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let $f(x)=x^2$, then $f(f(x))=x^4$, so $x^4$ is a continuous function from $\Bbb R$ to $\Bbb R$ which can be obtained as $f\circ f$ for a continuous $f\colon \Bbb R\to \Bbb R$. general example: for $f(x)=|x|^{\sqrt{d}}$, $d>0$, we have $(f\circ f)(x)=|x|^d$.

I wonder if it is possible to find for example continuous functions $f\colon \Bbb R\to \Bbb R$ such that

$f(f(x))=x^2+1$

$f(f(x))=x^3$

$f(f(x))=\sin x$

$f(f(x))=\arcsin x$

$f(f(x))=\ln x$

$f(f(x))=e^x$

Is it known?

I don't know which theory can be applied to such problem, I formulated these questions while walking with my dog :-)

Are there any general results or counterexamples?

Gregory Grant
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larry01
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1 Answers1

2

A few illustrative examples:

$f(f(x))=x^2+1$: It suffices to find $g\colon [0,\infty)\to[0,\infty)$ with $g(g(x))=x^2+1$ and let $f(x)=g(|x|)$. Note that $g\circ g$ is strictly increasing and has no fixed point. Therefore the sequence $x_0=0, x_1=\frac12$, $x_n=x_{n-2}^2+1$ is strictly increasing and diverges to $\infty$. Assume we have defined continuous $g_n\colon [x_0,x_n]\to \mathbb R$ such that $g_n$ is a bijection $[x_{k-1},x_k]\to [x_k,x_{k+1}]$ for $1\le k\le n$ and such that $g_n(g_n(x))=x^2+1$ for all $x\in[x_0,x_{n-1}]$. Then we can define $g_{n+1}\colon [x_0,x_{n+1}]\to \mathbb R$ with the corresponding property by letting $$g_{n+1}(x)=\begin{cases}g_n(x)&x\in[x_0,x_n]\\ g_n^{-1}(x)^2+1&x\in[x_n,x_{n+1}]\end{cases} $$ In summary this defines the $g$ we desired (if we start, e.g., with $g_1(x)=x+\frac12$).

$f(f(x))=x^3$: Your trick for $|x|^d$ can be adjusted: Let $f(x)=\operatorname{sgn}(x)\cdot |x|^{\sqrt 3}$.

Hagen von Eitzen
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