Trace operator ($T:H^1(\Omega )\to L^2(\partial \Omega )$)

On a bounded Lipschitz domain, the Sobolev trace operator is the continuous linear extension to $H^1(\Omega )$ of the boundary restriction of smooth functions. Continuity and density make this extension unique; its existence is the Sobolev trace theorem. The kernel prescribes a homogeneous Dirichlet boundary condition. Vanishing Sobolev trace on only part of the boundary instead gives the form domain for a mixed boundary condition, with the natural Neumann boundary condition on the remainder. On an assembly of congruent tiles, taking a Sobolev trace commutes with a constant matrix acting on the tile restrictions. This lets a transplantation theorem be proved on quadratic form domains even when corner eigenfunctions have singular classical derivatives.

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 $\Vert u_k{\Vert }_p=k^{-1/p}\Vert \varphi {\Vert }_p\Vert \psi {\Vert }_p\to 0$.