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Journal of Function Spaces and Applications
Volume 2012 (2012), Article ID 184186, 12 pages
http://dx.doi.org/10.1155/2012/184186
Research Article

On the Frame Properties of Degenerate System of Sines

Institute of Mathematics and Mechanics of NAS of Azerbaijan, B. Vahabzade 9, 1141 Baku, Azerbaijan

Received 27 June 2012; Accepted 17 August 2012

Academic Editor: Maria Isabel Berenguer

Copyright © 2012 Bilal Bilalov and Fatima Guliyeva. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Systems of sines with degenerate coefficients are considered in this paper. Frame properties of these systems in Lebesgue spaces are studied.

1. Introduction

Basis properties of classical system of exponents ( is the set of all integers) in Lebesgue spaces , , are well studied in the literature (see [14]). Bari in her fundamental work [5] raised the issue of the existence of normalized basis in which is not Riesz basis. The first example of this was given by Babenko [6]. He proved that the degenerate system of exponents with forms a basis for but is not Riesz basis when . This result has been extended by Gaposhkin [7]. In [8], the condition on the weight was found which make the system forms a basis for the weight space with a norm . Basis properties of a degenerate system of exponents are closely related to the similar properties of an ordinary system of exponents in corresponding weight space. In all the mentioned works, the authors consider the cases when the weight or the degenerate coefficient satisfies the Muckenhoupt condition (see, e.g., [9]). It should be noted that the above stated is true for the systems of sines and cosines, too.

Basis properties of the system of exponents and sines with the linear phase in weighted Lebesgue spaces have been studied in [1012]. Those of the systems of exponents with degenerate coefficients have been studied in [13, 14]. Similar questions have previously been considered in papers [1518].

In this work, we study the frame properties of the system of sines with degenerate coefficient in Lebesgue spaces, when the degenerate coefficient, generally speaking, does not satisfy the Muckenhoupt condition.

2. Needful Information

To obtain our main results, we will use some concepts and facts from the theory of bases.

We will use the standard notation. will be the set of all positive integers; will mean “there exist(s)”; will mean “it follows”; will mean “if and only if”; will mean “there exists unique”; or will stand for the set of real or complex numbers, respectively; is Kronecker symbol, .

Let be some Banach space with a norm . Then will denote its dual with a norm . By , we denote the linear span of the set , and will stand for the closure of .

System is said to be uniformly minimal in if

System is said to be complete in if . It is called minimal in if , for all .

The following criteria of completeness and minimality are available.

Criterion 1 (Hahn-Banach theorem). System is complete in if for all .

Criterion 2 (see [19]). System is minimal in it has a biorthogonal system that is, , for all .

Criterion 3. Complete system is uniformly minimal in , where is a system biorthogonal to it.
System is said to be a basis for if for all , .
If system forms a basis for , then it is uniformly minimal.

Definition 2.1 (see [20, 21]). Let be a Banach space and a Banach sequence space indexed by . Let , . Then is an atomic decomposition of with respect to if(i) , for all ;(ii) : (iii) , for all .

Definition 2.2 (see [20, 21]). Let be a Banach space and a Banach sequence space indexed by . Let and be a bounded operator. Then is a Banach frame for with respect to if(i) , for all;(ii) : (iii) , for all .

It is true the following.

Proposition 2.3 (see [20, 21]). Let be a Banach space and a Banach sequence space indexed by . Assume that the canonical unit vectors constitute a basis for and let and be a bounded operator. Then the following statements are equivalent: (i)  is a Banach frame for with respect to .(ii)  is an atomic decomposition of with respect to .

More details about these facts can be found in [2023].

3. Completeness and Minimality

We consider a system of sines with a degenerate coefficient where are points of degeneration and .

The notation , means that the inequality holds in sufficiently small neighborhood of the point with respect to the functions and . Thus, it is clear that and for . Proceeding from these relations, we immediately obtain that the inclusion , , is true if and only if the following relations hold In what follows, we will always suppose that this condition is satisfied. Assume that the function is otrhogonal to the system , that is where is a complex conjugate. By , we denote the Banach space of functions which are continuous on with a sup-norm and vanish at the ends of the interval . It is absolutely clear that . As the system of sines is complete in , we obtain from the relations (3.4) that a.e. on , and, consequently, a.e. on . This proves the completeness of system (3.3) in .

