f is constant?
Betelgeuse
Assume also that I know that f'(x)=0 for
all x in A, where A is an open dense subset of (a,b).
Can I conclude that f is a constant function?
Thanks in advance.
Jenny
Justin Young
"Betelgeuse" <betel...@rigel.bellatrix> wrote in message
news:42225E11...@rigel.bellatrix...
The Cantor function is a function that is continuous, differentiable,
increasing, non-constant, and the derivative is zero everywhere on the
complement of the Cantor set. If we restrict the function to (0,1), then
(0,1)-(cantor set) is open and dense.
David C. Ullrich
No, the function commonly known as the Cantor function is not
differentiable (at every point).
************************
David C. Ullrich
Justin Young
"David C. Ullrich" <ull...@math.okstate.edu> wrote in message
news:106621pdesktp2u0g...@4ax.com...
I thought that part was a little fishy, but some web site claimed it was
differentiable on [0,1] and proceeded to take the integral of f' over that
interval.
Betelgeuse
Justin Young wrote:
it is indeed a wrong claim! The difficulty here is exactly the "everywhere"
Regards,
Jenny
Dave Seaman
It is differentiable almost everywhere on [0,1] and the integral of its
derivative is 0.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Ioannis
>>No, the function commonly known as the Cantor function is not
>>differentiable (at every point).
>>
>
>
> I thought that part was a little fishy, but some web site claimed it was
> differentiable on [0,1] and proceeded to take the integral of f' over that
> interval.
It's not differentiable on at least one point. Take x=1/3, for example:
lim_{x->1/3+}f'(x)=0, (the Cantor set is zero from 1/3 to 2/3), while
from the left side of 1/3, the Cantor set at each step of its
construction is the line:
(y-1/2)/(x-1/3)=(2^n-1)/2^(n+1)/(1/3-1/3^(n+1)).
The right side tends to 3/2 as n->+oo, so left of 1/3 the Cantor
function approaches the linear function
(y-1/2)/(x-1/3)=3/2, consequently,
lim_{x->1/3-}f'(x)=lim_{x->1/3-}3/2(x-1/3)+1/2=1/2=/=0,
so it's not differentiable at x=1/3.
That is, unless I've made a mistake somewhere.
--
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
------------------------------------------
Eventually, _everything_ is understandable
Ioannis
[snip]
> That is, unless I've made a mistake somewhere.
Damn typos! The *slope* of the limit function is what matters. (Not the
limit function itself). So it should be lim_{x->1/3-}f'(x)=3/2=/=0, so
the Cantor function is not differentiable at x=1/3.
Incidentally, the same argument can be repeated for all corners of the
Cantor set, with the expression for the slope changed, to show
non-differentiability on a set of measure 0.
ArtflDodgr
Betelgeuse <betel...@rigel.bellatrix> wrote:
> Let f:(a,b)->R be differentiable in every point.
> Assume also that I know that f'(x)=0 for
> all x in A, where A is an open dense subset of (a,b).
> Can I conclude that f is a constant function?
Hint: Use the definition of f'(x) to deduce that if
f'(x_0) > 0 then f is strictly increasing in a neighborhood of x_0;
likewise if f'(x_0) < 0 then f is strictly decreasing in a neighborhood
of x_0.
--
A.
Daniel Grubb
>> Let f:(a,b)->R be differentiable in every point.
>> Assume also that I know that f'(x)=0 for
>> all x in A, where A is an open dense subset of (a,b).
>> Can I conclude that f is a constant function?
>Hint: Use the definition of f'(x) to deduce that if
>f'(x_0) > 0 then f is strictly increasing in a neighborhood of x_0;
>likewise if f'(x_0) < 0 then f is strictly decreasing in a neighborhood
>of x_0.
Not generally true. There are differentiable functions with the property that
{x:f'(x)=0}, {x:f'(x)<0}, and {x:f'(x)>0} are *all* dense. Furthermore,
the derivative can be chosen to be a bounded function.
--Dan Grubb
ArtflDodgr
gr...@lola.math.niu.edu (Daniel Grubb) wrote:
Indeed, forget my hint.
--
A.
poopd...@gmail.com
Nice problem. I take it this is for a course in real analysis?
Hint: Determine whether or not f' must be a continuous function of x.
If so, show that since f'(x) = 0 on a dense subset, it must be 0
everywhere. If you haven't proved it already, show that f'(x) = 0
everywhere implies that f is constant.
