TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 359, Number 4, April 2007, Pages 1829–1849 S 0002-9947(06)03950-X Article electronically published on November 22, 2006
COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS AND THE LAMBERT W FUNCTIONS
ANDRE´ BOIVIN AND HUALIANG ZHONG
Abstract. We study some of the properties of the solution system {eiλnt} of the delay-differential equation y(t)=ay(t − 1). We first establish some general results on the stability of the completeness of exponential systems in L2 and then show that the solution system above is always complete, but is not an unconditional basis in L2(−1/2, 1/2).
1. Introduction
It is well known that the solutions of differential-difference equations can be iλnt expressed as (an infinite) sum of exponentials cne where each iλn is a char- acteristic root of the equation. For a “nice” initial condition g(t), say, belonging to C0,orC1, the summation is known to converge pointwise to the solution it represents (see, for example, Bellman and Cooke [2]). In an attempt to better un- derstand this representation, we were led to study the structure (e.g. completeness, iλ t basis or frame properties) of the exponential systems {e n } when the iλn have a distribution similar to that of the characteristic roots of y (t)=ay(t − 1). The study of exponential systems, often referred to as the theory of nonharmonic Fourier series (see [23, 26]), has its origin in the classical works of R. Paley and N. Wiener [15] and N. Levinson [13]. One of the famous early results in the theory is int ∞ 2 that the basis property of the trigonometric system {e }−∞ is stable in L (−π,π) iλ t ∞ 2 in the sense that the system {e n }−∞ will always form a Riesz basis for L (−π,π) | − |≤ 1 if λn n L< 4 . M.I. Kadec’ [9], and R.M. Redheffer and R.M. Young [17] 1 have shown 4 to be optimal. Many (but not all) of the subsequent results on the completeness, frames, basis or interpolation properties of exponential systems required that the λn’s be located in a strip parallel to the real axis (e.g. [26]) or nearby the zeroes of a function of sine type (e.g. [1]). These results did not apply to our case since for the sequences we wish to consider, the λn’s are located in a curvilinear strip. The few known results that allowed for a (slow) growth of the imaginary part of the λn did not apply directly to our setting either and first needed to be extended. One of the main purposes of this paper is to provide such an extension. The paper is organized as follows. Section 2 contains basic definitions and nota- tions, and a discussion of the characteristic roots of the equation y (t)=ay(t − 1)
Received by the editors July 4, 2003 and, in revised form, February 4, 2005. 2000 Mathematics Subject Classification. Primary 42C15, 42C30, 34K07; Secondary 30B50. The first author was partially supported by a grant from NSERC of Canada.
c 2006 American Mathematical Society Reverts to public domain 28 years from publication 1829
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1830 A. BOIVIN AND H. ZHONG
in terms of the Lambert W functions, including a result on the asymptotic dis- tributions of the roots. In Sections 3 and 4, we present and prove two theorems regarding the stability of the completeness property in L2 of exponential systems under perturbations within a curvilinear strip. Previously imposed conditions in theorems of A.M. Sedletskii [18], and N. Fujii, A. Nakamura and R. Redheffer [6] and A.M. Sedletskii [23, §5.4, Theorem 5] have been relaxed. In Section 5, we apply Theorem 3.2 to special sequences related to the Lambert W functions Wn(a) 2 − 1 1 introduced in Section 2. In particular, we show the completeness in L ( 2 , 2 )of the system {eWn(a)t}. In Section 6, we show that the radius of completeness of {− } 1 the sequence iWn(a) is equal to 2 , and thus that the main results of Section 5 cannot be derived from the powerful theorem of Beurling and Malliavin [3, 10].
2. Preliminaries In this paper, we denote by C (respectively by R) the set of all complex (re- spectively real) numbers. Z denotes the set of all integers, and and mean summation and multiplication, respectively, through all the integers except 0. Un- less otherwise specified, all sequences considered in this paper will be indexed by the integers from −∞ to ∞. We say that an entire function f(z)isofexponential type γ if there is a constant A>0 such that
|f(z)|≤Aeγ|z|.
