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>Functions that do not have a derivative at all can certainly have that property. Perhaps you were confused by the notation "dx", suggesting a differential, for

>Functions that do not have a derivative at all can certainly have that property. Perhaps you were confused by the notation "dx", suggesting a differential, for the displacement, rather than the conventional Δx, but the sentence makes no sense if we try to read it as a statement about differentials. Possibly your correspondent is not a Greek Keyboard Achiever. Moreover, the functions they are discussing, while indeed not differentiable, are continuous and differentiable almost everywhere, so you can also prove Lipschitz continuity if you can prove a bound on their gradients at the points where they do exist. The Lipschitz property is not a property of functions with a derivative. I really do dislike this kind of drive by commenting, where you ignore an entire conversation and pretend someone does not understand what a gradient is.

kragen 1y ago
Functions with a derivative can definitely have the Lipschitz property. The “examples” section of https://en.m.wikipedia.org/wiki/Lipschitz_continuity https://en.m.wikipedia.org/wiki/Lipschitz_continuity begins with a subsection entitled, “Lipschitz continuous functions that are everywhere differentiable”. Most of the examples given there are differentiable in varying ways. (Unqualified “differentiable” normally means “everywhere differentiable”.) You're also contradicting your previous comment at https://news.ycombinator.com/item?id=44145078 https://news.ycombinator.com/item?id=44145078. I was trying very hard to ascribe some sort of sensible interpretation to your comments, but due to your aggression, now I regret having done so.