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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.

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2 Answers 2

up vote 3 down vote accepted

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

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

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 $T = T_m$.

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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

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