Bounding integral on boundary

Question

Let $\Omega$ be bounded domain with $C^1$-boundary, $g \in L^2(\partial\Omega)$ and $u \in W_2^1(\Omega)$. why $F(u)$ defined by following is bounded linear on $W_2^1(\Omega)$?

$$F(u)=\int_{\partial \Omega}ug dS.$$

Answer 1

The essential steps are the trace theorem and Hölder's inequality.

I take the version found in Evans 'Partial Differential Equations'

Trace Theorem: Assume $\Omega$ is bounded and $\partial \Omega$ is $C^1$. Then there exists a bounded linear operator $$T:W^1_p(\Omega) \to L^p(\partial \Omega)$$ such that $Tu=u|_{\partial \Omega}$ if $u \in W^1_p(\Omega) \cap C(\overline{\Omega})$ and $\|Tu\|_{L^p(\partial \Omega)} \leq C \|u\|_{W^1_p(\Omega)}$ for each $u \in W^1_p(\Omega)$.

We have given a functional $F:W^1_2(\Omega) \to \mathbb{R}$ such that

$$F(u)=\int_{\partial \Omega} Tu \, g \, \text{d}S$$

for a $g \in L^2(\partial \Omega)$. Notice that I have written a $T$ before $u$ in the integral since $u \in W^1_2(\Omega)$ itself is not defined on the boundary; but $Tu$ is. Sadly $T$ is often left out and one has to interpret $u$ in the right way.

Now using Hölder's inequality in the first step and the trace theorem in the second step yields

$$|F(u)|\leq \|Tu\|_{L^2(\partial \Omega)} \|g\|_{L^2(\partial \Omega)} \leq C \|u\|_{W^1_2(\Omega)} \|g\|_{L^2(\partial \Omega)}$$

and thus $F$ acts linearly and continuously on $W^1_2(\Omega)$.