Every Banach space is isometric to a subspace of $C(X)$ for some compact space $X$

Question

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 $X = (B^*)_1$.

Answer 1

The definition is right. The function $\beta$ evaluates on elements of $B$. Given $f\in B$ (this is a terrible choice in notation, to use $f$ for a point in an argument in which there are functions involved), $\beta f$ should be a function on $X$; so we need to say what its value is on some $\phi\in X$, and this value is $\phi(f)$.

Answer 2

The definition is as follows : For each $f \in B$, define $\beta(f) :X \to \mathbb{C}$ by $$ \beta(f)(\phi) = \phi(f) $$ For instance, if $B = L^1(\mathbb{R})$, then $X$ is the unit ball of $L^{\infty}$. In particular, it contains things like $$ \phi_x : \mathbb{R}\to \mathbb{C} \text{ given by } t \mapsto e^{itx} $$ and $$ \beta(f)(\phi_x) := \phi_x(f) = \int f(t)e^{itx}dt $$ Although this is not quite accurate, you can think of $\beta(f)$ as some sort of Fourier transform of $f$.

Answer 3

You have a Banach space $B$ and want to represent it as a space of functions. We can do it step by step.

1. 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 $B^*$.

2. If you give $B^*$ the weak-$^*$ topology, each $\hat{x}$ is a continuous function on $B^*$.

3. 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.

4. 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 $B^{**}$.