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In a course of functional analysis, I came across the following exercise which asks me to prove the existence of a specific space under which an isometric embedding exists. I do not seem to get which space it is.

The exact exercise is as follows.

Let $(E, \Vert \cdot \Vert)$ be a normed vector space over $\mathbb{K}$. Prove that there exists a compact topological space $X$ which also possesses the Hausdorff property and satisfies a linear isometric embedding $$e: (E, \Vert \cdot \Vert) \hookrightarrow (C_{\mathbb{K}}(X), \Vert \cdot \Vert_{\infty}).$$ Also show that if $E$ is complete, then $E$ is isometrically isomorph to a closed subspace of $C_{\mathbb{K}}(X)$.

Since I do not know which $X$ is meant I also cannot figure out the second partial task.

Any suggestions or hints are appreciated.

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Hint: Let $B$ be the closed unit ball in $E'$ (the topological dual of $E$). Then $B$ is compact with respect to the weak topology (that's what the Banach-Alaoglu theorem states). Consider the map$$\begin{array}{rccc}\Psi\colon&E&\longrightarrow&C_{\mathbb{K}}(B)\\&x&\mapsto&\left(\begin{array}{ccc}B&\longrightarrow&\mathbb{K}\\\alpha&\mapsto&\alpha(x)\end{array}\right)\end{array}.$$

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  • $\begingroup$ What is $\alpha$ exactly? $\endgroup$ – Taufi Dec 5 '17 at 23:08
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    $\begingroup$ @Taufi An element of the closed unit ball in $E'$. In particular, $\alpha$ is a linear map from $E$ into $\mathbb K$. $\endgroup$ – José Carlos Santos Dec 5 '17 at 23:11
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Hint: Consider the topological dual space $E^*$ with the weak-* topology, and take $$X:=\{f\in E^*\mid\|f\|\le 1\}\,.$$

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