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The function $f : \mathbb{R} \to \mathbb{R}$ is continuous . Also $0\lt x_1 \lt x_2 \lt x_3 \lt x_4 \ $ and $f(x_1) = x_2 \ , \ f(x_2) = x_3 \ , \ f(x_3) = x_4 \ , \ f(x_4) = x_1 $ . Prove that there is a $m \in \mathbb{R}$ so that $f(f(m)) = m$ .

My Try : Let $g(x) = f(x) - x $ and it is continuous over $\mathbb{R} $ . Furthermore , we have $g(x_1) \ , \ g(x_2) \ , \ g(x_3) \gt 0 $ and $g(x_4) \lt 0$ . The result is $f(c) = c$ for some $c$ between $x_3$ and $x_4$ . Applying $f$ to both sides yields to $f(f(c)) = f(c) = c$ .

Is my solution correct ? Also write other solutions . Thanks in advance !

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  • $\begingroup$ I think you've got the cleanest solution (my idea was the same; it boils down to using the intermediate value theorem). You actually prove a stronger result: there exists $m\in\Bbb R$ such that $f(m)=m$. This automatically means $m$ is a fixed point for any number of compositions of $f$. $\endgroup$
    – Clayton
    Commented Oct 2, 2018 at 20:06
  • $\begingroup$ @Clayton Thanks for your reply . If we have three $x_i$'s instead of four , it will work again . So the number of $x_i$'s is irrelevant to question ? $\endgroup$
    – S.H.W
    Commented Oct 2, 2018 at 20:14
  • $\begingroup$ Correct. All you really need are two points, $x_1<x_2$ with the conditions $f(x_1)=x_2$ and $f(x_2)=x_1$. This will cause $g(x)=f(x)-x$ to have the properties $g(x_2)<0<g(x_1)$, so there is a root of $g(x)$ in this interval. $\endgroup$
    – Clayton
    Commented Oct 2, 2018 at 20:19
  • $\begingroup$ Okay , thanks a lot . $\endgroup$
    – S.H.W
    Commented Oct 2, 2018 at 20:21

1 Answer 1

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Yes, the solution is correct.

In fact, you show something stronger, using a weaker hypothesis: all you need is a continuous function on an interval $[x_1,x_2]$ such that $f(x_1)\geq x_1$ and $f(x_2)\leq x_2$ to conclude that $f$ has a fixed point somewhere in $[x_1,x_2]$.

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  • $\begingroup$ Also it can be done, by finding such $m,n$ that $f(m)=n,f(n)=m$ since, $f$ oscillates and continuous, we must have such $m,n$. $\endgroup$
    – A learner
    Commented Jul 29, 2020 at 15:27
  • $\begingroup$ @Subhajit: I'm not sure how you would prove that without proving that $f$ has a fixed point. $\endgroup$
    – tomasz
    Commented Jul 29, 2020 at 21:33

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