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\begin{document}
\setcounter{page}{121}
\noindent\parbox{2.95cm}{\includegraphics*[keepaspectratio=true,scale=0.125]{AFA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
\qquad Ann. Funct. Anal. 3 (2012), no. 1, 121--127\\
{\footnotesize \qquad \textsc{\textbf{$\mathscr{A}$}nnals of
\textbf{$\mathscr{F}$}unctional
\textbf{$\mathscr{A}$}nalysis}\\
\qquad ISSN: 2008-8752 (electronic)\\
\qquad URL:
\textcolor[rgb]{0.00,0.00,0.99}{www.emis.de/journals/AFA/}
}\\[.5in]}
\title[Compactness in complex interpolation]{On compactness in
complex interpolation}
\author[J. Voigt]{J\"urgen Voigt}
\address{Fachrichtung Mathematik, Technische Universit\"at Dresden,
01062 Dresden, Germany.}
\email{\textcolor[rgb]{0.00,0.00,0.84}{juergen.voigt@tu-dresden.de}}
\dedicatory{{\rm Communicated by D. Werner }}
\subjclass[2010]{Primary 46B70; Secondary 47B07.}
\keywords{Complex interpolation, compact operator function.}
\date{Received: 4 January 2012; Accepted: 7 February 2012.}
\begin{abstract}
We show that, in complex interpolation, an operator function that is compact
on one side of the interpolation scale will be compact for all proper
interpolating
spaces if the right hand side $(Y^0,Y^1)$ is reduced to a single space.
A corresponding result, in restricted generality, is shown if the left hand side
$(X^0,X^1)$ is reduced to a single space.
These results are derived from the fact that a holomorphic operator valued
function on an open subset of $\C$ which
takes values in the compact operators on part of the boundary is in fact compact
operator valued.
\end{abstract} \maketitle
\section*{Introduction}
\noindent
Let $(X^0,X^1)$ and $(Y^0,Y^1)$ be %two
interpolation pairs of Banach spaces, and let $A\colon X_\Sigma\to
Y_\Sigma$ be a linear operator such that
$A\restrict_{X^j}\in\cL(X^j,Y^j)$ ($j=0,1$) and such that
$A\restrict_{X^0}\colon X^0\to Y^0$ is compact. (For notation
concerning interpolation we refer to Section \ref{comp-int}.) For
many important cases it is known that then the interpolating
operator $A\restrict_{X_t}\colon X_t\to Y_t$ resulting from complex
interpolation is compact for all $t\in(0,1)$. In particular, for the
case that the pair $(Y^0,Y^1)$ obeys a certain approximation
hypothesis, this property was shown by Persson \cite{per64}.
However, whether this is true in general is an open problem since
the fundamental paper of Calder\'on \cite{cal64}. For a survey on
developments and the state of the art concerning this problem we
refer to \cite{c-k-m96, sch00, cw-ja06, cwi10} and the references
therein.
In the particular cases where $X^0=X^1$ or $Y^0=Y^1$, it was shown already in
\cite{li-pe64} that the answer is positive, without further requirements on the
spaces.
It is the purpose of the present paper to present an analogous result for the
case that the operator $A$ is replaced by a suitable operator function $A\colon
S\to\cL(X_\Delta,Y_\Sigma)$, where $S$ is the strip
\[
S:=\sset{z\in\C}{0\le\Re z\le1}\spe.
\]
Assume that $A(z)\colon X^0\to Y^0$ is compact for all $z$ in some
subset of the imaginary axis, having positive measure.
If $Y^0=Y^1=:Y$, then we show that $A(z)$ is compact as
an operator in $\cL(X_{\Re z},Y)$, for all $z\in \interior S$
(Theorem~\ref{mainres}). Similarly, if $X^0=X^1:=X$, the dual of $X$ is
separable, and one of the spaces $Y^0,Y^1$ is reflexive, then we show that
$A(z)$ is compact as an operator in $\cL(X,Y_{\Re z})$, for all $z\in \interior
S$
(Theorem~\ref{mainres2}).
As a preliminary result we show that holomorphic operator valued functions on
open subsets of $\C$ which
take values in the compact operators on part of the boundary are compact
operator valued (Theorem~\ref{holcomp}).
