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\begin{document}
\title[Continuity of the norm]{Continuity of the norm of a composition operator}
\author{David B. Pokorny}
\address{Mathematics Department, University of California, Berkeley, CA 94720}
\email{davebrok@soda.csua.berkeley.edu}
\urladdr{www.csua.berkeley.edu/\symbol{126}davebrok}
\author{Jonathan E. Shapiro}
\address{Mathematics Department, California Polytechnic State University, San Luis
Obispo, CA 93407}
\email{jshapiro@calpoly.edu}
\urladdr{http://www.calpoly.edu/\symbol{126}jshapiro}
\thanks{We would like to thank Joel Shapiro for his helpful comments during the
preparation of this work. \ This material is based upon work supported by the
National Science Foundation under Grant No. 0097392.}
\date{}
\subjclass{Primary 47-A30}
\keywords{composition operator, norm}
\maketitle
\begin{abstract}
We explore the continuity of the map which, given an analytic self-map of the
disk, takes as its value the norm of the associated composition operator on
the Hardy space $H^{2}$. We also examine the continuity the functions which
assign to a self-map of the disk the Hilbert-Schmidt norm or the essential
norm of the associated composition operator and show these to be
discontinuous. \ Additionally, we characterize when the norm of a composition
operator is minimal.
\end{abstract}
\section{Introduction}
Let $\mathbb{D}$ denote the unit disc of the complex plane and $\phi$ be a
holomorphic function on $\mathbb{D}$ with $\phi(\mathbb{D})\subseteq
\mathbb{D}$. The equation $C_{\phi}f=f\circ\phi$ defines a \textit{composition
operator} on the Hardy Space $H^{2}$. For any such $\phi$, $C_{\phi}$ is a
bounded operator (see \cite[pg. 117]{CCBM}). \ It is known that if $\phi
_{n}\rightarrow\phi$ weakly in $H^{1}$ then $C_{\phi_{n}}\rightarrow C_{\phi}$
weakly. If $\phi_{n}\rightarrow\phi$ in $H^{1} $ then $C_{\phi_{n}}\rightarrow
C_{\phi}$ pointwise or strongly. These results and others are found in
\cite{HJS}. It is also known that $\phi_{n}\rightarrow\phi$ in some sense does
not in general imply that $\Vert C_{\phi}-C_{\phi_{n}}\Vert\rightarrow0$.
\ However, we show that in several cases, it does imply that $\left\|
C_{\phi_{n}}\right\| \rightarrow\left\| C_{\phi}\right\| $.
Let $ASM(\mathbb{D})$ be the set of all analytic self-maps of the disk,
considered as a subset of $H^{1}$ or $H^{\infty}$. We will denote this set
$ASM(\mathbb{D})^{1}$ or $ASM(\mathbb{D})^{\infty}$ respectively, when we wish
to make a distinction. Recall that the $H^{\infty}$ norm is given by
$\Vert\phi\Vert_{\infty}=\sup_{|z|<1}|\phi(z)|$, and the $H_{1}$ norm is given
by
\[
\Vert\phi\Vert_{1}=\int_{0}^{2\pi}|\phi(e^{i\theta})|\ \frac{d\theta}{2\pi}.
\]
Let $N\colon ASM(\mathbb{D})\rightarrow\mathbf{R}$ be given by $N(\phi)=\Vert
C_{\phi}\Vert$. We will use $N_{1}\colon ASM(\mathbb{D})^{1}\rightarrow
\mathbf{R}$ and $N_{\infty}\colon ASM(\mathbb{D})^{\infty}\rightarrow
\mathbf{R}$ to distinguish the topologies on the domain ($N\equiv N_{1}\equiv
N_{\infty}$). Stated in a different way, we are asking:
\emph{For which functions }$\phi\in ASM(\mathbb{D})$\emph{\ is }$N_{1}%
$\emph{\ or }$N_{\infty}$\emph{\ continuous?}
In addition to questions of the convergence of $\Vert C_{\phi_{n}}\Vert$ to
$\Vert C_{\phi}\Vert$, we will consider similar questions with the operator
norm replaced with either the essential norm $\Vert\ \Vert_{e}$ or the
Hilbert-Schmidt norm $\Vert\ \Vert_{HS}$. Recall that $\Vert T\Vert_{e}$ is
the distance from $T$ to the subspace of compact operators in the operator
norm topology and
\[
\Vert T\Vert_{HS}^{2}=\sum_{n=1}^{\infty}\Vert Te_{n}\Vert^{2}%
\]
where $\{e_{n}\}_{1}^{\infty}$ is a complete orthonormal basis. We will also
often suppress the subscript and write $\Vert f\Vert=\Vert f\Vert_{2}$ for the
$H^{2}$ norm of an analytic function in $\mathbb{D}$.
