# On the Frame Properties of Degenerate System of Sines

Authors: Bilal Bilalov; Fatima Guliyeva
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

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


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## Body

## 1. Introduction

Basis properties of classical system of exponents{eint}n∈ℤ (ℤ is the set of all integers) in Lebesgue spaces Lp(-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 L2 which is not Riesz basis. The first example of this was given by Babenko [6]. He proved that the degenerate system of exponents {|t|αeint}n∈ℤ with |α|<(1/2) forms a basis for L2(-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 {eint}n∈ℤ forms a basis for the weight space Lp,ρ(-π,π) with a norm ∥f∥p,ρ=(∫-ππ‍|f(t)|pρ(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].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.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; δnk is Kronecker symbol, δk={δkn}k∈ℕ.LetX 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{xn}n∈ℕ⊂X is said to be uniformly minimal in X if ∃δ>0(2.1)inf∀u∈L[{xn}n≠k]∥xk-u∥X≥δ∥xk∥X,∀k∈ℕ.System{xn}n∈ℕ⊂X is said to be complete in X if L[{xn}n∈ℕ]¯=X. It is called minimal in X if xk∉L[{xn}n≠k]¯, for all k∈ℕ.The following criteria of completeness and minimality are available.Criterion 1 (Hahn-Banach theorem).
System{xn}n∈ℕ⊂X is complete in X if f(xn)=0, for all n∈ℕ,f∈X*⇒f=0.Criterion 2 (see [19]).
System{xn}n∈ℕ⊂X is minimal in X⇔ it has a biorthogonal system {fn}n∈ℕ⊂X*,that is, fn(xk)=δnk, for all n,k∈ℕ.Criterion 3.
Complete system{xn}n∈ℕ⊂X is uniformly minimal in X⇔supn∥xn∥X∥yn∥X*<+∞, where {yn}n∈ℕ⊂X* is a system biorthogonal to it.
System{xn}n∈ℕ⊂X is said to be a basis for X if for all x∈X, ∃!{λn}n∈ℕ⊂K:x=∑n=1∞‍λnxn.
If system{xn}n∈ℕ⊂X forms a basis for X, then it is uniformly minimal.Definition 2.1 (see [20, 21]).
LetX be a Banach space and 𝒦 a Banach sequence space indexed by ℕ. Let {fk}k∈ℕ⊂X, {gk}k∈ℕ⊂X*. Then ({gk}k∈ℕ,{fk}k∈ℕ) is an atomic decomposition of X with respect to 𝒦 if(i)
{gk(f)}k∈ℕ∈𝒦, for all f∈X;(ii)
∃A,B>0:
(2.2)A∥f∥X≤∥{gk(f)}k∈ℕ∥𝒦≤B∥f∥X,∀f∈X;(iii)
f=∑k=1∞‍gk(f)fk, for all f∈X.Definition 2.2 (see [20, 21]).
LetX be a Banach space and 𝒦 a Banach sequence space indexed by   ℕ. Let {gk}k∈ℕ⊂X* and S:𝒦→X be a bounded operator. Then ({gk}k∈ℕ,S) is a Banach frame for X with respect to 𝒦 if(i)
{gk(f)}k∈ℕ∈𝒦, for allf∈X;(ii)
∃A,B>0:
(2.3)A∥f∥X≤∥{gk(f)}k∈ℕ∥𝒦≤B∥f∥X,∀f∈X;(iii)
S[{gk(f)}k∈ℕ]=f, for all f∈X.It is true the following.Proposition 2.3 (see [20, 21]).
LetX be a Banach space and 𝒦 a Banach sequence space indexed by   ℕ. Assume that the canonical unit vectors {δk}k∈ℕ constitute a basis for 𝒦 and let {gk}k∈ℕ⊂X* and S:𝒦→X be a bounded operator. Then the following statements are equivalent: (i)
({gk}k∈ℕ,S) is a Banach frame for X with respect to 𝒦.(ii)
({gk}k∈ℕ,{S(δk)}k∈ℕ) is an atomic decomposition of X with respect to 𝒦.More details about these facts can be found in [20–23].