Now consider the minimality of system (3.3) in . It is clear that if and only if It is easily seen that the system is biorthogonal to . So the following theorem is true.

Theorem 3.1. System is complete in , , if the relations (3.3) hold. Besides, it is minimal in if both (3.3) and (3.5) hold. Consequently, system is complete and minimal in if the following relations hold:

It is known that (see, e.g., [10, 11]) if , then system forms a basis for , . Let , where either or . In the sequel, we will suppose that the condition (3.6) is satisfied for . We have On the other hand where is some constant (in what follows will denote constants that may be different from each other), is such that does not contain the points . Let us show that . We have where . Thus It is absolutely clear that Taking into account this relation, we obtain where . Consequently It follows immediately that .

Regarding biorthogonal system we get Choose as small as the interval does not contain the points . Consequently where is some constant. We have First we consider the case . In this case, for sufficiently great , we have Let . Consequently and, as a result where are some constants. So we obtain that for , . Consequently, in this case we have Then it is known that (see, e.g., [22]) the system is not uniformly minimal and, besides, does not form a basis for .

Consider the case . Without limiting the generality, we will suppose that . In this case, with regard to the biorthogonal system we have Taking sufficiently small , we obtain As , then, in the absolutely same way as in the previous case, we get On the other hand, where . Similarly to the previous case again, we get As a result we obtain Thus, the following theorem is true.

Theorem 3.2. Let ; , and , where , . Then the system is complete and minimal in , but does not form a basis for it.

4. Defective Case

Here, we consider the defective system of sines , where , is some number. It follows directly from Theorem 3.2 that if the condition holds, then the system is minimal but not complete in . Assume . Let . Consider the completeness of system in . Suppose that is orthogonal to the system, that is, As and system is complete and minimal in , from (4.2) we get It is clear that as . Consequently, . As , then if and only if , and, consequently, . The similar result is true for . Thus, if , and , then the system is complete in . Now we consider the minimality of this system. Let We have Let us show that . In fact and, consequently From these relations, we immediately find that . As a result, . Then the relations (4.5) imply the minimality of system in . Similar result is true for . In the end, we obtain that if , then the system has a defect equal to 1.

Consider the case when , where . We look at the system , where , are some numbers. Let cancel this system out, that is, Using the previous reasoning, we find that for some constants , the following is true: Using representations we obtain ( is some constant), where it can be easily seen that and Thus, if and only if . Assume ; . As , it is clear that and . Suppose . We have It follows directly that for sufficiently small , we have where is some constant depending only on and . As a result, .

Consequently, . Moreover, it is not difficult to derive that . Thus, we obtain the following system for and : It is clear that . And, consequently, . As a result, , which, in turn, implies that the system is complete in . Let us show that it is also minimal in . Assume . Consider the system + , for all , where Simple calculations give the following representation: where , for all  are some constants. We obtain directly from this representation that . On the other hand, Thus, if , then the system is complete and minimal in , and, as a result, the system has a defect equal to 2. It is easy to see that the similar result is true if with . Continuing this way, we obtain that if , where , then the system is complete and minimal in , where , with .

Consider the basicity of system (i.e., the case of ) in . Similar to the case of , it can be proved that Concerning biorthogonal system, we have where the interval () does not contain the points . As , for , it is clear that Taking this circumstance into account, we have Consider the case of : where . Consequently, . Let . In this case we have and, consequently . As a result, we get that for , the system does not form a basis for . Assume that in this case, the system is a frame in , that is any function from can be expanded with respect to this system. As it does not form a basis for , zero has a non trivial decomposition, that is, where . As the system is complete and minimal in , it is clear that . Consequently, . It follows directly that the arbitrary element can be expanded with respect to the system . But this is impossible. Similar result is true for .

Proceeding in an absolutely similar way as we did in the previous case, we can prove that for , the system is complete and minimal in , but does not form a basis for it. Consequently, system has a defect equal to . The fact that it is not a frame in in this case too is proved as follows. Let . Assume that the system is a frame in . Then zero has a non trivial decomposition: . It is clear that , and let . It follows directly that the system is a frame in . The further reasoning is absolutely similar to the case of . This scheme is applicable for for all . Thus, we have proved the following main theorem.

Theorem 4.1. Let the following necessary condition be satisfied
Then the system is a frame (basis) in if and only if . Moreover, for , , it has a defect equal to , where .

Acknowledgment

The authors are thankful to the referees for their valuable comments.

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