Expanding the reciprocal of an entire function with zeros in a strip in a Kreǐn series

This article has been downloaded from IOPscience. Please scroll down to see the full text article.

2011 Sb. Math. 202 1853

(http://iopscience.iop.org/1064-5616/202/12/A06)

Download details:

IP Address: 171.67.34.205

The article was downloaded on 21/02/2013 at 17:29

Please note that terms and conditions apply.

View the table of contents for this issue, or go to the journal homepage for more

Home Search Collections Journals About Contact us My IOPscience

Sbornik : Mathematics 202:12 1853–1871 c⃝ 2011 RAS(DoM) and LMS

Matematicheskiı Sbornik 202:12 137–156 DOI 10.1070/SM2011v202n12ABEH004210

Expanding the reciprocal of an entire functionwith zeros in a strip in a Kreın series

V. B. Sherstyukov

Abstract. If L(λ) is an entire function with only simple zeros, all of whichlie in some strip, the problem of representing L−1(λ) by a partial fractionseries is solved.

Bibliography: 26 titles.

Keywords: entire function, partial fraction series.

§ 1. Introduction

The progress made in the theory of Hermitian operators and the moment problem(see, for example, [1], [2]) prompted M. Kreın to introduce a certain class of entirefunctions and start a systematic investigation of them. This proved to be quiteuseful in applications (see [3]). We shall say that an entire function L(λ) whichonly has real simple zeros λk = 0, k = 1, 2, . . . , is in the Kreın class if

1L(λ)

= R(λ) +∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

+λ

λ2k

+ · · ·+ λp−1

λpk

}

= R(λ) + λp∞∑k=1

1L′(λk)λ

pk

1λ− λk

(1.1)

for some integer p > 0, where∞∑k=1

1|λk|p|L′(λk)|

<∞, (1.2)

and R(λ) is a polynomial. Note that if we replace p by p+1 in (1.2), then this will beequivalent to the condition that the series (1.1) converges absolutely and uniformlyon each compact subset of the complex plane not containing zeros of L(λ).

It was proved in [3] (Theorem 4) that each function in the Kreın class has expo-nential type and, in addition,∫ +∞

−∞

ln+ |L(x)|1 + x2

dx <∞. (1.3)

This research was carried out with the support of the Russian Foundation for Basic Research(grant no. 09-01-00255-a), the Departmental Analytic Target Programme “Development of Scien-tific Potential in Higher Education” (grant no. 2.1.1/6827), in the framework of state contractsП 268, П 795 and П 943 of the Federal Agency of Education and the Federal Target Programme“Scientific and Scientific-Pedagogical Personnel of Innovative Russia” (grant no. П1109).

AMS 2010 Mathematics Subject Classification. Primary 30B50, 30D10.

1854 V.B. Sherstyukov

Entire functions of exponential type (EFETs) satisfying (1.3), form the well-known Cartwright class. Each function in the Cartwright class has completelyregular growth and its indicator diagram is an interval of the imaginary axis (see [3],and [4], Ch. V, § 6). It was noted in [5] that for functions in the Kreın class thisinterval contains the point λ = 0.

The condition that the zeros λk, k = 1, 2, . . . , be real is essential in Theorem 4of [3], which was mentioned above. We can show this using a simple examplefrom [3], § 4: if an entire function L(λ) of positive exponential type has an expansion(1.1), (1.2) with p = 0 and R(λ) ≡ 0, then we also have the expansion

1L(λ2)

=∞∑k=1

ck

2√λk

{1

λ−√λk

− 1λ+

√λk

},

although L(λ2) does not have exponential type any longer.Recall that the class A (notation introduced by B. Ya. Levin) is the standard

name for the set of entire functions, introduced by N. I. Akhiezer [6] in 1946, whosezeros satisfy ∑

λk =0

∣∣∣∣Im 1λk

∣∣∣∣ <∞.

In view of the above example and the remark on condition (1.2), the following result,which improves on Theorem 4 in [3] is of interest. It concerns representations forfunctions in the class A (see [3], Theorem 5); another proof can be found in [4],Ch. V, § 6, Theorem 13.

Kreın’s theorem. Let L(λ) be an entire function with simple zeros λk, k =1, 2, . . . , such that

∞∑k=1

∣∣∣∣Im 1λk

∣∣∣∣ <∞.

If in addition,1

L(λ)=

∞∑k=1

ckλ− λk

,

∞∑k=1

∣∣∣∣ ckλk∣∣∣∣ <∞, (1.4)

then L(λ) is a function of exponential type and (1.3) holds.

Of course, this result also holds for functions which can be expanded in a seriesof the form (1.1) because we can reduce this more general case to an expansion(1.4) by multiplying L(λ) by a suitable polynomial (see details in [3]).

Various properties of the functions having representations (1.1) or (1.4) havebeen investigated and used by several authors in function theory and harmonicanalysis, operator theory and differential equations. Besides the papers [1], [2] andthe monograph [4], which we have already mentioned, we can point to the investi-gations by Keldysh and Ostrovskiı ([7], Ch. V, § 6 and Ch. VI, § 2), de Branges [8],Korobeınik [9], [10], Borichev and Sodin [11] and Maergoız [12] (§ 6) and [13]. Forinstance, in [9] Korobeınik looked at the solution space of a differential equationof infinite order with symbol a(λ) ∈ [0, 1] having an expansion in the form (1.4).In [11] the authors reveal the role of the Kreın class in the analysis of functions of

Expanding in a Kreın series 1855

bounded type and Akhiezer-Levin sets, related to Bernstein’s problem of weightedapproximation on the real axis. Some recent results (see [12]) concern representingthe reciprocal of an entire function of a special kind occurring in the theory ofanalytic proximate orders by a series of partial fractions.

