Let $\Omega$ be an open set of $\mathbb{R}^n$ and $u\colon\Omega\longrightarrow\mathbb{R}$ be a Lipschitz function. By Rademacher Theorem I know that $u$ is differentiable almost everywhere in $\Omega$. Now, my book says that from this fact, it follows that
$$
\|Du\|_{L^{\infty}(\Omega)}\leq\operatorname{Lip}(u, \Omega),\qquad (\star)
$$
where $\operatorname{Lip}(u, \Omega)$ is the Lipschitz constant of $u$ in $\Omega$. I do not understand why inequality $(\star)$ holds. Can someone help me?
Thank You
Best Answer
Let $L:=\text{Lip}(u,\Omega)$. If the inequality were false, then there would exist $x\in \Omega$ such that $|Du(x)|\geq L+\varepsilon$, for some $\varepsilon > 0 $. In particular, $Du(x)\neq 0$. By definition of $Du$, there is $\delta >0$ such that for all $0<t\leq \delta $, $|h|=1$ we have $$\left|\frac{f(x+th)-f(x)}{t}-Du(x)\cdot h\right|\leq \varepsilon/2 $$ In particular, $$\left|\frac{f(x+th)-f(x)}{t}\right|\geq |Du(x)\cdot h|-\varepsilon/2 $$ now choose $h=\frac{Du(x)}{|Du(x)|}$ (recall that $Du(x)\neq 0$). Then we get $$\left|f(x+th)-f(x)\right|\geq (|Du(x)|-\varepsilon/2)t\geq (L+\varepsilon/2)t $$
which contradicts the fact that $L$ is the Lipschitz constant (recall that $|h|=1$).