(For exactness, you might actually have to prove that if g is a
continuous function and is constant on a dense subset, it is constant
everywhere. f' continuous and 0 is a special case of this. While
you're doing this, use the triangle inequality to set up your
delta-epsilon argument)
'cid 'ooh
Justin Young
>> I thought that part was a little fishy, but some web site claimed it was
>> differentiable on [0,1] and proceeded to take the integral of f' over
>> that
>> interval.
>
> It is differentiable almost everywhere on [0,1] and the integral of its
> derivative is 0.
how do you take an integral of a function on [0,1] that isn't even defined
on [0,1]?
Robert Israel
Justin Young <x_sta...@hotmail.com> wrote:
No problem if you're using the Lebesgue integral, as long as it's defined
almost everywhere (and is integrable).
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Justin Young
"Robert Israel" <isr...@math.ubc.ca> wrote in message
news:d005u0$isd$1...@nntp.itservices.ubc.ca...
> In article <5JMUd.61853$H05....@twister.nyroc.rr.com>,
> Justin Young <x_sta...@hotmail.com> wrote:
>
>>>> I thought that part was a little fishy, but some web site claimed it
>>>> was
>>>> differentiable on [0,1] and proceeded to take the integral of f' over
>>>> that
>>>> interval.
>
>>> It is differentiable almost everywhere on [0,1] and the integral of its
>>> derivative is 0.
>
>>how do you take an integral of a function on [0,1] that isn't even defined
>>on [0,1]?
>
> No problem if you're using the Lebesgue integral, as long as it's defined
> almost everywhere (and is integrable).
>
(why does this seem like an ad campaign?)
I suppose the rub is that the fundamental theorem fails in a sense. I was
thinking of the Riemann integral of course.
Betelgeuse
> Betelgeuse wrote:
> > Let f:(a,b)->R be differentiable in every point.
> > Assume also that I know that f'(x)=0 for
> > all x in A, where A is an open dense subset of (a,b).
> > Can I conclude that f is a constant function?
> > Thanks in advance.
> > Jenny
>
> Nice problem. I take it this is for a course in real analysis?
No. I do not think the real analysis average (or even good) student
could solve it!
>
>
> Hint: Determine whether or not f' must be a continuous function of x.
> If so, show that since f'(x) = 0 on a dense subset, it must be 0
> everywhere. If you haven't proved it already, show that f'(x) = 0
> everywhere implies that f is constant.
>
> (For exactness, you might actually have to prove that if g is a
> continuous function and is constant on a dense subset, it is constant
> everywhere. f' continuous and 0 is a special case of this. While
> you're doing this, use the triangle inequality to set up your
> delta-epsilon argument)
>
> 'cid 'ooh
Actually, f' need not be continuous. Otherwise it is trivial.
My feeling is that there is an example of a nonconstant
function satisfying the required conditions, but the example
(provided that it exists) is far from being easy (at least, for me).
What is true is that such an f cannot be absolutely continuous.
David C. Ullrich
<x_sta...@hotmail.com> wrote:
>
>"David C. Ullrich" <ull...@math.okstate.edu> wrote in message
>news:106621pdesktp2u0g...@4ax.com...
>> On Mon, 28 Feb 2005 00:39:51 GMT, "Justin Young"
>> <x_sta...@hotmail.com> wrote:
>>
>>>
>>>"Betelgeuse" <betel...@rigel.bellatrix> wrote in message
>>>news:42225E11...@rigel.bellatrix...
>>>> Let f:(a,b)->R be differentiable in every point.
>>>> Assume also that I know that f'(x)=0 for
>>>> all x in A, where A is an open dense subset of (a,b).
>>>> Can I conclude that f is a constant function?
>>>> Thanks in advance.
>>>> Jenny
>>>>
>>>
>>>The Cantor function is a function that is continuous, differentiable,
>>>increasing, non-constant, and the derivative is zero everywhere on the
>>>complement of the Cantor set. If we restrict the function to (0,1), then
>>>(0,1)-(cantor set) is open and dense.
>>
>> No, the function commonly known as the Cantor function is not
>> differentiable (at every point).
>>
>
>I thought that part was a little fishy, but some web site claimed it was
>differentiable on [0,1] and proceeded to take the integral of f' over that
>interval.
Oh. If some web site says it's differentiable then it
must be so.
************************
David C. Ullrich
David C. Ullrich
>
>Betelgeuse wrote:
>> Let f:(a,b)->R be differentiable in every point.
