TITLE: Every Banach space is isometric to a subspace of $C(X)$ for some compact space $X$
QUESTION [8 upvotes]: The book of Douglas says on page 12:
Theorem (Banach): Every Banach space $B$ is isometrically isomorphic to a closed subspace of $C(X)$ for some compact Hausdorff space $X$.
Proof: Let $X$ be $(B^*)_1$ with the w*-topology and define $\beta$: $B \rightarrow C(X)$ by
$(\beta f)(\phi) = \phi(f)$ with $\phi \in (B^*)_1$ and $f \in B$.
Next he prove the linearity of $\beta$ and the fact that $\beta$ is an isomorphism using the Hahn-Banach theorem
My question is:
What is $\beta$? I don't understand much of its definition.
Isn't supposed to be a function from $B \rightarrow C(X)$? Here it seems to me a function $B \rightarrow X$ because it's equal to $\phi(f)$ that is a function in $X = (B^*)_1$.

REPLY [8 votes]: You have a Banach space $B$ and want to represent it as a space of functions. We can do it step by step.

A natural choice is to take the candidate domain to be $B^*$: taking $x \in B$ to the function $\hat{x}(\phi) = \phi(x)$ gives you a vector space map $\tau$ from $B$ to scalar-valued linear functions on $B^*$.
If you give $B^*$ the weak-$^*$ topology, each $\hat{x}$ is a continuous function on $B^*$.
If you want the domain to be compact, then restrict $\hat{x}$ to the unit ball $(B^*)_1$ in $B^*$, which is compact by Banach-Alaoglu.
On this now compact domain, it's natural to give continuous functions on $(B^*)_1$ the uniform norm $\|\cdot \|_{\infty}$. The map $\tau$ is now a map between Banach spaces. Is it injective? Is it isometric? Hahn-Banach says yes to both.  

This make $\tau$ a "representation" of $B$ as (continuous) affine functions on a compact convex set. If you don't restrict to $(B^*)_1$, you get the embedding from $B$ to $B^{**}$.