Trace operator

Remarks

The operator exists and is unique. For $u$ smooth up to the boundary, the divergence theorem gives $\norm{u|_{\partial U}}_{L^p(\partial U)} \le C \norm{u}_{W^{1,p}(U)}$ with $C$ depending on $p$ and $U$. Such functions are dense in $W^{1,p}(U)$, so restriction extends uniquely to a continuous linear map on all of $W^{1,p}(U)$.

Example: on $U = (0, 1)$ the boundary is $\set{0, 1}$, and every $u \in W^{1,p}(0, 1)$ has a continuous representative. Its trace is the pair of values $(u(0), u(1))$ of that representative.

Non-example: on $U = (0, 1)$, restriction to the boundary is not bounded for the $L^p$ norm alone. The functions $u_k(x) = (1 - x)^k$ satisfy $u_k(0) = 1$ while $\norm{u_k}_{L^p(0,1)} \to 0$, so no bounded map $L^p(0, 1) \to \R^2$ extends restriction. This is why the definition uses $W^{1,p}(U)$.

The kernel of $T$ is $W^{1,p}_0(U)$. For $1 < p < \infty$, the image of $T$ is the fractional Sobolev space $W^{1 - 1/p, p}(\partial U)$, a result of Gagliardo.