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 $0$.
Function satisfiyng Hölder condition but not Lipschitz condition.
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real-analysis
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1Dear 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. – 2012-05-12
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0I dont know how to prove the first inequallity for Hölder condition, i dont know many useful inequallities. The last part its ok. – 2012-05-12
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0The inequality is here: http://math.stackexchange.com/questions/144434/an-inequality-x1-n-y1-n-leq-cx-y1-n – 2012-05-13
1 Answers
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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
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0Ty. Mean value theorem Right? The first part I cant do. I dont know how to prove inequallities =/. – 2012-05-12