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Let $A$ a uniformly elliptic matrix, i.e. there is $\Lambda >0$ s.t. $$\Lambda ^{-1}|\xi|^2\leq A\xi\cdot \xi\leq \Lambda |\xi|^2,$$ for all $\xi\in \mathbb R^d$.

Prove that for all $u,v\in H=\{u\in H^1(\Omega )\mid \int_\Omega u=0\} $ where $\Omega $ is a bounded open set of $\mathbb R^d$ that $$|a(u,v)|\leq C\int_\Omega |\nabla u||\nabla v|$$ if $$a(u,v)=\int_\Omega A(x)\nabla u\cdot \nabla v.$$

I'm sure it's not complicate, but I have $$|a(u,v)|\leq \int_{\Omega }|A(x)\nabla u\cdot \nabla v|,$$ but how can I use the uniformly ellipticity ? It would be fine if instead of $\nabla v$ I would have $\nabla u$). And what mean $A(x)$ that $A$ is a matrix that depend on $x$ ? (i.e. the entries are functions that depend on $x$) ?

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By Cauchy-Schwartz, $$|A(x)\nabla u\cdot \nabla v|\leq |A(x)\nabla u||\nabla v|.$$ Remark that $|A(x)\nabla u|^2\leq C|\nabla u|$ (why ?) and conclude.

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  • $\begingroup$ You have an extra power of two, should be $|A(x) \nabla u|$. $\endgroup$ – fourierwho Dec 24 '17 at 19:42
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Note that for a symmetric positive definite matrix A we have $$ \|A\| =\sup_{\|x\|_2=1}\{\|Ax\|_2\}= \sup_{\|x\|_2=1}\{|\langle Ax,x\rangle_2|\}$$ Therefore, since for all $x\in \Bbb R^d$ $$|Ax\cdot x| =\left|\sum_{i,j=1}^{d}A_{ij}x_ix_j \right|\le C\|x\|_2$$

We automatically for all $x\in \Bbb R^d$ we get

$$\|Ax\|_2\le C\|x\|_2$$

Now replacing $x_i= \partial_iu $ and making use of by Cauchy Schwartz inequality we have

$$|A\nabla u\cdot \nabla v| \le\|A\nabla u\|_2\| \nabla v\|_2\le C \|\nabla u\|_2\| \nabla v\|_2$$

Integrating both side and applying Cauchy Schwartz again we obtain: $$\int_\Omega|A\nabla u\cdot \nabla v| \le C\int_\Omega \|\nabla u\|_2\| \nabla v\|_2\le C\|\nabla u\|_{L^2(\Omega)}\| \nabla v\|_{L^2(\Omega)}$$

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