# Non-constant entire functions

Question: If $g$ is a non-constant entire function does it follow that $G_1(z)=g(z)-g\left(z+e^{g(z)}\right)$ is non-constant?

The reason I care is it would imply Prop 3 below, which in turn implies Prop 1, giving a proof that might seem better motivated than the proof here.

It seems at least plausible. The only thought I've had is a little vague: $-G_1(z)$ is something somewhat like $g'(z)e^{g(z)}$. And $g'e^g$ cannot be constant: If $g'e^g=c$ then $e^g=cz+d$. So $cz+d$ has no zero, hence $c=0$, and it follows that $g$ is constant. The problem is that the size of $g'(z)e^{g(z)}+G_1(z)$ depends on the size of $e^g$ and also on the size of $g''$...

Context: The other day at the link above I learned something totally new to me:

Prop 1. Suppose $f$ is entire. Then $f\circ f$ has no fixed point if and only if $f$ is a nontrivial translation ($f(z)=z+b$, $b\ne0$.)

The obvious question is when $f$ has a fixed point - this is trivial:

Prop 2. The entire function $f$ has no fixed point if and only if $f(z)=z+e^{g(z)}$ for some entire function $g$.

(Proof: $f$ has no fixed point if and only if $f(z)-z$ has no zero...)

It occurred to me to try to use Prop 2 to prove Prop 1: Suppose $f\circ f$ has no fixed point. Then $f$ has no fixed point, so $f(z)=z+e^{g(z)}$, and now we have something to work with in investigating when $f\circ f$ has a fixed point...

It turns out that to get to Prop 1 from Prop 2 we need this:

Prop 3 If $g$ is a non-constant entire function then $G_2(z)=e^{g(z)}+e^{g\left(z+e^{g(z)}\right)}$ has a zero.

Proving two assertions above:

If the answer to the question is yes then Prop 3 follows: If $G_1$ is non-constant then little Picard shows that there exist $k\in\Bbb Z$ and $z\in\Bbb C$ with $G_1(z)=(2k+1)\pi i$: this shows that $G_2(z)=0$.

Prop 3 implies Prop 1: Suppose $f\circ f$ has no FP Then $f$ has no FP, so $f(z)=z+e^{g(z)}$. But now $f(f(z))=z+G_2(z)$, so saying $f\circ f$ has no FP is saying exactly that $G_2$ has no zero. So Prop 3 implies that $g$ is constant, hence $f$ is a non-trivial translation.

In fact the two are equivalent (and hence Prop 3 is in fact true):

Prop 1 implies Prop 3: Say $g$ is non-constant and let $f(z)=z+e^{g(z)}$. Then $f$ is not a translation, so $f\circ f$ has a FP, hence as before $G_2$ has a zero.

My "might seem better motivated" might be worth justifying, given the bizarre appearance of Prop 3. It's really perfectly natural: If we're investigating when $f\circ f$ has a FP it's natural to first wonder when $f$ has a FP. Then Prop 2 is trivial, and given Prop 2 the question of when $f\circ f$ has a FP leads naturally to Prop 3, since $f(f(z))=z+G_2(z)$. (And now saying $G_2$ has a zero is clearly equivalent to saying $G_1(z)=(2k+1)\pi i$ for some $z$, and to prove that we only need to show that $G_1$ is non-constant, hence the question...) Perfectly clear, qed.

The answer is: yes, the property holds. I.e., there is no non-constant, entire $g$ such that $g(z)-g(z+e^{g(z)})$ is constant.

The proof goes by contradiction, in a couple of claims:

Claim 1: If $g(z)-g(z+e^{g(z)}) = a,$ then $a \ne (2k+1) i\pi$, for any $k \in \mathbb{Z}$.

Proof: let $z = w + e^{g(w)}.$ Then, from the property, it follows that

$$g(w) - g(w+(1+e^{-a})e^{g(w)}) = 2a.$$

If $a = (2k+1)i\pi$ for some $k \in \mathbb{Z},$ then we get that $a=0$ above, a contradiction.

Claim 2: $g’ e^g$ is non-constant.

This has been proved in the original post; explicitly, if it were constant, then we would have $e^g = rz + b,$ a contradiction.

Claim 3: $g’$ satisfies both $$g’ = (1+g’ e^g) g’(z+e^g)$$ and $$g’ = (1+(1+e^{-a})g’ e^g) g’(z+ (1+e^{-a})e^g).$$

Proof: The identity we obtained in the proof of Claim 1 is still valid; differentiate it, along with the original identity, and you obtain the result.

Now, for the ending of the proof, we know that either $g’e^g = -1$ or $g’e^g=-\frac{1}{1+e^{-a}}$ has a solution. Here, we used Claim 1 to prove that $e^a+1 \ne 0,$ and little Piccard, together with Claim 2. Let $z_0$ be a complex number that satisfies either of those equalities. Plugging in $z_0$ in the expressions from Claim 3, we obtain a contradiction: indeed, we would have that $g’(z_0) =0$, which contradicts directly the fact that $g’(z_0) e^{g(z_0)} \in \{-1, -\frac{1}{1+e^{-a}}\}$.This finishes the proof.

• This seems right. Alas I don't "understand" it in the sense I'd hoped to - oh well... – David C. Ullrich Aug 30 '18 at 14:30