>> Assume also that I know that f'(x)=0 for
>> all x in A, where A is an open dense subset of (a,b).
>> Can I conclude that f is a constant function?
>> Thanks in advance.
>> Jenny
>
>Nice problem. I take it this is for a course in real analysis?
>
>Hint: Determine whether or not f' must be a continuous function of x.
Uh, the word "hint" isn't all that appropriate, since this
suggestion doesn't work (because f' certainly need not be
continuous.)
>If so, show that since f'(x) = 0 on a dense subset, it must be 0
>everywhere. If you haven't proved it already, show that f'(x) = 0
>everywhere implies that f is constant.
>
>(For exactness, you might actually have to prove that if g is a
>continuous function and is constant on a dense subset, it is constant
>everywhere. f' continuous and 0 is a special case of this. While
>you're doing this, use the triangle inequality to set up your
>delta-epsilon argument)
>
>'cid 'ooh
************************
David C. Ullrich
David C. Ullrich
That's the way it seems to me, too.
>What is true is that such an f cannot be absolutely continuous.
Why not?
************************
David C. Ullrich
Dave Seaman
The integral is in the sense of Lebesgue. It's enough that the function
be measurable and defined almost everywhere on the interval (everywhere
except on a set of measure zero).
Dave Seaman
The fundamental theorem does not fail. In fact, the Lebesgue version of
the FTOC is stronger than the Riemann version.
It's also a theorem that a function is Riemann integrable iff it is
continuous almost everywhere, which would seem to indicate that the
Cantor function is Riemann integrable.
The World Wide Wade
ArtflDodgr <artfl...@aol.com> wrote:
A counterexample to this is f(x) = x + 2x^2*sin(1/x) at x_0 = 0.
We have f'(0) = 1, but f ping-pongs between x - 2x^2 and x + 2x^2
in such a way that f is strictly decreasing on a sequence of
intervals going to 0.
Robert Israel
Dave Seaman <dse...@no.such.host> wrote:
>The fundamental theorem does not fail. In fact, the Lebesgue version of
>the FTOC is stronger than the Riemann version.
There are several Lebesgue versions. One is: if f is differentiable
everywhere on [a,b] and f' is in L^1[a,b], then
f(b) - f(a) = int_a^b f'(x) dx
Another is
If f is absolutely continuous on [a,b] then f' exists almost
everywhere on [a,b] and
f(b) - f(a) = int_a^b f'(x) dx
Neither of those versions applies where f is the Cantor function, since
the Cantor function is not differentiable everywhere and is not absolutely
continuous on [0,1].
>It's also a theorem that a function is Riemann integrable iff it is
...bounded and...
>continuous almost everywhere, which would seem to indicate that the
>Cantor function is Riemann integrable.
The Cantor function is continuous, so of course it's Riemann integrable.
But that's irrelevant, since we're talking about integrating f', not f.
The derivative of the Cantor function is not everywhere defined, and
therefore not Riemann integrable.
Robert Israel
And in your article of 30 Aug 2002 in the thread "Increasing function
whose derivative is densely often equal to 0"
<http://groups-beta.google.com/group/sci.math/msg/bd4d9edf09f098bb>
you pointed to an example in Stromberg "An Introduction to Classical
Real Analysis" of a non-constant differentiable function whose derivative
is 0 on a dense set. Unfortunately not on a dense _open_ set (and I
don't see how to patch Stromberg's example to make it so).
Dave Seaman
> In article <d00b3u$6rt$2...@mailhub227.itcs.purdue.edu>,
> Dave Seaman <dse...@no.such.host> wrote:
>>It's also a theorem that a function is Riemann integrable iff it is
> ...bounded and...
Yes.
>>continuous almost everywhere, which would seem to indicate that the
>>Cantor function is Riemann integrable.
> The Cantor function is continuous, so of course it's Riemann integrable.
> But that's irrelevant, since we're talking about integrating f', not f.
> The derivative of the Cantor function is not everywhere defined, and
> therefore not Riemann integrable.
But that seems like an artificial distinction to me, since f' exists a.e.
and there is an extension to a function g:[0,1] -> R that is bounded and
continuous a.e. and therefore Riemann integrable. Morever, the extension
can be done pretty much arbitarily (so long as g remains bounded) and it
will have no effect on the value of the integral.
W. Dale Hall
That's something I'd like to see. Can you post a f'rinstance?
It appears that this would force f' to be discontinuous wherever
it's nonzero, kinda-sorta-like that function defined to be zero
on the irrationals, and equal to 1/q where x = p/q in lowest terms.
