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

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The same question was posted here:… – 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

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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$|u| _{L^2(U)} - |(u)_U| _{L^2(U)} = |u| _{L^2(U)} - |(u)_U| _{L^2(A^c)}$. Why this is equal? I don't see how this equality is true? – Sherry Feb 22 '15 at 7:10
@Sherry I answered this same question here. – ᴡᴏʀᴅs Feb 22 '15 at 7:22
@FamousBlueRaincoat My teacher gives me hint like this, but I don't understand how the hint is related to the problem: – Sherry Feb 22 '15 at 23:03
@FamousBlueRaincoat The hint is suppose $w$ is a smooth function on the ball $B$ and $S$ is a set of positive measure. For $x\in B$ and $y\in S$, $|w(x)-w(y)|\le C\int_0^1 |Dw(tx+(1-t)y)|dt$. and so $\int\int|w(x)-w(y)|dydx\le C\int_0^1 \int \int |Dw(tx+(1-t)y)|dydxdt$. Then split the integration in the $t$ variable into two parts $0 \le t \le \frac{1}{2}$ and $\frac{1}{2} \le t \le 1$ and estimate. In general, use approximation of $u$ by smooth functions. Could you please help me with some further explanation about how the hint is related to the problem? Thanks! – Sherry Feb 22 '15 at 23:08

Ok, this question can be proved by using general version of Poincare inequality.

This is Theorem 12.23 in Leoni's book. Let me copy it here:

Let $1\leq p<\infty$ and let $\Omega\subset\mathbb R^N$ be a connected extension domain for $W^{1,p}(\Omega)$ with finite measure. Let $E\subset \Omega$ be a lebesgue measureable set with positive measure. Then there exists a constant $C=C(p,\Omega,E)>0$ such that for all $u\in W^{1,p}(\Omega)$, $$ \int_\Omega|u(x)-u_E|^pdx\leq C\int_\Omega |\nabla u(x)|^pdx,\tag 1 $$ where $$ u_E:=\frac{1}{|E|}\int_E u(x)dx $$

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 $(1)$.

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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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