What else can we conclude about $f$ given this information?

Question

Consider a function $f: \mathbb{R} \mapsto \mathbb{R} $, with the following property:

Consider the sequence $x, f(x), f(f(x)), f(f(f(x))), ...$.
This sequences converges for all values of $x$.

For completeness, this sequence is $a_0=x, a_{n+1}=f(a_n)$.
For all $x \in \mathbb{R}$, there exists a $y \in \mathbb{R}$ such that $y = \displaystyle \lim_{m \to \infty} a_m$.

For example, $f$ might be a constant function. It might even be, $f(x)=x/2$ or $x/100$.

Is there anything known about continuïty? Because otherwise I think one could construct some ugly functions that satisfy — Vincent Batens, Jan 04 '24 at 02:24
@VincentBatens Yeah I think $f$ can be considered continous. — whoisit, Jan 04 '24 at 23:14

Siong Thye Goh · Answer 1 · 2024-01-04T10:29:48.833

0

We are certain that we can't find cycles $C=\{a_0, a_1, \ldots, a_{n-1}\}$ where all the elements are distinct and the lengths more than $1$ such that $a_{i+1}=f(a_i), $ where $i$ takes values from $0$ to $n-2$ and $a_0 = f(a_{n-1})$.

Remark:

This function need not have a fixed point as we can see from the counterexample in the comment.

edited Jan 04 '24 at 10:29

answered Jan 04 '24 at 07:01

Siong Thye Goh

149,520
20
88
149

3

if f(x)=x/2 except f(0)=-1. This function follows the rules, but has no fixed point - is that right? Then that contradicts your first claim. – whoisit Jan 04 '24 at 08:41
you are right, thanks for the counterexample. – Siong Thye Goh Jan 04 '24 at 10:18

What else can we conclude about $f$ given this information?

1 Answers1