TITLE: Fourier transform of convolution for $L^2$ functions
QUESTION [8 upvotes]: 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...

REPLY [3 votes]: One might argue that this is only a partial argument.
I understand the question in the following way: 
1) is  $f \ast g$ well defined: Yes, certainly, and since an easy application of Cauchy-Schwartz yields 
$$ \| f \ast g\|_\infty \leq \|f\|_2 \|g\|_2$$
we certainly have a bounded function. Since 
$$ \| T_x(f \ast) g \|_\infty = \|(T_xf -f) \ast g\|_\infty 
\leq \|T_x f -f\|_2 \|g\|_2\to 0  
\quad \mbox{for} \,\,  |x| \to 0$$ 
because the shift is continuous in the $L^2$-norm (prove it for compactly supported continuous functions, then by density you get it for every $f \in L^2$) we find that $f \ast g$ is in fact uniformly contiuous. Using the density of compactly supported (continuous) fucntion in $L^2(R)$ once more and the fact that the convolution product of two such function is thus a compactly supported function implies that $f \ast g \in C^0(R)$, i.e. 
$$\lim_{x \to \infty} f \ast g (x)  = 0 . $$ 
2) is the Fourier transform of $f \ast g$ defined (and there one has to say: only in the distributional setting, AND in this distributional setting there is an inverse to the Fourier transform, in the realm of tempered distributions!).
This makes the (good) argument given above a (non-trivial) valid statement.