Dirichlet's nowhere differentiable function
woodland
Here's Dirichlet's function f:(0,1)-->(0,1) defined by
f(x) = 1/q , if x=p/q in lowest terms with p,q in N
= 0 , if x irrational
I have to prove that this is differentiable nowhere in (0,1).
I know one way. Let 'c' be in (0,1). Then I have to come
up with two sequences x_n and t_n in (0,1)\{c} that converge
to 'c' but their corresponding difference quotients approach
two different values. But I have difficulty coming up with
such innovative sequences. Can you give some ideas?
regards
wood
G.E. Ivey
If c is irrational, so f(c)= 0, let {a_n} be a sequence of rational numbers converging to c. Then the sequence of difference quotients is (f(a_n)- 0)/a_n= 0.
Let {b_n} be a sequence of rational numbers converging to 0, with b_n= p_n/q_n reduced to lowest terms. Then
(f(b_n)-0)/b_n= (p_n/q_n)/(1/q_n)= p_n. What can you say about the limit of that sequence?
David C. Ullrich
wrote:
If c is rational then f is not even continuous at c.
So suppose that c is irrational. A sequence of irrationals
tending to c shows that if f'(c) exists it must be 0.
But suppose that p is a large prime, and choose q so
that q/p is as close as possible to c. Then it follows
that |q/p - c| <= ___, and hence that |f(q/p) - f(c)|
>= ___.
>regards
>wood
************************
David C. Ullrich
David C. Ullrich
<georg...@gallaudet.edu> wrote:
>First, you might note that lim(x->c) f(x)= 0 for all c so this function (more commonly called the "modified" Dirichlet function) is continuous only for c irrational- f is obviously not differentiable for c rational.
>
> If c is irrational, so f(c)= 0, let {a_n} be a sequence of rational numbers converging to c. Then the sequence of difference quotients is (f(a_n)- 0)/a_n= 0.
Actually (f(a_n)- 0)/(a_n-c) = 0.
> Let {b_n} be a sequence of rational numbers converging to 0,
Typo for "converging to c"? We've already established that f is
discontinuous at 0...
>with b_n= p_n/q_n reduced to lowest terms. Then
>(f(b_n)-0)/b_n= (p_n/q_n)/(1/q_n)= p_n. What can you say about the limit of that sequence?
Actually (f(b_n)-0)/b_n = 1/p_n, not that I see how that's relevant.
You seem to be suggesting that if c is irrational and b_n is _any_
sequence of rationals tending to c then (f(b_n) - f(c))/(b_n-c) fails
to approach 0. This is not true, there exist such sequences for which
the quotient does tend to 0. There also exist such sequences for which
the quptient does not tend to 0, but they need to be chosen a little
more carefully.
************************
David C. Ullrich
Rob Johnson
Indeed. Given an irrational c, let p_n be the n^th prime, and define
2 2
a = [c p ] + 1 + p if gcd([c p ] + 1, p ) = 1
n n n n n
2
= [c p ] + 2 + p otherwise
n n
2
b = p
n n
Then,
a_n f(a_n/b_n) - f(c) 1/p_n^2 1
lim --- = c and ----------------- < ------- = ---
n->oo b_n a_n/b_n - c 1/p_n p_n
Thus, {a_n/b_n} is a sequence of rationals converging to c, with the
differnce quotient tending to 0.
On the other hand, given an irrational c, it is a simple corollary of
the Lemma at <http://www.whim.org/nebula/math/approx.html> that there
are integer sequences {p_n} and {q_n}, with q_n increasing, p_n and q_n
coprime, and |c - p_n/q_n| < 1/q_n^2. This yields a rational sequence,
{p_n/q_n}, converging to c for which the difference quotient is bigger
than q_n.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
julia...@gmail.com
irrational, and f(x) = 1/q^k if x=p/q, with (p,q)=1. For what values
of k is f differentiable at some points? If we want f to be
differentiable at a given irrational x, how does the value of k that we
need depend on x?
Robert Low
You do come up with some interesting and instructive puzzles for us.
Where do you get them from?
Dave L. Renfro
>> Here's another question: First fix k>0. Then,
>> define f(x) = 0 if x is irrational, and f(x) = 1/q^k
>> if x=p/q, with (p,q)=1. For what values of k is
>> f differentiable at some points? If we want f to be
>> differentiable at a given irrational x, how does the
>> value of k that we need depend on x?
Robert Low wrote:
> You do come up with some interesting and instructive
> puzzles for us. Where do you get them from?
