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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 Poincare 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 $u\in W^{1,1}(U)$ satisfies $|\{x\in U\ |\ u(x)=0\}|>\alpha$.

As Henry pointed out, this was asked before but no answer was given. I hope someone will kindly give an answer this time. Thank you very much.

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The same question was posted here: math.stackexchange.com/questions/235296/… –  Henry T. Horton Nov 29 '12 at 23:55
    
I should have mentioned the post but I am aware of it. It's closed without answer. Thank you for pointing out that. –  Zheng Nov 30 '12 at 1:10

3 Answers 3

Read the proof of Poincare inequality in Evans' book carefully. At some point one conclude that $u-\tilde{u}$ is constant. That is where you will need your $\alpha$ in your problem. Hope this will help you solve the problem.

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a hint: write the poincaré inequality and study the integral : $$ \frac{\int_{U} |u(x)| dx}{|U|} = \frac{\int_{U - \{ x \in U ; u(x) = 0\}} |u(x)| dx}{|U|} $$

Use Holder in the last integral .

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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.

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