In Section \ref{minnormsec}, we show that the well known lower bound for the
norm of a composition operator $C_{\phi}$, equation (\ref{bounds}), is
attained (when $\phi(0)\neq0$) only when $\phi$ is a constant function. \ This
appears here as Theorem \ref{minnorm}. \ C. Hammond, in \cite{CH}, has
recently obtained this same theorem with a different proof.
\section{Hilbert-Schmidt Operators}
Let $ASM(\mathbb{D})_{HS}^{\infty}$ denote the set of analytic self-maps of
$\mathbb{D}$ that induce Hilbert-Schmidt composition operators, and give this
set the $H^{\infty}$ norm. Suppose that $\phi_{n}\rightarrow\phi$ are all in
$ASM(\mathbb{D})_{HS}^{\infty}$. Does it then follow that $\Vert C_{\phi_{n}%
}\Vert_{HS}\rightarrow\Vert C_{\phi}\Vert_{HS}$? In general, the answer is no,
as the example below demonstrates. Let
\[
\phi(z)={\frac{1}{1+\left( \frac{1-z}{1+z}\right) ^{1/2}}}%
\]
and
\[
\phi_{n}(z)={\frac{1}{1-\left( \frac{1-{\frac{n}{n+1}}}{2}\right)
^{1/2}+\left( \frac{1-{\frac{n}{n+1}}z}{1+z}\right) ^{1/2}}}%
\]
where the real part of $z^{1/2}$ is always positive.
We approach the Hilbert-Schmidt norms of composition operators by using the
formulas (as in \cite[Sec. 3.3]{CCBM})
\[
\left\langle C_{\phi},C_{\psi}\right\rangle _{HS}={\int_{0}^{2\pi}{\frac
{1}{1-\phi(e^{i\theta})\overline{\psi(e^{i\theta})}}}\ }\frac{{d\theta}}{2\pi
}\text{ }%
\]
and
\[
\Vert C_{\phi}\Vert_{HS}^{2}={\int_{0}^{2\pi}{\frac{1}{1-\left|
\phi(e^{i\theta})\right| ^{2}}}\ }\frac{{d\theta}}{2\pi}\text{.}%
\]
To prove that $C_{\phi}$ is Hilbert-Schmidt, we will show by direct
computation that the integral
\[
\int_{0}^{\pi}{\frac{1}{1-|\phi(e^{i\theta})|^{2}}}\frac{d\theta}{2\pi}%
\]
is finite. Since $\phi(e^{-i\theta})=\overline{\phi(e^{i\theta})}$ this will
show that $\Vert C_{\phi}\Vert_{HS}$ is finite. First note that for
$0<\theta<\pi$,
\[
{\frac{1-e^{i\theta}}{1+e^{i\theta}}}={\frac{\tan(\theta/2)}{i}},
\]
and if we set $t=\tan(\theta/2)$,
\[
\phi(e^{i\theta})={\frac{1}{1+\sqrt{\frac{t}{i}}}}={\frac{1}{1+{\frac{\sqrt
{t}}{\sqrt{2}}}-i{\frac{\sqrt{t}}{\sqrt{2}}}}}%
\]
so
\[
|\phi(e^{i\theta})|^{2}={\frac{1}{1+\sqrt{2t}+t}}%
\]
and thus
\[
{\frac{1}{1-|\phi(e^{i\theta})|^{2}}}={\frac{1}{1-{\frac{1}{1+\sqrt{2t}+t}}}%
}=1+{\frac{1}{\sqrt{2t}+t}}.
\]
It can be shown using basic calculus that the integral
\[
\int_{0}^{\pi}\left( 1+{\frac{1}{\sqrt{2\tan(\theta/2)}+\tan(\theta/2)}%
}\right) \frac{d\theta}{2\pi}%
\]
converges. Each of the maps $\phi_{n}$ has an angular derivative at 1 because
each $\phi_{n}$ is in fact analytic in a disc centered at 1. Thus none of the
$C_{\phi_{n}}$ are compact and thus none have finite Hilbert-Schmidt norm.
For each $n$, choose ${\frac{n}{n+1}} < r_{n}<1$ such that $\|C_{r_{n}\phi
_{n}}\|_{HS}>n$. Note that each $r_{n}\phi_{n}$ induces a Hilbert-Schmidt
operator. Now $\|C_{r_{n}\phi_{n}}\|_{HS}\rightarrow\infty$ while $r_{n}%
\phi_{n}\rightarrow\phi$ in $H^{\infty}$.
