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

Authors: Aydin Sh. Shukurov
Journal: Journal of Function Spaces
 (2017)

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## 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{ω(t)φn(t)}, where {φn(t)} is an exponential or trigonometric (cosine or sine) systems, becomes complete in the corresponding Lebesgue space Lp(-π,π) or Lp(0,π), respectively, whenever {ω(t)φn(t)} belongs to the corresponding Lebesgue space for all indices n (under the evident natural condition mes⁡{t:ω(t)=0}=0). It is also shown that the same conclusion does not remain valid for, in general, any complete or complete orthonormal system {φn(t)}. Besides it, the largest class of functions ω(t) for which the system {ωtsin⁡nt}n∈N is complete in Lp(0,π) space is determined.


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## Body

## 1. Introduction

The basis properties (completeness, minimality, and Schauder basicity) of systems of the form{ω(t)φn(t)}, where {φn(t)} 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 {|t|α·eint}n∈Z, where |α|<1/2 and α≠0, answering in the affirmative a question of Bari [16] on the existence of normalized basis for L2(-π,π) that is not a Riesz basis [2]. It is also known that, in general, the system {|t|α·eint}n∈Z is a basis in Lp(-π,π),1<p<∞ space if and only if -1/p<α<1/q, where 1/p+1/q=1 (it is, e.g., a direct consequence of the fact that the weight |t|α satisfies the Muckenhoupt condition if and only if -1<α<p-1 (see, e.g., [6])).Later, basicity properties (completeness, minimality, and Schauder basicity) of systems of more general form{∏j=1rt-tjαjeint}n∈Z and {∏j=1rt-tjαjcos⁡nt}n∈Z+ and {∏j=1rt-tjαjsin⁡nt}n∈N, where r≥1, 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 functionsω(t) for which the system {ω(t)φn(t)}, where {φn(t)} is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space Lp(-π,π) or Lp(0,π), respectively.It is easy to see that the relationωtcos⁡nt∈Lp(0,π),n∈N or ω(t)eint∈Lp(-π,π),n∈N is possible if and only if ω(t)∈Lp(0,π) or ω(t)∈Lp(-π,π), respectively. This observation shows the validity of the following fact.Proposition 1 (see [17]).
Letω(t) be any measurable function on (0,π)(on (-π,π)) such that the relations ωtcos⁡nt∈Lp(0,π),n∈N(ω(t)eint∈Lp(-π,π),n∈Z) and mes{t:ω(t)=0}=0 hold. Then the system {ωtcos⁡nt}n∈N({ω(t)eint}) is complete in the space Lp(0,π)(Lp(-π,π)).Indeed, if a functionf(t)∈Lq(0,π)(f(t)∈Lq(-π,π)) is orthogonal to the system {ωtcos⁡nt}n∈N({ω(t)eint}) then ∫0πftωtcos⁡ntdt=0 for all n∈N (∫-ππf(t)ω(t)eintdt=0 for all n∈Z). By the above observation, we have f(t)ω(t)∈L1(0,π) (f(t)ω(t)∈L1(-π,π)). 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 f(t)ω(t)=0 a.e. Using this equality and the condition mes{t:ω(t)=0}=0, we arrive at the equality f(t)=0a.e., which shows the completeness of the mentioned system.This proposition shows that the largest set of functionsω(t) for which the system {ω(t)eint}n∈Z (or {ωtcos⁡nt}n∈N) is complete in the space Lp(-π,π) (or Lp(0,π)) is the set of all functions ω(t) from Lp(-π,π) (or Lp(0,π)), for which mes{t:ω(t)=0}=0.Note that the observation similar to one that is given before the proposition is not valid for the sine system; more precisely,ωtsin⁡nt∈Lp(0,π),n∈N does not always imply ω(t)∈Lp(0,π). Therefore, the scheme used to prove the above proposition does not work in the case of the system {ωtsin⁡nt}n∈N. 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{ω(t)φn(t)}, where {φn(t)} is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space Lp(-π,π) or Lp(0,π), respectively, whenever mes{t:ω(t)=0}=0 and {ω(t)φn(t)} belongs to the corresponding Lebesgue space for all indices n. This conclusion may give rise to the impression that if the function ω(t) is such that mes{t:ω(t)=0}=0 and ω(t)φn(t)∈Lp(a,b) for all indices n, where {φn(t)} is any complete system in Lp(a,b), then the system {ω(t)φn(t)} is also complete in the space Lp(a,b). The arguments given below show that, in general, this is not true.


