Content-Type: multipart/mixed; boundary="-------------0511251232275" This is a multi-part message in MIME format. ---------------0511251232275 Content-Type: text/plain; name="05-404.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="05-404.keywords" Jacobi matrices, inverse spectral theory ---------------0511251232275 Content-Type: application/x-tex; name="wedersilva1v15.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="wedersilva1v15.tex" \documentclass[12pt]{article} \usepackage{hyperref} \usepackage{amssymb,amsmath,amsthm} \usepackage{enumerate} \setlength{\textwidth}{165mm} \setlength{\textheight}{218mm} \setlength{\topmargin}{6mm} %To get 1 inch margins use the following \setlength{\oddsidemargin}{-0mm} \setlength{\evensidemargin}{0mm} \setlength{\textwidth}{165mm} \setlength{\textheight}{228mm} \setlength{\topmargin}{-15mm} \newcommand{\sss}{\setcounter{equation}{0}} % The correct enumeration of equations \renewcommand{\theequation}{\arabic{section}.\arabic{equation}} % Theorem like environments \newtheorem{theorem}{THEOREM}[section] \renewcommand{\thetheorem}{\arabic{section}.\arabic{theorem}} \newtheorem{lemma}[theorem]{LEMMA} \renewcommand{\thelemma}{\arabic{section}.\arabic{lemma}.} \newtheorem{corollary}[theorem]{COROLLARY} \renewcommand{\thecorollary}{\arabic{section}.\arabic{lemma}.} \theoremstyle{definition} \newtheorem{remark}[theorem]{REMARK} \renewcommand{\theremark}{\arabic{section}.\arabic{remark}} \newcommand{\ere}{ {\mathbb R}} \newcommand{\ZETA}{{\mathbb Z}} \newcommand{\CE}{{\mathbb C}} \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\norm}[1]{\lVert#1\rVert} \newcommand{\I}{{\rm i}} %%%%macros-weder%%%%%% \def\cprime{$'$} \def\beq{\begin{equation}} \def\ene{\end{equation}} \def \ds {\displaystyle} \newcommand{\bull}{\hfill $\Box$} %% Mathematical operators \DeclareMathOperator{\re}{Re} \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\dom}{Dom} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator*{\res}{Res} \begin{document} \baselineskip=17 pt \parskip 6 pt \title{On the Two Spectra Inverse Problem for Semi-Infinite Jacobi Matrices \thanks{ Mathematics Subject Classification(2000): 47B36, 49N45,81Q10,47A75, 47B37, 47B39.} \thanks{ Research partially supported by Universidad Nacional Aut\'onoma de M\'exico under Project PAPIIT-DGAPA IN 105799, and by CONACYT under Project P42553­F.}} \author{Luis O. Silva and Ricardo Weder\thanks{Fellow Sistema Nacional de Investigadores.} \\[2mm] Instituto de Investigaciones en Matem\'aticas Aplicadas y en Sistemas \\Universidad Nacional Aut\'onoma de M\'exico \\ Apartado Postal 20-726, M\'exico D.\,F. 01000 \\[3mm] \texttt{silva@leibniz.iimas.unam.mx} \\\texttt{weder@servidor.unam.mx}} \date{} \maketitle \begin{center} \begin{minipage}{5.75in} \centerline{{\bf Abstract}} \bigskip We present results on the unique reconstruction of a semi-infinite Jacobi operator from the spectra of the operator with two different boundary conditions. This is the discrete analogue of the Borg-Marchenko theorem for Schr{\"o}dinger operators in the half-line. Furthermore, we give necessary and sufficient conditions for two real sequences to be the spectra of a Jacobi operator with different boundary conditions. \end{minipage} \end{center} \newpage %%%%%%%%%%%%%%%% \section{Introduction}\sss \label{sec:intro} In the Hilbert space $l_2(\mathbb{N})$ let us single out the dense subset $l_{fin}(\mathbb{N})$ of sequences which have a finite number of non-zero elements. Consider the operator $J$ defined for every $f=\{f_k\}_{k=1}^\infty$ in $l_{fin}(\mathbb{N})$ by means of the recurrence relation \begin{align} \label{eq:recurrence-coordinates} (Jf)_k &:= b_{k-1}f_{k-1} + q_k f_k + b_k f_{k+1}\quad k \in \mathbb{N} \setminus \{1\}\\ \label{eq:initial-coordinates} (Jf)_1 &:= q_1 f_1 + b_1 f_2\,, \end{align} where, for every $n\in\mathbb{N}$, $b_n$ is positive, while $q_n$ is real. $J$ is symmetric, therefore closable, and in the sequel we shall consider the closure of $J$ and denote it by the same letter. Notice that we have defined the Jacobi operator $J$ in such a way that \begin{equation} \label{eq:jm-0} \begin{pmatrix} q_1 & b_1 & 0 & 0 & \cdots \\[1mm] b_1 & q_2 & b_2 & 0 & \cdots \\[1mm] 0 & b_2 & q_3 & b_3 & \\ 0 & 0 & b_3 & q_4 & \ddots\\ \vdots & \vdots & & \ddots & \ddots \end{pmatrix} \end{equation} is the matrix representation of $J$ with respect to the canonical basis in $l_2(\mathbb{N})$ (we refer the reader to \cite{MR1255973} for a discussion on matrix representation of unbounded symmetric operators). It is known that the symmetric operator $J$ has deficiency indices $(1,1)$ or $(0,0)$ \cite[Chap. 4, Sec. 1.2]{MR0184042} and \cite[Corollary 2.9]{MR1627806}. In the case $(1,1)$ we can always define a linear set $D(g)\subset\dom (J^*)$ parameterized by $g\in\mathbb{R}\cup\{+\infty\}$ such that \begin{equation*} J^*\upharpoonright D(g)=:J(g) \end{equation*} is a self-adjoint extension of $J$. Moreover, for any self-adjoint extension (von Neumann extension) $\widetilde{J}$ of $J$, there exists a $\widetilde{g}\in\mathbb{R}\cup\{+\infty\}$ such that \begin{equation*} J(\widetilde{g})=\widetilde{J}\,, \end{equation*} \cite[Lemma 2.20]{MR1711536}. We shall show later (see the Appendix) that $g$ defines a boundary condition at infinity. To simplify the notation, even in the case of deficiency indices $(0,0)$, we shall use $J(g)$ to denote the operator $J=J^*$. Thus, throughout the paper $J(g)$ stands either for a self-adjoint extension of the non-selfadjoint operator $J$, uniquely determined by $g$, or for the self-adjoint operator $J$. In what follows we shall consider the inverse spectral problem for the self-adjoint operator $J(g)$. It turns out that if $J\ne J^*$ (the case of indices $(1,1)$), then for all $g\in\mathbb{R}\cup\{+\infty\}$ the Jacobi operator $J(g)$ has discrete spectrum with eigenvalues of multiplicity one, i.e., the spectrum consists of eigenvalues of multiplicity one that can accumulate only at $\pm \infty$, \cite[Lemma 2.19]{MR1711536}. Throughout this work we shall always require that the spectrum of $J(g)$, denoted $\sigma(J(g))$, be discrete, which is not an empty assumption only for the case $J(g)=J$. Notice that the discreteness of $\sigma(J(g))$ implies that $J(g)$ has to be unbounded. For the Jacobi operators $J(g)$ one can define boundary conditions at the origin in complete analogy to those of the half-line Sturm-Liouville operator (see the Appendix). Different boundary conditions at the origin define different self-adjoint operators $J_h(g)$, $h\in\mathbb{R}\cup\{+\infty\}$. $J_0(g)$ corresponds to the Dirichlet boundary condition, while the operator $J_\infty(g)$ has Neumann boundary condition. If $J(g)$ has discrete spectrum, the same is true for $J_h(g)$, $\forall h\in\mathbb{R}\cup\{+\infty\}$ (for the case of $h$ finite see Section 2 and for $h=\infty$, Section 4). In this work we prove that a Jacobi operator $J(g)$ with discrete spectrum is uniquely determined by $\sigma(J_{h_1}(g))$, $\sigma(J_{h_2}(g))$, with $h_1,h_2\in\mathbb{R}$ and $h_1 \ne h_2$, and either $h_1$ or $h_2$. If $h_1$, respectively, $h_2$ is given, the reconstruction method also gives $h_2$, respectively, $h_1$. Saying that $J(g)$ is determined means that we can recover the matrix (\ref{eq:jm-0}) and the boundary condition $g$ at infinity, in the case of deficiency indices $(1,1)$. We will also establish (the precise statement is in Theorem \ref{thm:suff-cond-two-gen}) that if two infinite real sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ that can accumulate only at $\pm \infty$ satisfy \begin{enumerate}[\ a)] \item $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ interlace, i.