If it is known that $\sum_{i,j=1}^{n}a_{ij}\xi_i\xi_j\geq \alpha^2|\xi|^2$, where $\xi=(\xi_1,\xi_2,...,\xi_n)\in\mathbb{R}^n$ then can it be said that $\sum_{i,j=1}^{n}a_{ij}\frac{\partial u}{\partial x_i}\frac{\partial u}{\partial x_j}$ is an equivalent norm to the usual norm of the Sobolev space $W_0^{1,2}(\Omega)$?. Note that $a_{ij}$ are all bounded measurable functions.
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1$\begingroup$ Is $(a^{-1})_{ij}$ also bounded and measurable? $\endgroup$– Igor KhavkineCommented Sep 28, 2016 at 7:10
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$\begingroup$ No information is available on the $1/a_{ij}$'s though. $\endgroup$– AlexanderCommented Sep 28, 2016 at 7:32
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3$\begingroup$ Yes, since also $\sum_{i,j=1}^{n}a_{ij}\xi_i\xi_j\leq \beta^2|\xi|^2$ for large enough $\beta$ depending on $n$ and suprema of $a_{ij}$'s. $\endgroup$– Fedor PetrovCommented Sep 28, 2016 at 8:12
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1$\begingroup$ Ah, yes, Fedor is right. Since the inequality you gave actually bounds $a_{ij}$ from below, the operator norm of $(a^{-1})_{ij}$ (the components of the inverse matrix, not the inverses of the matrix elements) should be bounded by $1/\alpha^2$ (and maybe a constant depending on $n$). $\endgroup$– Igor KhavkineCommented Sep 28, 2016 at 8:30
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3$\begingroup$ At first, you miss $\int$ in your formula for the norm. At second, your norm is Hilbert norm for a positive-definite inner product $\langle u,v\rangle=\int \sum_{i,j=1}^{n}a_{ij}\frac{\partial u}{\partial x_i}\frac{\partial v}{\partial x_j}$. This guarantees the triangle inequality. $\endgroup$– Fedor PetrovCommented Sep 28, 2016 at 9:28
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