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?
Functional Analysis – Reflexivity of Space of Bounded Functions
banach-spacesfunctional-analysis
Related Solutions
Let $(B(X), \|\cdot\|_\infty)$ be the space of bounded real-valued functions with the sup norm. This space is complete.
Proof: We claim that if $f_n$ is a Cauchy sequence in $\|\cdot\|_\infty$ then its pointwise limit is its limit and in $B(X)$, i.e. it's a real-valued bounded function:
Since for fixed $x$, $f_n(x)$ is a Cauchy sequence in $\mathbb R$ and since $\mathbb R$ is complete its limit is in $\mathbb R$ and hence the pointwise limit $f(x) = \lim_{n \to \infty } f_n(x)$ is a real-valued function. It is also bounded: Let $N$ be such that for $n,m \geq N$ we have $\|f_n - f_m\|_\infty < \frac{1}{2}$. Then for all $x$
$$ |f(x)| \leq |f(x) - f_N(x)| + |f_N(x)| \leq \|f - f_N \|_{\infty} + \|f_N \|_{\infty}$$
where $\|f - f_N \|_{\infty} \leq \frac12$ since for $n \geq N$, $ |f_n(x) - f_N(x)| < \frac12$ for all $x$ and hence $|f(x) - f_N(x)| = |\lim_{n \to \infty} f_n(x) - f_N(x)| = \lim_{n \to \infty} |f_n(x) - f_N(x)| \color{\red}{\leq} \frac12$ (not $<$!) for all $x$ and hence $\sup_x |f(x) - f_N(x)| = \|f-f_N\|_\infty \leq \frac12$.
To finish the proof we need to show $f_n$ converges in norm, i.e. $\|f_N - f\|_\infty \xrightarrow{N \to \infty} 0$:
Let $\varepsilon > 0$. Let $N$ be such that for $n,m \geq N$ we have $\|f_n-f_m\|_\infty < \varepsilon$. Then for all $n \geq N$
$$ |f(x) - f_n(x)| = \lim_{m \to \infty} |f_m(x) - f_n(x)| \leq \varepsilon $$
for all $x$ and hence $\|f- f_n\|_\infty \leq \varepsilon$.
Since my comment seems to have caused some confusion, here's the argument (the answer is yes, of course):
Let $(f_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $B_{b}(\Omega,\Sigma)$, that is to say:
For every $\varepsilon \gt 0$ there exists $N(\varepsilon)$ such that for all $m,n \geq N$ we have $\|f_n - f_m\|= \sup_{x \in \Omega}{|f_n(x) - f_m(x)|} \lt \varepsilon$.
For each $x \in \Omega$ we have $|f_n(x) - f_m(x)| \leq \|f_n - f_m\|$. It follows that $(f_n(x))_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Therefore $f(x) = \lim_{n \to \infty}{f_n(x)}$ exists (i.e. there is a function $f: \Omega \to \mathbb{R}$ such that $f_n \to f$ pointwise).
The pointwise limit of a sequence of measurable functions is measurable: For all $a \in \mathbb{R}$ we have $$ \begin{align*} \{x \in \Omega\,:\,f(x) \leq a\} & = \left\{ x \in \Omega \,:\, (\forall k \in \mathbb{N}) \, (\exists m \in \mathbb{N}) \, (\forall n \geq m)\quad f_n(x) \leq a+ \frac{1}{k} \right\} \\ & = \bigcap_{k = 1}^\infty \bigcup_{m = 1}^\infty \bigcap_{n = m}^\infty \left\{x \in \Omega \,:\, f_n(x) \leq a + \frac{1}{k}\right\} \in \Sigma. \end{align*} $$ It follows that $f$ is measurable.
For $m,n \geq N(\varepsilon)$ we have for all $x \in \Omega$ that $|f_n(x) - f_m(x)| \lt \varepsilon$. Letting $n \to \infty$ we see that $$ |f(x) - f_m(x)| \leq \varepsilon \quad\text{ for all } x \in \Omega\text{ whenever } m \geq N(\varepsilon). $$ It follows that
for all $x \in \Omega$ we have $|f(x)| \leq |f_m(x)| + \varepsilon \leq \|f_m\| + \varepsilon$, so that $f$ is bounded and by point 2 we already know that $f$ is measurable, so $f \in B_b(\Omega,\Sigma)$.
If $m \geq N(\varepsilon)$ we have $\|f - f_m\| \leq \varepsilon$ so that $f$ is the limit of $(f_n)$ in $B_b(\Omega,\Sigma)$.
Best Answer
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 $x \mapsto \sum a_n f_n(x)$.