If $u,v$ are a couples of distributions with compact supports,namely $u,v\in S^{'} $,then How to prove that $$\mathcal{F}(u\ast v)=\mathcal{F}(u)\mathcal{F}(v)$$ $\mathcal{F} $ denote the Fourier transform.
My attempt:I have known that $$(\mathcal{F}(u),v)=(u,\mathcal{F}(v)),u\in S',v\in S$$ $$(u\ast v,w)=(u,p),p(x)=(v,\tau_{-x}w)$$ However,I don't know how to define the product of distributions.
If $u,v$ have compact support, then so does $u*v$. Also, $\hat{u},\hat{v}$ are ordinary functions given by $$ \hat{u}(\xi) = \langle u(x), e^{-ix\xi}\rangle \\ \hat{v}(\xi) = \langle u(y), e^{-iy\xi}\rangle $$
Therefore, $$ (\hat{u}\hat{v})(\xi) = \langle u(x), e^{-ix\xi} \rangle \langle v(y), e^{-iy\xi} \rangle = \langle u(x), \langle v(y), e^{-iy\xi} \rangle e^{-ix\xi} \rangle = \langle u(x), \langle v(y), e^{-iy\xi} e^{-ix\xi} \rangle \rangle = \langle u(x), \langle v(y), e^{-i(x+y)\xi} \rangle \rangle = \langle (u*v)(t), e^{-it\xi} \rangle \rangle = \widehat{u*v}(\xi) . $$