Fourier transform of convolution for $L^2$ functions

Question

If $f,g\in L^1(\mathbb{R})$, it is not hard to show by definition that $$(\hat{f\ast g)}(t)=\hat{f}(t)\hat{g}(t).$$ But what about if $f,g\in L^2(\mathbb{R})$? The Fourier transform on $L^2(\mathbb{R})$ is defined in as an extension of the Fourier transform in the Schwartz class. Then it's hard to work with the definition...

Answer 1

For $f,g\in L^2$, the convolution $f*g$ belongs to $C_0(\mathbb R)$. Taking the Fourier transform of a $C_0$ function is problematic: such transforms are generally not functions, but merely distributions (as Jose27 points out).

It is better to work from right to left. The product $\hat f \hat g$ is in $L^1$. Apply the inverse Fourier transform to it (essentially same as Fourier transform, but with opposite sign in the exponential, and maybe different normalization). The result is a $C_0$ function. You want to show that this function is $f*g$.

Argue by density. For $f,g\in \mathcal S$ the result is known. The map $(f,g)\mapsto f*g$ is a continuous map from $L^2\times L^2$ into $C_0$. Also, $(f,g)\mapsto \hat f \hat g$ is continuous from $L^2\times L^2$ to $L^1$. The inverse Fourier transform is continuous from $L^1$ to $C_0$. The conclusion follows.

In the preceding paragraph, $C_0$ can be replaced by $L^\infty$ throughout (the norm is the same), saving you the trouble of proving that $f*g$ vanishes at infinity.