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\def\ref#1#2#3#4#5#6{#1, {\it #2,} #3 {\bf #4} (#5), #6.}
%References
\def\akh {1} % Akhiezer
\def\ag {2} % Akhiezer-Glazman
\def\as {3} % Alonso-Simon
\def\bgm {4} % Baker and Graves-Morris
\def\ds {5} % Dunford-Schwarz
\def\gs {6} % Gesztesy-Simon
\def\gdl {7} % Gil de Lamadrid
\def\ham {8} % Hamburger
\def\im {9} % Ismail-Masson
\def\ir {10} % Ismail-Rahman
\def\jl {11} % Jitomirskaya-Last
\def\kato {12} % Kato
\def\katz {13} % Katznelson
\def\kei {14} % Kreich
\def\krch {15} % Krein (1951)
\def\krsti {16} % Krein (1952-Dokl.)
\def\krstu {17} % Krein (1952-Priklad.)
\def\krkol {18} % Krein (1945)
\def\land {19} % Landau
\def\lang {20} % Langer
\def\liv {21} % Livsic
\def\loe {22} % Loeffer et al.
\def\mmc {23} % Masson-McClary
\def\naI {24} % Naimark (Self-adjoint extension)
\def\naII {25} % Naimark (1940)
\def\naIII {26} % Naimark (1943)
\def\naIV {27} % Naimark (1946)
\def\nel {28} % Nelson
\def\nev {29} % Nevanlinna
\def\nusI {30} % Nussbaum (1965)
\def\nusII {31} % Nussbaum (1969)
\def\pea {32} % Pearson
\def\rsI {33} % Reed-Simon I
\def\rsII {34} % Reed-Simon II
\def\st {35} % Shohat-Tamarkin
\def\simap {36} % Simon (1970)
\def\simjfa {37} % Simon (1978)
\def\sti {38} % Stieltjes
\def\sto {39} % Stone
\topmatter
\title The Classical Moment Problem as a Self-Adjoint
Finite Difference Operator
\endtitle
\rightheadtext{The Classical Moment Problem}
\author Barry Simon$^*$
\endauthor
\leftheadtext{B.~Simon}
\affil Division of Physics, Mathematics, and Astronomy \\
California Institute of Technology \\
Pasadena, CA 91125
\endaffil
\date November 14, 1997
\enddate
\thanks$^*$ This material is based upon work supported by the
National Science Foundation under Grant No.~DMS-9401491. The
Government has certain rights in this material.
\endthanks
\thanks To appear in {\it{Advances in Mathematics}}
\endthanks
\abstract This is a comprehensive exposition of the classical
moment problem using methods from the theory of finite difference
operators. Among the advantages of this approach is that the
Nevanlinna functions appear as elements of a transfer matrix
and convergence of Pad\'e approximants appears as the strong
resolvent convergence of finite matrix approximations to a
Jacobi matrix. As a bonus of this, we obtain new results on the
convergence of certain Pad\'e approximants for series of Hamburger.
\endabstract
\endtopmatter
\document
\vskip 0.1in
\flushpar{\bf \S 1. Introduction}
\vskip 0.1in
The classical moment problem was central to the development of
analysis in the period from 1894 (when Stieltjes wrote his
famous memoir [\sti]) until the 1950's (when Krein completed his
series on the subject [\krch, \krsti, \krstu]). The notion of
measure (Stieltjes integrals), Pad\'e approximants, orthogonal
polynomials, extensions of positive linear functionals
(Riesz-Markov theorem), boundary values of analytic functions,
and the Herglotz-Nevanlinna-Riesz representation theorem all
have their roots firmly in the study of the moment problem.
This expository note attempts to present the main results of
the theory with two main ideas in mind. It is known from early
on (see below) that a moment problem is associated to a certain
semi-infinite Jacobi matrix, $A$. The first idea is that the basic
theorems can be viewed as results on the self-adjoint extensions
of $A$. The second idea is that techniques from the theory of
second-order difference and differential equations should be
useful in the theory.
Of course, neither of these ideas is new. Both appear, for
example, in Stone's treatment [\sto], in Dunford-Schwartz [\ds],
and in Akhiezer's brilliant book on the subject [\akh]. But,
normally, (and in [\akh], in particular), these ideas are presented
as applications of the basic theory rather than as the central
tenets. By placing them front and center, many parts of the
theory become more transparent --- in particular, our realization
of the Nevanlinna matrix as a transfer matrix makes its
properties (as an entire function of order 1 and minimal type)
evident (see Theorem~4.8 below).
Two basic moment problems will concern us here:
\example{Hamburger Moment Problem} Given a sequence
$\gamma_0, \gamma_1,\gamma_2, \dots$ of reals, when is there a
measure, $d\rho$, on $(-\infty, \infty)$ so that
$$
\gamma_n = \int_{-\infty}^\infty x^n\, d\rho(x) \tag 1.1
$$
and if such a $\rho$ exists, is it unique? We let $\Cal M_H (\gamma)$
denote the set of solutions of (1.1).
\endexample
\example{Stieltjes Moment Problem} Given a sequence $\gamma_0,
\gamma_1, \gamma_2, \dots$ of reals, when is there a measure,
$d\rho$ on $[0,\infty)$ so that
$$
\gamma_n = \int_0^\infty x^n\, d\rho(x) \tag 1.2
$$
and if such a $\rho$ exists, is it unique? We let $\Cal M_S (\gamma)$
denote the set of solutions of (1.2).
\endexample
\vskip 0.1in
We will not attempt a comprehensive historical attribution of
classical results; see Akhiezer [\akh] and Shohat-Tamarkin
[\st] for that.
We will always normalize the $\gamma$'s so that $\gamma_0 =1$.
By replacing $\gamma_n$ by $\gamma_n/\gamma_0$, we can reduce
general $\gamma$'s to this normalized case. We will also demand
that $d\rho$ have infinite support, that is, that $\rho$ not be a
pure point measure supported on a finite set. This eliminates
certain degenerate cases.
One immediately defines two sesquilinear forms, $H_N$ and $S_N$,
on $\Bbb C^N$ for each $N$ by
$$\align
H_N (\alpha,\beta) &= \sum \Sb n=0,1,\dots, N-1 \\
m=0,1,\dots, N-1 \endSb \bar\alpha_n \beta_m \gamma_{n+m}
\tag 1.3 \\
S_N(\alpha,\beta) &= \sum \Sb n=0,1,\dots, N-1 \\
m=0,1,\dots, N-1 \endSb \bar\alpha_n \beta_m \gamma_{n+m+1}
\tag 1.4
\endalign
$$
and corresponding matrices $\Cal H_N$ and $\Cal S_N$ so that
$H_N (\alpha,\beta)=\langle \alpha, \Cal H_N \beta\rangle$
and $S_N(\alpha,\beta)=\langle \alpha, \Cal S_N \beta\rangle$
in the usual Euclidean inner product. Our inner products are
linear in the second factor and anti-linear in the first.
A standard piece of linear algebra says:
\proclaim{Lemma 1.1} An $N\times N$ Hermitean matrix $A$ is
strictly positive definite if and only if each submatrix
$A^{[1,J]} = (a_{ij})_{1\leq i, j\leq J}$ has $\det(A^{[1,J]})
>0$ for $J=1,2,\dots, N$.
\endproclaim
\demo{Proof} If $A$ is strictly positive definite, so is each
$A^{[1,J]}$, so their eigenvalues are all strictly positive and
so their determinants are all strictly positive.
For the converse, suppose that $A^{[1,N-1]}$ is positive
definite. By the min-max principle, the $(N-1)$ eigenvalues of
$A^{[1,N-1]}$ interlace the $N$ eigenvalues of $A^{[1,N]}
\equiv A$. If all the eigenvalues of $A^{[1,N-1]}$ are positive,
so are the $N-1$ largest eigenvalues of $A$. If the
$N^{\text{\rom{th}}}$ eigenvalue were non-positive, $\det(A)
\leq 0$. Thus, $A^{[1,N-1]}$ strictly positive definite and
$\det(A^{[N,N]}) >0$ imply $A^{[1,N]}$ is strictly positive
definite. An obvious induction completes the proof. \qed
\enddemo
This immediately implies
\proclaim{Proposition 1.2} $\{H_N\}^\infty_{N=1}$ are strictly
positive definite forms if and only if $\det(\Cal H_N)>0$ for
$N=1,2,\dots$. Similarly, $\{S_N\}^\infty_{N=1}$ are strictly
positive definite forms if and only if $\det(\Cal S_N)>0$ for
$N=1,2,\dots$.
\endproclaim
Suppose that the $\gamma_n$ obey (1.1). Then by an elementary
calculation,
$$\align
\int \biggl| \sum_{n=0}^{N-1} \alpha_n x^n\biggr|^2\, d\rho(x)
&= H_N (\alpha,\alpha) \tag 1.5 \\
\int x \biggl| \sum_{n=0}^{N-1} \alpha_n x^n\biggr|^2 \,
d\rho(x) &= S_N (\alpha, \alpha). \tag 1.6
\endalign
$$
Taking into account that if $\int |P(x)|^2\, d\rho(x)=0$, $\rho$
must be supported on the zeros of $P$, we have:
\proclaim{Proposition 1.3} A necessary condition that
{\rom{(1.1)}} holds for some measure $d\rho$ on \linebreak
$(-\infty, \infty)$ with infinite support is that each
sesquilinear form $H_N$ is strictly positive definite. A
necessary condition that there be a $d\rho$ supported on
$[0,\infty)$ is that each $H_N$ and each $S_N$ be strictly
positive definite.
\endproclaim
Suppose now that each $H_N$ is strictly positive definite. Let
$\Bbb C[X]$ be the family of complex polynomials. Given $P(X) =
\sum_{n=0}^{N-1} \alpha_n X^n$, $Q(X)=\sum_{n=0}^{N-1} \beta_n
X^n$ (we suppose the upper limits in the sums are equal by
using some zero $\alpha$'s or $\beta$'s if need be), define
$$
\langle P,Q\rangle = H_N (\alpha,\beta). \tag 1.7
$$
This defines a positive definite inner product on $\Bbb C[X]$,
and, in the usual way, we can complete $\Bbb C[X]$ to a Hilbert
space $\Cal H^{(\gamma)}$ in which $\Bbb C[X]$ is dense.
$\Bbb C[X]$ can be thought of as abstract polynomials or as
infinite sequences $(\alpha_0, \alpha_1,\dots, \alpha_{N-1}, 0,
\dots)$ which are eventually $0$ via $\alpha\sim \sum_{j=0}^{N-1}
\alpha_j X^j$.
We will start using some basic facts about symmetric and
self-adjoint operators on a Hilbert space --- specifically, the
spectral theorem and the von Neumann theory of self-adjoint
extensions. For an exposition of these ideas, see Chapters~VII,
VIII, and X of Reed-Simon, volumes~I and II [\rsI, \rsII] and
see our brief sketch at the start of Section~2.
Define a densely defined operator $A$ on $\Cal H^{(\gamma)}$ with
domain $D(A)=\Bbb C[X]$ by
$$
A[P(X)]=[XP(X)]. \tag 1.8
$$
In the sequence way of looking at things, $A$ is the right shift,
that is, $A(\alpha_0, \alpha_1, \dots, \alpha_N, 0, \dots)
\mathbreak = (0, \alpha_0, \alpha_1, \dots, \alpha_N, 0, \dots)$.
This means that
$$
\langle P,A[Q]\rangle = S_N (\alpha,\beta) \tag 1.9
$$
and, in particular,
$$
\langle 1, A^n 1\rangle = \gamma_n. \tag 1.10
$$
Since $\Cal H_N$ and $\Cal S_N$ are real-symmetric matrices, $A$
is a symmetric operator, that is, $\langle A[P],Q\rangle =
\langle P,A[Q]\rangle$. Moreover, if we define a complex
conjugation $C$ on $\Bbb C[X]$ by $C(\sum_{n=0}^{N-1} \alpha_n
X^n) = \sum_{n=0}^{N-1} \bar\alpha_n X^n$, then $CA = AC$.
It follows by a theorem of von Neumann (see Corollary~2.4 in
Section~2) that $A$ has self-adjoint extensions.
If each $S_N$ is positive definite, then $\langle P,A[P]\rangle
\geq 0$ for all $P$, and it follows that $A$ has a non-negative
self-adjoint extension $A_F$, the Friedrichs extension (discussed
further in Section~3). We thus see:
\proclaim{Proposition 1.4} If all $H_N$ are positive definite,
then $A$ has self-adjoint extensions. If all $S_N$ are positive
definite, then $A$ has non-negative self-adjoint extensions.
\endproclaim
Let $\tilde A$ be a self-adjoint extension of $A$. By the spectral
theorem, there is a spectral measure $d\tilde\mu$ for $\tilde A$
with vector $[1]\in\Cal H^{(\gamma)}$, that is, so that for
any bounded function of $\tilde A$,
$$
\langle 1, f(\tilde A) 1\rangle = \int f(x)\,
d\tilde\mu(x). \tag 1.11
$$
Since $1\in D(A^N)\subset D(\tilde A^N)$, (1.11) extends to
polynomially bounded functions and we have by (1.10) that,
$$
\gamma_N = \int x^N \, d\tilde\mu(x).
$$
We see therefore that a self-adjoint extension of $A$ yields a
solution of the Hamburger moment problem. Moreover, a non-negative
self-adjoint extension has $\text{supp}(d\tilde\mu) \subset
[0,\infty)$ and so yields a solution of the Stieltjes moment
problem. Combining this with Propositions~1.2, 1.3, and 1.4, we
have the first major result in the theory of moments.
\proclaim{Theorem 1 (Existence)} A necessary and sufficient
condition for there to exist a measure $d\rho$ with infinite
support obeying {\rom{(1.1)}} is that $\det(\Cal H_N)>0$ for
$N=1,2, \dots$. A necessary and sufficient condition that also
$d\rho$ be supported on $[0,\infty)$ is that both $\det(\Cal H_N)
>0$ and $\det(\Cal S_N)>0$ for $N=1,2,\dots$.
\endproclaim
Historically, existence was a major theme because the now
standard tools on existence of measures were invented in the
context of moment problems. We have settled it quickly, and the
bulk of this paper is devoted to uniqueness, especially the
study of cases of non-uniqueness. Non-uniqueness only occurs
in somewhat pathological situations, but the theory is so elegant
and beautiful that it has captivated analysts for a century.
Henceforth, we will call a set of moments
$\{\gamma_n\}^\infty_{n=0}$ with $\det(\Cal H_N)>0$ for all
$N$ a set of {\it{Hamburger moments}}. If both $\det(\Cal H_N)>0$
and $\det(\Cal S_N)>0$ for all $N$, we call them a set of
{\it{Stieltjes moments}}.
We will call a solution of the moment problem which comes from
a self-adjoint extension of $A$ a {\it von~Neumann solution}.
The name is in honor of the use below of the von~Neumann
theory of self-adjoint extensions of densely-defined symmetric
operators. This name, like our use of Friedrichs solution and
Krein solution later, is not standard, but it is natural from
the point of view of self-adjoint operators. As far as I know,
neither von~Neumann nor Friedrichs worked on the moment problem
per se. While Krein did, his work on the moment problem was not
in the context of the Krein extension we use to construct what
we will call the Krein solution. What we call von~Neumann solutions,
Akhiezer calls $N$-extremal and Shohat-Tamarkin call extremal.
This last name is unfortunate since we will see there exist many
solutions which are extreme points in the sense of convex set
theory, but which are not von~Neumann solutions (and so, not
extremal in the Shohat-Tamarkin sense).
Given the connection with self-adjoint extensions, the following
result is reasonable (and true!):
\proclaim{Theorem 2 (Uniqueness)} A necessary and sufficient
condition that the measure $d\rho$ in {\rom{(1.1)}} be unique
is that the operator $A$ of {\rom{(1.8)}} is essentially
self-adjoint \rom(i.e., has a unique self-adjoint extension\rom).
A necessary and sufficient condition that there be a unique measure
$d\rho$ in {\rom{(1.1)}} supported in $[0,\infty)$ is that $A$
have a unique non-negative self-adjoint extension.
\endproclaim
This result is surprisingly subtle. First of all, it is not
obvious (but true, as we will see) that distinct self-adjoint
extensions have distinct spectral measures $d\tilde\mu$, so
there is something to be proven before multiple self-adjoint
extensions imply multiple solutions of the moment problem. The
other direction is even less clear cut, for not only is it not
obvious, it is false that every solution of the moment problem
is a von~Neumann solution (Reed-Simon [\rsII] has an incorrect
proof of uniqueness that implicitly assumes every solution
comes from a self-adjoint extension). As we will see, once
there are multiple solutions, there are many, many more
solutions than those that come from self-adjoint extensions in
the von~Neumann sense of looking for extensions in
$\overline{D(A)}$. But, as we will see in Section~6, there is
a sense in which solutions are associated to self-adjoint
operators in a larger space.
We also note we will see cases where the Stieltjes problem has
a unique solution but the associated Hamburger problem does not.
The Hamburger part of Theorem~2 will be proven in Section~2
(Theorems~2.10 and 2.12); the Stieltjes part will be proven in
Sections~2 and 3 (Theorems~2.12 and 3.2). If there is a unique
solution to (1.1), the moment problem is called {\it{determinate}};
if there are multiple solutions, it is called {\it{indeterminate}}.
It is ironic that the English language literature uses these
awkward terms, rather than determined and undetermined. Stieltjes
was Dutch, but his fundamental paper was in French, and the names
have stuck. Much of the interesting theory involves analyzing
the indeterminate case, so we may as well give some examples that
illuminate non-uniqueness.
\example{Example 1.1} Let $f$ be a non-zero $C^\infty$ function
on $\Bbb R$ supported on $[0,1]$. Let $g(x) = \hat f(x)$, with
$\hat f$ the Fourier transform of $f$. Then
$$
\int_{-\infty}^\infty x^n g(x)\, dx = \sqrt{2\pi} \,
(-i)^n\, \frac{d^n f}{dx^n}\, (0)=0.
$$
Let $d\rho_1 (x) = (\text{Re}\, g)_+ (x)\, dx$, the positive part
of $\text{Re}\, g$, and let $d\rho_2 = (\text{Re}\, g)_- (x)\, dx$.
By the above,
$$
\int_{-\infty}^\infty x^n d\rho_1 (x) =
\int_{-\infty}^\infty x^n\, d\rho_2 (x)
$$
for all $n$. Since $d\rho_1$ and $d\rho_2$ have disjoint essential
supports, they are unequal and we have non-unique solutions of the
moment problem. (We will see eventually that neither is a
von~Neumann solution.) The moments from $\rho_1, \rho_2$ may not
be normalized. But we clearly have non-uniqueness after
normalization.
\endexample
This non-uniqueness is associated to non-analyticity in a Fourier
transform and suggests that if one can guarantee analyticity,
one has uniqueness. Indeed,
\proclaim{Proposition 1.5} Suppose that
$\{\gamma_n\}^\infty_{n=0}$ is a set of Hamburger moments and
that for some $C,R >0$,
$$
|\gamma_n |\leq CR^n n! \tag 1.12a
$$
Then the Hamburger moment problem is determinate.
If $\{\gamma_n\}^\infty_{n=0}$ is a set of Stieltjes moments
and
$$
|\gamma_n|\leq CR^n (2n)! \tag 1.12b
$$
then the Stieltjes moment problem is determinate.
\endproclaim
\demo{Proof} Let $d\rho$ obey (1.1). Then $x^{2n}\in L^1
(\Bbb R, d\rho)$, and by the monotone convergence theorem,
$$\align
\int_{-\infty}^\infty \text{cosh} \biggl(\frac{x}{2R}\biggr)\,
d\rho(x) &= \lim_{N\to\infty} \int_{-\infty}^\infty \sum_{n=0}^N
\biggl(\frac{x}{2R}\biggr)^{2n}
\frac{1}{(2n)!}\, d\rho(x)\\
&\leq C\, \lim_{N\to\infty} \sum_{n=0}^N \biggl(\frac12\biggr)^{2n}
= \frac43 C < \infty.
\endalign
$$
Thus, $e^{\alpha x} \in L^1 (\Bbb R, d\rho(x))$ for $|\alpha| <
\frac1{2R}$. It follows that $F_\rho (z) \equiv \int e^{izx}
d\rho(x)$ has an analytic continuation to $\{ z\mid\, |\text{Im}\,
z| < \frac1{2R}\}$. If $\mu$ is a second solution of (1.1),
$F_\mu (z)$ is also analytic there. But $-i^n \frac{d^n F_\rho}{dz}
(0)= \gamma_n$, so the Taylor series of $F_\rho$ and $F_\mu$ at $z=0$
agree, so $F_\rho = F_\mu$ in the entire strip by analyticity.
This implies that $\mu = \rho$ by a variety of means. For example,
in the topology of bounded uniformly local convergence, (i.e.,
$f_n\to f$ means $\sup \|f_n\|_\infty <0$ and $f_n(x)\to f(x)$
uniformly for $x$ in any $[-\kappa, \kappa]$), linear combinations
of the $\{ e^{iyx}\mid y\in\Bbb R\}$ are dense in all bounded
continuous functions. Or alternatively, $G_\mu (z) = \int
\frac{d\mu(x)}{x-z} = i\int_{-\infty}^0 e^{-iyz}F_\mu (y)\, dy$
for $\text{Im}\, z>0$, and $\mu$ can be recovered as a boundary
value of $G_\mu$.
The $(2n)!$ Stieltjes result follows from the $n!$ result and
Proposition~1.6 below. \qed
\enddemo
We will generalize Proposition~1.5 later (see Corollary~4.5).
\proclaim{Proposition 1.6} Let $\{\gamma_n\}^\infty_{n=0}$ be a
set of Stieltjes moments. Let
$$\alignat2
\Gamma_{2m} &= \gamma_m, \qquad && m=0,1,\dots \\
\Gamma_{2m+1} &= 0, \qquad && m=0,1,\dots .
\endalignat
$$
If $\{\Gamma_j\}^\infty_{j=0}$ is a determinate Hamburger problem,
then $\{\gamma_n\}^\infty_{n=0}$ is a determinate Stieljes
problem.
\endproclaim
\demo{Proof} Let $d\rho$ solve the Stieltjes problem. Let
$$
d\mu(x) = \tfrac12 [\chi_{[0,\infty)} (x)\, d\rho (x^2)
+\chi_{(-\infty, 0]}(x)\, d\rho (x^2)].
$$
Then the moments of $\mu$ are $\Gamma$. Thus uniqueness for the
$\Gamma$ problem on $(-\infty, \infty)$ implies uniqueness for
the $\gamma$ problem on $[0,\infty)$. \qed
\enddemo
\remark{Remark} We will see later (Theorem~2.13) that the
converse of the last assertion in Proposition~1.6 is true. This
is a more subtle result.
\endremark
\example{Example 1.2} This example is due to Stieltjes [\sti].
It is interesting for us here because it is totally explicit and
because it provides an example of non-uniqueness for the Stieltjes
moment problem. Note first that
$$
\int_0^\infty u^k u^{-\ln u} \sin(2\pi \ln u)\, du = 0.
$$
This follows by the change of variables $v=-\frac{k+1}{2} +
\ln u$, the periodicity of $\sin(\,\cdot\,)$, and the fact that
sin is an odd function. Thus for any $\theta\in [-1,1]$,
$$
\frac{1}{\sqrt\pi} \int_0^\infty u^k u^{-\ln u} [1+\theta
\sin(2\pi \ln u)]\, du = e^{\frac14(k+1)^2}
$$
(by the same change of variables) so $\gamma_k =
e^{\frac14 (k+1)^2}$ is an indeterminate set of Stieltjes
moments. Notice for $\theta\in (-1,1)$, if $d\rho_\theta (u)$
is the measure with these moments, then $\sin(2\pi \ln (u))/1+
\theta \sin(2\pi \ln (u))$ is in $L^2 (d\rho_\theta)$
and orthogonal to all polynomials, so $d\rho_\theta (u)$ is
a measure with all moments finite but where the polynomials are
not $L^2$ dense. As we shall see, this is typical of solutions of
indeterminate moment problems which are not von~Neumann solutions.
\endexample
Looking at the rapid growth of the moments in Example~1.2, you
might hope that just as a condition of not too great growth
(like Proposition~1.5) implies determinacy, there might be a
condition of rapid growth that implies indeterminacy. But that
is false! There are moments of essentially arbitrary rates of
growth which lead to determinate problems (see the remark
after Corollary~4.21 and Theorem~6.2).
\example{Example 1.3} There is a criterion of Krein [\krkol]
for indeterminacy that again shows the indeterminacy of
Example~1.2, also of the example of the moments of $\exp
(-|x|^\alpha)\, dx$ (on $(-\infty, \infty)$) with $\alpha <1$ and
also an example of Hamburger [\ham], the moments of
$$
\chi_{[0,\infty)}(x) \exp \biggl( - \frac{\pi\sqrt{x}}
{\ln^2 x + \pi^2}\biggr)\, dx.
$$
\endexample
\proclaim{Proposition 1.7 (Krein [\krkol])} Suppose that
$d\rho(x) = F(x)\, dx$ where $0\leq F(x) \leq 1$ and either
\rom{(i)} $\text{\rom{supp}}(F)= (-\infty, \infty)$ and
$$
\int_{-\infty}^\infty - \frac{\ln (F(x))}{1+x^2}\, dx < \infty
\tag 1.13a
$$
or
\rom{(ii)} $\text{\rom{supp}}(F) = [0,\infty)$ and
$$
\int_0^\infty -\frac{\ln (F(x))}{(1+x)}\, \frac{dx}
{\sqrt x} < \infty. \tag 1.13b
$$
Suppose also that for all $n$:
$$
\int_{-\infty}^\infty |x|^n F(x)\, dx < \infty.
\tag 1.13c
$$
Then the moment problem \rom(Hamburger in case {\rom{(i)}},
Stieltjes in case {\rom{(ii))}} with moments
$$
\gamma_n = \frac{\int x^n F(x)\, dx}{\int F(x)\, dx}
$$
is indeterminate.
\endproclaim
\remark{Remarks} 1. Hamburger's example is close to borderline
for (1.13b) to hold.
2. Since $\int_{-\infty}^\infty x^{2n} \exp(-|x|^\alpha)\, dx
= 2\alpha^{-1} \Gamma (\frac{2n+1}{\alpha})\sim (\frac{2n}
{\alpha})!$ and (1.13a) holds for $F(x) = \exp(-|x|^\alpha)$
if $\alpha < 1$, we see that there are examples of Hamburger
indeterminate moment problems with growth just slightly faster
than the $n!$ growth, which would imply by Proposition~1.5 that
the problem is determinate. Similarly, since $\int_0^\infty
x^n \exp(-|x|^\alpha)\, dx = \alpha^{-1} \Gamma (\frac{n+1}
{\alpha})\sim (\frac{n}{\alpha})!$ and (1.13b) holds for $F(x)
=\exp (-|x|^\alpha)$ if $\alpha < \frac12$, we see there are
examples of Stieltjes indeterminate problems with growth just
slightly faster than the $(2n)!$ of Proposition~1.5.
3. Since $F(x) = \frac12 e^{-|x|}$ has moments $\gamma_{2n} =
(2n)!$ covered by Proposition~1.5, it is a determinate problem. The
integral in (1.13a) is only barely divergent in this case. Similarly,
$F(x) = \chi_{[0,\infty)}(x) e^{-\sqrt x}$ is a Stieltjes
determinate moment problem by Proposition~1.5 and the integral
in (1.13b) is barely divergent. This says that Krein's conditions
are close to optimal.
4. Krein actually proved a stronger result by essentially the
same method. $F$ need not be bounded (by a limiting argument
from the bounded case) and the measure defining $\gamma_n$ can
have an arbitrary singular part. Moreover, Krein proves (1.13a)
is necessary and sufficient for $\{e^{i\alpha x}
\}_{0\leq\alpha<\infty}$ to not be dense in $L^2
(\Bbb R, F(x)\, dx)$.
5. Krein's construction, as we shall see, involves finding a
bounded analytic function $G$ in $\Bbb C_+ \equiv \{z\in\Bbb C
\mid\text{Im}\, z>0\}$ so that $|G(x+i0)| \leq F(x)$. From this
point of view, the fact that $F(x)$ cannot decay faster than
$e^{-|x|}$ is connected to the Phragm\'en-Lindel\"of principle.
6. The analog of Krein's result for the circle instead of
the half plane is due to Szego.
\endremark
\demo{Proof} Suppose we can find $G$ with $\text{Re}\, G
\not\equiv 0$ with $|G(x)|\leq F(x)$ so that
$$
\int x^n G(x)\, dx = 0.
$$
Then both $\frac{F(x)\, dx}{\int F(x)\, dx}$ and
$\frac{[F(x)+\text{Re}\, G(x)]\, dx}{\int F(x)\, dx}$ solve the
$\gamma_n$ moment problem, showing indeterminacy.
In case (i), define for $z\in\Bbb C_+$ (see (1.19) and the
definition of Herglotz function below):
$$
Q(z) = \frac1{\pi} \int \biggl( \frac1{x-z} - \frac{x}{1+x^2}
\biggr) [-\ln (F(x))]\, dx,
$$
which is a convergent integral by (1.13a). Then, $\text{Im}\,
Q(z) \geq 0$ and by the theory of boundary values at such
analytic functions [\katz], $\lim_{\varepsilon\downarrow 0}
Q(x+i\varepsilon) \equiv Q(x+i0)$ exists for a.e.~$x\in\Bbb R$
and $\text{Im}\, Q(x+i0)=-\ln F(x)$.
Let $G(z) = \exp (iQ(z))$ for $z\in\Bbb C_+$ and $G(x) = \exp
(iQ(x+i0))$. Then the properties of $Q$ imply $|G(z)| \leq 1$,
$\lim_{\varepsilon\downarrow 0} G(x+i0)= G(x)$ for a.e.~$x$
in $\Bbb R$, and $|G(x)| = F(x)$. A standard contour integral
argument then shows that for any $\varepsilon >0$ and $n>0$,
$$
\int x^n (1-i\varepsilon x)^{-n-2} G(x)\, dx =0.
$$
Since (1.13c) holds and $|G(x)|\leq F(x)$, we can take
$\varepsilon\downarrow 0$ and obtain $\int x^n G(x)\, dx =0$.
In case (ii), define
$$
Q(z) = \frac{2z}{\pi} \int_0^\infty - \frac{\ln F(x^2)}
{x^2 - z^2}\, dx.
$$
Again, the integral converges by (1.13b) and $\text{Im}\,
Q(z) >0$. Moreover, $Q(-\bar z) = -\overline{Q(z)}$. We have for
a.e.~$x\in\Bbb R$, $\text{Im}\, Q(x+i0) = -\ln F(x^2)$. Thus,
if $H(z) = \exp (iQ(z))$ for $z\in\Bbb C_+$ and $H(x) = \exp
(iQ(x+i0))$ for $x\in\Bbb R$ and $G(x) = \text{Im}\, H(\sqrt x)$
for $x\in [0,\infty)$, then, as above,
$$
\int_{-\infty}^\infty x^{2n+1} H(x)\, dx =0,
$$
so since $H(-x) = \overline{H(x)}$, $\int_0^\infty
x^{2n+1} G(x^2)\, dx = 0$ or $\int_0^\infty y^n
G(y)\, dy =0$. Since $|H(x)| = F(x^2)$, we have that $|G(x)|
\leq F(x)$. \qed
\enddemo
Additional examples of indeterminate moment problems (specified
by their orthogonal polynomials) can be found in [\im, \ir].
One point to note for now is that in all these examples, we have
non-uniqueness with measures $d\rho$ of the form $d\rho (x) =
G(x)\, dx$ (i.e., absolutely continuous). This will have
significance after we discuss Theorem~5 below.
To discuss the theory further, we must look more closely at the
operator $A$ given by (1.8). Consider the set $\{ 1,X,X^2,
\dots\}$ in $\Cal H^{(\gamma)}$. By the strict positivity of
$H_N$, these elements in $\Cal H^{(\gamma)}$ are linearly
independent and they span $\Cal H^{(\gamma)}$ by construction.
Thus by a Gram-Schmidt procedure, we can obtain an orthogonal
basis, $\{P_n(X)\}^\infty_{n=0}$, for $\Cal H^{(\gamma)}$. By
construction,
$$\alignat2
P_n (X) &= c_{nn}X^n + \text{ lower order}, \qquad &&
\text{with }c_{nn} > 0 \tag 1.14a \\
\langle P_n, P_m\rangle &= 0, \qquad && m=0,1,2,\dots, n-1
\tag 1.14b \\
\langle P_n, P_n\rangle &= 1. \tag 1.14c
\endalignat
$$
These are, of course, the well-known orthogonal polynomials for
$d\rho$ determined by the moments $\{\gamma_n\}^\infty_{n=0}$.
Note that often the normalization condition (1.14c) is
replaced by $c_{nn}\equiv 1$, yielding a distinct set of
``orthogonal" polynomials. There are explicit formulae for the
$P_n(X)$ in terms of determinants and the $\gamma$'s. We
discuss them in Appendix~A.
By construction, $\{P_j\}^n_{j=0}$ is an orthonormal basis for
the polynomials of degree $n$. The realization of elements of
$\Cal H^{(\gamma)}$ as $\sum_{n=0}^\infty \lambda_n P_n(X)$
with $\sum_{n=0}^\infty |\lambda_n|^2 <\infty$ gives a
different realization of $\Cal H^{(\gamma)}$ as a set of
sequences $\lambda = (\lambda_0, \dots)$ with the usual
$\ell^2 (\{0,1,2,\dots \})$ inner product. $\Bbb C[X]$
corresponds to these $\lambda$'s with $\lambda_n=0$ for $n$
sufficiently large. But we need to bear in mind this change
of realization as sequences.
Note that $\text{span}[1,\dots, X^n]=\text{span}[P_0, \dots,
P_n(X)]$. In particular, $XP_n(X)$ has an expansion in $P_0,
P_1, \dots, P_{n+1}$. But $\langle XP_n, P_j\rangle =
\langle P_n, XP_j\rangle =0$ if $j0$ and
$$
c_n = \langle P_{n+1}, XP_n\rangle = \langle P_n, XP_{n+1}
\rangle = a_n.
$$
(1.15) thus becomes
$$
XP_n(X) = a_n P_{n+1}(X) + b_n P_n(X) + a_{n-1} P_{n-1}(X).
\tag 1.15$^\prime$
$$
Since $\{P_n(X)\}$ are an orthonormal basis for
$\Cal H^{(\gamma)}$, this says that in this basis, $A$ is given
by a tridiagonal matrix, and $D(A)$ is the set of sequences of
finite support.
Thus, given a set $\{\gamma_n\}^\infty_{n=0}$ of Hamburger
moments, we can find $b_0, b_1, \dots$ real and $a_0, a_1, \dots$
positive so that the moment problem is associated to self-adjoint
extensions of the Jacobi matrix,
$$
A=\pmatrix
b_0 & a_0 & 0 & 0 & \dots \\
a_0 & b_1 & a_1 & 0 & \dots \\
0 & a_1 & b_2 & a_2 & \dots \\
0 & 0 & a_2 & b_3 & \dots \\
\dots & \dots & \dots & \dots & \dots
\endpmatrix . \tag 1.16
$$
There are explicit formulae for the $b_n$'s and $a_n$'s in
terms of the determinants of the $\gamma_n$'s. We will discuss
them in Appendix A.
Conversely, given a matrix $A$ of the form (1.16), we can find
a set of moments for which it is the associated Jacobi matrix.
Indeed, if $\delta_0$ is the vector $(1,0, \dots, 0, \dots )$ in
$\ell_2$, then
$$
\gamma_n = (\delta_0, A^n \delta_0).
$$
Thus the theory of Hamburger moments is the theory of
semi-infinite Jacobi matrices.
So far we have considered the $P_n (X)$ as abstract polynomials,
one for each $n$. It is useful to turn this around. First,
replace the abstract $X$ by an explicit complex number. For
each such $z$, consider the semi-infinite sequence $\pi(z) =
(P_0(z), P_1(z), P_2(z),\dots)$. (1.15$^\prime$) now becomes
(with $P_{-1}(z)$ interpreted as $0$):
$$
a_n P_{n+1}(z) + (b_n -z) P_n(z) + a_{n-1} P_{n-1}(z) = 0,
\qquad n\geq 0 \tag 1.17
$$
so $P_n(z)$ obeys a second-order difference equation. If, for
example, $\pi(x_0)\in\ell^2$, then $x_0$ is formally an eigenvalue
of the Jacobi matrix $A$. (Of course, $\pi$ is never a finite
sequence because it obeys a second-order difference equation and
$\pi_0(z)=1$. Thus, $\pi\notin D(A)$. It may or may not happen
that $\pi\in D(\bar A)$, so the formal relation may or may not
correspond to an actual eigenvalue. It will always be true,
though, as we shall see, that if $\pi(x_0)\in\ell^2$, then
$(A^*-x_0)\pi(x_0)=0$.)
For convenience, set $a_{-1}=1$. For any $z\in \Bbb C$, the
solutions of the equation
$$
a_n u_{n+1} + (b_n-z) u_n + a_{n-1} u_{n-1} = 0,
\qquad n \geq 0 \tag 1.18
$$
are two-dimensional, determined by the initial data $(u_{-1},
u_0)$.
$P_n (z)$ corresponds to taking
$$
u_{-1} =0, \qquad u_0 = 1.
$$
There is a second solution, $Q_n (z)$, taking initial
conditions
$$
u_{-1} = -1, \qquad u_0 = 0.
$$
We will also define $\xi(z)=(Q_0(z), Q_1(z), \dots)$. It can be
seen by induction that for $n\geq 1$, $Q_n(X)$ is a polynomial of
degree $n-1$. We will see later (Proposition~5.16) that the $Q$'s
are orthogonal polynomials for another moment problem.
As we will see, this normalization is such that if some
combination $\eta (z) \equiv t\pi(z) + \xi(z)\in\ell^2$, then
$(A^* -z) \eta = \delta_0$ and $\langle \delta_0, \eta\rangle
=t$. Here $\delta_n$ is the Kronecker vector with $1$ in
position $n$ and zeros elsewhere. We will have a lot more to say
about $Q_n(z)$ in Section~4.
There is a fundamental result relating the solution vectors
$\pi,\xi$ to determinacy of the moment problem. In Section~4
(see Proposition~4.4 and Theorem~4.7), we will prove
\proclaim{Theorem 3} Fix a set of moments and associated
Jacobi matrix. Then the following are equivalent:
\roster
\item"\rom{(i)}" The Hamburger moment problem is indeterminate.
\item"\rom{(ii)}" For some $z_0\in\Bbb C$ with $\text{\rom{Im}}
\, z_0 \neq 0$, $\pi(z_0)\in\ell_2$.
\item"\rom{(iii)}" For some $z_0\in\Bbb C$ with
$\text{\rom{Im}}\,z_0 \neq 0$, $\xi(z_0)\in\ell_2$.
\item"\rom{(iv)}" For some $x_0\in\Bbb R$, both $\pi(x_0)$
and $\xi(x_0)$ lie in $\ell_2$.
\item"\rom{(v)}" For some $x_0\in\Bbb R$, both $\pi(x_0)$ and
$\frac{\partial\pi}{\partial x}(x_0)$ lie in $\ell_2$.
\item"\rom{(vi)}" For some $x_0\in\Bbb R$, both $\xi(x_0)$
and $\frac{\partial\xi}{\partial x}(x_0)$ lie in $\ell_2$.
\item"\rom{(vii)}" For all $z\in\Bbb C$, both $\pi(z)$ and
$\xi(z)$ lie in $\ell_2$.
\endroster
\endproclaim
\remark{Remarks} 1. Appendix A has explicit determinantal
formulae for $\sum_{n=0}^N |P_n (0)|^2$ and $\sum_{n=0}^N
|Q_n (0)|^2$ providing ``explicit" criteria for determinacy in
terms of limits of determinants.
2. Theorem~3 can be thought of as a discrete analog of Weyl's
limit point/limit circle theory for self-adjoint extensions of
differential operators; see [\rsII].
\endremark
This implies that if all solutions of (1.18) lie in $\ell^2$
for one $z\in\Bbb C$, then all solutions lie in $\ell^2$ for all
$z\in\Bbb C$.
A high point of the theory, discussed in Section~4, is an
explicit description of all solutions of the moment problem in
the indeterminate case in terms of a certain $2\times 2$
matrix valued entire analytic function. It will suffice to
only vaguely describe the full result in this introduction.
\definition{Definition} A {\it{Herglotz function}} is a function
$\Phi(z)$ defined in $\Bbb C_+ \equiv\{z\in\Bbb C\mid\text{Im}\,
z>0\}$ and analytic there with $\text{Im}\,\Phi(z) >0$ there.
\enddefinition
These are also sometimes called Nevanlinna functions. It is a
fundamental result (see, e.g., [\ag]) that given such a $\Phi$,
there exists $c\geq 0$, $d$ real, and a measure $d\mu$ on $\Bbb R$
with $\int\frac{d\mu(x)}{1+x^2} <\infty$ so that either $c\neq
0$ or $d\mu \neq 0$ or both, and
$$
\Phi(z) = cz + d + \int \biggl[ \frac{1}{x-z}
-\frac{x}{1+x^2}\biggr] \, d\mu (x). \tag 1.19
$$
\proclaim{Theorem 4} The solutions of the Hamburger moment
problem in the indeterminate case are naturally parametrized by
Herglotz functions together with the functions $\Phi(z)= t\in
\Bbb R \cup\{\infty\}$. This later set of solutions are the
von~Neumann solutions.
If the Stieltjes problem is also indeterminate, there is
$t_0 >0$ so that the solutions of the Stieltjes problem are
naturally parametrized by Herglotz functions, $\Phi(z)$, which
obey $\Phi(x+i0)\in(t_0,\infty)$ for $x\in(-\infty,0)$ together
with $[t_0,\infty) \cup\{\infty\}$. The later set of solutions
are the von~Neumann solutions.
\endproclaim
We will prove Theorem~4 in Section~4 (see Theorems~4.14 and
4.18). From the explicit Nevanlinna form of the solutions, one
can prove (and we will in Section~4; see Theorems~4.11 and 4.17)
\proclaim{Theorem 5} In the indeterminate case, the von~Neumann
solutions are all pure point measures. Moreover, for any $t\in
\Bbb R$, there exists exactly one von~Neumann solution,
$\mu^{(t)}$, with $\mu^{(t)}(\{t\}) >0$. Moreover, for any other
solutions, $\rho$, of the moment problem, $\rho(\{t\}) <\mu^{(t)}
(\{t\})$.
\endproclaim
\remark{Remark} The parametrization $\mu^{(t)}$ of this theorem
is inconvenient since $t\mapsto \mu^{(t)}$ is many to one (in
fact, infinity to one since each $\mu^{(t)}$ has infinite support).
We will instead use a parametrization $t\mapsto \mu_t$ given by
$$
\int \frac{d\mu_t (x)}{x}=t,
$$
which we will see is a one-one map of $\Bbb R\cup \{\infty\}$
to the von~Neumann solutions.
\endremark
\example{Examples 1.1 and 1.2 revisited} As noted, the
explicit non-unique measures we listed were absolutely
continuous measures. But there are, of necessity, many pure
point measures that lead to the same moments. Indeed, the
measures $\mu_t$ associated to the von~Neumann solutions are each
pure point. As we will see, there are many other pure point
measures with the given moments. If $\nu$ is a Cantor measure,
then it can be proven that $\int \mu_t\, d\nu(t)$ will be a
singular continuous measure with the given moments. In the
indeterminate case, the class of measures solving (1.1) is always
extremely rich.
Given a set of moments $\{\gamma_n\}^\infty_{n=0}$ and $c\in
\Bbb R$, one can define a new set of moments
$$
\gamma_n (c) = \sum_{j=0}^n \binom{n}{j} c^j \gamma_{n-j}.
\tag 1.20
$$
For the Hamburger problem, the solutions of the
$\{\gamma_n\}^\infty_{n=0}$ and each
$\{\gamma_n(c)\}^\infty_{n=0}$ problem are in one-one
correspondence. If $\mu$ solves the $\{\gamma_n\}^\infty_{n=0}$
problem, then $d\rho(x-c)=d\rho_c (x)$ solves the $\{\gamma_n
(c)\}^\infty_{n=0}$ problem, and vice-versa. But translation
does not preserve the condition $\text{supp}(\rho)\subset[0,
\infty)$, so that as $c$ decreases, the set of solutions of the
Stieltjes problem shrinks. We will show that for any indeterminate
Stieltjes moment problem, there is always a $c_0$, so that for
$c>c_0$, $\{\gamma_n(c)\}^\infty_{n=0}$ is an indeterminate
Stieltjes problem. For $c0$, $f^{[N,N]}
(x)$ is monotone decreasing in $N$ and $f^{[N-1,N]}(x)$ is
monotone increasing.
\endproclaim
In terms of the language of Section~3, $f_-$ is associated to
the Friedrichs solution and $f_+$ to the Krein solution. For an
interesting application of Theorem~6, see [\loe, \simap]. In
Section~5, we will also discuss series of Hamburger and prove
the new result that the $f^{[N,N]}(z)$ Pad\'e approximants always
converge in that case.
