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Show that a continuous mapping f from $[0,1]$ to $[0,1]$ which satisfies $f(f(x)) = x$ for all $x\in [0,1]$ and for which $f(x) \neq x$ for at least one $x\in [0,1]$ must have exactly one fixed point.

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What have you tried? Any Ideas that you have had in order to show some initiative? –  CBenni Dec 21 '12 at 13:28
    
Hint: $f$ must be $1-1$. Show that is must be either strictly increasing or strictly decreasing. Then show it can't be increasing. –  Thomas Andrews Dec 21 '12 at 13:30
    
Thanks for your help –  salar_ve Dec 22 '12 at 15:39

2 Answers 2

It is not true. The identity $f$ given by $f(x)=x$ for all $x\in[0,1]$ satisfies your assumption and has a continuum of fixed points.

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There should $\exists x\in[0,1]\;f(x)\neq x$ –  CBenni Dec 21 '12 at 13:27

Since $f(0)-0\geq0$ and $f(1)-1\leq0$, by the IVT, there must be at least one fixed point. Since $f$ is its own left-sided inverse, the function must be injective, hence it is either strictly increasing or strictly decreasing.

If it is decreasing, then everything is okay (indeed, the function $f(x)=-x+1$ has the properties, and even more generally, $f(x)=-x+(1-c)$). I claim that it cannot be increasing, therefore it must be decreasing. Since we know there is at least one $x$ where $f(x)\neq x$, we know either $f(x)>x$ or $f(x)<x$. Since strictly increasing functions preserve the inequality, we have $f\big(f(x)\big)>f(x)>x$ or vice versa, and in particular, the composition is not the identity. Thus, the function must be decreasing.

Now, since the function is strictly decreasing, it cannot have more than one fixed point, thus we have proven the conclusion.

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Thanks for your help I need help for this question Does a continuous mapping f : R ⟶ R which satisfies f(f(x)) =x for each x ∈ R necessarily have a fixed point? –  salar_ve Dec 22 '12 at 15:42
    
From above, we know that it must be injective and decreasing. Thinking about this, can you say whether or not it must cross the line $y=x$ (i.e., it has a fixed point)? –  Clayton Dec 22 '12 at 16:36

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