\documentclass[12pt]{article}
\usepackage{amsmath,amsfonts,fullpage}

\def\C{{\mathbb C}}
\def\e{{\mathrm e}}
\def\d{{\mathrm d}}
\def\i{{\mathrm i}}
\def\interior{\operatorname{int}}
\def\dist{\operatorname{dist}}
\def\Ker{\operatorname{Ker}}
\def\Range{\operatorname{Range}}
\def\N{{\mathbb N}}
\def\R{{\mathbb R}}
\def\T{{\mathbb T}}
\def\Z{{\mathbb Z}}
\def\L{{\mathcal L}}
\def\K{{\mathcal K}}
\def\D{{\mathcal D}}

\begin{document}

\title{Functional Analysis}
\author{Homework 9}
\date{\today}
\maketitle

\begin{enumerate}

\item Let $E,F$ be Banach spaces and let $A \colon \D(A) \subset E \to
F$ be a linear, closed, densely defined, unbounded operator. 

Show that $A$ is invertible if and only if
\begin{itemize}
\item[(i)] There exists $\alpha>0$ such that for all $u \in \D(A)$
\[
  \lVert Au \rVert_F \geq \alpha \, \lVert u \rVert_E \,.
\]
\item[(ii)] $\Range A$ is dense in $F$.
\end{itemize}
Then, moreover, $\lVert A^{-1} \rVert_{\L(F,E)} \leq \dfrac1\alpha$.

\item Let $H = \ell^2(\N)$.  
\begin{enumerate}
\item Show that the operator defined via $A e_n
= n^{-1} \, e_n$ for $n \in \N$, where $\{e_n\}$ is the canonical
basis in $\ell^2$, is compact.  
\item What is $\sigma(A)$?
\item Let $R$ be the right shift operator on $\ell^2$, i.e.\ $R(a_1,
a_2, \dots) = (0, a_1, a_2, \dots)$.  What is $\sigma(RA)$?
\end{enumerate}

\item 
\begin{enumerate}
\item Let $E,F$ be Banach spaces and $A \colon \D(A) \subset E \to F$
be a linear, closed, densely defined, and invertible unbounded
operator.  Show that if $B \in \L(E,F)$ such that $\lVert A^{-1}B
\rVert_{\L(E)}<1$, then $A+B$ is invertible with
\[
  (A+B)^{-1} = (I + A^{-1}B)^{-1} \, A^{-1} \,.
\]

\emph{Note:} Use Neumann series.  We sketched the proof in class.
\item Now take $F=E$.  Use part (a) to show that the \emph{resolvent
map} 
\[
  \lambda \mapsto (A - \lambda I)^{-1}
\]
is a continuous map from $\rho(A) \subset \C \to \L(E)$.  

\item Show further that the resolvent map is differentiable (hence
analytic) on $\rho(A)$.

\emph{Hint:} Note that
\[
  \frac{(A-\lambda I)^{-1}(A-\lambda_0I)^{-1}}{\lambda-\lambda_0}
  = (A-\lambda_0 I)^{-1} \, (A-\lambda I)^{-1} \,.
\]
(Why?)
\end{enumerate}

\item Let $E,F$ be Banach spaces and 
\begin{align*}
  A & \colon \D(A) \subset E \to F \,, \\
  A^\dagger & \colon \D(A^\dagger) \subset F^* \to E^*
\end{align*}
be two linear, closed, densely defined, unbounded operators such that
\[
  \langle f, Au \rangle_{F^*,F} 
  = \langle A^\dagger f, u \rangle_{E^*,E} 
\]
for all $f \in \D(A^\dagger)$ and $u \in \D(A)$.

Now suppose there exist
\begin{align*}
  G & \colon F \to \D(A) \subset E \,, \\
  G^\dagger & \colon E^* \to \D(A^\dagger) \subset F^*
\end{align*}
such that
\begin{equation*}
  AG = I_F \qquad\text{and}\qquad A^\dagger G^\dagger = I_{E^*} \,.
  \tag{*}
\end{equation*}
Then $A^\dagger = A^*$.

\emph{Hint:} The issue here is whether $\D(A^\dagger)=\D(A^*)$.  Note
that $f \in \D(A^*)$ if and only if there exists $g \in E^*$ such that 
\[
  \langle f,Au \rangle_{F^*,F} = \langle g,u \rangle_{E^*,E} \,.
\]

\emph{Note:} The above criterion is useful because, on the one hand,
$\D(A^*)$ is often difficult to establish directly.  On the other
hand, the ``formal adjoint'' $A^\dagger$ and the respective inverses
can often be obtained by explicit computation.  So the proof that the
``natural domain'' of the formal adjoint, $\D(A^\dagger)$, already
exhausts all of $\D(A^*)$ reduces to the verification of the two
identities (*).  See next question.

\item Consider $A=-\partial_{xx}$ on $L^2([0,1])$.  
\begin{enumerate}
\item Show that $A$ is self-adjoint with
\[
  \D(A) = \{ u \in L^2([0,1]) \colon \partial_{xx}u \in L^2([0,1]), 
  u(0)=u(1)=0 \} \,.
\]
\item Show that $A$ is symmetric ($L^*$ is an extension of $L$), but
not self-adjoint with
\[
  \D(A) = \{ u \in L^2([0,1]) \colon \partial_{xx}u \in L^2([0,1]), 
  u(0)=\partial_xu(0)=u(1)=\partial_xu(1)=0 \} \,.
\]
\end{enumerate}
Describe the point, continuous, and residual spectrum in each case.

\end{enumerate}
\end{document}
