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Does the function $f(x)=\sqrt{x}\sin(1/x),x\in(0,1],f(0)=0,$ satisfy the uniform Lipschitz condition $|f(x)-f(y)|0$?

Any help is appreciated. Thanks

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    Such a condition is usually referred to as Holder continuity, not Lipschitz (which would be stronger).2014-05-04

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no, take $x =\frac{1}{2n\pi +\frac{\pi}{2}}$ and $y =\frac{1}{2n\pi -\frac{\pi}{2}}$. Then $\vert f(x)-f(y)\vert$ behave like $\frac{1}{\sqrt{n}}$ and $\vert x-y \vert^\frac{1}{2}$ behaves like $\frac{1}{n}$.

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    I think you are right. Thanks a lot for your help.2011-11-28