Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $U$ be a convex open set in $\mathbb{R}^n$ and $f:U\longrightarrow \mathbb{R}$ such that there is $M>0$ with the following property:

$\forall x\in U , \exists r >0: \forall a,b \in B(x,r),|f(a)-f(b)|\le M\cdot ||a-b||$ then it holds that

$\forall x,y\in U,|f(x)-f(y)|\le M\cdot ||x-y||. $

share|cite|improve this question
As I already commented on a previous question of yours: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Prove..."), not a request for help, so please consider rewriting it. – Zev Chonoles Jul 8 '12 at 6:35

Hint: For any pair $x,y\in U$ cover the compact set $\{ty+(1-t)x: t\in [0,1]\}\subset U$ with finitely many open balls $B_1,\ldots,B_n$ such that $f$ is Lipschitz on each $B_i$. Let $$X=\{t: t\in \text{ more than one } B_i\}$$ which is a finite union of intervals, and let $\epsilon$ be the sum of their lengths. Note that $\epsilon$ can be made arbitrarily small. Apply the triangle inequality, and let $\epsilon\to 0$.

Remark: This actually works for any path-connected $U$, by replacing the set $\{ty+(1-t)x: t\in [0,1]\}$ with $\{p(t):t\in [0,1]\}$ for some path $p$ connecting $x$ and $y$.

share|cite|improve this answer
Re: the remark - path connected is not enough, you need some control over the length of connecting curves. The technical term is quasiconvex. – user31373 Jul 9 '12 at 13:00
Actually, the answer is a bit too involved - since the set is assumed convex, no covers are needed - straight line path is available – user31373 Jul 9 '12 at 13:02

Locally Lipschitz functions are not (globally Lipschitz) if we agree that a function $f:X \to Y$ is locally Lipschitz provided any $x \in X$ admits a neighbourhood $U_x$ such that its restriction $f|U_x$ is (globally) Lipschitz (see, e.g. In that case the Lipschitz constant of $f|U_x$ does depend on $x$, which is not the case in the definition you gave. With the usual definition of "locally Lipschitz" one can show that any differentiable function $f:(a,b)\to \mathbb{R}$ for which $f'$ is unbounded on $(a,b)$ and bounded on compact subsets of $(a,b)$ is locally Lipschitz but not (globally) Lipschitz. Functions satisfying the property you mention are more than just locally Lipschitz. For instance a $C^1$ function $f:(a,b) \to \mathbb{R}$ satisfies "your property" iff $\sup_{a<x<b}|f'(x)| \le M$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.