About the trace inequality in Sobolev space

Question

Given $\Omega$ is a region with Lipschitz domain and $1\leq p\leq +\infty$. Prove that there exists a constant $C>0$ such that $$ \Vert u\Vert_{L^p(\partial\Omega)} \leq C \Vert u\Vert_{L^p(\Omega)}^{1-\tfrac{1}{p}} \Vert u\Vert_{H^1(\Omega)}^{\tfrac{1}{p}},\quad\forall u\in W^{1,p}(\Omega) $$ I have no idea how to prove this, can somebody give me a hint :(

Answer 1

As usual the idea is to reduce matters to the simplest case: By working locally you can look at the case of $\Omega$ being the domain above the graph of a Lipschitz function. As a toy model, lets look at the case this function is identically zero, i.e. $\Omega=\mathbb{R}^n_+$ is the upper half space.

Write $(x,t)\in \mathbb{R}^{n-1}\times (0,\infty)$ for coordinates in $\Omega$, and so for any $f\in C^\infty_c(\mathbb{R}^n)$ we have by the fundamental theorem of calculus $$ \int_{\mathbb{R}^{n-1}}|f(x,0)|^p\,dx = -\int_{\mathbb{R}^{n-1}} \int_0^\infty \partial_t(|f(x,t)|^p)\, dt \, dx = p \int_{\mathbb{R}^n_+} |f(x,t)|^{p-2} f(x,t)\partial_tf(x,t)\, dt dx. $$ Taking absolute values and usign Hölder's inequality gives $$ \int_{\mathbb{R}^{n-1}}|f(x,0)|^p\,dx \leq \int_{\mathbb{R}^n_+} |f(x,t)|^{p-1}|\partial_t f(x,t)|\, dtdx \leq \| f\|_{L^p(\mathbb{R}^n_+)}^{p-1}\| \partial_t f\|_{L^p(\mathbb{R}^n_+)}, $$
and this is a stronger version of your inequality.

For the general case of a Lipschitz graph you want to do the same thing, though it gets a bit ugly notation wise due to the weight $\sqrt{1+|\nabla \varphi(x)|^2}$ that appears in the surface measure, where $\varphi$ is your Lipschitz function. Still, I'd encourage you to try it out on your own.