As a by-product of the proof of Theorem~6, we will find
\proclaim{Theorem 7} Let $\{\gamma_n\}^\infty_{n=0}$ be a set
of Stieltjes moments. Then the Stieltjes problem is indeterminate
if and only if
$$
\sum_{n=0}^\infty |P_n (0)|^2 < \infty \tag 1.26
$$
and
$$
\sup_n \biggl| \frac{Q_n(0)}{P_n(0)}\biggr| < \infty . \tag 1.27
$$
\endproclaim
\proclaim{Theorem 8} Let $\{\gamma_n\}_{n=0}^\infty$ be a set
of Stieltjes moments. Then the Stieltjes problem is determinate
while the Hamburger problem is indeterminate if and only if
$$
\sum_{n=0}^\infty |Q_n (0)|^2 < \infty \tag 1.28
$$
and
$$
\lim_{n\to\infty} \biggl| \frac{Q_n(0)}{P_n(0)}\biggr| = \infty .
\tag 1.29
$$
\endproclaim
We will see that (1.26)/(1.27) are equivalent to a criterion of
Stieltjes.
Theorem~6 is proven in Section~5 as parts of Theorem~5.2,
Proposition~5.6, Proposition~5.8, and Proposition~5.11. Theorem~7
is Theorem~5.21 and Theorem~8 is Proposition~5.22.
In Section~6, which synthesizes and extends results in
Akhiezer [\akh], we will study when the closure of the
polynomials has finite codimension, and among our results proven
as part of Theorem~6.4 is
\proclaim{Theorem 9} Let $\rho$ lie in $\Cal M^H (\gamma)$ and
let $\Cal H_0$ be the closure of the set of polynomials in
$\Cal H_\rho \equiv L^2 (\Bbb R, d\rho)$. Then $\Cal H^\bot_0$
has finite dimension if and only if the Herglotz function, $\Phi$,
in the Nevanlinna parametrization of Theorem~4 is a real rational
function \rom(i.e., a ratio of real poynomials\rom). Equivalently,
if and only if the measure $\mu$ of {\rom{(1.19)}} has finite support.
\endproclaim
In fact, we will see that $\dim(\Cal H^\bot_0)$ is the degree of
the rational function $\Phi$.
One consequence of this will be the following, proven in
Appendix~B (see Theorems~B.1 and B.4).
\proclaim{Theorem 10} $\Cal M^H(\gamma)$ is a compact convex
set \rom(in the weak topology as a subset of the dual of $C(\Bbb R
\cup\{\infty\})$\rom). Its extreme points are dense in the set.
\endproclaim
\remark{Remark} Do not confuse this density with the fact that
the Krein-Millman theorem says that the convex combinations of
the extreme points are dense. Here we have density without
convex combinations, to be compared with the fact that in many
other cases (e.g., probability measures on a compact Hausdorff
space), the extreme points are closed.
\endremark
\vskip 0.1in
Here is a sketch of the contents of the rest of this paper. In
Section~2, we review von Neumann's theory of self-adjoint
extensions and use it to prove the uniqueness result (Theorem~2)
in the Hamburger case. In Section~3, we review the
Birman-Krein-Vishik theory of semi-bounded self-adjoint
extensions and use it to prove the uniqueness result (Theorem~3)
in the Stieltjes case. In Section~4, we turn to a detailed
analysis of the polynomials $P$ and $Q$ in terms of transfer
matrices for the difference equation associated to the Jacobi
matrix and prove Theorems~3, 4, and 5. In Section~5, we discuss
Pad\'e approximations and prove Theorems~6, 7, and 8. In Section~6,
we discuss solutions, $\rho$, where the closure of the polynomials
has finite codimension in $L^2 (\Bbb R, d\rho)$. Appendix~A is
devoted to a variety of explicit formulae in terms of
determinants of moments, and Appendix~B to the structure of
$\Cal M^H(\gamma)$ as a compact convex set. Appendix~C summarizes
notation and some constructions.
\vskip 0.3in
\flushpar{\bf \S 2. The Hamburger Moment Problem as a
Self-Adjointness Problem}
\vskip 0.1in
Let us begin this section with a brief review of the von
Neumann theory of self-adjoint extensions. For further details,
see [\ag, \ds, \rsI, \rsII]. We start out with a densely defined
operator $A$ on a Hilbert space $\Cal H$, that is, $D(A)$ is a
dense subset of $\Cal H$ and $A:D(A)\to\Cal H$ a linear map. We
will often consider its graph $\Gamma(A)\subset\Cal H \times
\Cal H$ given by $\Gamma(A)=\{(\varphi, A\varphi)\mid\varphi\in
D(A)\}$. Given operators $A,B$, we write $A\subset B$ and say $B$
is an {\it{extension}} of $A$ if and only if $\Gamma(A)\subset
\Gamma(B)$.
One defines a new operator $A^*$ as follows: $\eta\in D(A^*)$
if and only if there is a $\psi\in\Cal H$ so that for all $\varphi
\in D(A)$, $\langle \psi,\varphi\rangle = \langle\eta, A\varphi
\rangle$. We set $A^*\eta =\psi$. In essence, $A^*$, called
the {\it{adjoint}} of $A$, is the maximal object obeying
$$
\langle A^*\eta, \varphi\rangle = \langle \eta, A\varphi\rangle
\tag 2.1
$$
for all $\eta\in D(A^*)$, $\varphi\in D(A)$.
An operator is called {\it{closed}} if and only if $\Gamma(A)$ is
a closed subset of $\Cal H\times\Cal H$ and it is called
{\it{closable}} if and only if $\overline{\Gamma(A)}$ is the
graph of an operator, in which case we define $\bar A$, the
{\it{closure}} of $A$, by $\Gamma(\bar A)=\overline{\Gamma(A)}$.
Thus, $D(\bar A)=\{\varphi\in\Cal H\mid\exists \varphi_n\in D(A),
\text{ so that } \varphi_n\to\varphi \text{ and } A\varphi_n
\text{ is Cauchy}\}$ and one sets $A\varphi = \lim A\varphi_n$.
Adjoints are easily seen to be related to these notions: $A^*$
is always closed; indeed,
$$
\Gamma (A^*)=\{(A\varphi, -\varphi)\in \Cal H \times \Cal H
\mid \varphi\in D(A)\}^\bot . \tag 2.2
$$
$A^*$ is densely defined if and only if $A$ is closable, in
which case (by (2.2)), $\bar A = (A^*)^*$.
An operator is called {\it{symmetric}} if $A\subset A^*$
(equivalently, if $\langle\varphi, A\psi\rangle = \langle A
\varphi, \psi\rangle$ for all $\varphi,\psi\in D(A)$),
{\it{self-adjoint}} if $A=A^*$, and {\it{essentially self-adjoint}}
if $\bar A$ is self-adjoint (equivalently, if $A^*$ is self-adjoint).
Notice that symmetric operators are always closable since $D(A^*)
\supset D(A)$ is dense. Note also that if $A$ is symmetric, then
$A^*$ is symmetric if and only if $A$ is essentially self-adjoint.
Von Neumann's theory solves the following fundamental question:
Given a symmetric operator $A$, when does it have self-adjoint
extensions, are they unique, and how can they be described? If
$B$ is a self-adjoint extension of $A$, then $B$ is closed, so
$\bar A \subset B$. Thus, looking at self-adjoint extensions
of $A$ is the same as looking at self-adjoint extensions of
$\bar A$. So for now, we will suppose $A$ is a closed symmetric
operator.
Define $\Cal K_\pm = \ker(A^* \mp i)$, that is, $\Cal K_+ =
\{\varphi\in\Cal H \mid A^*\varphi = i\varphi\}$. They are called
the {\it deficiency subspaces}. Using $\|(A\pm i)\varphi\|^2 =
\|A\varphi\|^2 + \|\varphi\|^2$, it is easy to see that if $A$
is closed, then $\text{Ran}(A\pm i)$ are closed, and since
$\ker(A^*\mp i)=\text{Ran}(A\pm i)^\bot$, we have
$$
\text{Ran}(A\pm i) = \Cal K^\bot_\pm . \tag 2.3
$$
Let $d_\pm = \dim(\Cal K_\pm)$, the {\it{deficiency indices}}
of $A$.
Place the graph norm on $D(A^*)$, that is,
$\|\varphi\|^2_{A^*} = \langle \varphi, \varphi\rangle +
\langle A^*\varphi, A^*\varphi\rangle$. This norm comes from an
inner product, $\langle \varphi,\psi\rangle_{A^*} = \langle
\varphi,\psi\rangle + \langle A^*\varphi, A^*\psi\rangle$.
\proclaim{Proposition 2.1} Let $A$ be a closed symmetric operator.
Then
$$
D(A^*) = D(A) \oplus \Cal K_+ \oplus \Cal K_- , \tag 2.4
$$
where $\oplus$ means orthogonal direct sum in the $\langle \,
\cdot\, , \,\cdot\,\rangle_{A^*}$ inner product.
\endproclaim
\demo{Proof} If $\varphi\in\Cal K_+$ and $\psi\in\Cal K_-$,
then $\langle\varphi, \psi\rangle_{A^*} = \langle\varphi,\psi
\rangle + \langle i\varphi, -i\psi\rangle =0$ so $\Cal K_+
\bot_{A^*} \Cal K_-$. If $\varphi\in D(A)$ and $\psi\in
\Cal K_\pm$, then $\langle A^*\varphi, A^*\psi\rangle =
\langle A\varphi, \pm i\psi\rangle = \langle \varphi, \pm i
A^* \psi\rangle = -\langle \varphi, \psi\rangle$, so $D(A)
\bot_{A^*} \Cal K_+ \oplus \Cal K_-$.
Let $\eta\in D(A^*)$. By (2.3), $\text{Ran}(A+i)+ \Cal K_+ =
\Cal H$ so we can find $\varphi\in D(A)$ and $\psi\in \Cal K_+$,
so $(A^* +i)\eta = (A+i)\varphi + 2i\psi$. But then $(A^* + i)
[\eta - \varphi -\psi] =0$, so $\eta - \varphi - \psi\in\Cal K_-$,
that is, $\eta\in D(A) + \Cal K_+ + \Cal K_-$. \qed
\enddemo
\proclaim{Corollary 2.2} \rom{(i)} Let $A$ be a closed symmetric
operator. Then $A$ is self-adjoint if and only if $d_+ = d_- =0$.
\rom{(ii)} Let $A$ be a symmetric operator. Then $A$ is essentially
self-adjoint if and only if $d_+ = d_- =0$.
\endproclaim
If $A\subset B$, then $B^* \subset A^*$, so if $B$ is symmetric,
then $A\subset B\subset B^* \subset A^*$. Thus, to look for
symmetric extensions of $A$, we need only look for operators $B$
with $A\subset B\subset A^*$, that is, for restriction of $A^*$ to
$D(B)$'s with $D(A)\subset D(B)$. By Proposition~2.1, every
such $D(B)$ has the form $D(A)+S$ with $S\subset\Cal K_+ +
\Cal K_-$. On $D(A^*)\times D(A^*)$, define the sesquilinear form
(sesquilinear means linear in the second factor and anti-linear
in the first)
$$
Q(\varphi,\psi) = \langle\varphi, A^*\psi\rangle_{\Cal H} -
\langle A^* \varphi,\psi\rangle_{\Cal H}.
$$
\proclaim{Proposition 2.3} Let $A$ be a closed symmetric
operator. Then
\roster
\item"\rom{(i)}" The operators $B$ with $A\subset B\subset A^*$
are in one-one correspondence with subspaces $S$ of $\Cal K_+
+\Cal K_-$ under $D(B)=D(A)+S$.
\item"\rom{(ii)}" $B$ is symmetric if and only if $Q\restriction
D(B)\times D(B)=0$.
\item"\rom{(iii)}" $B$ is symmetric if and only if $Q\restriction
S\times S=0$.
\item"\rom{(iv)}" $B$ is closed if and only if $S$ is closed in
$\Cal K_+ \oplus \Cal K_-$ in $D(A^*)$ norm.
\item"\rom{(v)}" $\varphi\in D(A^*)$ lies in $D(B^*)$ if and
only if $Q(\varphi,\psi)=0$ for all $\psi\in D(B)$.
\item"\rom{(vi)}" Let $J:\Cal K_+ \oplus \Cal K_- \to \Cal K_+
\oplus \Cal K_-$ by $J(\varphi, \psi) = (\varphi, -\psi)$.
If $D(B) = D(A)+S$, then $D(B^*)=D(A)+J[S]^\bot$, where $^\bot$
is in $\Cal K_+ \oplus \Cal K_-$ in the $D(A^*)$ norm.
\item"\rom{(vii)}" $\Cal K_+(B) = \Cal K_+ \cap S^\bot$,
$\Cal K_-(B) = \Cal K_- \cap S^\bot$ \rom(with $^\bot$ in the
$\langle \, , \, \rangle_{A^*}$ inner product\rom).
\endroster
\endproclaim
\demo{Proof} We have already seen (i) holds and (ii) is obvious.
(iii) holds since if $\varphi\in D(A)$ and $\psi\in D(A^*)$, then
$Q(\varphi, \psi)=0$ by definition of $A^*$ and symmetry of $A$.
Thus, if $\varphi_1, \varphi_2 \in D(A)$ and $\psi_1, \psi_2 \in
S$, then $Q(\varphi_1 + \psi_1, \varphi_2 + \psi_2) = Q(\psi_1,
\psi_2)$. (iv) is immediate if one notes that $\Gamma (B) =
\Gamma(A)\oplus \{(\varphi, A^*\varphi)\mid \varphi\in S\}$
with $\oplus$ in $\Cal H\times\Cal H$ norm and that the
$\Cal H\times\Cal H$ norm on $\{(\varphi, A^*\varphi)\mid
\varphi\in S\}$ is just the $D(A^*)$ norm. (v) follows from the
definition of adjoint.
To prove (vi), let $\eta = (\eta_1, \eta_2)$, $\varphi =
(\varphi_1, \varphi_2)\in \Cal K_+ \oplus\Cal K_-$. Then, direct
calculations show that
$$\align
Q(\eta, \varphi) &= 2i [\langle\eta_1, \varphi_1\rangle_{\Cal H}
- \langle \eta_2, \varphi_2\rangle_{\Cal H}] \tag 2.5 \\
\langle \eta, \varphi\rangle_{A^*} &= 2[\langle \eta_1,
\varphi_1\rangle_{\Cal H} + \langle\eta_2, \varphi_2
\rangle_{\Cal H}]. \tag 2.6
\endalign
$$
Thus, $Q(\eta, \varphi)=0$ if and only if $\eta\bot J\varphi$.
(v) thus implies (vi).
To prove (vii), note that by (vi), $\Cal K_\pm(B) = \Cal K_\pm
\cap D(B^*)=\Cal K_\pm\cap J[S]^\bot =\Cal K_\pm \cap S^\bot$
since $\varphi\in\Cal K_\pm$ lies in $J[S]^\bot$ if and only
if $J\varphi$ lies in $S^\bot$ if and only if $\varphi$ lies
in $S^\bot$ by $J\varphi = \pm\varphi$ if $\varphi\in\Cal K_\pm$.
\qed
\enddemo
With these preliminaries out of the way, we can prove
von~Neumann's classification theorem:
\proclaim{Theorem 2.4} Let $A$ be a closed symmetric operator.
The closed symmetric extensions $B$ of $A$ are in one-one
correspondence with partial isometries of $\Cal K_+$ into
$\Cal K_-$, that is, maps $U:\Cal K_+ \to \Cal K_-$ for which
there are closed subspaces $\Cal U_+\subset\Cal K_+$ and $\Cal U_-
\subset\Cal K_-$ so that $U$ is a unitary map of $\Cal U_+ \to
\Cal U_-$ and $U\equiv 0$ on $\Cal U^\bot_+$. If $B$
corresponds to $U$, then $D(B) =D(A) + \{\varphi+U\varphi\mid
\varphi\in \Cal U_+\}$. Moreover, $\Cal K_\pm(B) = \Cal K_\pm
\cap \Cal U^\bot_\pm$ \rom($\Cal H$ orthogonal complement\rom).
In particular, $A$ has self-adjoint extensions if and only if
$d_+ =d_-$ and then the self-adjoint extensions are in one-one
correspondence with unitary maps $U$ from $\Cal K_+$ to $\Cal K_-$.
\endproclaim
\demo{Proof} Let $B$ be a symmetric extension of $A$. Let
$\varphi\in D(B)\cap (\Cal K_+ \oplus\Cal K_-)$, say $\varphi
= (\varphi_1 + \varphi_2)$. By (2.5), $Q(\varphi,\varphi) =
2i(\|\varphi_1\|^2 - \|\varphi_2\|^2)$, so $B$ symmetric
implies that $\|\varphi_1\| = \|\varphi_2\|$, so $D(B)\cap
(\Cal K_+ \oplus\Cal K_-)=\{\text{some }(\varphi_1, \varphi_2)\}$,
where $\|\varphi_1\| = \| \varphi_2\|$. Since $D(B)$ is a subspace,
one cannot have $(\varphi_1, \varphi_2)$ and $(\varphi_1,
\varphi'_2)$ with $\varphi_2 \neq \varphi'_2$ (for then
($0, \varphi_2 - \varphi'_2)\in D(B)$ so $\| \varphi_2
-\varphi'_2\|=0$). Thus, $D(B)\cap (\Cal K_+ \oplus\Cal K_-)$ is
the graph of a partial isometry. On the other hand, if $U$ is a
partial isometry, $Q(\varphi +U\varphi, \psi + U\psi) = 2i
(\langle \varphi, \psi\rangle - \langle U\varphi, U\psi\rangle)
=0$ so each $U$ does yield a symmetric extension.
That $\Cal K_\pm(B)=\Cal K_\pm \cap \Cal U^\bot_\pm$ follows
from (vii) of the last proposition if we note that if $\varphi
\in\Cal K_+$, then $\varphi\perp_{A^*}\{(\psi+U\psi) \mid
\psi\in\Cal U_+\}$ if and only if $\varphi\perp_{A^*}\{\psi\mid
\psi\in\Cal U_+\}$ if and only if $\varphi\perp_\Cal H \Cal U_+$
(by (2.6)).
Thus $B$ is self-adjoint if and only if $U$ is a unitary from
$\Cal K_+$ to $\Cal K_-$, which completes the proof. \qed
\enddemo
Recall that a map $T$ is {\it anti-linear} if $T(a\varphi +
b\psi)=\bar a T(\varphi) + \bar b T(\psi)$ for $a,b\in
\Bbb C$, $\varphi,\psi\in\Cal H$; that $T$ is {\it anti-unitary}
if it is anti-linear, a bijection, and norm-preserving; and that
a {\it complex conjugation} is an anti-unitary map whose
sequence is $1$.
\proclaim{Corollary 2.5} Let $A$ be a symmetric operator. Suppose
there exists a complex conjugation $C:\Cal H\to\Cal H$ so that
$C:D(A)\to D(A)$ and $CA\varphi = AC\varphi$ for all $\varphi
\in D(A)$. Then $A$ has self-adjoint extensions. If $d_+ =1$,
every self-adjoint extension $B$ is real, that is, obeys $C:
D(B) \to D(B)$ and $CB\varphi=BC\varphi$ for all $\varphi
\in D(B)$. If $d_+ \geq 2$, there are non-real self-adjoint
extensions.
\endproclaim
\demo{Proof} $C$ is an anti-unitary from $\Cal K_+$ to
$\Cal K_-$ so $d_+ = d_-$.
We claim that if $B$ is self-adjoint and is associated to $U:
\Cal K_+\to\Cal K_-$, then $B$ is real if and only if $CUCU=1$
for it is easy to see that $C:D(A^*)\to D(A^*)$ and $CA^*\varphi
= A^*C\varphi$, so $B$ is real if and only if $C$ maps $D(B)$ to
itself. Thus, $C(\varphi +U\varphi)$ must be of the form $\psi +
U\psi$. Since $\psi$ must be $CU\varphi$, this happens if and only
if $C\varphi =UCU\varphi$, that is, $CUCU\varphi=\varphi$. This
proves our claim.
If $d_+=1$ and $\varphi\in\Cal K_+$, $CU\varphi=e^{i\theta}
\varphi$ for some $\theta$. Then $(CUCU)\varphi = Ce^{i\theta}
U\varphi=e^{-i\theta}CU\varphi=\varphi$ showing that every
self-adjoint extension is real.
If $d_+ \geq 2$, pick $\varphi,\psi\in\Cal K_+$ with $\varphi
\perp\psi$ and let $U\varphi =C\psi$, $U\psi=iC\varphi$.
Then $CUCU\varphi =CUCC\psi=CU\psi = CiC\psi=-i\varphi$,
so $CUCU\not\equiv 1$. Thus, the $B$ associated to this $U$
will not be real. \qed
\enddemo
Next we analyze a special situation that will always apply to
indeterminate Hamburger moment problems.
\proclaim{Theorem 2.6} Suppose that $A$ is a closed symmetric
operator so that there exists a complex conjugation under which
$A$ is real. Suppose that $d_+ =1$ and that $\ker(A)=\{0\}$,
$\dim\ker(A^*)=1$. Pick $\varphi\in\ker(A^*)$, $C\varphi=
\varphi$, and $\eta\in D(A^*)$, not in $D(A)+\ker(A^*)$. Then
$\langle \varphi, A^*\eta\rangle\neq 0$ and $\psi = \{\eta -
[\langle\eta, A^*\eta\rangle/\langle\varphi, A^*\eta\rangle]
\varphi\}/ \langle\varphi, A^*\eta\rangle$ are such that in
$\varphi,\psi$ basis, $\langle \,\cdot\, , A^*\,\cdot\,\rangle$
has the form
$$
\langle\,\cdot\, , A^*\,\cdot\,\rangle =
\pmatrix 0 & 1 \\ 0 & 0 \endpmatrix . \tag 2.7
$$
The self-adjoint extensions, $B_t$, can be labelled by a
real number or $\infty$ where
$$\alignat2
D(B_t) &= D(A) + \{\alpha(t\varphi + \psi) \mid \alpha\in
\Bbb C\} \qquad && t\in\Bbb R \\
&= D(A) + \{\alpha\varphi \mid \alpha\in\Bbb C \} \qquad &&
t=\infty.
\endalignat
$$
The operators $B_t$ are independent of which real $\psi$ in
$D(A^*)\backslash D(A)$ is chosen so that {\rom{(2.7)}} holds.
\endproclaim
\demo{Proof} If $\langle\varphi, A^*\eta\rangle =0$, then the
$\langle\,\cdot\, ,A^*\,\cdot\,\rangle$ matrix with basis
$\varphi, \eta$ would have the form $\left( \smallmatrix a & 0 \\
0 & 0 \endsmallmatrix\right)$ with $a\in\Bbb R$, in which case
$Q\equiv 0$ on the span of $\varphi$ and $\eta$, which is
incompatible with the fact that $d_+=1$. Thus, $\langle\varphi,
A^*\eta\rangle \neq 0$ and (2.7) holds by an elementary calculation.
Once (2.7) holds, it is easy to see that the subspaces $S$ of
$D(A^*)$ with $D(A)\subset S$, $\dim (S/D(A))=1$, and $Q
\restriction S\times S=0$ are precisely the $D(B_t)$.
If $\tilde\psi$ is a second $\psi$ for which (2.7) holds,
then $Q(\psi -\tilde\psi, \rho)=0$ for all $\rho\in D(A^*)$,
so $\psi-\tilde\psi\in D(A)$ and the $D(B_t)$'s are the same
for $\psi$ and $\tilde\psi$. \qed
\enddemo
\remark{Remarks} 1. It is easy to see that if $\ker(A)=\{0\}$,
then $\dim\ker(A^*)\leq d_+$, so if $d_+=1$ and $\ker(A)=\{0\}$,
$\dim\ker(A^*)$ is either $0$ or $1$. The example of $A =
-\frac{d^2}{dx^2}$ on $L^2 (0,\infty)$ with $D(A)=C^\infty_0
(0,\infty)$ shows that it can happen that $A$ is real, $d_+=1$,
but $\dim\ker(A^*)=0$.
2. One example where this formalism is natural is if $A\geq
\alpha >0$ (for $\alpha\in\Bbb R$), in which case $\dim\ker(A^*)
=d_+$. One takes $\eta= A^{-1}_F \varphi$, where $A_F$ is the
Friedrichs extension. We will discuss this further in Section~3.
A second example, as we will see, is the indeterminate moment
problem, in which case one can take $\varphi = \pi (0)$,
$\psi = \xi (0)$.
\endremark
\vskip 0.1in
We can now turn to the analysis of the symmetric operator $A$ on
$\Cal H^{(\gamma)}$ described in Section~1. In terms of the
explicit basis $P_n (X)$ in $\Cal H^{(\gamma)}$, we know $A$
has the Jacobi matrix form (1.16). We will first explicitly
describe $A^*$ and the form $Q$. Given any sequence $s=
(s_0, s_1, \dots)$, we define a new sequence $\Cal F(s)$ (think
of $\Cal F$ for ``formal adjoint") by
$$
\Cal F(s)_n = \cases b_0 s_0 + a_0 s_1 & \text{if } n=0 \\
a_{n-1} s_{n-1} + b_ns_n + a_n s_{n+1} & \text{if } n\geq 1.
\endcases
$$
Given any two sequences, $s_n$ and $t_n$, we define their
Wronskian,
$$
W(s,t)(n) = a_n(s_{n+1}t_n - s_n t_{n+1}). \tag 2.8
$$
The Wronskian has the following standard property if
$$
a_n s_{n+1} + c_n s_n + a_{n-1} s_{n-1} = 0 \tag 2.9
$$
and
$$
a_n t_{n+1} + d_n t_n + a_{n-1} t_{n-1} =0. \tag 2.10
$$
Then multiplying (2.9) by $t_n$ and subtracting (2.10)
multiplied by $s_n$, we see that
$$
W(s,t)(n) - W(s,t)(n-1) = (d_n -c_n)s_n t_n. \tag 2.11
$$
In particular, if $d=c$, then $W$ is constant.
\proclaim{Theorem 2.7} Let $A$ be the Jacobi matrix {\rom{(1.16)}}
with $D(A) =\{s\mid s_n =0$ for $n$ sufficiently large$\}$. Then
$$
D(A^*) =\{s\in\ell_2\mid \Cal F(s)\in\ell_2\} \tag 2.12
$$
with
$$
A^* s=\Cal F(s) \tag 2.13
$$
for $s\in D(A^*)$. Moreover, if $s,t\in D(A^*)$, then
$$
\lim_{n\to\infty} W(\bar s,t)(n) = \langle A^* s,t\rangle
-\langle s, A^* t\rangle. \tag 2.14
$$
\endproclaim
\demo{Proof} Since the matrix (1.16) is a real-symmetric matrix,
it is easy to see that for any $t\in D(A)$, and any sequences
$$
\langle s,At\rangle = \langle \Cal F(s),t\rangle. \tag 2.15
$$
Since $t$ and $At$ are finite sequences, the sum in $\langle
\, \cdot\, , \, \cdot\,\rangle$ makes sense even if $s$ or
$\Cal F(s)$ are not in $\ell_2$. Since $D(A)$ is dense in $\ell_2$,
(2.15) says that $s\in\ell_2$ lies in $D(A^*)$ precisely if
$\Cal F(s)\in \ell_2$ and then (2.13) holds. That proves the
first part of the theorem.
For the second, we perform a calculation identical to that
leading to (2.11):
$$\align
\sum_{n=0}^N \left[\, \overline{\Cal F s_n}\, t_n -
\overline{s_n}\, (\Cal Ft_n)\right] &= W(\bar s,t)(0) +
\sum_{n=1}^N W(\bar s,t)(n)-W(\bar s,t)(n-1) \\
&= W(\bar s,t)(N). \tag 2.16
\endalign
$$
If $s,t\in D(A^*)$, then the left side of (2.16) converges to
$\langle A^* s,t\rangle - \langle s, A^* t\rangle$ as $N\to
\infty$, so (2.14) holds. \qed
\enddemo
\proclaim{Lemma 2.8} If $\varphi\in D(A^*)$, $(A^*-z)\varphi
=0$ for some $z$ and $\varphi_0 =0$, then $\varphi\equiv 0$.
\endproclaim
\demo{Proof} Essentially, this is so because $(A^*-z)s =0$ is a
second-order difference equation with $s_{-1}\equiv 0$, so
solutions are determined by $s_0$. Explicitly, suppose $A^* s=zs$
and $s_0 =0$. Then
$$\alignat2
s_{n+1} &= a^{-1}_n [(-b_n +z)s_n - a_{n-1} s_{n-1}], \qquad
&& n\geq 1 \\
&= a^{-1}_0 [(-b_0 +z) s_0], \qquad && n=0.
\endalignat
$$
By a simple induction, $s\equiv 0$. \qed
\enddemo
\proclaim{Corollary 2.9} The operator $A$ associated to a Hamburger
moment problem always has either deficiency indices $(1,1)$ or
deficiency indices $(0,0)$.
\endproclaim
\demo{Proof} We have already seen that $d_+ =d_-$ so it suffices
to show that $\ker(A^* -i)$ has dimension at most $1$. By Lemma~2.8,
the map from $\ker(A^* -i)$ to $\Bbb C$ that takes $s\mapsto s_0$
is one-one, so $\ker(A^* -i)$ is of dimension $0$ or $1$. \qed
\enddemo
We are now ready to prove the more subtle half of the Hamburger
part of Theorem~2.
\proclaim{Theorem 2.10 (one quarter of Theorem~2)} Let $A$ be
essentially self-adjoint. Then the Hamburger moment problem has
a unique solution.
\endproclaim
\demo{Proof} Pick $z$ with $\text{Im}\, z>0$ and a measure $\rho$
obeying (1.1). Since $A$ is essentially self-adjoint, $(A-i)
[D(A)]$ is dense in $\ell^2$, and by an identical argument,
$(A-z)[D(A)]$ is dense in $\ell^2$. Thus, there exists a
sequence $R^{(z)}_n (X)$ of polynomials so that
$$
\|(X-z)R^{(z)}_n (X)-1\| \to 0
$$
in $\Cal H^{(\gamma)}$, and thus
$$
\int [(x-z) R^{(z)}_n (x) -1]^2\, d\rho(x) \to 0
$$
since $d\rho(x)$ realizes $\Cal H^{(\gamma)}$ on polynomials.
Now $\frac1{x-z}$ is bounded for $x\in\Bbb R$ since $\text{Im}
\, z>0$. Thus
$$
\int \biggl|R^{(z)}_n (x) - \frac{1}{x-z}\biggr|^2\,
d\rho(x)\to 0.
$$
It follows that
$$
G_\rho (x) \equiv \int\frac{d\rho(x)}{x-z} = \lim_{n\to\infty} \int
R^{(z)}_n(x)\, d\rho(x)
$$
is independent of $\rho$. Since $G_\rho(x)$ determines $\rho$
(because $\rho(\{a\})=\lim_{\varepsilon\downarrow 0} \varepsilon \,
\text{Im}\, G_\rho (a+i\varepsilon)$ and $\rho ((a,b)) + \rho
([a,b])=\frac2{\pi} \lim_{\varepsilon\downarrow 0} \int_a^b
\text{Im}\, G_\rho (y+i\varepsilon)\, dy$), all $\rho$'s must
be the same. \qed
\enddemo
\proclaim{Theorem 2.11} Let $A$ be a Jacobi matrix with $D(A)$
the sequences of finite support. Suppose $A$ is not essentially
self-adjoint and $B,F$ are distinct self-adjoint extensions of
$A$. Then
$$
\biggl( \delta_0, \frac{1}{B-i}\, \delta_0\biggr) \neq
\biggl( \delta_0, \frac{1}{F-i}\, \delta_0\biggr). \tag 2.17
$$
\endproclaim
\remark{Remarks} 1. The proof shows for any $z\in\Bbb C\backslash
\Bbb R$, $(\delta_0, (B-z)^{-1}\delta_0) \neq (\delta_0 (F-z)^{-1}
\delta_0)$. It also works for $z\in\Bbb R$ as long as $z$ is in
the resolvent set for both $B$ and $F$.
2. In Section~4, we will have a lot more to say about the
possible values of $(\delta_0, \frac1{B-z}\delta_0)$ as $B$
runs through the self-adjoint extensions of $A$.
\endremark
\demo{Proof} We first claim that $\delta_0 \notin\text{Ran}(A-i)$.
For suppose on the contrary that there exists $\eta\in D(A)$ and
$\delta_0 = (A-i)\eta$ and that $(A^* +i)\varphi =0$. Then
$(\delta_0, \varphi) = ((A-i)\eta, \varphi) = (\eta, (A^* +i)
\varphi)=0$. By Lemma~2.8, $\varphi\equiv 0$. Thus, if $\delta_0
\in\text{Ran}(A-i)$, then $\ker(A^* +i)=\{0\}$, so $A$ is
essentially self-adjoint. By hypothesis, this is false, so
$\delta_0 \notin \text{Ran}(A-i)$.
Thus, $(B-i)^{-1}\delta_0$ and $(F-i)^{-1}\delta_0$ are in
$D(A^*)\backslash D(A)$. Since $\dim(D(B)/D(A)) =\dim(D(F)/D(A))
=1$, if these vectors were equal, $D(B)$ would equal $D(F)$, so
$B=A^*\restriction D(B)$ would equal $F=A^*\restriction D(F)$.
Thus, $(B-i)^{-1}\delta_0 \neq (F-i)^{-1}\delta_0$.
Let $\eta = (B-i)^{-1}\delta_0 - (F-i)^{-1}\delta_0$. Then
$$
(A^* -i)\eta = (A^* -i)(B-i)^{-1}\delta_0 - (A^* -i)
(F-i)^{-1}\delta_0 = \delta_0 - \delta_0 =0 ,
$$
so $\eta\neq 0$ implies $\eta_0 \neq 0$ by Lemma~2.8. Thus,
$(\delta_0, \eta) = (\delta_0, (B-i)^{-1}\delta_0) -
(\delta_0, (F-i)^{-1}\delta_0)\neq 0$. \qed
\enddemo
As a corollary of Theorem~2.11, we have
\proclaim{Theorem 2.12 (two quarters of Theorem~2)} A Hamburger
moment problem for which $A$ is not essentially self-adjoint
is indeterminate. A Stieltjes moment problem for which
$A$ has multiple non-negative self-adjoint extensions is
indeterminate.
\endproclaim
\demo{Proof} Theorem~2.11 implies that distinct self-adjoint
extensions lead to distinct spectral measures since
$$
(\delta_0, (B-i)^{-1}\delta_0) = \int \frac{d\mu^B (x)}{x-i} \, ,
$$
where $\mu^B$ is the solution to (1.1) associated to $B$.
Positive self-adjoint extensions yield solutions of the
Stieltjes moment problem. \qed
\enddemo
With this machinery available, we can prove the converse of
Proposition~1.6:
\proclaim{Theorem 2.13} Let $\{\gamma_n\}^\infty_{n=0}$ be a set
of moments for which the Stieltjes problem has solutions. Let
$$\alignat2
\Gamma_{2m} &= \gamma_m, \qquad && m=0,1,\dots \tag 2.18a \\
\Gamma_{2m+1} &= 0, \qquad && m=0,1,\dots. \tag 2.18b
\endalignat
$$
Then $\{\Gamma_j\}^\infty_{j=0}$ is a determinate Hamburger
problem if and only if $\{\gamma_n\}^\infty_{n=0}$ is a
determinate Stieltjes problem.
\endproclaim
\demo{Proof} The proof of Proposition~1.6 shows there is a
one-one correspondence between solutions of the Stieltjes problem
for $\gamma$ and those solutions $d\rho$ of the Hamburger
problem for $\Gamma$ with $d\rho(-x)=d\rho(x)$. We will call
such solutions symmetric. Thus it suffices to show that if the
Hamburger problem for $\Gamma$ is indeterminate, then there are
multiple symmetric solutions.
Let $U$ map $\Bbb C(X)$ to itself by $U[P(X)]=P(-X)$. By
$\Gamma_{2m+1}=0$, $U$ is unitary, and so extends to a unitary
map from $\Cal H^{(\Gamma)}$ to itself. Clearly, $U$ maps $D(A)
=\Bbb C(X)$ to itself and $U\!AU^{-1}=-A$.
Thus, $U$ maps $D(A^*)$ to itself and $U\!A^*U^{-1}=-A^*$. This
means that $U$ maps $\Cal K_+$ to $\Cal K_-$.
Let $C$ be the complex conjugation under which $A$ is real. Then
$UC=CU$ is an anti-unitary map of $\Cal K_+$ to itself. So if
$\varphi$ is a unit vector in $\Cal K_+$, then $UC\varphi =
e^{i\theta}\varphi$ for some real $\theta$. Thus, $UC
(e^{i\theta/2}\varphi) = e^{-i\theta/2}UC\varphi = e^{i\theta/2}
\varphi$ so there exists $\psi\neq 0$ in $\Cal K_+$ with
$UC\psi=\psi$.
In particular, since $UC=CU$, $U(\psi\pm C\psi) = \pm(\psi \pm
C\psi)$ and therefore, if $B=A^*\restriction D(B)$ and $F=
A^*\restriction D(F)$, where $D(B)=D(A) + [\psi + C\psi]$ and
$D(F)=D(A)+[\psi - C\psi]$, then $U$ leaves both $D(B)$ and
$D(F)$ invariant, and so $B$ and $F$ are distinct extensions
with $U\!BU^{-1}=-B$ and $U\!FU^{-1}=-F$. Their spectral measures
thus yield distinct symmetric extensions. \qed
\enddemo
\remark{Remark} Since $U(\psi + e^{i\theta}C\psi)=e^{i\theta}
(\psi + e^{-i\theta}C\psi)$, we see that if $\theta\neq 0,\pi$,
then $U$ does not leave $D(B_\theta)$ invariant. Thus among
von~Neumann solutions of the $\Gamma$ problem, exactly two are
symmetric.
\endremark
As a consequence of our identification of $A^*$, we can prove
one small part of Theorem~3. Recall that $\pi(x_0)$ is the
sequence $\{P_n(x_0)\}^\infty_{n=0}$ and $\xi(x_0)$ is the
sequence $\{Q_n(x_0)\}^\infty_{n=i}$.
\proclaim{Theorem 2.14} Let $\{\gamma_n\}^\infty_{n=0}$ be a set
of Hamburger moments. If either of the following holds, then
the Hamburger moment problem is indeterminate.
\roster
\item"\rom{(i)}" For some $x_0\in\Bbb R$, $\pi(x_0)$ and
$\frac{\partial\pi}{\partial x} (x_0)$ lie in $\ell_2$.
\item"\rom{(ii)}" For some $x_0\in\Bbb R$, $\xi(x_0)$ and
$\frac{\partial\xi}{\partial x}(x_0)$ lie in $\ell_2$.
\endroster
\endproclaim
\remark{Remark} Since $P_n(z)$ is a polynomial, $\frac{\partial\pi}
{\partial x}$ makes sense as a sequence.
\endremark
\demo{Proof} In terms of the formal adjoint $\Cal F$, we have
$$\align
\Cal F(\pi(x_0)) &= x_0 \pi(x_0) \tag 2.19 \\
\Cal F \bigg(\frac{\partial\pi}{\partial x}\, (x_0) \biggr)
&= \pi(x_0) + x_0 \, \frac{\partial \pi}{\partial x}\,
(x_0) \tag 2.20 \\
\Cal F (\xi(x_0)) &= \delta_0 + x_0 \xi(x_0) \tag 2.21 \\
\Cal F \biggl( \frac{\partial\xi}{\partial x}\, (x_0)\biggr)
&= \xi(x_0) + x_0 \, \frac{\partial\xi}{\partial x_0}\, .
\tag 2.22
\endalign
$$
Thus, if (i) holds, we conclude $\pi$, $\frac{\partial\pi}
{\partial x}$ are in $D(A^*)$ and
$$
(A^* -x_0) \pi(x_0)=0, \qquad \left. (A^* - x_0) \frac{\partial\pi}
{\partial x}\right|_{x=x_0} = \pi(x_0).
$$
If $A^*$ were self-adjoint, then
$$
\| \pi(x_0)\|^2 = \bigg\langle \left. \pi, (A^*-x_0)
\frac{\partial\pi}{\partial x}\right|_{x=x_0}\bigg\rangle =
\biggl\langle \left. (A^* - x_0) \pi, \frac{\partial\pi}
{\partial x}\right|_{x=x_0} \bigg\rangle =0,
$$
which is impossible since $P_0 (x_0) =1$. Thus, $A^*$ is not
self-adjoint and the problem is indeterminate.
If (ii) holds, $\xi, \frac{\partial\xi}{\partial x}$ are in
$D(A^*)$ and
$$
(A^* - x_0) \xi(x_0) = \delta_0, \qquad \left. (A^* - x_0)
\frac{\partial\xi}{\partial x}\right|_{x=x_0} = \xi(x_0).
$$
If $A^*$ were self-adjoint, then
$$\align
\|\xi(x_0)\|^2 &= \bigg\langle \left. \xi(x_0), (A^* - x_0)
\frac{\partial\xi}{\partial x}\right|_{x=x_0} \bigg\rangle \\
&= \bigg\langle\left. (A^* - x_0) \xi(x_0), \frac{\partial \xi}
{\partial x}\right|_{x=x_0} \bigg\rangle \\
&= \left. \frac{\partial Q_0(x)}{\partial x} \right|_{x=x_0}
=0
\endalign
$$
since $Q_0(x_0)\equiv 0$. This is impossible since $Q_1(x_0)\neq 0$
and so again, the problem is indeterminate. \qed
\enddemo
While we have used the theory of essential self-adjointness
to study the moment problem, following Nussbaum [\nusI], one can
turn the analysis around:
\definition{Definition} Let $B$ be a symmetric operator on a
Hilbert space $\Cal H$ and let $\varphi\in C^\infty (B)=
\cap_n D(B^n)$. $\varphi$ is called {\it{a vector of uniqueness
for}} $B$ if and only if the Hamburger moment problem for
$$
\gamma_n = \frac{(\varphi, B^n \varphi)}{\|\varphi\|^2} \tag 2.23
$$
is determinate.
\enddefinition
\proclaim{Theorem 2.15 (Nussbaum [\nusI])} Let $B$ be a
\rom(densely defined\rom) symmetric operator on a Hilbert space
$\Cal H$. Suppose that $D(B)$ contains a dense subset of vectors
of uniqueness. Then $B$ is essentially self-adjoint.
\endproclaim
\demo{Proof} Let $\Cal U$ be the vectors of uniqueness for $B$.
Let $\varphi\in\Cal U$ and let $A_\varphi$ be the restriction
of $B$ to the closure of the space spanned by
$\{B^n\varphi\}^\infty_{n=0}$. $A_\varphi$ is unitarily
equivalent to the Jacobi matrix for the moment problem (2.23).
Thus, $A_\varphi$ is essentially self-adjoint on $\Cal H_\varphi$,
the closure of the span of $\{B^n\varphi\}^\infty_{n=0}$.
Therefore, $\overline{(A_\varphi +i)[D(A_\varphi)]} =
\Cal H_\varphi$ and, in particular, $\varphi \in
\overline{(A_\varphi +i)[D(A_\varphi)]} \subset
\overline{(B+i)[D(B)]}$. It follows that $\Cal U\subset
\overline{(B+i)[D(B)]}$ and thus, $\overline{\text{Ran}(B+i)} =
\Cal H$. Similarly, $\overline{\text{Ran}(B-i)}=\Cal H$, so
$B$ is essentially self-adjoint by Corollary~2.2. \qed
\enddemo
The motivation for Nussbaum was understanding a result of
Nelson [\nel].
\definition{Definition} Let $B$ be a symmetric operator.
$\varphi\in C^\infty$ is called
\roster
\item"{}" {\it{analytic}} if $\| B^n \varphi\| \leq CR^n n!$
\item"{}" {\it{semi-analytic}} if $\|B^n \varphi\| \leq
CR^n (2n)!$
\item"{}" {\it{quasi-analytic}} if $\sum_{n=1}^\infty
\|B^{2n}\varphi\|^{1/2n} <\infty$.
\item"{}" {\it{Stieltjes}} if $\sum_{n=1}^\infty
\|B^n \varphi\|^{1/2n}<\infty$.
\endroster
\enddefinition
\proclaim{Corollary 2.16} If $B$ is a symmetric operator, then
any analytic or quasi-analytic vector is a vector of uniqueness.
If $B$ is also bounded from below, then any semi-analytic or
Stieltjes vector is a vector of uniqueness. In particular,
\roster
\item"\rom{(i)}" {\rom{(Nelson [\nel])}} If $D(B)$ contains a
dense set of analytic vectors, then $B$ is essentially self-adjoint.
\item"\rom{(ii)}" {\rom{(Nussbaum [\nusII], Masson-McClary
[\mmc])}} If $D(B)$ contains a dense set of semi-analytic vectors
and is bounded from below, then $B$ is essentially self-adjoint.
\item"\rom{(iii)}" {\rom{(Nussbaum [\nusI])}} If $D(B)$ contains
a dense set of quasi-analytic vectors, then $B$ is essentially
self-adjoint.
\item"\rom{(iv)}" {\rom{(Nussbaum [\nusII], Masson-McClary
[\mmc])}} If $D(B)$ contains a dense set of Stieltjes vectors
and is bounded from below, then $B$ is essentially self-adjoint.