\,e., between two elements of a sequence there is one and only one element of the other. Thus, we assume below that $\lambda_k<\mu_k<\lambda_{k+1}$. \item The series $\sum_k(\mu_k-\lambda_k)$ converges, so \begin{equation*} 0<\sum_k(\mu_k-\lambda_k)=:\Delta < \infty\,. \end{equation*} \item The sequence of positive numbers $\{\tau_n\}_n$ defined by \begin{equation*} \tau_n^{-1}:= \frac{\mu_n-\lambda_n}{\Delta} \prod_{k\ne n} \frac{\mu_k-\lambda_n}{\lambda_k-\lambda_n}\,, \end{equation*} is such that \begin{equation*} \abs{\sum_k\frac{\lambda_k^m}{\tau_k}}<\infty\,, \qquad\text{for } m=0,1,2,\dots \end{equation*} \item The function $\rho(t):=\sum_{\lambda_k\le t}\tau_k^{-1}$ is such that the set of all polynomials is dense in $L_2(\mathbb{R},d\rho)$. \end{enumerate} Then, for any real number $h_1$, there exists a unique Jacobi operator $J$, a unique $h_2>h_1$, and if $J \ne J^\ast$, a unique $g\in\mathbb{R}\cup\{+\infty\}$, such that $\sigma(J_{h_2}(g))=\{\lambda_k\}_k$ and $\sigma(J_{h_1}(g))=\{\mu_k\}_k$. Moreover, we show that if the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are the spectra of a Jacobi operator $J(g)$ with two different boundary conditions $h_10\,. \end{equation*} Using the Neumann expansion for the resolvent (cf.\cite[Chap. 6, Sec. 6.1]{MR1711536}) \begin{equation*} (J_h(g)-\zeta I)^{-1}= -\sum_{k=0}^{N-1}\frac{(J_h(g))^k}{\zeta^{k+1}} +\frac{(J_h(g))^N}{\zeta^{N}} (J_h(g)-\zeta I)^{-1}\,, \end{equation*} where $\zeta\in\mathbb{C}\setminus\sigma(J(g))$, one can easily obtain the following asymptotic formula \begin{equation} \label{eq:m-asympt} m_h(\zeta,g)=-\frac{1}{\zeta}-\frac{q_1-h}{\zeta^2} -\frac{b_1^2+(q_1-h)^2}{\zeta^3} +O(\zeta^{-4})\,, \end{equation} as $\zeta\to\infty$ ($\im \zeta\ge \epsilon$, $\epsilon>0$). An important result in the theory of Jacobi operators is the fact that $m(\zeta,g)$ completely determines $J(g)$ (the same is of course true for the pair $m_h(\zeta,g)$ and $J_h(g)$). There are two ways for recovering the operator from the Weyl $m$-function. One way consists in obtaining first $\rho(t)$ from $m(\zeta,g)$ by means of the inverse Stieltjes transform (cf. \cite[Appendix B]{MR1711536}), namely, \begin{equation*} \rho(b)-\rho(a)=\lim_{\delta\downarrow 0}\lim_{\epsilon\downarrow 0} \frac{1}{\pi}\int_{a+\delta}^{b+\delta}\, \left(\im m(x+\I\epsilon,g)\right) dx\,. \end{equation*} The function $\rho$ is such that all the moments of the corresponding measure are finite \cite{MR0184042,MR1627806}. Hence, all the elements of the sequence $\{t^k\}_{k=0}^\infty$ are in $L_2(\mathbb{R},d\rho)$ and one can apply, in this Hilbert space, the Gram-Schmidt procedure of orthonormalization to the sequence $\{t^k\}_{k=0}^\infty$. One, thus, obtains a sequence of polynomials $\{P_k(t)\}_{k=0}^\infty$ normalized and orthogonal in $L_2(\mathbb{R},d\rho)$. These polynomials satisfy a three term recurrence equation \cite{MR1627806} \begin{align} \label{eq:favard-system1} tP_k(t) &= b_{k-1}P_{k-1}(t) + q_k P_k(t) + b_k P_{k+1}(t) \quad k \in \mathbb{N} \setminus \{1\}\\ \label{eq:favard-system2} tP_1(t) &= q_1 P_1(t) + b_1 P_2(t)\,, \end{align} where all the coefficients $b_k$ ($k\in\mathbb{N}$) turn out to be positive and $q_k$ ($k\in\mathbb{N}$) are real numbers. The system (\ref{eq:favard-system1}) and (\ref{eq:favard-system2}) defines a matrix which is the matrix representation of $J$. We shall refer to this procedure for recovering $J$ as the method of orthogonal polynomials. The other method for determining $J$ from $m(\zeta,g)$ was developed in \cite{MR1616422} (see also \cite{MR1643529}). It is based on the asymptotic behavior of $m(\zeta,g)$ and the Ricatti equation \cite{MR1643529}, \begin{equation} \label{eq:ricatti} b_n^2 m^{(n)}(\zeta,g)= q_n-\zeta-\frac{1}{m^{(n-1)}(\zeta,g)}\,,\quad n\in\mathbb{N}\,, \end{equation} where $m^{(n)}(\zeta,g)$ is the Weyl $m$-function of the Jacobi operator associated with the matrix (\ref{eq:jm-0}) with the first $n$ columns and $n$ rows removed. After obtaining the matrix representation of $J$, one can easily obtain the boundary condition at infinity which defines the domain of $J(g)$ in the non-selfadjoint case. Indeed, take an eigenvalue of $J(g)$, i.\,e., a pole of $m(\zeta,g)$. Since the corresponding eigenvector $f(\lambda)=\{f_k(\lambda)\}_{k=1}^\infty$ is in $\dom (J(g))$, it must be that \begin{equation*} \lim_{n\to\infty}W_n\bigl(v(g),f(\lambda)\bigr)=0\,. \end{equation*} This implies that either $\lim_{n\to\infty} W_n\bigl(\{Q_{k-1}\}_{k=1}^\infty,f(\lambda)\bigr)=0$, which means that $g=+\infty$, or \begin{equation*} g=-\frac {\lim_{n\to\infty}W_n\bigl(\{P_{k-1}(0)\}_{k=1}^\infty,f(\lambda)\bigr)} {\lim_{n\to\infty}W_n\bigl(\{Q_{k-1}(0)\}_{k=1}^\infty,f(\lambda)\bigr)} \,. \end{equation*} If the spectrum of $J_h(g)$ is discrete, say $\sigma(J_h(g))=\{\lambda_k\}_k$, the function $\rho(t)$ defined by (\ref{eq:spectral-measure}) can be written as follows \begin{equation*} \rho(t)=\sum_{\lambda_k\le t}\frac{1}{\alpha_k}\,, \end{equation*} where the coefficients $\{\alpha_k\}_k$ are called the normalizing constants and are given by \begin{equation} \label{eq:def-normalizing} \alpha_n=\sum_{k=0}^\infty\abs{P_k(\lambda_n)}^2\,. \end{equation} Thus, $\sqrt{\alpha_n}$ equals the $l_2$ norm of the eigenvector $f(\lambda_n)=\{P_k(\lambda_n)\}_{k=0}^\infty$ corresponding to $\lambda_n$, normalized in such a way that $f_1(\lambda_n)=1$. Clearly, \begin{equation} \label{eq:sum-normalizing-constants} 1=\langle e_1,e_1\rangle=\int_\mathbb{R}d\rho= \sum_{k}\frac{1}{\alpha_k}\,. \end{equation} The Weyl $m$-function in this case is given by \begin{equation} \label{eq:m-normalizing} m_h(\zeta,g)=\sum_{k}\frac{1}{\alpha_k(\lambda_k-\zeta)}\,. \end{equation} From this we have that \begin{equation*} (\lambda_n-\zeta)m_h(\zeta,g) = (\lambda_n-\zeta)\sum_k\frac{1}{\alpha_k(\lambda_k-\zeta)}= \sum_{k\ne n}\frac{\lambda_n-\zeta}{\alpha_k(\lambda_k-\zeta)} + \frac{1}{\alpha_n}\,. \end{equation*} Therefore, \begin{equation} \label{eq:normalizing-const-formula} \alpha_n^{-1}=\lim_{\zeta\to\lambda_n}(\lambda_n-\zeta)m(\zeta,g) =-\res_{\zeta=\lambda_n}m(\zeta,g)\,. \end{equation} Let us now introduce an appropriate way for enumerating sequences that we shall use. Consider a pair of infinite real sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ that have no finite accumulation points and that interlace, i.\,e., between two elements of one sequence there is one and only one element of the other. We use $M$, a subset of $\mathbb{Z}$ to be defined below, for enumerating the sequences as follows \begin{equation} \label{eq:enum-zeros-poles} \forall k\in M\quad \lambda_k<\mu_k<\lambda_{k+1}, \end{equation} where \begin{enumerate}[\ a)] \item If $\inf_k\{\lambda_k\}_k=-\infty$ and $\sup_k\{\lambda_k\}_k=\infty$, \begin{equation} \label{eq:unbounded-case} M:=\mathbb{Z}\quad\text{and we require}\quad \mu_{-1} < 0 < \lambda_1\,. \end{equation} \item If $0<\sup_k\{\lambda_k\}_k<\infty$, \begin{equation} \label{eq:bounded-above-positive} M:= \{k\}_{k=-\infty}^{k_{\max}}\,, \ (k_{\max}\ge 1)\quad\text{and we require}\quad \mu_{-1} < 0 < \lambda_1\,. \end{equation} \item If $\sup_k\{\lambda_k\}_k \leq 0$, \begin{equation} \label{eq:bounded-above-0} M:=\{k\}_{k=-\infty}^0\,. \end{equation} \item If $\inf_k\{\mu_k\}_k\geq 0$, \begin{equation} \label{eq:bounded-below-0} M:=\{k\}_{k=0}^\infty\,. \end{equation} \item If $-\infty<\inf_k\{\mu_k\}_k < 0$, \begin{equation} \label{eq:bounded-below-neg} M:=\{k\}_{k=k_{\min}}^{\infty}\,, \ (k_{\min}\le -1)\quad\text{and we require}\quad \mu_{-1} < 0 < \lambda_1\,. \end{equation} \end{enumerate} Notice that, by this convention for enumeration, the only elements of $\{\lambda_k\}_{k\in M}$ and $\{\mu_k\}_{k\in M}$ allowed to be zero are $\lambda_0$ or $\mu_0$. \section{Boundary conditions being rank one perturbations}\sss \label{sec:sep-conditions} In this section we consider a pair of operators $J_{h_1}(g)$ and $J_{h_2}(g)$, where $h_1,h_2\in \mathbb{R}$, that is, rank one perturbations of the Jacobi operator $J(g)$. \subsection{Recovering the matrix from two spectra} \label{sec:recov-matr-from-1} Let $g\in\mathbb{R}\cup\{+\infty\}$ be fixed. Since $J_h(g)$ is a rank one perturbation of $J(g)$, the domain of $J(g)$ coincides with the domain of $J_h(g)$ for all $h\in\mathbb{R}$. Moreover, as the perturbation is analytic in $h$, the multiplicity-one eigenvalues, $\lambda_k(h)$, and the corresponding eigenvectors, are analytic functions of $h$ \cite{ka}. \begin{lemma} \label{lem:halilova1} Let $\{\lambda_k(h)\}_k$ be the set of eigenvalues of $J_h(g)$ ($h\in\mathbb{R}$). For a fixed $k$ the following holds \begin{equation} \label{eq:lem-halilova} \frac{d}{dh}\lambda_k(h)=-\frac{1}{\alpha_k(h)}\,, \end{equation} where $\alpha_k(h)$ is the normalizing constant corresponding to $\lambda_k(h)$. \end{lemma} \begin{proof} For the sake of simplifying the formulae, we write $J_h$ and $\lambda(h)$ instead of $J_h(g)$ and $\lambda_k(h)$, respectively ($k$ is fixed). Let us denote by $f(h)$ the eigenvector of $J_h$ corresponding to $\lambda(h)$. Take any $\delta>0$, taking into account that $\dom (J_{h+\delta})=\dom (J_h)$ and that $J_h$ is symmetric for any $h\in\mathbb{R}$, we have that \begin{gather*} (\lambda(h+\delta)-\lambda(h)) \langle f(h+\delta),f(h)\rangle=\\ \langle J_{h+\delta}f(h+\delta),f(h)\rangle - \langle f(h+\delta),J_{h}f(h)\rangle=\\= \langle (J_{h+\delta}-J_{h}+J_{h})f(h+\delta),f(h)\rangle - \langle f(h+\delta),J_{h}f(h)\rangle =\\= \langle (J_{h+\delta}-J_{h})f(h+\delta),f(h)\rangle = -\delta\,. \end{gather*} Therefore, \begin{equation*} \lim_{\delta\to 0}\frac{\lambda(h+\delta)-\lambda(h)}{\delta}= -\lim_{\delta\to 0}\frac{1}{\langle f(h+\delta),f(h)\rangle}= -\frac{1}{\alpha_k(h)}\,. \end{equation*}\\ \end{proof} The cornerstone of our analysis below is again the Weyl $m$-function. Let us establish the relation between $m_h(\zeta,g)$ and $m(\zeta,g)$. Consider the second resolvent identity \cite{MR566954}: \begin{equation} \label{eq:second-resolvent-identity} (J_h(g)-\zeta I)^{-1}-(J(g)-\zeta I)^{-1}= (J(g)-\zeta I)^{-1}(J(g)-J_h(g))(J_h(g)-\zeta I)^{-1}\,, \end{equation} where $\zeta\in\mathbb{C}\setminus\{\sigma(J(g))\cup\sigma(J_h(g))\}$. Then, for $h\in\mathbb{R}$, \begin{equation*} \begin{split} m_h(\zeta,g)-m(\zeta,g)&= \langle\left((J_h(g)-\zeta I)^{-1}-(J(g)-\zeta I)^{-1}\right)e_1,e_1\rangle \\ &= \big\langle(J(g)-\zeta I)^{-1} (h\langle\cdot,e_1\rangle e_1) (J_h(g)-\zeta I)^{-1}e_1,e_1\big\rangle \\ &= \big\langle h\langle(J_h(g)-\zeta I)^{-1}e_1,e_1\rangle (J(g)-\zeta I)^{-1} e_1, e_1\big\rangle \\ &= h m_h(\zeta,g)m(\zeta,g)\,. \end{split} \end{equation*} Hence, \begin{equation} \label{eq:m-two-boundaries} m_h(\zeta,g)=\frac{m(\zeta,g)}{1-hm(\zeta,g)}\,. \end{equation} \begin{remark} \label{rem:zeros-poles-m-h} If $J(g)$ has discrete spectrum, then $m(\zeta,g)$ is meromorphic and, by (\ref{eq:m-two-boundaries}), so is $m_h(\zeta,g)$. The poles of $m_h(\zeta,g)$ are the eigenvalues of $J_h(g)$. Since the poles of the denominator and numerator in (\ref{eq:m-two-boundaries}) coincide, assuming that $h\ne 0$, the poles of $m_h(\zeta,g)$ are given by the zeros of $1-hm(\zeta,g)$ and the zeros of $m_h(\zeta,g)$ by the zeros of $m(\zeta,g)$. Thus, $J_{h_1}(g)$ and $J_{h_2}(g)$ have different eigenvalues, provided that $h_1\ne h_2$. \end{remark} \begin{theorem} \label{uniq.1} \label{th:recon-by-levin-gen} Consider the Jacobi operator $J(g)$ with discrete spectrum. The sequences $\{\mu_k\}_k=\sigma(J_{h_1}(g))$ and $\{\lambda_k\}_k=\sigma(J_{h_2}(g)), h_1 \ne h_2$, together with $h_1$ (respectively, $h_2$) uniquely determine the operator $J$, $h_2,$ (respectively, $h_1$) and, if $J\ne J^*$, the boundary condition $g$ at infinity. \end{theorem} \begin{proof} Without loss of generality we can assume that $h_1 < h_2$. Consider the Weyl $m$-function $m(\zeta,g)$ of the operator $J(g)$. Let us define the function \begin{equation} \label{eq:teschl-function} \mathfrak{m}(\zeta,g)=\frac{m_{h_2}(\zeta,g)}{m_{h_1}(\zeta,g)}\,, \qquad\zeta\in\mathbb{C}\setminus\mathbb{R}\,. \end{equation} Notice first that the zeros of $\mathfrak{m}(\zeta,g)$ are the eigenvalues of $J_{h_1}$ while the poles of $\mathfrak{m}(\zeta,g)$ are the eigenvalues of $J_{h_2}$. This follows from Remark~\ref{rem:zeros-poles-m-h} and (\ref{eq:teschl-function}). Let us now show that $\mathfrak{m}(\zeta,g)$ is a Herglotz or an anti-Herglotz function. Indeed, since $m(\zeta,g)$ is Herglotz, then \begin{equation} \label{eq:m-h-herglotz} \mathfrak{m}(\zeta,g)=\frac{1-h_1m}{1-h_2m}= 1+\frac{-1}{\frac{h_2}{h_2-h_1}+ \frac{-1}{(h_2-h_1)m(\zeta,g)}}\,. \end{equation} Therefore, $\mathfrak{m}(\zeta,g)$ is Herglotz or anti-Herglotz depending on the sign of $h_2-h_1$. Recall that if a function $f$ is Herglotz, then, $-\frac{1}{f}$ is also Herglotz. Since $h_2-h_1>0$, $\mathfrak{m}(\zeta,g)$ is a Herglotz function. Thus, the zeros $\{\mu_k\}_k$ of $ \mathfrak{m}(\zeta,g)$ and its poles $\{\lambda_k\}_k$ interlace. Let us use the convention (\ref{eq:enum-zeros-poles})--(\ref{eq:bounded-below-neg}) for enumerating the zeros and poles of $\mathfrak{m}(\zeta,g)$. By this convention, if the sequence $\{\lambda_k\}_k$ (or $\{\mu_k\}_k$) is bounded from below, the least of all zeros is greater than the least of all poles, while, if $\{\lambda_k\}_k$ is bounded from above, the greatest of all poles is less that the greatest of all zeros. It is easy to verify, using for instance (\ref{eq:lem-halilova}), that this is what we have for the zeros and poles of $\mathfrak{m}(\zeta,g)$ when $J(g)$ is semi-bounded. According to \cite[Chap. 7, Sec.1, Theorem 1]{MR589888}, the meromorphic Herglotz function $\mathfrak{m}(\zeta,g)$, with its zeros and poles enumerated as convened, can be written as follows \begin{equation} \label{eq:levin-herglotz-gen} \mathfrak{m}(\zeta,g)= C \frac{\zeta-\mu_0}{\zeta-\lambda_0} \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}\,,\qquad C>0\,, \end{equation} where the prime in the infinite product means that it does not include the factor $k=0$. From the asymptotic behavior of $m(\zeta,g)$, given by (\ref{eq:m-asympt}), one easily obtains that, as $\zeta\to\infty$ with $\im \zeta\ge\epsilon$ ($\epsilon>0$), \begin{equation} \label{eq:M-asympt} \mathfrak{m}(\zeta,g)=1+(h_1-h_2)\zeta^{-1} +(h_1-h_2)(q_1 -h_2)\zeta^{-2}+O(\zeta^{-3})\,. \end{equation} Therefore, \begin{equation*} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \mathfrak{m}(\zeta,g)=1\,. \end{equation*} Then, using (\ref{eq:levin-herglotz-gen}), we have \begin{equation} \label{eq:c-deter-2-spect-gen} C^{-1}=\lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1} \,,\qquad \epsilon>0\,. \end{equation} Thus, $\mathfrak{m}(\zeta,g)$ is completely determined by the spectra $\sigma(J_{h_1}(g))$ and $\sigma(J_{h_2}(g))$. Having found $\mathfrak{m}(\zeta,g)$, we can determine $h_2,$ respectively, $h_1$, by means of (\ref{eq:M-asympt}). Hence, from (\ref{eq:m-h-herglotz}) one obtains $m(\zeta,g)$ and, using the methods introduced in the preliminaries, $J$ is uniquely determined. In the case when $J\ne J^*$, we can also find the boundary condition $g$ at infinity as indicated in Section 2.\\ \end{proof} In \cite{MR1643529} (see also \cite{MR0190761}) it is proven that the discrete spectra of $J_{h_1}(g)$ and $J_{h_2}(g)$, together with $h_1$ and $h_2$ uniquely determine $J$ and the boundary condition $g$ in the $(1,1)$ case. Our result shows that it is not necessary to know both $h_1$ and $h_2$, one of them is enough. It turns out that if one knows the spectra $\sigma(J_{h_1}(g))$ and $\sigma(J_{h_2}(g))$ together with $q_1$, the first element of the matrix's main diagonal, it is possible to recover uniquely the matrix, the boundary conditions $h_1$, $h_2$ and the boundary condition at infinity, $g$, if any. Indeed, the term of order $\zeta^{-1}$ in the asymptotic expansion of $\mathfrak{m}(\zeta,g)$ (\ref{eq:M-asympt}) determines $h_1-h_2$. Since the coefficient of $\zeta^{-2}$ term is $(h_1-h_2)(q_1-h_2)$, if we know $q_1$ one finds $h_2$, and then $h_1$. \subsection{Necessary and Sufficient conditions } \label{sec:suff-cond-two-gen} \begin{theorem} \label{thm:suff-cond-two-gen} Given $h_1\in\mathbb{R}$ and two infinite sequences of real numbers $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ without finite points of accumulation, there is a unique real $ h_2 > h_1$, a unique operator $J(g)$, and if $J \ne J^\ast$ also a unique $g \in \mathbb R \cup \{+\infty\}$, such that, $\{\mu_k\}_k =\sigma(J_{h_1}(g))$ and $\{\lambda_k\}_k =\sigma(J_{h_2}(g))$ if and only if the following conditions are satisfied. \begin{enumerate}[\ a)] \item $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ interlace and, if $\{\lambda_k\}_k$ is bounded from below, $\min_k\{\mu_k\}_k>\min_k\{\lambda_k\}_k$, while if $\{\lambda_k\}_k$ is bounded from above, $\max_k\{\lambda_k\}_k<\max_k\{\mu_k\}_k$. So we use below the convention (\ref{eq:enum-zeros-poles})--(\ref{eq:bounded-below-neg}) for enumerating the sequences. \label{interlace-sufficient} \item The following series converges \begin{equation*} \sum_{k\in M}(\mu_k-\lambda_k)=\Delta<\infty\,. \end{equation*}\label{sum-spectr-sufficient} \item The sequence of numbers $\{\tau_n\}_{n\in M}$ defined by \begin{equation} \label{eq:tau-def-1} \tau_n^{-1}:= \frac{\mu_n-\lambda_n}{\Delta} \prod_{\substack{k\in M\\k\ne n}} \frac{\mu_k-\lambda_n}{\lambda_k-\lambda_n}\,, \quad \forall n\in M \end{equation} is such that \begin{equation*} \abs{\sum_{k\in M}\frac{\lambda_k^m}{\tau_k}}<\infty\,, \quad\text{for }m=0,1,2,\dots \end{equation*} \label{finite-moments-sufficient} \item The function $\rho(t):=\sum_{\lambda_k\le t}\tau_k^{-1}$ is such that the set of all polynomials is dense in $L_2(\mathbb{R},d\rho)$.