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In this question, Is there an analytic function with $f(z)=f(e^{iz})$?, it was settled that there exists no non-constant analytic function $f$ such that $f=f \circ g$, where $g(z)=e^{iz}$. Below is an alternate attempt to prove it.

$\textbf{Proof}: ~~$Suppose $f'(z_0)\not = 0$ for some $z_0 \in \mathbb{C}$. Then, $f$ is locally injective around $z_0$, which implies $z=e^{iz}$ locally around $z_0$. Clearly this is impossible, since the zero set of $z \to z-e^{iz}$ can atmost be countable.

However, these approach seems to prove the following - Suppose $g$, $h $ are analytic, then for any non-constant analytic $f$, $f \circ g \not= f \circ h $ unless $g=h$.

This is not true. Am I missing something?

Community
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Hmm.
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1 Answers1

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It's wrong. It's perfectly possible to have $f\circ g=f\circ h$ with $h\neq g$; take for example $f:z\to z^2$ and $h=-g$, or $f:z\to e^z$ and $h=g+2ik\pi$ with $k\in \mathbb{Z}\setminus\{0\}$.

The problem in your argument is that you consider a point $z_0$ at the neighborhood of which $f$ is injective, but there is no reason to assume that $e^{iz_0}=z_0$...

Arnaud D.
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  • Ahh yes of course. Silly mistake on my part ! Thank you. – Hmm. Jun 20 '16 at 11:04
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    But of course, if $e^{iz_0}$ lands inside that neighbourhood of $z_0$ on which $f$ is injective, then my arguments will go through. In other words, we can look for the fixed point of $z \to e^{iz}$ to begin with. This motivates the answer to the original question to a great extent (at least for me). Thank you once again. – Hmm. Jun 20 '16 at 11:09
  • @Hmm. Yes I realized that when I looked the other question after writing my answer. It explains why it works in this very precise case, but not in my counterexamples : $z\mapsto z+2i\pi$ has no fixed points, and $z\mapsto -z$ has only one, but $z\mapsto z^2$ is not locally injective at $0$... – Arnaud D. Jun 20 '16 at 11:14