\endroster
\endproclaim
\demo{Proof} The first two assertions follow Proposition~1.5,
Corollary~3.4, and Corollary~4.5. Theorem~2.14 completes the
proof. \qed
\enddemo
\vskip 0.3in
\flushpar{\bf \S 3. The Stieltjes Moment Problem as a
Self-Adjointness Problem}
\vskip 0.1in
One goal in this section is to prove the remaining part of
Theorem~2, namely, that if the operator $A$ has a unique
non-negative self-adjoint extension, then the Stieltjes problem
is determinate. In the indeterminate case, we will also introduce
two distinguished non-negative von~Neumann solutions: the
Friedrichs and Krein solutions.
The name Friedrichs and Krein for the associated self-adjoint
extensions is standard. We will naturally carry over the names
to the measures $d\mu_F$ and $d\mu_K$ that solve the Stieltjes
problem. Those names are not standard. Krein did important work
on the moment problem, but not in the context of what we will
call the Krein solution because of his work on self-adjoint
extensions.
We begin with a very brief summary of the Birman-Krein-Vishik
theory of extensions of strictly positive operators (see [\as]
for a complete exposition). Suppose $A$ is a closed symmetric
operator and for some $\alpha >0$, $(\varphi, A\varphi) \geq\alpha
\| \varphi\|^2$ for all $\varphi\in D(A)$. One closes the quadratic
form analogously to closing the operator, that is, $\varphi\in
Q(A_F)$ if and only if there is $\varphi_n\in D(A)$ so that
$\varphi_n \to \varphi$ and $(\varphi_n - \varphi_m, A(\varphi_n
-\varphi_m))\to 0$ as $n\to\infty$. One then sets $(\varphi,
A_F \varphi)=\lim(\varphi_n, A\varphi_n)$. The quadratic form
$(\varphi, A_F \varphi)$ on $Q(A_F)$ is closed, that is,
$Q(A_F)$ is complete in the norm $\sqrt{(\varphi, A_F \varphi)}$.
Not every quadratic form abstractly defined is closable (e.g.,
on $L^2(\Bbb R, dx)$, the form with form domain $C^\infty_0
(\Bbb R)$ given by $\varphi\mapsto |\varphi(0)|^2$ is not
closable), but it is not hard to see that if a quadratic form
comes from a non-negative symmetric operator, it is closable
(see Theorem~X.23 in [\rsII] or [\kato], Chapter~VI).
While a closed symmetric operator need not be self-adjoint, it
is a deep result (see [\kato, \rsII]) that a closed positive quadratic
form is always the quadratic form of a self-adjoint operator. It
follows that $A_F$ is the quadratic form of a self-adjoint
operator, the {\it{Friedrichs extension}} of $A$. It follows that
$\inf \,\text{spec}(A_F)=\alpha >0$.
Define $N=\ker(A^*)$ which is closed since $A^*$ is a closed
operator. Then $D(A^*)=D(A)+N + A^{-1}_F [N]$, where $+$ means
independent sum. In particular, $N\neq \{0\}$ if $A$ is not
self-adjoint (which we suppose in the rest of this paragraph).
The {\it{Krein extension}} is the one with $D(A_K)=D(A)+N$. Then
$Q(A_K)=Q(A_F) + N$. Since $N\subset D(A_K)$, we have that
$\inf\,\text{spec}(A_K)=0$ so $A_K \neq A_F$. The set of
non-negative self-adjoint extensions of $A$ is precisely the
set of self-adjoint operators $B$ with $A_K \leq B \leq A_F$
(where $0\leq C\leq D$ if and only if $Q(D) \subset Q(C)$ and
$(\varphi, C\varphi)\leq (\varphi, D\varphi)$ for all $\varphi\in
Q(D)$; equivalently, if and only if $(D+1)^{-1}\leq (C+1)^{-1}$
as bounded operators). These $B$'s can be completely described
in terms of closed but not necessarily densely defined quadratic forms,
$C$, on $N$. One can think of $C=\infty$ on $Q(C)^\bot$. $A_F$
corresponds to $C\equiv\infty$ (i.e., $Q(C)=\{0\}$) and $A_K$ to
$C\equiv 0$. In general, $Q(A_C) = Q(A_F) + Q(C)$ and if $\varphi
\in Q(A_F)$ and $\eta\in Q(C)$, then $(\varphi + \eta, A_C
(\varphi + \eta)) = (\varphi, A_F \varphi)+(\eta, C\eta)$.
Two aspects of Jacobi matrices (defined on sequences of finite
support) make the theory special. First, $\dim(N)\leq 1$.
Second, by Theorem~5 (which we will prove in the next section
without using any results from the present section), if $A$ is
not essentially self-adjoint, $A_F$ has a discrete spectrum with
the property that if $t\in\text{spec}(A_F)$, then $\mu_F
(\{t\}) > \mu (\{t\})$ for any other solution, $\mu$, of the
Hamburger moment problem. Moreover, if $t\in\text{spec}(A_F)$,
$t\notin\text{spec}(B)$ for any other self-adjoint extension,
$B$.
\proclaim{Proposition 3.1} Let $A$ be a Jacobi matrix defined
on the sequences of finite support, and suppose that $A$ has
deficiency indices $(1, 1)$ and that $A$ is non-negative
\rom(i.e., that the moment problem associated to $A$ has at
least some solutions of the Stieltjes moment problem\rom).
Then $A$ has a unique non-negative self-adjoint extension if
and only if $0$ is an eigenvalue in the spectrum of $A_F$.
\endproclaim
\demo{Proof} Let $\alpha=\inf \,\text{spec}(A_F)$. By hypothesis,
$\alpha \geq 0$, and by Theorem~5, $\alpha$ is a discrete,
simple eigenvalue of $A_F$. We want to show that if $\alpha >0$,
then $A$ has additional non-negative self-adjoint extensions; and
contrary-wise, if $\alpha =0$, then it has no additional
non-negative self-adjoint extensions.
If $\alpha >0$, $A\geq \alpha >0$, then $A$ has a Krein extension
distinct from the $A_F$, so there are multiple non-negative
self-adjoint extensions.
Suppose $\alpha =0$ and that $B$ is some non-negative self-adjoint
extension. Since $A_F$ is the largest such extension, $0\leq B
\leq A_F$, so
$$
(A_F +1)^{-1} \leq (B+1)^{-1} \leq 1 . \tag 3.1
$$
Since $\alpha =0$, $\|(A_F +1)^{-1}\| =1$, so (3.1) implies that
$\|(B+1)^{-1}\|=1$. Since $B$ has discrete spectrum by Theorem~5,
this means $0$ is an eigenvalue of $B$. But, by Theorem~5 again,
$A_F$ is the unique self-adjoint spectrum with a zero eigenvalue.
Thus, $B=A_F$, that is, there is a unique non-negative
self-adjoint extension. \qed
\enddemo
\proclaim{Theorem 3.2 (last quarter of Theorem~2)} Suppose that
$A$ is a Jacobi matrix on the vectors of finite support and that
$A$ is non-negative. If $A$ has a unique non-negative self-adjoint
extension, then the associated Stieltjes moment problem is
determinate.
\endproclaim
\demo{Proof} Clearly, if $A$ is essentially self-adjoint, then
the Hamburger problem is determinate and so, a fortiori, the
Stieltjes problem is determinate. Thus, we need only consider the
case where $A$ has multiple self-adjoint extensions, but only one
that is non-negative.
Then $A$ has deficiency indices $(1,1)$. By Proposition~3.1,
$\alpha =\inf\,\text{spec}(A_F)=0$. Moreover, by Theorem~5, $0$
is a pure point of $\mu_F$. Let $\varphi = P_{\{0\}}(A_F)\delta_0$.
By Lemma~2.8, $\varphi\neq 0$. Then there exists $\varphi_n\in D(A)$
so $\varphi_n\to\varphi$ with $\|\varphi_n\|=\|\varphi\|$ and
$(\varphi_n, A_F \varphi_n)\to (A_F \varphi, \varphi)=0$. If
$\mu_F (\{0\}) =\tau$, then $\|\varphi\|^2 = \langle \varphi,
\delta_0\rangle = \tau$ so $\|\varphi_n\|=\sqrt\tau$ and
$\langle \varphi_n, \delta_0\rangle \to\tau$.
Suppose $\rho$ is some solution of the Stieltjes moment problem.
Since $\varphi_n\to\varphi$ in $L^2 (d\mu_F)$, $\varphi_n$ is
Cauchy in $L^2(d\rho)$, and so $\varphi_n \to f$ in $L^2(d\rho)$
and $\|f\|=\sqrt\tau$. Since $\int x\varphi_n(x)^2\, d\rho
\to 0$, we conclude that $\int xf^2\, d\rho =0$. Since $\rho$
is a measure supported on $[0,\infty)$, we conclude $f(x)=0$ for
$x\neq 0$, and thus $\rho$ has a pure point at zero also. Since
$(\varphi_n, \delta_0)\to\tau$, we see that $\int f\, d\rho=\lim
\int \varphi_n(x)\, d\rho(x)=\lim (\varphi_n, \delta_0)=\tau$.
But $\int f^2 \, d\rho = \|f\|^2 =\tau$ and $\int f\,d\rho=\tau$.
Thus $f(0)=1$ and so $\rho(\{0\})=\tau$. But Theorem~5 asserts
that any solution of the Hamburger problem distinct from $\mu_F$
has $\mu_F (\{0\}) <\tau$. We conclude that $\rho=\mu_F$, that
is, that the solution of the Stieltjes problem is unique. \qed
\enddemo
We will call the solution of the Stieltjes problem associated to
the Friedrichs extension of $A$ the {\it Friedrichs solution}.
If the Stieltjes problem is indeterminate, then $A\geq \alpha >0$
by Proposition~3.1 and Theorem~3.2. Moreover, since the Hamburger
problem is indeterminate, $N=\ker(A)=\text{span}[\pi(0)]$ has
dimension $1$, so the Krein extension exists and is distinct from
the Friedrichs extension. We will call its spectral measure the
{\it Krein solution} of the Stieltjes problem.
As discussed in the introduction, every indeterminate Stieltjes
problem has associated to it an example of a determinate
Stieltjes problem where the associated Hamburger problem is
indeterminate. For by Proposition~3.1 and Theorem~3.2, the
indeterminate Stieltjes problem is associated to an $A$ where
the bottom of the spectrum of $A_F$ is some $f_0 >0$. The
moment problem $\{\gamma_m (-f_0)\}$ with $\gamma_m (-f_0)=
\sum_{j=0}^m \binom{m}{j} \gamma_j (-f_0)^{m-j}$ will be
determinate Stieltjes (since the associated $A_F$ is the old
$A_F - f_0$) but indeterminate Hamburger.
We summarize with
\proclaim{Theorem 3.3} Suppose that $\{\gamma_n\}^\infty_{n=0}$
is a set of Stieltjes moments. Then, if $\gamma$ is Hamburger
indeterminate, there is a unique $c_0 \leq 0$ so that $\gamma
(c_0)$ is Stieltjes determinate. Moreover, $\gamma(c)$ is
Stieltjes indeterminate if $c>c_0$ and $\gamma(c)$ are not
Stieltjes moments if $c0$. We noted that
with $\{a_n\}^\infty_{n=0}$, $\{b_n\}^\infty_{n=0}$, and $a_{-1}
\equiv 1$, the elements of the Jacobi matrix associated to
$\gamma$, the sequence $u_n = P_n(z)$; $n=0,1,2,\dots$ obeys
$$
a_n u_{n+1} + (b_n -z) u_n + a_{n-1} u_{n-1} =0 \tag 4.1
$$
with
$$
u_{-1} =0, \qquad u_0 =1 . \tag 4.2
$$
We also introduced the sequence of polynomials $Q_n (X)$ of
degree $n-1$ by requiring that $u_n = Q_n(z)$ solve (4.1) with
the initial conditions
$$
u_{-1} = -1, \qquad u_0 =0 . \tag 4.3
$$
Notice that the Wronskian $W(Q_\cdot (z), P_\cdot(z))(-1) = a_{-1}
[Q_0(z) P_{-1}(z) - Q_{-1}(z) P_0(z)]=1$. Thus, by (2.11) we
have:
\proclaim{Proposition 4.1}
$$
a_{k-1} [Q_k(z) P_{k-1}(z) -
Q_{k-1}(z) P_k(z)]\equiv 1
$$
\endproclaim
The following is a useful formula for the $Q$'s in terms of the
$P$'s. Note that since $\frac{X^n - Y^n}{(X-Y)} =
\sum_{j=0}^{n-1} X^j Y^{n-1-j}$ for any polynomial $P$,
$\frac{P(X)-P(Y)}{X-Y}$ is a polynomial in $X$ and $Y$.
\proclaim{Theorem 4.2} For $n\geq 0$,
$$
E_X \biggl(\frac{P_n(X) - P_n(Y)}{X-Y}\biggr) = Q_n (Y).
$$
\endproclaim
\demo{Proof} $\frac{[P_n(X)-P_n(Y)]}{(X-Y)}$ is a polynomial in
$X$ and $Y$ of degree $n-1$, so we can define polynomials
$$
R_n(Y)\equiv E_X \biggl(\frac{P_n(X)-P_n(Y)}{X-Y}\biggr).
$$
Subtract the equations,
$$\align
a_n P_{n+1}(X) + b_n P_n(X) + a_{n-1} P_{n-1}(X) &= XP_n(X) \\
a_n P_{n+1}(Y) + b_n P_n (Y) + a_{n-1} P_{n-1}(Y) &= YP_n(Y)
\endalign
$$
and see that for $n=0,1,2,\dots$ with $R_{-1}(Y)\equiv 0$:
$$\align
a_n R_{n+1}(Y) + b_n R_n(Y) + a_{n-1} R_{n-1}(Y) &=
E_X\biggl(\frac{XP_n(X)-YP_n(Y)}{X-Y}\biggr) \\
&= E_X \biggl(\frac{YP_n(X) - YP_n(Y)}{X-Y} + P_n (X)\biggr)\\
&= YR_n (Y) + \delta_{n0}.
\endalign
$$
Replace the abstract variable $Y$ with a complex variable $z$ and
let $\tilde R_n (z) = R_n(z)$ if $n\geq 0$ but $\tilde R_{-1}(z)
=-1$. Then
$$
a_n \tilde R_{n+1}(z) + b_n \tilde R_n (z) + a_{n-1}
\tilde R_{n-1}(z) = z\tilde R_n (z)
$$
(since the $\delta_{n0}$ term becomes $-a_{n-1}\tilde R_{n-1}(z)$
for $n=0$). Moreover,
$$
\tilde R_{-1}(z)=-1, \qquad \tilde R_0 (z) = E_X \biggl(
\frac{1-1}{X-Y}\biggr) =0
$$
so by uniqueness of solutions of (4.1) with prescribed initial
conditions, $\tilde R_n (z) = Q_n (z)$, that is, for $n\geq 0$,
$R_n =Q_n$, as claimed. \qed
\enddemo
One consequence of Theorem~4.2 is that $P_n(X)$ and $Q_n(X)$
have the same leading coefficient, that is, if $P_n(X) = c_{nn}
X^n + \cdots$, then $Q_n(X) = c_{nn} X^{n-1} + \cdots$.
The following calculation is part of the standard theory [\akh]
and plays a critical role in what follows:
\proclaim{Theorem 4.3} Let $\rho$ solve the moment problem and
assume $z\in\Bbb C\backslash\Bbb R$. Set $\zeta = G_\rho (z)
\equiv \int\frac{d\rho(x)}{x-z}$. Then
$$
\langle (x-z)^{-1}, P_n (x)\rangle_{L^2 (d\rho)} =
Q_n (z) + \zeta P_n (z). \tag 4.4
$$
In particular,
$$
\sum_{n=0}^\infty |Q_n(z) +
\zeta P_n(z)|^2 \leq \frac{\text{\rom{Im}}\,\zeta}
{\text{\rom{Im}}\, z} \tag 4.5
$$
with equality if $\rho$ is a von~Neumann solution.
\endproclaim
\remark{Remark} We will eventually see (Proposition~4.15 and
its proof) that the ``if" in the last sentence can be replaced
by ``if and only if."
\endremark
\demo{Proof}
$$
\int \frac{P_n(x)}{x-z}\, d\rho(x) = \zeta P_n(z) + \int
\frac{P_n(x) - P_n(z)}{x-z}\, d\rho(x) = \zeta P_n(z) +
Q_n(z)
$$
by Theorem~4.2. This is just (4.4). (4.5) is just Parseval's
inequality for the orthonormal set $\{P_n(x)\}^\infty_{n=0}$ in
$L^2(d\rho)$ if we note that
$$
\int\frac{d\rho(x)}{|x-z|^2} = \frac{1}{z-\bar z} \int d\rho(x)
\biggl[ \frac{1}{x-z} - \frac{1}{\bar x - \bar z}\biggr]
=\frac{\text{Im}\, \zeta}{\text{Im}\,z}\,.
$$
If $\rho$ comes from a self-adjoint extension, by construction
of $\Cal H^{(\gamma)}$, $\{P_n(x)\}^\infty_{n=0}$ is an
orthonormal basis for $L^2(d\rho)$, so equality holds in (4.5).
\qed
\enddemo
\remark{Remark} In $\Cal H^{(\gamma)}$, thought of as limits
of polynomials, the vector with components $Q_n (z) + \zeta
P_n(z)$ is represented by
$$
\sum_n [Q_n(z) + \zeta P_n(z)] P_n(X).
$$
This product of $P$'s may seem strange until one realizes that
$\sum_{n=1}^N P_n (y)P_n(x)$ is a reproducing kernel in $L^2
(d\rho)$ for polynomials of degree $N$. This links the construction
to ideas in Landau [\land].
\endremark
Given Theorem~2, the following proves the equivalence of parts
(i), (ii), (iii), and (vii) for $z\in\Bbb C\backslash\Bbb R$ of
Theorem~3.
\proclaim{Proposition 4.4} Suppose that $z_0\in\Bbb C
\backslash\Bbb R$. Then the following are equivalent:
\roster
\item"\rom{(i)}" The Jacobi matrix $A$ is not essentially
self-adjoint.
\item"\rom{(ii)}" $\pi(z_0) = \{P_n(z_0)\}^\infty_{n=0}$ is
in $\ell^2$.
\item"\rom{(iii)}" $\xi(z_0) = \{Q_n(z_0)\}^\infty_{n=0}$
is in $\ell^2$.
\item"\rom{(iv)}" Both $\pi(z_0)$ and $\xi(z_0)$ are in
$\ell^2$.
\endroster
Moreover, when any of these conditions holds, there is a closed
disk of positive radius, $\Cal D(z_0)$ , in the same half-plane
as $z_0$ so that for any solution $\rho$ of the moment problem,
$\zeta=G_\rho (z_0)\in\Cal D(z_0)$. The values of $\zeta$
when $\rho$ is a von~Neumann solution lie on $\partial
\Cal D(z_0)$ and fill this entire boundary. Every point in
$\Cal D(z_0)$ is the value of $G_\rho(z_0)$ for some
$\rho$ solving the moment problem.
In addition, if these conditions hold, $\pi(z_0)$ and $\xi(z_0)$
lie in $D(A^*)\backslash D(A)$.
\endproclaim
\remark{Remarks} 1. One can show (using (2.11)) that the center
of the disk is
$$
\lim_{n\to\infty} - \frac{Q_n(z_0)\,
\overline{P_{n-1}(z_0)} - Q_{n-1}(z_0)\,
\overline{P_n(z_0)}}{P_n(z_0) \overline{P_{n-1}(z_0)} -
P_{n-1}(z_0)\, \overline{P_n(z_0)}}
$$
and the radius is
$$
\frac{1}{2 |\text{Im}\, z_0|} \,\, \frac{1}{\|\pi(z_0)\|^2}
\tag 4.6
$$
but these explicit formulae will play no role in our discussions.
2. We will see later (Theorem~4.14) that if $\rho$ is not a
von~Neumann solution, then $\zeta\in\Cal D(z_0)^{\text{\rom{int}}}$.
\endremark
\demo{Proof} Since $A$ is real, $A$ fails to be essentially
self-adjoint if and only if there is a non-zero solution of
$(A^*-z_0)u=0$. By Theorem~2.6 and the unique solubility of
second-order difference operators given $u_{-1}=0$ and $u_0$,
every such solution has $u_n = u_0 P_n(z_0)$ so (i) is equivalent
to (ii). Let $\zeta = G_\rho(z_0)$ for some von~Neumann solution.
Then $\xi(z_0) + \zeta \pi(z_0)\in\ell^2$, so (ii) is equivalent
to (iii) or (iv).
If (i)--(iv) hold, then (4.5) has the form
$$
a|\zeta|^2 + b\zeta + \bar b\bar\zeta + c \leq 0,
$$
where $a=\|\pi(z_0)\|^2$, $c=\|\xi(z_0)\|^2$, and $b=2
\langle\xi(z_0), \pi(z_0)\rangle - \frac{i}{2\, \text{Im}\, z_0}$.
The set where this inequality holds is always a disk, $\Cal D(z_0)$,
although a priori, it could be empty depending on the values of
$a,b,c$. However, by Theorem~2.11, we know that there are multiple
$\zeta$'s obeying (4.5) so the disk must have strictly positive
radius. By Theorem~4.3, $\zeta$'s for von~Neumann solutions obey
equality in (4.5) and so lie in $\partial\Cal D(z_0)$.
The self-adjoint extensions are parametrized by a circle
(unitaries from $\Cal K_+$ to $\Cal K_-$) in such a way that
the map from the parametrization to values of $(\delta_0,
(B_t-z_0)^{-1}\delta_0)$ is continuous. By Theorem~2.11, this map
is one-one. Every one-one continuous map of the circle to itself
is surjective, so all of $\partial\Cal D(z_0)$ occurs.
Given a point $\zeta$ in $\Cal D(z_0)^{\text{\rom{int}}}$, find
$\zeta_0$ and $\zeta_1$ in $\partial\Cal D(z_0)$ and $\theta\in
(0,1)$ so $\zeta = \theta\zeta_0 + (1-\theta)\zeta_1$. If
$\mu_0, \mu_1$ are the von~Neumann solutions that have
$G_{\mu_i} (z_0) = \zeta_i$ ($i=0,1)$, then $\rho =\theta\mu_0 +
(1-\theta)\mu_1$ is a measure solving the moment problem with
$G_\rho (z_0)=\zeta$.
If (iv) holds, then $\pi(\bar z_0), \xi(z_0)\in D(A^*)$.
By (2.14) and Proposition~1.4, $\langle A^*\pi(\bar z_0),
\xi(z_0)\rangle \mathbreak - \langle \pi(\bar z_0), A^*
\xi(z_0)\rangle =-1$, so neither $\pi(\bar z_0)$ nor $\xi(z_0)$
lies in $D(A)$. \qed
\enddemo
\remark{Remark} The Cayley transform way of looking at
self-adjoint extensions says that for any $z_0 \in \Bbb C
\backslash\Bbb R$,
$$
(B_t - \bar z_0)(B_t - z_0)^{-1} = V(z_0) +
e^{i\theta(z_0, t)} X(z_0),
$$
where $V(z_0)$ is the $t$-independent map $(\bar A - \bar z_0)
(\bar A -z_0)^{-1}$ from $\overline{\text{Ran}(A-z_0)}$ to
$\overline{\text{Ran}(A-\bar z_0)}$ and $X(z_0)$ is any
isometry from $\ker(A^* - \bar z_0)$ to $\ker(A^* -z_0)$
extended to $\Cal H$ as a partial isometry. $\theta$ is a
parametrization depending on $z_0, t$ (and the choice of
$X(z_0)$), but the theory guarantees that as $t$ varies through
all possible values, $e^{i\theta}$ runs through the full circle
in an injective manner. Now $(B_t - \bar z_0)(B_t - z_0)^{-1}
=1 + 2i(\text{Im}\, z_0)(B_t - z_0)^{-1}$, so
$$
(\delta_0, (B_t - z_0)^{-1}\delta_0) =
[2i(\text{Im}\, z_0)]^{-1} [-1 + (\delta_0, V(z_0)\delta_0)
+e^{i\theta(z_0, t)} (\delta_0, X(z_0)\delta_0)]
$$
is seen directly to be a circle. Since one can take $X(z_0)$ to be
$$
(\delta_n, X(z_0)\delta_m) = \frac{P_m (\bar z_0) P_n(z_0)}
{\sum_{j=0}^\infty |P_j (z_0)|^2}
$$
and $P_0(z)\equiv 1$, we even see that the radius of the circle
is given by (4.6).
\endremark
\proclaim{Corollary 4.5} If a Hamburger problem is indeterminate,
then $\sum_{n=0}^\infty |a_n|^{-1} <\infty$. In particular, if
$\sum_{n=0}^\infty |a_n|^{-1} =\infty$, \rom(e.g., $a_n \equiv
1$\rom), then the Hamburger problem is determinate. If
$$
\sum_{n=1}^\infty \gamma^{-1/2 n}_{2n} = \infty,
$$
then the Hamburger problem is determinate. If
$$
\sum_{n=1}^\infty \gamma^{-1/2 n}_n = \infty
$$
for a set of Stieltjes moments, that problem is both Stieltjes
and Hamburger determinate.
\endproclaim
\remark{Remarks} 1. The last pair of assertions is called
Carleman's criterion. It generalizes Proposition~1.5 and proves
uniqueness in some cases where growth doesn't imply analyticity of
the Fourier transform.
2. See Corollary~5.24 for the Stieltjes analog of the first
assertion.
\endremark
\demo{Proof} By Proposition~4.1 and the Schwarz inequality,
$$
\sum_{n=0}^N |a_n|^{-1} \leq 2 \biggl( \sum_{n=0}^{N+1}
|P_n (z)|^2\biggr)^{1/2} \bigg( \sum_{n=0}^{N+1} |Q_n (z)|^2
\biggr)^{1/2}.
$$
Thus, divergence of $\sum_{n=0}^\infty |a_n|^{-1}$ implies
that either $\pi$ or $\xi$ (or both) fail to be $\ell_2$, and so
determinacy.
Consider Carleman's criterion in the Hamburger case. By induction
in $n$ and (4.1) starting with $P_0(x)=1$, we see that
$$
P_n(x) = (a_1 \dots a_n)^{-1} x^n + \text{lower order},
$$
so $\langle(a_1 \dots a_n)^{-1}x^n, P_n(x)\rangle =1$ and
thus, by the Schwartz inequality,
$$
1 \leq \gamma_{2n} (a_1 \dots a_n)^{-2}
$$
hence
$$
\gamma^{-1/2n}_{2n} \leq (a_1 \dots a_n)^{-1/n}.
$$
Therefore our result follows from the divergence criteria proven at
the start of the proof and the inequality
$$
\sum_{j=1}^n (a_1 \dots a_j)^{-1/j} \leq 2e \sum_{j=1}^n a^{-1}_j
$$
(which is actually known to hold with $e$ in place of $2e$).
To prove this, note first that $1+x \leq e^x$ so $(1+n^{-1})^n
\leq e$ so using induction, $n^n \leq e^n n!$, so by the
geometric-arithmetic mean inequality,
$$\align
(a_1 \dots a_j)^{-1/j} &= (a^{-1}_1 2a^{-1}_2 \dots
ja^{-1}_j)^{1/j} (j!)^{-1/j} \\
&\leq \frac{e}{j^2} \sum_{k=1}^j ka^{-1}_k.
\endalign
$$
Thus,
$$
\sum_{j=1}^n (a_1 \dots a_j)^{-1/j} \leq e \sum_{k=1}^n
a^{-1}_k \biggl( \sum_{j=k}^n \frac{k}{j^2}\biggr) \leq
2e \sum_{j=1}^n a^{-1}_j
$$
since
$$
\sum_{j=k}^\infty \frac{k}{j^2} \leq \frac{1}{k} + k
\int_k^\infty \frac{dy}{y^2} \leq 2.
$$
By Proposition~1.6, Carleman's criterion in the Hamburger case
means that if a set of Stieltjes moments obeys $\sum_{n=1}^\infty
(\gamma_n)^{-1/2n}=\infty$, then it is Stieltjes determinate. Thus
by Theorem~3.3, it suffices to show that
$$
\sum_{n=1}^\infty (\gamma_n (c))^{-1/2n} \geq D(c)
\sum_{n=1}^\infty \gamma_n^{-1/2n}
$$
for $c>0$ and $D(c) < \infty$ to conclude Hamburger determinacy.
Note first that $\frac{\gamma_{j+1}}{\gamma_j} \geq
\frac{\gamma_j}{\gamma_{j-1}}$ by the Schwarz inequality. Thus,
if $\alpha = \frac{\gamma_0}{\gamma_1}$, we have that
$$
\gamma_{n-j} \leq \gamma_n \alpha^j.
$$
Therefore,
$$
\gamma_n (c) \leq \gamma_n \sum_{j=0}^n \binom{n}{j} c^j
\alpha^j = (1 + c\alpha)^n \gamma_n
$$
and so
$$
\sum_{n=1}^\infty (\gamma_n (c))^{-1/2n} \geq
(1 + c\alpha)^{-1/2} \sum_{n=1}^\infty \gamma_n^{-1/2n}. \qed
$$
\enddemo
To complete the proof of Theorem~3, we need to show that if every
solution of $[(A^* -z_0)u]_n =0$ ($n\geq 1$) is $\ell^2$ for a
fixed $z=z_0$, the same is true for all $z\in\Bbb C$ and that
$\frac{\partial\pi}{\partial x}$, $\frac{\partial\xi}{\partial x}$
are in $\ell^2$. We will do this by the standard method of
variation of parameters. So we write a general solution of
$[(A^*-z)u]_n =0$ ($n\geq 1$) in terms of $P_n(z_0)$ and
$Q_n(z_0)$ using
$$
\binom{u_{n+1}}{u_n} \equiv \alpha_n
\binom{P_{n+1}(z_0)}{P_n (z_0)} + \beta_n
\binom{Q_{n+1}(z_0)}{Q_n (z_0)} . \tag 4.7
$$
Since $W(P_\cdot(z_0), Q_\cdot(z_0)) = -1$, the two vectors on
the right of (4.7) are linearly independent, and so they span
$\Bbb C^2$ and $(\alpha_n, \beta_n)$ exist and are unique.
Indeed, by the Wronskian relation:
$$\align
\alpha_n &= W(Q_\cdot(z_0), u)(n) \tag 4.8a \\
\beta_n &= -W (P_\cdot(z_0), u)(n). \tag 4.8b
\endalign
$$
A straightforward calculation using the fact that $P,Q$ obey
(4.1) with $z=z_0$, (2.11), and (4.8) shows that (4.1) at $n$
is equivalent to
$$
\binom{\alpha_n}{\beta_n} = [1 + (z-z_0) S(n, z_0)]
\binom{\alpha_{n-1}}{\beta_{n-1}}, \tag 4.9
$$
where
$$
S(n,z_0) = \pmatrix -Q_n (z_0) P_n(z_0) & -Q_n (z_0) Q_n (z_0) \\
P_n (z_0) P_n (z_0) & P_n (z_0) Q_n (z_0) \endpmatrix .
\tag 4.10
$$
For example, using (2.11) and (4.7) for $n\to n-1$:
$$\align
\alpha_n &= W(Q_\cdot (z_0), u)(n) \\
&= W(Q_\cdot (z_0), u)(n-1) + (z_0-z) Q_n (z_0)u_n \\
&= \alpha_{n-1} - (z-z_0) Q_n (z_0) \{\alpha_{n-1} P_n (z_0)
+ \beta_{n-1} Q_n (z_0)\}.
\endalign
$$
Notice that
$$
S^2 = \det(S) = \text{Tr}(S)=0. \tag 4.11
$$
The following is obvious and explains why $\ell^2$ solutions are
so natural:
If $\pi(z_0) = \{P_n (z_0)\}^\infty_{n=0}$ and $\xi(z_0) =
\{Q_n (z_0)\}^\infty_{n=0}$ are {\it both} in $\ell^2$, then
$$
\sum_{n=0}^\infty \|S(n,z_0)\| <\infty. \tag 4.12
$$
\proclaim{Lemma 4.6} Let $A_n$ be a sequence of matrices with
$\sum_{n=0}^\infty \|A_n\|<\infty$. Let $D_N(z) = (1+zA_N)
(1+zA_{N-1})\dots (1+zA_0)$. Then $D_\infty (z) = \lim D_N (z)$
exists for each $z\in\Bbb C$ and defines an entire function of $z$
obeying
$$
\|D_\infty (z)\| \leq c_\varepsilon \exp(\varepsilon |z|)
$$
for each $\varepsilon >0$.
\endproclaim
\demo{Proof} Notice first that
$$
\| (1+B_N)\dots (1+B_0)\| \leq \prod_{j=0}^N (1+\|B_j\|)
\leq \exp \biggl(\sum_{j=0}^N \|B_j\|\biggr) \tag 4.13
$$
and that
$$
\| (1+B_N)\dots (1+B_0) - 1 \| \leq \prod_{j=0}^N
(1+\|B_j\|) - 1 \leq \exp \biggl( \sum_{j=0}^N \|B_j\|
\biggr) - 1. \tag 4.14
$$
>From (4.14), we see that
$$\align
\| D_{N+j}(z) - D_N(z)\| &\leq \biggl[ \exp \biggl( |z|
\sum_{N+1}^{N+j} \| A_j \|\biggr) -1 \biggr] \| D_N(z)\| \\
&\leq \biggl[ \exp \biggl( |z| \sum_{N+1}^\infty \|A_j\|
\biggr) -1 \biggr] \exp\biggl( |z| \sum_{j=0}^N \|A_j\|
\biggr),
\endalign
$$
from which it follows that $D_N (z)$ is Cauchy uniformly for
$z$ in balls of $\Bbb C$. Thus, $D_\infty$ exists and is entire
in $z$. By (4.13),
$$
\|D_\infty (z)\| \leq \prod_{j=0}^N (1 + |z| \|A_j\|)
\exp \biggl( |z| \sum_{j=N+1}^\infty \|A_j\|\biggr),
$$
so given $\varepsilon$, choose $N$ so $\sum_{j=N+1}^\infty
\|A_j\|\leq\frac{\varepsilon}{2}$ and use the fact that the
polymonial $\prod_{j=0}^N (1+|z| \|A_j\|)$ can be bounded by
$c_\varepsilon \exp(\frac12\varepsilon |z|)$. \qed
\enddemo
Given Proposition~4.4 and Theorem~2.14, the following completes
the proof of Theorem~3:
\proclaim{Theorem 4.7} Let $A$ be a Jacobi matrix and consider
solutions of {\rom{(4.1)}} for $z\in\Bbb C$. If for some $z_0
\in\Bbb C$, all solutions are in $\ell^2$, then that is true for
each $z\in\Bbb C$. Moreover, $\pi(z)$ and $\xi(z)$ are analytic
$\ell_2$-valued functions so, in particular, $\frac{\partial\pi}
{\partial z}, \frac{\partial\xi}{\partial z}\in\ell_2$ for all
$z$. The disk $\Cal D(z_0)$ varies continuously as $z$ runs through
$\Bbb C_+$.
\endproclaim
\demo{Proof} Define
$$
T(n,-1; z, z_0) = (1+(z-z_0) S(n,z_0)) \dots
(1+(z-z_0)S(0,z_0)).
$$
By (4.12) and Lemma~4.6,
$$
\sup_n \|T(n, -1,; z, z_0) \| \leq \exp \biggl( |z-z_0|
\sum_{j=0}^\infty \|S(j,z)\|\biggr) < \infty.
$$
Thus for any initial $\binom{\alpha_{-1}}{\beta_{-1}}$, if
$u$ has the form (4.7), then $\sup_n \|\binom{\alpha_n}
{\beta_n}\|\leq \mathbreak \sup_n \| T(n,-1; z, z_0)\|
\| \binom{\alpha_{-1}}{\beta_{-1}}\| \equiv C$. Thus, $u_n
= \alpha_n P_n(z_0) + \beta_n Q_n (z_0)$ has $|u_n|^2 \leq
C^2 [P_n(z_0)^2 + Q_n (z_0)^2]$ which is in $\ell_1$. Since
$\pi(z)$ is associated to $T(n, -1; z, z_0)\binom{1}{0}$ and
$\xi(z)$ is associated to $T(n,-1; z,z_0)\binom{0}{1}$, the
claimed analyticity holds because $\sup_n \|T(n, -1; z, z_0)\|$
is bounded as $z$ varies in bounded sets.
Continuity of $\Cal D(z_0)$ follows from the formula for $\partial
\Cal D(z_0)$, viz.~$w\in\partial\Cal D(z_0)$ if and only if
$$
\| \pi(z_0)\|^2 |w|^2 + 2\, \text{Re}\biggl[\biggl(
2\langle \xi(z_0), \pi(z_0)\rangle - \frac{i}
{2\,\text{Im}\, z_0} \biggr) w \biggr] + \|\xi(z_0)\|^2 =0.
\qed
$$
\enddemo
By Theorem~3 and Lemma~4.6, if the Hamburger problem is
indeterminate,
$$
T(\infty, -1; z, z_0) \equiv \lim_{n\to\infty} T(n,-1;
z, z_0) \tag 4.15
$$
exists. We define four functions $A(z)$, $B(z)$, $C(z)$, $D(z)$
by
$$
T(\infty, -1; z,z_0=0) = \pmatrix -B(z) & -A(z) \\
D(z) & C(z) \endpmatrix . \tag 4.16
$$
The {\it Nevanlinna matrix} is defined by
$$
N(z) = \pmatrix A(z) & C(z) \\
B(z) & D(z) \endpmatrix . \tag 4.17
$$
\proclaim{Theorem 4.8}
\roster
\item"\rom{(i)}" Each of the functions $A$, $B$, $C$, $D$ is
an entire function obeying
$$
|f(z)| \leq c_\varepsilon \exp (\varepsilon |z|)
$$
for each $\varepsilon >0$.
\item"\rom{(ii)}" For $z$ small,
$$
B(z) = -1 + O(z) \tag 4.18a
$$
and
$$
D(z) = \alpha z + O(z^2) \tag 4.18b
$$
with $\alpha >0$.
\item"\rom{(iii)}" $AD-BC\equiv 1$.
\endroster
\endproclaim
\demo{Proof} (i) follows from Lemma~4.6. (ii) follows if we note
that
$$
T(\infty, -1; z,z_0=0) = 1 + z\sum_{n=0}^\infty S(n, z_0=0) +
O(z^2).
$$
(4.18a) is immediate and (4.18b) follows since (4.10) says that
$\alpha = \sum_{n=0}^\infty P_n (0)^2 >0$. (iii) holds since
by (4.11), $\det T =1$. \qed
\enddemo
For our purposes, the formula for $A,B,C,D$ as matrix elements of
an infinite product is sufficient, but there is an infinite sum
formula connected to standard equations for perturbed solutions
(see, e.g., [\pea, \jl]) that lets us make contact with the more
usual definitions [\akh].
\proclaim{Theorem 4.9} If the Hamburger problem is indeterminate,
then,
$$\align
A(z) &= z\sum_{n=0}^\infty Q_n(0) Q_n(z) \\
B(z) &= -1 + z\sum_{n=0}^\infty Q_n(0) P_n(z) \\
C(z) &= 1 + z\sum_{n=0}^\infty P_n(0) Q_n(z) \\
D(z) &= z \sum_{n=0}^\infty P_n(0) P_n(z).
\endalign
$$
\endproclaim
\demo{Proof}
$$\align
T(n, -1; z,z_0 =0) &= (1+z S(n, z_0=0)) \dots (1+z S(0,z_0=0)) \\
&= 1 + \sum_{j=0}^n zS(j, z_0=0) \prod_{k=0}^{j-1} (1+z S(k,z_0=0))
\endalign
$$
so it suffices to show that
$$\split
S(j,z_0=0) (1 +z S(j-1, z_0&=0)) \dots (1+zS(0,z_0=0)) \\
&\quad =
\pmatrix -Q_j(0) P_j(z) & -Q_j (0) Q_j(z) \\
P_j(0) P_j(z) & P_j(0) Q_j(z) \endpmatrix
\endsplit . \tag 4.19
$$
By definition,
$$
(1+zS(j-1, z_0=0)) \dots (1+zS(0,z_0=0)) =
\pmatrix \alpha^{(1)}_{j-1} & \alpha^{(2)}_{j-1} \\
\beta^{(1)}_{j-1} & \beta^{(2)}_{j-1} \endpmatrix ,
$$
where
$$
\alpha^{(1)}_{j-1} \binom{P_j(0)}{P_{j-1}(0)} +
\beta^{(1)}_{j-1} \binom{Q_j(0)}{Q_{j-1}(0)} =
\binom{P_j(z)}{P_{j-1}(z)}
$$
and
$$
\alpha^{(2)}_{j-1} \binom{P_j(0)}{P_{j-1}(0)} +
\beta^{(2)}_{j-1} \binom{Q_j(0)}{Q_{j-1}(0)} =
\binom{Q_j(z)}{Q_{j-1}(z)}.
$$
Thus by (4.10),
$$\align
\text{LHS of (4.19)} &= \pmatrix
-Q_j(0)[\alpha^{(1)}_{j-1} P_j(0) + \beta^{(1)}_{j-1} Q_j(0)]
& -Q_j(0) [\alpha^{(2)}_{j-1} P_j(0) + \beta^{(2)}_{j-1} Q_j (0)] \\
P_j(0)[\alpha^{(1)}_{j-1} P_j(0) + \beta^{(1)}_{j-1} Q_j(0)]
& P_j(0) [\alpha^{(2)}_{j-1} P_j(0) + \beta^{(2)}_{j-1}
Q_j(0)] \endpmatrix \\
&= \text{RHS of (4.19)}. \qed
\endalign
$$
\enddemo
$\pi(0) = \{P_n(0)\}$ and $\xi(0)=\{Q_n(0)\}$ span $D(A^*)/D(A)$
by Proposition~4.4. Notice that $A^* \pi(0)=0$ while $A^*\xi(0)=
\delta_0$, so $\langle \pi(0), A^* \xi(0)\rangle =1$ and (2.7)
holds. Thus, we can parametrize the self-adjoint extensions $B_t$
of $A$ by
$$
D(B_t) = D(A) + \{\alpha(t\pi(0)+\xi(0))\mid \alpha\in
\Bbb C\} \tag 4.20a
$$
if $t<\infty$ and
$$
D(B_\infty) = D(A) + \{\alpha \pi(0)\}. \tag 4.20b
$$
This is equivalent to defining $t$ by
$$
t=(\delta_0, B^{-1}_t \delta_0). \tag 4.20c
$$
\proclaim{Theorem 4.10} For each $t\in\Bbb R\cup \{\infty\}$ and
$z\in\Bbb C\backslash\Bbb R$,
$$
(\delta_0, (B_t -z)^{-1}\delta_0) = -\frac{C(z)t+A(z)}
{D(z)t +B(z)}\, . \tag 4.21
$$
\endproclaim
\demo{Proof} Let us sketch the proof before giving details:
$$
T(-1, \infty; z, z_0=0) \equiv T(\infty, -1; z, z_0=0)^{-1}.
\tag 4.22
$$
Then
$$
\binom{\alpha_{-1}}{\beta_{-1}} \equiv T(-1,\infty; z, z_0=0)
\binom{t}{1} \tag 4.23
$$
is such that the $u_n$ associated to $\binom{\alpha_{-1}}
{\beta_{-1}}$ obeys (4.1) and is asymptotically $t\pi (0) +
\xi(0)$ and so in $D(B_t)$. $(B_t-z)^{-1} \delta_0$ will
thus be $-\frac{u_t}{u_t (-1)}$ and $(\delta_0, (B_t -z)^{-1}
\delta_0)$ will be $-\frac{u_t(0)}{u_t(-1)}$. But $u_t(0) =
\alpha_{-1}$ and $u_t (-1) = -\beta_{-1}$ so $(\delta_0,
B_t-z)^{-1}\delta_0)$ will be $\frac{\alpha_{-1}}{\beta_{-1}}$,
which is given by (4.21).
Here are the details. If $\varphi\in D(A^*)$, then by (2.14),
$\varphi\in D(B_t)$ if and only if
$$
\lim_{n\to\infty} W(\varphi, t\pi(0) + \xi(0))(n)=0.
\tag 4.24
$$
Suppose $u$ solves (4.1) for a given $z$. Then $u\in D(A^*)$
and $u$ has the form (4.7), where by (4.15), $\lim_{n\to\infty}
\binom{\alpha_n}{\beta_n} = \binom{\alpha_\infty}{\beta_\infty}$
exists. Clearly,
$$
W(u, t\pi(0) + \xi(0))(n) = -\alpha_n + \beta_n t
$$
so (4.24) holds if and only if
$$
\alpha_\infty = t\beta_\infty.
$$
Thus, if $T(-1, \infty; z, z_0=0)$ is given by (4.22), then
$\binom{\alpha_{-1}}{\beta_{-1}}$ given by (4.23) is initial data
for a solution $u$ of (4.1) that has $u\in D(B_t)$. A solution
$u$ of (4.1) has
$$
(A^* -z)u = -u_{-1} \delta_0
$$
and thus, if $u$ is associated to the data in (4.23),
$$
(B_t -z)u = -u_{-1}\delta_0
$$
and
$$
(\delta_0, (B_t -z)^{-1} \delta_0) = -\frac{u_0}{u_{-1}}\, .
$$
But
$$
\binom{u_0}{u_{-1}} = \alpha_{-1}
\binom{P_0(0)}{P_{-1}(0)} + \beta_{-1}
\binom{Q_0(0)}{Q_{-1}(0)} = \binom{\alpha_{-1}}{-\beta_{-1}}
$$
so
$$
(\delta_0, (B_t -z)^{-1}\delta_0) = \frac{\alpha_{-1}}
{\beta_{-1}}\, .
$$
Since $T(\infty, -1; z, z_0=0)$ has the form (4.16) and has
determinant $1$, its inverse has the form
$$
\pmatrix C(z) & A(z) \\
-D(z) & -B(z) \endpmatrix
$$
and so $\alpha_{-1} = C(z)t + A(z)$, $\beta_{-1} = -D(z)t -
B(z)$. Thus (4.21) is proven. \qed
\enddemo
\remark{Remark} Our convention for the parameter $t$ differs from
that in Akhiezer [\akh], which is the traditional one! If $s=
-t^{-1}$, then
$$
-\frac{Ct+A}{Dt+B} = -\frac{As - C}{Bs-D}\, .
$$
The parameter $s$ is what he calls $t$. Our $\Phi(z)$ later and
his $\Phi(z)$, which I will call $\Psi(z)$, are related by $\Phi(z)
= -\Psi(z)^{-1}$. Since this transformation takes Herglotz functions
to themselves, we both deal with Herglotz functions. See [\im] for
an interesting alternate reparametrization.