What about $\phi$ such that $\Vert\phi\Vert_{\infty}<1$? Is the map
$N_{HS}(\phi)=\Vert\phi\Vert_{HS}$ continuous at such points? Suppose that
$\Vert\phi\Vert_{\infty}<1$ and $\phi_{n}\rightarrow\phi$ in $ASM(\mathbb{D}%
)_{HS}^{\infty}$. There exists an $r<1$ such that for sufficiently large $n$,
$\Vert\phi_{n}\Vert_{\infty}n$. We now have $\Vert C_{r_{n}\eta_{n}}\Vert_{HS}%
\rightarrow\infty$ and $r_{n}\eta_{n}\rightarrow0$ in $H^{1}$ as
$n\rightarrow\infty$.
\textit{Remark:} It is easy to check that if $\phi\in ASM(\mathbb{D})$
satisfies $\Vert C_{\phi}\Vert_{HS}=\infty$ then $\Vert C_{r\phi}\Vert_{HS}$
is unbounded for $r\in(0,1)$.
\section{Continuity of the norm}
We now address the first question raised in the introduction. Is $N_{1}$ or
$N_{\infty}$ continuous? As the $H^{\infty}$ topology is stronger than $H^{1}$
topology, a map $\phi$ at which $N_{1}$ is continuous is a map at which
$N_{\infty}$ is continuous.
\begin{theorem}
The map $N_{1}$ (and thus $N_{\infty}$) is continuous at $\phi$ if either
$\phi(0)=0$ or $\phi$ is an inner map. The map $N_{\infty}$ is continuous at
$\phi$ which satisfy $\phi(\mathbb{D})\subseteq r\mathbb{D}$ for some $r<1$.
\end{theorem}
\begin{proof}
The following estimate for the norm of a composition operator is well-known
(for instance in \cite[Cor. 3.7]{CCBM}):
\begin{equation}
{\frac{1}{\sqrt{1-|\phi(0)|^{2}}}}\leq\Vert C_{\phi}\Vert\leq{\frac
{1+|\phi(0)|}{\sqrt{1-|\phi(0)|^{2}}}}\label{bounds}%
\end{equation}
Furthermore, it is noted in \cite{JHS1} that the upper bound is attained if
$\phi$ is inner.
If $\phi(0)=0$, then $\Vert C_{\phi}\Vert=1$ and $\phi_{n}(0)\rightarrow0,$ so
the bounds given above force $\lim_{n\rightarrow\infty}\Vert C_{\phi_{n}}%
\Vert=1$.
Let $T\in\mathcal{L}(H^{2})$ be a bounded linear operator. Define $\Vert
T\Vert_{(m)}=\Vert TK_{m}\Vert$ where $K_{m}$ is the orthogonal projection of
$H^{2}$ onto the subspace spanned by $\{1,z,z^{2},\dots,z^{m-1}\}$. Then
\[
\Vert T\Vert_{(n)}\leq\Vert T(1)\Vert+\Vert T(z)\Vert+\dots+\Vert
T(z^{m-1})\Vert,
\]
so if $\phi_{n}\rightarrow\phi$ in $H^{1}$ (and thus in $H^{2}$), one obtains
\[
\Vert C_{\phi_{n}}-C_{\phi}\Vert_{(n)}\leq\Vert\phi_{n}-\phi\Vert_{2}%
+\Vert\phi_{n}^{2}-\phi^{2}\Vert_{2}+\dots+\Vert\phi_{n}^{m-1}-\phi^{m-1}%
\Vert_{2}.
\]
Since
\[
\Vert\phi_{n}^{k}-\phi^{k}\Vert_{2}\leq\Vert\phi_{n}-\phi\Vert_{2}\Vert
\phi_{n}^{k-1}+\cdots+\phi^{k-1}\Vert_{\infty}\leq k\Vert\phi_{n}-\phi
\Vert_{2},
\]
we have $\Vert C_{\phi_{n}}-C_{\phi}\Vert_{(m)}\leq m^{2}\Vert\phi_{n}%
-\phi\Vert_{2}$. Thus $\phi\mapsto\Vert C_{\phi}\Vert_{(m)}$ is a continuous
map on $ASM(\mathbb{D})^{1}$ for all $m$. As $\lim_{m\rightarrow\infty}\Vert
C_{\phi}\Vert_{(m)}=\Vert C_{\phi}\Vert$ is an increasing limit for all
$\phi\in ASM(\mathbb{D})^{1}$, $N_{1}(\phi)$ is lower semi-continuous for all
$\phi$. In other words, if $\phi$ is given and $\varepsilon>0$ then there
exists $\delta>0$ such that $\Vert\phi-\psi\Vert\leq\delta$ implies $\Vert
C_{\psi}\Vert\geq\Vert C_{\phi}\Vert-\varepsilon$. Note that if $\phi
_{n}\rightarrow\phi$ in $H^{1}$ then $\phi_{n}(0)\rightarrow\phi(0)$. This can
be seen using properties of subharmonic functions (see \cite[Th. 2.6 and
1.4]{PD}). For each $n$, $g_{n}(z)=\phi_{n}(z)-\phi(z)$ is a bounded analytic
function in $D$ and $|g_{n}|$ is subharmonic so
\[
\Vert g_{n}\Vert_{1}=\lim_{r\rightarrow1^{-}}\int_{0}^{2\pi}|g_{n}%
(re^{i\theta})|\ \frac{d\theta}{2\pi}\geq|g_{n}(0)|.
