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I have a question. When is the Fourier transform of a tempered distribution a function? I guess if the FT is a function, itself must also be a function. But I don't know how to go further. Thanks for any help!

user146507
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1 Answers1

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Your guess is incorrect. Dirac's $\delta$ is a tempered distribution, and its Fourier transform is the function $\hat\delta(\xi)=1$. More generally, the Fourier transform of any bounded measure is a bounded function. Then it is easy to see that the same is true of its derivatives. For instance $\widehat{\delta'}(\xi)=i\,\xi$.

Julián Aguirre
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  • You are right! Thanks for correcting me. Is there any if and only if condition? – user146507 Mar 21 '16 at 14:17
  • No, as far as I know. – Julián Aguirre Mar 21 '16 at 14:56
  • Okay. Thanks! What if the tempered distribution is a weak solution of heat equation with Lu=f, L is the heat operator and f is in some sobolev space and the initial value of u is also in some sobolev space? Can we get the conclusion that the FT of that tempered distribution is a function? I know it is beyond my original question but thanks for reading it. – user146507 Mar 21 '16 at 15:02