Let $X$ be a normed space. $F \subset X$ is compact in $X$ if every open cover of $F$ contains a finite subcover, that is, for every family $\set{G_{\alpha}}$ of open sets in $X$ such that $F \subset\bigcup_{\alpha} G_{\alpha}$ there exists $\set{G_{\alpha_1}, \cdots, G_{\alpha_n}}\subset\set{G_{\alpha}}$ such that $F \subset\bigcup_{k =1}^n G_{\alpha_k}$.
\end{defi}
\begin{thm}
$F$ is compact in $X$ if and only if every sequence $\anyseqdef{F}$ has a subsequence that is convergent in $F$.
\end{thm}
\begin{proof}
\noproof
\end{proof}
\begin{defi}
A set $F \subset X$ is said to be relatively compact if $\closure{F}$ is compact.
Every bounded set in a finite-dimensional normed space is relatively compact.
\end{defi}
\begin{defi}
Let $X$ and $Y$ be normed spaces. An operator $T: X \rightarrow Y$ is called a compact linear operator if $T$ is linear and if for every bounded subset $M \subset X$ the image $T(M)$ is relatively compact.
\end{defi}
\begin{thm}[Compactness Criterion]
Let $X$ and $Y$ be normed spaces and $T: X \rightarrow Y$ a linear operator.
Then $T$ is compact if and only if it maps every bounded sequence $\anyseqdef{X}$ onto a sequence $\anyseqdef[Tx]{Y}$ that has a convergent subsequence, that is
\[
\forall\anyseqdef{X} ~\exists\left(Tx_{n_k}\right) \subset Y: \quad Tx_{n_k}\conv{k \rightarrow\infty} y \in Y
\]
\end{thm}
\begin{proof}
\noproof
\end{proof}
\begin{thm}\label{thm:23.6}
If $T: X \rightarrow Y$ is bounded and $\Im T = T(X)$ is finite-dimensional, then $T$ is compact.
\end{thm}
\begin{eg}
Consider $X = Y = l^2$ over the field $\field$. The operator $T$ defined by
is a three-dimensional subspace of $l^2$. By \Cref{thm:23.6}$T$ is compact.
\end{eg}
\begin{thm}\label{thm:23.8}
Let $\seq{T}$ be a sequence of compact linear operators from a normed space $X$ to a Banach space $Y$. If $T_n \rightarrow T$ in $B(X, Y)$ then $T$ is compact.
is also unbounded. We will not discuss what $\domain(T)$ is at this point, however we will do so later.
Here we will only remark that all continuously differentiable functions with compact support and Hermite polynomials belong to $\domain(T)$.
\end{eg}
\begin{eg}
Let $H$ be a complex Hilbert space. Let $T: \domain(T)\rightarrow H$ be a densely defined linear operator.
The adjoint operator $T^*: \domain(T^*)\rightarrow H$ of $T$ is defined as follows. The domain $\domain(T^*)$ of $T^*$ consists of all $y \in H$ such that $\exists y^*\in H$ satisfying
\[
\innerproduct{Tx}{y} = \innerproduct{x}{y^*}, \quad\forall x \in\domain(T)
\]
For each such $y \in\domain(T^*)$ define $T^*y := y^*$. Remark that $\domain(T^*)$ is not necessarily equal to $H$.
Since $\domain(T)$ is dense in $H$, for every $y \in\domain(T^*)$ there exists a unique $y^*$ satisfying the above equation.
Before we discuss the properties of adjoint operators, we will first take a look at the extension of a linear operator. Consider again the differentiation operator
$T_1$ and $T_2$ are different operators, but $\domain(T_1)\subset\domain(T_2)$ and $T_1= T_2\vert_{\domain(T_1)}$.
\end{eg}
\begin{defi}
An operator $T_2$ is said to be an extension of another operator $T_1$ if $\domain(T_1)\subset\domain(T_2)$ and $T_1= T_2\vert_{\domain(T_1)}$.
