Let $\alpha > 0$. Prove that there exists a constant $C > 0$ that depends only on $\alpha, U, n$ such that $|| u ||_{L_2} \leq C ||\nabla u ||\ \forall\ u \in\ H^1(U)$ such that $|M| \geq \alpha$, where $M = \{x \in U : u(x) = 0\}$
My attempt: I know that under these conditions, $||u - (u)_{U}||_{L_2} \leq C ||\nabla u||_{L_2}$, for some constant $C$, (where $(u)_{U} = \frac{1}{|U|} \int_{U} u\ dx$), so I substracted and added $(u)_{U}$.
$|| u - (u)_{U} + (u)_{U}||^2_{L_2} = $
$= \int_U (u - (u)_{U})^2 dx + 2 \int_U (u - (u)_{U}) (u)_{U}\ dx+ \int_U ((u)_{U})^2\ dx =: (I) + (II) + (III)$
I already know how to bound the first term, and I would like to show that the second plus the third term is less than zero, but it's not. Some things cancel, and I get $(I) + (II) = \frac{1}{|U|} (\int_U u)^2$, which is positive.
I know somewhere in the middle I should replace $U$ for $M$,but I get the same result if I do that. What am I doing wrong? Is this the wrong approach? I tried other things too, but I didn't get anywhere. Thanks!
Let me recall Theorem 12.23 in Leoni's book:
What you need to do is set $E:=\{x\in\Omega,\,u(x)=0\}$ and notice that $|E|>0$. Then $u_E=0$ and we done by $(1)$.
The Theorem itself is proved by using contradiction. Try to prove it before you go ahead and look at this book. The proof is shoot.