Why are translation invariant operators on $L^2$ multiplier operators

Question

For $m \in L^\infty$, we can define the multiplier operator $T_m \in L(L^2,L^2)$ implicitly by

$\mathcal F (T_m f)(\xi) = m(\xi) \cdot (\mathcal F T_m)(\xi)$

where $\mathcal F$ is the Fourier transform. It is obvious from the defintion that $T_m$ commutes with translations.

How can you show the converse, i.e. every translation invariant $T \in L(L^2,L^2)$ is induced by a multiplier $m_T$? I have no idea how this might work.

How is it obvious that $T_m$ commutes with translations? – Brozovic Nov 07 '19 at 14:04 — Brozovic, Nov 07 '19 at 14:04

score 9 · Answer 1 · answered Apr 04 '11 at 09:10

9

Here's a sketch proof, which I think gets to the core reason why this is true. Remember that the fourier transform takes translations to multiplication by characters. So if $T$ commutes with translation, $\hat T = \mathcal{F} T \mathcal{F}^{-1}$ will commute with multiplication by characters. So $\hat T$ commutes with all operators which are in the closure of the linear span of the operators given by multiplication by characters. That is, $\hat T$ commutes with multiplication by any continuous function. It's not so hard to then show that $\hat T$ must itself by multiplication by some $m\in L^\infty$; under your definition, this means precisely that $T = T_m$.

answered Apr 04 '11 at 09:10

Matthew Daws

2,825

Could you explain how you pass to $\hat T$ commuting with the closed linear span of the characters? I can't seem to get the limits to work out right. – Potato Feb 13 '13 at 22:34
1

If the group is non-compact (for example, $\mathbb{R}$), then the uniform closure of the linear span of the characters yields only almost periodical functions (not all bounded uniformly continuous functions). It would be interesting to see a simple solution of the original problem for this case. – Egor Maximenko Feb 22 '17 at 20:50
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@Egor I think a weak topology (any of the operator topologies would work) comes to the rescue. If an operator T commutes with translation operators, it commutes with any operator in the WOT closure of the translations. The same is true after taking the Fourier transform, and so we can look at the WOT closure of the characters (and not just the uniform closure). That's enough. – Matthew Daws Feb 22 '17 at 21:21
@Matthew Dows Good point! Thank you very much for the idea! – Egor Maximenko Feb 23 '17 at 04:45

score 3 · Accepted Answer · edited May 02 '20 at 08:50

3

This is Theorem 3.16 in the first chapter of Introduction to Fourier Analysis on Euclidean Spaces, by E.M. Stein and G. Weiss.

edited May 02 '20 at 08:50

Calvin Khor

34,903

answered Mar 29 '11 at 12:56

Julián Aguirre

76,354

perfect, thank you very much. – shuhalo Mar 29 '11 at 14:50

Why are translation invariant operators on $L^2$ multiplier operators

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