\section{On compactness for holomorphic operator functions}
\begin{theorem}\label{holcomp}
Let $X,Y$ be Banach spaces, and let $Z\subseteq Y'$ be a separating set for
$Y$.
Let $\Omega:=\set{z\in\C}{|z|<1,\ \Im z>0}$,
\[
T\colon \Omega\cup(-1,1)\to\cL(X,Y)
\]
bounded, $z\mapsto y'(T(z)x)$ continuous for all
$y'\in Z$, $x\in X$, and $T\restrict_\Omega$ holomorphic.
Assume that $T(z)$ is a compact operator for all $z\in E$, where
$E\subseteq(-1,1)$
is a set of positive measure, and that
$T\restrict_{(-1,1)}$ is strongly measurable.
Then $T(z)$ is compact for all $z\in\Omega$.
\end{theorem}
\begin{proof} %[Proof of Theorem \ref{holcomp}]
Without restriction we may assume that $X$ is separable.
Also, we may assume that $T$ has an extension to
$\closure\Omega$ such that $z\mapsto y'(T(z)x)$ is continuous on
$\closure\Omega$ for all $y'\in Z$, $x\in X$.
(Replace $T$ by $z\mapsto T(\alpha z)$, for suitable $\alpha\in(0,1)$.)
Let $D:=B_\C(0,1)$ be the open unit disc in
$\C$, let $C:=\partial D$ be the unit circle, and let $\mu$ be the normalised
line measure on $C$.
There exists a homeomorphism $\kappa\colon\overline\Omega\to\overline D$, such
that $\kappa\colon\Omega\to D$ is biholomorphic and such that the setup is
transformed into the situation that
\[
T\colon\overline D\to\cL(X,Y)
\]
is bounded, $z\mapsto y'(T(z)x)$ is continuous on $\overline D$, for all $y'\in
Z$ and $x\in X$, that $T\restrict_ D$ is holomorphic, that $T(\xi)$ is a compact
operator for all $\xi\in E$, where $E\subseteq C$ is a set of positive measure,
and
that $T\restrict_ C$ is strongly measurable.
Let $p\colon D\times C\to(0,\infty)$ be the $2$-dimensional Poisson kernel,
\[
p(z,\xi)=\frac{1-|z|^2}{|z-\xi|^2}\spe.
\]
For $z\in D$, Poisson's formula implies that
\[
T(z)=\int_C T(\xi)p(z,\xi)\,d\mu(\xi)\spe.
\]
Here, the integral is a strong integral, and the equality is obtained by
inserting $x\in X$ and applying
functionals $y'\in Z$. We split the integral as $T(z)=T_1(z)+T_2(z)$, with
\begin{align*}
T_1(z)&:=\int_E T(\xi)p(z,\xi)\,d\mu(\xi)\spe,\\
T_2(z)&:=\int_{C\setminus E} T(\xi)p(z,\xi)\,d\mu(\xi)\qquad(z\in D)\spe.
\end{align*}
Then $T_1(z)$ is a compact operator, by the strong convex
compactness property of the compact operators;
cf.\ \cite[Theorem 1.3]{voi92b}. We estimate $T_2(z)$ by
\[
\|T_2(z)\|\le\int_{C\setminus E}\|T(\xi)\|p(z,\xi)\,d\mu(\xi)
\]
(noting that $\xi\mapsto\|T(\xi)\|$ is measurable, by the separability of $X$).
Now it follows that $\|T_2(r\xi)\|\to 0$ as $r\to1\llim$, for $\xi\in E$ a.e.;
see, e.g., \cite[Corollary 2 of Theorem 1.2]{dur70}.
Let $\eta\in\cL(X,Y)'$ be vanishing on the compact operators, and define
\[
\phi(z):=\eta(T(z))\qquad(z\in D)\spe.
\]
Then $\phi$ is bounded and holomorphic, and the
previous considerations show that $\phi(r\xi)=\eta(T_2(r\xi))\to0$
($r\to1\llim$)
for $\xi\in E$ a.e. This implies that
$\eta(T(z))=\phi(z)=0$ ($z\in D$), by \cite[Theorem 17.18]{rud66}.
