Theorem 10.3.2 (Linear Extension Theorem (Normed)). Let $E$ be a normed vector space over $K \in \RC$, $F$ be a Banach space 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 $\ol T|_{D} = T$.
$\normn{\ol T}_{L(E; F)}= \normn{T}_{L(D; F)}$.
For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
Proof. (1), (U): By Theorem 8.5.5.
(2): Since $\ol{T}$ is continuous, the function
\[N: E \to \real \quad x \mapsto \normn{\ol T}_{F}- \norm{T}_{L(D; F)}\cdot \norm{x}_{E}\]
is continuous, so $\bracs{N \le 0}\supset D$ is closed. By density of $D$, $\bracs{N \le 0}= E$. Therefore $\normn{\ol T}_{L(E; F)}= \normn{T}_{L(D; F)}$.$\square$