In [14], the author describes the sets of real numbers that can be the zero sets ofreal functions in the Kreın class in terms of special conformal maps. We emphasizethat this description is pretty intricate and because of its specific features it cannotreally be used to find manageable tests for the representability of the reciprocal ofan entire function by a Kreın-type series. On the other hand, obtaining such testsis an important problem, which arises from both the intrinsic needs of the theory ofentire functions and the many applications of functions in the Kreın class that havebeen discovered. Many authors have displayed interest in this problem: de Branges,Koosis, Pedersen, Bakan amongst others. In this section we present several knownresults. For example, in 1959 de Branges showed that if an EFET L(λ) with onlysimple real zeros Λ = (λk)∞k=1 is in the Cartwright class and satisfies

|L(iy)| → +∞, y → ±∞,∑λk =0

1|L′(λk)| |λk|

<∞, (1.5)

then1

L(λ)=

∞∑k=1

1L′(λk)(λ− λk)

, λ ∈ C \ Λ (1.6)

(see [15], Lemma 2 for G ≡ 1). Koosis [16], pp. 204, 205 showed that we cannotdrop the condition that L is in the Cartwright class in de Branges’ lemma whilstretaining the other assumptions and the conclusion. In [17], Theorem 6.6 Pedersenshowed that a function of minimal exponential type with simple real zeros whichsatisfies (1.5) has an expansion (1.6). The most complete results on the inversion ofKreın’s theorem before this author’s papers [5] and [18] are due to Bakan [19]–[21]and have been published in the last decade. To state these results we borrow theirnotation; we shall also leave out the trivial case when the set Λ is empty or finite.

For an entire function L with simple zeros, which form a set Λ = (λk)∞k=1, let

dL := inf{q ∈ Z :

∑λk =0

1|L′(λk)| |λk|q+1

<∞}, χΛ(0) =

{1 if 0 ∈ Λ,0 if 0 /∈ Λ.

If dL < +∞, then for each integer p > max{0, dL} we can introduce the entirefunction

∆pL(λ) :=

1L(λ)

− χΛ(0)L′(0)λ

−∑λk =0

λp

λpkL′(λk)(λ− λk)

, λ ∈ C.

In [19], Theorem 3.1 and [20], Theorem 3.1, Bakan proved a new version of Kreın’stheorem, which we reproduce in full.

Bakan’s Theorem (I). Let L be a real entire (nonconstant) function with onlysimple real zeros Λ = (λk)∞k=1 such that dL < +∞. Then the following results areequivalent.

1856 V.B. Sherstyukov

1) There exists a nonnegative integer p > dL such that the entire function ∆pL

is a polynomial.2) L is an entire function of exponential type satisfying (1.3) (in other words,

L is a function in the Cartwright class).3) If Λ is a semibounded subset of the real axis, then L has exponential type

zero, while if Λ is not bounded either above or below, then L is an entirefunction of positive exponential type.

Clearly, for functions that are real on the real axis the implication 1) =⇒ 2) coin-cides with Kreın’s theorem. The proof of 2) =⇒ 1), in fact, reduces to a referenceto de Branges’ lemma. As for the implication 3) =⇒ 1), for functions of minimalexponential type it was established in Pedersen’s theorem mentioned above. Thus,the following points in Bakan’s Theorem (I) are new relative to the previous results:first, the implication 1) =⇒ 3), supplementing Kreın’s theorem; second, the resultstating that the reciprocal of a real EFET L with Λ ⊂ R and dL < +∞ expandsin a Kreın series. In our opinion, this is the most interesting result. Its proof isbased on Theorem 3.2 in [19] and on [20], and has been applied to topical problemsrelating to the compactness of families of entire functions (see [21]). For it to hold itis essential that L be real. In this connection we state a result recently announcedin [22] (Theorem 1), where L is not assumed to be real on R.

Let q ∈ Z. We say that a transcendental entire function L with only simplezeros Λ = (λk)∞k=1 such that dL < +∞ belongs to the class K q if for q 6 0 wehave (1.6) and for q > 1 there exists a polynomial P of degree q with only simplezeros M = (µj)

qj=1, which lie in C \ Λ, such that the following expansion holds for

λ ∈ C \ (Λ ∪M):

1L(λ)P (λ)

=q∑j=1

1P ′(µj)L(µj)(λ− µj)

+∞∑k=1

1P (λk)L′(λk)(λ− λk)

.

Bakan’s Theorem (II). Let q ∈ Z and suppose that dL < +∞. Then L belongsto the class A ∩ K q if and only if L is in the Cartwright class and there exista set E of relative measure zero and an integer N > 0 such that |y|N |L(iy)| → +∞as y → ±∞, |y| ∈ R+ \ E.

Kreın’s theorem, the results due to de Branges and Koosis and Bakan’s Theorems(I) and (II) could suggest that in describing the characteristic features of a (notnecessarily real) entire function with real zeros whose reciprocal has an expansionin a series of partial fractions we cannot get along without using the Cartwrightclass and some lower bounds for the growth of the function, on the imaginary axissay. However, it turns out that for functions with real zeros (and even for functionswith zeros in a strip; see [18]) only the existence of an upper bound for the growthof its absolute value and the fact that the indicator diagram contains the pointλ = 0 (see [5]) are crucial for this range of questions.

The author’s paper [18] focused on the properties of the function which areresponsible for the existence of an expansion in principle. More precisely, we foundnecessary and sufficient conditions, minimal in a certain sense, for the reciprocalof an entire function with zeros in a strip to have an expansion in a Kreın-typeseries. In this paper, we solve an ‘individual’ problem for such a function, namely

Expanding in a Kreın series 1857

expanding it in a series (1.1) for any p > 0. The corresponding results are new;they generalize and refine the author’s previous results in [5] and make up § 2. Themethods of our proofs are distinct from the ones in [19], [20]; they are based onideas from [5] and [18].

The author is pleased to acknowledge A. Yu. Popov’s interest in this work andour valuable discussions of its results.

§ 2. The main result

The proof of the central result of this section is based on an auxiliary result closeto a lemma due to Sedletskiı; see [23], Ch. 5, § 5.3.

