\section{Compact Operators}
\label{section:compact-operator}

\begin{definition}[Compact Operator]
\label{definition:compact-operator}
    Let $E, F$ be locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then $T$ is \textbf{compact} if there exists $U \in \cn_E(0)$ such that $T(U)$ is relatively compact in $F$. 

    The set $\mathcal{K}(E; F)$ is the \textbf{space of compact operators} from $E$ to $F$. 
\end{definition}

\begin{proposition}
\label{proposition:compact-normed-complete}
    Let $E$ be a normed space over $K \in \RC$ and $F$ be a complete Hausdorff topological vector space over $K$, then $\mathcal{K}(E; F)$ is a closed subspace of $L_b(E; F)$. 
\end{proposition}
\begin{proof}
    Let $T \in \ol{\mathcal{K}(E; F)}$ and $U \in \cn_F(0)$ be circled, then there exists $S \in \mathcal{K}(E; F)$ such that $Sx - Tx \in U$ for all $x \in B_E(0, 1)$. Since $S$ is compact, there exists $Y \subset F$ finite with $S(B_E(0, 1)) \subset Y + U$. In which case, $T(B_E(0, 1)) \subset Y + U + U = Y + 2U$. Therefore $T(B_E(0, 1))$ is totally bounded, and as $F$ is complete, relatively compact in $F$. 
\end{proof}

\begin{theorem}[Schauder]
\label{theorem:compact-adjoint}
    Let $E, F$ be normed vector spaces over $K \in \RC$ with $F$ being complete and $T \in L(E; F)$, then $T$ is compact if and only if $T^* \in L(F^*; E^*)$ is compact. 
\end{theorem}
\begin{proof}
    ($\Rightarrow$): Let $\cf \subset F^*$ be bounded, then $\cf$ is equicontinuous and hence relatively compact in the weak* topology by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Since $F$ is complete, $\ol{T(B_E(0, 1))}$ is compact. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is relatively compact with respect to the topology of uniform convergence on $\ol{T(B_E(0, 1))}$. Thus $T^*(\cf) = \bracsn{\phi \circ T|\phi \in \cf}$ is relatively compact with respect to the topology of uniform convergence on $B_E(0, 1)$. In other words, $T^*(\cf)$ is relatively compact in $E^*$, and $T^*$ is a compact operator. 

    ($\Leftarrow$): By \autoref{proposition:operator-space-completeness}, $E^*$ is complete. The preceding case then implies that $T^{**} \in L(E^{**}; F^{**})$ is compact. As such, its restriction to $E$, being identified with $T$, is also compact. 
\end{proof}

\begin{theorem}[Gantmacher]
\label{theorem:weakly-compact-biadjoint}
    Let $E, F$ be Banach spaces over $K \in \RC$, and $T \in L(E; F)$, then the following are equivalent:
    \begin{enumerate}
        \item $T(B_E(0, 1))$ is relatively $\sigma(F, F^*)$-compact. 
        \item $T^{**}(E^{**}) \subset F \subset F^{**}$. 
    \end{enumerate}
\end{theorem}
\begin{proof}
    Let $B_E$ be the closed unit ball of $E$, and $B_{E^{**}}$ be the closed unit ball of $E^{**}$. By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}, $B_E$ is $\sigma(E^{**}, E^*)$-dense in $B_{E^{**}}$. Since $T^{**}$ is a $\sigma(E^{**}, E^*)$-$\sigma(F^{**}, F^*)$-continuous extension of $T$, 
    \[
    T^{**}(B_{E^{**}}) = T^{**}\paren{\ol{B_E}^{\sigma(E^{**}, E^*)}} \subset \ol{T(B_E)}^{\sigma(F^{**}, F^*)}
    \]

    by \autoref{proposition:closure-of-image}. On the other hand, $B_{E^{**}}$ is $\sigma(E^{**}, E^*)$-compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Hence \autoref{proposition:compact-extensions} implies that $T^{**}(B_{E^{**}})$ is $\sigma(F^{**}, F^*)$-closed, so
    \[
    T^{**}(B_{E^{**}}) = T^{**}\paren{\ol{B_E}^{\sigma(E^{**}, E^*)}} = \ol{T(B_E)}^{\sigma(F^{**}, F^*)}
    \]

    The above equality shows that the following five statements are equivalent:
    \begin{enumerate}[label=(\roman*)]
        \item $T(B_E)$ is relatively $\sigma(F, F^*)$-compact. 
        \item $\ol{T(B_E)}^{\sigma(F^{**}, F^*)} = \ol{T(B_E)}^{\sigma(F, F^*)}$. 
        \item $\ol{T(B_E)}^{\sigma(F^{**}, F^*)} \subset F$. 
        \item $T^{**}(B_{E^{**}}) \subset F$. 
        \item $T^{**}(E^{**}) \subset F$. 
    \end{enumerate}

    where (i) is equivalent to (1), and (v) is equivalent to (2). 
\end{proof}