Representation of the reciprocal of an entire function by series of partial fractions and

exponential approximation

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Sbornik : Mathematics 200:3 455–469 c© 2009 RAS(DoM) and LMS

Matematicheskiı Sbornik 200:3 147–160 DOI 10.1070/SM2009v200n03ABEH004004

Representation of the reciprocal of an entire functionby series of partial fractions and exponential approximation

V.B. Sherstyukov

Abstract. Conditions under which the reciprocal 1/L(λ) of an entire func-tion with simple zeros λk can be represented as a series of partial fractionsck/(λ − λk), k = 1, 2, . . . , are investigated. The possibility of such a rep-resentation is characterized, as is conventional, in terms of a particular‘asymptotically regular’ behaviour of the function L(λ). Applications tocomplete systems of exponentials on a line interval and to representativesystems of exponentials in a convex domain are considered.

Bibliography: 18 titles.

Keywords: entire function, series of partial fractions, representative sys-tems of exponentials.

§ 1. The setting of the problem. The statement of the main results

In 1947, M. Kreın’s paper [1] appeared; one interesting class of entire functions,which arose in order to study the spectral theory of Hermitian operators and themoment problem (see [2], [3]), was introduced and studied. A function L(λ) havingonly real simple zeros λk, k = 1, 2, . . . , is said to belong to the aforementionedKreın class if, for some p > 0,

∞∑k=1

1|λk|p |L′(λk)|

<∞, (1)

1L(λ)

= R(λ) +∞∑k=1

1L′(λk)

{1

λ− λk+

1λk

+λ

λ2k

+ · · ·+ λp−1

λpk

}, (2)

where R(λ) is a polynomial.Kreın’s theorem (see [1], Theorem 4) states that if conditions (1) and (2) are

satisfied, then L(λ) is an entire function of exponential type (e.f.e.t.) and∫ +∞

−∞

ln+ |L(x)|1 + x2

dx <∞. (3)

As usual, ln+ a := 12 (|ln a| + ln a), a > 0. In view of (1), expansion (2) can be

restated in an equivalent form (see [1]):

1L(λ)

= R(λ) + λp∞∑k=1

ckλ− λk

,∞∑k=1

|ck| <∞,

This research was carried out with the financial support of the Departmental Analytic TargetedProgramme “Development of Scientific Potential in Higher Education” (grant no. 2.1.1/6827).

AMS 2000 Mathematics Subject Classification. Primary 30D20; Secondary 30B99.

456 V.B. Sherstyukov

which for R(λ) ≡ 0 and p = 0 assumes the form

1L(λ)

=∞∑k=1

ckλ− λk

,∞∑k=1

|ck| <∞, (4)

and shows that a function in the ‘Kreın class’ is of completely regular growth andits indicator diagram is a closed interval on the imaginary axis containing the origin(see [1], [4], Ch. V, § 6).

The following is a more general result (see [1], Theorem 5), for a different prooffrom the original, we refer the reader to [4], Ch. V, § 6, Theorem 13.

Kreın’s Theorem. Let L(λ) be an entire function which has simple zeros λk,k = 1, 2, . . . , satisfying

∞∑k=1

|Imλk||λk|2

<∞. (5)

If, in addition,

1L(λ)

=∞∑k=1

ckλ− λk

,

∞∑k=1

∣∣∣∣ ckλk∣∣∣∣ <∞, (6)

then L(λ) is a function of at most exponential type, and this function satisfies (3).

Entire functions whose zeros satisfy condition (5) on their closeness to the realaxis (this condition may be written as

∑∞k=1

∣∣Im 1λk

∣∣ <∞) are usually called func-tions of class A (see [4], Ch. V). For further extensions of Kreın’s theorem, referencemay be made to [5], Ch. VI, § 2 and to [6]. Various properties of functions admit-ting a representation of the form of (2), (4) or (6) have been studied and used inmany papers on the theory of functions and differential equations (see, for instance,[5], Ch. V, § 6, [6], [7], [8], § 6, [9] and [10] in addition to the previously mentionedpapers [2], [3] and the book [4]). Even so, the important question of conditionsdetermining when the reciprocal of an entire function (or, more generally, a mero-morphic function) can be expanded in a series of partial fractions does not seemto have been explored sufficiently. Neither Mittag-Leffler’s classical theorem norCauchy’s method for expanding a meromorphic function in partial fractions pro-vide a complete answer here. For this reason the following problem was posed byus in [9].

Given an entire function L(λ) of class A with simple zeros λk, k = 1, 2, . . . , finda criterion for its reciprocal 1/L(λ) to be represented as a series (6). In [9] thisproblem was solved, but only for a function with real zeros, and a necessary andsufficient condition for the existence of a Kreın-type expansion was stated thereonly in terms of the sequence {L′(λk)}∞k=1. In this paper we examine this problemin a more general setting.

