Theorem 8.5.5 (Linear Extension Theorem (TVS)). Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_{D} = T$.
For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
Proof. By (3) of Definition 8.5.1, $T \in UC(D; F)$. By Theorem 5.5.6, there exists a unique $\ol T \in C(E; F)$ such that $\ol T|_{D} = T$.
It remains to show that $\ol T$ is linear. Since $\ol T$ is continuous, the maps
and
are continuous. Thus $\bracs{A = 0}\supset D \times D$ and $\bracs{M = 0}\supset K \times D$ are both closed. By density of $D$, $\bracs{A = 0}= E \times E$ and $\bracs{M = 0}= K \times E$. Therefore $T$ is linear.$\square$