Cfrgtkky

I want to prove that the equation
$$u -Delta u=f$$

with $f in L^2 (mathbb{R}^n)$ admits a solution in $H^2 (mathbb{R}^n)$ with $n=1,3$.

Taking Fourier transforms and resolving, I get:

$$u(x)=left( frac {e^{-|y|}}{2} ast f right) (x)$$

when $n=1$ and

$$u(x)=left( frac {e^{-|y|}}{4 pi |y|} ast f right) (x)$$

when $n=3$. Do these functions belong to $H^2 (mathbb{R}^n)$? And how do I prove it?
(I don’t know if I should use the $H$ definition with approximations or the $W$ definition with weak derivatives)

EDIT: I just realized (don’t know why it took so much) that the problem I had can be solved by using Young inequality. For some reason I kept thinking I needed the derivative of $frac{e^{-|x|}}{|x|}$ to be in $L^2$

Fourier transforming, we see that the equation is equivalent to
$$
(1+|xi|^2)hat{u}(xi) = hat{f}(xi). $$
In particular,
$$
|u|_{H^2}^2=int_{mathbb R^n} |hat{u}(xi)(1+|xi|^2)|^2, dxi=int_{mathbb R^n} |hat{f}(xi)|^2, dxi, $$
and $hat{f}in L^2(mathbb R^n)$ by Plancherel's theorem, so the integral in the right-hand side is finite. Thus, $uin H^2$.