# Sums of entire surjective functions

Suppose $$(f_n)_n$$ is a countable family of entire, surjective functions, each $$f_n:\mathbb{C}\to\mathbb{C}$$. Can one always find complex scalars $$(a_n)_n$$, not all zero, such that $$\sum_{n=1}^{\infty} a_n f_n$$ is entire but not-surjective? In fact, I am interested in this question under the additional assumption that $$(f_n)_n$$ are not polynomials.

• I'm not sure I understand the question. Are there some extra assumptions on the $f_n$ or on the $a_n$ ? For instance, assume that $f_1=-f_2$, then $a_1=a_2=1$, $a_n=0$, $n\geq 3$, will yield $\sum a_nf_n=0$... – M. Dus Mar 22 at 21:14
• @M.Dus: I suppose OP asks whether one can always find such scalaras. I also guess that the sum is supposed to be entire, not surjective and non-constant. – Mateusz Kwaśnicki Mar 22 at 21:18
• @Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective. – Alexandre Eremenko Mar 22 at 21:21
• Maybe $(f_n)$ denotes an infinite sequence of functions? – Nik Weaver Mar 22 at 21:31
• @AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family. – user137377 Mar 22 at 23:17

One expects there to be no such $$a_n$$ in general, because the "typical" entire functions is surjective (those that aren't are of the special form $$z \mapsto c + \exp g(z)$$). An explicit example is $$f_n(z) = \cos z/n$$: any convergent linear combination $$f = \sum_n a_n f_n$$ is of order $$1$$, so if $$f$$ is not surjective then $$g$$ is a polynomial of degree at most $$1$$; but $$f$$ is even, so must be constant, from which it soon follows that $$a_n=0$$ for every $$n$$.

The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)

For example, all non-constant functions of order less than $$1/2$$ are surjective. This follows from an old theorem of Wiman that for such function $$f$$ there exists a sequence $$r_k\to\infty$$ such that $$\min_{|z|=r_k}|f(z)|\to\infty$$ as $$k\to \infty.$$ And of course linear combinations of functions of order less than $$1/2$$ are of order less than $$1/2$$.

Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $$\Lambda$$ be a sequence of integers $$n_k$$ which grows sufficiently fast, for example, such that $$n_k/k\to\infty$$, and consider the class of entire functions of the form $$f(z)=\sum_{n\in\Lambda}c_nz^n.$$ It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.

Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series, Proc. LMS, 1970, 21 525-539.

• Didn't the OP ask about infinite sums? The exponential function is the sum of its power series, and it is not surjective, so if $f_n(z) = z^n$, the answer is yes. – Mateusz Kwaśnicki Mar 23 at 8:25
• Do we really need a theorem of Wiman (or anyone else) to verify your claim that non-constant functions $f$ of order $<1$ are surjective? Doesn't this just follow from the Hadamard factorization, which implies that if $f-c$ doesn't have a zero, then $f-c\equiv d$. – Christian Remling Mar 23 at 17:40
• @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion. – Alexandre Eremenko Mar 23 at 20:52
• @AlexandreEremenko: Ok, thanks. I was just thinking that everyone knowns Hadamard factorization while you are probably the only one who is familiar with Wiman's theorem, but of course it doesn't matter... – Christian Remling Mar 25 at 14:53
• @Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-) – Alexandre Eremenko Mar 25 at 18:46