Bound L^2 norm of gradient by L^infinity norm

For $u\in H^1_{loc}(\mathbb{R}^2)$ a weak solution to

$$-div(a\cdot \nabla u) = 0$$

with $a_{ij}$ constant and strongly ellipctic, we showed that

$$\int_{B(x_0,s)} |\nabla u|^2 dx \leq \left(\frac{2s}{r}\right)^\alpha \int_{B(x_0,r)} |\nabla u|^2 dx $$

for some constant $\alpha > 0$.

Our goal is now to show a version of Liouvilles theorem i.e. that if in addition $u\in L_\infty(\mathbb{R}^2)$, $u$ must already be constant.

My idea is to bound $\int_{B(x_0,r)} |\nabla u|^2 dx$ somehow and then let $r\to\infty$. This way we would get for all $x_0$ and $s$ that $u$ is constant on $B(x_0,s)$ etc.

The problem is now to bound the integral. I thought to this end we want to show

$$||\nabla u ||_{L^2(V)} \leq C$$

for some constant $C$ and all compact $V$, where I assumed that it would be something like $c\cdot ||u||_\infty$ for some other constant $c$. But as pointed out in the answers, there are counterexamples.

It feels like I‘m mentally blocked (or as we would say in Germany „ich hab ein Brett vorm Kopf“).

Thanks for the help!

Solution 1:

Such an estimate is not necessarily true.

Consider the function $u(x) = \cos x^2$ for $x \in \mathbb{R}$. Then $\|u\|_\infty = 1$, but $\|\nabla u\|_{L^2(-M,M)} \sim M^{3/2}$ as $M$ becomes large.

Solution 2:

You have the right idea: Remember the basic energy/Caccioppoli inequality, which is the first step in proving the inequality that you have (I'll omit the centers of the balls, they can be whatever): $$ \int_{B_r} |\nabla u|^2\, dx \lesssim r^{-2}\int_{B_{2r}} |u|^2\, dx. $$ Now just bound the last integral with the $L^\infty$ norm of $u$. As you'll see the fact that you're in dimension $n=2$ is crucial for this!

I'm not yet sure how to fit the counterexample, but my guess is that that function doesn't satisfy an elliptic equation on the line.