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Re: singular differentiable strictly increasing function
Posted:
May 3, 2012 2:37 AM
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On May 2, 9:27 pm, Tonico <Tonic...@yahoo.com> wrote:
> On May 3, 4:57 am, Gus Gassmann <horand.gassm...@googlemail.com>
> wrote:
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>
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> > On May 2, 10:37 pm, Tonico <Tonic...@yahoo.com> wrote:
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> > > On May 2, 9:30 pm, boyandshark <amitgan...@gmail.com> wrote:
>
> > > > Looking around at various sources, I could not quite find an answer to
> > > > the following issue:.
>
> > > > Suppose we have a differentiable real value function that is strictly
> > > > increasing over its domain. Do these conditions this imply that the
> > > > derivative is almost everywhere non-zero? I know there exist counter-
> > > > examples to this claim if the function is only almost everywhere
> > > > differentiable - hence I was wondering if it was possible that the
> > > > strengthening to everywhere differentiable might remove these
> > > > pathologies?
>
> > > > Thanks immensely for any feedback.
>
> > > Not only different from zero: if a differentiable (on its domain)
> > > function (one real variable and real valued one) is strictly
> > > increasing in its domain then its derivative is actually positive.
>
> > This is certainly not true: What about f(x) = x^3?-
>
> Indeed, inflexion points are problematic, and checking the example one
> could ask whether there can be "lots of these points" as to make the
> derivative to be zero in aset with non-zero measure.
One could ask that. Indeed, I thought that was the original poster's
question.
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