> [!theorem] > > Let $p \in [1, \infty)$, $U \subset \real^n$ be bounded and [[Open Set|open]] such that $\partial U$ is [[Differentiable Boundary|of class]] $C^1$. Then there exists a [[Bounded Linear Map|linear operator]] > $ > T: W^{1, p}(U) \to L^p(\partial U) > $ > such that > 1. $Tf = f|_{\partial U}$ for any $f \in UC(U)$. > 2. $T$ is bounded with respect to the [[Sobolev Space|Sobolev norm]]. > > known as the **trace** on $\partial U$. > > *Proof*. Let $f \in UC^1(u)$ be a [[Space of Uniformly Continuously Differentiable Functions|UC]] function, and we extend $f$ to $\partial U$ by taking limits. Let $x \in \partial U$, then by the [[Local Trace Theorem]], there exists $C_x \ge 0$ and $V_x \in \cn^o(x)$, independent of $f$, such that $\norm{f|_{\partial U \cap V_x}}_p \le C_x\norm{f}_{W^{1, p}(U)}$. > > By [[Compactness|compactness]], the exists a finite subcover $\bracs{V_j}_1^k$ of $\partial U$. From here, > $ > \norm{f|_{\partial U}} \le \sum_{j = 1}^k\norm{f|_{\partial U \cap V_j}}_p \le \sum_{j = 1}^kC_j \norm{f}_{W^{1, p}(U)} > $ > > This defines the trace operator on $UC^1(U)$. By the [[Approximation of Sobolev Space by Smooth Functions|density of UC functions]] and the [[Linear Extension Theorem|linear extension theorem]], $T$ admits a unique, norm-preserving extension to $W^{1, p}(U)$. > > If $f \in UC(U)$, then there exists a sequence $\seq{f_n} \subset UC^1(U)$ that converges uniformly to $f$. Therefore $Tf = f|_{\partial U}$ for any $f \in UC(U)$.