I was reading a theorem about the trace of a function (taken from Evans' book on PDEs). In the proof, I don't understand the last paragraph. Why the uniform convergence of $u_m$ to $u$ implies $Tu=u|_{\partial U}$ for a function in $W^{1,p}\cap C(\overline{U})$?
THEOREM 1 (Trace Theorem). Assume $U$ is bounded and $\partial U$ is $C^1$. Then there exists a bounded linear operator $$T : W^{1,p}(U) \rightarrow L^p(\partial U)$$ such that
$\quad$(i) $Tu=u|_{\partial U}$ if $u \in W^{1,p}(U) \cap C(\bar{U})$
and
$\quad$(ii) $$\|Tu\|_{L^p(\partial U)} \le C \| U \|_{W^{1,p} (U)},$$ for each $u \in W^{1,p}(U)$, with the constant $C$ depending only on $p$ and $U$.
DEFINITION. We call $Tu$ the trace of $u$ on $\partial U$.
Proof. 1. Assume first $u \in C^1(\bar{U})$. As in the first part of the proof of Theorem 1 in ยง5.4 let us also intially suppose $x^0 \in \partial U$ and $\partial U$ is flat near $x^0$, lying in the plane $\{x_n=0\}$. Choose an open ball $B$ as in the previous proof and let $\hat{B}$ denote the concentric ball with radius $r/2$.
$\quad$Select $\zeta \in C_c^\infty(B)$, with $\zeta \ge 0$ in $B$, $\zeta \equiv 1$ on $\hat{B}$. Denote by $\Gamma$ that portion of $\partial U$ within $\hat{B}$. Set $x'=(x_1,\ldots,x_{n-1}) \in \mathbb{R}^{n-1} = \{x_n=0\}$. Then
\begin{align}
\int_\Gamma |u|^p \, dx' &\le \int_{\{x_n=0\}} \zeta |u|^p \, dx' \\
&= -\int_{B^+} (\zeta |u|^p)_{x_n} \, dx \\
&= -\int_B^+ |u|^p \zeta_{x_n} + p|u|^{p-1} (\text{sgn} u)u_{x_n} \zeta \, dx \\
&\le C \int_{B^+} |u|^p + |Du|^p \, dx,
\end{align}
If $x^0\in\partial U$, but $\partial U$ is not flat near $x^0$, we as usual straighten out the boundary near $x^0$ to obtrain the setting above. Applying estimate (1) and changing variables, we obtain the bound $$\int_{\Gamma}|u|^pdS\le C\int_U|u|^p+ |Du|^pdx,$$ where $\Gamma$ is some open subset of $\partial U$ containing $x^0$.
Since $\partial U$ is compact, there exists finitely many points $x_i^0\in\partial U$ and open subsets $\Gamma_i\subset \partial U (i=1,…,N)$ such that $\partial U =\bigcup_{i=1}^N \Gamma_i$ and $$\|u\|_{L^p(\Gamma_i)}\le\|u\|_{W^{1,p}(U)}\quad(i=1,\ldots,N).$$
Consequently, if we write $$Tu:=u|_{\partial U},$$ then $$\|Tu\|_{L^p(\partial U)}\le C\|u\|_{W^{1,p}(U)}$$ for some appropriate constant $C$, which does not depend on $u$.
- Inequality (2) holds for $u\in C^1(\overline{U})$. Assume now $u\in W^{1,p}(U)$. Then there exists functions $u_m\in C^{\infty}(\overline{U})$ converging to $u$ in $W^{1,p}(U)$. According to (2) we have $$\|Tu_m-Tu_l\|_{L^p(\partial U)}\le C\|u_m-u_l\|_{W^{1,p}(U)};$$ so that $\{Tu_m\}_{m=1}^{\infty}$ is a Cauchy sequence in $L^p(\partial U)$. We define $$Tu:=\lim\limits_{m\to\infty}Tu_m,$$ the limit taken in $L^p(\partial U)$. According to (3) this definition does not depend on the particular choice of smooth functions approximating $u$.
Finally if $u\in W^{1,p}\cap C(\overline{U})$, we note that the function $u_m\in C^{\infty}(\overline{U})$ constructed in the proof of Theorem 3 in paragraph 5.3.3 converges uniformly to $u$ on $\overline{U}$. Hence $Tu=u|_{\partial U}$.
Best Answer
For every $u \in W^{1,p}(U)$, the trace $Tu$ is defined as the limit in $L^p(\partial U)$ of the functions $Tu_m:=u_m|_{\partial U}$ where $u_m\in C^{\infty}(\overline{U})$ converge to $u$ in $W^{1,p}(U)$ (and the limit is shown to be independent of the choice of the approximating sequence $\{u_m\}$.). Now if $u_m$ converge uniformly to $u$ in $\bar{U}$ (which is the case for $u\in W^{1,p}(U) \cap C(\bar{U})$), then the uniform convergence also holds on $\partial U$, so (since the measure used on $ \partial U$ is finite) $u_m|_{\partial U}\to u|_{\partial U}$ in $L^p(\partial U)$ as well. Uniqueness of limits in $L^p(\partial U)$ then guarantees that $u|_{\partial U}=Tu$ as members of $L^p(\partial U)$.