6

Let $f$ be an entire function such that $f\circ f$ has no fixed points. Prove that $f$ is a translation $$z\mapsto f(z)=z+b \qquad (b\neq 0)$$

Firstly, we prove that there exists a constant $c\in \mathbb{C}\backslash \{0,1\}$ such that $$f(f(z))-z=c(f(z)-z) $$ applying Picard's little theorem. If $c=0$, then $f(f(z))=z$, so $f\circ f$ has a fixed point (absurd). If $c=1$, then $f(f(z))=f(z)$, so $f$ is the identity $f(z)=z$ and of course it has fixed point (absurd). Then, $$F(z)=\frac{f(f(z))-z}{(f(z)-z)}$$ is an entire function which does not take the values 0 and 1 so, by Picard's little theorem, it must be constant.

Also, I've proved that $f'\circ f$ is a constant function. Let's see this. Differentiating $$f(f(z))-z=c(f(z)-z) $$ we have $$f'(z)f'(f(z))-1=cf'(z)-c$$ $$f'(z)[f'(f(z))-c]=1-c$$ Again, the entire function $$G(z)=f'(z)[f'(f(z))-c]$$ does not take the values $0$ and $1$ so, by Picard's Little Theorem, then is constant.

However, I don't know how to prove this problem. Any help would be appreciate.

user326159
  • 3,322

1 Answers1

1

Note there may or may not be considerable overlap between what's below and what's in the current version of the question; if so it's because the OP was revising the question while I was typing the answer. What actually happened: In the original version of the question he or she asserted that $f'\circ f$ was constant. I showed how the result followed from that and asked how to show that $f'\circ f$ was constant. He gave a few not quite right arguments to that effect while I was concocting my own proof. He gets credit for $f(f(z))-z=c(f(z)-z)$ and for saying that it follows that $f'\circ f$ is constant...

Hmm. Took me a minute to see how to prove the first thing you say you've proved. I don't see yet how to show that $f'\circ f$ is constant. But if that's correct you're done: Since $f$ has no fixed point $f$ is nonconstant; so the range of $f$ is dense, hence $f'(f(z))=k$ for all $z$ implies $f'(z)=k$ for all $z$.

How do you show $f'\circ f$ is constant?

Ah, here's how you show that: First, if $c=0$ then $f\circ f$ has lots of fixed points. So $c\ne0$. Differentiating the first identity shows that $$f'(z)(f'(f(z))-c)=1-c.$$

Hence $f'\circ f$ cannot take the values $0$ or $c$. If $f'\circ f(w)=c$ then $c=1$, so $f(f(z))=f(z)$, for all $z$,, so $f(z)$ is a fixed point of $f$ and hence of $f\circ f$. And if $f'\circ f(w)=0$ then in particular $f'$ has a zero, so it follows again that $c=1$.

So Picard shows that $f'\circ f$ is constant (since $c\ne0$).

For the benefit of anyone confused by the proof in the OP that $f(f(z))-z=c(f(z)-z)$: since $f$ has no fixed point, $$F(z)=\frac{f(f(z))-z}{f(z)-z}$$ is entire. If $F(z)=0$ then $f\circ f$ has a fixed point,, while $F(z)=1$ implies $f(f(z))=f(z)$, so $f$ has a fixed point. So Picard shows $F$ is constant.

David C. Ullrich
  • 93,621