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Re: singular differentiable strictly increasing function
Posted:
May 3, 2012 9:28 PM
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On May 3, 6:42 pm, Tonico <Tonic...@yahoo.com> wrote:
> On May 4, 1:55 am, Butch Malahide <fred.gal...@gmail.com> wrote:
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> > On May 3, 4:40 am, Tonico <Tonic...@yahoo.com> wrote:
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> > > On May 3, 9:37 am, Butch Malahide <fred.gal...@gmail.com> wrote:
> > > > On May 2, 9:27 pm, Tonico <Tonic...@yahoo.com> wrote:
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> > > > > On May 3, 4:57 am, Gus Gassmann <horand.gassm...@googlemail.com>
> > > > > wrote:
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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:
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> > > > > > > > Looking around at various sources, I could not quite find an answer to
> > > > > > > > the following issue:.
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> > > > > > > > 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?
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> > > > > > > > Thanks immensely for any feedback.
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> > > > > > > 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.
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> > > > > > This is certainly not true: What about f(x) = x^3?-
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> > > > > 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.
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> > > > One could ask that. Indeed, I thought that was the original poster's
> > > > question.
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> > > Think again. The OP was whether an strictly ascending function has a
> > > non-zero derivative, and I wrongly answered yes. When the inflexion
> > > points came up I asked what I asked about the inflexion points, which
> > > is NOT the OP.
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> > I see.
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> > OP asked: must the derivative of such a function be almost everywhere
> > nonzero?
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> > You asked: can the derivative of such a function be zero in a set with
> > non-zero measure?
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> > Completely different questions. My bad. Just guessing here, but I'd
> > say the answer to the first question is "no" but the answer to the
> > second is "yes".-
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> And also I was asking about the ammount of inflexion points that could
> make the derivative vanish without the function ceasing to be strictly
> increasing.
The OP's function was only assumed to be differentiable. But let's say
it's twice differentiable, so we can talk about inflexion points.
Surely, the set of inflexion points is at most countable?
Suppose x is an inflexion point of f'', that is, f'' changes sign at
x. So there is an interval (a, b) about x such that, either f'' is
positive on (a, x) and negative on (x, b), or else f'' is negative on
(a, x) and positive on (x, b). In either case, the inflexion point x
is an isolated zero of f''. Hence, the set of all inflexion points is
a discrete subset of the real line, and therefore countable. Right?
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