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\vglue 2truecm

\centerline {\titre SHARP DETERMINANTS AND KNEADING OPERATORS}
\bigskip \smallskip
\centerline {\titre FOR HOLOMORPHIC MAPS}
\vglue 1.5 truecm 
\centerline {\douze 
V. Baladi\footnote{$^1$}{{\huit Section de Math\'ematiques, Universit\'e de
Gen\`eve, CH-1211 Gen\`eve 24, Switzerland,\hfill\break 
$\phantom{XX}$
baladi@sc2a.unige.ch (on
leave from CNRS, UMR 128, ENS Lyon, France)
}},  A. Kitaev\footnote{${}^{2}$}{{\huit L.D. Landau Institute for
Theoretical Physics, Moscow, Russia, kitaev@itp.ac.ru}}, 
D. Ruelle\footnote{${}^{3}$}{{\huit I.H.E.S., 91440 Bures-sur-Yvette,
France}}, and  S. Semmes\footnote{${}^{4}$}{{\huit Rice University, Dept of
Mathematics, Houston, TX 77251, USA}} } 
\vglue 1 truecm 
\centerline{October 1995}
\vglue 1.5 truecm

In an earlier paper [1] two of us studied {\it generalized transfer
operators} $\M$ associated to maps of intervals of $\R$. In particular the
analyticity of a {\it sharp determinant} $\Det^\#(1-z \M)$ was analyzed on the
basis of spectral properties of $\M$ and by relating $\Det^\#(1-z \M)$ to the
determinant of $1 + \DD(z)$ for a suitable {\it kneading operator} $\DD(z)$.
\medskip 
It is a natural idea to try to replace monotone maps of intervals of $\R$ by
holomorphic diffeomorphisms of domains of $\C$ and to transpose the results of
[1] to this new situation. This program of transposition to the complex has
been carried through in part, and the present preprint shows our results. In a
suitable function-theoretical setting one can analyze the spectral properties
of the generalized transfer operator $\M$ and relate formally the sharp
determinant $\Det^\#(1-z\M)$ to the determinant of $1 + \DD(z)$ where $\DD(z)$
is a kneading operator. Unfortunately the trace of $\DD(z)$ might diverge, and
only regularized determinants like $\Det_3(1+ \DD(z))$ make sense a priori. The
present preprint leaves unfinished the task of either bounding $\Tr \DD(z)$ or
making suitable subtractions in the definition of $\Det^\#(1- z \M)$.
Nevertheless it has appeared useful to record the existing non trivial results
obtained, in view of their later utilisation.
\medskip
The present preprint is a result of extensive discussions between the authors.
It consists of three parts. The first part contains an identity between formal
power series. The second part is about spectral properties, and the third part
(traces and determinants) concerns bounds on $\Tr \DD(z)^2$ and higher traces.




\bigskip\bigskip

\noindent{\titre 1. AN IDENTITY BETWEEN POWER SERIES.}
\medskip

\noindent{\bf 1.1. Sharp trace and sharp determinants for transfer
operators and related operators.}

\medskip

Let $E$ be a compact subset of the complex plane which we assume
fixed throughout.
We shall consider the algebra
$\AA$ of {\it transfer operators}
$\MM$ defined  on the vector space 
of complex measures by
$$
\MM \Phi(x) = \sum_{\omega\in \Omega} 
g_\omega(x)\, \Phi(\psi_\omega(x))\, ,\eqno(1.1)
$$
where $\Omega$ is a finite set, 
each $g_\omega : \C \to \C$ is $C^\infty$
with nonempty compact support,
and each $\psi_\omega$ is a holomorphic diffeomorphism from
an open set $\Lambda_\omega \subseteq E$ onto its image
$\psi_\omega \Lambda_\omega \subset E$,
with ${\rm supp} \, g_\omega \subseteq \Lambda_\omega$.
(At the cost of additional
assumptions the smoothness requirement on
$g_\omega$ could be somewhat weakened
--- see also the last subsection of this section ---
and the setting
may be generalized to the case where $\Omega$ is countable,
or where $\sum_\omega$ is replaced by an integral.)
In later sections, we shall  restrict $\MM$
to   Banach subspaces of the space $\BB_{00}(E)$
of  measures 
with support in $E$. 
Note for the moment that $\MM$ in fact maps the space
of measures into $\BB_{00}(E)$.

\smallskip

We start by the important observation that the representation
(1.1) of an operator $\MM \in \AA$ is essentially unique
(because the $\psi_\omega$ are holomorphic).
To make this remark more precise, we say that
a representation 
$\MM\Phi = \sum_{i\in \II} \sum_{\omega \in \Omega_i} 
           g_{\omega} \, \Phi \circ \psi_{\omega}$
is finer than 
$\MM\Phi = \sum _{i\in \II} g_i \, \Phi \circ \psi_i$
if each $\psi_{\omega}$ is a restriction of the corresponding $\psi_i$ to some 
subset, and $\sum_{\omega \in \Omega_i} g_{\omega} = g_i$. Two representations
are called equivalent if there is a third one which is finer than both.
The precise claim is now that any two 
representations
of $\MM\in \AA$ are equivalent.
(To prove this, write the difference of the two representations
and then use the fact that it represents the zero operator on $\BB_{00}(E)$
in particular on $\Phi$
which are Dirac measures,
noting that whenever 
$\psi_\omega$ and $\psi'_{\omega'}$ do not coincide on
some open connected set,
the set of $x\in (\Lambda_\omega \cap  \Lambda'_{\omega'})$ 
with $\psi_\omega(x)=\psi'_{\omega'}(x)$ is finite,
and that when they do coincide a partition of unity may
be used to produce an associated refinement.)

\smallskip

We define the {\it sharp trace} of $\MM \in  \AA$ to be the
Cauchy principal value of the integral 
$$
\Tr^\# \MM =
\sum_{\omega\in \Omega}
\int_\C \bar \partial (g_\omega(x)) \,  
\sigma(\psi_\omega(x) -x) \, dx \, \, ,
\eqno (1.2)
$$
where $\bar \partial f(x) = 
1/2( {\partial \over \partial x_1}
f(x_1+ix_2) + i {\partial \over \partial x_2} f(x_1+ix_2))$
in the sense of distributions, 
the measure $dx$ is Lebesgue measure on the complex
plane, and
$$
\sigma(x) = \left \{ \matrix { 
{1\over \pi\,  x} & x\ne 0\hfill        \cr
 0 \hfill               &\hfill x=0 \, .\cr
}
\right .
$$
(We shall also use the notation 
$\partial f(x) =
1/2( {\partial \over \partial x_1}
f(x_1+ix_2) - i {\partial \over \partial x_2} f(x_1+ix_2))$.)
Observe that the terms in (1.1) with $\psi_\omega$
the identity do not contribute to the sharp trace.
Obviously, two equivalent representations of $\MM$ produce the
same value for the sharp trace, so that $\Tr^\#\MM$ does not depend
on the choice of a representaion by the above remarks.


Note that if  $\MM$ is such that for all $\omega$
$$
\partial \psi_\omega(x)\ne 1\quad { \rm whenever }\quad \psi_\omega(x)=x
\quad {\rm for } \quad x\in 
{\rm supp}\, g_\omega \, , \eqno (1.3)
$$
it would suffice to assume that the $g_\omega$ are $C^1$ in order
to define the sharp trace. 
(Warning: Property (1.3) is not preserved 
by taking powers $\MM^n$ of $\MM$.) See  the end of this section 
for an explicit
formula for the sharp trace as a sum
over fixed points of the $\psi_\omega$ when 
the ``simple fixed points property'' (1.3) holds.


\medskip
We now extend the domain of definition of the sharp trace.
For this, we introduce the linear operator
$$
S \Phi (x) = \int_\C \sigma(x-y)\,  \Phi(y)\, dy\, ,\eqno (1.4)
$$
which sends distributions with compact support
to distributions,
measures with compact support to  measures,
and $C^\infty$ functions with compact
support to $C^\infty$ functions.
Note for further use the property that
$\bar \partial S$ is the identity
map on compactly supported distributions, and in particular on measures
supported in $E$.
(To prove this,
use that  $ \bar \partial \sigma$ is the Dirac
mass at $0$, a proof of this well-known equality may
be found, e.g., in [3, p. 34].) A consequence of $\bar \partial S \Phi = \Phi$ is that
$S$ maps measures with compact support to  measures without atoms.
Observe that $\MM \in \AA$ maps any atomless measure to an
atomless measure in $\BB_{00}(E)$.

We write $\AA^S$ for the
algebra of operators which are linear combinations of
finite alternating products of transfer
operators  $\AA$ and  operators
$S$,  with at least one factor
$S$, and denote by
$\AA^S_L$, respectively
$\AA^S_R$ those elements of
$\AA^S$  such that the leftmost (respectively rightmost)
factor is different
from $S$.
Operators in $\AA^S_L$ act on $\BB_{00}(E)$,
and operators in $\AA^S_R$, or more generally $\AA^S$,
map $\BB_{00}(E)$ to 
measures. 
\bigskip \noindent
{\bf Lemma 1.1.} {\it For an operator $\KK$ in $\AA^S$  
there is a unique kernel $\KK_{xy}:\C \times \C\to \C$
such that:} 

\medskip
\itemitem{(1.5 a)} {\it $\KK \Phi(x) = \int_\C \KK_{xy}\,  \Phi(y) \, dy$
for all $\Phi \in \BB_{00}(E)$,}
\smallskip
\itemitem {(1.5 b)}
{\it $\KK_{xy}$ is a finite sum  of terms 
$h(x)\cdot\tilde  h(y)  \cdot \sigma(\psi x -\tilde \psi y)$
where $h$, $\tilde h$ are $C^\infty$  functions,  
$\psi$, $\tilde \psi$ are local holomorphic diffeomorphisms on
open sets $\Lambda$, respectively $\tilde \Lambda$,
containing the supports of $h$, respectively $\tilde h$,
and $h$, respectively $\tilde h$,
has compact support if $\KK \in \AA^S_L$, respectively $\AA^S_R $.}

\bigskip
\noindent
{\bf Proof of Lemma 1.1:} We consider a summand in  $\KK$ and prove the result
by induction on the number of factors in the summand,
starting the decomposition from the right.
The existence of a representation (1.5.a.b)
is clear for  $S$, and also
for $S \MM$ and $\MM S$ with $\MM \in \AA$, since
$$
\eqalign
{
 S \MM \Phi(x)&=
\int_\C \sigma(x-y)  \sum_{\omega\in \Omega} 
g_\omega(y)\, \Phi(\psi_\omega(y))\, dy\cr
&=\int_\C \sum_{\omega\in \Omega} \chi_{\psi_\omega(\Lambda_\omega)}
\cdot 
{\sigma(x- \psi_\omega^{-1} (z)) g_\omega(\psi_\omega^{-1}(z))
\, |\partial (\psi_\omega^{-1}) (z)|^2}
\, \Phi(z)\, dz\cr
\MM S\Phi(x) &= \sum_{\omega\in \Omega} 
g_\omega(x)\, \int_\C \sigma(\psi_\omega(x) -y)\, \Phi(y)\, dy \, .
}\eqno (1.6)
$$
The kernel representation property (1.5.a.b) is preserved 
when multiplying on the left by $\MM\in \AA$:
$$
\MM \KK \Phi(x) =  \int_\C \, 
\sum_{\omega\in \Omega}  g_\omega(x)\, \KK_{\psi_\omega(x) y} \Phi(y)\, dy \,
.\eqno (1.7) $$
Multiplication to the left by $S$ is slightly more delicate.
By Fubini, we have
$$
S \KK  \Phi (x) = \int_\C \int_\C \sigma(x-y) \, \KK_{yz} \,  \Phi(z) \, dz\, dy=
\int_\C \int_\C \sigma(x-y) \, \KK_{yz} \, dy\,  \Phi(z)  \, dz
\, . \eqno(1.8)
$$
Let us now study the kernel $\int_\C \sigma(x-y) \, \KK_{yz} \, dy$
of  $S \KK$
defined by  (1.8), using the induction
assumption (1.5.a.b)
on $\KK_{xy}$.
First note   that, since
$\bar \partial \sigma$ is the Dirac measure at zero,
we have 
for any holomorphic diffeomorphism $\Psi$:
$$
\eqalign
{
\bar \partial (\sigma \circ \Psi) (x) &= 
\delta_0 \circ \psi \cdot \overline {\partial \Psi} (x)\cr
&=\delta_{\Psi^{-1}(0)}
\,  | \partial (\Psi^{-1})(0)|^2\, 
\overline {\partial  \Psi} (\Psi^{-1}(0)) \cr
&=
{\delta_{\Psi^{-1}(0)}  \over  \partial \Psi(\Psi^{-1}(0)) }\, . \cr
}\eqno(1.9)
$$
Therefore, for any $C^\infty$ functions $h$, $\tilde h$, 
with $h$ compactly supported 
(recall that no composition $S^2$ is  allowed),
any local holomorphic diffeomorphism $\psi:\Lambda \to \C$, and any
points $u\in \psi(\Lambda)$, $x \in \Lambda$, with
$\psi (x)\ne u$, we get, since $h=\bar \partial S h$
$$
\eqalign
{
\int_\C h(y) \, \sigma(x-y) \, \sigma(\psi(y)&-u) \, dy
=\cr
&(Sh)(x)\, \sigma(\psi(x)-u)
- {(Sh)(\psi^{-1}(u)) \, \sigma(x -\psi^{-1}(u))
\over \partial \psi (\psi^{-1}(u)) }\, .
}\eqno(1.10)
$$
To check that (1.10) makes sense, we observe 
that since $S h$ is holomorphic outside of the support of
$h$, and since $Sh(z)$ goes to zero
as $|z| \to \infty$  by construction, $Sh$ must have
support contained in the support of $h$ (and in particular in $\Lambda$).
Formula (1.10)  for $u=\tilde \psi(z)$
defines a summand of the new kernel $(S\KK)_{xz}$
except on a set of points $(x,z)$
of zero two-dimensional complex Lebesgue measure:  
for fixed $z$ ($\psi$, $\tilde \psi$),
there are at most
finitely many  $x$ such that $\psi (x) =\tilde \psi (z)$
(and similarly, at most finitely many such $z$ for each $x$,
$\psi$, $\tilde \psi$).
Note also that the factor $\tilde h (z)$ implicit
in both sides of (1.10) has compact support if $\KK \in \AA^S_R$. 

We {\it define} the
left-hand side of (1.10) to be zero when $\psi(x)=u$,
in other words we extend formula (1.10) to such points $x$, $u$.
(Note that the Cauchy principal value of
the left-hand-side of (1.10) does {\it not} always vanish
when $\psi(x)=u$. An easy counterexample can be obtained
with $\psi$ the identity map, assuming that $S \, \partial h$
does not vanish at $x$.)
Our  definition is legitimate
in that it is consistent with the action of
$S \KK$ on $\BB_{00}(E)$ (because the ambiguity
only concerns a set of two-dimensional complex
Lebesgue measure zero). 

(Although we shall not need this,
we note that when applying (1.10) inductively at an intermediate
$S$ factor, the choice for the case $\psi(x)=u$ we just made
is unconsequential for the kernel since the
concerned set  is of zero measure and will be washed
out by next integration.)

By induction we thus prove the existence of a representation
(1.5.a.b) for $\KK \in \AA^S$.
If $\KK$, $\KK' \in \AA^S$
and $\KK \Phi =\KK'\Phi$ for all $\Phi \in \BB_{00}(E)$
then the  kernels
$\KK_{xy}$ and $\KK'_{xy}$ in (1.5) 
must coincide as functions on $\C \times \C$.
(Using $\Phi$ the Dirac mass at an arbitrary point $y_0$ one sees that
the difference of the densities of the image measures
$\PP_{y_0}(x)=\KK_{xy_0}-\KK'_{x y_0}$ vanishes for Lebesgue almost all $x$.
Since $\PP_{y_0}$ is $C^\infty$ except at finitely many points where it
vanishes, it must vanish identically.)\ \cube
\medskip
If $\KK \in \AA^S_L + \AA^S_R$ has kernel
$\KK_{xy}$ as in (1.5.a.b)
then  the function $\KK_{xx}$ 
is a finite sum of terms $\hat h(x)  \sigma(\Psi (x) )$
where $\hat h(x)$ is $C^\infty$, with compact support, 
and  $\Psi(x)$ is holomorphic (and therefore vanishes
to finite order if not identically).
In particular the Cauchy principal value
of $\int_\C \KK_{xx} \, dx$ is well defined for $\KK \in \AA^S_L 
+ \AA^S_R$.
Therefore, we may define the {\it sharp trace} of $\KK \in \AA^S_L+\AA^S_R$
to be this Cauchy principal value:
$$
\Tr^\# \KK=\int_\C \KK_{xx} \, dx \, .\eqno (1.11)
$$ 

Before we proceed with
our extension of the domain of the
definition of the sharp trace,
we note that if $\MM + \KK$, with
$\MM \in \AA$ and $\KK \in \AA^S_L+\AA^S_R$,
vanishes as an operator mapping $\BB_{00}(E)$ to measures,
then $\MM$, and therefore also $\KK$, must vanish.
(To show this,  consider the Dirac measure $\delta_{x_0}$
at $x_0$ in $\C$.
If $\MM \ne 0$, then the measure
$\MM \delta_{x_0}$ has atoms for $x_0$ well-chosen 
in function of a given representation (1.1). However,
$\KK \delta_{x_0}$ is an atomless measure 
for any $x_0$ by
above considerations.)
We now extend the definition
linearly to $\AA'=\AA\oplus (\AA^S_L+\AA^S_R)$, using (1.2).
By the uniqueness
statement in Lemma 1.1,
we have that $\Tr^\#\KK$ only depends on $\KK \in \AA'$ as an operator
mapping $\BB_{00}(E)$ to measures.

Although $\AA'$ is not an algebra,
both $\AA'_L=\AA \oplus \AA^S_L$ and $\AA'_R=\AA \oplus \AA^S_R$ are algebras.
In particular, if $\KK \in \AA'_L$ (or $\AA'_R$), then
$\KK^m \in \AA'_L$ (respectively $\AA'_R$) for all $m \ge 1$.


\medskip

 
We note the following ``almost-trace properties''
of $\Tr^\#$:
\bigskip \noindent
{\bf Lemma 1.2.}\smallskip

\item
{(1)} {\it Let $\KK$ belong to $\AA^S$.
Then $\Tr^\#  \KK \MM = \Tr^\# \MM \KK $ for any $\MM\in \AA$.}
\smallskip
\item {(2)} {\it Let $\KK$ belong to
$\AA^S_L\cap\AA^S_R$.
Then 
$
\Tr^\#  S \KK  = \Tr^\# \KK S  
$.}

\bigskip


\noindent In Lemma 1.6, we shall see that that 
$\Tr^\# \MM_1 \MM_2=\Tr^\# \MM_2 \MM_1$
for  $\MM_1$, $\MM_2 \in \AA$.

\medskip
\noindent {\bf Proof of Lemma 1.2:} For the first claim, we use (1.6--1.7) 
(and the analogue of (1.7) for $\KK\MM$)
and write
$$
\eqalign
{
\Tr^\# \KK\MM
&= \int_\C \sum_{\omega\in \Omega}  \chi_{\psi_\omega(\Lambda_\omega)}
{\KK_{x \psi_\omega^{-1} x}\,  g_\omega(\psi_\omega^{-1}(x))
\, |\partial (\psi_\omega^{-1}) (x)|^2}\, dx \cr
&=
\int_\C 
\sum_{\omega\in \Omega} \KK_{\psi_\omega( y) y} \,
g_\omega(y) \, dy= \Tr^\# \MM \KK \, .
}
$$
In the above equalities, we used the fact that the standard change
of variables formula holds for Cauchy principal values,
whenever the change of variables is holomorphic.
Since antiholomorphic maps are harmonic,
this can be proved by induction on $m \ge 2$
as follows ($g$ is a compactly supported $C^\infty$ function,
and $\psi^{-1}$ is a holomorphic diffeomorphism on the support
of $g$):
$$
\eqalign
{
(m-1) \int_\C {1\over y^{m}}\, g(y) \, dy &=
\int_\C {1\over  y^{m-1}} \, \partial g(y) \, dy
\cr
&= 
\int_\C {1\over  (\psi x)^{m-1}} \, \partial g(\psi x) 
\partial \psi(x) \, \bar \partial \bar \psi (x) \, dx\cr
&= 
\int_\C {1 \over  (\psi x)^{m-1}} 
\biggl [ \partial g(\psi x) \,  \partial \psi(x) \, \bar \partial \bar \psi (x) 
+ g(\psi x) \, \partial \bar \partial \bar \psi (x) \biggr ] \, dx\cr
&=
(m-1) \int_\C {\partial \psi (x) \over (\psi x)^m} 
\, g(\psi x) \bar \partial \bar \psi (x)\,  dx \, .\cr
}
$$


For the second claim, use (1.8) 
(and the equivalent formula for $\KK S$) to get 
$$
\Tr^\#  S \KK  =\int_\C \int_\C  \sigma(x-y) \, \KK_{yx} \,dy\, dx \qquad
\Tr^\# \KK S=\int_\C \int_\C \KK_{xy} \, \sigma(y-x) \, dy\, dx\, ,
$$
where it is implicit (from the definition
given after (1.10))
that we have suppressed from $\KK_{xy}$ 
all terms of
the form $h(x)\tilde h(y) \sigma(\psi x -\tilde\psi y)$
with $\psi$ and $\tilde \psi$ identical on some open set, taking
finer presentations if necessary
(note that the suppressed terms are the same for both traces).
To finish, apply Fubini. \cube
\medskip

We can obviously
extend additively the sharp trace
from the vector space $\AA'$ to the vector space
$\AA''=\AA''_L + \AA''_R$
with $\AA''_M=\AA[[z]]\oplus \AA^S_M[[z]]$ 
where $\AA''_M[[z]]$ is the algebra
of formal power series with coefficients in $\AA^S_M$
for $M=L,R$. We use the same notation for this extension
$$
\Tr^\# : \AA'' \to \C [[z]] \, .
$$


We  define
the {\it sharp determinant} of $\KK(z)\in z\AA''_L$  or $z\AA''_R$
by the following formal power series:
$$
\Det^\# (1-\KK(z)) =
\exp -\sum_{m=1}^\infty {1 \over m}
\Tr^\# (\KK(z))^m \in 1+ \C [[z]] \, . \eqno (1.12)
$$


\bigskip
\noindent{\bf 1.2. The kneading operator $\DD(z)$ and the main identity.}
\medskip

We associate to $\MM\in  \AA$ an operator 
$\NN=\NN(\MM) \in  \AA$ defined by
$$
\NN \Phi (x) = \sum_{\omega \in \Omega}   \bar \partial g_\omega(x) \,
\Phi(\psi_\omega (x)) \, ,\eqno (1.13)
$$
which sends measures to measures supported in $E$.
(Note that $\NN(\MM)$ only depends on $\MM$ as an operator
on $\BB_{00}(E)$.)

