Today the 17 October 2019 I discussed a very remarkable fixed point theorem discovered by the Ukrainian mathematician Oleksandr Micholayovych Sharkovsky.
We recall that a periodic point of period for a function
is a point
such that
. With this definition, a periodic point of period
is also periodic of period
for every
which is a multiple of
. If
but
for every
from 1 to
, we say that
is the least period of
.
Theorem 1. (Sharkovsky’s “little” theorem) Let be an interval and let
be a continuous function su. If
has a point of least period 3, then it has points of arbitrary least period; in particular, it has a fixed point.
Note that no hypothesis is made on being open or closed, bounded or unbounded.
Our proof of Sharkovsky’s “little” theorem follows the one given in (Sternberg, 2010), and could even be given in a Calculus 1 course: the most advanced result will be the intermediate value theorem.
Lemma 1. Let be a compact interval of the real line, let
be a continuous function. Suppose that for some compact interval
it is
. Then
has a fixed point in
.
Proof. Let and
be the minimum and the maximum of
in
, respectively. As
, it is
and
. Choose
such that
and
. Then
is nonpositive at
and nonnegative at
. By the intermediate value theorem applied to
,
must have a fixed point in the closed and bounded interval (possibly reduced to a single point) delimited by
and
, which is a subset of
.
Lemma 2. In the hypotheses of Lemma 1, let be a closed and bounded interval contained in
. Then there exists a closed and bounded subinterval
of
such that
.
Proof. Let . We may suppose
, otherwise the statement is trivial. Let
be the largest such that
. Two cases are possible.
- There exists
such that
. Let
be the smallest such
, and let
. Then surely
, but if for some
we had either
or
, then by the intermediate value theorem, for some
we would also have either
or
, against our choice of
and
.
for every
. Let then
be the largest
such that
, and let
. Then
for reasons similar to those of the previous point.
Proof of Sharkovsky’s “little” theorem. Let be such that
,
, and
. Up to cycling between these three values and replacing
with
, we may suppose
. Fix a positive integer
: we will prove that there exists
such that
and
for every
.
Let and
be the “left” and “right” side of the closed and bounded interval
: then
and
by the intermediate value theorem. In particular,
, and Lemma 1 immediately tells us that
has a fixed point
in
. Also,
, so
also has a point of period 2 in
, again by Lemma 1: call it
. This point
cannot be a fixed point, because then it would also belong to
as
, but
which has period 3. As we can obviously take
, we only need to consider the case
.
By Lemma 2, there exists a closed and bounded subinterval of
such that
. In turn, as
, there also exists a closed and bounded subinterval
of
such that
, again by Lemma 2: but then,
. By iterating the procedure, we find a sequence of closed and bounded intervals
such that, for every
,
and
.
We stop at and recall that
: we are still in the situation of Lemma 2, with
in the role of
. So we choose
as a closed and bounded subinterval not of
, but of
, such that
. In turn, as
, there exists a closed and bounded subinterval
of
such that
. Following the chain of inclusions we obtain
. By Lemma 1,
has a fixed point
in
, which is a periodic point of period
for
.
Can the least period of for
be smaller than
? No, it cannot, for the following reason. If
has period
, then so has
, and in addition
is divisible by
. But
while
for every
: consequently, if
has period
, then
. But this is impossible, because
by construction as
, while
.
Theorem 1 is a special case of a much more general, and complex, result also due to Sharkovsky. Before stating it, we need to define a special ordering on positive integers.
Definition. The Sharkovsky ordering between positive integers is defined as follows:
- Identify the number
, with
odd integer, with the pair
.
- Sort the pairs with
in lexicographic order.
That is: first, list all the odd numbers, in increasing order; then, all the doubles of the odd numbers, in increasing order; then, all the quadruples of the odd numbers, in increasing order; and so on.
For example,and
- Set
for every
and
.
That is: the powers of 2 follow, in the Sharkovskii ordering, any number which has an odd factor.
For example,.
- Sort the pairs of the form
—i.e., the powers of 2—in reverse order.
The set of positive integer with the Sharkowsky ordering has then the form:
Note that is a total ordering.
Theorem 2. (Sharkovsky’s “great” theorem) Let be an interval on the real line and let
be a continuous function.
- If
has a point of least period
, and
, then
has a point of least period
. In particular, if
has a periodic point, then it has a fixed point.
- For every
integer it is possible to choose
and
so that
has a point of minimum period
and no points of minimum period
for any
. In particular, there are functions whose only periodic points are fixed.
Bibliography:
- Keith Burns and Boris Hasselblatt. The Sharkovsky theorem: A natural direct proof. The American Mathematical Monthly 118(3) (2011), 229–244. doi:10.4169/amer.math.monthly.118.03.229
- Robert L. Devaney, An Introduction to Chaotic Dynamical Systems, Second Edition, Westview Press 2003.
- Shlomo Sternberg, Dynamical Systems, Dover 2010.