Frame Properties of a Part of the System of Exponents in Hardy Weighted Classes
We consider a part of the classic system of exponents, study its frame property in Hardy weighted classes when, generally speaking the weight may not satisfy the Muckenhoupt condition.
Keywords: system of exponents, Hardy weighted class
American Journal of Mathematical Analysis, 2013 1 (1),
Received December 25, 2012; Revised January 30, 2013; Accepted February 25, 2013Copyright © 2013 Science and Education Publishing. All Rights Reserved.
Cite this article:
- Muradov, Togrul R.. "Frame Properties of a Part of the System of Exponents in Hardy Weighted Classes." American Journal of Mathematical Analysis 1.1 (2013): 1-7.
- Muradov, T. R. (2013). Frame Properties of a Part of the System of Exponents in Hardy Weighted Classes. American Journal of Mathematical Analysis, 1(1), 1-7.
- Muradov, Togrul R.. "Frame Properties of a Part of the System of Exponents in Hardy Weighted Classes." American Journal of Mathematical Analysis 1, no. 1 (2013): 1-7.
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The basis properties of the classic system of exponents (are integers) in Lebesgue spaces , , have been studied well and elucidated in the known monographs [1, 2, 3, 4]. In his basic work  N.K.Bari has put a question on existence of the normed basis in , that is not the Riesz basis. Babenko showed the first example . He proved that the degenerating system of exponents for forms a basis in , but is not the Riesz basis for . V.F.Gaposhkin generalized these results . A condition on the weight , when the system forms a basis in the weighted space with the norm was found in . The basis properties of the degenerate system of exponents are closely connected with similar properties of the ordinary system of exponents in the appropriate weighted space. In all enumerated works the cases when the weight or the degenerating coefficient satisfies the Muckenhoupt condition (see for example ). It should be noted that these arguments are valid for the system of sines and cosines as well.
Basis properties of the system of exponents and sines with linear phase in Lebesque weighted spaces were studied in [10, 11, 12]. The same properties for the system of exponents with degenerating coefficients were studied in [13, 14].
It should be noted that the solution of a lot of problems of mathematical physics and mechanics (see for instance [10, 11] ) by the Fourier method requires to study frame properties of the system of the form ( are natural number) in weighted spaces of functions. This issue is closely connected with studying the similar issue for the system of exponents . One of the methods for studying the basis properties of this system is the method of a boundary value problem of theory of analytic functions. This methods requires to study similar properties of a part of the considered system in Hardy weighted classes. The present paper is devoted to studying frame properties of a “half” of this system in Hardy weighted classes. Similar issues for the system of sines, cosines and exponents were studied in [15, 16, 17].
2. Necessary Information
By obtaining main results we’ll use the following notion and facts from the frame theory. Accept standard denotation. denotes natural numbers; means there exists; means it follows; stands for if and only if; means exists uniquely; denote real numbers or are complex numbers.
Let be some Banach space with the norm , be a space conjugated to its with appropriate norm . We’ll denote by a linear span of the set , by the closure of .
The system is said to be uniformly-minimal in , if .
The system is said to be complete in , if . It is called minimal in , if .
The following criteria of completeness and minimality hold.
Criterion 1. The system is complete , if .
Criterion 2. The system is minimal in it has a biorthogonal system i.e. is Kronecker’s symbol.
Criterion 3. The complete system is uniformly-minimal in , where is a system biorthogonal to it.
The system is called a basis in , if for , .
Cite some facts on frames.
Definition 1. Let be a Banach space and a Banach sequence space indexed by . Let . Then is an atomic decomposition of with respect to if :
Definition 2. 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 :
It is true the following
Proposition 1. 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 .
It the system forms a basis in , then it is uniformly-minimal.
, means that for a sufficiently small vicinity of the point it holds .
3. Hardy Weighted Classes
Denote by an ordinary Hardy class of functions analytic interior to a unit circle with the norm
Let be non-tangential boundary values of the function on a unique circle . It is known that it holds
Let be some weight function on . Assume , where is the Lebesgue space of functions with the space . Determine in the norm
and denote the appropriate space by . It is easy to establish the validity of the following lemma.
Lemma 1. Let . Then the norm , with respect to norm (1) in Banach.
