This question is taken out of the text book High Dimensional Probability by Roman. I am unsure from the context if it means locally or globally Lipschitz. But the question is as follows:
Every differentiable function $f:\mathbb{R}^n\rightarrow{\mathbb{R}}$ is Lipschitz, and
$\|f\|_{Lip}\le{â„âfâ„_â}$ .
I have shown that if a function is continuously differentiable, it is locally Lipchitz. But since the question uses $\|f\|_{Lip}$ I believe I am to show that if a function is only differentiable, it is globally Lipchitz.
Any hints / clues?
Best Answer
Roman Vershynin has made the final draft of his book High-Dimensional Probability freely downloadable from here (click on the link labelled "Your Copy" on the left side of the page).
I don't believe the statement of exercise $5.1.2$(b) on p.$106$ is correct as given. It should read:
"Every differentiable function $\ f:\mathbb{R}^n \rightarrow\mathbb{R}\ $ with bounded gradient is Lipschitz, and $$ \|f\|_\text{Lip}\le \sup_\limits{x\in\mathbb{R}^n}\|\nabla f(x)\|_2." $$ As indicated by mathcounterexamples.net in the comments, this version of the statement can be proved by appealing to the mean-value theorem: \begin{align} \left|f\left(x_2\right)-f\left(x_1\right)\right|&=\left|\left\langle\nabla f\left(\lambda x_1 + (1-\lambda)x_2\right),x_2-x_1\right\rangle\right|\\ &\hspace{4em}\text{for some }\lambda\in(0,1)\\ &\le \left\| \nabla f\left(\lambda x_1 + (1-\lambda\right)x_2)\right\|\,\left\|x_2-x_1\right\|\\ &\le \sup_\limits{x\in\mathbb{R}^n}\|\nabla f(x)\| \left\|x_2-x_1\right\|\ . \end{align}