Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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 $T = T_m$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.