# Space of bounded functions is reflexive if the domain is finite

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

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
Where are you using the compactness of $X$? – J.G. Oct 1 '15 at 16:30
I'm not sure that this is exactly easier. Proving that $\ell^1$ is not reflexive is actually a bit challenging, and it requires the axiom of choice. Proving that $c_0$ is not reflexive is a lot easier and can be done explicitly. – Nate Eldredge Nov 4 '15 at 21:46
@NateEldredge: Both answers use Urysohn and that a closed subspace of a reflexive space is reflexive; the usual proof of the latter fact uses Hahn-Banach (it is consistent that $\ell^\infty$ is reflexive, but, as you note, its closed subspace $c_0$ is never reflexive). Once you know that fact, it is not hard to deduce "$X$ reflexive iff $X^\ast$ is reflexive". But this also shows that $\ell^1 = (c_0)^\ast$ is not reflexive without needing to "construct" a Banach limit. I would say both answers are about on the same level of difficulty. – user160629 Jun 12 at 0:30
@J.G. If you want to prove that $\delta_n \neq \delta_m$, you need some form of Urysohn's lemma (you need to find a function $f: X \to [0,1]$ such that $f(x_n) = 0$ and $f(x_m) = 1$). – user160629 Jun 12 at 0:33

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

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