Distribution of Zeros of Exponential-Type Entire Functions with Constraints on Growth along a Line
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ribution of Zeros of Exponential-Type Entire Functions with Constraints on Growth along a Line A. E. Salimova1* and B. N. Khabibullin1** 1
Bashkir State University, Ufa, 450074 Russia
Received November 13, 2019; in final form, November 13, 2019; accepted March 18, 2020
Abstract—Let g = 0 be an entire function of exponential type in the complex plane C, and let Z = {zk }k=1,2,... be a sequence of points in C. We give a criterion for the existence of an entire function f = 0 of exponential type which vanishes on Z and satisfies the constraint ln |f (iy)| ≤ ln |g(iy)| + o(|y|),
y → ±∞.
Our results generalize and develop joint results of P. Malliavin and L. A. Rubel. Applications to multipliers for entire functions of exponential type, to analytic functionals and their convolutions in the complex plane, and to the completeness problem for exponential systems in spaces of locally analytic functions on compact spaces in terms of the widths of these spaces are given. DOI: 10.1134/S0001434620090308 Keywords: zeros of entire function, multiplier, analytic functional, convolution, completeness of exponential systems.
1. MAIN RESULTS We say that an entire function f in the complex plane C vanishes on a sequence Z = {zk }k=1,2,... of points on the complete plane C (and write f (Z) = 0) if, for each point z ∈ C, the multiplicity of the zero of f at z is not less than the number of occurrences of this point z in the sequence Z. An entire function f is called an entire function of exponential type (e.f.e.t) if its type typef := lim sup z→∞
ln |f (z)| |z|
(1.1)
is finite. Given an e.f.e.t. f , the 2π-periodic function hf (θ) := lim sup r→+∞
ln |f (reiθ )| , r
θ ∈ R,
is called its growth indicator [1], [2]. A sequence Z = {zk } ⊂ C is separated from the imaginary axis iR if |Re zk | |Im zk | >0 ⇐⇒ x→b
MATHEMATICAL NOTES
Vol. 108 No. 4 2020
(2.8)
584
SALIMOVA, KHABIBULLIN
then the increasing function mlh [nmrh [f ]] multiplied by 1/n satisfies the conditions 1 lh rh f (R) ≤ Am r, R; m [nm [f ]] for all a ≤ r < R < +∞, n 1 (mlh [nmrh [f ]])(R) = 0. lim b>R→b n(R)
(2.9A) (2.90 )
Proof. Let us prove the first part of the theorem. The function Am (a, R; mrh [f ]) is decreasing in R ∈ (a, b) by Proposition 2.1. Inequality (2.7A) is derived from the chain (2.5)
Am (r, R; f ) ≤ Am (r, R; mrh [f ]) ≤ sup Am (r, R; mrh [f ]) = Am (a, R; mrh [f ]), r∈[a,R]
where the last relation is equality (2.3) of Proposition 2.1. Let c > 0. It follows from the second relation in (2.6) that mrh [f ](x) = o(1) for b > x → b and there exists a number Rc ∈ [a, b) such that mrh [f ](x) ≤ c
for x ∈ [Rc , b).
(2.10)
For R > Rc , by the definition (2.1) of integral mean, we have ˆ Rc ˆ R 1 rh + mrh [f ] dm Am ((a, R; m [f ]) = m(R) − m(a) a Rc ≤
(m(Rc ) − m(a)) sup[a,Rc ] mrh [f ] m(R) − m(Rc ) +c . m(R) − m(a) m(R) − m(a)
Therefore, for b > R → b, the first and the third relation in (2.6) imply lim sup Am (a, R; mrh [f ]) ≤ c. b>R→b
Since c > 0 is arbitrary, this gives (2.70 ). Let us prove the second part. According to (2.4)
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