I've been trying to prove the following for a while, out of curiosity. For any (possibly discontinuous, etc) function $f:A\to A$ such that $$ f^n(x) = f(x),\quad \forall x\in A, n\in \mathbb{N} $$ where $f^n$ is the $n$-repeated application of the function $f$ on its argument, then $f$ is either the identity mapping or constant everywhere.

Is this statement true? It feels like it might be, at a first glance, but I have little intuition for these types of statements, so it's unclear if there exists a weird counterexample from adding Choice into the mix.

For now, we can say, for example, that $A = \mathbb{R}$ (though I'm quite curious about the general case as well).

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    $\begingroup$ Take $f(1) = 1$ and $f(x) = 0$ for $x \neq 1$ for example. $\endgroup$
    – Cauchy
    Aug 17, 2017 at 15:33
  • $\begingroup$ Note that $f^n(x) = f(x)$ for all $x \in A, n \in \Bbb N$ is equivalent to simply saying that $f(f(x)) = f(x)$ for all $x \in A$. $\endgroup$ Aug 17, 2017 at 15:43

2 Answers 2


The answer is no. Take $A = \Bbb R$ and $f(x) = |x|$.

In general, we can say that if $f(f(x)) = f(x)$, then $f$ is equal to the identity over the image of $f$. In particular: consider any $y$ in the image of $f$. Then $y = f(x)$ for some $x \in A$, and we have $$ f(y) = f(f(x)) = f(x) = y $$ If $f$ is surjective, then the image of $f$ is $A$, and so $f$ must be the identity map over $A$.

Note that if $f$ is injective, then there exists a function $g(x)$ such that $g(f(x)) = x$, and we have $$ f(f(x)) = f(x) \quad \text{for all }x \in A \implies \\ g(f(f(x)) = g(f(x)) \quad \text{for all }x \in A\implies \\ f(x) = x \quad \text{for all }x \in A $$

  • $\begingroup$ Ah, thank you! Much easier than expected. I do wonder, are there any additional conditions we can add on $f$ for which this would imply that it is either constant or the identity? It's unclear to me that this would be the case. $\endgroup$ Aug 17, 2017 at 15:34
  • $\begingroup$ Ah, brilliant; the latter one was a condition I also managed to scrap together, but didn't formalize. Thank you! $\endgroup$ Aug 17, 2017 at 15:40

Just take $f$ to be a linear projection, it neither a constant nor the identity.


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