# $f(f(f(x))) = x$. Prove or disprove that f is the identity function [duplicate]

Let $f$ be a continuous function on $\mathbb R$ satisfying the relation $$f(f(f(x))) = x\ \text{for all}\ x \in \mathbb R$$ Prove or disprove that $f$ is the identity function.

I tried taking the derivative. From the derivative, I'm not sure about it, but I concluded it had to be of degree 1 if it is a polynomial since if it'd have been of degree 2 or higher.. there needed to be terms of $x$ in the derivative.. which are not there.

• Hint : $x,f(x),f (f(x))$ must be different. Now use the intermediate value theorem for $f$ in a clever fashion. Interestingly enough, something similar can be done for functions continuous on the unit circle. Apr 30, 2017 at 4:11
• Hello, Would you mind showing us what you have tried, or what your think is the answer. It would help immensely in showing you how and why it is what it is. Apr 30, 2017 at 4:11
• There are ways to disprove. Try some functions with undetermined coefficients. Apr 30, 2017 at 4:14
• @GeorgeLaw thanks for pointing that out.. I forgot that if I took the derivative.. I was assuming that f is differentiable.. but couldn't we at least assume in general that f has to be differentiable over some of the domain? Apr 30, 2017 at 4:20
• Apr 30, 2017 at 4:27

You can prove that $f(x) \equiv x$ in four fairly simple steps:

1. Show that $f$ is one-to-one. Assume $f(x) = f(y)$ and show that this implies $x=y$ by applying $f$ two times to each side of the equation.
2. Show that a continuous function that is one-to-one has to be strictly increasing or decreasing. This follows for example by the mean-value theorem.
3. Show that $f$ cannot be strictly decreasing. If it is then $f(x) < f(y)$ for $x > y$. Now apply $f$ two times show that this gives us a contradiction ($x<y$).
4. Show that if $f$ is strictly increasing then $f(x) = x$. Assume 1) $f(x) > x$ and 2) $f(x) < x$ and try to derive a contradiction in both these cases (again by applying $f$ to the inequality using that $f$ is increasing).

All these points apart from the first two are covered in the answer to this question.

The only solution is $f(x)=x$. First, we note that if we have two functions $f:X\to Y$ and $g:Y\to X$ such that $f\circ g$ and $g\circ f$ are identity maps, then $f$ and $g$ must be bijective functions. Now take $g=f\circ f$.

Further, because $f$ is continuous and bijective, it must be monotonic. Now, you have two cases, $f$ is increasing, or $f$ is decreasing. If $f$ is increasing and $f(x)>x$, then $x=f(f(f(x)))>f(f(x)>f(x)>x$, which is a contradiction. The inequalities are reversed if $f(x)<x$. If $f$ is decreasing, we have that if $x<f(x)$, then $f(x)>f(f(x))$, so $f(f(x))<f(f(f(x))=x$, and hence $x>f(x)$, which is a contradiction, and similarly if $f(x)<x$.

However, a harder and more interesting problem is asking if you can have a point with period 3. This is impossible for polynomials with integer coefficients, but is possible for continuous functions, but when it happens, things are messy. As shown in the paper "Period three implies chaos", if you have a point of period 3, you will have a point of every other possible order.

• The Last result @Aaron had mentioned is the very interesting theorem named SARKOVSKII's theorem
– SJA
Dec 12, 2020 at 4:53
• Actually, Sharkovsii’s theorem is a refinement of this which says exactly which periodicities imply which other periodicities. There’s is a well ordering of the integers for this, 3 being the first element. en.wikipedia.org/wiki/Sharkovskii's_theorem Dec 12, 2020 at 4:57
• @ Aaron can you please refer a good book of discrete dynamical system. In particularly where structural stability is discussed in a nice way.. Because to prove a map structurally stable , I find it quite difficult. Actually I follow Robert Devaney's book . Please suggest me a good reference.
– SJA
Dec 12, 2020 at 5:16
• @SJA I do not know of any references. I have never read any books on dynamical systems, just a random paper I found on Sharkovskii's theorem. Dec 12, 2020 at 5:52