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Let $C_b(X)$ be a space of bounded continuous functions on a locally compact space $X$ equipped with the supremum norm. How to show that $C_b(X)$ is reflexive if and only if $X$ is finite?

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Do you know what the dual of $C_b(X)$ is? –  Christopher A. Wong Dec 4 '12 at 2:18
This can be reduced to the case where $X$ is compact. $C_b(X)\cong C(\beta X)$. –  Jonas Meyer Dec 4 '12 at 5:31
Why do you accept an incomplete answer when you received three full answers? –  Martin Dec 19 '12 at 21:19

4 Answers 4

up vote 2 down vote accepted

This is only a partial answer. I assume that $X$ have finite amount of connected components.

If $X$ is finite then $C_b(X)$ is finite dimensional and the result follows. If $X$ is infinite then $C_b(X)$ and $c_b(X)^{**}$ are infinite dimensional. Assume $C_b(X)$ is reflexive. Since $C_b(X)^{**}$ is the dual space, then by Banach–Alaoglu theorem its unit ball of $C_b(X)^{**}$ is weak-$^*$ compact. Hence by Krein–Milman theorem this unit ball is closed convex hull of its extreme points. Since $C_b(X)$ is reflexive, then $C_b(X)^{**}$ is isometricaly isomorphic to $C_b(X)$. As the consequence extreme points of unit ball of $C_b(X)^{**}$ is one-to-one correspondence with exteme points of unita ball of $C_b(X)$. One can check that there finitely many extreme points in the unit ball of $C_b(X)$. They are of the form $\sum\limits_{i=1}^n\alpha_i{1_{S_i}}$ whre $\{S_i\}_{i=1}^n$ are connected component of $X$ and $|\alpha_i|=1$ for all $i$. Thus unit ball of $C_b(X)^{**}$ is closed convex hull of finite amount of points! This implies that $C_b(X)^{**}$ is finite dimensional. Contradiction, hence $C_b(X)$ is not reflexive.

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Isn't any $f$ with $|f|\equiv 1$ an extreme point in $C_b(X)$? –  Jonas Meyer Dec 4 '12 at 5:07
@JonasMeyer You are right. I thought of $C([0,1])$ when was writing this answer. –  Norbert Dec 4 '12 at 5:11
Still, in $C[0,1]$, $f$ is an extreme point in the unit ball if and only if $|f(x)|=1$ for all $x\in[0,1]$. –  Jonas Meyer Dec 4 '12 at 5:13
@JonasMeyer I consider real case. Well to answer this question one may apply a sledgehammer from this answer –  Norbert Dec 4 '12 at 5:13
Oh yeah, I should have realized that. :) That could handle many cases, then. Thanks. –  Jonas Meyer Dec 4 '12 at 5:15

It is clear that for finite $X$ the space $C_b(X)$ is reflexive because it is finite-dimensional.

Assume $X$ is infinite. Choose a sequence of distinct points $(x_n) \subset X$. The functionals $\delta_n (f) = f(x_n)$ yield an isometric embedding $(a_n) \mapsto \sum a_n \delta_n$ of $\ell^1$ into $C_b(X)^\ast$. This gives a closed non-reflexive subspace of $C_b(X)^\ast$, so $C_b(X)^\ast$ is not reflexive and hence $C_b(X)$ is not reflexive either.

The argument I gave is basically an easier (dual) version of Danny Leung's answer. The fact that $c_0$ embeds isometrically into $C_b(X)$ is a simple application of Urysohn's lemma: choose a sequence of pairwise distinct points in $X$ (which is possible since $X$ is infinite. Use Urysohn's lemma to construct a sequence of functions $f\colon X \to [0,1]$ such that $f_n(x_m) = \delta_{mn}$ with pairwise disjoint supports by induction. To embed $c_0$ isometrically into $C_b(X)$ send a sequence $a_n \in c_0$ to the continuous function $x \mapsto \sum a_n f_n(x)$.

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I didn't see the question about the dual space of $C_b(X)$ in the comments. It has no better description than the space of Radon measures on $\beta X$ via the identification of $C_b(X) = C(\beta X)$ (Stone–Čech compactification + Riesz representation theorem). –  Martin Dec 4 '12 at 18:09

I assume here that $X$ is Hausdorff.

Let $M$ denote the set of all finite, signed, countably additive Borel measures on $X$; we know that $M$ is a closed subspace of $C_b(X)^*$.

Suppose first that $X$ is not discrete; then there exists a bounded, measurable, discontinuous $f : X \to \mathbb{R}$. (For example, if $U$ is open and not closed, set $f = 1_U$.) The map $\mu \mapsto \int f\,d\mu$ is a continuous linear functional on $M$, so by Hahn-Banach it has a continuous extension to a $\xi \in C_b(X)^{**}$. But there can be no $g \in C_b(X)$ with $\xi(\ell) = \ell(g)$ for all $\ell \in C_b(X)^*$, by considering that if we take $\ell = \ell_x$ to be a point mass at an arbitrary $x$, this forces $g(x) = \ell_x(g) = \xi(\ell_x) = f(x)$, but $f$ is not continuous. (Here we used the assumption that $X$ was Hausdorff, so that singleton sets are closed, hence Borel, and "point masses" work the way you think they do.)

Next suppose that $X$ is discrete and infinite. Then the compactly (i.e. finitely) supported functions $C_c(X)$ are a non-dense subset of $C_b(X)$ (constant functions are not in their closure), so by Hahn-Banach there is a nonzero linear functional $\ell \in C_b(X)^*$ with $\ell(g) = 0$ for all $g \in C_c(X)$. Clearly $\ell \notin M$, so $M$ is not dense in $C_b(X)^*$, and by Hahn-Banach again there exists a nonzero $\xi \in C_b(X)^{**}$ with $\xi(\mu) = 0$ for $\mu \in M$. On the other hand, if $g \in C_b(X)$ and $\int g\,d\mu = 0$ for all $\mu \in M$, we have $g=0$, so $\xi$ cannot correspond to an element of $C_b(X)$.

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If $X$ is equipped with discrete topology, then there doesn't exist a discontinuous function. –  Guillermo Jun 23 '14 at 18:59

If X is infinite, then $C_b(X)$ contains a copy of $c_0$.

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