If $f(x) \in L^2(\mathbb R^d)$ then $f(x-y) \in L^2(\mathbb R^d \times \mathbb R^d)$

Question

I'm very interested to know if the above statement is true.

I know that if $f(x)$ is measurable then $f(x-y)$ is measurable but that doesn't say much regarding integrability, and certainly not squared integrability.

Is the statement true?

Asking because I saw someone make the statement that $\frac{1}{1+|x-y|^2}$ is $L^2$ because $\frac{1}{1+|z|^2}$ is $L^2$ and apparently $\iint \frac{1}{1+|x-y|^2} dxdy\leq \int\frac{1}{1+|z|^2}dz$

The author didn't see fit to include a proof.

Answer 1

No. We have $$ \int\int |f(x-y)|^2 dxdy=\int \int |f(x)|^2 dx dy=\infty, $$ unless $f$ is identically 0.

Answer 2

It doesn't seem true. For $d=1$, if $f(x)=I_{[0,1]}(x)$ is a rectangle over the unit interval, then $f\in L^2(R)$. OTOH $g(x,y):=f(x-y)$ as a function of $(x,y)$ is an infinite rectangular tube, so is not integrable over $R^2$, nor is $g^2$, which equals $g$.