Let $H=(H,(\cdot,\cdot))$ be a Hilbert space and $G\subset H$ equiped with a norm of $H$. Let $F=(F,||\cdot||_{F})$ a Banach space and $S:G \longrightarrow F$ a bounded linear operador. Prove that, there exists a bounded linear operator $T: H \longrightarrow F$ such that it extends $S$ and $||T||_{\mathcal{L}(H,F)}=||S||_{\mathcal{L}(G,F)}$.
I have already been able to prove that such $ T $ exists by taking $ T = \tilde{S} \circ P_\bar {G}$, where $ \tilde{S}: \bar{G} \longrightarrow F $ is the extension of $ S $ (Theorem 2.7.11, Kreyszig) and $ P_\bar{G} : H \longrightarrow \bar{G} $ is the projection on the closed convex $ \bar{G} $.
I could not prove that $||T||_{\mathcal{L}(H,F)}=||S||_{\mathcal{L}(G,F)}$.
Best Answer
Hint: prove two inequalities.
The inequality $\|T\|_{L(H,F)} \ge \|S\|_{L(G,F)}$ is more or less trivial, when you note that for $g \in G$ we have $Tg = Sg$.
For the inequality $\|T\|_{L(H,F)} \le \|S\|_{L(G,F)}$, show:
$\|\bar{S}\|_{L(\bar{G}, F)} \le \|S\|_{L(G,F)}$, using the continuity of $\bar{S}$ and the fact that $G$ is dense in $\bar{G}$
$\|P_{\bar{G}}\|_{L(H, \bar{G})} \le 1$.
(Actually 1 and 2 are both equalities, but you don't need their other directions.)