Lemma $9.2$: Let $u\in W_0^{1,1}(\Omega^+),f\in L^p(\Omega^+),1<p<\infty$ satisfy $\Delta u=f$ weakly in $\Omega^+$ with $u=0$ near $(\partial \Omega)^+$. then $u\in W^{2,p}(\Omega^+)\cap W_0^{1,p}(\Omega^+)$ and $$||D^2u||_{L^p(\Omega^+)}\leq C||f||_{L^p(\Omega^+)}.$$
The proof goes like this:
$1.$ Extend $u,f$ to whole of $\mathbb R^n$ to show that the new $u$ solves $\Delta u=f$ weakly in $\mathbb R^n$. And new $u\in W^{1,1}(\mathbb R^n)$ (why not in $W_0^{1,1}$?) and $f\in L^p(\mathbb R^n)$.
$2.$ The regularisation $u_h\in C_c^{\infty}(\mathbb R^n)$ and $\Delta u_h=f_h$ in $\mathbb R^n$. Using the corollary $9.10$, we get $$\label{1}||D^2u_h||_p\leq C||f_h||_p\tag{1}$$ $3.$ Then they take $h\to0$ to get $u_h\to u$ in $W^{2,p}(\mathbb R^n)$. Since $u_h(x',0)=0$, we get $u\in W^{1,p}_0(\Omega+)$.
I have doubts with step 3. I have a few questions:
- How are they getting the convergence in $W^{2,p}$? I don't understand why $u\in L^p$? I understand that the sequence $\{D^2u_h\}_h$ converges to some $v$ in $L^p$ because of $\ref{1}$. But why is it $D^2u$?
- How does $u_h(x',0)=0$ imply $u\in W^{1,p}_0(\Omega+)$? Can we say $u\in W_0^{2,p}$?