Since this is true for any $\eta$ as above, the Hahn-Banach
theorem implies that $T(z)$ is compact for all $z\in D$.
\end{proof}
\begin{remarks}\label{rems1}
(a) In the proofs of Theorems~\ref{mainres} and \ref{mainres2}
we will apply Theorem~\ref{holcomp}
with
the interior of the strip $S$ instead of the set $\Omega$. The transformation to
this situation is achieved by a suitable biholomorphic mapping.
(b) If, in the hypothesis of Theorem~\ref{holcomp}, one assumes more strongly
that $T$ is operator norm continuous on $\Omega\cup(-1,1)$,
then the proof is easier,
as one can use the last step of the proof directly: Proceeding as above one
obtains a function $\phi$ which is continuous on $\overline D$ and vanishes on
the set $E\subseteq C$ of positive measure. This implies that $\phi=0$, by
\cite[Theorem 17.18]{rud66}. It is the main point of
Theorem~\ref{holcomp} to deal with a less restrictive
hypothesis for $T$ concerning the behaviour on $(-1,1)$.
(c) The hypotheses of Theorem~\ref{holcomp} are somewhat reminiscent of the
setting used in \cite[9.6]{cal64}.
(d) If $Z\subseteq Y'$ is an \emph{almost norming} subspace (i.e.,
\[
\|y\|_Z:=\sup\set{|y'(y)|}{y'\in Z,\ \|y'\|\le1}\qquad(y\in Y)\spe,
\]
defines a norm $\|\cdot\|_Z$ which is equivalent to the norm on $Y$), then
the hypothesis that $T\restrict_{(-1,1)}$ is strongly measurable can
be replaced by the requirement that $T(\cdot)x$ is essentially
separably valued on
$(-1,1)$, for all $x\in X$. This is a consequence of a slight strengthening of
Pettis' theorem on measurability as described in
\cite[Section II.1, Corollary 4]{di-uh77}.
\end{remarks}
\section{Compact interpolation of operator functions}
\label{comp-int}
\noindent
In this section we assume that $(X^0,X^1)$ and $(Y^0,Y^1)$ are
interpolation pairs of Banach
spaces (cf. \cite[Sect.\,2.3]{be-lo76}, \cite[Introduction]{lun09}).
We recall the notation $X_\Delta:=X^0\cap X^1$, $X_\Sigma:=X^0+X^1$, and we
assume that $\check X$ is a dense subspace of $X_\Delta$. For the
definition of the complex interpolation spaces we recall the space
\begin{align*}
\cF_0(X):=\bigl\{f\in C_0(S;X_\Sigma);&\ f\text{ holomorphic on }\interior S,\\
&(s\mapsto f(j+\imu s))\in C_0(\R;X^j)\ (j=0,1)\bigr\}\spe,
\end{align*}
with the norm given by
\[
\tripnorm f\tripnorm:=\sup\set{\|f(j+\imu s)\|_j}{s\in\R,\ j=0,1}\spe.
\]
For $z\in S$, the interpolation space $X_z$ is defined by
\[
X_z:=\set{f(z)}{f\in\cF_0(X)}\spe,
\]
with norm
\[
\|x\|_z:=\inf\set{\tripnorm f\tripnorm}{f\in\cF_0(X),\ f(z)=x}\spe.
\]
(Usually, the spaces $X_t$ are only considered for $0\le t\le1$; note that
evidently $X_z=X_{\Re z}$ for all $z\in S$.) Finally, we define
\[
A_0(S):=\set{\phi\in C_0(S)}{\phi\restrict_{\interior S}\text{
holomorphic}}\spe,
\]
and we recall that
\[
\check\cF_0(X,\check X):=\lin\set{\phi x}{\phi\in A_0(S),\ x\in\check X}
\]
is a dense subspace of $\cF_0(X)$; cf. \cite{cal64} or \cite[Lemma
4.2.3]{be-lo76}.
