Increasing function whose derivative is densely often equal to 0
Alan Stern
of Calculus, I came up with this problem: Find a differentiable,
monotonically increasing function whose derivative is equal to 0 on a
dense set. We might even require that the derivative be bounded.
Does anyone know of any simple examples of this sort of thing?
Alan Stern
Ignacio Larrosa Cañestro
news:a879975b.02082...@posting.google.com...
If the function can be no differenciable in a numerable set of points,
the Cantor function o 'Devil Staircase' is that you are looking for.
--
Saludos,
Ignacio Larrosa Cañestro
A Coruña (España)
ilar...@mundo-r.com
David C. Ullrich
>While thinking about some borderline cases for the Fundamental Theorem
>of Calculus, I came up with this problem: Find a differentiable,
>monotonically increasing function whose derivative is equal to 0 on a
>dense set.
As has been pointed out, if the function is only required to be
differentiable on a dense set then there is a standard example
known by various names like "Devil's Staircase" or "Cantor-Lebesgue
function". If the function is required to be differentiable
everywhere and have f' = 0 on a dense set I'm pretty sure there
are still examples, but they're not as well known or easy to
describe. (Note that if the function is supposed to be differentiable
everywhere and have f' = 0 "almost everywhere" in the sense of
Lebesgue measure then there is no such example. But saying f' = 0
on a dense set is a much weaker condition.)
Let's say that an increasing function is "almost differentiable"
is at every point either it is differentiable or it has
derivative = + infinity (there may be a standard term for this.)
Suppose you could construct g like so:
(i) g is increasing
(ii) g is almost differentiable everywhere
(iii) g' > 0 everywhere (allowing the value +infinity)
(iv) the set of g(x) such that g'(x) = + infinity is dense.
Then the inverse of g will be a differentiable function
(by (iii)) with f' = 0 on a dense set (by (iv)). It's
easier to say how to construct g than say how to
construct f directly; g will be an infinite sum,
while the corresponding construction for f would
involve taking parts of f, separating them
horizontally and putting new stuff in the resulting
gaps (if it's not clear what that meant, well I
_said_ that the construction of g was easier to
describe...)
You can construct a g satisfying all the conditions
above like so:
g = g_0 + g_1 + g_2 + ...,
where g_0(x) = x for all x. Each g_n will satisfy
g_n'(x) = 0 for all x, _except_ that g_n'(x) > 0 for
x very near a point x_n, with g_n'(x_n) = + infinity.
It's clear then that if G_n = g_0 + ... + g_n
then G_n satisfies (i), (ii), and (iii), and
will also satisfy G_n'(x_j) = + infinity for
j = 1, ... n.
You have to choose the g_n very carefully, so
that when you pass to the limit you get a
function satisfying all four properties.
Having chosen g_1, ... g_n, it seems clear
to me that it's possible to give constraints
that g_{n+1} must satisfy, and then choose
the interval where g_{n+1} > 0 so tiny that
these constraints will be satisfied, in
such a way that everything works.
I'm "certain" I could do that given enough
time and space to write down the details.
I think there's an example in a certain
book at the office - if I think if it I'll
try to look it up.
>We might even require that the derivative be bounded.
If f is differentiable everywhere and f' is bounded then
f satisfies the fundamental theorem of calculus, at least
if you take the Lebesgue integral of f' instead of the
Riemann integral.
Alan Stern
> On 27 Aug 2002 11:48:33 -0700, st...@rowland.org (Alan Stern) wrote:
>
> >While thinking about some borderline cases for the Fundamental Theorem
> >of Calculus, I came up with this problem: Find a differentiable,
> >monotonically increasing function whose derivative is equal to 0 on a
> >dense set.
>
> is at every point either it is differentiable or it has
> derivative = + infinity (there may be a standard term for this.)
> Suppose you could construct g like so:
>
> (i) g is increasing
> (ii) g is almost differentiable everywhere
> (iii) g' > 0 everywhere (allowing the value +infinity)
> (iv) the set of g(x) such that g'(x) = + infinity is dense.
>
> Then the inverse of g will be a differentiable function
> (by (iii)) with f' = 0 on a dense set (by (iv)). It's
> easier to say how to construct g than say how to
> construct f directly; g will be an infinite sum,
> while the corresponding construction for f would
> involve taking parts of f, separating them
> horizontally and putting new stuff in the resulting
> gaps (if it's not clear what that meant, well I
> _said_ that the construction of g was easier to
> describe...)
In my original construction, I found an example for f very much in the
way you described, and it was indeed messy. It does seem that it
would be simpler to find the appropriate conditions for building g.
And I guess there aren't any simple or well-known examples of this
sort of thing.
