# FT of a tempered distribution is a function

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!

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 $\widehat{\delta'}(\xi)=i\,\xi$.