Turning to the statement of the main results we recall that a subset of the realaxis is said to be relatively dense with respect to measure (see [11]) if there existpositive numbers l and δ such that every line interval [x, x + l], x ∈ R, containsa part of this set with Lebesgue measure greater than δ. This definition can becarried over in a natural way (see [10]) from the real axis to an arbitrary line in C.

Representation of the reciprocal of an entire function 457

Let L(λ) be an entire function with simple zeros Λ = {λk}∞k=1, γ a line throughthe origin, R = {rk}∞k=1 some sequence of positive numbers, and let

U =∞⋃k=1

{λ : |λ− λk| < rk}.

A pair (γ,R) is said to be compatible with the function L if the following threeconditions are simultaneously satisfied:

the set S = γ \ U is relatively dense on γ with respect to measure; (7)

∃α > 0 :1

L′(λk)rk= O(|λk|α), k →∞; (8)

∃β > 0 :1

L(λ)= O(|λ|β), λ ∈ S. (9)

Theorem 1. A necessary and sufficient condition for an e.f.e.t. L(λ) with (all)simple zeros Λ = {λk}∞k=1 to admit a Kreın-type expansion

1L(λ)

= Q(λ) +∞∑k=1

(ak

λ− λk+ Pk(λ)

)(10)

(the series converges absolutely and uniformly in C \ Λ; Q(λ) and the Pk(λ), k =1, 2, . . . , are polynomials of bounded degree) is that L has a non-negative indicatorand there exists a pair (γ,R) compatible with L.

It is natural to expect that the statement of the theorem will be a little simpli-fied if some additional apriori information about the location of the zeros of thefunction L(λ) is available. For example, it is very likely that under condition (5)the possibility of expanding of 1/L(λ) in a series (10) can be expressed only interms of estimates for the growth rate of the sequence {L′(λk)}∞k=1. As of now, thisconjecture has been confirmed only for an entire function all of whose zeros lie insome horizontal strip (clearly, such a function is in the class A).

Theorem 2. Let L(λ) be an entire function with simple zeros Λ = {λk}∞k=1 lying inthe strip |Imλ| 6 h, h > 0. The reciprocal 1/L(λ) can be expanded in a series (10)if and only if the following conditions are satisfied simultaneously :

i) L(λ) is an e.f.e.t. whose indicator diagram contains the origin ;

ii) ∃α > 0 :1

L′(λk)= O(|λk|α) as k →∞.

§ 2. The proof of the main results

For brevity in the proofs of Theorems 1 and 2 we start by proving an auxiliaryresult.

Lemma 1. Suppose that an entire function L(λ) admits an expansion (10) ina Kreın-type series. Then the asymptotic estimate

1L′(λk)

= O(|λk|α), k →∞,

holds for some α > 0.

458 V.B. Sherstyukov

Proof. Suppose that the meromorphic function 1/L(λ) can be expanded in a series(10) which converges absolutely and uniformly in the domain C \ Λ and in whichthe degrees of all the polynomials Pk(λ) do not exceed some positive integer p. Inparticular, this gives ak = 1/L′(λk), k = 1, 2, . . . . We invoke a trick borrowedfrom [1] and multiply the function L(λ) by a suitable polynomial T (λ) such thatits degree m exceeds the maximum of p and the degree of the polynomial Q(λ) byat least 2. Putting L1(λ) = L(λ)T (λ) this gives

1L1(λ)

=m∑j=1

bjλ− µj

+∞∑k=1

ckλ− λk

,∞∑k=1

|ck| |λk|2 <∞. (11)

It is worth noting that it is possible to prove Lemma 1 through the use of a lower-order polynomial, thus obtaining an expansion of the form (4); however in thesubsequent analysis it will be convenient to use (11). Direct calculation shows that

ck =1

L′(λk)T (λk), k = 1, 2, . . . .

and so, from the restriction imposed on the coefficients in the expansion (11) it fol-lows that

∃C > 0 :|λk|2

|L′(λk)|6 C|T (λk)|, k = 1, 2, . . . ,

whence the lemma.

Proof of Theorem 1. Sufficiency. Let L(λ) be an e.f.e.t. with simple zeros Λ ={λk}∞k=1. Suppose that for L(λ) there exists a compatible pair (γ,R). We choosea polynomial T (λ) of degreem with simple zeros {µj}mj=1 such that {µj}mj=1∩Λ = ∅,where m = max

{[α]+3, [β]+1

}, and α and β are given by conditions (8) and (9) for

a pair (γ,R) compatible with L(λ), and where the symbol [x] denotes the integerpart of x. Consider the function L1(λ) = L(λ)T (λ). For all k = 1, 2, . . . andj = 1, 2, . . . ,m we have L′1(λk) = L′(λk)T (λk) and L′1(µj) = L(µj)T ′(µj). Theauxiliary function

ϕ(λ) =1

L1(λ)−

∞∑k=1

1L′1(λk)(λ− λk)

−m∑j=1

1L′1(µj)(λ− µj)

is easily seen to be an e.f.e.t. Using condition (7) we estimate |ϕ(λ)| from above onthe set S, which is relatively dense with respect to measure on the line γ. Lettingλ→∞ we clearly have

m∑j=1

1L′1(µj)(λ− µj)

→ 0.

