Entire function $f$ such that $f(z)=f(P(z))$, where $P$ is a polynomial of degree at least $2$.

Question

Let $f$ be a entire function and $P$ be a polynomial of degree at least $2$.

If $$f(z)=f(P(z)),\quad \forall z\in\mathbb C,$$ Is the entire function $f$ constant function?

My thought: If $f$ is not constant, then $f$ must be transcendental entire function. If the sequence $\{z_n=P^n(z)\}$ iterated by $P$ has a limit point $w_0$ in $\mathbb C$, then $$f(z)=f(z_1)=\cdots=f(z_n)=\cdots,\quad \lim_{n\to\infty}f(z_n)=f(w_0),$$ then $f(z)\equiv w_0$ by Identity theorem. But this is not always ture. For enample, if $P(z)=z^2+1$, take $z=1$, then $z_n\to\infty$.

Also if $$f(z)=f(e^z),\quad \forall z\in\mathbb C,$$ can we have $f$ is constant function?

Any hints and help will welcome.

Answer 1

$P$ is a polynomial of degree at least two, so $P(z)/z \to\infty$ for $z \to \infty$ and therefore exists $R > 0$ such that $|P(z)| > |z|$ for $|z| \ge R$.

Now choose $w$ with $|w| = R$ and $|f(w)| = \max \{ |f(z)| : |z| = R \}$ and choose $z_1$ with $P(z_1) = w$. Then $|z_1| < R$ and $$ |f(z_1)| = |f(P(z_1))| = |f(w)| $$ so that $f$ is constant by the maximum modulus principle.

Answer 2

Let's note that if $\deg P \ge 2$ then for large enough $|w|=R$ the equation $P(z)=w$ has a solution $|z_w|<R$ since if $P(z)=az^n+..., a \ne 0, n \ge 2$ we have $|P(z)/z^n| \to |a|, |z| \to \infty$ so $|P(z)| \ge |a||z|^n/2, |z| \ge R(P)$ large enough, hence $|P(z)| \ge |a|R^n/2 >R, |z| \ge R$ large enough so for $|w|=R$ and $P(z)=w$ we must have $|z|<R$

But if now we take $w$ which attains maximum modulus of $f$ on the disc of radius $R$ as above we get a $|z|<R$ st $P(z)=w$ so $f(z)=f(w)$ and that implies $f$ constant.

A similar argument works for $f(z)=f(e^z)$ noting that $|w|=R >1$ then $|\log w|^2 \ge \log^2R+\pi^2$ when we choose the principal branch (and $\log (-R)=\log R +i \pi$ say) so for $|w|=R$ large enough we find $|z|< R, e^z=w$ and the maximum modulus gives $f$ constant as above

Answer 3

We can restrict around a fixed point $z$ of $P(z)$, which can be assumed to be $z=0$. So now look at

$$f(z) = f(P(z))$$

as germs of holomorphic functions around $0$. IF $f$ were not constant, then $f(z) =c+ a z^m + \textrm{h.o.t}$, so $f(z) = c + g(z)^m$, and we get

$$g(z)^m = g(P(z))^m$$

and from here

$$\omega g(z) = g(P(z))$$

where $\omega^m = 1$. Composing with $P(z)$ succesively we get $$g(z) = g(P^m(z))$$

But note that $g$ is invertible, so from the above we get

$$z = P^m(z)$$ not possible.

($P^m(z)$ means $P\circ P\circ \cdot \circ P(z)$)

$\bf{Added:}$ The same reasoning would apply for the problem $f(z) = f(P(z))$, when $P(z)$ has some fixed points, but is not linear, for instance $P(z) = e^z$.

Answer 4

I'd like to add to all the good existing answers that in fact, one can completely answer the question which entire functions satisfy $$f(z) = f(P(z))$$ for any complex polynomial $P$. For details see https://math.stackexchange.com/a/4915599/96384. In most cases, only constant functions satisfy this:

1. $f$ is constant unless: $P(z) = az+b$ and there is some $k \in\{1,2,\dots\}$ with $a^k=1$ i.e. $a$ is a root of unity.

In this remaining case, we encounter two distinctly different classes of exceptions:

2. In case $a=1$, we are asking about periodic entire functions $f(z) = f(z+b)$ with period $b\neq 0$ (of course for $b=0$ the condition is empty). These are well-known to be in one-to-one correspondence with analytic functions $g$ on the punctured plane $\mathbb C \setminus \{0\}$, via defining $g(w) = f\left(\dfrac{b}{2\pi i}\log(w)\right)$; that is, $f$ can be written as a "Laurent series in $e^{\frac{2\pi i}{b}}$", $$f(z)=\sum_{n=-\infty}^\infty a_n e^{\frac{2\pi i}{b} n}$$ for certain $a_n$. Compare e.g. Show this formula holds for an analytic, periodic function in each half plane or Periodic holomorphic function on the right half-plane or Entire "periodic" function and links from there about more statements to be made once one imposes growth conditions.

3. In case $a^k=1$ for a minimal $k\ge 2$, i.e. $a$ is a primitive $k$-th root of unity, the following interesting thing happens: $P(z)=az+b$ has the fixed point $p:= \frac{b}{1-a}$, and a quick calculation translates the condition $f(z) = f(P(z))$ into saying that the Taylor Series of $f$ around $p$ is of the form $$f(z)=\sum_{m\ge 0} a_{mk}(z-p)^{mk}$$ i.e. it contains only terms where the exponent is a multiple of $k$. Or in other words, it is a "Taylor Series in $(z-p)^k$". E.g. for $p=0$, $k=2$, these are the "even" functions $f(z)= f(-z)$.