## 2. Auxiliary Facts

Lemma 2.
Letω(t) be any measurable function on (0,π). The relation ωtsin⁡nt∈Lp(0,π),∀n∈N is possible if and only if t(t-π)ω(t)∈Lp(0,π).Proof.
Necessity. Let the relation ωtsin⁡nt∈Lp(0,π),n∈N hold. Write the function ωtsin⁡t in the following form: (1)ωtsin⁡t=tt-πωt·sin⁡ttt-π.
Consider an auxiliary function(2)Φt=sin⁡ttt-π,ift∈0,π;-1π,ift=0,π.
It is evident that the functionΦ(t) is continuous and never vanishes at the segment [0,π]. Therefore, there is a positive number m such that Φt>m for all t∈[0,π]. Using these inequalities in (1), we obtain the following estimation: (3)ωtsin⁡t≥m·tt-πωt,∀t∈0,π.
This estimation impliest(t-π)ω(t)∈Lp(0,π) since ωtsin⁡t∈Lp(0,π) by the condition of the Lemma. The necessity part of the lemma is proven.
Sufficiency. Assume that t(t-π)ω(t)∈Lp(0,π). Take an arbitrary natural number n. Write the function ωtsin⁡nt in the following form: (4)ωtsin⁡nt=tt-πωt·sin⁡nttt-π.
Consider an auxiliary function(5)Φt=sin⁡nttt-π,ift∈0,π;-nπ,ift=0;n·-1nπ,ift=π.
It is evident that the functionΦ(t) is continuous on [0,π]. Therefore, there is a number M such that Φt<M for all t∈[0,π]. Using these inequalities in (4), we obtain the following estimation: (6)ωtsin⁡nt≤M·tt-πωt,∀t∈0,π.
Sincet(t-π)ω(t)∈Lp(0,π), the last estimation implies ωtsin⁡nt∈Lp(0,π). The lemma is proven.Taking into account that the proof of the necessity part of this lemma relies only on the relationωtsin⁡t∈Lp(0,π) and using the sufficiency part of the same lemma we obtain the validity of the following.Lemma 3.
Letω(t) be any measurable function on (0,π). The relation ωtsin⁡nt∈Lp(0,π),∀n∈N is possible if and only if ωtsin⁡t∈Lp(0,π).


## 3. Main Results

The main aim of this note is to prove the following.Theorem 4.
Letω(t) be any measurable function on (0,π) such that(7)1mes⁡t:ωt=0=0,2ωtsin⁡nt∈Lp0,π,∀n∈N.
Then the system{ωtsin⁡nt}n∈N is complete in Lp(0,π) space.Proof.
Let the functionf(t)∈Lq(0,π) be orthogonal to the system {ωtsin⁡nt}n∈N: (8)∫0πftωtsin⁡ntdt=0,∀n∈N.Then the equalities (9)∫0πftωtsin⁡n+1tdt=0,(10)∫0πftωtsin⁡n-1tdt=0hold for all n=0,1,2,….. Subtracting (10) from (9) we obtain (11)∫0πftωtsin⁡tcos⁡ntdt=0,for all n=0,1,2,….. Note that ωtsin⁡t∈Lp(0,π) (by the condition of the theorem). Therefore ftωtsin⁡t∈L1(0,π) and the fact that the Fourier coefficients of a summable function with respect to the cosine system is unique (along with (11)) imply that ftωtsin⁡t=0 a.e. on (0,π). The last relation implies f(t)=0 a.e. on (0,π) since mes{t:ω(t)=0}=0.
The theorem is proven.It should be mentioned that the proposition given above and the analogous result for sine system show that the system{ω(t)φn(t)}, where {φn(t)} is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space Lp(-π,π) or Lp(0,π), respectively, whenever mes{t:ω(t)=0}=0 and {ω(t)φn(t)} belongs to the corresponding Lebesgue space for all indices n. This conclusion may give rise to the impression that if the function ω(t) is such that mes{t:ω(t)=0}=0 and ω(t)φn(t)∈Lp(a,b) for all indices n, where {φn(t)} is any complete system in Lp(a,b), then the system {ω(t)φn(t)} is also complete in the space Lp(a,b). The arguments given below show that, in general, this is not true.Theorem 5.
There is a complete system{φn(t)} in Lp(-π,π) space and a measurable function ω(t) such that mes{t:ω(t)=0}=0 and ω(t)φn(t)∈Lp(a,b) for all indices n but the system {ω(t)φn(t)} is not complete in Lp(-π,π).Proof.
Define the functionφn(t) as follows: (12)φnt=teintfor all n∈Z/{0}; in other words, let {φn(t)} be the system {teint}n∈Z/{0}. Take ω(t)≡1/t. Then mes{t:ω(t)=0}=0 and {φn(t)} is complete in L2(-π,π) (see, e.g., [9] or [10]). But, in this case, the system {ω(t)φn(t)} coincides with {eint}n∈Z/{0} which, obviously, is not complete in L2(-π,π). The theorem is proven.Applying the Schmidt’s orthogonalization process to the system given in Theorem5 we obtain immediately the validity of the following fact.Theorem 6.
There is a complete orthonormal system{φn(t)} in Lp(-π,π) space and a measurable function ω(t) such that mes{t:ω(t)=0}=0 and ω(t)φn(t)∈Lp(-π,π) for all indices n, but the system {ω(t)φn(t)} is not complete in Lp(-π,π).Using Lemmas2 and 3 from the previous section, we obtain the following descriptions of the largest set of functions ω(t) for which the system {ωtsin⁡nt}n∈N is complete in Lp(0,π).Theorem 7.
The system{ωtsin⁡nt}n∈N is complete in Lp(0,π) if and only if t(t-π)ω(t)∈Lp(0,π) and mes{t:ω(t)=0}=0.Theorem 8.
The system{ωtsin⁡nt}n∈N is complete in Lp(0,π) if and only if ωtsin⁡t∈Lp(0,π) and mes{t:ω(t)=0}=0.



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