[proofplan] We prove the Rellich-Kondrachov compactness theorem by extracting a convergent subsequence from any bounded sequence in $W^{1,p}(U)$. The argument proceeds in five stages: extend the sequence to $\mathbb{R}^n$ with compact support using the Extension Theorem; mollify the extended sequence and apply the Arzela-Ascoli Theorem to obtain uniform convergence at each fixed mollification scale; estimate the $L^1$ mollification error uniformly in the sequence index; extract a diagonal subsequence that is Cauchy in $L^1(U)$ via Cantor's diagonal argument; and finally interpolate between $L^1$ convergence and the uniform $L^{p^*}$ bound from the [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61) to upgrade convergence to $L^q(U)$ for any $q < p^*$. [/proofplan] [step:Extend the bounded sequence to $\mathbb{R}^n$ with compact support] Let $(u_m)_{m=1}^\infty$ be a sequence in $W^{1,p}(U)$ satisfying $\|u_m\|_{W^{1,p}(U)} \le M$ for all $m \in \mathbb{N}$. Since $\partial U$ is $C^1$, the Extension Theorem provides a bounded linear operator $E: W^{1,p}(U) \to W^{1,p}(\mathbb{R}^n)$ with $\|Eu_m\|_{W^{1,p}(\mathbb{R}^n)} \le C_E M$. Let $V \supset \overline{U}$ be a bounded open set with $\operatorname{dist}(\partial U, \partial V) > 2$. Choose a cutoff function \begin{align*} \zeta : \mathbb{R}^n &\to [0,1] \\ x &\mapsto \zeta(x) \end{align*} with $\zeta \in C_c^\infty(\mathbb{R}^n)$, $\zeta \equiv 1$ on $U$, and $\operatorname{supp}(\zeta) \subset V$. Define $\overline{u}_m = \zeta \cdot Eu_m$. Then $\overline{u}_m \in W^{1,p}(\mathbb{R}^n)$ with $\operatorname{supp}(\overline{u}_m) \subset V$, $\overline{u}_m|_U = u_m$, and $\|\overline{u}_m\|_{W^{1,p}(\mathbb{R}^n)} \le M'$ for a constant $M' = M'(C_E, \|\zeta\|_{C^1}, M)$ independent of $m$. [guided] The extension is necessary because the GNS inequality and mollification arguments require functions defined on all of $\mathbb{R}^n$. The cutoff $\zeta$ ensures compact support inside a fixed bounded set $V$, which is needed for the Arzela-Ascoli argument in the next step. Since $\partial U$ is $C^1$, the Extension Theorem yields a bounded linear operator $E: W^{1,p}(U) \to W^{1,p}(\mathbb{R}^n)$ with $Eu_m|_U = u_m$ and $\|Eu_m\|_{W^{1,p}(\mathbb{R}^n)} \le C_E \|u_m\|_{W^{1,p}(U)} \le C_E M$. However, $Eu_m$ may have support spreading across all of $\mathbb{R}^n$. To localise, we multiply by a smooth cutoff $\zeta \in C_c^\infty(\mathbb{R}^n)$ with $\zeta \equiv 1$ on $U$ and $\operatorname{supp}(\zeta) \subset V$. The product rule for [weak derivatives](/page/Weak%20Derivative) gives \begin{align*} \nabla(\zeta \cdot Eu_m) = \nabla\zeta \cdot Eu_m + \zeta \cdot \nabla(Eu_m), \end{align*} so $\|\overline{u}_m\|_{W^{1,p}} \le (\|\nabla\zeta\|_{L^\infty} + \|\zeta\|_{L^\infty}) \|Eu_m\|_{W^{1,p}} \le M'$ where $M' = (1 + \|\nabla\zeta\|_{L^\infty}) C_E M$. The key property is that $\overline{u}_m|_U = u_m$ (since $\zeta = 1$ on $U$), so any convergence of $\overline{u}_m$ on $U$ is convergence of $u_m$ itself. [/guided] [/step] [step:Mollify and apply the Arzela-Ascoli Theorem to extract uniformly convergent subsequences] Fix $\varepsilon > 0$. Let $\eta \in C_c^\infty(B(0,1))$ be the [standard mollifier](/page/Standard%20Mollifier) with $\int_{\mathbb{R}^n} \eta \, d\mathcal{L}^n = 1$, and define $\eta_\varepsilon(x) = \varepsilon^{-n} \eta(x/\varepsilon)$. The