The Hardy weighted class of functions analytic exterior to a unique circle and having a pole at the point at infinity of order no higher than , is determined similarly. Consequently
where are non-tangential boundary values on a unit circle of the function exterior to a unit circle. The following lemma is valid.
Lemma 2. Let . Then the space is Banach.
Let be a unit ball and be a unit circle. Denote the contraction of the classes and to a unit circle by and , i.e., .
In what follows we should pay attention to the following easily provable lemma.
Lemma 3. Let . Then, if , then , where
Indeed, let . It is clear that is analytic in and
From it follows that . Let be some number. Assume и . Having applied the Hölder’s inequality, we get
and choose from the conditions . This is always possible since , as . Then from relation (2) and from it follows that . The belonging of the function to the space follows from the Smirnov theorem (see for instance ).
Consider the part of the system of exponents. Assume
where (are real numbers). It is seen that the system belongs to the space iff . Indeed, for each fixed the function is the contraction of to . Recall the class of weights satisfying the Muckenhoupt condition. Denote by a class of weights , satisfying the Muckenhoupt condition, i.e.
where is an arbitrary subinterval of the segment , is a Lebesgue measure. Consider the case . This case holds iff . In this case the system forms a basis in . Take . Consequently, . As a result, it is clear that
where are non-tangential boundary values of on . Hence it directly follows that expands in series in the system in . And this expansion is obviously unique. Thus, the following theorem is valid.
Theorem 1. Let the following inequality hold
Then the system of exponents forms a basis in .
The following theorem is proved in the same way.
Theorem 2. If conditions (3), are fulfilled, the system of exponents forms a basis in .
Find a system biorthogonal to ... Take , where is the number conjugated to . This element generates the functional by the expression
This directly follows from the estimation
Then the system biorthogonal to is . Since it follows from (3) that this system belongs to . We’ll need the following
Lemma 4. Let . Then is continuously embedded into , if , i.e.
If suffices to establish validity of this lemma for . Let . We have
where . Applying the Hölder’s inequality we get
Hence it directly follows that . It follows from that . Consequently, by definition и, where . The lemma is proved.
5. Completeness. Minimality
We’ll consider the case when . Let the following hold
Assume . It follows from condition (4) that . Then by Theorem 1 the system forms a basis in . Assume that the functional annihilates the system , i.e. . By Lemma 1 and so . From the completeness of the system in we get, . As a result, we get that the system is complete in . Assume . Show that by fulfilling condition (4) the system is minimal in . Consider the system
Taking into account the relation , we get that on it holds the representation
It follows from conditions (3) that system (5) belongs to the space . Then from relations (6) we get that it is biorthogonal to the system and as a result is minimal in .
Consider the completeness of the system in . Let the functional annihilate the system
As it was established, the system forms a basis in , where . It follows from that . From the uniqueness of the biorthogonal system to the basis and from relations (7) we get that the functional is realized by the function
where here in the sequel denotes an absolute constant.
It is obvious that generates some bounded functional on . Consequently,
In the other hand, the system forms a basis in where . It follows from that . From the uniqueness of the biorthonormed system to the basis and from relations (7) we get .
It is easy to note that relation (8) holds also for negative values . As it follows from the results of , in this case the system of exponents is complete in . Then it is clear that , i.e. . As a result, we get that the system is complete and minimal in . Show that in this case the system doesn’t form a basis in . Verify the fulfillment of the inequality
where . We have . Consequently, for the basicity of with respect to the biorthogonal system we get the condition
Let the segment don’t contain the points . Then it is clear that
For sufficiently large we have . As a result
If , then hence we immediately get that as and so . Then the system is not uniformly minimal in and at the same time doesn’t form a basis in it. Consequently, in this case, the system has a deficiency equal 1.
Consider the case . In this case we have
For it holds
Let . Take . As a result
Using the elementary identities
we easily get the expression
where . Consequently
From the convergence of the integrals hence it directly follows that . From the previous reasonings we get that the system doesnt form a basis in . Thus, the following is valid.
Theorem 3. If , then the system forms a basis in . For the system is complete and minimal in , but doesn’t form a basis in it.
The similar result holds in , as well , i.e.
Theorem 4 . Let . Then the system is complete and minimal in , but doesn’t form a basis in it . For , the system is complete and minimal in , but doesn’t form a basis in it.
This work was supported by the Science Foundation of State Oil Company of Azerbaijan Republic.
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