TITLE: Is $L^2(\mathbb{R})$ with convolution a Banach Algebra?
QUESTION [11 upvotes]: Is $L^2(\mathbb{R})$ a Banach algebra, with convolution?
I am pretty sure the answer is no, because I think that
$f,g \in L^2(\mathbb{R})$ does not imply that $f*g \in L^2(\mathbb{R})$. However, I can't seem to find a counterexample, or a proof that this is not the case, using Fourier Transform for example.
Can someone give a hint?
Thank you

REPLY [5 votes]: As seen in the other answers $L^2(\mathbb{R})$ is not a Banach algebra. Instead we may consider a weighted $L^2$-space, let $w$ be a positive continuous function defined on $\mathbb{R}$ and $L^2_w= L^2_w(\mathbb{R})$ be the space of all measurable functions $f$ such that
$fw\in L^2(\mathbb{R})$ - the norm is defined by
$$
\|f\|_{L^2_w}=\|fw\|_{L^2}.
$$
It is an exercise to show that $L^2_w$ is a Banach space. If we defined convolution as in the non-weighted space, that is
$$f*g(x)=\int_\mathbb{R} f(y) g(x-y)dy\qquad\text{(note no weight here)} $$
we will get a Banach algebra if $w^{-2}*w^{-2}\le w^{-2}$. Unfortunately, I do not know if this condition can be improved. See this thread for more details, and a proof of the last statement:  When is the weighted space $\ell^p(\mathbb{Z},\omega)$ a Banach algebra ($p>1$)?