

\documentclass[11pt]{amsart}
\baselineskip=7.0mm
\usepackage{amsmath}
\usepackage{amssymb}

\theoremstyle{plain}


%\newtheorem{theorem}{Theorem}[chapter]
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{prop}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{xca}[theorem]{Exercise}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}

%\numberwithin{equation}{section}


\setlength{\textwidth}{5.8in} \setlength{\textheight}{8.0in}
\hoffset=-0.45truein \voffset=0.1truein


\begin{document}

\newcommand{\x}{\mathcal X}
\newcommand{\s}{\mathcal S}
\newcommand{\xx}{\mathcal X^*}
\newcommand{\xxx}{\mathcal X^{**}}
\newcommand{\y}{\mathcal Y}
\newcommand{\m}{\mathcal M}
\newcommand{\h}{\mathcal H}
\newcommand{\nin}{\noindent}
%\define\hp{\overset\rightharpoonup\to}
%\define\rp{\rightharpoonup}
%\def\intt{\int\!\!\!\!\!\!{-}}




\title{PROJECT: OPERATORS IN HILBERT SPACES}
\maketitle

Prove all the following lemmas, theorems and examples.

\section{Compact Operators}

In the following, all spaces $\x$, $\y$ and $\mathcal{Z}$ are
assumed to be Hilbert spaces.

\begin{lemma} Let $A:
\x \to \y$ be a linear operator. Then the following statements are
equivalent.

(1) ${A(B_1)}$ is sequentially compact in $\y$ where $B_1 = \{
x\in \x; \|x\| \leq 1\}$.

(2) ${A(B)}$ is sequentially compact in $\y$ where $B$ is any
bounded set in $\x$.

(3) For any bounded sequence $\{x_n\}$ in $\x$, $\{A x_n\}$ has a
convergent subsequence in $\y$.
\end{lemma}

\begin{definition}  Such a linear operator is called compact.  Let
$\mathcal{C}(\x,\y)=$ all compact linear operators from $\x$ to
$\y$.\end{definition}

\begin{lemma} There hold the following results.

(1) $\mathcal{C}(\x,\y) \subseteq \mathcal{B}(\x,\y)$.

(2) $\mathcal{C}(\x,\y)$ is a linear subspace and closed in
$\mathcal{B}(\x,\y)$.

(3) If $A \in \mathcal{C}(\x,\y), B \in
\mathcal{B}(\y,\mathcal{Z})$, then $BA \in
\mathcal{C}(\x,\mathcal{Z})$.  If $A\in \mathcal{B}(\x,\y), B\in
\mathcal{C}(\y,\mathcal{Z})$, then $BA \in
\mathcal{C}(\x,\mathcal{Z})$.

(4) Let $A\in \mathcal{C}(\x,\y)$ and $\m$ be a closed subspace in
$\x$. Then $A|\m\in \mathcal{C}(\m,\y)$.

(5) Let $A\in \mathcal{C}(\x,\y)$. Then
$\overline{\mathcal{R}(A)}$ is separable.\end{lemma}


\begin{theorem}\label{com.1}  Let
$A\in\mathcal{B}(\x, \y)$.  Then $A\in \mathcal{C}(\x, \y)$ if and
only if $A^* \in \mathcal{C}(\y, \x)$.\end{theorem}

Hint: If $A\in \mathcal{C}(\x, \y)$, then $AA^*\in \mathcal{C}(\y,
\y)$.

Now we provide several methods to verify whether a bounded linear
operator is compact.

\begin{definition}  An operator $A \in \mathcal{B}(\x,\y)$ is completely
continuous if for any $x_n \rightharpoonup x$ weakly in $\x$ there
holds $Ax_n \to Ax$ in norm in $\y$.\end{definition}

\begin{lemma} Let $A \in \mathcal{B}(\x,\y)$. Then
$A \in \mathcal{C}(\x,\y)$ if and only if $A$ is completely
continuous. \end{lemma}



\begin{definition}  A linear operator $A: \x \to \y$ has a finite
rank if $\mathcal{R}(A)$ is finite dimensional.  Let
$\mathcal{C}_0(\x, \y) =$ all linear operators of finite
rank.\end{definition}

Operators in $\mathcal{C}_0(\x,\y)$ have a simple formula.


\begin{lemma}  Let
$A\in \mathcal{B}(\x, \y)$. Then $A\in \mathcal{C}_0(\x, \y)$ if
and only if $A^* \in \mathcal{C}_0(\y, \x)$.\end{lemma}

Hint: Let $A\in \mathcal{C}_0(\x, \y)$. Then there exist $\{x_1 ,
\cdots , x_n \}\subset \x $ and $\{y_1 , \cdots , y_n \}\subset \y
$ such that
$$
Ax = \sum_{i=1}^n(x, x_i)y_i\text{ for any } x \in \x.
$$


\begin{lemma} (1) $\mathcal{C}_0(\x, \y)$ is a linear space
and $\mathcal{C}_0(\x, \y) \subseteq \mathcal{C}(\x, \y)$.

