Let $\Omega\subset \mathbb{R}^n$ an open bounded domain. Let $\kappa:\Omega \to \mathbb{R}$ a continuous function which $\beta\leq \kappa(x)\leq M \quad \forall x \in \Omega$, where $0<\beta,M.$ Define $$S:=\{v \in H_0^1(\Omega):\quad \|v\|_{H^1(\Omega)}\leq 1\}.$$ Show that for any $u \in H_{0}^1(\Omega)$ exists a unique $g \in S$ such that $$\int_{\Omega}\kappa(x)|\nabla u(x)-\nabla g(x)|^2dx=\min_{v \in S} \int_{\Omega}\kappa(x)|\nabla u(x)-\nabla v(x)|^2dx.$$
My attempts: Of course S is closed and convex subset of $H_0^1(\Omega)$. According to the Approximation Theorem for Hilbert Spaces there is a unique $g \in S$ $$\| u-g\|^2_{H^1(\Omega)}=\min_{v\in S}\int_{\Omega}|u(x)-v(x)|^2+\int_{\Omega}|\nabla u(x)-\nabla v(x)|^2dx$$ and $\langle u-g,v\rangle_{H^1(\Omega)} = 0 \quad \forall v \in V $. Now we can employ another inner product such as: $$\langle u,v \rangle_\kappa := \int_{\Omega} \frac{1}{\kappa(x)}\nabla u(x)\nabla v(x)dx+\int_{\Omega} \kappa(x)u(x)v(x)dx$$ and $\|u\|_{\kappa}=\langle u,u \rangle^{1/2}_\kappa$. Of course, $\min\{\frac{1}{M},\beta\}\|u\|^2_{H^1(\Omega)}\leq \|u\|^2_\kappa \leq \max\{\frac{1}{\beta},M\}\|u\|^2_{H^1(\Omega)}$, then $H^1_0(\Omega)$ is still a Hilbert space and $S$ remains closed and convex. So we can say that exist a unique $g \in S$. Such that,
$$\| u-g\|^2_{\kappa}=\min_{v\in S}\int_{\Omega}\frac{1}{\kappa(x)}|u(x)-v(x)|^2+\int_{\Omega}\kappa(x)|\nabla u(x)-\nabla v(x)|^2dx$$ Where $\langle u-g,v \rangle_\kappa = 0 \quad \forall v \in S$. There is a way to get conclusion from here or we need to find another approach?