# “Poincaré” inequality for $H^1$

I have to show the following: Let $U\subset \mathbb{R}^n$ be nice (i.e. bounded, open and boundary of class $C^1$). Further there's a function

$$f:H^1(U) \to \mathbb{R}^n$$

which is continuous and satisfies: $r\in \mathbb{R} \wedge f(r\mathbf1) = 0\Rightarrow r=0$ and $\forall \lambda\ge 0, u\in H^1(U): f(\lambda u)=\lambda f(u)$. Now I want to show a certain type of a Poincaré inequality, i.e.

$$\|u\|_{L^2}^2 \le C(U,f)(\|\nabla u\|^2_{L^2}+|f(u)|^2) \hspace{12pt} \forall u \in H^1(U)$$

The hint was: Prove by contradiction and use Rellich-Kondrachov embedding theorem.

First of all, I know the proof for a Poincaré type inequality for a closed subspace of $H^1$ which does not contain the non zero constant functions.

This is what I did so far:

Suppose not, then there are $c_k \to \infty$ such that $0\not= u_k\in H^1(U)$ with

$$c_k(\|\nabla u_k\|^2_2+|f(u_k)|^2) < \|u_k\|^2_2$$

As in the proof of the above mentioned theorem, I assume $\|u_k\|_2=1$. Now we have

$$c_k(\|\nabla u_k\|_2^2 +|f(u_k)|^2)<1$$

Since the term in the bracket is positive, we have $(\|\nabla u_k\|_2^2 +|f(u_k)|^2)\to 0$ as $c_k\to\infty$, now my first question:

Can I deduce from here, that $(\|\nabla u_k\|_2^2 +|f(u_k)|^2)$ must be bounded, therefore $u_k$'s would be bounded in $H^1$. Is this obvious or how could I show that?

Hence due to Rellich-Kondrachov embedding theorem, there's a subsequence $u_{k_l}$ which is Cauchy in $L^2$ and therefore converges in $L^2$ to $u\in L^2$. Since $\|u_k\|_2=1$ we have $u\not=0$.

Now I should show that from $(\|\nabla u_k\|_2^2 +|f(u_k)|^2)\to 0$ it follows that $u_k \to 0$. This would be my contradiction. I guess here I have to use the property of $f$, since I have them not used yet. A hint would be much appreciated.

There's a second part of this exercises and unfortunately I do not know how to start:

$$H_\Gamma:=\{u\in H^1(U);u=0 \mbox{ on } \Gamma\subset \partial U\}$$

and the volume of $\Gamma$ with respect to the surface measure is strictly positive. Then there's a constant $C>0$ not depending on $u$ such that:

$$\|u\|_{H^1}\le C \|\nabla u\|_2$$

Thanks a lot for your help.

math

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For the first part, if $U$ is assumed connected, I think I have a hint. First, show that $u$ is in fact in $H^1(U)$ and that $\nabla u_k\to \nabla u$, and it gives you that $u$ is constant and by continuity of $f$ that $f(u)=0$. So by the property $u=0$, which yields a contradiction since $||u||_{L^2}=1$. For the second problem, I don't have a simple solution, maybe using the definition of the integration over the boundary (which charts). –  Davide Giraudo Mar 31 '12 at 13:49
Is connectedness contained in the nice properties of $U$? –  Davide Giraudo Apr 1 '12 at 9:25
I guess we can assume that $U$ is connected. –  math Apr 1 '12 at 12:17
@ Davide Giraudo: Any suggestions how to show these two things: 1. $u\in H^1(U)$, here I have to show the following: $\int u\nabla \phi=-\int g \phi$ for all $\phi \in C^\infty_c$ and where $g$ is a function in $L^2$. Second I have to show: $\int(\nabla u_k -\nabla u)^2 dx =0$ as $k\to \infty$. How to show this? In fact, I don't know how to handle the square under the integral sign. Thanks again! –  math Apr 2 '12 at 7:20

If the inequality wasn't true, then for each $k\in\mathbb Z_{\geq 1}$ we would be able to find $u_k\in H^1(U)$ such that $\lVert u_k\rVert_{L^2}=1$ and $k^{—1}\geq \lVert \nabla u_k\rVert^2_{L^2}+|f(u_k)|^2$. Since $\lVert u_k\rVert_{H^1}^2\leq 2$, we can, by Rellich-Kondrachov embedding theorem, find a subsequence $v_k=u_{\varphi(k)}$, where $\varphi\colon\mathbb Z_{\geq 1}\to \mathbb Z_{\geq 1}$ is strictly increasing and $v_k$ converges weakly to $u$ in $L^2(U)$. If $\psi\in\mathcal D(U)$ and $j\in \{1,\ldots,n\}$, we have $$\int\frac{\partial v_k}{\partial x_j}\psi dx=-\int v_k\frac{\partial \psi}{\partial x_j}dx$$ and since $v_k$ converges weakly to $u$, we have $\lim_{k\to\infty}\int v_k\frac{\partial \psi}{\partial x_j}dx=\int u\frac{\partial \psi}{\partial x_j}dx$, and since $\lVert \nabla v_k\rVert_{L^2}\to 0$, we have that $\lim_{k\to\infty}\int\frac{\partial v_k}{\partial x_j}\psi dx=0$, so for all test function $\psi$: $\int u\frac{\partial \psi}{\partial x_j}dx=0$, and by connectedness $u$ is constant. Now using the properties of $f$ it should work.