\endremark
\vskip 0.1in
We turn next to half of Theorem~5. We will prove that each
$B_t$ has point spectrum with any two disjoint. The condition
that $\mu_t (\{t\}) >\rho (\{t\})$ for any other solution of the
moment problem will wait until after we have proven Theorem~4.
\proclaim{Theorem 4.11 (half of Theorem 5)} Suppose the
Hamburger problem is indeterminate. Each $B_t$ has pure point
spectrum only. The eigenvalues of the different $B_t$'s are
distinct and every $x\in\Bbb R$ is an eigenvalue of some $B_t$.
If $x$ is an eigenvalue of $B_t$, then
$$
\mu_t (\{x\}) = \frac{1}{\sum_{n=0}^\infty |P_n (x)|^2}\, .
\tag 4.25
$$
Moreover, if $\lambda_n (B_t)$ are the eigenvalues of $B_t$
\rom(ordered in some convenient way\rom), then for any $p>1$,
$$
\sum_n |\lambda_n (B_t)|^{-p} <\infty. \tag 4.26
$$
\endproclaim
\demo{Proof} Let $\mu_t$ be the spectral measure for $B_t$. By
Theorem~4.10,
$$
\int\frac{d\mu_t (x)}{x-z} = -\frac{C(z)t+A(z)}{D(z)t+B(z)}
$$
is a meromorphic function of $z$, since $A,B,C,D$ are entire.
That implies that $\mu_t$ is pure point only and these points
are given by solutions of
$$
D(z)t + B(z) =0.
$$
Since $A,B,C,D$ are real when $z=x$ is real, for any $x$,
$x\in\text{spec}(B_t)$ for $t=-\frac{B(x)}{D(x)}$. Since $AD-BC
\equiv 1$, $B$ and $D$ have no simultaneous zeros, so the
eigenvalues are distinct for different $t$'s. (4.26) is a
standard result on the zeros of an entire function like
$D(z)t+ B(z)$, which obeys (i) of Theorem~4.8.
To prove (4.25), note that if $x$ is an eigenvalue of $B_t$,
then the corresponding normalized eigenvector $\varphi$
obeys $A^*\varphi = x\varphi$. It follows that $\varphi_n
= \varphi_0 P_n (x)$ and then by normalization, that
$\varphi^2_0 = 1/\sum_{n=0}^\infty |P_n(x)|^2$. Thus, $\mu_t
(\{x\}) = \varphi^2_0$ is given by (4.25). \qed
\enddemo
Now define the map $F(z): \Bbb C\cup\{\infty\} \to \Bbb C
\cup\{\infty\}$ by
$$
F(z)(w) = -\frac{C(z)w + A(z)}{D(z)w + B(z)}\, . \tag 4.27
$$
$F(z)$ as a fractional linear transformation is one-one and
onto, and takes $\Bbb R\cup\{\infty\}$ onto a circle or straight
line. By Proposition~4.4 and (4.21), $F(z) [\Bbb R\cup\{\infty\}]$
is precisely the circle $\partial\Cal D(z)$ when $\text{Im}\, z
\neq 0$.
\proclaim{Proposition 4.12} If $\text{\rom{Im}}\, z>0$, $F(z)$
maps the upper half-plane $\Bbb C_+ = \{z\in\Bbb C\mid
\text{\rom{Im}}\, z>0\}$ onto the interior of the disk
$\Cal D(z)^{\text{\rom{int}}}$.
\endproclaim
\demo{Proof} For each $z\in\Bbb C\backslash\Bbb R$, $F(z)$ must
map $\Bbb C_+$ onto either $\Cal D(z)^{\text{\rom{int}}}$ or
$\Bbb C\backslash\Cal D(z)$ since $\Bbb R$ is mapped to
$\partial\Cal D(z)$ and a fractional linear transformation
maps $\Bbb C\cup\{\infty\}$ to itself bijectively. Thus it
suffices to show $F(z)[i]\in\Cal D(z)^{\text{\rom{int}}}$ for
all $z$ with $\text{Im}\,z>0$. But $F(z)[i]$ moves analytically,
and so continuously. It can only move from
$\Cal D(z)^{\text{\rom{int}}}$ to $\Bbb C\backslash\Cal D(z)$ by
lying on $\partial\Cal D(z)$, which is impossible if $\text{Im}\,
z>0$ since then $\partial\Cal D(z)=F(z)[\Bbb R\cup\{\infty\}]$.
Thus it suffices to show $F(z)[i]\in\Cal D(z)^{\text{\rom{int}}}$
for $z=ix$ with $x>0$ and small.
Now, $F(z)[i]\in\Cal D(z)^{\text{\rom{int}}}$ if and only if
$F(z)$ takes the lower half-plane to $\Bbb C\cup\{\infty\}
\backslash\Cal D(z)$, which holds if and only if $F(z)^{-1}
[\infty]$ is in the lower half-plane. Thus it suffices to show
for $x$ small and positive that $F(ix)^{-1}[\infty]\in\Bbb C_-$.
$F(ix)^{-1}[\infty]$ is that $w$ with $D(ix)w+B(ix)=0$, that is,
$w=-\frac{B(ix)}{D(ix)}$. But by (4.18), $B(ix) =-1 + O(x)$ and
$D(ix) = i\alpha x + O(x^2)$ with $\alpha >0$, so $w=-i
\alpha^{-1} x^{-1} + O(1)$ lies in $\Bbb C_-$, which suffices
to prove the proposition. \qed
\enddemo
We now follow the classical argument of Nevanlinna [\nev] to
obtain Theorem~4. We need to relate solutions of (1.1) to
asymptotics of the Stieltjes transform of $\rho$ following
Hamburger [\ham] and Nevanlinna [\nev].
\proclaim{Proposition 4.13} Let $\Cal M^H(\gamma) = \{\rho \mid
\rho\text{\rom{ obeys (1.1)}}\}$. Let $G_\rho(z) = \int
\frac{d\rho(x)}{x-z}$. Then for any $N$ as $y\to\infty$,
$$
y^{N+1} \biggl[G_\rho (iy) + \sum_{n=0}^N (-i)^{n+1}
y^{-n-1} \gamma_n\biggr] \to 0 \tag 4.28
$$
uniformly for $\rho$ in $\Cal M^H(\gamma)$. Conversely, if $G$
is a function on $\Bbb C_+$ with $\text{\rom{Im}}\, G(z)>0$ there
and so that {\rom{(4.28)}} holds for each $N$, then $G(z) =
G_\rho (z)$ for some $\rho$ in $\Cal M^H(\gamma)$.
\endproclaim
\demo{Proof} By the geometric series with remainder, the left side
of (4.28) is
$$
-(-i)^{N+1} y^{N+1} \int \frac{d\rho(x)}{x-iy} \,\,
\frac{x^{N+1}}{y^{N+1}} \equiv R_N(\rho) \tag 4.29
$$
so, using $|x-iy|^{-1}\leq y^{-1}$,
$$
|R_N(\rho)| \leq y^{-1} \int |x|^{N+1} d\rho(x)
\cases = \gamma_{N+1} y^{-1} & \text{if $N$ is odd} \\
\leq \frac12 [\gamma_N + \gamma_{N+2}]y^{-1} & \text{if $N$ is even}.
\endcases
$$
This shows that (4.28) holds and the convergence is uniform for
$\rho\in \Cal M^H(\gamma)$.
For the converse, first use the Herglotz representative theorem
which says that if $G$ maps $\Bbb C_+$ to $\Bbb C_+$, then
for some measure $d\rho$, some $c\geq 0$ and some real $d$:
$$
G(z) = cz + d + \int d\rho(x) \biggl[ \frac{1}{x-z} -
\frac{x}{1+x^2}\biggr] , \tag 4.30
$$
where, a priori, $\rho$ only obeys
$$
\int\frac{d\rho(x)}{1+ x^2} < \infty.
$$
By (4.28),
$$
y[G(iy)] \to i\gamma_0. \tag 4.31
$$
If (4.30) holds, then $y^{-1}G(iy)\to ic$, so $c=0$. Since
$c=0$, (4.30) says
$$
y\text{ Im}\, G(iy) = \int\frac{y^2}{x^2 + y^2}\, d\rho(x),
$$
so (4.31) and the monotone convergence theorem implies that
$\int d\rho(x)=\gamma_0$. Once this is true, (4.30) implies that
as $y\to\infty$,
$$
\text{Re}\, G(iy)\to d - \int d\rho(x)\, \frac{x}{1+x^2}
\tag 4.32
$$
so (4.31) implies the right side of (4.32) is zero, and thus
(4.30) becomes
$$
G(z) = \int \frac{d\rho(x)}{x-z}\, . \tag 4.33
$$
We will now prove inductively that $\rho$ obeys (1.1), that is,
$\rho\in \Cal M^H(\gamma)$. Suppose that we know that (1.1)
holds for $n=0,1,\dots, 2M-2$. (4.28) for $N=2M$ then implies
that
$$
\int\frac{(iy)^2 x^{2M-1}}{x-iy}\, d\rho(x) + iy\gamma_{2M-1}
\to -\gamma_{2M}.
$$
Taking real and imaginary parts, we see that
$$\align
\gamma_{2M-1} &= \lim_{y\to\infty} \int \frac{y^2 x^{2M-1}}
{x^2 + y^2}\, d\rho(x) \tag 4.34 \\
\gamma_{2M} &= \lim_{y\to\infty} \int \frac{y^2 x^{2M}}
{x^2 + y^2}\, d\rho(x). \tag 4.35
\endalign
$$
(4.35) and monotone convergence implies that $\int x^{2M}\,
d\rho(x)<\infty$, so that dominated convergence and (4.34),
(4.35) imply that (1.1) holds for $n=0,1,2,\dots, 2M$. \qed
\enddemo
Let $\Cal F$ be the set of analytic maps from $\Bbb C_+$ to
$\bar\Bbb C_+ \cup\{\infty\}$. By the open mapping theorem for
analytic functions, $\Phi \in\Cal F$ either maps $\Bbb C_+$ to
$\Bbb C_+$ or $\Phi$ has a constant value in $\Bbb R\cup
\{\infty\}$.
\proclaim{Theorem 4.14 (Nevanlinna [\nev])} Let $\gamma$ be a
set of indeterminate Hamburger moments. Then there is a one-one
correspondence between solutions, $\rho$, of the moment problem
and functions $\Phi\in\Cal F$ given by
$$
G_\rho \equiv \int \frac{d\rho(x)}{x-z} =
-\frac{C(z)\Phi(z) + A(z)}{D(z) \Phi(z) + B(z)}\, . \tag 4.36
$$
In particular, if $\rho$ is not a von~Neumann solution,
$G_\rho(z) \in D(z)^{\text{\rom{int}}}$ for all $z\in\Bbb C_+$.
\endproclaim
\remark{Remarks} 1. By (4.21), the von~Neumann solutions $\mu_t$
correspond precisely to the case $\Phi(z)\equiv t\in\Bbb R \cup
\{\infty\}$.
2. We will call the Herglotz function associated to some
$\rho\in\Cal M^H(\gamma)$ the {\it Nevanlinna function} of
$\rho$ and denote it by $\Phi_\rho$. Conversely, given $\Phi$
(and $\gamma$), we denote by $\rho_\Phi$ the associated solution
of the moment problem.
3. In Section~6, we will discuss the $\rho_\Phi$'s associated to
the case where the measure $\mu$ in the Herglotz representation
(1.19) is a measure with finite support.
\endremark
\demo{Proof} Let $\Phi\in\Cal F$ and let $G(z)$ denote the right
side of (4.36). In terms of the map $F(z)$ of (4.27), $G(z) =
F(z)(\Phi(z))$, so by Proposition~4.12, $G(z)\in\Cal D(z)$. By
the uniformity in Proposition~4.13, the fact that the $\mu_t\in
\Cal M^H(\gamma)$ and that $\{G_{\mu_t}(z)\mid t\in\Bbb R\cup
\{\infty\} \}=\partial\Cal D(z)$, this implies that $G(z)$ obeys
(4.28). Thus by Proposition~4.13, $G(z)=G_\rho(z)$ for some $\rho
\in \Cal M^H(\gamma)$.
Conversely, let $\rho\in\Cal M^H(\gamma)$ and consider
$G_\rho(z)$. By Theorem~4.3, Proposition~4.4 and Proposition~4.12,
$\Phi(z) \equiv F(z)^{-1} (G_\rho (z))\in \bar\Bbb C_+ \cup
\{\infty\}$, and so $\Phi\in\Cal F$. Thus, $G_\rho(z) = F(z)
(\Phi(z))$, that is, (4.36) holds.
If $\rho$ is not a von~Neumann solution, then $\text{Im}\,
\Phi_\rho (z) >0$, so $G_\rho(z) = F(z)(\Phi_\rho(z))$ is in
the interior of $\Cal D(z)$. \qed
\enddemo
\proclaim{Proposition 4.15} Let $\rho\in\Cal M^H(\gamma)$. If
$\rho$ is a von~Neumann solution, then the polynomials are
dense in $L^2(\Bbb R, d\rho)$ and for each $z_0\in\Bbb C_+$, there
is no other $\mu$ in $\Cal M^H(\gamma)$ with $G_\mu (z_0)=
G_\rho(z_0)$. If $\rho$ is not a von~Neumann solution, then the
polynomials are not dense in $L^2 (\Bbb R, d\rho)$ and for each
$z_0\in \Bbb C_+$, there are distinct $\mu$'s in $\Cal M^H(\gamma)$
with $G_\mu (z_0)=G_\rho(z_0)$.
\endproclaim
\demo{Proof} If $\rho$ is a von~Neumann solution, the polynomials
are dense by construction and the uniqueness result holds since
$G_\mu(z_0)=G_\rho(z_0)$ if and only if $\Phi_\mu (z_0) =
\Phi_\rho(z_0)$ and $\Phi_\rho(z_0)$ is then real.
Conversely, suppose $\rho$ is not a von~Neumann solution. Then,
$G_\rho(z_0)\in D(z)^{\text{\rom{int}}}$, so by the proof of
Theorem~4.3, $(x-z_0)^{-1}$ is not in the $L^2$ closure of the
polynomials. Thus, the polynomials are not dense. If $\Phi_\rho$
is not a constant, set $\Psi(z) = -|\Phi_\rho(z_0)|^2 \Phi_\rho
(z)^{-1} +2\text{Re}\, \Phi_\rho (z_0)$. If $\Phi_\rho(z)$ is
a constant, $q$ (of necessity with $\text{Im}\, q>0$), set
$\Psi(z) = cz+d$ with $c=\frac{\text{Im}\, q}{\text{Im}\, z_0}$
($>0$) and $d=\text{Re}\, q -c\,\text{Re}\, z_0$. In either case,
$\Psi \neq \Phi$, $\Psi$ is Herglotz, and $\Psi(z_0) = \Phi(z_0)$.
Thus, if $\mu = \rho_\Psi$, we have $G_\mu (z_0) = F(z_0)
(\Psi(z_0)) = F(z_0)(\Phi(z_0)) = G_\rho(z_0)$. \qed
\enddemo
To complete the proof of Theorem~5, we need the following fact
about Herglotz functions:
\proclaim{Theorem 4.16} Let $\Phi(z)$ be a Herglotz function so
$$
\Phi(z) = cz+ d + \int d\mu(x) \biggl[\frac{1}{x-z} -
\frac{x}{1+x^2}\biggr]
$$
with $d$ real, $c\geq 0$, and either $c>0$ or $d\mu\neq 0$.
Suppose for some $x_0$, $\Phi(x_0 + i\varepsilon)\to t\in
\Bbb R$. Then either $|\frac{[\Phi(x_0 +i\varepsilon)-t]}
{i\varepsilon}|\to\infty$ or else $\int \frac{d\mu(x)}
{(x-x_0)^2}<\infty$
and
$$
\lim_{\varepsilon\downarrow 0} \frac{\Phi(x_0 + i\varepsilon)
-t}{i\varepsilon} = c+ \int\frac{d\mu(x)}{(x-x_0)^2}\, .
\tag 4.37
$$
\endproclaim
\remark{Remark} Do not confuse the $\mu$ in the Herglotz
representation for $\Phi$ and the measure $\rho$ of (4.36).
In particular, $\mu$ is allowed to have finite support even
though we are suppposing that $\rho$ does not.
\endremark
\demo{Proof} Note first that
$$
\frac{\text{Im}\, \Phi(x_o + i\varepsilon)}{\varepsilon} =
c + \int \frac{d\mu(x)}{(x-x_0)^2 + \varepsilon^2}\, .
\tag 4.38
$$
On the other hand,
$$
\text{Im}\, \frac{\partial\Phi}{\partial z}\, (x_0 + i\varepsilon)
=\int\frac{d\mu(x)\,2\varepsilon (x-x_0)}
{[(x-x_0)^2 + \varepsilon^2]^2}\, . \tag 4.39
$$
If $\int\frac{d\mu(x)}{(x-x_0)^2}=\infty$, then (4.38) implies
$\text{Im}\,[\frac{\Phi(x_0 +i\varepsilon)-t}{\varepsilon}]
\to\infty$ , so if the limit is finite, then $\int
\frac{d\mu(x)}{(x-x_0)^2}<\infty$. (4.39) and the dominated
convergence theorem implies that $\text{Im}\, \frac{\partial\Phi}
{\partial z} (x_0 + i\varepsilon)\to 0$ so that $\frac{[\text{Re}\,
\Phi(x_0 + i\varepsilon)-t]}{i\varepsilon} = \frac{1}{\varepsilon}
\int_0^\varepsilon \text{Im}\, \frac{\partial\Phi}{\partial z}
(x_0 + iy)\, dy \to 0$. This and (4.38) implies (4.37). \qed
\enddemo
\proclaim{Theorem 4.17 (end of Theorem~5)} Let $\gamma$ be an
indeterminate moment problem. Let $\rho\in \Cal M^H(\gamma)$
correspond to a $\Phi$ which is not a constant in $\Bbb R \cup
\{\infty\}$ \rom(so that $\rho$ is not a von~Neumann solution\rom).
Suppose $\alpha\equiv\rho(\{x_0\}) >0$ for some point $x_0\in
\Bbb R$. Then there is a von~Neumann solution $\mu_t$ so that
$\mu_t (\{x_0\})>\alpha$.
\endproclaim
\demo{Proof} We will suppose that $D(x_0)\neq 0$. If $D(x_0)
=0$, then $B(x_0)\neq 0$ and we can use
$$
-\frac{C(z)\Phi(z)+A(z)}{D(z)\Phi(z)+B(z)} =
-\frac{A(z) (-\Phi(z))^{-1} - C(z)}{B(z)(-\Phi(z))^{-1}
- D(z)}
$$
and $-\Phi(z)^{-1}$ in place of $\Phi(z)$ to give an identical
argument. Define $t=-\frac{B(x_0)}{D(x_0)}\in\Bbb R$. Since
$AD-BC=1$,
$$
C(x_0)t + A(x_0) = \frac{1}{D(x_0)} \tag 4.40
$$
is non-zero and has the same sign as $D$.
$$
(-i\varepsilon) G_\rho (x_0+i\varepsilon) = (i\varepsilon)
\biggl[ \frac{C(x_0 + i\varepsilon) \Phi(x_0 + i\varepsilon)
+ A(x_0 + i\varepsilon)}{D(x_0+i\varepsilon)\Phi(x_0+i\varepsilon)
+B(x_0+i\varepsilon)} \biggr] \to \alpha.
$$
This is only possible if $G_\rho (x_0+i\varepsilon) \to\infty$,
which requires that $\Phi(x_0+i\varepsilon)\to t$. Then
$$
C(x_0+i\varepsilon) \Phi(x_0+i\varepsilon) +A(x_0+i\varepsilon)
\to C(x_0)t +A(x_0)
$$
while
$$
\frac{D(x_0+i\varepsilon) \Phi(x_0+i\varepsilon) +
B(x_0+i\varepsilon)}{i\varepsilon} \to D'(x_0) t + B'(x_0) +
D(x_0) \lim_{\varepsilon\downarrow 0} \frac{\Phi(x_0+i\varepsilon)
-t}{i\varepsilon}\, .
$$
Thus using (4.40),
$$
\alpha = \frac{1}{D(x_0)^2 \lim_{\varepsilon\downarrow 0}
\frac{\Phi(x_0+i\varepsilon) -t}{i\varepsilon} + D(x_0)
[D'(x_0)t + B'(x)]}\, .
$$
On the other hand, taking $\Phi(x)\equiv t$ to describe $\mu_t$,
$$
\mu_t (\{x_0\}) = \frac{1}{D(x_0) [D'(x_0)t + B'(x_0)]}
$$
must be positive. It follows from Lemma~4.16 that $\int
\frac{d\mu(x)}{(x-x_0)^2}<\infty$ and that
$$
\alpha^{-1} > \mu_t (\{x_0\})^{-1}. \qed
$$
\enddemo
Finally, we turn to the Nevanlinna parametrization in the
indeterminate Stieltjes case. As we have seen, among the $B_t$
there are two distingtuished ones, $B_F$ and $B_K$. The Krein
extension has $\ker(A^*)\subset D(B_K)$ so $\pi(0)\in D(B_K)$,
which means $t=\infty$ in the notation of (4.20). The Friedrichs
extension is $B_F=B_{t_0}$ where $t_0 = (\delta_0, A^{-1}_F
\delta_0)$.
\proclaim{Theorem 4.18 (Stieltjes case of Theorem 4)} If the
Stieltjes problem for $\gamma$ is indeterminate, its solutions
obey {\rom{(4.36)}} for precisely those $\Phi$ that either are
a constant in $[t_0, \infty)\cup\{\infty\}$ or that obey $\Phi(z)$
is analytic in $\Bbb C\backslash [0,\infty)$ with $\Phi(x) \in
[t_0,\infty)$ if $x\in (-\infty, 0)$. Here $t_0 = (\delta_0,
A^{-1}_F \delta_0)$.
\endproclaim
\remark{Remark} Such $\Phi$'s are precisely those of the form
$$
\Phi(z) = d + \int_0^\infty \frac{d\mu(x)}{x-z}\, , \tag 4.41
$$
where $d\geq t_0$ and $\int_0^\infty \frac{d\mu(x)}{x+1}<\infty$.
\endremark
\demo{Proof} Clearly, the solutions of the Stieltjes problem are
precisely the solutions $\rho$ of the Hamburger problem with
$\lim_{\varepsilon\downarrow 0} G_\rho (-y + i\varepsilon)$ real
for each $y\in (0,\infty)$. Since the $B_t$'s obey $t=(\delta_0,
B^{-1}_t \delta_0)$ and $B_K < B_t < B_F$ means $t_0 < (\delta_0,
B^{-1}_t \delta_0)<\infty$, we know that for $y\in (-\infty,0)$,
$F(y)$ maps $[t_0, \infty)$ into $(0,\infty)$ ($F(\,\cdot\,)$ is
given by (4.27)). It follows that if $\Phi$ obeys $\Phi(x)\in
[t_0, \infty)$ for $x\in (-\infty, 0)$, then $G_\rho (y+i
\varepsilon)$ has a limit in $(0,\infty)$ on $(-\infty, 0)$,
that is, $\Phi$ defines a solution of the Stieltjes moment
problem. The converse follows from this argument and the next
theorem. \qed
\enddemo
\proclaim{Theorem 4.19} Let $d\mu_F$ and $d\mu_K$ be the
Friedrichs and Krein solutions of an indeterminate Stieltjes
moment problem and let $d\rho$ be another solution. Then for
all $y\in (0,\infty)$,
$$
\int_0^\infty \frac{d\mu_F (x)}{x+y} \leq
\int_0^\infty \frac{d\rho(x)}{x+y} \leq
\int_0^\infty \frac{d\mu_K (x)}{x+y}\, . \tag 4.42
$$
\endproclaim
This will be proven below as part of Theorem~5.2.
\proclaim{Corollary 4.20} If $\rho\in\Cal M^S(\gamma)$ and
$d\rho\neq d\mu_F$, then for $y\in [0,\infty)$,
$$
\int_0^\infty \frac{d\mu_F(x)}{x+y} < \int_0^\infty
\frac{d\rho(x)}{x+y}\,. \tag 4.43
$$
\rom(Note that the inequality is strict.\rom)
\endproclaim
\demo{Proof} For each $y\in (0,\infty)$, $F(-y)(\, \cdot\,)$ is
a strictly monotone map of $[t_0, \infty)$ to $[G_{\mu_F}(-y),
\mathbreak G_{\mu_K}(-y))$ by the proof of Theorem~4.18 and the
fact that any fractional linear map of $\Bbb R$ to $\Bbb R$ is
monotone in any region where it is finite. By a limiting argument,
the same is true for $y=0$. Thus, (4.43) follows from the relation
of Nevanlinna functions $\Phi_\rho(y) > \Phi_{\mu_F}(y) \equiv
t_0$. This is immediate from (4.41). \qed
\enddemo
As a corollary of this, we see
\proclaim{Corollary 4.21} Let $\gamma$ be a set of indeterminate
Stieltjes moments and let $d\mu_F$ be its Friedrichs solution. Let
$r=\int x^{-1} d\mu_F(x)$ and let
$$
\tilde\gamma_0 = 1; \qquad \tilde\gamma_n = r^{-1}\gamma_{n-1},
\qquad n=1,2,\dots .
$$
Then $\{\tilde\gamma_j\}^\infty_{j=0}$ is a determinate
Hamburger moment problem.
\endproclaim
\remark{Remark} This lets us produce Stieltjes determinate
moment problems with essentially arbitrary rates of growth for
$\gamma_n$. By using Theorem~2.13, we obtain determinate
Hamburger problems with arbitrary fast growth also.
\endremark
\demo{Proof} $r^{-1} x^{-1} d\mu_F\equiv d\lambda_0(x)$ is a
Stieltjes measure whose moments are $\tilde\gamma$. Let $d\lambda(x)$
be another such Stieltjes measure and let $d\rho(x)=rx\, d\lambda(x)$.
Then the moments of $d\rho$ are $\gamma$ and $\int x^{-1} d\rho(x)
=r$. So, by Corollary~4.20, $\rho=\mu_F$. Thus, $\lambda =
\lambda_0$. Thus, $\lambda_0$ is the unique solution of the Stieltjes
problem. But $0\not\in\text{supp}(d\lambda_0)$. Thus by
Proposition~3.1, the Hamburger problem is also determinate.
\qed
\enddemo
\vskip 0.3in
\flushpar{\bf \S 5. Pad\'e Approximants, Continued Fractions, and
Finite Matrix Approximations}
\vskip 0.1in
In this section, we will primarily consider aspects of the
Stieltjes moment problem. We will discuss finite matrix
approximations which converge to the Friedrichs and Krein
extensions of $A$ and, as a bonus, obtain Theorem~4.19 and
Theorems~5, 6, 7, and 8. We will see that the convergence of the
Pad\'e approximants (equivalently, continued fractions of
Stieltjes) can be reinterpreted as strong resolvent convergence
of certain finite matrix approximations to $A_F$ and $A_K$.
Before we start, we note that our continued fraction expressions
are related to but distinct from those of Stieltjes. Our basic
object (the approximations to $A_F$) are of the form:
$$
\cfrac1 \\
-z+b_0 - \cfrac{a^2_0} \\
-z+b_1 - \cfrac{a^2_1} \\
-z+b_2 + \cdots \endcfrac \tag 5.1a
$$
with two terms (an affine polynomial) at each stage while
Stieltjes' are of the form
$$
\cfrac1 \\
c_1 w + \cfrac1 \\
c_2 + \cfrac1 \\
c_3 w+\cdots \endcfrac \tag 5.1b
$$
with a single term in each step and alternate constant and linear
terms. This gives him twice as many continued fractions, so his
alternate between the two sets of rational approximations we get.
We will say more about this later.
Until Theorem~5.29 below, we will suppose
$\{\gamma_n\}^\infty_{n=0}$ is a set of Stieltjes moments.
Given the Jacobi matrix (1.16), we will consider two $N\times N$
approximations. The first is obvious:
$$
A^{[N]}_F = \pmatrix
b_0 & a_0 & {} & {} & {} \\
a_0 & b_1 & {} & {} & {} \\
{} & {} & \ddots & {} & {} \\
{} & {} & {} & b_{N-2} & a_{N-2} \\
{} & {} & {} & a_{N-2} & b_{N-1} \endpmatrix . \tag 5.2a
$$
The second differs by the value of the $NN$ coefficient:
$$
A^{[N]}_K = \pmatrix
b_0 & a_0 & {} & {} & {} \\
a_0 & b_1 & {} & {} & {} \\
{} & {} & \ddots & {} & {} \\
{} & {} & {} & b_{N-2} & a_{N-2} \\
{} & {} & {} & a_{N-2} & b_{N-1}-\alpha_{N-1} \endpmatrix ,
\tag 5.2b
$$
where $\alpha_{N-1}$ is chosen so that $A^{[N]}_K$ has a zero
eigenvalue. That such an $\alpha_{N-1}$ exists is the content of
\proclaim{Lemma 5.1} There is a unique $\alpha_{N-1}>0$ so that
{\rom{(5.2b)}} has a zero eigenvalue. $A^{[N]}_K \geq 0$ and
$\alpha_{N-1}$ obeys
$$
(b_{N-1} - \alpha_{N-1}) P_{N-1}(0) + a_{N-2} P_{N-2}(0)=0
\tag 5.3
$$
and
$$
\alpha_{N-1} = -a_{N-1}\, \frac{P_N(0)}{P_{N-1}(0)}\, . \tag 5.4
$$
Moreover,
$$
(b_N-\alpha_N)(\alpha_{N-1}) -a^2_{N-1} = 0. \tag 5.5
$$
\endproclaim
\demo{Proof} Any solution of $A^{[N]}_K u=0$ must obey the
eigenfunction equation (4.1) and so be a multiple of $P_j(0)$,
that is,
$$
u_j = P_j(0), \qquad j=0,1,\dots,N-1.
$$
This obeys the condition at the lower right of the matrix if
and only if (5.3) holds.
If $P_{N-1}(0)$ were $0$, then $A^{[N-1]}_F$ would have a zero
eigenvalue, which is impossible since $A^{[N-1]}_F$ is strictly
positive definite (since the form $S_{N-1}$ is strictly positive
definite). Thus, $P_{N-1}(0)\neq 0$ and so (5.3) has a unique
solution. Since $A^{[N]}_F$ is positive definite, this solution
must have $\alpha_{N-1}>0$.
The eigenfunction equation for $P_N(0)$ then yields (5.4). (5.3)
for $N\to N+1$ and (5.4) yield two formulas for $\frac{P_N(0)}
{P_{N-1}(0)}$. Setting these to each other yields (5.5). \qed
\enddemo
The first main result of this section, towards which we are
heading, is
\proclaim{Theorem 5.2 (includes Theorem 4.19)} Let
$$
f^+_N(z)\equiv (\delta_0, (A^{[N]}_K-z)^{-1}\delta_0) \tag 5.6
$$
and
$$
f^-_N(z) \equiv (\delta_0, (A^{[N]}_F -z)^{-1}\delta_0). \tag 5.7
$$
Then for $x\in (0,\infty)$,
$$
f^+_N(-x) \geq f^+_{N+1}(-x) \text{ converges to } (\delta_0,
(A_K +x)^{-1}\delta_0)\equiv f^+ (-x)
$$
while
$$
f^-_N (-x) \leq f^-_{N+1} (-x) \text{ converges to }
(\delta_0, (A_F +x)^{-1}\delta_0) \equiv f^-(-x).
$$
The convergence holds for any $z\in\Bbb C\backslash [0,\infty)$
and is uniform on compacts. Moreover, for any solution $\rho$ of
the moment problem and $x>0$,
$$
f^- (-x) \leq \int\frac{d\rho(y)}{x+y} \leq f^+ (-x). \tag 5.8
$$
\endproclaim
\remark{Remarks} 1. Notice that (5.8) is (4.42).
2. Let $A^{[N]}(\beta)$ be (5.2b) with $\alpha_{N-1}$
replaced by $-\beta$. Then $A^{[N]}_K$ is quite a natural
approximation: It is that $A^{[N]} (\beta)$ which is minimal
among the non-negative matrices --- reasonable to capture $A_K$,
the minimal non-negative self-adjoint extension. From this point
of view, $A^{[N]}_F$ seems less natural. One would instead want
to consider $\lim_{\beta\to\infty} A^{[N]}(\beta)$. In fact, one
can show this limit is ``essentially" $A^{[N-1]}_F$ in the sense
that
$$
\lim_{\beta\to\infty} (\delta_0, (A^{[N]}(\beta)-z)^{-1}\delta_0)
=(\delta_0, (A^{[N-1]}_F -z)^{-1} \delta_0).
$$
Thus, the $A^{[N]}(\beta)$ interpolate between $A^{[N]}_K$
and $A^{[N-1]}_F$.
\endremark
\proclaim{Proposition 5.3} Let $\tilde A^{[N]}_K$ be the operator
on $\ell^2$ which is $A^{[N]}_K \oplus 0$. Then $\tilde A^{[N]}_K$
converges to $A_K$ in strong resolvent sense.
\endproclaim
\remark{Remark} For self-adjoint operators $\{A_n\}^\infty_{n=1}$
and $A_\infty$, we say $A_n$ converges to $A_\infty$ in strong
resolvent sense if and only if for all $z\in\Bbb C\backslash
\Bbb R$, $(A_n -z)^{-1}\varphi \to (A_\infty - z)^{-1}\varphi$
for all vectors $\varphi$. It is a basic theorem (see Reed-Simon
[\rsI]) that it suffices to prove this for a single $z\in
\Bbb C\backslash\Bbb R$. The same proof shows that if all
$A_n$ and $A_\infty$ are non-negative, one then also has
convergence for $z\in(-\infty, 0)$. Moreover, one has
convergence uniformly for $z$ in compact subsets of $\Bbb C
\backslash [0,\infty)$.
\endremark
\demo{Proof} Suppose first that the Hamburger problem is
indeterminate. Thus, $\pi(0)=\mathbreak (P_0(0),\dots) \in
\ell_2$. Let $P^{[n]}(0)$ be the vector $(P_0(0), \dots,
P_{n-1}(0), 0,0,\dots )$. Then $A^{[n]}_K P^{[n]} \mathbreak
=0$ so $(A^{[n]}_K-z)^{-1} P^{[n]}=-z^{-1} P^{[n]}$ for any
$z\in\Bbb C\backslash\Bbb R$. Since $\| (A^{[n]}_K -z)^{-1}\|
\leq |\text{Im}\,z|^{-1}$ for any such $z$, and $\|P^{[n]} -
\pi (0)\| \to 0$, we conclude that
$$
(A^{[n]}_K -z)^{-1} \pi (0) \to -z^{-1} \pi (0) =
(A_K -z)^{-1} \pi (0) \tag 5.9
$$
since $A_K \pi (0) =0$.
Suppose $\varphi = (A-z)\eta$ for $\eta\in D(A)$, that is, a
finite sequence. Then $(A_K -z)\eta = \varphi = (A^{[n]}_K -z)
\eta$ for $n$ large, and thus $\lim_{n\to\infty} (A^{[n]}_K
-z)^{-1} \varphi = \eta= (A_K -z)^{-1} \varphi$. Thus, $(A^{[n]}_K
-z)^{-1}\varphi \to (A-z)^{-1}\varphi$ for $\varphi\in
\overline{\text{Ran}(A-z)}$. In the determinate case, that is
all $\varphi$'s; while in the indeterminate case, we claim that
$\overline{\text{Ran}(A-z)} + [\pi(0)]$ is all of $\ell^2$
so (5.9) completes the proof.
That leaves the proof of our claim that $\overline{\text{Ran}
(A-z)} + [\pi(0)]$ is $\ell^2$. We first note that $\pi(0)\notin
D(\bar A)$ by Proposition~4.4. $(A_K -z)^{-1}
(\,\overline{\text{Ran}(A-z)} + [\pi(0)]) = D(\bar A) + [\pi(0)]$
since $(A_K - z)^{-1})\pi(0) =-z^{-1}\pi(0)$. But
$D(A_K)/D(\bar A)$ has $\dim 1$ and $\pi(0)\notin D(\bar A)$ so
$(A_K-z)^{-1} (\, \overline{\text{Ran}(A-z)} + [\pi(0)])=
D(A_K)$ and thus, since $\text{Im}\, z\neq 0$,
$\overline{\text{Ran}(A-z)} + [\pi(0)]=\ell_2$, as claimed. \qed
\enddemo
\proclaim{Proposition 5.4} Let $\Tilde{\Tilde A}^{[N]}_K$ be the
$(N+1)\times (N+1)$ matrix which is $A^{[N]}_K \oplus 0$,
that is, it has zeros in its last row and column. Then
$$
\Tilde{\Tilde A}^{[N]}_K \leq A^{[N+1]}_K. \tag 5.10
$$
In particular, for $x>0$,
$$
(\delta_0, (A^{[N+1]}_K + x)^{-1}\delta_0) \leq
(\delta_0, (\Tilde{\Tilde A}^{[N]}_K + x)^{-1} \delta_0) =
(\delta_0, (A^{[N]}_K + x)^{-1}\delta_0).
$$
\endproclaim
\remark{Remark} Propositions~5.3 and 5.4 prove the part of
Theorem~5.2 involving monotonicity of $f^+_N$ and its
convergence.
\endremark
\demo{Proof} $B\equiv A^{[N+1]}_K - \Tilde{\Tilde A}^{[N]}_K$ is
an $(N+1)\times (N+1)$ matrix which has all zeros except for a
$2\times 2$ block in its lower right corner. This block is
$$
\pmatrix \alpha_{N-1} & a_{N-1} \\
a_{N-1} & b_N - \alpha_N \endpmatrix ,
$$
which is symmetric with a positive trace and determinant zero
(by (5.5)). Thus, the block is a positive rank one operator so
$B$ is positive, proving (5.10). \qed
\enddemo
The monotonicity of the $A^{[N]}_F$'s and their convergence is
a special case of monotone convergence theorems for forms.
Consider non-negative closed quadratic forms whose domain may not
be dense. Each such form $q_B$ with $D(q_B)$ is associated to a
non-negative self-adjoint operator $B$ on $\overline{D(q_B)}$.
We define $(B -x)^{-1}$ to be $0$ on $D(q_B)^\bot$. This
essentially sets $B=\infty$ on $D(q_B)^\bot$ consistent with the
intention that $D(q_B)$ is the set of $\varphi$ for which
$q_B(\varphi) <\infty$.
Given two forms $q_B$ and $q_C$, we say $q_B \leq q_C$ if
and only if $D(q_B)\supseteq D(q_C)$ (intuitively $q_C(\varphi)
< \infty$ implies $q_B(\varphi) <\infty$) and for all $\varphi
\in q_C$, $q_C(\varphi)\geq q_B(\varphi)$. It is a fundamental
result [\kato, \simjfa] that if $q_B \leq q_C$, then for all
$x\geq 0$ and all $\varphi$,
$$
(\varphi, (C+x)^{-1}\varphi) \leq (\varphi, (B+x)^{-1}\varphi).
\tag 5.11
$$
One monotone convergence theorem for forms [\kato, \simjfa]
says the following: Let $\{q_{A_n}\}^\infty_{n=1}$ be a sequence
of quadratic forms with $q_{A_n} \geq q_{A_{n+1}}$. Then $(A_n
+x)^{-1}$ converges to the resolvent, $(A_\infty +x)^{-1}$, of an
operator $A_\infty$. It is the operator associated to the form
$q_{A_\infty}$ defined to be the closure of $\tilde q_{A_\infty}$
where $D(\tilde q_{A_\infty}) = \cup D(q_{A_n})$ with
$\tilde q_{A_\infty}(\varphi) = \lim_{n\to\infty} q_{A_n}
(\varphi)$ provided that the form $\tilde q_{A_\infty}$ is
closable.
We can apply this to the $A^{[N]}_F$. Let $q^{[N]}_F$ be defined
to be the form with $D(q^{[N]}_F)=\{(\varphi_0, \dots,
\varphi_{N-1}, 0, \dots, 0, \dots)\}$ and $q^{[N]}_F (\varphi)
= (\varphi, A\varphi)$. Then $D(q^{[N]}_F)\subset D(q^{[N+1]}_F)$
and $q^{[N]}_F (\varphi) = q^{[N+1]}_F (\varphi)$ if $\varphi\in
D(q^{[N]}_F)$. Thus, $q^{[N]}_F \geq q^{[N+1]}_F$ (in essence,
$\tilde A^{[N]}_F = A^{[N]}_F \oplus \infty$ on $\Bbb C^{N+1}$).
The monotone convergence theorem applies. $A_\infty$ is just the
closure of $\varphi\mapsto (\varphi, A\varphi)$ on finite sequences,
that is, the Friedrichs extension $A_F$, so we have the
convergence and monotonicity part of Theorem~5.2 for $f^-_N$
(using (5.11)):
\proclaim{Proposition 5.5} Let $x>0$. Then $(\delta_0,
(A^{[N+1]}_F +x)^{-1} \delta_0) \geq (\delta_0, (A^{[N]}_F + x)^{-1}
\delta_0)$ and this converges to $(\delta_0, (A_F+x)^{-1}\delta_0)$
as $N\to\infty$.
\endproclaim
To head towards the remaining part of Theorem~5.2, viz.~(5.8),
we need expressions for $(\varphi, (A^{[N]}_F -z)^{-1}\varphi)$
and $(\varphi, (A^{[N]}_K -z)^{-1}\varphi)$, which are of
independent interest.
\proclaim{Proposition 5.6}
$$
(\delta_0, (A^{[N]}_F -z)^{-1}\delta_0)=
-\frac{Q_N(z)}{P_N(z)}\, \tag 5.12
$$
\endproclaim
\demo{Proof} The only possible eigenfunctions of $A^{[N]}_F$ are
the vectors $\pi^{[N]} (z) \equiv (P_0(z), \dots, \mathbreak
P_{N-1}(z))$. By the eigenfunction equation (4.1), this is an
eigenfunction of $A^{(N)}_F$ if and only if $P_N(z)=0$. Thus, the
$N$ poles of $(\delta_0, (A^{[N]}_F -z)^{-1}\delta_0)$ are
precisely the $N$-zeros of $P_N(z)$. The zeros are distinct
because the eigenvalues are simple (since $\pi^{[N]}(z)$ is the
only potential eigenfunction, since eigenfunctions must have
$u_0\neq 0$).
Now let the $B^{[N]}_F$ be the $(N-1)\times (N-1)$ matrix obtained
by removing the top row and left column of $A^{[N]}_F$. By
Cramer's rule, $(\delta_0, (A^{[N]}_F -z)^{-1}\delta_0) =
\frac{\det (B^{[N]}_F -z)}{\det (A^{[N]}_F-z)}$, so the $N-1$
zeros of $(\delta_0, (A^{[N]}_F -z)^{-1}\delta_0)$ are precisely
the $N-1$ eigenvalues of $B^{[N]}_F$. The only possible
eigenfunctions of $B^{[N]}_F$ are $\xi(z)^{[N-1]} = (Q_1(z),
\dots, Q_{N-1}(z))$ (since $Q_0(z)=0$, $Q_1(z)=\frac{1}{a_0}$).
Thus, the eigenvalues of $B^{[N]}_F$ are precisely $z$'s with
$Q_N(z)=0$. It follows that
$$
(\delta_0, (A^{[N]}_F -z)^{-1}\delta_0) = \frac{d_N Q_N(z)}
{P_N(z)}\, ,
$$
and we need only show that $d_N \equiv -1$.
Since $(\delta_0, (A^{[N=1]}_F -z)^{-1}\delta_0) = (b_0 -z)^{-1}$
and $Q_1(z) = \frac{1}{a_0}$, $P_1(z) = \frac{(z-b_0)}{a_0}$,
we see that $d_1 =-1$. On the other hand, Proposition~4.1 implies
that
$$
\frac{Q_k (z)}{P_k (z)} - \frac{Q_{k-1}(z)}{P_{k-1}(z)} =
\frac{1}{a_{k-1} P_k (z) P_{k-1}(z)} = o\biggl(\frac{1}{z}\biggr).
$$
Moreover, $(\delta_0, (A-z)^{-1})\delta_0) = -z^{-1} +O(z^{-2})$,
so $d_k = d_{k-1}$, that is, $d_N =-1$ for all $N$. \qed
\enddemo
\remark{Remarks} 1. The proof shows that for suitable $c_N$
(indeed, an induction shows that $c_N = (-1)^N (a_0 \dots
a_{N-1})^{-1}$), we have
$$
P_N(z) = c_N \det(A^{[N]}_F -z) \tag 5.13
$$
and
$$
Q_N(z) = -c_N \det(B^{[N]}_F -z). \tag 5.14
$$
2. Since $A^{[N]}_F$ is strictly positive definite, (5.13) shows
once more that $P_N(0)\neq 0$ in the Stieltjes case. Since
$B^{[N]}_F$ is also strictly positive definite (as a submatrix
of $A^{[N]}_F$), (5.14) shows that $Q_N(0)\neq 0$ also in the
Stieltjes case.
