# 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 $$f(x)$$.

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

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,

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

• Sorry if I haven't understood correctly, but in all your examples $f_{\infty}(x)$ is a constant function right? – Mathphile Oct 8 '19 at 4:15
• 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. – Sangchul Lee Oct 8 '19 at 4:17
• +1 I see. Can you find any continuous example? Sorry for all the confusion. I should have been more specific in my question. – Mathphile Oct 8 '19 at 4:21
• @Mathphile, No worries. I updated my answer. – Sangchul Lee Oct 8 '19 at 4:50
• nice! I'll mark as answer after some time just in case if I get any other good examples. – Mathphile Oct 8 '19 at 4:57

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 $$f(x)$$

• +1 Yes, I see what you mean. Although your answer is correct, I was looking for some function that keeps changing as $x$ changes. – Mathphile Oct 8 '19 at 3:53
• 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 $f(x)=x$ – Isaac Browne Oct 8 '19 at 4:00
• How did you come up with the example of $\arctan (2x)$ btw? – Mathphile Oct 8 '19 at 17:31
• 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 $\arctan(2x)$. – Isaac Browne Oct 8 '19 at 17:57
• Ah, so a function like $f(x)=x^{1/x}$ should work as well right? – Mathphile Oct 8 '19 at 19:02