Finally, we  define the {\it formal kneading operator}
$\DD=\DD(z)\in \AA^S_L[[z]]$ associated to $\MM\in  \AA$ by
$$
\DD(z)  = z \, \NN \, (1-z \MM)^{-1} \, S 
 = \sum_{k=1}^\infty z^{k} \, \NN \, \MM^{k-1} \,  S\, . 
\eqno (1.14)
$$


{\it The point is that for 
$|z|^{-1}$ larger than the
spectral radius of $\MM$ on
$\BB_{00}(E)$, we may view $\DD(z)$ as
acting on  $\BB_{00}(E)$,   
and we shall see in Section 3, restricting
to a Banach subspace of $\BB_{00}(E)$,
that the operator $\DD^2(z)$ is almost trace-class
so that one should be able to relate
its sharp determinant to a classical regularized determinant.}
We hope that this informal comment shows the importance of
the main result of this section:
\bigskip \noindent
{\bf Proposition 1.3.} {\it For any $\MM$ in $\AA$, 
and $\DD(z)$ defined by} (1.13--1.14),
{\it we have the following identity between formal power series}
$$
\Det^\# (1+ \DD(z)) 
={1 \over \Det^\# (1-z\MM)}
\, . \eqno (1.15)
$$
\bigskip
One way to show Proposition 1.3 would be to adapt
the proof of the formally identical result 
on transfer operators acting on functions
of bounded variation on  $\R$ in 
[1, Proposition 3.1] (using  the fact that $1/(\pi x)$ is the complex analogue
of the function $(1/2)\sgn \, (x)$ used there).
We shall give here a more streamlined proof.
We first need a lemma and its corollary:


\bigskip \noindent {\bf Lemma 1.4.} {\it For any $\MM$ in $ \AA$ and the associated
operator $\NN(\MM)\in\AA$, we have}
 $$
\Tr^\# \MM = \Tr^\# \NN S
=\Tr^\# S \NN \, .
$$
\bigskip 
\noindent
{\bf Proof of Lemma 1.4:} The second equality is a consequence of the first
claim of Lemma 1.2, so that
it suffices to check the first one.
By definition
$$
\eqalign
{
\NN S \Phi(x)&=
\sum_{\omega\in \Omega} \int_\C    \bar \partial g_\omega(x) \,
\sigma(\psi_\omega x-y ) \, \Phi(y) \, dy \, .\cr
}
$$
Therefore
$$
\Tr^\# \NN S
= \sum_{\omega\in \Omega}
\int_\C \bar \partial g_\omega(x) \, \sigma(\psi_\omega x-x )\, dx =
\Tr^\# \MM\, .\ \cube
$$
\bigskip \noindent
{\bf Corollary 1.5.} {\it For any $\MM \in \AA$ and the associated operator
$\NN(\MM)\in \AA$, the operator $ \widetilde \MM=\MM-S \NN \in \AA^S_R$ has the
property that $\Tr^\#  (\widetilde \MM)^\ell = 0$ for all integer $\ell \ge 1$.}
\bigskip
\noindent
{\bf Proof of Corollary 1.5:} We only need to check that
$$
\widetilde {\MM^\ell}
\widetilde \MM  = \widetilde {\MM ^{\ell+1}}\, .\eqno (1.16)
$$
in the sense of operators from $\BB_{00}(E)$ to  measures,
for all $\ell \ge 1$.
(Indeed, using the 
representation (1.5)
for $(\widetilde \MM)^\ell\in \AA^S_R$ and
$\widetilde {\MM^\ell}\in \AA^S_R$ , it follows that
$\Tr^\# (\widetilde \MM)^\ell = \Tr^\#
\widetilde {\MM^\ell}$ which vanishes by Lemma 1.4
applied to $\MM^\ell$.)
We write $\NN_\ell$ for the transfer operator associated
to $\MM^\ell$, so that $\widetilde{ \MM^\ell}=\MM^\ell -S \NN_\ell$. 
We also associate to $\MM\in \AA$ an
operator $\MM_{0,1}\in \AA$ defined by
$$
\MM_{0,1} \Phi (x) = \sum_{\omega \in \Omega}
g_\omega(x) \, \overline {\partial \psi_\omega} (x)\,  \Phi( \psi_\omega(x)) 
\,  , \eqno(1.17)
$$
noting that for $\ell \ge 1$ the chain rule implies
$(\MM^\ell  )_{0,1}= ( \MM_{0,1})^\ell$ (we
simply write $\MM^\ell_{0,1}$).
Since each $\psi_\omega$ is
holomorphic we have by the
Leibniz rule, and because
$\bar \partial S=S\bar \partial$ is the identity on 
compactly supported distributions, for all $\ell \ge 1$:
$$
\widetilde {\MM^\ell}=\MM^\ell -S\NN_\ell
=S\MM_{0,1}^\ell \bar \partial \, , \eqno(1.18)
$$
in the sense of operators from $\BB_{00}(E)$ to
measures.
Using again that $\bar \partial S$ is the identity
on distributions with compact support, we thus get
$$
\eqalign
{
\widetilde {\MM^\ell} \widetilde \MM &=
S\MM_{0,1}^\ell \bar \partial S\MM_{0,1} \bar \partial
=S\MM_{0,1}^\ell  \MM_{0,1} \bar \partial
= \widetilde {\MM ^{\ell+1}}\, .\ \cube
}
$$
\bigskip \noindent
{\bf Proof of Proposition 1.3:} Let $M_i=L$ or $R$, $i=1, 2,3$.
We first note that whenever $\KK_1$ and $\KK_2$ are two elements
of $\AA''_{M_1}\cup \{S\}$ with 
$\KK_1\KK_2 \in \AA''_{M_2}$, $\KK_2\KK_1 \in \AA''_{M_3}$,
and
$\Tr^\# (\KK_1 \KK_2 )^m= \Tr^\# (\KK_2 \KK_1)^m$ 
for all $m \ge 1$ then
$$
\Det^\# (1+\KK_1\KK_2) =\Det^\#(1+ \KK_2\KK_1)\, . \eqno (1.19) 
$$
If, additionally,
$\KK(\vec \jmath)=\KK_1^{j_m} \KK_2^{j_{m-1}} \cdots \KK_2^{j_0}$
is in $\AA''_L$ or $\AA''_R$ for all integer $m \ge 1$,
and $j_\ell \ge 0$ for $0\le \ell \le m$,
with $\sum_{0\le \ell\le m} j_{\ell} \ge 1$,
and if $\Tr^\# \KK(\vec \jmath)$ is invariant under
circular permutations of $\vec \jmath$, then
$$
\Det^\# (1+\KK_1) \, \Det^\#(1+\KK_2) = 
\Det^\# ((1+\KK_1)\, (1+\KK_2))  \, . \eqno (1.20) 
$$
(See, e.g., [5, Appendix A] for a proof.)
We now have by  Lemma 1.2, (1.19--1.20)
and Corollary 1.5:
$$
\eqalign
{
\Det^\#(1+\DD(z))  \, \Det^\#(1-z\MM)&=
\Det^\#(1+ z \NN(1-z\MM)^{-1} S)\, \Det^\#(1-z\MM)\cr
&=\Det^\#(1+ z S\NN(1-z\MM)^{-1} )\, \Det^\#(1-z\MM)\cr
&=\Det^\#(1 -z\MM + z S\NN) \cr
&=\Det^\#(1-z \widetilde \MM) =1 \, .\  \cube
}
$$
\smallskip

As a consequence of our computations, we are able to prove:
\bigskip \noindent
{\bf Lemma 1.6.} {\it For any $\MM$, $\PP$
in $\AA$
we have $\Tr^\# (\MM\PP)= \Tr^\# (\PP \MM)$.}
\bigskip \noindent
{\bf Proof of Lemma 1.6:} We shall use Lemma 1.4, which tells us that
$$
\Tr^\#(\MM\PP)=\Tr^\#S\NN(\MM\PP) \quad {\rm and }\quad
\Tr^\#(\PP\MM)=\Tr^\#S\NN(\PP\MM)\, ,
$$
with $\NN(\QQ)$ as in (1.13).
By definition and the Leibniz rule we have
$$
\NN(\MM\PP) =\NN(\MM) \PP + \MM_{0,1} \NN(\PP) \, ,\eqno (1.21)
$$
with $\MM_{0,1}$ defined in (1.17).
Now, using once more that $\bar \partial S$ is the identity
on $\BB_{00}(E)$ to apply (1.18), we get the following equality
between operators from $\BB_{00}(E)$ to measures:
$$
\eqalign
{
S \MM_{0,1} \NN(\PP) &=
S\MM_{0,1} \bar \partial S \NN(\PP)\cr
&=(\MM -S\NN(\MM)) \, S \NN (\PP)\cr
&= \MM S \NN(\PP) -S\NN(\MM) S \NN(\PP) \, .
}\eqno (1.22)
$$
Putting (1.21) and (1.22) together, and applying
Lemma 1.2  (1) yields
$$
\eqalign
{
\Tr^\#S\NN(\MM\PP)&=
\Tr^\# S\NN(\MM) \PP+\Tr^\# \MM S \NN(\PP) -\Tr^\#S\NN(\MM) S \NN(\PP) \cr
&=\Tr^\# \PP S\NN(\MM) +\Tr^\# S\NN(\PP)\MM  -\Tr^\#S\NN(\PP) S \NN(\MM) \cr
&=\Tr^\#S\NN(\PP\MM)\, ,
}
$$
as desired.
\ \cube
\bigskip
\noindent{\bf 1.3. The kneading operators $\DD_{(r)}$.}
\medskip
We introduce yet another linear operator sending compactly supported distributions
(or measures)  to distributions (measures),
$$
S_{(r)} \Phi (x) = 
\int_\C  \sigma_r(x-y)\,  \Phi(y)\, dy\, ,
$$
for integer $r \ge 0$, where we 
set $\sigma_0(x)=\sigma(x)$, $\sigma_1(x)=\log(|x|)/\pi$ for $x \ne 0$,
and $\sigma_1(0)=0$,
and  generally 
$\sigma_r$ for $r\ge 2$ a solution of $\partial \sigma_r =\sigma_{r-1}$
which we choose to be continuous
and zero at the origin 
(for example, $\sigma_2(x) = x(\log |x|-1)$). 
By definition $\partial^r \sigma_r=\sigma$.
Note also that $\sigma_0 \chi_E$ is in $L^p$
for all $p< 2$, $\sigma_1 \chi_E$ is in $L^q$ for all
$q < \infty$, where $\chi_E$ denotes the characteristic function
of $E$, and $\sigma_r$ is continuous for all $r \ge 2$.
Introducing the notation
$$
\bar S^r \Phi = \bar \partial S_{(r)} \Phi\, ,\eqno (1.23)
$$
for $\Phi \in \BB_{00}(E)$, and integer
$r \ge 0$ it is not difficult
to check that $\partial  ^r \bar S^r=\bar S^r \partial^r$
is the identity on $\BB_{00}(E)$.

For $\MM \in \AA$
we now define the associated operators $\DD_{(r)}(z) \in 
z \partial ^{r} \AA[[z]]   S_{(r)}$ 
for integers $r \ge 0$,
following [6], by
$$
\DD_{(r)}(z) = \partial^{r} \DD(z) \bar S^r =
z \partial^{r}\, \NN \, (1-z\MM)^{-1} \,  S_{(r)} \, .\eqno (1.24)
$$
(Recall that $\NN=\NN(\MM)$ was defined in (1.13).)
Clearly, $\DD_{(0)}(z)=\DD(z)$. 

By definition of $S_{(r)}$,
if $z^{-1}$ is outside of the spectrum
of $\MM$ acting on the space $\BB_{r1}$ of distributions
$\Phi$ and
such that $\bar \partial \partial^r \Phi$ is a measure
(use that $\sigma_r \in \BB_{r1}$),
the operator $\DD_{(r)}(z)$ has a chance to be well-behaved
when acting on a suitable Banach subspace
of $\BB_{00}(E)$. 
The main point (see Lemma 1.7 below)
is that the sharp determinant is preserved by the modifications
(1.24), where we first need to extend the sharp
determinant to $\partial ^{r} \AA[[z]]  S_{(r)}$,
i.e., to define the sharp trace of
powers of operators in $\partial ^{r} \AA[[z]]  S_{(r)}$. 
To do this, we first
write $\KK=\partial ^{r} \MM  S_{(r)}$
(for $\MM\in \AA$) in kernel form:
$$
\eqalign
{
\KK\Phi(x)&=
\partial ^{r} \MM  S_{(r)} \, \Phi(x)\cr
&=\partial^r_x \biggl ( \sum_\omega g_\omega(x)
\int_\C  \sigma_r(\psi_\omega x-y)\,  \Phi (y)\, dy \biggr )\cr
&=
\int_\C \sum_\omega 
\sum_{s=0}^r {r \choose s} \bigl ( \partial^s_x  g_\omega \bigr )(x) \, 
\partial^{r-s}_x \bigl ( \sigma_r(\psi_\omega x-y) \bigr )
\,  \Phi (y)\, dy  \cr
&=\int \KK_{xy} \Phi(y) \, dy \, ,
}\eqno (1.25)
$$
with $\KK_{xy}$ a finite sum  of terms  
$h(x)  \cdot \sigma_t(\psi x - y)$,
for $0\le t \le r$,
where $h$ is a    $C^\infty$  function with compact
support, and
$\psi$ is a local holomorphic diffeomorphism on
an open set $\Lambda$,
containing the supports of $h$.
When constructing inductively
a kernel $\LL_{xz}$ for $\LL$  a finite product of operators of the
type (1.25), we are confronted with integrals
of the type
$$
\int_\C h(y) \sigma_r (u-y) \, \sigma_s ( \psi y-v)
\, dy\eqno (1.26)
$$
where 
$h$ is a    $C^\infty$  function with compact
support, and
$\psi$ a local holomorphic diffeomorphism on
an open set $\Lambda$
containing the support of $h$,
and the integers $r$ and $s$ are both nonnegative.
If $r=s=0$, we {\it define}
the new kernel by the same rule as in (1.10), thus
collapsing the two $\sigma_0$ in (1.26) and
producing two terms with just one $\sigma_0$ factor.
If $r\cdot s=0$ but $r+s > 0$, 
since $\bar S^t h$ is a well defined
$C^\infty$ function supported in $E$ for $t\ge 0$
(just conjugate the arguments following (1.10)) we may
$\partial_y$-integrate by parts in (1.26)
(using that $\partial^t \bar S^t$
is the identity) and thus choose on which side (left or right) we want
the $\sigma_0$ factor to be. Applying these remarks,
we end up for $\LL_{xz}$ with a sum of terms of two types: either
a single 
$$
h_0(x) \,  h_1(z) \, \sigma_0(\psi_1(x)-\tilde \psi_1(z))\eqno (1.27)
$$
or an  integral
$$
\eqalign
{
h_0(x) \, h_{j+1}(z)\,
\int_{\C\times \cdots \times \C}&
h_1(y_1)\,  \sigma_{s_1}  ( \psi_1 x- \tilde \psi_1 y_1) \,  
h_2(y_2) \, \sigma_{s_2}  ( \psi_2 y_1-\tilde \psi_2 y_2)\cr
&\cdots  
\, \sigma_{s_j}  (\psi_j y_{j-1} -\tilde \psi_j z) 
\, dy_1 \cdots dy_{j-1} 
}
\eqno (1.28)
$$
where $j \ge 2$, and all the $s_i$ except perhaps the second
one $s_2$
are strictly positive
(the $h_i$ are $C^\infty$
with compact support, and the 
$\psi_i$, $\tilde \psi_i$ local holomorphic diffeomorphisms
on open sets as usual). 
Because of the properties of the $\sigma_i$,  the Cauchy principal value 
of $\int \LL_{xx}\, dx$
is thus well defined and we may set $\Tr^\# \LL$ to be this value.
The result can  now be stated:  

\bigskip \noindent
{\bf Lemma 1.7.} {\it For $\MM \in \AA$ and
$\DD_{(r)}(z)$ defined by (1.24),
we have for all integers $r \ge 0$ 
$\Det^\#(1+\DD_{(r)}(z)) = \Det^\#(1+\DD(z))$.}
\bigskip
\noindent
{\bf Proof of Lemma 1.7:}
By  definition of the sharp determinant, 
we must check that
for any $\PP  \in \AA^S_R\cap \AA^S_L$ and integer $r \ge 1$
$$
\Tr^\# \PP S = \Tr^\# \partial ^{r} \PP S \bar S^r \, .\eqno (1.29)
$$
Clearly, it suffices to verify that for all
operators $\KK=\partial^s \PP S \bar S^{s-1}$ with $\PP\in \AA^S_R\cap
\AA^S_L$
and integer $s \ge 1$ we have
$$
\Tr^\# \KK \bar S =\Tr^\# \bar S \KK   .\eqno (1.30)
$$
(Both sides of (1.30) are well defined
since 
$\KK\bar S=\partial^{s} \PP S \bar S^{s}$
and $\bar S \KK =\partial^{s-1} \PP S \bar S^{s-1}$.)
We first observe that since
$\bar  S\partial= \partial \bar S$ is the identity on $\BB_{00}(E)$, 
we have
$$
\bar S \Phi(x) = \int_{\C} \bar \sigma(x-y) \, \Phi(y) \, dy \, ,
$$
with $\bar \sigma(x) = \overline{\sigma(x)}$
having the property that $\partial \bar \sigma$
is the Dirac mass at $0$.
By (1.25--1.28)
we know that the operator $\KK$ appearing in (1.30)
acting on $\BB_{00}(E)$
can be written  in kernel form
with $\KK_{xy}$ a finite sum  of terms
$$
h(x) \cdot \tilde h(y) \cdot \sigma_{-1}(\psi x - \tilde \psi y)\eqno (1.31)
$$
where $\sigma_{-1} = \partial \sigma$
(apply $\partial_x$ to terms (1.27))
and terms
$$
\int_{\C\times\cdots\times \C} 
h_1(y_1) \, \sigma_{s_1-1} (\psi_1 x-\tilde \psi_1y_1)  
\cdots  \sigma_{s_j} (\psi_j y_{j-1} -\tilde \psi_j z) 
\, dy_1 \cdots dy_{j-1}\eqno (1.32)
$$
with $s_1-1, s_2 \ge 0$ and the other $s_i > 0$
(apply $\partial_x$ to (1.28)).
If (1.32) contains two $\sigma_0$ factors we proceed
as described above to define its value.
Using this kernel $\KK_{xy}$, and (1.30)
we write
$$
\KK \bar S \Phi(x)=\int_\C\int_\C
\KK_{xy} \bar \sigma(y-z) \, \Phi(z) \, dz \, dy
\quad
\bar S \KK \Phi(x)=\int_\C\int_\C
\bar \sigma(x-y) \KK_{yz} \, \Phi(z) \, dz \, dy\, .\eqno (1.33)
$$
We claim that we can use the expression
$\int \KK_{xy} \, \bar \sigma(y-z) \, dy$ from (1.33)
for the kernel of $\KK\bar S$ without problems
(and similarly for $\bar S\KK$). 
The only summands of $\KK_{xy}$ which require some care
are those of the form (1.31), for which we
may legitimately use (since $\partial \bar \sigma$ is the
Dirac mass at the origin)
$$
\eqalign
{
\int \tilde h(y) &\, \sigma_{-1} (\psi x -\tilde \psi y) \, 
\bar \sigma(y-z) \, dy
=\cr
&={\tilde h (z)  \over \tilde \psi'(z)} \, \sigma_0(\psi x -\tilde\psi z) 
+\int 
{\partial \tilde h(y)  \tilde \psi'(y) -\tilde h(y) \tilde \psi''(y)
\over (\tilde \psi'(y))^2}
\, \sigma_0( \psi x -\tilde \psi y)\,  \bar \sigma(y-z) \, dy \, .
}
$$
(A similar computation exists for $\bar S \KK$.)
Finally, 
we just apply Fubini and get:
$$
\Tr^\#\KK \bar S =
\int_{\C} 
\int_\C \KK_{xy} \bar \sigma(y-x) \, dy \, dx
=
\int _\C 
\int_\C \bar \sigma(x-y) \KK_{yx}\, dy \, dx = \Tr^\# \bar S \KK \, ,
$$
as required.
\ \cube
\bigskip \noindent
{\bf 1.4. The half-adjoint, a functional equation, and the kneading
operator $ \widehat \DD(z)$.}
\medskip
We associate with $\MM$ 
another operator in $\AA$ 
(the {\it half-adjoint} of $\MM$) defined by
$$
\widehat \MM \Phi(x) = \sum_{\omega \in \Omega}{g_\omega( \psi_\omega^{-1}x) 
\over \partial \psi_\omega(\psi_\omega^{-1} (x))}\, 
\Phi(\psi^{-1}_\omega x) \, .\eqno (1.34)
$$
(We assume that ${\rm supp}\, g_\omega \subset \psi_\omega \Lambda_\omega$.)
Note that $\widehat \MM$ is independent of the
representation of $\MM$ of the form (1.1).
Clearly $\widehat{\phantom{\MM}}$
is an involution. Since each $\psi_\omega$ is a local holomorphic
diffeomorphism,
$$
\eqalign
{
\Tr^\# \widehat \MM &=
\int_\C \bar \partial \left (
{g_\omega  \over \partial \psi_\omega} \circ \psi_\omega^{-1} \right )
\, \sigma (\psi_\omega^{-1} (x)-x) \, dx\cr
&=\int_\C \bar \partial \left (
{g_\omega  \over \partial \psi_\omega} \right ) (\psi_\omega^{-1} (x)) 
\, \overline {\partial \psi_\omega^{-1}} (x) \,
\, \sigma (\psi_\omega^{-1} (x)-x) \, dx\cr
&=\int_\C
{\bar \partial  g_\omega  \over \partial \psi_\omega}  (\psi_\omega^{-1} (x))
\,\overline {\partial \psi_\omega^{-1}} (x) \,
\, \sigma (\psi_\omega^{-1} (x)-x) \, dx\cr
&=\int_\C
{\bar \partial  g_\omega  \over \partial \psi_\omega}  (y) \,
\overline {\partial \psi_\omega^{-1}} (\psi_\omega (y)) \,
|\partial \psi_\omega (y)|^2
\, \sigma (y-\psi_\omega y) \, dy\cr
&=-\Tr^\# \MM \, .
}\eqno (1.35)
$$
Since $\widehat{\MM_1 \MM_2} = \widehat \MM_2 \widehat \MM_1$
we have $\Tr^\# (\widehat \MM)^n =
-\Tr^\# \MM^n$ for all integers $n\ge 1$. 
We therefore obtain the functional equation
$$
\Det^\# (1-z\MM) \, \Det^\# (1-z\widehat \MM)=1 \, .\eqno (1.36)
$$
Replacing $\MM$
by $\widehat \MM$ and $\NN( \MM)$
by $\NN(\widehat \MM)$ in (1.9)
we may define an operator $\widehat \DD(z)\in \AA''$.
Using the functional
equation (1.36)
and the above properties
of $\widehat{\phantom{\MM}}$,
we may now write a more complete version of Proposition 1.3:
$$
\Det^\# (1+ \DD(z)) = \Det^\# (1-z\widehat \MM) 
={1 \over \Det^\# (1-z\MM)}
={1 \over \Det^\# (1+ \widehat \DD(z))}
\, .  \eqno (1.37)
$$
We also define operators $\widehat \DD_{(r)}(z)$ for
integers $r\ge 0$
by $z \partial^r \NN(\widehat \MM) (1-z\widehat \MM)^{-1} S_{(r)}$
and note as in Lemma 1.7 that 
$\Det^\#(1+\widehat \DD_{(r)}(z))=\Det^\#(1+\widehat\DD(z))$ for all $r$.