Alan Stern
David C. Ullrich
Possibly not simple, but I do believe that there are examples
that are well-known, at least in the sense that they appear
in papers and books. When I got to the office I forgot about
this - distracted by silliness like classes, students and
meetings. Before I hit the Send button I'm gonna leave myself
a note to look it up tomorrow...
>Alan Stern
Gerry Myerson
st...@rowland.org (Alan Stern) wrote:
=> Find a differentiable, monotonically increasing function whose
=> derivative is equal to 0 on a dense set.
The first place to look for this sort of thing is Gelbaum & Olmsted,
Counterexamples in Analysis. Chapter 8, #30 is "A continuous strictly
monotonic function with a vanishing derivative almost everywhere."
--
Gerry Myerson (ge...@mpce.mq.edi.ai) (i -> u for email)
Virgil
> > On 27 Aug 2002 11:48:33 -0700, st...@rowland.org (Alan Stern) wrote:
> >
> > >While thinking about some borderline cases for the Fundamental Theorem
> > >of Calculus, I came up with this problem: Find a differentiable,
> > >monotonically increasing function whose derivative is equal to 0 on a
> > >dense set.
The closest I have seen to this is in "Counterexamples in Analysis",
Gelbaum and Oemstead, copyright 1964, Holden-Day, Inc.
Chapter 8, example 15, gives a function described as "A Continuous
Monotonic function with a vanishing derivative almost everywhere."
As might be expected, it is based on the Cantor set (of recursively
deleted middle thirds).
Ioannis
[snip]
> Chapter 8, example 15, gives a function described as "A Continuous
> Monotonic function with a vanishing derivative almost everywhere."
>
> As might be expected, it is based on the Cantor set (of recursively
> deleted middle thirds).
And for those that want to take a peak at the "Devil's Staircase", here
are some successively better approximations done years ago:
<http://users.forthnet.gr/ath/jgal/math/cantor.html>
--
Ioannis
http://users.forthnet.gr/ath/jgal/
____________________________________________
"You cannot go against Nature, because going
against Nature is part of Nature too".
David C. Ullrich
<ge...@mpce.mq.edi.ai.i2u4email> wrote:
>In article <a879975b.02082...@posting.google.com>,
>st...@rowland.org (Alan Stern) wrote:
>
>=> Find a differentiable, monotonically increasing function whose
>=> derivative is equal to 0 on a dense set.
>
>The first place to look for this sort of thing is Gelbaum & Olmsted,
>Counterexamples in Analysis. Chapter 8, #30 is "A continuous strictly
>monotonic function with a vanishing derivative almost everywhere."
As (ahem) already noted, that cannot be an example of the
desired sort: If f is continuous, increasing, and has f'=0
almost everywhere then it cannot be differentiable
(everywhere).
David Bernier
For x in R, let frac(x) be x minus the largest integer less than or
equal to x. So frac(pi) = 3, frac(7) = 0, and so on.
Given x in R, express frac(x) in base 6. This leads to two ways
of writing 1/6: 0.1 [base 6] and 0.055555555555... [base 6].
For definiteness, let's exclude base 6 expansions with an infinite
tail of consecutive 5's.
We can use frac(x) in base 6 to code for a random walk on the 3-D lattice
ZxZxZ where we start at (0,0,0) and
1 means move UP
2 means move DOWN
3 means move LEFT
4 means move RIGHT
5 means move FORWARD
0 means move BACWARDS.
Then, any given x in R codes for a well-defined random walk on ZxZxZ through
the encoding of frac(x) in base 6.
Let A= {x in R: the random walk determined as above returns to the
origin either never or an even number of times}
Let B= R\A.
Then both A and B have strictly positive Lebesgue measure, and both
A and B are dense in R. The fact that both A and B have Lebesgue
measure >0 follows from the fairly well-known fact that in a random walk
on ZxZxZ, the probability of ever returning to the origin is >0 and <1.
Both A and B are dense in R, roughly because to know whether the number
of returns to the origin is odd or even, we need to wait till the
random walk is "finished".
If we let chi_B be the characteristic function of B, and define
f(y) = int_{-oo, y} { chi_B(t) dt } [ Lebesgue integration ],
then I would guess that f is strictly increasing, differentiable, and
that f'(y) = 0 for any y in A.
That's as far as I've got, so there are still things to prove...
David Bernier
Robert Israel
David Bernier <ez...@yahoo.com> wrote:
>st...@rowland.org (Alan Stern) wrote in message
>news:<a879975b.02082...@posting.google.com>...
>> While thinking about some borderline cases for the Fundamental Theorem
>> of Calculus, I came up with this problem: Find a differentiable,
>> monotonically increasing function whose derivative is equal to 0 on a
>> dense set. We might even require that the derivative be bounded.
>>
>> Does anyone know of any simple examples of this sort of thing?