Further, for λ ∈ S,

∞∑k=1

1|L′1(λk)| |λ− λk|

6∞∑k=1

1|L′(λk)| |T (λk)|rk

6 B∞∑k=1

|λk|α

|T (λk)|,

Representation of the reciprocal of an entire function 459

where the constant B is independent of λ, and the last series is convergent due tothe choice of m and the fact that

∑∞k=1 1/|λk|2 <∞. Finally, if λ ∈ S is sufficiently

large in modulus, then the estimate

1|L1(λ)|

=1

|L(λ)| |T (λ)|6 C

|λ|β

|T (λ)|

holds with the constant C in condition (9) independent of λ. Thus, the polynomialT (λ) is chosen so that the function ϕ(λ) is bounded on S. By the main lemmaof [11] it follows that ϕ(λ) is bounded on the line γ, and hence (see, for instance,[4], Ch. V, § 4), ϕ(λ) is of completely regular growth. Putting

ψ(λ) =∞∑k=1

L1(λ)L′1(λk)(λ− λk)

+m∑j=1

L1(λ)L′1(µj)(λ− µj)

,

we arrive at the relation 1 = ψ(λ) + L1(λ)ϕ(λ), in which ψ(λ) is an e.f.e.t. whoseindicator hψ(θ) does not exceed the indicator hL1(θ) of the function L1(λ). Sinceϕ(λ) is of completely regular growth, for each θ ∈ [0, 2π) the indicators of the func-tions L1(λ)ϕ(λ), L1(λ) and ϕ(λ) satisfy hL1ϕ(θ) = hL1(θ) + hϕ(θ). Consequently,since hL1(θ) is non-negative, it follows that hL1(θ) + hϕ(θ) 6 hL1(θ) for all θ, andhence ϕ(λ) is of exponential type zero. As this function is bounded on the line γ,it is constant using known results of Phragmen-Lindelof type. Moreover, from theabove estimates it follows that ϕ(λ) → 0 as λ→∞, λ ∈ S. This gives ϕ ≡ 0 and

1L1(λ)

=∞∑k=1

1L′1(λk)(λ− λk)

+m∑j=1

1L′1(µj)(λ− µj)

,

and the series converges absolutely and uniformly on every compact set not contain-ing zeros of the function L1(λ) because

∑∞k=1 1/|L′1(λk)| < ∞. Now (10) follows

relatively easily:

1L(λ)

=∞∑k=1

T (λ)L′(λk)T (λk)(λ− λk)

+m∑j=1

T (λ)L(µj)T ′(µj)(λ− µj)

=∞∑k=1

1L′(λk)

{1

λ− λk+T ′(λk)T (λk)

+ · · ·+ T (m)(λk)T (λk)m!

(λ− λk)m−1

}

+m∑j=1

1L(µj)

{1 +

T ′′(µj)2T ′(µj)

(λ− µj) + · · ·+ T (m)(µj)T ′(µj)m!

(λ− µj)m−1

}

= Q(λ) +∞∑k=1

(1

L′(λk)(λ− λk)+ Pk(λ)

);

here Q(λ) are Pk(λ), k = 1, 2, . . . , are polynomials of degree m− 1 at most.Necessity. Consider an arbitrary line γ through the origin and define

R = {rk}∞k=1, rk =1

|λk|2, k = 1, 2, . . . .

460 V.B. Sherstyukov

From (10), by Lemma 1 we obtain condition (8) as a corollary to the represent-ation (11). It also follows from (11) that hL(θ) > 0. Since L(λ) is of exponentialtype, we have

∑∞k=1 rk <∞, and so the set of discs Uk = {λ : |λ−λk| < rk} has zero

line density (that is, it is a C0-set), and the set S = γ \⋃∞k=1 Uk is relatively dense

with respect to measure on γ. The pair (γ,R) is compatible with L(λ) because (11)in combination with (8) implies the bound

∃B > 0 : ∀λ ∈ S 1|L(λ)|

6 B|T (λ)|.

The proof of Theorem 1 is complete.

Corollary 1. Let L(λ) be an e.f.e.t. of completely regular growth with simple zerosΛ = {λk}∞k=1 and with a positive indicator hL(θ). Then the following bound is nec-essary and sufficient for the representation (10) to hold :

∃α > 0 :1

L′(λk)= O(|λk|α), k →∞.

