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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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