It doesn't seem that integrating such an example would work, since
its Riemann integral doesn't exist, and the Lebesgue integral ignores
the set of measure zero where the discontinuities belong. I imagine
you could take a sum of differentiable functions where you control
the derivatives of the partial sums to have this behavior, and at
the same time preserve differentiability in the limit.
I'm guessing an ASCII-art picture of the graph is just way out of
the question, so some outline of the construction would be edifying.
Thanks,
Dale (not an analyst, and not pretending to have been one).
David C. Ullrich
<dse...@no.such.host> wrote:
>On 1 Mar 2005 00:27:57 GMT, Robert Israel wrote:
>> In article <d00b3u$6rt$2...@mailhub227.itcs.purdue.edu>,
>> Dave Seaman <dse...@no.such.host> wrote:
>
>>>It's also a theorem that a function is Riemann integrable iff it is
>
>> ...bounded and...
>
>Yes.
>
>>>continuous almost everywhere, which would seem to indicate that the
>>>Cantor function is Riemann integrable.
>
>> The Cantor function is continuous, so of course it's Riemann integrable.
>> But that's irrelevant, since we're talking about integrating f', not f.
>> The derivative of the Cantor function is not everywhere defined, and
>> therefore not Riemann integrable.
>
>But that seems like an artificial distinction to me, since f' exists a.e.
>and there is an extension to a function g:[0,1] -> R that is bounded and
>continuous a.e. and therefore Riemann integrable. Morever, the extension
>can be done pretty much arbitarily (so long as g remains bounded) and it
>will have no effect on the value of the integral.
Good point, except that the derivative is _not_ bounded. (Unless you
mean something very different from what I've assumed we're meaning
by "Cantor function"...)
************************
David C. Ullrich
David C. Ullrich
>In article <0ec721pqqufsrsh6t...@4ax.com>,
>David C. Ullrich <ull...@math.okstate.edu> wrote:
>>On Mon, 28 Feb 2005 22:42:09 GMT, Betelgeuse
>><betel...@rigel.bellatrix> wrote:
>
>>>Actually, f' need not be continuous. Otherwise it is trivial.
>>>My feeling is that there is an example of a nonconstant
>>>function satisfying the required conditions, but the example
>>>(provided that it exists) is far from being easy (at least, for me).
>
>>That's the way it seems to me, too.
>
>And in your article of 30 Aug 2002 in the thread "Increasing function
>whose derivative is densely often equal to 0"
><http://groups-beta.google.com/group/sci.math/msg/bd4d9edf09f098bb>
>you pointed to an example in Stromberg "An Introduction to Classical
>Real Analysis" of a non-constant differentiable function whose derivative
>is 0 on a dense set. Unfortunately not on a dense _open_ set (and I
>don't see how to patch Stromberg's example to make it so).
Simple<g>: Take a dense sequence (a_n) where g' = 0, and
define f by sort of sliding bits of the graph around so
that each a_n gets expanded to an open interval on which
f is constant. Of course then we have to show that f is
differentiable at the _other_ points...
(But seriously: I wasn't going to look up that example,
but evidently you have. Iirc it proceeds by first
constructing the inverse function, making sure it
has infinite derivative at a lot of places. Have you
tried doing the same thing except with an inverse
non-function, something whose graph contains a bunch
of vertical segments?)
>Robert Israel isr...@math.ubc.ca
>Department of Mathematics http://www.math.ubc.ca/~israel
>University of British Columbia Vancouver, BC, Canada
************************
David C. Ullrich
Dave Seaman
I suspect that's where the difference lies. The function I mean by
"Cantor function" is the one described in Royden (exercise 46 on p. 48 of
the 2nd ed.). That function has the properties f(0) = 0, f(1) = 1, f is
monotone, and f is constant on each interval contained in the complement
of the Cantor set. Therefore f' is zero wherever it is defined.
Dave Seaman
> I suspect that's where the difference lies. The function I mean by
> "Cantor function" is the one described in Royden (exercise 46 on p. 48 of
> the 2nd ed.). That function has the properties f(0) = 0, f(1) = 1, f is
> monotone, and f is constant on each interval contained in the complement
> of the Cantor set. Therefore f' is zero wherever it is defined.
My "therefore" in that last statement is probably unjustified, since I
have not adequately characterized the Cantor function. Perhaps it will
help if I add that f(x) = 1/2 for x in (1/3,2/3), f(x) = 1/4 for x in
(1/9,2/9), f(x) = 3/4 for x in (7/9,8/9), etc.
