Convolution Identity for Schwartz Space

Question

Let $f_1, f_2$ be Schwartz functions on the real line.

I'd like to show that $$(f_1 \star f_2)^{\wedge}= (2\pi)^{d/2} (\hat f_1 \hat f_2).$$

I use the identity $(2\pi)^{d/2} (f_1 f_2)^{\wedge} = \hat f_1 \star \hat f_2$. Equivalently, $(2\pi)^{d/2} (f_1 f_2) = (\hat f_1 \star \hat f_2)^{\vee}$.

Here is my attempt: \begin{align*} (f_1 \star f_2)^{\wedge} &= \mathcal{F}^{-1} (\mathcal{F}^2 (f_1 \star f_2))\\ &= (R(f_1 \star f_2))^{\vee}\\ &=^{?} (R(f_1) \star R(f_2))^{\vee}\\ &= ({\hat {\hat f_1}} \star {\hat {\hat f_2}} )^{\vee}\\ &= (2\pi)^{d/2} (\hat f_1 \hat f_2) \end{align*}

I can't prove, however, that $R(f_1 \star f_2) = R(f_1) \star R(f_2)$. I've tried writing the convolution in terms of the integral and doing various changes of coordinate, but nothing works. I'm beginning to suspect that reflection does not distribute over convolution.

Any help is appreciated!

Answer 1

I gather that $R(f)(x)=f(-x)$? If so this is clear - it's hard to imagine what changes of coordinate you tried that didn't work. Since $\int\phi(-y)\,dy=\int\phi(y)\,dy$ we have $$\begin{aligned}R(f)*R(g)(x)&=\int R(f)(x-y)R(g)(y)\,dy \\&=\int f(y-x)\,g(-y)\,dy\\&=\int f(-y-x)g(y)\,dy=f*g(-x).\end{aligned}$$