If $x_{n+1}=f(x_n)$ and $x_{n+1}-x_n\to 0$, then $\{x_n\}$ converges

Question

Let $f:[0,1] \rightarrow [0,1]$be a continuous function. Choose any point $x_0 \in [0,1]$ and define a sequence recursively by $x_{n+1}=f(x_n)$. Suppose $\lim_{n \rightarrow \infty}x_{n+1}-x_n =0$, does this sequence converge?

Answer 1

The answer is YES - I have fixed the proof of Hongyi:

Let $K$ be the set of sub-sequential limits of the sequence $\{x_n\}$. Then $K$ is compact, and since $x_{n+1}-x_n\to 0$, $K$ is also connected. (This requires some more work.) Hence $K$ is of the form $$ K=[a,b]\subset [0,1]. $$ If $a=b$, then we are done, since this means that the sequence $\{x_n\}$ has only one sub-sequential limit, and hence converges.

We shall next show that $a<b$ implies that $f(x)=x$ on $[a,b]$, and hence the sequence is eventually constant. This contradicts the fact that its sub-sequential limits are all the points of $[a,b]$.

Assume that $f(x)\not\equiv x$ in $[a,b]$, and let $x_0\in(a,b)$ with $f(x_0)>x_0$. (The case $f(x_0)<x_0$ is treated similarly.) Then there exist $h>0$, such that $$ f(x)-x\ge 0\quad\text{whenever}\quad x\in [x_0-h,x_0+h]\subset (a,b). $$ Since $b$ is a limit point of $\{x_n\}$, there exists $x_{n_0}\in (x_0+h,1]$, and the $n_0$ can be picked so that $$ \lvert x_{n+1}-x_{n}\rvert <h, \quad \text{when}\quad n\ge n_0. $$ This means that, if $x_{n_0}$ is very close to $b$, then $x_n$, $n\ge n_0$, CAN NOT get smaller than $x_0$, since in the whole interval $[x-h,x+h]$, $f(x_n)\ge x_n$. Thus, $a$ can not be a limit point.

Answer 2

Yes. Let $I = \lim_{k \rightarrow \infty} I_k$ where $I_k$ is the closure of $\{x_k,x_{k+1},...\}$. Note that because $x_{k+1}-x_k \rightarrow 0$, $I$ is connected, thus either a singleton or an interval. If $I$ is a singleton, we are done. If $I$ is an interval $(x_-,x_+)$, then $f(x)=x$ on this interval. It then follows, unless $x_k$ is constant for sufficiently large $k$, that $x_k$ is strictly monotone in $k$; and thus that the sequence must converge to some point within $I$.

Answer 3

I'm fairly sure it need not converge.

Let $$H_n = \sum_{i=1}^n \frac{1}{i}$$

Define $y_n = H_n \pmod{1}$ if $\lfloor H_n \rfloor$ is even, and $1-H_n \pmod{1}$ otherwise. This sequence oscillates between 0 and 1 indefinitely, so it doesn't converge; although $y_{n+1} - y_n$ tends to $0$. It is the case that the difference between distinct harmonic numbers is never an integer, according to Wikipedia, so this sequence never duplicates any value.

We just need to show that we can define continuous $f: [0,1] \to [0,1]$ such that $f(x_n) = x_{n+1}$. I've convinced myself of this with some rather dubious reasoning about infinitesimals.