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It is well-known that a holomorphic function $f:\mathbb{C}\to \mathbb{C}$ which is injective must be of the form $f(z)=az+b$ for some $a,b\in \mathbb{C}$ with $a\neq 0$. See here for a proof.

My question is whether a similar thing holds true for real-analytic functions. In this situation, an example such as $f(x)=x^3$ shows that being analytic and bijective doesn't imply that the inverse is necessarily analytic. With this in mind, I'll state the question as follows: Suppose there exists a real-analytic $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is a bijection and $f^{-1}$ is also real-analytic. Is it the case that there must exist $a,b\in \mathbb{R}$ such that $a\neq 0$ and $f(x)=ax+b$ ?

My intuition says that there could be something like $\arctan(x)$ which satisfies all the requirements except that the domain is $(-\pi/2,\pi/2)$ rather than all of $\mathbb{R}$.

TheEmptyFunction
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    Consider $x^3+x$. – Moishe Kohan Sep 06 '23 at 16:42
  • Consider the set $S={ f_n(x): n \in \mathbb{N}},$ where $f_n(x)=x^{2n+1}$ for each $n \in \mathbb{N}.$ Each element in $S$ satisfies your condition. None of the element of $S$ is linear. – MathRookie2204 Sep 06 '23 at 17:05
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    @MathRookie2204 OP already accounted for your counterexample by talking about $x^3$; none of the functions you wrote have analytic inverses (the origin being the troublesome point). The example above (or more generally $x^{2n+1}+x$) do indeed have analytic inverses but aren’t affine. – peek-a-boo Sep 06 '23 at 17:30
  • Thanks for pointing out @peek-a-boo – MathRookie2204 Sep 07 '23 at 03:55

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