\endremark
(5.13)--(5.14) imply some facts about the zeros of $P_N$ and
$Q_N$:
\proclaim{Proposition 5.7} All the zeros of each $P_N(z)$ and
each $Q_N(z)$ are real. Moreover, there is exactly one zero of
$Q_N(z)$ and one zero of $P_{N-1}(z)$ between any pair of
successive zeros of $P_N(z)$.
\endproclaim
\demo{Proof} The first assertion follows from the fact that the
eigenvalues of a real-symmetric matrix are all real. The second
assertion follows from the fact that if $X$ is an $N\times N$
real-symmetric matrix and $Y$ is an $N-1 \times N-1$ submatrix,
then by the min-max principle, there is exactly one eigenvalue
of $Y$ between any pair of eigenvalues of $X$. \qed
\enddemo
\remark{Remark} Since the $Q_N$'s are orthogonal polynomials for
another moment problem (see Proposition~5.16), between any two
successive zeros of $Q_N(z)$, there is a single zero of $Q_{N-1}
(z)$.
\endremark
Now define
$$\align
M_N(z) &= P_N(z) - P_N(0)\, \frac{P_{N-1}(z)}{P_{N-1}(0)} \tag 5.15 \\
N_N(z) &= Q_N(z) - P_N(0)\, \frac{Q_{N-1}(z)}{P_{N-1}(0)}\tag 5.16
\endalign
$$
since $A^{[j]}_F$ is strictly positive, $0$ is never an eigenvalue
and $P_j(0)\neq 0$ for all $j$, so $M$ is well-defined. Notice
$M_N(0)=0$.
\proclaim{Proposition 5.8}
$$
(\delta_0, (A^{[N]}_K -z)^{-1}\delta_0) = -\frac{N_N(z)}
{M_N(z)} \tag 5.17
$$
\endproclaim
\demo{Proof} Let $c_N$ be the constant in (5.13) and let
$B^{[N]}_K$ be $A^{[N]}_K$ with the top row and left-most
column removed. We will prove that
$$
M_N(z) = c_N\det(A^{[N]}_K-z); \qquad N_N(z)=-c_N \det
(B^{[N]}_K -z), \tag 5.18
$$
from which (5.17) holds by Cramer's rule.
Clearly,
$$
\det (A^{[N]}_K -z) = \det(A^{[N]}_F -z) - \alpha_{N-1}
\det(A^{[N-1]}_F -z)
$$
so
$$
c_N \det(A^{[N]}_K -z) = P_N(z) - \beta_N P_{N-1}(z),
$$
where
$$
\beta_N = \alpha_{N-1} \, \frac{c_N}{c_{N-1}}\, .
$$
But $\det(A^{[N]}_K) =0$ by construction. So $P_N(0) - \beta_N
P_{N-1}(0)=0$ and thus, $c_N \det (A^{[N]}_K -z) = M_N (z)$.
In the same way,
$$
c_N \det(B^{[N]}_K -z) = -Q_N(z) + \beta_N Q_{N-1}(z) = -N_N (z).
\qed
$$
\enddemo
Because of our convergence theorem, we have
\proclaim{Corollary 5.9} Let $\{\lambda^{[N]}_i\}^N_{i=1}$
be the zeros of $P_N(z)$ and let $\nu^{[N]}_i = -
\frac{Q_N (\lambda^{[N]}_i)}{P'_N (\lambda^{[N]}_i)}$. Then
$\sum_{i=1}^N \nu^{[N]}_i \delta (\lambda - \lambda^{[N]}_i)$
converges to $d\mu_F (\lambda)$, the Friedrichs solution
of the moment problem. A similar formula holds for the Krein
solution, $d\mu_K(\lambda)$, with $P,Q$ replaced by $M,N$.
\endproclaim
The following concludes the proof of Theorem~5.2. It extends
the calculation behind (4.4):
\proclaim{Theorem 5.10} Let $d\rho$ solve the Stieltjes moment
problem. Then for $x> 0$ and $N\geq 1$,
$$
-\frac{Q_N(-x)}{P_N(-x)} \leq \int \frac{d\rho(y)}{x+y}
\leq -\frac{N_N(-x)}{M_N (-x)}\, . \tag 5.19
$$
\endproclaim
\demo{Proof} $\frac{P_N (y) - P_N(z)}{y-z}$ is a polynomial in
$y$ of degree $N-1$ so
$$
\int d\rho(y) P_N(y)\biggl[ \frac{P_N(y)-P_N(z)}{y-z}\biggr]
=0. \tag 5.20
$$
On the other hand, (4.4) says that
$$
\int d\rho(y)\, \frac{P_N(y)}{y-z} = P_N(z) \int\frac{d\rho(y)}
{y-z} + Q_N(z). \tag 5.21
$$
Thus for $z=-x$ with $x>0$,
$$
0\leq \int\frac{d\rho(y) P^2_N(y)}{y+x} = P_N(-x) \int
\frac{d\rho(y)P_N(y)}{y+x} = P_N (-x)^2 \int \frac{d\rho(y)}
{x+y} + Q_N (-x) P_N (-x).
$$
Dividing by $P_N(-x)^2$, we obtain the left-most inequality in
(5.19).
Similarly, since $M_N(0)=0$, $\frac{M_N(z)}{z}$ is a polynomial
of degree $N-1$ and so $[y^{-1}M_N(y) -z^{-1} M_N(z)]/y-z$
is a polynomial of degree $N-2$ in $y$, which is orthogonal to
$P_N (y)$ and $P_{N-1}(y)$ and so to $M_N(y)$. Thus
$$
\int d\rho(y)\, \frac{M_N(y)}{y-z} \biggl[ \frac{M_N(y)}{y}
-\frac{M_N(z)}{z}\biggr] =0. \tag 5.22
$$
Since for $N\geq 0$,
$$
Q_N(z) = \int d\rho(y)\, \frac{P_N(y)-P_N(z)}{y-z} \tag 5.23
$$
for each $z$, we see that for $N\geq 1$,
$$
N_N(z) = \int d\rho(y) \, \frac{M_N (y)- M_N(z)}{y-z}\, .
\tag 5.24
$$
Therefore, for $x>0$,
$$\align
0\leq \int \frac{d\rho(y)}{y+x}\,\, \frac{M_N(y)^2}{y} &=
\frac{M_N(-x)}{(-x)} \int d\rho(y)\, \frac{M_N(y)}{y+x}
\qquad \text{by (5.22))} \\
&=\frac{M_N(-x)^2}{(-x)} \int \frac{d\rho(y)}{x+y} +
\frac{M_N(-x)N_N(-x)}{(-x)}\, .
\endalign
$$
Dividing by $\frac{M_N(-x)^2}{x}$, we obtain the second half
of (5.19). \qed
\enddemo
Next, we turn to Theorem~6 and the connection of Pad\'e
approximants to these finite matrix approximations. Given a formal
power series $\sum_{n=0}^\infty \kappa_n z^n$, we define the
Pad\'e approximants $f^{[N,M]}(z)$ as follows (see Baker and
Graves-Morris [\bgm] for background on Pad\'e approximants): We
seek a function $f^{[N,M]}(z)$ so that
$$
f^{[N,M]}(z) = \frac{A^{[N,M]}(z)}{B^{[N,M]}(z)}\, , \tag 5.25
$$
where $A^{[N,M]}(z)$ is a polynomial of degree at most $N$,
$B^{[N,M]}(z)$ is a polynomial of degree at most $M$, as $z\to 0$,
$$
f^{[N,M]}(z) - \sum_{j=0}^{N+M} \kappa_j z^j = O(z^{N+M+1})
\tag 5.26
$$
and
$$
B^{[N,M]}(0)=1. \tag 5.27
$$
There is at most one solution, $f$, for these equations since if
$\tilde A, \tilde B$ are another pair of such polynomials, by
(5.26), $\tilde A B-A\tilde B = O(z^{N+M+1})$ must be zero since
$\tilde A B-A\tilde B$ is a polynomial of degree $N+M$. Thus,
$$
A^{[N,M]}(z) \tilde B^{[N,M]}(z) - \tilde A^{[N,M]}(z)
B^{[N,M]}(z) = 0. \tag 5.28
$$
This implies $A/B = \tilde A/\tilde B$, showing $f$ is uniquely
determined as an analytic function. If
$$
\deg A=N, \qquad \deg B=M, \qquad A,B \text{ relatively prime},
\tag 5.29
$$
then (5.28) shows $\tilde A = A$ and $\tilde B=B$, so $A$ and
$B$ are uniquely determined. It can be shown ([\bgm]) that if
$$
\det ((\kappa_{N-M+i+j-1})_{1\leq i,j\leq M})\neq 0 \tag 5.30
$$
then $A,B$ exist and obey (5.29).
There are degenerate cases where a Pad\'e approximant exists,
but $A,B$ are not unique (e.g., for $\sum_{n=0}^\infty \kappa_n
z^n = 1+z$, $f^{[2,1]}(z) = 1+z$ can be written as $A(z) =
(1+z)(1+\alpha z)$, $B(z) = (1+\alpha z)$ for any $\alpha$).
In any event, if a solution of (5.25)--(5.27) exists, we say the
$[N,M]$ Pad\'e approximant exists and denote it by $f^{[N,M]}(z)$.
In the context of Theorem~6, we are interested in the Pad\'e
approximants for the Taylor series of
$$
f(z) = \int_0^\infty \frac{d\rho(x)}{1+zx}\, , \tag 5.31
$$
where $\rho$ is a measure with finite moments. Without loss,
normalize $\rho$ so $\int d\rho =1$. Recall the context of
Theorem~6. We have
$$
\kappa_n = (-1)^n \int_0^\infty x^n\, d\rho(x)
$$
and want to formally sum near $z=0$:
$$
f(z) \sim \sum_{n=0}^\infty \kappa_n z^n.
$$
If we define
$$\gamma_n = \int_0^\infty x^n\, d\rho(x),
$$
then near $w=\infty$,
$$
G_\rho (w) \equiv \int_0^\infty \frac{d\rho(x)}{x-w} \sim
-\sum_{n=0}^\infty \gamma_n w^{-n-1}. \tag 5.32
$$
Thus formally,
$$
f(z) = \frac{1}{z}\, G_\rho \biggl(-\frac{1}{z}\biggr). \tag 5.33
$$
We begin by noting:
\proclaim{Proposition 5.11} As $|w|\to\infty$,
$$\align
-\frac{Q_N(w)}{P_N(w)} &= -\sum_{j=0}^{2N-1} \gamma_j w^{-j-1}
+O(w^{-2N-1}) \tag 5.34 \\
-\frac{N_N(w)}{M_N(w)} &= -\sum_{j=0}^{2N-2} \gamma_j
w^{-j-1} + O(w^{-2N}) \tag 5.35
\endalign
$$
\endproclaim
\demo{Proof} Let $A$ be the Jacobi matrix for the moment problem
$\{\gamma_n\}^\infty_{n=0}$. Then
$$
\gamma_n = \langle \delta_0, A^n \delta_0 \rangle. \tag 5.36
$$
On the other hand, we have an expansion converging near infinity
for
$$
\langle \delta_0, (A^{[N]}_F -w)^{-1}\delta_0\rangle =
-\sum_{j=0}^\infty (\delta_0, (A^{[N]}_F)^j \delta_0)w^{-j-1},
$$
so (5.34) follows from
$$
\langle \delta_0, (A^{[N]}_F)^j \delta_0\rangle =
\langle \delta_0, A^j \delta_0\rangle , \qquad j=0,1,\dots, 2N-1. \tag 5.37
$$
To prove (5.37), note that
$$
(A^{[N]}_F)^j \delta_0 = A^j_F \delta_0 \tag 5.38
$$
if $j=0,\dots, N-1$ and for $B$ symmetric,
$$
\langle \delta_0, B^{2j}\delta_0\rangle = \|B^j\delta_0\|
$$
and that
$$
\langle \delta_0, B^{2j+1}\delta_0\rangle = \langle B^j\delta_0, B(B^j
\delta_0)\rangle. \tag 5.39
$$
To obtain (5.35), note that if $A^{[N]}_F$ is replaced by
$A^{[N]}_K$, (5.38) still holds for $j=0,\dots, N-1$, but
(5.39) fails for $j=N-1$ and so the analog of (5.37) only holds
for $j=0,\dots, 2N-2$. \qed
\enddemo
\remark{Remark} While (5.37) holds for $j=0,1,2,\dots, 2N-1$,
it never holds for $j=2N$. For if it did, we would have for any
polynomial, $R$, of degree $2N$ that
$$
\langle \delta_0, R(A^{[N]}_F)\delta_0\rangle =
\langle \delta_0, R(A)\delta_0\rangle .
$$
But this is false for $R(x) = x^2 P_{N-1}(x)^2$ since
$$
\langle \delta_0, R(A)\delta_0\rangle =
\|A\delta_{N-1}\|^2 = a^2_{N-1} + a^2_{N-2} + b^2_{N-1},
$$
while
$$
\langle \delta_0, R(A^{[N]}_F)\delta_0\rangle =
\|A^{[F]}_N \delta_{N-1}\|^2 =a^2_{N-2} + b^2_{N-1}.
$$
\endremark
\demo{Proof of Theorem 6} By (5.33) and (5.34) as $z\to 0$,
$$
-\frac{Q_N(-\frac{1}{z})}{zP_N(-\frac{1}{z})} =
\sum_{j=0}^{2N-1} \kappa_j z^j +O(z^{2N}). \tag 5.40
$$
Now $z^{N-1} Q_N(-\frac{1}{z})$ is a polynomial of degree $N-1$
since $Q_N(0)\neq 0$ and $z^N P_N(-\frac{1}{z})$ is a polynomial
of degree $N$ since $P_N(0)\neq 0$. Moreover, $\lim_{z\to 0}
z^N P_N(-\frac{1}{z})\neq 0$ since $P_N$ has degree $N$. Thus
(5.40) identifies $f^{[N-1,N]}$ and $-z^{N-1} Q_N(-\frac{1}{z})
/ z^N P_N (-\frac{1}{z})$. Similarly, noting that since $M_N(0)
=0$ (but $M'_N(0)\neq 0$ since $A^N_K$ has simple eigenvalues),
$z^N M_N(-\frac{1}{z})$ is a polynomial of degree $N-1$, and we
identify $f^{[N-1, N-1]}(z)$ and $-z^{N-1}N_N(-\frac{1}{z}) /
z^N M_N(-\frac{1}{z})$. Theorem~5.2 thus implies Theorem~6. \qed
\enddemo
\remark{Remarks} 1. Since $P_N$ and $Q_N$ are relatively prime
(by Proposition~5.7) and similarly for $M_N$, $N_N$, we are in
the situation where the numerator and denominator are uniquely
determined up to a constant.
2. Suppose $\gamma_n \leq Ca^n$ so $\rho$ is unique and supported
on some interval $[0, R]$. Then if $R$ is chosen as small as
possible, $\sum_{n=0}^\infty \kappa_n z^n$ has radius of
convergence $R^{-1}$, but the Pad\'e approximants converge in
the entire region $\Bbb C\backslash (-\infty, -R^{-1}]$!
\endremark
Before leaving our discussion of Pad\'e approximants for series
of Stieltjes, we will consider the sequences $f^{[N+j, N]}(z)$
for $j$ fixed. We will show they each converge as $N\to\infty$
with limits that are all distinct (as $j$ varies) if the
Stieltjes moment problem is indeterminate and all the same if
a certain set of Stieltjes moment problems are all determinate.
Given a set of moments $\{\gamma_j\}^\infty_{j=1}$ and $\ell=1,2,
\dots$, define for $j=0,1,\dots$
$$
\gamma^{(\ell)}_j \equiv \frac{\gamma_{j+\ell}}{\gamma_\ell}.
\tag 5.41
$$
\proclaim{Proposition 5.12} Suppose that $\{\gamma_j
\}^\infty_{j=0}$ is a set of Stieltjes \rom(resp.~Hamburger\rom)
moments. Then $\{\gamma^{(\ell)}_j\}^\infty_{j=0}$ is the set of
moments for a Stieltjes \rom(resp.~Hamburger\rom) problem for
$\ell = 1,2,\dots$ \rom(resp.~$\ell=2,4,6,\dots$\rom). This
problem is indeterminate if the original problem is.
\endproclaim
\demo{Proof} We will consider the Stieltjes case. Let $\rho\in
\Cal M^S(\gamma)$. Then $\gamma^{-1}_\ell x^\ell\, d\rho(x)
\equiv d\rho^{(\ell)}(x)$ solves the moment problem for
$\gamma^{(\ell)}$. If $\rho_1, \rho_2 \in\Cal M^S(\gamma)$,
then $\rho^{(\ell)}_1, \rho^{(\ell)}_2$ are both in $\Cal M^S
(\gamma^{(\ell)})$. If $\rho^{(\ell)}_1 = \rho^{(\ell)}_2$, then
$\rho_1 - \rho_2$ is supported at zero, which means it is zero
since $\int d\rho_1(x)=\gamma_0=\int d\rho_2 (x)$. Thus, if the
$\gamma$ problem is indeterminate, so is the $\gamma^{(\ell)}$
problem. \qed
\enddemo
\example{Example} As Corollary~4.21 shows, there can be determinate
moment problems so that $\gamma^{(1)}$ is indeterminate (in the
language of that corollary, $\tilde\gamma^{(1)}=\gamma$). Thus,
the converse of the last statement in Proposition~5.12 fails.
\endexample
As an aside, we present a proof of a criterion for indeterminacy
of Hamburger [\ham], especially interesting because of a way of
rewriting it in terms of a single ratio of determinants (see
Theorem~A.7). Note this aside deals with the Hamburger problem.
\proclaim{Proposition 5.13} A necessary and sufficient condition
for a set of Hamburger moments to be indeterminate is that
\roster
\item"\rom{(i)}" $\sum_{j=0}^\infty |P_j (0)|^2 < \infty$ and
\item"\rom{(ii)}" $\sum_{j=0}^\infty |P_j^{(2)}(0)|^2 < \infty$
\endroster
where $P^{(2)}_j(x)$ are the orthogonal polynomials for the
$\gamma^{(2)}$ moment problem.
\endproclaim
\demo{Proof} If $\{\gamma_j\}^\infty_{j=0}$ is indeterminate by
Proposition~5.12, so is $\{\gamma_j^{(2)}\}^\infty_{j=0}$, and
then (i), (ii) follow by Theorem~3. The converse is more
involved. The intuition is the following: If $d\rho$ is an even
measure, then $P^{(2)}_{2j}(x)=\sqrt{\gamma_2} \frac{P_{2j+1}(x)}
{x}$, so $P^{(2)}_m(0)=\sqrt{\gamma_2}\frac{\partial P_{m+1}}
{\partial x} (0)$ (both sides are zero if $m$ is odd). So (ii)
is equivalent to $\sum_{j=0}^\infty |\frac{\partial P_j(0)}
{\partial x}|^2 <\infty$ and condition (v) of Theorem~3 says that
(i), (ii) imply indeterminacy. Our goal is to prove in general
that when (i) holds, $P^{(2)}_j(0)$ and $\frac{\partial P_{j+1}}
{\partial x}(0)$ are essentially equivalent.
Let $S_n(x) = P^{(2)}_n (x)$, $\eta_n = \frac{\partial P_n}
{\partial x}(0)$, and define
$$\align
\alpha_n &= \int x S_n(x)\, d\rho(x) \tag 5.42 \\
\beta_n &= \int [x S_n (x)][P_{n+1}(x)]\, d\rho(x), \tag 5.43
\endalign
$$
where $\rho$ is any solution of the $\gamma$ moment problem.
Since $S_n$ is an orthogonal polynomial for $d\rho^{(2)}$, we
have
$$
\int [x S_n(x)]x^j\, xd\rho(x)=0, \qquad j=0,1,\dots, n-1
$$
and thus
$$
\int x S_n(x) P_j(x)\, d\rho(x) = \alpha_n P_j (0),
\qquad j=0,1,\dots, n
$$
so the orthogonal expansion for $xS_n(x)$ is
$$
xS_n (x) = \beta_n P_{n+1}(x) + \alpha_n \sum_{j=0}^n
P_j (0) P_j(x). \tag 5.44
$$
Since $xS_n(x)$ vanishes at zero,
$$
\beta_n P_{n+1}(0) + \alpha_n \sum_{j=0}^n |P_j (0)|^2 = 0.
\tag 5.45
$$
Since
$$
\int |xS_n (x)|^2 \, d\rho(x) = \gamma_2 \int S_n(x)^2\,
d\rho^{(2)}(x) = \gamma_2,
$$
we have by (5.44) that
$$
\gamma_2 = \beta^2_n + \alpha^2_n \sum_{j=0}^n |P_j(0)|^2.
\tag 5.46
$$
Moreover, taking derivatives of (5.44) at $x=0$,
$$
S_n(0) = \beta_n \eta_{n+1} + \alpha_n \sum_{j=0}^n
P_j(0)\eta_j. \tag 5.47
$$
By (5.46), $\beta_n$ is bounded, so by (5.45),
$$
|\alpha_n| \leq C_1 |P_{n+1}(0)|. \tag 5.48
$$
By hypothesis, $\sum_{j=0}^\infty |P_j (0)|^2 <\infty$, so
$P_{n+1}(0)\to 0$ and thus by (5.46), $\beta_n \to \sqrt{\gamma_2}$
as $n\to\infty$. Since $\beta_n >0$ for all $n$ (this follows from
the fact that both $xS_n(x)$ and $P_{n+1}(x)$ have positive
leading coefficient multiplying $x^{n+1}$), we see that
$\beta^{-1}_n$ is bounded. Using (5.48), the Schwarz inequality
on the sum in (5.47), and $\sum_{j=0}^\infty |P_j(0)|^2 <
\infty$ again, we see that
$$\align
|\eta_{n+1}|^2 &\leq C\biggl(|S_n(0)|^2 + |P_{n+1}(0)|^2
\sum_{j=0}^n |\eta_j|^2\biggr) \\
&\leq C (|S_n(0)|^2 + |P_{n+1}(0)|^2) \biggl( 1 + \sum_{j=0}^n
|\eta_j|^2\biggr).
\endalign
$$
It follows that
$$
\biggl[ 1 + \sum_{j=0}^{n+1} |\eta_j|^2\biggr] \leq [1+ C
(|S_n(0)|^2 + |P_{n+1}(0)|^2)]\biggl[ 1 + \sum_{j=0}^n
|\eta_j|^2\biggr]
$$
and so by induction that
$$
\sup_{1 \leq n < \infty} \sum_{j=0}^{n+1} |\eta_j|^2 \leq
\sup_{1 \leq n < \infty} \prod_{j=1}^{n+1} [1+ C(|S_{j-1}(0)|^2
+ |P_j (0)|^2)] <\infty
$$
since $\sum_{j=0}^\infty |P_j(0)|^2 + \sum_{j=0}^\infty |S_j
(0)|^2 < \infty$ by hypothesis. Thus, if (i) and (ii) hold,
$\eta_n \in\ell_2$ and thus, the problem is indeterminate by
Theorem~3. \qed
\enddemo
\proclaim{Theorem 5.14} Let $\{\gamma_n\}^\infty_{n=0}$ be a
set of Stieltjes moments. Fix $\ell \geq 1$ and let $P^{(\ell)}_N
(z)$, $Q^{(\ell)}_N(z)$, $M^{(\ell)}_N(z)$, and $N^{(\ell)}_N
(z)$ be the orthogonal polynomials and other associated
polynomials for the $\gamma^{(\ell)}$ moment problem. Let
$f^{[N,M]}(z)$ be the Pad\'e approximants for the series of
Stieltjes $\sum_{j=0}^\infty \kappa_j z^j$ where $\kappa_j =
(-1)^j \gamma_j$. Then
$$
f^{[N+\ell-1, N]}(z) = \sum_{j=0}^{\ell-1} (-1)^j \gamma_j
z^j + (-1)^\ell \gamma_\ell z^\ell
\biggl[\frac{Q^{(\ell)}_N (-\frac{1}{z})}
{zP^{(\ell)}_N (-\frac{1}{z})}\biggr] \tag 5.49
$$
and
$$
f^{[N+\ell-1, N-1]}(z) = \sum_{j=0}^{\ell-1} (-1)^j \gamma_j
z^j + (-1)^\ell \gamma_\ell z^\ell
\biggl[\frac{N^{(\ell)}_N (-\frac{1}{z})}
{zM^{(\ell)}_N (-\frac{1}{z})}\biggr]. \tag 5.50
$$
In particular:
\roster
\item"\rom{(1)}" For each $\ell$, $(-1)^\ell f^{[N+\ell-1, N]}
(x)$ is monotone increasing to a finite limit for all $x\in
(0,\infty)$.
\item"\rom{(2)}" For all $z\in\Bbb C\backslash (-\infty, 0]$,
$\lim_{N\to\infty} f^{[N+\ell -1, N]}(z) \equiv f_\ell(z)$
exists.
\item"\rom{(3)}" The $\gamma^{(\ell)}$ moment problem is
\rom(Stieltjes\rom) determinate if and only if $f_\ell (z) =
f_{\ell+1}(z)$.
\item"\rom{(4)}" If the $\gamma^{(\ell)}$ moment problem is
determinate, then $f_0(z) = f_1(z) = \cdots = f_{\ell+1}(z)$.
\item"\rom{(5)}" If the $\gamma^{(\ell)}$ moment problem is
indeterminate, then as $x\to\infty$,
$$
f_{\ell+1}(x) = (-1)^\ell \gamma_\ell \mu^{(\ell)}_K
(\{0\}) x^\ell + O(x^{\ell-1}), \tag 5.51
$$
where $\mu^{(\ell)}_K$ is the Krein solution of the
$\gamma^{(\ell)}$ problem. In particular if $\ell >1$,
$-f_{\ell +1}(x)$ is not a Herglotz function \rom(as it is
if the problem is determinate\rom).
\endroster
\endproclaim
\remark{Remarks} 1. Thus, it can happen (e.g., if $\gamma$ is
determinate but $\gamma^{(1)}$ is not (cf.~Corollary~4.21)
that $f^{[N-1, N]}$ and $f^{[N,N]}$ have the same limit, but
that $f^{[N+1, N]}$ has a different limit.
2. If the $\{\gamma_j\}^\infty_{j=0}$ obey a condition that
implies uniqueness and which is invariant under $\gamma_j \to
\gamma_{j+1}$ (e.g., the condition of (1.12b) of Proposition~1.5),
then all $\gamma^{(\ell)}$ are determinate; see Theorem~5.19
below.
3. Even for $\ell=1$ and consideration of $f^{[N-1,N]}$ and
$f^{[N,N]}$, we have something new --- the remarkable fact that
$xd\mu_K(x)/\gamma^{-1}_1$ is $d\mu^{(1)}_F(x)$. Multiplication
by $x$ kills the point measure at $x=0$ and produces a measure
supported on some set $[R,\infty)$ with $R>0$. We also see that
monotonicity for $f^-_N$ implies monotonicity of $f^+_N$. The
direction of monotonicity flips because of the minus sign
($(-1)^\ell = -1$) in (5.49).
4. In particular, we have $P^{(1)}_N(x) = c_N x^{-1} M_{N+1}
(x)$, that is, for any $\ell\neq m$, \linebreak $\int x^{-1}
M_m(x) M_\ell(x)\, d\rho(x) = 0$, something that can be checked
directly.
5. This theorem implies that all the $f^{[N,M]}$ with $N\geq
M-1$ exist and their denominators obey a three-term recursion
relation.
6. The connection in Remark~3 extended to $xd\mu^{(\ell)}_K
(x) = c_\ell d\mu^{(\ell+1)}_F (x)$ implies that if any
$\gamma^{(\ell)}$ is indeterminate, then $d\mu_K(x)$ is a
point measure. Thus if $\kappa_n = (-1)^n \int_0^\infty
x^n \, d\rho(x)$ where $\rho$ is associated to a determinate
Stieltjes problem and $d\rho$ is not a point measure, then all
the $f_\ell (z)$'s, $\ell=0,1,2,\dots$, are equal.
\endremark
\demo{Proof} It is easy to check that the right side of (5.49)
and (5.50) are ratios of polynomials whose degrees are at most
of the right size. (Because $P^{(\ell)}_N(0)\neq 0 \neq
M^{(\ell)'}_N(0)$, it is easy that the denominators are always
precisely of the right size.) Since $M^{(\ell)}_N (z)$ and
$P^{(\ell)}_N(z)$ are of degree $n$, the rationalized
denominators do not vanish at $\frac1{z}=0$. By Proposition~5.11
for the $\gamma^{(\ell)}$ problem, they have the proper
asymptotic series to be $f^{[N+\ell-1, N]}(z)$ and
$f^{[N+\ell-1, N-1]}(z)$, respectively. That proves (5.49)
and (5.50).
Assertions (1), (2), (3) are then just Theorem~6 for the
$\gamma^{(\ell)}$ moments. To prove assertion (4), note that
if $\gamma^{(\ell)}$ is determined by Proposition~5.12, so
are $\gamma^{(\ell-1)}, \gamma^{(\ell-2)}, \dots, \gamma^{(1)},
\mathbreak \gamma$, and so by (3), we have $f_{\ell+1} = f_\ell
= \cdots = f_0$.
To prove assertion (5), note that if the $\gamma^{(\ell)}$
moment problem is indeterminate, $d\mu^{(\ell)}_K(\{0\})
\mathbreak >0$. Thus,
$$
\lim_{x\to\infty} \int \frac{d\mu^{(\ell)}_K(y)}{1+xy}
= \mu^{(\ell)}_K \{0\} > 0. \tag 5.52
$$
By (5.50),
$$
f_{\ell+1}(x) = \sum_{j=0}^{\ell-1} (-1)^j \gamma_j x^j +
(-1)^\ell \gamma_\ell x^\ell \int_0^\infty
\frac{d\mu^{(\ell)}_K(y)}{1+xy}\, ,
$$
so (5.52) implies (5.51). \qed
\enddemo
To deal with the sequences $f^{[N-1+\ell, N]}(z)$ with $\ell <0$,
we need to introduce yet another modified moment problem
associated to a set of Hamburger (or Stieltjes) moments. To
motivate what we are looking for, let $\rho\in\Cal M^H(\gamma)$
and let $G_\rho(z)$ be the associated Stieltjes transform. Then
$-G_\rho(z)^{-1}$ is a Herglotz function which has an asymptotic
series to all orders as $z\to i\infty$. Since $G_\rho(z) \sim
-\frac1{z} (1+\gamma_1 z^{-1} + \gamma_2 z^{-2} + O(z^{-3}))$,
$$
-G_\rho(z)^{-1} \sim z-\gamma_1 - (\gamma_2 - \gamma^2_1)
z^{-1} +O(z^{-2})
$$
so (by the proof of Proposition~4.13) the Herglotz representation
of $-G_\rho(z)^{-1}$ has the form
$$
-G_\rho(z)^{-1} = z - \gamma_1 + (\gamma_2 - \gamma^2_1)
\int \frac{d\tilde\rho(x)}{x-z}\, , \tag 5.53
$$
where $d\tilde\rho$ is a normalized measure. Its moments,
which we will call $\gamma^{(0)}_j$, only depend on the
asymptotic series for $G_\rho(z)$, and so only on the original
moments. We could do everything abstractly using (5.53), but
we will be able to explicitly describe the relation between the
moment problems.
The key is the following formula, a Ricatti-type equation well
known to practitioners of the inverse problem [\gs] (which we will,
in essence, prove below):
$$
-m_0 (z)^{-1} = z-b_0 + a^2_0 m_1 (z), \tag 5.54
$$
where $m_0(z) = \langle\delta_0, (A-z)^{-1}\delta_0\rangle$ is
just $G_\rho(z)$ for a spectral measure, and $m_1(z) =
\langle \delta_0, (A^{[1]} -z)^{-1}\delta_0\rangle$, where $A^{[1]}$
is obtained from $A$ by removing the top row and left-most column. (
5.54) is just (5.53) if we note that $b_0 = \gamma_1$ and $a^2_0 =
\gamma_2 - \gamma^2_1$. Thus, we are led to define $\gamma^{(0)}_j$
as the moments associated to the Jacobi matrix $A^{[1]}$ obtained
by removing the top row and left-most column, that is,
$$
A^{[1]} = \pmatrix
b_1 & a_1 & 0 & \dots & \dots \\
a_1 & b_2 & a_2 & \dots & \dots \\
0 & a_2 & b_3 & a_3 & \dots \\
\vdots & \vdots & \vdots & \vdots & \vdots
\endpmatrix .
$$
\proclaim{Proposition 5.15} Let $\{\gamma_j\}^\infty_{j=0}$
be a set of Hamburger moments. Then the $\{\gamma^{(0)}_j
\}^\infty_{j=0}$ Hamburger problem is determinate if and only
if $\{\gamma_j\}^\infty_{j=0}$ is.
\endproclaim
\demo{Proof} Let $\tilde A^{[1]}$ be $A^{[1]}$ with a row of
zeros added at the top and column of zeros on the left. Since
$\tilde A^{[1]}=0 \oplus A^{[1]}$, $\tilde A^{[1]}$ is
essentially self-adjoint if and only if $A^{[1]}$ is. $A -
\tilde A^{[1]}$ is a matrix with three non-zero elements and
essential self-adjointness is provided by bounded perturbations.
Thus, $A$ is essentially self-adjoint if and only if $A^{[1]}$
is. By Theorem~2, we have the equivalence of determinacy for the
Hamburger problem. \qed
\enddemo
\remark{Remark} As we will see shortly, this result is not true
in the Stieltjes case.
\endremark
Let $P^{(0)}_N(x)$, $Q^{(0)}_N(x)$, and $f^{(0)}_N(z) = -
\frac{Q^{(0)}_N(z)}{P^{(0)}_N(z)}$ be the polynomials and
finite matrix approximation for the $\gamma^{(0)}$ problem. Then
\proclaim{Proposition 5.16}
$$\alignat2
&\text{\rom{(i)}} \qquad && P^{(0)}_N(x) = a_0 Q_{N+1}(x)
\tag 5.56 \\
&\text{\rom{(ii)}} \qquad && Q^{(0)}_N(x) = -\frac1{a_0} \,
P_{N+1}(x) + \frac{x-b_0}{a_0}\, Q_{N+1}(x) \tag 5.57 \\
&\text{\rom{(iii)}} \qquad && -f^-_{N+1} (z)^{-1} = z -b_0 +
a^2_0 f^{(0)}_N(z) \tag 5.58
\endalignat
$$
\endproclaim
\remark{Remarks} 1. (i) implies that the $Q_N$'s are orthogonal
polynomials for some measure.
2. (5.58) is (5.54).
\endremark
\demo{Proof} For each fixed $x$, $P^{(0)}_N(x), Q^{(0)}_N(x)$
obey the same difference equation as $P_{N+1}(x)$, $Q_{N+1}(x)$
so we need (5.56), (5.57) at the initial points $N=-1,0$ where
it is required that
$$
P^{(0)}_{-1}(x) = 0, \quad P^{(0)}_0(x)=1; \qquad
Q^{(0)}_{-1}(x)=-\frac{1}{a_0}, \quad Q^{(0)}_0(x) =0
$$
($Q^{(0)}_{-1}(x) = -\frac1{a_0}$ since we don't make the
convention $a^{(0)}_{-1} = 1$, but rather $a^{(0)}_{-1} = a_0$
consistent with $a^{(0)}_n=a_{n+1}$). This is consistent with
(5.56) if we note that
$$
Q_0 (x) = 0, \quad Q_1(x) = \frac1{a_0}; \qquad
P_0 (x) = 1, \quad P_1 (x) = \frac{(x-b_0)}{a_0}\, .
$$
This proves (i), (ii).
They in turn imply
$$
a^2_0\, \frac{Q^{(0)}_N(x)}{P^{(0)}_N(x)} = x-b_0 -
\frac{P_{N+1}(x)}{Q_{N+1}(x)}\, ,
$$
which, by (5.12), proves (iii). \qed
\enddemo
\proclaim{Proposition 5.17} Let $\{\gamma_j\}^\infty_{j=0}$ be
a set of Stieltjes moments. If the Hamburger problem is
indeterminate, then the $\gamma^{(0)}$ Stieltjes problem is
indeterminate \rom(even if the $\gamma$ Stieltjes problem is
determinate\rom).
\endproclaim
\demo{Proof} As we will see below (Proposition 5.22 $\equiv$
Theorem~8), a set of Stieltjes moments which is Hamburger
indeterminate is Stieltjes determinate if and only if $L\equiv
\lim_{N\to\infty} [-\frac{Q_N(0)}{P_N (0)}]$ is infinite. By
(5.58),
$$
a^2_0 L^{(0)} = b_0 - L^{-1}.
$$
Since $L>0$, $L^{(0)}$ is never infinite. \qed
\enddemo
\remark{Remarks} 1. Thus, if $\gamma$ is Hamburger indeterminate
but Stieltjes determinate, $\gamma$ and $\gamma^{(0)}$ have
opposite Stieltjes determinacy.
2. The spectral way to understand this result is to note that if
$\gamma$ is Hamburger indeterminate, the Friedrichs extensions of
$A$ and $A^{[1]}$ have interlacing eigenvalues. Given that $A \geq
0$, $A^{[1]}$ must be strictly positive.
\endremark
For $\ell <0$ and integral, we let
$$
\gamma^{(\ell)}_j = [\gamma^{(0)}]^{(-\ell)}_j =
\frac{\gamma^{(0)}_{j-\ell}}{\gamma^{(0)}_{-\ell}}\, .
$$
Then:
\proclaim{Theorem 5.18} Let $\{\gamma_n\}^\infty_{n=0}$ be a set
of Stieltjes moments. Fix $\ell \leq 0$ and let $P^{(\ell)}_N(z)$,
$Q^{(\ell)}_N(z)$, $M^{(\ell)}_N(z)$, and $N^{(\ell)}_N(z)$ be
the orthogonal polynomials and other associated polynomials for
the $\gamma^{(\ell)}$ moment problem. Let $f^{[N,M]}(z)$ be the
Pad\'e aproximants for the series of Stieltjes $\sum_{j=0}^\infty
\kappa_j z^j$ where $\kappa_j = (-1)^j \gamma_j$. Then
$$
f^{[N+\ell-1,N]}(z) = \biggl\{ 1-\gamma_1 z -
\sum_{j=0}^{-\ell-1} (-1)^j \gamma^{(0)}_j z^{j+2} + (-1)^{\ell+1}
\gamma^{(0)}_{-\ell} z^{-\ell +2} \biggl[\frac{Q^{(\ell)}_{N+\ell-1}
(-\frac{1}{z})}{zP^{(\ell)}_{N+\ell-1} (-\frac1{z})}\biggr]
\biggr\}^{-1} \tag 5.59
$$
and
$$
f^{[N+\ell-1,N+1]}(z) = \biggl\{ 1-\gamma_1 z -
\sum_{j=0}^{-\ell-1} (-1)^j \gamma^{(0)}_j z^{j+2} + (-1)^{\ell+1}
\gamma^{(0)}_{-\ell} z^{-\ell +2} \biggl[\frac{N^{(\ell)}_{N+\ell-1}
(-\frac{1}{z})}{zM^{(\ell)}_{N+\ell} (-\frac1{z})}\biggr]
\biggr\}^{-1}. \tag 5.60
$$
In particular,
\roster
\item"\rom{(1)}" For each $\ell$, $(-1)^\ell f^{[N+\ell-1, N]}
(x)$ is monotone increasing to a finite limit for all $x\in
(0,\infty)$.
\item"\rom{(2)}" For all $z\in\Bbb C\backslash (-\infty, 0]$,
$\lim_{N\to\infty} f^{[N+\ell -1, N]}(z) \equiv f_\ell(z)$
exists.
\item"\rom{(3)}" The $\gamma^{(\ell)}$ moment problem is
Stieltjes determinate if and only if $f_\ell(z) = f_{\ell-1}(z)$.
\item"\rom{(4)}" If the $\gamma^{(\ell)}$ moment problem is
determinate, then $f_0(z) = f_{-1}(z) = \cdots = f_\ell(z)
= f_{\ell-1}(z)$.
\item"\rom{(5)}" If the $\gamma^{(\ell)}$ moment problem is
indeterminate, then as $x\to\infty$,
$$
f_{\ell-1}(x) = (-1)^{\ell+1} (\gamma^{(0)}_{-\ell})^{-1}
x^{\ell-2} \mu^{(\ell)}_K (\{0\})^{-1} + O(x^{\ell-3}),
$$
where $\mu^{(\ell)}_K$ is the Krein solution of the
$\gamma^{(\ell)}$ problem. In particular, $-f_{\ell-1}(x)$ is
not a Herglotz function \rom(as it is if the problem is
determinate\rom).
\endroster
\endproclaim
\demo{Proof} By (5.58), we have a relation between formal power
series:
$$
\biggl( \sum_{j=0}^\infty (-1)^j \gamma_j z^j\biggr)^{-1} =
1 + \gamma_1 z - (\gamma_2 - \gamma^2_1) z^2 \biggl[
\sum_{j=0}^\infty (-1)^j \gamma^{(0)}_j z^j\biggr], \tag 5.61
$$
from which one obtains
$$
f^{[M,N+2]}(z) = (1+\gamma_1 z - (\gamma_2 - \gamma^2_1) z^2
f^{(0)[N,M]}(z))^{-1}.
$$
Thus, (5.59) is just (5.49) and (5.60) is (5.50). The
consequence (1)--(5) follows as in the proof of Theorem~5.14.
\qed
\enddemo
We summarize and extend in the following:
\proclaim{Theorem 5.19} Let $\sum_{j=0}^\infty \kappa_j z^j$
be a series of Stieltjes. Then for each $\ell = 0, \pm 1, \pm 2,
\dots$, $f_\ell(z) = \lim_{N\to\infty} f^{[N+\ell, N]}(z)$
exists uniformly for $z$ in compact subsets of $\Bbb C
\backslash (-\infty, 0]$. Moreover,
\roster
\item"\rom{(1)}" If any two $f_\ell$'s are different, then all of
them are meromorphic.
\item"\rom{(2)}" If the $|\kappa_j|=\gamma_j$ moment problem
is determinate and the measure solving the moment problem is not
a discrete point measure, then all $f_\ell$'s are equal.
\item"\rom{(3)}" If $|\kappa_j|\leq C^j (2j)!$, then all
$f_\ell$'s are equal.
\endroster
\endproclaim
\demo{Proof} The first assertion is Theorem~5.14 for $\ell \geq
1$ and Theorem~5.18 for $\ell\leq 0$. If some $f_\ell \neq
f_{\ell+1}$, then some $\gamma^{(\ell)}$ are indeterminate, so
the corresponding $d\rho_F$'s are pure point and $f_\ell$'s
meromorphic. This proves (1). Under the hypothesis of (2),
$f_0$ is not meromorphic, so by (1), all $f$'s are equal. To
prove (3), an induction using (5.61) proves $|\gamma^{(0)}_j|
\leq \tilde C^j (2j)!$, so by Proposition~1.5, all the
$\gamma^{(\ell)}$ moment problems are determinate. \qed
\enddemo
This completes our discusion of Pad\'e approximants for
series of Stieltjes. We return to consequences of Theorem~5.2
for the study of the Stieltjes moment problem.
\vskip 0.1in
As we have seen for any $x>0$, $-\frac{Q_N(-x)}{P_N(-x)}$ is
monotone increasing and $-\frac{xN_N (-x)}{M_N(-x)}$ is monotone
decreasing. By taking limits (recall $M_N(0)=0$), we see that
$-\frac{Q_N(0)}{P_N(0)}$ is monotone increasing and
$\frac{N_N(0)}{M'_N(0)}$ is decreasing.
\proclaim{Proposition 5.20}
$$\alignat2
&\text{\rom{(i)}} \qquad && \lim_{N\to\infty} - \frac{Q_N(0)}
{P_N(0)} = \int y^{-1}\, d\mu_F (y) \\
&\text{\rom{(ii)}} \qquad && \lim_{N\to\infty} \frac{N_N(0)}
{M'_N(0)} = \mu_K (\{0\}),
\endalignat
$$
where $\mu_F$ \rom(resp.~$\mu_K$\rom) is the Friedrichs
\rom(resp.~Krein\rom) solution.
