On fourier transform of convolution

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I'm studying on the book Introduction to the Theory of Distributions, by F.G. Friedlander & M. Joshi.

Theorem. ${\scr F}(u * v) = {\scr F} (u) {\scr F}(v) \qquad \forall \, u, v \in {\scr E}'(\mathbb{R}^n) $

Here ${\scr F}$ is the Fourier transform operator, $*$ is a convolution and ${\scr E}'$ is the space of all compactly supported distributions.

I'm trying to prove this, but I'm a bit stuck.

Remark. Let $u \in {\scr E}'(\mathbb {R}^n)$. Then for every smooth test function $f(x) \in \mathscr{E}(\mathbb{R}^n)$, one has $${\scr F}(u(f))= u({\scr F}(f)) = \langle u, \mathscr{F}(f)\rangle = \langle u, \int_{\mathbb{R}^n} e^{ik\cdot x} f(x)\, dx \rangle = \langle u, \langle w_f, e^{i k \cdot x} \rangle \rangle =: (u \otimes w_f)(e^{ik\cdot x})$$

Where $w_f$ is the distribution generated by $f(x)$ via integration.

By definition $(u * v)(f) := (u \otimes v)(f(x+y)) = u(v(f(x+y))) = \langle u, \langle v, f(x+y) \rangle \rangle $.

So in this case, ${\mathscr F}(u *v)= \langle u, \langle v, {\mathscr F}(f(x+y)) \rangle = \langle u, ( v \otimes w_f)(e^{ik\cdot (x+y)})\rangle$

...and now?

The goal should be writing something like $[(u \otimes w_f)(e^{ik\cdot x})] \ [(v \otimes w_f)(e^{ik\cdot y})] = {\mathscr F}(u) {\mathscr F}(v) $

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I will write $\hat{u}$ instead of $\mathscr{F}u.$

First two lemmas that you can try to prove:

  1. If $u,v\in\mathcal{E}'$ then $u*v\in\mathcal{E}'$.
  2. If $u\in\mathcal{E}'$ then $\hat{u}\in C^\infty$ and $\hat{u}(\xi)=\langle u(x), e^{-i\xi x} \rangle.$

Also note that $\langle u\otimes v, \phi\otimes\psi\rangle = \langle u, \phi \rangle \, \langle v, \psi \rangle.$

Now the main proof: $$ \widehat{u*v}(\xi) = \langle (u*v)(z), e^{-i\xi z} \rangle = \langle u(x)v(y), e^{-i\xi (x+y)} \rangle = \langle u(x)v(y), e^{-i\xi x} e^{-i\xi y} \rangle \\ = \langle u(x), e^{-i\xi x} \rangle \langle v(y), e^{-i\xi y} \rangle = \hat{u}(\xi) \, \hat{v}(\xi). $$