Lemma 2.1. Let Λ = (λk)∞k=1 be a sequence of finite upper density with a uniquelimit point at infinity that lies in a horizontal strip | Imλ| 6 h. Also assume thatµk := Reλk = 0 for all k. Then the infinite product

π(λ) :=∞∏k=1

(1− λ

λk

)(1− λ

µk

)−1

has the following estimate on the imaginary axis :

infµ∈R

|π(iµ)| > 0.

Proof. Certainly, the infinite product π(λ) converges outside the strip in question.Setting λk = µk + iνk, |νk| 6 h, k = 1, 2, . . . , we see that

|π(iµ)|2 =∞∏k=1

∣∣∣∣ iµ− λkiµ− µk

∣∣∣∣2 · ∣∣∣∣µkλk∣∣∣∣2 =

∞∏k=1

(µ− νk)2 + µ2k

µ2 + µ2k

· µ2k

µ2k + ν2

k

>∞∏k=1

(|µ| − h)2 + µ2k

µ2 + µ2k

· µ2k

µ2k + h2

, µ ∈ R, |µ| > h.

By assumption there exists A > 0 such that for all t > 0 we have n(t) 6 At, wheren(t) is the counting function of the sequence (µk)∞k=1. Hence

ln∞∏k=1

µ2k + h2

µ2k

=∞∑k=1

ln(

1 +h2

µ2k

)6 πAh,

which shows that∞∏k=1

µ2k

µ2k + h2

> e−πAh.

1858 V.B. Sherstyukov

Now integrating by parts, for |µ| > 2h we obtain

ln∞∏k=1

µ2k + µ2

µ2k + (|µ| − h)2

=∞∑k=1

ln(

1 +µ2 − (|µ| − h)2

µ2k + (|µ| − h)2

)

6 h(2|µ| − h)∞∑k=1

1µ2k + (|µ| − h)2

= h(2|µ| − h)∫ +∞

0

dn(t)t2 + (|µ| − h)2

= h(2|µ| − h)(

n(t)t2 + (|µ| − h)2

∣∣∣∣+∞0

+∫ +∞

0

2tn(t) dt(t2 + (|µ| − h)2)2

)= h(2|µ| − h)

∫ +∞

0

2tn(t) dt(t2 + (|µ| − h)2)2

6 Ah(2|µ| − h)∫ +∞

0

2t2 dt(t2 + (|µ| − h)2)2

= Ah(2|µ| − h)∫ +∞

0

dt

t2 + (|µ| − h)2=πAh(2|µ| − h)

2(|µ| − h)6

3πAh2

.

Hence∞∏k=1

(|µ| − h)2 + µ2k

µ2 + µ2k

> e−3πAh/2, |µ| > 2h.

Finally,

|π(iµ)|2 > e−5πAh/2, |µ| > 2h,

which gives us the required estimate on the whole of the imaginary axis since theinfinite product π(λ) converges and does not vanish on this axis.

We also require the following elementary lemma.

Lemma 2.2. Let L(λ) be an entire function with simple zeros forming a set Λ =(λk)∞k=1, 0 /∈ Λ, and assume that

1L(λ)

= Qp(λ) +∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

+ · · ·+ λp

λp+1k

}(2.1)

for some integer p > 0, where the series converges absolutely on the set C \ Λ andQp(λ) is a polynomial of degree at most p. Then

Qp(λ) =1

L(0)− L′(0)L2(0)

· λ+ · · ·+ 1p!

(1

L(λ)

)(p)

(0) · λp.

Proof. Since the series in (2.1) converges uniformly on compact subsets of C \ Λ,we find the coefficients of Qp(λ) in the standard way by differentiating (2.1) therequired number of times.

The main result giving criteria which relax the conditions known previously forthe reciprocal of an entire function with simple zeros to be expandable in a Kreınseries is given in the following statement.

Expanding in a Kreın series 1859

Theorem 2.1. Let L(λ) be an entire function with simple zeros forming a sequenceΛ = (λk)∞k=1, 0 /∈ Λ, such that all the λk are simple and lie in a strip Π. Thenthe reciprocal 1

L(λ) expands for some nonnegative p in a Kreın series convergingabsolutely and uniformly on compact subsets of C \ Λ:

1L(λ)

=1

L(0)− L′(0)L2(0)

· λ+ · · ·+ 1p!

(1

L(λ)

)(p)

(0) · λp

+∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

+ · · ·+ λp

λp+1k

}(2.2)

if and only if the following conditions hold simultaneously :i) L(λ) has exponential type and its indicator diagram is a line segment ortho-

gonal to the boundary of Π and containing λ = 0;ii)

∑∞k=1

1|L′(λk)| |λk|p+2 < +∞.

Proof. By assumption, Λ lies in a strip. We can always put it in a wider strip whichis symmetric relative to λ = 0. Rotating the plane through a suitable angle in thestandard way we can assume without loss of generality that

Π = Πh := {λ ∈ C : | Imλ| 6 h}, h > 0,

and the line segment in i) lies on the imaginary axis.Necessity. Let L(λ) be an entire function with simple zeros forming a sequenceΛ = (λk)∞k=1, | Imλk| 6 h, 0 /∈ Λ, which has the representation (2.2). The conver-gence mode of (2.2) ensures ii).

Now we verify that L(λ) is a function of exponential type. We use the argumentdeveloped by Kreın [3], for functions with real zeros. We write (2.2) as

1L(λ)

= Qp(λ) + λp+1f(λ),

and shall show that the function

f(λ) :=∞∑k=1

ckλ− λk

,

∞∑k=1

∣∣∣∣ ckλk∣∣∣∣ <∞,

which occurs in this representation with coefficient λp+1, belongs to the Nevanlinnaclass N in both half-planes, Π+

h ={λ∈C : Imλ>h

}and Π−h =

{λ∈C : Imλ<−h

}.

This means that in each half-plane the function ln+ |f(λ)| has a harmonic majorant.We shall carry out the proof in the half-plane Π+

h . We shall use facts about thestructure of Nevanlinna classes which can be found in [3].

