When is $C(X)$ reflexive?

Question

Let $X$ be a compact Hausdorff space. What are sufficient and necessary conditions on $X$ under which $C(X)$ would be a reflexive Banach space.

Is there a non reflexive Banach space $C(X)$ such that $C(X)$ is isomorphic to its bidual?

Answer 1

$C(X)$ is reflexive if and only if $X$ is finite (in which case $C(X)$ is finite-dimensional). Otherwise, you will find a seqeunce of norm-one disjointly supported functions. These functions span an (isometric) copy of $c_0$, which is certainly not reflexive.

$C(X)$ is never isomorphic to $C(X)^{**}$ unless $X$ is finite. This is because $$C(X)^{**}\cong (\bigoplus_{i\in I} L_\infty(\mu_i))_{\ell_\infty(I)},$$ where $(\mu_i)_{i\in I}$ is a maximal family of pairwise singular probability measures on $X$, so if there were an isomorphism $C(X)\to C(X)^{**}$ we could iterate taking even duals and we would run quickly into cardinality troubles.

Answer 2

Theorem. Let $X$ be a compact Hausdorff space. Then $C(X)$ is reflexive iff $X$ is finite.

Proof (elementary, I hope): A finite Hausdorff space $X$ is clearly discrete, so if $X$ is finite, then $C(X)$ is isomorphic to $\mathbb{K}^X$ (where $\mathbb{K}=\mathbb C$ or $\mathbb{K}=\mathbb R$), hence reflexive.

Assume then that $X$ is infinite. By the Banach-Alaoglu theorem [Rudin FA: 3.15], it suffices to show that the weak unit ball $U$ of $C(X)$ is not compact, e.g., that $\mu(U)$ is not compact by some $\mu\in C(X)^*$, as "Continuous images of compact sets are compact" [Google]. (Indeed, then $C(X)$ cannot be a dual space, by Banach-Alaoglu.)

But $C(X)^*=rca(X)$. https://en.wikipedia.org/wiki/Continuous_functions_on_a_compact_Hausdorff_space

$\mu=\sum_k 2^{-k} (-1)^k 1_{x_k}\in rca(X)$ has supremum $1$ on $U$ but no maximum, so we are done. (Here $1_x$ is the unit mass at $x$.)

Indeed, let $A_n=1,3,...,2n-1$, $B_n=2,4,...,2n$ (both closed, being finite unions of singletons), and let $f_n(A_n)=-1$, $f_n(B_n)=1$, $f_n\in C(X,[-1,1])$, by https://en.wikipedia.org/wiki/Urysohn%27s_lemma

Then $f_n\in U$ and $\mu(f_n)\to 1$, as $n\to\infty$.

But if $g\in U$, $\mu(g)=1$, then $g(x_n)=(-1)^n$. Hence, then $g(x_n)$ is not a Cauchy sequence, violating $g(x_n)\to g(x)$ (i.e., $g$ is discontinuous), a contradiction, so $1\not\in \mu(U)$, QED.