How to prove $\int_{\Omega}\frac{1}{\kappa}uv\ +\ \int_{\Omega}\kappa\nabla u\cdot\nabla v$ is an inner product in $H^1$ I need help with this problem from my homework

Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ and let $\kappa : \Omega\rightarrow\mathbb{R}$ be a continuous function, such that there's constants $M, \beta > 0$, such that $\beta \leq \kappa(x) \leq M$ for each $x\in\Omega$. Now, consider
$$\langle u, v\rangle_{\kappa}\ :=\ \int_{\Omega}\frac{1}{\kappa}uv\ +\ \int_{\Omega}\kappa\nabla u\cdot\nabla v,\;\;\; \forall\ u,v\in H^1(\Omega),$$
and show that


*

*$\langle \cdot, \cdot\rangle_{\kappa}$ is an inner product of $H^1(\Omega) = \{v\in L^2(\Omega) : \nabla v\in L^2(\Omega)\}$,

*$\|\cdot\|_{\kappa} = \sqrt{\langle \cdot, \cdot\rangle_{\kappa}}$, is equivalent with the norm
$$\|u\|_{H^1(\Omega)}\ = \ \left(\int_{\Omega}u^2\ + \int_{\Omega}\nabla u\cdot\nabla u\right)^{1/2}.$$



Please somebody can help me. Thanks in advance.
 A: This is a community-wiki answer trying to remove this question from the unanswered queue.


Q1: Show that $\langle \cdot, \cdot\rangle_{\kappa}$ is an inner product of $H^1(\Omega) = \{v\in L^2(\Omega) : \nabla v\in L^2(\Omega)\}$.

Symmetric: $\langle u, v\rangle_{\kappa} = \langle v, u\rangle_{\kappa}$ follows from dot product and scalar product is symmetric.
Bi-linearity: $\langle au+bw, v\rangle_{\kappa} = 
a\langle u, v\rangle_{\kappa}+b\langle w, v\rangle_{\kappa}$ follows from the linearity of the integration, as well as the linearity of the $\nabla$ operator.
Positive definite: 
$$\langle u, u\rangle_{\kappa} = \int_{\Omega}\frac{1}{\kappa}u^2\ +\int_{\Omega}\kappa|\nabla u|^2 \geq 0.$$
Now $$\langle u, u\rangle_{\kappa}=0 \implies 
0 \geq \int_{\Omega}\frac{1}{\kappa}u^2 \geq \frac{1}{M}\int_{\Omega}u^2 \implies \|u\|_{L^2(\Omega)} = 0,$$
and this tells us that $u = 0$ a.e.
Hence it is an inner-product.


Q2: Show that $\|\cdot\|_{\kappa} = \sqrt{\langle \cdot, \cdot\rangle_{\kappa}}$, is equivalent with the norm $\|\cdot\|_{H^1(\Omega)}$.

First for any $u\in H^1(\Omega)$:
$$\|u\|_{\kappa}^2 =\int_{\Omega}\frac{1}{\kappa}u^2 + \int_{\Omega}\kappa|\nabla u|^2 .$$
Hence
$$\|u\|_{\kappa}^2 \leq \frac{1}{\beta}\int_{\Omega}u^2 + M\int_{\Omega}|\nabla u|^2 
\leq \max\{\beta^{-1},M\}\|u\|_{H^1(\Omega)}^2, $$
as well as
$$\|u\|_{\kappa}^2  \geq \frac{1}{M}\int_{\Omega}u^2 + \beta\int_{\Omega}|\nabla u|^2 
\geq \min\{\beta,M^{-1}\}\|u\|_{H^1(\Omega)}^2. $$
This is 
$$
c\|u\|_{H^1(\Omega)} \leq \|u\|_{\kappa}\leq C\|u\|_{H^1(\Omega)},
$$
which is the equivalence.
