Letter to the Editor | Open Access

# Comment on “On the Frame Properties of Degenerate System of Sines”

**Academic Editor:**Yoshihiro Sawano

#### Abstract

The proof of Theorem 3.1 of the paper “On the Frame Properties of Degenerate System of Sines” (see (Bilalov and Guliyeva, 2012)) published earlier in this journal contains a gap; the reasoning given there to prove this theorem is not enough to state the validity of the mentioned theorem. To overcome this shortage we state the most general fact on the completeness of sine system which implies in particular the validity of this fact. It is shown in this note that the system , where is an exponential or trigonometric (cosine or sine) systems, becomes complete in the corresponding Lebesgue space or , respectively, whenever belongs to the corresponding Lebesgue space for all indices (under the evident natural condition ). It is also shown that the same conclusion does not remain valid for, in general, any complete or complete orthonormal system . Besides it, the largest class of functions for which the system is complete in space is determined.

#### 1. Introduction

The basis properties (completeness, minimality, and Schauder basicity) of systems of the form , where is an exponential or trigonometric (cosine or sine) system, are investigated in several papers (see, e.g., [1–15]). For example, Babenko gave an example , where and , answering in the affirmative a question of Bari [16] on the existence of normalized basis for that is not a Riesz basis [2]. It is also known that, in general, the system is a basis in space if and only if , where (it is, e.g., a direct consequence of the fact that the weight satisfies the Muckenhoupt condition if and only if (see, e.g., [6])).

Later, basicity properties (completeness, minimality, and Schauder basicity) of systems of more general form and and , where , have been investigated in several papers (see, e.g., [1, 3–15]).

The aim of this note is the determination of the largest class of functions for which the system , where is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space or , respectively.

It is easy to see that the relation or is possible if and only if or , respectively. This observation shows the validity of the following fact.

Proposition 1 (see [17]). *Let be any measurable function on on such that the relations and hold. Then the system is complete in the space .*

Indeed, if a function is orthogonal to the system () then for all ( for all ). By the above observation, we have (). This fact and the fact that the Fourier coefficients of a summable function with respect to the cosine (with respect to the exponential) system are uniquely determined and imply that a.e. Using this equality and the condition , we arrive at the equality a.e., which shows the completeness of the mentioned system.

This proposition shows that the largest set of functions for which the system (or ) is complete in the space (or ) is the set of all functions from (or ), for which .

Note that the observation similar to one that is given before the proposition is not valid for the sine system; more precisely, does not always imply . Therefore, the scheme used to prove the above proposition does not work in the case of the system . This feature of sine system was overlooked in the proof of one of the results of the paper [1]; but the results of this note show that the statement of the mentioned result from [1] is true.

It should be mentioned that the proposition given above and the analogous result for sine system which will be proven in this note show that the system , where is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space or , respectively, whenever and belongs to the corresponding Lebesgue space for all indices . This conclusion may give rise to the impression that if the function is such that and for all indices , where is any complete system in , then the system is also complete in the space . The arguments given below show that, in general, this is not true.

#### 2. Auxiliary Facts

Lemma 2. *Let be any measurable function on . The relation is possible if and only if .*

*Proof. **Necessity*. Let the relation hold. Write the function in the following form: Consider an auxiliary functionIt is evident that the function is continuous and never vanishes at the segment . Therefore, there is a positive number such that for all . Using these inequalities in (1), we obtain the following estimation: This estimation implies since by the condition of the Lemma. The necessity part of the lemma is proven.*Sufficiency*. Assume that . Take an arbitrary natural number . Write the function in the following form: Consider an auxiliary functionIt is evident that the function is continuous on . Therefore, there is a number such that for all . Using these inequalities in (4), we obtain the following estimation: Since , the last estimation implies . The lemma is proven.

Taking into account that the proof of the necessity part of this lemma relies only on the relation and using the sufficiency part of the same lemma we obtain the validity of the following.

Lemma 3. *Let be any measurable function on . The relation is possible if and only if .*

#### 3. Main Results

The main aim of this note is to prove the following.

Theorem 4. *Let be any measurable function on such that** Then the system is complete in space.*

*Proof. *Let the function be orthogonal to the system : Then the equalities hold for all . Subtracting (10) from (9) we obtain for all . Note that (by the condition of the theorem). Therefore and the fact that the Fourier coefficients of a summable function with respect to the cosine system is unique (along with (11)) imply that a.e. on . The last relation implies a.e. on since .

The theorem is proven.

It should be mentioned that the proposition given above and the analogous result for sine system show that the system , where is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space or , respectively, whenever and belongs to the corresponding Lebesgue space for all indices . This conclusion may give rise to the impression that if the function is such that and for all indices , where is any complete system in , then the system is also complete in the space . The arguments given below show that, in general, this is not true.

Theorem 5. *There is a complete system in space and a measurable function such that and for all indices but the system is not complete in .*

*Proof. *Define the function as follows: for all ; in other words, let be the system . Take . Then and is complete in (see, e.g., [9] or [10]). But, in this case, the system coincides with which, obviously, is not complete in . The theorem is proven.

Applying the Schmidt’s orthogonalization process to the system given in Theorem 5 we obtain immediately the validity of the following fact.

Theorem 6. *There is a complete orthonormal system in space and a measurable function such that and for all indices , but the system is not complete in .*

Using Lemmas 2 and 3 from the previous section, we obtain the following descriptions of the largest set of functions for which the system is complete in .

Theorem 7. *The system is complete in if and only if and .*

Theorem 8. *The system is complete in if and only if and .*

#### Competing Interests

The authors state that there are no competing interests related to this paper.

#### Acknowledgments

The author is grateful to Professor B. T. Bilalov for drawing his attention to this question and encouraging discussion. The author also thanks N. J. Guliyev for his useful assistance. This work was supported by the Science Development Foundation under the President of the Republic of Azerbaijan, Grant N EIF/GAM-3-2014-6(21)-24/05/1.

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#### Copyright

Copyright © 2017 Aydin Sh. Shukurov. 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.