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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 $f(x)$.

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  • $\begingroup$ Do you know Banach‘s Fixed Point Theorem? $\endgroup$ – Qi Zhu Oct 8 at 3:33
  • $\begingroup$ @QiZhu I am not familiar with it. $\endgroup$ – Mathphile Oct 8 at 3:41
  • $\begingroup$ What about $f(x)=x$? $\endgroup$ – Robert Israel Oct 8 at 4:19
  • $\begingroup$ @RobertIsrael I think that may be the only example. Can you find any other example? $\endgroup$ – Mathphile Oct 8 at 4:25
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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

$\hspace{4em}$Graph of f

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

$\hspace{7.5em}$Graph of f

This function is constructed in such a way that the function iteration stabilizes within finitely many steps on each bounded interval:

$\hspace{7.5em}$Graph of f_n's

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.

$$ f(x) = \begin{cases} x, & -1 \leq x \leq 1, \\ 2-x, & 1 < x < 2, \\ x-2, & x \geq 2, \\ -2-x, & -2 < x < -1, \\ x+2, & x \leq -2. \end{cases} $$

$\hspace{7.5em}$Graph of f

Then its $n$-fold compositions look like

$\hspace{7.5em}$Graph of f_n's


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,

    $$ f_n(x) \to \begin{cases} (2k+1)\pi, & \text{if $k$ is an integer and $2k\pi < x < (2k+2)\pi$} \\ 2k\pi, & \text{if $k$ is an integer and $x = 2k\pi$} \end{cases} $$

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  • $\begingroup$ Sorry if I haven't understood correctly, but in all your examples $f_{\infty}(x)$ is a constant function right? $\endgroup$ – Mathphile Oct 8 at 4:15
  • $\begingroup$ The answer was written based on the original example, so some examples (including all examples in #1) have constant limit. But the last two examples are non-constant. $\endgroup$ – Sangchul Lee Oct 8 at 4:17
  • $\begingroup$ +1 I see. Can you find any continuous example? Sorry for all the confusion. I should have been more specific in my question. $\endgroup$ – Mathphile Oct 8 at 4:21
  • $\begingroup$ @Mathphile, No worries. I updated my answer. $\endgroup$ – Sangchul Lee Oct 8 at 4:50
  • $\begingroup$ nice! I'll mark as answer after some time just in case if I get any other good examples. $\endgroup$ – Mathphile Oct 8 at 4:57
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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 $f(x)$

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    $\begingroup$ +1 Yes, I see what you mean. Although your answer is correct, I was looking for some function that keeps changing as $x$ changes. $\endgroup$ – Mathphile Oct 8 at 3:53
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    $\begingroup$ Interesting. Well if you want everything to be continuous, I don't think that is possible unless $f(x)=x$ (over some small interval), as the values must converge to fixed points of $f$, that is points where $f(x)=x$ $\endgroup$ – Isaac Browne Oct 8 at 4:00
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    $\begingroup$ How did you come up with the example of $\arctan (2x)$ btw? $\endgroup$ – Mathphile Oct 8 at 17:31
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    $\begingroup$ I wanted a function which had slope less than $1$ at the desired fixed points and as $x\to\infty$. Arctangent satisfies the latter condition, as it is in fact bounded and increasing!. But, $\arctan(x)$ doesn't work b/c it only has one intersection with the line $y=x$, so I changed it to $\arctan(2x)$. $\endgroup$ – Isaac Browne Oct 8 at 17:57
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    $\begingroup$ Ah, so a function like $f(x)=x^{1/x}$ should work as well right? $\endgroup$ – Mathphile Oct 8 at 19:02

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