# On the Completeness of  the System{zλnlogjz}

Authors: Xiangdong Yang
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

LetLa2(Ω)
denote a Hilbert space consisting of analytic functions on an unbounded domain Ω
located outside an angle domain with vertex at the origin. We obtain a completeness theorem for the system MΛ={zλnlogjz,j=0,1,…,mn-1}n=1∞, in La2(Ω).


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## Body

## 1. Introduction

LetΩ denote a domain in the complex z plane. Let La2(Ω) denote the space consisting of all functions f analytic in Ω with
(1.1)∬Ω|f(z)|2dm<∞,z=x+iy,
where dm is the area element in the z plane (i.e., dm=dxdy=rdrdθ for z=reiθ). It is well known that, with the inner product (1.2)〈f,g〉=∬Ωf(z)g(z)¯dm,
and the norm ∥f∥=〈f,f〉1/2, La2(Ω) is a Hilbert space (see, e.g., [1, Chapter  1]).We say a system{hn}⊂La2(Ω) is complete in La2(Ω) if its linear span is dense in La2(Ω) (see, e.g., [2–7]). The following lemma provides an elementary fact on completeness (see, e.g., [1, Chapter   1] and [3, Lemma  1.1]).Lemma 1.1.
A necessary and sufficient condition for the system{hn} to be complete in La2(Ω) is that for any f∈La2(Ω), if 〈f,hn〉=0 for all hn, then f(z)≡0.ADzhrbasian domain (see, e.g., [8]) is an unbounded simply connected domain satisfying the following conditions:ConditionΩ(I). For r>0, let σ(r) denote the linear measure of the intersection of the circle |z|=r and Ω. There exists r0>0 such that, for r>r0, (1.3)σ(r)≤e-α(r),
where α(r)>0 satisfies(1.4)α(r)=α(r0)+∫r0rφ(t)tdt
with φ(r)≥0 and φ(r)↑∞ as r→∞.ConditionΩ(II). The complement of Ω consists of m unbounded simply connected domains Ωj(j=1,2,…,m), each containing an angle domain Δj with opening π/βj,βj>1/2.For a Dzhrbasian domain satisfying ConditionΩ(I) and Condition Ω(II), Dzhrbasian proved that if (1.5)∫∞α(r)r1+ϑdr=+∞
(here ∫∞ means that the lower limit of the integral is sufficiently large), where(1.6)ϑ=max⁡{β1,…,βm},
then the polynomial system {1,z,z2,…} is complete in La2(Ω).Motivated by the result of Dzhrbasian, Shen [4–7] studied the completeness of the system {zλn} in La2(Ω), where {λn} is a sequence of complex numbers satisfying(1.7)theλnarealldistinctandlim⁡n→∞|λn|=∞,(1.8)lim⁡n→∞n|λn|=D,0<D<∞,(1.9)Reλn>0,|Im⁡λn|≤A
for some constant A. Shen also supposed that Ω is a Dzhrbasian domain with the vertex of Δ1 at the origin (hence 0 is outside of Ω). The following result was obtained in [4, 5].Theorem A.
Assume that the sequence{λn} satisfy (1.7), (1.8), and (1.9), and  Ω is a Dzhrbasian domain which satisfies  Ω(I),  Ω(II). If
(1.10)2β1(1-D)<1,
and, for some ɛ0,
(1.11)∫∞α(r)r1+ηdr=∞,
where
(1.12)η=max⁡{ϑ,11/β1-2(1-D)+ɛ0},
then the system {zλn} is complete in La2(Ω).An improved version of Shen's result is given in [3]. Let Ω be a Dzhrbasian domain with the added requirement of(1.13)Δ1={z:|arg(z)-π|<π2ϖ},
where ϖ>1/2 is some constant. In [3], for such a domain, results on the completeness on {zλn} in La2(Ω) were obtained, assuming {zλn} is a sequence of complex numbers satisfying (1.7) and (1.8), but (1.9) is replaced by the more general condition(1.14)|arg(λn)|<β<π2,
thus allowing ℑ(λn)→0 as n→∞.More accurately, the main result in [3] is described as follows.Theorem B.
Assume that the sequence{λn} satisfy (1.7), (1.8), and (1.14), and  Ω is a Dzhrbasian domain which satisfies  Ω(I),  Ω(II), and (1.13). Moreover, assume that
(1.15)2ϖ(1-Dcos⁡β)<1.
Let
(1.16)η=max⁡{ϑ,1h+ɛ0},
where ϑ is defined in (1.6), ɛ0 is some positive number, and
