Prove that $a(u,v)=\int_\Omega A\nabla u\cdot \nabla v$ is continuous if $A$ uniformly elliptic. 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 ?

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:
$$\int_\Omega|A\nabla u\cdot \nabla v| \le C\int_\Omega \|\nabla u\|_2\| \nabla v\|_2\le C\|\nabla u\|_{L^2(\Omega)}\| \nabla v\|_{L^2(\Omega)}$$