\eject \noindent
{\bf 1.5. Simple fixed points: a formula for the sharp trace
of a transfer operator.}

\medskip

We assume for a moment that condition (1.3)
holds for $\MM$, but relax the smoothness
assumption on the $g_\omega$, requiring only that
they are $C^1$ with support
in $E$. With these assumptions, each
$\bar \partial g_\omega (x) \, \sigma(\psi_\omega(x)-x)$ is
in fact integrable.
Recalling (1.9), and observing that there are only finitely
many points $x\in \C$ with $\psi_\omega(x)=x$
(because  (1.3) excludes the case
where $\psi_\omega$ is the identity map), we get the explicit formula:
$$
\Tr^\# \MM = -
\sum_{\omega\in \Omega}
\sum_{x : \psi_\omega(x) =x} 
{g_\omega(x) \over   \partial \psi_\omega(x) - 1}  \, .\eqno (1.38)
$$



\smallskip

When (1.3) does not
hold, i.e. when 
$\partial \psi_\omega (x_0)=1$ for some $x_0=\psi_\omega(x_0)$
in the support of $g_\omega$, then either $\psi_\omega$
is the identity,
or the tangency is of finite order, 
i.e., $\partial^k \psi_\omega(x_0) \ne 0$
for some $k \ge 2$.
In the second case,
one may thus use $\partial$ integration by parts to obtain
a formula for the principal
value of $\int \bar \partial g_\omega (x) \, \sigma(\psi_\omega
x-x) \, dx$. The formula will involve in general
the derivatives
of $\partial^j g_\omega$ at $x_0$ of order $0\le j\le k-1$. We just consider a simple example,
where $\psi_\omega(x) = x + \alpha x^2$ with $\alpha \in \C$
($\alpha \ne 0$) and $g_\omega$ is $C^2$. Then:
$$
\eqalign
{
\int_\C \bar \partial g_\omega (x) \, \sigma(\psi_\omega x-x) \, dx&=
\int_\C \bar \partial g_\omega (x) \, \sigma(\alpha x^2) \, dx 
=\int_\C \partial \bar \partial g_\omega (x) \, \sigma(\alpha x) \, dx\cr
&=\int_\C \bar \partial  \partial g_\omega (x) \, \sigma(\alpha x) \, dx
= -{\partial  g_\omega (0) \over \alpha }\, .
}
$$



\bigskip
\bigskip

\noindent{\titre 2. SPECTRAL PROPERTIES}

\bigskip \noindent 
{\bf 2.1. The spaces $ \bf \B_{KL}$.}
\medskip For $K,L \ge 0$, let $\B_{KL}$ be the Banach space of distributions
$\Phi$ with support in a fixed bounded set $B \subset \C$ and 
$\partial^K\, \bar
\partial^L \Phi$ = a measure. We write $\Vert \Phi \Vert_{KL} =$ mass of 
$\vert
\partial^K\, \bar
\partial^L \Phi \vert$. We shall generally use a functional notation
for the elements $\Phi$ of $\B_{KL}$. 
Note that $\partial^K \, \bar \partial^L$ is a
canonical isomorphism of $\B_{KL}$ to a subspace of the space of measures with
support in $B$.
\medskip If $\d_0$ is the unit mass at $0 \in \C$, the equation 
$\partial^u \bar
\partial^v \varphi = \d_0$ has a fundamental solution $\varphi_{uv}$ such that $\varphi_{01}
(z) = {1 \over \pi z},\ \varphi_{10} (z) = {1 \over \pi \overline z},\ \varphi_{02}(z) =
{\overline z \over \pi z},\ \varphi_{11} (z) = {1 \over \pi} \log |z|,\ \varphi_{20} (z) = {z
\over \pi z}, \ldots \varphi_{21} (z) = {z \over \pi} (\log |z|+1)$, etc. Therefore 
$$ 
\matrix {
 \B_{KL} \subset L^p(B)                                           \hfill      
& \hbox {\rm for all} \  p < 2 \ \if \  K+L=1         \hfill    \cr 
 \B_{KL} \subset L^\infty(B)                                      \hfill 
& \hbox{\rm if}\  K+L=2 \  \and \ KL=0                 \hfill    \cr 
 \B_{KL} \subset L^q(B)                                           \hfill 
&   \hbox{\rm for all} \ q < \infty \ \if \  K=L=1  \hfill    \cr  
 \B_{KL} \subset \CC_0 (B)                                        \hfill 
& \hbox{\rm (continuous functions on}\ \C \  \hbox{\rm
vanishing outside of} \ B) \  \if  \ K+L \ge 3. 										\hfill				\cr 
}$$
   
In fact one can prove that $\B_{20} \cap \B_{11} \cap \B_{02}
\subset \CC_0(B)$ [2]. \medskip Note that $\B_{K'L'} \subset
\B_{KL}$ if $K' \ge K,\ L' \ge L$. Also, since 
$\partial^u \bar \partial^v \B_{KL}
\subset \B_{K-u, L-v}$ for\break $u \le K,\  v \le L$, we have ready 
information on
spaces in which these derivatives lie. 
\bigskip \noindent {\bf 2.2. The operators $ \bf \M_{k
\ell}, \widehat \M_{k \ell}$.} 
\medskip For $k,\ell \in \Z$, let 
$$\eqalign {
\M_{k \ell} \Phi &= \sum_{ \in \O} g_\o . (\psi'_\o)^k. (\bar \psi'_\o)^\ell
. (\Phi \circ \psi_\o) \cr
\widehat \M_{k \ell} \Phi &= \sum_{\o \in \O} (g_\o \circ \psi_\o^{-1}) . (\psi'_\o
\circ \psi_\o^{-1})^k . (\bar \psi'_\o \circ \psi^{-1}_\o)^\ell . (\Phi \circ
\psi^{-1}_\o). \cr
}$$
We assume that $\O$ is finite, $g_\o \in \CC_0(B)$, and that $\psi_\o$ is an
invertible holomorphic map $\Lambda_\o \to \psi_\o \Lambda_\o$ with open $\Lambda_\o
\supset \supp g_\o$ and $\Lambda_\o,\ \psi_\o \Lambda_\o \subset B$. [At the
cost of making more complicated assumptions below, one could take 
$\O$ countable infinite, or
replace $\sum\limits_{\o}$ by an integral, or assume only that $g_\o$ vanishes
outside $\Lambda_\o$].
\medskip Let us say that the family $(g_\o)$ is $\B_{KL}$ {\it adapted} if
$g_\o \in \B_{KL}$ and $K+L \ge 3$. If $K+L=2$ we require 
$\partial^K \, \bar \partial L
g_\o \in L^r$ for some $r > 1$. If $K + L = 1$ we 
require $\partial^K \bar
\partial^L g_\o \in L^r$ for some $r > 2$. 
Note that if $K' \le K,\ L' \le L$ and
$(g_\o)$ is $\B_{KL}$ adapted, then $(g_\o)$ is also $\B_{K'L'}$ adapted.
\medskip 
With the above definition, if $(g_\o)$ is $\B_{KL} $ 
adapted then $\M_{k \ell},\
\widehat \M_{k \ell} : \B_{KL} \to \B_{KL}$ are bounded operators for all
$k,\ell$. [This is readily checked by expanding $\partial^K \, 
\bar \partial^L \M_{KL}
\Phi$ and using the properties of the spaces $\B_{KL}$ obtained in 
Section 2.1].
\medskip 
We may also let $\M_{k \ell} $ act on $\CC_0 (B)$; we let  $R_{k \ell},
\widehat
R_{k \ell}$ be the corresponding spectral radii: 
$$\eqalign {
R_{k \ell} &= \lim_{m \to \infty} \bigl ( \Vert (\M_{k \ell})^m \Vert_\infty
\bigr )^{1/m} \cr
\widehat R_{k \ell} &= \lim_{m \to \infty} \bigl ( \Vert (\widehat \M_{k \ell})^m
\Vert_\infty \bigr )^{1/m}. \cr
}$$
\medskip Let us assume that $(g_\o)$ is $\B_{KL}$ adapted and $\Phi \in
\B_{KL}$. We may write 
$$\partial^K \, \bar \partial^L(\M_{k \ell} \Phi) = \M^* \partial^K\, \bar
\partial^L \Phi +
\M' \partial^K\, \bar
\partial^L \Phi$$
where $\M^*$ and $\M'$ act on measures. The operator $\M^*$ is given in
functional notation by 
$$(\M^* \Phi)(x) = \sum_\o g_\o \cdot (\psi'_\o)^{k+K} 
(\bar  \psi'_\o)^{\ell +L}
(\Psi \circ \psi_\o)$$
i.e., formally, $\M^* = \M_{k+K, \ell+L}$. We turn now to the definition of
$\M'$. If we act repeatedly on the measure $\Psi$ by convolution with the
fundamental solution $\varphi_{01}$ or $\varphi_{10}$ and multiplication by 
a smooth
function $X$ with compact support and equal to 1 on $B$ we obtain terms
$\Psi_{uv}$ with $u \le K,\ v \le L$ such that when $\Psi = 
\partial^K \, \bar \partial^L \Phi$, the $\Psi_{uv}$ are the lower order 
derivatives $\partial^{u} \, \bar \partial^v \Phi$. We obtain $\M' \Psi$ by adding the $\Psi_{uv} \circ \psi_\o$
multiplied by appropriate derivatives of $g_\o,\ \psi'_\o,\ \bar \psi'_\o$ and
summing over $\o$. For $K+L-u-v = 1,2,\ge 3$ the maps $\Psi \to \Psi_{uv}$ are
found to be compact from measures to $L^p$ (with $p < 2$), $L^q$ (with $q <
\infty$) or $\CC_0$ ($\supp X$); from this it follows that $\M'$ (acting on
measures) is compact.
\medskip
Similarly
$$\partial^K \, \bar \partial^L(\widehat \M_{k \ell} \Phi) = 
\widehat \M^* \partial^K \, \bar
\partial^L \Phi + \widehat \M' \partial^K\, \bar
\partial^L \Phi$$
where $\M^*$ and $\M'$ act on measures; $\widehat \M^*$ is given formally by $\widehat
\M^* = \widehat \M_{k-K, \ell-L}$ and $\widehat \M'$ is compact.
\medskip \noindent 
{\bf 2.3. Theorem.} {\it Let $(g_\o)$ be $\B_{KL}$ adapted. 
Then the essential spectral radius of $\M_{k \ell}$  (resp. $\widehat \M_{k
\ell}$)  acting on $\B_{KL}$ is $\le \widehat R_{k + K-1, \ell +L-1}$ (resp.
$\le R_{k-K+1, \ell-L+1}$).}
\medskip Indeed, a direct computation shows that $\M^*$ 
(respectively
$\widehat \M^*$) is
the adjoint of $\widehat \M_{k+K-1, \ell+L-1}$ (resp. $\M_{k-K+1,\ell-L+1}$) acting
on continuous functions. Therefore the spectral radius of $\M^*$ (resp. $\widehat
\M^*$) is $\le \widehat R_{k+K-1,\ell + L-1} $ 
(resp. $\le R_{k-K+1,\ell-L+1}$). The
theorem results from the fact that a compact perturbation does not change the
essential spectral radius, and from the fact that 
$\partial^K \, \bar \partial^L$ maps
$\B_{KL}$ isometrically into the measures, replacing $\M_{k \ell}$ (resp.
$\widehat \M_{k \ell}$) by $\M^* + \M'$ (resp. $\widehat \M^* + 
\widehat \M'$). \cube

\medskip\noindent {\bf 2.4. Theorem.} {\it Let $(g_\o)$ be $\B_{KL}$ 
adapted and $K+L
\ge 2$.}
\medskip (a) {\it The spectral radius of $\M_{k \ell}$ (resp. 
$\widehat \M_{k \ell}$) is $\le \max (\widehat R_{k+K-1,\ell+L-1}, 
R_{k \ell})$ (resp. $\le \max
(R_{k-K+1,\ell -L+1}, \widehat R_{k \ell})$.}
\medskip (b) {\it The generalized eigenfunctions corresponding to eigenvalues
$\lambda$ satisfying 
$| \lambda | >\break \widehat R_{k+K-1,\ell + L-1}$ 
(resp. $> R_{k-K+1,\ell
-L+1})$ are continuous.}
\medskip Let $\rho$ be the spectral radius of $\M_{k \ell}$ acting on $\B_{KL}$.
It suffices to prove (a) under the assumption $\rho > \widehat R_{k+K-1, \ell +
L-1}$. This assumption implies that $\M_{k \ell}$ has an eigenvalue $\lambda$ with
$|\lambda | = \rho$ and a corresponding eigenfunction $\Phi \in \B_{KL}$. If (b)
holds then $\Phi$ is bounded hence $\rho = | \lambda| \le R_{k \ell}$, proving (a).
\medskip Since $\B_{KL} \subset \CC_0 (B)$ when $K+L \ge 3$, it suffices to prove
(b) for $K+L=2$. Let $\lambda$, with $| \lambda | 
> \widehat R_{k+K-1,\ell + L-1}$, be
an eigenvalue of $\M_{k \ell}$ acting on $\B_{KL}$, and $\Phi$ be a
corresponding eigenfunction. We have thus 
$$(\M^* - \lambda) \partial^K\, \bar
\partial^L \Phi = - \M' \partial^K\, \bar
\partial^L \Phi$$
and the right hand side is in $L^r(B)$ for some $r > 1$ 
because $\M_{k \ell}$
is $\B_{KL}$ adapted. The spectral radius of $\M^*$ acting on $L'$ is 
$\le
\widehat R_{k+K-1,\ell +L-1}$ (as noted in the proof of theorem 2.3) hence $< |
\lambda |$, and $\M^*$ is also bounded on $L^\infty$. Using the Riesz-Thorin
interpolation theorem we therefore see that the spectral radius of $\M^*$
acting on $L^r$ becomes $< | \lambda |$ if $r$ is sufficiently close to 1. We
can now conclude that 
$$\partial^K\, \bar
\partial^L \Phi = - (\M^* - \lambda)^{-1} \M' \partial^K \, 
\bar \partial^L
\Phi$$
is in $L^r$, and $\Phi$ is thus in $\CC_0(B)$. Generalized eigenfunctions are
treated in similar manner, and this completes the verification of (b).
\medskip The case of $\widehat \M_{k \ell}$ is handled by the same arguments.
\cube
\medskip \noindent {\bf Remark.} Let $(g_\o)$ be $\B_{KL}$ adapted, with $K+L=1$. If
$\Phi$ is a (generalized) eigenfunction of $\M_{k \ell}$ to an eigenvalue $\lambda$ with $|
\lambda | > \widehat R_{k+K-1,\ell +L-1}$, we have as above 
$\partial^K \, \, \bar \partial^L
\Phi \in L^r$ for some $r > 1$. Therefore $\Phi \in L^p$ for some $p > 2$.
\medskip \noindent {\bf 2.5. Theorem.} {\it Let $g_\o \ge 0,\ (g_\o)$ be $\B_{KL}$
adapted with $K+L \ge 2$, and $\widehat R_{k+K-1,k+L-1} < R_{kk}$ (resp. $R_{k-K+1,k-L+1}
< \widehat R_{kk}$). Then $R_{kk}$ is an eigenvalue of $\M_{kk}$ acting on
$\B_{KL}$ (resp. $\widehat R_{kk}$ is an eigenvalue of $\widehat \M_{kk}$ 
acting on
$\B_{KL}$) and there is a corresponding eigenfunction $\Phi \ge 0$.}
\medskip 
A very similar result has been proved in a related situation [4]. 
We give here
again a complete proof.
\medskip We shall use the notation $\M = \M_{kk},\ R = R_{kk},\ \Vert \cdot
\Vert = \Vert \cdot \Vert_{KL}$. We can take $\chi$ smooth such that $\chi$
takes values in $[0,1]$, has support in $B$, and is 1 on the $\psi_\o \supp
g_\o$. 
\medskip
The spectral radius $\rho$ of $\M$ acting on $\B_{KL}$ is the same as the
spectral radius of the operator $\M^* + \M'$ acting on the space 
$\partial^K \, \bar
\partial^L \B_{KL}$ of measures (the operators $\M^*,\ \M'$ are defined as
earlier). Therefore 
$$ \displaylines {
\hbox {\rm spectral radius of}\ \M^* + \M'\cr
 \hbox{\rm acting on}\ \partial^K \, \bar
\partial^L \B_{KL} \cap L^1 \le \rho . \cr}
$$
Writing 
$$\eqalign {
\s &= \  \build{\rm lim \, sup}_{m \to \infty}^{} \Vert \M^m \chi \Vert^{1/m} \cr
&= \  \build{\rm lim \, sup}_{m \to \infty}^{} \Bigl ( 
\left\Vert (\M^* + \M')^m \partial ^K
\, \bar \partial^L \chi \right\Vert_1 \Bigr)^{1/m} \cr
}$$
we also have $\s \le \rho$.
\medskip 
Because $(g_\o)$ is $\B_{KL}$ adapted, $\M'$ is bounded $L^1 \to L^r$ for 
some
$r > 1$ and so $\M^* + \M'$ is bounded $L^r \to L^r$. Therefore, given $\ve >
0$, we can find (using the H\"older inequality) $s > 1$ such that 
$$\build {\rm lim \, sup}_{m \to \infty}^{} \left ( 
\left \Vert (\M^* +\M')^m \partial^K
\, \bar \partial^L \chi \right \Vert_s \right )^{1/m} < \s + \ve.$$
We have (using the positivity of the $g_\o$)
$$\eqalign {
R &= \lim_{m \to \infty}\ (\hbox{\rm norm of}\ \M^m\ 
\hbox{\rm acting on}\
L^\infty)^{1/m} \cr
&= \lim_{m \to \infty} (\Vert \M^m \chi \Vert_\infty)^{1/m} \cr
&\le \  \build{\rm lim \, sup}_{m \to \infty}^{} (\const. 
\left \Vert \partial^K \, \bar
\partial^L (\M^m \chi) \right \Vert_s )^{1/m} \cr
&= \  \build{\rm lim \, sup}_{m \to \infty}^{} \left ( \left \Vert (\M^* + \M')^m
\partial^K \, \bar  \partial^L \chi \right \Vert_s \right )^{1/m} \cr
& < \s + \ve .\cr
}$$
Hence $R \le \s \le \rho$ and, if $R > \widehat R_{k+K-1,k+L-1}$, 
Theorem 2.4 gives
$\rho \le R$. Therefore $\s = \rho = R$ and 
$$\build{\rm lim \, sup}_{m \to \infty}^{} \Vert \M^m \chi \Vert^{1/m} = R.$$
Replacing everywhere lim sup by lim inf we see that 
$$\lim_{m \to \infty} \Vert \M^m \chi \Vert^{1/m} = R \eqno (1)$$
\medskip
We may write 
$$\chi = \Psi + \sum_j \Psi_j \eqno (2)$$
where, for each $j,\ \lambda_j$ is an eigenvalue of $\M$ (acting on $\B$) 
with $|
\lambda_j| = R$ and $\Psi_j$ is in the corresponding generalized eigenspace;
$\Psi$ is such that 
$$\lim_{m \to \infty} {\Vert \M^m \Psi \Vert \over \widetilde \lambda^m} = 0$$
with $0 < \widetilde \lambda < R$. In view of (1) the $\Psi_j$ do not all vanish.
Write the restriction of $\M$ to the generalized eigenspaces corresponding to
the $\lambda_j$ in Jordan normal form: it is then readily seen that there is an
integer $k \ge 0$ such that 
$$\lim_{m \to \infty} {1 \over \lambda^m_j m^k}\, \M^m \Psi_j = \Phi_j$$
and
$$\M \Phi_j = \lambda \Phi_j$$
for all $j$, and $\Phi_j \not = 0$ for some $j$. From (2) we get
$$0 \le {\M^m \chi \over R^m m^k}\, = \, {\M^m \Psi \over R^m m^k}\, + \sum_j
\left ( {\lambda_j \over R} \right )^m\, {\M^m \Psi_j \over \lambda^m_j m^k}.$$
Therefore
$$\sum_j \left ({\lambda_j \over R} \right ) ^m \Phi_j \ge - \varphi_m \eqno (3)$$
where both sides of this inequality are real functions $\in \B$ and 
$$\varphi_m = {\M^m \Psi \over R^m m^k} + \sum_j \left ( {\lambda_j \over R} \right
)^m \left ( {\M^m \Psi_j \over \lambda_j^m m^k} - \Phi_j \right ).$$
Note that $\varphi_m \to 0$ in $\B$, hence in $L^q$ for $q < \infty$.
\medskip Let $\langle \quad \rangle_m$ denote the average $\lim\limits_{M \to +
\infty} {1 \over M} \sum\limits_{m=0}^{M-1}$, and write $\lambda_0 = R$ (with
$\Psi_0 = \Phi_0 = 0$ if $\lambda_0$ is not an eigenvalue of $\M$). For arbitrary
real $\a,\b$, we have 
$$\displaylines {
\left\langle \bigl ( 1+ \sin (m \, \a + \b)\bigr ) \sum_j \left ({\lambda j
\over R} \right )^m \Phi_j \right\rangle_m \cr
\ge - \left \langle \bigl ( 1 + \sin (m \,  \a + \b) \bigr ) \varphi_m \right
\rangle_m \cr
}
$$ 
hence 
$$\Phi_0 + \left \langle \sin (m \a + \b) \sum_j \left ( {\lambda_j \over R} \right )^m
\Phi_j \right \rangle_m \ge 0.$$
If we had $\Phi_0 = 0$, taking $\b = 0,\ {\pi \over 2},\ \pi,\ {3 \pi \over
2}$ would give
$$\left \langle e^{im \a} \sum_j \left ( {\lambda_j \over R} \right )^m \Phi_j
\right\rangle_m = 0$$
implying that $\Phi_j = 0$ for all $j$ contrary to our assumptions. Therefore
$\lambda_0 = R$ is an eigenvalue and (taking $\b = 0,\pi)$
$$\Phi_0 \pm \left \langle \sin m \, \a \sum_j \left ( {\lambda_j \over R} \right )^m \Phi
j \right \rangle_m \ge 0$$
so that $\Phi_0 \ge 0$, and $\Phi_0$ is not identically $0$. \cube 




