Is there some function $f(x)$ for which $f_{\infty}(x)$ is a continuous non-constant function that converges? let $f_1(x)=f(x)$
$f_2(x)=f(f(x))$
$f_3(x)=f(f(f(x)))$ and so on...

Is there some function $f(x)$ for which $f_{\infty}(x)$ is a continuous non-constant function that converges? It is okay if the $f_{\infty}(x)$ you find has a finite radius of convergence.


The only example that comes to my mind would be like,
$$f_n(x)= ^nx$$
where $^nx$ is the tetration, as $^{\infty}x$ converges for $e^{-e} \lt x \le e^{1/e}$. Although in this example I am not sure how we could define $f(x)$. 
 A: New Answer. We examine a couple of examples such that $f_{\infty}$ exists as a continuous function.

Example 1. Consider the following saw-tooth function:
$$ f(x) = \begin{cases}
x, & \text{if } -1 \leq x \leq 1, \\
-f(x-2), & \text{for all $x$}
\end{cases} $$
Its graph looks like
$\hspace{4em}$
Then it is easy to check that $ f_n(x) = f(x)$ for all $n\geq 1$, and so, $f_{\infty}(x) = f(x)$. This is obviously a continuous function.



Example 2. For a less trivial example, consider
$$ f(x) = \begin{cases}
x, & \text{if $|x|\leq \pi$}, \\
\sin(x)+\frac{x+\pi}{2}, & \text{if $x > \pi$}, \\
\sin(x) + \frac{x-\pi}{2}, & \text{if $x < -\pi$}.
\end{cases} $$
Its graph looks like
$\hspace{7.5em}$
This function is constructed in such a way that the function iteration stabilizes within finitely many steps on each bounded interval:
$\hspace{7.5em}$
So, for each interval $[a, b]$, there exists $N$ such that $f_{\infty} \equiv f_n$ on $[a, b]$ for all $n \geq N$, and so, $f_{\infty}$ is continuous.



Example 3. Perhaps the following example better explains the mechanism of Example 2.
$$ f(x) = \begin{cases}
x, & -1 \leq x \leq 1, \\
2-x, & 1 < x < 2, \\
x-2, & x \geq 2, \\
-2-x, & -2 < x < -1, \\
x+2, & x \leq -2.
\end{cases} $$
$\hspace{7.5em}$
Then its $n$-fold compositions look like
$\hspace{7.5em}$


Old Answer. There is a plethora of examples.
1. Suppose that $I \subseteq \mathbb{R}$ is a closed interval and $f : I \to I$ is a contraction, meaning that there exists a constant $0 \leq c < 1$ for which $|f(x) - f(y)| \leq c|x - y|$ for all $x, y \in I$. Then by the contraction mapping theorem (a.k.a. Banach fixed point theorem), there exists a unique $x_0$ such that $f_n(x) \to x_0$ for all $x \in I$. Here are some examples.

*

*$f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = cx$ for some $-1 < c < 1$. Then $f_n(x) \to 0$.


*$f : [0, \infty) \to [0, \infty)$ given by $f(x) = \frac{1}{2+x}$. Then we can check that $|f(x)-f(y)| \leq \frac{1}{4}|x - y|$. So there exists $x_0 \in [0, \infty)$ such that $f_n(x) \to x_0$ for any $x \in [0, \infty)$. Such $x_0$ can be located by solving $f(x_0) = x_0$, which in turn yields $x_0 = \sqrt{2}-1$.


*$f : [0, 1] \to [0, 1]$ given by $f(x) = \cos (x)$. Then $|f(x)-f(y)|\leq \sin(1)|x - y|$ holds, and so, there exists a unique $x_0 \in [0, 1]$ such that $f_n(x) \to x_0$. This $x_0$ is the solution of $\cos(x_0) = x_0$ on $[0, 1]$, but this number seems to have no closed form.
2. Suppose that $I \subset \mathbb{R}$ is a closed bound interval and $f : I \to I$ is continuous and non-decreasing. Then $f_n(x)$ always converges as $n\to\infty$. The limit can be located by the following criteria: (1) If $f(x) > x$, then $f_n(x)$ converges to the smallest fixed point of $f$ larger than $x$. (2) If $f(x) < x$, then $f_n(x)$ converges to the largest fixed point of $f$ less than $x$. (3) If $f(x) = x$, then $f_n(x) \to x$.

*

*$f : [0, 1] \to [0, 1]$ given by $f(x) = \sin(x)$. Since $\sin(x) < x$ for all $x > 0$, it follows that $f_n(x) \to 0$.


*$f : [0, 4] \to [0, 4]$ given by $f(x) = \sqrt{2x}$. Using the above criteria, we can check that $f_n(0) \to 0$ and $f_n(x) \to 2$ for all $0 < x \leq 4$.


*$f : [-n\pi, n\pi] \to [-n\pi, n\pi]$ where $n$ is a positive integer and $f(x) = x+\sin(x)$. Then $f'(x) = 1+\cos(x) \geq 0$ and so $f$ is increasing. Moreover,
$$ f_n(x) \to \begin{cases}
(2k+1)\pi, & \text{if $k$ is an integer and $2k\pi < x < (2k+2)\pi$} \\
2k\pi, & \text{if $k$ is an integer and $x = 2k\pi$}
\end{cases} $$
A: These definitely exist! Consider for example
$$f(x) = \arctan(2x)$$
Then for positive $x$, this converges to a positive value, and for negative $x$, it approaches a negative value. The values are namely the fixed points of $f(x)$