I don't know where the previous poster saw this,
but it's fairly well known. At one time (before the
mid 1960's) it was less well known and re-discovered
several times. Below is a chronological list of the
literature on this issue that I've come across. I'm
not including textbooks because I haven't tried making
a systematic search for this particular topic, aside
from collecting the papers I've encountered about it
into one notebook binder.
Note: Because I ran out of time, I didn't say anything
about the last few papers. There also might be some
that I have cited elsewhere in a manuscript that I
was working on about a year ago, but I don't have
time to check now. I also didn't get a chance to
proof-read what I wrote below. (This note was written
just before posting this, as I was rushing out the
door to get to work on time!)
[1] W. D. A. Westfall, "A class of functions having
a peculiar discontinuity", Bulletin of the American
Mathematical Society 15 (1908-09), 225-226.
[Submission date: 12 December 1908]
Westfall considers Ruler-like functions of the
form f(p/q) /= 0 when p and q are co-prime,
f(x) = 0 for x irrational, and the limit as
q --> oo of f(p/q) is zero. As examples, he
gives f(p/q) = (1/q)^n, 1/(q!), and (1/q)^q.
In a footnote he says the (1/q)^n example was
used by Osgood in class lectures, but Westfall
doesn't say what Osgood knew and/or proved about
it. Westfall uses the Liouville algebraic
n'th order irrationality result to show that
their exist irrational points of non-differentiability
in every interval.
[2] Franz Lukács, "Eine unstetige und differenzierbare
Fundtion", Mathematische Annalen 70 (1911), 561-562.
[Submission date: 1 November 1910]
Appears to prove the same thing that Westfall did
(in the generality that Westfall did, with the
specific example f(p/q) = (1/q)^q being given).
[3] T. Hayashi, "Eine stetige und nicht-differenzierbare
function", Tôhoku Mathematical Journal 1 (1911-12),
140-142. [No submission date given.]
Mentions Lukács' paper. I'm not sure if Hayashi
is filling in some gaps from Lukács or extending
the results from Lukács in some way. The paper is
in German, which I can't read. (Lukács' paper is
also in German, but in that case it was easy to
figure out what Lukács was doing. In this case,
since Hayashi already knows of Lukács' paper,
the issue of what Hayashi is doing is not as
immediately apparent to me.)
[4] Bohus Jurek, "Sur la dérivabilité des fonctions à
variation bornée", Casopis Pro Pestování Matematiky
65 (1935), 8-27.
It appears that he proves some general results
concerning the zero Hausdorff h-measure of
sets of non-differentiability for bounded
variation functions such that the sum of the
h-values of the countably many jump discontinuities
is finite (special case: h(t) = t^a for a fixed
0 < a < 1). General "h-versions" of the ruler
function seem to appear as examples, and V. Jarnik's
more precise results about the Hausdorff dimension
of Liouville-like Diophantine approximation results
is used.
[5] J. W. Gaddum, "Solution to Monthly Problem E 935",
American Mathematical Monthly 58 (1951), 340.
C. D. Olds posed the problem of giving an example
of a function which is discontinuous in an every-
where dense set that is also differentiable in an
everywhere dense set. The editors note that several
solutions offered the ruler version f(p/q) = (1/q)^q,
along with the observation (presumably from Liouville's
approximation theorem, but this isn't actually stated
in the published text) that this function is not
differentiable at every algebraic irrational.
(Gaddum's solution is more elementary.)
[6] Simon Cetkovic, "Sur la différentiabilité des deux
familles des fonctions réelles", Bull. Soc. Math.
Phys. Serbie 4 (1952), 53-57.
MR 14,855b (E. Hewitt): "Let A and B be complementary
sets of natural numbres, each containing an infinite
number of primes. Let S be the set of real numbers
which can be written in the form p/q, where p is an
integer, q in B, andpa and q are relatively prime.
Let F(x) = 0 on S' and let F(p/q) = q^(-alpha) for
p/q in S, alpha being an arbitrary positive real
number. Then, if 0 < alpha <= 1, F is nowhere
differentiable but is continuous on an everywhere
dense set; if 1 < alpha, then F is discontinuous
on a dense set but is also differentiable on a
dense set."
[7] Simon Cetkovic, "La relation entre l'ordre des
nombres algébriques et la différenciabilité d'une
famille des fonctions", Bull. Soc. Math. Phys.
Serbie 5 (1953), 91-92.
MR 15,693d (no reviewer cited): "Après avoir défini
une famille de fonctions de la manière suivante:
F(x,b) = 0, si x est un nombre irrationel,
F(x,b) = (1/q)^b, si x est un nombre rationnel
de la forme p/q, p et q étant des entiers sans
facteurs communs avec q /= 0 et où b est une
paramètre arbitraire, l'auteur démontre: La
fonction F(x,b) a une dérivée aux tous points
irrationnels et algébriques de l'ordre <= n
pour n < b, quoique elle ne soit pas continue
aux points d'un ensemble de points partout dense."
