Lax Milgram Application I found the next exercise in a book. I'm stuck at proving the coercivity of the bilinear functional in the variational formulation of the problem.

Suppose that $\Omega$ is a regular $\mathcal{C}^1$ bounded open set. Prove the existence and uniqueness for the following problem with Fourier limit conditions
  $$ \begin{cases} -\Delta u&=f & \text{in }\Omega \\ \frac{\partial u}{\partial n}+u&=g & \text{in } \partial \Omega \end{cases}$$
  where $f \in L^2(\Omega)$ and $g$ is the trace on $\partial \Omega$ of a function in $H^1(\Omega)$.

Can you please give me a hint on how to prove the coercivity? Thank you.
 A: The argument is basically the same as a proof of the Poincare inequality in $H_0^1$: Assume that $a$ (as in Davide's comment) is not coercive, then we can take $u_n\in H^1$ with $\| u_n\|_{H^1}=1$ such that 
$$\| \nabla u_n\|_{L^2(\Omega)}^2 + \| u_n \|_{L^2(\partial\Omega)}^2 \leq n^{-1} \| u_n\|_{H^1}^2=n^{-1}$$
First notice that the inequality gives that $\nabla u_n, u_n|_{\partial\Omega} \to 0$ strongly in the corresponding $L^2$ spaces. By basic results on Sobolev spaces, if we can prove that there is $u\in L^2(\Omega)$ with $u_n \to u$ strongly in this last space, we could conclude two things: $u\in H^1$ and $u$ is constant on each connected component of $\Omega$. So $u$ is locally constant, by smoothness of the domain each connected component must have a nontrivial piece of the boundary, so if $u\neq 0$ in a component we would have a non-zero trace, which is absurd since the trace is a bounded linear operator and $u_n|_{\partial\Omega} \to 0$. We conclude that $u=0$ on $\Omega$. This is of course a contradiction since $u_n\to u$ in $H^1$ and $\|u_n\|_{H^1}=1$.
Now all we have to do is show such an $u$. By reflexivity there is $u\in H^1$ with (really a subsequence) $u_n\rightharpoonup u$ in $H^1$, and by Rellich's theorem $u_n \to u$ in $L^2$. 
