Entire Functions with Separated Zeros and 1-Points

We consider transcendental entire functions of finite order for which the zeros and 1-points are in disjoint sectors. Under suitable hypotheses on the sizes of these sectors we show that such functions must have a specific form, or that such functions do not exist at all.


Introduction and Results
Our starting point is the following result of Biernacki [3, p. 533]. Theorem A There is no transcendental entire function of finite order for which the zeros accumulate in one direction and the 1-points accumulate in a different direction.
Here we say that a set {a n } of complex numbers accumulates in one direction if there exists a ray such that for every open sector bisected by this ray all but finitely many a n lie in this sector.
We will prove the following generalizations of Theorem A. Theorem 1.1 Let S 0 and S 1 be closed sectors in C satisfying S 0 ∩ S 1 = {0}. Let θ j denote the opening angle of S j and suppose that min{θ 0 , θ 1 } < π 2 and max{θ 0 , θ 1 } < π.
Then there is no transcendental entire function of finite order for which all but finitely many zeros are in S 0 while all but finitely many 1-points are in S 1 .
Theorem 1.2 Let S be a closed sector in C of opening angle less than π/3 and let H be a closed half-plane intersecting S only in 0. Let f be a transcendental entire function of finite order. Suppose that all but finitely many zeros of f are in S while all but finitely many 1-points are in H . Then f has the form f (z) = P(z)e az where P is a polynomial and a ∈ C.
The following examples show that the hypotheses of Theorems 1.1 and 1.2 are sharp. We will verify in Sect. 2 that these examples have the stated properties.
The exponential function shows that the condition max{θ 0 , θ 1 } < π in Theorem 1.1 cannot be relaxed to max{θ 0 , θ 1 } ≤ π . This is also shown by the following example, which has infinitely many zeros and 1-points. This example also shows that the conclusion of Theorem 1.2 does not hold if S has opening angle equal to π/3.

Example 1.2 Let a and b be defined by
Then all but finitely many zeros of f are in {z : | arg z| ≤ π/6} while all but finitely many 1-points of f are in {z : Re z ≤ 0}.
where p and q are polynomials and c ∈ C.
The conclusion of Theorem 1.3 does not hold if one of the sectors S 0 and S 1 has opening angle greater than π . Similarly, the half-plane in Theorem 1.2 cannot be replaced by a sector of opening angle greater than π . For example, if 0 < ρ < 1 and (a n ) is a sequence of positive real numbers satisfying a n ∼ n 1/ρ as n → ∞, then is an entire function with positive zeros for which the 1-points accumulate at the rays arg z = ±π(1 − 1/2ρ); see, e.g., [8,Sect. 2.5] for the asymptotics of canonical products with positive zeros. The restriction on the order is essential in all stated theorems. In fact, for any two distinct directions there exists an entire function f of infinite order such that the zeros of f accumulate in one of these directions while the 1-points accumulate in the other direction; see, e.g., [2,Thm. 5].
On the other hand, a classical result of Edrei [5] says that if the zeros and 1-points of an entire function f lie on finitely many rays, then f has finite order.

Examples
We consider functions f of the form (1.1). Let d := deg(q) and let A be the leading coefficient of q so that q(z) ∼ Az d as z → ∞. For k ∈ {1, 2, . . . , d} we put It is easy to see that the limits exist. For ε > 0 we then have, as |z| → ∞, Here we have put φ d+1 = φ 1 + 2π .
The following result is implicit in [2]. For completeness, we include a proof.
Lemma 3.1 Let u be subharmonic function in a neighborhood of 0 and u(0) = 0. Let α ∈ (0, π] and suppose that u(z) < 0 for | arg z| < α. Then α ≤ π/2, and there exists c > 0 and r 0 > 0 such that Proof Let r 1 > 0 be such that the closed disk of radius r 1 around 0 is in the domain of u. Then there exists a harmonic majorant h of u in the (truncated) sector S α := {z : | arg z| < α, |z| < r 1 } such that h(re ±iα ) = 0 for r ∈ (0, r 1 ). Let r 2 := r π/(2α) 1 and S π/2 := {z : where the principal branch of the root is used, is negative and harmonic in S π/2 , and zero on the vertical part of the boundary. By the reflection principle v extends to a harmonic function in the disc of radius r 2 around 0, and ∇v(0) = 0. Therefore v(z) = −c 0 Re z + O(z 2 ) near zero for some c 0 > 0. Thus To show that α ≤ π/2, let S α = {z : α < arg z < 2π − α, |z| < r 1 } be the complement of S α in the disk of radius r 1 and let h 1 be a harmonic majorant of u in S α which satisfies h 1 (re ±iα ) = 0 for r ∈ (0, r 1 ). The same argument as before shows that h 1 (z) = O(z π/(2β) ) as z → 0, where β := π − α. Thus there exists a constant Together with (3.1) this yields that We conclude that β ≥ α and thus α ≤ π/2.
In the case that α = π/2 we will use the following result.