I haven't actually proved that this function is differentiable only on
the complement of the Cantor set, but that's what my comments were based
on.
David C. Ullrich
<dse...@no.such.host> wrote:
Hmm, actually I guess you're right - I had sort of a different
notion of derivative in mind, not the one we're actually talking
about here.
(Loosely speaking, "has a bounded derivative" is identified
in my mind with "satisfies a Lip_1 condition". The two are
equivalent if we're talking about the distribution derivative;
that's a bounded function iff f is Lipschitz, and the Cantor
function is certainly not Lipschitz. So saying it has a
bounded derivative sounds immediately wrong to me, but no,
in the sense we're talking about here it may well be that
the derivative is 0 everywhere it's defined.)
************************
David C. Ullrich
Betelgeuse
"David C. Ullrich" wrote:
>
> >What is true is that such an f cannot be absolutely continuous.
>
> Why not?
oops! you are right. It could as well be AC.
Regards,
Jenny
Robert Israel
Dave Seaman <dse...@no.such.host> wrote:
>I suspect that's where the difference lies. The function I mean by
>"Cantor function" is the one described in Royden (exercise 46 on p. 48 of
>the 2nd ed.). That function has the properties f(0) = 0, f(1) = 1, f is
>monotone, and f is constant on each interval contained in the complement
>of the Cantor set. Therefore f' is zero wherever it is defined.
I don't think the "therefore" is at all obvious, though it may be true
for the case of the Cantor set.
It's certainly false for some functions with the same properties except
that the Cantor set is replaced by another set that is homeomorphic to it.
Thus there are increasing homeomorphisms g and h of [0,1] to itself such
that g(f(h(x))) has a nonzero derivative at, say, x = 1/2, because
we can make, say, x <= g(f(h(x))) <= x + (x-1/2)^2.
Dave Seaman
> In article <d01q9n$kvs$1...@mailhub227.itcs.purdue.edu>,
> Dave Seaman <dse...@no.such.host> wrote:
>>I suspect that's where the difference lies. The function I mean by
>>"Cantor function" is the one described in Royden (exercise 46 on p. 48 of
>>the 2nd ed.). That function has the properties f(0) = 0, f(1) = 1, f is
>>monotone, and f is constant on each interval contained in the complement
>>of the Cantor set. Therefore f' is zero wherever it is defined.
> I don't think the "therefore" is at all obvious, though it may be true
> for the case of the Cantor set.
Agreed, as I stated elsewhere. That was a careless use of "therefore".
> It's certainly false for some functions with the same properties except
> that the Cantor set is replaced by another set that is homeomorphic to it.
> Thus there are increasing homeomorphisms g and h of [0,1] to itself such
> that g(f(h(x))) has a nonzero derivative at, say, x = 1/2, because
> we can make, say, x <= g(f(h(x))) <= x + (x-1/2)^2.
Right. I was basically saying that in the case of the standard Cantor
function, (a) the derivative is zero on the complement of the Cantor set,
and (b) the derivative is undefined on the Cantor set itself. I am
certain of (a), and I think (b) is also correct.
Robert Israel
Dave Seaman <dse...@no.such.host> wrote:
>On 2 Mar 2005 00:48:20 GMT, Robert Israel wrote:
>> In article <d01q9n$kvs$1...@mailhub227.itcs.purdue.edu>,
>> Dave Seaman <dse...@no.such.host> wrote:
>Right. I was basically saying that in the case of the standard Cantor
>function, (a) the derivative is zero on the complement of the Cantor set,
>and (b) the derivative is undefined on the Cantor set itself. I am
>certain of (a), and I think (b) is also correct.
Yes. Note that if f is the standard Cantor function, for any point
e in the Cantor set and any positive integer n, there are b,c with
b <= e <= c and c-b = 3^(-n) such that f(c) - f(b) = 2^(-n), so
that (f(c)-f(b))/(c-b) = (3/2)^n. Thus there is some y with
0 < |y-e| <= 3^(-n) and |f(y) - f(e)|/|y-e| >= (3/2)^n (in fact,
y can be one of b and c).
AGUIRRE ESTIBALEZ Julian
> In article <0ec721pqqufsrsh6t...@4ax.com>,
> David C. Ullrich <ull...@math.okstate.edu> wrote:
> >On Mon, 28 Feb 2005 22:42:09 GMT, Betelgeuse
> ><betel...@rigel.bellatrix> wrote:
>
> >>Actually, f' need not be continuous. Otherwise it is trivial.
