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In this https://web.abo.fi/fak/mnf/mate/kurser/fourieranalys/chap3.pdf Section 3.7: Convolutions (Page 79) lecture notes, it is said that the fourier transform of a convolution of distributions $\mathcal{F}(f*g)$ can be defined if the fourier transform of $g$ is in the space of tempered test functions, i.e. $\mathcal{F}(g)\in C^\infty_{pol}$. Where the tempered test functions are smooth functions whose derivatives grow at most polynomially.

The assumptions are that $f$ and $g$ are tempered distributions.

Now I am wondering, since we define distributions on test functions as integrals anyways, couldn't we relax the assumption an say, that $\mathcal{F}(g)$ must be smooth only up to a null set (and fulfill the growing condition)?

I.e. it may have discontinuities or kinks at discrete points.

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  • $\begingroup$ Could you please give a more specific reference to the linked lecture notes (section, page, and theorem, lemma, or formula number)? Are there any assumptions on $f$ as well as $g$? $\endgroup$
    – Steven Clark
    Commented Mar 19 at 15:36
  • $\begingroup$ @StevenClark Section 3.7: Convolutions (Page 79). The assumptions are that $f$ and $g$ are distributions. $\endgroup$
    – Confuse-ray30
    Commented Mar 19 at 16:14
  • $\begingroup$ I don't believe the convolution theorem is limited to smooth tempered distributions or smooth Schwartz functions. Assuming $f(x)=e^{-x^2}$, I believe the convolution theorem applies to the convolution $[f(y) * g(y)](x)$ when $g(x)=\frac{1}{4} (\text{sgn}(1-x)+\text{sgn}(x+1))$ and also when $g(x)=e^{-|x|}$. I can provide more details in an answer if you're interested in these two particular examples. $\endgroup$
    – Steven Clark
    Commented Mar 20 at 18:57
  • $\begingroup$ @StevenClark Sure, since both of them clearly are not smooth, but the convolution can be defined in that way nonetheless, it is a nice example. The only thing that bugs me about this is, that maybe it works only because $f=e^{-x^2}$ is a rapidly falling function, which of course induces a distribution as well, but is still a very nicely behaved function. Imagine instead $f=1/(x^2+a)$, which induces a distribution, but is not nice at all. $\endgroup$
    – Confuse-ray30
    Commented Mar 21 at 12:28
  • $\begingroup$ @StevenClark also, do you perhaps mean that e.g. $\mathcal{F}(g)=e^{-|x|}$? Because the theorem says that the fourier transform of $g$ must be smooth, not necessarily $g$ itself. $\endgroup$
    – Confuse-ray30
    Commented Mar 21 at 12:32

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