I was looking at this question and attempting to come up with my own answer when I "proved" a statement which seems far too strong: if $f\in L^2(\mathbb R^n)$, then the unique solution to $u-\Delta u=f$ is in $L^\infty(\mathbb R^n)$. In particular, this would imply $H^2(\mathbb R^n)\subset L^\infty(\mathbb R^n)$. Could someone look at my working and see where my mistake is? (Or else verify that the statement is true, although I doubt this is the case.)
Using the convention that the Fourier transform is given by $$\mathcal Fg(t)=\int g(x)e^{-2\pi ix\cdot t}\ \mathrm dx$$ for $g\in L^1(\mathbb R^n)\cap L^2(\mathbb R^n)$, one can easily show that the unique solution $u$ to $u-\Delta u=f$ is $$u=\mathcal F^{-1}\left(\frac{\mathcal Ff(t)}{1+4\pi^2|t|^2}\right).$$ Let $\varphi\in C^\infty_c(\mathbb R^n)$ be non-negative with $\int\varphi=1$ and define $\varphi_k(x)=k^{-n}\varphi(kx)$. Then $f_k:=f*\varphi_k\in C^\infty_c(\mathbb R^n)$, so in particular $f_k\in\mathcal S$ where $\mathcal S$ is the space of Schwarz functions. Using the fact that $\mathcal F:\mathcal S\rightarrow\mathcal S$ is a bijection, $\mathcal Ff_k\in\mathcal S$ and hence $$g_k:=\frac{\mathcal Ff_k(t)}{1+4\pi^2|t|^2}\in\mathcal S.$$ It follows that both $u_k:=\mathcal F^{-1}g_k$ and its Fourier transform are integrable, so by the Fourier inversion theorem $u_k(x)=\int g_k(t)e^{2\pi ix\cdot t}\ \mathrm dt$. Now since $\mathcal Ff_k=\mathcal F(f*\varphi_k)=(\mathcal Ff)(\mathcal F\varphi_k)$ and $\|\mathcal F\varphi_k\|_\infty\leq\|\varphi_k\|_1=1$, it follows that $$\|u_k\|_\infty\le\int\frac{|\mathcal Ff(t)|}{1+4\pi^2|t|^2}\ \mathrm dt\le\|\mathcal Ff\|_2\left(\int\frac1{(1+4\pi^2|t|^2)^2}\ \mathrm dt\right)^{1/2}.$$ Since $(\varphi_k)$ is an approximate inentity of convolution, $f_k\to f$ in $L^2$. All the linear maps involved are $L^2$-bounded so $u_k\to u$ in $L^2$ and hence almost everywhere along a subsequence. Hence we conclude $$\|u\|_\infty\le\|\mathcal Ff\|_2\left(\int\frac1{(1+4\pi^2|t|^2)^2}\ \mathrm dt\right)^{1/2}<\infty$$ and so $u\in L^\infty(\mathbb R^n)$.