> >>My feeling is that there is an example of a nonconstant
> >>function satisfying the required conditions, but the example
> >>(provided that it exists) is far from being easy (at least, for me).
f' need not be continuos, but satisfies the intermediate values condition,
that is, if f' takes two different values, it takes all values in between
(if I remember correctly, there is a proof of this fact in Rudin's
Principles of Mathematical Analysis).
This, together with the fact that an open set in R is the countable
union of disjoint intervals, proves that f must in fact be constant.
Proof: The open set where f'=0 is the union of open intervals (a_n,b_n),
that we assume ordered so that b_n<=a_{n+1}. Since it is dense in the
interval where f is defined, it must be that b_n=a_{n+1}. Thus, f' has at
most countably many discontinuities, ant therefore it takes at most
countably many different values. Lets say that f'(x)>0 for some x. Then f'
must take all the values in the interval [0,f'(x)], wich is uncountable.
Julian Aguirre
Universidad del Pais Vasco
Daniel Grubb
>> Not generally true. There are differentiable functions with the property that
>> {x:f'(x)=0}, {x:f'(x)<0}, and {x:f'(x)>0} are *all* dense. Furthermore,
>> the derivative can be chosen to be a bounded function.
>>
>> --Dan Grubb
>That's something I'd like to see. Can you post a f'rinstance?
There is an example in Stromberg's 'An Introduction to Classical Real
Analysis'. I think that Ullrich and Israel have been discussing it.
--Dan Grubb
Daniel Grubb
>>Right. I was basically saying that in the case of the standard Cantor
>>function, (a) the derivative is zero on the complement of the Cantor set,
>>and (b) the derivative is undefined on the Cantor set itself. I am
>>certain of (a), and I think (b) is also correct.
>Yes. Note that if f is the standard Cantor function, for any point
>e in the Cantor set and any positive integer n, there are b,c with
>b <= e <= c and c-b = 3^(-n) such that f(c) - f(b) = 2^(-n), so
>that (f(c)-f(b))/(c-b) = (3/2)^n. Thus there is some y with
>0 < |y-e| <= 3^(-n) and |f(y) - f(e)|/|y-e| >= (3/2)^n (in fact,
>y can be one of b and c).
Hmmm... If we take an evenly divided Cantor set with strictly
positive measure, this argument fails. Let the proportionality factors
at each stage be a_1, a_2,.... For such a set, the corresponding
function would have b and c with c-b=a_1a_2...a_n and f(c)-f(b)=1/2^n,
so the difference quotient is 1/(2^n a_1 a_2...a_n) which goes to
{\lambda(C)}^(-1). This function will also have derivative 0 on a dense
open set.
Unfortunately, this argument shows that the function is definitely not
differentiable at the endpoints since the one-sided derivatives are 0
and {\lambda(C)}^(-1). It's not clear whether the one-sided derivatives
exist other places. At the very least, this example makes me more confident
a counterexample exists for the OP.
--Dan Grubb
Daniel Grubb
>Hmmm... If we take an evenly divided Cantor set with strictly
>positive measure, this argument fails. Let the proportionality factors
>at each stage be a_1, a_2,.... For such a set, the corresponding
>function would have b and c with c-b=a_1a_2...a_n and f(c)-f(b)=1/2^n,
>so the difference quotient is 1/(2^n a_1 a_2...a_n) which goes to
>{\lambda(C)}^(-1). This function will also have derivative 0 on a dense
>open set.
Actually, the derivative exists a.e. and is the characteristic function
of C divided by the measure of C. Now, if there was a way of moderating
the one-sided difference quotients to get 0 at the endpoints we might
have an example.
--Dan Grubb
Dave Seaman
On 2 Mar 2005 09:39:22 -0800, scattered wrote:
> AGUIRRE ESTIBALEZ Julian wrote:
> (snip)
>> Proof: The open set where f'=0 is the union of open intervals
> (a_n,b_n),
>> that we assume ordered so that b_n<=a_{n+1}.
> (snip)
> Why assume you can do that? You can't if you have say an open covering
> of the rationals in (0,1) of total length 1/2. The complement of an
> open dense set need not be countable.
The difference is that you can't do that with disjoint intervals.
> The problem is that, given (a_n,b_n), b_n might be a cluster point of
> the a_k, so that there is no "next" interval
Yes, that can happen if the intervals are open. It doesn't happen if the
intervals are half-open to the right.