Obviously, the function f(λ) can be represented as

f(λ) = f1(λ)− f2(λ) + if3(λ)− if4(λ),

where each of the functions fm(λ) is not identically equal to zero and is representedby a series ∑

k∈Jm

ρk,mλ− λk

, ρk,m > 0, Jm ⊂ N, m = 1, 2, 3, 4.

1860 V.B. Sherstyukov

It is known that a sufficient condition for an analytic function in a simply connecteddomain G ⊂ C to belong to the class N in G is that its imaginary part has constantsign in the domain. For λ ∈ Π+

h and ρk > 0 we have

Im∑k∈Jm

ρk,mλ− λk

=∑k∈Jm

ρk,m Im1

λ− λk=

∑k∈Jm

ρk,m Im(λk − λ)|λ− λk|2

<∑k∈Jm

ρk,m Im(λk − h)|λ− λk|2

6 0.

Hence the functions fm(λ), m = 1, 2, 3, 4, —and therefore also f(λ) —belong to theclass N in Π+

h . Hence it follows from the representation (2.2) that 1L(λ) and L(λ)

belong to the class N in Π+h . In a similar way we can verify that L(λ) belongs to

the class N in Π−h because for λ ∈ Π−h and ρk,m > 0 we have

Im∑k∈Jm

ρk,mλ− λk

>∑k∈Jm

ρk,m Im(λk + h)|λ− λk|2

> 0.

Now by Theorem 2 in [3] as applied to the half-plane Π+h , there exists a real constant

σ1 such that for each θ ∈ (0, π)

limr→∞

ln |L(ih+ reiθ)|r

= σ1 sin θ

uniformly in θ ∈ [δ, π − δ] for each δ ∈ (0, π). From this we obtain a limit relationfor |L(λ)| on rays emanating from λ = 0. Namely, the limits

limr→∞

ln |L(reiθ)|r

= σ1 sin θ, θ ∈ (0, π),

exist. A similar relation also holds for θ ∈ (−π, 0) with some constant σ2 (indepen-dent of θ). Consequently, the function L has exponential type and its indicator isgiven by

hL(θ) =

{σ1 sin θ for θ ∈ [0, π],σ2 sin θ for θ ∈ (−π, 0).

Hence the indicator diagram D(L) of L is an interval of the imaginary axis.Thus, to deduce i) it remains to verify that 0 ∈ D(L). In view of ii), we shall

transform (2.2) as follows:

1L(λ)

= Qp(λ) + λp+1∞∑k=1

1L′(λk)λ

p+1k

1λ− λk

= Qp(λ) + λp+2∞∑k=1

1L′(λk)λ

p+2k

(1

λ− λk− 1λ

)

= Qp(λ)− λp+1∞∑k=1

1L′(λk)λ

p+2k

+ λp+2∞∑k=1

1L′(λk)λ

p+2k

1λ− λk

= Tp+1(λ) + λp+2∞∑k=1

1L′(λk)λ

p+2k

1λ− λk

,

Expanding in a Kreın series 1861

where Tp+1(λ) is a polynomial of degree at most p + 1. Estimating |L(λ)| on theimaginary axis, for each µ ∈ R, |µ| > h, we find

1|L(iµ)|

6 |Tp+1(iµ)|+ |µ|p+2

|µ| − h

∞∑k=1

1|L′(λk)| |λk|p+2

.

This shows that hL(±π2 ) > 0 and therefore 0 ∈ D(L).

Sufficiency. Let L(λ) be an EFET with simple zeros Λ = (λk)∞k=1 0 /∈ Λ, lying inthe strip | Imλ| 6 h which satisfies condition ii) and let D(L) be an interval on theimaginary axis containing the origin. We claim that we have an expansion (2.2).We prove this by induction on p > 0.

For p = 0 condition ii) is the requirement that the series

∞∑k=1

1|L′(λk)| |λk|2

converges. We must verify that

1L(λ)

=1

L(0)+

∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

}in this case.

Assuming in what follows that L(λ) has no purely imaginary zeros, we write outits Hadamard-Weierstrass representation:

L(λ) = eaλ+b∞∏k=1

(1− λ

λk

)eλ/λk , a, b ∈ C.

Setting

λk = µk + iνk, k = 1, 2, . . . , d = a+∞∑k=1

(1λk

− 1µk

),

π(λ) =∞∏k=1

(1− λ

λk

)(1− λ

µk

)−1

, L1(λ) = π(λ)eλRe d∞∏k=1

(1− λ

µk

)eλ/µk ,

we obtainL(λ) = L1(λ)eiλ Im d+b. (2.3)

In our case we require Lemma 2.1 to obtain a lower estimate of |π(λ)| on theimaginary axis; in combination with the obvious inequality∣∣∣∣ ∞∏

k=1

(1− iµ

µk

)eiµ/µk

∣∣∣∣ > 1, µ ∈ R,

this yields the following result:

∃D > 0 : |L1(iµ)| > D ∀µ ∈ R. (2.4)

1862 V.B. Sherstyukov

In view of the constraints ii) and |νk| 6 h, k = 1, 2, . . . , from (2.3) we obtain

∞∑k=1

1|L′1(λk)| |λk|2

< +∞. (2.5)

We start by showing that

1L1(λ)

=1

L1(0)+

∞∑k=1

1L′1(λk)

{1

λ− λk+

1λk

}. (2.6)

Let

ϕ(λ) :=1λ

(1

L1(λ)− 1L1(0)

)−

∞∑k=1

1L′1(λk)λk(λ− λk)

be an auxiliary function; it is analytic in C \ ({0} ∪ Λ), with residues at λ = 0and λ ∈ Λ equal to zero, so ϕ(λ) is an entire function. We claim that ϕ(λ) hasexponential type. To show this we represent ϕ(λ) as the ratio of two entire functions,

L1(0)− L1(λ)− L1(0)∞∑k=1

λL1(λ)L′1(λk)λk(λ− λk)

and λL1(0)L1(λ).

