How to deduce the uniform ellipticity condition from an integral condition Let $\Omega$ be a bounded open set, and let $A$ be a $N\times N$ symmetric matrix with entries in $L^\infty(\Omega, \mathbb{R})$, such that for some positive $\lambda$ and $\Lambda$ the following inequalities hold for every $u \in W^{1,2}_0(\Omega, \mathbb{R})$:
$$ 
\lambda \int \limits_\Omega \! | \nabla u(x) |^2 \, \mathrm{d}x \leq \int \limits_\Omega \! A(x) \nabla u(x) \cdot \nabla u(x) \, \mathrm{d} x \leq \Lambda \int \limits_\Omega \! | \nabla u(x) |^2 \, \mathrm{d}x
$$
Is there a way to deduce that $A$ satisfies the following pointwise inequalities
$$
\lambda | \xi |^2 \leq A(x) \xi \cdot \xi \leq \Lambda |\xi|^2
$$
for almost every $x \in \Omega$ and for all $\xi \in \mathbb{R}^N$?
The first strategy I tried to adopt was to choose a function in $W^{1,2}_0(\Omega)$ whose gradient is essentially a fixed vector $\xi \in \mathbb{R}^N$. However, to choose it in such a way that it belongs to $W^{1,2}_0(\Omega)$ it must vanish near the boundary. Hence I chose
$$
u(x) := (\xi \cdot x ) \zeta(x)
$$
where $\zeta$ is a cutoff function such that $\zeta \equiv 1$ on $\omega \subset \subset \Omega$ and with support in $\Omega$. Maybe, if it is convenient, we can also control $|\nabla \zeta|$ when $\partial{\omega}$ is "near" $\partial{\Omega}$. But I don't see how to obtain the pointwise estimates from the intagral ones!
Second strategy. I tried to argue by contradiction. Assume, for example, that the inequality
$$
\lambda |\xi|^2 \leq A(x) \xi \cdot \xi
$$
doesn't hold for almost every $x \in \Omega$ for some $\xi \in \mathbb{R}^N$. This means that the set
$$
\omega := \{ x \in \Omega \ | \ \lambda |\xi|^2 > A(x) \xi \cdot \xi \}
$$
has positive measure. If I was able to construct a function in $W^{1,2}_0(\Omega)$ whose gradient is $\xi$ on $\omega$ and $0$ in $\Omega \setminus \omega$, I would conclude integrating, contradictiong the first intregral inequality. Though such function cannot be constructed in that precise way, maybe I can argue as in the first strategy using suitable cutoff functions, but my tries failed.
Any suggestions?
 A: A sketch of what you should do. 


*

*Show that if $g(x)$ is a $L^\infty$ function such that $\int |g(x) \phi(x)|\mathrm{d}x \leq \Lambda \int |\phi(x)|\mathrm{d}x$ for every $\phi\in C^\infty_0$ then $|g(x)| \leq \Lambda$. (Hint, suppose not, let $S = \{ |g(x)| > \Lambda\}$ has positive measure. Let $\phi_{\eta,\epsilon}(x)$ be a mollified (with parameter $\eta$) version of the indicator function $\mathbf{1}_{S\cap \Omega_\epsilon}$ where $\Omega_\epsilon$ is the subset of $\Omega$ that is at least $\epsilon$ distance from the boundary. Take $\eta,\epsilon$ sufficiently small. For small $\epsilon$ we guarantee that $S\cap \Omega_\epsilon$ has positive measure. For small $\eta$ we guarantee that $\mathrm{supp}(\phi_{\eta,\epsilon})$ has arbitrarily small measure.)

*Similarly work with the lower bound $\lambda$. 

*Choose your test function $u$ in your original problem to be something of the form 
$$ u_{\eta,\epsilon,k}(x) = \phi_{\eta,\epsilon}(x) \exp(i k x\cdot\xi) \in C^\infty_0(\Omega) \tag{*}$$
and take $k \to \infty$ (sufficiently large depending on the choice of $\eta$ and $\epsilon$ so that the term corresponding to $k^2 A\xi\cdot\xi$ "swamps" the contribution when the derivative falls on $\phi_{\eta,\epsilon}$). 
Note that this is more akin to your second strategy. 

A short word on the form (*): one should think of the scaling in $\eta$ as sharpening the support of the function $u$. But by the uncertainty principle this will necessarily broaden the Fourier support. We compensate this by doing a translation on the Fourier side (corresponding to increasing $k$, a phase shift in physical space). The end result is that while the size of the Fourier support is huge, we can make it concentrated in an arbitrarily small angular neighborhood of the ray corresponding to $\xi$. (Remember, a small angle can still give lots of room if you go far enough away from the origin.) 
That we can use this method of proof hinges on the fact that $A\xi\cdot \xi$ is homogeneous in scale for $\xi$, and the only part that is important is the direction!