\label{density-poly-sufficient} \end{enumerate} \end{theorem} \begin{proof} We first prove that if $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are the spectra of $J_{h_2}(g)$ and $J_{h_1}(g)$, with $ h_2 > h_1$, then \emph{\ref{interlace-sufficient}}), \emph{\ref{sum-spectr-sufficient}}), \emph{\ref{finite-moments-sufficient}}), and \emph{\ref{density-poly-sufficient}}) hold true. The condition \emph{\ref{interlace-sufficient}}) follows directly from the proof of the previous theorem. To prove that \emph{\ref{sum-spectr-sufficient}}) holds, observe that (\ref{eq:lem-halilova}) implies \begin{equation*} \mu_k-\lambda_k= \int_{h_1}^{h_2}\frac{dh}{\alpha_k(h)}\,. \end{equation*} Consider a sequence $\{M_n\}_{n=1}^\infty$ of subsets of $M$, such that $M_n\subset M_{n+1}$ and $\cup_nM_n=M$, then, using (\ref{eq:sum-normalizing-constants}), we have \begin{equation*} s_n:=\sum_{k\in M_n}(\mu_k-\lambda_k)= \sum_{k\in M_n}\int_{h_1}^{h_2}\frac{dh}{\alpha_k(h)}= \int_{h_1}^{h_2}\sum_{k\in M_n}\frac{dh}{\alpha_k(h)}\le h_2-h_1\,. \end{equation*} The sequence $\{s_n\}_{n=1}^\infty$ is then convergent and clearly \begin{equation*} \sum_{k\in M}(\mu_k-\lambda_k)=\lim_{n\to\infty}s_n=h_2-h_1\,. \end{equation*} Thus, $\Delta=h_2-h_1$. The convergence of the series in \emph{\ref{sum-spectr-sufficient}}) allows us to write (\ref{eq:c-deter-2-spect-gen}) as follows \begin{equation*} C^{-1}= \sideset{}{'}\prod_{k\in M}\frac{\lambda_k}{\mu_k} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \sideset{}{'}\prod_{k\in M} \frac{\mu_k-\zeta}{\lambda_k-\zeta} \,,\qquad \epsilon>0\,. \end{equation*} Now, using again \emph{\ref{sum-spectr-sufficient}}), it easily follows that for any $\epsilon>0$ \begin{equation*} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \sideset{}{'}\prod_{k\in M} \frac{\mu_k-\zeta}{\lambda_k-\zeta}= \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \sideset{}{'}\prod_{k\in M} \left(1+\frac{\mu_k-\lambda_k}{\lambda_k-\zeta}\right)=1\,. \end{equation*} Thus, $C=\sideset{}{'}\prod_{k\in M}\mu_k/\lambda_k$ and by (\ref{eq:levin-herglotz-gen}), \begin{equation} \label{eq:short} \mathfrak{m}(\zeta,g)=\prod_{k\in M} \frac{\mu_k-\zeta}{\lambda_k-\zeta}\,. \end{equation} Let us now find formulae for the normalizing constants in terms of the sets of eigenvalues for different boundary conditions. By (\ref{eq:normalizing-const-formula}), \begin{equation*} \alpha_n^{-1}(h_2,g)=\lim_{\zeta\to\lambda_n} (\lambda_n-\zeta)m_{h_2}(\zeta,g)\,. \end{equation*} Using the second resolvent identity, as we did to obtain (\ref{eq:m-two-boundaries}), we have that \begin{equation*} m_{h_1}(\zeta,g)= \frac{m_{h_2}(\zeta,g)}{1-(h_1-h_2)m_{h_2}(\zeta,g)}\,. \end{equation*} Therefore, \begin{equation} \label{eq:m-gen-m-h} \mathfrak{m}(\zeta,g)= \frac{m_{h_2}(\zeta,g)}{m_{h_1}(\zeta,g)} = 1-(h_1-h_2)m_{h_2}\,, \qquad\zeta\in\mathbb{C}\setminus\mathbb{R}\,. \end{equation} Then, the normalizing constants are given by \begin{equation*} \alpha_n^{-1}(h_2,g)=\lim_{\zeta\to\lambda_n} (\lambda_n-\zeta)\frac{\mathfrak{m}(\zeta,g) -1}{h_2-h_1}= \frac{1}{h_2-h_1}\lim_{\zeta\to\lambda_n} (\lambda_n-\zeta)\mathfrak{m}(\zeta,g)\,. \end{equation*} Now, \begin{equation} \label{eq:res-m-computation} \begin{split} \lim_{\zeta\to\lambda_n}(\lambda_n-\zeta)\mathfrak{m}(\zeta,g)&= \lim_{\zeta\to\lambda_n}(\lambda_n-\zeta) \prod_{k\in M} \frac{\mu_k-\zeta}{\lambda_k-\zeta}=\\ &=(\mu_n-\lambda_n) \prod_{\substack{k\in M\\k\ne n}} \frac{\mu_k-\lambda_n}{\lambda_k-\lambda_n}\,. \end{split} \end{equation} Hence, \begin{equation} \label{eq:n-normalizing-spect-gen} \alpha_n^{-1}(h_2,g)= \frac{\mu_n-\lambda_n}{h_2-h_1} \prod_{\substack{k\in M\\k\ne n}} \frac{\mu_k-\lambda_n}{\lambda_k-\lambda_n}\,. \end{equation} Notice that, since $\Delta=h_2-h_1$, it follows from (\ref{eq:n-normalizing-spect-gen}) that $\tau_n=\alpha_n$ for all $n\in M$. Thus, \emph{\ref{density-poly-sufficient}}) and \emph{\ref{finite-moments-sufficient}}) follow from the fact that the measure corresponding to the spectral function $\rho$ of the self-adjoint extension $J_{h_2}(g)$ is $N$-extremal and all its moments are finite \cite{MR0184042}, \cite[Proposition 4.15]{MR1627806}. We now prove that conditions \emph{\ref{interlace-sufficient}}), \emph{\ref{sum-spectr-sufficient}}), \emph{\ref{finite-moments-sufficient}}), and \emph{\ref{density-poly-sufficient}}) are sufficient. Let $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ be sequences as in \emph{\ref{interlace-sufficient}}) and \emph{\ref{sum-spectr-sufficient}}). Then, \begin{equation} \label{eq:prod-posit-gen} 0<\prod_{\substack{k\in M\\k\ne n}} \frac{\mu_k-\lambda_n}{\lambda_k-\lambda_n}<\infty\,. \end{equation} The convergence of this product allows us to define the sequence of numbers $\{\tau_n\}_{n\in M}$. Observe that for all $n\in M$, $\tau_n>0$. Indeed, $\Delta>0$ and (\ref{eq:enum-zeros-poles})--(\ref{eq:bounded-below-neg}) yield $\mu_n-\lambda_n>0$ for all $n\in M$. Thus, taking into account (\ref{eq:prod-posit-gen}), we obtain \begin{equation} \label{eq:normalizing-posit-gen} \tau_n>0\,,\qquad\forall\, n\in M\,. \end{equation} Let us now define the function \begin{equation} \label{eq:rho-thru-normalizing-def-gen} \rho(t)=\sum_{\lambda_k\le t}\frac{1}{\tau_k}\,,\quad t\in\mathbb{R}\,. \end{equation} Since (\ref{eq:normalizing-posit-gen}) holds, $\rho$ is a monotone non-decreasing function and has an infinite number of points of growth. Notice also that $\rho$ is right continuous. Now, we want to show that for the measure corresponding to $\rho$ all the moments are finite and \begin{equation} \label{eq:measure-normal-gen} \int_{\mathbb{R}}d\rho(t)=1\,. \end{equation} The fact that the moments are finite follows directly from condition \emph{\ref{finite-moments-sufficient}}). Indeed, \begin{equation*} \int_{\mathbb{R}}t^md\rho(t) = \sum_{k\in M}\frac{\lambda_k^m}{\tau_k}\,. \end{equation*} We show next that (\ref{eq:measure-normal-gen}) holds true. Given the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ satisfying \emph{\ref{interlace-sufficient}}) and \emph{\ref{sum-spectr-sufficient}}) we can define the function \begin{equation} \label{eq:m-tilde-def-gen} \widetilde{\mathfrak{m}}(\zeta)= \prod_{k\in M} \frac{\mu_k-\zeta}{\lambda_k-\zeta}\,. \end{equation} Taking into account (\ref{eq:tau-def-1}), one obtains that \begin{equation*} \res_{\zeta=\lambda_n} (\widetilde{\mathfrak{m}}(\zeta) -1) =-\frac{\Delta}{\tau_n}\,. \end{equation*} In view of \emph{\ref{sum-spectr-sufficient}}), we easily find that \begin{equation} \label{eq:m-1-to-0} \begin{split} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} (\widetilde{\mathfrak{m}}(\zeta) -1) &= \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \prod_{k\in M} \frac{\mu_k-\zeta}{\lambda_k-\zeta} -1 =\\&= \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \prod_{k\in M}\left(1+ \frac{\mu_k-\lambda_k}{\lambda_k-\zeta}\right) - 