Since L1(λ) is an EFET, it is sufficient to show that

ψ(λ) :=∞∑k=1

L1(λ)L′1(λk)λk(λ− λk)

is an EFET and to use a corollary to the theorem on categories (see [4], Ch. I, § 9).For a suitable upper estimate for |ψ(λ)| we require the following notation for theconstants (see (2.5)):

A1 :=∞∑k=1

1|L′1(λk)| |λk|2

, A2 :=∞∑k=1

1|L′1(λk)| |λk|3

,

A3 :=∞∑k=1

1|L′1(λk)| |λk|4

.

Furthermore, let

U :=∞⋃k=1

Uk, Uk :={λ ∈ C : |λ− λk| < |λk|−2

}.

We give several estimates for λ /∈ U . First,∣∣∣∣ ∞∑k=1

1L′1(λk)λ

3k(λ− λk)

∣∣∣∣ =∣∣∣∣ ∞∑k=1

λ

L′1(λk)λ4k(λ− λk)

−∞∑k=1

1L′1(λk)λ

4k

∣∣∣∣6 |λ|

∞∑k=1

1|L′1(λk)| |λk|4|λ− λk|

+A3 6 A1|λ|+A3.

Expanding in a Kreın series 1863

Second, ∣∣∣∣ ∞∑k=1

1L′1(λk)λ

2k(λ− λk)

∣∣∣∣ =∣∣∣∣ ∞∑k=1

λ

L′1(λk)λ3k(λ− λk)

−∞∑k=1

1L′1(λk)λ

3k

∣∣∣∣6 |λ|

∣∣∣∣ ∞∑k=1

1L′1(λk)λ

3k(λ− λk)

∣∣∣∣ +A2 6 A1|λ|2 +A3|λ|+A2.

Finally, from the above estimates for the same λ we obtain

|ψ(λ)| = |L1(λ)|( ∞∑k=1

λ

L′1(λk)λ2k(λ− λk)

−∞∑k=1

1L′1(λk)λ

2k

)6 |L1(λ)|

(A1|λ|3 +A3|λ|2 +A2|λ|+A1

).

Hence there exists M > 0 such that

|ψ(λ)| 6 M |λ|3|L1(λ)|, λ /∈ U.

The set U is a union of discs whose radii have a finite sum because the convergenceexponent of the sequence Λ does not exceed the order of L1(λ). Using the mainresult in [24] we can cover U by disjoint discs forming a set of linear density zero.Applying the maximum modulus principle to ψ(λ) in these discs, from the estimatefor |ψ(λ)| obtained above for λ /∈ U , since the indicator of L1(λ) is continuous, wesee that ψ(λ) has exponential type and its indicator does not exceed hL1(θ).

Now we find estimates for the growth of the EFET ϕ(λ) on the imaginary axis.By (2.5), for any small ε > 0 there exists an index N such that

∞∑k=N+1

1|L′1(λk)| |λk|2

<ε

2√

2,

and furthermore, |µk| > h, k > N . Now if µ ∈ R, |µ| > h, and k > N , then

|iµ− λk| > |µk| >1√2

√h2 + µ2

k >1√2

√ν2k + µ2

k =|λk|√

2.

Hence for all sufficiently large absolute values of µ∣∣∣∣ ∞∑k=1

1L′1(λk)λk(iµ− λk)

∣∣∣∣ 6N∑k=1

1|L′1(λk)| |λk| |iµ− λk|

+∞∑

k=N+1

1|L′1(λk)| |λk| |iµ− λk|

<ε

2+√

2∞∑

k=N+1

1|L′1(λk)| |λk|2

< ε.

Taking (2.4) into account we see that as µ→ ±∞,

ϕ(iµ) → 0.

By Theorem 11 in [4], Ch. V, § 4, ϕ(λ) has completely regular growth. It followsfrom the definition of this function that

ϕ(λ)λL1(λ) = 1− L1(λ)L1(0)

− λψ(λ). (2.7)

1864 V.B. Sherstyukov

Since the indicator diagram D(L1) can be obtained from D(L) by a translationalong the imaginary axis and as (2.4) holds, D(L1) is an interval on the imaginaryaxis containing the origin. Now by (2.7), since the indicator hL1(θ) is nonnegative,hψ(θ) 6 hL1(θ) and ϕ(λ) has completely regular growth, it follows that ϕ ∈ [1, 0].But if such a function vanishes on the imaginary axis, it must vanish identically.Thus, we have proved that the expansion (2.6) exists. By Kreın’s theorem L1(λ) isa function of completely regular growth, therefore the function L(λ) is too.

Now we must verify that we have an expansion of the form (2.6) for L(λ).We consider three cases which are possible under condition i).If λ = 0 is not an end-point of the interval coinciding with the indicator diagram

of L(λ), then |L(iµ)| → ∞ as µ→ ±∞. In this case 1L(λ) is represented by a Kreın

series (2.6), and this can be demonstrated using the arguments we have appliedto L1(λ).

Now let λ = 0 be an end-point of the indicator diagram of L(λ); for definiteness,let [−iδ, 0], δ > 0, be the indicator diagram. Then for each ε ∈ (0, δ) we canrepeat the above arguments for the function L(λ)e−iελ (the same estimate for thederivative at the λk holds because these points lie in a horizontal strip). Then weobtain an expansion into partial fractions:

eiελ

L(λ)=

1L(0)

+∞∑k=1

eiελk

L′(λk)

{1

λ− λk+

1λk

}. (2.8)

Since the series on the right-hand side of (2.8) converges uniformly in ε for fixed λ(again, we use the inequalities |νk| 6 h, k = 1, 2, . . . ), we can pass to the limit asε→ +0 termwise and establish the required relation

1L(λ)

=1

L(0)+

∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

}.

