Re: differentiability of function

In article <bfjks1pffbm3bst56qugl1grig6ipldn4c@xxxxxxx>,
David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:

> On Sat, 14 Jan 2006 18:19:03 -0800, The World Wide Wade
> <waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote:
>
> >In article <gv7fs1d7c4dr0lo3e6fa235v4815p1r2v6@xxxxxxx>,
> > David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:
> >
> >> >Let f: Rn -> R be a function, let f ' (x;h) denote directional
> >> >derivative of f at x in direction h, and let U be an open subset in Rn.
> >> >
> >> >
> >> >Suppose that for every x from U the following conditions hold:
> >> >1. f '(x;-h) = f '(x;h) for every h from Rn
> >> >2. the function h -> f '(x;h) is Lipschitz
> >> >
> >> >Can we find an example of function which satisfies the previous
> >> >assumptions, but which is not differentiable at least at one point of U
> >> >?
> >>
> >> Certainly. For example, in the plane: Let S be the unit circle.
> >> Let g be any smooth function on S such that g(-x) = -g(x) for
> >> all x in S, and then define f by f(rx) = r g(x) for x in S,
> >> r >= 0. That gives f'(0,h) = f(h), so g smooth implies that
> >> f'(0,h) is Lipschitz. But for f to be differentiable at 0
> >> would reguire that f'(0,h) depend linearly on h; that doesn't
> >> happen unless you got very unlucky in your choice of g.
> >
> >Hmmm ... What he wrote was f'(x;-h) = f'(x;h). Of course f(x) =
> >|x| satisfies that and is nondifferentiable at 0. That would be
> >kind of a silly problem. He probably meant f'(x;-h) = -f'(x;h) as
> >you interpreted it.
>
> Oops. Yes, I misread that, doubtless because indeed the
> version he wrote doesn't make much sense.

Actually, my example didn't work for the "as is" problem, because
he says "for all x". So the answer to the question as stated is
that any f satisfying 1. (for all x in U) is either constant on
(connected components of) U or nondifferentiable at infinitely
many points of U.

This suggests the question: If f maps (a,b) into R and E = {x in
(a,b) : f'(x,1) = f'(x,-1) <> 0}, how big can E be?
.

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