mollified functions are \begin{align*} v_{\varepsilon,m} : \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto (\eta_\varepsilon * \overline{u}_m)(x) = \int_{\mathbb{R}^n} \eta_\varepsilon(x - y) \, \overline{u}_m(y) \, d\mathcal{L}^n(y). \end{align*} Each $v_{\varepsilon,m}$ is smooth with $\operatorname{supp}(v_{\varepsilon,m}) \subset V_\varepsilon := \{x : \operatorname{dist}(x, V) < \varepsilon\}$. We verify uniform boundedness: since $V$ has finite measure, Holder's inequality gives $\|\overline{u}_m\|_{L^1(V)} \le \mathcal{L}^n(V)^{1 - 1/p} \|\overline{u}_m\|_{L^p(V)} \le \mathcal{L}^n(V)^{1-1/p} M'$. Then \begin{align*} \|v_{\varepsilon,m}\|_{L^\infty} \le \|\eta_\varepsilon\|_{L^\infty} \|\overline{u}_m\|_{L^1} \le \varepsilon^{-n} \|\eta\|_{L^\infty} \mathcal{L}^n(V)^{1-1/p} M'. \end{align*} We verify [equicontinuity](/page/Equicontinuity): $\nabla v_{\varepsilon,m} = (\nabla\eta_\varepsilon) * \overline{u}_m$, so \begin{align*} \|\nabla v_{\varepsilon,m}\|_{L^\infty} \le \|\nabla\eta_\varepsilon\|_{L^\infty} \|\overline{u}_m\|_{L^1} \le \varepsilon^{-n-1} \|\nabla\eta\|_{L^\infty} \mathcal{L}^n(V)^{1-1/p} M'. \end{align*} This uniform Lipschitz bound (depending on $\varepsilon$ but not on $m$) together with uniform pointwise boundedness satisfies the hypotheses of the Arzela-Ascoli Theorem on the [compact](/page/Compact%20Space) set $\overline{V}$. Therefore, for each fixed $\varepsilon > 0$, the sequence $(v_{\varepsilon,m})_{m=1}^\infty$ admits a subsequence converging uniformly on $\overline{V}$. Since $\overline{V}$ has finite measure, uniform convergence implies $L^1(V)$ convergence. [guided] Mollification replaces the $W^{1,p}$ functions $\overline{u}_m$ with smooth approximations $v_{\varepsilon,m} = \eta_\varepsilon * \overline{u}_m$ at the cost of an approximation error controlled by $\varepsilon$. At any fixed mollification scale $\varepsilon > 0$, the smoothed sequence lives in a uniformly bounded, [equicontinuous](/page/Equicontinuity) family, so the Arzela-Ascoli Theorem extracts a uniformly convergent subsequence. The two estimates that verify the Arzela-Ascoli hypotheses are as follows. **Uniform pointwise boundedness.** For each $x \in \mathbb{R}^n$, Young's convolution inequality gives \begin{align*} |v_{\varepsilon,m}(x)| = |(\eta_\varepsilon * \overline{u}_m)(x)| \le \|\eta_\varepsilon\|_{L^\infty(\mathbb{R}^n)} \|\overline{u}_m\|_{L^1(\mathbb{R}^n)}. \end{align*} The mollifier satisfies $\|\eta_\varepsilon\|_{L^\infty} = \varepsilon^{-n} \|\eta\|_{L^\infty}$. For the $L^1$ norm, since $\overline{u}_m$ is supported in the bounded set $V$ with $\mathcal{L}^n(V) < \infty$, Holder's inequality with exponents $p$ and $p/(p-1)$ yields \begin{align*} \|\overline{u}_m\|_{L^1(\mathbb{R}^n)} = \|\overline{u}_m\|_{L^1(V)} \le \mathcal{L}^n(V)^{1 - 1/p} \|\overline{u}_m\|_{L^p(V)} \le \mathcal{L}^n(V)^{1-1/p} M'. \end{align*} Combining: $\|v_{\varepsilon,m}\|_{L^\infty} \le \varepsilon^{-n} \|\eta\|_{L^\infty} \mathcal{L}^n(V)^{1-1/p} M'$. This bound depends on $\varepsilon$ but is uniform in $m$. **Equicontinuity.