Now assume that the indicator diagram of L(λ) reduces to the single point λ = 0,so that L(λ) has exponential type zero. Conditions (2.3) and (2.4) show that forD1 = DeRe b > 0 we have

|L(iµ)| > D1e−µ Im d, µ ∈ R.

However, a function of exponential type zero can satisfy the last inequality only ifIm d = 0. Hence L(λ) = ebL1(λ), and we have already derived the expansion (2.6)for 1

L1(λ) . This completes the proof of sufficiency for p = 0.

It remains to make the induction step. Assume that for some p ∈ N each EFETL(λ) with simple zeros

Λ = (λk)∞k=1 ⊂{λ ∈ C : | Imλ| 6 h

}\ {0} such that

∞∑k=1

1|L′(λk)| |λk|p+1

< +∞,

Expanding in a Kreın series 1865

and with the property 0 ∈ D(L) ⊂ iR has the representation

1L(λ)

=1

L(0)− L′(0)L2(0)

· λ+ · · ·+ 1(p− 1)!

(1

L(λ)

)(p−1)

(0) · λp−1

+∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

+ · · ·+ λp−1

λpk

}.

Then we look at an EFET L(λ) where the indicator diagram and zeros have thesame properties, but

∞∑k=1

1|L′(λk)| |λk|p+2

< +∞.

Then for fixed µ /∈ Λ∪{0} the function Lµ(λ) := L(λ)(λ−µ) satisfies the inductiveassumptions and therefore

1Lµ(λ)

=1

Lµ(0)−L′µ(0)L2µ(0)

· λ+ · · ·+ 1(p− 1)!

(1

Lµ(λ)

)(p−1)

(0) · λp−1

+∞∑k=1

1L′µ(λk)

{1

λ− λk+

1λk

+ · · ·+ λp−1

λpk

}+

1L(µ)

{1

λ− µ+

1µ

+ · · ·+ λp−1

µp

}.

Hence

1L(λ)

= (λ− µ){

1Lµ(0)

−L′µ(0)L2µ(0)

· λ+ · · ·+ 1(p− 1)!

(1

Lµ(λ)

)(p−1)

(0) · λp−1

}+

λp

µpL(µ)+

∞∑k=1

(λ− µ)λp

L′(λk)λpk(λ− λk)(λk − µ)

.

Next we make some transformations:∞∑k=1

(λ− µ)λp

L′(λk)λpk(λ− λk)(λk − µ)

= λp+1∞∑k=1

λk

L′(λk)λp+1k (λ− λk)(λk − µ)

− λpµ

∞∑k=1

1L′(λk)λ

pk(λ− λk)(λk − µ)

= λp+1∞∑k=1

1L′(λk)λ

p+1k (λ− λk)

+ λpµ

∞∑k=1

1L′(λk)λ

p+1k (λk − µ)

.

We shall use the notation Qp,µ(λ) for the polynomial

(λ− µ){

1Lµ(0)

−L′µ(0)L2µ(0)

· λ+ · · ·+ 1(p− 1)!

(1

Lµ(λ)

)(p−1)

(0) · λp−1

}+ λp

{1

µpL(µ)+ µ

∞∑k=1

1L′(λk)λ

p+1k (λk − µ)

}

1866 V.B. Sherstyukov

which has degree at most p. Then we can write

1L(λ)

= Qp,µ(λ) + λp+1∞∑k=1

1L′(λk)λ

p+1k (λ− λk)

,

∞∑k=1

1|L′(λk)| |λk|p+2

< +∞.

Using Lemma 2.2 and the principle of mathematical induction we see that theconditions of the theorem are sufficient for each integer p > 0, so the proof iscomplete.

The next two theorems are proved in a completely similar way.

Theorem 2.2. Let L(λ) be an entire function with simple zeros lying in somestrip and forming a sequence {0} ∪Λ, Λ = (λk)∞k=1. Then the function 1

L(λ) can berepresented by a Kreın series

1L(λ)

= − L′′(0)2(L′(0))2

+ · · ·+ 1(p+ 1)!

(λ

L(λ)

)(p+1)

(0) · λp +1

L′(0)1λ

+∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

+ · · ·+ λp

λp+1k

}(2.9)

converging absolutely and uniformly on each compact subset of C \ ({0}∪Λ), wherep > 0 is an integer, if and only if conditions i) and ii) in Theorem 2.1 hold simul-taneously.

Theorem 2.3. Let L(λ) be the same function as in Theorem 2.1 (or Theorem 2.2).Then the reciprocal 1

L(λ) can be expanded in a series

∞∑k=1

1L′(λk)

1λ− λk

(in a series1

L′(0)1λ

+∞∑k=1

1L′(λk)

1λ− λk

,

respectively) which converges absolutely and uniformly on each compact subset notcontaining zeros of L(λ) if and only if, in addition to the absolute convergence ofthe series

∑∞k=1

1L′(λk)λk

, condition i) is satisfied.

An analysis of the arguments in the proof of Theorem 2.1 gives us the followingresults.

Corollary 2.1. Let L(λ) be a function of exponential type zero which has simplezeros all of which lie in a strip. Then a representation of the form (2.2) (if L(0) = 0)or (2.9) (if L(0) = 0) holds if and only if condition ii) is fulfilled.

It is claimed in [22], Theorem 2 that a result analogous to Corollary 2.1 alsoholds without any restrictions on the position of the zeros of L(λ). They state thatthe scheme of the proof is due to Sodin and is based on the Carleman-Tsuji-Heinsconvexity formula.

Expanding in a Kreın series 1867

Corollary 2.2. Let L(λ) be an entire function all of whose zeros λk are simpleand lie in a strip. Then the following conditions are equivalent :

1) the representation

1L(λ)

=∞∑k=1

ckλ− λk

,

∞∑k=1

|ck| <∞

holds ;2) L has exponential type and the indicator diagram D(L) is a line segment

orthogonal to the boundary of the strip; 0 ∈ D(L) and

∞∑k=1

1|L′(λk)|

<∞.