The totality of all entire functions of exponential type at most π that are square integrable on the real axis is known as the Paley-Wiener space (see [26]) which ∞ is a Hilbert space with respect to the inner product (f,g)= −∞ f(x)g(x)dx. Asystem{eiλnt} of complex exponentials is closed in Lp(−γ,γ), 1 ≤ p<∞,if every f ∈ Lp(−γ,γ) can be approximated in Lp norm by (finite) linear combinations of the functions eiλnt. Asystem{eiλnt} of complex exponentials is complete in Lp(−γ,γ), 1 ≤ p<∞, if the relations γ f(t)eiλntdt =0 −γ
p for all n with f ∈ L imply that f = 0 a.e. In this case {λn} is called a complete sequence. q p 1 1 Duality shows that closure in L is equivalent to completeness in L if p + q =1, and 1 Proposition A (N. Levinson, [13]). For the system {eiλnt} to be incomplete in C(−γ,γ) (or in Lp(−γ,γ), 1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1831 in the form γ (2.1) f(z)= eiztdω(t), −γ where ω(t) is of bounded variation on (−γ,γ) (or ω(t) ∈ Lq(−γ,γ), 1/p+1/q =1). Note 2.1. Equation (2.1) defines a function of exponential type at most γ and thus, when p =2andγ = π, the Plancherel Theorem shows that f in fact belongs to the Paley-Wiener space. Definition 2.2. Acompletesystem{eiλnt} in Lp that ceases to be complete when any one of its element is removed is said to be an exact complete system. When exactly m elements have to be removed (or added) in order that the new system be exact, then the excess Ep(λ) of the system is m (or −m). Definition 2.3. A sequence {f1,f2, ...} in an infinite-dimensional Banach space X is said to be a Schauder basis for X if for each f ∈ X, there is a unique sequence { } ∞ of scalars c1,c2,... such that f = n=1 cnfn, i.e. n f − cifi → 0asn →∞. i=1 Henceforth, the term basis will always mean a Schauder basis. To study the completeness of the solution system of the delay-differential equa- tion y (t)=ay(t−1), where a is a nonzero fixed real number, we need to understand the zeroes of its characteristic function g(z)=z −ae−z as a function of the variable a as well as their asymptotic distribution. − 1 − 1 Note that when a = e , all the zeroes of g(z)aresimpleandthatwhena = e , − − 1 there is a double zero at z = 1. Assuming first that a = e and that g(z)=0, iφ and writing z as z = η1 + iη2 = re (−π<φ≤ π), we get that (2.2) η1 =log|a|−log r and η2 =arga +2πn − φ for some n ∈ Z. Now assume that a is real. It is known (see [25] or [24]) that: − 1 ≤ ≤ (1) If a does not satisfy e a 0, then there is a zero zn corresponding to each integer n ∈ Z. − 1 (2) If a satisfies e License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1832 A. BOIVIN AND H. ZHONG Proposition B (S. Verblunsky, [24]). Suppose Wn(a) is the Lambert W function ∈ |a| for n Z,andα = 2π .Then i) when a>0, for all n =0 , we have α 1 log |n| 2 W (a)={log + sign(n)+O } n |n| 4n n α 1 log | | log |n| + i{2π(n − sign(n)) + n + O }, 4 2πn n2 and W0(a) > 0; − 1 ∈ \{− } ii) when e and −1 (1) if a>0, W0(a)isreal,andWn(a)=W−n(a) for all n ≥ 1; − 1 ≤ ≥ (2) if e a<0, only W0(a)andW−1(a) are real, and for all n 1, we have Wn(a)=W−n−1(a); − 1 (3) if a< e ,noneoftheWn(a) are real, and we have that Wn(a)=W−n−1(a) for all n ≥ 0. For convenience, we modify the sequence and rescale the Wn(a) in the following ways: When a>0, we set i (2.4a) V (a)=− W (a)forn ∈ Z. n 2π n When a<0, we set ⎧ ⎨ − i 2π Wn(a),n>0, (2.4b) Vn(a)=⎩ 0,n=0, − i 2π Wn−1(a),n<0. Furthermore we set ρn(a)=ReVn(a)andσn(a)=ImVn(a). It follows that when a<0, {Vn(a)} has “one element less” than {Wn(a)}.Butit gives us a nice symmetrical index on both sequences {ρn(a)} and {Vn(a)}. Indeed, from the definition of Vn(a) and Proposition B, we immediately obtain: Lemma 2.4. Define Vn(a) and ρn as above. Then Vn(a)=−V−n(a) and ρn(a)= −ρ−n(a) for all integers n and (nonzero) real a. Moreover Vn(a) has the following asymptotic property. