# Prove that there is a $m \in \mathbb{R}$ so that $f(f(m)) = m$

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 !

• 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 $f$. Oct 2 '18 at 20:06
• @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 ? Oct 2 '18 at 20:14
• 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. Oct 2 '18 at 20:19
• Okay , thanks a lot . Oct 2 '18 at 20:21

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 $$[x_1,x_2]$$.
• 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 $m,n$. Jul 29 '20 at 15:27
• @Subhajit: I'm not sure how you would prove that without proving that $f$ has a fixed point. Jul 29 '20 at 21:33