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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

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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 $|h|=1$).

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    $\begingroup$ Thank You for your attempt. Maybe there is another way to prove the inequality. Indeed, where exists, $\frac{\partial u}{\partial x_i}=\lim_{t\rightarrow0}\frac{u(x+te_i)-u(x)}{t}$, from which it follows that $\|Du\|_{L^{\infty}(\Omega)}\leq\operatorname{Lip}(u, \Omega)$. Is it right? $\endgroup$ – Jeji Jun 5 '18 at 14:49
  • $\begingroup$ As $|Du|$ we choose $|Du|=\max_{i=1,\ldots,n}|u_{x_i}|$, which is equivalent to the Euclidean norm $\endgroup$ – Jeji Jun 5 '18 at 14:53
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    $\begingroup$ Yes, that should be the 'limit version' of my proof, I basically made that limit explicit with $\varepsilon$s and $\delta$s, aside from that nothing is different I would say. For sure taking the limit directly as you did is more efficient though. $\endgroup$ – Lorenzo Quarisa Jun 5 '18 at 14:55
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    $\begingroup$ Also I took the limit from an arbitrary direction, but since the norm of the gradient (provided it exists, and we know it does) only depends on the partial derivatives alongside the axes, your approach is equivalent $\endgroup$ – Lorenzo Quarisa Jun 5 '18 at 14:57

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