\endproclaim
\demo{Proof} By Theorem~5.2 and Proposition~5.6, for $x>0$,
$$
-\frac{Q_N(-x)}{P_N(-x)} \leq \int (y+x)^{-1}\, d\mu_F (y),
$$
so taking $x$ to zero and $N$ to infinity, we see that
$$
\lim_{N\to\infty} -\frac{Q_N(0)}{P_N(0)} \leq \int y^{-1} \,
d\mu_F (y).
$$
On the other hand, since $-\frac{Q_N(-x)}{P_N(-x)}=(\delta_0,
(A^{[N]}_F + x)^{-1}\delta_0)$ is monotone increasing as $x$
decreases, we have for each $N$ and $x>0$,
$$
-\frac{Q_N(-x)}{P_N(-x)} \leq \lim_{N\to\infty}
-\frac{Q_N(0)}{P_N(0)}\, ,
$$
so taking $N$ to infinity for fixed $x>0$ and using Theorem~5.2,
$$
\int (x+y)^{-1} \, d\mu_F (y) \leq \lim_{N\to\infty}
-\frac{Q_N(0)}{P_N(0)}\, .
$$
Taking $x$ to zero, we see that
$$
\int y^{-1} \, d\mu_F (y) \leq \lim_{N\to\infty}
-\frac{Q_N(0)}{P_N(0)}\, ,
$$
so (i) is proven.
The proof of (ii) is similar. By Theorem~5.2 and Proposition~5.8,
$$
-\frac{xN_N(-x)}{M_N (-x)} \geq \int \frac{x}{y+x}\,
d\mu_K (y) \geq \mu_K (\{0\}),
$$
so taking $x$ to zero and $N$ to infinity,
$$
\lim_{N\to\infty} \frac{N_N(0)}{M'_N(0)} \geq \mu_K (\{0\}).
$$
On the other hand, since $-\frac{xN_N(-x)}{M_N(-x)} = (\delta_0,
x(A^{[N]}_K + x)^{-1} \delta_0)$ is monotone decreasing as $x$
decreases, we have for each $N$ and $x>0$,
$$
-\frac{xN_N(-x)}{M_N(-x)} \geq \lim_{N\to\infty}
\frac{N_N(0)}{M'_N(0)}\, ,
$$
so taking $N$ to infinity for fixed $x>0$ and using Theorem~5.2,
$$
\int \frac{x}{x+y}\, d\mu_K(y) \geq \lim_{N\to\infty}
\frac{N_N(0)}{M'_N(0)}\, .
$$
Taking $x$ to zero, we see that
$$
\mu_K (\{0\}) \geq \lim_{N\to\infty} \frac{N_N(0)}
{M'_N(0)}\, ,
$$
so (ii) is proven. \qed
\enddemo
This leads us to define (note that we use $\frac{M'_N(0)}{N_N(0)}$,
not $\frac{N_N(0)}{M'_N(0)}$ so $M,L\in (0,\infty)\cup \{\infty\}$):
$$\align
L &= \lim_{N\to\infty} -\frac{Q_N(0)}{P_N(0)} \tag 5.62 \\
M &= \lim_{N\to\infty} \frac{M'_N(0)}{N_N(0)}, \tag 5.63
\endalign
$$
so Proposition~5.20 says that
$$\align
L &= \int y^{-1}\, d\mu_F (y) \\
M^{-1} &= \mu_K (\{0\}).
\endalign
$$
By (4.25), $\mu_K (\{0\}) = 1/\sum_{n=0}^\infty |P_N(0)|^2$, so
$$
M=\sum_{n=0}^\infty |P_n(0)|^2 . \tag 5.64
$$
\demo{Note} (4.25) was only proven in the indeterminate case,
but the argument applies in the determinate case also. If $x$ is
an eigenvalue of a solution $\mu$ of the moment problem associated
to a self-adjoint extension, then $\mu (\{x\}) = 1/
\sum_{n=0}^\infty |P_n(x)|^2$.
\enddemo
\proclaim{Theorem 5.21 ($\equiv$ Theorem 7)} Let $\{\gamma_n
\}^\infty_{n=0}$ be the moments of a Stieltjes problem. Then the
problem is indeterminate if and only if
$$
L < \infty \qquad \text{\rom{and}} \qquad M<\infty.
$$
Equivalently, the problem is determinate if and only if
$$
L=\infty \qquad \text{\rom{or}} \qquad M=\infty.
$$
\endproclaim
\demo{Proof} If $L<\infty$ and $M<\infty$, then $\int y^{-1}
d\mu_F (y) <\infty$, while $\mu_K(\{0\}) >0$, so clearly,
$\mu_F \neq \mu_K$ and the problem is indeterminate. Conversely,
if the problem is indeterminate, then by Proposition~3.1, $\alpha
=\inf \, \text{spec}(A_F)>0$, so $\int y^{-1} \, d\mu_F(y) \leq
\alpha^{-1} < \infty$ and $L<\infty$. Moreover, since the
Stieltjes problem is indeterminate, so is the Hamburger problem;
and thus by (5.64), $M<\infty$. \qed
\enddemo
\proclaim{Theorem 5.22 ($\equiv$ Theorem 8)} Let $\{\gamma_n
\}^\infty_{n=0}$ be a set of Stieltjes moments. Then the
Stieltjes problem is determinate while the Hamburger problem is
indeterminate if and only if
$$
\sum_{n=0}^\infty |Q_n(0)|^2 < \infty \tag 5.65
$$
and $L=\infty$.
\endproclaim
\demo{Proof} For a set of Stieltjes moments $-\frac{Q_n(0)}
{P_n(0)}$ is positive and monotone increasing, so $|Q_n(0)| \geq
\frac{|Q_1(0) P_n(0)|}{|P_1(0)|}$. Since $Q_1(0)\neq 0$, we see
that (5.65) implies also that $M<\infty$. Thus, (5.65) is
equivalent to the Hamburger problem being indeterminate. Given
that $M<\infty$, Theorem~5.21 says that determinacy of the
Stieltjes problem is equivalent to $L=\infty$. \qed
\enddemo
As our next topic, we will further examine the conditions $M<
\infty$ and $L<\infty$ to see they are conditions of Stieltjes
and Krein in a different form. Define
$$
\ell_n = -\frac{Q_n(0)}{P_n(0)} +
\frac{Q_{n-1}(0)}{P_{n-1}(0)}\, , \qquad n\geq 1 \tag 5.66
$$
and
$$\align
m_n &= \frac{M'_n(0)}{N_n(0)} -
\frac{M'_{n-1}(0)}{N_{n-1}(0)}\, , \qquad n\geq 2 \tag 5.67a \\
m_1 &= \frac{M'_1 (0)}{N_1 (0)}\, . \tag 5.67b
\endalign
$$
By the monotonicity properties of this section, $\ell_n >0$,
$m_n >0$. By definition and $Q_0(0) =0$,
$$\alignat2
-\frac{Q_N(0)}{P_N(0)} &= \sum_{n=1}^N \ell_n, \qquad &&
L=\sum_{n=1}^\infty \ell_n \\
\frac{M'_N(0)}{N_N(0)} &= \sum_{n=1}^N m_n, \qquad &&
M=\sum_{n=1}^\infty m_n.
\endalignat
$$
\proclaim{Proposition~5.23} For $N\geq 1$,
$$\align
m_N &= |P_{N-1}(0)|^2 \tag 5.68 \\
\ell_N &= -[a_{N-1} P_N(0) P_{N-1}(0)]^{-1}. \tag 5.69
\endalign
$$
\endproclaim
\remark{Remark} Since $A^{[N]}_F >0$, $P_N(z)$ has no zeros on
$(-\infty, 0]$ and thus, $P_N(z)$ has the same sign near
$-\infty$ and $0$. But $P_N(z)=c_N z^N + \text{ lower order}$
with $c_N >0$, so $(-1)^N P_N(0)>0$ (cf.~(5.13)). Thus, $P_N(0)
P_{N-1}(0)<0$ and the minus sign in (5.69) is just what is needed
to ensure that $\ell_N >0$.
\endremark
\demo{Proof} $A^{[N]}_K$ has $\{ P_0(0), \dots, P_{N-1}(0)\}$ as
its eigenfunction with eigenvalue zero. Thus, $\lim_{x\to 0}
x(\delta_0, (A^{[N]}_K + x)^{-1} \delta_0) = |P_0(0)|^2 /
\sum_{j=0}^{N-1} |P_j (0)|^2$. Since $P_0(0)=1$, we see that
$$
\frac{M'_N(0)}{N_N(0)} = \sum_{j=0}^{N-1} |P_j (0)|^2, \tag 5.70
$$
which implies (5.68). (5.69) follows immediately from
Proposition~4.1. \qed
\enddemo
\proclaim{Corollary 5.24} If $A$ given by {\rom{(1.16)}} is the
Jacobi matrix associated to an indeterminate Stieltjes moment
problem, then
$$
\sum_{n=0}^\infty a^{-1/2}_n < \infty.
$$
In particular, if $\sum_{n=0}^\infty a^{-1/2}_n = \infty$ for the
Jacobi matrix associated to a Stieltjes problem, then the problem
is determinate.
\endproclaim
\demo{Proof} If the problem is indeterminate, by Proposition~5.23
and Theorem~5.21, \linebreak $a^{-1/2}_n |P_n(0)
P_{n-1}(0)|^{-1/2}$ and $|P_n(0) P_{n-1}(0)|^{1/2}$ both lie in
$\ell^2$, so their product lies in $\ell^1$. \qed
\enddemo
Now we will define, following Stieltjes and Krein, functions
$$\alignat2
U_n(x) &= \frac{P_n(-x)}{P_n(0)} \, , \qquad && n \geq 0
\tag 5.71 \\
V_n(x) &= -\frac{Q_n(-x)}{P_n(0)}\, , \qquad && n\geq 0
\tag 5.72 \\
G_n(x) &= -a_{n-1} M_n(-x) P_{n-1}(0), \qquad && n\geq 1
\tag 5.73 \\
H_n(x) &= a_{n-1} N_n (-x) P_{n-1}(0), \qquad && n\geq 1.
\tag 5.74
\endalignat
$$
We claim that
\proclaim{Proposition 5.25}
\roster
\item"\rom{(i)}" $U_n(0)=1$
\item"\rom{(ii)}" $G_n(0) = 0$
\item"\rom{(iii)}" $H_n(0) = 1$
\item"\rom{(iv)}" $U_n(x) H_n(x) - V_n(x) G_n(x) =1
\text{\rom{ for }} n\geq 1$
\item"\rom{(v)}" $ f^+_n(-x) - f^-_n (-x) =
\frac{1}{U_n(x) G_n(x)} \text{\rom{ for }} n\geq 1$
\endroster
\endproclaim
\demo{Proof} (i) is immediate from the definition (5.71), (ii) from
$M_n(0)=0$, and (iii) follows from Proposition~4.1 and the
definition (5.16) of $N_N$ (and explains why we multiply $M$ and
$N$ by $a_{n-1}P_{n-1}(0)$). To prove (iv), we note that for some
constants $\alpha_n, \beta_n, \gamma_n$,
$$\alignat3
U_n(x) &= \alpha_n P_n(-x), & \qquad V_n(x)
&= -\alpha_n Q_n (-x) \\
G_n(x) &= \beta_n P_n(-x) + \gamma_n P_{n-1}(-x), & \qquad
H_n(x) &= -\beta_n Q_n (-x) - \gamma_n Q_{n-1}(-x),
\endalignat
$$
so
$$
U_n(x) H_n(x) - V_n(x) G_n(x) = -\alpha_n \gamma_n
[P_n (-x) Q_{n-1}(-x) - Q_n (-x) P_{n-1} (-x)]
$$
is constant as $x$ is varied for fixed $n$. But at $x=0$, by
(i)--(iii) this combination is $1$. (v) follows from (iv) and
the definitions. \qed
\enddemo
The following says that for $x>0$, $U,G$ can be associated with
the equation of motion of a string of beads (see [\akh]). This
is the starting point of deep work of Krein [\krstu].
\proclaim{Theorem 5.26}
\roster
\item"\rom{(i)}" $U_n(x) - U_{n-1}(x) = \ell_n G_n (x), \quad
n\geq 1$
\item"\rom{(ii)}" $G_{n+1}(x) - G_n(x) = m_{n+1}x U_n(x),
\quad n\geq 1$
\item"\rom{(iii)}" $G_1 (x) = m_1 x \quad U_0(x)=1$
\endroster
\endproclaim
\demo{Proof} (i) By definition of $U_n$,
$$
U_n (x) - U_{n-1}(x) = \frac{M_n (-x)}{P_n(0)} = \ell_n G_n(x)
$$
by (5.69) and (5.73).
(ii) We have for $n\geq 1$,
$$
a_n P_{n+1} (-x) + b_n P_n (-x) + a_{n-1} P_{n-1} (-x) =
-x P_n (-x). \tag 5.75
$$
(5.75) for $x=0$ implies that
$$
b_n = -a_n P_{n+1}(0) P_n(0)^{-1} -a_{n-1} P_{n-1}(0)
P_n(0)^{-1}, \tag 5.76
$$
which we can substitute into (5.75) to obtain
$$
a_n M_{n+1} (-x) - a_{n-1} P_{n-1}(0) P_n(0)^{-1} M_n(-x) =
-x P_n(x).
$$
Multiplying by $-P_n(0)$ and using the definitions (5.71)/(5.73),
we see
$$
G_{n+1}(x) - G_n (x) = xP_n(0)^2 U_n(x)
$$
so (5.68) implies the result.
(iii) $U_0(x)$ is a constant so $U_0(0)=1$ implies $U_0(x)\equiv
1$. $G_1$ is a linear polynomial with $G_1 (0)=0$, so $G_1(x) =
G'_1 (0)x$. For any $n$, $G'_n (0)=\lim_{x\downarrow 0}
\frac{x^{-1}G_n(x)}{H_n(x)} = \sum_{j=1}^n m_j$. \qed
\enddemo
We can use these equations to get further insight into
Theorem~5.2. First, we obtain explicit bounds on the rate of
convergence of $f^+_N(-x) - f^-_N(-x)$ to zero in the
determinate case, where either $\sum m_j$ or $\sum \ell_j$ or
both diverge.
\proclaim{Theorem 5.27} For $x>0$,
$$
f^+_N (-x) - f^-_N (-x) \leq \biggl[m_1 x^2 \biggl(
\sum_{j=1}^n m_j \biggr) \biggl(\sum_{j=1}^n \ell_j\biggr)
\biggl]^{-1}.
$$
\endproclaim
\demo{Proof} It follows from Theorem~5.26 and induction that
$U_n(x)$ and $G_n(x)$ are non-negative for $x>0$ and then that
they are monotone in $n$. In particular, $U_n(x)\geq 1$, $G_n(x)
\geq m_1 x$ from which, by (i), (ii),
$$\align
U_n(x) &\geq m_1 \biggl( \sum_{j=1}^n \ell_j \biggr) x \\
G_n(x) & \geq \biggl(\sum_{j=1}^n m_j\biggr) x.
\endalign
$$
(v) of Proposition~5.25 then completes the proof. \qed
\enddemo
And we obtain directly that in the indeterminate case, $f^\pm$
are meromorphic functions.
\proclaim{Theorem 5.28} {\rom{(i)}} For any $z\in\Bbb C$, we
have that for $n\geq 1$,
$$\align
|U_n(z)| &\leq \prod_{j=1}^n (1+\ell_j) \prod_{j=1}^n
(1+m_j |z|) \\
|G_n(z)| &\leq \prod_{j=1}^{n-1} (1+\ell_j) \prod_{j=1}^n
(1+m_j |z|).
\endalign
$$
{\rom{(ii)}} If $M<\infty$ and $L<\infty$, then for all $z\in
\Bbb C$, $U_n(z), V_n(z), G_n(z), H_n(z)$ converge to functions
$U_\infty(z), V_\infty(z), G_\infty(z), H_\infty(z)$, which
are entire functions obeying
$$
|f(z)| \leq C_\varepsilon \exp(\varepsilon(z))
$$
for each $\varepsilon >0$.
{\rom{(iii)}} $f^-(z) = \frac{V_\infty (-z)}{U_\infty (-z)}$,
$f^+(z) =\frac{H_\infty (-z)}{G_\infty(-z)}$\, .
{\rom{(iv)}} $f^+(z) - f^- (x) = \frac{1}{U_\infty (-z)
G_\infty(-z)}$\, .
\endproclaim
\remark{Remark} In terms of the Nevanlinna functions $A,B,C,D$,
one can see (using the fact that the Friedrichs solution is
associated to $B_t$ with $t=L$) that $G_\infty (-z) = -D(z)$,
$H_\infty (-z) = C(z)$, $U_\infty (-z) = -B(z) - LD(z)$, and
$V_\infty(-z) = A(z) +LC(z)$, where $L=\sum_{j=1}^\infty \ell_j$.
\endremark
\demo{Proof} (i) follows by an elementary induction from
Theorem~5.26. Similarly, one follows the proof of that theorem to
show that $V_n, H_n$ obey
$$\alignat2
&V_n(x) - V_{n-1}(x) = \ell_n H_n(x), \qquad && n\geq 1 \\
&H_{n+1}(x) - H_n (x) = m_{n+1}x V_n(x), \qquad && n\geq 1 \\
&H_1(x) =1 \qquad V_0(x)=0
\endalignat
$$
and obtains inductively that
$$\align
|V_n(z)| &\leq \prod_{j=1}^n(1+\ell_j) \prod_{j=2}^n
(1+m_j |z|) \\
|H_n(z)| &\leq \prod_{j=1}^{n-1} (1+\ell_j) \prod_{j=2}^n
(1+m_j |z|).
\endalign
$$
Thus, if $L,M <\infty$, we first see that $U,V,H,G$ are bounded
and then, since $\sum_{j\geq n} \ell_j \to 0$, $\sum_{j\geq n}
m_j \to 0$, that each sequence is Cauchy as $n\to\infty$. The
$C_\varepsilon$ bound is easy for products $\prod_{j=1}^\infty
(1+m_j |z|)$ with $\sum_1^\infty m_j <\infty$. (iii), (iv) are
then immediate from the definitions. \qed
\enddemo
To make the link to Stieltjes' continued fractions, we need
to note the relation between $a_n$, $b_n$, $\ell_n$, and $m_n$.
Immediately from Proposition~5.23, we have
$$
a_n = \bigl( \ell_{n+1} \sqrt{m_{n+1}m_{n+2}}\,\bigr)^{-1},
\qquad n\geq 0 \tag 5.77
$$
and by (5.76),
$$\align
b_n &= m^{-1}_{n+1} (\ell^{-1}_n + \ell^{-1}_{n+1}),
\qquad n \geq 1 \tag 5.78 \\
b_0 &= m^{-1}_1 \ell^{-1}_1. \tag 5.79
\endalign
$$
(Parenthetically, we note that these equations can be used
inductively to define $m_j, \ell_j$ given $a_j, b_j$. We have
$m_1 =1$, so (5.79) gives $\ell_1$. Given $\ell_1, \dots, \ell_j$
and $m_1, \dots, m_j$, we can use $a_{j-1}$ and (5.77) to find
$m_{j+1}$ and then $b_j$ and (5.78) to get $\ell_{j+1}$.)
Stieltjes' continued fractions are of the form (5.1b). Let
$$
d_0 = c^{-1}_1, \qquad d_n = (c_n c_{n+1})^{-1} \tag 5.80
$$
for $n\geq 1$ so (5.1b) becomes
$$
\cfrac d_0 \\
w + \cfrac d_1 \\
1 + \cfrac d_2 \\
w+\dots \endcfrac
$$
Now use the identity
$$
w +
\cfrac \beta_1 \\
1 + \cfrac \beta_2 \\
f(w) \endcfrac = w+\beta_1 - \frac{\beta_1\beta_2}{\beta_2 +
f(w)}
$$
to see that (5.1b) has the form (5.1a) if $w=-z$, $d_0=1$, and
$$\align
b_0 &= d_1 \tag 5.81a \\
b_n &= d_{2n+1} + d_{2n} \tag 5.81b \\
a^2_n &= d_{2n+1} d_{2n+2}. \tag 5.81c
\endalign
$$
Thus, (5.80) and (5.81) are consistent with (5.77)--(5.79) if
and only if
$$
c_{2j-1} = m_j, \qquad c_{2j} = \ell_j. \tag 5.82
$$
Thus, we have seen that (5.1a) is (5.1b) if $w=-z$ and the $c$'s
are given by (5.82). Stieltjes' criterion that depends on whether
or not $\sum_{j=1}^\infty c_j < \infty$ is equivalent to $L<\infty$
and $M<\infty$.
The connection (5.82) is not surprising. In (5.1b) if $w=0$,
the continued fraction formally reduces to $c_2 + c_4 + \cdots$
consistent with $\sum_{j=1}^\infty \ell_j = -\lim_{N\to\infty}
\frac{Q_N(0)}{P_N(0)}$. On the other hand, if we multiply by $w$,
the continued fraction formally becomes
$$
\cfrac 1 \\
c_1 + \cfrac 1 \\
c_2 w + \cfrac 1 \\
c_3 + \cdots \endcfrac
$$
which, formally at $w=0$, is $(c_1 + c_3 + \cdots)^{-1}$, consistent
with $\sum_{j=1}^\infty m_j = \lim_{N\to\infty}
\lim_{x\downarrow 0} \mathbreak [-\frac{xN_N(-x)}{M_N(-x)}]$.
We conclude this section by discussing Pad\'e approximants
for the Hamburger analog of series of Stieltjes. A series of
Hamburger is a formal power series $\sum_{j=0}^\infty \kappa_j
z^j$ with
$$
\kappa_j = (-1)^j \int_{-\infty}^\infty x^j \, d\rho(x)
$$
for some measure $\rho$. We will begin with the ``principal"
Pad\'e approximants $f^{[N-1,N]}(z)$:
\proclaim{Theorem 5.29} Let $\sum_{j=0}^\infty \kappa_j x^j$
be a series of Hamburger. Then:
\roster
\item"\rom{(i)}" The $f^{[N-1, N]}(z)$ Pad\'e approximants
always exist and are given by
$$
f^{[N-1, N]}(z) = (\delta_0, (1+zA^{[N]}_F)^{-1} \delta_0)
= -\frac{z^{N-1} Q_N(-\frac1{z})}{z^N P_N (\frac1{z})}\, .
\tag 5.83
$$
\item"\rom{(ii)}" If the associated Hamburger moment problem is
determinate, then for any $z$ with $\text{\rom{Im}}\, z\neq 0$,
$$
\lim_{N\to\infty} f^{[N-1, N]}(z) =
-\frac{z^{N-1} Q_N(-\frac1{z})}{z^N P_N (\frac1{z})}
\tag 5.84
$$
exists and equals $\int \frac{d\rho(x)}{1+zx}$ for the unique
solution, $\rho$, of the moment problem.
\item"\rom{(iii)}" The sequence $f^{[N-1, N]}(z)$ is pre-compact
in the family of functions analytic in $\Bbb C_+$ \rom(in the
topology of uniform convergence on compact sets\rom).
\item"\rom{(iv)}" Any limit of $f^{[N-1, N]}(z)$ is of the form
$\int \frac{d\rho(x)}{1+zx}$ where $\rho$ is a von~Neumann
solution of the Hamburger problem.
\endroster
\endproclaim
\demo{Proof} (i) The proof that (5.12) holds (Proposition~5.6)
is the same as in the Stieltjes case, and from there we get
(5.34) as in the Stieltjes case. Since $\lim_{z\to 0} z^N P_N
(\frac1{z})\neq 0$, we see $f^{[N-1,N]} (z)$ exists and is given
by (5.83).
(iii) Since
$$
|zf^{[N-1, N]}(z)| = |(\delta_0, (A^{[N]}_F + z^{-1})^{-1}
\delta_0)| \leq |\text{Im}(z^{-1})|^{-1},
$$
we see that
$$
|f^{[N-1,N]}(z)| \leq |z| \, |\text{Im}(z)|^{-1}
$$
is bounded on compacts. By the Weierstrass-Vitali theorem, such
functions are pre-compact.
(ii), (iv) By the proof of Proposition~4.13, if $d\rho$ is any
measure whose first $M+2$ moments are $\gamma_0, \gamma_1,
\dots, \gamma_{M+2}$, then
$$
\biggl| y^{M+1} \biggl[G_\rho (iy) + \sum_{n=0}^M (-i)^{n+1}
y^{-n-1} \gamma_n \biggr]\biggr| \leq ay^{-1} \tag 5.85
$$
with $a$ only depending on $\{\gamma\}$ (either $a=\gamma_{M+1}$
or $a=\frac12 [\gamma_M + \gamma_{M+2}]$). Since $\langle\delta_0,
(A^{[N]}_F)^j \delta_0\rangle \mathbreak =\gamma_j$ for $j\leq 2N-1$,
we see that for $M$ fixed, (5.85) holds uniformly for $N$ large if
$G_N(z) =-z^{-1} f^{[N-1, N]}(-\frac1{z})$, so any limit point
of the $G$'s obeys (5.85) and is Herglotz. Thus by Proposition~4.13,
it is of the form $G_\rho(z)$ with $\rho\in\Cal M^H(\gamma)$.
Thus the limit for $f^{[N-1,N]}(z)$ as $N\to\infty$, call it
$f(z)$, has the form
$$
f(z) = \int \frac{d\rho(x)}{1+xz}
$$
with $\rho\in\Cal M^H(\gamma)$. In the determinate case, $\rho$
must be the unique solution, so all limit points are equal. Thus
compactness implies convergence, and we have proven (ii).
In the indeterminate case, we note that $\delta_j = P_j
(A^{[N]}_F )\delta_0$ for $j=0,\dots, N-1$ so
$$
(A^{[N]}_F -z)^{-1}\delta_0 = \sum_{j=0}^{N-1} \langle P_j
(A^{[N]}_F)\delta_0, (A^{[N]}_F -z)^{-1}\delta_0\rangle \delta_j.
$$
As in the proof of Theorem~4.3, we conclude that (with $G_N(z)
=\langle\delta_0, (A^{[N]}_F -z)^{-1}\delta_0\rangle$),
$$
\sum_{j=0}^{N-1} |Q_j (z) + G_N(z) P_j (z)|^2 =
\frac{\text{Im}\, G_N(z)}{\text{Im}\, z}\, .
$$
Thus, since $Q,P \in \ell_2$ if $G_N(z) \to G(z)$ through a
subsequence, the limit $\zeta = G(z)$ obeys equality in (4.5).
By Theorem~4.14, any such solution of the moment problem is a
von~Neumann solution. \qed
\enddemo
We want to show that in many cases, $\lim f^{[N-1, N]}(z)$
will not exist. Consider a set of Stieltjes moments,
$\{\gamma_n\}^\infty_{n=0}$ and the associated even Hamburger
problem with moments
$$
\Gamma_{2n} = \gamma_n, \qquad \Gamma_{2n+1} =0. \tag 5.86
$$
Given $\rho\in\Cal M^S(\gamma)$, let $\tilde\rho\in\Cal M^H
(\Gamma)$ be given by
$$
d\tilde\rho(x) = \tfrac12 (\chi_{[0,\infty)}(x)\, d\rho(x^2)
+ \chi_{(-\infty, 0]}(x)\, d\rho(x^2)) \tag 5.87
$$
so that Theorem~2.12 says that this sets up a one-one
correspondence between even integrals in $\Cal M^H(\Gamma)$
and $\Cal M^S(\gamma)$. Let $\tilde P_n, \tilde Q_n$ be the
orthogonal polynomials for $\Gamma$ and $P_n, Q_n, M_n, N_n$
for $\gamma$. Define $d_n$ by $d^2_n \int M_n(x)^2 x^{-1}\,
d\rho(x)=1$. Then we claim that
$$\align
G_{\tilde\rho}(z) &= zG_\rho(z^2) \tag 5.88 \\
\tilde P_{2n}(x) &= P_n(x^2) \tag 5.89 \\
\tilde P_{2n-1}(x) &= \frac{d_n M_n(x^2)}{x} \tag 5.90 \\
\tilde Q_{2n}(x) &= xQ_n (x^2) \tag 5.91 \\
\tilde Q_{2n-1}(x) &= d_n N_n (x^2). \tag 5.92
\endalign
$$
To prove (5.89) and (5.90), we note that the right sides have
the correct degree, and since $P_n(x^2)$ is even and
$\frac{M_n(x^2)}{x}$ is odd, they are $d\tilde\rho$ orthogonal
to each other. Easy calculations prove the $d\tilde\rho$
orthogonality of two $P_n(x^2)$'s and two $\frac{M_n(x^2)}{x}$'s.
The formulae for $\tilde Q$ and $G_{\tilde\rho}$ follow by
direct calculations using Theorem~4.2 for $Q$.
\proclaim{Theorem 5.30} Let $\gamma, \Gamma$ be related by
{\rom{(5.86)}} and let $d\tilde\rho^{(F)}, d\tilde\rho^{(K)}$
be the images of $d\rho_F, d\rho_K$ \rom(for $\gamma$\rom)
under the map $\rho \mapsto \tilde\rho$ given by {\rom{(5.87)}}.
Let $\tilde f^{[N,M]}(z)$ be the Pad\'e approximants for
$\sum_{n=0}^\infty (-1)^n \Gamma_n z^n$. Then
$$\align
\lim_{N\to\infty} \tilde f^{[2N-1, 2N]}(z) &= \int
\frac{d\tilde\rho^{(F)}(x)}{1+xz} \tag 5.93 \\
\lim_{N\to\infty} \tilde f^{[2N, 2N+1]}(z) &= \int
\frac{d\tilde\rho^{(K)}(x)}{1+xz}\, . \tag 5.94
\endalign
$$
In particular, if the $\Gamma$ problem is indeterminate, $\lim
f^{[N-1, N]}(z)$ does not exist.
\endproclaim
\demo{Proof} This follows directly from Theorem~6 for the $\gamma$
problem and (5.88)--(5.92). \qed
\enddemo
Of course, in the special case, the non-convergence of
$f^{[N-1, N]}(z)$ is replaced by something almost as nice since
the even and odd subsequences separately converge. But there is
no reason to expect this to persist if $\Gamma$ is not even and
indeed, simple numerical examples ([\kei]) illustrate
non-convergence is the rule. One can understand why this happens.
In the indeterminate case, there is a one-parameter family of
extensions, each with discrete spectrum. There is, in general,
nothing to tack down $A^{[N]}_F$ so it wanders. In the even case,
two self-adjoint extensions are even, one with an eigenvalue at
$0$ and one without, and the $A^{[N]}_F$ for odd $N$ and even
$N$ track these. This suggests that $A^{[N]}_K$, which is tacked
down to have a zero eigenvalue, should converge in strong
resolvent sense and the corresponding $f^{[N,N]}$'s should
converge. This is true and is not a previously known result:
\proclaim{Theorem 5.31} Let $\{\gamma_n\}^\infty_{n=0}$ be a
set of Hamburger moments and let $\sum_{n=0}^\infty (-1)^n
\gamma_n z^n$ be the associated series of Hamburger. Then the
$f^{[N,N]}(z)$ Pad\'e approximants exist if and only if
$$
P_N(0)\neq 0. \tag 5.95
$$
Moreover, {\rom{(5.95)}} holds if and only if there exists
$\alpha_N$ so that the matrix $A^{[N+1]}_K$ given by {\rom{(5.2b)}}
has a zero eigenvalue, and then
$$
f^{[N,N]}(z) = (\delta_0, (1+zA^{[N+1]}_K)^{-1}\delta_0).
\tag 5.96
$$
If $N\to\infty$ through the sequence of all $N$'s for which
{\rom{(5.95)}} holds, then
$$
\lim \Sb N\to \infty \\ N:P_N(0)\neq 0 \endSb
f^{[N,N]}(z) = \int \frac{d\rho(x)}{1+xz}\, , \tag 5.97
$$
where $d\rho$ is the unique solution of the moment problem in
case it is determinate and the unique von~Neumann solution with
$\mu (\{0\})>0$ if the moment problem is indeterminate.
\endproclaim
\remark{Remarks} 1. If $P_N(0)=0$, then, by the second-order
equation $P_N(0)$ obeys, we have that $P_{N+1}(0)\neq 0 \neq
P_{N-1}(0)$, so (5.95) can fail at most half the time. Typically,
of course, it always holds (e.g., under a small generic
perturbation of the Jacobi matrix, one can prove that it will
hold even if initially it failed).
2. If the $\gamma$'s are a set of even moments, then (5.95) holds
exactly for $N$ even and
$$
f^{[2M, 2M]}(z) = \frac{z^{2M} Q_{2M+1} (-\frac1{z})}
{z^{2M+1} P_{2M+1} (-\frac1{z})}\, ,
$$
so we get convergence of those measures we called $d\tilde\rho$
above.
3. The non-existence of $f^{[N,N]}$ is using the Baker definition
that we have made. There is a classical definition which would
define $f^{[N,N]}(z)$ even if $P_N(0)=0$, (but (5.26) and (5.27)
would both fail!). Under this definition, $f^{[N,N]}(z)$ would
be $f^{[N-1, N-1]}(z)$ when $P_N(0)=0$ and (5.97) would hold
using all $N$ rather than just those $N$ with $P_N(0)\neq 0$.
\endremark
\demo{Proof} Define
$$\align
\widetilde M_N(z) &= P_{N-1}(0) P_N(z) - P_N(0) P_{N-1}(z)
\tag 5.98 \\
\tilde N_N(z) &= P_{N-1}(0) Q_N(z) - P_N(0)P_{N-1}(z).
\tag 5.99
\endalign
$$
Then by the Wronskian calculations (see Proposition~4.1); also
see (5.70)),
$$\align
\tilde N_N(0) &= \frac{1}{a_{N-1}} \neq 0 \tag 5.100 \\
\widetilde M'_N(0) &= \sum_{j=0}^{N-1} \frac{P_j(0)^2}
{a_{N-1}} \neq 0 \tag 5.101
\endalign
$$
and, of course,
$$
\widetilde M_N(0)=0. \tag 5.102
$$
Define
$$\align
A_N(z) &= z^N \tilde N_{N+1} \bigl(-\tfrac{1}{z}\bigr)
\tag 5.103 \\
B_N(z) &= z^{N+1} \widetilde M_{N+1} \bigl( -\tfrac{1}{z}\bigr).
\tag 5.104
\endalign
$$
By (5.100), $\deg A_N=N$ and by (5.101/5.102), $\deg B_N = N$.
Moreover, by the definition (5.98),
$$
B_N(0) = 0 \Longleftrightarrow P_N(0)=0 \tag 5.105
$$
and
$$
B_N(0) = 0 \Longrightarrow P'_N(0)\neq 0 \tag 5.106
$$
since $P_N(0)=0 \Rightarrow P_{N-1}(0)\neq 0$.
Moreover, we claim that
$$
A_N(z) - B_N(z) \sum_{j=0}^{2N} \gamma_j z^j =
O(z^{2N+1}). \tag 5.107
$$
This can be seen either by algebraic manipulation as in the
remark following the proof of Theorem~A.5 or by noting that
we have proven it in (5.39) for Stieltjes measures. So if
$\rho\in\Cal M^H(\gamma)$ and $\mu_0$ is a Stieltjes measure
and $\rho_t = (1-t)\mu_0 + t\rho$, then $B_N(-z), A_N(-z)$
are real analytic functions of $t$ and (5.107) is true for
all small $t$ in $(0,1)$ since $A^{[N+1]}_F (t) >0$ for small
$t$ (and fixed $N$) and that is all the proof of (5.106)
depends on.
If $P_N(0)\neq 0$, $B_N(0)\neq 0$, so $B^{[N,N]}(z) =
\frac{B_N(z)}{B_N(0)}$ and $A^{[N,N]}(z) = \frac{A_N(z)}
{B_N(0)}$ yield denominator and numerator of $f^{[N,N]}(t)$;
so $f^{[N,N]}$ exists.
Moreover, the proof of Lemma~5.1 shows that an $\alpha_N$ exists
giving $A^{[N+1]}_K$ a zero eigenvalue if and only if $P_N(0)
\neq 0$. In that case, the proof of Proposition~5.8 is valid,
and we have that (5.17) holds. Noting that $\frac{N_N}{M_N} =
\frac{\tilde N_N}{\widetilde M_N}$, we see that (5.96) is then
valid.
On the other hand, suppose that $P_N(0)=0$ and that $f^{[N,N]}
(z)$ exists. Then by (5.107), we have that
$$
A_N(z) B^{[N,N]} - B_N(z) A^{[N,N]}(z) = O(z^{2N+1})
$$
and so it is zero since the left side of this is of degree $2N$.
Since $P_N(0) =0$, $\widetilde M_{N+1}(z) = -P_{N+1}(0) P_N(z)
=c\widetilde M_N(z)$ where $c=-P_{N+1}(0)\mathbreak
P_{N-1}(0)^{-1}$, and similarly, $\tilde N_{N=1}(z) =
c\tilde N_N(z)$. It follows that
$$
A_N(z) = cz A_{N-1}(z), \qquad B_N(z) = czB_{N-1}(z).
$$
Thus, if $P_N(0)=0$ but $f^{[N,N]}(z)$ exists, then
$$
f^{[N,N]}(z) = f^{[N-1, N-1]}(z),
$$
that is, the $[N-1, N-1]$ Pad\'e approximant would need to have an
error of order $O(z^{2N+1})$.
Since $P_N(0)=0$, $A^{[N]}_K = A^{[N]}_F$ and so we would have
that $\langle\delta_0, (A^{[N]}_F)^j \delta_0\rangle = \gamma_j$
for $j=0,1,\dots, 2N$. But as noted in the remark following
the proof of Proposition~5.11, this can never happen. We
conclude that if $P_N(0)=0$, $f^{[N,N]}$ must fail to exist.
To complete the proof, define $A_K$ to be $\bar A$ if the
problem is determinate and the unique von~Neumann extension
with eigenvalue zero if the problem is indeterminate (so $A_K
\pi(0)=0$). Then the proof of Proposition~5.3 goes through without
change and shows that $A^{[N]}_K$ converges to $A_K$ in strong
resolvent sense. This implies (5.97). \qed
\enddemo
As for other $f^{[N+\ell -1, N]}(z)$ for series of Hamburger, we
can only define $\gamma^{(j)}$ moment problems for $j$ an even
integer (positive or negative). This means that we can use
(5.49)/(5.59) if $\ell$ is an even integer and (5.50)/(5.60) if
$\ell$ is an odd integer. The result is the following:
\proclaim{Theorem 5.32} Let $\{\gamma_n\}^\infty_{n=0}$ be a
set of Hamburger moments and let $\sum_{n=0}^\infty (-1)^n
\gamma_n z^n$ be the associated series of Hamburger. Then:
\roster
\item"\rom{(i)}" If $\ell$ is any odd integer, the Pad\'e
approximants $f^{[N+\ell-1, N]}(z)$ exist if and only if
$P^{(j)}_M(0)\neq 0$ where for $\ell >0$, $j=\ell-1$, and $M=N$
and if $\ell <0$, $j=\ell +1$ and $M=N+\ell -1$. The functions
that exist converge to a finite limit $f_\ell (z)$.
\item"\rom{(ii)}" If $\ell$ is any even integer, the Pad\'e
approximants $f^{[N+\ell -1, N]}(z)$ all exist and lie in a
compact subset in the topology of uniform convergence on
compact subsets of $\Bbb C_+$. Any limit point, $f_\ell (z)$,
is associated to a von~Neumann solution, $\rho^{(\ell)}$,
of the $\gamma^{(\ell)}$ moment problem via
$$
f_\ell (z) = \sum_{j=0}^{\ell-1} (-1)^j \gamma_j z^j +
(-1)^\ell \gamma_\ell z^\ell \int_{-\infty}^\infty
\frac{d\rho^{(\ell)} (x)}{1+xz}
$$
if $\ell >0$ and
$$
f_\ell (z) = \biggl\{ 1 - \gamma_1 z - \sum_{j=0}^{-\ell-1}
(-1)^j \gamma^{(0)}_j z^{j+2} + (-1)^{\ell +1}
\gamma^{(0)}_{-\ell} z^{-\ell + 2} \int_{-\infty}^\infty
\frac{d\rho^{(\ell)}(x)}{1+xz} \biggr\}^{-1}
$$
if $\ell <0$. In particular, if the $\gamma^{(\ell)}$ problem
is determinate, $f^{[N+\ell -1, N]}(z)$ is convergent.
\item"\rom{(iii)}" Let $\ell$ be an even integer. If the
$\gamma^{(\ell)}$ problem is determinate \rom(as a Hamburger
problem\rom), then for $\ell\geq 0$, $f_{\ell + 1}(z) =
f_\ell (z) = \cdots = f_0(z) = f_{-1}(z)$, and for $\ell \leq 0$,
$f_{\ell -1}(z) = f_\ell (z) =\cdots = f_0(z) = f_1(z)$. In
particular, if $|\gamma_j| \leq C^{j+1} j!$, then all
$f^{[N+\ell -1, N]}(z)$ converges to the same
$\ell$-independent limits as $N\to\infty$.
\endroster
\endproclaim
\vskip 0.3in
\flushpar{\bf \S 6. Solutions of Finite Order}
\vskip 0.1in
In this section, we will continue the analysis of the
indeterminate Hamburger moment problem. Throughout,
$\{\gamma_n \}^\infty_{n=0}$ will be a set of indeterminate
Hamburger moments and $\rho$ will be a solution of (1.1), the
moment problem for $\gamma$.
We have picked out the von~Neumann solutions as coming from
self-adjoint extensions of the Jacobi matrix, $A$. But in a
certain sense, every $\rho$ does come from a self-adjoint
extension of $A$! For let $\Cal H_\rho = L^2 (\Bbb R, d\rho)$,
let $D(B)=\{f\in\Cal H_\rho \mid \int x^2 |f(x)|^2\, d\rho(x)
<\infty\}$, and let
$$
(Bf)(x) = xf(x).
$$
Let $\Bbb C[X]$ be the set of polynomials in $x$ which lies in
$\Cal H_\rho$ since all moments are finite, and let $\Cal H_0 =
\overline{\Bbb C[X]}$. Corollary~4.15 says that $\Cal H_0 =
\Cal H_\rho$ if and only if $\rho$ is a von~Neumann solution.
Let $A_\rho = B\restriction \Bbb C[X]$. Then to say $\rho$ is
a solution of the Hamburger moment problem is precisely to say
that $A_\rho$ is unitarily equivalent to the Jacobi matrix, $A$,
associated to $\gamma$ under the natural map (so we will drop
the subscript $\rho$). In that sense, $B$ is a self-adjoint
extension of $A$, but not in the von~Neumann sense.
$B$ is minimal in the sense that there is no subspace of
$\Cal H_\rho$ containing $\Cal H_0$ and left invariant by all
bounded functions of $B$ (equivalently, left invariant by all
$(B-z)^{-1}$, $z\in\Bbb C\backslash\Bbb R$ or by all
$e^{iBs}$, $s\in\Bbb R$). So (if $\rho$ is not a von~Neumann
extension) we are in the strange situation where $D(B)\cap
\Cal H_0$ is dense in $\Cal H_0$, $B[D(B)\cap\Cal H_0]\subset
\Cal H_0$, but $\Cal H_0$ is not invariant for the resolvents
of $B$!
It is not hard to see that the set of all solutions of the
moment problem is precisely the set of all self-adjoint
``extensions" of $A$ in this extended sense, which are minimal
($\Cal H_0$ is cyclic for $B$) and modulo a natural unitary
equivalence. This is a point of view originally exposed by
Naimark [\naI, \naII, \naIII, \naIV] and developed by Livsic
[\liv], Gil de Lamadrid [\gdl], and Langer [\lang]. See the
discussion of Naimark's theory in Appendix~1 of [\ag].
In the language of Cayley transforms, the extensions associated
to $\rho$'s that we will call of order at most $n$ below are
parametrized by unitary maps, $U$, from $\Cal K_+\oplus\Bbb C^n$
to $\Cal K_- \oplus\Bbb C^n$ with $U_1, U_2$ equivalent if and
only if there is a unitary map $V:\Bbb C^n \to \Bbb C^n$ so $U_1
(1\oplus V) = (1\oplus V)U_2$. From this point of view, these
extensions are then parametrized by the variety of conjugacy
classes of $\Cal U(n+1)$ modulo a $\Cal U(n)$ subgroup. This has
dimension $(n+1)^2 - n^2 = 2n +1$. Our parametrization below in
terms of a ratio of two real polynomials of degree at most $n$ also
is a variety of dimension $2n+1$, showing the consistency of the
parametrization. (The $\rho$'s of exact order $n$ will be a
manifold.)
Given any $\rho$, we define the {\it{order}} of $\rho$,
$$
\text{ord}(\rho) = \dim(\Cal H_\rho/\Cal H_0)
$$
so the von~Neumann solutions are exactly the solutions of
order $0$. Our main result in this section will describe all
solutions of finite order which turn out to correspond
precisely to those $\rho$'s whose Nevanlinna function
$\Phi_\rho$ are ratios of real polynomials.
They will also be $\rho$'s for which $d\rho / \prod_{i=1}^n
|x-z_i|^2$ is the measure of a determinate moment problem. We
will therefore need some preliminaries about such problems.