## 3. Completeness and Minimality

We consider a system of sines(3.1){Snν}n∈ℕ≡{ν(t)sinnt}n∈ℕ,
with a degenerate coefficient ν(3.2)ν(t)=∏k=0r‍|t-tk|αk,
where 0=t0<t1<⋯<tr=π are points of degeneration and {αk}0r⊂R.The notationf~g,t→a, means that the inequality 0<δ≤|f(t)/g(t)|≤δ-1<+∞ holds in sufficiently small neighborhood of the point t=a with respect to the functions f and g. Thus, it is clear that sinnt~t,t→0 and sinnt~π-t,t→π for alln∈ℕ. Proceeding from these relations, we immediately obtain that the inclusion {Snν}n∈ℕ⊂Lp(0,π), 1≤p<+∞, is true if and only if the following relations hold
(3.3)α0;αr∈(-1p-1,+∞),{αk}1r-1⊂(-1p,+∞).
In what follows, we will always suppose that this condition is satisfied. Assume that the function f∈Lq(0,π)((1/p)+(1/q)=1) is otrhogonal to the system {Snν}n∈ℕ, that is
(3.4)∫0π‍ν(t)sinntf¯(t)dt=0,∀n∈ℕ,
where (·¯) is a complex conjugate. By C0[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∈L1(0,π)⊂C0*[0,π]. As the system of sines {sinnt}n∈ℕ is complete in C0[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 Lp(0,π).Now consider the minimality of system (3.3) in Lp(0,π). It is clear that {Snν-1}n∈ℕ⊂Lq(0,π) if and only if
(3.5)α0;αr∈(-∞,1q+1),{αk}1r-1⊂(-∞,1q).
It is easily seen that the system {(2/π)Snν-1}n∈ℕ is biorthogonal to {Snν}n∈ℕ. So the following theorem is true.Theorem 3.1.
System{Snν}n∈ℕ is complete in Lp(0,π), 1≤p<+∞, if the relations (3.3) hold. Besides, it is minimal in Lp(0,π) if both (3.3) and (3.5) hold. Consequently, system {Snν}n∈ℕ is complete and minimal in Lp(0,π) if the following relations hold:
(3.6)α0;αr∈(-1p-1,1q+1),{αk}1r-1⊂(-1p,1q).It is known that (see, e.g., [10, 11]) if {αk}0r⊂(-(1/p),(1/q)), then system {Sn(ν)}n∈ℕ  forms a basis for Lp(0,π), 1<p<+∞. Let β∈[(1/q),(1/q)+1), where either β=α0 or β=αr. In the sequel, we will suppose that the condition (3.6) is satisfied for {αk}1r-1. We have
(3.7)∥Snν∥pp=∫0π‍|ν(t)|p|sinnt|pdt≤∫0π‍|ν(t)|pdt<+∞,∀n∈ℕ.
On the other hand
(3.8)∥Snν∥pp≥∫0δ‍|ν(t)|p|sinnt|pdt≥c∫0δ‍tα0p|sinnt|pdt=cn1+α0p∫0nδ‍tα0p|sint|pdt,
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 {αk}1r-1. Let us show that infn∈ℕ∥Snν∥p>0. We have (α=α0p+1)(3.9)∥Snν∥pp≥cnα∑k∈Mn‍∫kπ+(π/4)(k+1)π-(π/4)‍tα-1|sint|pdt≥cnα∑k∈Mn‍∫kπ+(π/4)(k+1)π-(π/4)‍tα-1dt,
