# Lipschitz continuity proof given boundedness of a derivate (not a derivative)

• Nov 27th 2010, 12:11 PM
hatsoff
Lipschitz continuity proof given boundedness of a derivate (not a derivative)
Hey guys. Tough problem here (tough for me anyway...)

Quote:

Suppose the derivate $\displaystyle D^+$ of a function $\displaystyle f$ on $\displaystyle [a,b]\subseteq\overline{\mathbb{R}}$ is bounded, where

$\displaystyle D^+=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}$.

Show that $\displaystyle f$ is Lipschitz continuous.
I can show that $\displaystyle f$ is of bounded variation, and therefore is differentiable almost everywhere. I don't know if that's a promising approach, however, nor even if it is, how to finish the proof.

One other possible avenue is this: The proof that a function is Lipschitz if its derivative is bounded uses the mean value theorem. Is there maybe some variation of the mean value theorem for $\displaystyle D^+$ which I could use in this case?

Any help would be much appreciated!
• Nov 28th 2010, 12:22 PM
Opalg
That result looks false to me. Suppose that $\displaystyle f$ is the function

$\displaystyle f(x) = \begin{cases}0&(x<0),\\1&(x\geqslant0).\end{cases}$

Then $\displaystyle f$ has $\displaystyle D^+$ equal to 0 everywhere (in the interval [–1,1) say), but it isn't even continuous, let alone Lipschitz continuous.
• Nov 28th 2010, 12:53 PM
hatsoff
Quote:

Originally Posted by Opalg
That result looks false to me. Suppose that $\displaystyle f$ is the function

$\displaystyle f(x) = \begin{cases}0&(x<0),\\1&(x\geqslant0).\end{cases}$

Then $\displaystyle f$ has $\displaystyle D^+$ equal to 0 everywhere (in the interval [–1,1) say), but it isn't even continuous, let alone Lipschitz continuous.

Yeah, that's what I was thinking, too, but I didn't have the confidence to say so. Okay, then. Thanks!
• Nov 29th 2010, 12:47 AM
Opalg
It occurs to me that you may have got the definition of $\displaystyle D^+$ a bit wrong. Suppose that it should be $\displaystyle D^+ = \limsup_{h\to0}\frac{f(x+h)-f(x)}h$. In other words, the "+" in $\displaystyle D^+$ indicates that the lim (in the definition of a derivative) has become a limsup, but it is taken in both directions, not just from the right.

With that definition, it seems plausible that if $\displaystyle D^+$ is bounded then f should be Lipschitz continuous. But I don't offhand see how to prove that.