1=0\,. \end{split} \end{equation} Thus, on the basis of \v{C}ebotarev's theorem on the representation of meromorphic Herglotz functions \cite[Chap. VII, Sec.1 Theorem 2]{MR589888}, one obtains \begin{equation} \label{eq:m-thru-normalizing-gen} \widetilde{\mathfrak{m}}(\zeta) - 1 = \sum_{k\in M}\frac{\Delta}{(\lambda_k-\zeta)\tau_k}\,. \end{equation} We now define the function $\widetilde{m}(\zeta):= \frac{\widetilde{\mathfrak{m}}(\zeta)-1}{\Delta}$. Then, (\ref{eq:m-thru-normalizing-gen}) yields \begin{equation} \label{eq:m-normalizing-gen} \widetilde m(\zeta) = \sum_{k\in M}\frac{1}{\tau_k(\lambda_k-\zeta)}\,. \end{equation} We next show that \begin{equation*} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \zeta\widetilde m(\zeta) = -1\,. \end{equation*} Indeed, \begin{equation*} \begin{split} \frac{\widetilde{\mathfrak{m}}(\zeta)}{\Delta}&= \frac{1}{\Delta}\prod_{k\in M} \frac{\mu_k-\zeta}{\lambda_k-\zeta}=\\ &=\frac{1}{\Delta} \exp\left\{\sum_{k\in M}\ln\left( \frac{\mu_k-\zeta}{\lambda_k-\zeta}\right)\right\}=\\ &=\frac{1}{\Delta} \exp\left\{\sum_{k\in M}\ln\left( 1+ \frac{\mu_k-\lambda_k}{\lambda_k-\zeta}\right)\right\}=\\ &=\frac{1}{\Delta} \exp\left\{\sum_{k\in M}\sum_{p=1}^\infty(-1)^{p-1}\left( \frac{\mu_k-\lambda_k}{\lambda_k-\zeta}\right)^p\right\}\,. \end{split} \end{equation*} Thus, as $\zeta\to\infty$ with $\im\zeta\ge\epsilon$ ($\epsilon>0$), \begin{equation*} \frac{\widetilde{\mathfrak{m}}(\zeta)}{\Delta} =\frac{1}{\Delta}+ \frac{1}{\Delta} \sum_{k\in M}\frac{\mu_k-\lambda_k}{\lambda_k-\zeta} + O(\zeta^{-2})\,. \end{equation*} Then, \begin{equation*} \begin{split} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \zeta\widetilde m(\zeta)&= \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \zeta \frac{1}{\Delta} \sum_{k\in M}\frac{\mu_k-\lambda_k}{\lambda_k-\zeta}=\\&= -\frac{1}{\Delta}\sum_{k\in M}(\mu_k-\lambda_k)=-1\,. \end{split} \end{equation*} Also, from (\ref{eq:m-normalizing-gen}) one has \begin{equation*} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \zeta\widetilde m(\zeta) =-\sum_{k\in M}\frac{1}{\tau_k}\,. \end{equation*} Therefore, \begin{equation*} 1=\sum_{k\in M}\frac{1}{\tau_k}=\int_\mathbb{R}d\rho(t)\,. \end{equation*} Having found a function $\rho$ with infinitely many growing points and such that (\ref{eq:measure-normal-gen}) is satisfied and all the moments exist, one can obtain, applying the method of orthogonal polynomials (see Section 2), a tridiagonal semi-infinite matrix. Let us denote by $\widehat{J}$ the operator whose matrix representation is the obtained matrix. By what has been explained before, this operator is closed and symmetric. Now, if $\widehat{J}=\widehat{J}^*$, we know that $\rho(t)=\langle E(t)_{ \widehat{J}}\, e_1,e_1\rangle$, where $E_{\widehat{J}}(t)$ is the spectral decomposition of the self-adjoint Jacobi operator $\widehat{J}$. We define $h_2:=\Delta+h_1$ and $J:= \widehat{J}+ h_2 \langle\cdot,e_1\rangle\, e_1$. Then, obviously, $\widehat{J}= J_{h_2}$. If $\widehat{J}\ne \widehat{J}^*$, the Stieltjes transform of $\rho$ is the Weyl $m$-function $m_{h_2}(\zeta,g)$ of some self-adjoint extension of $\widehat{J}$ that we denote by $\widetilde{J}$. This is true only because the polynomials are dense in $L_2(\mathbb{R},d\rho)$. Indeed, the density of polynomials in $L_2(\mathbb{R},d\rho)$ means that the measure associated with $\rho$ is $N$-extremal \cite{MR0184042}. This in its turn implies that $m_{h_2}(\zeta,g)$ lies on the Weyl circle, and then, it is the Weyl $m$-function of some self-adjoint extension of $\widehat{J}$ \cite{MR0184042}, \cite[Proposition 4.15]{MR1627806}. We define, $J:= \widehat{J}+ h_2 \langle\cdot,e_1\rangle\, e_1$. Then, $ \widetilde{J}+ h_2 \langle\cdot,e_1\rangle\, e_1$ is a self-adjoint extension of $J$ and hence, $ \widetilde{J}+ h_2 \langle\cdot,e_1\rangle\, e_1= J(g)$ for some unique $g \in \mathbb R \cup \{\infty\}$. Furthermore, we obviously have that, $\widetilde{J}= J_{h_2}(g)$. We uniquely reconstruct $m(\zeta,g)$ from $m_{h_2}(\zeta,g)$ using (\ref{eq:m-two-boundaries}) and then, we uniquely reconstruct $g$ as explained in Section 2. Notice that we have \begin{equation*} m_{h_2}(\zeta,g)=\int_\mathbb{R}\frac{d\rho(t)}{t-\zeta} =\widetilde{m}(\zeta)\,. \end{equation*} It remains to show that $\sigma(J_{h_2}(g))=\{\lambda_k\}_k$ and $\sigma(J_{h_1}(g))=\{\mu_k\}_k$. To this end consider the function $\mathfrak{m}(\zeta,g)$ for the pair $J_{h_2}$ and $J_{h_1}$: \begin{equation*} \mathfrak{m}(\zeta,g)= \frac{m_{h_2}(\zeta,g)}{m_{h_1}(\zeta,g)}\,, \qquad\zeta\in\mathbb{C}\setminus\mathbb{R}\,. \end{equation*} Let the sequence $\{\gamma_k\}_k$ denote the spectrum of $J_{h_1}$. Then, arguing as in the proof of (\ref{eq:short}) we obtain that \begin{equation*} \mathfrak{m}(\zeta,g)=\prod_{k\in M} \frac{\gamma_k-\zeta}{\lambda_k-\zeta}\,. \end{equation*} Since we have already proven that \emph{\ref{interlace-sufficient}}) and \emph{\ref{sum-spectr-sufficient}}) are necessary conditions, we have that \begin{equation*} \sum_{k\in M}(\gamma_k-\lambda_k)=\Delta<\infty\,. \end{equation*} Then, as in the proof of (\ref{eq:m-1-to-0}), it follows that \begin{equation*} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} (\mathfrak{m}(\zeta) -1)=0\,. \end{equation*} Hence by \v{C}ebotarev's theorem \cite[Chap. VII, Sec.1 Theorem 2]{MR589888}, \begin{equation*} \mathfrak{m}(\zeta,g)=1+\sum_{k\in M} \frac{h_2-h_1}{(\lambda_k-\zeta)\alpha_k(h_2,g)}\,, \end{equation*} where we compute the residues of $\mathfrak{m}(\zeta)$ as in (\ref{eq:res-m-computation}). Thus, since $\alpha_k(h_2,g)=\tau_k$, $\forall k\in M$, \begin{equation*} \mathfrak{m}(\zeta,g)= 1+\sum_{k\in M} \frac{\Delta}{(\lambda_k-\zeta)\tau_k}= \widetilde{\mathfrak{m}}(\zeta,g)\,. \end{equation*} But $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are the poles and zeros of $\widetilde{\mathfrak{m}}(\zeta,g)$ and then, the eigenvalues of $J_{h_2}(g)$ and $J_{h_1}(g)$, respectively.\\ \end{proof} \begin{remark} \label{rem:1-1} We draw the reader's attention to the fact the matrix associated with the function $\rho$, constructed in the proof of the previous theorem, may have deficiency indices $(1,1)$ \cite{MR0184042,MR1627806,MR1711536}. If we drop the condition of the density of polynomials in $L_2(\mathbb{R},d\rho)$, then the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ correspond to the spectra of some generalized extensions of $J_{h_2}(g)$ and $J_{h_1}(g)$, respectively (see \cite{MR1627806}). The generalized extensions of symmetric operators, which are not von Neumann extensions, were first introduced by Naimark (see Appendix I in \cite{MR1255973} on Naimark's theory). \end{remark} In \cite{MR0221315} the case of Jacobi operators bounded from below is considered. A uniqueness result is proven, and some sufficient conditions for a pair of sequences to be the spectra of a Jacobi operator with different boundary conditions are given. \section{Dirichlet-Neumann conditions}\sss \label{sec:d-n-cond} \subsection{Recovering the matrix from two spectra} \label{sec:recov-matr-from} In this section we shall consider the pair of Jacobi operators $J_0(g)=J(g)$ and $J_\infty(g)$. Here, as before, we keep the convention of writing $J(g)$ even if $J=J^*$. The matrix representation of $J_\infty(g)$ corresponds to the matrix representation of $J(g)$ with the first column and row removed. From the Ricatti equation (\ref{eq:ricatti}), taking into account that $m^{(0)}(\zeta,g)=m(\zeta,g)$ and $m^{(1)}(\zeta,g)=m_\infty(\zeta,g)$, we have \begin{equation} \label{eq:relation-weyl-functions} m_\infty(\zeta,g)=-\frac{1}{b_1^2}\left((\zeta-q_1) + \frac{1}{m(\zeta,g)}\right) \end{equation} As before, we assume that the spectrum of $J=J(g)$ is discrete. If $m(\zeta,g)$ is a meromorphic function, then, by (\ref{eq:relation-weyl-functions}), $m_\infty(\zeta,g)$ is also meromorphic and the spectrum of $J_\infty(g)$ is discrete. The poles of $m(\zeta,g)$ are the eigenvalues of $J(g)$, while the zeros of $m(\zeta,g)$ are the eigenvalues of $J_\infty(g)$. Since $m(\zeta,g)$ is always a Herglotz function, under