2

If $f$,$g$ are entire functions such that and $g′(z)=f(g(z))$ everywhere, then show that $f$ is linear or $g$ is constant.

This claim was mentioned without proof in this related question (Applications of the Little and Great Theorems of Picard). I am unable to prove this.

One way might be to show that $f$ is injective by using Picard's Little Theorem - but I have no idea how to proceed.

Agile_Eagle
  • 2,922

1 Answers1

2

Assume that there is such a solution with $g(a)=g(b)$. Then $h(z)=g(z-a+b)$ is such that $h(a)=g(a)$ and $h’=f(h), g’=f(g)$, thus $h=g$ so $g$ is $b-a$-periodic.

In particular, we can assume that $g$ is linear or that its set of periods is $2i\pi\mathbb{Z}$. In the first case, $f$ is constant (unless $g$ is).

In the second case, $g(z)=h(e^z)$, and $h$ is injective. But $h$ has singularities at $0$ and $\infty$, thus $h$ is meromorphic on $\mathbb{CP}^1$ (by eg Casorati-Weierstrass, or Picard’s theorem) and injective on that projective line minus two points, so it is a rational fraction of degree $1$ defined on $\mathbb{C}^{\times}$, so that $g(z)=C+De^{\pm z}$, and $g’(z)=\pm D e^{\pm z}$, so that $f(t)=\pm (t-C)$ is linear.

Aphelli
  • 34,439
  • 1
    Should it be $h(z)=g(z-a+b)$? – Martin R Apr 17 '22 at 16:30
  • 1
    @MartinR Thank you for pointing this out! I corrected. – Aphelli Apr 17 '22 at 16:33
  • @Mindlack Why does periodicity imply $g = h(e^z)$ for some injective $h$? – MathPanda Apr 17 '22 at 18:00
  • @MathPanda: my claim is more subtle. For any complex number $d$, the function $g(z+d)-g(z)$ has a zero iff it is zero. So let $\Lambda$ be the group of periods of $g$. Holomorphic functions have isolated zeros, so $\Lambda$ must be a discrete subgroup. As $g$ is entire, $\Lambda$ cannot have rank two, so $\Lambda=\alpha\mathbb{Z}$ for some $\alpha$. If $\alpha=0$, then $g$ is injective (hence linear), otherwise up to scaling we may assume $\alpha=2i\pi$ and thus $g(z)=h(e^z)$. If $u=e^x$, $v=e^y$ and $h(u)=h(v)$ so $g(x)=g(y)$ so $x-y \in \Lambda$ so that $u=v$. – Aphelli Apr 17 '22 at 18:59