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If $f: \mathbb{R} \to \mathbb{R}$ and $f$ is strictly increasing, show that $f(x) = x$ if $f(f(f(x))) = x$.

So this compulsorily ESTABLISHES that $f(x) = x$ only, and no other solution. So, merely substituting $f(x) = x$ and hence showing the given equality holds will not earn any credit.

I was proceeding via inverses, but then I got confused with the notation. And for that, the problem got tricky.

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  • $\begingroup$ Where exactly did you stumble? We won't do the problem for you, just help you over the bump. $\endgroup$ – vonbrand Mar 19 '14 at 14:30
  • $\begingroup$ I think the elegance of the solution. $\endgroup$ – Landon Carter Mar 19 '14 at 14:36
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Suppose that $f(x_0)>x_0$ for some $x_0$. Then $$ f(f(f(x_0)))>f(f(x_0))>f(x_0)>x_0, $$ contradicting the hypothesis that $f(f(f(x_0)))=x_0$. We get a similar contradiction if $f(x_0)<x_0$. So the only option is that $f(x)=x$ for all $x$.

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  • $\begingroup$ You mean "we get a similar contradiction of $f(x_0) < x_0$" I think. $\endgroup$ – Frank Mar 19 '14 at 14:26
  • $\begingroup$ Thanks a lot! This was really helpful. $\endgroup$ – Landon Carter Mar 19 '14 at 14:40
  • $\begingroup$ @Frank: indeed. Thanks! $\endgroup$ – Martin Argerami Mar 19 '14 at 14:51
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If f is strictly increasing, then if x>y, f(x)>f(y). Suppose f(f(f(x)))=f(x).

Case 1: Suppose $f(x) < x$. So, $x>f(x)$. Thus, $f(x)>f(f(x))$ since $f$ is strictly increasing. It then would follow that $f(f(f(x)))>f(f(x))$, since $f(f(f(x)))=f(x)$. But, since f is strictly increasing we can also infer that $f(f(x))>f(f(f(x)))$. Thus we have a contradiction and it is not the case that $f(x) < x$.

Case 2: Suppose $f(x)>x$. The reasoning of this case comes as similar to that of case 1. So, it does not hold that $f(x)>x$.

By trichotomy it follows that $f(x)=x$

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  • $\begingroup$ Thank you. The elegance was eluding me. $\endgroup$ – Landon Carter Mar 19 '14 at 14:37
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    $\begingroup$ +1 for "trichotomy". People have lost classy talk these days. $\endgroup$ – Guy Mar 19 '14 at 14:39

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