On the Frame Properties of Degenerate System of Sines

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


Introduction
Basis properties of classical system of exponents {e int } n∈Z Z is the set of all integers in Lebesgue spaces L p −π, π , 1 ≤ p < ∞, are well studied in the literature see 1-4 .Bari in her fundamental work 5 raised the issue of the existence of normalized basis in L 2 which is not Riesz basis.The first example of this was given by Babenko 6 .He proved that the degenerate system of exponents {|t| α e int } n∈Z with |α| < 1/2 forms a basis for L 2 −π, π but is not Riesz basis when α / 0. This result has been extended by Gaposhkin 7 .In 8 , the condition on the weight ρ was found which make the system {e int } n∈Z forms a basis for the weight space L p,ρ −π, π with a norm f p,ρ π −π |f t | p ρ t dt 1/p .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 10-12 .Those of the systems of exponents with degenerate coefficients have been studied in 13, 14 .Similar questions have previously been considered in papers 15-18 .

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.N 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"; K ≡ R or K ≡ C will stand for the set of real or complex numbers, respectively; Let X be some Banach space with a norm • X .Then X * will denote its dual with a norm • X * .By L M , we denote the linear span of the set M ⊂ X, and M will stand for the closure of M.
System {x n } n∈N ⊂ X is said to be uniformly minimal in The following criteria of completeness and minimality are available.
If system {x n } n∈N ⊂ X forms a basis for X, then it is uniformly minimal.
Definition 2.1 see 20, 21 .Let X be a Banach space and K a Banach sequence space indexed by N.
ii ∃A, B > 0: Definition 2.2 see 20, 21 .Let X be a Banach space and K a Banach sequence space indexed by N. Let {g k } k∈N ⊂ X * and S : K → X be a bounded operator.Then {g k } k∈N , S is a Banach frame for X with respect to ii ∃A, B > 0: It is true the following.ii {g k } k∈N , {S δ k } k∈N is an atomic decomposition of X with respect to K.
More details about these facts can be found in 20-23 .

Completeness and Minimality
We consider a system of sines sufficiently small neighborhood of the point t a with respect to the functions f and g.Thus, it is clear that sin nt ∼ t, t → 0 and sin nt ∼ π − t, t → π for all n ∈ N. Proceeding from these relations, we immediately obtain that the inclusion {S ν n } n∈N ⊂ L p 0, π , 1 ≤ p < ∞, 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 f ∈ L q 0, π 1/p 1/q 1 is otrhogonal to the system {S ν n } n∈N , that is where • is a complex conjugate.By C 0 0, π , we denote the Banach space of functions which are continuous on 0, π with a sup-norm and vanish at the ends of the interval 0, π .It is absolutely clear that νf ∈ L 1 0, π ⊂ C * 0 0, π .As the system of sines {sin nt} n∈N is complete in C 0 0, π , we obtain from the relations 3.4 that ν t f t 0 a.e. on 0, π , and, consequently, f t 0 a.e. on 0, π .This proves the completeness of system 3.3 in L p 0, π .Now consider the minimality of system 3.
It is easily seen that the system 3 and 3.5 hold.Consequently, system {S ν n } n∈N is complete and minimal in L p 0, π if the following relations hold: It is known that see, e.g., 10, 11 if {α k } r 0 ⊂ − 1/p , 1/q , then system {S n ν } n∈N forms a basis for L p 0, π , 1 < p < ∞.Let β ∈ 1/q , 1/q 1 , where either β α 0 or β α r .In the sequel, we will suppose that the condition 3.6 is satisfied for {α k } r−1 1 .We have On the other hand where c > 0 is some constant in what follows c will denote constants that may be different from each other , δ > 0 is such that 0, δ does not contain the points {α k } r−1 1 .Let us show that inf n∈N S ν n p > 0. We have α α 0 p 1 where Taking into account this relation, we obtain where

3.13
It follows immediately that inf n S ν n p > 0. Regarding biorthogonal system we get S −ν n q q π 0 ν −q t |sin nt| q dt.

3.14
Choose ε > 0 as small as the interval 0, ε does not contain the points {α k } r−1 1 .Consequently where m > 0 is some constant.We have |sin t| q t α 0 q dt.

3.16
First we consider the case α 0 ∈ 1/q , 1/q 1 .In this case, for sufficiently great n, we have and, as a result

3.19
where c i are some constants.So we obtain that for β ∈ 1/q , 1/q 1 , sup n S −ν n q ∞.Consequently, in this case we have Then it is known that see, e.g., 22 the system {S n ν } n∈N is not uniformly minimal and, besides, does not form a basis for L p .Consider the case β ∈ − 1/p − 1, − 1/p .Without limiting the generality, we will suppose that β α 0 .In this case, with regard to the biorthogonal system we have

