#### Abstract

Let denote a Hilbert space consisting of analytic functions on an unbounded domain located outside an angle domain with vertex at the origin. We obtain a completeness theorem for the system , in .

#### 1. Introduction

Let denote a domain in the complex plane. Let denote the space consisting of all functions analytic in with where is the area element in the plane (i.e., for ). It is well known that, with the inner product and the norm , is a Hilbert space (see, e.g., [1, Chapter  1]).

We say a system is complete in if its linear span is dense in (see, e.g., ). 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 to be complete in is that for any , if for all , then .

A Dzhrbasian domain (see, e.g., ) is an unbounded simply connected domain satisfying the following conditions:

Condition . For , let denote the linear measure of the intersection of the circle and . There exists such that, for , where satisfies with and as .

Condition . The complement of consists of unbounded simply connected domains , each containing an angle domain with opening .

For a Dzhrbasian domain satisfying Condition and Condition , Dzhrbasian proved that if (here means that the lower limit of the integral is sufficiently large), where then the polynomial system is complete in .

Motivated by the result of Dzhrbasian, Shen  studied the completeness of the system in , where is a sequence of complex numbers satisfying for some constant . Shen also supposed that is a Dzhrbasian domain with the vertex of at the origin (hence 0 is outside of ). The following result was obtained in [4, 5].

Theorem A. Assume that the sequence satisfy (1.7), (1.8), and (1.9), and is a Dzhrbasian domain which satisfies , . If and, for some , where then the system is complete in .

An improved version of Shen's result is given in . Let be a Dzhrbasian domain with the added requirement of where is some constant. In , for such a domain, results on the completeness on in were obtained, assuming is a sequence of complex numbers satisfying (1.7) and (1.8), but (1.9) is replaced by the more general condition thus allowing as .

More accurately, the main result in  is described as follows.

Theorem B. Assume that the sequence satisfy (1.7), (1.8), and (1.14), and is a Dzhrbasian domain which satisfies , , and (1.13). Moreover, assume that Let where is defined in (1.6), is some positive number, and If then the system is complete in .

Remark 1.2. The in (1.17) is well defined, for reference we refer to [3, Remark  4].

In this paper, motivated by the work in , we will investigate the completeness of the system in , where is a Dzhrbasian domain with the added requirement of satisfying (1.13). The system is associated with the multiplicity sequence , that is, a sequence where are complex numbers with wherever , and each having multiplicity equal to . The sequence satisfies (1.7), (1.14) and also where is the so-called counting function of the sequence . We note that when for all , the above relation is equivalent to (1.8). To describe even further the sequence , thus the system as well, we need some definitions from . We denote by the class of all complex sequence , satisfying the following properties: , for one has that for some constant and . The following definition is from .

Definition 1.3. Let the sequence and , be real positive numbers such that . We say that a sequence belongs to the class if for all n we have and for all one of the following holds: , .
We may write B in the form of a multiplicity sequence , by grouping together all those terms that have the same modulus and ordering them so that . This form of B is called as reordering (see ).
We prove the elementary fact.

Lemma 1.4. Suppose is a Dzhrbasian domain such that Conditions (I), Condition (II), and (1.13) are satisfied. Moreover, suppose is a sequence of complex numbers which is a reordering of of a sequence such that as , satisfying (1.14). Then .

Proof. Due to the definition of the domain , the principal branch of , that is, , is well defined on . Thus, is an analytic function in . Let and . Considering and whenever , there exists some positive constant such that for . Since as , for sufficiently large which is denoted by , we have . Without loss of generality, we can suppose . By (1.4), we have Thus, we have

The main result of this paper is as follows.

Theorem 1.5. Suppose that is a Dzhrbasian domain such that Conditions (I), (II), and (1.13) are satisfied. Moreover, suppose that is a sequence of complex numbers which is a reordering of of a sequence such that as , satisfying (1.14). If where is defined in (1.16), is defined in (1.6), and is defined in (1.17), then the system is complete in .

The paper is organized as follows. In Section 2, 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 where denotes the multiplicity of the term and the integral

For sufficiently small , denote Under the Conditions (I), (II), and Definition 1.3, by , we can get the following estimates which will play an important role in the proof of Theorem 1.5.

Lemma 2.1. Given , where is a constant which depends only on .

Lemma 2.2. There exists a sequence with ( is some sufficiently small positive number) such that, for , and, for , where A is a constant independent of and , while is a small positive number satisfying

Let and denote the image of in the plane by . It follows from Condition and (1.13) that must be located inside the strip Denote Suppose that from which follows, choosing sufficiently small so that It is obvious that if and , then , that is, in (2.3). Thus for any , we can define a function for by

Remark 2.3. By Lemma  2.6 in , when is fixed is analytic for ; when is fixed, is both measurable and bounded for . Thus, it is not hard to prove that is analytic and bounded in (see [11, Chapter 10, Exercise 16; 1, Section 3] and [3, page 8]).

The following lemma will be crucial in our proof of Theorem 1.5.

Lemma 2.4. If for , where is defined by (2.12), then

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. Let . If is analytic and bounded in the half-plane and then .

We also need a result of M. M. Dzhrbasian (see [13, Section 10, Lemma  1]).

Lemma 2.6. Suppose be given as in (1.4), let Then there exists some constant such that for sufficiently large

#### 3. Proof of Theorem 1.5

Proof. Let us fix some notations. Throughout this section, will denote positive constants, and it may be different at each occurrence.
To prove Theorem 1.5, it suffices to show that if and then . We claim that, by letting be the function as in (2.12), we only need to prove for . Indeed by Lemma 2.4, it follows that (2.13) is satisfied, that is . Since (1.24) holds, by Dzhrbasian's result the system is complete in which means . Our claim is now justified.
For , let be the sequence defined in Lemma 2.2, with where is a sufficiently small positive number. Then Since we have Hence, for , . By (2.5) and (2.6) in Lemma 2.2, we have where is a constant independent of and . Hence, for , By Schwarz'z inequality and, by Condition , we have the estimate where is some positive constant independent of and . Thus, by , we have for every . Hence,
Let Then If we let then it follows from Lemma 2.6 that there is some constant so that Combining (3.12) and (3.14) shows that for In order to use Lemma 2.5, we transform the domain into the upper half-plane . First, let , is then transformed into an angle , where Let . The above angle domain is transformed into the right half-plane . Finally, let ; the right half-plane is then transformed into the upper half-plane .
More accurately, we have
Define ; it is obvious that is analytic and bounded in the upper half-plane . By (3.15), for and sufficiently large, we have where is some positive constant independent of , is given by (3.16), and Let in (2.7), then
Denote By (3.18), for and sufficiently large, we have It is obvious that can be chosen such that .
Denote By (3.22), for and sufficiently large, we have Since , choosing sufficiently small yields where is defined in (1.16). Thus, by (3.24), where is some positive constant independent of . Thus, by (1.24), we have Hence Let mean that the upper limit of the integral is a negative number with sufficiently large magnitude. Similarly, we have Hence By Remark 2.3, we know that for every finite closed interval , thus and, by Lemma 2.5, .

#### Acknowledgments

The author gratefully acknowledges the help of Prof. E. Zikkos to improve the original version of the paper. The author also thanks the referees for improvement of the paper. This paper supported by Natural Science Foundation of Yunnan Province in China (Grant no. 2009ZC013X) and Basic Research Foundation of Education Bureau of Yunnan Province in China (Grant no. 09Y0079).