Added corollaries.

src/fa/norm/ap.tex

@@ -99,6 +99,7 @@
         \item $E$ has the approximation property. 
         \item For any Banach space $F$, the closure of $F^* \otimes E$ in $L(F; E)$ is $\mathcal{K}(F; E)$.
         \item For any Banach space $F$, the canonical map $F^* \wh \otimes_\pi E \to L(F; E)$ is injective. 
+        \item The canonical map $E^* \wh \otimes_\pi E \to L(E; E)$ is injective.
     \end{enumerate}
 
     and the following are equivalent:
@@ -144,7 +145,7 @@
     As the above holds for all $\eps > 0$, $\dpn{T, S}{F^* \wh \otimes_\pi E} = 0$. Therefore $T = 0$ as an element of $F^* \wh \otimes_\pi E$. 
 
 
-    $\neg$ (1) $\Rightarrow$ $\neg$ (3): Suppose that $E$ suffers from a lack of the approximation property, then $\text{Id}$ is not in the closure of $E^* \otimes E$ in $L_c(E; E)$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in L_c(E; E)^*$ such that $\dpn{\text{Id}, \phi}{L_c(E; E)} = 1$, but $\dpn{T, \phi}{L_c(E; E)} = 0$ for all $T \in E^* \otimes E$. 
+    $\neg$ (1) $\Rightarrow$ $\neg$ (4): Suppose that $E$ suffers from a lack of the approximation property, then $\text{Id}$ is not in the closure of $E^* \otimes E$ in $L_c(E; E)$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in L_c(E; E)^*$ such that $\dpn{\text{Id}, \phi}{L_c(E; E)} = 1$, but $\dpn{T, \phi}{L_c(E; E)} = 0$ for all $T \in E^* \otimes E$. 
 
     By \autoref{lemma:compact-operator-topology-banach-dual}, there exists a null sequence $\seq{x_n} \subset E$ and $\seq{\psi_n} \in l^1(\natp; E^*)$ such that for each $T \in L(E; E)$, 
     \[
@@ -197,3 +198,41 @@
     Finally, since $A$ is compact, \hyperref[Goldstine's Theorem]{theorem:goldstine-weak} and the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} allow assuming without loss of generality that $T^*$ takes the form of an element of $E_U \otimes E^*$ on $(E^*)_A$. In which case, $T^*$ indeed corresponds to an element of $E^{**} \otimes E^*$ such that $\norm{T^*\phi - \phi}_{E^*} \le \eps$ for all $\phi \in A$. 
 \end{proof}
 
+\begin{corollary}
+\label{corollary:approximation-property-dual}
+    Let $E$ be a Banach space over $K \in \RC$. If $E^*$ has the approximation property, then so does $E$. 
+\end{corollary}
+\begin{proof}
+    By (3) of \autoref{theorem:approximation-property-dual}, for any Banach space $F$, the canonical map from $F^{*} \wh \otimes_\pi E^*$ to $L(F; E^*)$ is injective. Since $L(F; E^*)$ is canonically isomorphic to $L(E; F^*)$, the canonical map from $F^* \otimes_\pi E^*$ to $L(E; F^*)$ is then injective. 
+
+    Now, let $F := E^*$, then the above yields an injection $E^{**} \wh \otimes_\pi E^*$ to $L(E; E^{**})$. Let $T \in E \wh \otimes_\pi E^*$. By \autoref{theorem:metrisable-tensor-product}, there exists $\seq{x_n} \subset E$ and $\seq{\phi_n} \subset E^*$ such that $\sum_{n \in \natp}\norm{x_n}_{E}\norm{\phi_n}_{E^*} < \infty$ and $T = \sum_{n =1}^\infty x_n \otimes \phi_n$. As an operator, for each $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty x_n \dpn{x, \phi_n}{E} \in E
+    \]
+
+    Therefore the restriction of the canonical map $E^{**} \wh \otimes_\pi E^* \to L(E; E^{**})$ to $E \wh \otimes_\pi E^*$ yields an injection into $L(E; E)$. By (4) of \autoref{theorem:approximation-property-dual}, $E$ has the approximation property. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:approximation-property-nuclear}
+    Let $E$ and $F$ be Banach spaces over $K \in \RC$. If $E^*$ or $F$ has the approximation property, then the canonical map
+    \[
+    E^* \otimes_\pi F \to N(E; F) \quad \braks{\sum_{j = 1}^n \phi_j \otimes y_j}(x) = \sum_{j = 1}^n y_j \dpn{x, \phi_j}{E}
+    \]
+
+    extends to an isometric isomorphism. 
+\end{corollary}
+\begin{proof}
+    If $F$ has the approximation property, then the canonical map $E^* \wh \otimes_\pi F \to N(E; F)$ is injective by (3) of \autoref{theorem:approximation-property-dual}. 
+
+    If $E^*$ has the approximation property, then the canonical map
+    \[
+    F^{**} \wh \otimes E^{*} \to N(F^*; E^*) \iso N(E; F^{**})
+    \]
+    
+    is injective. Restricting to $F \wh \otimes E^*$ yields an injection into $N(E; F^{**})$. 
+\end{proof}
+
+
+
+