Sobolev trace theorem

This theorem describes the continuous extension of the Trace operator. For $s>1/2$, restriction to the coordinate hyperplane extends uniquely to a bounded linear mapIn Fourier variables, Cauchy--Schwarz bounds the integral over the normal frequency because $(1+t^2{)}^{-s}$ is integrable exactly when $s>1/2$.

On a bounded smooth domain, the Sobolev trace theorem can be strengthened toFor smooth $u$, extend the outward unit normal to a smooth vector field $X$ near the closure of $U$ with $X\cdot \nu =1$ on its boundary. The divergence theorem gives ${\int }_{\partial U}{u}^2={\int }_U(\mathrm{div}X)u^2+2uX\cdot Du$, and the Cauchy-Schwarz inequality gives the estimate. Density extends it to $H^1(U)$. A bound containing only the product $\Vert u{\Vert }_2\Vert Du{\Vert }_2$ is false for nonzero constant functions. Equivalently, for every $\epsilon >0$, the Young inequality gives $\Vert Tu{\Vert }_2^2\le \epsilon \Vert Du{\Vert }_2^2+C_{\epsilon }\Vert u{\Vert }_2^2$.

For $1\le p<\infty$, boundary restriction on smooth functions extends uniquely to a bounded linear map $T:W^{1,p}({\mathbb{R}}_{+}^n)\to L^p({\mathbb{R}}^{n-1})$. The estimate follows by applying the fundamental theorem of calculus on normal line segments and then integrating over the boundary variables.

For $1\le p<\infty$, no bounded map from $L^p({\mathbb{R}}^n)$ to $L^p({\mathbb{R}}^{n-1})$ can agree with restriction on continuous functions. Indeed,has a fixed nonzero trace while $\Vert u_k{\Vert }_p=k^{-1/p}\Vert \varphi {\Vert }_p\Vert \psi {\Vert }_p\to 0$.