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Exotic functions (elementary)

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Dave L. Renfro

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Oct 5, 2006, 3:54:59 PM10/5/06
to
I've recently made a couple of posts to another
group (not usenet) that I thought might be of
interest to some people in sci.math, and so
I've merged those posts together to form this one.

http://mathforum.org/kb/thread.jspa?threadID=1463266

Incidentally, neither of the properties being considered
(unbounded in every interval, dense graph) can occur for
a Baire one function. However, the first example I give
is a Baire two function, and I'm pretty sure it's possible
(use condensation of singularities?) to obtain a Baire two
function whose graph is dense in the plane.


TABLE OF CONTENTS

1. A FUNCTION UNBOUNDED IN EVERY INTERVAL

2. A FUNCTION WHOSE GRAPH IS DENSE IN THE PLANE


-----------------------


1. A FUNCTION UNBOUNDED IN EVERY INTERVAL

I had a little free time, so I thought I'd
discuss a function, related to the ruler function

<http://mathforum.org/kb/thread.jspa?messageID=4594375>,

whose behavior illustrates a possibility that some
might consider impossible for a function to have.

We define f from the reals to the reals as follows.

f(x) = 0 if x is irrational or x=0.

f(p/q) = q if x is rational and x = p/q in lowest
terms and x has the same sign as p (i.e. q > 0).

Then f has the property of being unbounded in
every interval. To see this, consider an arbitrary
interval J. Then, for each integer n > 0 (no matter
how large), there exist infinitely many reduced-form
fractions p/q in J such that p > n. This is easy
to see if you consider "grids" on the number line
formed by marking the points

..., -3/q, -2/q, -1/q, 0, 1/q, 2/q, 3/q, ...,

first for q = M+1, then for q = M+2, then for
q = M+3, and so on. If the interval J is very
short, the first few of these grids might not
produce any points in J, but once q gets large
enough, the grids will start producing points
in J. Of course, only those p/q's where p and q
are relatively prime will matter, but this is
easily taken care of by just considering those
integers q > M such that q is a prime number.

Note that the value of f is bounded at each point.
[All I'm saying is that for each real number r,
|f(r)| < oo, which is a no-brainer.] However, for
each point r, the limiting behavior of f at x=r
is "unbounded". By this, I mean that f is unbounded
in every neighborhood of x=r.

Incidentally, f is unbounded in only one way,
namely from above. However, if we replace
"f(p/q) = q" with "f(p/q) = [(-1)^p] * q",
then I believe we'll get a function whose graph
is unbounded, both from above and from below,
in every interval.

As badly behaved as this function is (it's worse
than being discontinuous at each point, because
a function can be discontinuous at each point and
still be bounded on the entire real line), its
graph is relatively sparse in the plane. For
example, the graph of this function is certainly
a subset of the union of all horizontal lines
having an integer y-intercept, and even this
larger set misses a lot of regions with positive
area in the plane (e.g. the interior of the circle
with center (1, 1/2) of radius 1/2).


-----------------------


2. A FUNCTION WHOSE GRAPH IS DENSE IN THE PLANE

Now I'll show that we can define a function g,
much worse than this, so that the graph of g is
dense in the plane. This means that the interior
of *every* circle in the plane will contain points
belonging to the graph of g.

Let D_1, D_2, D_3, ... be an infinite sequence of
sets of nonzero real numbers such that each pair
of the sets has empty intersection and each of the
sets is a dense subset of the real numbers. See below
for several ways to obtain such a sequence of sets.

Write the rational numbers as a sequence r_1, r_2,
r_3, ... We can do this because the rational numbers
form a countable set. Indeed, there are many explicit
methods for carrying this out -- google phrases such
as "rationals are countable", "list the rationals",
"countability of the rationals", etc.

For each n = 1, 2, 3, ..., let L_n be the horizontal
"line" L_n = { (x, r_n) : x belongs to D_n }.

That is, L_n is the set of all points in the plane
whose x-coordinates are numbers chosen from D_n and
whose y-coordinates are the (fixed) number r_n.