AGUIRRE ESTIBALEZ Julian
> (Please ignore my other message if it appears.)
>
> On 2 Mar 2005 09:39:22 -0800, scattered wrote:
>
> > AGUIRRE ESTIBALEZ Julian wrote:
> > (snip)
> >> Proof: The open set where f'=0 is the union of open intervals
> > (a_n,b_n),
> >> that we assume ordered so that b_n<=a_{n+1}.
> > (snip)
>
> > Why assume you can do that? You can't if you have say an open covering
> > of the rationals in (0,1) of total length 1/2. The complement of an
> > open dense set need not be countable.
I had forgotten about this example.
> The difference is that you can't do that with disjoint intervals.
>
> > The problem is that, given (a_n,b_n), b_n might be a cluster point of
> > the a_k, so that there is no "next" interval
I had allready seen this posibility, but (wrongly) thougt that it did not
affect my argument. I do not remember who said "the prestige of a
mathematician is proportional to the number of wong proofs she has given".
> Yes, that can happen if the intervals are open. It doesn't happen if the
> intervals are half-open to the right.
> Dave Seaman
> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
> <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Julian Aguirre
Universidad del Pais Vasco
Dave Seaman
> Dave Seaman wrote:
>> The difference is that you can't do that with disjoint intervals.
> Sure you can. The rationals in (0,1) can be covered by an open set of
> measure 1/2, as an open set it will be a coutable union of disjoint
> open intervals. Am I missing something?
You're right. Sorry.
David C. Ullrich
wrote:
Something like this seems like an obvious approach - at least it's the
first thing I think of. Never quite saw how to fix it up to be
differentiable at every point, though (I don't think that the
endpoints are the only problem...)
>--Dan Grubb
************************
David C. Ullrich
Gerry Myerson
AGUIRRE ESTIBALEZ Julian <mtpa...@lg.ehu.es> wrote:
> I had allready seen this posibility, but (wrongly) thougt that it did not
> affect my argument. I do not remember who said "the prestige of a
> mathematician is proportional to the number of wong proofs she has given".
It seems that in A Mathematician's Miscellany, Littlewood quotes
Besicovitch to the effect that "A mathematician's reputation rests on
the number of bad proofs he has given," but I don't know where in the
book it might be.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Robert B. Israel
> On 1 Mar 2005 00:40:31 GMT, isr...@math.ubc.ca (Robert Israel) wrote:
>
> >In article <0ec721pqqufsrsh6t...@4ax.com>,
> >David C. Ullrich <ull...@math.okstate.edu> wrote:
> >>On Mon, 28 Feb 2005 22:42:09 GMT, Betelgeuse
> >><betel...@rigel.bellatrix> wrote:
> >
> >>>Actually, f' need not be continuous. Otherwise it is trivial.
> >>>My feeling is that there is an example of a nonconstant
> >>>function satisfying the required conditions, but the example
> >>>(provided that it exists) is far from being easy (at least, for me).
> >
> >>That's the way it seems to me, too.
> >
> >And in your article of 30 Aug 2002 in the thread "Increasing function
> >whose derivative is densely often equal to 0"
> ><http://groups-beta.google.com/group/sci.math/msg/bd4d9edf09f098bb>
> >you pointed to an example in Stromberg "An Introduction to Classical
> >Real Analysis" of a non-constant differentiable function whose derivative
> >is 0 on a dense set. Unfortunately not on a dense _open_ set (and I
> >don't see how to patch Stromberg's example to make it so).
>
> Simple<g>: Take a dense sequence (a_n) where g' = 0, and
> define f by sort of sliding bits of the graph around so
> that each a_n gets expanded to an open interval on which
> f is constant. Of course then we have to show that f is
> differentiable at the _other_ points...
Aye, there's the rub. Too much sliding around near
points where the derivative is nonzero is likely to lead to
trouble. Suppose you have a point b and some epsilon > 0
such that for some sequence of positive numbers delta_n -> 0,
(b-delta_n,b+delta_n) contains an interval of length at least
epsilon*delta_n on which f' = 0. Then either f'(b) = 0 or
f is not differentiable at b.
> (But seriously: I wasn't going to look up that example,
> but evidently you have. Iirc it proceeds by first
> constructing the inverse function, making sure it
> has infinite derivative at a lot of places. Have you
> tried doing the same thing except with an inverse
> non-function, something whose graph contains a bunch
> of vertical segments?)
