sequence

[alberto]

Hi
Let (a_n) be a sequence which converges to a and let f be continuous
map such that
f^n(a_n) = a_n for all n. Is it true that a is a limit point of the
sequence
(f^n(a))?

[William Elliot]

It's true if f has Lipschitz constant of 1 or less, ie
for all x,y, d(f(x),f(y)) <= d(x,y).

Then we have the stronger

(aj)_j -> a, (f^j(aj))_j -> b implies (f^j(a))_j -> b.

Proof. d(f^j(a),b) < d(f^j(a),f^j(aj)) + d(f^j(aj),b)
<= d(a,aj) + d(f^j(aj).b)

----

[Ilmari Karonen]

[Corrected version after moderator comments.]

No. A counterexample would be the tent map, f(x) = 1-|2x-1| on the
interval [0,1], with a_n = 1/(1+1/2^n). Then a_n is a fixed point of
the n-th iterate of f, with a_n -> 1 as n -> infinity, but f^n(1) = 0
for all n.

(If by f^n you meant exponentiation rather than iteration, the same
tent map still provides a counterexample, except this time with a_n =
2^(-n/(n-1)). Then (f(a_n))^n = a_n, a_n -> 1/2, but (f(1/2))^n = 1
for all n.)

--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.

[se...@in.com]

The reason for the negative answer is that the limit set of the orbits
of the a_n‰??s need not be formed by recurrent points, although it is
expected to have some cyclic structure (at least in the compact
case).
Here is another counterexample. Take the space C of all double-sided
sequences of zeroes and ones indexed by the integers
x = (...x(-1), x(0), x(1)..., x(k),...), x(k) in {0,1} with the usual
metric
d(x,y) = Sum |x(k) - y(k)|/2^|k| , the sumtaken over the integers k in
Z.
This is a compact space homeomorphic to the Cantor set. Let f:C-->C
denote the left shift in C. Take now the sequence a_n of points in C
defined by a_n(m)=1 if n|m and a_n(m)=0 otherwise. For example
a_2=(...0101010...), a_3=(...00100100100...) etc. ; so f^n(a_n)=a_n.
Then a_n --> a, where a=(...000010000...) and clearly f^n(a)-->(...
000000...), so a is not a limit point of f^n(a).

Simeon