# 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

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

• For the inner product part, just try to check the requirements of the definition, one by one. Where do you get stuck? And for the second part, write up the sort of inequalities (there are two) that you would need to prove to show the equivalence of the norms. Square both sides to get rid of the square roots, and try to use what you know about $\kappa$ to prove the inequalities. – Harald Hanche-Olsen Apr 25 '13 at 21:56
• In the last question I told you what to do... Also, you don't need that the norms are equivalent for your last question if you show it's an inner product. – toypajme May 4 '13 at 22:25

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.