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In relation to this question , To prove : If $f^n$ has a unique fixed point $b$ then $f(b)=b$ , if $f: \mathbb R \to \mathbb R $ is a continuous function such that for some $n_o \in \mathbb N$ the $n_o$th iterate of $f$ has a fixed point , then is it true that $f$ has a fixed point ?

Is it possible that if $b$ is a fixed point of $f^{n_o}$ then $b$ must also be a fixed point of $f$?

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Suppose a fixed point $b$ of $f^{n_0}$ is not a fixed point of $f$. Then there is a smallest $k > 1$ such that $f^{k}(b) = b$ (this $k$ must be a divisor of $n_0$, but that's not important here). Then $b, f(b),\dotsc, f^{k-1}(b)$ are distinct points.

Let $f^m(b)$ be the largest among these points.

Since $f^{m-1}(b) < f^m(b)$ and $f^m(b) > f^{m+1}(b)$, the continuous function $g(x) = f(x) - x$ attains a positive value in $f^{m-1}(b)$, and a negative value in $f^m(b)$, hence $g$ has a zero in the interval $\bigl(f^{m-1}(b),f^m(b)\bigr)$, that is, $f$ has a fixed point in that interval.

There are conditions that force $b$ to be a fixed point of $f$, for example monotonicity, or contractiveness.

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Let $f$ be a function on the set $\{-1,1\}$ given by $f(1) = -1$ and $f(-1)=1$. $f^2$ has two fixed points $f^2(\pm 1) = \pm 1$, but $f$ does not have any.

So this idea does not hold in this simple case.


Otherwise examine $f(x) = -x$ which has only one fixed point at $x=0$. $f^2(x)=x$ and every point is a fixed point. This tells us that if $b$ is a fixed point for $f^2$, then it is not necessarily a fixed point for $f$.

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    $\begingroup$ I said $f$ is defined and continuous on the whole real line $\endgroup$ – Souvik Dey Jul 2 '14 at 14:09
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    $\begingroup$ True, but this function is not from $\mathbb R$ into $\mathbb R$ so this doesn't answer the question. $\endgroup$ – leo Jul 2 '14 at 14:10
  • $\begingroup$ Not hard to fix. Same example really. $\endgroup$ – Joel Jul 2 '14 at 14:13
  • $\begingroup$ It's not fixed. A counterexample would be $f$ such that $f^n$ has fixed points but $f$ does not. Your $f$ has a fixed point. $\endgroup$ – Kyle Jul 2 '14 at 14:16
  • $\begingroup$ There were two questions asked by the op. This addresses the second. $\endgroup$ – Joel Jul 2 '14 at 14:17

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