Stone-Weierstrass analogue for $L^p$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T19:24:31Z http://mathoverflow.net/feeds/question/96006 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96006/stone-weierstrass-analogue-for-lp Stone-Weierstrass analogue for $L^p$ Fedor Petrov 2012-05-04T18:26:06Z 2012-05-04T21:34:33Z <p>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 $1\leq p &lt; \infty$? </p> http://mathoverflow.net/questions/96006/stone-weierstrass-analogue-for-lp/96025#96025 Answer by Mikael de la Salle for Stone-Weierstrass analogue for $L^p$ Mikael de la Salle 2012-05-04T21:34:33Z 2012-05-04T21:34:33Z <p>Yes (I assume that the measure is finite). Here is a proof that uses the von Neumann bicommutant theorem (or rather <a href="http://en.wikipedia.org/wiki/Kaplansky_density_theorem" rel="nofollow">Kaplansky's density theorem</a>).</p> <p>See <code>$A \subset L^\infty(X,\mu) \subset B(L^2(X,\mu))$</code> where <code>$L^\infty$</code> acts on $L^2$ by pointwise multiplication. Then the assumption that $A$ separates points is exactly that the commutant of $A$ is <code>$L^\infty(X,\mu)$</code>, so that the bicommutant of $A$ is <code>$L^\infty(X,\mu)$</code>. Therefore, by Kaplansky's density theorem, any <code>$f \in L^\infty$</code> with <code>$\|f\|_\infty \leq 1$</code> belongs to the strong operator topology closure of <code>$\{g \in A, \|g\|_\infty\leq 1\}$</code>. Equivalently, there is a net <code>$g_\alpha \in A$</code> with <code>$\|g_\alpha\|_\infty \leq 1$</code> such that, for every <code>$\xi \in L^2$</code>, <code>$\|g_\alpha \xi - f \xi\|_2\to 1$</code>. In particular (using that the constant function $1$ belongs to $L^2$), <code>$\|g_\alpha - f\|_2 \to 0$</code>. But by this implies that for every $1\leq p &lt; \infty$, <code>$\|g_\alpha - f\|_p \to 0$</code>~: if $p&lt;2$ this is because the $L^p \subset L^2$ (the measure is finite), whereas if $p>2$ this is the inequality <code>$\| \cdot \|_p \leq \|\cdot \|_\infty^\theta \|\cdot \|_2^{1-\theta}$</code> for $\theta=1-2/p>0$.</p> <p>This proves that the <code>$\| \cdot \|_p$</code>-closure of $A$ contains <code>$L^\infty$</code>, and hence it is $L^p$.</p>