On the closedness of $L^2$ under convolution

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

Solution 1:

Since the Fourier Transform of the product of two functions is the same as the convolution of their Fourier Transforms, and the Fourier Transform is an isometry on $L^2$, all we need find is an $L^2$ function that when squared is no longer an $L^2$ function. Take the function $$ f(x)=e^{-x^2}|x|^{-1/3} $$ $f\in L^2(\mathbb{R})$, yet $f^2\not\in L^2(\mathbb{R})$. Thus, $\hat{f}\in L^2(\mathbb{R})$, yet $\hat{f}*\hat{f}\not\in L^2(\mathbb{R})$.

Exposition:

The reason that it is hard to come up with an explicit example without using the Fourier Transform, is that the $L^2$ functions involved in the convolution do not decay at $\infty$ quickly enough to be integrable; that is, the convolution requires cancellation to evaluate. The $\hat{f}$ given above is not in $L^1$ (if it were, then $f$ would be bounded), so trying to compute the convolution with itself would be extremely difficult.