It is a direct consequence of Fubini's theorem that if $f,g \in L^1(\mathbb{R})$, then the convolution $f *g$ is well defined almost everywhere and $f*g \in L^1(\mathbb{R})$. Thus, $L^1(\mathbb{R})$ is closed under convolution, and it is a Banach algebra without unit since we have the inequality
$$\|f*g\|_{1} \leq \|f\|_1 \|g\|_1 \qquad (f,g \in L^1(\mathbb{R})).$$ Now, it follows from Hölder's inequality that if $f,g \in L^2(\mathbb{R})$, then $f*g$ is bounded.
My question is the following : Does $f*g$ necessarily belongs to $L^2(\mathbb{R})$? In other words, is $L^2(\mathbb{R})$ closed under convolution?
Since a quick google search seem to result in a negative answer, I also ask the following question :
Can you give an explicit example of two functions $f, g \in L^2(\mathbb{R})$ such that $f*g \notin L^2(\mathbb{R})$?
Thank you, Malik
, $1/p+1/p'=1$. Thus, the answers in Steven's link do answer your question. An _explicit_ example is $f(x)=g(x)={}_1F_2(\frac38;\frac12,\frac{11}{8};-\frac{x^2}{2})$.
– 2012-03-09