\bigskip\bigskip
\noindent
{\titre 3. TRACES AND DETERMINANTS -- EXISTENCE.}

\medskip

	In this section we want to justify the existence of the traces
and determinants of some of the operators which arise in this paper.
We begin with some attention to the function spaces on which we shall
work.

	Given a compact set $E \subseteq {\bf C}$, let
${\cal B}(E) = {\cal B}_{0,2}(E)$ denote the space of distributions $f$ on 
${\bf C}$ such that  ${\rm supp} \, f \subseteq E$
and $\overline{\partial}^2 f$ (taken in the sense of distributions)
is a finite measure on ${\bf C}$.  We define the ${\cal B}$-norm of $f$ by
$$
	\|f\|_{\cal B} = \int_{\bf C} |\overline{\partial}^2 f|,
								\eqno (3.1)
$$
and we use this norm for all the spaces ${\cal B}(E)$.

\medskip
\noindent
{\bf Lemma 3.2.}  {\it If $f \in {\cal B}(E)$, then $f$ is in fact represented
by a bounded measurable function (which we also denote by $f$) and we
have that
$$
	\|f\|_\infty \le {2 \over \pi} \int_{\bf C} |\overline{\partial}^2 f|.
								\eqno (3.3)
$$
}

\medskip

	To see this we first recall that 
$$
	\overline{\partial} \, ({ 1 \over x}) = \pi \, \delta_0 
		\quad\hbox{ and }\quad
	\overline{\partial}^2 ({\overline{x} \over x}) = \pi \, \delta_0
								\eqno (3.4)
$$
in the sense of distributions on ${\bf C}$,
where $\delta_0$ denotes the Dirac mass at the origin.	The first equation 
is well-known, and the second equation follows from the first.
(Note that ${ 1 \over x}$ is locally integrable on ${\bf C}$.)

	Let $f \in {\cal B}(E)$ be given, and define $F$ by
$$
	F(x) = \int_{\bf C} {1 \over \pi} 
		\,{\overline{x}-\overline{y}  \over  x-y } 
			\,\,\overline{\partial}^2 f(y).		\eqno (3.5)
$$
This is a bounded measurable function on ${\bf C}$, since 
$\overline{\partial}^2 f$ is a finite measure, and we have that
$$
	\|F\|_\infty \le {1 \over \pi}  \int_{\bf C} |\overline{\partial}^2 f|.
								\eqno (3.6)
$$
We also have that
$\overline{\partial}^2 F = \overline{\partial}^2 f$ (in the sense
of distributions), because of (3.4).  
Thus $F - f$ is holomorphic, and it is therefore
constant, since $F$ is bounded and $f$ has compact support.  
This constant is bounded by $\|F\|_\infty$, and  (3.3) follows.
This proves Lemma 3.2.

\medskip
\noindent
{\bf Lemma 3.7.}  {\it If $E \subseteq {\bf C}$ is compact and 
$f \in {\cal B}(E)$, then $\overline{\partial} f$ (taken in the sense
of distributions) is an integrable function on ${\bf C}$ which is
given by
$$
	\overline{\partial} f(x) = \int_{\bf C} {1 \over \pi} \, {1 \over x-y} 
			\,\,\overline{\partial}^2 f(y).
								\eqno (3.8)
$$
Also, $f \mapsto \overline{\partial} f$ defines a bounded linear
operator from ${\cal B}(E)$ into $L^p(E)$ for all $p < 2$.}

\medskip

	Indeed, if $f \in {\cal B}(E)$ and $F$ is defined by (3.5),
then we saw above that $F-f$ is constant.  Thus $\overline{\partial} f
= \overline{\partial} F$.  This permits us to derive (3.8) from
(3.5) by standard arguments in distribution theory.  Of course
$\overline{\partial} f \equiv 0$ on ${\bf C}
\backslash E$, since ${\rm supp} \, f \subseteq E$.  The last part,
about $\overline{\partial} f \in L^p(E)$, follows from the fact
that ${1 \over x} \in L^p_{loc}$ for all $p < 2$.  This proves
Lemma 3.7.

	In the next lemmas we accumulate some facts about linear operators
acting on ${\cal B}(E)$.

\medskip
\noindent
{\bf Lemma 3.9.} {\it Let $E$ be a compact subset of ${\bf C}$.
Suppose that 
$$
\eqalign{
	& g \hbox{ is a continuous function on } {\bf C},		 \cr
	& \hbox{the second derivatives of $g$ are locally integrable,}   \cr
	& \hbox{and } \nabla g \in L^p_{loc} \hbox{ for some } p > 2.	 \cr
}
								\eqno (3.10)
$$
(All these derivatives of $g$ are taken in the sense of distributions.)
Then $f \mapsto g \, f$
defines a bounded linear operator on ${\cal B}(E)$.  If the support of
$g$ is contained in some compact set $W \subseteq {\bf C}$, then
$f \mapsto g \, f$ defines a bounded linear operator from ${\cal B}(E)$
into ${\cal B}(W)$.}

\medskip

	According to the Leibniz rule we have that
$$
	\overline{\partial}^2  (gf) = (\overline{\partial}^2  g) \, f
		+ 2 (\overline{\partial} \, g) \, (\overline{\partial} \, f)
			+ g \, (\overline{\partial}^2 f).
								\eqno (3.11)
$$
Since we are working with distributional derivatives we should be a
little bit careful.  This identity would be a standard tautology from
distribution theory if $g$ were $C^\infty$.  In general the right hand
side makes sense as a measure because of (3.10) and Lemmas 3.2 and
3.7, and it can be derived from the case where $g$ is $C^\infty$ by a
standard approximation argument.

	Lemma 3.9 follows easily from the identity (3.11) and the
bounds in Lemmas 3.2 and 3.7. 

	Recall that if $\eta \in L^1({\bf C})$ and if $h \in L^p({\bf C})$
for some $p, 1 \le p \le \infty$, then $\eta * h \in L^p({\bf C})$
and
$$
	\|\eta * h\|_p \le \|\eta\|_1 \, \|h\|_p.
$$
Here $\eta * h$ denotes convolution between $\eta$ and $h$.  The next
lemma provides an analogous result for the ${\cal B}$ spaces.


\medskip
\noindent
{\bf Lemma 3.12.} {\it Suppose that $g$ satisfies (3.10) and has
support contained in the compact set $W$, and let $\theta$ be a
locally integrable function on ${\bf C}$.  Then $f \mapsto g \,
(\theta * f)$ defines a bounded linear operator from ${\cal B}(E)$
into ${\cal B}(W)$ for every compact set $E \subseteq {\bf C}$.  This
operator is also compact, and in fact it can be approximated in the
operator norm by finite rank operators of the form $f \mapsto \sum_i g
\, a_i \langle f, b_i \rangle$, where $\{a_i\}, \{b_i\}$ are 
finite sequences of polynomials, and $\langle \cdot, \cdot
\rangle$ denotes the standard pairing of functions on ${\bf C}$.}

\medskip

	Consider first the part about the boundedness of the operator.
The main point is that convolution commutes with differentiation, so
that $\overline{\partial}^2 \theta * f = \theta *
(\overline{\partial}^2 f)$, etc.  This permits us to control
$\theta*f$ and its first two $\overline{\partial}$ derivatives in
terms of $f$ and its first two $\overline{\partial}$ derivatives, and
then we can get $g \, (\theta * f)$ into ${\cal B}(W)$ using the same
argument as in Lemma 3.9.

	Now consider the second part, about compactness.  Given any
compact set $K$ in ${\bf C}$, we can approximate $\theta$ by
polynomials in $L^1(K)$.  By choosing $K$ correctly we can approximate
our operator $f \mapsto g \, (\theta * f)$ in the operator norm ${\cal
B}(E) \to {\cal B}(W)$ by operators of the same form, but with
$\theta$ replaced by polynomials.  This uses the fact that if $\eta
\in L^1_{loc}$, then the operator norm of $f \mapsto g \, (\eta * f)$,
as an operator from ${\cal B}(E)$ into ${\cal B}(W)$, is controlled by
the $L^1(K)$ norm of $\eta$ if we choose $K$ correctly.  This fact can
be checked using the argument of the preceding paragraph.  A simple
computation shows that $f \mapsto g \, (\theta * f)$ is a finite rank
operator of the desired form when $\theta$ is a polynomial.

	This proves Lemma 3.12.

	Let us call a linear operator acting on functions on ${\bf C}$ a
transfer operator if it can be expressed in the form
$$
	{\cal M} \Phi (x) = 
	     \sum_{\omega \in \Omega} g_\omega(x) \, \Phi \circ \psi_\omega(x).
								\eqno (3.13)
$$
As before we assume that $\Omega$ is finite, that each $g_\omega$ is
at least a continuous function with compact support, 
that each $\psi_\omega$ is a holomorphic
diffeomorphism from an open set $\Lambda_\omega \subseteq {\bf C}$ onto
its image, and that $\Lambda_\omega \supseteq {\rm supp} \, g_\omega$.
Although $\Phi \circ \psi_\omega(x)$ is not
defined for all $x$, it is defined on the support of $g_\omega$, and we
can simply interpret $\Phi \circ \psi_\omega(x)$ to be $0$ when $x$ does
not lie in the domain of $\psi_\omega$.  

	Note that these transfer operators are well-defined on
$L^1_{loc}$ functions, for instance, because the $\psi_\omega$'s are
diffeomorphisms (on their domains) and preserve sets of Lebesgue
measure $0$ in particular.  We shall frequently use the fact that
transfer operators are bounded on $L^\infty$.

\medskip
\noindent
{\bf Lemma 3.14.} {\it Let $E$ and $W$ be compact subsets of ${\bf
C}$.  If ${\cal M}$ is a transfer operator as in (3.13), if each
$g_\omega$ satisfies (3.10), and if ${\rm supp} \, g_\omega \subseteq
W$ for each $\omega$, then ${\cal M}$ defines a bounded linear
operator from ${\cal B}(E)$ into ${\cal B}(W)$.}

\medskip

	In order to deal with this in a reasonable way it is helpful
to establish first a technical fact about membership in ${\cal B}(W)$.

\medskip
\noindent
{\bf Sublemma 3.15.} {\it An integrable function $f$ on ${\bf C}$
lies in ${\cal B}(W)$ if and only if there is a sequence of smooth
functions $\{f_j\}$ such that $f_j \to f$ in the $L^1$ norm, $\sup_j
\|f_j\|_{\cal B} < \infty$, and the supports of the $f_j$'s shrink to
a subset of $W$, in the sense that for each open set $U \supseteq W$
there is an $l$ such that ${\rm supp} \, f_j \subseteq U$ when $j >
l$.}

\medskip

	The proof of this is standard.  If $f \in {\cal B}(W)$, then
we can get a sequence $\{f_j\}$ as above by convolving $f$ with an
approximation to the identity with supports shrinking to $\{0\}$.
The resulting functions will have supports shrinking to a subset
of $W$, and the ${\cal B}$ norms will remain bounded because
derivatives commute with convolution, and because the convolution
of an $L^1$ function with a measure is again an $L^1$ function
whose $L^1$ norm is bounded by the product of the $L^1$ norm of
the original function and the total variation of the measure.
Conversely, suppose that $f$ is integrable and that there exists an
approximating sequence $\{f_j\}$ as in the sublemma.  Then we have
that $\overline{\partial}^2 f_j \to \overline{\partial}^2 f$ in the
sense of distributions.  Our assumptions on $\{f_j\}$ imply a uniform
mass bound on $\overline{\partial}^2 f_j$, and hence
$\overline{\partial}^2 f$ must be a finite measure.  Since $f$ is
certainly supported in $W$ we get that $f \in {\cal B}(W)$, as
desired.  This proves Sublemma 3.15.

	Let us come back now to the proof of Lemma 3.14.  If we
assume that $\Phi$ is smooth, or if we compute formally, then we can get
that $\|{\cal M} \Phi\|_{\cal B} \le C \|\Phi\|_{\cal B}$ for some constant
$C$ which does not depend on $\Phi$.  This uses a Leibniz computation
like (3.11) and our assumptions on the $g_\omega$'s to reduce the
problem to one of having estimates for $\Phi \circ
\psi_\omega$ and its $\overline{\partial}$ derivatives on the support
of $g_\omega$, in the same way as in the proof of Lemma 3.9.  The
holomorphicity of the $\psi_\omega$'s permits us to compute
$\overline{\partial}$ derivatives of $\Phi \circ \psi_\omega$ in terms of
$\overline{\partial}$ derivatives of $\Phi$, with extra multiplicative
factors coming from the derivatives of the $\psi_\omega$'s.
Composition with $\psi_\omega$ and multiplication by its derivatives
do not disturb integrability properties, at least if we remain within
the compact subset ${\rm supp} \, g_\omega$ of the domain of
$\psi_\omega$, as we do here.  Thus we can bound $\|{\cal M} \Phi\|_{\cal
B} = \|\overline{\partial}^2 {\cal M} \Phi\|_1$ in terms of $\|\Phi\|_{\cal
B}$, using also Lemmas 3.2 and 3.7.  Once we have this bound for all
smooth functions $\Phi$ with compact support we can use Sublemma 3.15 to
get that ${\cal M}$ actually maps ${\cal B}(E)$ into ${\cal B}(W)$.
This proves Lemma 3.14.

	Another operator that we shall be interested in is the operator
$S$ defined by
$$
	S f(x) = \int_{\bf C} {1 \over \pi} \, {1 \over x-y} \, f(y) \, dy.
								\eqno (3.16)
$$
This operator sends integrable functions with compact support to locally
integrable functions on ${\bf C}$, or, more generally, it sends distributions
with compact support to distributions.  It is the inverse to 
$\overline{\partial}$, in the sense that
$$
	\overline{\partial} \, (S f) = f \quad\hbox{ and } \quad
		S (\overline{\partial} \, f) = f		\eqno (3.17)
$$
for all distributions $f$ on ${\bf C}$ with compact support.  These
equations come down to the first formula in (3.4), by the standard
tricks in distribution theory.  (That is, the distributional interpretation
of (3.17) converts these equations, via duality, to their counterparts for 
test functions, which can then be reduced to (3.4).)

	We shall also want to have a nice one-parameter family of
regularizations of $S$.  Fix a $C^\infty$ function $\nu$ on ${\bf C}$,
once and for all, with 
$$
	\nu(x) = 1 \,\hbox{ when }\, |x| \ge 1 \quad\hbox{ and }\quad
			\nu(x) = 0 \,\hbox{ when }\, |x| \le {1 \over 2}.
								\eqno (3.18)
$$
Define an operator $S_t$ for $t>0$ by
$$
	S_t f(x) = \int_{\bf C} {1 \over \pi} \, {1 \over x-y} \, 
				\nu({x-y \over t}) \, f(y) \, dy.
								\eqno (3.19)
$$
Again this operator acts on functions or distributions with compact support.
Notice that 
$$
\eqalign{
	{1 \over x-y} \, \nu({x-y \over t}) & = {1 \over x-y} 
			\qquad\hbox{ when } |x-y| \ge t, 	\cr
					    & = 0
			\,\qquad\qquad\hbox{ when } |x-y| \le t/2.
}
					 			\eqno (3.20)
$$
Thus $S_t$ approximates $S$ but its kernel has no singularity.

	In practice one should think of $t$ as being small, or at least
not large, and for simplicity we shall restrict our attention to $t \le 1$.

\medskip
\noindent
{\bf Lemma 3.21.} {\it Suppose that $g$ satisfies (3.10) and has
support contained in the compact set $W$.  Then $f \mapsto g \, (S f)$
and $f \mapsto g \, (S_t f)$ define bounded linear operators from
${\cal B}(E)$ into ${\cal B}(W)$ for every compact set $E \subseteq
{\bf C}$ and for every $t \in (0,1]$.  Moreover, the second operator
converges to the first one as $t \to 0$ in operator norm (defined
using $\|\cdot\|_{\cal B}$).  In fact, the operator norm of $g S - g S_t$
is $O(t)$.}

\medskip

	The boundedness of these operators follows from Lemma 3.12,
because they are given in terms of convolutions of locally integrable
functions.  The convergence in the operator norm follows from the
fact that ${1 \over x} \, \nu({x \over t})$ converges to ${1 \over x}$
as $t \to 0$ in $L^1$ on any compact subset of ${\bf C}$.  In fact 
$$
	\int_{\bf C} |{1 \over x} - {1 \over x} \, \nu({x \over t})| \, dx
		= \hbox{ constant} \cdot t,	
$$
as one can easily check.  This implies that $g S - g S_t$ has norm $\,
= O(t)$, because the norm of the operator described in Lemma 3.12 is
controlled by the $L^1$ norm of the function $\theta$ given there.
This proves Lemma 3.21.

	One of our main goals in this section is to show that certain
operators constructed from $S$ are trace class.  Let us recall the
definition.

	If $X$ and $Y$ are Banach spaces, then we say that a linear
operator $T : X \to Y$ is trace class (or nuclear) if it can be represented
as a sum of rank one operators,
$$
	T u = \sum_j \lambda_j \, w_j \langle u, v_j \rangle,	\eqno (3.22)
$$
where the $w_j$'s are elements of the unit ball of $Y$, the $v_j$'s
are elements of the unit ball of the dual space of $X$,
$\langle \cdot, \cdot \rangle$ denotes the pairing between
$X$ and its dual space, and the
$\lambda_j$'s are scalars which satisfy $\sum_j |\lambda_j| < \infty$.
The infimum of $\sum_j |\lambda_j|$ over all such representations
of $T$ is called the trace norm of $T$.

	Note that the composition of a bounded operator and a trace class
operator (in either order) is trace class.

	Unfortunately operators like $g \, S$ are not trace class, the
singularity in the kernel is too strong.  We shall see that products
of $S$'s and commutators with $S$'s can give rise to trace class
operators in a natural way.  Or almost anyway; we shall see that
products of $S_t$'s and commutators with $S_t$'s give rise to trace
class operators whose trace norm is bounded uniformly in $t$.

	Let us begin with a general criterion for an integral operator
to be trace class.