\]
If, on the other hand, $\phi$ is inner, then
\[
\lim_{n\rightarrow\infty}\Vert C_{\phi_{n}}\Vert\leq\lim_{n\rightarrow\infty
}{\frac{1+|\phi_{n}(0)|}{\sqrt{1-|\phi_{n}(0)|^{2}}}}={\frac{1+|\phi
(0)|}{\sqrt{1-|\phi(0)|^{2}}}}=\Vert C_{\phi}\Vert.
\]
Since $N_{1}(\phi)$ is lower semi-continuous, this gives
\[
\lim_{n\rightarrow\infty}\Vert C_{\phi_{n}}\Vert=\Vert C_{\phi}\Vert.
\]
Suppose now that $\phi_{n}\rightarrow\phi$ in $H^{\infty}$ and $\phi
(\mathbb{D})\subseteq r\mathbb{D}$ for some $r<1$. In this case, there exists
an $r^{\prime}<1$ such that $|\phi_{n}(z)|0$.
\end{proof}
The converse of this theorem remains open. For reference, define the maps
$N_{1,e}$ on $ASM(D)^{1}$ and $N_{\infty,e}$ on $ASM(D)^{\infty}$ by
$N_{1,e}(\phi)=N_{\infty,e}(\phi)=\Vert C_{\phi}\Vert_{e}$.
\section{Composition operators of minimal norm\label{minnormsec}}
Consider the upper and lower bounds for the norm of a composition operator on
$H^{2}$ that appear in equation (\ref{bounds}). It is known precisely when
$\Vert C_{\phi}\Vert$ achieves this upper bound (see \cite{JHS1}). Namely, if
$\phi(0)=0$, then $\Vert C_{\phi}\Vert=1$, and if $\phi(0)\not =0$ then $\Vert
C_{\phi}\Vert=\sqrt{\frac{1+|\phi(0)|}{1-|\phi(0)|}}$ if and only if $\phi$ is inner.
Suppose that $\phi$ is now a constant function. A straightforward calculation
shows that $\Vert C_{\phi}\Vert=\sqrt{\frac{1}{1-|\phi(0)|^{2}}}$. Does this
equation characterize the constant analytic self-maps of $D$? \ The answer is yes:
\begin{theorem}
\label{minnorm}Suppose that $\phi$ is an analytic self-map of $D$,
$\phi(0)\not =0$, and $\phi$ is not constant. Then
\[
\Vert C_{\phi}\Vert>\sqrt{\frac{1}{1-|\phi(0)|^{2}}}.
\]
\end{theorem}
\begin{proof}
Let $\phi(0)=a$ and without loss of generality assume that $a$ is real and
positive. Since non-constant analytic maps are open, there is a real $b>a$
such that $\phi(z)=b$ for some $z\in D$. Let $x>0$ and observe that
\[
{\frac{\Vert C_{\phi}^{\ast}(K_{0}+xK_{z})\Vert}{\Vert K_{0}+xK_{z}\Vert}}%
\leq\Vert C_{\phi}\Vert
\]
where $K_{t}(w)={\frac{1}{1-\overline{t}w}}$ is the reproducing kernel at
$t\in D$. We have
\begin{align*}
{\frac{\Vert C_{\phi}^{\ast}(K_{0}+xK_{z})\Vert^{2}}{\Vert K_{0}+xK_{z}%
\Vert^{2}}} & ={\frac{{\frac{1}{1-a^{2}}}+{\frac{2x}{1-ab}}+{\frac{x^{2}%
}{1-b^{2}}}}{1+2x+{\frac{x^{2}}{1-|z|^{2}}}}}\\
& ={\frac{1}{1-a^{2}}}\left( {\frac{1+2x{\frac{1-a^{2}}{1-ab}}+x^{2}%
{\frac{1-a^{2}}{1-b^{2}}}}{1+2x+{\frac{x^{2}}{1-|z|^{2}}}}}\right) .
\end{align*}
Now notice that since $b>a>0$, we have ${\frac{1-a^{2}}{1-ab}}>1$. If $x$
satisfies $0