Proof. Since $\phi \in L_{c}(E; F)^{*}$, there exists $A \subset E$ compact and $\alpha > 0$ such that $|\dpn{T, \phi}{L_c(E; F)}| \le \alpha\sup_{x \in A}\norm{Tx}_{F}$ for all $T \in L(E; F)$. After rescaling $A$, assume without loss of generality that $\alpha = 1$, so that $|\dpn{T, \phi}{L_c(E; F)}| \le \sup_{x \in A}\norm{Tx}_{F}$ for all $T \in L(E; F)$.

By Lemma 12.3.7, there exists $\seq{x_n}\subset E$ with $\limv{n}x_{n} = 0$ and $A \subset \ol{\conv}(\seq{x_n})$. Since $E$ is complete, Mazur’s Theorem implies that $\ol{\conv}(\seq{x_n})$ is compact as well. Thus for each $T \in L(E; F)$,

In particular,

As $\seq{x_n}$ is a null sequence in $E$, $\seq{Tx_n}\in c_{0}(\natp; F)$ for each $T \in L(E; F)$. Let $L = \bracs{\seq{Tx_n}|T \in L(E; F)}$, then $L$ is a subspace of $c_{0}(\natp; F)$. By the above estimate, $\phi$ factors through $L$ as follows:

The Hahn-Banach Theorem then yields an extension $\Phi$ of $\wh \phi$ as shown below:

By Theorem 15.5.5, there exists $\seq{\psi_n}\in l^{1}(\natp; F^{*})$ such that for each $y \in c_{0}(\natp; F)$, $\dpn{y, \Phi}{c_0(\natp; F)}= \sum_{n = 1}^{\infty} \dpn{y_n, \psi_n}{F}$. In particular, for each $T \in L(E; F)$,

$\square$