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 $f:[0,+\infty)\to[0,+\infty)$ be defined by $f(x)=x^{1/n}$ where $n\in\mathbb{N}$. Show that $f$ satisfies $|f(x)-f(y)|\leq 2^{(n-1)/n}|x-y|^{1/n}$. Prove that $f$ isn't Lipschitz in any interval cointaining $0$.

share|cite|improve this question
Dear Gastón, in order to get some help for questions, it could be good to show what you have tried so far and, in the future, try to write questions not in an "exercise" fashion, but showing what was your effort so far. – matgaio May 12 '12 at 2:29
I dont know how to prove the first inequallity for Hölder condition, i dont know many useful inequallities. The last part its ok. – Gastón Burrull May 12 '12 at 3:27
The inequality is here:… – Gastón Burrull May 13 '12 at 2:24
up vote 2 down vote accepted

Note that "$f'(0)=+\infty$".
Formally, given $\epsilon>0$, you can find $\delta>0$ such that $x\in[0,0+\delta)$ implies $|f(x)-f(0)|>\epsilon|x-0|$, and in particular, $f$ cannot be Lipschits

share|cite|improve this answer
Ty. Mean value theorem Right? The first part I cant do. I dont know how to prove inequallities =/. – Gastón Burrull May 12 '12 at 4:02

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.