Q:

Evans PDE Problem 5.15: Poincaré inequality for functions with large zero set

I have trouble proving the following problem (Evans PDE textbook 5.10. #15). Could anyone kindly help me solving the problem? I know that I should somehow use Poincaré inequality but I still cannot solve it. 
Fix $\alpha>0$ and let $U=B^0(0,1)\subset \mathbb{R}^n$. Show there exists a constant $C$ depending only on $n$ and $\alpha$ such that
$$
\int_U u^2 dx \le C\int_U|Du|^2 dx, 
$$
provided that $u\in W^{1,2}(U)$ satisfies $|\{x\in U\ |\ u(x)=0\}|>\alpha$. 

A:

Setting: $A\equiv \{x\in U: u(x)=0\}$.
By using the Poincaré-Witinger inequality, we have
\begin{eqnarray*}
C(n)\times \|\nabla u\| _{L^2(U)}
&\ge&
\|u - (u)_U\| _{L^2(U)}
\\
&\ge &
\|u\| _{L^2(U)} - \|(u)_U\| _{L^2(U)}
\\
&=&
\|u\| _{L^2(U)} - \|(u)_U\| _{L^2(A^c)}
\\
&=&
\|u\| _{L^2(U)} - \left|(u)_U\right| \times \left|A^c \right|^{1/2}
\\
&\ge&
\|u\| _{L^2(U)} - {1\over |U|} \times\left[ \int_{U} |u| dx\right] \times  |A^c|^{1/2}
\\
&=&
\|u\| _{L^2(U)} - {1\over |U|}\times \left[ \int_{A^c} |u| dx\right]\times |A^c|^{1/2}
\end{eqnarray*}
By Holder inequality, this last factor is equal to the largest
\begin{eqnarray*}
&\ge&
\|u\| _{L^2(U)} - {1\over |U|}\times \left|A^c\right|^{1/2} \times \|u\|_{L^2(U)} \times|A^c|^{1/2}
\\
&=&
\|u\| _{L^2(U)} - {1\over |U|}\times \left|A^c\right| \times \|u\|_{L^2(U)}
\\
&\stackrel{\boxed{-|A^c|\ge \alpha- |U|}}{\ge}&
\|u\| _{L^2(U)} + {1\over |U|}\times (\alpha - |U|)\times \|u\|_{L^2(U)}
\\
&=&
\left[1+ {1\over |U|} \  \left(\alpha - |U|\right)\right]\times \|u\|_{L^2(U)}
\end{eqnarray*}
Since $\alpha>0 \Rightarrow |U|>|U|-\alpha \Rightarrow 1>{|U|-\alpha \over |U|} \Rightarrow 1+ {\alpha -|U| \over |U|}>0.$
Denoting the constant $C_1(\alpha) \equiv 1+ {\alpha -|U| \over |U|} >0$, inequality obtained above we obtain that
$$
\|u\|_{L^2(U)} \le {C(n)\over C_1(\alpha)} \times \|\nabla u\|_{L^2(U)},
$$
concluding the proof.$_\blacksquare$
I hope I have helped. 
A hug.