Given $z_1, \dots, z_n \in\Bbb C_+$, by using partial fractions
we can write for $x$ real,
$$
\frac{x^m}{\prod_{i=1}^n |x-z_i|^2} = P_{m-2n} (x; z_1, \dots,
z_n) + \sum_{i=1}^n \biggl[ \frac{z^m_i}{z_i - \bar z_i} \,
\frac{1}{x-z_i} + \frac{\bar z^m_i}{\bar z_i - z_i} \,
\frac{1}{x-\bar z_i}\biggr] \tag 6.1
$$
where $P_{m-2n}$ is a polynomial in $x$ of degree $m-2n$. To
see (6.1), write $\prod_{i=1}^n |x-z_i|^2 = \prod_{i=1}^n
(x-z_i)(x-\bar z_i)$ and analytically continue. The sum in
(6.1) comes from computing the residues of the poles of these
functions. Define
$$
\Gamma^{(0)}_m (z_1, \dots, z_n) = E^\gamma_x (P_{m-2n}
(x; z_1, \dots, z_n)), \tag 6.2
$$
where, as usual, $E^\gamma_x$ is the expectation with respect
to any solution of the moment problem. For $\zeta_1, \dots,
\zeta_n\in\Bbb C_+$, let
$$
\Gamma_m (z_1, \dots, z_n; \zeta_1, \dots, \zeta_n) =
\Gamma^{(0)}_m (z_1, \dots, z_n) + \sum_{i=1}^n \biggl[
\frac{z^m_i}{z_i - \bar z_i}\, \zeta_i +
\frac{\bar z^m_i}{\bar z_i - z_i}\, \bar\zeta_i \biggr].
\tag 6.3
$$
\proclaim{Theorem 6.1} Fix $z_1, \dots, z_n$ and $\zeta_i,
\dots, \zeta_n$ in $\Bbb C_+$. There is a one-one correspondence
between $\nu\in\Cal M^H (\Gamma(z_1, \dots, z_n; \zeta_1,
\dots, \zeta_n))$ and those $\rho$ in $\Cal M^H(\gamma)$ which
obey
$$
\int \frac{d\rho(x)}{x-z_i} = \zeta_i, \qquad i=1,\dots, n
\tag 6.4
$$
under the association
$$
d\rho(x) \leftrightarrow d\nu (x) = \prod_{i=1}^n |x-z_i|^{-2}
\, d\rho(x). \tag 6.5
$$
That is, $\rho$ has moments $\{\gamma_m\}^\infty_{m=0}$ with
subsidiary conditions {\rom{(6.4)}} if and only if $\nu$
given by {\rom{(6.5)}} has the moments $\{\Gamma_m (z_1,
\dots, z_n; \zeta_1, \dots, \zeta_n)\}^\infty_{m=0}$.
\endproclaim
\demo{Proof} This is pure algebra. The functions
$\{x_m\}^\infty_{m=0}$, $\{\frac{1}{x-z_i}\}^n_{i=1}$ are
linearly independent as analytic functions (finite sums only).
(6.1) says that each function $x^m / \prod_{i=1}^n (x-z_i)
(x-\bar z_i)\equiv Q_m(x)$ is in the span, $\Cal S$, of these
functions. On the other hand, since $\prod_{i=1}^n (x-z_i)
(x-\bar z_i)\equiv L(x)$ is a polynomial in $x$, and $x^\ell Q_m
(x) = Q_{\ell+m}(x)$, $L(x) Q_m(x)$ is in the span of the
$\{Q_m(x)\}$. Similarly, since $\frac{L(x)}{(x-z_i)}$ is a
polynomial, $L(x) Q_0(x) \frac{1}{(x-z_i)}$ lies in that span.
Thus, $\{Q_m(x)\}$ is also a basis of $\Cal S$. It follows that
there is a one-one correspondence between assignments of numbers
$Q_m(x)\mapsto \Gamma_m$ and of assignments $x^m \mapsto \gamma_m$,
$\frac{1}{x-z_i}\mapsto\zeta_i$, $\frac{1}{x-\bar z_i}\mapsto
\lambda_i$ since each defines a linear functional on $\Cal S$.
For the linear functional to be real in $\Gamma_m$ representation,
all $\Gamma_m$ must be real. For reality in $(\gamma,\zeta,\lambda)$
representation, we must have $\gamma_m$ real and $\lambda_i =
\bar\zeta_i$. (6.3) is just an explicit realization of the map
from $\gamma,\zeta, \bar\zeta$ to $\Gamma$. Thus, a real $d\rho$
has moments $\gamma_m$ with $\zeta_i$ conditions if and only if
$\int x^m \prod_{i=1}^n |x-z_i|^{-2}\, d\rho(x)=\Gamma_m$ for
all $m$. Since (6.5) shows $\nu$ is positive if and only if
$\rho$ is, we see that the claimed equivalence holds. \qed
\enddemo
As an aside of the main theme of this section, we can construct
determinate Hamburger moments quite close to indeterminate ones.
The result is related to Corollary~4.21.
\proclaim{Theorem 6.2} Let $\{\gamma_n\}^\infty_{n=0}$ be a set
of Hamburger moments. Then for any $a\in (0,\infty)$, there is
a $c_a$ so that the moment problem with moments
$$\align
\tilde\gamma_{2n+1} &= 0 \\
\tilde\gamma_{2n} &= \gamma_{2n} -a^2 \gamma_{2n-2} +
a^4 \gamma_{2n-2} + \cdots + (-1)^n a^{2n} \gamma_0 +
(-1)^{n+1} c_a a^{2n+2} \tag 6.6
\endalign
$$
is a determinate moment problem.
\endproclaim
\demo{Proof} Let $\rho_1$ be any solution for the Hamburger
problem. Since $\frac12 [d\rho_1 (x) + d\rho_1 (-x)]=d\tilde\rho$
has moments
$$
\int x^m\, d\tilde\rho(x) = \cases 0 & m\text{ odd} \\
\gamma_{2m} & m\text{ even}
\endcases \quad \equiv \gamma'_{2m},
$$
we can suppose $\gamma_{2m+1}=0$. Let $\rho$ be a von~Neumann
solution to the $\gamma'_m$ moment problem which is invariant
under $x\to -x$. If the $\gamma'_m$ problem is determinate, then
the unique solution is invariant. Otherwise, this follows from the
proof of Theorem~2.13. Let $d\mu(x) = \frac{d\rho(x)}{1+a^2 x^2}$.
By Corollary~4.15 and Theorem~6.1, $\rho$ is the unique solution
of the $\gamma'$ moment problem with $\int \frac{d\rho(x)}
{x-ia^{-1}}\equiv\zeta_{a^{-1}}$. Thus, the $\Gamma_m (ia^{-1},
\zeta_{a^{-1}})$ problem is determinate. If $c_a = a^{-1}\,
\text{Im}\, \zeta_{a^{-1}}$, then a calculation using the
geometric series with remainder shows that $\Gamma_m (-ia^{-1},
\zeta_{a^{-1}})=\tilde\gamma_m$ given by (6.6). \qed
\enddemo
\vskip 0.1in
Returning to the main theme of this section, we next examine
when a Herglotz function is a real rational function. Define for
$z\in\Bbb C_+$,
$$
\varphi_z (x) \equiv \frac{1+xz}{x-z} \tag 6.7a
$$
viewed as a continuous function on $\Bbb R \cup\{\infty\}$ with
$$
\varphi_z (\infty) = z. \tag 6.7b
$$
If $d\sigma (x) = [\frac{d\mu(x)}{1+x^2}] + c\delta_\infty$ as
a finite measure on $\Bbb R\cup\{\infty\}$, then (1.19) can be
rewritten as
$$
\Phi(z) = d + \int_{\Bbb R\cup\{\infty\}} \varphi_z (x)\,
d\sigma(x). \tag 6.8
$$
If $\sigma(\{\infty\}) = 0$ and $\int |x|\, d\sigma(x) < \infty$,
define $\tilde d = d-\int x\,
d\sigma(x)$.
\proclaim{Proposition 6.3} A Herglotz function is a ratio of two
real polynomials if and only if the representation {\rom{(6.8)}}
has a $\sigma$ with finite support. If $\sigma$ has exactly $N$
points in its support, there are three possibilities for the
degrees of $P,Q$ in $\Phi (z) = \frac{P(z)}{Q(z)}$ with $P,Q$
relatively prime polynomials:
\roster
\item"\rom{(i)}" $(\sigma(\{\infty\})=0, \tilde d=0), \quad
\deg P=N-1, \ \deg Q=N$
\item"\rom{(ii)}" $(\sigma(\{\infty\})=0, \tilde d\neq 0), \quad
\deg P=\deg Q=N$
\item"\rom{(iii)}" $(\sigma(\{\infty\})\neq 0), \quad
\deg P = N, \ \deg Q=N-1$
\endroster
In all cases, we say $N=\max(\deg(P), \deg(Q))$ is the degree of
$\Phi$.
\endproclaim
\demo{Proof} Elementary. \qed
\enddemo
\remark{Remark} The set of $\sigma$'s with exactly $N$ pure
points is a manifold of dimension $2N$ ($N$ points, $N$ weights).
$d\in\Bbb R$ is another parameter so the set of such $\Phi$'s is
a manifold of dimension $2N+1$.
\endremark
As a final preliminary, for any $z\in\Bbb C\backslash\sigma(B)
\equiv \Bbb C \backslash\text{ supp}(d\rho)$ and $n=1,2,\dots$,
we introduce the functions $e_n(z)$ on $\text{supp}(d\rho)$,
$$
e_n(z)(x) = \frac{1}{(x-z)^n},
$$
thought of as elements of $\Cal H_\rho = L^2 (\Bbb R, d\rho)$.
The main result of this section is:
\proclaim{Theorem 6.4} Let $\gamma$ be a set of indeterminate
Hamburger moments and let $\rho_0 \in \Cal M^H(\gamma)$. Fix
$N\in \{0,1,\dots\}$. Then the following are equivalent:
\roster
\item"\rom{(1)}" $\rho_0$ has order at most $N$.
\item"\rom{(2)}" For some set of distinct $z_1, \dots, z_N \in
\Bbb C\backslash\Bbb R$, $\Cal H_0 \cup \{e_1 (z_j)\}^N_{j=1}$
span $\Cal H_{\rho_0}$.
\item"\rom{(3)}" For any set of distinct $z_1, \dots, z_N\in
\Bbb C\backslash\Bbb R$, $\Cal H_0\cup\{e_1 (z_j)\}^N_{j=1}$
span $\Cal H_{\rho_0}$.
\item"\rom{(4)}" For some $z_0 \in \Bbb C\backslash\Bbb R$,
$\Cal H_0 \cup \{e_j (z_0)\}^N_{j=1}$ span $\Cal H_{\rho_0}$.
\item"\rom{(5)}" For any $z_0\in\Bbb C\backslash\Bbb R$,
$\Cal H_0 \cup \{e_j (z_0)\}^N_{j=1}$ span $\Cal H_{\rho_0}$.
\item"\rom{(6)}" For some set of $z_0, z_1, \dots, z_N \in
\Bbb C_+$ and $\zeta_j = G_{\rho_0}(z_j)$, there is no other
$\rho\in\Cal M^H(\gamma)$ with $G_\rho (z_j)=\zeta_j$.
\item"\rom{(7)}" For all sets of $z_0, z_1, \dots, z_N \in
\Bbb C_+$ and $\zeta_j = G_{\rho_0}(z_j)$, there is no other
$\rho\in\Cal M^H(\gamma)$ with $G_\rho(z_j)=\zeta_j$.
\item"\rom{(8)}" For some $z_0, z_1, \dots, z_N \in \Bbb C_+$
and $\zeta_j = G_{\rho_0}(z_j)$, the $\Gamma (z_0, \dots,
z_N; \zeta_0, \dots, \zeta_N)$ moment problem is determinate.
\item"\rom{(9)}" For any $z_0, z_1, \dots, z_N \in \Bbb C_+$
and $\zeta_j = G_{\rho_0}(z_j)$, the $\Gamma (z_0, \dots,
z_N; \zeta_0, \dots, \zeta_N)$ moment problem is determinate.
\item"\rom{(10)}" For some $z_1, \dots, z_N\in\Bbb C_+$ and
$\zeta_j = G_{\rho_0}(z_j)$, $\rho_0$ is a von~Neumann solution
of the $\Gamma (z_1, \dots, z_N; \zeta_1, \dots, \zeta_N)$
moment problem.
\item"\rom{(11)}" For any $z_1, \dots, z_N \in\Bbb C_+$ and
$\zeta_j = G_{\rho_0}(z_j)$, $\rho_0$ is a von~Neumann solution
of the $\Gamma (z_1, \dots, z_N; \zeta_1, \dots, \zeta_N)$
moment problem.
\item"\rom{(12)}" The Nevanlinna function $\Phi$ of $\rho_0$
is a rational function of degree at most $N$.
\endroster
In particular, the Nevanlinna function $\Phi$ of $\rho_0$ has
degree $N$ if and only if $\rho_0$ has order $N$. Moreover,
if one and hence all those conditions hold, $\rho_0$ is an
extreme point in $\Cal M^H(\gamma)$ and is also a pure point
measure.
\endproclaim
\remark{Remark} The measures obeying (1)--(12) are what Akhiezer
calls canonical solutions of order $N$.
\endremark
\proclaim{Lemma 6.5} If $\varphi\in \Cal H_0$ and $z\notin
\sigma(B)$, then
$$
(B-z)^{-1} \varphi \in \Cal H_0 + [e_1(z)].
$$
\endproclaim
\demo{Proof} Since $(B-z)^{-1}$ is bounded and $\Cal H_0 +
[e_1(t)]$ is closed, it suffices to prove this for $\varphi(x)
=P(x)$, a polynomial in $x$. But then
$$\align
(B-z)^{-1}\varphi &= (x-z)^{-1} P(x) = P(z)(x-z)^{-1} +
\frac{(P(x)-P(z))}{x-z} \\
&= P(z) e_1(z) + R(x)
\endalign
$$
for a polynomial $R$, that is, $(B-z)^{-1}\varphi \in \Cal H_0
+e_1(z)$. \qed
\enddemo
\proclaim{Lemma 6.6} Supose that $z_0, z_1, \dots, z_\ell \in
\Bbb C\backslash\sigma(B)$ are distinct. If $e_m(z_0)$ is in the
span of $\Cal H_0 \cup \text{\rom{span}}[\{e_j(z_0)\}^{m-1}_{j=1}]
\cup \text{\rom{span}}[\{e_1 (z_k)\}^\ell_{k=1}]$, then so is
$e_{m+1}(z_0)$.
\endproclaim
\remark{Remark} $\ell$ can be zero.
\endremark
\demo{Proof} By hypothesis, for some $\varphi\in\Cal H_0$ and
$\{\alpha_j\}^{m-1}_{j=1}$ and $\{\beta_k\}^\ell_{k=1}$ in
$\Bbb C$,
$$
e_m (z_0) = \varphi + \sum_{j=1}^{m-1} \alpha_j e_j (z_0)
+ \sum_{k=1}^\ell \beta_k e_1(z_k).
$$
Apply $(B-z_0)^{-1}$ to this using $(B-z_0)^{-1} e_j (z_0) =
e_{j+1} (z_0)$, Lemma~6.5, and $(B-z_0)^{-1} e_1 (z_k) =
(z_0 - z_k)^{-1} [e_1 (z_0) - e_1 (z_k)]$ and we see that
$e_{m+1}(z_0)$ is in the requisite span. \qed
\enddemo
\proclaim{Proposition 6.7} Let $\ell = \text{\rom{ord}}(\rho)
<\infty$. Then for any set $\{z_i\}^\ell_{i=1}$ of $\ell$
distinct points in $\Bbb C\backslash\sigma(B)$, $\Cal H_0 \cup
\{e_1(z_i)\}^\ell_{i=1}$ span $\Cal H_\rho$.
\endproclaim
\demo{Proof} By the Stone-Weierstrass theorem, linear combinations
of $\{(x-z)^{-1} \mid z\in\Bbb C\backslash \sigma(B)\}$ are
dense in the continuous functions on $\sigma(B)$ vanishing at
infinity, and so they are dense in $L^2 (\Bbb R, d\rho)$. It
follows that we can find $\{z_i\}^\ell_{i=1}$ so that $\Cal H_0
\cup \{e_1 (z_i)\}^\ell_{i=1}$ span $\Cal H_\rho$.
Pick $z_0 \in\Bbb C\backslash\sigma(B)$ and let $W = (B-z_\ell)
(B-z_0)^{-1}$. Then $W$ is bounded with bounded inverse, so it
maps dense subspaces into dense subspaces. By Lemma~6.5, $W
[\Cal H_0] \subset \Cal H_0 + [e_1 (z_0)]$. For $i=1, \dots,
\ell-1$, $We_1 (z_i) = e_1 (z_i) + (\frac{z_0 - z_\ell}
{z_0 -z_i}) [e_1(z_0) - e_1(z_i)]$ and $We_1 (z_\ell) = e_1 (z_0)$.
Thus, $W$ maps the span of $\{e_1 (z_i)\}^\ell_{i=1}$ into
$\{e_1 (z_i)\}^{\ell-1}_{i=0}$. So $\Cal H_0 \cup
\{e_1 (z_i)\}^{\ell-1}_{i=0}$ span $\Cal H_\rho$. By successive
replacement, we can move the $z_i$'s to an arbitrary set of
distinct points. \qed
\enddemo
\remark{Remark} Proposition~6.7 shows (1)--(3) of Theorem~6.4
are equivalent.
\endremark
\proclaim{Proposition 6.8} Suppose that $\ell =\text{\rom{ord}}
(\rho)<\infty$. Then for any $z_0\in\Bbb C\backslash\sigma(B)$,
$\Cal H_0 \cup \{e_j (z_0)\}^\ell_{j=1}$ span $\Cal H_\rho$.
\endproclaim
\demo{Proof} By hypothesis, there must be some dependency
relation
$$
\varphi + \sum_{j=1}^{\ell + 1} \alpha_j e_j (z_0)=0 \tag 6.10
$$
for $\varphi\in\Cal H_0$ and some $(\alpha_1, \dots,
\alpha_{\ell+1})\neq 0$. Let $k+1 = \max\{j\mid \alpha_j \neq 0\}$.
Then solving {\rom{(6.10)}} for $e_{k+1} (z_0)$, we see that
$e_{k+1} (z_0)$ lies in the span of $\Cal H_0 \cup \{e_j (z_0)
\}^k_{j=1}$; and so by induction and Lemma~6.6, all $e_m
(z_0)$ lie in this span and so in the span of $\Cal H_0 \cup
\{e_j (z_0)\}^\ell_{j=1}$. $e_1(z)$ is an analytic function in
$\Bbb C\backslash\Bbb R$ with Taylor coefficients at $z_0$
equal to $e_n (z_0)$, so $e_1(z)$ lies in the span of $\Cal H_0
\cup \{e_j (z_0)\}^\ell_{j=1}$ for $z$ in the same half plane
as $z_0$.
By Proposition~6.7, these $e_1$'s
together with $\Cal H_0$ span $\Cal H_\rho$. \qed
\enddemo
\remark{Remark} Proposition~6.8 shows that (1), (4), (5) of
Theorem~6.4 are equivalent.
\endremark
\proclaim{Proposition 6.9} Fix distinct points $z_1, \dots,
z_\ell \in\Bbb C\backslash\Bbb R$. Let
$$
d\mu (x) = \prod_{i=1}^\ell |x-z_i|^{-2}\, d\rho(x).
$$
Then the polynomials are dense in $L^2(\Bbb R, d\mu)$ if and
only if $\Cal H_0 \cup \{e_1 (z_j)\}^\ell_{j=1}$ span
$\Cal H_\rho$.
\endproclaim
\demo{Proof} Let $U$ be the unitary map from $\Cal H_\rho$ to
$\Cal H_\mu$ given by
$$
(Uf)(x) = \prod_{i=1}^\ell (x-z_i) f(x).
$$
Then $U$ maps the span of $\Bbb C[X]\cup\{e_1(z_i)\}^\ell_{i=1}$
onto $\Bbb C[X]$ for $P(x)$ is a polynomial of degree $m$ if
and only if $\prod_{i=1}^\ell (x-z)^{-1} P(x)$ is a linear
combination of a polynomial of degree $\max(0, m-\ell)$ and
$\{(x-z_i)^{-1}\}^\ell_{i=1}$. Thus, the density assertions
are equivalent. \qed
\enddemo
\remark{Remark} Proposition~6.9 and Corollary~4.15 show the
equivalence of (2), (3), (10), and (11) of Theorem~6.4 (given
that we already know that (2) and (3) are equivalent).
\endremark
\proclaim{Proposition 6.10} If $\text{\rom{ord}}(\rho_0)$ is
finite, then $\rho_0$ is a pure point measure with discrete
support.
\endproclaim
\demo{Proof} Let $\ell = \text{ord}(\rho_0)$. Pick distinct
$\{z_i\}^\ell_{i=1}$ in $\Bbb C_+$. Then by Propositions~6.7 and
6.9, the polynomials are dense in $L^2 (\Bbb R, d\mu_\ell)$
where for $k=0,1,\dots, \ell$, we set
$$
d\mu_k (x) = \prod_{j=1}^k |x-z_i|^{-2} \, d\rho(x).
$$
Let $k_0$ be the smallest $k$ for which the polynomials are
dense in $L^2 (\Bbb R, d\mu_k)$. If $k_0 =0$, then $d\rho$
is a von~Neumann solution of the $\gamma$ problem. If $k_0 >0$,
then $d\mu_{k_0 -1}$ is not a von~Neumann solution of its
moment problem. So by Proposition~4.15 and Theorem~6.1, the
moment problem for $\Gamma^{(0)}_m \equiv \Gamma_m (z_1, \dots,
z_{k_0 -1}, z_{k_0}; \zeta_1, \dots, \zeta_{k_0-1}, \zeta_k)$
is indeterminate. But since the polynomials are dense in $L^2
(\Bbb R, d\mu_k)$, $d\mu_{k_0}$ is a von~Neumann solution
of an indeterminate problem.
Either way, $d\mu_{k_0}$ is a von~Neumann solution of an
indeterminate problem. It follows Theorem~4.11 that
$d\mu_{k_0}$ and so $d\rho$ is a discrete point measure. \qed
\enddemo
As a final preliminary to the proof of Theorem~6.4, we need a
known result about interpolation of Herglotz functions. Given
$z_1, \dots, z_n; w_1, \dots, w_n\in C_+$ with the $z$'s distinct,
define the $n\times n$ matrix $D$ by
$$
D_{ij}(z_1, \dots, z_n; w_1, \dots, w_n) = \frac{w_i -\bar w_j}
{z_i - \bar z_j}\, . \tag 6.11
$$
\proclaim{Theorem 6.11} Pick $z_1, \dots, z_n \in\Bbb C_+$ with
$n\geq 1$. There exists a Herglotz function $\Phi$ with
$$
\Phi (z_i) = w_i, \qquad i=1,\dots, n \tag 6.12
$$
if and only if $D(z_1, \dots, z_n; w_1, \dots, w_n)$ is a \rom(not
necessarily strictly\rom) positive definite matrix. Moreover, the
following are equivalent, given such a $\Phi$\rom:
\roster
\item"\rom{(1)}" $\det\,D(z_1, \dots, z_n; w_1, \dots, w_n)
=0$.
\item"\rom{(2)}" There is a unique Herglotz function $\Phi$
obeying {\rom{(6.12)}}.
\item"\rom{(3)}" $\Phi$ is a real rational polynomial of degree
at most $n-1$.
\endroster
\endproclaim
\remark{Remark} Since (3) is independent of the choice of
$\{z_i\}^n_{i=1}$, so are (1), (2).
\endremark
We will sketch a proof of this result (fleshing out some
arguments in [\akh]) below.
\demo{Proof of Theorem 6.4} As already noted, Propositions~6.7,
6.8, and 6.9 show that (1), (2), (3), (4), (5), (10), and (11)
are equivalent. Theorem~6.1 and Corollary~4.15 prove the
equivalence of (10) and (11) with (6), (7), (8), and (9). By
Theorems~4.14 and 6.11, (6) and (12) are equivalent.
Since $\text{ord}(\rho)=N$ is equivalent to $\text{ord}(\rho)
\geq N$ and the negation of $\text{ord}(\rho)\leq N-1$, we see
that $\Phi_{\rho_0}$ has degree precisely equal to $\text{ord}
(\rho_0)$. That $\rho_0$ is then an extreme point is proven in
Appendix~B. \qed
\enddemo
We conclude this section by proving Theorem~6.11.
\proclaim{Proposition 6.12} Let $\Phi$ be a Herglotz function
and let $w_i = \Phi (z_i)$, $i=1,\dots, n$ for distinct $z_1,
\dots, z_n \in\Bbb C_+$. Let $D$ be given by {\rom{(6.11)}}.
Then $D(z_1, \dots, z_n; w_1, \dots, w_n)$ is \rom(not
necessarily strictly\rom) positive definite and $\det\, D=0$
if and only if $\Phi$ is a real ratio of polynomials of degree
$n-1$ or less.
\endproclaim
\demo{Proof} Using the Herglotz representation (1.19), we get a
representation of $D_{ij}$,
$$
\frac{w_i-\bar w_j}{z_i - \bar z_j} = c + \int d\mu(x)\,
\frac{1}{x-z_i}\, \frac{1}{x-\bar z_j}\, ,
$$
from which we obtain for $\alpha\in\Bbb C^N$:
$$
\sum_{i,j=1}^n \bar\alpha_i \alpha_j D_{ij} =
c \, \biggl| \sum_{i=1}^n \alpha_j \biggr|^2 +
\int_{-\infty}^\infty d\mu(x) \biggl| \sum_{i=1}^n
\frac{\alpha_i}{x_i - z_i} \biggr|^2, \tag 6.13
$$
proving the positivity.
Suppose $\det(D) =0$. Then there is a non-zero $\alpha\in
\Bbb C^n$ so that the right side of (6.13) is $0$. Note that
$\sum_{i=1}^n \frac{\alpha_i}{(x - z_i)} = \prod_{i=1}^n
(x-z_i)^{-1} Q(x)$, where $Q$ is a polynomial of degree $n-1$ if
$\sum_{i=1}^n \alpha_i \neq 0$ and of degree at most $n-2$ if
$\sum_{i=1}^n \alpha_i =0$. Thus, for the right side of (6.13)
to vanish for some non-zero $\alpha$ in $\Bbb C^n$, either
$c\neq 0$ and $\mu$ is supported at $n-2$ or fewer points, or
else $c= 0$ and $\mu$ is supported at $n-1$ or fewer points.
Either way, by Proposition~6.3, $\Phi$ is a real rational
function of degree at most $n-1$.
Conversely, suppose $\Phi$ is a real rational function of degree
precisely $n-1$. Then either $\mu$ is supported at $n-1$ points
(say, $x_1, \dots, x_{n-1}$) or $c\neq 0$ and $\mu$ is supported
at $n-2$ points (say, $x_1, \dots, x_{n-2}$). The map $\psi :
\Bbb C^n \to \Bbb C^{n-1}$ given by
$$
\psi_j (\alpha) = \sum_{i=1}^n \frac{\alpha_i}{x_j - z_i}
$$
(with $\psi_{n-1}(\alpha) = \sum_{i=1}^n \alpha_i$ if $c\neq 0$)
has a non-zero kernel by dimension counting. Thus, the right side
of (6.13) is zero for some $\alpha\in\Bbb C^N$. If degree $\Phi$
is smaller than $n-1$, we can find $(\alpha_1, \dots, \alpha_\ell)
\in\Bbb C^\ell$ so that $\sum_{i,j=1}^\ell \bar\alpha_i \alpha_j
D_{ij}=0$, which still implies that $\det(D)=0$. \qed
\enddemo
\proclaim{Proposition 6.13} Let $z_1, \dots, z_n; w_1, \dots,
w_n$ lie in $\Bbb C_+$ with the $z_i$ distinct. Suppose the
matrix $D_{ij}$ given by {\rom{(6.11)}} is \rom(not necessarily
strictly\rom) positive definite. Then there exists a Herglotz
function with $\Phi (z_i) = w_i$, $i=1,\dots, n$.
\endproclaim
\demo{Proof} This is a fairly standard use of the Hahn-Banach
theorem. We will use the representation (6.8) for $\Phi$ and
the function $\varphi_z$ of (6.7). Begin by noting that without
loss of generality, we can suppose $z_n =i$ (by mapping $\Bbb C_+
\to \Bbb C_+$ with a linear map that takes the original $z_n$ to
$i$) and that $\text{Re}\, w_n =0$ (by adjusting the constant
$d$ in (6.8)). Indeed, since
$$
\varphi_i (x) \equiv i, \tag 6.14
$$
this choice $\text{Re}\, \Phi(i)=0$ is equivalent to $d=0$ in the
representation (6.8).
Let $V$ be the $2n-1$-dimensional subspace of $C(\Bbb R\cup
\{\infty\})$, the real-valued continuous functions on $\Bbb R\cup
\{\infty\}$ spanned by $1$, $\{\text{Re}\, \varphi_{z_j}(\,\cdot\,)
\}^{n-1}_{j=1}$ and $\{\text{Im}\, \varphi_{z_j}(\, \cdot \,)
\}^{n-1}_{j=1}$. Any $f\in V$ can be written:
$$
f= A_n + \sum_{j=1}^{n-1} A_j \varphi_j(x) + \bar A_j \,
\overline{\varphi_j(x)} \tag 6.15
$$
with $A_1, \dots, A_{n-1}\in\Bbb C$ and $A_n \in\Bbb R$. Define
a linear functional $f:V\to\Bbb R$ by
$$
\ell (f) = A_n \, \text{Im}\, w_n + \sum_{j=1}^{n-1} A_j
w_j + \bar A_j \bar w_j \tag 6.16
$$
if $f$ has the form (6.15).
We will prove shortly that
$$
f(x) > 0 \qquad \text{for all } x \Rightarrow \ell(f)>0. \tag 6.17
$$
Assuming this, we let $X = \{f\in C(\Bbb R\cup\{\infty\} \mid
f(x) >0 \text{ for all }x\}$ and $Y=\{f\in V \mid \ell(f) =0\}$.
Since $X$ is open and $Y$ closed, by the separating hyperplane
version of the Hahn-Banach theorem (see [\rsI]), there is a
linear functional $L$ on $\Bbb C(\Bbb R\cup\{\infty\})$ so
$L>0$ on $X$ and $L\leq 0$ on $Y$. If $L$ is normalized so
$L(1) = \text{Im}\, w_n$, it is easy to see that $L$ extends
$\ell$, and so defines a measure $\sigma$ on $\Bbb R\cup
\{\infty\}$ with
$$
\Phi(z_i) \equiv \int \varphi_{z_i}(x)\, d\sigma (x) = w_i,
\qquad i=1,\dots, n.
$$
Thus, if (6.17) holds, we have the required $\Phi$.
Since $f(x) >0$ on the compact set $\Bbb R\cup\{\infty\}$ implies
$f\equiv\varepsilon + g$ with $g\geq 0$ and $\varepsilon >0$
(by hypothesis, $1\in V$) and $\text{Im}\, w_n >0$, we need
only show $f\geq 0$ implies $\ell(f)\geq 0$.
If $f$ has the form (6.15), we can write
$$
f(x) = \frac{Q(x)}{\prod_{j=1}^{n-1} |x-z_j|^2}\, ,
$$
where $Q$ is a polynomial. $f\geq 0$ implies $Q\geq 0$. Any real
non-negative polynomial has roots either in complex conjugate
pairs or double real roots, so $Q$ must have even degree and
have the form
$$
Q(x) = c^2 \prod_{k=1}^\ell |x-y_k|^2
$$
for suitable $y_1, \dots, y_\ell \in \Bbb R\cup\Bbb C_+$. Thus,
with $R(x) = c \prod_{k=1}^\ell (x-y_k)$,
$$
f = |h|^2 \qquad \text{with}\qquad h(x) = \frac{R(x)}
{\prod_{j=1}^{n-1} (x-z_j)}\, .
$$
Since $f$ is bounded on $\Bbb R\cup\{\infty\}$, $\deg(R)\leq n-1$,
and thus $h$ has the form
$$
h(x) = \beta_n + \sum_{j=1}^{n-1} \beta_j (x-z_j)^{-1}. \tag 6.18
$$
Since $(x-z_j)^{-1} = (z_j -i)^{-1} [\frac{(x-i)}{(x-z_j)}-1]$,
we can rewrite (6.18) as
$$
h(x) = \sum_{j=1}^n \alpha_j\, \frac{x-i}{x-z_j}
$$
for real $x$. Thus,
$$
f(x) = |h(x)|^2 = \sum_{j=1}^n \bar\alpha_i \alpha_j
\frac{x^2 +1}{(x-\bar z_j)(x-z_j)} = \sum_{j=1}^n \bar\alpha_i
\alpha_j \biggl[ \frac{\varphi_{z_j}(x) -
\overline{\varphi_{z_i}(x)}}{z_j - \bar z_i}\biggr]
$$
so that
$$
\ell (f) = \sum_{i,j=1}^n \bar\alpha_i \alpha_j D_{ij},
$$
which is non-negative by hypothesis. \qed
\enddemo
\demo{Proof of Theorem 6.11} The first assertion is a direct
consequence of Propositions 6.12 and 6.13. Proposition~6.12
shows the equivalence of (1) and (3).
To prove (1) $\Rightarrow$ (2), suppose (1) holds and, as in the
last proof, we can suppose that $z_n=i$ and $\text{Re}\, w_n =0$.
Then by our proof of Proposition~6.12, the set of points where
the measure $\sigma$ of (6.8) is supported is determined by the
$\alpha$ with $\sum_i D_{ij} \alpha_j =0$ as an $n-1$ point set
$x_1, \dots, x_{n-1}$ (with $x_{n-1}=\infty$ allowed). Thus, any
such $\Phi$ is of the form $P/Q$ where $\deg(P)\leq n-1$ and
$Q(z)=\prod_{j=1}^{n-1} (z-x_j)$ (if $x_{n-1}=\infty$, the
product only goes from $1$ to $n-2$), If $\Phi_1$ and $\Phi_2$
are two solutions of (6.12), they have the same $Q$ but they
could be distinct $P$'s, say $P_1$ and $P_2$. But by (6.12),
$P_1(z) - P_2(z)$ vanishes at $z_1, \dots, z_n$. Since $P_1 -
P_2$ is a polynomial of degree $n-1$, this is only possible if
$P_1 - P_2 =0$, that is, $\Phi_1 = \Phi_2$. Thus, (1)
$\Rightarrow$ (2).
For the converse, suppose the determinant
$$
D(z_1, \dots, z_n; w_1, \dots, w_n) >0. \tag 6.19
$$
If $n=1$, since $z,w\in\Bbb C_+$, it is easy to see there are
multiple $\Phi$'s obeying (6.12). (For example, as usual we can
consider the case $z=w=i$ and then that $\Phi(z) = z$ or $\Phi
(z) = -z^{-1}$.) So we suppose $n\geq 2$.
Consider the function $g(w) = \det[D(z_1, \dots, z_n; w_1,
\dots, w_{n-1}, w)]$ where we vary $w_n$. $g(w)$ is of the
form $c|w|^2 + dw + \bar d\bar w + e$ with $c<0$, $e$ real,
and $d$ complex. ($c$ is strictly negative since it is $-\det
(D(z_1, \dots, z_{n-1}; w_1, \dots, w_{n-1})$ and we are supposing
(6.19) and $n\geq 2$.) Thus, the set $g(w)\geq 0$ is a disk and
since $g(w_n) >0$, $w_n$ is in its interior. Let $w_0$ be the
center of this disk, $R$ its radius, and let $w(\theta) = w_0
+ Re^{i\theta}$. Since we have proven (1) $\Rightarrow$ (2), there
is a unique $\Phi$ with
$$
\Phi_\theta (z_i) = w_i, \quad i=1,\dots, n-1; \qquad
\Phi_\theta (z_n) = w(\theta) \tag 6.20
$$
and it is a rational function of degree at most $n-1$.
As usual, we can suppose $z_1 = w_1 = i$ so each $\Phi_\theta
(z)$ has the form
$$
\Phi_\theta (z) = \int \frac{1+xz}{x-z}\, d\sigma_\theta (x),
\tag 6.21
$$
where $d\sigma_\theta$ is a probability measure on $\Bbb R
\cup \{\infty\}$. If $\theta_k \to \theta_\infty$, and
$d\sigma_{\theta_k}\to d\rho$, then the Herglotz function
associated to $d\rho$ obeys (6.20) for $\theta_\infty$ and
so it must be $\Phi_{\theta_\infty}$. It follows (since the
probability measures are compact) that $d\rho =
d\sigma_{\theta_\infty}$ and thus $\{d\sigma_\theta\}$ is closed
and $d\sigma_\theta \mapsto \Phi_\theta$ is continuous. Since a
continuous bijection between compact sets has a continuous
inverse, $\theta \mapsto d\sigma_\theta$ is continuous.
Since $\Phi_\theta$ is unique, $d\sigma_\theta$ is a pure point
measure with at most $n-1$ pure points. Since the function is
not determined by $(z_1, \dots, z_{n-1})$, there must be exactly
$n-1$ points. It follows that the points in the support must vary
continuously in $\theta$.
Note next that for any $\theta$, we can find $\theta'$ and
$t_\theta \in (0,1)$ so that
$$
w_n = t_\theta w(\theta) + (1-t_\theta)w(\theta').
$$
Suppose that there is a unique $\Phi$ obeying (6.12). It follows
that for any $\theta$,
$$
\Phi(z) = t_\theta \Phi_\theta (z) + (1-t_\theta) \Phi_{\theta'}
(z) \tag 6.22
$$
and thus $\Phi(z)$ has a representation of the form (6.21) with
$d\sigma$ a point measure with at most $2n-2$ pure points. By
(6.22) again, each $d\sigma_\theta$ must be supported in that
$2n-2$ point set, and then by continuity, in a fixed $n-1$
point set. But if $d\sigma_\theta$ is supported in a fixed
$\theta$-independent $n-1$ point set, so is $d\sigma$, and
thus (6.19) fails. We conclude that there must be multiple
$\Phi$'s obeying (6.12). \qed
\enddemo
\vskip 0.3in
\flushpar {\bf Appendix A: The Theory of Moments and
Determinantal Formulae}
\vskip 0.1in
The theory of moments has a variety of distinct objects constructed
in principle from the moments $\{\gamma_n\}^\infty_{n=0}$:
the orthogonal polynomials $\{ P_n(x)\}^\infty_{n=0}$, the
associated polynomials $\{Q_n(x)\}^\infty_{n=0}$, the sums
$\sum_{j=0}^n P_j(0)^2$ and $\sum_{j=0}^n Q_j (0)^2$, the
Jacobi matrix coefficients, and the approximations
$-\frac{Q_n (x)}{P_n(x)}$ and $-\frac{N_n(x)}{M_n(x)}$. It turns
out most of these objects can be expressed as determinants. These
formulae are compact and elegant, but for some numerical
applications, they suffer from numerical round-off errors in
large determinants.
We have already seen two sets of determinants in Theorem~1.
Namely, let $\Cal H_N$ be the $N\times N$ matrix,
$$
\Cal H_N = \pmatrix
\gamma_0 & \gamma_1 & \dots & \gamma_{N-1}\\
\gamma_1 & \gamma_2 & \dots & \gamma_N \\
\vdots & \vdots & {} & \vdots \\
\gamma_{N-1} & \gamma_N & \dots & \gamma_{2N-2}
\endpmatrix \tag A.1a
$$
and $\Cal S_N$, the matrix
$$
\Cal S_N = \pmatrix
\gamma_1 & \gamma_2 & \dots & \gamma_N\\
\gamma_2 & \gamma_3 & \dots & \gamma_{N+1} \\
\vdots & \vdots & {} & \vdots \\
\gamma_N & \gamma_{N+1} & \dots & \gamma_{2N-1}
\endpmatrix \tag A.1b
$$
and define
$$
h_N = \det (\Cal H_N), \qquad s_N = \det(\Cal S_n) \tag A.2
$$
(for comparison, $D_N$ from [\akh] is our $h_{N+1}$). We will use
$h_N(\gamma)$ if we need to emphasize what moments $\{\gamma_n
\}^\infty_{n=0}$ are involved. Thus, Theorem~1 says $h_N >0$ for
all $N$ is equivalent to solubility of the Hamburger problem and
$h_N >0$, $s_N>0$ for all $N$ is equivalent to solubility of the
Stieltjes problem.
There is an interesting use of determinants to rewrite $h_N$ and
$s_N$ in terms of the moment problem that makes their positivity
properties evident. Suppose $d\rho$ obeys (1.1). Then
$$\align
h_N &= \int \det ((x^{a+b}_a)_{0\leq a,b\leq N-1})
\prod_{a=0}^{N-1} d\rho(x_a) \\
&= \int \biggl[ \prod_{a=1}^n x^a_a\biggr] \det
((x^b_a))_{0\leq a,b\leq N-1} \prod_{a=0}^{N-1} d\rho(x_a).
\endalign
$$
Permuting over indices, we see that in $\prod_{a=1}^n (x^a_a)
\det (x^b_a)$, we can replace $x_a$ by $x_{\pi(a)}$ for any
permutation, $\pi$. Since $\det (x^b_{\pi(a)})=(-1)^\pi
\det(x^b_a)$, we see that
$$
h_N = (N!)^{-1} \int [\det((x^b_a)_{0\leq a,b\leq N-1})]^2
\prod_{a=0}^{N-1} d\rho(x_a).
$$
Recognizing the Vandermonde determinant, we have
$$
h_N = (N!)^{-1} \int \prod_{0\leq a < b\leq N-1} (x_a - x_b)^2
\prod_{a=0}^{N-1} d\rho(x_a). \tag A.3a
$$
Similarly,
$$
s_N = (N!)^{-1} \int x_0 \dots x_{N-1}
\prod_{0\leq a < b \leq N-1} (x_a - x_b)^2
\prod_{a=0}^{N-1} d\rho(x_a).
\tag A.3b
$$
The most basic formula is:
\proclaim{Theorem A.1} $P_n(x)$ is given by
$$
P_n(x) = \frac{1}{\sqrt{h_n h_{n+1}}}\,
\det \pmatrix
\gamma_0 & \gamma_1 & \dots & \gamma_n \\
\gamma_1 & \gamma_2 & \dots & \gamma_{n+1} \\
\vdots & \vdots & {} & \vdots \\
\gamma_{n-1} & \gamma_n & \dots & \gamma_{2n-1} \\
1 & x & \dots & x^n
\endpmatrix. \tag A.4
$$
\endproclaim
\demo{Proof} Let $S_n(x)$ be the determinant on the right side
of (A.4). Then for any solution $\rho$ of the moment problem, we
have that $\int S_n(x) x^j\, d\rho(x)$ is given by the same
determinant, but with the last row replaced by $\gamma_j \
\gamma_{j+1} \ \dots \ \gamma_{j+n}$. It follows that
$$\alignat2
\int x^j S_n (x)\, d\rho(x) &=0, \qquad && j=0,1,\dots, n-1
\tag A.5 \\
&= h_{n+1} \qquad && j=n.
\endalignat
$$
In particular, since
$$
S_n (x) = h_n x^n + \text{ lower order}, \tag A.6
$$
we see that
$$
\int S_n (x)^2\, d\rho(x) = \ h_n h_{n+1}. \tag A.7
$$
>From (A.5)--(A.7), it follows that $P_n(x) = \frac{S_n (x)}
{\sqrt{h_n h_{n+1}}}$. \qed
\enddemo
>From (A.4), we can deduce formulae for the coefficients
$a_n, b_n$ of a Jacobi matrix associated to the moments
$\{\gamma_n\}^\infty_{n=0}$. Let $\tilde h_n$ be the $n\times n$
determinant obtained by changing the last column in $\Cal H_N$:
$$
\tilde h_n = \det \pmatrix
\gamma_0 & \gamma_1 & \dots & \gamma_{n-2} & \gamma_n \\
\vdots & \vdots &{} & \vdots & \vdots \\
\gamma_n & \gamma_{n+1} & \dots & \gamma_{2n-3} &
\gamma_{2n-1}
\endpmatrix.
$$
Thus (A.4) implies that
$$
P_n(x) = \sqrt{\frac{h_n}{h_{n+1}}} \, \biggl[ x^n -
\frac{\tilde h_n}{h_n}\, x^{n-1} + \text{ lower order}
\biggr] . \tag A.8
$$
\proclaim{Theorem A.2} For $n\geq 0$,
$$\alignat2
&\text{\rom{(i)}} \qquad && a_n = \biggl(\frac{h_n h_{n+2}}
{h^2_{n+1}}\biggr)^{1/2} \\
&\text{\rom{(ii)}} \qquad && \sum_{j=0}^n b_j =
\frac{\tilde h_{n+1}}{h_{n+1}}\, .
\endalignat
$$
\endproclaim
\demo{Proof} By the definition of $a_n$ and $b_n$, we have that
$$
xP_n (x) = a_n P_{n+1}(x) + b_n P_n (x) + a_{n-1}
P_{n-1}(x). \tag A.9
$$
Identifying the $x^{n+1}$ and $x^n$ terms in (A.9) using (A.8),
we see that
$$\gather
\sqrt{\frac{h_n}{h_{n+1}}} = a_n \sqrt{\frac{h_{n+1}}
{h_{n+2}}} \tag A.10 \\
\sqrt{\frac{h_n}{h_{n+1}}} \biggl( - \frac{\tilde h_n}{h_n}
\biggr) = a_n \sqrt{\frac{h_{n+1}}{h_{n+2}}} \biggl( -
\frac{\tilde h_{n+1}}{h_{n+1}} \biggr) + b_n
\sqrt{\frac{h_n}{h_{n+1}}}\, . \tag A.11
\endgather
$$
(A.10) implies (i) immediately, and given (A.10), (A.11) becomes
$$
b_n = \frac{\tilde h_{n+1}}{h_{n+1}} - \frac{\tilde h_n}
{h_n}\, ,
$$
which implies (ii) by induction if we note the starting point
comes from looking at the constant term in $x P_0(x) =x = a_0
P_1 (x) + b_0 P_0 (x)$, which implies that $b_0 =
\frac{\tilde h_1}{h_1} = \frac{\gamma_1}{\gamma_0}$. \qed
\enddemo
\remark{Remark} (ii) has an alternate interpretation. (A.8) says
that $\frac{\tilde h_{n+1}}{h_{n+1}}$ is the sum of the $n+1$
roots of $P_{n+1}(x)$. But $P_{n+1}(x)$ is a multiple of the
determinant of the Jacobi matrix $A^{[n+1]}_F$. So the sum of
the roots is just the trace of $A^{[n+1]}_F$, that is,
$\sum_{j=0}^n b_j$.
