Characterization of $L^2$ operators that commute with multiplication by continuous functions

Question

Suppose $T$ is a bounded operator from $L^2$ to $L^2$ that commutes with pointwise multiplication by any continuous function $g$. That is, if $f\in L^2$,

$$gT(f)=T(gf).$$

Why must $T$ necessarily be multiplication by a $L^\infty$ function?

I am asking this question in order to understand the answer of Matthew Daws here.

Norbert · Accepted Answer · 2013-02-11T07:44:57.933

4

I assume that underlying measure space is $\sigma$-finite

If underlying measure space is finite just set $f=1$ and recall that all continuous functions (denoted by $g$) are dense in $L_2$.

To show that $T(1)$ is in $L_\infty$ you need to note that $$ \Vert g T(1)\Vert_2=\Vert T(g)\Vert_2\leq \Vert T\Vert\Vert g\Vert_2 $$ and use the following fact: If $\Vert gh\Vert_2\leq C\Vert g\Vert_2$ for all $g\in L_2$, then $h\in L_\infty$ and what is more $\Vert h\Vert_\infty\leq C$.

Otherwise decompose it onto disjoint union of finite measurable subsets and reduce your problem to the previous case.

edited Feb 11 '13 at 07:44

answered Feb 09 '13 at 19:52

Norbert

56,803

So this shows that $T$ is multiplication by some $m$. How does one show this $m$ is in $L^\infty$? – Potato Feb 11 '13 at 07:38
@Potato see edits – Norbert Feb 11 '13 at 07:45
Thank you! $ $ $ $ $ $ $ $ – Potato Feb 11 '13 at 08:02
@Norbert: That "following fact" is the crux of the problem; not giving its proof is like not proving anything. – Alex M. Oct 08 '21 at 10:47
It is a well-known fact. I just reduced the original problem to it. You can add the proof for the "following fact" if you wish. – Norbert Oct 08 '21 at 15:59

Characterization of $L^2$ operators that commute with multiplication by continuous functions

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