The following question is an exercise from Evans' textbook Partial Differential Equations.
Given $\alpha > 0$ and $U = B_0(0,1)$, show that there exists a constant $C$, depending only on $n$ and $\alpha$, such that $$\int_U u^2 dx \leq C \int_U |\nabla u|^2 dx$$ provided $|\{x\in U | u(x) = 0\}| \equiv |M| \geq \alpha > 0$ and $u\in H^1(U)$.
This question has been asked before on this website. However, I am writing to verify my proposed solution. Here it is:
Proof:
Define $\alpha' = |\{x\in U | u(x) = 0\}| \equiv |M|$. By assumption, $\alpha' \geq \alpha > 0$. Let $u \in H^1(U)$. If $\alpha' = |U|$, then $|M^c| = 0$. Hence $u$ is non-zero on a set of measure zero. The inequality is true trivially. So WLOG assume that $\alpha' \neq |U|$. Since $U$ is an open, connected set with $C^1$ boundary, Poincare's inequality implies that:
$$ ||u - u_U ||_{L^2(U)} \leq C || Du ||_{L^2(U)}. $$
for some constant $C > 0$. We directly manipulate the inequality above. First note that:
$$ C || Du ||_{L^2(U)} \geq || u - u_U ||_{L^2(U)} \geq || u ||_{L^2(U)} - || u_U ||_{L^2(U)} = || u ||_{L^2(U)} - | u_U | |U|^{1/2} \geq ||u||_{L^2(U)} - \frac{1}{|U|^{1/2}} \int_{U} |u(x)| dx. $$
The first inequality follows from Poincare's inequality. The second inequality follows from the triangle inequality. The third equality follows by the definition of the $L^2$ norm. The fourth inequality is a simple estimate using the definition of $u_U$. Continuing with the manipulations, we have:
$$ ||u||_{L^2(U)} - \frac{1}{|U|^{1/2}} \int_{U} |u(x)| dx = || u ||_{L^2(U)} - \frac{1}{|U|^{1/2}} \int_{M^c} |u(x)| dx \geq || u ||_{L^2(U)} - \frac{|M^c|^{1/2}}{|U|^{1/2}} ||u||_{L^2(U)}. $$
The first equality follows since $u \equiv 0$ on $M$. The second inequality follows from Holder's inequality and noting that $||u||_{L^2(U)} \geq ||u||_{L^2(M^c)}$. Noting that:
$$ || u ||_{L^2(U)} - \bigg ( \frac{|M^c|}{|U|} \bigg )^{1/2} ||u||_{L^2(U)} = || u ||_{L^2(U)} - \bigg ( \frac{|U| - |M| }{|U|} \bigg )^{1/2} ||u||_{L^2(U)} = \bigg ( 1 - \sqrt{ 1 - \frac{|M| }{|U|} } \bigg ) ||u||_{L^2(U)} $$
All in all, we have:
$$ ||u||_{L^2(U)} \leq \frac{C }{ \bigg ( 1 - \sqrt{ 1 - \frac{|M| }{|U|} } \bigg ) } ||Du||_{L^2(U)} $$
Taking squares on both sides since both sides are non-negative yields the desired result.
Please feel free to critique the proof.