\endremark
>From (A.4) and Theorem~4.2, (i.e., $Q_n(x) = E_X
(\frac{P_n(X) - P_n(Y)}{X-Y})$), we immediately get
\proclaim{Theorem A.3}
$$
Q_n(x) = \frac{1}{\sqrt{h_n h_{n+1}}}\,
\det \pmatrix
\gamma_0 & \gamma_1 & \dots & \gamma_n \\
\gamma_1 & \gamma_2 & \dots & \gamma_{n+1} \\
\vdots & \vdots & {} & \vdots \\
\gamma_{n-1} & \gamma_n & \dots & \gamma_{2n-1} \\
R_{n,0}(x) & R_{n,1}(x) & \dots & R_{n,n}(x)
\endpmatrix \tag A.12
$$
where
$$\align
R_{n,j}(x) &= \sum_{k=0}^{j-1} \gamma_{j-1-k} x^k, \qquad
j \geq 1 \tag A.13a \\
R_{n,j=0} (x) &= 0. \tag A.13b
\endalign
$$
\endproclaim
\demo{Proof} $\frac{x^j - y^j}{x-y} = \sum_{k=0}^{j-1}
y^{j-1-k}x^k$ so $E_X (\frac{x^j - y^j}{x-y}) = R_{n,j}(x)$. \qed
\enddemo
\remark{Remark} By (A.4), $P_n(x) \sum_{k=0}^{2n-1} \gamma_k
x^{-k-1}$ has the same form as (A.4) but with the bottom row
replaced by $S_{n,0}(x) \dots S_{n,n}(x)$ where
$$\align
S_{n,j}(x) &= \sum_{k=0}^{2n-1} \gamma_k x^{j-k-1} =
\sum_{\ell = -(2n-j)}^{j-1} \gamma_{j-1-\ell} x^\ell \\
&= R_{n,j}(x) + x^{-1} \gamma_j + x^{-2} \gamma_{j+1} +
\cdots + x^{-n} \gamma_{j+n-1} + O(x^{-n-1}).
\endalign
$$
Recognizing $x^{-1} \gamma_j + x^{-2} \gamma_{j+1} + \cdots +
x^{-n-1} \gamma_{j+n-1}$ as $x^{-1}$ times the first row of
the matrix in (A.4) plus $x^{-2}$ times the second row plus
\dots, we conclude that
$$
P_n(x) \biggl(\sum_{k=0}^{2n-1} \gamma_k x^{-k-1}\biggr) =
Q_n(x) + O(x^{-n-1})
$$
and thus
$$
-\frac{Q_n(x)}{P_n(x)} = -\sum_{k=0}^{2n-1} \gamma_k
x^{-k-1} + O(x^{-2n-1})
$$
consistent with the $f^{[N-1,N]}$ Pad\'e formula (5.28).
\endremark
We saw the quantity $L$ of (5.29) is important. Here is a
formula for it.
\proclaim{Theorem A.4} $L=\lim_{n\to\infty} - \frac{Q_n(0)}
{P_n(0)}$ and
$$
-\frac{Q_n(0)}{P_n(0)} = \frac{t_n}{s_n}\, , \tag A.14a
$$
where $s_n = \det(\Cal S_N)$ and
$$
t_n = -\det \pmatrix
0 & \gamma_0 & \gamma_1 & \dots & \gamma_n \\
\gamma_0 & \gamma_1 & \gamma_2 & \dots & \gamma_{2n+1} \\
\vdots & \vdots & \vdots & {} & \vdots \\
\gamma_n & \gamma_{n+1} & \gamma_{n+2} & \dots & \gamma_{2n-1}
\endpmatrix . \tag A.14b
$$
\endproclaim
\demo{Proof} Follows by putting $x=0$ in our formula for $P_n(x)$
and $Q_n(x)$. \qed
\enddemo
Next we have explicit formulae for $M_n(x)$ and $N_n(x)$, the
polynomials introduced for Section~5 (see (5.15) and (5.16)).
\proclaim{Theorem A.5}
$$\align
M_n(x) &= \frac{1}{s_{n-1}} \sqrt{\frac{h_n}{h_{n+1}}}
\pmatrix
\gamma_1 & \gamma_2 & \dots & \gamma_n \\
\gamma_2 & \gamma_3 & \dots & \gamma_{n+1} \\
\vdots & \vdots & {} & \vdots \\
\gamma_{n-1} & \gamma_{n-2} & \dots & \gamma_{2n-2} \\
x & x^2 & \dots & x^n
\endpmatrix \tag A.15 \\
N_n(x) & = \frac{1}{s_{n-1}} \sqrt{\frac{h_n}{h_{n+1}}}
\pmatrix
\gamma_1 & \gamma_2 & \dots & \gamma_n \\
\gamma_2 & \gamma_3 & \dots & \gamma_{n+1} \\
\vdots & \vdots & {} & \vdots \\
\gamma_{n-1} & \gamma_{n-2} & \dots & \gamma_{2n-2} \\
R_{n,1}(x) & R_{n,2}(x) & \dots & R_{n,n}(x)
\endpmatrix \tag A.16
\endalign
$$
where $R_{i,j}(x)$ is given by {\rom{(A.13)}}.
\endproclaim
\remark{Remarks} 1. $P_n$ is given by a $(n+1)\times (n+1)$
matrix. To get $M_n$, which has an $n\times n$ matrix, we drop
the first column and next to last row.
2. It is interesting that (A.15) does not obviously follow from
the basic definition (5.15).
\endremark
\demo{Proof} The right side of (A.15) (call it $\widetilde M_n(x)$)
has the following properties:
\roster
\item"\rom{(i)}" It is a polynomial of degree $n$.
\item"\rom{(2)}" It obeys $\widetilde M_n(0)=0$.
\item"\rom{(3)}" It obeys $E_x (x^j \widetilde M_n(x)) =0$ for
$j=0,1,\dots, n-2$ since the corresponding matrix has two equal
rows.
\item"\rom{(4)}" It obeys $\widetilde M_n(x) = \sqrt{h_n/h_{n+1}}
\, x^n + \text{ lower order}$ so it has the same highest degree
term as $P_n(x)$.
\endroster
These properties uniquely determine $M_n(x)$ so (A.15) is proven.
(A.16) then follows from (5.24). \qed
\enddemo
\remark{Remark} By mimicking the argument following Theorem~A.3,
one finds that
$$
M_n(x) \sum_{k=0}^{2n-2} \gamma_k x^{-k-1} = N_n(x) +
O(x^{-n})
$$
so that
$$
-\frac{N_n(x)}{M_n(x)} = -\sum_{k=0}^{2n-2} \gamma_k
x^{-k-1} + O(x^{-2n}),
$$
consistent with the $f^{[N-1, N-1]}$ Pad\'e formula used in the
proof of Theorem~6 (at the end of Section~5).
\endremark
Remarkably, there are also single determinant formulae for
$\sum_{j=0}^n P_j(0)^2$ and $\sum_{j=0}^n \mathbreak Q_j(0)^2$.
\proclaim{Theorem A.6} We have that
$$\align
\sum_{j=0}^n P_j (0)^2 &= \frac{v_n}{h_{n+1}} \tag A.17 \\
\sum_{j=0}^n Q_j (0)^2 &= - \frac{w_{n+2}}{h_{n+1}}\, ,
\tag A.18
\endalign
$$
where $v_n$ is given by the $n\times n$ determinant,
$$
v_n = \det \pmatrix
\gamma_2 & \dots & \gamma_{n+1} \\
\gamma_3 & \dots & \gamma_{n+2} \\
\vdots & {} & \vdots \\
\gamma_{n+1} & \dots & \gamma_{2n} \\
\endpmatrix \tag A.19
$$
and $w_{n+2}$ by the $(n+2)\times (n+2)$ determinant,
$$
w_n = \det \pmatrix
0 & 0 & \gamma_0 & \gamma_1 & \dots & \gamma_{n-1} \\
0 & \gamma_0 & \gamma_1 & \gamma_2 & \dots & \gamma_n \\
\gamma_0 & \gamma_1 & \gamma_2 & \gamma_3 & \dots & \gamma_{n+1} \\
\vdots & \vdots & \vdots & \vdots & {} & \vdots \\
\gamma_{n-1} & \gamma_n & \gamma_{n+1} & \gamma_{n+2} & \dots &
\gamma_{2n}
\endpmatrix \tag A.20
$$
\endproclaim
\remark{Remark} Thus by Theorems~3 and 7, indeterminacy for
both the Hamburger and Stieltjes problems can be expressed in
terms of limits of ratios of determinants. In the Hamburger
case, we need (A.17) and (A.18) to have finite limits in order
that the problem be indeterminate. In the Stieltjes case, (A.17)
and (A.14) must have finite limits for indeterminacy to hold.
\endremark
\demo{Proof} We actually prove a stronger formula. Let
$$
B_N(x,y)= \sum_{n=0}^N P_n(x) P_n(y) . \tag A.21
$$
We will show that
$$
B_n (x,y) = -h^{-1}_{n+1} \det \pmatrix
0 & 1 & x & \dots & x^N \\
1 & \gamma_0 & \gamma_1 & \dots & \gamma_N \\
y & \gamma_1 & \gamma_2 & \dots & \gamma_{N+1} \\
\vdots & \vdots & \vdots & {} & \vdots \\
y^N & \gamma_N & \gamma_{N+1} & \dots & \gamma_{2N}
\endpmatrix \tag A.22
$$
(A.17) then follows by setting $x=y=0$ and noting that if $C =
\{C_{ij}\}_{1\leq i \leq j \leq N}$ is an $N\times N$ matrix
with $C_{11}=0$, $C_{1j} = \delta_{2j}$, and $C_{i1} =
\delta_{i2}$, then
$$
\det ((C_{ij})_{1\leq i,j\leq N}) = -\det ((C_{ij})_{3\leq
i,j\leq N}).
$$
(A.18) then follows from $Q_j (z) = E_x
\bigl(\frac{[P_j (x) - P_j (z)]}{(x-z)}\bigr)$, since
$$
\sum_{n=0}^N Q_n(0)^2 = E_x E_y ([B_N (x,y) - B_N (x,0) -
B_N (0,y) + B_N (0,0)] x^{-1} y^{-1}).
$$
Thus we need only prove (A.22). To do this, let $\tilde B_N(x,y)$
be the right side of (A.22). Consider $E_x (\tilde B_N(x,y) x^j)$
for $0\leq j \leq N$. This replaces the top row in the
determinant by $(0 \ \gamma_j \ \gamma_{j+1} \ \dots \
\gamma_{N+j})$. The determinant is unchanged if we subtract the
row $(y^j \ \gamma_j \ \dots \ \gamma_{N+j})$ from this row.
That is, $E_x (\tilde B_N(x,y)x^j)$ is given by the determinant
with top row $(-y^j \ 0 \ 0 \ \dots \ 0)$. Thus,
$$
E_x (\tilde B_N (x,y) x^j) = -h^{-1}_{n+1} (-y^j)
h_{n+1} = y^j.
$$
So, $\tilde B_N(x,y)$ is a reproducing kernel. For any polynomial
$P(x)$ of degree $N$ or smaller, $E_x (\tilde B_N(x,y) P(x)) =
P(y)$. But
$$
\tilde B_N (x,y) = \sum_{j=0}^N E_x (\tilde B_N (x,y)
P_j (x)) P_j(x) = B_N (x,y)
$$
since $\{P_j(x)\}^N_{j=0}$ is an orthonormal basis in the
polynomials of degree $N$ or less. Thus (A.22) is proven. \qed
\enddemo
In terms of the $\gamma^{(\ell)}$ moment problems (with moments
$\gamma^{(\ell)}_j = \frac{\gamma_{\ell+j}}{\gamma_\ell}$), we
recognize $v_n$ as $(\gamma_2)^{n-1} h_{n-1}(\gamma^{(2)})$.
By (A.17) for the $\gamma^{(2)}$ problem,
$$
\sum_{j=0}^{n-1} P^{(2)}_j (0) = \frac{v_{n-1} (\gamma^{(2)})}
{h_n (\gamma^{(2)})}\, .
$$
But $h_n (\gamma^{(2)}) = v_n (\gamma)(\gamma_2)^{-n}$ and
$v_{n-1} (\gamma^{(2)}) = \gamma^{1-n}_2 y_{n-1}(\gamma)$ where
$$
y_{n-1} = \pmatrix
\gamma_4 &\dots & \gamma_{n+2} \\
\gamma_5 & \dots & \gamma_{n+3} \\
\vdots & {} & \vdots \\
\gamma_{n+1} & \dots & \gamma_{2n}
\endpmatrix . \tag A.23
$$
Thus, using Proposition~5.13:
\proclaim{Theorem A.7}
$$
\gamma^{-1}_2 \biggl( \sum_{j=0}^n P_j (0)^2 \biggr)
\biggl( \sum_{j=0}^{n-1} P^{(2)}_j (0)^2\biggr) =
\frac{y_{n-1}}{h_{n+1}}\, .
$$
In particular, the Hamburger problem is determinate if and only
if $\lim_{n\to\infty} \frac{y_{n-1}}{h_{n+1}} = \infty$.
\endproclaim
So we have a simple ratio of determinants to determine
determinacy.
\vskip 0.3in
\flushpar {\bf Appendix B: The Set of Solutions of the
Moment Problem as a Compact Convex Set}
\vskip 0.1in
In this appendix, we will prove (following [\akh]) that
$\Cal M^H (\gamma)$ and $\Cal M^S(\gamma)$ are compact convex
sets whose extreme points are dense. Each set is a subset of
$\Cal M_+(\Bbb R\cup\{\infty\})$, the set of measures on the
compact set $\Bbb R\cup\{\infty\}$. This set, with the condition
$\int d\rho(x) = \gamma_0$, is a compact space in the topology
of weak convergence (i.e., $\int f(x)\, d\rho_n(x) \to \int f(x)\,
d\rho(x)$ for continuous functions $f$ on $\Bbb R\cup\{\infty\}$).
\proclaim{Theorem B.1} $\Cal M^H(\gamma)$ and $\Cal M^S(\gamma)$
are closed in the weak topology and so are compact convex sets.
\endproclaim
\remark{Remark} Since the $x^n$'s are unbounded, this does not
follow from the definition of the topology without some additional
argument.
\endremark
\demo{Proof} By Propositions~4.4 and 4.13, $\mu\in\Cal M^H
(\gamma)$ if and only if
$$
\int (x-z_0)^{-1}\, d\mu(x) \in D(z_0) \tag B.1
$$
for all $z_0\in\Bbb C$. The set of $\mu$'s that obey (B.1) for
a fixed $z_0$ is closed since $D(z_0)$ is closed and
$(x-z_0)^{-1}$ is in $C(\Bbb R\cup\{\infty\})$. Thus, the
intersection over all $z_0$ is closed. $\Cal M^S(\gamma) =
\Cal M^H(\gamma) \cap \{\mu\mid \int f(x)\, d\mu(x)=0 \text{ if }
\text{supp}\, f\subset (-\infty, 0)\}$ is an intersection of
closed sets. \qed
\enddemo
\remark{Remark} To get compact sets of measures, we need to consider
$\Bbb R\cup\{\infty\}$ rather than $\Bbb R$. Thus a priori, the integral
in (B.1) could have a point at infinity giving a constant term as
$z_0 = iy$ with $y\to\infty$. Since $D(z_0)\to\{0\}$ as $z_0 = iy$
with $y\to\infty$, this term is absent.
\endremark
\proclaim{Theorem B.2 (Naimark)} $\mu\in\Cal M^H(\gamma)$
\rom(resp.~$\Cal M^S(\gamma)$\rom) is an extreme point if and
only if the polynomials are dense in $L^1 (\Bbb R, d\rho)$.
\endproclaim
\remark{Remark} Compare with density in $L^2 (\Bbb R, d\rho)$
which picks out the von~Neumann solutions.
\endremark
\demo{Proof} This is a simple use of duality theory. The
polynomials fail to be dense if and only if there exists a
non-zero $F\in L^\infty (\Bbb R, d\rho)$ so that
$$
\int x^n F(x)\, d\rho(x) =0 \tag B.2
$$
for all $n$. We can suppose that $\| F\|_\infty = 1$, in which
case $d\rho_\pm = (1\pm F)\, d\rho$ both lie in $\Cal M^H
(\gamma)$ with $\rho=\frac12 (\rho_+ + \rho_-)$, so $\rho$ is not
extreme.
Conversely, suppose $\rho=\frac12(\rho_+ + \rho_-)$ with
$\rho_+ \neq \rho_-$ and both in $\Cal M^H(\gamma)$. Then
$\rho_\pm \leq 2\rho$ so $\rho_+$ is $\rho$-absolutely
continuous, and the Radon-Nikodyn derivative, $\frac{d\rho_+}
{d\rho}$, obeys $\| \frac{d\rho_+}{d\rho}\|_\infty \leq 2$.
Let $F=1-\frac{d\rho_+}{d\rho}$, so $\|F\|_\infty \leq 1$ and
$F\neq 0$ since $\rho_+ \neq \rho_-$. Then
$$
\int F(x) x^n \, d\rho = \int x^n \, d\rho - \int x^n \,
d\rho_+ = 0.
$$
Thus, we have proven the result for $\Cal M^H(\gamma)$. Since
$\Cal M^S(\gamma) = \{\rho\in \Cal M^H(\gamma) \mid \int f(x)
\, d\rho(x)=0 \text{ for $f$ in } C(\Bbb R\cup\{\infty\})
\text{ with support in } (-\infty, 0)\}$, $\Cal M^S (\gamma)$
is a face of $\Cal M^H (\gamma)$, so extreme points of $\Cal M^S
(\gamma)$ are exactly those extreme points of $\Cal M^H(\gamma)$
that lie in $\Cal M^S(\gamma)$. \qed
\enddemo
\proclaim{Theorem B.3} Let $\rho\in \Cal M^H(\gamma)$ have
$\text{\rom{ord}}(\rho)<\infty$. Then $\rho$ is an extreme point.
\endproclaim
\demo{Proof} By Theorem~6.4, for some $z_1, \dots, z_n \in
\Bbb C_+$, we have that the polynomials are dense in $L^2 (\Bbb R,
\prod_{j+1}^n |x-z_j|^{-2}d\rho)$. For any polynomially bounded
continuous function
$$
\int |f(x)|\, d\rho(x) \leq \biggl( \int |f(x)|^2 \prod_{j=1}^n
|x-z_j|^{-2}\, d\rho(x) \biggr)^{1/2}
\biggl(\int \prod_{j=1}^n (x-z_j)^2 \, d\rho(x)\biggr)^{1/2}
$$
by the Schwarz inequality. It follows that the identification
map is continuous from $L^2 (\Bbb R, \prod_{j=1}^n (x-z)^{-2}
\, d\rho)$ into $L^1 (\Bbb R, d\rho)$, so the polynomials are
dense in $L^1$-norm in the continuous functions, and so in
$L^1 (\Bbb R, d\rho)$. By Theorem~B.2, $\rho$ is an extreme
point. \qed
\enddemo
\proclaim{Theorem B.4} The extreme points are dense in $\Cal M^H
(\gamma)$ \rom(and in $\Cal M^S(\gamma)$\rom).
\endproclaim
\demo{Proof} The finite point measures are dense in the finite
measures on $\Bbb R\cup\{\infty\}$. Thus, by the Herglotz
representation theorem in form (6.8), if $\Phi$ is a Herglotz
function, there exist real rational Herglotz functions $\Phi_n$
so that $\Phi_n(z) \to \Phi(z)$ for each $z\in\Bbb C_+$.
Now let $\rho\in \Cal M^H(\gamma)$ and let $\Phi_\rho$ be the
Nevanlinna function of $\rho$. Let $\Phi_n$ be as above and
$\rho_n = \rho_{\Phi_n}$. Then by (4.35), $G_{\rho_n}(z) \to
G_\rho (z)$ for each $z\in\Bbb C_+$ and so $\rho_n \to \rho$
weakly. By Theorems~6.4 and B.3, each such $\rho_n$ is an
extreme point.
For the Stieltjes case, we need only note that by Theorem~4.18
and the remark after it, if $\rho\in \Cal M^S(\gamma)$, then
the approximating $\Phi_n$'s can be chosen so that $\rho_{\Phi_n}
\in\Cal M^S (\gamma)$. \qed
\enddemo
\proclaim{Theorem B.5} For any indeterminate set of Hamburger
moments $\{\gamma_n\}^\infty_{n=0}$, $\Cal M^H(\gamma)$ has
extreme points $\rho$ with $\text{\rom{ord}}(\rho)=\infty$.
\endproclaim
\demo{Proof} We first pick positive $\alpha_j$ strictly decreasing
so that for any $\rho\in\Cal M^H(\gamma)$ we have that
$$
\sup_n \int \prod_{j=1}^n (1+\alpha^2_j x^2)^2\, d\rho (x)
\leq 2. \tag B.3
$$
We can certainly do this as follows: Since the integral only
depends on the moments $\gamma_j$, we need only do it for some
fixed $\rho_0 \in\Cal M^H(\gamma)$. Since
$$
\lim_{\alpha_1 \downarrow 0} \int (1+\alpha^2_1 x^2)^2\,
d\rho(x)= 1,
$$
we can pick $\alpha_1 >0$ so
$$
\int (1+\alpha^2_1 x^2)^2 \, d\rho(x) <2.
$$
We then pick $\alpha_2, \alpha_3, \dots$ inductively so the
integral is strictly less than 2. Then the $\sup$ in (B.3) is
bounded by 2. The product must be finite for a.e.~$x$ w.r.t.
$d\rho$, so
$$
\sum_{j=1}^\infty \alpha^2_j < \infty. \tag B.4
$$
So if
$$
g_n(x) = \prod_{j=1}^n (1+\alpha^2_j x^2),
$$
then for any real $x$,
$$
g(x) = \lim_{n\to\infty} g_n(x)
$$
exists, and by (B.3),
$$
\int g(x)^2\, d\rho(x) < \infty
$$
for any $\rho\in\Cal M^H(\gamma)$.
Now pick $\zeta_1, \zeta_2, \dots$ in $\Bbb C_+$ and
$\rho_1, \rho_2, \dots$ in $\Cal M^H(\gamma)$ inductively so
$\rho_k$ obeys
$$
\int (x-i\alpha^{-1}_j)^{-1} \, d\rho_k (x) = \zeta_j,
\qquad j=1,\dots, k \tag B.5
$$
and then $\zeta_{k+1}$ is the middle of the disk of allowed values
for $\int (x-i\alpha^{-1}_{j+1})^{-1}\, d\rho(x)$ for those $\rho$
in $\Cal M^H(\gamma)$ which obey (B.5). Pick a subsequence of the
$\rho_j$'s which converges to some $\rho_\infty \in \Cal M^H
(\gamma)$. Then $\rho_\infty$ obeys (B.5) for all $k$.
Let
$$
\Cal M_k = \{\rho\in\Cal M^H(\gamma)\mid \rho
\text{ obeys (B.5)}\}
$$
and
$$
\Cal M_\infty = \cap \Cal M_k.
$$
Define for all $n,m$,
$$
\Gamma^{(n)}_m = \int x^m \prod_{j=1}^n (1+\alpha^2_j x^2)^{-1}
\, d\rho_n (x).
$$
By Theorem~6.1, $\mu\in\Cal M^H(\Gamma^{(n)})$ if and only if
$d\rho \equiv g_n\, d\mu$ lies in $\Cal M_k$. $\Gamma^{(n)}_m$
is decreasing to $\Gamma^{(\infty)}_m$ and
$$
\Gamma^{(\infty)}_m = \int x^m g(x)^{-1}\, d\rho(x)
$$
by a simple use of the monotone convergence theorem.
We claim that $\mu\in\Cal M^H(\Gamma^{(\infty)})$ if and only if
$d\rho = g\, d\mu$ lies in $\Cal M_\infty$. For
$$
\gamma_n = \lim_{n\to\infty} \int x^m g(x)^{-1} g_n(x)\,
d\rho(x)
$$
and the right side only depends on the moments
$\Gamma^{(\infty)}_m$ and similarly for calculation of $\int
(x-i\alpha_j)^{-1}\, d\rho(x)$.
Now let $\mu$ be a von~Neumann solution of the $\Gamma^{(\infty)}$
moment problem and $d\rho=g\,d\mu$. Since $\int g^2\, d\rho <
\infty$ by construction, the proof of Theorem~B.3 shows that $\rho$
is an extreme point. On the other hand, since $\rho$ obeys (B.5)
and $\zeta_j$ is not in the boundary of its allowed circle,
$\text{ord}(\rho) = \infty$. \qed
\enddemo
\vskip 0.3in
\flushpar {\bf Appendix C: Summary of Notation and Constructions}
\vskip 0.1in
Since there are so many objects and constructions associated to
the moment problem, I am providing the reader with this summary of
them and where they are discussed in this paper.
\definition{C1 \quad Structure of the Set of Moments}
\vskip 0.1in
$\{\gamma_n\}^\infty_{n=0}$ is called a set of {\it{Hamburger
moments}} if there is a positive measure $\rho$ on $(-\infty,
\infty)$ with $\gamma_n = \int x^n\, d\rho(x)$ and a set of
{\it{Stieltjes moments}} if there is a $\rho$ with support in
$[0,\infty)$. We take $\gamma_0=1$ and demand $\text{supp}(\rho)$
is not a finite set. Theorem~1 gives necessary and sufficient
conditions for existence. If there is a unique $\rho$, the
problem is called {\it{determinate}}. If there are multiple
$\rho$'s, the problem is called {\it{indeterminate}}. If a set of
Stieltjes moments is Hamburger determinate, it is a fortiori
Stieltjes determinate. But the converse can be false (see the
end of Section~3).
$\Cal M^H(\gamma)$ (resp.~$\Cal M^S(\gamma)$) denotes the set
of all solutions of the Hamburger problem (resp.~all $\rho\in
\Cal M^H(\gamma)$ supported on $[0,\infty)$). They are compact
convex sets (Theorem~B.1) whose extreme points are dense
(Theorem~B.4). Indeterminate Hamburger problems have a
distinguished class of solutions we have called {\it{von~Neumann
solutions}} (called $N$-extremal in [\akh] and extremal in
[\st]). They are characterized as those $\rho\in\Cal M^H
(\gamma)$ with the polynomials dense in $L^2 (\Bbb R, d\rho)$.
They are associated to self-adjoint extensions $B_t$ of the
Jacobi matrix associated to $\gamma$ (see {\bf{C4}} below for
the definition of this Jacobi matrix). The parameter $t$ lies in
$\Bbb R\cup \{\infty\}$ and is related to $B_t$ and the solution
$\mu_t\in \Cal M^H(\gamma)$ by
$$
t = (\delta_0, B^{-1}_t \delta_0) = \int x^{-1} d\mu_t(x).
\tag C.1
$$
In the indeterminate Stieltjes case, there are two distinguished
solutions: $\mu_F$, the {\it{Friedrichs solution}}, and $\mu_K$,
the {\it{Krein solution}}. Both are von~Neumann solutions of the
associated Hamburger problem and are characterized by $\inf
[\text{supp}(\mu_F)] > \inf[\text{supp}\,\rho]$ for any other
$\rho\in\Cal M^H(\gamma)$ and by $\mu_K(\{0\})>0$ and $\mu_K
(\{0\}) >\rho(\{0\})$ for any other $\rho\in\Cal M^H(\gamma)$
(see Theorems~3.2 and 4.11 and 4.17). All solutions of the
Stieltjes problem lie between $\mu_F$ and $\mu_K$ in the sense
of (4.42) (see Theorem~5.10).
We define
$$
\Bbb C_+ = \{ z\in\Bbb C \mid\text{Im}\, z>0\}.
$$
For any probability measure $\rho$, we define its {\it{Stieltjes
transform}} as a function of $\Bbb C\backslash\text{supp}(\rho)$
by
$$
G_\rho(z) = \int \frac{d\rho(x)}{x-z}\, .
$$
These functions map $\Bbb C_+$ to $\Bbb C_+$ and have an
asymptotic series
$$
G_\rho (z) \sim -z^{-1} \biggl( \sum_{j=0}^\infty
\gamma_j z^{-j}\biggr) \tag C.2
$$
as $|z| \to \infty$ with $\min(|\text{Arg}(z)|, |\text{Arg}(-z)|)
>\varepsilon$. (C.2) characterizes $\Cal M^H(\gamma)$ and the
asymptotics are uniform over $\Cal M^H(\gamma)$ (Proposition~4.13).
For an indeterminate moment problem, we defined $\text{ord}(\rho)$,
the {\it{order}} of $\rho$, to be the codimension of the closure
of the polynomials in $L^2 (\Bbb R, d\rho)$. The solutions of
finite order are dense in $\Cal M^H(\gamma)$ and in $\Cal M^S
(\gamma)$, and each is an extreme point. Solutions of finite order
(and, in particular, the von~Neumann solutions) are discrete pure
point measures; equivalently, their Stieltjes transforms are
meromorphic functions (Theorems~4.11 and 6.4).
\enddefinition
\vskip 0.1in
\definition{C2 \quad Nevanlinna Parametrization}
\vskip 0.1in
We used $\Cal F$ to denote the analytic maps of $\Bbb C_+$ to
$\bar\Bbb C_+$ where the closure is in the Riemann sphere. The
open mapping theorem implies that if $\Phi\in\Cal F$, either
$\Phi(z) \equiv t$ for some $t\in\Bbb R\cup\{\infty\}$ or else
$\Phi$ is a Herglotz function which has a representation of
the form (1.19).
Associated to any indeterminate moment problem are four entire
functions, $A(z)$, $B(z)$, $C(z)$, $D(z)$, obeying growth
condition of the form $|f(z)|\leq c_\varepsilon \exp(\varepsilon
|z|)$. We defined them via the transfer matrix relation (4.16),
but they also have explicit formulae in terms of the orthogonal
polynomials, $P$ and $Q$ (defined in {\bf{C4}} below)---these
are given by Theorem~4.9.
We define a fractional linear transformation $F(z) : \Bbb C \cup
\{\infty\} \to \Bbb C\cup\{\infty\}$ by
$$
F(z)(w) = -\frac{C(z)w + A(z)}{D(z)w + B(z)}\, . \tag C.3
$$
For indeterminate Hamburger problems, there is a one-one
correspondence between $\rho\in\Cal M^H(\gamma)$ and $\Phi\in
\Cal F$ given by
$$
G_\rho(z) = F(z)(\Phi(z))= - \frac{C(z)\Phi(z) + A(z)}
{D(z) \Phi(z) + B(z)} \tag C.4
$$
for $z\in\Bbb C_+$ (Theorem~4.14). $\Phi$ is called the Nevanlinna
function of $\rho$, denoted by $\Phi_\rho$. The von~Neumann
solutions correspond precisely to $\Phi(z) = t$ where $t$ is
given by (C.1) (Theorem~4.10). The solutions of finite order
correspond precisely to the case where $\Phi$ is a ratio of real
polynomials (Theorem~6.4).
Given a set of Stieltjes moments, the $\rho\in\Cal M^S(\gamma)$
are precisely those $\rho\in\Cal M^H(\gamma)$ whose Nevanlinna
function has the form (Theorem~4.18)
$$
\Phi(z) = d_0 + \int_0^\infty \frac{d\mu(x)}{x-z}
$$
with $\int d\mu(x) < \infty$ and $d_0 \geq t_F = \int d\rho_F(x)
= (\delta_0, A^{-1}_F\delta_0)$.
For $z\in\Bbb C_+$, the image of $\bar\Bbb C_+ \cup \{\infty\}$
under $F(z)$ is a closed disk denoted by $\Cal D(z)$. Von~Neumann
solutions $\rho$ have $G_\rho(z)\in\partial\Cal D(z)$ for all $z$
while other solutions have $G_\rho(z)\in\Cal D(z)^{\text{\rom{int}}}$
for all $z$ (Theorem~4.3, Proposition~4.4, and Theorem~4.14).
\enddefinition
\vskip 0.1in
\definition{C3 \quad Derived Moment Problems}
\vskip 0.1in
Any set of moments has many families of associated moments. For
each real $c$, $\gamma(c)$ is defined by
$$
\gamma_n (c) = \sum_{j=0}^n \binom{n}{j} c^j \gamma_{n-j}.
\tag C.5
$$
There is a simple map ($\rho\mapsto \rho (\,\cdot\, -c)$) that
sets up a bijection between $\Cal M^H(\gamma)$ and $\Cal M^H
(\gamma(c))$ and, in particular, $\gamma$ is Hamburger
determinate if and only if $\gamma(c)$ is Hamburger determinate.
But the analog is not true for the Stieljes problem. Indeed,
(end of Section~3), if $\gamma$ is a set of indeterminate
Stieltjes moments and $c_F = -\inf\,\text{supp}(d\mu_F)$, then
$\gamma(c)$ is a set of Stieltjes moments if and only if $c
\geq c_F$, and it is Stieltjes determinate if and only if $c =
c_F$. The orthogonal polynomials for $\gamma(c)$ and $\gamma$
are related via $P^{(\gamma(c))}_N(z) = P_N(z-c)$.
Given a set of Stieltjes moments $\{\gamma_n\}^\infty_{n=0}$,
one defines Hamburger moments $\{\Gamma_n\}^\infty_{n=0}$ by
$$
\Gamma_{2m} = \gamma_m, \qquad \Gamma_{2m+1} = 0. \tag C.6
$$
There is a simple map ($d\rho\mapsto d\mu(x) = \frac12
\chi_{[0,\infty)}(x)\, d\rho(x^2) + \frac12 \chi_{(-\infty, 0]}
(x)\, d\rho(x^2)$) that sets up a bijection between $\Cal M^H
(\gamma)$ and those $\mu\in\Cal M^H(\Gamma)$ with $d\mu (-x)
=d\mu(x)$. Then $\gamma$ is Stieltjes determinate if and only
if the $\Gamma$ problem is Hamburger determinate (Theorem~2.13).
There are simple relations between the orthogonal polynomials
for $\gamma$ and those for $\Gamma$ ((5.89)--(5.92)).
For $\ell=1,2,\dots$, define
$$
\gamma^{(\ell)}_j = \frac{\gamma_{\ell+j}}{\gamma_\ell}\, .
\tag C.7
$$
If $\gamma$ is a set of Stieltjes moments, so are $\gamma^{(\ell)}$
for all $\ell$, and if $\gamma$ is a set of Hamburger moments, so
are $\gamma^{(\ell)}$ for $\ell = 2,4,\dots$. The map $\rho \mapsto
d\rho^{(\ell)} = x^\ell \frac{d\rho(x)}{\gamma_\ell}$ is a map
from $\Cal M^H(\gamma)$ to $\Cal M^H(\gamma^{(\ell)})$ which is
injective. But in the indeterminate case, it is {\it{not}}, in
general, surjective. If $\gamma$ is indeterminate, so is
$\gamma^{(\ell)}$, but the converse may be false for either the
Hamburger or Stieltjes problems (Corollary~4.21). For general
$\ell$, the connection between the orthogonal polynomials for
$\gamma$ and $\gamma^{(\ell)}$ is complicated (see the proof of
Proposition~5.13 for $\ell=2$), but (see the remarks following
Theorem~5.14)
$$
P^{(1)}_N(z) = \frac{d_N [P_N(0) P_{N+1}(z) - P_{N+1}(0) P_N(z)]}
{z}\, .
$$
Given a set of Hamburger moments $\{\gamma_j\}^\infty_{j=0}$,
one defines a new set $\{\gamma^{(0)}_j\}^\infty_{j=0}$ by the
following formal power series relation:
$$
\biggl[ \sum_{n=0}^\infty (-1)^n \gamma_n z^n \biggr]^{-1}
= 1 - \gamma_1 z + (\gamma_2 - \gamma^2_1) z^2
\biggl( \sum_{n=0}^\infty (-1)^n \gamma^{(0)}_n z^n \biggr).
\tag C.8
$$
This complicated formula is simple at the level of Jacobi
matrices. If $\{a_n\}^\infty_{a=0}$, $\{b_n\}^\infty_{n=0}$
(resp.~$\{a^{(0)}_n\}^\infty_{n=0}$, $\{b^{(0)}_n
\}^\infty_{n=0}$) are the Jacobi matrix coefficients associated
to $\gamma$ (resp.~$\gamma^{(0)}$), then
$$
a^{(0)}_n = a_{n+1}, \qquad b^{(0)}_n = b_{n+1}.
$$
The map $\rho\mapsto \tilde\rho$ given by (5.53) sets up a
bijection between $\Cal M^H(\gamma)$ and $\Cal M^H(\gamma^{(0)})$,
and, in particular, $\gamma$ is a Hamburger determinate if and
only if $\gamma^{(0)}$ is (Proposition~5.15). This is not, in
general, true in the Stieltjes case (see Proposition~5.17). There
is a simple relation between the orthogonal polynomials for
$\gamma$ and for $\gamma^{(0)}$; see Proposition~5.16. In
particular,
$$
P^{(0)}_N(z) = a_0 Q_{N+1}(z). \tag C.9
$$
For $\ell= -1, -2, \dots$, we defined
$$
\gamma^{(\ell)} = (\gamma^{(0)})^{(-\ell)} =
\frac{\gamma^{(0)}_{j-\ell}}{\gamma^{(0)}_{-\ell}}\, .
\tag C.10
$$
\enddefinition
\vskip 0.1in
\definition{C4 \quad Associated Polynomials}
Given a set of Hamburger moments, the fundamental orthogonal
polynomials $P_N(z)$ are defined by requiring for any $\rho\in
\Cal M^H(\gamma)$ that
$$
\int P_j (x) P_\ell(x)\, d\rho(x) = \delta_{j\ell} \tag C.11
$$
and $P_N(x) = c_{NN} x^N + \text{ lower order terms}$ with
$c_{NN} >0$. They obey a three-term recursion relation
$$
xP_N(x) = a_N P_{N+1}(x) + b_N P_N(x) + a_{N-1} P_{N-1}(x).
\tag C.12
$$
The Jacobi matrix $A$ associated to $\gamma$ is the tridiagonal
matrix given by (1.16).
The associated polynomials $Q_N(x)$ are of degree $N-1$ and
defined to obey (C.12) but with the starting conditions,
$Q_0(x)=0$, $Q_1(x) = \frac1{a_0}$. They are related to $P_N$
by (Theorem~4.2)
$$
Q_N(x) = \int \frac{P_N(x) - P_N(y)}{x-y}\, d\rho(y) \tag C.13
$$
for any $\rho\in\Cal M^H(\gamma)$. As noted in (C.9), they are
up to constants, orthogonal polynomials for another moment problem,
namely, $\gamma^{(0)}$.
In the Stieltjes case, we defined polynomials $M_N$ by
$$
M_N(x) = P_N(x) - \frac{P_N(0)P_{N-1}(x)}{P_{N-1}(0)}\, .
\tag C.14
$$
They vanish at $x=0$ so $\frac{M_N(x)}{x}$ is also a
polynomial. It is of degree $N-1$ and is up to a constant, the
principal orthogonal polynomial of the moment problem
$\gamma^{(1)}$. $N_N(x)$ is defined analogously to (C.13) with
$P_N$ replaced by $M_N$. In Section~5, it was useful to change
the normalizations and define multiples of $P_N(-x)$, $Q_N(-x)$,
$M_N(-x)$, and $N_N(-x)$ as functions $U_N(x)$, $V_N(x)$,
$G_N(x)$, $H_N(x)$ given by (5.71)--(5.74).
\enddefinition
\vskip 0.1in
\definition{C5 \quad Pad\'e Approximants and Finite Matrix
Approximations}
\vskip 0.1in
If $\{\gamma_n\}^\infty_{n=0}$ is a set of Hamburger
(resp.~Stieltjes) moments, the formal power series
$\sum_{n=0}^\infty (-1)^n \gamma_n z^n$ is called a
{\it{series of Hamburger}} (resp.~{\it{a series of Stieltjes}}).
Formally, it sums to
$$
\int \frac{d\rho(x)}{1+xz}
$$
if $\rho\in\Cal M^H(\gamma)$. The Pad\'e approximants
$f^{[N,M]}(z)$ to these series, if they exist, are defined by
(5.25)--(5.27) as the rational function which is a ratio of a
polynomial of degree $N$ to a polynomial of degree $M$, whose
first $N+M+1$ Taylor coefficients are $\{(-1)^n\gamma_n
\}^{N+M}_{n=0}$.
For a series of Stieltjes, for each fixed $\ell = 0,\pm 1, \dots$,
$$
\lim_{N\to\infty} f^{[N+\ell-1, N]}(z) \equiv f_\ell (z)
$$
exists for all $z\in\Bbb C\backslash (-\infty, 0)$ and defines
a function analytic there. Indeed, for $x\in [0,\infty)$,
$(-1)^\ell f^{[N+\ell -1, N]}(x)$ is monotone increasing
(Theorem~6, Theorem~5.14, and Theorem~5.18). Moreover,
$$
f_0(z) = \int \frac{d\rho_F(x)}{1+xz}\, , \qquad
f_1(z) = \int \frac{d\rho_K (x)}{1+xz}\, ,
$$
where $\rho_F$, $\rho_K$ are the Friedrichs and Krein solutions
of the moment problem. In particular, the moment problem is
Stieltjes determinate if and only if $f_0 = f_1$.
There is a connection between $f^{[N+\ell-1,N]}(z)$ and the
moment problems $\gamma^{(\ell)}$ and $\gamma^{(\ell-1)}$ (if
$\ell < 0$, $\gamma^{(\ell)}$ and $\gamma^{(\ell+1)}$) given
in Theorem~5.14 and Theorem~5.18. In particular, if $\ell >0$ and
$\gamma^{(\ell)}$ is Stieltjes determinate, then $f_{-1}(z) =
f_0(z) = \cdots = f_{\ell+1}(z)$, and if $\ell <0$ and
$\gamma^{(\ell)}$ is Stieltjes determinate, then $f_1(z) =
f_0(z) = \cdots = f_\ell(z) = f_{\ell-1}(z)$.
The situation for series of Hamburger is more complicated, but,
in general, \linebreak $f^{[N+\ell-1, N]} (z)$ converges for
$z\in\Bbb C_+$ and $\ell = \pm 1, \pm 3, \dots$ (Theorem~5.31 and
Theorem~5.32; see Theorem~5.31 for issues of existence of the
Pad\'e approximant).
There are connections between the Pad\'e approximants and the
polynomials $P,Q,M,N$ as well as two finite matrix approximations
$A^{[N]}_F$ and $A^{[N]}_K$ to the Jacobi matrix, $A$. $A^{[N]}_F$
is the upper right $N\times N$ piece of $A$. $A^{[N]}_K$ differs
by adjusting the $NN$ matrix element of $A^{[N]}_F$ so $\det
(A^{[N]}_K)=0$. Then
$$\align
f^{[N-1, N]}(z) &= \langle\delta_0, (1+zA^{[N]}_F )^{-1} \delta_0\rangle
= -\frac{z^{N-1}Q_N (-\frac1{z})}{z^N P_N(-\frac1{z})} \\
f^{[N,N]}(z) &= \langle\delta_0, (1+zA^{[N+1]}_K)^{-1} \delta_0\rangle =
-\frac{z^N N_{N+1} (-\frac1{z})}{z^{N+1}M_{N+1} (-\frac1{z})}\, .
\endalign
$$
The connection to the continued fractions of Stieltjes is
discussed after Theorem~5.28.
\enddefinition
\vskip 0.1in
\definition{C6 \quad Criteria for Determinacy}
\vskip 0.1in
Criteria for when a Hamburger problem is determinate can be
found in Proposition~1.5, Theorem~3, Corollary~4.5,
Proposition~5.13, Theorem~A.6, and Theorem~A.7.
Criteria for when a Stieltjes problem is determinate can be
found in Proposition~1.5, Theorem~7, Corollary~4.5,
Theorem~5.21, Corollary~5.24, Theorem~A.4, and Theorem~A.6. In
particular, if one defines
$$\align
c_{2j} &= \ell_j = -[a_{j-1} P_j (0) P_{j-1}(0)]^{-1} \\
c_{2j-1} &= m_j = |P_{j-1}(0)|^2
\endalign
$$
(the $c$'s are coefficients of Stieltjes continued fractions;
the $m$'s and $\ell$'s are natural parameters in Krein's theory),
then the Stieltjes problem is determinate if and only if
(Theorem~5.21 and Proposition~5.23) $\sum_{j=0}^\infty c_j =
\infty$.
\enddefinition
\vskip 0.3in
\definition{Acknowledgments} I would like to thank George Baker,
Percy Deift, Rafael del Rio, Fritz Gesztesy, Mourad Ismail,
Uri Keich, and Sasha Kiselev for valuable comments.
\enddefinition
\vskip 0.3in
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\enddocument