I am confused about a step in the proof in Gilbarg-Trudinger, when they solve the Dirichlet problem using $L^p$-estimates in chapter 9.
First, the setting: Let $\Omega\subset \mathbb R^n$ has $C^{1,1}$ boundary, $1<p<\infty$, $u\in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)$, $$L = a^{ij} D_{ij} + b^i D_iu + cu = f$$ an elliptic equation with $a^{ij} \ge \lambda I$, $a^{ij}\in C^0(\overline\Omega)$, $b^i, c\in L^\infty$, $|a^{ij}|, |b^i|, |c|\le \Lambda$ and $f\in L^p$.
Then they prove the following lemma:
Lemma 9.17 (p.242) If in pacticular $c\le 0$, then there is a constant $C$ so that $$\tag{1} \|u\|_{L^p} \le C \| Lu\|_{L^p}$$ for all $u\in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)$.
The lemma is proved using a (standard) contradiction argument. In particular it is not clear how $C$ depends on $L$.
Now, my question: in the next page they apply the lemma in the following situation: Let $a^{ij}_m \in C^1(\overline\Omega)$ and $a^{ij}_m\to a^{ij}$ in $C^0(\overline \Omega)$ as $m\to \infty$. Let $$L_m = a^{ij}_m D_{ij} + b^i D_i +c$$ Let $u_m \in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)$ such that $L_mu_m = f\in L^p$. Now they claim:
... we infer from Lemma 9.17 that the sequence $\{u_m\}$ is bounded in $W^{2,p}(\Omega)$.
I guess they are using the $L^p$-estimates
$$\|u_m\|_{W^{2,p}} \le C( \| u_m\|_{L^p} + \|f\|_{L^p})$$
together with $(1)$ that $\|u_m\|_{L^p}\le C\| L_m u_m\|_{L^p} = C\|f\|_{L^p}$. But this is assuming that $C$ can be chosen independent of $L_m$.
Am I getting it wrong? Do they mean something else?
I don't know if you can prove a stronger variant of lemma 9.17, but you can adapt the argument to prove theorem 9.15. This exploits the fact that $a^{ij}_m$ convergences uniformly to $a^{ij}$ on $\Omega$ in the proof.
We want $u_m$ to be bounded in $L^p(\Omega).$ If not, by passing to a subsequence and setting $v_m = u_m / \lVert u_m \rVert_{L^p(\Omega)}$ and $\tilde f_m = f_m / \lVert u_m \rVert_{L^p(\Omega)}$ we get $L_mv_m = \tilde f_m$ weakly, $\lVert v_m \rVert_{L^p(\Omega)} = 1$ and $\lVert\tilde f_m\rVert_{L^p(\Omega)} \rightarrow 0.$
Since $a^{ij}_m$ converges to $a^{ij}$ uniformly on $\Omega,$ passing to a subsequence such that $v_m \rightharpoonup v$ weakly in $W^{2,p}(\Omega)$ we can infer that $L_mv_m$ converges to $Lv.$ This follows by writing, $$ \int_{\Omega} g\,a_m^{ij}D^{ij}v_m = \int_{\Omega} \left(g\,(a_m^{ij}-a^{ij})D^{ij}v_m + g\,a^{ij}D^{ij}(v_m - v)\right) $$ where $g \in L^{p/(p-1)}(\Omega)$ is test function and letting $m \rightarrow \infty.$ Thus $Lv=0$ weakly with $\lVert v \rVert_{L^p(\Omega)}=1,$ which gives the desired contradiction.