\medskip
\noindent
{\bf Lemma 3.23.} {\it Let $X$ be a Banach space and let $E$ be a compact
subset of ${\bf C}$.  Suppose that $H : E \to X$ is continuous (using the
norm topology for $X$).  Define
an operator $T : L^\infty(E) \to X$ by
$$
	T f = \int_{\bf C} H(y) \, f(y) \, dy.			\eqno (3.24)
$$
Then $T$ is trace class, and 
$$
	 \hbox{\rm trace norm } T \le |E| \cdot \sup_{y \in E} \|H(y)\|_X,
								\eqno (3.25)
$$
where $|E|$ denotes the Lebesgue measure of $E$.}

\medskip

	This result is pretty standard, but let us prove it for the
sake of completeness.  We need to approximate $T$ by finite rank
operators, and so we approximate the integral in (3.24) by Riemann
sums.

	Incidentally, $H$ is uniformly continuous on $E$, since $E$ is 
compact, and this ensures that there is no funny business in defining
the integral in (3.24).  This will be made explicit in the argument
that follows.

	Fix a square $Q_0$ in ${\bf C}$ which contains $E$.  For each $j
= 0, 1, 2, \ldots$ let $\Delta(j)$ denote the obvious decomposition of
$Q_0$ into $2^{2j}$ subsquares of $Q_0$, each of size $2^{-j}$ times the
size of $Q_0$, with sides parallel to the same axes as for $Q_0$, and with
the elements of $\Delta(j)$ having disjoint interiors.  Thus when
$j=0$ we simply get $Q_0$ back again, when $j=1$ we get the usual
decomposition of $Q_0$ into four pieces, etc.

	If $Q$ is any square in ${\bf C}$, let $\gamma(Q)$ denote its center.

	Given $j \ge 0$ define $T_j : L^\infty(E) \to X$ by
$$
	T_j f(x) = \sum_{Q \in \Delta(j)} \,  H(\gamma(Q)) \,\,
				\int_{Q \cap E} f(y) \, dy.	
								\eqno (3.26)
$$
Let us check that this finite-rank operator satisfies
$$
	\hbox{trace norm } T_j \le |E| \cdot \sup_{y \in E} \|H(y)\|_X.
								\eqno (3.27)
$$
This comes down to the observation that
$$
\eqalign{
	f \mapsto \int_{Q \cap E} f(y) \, dy \quad
		& \hbox{is a bounded linear functional on } L^\infty(E) \cr
		& \hbox{with norm } \le |Q \cap E|.			\cr
}
								\eqno (3.28)
$$

	Next let us check that
$$
	\lim_{j,k \to \infty} \,\, \hbox{trace norm } (T_k - T_j) = 0.	
								\eqno (3.29)
$$
Let $j$ and $k$ be large and given, with $k > j$.  Notice that every
$Q \in \Delta(k)$ has a unique
``ancestor'' $\widehat {Q}(j)$ in $\Delta(j)$, where 
$Q \subseteq \widehat {Q}(j)$
but the interior of $Q$ is disjoint from all other elements of $\Delta(j)$.
In other words the elements of $\Delta(k)$ can be obtained by subdividing the
elements of $\Delta(j)$.  This observation leads us to the formula
$$
\eqalign{
	(T_k - T_j)f & = \sum_{Q \in \Delta(k)} \,  H(\gamma(Q)) \,\,
				\int_{Q \cap E} f(y) \, dy
		       - \sum_{R \in \Delta(j)} \,  H(\gamma(R)) \,\,
				\int_{R \cap E} f(y) \, dy		\cr
		     & = \sum_{Q \in \Delta(k)} \,  
			    (H(\gamma(Q)) - H(\gamma(\widehat{Q}(j))) \,\,
				 \int_{Q \cap E} f(y) \, dy.		\cr
}
								\eqno (3.30)
$$
This formula uses the fact that every element $R$ of $\Delta(j)$
is the disjoint union of its descendants in $\Delta(k)$ (so that
$\int_{R \cap E} f$ equals the sum of the corresponding integrals
for the descendants $Q$ of $R$ in $\Delta(k)$).  On the other hand the
(uniform) continuity of $H$ implies that
$$
	\lim_{j,k \to \infty} \,\, \max_{Q \in \Delta(k)} 
		\|H(\gamma(Q)) - H(\gamma(\widehat{Q}(j))\|_X
			= 0.					\eqno (3.31)
$$
It is easy to derive (3.29) from (3.30), (3.31), and (3.28).

	Thus $\{T_j\}$ is a Cauchy sequence with respect to the trace norm.
Of course $\{T_j\}$ converges to $T$ in the operator norm.  Standard arguments
imply that $T$ is trace class and that
$$
	\hbox{trace norm } T \le \liminf_{j \to \infty} 
						\,\, \hbox{trace norm } T_j
				\le |E| \cdot \sup_{y \in E} \|H(y)\|_X.
								\eqno (3.32)
$$

	This completes the proof of Lemma 3.23.

	In practice we shall want to apply Lemma 3.23 with
$L^\infty(E)$ replaced with ${\cal B}(E)$, and we can do that freely
because of Lemma 3.2.  We shall also normally take $X$ to be some
${\cal B}$ space.

	Now we want to look at operators constructed from $S$, starting
with the very smooth $S_t$'s.

\medskip
\noindent
{\bf Lemma 3.33.} {\it Let $E$ and $W$ be compact subsets of ${\bf
C}$, and suppose that $g$ is supported in $W$ and satisfies (3.10).
Then $f \mapsto g \, (S_t f)$ defines a trace class operator from
$L^\infty(E)$ into ${\cal B}(W)$ for every $t \in (0,1]$, and the trace norm
is $O(t^{-1})$.}

\medskip

	The hypothesis on $g$ is not sharp, but it is adequate.  We shall
not try to get the sharpest results here.

	As above, we can replace $L^\infty(E)$ with ${\cal B}(E)$
for free, because of Lemma 3.2.

	Let us first observe that we can reduce to the case where $g$
is smooth.  Indeed, let $\chi$ be a smooth function with support
contained in some compact set $W'$ and with $\chi \equiv 1$ on $W$.
Then we can view $g \, S_t : L^\infty (E) \to {\cal B}(W)$ as the
composition of $\chi \, S_t : L^\infty (E) \to {\cal B}(W')$ and the
operator of multiplication by $g$, which defines a bounded operator
from ${\cal B}(W')$ into ${\cal B}(W)$ (Lemma 3.9).  This permits 
us to reduce to the case where $W, g$ are replaced with $W', \chi$.
Thus we may assume that $g$ is smooth.

	We want to apply Lemma 3.23.  We can write $g \, S_t$ explicitly as
$$
	(g \, S_t) f(x)  = {1 \over \pi} \int_{\bf C} g(x) \, {1 \over x-y} \, 
				\nu({x-y \over t}) \, f(y) \, dy.
								\eqno (3.34)
$$
The mapping
$$
	y \to g(x) \, {1 \over x-y} \, \nu({x-y \over t})	\eqno (3.35)
$$
defines a continuous map from ${\bf C}$ into ${\cal B}(W)$.  This is
easy to check, using the smoothness of ${1 \over x-y} \, \nu({x-y \over t})$.
Therefore $g \, S_t$ defines a trace
class map from $L^\infty(E)$ into ${\cal B}(W)$.  We also get that
$$
\eqalign{
	\hbox{trace norm } g \, S_t \, & \le \, |E| \cdot 
	\sup_{y \in E} \, \| 
	    \, g(x) \, {1 \over x-y} \, \nu({x-y \over t}) \, \|_{\cal B}
									\cr
	& \le \, |E| \cdot 
	\sup_{y \in E} \, \int_{\bf C} |\, \overline{\partial}^2_x 
		\{ \, g(x) \, {1 \over x-y} \, \nu({x-y \over t})\}| \, dx 
									\cr
}
								\eqno (3.36)
$$
We want to show that the right hand side is $O(t^{-1})$.  

	Notice that
$$
\eqalign{
	|\, \overline{\partial}^2_x 
		\{ \, g(x) \, {1 \over x-y} \, \nu({x-y \over t})\}|
	& = {1 \over |x-y|} |\, \overline{\partial}^2_x 
		\{ \, g(x) \, \nu({x-y \over t})\}|			\cr
	& \le {2 \over t} \, |\, \overline{\partial}^2_x 
		\{ \, g(x) \, \nu({x-y \over t})\}|,			\cr
}
								\eqno (3.37)
$$
since ${1 \over x-y}$ is holomorphic in $x$ away from the pole $x = y$,
and because $\nu({x-y \over t})$ vanishes when $|x-y| \le t/2$, by (3.18).
Therefore
$$
	\hbox{trace norm } g \, S_t \, \le \, |E| \cdot 
	\sup_{y \in E} \, \int_{\bf C} {2 \over t} \, 
		|\, \overline{\partial}^2_x 
		\{ \, g(x) \, \nu({x-y \over t})\}| \, dx.
								\eqno (3.38)
$$
Thus it suffices to show that
$$
	\sup_{0 < t \le 1} \sup_{y \in E} \, \int_{\bf C} \, 
		|\, \overline{\partial}^2_x 
		 	\{ \, g(x) \, \nu({x-y \over t})\}| \, dx
					< \infty.		\eqno (3.39)
$$

	Let us first make some preliminary observations.  Using (3.18)
and calculus we get that
$$
\eqalignno{
	& \,\,\,\overline{\partial}_x \, \nu({x-y \over t}) = 
		\overline{\partial}^2_x \, \nu({x-y \over t}) = 0 
		\quad\hbox{ when } |x-y| > t \hbox{ or } |x-y| < {t \over 2},
								& (3.40) \cr
	& \sup_{x,y \in {\bf C}} |\overline{\partial}_x \, \nu({x-y \over t})|
 		= \sup_{x,y \in {\bf C}} \, t^{-1} \, 
				|(\overline{\partial} \nu)({x-y \over t})|
			= O(t^{-1}), \qquad\hbox{ and}		& (3.41) \cr
	& \sup_{x,y \in {\bf C}} 
		|\overline{\partial}_x^2 \, \nu({x-y \over t})|
 		= \sup_{x,y \in {\bf C}} \, t^{-2} \,
				|(\overline{\partial}^2 \nu)({x-y \over t})|
			= O(t^{-2}).				& (3.42) \cr
}
$$
Hence
$$
\eqalignno{
	& \int_{\bf C} |\overline{\partial}_x \, \nu({x-y \over t})|
			= O(t) \quad\hbox{ and}			& (3.43) \cr
	& \int_{\bf C} |\overline{\partial}^2_x \, \nu({x-y \over t})|
			= O(1).					& (3.44) \cr
}
$$

	To prove (3.39) one simply uses the Leibniz formula and
remembers that $g$ is smooth and has compact support.  When both
$\overline{\partial}_x$'s land on $g$ the integral is bounded because
$\nu$ is bounded.  When one $\overline{\partial}_x$ lands on $g$ and
one lands on the $\nu$ we can control the integral using (3.43). When
both $\overline{\partial}_x$'s land on $\nu$ we use (3.44).

	This proves (3.39), and Lemma 3.33 follows.

\medskip
\noindent
{\bf Lemma 3.45.}  {\it Let $E$ and $W$ be compact subsets of ${\bf C}$,
and let $g$ and $h$ be functions on ${\bf C}$ such that $g$ satisfies
(3.10), ${\rm supp} \, g \subseteq W$, and 
$$
	h \in C^1 \quad \hbox{ and } \quad 
		\overline{\partial}^2 h \in L^p_{loc}
		\,\,\hbox{ for some } p > 2.			\eqno (3.46)
$$
Then the operator $g \, [h, S_t] : L^\infty(E) \to {\cal B}(W)$
is trace class for each $t \in (0,1]$, where we identify $g$ and $h$ with their
corresponding multiplication operators, and 
$$
	\sup_{0 < t \le 1} \, \hbox{\rm trace norm } g \, [h, S_t] < \infty.
								\eqno (3.47)
$$
}

	To prove this we first observe that we may take $g$ to be
smooth.  This follows from the same argument as in the corresponding
step of the proof of Lemma 3.33.

	To show that $g \, [h, S_t]$
is trace class we apply Lemma 3.23.  We need to check that
$$
	y \to g(x) \, (h(x) - h(y)) \, {1 \over x-y} \, \nu({x-y \over t})
								\eqno (3.48)
$$
defines a continuous map from ${\bf C}$ into ${\cal B}(W)$, and we need to
bound its supremum norm.  The factor
of $g(x)$ already ensures that these functions of $x$ are supported in
$W$, and so it suffices to show that
$$
	y \to \overline{\partial}^2_x \{ \, g(x) \, (h(x) - h(y)) \, 
					{1 \over x-y} \, \nu({x-y \over t})\}
								\eqno (3.49)
$$
defines a continuous map from ${\bf C}$ into the finite measures on ${\bf C}$
(continuity with respect to the total variation norm on the space of
finite measures on ${\bf C}$),
and to bound the total variations of these measures.  In this case these
measures are given by integrable functions, because the $\nu$ kills the 
singularity in ${1 \over x-y}$ and because of our assumptions (3.46) on $h$,
and the continuity of (3.49) as a map into $L^1({\bf C})$ is immediate.
The issue is to control the $L^1$ norms, i.e., to prove that
$$
	\sup_{0 < t \le 1} \, \sup_{y \in E} \,
		\int_W |\overline{\partial}^2_x \{ \, g(x) \, (h(x) - h(y))
				\, {1 \over x-y} \, \nu({x-y \over t})\}| \, dx
		< \infty.					\eqno (3.50)
$$

	The computations are neither exciting nor difficult, but let
us be a little bit careful.  Notice first that we can pull the ${1 \over x-y}$
outside the $\overline{\partial}^2_x$, because ${1 \over x-y}$ is holomorphic
in $x$ away from $y$ and because $\nu({x-y \over t})$ vanishes for $x$
in a neighborhood of $y$ (by (3.18)).  Thus we can reduce to showing that
$$
	\sup_{0 < t \le 1} \, \sup_{y \in E} \,
	    \int_W \, |{1 \over x-y} \, \overline{\partial}^2_x 
		\{ \, g(x) \, (h(x) - h(y)) \, \nu({x-y \over t})\}| \, dx
		< \infty.					\eqno (3.51)
$$

	At this stage one simply has to use the Leibniz formula and treat
the various terms.  Consider first the term
$$
	{1 \over x-y} \, \{ \, g(x) \, (\overline{\partial}^2_x h(x)) \, 
					\nu({x-y \over t})\}	\eqno (3.52)
$$
Since $g$ and $\nu$ are bounded this is controlled by our assumption 
(3.46), H\"older's inequality, and the fact that ${1 \over u}$ lies
in $L^q_{loc}({\bf C})$ for all $q < 2$.  Next consider
$$
	{1 \over x-y} \, (\overline{\partial}_x h(x)) \,
		\overline{\partial}_x \{ \, g(x) \, \nu({x-y \over t})\}.
								\eqno (3.53)
$$
We can ignore $\overline{\partial}_x h(x)$, because it is bounded on
the compact set $W$ by (3.46).  Thus we are faced with estimating
$$
	\int_W \, |{1 \over x-y} 
		\{(\overline{\partial}_x g(x)) \, \nu({x-y \over t})
		  + g(x) \, \overline{\partial}_x \nu({x-y \over t}) \}| \, dx.
								\eqno (3.54)
$$
The $\overline{\partial}_x g(x)$ term is easily controlled because $g$
is smooth, so that $\overline{\partial}_x g(x)$ is bounded on $W$,
and because $\nu$ is bounded.  For the remaining
piece we use the boundedness of $g$ and (3.40), (3.41) to get that
$$
	\sup_{t>0} \, \sup_{y \in E} \, \int_W \, |{1 \over x-y} \, 
		\overline{\partial}_x \nu({x-y \over t})| \, dx < \infty.
								\eqno (3.55)
$$
Thus we conclude that (3.54) is bounded in $t$.

	It remains to show that
$$
	\sup_{t>0} \, \sup_{y \in E} \, \int_W \, |{h(x) - h(y) \over x-y} \, 
	   \overline{\partial}^2_x \{ \, g(x) \, \nu({x-y \over t})\}| \, dx
		< \infty.					\eqno (3.56)
$$
Notice first that
$$
	\sup_{x \in W} \sup_{y \in E} |{h(x) - h(y) \over x-y}| < \infty,
								\eqno (3.57)
$$
since we are assuming that $h$ is $C^1$.  Therefore (3.56) reduces to
$$
	\sup_{t>0} \, \sup_{y \in E} \, \int_W \, 
	  |\overline{\partial}^2_x \{ \, g(x) \, \nu({x-y \over t})\}| \, dx
		< \infty.					\eqno (3.58)
$$
Using Leibniz again we are faced with three terms.  The first is
$(\overline{\partial}^2_x g(x)) \, \nu({x-y \over t})$.  This is
bounded on $W$ since $g$ is smooth and $\nu$ is bounded, and hence the
integral is bounded.  The second term is $(\overline{\partial}_x g(x))
\, (\overline{\partial}_x \nu({x-y \over t}))$.  The integral of this
is bounded because of (3.43) and the boundedness of
$\overline{\partial} g$.  Similarly, we can control the contribution
of the $g(x) \, \overline{\partial}^2_x \nu({x-y \over t})$ term to
(3.58) using (3.44).  Thus we get (3.58), and hence (3.56).

	This proves (3.50), as desired, and Lemma 3.45 follows.

	It would be nice if the commutator between $S_t$ and a
transfer operator had bounded trace norm, at least under suitable
conditions.  There is a result like this, but unfortunately we cannot
quite take the commutator, we have to convert one transfer operator
into another, and we also get an annoying error term that comes from
the truncation in $S_t$.  

	Given a transfer operator ${\cal M}$ as in (3.13), define
${\cal M}_{0,1}$ by
$$
	{\cal M}_{0,1} \Phi (x) = 
	     \sum_{\omega \in \Omega} g_\omega(x) \, \overline{\psi}'_\omega(x)
						\, \Phi \circ \psi_\omega(x).
								\eqno (3.59)
$$


\medskip
\noindent
{\bf Lemma 3.60.} {\it Suppose that $g_\omega, \psi_\omega$, etc.,
are as in the paragraph after (3.13), and suppose also that each
$g_\omega$ satisfies (3.46).  Let $E$ and $W$ be compact sets in
${\bf C}$, and let $\chi$
be a smooth function on ${\bf C}$ which is supported in $W$.  Then
the operator $\chi \, ({\cal M} \, S_t - S_t \, {\cal M}_{0,1}) :
{\cal B}(E) \to {\cal B}(W)$ is trace class for every $t \in (0,1]$,
and we can write it as $T_t + U_t$, where $T_t$ and $U_t$ are 
trace class operators from ${\cal B}(E)$ into ${\cal B}(W)$
which satisfy
$$
\eqalignno{
	& \sup_{0 < t \le 1} \, \hbox{\rm trace norm } T_t < \infty	
			\qquad\hbox{ and}			& (3.61) \cr
	& \hbox{\rm operator norm } U_t = O(t).			& (3.62) \cr
}
$$
}

	We are not asserting a uniform bound on the trace norms
of the error terms $U_t$.  The proof will give a bound of $O(t^{-1})$,
but we shall not need this.  These error terms are an unfortunate
consequence of our truncations, they are not truly natural.  In our
applications of Lemma 3.60 these error terms will not be so bad,
because there will be another factor of $S_t$, and the good estimate
for the operator norm of the error term in (3.62) will balance out
the bad estimate for the trace norm in Lemma 3.33.

	Before we begin the proof of Lemma 3.60 in earnest let us
dispense with some preliminary reductions.

	We may as well assume that $\Omega$ has only one element
$\omega$, so that we can get rid of the sum in our transfer operator.

	Let us show that we can replace the function $g_\omega$ with
something smooth.  Let $h$ be a smooth function with compact support
such that $h \equiv 1$ on ${\rm supp} \, g_\omega$ and ${\rm supp} \,
h \subseteq \Lambda_\omega$ (the domain of $\psi_\omega$).  We may as
well assume that $E \supseteq {\rm supp} \, h$, since otherwise we can
simply enlarge $E$, which strengthens the conclusion of the lemma.

	Let us write $\psi$ for
$\psi_\omega$, to simplify the notation.  Define a transfer
operator ${\cal N}$ by
$$
	{\cal N} \, \Phi (x) = h(x) \, \Phi \circ \psi(x).	\eqno (3.63)
$$
Also, as in (3.59), define ${\cal N}_{0,1}$ by
$$
	{\cal N}_{0,1} \, \Phi (x) = 
		h(x) \, \, \overline{\psi}'(x) \, \Phi \circ \psi(x).
								\eqno (3.64)
$$
Thus ${\cal M} = g_\omega \, {\cal N}$ and 
${\cal M}_{0,1} = g_\omega \, {\cal N}_{0,1}$.
We can rewrite our operator as
$$
	\chi \, ({\cal M} \, S_t - S_t \, {\cal M}_{0,1}) =
		g_\omega \, \chi \, ({\cal N} \, S_t - S_t \, {\cal N}_{0,1})
		+ \chi \, [g_\omega,S_t] \, {\cal N}_{0,1}.	\eqno (3.65)
$$
The last term is trace class, with bounded norm, because of Lemma 3.45
(with $g_\omega$ playing the role that $h$ did there), and because
the transfer operator ${\cal N}_{0,1}$ is bounded on $L^\infty$.
Thus we need only show that the first term on the
right hand side satisfies the conclusions of the lemma.  For this we may forget
about the $g_\omega$ in front, because multiplication by it defines a bounded
linear operator on ${\cal B}(W)$, by Lemma 3.9.  

	In summary, it suffices to show that the operator
$\chi \, ({\cal N} \, S_t - S_t \, {\cal N}_{0,1}): 
{\cal B}(E) \to {\cal B}(W)$ is trace class, and that we can write it as
a sum of two trace class operators, where one has uniformly bounded
trace norm and the other has operator norm which is $O(t)$.


