(This question is a duplicate from here)
We start with the linear elliptic PDE
$$ -\operatorname{div}(A\nabla u)=f \quad\text{in}\ \Omega,\\ u=0 \quad\text{on}\ \partial\Omega $$ We assume that $\Omega\subset\mathbb{R}^3$ is a smooth domain, that $A\colon\Omega\to\mathbb{R}^3$$A\in C^\alpha(\overline{\Omega})$ (for some $\alpha>0$) is bounded anda symmetric, positive definite matrix-valued function (there is $c_A>0$ such that $A\xi\cdot\xi>c_\alpha|\xi|^2$$A(x)\xi\cdot\xi>c_\alpha|\xi|^2$ for all $\xi\in\mathbb{R}^3$ and all $x\in\Omega$), and a function $f\in L^s(\Omega)$ for some $s>2$$f\in L^\infty(\Omega)$.
For some $\delta>0$ and some continuoussymmetric $B\colon\overline{\Omega}\to\mathbb{R}^3$$B\in C^\alpha(\overline{\Omega})$, we also look at the perturbed problem
$$ -\operatorname{div}((A+\delta B)\nabla v)=f \quad\text{in}\ \Omega,\\ v=0 \quad\text{on}\ \partial\Omega. $$
My question: If I know, that I have bounds on the gradient of $u$, like
$$ \|\nabla u\|_{L^\infty(\Omega)}\leq C(C_P,|\Omega|,c_\alpha,\|f\|_{L^s(\Omega)}), $$$$ \|\nabla u\|_{L^\infty(\Omega)}\leq C(C_P,|\Omega|,c_\alpha,\|f\|_{L^\infty(\Omega)}), $$
can I conclude something similar for $v$ (for sufficiently small $\delta$), like
$$ \|\nabla v\|_{L^\infty(\Omega)}\leq C(C_P,|\Omega|,c_\alpha,\|f\|_{L^s(\Omega)},\delta, \|B\|_{L^\infty(\Omega)})? $$$$ \|\nabla v\|_{L^\infty(\Omega)}\leq C(C_P,|\Omega|,c_\alpha,\|f\|_{L^\infty(\Omega)},\delta, \|B\|_{C^\alpha(\overline{\Omega})})? $$ In these estimates, $C_P$ denotes the Poincare inequality constant.
I tried to prove this estimate via difference quotients and cut-offs but somehow got stuck. If it holds, what technique would you use to prove this? And if it is not possible, why could this estimate fail?
