I would like to find the set of continuous functions $f_n(x)$, where $f_n(x):\mathbb{R}\to \mathbb{R}$ satisfies $$f_n(f_n(f_n(f_n...(x)))) = x$$ where there are $n$ iterations of $f(x)$. For example $f_1(x)$ would be the solution to $f_1(x)=x$. $f_2(x)$ would be the solution to $f_2(f_2(x)) = x$.

For $f_1(x)$, the only solution is $f_1(x)=x$. For $f_2(x)$, the solutions are involutions.

For $f_3(x)$, the only answer is $f_3(x)=x$. For all other $f_n(x)$, one solution is $f_n(x) = x$.

My question: For $n \ge 3$, is $f_n(x) = x$ the only solution? If not, what are the solutions?

Edit: @MattSamuel said that any involution works for an even $n$. This is because $f_n(f_n(x))$ can be replaced with $x$. For example, $$f_2(f_2(f_2(f_2(f_2(f_2(x)))))) = f_2(f_2(f_2(f_2(x)))) = f_2(f_2(x)) = x$$ However, this does not necessarily mean that involutions are the only set of solutions for $f_{2k}(x)$.

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    $\begingroup$ Involutions work for any even number of compositions. $\endgroup$ Commented Jul 27, 2019 at 19:48
  • $\begingroup$ @MattSamuel Wow, that is true. Can functions that aren't involutions work for an even number of compositions? $\endgroup$ Commented Jul 27, 2019 at 19:51
  • $\begingroup$ Probably, but I don't know any off hand. $\endgroup$ Commented Jul 27, 2019 at 19:52
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    $\begingroup$ @PackSciences, only real functions. $\endgroup$ Commented Jul 27, 2019 at 19:58
  • 1
    $\begingroup$ Ok, good you edited the question $\endgroup$ Commented Jul 27, 2019 at 19:58

1 Answer 1


There are no monotone solution for odd $n$ other then $f_n(x) = x$. There are also no continuous solutions as any continuous injection is monotone.

There are many discontinuous solutions for any $n$. Represent any real number $x$ as $x = \lfloor x \rfloor + \{x\}$ (integer and floor part), so we now have bijection $\mathbb{R} \leftrightarrow [0, 1) \times \mathbb{Z}$. Choose your favorite bijection $g_n: \mathbb Z \leftrightarrow \mathbb Z$ of order $n$ - for example, $g_n(i) = (i + 1) \mod n + \lfloor\frac i n\rfloor$ - split $\mathbb{Z}$ into segments of length $n$ and rotate any segment. Now define $f_n(x) = \{x\} + g_n(\lfloor x \rfloor)$. It's even continuous everywhere but in points $n - 1 + kn$.

(for simplicity, I'll denote $f^n$ to be $n$-th iteration of $f$ - we will not need powers here)

The only continuous solutions are involutions - answer you linked can be extended to proof it. $f$ have to be monotonic - if it isn't - say we have $f(x) > f(y) > f(z)$ while $x > z > y$ - then it's not injective, as there is point in $q \in [z, x]$ s.t. $f(q) = f(y)$, so we will have $f^n(q) = f^n(y)$ but $q \neq y$.

If $f$ is strictly increasing, then $f(x) = x$ by @Najib's argument.

If $f$ is strictly decreasing, then $f$ has a single fixed point $x_0$. We have $f(x_0 + a) < f(x_0) = x_0$ and $f(x_0 - a) > f(x_0) = x_0$ for positive $a$. $g =f\circ f$ is continuous and injective - so monotonic. As $f(x_0 + 1) < f(x_0)$, we have $f(f(x_0 + 1)) > x_0$, so $g$ is increasing. If $g(x) \neq x$ for some $x$, we will again have $f^n(x) \neq x$. So $g(x) = x$. And thus $f$ is involution.

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    $\begingroup$ Added part about continuous functions - involutions are the only answer. $\endgroup$
    – mihaild
    Commented Jul 27, 2019 at 21:50
  • $\begingroup$ Just to make sure I understand correctly, there are no other solutions for odd $n$ other than $f_n(x) = x$ and no other even solutions for even $n$ other than involutions? $\endgroup$ Commented Jul 28, 2019 at 17:46
  • $\begingroup$ Yes. Also, the only involutions on $\mathbb{R}$ are $f(x) = x$ and $f(x) = a - x$. $\endgroup$
    – mihaild
    Commented Jul 28, 2019 at 18:53
  • $\begingroup$ That doesn't sound right. Isn't $1/x$ an involution too? $\endgroup$ Commented Jul 28, 2019 at 19:48
  • $\begingroup$ Sorry, the only continuous involutions. $1 / x$ (defined in $0$ as $0$) isn't continuous. $\endgroup$
    – mihaild
    Commented Jul 28, 2019 at 20:06

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