TITLE: $ W_0^{1,p}$ norm bounded by norm of Laplacian
QUESTION [7 upvotes]: Let $f\in W_0^{1,p}(U)$, for $U$ a bounded domain and $p < n/(n-1)$. I am trying to prove that there is an inequality of the form
$$\|f\|_{W^{1,p}} \leq C \int_{\Omega} |\Delta f| $$
where $\Delta$ is the Laplacian. I tried applying the Sobolev embedding theorem, followed by the inequality
$$\|D^2 f\|_p \leq C \|\Delta f\|_p$$
but this does not hold for $p = 1$. I also tried using elliptic estimates, also to no avail. Does anyone have any suggestions? 
EDIT: Perhaps it is possible to deduce this from the $L^p$ inequality 
$$\|D^2 f\|_{L^p} \leq C \|\Delta f\|_{L^p}$$
for $u\in W_0^{2,p}(\Omega)$, $1 < p < \infty$?

REPLY [2 votes]: The domain (variously called $U$ and $\Omega$ in the question) is not very important here. Since $f\in W^{1,p}_0(U)$, the extension of $f$ by zero belongs to $W^{1,p}_0(\mathbb R^n)$. Let $k(x)=c_n|x|^{2-n}$ be the fundamental solution of $\Delta$ on $\mathbb R^n$. The function $g=\Delta f$ belongs to $L^1(\mathbb R^n)$ by assumption. Therefore, the convolution $g*k$ vanishes at infinity. Since $f-g*k$ is harmonic, we have $f=g*k$. 
Next, differentiation $\nabla f = g*(\nabla k)$ is legitimate because $\nabla k$ is also an integrable kernel (we do not get into singular integrals here). Most importantly, $|\nabla k|\approx |x|^{1-n}\in L^p$ in the specified range of $p$. Young's inequality for convolution yields $\|\nabla f\|_p\le \|g\|_1 \|\nabla k\|_p$. $\quad\Box$