[proofplan]
We reduce the problem of extracting convergent subsequences in $C(\bar{U})$ to the Arzela-Ascoli theorem by showing that any bounded sequence in $W^{1,p}(U)$ is uniformly bounded and equicontinuous on $\bar{U}$. The key tool is [Morrey's Inequality](/theorems/66), which converts an $L^p$ bound on derivatives (with $p > n$) into a Holder norm bound with exponent $\gamma = 1 - n/p$. We first extend functions to $\mathbb{R}^n$ using the [Extension Theorem](/theorems/77), apply Morrey's inequality to the extensions, and then invoke Arzela-Ascoli on $\bar{U}$.
[/proofplan]
[step:Extend the bounded sequence from $W^{1,p}(U)$ to $W^{1,p}(\mathbb{R}^n)$]
Let $\{u_m\}_{m=1}^{\infty} \subset W^{1,p}(U)$ be a bounded sequence with $\|u_m\|_{W^{1,p}(U)} \leq M$ for all $m$. Since $\partial U$ is $C^1$, the [Extension Theorem](/theorems/77) provides a bounded linear operator
\begin{align*}
E: W^{1,p}(U) \to W^{1,p}(\mathbb{R}^n)
\end{align*}
satisfying $Eu_m = u_m$ on $U$ and $\operatorname{supp}(Eu_m) \subseteq V$ for some fixed bounded open set $V \supset \bar{U}$. The boundedness of $E$ gives
\begin{align*}
\|Eu_m\|_{W^{1,p}(\mathbb{R}^n)} \leq C_E \|u_m\|_{W^{1,p}(U)} \leq C_E M,
\end{align*}
where $C_E = \|E\|_{\mathcal{L}(W^{1,p}(U), W^{1,p}(\mathbb{R}^n))}$ is the operator norm of the extension.
[guided]
Why is the extension step necessary? [Morrey's Inequality](/theorems/66) is stated for functions in $W^{1,p}(\mathbb{R}^n)$, not for functions on a domain $U$. Without extension, we cannot directly apply the global estimate. The [Extension Theorem](/theorems/77) provides a bounded linear operator
\begin{align*}
E: W^{1,p}(U) \to W^{1,p}(\mathbb{R}^n)
\end{align*}
satisfying $Eu_m = u_m$ on $U$ and $\operatorname{supp}(Eu_m) \subseteq V$ for a fixed bounded open set $V \supset \bar{U}$. The $C^1$ boundary hypothesis on $U$ is consumed here: it ensures the existence of such an extension operator. Domains with irregular boundaries (e.g., cusps) may fail to admit a bounded extension operator, and the compact embedding can fail in those cases.
The boundedness of $E$ gives $\|Eu_m\|_{W^{1,p}(\mathbb{R}^n)} \leq C_E\|u_m\|_{W^{1,p}(U)} \leq C_E M$, so the extended sequence is bounded in $W^{1,p}(\mathbb{R}^n)$. The support condition $\operatorname{supp}(Eu_m) \subseteq V$ is a technical convenience: the extended functions are compactly supported, so there is no issue with integrability at infinity.
[/guided]
[/step]
[step:Apply Morrey's inequality to obtain uniform Holder bounds]
[claim:Uniform boundedness and equicontinuity]
The sequence $\{u_m\}$ is uniformly bounded and equicontinuous on $\bar{U}$.
[/claim]
[proof]
Since $p > n$, [Morrey's Inequality](/theorems/66) applies to each $Eu_m \in W^{1,p}(\mathbb{R}^n)$. The inequality states that there exists a constant $C = C(n, p) > 0$ such that
\begin{align*}
\|Eu_m\|_{C^{0,\gamma}(\mathbb{R}^n)} \leq C \|Eu_m\|_{W^{1,p}(\mathbb{R}^n)} \leq C \cdot C_E M =: M',
\end{align*}
where $\gamma = 1 - n/p > 0$ (the positivity of $\gamma$ requires $p > n$). Restricting to $\bar{U}$, on which $Eu_m = u_m$:
\begin{align*}
\|u_m\|_{C^{0,\gamma}(\bar{U})} \leq M' \quad \text{for all } m.
