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).

  • $\begingroup$ Note that this holds for any $\lambda > 0$ (with $C$ depending on $\lambda$). $\endgroup$ – gerw Dec 13 '13 at 8:09

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.