The derivative of a map is Lipschitz?

Question

If $f: U \rightarrow \mathbb{R}^m$ is a map of class $C^2$ defined on a compact $U \subset \mathbb{R}^n$ with non-empty interior, is it true that the map $Df : U \rightarrow L(\mathbb{R}^n, \mathbb{R}^m)$ is Lipschitz? (Where, of course, $L(\mathbb{R}^n, \mathbb{R}^m)$ is a normed vector space with the operator norm).

This seems a tricky question, because the linear map $Df(x)$ is, of course Lipschitz, but I don't know if this is true for the map above.

@copper.hat usually by assuming that there's a function $g$ defined on an open set containing $U$ which is differentiable on $U$ and coincides with $f$ everywhere on $U$. — , Apr 30 '17 at 21:34

score 0 · Answer 1 · answered Apr 30 '17 at 21:02

0

At least there is a simple proof if $U$ is also convex.

Hint: $$ Df(x+h) - Df(x) = \int_0^1 Df^2(x+th)h dt. $$

answered Apr 30 '17 at 21:02

user251257

9,229

The derivative of a map is Lipschitz?

1 Answers1