It looked complicated to me - I was hoping you'd do it. :)
David Bernier
An open dense set has positive Lebesgue measure. If the condition
on the set where f' is zero is relaxed to just "positive Lebesgue
measure", then I'm 80% sure that there exists
an f: [0,1] -> [0,1] which is continuous,
strictly increasing, bijective, differentiable
and where the set where f' is 0 has positive
Lebesgue measure.
David Bernier
Robert Israel
David Bernier <davi...@videotron.ca> wrote:
>An open dense set has positive Lebesgue measure. If the condition
>on the set where f' is zero is relaxed to just "positive Lebesgue
>measure", then I'm 80% sure that there exists
>an f: [0,1] -> [0,1] which is continuous,
>strictly increasing, bijective, differentiable
>and where the set where f' is 0 has positive
>Lebesgue measure.
Let E be any nowhere-dense perfect subset of [0,1], {V_n = (a_n, b_n)} an
enumeration of the open intervals forming its complement.
Let g_n be a C^infinity function that is 0 on [0,a_n], 1 on [b_n, 1] and
strictly increasing on V_n.
Take f = sum_{n=1}^infinity a_n g_n, where all
a_n > 0. If a_n decreases rapidly enough, the series can be made
to converge in C^k[0,1] for all k: e.g. take
0 < a_n < 2^{-n}/max{||g_n^(j)||_infinity, 0 <= j <= n}.
Then f is C^infinity, strictly increasing, and all its derivatives are
0 on E.
David C. Ullrich
wrote:
Ok, had some free time Friday to look it up, and as of
this morning I've convinced myself that there's no
straightforward generalization of the argument that's
going to work here - if it's possible to make the
argument work one of the key lemmas will have to be
replaced by something much more subtle regarding
rates of convergence blah blah to get differentiability.
If f : R -> R is increasing, (ie non-decreasing) let's
say
Df(x) = lim_{h -> 0} (f(x+h) - f(x))/h,
where +infinity is allowed as a limit, and say
f is "i-diff" at x if Df(x) exists.
Say
Mf(x) = sup_{h <> 0} (f(x+h) - f(x))/h.
The key lemma, not explicitly stated in the proof
in the book but what really makes it work, is this:
Key Lemma: Suppose that each f_n is increasing and
is i-diff everywhere. Suppose that
f(x) = sum f_n(x)
converges (to something finite) everywhere. Then
(i) if sum Df_n(x) = infinity then Df(x) = infinity
(ii) if sum Mf_n(x) < infinity then Df(x) = sum Df_n(x).
(Proof: (i) is clear since all the f_n are increasing.
For (ii), let epsilon > 0 and choose N so that
sum{n>N} Mf_n(x) < epsilon...)
The hypothesis sum Mf_n(x) < infinity in (ii) is
crucially important - it's what lets us say the
derivative of the sum is the sum of the derivatives.
Now the key lemma is applied in the book like so:
If phi(x) = x^(1/3) then it's actually true that
(*) M phi <= c D phi
everywhere. So we let f be a sum of phi_n = constants times
translates and dilates of phi, and it follows that
Df = sum D phi_n everywhere, because sum M phi_n is finite
at every point where sum D phi_n is finite.
We wouldn't a priori _need_ (*) here, but we do need
sum M phi_n < infinity at every point where sum D phi_n
is finite, and this is not going to be true for any
obvious reason if phi_n is a random dilate/translate
of phi, _unless_ (*) holds (if c_n -> infinity then
there exists a_n > 0 with sum a_n < infinity but
sum c_n a_n = infinity, and we have to rule that
out somehow, or replace the Key Lemma with something
else.) For a little while I thought we didn't need
(*), just something else to ensure that sum M phi_n
is finite at every point where sum D phi_n is finite,
but there is no such condition (as long as we're just
talking about "random" translates and dilates...)
(*) is much more essential than I realized at first.
Now say we try to modify the argument to get something
with derivative vanishing on a dense open set. We
use an argument as above, except that phi is not
quite a function, it's an increasing function
except its graph contains a vertical segment.
Never mind precise definitions, since this is not
going to work anyway:
Say the graph of phi contains the segment from
(0,0) to (0,1). Then for small positive x we're
going to have M phi(x) >= 1/x. So if (*) holds
we have D phi(x) >= c/x, which is impossible since
int_0^1 1/x = infinity.
************************
David C. Ullrich
David C. Ullrich
Sure, that's much easier.
>David Bernier
************************
David C. Ullrich