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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 $\|f\|_\infty$ = $\sup_{x \in X} |f(x)|$.

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1 Answer 1

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.

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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 $V$. –  GEdgar May 2 '12 at 21:52
    
You're right, the phrasing was rather bad. I hope it is clearer now. Thanks! –  t.b. May 2 '12 at 22:04
    
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 $\Vert v\Vert_C = \sup_{f\in B_{V^*}} |v(f)|= \sup_{f\in B_{V^*}} |f(v)|=\Vert v\Vert_V$. –  David Mitra May 2 '12 at 22:05
    
@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. –  t.b. May 2 '12 at 22:08

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