Theorem 5 of pag 283 of the text "Partial Differential Equations - Second Edition (Lawrence C. Evans)" states
Let $U$ be a bounded, open subset of $\mathbb{R}^n$, and suppose $\partial U$ is $C^1$. Assume $n < p \leq \infty$ and $u \in W^{1,p}(U)$.
Then $u$ has a version $u^* \in C^{0,\gamma}(\overline{U})$, for $\gamma := 1 - \frac{n}{p}$, with the estimate
\begin{equation} ||u^*||_{C^{0,\gamma}(\overline{U})} \leq C||u||_{W^{1,p}(U)}\end{equation}
The costant $C$ depends only on $p,n$ and $U$
Where $u^*$ is a version of $u$ means that $u^*(x) = u(x)$ almost everywhere for $x \in U$, and
\begin{equation}||u^*||_{C^{0,\gamma}(\overline{U})} := sup_{x \in \overline{U}}{|u^*(x)|} + sup_{\substack{x,y \in U \\ x \neq y}}{\frac{|u^*(x)-u^*(y)|}{|x - y|^\gamma}}\end{equation}
The book proves the theorem in the case $p < \infty$ and says "The case $p = \infty$ is easy to prove directly".
I'm only interested in the case $p = \infty$.
I only managed to prove that there is a costant $C$ and that $u$ has a version $u^*$ such that
$$\sup_{x \in \overline{U}}{|u^*(x)|} \leq C||u||_{W^{1,\infty}(U)}$$
I don't understand how to prove that
$$ sup_{\substack{x,y \in \overline{U} \\ x \neq y}}{\frac{|u^*(x)-u^*(y)|}{|x - y|}} \leq C||u||_{W^{1,\infty}(U)}$$
Note that there is no hypothesis about the connection of $U$.
My question is : How can I prove the teorem in an easy way in the case $p = \infty$ ?
Here is my attempt. Since $U$ is bounded, we have $W^{1, \infty} \subset W^{1, q}$ for all $q$, so we already know that $u$ is continuous. Denote by $u_1$ an extension of $u$ to $\mathbb{R}^n$ that satisfies $\|u_1\|_{W^{1, \infty}} \leq C\|u\|_{W^{1, \infty}}$. Let $\eta \geq 0$ be the "standard mollifier". For $h > 0$, let $u_{h} = \eta_h * u_1$. Let $x, y \in \mathbb{R}^n$ be arbitrary. We have by the fundamental theorem of calculus and $Du_h = \eta_h * Du_1$ that $$|u_h(x) - u_h(y)| \leq \|Du_{h}\|_{L^\infty}|x - y| \leq \|Du_1\|_{L^\infty}|x - y|.$$ Since $u$ is continuous on $U$, we have $u_{h} \to u$ locally uniformly on $U$, so if $x, y \in U$ we can take $h \searrow 0$ to get $$|u(x) - u(y)| \leq \|Du_1\|_{L^\infty}|x - y| \leq C\|Du\|_{L^\infty}|x - y|.$$ This shows that $u$ is in fact Lipschitz, as desired.