From: ikastan@sol.uucp (Ilias Kastanas)
Subject: Re: Cycles of continuous functions
Date: 19 Sep 1999 10:37:26 GMT
Newsgroups: sci.math
Keywords: Entire functions with no 2-cycles (f(f(x))=x)

In article <7rrc01$gep$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
@In article <130919990735251590%edgar@math.ohio-state.edu.nospam>,
@ "G. A. Edgar" <edgar@math.ohio-state.edu.nospam> writes:
@
@> Sarkovskii is for functions R -> R but not for C -> C, right?
@
@Here's an interesting complex analysis exercise: 
@
@Show that the only entire functions that have no fixed points and no 
@2-cycles are the translations z -> z+a (for a <> 0).
@
@I'll post a solution some time next week if nobody does it sooner.
@
@Robert Israel                                israel@math.ubc.ca
@Department of Mathematics        http://www.math.ubc.ca/~israel 
@University of British Columbia            
@Vancouver, BC, Canada V6T 1Z2




	If the entire function f(z) has no fixed points,  f(z) - z  is
entire and never  = 0.  Then  (f(f(z)) - f(z))/(f(z) - z)  is also entire,
never  = 0,  and  (f  having no 2-cycles) never  = -1  either.  Hence it
is constant,  = c  say, and thus

	|f(f(z)) - f(z)|  =  |c| |f(z) - z|.

	Suppose  f(z) - z  isn't constant;  then it cannot be a polynomial...
it is a transcendental entire function, and so is  f.  So for every  N > 0
there is an  R > N  and a  z,  |z| = R  with  f(z)  in the unit disk  D
(|f(z)| < 1).  For such  z's,  then,  |f(f(z)) - f(z)|  can be made as
large as desired;  the image of  D  under  f  isn't bounded -- contradicting
the compactness of  D.

	Therefore  f(z) - z = a non-0 constant.


	(Hmm...  I thought this would qualify for the "big hammer" thread...
aircraft carrier against a sailboat.   But I'm not using Great Picard,
after all;  just little Picard, and Casorati-Weierstrass).



						Ilias
==============================================================================

From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Cycles of continuous functions
Date: 21 Sep 1999 18:40:39 GMT
Newsgroups: sci.math

In article <7s2ed6$kou$1@hades.csu.net>,
 ikastan@sol.uucp (Ilias Kastanas) writes:
> In article <7rrc01$gep$1@nntp.itservices.ubc.ca>,
> Robert Israel <israel@math.ubc.ca> wrote:

> @Show that the only entire functions that have no fixed points and no 
> @2-cycles are the translations z -> z+a (for a <> 0).

> 	(Hmm...  I thought this would qualify for the "big hammer" thread...
> aircraft carrier against a sailboat.   But I'm not using Great Picard,
> after all;  just little Picard, and Casorati-Weierstrass).

I think little Picard (at least) is the natural tool for this kind of
problem.

Here's a bit of a generalization.

Theorem: If f and g are entire functions such that f, g and f \circ g have
no fixed points, then f and g are both translations.

Proof: Consider (f(g(z))-z)/(g(z)-z).  This is analytic, and never takes 
values 0 or 1, so by little Picard it's constant.  Thus for some constant c
(not equal to 0 or 1),

f(g(z)) - c g(z) = (1-c) z

Now the right side is one-to-one, so g must be one-to-one, and the only
one-to-one entire functions are translations (well, that's a consequence
of Casorati-Weierstrass too).  Therefore g is a translation.
If g(z) = z + b, then f(z+b) - c(z+b) is one-to-one, and so f is also
a translation.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2