where Mn≡{k≥0:(k+1)π-(π/4)≤nδ}. Thus
(3.10)∥Snν∥pp≥cnα∑k∈Mn‍∫4k+14k+3‍tα-1dt.
It is absolutely clear that
(3.11)∫4k-14k+1‍tα-1dt≤∫4k+14k+3‍tα-1dt,∀k∈Mn.
Taking into account this relation, we obtain
(3.12)∥Snν∥pp≥cnα∑k∈Mn‍[∫4k+14k+3‍tα-1dt+∫4k-14k+1‍tα-1dt]≥cnα∫0λn‍tα-1dt=c(λnn)α,
where λn≥(nδ/π)-2. Consequently
(3.13)∥Snν∥pp≥c(δπ-2n)α→c(δπ)α>0,n→∞.
It follows immediately that infn∥Snν∥p>0.Regarding biorthogonal system we get(3.14)∥Sn-ν∥qq=∫0π‍ν-q(t)|sinnt|qdt.
Choose ε>0 as small as the interval [0,ε] does not contain the points {αk}1r-1. Consequently
(3.15)∥Sn-ν∥qq=m∫0ε‍t-α0q|sinnt|qdt,
where m>0 is some constant. We have
(3.16)∥Sn-ν∥qq≥mnα0q-1∫0nε‍|sint|qtα0qdt.
First we consider the case α0∈((1/q),(1/q)+1). In this case, for sufficiently great n, we have
(3.17)∥Sn-ν∥qq≥mnα0q-1∫01‍|sint|qtα0qdt→∞,n→∞.
Let α0=(1/q). Consequently
(3.18)∥Sn-ν∥qq≥m∫0nε‍|sint|qtdt,
and, as a result
(3.19)supn∥Sn-ν∥qq≥m∫1+∞‍|sint|qtdt≥m∑k=1∞‍∫kπ+(π/4)(k+1)π-(π/4)‍|sint|qtdt≥mc1∑k=1∞‍1(k+1)π-(π/4)∫kπ+(π/4)(k+1)π-(π/4)‍1dt=c2∑k=1∞‍1(k+1)π-(π/4)=+∞,
where ci are some constants. So we obtain that for β∈[(1/q),(1/q)+1), supn∥Sn-ν∥q=+∞. Consequently, in this case we have
(3.20)supn∥Snν∥p∥Sn-ν∥q=+∞.
Then it is known that (see, e.g., [22]) the system {Sn(ν)}n∈ℕ is not uniformly minimal and, besides, does not form a basis for Lp.Consider the caseβ∈(-(1/p)-1,-(1/p)]. Without limiting the generality, we will suppose that β=α0. In this case, with regard to the biorthogonal system we have
(3.21)∥Sn-ν∥qq=∫0π‍ν-q(t)|sinnt|qdt≤∫0π‍ν-q(t)dt<+∞,∀n∈ℕ.
Taking sufficiently small δ>0, we obtain
(3.22)∥Sn-ν∥qq≥∫0δ‍ν-q(t)|sinnt|qdt≥c∫0δ‍t-α0q|sinnt|qdt.
As α0q<0, then, in the absolutely same way as in the previous case, we get
(3.23)infn∥Sn-ν∥q>0.
On the other hand,
(3.24)∥Snν∥pp=∫0π‍νp(t)|sinnt|pdt≥c∫0δ‍|sinnt|pt-α0pdt,
where α0p≤1. Similarly to the previous case again, we get
(3.25)∫0δ‍tα0p|sinnt|pdt→+∞,n→∞.
As a result we obtain
(3.26)supn∥Snν∥p∥Sn-ν∥q=+∞,forβ∈(-1p-1,-1p].
Thus, the following theorem is true.Theorem 3.2.
Let{α0;αr}⊂(-(1/p)-1,-(1/p)+2); {α0;αr}⋂‍Mp(0)≠∅, and Ar⊂(-(1/p),(1/q)), where Mp(0)≡(-(1/p)-1,-(1/p)]⋃‍[-(1/p)+1,-(1/p)+2), Ar≡{αk}1r-1. Then the system {Snν}n∈ℕ is complete and minimal in Lp(0,π),1≤p<+∞, but does not form a basis for it.


