# Convolution Theorem for Distributions

I am searching for a version of the convolution theorem for functions (i.e. $\hat{f \cdot g} = \hat f \star \hat g$) that also applies to Distributions/tempered Distributions. Is it automatically true, if $f, g$ satisfy Hörmanders Criterion (i.e. $(x,k) \in WF(f) \Rightarrow (x,-k) \notin WF(f)$).

I feel, like it should be especially true, if the sum of the supports of the Fourier transforms (in the case of tempered Distributions) does not contain 0, i.e. $k \in supp(\hat f) \Rightarrow -k \notin supp(\hat g)$

I am especially interested in the case of the massiv two-point"function" in 2D $f(t,x) = g(t,x) = \Delta_m(t,x) = \int \delta(\omega^2-k^2-m^2) \Theta (\omega) e^{-i\omega t + i kx} \,d\omega \,dk$.

Any pointers to resources, or an idea on how to proof that would appreciated, since I only found statements about the product of Distributions with testfunctions and their convolution. Or does an argument like "The convolution theorem applies, if $f$ is a Distribution and $g$ a test function. So if $f$ and $g$ are Distributions and $f \cdot g$ and $\hat f \star \hat g$ exist, then we must automatically have $\hat{f \cdot g} = \hat f \star \hat g$"?

edit I think, I have an answer. If anyone can confirm, that my reasoning is correct, that would be great.

So we have the following setup: $f \in \mathcal{S}', \mathcal{S} \ni g_n \to g \in \mathcal{S}, \phi \in \mathcal{S}$.

We first show that $\left<\hat{f g_n}, \phi\right> = \left<\hat f \star \hat g_n, \phi\right>$ for all $g_n \in \mathcal{S}$: $$\left<\hat{f g_n},\phi\right> = \left<{f g_n},\hat\phi\right> = \left<f,g_n\hat\phi\right> = \left<f,\hat{\hat g_n\star\phi(-\cdot)}\right> = \left<\hat f,\hat g_n(-\cdot)\star\phi\right> = \left<\hat f \star \hat g_n,\phi\right>$$ by the definition of multiplication with testfunctions, convolution with testfunctions and fourier transform of Distributions.

If we take now the limit $n \to \infty$ on one side and it exists for all testfunctions $\phi$ (it must not in general!) the limit on the other side exists too for all testfunctions $\phi$ and we have shown, that if the multiplication of two distributions works (e.g. by Hörmanders Theorem) the convolution theorem also applies to their fourier transforms and vice versa.

• You know (till now) nothing about the coodinate-wise continuity of the multiplication. So why should $fg_n$ converge to $fg$, even if the latter exists by the Hörmander criterion? – Vobo Aug 11 '18 at 16:06
• That is right, in fact I just came up with an counter-example: Take $g_n (x) = e^{-x^2}\cdot \cos(n\cdot x)$ and $f = \delta$. Then as a distribution we have $g_n \to 0$, but $g_n \cdot f = \delta ~\forall n$. So I have to think a bit harder here... – Jan Lukas Bosse Aug 14 '18 at 7:15
• I am pretty sure, "Product of distributions" by M. Oberguggenberger, J. reine angew. Math. 365 (1986) will be helpful ... – Vobo Aug 14 '18 at 19:23
• Yup, that looks really helpful! Thank you for pointing out this article. – Jan Lukas Bosse Aug 15 '18 at 7:56