Lemma 3.2 Let u be subharmonic in C.
Suppose that u(0) = 0 and that there exist ρ > 0 and K > 0 such that u(z) ≤ K |z| ρ for all z ∈ C. If u is negative in some half-plane, then ρ = 1.
In the proof of Lemma 3.2 we will use the following version of the Phragmén-Lindelöf theorem [14, Cor. 2.3.8].

Proof of Lemma 3.2
Without loss of generality we may assume that u(z) < 0 for Re z < 0. Lemma 3.1 and the hypotheses that u(0) = 0 and u(z) ≤ K |z| ρ yield that if r is sufficiently small, then This yields that ρ ≤ 1. Hence (3.2) holds with A = B = K . Jensen's formula (see [13,Sect. 7.2] or [9, Sect. 3.9]) yields that The order ρ of u is defined by Note that if u = log | f | for some entire function f , and if M(r ) denotes the maximum modulus of f , then we have B(r , u) = log M(r ). Thus the order of the subharmonic function u coincides with that of the entire function f .
If u has finite order, then there exists a non-negative integer q satisfying q ≤ ρ ≤ q + 1 such that Using (3.5) we also see that Moreover, integration by parts shows that the latter condition is equivalent to The following result [9,Thm. 4.2] is the subharmonic version of the Hadamard factorization theorem.

Lemma 3.4
Let u be subharmonic of finite order ρ with Riesz measure μ. Let q be the minimal integer such that (3.6) holds and let R > 0. Then u can be written in the form and h is a harmonic polynomial of degree at most ρ.