(1.17)h=max⁡0<δ<Dcos⁡β-1+1/2ϖ2δD2sin2β+δ2(Dcos⁡β-1+12ϖ-δ).
If
(1.18)∫∞α(r)r1+ηdr=+∞,
then the system {zλn} is complete in La2(Ω).Remark 1.2.
Theh in (1.17) is well defined, for reference we refer to [3, Remark  4].In this paper, motivated by the work in [3–7], we will investigate the completeness of the system MΛ={zλnlogjz,j=0,1,…,mn-1}n=1∞ in La2(Ω), where Ω is a Dzhrbasian domain with the added requirement of Δ1 satisfying (1.13). The system MΛ is associated with the multiplicity sequence Λ={λn,mn}n=1∞, that is, a sequence where {λn} are complex numbers with λn≠λm wherever n≠m, and each λn having multiplicity equal to mn. The sequence Λ satisfies (1.7), (1.14) and also(1.19)lim⁡nΛ(t)t=D,0<D<∞,
where nΛ(t)=∑|λn|≤tmn is the so-called counting function of the sequence Λ. We note that when mn=1 for all λn, the above relation is equivalent to (1.8). To describe even further the sequence Λ, thus the system MΛ as well, we need some definitions from [9]. We denote by L(c,D) the class of all complex sequence A={an}, |an|≤|an+1| satisfying the following properties: (1)n/|an|→D≥0,(2) for n≠k one has that |an-ak|≥c|n-k| for some constant c and (3)sup|arg(an)|<π/2. The following definition is from [9].Definition 1.3.
Let the sequenceA∈L(c,D) and a, b be real positive numbers such that a+b<1. We say that a sequence B={bn}n=1∞ belongs to the class Aa,b if for all n∈ℕ we have
(1.20)bn∈{z:|z-an|≤ana},
and for all k≠n one of the following holds: (i)
bk=bn,(ii)
|bk-bn|≥max{e-|ak|b,e-|an|b}.
We may writeB in the form of a multiplicity sequence Λ={λn,mn}n=1∞, by grouping together all those terms that have the same modulus and ordering them so that |λn|<|λn+1|. This form of B is called as {λ,m}  reordering  (see [9]).
We prove the elementary fact.Lemma 1.4.
SupposeΩ is a Dzhrbasian domain such that Conditions Ω (I), Condition Ω (II),  and (1.13) are satisfied.  Moreover, suppose Λ={λn,mn}n=1∞ is a sequence of complex numbers which is a {λ,m} reordering of B={bn}∈Aa,b of a sequence A={an}∈L(c,D) such that arg(an)→0 as n→∞, satisfying (1.14). Then zλnlogmnz∈La2(Ω).Proof.
Due to the definition of the domainΔ1, the principal branch of logz, that is, logz, is well defined on Ω. Thus, (logjz)(zλn)=logjzexp{λnlogz} is an analytic function in Ω. Let z=reiθ and λn=|λn|eiθn. Considering |θn|<β<π/2 and |θ|<π-π/2ϖ whenever z∈Ω, there exists some positive constant A such that for z∈Ω. 							(1.21)|zλnlog⁡mnz|<(Ar)|λn|+mn.
Since φ(r)↑∞ as r→∞, for r sufficiently large which is denoted by r1, we have φ(r)>2(|λn|+mn)+2. Without loss of generality, we can suppose r1>r0. By (1.4), we have
(1.22)e-α(r)r2(|λn|+mn)<exp{-(2(|λn|+mn)+2)}r2(|λn|+mn)=r12(|λn|+mn)r42.
Thus, we have
(1.23)∬Ω|zλnlog⁡mnz|2dxdy≤∫0r02πr(Ar)2(|λn|+mn)dr+∫r0∞σ(r)(Ar)2(|λn|+mn)dr≤2πA2(|λn|+mn)r02(|λn|+mn)+22(|λn|+mn)+2+∫r0∞e-α(r)(Ar)2(|λn|+mn)dr<∞.The main result of this paper is as follows.Theorem 1.5.
Suppose thatΩ is a Dzhrbasian domain such that Conditions Ω (I), Ω (II), and (1.13) are satisfied. Moreover, suppose that Λ={λn,mn}n=1∞ is a sequence of complex numbers which is a {λ,m} reordering of B={bn}∈Aa,b of a sequence A={an}∈L(c,D) such that arg(an)→0 as n→∞, satisfying (1.14). If
(1.24)∫∞α(r)r1+ηdr=+∞,
where η is defined in (1.16), ϑ is defined in (1.6), and h is defined in (1.17), then the system MΛ is complete in La2(Ω).The paper is organized as follows. In Section2, crucial lemmas in proving Theorem 1.5 will be presented. In Section 3, the completeness theorem above will be proved.


