> [!theorem] > > Let $d \in \nat$, $d < p \le \infty$, $U \subset \real^d$ be an [[Open Set|open set]] such that either $U$ is bounded [[Open Set|open set]] with $\partial U$ being [[Differentiable Boundary|of class]] $C^1$ or $U = \real^d$, and $W^{1, p}(U)$ be the [[Sobolev Space|Sobolev space]] on $U$. Let $\gamma = 1 - d/p$, then > 1. On the [[Space of Continuously Differentiable Functions|continuously differentiable]] functions, the identity $I: C^1(U) \cap W^{1, p}(U) \to C^{0, \gamma}(U)$ is a [[Bounded Linear Map|bounded linear map]] into the [[Hölder Space]]. > 2. $I$ [[Linear Extension Theorem|extends uniquely]] into a bounded linear operator $W^{1, p}(U) \to C^{0, \gamma}(U)$, such that $If = f$ [[Almost Everywhere|a.e.]] for all $f \in W^{1, p}(U)$. In other words, for every $f \in W^{1, p}(U)$, there exists $f^* \in C^{0, \gamma}(U)$ such that $f = f^*$ a.e. > 3. For every $f \in W^{1, p}_{\text{loc}}(U)$, $f$ is [[Derivative|differentiable]] a.e., with its derivative equal to the weak derivative a.e. > > *Proof*. First suppose that $U = \real^d$. By the inequalities in the section below, $I$ is a bounded linear map into the Hölder space on $\real^d$. > > By the [[Linear Extension Theorem]], $I$ admits a unique extension to $W^{1, p}(U)$. Let $f \in W^{1, p}(U)$ and $\seq{f_n} \in C^1(U) \cap W^{1, p}(U)$ such that $f_n \to f$ in $W^{1, p}$ and almost everywhere. On the other hand, since $f_n = If_n \to If$ in $C^{0, \gamma}$, $f_n \to If$ [[Uniform Convergence|uniformly]] and pointwise, so $If = f$ almost everywhere. > > Now suppose that $U$ is bounded with $C^1$ boundary. By the [[Extension Theorem for Sobolev Spaces|Extension Theorem]], we can express the embedding > $ > \begin{CD} > W^{1, p}(U) @>\text{extend}>> W^{1, p}(\real^d) @>{I}>> C^{0, \gamma}(\real^d) @>{\text{restrict}}>> C^{0, \gamma}(U) > \end{CD} > $ > as the above composition, which preserves all of the discussed properties. > > Let $f \in W^{1, p}_{\text{loc}}(U) \cap C^1(U)$, then there exists $C \ge 0$ such that for any $x, y \in U$, > $ > \abs{f(x) - f(y)} \le C\abs{x - y}^{1 - d/p} \cdot \norm{Df}_{L^p(B(x, 2\abs{x - y}))} > $ > which extends to the continuous version of $f$ for all $f \in W^{1, p}_{\text{loc}}(U)$ by continuity. In particular, using the above on $g = f - Df(x_0)$ yields that > $ > \begin{align*} > \abs{f(x) - f(y) - Df(x)(x - y)} &\le C\abs{x - y}^{1 - d/p} \cdot \norm{Df - Df(x_0)}_{L^p(B(x, 2\abs{x - y}))} \\ > &\le C\abs{x - y} \cdot \braks{\frac{\int_{B(x, 2\abs{x - y})} \abs{Df - Df(x_0)}^p}{m(B(x, 2\abs{x - y}))}}^{1/p} > \end{align*} > $ > where the second term goes to $0$ [[Almost Everywhere|a.e.]] as $\abs{x - y} \to 