A question about continuity of the derivative
Stephen J. Herschkorn
which is differentiable everywhere but continuously differentiable nowhere?
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
Benjamin Lotto
<hers...@rutcor.rutgers.edu> wrote:
> Does there exist a function in R^(0,1) (domain is the open interval)
> which is differentiable everywhere but continuously differentiable nowhere?
Yes, although I can't recall the details.
I'm sure there's an example of this sort in "Counterexamples in
Analysis," which is about to be republished by Dover (if amazon.com is
to be trusted).
There is no easy example, since the derivative has to have the
intermediate value property.
--
(Remove spots to reply.)
Fred Galvin
> Does there exist a function in R^(0,1) (domain is the open
> interval) which is differentiable everywhere but continuously
> differentiable nowhere?
No. Let me quote the relevant facts from Casper Goffman's _Real
Analysis_.
THEOREM 2. [p. 117] If the derivative f'(x) of f(x) exists, then f'(x)
is the limit of a sequence of continuous functions.
Proof. . . . g_n(x) = n{f(x + 1/n) - f(x)} is continuous . . .
THEOREM 7. [p. 110] The set of points of discontinuity of a function
which is the limit of a convergent sequence of continuous functions is
of the first category.
Exercise 2.3 [p. 120] Show that if the derivative of f(x) exists its
set of points of discontinuity is of the first category; and that,
conversely, for every set S of the first category there is a
derivative which is discontinuous at every point of S.
--
It takes steel balls to play pinball.
Kamlesh Parwani
If I'm understanding this question correctly, it asks, is there a
function that has a derivative, but the derivative is continuous
nowhere?
Construct a function that is exactly equal to 1 on a dense set of
positive measure and zero on the complement, which is also dense on a
set of positive measure. Suppose one can actually do this. Call this
function g. Integrate this function to get another continuous
function,monotone increasing, called f.
f(x)=integral of g from 0 to x.
Now f is differentiable because g is it's derivative, which is nowhere
continuous.
So the hard part is to construct g.
Let A be a set that is the union of the rationals, in (0,1), and a fat
cantor set, say of measure 1/2. This is certainly dense because it
has the rationals. It's complement can be described as the
irrationals minus a fat cantor set, which is nowhere dense. So the
complement is also dense and has positive measure.
Let g be defined as 1 on A and 0 on the complement of A.
Let me know if this is works. I'm sure that I have answered some
question--it may not be the one that you were asking.
Terry E
"Kamlesh Parwani" <for...@math.northwestern.edu> wrote in message
news:514a6bf.02121...@posting.google.com...
> Benjamin Lotto <benlot...@hvc.rr.removespot.com> wrote in message
news:<151220021815364162%benlot...@hvc.rr.removespot.com>...
> > In article <3DFD05AD...@rutcor.rutgers.edu>, Stephen J. Herschkorn
> > <hers...@rutcor.rutgers.edu> wrote:
> >
> > > Does there exist a function in R^(0,1) (domain is the open interval)
> > > which is differentiable everywhere but continuously differentiable
nowhere?
> >
> > Yes, although I can't recall the details.
> >
> > I'm sure there's an example of this sort in "Counterexamples in
> > Analysis," which is about to be republished by Dover (if amazon.com is
> > to be trusted).
> >
> > There is no easy example, since the derivative has to have the
> > intermediate value property.
>
> If I'm understanding this question correctly, it asks, is there a
> function that has a derivative, but the derivative is continuous
> nowhere?
>
> Construct a function that is exactly equal to 1 on a dense set of
> positive measure and zero on the complement, which is also dense on a
> set of positive measure. Suppose one can actually do this. Call this
> function g. Integrate this function to get another continuous
> function,monotone increasing, called f.
>
> f(x)=integral of g from 0 to x.
>
for it to be Lebesgue intergable?
Fred Galvin
> Benjamin Lotto <benlot...@hvc.rr.removespot.com> wrote in message
> news:<151220021815364162%benlot...@hvc.rr.removespot.com>...
