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Let $\Omega$ be an open set with boundary $\partial\Omega$. Let $u \in H^1(\Omega)$.

There exists a $\lambda \in \mathbb{R}$ such that $$\int_\Omega |\nabla u |^2 + \lambda\int_{\partial\Omega}u^2 \geq C\lVert u \rVert^2_{H^1(\Omega)}$$ for some constant $C$.

I don't understand why this inequality is true. I thought maybe there is something to do with the right inverse of the map trace being continuous but I am not sure if this is correct. Help appreciated.

Some additional info about $u$:

For $v \in H^{\frac 1 2}(\partial\Omega)$, $u$ is the solution of $-\Delta u = 0$ on $\Omega$ with $u = v$ on $\partial \Omega$.

(I saw this in page 135 of Lions' Quelques methodes... book).

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  • $\begingroup$ Note that this holds for any $\lambda > 0$ (with $C$ depending on $\lambda$). $\endgroup$ – gerw Dec 13 '13 at 8:09
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Fix $\lambda>0$ and suppose ad absurdum that there is a sequence $u_n\in H^1$ such that $$\int_\Omega|\nabla u_n|^2+\lambda\int_{\partial\Omega}u_n^2<\frac{1}{n}\|u_n\|_{1,2}\tag{1}$$

If we dive the above expression by $\|u_n\|_{1,2}$ and denote by $v_n=\frac{u_n}{\|u_n\|_{1,2}}$ we get that $$\int_\Omega |\nabla v_n|^2+\lambda\int_{\partial\Omega}v_n^2<\frac{1}{n},\ \ \|v_n\|_{1,2}=1~~ \forall n\tag{2}$$

We conclude from $(2)$ that $\int_\Omega |\nabla u_n|^2$ and $\int_{\partial\Omega}{v_n}^2$ converge to zero and $\int_\Omega u_n^2\to 1$. Assume without loss of generality that $u_n\rightharpoonup u$ in $H^1$, where $\rightharpoonup$ denotes weak convergence. Assume also without loss of generality that $u_n\to u$ in $L^2$.

Note that $\|\nabla u\|_2=0$ which implies that $u$ is constant a.e. On the other hand $\int_{\partial\Omega} u_n^2\to\int_{\partial\Omega}u^2$, hence, $\int_{\partial\Omega}u^2=0$ which implies that the trace of $u$ is zero. Because $u$ is constant we conclude that $u$ is zero in the whole $\Omega$.

To finish, note that as $\int_\Omega u_n^2\to 1$, we must have $\int_\Omega u^2=1$ which is an absurd.

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  • $\begingroup$ Thank you Tomas. May I ask how this proof occurred to you and if you have a source for this? Thanks $\endgroup$ – soup Dec 16 '13 at 15:44
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    $\begingroup$ You are welcome @soup. In fact, this is a standard argument in PDE's. See for example here math.stackexchange.com/questions/601996/… or here: math.stackexchange.com/questions/361423/… This argument is called a argument of compacity and uses the fact that the problem is homogeneous. $\endgroup$ – Tomás Dec 16 '13 at 15:47
  • $\begingroup$ Thanks, I never that argument. Excellent answers there. $\endgroup$ – soup Dec 16 '13 at 18:07
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    $\begingroup$ @soup This mimics the proof of the Poincaré inequality. See page 290 of Evans for this exact argument. One should also note that the fact we can consider a weakly convergent subsequence comes from the Rellich-Kondrachov Compactness Theorem. $\endgroup$ – AmorFati May 29 '17 at 10:24

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