Let $\Omega \subseteq \mathbb{R}^n$ be an open set, and let's consider $\gamma \in L^{\infty}(\Omega)$, $u\in H^1_{\text{loc}}(\Omega)$.
Here $H^1_{\text{loc}}(\Omega)=\{f: \Omega\rightarrow \mathbb{R}: f \in L^2(\Omega) , D_jf \in L^2_{\text{loc}}(\Omega)\}$, where $D_jf$ is the partial derivative $\frac{\partial}{\partial x_j}$, in the weak sense.
Suppose $u$ weakly solves $D_j(\gamma D_iu)=0$, i.e. $\forall \phi \in C_0^{\infty}(\Omega)$ one has $\int_{\Omega}\gamma D_iuD_j\phi=0$.
Can I conclude that $\forall \phi \in H_0^{1}(\Omega)$ we have $\int_{\Omega}\gamma D_iuD_j\phi=0$?
My try: $\phi \in H^1_0(\Omega) \iff$ there is a sequence $\phi_n\in C_0^{\infty}(\Omega)$ that converges to $\phi$ in $H_1(\Omega).$ Therefore I can write $\int_{\Omega}\gamma D_iuD_j\phi_n=0$. This integral makes sense, since $\int_{\Omega}=\int_{K_n}$, where $K_n$ is the compact support of $\phi_n$. Therefore $\gamma \in L^{\infty}(K_n) + D_iu \in L^2(K_n) \implies \gamma D_iu \in L^2(K_n)$.
Since $D_j\phi_n$ also is in $L^2(K_n)$ the integral $\int_{\Omega}\gamma D_iuD_j\phi_n$ exists, and is equal to $0$. Therefore, $\forall n$ we have $(\gamma D_iu,D_j \phi_n)_{L^2(\Omega)}=0$, and since $\phi_n \rightarrow \phi$ in $H^1(\Omega)$ we get $D_j \phi_n \rightarrow D_j \phi$ in $L^2(\Omega)$ and by the continuity of the inner product $0=(\gamma D_iu,D_j \phi)_{L^2(\Omega)}=\int_{\Omega}\gamma D_iuD_j\phi$.
Is everything correct?