3
$\begingroup$

Let $C(K)=$ all continuous complex valued functions on compact Hausdroff $K$. 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 it helps?

Please anyone's help would be appreciated.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Toeplitz Nov 3 '15 at 22:37
  • $\begingroup$ @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. $\endgroup$ – user99914 Nov 3 '15 at 22:49
  • $\begingroup$ I am still confused why one point set is open. How you are relating with u. Please write in detail, sorry to bother you $\endgroup$ – Toeplitz Nov 3 '15 at 23:15
  • 1
    $\begingroup$ That $u$ is continuous, so $\{x\}= u^{-1} (0.5,1.5)$ is open. @Toeplitz $\endgroup$ – user99914 Nov 3 '15 at 23:16
  • 2
    $\begingroup$ 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 $\endgroup$ – user99914 Nov 4 '15 at 5:06
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What if K is not metric space? $\endgroup$ – Toeplitz Nov 3 '15 at 21:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.