> > In article <3DFD05AD...@rutcor.rutgers.edu>, Stephen J. Herschkorn
> > <hers...@rutcor.rutgers.edu> wrote:
> >
> > > Does there exist a function in R^(0,1) (domain is the open interval)
> > > which is differentiable everywhere but continuously
> > >differentiable nowhere?
> >
> > Yes, although I can't recall the details.
No, there is no such function. This question was answered in my
earlier posting.
> > I'm sure there's an example of this sort in "Counterexamples in
> > Analysis," which is about to be republished by Dover (if amazon.com is
> > to be trusted).
> >
> > There is no easy example, since the derivative has to have the
> > intermediate value property.
>
> If I'm understanding this question correctly, it asks, is there a
> function that has a derivative, but the derivative is continuous
> nowhere?
Yes, that was the question.
> Construct a function that is exactly equal to 1 on a dense set of
> positive measure and zero on the complement, which is also dense on a
> set of positive measure.
Such a function cannot be a derivative, because it does not have the
intermediate value property.
> Suppose one can actually do this.
That's easily done. Define g(x) = 1 if x is either rational and
greater than 1/2 or else irrational and less than 1/2; and define g(x)
= 0 elsewhere. Then g(x) = 1 on a dense set of positive measure, and
g(x) = 0 on a dense set of positive measure. Of course, g(x) is not a
derivative.
Robert Israel
Kamlesh Parwani <for...@math.northwestern.edu> wrote:
>Benjamin Lotto <benlot...@hvc.rr.removespot.com> wrote in message
>news:<151220021815364162%benlot...@hvc.rr.removespot.com>...
>> In article <3DFD05AD...@rutcor.rutgers.edu>, Stephen J. Herschkorn
>> <hers...@rutcor.rutgers.edu> wrote:
>>
>> > Does there exist a function in R^(0,1) (domain is the open interval)
>> > which is differentiable everywhere but continuously differentiable nowhere?
>function that has a derivative, but the derivative is continuous
>nowhere?
Correct. And the answer is no, as Fred Galvin said.
>Construct a function that is exactly equal to 1 on a dense set of
>positive measure and zero on the complement, which is also dense on a
>set of positive measure. Suppose one can actually do this. Call this
>function g. Integrate this function to get another continuous
>function,monotone increasing, called f.
>f(x)=integral of g from 0 to x.
>Now f is differentiable because g is it's derivative, which is nowhere
>continuous.
No, f is not differentiable everywhere, and where it is differentiable
the derivative is not necessarily equal to g.
f'(x) is the limit as h -> 0 (both sides!) of 1/h int_x^{x+h} g(t) dt.
In general there's no reason for this limit to exist everywhere, or
to be equal to g(x) where it does exist. The Lebesgue differentiation
theorem says it is g(x) almost everywhere, but the question asked for
"everywhere".
>So the hard part is to construct g.
>Let A be a set that is the union of the rationals, in (0,1), and a fat
>cantor set, say of measure 1/2. This is certainly dense because it
>has the rationals. It's complement can be described as the
>irrationals minus a fat cantor set, which is nowhere dense. So the
>complement is also dense and has positive measure.
>Let g be defined as 1 on A and 0 on the complement of A.
The fact that g is 1 on the rationals has _no_ effect on the integral
of g on any interval, and so is quite irrelevant. The derivative of
this function will be 0 outside of your "fat cantor set" (and therefore
will be continuous there).
You _can_ construct a set A such that A and its complement have positive
measure in each interval of positive length, using the union of a sequence
of "fat cantor sets". This leads to a function f such that f' = 1 almost
everywhere on A and 0 almost everywhere on its complement, so in
particular f' is nowhere continuous. But, alas, f' does not exist
everywhere.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Miro
> Does there exist a function in R^(0,1) (domain is the open interval)
> which is differentiable everywhere but continuously differentiable nowhere?