It's true f is strictly increasing; unfortunately, the other guesses are
false.
Let p_0 be the probability that the walk returns to the origin given that
it starts at the origin. This is also the probability that the walk
hits the origin given that it starts one step away from the origin.
Let p_2 < p_0 be the probability that the walk hits the origin, given
that it starts at (2,0,0). Thus the probability that the walk hits
the origin an odd number of times, given that it starts at (1,0,0), is
p_0 (1 - p_0) + p_0^3 (1 - p_0) + ... = p_0/(1+p_0), and the probability
that it hits the origin an odd number of times, given that it starts at
(2,0,0), is p_2/(1+p_0). Call these numbers r_1 and r_2 respectively.
Choose y_0 encoding a random walk that never hits the origin (so y_0 is in
A), but is at (1,0,0) infinitely many times and at (2,0,0) infinitely many
times. If n is one of the times it is at (1,0,0), there is an interval
[a_n, a_n+6^(-n)) containing y_0, consisting of all the numbers encoding
walks whose first n steps are the same as for y_0. In particular, these
walks are at (1,0,0) at time n. The fact that the probability of hitting
the origin an odd number of times starting at (1,0,0) is r_1 means that
(f(a_n + 6^(-n))-f(a_n))/6^(-n) = r_1. Similarly, if m is one of the
times the walk for y_0 is at (2,0,0), there is an interval
[a_m, a_m+6^(-m)) containing y_0 such that (f(a_m + 6^(-m))-f(a_m))/6^(-m)
= r_2. We conclude that f is not differentiable at y_0.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
David C. Ullrich
Incredibly, my recollection of what was on a certain page of
a certain book was correct. See Exercise 6(e) in Chapter 4
of Stromberg "An Introduction to Classical Real Analysis"
for a clean exposition of an example (possibly not quite
"simple").
>Alan Stern
Herman Rubin
Gerry Myerson <ge...@mpce.mq.edi.ai.i2u4email> wrote:
>In article <a879975b.02082...@posting.google.com>,
>st...@rowland.org (Alan Stern) wrote:
>=> Find a differentiable, monotonically increasing function whose
>=> derivative is equal to 0 on a dense set.
>The first place to look for this sort of thing is Gelbaum & Olmsted,
>Counterexamples in Analysis. Chapter 8, #30 is "A continuous strictly
>monotonic function with a vanishing derivative almost everywhere."
This is not the question; such a function is easy
to produce. If a function has the property of
differentiability and its derivative vanishes
almost everywhere, it must be constant.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Herman Rubin
>> Alan Stern
There are easier ways to construct such sets; I have given
one here using the binary expansion, and easier to compute.
>If we let chi_B be the characteristic function of B, and define
>f(y) = int_{-oo, y} { chi_B(t) dt } [ Lebesgue integration ],
>then I would guess that f is strictly increasing, differentiable, and
>that f'(y) = 0 for any y in A.
This does not work. The derivative of f is 0 or 1 almost
everywhere, but does not exist everywhere.
David Bernier
> In article <50dce0c5.02082...@posting.google.com>,
> David Bernier <ez...@yahoo.com> wrote:
> >st...@rowland.org (Alan Stern) wrote in message
> >news:<a879975b.02082...@posting.google.com>...
> >> While thinking about some borderline cases for the Fundamental Theorem
> >> of Calculus, I came up with this problem: Find a differentiable,
> >> monotonically increasing function whose derivative is equal to 0 on a
> >> dense set. We might even require that the derivative be bounded.
> >>
> >> Does anyone know of any simple examples of this sort of thing?
>
> >For x in R, let frac(x) be x minus the largest integer less than or
> >equal to x. So frac(pi) = 3, frac(7) = 0, and so on.
>
> >Given x in R, express frac(x) in base 6. This leads to two ways
> >of writing 1/6: 0.1 [base 6] and 0.055555555555... [base 6].
> >For definiteness, let's exclude base 6 expansions with an infinite
> >tail of consecutive 5's.
> It's true f is strictly increasing; unfortunately, the other guesses are
> false.
[...]
Thanks for clearing this up. I've been reading your disproof;
while I'm not done reading, it looks quite interesting.
David Bernier
David Bernier
[...]
> There are easier ways to construct such sets; I have given
> one here using the binary expansion, and easier to compute.
I was previously unaware of your construction.
> >If we let chi_B be the characteristic function of B, and define
> >f(y) = int_{-oo, y} { chi_B(t) dt } [ Lebesgue integration ],
> >then I would guess that f is strictly increasing, differentiable, and
> >that f'(y) = 0 for any y in A.
>
> This does not work. The derivative of f is 0 or 1 almost
> everywhere, but does not exist everywhere.
Ok.
David Bernier