Proof, [Corollary III.7.1.1-III.7.1.3, SW99]. Let $\seq{\phi_n}\subset F^{*}$ be an equicontinuous sequence, $B \in B(G)$ be convex and circled, $\seq{y_n}\subset B$, and $\seq{\lambda_n}\subset K$ such that

1. (a)

   The auxiliary space $G_{B}$ is a Banach space.

2. (b)

   For each $x \in F$, $Sx = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{F}$.

3. (c)

   $\sum_{n \in \natp}|\lambda_{n}| < \infty$.

(1): Let $U = \bigcap_{n \in \natp}\phi_{n}^{-1}(B_{K}(0, 1))$, then since $\seq{\phi_n}$ is equicontinuous, $U$ is a convex and circled neighbourhood of $0$ in $F$. Given that $G_{B}$ is complete, $S$ is the following composition of continuous maps:

By Tychonoff’s Theorem, $\overline{B_K(0,1)}^{\natp}$ is compact. Since $S(U)$ is contained in its image in the above diagram, $S(U)$ is relatively compact.

(2): Since $T \in L(E; F)$, $\seq{\phi_n \circ T}\subset E^{*}$ is equicontinuous. Thus for any $x \in E$,

and $S \circ T \in N(E; G)$.

(3): Using (1), assume without loss of generality that $B$ is also compact. In which case, $R(B)$ is a convex, circled, and compact set in $H$ containing $0$. Thus $H_{R(B)}$ is a Banach space. For each $x \in F$,

and $R \circ S \in N(F; H)$.

(4): By the linear extension theorem, such an extension exists and is unique. Moreover, $\seq{\phi_n}\subset F^{*}$ extend into an equicontinuous family $\bracsn{\wh \phi_n}_{1}^{\infty} \subset \wh F^{*}$. Since $G_{B}$ is complete, the extension $\wh S \in L(\wh F; G)$ takes the form

Therefore $\wh S \in N(\wh F; G)$.$\square$