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1833 When a>0, ⎧ ⎨⎪ { − 1 − (1) } { 1 | | } n 4 εn (a) + i 2π log n + O(1) ,n>0, Vn(a)=⎪ ⎩ { 1 (1) } { 1 | | } n + 4 + εn (a) + i 2π log n + O(1) ,n<0. When a<0, ⎧ ⎨⎪ { 1 − (2) } { 1 | | } n + 4 εn (a) + i 2π log n + O(1) ,n>0, Vn(a)=⎪ ⎩ { − 1 (2) } { 1 | | } n 4 + εn (a) + i 2π log n + O(1) ,n<0. (j) log |n| (j) Here εn (a)=O( |n| ) and εn (a) > 0, j =1, 2, for sufficiently large n depending on a. We end this section by noting that the above rescaling of Wn(a)isofnoconse- quence for the theory that follows. Indeed we have: Lemma 2.5. iλ t p (1) Let µn = σλn, σ>0.If{e n } is complete in L (−γ,γ), 1 { iµnt} p − γ γ Proof. Let µn = σλn.If e is not complete in L ( σ , σ ), then by Proposi- tion A, there exists an entire function g of the form γ σ g(z)= eiztdω(t), − γ σ q − γ γ such that g(µn)=0whereω(t)isinL ( σ , σ )and1/p +1/q =1. Substituting σw for z in the above integral, we obtain γ σ γ iσwt iws g1(w)=g(σw)= e dω(t)= e dω1(s), − γ − σ γ q where ω1(s)=ω(s/σ)isinL (−γ,γ), and s = σt.Sog1 is an entire function sat- iλnt p isfying g1(λn)=g(µn) = 0. By Proposition A, {e } is incomplete in L (−γ,γ), which is a contradiction. The proof of (2) is the same as that for (1) provided this time we substitute z +τ for z in the integral. 3. Stability of completeness – Part I iλ t 2 N. Levinson (see [13]) showed that {e n } is complete in L (−π,π)if|λn − n|≤ 1 1 4 , and incomplete if λn = n+(4 + )sign(n)forn = 0 and arbitrary constant >0, iλ+t iλ−t and λ0 =0. So it is not too surprising that the two systems {e n } and {e n } have received a lot of attention, where the λ+ and λ− are defined by n n n +sign(n) 1 ,n∈ Z/{0}, (3.1) λ+ = 4 n 0,n=0, License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1834 A. BOIVIN AND H. ZHONG and n − sign(n) 1 ,n∈ Z/{0}, (3.2) λ− = 4 n 0,n=0. See for example [17, 18, 20, 22, 26, 27, 28]. In 1940, N. Levinson gave the following sufficient condition for completeness. Proposition C (N. Levinson, [13]). Suppose that {λn} is a sequence of complex | |≤ numbers, that nλ(t) denotes the number of points λk in the disc z t and let R nλ(t) { } N(R)= 0 t dt. If the sequence λn satisfies 2γ 1 lim sup{N(R) − R + ln R} > −∞, R→∞ π p iλnt p then the system {e } is complete in L (−γ,γ)(1 Remark 3.1. It follows from Proposition C that if the sequence {λn} is either − { +} { } { iλnt} λn or λn as defined by (3.1) and (3.2), then the system e is complete in L2(−π,π). We note also that although 1/(2p) is the best possible constant, the condition above can still be relaxed if the constant is replaced by a sequence depending on n. Theorem A (A.M. Sedletskii, [18]). Let {λn} and {µn} be two real sequences and 2 0 <γ<∞. Then the excess E2(λ) is equal to the excess E2(µ) in L (−γ,γ) if either one of the following conditions holds: s (1) For some 0 Theorem 3.2. Let 0 <γ<∞,andlet{λn} and {µn} be two sequences of distinct complex numbers such that (3.3) |λn − µn|≤φ(|n|), | | φ(|n|) ∞ where φ( n ) is nonincreasing and tends to 0,and |n| < . Then the excesses of the two exponential systems {eiλnt} and {eiµnt} in L2(−γ,γ) are equal, that is, E2(λ)=E2(µ). For the proof, we will need the following two results. Proposition D (N. Levinson, [12, Proof of Theorem III], [13, Theorem VIII]). Suppose f(z) is an entire function of exponential type a and {λn} is its zero set. Let n+(r) and n−(r) be the counting functions for its zeroes in the right half disc {z ∈ C : |z| License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1835 then for some constant C0, the following statements hold: n+(r) n−(r) (1) lim = lim = C0. r→∞ r r→∞ r n1(r) n2(r) (2) lim = lim = C0. r→∞ r r→∞ r Remark 3.3. The function f defined by (2.1) in Proposition A is bounded on the real axis, and thus such an f satisfies the hypotheses of Proposition D. But by Bernstein’s theorem (see [4, Notes for Chapter 3]), such an f must be of nonzero type. The proof of Proposition D then yields in this case that C0 > 0. Proposition E (A.M. Sedletskii, [21]). { }∞ a) Suppose G(z) is in the Paley-Wiener space and its zero set is zn n=0. Then lim 1 exists. r→∞ zn 0<|zn| Proof of Theorem 3.2. To prove the theorem, we need only to show that {µn} is not 2 2 a complete sequence in L (−γ,γ)if{λn} is assumed to be incomplete in L (−γ,γ). Without loss of generality, we can assume γ = π. 2 So assume that {λn} is incomplete in L (−π,π). Then by Proposition A, there is an entire function F (z) of the form (2.1) in the Paley-Wiener space such that the sequence {λn} is a subset of its zeroes. Noting that for any zero point β of F (z), F (z)/(z − β) belongs to the Paley-Wiener space (see [13], [11, Appendix III, Theorem 2]), we can assume without loss of generality that {λn} is exactly the zero set of F and that all the zeroes are simple. Since any finite number of terms can be replaced without altering the complete- ness (see [13]), considering Proposition D, we can assume that µ0 = λ0 =0and Re λn < 0forn<0andReλn ≥ 0forn>0. We know from Proposition E that 1 exists, and the Hadamard factorization λn of F (z)givesthat z F (z)=Beazz (1 − ). λn Let λn = ρn + iσn. We first show that there exists a δ>0 such that δ |λ | (3.4) ≤ n ≤ 2δ 2 |n| License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1836 A. BOIVIN AND H. ZHONG and δ |ρ | (3.5) ≤ n ≤ 8δ 8 |n| for |n| large enough. Indeed, from Proposition D and Remark 3.3, we have that + − C0 n (|λn|) C0 n (|λn|) < < 2C0 and < < 2C0 2 |λn| 2 |λn| + for sufficiently large n. From our assumptions on the λn’s, it follows that n (|λn|)= − n (|λ−n|) ≈|n|. Thus, for large enough n, 2 |λ | 1 ≥ n ≥ , C0 |n| 2C0 and (3.4) follows with δ =1/C0. C0 ≤ nρn ≤ Similarly, from (2) of Proposition D, we have 2 2C0. Combining 2 |λn| with (3.4), we get (3.5). Now from the hypotheses of the theorem and Proposition E, it follows that µ λ − µ 2φ(|n|) lim 1 − n = lim n n ≤ lim =0 →±∞ →±∞ →∞ n λn n λn n δ|n| and − | | 1 λn µn 1 ≤ 4φ( n ) 1 ∞ = + 2 2 + < . µn λnµn λn δ n λn We now set z F ∗(z)=Beazz (1 − ). µn To reach our goal we will apply the second part of Proposition E with h = 0 and will show that |F ∗(x)|≤const|F (x)|, x ∈ R. By Proposition F below (see Section 5) we can assume |σn|≥1 for all n, and since by hypothesis the sequence {φ(|n|)} must be bounded, Proposition F again allows us to change the imaginary part of µn to agree with that of λn.Sowemaysetµn = λn +τn where τn is real. It follows that for x ∈ R, 2 − 2 2 2 2 x (ρn x + τn) + σn λn x 1 − = 1 − µ (ρ − x)2 + σ2 µ λ n n n n n − 2 2 λn µn x =(1+ηn) 1+ 1 − µn λn | − 2| { − 2 2 } ≤ ≤ where ηn = 2τn(ρn x )+τn / (ρn x) + σn . Note that when 0 u b0 < 1, uk u we have log(1 + u)= ≤ . So log(1 + un)isconvergentif un is k 1−b0 convergent. |λn| δ Since | | ≥ and |λn − µn|≤φ(|n|), the hypothesis on φ implies that n 2 − | − | |τ 2 | | λn µn | | |≤ 2τn(ρn x) n log 1+ converges. Next, since ηn − 2 2 + − 2 2 , to show µn (ρn x) +σn (ρn x) +σn that |ηn| is bounded on the positive real axis and that the bound is independent of x, we consider three cases: x | − |≥| | 1) When ρn < 2 , x ρn ρn = O(n). So from the hypothesis on φ and (3.5), we have that 2φ(|n|) φ2(|n|) |η |≤ + License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1837 x ≤ 2) When 2 ρn where n(t) is the counting function of ρn in the interval (0,t). Since φ(t) is nonincreasing, and n(t)/t is bounded uniformly, as shown by (3.5), we have that both I2 and I4 are uniformly bounded. For I1, again since φ(t)is nonincreasing, we have that − − x x 1 x t 2 s 2φ( 8δ ) 2φ( 8δ ) I1 ≤ dn(t) ≤ dn(s). x x − t s 2 1 φ(|n|) ∞ Since |n| < ,wegetthatI1 is uniformly bounded. Similarly, since φ(x)is nonincreasing, it follows that x−1 x−t x φ( 8δ ) I3 ≤ φ( ) dn(t), 16δ x x − t 2 which is uniformly bounded. 3) When ρn >x+1, | − | | 2 | | |≤ 2φ(n)(ρn x) φ (n) ηn 2 + 2 . (ρn − x) +1 (ρn − x) +1 ρn>x+1 ρn>x+1 ρn>x+1 Similar to the above discussion, we only need to establish a uniform bound for |φ(n)| I5 = .Since ρn>x+1 (ρn−x) ∞ φ( t ) ∞ φ( 1 (u + x)) ≤ 8δ ≤ 8δ I5 − dn(t) dn(u) x+1 t x 1 u and since φ(n) does converge and φ(x) is nonincreasing for x>0, we get that n ∞ φ( 1 (u)) ≤ 8δ I5 1 u dn(u), which is uniformly bounded. Thus we get |F ∗(x)|≤const|F (x)| for x>0. The same argument can be applied to the case of x<0. So |F ∗(x)/F (x)| is uniformly bounded on the real axis, and this completes the proof. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1838 A. BOIVIN AND H. ZHONG 4. Stability of completeness – Part II In the last section, we have discussed and compared the excesses of two complex sequences which were close to each other. In this section, we consider some cases where the corresponding elements of the two sequences are some fixed distance apart. In 1999, N. Fujii, A. Nakamura and R. Redheffer published the following result. Theorem B ([6]). Let {µn} be a complex sequence such that |µn − n|≤c for −∞ λn = µn + a and λ−n = µ−n − b for n>0, where a ≥ 0 and b ≥ 0 are constants. Then E2(λ) ≤ E2(µ) on the interval (−π,π). This theorem is in fact a special case of a more general result obtained previously by A.M. Sedletskii ([20], [23, §5.4, Theorem 5]). We will extend the condition |µn − n|≤c to | Re µn − n|≤c, and prove the following result. Theorem 4.1. Assume that {λn} and {µn} are two