3.22
As α 0 q < 0, then, in the absolutely same way as in the previous case, we get inf n S −ν n q > 0.

3.23
On the other hand,

3.25
As a result we obtain

3.26
Thus, the following theorem is true.

Defective Case
Here, we consider the defective system of sines {S ν n } n∈N k 0 , where N k 0 ≡ N \ {k 0 }, k 0 ∈ N is some number.It follows directly from Theorem 3.2 that if the condition Consider the completeness of system {S ν n } n∈N k 0 in L p 0, π .Suppose that f ∈ L q 0, π is orthogonal to the system, that is, As νf ∈ L 1 ⊂ C * 0 0, π and system {sin nt} n∈N is complete and minimal in C 0 0, π , from 4.2 we get It is clear that ν −1 t ∼ t −α 0 , sin k 0 t ∼ t as t → 0. Consequently, f ∼ t −α 0 1 , t → 0. As q −α 0 1 ≤ −1, then f ∈ L q 0, π if and only if c 0, and, consequently, f 0. The similar result is true for α r ∈ M 1 p .Thus, if α 0 ; α r > − 1/p − 1, and max{α 0 ; α r } ∈ M 1 p , then the system {S ν n } n∈N k 0 is complete in L p 0, π .Now we consider the minimality of this system.Let We have Let us show that {ϑ n } n∈N k 0 ⊂ L q 0, π .In fact and, consequently From these relations, we immediately find that ϑ n t ∼ t 3−α 0 , t → 0. As a result, {ϑ n } n∈N k 0 ⊂ L q 0, π .Then the relations 4.5 imply the minimality of system {S ν n } n∈N k 0 in L p 0, π .
Similar result is true for α r ∈ M 1 p .In the end, we obtain that if max{α 0 ; α r } ∈ M 1 p , then the system {S ν n } n∈N has a defect equal to 1. Consider the case when max{α 0 ; α r } α 0 ∈ M 2 p , where M 2 p ≡ 1/q 3, 1/q 5 .We look at the system Using the previous reasoning, we find that for some constants c 1 , c 2 , the following is true: Using representations we obtain c / 0 is some constant , where it can be easily seen that g 2 ∈ L q and ∈ L q and t 3−α 0 / ∈ L q .Suppose b 1 / 0. We have

4.13
Journal of Function Spaces and Applications 9 It follows directly that for sufficiently small δ > 0, we have where c δ > 0 is some constant depending only on δ and b 2 .As a result, f / ∈ L q .Consequently, b 1 0.Moreover, it is not difficult to derive that b 2 0. Thus, we obtain the following system for c 1 and c 2 :

4.15
It is clear that det where the interval 0, ε ε > 0 does not contain the points Taking this circumstance into account, we have Consider the case of α 0 ∈ 1/q 1, 1/q 3 : where α −α 0 q q 1 < 0. Consequently, sup n S −ν n q ∞.Let α 0 1/q 1.In this case we have and, consequently sup n S −ν n q ∞.As a result, we get that for α 0 ∈ M 1 p , the system {S ν n } n∈N {n 1 } does not form a basis for L p .Assume that in this case, the system {S ν n } n∈N is a frame in L p , that is any function from L p can be expanded with respect to this system.As it does not form a basis for L p , zero has a non trivial decomposition, that is, where ∃n 0 ∈ N : a n 0 / 0. As the system {S ν n } n∈N {n 1 } is complete and minimal in L p , it is clear that a n 1 / 0. Consequently, S ν n 1 n / n 1 a n /a n 1 S ν n .It follows directly that the arbitrary element can be expanded with respect to the system {S ν n } n∈N {n 1 } .But this is impossible.Similar result is true for max{α 0 ; α r } ∈ M 1 p .
Proceeding in an absolutely similar way as we did in the previous case, we can prove that for max{α 0 ; α r } ∈ M k p , the system {S ν n } n∈N {n k } is complete and minimal in L p , but does not form a basis for it.Consequently, system {S ν n } n∈N has a defect equal to k .The fact that it is not a frame in L p in this case too is proved as follows.Let k 2 : {n k } ≡ {n 1 ; n 2 }.Assume that the system {S n } n∈N is a frame in L p .Then zero has a non trivial decomposition: 0 ∞ n 1 a n S ν n .It is clear that |a n 1 | |a n 2 | > 0, and let a n 1 / 0. It follows directly that the system {S n } n / n 1 is a frame in L p .The further reasoning is absolutely similar to the case of k 1.This scheme is applicable for for all k ∈ N. Thus, we have proved the following main theorem.Theorem 4.1.Let the following necessary condition be satisfied 4.26 Then the system {S ν n } n∈N is a frame (basis) in L p if and only if α 0 ; α r ∈ − 1/p , 1/q .Moreover, for max{α 0 ; α r } ∈ M k p , k ∈ N, it has a defect equal to k , where M k p ≡ 1/q 2k, 1/q 2 k 1 .

Proposition 2.3 see
20, 21 .Let X be a Banach space and K a Banach sequence space indexed by N. Assume that the canonical unit vectors {δ k } k∈N constitute a basis for K and let {g k } k∈N ⊂ X * and S : K → X be a bounded operator.Then the following statements are equivalent: i {g k } k∈N , S is a Banach frame for X with respect to K.
} n∈N has a defect equal to 2. It is easy to see that the similar result is true if α r ∈ M 2 p with α 0 ≤ α r .Continuing this way, we obtain that if β max{α 0 ; α r } α 0 ∈ M } n∈N {n k } is complete and minimal in L p , where {n k } {n 1 ; . ..; n k } ⊂ N, n i / n j with i / j.Consider the basicity of system {S ν n } n∈N {n 1 } i.e., the case of k 1 in L p .Similar to the case of M π 0