## 2. Preliminary Lemmas

We consider the function(2.1)G(z)=∏n=1∞(1-z2λn2)μn,
where μn denotes the multiplicity of the term 1-z2/λn2 and the integral(2.2)K(s)=-12πi∫-∞+∞e-iysG(iy)dy,s=u+iv.For sufficiently smallδ>0, denote(2.3)Bδ={s=u+iv:|v|≤πDcos⁡β-δπ}.
Under the Conditions Ω (I), Ω (II), and Definition 1.3, by [10], we can get the following estimates which will play an important role in the proof of Theorem 1.5.Lemma 2.1.
Givenɛ>0,
(2.4)1|K(iy)|≤A(ɛ)e(-πDcos⁡α+ɛ)|y|,
where A(ɛ) is a constant which depends only on ɛ.Lemma 2.2.
There exists a sequence{tk} with k≥tk≥(1-λ)k (λ is some sufficiently small positive number) such that, for s=u+iv∈Bδ,Res=u≥0,
(2.5)|K(s)-∑|λn|<tk∑m=0mn-1an,msme-λns|≤Atke-utksin(μπ),
and, for s=u+iv∈Bδ,Res=u≤0,
(2.6)|K(s)-∑|λn|<tk∑m=0mn-1an,msme-λns|≤Atkeutk,
where A is a constant independent of s and tk, while μ is a small positive number satisfying
(2.7)tan(μπ)<δDsinβ.Letz=eξ,ξ=ξ1+iξ2 and denote the image of Ω in the ξ plane by Ω′. It follows from Condition Ω(II) and (1.13) that Ω′ must be located inside the strip (2.8)Bξ={ξ=ξ1+iξ2:|ξ2|<π(1-12ϖ)}.
Denote (2.9)Bϖ={s=u+iv:|v|<πDcos⁡β-π(1-12ϖ)},Bϖδ={s=u+iv:|v|≤πDcos⁡β-δπ-π(1-12ϖ)}.
Suppose that (2.10)2ϖ(1-Dcos⁡β)<1
from which πDcosβ-π(1-1/2ϖ)>0 follows, choosing δ sufficiently small so that(2.11)0<δ<Dcos⁡β-1+12ϖ.
It is obvious that if s∈Bϖδ and ξ∈Ω′, then |Im(s-ξ)|<πDcosβ-δπ, that is, s-ξ∈Bδ in (2.3). Thus for any f(z)∈La2(Ω), we can define a function for s∈Bϖδ by(2.12)F(s)=∫Ω′f(eξ)¯|eξ|2K(s-ξ)dξ1dξ2,ξ=ξ1+iξ2.Remark 2.3.
By Lemma  2.6 in [10], when ξ∈Ω′ is fixed K(s-ξ) is analytic for s∈Bϖδ; when s∈Bϖδ is fixed, K(s-ξ) is both measurable and bounded for ξ∈Ω′. Thus, it is not hard to prove that F(s) is analytic and bounded in Bϖδ (see [11, Chapter 10, Exercise 16; 1, Section 3] and [3, page 8]).The following lemma will be crucial in our proof of Theorem1.5.Lemma 2.4.
If fors∈Bϖδ, F(s)≡0 where F(s) is defined by (2.12), then
(2.13)∬Ωf(z)¯zndxdy=0,n=0,1,2,….Proof.
See [3, Lemma 2.4].We end this section by presenting two more lemmas. The first one is the so-called Carleman's Theorem (see [12, page 103]).Lemma 2.5.
Letlog-r=max{-logr,0}. If g(w) is analytic and bounded in the half-plane Im(w)≥0 and
(2.14)∫-∞+∞log⁡-|g(t)|1+t2dt=∞,
then g(w)≡0.We also need a result of M. M. Dzhrbasian (see [13, Section 10, Lemma  1]).Lemma 2.6.
Supposeα(r) be given as in (1.4), let
(2.15)Mn=∫r0∞e-α(r)rndr,Φ(r)=sup⁡n≥1rnM2n.
Then there exists some constant A>0 such that for r sufficiently large
(2.16)log⁡Φ(r)≥Aα(r).