complex sequences such that λ0 = µ0 and λn = µn + a and λ−n = µ−n − b for n>0, where a ≥ , b ≥ are constants. Suppose µ β iσ and σ ψ |n| ,where 0 0 n = n + n n = ( ) ≤ ψ2(|n|) ∞ ψ(x) is increasing and satisfies ψ(2x) 2ψ(x) for x>x0 > 0 and n2 < . If | Re µn −n|≤c for all integers n,thenE2(λ) ≤ E2(µ) on the interval (−π,π). ∞ Proof. Suppose {µn}−∞ is exact. To prove Theorem 4.1, it suffices to show that {λn}n∈Z\{0} is incomplete. From Proposition A we know that there is an en- tire function F1(z) vanishing on the sequence {µn}n∈Z\{0} such that F1(x) ∈ 2 L (−∞, ∞). From the condition on the sequence {λn},wehaveλn = αn + iσn, where αn and σn are real, and αn = βn + a, α−n = β−n − b, n > 0. It follows from the conditions |βn − n|≤c and σn = ψ(|n|)that ∞ ∞ 1 1 (µ − n)+(µ− + n) + = n n µ µ− µ µ− n=1 n n n=1 n n ∞ 2c 2σ ≤ ( + n ) < ∞. |µ ||µ− | |µ ||µ− | n=1 n n n n Thus by Hadamard’s factorization theorem, we have ∞ az z z F1(z)=Ke (z − µ0) (1 − )(1 − ) µ− µ n=1 n n for some constant a. As in the proof of Theorem 3.2, we may assume that µ0 = λ0 =0and Re µn < 0forn<0andReµn ≥ 0forn ≥ 0. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1839 Following the argument presented in the proof of Theorem 3.2, we set ∞ az z z G1(z)=Ke z (1 − )(1 − ), β− β n=1 n n ∞ az z z G2(z)=Ke z (1 − )(1 − ), α− α n=1 n n ∞ az z z F2(z)=Ke z (1 − )(1 − ). λ− λ n=1 n n Then from Proposition E, it suffices to show that for some constant C (4.1) |F2(x − i)|≤C|F1(x − i)| on the x-axis. ψ2(|n|) ∞ | − |≤ From the hypotheses of the theorem, we have that n2 < and αn n 2 σn const, so α2 converges. It follows that n 2 αn −1 σn =exp log(1 + 2 ) λn 2 αn converges to a nonzero constant A.Consequently, ∞ − z − z ∞ F (z) (1 − )(1 ) α− α (z − λ− )(z − λ ) 2 λ n λn n n n n = z z = G2(z) (1 − )(1 − ) λ−n λn (z − α−n)(z − αn) n=1 α−n αn n=1 = A|φ−(z)||φ+(z)|, ∞ − ∞ z λ−n z−λn where φ−(z)= and φ+(z)= .Notethat n=1 z−α−n n=1 z−αn ∞ − − ∞ 2 (x α−n) (σ−n +1)i σ−n +2σ−n | − − | φ (x i) = = 1+ 2 (x − α−n) − i (x − α−n) +1 n=1 n=1 2 σ +2σ− 1 | −n n | =exp log 1+ 2 , 2 (x − α−n) +1 F2(x−i) which converges. Furthermore |φ−(x − i)|→1and − /|φ+(x − i)|→A as G2(x i) F2(x−i) x → +∞. Similarly, we have |φ+(x − i)|→1and /|φ−(x − i)|→A as G2(x−i) x →−∞. Next consider F1(z)/G1(z). Similar to the above discussion, we have ∞ (1 − z )(1 − z ) µ−n µn = B|ψ−(z)||ψ (z)|, (1 − z )(1 − z ) + n=1 β−n βn where ∞ ∞ − ∞ − β−n βn z µ−n z µn B = ,ψ−(z)= and ψ+(z)= . µ− µ z − β− z − β n=1 n n n=1 n n=1 n Repeating the argument above, we get that |ψ−(x − i)|→1asx → +∞ and |ψ+(x − i)|→1asx →−∞. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1840 A. BOIVIN AND H. ZHONG Consequently, − F1(x i) /|ψ+(x − i)|→B as x → +∞ G1(x − i) and − F1(x i) /|ψ−(x − i)|→B as x →−∞. G1(x − i) Note that when n>0, λn = µn + a,andwhenn<0, λn = µn − b.Sowehavethat ∞ ∞ x − i − λ (x − i − a) − µ φ (x − i)= n = n = ψ (x − a − i) + x − i − α (x − i − a) − β + n=1 n n=1 n and that φ−(x − i)=ψ−(x + b − i). To prove (4.1) for all x ∈ R, we consider the three cases: x<−M, x>M and −M ≤ x ≤ M for some sufficiently large positive M.Firstforx<−M,wehave that − F2(x i) |F2(x − i)| = |G2(x − i)|≤2A|φ−(x − i)||G2(x − i)| G2(x − i) =2A|ψ−(x − i + b)||G (x − i)| 2 − 4A F1(x i + b) ≤ |G2(x − i)|. B G1(x − i + b) − From Lemma 4.2 below, it is sufficient to show that | G2(x i) | is uniformly bounded. G1(x−i+b) Note that − ∞ (1 − x−i )(1 − x−i ) ∞ − − G2(x i) α−n αn β−nβn x i αn = x−i+b x−i+b = G1(x + b − i) (1 − )(1 − ) α−nαn x − i − βn + b n=1 β−n βn n=1 ∞ = L(n, x), n=1 where [α− α +(bα − aα− ) − ab][x − i − α ] L(n, x)= n n n n n α−nαn(x − i − αn + a + b) α− α (x − i − α )+(bα − aα− − ab)(x − i − α ) = n n n n n n α−nαn(x − i − αn + a + b) {−(a + b)α α− +(bα − aα− − ab)(−α )} =1+ n n n n n α−nαn(x − i − αn + a + b) (x − i)(bα − aα− − ab) + n n . α−nαn(x − i − αn + a + b) Set αn = n + cn for all n ∈ Z.ThenL(n, x) can be written as (x − i)(n(a + b)+(bc − ac− − ab)) L(n, x)=1+ n n α−nαn(x − i − αn + a + b) 2 n(−bc− − bc + ab)+b(c c− + c + ac ) + n n n n n n α−nαn(x − i − αn + a + b) (x − i)K(1) + K(2) =1+ n n , α−n(x − i − αn + a + b) License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1841 (1) (2) where both Kn and Kn are real and can be bounded with a bound independent (1) of n.Moreover,Kn → (a + b) > 0asn →∞. Noting that the denominator of the 2 last expression contains a term in n , it follows that