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.
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$\begingroup$ 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 $\sum a_nf_n=0$... $\endgroup$– M. DusCommented Mar 22, 2019 at 21:14
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1$\begingroup$ @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. $\endgroup$– Mateusz KwaśnickiCommented Mar 22, 2019 at 21:18
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$\begingroup$ @Mateusz Kwasnicki: this does not help. Suppose they are all polynomials. If the linear combination is not constant is must be surjective. $\endgroup$– Alexandre EremenkoCommented Mar 22, 2019 at 21:21
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1$\begingroup$ Maybe $(f_n)$ denotes an infinite sequence of functions? $\endgroup$– Nik WeaverCommented Mar 22, 2019 at 21:31
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$\begingroup$ @AlexandreEremenko I made some edits, I hope it is more clear now. Indeed, the question is if this is true for any such family. $\endgroup$– user137377Commented Mar 22, 2019 at 23:17
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2 Answers
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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 $n$.
answered Mar 22, 2019 at 23:41
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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.
answered Mar 23, 2019 at 1:37
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1$\begingroup$ 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. $\endgroup$ Commented Mar 23, 2019 at 8:25
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1$\begingroup$ 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 $f-c\equiv d$. $\endgroup$ Commented Mar 23, 2019 at 17:40
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$\begingroup$ @Christian Remling: Hadamard factorization is fine of course. On my opinion, Wiman's theorem is somewhat simpler, but this is a question of opinion. $\endgroup$ Commented Mar 23, 2019 at 20:52
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$\begingroup$ @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... $\endgroup$ Commented Mar 25, 2019 at 14:53
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$\begingroup$ @Christian Remling: I agree with the first part ("everyone...") but the second part ("only one...") is certainly an exaggeration:-) $\endgroup$ Commented Mar 25, 2019 at 18:46