# Make a function that is not of bounded variation on $[a,b]$

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)|

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!!

## 1 Answer

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.

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