Let $f \in L^2(\mathbb R^n)$ and $u \in \mathcal S'(\mathbb R^n)$, where $\mathcal S'$ denote the tempered distribution space. I want to show that if $$ u - \Delta u = f \ \text{ in }\ \mathcal S', $$ then $u \in H^2(\mathbb R^n)$, which means, $$ \| (1+|\xi|^2) \hat u(\xi)\|_{L^2} = \int_{\mathbb R^n} \big| (1+|\xi|^2) \hat u(\xi) \big|^2 d\xi < \infty. $$ I have shown that $\mathcal F[(1+|\xi|^2) \varphi(x)] = (1-\Delta)\hat\varphi$ for any $\varphi \in \mathcal S$. Thus $$ \langle (1+|\xi|^2) \hat u, \varphi \rangle = \langle \mathcal F[(1-\Delta) u], \varphi \rangle = \langle \hat f, \varphi \rangle, \quad \forall \varphi \in \mathcal S. $$ But I have no idea how to show the rest.
If, at least, $\hat u$ is a locally integrable function, we can conclude that $\| (1+|\xi|^2) \hat u(\xi)\|_{L^2} = \| \hat f \|_{L^2} < \infty$, since $(1+|\xi|^2) \hat u(\xi) = \hat f$ a.e. by the fundamental lemma of calculus of variations. But for $u$ that is a tempered distribution, how to deal with it?
Consider a smooth function $\psi$ which is supported in $B(0,1)$, it is positive and has integral $1$, and set $\psi_m(x)=m\psi(mx)$. Consider now the convolutions to two sides of the equation, $$ u_m -\Delta u_m = f_m, $$ where $u_m = u*\psi_m$, and $f_m = f*\psi_m$. We have $$ f_m \to f, \qquad \text{in $L^2$ and in $\mathcal{S}'$} $$ and $$ u_m \to u, \qquad \text{in $\mathcal{S}'$}. $$ Since $\|u_m\|_{H^2} \le \|f_m\| \le C(\|f\| + 1)$, we have $u_m \to u$ in $L^2$, and in $H^2$ weakly. (If necessary, we choose a subsequence of $\{m\}$.)
The problem is similar as the following answer.
Relation between distribution and $L^p$