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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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  • $\begingroup$ Thanks Daw...can you throw some light on it or at least refer an article/book?. I was in fact struggling to show that it is a norm firstly. Sorry for the naivety. $\endgroup$ – Alexander Sep 28 '16 at 6:33
  • $\begingroup$ the triangle inequality part only. $\endgroup$ – Alexander Sep 28 '16 at 6:33
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The norm is actually defined as $$ \|u\|:= \left( \sum_{i,j} \int_\Omega a_{ij} u_{x_i} u_{x_j} \right)^{1/2}. $$ Wlog we can assume that $a_{ij}=a_{ji}$. Otherwise, we replace $a_{ij}$ in the definition of the norm by $\frac12(a_{ij}+a_{ji})$. In order to that this is a norm, we show that it is induced by the scalar product: $$(u,v):= \sum_{i,j} \int_\Omega a_{ij} u_{x_i} v_{x_j}.$$ Due to the assumptions, $(u,v)$ is well-defined for all $u,v\in W_0^{1,2}(\Omega)$. Now, $(\cdot,\cdot)$ is obviously linear in both arguments and symmetric. Moreover, $(u,u)\ge0$, and $(u,u)=0$ iff $u=0$. Hence $(\cdot,\cdot)$ is a scalar product, and the induced norm a norm.

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