The function g, and here I'm identifying the function
with its graph, consists of all the points in any
of the sets L_n, along with every point (if any) of
the form (x',0), where x' is a real number that doesn't
belong to any of the sets D_1, D_2, D_3, ...

g is function -- It is easy to see that g satisfies
the vertical line test by using the fact that the sets
D_1, D_2, ... are pairwise disjoint and the fact that
none of these sets contains 0. Also, the domain of g
is the set of all real numbers. This is because each
real number is either in one of the "D-sets" (in which
case g contains an ordered pair whose x-coordinate is
that real number) or it isn't in any of the "D-sets"
(in which case the value of g at that real number
is 0).

g is dense in the plane -- It suffices to show that
the (possibly) smaller set consisting of all the
horizontal "lines" L_1, L_2, ... is dense in the
plane. To see this, we need to show that each
rectangle [a,b] x [c,d] in the plane contains
in its interior at least one point belonging to
these "L-sets". Because the rational numbers are
dense, there exists a rational number, call it
r_k (i.e. let k be the index for one such rational
number in our listing of the rational numbers as
r_1, r_2, ...), such that c < r_k < d. Because the
set D_k is dense, there exists a number s in D_k
such that a < s < b. Then (s, r_k) belongs to L_k,
and hence to the union of the horizontal "lines",
and (s, r_k) lies in the interior of the rectangle
[a,b] x [c,d].


HOW TO OBTAIN THE SETS D_1, D_2, D_3, ...

One way is to let D_1 be the set of all nonzero
rational numbers whose reduced-fraction form
has a denominator that is some power of 2,
D_2 be the set of all nonzero rational numbers
whose reduced-fraction form has a denominator
that is some power of 3, ..., D_n be the
set of all nonzero rational numbers whose
reduced-fraction form has a denominator that
is some power of the n'th prime number, ...

A 2'nd way is to let D_1 be the set of all
nonzero rational numbers whose reduced-fraction
form has a prime denominator, D_2 be the set
of all nonzero rational numbers whose reduced-
fraction form has a denominator that is a product
of two prime numbers, ..., D_n be the set of
all nonzero rational numbers whose reduced-fraction
form has a denominator that is a product of n
prime numbers, ...

A 3'rd way is to let
D_1 = { t + sqrt(2): t is a rational number}
D_2 = { t + sqrt(3): t is a rational number}
.
.
.
D_n = { t + sqrt(p_n): t is a rational number},
where p_n is the n'th prime number,
.
.
.

A 4'th way is to let
D_1 = { t*sqrt(2): t is a nonzero rational number}
D_2 = { t*sqrt(3): t is a nonzero rational number}
.
.
.
D_n = { t*sqrt(p_n): t is a nonzero rational number},
where p_n is the n'th prime number,
.
.
.

The 4'th way is probably the easiest to verify the
properties for. To show D_n is dense, let a,b be
two real numbers with a < b. Choose t' to be a
nonzero rational number such that t' lies between
a / sqrt(p_n) and b / sqrt(p_n). [We can do this
because the rational numbers, and hence the nonzero
rational numbers as well, are dense in the reals.]
Then t'*sqrt(p_n) lies between a and b. To show that
the sets are pairwise disjoint, suppose t*sqrt(p)
belongs to one of the sets and t'*sqrt(p') belongs
to another one of the sets. Now it might be the case
that t = t', but since the two numbers are taken
from different "D-sets", we must have p not equal
to p'. Could these two numbers be equal? No, because
if t*sqrt(p) = t'*sqrt(p'), we'd have t/t' = sqrt(p'/p),
a contradiction, since t/t' is a rational number
and sqrt(p'/p) is an irrational number (because
the ratio of two different prime numbers can't be
a perfect square).


-----------------------


Dave L. Renfro

Gerry Myerson

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Oct 5, 2006, 8:53:57 PM10/5/06
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In article <1160078099.4...@i42g2000cwa.googlegroups.com>,

"Dave L. Renfro" <renf...@cmich.edu> wrote:

> 1. A FUNCTION UNBOUNDED IN EVERY INTERVAL
>
> 2. A FUNCTION WHOSE GRAPH IS DENSE IN THE PLANE

There are also examples in Gelbaum & Olmsted, Counterexamples
in Analysis.

A more-or-less natural example of #2 arises in Number Theory,
in the study of the Dedekind sum.

s(h, k) = sum, v = 1 to k - 1, of ((v/k)) ((hv/k)),
where ((x)) is 0 if x is an integer, {x} - 1/2 otherwise.

It's known that s(ah, ak) = s(h, k) for positive a,
so, if x = h/k, we can define s from Q to Q by s(x) = s(h, k).