In this case we write $T_1\subset T_2$.
\end{defi}
\begin{thm}
Let $T: \domain(T)\rightarrow H$ be a linear operator, where $\domain(T)\subset H$. Then
\begin{enumerate}[(i)]
\item$T$ is closed if and only if $x_n \longrightarrow x, ~x_n \in\domain(T)$ and $Tx_n \longrightarrow y$ imply $x \in\domain(T)$ and $Tx = y$.
\item If $T$ is closed and $\domain(T)$ is closed, then $T$ is bounded.
\item Let $T$ be bounded. Then $T$ is closed if and only if $\domain(T)$ is closed.
\end{enumerate}
\end{thm}
\begin{proof}
\noproof
\end{proof}
\begin{thm}
Let $T$ be a densely defined operator on $H$. Then the adjoint operator $T^*$ is closed.
\end{thm}
\begin{proof}
\noproof
\end{proof}
\begin{defi}
If a linear operator $T$ has an extension $T_1$ which is a closed linear operator, then $T$ is said to be closable.
If $T$ is closable, then there exists a minimal closed operator $\closure{T}$ satisfying $T \subset\closure{T}$. The operator $\closure{T}$ is said to be the closure of $T$.
\end{defi}
\begin{thm}
Let $T: \domain(T)\rightarrow H$ be a densely defined linear operator. If $T$ is symmetric, its closure $\closure{T}$ exists and is unique.
\end{thm}
\begin{proof}
\noproof
\end{proof}
\begin{thm}
Let $U: H \rightarrow H$ be a unitary operator. Then there exists a spectral family $\set{E_{\theta}}_{\pi}$ on $[-\pi, \pi]$ such that
\[
U = \int_{-\pi}^{\pi} e^{i\theta}\dd{E_{\theta}}
\]
where the integral is understood in the sense of uniform operator convergence.
\end{thm}
\begin{hproof}
One can show that there exists a bounded self-adjoint linear operator $S$ with $\sigma(S)\subset[-\pi, \pi]$ such that
\begin{equation}
U = e^{iS} = \cos S + i\sin S
\end{equation}
Let $\set{E_{\theta}}$ be a spectral family for $S$ on $[-\pi, \pi]$. Then
\[
S = \int_{-\pi}^{\pi}\theta\dd{E_{\theta}}
\]
Hence
\[
U = e^{iS} = \int_{-\pi}^{\pi}\cos\theta\dd{E_{\theta}} + i \int_{-\pi}^{\pi}\sin\theta\dd{E_{\theta}} = \int_{-\pi}^{\pi} e^{i\theta}\dd{E_{\theta}}
\]
\end{hproof}
\begin{defi}
Let $T: \domain(T)\rightarrow H$ be a self-adjoint linear operator, where $\domain(T)$ is dense in $H$ and $T$ may be unbounded.
Define a new operator
\[
U = (T - iI)(T + iI)^{-1}
\]
called the Cayley transform of $T$. It is defined on the entire Hilbert space since we know that $-i \not\in\sigma(T)\subset\realn$. One can also check that it is unitary and
\[
T = i(I + U)(I - U)^{-1}
\]
\end{defi}
\begin{thm}[Spectral Representation for Unbounded Self-Adjoint Operators]
Let $T: \domain(T)\rightarrow H$ be a self-adjoint linear operator and let $\domain(T)$ be dense in $H$.
Let $U$ be the Cayley transform of $T$ and $\set{\tilde{E}_{\theta}}$ a spectral family in the spectral representation for $-U$. Then
\[
T = \int_{-\pi}^{\pi}\tan\frac{\theta}{2}\dd{\tilde{E}_{\theta}} = \int_{-\infty}^{\infty}\lambda\dd{E_{\lambda}}
\]
where $E_{\lambda}=\tilde{E}_{2\arctan\lambda}, ~\lambda\in\realn$.
