> [!theoremb] Theorem > > Let $d \in \nat$, $1 \le p < \infty$, $p^*$ be its [[Sobolev Conjugate|Sobolev conjugate]] and $\cf \subset W^{1, p}(\real^d)$ be a family of functions in the [[Sobolev Space|Sobolev space]] such that > 1. $M = \sup_{f \in \cf}\norm{f}_{W^{1,p}(\real^d)}$. > 2. There exists $K \subset \real^d$ [[Compactness|compact]] such that $\supp{f} \subset K$ for all $f \in \cf$. > > then $\cf$ is [[Totally Bounded|totally bounded]] in $L^q$ for every $q < p^*$. Moreover, for any bounded open set $U \subset \real^d$ with $\partial U \in C^1$, the inclusion > $ > W^{1, p}(U) \to L^q(U) > $ > is compact. # Proof Let $\cf \subset W^{1, p}(\real^d)$ be a family of functions satisfying the requirements of the theorem. There are no sufficiently nice subspaces of $W^{1, p}(\real^d)$ where the compactness result can be applied directly, so we will take families that approximate $\cf$. Let $\phi \in C_c^\infty(\real^d)$ be a [[Mollifier|mollifier]]. For each $t > 0$, define $ \cf_t = \bracs{f * \phi_t: f \in \cf} $ ### The Compactness Lemma These families must satisfy two conditions for the compactness to hold. > [!theorem] > > Let $(X, d)$ be a [[Metric Space|metric space]] and $E \subset X$. Suppose that for every $R > 0$, there exists $E_R \subset X$ such that: > 1. $E_R$ is totally bounded. > 2. For any $x \in E$, there exists $y \in E_R$ such that $d(x, y) < R/2$. > > then $E$ is also totally bounded. > > *Proof*. Let $R > 0$, then there exists $\seqf{x_j} \subset X$ such that $E_R \subset \bigcup_{i = 1}^n B(x_j, R/2)$. Let $x \in E$, then there exists $y \in E_R$ such that $d(x, y) < R/2$, and $1 \le j \le n$ such that $d(x_j, y) < R/2$. Therefore $E \subset \bigcup_{i = 1}^n B(x_j, R)$. As such a finite covering holds for all $R > 0$, $E$ is totally bounded. ### Totally Bounded > [!theorem] > > For each $t > 0$, $\cf_t$ is totally bounded. > > *Proof*. Since $\cf$ has a common [[Compactly Supported|compact support]], so does $\cf_t$. Using the [[Arzelà-Ascoli Theorem]], it's sufficient to show that $\cf_t$ is pointwise bounded and equicontinuous. Firstly, > $ > \norm{f * \phi_t}_u \le \norm{\phi_t}_q\norm{f}_p \le \norm{\phi_t}_q \cdot M > $ > for all $f \in \cf$, so $\cf_t$ is uniformly bounded. On the other hand, for any $x, y \in \real^d$, > $ > \begin{align*} > \abs{f * \phi_t(x) - f * \phi_t(y)} &\le \int_{\real^d}\abs{\phi_t(x - z) - \phi_t(y - z)} \cdot f(z)dz \\ > &\le \norm{\tau_{x - y}\phi_t - \phi_t}_q \cdot \norm{f}_p \\ > &\le \sup_{\abs{z} \le \abs{x - y}}\norm{\tau_z \phi_t - \phi_t}_q \cdot M > \end{align*} > $ > where $\sup_{\abs{z} \le \abs{x - y}}\norm{\tau_z \phi_t - \phi_t}_q \to 0$ as $\abs{x - y} \to 0$, so $\cf_t$ is uniformly equicontinuous. ### Distance > [!theorem] > > Let $f \in W^{1, p}(\real^d)$ be of compact support, then > $ > \norm{f * \phi_t - f}_1 \le t\norm{\phi}_1\norm{Df}_{L^1} > $ > *Proof*. First suppose that $f \in C_c^\infty(\real^d)$ and let $x \in \real^d$, then > $ > \begin{align*} > \abs{f * \phi_t(x) - f(x)} &\le \int_{B(0, t)}\abs{\phi_t(y)}\abs{f(x - y) - f(x)}dy \\ > &\le t\int_{B(0, t)}\abs{\phi_t(z)}\int_{0}^1\abs{Df(x -sy)}dsdy \\ > \int_{\real^d}\abs{f * \phi_t(x) - f(x)}dx &\le t\int_{B(0, t)}\abs{\phi_t(z)}\int_{0}^1\abs{Df(x -sy)}dxdsdy \\ > &\le t\norm{Df}_1 \norm{\phi_t}_1 > \end{align*} > $ > > If $f$ is arbitrary, taking an approximating sequence yields the desired result. > [!theorem] > > Let $f \in W^{1, p}(\real^d)$ be of compact support, then > $ > \norm{f * \phi_t - f}_{q} \le C_{p, d}\norm{\phi}_1\norm{f}_{W^{1, p}} m(\supp{f})^{\lambda p/(p - 1)} \cdot t^{\lambda} > $ > where > $ > \lambda = \frac{1/q - 1/p^*}{1 - 1/p^*} > $ > *Proof*. By $L^p$ [[Intermediate Interpolation of Lp Spaces|interpolation]], > $ > \norm{f * \phi_t - f}_q \le \norm{f * \phi_t - f}_1^\lambda \cdot \norm{f * \phi_t - f}_{p^*}^{1 - \lambda} > $ > By the [[Gagliardo-Nirenberg-Sobolev Inequality]] and [[Young's Inequality]], > $ > \norm{f * \phi_t - f}_{p^*} \le C_{p, d}\norm{f * \phi_t - f}_{W^{1, p}} \le C_{p, d}\norm{\phi}_1 \norm{f}_{W^{1, p}} > $ > By the inclusion of $L^p$ spaces on finite measure spaces, > $ > \begin{align*} > \norm{f * \phi_t - f}_q &\le (t\norm{\phi}_1\norm{Df}_{L^1})^{\lambda}(C_{p, d}\norm{\phi}_1 \norm{f}_{W^{1, p}})^{1 - \lambda} \\ > &= C_{p, d} \norm{\phi}_1 \cdot \norm{Df}_{L^1}^\lambda \norm{f}_{W^{1, p}}^{1 - \lambda} \cdot t^\lambda \\ > &\le C_{p, d}\norm{\phi}_1 \cdot m(\supp{f})^{\lambda p/(p-1)} \norm{f}_{W^{1, p}} \cdot t^\lambda > \end{align*} > $ ### Conclusion > [!theorem] > > $\cf$ is totally bounded in $L^q$. > > *Proof*. It's sufficient to verify condition $(2)$ of the compactness lemma. Since $\cf$ is bounded, we have that > $ > \norm{f * \phi_t - f}_q \le C_{p, d}M\norm{\phi}_1 m(K)^{\lambda p/(p - 1)} \cdot t^\lambda > $ > Condition $(2)$ for any $R > 0$ can be satisfied by $\cf_t$ for sufficiently small $t > 0$.