Proof of Theorem 2. Necessity. Suppose that representation (10) holds. ApplyingLemma 1 gives condition ii). Since all the zeros of L(λ) lie in a horizontal strip,it follows that L ∈ A, and so, by Kreın’s theorem, this function is of exponentialtype, and its indicator diagram lies on the imaginary axis. To verify condition i)it remains to observe, using Theorem 1, that the indicator diagram contains theorigin. However, since multiplication by a polynomial leaves the indicator diagramof a function unchanged, instead of (11) we can use the following representationwithout loss of generality:

1L(λ)

=∞∑k=1

ckλ− λk

,

∞∑k=1

|ck| <∞ .

If µ ∈ R, |µ| > h, then

1|L(iµ)|

6∞∑k=1

|ck||iµ− λk|

6∞∑k=1

|ck||µ− Imλk|

61

|µ| − h

∞∑k=1

|ck| <∞,

and so hL(±π/2) > 0, proving the necessity.Sufficiency. Suppose that an entire function L(λ), with zeros in the strip |Imλ| 6 h,satisfies conditions i) and ii). Assuming in what follows that iR ∩ Λ = ∅ we writethe Hadamard-Weierstrass representation for L(λ):

L(λ) = eaλ+b∞∏k=1

(1− λ

λk

)eλ/λk , a, b ∈ C.

Putting

µk = Reλk, k = 1, 2, . . . ,

d = a+∞∑k=1

(1λk

− 1µk

), χ(λ) =

∞∏k=1

(1− λ

λk

)(1− λ

µk

)−1

,

L1(λ) = χ(λ)eλRe d∞∏k=1

(1− λ

µk

)eλ/µk ,

Representation of the reciprocal of an entire function 461

this givesL(λ) = L1(λ)eiλ Im d+b. (12)

The subsequent analysis relies on a lemma of Sedletskiı (see [12], Ch. 5, § 5.3).

Lemma 2. Suppose that sequences {λk}∞k=1 and {µk}∞k=1 lie in the horizontal strip|Imλ| 6 h < +∞ without condensation. Suppose also that Reλk = Reµk for all k.Then ∣∣∣∣ ∏′ 1− λ/λk

1− λ/µk

∣∣∣∣ � 1, |Imλ| > H > h

(the sign ′ indicates that the infinite product does not contain factors with λk,µk = 0, and the notation � 1 means that this infinite product is convergent outsidethe strip |Imλ| < H and is bounded away from zero there).

In our case we need Sedletskiı’s lemma to estimate |χ(λ)| from below on theimaginary axis. In combination with the evident inequality∣∣∣∣ ∞∏

k=1

(1− iµ

µk

)eiµ/µk

∣∣∣∣ > 1, µ ∈ R,

this gives∃D > 0 : |L1(iµ)| > D ∀µ ∈ R : |µ| > H > h. (13)

In view of condition ii) and the restriction |Imλk| 6 h, k = 1, 2, . . . , by (12) weobtain

1L′1(λk)

= O(|λk|α), k →∞. (14)

Conditions (13) and (14) enable us to apply Theorem 1 to the function L1(λ), andso its reciprocal 1/L1(λ) expands in a series of the form (10). By Kreın’s theorem,L1(λ) is a function of completely regular growth, hence so is the function L(λ).This also shows that the indicator diagram of the function L(λ) is an intervalon the imaginary axis. Consider the three possible cases implied by condition i).If the origin is not an end-point of the interval representing the indicator diagramof the function L(λ), then |L(iµ)| → ∞ as µ → ±∞. In this case it follows fromTheorem 1 that 1/L(λ) is representable by a Kreın-type series (10) because the pair(γ,R) is compatible with L for γ = iR, R = {rk}∞k=1, and rk = 1, k = 1, 2, . . . .

Suppose now that the indicator diagram of L(λ) has the origin as one of itsend-points; for definiteness assume that this diagram is [−iδ, 0], δ > 0. Then foreach ε ∈ (0, δ) the ‘if’ part of Theorem 1 is effective for the function L(λ)e−iελ.For brevity we may assume (without loss of generality in our arguments) that incondition ii) α > 2 and that the resulting expansion in partial fractions is as follows:

eiελ

L(λ)=

∞∑k=1

ckeiελk

λ− λk,

∞∑k=1

|ck| <∞. (15)

The series on the right-hand side of (15) is uniformly convergent with respect to εfor fixed λ (here we make use of the inequalities |Imλk| 6 h, k = 1, 2, . . . ) and sowe can pass termwise to the limit as ε→ +0, which yields

1L(λ)

=∞∑k=1

ckλ− λk

,

∞∑k=1

|ck| <∞.

