Interior estimates for $H^1$-norm of solutions of divergence-type elliptic PDE Let $u\in H^1(\Omega)$ be a weak solution of $Lu=f$, where the coefficients of $L$ satisfy $a^{ij}{\xi}_i {\xi}_j \ge \lambda {|\xi|}^2$ for any $x\in \Omega$, $\xi \in {\mathbb{R}}^n$ and ${\|a^ij\|}_{\infty},{\|b^ij\|}_{\infty},{\|c^ij\|}_{\infty} \le \Lambda$, where both $\lambda$ and $\Lambda$ are positive constants. Assume that $f\in L^2(\Omega)$. Derive an estimate for ${\|u\|}_{H^1(\Omega')}$ for any open subset $\Omega'\Subset\Omega$. 

I attempted to prove ${\|u\|}_{H^1(\Omega')}\le C({\|u\|}_{\infty}+{\|f\|}_{\infty})$ but that is wrong. The answer says ${\|u\|}_{H^1(\Omega')}\le C({\|u\|}_2+{\|f\|}_2)$, where $C=C(n,\lambda,\Lambda,\textrm {dist}(\Omega', \partial\Omega))$ and I cannot work that out. 
 A: This is a typical energy estimate for an elliptic PDE. Recall that the divergence-type operator
$$
Lu = \sum_{i} \left(D_i \left(\sum_j a^{ij}D_ju + b^i u\right) + c^i D_i u \right)
\tag1$$
is associated with a bilinear form
$$
\mathcal L(u,v) = \int_\Omega \left\{\sum_{i} \left(\sum_j a^{ij}D_ju + b^i u\right)D_i v -( c^i D_i u )v \right\}
\tag2$$
The weak form of the equation $Lu=f$ is $$\mathcal L(u,v) + \int_{\Omega} fv = 0 \tag3$$ for every test function $v$, which can be taken from $H^1_0(\Omega)$ (the test function must vanish on the boundary). 
Pick a smooth cut-off function $\varphi$ that has compact support in $\Omega$ and is identically $1$ on $\Omega'$. This can be done so that $|D\varphi|\le C/\operatorname{dist}(\Omega', \partial \Omega)$. Put $v=u\varphi$ in (3) and write out all of the terms. The main one is 
$$
\int_\Omega \varphi \sum_{ij}a^{ij}D_juD_iu  \ge \lambda \int_{\Omega'} |Du|^2
$$
The rest involve at most one derivative of $u$, and are handled with weighted Cauchy-Schwarz inequality, 
$$2|ab|\le \epsilon a^2 + \epsilon^{-1}b^2$$
so that any $|Du|^2$ you get has tiny coefficient, which will be dominated by $\lambda$. E.g., 
$$
\left| \int_\Omega u D_j u \right| \le  \frac{\epsilon^{-1}}{2} \int_\Omega u^2   + \frac{\epsilon}{2} \int_\Omega  |Du|^2  
$$
When the dust settles, (3) leads to 
$$
 \int_{\Omega'} |Du|^2 \le C \int_{\Omega} (u^2 + f^2)
$$
as claimed.
