A functional equation concerning analytic functions

Question

Let $P$ be a polynomial; we ask about the existence of a non-constant analytic function $f : \Bbb{C}\longrightarrow \Bbb{C}$ 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.

Answer 1

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 $1$.

Answer 2

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).

Answer 3

An elementary argument based on the identity principle for power series.

If $\text{deg}(P)>1$, it has a fixed point $z_0$. We can assume $z_0=0$ (we can replace $f$ with $F(z):=f(z_0+z)$ and $p$ with $P(z):=p(z_0+z)-z_0$, so $F(P(z))=F(z)$ and $P(0)=0$). Then the conclusion is true even in a more general situation; in fact, just by comparing the power series expansions we have:

Let $f\in\mathbb{C}[[z]]$ and $p\in z\mathbb{C}[[z]]$ satisfy $f(z)=f(p(z))$. Then either

• $f$ is constant,

  or

• There are $n\in\mathbb N$ and $\lambda \in\mathbb C$ such that $p(z)=\lambda z$, $\lambda^n=1$ and $f\in\mathbb C[[z^n]]$.

$$\sim *\sim$$

All details: Write $f:= \sum_{k\ge0}f_kz^k$ and $p:= \sum_{k\ge1}p_kz^k$. Assume $f$ is not constant, and let $m$ be the minimum index of nonzero coefficients of $f$. So $\bf(*)$ we have $f=f_0+f_mz^m+O(z^{m+1})$ with $f_m\neq0$ and $p=p_1z+O(z^2)$. Hence $$f(p(z))=f_0+f_mp_1^mz^m +O(z^{m+1}),$$ which compared with $f(z)=f_0+f_mz^m+O(z^{m+1})$ implies that $ p_1^m=1$, so $\lambda:=p_1$ is some primitive $n$-th root of unity for some $n$. To prove $p$ is linear we can assume w.l.o.g $p_1=1$: we may replace $p$ with the $m$-fold composition $P:=p^{(m)}$, that has linear coefficient $p_1^m=1$ (and if $P$ is linear so is $p$). So assuming $p_1=1$ we prove that $p$ is linear showing by complete induction that $p_r=0$ for all $r \ge2$. Assume $p_j=0$ for all $1<j<r$. Next, by the Faà di Bruno formula: $$ f(p(z)) =\sum_{k\in\mathbb{N},\, j\in\mathbb{N}_+^k} f_kp_{j_1}\dots p_{j_k} z^{j_1+\dots+j_k},$$ we have in particular
$$f(p(z))-f(z)=mf_m p_rz^{m+r-1}+O(z^{m+r}),$$ which implies $p_r=0$, and proves that is linear. Finally, since $p(z)=\lambda z$, we have $f(p(z)) -f(z)= \sum_{k\ge0}f_k(\lambda^k-1)z^k=0$ while $\lambda^k\neq1$ for $k\notin n\mathbb Z$, and this implies $f\in\mathbb{C}[[z^n]].\qquad$ $\square$

$\bf(*)$ in a context of formal power series here $O(z^m)$ has the formal meaning of "any element of the ideal $z^{m}\mathbb{C}[[z]]$".