Infinite composition of function that has a fixed point

Question

I have faced an interesting question working with functions that have a fixed point(i.e such $f$ that $\exists$ $x: f(x)=x$ )

So I asked myself quite a gerenal question that I did't find easy to answer, namely:

Let ($\mathrm X$,$\mathscr P $) be a metric space and $f$ : $\mathrm X$ $\to$ $\mathrm X$ is a function with the only fixed point, i.e. $$\exists ! x_0: f(x_0)=x_0$$

And let $y \in \mathrm X$-another point from $\mathrm X$. Is it true, that if $ \exists \lim_{n\to \infty}f^n$($y$), then this limit must be $x_0$? (When I write $f^n(y)$ I actually mean $(f(f(f(f...(f(y))- n$ $times$ )

It's qiute easy to build an example when such limit doesn't exist, but if it does, I have a hypothesis that it must be $x_0$ but have no idea of proving it.

Answer 1

Suppose first that $f$ is continuous. Then write $$f^{n+1}(y) = f(f^n(y))$$

The left part tends to the limit $l$ of $(f^n(y))_{n \in \mathbb{N}}$, and by continuity of $f$, the second part tends to $f(l)$. So you get $l=f(l)$, so if $f$ has only one fixed point, $l$ is indeed equal to this fixed point.

If $f$ is not continuous, can you build a counter-example ?

Answer 2

It does not. Let

$$f(x)=\begin{cases} 1 & \mathrm{if\ }x=0 \\ x^2 & \mathrm{otherwise},\end{cases}$$

and consider value $y=\frac{1}{2}$, for example.

I believe your property should hold if $f$ is continuous, however.