7
$\begingroup$

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?

This question was asked before, but did not receive an answer at the time.

When are entire functions surjective?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
19
$\begingroup$

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).

Improve this answer
$\endgroup$
16
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @Nomas: Thanks I corrected the misprint. $\endgroup$
    – Alexandre Eremenko
    Commented Jul 29, 2021 at 0:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .