Maximum of gradient of convex functions [closed]

Question

The question comes from the page 472, Elliptic partial differential equations of second order/ David Gilbarg, Neil S. Trudinger. In one dimension it's obviously true, but it seems more involved in higher dimensions. Even though the authors consider it trivial but I didn't find any references for that. Can anybody give a suggestion?

Let $f:B\left(0,2\right)\subseteq\mathbb{R}^{n}\rightarrow\mathbb{R}$ be a smoothly convex function. Prove that $$\underset{\overline{B}\left(0,1\right)}{\max}\left|\nabla f\right|=\underset{\partial B\left(0,1\right)}{\max}\left|\nabla f\right|.$$

Thank you.

Answer 1

Thanks Willie's suggestion, I give my answer as follow:

Assume $p\in B\left(0,1\right)$ so that $\left|\nabla\left(p\right)\right|=\underset{\overline{B}\left(0,1\right)}{\max}\left|\nabla f\right|$. Set $g\left(t\right):=f\left(p+\nabla f\left(p\right)t\right)$ then $g:\left(-\delta,\beta\right)\rightarrow\mathbb{R}$ is convex. It's easy to see that there is $t_{0}>0$ such that $p+t_{0}\nabla f\left(p\right)\in\partial B\left(0,1\right)$. Therefore, $g'\left(0\right)\leq g'\left(t_{0}\right)$, which leads to $$\left|\nabla f\left(p\right)\right|\leq\left|\nabla f\left(p+t_{0}\nabla f\left(p\right)\right)\right|$$, we get the desired result.