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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 $|f(x)-f(y)|<M|x-y|^{1/2},M>0$?

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). –  anonymous May 4 at 9:52
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1 Answer 1

up vote 3 down vote accepted

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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>Hello, Paul, Thanks a lot for your hint. But even with your special case, there should still exist a $M$ s.t. the inequality is satisfied. –  Sam Nov 27 '11 at 19:03
    
i edit my answer to be more precise –  Paul Nov 27 '11 at 19:13
    
I think you are right. Thanks a lot for your help. –  Sam Nov 28 '11 at 1:47
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