our assumption on the discreteness of $\sigma(J(g))$, $m(\zeta,g)$ is a meromorphic Herglotz function. This implies that $\sigma(J(g))$ and $\sigma(J_\infty(g))$ are interlaced, that is, between two successive eigenvalues of one operator there is exactly one eigenvalue of the other. Let the sequence $\{\lambda_k\}_k$ denote the eigenvalues of $J(g)$ (the poles of $m(\zeta,g)$). Furthermore, $\{\mu_k\}_k$ will stand for the eigenvalues of $J_\infty(g)$ (the zeros of $m(\zeta,g)$). It is worth remarking that, in contrast to the case of boundary conditions being rank one perturbations, here our convention for enumerating the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ does not work in the case when $J(g)$ is semi-bounded from above. Indeed, it follows from the mini-max principle \cite{rs} that if $J(g)$ is bounded from below the smallest of all poles is less than the smallest of all zeros of $m(\zeta,g)$, and if $J(g)$ is bounded from above, the min-max principle applied to $-J(g)$ implies that the greatest of all zeros is less than the greatest of all poles of $m(\zeta,g)$. So let us consider first the case when $J(g)$ is not semi-bounded or semi-bounded from below and enumerate the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ by (\ref{eq:enum-zeros-poles})--(\ref{eq:bounded-below-neg}). Then, by the same theorem we used to obtain (\ref{eq:levin-herglotz-gen}) \cite{MR589888}, $m(\zeta,g)$ can be written as follows \begin{equation} \label{eq:levin-herglotz} m(\zeta,g) = C \frac{\zeta-\mu_0}{\zeta-\lambda_0} \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}\,,\qquad C>0\,, \end{equation} where, as before, the prime in the infinite product means that it does not include the factor $k=0$. If $J(g)$ is bounded from above, then we are still able to use (\ref{eq:enum-zeros-poles})--(\ref{eq:bounded-above-0}) for enumerating the zeros and poles of the meromorphic Herglotz function $-\frac{1}{m(\zeta,g)}$. Thus, \begin{equation} \label{eq:levin-herglotz-inv} -\frac{1}{m(\zeta,g)}=\widetilde{C} \frac{\zeta-\lambda_0}{\zeta-\mu_0} \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\lambda_k}\right) \left(1-\frac{\zeta}{\mu_k}\right)^{-1}\,,\qquad \widetilde C>0\,. \end{equation} Notice that, since we have enumerated zeros and poles of $-\frac{1}{m(\zeta,g)}$ by our convention, we have now \begin{equation} \label{eq:enum-zeros-poles-alt} \forall k\in M\,,\quad\mu_k<\lambda_k<\mu_{k+1}\,, \end{equation} and \begin{enumerate}[\ a)] \item if $ 0 < \sup_k\{\mu_k\}_k< \infty$, \begin{equation} \label{eq:bounded-above-pos-alt} M:= \{k\}_{k=-\infty}^{k_{\max}}\,, \ (k_{\max}\ge 1)\quad\text{requiring}\quad \lambda_{-1} < 0 < \mu_1\,, \end{equation} \item if $\sup_k\{\mu_k\}_k\le 0$, \begin{equation} \label{eq:bounded-above-0-alt} M:=\{k\}_{k=-\infty}^0\,. \end{equation} \end{enumerate} Here again $\lambda_0$ or $\mu_0$ are the only ones allowed to be zero. Equations (\ref{eq:levin-herglotz}) and (\ref{eq:levin-herglotz-inv}) can be written in one formula \begin{equation} \label{eq:levin-herglotz-yes-not} m(\zeta,g) = K \frac{\zeta-\mu_0}{\zeta-\lambda_0} \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}\,, \end{equation} where, if $J(g)$ is not semi-bounded from above, $K=C$ and $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are enumerated by (\ref{eq:enum-zeros-poles}), (\ref{eq:unbounded-case}), (\ref{eq:bounded-below-0}), and (\ref{eq:bounded-below-neg}), while $K=-\widetilde{C}^{-1}$ and $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are enumerated by (\ref{eq:enum-zeros-poles-alt})--(\ref{eq:bounded-above-0-alt}) if $J(g)$ is semi-bounded from above. We give now, for the reader's convenience, a simple proof of a theorem that was proven by Fu and Hochstadt \cite{MR49:9676} for regular Jacobi operators (a regular Jacobi matrix is defined in \cite{MR49:9676}), and by Teschl \cite{MR1643529} in the general case. \begin{theorem}(Fu and Hochstadt, Teschl) \label{uniq.2} Consider the Jacobi operator $J(g)$ with discrete spectrum. The sequences $\{\lambda_k\}_k=\sigma(J(g))$ and $\{\mu_k\}_k=\sigma(J_\infty(g))$ uniquely determine the operator $J$ and, if $J\ne J^*$, the boundary condition, $g$, at infinity. \end{theorem} \begin{proof} From (\ref{eq:m-asympt}) we know that \begin{equation*} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \zeta m(\zeta,g)=-1\,,\qquad \epsilon>0\,. \end{equation*} Then, if $J(g)$ is not semi-bounded from above, (\ref{eq:levin-herglotz}) yields \begin{equation} \label{eq:c-deter-2-spect} C^{-1}=-\lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}}\zeta \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}\,, \qquad \epsilon>0\,, \end{equation} where $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are enumerated by (\ref{eq:enum-zeros-poles}), (\ref{eq:unbounded-case}), (\ref{eq:bounded-below-0}), and (\ref{eq:bounded-below-neg}). On the other hand, in the semi-bounded from above case (\ref{eq:levin-herglotz-inv}) implies \begin{equation} \label{eq:tilde-c-deter-2-spect} \widetilde{C}^{-1}= \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}} \frac{1}{\zeta} \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\lambda_k}\right) \left(1-\frac{\zeta}{\mu_k}\right)^{-1}\,, \qquad \epsilon>0\,. \end{equation} where $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are enumerated by (\ref{eq:enum-zeros-poles-alt})--(\ref{eq:bounded-above-0-alt}). Thus, in any case, one can find $K$, the constant in (\ref{eq:levin-herglotz-yes-not}), from the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$. Therefore, the spectra $\sigma(J(g))$ and $\sigma(J_\infty(g))$ uniquely determine $m(\zeta,g)$. Having found $m(\zeta,g)$ we can, using the methods introduced in Section \ref{sec:preliminaries}, determine $J$ and, in the case when $J\ne J^*$, also find uniquely the boundary condition at infinity, $g$.\\ \end{proof} \begin{remark} \label{rem:K-expression} It turns out that, by (\ref{eq:c-deter-2-spect}) and (\ref{eq:tilde-c-deter-2-spect}), $K$ can be written as \begin{equation} \label{eq:K-gen-expression} K^{-1}=-\lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}}\zeta \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}\,, \qquad \epsilon>0\,, \end{equation} where the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ have been enumerated by (\ref{eq:enum-zeros-poles}), (\ref{eq:unbounded-case}), (\ref{eq:bounded-below-0}), and (\ref{eq:bounded-below-neg}), when $J(g)$ is not semi-bounded from above and by (\ref{eq:enum-zeros-poles-alt})--(\ref{eq:bounded-above-0-alt}), otherwise. \end{remark} In what follows the Weyl $m$-function will be written through (\ref{eq:levin-herglotz-yes-not}) with $K$ given by (\ref{eq:K-gen-expression}). From (\ref{eq:levin-herglotz-yes-not}) one can obtain straightforward formulae for the normalizing constants (\ref{eq:def-normalizing}) in terms of the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$. Indeed, when $n\ne 0$ \begin{equation*} \begin{split} \lim_{\zeta\to\lambda_n}(\lambda_n-\zeta)m(\zeta,g)&= \lim_{\zeta\to\lambda_n}(\lambda_n-\zeta) K \frac{\zeta-\mu_0}{\zeta-\lambda_0} \sideset{}{'}\prod_{k\in M}\frac{ 1-\frac{\zeta}{\mu_k}} {1-\frac{\zeta}{\lambda_k}}=\\[3mm] &=K\frac{\lambda_n}{\mu_n}(\mu_n-\lambda_n) \frac{\lambda_n-\mu_0}{\lambda_n-\lambda_0} \sideset{}{'}\prod_{\substack{k\in M\\k\ne n}}\frac{ 1-\frac{\lambda_n}{\mu_k}} {1-\frac{\lambda_n}{\lambda_k}}\ . \end{split} \end{equation*} Formulae (\ref{eq:normalizing-const-formula}) and (\ref{eq:K-gen-expression}) then give \begin{equation} \label{eq:n-normalizing-spect} \alpha_n^{-1}=-\frac{ \frac{\lambda_n}{\mu_n}(\mu_n-\lambda_n)\frac{\lambda_n-\mu_0} {\lambda_n-\lambda_0} \sideset{}{'}\prod_{\substack{k\in M\\k\ne n}} \left(1-\frac{\lambda_n}{\mu_k}\right) \left(1-\frac{\lambda_n}{\lambda_k}\right)^{-1}} {\lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}}\zeta \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}}\,, \quad n\ne 0\,. \end{equation} Analogously, \begin{equation} \label{eq:0-normalizing-spect} \alpha_0^{-1}= -\frac{(\mu_0-\lambda_0) \sideset{}{'}\prod_{k\in