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Let $C(K)$ be the set of all continuous complex valued functions on a compact Hausdorff space $K$. Is it true that $K$ must be finite if $C(K)$ is reflexive? To me it seems true, but I don't know how to prove it. As $C(K)$ is reflexive then we have canonical isometry onto $C(K)^{**}$. How does that help?

ViktorStein
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Toeplitz
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3 Answers3

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The dual space of $C(K)$ is the space of Radon measure $\nu$ on $K$. For every bounded Borel function $u$ on $K$, one can define $\bar u \in C(K)^{**}$ by

$$\bar u (\nu) = \int_K u d\nu,\ \ \ \ \forall \nu \in C(K)^*. $$

Note that the canonical embedding $\Phi: C(K) \to C(K)^{**}$ is given by $$\Phi f(\nu) = \nu(f) = \int_K fd\nu,$$ thus if $\Phi$ is surjective, for all bounded Borel measurable $u$ there is $f\in C(K)$ so that

$$\bar u = \Phi f \Rightarrow \int_K u d\nu = \int_K f d\nu$$

for all Radon measure $\nu$. In particular for the Dirac measure $\nu = \delta_x$, $x\in K$, we have $$u(x) = f(x).$$

Thus all bounded Borel measurable functions are indeed continuous. In particular, all one point sets are open and so $K$ is discrete. By compactness, $K$ is finite.

  • I am confused in last line. ya it is true that one point sets are closed since K is Hausdorff. But how you conclude K is discrete. What is connection with this by showing all $u$ are continuous. – Toeplitz Nov 03 '15 at 22:37
  • @Toeplitz : Sorry. I meant to say one point set are open, as now the function $u$ with $u(x) = 1$ and zero otherwise is continuous. –  Nov 03 '15 at 22:49
  • I am still confused why one point set is open. How you are relating with u. Please write in detail, sorry to bother you – Toeplitz Nov 03 '15 at 23:15
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    That $u$ is continuous, so ${x}= u^{-1} (0.5,1.5)$ is open. @Toeplitz –  Nov 03 '15 at 23:16
  • $u$ is now continuous, then why $u(x)= 1$ and zero? – Toeplitz Nov 03 '15 at 23:22
  • you proved $u$ is continuous function on K, What you mean by $u(x)=1$ and zero otherwise continuous, please explain me little bit. After this I understand how ${x}$ is open. – Toeplitz Nov 04 '15 at 00:20
  • $u(y) = \begin{cases} 1 & \text{if }y=x, \ 0 &\text{if } y\neq x\end{cases}$ is a bounded Borel function, thus is continuous. I set such a $u$. @Toeplitz –  Nov 04 '15 at 03:18
  • I guess wherever u wrote $X$ is $K$, right? I am confused, how $u$ and $\bar u$ is related. What is $\bar u$ mean? – Toeplitz Nov 04 '15 at 04:30
  • Yes, all $X$ should be $K$. Edited. $\bar u$ is an element in $C(K)^{**}$ defined by that integration. @Toeplitz –  Nov 04 '15 at 04:32
  • you are saying that if we define $\bar u$ as above then $\bar u$ is linear and bounded so $\bar u \in C(K)^{**}$, right? – Toeplitz Nov 04 '15 at 04:57
  • Yes. @Toeplitz ${}{}$ –  Nov 04 '15 at 04:58
  • is it easy or difficult to prove dual of $C(K)$ is space of Random measure $\nu$ on K? I don't know proof, I am just accepting as grant because you said. – Toeplitz Nov 04 '15 at 05:04
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    Oh, I would not say that the proof is easy, but this fact is well known and can be found in Theorem 2.14 of Rudin's Real and complex analysis (and also chapter 6). @Toeplitz –  Nov 04 '15 at 05:06
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    That's called Riesz representation theorem. –  Nov 04 '15 at 05:08
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One simple case is when K is a metric space. Since it is compact, it is separable. Choose a dense set $\{y_n\} $ of K, them the functions $x \mapsto d (x,y_n) $ generate a separable subalgebra of C(K) that is dense by Stone weierstrass. Hence C(K) is separable.

However, for each $x\in K $, the evaluation maps $ev_x $ are bounded linear functionals at a distance 1 from each other. Thus if K is infinite, then $C(K)^{\ast} $ is not separable, and so C(K) cannot be reflexive.

OTOH, if K is finite, the C(K) is finite dimensional and hence reflexive.

Prahlad Vaidyanathan
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There is another different answer to you question in the special case that you allow $C(K)$ for $K = \mathbb T$, the torus, to be equipped with the weak topology.

The following is taken from the article Exact Support Recovery for Sparse Spikes Deconvolution by Vincent Duval and Gabriel Peyré, from subsection 2.1 Topology of Radon Measures:

$M(\mathbb T)$ can be defined as the dual space of $C(\mathbb T)$ endowed with the uniform norm $\| \cdot \|_{\infty}$. It is naturally a Banach space when endowed with the dual norm (also called total variation norm) $$ \| m \|_{\text{TV}} := \sup_{\substack{f \in C(\mathbb T) \\ \| f \|_{\infty} = 1}} \int_{\mathbb T} f dm. $$ In that case the dual of $M(\mathbb T)$ is a complicated space, and it is strictly larger than $C(\mathbb T)$, as $C(\mathbb T)$ is not reflexive. However, if we endow $M(\mathbb T)$ with its weak$^*$ topology (i.e. the coarsest topology such that elements of $C(\mathbb T)$ defined continuous linear functionals on $M(\mathbb T)$), then $M(\mathbb T)$ is a locally convex space whose dual is $C(\mathbb T)$. Thus if we endow $C(\mathbb T)$ with its weak and $M(\mathbb T)$ with its weak$^*$-topology, both have symmetrical roles, on is the dual of the other, and conversely.

ViktorStein
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