Norm of the gradient of a Lipschitz function

Question

Let $f:\mathbb{R}^n\mapsto \mathbb{R}$ be a differentiable Lipschitz function with Lipschitz constant $K$. Is it true that, ${\|\nabla f(x)\|}_2\leq K$, for all $x\in\mathbb{R}^n$, where $\nabla f(x)$ denotes vector of partial derivatives of $f$ at $x$. I get an upper bound, but cannot make it free of $n$. Thank you.

Answer 1

By differentiability $$ \nabla f(x)^Th = f(x+h)-f(x) + o(h), $$ so $$ |\nabla f(x)^Th| \le |f(x+h) - f(x)| + |o(h)| \le K \|h\|_2 + |o(h)|. $$ Fix $h$ with $\|h\|_2=1$, replace $h$ in the above formula by $th$, $t>0$, then divide by $t$ to get $$ |\nabla f(x)^Th| \le K + t^{-1} |o(th)|, $$ now passing to the limit $t\searrow0$ proves $$ |\nabla f(x)^Th| \le K $$ for all $h$ with $\|h\|_2=1$, which implies the claim.