Hickerson (Continued fractions and density results for Dedekind sums,
J Reine Angew Math 290, 1977, 113-116, MR 55 #12611) proved that
the graph of s is dense in the plane.

More recently, Nick Phillips and I (Lines full of Dedekind sums,
Bull London Math Soc 36, 2004, 547-552) proved that, with one
exception, every line through the origin with rational slope goes
through infinitely many points on the graph of s(x). We suspect
that such points are dense on these lines, but we only prove this
for the line y = x.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

Dave L. Renfro

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Oct 6, 2006, 12:07:59 PM10/6/06
to
Dave L. Renfro wrote (in part):

>> 1. A FUNCTION UNBOUNDED IN EVERY INTERVAL
>>
>> 2. A FUNCTION WHOSE GRAPH IS DENSE IN THE PLANE

Gerry Myerson wrote (in part):

> There are also examples in Gelbaum & Olmsted, Counterexamples
> in Analysis.
>
> A more-or-less natural example of #2 arises in Number Theory,
> in the study of the Dedekind sum.

Thanks for the reference. I'll look it up sometime and keep
a record of its existence. I rarely look at number theory
papers, unless they appear to involve something I'm marginally
interested in (such as normal or transcendental numbers,
especially when exceptional set behavior is involved in
some way -- Baire category, Hausdorff dimension, etc.),
so I probably passed right over this article when that
issue of Bull. London Math. Soc. came out.

I wrote my examples for a listgroup devoted to AP-calculus
(high school calculus in the U.S. designed to give students
an opportunity to later receive college credit), which is
why I gave all the proofs at the level and detail I gave
them. Two days ago, due to some issues raised in the
thread "Sequence of continuous functions unbounded on
irrationals" and some discussion in the AP-calculus list
group about strange (for them) functions, I wondered if
there was a fairly simple way to give a function that was
unbounded on every interval and a function whose graph
was dense in the plane. After a few minutes, I came up
with the f(p/q) = q example and wrote it up. I had an
idea for how to go about getting a function whose graph
was dense in the plane (construct it by a union of
horizontal lines, full of holes, that had disjoint
projections onto the x-axis), but I decided to hold
off posting that example 2 days ago. (For one thing,
there was something I needed to take care of at work
by the time I had finished writing up the f(p/q) = q
example.) Since I again had some free time yesterday
(and today, as should now be obvious), I decided to
go ahead and see if I could write up the dense graph
example in such a way that an interested high school
calculus teacher or a reasonably good high school
student could follow the proofs.

I didn't know my first example was in Gelbaum/Olmsted,
or I would have cited it. I knew it couldn't be all that
original (I've seen many variations on "f(p/q) = some
function involving the integers p and/or q" in the
literature), but I didn't realize the *exact* example
that I came up with and for the reason I wanted it
for was in Gelbaum/Olmsted.

For the record, it's in Chapter 2, Example 5, p. 23 of
the 1992 Dover edition of Gelbaum/Olmsted.

As for functions whose graphs are dense in the plane,
I've known of some more advanced ways to get this. The
most common ways are (A) nonlinear additive functions,
and (B) exotic Darboux functions. In Gelbaum/Olmsted,
these functions are given in (or easily constructed using):
Chapter 2, Example 26, p. 33; Chapter 8, Example 27,
pp. 104-105; Chapter 10, Example 9, pp. 134-135.

(A) Recall that f:R --> R is additive if f(x+y) = f(x) + f(y)
for all real numbers x and y.

Theorem: If f is a nonlinear additive function, then the graph
of f is dense in R^2.

Proof: Pick b so that f(b) differs from b*f(1). Define g by

g(x) = [f(x) - x*f(1)] / C,

where C is the constant [f(b) - b*f(1)].

An easy computation shows that g is additive. Moreover,
since g(1) = 0, it follows that g is zero for every
rational number.

We will first show that the graph of g is dense in R^2.
To this end, let (u,v) be a point in R^2 and let delta
be a positive number. Choose a rational number s whose
distance from v is less than (1/2)*delta, and then choose
a rational number r whose distance from u-sb is less than
(1/2)*delta. Then (r+sb, s) belongs to the interior of the
delta-by-delta square centered at (u,v). Moreover, a short
computation using the fact that g is Q-linear shows that
(r+sb, s) belongs to the graph of g. Therefore, the graph
of g is dense in R^2.