## 4. Defective Case

Here, we consider the defective system of sines{Snν}n∈ℕ(k0), where ℕ(k0)≡ℕ∖{k0}, k0∈ℕ is some number. It follows directly from Theorem 3.2 that if the condition
(4.1){α0;αr}⋂‍Mp(0)≠∅;Ar⊂(-1p,1q),
holds, then the system {Snν}n∈ℕ(k0) is minimal but not complete in Lp(0,π),1≤p<+∞. Assume Mp(1)≡[(1/q)+1,(1/q)+3). Let α0∈Mp(1). Consider the completeness of system {Snν}n∈ℕ(k0) in Lp(0,π). Suppose that f∈Lq(0,π) is orthogonal to the system, that is,
(4.2)∫0π‍ν(t)sinntf(t)¯dt=0,∀n∈ℕ(k0).
As νf∈L1⊂C0*[0,π] and system {sinnt}n∈ℕ is complete and minimal in C0[0,π], from (4.2) we get
(4.3)ν(t)f(t)=csink0t⇒f(t)=csink0tν(t).
It is clear that ν-1(t)~t-α0,sink0t~t as t→0. Consequently, f~t-α0+1,t→0. As q(-α0+1)≤-1, then f∈Lq(0,π) if and only if c=0, and, consequently, f=0. The similar result is true for αr∈Mp(1). Thus, if α0;αr>-(1/p)-1, and max{α0;αr}∈Mp(1), then the system {Snν}n∈ℕ(k0) is complete in Lp(0,π). Now we consider the minimality of this system. Let
(4.4)ϑn(t)=ν-1(t)[sinnxn-sink0xk0],n∈ℕ(k0).
We have
(4.5)〈Skν,ϑn〉=∫0π‍Skν(t)ϑn(t)¯dt=π2nδkn,∀k,n∈ℕ(k0).
Let us show that {ϑn}n∈ℕ(k0)⊂Lq(0,π). In fact
(4.6)sinntn=t-n2t36+o(t3),t→0,sink0tk0=t-k02t36+o(t3),t→0,
and, consequently
(4.7)sinntn-sink0tk0=16(k02-n2)t3+o(t3),t→0.
From these relations, we immediately find that ϑn(t)~t3-α0,t→0. As a result, {ϑn}n∈ℕ(k0)⊂Lq(0,π). Then the relations (4.5) imply the minimality of system {Snν}n∈ℕ(k0) in Lp(0,π). Similar result is true for αr∈Mp(1). In the end, we obtain that if max{α0;αr}∈Mp(1), then the system {Snν}n∈ℕ has a defect equal to 1.Consider the case whenmax{α0;αr}=α0∈Mp(2), where Mp(2)≡[(1/q)+3,(1/q)+5). We look at the system {Snν}n∈ℕ(k1;k2), where ℕ(k1;k2)≡ℕ∖{k1;k2},k1;k2∈ℕ,k1≠k2, are some numbers. Let f∈Lq(0,π) cancel this system out, that is,
(4.8)〈Snν,f〉=0,∀n∈ℕ(k1;k2).
Using the previous reasoning, we find that for some constants c1,c2, the following is true:
(4.9)f(t)=c1ν-1(t)sink1t+c2ν-1(t)sink2t.
Using representations
(4.10)sinkit=kit-ki36t3+o(t4),t→0,i=1,2,
we obtain
(4.11)f(t)~ct-α0g1(t)+g2(t),t→0,
(c≠0 is some constant), where it can be easily seen that g2∈Lq and
(4.12)g1(t)=(c1k1+c2k2)t-16(c1k13+c2k23)t3.
Thus, f∈Lq if and only if t-α0g1(t)∈Lq. Assume b1=c1k1+c2k2; b2=-(1/6)(c1k13+c2k23). As (1-α0)q≤-1(3-α0)q≤-1, it is clear that t1-α0∉Lq and t3-α0∉Lq. Suppose b1≠0. We have
(4.13)|t-α0g1(t)|=|b1|t1-α0|1+b1-1b2t2|.
It follows directly that for sufficiently small δ>0, we have
(4.14)|t-α0g1(t)|≥cδ|b1|t1-α0,∀t∈(0,δ),
where cδ>0 is some constant depending only on δ and b2. As a result, f∉Lq.Consequently,b1=0. Moreover, it is not difficult to derive that b2=0. Thus, we obtain the following system for c1 and c2:
(4.15)c1k1+c2k2=0,c1k13+c2k23=0.
It is clear that det|k1k2k13k23|≠0. And, consequently, c1=c2=0. As a result, f=0, which, in turn, implies that the system {Snν}n∈ℕ(k1;k2) is complete in Lp. Let us show that it is also minimal in Lp. Assume γn(k)=(1/6)(k2-n2). Consider the system ϑn(k1;k2)(t)≡[an(k1;k2)sinnt-(1/k1γn(k1))sink1t + (1/k2γn(k2))sink2t]ν-1(t), for all n∈ℕ(k1;k2), where
(4.16)an(k1;k2)=1n(1γn(k1)-1γn(k2))≠0.
Simple calculations give the following representation:
