Let $\Omega$ be a bounded smooth region in $R^n$ and $u$ satisfy
$-\Delta u+a(x)u=f, \ \ u|_{\partial \Omega}=0$,
where $a(x)\geq 0$ and $f(x)$ are smooth functions. I wonder if the following estimate holds
$||a(x)u||_{H^1(\Omega)} \leq C ||a||\ \||f||_{H^1(\Omega)}$,
where $C$ is independent of $u$ and $a(x)$. What norm of $a(x)$ should appear in the right hand side of the above inequality?
I disagree with the accepted answer. You do not need a Lipschitz bound on $a$. I assume for simplicity that $a\in L^\infty(\Omega)$.
Step 1. The solution $u$ exists and is unique in $H^1_0(\Omega)$, by Lax-Milgram for example.
Step 2. Write $g:=f-au$. Then $u$ satisfies $-\Delta u = g$ in $\Omega$, so provided $\Omega$ is $C^1$ for example, $u\in H^2(\Omega)\cap H^1_0(\Omega)$. So
2.a If $n$ the dimension of the ambient space verifies $n<4$ then $u\in C^{0,1/4}(\Omega)$ and therefore $$ au \in H^{1}(\Omega) \mbox{ when } a\in H^1(\Omega). $$ 2.b If $n>4$ then $u\in L^{\frac{2n}{n-4}}(\Omega)$ and therefore $$ au \in H^{1}(\Omega) \mbox{ when } a\in W^{1,n/2}(\Omega). $$ 2.c If $n=4$, $a\in W^{1,2+\epsilon}(\Omega)$ with $\epsilon>0$ is enough.
3.a. Suppose $n \geq 6$, and $a\in H^{1}\cap L^\infty$. If we allow $f\in H^1$, as suggested in the question, then $f\in L^{\frac{2n}{n-2}}$, $au \in L^{\frac{2n}{n-4}}$ so $u \in W^{2,\frac{2n}{n-4}}$ and as $\frac{2}{n-4}< 2$, $u$ is Holder, therefore bounded and we are back to 2.a
3.b and 3.c, n=5 and n=4, argue as in 3.a but it bootstrap twice instead of once.
Therefore in all dimension, $a\in H^{1}\cap L^\infty$ is enough.
Writing $$ \langle-\Delta u + au, u\rangle_{L^2(\Omega)}=\langle f, u\rangle_{L^2(\Omega)}, $$ using the Dirichlet boundary condition, you get $$ \Vert \nabla u\Vert_{L^2(\Omega)}^2+\langle au, u\rangle_{L^2(\Omega)} \le \Vert f\Vert_{L^2(\Omega)}\Vert u\Vert_{L^2(\Omega)}, $$ so that you control the $H^1$ norm of $u$ by the $L^2$ norm of $f$. To get a control of the $H^1$ norm of $au$, you will need some Lipschitz norm of $a$.