Proofs of the Theorems
Proof of Theorem 1. 3 Let θ j be the opening angle of S j . Since only one of the two sectors can have opening angle equal to π , we may assume without loss of generality that θ 0 < π. We may also assume that f (0) / ∈ {0, 1}. First we show that the genus of f is at least 1. Suppose that this is not the case so that f is of genus 0. Then f has the form with all but finitely many a k contained in S 0 and all but finitely many b k contained in S 1 . The Gauss-Lucas theorem and Hurwitz's theorem imply that all zeros of f are contained in the convex hull of the a k as well as in the convex hull of the b k . Thus f has only finitely many zeros. Since we assumed that f has genus 0, this contradicts our hypothesis that f is transcendental. Hence f has genus at least 1.
Let ρ be the order of f . Since f has genus at least 1 we have ρ ≥ 1. Let M(r ) denote the maximum modulus of f . We proceed as in [2] and note that since f is of finite order, there exists a sequence (r k ) tending to ∞ such that log M(tr k ) = O(log M(r k )) as k → ∞, The result of Drasin and Shea [4] says that and if λ ∈ R satisfies ρ * ≤ λ ≤ ρ * , then log M(r ) has a sequence of Pólya peaks of order λ; that is, there exists a sequence (r k ) tending to ∞ such that if ε > 0, then provided k is large enough.
For a sequence (r k ) satisfying (4.1) we consider, as in [2], the subharmonic functions Arguing as in [ exist and are subharmonic in C. Here the convergence is in the Schwartz space D . This implies that we also have convergence in L 1 loc . Moreover, the limits in (4.3) have the following properties: If (r k ) is a sequence of Pólya peaks of order λ > 0, then we also have We refer to [2, p. 97] for the deduction of these properties.
Suppose first that (r k ) is a sequence of Pólya peaks of order λ > 0 so that (a)-( f ) hold. Note that such a sequence exists by (4.2) since ρ ≥ 1.
Let P be the set where one of the functions u and v is positive. In view of (a) this set coincides with the set where both functions are positive. Let z 0 ∈ P. Then z 0 = 0 by (e) and since S 0 ∩ S 1 = {0} we deduce from (c) that one of the functions u and v is harmonic in some neighborhood of z 0 . In particular, it is continuous in z 0 and thus positive in some (possibly smaller) neighborhood of z 0 . Hence P is open.
Let N be the set of points where at least one of the functions u and v is negative. Since subharmonic functions are upper semicontinuous, N is also open.
Thus the complement E := C\(P ∪ N ) = {z : u(z) = v(z) = 0} is closed. We will show that E has no interior. Suppose to the contrary that E has an interior point z 0 . Without loss of generality we may assume that z 0 / ∈ S 1 . By (c), the function v is harmonic in C\S 1 , which contains a neighborhood of z 0 . It follows that v(z) = 0 for all z ∈ C\S 1 . By (a) we have u(z) ≤ 0 for all z ∈ C\S 1 . If u(z 1 ) < 0 for some z 1 ∈ C\S 1 , then u(z) < 0 for all z ∈ C\S 1 by the maximum principle. This is a contradiction since u(z 0 ) = 0 and we assumed that z 0 ∈ C\S 1 . Hence u(z) = 0 for all z ∈ C\S 1 . Since u is harmonic in C\S 0 by (c), we find that u(z) = 0 for all z ∈ C, a contradiction. Thus E has no interior.
Our next goal is to show that either N ⊂ S 0 ∪ S 1 or N ⊃ C\S 1 , where the latter case can occur only if θ 1 = π . To this end, let Q be a component of N such that u(z) < 0 for z ∈ Q. We will show that ∂ Q ⊂ S 1 . In order to do so, suppose that z 0 ∈ ∂ Q\S 1 . Then v is harmonic in some neighborhood V of z 0 . By (a) and (b) we have v(z) = 0 for z ∈ Q. It follows that v(z) = 0 for z ∈ V . On the other hand, V also contains a point z 1 such that u(z 1 ) > 0. Thus v(z 1 ) = u(z 1 ) > 0 by (a). This is a contradiction. Thus ∂ Q ⊂ S 1 . If θ 1 < π, then Q cannot contain C\S 1 by Lemma 3.1. Hence Q ⊂ S 1 if θ 1 < π. But if θ 1 = π , then it is also possible that Q ⊃ C\S 1 .
Similarly, if Q is a component of N such that v(z) < 0 for z ∈ Q, then ∂ Q ⊂ S 0 . Since we assumed that θ 0 < π, this implies together with Lemma 3.1 that Q ⊂ S 0 . Overall we see that N ⊂ S 0 ∪ S 1 or N ⊃ C\S 1 , as claimed above. In the latter case, u(z) < 0 for z ∈ C\S 1 .
We first consider the case that N ⊂ S 0 ∪ S 1 . It follows that u(z) ≥ 0 and v(z) ≥ 0 for z ∈ C\(S 0 ∪ S 1 ). Since both u and v are harmonic in C\(S 0 ∪ S 1 ) we actually have u(z) > 0 and v(z) > 0 for z ∈ C\(S 0 ∪ S 1 ) by the minimum principle. Hence u(z) = v(z) for z ∈ C\(S 0 ∪ S 1 ) by (a). Thus the function is well-defined and harmonic in C\{0}. By the removable singularity theorem [1,Thm. 2.3] it is harmonic in C. Hence w has the form w = Re g for some entire function g. Since w(0) = 0 we may choose g such that g(0) = 0. Since (r k ) is a sequence of Pólya peaks of order λ > 0, we can deduce from ( f ) that w(z) = Re g(z) ≤ |z| λ . (4.5) This implies that g is a polynomial. In fact, λ is a positive integer and we have for some c ∈ C. Moreover, |c| = 1 by (d).
