A question about a Sobolev space trace inequality (don't understand why it is true)

Question

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).

Answer 1

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.