# Relation between $L^{\infty}$ norm of the gradient and Lipschitz constant

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

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 $|h|=1$).
• 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? – Jeji Jun 5 '18 at 14:49
• As $|Du|$ we choose $|Du|=\max_{i=1,\ldots,n}|u_{x_i}|$, which is equivalent to the Euclidean norm – Jeji Jun 5 '18 at 14:53
• 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. – Lorenzo Quarisa Jun 5 '18 at 14:55