6
$\begingroup$

I want to prove that for Banach space V there is a compact topological space $X$ so that $V$ is isometrically isomorphic to a closed subspace of $C(X)$-continuous function on a (compact) topological space X, equipped with the supremum norm $\|f\|_\infty$ = $\sup_{x \in X} |f(x)|$.

|cite|improve this question
$\endgroup$
5
$\begingroup$

Consider the canonical isometric embedding $\operatorname{ev}: V \to V^{\ast\ast}$.

Let $X$ be the unit ball of $V^\ast$, which is compact in the weak$^\ast$-topology by Alaoglu's theorem. By definition we have $\operatorname{ev}_v(\varphi) = \varphi(v)$ for a linear functional $\varphi \in X$ and $\operatorname{ev}_v$ is a continuous function on $X$. Finally observe that the sup-norm of $\operatorname{ev}_v$ on $X$ is the same as its norm as element of $V^{\ast\ast}$ and thus $v \mapsto \operatorname{ev}_v \in C(X)$ is isometric.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Do you think Tom understood that? Given Banach space $V$, we take $X$ to be the unit ball of $V^*$ in its weak* topology. $X$ is a compact Hausdorff space. The embedding of $V$ into $C(X)$ is defined as follows: element $v$ of $V$ corresponds to function $f$ on $X$, where $f(x) := x(v)$. Elements of $X$ are linear functionals on $V$. $\endgroup$ – GEdgar May 2 '12 at 21:52
  • $\begingroup$ You're right, the phrasing was rather bad. I hope it is clearer now. Thanks! $\endgroup$ – t.b. May 2 '12 at 22:04
  • $\begingroup$ Perhaps I'm missing something, but why mention $V^**$? The norm of $g\in C(X)$ is $\sup_{f\in B_{V^*}} |g(f)|$. For $v\in V$, this gives $\Vert v\Vert_C = \sup_{f\in B_{V^*}} |v(f)|= \sup_{f\in B_{V^*}} |f(v)|=\Vert v\Vert_V$. $\endgroup$ – David Mitra May 2 '12 at 22:05
  • $\begingroup$ @David: the last equality $\sup_{f \in B_{V^\ast}} |f(v)| = \|v\|_V$ expresses that the embedding $V \to V^{\ast\ast}$ is isometric (Hahn-Banach). I prefer to put it this way, but of course you can avoid it, if you want. $\endgroup$ – t.b. May 2 '12 at 22:08
  • $\begingroup$ @t.b.: A small doubt. How do we show that $ev_{V}$ is a closed subspace of $(C(X),\|.\|_\infty)$ ? $\endgroup$ – pikachuchameleon Mar 17 '16 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.