Fourier Transform and Convolution with functions in $L^2$

Question

Let $f,g\in L^2(\mathbb{R})$, assume $f*g\in L^2$, do we have the following formula: $ \mathcal{F}(f*g)=\mathcal{F}(f)\cdot \mathcal{F}(g) $ where $\mathcal{F}$ is the (extended) Fourier transform?

Here's what bother me a little bit:

1. Normally, $f*g\notin L^2$, it's in $L^\infty$ by Young's inequality. So what's the integrability condition on $f,g$ so that this is true?

2. The formula is of course trivial if $f,g\in L^1$, it's also easy to get the result if one of them is in $L^1$ and the other in $L^2$ by density argument. Now that both of them are in $L^2$, I'm not sure how the density argument could apply. Essentially, I think it boils down to my first question. Can you approxiamte $f*g$ by $f_n*g_n$ in the $L^2$ norm with $f_n,g_n\in L^1$? (One can of course do that individually, how about the convolution?)

Thanks for the help!

Answer 1

$\newcommand\F{\mathcal F}$ $\newcommand\Fi{\mathcal F^{-1}}$ $\newcommand\S{\mathcal S}$ I believe the formula is correct. It seems like there should be a more elementary argument, but what seems like the easiest proof is to show that it's true for every $f,g\in L^2$, if we interpret $\F(f*g)$ in the sense of tempered distributions.

So. Say $\F:\S'\to\S'$ is the tempered-distribution Fourier transform.

First note that of course this is the same as the classical Fourier, or better Fourier-Placherel, transform when applied to an $L^2$ function:

If $f\in L^2$ then $\F(f)=\hat f$.

Proof: If $\phi\in\S$ then $\langle\F(f),\phi\rangle=\int f\hat\phi=\int\hat f\phi$.

Lemma. If $(f_n)$ is uniformly bounded and $f_n\to f$ almost everywhere then $f_n\to f$ in $\S'$.

Proof: If $\phi\in S$ then dominated convergence shows that $\int f_n\phi\to\int f\phi$.

Proposition. If $f,g\in L^2$ then $\F(f*g)=\F(f)\F(g)$.

Proof. Say $f_n,g_n\in\S$ with $f_n\to f$ and $g_n\to g$ in $L^2$. Then $$\Fi(\F(f_n)\F(g_n))=f_n*g_n.$$Since $\F(f_n) \to\F(f)$ in $L^2$ and similarly for $g$, $\F(f_n)\F(g_n)\to\F(f)\F(g)$ in $L^1$. Hence $\F(f_n)\F(g_n)\to\F(f)\F(g)$ in $\S'$, and since $\Fi$ is continuous on $\S'$ it follows that $$f_n*g_n=\Fi(\F(f_n)\F(g_n))\to\Fi(\F(f)\F(g))$$in $\S'$. But $||f_n*g_n||_\infty\le||f_n||_2||g_n||_2$, hence $f_n*g_n$ is uniformly bounded; since $f_n*g_n\to f*g$ pointwise (in fact uniformly) the lemma shows that $f*g=\Fi(\F(f)\F(g))$.