462 V.B. Sherstyukov

Suppose finally that the indicator diagram of the function L(λ) consists of theorigin alone, that is, L(λ) is of exponential type zero. Conditions (12) and (13)show that the inequality

|L(iµ)| > D1e−µ Im d ∀µ ∈ R : |µ| > H > h

holds with D1 = DeRe b > 0. But a function of exponential type zero satisfies thelast inequality only for d = 0. Hence L(λ) = ebL1(λ); and expansion (10) has beenalready obtained for 1/L1(λ). This completes the proof of Theorem 2.

An analysis of the above arguments leads us to the following results.

Corollary 2. Suppose that an e.f.e.t. L(λ) has simple zeros Λ = {λk}∞k=1 lying inthe strip |Imλ| 6 h. Suppose also that, for some α > 0,

1L′(λk)

= O(|λk|α), k → +∞.

Then ∫ +∞

−∞

ln+ |L(x)|1 + x2

dx <∞.

Consequently, L has completely regular growth, and the indicator diagram of L isan interval on the imaginary axis (possibly degenerating to a point).

Corollary 3. Suppose L(λ) is of zero exponential type and has simple zeros Λ ={λk}∞k=1, |Imλk| 6 h, k = 1, 2, . . . . A necessary and sufficient condition thatrepresentation (10) be valid is that condition ii) of Theorem 2 hold.

§ 3. Completeness on a line interval

One of many possible ways in which Kreın-type expansions may be applied isto investigate the approximative properties of systems of exponentials in variousfunction spaces. As usual, C[a, b] denotes the Banach space of all continuous func-tions on the interval [a, b] with the supremum norm ‖f‖[a,b] := supx∈[a,b] |f(x)|;and Lp(a, b), 1 6 p < ∞, denotes the (quotient) space of all measurable functionson (a, b), whose absolute value raised to the power p is integrable. It becomesa Banach space under the norm

‖f‖p,(a,b) :=(∫ b

a

|f(x)|p dx)1/p

.

An arbitrary sequence of pairwise distinct points Λ = {λk}∞k=1 ⊂ C generates thesystem of exponentials EΛ = {eλkz}∞k=1. A system of exponentials is said to becomplete on [a, b] if it is complete in the space C[a, b]; that is, when the closure ofthe span of this system coincides with C[a, b]. The question of the completenessof systems of exponentials on an interval has a long history; a partial account maybe found in the monograph [12] and the survey article [13]. We state some results,which will be used below, and which are closely related to the subject in question.We put in the class C (the Cartwright class) all the e.f.e.t. L(λ) for which the inte-

gral∫ +∞

−∞

ln+ |L(x)|1 + x2

dx exists; or, equivalently, condition (3) in the introduction

Representation of the reciprocal of an entire function 463

is satisfied. One of the main results bordering on the theory of completeness ofexponential systems on a line segment is the famous Beurling-Malliavin multipliertheorem (see [14]), from which the condition for completeness in terms of e.f.e.t.easily follows (see [13], Ch. 2, § 2.1, Theorem 2.1.12). We state this result in a some-what weaker form: if L is an e.f.e.t. of class C with indicator hL(θ) and simple zerosΛ = {λk}∞k=1, then the system EiΛ is complete both on [a, b] and in Lp(a, b), pro-vided that b − a < hL(π2 ) + hL(−π

2 ); the system fails to be complete on [a, b] andin Lp(a, b) when b − a > hL(π2 ) + hL(−π

2 ). This is to say, the supremum ρ(Λ) ofall the half-lengths (radii) of the line intervals on which the system EiΛ is complete(the completeness radius of the sequence Λ) is 1

2

(hL(π2 ) + hL(−π

2 )). Combining

this with Kreın’s theorem gives the following result.

Proposition. Suppose that L(λ) is a function of class A with simple zeros Λ ={λk}∞k=1 whose reciprocal admits an expansion in partial fractions of the form (10).Then the completeness radius of the sequence Λ is half the length of the indicatordiagram of the function L.

Taking account of Corollary 2 we obtain the following result.

Corollary 4. Suppose that all the zeros λk of an e.f.e.t. L(λ) are simple and liein the strip |Imλ| 6 h. Suppose also that |L′(λk)| > M |λk|−α for some M , α > 0and for all k ∈ N. Then ρ(Λ) = 1

2

(hL(π2 ) + hL(−π

2 )).

It is worth noting that for the special function L(λ) = λ∏′(1− λ

λk

)eλ/λk , which

generates the system {e−ikz}k∈Z\Λ, Λ = {λk}⊂Z, λ0 = 0, the condition that L(λ)can be expanded in partial fractions is equivalent to enhanced (in a certain sense)completeness of Ei(Z\Λ) in the class L2(−π, π) (see [9] for details).