Well, some people (mainly from applied sciences) don't understand
"continuous differentiability at x" as having f' continuous at x, but
rather having f'
continuous in some neighborhood of x. Just for sure this comment:
As was answered by Fred Galvin, any function f' that is a (finite)
derivative (everyvhere!) of some (necessarily continuous) function f
has a dense set of continuity points. But f' need not be continuous on
any subinterval, in fact
it can be discontinuous almost everywhere. It is even possible that
f' is positive on a dense set and at the same time f' is negative on
another
dense set. Necessarily, the zero set of f' is then dense as well, by
intermediate value property. In fact, all continuity points of f' then
belong
to the zero set of f' (which can be at the same time of Lebesgue
measure zero).
Moreover, both sets, the one where f' is positive, and the one where
f' is negative, can be of positive measure in every subinterval...
So, even if necessarily f' is continuous at points of a dense set, at
the same
time f' can be highly discontinuous.
Miro
Dave L. Renfro
[sci.math Dec 15 2002 6:50:26:000PM]
http://mathforum.org/discuss/sci.math/m/466167/466167
wrote
> Does there exist a function in R^(0,1) (domain is the open
> interval) which is differentiable everywhere but continuously
> differentiable nowhere?
No. The continuity set of a derivative is dense. In fact, it
has cardinality c in every interval. On the other hand, the
discontinuity set of a derivative can have any of the following
properties --->
1. dense in the reals
2. cardinality c in every interval
3. positive measure (hence, not Riemann integrable)
4. positive measure in every interval
5. full measure in every interval (i.e. measure zero complement)
6. have a Hausdorff dimension zero complement
A sci.math post of mine from a couple of months ago addresses
this topic. A copy of this post is pasted below. This post
also includes a link to another post that contains a large
number of references.
Note that your question in the thread "Measure zero and first
category" at
http://mathforum.org/discuss/sci.math/t/466275
is also taken care of below (in a "big hammer" way).
Dave L. Renfro
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http://mathforum.org/discuss/sci.math/m/451663/453478
Zdislav V. Kovarik <kov...@mcmail.cis.mcmaster.ca>
[sci.math Oct 23 2002 3:20:30:000PM]
http://mathforum.org/discuss/sci.math/m/451663/452331
wrote
> Am I missing something?
> The "everywhere existing derivative" is a pointwise limit of a
> sequence of continuous functions (f(x+1/n)-f(x))/(1/n), with
> slight adjustments at the endpoints.
> Such functions (of "first Baire class") are known to have
> plenty of points of continuity (a set of second Baire category).
Although this matter seems to have been taken care of already
in this thread (I've been away from sci.math a bit lately, working
on other things), those interested in a more complete story of
the continuity set of a derivative will want to see the following
sci.math post of mine --->>>
HISTORICAL ESSAY ON CONTINUITY OF DERIVATIVES [Jan. 23, 2000]
http://mathforum.org/discuss/sci.math/t/251588
Briefly, a subset D of the reals is a discontinuity set for some
bounded derivative if and only if D is an F_sigma first category
set. Moreover, derivatives for which D is large are plentiful in
the sense of Baire category (use the sup norm on the collection
of bounded derivatives) -- For most bounded derivatives, the set D
is the complement of a measure zero set. Note that this is much
stronger than simply saying that D has positive measure (i.e. that
the derivative isn't Riemann integrable). Note also that D and
the complement of D, for any of these Baire-typical bounded
derivatives, gives a partition of the reals into a measure zero set
and a first category set. In 1993 Bernd Kirchheim strengthened
this by proving that given any Hausdorff measure function h,
the set D is the complement of a set that has Hausdorff h-measure
zero for most bounded derivatives. (I seem to have left out
Kirchheim's result in the above essay.)
For those that are interested, note that I've included some
additional information in the references section of this post.
While I'm here, I suppose I may as well plug some of my other
more extensive posts concerning real analysis pathology --->>>
HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS [May 1, 2000]
http://mathforum.org/discuss/sci.math/t/267778
ESSAY ON NON-DIFFERENTIABILITY POINTS OF MONOTONE FUNCTIONS
[Nov. 4, 2000]
http://mathforum.org/discuss/sci.math/t/301615
ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS [May 9 and 20, 2002]
http://mathforum.org/discuss/sci.math/t/410034
Dave L. Renfro
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