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 $1\leq p < \infty$?

  • $\begingroup$ $L^1$ is a Banach algebra. The spectral calculus is probably a mess, but does this not help. Similar $L^\infty$. Then you can interpolate? $\endgroup$ – Marc Palm May 4 '12 at 19:16
  • $\begingroup$ 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 $ba \neq L^p$. $\endgroup$ – Marc Palm May 4 '12 at 19:29
  • $\begingroup$ 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. $\endgroup$ – Marc Palm May 4 '12 at 19:36
  • $\begingroup$ L^1 is not a Banach algebra for pointwise product. $\endgroup$ – Yemon Choi May 4 '12 at 19:49
  • $\begingroup$ @Yemon Choi: I am well aware of that fact, I consider the interval as a circle, so I have $L^1$ of group, sorry that I was not clear about that. $\endgroup$ – Marc Palm May 5 '12 at 9:38

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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    $\begingroup$ 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 $L_p(K,\nu)$. $\endgroup$ – Bill Johnson May 5 '12 at 16:12
  • $\begingroup$ @Bill: I am so illiterate, sorry, what is a "Stone space" here? $\endgroup$ – Fedor Petrov May 7 '12 at 19:15
  • $\begingroup$ Presumably, the maximal ideal space or, equivalently, the Stone space of the Boolean algebra $\wp(X)$ modulo $\mu$-null sets. $\endgroup$ – Jan Veselý May 9 '12 at 22:57
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    $\begingroup$ Then how do we conclude that Stone Weierstrass condition holds from the separation property? $\endgroup$ – Fedor Petrov May 12 '12 at 21:22
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    $\begingroup$ Bill, the Stone-Weïerstrass Theorem does not apply here because the condition of separability does not include all points in K. In addition, the density Theorem proposed by Mikael is valid for the weak*-topology of L^\infty, not for the norm topology. $\endgroup$ – Jean Bellissard Jun 9 '17 at 17:21

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