§ 4. Representative systems of exponentials

Let G be a bounded convex domain in C with support function h(−θ); letA(G) be the space of all analytic functions in G endowed with the topology ofuniform convergence on compact subsets of G; let Λ = {λk}∞k=1 be a sequenceof pairwise distinct complex numbers with a unique limit point at infinity; and letEΛ = {eλkz}∞k=1. In accordance with the general definition (see [15]), a sequence EΛ

is called an absolutely representative system (ARS) in A(G) if any function f(z) ∈A(G) can be represented (not necessarily uniquely) as the sum of a series

f(z) =∞∑k=1

ckeλkz, ck ∈ C, k = 1, 2, . . . , (16)

which converges absolutely relative to the topology τG of the space A(G). Also,a sequence EΛ is said to be an effectively absolutely representative system (EARS)in A(G) if for each function f(z) ∈ A(G) there exists a method of finding thecoefficients of at least one expansion of f(z) in a series (16). The first resultsconcerning ARS and EARS of exponentials were obtained by Leont’ev (see [16]),and we shall state one of them below. To this end we take an e.f.e.t. L(λ) withgrowth indicator hL(θ) = h(θ) and with simple zeros λk, k = 1, 2, . . . , and makeup the system of functions

ψn(t) =1

L′(λn)

∫ ∞

0

L(λ)λ− λn

e−λt dλ, n = 1, 2, . . . (17)

464 V.B. Sherstyukov

(integration proceeds over the ray arg λ = ϕ). In [16], Ch. IV, § 1 it is provedthat all the ψn(t) are regular outside G and that the system (17) is biorthogonalto {eλkz}∞k=1. Next, corresponding to each analytic function f(z) on G (that is,f(z) ∈ A(G)) is the series

f(z) ∼∞∑k=1

ckeλkz, ck =

12πi

∫Γ

f(t)ψk(t) dt, k > 1, (18)

where Γ is a contour enclosing G, and f(z) is analytic on it and inside it. We usethe symbol [1;h(θ)) to denote the set of all e.f.e.t. whose indicators are less thanh(θ); also we let [1; 0] be the class of entire functions of exponential type zero.

Leont’ev’s Theorem (see the Appendix to [16]). A necessary and sufficient con-dition for the series (18) to converge in G to the corresponding function f(z) (what-ever f(z) ∈ A(G) may be) is as follows :

Φ(λ) =∞∑k=1

Φ(λk)L′(λk)

L(λ)λ− λk

∀Φ ∈ [1;h(θ)). (19)

The book [16] also contains a number of other conditions, equivalent to (19),for the series (18) to converge (absolutely, relative to the topology τG) preciselyto f(z). In particular, this will be so if L has completely regular growth and

limk→∞

{1|λk|

ln1

|L′(λk)|+ h(arg λk)

}6 0 (20)

(in the terminology of [17], L is an (1, h(θ))-interpolating function).Leont’ev’s theorem furnishes an example of an EARS in the space (A(G), τG).

From the results of [16] and [15] it follows that a system of exponentials withexponents at the zeros of an (1, h(θ))-interpolating function also forms an EARSin the space A(G). We shall show that setting Φ(λ) ≡ 1 in condition (19) ensuresthe convergence of the series (18) in G to the corresponding function f(z). It isconvenient for the statements of the results below to introduce some classes of entirefunctions. Let

M(Λ;h) :={L(λ) is an e.f.e.t. with hL(θ) 6 h(θ) : all λk are simple zeros of L

}.

If L ∈M(Λ;h), then L(λ) may have zeros of arbitrary multiplicity different from λk.Also let

M(Λ;h) :={L ∈M(Λ;h) : the set v(L) \ Λ has density at most zero

},

where v(L) denotes the zero set of the function L.

Korobeınik’s Theorem (see [15], Ch. III, § 1, Theorem 6 for ρ = 1). Suppose thatM(Λ;h) 6= ∅ and that L(λ) ∈M(Λ;h). Then the system EΛ is an ARS in A(G) ifand only if there exists a function D(λ) ∈ [1; 0] \ {0} such that

D(λ)eλz =∞∑k=1

L(λ)λ− λk

D(λk)eλkz

L′(λk), z ∈ G, λ ∈ C (21)

(the series converges absolutely relative to the topology τG).

Representation of the reciprocal of an entire function 465

Theorem 3 (see [10], Theorem 6). A necessary and sufficient condition for a sys-tem EΛ to be an ARS in A(G) is that Λ contains a subsequence Λ′ ⊂ Λ satisfyingthe following condition : there exists a function L ∈ M(Λ′;h) with indicator h(θ)and lower indicator hL,−(θ) such that the inequality hL,−(θ0) + hL,−(π + θ0) > 0holds for some θ0 ∈ [0, 2π) and, in addition,

limk→∞λk∈Λ′

{1|λk|

ln |L′(λk)| − h(arg λk)}

= 0.