	We need to compute the kernel of
$\chi \, ({\cal N} \, S_t - S_t \, {\cal N}_{0,1})$.  By definitions we have
that
$$
	S_t \, {\cal N}_{0,1} f(x) = {1 \over \pi} \int_{\bf C} {1 \over x-y} 
			\, \nu({x-y \over t}) \, 
			h(y) \, \overline{\psi}'(y)
						\, f(\psi(y)) \, dy.
								\eqno (3.66)
$$
We can make a change of variables to get
$$
   S_t \, {\cal N}_{0,1} f(x) = 
		{1 \over \pi} \int_{\bf C} {1 \over x-\psi^{-1}(y)}
			\, \nu({x- \psi^{-1}(y) \over t})
			\, h(\psi^{-1}(y)) \,
				(\psi^{-1})'(y) \, f(y) \, dy.
								\eqno (3.67)
$$
(Don't forget about the jacobian.)
We are abusing our notation somewhat here; when 
$y \notin \psi({\rm supp} \, h)$ one should interpret $h(\psi^{-1}(y))$
and the whole integrand as being $0$, and we shall follow this convention
throughout these computations.  We can now write 
$$
	(\chi \, ({\cal N} \, S_t - S_t \, {\cal N}_{0,1}))f(x) =
		{1 \over \pi} \int_{\bf C} K(x,y) \, f(y) \, dy,	
								\eqno (3.68)
$$
where
$$
\eqalign{
    K(x,y) = \,\, \chi(x) \, \{h(x) \, & {1 \over \psi(x) - y}
			\,\,\, \nu({\psi(x)-y \over t}) 		\cr
		    - & { (\psi^{-1})'(y) \over x-\psi^{-1}(y)} 
			\, \nu({x- \psi^{-1}(y) \over t})
			\, h(\psi^{-1}(y)) \}.				\cr
}
								\eqno (3.69)
$$

	It is easy to see that
$$
	y \to K(x,y)						\eqno (3.70)
$$
is a continuous map from $E$ into ${\cal B}(W)$.  Indeed, we have that
$K(x,y) = 0$ when $x \notin W$, since $\chi$ is assumed to be supported
in $W$, and we also have that $K(x,y)$ is smooth, since
$\chi$ and $h$ are smooth and the $\nu$
kills the singularity in ${1 \over x-y}$.  Therefore our operator
$\chi \, ({\cal N} \, S_t - S_t \, {\cal N}_{0,1})$ is trace class, by
Lemma 3.23.  From Lemma 3.23 we also get the estimate
$$
\eqalign{
	\hbox{\rm trace norm } 
		\chi \, ({\cal N} \, S_t - S_t \, {\cal N}_{0,1})
& \le \, |E| \, {1 \over \pi} \,\, \sup_{y \in E} \, \, \|K(\cdot,y)\|_{\cal B}
									\cr
& \le \, |E| \, {1 \over \pi} \,\, \sup_{y \in E} \, \int_W 
			|\overline{\partial}^2_x K(x,y)| \, dx.		\cr
}
								\eqno (3.71)
$$
Unfortunately we shall not get a uniform bound (in $t$) for
$$
	\sup_{y \in E} \, \int_W 
			|\overline{\partial}^2_x K(x,y)| \, dx.	\eqno (3.72)
$$
The problem stems from the two
different ways in which $\nu$ appears in (3.69).  Before we get to the 
heart of this we should deal with some preliminary issues.  We should
do some bookkeeping concerning the singularity of $K(x,y)$.

	Let $\Lambda = \Lambda_\omega$ denote the domain of $\psi$, an open
set which contains the compact set ${\rm supp} \, h$.
Choose $r > 0$ so that 
$$
	u \in \Lambda \quad\hbox{ when } v \in {\rm supp} \, h 
		\hbox{ and } |u - v| \le 10 r.			\eqno (3.73)
$$
Of course $r$ does not depend on $t$.  Let $H_i, M_i$, $i = 1,2,3$, denote
the compact sets defined by
$$
       H_i = \{u \in {\bf C}: \hbox{dist} \, (u, {\rm supp} \, h) \le i \, r\},
		\qquad M_i = \psi(H_i).
								\eqno (3.74)
$$
We can find a constant $s > 0$ (depending
on $r$ but not $t$) so that $s < r$,
$$
\eqalignno{
	& x \in {\rm supp} \, h \quad\enspace \, \hbox{ and } \, 
			|\psi(x) - y| \le s 
		\qquad \hbox{ imply } \, y \in  M_1, 	
				\quad\hbox{ and}		& (3.75) \cr
	& y \in \psi({\rm supp} \, h) \, \hbox{ and } \, 
		|x- \psi^{-1}(y)| \le s \quad \hbox{ imply } 
		x \in H_1.					& (3.76) \cr
}
$$
(Actually, (3.76) follows from (3.73), since $s < r$.)

	The point of this parameter $s$ is that we do not need to worry
about $t \ge s$ and we have some useful localizations when $t < s$.
The first assertion is made precise by the observation that
$$
	\sup_{s \le t \le 1} \, \sup_{y \in E} \, \int_W 
		|\overline{\partial}^2_x K(x,y)| \, dx < \infty.	
								\eqno (3.77)
$$
Indeed, as long as $t$ is bounded away from $0$, everything in (3.69)
is smooth, with uniform estimates.  Thus we conclude that (3.71) is bounded
for $t \ge s$, and so we automatically have the decomposition that we seek
when $t \ge s$ (with no error term).  
>From now on we restrict ourselves to $t < s$.

	Let $\Gamma$ denote the operator $\pi \, \chi \, ({\cal N} \, S_t -
S_t \, {\cal N}_{0,1})$.  In the argument that follows we shall successively
decompose $\Gamma$ into pieces, peeling away the simplest terms until we
get to the main part.

	Let $\theta(y)$ be a smooth function on ${\bf C}$ which satisfies
$\theta \equiv 1$ on $M_1$ and ${\rm supp} \, \theta \subseteq M_2$.
Let us split $\Gamma$ into $\Gamma_1 + \Gamma_2$, where
$\Gamma_1 (f) = \Gamma (\theta f)$ and
$\Gamma_2 (f) = \Gamma ((1-\theta) f)$.  Let us check that
$\Gamma_2 : {\cal B}(E) \to {\cal B}(W)$ is trace class and that
$$
	\sup_{0 < t < s} \hbox{ trace norm } \Gamma_2 < \infty.	\eqno (3.78)
$$
The kernel of $\Gamma_2$ 
is just $K(x,y) \, (1 - \theta(y))$.  Lemma 3.23 implies that 
$\Gamma_2$ is trace class (since $K(x,y)$ is smooth)
and that (3.78) will follow if we can prove that
$$
	\sup_{0 < t < s} \, \sup_{y \in E} \, \int_W 
		|\overline{\partial}^2_x K(x,y) 
					\, (1 - \theta(y))| \, dx < \infty.
								\eqno (3.79)
$$
The formula (3.69) for $K(x,y)$ implies that
$$
	\overline{\partial}^2_x K(x,y) \, (1 - \theta(y))
		= \overline{\partial}^2_x 
		\{\chi(x) \, h(x) \,  {1 \over \psi(x) - y}
			\,\,\, \nu({\psi(x)-y \over t})\} \, (1 - \theta(y)),
								\eqno (3.80)
$$
because $h(\psi^{-1}(y))$ vanishes on the support of $1 - \theta$.
(Remember our convention from (3.67).)  The right side of (3.80) 
will be different from $0$ only when $x \in {\rm supp} \, h$ and
$y \notin M_1$.
For these $x$, $y$ we have that $|\psi(x)-y| > s$, because of (3.75). 
This means that $|\psi(x)-y| > t$ for the $t$'s 
that we are considering, and so 
$\nu({\psi(\cdot)-y \over t}) \equiv 1$ on a neighborhood of such an $x$,
by (3.18).  Thus we may ignore it in (3.80).  We conclude that (3.80)
remains bounded when $x \in {\rm supp} \, h$, $y \notin M_1$, 
and $t < s$, since 
$|\psi(x)-y| > s$ in this case, and this keeps us away from the singularity.  
This proves (3.79) and therefore (3.78).

	In order to finish the proof of Lemma 3.60, it suffices to
show that 
$$
\eqalign{
	& \hbox{we can decompose } \Gamma_1 : {\cal B}(E) \to {\cal B}(W) 
		\hbox{ into a sum of two operators}			  \cr
	& \hbox{for each } 0 < t < s, 	 
		\hbox{ where one is trace class with bounded trace norm}  \cr
	& \hbox{and the other has operator norm  } = O(t).  	  	  \cr
}
								\eqno (3.81)
$$
(Remember the reductions which preceded (3.66).)

	Let us split $\Gamma_1$ into two more pieces, corresponding
to $x$ near or far from ${\rm supp} \, h$.  
Let $\eta(x)$ be a smooth cut-off function on ${\bf C}$
which satisfies $\eta \equiv 1$ on a neighborhood of
$H_1$ and ${\rm supp} \, \eta \subseteq H_2$.
Define operators $\Gamma_{11}$ and $\Gamma_{12}$ by 
$$
	\Gamma_{11} (f) = \eta \Gamma_1(f) = \eta \Gamma(\theta f), 
		\quad \Gamma_{12} (f) = (1 - \eta) \, \Gamma_1(f) 
					= (1 - \eta) \, \Gamma(\theta f).
								\eqno (3.82)
$$
Thus $\Gamma_1 = \Gamma_{11} + \Gamma_{12}$.

	Let us check that $\Gamma_{12}$ defines a trace class operator
from ${\cal B}(E)$ into ${\cal B}(W)$ with trace norm which is
uniformly bounded in $t$, $0 < t < s$.  The kernel of
$\Gamma_{12}$ is just $(1 - \eta(x)) \, K(x,y) \, \theta(y)$.
This kernel is smooth, and so we can apply Lemma
3.23 to conclude that it is trace class.  To bound the trace norm it
suffices to show that
$$
	\sup_{0 < t < s} \, \sup_{y \in E} \, \int_{W}
	 |\, \overline{\partial}^2_x \{(1 - \eta(x)) \, K(x,y) \, \theta(y)\}| 
				\, dx < \infty.			\eqno (3.83)
$$
The $x$'s that are relevant for this integral all lie outside $H_1$
(since $\eta \equiv 1$ on a neighborhood of $H_1$), and in particular 
they all lie outside the support of $h$.  For these $x$'s we have that
$$
\eqalign{
	& \overline{\partial}^2_x \{(1 - \eta(x)) \, K(x,y) \, \theta(y)\} =
									\cr
	& \qquad\qquad
		- \overline{\partial}^2_x \{(1 - \eta(x)) \, \chi(x) \,\, 
		     { (\psi^{-1})'(y) \over x-\psi^{-1}(y)} 
			\, \nu({x- \psi^{-1}(y) \over t})
			\, h(\psi^{-1}(y)) \, \theta(y)\}.		\cr
}
								\eqno (3.84)
$$
This quantity vanishes unless $y \in \psi({\rm supp} \, h)$, because
of the $h(\psi^{-1}(y))$.  For these $x$'s and $y$'s -- i.e., $x \notin H_1$,
$y \in \psi({\rm supp} \, h)$ -- we have that $|x- \psi^{-1}(y)| > s$, 
because of (3.76).  Thus $|x- \psi^{-1}(y)| > t$
for the $t$'s that are relevant to (3.83), and so $\nu({x-
\psi^{-1}(y) \over t}) = 1$ for these $x$'s and $y$'s and on
neighborhoods of these $x$'s and $y$'s, because of (3.18).  This
means that we can ignore the $\nu$ part of (3.84).  What remains in
(3.84) is uniformly bounded for the relevant $x$'s and $y$'s, because
of our bound $|x- \psi^{-1}(y)| > s$ and the smoothness of the various
functions.  Therefore (3.83) holds, and we conclude that
$\Gamma_{12}$ defines a trace class operator with uniformly bounded
norm.

	We have now reduced Lemma 3.60 to the problem of showing that
$\Gamma_{11} : {\cal B}(E) \to {\cal B}(W)$ can be decomposed into a
sum of two operators for each $0 < t < s$, where one is trace class
with bounded trace norm and the other has operator norm which is
$O(t)$.

	The kernel associated to $\Gamma_{11}$ is given by $\eta(x) \,
K(x,y) \, \theta(y)$.  Define new kernels $J_i(x,y)$, $i=1, 2, 3$, by
$$
\eqalignno{
	& \quad J_1(x,y) = \, \eta(x) \, \chi(x) \, h(x)
			\, {1 \over \psi(x) - y}
		\{\nu({\psi(x)-y \over t}) - \nu({x- \psi^{-1}(y) \over t})\}
					 \, \theta(y),
								& (3.85)  \cr
	& \quad J_2(x,y) = \, \eta(x) \, \chi(x) \, h(x) \, 
	    \{{1 \over \psi(x) - y} - { (\psi^{-1})'(y) \over x-\psi^{-1}(y)}\}
		\, \nu({x- \psi^{-1}(y) \over t})  \, \theta(y),
								& (3.86) \cr
	& \quad J_3(x,y) =  \, \eta(x) \, \chi(x) \, 
			\{h(x) - h(\psi^{-1}(y)) \} \,
			{ (\psi^{-1})'(y) \over x-\psi^{-1}(y)}
			\, \nu({x- \psi^{-1}(y) \over t}) \, \theta(y).
								& (3.87) \cr
}
$$
We can pretend that $\psi^{-1}(y)$ is always defined in these formulae, 
because $\theta$ is supported in $M_2 \subseteq \psi(\Lambda)$.
One can compute directly from (3.69) that
$$
	\eta(x) \, K(x,y) \, \theta(y) = J_1(x,y) + J_2(x,y) + J_3(x,y).
								\eqno (3.88)
$$
These kernels $J_i(x,y)$ are all smooth and they all vanish when $x \in {\bf C}
\backslash W$, and therefore they define trace class operators
from ${\cal B}(E)$ into ${\cal B}(W)$, by Lemma 3.23.  
It remains to get estimates.
It turns out that the trace norms of the operators which correspond to
$J_2(x,y)$ and $J_3(x,y)$ are bounded, while $J_1(x,y)$ is our long-awaited
error term.

	Let us first check that the trace norm of the operator corresponding
to $J_2(x,y)$ is bounded.  In view of Lemma 3.23 it suffices to show that
$$
	\sup_{0 < t < s} \,\, \sup_{y \in E}
		\int_{W} | \, \overline{\partial}^2_x J_2(x,y)| \, dx
			< \infty.				\eqno (3.89)
$$
The main point for this term is that
$$
	{1 \over \psi(x) - y} - { (\psi^{-1})'(y) \over x-\psi^{-1}(y)}
								\eqno (3.90)
$$
is smooth (and even holomorphic) for $x \in \Lambda$ and $y \in \psi(\Lambda)$.
To see this it is a little more pleasant to set $y = \psi(z)$, so that
(3.90) becomes
$$
	{1 \over \psi(x) - \psi(z)} - { \,\,\,\psi'(z)^{-1} \over x-z}.
								\eqno (3.91)
$$
This uses also the identity $(\psi^{-1})'(\psi(z)) = \psi'(z)^{-1}$.
The smoothness (and holomorphicity) of (3.91) is a standard exercise.
(The poles cancel.)  For (3.89) we only care about $x$'s in 
${\rm supp} \, \eta$, which is contained in the compact subset $H_2$ of 
$\Lambda$, and we only care about the $y$'s in ${\rm supp} \, \theta$, 
which is a compact subset of $\psi(\Lambda)$.  Thus (3.89) only involves
(3.90) on a compact subset of its domain.  We conclude that we can
write $J_2(x,y)$ as
$$
	J_2(x,y) = L(x,y) \, \nu({x- \psi^{-1}(y) \over t}),	\eqno (3.92)
$$
where $L(x,y)$ is smooth and has compact support, and where $L(x,y)$
also does not depend on $t$.  It is not hard to verify (3.89) using
(3.92), the Leibniz rule, the boundedness of $\nu$, and the estimates
(3.43) and (3.44).  Thus the operator that corresponds to $J_2(x,y)$
has bounded trace norm.

	To check that the trace norm of the operator that
corresponds to $J_3(x,y)$ is bounded it suffices to show that
$$
	\sup_{0 < t < s} \,\, \sup_{y \in E}
		\int_{W} | \, \overline{\partial}^2_x J_3(x,y)| \, dx
			< \infty,				\eqno (3.93)
$$
because of Lemma 3.23.  We can rewrite this as
$$
	\sup_{0 < t < s} \,\, \sup_{y \in E \cap M_2}
		\int_{W} | \, \overline{\partial}^2_x J_3(x,y)| \, dx
			< \infty,				\eqno (3.94)
$$
since ${\rm supp} \, \theta \subseteq M_2$.  Set
$$
	N(x,z) = \, \eta(x) \, \chi(x) \, \{h(x) - h(z) \} \,
		    { 1 \over x- z} \, \nu({x- z \over t}) \, \theta(\psi(z)).
								\eqno (3.95)
$$
In order to prove (3.94) it suffices to show that
$$
	\sup_{0 < t < s} \,\, \sup_{z \in H_2}
		\int_{W} | \, \overline{\partial}^2_x N(x,z)| \, dx
			< \infty.				\eqno (3.96)
$$
Let us check this.  In passing from $J_3(x,y)$ to $N(x,z)$ we have
made two changes.  The first is that we dropped the $(\psi^{-1})'(y)$,
which does not matter because it is bounded and pulls through the
$\overline{\partial}^2_x$ (since it does not depend on $x$).  The
second change was to replace $y$ with $\psi(z)$.  This is again
trivial from the perspective of $x$ and $\overline{\partial}^2_x$, and
we accomodated it in (3.96) by taking the supremum over $z \in H_2$
instead of $y \in E \cap M_2$.  (Remember from (3.74) that $M_2 =
\psi(H_2)$.)  Thus (3.96) will imply (3.94).  To prove
(3.96) one can use exactly the same argument as used to prove
(3.50).  In fact (3.96) is practically the same as (3.50), except
that we have different functions now.  The present situation is a
little simpler, because $\eta$, $\chi$, and $h$ are smooth.  At any
rate we get (3.96) and hence (3.93), and we conclude that the
operator which corresponds to $J_3(x,y)$ has bounded trace norm.

	It remains to deal with the operator $J_1$ given by
$$
	J_1 f(x) = \int_{\bf C} J_1(x,y) \, f(y) \, dy.		\eqno (3.97)
$$
This will give us the error term $U_t$ described in Lemma 3.60, and
we want to show that it has norm $= O(t)$ as an operator from
${\cal B}(E)$ into ${\cal B}(W)$.  Let us begin by rewriting (3.85)
as
$$
	J_1(x,y) = \, \alpha(x)
			\, {1 \over \psi(x) - y}
		\{\nu({\psi(x)-y \over t}) - \nu({x- \psi^{-1}(y) \over t})\}
					 \, \theta(y),
								\eqno (3.98)
$$
where $\alpha(x) = \eta(x) \, \chi(x) \, h(x)$.  Thus $\alpha(x)$ is a
smooth function with ${\rm supp} \, \alpha \subseteq W \cap {\rm supp} \, h$.

	Set $\mu = \nu - 1$, where $\nu$ is as in (3.18).  Thus $\mu$ is
a $C^\infty$ function on ${\bf C}$ such that
$$
	\mu(x) = 0 \,\hbox{ when }\, |x| \ge 1 \quad\hbox{ and }\quad
			\mu(x) = 1 \,\hbox{ when }\, |x| \le {1 \over 2}.
								\eqno (3.99)
$$
We can rewrite (3.98) as
$$
	J_1(x,y) = \, \alpha(x)
			\, {1 \over \psi(x) - y}
		\{\mu({\psi(x)-y \over t}) - \mu({x- \psi^{-1}(y) \over t})\}
					 \, \theta(y).
								\eqno (3.100)
$$
This is more convenient for localizing.

	Let us now split the two $\mu$ terms apart.  Set
$$
\eqalignno{
	& J_{11}(x,y) = \, \alpha(x)
			\, {1 \over \psi(x) - y} \,
		\mu({\psi(x)-y \over t})
					 \, \theta(y)		& (3.101) \cr
	& J_{12}(x,y) = \, \alpha(x)
			\, {1 \over \psi(x) - y} \,
		\mu({x- \psi^{-1}(y) \over t})
					 \, \theta(y),		& (3.102) \cr
}
$$
and let $J_{11}$ and $J_{12}$ denote the corresponding operators, as in
(3.97), so that $J_1 = J_{11} - J_{12}$.

\medskip
\noindent
{\bf Sublemma 3.103.}  {\it $J_{11} : {\cal B}(E) \to {\cal B}(W)$
has operator norm $\, = O(t)$.}

\medskip

	To prove this we want to first peel off the $\psi$ and 
the $\alpha$ into a
separate operator.  Define a transfer operator ${\cal L}$ by
$$
	{\cal L} f(x) = \alpha(x) \, f(\psi(x)).		\eqno (3.104)
$$
Remember that ${\rm supp} \, \alpha \subseteq {\rm supp} \, h$ is a compact
subset of the domain $\Lambda$ of $\psi$, so that (3.104) makes sense.
Set $b_t(w) = \, {1 \over w} \, \mu({w \over t})$,
and define the multiplication operator $\Theta$ and the convolution
operator $B_t$ by $\Theta (f) = \theta f$ and
$B_t f = b_t * f$.  We have that
$$
	J_{11} = {\cal L} \circ B_t \circ \Theta,		\eqno (3.105)
$$
as one can compute from (3.101).

