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Let $\Omega \subset \mathbb R^d$ smooth, bounded and connected domain. Let $A\in \mathbb R^{d\times d}$ symetric and uniformly elliptic, i.e. there is $C>0$ such that $$C^{-1}\|x\|^2\leq Ax\cdot x\leq C\|x\|^2.$$

How can I prove that $$a(u,v)=\int_\Omega A \nabla u\cdot \nabla v,$$ continuous ? I know that $$|a(u,v)|\leq \int_\Omega \|A\nabla u\|\|\nabla v\|.$$ I suppose that $|A\nabla u|\leq C|\nabla u|$, but I can't prove it since $|A\nabla u|$ is not $A\nabla u\cdot \nabla u$. I also tried as $$|A\nabla u|^2=A^2\nabla u\cdot \nabla u,$$ but is also $A^2$ uniformly elliptic ? If yes how can I prove it ? If no, how can I conclude ?

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  • $\begingroup$ continuous for what norm $\endgroup$ – user503348 Jan 3 '18 at 9:55
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If $A$ is a symmetric matrix, then $Ax \cdot x \le C \|x\|^2$ implies that all eigenvalues of $A$ are bounded from above by $C$. Hence, $\|A x\| \le C \|x\|$.

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  • $\begingroup$ Thank you for your answer, but I still don't understand why $\|Ax\|\leq C\|x\|$ with your argument of eigenvalue. Could you please develop ? $\endgroup$ – user386627 Jan 3 '18 at 9:52
  • $\begingroup$ You just need to expand $x$ in an orthonormal basis of eigenvectors of $A$. $\endgroup$ – gerw Jan 3 '18 at 10:32
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Note that for a symmetric 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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  • $\begingroup$ I agree that $\|A\|=\sup\{|<Ax,x>|: \|x\|_2\leq 1\}$ but why it's equal at $\sup_{\|x\|\leq 1}\|Ax\|_2$ ? $\endgroup$ – user380364 Jan 21 '18 at 14:05

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