## 3. Proof of Theorem1.5

Proof.
Let us fix some notations. Throughout this section,A will denote positive constants, and it may be different at each occurrence.
To prove Theorem1.5, it suffices to show that if f∈La2(Ω) and
(3.1)〈f(z),zλnlog⁡jz〉=0,j=0,1,2,…,mn-1,n=1,2,…,
then f(z)≡0. We claim that, by letting F(s) be the function as in (2.12), we only need to prove F(s)≡0 for s∈Bϖδ. Indeed by Lemma 2.4, it follows that (2.13) is satisfied, that is 〈f(z),zn〉=0,n=0,1,…. Since (1.24) holds, by Dzhrbasian's result the system {zn} is complete in La2(Ω) which means f(z)≡0. Our claim is now justified.
Fors∈Bδ, let {tk} be the sequence defined in Lemma 2.2, with k≥tk≥(1-λ)k where λ is a sufficiently small positive number. Then
(3.2)F(s)=∬Ω′f(eξ)¯|eξ|2K(s-ξ)dξ1dξ2,ξ=ξ1+iξ2=∬Ω′f(eξ)¯|eξ|2[K(s-ξ)-∑|λn|<tk∑m=0mn-1an,m(s-ξ)me-λn(s-ξ)]dξ1dξ2+∬Ω′f(eξ)¯|eξ|2(∑|λn|<tk∑m=0mn-1an,m(s-ξ)me-λn(s-ξ))dξ1dξ2=:F1,k(s)+F2,k(s).
Since
(3.3)〈f(z),zλnlog⁡jz〉=0,j=0,1,2,…,mn-1,n=1,2,…,
we have
(3.4)F2,k(s)=0.
Hence, for s=u+iv∈Bδ, F(s)=F1,k(s). By (2.5) and (2.6) in Lemma 2.2, we have
(3.5)|F(s)|=|F1,k(s)|≤Atk(e-utksin(μπ)∬Ω′⋂{Re(s-ξ)≥0}|f(eξ)¯||eξ|2|eξ|tksin(μπ)dξ1dξ2+e-utk∬Ω′⋂{Re(s-ξ)≤0}|f(eξ)¯||eξ|2|eξ|tkdξ1dξ2),
where A is a constant independent of k and s. Hence, for Res=u≥0,
(3.6)|F(s)|≤Atk(∬Ω|f(z)||z|tkdxdy|es|tksin⁡(μπ)+∬Ω|f(z)||z|tkdxdy|es|tk)≤Atk∬Ω|f(z)||z|tkdxdy|es|tksin⁡(μπ).
By Schwarz'z inequality
(3.7)|F(s)|≤Atk(∬Ω|f(z)|2dxdy)1/2(∬Ω|z|2tkdxdy)1/2|es|tksin(μπ),
and, by Condition Ω(I), we have the estimate
(3.8)∬Ω|z|2tkdxdy≤Atk∫r0∞e-α(r)r2tkdr,
where A is some positive constant independent of k and s. Thus, by k≥tk≥(1-λ)k, we have
(3.9)|F(s)|≤Atk(∫r0∞e-α(r)r2tkdr)1/2|es|tksin(μπ)≤Ak(∫r0∞e-α(r)r2kdr)1/2|es|(1-λ)ksin(μπ),
for every k=1,2,…. Hence,
(3.10)|F(s)|≤inf⁡k≥1{Ak(∫r0∞e-α(r)r2kdr)1/2|es|(1-λ)ksin(μπ)}=inf⁡k≥1{(∫r0∞e-α(r)r2kdr)1/2((1/A)|es|(1-λ)sin(μπ))k}.
Let(3.11)t=1A|es|(1-λ)sin(μπ),Mn=∫r0∞e-α(r)rndr.
Then
(3.12)|F(s)|≤inf⁡n≥1{M2ntn}.
If we let
(3.13)Φ(t)=sup⁡n≥1tnM2n,
then it follows from Lemma 2.6 that there is some constant q>0 so that
