Proof. By (3) of Theorem 13.8.5, 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 $F^{*} \wh \otimes_{\pi} E^{*}$ to $L(E; F^{*})$ is then injective.

Now, let $F := E^{*}$, then the above yields an injection from $E^{**}\wh \otimes_{\pi} E^{*}$ to $L(E; E^{**})$. Let $T \in E \wh \otimes_{\pi} E^{*}$. By Theorem 12.11.4, 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 $x \in E$,

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 Theorem 13.8.5, $E$ has the approximation property.$\square$