[8] I. J. Schoenberg, "The integrability of certain
functions and related summability methods", American
Mathematical Monthly 66 (1959), 361-375.
After observing that the usual ruler function is
Riemann integrable (it has countably many points
of discontinuity), Schoenberg proves the following
theorem: Let {b_n} be a sequence of real numbers
and define the ruler-like function f by putting
f(p/q) = b_q. Then f is Riemann integrable on
[0,1] if and only if b_n --> 0 as n --> oo.
Schoenberg does this in the first two pages.
There is quite a bit more which is too technical
for me to easily describe now.
[9] I. J. Schoenberg, "The integrability of certain
functions and related summability methods II",
American Mathematical Monthly 66 (1959), 562-563.
Some follow-up literature results that has come
to his attention, including the paper by Lukács
(but none of the others listed above).
[10] Solomon Marcus, "Sur les propriétés différentielles
des fonctions dont les points de continuité
forment un ensemble frontiére partout dense",
Annales Scientifiques de l'École Normale Supérieure
(3) 79 (1962), 1-21.
A very thorough survey of this topic making use of
Lebesgue measure and Baire category to describe
the size of the exceptional sets involved. Many
variations of the ruler function are extensively
analyzed, and most of the references listed above
are cited (but not Jurek or Westfall).
[11] Gerald J. Porter, "On the differentiability of a
certain well-known function", American Mathematical
Monthly 69 (1962), 142.
A short proof using decimal expansion ideas that
the usual ruler function is not differentiable
at each point.
[12] G. A. Heuer, "Functions continuous at irrationals
and discontinuous at rationals", abstract of talk
given on 2 November 1963 at North Dakota State
University, American Mathematical Monthly 71
(1964), 349.
Two results are stated, which were published later
by Heuer.
[13] M. D. Mavindurve, "Solution to Monthly Problem 5181",
American Mathematical Monthly 72 (1965), 326-327.
Solves this problem posed by E. J. Burr: Let f
be ruler-like, with f(p/q) = b_q, where {b_n}
is a fixed sequence of real numbers. Find, or
prove the non-existence of, sequences {b_n} with
b_n --> 0 such that (a) f' exists nowhere,
(b) f' exists for some x, (c) f' exists for all
irrational x.
[14] G. A. Heuer, "Functions continuous at the irrationals
and discontinuous at the rationals", American
Mathematical Monthly 72 (1965), 370-373.
[15] G. A. Heuer, "A property of functions discontinuous
on a dense set", American Mathematical Monthly
73 (1966), 378-379.
[16] E. M. Beesley, "Lipschitzian points", American
Mathematical Monthly 79 (1972), 603-608.
[17] Alec Norton, "Continued fractions and differentiability
of functions", American Mathematical Monthly 95
(1988), 639-643.
[18] Richard Darst and Gerald Taylor, ???, American
Mathematical Monthly 103 (1996), 415-416.
[See also notes by the editor in 104 (1997),
p. 92.]
[19] Marc Frantz, "Two functions whose powers make
fractals", American Mathematical Monthly 105
(1998), 609-617. [See also "Editor's Endnotes
in 106 (1999), p. 284.]
[20] William Dunham, "Nondifferentiability of the
ruler function", Mathematics Magazine 76 (2003),
140-142.
Dave L. Renfro
David C. Ullrich
wrote:
>In article <iid1n1hs5o8tmveqt...@4ax.com>,
>David C. Ullrich <ull...@math.okstate.edu> wrote:
>>On Tue, 08 Nov 2005 08:47:35 EST, "G.E. Ivey"
>><georg...@gallaudet.edu> wrote:
>>
>>>[...]
Which gives more than the problem asks for, in fact the "upper
derivative" is infinite. We should maybe note that the original
problem is much easier than this, only requires |c - p_n/q_n|
< 1/q_n.
>Rob Johnson <r...@trash.whim.org>
>take out the trash before replying
************************
David C. Ullrich
David C. Ullrich
>Here's another question: First fix k>0. Then, define f(x) = 0 if x is
>irrational, and f(x) = 1/q^k if x=p/q, with (p,q)=1. For what values
>of k is f differentiable at some points?
If k <= 2 then the argument that Rob Johnson gave shows that
it's nowhere differentiable (if k < 2 then in fact the
"upper derivative" is infinite everywhere; not sure about
that if k = 2.) If k > 2 it's not hard to show that f is
differentiable almost everywhere.
>If we want f to be
>differentiable at a given irrational x, how does the value of k that we
>need depend on x?
