12.3 Linear Maps

Proposition 12.3.1.label Let $E, F$ be normed vector spaces, then the topology on $L_{b}(E; F)$ is induced by the operator norm

\[\norm{\cdot}_{L(E; F)}: L(E; F) \to [0, \infty) \quad T \mapsto \sup_{\substack{x \in E \\ \norm{x}_E = 1}}Tx\]

Proof. By Proposition 11.8.1.$\square$

Theorem 12.3.2 (Linear Extension Theorem (Normed)).label 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:

  1. (1)

    There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_{D} = T$.

  2. (2)

    $\normn{\ol T}_{L(E; F)}= \normn{T}_{L(D; F)}$.

  3. (U)

    For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.

Proof. (1), (U): By Theorem 10.5.5.

(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 $\normn{\ol T}_{L(E; F)}= \normn{T}_{L(D; F)}$.$\square$