Let $U$ be an open bounded subset of ${\mathbb{R}}^n$ and consider $a_{ij}(x)$ to be a matrix so that there exists a constant $C>0$ such that for all $\xi \in {\mathbb{R}}^n$ and $x\in U$ we have $${\sum}_{i, j =1}^n a_{ij}(x){\xi}_{i} {\xi}_{j} \geq C |\xi |^2.$$ Also, for $u, v \in H_{0}^{1}(u)$ we have $$B(u, v) = {\int}_{U}{\sum}_{i, j =1}^n a_{ij}(x)\frac{\vartheta u}{\vartheta x_i}\frac{\vartheta v}{\vartheta x_j} dx .$$ How can I prove that If $a_{ij}(x) \in L^{\infty}(U)$, then there exist constants $\alpha, \beta$ such that $$\beta \|u\|_{H_{0}^{1}}^2 \leq B(u, u) \leq \alpha \|u\|_{H_{0}^{1}(U)}^{2}$$ for all $u \in H_{0}^{1}(U)$?
I thought about working considering the Neumann problem, but it did not led me anywhere.
Any help would be very much appreciated it.
Since $a_{ij}$ are bounded and we have, $|ab|\le \frac{a^2+b^2}{2}$
$$|B(u, u) |= |\int_{U}{\sum}_{i, j =1}^n a_{ij}(x)\frac{\vartheta u}{\vartheta x_i}\frac{\vartheta u}{\vartheta x_j} dx|\le K {\int}_{U}{\sum}_{i, j =1}^n|\frac{\vartheta u}{\vartheta x_i}\frac{\vartheta u}{\vartheta x_j}| dx \\\le \frac{K}{2}\left( {\int}_{U}{\sum}_{i, j =1}^n|\frac{\vartheta u}{\vartheta x_i}|^2 dx+ {\int}_{U}{\sum}_{i, j =1}^n|\frac{\vartheta u}{\vartheta x_j}|^2 dx\right)\\= Kn{\int}_{U}\sum_{i=1}^n|\frac{\vartheta u}{\vartheta x_i}|^2 dx \le \color{red}{Kn\|u\|_{H_0^1(U)}^2}$$
Since $U$ is bounded, By, Poincare-friedrichs inequality we have, $$\|\nabla u\|^2_{L^2(U)}\ge k\| u\|^2_{L^2(U)} $$
Then we have, $$\| u\|^2_{H_0^1(U)}=\|\nabla u\|^2_{L^2(U)} + \| u\|^2_{L^2(U)} \le (1+k^{-1})\|\nabla u\|^2_{L^2(U)}$$
Using, this, if we take $\xi_i =\frac{\vartheta u}{\vartheta x_i} $. The ellipticity gives that
$$B(u,u)={\int}_{U}{\sum}_{i, j =1}^n a_{ij}(x)\frac{\vartheta u}{\vartheta x_i}\frac{\vartheta u}{\vartheta x_j} dx \ge C \int_{U}\sum_{i=1}^n |\frac{\vartheta u}{\vartheta x_i}|^2 \\= C\|\nabla u\|^2_{L^2(U)} \ge\color{blue}{ \frac{C}{1+k^{-1}}\| u\|^2_{H_0^1(U)}}$$ Done!!!