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A friend of mine gave this question to me :

Find the value of $\int_0^1f(x)dx$ if $f$ is a real, non-constant, differentiable function satisfying $$f(f(x))=1-x \space\space\space\space\space\space\forall x\in(0,1)$$

The only thing that came to my mind was to form a differential equation with the given equation and find a solution but that was a dead end. If someone could give me a hint on how to tackle this problem it would be great.

EDIT: All this time wonghang and John Ma's answers had me convinced that this problem was not well thought-out. But like Sanket has pointed out, if we disregard the differentiable assumption for a while, does there now exist an $f$ satisfying the given conditions? Can someone find an example? I tried to produce one using some variants of the absolute value function but to no avail.

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  • $\begingroup$ And to think #ff69b4 was able to come up with the correct answer for this... $\endgroup$
    – AvZ
    Nov 17, 2015 at 17:06
  • $\begingroup$ @AvZ welcome back to city 17 $\endgroup$
    – najayaz
    Nov 17, 2015 at 17:34
  • $\begingroup$ Is it safer here? I hoped to see the last of Dr. Breen in City 14 $\endgroup$
    – AvZ
    Nov 17, 2015 at 17:37
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    $\begingroup$ If it isn't the man himself... The rise of #ff69b4 $\endgroup$
    – AvZ
    Nov 21, 2015 at 17:32

3 Answers 3

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I think $f(x)$ does not exist at all. $f(f(x))$ is decreasing on $(0,1)$.

$f(x)$ is either increasing or decreasing on $(0,1)$. (non-constant, real, differentiable)

If $f(x)$ is increasing, then $f(f(x))$ is increasing. If $f(x)$ is decreasing, then $f(f(x))$ is increasing.

There is no chance for $f(f(x))$ to be a decreasing function like $1-x$.

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  • $\begingroup$ Why is f either strictly increasing or decreasing? $\endgroup$
    – nkm
    Oct 30, 2015 at 6:46
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    $\begingroup$ @NathanielMayer By $-1 = f'(f(x)) \cdot f'(x)$. Then $f'$ can't be zero. Then it's either $>0$ or $<0$ by Darboux theorem. $\endgroup$
    – user99914
    Oct 30, 2015 at 6:47
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    $\begingroup$ @NathanielMayer As $f^2$ is strictly decreasing, $f^2$ is injective and so must be $f$. By the intermediate value theorem, a continuous injective function is either strictly increasing or strictly decreasing. $\endgroup$
    – k.stm
    Oct 30, 2015 at 6:48
  • $\begingroup$ I think one can also show that any such possible $f$ would have to be of the form $a+bx$ and reach a contradiction this way. $\endgroup$
    – gary
    Oct 30, 2015 at 7:04
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    $\begingroup$ In fact, this answer assumes that $f:[0,1]\to[0,1]$, which is not necessarily true. Take a look at the answer of John Ma. $\endgroup$ Oct 30, 2015 at 8:24
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The given function is not possible, but if we assume the given function is not differentiable which should have been mentioned then answer would have been $\frac 12$.

$$f(f(x)) =1-x\implies f(1-x)=1-f(x)$$ $$\int_0^1 f(x)dx=1-\int_0^1f(1-x)dx=1-\int_0^1 f(x)dx\implies \int_0^1 f(x)dx=\frac 12$$

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  • $\begingroup$ Astute observation (+1) $\endgroup$
    – AvZ
    Nov 21, 2015 at 17:35
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Another point of view:

For all $C > 1$, let $f$ be a differentiable function on $(0,C+1)$ such that $$f(x) = \begin{cases} C+x & \text{if }x \in (0,1) \\ C+1 - x & \text{if }x \in (C, C+1).\end{cases}$$

Then for all $x\in (0,1)$, $f(f(x)) = f(C+x) = C+1 - (C+x) = 1-x$. But

$$\int_0^1 f(x) dx = C + \frac 12$$

can be any number bigger than $\frac 32$.

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