M} \left(1-\frac{\lambda_0}{\mu_k}\right) \left(1-\frac{\lambda_0}{\lambda_k}\right)^{-1}} {\lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}}\zeta \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}} \,. \end{equation} \subsection{Necessary and Sufficient Conditions } \label{sec:suff-cond-two} The following result establishes necessary and sufficient conditions for two given sequences of real numbers to be the spectra of $J(g)$ and $J_\infty(g)$. \begin{theorem} \label{thm:dn-suff-cond-two-gen} Given two infinite sequences of real numbers $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ without finite points of accumulation, there is a unique operator $J(g)$, and if $J\ne J^\ast$ also a unique $ g \in \mathbb R \cup \{+\infty\}$, such that $\{\lambda_k\}_k=\sigma(J(g))$ and $\{\mu_k\}_k=\sigma(J_\infty(g))$ if and only if the following conditions are satisfied. \begin{enumerate}[\ a)] \item $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ interlace and, if $\{\lambda_k\}_k$ is bounded from below, $\min_k\{\mu_k\}_k>\min_k\{\lambda_k\}_k$, if $\{\lambda_k\}_k$ is bounded from above, $\max_k\{\lambda_k\}_k>\max_k\{\mu_k\}_k$. So we use below the convention (\ref{eq:enum-zeros-poles}), (\ref{eq:unbounded-case}), (\ref{eq:bounded-below-0}), and (\ref{eq:bounded-below-neg}) for enumerating the sequences when $J(g)$ is not semi-bounded from above and (\ref{eq:enum-zeros-poles-alt})--(\ref{eq:bounded-above-0-alt}) otherwise.\label{dn-interlace-suff} \item The limit \begin{equation} \label{eq:limit-over-imaginary} \lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}}\I\xi \sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1} \end{equation} is finite and negative when the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are not bounded from above, and it is finite and positive otherwise.\label{dn-limit-sufficient} \item Let $\{\tau_n\}_{n\in M}$ be defined by \begin{equation*} \tau_n^{-1}=-\frac{ \frac{\lambda_n}{\mu_n}(\mu_n-\lambda_n)\frac{\lambda_n-\mu_0} {\lambda_n-\lambda_0} \sideset{}{'}\prod_{\substack{k\in M\\k\ne n}} \left(1-\frac{\lambda_n}{\mu_k}\right) \left(1-\frac{\lambda_n}{\lambda_k}\right)^{-1}} {\lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}}\I\xi \sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1}}\,, \quad \forall n\in M\,,\ n\ne 0\,, \end{equation*} and \begin{equation*} \tau_0^{-1}= -\frac{(\mu_0-\lambda_0) \sideset{}{'}\prod_{k\in M} \left(1-\frac{\lambda_0}{\mu_k}\right) \left(1-\frac{\lambda_0}{\lambda_k}\right)^{-1}} {\lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}}\I\xi \sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1}}. \end{equation*} The sequence $\{\tau_n\}_{n\in M}$ is such that \begin{equation*} \abs{\sum_{k\in M}\frac{\lambda_k^m}{\tau_k}}<\infty\,, \quad\text{for } m=0,1,2\dots \end{equation*}\label{dn-finite-moments-sufficient} \item The function $\rho(t)=\tau_k^{-1}\delta(t-\lambda_k)$ is such that the set of all polynomials is dense in $L_2(\mathbb{R},d\rho)$.\label{dn-density-sufficient} \end{enumerate} \end{theorem} \begin{proof} We begin the proof by showing that the sequences $\sigma(J(g))=\{\lambda_k\}_k$ and $\sigma(J_h(g))=\{\mu_k\}_k$ satisfy \emph{\ref{dn-interlace-suff}}), \emph{\ref{dn-limit-sufficient}}), \emph{\ref{dn-finite-moments-sufficient}}), and \emph{\ref{dn-density-sufficient}}). Since the Weyl $m$ function is Herglotz, the eigenvalues of $J(g)$ and $J_\infty(g)$ interlace as indicated in \emph{\ref{dn-interlace-suff}}). To prove that \emph{\ref{dn-limit-sufficient}}) holds, consider first the case when $J(g)$ is not semi-bounded or only bounded from below, then (\ref{eq:c-deter-2-spect}) yields \emph{\ref{dn-limit-sufficient}}). If $J(g)$ is semi-bounded from above, (\ref{eq:tilde-c-deter-2-spect}) implies \emph{\ref{dn-limit-sufficient}}). On the basis of (\ref{eq:n-normalizing-spect}) and (\ref{eq:0-normalizing-spect}), $\forall n\in M$, $\tau_n$ coincides with the normalizing constant $\alpha_n$. Moreover \emph{\ref{dn-density-sufficient}}) and \emph{\ref{dn-finite-moments-sufficient}}) follow from the fact that the measure corresponding to $\rho$ is the spectral measure of the self-adjoint extension $J(g)$ and therefore, it is $N$-extremal (see \cite{MR0184042}) and all its moments are finite \cite{MR0184042,MR1627806}. Let us now suppose that we are given two real sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ that satisfy \emph{\ref{dn-interlace-suff}}). It can be shown that \begin{equation} \label{eq:prod-posit} 0<\sideset{}{'}\prod_{\substack{k\in M\\k\ne n}} \left(1-\frac{\lambda_n}{\mu_k}\right) \left(1-\frac{\lambda_n}{\lambda_k}\right)^{-1}<\infty\,. \end{equation} Indeed, the convergence of the infinite product follows from \emph{\ref{dn-interlace-suff}}) and is part of the Theorem 1 in \cite[Chap. 7, Sec.1]{MR589888} used to obtain (\ref{eq:levin-herglotz-gen}). We give here, for the reader's convenience, some details. The product in (\ref{eq:prod-posit}) converges if and only if \begin{equation*} \sideset{}'\sum_{\substack{k\in M\\k\ne n}}\left\{ \left(1-\frac{\lambda_n}{\mu_k}\right) \left(1-\frac{\lambda_n}{\lambda_k}\right)^{-1} -1\right\}=\lambda_n\sideset{}'\sum_{\substack{k\in M\\k\ne n}} \left(\frac{1}{\lambda_k}-\frac{1}{\mu_k}\right) \left(1-\frac{\lambda_n}{\lambda_k}\right)^{-1}<\infty\,, \end{equation*} where prime means that the summand $k=0$ is excluded. Thus, we have to prove that \begin{equation*} \sideset{}'\sum_{k\in M} \left(\frac{1}{\lambda_k}-\frac{1}{\mu_k}\right)<\infty\,. \end{equation*} It will suffice to consider that in \emph{\ref{dn-interlace-suff}}) the sequences are ordered by (\ref{eq:enum-zeros-poles}) with $M$ given by (\ref{eq:unbounded-case}). For any $k\in\mathbb{N}$, (\ref{eq:enum-zeros-poles}) implies \begin{equation*} 0< \left(\frac{1}{\lambda_k}-\frac{1}{\mu_k}\right)< \left(\frac{1}{\lambda_k}-\frac{1}{\lambda_{k+1}}\right)\,, \quad\forall k\in\mathbb{N}\,. \end{equation*} Clearly, $\sum_{k\in\mathbb{N}}\left(\frac{1}{\lambda_k} -\frac{1}{\lambda_{k+1}}\right)$ is convergent. Analogously, it can be proven that \begin{equation*} \sum_{k\in\mathbb{N}}\left(\frac{1}{\lambda_{-k}} -\frac{1}{\mu_{-k}}\right)<\infty\,. \end{equation*} Having established the convergence of the the product in (\ref{eq:prod-posit}), its positivity follows easily. We have, therefore, a sequence of real numbers $\{\tau_k\}_{k\in M}$ and let us now show that $\tau_n>0$, $\forall n\in M$. First notice that (\ref{eq:enum-zeros-poles}), (\ref{eq:unbounded-case}), (\ref{eq:bounded-below-0}), and (\ref{eq:bounded-below-neg}), yield \begin{equation*} \frac{\lambda_n}{\mu_n}(\mu_n-\lambda_n)\frac{\lambda_n-\mu_0} {\lambda_n-\lambda_0}>0\quad (n\ne 0)\quad\text{ and }\quad\mu_0-\lambda_0>0\,. \end{equation*} On the other hand (\ref{eq:enum-zeros-poles-alt})--(\ref{eq:bounded-above-0-alt}) imply \begin{equation*} \frac{\lambda_n}{\mu_n}(\mu_n-\lambda_n)\frac{\lambda_n-\mu_0} {\lambda_n-\lambda_0}<0\quad (n\ne 0) \quad\text{ and }\quad\mu_0-\lambda_0<0\,. \end{equation*} From these last inequalities, taking into account (\ref{eq:prod-posit}) and condition \emph{\ref{dn-limit-sufficient}}) we obtain \begin{equation} \label{eq:normalizing-posit} \tau_n>0\,,\qquad\forall\, n\in M\,. \end{equation} Let us now define the function \begin{equation} \label{eq:rho-thru-normalizing-def} \rho(t)=\sum_{\lambda_k\le t}\frac{1}{\tau_k}\,,\qquad\forall t\in\mathbb{R}\,. \end{equation} In view of (\ref{eq:normalizing-posit}), $\rho$ is a monotone non-decreasing function and has an infinite number of points of growth. Now, we want to show that, for the measure corresponding to $\rho$, all the moments are finite and \begin{equation} \label{eq:measure-normal} \int_{\mathbb{R}}d\rho(t)=1\,. \end{equation} The fact that the moments are finite follows directly from condition \emph{\ref{dn-finite-moments-sufficient}}). Indeed, \begin{equation*} \int_{\mathbb{R}}t^md\rho(t) = \sum_{k\in M}\frac{\lambda_k^m}{\tau_k}\,. \end{equation*} We show next that (\ref{eq:measure-normal}) is true. Given the sequences $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ satisfying \emph{\ref{dn-interlace-suff}}) and \emph{\ref{dn-limit-sufficient}}), we can define the function \begin{equation} \label{eq:m-tilde-def} \widetilde{m}(\zeta):= -\frac{\frac{\zeta-\mu_0}{\zeta-\lambda_0} \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}} {\lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}}\I\xi \sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1}}\,. \end{equation} Now, arguing as in the proof of (\ref{eq:n-normalizing-spect}) and (\ref{eq:0-normalizing-spect}), we obtain \begin{equation*} \res_{\zeta=\lambda_n}\widetilde m(\zeta)=-\tau_n^{-1}\,. \end{equation*} On the other hand, \begin{equation*} \lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}} \widetilde m(\I\xi) = -\lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}} \frac{\sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1}} {\I\xi \sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1}}=0\,. \end{equation*} Thus, using again \v{C}ebotarev's theorem \cite{MR589888} we find that \begin{equation} \label{eq:m-thru-normalizing} \widetilde m(\zeta) = \sum_{k\in M}\frac{1}{\tau_k(\lambda_k-\zeta)}\,. \end{equation} It follows from (\ref{eq:m-tilde-def}) that \begin{equation*} \lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}} \I\xi\widetilde m(\I\xi) = -\lim_{\substack{\xi\to\infty \\\xi\in\mathbb{R} }} \frac{\I\xi\sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1}} {\I\xi \sideset{}{'}\prod_{k\in M} \left(1-\frac{\I\xi}{\mu_k}\right) \left(1-\frac{\I\xi}{\lambda_k}\right)^{-1}}=-1\,. \end{equation*} Also from (\ref{eq:m-thru-normalizing}) one has \begin{equation*} \lim_{\substack{\xi\to\infty \\ \xi\in\mathbb{R}}} \I\xi\widetilde m(\I\xi) =-\sum_{k\in M}\frac{1}{\tau_k}\,. \end{equation*} Therefore, \begin{equation*} 1=\sum_{k\in M}\frac{1}{\tau_k}=\int_\mathbb{R}d\rho(t)\,. \end{equation*} We have found a function $\rho(t)$ with infinitely many growing points, such that all the moments exists for the corresponding measure and (\ref{eq:measure-normal}) holds. Therefore one can obtain, applying the method of orthogonal polynomials (see Section 2), a tridiagonal semi-infinite matrix. Let us denote by $J$ the operator whose matrix representation is the obtained matrix. As was mentioned before, $J$ is symmetric and closed. Now, if $J=J^*$, we know that $\rho(t)=\langle E_{J}(t)e_1,e_1\rangle$, where $E_{J}(t)$ is the spectral decomposition of the self-adjoint Jacobi operator $J$. If $J\ne J^*$, then the Stieltjes transform of $\rho(t)$ is the Weyl $m$-function $m(\zeta,g)$ of some self-adjoint extension of $J$ with boundary conditions at infinity given by $g$, that is, \begin{equation*} m(\zeta,g)=\int_\mathbb{R}\frac{d\rho(t)}{t-\zeta}\,. \end{equation*} This last assertion is true only because the polynomials are dense in $L_2(\mathbb{R},d\rho)$. Indeed, the density of polynomials in $L_2(\mathbb{R},d\rho)$ means that $\rho$ is $N$-extremal \cite{MR0184042}. This in its turn implies that $m(\zeta,g)$ lies on the Weyl circle, and then it is the Weyl $m$-function of some self-adjoint extension $J(g)$ \cite{MR0184042}, \cite{MR1627806}. It remains to show that $\sigma(J(g))=\{\lambda_k\}_k$ and $\sigma(J_\infty(g))=\{\mu_k\}_k$. So we start from (\ref{eq:rho-thru-normalizing-def}) and find the Weyl $m$-function of $J(g)$ using (\ref{eq:m-thru-normalizing}) \begin{equation*} m(\zeta,g)=\int_\mathbb{R}\frac{d\rho(t)}{t-\zeta}= \sum_{k\in M}\frac{1}{\tau_k(\lambda_k-\zeta)}= \widetilde m(\zeta)\,. \end{equation*} But $\{\lambda_k\}_k$ and $\{\mu_k\}_k$ are the poles and zeros of $\widetilde m$ and then the eigenvalues of $J(g)$ and $J_\infty(g)$, respectively.\\ \end{proof} For Jacobi operators semi-bounded from below, necessary and sufficient conditions are given in \cite{MR499269}. Note that Remark \ref{rem:1-1} can also be made here. It is worth mentioning that, from (\ref{eq:c-deter-2-spect}) and (\ref{eq:tilde-c-deter-2-spect}), it follows that, when \emph{\ref{dn-limit-sufficient}}) is seen as a necessary condition, one could write \begin{equation*} \lim_{\substack{\zeta\to\infty \\ \im \zeta\ge\epsilon}}\zeta \sideset{}{'}\prod_{k\in M} \left(1-\frac{\zeta}{\mu_k}\right) \left(1-\frac{\zeta}{\lambda_k}\right)^{-1}\,, \quad\epsilon>0\,, \end{equation*} instead of (\ref{eq:limit-over-imaginary}). \section*{Appendix: Boundary conditions for Jacobi operators}\sss \setcounter{section}{1} \renewcommand{\theequation}{\Alph{section}.\arabic{equation}} The difference expression $\gamma$ defined by (\ref{eq:recurrence-spectral}) and (\ref{eq:initial-spectral}) can be written together in one equation with the help of some conditions. Indeed, consider the difference expression $\tilde\gamma$ given by \begin{equation} \label{eq:general-difference} (\tilde\gamma f)_k = b_{k-1}f_{k-1} + q_k f_k + b_k f_{k+1}\,,\quad k \in \mathbb{N}\quad (b_0=1)\,. \end{equation} Clearly, $\gamma f$ is equal to $\tilde\gamma f$ provided that \begin{equation} \label{eq:Dir-bound-cond} f_0=0\,. \end{equation} This requirement can be considered as a boundary condition for the difference equation (\ref{eq:general-difference}). Notice that, although $f_0$ is not an element of the sequence $\{f_k\}_{k=1}^\infty$, it can be used to introduce boundary conditions for (\ref{eq:general-difference}) which turn out to be completely analogous to the boundary conditions at the origin for the Sturm-Liouville operator on the semi-axis. We shall refer to (\ref{eq:Dir-bound-cond}) as the Dirichlet boundary condition. Thus, $J$ is the closure of the operator which acts on sequences of $l_{fin}(\mathbb{N})$ by (\ref{eq:general-difference}) with the Dirichlet boundary condition (\ref{eq:Dir-bound-cond}). Suppose that the deficiency indices of $J$ are $(1,1)$ and consider now the following solution of (\ref{eq:recurrence-spectral}) \begin{equation*} \tilde{v}_k(\beta):=Q_{k-1}(0)\cos\beta + P_{k-1}(0)\sin\beta\,,\quad\beta\in[0,\pi)\,. \end{equation*} Let us define the set \begin{equation} \label{eq:set-extensions} \bigl\{f=\{f_k\}_{k=1}^\infty\in l_2(\mathbb{N}): \tilde\gamma f\in l_2(\mathbb{N}),\, \lim_{n\to\infty}W_n(\tilde{v}(g),f)=0\bigr\}\,. \end{equation} Notice that $D(g)$, defined by (\ref{eq:beta-extensions-domain}), coincides with (\ref{eq:set-extensions}) as long as $g=\cot \beta$. As pointed out in Section 2, the domain of every self-adjoint extension of $J$ is given by (\ref{eq:set-extensions}) for some $\beta$, and different $\beta$'s define different self-adjoint extensions \cite{MR1711536}. Let us denote these self-adjoint extensions by $J(g)$, as we did in Section 2, bearing in mind that $g=\cot \beta$. The condition \begin{equation} \label{eq:boundary-infinity} \lim_{n\to\infty} W_n(\tilde{v}(g),f)=0\,, \quad\ f\in\dom (J^*) \end{equation} determining the restriction of $J^*$ is considered to be a boundary condition at infinity. In analogy with the case of Sturm-Liouville operators one can define general boundary conditions at zero for the difference expression (\ref{eq:general-difference}). To this end, consider the operator $J(\alpha,g)$ defined by the difference expression (\ref{eq:general-difference}) with boundary condition at infinity (\ref{eq:boundary-infinity}) if necessary, and boundary condition ``at the origin'' \begin{equation} \label{eq:boundary-origin} f_1\cos\alpha + f_0\sin\alpha = 0\,,\qquad \alpha\in [0,\pi)\,. \end{equation} Thus, if $\alpha\in (0,\pi)$, \begin{equation*} J(\alpha,g)=J(g)-\cot\alpha\langle\cdot,e_1\rangle e_1\,. \end{equation*} Therefore $J(\alpha,g)=J_h(g)$, provided that $h=\cot\alpha$. When $\alpha=0$, from (\ref{eq:boundary-origin}), one has $f_1=0$ and (\ref{eq:general-difference}) is used to define the action of the operator for $k\ge 2$. $J(0,g)$ is said to be operator $J(g)$ with Neumann boundary condition. For this case we have that $J(0,g)$ is equal to $J_\infty(g)$. \\[5mm] \textbf{\large Acknowledgments}\hspace{3mm} We thank Rafael del Rio for a hint on the literature. \begin{thebibliography}{10} \bibitem{MR0184042} N.~I. 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