Prove these two functions are in $H^2(\mathbb{R}^n)$

Question

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 $L^2$

Answer 1

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