0$ by the [[Lebesgue Differentiation Theorem]]. # The Inequalities Let $f \in C^1(\real^d) \cap W^{1, p}(\real^d)$. > [!theorem] > > There exists $C_{p, d} \ge 0$, independent of $f$, such that $\norm{f}_u \le \norm{f}_{W^{1, p}}$. > > *Proof*. Let $x \in \real^d$, then using a lemma about the [[Lebesgue Set|Lebesgue set]], > $ > \begin{align*} > \abs{f(x)} &= C_d\int_{B(x, 1)} \abs{f(x)}dy \\ > &\le C_d\braks{\int_{B(x, 1)} \abs{f(y) - f(x)}dy + \int_{B(x, 1)} \abs{f(y)}dy} \\ > &\le C_d \int_{B(x, 1)}\frac{\abs{Df(y)}}{\abs{x - y}^{d - 1}}dy + C_d\norm{f}_{L^1(B(x, 1))} > \end{align*} > $ > By [[Interpolation of Finite Lp Spaces|Lyapunov's inequality]], there exists $C_p \ge 0$ such that $\norm{f}_{L^1(B(x, 1))} \le C_p \norm{f}_p$. On the other hand, let $q$ be the Hölder conjugate of $p$, then by [[Hölder's Inequality]], > $ > \int_{B(x, 1)}\frac{\abs{Df(y)}}{\abs{x - y}^{d - 1}}dy \le \norm{Df}_p \cdot \braks{\int_{B(0, 1)}\frac{1}{\abs{y}^{(d - 1)q}}dy}^{1/q} > $ > where $q = p/(p - 1)$ means that $p(d-1)/(p - 1) < d$, so > $ > \int_{B(0, 1)}\frac{1}{\abs{y}^{p(d-1)/(p-1)}}dy < \infty > $ > Collecting the constants yields the desired result. > [!theorem] > > Let $\gamma = 1 - d/p$, then there exists $C_{p, d} \ge 0$, independent of $f$, such that $[f]_\gamma \le C \norm{f}_{W^{1, p}}$. > > *Proof*. Let $x, y \in \real^d$ with $r = \abs{x - y}$, and $W = B(x, r) \cap B(y, r)$, then > $ > \begin{align*} > \abs{f(x) - f(y)} &= \frac{1}{m(W)}\int_W\abs{f(x) - f(y)}dz \\ > &\le \frac{1}{m(W)}\braks{\int_W\abs{f(x) - f(z)}dz + \int_W\abs{f(y) - f(z)}dz} > \end{align*} > $ > Using the same lemma on the Lebesgue set, and applying [[Hölder's Inequality]] yields > $ > \begin{align*} > \frac{1}{m(W)}\int_W\abs{f(x) - f(z)}dz &\le C_d\int_{B(x, r)}\frac{\abs{Df(z)}}{\abs{z - x}^{d - 1}}dz \\ > &\le C_d\norm{Df}_p \braks{\int_{B(0, r)}\frac{1}{\abs{z}^{p(d - 1)/(p - 1)}}dz}^{(p-1)/p} > \end{align*} > $ > where using [[Polar Integration]], > $ > \begin{align*} > \int_{B(0, r)}\frac{1}{\abs{z}^{p(d - 1)/(p - 1)}}dz &= \int_0^r \frac{s^{d - 1}}{s^{p(d-1)/(p-1)}}\int_{\partial B(0, 1)}d\sigma(x)ds \\ > &= C_d \int_0^rs^{d - 1 - p(d - 1)/(p - 1)}ds \\ > &= C_dr^{d - p(d - 1)/(p - 1)} \\ > \braks{\int_{B(0, r)}\frac{1}{\abs{z}^{p(d - 1)/(p - 1)}}dz}^{(p-1)/p} &= C_d r^{[d - p(d - 1)/(p - 1)] \cdot (p-1)/p} > \end{align*} > $ > where > $ > \begin{align*} > \paren{d - \frac{p(d-1)}{p-1}} \cdot \frac{p - 1}{p} &= \frac{d(p-1)}{p} - d + 1 \\ > &= d - \frac{d}{p}+1 = 1 - \frac{d}{p} > \end{align*} > $ > Therefore > $ > \frac{1}{m(W)}\int_W\abs{f(x) - f(z)}dz \le C_dr^\gamma > $ > Applying the same argument to the second term gives the desired result.