\end{align*}
Unpacking the Holder norm, this encodes two properties:
**Uniform boundedness:** $\|u_m\|_{L^{\infty}(\bar{U})} \leq M'$ for all $m$.
**Equicontinuity:** For all $x, y \in \bar{U}$ and all $m$:
\begin{align*}
|u_m(x) - u_m(y)| \leq M' |x - y|^{\gamma}.
\end{align*}
Given $\varepsilon > 0$, choosing $\delta = (\varepsilon / M')^{1/\gamma}$ ensures $|u_m(x) - u_m(y)| < \varepsilon$ whenever $|x - y| < \delta$, uniformly in $m$.
[/proof]
[guided]
[Morrey's Inequality](/theorems/66) is the engine of the entire proof. It converts an integral bound ($\|u\|_{W^{1,p}} \leq M'$) into a pointwise regularity estimate. Specifically, for each $Eu_m \in W^{1,p}(\mathbb{R}^n)$ with $p > n$:
\begin{align*}
\|Eu_m\|_{C^{0,\gamma}(\mathbb{R}^n)} \leq C\|Eu_m\|_{W^{1,p}(\mathbb{R}^n)} \leq C \cdot C_E M =: M',
\end{align*}
where $\gamma = 1 - n/p > 0$. Restricting to $\bar{U}$ (where $Eu_m = u_m$), the Holder norm bound encodes:
**Uniform boundedness:** $\|u_m\|_{L^\infty(\bar{U})} \leq M'$ for all $m$.
**Equicontinuity:** For all $x, y \in \bar{U}$ and all $m$:
\begin{align*}
|u_m(x) - u_m(y)| \leq M'|x - y|^{\gamma}.
\end{align*}
Given $\varepsilon > 0$, choosing $\delta = (\varepsilon/M')^{1/\gamma}$ ensures $|u_m(x) - u_m(y)| < \varepsilon$ whenever $|x - y| < \delta$, uniformly in $m$.
The Holder exponent $\gamma = 1 - n/p$ is positive precisely because $p > n$. As $p \to n^+$, the exponent $\gamma \to 0^+$ and the Holder control degenerates. When $p = n$, Morrey's inequality fails entirely (one gets BMO estimates instead), and for $p < n$ the relevant inequality is Gagliardo-Nirenberg-Sobolev, which gives $L^{p^*}$ integrability rather than pointwise control. The fact that $M'$ is independent of $m$ is what gives *uniform* boundedness and *equi*continuity.
[/guided]
[/step]
[step:Extract a uniformly convergent subsequence via the Arzela-Ascoli theorem]
[claim:Convergent subsequence in $C(\bar{U})$]
There exists a subsequence $\{u_{m_j}\}$ that converges uniformly on $\bar{U}$.
[/claim]
[proof]
The set $\bar{U}$ is compact (since $U$ is bounded, $\bar{U}$ is closed and bounded in $\mathbb{R}^n$, hence compact by the Heine-Borel theorem). By the previous step, the family $\{u_m\}_{m=1}^{\infty}$ is uniformly bounded and equicontinuous in $C(\bar{U})$. The Arzela-Ascoli theorem guarantees the existence of a subsequence $\{u_{m_j}\}_{j=1}^{\infty}$ and a function $u \in C(\bar{U})$ such that
\begin{align*}
\|u_{m_j} - u\|_{C(\bar{U})} = \sup_{x \in \bar{U}} |u_{m_j}(x) - u(x)| \to 0 \quad \text{as } j \to \infty.
\end{align*}
[/proof]
[/step]
[step:Conclude compactness of the embedding]
Since every bounded sequence in $W^{1,p}(U)$ has a subsequence converging in $C(\bar{U})$, the embedding $W^{1,p}(U) \hookrightarrow C(\bar{U})$ is compact:
\begin{align*}
W^{1,p}(U) \subset\subset C(\bar{U}).
\end{align*}
[/step]