Using the fact that the graph of g is dense in R^2,
it is not difficult to show that the graph of
f(x) = C*g(x) + x*f(1) is also dense in R^2.

This proof can be found in the following reference. The result
itself is much older, and I also believe this specific proof
is not original with Wilansky, but I'm not certain about this
last statement.

Albert Wilansky, "Additive functions", pp. 97-124 in "Lectures
On Calculus" (edited by Kenneth O. May), Holden-Day, 1967.

In fact, each nonlinear additive function bilaterally majorizes
every Lebesgue measurable function on every set of positive
Lebesgue measure. Note that simply being unilaterally unbounded
on every set of positive measure is enough (Lusin's theorem
in measure theory and the fact that continuous functions are
bounded on compact sets) to force every restriction of a
nonlinear additive function to a set of positive measure
to be a nonmeasurable function.

(B) Darboux (1875) proved that derivatives satisfy the
intermediate value theorem (i.e. are Darboux functions)
and he gave an example of a derivative that is discontinuous
on a certain countable dense subset of the rationals.
I don't know if Darboux's function is unbounded on every
interval, but I doubt it would be difficult to modify
his construction to get this property, in which case
we'd have a function whose graph is dense (and then some)
in the plane.

Lebesgue (p. 90 of his 1904 book) gave an example of
a Darboux function that is discontinuous everywhere
in [0,1] (see Ilias Kastanas' post below for the details)
which also can be easily modified to give a function
whose graph is dense (and then some) in the plane.

"Non zero Lebesgue measure" (Ilias Kastanas, 8 January 1996)
http://groups.google.com/group/sci.math/msg/502fde0541cc97a0

Incidentally, nonlinear additive functions can map onto
the rational numbers, and hence nonlinear additive functions
can fail to be Darboux. (Note that my example of a function
that has a dense graph also maps onto the rationals.)

To continue the theme of pathological Darboux functions,
note that (easy modifications of) Darboux's and Lebesgue's
examples automatically have the property that they assume
every real number infinitely often in every interval,
since every interval contains infinitely many pairwise
disjoint subintervals.

Israel Halperin (1950) gave an example of a function
f:R --> R that assumes every real number in each nonempty
perfect set, and another example (I think his other one
is a different function, but I'm not sure) of a function
g:R --> R that assumes every real number c-many times
in each nonempty perfect set. Using the fact that every
nonempty perfect set contains c-many pairwise disjoint
perfect subsets (for a proof, see my 2 July 2001 post
below), Solomon Marcus (1960) observed that any function
having the property of Halperin's 1'st example automatically
has the property of Halperin's 2'nd example. Full citations
of Halperin's and Marcus' papers can be found in my
4 December 2000 post below.

"uncountability of the reals" (Dave L. Renfro, 2 July 2001)
http://groups.google.com/group/sci.math/msg/86ef3b7928eff8e9

"Strange function" (Dave L. Renfro, 4 December 2000)
http://groups.google.com/group/sci.math/msg/712092d7b9852455

Finally, for those who are interested, below are 4 books
that those who enjoy the kinds of things in Gelbaum/Olmsted's
"Counterexamples in Analysis" would want to know about.
I've listed them from lowest to highest for to the background
they assume the reader has.

Ralph P. Boas, "A Primer of Real Functions", 4'th edition
(revised and updated by Harold P. Boas), Carus Mathematical
Monographs 13R, The Mathematical Association of America, 1996.

Arnoud C. M. Van Rooij and Wilhelmus H. Schikhof, "A Second
Course in Real Functions", Cambridge University Press, 1982.

John C. Morgan II, "Point Set Theory", Pure and Applied
Mathematics #131, Marcel Dekker, 1989.

Alexander B. Kharazishvili, "Strange Functions in Real
Analysis", Marcel Dekker, 2000.


Dave L. Renfro

Michael Press

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Oct 7, 2006, 12:43:52 AM10/7/06
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In article
<1160078099.4...@i42g2000cwa.googlegroups.com>,
"Dave L. Renfro" <renf...@cmich.edu> wrote:

> HOW TO OBTAIN THE SETS D_1, D_2, D_3, ...

How about this?

Let v be transcendental. D_n = Q + v^n.
If x in D_m and y in D_n and x=y then there exists
b in Q st v^m - v^n + b = 0. Since v is not a root of
a polynomial with rational coefficients we have m=n,
and b = 0. The D_n are obviously dense in R.

--
Michael Press

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