** Differentiating under the convolution integral, $\nabla v_{\varepsilon,m}(x) = \int_{\mathbb{R}^n} \nabla_x \eta_\varepsilon(x - y) \, \overline{u}_m(y) \, d\mathcal{L}^n(y) = (\nabla\eta_\varepsilon) * \overline{u}_m(x)$, so \begin{align*} \|\nabla v_{\varepsilon,m}\|_{L^\infty(\mathbb{R}^n)} \le \|\nabla\eta_\varepsilon\|_{L^\infty} \|\overline{u}_m\|_{L^1} \le \varepsilon^{-(n+1)} \|\nabla\eta\|_{L^\infty} \mathcal{L}^n(V)^{1-1/p} M'. \end{align*} Setting $L_\varepsilon = \varepsilon^{-(n+1)} \|\nabla\eta\|_{L^\infty} \mathcal{L}^n(V)^{1-1/p} M'$, the mean value inequality gives $|v_{\varepsilon,m}(x) - v_{\varepsilon,m}(y)| \le L_\varepsilon |x - y|$ for all $x, y \in \mathbb{R}^n$ and all $m$, establishing equicontinuity with Lipschitz modulus $L_\varepsilon$ uniform in $m$. The Arzela-Ascoli Theorem applies on the [compact](/page/Compact%20Space) set $\overline{V}$ (compact because $V$ is bounded and $\overline{V}$ is a closed bounded subset of $\mathbb{R}^n$): for each fixed $\varepsilon > 0$, the sequence $(v_{\varepsilon,m})_{m=1}^\infty$ admits a subsequence converging uniformly on $\overline{V}$. Uniform convergence on a finite-measure set implies $L^1$ convergence: \begin{align*} \|v_{\varepsilon,m_i} - v_{\varepsilon,m_j}\|_{L^1(V)} \le \mathcal{L}^n(V) \|v_{\varepsilon,m_i} - v_{\varepsilon,m_j}\|_{L^\infty(\overline{V})} \to 0. \end{align*} [/guided] [/step] [step:Estimate the $L^1$ mollification error uniformly in the sequence index] [claim:Mollification error] There exists $C > 0$ independent of $m$ and $\varepsilon$ such that \begin{align*} \|v_{\varepsilon,m} - \overline{u}_m\|_{L^1(V)} \le C\varepsilon M' \end{align*} for all $m \in \mathbb{N}$ and $\varepsilon > 0$. [/claim] [proof] Using the substitution $y \mapsto z = y/\varepsilon$ in the convolution and the normalisation $\int \eta \, d\mathcal{L}^n = 1$: \begin{align*} v_{\varepsilon,m}(x) - \overline{u}_m(x) = \int_{B(0,1)} \eta(z) \left( \overline{u}_m(x - \varepsilon z) - \overline{u}_m(x) \right) d\mathcal{L}^n(z). \end{align*} By the Fundamental Theorem of Calculus along the segment from $x$ to $x - \varepsilon z$, parameterised by $t \in [0, 1]$: \begin{align*} \overline{u}_m(x - \varepsilon z) - \overline{u}_m(x) = -\varepsilon \int_0^1 \nabla\overline{u}_m(x - t\varepsilon z) \cdot z \, d\mathcal{L}^1(t). \end{align*} Taking absolute values, integrating over $V$ with respect to $\mathcal{L}^n(x)$, and applying Fubini's Theorem to exchange the order of integration: \begin{align*} \|v_{\varepsilon,m} - \overline{u}_m\|_{L^1(V)} &\le \varepsilon \int_{B(0,1)} \eta(z) |z| \int_0^1 \int_V |\nabla\overline{u}_m(x - t\varepsilon z)| \, d\mathcal{L}^n(x) \, d\mathcal{L}^1(t) \, d\mathcal{L}^n(z). \end{align*} By translation invariance of the Lebesgue measure, $\int_V |\nabla\overline{u}_m(x - t\varepsilon z)| \, d\mathcal{L}^n(x) \le \|\nabla\overline{u}_m\|_{L^1(\mathbb{R}^n)}$. Since $V$ has finite measure and $\overline{u}_m$ is supported in $V$, Holder's inequality gives $\|\nabla\overline{u}_m\|_{L^1(\mathbb{R}^n)} \le \mathcal{L}^n(V)^{1-1/p} \|\nabla\overline{u}_m\|_{L^p(\mathbb{R}^n)} \le \mathcal{L}^n(V)^{1-1/p} M'$. Since $\int_{B(0,1)} \eta(z) |z| \, d\mathcal{L}^n(z) \le 1$ and $\int_0^1 d\mathcal{L}^1(t) = 1$: \begin{align*} \|v_{\varepsilon,m} - \overline{u}_m\|_{L^1(V)} \le \varepsilon \, \mathcal{L}^n(V)^{1-1/p} M'. \end{align*} Setting $C = \mathcal{L}^n(V)^{1-1/p}$ completes the claim. [/proof] [/step] [step:Extract a diagonal subsequence that is Cauchy in $L^1(U)$] Define a decreasing sequence of scales $\varepsilon_k = 1/k$ for $k \in \mathbb{N}$. At scale $\varepsilon_1 = 1$, apply the Arzela-Ascoli extraction from the second step to obtain a subsequence $(\overline{u}_{\sigma_1(m)})_{m=1}^\infty$ such that $(v_{1, \sigma_1(m)})_{m=1}^\infty$ converges in $L^1(V)$. At scale $\varepsilon_2 = 1/2$, extract a further subsequence $\sigma_2$ from $\sigma_1$ such that $(v_{1/2, \sigma_2(m)})$ converges in $L^1(V)$. Continuing inductively, at stage $k$, extract $\sigma_k$ from $\sigma_{k-1}$ such that $(v_{1/k, \sigma_k(m)})$ converges in $L^1(V)$. Define the diagonal subsequence $w_m = \overline{u}_{\sigma_m(m)}$. We verify that $(w_m)$ is Cauchy in $L^1(V)$. Let $\delta > 0$. Choose $k$ large enough that $C M' / k < \delta / 3$. By the mollification error estimate: \begin{align*} \|v_{1/k, \sigma_m(m)} - w_m\|_{L^1(V)} \le \frac{CM'}{k} < \frac{\delta}{3} \quad \text{for all } m. \end{align*} Since $(v_{1/k, \sigma_m(m)})_{m \ge k}$ is a subsequence of the $k$-th extracted sequence (by the nesting property of the diagonal), it converges in $L^1(V)$ and hence is Cauchy. Choose $N$ such that for all $i, j \ge N$: \begin{align*} \|v_{1/k, \sigma_i(i)} - v_{1/k, \sigma_j(j)}\|_{L^1(V)} < \frac{\delta}{3}. \end{align*} By the triangle inequality: \begin{align*} \|w_i - w_j\|_{L^1(V)} \le \|w_i - v_{1/k, \sigma_i(i)}\|_{L^1(V)} + \|v_{1/k, \sigma_i(i)} - v_{1/k, \sigma_j(j)}\|_{L^1(V)} + \|v_{1/k, \sigma_j(j)} - w_j\|_{L^1(V)} < \delta. \end{align*} Since $U \subset V$ and $w_m|_U = u_{\sigma_m(m)}$, the restricted sequence is Cauchy in $L^1(U)$. [guided] The diagonal argument combines two ingredients: at each fixed scale $\varepsilon_k$, the mollified sequence has a convergent subsequence (from Arzela-Ascoli), but different scales produce different subsequences. Cantor's diagonal trick selects a single subsequence that works for all scales simultaneously. At stage $k$, we have a subsequence $\sigma_k$ such that $(v_{1/k, \sigma_k(m)})_m$ converges in $L^1(V)$. The nesting $\sigma_k \subset \sigma_{k-1} \subset \cdots$ ensures that the tail of the diagonal sequence $(w_m)_{m \ge k} = (\overline{u}_{\sigma_m(m)})_{m \ge k}$ is a subsequence of the $k$-th extraction. Therefore $(v_{1/k, \sigma_m(m)})_{m \ge k}$ converges in $L^1(V)$. To show $(w_m)$ is Cauchy in $L^1(V)$, we use the triangle inequality with the mollified sequence as an intermediary. The three terms are: 1. $\|w_i - v_{1/k, \sigma_i(i)}\|_{L^1} \le CM'/k$ (mollification error, from the claim above), 2. $\|v_{1/k, \sigma_i(i)} - v_{1/k, \sigma_j(j)}\|_{L^1} < \delta/3$ (Cauchy property of the $k$-th mollified sequence for large $i, j$), 3. $\|v_{1/k, \sigma_j(j)} - w_j\|_{L^1} \le CM'/k$ (mollification error again). Choosing $k$ so that $CM'/k < \delta/3$ makes the total less than $\delta$. Since $w_m|_U = u_{\sigma_m(m)}$, the original sequence has a subsequence Cauchy in $L^1(U)$. [/guided] [/step] [step:Interpolate between $L^1$ and $L^{p^*}$ to upgrade convergence to $L^q(U)$] Since $1 \le q < p^*$, there exists a unique $\theta \in (0, 1]$ satisfying \begin{align*} \frac{1}{q} = \frac{\theta}{1} + \frac{1 - \theta}{p^*}. \end{align*} The $L^q$ interpolation inequality gives, for any $i, j \in \mathbb{N}$: \begin{align*} \|w_i - w_j\|_{L^q(U)} \le \|w_i - w_j\|_{L^1(U)}^\theta \, \|w_i - w_j\|_{L^{p^*}(U)}^{1 - \theta}. \end{align*} By the [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61) (extended to bounded domains via the Extension Theorem), the continuous embedding $W^{1,p}(U) \hookrightarrow L^{p^*}(U)$ yields $\|u_m\|_{L^{p^*}(U)} \le C_* M$ for all $m$, where $C_* = C_*(n, p, U)$. Therefore \begin{align*} \|w_i - w_j\|_{L^{p^*}(U)} \le \|w_i\|_{L^{p^*}(U)} + \|w_j\|_{L^{p^*}(U)} \le 2C_* M. \end{align*} Substituting: \begin{align*} \|w_i - w_j\|_{L^q(U)} \le (2C_* M)^{1-\theta} \, \|w_i - w_j\|_{L^1(U)}^\theta. \end{align*} Since $(w_m)$ is Cauchy in $L^1(U)$ by the previous step and $\theta > 0$, the right-hand side tends to $0$ as $i, j \to \infty$. Hence $(w_m)$ is Cauchy in $L^q(U)$. Since $L^q(U)$ is a [complete](/page/Complete%20Metric%20Space) [Banach space](/page/Banach%20Space), the subsequence converges in $L^q(U)$. This proves that every bounded sequence in $W^{1,p}(U)$ has a subsequence converging in $L^q(U)$, establishing the compact embedding $W^{1,p}(U) \hookrightarrow\hookrightarrow L^q(U)$. [guided] The interpolation step is where the argument closes. We have two pieces of information about the subsequence $(w_m)$: 1. It is Cauchy in $L^1(U)$ (from the diagonal argument), so $\|w_i - w_j\|_{L^1(U)} \to 0$ as $i, j \to \infty$. 2. It is uniformly bounded in $L^{p^*}(U)$ (from the GNS inequality), with $\|w_m\|_{L^{p^*}(U)} \le C_* M$ for all $m$. The first property gives convergence in a weak topology but not yet in $L^q$ for $q > 1$. The second gives no convergence at all, only boundedness. The $L^q$ interpolation inequality bridges the gap by combining both pieces of information. For $1 \le q < p^*$, write $q$ as a weighted harmonic mean of $1$ and $p^*$: there exists a unique $\theta \in (0, 1]$ satisfying \begin{align*} \frac{1}{q} = \frac{\theta}{1} + \frac{1 - \theta}{p^*}. \end{align*} Solving for $\theta$ explicitly: \begin{align*} \theta = \frac{1/q - 1/p^*}{1 - 1/p^*} = \frac{p^* - q}{q(p^* - 1)}. \end{align*} The condition $q < p^*$ ensures $1/q > 1/p^*$, so $\theta > 0$. The condition $q \ge 1$ ensures $1/q \le 1$, so $\theta \le 1$. With this exponent, Holder's inequality applied to $|w_i - w_j|^q = |w_i - w_j|^{q\theta} \cdot |w_i - w_j|^{q(1-\theta)}$ with conjugate exponents $1/\theta$ and $1/(1-\theta)$ yields the $L^q$ interpolation inequality: \begin{align*} \|w_i - w_j\|_{L^q(U)} \le \|w_i - w_j\|_{L^1(U)}^\theta \, \|w_i - w_j\|_{L^{p^*}(U)}^{1-\theta}. \end{align*} Now we bound the $L^{p^*}$ factor. By the [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61), applied to the extended functions via the Extension Theorem, $\|w_m\|_{L^{p^*}(U)} \le C_* M$ for all $m$. The triangle inequality gives \begin{align*} \|w_i - w_j\|_{L^{p^*}(U)} \le \|w_i\|_{L^{p^*}(U)} + \|w_j\|_{L^{p^*}(U)} \le 2C_* M. \end{align*} Substituting into the interpolation inequality: \begin{align*} \|w_i - w_j\|_{L^q(U)} \le (2C_* M)^{1-\theta} \, \|w_i - w_j\|_{L^1(U)}^\theta. \end{align*} The prefactor $(2C_* M)^{1-\theta}$ is a fixed constant. Since $(w_m)$ is Cauchy in $L^1(U)$ (from the diagonal argument), $\|w_i - w_j\|_{L^1(U)} \to 0$ as $i, j \to \infty$. Since $\theta > 0$, the factor $\|w_i - w_j\|_{L^1(U)}^\theta \to 0$ as well, proving that $(w_m)$ is Cauchy in $L^q(U)$. Completeness of $L^q(U)$ then gives convergence $w_m \to u$ in $L^q(U)$ for every $1 \le q < p^*$. [/guided] [/step]