In addition to the previous assumptions we assume that we are given a
family $(A(z))_{z\in S}$ of linear mappings $A(z)\colon\check X\to Y_\Sigma$,
with the following properties:
\begin{enumerate}
\item[(i)] For all $x\in\check X$ the function $A(\cdot)x\colon S\to Y_\Sigma$
is continuous, bounded, and holomorphic on $\interior S$;
\item[(ii)] for $x\in\check X$, $j=0,1$, the function $\R\ni s\mapsto A(j+\imu
s)x\in Y^j$ is continuous, and
\[
M_j:=\sup\set{\|A(j+\imu s)x\|_{Y^j}}{s\in\R,\ x\in\check X,\ \|x\|_{X^j}\le1}
<\infty\spe.
\]
\end{enumerate}
If these conditions are satisfied, then, for all $z\in S$, the operator $A(z)$
has a unique extension to an operator $A(z)\in\cL(X_z,Y_z)$, and this operator
satisfies $\|A(z)\|_{\cL(X_z,Y_z)}\le M_0^{1-\Re z}M_1^{\Re z}$. We refer to
\cite{cw-ja84} (or \cite[Theorem 2.1]{voi92a}, \cite[Theorem 2.7]{lun09})
for the proof of this statement.
\begin{theorem}\label{mainres}
In the previous setup, assume that $Y^0=Y^1=:Y$, and assume additionally that
there exists a subset $E\subseteq\R$ of positive measure such that
$A(\imu s)\in\cL(X_0,Y)$ is a compact operator, for all $s\in E$.
Then the operator $A(z)\in\cL(X_z,Y)$ is compact for all $z\in \interior S$.
\end{theorem}
\begin{proof}
The hypotheses imply that for $f\in\check\cF(X,\check X)$, the function
$z\mapsto A(z)f(z)$ belongs to $\cF_0(Y):=\cF_0((Y,Y))$. Moreover, the mapping
\[
f\mapsto A(\cdot)f(\cdot)
\]
is continuous with respect to the norms on $\cF_0(X)$ and $\cF_0(Y)$, which
implies that it has a unique extension to $\cF_0(X)$. As $A(z)\in\cL(X_z,Y)$ for
all $z\in S$, this extension is given pointwise by $A(z)f(z)$ ($f\in\cF_0(X)$).
This shows that the function $z\mapsto A(z)f(z)$ belongs to $\cF_0(Y)$, for
all $f\in\cF_0(X)$.
Now we define the operator function $T\colon S\to\cL(\cF_0(X),Y)$ by
\[
T(z)f:= A(z)f(z) \qquad (f\in\cF_0(X)\spe,\ z\in S)\spe.
\]
Because $\cF_0(X)\ni f\mapsto f(z)\in X_z$ is a continuous linear operator, the
hypothesis of the theorem implies that $T(\imu s)$ is compact for all
$s\in E$. Further, the properties mentioned above imply that $T$ is
holomorphic on $\interior S$, strongly continuous on $S$, and $T(z)$ tends
strongly to $0$ as $|z|\to\infty$.
Therefore the application of Theorem \ref{holcomp} yields that $T(z)$ is
compact for all $z\in\interior S$. For $z\in S$, the operator $\cF_0(X)\ni
f\mapsto f(z)\in X_z$ is a quotient map (in particular, maps the unit ball of
$\cF_0(X)$ onto the unit ball of $X_z$), and therefore the definition of $T(z)$
implies that $A(z)\in\cL(X_z,Y)$ is compact for all $z\in\interior S$.
\end{proof}
\begin{remarks}
(a) For the case that the pair $(Y^0,Y^1)$ obeys an approximation hypothesis,
Calder\'on \cite[10.4]{cal64} presented a result corresponding to
Theorem~\ref{mainres}. We just mention another noteworthy difference
between the hypotheses in the results: In \cite{cal64} the function $A$ is
required to be operator norm continuous (in a certain sense) on $S$.
(b)
One might ask whether, analogously to the case of a single operator, one could
also treat the case that $X^0=X^1$ whereas the interpolation pair $(Y^0,Y^1)$ is
not reduced to a single space.
This would be highly desirable. Indeed, a glance at the proof of
Theorem~\ref{mainres} shows that this would mean that one could also treat the
general case where both interpolation pairs are non-trivial. In particular, this
would solve the long-standing open problem mentioned above, concerning a single
operator. We did not see how to apply Theorem~\ref{holcomp} to this case, in the
general setup. In Theorem~\ref{mainres2} below we present a result of this kind,
under more restrictive hypotheses.
