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If I have an entire function give as a power series $f(z)=\sum_{i=0}^{\infty}a_iz^i$, is there a way/technique to check if the function is surjective? Weierstarss factorization theorem gives that $f$ is not surjective if and only if $f=\exp(g)+c$ for some $\require{cancel}\cancel{surjective}$ entire function $g$. However I don't know how to check this. Is there, for example, some known result that gives the surjectivity of $f$ depending on the behavior (rate of convergence?) of the coefficients?

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When are entire functions surjective?

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Your statement that if $f$ is not surjective, then $f=e^g+c$ where $g$ is surjective is wrong: $g$ does not have to be surjective. Example: $f(z)=e^{e^z}$. By the way, this example permits an infinite iteration: there is an infinite sequence of entire functions $f_n$ such that all of them are zero-free, and $e^{f_{n+1}}=f_n$.

There are various results which imply that $f$ is surjective, you have to be more specific, what kind of criterion you want. In terms of growth of coefficients, if $$\rho:=\limsup_{n\to\infty}\frac{n\log n}{-\log|a_n|}<1$$ then $f$ is surjective. If this is equal to $1$, but $$\sigma:=\limsup_{n\to\infty}n|a_n|^{1/n}=0,$$ then $f$ is surjective.

Moreover, if $f$ is not surjective then either $\rho$ is a positive integer and $\sigma$ is non-zero and finite, or $\rho=\infty$. This gives a pretty strong sufficient condition of surjectivity in terms of coefficients.

Another types of conditions can be obtained when you know something about $f$ beyond the coefficients. For example, if $f$ has infinitely many zeros, but not too many: $$\limsup_{r\to\infty}\frac{\log n(r)}{\log r}<\rho,$$ where $n(r)$ is the number of zeros in the disk $|z|\leq r$, then $f$ is surjective. This can be very much refined, if needed.

One can give very many other sufficient conditions in terms of coefficients, for example if many of the coefficients are $0$ (gap series), and in other terms. For example, one can combine growth conditions with gap conditions. But the only reasonable necessary and sufficient condition is that $f=e^g+c$ with some entire $g$.

Reference: B. Ya. Levin books on entire functions (any of the two of them).

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  • $\begingroup$ This is exactly the type of conditions I was looking for, thank you. From what I can understand, the first conditions mean that $|a_n|$ does not go to $0$ too fast. However in my case $\rho=\infty$, I will look in the reference you kindly suggested, perhaps there will be something helpful there. By the way, if $f$ has finitely, but non-zero, many zeros (or any other value), Picard implies that it must be surjective right? So another way would be to show that it takes a certain value finitely many times. $\endgroup$
    – Markus
    Commented May 28, 2017 at 4:26
  • $\begingroup$ My second criterion generalizes the case of finitely many zeros. Speaking of $\rho=\infty$, it is more difficult, but if you give the exact statement that you need I may try to help. $\endgroup$
    – Alexandre Eremenko
    Commented May 28, 2017 at 5:15
  • $\begingroup$ The (family of) series I am looking at is a bit strange. Consider a $n\times n$ matrix $A$, two length $1$ column vectors $x$ and $y$ and assume the function $f(z)=\sum z^n x^{T}A^n y$ is entire for any choice of such $x$ and $y$ ($A$ is fixed). I want to conclude that $f$ is surjective or constant. Probably it cannot be done in general, as the result will depend on the choice of $x$, $y$, and $A$. $\endgroup$
    – Markus
    Commented May 28, 2017 at 15:13
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    $\begingroup$ There is a small typo in the formula for $\sigma$, it should be $\limsup_{n\to\infty}n|a_n|^{1/n}$. $\endgroup$
    – Martin R
    Commented Jul 28, 2021 at 17:23
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    $\begingroup$ @Nomas: Thanks I corrected the misprint. $\endgroup$
    – Alexandre Eremenko
    Commented Jul 29, 2021 at 0:02

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