The last theorem refines the corresponding criterion due to Abanin (see [18]), inwhich instead of the condition on the lower indicator, the complete regularity ofthe growth of L was assumed. At present it is not clear whether we can drop thiscondition completely. It is possible that the results obtained here will be a step inthe direction of solving this problem. At any rate, they seem to throw into relieffor the first time the important role in the theory of ARS of exponentials that isplayed by expansions of the generating function in partial fractions.

Theorem 4. Let G be a convex domain in C containing the origin, and let h(−θ)be the support function of G. Also let L(λ) be an e.f.e.t. with indicator h(θ) andwith simple zeros λk, k = 1, 2, . . . . Then the following conditions are equivalent :

1) the series (18) corresponding to an arbitrary function f(z) ∈ A(G) is abso-lutely convergent to f(z) in A(G);

2)1

L(λ)=

∞∑k=1

1L′(λk)(λ− λk)

, where{

1L′(λk)

}∞

k=1

satisfies (20);

3) the meromorphic function ϕ(λ) :=∞∑k=1

1L′(λk)(λ− λk)

has no zeros (the

coefficients1

L′(λk)of the series satisfy condition (20)).

Proof. We need to prove the implications 2) =⇒ 1) and 3) =⇒ 2), however 3) readilyfollows from 2), and 1) =⇒ 2) holds by Leont’ev’s theorem. Now we assume thatthe reciprocal 1/L(λ) can be represented by the series

∞∑k=1

1L′(λk)(λ− λk)

,

where condition (20) is satisfied. In [6] it is shown that L has a completely regulargrowth. This immediately implies (19) (as we observed above after the formulationof Leont’ev’s theorem). Let us prove the implication 3) =⇒ 2). From condition 3)it readily follows that

ψ(λ) := ϕ(λ)L(λ) =∞∑k=1

L(λ)L′(λk)(λ− λk)

is an e.f.e.t. without zeros and with hψ(θ) 6 h(θ). Consequently, there exist a ∈ Cand A ∈ C \ {0} such that ψ(λ) = Aeaλ and we have ψ(λk) = 1 for all k ∈ N;that is, eaλk = 1/A for all k. This gives Re(aλk) = − ln |A|, k = 1, 2, . . . . For a 6= 0the above equalities mean that the points λk line on one line. We proved in [6] that

466 V.B. Sherstyukov

in this case it follows from (20) that the indicator diagram of the function L(λ) hasno interior points. We have reached a contradiction, hence a = 0. But then A = 1and ψ(λ) ≡ 1, that is,

1L(λ)

=∞∑k=1

1L′(λk)(λ− λk)

.

This proves condition 2), and Theorem 4 now follows.

Now we shall improve upon Korobeınik’s theorem for a domain G 3 0 by show-ing that it suffices to take z = 0 in relation (21), and shall also supplement theAbanin-Sherstyukov criterion cited above by the corresponding statement aboutthe expansion in partial fractions.

Theorem 5. Let L ∈M(Λ;h) and 0 ∈ G. Then the system EΛ is an ARS in A(G)if and only if the expansion

D(λ)L(λ)

=∞∑k=1

D(λk)L′(λk)(λ− λk)

(22)

holds with some function D(λ) ∈ [1; 0] distinct from identical zero and also

limk→∞

{1|λk|

ln∣∣∣∣D(λk)L′(λk)

∣∣∣∣ + h(arg λk)}

6 0. (23)

Proof. The necessity is immediate. If in Theorem 5 the system EΛ is an ARSin A(G), then the expansion (21) in Korobeınik’s criterion holds. In particular,for z = 0 ∈ G we arrive at the expansion (22) and condition (23) is equivalent tothe absolute convergence of the series on the right-hand side of (21) relative to thetopology τG with some fixed λ /∈ Λ.

Conversely, suppose that condition (23) is fulfilled and we have the expan-sion (22); suppose also that v(L) \ Λ 6= ∅ and that λ0 ∈ v(L) \ Λ. It is easilyverified that the function D(λ) vanishes at the point λ0; this gives v(L)\Λ ⊆ v(D).Consequently, the set v(L) \ Λ has density zero and L ∈ M(Λ;h). Next considerthe auxiliary function

ϕ(λ, z) := D(λ)eλz −∞∑k=1

D(λk)L′(λk)

L(λ)λ− λk

eλkz ,

which, for fixed λ ∈ C, is analytic in the domain G (see (23)) and also is an e.f.e.t.(in λ, for fixed z ∈ G) with indicator not exceeding h(θ). The important thingis that ϕ(µ, z) = 0 for all µ ∈ v(L) and all z ∈ G. The last relation shows thereforethat, for fixed z ∈ G, the function