	Set $E' = \{u \in {\bf C} : {\rm dist} (u,E) \le  s\}$.
Note that
$$
	\hbox{supp } B_t f \subseteq E' \quad\hbox{ when }\quad
		\hbox{supp } f \subseteq E.			\eqno (3.106)
$$
This follows
from (3.99) and the fact that we are restricting ourselves to $t < s$
(since (3.77)).  The main point now is that $B_t$ maps ${\cal B}(E)$ into
${\cal B}(E')$ with norm $\le \|b_t\|_1$.  Indeed, as in Lemma 3.12,
we have that 
$\overline{\partial}^2 (b_t * f) = b_t * (\overline{\partial}^2 f)$,
and so our assertion about $B_t : {\cal B}(E) \to {\cal B}(E')$ reduces
to (3.106) and the standard fact that the convolution of an $L^1$ function
with a finite measure is an $L^1$ function, with the norm of the result
being less than or equal to the product of the norms of the input.
On the other hand we have that
$$
	\|b_t\|_1 = t \, \|b_1\|_1,				\eqno (3.107)
$$
by an easy scaling argument, and $\|b_1\|_1 < \infty$ by direct computation.
Thus we conclude that $B_t : {\cal B}(E) \to {\cal B}(E')$ has norm
$\, = O(t)$.

	To finish the proof of Sublemma 3.103 it suffices to observe
that $\Theta : {\cal B}(E) \to {\cal B}(E)$ is bounded, because of
Lemma 3.9, and that ${\cal L} : {\cal B}(E') \to {\cal B}(W)$ is bounded,
because of Lemma 3.14 (since ${\rm supp} \, \alpha \subseteq W$).

	This completes the proof of Sublemma 3.103.

	It remains to show that $J_{12}$ has norm $\, = O(t)$ as an
operator from ${\cal B}(E)$ to ${\cal B}(W)$.  The argument will be similar
in spirit to the proof of Sublemma 3.103, but the details will be messier
because the formulae are less convenient.

	Let us first work towards peeling off the $\alpha$ from $J_{12}$.
Define $Q(x,y)$ by
$$
	Q(x,y) = \, {1 \over \psi(x) - y} \,
		\mu({x- \psi^{-1}(y) \over t})
					 \, \theta(y)		\eqno (3.108)
$$
when $(x,y) \in \Lambda \times \psi(\Lambda)$, $Q(x,y) = 0$ otherwise.
Remember that $\theta(y) \ne 0$ implies that $y \in M_2$, a compact subset
of $\psi(\Lambda)$, and notice that $y \in M_2 = \psi(H_2)$ and
$\mu({x- \psi^{-1}(y) \over t}) \ne 0$ imply that $|x - \psi^{-1}(y)| < t$
and hence $x \in H_3$.  (See (3.74), and remember that $t < s < r$.)   
Thus $Q(x,y)$ is smooth, 
since $H_3$ is a compact subset of $\Lambda$.  For the record, we have that
$$
	Q(x,y) \ne 0 \quad \hbox{ implies }
			y \in M_2 \hbox{ and } x \in H_3.	\eqno (3.109)
$$

	Define an operator $Q$ by
$$
	Q f(x) = \int_{\bf C} Q(x,y) \, f(y) \, dy.		\eqno (3.110)
$$
Thus ${\rm supp} \, Qf \subseteq H_3$ for all $f$, by (3.109).  Since
multiplication by $\alpha$ defines a bounded operator from 
${\cal B}(H_3)$ into ${\cal B}(W)$, by Lemma 3.9 (and the fact that
${\rm supp} \, \alpha \subseteq W$), we are reduced to showing that
$Q : {\cal B}(E) \to {\cal B}(H_3)$ has operator norm $\, = O(t)$.
Since we know ${\rm supp} \, Qf \subseteq H_3$ already, it suffices to
show that
$$
	\overline{\partial}^2 \circ Q : {\cal B}(E) \to L^1(H_3)
		\quad\hbox{ has operator norm } = O(t).		\eqno (3.111)
$$

	We would like to move the $\psi^{-1}$ from inside $Q(x,y)$,
essentially by making the change of variables $y = \psi(z)$.
Define $P(x,z)$ by
$$
	P(x,z) = \, {1 \over \psi(x) - \psi(z)} \,
		\mu({x- z \over t})
		\, \theta(\psi(z)) \, |\psi'(z)|^2		\eqno (3.112)
$$
when $(x,z) \in \Lambda \times \Lambda$, $P(x,z) = 0$ otherwise.
Thus $P(x,z) = Q(x,\psi(z)) \, |\psi'(z)|^2$, modulo technicalities.
Again this is a smooth function, and we have that
$$
	P(x,z) \ne 0 \quad \hbox{ implies that }
			z \in H_2 \hbox{ and } x \in H_3,	\eqno (3.113)
$$
by (3.109).  We can rewrite (3.110) as
$$
	Q f(x) = \int_{\bf C} P(x,z) \, f(\psi(z)) \, dz.	\eqno (3.114)
$$

	For the purpose of computing distributional
$\overline{\partial}$ derivatives the
${1 \over \psi(x) - \psi(z)}$ singularity in $P(x,z)$ and its counterpart
in $Q(x,y)$ are slightly annoying.  These singularities are integrable (see
also (3.124) and (3.125) below), 
and so there is no problem with using these kernels to define our operators,
but we should be slightly careful with differentiating them.  To avoid
any problem let us approximate them by more regular kernels.  Given
$\epsilon > 0$, $\epsilon < t$, set
$$
	P_\epsilon (x,z) = \, {1 \over \psi(x) - \psi(z)} \,
		\{\mu({x- z \over t}) - \mu({x- z \over \epsilon})\}
		\, \theta(\psi(z)) \, |\psi'(z)|^2		\eqno (3.115)
$$
when $(x,z) \in \Lambda \times \Lambda$, $P_\epsilon(x,z) = 0$ otherwise.
The point here is that $P_\epsilon (x,z) = 0$ when $|x-z| < \epsilon/2$,
because of (3.99), and so we have killed the singularity.  Define
operators $Q_\epsilon$ by
$$
	Q_\epsilon f(x) = \int_{\bf C} P_\epsilon(x,z) \, f(\psi(z)) \, dz.
								\eqno (3.116)
$$
We have that
$$
	Q_\epsilon f \to Q f \quad \hbox{ as } \epsilon \to 0
								\eqno (3.117)
$$
in the $L^\infty$ norm for all bounded functions $f$, for instance.
(This uses the integrability of the singularity in $P(x,y)$.)

	Let us compute $\overline{\partial}^2 \circ Q_\epsilon$.  We have that
$$
	\overline{\partial}^2_x P_\epsilon(x,z) = \overline{\partial}^2_z 
		\bigl({1 \over \psi(x) - \psi(z)} \,
		\{ \mu({x- z \over t}) - \mu({x- z \over \epsilon})\}\bigr)
			\, \theta(\psi(z)) \, |\psi'(z)|^2	\eqno (3.118)
$$
on $\Lambda \times \Lambda$. Notice that we are converting $x$
derivatives into $z$ derivatives. We have used here
the fact that ${1 \over \psi(x) - \psi(z)}$ is holomorphic away from the
singularity. Therefore
$$
\eqalign{
	& \overline{\partial}^2_x Q_\epsilon f(x) =			\cr
	& \qquad  \int_{\bf C} {1 \over \psi(x) - \psi(z)} \,
		\{ \mu({x- z \over t}) - \mu({x- z \over \epsilon})\} \,
     \overline{\partial}^2_z \{\theta(\psi(z)) \, |\psi'(z)|^2 \, f(\psi(z)) \}
				\, dz					\cr
}
								\eqno (3.119)
$$
for $x \in \Lambda$ and $f \in {\cal B}(E)$.  Let us be careful about what
this last $\overline{\partial}^2_z$ expression really means.  Define a
transfer operator ${\cal T}$ by
$$
	{\cal T} f(z) = \theta(\psi(z)) \, |\psi'(z)|^2 \, f(\psi(z)).
								\eqno (3.120)
$$
More precisely, we view $\theta(\psi(z))$ here as a smooth function defined
on all of ${\bf C}$, and not just the domain $\Lambda$ of $\psi$, by
setting it to be $0$ on ${\bf C} \backslash \Lambda$.  This is reasonable,
because ${\rm supp} \, \theta \subseteq M_2$ is a compact subset of 
$\psi(\Lambda)$, so that $\theta(\psi(z))$ vanishes off of the compact
subset $H_2$ of $\Lambda$.  With this interpretation, and similar
remarks for $|\psi'(z)|^2$, ${\cal T}$ is a transfer operator in the
sense of (3.13).  Thus ${\cal T}$ maps ${\cal B}(E)$ boundedly into
${\cal B}(H_2)$, by Lemma 3.14.  In particular, $\overline{\partial}^2
{\cal T}f$ is defined as a finite measure when $f \in {\cal B}(E)$, and
so we can rewrite (3.119) as
$$
	\overline{\partial}^2_x Q_\epsilon f(x) =	
		\int_{\bf C} {1 \over \psi(x) - \psi(z)} \,
		\{ \mu({x- z \over t}) - \mu({x- z \over \epsilon})\} \,
			\overline{\partial}^2_z ({\cal T}f)(z) \, dz.
								\eqno (3.121)
$$
Here $\overline{\partial}^2_z ({\cal T}f)(z) \, dz$ denotes the measure
$\overline{\partial}^2 {\cal T}f$, which may or may not be absolutely
continuous.  Note that this measure is supported in $H_2$.

	Before we take the limit as $\epsilon \to 0$ we need to record
an estimate.

\medskip
\noindent
{\bf Sublemma 3.122.}  {\it Suppose that $\sigma$ is a nonnegative measure
on $H_2$, and consider the function of $x$ defined on $\Lambda$ by
$$
	\int {1 \over |\psi(x) - \psi(z)|} \, |\mu({x- z \over t})| \,
					d\sigma(z).		\eqno (3.123)
$$
If $t < s$ (as usual), then this is an integrable function supported on
$H_3$ whose $L^1$ norm is bounded by a constant times $t \, \sigma (H_2)$.}

\medskip

	The condition on the support comes from (3.99) and (3.74), as usual.
For the $L^1$ bound we observe first that there
is a constant $C > 0$ such that
$$
	|\psi(x) - \psi(z)| \ge C^{-1} \, |x-z| 
		\quad \hbox{ when } x \in H_3, \, z \in H_2.	\eqno (3.124)
$$
Once we have this inequality the $L^1$ bound follows from Fubini's theorem
and the fact that
$$
	\int {1 \over |w|} \, |\mu({w \over t})| \, dw 
					= \hbox{ constant} \cdot t.
								\eqno (3.125)
$$
(Remember (3.99).)  This proves Sublemma 3.122.

	Let us check now that
$$
	\overline{\partial}^2_x Q f(x) =	
		\int_{\bf C} {1 \over \psi(x) - \psi(z)} \,
		   \mu({x- z \over t}) \,
			\overline{\partial}^2_z ({\cal T}f)(z) \, dz
								\eqno (3.126)
$$
on $\Lambda$ when $f \in {\cal B}(E)$.
The point is simply to send $\epsilon \to 0$ in (3.121).  
Sublemma 3.122 (applied with $t$ replaced with $\epsilon$) implies
that the right side of (3.121) converges to the right side of
(3.126) in $L^1(\Lambda)$.  On the other hand (3.117) implies
that the left side of (3.121) converges to the left side of
(3.126) on $\Lambda$ in the sense of distributions.  This
proves (3.126).

	Next we want to check that
$$
	\int_{\bf C} |\overline{\partial}^2_x Q f(x)| \, dx
		\le C \, t \, \|f\|_{\cal B}			\eqno (3.127)
$$
for some constant $C$ when $f \in {\cal B}(E)$.  We already know that
${\rm supp} \, Qf \subseteq H_3$, and that $H_3$ is a compact subset
of $\Lambda$, and so it is enough to consider the integral over $\Lambda$.
Thus we can use (3.126).  Sublemma 3.122 implies that the left
side of (3.127) is bounded by a constant times $t$ times the total
variation of the measure $\overline{\partial}^2_z ({\cal T}f)$.
The total variation of $\overline{\partial}^2_z ({\cal T}f)$ is bounded
by a constant times $\|f\|_{\cal B}$, since the transfer operator
${\cal T}$ maps ${\cal B}(E)$ boundedly into
${\cal B}(H_2)$, by Lemma 3.14.  This yields (3.127).

	This proves (3.111), which implies in turn that $J_{12}$ maps
${\cal B}(E)$ into ${\cal B}(W)$ with norm $\, = O(t)$.  This was the
last step in the proof of Lemma 3.60, and so the proof of Lemma 3.60
is now complete.

\medskip
\noindent
{\bf Lemma 3.128.} {\it Let $E, H$, and $W$ be compact subsets of
${\bf C}$, and let $g$ and $h$ be functions on ${\bf C}$ such that $g$
satisfies (3.10), $h$ satisfies (3.46), ${\rm supp} \, g \subseteq
W$, and ${\rm supp} \, h \subseteq H$.  Then the operator $g S_t h S_t
: L^\infty(E) \to {\cal B}(W)$ is trace class for each $t \in (0,1]$,
where we identify $g$ and $h$ with their corresponding multiplication
operators, and
$$
	\sup_{0 < t \le 1} \hbox{\rm trace norm } g S_t h S_t < \infty.
								\eqno (3.129)
$$
}

\medskip

	To prove this we begin with some preliminary reductions.  We
may as well assume that $g$ is smooth, since otherwise we can view $g
S_t h S_t$ as the composition of the operator of multiplication by $g$
and $g_0 S_t h S_t$, where $g_0$ is some smooth function such that
$g_0 \equiv 1$ on ${\rm supp} \, g$ and ${\rm supp} \, g_0 \subseteq
W_0$ for some compact set $W_0 \supseteq W$.  If we can prove the
lemma for smooth $g$'s, then we can prove it for $g_0 S_t h S_t$ (with
$W$ replaced by $W_0$), and then we can use the fact that
multiplication by $g$ defines a bounded operator from ${\cal B}(W_0)$
into ${\cal B}(W)$ (by Lemma 3.9) to get back to $g S_t h S_t$.  Thus
we may assume that $g$ is smooth.

	Set $F_i = \{x \in {\bf C} : {\rm dist} (x, E \cup W) \le i\}$
$i = 1,2$.
We may assume that $h$ is a smooth function which satisfies 
$$
	h \equiv 1 \quad\hbox{ on } F_2.  			\eqno (3.130)
$$
Indeed, if not, let $H_0$ be a compact
subset of ${\bf C}$ which contains $F_2$ and ${\rm supp} \, h$ in
its interior, and let $h_0$ be a smooth function which satisfies $h_0
\equiv 1$ on $F_2 \cup {\rm supp} \, h$ and ${\rm supp} \, h_0
\subseteq H_0$.  Then
$$
	g S_t h S_t = g S_t h \, h_0 S_t 
			= g \, [S_t, h] h_0 S_t + h \, g S_t h_0 S_t.
								\eqno (3.131)
$$
The operator $g \, [S_t, h] h_0 S_t$ is trace class, with 
$$
	\sup_{0 < t \le 1} \hbox{ trace norm } g \, [S_t, h] h_0 S_t < \infty,
								\eqno (3.132)
$$
because of Lemmas 3.45 and 3.21.  If we can prove the lemma for
$g S_t h_0 S_t$ (with $H$ replaced by $H_0$), then we shall know that
$h \, g S_t h_0 S_t$ is trace class with bounded norm, since multiplication
by $h$ defines a bounded operator on ${\cal B}(W)$.  This would then imply
our original $g S_t h S_t$ is trace class with bounded norm.  

	Thus we may assume that $h$ satisfies (3.130), and that $g$
and $h$ are both smooth.

	Next, let us check that $g (S-S_t) h S_t : L^\infty(E) \to
{\cal B}(W)$ is trace class with bounded norm.  We know from Lemma
3.33 that $h S_t : L^\infty(E) \to {\cal B}(H)$ is trace class with
norm $\, = O(t^{-1})$.  We also know from Lemma 3.21 that $g (S-S_t)
: {\cal B}(H) \to {\cal B}(W)$ is a bounded operator with norm $\, =
O(t)$.  Therefore $g (S-S_t) h S_t$ is trace class with bounded norm.

	Thus we are reduced to showing that $g S h S_t : L^\infty(E)
\to {\cal B}(W)$ is a trace class operator with bounded norm.

	We need to put $S$ in a more convenient form.  Define a convolution
operator $T$ by 
$$
	T f(x) = \int_{\bf C} {1 \over \pi}
		 \, {\overline{x} - \overline{y} \over x-y} \, f(y) \, dy.
								\eqno (3.133)
$$
This makes sense for integrable functions with compact support, for instance.
We have that
$$
	S = T \circ \overline{\partial}				\eqno (3.134)
$$
as operators acting on, say, smooth functions with compact support.  
This is not hard to check, using the fact that
$$
	\overline{\partial}({\overline{x} \over x}) = {1 \over x}
								\eqno (3.135)
$$
in the sense of distributions.  

	Since $h S_t$ maps $L^\infty(E)$ into smooth functions with
compact support, we have that 
$$
	g S h S_t (f) = g T \, \overline{\partial} (h S_t(f))
			= g T (\overline{\partial} h) S_t(f)
				+ g T h (\overline{\partial} \circ S_t) (f)
								\eqno (3.136)
$$
when $f \in L^\infty(E)$.  Here $\overline{\partial} h$ means multiplication
by the function $\overline{\partial} h$.


\medskip
\noindent
{\bf Sublemma 3.137.} {\it The operator $g T (\overline{\partial} h)
S_t : L^\infty(E) \to {\cal B}(W)$ is trace class with bounded trace norm.}

\medskip

	It suffices to show that $(\overline{\partial} h) S_t :
L^\infty(E) \to {\cal B}(H)$ is trace class with bounded norm, 
since $g T : {\cal B}(H) \to {\cal B}(W)$ is bounded, by Lemma 3.12.

	We can write $(\overline{\partial} h) S_t$ as
$$
	((\overline{\partial} h) S_t)(f) (x) 
		= \int_E J(x,y) \, f(y) \, dy,		\eqno (3.138)
$$
where 
$$
	J(x,y) = \overline{\partial} h(x) \, {1 \over \pi} \, 
			{ 1 \over x-y} \, \nu({x-y \over t}).
							\eqno (3.139)
$$
(See (3.19).)  This is a smooth function of $x$ and $y$, and $J(x,y) = 0$
when $x \notin H$, since ${\rm supp} \, h \subseteq H$.  
Thus $y \mapsto J(\cdot, y)$ defines a continuous
mapping from $E$ into ${\cal B}(H)$, and Lemma 3.23 implies that 
$(\overline{\partial} h) S_t : L^\infty(E) \to {\cal B}(H)$ is trace
class and that its trace norm is bounded by
$$
	|E| \cdot \sup_{y \in E} \|J(\cdot, y)\|_{\cal B} = 
		|E| \cdot \sup_{y \in E} \int_{\bf C} 
		|\overline{\partial}^2_x J(x,y)| \, dx < \infty.
							\eqno (3.140)
$$

	Let us check that
$$
	J(x,y) = \overline{\partial} h(x) \, {1 \over \pi} \, { 1 \over x-y}
							\eqno (3.141)
$$
when $t \le 1$ and $y \in E$, i.e., that we can drop the $\nu$.
This is trivial when $x \notin {\rm supp} \, \overline{\partial} h$,
since both sides of (3.141) then vanish.
If $y \in E$ and $x \in {\rm supp} \, \overline{\partial} h$, then
$|x-y| \ge 2$ because of (3.130).  This implies that $|x-y| \ge t$ and
hence that $\nu({x-y \over t}) = 1$ (by (3.18)), which gives (3.141) in this
case too.  Thus (3.141) holds when $t \le 1$ and $y \in E$.

	Using (3.141) we see that the (3.140) does not depend on $t$
when $t \in (0,1]$.  It is finite, since $J(x,y)$ is smooth, and so we
conclude that it is bounded for $t$ in this range.  This proves
Sublemma 3.137.
		
	Next we consider $g T h (\overline{\partial} \circ S_t)$.  

\medskip
\noindent
{\bf Sublemma 3.142.} {\it  $g T h (\overline{\partial} \circ S_t)(f) =
g T (\overline{\partial} \circ S_t)(f)$ when $f \in L^\infty(E)$
and $t \in (0,1]$.}

\medskip

	Set $E_t = \{x \in {\bf C}: {\rm dist}(x,E) \le t\}$.  Let us check
that 
$$
	\hbox{supp } (\overline{\partial} \circ S_t)(f) \subseteq E_t
		\quad\hbox{ when } f \in L^\infty(E).		\eqno (3.143)
$$
Indeed,
$$
	(\overline{\partial} \circ S_t) f(x) =
		\int_{\bf C} {1 \over \pi} \, 
		{ 1 \over x-y} \, (\overline{\partial} \nu)({x-y \over t})
			\,\, t^{-1} \, f(y) \, dy,		\eqno (3.144)
$$
by (3.19) and calculus.  We also have that
$$
	\overline{\partial} \nu (u) = 0 
		\qquad\hbox{ when } |u| \ge 1 \hbox{ or } |u| \le {1 \over 2},
								\eqno (3.145)
$$
because of (3.18).  This and (3.144) imply (3.143).

	On the other hand, $h \equiv 1$ on $E_t$ when $t \in (0,1]$, because
of (3.130).  This implies Sublemma 3.142.

	To finish the proof of Lemma 3.128 it remains to show that $g
T (\overline{\partial} \circ S_t) : L^\infty(E) \to {\cal B}(W)$ is
trace class with bounded trace norm.

	Define functions $k(x)$ and $b_t(x)$ on ${\bf C}$ by
$$
	k(x) = {1 \over \pi} \, {\overline{x} \over x} \quad\hbox{ and } \quad
		b_t(x) = {1 \over \pi} \, t^{-1} \, { 1 \over x} 
			\, (\overline{\partial} \nu)({x \over t}).
								\eqno (3.146)
$$
Thus $T$ and $\overline{\partial} \circ S_t$ are given by $Tf = k*f$
and $(\overline{\partial} \circ S_t)f = b_t * f$.  We also have that
$$
	b_t(x) = 0 \quad \hbox{ when } |x| \ge t 
		\qquad\hbox{ and } \qquad \|b\|_\infty \le C \, t^{-2}	
								\eqno (3.147)
$$
for some constant $C$.  These facts follow from (3.145) and the observation
that $\overline{\partial} \nu$ is bounded.

