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Suppose we have a compact set $K$. I know that the space $C(K,\mathbb{C})$ of continuous functions is complete with respect to the norm $\|f\| = \sup_{x\in K} |f(x)|$. Let $L^{\infty}$ be the space of bounded measurable functions (with the Borel subsets as $\sigma$-algebra). Is $C(K,\mathbb{C})$ pointwise dense in $L^{\infty}$ ?

Now, if this is true and suppose that $K$ is a compact subset of $\mathbb{C}$. I know from the Stone-Weierstrass Theorem that the polynomials on $K$ are uniformly dense in $C(K,\mathbb{C})$, so certainly pointwise dense. Are the polynomials on $K$ also dense in $C(K,\mathbb{C})$ ?

Any help would be appreciated.

KevinDL
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  • See also this related thread – t.b. Jun 07 '12 at 07:30
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    What do you mean by pointwise-dense? $C(K)$ is not dense in $L^\infty$ because $C(K)$ is complete in the $\sup$-norm. – AD - Stop Putin - Jun 07 '12 at 07:31
  • I mean that when you look at $C(K)$ and $L^{\infty}$ as subspaces of $K^{\mathbb{C}}$ with the product topology, does the closure of $C(K)$ contain $L^{\infty}$ ? – KevinDL Jun 07 '12 at 07:34
  • On the other hand, given a measureable function $f$ there is a sequence $f_n\in C(K)$ such that $\lim_{n\to\infty}f_n(x)=f(x)$ for almost all $x$. But I am sure this is not what you mean. – AD - Stop Putin - Jun 07 '12 at 07:37
  • No, that is exactly what I mean (although I just need essentially bounded measurable functions). Maybe I forgot to mention, but with $L^\infty$ I mean the essentially bounded functions modulo the ones which are essentially $0$ everywhere. – KevinDL Jun 07 '12 at 07:39
  • So, I understand that the question is whether the functions of Baire class $\le 1$ are dense in $L^{\infty}$ (http://en.wikipedia.org/wiki/Baire_function). I think this is true, because the characteristic function of any closed set $E$ is of Baire class $\le 1$; it is the pointwise limit of $(1-n\mathrm{dist},(x,E))^+$ as $n\to\infty$. –  Jun 07 '12 at 16:01
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    Re: "I know from the Stone-Weierstrass Theorem that the polynomials on $K$ are uniformly dense in $C(K,\mathbb C)$". -- This is not true in general; $K={z:|z|=1}$ is a counterexample. See Mergelyan's theorem. – ˈjuː.zɚ79365 Jun 19 '13 at 13:31

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