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

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

Hey guys. Tough problem here (tough for me anyway...)

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

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

Show that $f$ is Lipschitz continuous.
I can show that $f$ 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 $D^+$ which I could use in this case?

Any help would be much appreciated!

2. That result looks false to me. Suppose that $f$ is the function

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

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

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

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

Then $f$ has $D^+$ 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!

4. It occurs to me that you may have got the definition of $D^+$ a bit wrong. Suppose that it should be $D^+ = \limsup_{h\to0}\frac{f(x+h)-f(x)}h$. In other words, the "+" in $D^+$ 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 $D^+$ is bounded then f should be Lipschitz continuous. But I don't offhand see how to prove that.