(2) $\mathcal{C}_0(\x, \y)$ is dense in $\mathcal{C}(\x, \y)$,
i.e., $\overline{\mathcal{C}_0(\x,\y)} =
\mathcal{C}(\x,\y)$.\end{lemma}



\begin{example}  Let $\Omega$ be a bounded measurable subset in
$\mathbb{R}^n$ and $K\in \mathcal{L}^2 (\Omega \times \Omega)$.
Define $A: \mathcal{L}^2 (\Omega) \to \mathcal{L}^2 (\Omega)$ by
$$
Au (x) = \int_{\Omega} K(x, y) u (y) dy \text{ for any } u \in
\mathcal{L}^2 (\Omega).
$$
Then $A$ is a compact operator on $\mathcal{L}^2 (\Omega)$.
\end{example}








\section{Self-Ajoint Operators}

In this section we always assume that $\x$ is a
$\mathbb{C}$-Hilbert space.

\begin{definition}  $A\in \mathcal{B}(\x)$ is self-adjoint if
$A=A^*$, i.e.,  $ (Ax,y) = (x,Ay) \text{ for any }  x,y\in\x.
$\end{definition}

\begin{lemma}  Let $A\in \mathcal{B}(\x)$. Then $A$ is self-adjoint
if and only if
$(Ax, x)\in\mathbb{R}$ for any $x\in \x$.\end{lemma}

Hint: Consider $(A(x+cy), x+cy)$ for $c=1$ and $c=i$.

The preceding result is false if $\x$ is only assumed to be an
$\mathbb{R}$-Hilbert space.  Indeed for any operator $A$ in an
$\mathbb{R}$-Hilbert space, $(Ax,y)\in\mathbb{R}$.

\begin{lemma}  Let $A\in \mathcal{B}(\x)$ be self-adjoint. Then
$$
\|A\|=\sup\{|(Ax,x)|; \| x\|=1\}.
$$\end{lemma}



\begin{lemma}  Let $A\in \mathcal{B}(\x)$ be self-adjoint
and $\lambda\in\mathbb C$. Then $\lambda I-A$ is invertible in
$\mathcal{B}(\x)$ if and only if there exists a constant $c>0$
such that $$\|(\lambda I-T)x\|\ge c\|x\|\text{ for any }x\in\x.$$
\end{lemma}

For a self-adjoint operator $A\in\mathcal{B}(\x)$, we define
$$m_-=\inf_{x\in\x, \|x\|=1}(Ax, x),\quad
m_+=\sup_{x\in\x, \|x\|=1}(Ax, x).$$

\begin{theorem}  Let $A\in \mathcal{B}(\x)$ be self-adjoint.

(1) For any $\lambda\notin\mathbb{R}$ the operator $\lambda I-A$
is invertible in $\mathcal{B}(\x)$.

(2) For any $\lambda\in\mathbb R\setminus [m_-, m_+]$, $\lambda
I-A$ is invertible in $\mathcal{B}(\x)$.
\end{theorem}

\begin{lemma}  Let $A\in \mathcal{B}(\x)$ be self-adjoint
such that $(Ax, x)\ge 0$ for any $x\in\x$.  Then

(1) $|(Ax, y)|^2\le (Ax, x)(Ay, y)$ for any $x, y\in\x$.

(2) $\|Ax\|^2\le \|A\|(Ax, x)$ for any $x\in\x$.
\end{lemma}



\begin{cor}  Let $A\in \mathcal{B}(\x)$ be self-adjoint. Then
$m_+I-A$ and $m_-I-A$ are not invertible in $\mathcal{B}(\x)$.
\end{cor}

\begin{theorem} Let $A\in \mathcal{B}(\x)$ be self-adjoint.

(1) If there exists an $x_-\in\x$ with $\|x_-\|=1$ and $(Ax_-,
x_-)=m_-$, then $m_-$ is an eigenvalue of $A$ and $x_-$ is a
corresponding eigenvector.

(1) If there exists an $x_+\in\x$ with $\|x_+\|=1$ and $(Ax_+,
x_+)=m_+$, then $m_+$ is an eigenvalue of $A$ and $x_+$ is a
corresponding eigenvector.
\end{theorem}




\end{document}