%The result shown subsequently is as close as we can get, momentarily.
\end{remarks}
\begin{theorem}\label{mainres2}
In the setup of the beginning of this section, assume that $X^0=X^1=:X$. Further
assume that one of the spaces $Y^0,Y^1$
is reflexive, that there exists a subset $E\subseteq\R$ of positive measure such
that
$A(\imu s)\in\cL(X,Y^0)$ is a compact operator for all $s\in E$, and that the
function $\R\ni s\mapsto A(\imu s)'\in\cL(Y^{0\prime},X')$ is strongly
measurable.
Then the operator $A(z)\in\cL(X,Y_z)$ is compact for all $z\in \interior S$.
\end{theorem}
\begin{proof}
It is easy to see that, without restriction, one may assume that $Y_\Delta$ is
dense in $Y^0$ and $Y^1$, and thus $Y_j=Y^j$ ($j=0,1$).
>From \cite[9.5, 12.1 and 12.2]{cal64} we then know that the
space $Y_z$ is reflexive and that $(Y')_z=(Y_z)'$ (with the obvious notation),
for all $z\in\interior S$.
For all $x\in X$, $y'\in Y_0'\cap Y_1'=(Y')_\Delta=(Y_\Sigma)'$, we obtain that
the function
\[
z\mapsto\langle x,A(z)'y'\rangle=\langle A(z)x,y'\rangle
\]
is bounded and continuous on $S$ and holomorphic on $\interior S$. Therefore,
for all $f\in\check\cF_0(Y',(Y')_\Delta)$, the function
\[
z\mapsto\langle x,A(z)'f(z)\rangle
\]
belongs $C_0(S)$, is holomorphic on $\interior S$, and we have the estimate
\[
|\langle x,A(z)'f(z)\rangle|
\le\max\bigl\{M_0,M_1\bigr\}\|x\|\tripnorm f\tripnorm\spe.
\]
The denseness of $\check\cF_0(Y',(Y')_\Delta)$ in $\cF_0(Y')$ implies that the
last properties carry over to all $f\in\cF_0(Y')$. As in the proof of
Theorem~\ref{mainres} we now define $T\colon S\to\cL(\cF_0(Y'),X')$ by
\[
T(z)f:=A(z)'f(z)\qquad(f\in\cF_0(Y')\spe,\ z\in S)\spe.
\]
Then we can apply Theorem~\ref{holcomp} to the function $T$. Indeed, $X\subseteq
X''$
is a norming subspace for $X'$, the continuity and holomorphy assumptions hold
(for the holomorphy we refer to \cite[Chapter III, Remark 1.38]{kat66}),
$T(\imu s)\colon f\mapsto A(\imu s)'f(\imu s)$ is compact because $f\mapsto
f(\imu s)$ is
continuous and the dual operator $A(\imu s)'\in\cL(Y_0',X')$ of the compact
operator
$A(\imu s)\in\cL(X,Y_0)$ is compact for all $s\in E$, and the measurability of
$\R\ni s\mapsto A(\imu s)'f(\imu s)$, for $f\in\cF_0(Y')$, follows from the
strong measurability of $\R\ni s\mapsto A(\imu s)'$ and the continuity of
$\R\ni s\mapsto f(\imu s)$.
The application of Theorem~\ref{holcomp} then yields that $T(z)$ is compact for
all $z\in\interior S$, and as in the proof of Theorem~\ref{mainres} we conclude
that $A(z)'\in\cL((Y')_z,X')=\cL((Y_z)',X')$ is compact. By Schauder's theorem
we obtain that $A(z)\in\cL(X,Y_z)$ is a compact operator.
\end{proof}
\begin{remark}
In Theorem \ref{mainres2},
the assumption of strong measurability follows from the remaining hypotheses if
the dual $X'$ of $X$ is separable. This is a consequence of
Remark~\ref{rems1}(d).
\end{remark}
\bigskip \textbf{Acknowledgement.} The author is grateful to Michael
Cwikel for helpful and encouraging correspondence.
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\end{document}
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