ψ(λ, z) :=ϕ(λ, z)L(λ)

=D(λ)L(λ)

eλz −∞∑k=1

D(λk)L′(λk)

eλkz

λ− λk

is an e.f.e.t. in λ. We claim that ψ(λ, z) = 0 for all λ ∈ C and all z ∈ G. Let Ube the union of the discs Uk = {λ ∈ C : |λ − λk| < e−δ|λk|}, k = 1, 2, . . . , with

Representation of the reciprocal of an entire function 467

a finite sum of the radii for any δ > 0. Since 0 ∈ G, we may choose an interval Kon the imaginary axis which lies in G. Let hK(−θ) be its support function. In whatfollows we assume that δ > 0 is taken from the condition hK(θ) + 2δ < h(θ) forall θ ∈ [0, 2π). Because of (23) and the choice of δ, for all x ∈ R \ U and z ∈ K,by (22) we can write

|ψ(x, z)| 6∣∣∣∣D(x)L(x)

∣∣∣∣ +∞∑k=1

e(2δ−h(arg λk)+hK(arg λk))|λk| 6 M,

where the constant M is independent of the arguments of the function ψ. Thisenables us to apply the scheme developed in [10], Theorems 3 and 5. For fixedz ∈ K we obtain the following chain of implications:

|ψ(x, z)| is bounded on R \ U =⇒ |ψ(x, z)| is bounded on R=⇒ ψ(λ, z) is a function of completely regular growth =⇒ ψ(λ, z) ∈ [1; 0]=⇒ ψ(λ, z) ≡ cz = const.

It is easily seen that for all z ∈ K, ψ(x, z) → 0 as x→∞, x /∈ U , because

|ψ(x, z)| =∣∣∣∣ ∞∑k=1

D(λk)(exz − eλkz)L′(λk)(x− λk)

∣∣∣∣ 6 2∞∑k=1

e|λk|(δ−h(arg λk)+|z|)

|x− λk|.

Thus, cz = 0 for all z ∈ K, and therefore, by the uniqueness theorem, for all z ∈ G.A consequence of the above analysis is that ϕ(λ, z) = 0 for all z ∈ G and all λ ∈ C,whence for the same z and λ we obtain

D(λ)eλz =∞∑k=1

D(λk)L′(λk)

L(λ)λ− λk

eλkz.

By Korobeınik’s criterion, EΛ is an ARS in A(G), and hence the proof of Theorem 5is complete.

We shall complete our description of the exponents of absolutely representativesystems of exponentials in terms of expansions in partial fractions by establishingthe main result of this section.

Theorem 6. Let G be a bounded convex domain in C containing the origin. A nec-essary and sufficient condition for a system of exponentials EΛ to be an ARSin A(G) is that there exist Λ′ ⊂ Λ, L(λ) ∈ M(Λ′;h) and D(λ) ∈ [1; 0] \ {0} suchthat

D(λ)L(λ)

=∑λk∈Λ′

dkλ− λk

, (24)

limk→∞λk∈Λ′

{1|λk|

ln |dk|+ h(arg λk)}

6 0. (25)

Proof. It is clear that in the representation (24) we have dk = D(λk)/L′(λk) andλk ∈ Λ′.

468 V.B. Sherstyukov

Sufficiency. From (24) and (25) it follows by Theorem 5 that EΛ′ is an ARS in thespace A(G). A fortiori this holds for the system EΛ.

Necessity. Suppose that EΛ is an ARS in A(G). From Abanin’s theorem, cited above,we obtain the existence of a subsequence Λ′ ⊂ Λ and a function L(λ) ∈ M(Λ′;h)of completely regular growth with indicator h(θ). Moreover,

limk→∞λk∈Λ′

{1|λk|

ln |L′(λk)| − h(arg λk)}

= 0.

But then EΛ′ is also an ARS in A(G), and so the hypotheses of Theorem 5 aresatisfied, whence we obtain (24) and (25). The proof is complete.

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Representation of the reciprocal of an entire function 469

[17] Yu. F. Korobeınik, “Convolution equations in the complex domain”, Mat. Sb.127(169):2(6) (1985), 173–197; English transl. in Math. USSR-Sb. 55:1 (1986),171–194.

[18] A.V. Abanin, “Geometric criteria for representation of analytic functions by series ofgeneralized exponentials”, Dokl. Ross. Akad. Nauk 323:5 (1992), 807–810; English transl.,Dokl. Math. 45:2 (1992), 407–411.

V. B. Sherstyukov

Moscow Engineering Physics Institute

(National Nuclear Research University)

E-mail : shervb73@gmail.com

Received 12/NOV/07 and 31/JUL/08Translated by A. ALIMOV