At the end of the section of the Banach-Alaoglu theorem my professor wrote in his script:
"The B-A theorem can be used to prove that for every Banach space $X$ there exists a compact space $K$ and an isometric embedding $T: X \hookrightarrow C(K)$, i.e $T$ satiesfies $ ||Tx|| = ||x||$ but $T$ is not necessarily surjective. Therefore any Banach space can be realized as a subspace of the space of continuous functions on a compact space. Proving that is given as an exercise."
As it seems to follow directly from the Banach Alaoglu theorem I thought of using the closed unit ball in $X^{\star}$ with respect to the weak-$\star$ topology as $K$. When I am now looking at functions on this $K$ it seems to be related to elements in the bidual space, as we are mapping from $X^{\star}$. But I dont see why that should end up in an isometric embedding. What am I missing here and how could I prove the statement?
Best Answer
Let $K$ be the closed unit ball of $X^{*}$ with its weak* topology. This is a compact Hausdorff space. Provide $C(K)$ with the usual sup norm. Consider the map $\phi: X \to C(K)$ defined by $\phi (x) (x^{*})=x^{*}(x)$. This is a well-defined linear map and $\|\phi (x)\|_{C(K)} =\sup \{|x^{*}(x)|: \|x^{*}\| \leq 1\}=\|x\|$.
[Note that $|x^{*}(x)| \leq \|x^{*}\| \|x\| \leq \|x\|$ for all $x ^{*} \in K$. By Hahn Banach Theorem there exist $x^{*}$ such that $\|x^{*}\|=1$ and $|x^{*}(x)|=\|x\|$. Hence the supremum is exactly $\|x\|$.]