# Iterating a real continuous injective function having no fixed points.

Let $f: \mathbb R \rightarrow\mathbb R$ be a continuous injective function. If $f(x)≠x ,\forall x∈\mathbb R$ and there exists a positive integer $n$ such that $f^n(x)=x , \forall x∈\mathbb R$ , then how do we prove that $f^2(x)=x ,\forall x∈\mathbb R$ ?

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The conditions on $f$ should restrict it to be monotonically increasing and either $f(x) < x$ for all $x \in \mathbb{R}$ or $f(x) > x$ for all $x \in \mathbb{R}$. – Pratyush Sarkar Dec 22 '13 at 5:49
What do you mean by $f^n(x)$: do you mean $f(x)\cdot f(x)\cdots f(x)$ or $f\bigl(f(\cdots f(x)\cdots)\bigr)$? – Matemáticos Chibchas Dec 22 '13 at 5:57
@Matemáticos Chibchas:- $f^n$ is $f$ composed $n$ times. – Souvik Dey Dec 22 '13 at 5:58
If my conclusions above are correct, then does such an $n$ even exist? For example, if $f(x) > x$ for all $x \in \mathbb R$, then we have $f^n(x) > f^{n - 1}(x) > \dotsb > f(x) > x$ and hence $f^n(x) \neq x$ for any $x$ and $n$. – Pratyush Sarkar Dec 22 '13 at 6:06

I don't think any such $f$ exists, hence the statement is vacuously true.

Since $f$ is injective, it must be strictly monotonic.

Since $f(x) \neq x$ for all $x$, and $f$ is continuous, we must have (i) $f(x)<x$ for all $x$ or (ii) $f(x)>x$ for all $x$. In the first case, we see that $\lim_{x \to -\infty} f(x) = -\infty$, and in the second case, $\lim_{x \to \infty} f(x) = \infty$, hence $f$ must be increasing.

Combining, we see that $f$ is strictly increasing.

In Case (i), $f(x) < x$, hence $f(f(x)) < f(x)$, which gives $f^2(x) < x$.

In Case (ii), $f(x) > x$, hence $f(f(x)) > f(x)$, which gives $f^2(x) > x$.

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