That result looks false to me. Suppose that is the function
Then has equal to 0 everywhere (in the interval [–1,1) say), but it isn't even continuous, let alone Lipschitz continuous.
Hey guys. Tough problem here (tough for me anyway...)
I can show that 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.Suppose the derivate of a function on is bounded, where
.
Show that is Lipschitz continuous.
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 which I could use in this case?
Any help would be much appreciated!
It occurs to me that you may have got the definition of a bit wrong. Suppose that it should be . In other words, the "+" in 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 is bounded then f should be Lipschitz continuous. But I don't offhand see how to prove that.