Is the function $f(x)=x^{\gamma+1}$, where $x>0 $ and $\gamma<0$ Lipschitz continuous ? I am a bit confused !
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Have you tried to solve this with some examples of specific values of $\gamma$, like say $-\frac12$? – Jonas Meyer Aug 22 '14 at 04:37
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See http://math.stackexchange.com/questions/723821/prove-that-if-f-is-derivable-on-a-b-and-f-is-lipschitz-continuos-then-f, http://math.stackexchange.com/questions/214511/is-the-function-fx-x1-2-lipschitz-continuous?rq=1, http://math.stackexchange.com/q/214290/ – Jonas Meyer Aug 22 '14 at 04:42
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thank you very much for your suggestions – peter5 Aug 22 '14 at 04:52
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is we let $x\in[\alpha,\bar{a}]$ where $\alpha>0$ then due to the fact that the derivative is bounded the function is Lipschitz. Is that correct? – peter5 Aug 22 '14 at 05:13
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peter5: On $[\alpha,\infty)$ with $\alpha>0$, $f(x)=x^p$ with $p\leq 1$ is Lipschitz, and yes, boundedness of the derivative implies this, thanks to the mean value theorem. – Jonas Meyer Aug 22 '14 at 05:16
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so, the problem if i got it right is around 0 ? Thank you very much for your help – peter5 Aug 22 '14 at 05:21
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Yes. You're welcome. – Jonas Meyer Aug 22 '14 at 05:21