for any given x, the infinite product L(n, x) converges. Next we show that L(n, x) is uniformly bounded for x<−M: (1) (2) [(x − i)Kn + Kn ][x − αn + a + b + i] L(n, x)=1+ 2 α−n((x − αn + a + b) +1) (1) (2) (1) (1) (2) [(xKn + Kn )(x − αn + a + b)+Kn ]+i[Kn (αn − a − b)+Kn ] =1+ 2 α−n((x − αn + a + b) +1) (1) (1) − − (2) (1) (2) Kn Kn (αn a b)+Kn (xKn + Kn )+ − + i − =1+ (x αn+a+b) (x αn+a+b) α−n(x − αn + a + b + δn) (xK(1) + K(3)) − iK(4) =1+ n n n α−n(x − αn + a + b + δn) (−x)K(1) − K(3) K(4) = {1+ n n } + i n , α−n(αn − a − b − δn − x) α−n(αn − a − b − δn − x) 1 (3) (4) where δn = and Kn and Kn can be bounded with a bound independent x−αn+a+b (1) (3) of n and x.Letωn = α−n(αn − a − b − δn − x)andΓn =(−x)Kn − Kn .Then we have that 2 (4) 2 − − − (4) 2 −| |2 − Γn − Γn +(Kn ) Γn[ 2ωn Γn] (Kn ) 1 L(n, x) = 2 2 = 2 ωn (ωn) (ωn) (1) (3) (4) 2 Γn{(−x)(−2α−n − Kn )+[−2ωn − 2α−nx + Kn ]}−(Kn ) = 2 . (ωn) (1) Choose N>0, independently of x, such that when n>N,wehave−2α−n −Kn > (3) 0and−2ωn − 2α−nx + Kn > 1. Then for sufficiently large M>0andx<−M, − (4) 2 −| |2 ≥ ωn (Kn ) 1 L(n, x) 2 > 0. (ωn) It follows that when x<− M, |L(n, x )| < 1forn>Nand |L(n, x)| 4AC (M,N) |F (x − i)|≤ 0 |F (x − i + b)| 2 B 1 4AC (M,N) ≤ 0 |K(x, b)||F (x − i)| B 1 ≤ C1(M,N)|F1(x − i)| for every x satisfying x<−M,whereC(M,N), C0(M,N)andC1(M,N)are constants. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1842 A. BOIVIN AND H. ZHONG Similarly, for x>M we have (x − i)(n(a + b)+(bc − ac− − ab)) L(n, x)=1+ n n α−nαn(x − i − α−n − a − b) 2 n(−ac− − ac + ab)+b(c c− + c− + bc− ) + n n n n n n α−nαn(x − i − α−n − a − b) (x − i)K + K =1+ 1 2 , α−n(x − i − α−n − a − b) a+b where K1 satisfies 0 < 2 Lemma 4.2. Assume that {µn} is defined as in Theorem 4.1. Then for any given ∞ − constant b, K(x, b)= µn x+i+b is bounded unformly on the x-axis. n=−∞ µn−x+i Proof. From the definition of K(x, b), we can rewrite it as ∞ ∞ b b K(x, b)= (1 + )= (1 + ) µ − x + i µ − x + i n=−∞ n m=−∞ m+[x] ∞ 2 b b(µ + µ− − 2x +2i)+b =(1+ ) (1 + m+[x] m+[x] ). µ − x + i (µ − x + i)(µ− − x + i) [x] m=1 m+[x] m+[x] | − | − Since µn = βn +iσn with βn n = O(1), we have µm+[x] x = βm +iσm+[x] with − − βm =(βm+[x] x). From the restriction above on βm, we know that βm m = cm and |cm|≤c for m ∈ Z.Thus b K(x, b)=(1+ ) µ[x] − x + i ∞ b(c + c− + b)+ib(σ + σ− +2) × m m m+[x] m+[x] (1 + ). (β + i(σ + 1))(β− + i(σ− +1)) m=1 m m+[x] m m+[x] Next, we try to prove that for a sufficiently large x with x>0, K(x, b) is bounded x x uniformly. Choose x0 such that when x>x0, ψ(x) < 2 and ψ(x) < 2ψ( 2 ), then ∞ ψ(m +[x]) + ψ(|[x] − m|) K(x, b) ≤ C (1 + ) 2 2 2 2 | − | m=3 m + ψ (m +[x]) m + ψ ( [x] m ) ∞ = C (1 + M(m, x)). m=3 To prove the uniform boundedness of K(x, b), we only need to consider the following summation: [ψ( x)]−1 [x] ∞ ( + + )M(m, x)=S1 + S2 + S3. m=3 m=[ψ(x)] m=[x]+1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1843 1) When 3 ≤ m<|ψ(x)| and x>x0, we know that [ψ(x)]−1 [ψ(x)]−1 1 1 S1 ≤ 2 ≤ 2 ≤ 4. 2 2 | − | ψ( x ) m=3 m + ψ ( [x] m ) m=3 2 ≤ ≤ [x] ψ(m+x) ≤ 2) When ψ(x) m<[x], we have S2 2 [ψ(x)] m2 . Since ψ(2x) 2ψ(x) for x>x0,wegetthat [x] 1 1 1 S ≤ 2ψ(2x) ≤ 4ψ(x)( − ), 2 m2 [ψ(x)] [x] m=[ψ(x)] which is uniformly bounded. 3) When m ≥ [x]andx>x0,wehave ∞ ∞ ψ(m +[x]) ψ(2m) S ≤ 2 ≤ , 3 m2 m2 m=[x] m=[x] which is also uniformly bounded. ∞ ≤ Thus m=1 M(m, x) is uniformly bounded when x>x0.For0 If there exists xj →∞such that ψ([xj] − i)=o(ψ([xj])) for i =1, ..., [xj], then ψ([x ]+m) ψ([x ]) j ≥ j . 2 2 2 2 2 2 m + ψ ([xj]+m) m + ψ ([xj] − m) m + ψ ([xj])(m + o(ψ(xj))) Since ψ(xj)=o(xj), a straightforward evaluation yields that xj ψ(xj) du ≈ log(ψ(xj)), 2 2 1 u + ψ (x )u j which implies that the (1 + M(m, xj)) may increase with xj for some particular sequences {µn}. This explains why we had to impose some restriction on the growth of the imag- inary parts of the sequence µn.ItisanopenquestionwhetherTheorem4.1holds without any such restriction. Note that the condition imposed on ψ(x)inTheo- rem 4.1 is satisfied by many smoothly increasing functions such as (log x)α or xα 1 (0 <α< 2 ). 5. Completeness of the solution system In this section we wish to apply the result of Section 3 to the special sequences introduced in Section 2. In particular we will prove: Theorem 5.1. Suppose {Vn(a)} is defined by (2.4). Then the excess E2(V ) of iVn(a)t 2 {e } in L (−π,π) satisfies E2(V )=0if a<0 and E2(V )=1if a>0. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1844 A. BOIVIN AND H. ZHONG From (2.4) and Lemma 2.5, we then get: Corollary 5.2. Let Wn(a) be the characteristic roots of the equation y (t)= ay(t − 1),wherea ∈ R, a =0 . Then the solution system {eWn(a)t} is complete in 2 − 1 1 L ( 2 , 2 ) and its excess is equal to 1. We show in Section 6 that Corollary 5.2 (and thus Theorem 5.1) is best possible in the sense that the solution system {eWn(a)t} fails to be complete on any interval of length greater than one. It is known that for {eiλnt} to be an unconditional basis, it is necessary that the sequence {λ } satisfies the condition n − λk λm (5.1) inf > 0. m λ − λ k= m k m See for example [14]. For (5.1) to hold, it is necessary that the sequence satisfies − λk λn (5.2) ≥ a>0,k= n. λk − λn See [7, Chap VII, Theorem 1.1]. Since clearly λn = Vn(a) does not satisfy (5.2), we have: Wn(a)t Proposition 5.3. The system {e }n∈Z\{0} is not an unconditional basis in L2(−γ,γ), for any γ>0. Note that a complete description of unconditional bases in L2(−γ,γ)oftheform − eiλnt was given by A.M. Minkin [14]. For a direct proof that {eWn( 1/e)t}, n =0, 2 − 1 1 is not even a Schauder basis in L ( 2 , 2 ), see [29]. Before proving Theorem 5.1, we need to introduce some results. In 1977, Red- heffer proved the following. Proposition F (R.M. Redheffer, [16]). Let {λn} and {µn} be two sequences of complex numbers, and suppose that Re(λn)=Re(µn) and | Im λn − Im µn|≤const. Then the L2 excesses of the two exponential systems {eiλnt} and {eiµnt} are equal, i.e. E2(λ)=E2(µ). Sedletskii [19] pointed out that the above theorem fails to be true in L1(−π,π) or C(−π,π). Furthermore, in 1985 he constructed in L2(−π,π) an example which shows that, in general, the boundedness condition on the imaginary parts in the above proposition cannot be removed (see [21]). However he obtained the following two results. Proposition G (A.M. Sedletskii, [19]). Suppose {hn} is a real sequence, and E2 is the excess of the exponential system {ei(n+ihn)t} in L2(−π,π). 1) If for some α ∈ [0, ∞) |hn|≤α log |n| (|n|≥n0), ≤ { } 1 ≤ { } then E2 [απ]+1. If, in addition, απ < 2 ,thenE2 [απ].Here[x] and x denote, respectively, the integral and fractional parts of x. 2) If moreover h2 (5.3) n < ∞ n2 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1845 and |hn|≥α log |n| (|n|≥n0,α∈ [0, ∞)), then E2 ≥ [απ]. If, in addition, {απ}≥1/2,thenE2 ≥ [απ]+1. In particular, if condition (5.3) is satisfied and hn/log |n|→∞as n →∞,then ∞ { i(n+ihn)t} 2 − the system e n=−∞ has infinite excess in L ( π,π). Proposition H (A.M. Sedletskii, [21]). Assume that the points {λn} and {µn} lie in the curvilinear strip {z : |y|≤φ(|x|)},whereφ(x)(x ≥ 0) is a positive nondecreasing function such that 2 2 φ (|n|)/n < ∞ and Re λn =Reµn,n∈ Z. If | Im λn|≤|Im µn| for all n ∈ Z,thenE2(λ) ≤ E2(µ). { } 1 | | Theorem 5.4. Suppose λn is defined by (3.1), µn = λn+ihn with hn = 2π log n , iµnt and h0 =0.ThenE2(µ)=0, that is, {e } is complete and exact. The theorem above can be derived from a result of Sedletskii which he proves using estimates of the Mittag-Leffler function (see Sedletskii’s formula below). Here we present a new proof based on the Lambert W functions, and the approach used here will also apply for the proof of Theorem 5.1. First we give one more lemma. − 1 { (Wn( e )+1)t ∈ \{ − }} − 1 1 Lemma 5.5. The system e ,n Z 0, 1 is incomplete in C[ 2 , 2 ] p − 1 1 ∞ (or in L ( 2 , 2 ), 1 Proof of Theorem 5.4. Since from Remark 3.1, {eiλnt,n ∈ Z} is complete, by Proposition H, E2(µ) ≥ 0. Thus to prove the theorem, we only need to show that {ei(λn+ihn)t,n∈ Z \{0}} is incomplete. 1 (W (− 1 )+1)t From Lemma 2.5 and Lemma 5.5, we get that {e 2π n e ,k ∈ Z \{0, −1}} 2 − 1 { − 1 } ≥ is incomplete in L ( π,π). Set Vn = 2πi Wn( e )+1 , n 0, and Vn = 1 { − 1 } { iVnt ∈ { }} 2 − 2πi Wn−1( e )+1 , n<0. Then e ,n Z/ 0 is incomplete in L ( π,π). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1846 A. BOIVIN AND H. ZHONG { } { } { − 1 − Furthermore if we set Vn = ρn+iσn,then Vn and ρn coincide with Vn( e ) i } { − 1 } 2π and ρn( e ) , respectively, as defined before in Section 2. In addition, by Lemma 2.4 1 |σ − log |n|| ≤ const. n 2π i(ρ +i 1 log |n|)t 2 Thus {e n 2π ,n =0 } is incomplete in L (−π,π)byPropositionF. − 1 Since ρn = ρn( e ), and λn is defined by (3.1), then by Lemma 2.4 again, we have log |n| |ρ − λ | = O( ). n n |n| { i | |} { i | |} It follows that the two sequences ρn + 2π log n and λn + 2π log n satisfy the i(ρ + i log |n|)t hypotheses of Theorem 3.2. Thus the two exponential systems {e n 2π } i(λ + i log |n|)t i(λ + i log |n|)t and {e n 2π } have the same excess, which means that {e n 2π ,n∈ Z \{0}} is incomplete. This ends the proof of Theorem 5.4. We note that there is a small misprint in the paper [19] of A.M. Sedletskii from 1978. He actually proved the following formula. Sedletskii’s formula (A.M. Sedletskii, [19]). Suppose µn = n − β sign(n)+ iα log |n|, n ∈ Z \{0},whereα ≥ 0, β real. Define φ(z) by z φ(z)=z (1 − ). λn Then |φ(z)| = O(|z|απ+2β) on Im z = −1,and ⎧ ⎪ { } 1 ⎨[απ +2β] if απ +2β < 2 , E2(µ)= ⎩⎪ { }≥ 1 [απ +2β]+1 if απ +2β 2 , where [x] and {x} denote, respectively, the integral and fractional parts of x. The above formula immediately yields the following lemma. { } 1 | | Lemma 5.6. Suppose λn is defined by (3.2), µn = λn +ihn with hn = 2π log n , and h0 =0.ThenE2(µ)=[απ +2β]=1. Proof of Theorem 5.1. Suppose a<0, and µn is defined as in Theorem 5.4. Then from Lemma 2.4, we get that | Im Vn(a) − Im µn|≤const and | Re Vn(a) − Re µn|≤ n(a), which satisfies the conditions on φ(|n|) in Theorem 3.2. Since E2(µ)=0, following the approach to prove Theorem 5.4, it is easy to see that E2(Vn(a)) = 0 for a<0. Using Lemma 5.6, Theorem 3.2 and Proposition F, the same argument will prove that E2(Vn(a)) = 1 for a>0. 6. Radius of completeness In this section, using the theorem of A. Beurling and P. Malliavin (Theorem C below), we show that the radius of completeness of the sequence {Vn(a)} is π,and thus that radius of completeness of {−iWn(a)} is one-half. Recall that to an arbitrary sequence Λ = {λn} of complex numbers containing no repetition, we can always associate a number R ∈ [0, ∞] called the radius of License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPLETENESS OF SYSTEMS OF COMPLEX EXPONENTIALS 1847 completeness characterized by the following property (see [10, Chap. IV]): iλ t 2 The linear span of {e n }, λn ∈ Λ, is dense in L (−L, L) for all 2 L iλnt When L = RΛ, examples show that {e } may or may not be complete in 2 L (−L, L). We claim that for Λ = {Vn(a)}, RΛ is equal to π.Thefactthat {eiVn(a)t} is complete in L2(−π,π) is the content of our Theorem 5.1 above. Assume that Λ is a sequence of (strictly) positive real numbers, which may contain repetition but has no accumulation points other than ∞.LetnΛ (t)bethe number of points of Λ , counting repetition, in [0,t]. Fix D>0andlet n (τ) − n (t) O (Λ )={t>0: Λ Λ >Dfor at least one τ>t}. D τ − t O D(Λ ) is an open set, and thus can be written as the union of at most countably O ⊂ ∞ many open intervals, i.e. D(Λ )= k≥0(ak,bk) (0, ). Set OD(Λ ) = ∞ if one of the intervals (ak,bk) is infinite. Otherwise set 2 bk − ak OD(Λ ) = . ak k≥1 There exists a number DΛ > 0 such that OD(Λ ) < ∞, for all D>DΛ and OD(Λ ) = ∞ whenever 0 { 1 ∈ } Λ = λn = ; λn Λ, (λn) =0 , (1/λn) 1 unless | ( )| = ∞,inwhichcaseRΛ = ∞. λn=0 λn Proposition 6.1. Fix a ∈ R \{0} and let Λ={Vn(a),n∈ Z}.ThenRΛ = π. Proof. Note that | ( 1 )| < ∞, so by Theorem C, it suffices to prove that n Vn(a) 1 DΛ =1,whereΛ = {λ }n∈Z with λ =1/ { }. n n Vn(a) Fix a<0(fora>0, the case is similar). We will show that D =1,the Λ+ ∈ proof that DΛ− = 1 being essentially the same. Note that for λn Λ+,wehaveby Lemma 2.4, 1 2 | | 1 2 log n + O(1) λ = n + − (2)(a)+ 4π , n 4 n 1 − (2) n + 4 n (a) License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1848 A. BOIVIN AND H. ZHONG (2) log n where n (a)=O( n ), and thus 1 λ = n + + (a) n 4 n → →∞ with n(a) 0asn . We first show that D ≥ 1. Fix 0 Acknowledgements We would like to express our sincere gratitute to Professor R. Corless for bringing the problem of representation of the solutions of delay-differential equations to our attention and for many useful discussions, and to Professor A.M. Sedletskii and an unknown referee for their valuable comments. 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Box 5015, London, Ontario, Canada N6A 5K8 Current address: Department of Radiation Oncology, Virginia Commonwealth University, 401 College Street, Richmond, Virginia 23298 E-mail address: [email protected] License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use