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Let $A$ be a complex algebra of bounded measurable functions on the measure space $(X,\mu)$ (case of $[0,1]$ with Lebesgue measure is enough for me) closed under conjugation. Assume that $A$ separates points, i.e. there is no non-trivial measurable partition of $X$ such that each function in $A$ is constant on (almost every) part. Is it true that $A$ is dense in $L^p(X,\mu)$ for $1\leq p < \infty$?

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@Yemon: oh, of course. Thank you. –  Fedor Petrov May 4 '12 at 19:12
    
$L^1$ is a Banach algebra. The spectral calculus is probably a mess, but does this not help. Similar $L^\infty$. Then you can interpolate? –  plusepsilon.de May 4 '12 at 19:16
    
Here is another idea: $L^p$ and $L^q$ are dual for appropiate $p$ and $q$, it is certainly enough to show that for every functional is distinguiehed by elements of $A$. This does not help for $L^\infty$, where the dual is $ba \neq L^p$. –  plusepsilon.de May 4 '12 at 19:29
    
I meant in my last sentence, then one can do $L^\infty$ seperately, if it works for $ba$. I actually would believe that $L^\infty$ implies the other stuff, since it will imply it for $L^1$, since dense in $ba$, and for all other $L^p$ since the characteristic functions are in all $L^p$ and span, and interpolation takes care over convergence in the right notion. –  plusepsilon.de May 4 '12 at 19:36
    
<deep breath> L^1 IS NOT A BANACH ALGEBRA FOR POINTWISE PRODUCT –  Yemon Choi May 4 '12 at 19:49

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up vote 7 down vote accepted

Yes (I assume that the measure is finite). Here is a proof that uses the von Neumann bicommutant theorem (or rather Kaplansky's density theorem).

See $A \subset L^\infty(X,\mu) \subset B(L^2(X,\mu))$ where $L^\infty$ acts on $L^2$ by pointwise multiplication. Then the assumption that $A$ separates points is exactly that the commutant of $A$ is $L^\infty(X,\mu)$, so that the bicommutant of $A$ is $L^\infty(X,\mu)$. Therefore, by Kaplansky's density theorem, any $f \in L^\infty$ with $\|f\|_\infty \leq 1$ belongs to the strong operator topology closure of $\{g \in A, \|g\|_\infty\leq 1\}$. Equivalently, there is a net $g_\alpha \in A$ with $\|g_\alpha\|_\infty \leq 1$ such that, for every $\xi \in L^2$, $\|g_\alpha \xi - f \xi\|_2\to 1$. In particular (using that the constant function $1$ belongs to $L^2$), $\|g_\alpha - f\|_2 \to 0$. But by this implies that for every $1\leq p < \infty$, $\|g_\alpha - f\|_p \to 0$~: if $p<2$ this is because the $L^p \subset L^2$ (the measure is finite), whereas if $p>2$ this is the inequality $\| \cdot \|_p \leq \|\cdot \|_\infty^\theta \|\cdot \|_2^{1-\theta}$ for $\theta=1-2/p>0$.

This proves that the $\| \cdot \|_p$-closure of $A$ contains $L^\infty$, and hence it is $L^p$.

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Why go to the non commutative setting, Mikael? Just use the Stone Weierstrass theorem in $K(K)$ with $K$ the Stone space of $L_\infty(\mu)$. $\mu$ induces a measure $\nu$ on $K$ s.t. for all $p$, $L_p(\mu)$ is naturally identified with $L_p(K,\nu)$. –  Bill Johnson May 5 '12 at 16:12
    
@Bill: I am so illiterate, sorry, what is a "Stone space" here? –  Fedor Petrov May 7 '12 at 19:15
    
Presumably, the maximal ideal space or, equivalently, the Stone space of the Boolean algebra $\wp(X)$ modulo $\mu$-null sets. –  Jan Veselý May 9 '12 at 22:57
    
Then how do we conclude that Stone Weierstrass condition holds from the separation property? –  Fedor Petrov May 12 '12 at 21:22

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