************************
David C. Ullrich
Rob Johnson
I believe that for any k > 2, f is differentiable at almost all x.
However, for any k, there are some x at which f is not differentiable.
In fact, there are some x for which f is not differentiable at x for
any k. This depends on the sizes of the continuants of the continued
fraction for x.
Rob Johnson
That's true, and a little simpler to show, but it's nice to show the
sharper inequality. The 1/q_n^2 inequality shows that there are
sequences of rationals approaching any irrational with arbitrarily
small or large difference quotients. It also works up to the limit
of the exponents (2) in the problem, and that turned out to be useful.
David C. Ullrich
wrote:
>In article <1131522416.7...@g43g2000cwa.googlegroups.com>,
>julia...@gmail.com wrote:
>>Here's another question: First fix k>0. Then, define f(x) = 0 if x is
>>irrational, and f(x) = 1/q^k if x=p/q, with (p,q)=1. For what values
>>of k is f differentiable at some points? If we want f to be
>>differentiable at a given irrational x, how does the value of k that we
>>need depend on x?
>
>I believe that for any k > 2, f is differentiable at almost all x.
Hint: If E_q is a subset of [0,1] and the sum of the measure of
E_q is finite then almost every point lies in only finitely
many E_q.
Given eps > 0, if you choose E_q right this shows that the lim sup
of the difference quotients at x is < eps for almost all x.
>However, for any k, there are some x at which f is not differentiable.
>In fact, there are some x for which f is not differentiable at x for
>any k. This depends on the sizes of the continuants of the continued
>fraction for x.
>
>Rob Johnson <r...@trash.whim.org>
>take out the trash before replying
David C. Ullrich
Rob Johnson
Yes, that's the way it goes. Let E_q be the intervals around each p/q
of radius 1/q^{k/2+1}. The sum of the measures of the E_q is finite
(there are phi(q) < q fractions in [0,1] with q in the denominator, so
|E_q| < 2/q^{k/2}). Furthermore, the difference quotient between an x
outside E_q and p/q is less than 1/q^{k/2-1}.
As you pointed out, almost all x in [0,1] are in at most finitely many
E_q. Suppose x is in finitely many E_q. For any eps > 0, there is a
del > 0 so that for any p/q with |x-p/q| < del, 1/q^{k/2-1} < eps.
This means that the limit of the difference quotients for any rational
sequence converging to x is 0.
David C. Ullrich
wrote:
>In article <43724394....@text.giganews.com>,
>ull...@math.okstate.edu (David C. Ullrich) wrote:
>>On Wed, 09 Nov 2005 17:20:14 GMT, r...@trash.whim.org (Rob Johnson)
>>wrote:
>>
>>>In article <1131522416.7...@g43g2000cwa.googlegroups.com>,
>>>julia...@gmail.com wrote:
>>>>Here's another question: First fix k>0. Then, define f(x) = 0 if x is
>>>>irrational, and f(x) = 1/q^k if x=p/q, with (p,q)=1. For what values
>>>>of k is f differentiable at some points? If we want f to be
>>>>differentiable at a given irrational x, how does the value of k that we
>>>>need depend on x?
>>>
>>>I believe that for any k > 2, f is differentiable at almost all x.
>>
>>Hint: If E_q is a subset of [0,1] and the sum of the measure of
>>E_q is finite then almost every point lies in only finitely
>>many E_q.
>>
>>Given eps > 0, if you choose E_q right this shows that the lim sup
>>of the difference quotients at x is < eps for almost all x.
>>
>>>However, for any k, there are some x at which f is not differentiable.
>>>In fact, there are some x for which f is not differentiable at x for
>>>any k. This depends on the sizes of the continuants of the continued
>>>fraction for x.
>
>Yes, that's the way it goes. Let E_q be the intervals around each p/q
>of radius 1/q^{k/2+1}.
Huh, I guess that works too. What I had in mind was intervals of
radius 1/(eps * q^k).
>The sum of the measures of the E_q is finite
>(there are phi(q) < q fractions in [0,1] with q in the denominator, so
>|E_q| < 2/q^{k/2}). Furthermore, the difference quotient between an x
>outside E_q and p/q is less than 1/q^{k/2-1}.
>
>As you pointed out, almost all x in [0,1] are in at most finitely many
>E_q. Suppose x is in finitely many E_q. For any eps > 0, there is a
>del > 0 so that for any p/q with |x-p/q| < del, 1/q^{k/2-1} < eps.
>This means that the limit of the difference quotients for any rational
>sequence converging to x is 0.
>
>Rob Johnson <r...@trash.whim.org>
>take out the trash before replying
************************
David C. Ullrich