	Given $f \in L^\infty(E)$ we have that
$T (\overline{\partial} \circ S_t)(f) = k * b_t * f$, so that
$$
	(g T (\overline{\partial} \circ S_t))(f)(x) =
		\int_E g(x) \, (k * b_t)(x-y) \, f(y) \, dy.	\eqno (3.148)
$$
Note that $b_t$ and hence $k * b_t$ are smooth.  Remember also that
${\rm supp} \, g \subseteq W$.  We can use Lemma 3.23 to conclude that
$g T (\overline{\partial} \circ S_t) : L^\infty(E) \to {\cal B}(W)$ is
trace class with 
$$
\eqalign{
	\hbox{trace norm } g T (\overline{\partial} \circ S_t)
   & \le |E| \cdot \sup_{y \in E} \|g(\cdot) \, (k * b_t)(\cdot-y)\|_{\cal B}
									\cr
   & \le |E| \cdot \sup_{y \in E} 
	    \int_{\bf C} |\overline{\partial}^2_x \{g(x) \, (k * b_t)(x-y)\}|
						\, dx			\cr
   & = |E| \cdot \sup_{y \in E} 
		\int_W |\overline{\partial}^2_x \{g(x) \, (k * b_t)(x-y)\}|
						\, dx.			\cr
}
								\eqno (3.149)
$$
In this last equality we have used the fact that $g$ is supported in $W$.

	To estimate the integral in (3.149) we use the formula
$$
\eqalign{
	\overline{\partial}^2_x \{g(x) \, (k * b_t)(x-y) \}
		& = (\overline{\partial}^2 g)(x) \, (k * b_t)(x-y)
	+ 2 (\overline{\partial} g)(x) \, \overline{\partial}_x(k * b_t(x-y))
									\cr
	& \qquad\qquad\qquad + g(x) \, \overline{\partial}^2_x(k * b_t(x-y))
									\cr
		& = (\overline{\partial}^2 g)(x) \, (k * b_t)(x-y)
	+ 2 (\overline{\partial} g)(x) \, ((\overline{\partial} k) * b_t(x-y))
									\cr
	& \qquad\qquad\qquad + g(x) \, ((\overline{\partial}^2 k) * b_t(x-y)).
									\cr
}
								\eqno (3.150)
$$
These derivatives of $k$ should be taken in the sense of distributions, 
and then the convolution with $b_t$ makes everything smooth again,
since $b_t$ is smooth.

	Inserting the right side of (3.150) into the integral in (3.149)
we get three terms to estimate.  For the first we have that
$$
\eqalign{
	\int_W |(\overline{\partial}^2 g)(x) \, (k * b_t)(x-y)| \, dx
      & \le \|\overline{\partial}^2 g \|_\infty \, \|k * b_t\|_\infty \,\, |W|
									\cr
	& \le \|\overline{\partial}^2 g \|_\infty 
			\, \|k\|_\infty \, \|b_t\|_1 \,\, |W|.		\cr
}
								\eqno (3.151)
$$
Of course $\overline{\partial}^2 g$ is bounded (since $g$ is smooth)
and $k$ is bounded (by inspection).  We also have that $\|b_t\|_1$ is bounded,
uniformly in $t$, as one can check from (3.147).  Thus the contribution
of this term to the integral in (3.149) is bounded.

	For the second term on the right side of (3.150) we begin with the
observation that
$$
	\overline{\partial} k (x) = {1 \over \pi} \, {1 \over x}.
								\eqno (3.152)
$$
This is easy to check.  For the relevant integral we have that
$$
\eqalign{
	\int_W |(\overline{\partial} g)(x) \, 
		((\overline{\partial} k) * b_t(x-y))| \, dx
	& \le \|\overline{\partial} g\|_\infty \int_W
		|(\overline{\partial} k) * b_t(x-y)| \, dx		\cr
	& \le \|\overline{\partial} g\|_\infty \int_W \int_{\bf C}
		{1 \over \pi} \, {1 \over |u|} \, |b_t(x-y-u)| \, du dx.  \cr
}
								\eqno (3.153)
$$
Remember that $|x-y-u| \le t$ when $|b_t(x-y-u)| \ne 0$, because of
(3.147).  Let $Z$ be a compact set which is large enough so that
$u \in Z$ whenever $|x-y-u| \le 1$ for some $x \in W$ and $y \in E$.
Using Fubini's theorem we get that
$$
	\int_W \int_{\bf C} {1 \over |u|} \, |b_t(x-y-u)| \, du dx
		\le \|b_t\|_1 \int_Z {1 \over |u|} \, du.	\eqno (3.154)
$$
We have already seen that $\|b_t\|_1$ is bounded, uniformly in $t$, 
and this last integral over $Z$ is
also finite.  Thus we conclude that the contribution of the second term
on the right side of (3.150) to the right side of (3.149) is also
bounded, uniformly in $t$.

	The last term on the right side of (3.150) is 
$g(x) \, ((\overline{\partial}^2 k) * b_t(x-y))$.  For this we use
the fact that $\overline{\partial}^2 k$ is the Dirac delta function
at the origin, as in (3.4).  Thus 
$$
	g(x) \, ((\overline{\partial}^2 k) * b_t(x-y))
		= g(x) \, b_t(x-y).				\eqno (3.155)
$$
The relevant integral for (3.149) reduces to
$$
	\int_W |g(x) \, b_t(x-y)| \, dx 
		\le \|g\|_\infty \|b_t\|_1.			\eqno (3.156)
$$
Again this is bounded, uniformly in $t$.

	Thus we conclude from (3.149) and these estimates that
$$
	\sup_{0 < t \le 1} 
	\hbox{trace norm } g T (\overline{\partial} \circ S_t)
		< \infty.					\eqno (3.157)
$$
This completes the proof of Lemma 3.128, because of (3.136) and
Sublemmas 3.142 and 3.137.

	Now we come to the main result of this section.  For this we need to
first set some notation and assumptions.

	Let ${\cal M}$ be a transfer operator as in (3.13).  We
assume that $g_\omega, \psi_\omega$, and $\Lambda_\omega$ satisfy the
same conditions as in the paragraph just after (3.13).  Fix a compact
set $B \subseteq {\bf C}$, and assume that $\Lambda_\omega,
\psi_\omega(\Lambda_\omega) \subseteq B$ for each $\omega$.
We shall make the standing assumption that 
$$
	g_\omega \hbox{ and } \overline{\partial} g_\omega 
		\hbox{ satisfy (3.46) for each } \omega \in \Omega.
								\eqno (3.158)
$$
This ensures that ${\cal
M}$ defines a bounded operator on ${\cal B}(B)$, by Lemma 3.14.

	Fix a compact set $K$ contained in the interior of $B$ such that
$$
	K \supseteq \bigcup_{\omega \in \Omega} 
	   \{\hbox{supp } g_\omega \cup \psi_\omega(\hbox{supp } g_\omega)\}.
								\eqno (3.159)
$$
Let $K_1$ be another compact subset of the interior of $B$ which contains
$K$ in its interior, and fix a smooth cut-off function $\eta$ on ${\bf C}$
such that
$$
	\hbox{supp } \eta \subseteq K_1 \qquad \hbox{ and } \qquad
		\eta \equiv 1 \quad\hbox{ on a neighborhood of } K.
								\eqno (3.160)
$$
Note that
$$
\eqalignno{
	& \quad {\cal M}(f) \equiv 0 \qquad\hbox{ on } B \backslash K
								& (3.161) \cr
	& {\cal M} (\eta \, f) = \eta \, {\cal M}(f) = {\cal M}(f)
								& (3.162) \cr
}
$$
for all functions $f$.
Notice also that multiplication by $\eta$ defines a bounded operator on
${\cal B}(B)$, by Lemma 3.9.

	Fix a complex number $z$, our spectral parameter.  From now on we
shall assume that
$$
	I - z {\cal M} \quad\hbox{ is invertible on } {\cal B}(B).
								\eqno (3.163)
$$
We can automatically extend $(I - z {\cal M})^{-1}$ to the larger space
$$
	\{ f \in L^\infty({\bf C}): \eta \, f \in {\cal B}(B)\}
								\eqno (3.164)
$$
by the formula
$$
	(I - z {\cal M})^{-1} f 
		= (1-\eta) f + (I - z {\cal M})^{-1} (\eta \, f).
								\eqno (3.165)
$$
This is consistent with the original definition of $(I - z {\cal M})^{-1}$
on ${\cal B}(B)$ because of (3.162).  Note that the image of
${\cal B}(B)$ under $S$ or an $S_t$ is always contained in (3.164),
because of Lemma 3.21.

	Define the transfer operator ${\cal N}$ by
$$
	{\cal N} \Phi (x) = 
	     \sum_{\omega \in \Omega} (\overline{\partial} g_\omega)(x) 
				\, \Phi \circ \psi_\omega(x).
								\eqno (3.166)
$$
This enjoys the same sort of properties as ${\cal M}$ does -- like
(3.161) and (3.162) -- since ${\rm supp} \, \overline{\partial} g_\omega
\subseteq {\rm supp} \, g_\omega$.  Our assumption (3.158) implies that
${\cal N}$ is bounded on ${\cal B}(B)$, because of Lemma 3.14.

	Given integers $k$ and $l$ define a new transfer operator
${\cal M}_{k,l}$ by
$$
	{\cal M}_{k,l} \Phi (x) = 
	     \sum_{\omega \in \Omega} g_\omega(x) \, (\psi'_\omega(x))^k \,
		(\overline{\psi}'_\omega(x))^l \, \Phi \circ \psi_\omega(x).
								\eqno (3.167)
$$
Again these transfer operators satisfy the analogues of (3.161) and (3.162).
We define ${\cal N}_{k,l}$ in the same manner, replacing $g_\omega$
with $\overline{\partial} g_\omega$ as in (3.166).  Notice that
$$
	g_\omega \, (\psi'_\omega)^k \,	(\overline{\psi}'_\omega)^l
	\hbox{ and } \overline{\partial} g_\omega 
			\, (\psi'_\omega)^k \,	(\overline{\psi}'_\omega)^l
		\hbox{ satisfy (3.46) for each } \omega \in \Omega.
								\eqno (3.168)
$$
This is easy to check, using (3.158) and the holomorphicity and
hence smoothness of $\psi_\omega$ (and the fact that ${\rm supp} \, g_\omega$
is a compact subset of the domain of $\psi_\omega$).  A useful consequence
of this observation is that
$$
	\hbox{the } {\cal M}_{k,l}\hbox{'s and } 
		{\cal N}_{k,l}\hbox{'s are all bounded operators on }
			{\cal B}(B).				\eqno (3.169)
$$
This follows from Lemma 3.14.

	We are going to need to make the technical assumption that
$$
	I - z {\cal M}_{0,l} \quad\hbox{ is invertible on } {\cal B}(B)
		\hbox{ for at least one of } \, l = 1, -1.
								\eqno (3.170)
$$
Once we know that this operator is invertible on ${\cal B}(B)$ we can extend
it to (3.164) as before, using a formula like (3.165).

	Define the kneading operators ${\cal D}_t = {\cal D}_t(z)$ by
$$
	{\cal D}_t = {\cal N} (I - z {\cal M})^{-1} S_t.	\eqno (3.171)
$$
We can think of this initially as mapping elements of ${\cal B}(B)$ to
some functions on ${\bf C}$, i.e., $S_t$ maps ${\cal B}(B)$ into the
space (3.164), and then $(I - z {\cal M})^{-1}$ and ${\cal N}$ act
on this space.  However, ${\cal D}_t$ is actually a bounded operator 
on ${\cal B}(B)$.  To see this note that
$$
\eqalign{
	{\cal D}_t & = {\cal N} (I - z {\cal M})^{-1} S_t		\cr
		& = {\cal N} \, \eta \, (I - z {\cal M})^{-1} S_t
		= {\cal N} (I - z {\cal M})^{-1} \, \eta \, S_t,	\cr
}
								\eqno (3.172)
$$
where we are identifying $\eta$ with the associated multiplication operator.
This identity follows from the analogue of (3.162) for ${\cal N}$ and
the fact that $(I - z {\cal M})^{-1}$ commutes with multiplication by $\eta$
(because of (3.162)).  Once we have this identity we may conclude that
${\cal D}_t$ is a bounded operator on ${\cal B}(B)$, and even a trace class
operator, because $\eta \, S_t$ is a trace class operator on ${\cal B}(B)$,
by Lemma 3.33.  

\medskip
\noindent
{\bf Theorem 3.173.} {\it Let ${\cal M}, g_\omega, \Lambda_\omega$,
etc., be as above.  Assume in particular that (3.158), (3.163), and
(3.170) hold.  Then ${\cal D}_t$ is a trace class operator on ${\cal
B}(B)$ for each $t \in (0,1]$, and we have that}
$$
	\sup_{0 < t \le 1} \hbox{ trace norm } {\cal D}_t^2 < \infty.
								\eqno (3.174)
$$


\medskip

	It would be much nicer if the trace norms of the ${\cal
D}_t$'s themselves were bounded, but this does not seem to work.

	The proof of the theorem is fairly simple given all of our previous
lemmas.  We already noted above that the ${\cal D}_t$'s are trace class,
and so the only issue is the bound (3.174).  Using (3.172) we can write
out ${\cal D}_t^2$ as 
$$
	{\cal D}_t^2 = {\cal N} (I - z {\cal M})^{-1} \, \eta \, S_t \,
			{\cal N} (I - z {\cal M})^{-1} \, \eta \, S_t.
								\eqno (3.175)
$$
To prove (3.174) it suffices to show that
$$
	\sup_{0 < t \le 1} \hbox{ trace norm } \,\,
		\eta \, S_t \, {\cal N} (I - z {\cal M})^{-1} \, \eta \, S_t
					< \infty,
								\eqno (3.176)
$$
since ${\cal N} (I - z {\cal M})^{-1}$ is a bounded operator.

	The main point is that we have two $\eta \, S_t$'s, so that we
can try to use Lemma 3.128.  We have to commute an $\eta \, S_t$
around other operators, but we have a lot of lemmas for doing precisely
that.	


\medskip
\noindent
{\bf Lemma 3.177.}  {\it 
$(\eta \, S_t \, {\cal N} - {\cal N}_{0,-1} \, \eta \, S_t) \,
(I - z {\cal M})^{-1} \, \eta \, S_t$ is a trace class operator
on ${\cal B}(B)$ for $t \in (0,1]$, and it has bounded trace norm.}

\medskip

	To see this we apply Lemma 3.60, with ${\cal M}$ replaced
with ${\cal N}_{0,-1}$, so that ${\cal M}_{0,1}$ should be replaced
with ${\cal N}$.  We also take $E = W = B$ and $\chi = \eta$
in Lemma 3.60.  Note that the coefficient functions for
${\cal N}_{0,-1}$ satisfy the requirements of Lemma 3.60,
because of (3.168).  From Lemma 3.60 we get a decomposition
of $\eta ({\cal N}_{0,-1} S_t - S_t {\cal N}) : {\cal B}(B) \to {\cal
B}(B)$ into $T_t + U_t$, where $T_t$ is trace class with bounded norm
and $U_t$ has operator norm $\, = O(t)$.  Notice that 
$$
	{\cal N}_{0,-1} \, \eta \, S_t - \eta \, S_t {\cal N}
	= \eta ({\cal N}_{0,-1} S_t - S_t {\cal N}) = T_t + U_t,
								\eqno (3.178)
$$
because $\eta$ commutes with ${\cal N}_{0,-1}$ (as
in (3.162)).  Hence
$$
	(\eta \, S_t \, {\cal N} - {\cal N}_{0,-1} \, \eta \, S_t) \,
		(I - z {\cal M})^{-1} \, \eta \, S_t
	= - (T_t + U_t) \, (I - z {\cal M})^{-1} \, \eta \, S_t.
								\eqno (3.179)
$$
The term with the $T_t$ is trace class with bounded trace norm because
$T_t$ is and the other operators are uniformly bounded.  The term with
the $U_t$ is also trace class with bounded trace norm because $\eta \,
S_t$ is trace class with trace norm $O(t^{-1})$ (Lemma 3.33), 
because $U_t$ has operator norm $O(t)$, and because $(I - z {\cal
M})^{-1}$ is a bounded operator which does not depend on $t$.  This
proves Lemma 3.177.
	
	To prove (3.176) it now suffices to show that
$$
	\sup_{0 < t \le 1} \hbox{ trace norm } \,\,
		\eta \, S_t \, (I - z {\cal M})^{-1} \, \eta \, S_t
					< \infty,
								\eqno (3.180)
$$
because of Lemma 3.177 and because ${\cal N}_{0,-1}$ is a bounded operator
on ${\cal B}(B)$, as in (3.169).

	We want to commute an $\eta \, S_t$ around $(I - z {\cal
M})^{-1}$.  We can do this either from the left or from the right, and
this corresponds to the ambiguity in (3.170).  For the sake of
definiteness we commute it from the left.

	Assume that we can take $l = -1$ in (3.170).  We want to
analyze the operator 
$$
	\Gamma = \eta \, S_t \, (I - z {\cal M})^{-1} -
			(I - z {\cal M}_{0,-1})^{-1} \, \eta \, S_t.
								\eqno (3.181)
$$
We can write this as
$$
	\Gamma = (I - z {\cal M}_{0,-1})^{-1} \, \Gamma_0 
			\, (I - z {\cal M})^{-1},
								\eqno (3.182)
$$
where
$$
	\Gamma_0 = (I - z {\cal M}_{0,-1}) \, \eta \, S_t -
			\eta \, S_t \, (I - z {\cal M}).	\eqno (3.183)
$$
We can rewrite $\Gamma_0$ as
$$
	\Gamma_0 = 
	-z \{ {\cal M}_{0,-1} \, \eta \, S_t - \eta \, S_t \, {\cal M} \}.
								\eqno (3.184)
$$
We apply Lemma 3.60 again, with the ${\cal M}$ in Lemma 3.60 replaced
with ${\cal M}_{0,-1}$, so that the ${\cal M}_{0,1}$ in Lemma 3.60
corresponds to ${\cal M}$ here.  As before, we are using (3.168) 
to know that the hypotheses in Lemma 3.60 on the coefficient functions
are satisfied, and we take $E = W = B$ and $\chi = \eta$
in Lemma 3.60.  We conclude that
$$
	\eta \, ({\cal M}_{0,-1}  \, S_t - S_t \, {\cal M}) =
		T'_t + U'_t,					\eqno (3.185)
$$
where $T'_t$ is trace class on ${\cal B}(B)$ with uniformly bounded trace norm,
and $U'_t$ is a bounded operator on ${\cal B}(B)$ with norm $\, = O(t)$.
We also get
$$
	\Gamma_0 = - z (T'_t + U'_t), 				\eqno (3.186)
$$
because $\eta$ commutes with ${\cal M}_{0,-1}$, as in (3.162).  Using
this and (3.182) we get a decomposition
$$
	\Gamma =  T''_t + U''_t, 				\eqno (3.187)
$$
where $T''_t$ is trace class on ${\cal B}(B)$ with uniformly bounded 
trace norm, and $U''_t$ is a bounded operator on ${\cal B}(B)$ with 
norm $\, = O(t)$.

	Let us use this to prove (3.180).  We have that
$$
	\eta \, S_t \, (I - z {\cal M})^{-1} \, \eta \, S_t
		= \Gamma \, \eta \, S_t
			+ (I - z {\cal M}_{0,-1})^{-1} \, (\eta \, S_t)^2,
								\eqno (3.188)
$$
by algebra.  The last term is trace class with bounded trace norm,
because of Lemma 3.128 and the boundedness of $(I - z {\cal
M}_{0,-1})^{-1}$.  The first term on the right splits into $T''_t \,
\eta \, S_t + U''_t \, \eta \, S_t$.  As usual, $T''_t \, \eta \, S_t$
is trace class with bounded norm because $T''_t$ is, and $U''_t \,
\eta \, S_t$ is trace class with bounded norm because $\eta \, S_t$ is
trace class with norm $= O(t^{-1})$ and because $U''_t$ has operator
norm $\, = O(t)$.  This takes care of the right side of (3.188), and
so we conclude that the left side does have bounded trace norm, i.e.,
(3.180) holds.

	If $l=1$ in (3.170), then we perform a similar analysis of
$$
	(I - z {\cal M})^{-1} \, \eta \, S_t -
		\eta \, S_t \, (I - z {\cal M}_{0,1})^{-1}.	\eqno (3.189)
$$
That is, we use Lemma 3.60 to prove that this operator admits a decomposition
like (3.187), and then we use this decomposition to prove (3.180) in
practically the same manner as before, using also Lemma 3.128.

	This completes the proof of Theorem 3.173.
			


\vglue 1.2 truecm

\centerline {\bf References}
\vglue 0.9 truecm

\ref {1} {V. Baladi and D. Ruelle, {\it Sharp determinants}, IHES preprint,
1994, to appear Invent. Math.} \medskip

\ref {2} {B. Fuglede and L. Schwartz. Un nouveau th\'eor\`eme sur les
distributions. CRAS Paris 263A, 899-901 (1966)} \medskip


\ref{3}{F. Treves, {\it Basic linear partial differential
equations}, Academic Press, N.Y., 1975.}
\medskip


\ref{4}{D. Ruelle.
Spectral properties of a class of operators associated with maps in one
dimension. Ergod. Th. and Dynam. Syst. {\bf 11}, 757--767 (1991)}
\medskip

\ref{5}{D. Ruelle, {\it Functional determinants related to dynamical
systems and the thermodynamic formalism}, 
Lezioni Fermiane, IHES preprint, 1995.}
\medskip

\ref{6}{D. Ruelle, {\it Sharp zeta functions for
smooth interval maps}, IHES preprint, 1995.}


\bye