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A function $f$ ,defined on [a,b],is said to satisfy a uniform Lipschitz condition of order $\alpha > 0$ on [a,b] if there exists a constant $M>0$ such that $|f(x)-f(y)|<M|x-y|^{\alpha}$

Give an example of a function satisfying a uniform Lipschitz condition of order $\alpha<1$ on [a,b] such that f is not of bounded variation on [a,b].

Give an example of a function $f$ which is of bounded variation on [a,b],but which satisfies no uniform Lipschitz condition.

I think of function like $f=\sqrt{x}sin(\frac{1}{x})$,and like $f=xcos(\frac{\pi}{2x})$,but they are not work,I don’t know how to deal with $|\sqrt{x}sin(\frac{1}{x})-\sqrt{y}sin(\frac{1}{y})|$

Could you help me? I don’t know how to make a such function.

What do you think the problem when the problem ask you to give a specific example ?

Thank you very much!!

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The Weierstrass function is not of bounded variation since it is nowhere differentiable, but is Hölder continuous, i.e., it satisfies a Lipschitz condition of order $\alpha < 1$.

For the second part, a discontinuous step function (or any monotone or piecewise monotone function with a jump discontinuity) has bounded variation but cannot satisfy a Lipschitz condition.

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  • $\begingroup$ I got it ! Thank you very much!! $\endgroup$ – Rancho Aug 15 at 0:44
  • $\begingroup$ @Rancho: You're welcome. $\endgroup$ – RRL Aug 15 at 16:19

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