Added Gantmacher's theorem.

src/fa/norm/compact.tex

@@ -26,3 +26,34 @@
     ($\Leftarrow$): 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}
+