(4.17)ϑn(k1;k2)(t)~cnt5,t→0,
where cn≠0, for all n∈ℕ(k1;k2) are some constants. We obtain directly from this representation that {ϑn(k1;k2)}n∈ℕ(k1;k2)⊂Lq. On the other hand,
(4.18)〈Snν,ϑn(k1;k2)〉=an(k1;k2)δnk={≠0,n≠k,=0,n=k,∀n,k∈ℕ(k1;k2).
Thus, if α0∈Mp(2), then the system {Snν}n∈ℕ(k1;k2) is complete and minimal in Lp, and, as a result, the system {Snν}n∈ℕ has a defect equal to 2. It is easy to see that the similar result is true if αr∈Mp(2) with α0≤αr. Continuing this way, we obtain that if β=max{α0;αr}=α0∈Mp(k), where Mp(k)≡[-(1/p)+2k,-(1/p)+2(k+1)), then the system {Snν}n∈ℕ{n-k}is complete and minimal in Lp, where {n-k}={n1;…;nk}⊂ℕ, ni≠nj with i≠j.Consider the basicity of system{Snν}n∈ℕ{n-1} (i.e., the case of k=1) in Lp. Similar to the case of Mp(0), it can be proved that
(4.19)0<infn∈ℕ{n-1}∥Snν∥p≤supn∈ℕ{n-1}∥Snν∥p<+∞.
Concerning biorthogonal system, we have
(4.20)∫0π‍ν-q(x)|sinnxn-sink1xk1|qdx≥c∫0ε‍x-qα0|sinnxn-sinn1xn1|qdx,
where the interval [0,ε] (ε>0) does not contain the points {tk}1r-1. As (sinn1x/n1)-x~x3, for x→0, it is clear that (-qα0+3>-1)(4.21)∫0ε‍x-qα0|sinn1xn1-x|qdx<+∞.
Taking this circumstance into account, we have
(4.22)∥Sn-ν∥qq≥c[∫0ε‍x-qα0|sinnxn-x|qdx-∫0ε‍x-qα0|sinn1xn1-x|qdx].
Consider the case of α0∈((1/q)+1,(1/q)+3):
(4.23)∫0ε‍x-qα0|sinnxn-x|qdx=n-q∫0ε‍x-qα0|sinnx-nx|qdx=n-α∫0nε‍t-qα0|sinnt-t|qdt≥cn-α∫0nε‍t-qα0+qdt=cα(εα-n-α),
where α=-α0q+q+1<0. Consequently, supn∥Sn-ν∥q=+∞. Let α0=(1/q)+1. In this case we have
(4.24)∫0ε‍x-qα0|sinnxn-x|qdx≥c∫1nε‍t-1dt=clnnε,
and, consequently supn∥Sn-ν∥q=+∞. As a result, we get that for α0∈Mp(1), the system {Snν}n∈ℕ{n-1} does not form a basis for Lp. Assume that in this case, the system {Snν}n∈ℕ is a frame in Lp, that is any function from Lp can be expanded with respect to this system. As it does not form a basis for Lp, zero has a non trivial decomposition, that is,
(4.25)0=∑n=1∞‍anSnν,
where ∃n0∈ℕ:an0≠0. As the system {Snν}n∈ℕ{n-1} is complete and minimal in Lp, it is clear that an1≠0. Consequently, Sn1ν=∑n≠n1‍(an/an1)Snν. It follows directly that the arbitrary element can be expanded with respect to the system {Snν}n∈ℕ{n-1}. But this is impossible. Similar result is true for max{α0;αr}∈Mp(1).Proceeding in an absolutely similar way as we did in the previous case, we can prove that formax{α0;αr}∈Mp(k), the system {Snν}n∈ℕ{n-k}is complete and minimal in Lp, but does not form a basis for it. Consequently, system {Snν}n∈ℕ has a defect equal to (k). The fact that it is not a frame in Lp in this case too is proved as follows. Let k=2:{n-k}≡{n1;n2}. Assume that the system {Sn}n∈ℕ is a frame in Lp. Then zero has a non trivial decomposition: 0=∑n=1∞‍anSnν. It is clear that |an1|+|an2|>0, and let an1≠0. It follows directly that the system {Sn}n≠n1 is a frame in Lp. 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)α0;αr∈(-1p-1,+∞),{αk}1r-1⊂(-1p;1q).
Then the system{Snν}n∈ℕ is a frame (basis) in Lp if and only if α0;αr∈(-(1/p),(1/q)). Moreover, for max{α0;αr}∈Mp(k), k∈ℕ, it has a defect equal to (k), where Mp(k)≡[(1/q)+2k,(1/q)+2(k+1)).



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