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I was trying to argue by contradiction. Say there is a $x_0 \in [0,1]$ so that $f(x_0) \neq x_0$. WLOG, assume $f(x_0) < x_0$. Now, notice that $f \circ f$ is the inverse of $f$ by definition. so $f$ has to be monotonic. (increasing or decreasing).

Assume it is decreasing. That is, if $a<b \implies f(a) > f(b)$.

Now, we know there is some $\alpha \in [0,1]$ so that $f(\alpha) = \alpha $. No WLOG assume $\alpha > x_0$, then

$$ f(\alpha) < f(x_0) \implies \alpha < f(x_0) < x_0$$

This is a contradiction.

Is this a valid argument?

  • $\begingroup$ Your assumption that $f$ is decreasing is a little risky, given that $f(1) > f(0)$. $\endgroup$ – user296602 Oct 9 '18 at 23:17
  • $\begingroup$ When you already have two points $\alpha$ , namely 0 and 1, with $f(\alpha)=\alpha$ saying there exists $\alpha$ such that $f(\alpha)=\alpha$ is making things unnecessarily complicated right$ $\endgroup$ – Kabo Murphy Oct 9 '18 at 23:25

Claim. $f$ is strictly increasing. (I thought it is trivial but you all doubt it. Little confused.)

Proof. If not, say $x_1>x_2$ such that $f(x_1)\leq f(x_2)$, then since $f(0)\leq f(x_1)\leq f(x_2)\leq f(1)$, by intermediate value theorem of continuous function, we can find $\xi\in(0,x_2]$ such that $f(\xi)=f(x_1)$, therefore $$ f\circ f\circ f(\xi)=f\circ f\circ f(x_1), $$ destroying the condition $f\circ f\circ f(x)=x$. $\square$

Then, suppose there is $x_0\in(0,1)$ such that $x_0>f(x_0)$, then we have $$ x_0>f(x_0)>f(f(x_0))>f(f(f(x_0))), $$ contradictory. $x_0<f(x_0)$ likewise.

  • 1
    $\begingroup$ Why $f$ must be strictly increasing? It isnt given. $\endgroup$ – James Oct 10 '18 at 0:51
  • $\begingroup$ Given $f(0)=0$ and $f(1)=1$. Impossible for decreasing. $\endgroup$ – JRen Oct 10 '18 at 1:16
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    $\begingroup$ counterexample: $f(x)=\sin(\frac52\pi x)$... $\endgroup$ – AccidentalFourierTransform Oct 10 '18 at 1:37
  • $\begingroup$ @AccidentalFourierTransform Your $f$ has $f(f(f(1/5)))=1$. How could? $\endgroup$ – JRen Oct 10 '18 at 1:51
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    $\begingroup$ $f \circ f \circ f = Id$ implies $f$ is bijection. Bijection and continuous implies strictly monotonic. $\endgroup$ – N. S. Oct 10 '18 at 3:11

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