Trace inequality

Question

Could you please give me a hint on how to prove the following inequality

$$\|u\|_{L^2(\Gamma)}\le C\|u\|^{\frac12}_{L^2(\Omega)}\|u\|^{\frac12}_{H^1}, \quad \forall u\in H^1.$$,

Answer 1

Typically, you get an estimate like $$ \lVert u \rVert_{H^{s-1/2}(\Gamma)} \le C_s \, \lVert u \rVert_{H^{s}(\Omega)} $$ for $s > 1/2$ by embedding theorems (maybe some regularity of the boundary is needed). Here, $H^{s}(\Omega)$ is a fractional Sobolev space.

By interpolation theory, you get $$ \lVert u \rVert_{H^{s}(\Omega)} \le C\, \lVert u \rVert_{L^2(\Omega)}^{1-s}\, \lVert u \rVert_{H^1(\Omega)}^{s}. $$ Combining the estimates, we have $$\lVert u \rVert_{L^2(\Gamma)} \le C \, \lVert u \rVert_{H^{s-1/2}(\Gamma)} \le C \, C_s \, \lVert u \rVert_{L^2(\Omega)}^{1-s}\, \lVert u \rVert_{H^1(\Omega)}^{s}$$ for all $s > 1/2$. Maybe, one can trace the dependence of $C_s$ as $s \to 1/2$ to conclude something.