(3.14)Φ(t)>eqα(t).
Combining (3.12) and (3.14) shows that for ℜs≥0(3.15)|F(s)|≤e-Aα(r¯),r¯=1A|es|(1-λ)sin(μπ).
In order to use Lemma 2.5, we transform the domain Bδ into the upper half-plane Imz≥0. (i)
First, letz1=es, Bδ is then transformed into an angle |argz1|≤mπ, where
(3.16)m=Dcos⁡β-δ-1+12ϖ.(ii)
Letz2=z11/2m. The above angle domain is transformed into the right half-plane Rez2≥0.(iii)
Finally, letz=iz2; the right half-plane is then transformed into the upper half-plane Imz≥0.
More accurately, we have(3.17)|es|=|z1|=|z22m|=|(-iz)2m|=|z2m|,F(s)=F(log⁡z1)=F(log⁡z22m)=F(log⁡⁡(-iz)2m).
Defineg(z)=F(log(-iz)2m); it is obvious that g(z) is analytic and bounded in the upper half-plane Imz≥0. By (3.15), for Imz≥0 and |z| sufficiently large, we have
(3.18)|g(z)|≤e-Aα(A|z|2m(1-λ)sin(μπ))=e-Aα(A|z|m′),
where A is some positive constant independent of z, m is given by (3.16), and
(3.19)m′=2m(1-λ)sin(μπ)=2(Dcos⁡β-δ-1+12ϖ)(1-λ)sin(μπ).
Let tan(μπ)→δ/(Dsinβ) in (2.7), then
(3.20)sin(μπ)⟶δD2sin2β+δ2.
Denote(3.21)m′′=2δD2sin2β+δ2(Dcos⁡β-δ-1+12ϖ)(1-λ).
By (3.18), for Imz≥0 and |z| sufficiently large, we have
(3.22)|g(z)|≤e-A2α(A3|z|m′′).
It is obvious that δ can be chosen such that 0<δ<Dcosβ-1+1/2ϖ.
Denote(3.23)h′=max⁡0<δ<Dcos⁡β-1+1/2ϖm′′.
By (3.22), for Imz≥0 and |z| sufficiently large, we have
(3.24)|g(z)|≤e-A2α(A3|z|h′).
Since h′=h(1-λ), choosing λ sufficiently small yields
(3.25)1h′<1h+ɛ0,
where ɛ0 is defined in (1.16). Thus, by (3.24),
(3.26)∫∞log⁡|g(t)|t2dt≤-A∫∞α(w)w1+1/h′dw,
where A is some positive constant independent of w=cth′. Thus, by (1.24), we have
(3.27)∫∞log⁡|g(t)|t2dt=-∞.
Hence
(3.28)∫∞log⁡|g(t)|1+t2dt=-∞.
Let ∫-∞ mean that the upper limit of the integral is a negative number with sufficiently large magnitude. Similarly, we have
(3.29)∫-∞log⁡|g(t)|t2dt≤∫-∞-A2α(A3|t|h′)t2dt=∫∞-A2α(A3th′)t2dt=-∞.
Hence
(3.30)∫-∞log⁡|g(t)|1+t2dt=-∞.
By Remark 2.3, we know that
(3.31)∫ablog⁡|g(t)|1+t2dt<+∞
for every finite closed interval [a,b], thus
(3.32)∫-∞∞log⁡|g(t)|1+t2dt=-∞,
and, by Lemma 2.5, g(z)≡0.



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