In the above reasoning we can take any λ ∈ [ρ * , ρ * ] ∩ (0, ∞), but the conclusion gives that λ is a positive integer. Together with (4.2) we conclude that (4.7) Thus the only possible value for λ is λ = ρ. It follows from (4.7) that (4.1) is satisfied for any sequence (r k ) tending to ∞. We mention that this kind of argument appears first in [6,Sect. 7] and [7, p. 1209], and it was also used in [2, p. 100]. More precisely, (4.7) implies that for any δ > 0 there exist r 0 , t 0 > 0 such that This still yields (e). Still assuming that the set N where one of the functions u and v is negative is contained in S 0 ∪ S 1 , we again find that the function w defined by (4.4) is harmonic and of the form w = Re g for some entire function g. Instead of (4.5), which was obtained from ( f ), we now deduce from ( f ) that This implies that g is a polynomial of degree at most ρ + δ which has a zero of multiplicity at least ρ − δ at the origin. Choosing δ < 1 we again find that g has the form (4.6) with λ = ρ; that is, To summarize, every sequence tending to ∞ has a subsequence (r k ) such that Next we note that log | f | = Re(log f ) for a branch log f of the logarithm. Also, the derivative h of a holomorphic function h can be computed from its real part, via if h is holomorphic in {z : |z − a| ≤ r }. We can thus deduce from (4.9) that In particular, if T 1 is a closed subsector of C\S 1 , then f has only finitely many zeros in T 1 . Applying the same argument to (4.10) yields that f has only finitely many zeros in any closed subsector T 0 of C\S 0 . As we may choose T 0 and T 1 such that T 0 ∪ T 1 = C we conclude that f has only finitely many zeros in C. Since f and hence f have finite order this implies that f has the form f = pe q with polynomials p and q. Thus f has the form (1.1). It remains to consider the case that N ⊃ C\S 1 , with θ 1 = π and u(z) < 0 for z ∈ C\S 1 . We may assume without loss of generality that S 1 is the left half-plane and S 0 = {z : | arg z| ≤ π − ε} for some ε > 0. It follows from (e) and ( f ) that u satisfies the hypotheses of Lemma 3.2. This lemma then yields that ρ = 1. Since u is harmonic in C\S 0 by (c), Lemma 3.5 yields that u has the form u(z) = az + b. Since u(0) = 0 we have b = 0 and using (d) we see that |a| = 1 and in fact a = −1. Hence u(z) = − Re(z); that is, As before we can now deduce that the limits (4.3) exist not only if (r k ) is chosen as a sequence of Pólya peaks, but in fact for every sequence (r k ) tending to ∞. Once this is known, it is not difficult to see that the answer to the question of whether N ⊂ S 0 ∪ S 1 or N ⊃ C\S 1 does not depend on the choice of the sequence (r k ). Thus we only have to deal with the case that (4.11) holds for every sequence (r k ) tending to ∞. We will show that this implies that f has the form f (z) = e az+b with constants a and b. In particular, f has the form (1.1).
It follows from (4.11) that there is a curve γ tending to ∞ near the imaginary axis in both directions such that | f (z)| = 1 for z ∈ γ . Suppose that f has at least one zero. Then f is unbounded on the imaginary axis by Lemma 3.5, applied to the subharmonic function log | f |. Thus there exists a real sequence (y k ) such that as k → ∞. Without loss of generality we may assume that y k → +∞. Assuming that T k > 1 there exists x k > 0 such that z k := x k + iy k lies on the curve γ . We have x k = o(y k ) as k → ∞ by (4.11). As before it follows from (4.11) with r k = y k by differentiation that lim k→∞ y k f (y k z) f (y k z) log M(y k ) = −1 for z ∈ C\S 0 . (4.13) Put L k := (log M(y k ))/y k . It follows from (4.13) that provided k is large enough. Hence Let now γ k be the component of the intersection of γ with the disk {z : |z − z k | ≤ x k } that contains z k . Then f • γ k is a curve contained in the unit circle. We have f (z) = 0 the rays arg z = φ k + π/(2d) and arg z = φ k − π/(2d) is contained in T . Since the zeros or 1-points accumulate at these rays, this is a contradiction.
Proof of Theorem 1.2 Let again φ k and a k be given by (2.1) and (2.2), with d := deg(q). As in the proof of Theorem 1.1 we have a k ∈ {0, 1} for all k ∈ {1, . . . , d}. By hypothesis there exists a closed sector T of opening angle greater than π/3 which intersects H and S only in 0. This implies that T does not intersect any of the rays arg z = φ k ± π/(2d). It follows that π/d > π/3 and thus d < 3. Suppose that d = 2. Since the 1-points are contained in a half-plane we have a k = 0 for some k ∈ {1, 2}. This implies that the zeros accumulate at both rays arg z = φ k ± π/4. Hence there are infinitely many zeros not contained in S, a contradiction.
It follows that d = 1. This implies that f has the form given.
Funding Open Access funding provided by Projekt DEAL.
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.