Corollary 2.3. Let L(λ) be an even entire function with zeros

Λ = (±λk)∞k=1

which are simple and lie in a strip. Then the expansion

1L(λ)

=1

L(0)+

12

(1

L(λ)

)′′(0) · λ2 + · · ·+ 1

(2m)!

(1

L(λ)

)(2m)

(0) · λ2m

+∞∑k=1

2λ2m+2

λ2m+1k L′(λk)(λ2 − λ2

k), (2.10)

where the series is absolutely convergent in C \ Λ, holds if and only if L has expo-nential type and

∞∑k=1

1|L′(λk)| |λk|2m+3

<∞.

Corollary 2.4. Let L(λ) be an odd entire function all of whose zeros

{0} ∪ Λ, Λ = (±λk)∞k=1

are simple and lie in a strip. Then the expansion

1L(λ)

=12

(λ

L(λ)

)′′(0) · λ+ · · ·+ 1

(2m)!

(λ

L(λ)

)(2m)

(0) · λ2m−1

+1

L′(0)1λ

+∞∑k=1

2λ2m+1

λ2mk L′(λk)(λ2 − λ2

k), (2.11)

where the series is absolutely convergent in C \ ({0}∪Λ), holds if and only if L hasexponential type and

∞∑k=1

1|L′(λk)| |λk|2m+2

<∞.

1868 V.B. Sherstyukov

§ 3. Some applications of the main results

One of the most simple, geometric applications of Theorems 2.1–2.3 is a realsimplification of the known procedures for expanding particular functions into par-tial fractions, so that in fact we can immediately write out the required formulae.For instance, using (2.9)–(2.11) we can easily recover the following well knownexpansions:

1sinλ

=1λ

+∞∑k=1

(−1)k2λ

λ2 − π2k2,

1cosλ

=∞∑k=1

(−1)k(2k − 1)π

λ2 − (2k − 1)2π2/4,

1eλ − 1

= −12

+1λ

+∞∑k=1

2λλ2 + 4π2k2

.

Theorems on partial fraction expansions find deeper applications in the theoryof complete and representing systems of exponentials (see, for example, [18]).

To illustrate our results clearly we discuss partial fraction expansions of functionsin some special classes. The results below (Propositions 3.1 and 3.2) refine theauthor’s preliminary results in [5]. We first state two well-known facts required inwhat follows. The first is due to Levin (1949; for example, see [25]), and the secondis due to Sedletskiı (see [26], Lemma 6).

Theorem A. Let

L(λ) =∞∏k=1

(1− λ2

λ2k

), 0 < λk ↑ ∞, |λk − k| 6 d <∞.

Then for all λ ∈ C and some constant A > 0

|L(λ)| 6 A(1 + |λ|4d)eπ| Imλ|.

In addition, if the condition that the zeros are separated holds : infk =j |λk−λj | > 0,then there exists a constant B > 0 such that for each k > 1,

|L′(λk)| > B|λk|−4d.

Theorem B. Assume that the zeros of the canonical product

L(λ) = λ

∞∏k=1

(1− λ

λk

)satisfy

λk = kβ +O(1), β > 1.

Then the following asymptotic equation holds as k →∞:

L′(λk) = exp(πk cot

π

β+O(ln k)

).

Expanding in a Kreın series 1869

In particular, the result of Theorem B holds for an infinite product of the form

L0(λ) = λ

∞∏k=1

(1− λ

kβ

).

For instance, if β = 2 then we obtain the following formula illustrating Theorem B:

L0(λ) = λ ·∞∏k=1

(1− λ

k2

)=

√λ sinπ

√λ

π, L′0(k

2) =(−1)k

2, k ∈ N.

For β > 2 there exist a, γ > 0 such that

|L′(λk)| = exp(ak +O(ln k)) > eak · k−γ ,

and therefore∞∑k=1

1|L′(λk)|

< +∞.

On the other hand, if β ∈ (1, 2), then setting

π cotπ

β=: −a < 0

for brevity, for k > ko(a) we have the estimate

|L′(λk)| = e−ak+O(ln k) 6 e−bk

for some b > 0, so that condition ii) in Theorem 2.1 fails for all p > 0. HenceTheorems 2.2 and 2.3 give us the following result.

Proposition 3.1. For β > 2 the function in the assumptions of Theorem B hasthe expansion

1L(λ)

=1λ

+∞∑k=1

1L′(λk)

1λ− λk

in a series which converges absolutely and uniformly on each compact set not con-taining any zeros of L(λ). For 1 < β < 2 the reciprocal of the function L(λ) inTheorem B cannot be expanded in a series of the form (2.9).

Proposition 3.1 demonstrates that a function with (L′(λk))∞k=1 converging rapidlyto zero cannot be represented by a Kreın series. Another obstacle to such an expan-sion may be that 0 /∈ D(L). Here we can give an example due to Koosis. Let

L(λ) =∞∏k=1

(1− λ

2k

)eλ/2

k

= eλ∞∏k=1

(1− λ

2k

).

This function has only the simple real zeros λk = 2k, k ∈ N, and has exponentialtype because

∑∞k=1 2−k <∞. Clearly, D(L) = {1}. In addition,

∞∑k=1

1|L′(2k)|

<∞,

|L(iy)| → ∞ as y → ±∞,1

L(x)→∞ as x→ −∞.

1870 V.B. Sherstyukov

In spite of what we have obtained previously, this last property means that we donot even have the equality

1L(x)

=∞∑k=1

1L′(2k)(x− 2k)

, x ∈ R,

because the right-hand side converges to zero as x→ −∞.Now let L(λ) be the canonical product in Theorem A. The sequence (λk)∞k=1 has

density

limk→∞

k

λk= 1,

and the separation condition for the points in this sequence means that the functionin question has a regular zero set Λ = (±λk)∞k=1. It is well known that L(λ) hascompletely regular growth and D(L) is the interval [−πi, πi] on the imaginary axis.Moreover, the condition

∞∑k=1

1|L′(λk)| |λk|2m+3

<∞

holds with m = 1 + [2d− 1]. From Corollary 2.4 we obtain the following result.

