Advanced Analysis, Notes 14: Banach spaces (application: the Stone–Weierstrass Theorem revisited; structure of C(K))

by Orr Shalit

In this post we will use the Krein–Milman theorem together with the Hahn–Banach theorem to give another proof of the Stone–Weierstrass theorem. The proof we present does not make use of the classical Weierstrass approximation theorem, so we will have here an alternative proof of the classical theorem as well.

1. The Stone–Weierstrass Theorem

We prove the real valued version of the Stone–Weierstrass Theorem. The complex version follows readily (see this post for details).

Theorem 1 (Stone–Weierstrass Theorem): Let A be a closed subalgebra of C_\mathbb{R}(X) that contains the constant functions and separates points. Then A = C_\mathbb{R}(X)

Proof: Recall that C_\mathbb{R}(X)^* = M(X), the space of regular Borel signed measures on X. Recall that the norm of a measure \mu \in M(X) is given by its total variation. Equivalently, \|\mu \| = \mu^+(X) + \mu^-(X), where \mu = \mu^+ - \mu^- is the Hahn–Jordan decomposition of \mu.

Consider the closed unit ball (A^\perp)_1 of the annihilator A^\perp of A:

(A^\perp)_1 = \{\mu \in M(X): \|\mu\| \leq 1 , \int f d \mu = 0\} .

The fact that A is an algebra has the following consequence: if f \in A and \mu \in A^\perp then f d \mu \in (A^\perp)_1Indeed, for all g \in A we have

\int g (f d \mu ) = \int (fg) d \mu = 0.

Our goal is to show that (A^\perp)_1 = \{0\} because the Hahn–Banach Theorem (more precisely, Corollary 16 in Notes 6) then tells us that A = C_\mathbb{R}(X).

Suppose, for the sake of reaching a contradiction, that (A^\perp)_1 \neq \{0\}. Clearly, (A^\perp)_1 is a compact convex subset of M(X)_1. By the Krein–Milman Theorem, (A^\perp)_1 has an extreme point \nu. It is easy to see that the assumption (A^\perp)_1 \neq \{0\} forces \|\nu\| = 1.

We will now show that the support of \nu must be a single point. Assume that we have two distinct points x,y \in supp(\nu). Then by the assumptions on A, there is some f \in A such that f(x) = 0, f(y) = 1. Define g = \frac{f^2}{2\|f\|^2}. Then g \in A and 0 \leq g \leq 1/2. There is a neighborhood U of y such that g is positive on U. Thus, since y is in the support of \nu, \|g \nu\| >0. Likewise, \|(1-g) d \nu\| > 0.

We now form the measures \nu_1 = \frac{g}{\|g d\nu\|} d \nu and \nu_2 = \frac{(1-g)}{\|(1-g)d\nu\|} d \nu. We have noted above that \nu_1, \nu_2 \in A^\perp, and clearly \|\nu_i\| \leq 1. From the fact that 0 < g \leq 1/2 on U, it follows that neither of these measures is equal to \nu.

Recall that by the Hahn–Jordan decomposition theorem, there is a partition X = P \cup N of X into disjoint sets P, N such that \nu(E) \geq 0 if E \subseteq P and \nu(E) \leq 0 if E \subseteq N. Compute

\|(1-g) d\nu\| = \int_P (1-g) d\nu - \int_N (1-g) d\nu = \|\nu\| - \|g d\nu\| = 1 - \|g d\nu\|.

Putting t = \|g d\nu\|, we find \nu = t \nu_1 + (1- t)\nu_2. This is a contradicts the extremeness of \nu. The contradiction shows that the support of \nu must be a singleton \{x\}.

But if supp(\nu) = \{x\}, then \nu = c \delta_x for |c|=1. It follows that \int f \nu = \int f c \delta_x = c f(x) = 0 for all f \in A. But this is absurd, since 1 \in A. The contradiction implies that A^\perp = \{0\}, and the consequence is that A = C_\mathbb{R}(X).

2. The isometric structure of C(K)

The Banach spaces C(K) give a rich source of “concrete” examples of Banach spaces. It turns out that their role in the general theory is much larger than appears at first.

Exercise A: Every Banach space is isometric with a closed subspace of C(K), where K is some compact Hausdorff topological space.

However, the spaces consisting of all continuous functions on some compact Hausdorff space are far from exhausting the examples of Banach spaces.

We will now give an application of extreme points to the structure theory of the spaces C(K). Even though the objects of interest here are the “classical analytic” spaces C(K), I do not consider this as an application of functional analysis to analysis, since we are answering a question that arises only in the framework of functional analysis. This is more of an application of functional analysis to itself. The reader may rest assured that even though this is not a “proper application”, I still find the following theorem and the question that it answers super interesting.

If K and L are two compact Hausdorff spaces, and if h : K \rightarrow L is a homeomorphism, then it is clear that the map C(L) \rightarrow C(K) given by f \mapsto f \circ h is an isometric isomorphism. Also, if u \in C(K) is such that |u|=1, then f \mapsto uf is an isometric automorphism of C(K). It is quite remarkable that every isometric isomorphism arises as the composition of these two examples.

Theorem 2: Let K and L be two compact Hausdorff spaces. Suppose that there is an isometric isomorphism T : C(L) \rightarrow C(K). Then there exists a homeomorphism h : K \rightarrow L and u \in C(K) with |u|\equiv 1 such that for all f \in C(L) and all k \in K

(Tf)(k) = u(k) f(h(k)).

In particular, if C(K) and C(L) are isometric, then K and L are homeomorphic. Thus, the topology of K is completely determined by the structure of the normed space C(K).

Proof: Consider T^* : C(K)^* \rightarrow C(L)^* given by T^* (\rho) = \rho \circ T.

Exercise B: T^* is an isometric isomorphism.

It follows from the exercise that T^* takes the set of extreme points of (C(K)^*)_1 bijectively onto the set of extreme points of (C(L)^*)_1.

Exercise C: The extreme points of (C(K)^*)_1 are of the form c\delta_k, where |c|=1.

It follows that for every k \in K, there is some h(k) \in L and some |u(k)|=1 such that T^*(\delta_k) = u(k) \delta_{h(k)}. Spelling out the duality, we have \delta_k(Tf) = u(k) \delta_{h(k)}(f), or

(Tf)(k) = u(k) f(h(k)).

It remains to show that u and h are continuous. This will remain an exercise.