Entire Functions and Lacunary Values.

[WWGD]

#Hi All,
Let ## f: \mathbb C \rightarrow \mathbb C ## be entire, i.e., analytic in the whole Complex plane. By one of Picard's theorems, ##f ## must be onto , except possibly for one value, called the lacunary value.
Question: say ##0## is the lacunary value of ##f ##. Must ## f ## be of the form ##e^{g(z)}## , with ##g(z)## analytic?.
Clearly if ##g(z)## has no lacunary values, then ## e^{g(z)}## will have only ##0## as its lacunary value,
and if ##g(z) ## has only ##w## as its lacunary value, then this value will be assumed in ## w+i2\pi n ##, so ##0## will still be the lacunary value of ## e^{g(z)}##. Maybe we can consider composing functions with known lacunary values, but I don't see offhand how, since I don't know the lacunary values of general entire functions.
I considered using Weirstrass factorization thm, but it seems overkill and has led nowhere.
Thanks.

 

[micromass]

Theorem: Let ##D## be any elementary domain (for example ##D = \mathbb{C}##), let ##f:D\rightarrow \mathbb{C}## be an analytic function such that ##f(z) \neq 0## for each ##z\in D##, then there exist an analytic function ##h## such that ##f(z) = e^{h(z)}## for all ##z\in D##.

Proof: Let ##F## be a primitive of ##f^\prime / f##. Then put ##G(z) = \frac{e^{F(z)}}{f(z)}##. One checks easily that ##G^\prime (z) = 0## for all ##z\in D##. Therefore, ##e^{F(z)} = Cf(z)## for some nonzero constant ##C##. Since ##e^z## is surjective, we can write ##C = e^c##. Then ##h(z) = F(z) - c## does the trick.

 

[WWGD]

Thanks, that shows the side I am aware of, but does it show that all entire functions that do not hit zero are of this form? Sorry, I know I am being lazy and using composition of entire functions would give an answer, but I wondered if someone knew the answer offhand.

 

[mfb]

Thanks, that shows the side I am aware of, but does it show that all entire functions that do not hit zero are of this form?

They all can be expressed in that way. That's what the theorem says.

 

[micromass]

I don't get it, how does my answer not answer your question?

 

[WWGD]

Sorry for not being clear: I am looking for a common way of generating/describing all such functions, I would say with an iff rule. I know that every function _can_ be expressed that way. What I meant to say is: does that form exhaust all possible functions with lacunary value 1?. I am looking for something along the lines of ##f: \mathbb C \rightarrow \mathbb C ## is an automorphism (global diffeomorphism) iff ##f(z)= az+b ##. Yes, you are right that any function ##f(z)## with ##f(z)## entire and never ##0## does have a logarithm ## g(z) ## in a simply-connected region so that ## e^{g(z)}=f(z) ## , but what I should have asked is whether this exhausts all possible such functions. Hope I was clearer this time

 

[micromass]

Sorry, I really don't get it.

 

[WWGD]

Let me think it through and I will try to be more clear. Thanks.

 

[mfb]

"For every entire nonzero non-constant function f there exists g with [property]" is as exhaustive as it can get.
The other direction is also true:
For every non-constant entire g there is a non-constant nonzero entire f with f(z)=e^g(z).