Let $d \in \nat$, $U \subset \real^d$ be a bounded [[Open Set|open]] set with $\partial U \in C^1$, $1 \le p < \infty$, $k \in \nat$, and let $W^{k, p}(U)$ be the [[Sobolev Space|Sobolev space]]. > [!theorem] Sobolev Embedding Theorem (I) > > Suppose that $k/d < 1/p$. Let $1 \le l \le k$ and $q_l \in \real$ such that > $ > \frac{1}{q_l} = \frac{1}{p} - \frac{1}{d} > $ > then the inclusion map $\iota: W^{k, p}(U) \to W^{k - l, q}(U)$ is [[Bounded Linear Map|bounded]] for all $1 \le q \le q_l$, and [[Compact Operator|compact]] for all $1 \le q < q_l$. > > Moreover, for arbitrary $1 \le p \le \infty$, the inclusion $W^{1, p}(U) \to L^p(U)$ is always compact. > > *Proof*. Suppose that $k/d < 1/p$. For any $f \in W^{k, p}(U)$, by the [[Gagliardo-Nirenberg-Sobolev Inequality]] applied to all $D^\alpha f$ with $\abs{\alpha} \le k - 1$, the inclusion $W^{k, p} \to W^{l - 1, p^*}$ is bounded. Since > $ > \frac{1}{p^*} = \frac{1}{p} - \frac{1}{d} = \frac{1}{q_1} > $ > we have that $q_j^* = q_{j+1}$. Applying this inductively yields the desired embedding. For $q < q_l$, the inclusion is compact by the [[Kondrachov Compactness Theorem]]. > > Now suppose that $1 \le p \le \infty$ is arbitrary. If $p < d$, then $p^* > p$ and compactness follows from the Rellich-Kondrachov compactness theorem. If $p \ge d$, then the inclusion > $ > \begin{CD} > W^{1, p}(U) @>>> W^{1, q}(U) @>>> L^{q^*}(U) > \end{CD} > $ > is compact for all $1 \le q < d$. Since $q^* \to \infty$ as $q \upto d$, the above can be composed with inclusion into $L^p$ for $q$ sufficiently close to $d$. Therefore the inclusion $W^{1, p} \to L^p$ is compact. > [!theorem] Sobolev Embedding Theorem (II) > > Suppose that $k/d > 1/p$. Let > $ > \gamma \in \begin{cases} > \left(0, 1 + \fl{\frac{d}{p}} - \frac{d}{p} \right] &\frac{d}{p} \not\in \integer \\ > (0, 1) & \frac{d}{p} \in \integer > \end{cases} > $ > then there exists a bounded linear map > $ > I: W^{k, p}(U) \to C^{k - \fl{d/p} - 1, \gamma}(U) > $ > into the [[Hölder Space|Hölder space]] such that $D^\alpha (If) = D^\alpha f$ [[Almost Everywhere|a.e.]] for all $\alpha$ with $\abs{\alpha} \le k - \fl{d/p} - 1$ and $f \in W^{k, p}(U)$. > > *Proof*. If $d/p < 1$, then by [[Morrey's Inequality]] on $D^\alpha f$ with $\abs{\alpha} \le k - 1$ yields a bounded linear map > $ > I: W^{k, p}(U) \to C^{k - 1, 1 - d/p}(U) > $ > satisfying the desired criteria. > > Suppose that $d/p \not\in \integer$, let > $ > l = \fl{\frac{d}{p}} < \frac{d}{p} < k \quad \frac{l}{d} < \frac{1}{p} > $ > then the inclusion > $ > \iota: W^{k, p}(U) \to W^{k - l, pd/(d - pl)}(U) > $ > is bounded by the Sobolev Embedding Theorem (I). Composing with $I$ from above yields > $ > I: W^{k, p}(U) \to C^{k - \fl{d/p} - 1, 1 - d(d-pl)/(pd)} > $ > where > $ > \begin{align*} > 1 - \frac{d(d - pl)}{pd} &= 1 - \frac{d(d - p\fl{d/p})}{pd} = \frac{p - d + p\fl{d/p}}{p} \\ > &= 1 - \frac{d}{p} + \fl{\frac{d}{p}} > \end{align*} > $ > > Now suppose that $d/p \in \integer$. Let > $ > l = \frac{d}{p} - 1 < \frac{d}{p} < k \quad \frac{l}{d} < \frac{1}{p} > $ > then the inclusion > $ > \iota: W^{k, p}(U) \to W^{k - l, pd/(d - pl)}(U) = W^{k - d/p, d}(U) > $ > where > $ > \frac{pd}{d - pl} = \frac{pd}{d - d + p} = d > $ > Since $U$ has finite measure, the inclusion into $W^{k - d/p, q}(U)$ is bounded for all $1 \le q < d$. Composing with $I$ from above yields > $ > I: W^{k, p}(U) \to C^{k - d/p - 1, 1 - d/q^*}(U) > $ > Given that $q^* \to \infty$ as $q \to d$, the above inclusion is continuous for all exponents in $(0, 1)$.