Let $\Omega\subset\mathbb{R}^n$ be a bounded open set with boundary $\Gamma$ . Let $\alpha>0$ . Define $$a(u,v)=\int_\Omega\nabla u.\nabla v+\int_\Omega uv+\alpha\int_\Gamma uv$$ for $u,v\in H^1(\Omega)$ . Let $f\in L^2(\Omega)$ . Show that there exist unique $u\in H^1(\Omega)$ such that $$a(u,v)=\int_\Omega fv$$ for every $v\in H^1(\Omega)$ .
We would want to use Lax-Milgram . It is easy to see the continuity of $a(u,v)$ using trace theorem . But to show ellipticity , we should have $$a(u,u)=||u||_{H^1(\Omega)}^2+\alpha||u||_{L^2(\Gamma)}^2$$ the RHS greater or equal to some $C||u||_{H^1(\Omega)}^2$ . Trace theorem gives the reverse kind of inequality . How to get something like $||u||_{L^2(\Gamma)}^2\geq d||u||_{H^1(\Omega)}^2$ ? Any help is appreciated .
The thread daw linked shows a much stronger result, namely that you an still apply Lax-Milgram in the absence of the $uv$ term, but here the argument is much simpler.
Since you already have the full $H^1$-norm from the first two terms, it suffices to just saw the last term in non-negative (that is, the inequality you seek with $d=0$). Explicitly we have $$ a(u,u) = \lVert u \rVert_{H^1(\Omega)}^2 + \alpha \lvert u \rvert_{L^2(\Gamma)}^2 \geq \lVert u \rVert_{H^1(\Omega)}^2. $$ As you say the trace theorem is still needed for the upper bound, but you don't need it for the lower bound thanks to the special structure of the equation.