MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $P$ be a polynomial and suppose $f : \Bbb{C}\longrightarrow \Bbb{C}$ is a non-constant analytic function such for all $z \in \Bbb{C}, f(z) = f(P(z))$. Clearly when $P$ is linear we can find such $f$. what happens when $P$ is not linear ? Any suggestion would be helpful.

share|cite|improve this question
«Clearly» when $P$ has degree $1$ and a fixed-point which is either attractive or repulsive then every such $f$ is constant. This is the generic situation for polynomials of degree $1$ (especially linear ones). – Loïc Teyssier Jul 26 '13 at 14:11
The linear function $z\mapsto\exp(i\alpha\pi)z$, with $\alpha$ real and irrational, is not covered by Loic's comment but also admits no non-constant $f$ as in the question. – Andreas Blass Jul 26 '13 at 15:29
up vote 3 down vote accepted

Here is a more conventional proof:-) Let $M(r)=\max\{|f(z)|:|z|=r\}$. Maximum principle implies that this function is strictly increasing (unless $f$ is constant). This gives a contradiction because |P(z)|>|z| when $z$ is sufficiently large, and $P$ is of degree greater than $1$.

share|cite|improve this answer
Mine was more «dynamical» and yours «analytical». I don't know about «conventional» ;) – Loïc Teyssier Jul 27 '13 at 16:55

If the degree of $P$ is greater than $1$ then the Julia set $J$ of $P$ is a nonempty perfect compact set of $\mathbb C$, completely invariant by $P$. Obviously $f$ is constant on any orbit $(P^{\circ n}(z))_n$. It is well known that most orbits of points of $J$ are dense in $J$, so that $f$ must be constant on $J$ (and therefore constant everywhere).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.