> [!definition] > > Let $U \subset \real^d$ be [[Open Set|open]] and $f: U \to \complex$ be a [[Continuity|continuous]] function, then $f$ is **Lipschitz continuous** if there exists $C \ge 0$ such that > $ > \abs{f(x) - f(y)} \le C\abs{x - y} > $ > for all $x, y \in U$. The smallest such $C$ is known as the **Lipschitz constant** of $f$, denoted as $\text{Lip}(f)$. > [!theorem] > > Let $W^{1, \infty}(\real^d)$ be the [[Sobolev Space|Sobolev space]], then there exists a [[Space of Toplinear Isomorphisms|toplinear isomorphism]] > $ > I: W^{1, \infty}(\real^d) \to C^{0, 1}(\real^d) > $ > such that for any $f \in W^{1, \infty}(\real^d)$, $If = f$ [[Almost Everywhere|a.e.]], $\norm{If}_u = \norm{f}_{L^\infty}$, and $[If]_{C^{0, 1}} = \norm{Df}_\infty$. > > *Proof*. Firstly, for any $V \subset \subset \real^d$, there exists a bounded linear map > $ > I: W^{1, \infty}(\real^d) \to C^{0, \gamma}(V) > $ > such that $If(x) = f(x)$ for almost every $x \in U$ and $f \in W^{1, \infty}(\real^d)$. By gluing, there exists a bounded linear map > $ > I: W^{1, \infty}(\real^d) \to BC(\real^d) > $ > such that $If = f$ a.e. for all $f \in W^{1, \infty}(\real^d)$. To show the continuity of $I$, it's sufficient to verify that $[If]_{C^{0, 1}} \le \norm{Df}_\infty$. To this end, let $\phi \in C_c^\infty(\real^d)$ be a non-negative mollifier. For any $x, y \in \real^d$, there exists $V \subset \subset \real^d$ such that $x, y \in V$. From here, > $ > f * \phi_t = If * \phi_t \to If > $ > uniformly on $V$ as $t \to 0$ since $If|_{V}$ is uniformly continuous. On the other hand, $\norm{D(f * \phi_t)}_{L^\infty(V)} \le \norm{Df}_{L^\infty(\real^d)}$ for all $t > 0$, so > $ > \begin{align*} > \abs{(f * \phi_t)(x) - (f * \phi_t)(y)} &\le \abs{x - y}\int_0^1\abs{D(f * \phi_t)(x + s(y - x))}ds \\ > &\le \abs{x - y} \cdot \norm{D(f * \phi_t)}_{L^\infty(V)} \\ > &\le \abs{x - y} \cdot \norm{Df}_{L^\infty(\real^d)} > \end{align*} > $ > taking the limit yields that > $ > \abs{If(x) - If(y)} \le \abs{x - y} \cdot \norm{Df}_{L^\infty(\real^d)} > $ > > Now let $I^{-1}: C^{0, 1}(\real^d) \to W^{1, \infty}(\real^d)$ be the inclusion map. To show continuity, it's sufficient to show that $f \in W^{1, \infty}(\real^d)$ and $\norm{Df}_{L^\infty(\real^d)} \le [f]_{C^{0, 1}(\real^d)}$. For any $1 \le j \le d$, $h \in \real$, and $x \in \real^d$, > $ > \abs{D^h_jf(x)} = \abs{\frac{f(x + h) - f(x)}{h}} \le [f]_{C^{0, 1}(\real^d)} > $ > and $\abs{D^hf(x)} \le [f]_{C^{0, 1}(\real^d)}$. Since the [[Difference Quotient|difference quotient]] is bounded in $L^p(V)$ for any $V \subset \subset \real^d$ and $1 < p < \infty$, $f \in W^{1, p}_{\text{loc}}(\real^d)$ with $D_j^hf \to D_jf$ in $L^p_{\text{loc}}(\real^d)$ for any $1 < p < \infty$. For any $V \subset \subset \real^d$, there exists a sequence $\seq{h_n} \subset \real$ such that $h_n \to 0$ and $D^{h_n}f \to Df$ a.e. on $V$, so $\norm{Df}_{L^{\infty}(\real^d)} \le [f]_{C^{0, 1}(\real^d)}$.