Proposition 3.2. Assume that the canonical product L(λ) satisfies the initialassumptions of Theorem A. Then the reciprocal [L(λ)]−1 has representation (2.10)with m = 1 + [2d − 1]. In particular, if d < 1

2 , then this representation can bewritten as follows :

1L(λ)

= 1 +∞∑k=1

2λ2

λkL′(λk)(λ2 − λ2k).

Bibliography

[1] M.G. Krein, “On a remarkable class of Hermitian operators”, Dokl. Akad. NaukSSSR 44:5 (1944), 191–195; English transl. in C. R. (Dokl.) Acad. Sci. URSS(N.S.) 44 (1944), 175–179.

[2] H.L. Hamburger, “Hermitian transformations of deficiency-index (1, 1), Jacobimatrices and undetermined moment problems”, Amer. J. Math. 66:4 (1944),489–522.

[3] M.G. Kreın, “A contribution to the theory of entire functions of exponential type”,Izv. Akad. Nauk SSSR Ser. Mat. 11:4 (1947), 309–326. (Russian)

[4] B. Ja. Levin (B. Ya. Levin), Distribution of zeros of entire functions, Gostekhizdat,Moscow 1956; English transl., Amer. Math. Soc., Providence, RI 1964.

[5] V.B. Sherstyukov, “Expanding meroporphic functions of a special form into partialfractions”, Anal. Math. 33:1 (2007), 63–81. (Russian)

[6] N. I. Akhiezer, “Some properties of integral transcendent functions of exponentialtype”, Izv. Akad. Nauk SSSR Ser. Mat. 10:5 (1946), 411–428. (Russian)

[7] A.A. Goldberg and I. V. Ostrovskiı, Value distribution of meromorphic functions,Nauka, Moscow 1970; English transl., Transl. Math. Monogr., vol. 236, Amer.Math. Soc., Providence, RI 1992.

Expanding in a Kreın series 1871

[8] L. de Branges, Hilbert spaces of entire functions, Prentice-Hall, Englewood Cliffs,NJ 1968.

[9] Ju. F. Korobeinik (Yu. F. Korobeınik), “Boundary properties of analytic solutionsof differential equations of infinite order”, Mat. Sb. 115(157):3(7) (1981), 364–390;English transl. in Math. USSR-Sb. 43:3 (1982), 323–345.

[10] Yu. F. Korobeınik, Solvability of some general classes of linear operator equationsin the complex domain, Rostov State University Publishing House, Rostov-on-Don2005. (Russian)

[11] A. Borichev and M. Sodin, “Krein’s entire functions and the Bernsteinapproximation problem”, Illinois J. Math. 45:1 (2001), 167–185.

[12] L. S. Maergoiz, “Indicator diagram and generalized Borel-Laplace transforms forentire functions of a given proximate order”, Algebra i Analiz 12:2 (2000), 1–63;English transl. in St. Petersburg Math. J. 12:2 (2001), 191–232.

[13] L. S. Maergoiz, “On partial fraction expansion for meromorphic functions”, Mat.Fiz. Anal. Geom. 9:3 (2002), 487–492.

[14] A.A. Gol’dberg, “A class of entire functions”, Dokl. Math. 17:1 1976 39–41.

[15] L. de Branges, “The Bernstein problem”, Proc. Amer. Math. Soc. 10:5 (1959),825–832.

[16] P. Koosis, The logarithmic integral, vol. I, Cambridge Stud. Adv. Math., vol. 12,Cambridge Univ. Press, Cambridge 1988.

[17] C. Berg and H. L. Pedersen, “Nevanlinna matrices of entire functions”, Math.Nachr. 171:1 (1995), 29–52.

[18] V.B. Sherstyukov, “Representation of the reciprocal of an entire function byseries of partial fractions and exponential approximation”, Mat. Sb. 200:3 (2009),147–160; English transl. in Sb. Math. 200:3 (2009), 455–469.

[19] A.G. Bakan, “Polynomial approximation in Lp(R, dµ)”, Nats. Akad. Nauk Ukrain.Inst. Mat. Preprint 7 (1998), 1–45.

[20] A.G. Bakan, “Polynomial density in Lp(R, dµ) and representation of all measureswhich generate a determinate Hamburger moment problem”, Approximation,optimization and mathematical economics (Guadeloupe 1999), Physica, Heidelberg2001, pp. 37–46.

[21] A.G. Bakan, “A polynomial form of the de Branges conditions for the density ofalgebraic polynomials in the space C0

w”, Ukrain. Mat. Zh. 57:3 (2005), 305–319;English transl. in Ukr. Math. J. 57:3 (2005), 364–381.

[22] A.G. Bakan, “Expanding the reciprocal of an entire function into partialfractions”, Dopov. Nats. Akad. Nauk Ukr. Mat. Prirodozn. Tekh. Nauki , 2009, no. 2,11–13. (Russian)

[23] A.M. Sedletskiı, Classes of analytic Fourier transforms and exponentialapproximations, Fizmatlit, Moscow 2005. (Russian)

[24] I. F. Krasichkov-Ternovskiı, “A geometric lemma useful in the theory of entirefunctions and Levinson-type theorems”, Mat. Zametki 24:4 (1978), 531–546;English transl. in Math. Notes 24:4 (1978), 784–792.

[25] R.M. Redheffer, “Completeness of sets of complex exponentials”, Advances inMath. 24:1 (1977), 1–62.

[26] A.M. Sedletskii, “Annihilable systems of trigonometric systems”, Mat. Zametki34:2 (1983), 237–248; English transl. in Math. Notes 34:2 (1983), 601–608.

V.B. Sherstyukov

Moscow Engineering Physics Institute

(National Research Nuclear University)

E-mail : shervb73@gmail.com

Received 6/JUL/10 and 1/NOV/10Translated by N. KRUZHILIN