If $f(f(f(x)))+f(x)=2$ for all $0≤x≤2$, where $f(x)$ is a continuous function, then find $\int_0^2 f(x) dx$

Substitute $f(x)=2-t$ to get $$\begin{equation} f(f(2-t))=t \end{equation}$$ Take inverse on both sides and then integrate from $0$ to $2$ $$\int_0^2f(2-t)=\int_0^2f^{-1}(t)$$ On the other hand, substituting $f(f(x))=z$ in original equation gives $$f(z)=2-f^{-1}(z)$$ Putting the value of $f^{-1}(z)$ back we can solve for $\int_0^2 f(x) dx=2$

My solution gave me the right answer, but it's clearly wrong because 1) the range of $f(x)$ or $f(f(x))$ need not be a subset of the domain and 2) inverse of $f(x)$ may not exist. Also, the given data $0≤x≤2$ seems to be superfluous here. So what is the proper way of doing it?

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    $\begingroup$ This question looks like there are missing details: it is not specified the domain of $f$ (all we know is that the domain contains $[0;2]$), and the codomain as well. Maybe it is implicitly assumed that $f: [0;2] \to [0;2]$? $\endgroup$
    – Crostul
    Nov 10, 2021 at 9:34
  • $\begingroup$ @Crostul that does seem to clear most of the trouble, so I guess you're right! One thread that would be left I suppose is checking for the validity of taking the inverse; will we have to assume an inverse exists? Or maybe there's some slightly different procedure that doesn't require taking an inverse $\endgroup$
    – Amadeus
    Nov 10, 2021 at 15:44

1 Answer 1


$f$ is not invertible. Consider $$f(f(f(x)))=2-f(x)$$ and suppose $f$ is increasing on $[0,2]$. Then \begin{align*} f(f(f(0)))<f(f(f(2))),\quad 2-f(0)>2-f(2) \end{align*} Conversly, suppose $f$ is decreasing on $[0,2]$. Then \begin{align*} f(f(f(0)))>f(f(f(2))),\quad 2-f(0)<2-f(2) \end{align*} It leads contradiction, so $f$ cannot increase or decrease on $[0,2]$. There's a trivial solution $f(x)=1$ to the functional equation that has $$\int_0^2 f(x)dx=2$$ And I'm not sure that there exists other function satisfying the equation.


We can prove $f(x)=1$. Suppose $f$ is a function defined in closed interval $[0,2]$. If $y$ is in range of $f$, then $2-y$ is also in range of $f$ by $$f\circ f\circ f(x)=2-f(x).$$ Since continuous function maps closed interval to closed interval $$f([0,2])=[1-p,1+p]$$ for $0\leq p$. Then suppose $0<p$ and consider $f:[1-p,1+p]\to[1-p,1+p]$. Then following holds. $$f\circ f(x)=2-x$$ Note that $f$ is invertible since $$f\circ f\circ f\circ f(x)=x\quad\Rightarrow\quad f^{-1}(x)=f\circ f\circ f(x)$$ We already verified that $f$ is not invertible, so $p=0$, that is $f$ is constant.

  • $\begingroup$ Wow that was really great! So ig I got the correct answer just by chance, even though I made an incorrect assumption? And also, just to be clear, you also assumed that $f: [0;2] \to [0;2]$ is given like @Crostul suggested in his comment, right? $\endgroup$
    – Amadeus
    Nov 12, 2021 at 6:03
  • $\begingroup$ @Amadeus Yeah, I assumed $f:[0,2]\to [0,2]$. I think that this assumption is necessary because the condition only holds on $[0,2]$. $\endgroup$
    – suww
    Nov 12, 2021 at 6:37
  • $\begingroup$ Thanks for the clarification; btw, I noticed this only just now but isn't $f$ is invertible implied only for $x: [1-p,1+p]$? So it could still be not invertible on $x: [0,2]$ and not contradict our first result $\endgroup$
    – Amadeus
    Nov 12, 2021 at 6:51
  • $\begingroup$ @Amadeus The same argument can be applied to $[1-p,1+p]$ instead of $[0,2]$. Then $f:[1-p,1+p]\to [1-p,1+p]$ is not invertible and we get contradiction. $\endgroup$
    – suww
    Nov 12, 2021 at 7:30
  • $\begingroup$ Oh yeah I got it now, thanks! $\endgroup$
    – Amadeus
    Nov 12, 2021 at 7:55

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