Mathematical Problems in Engineering

Vol. 23, No. 3 (2000) 175–180 S0161171200001964

© Hindawi Publishing Corp.

LACUNARY STATISTICAL CONVERGENCE AND INCLUSION PROPERTIES BETWEEN LACUNARY METHODS

JINLU LI (Received 6 August 1998)

Abstract.A lacunary sequence is an increasing integer sequenceθ= {kr}such thatkr− k_r−1→ ∞asr → ∞. A sequence xis called s_θ-convergent toLprovided that for each ε >0, limr(1/(kr−kr−1)){the number ofkr−1< k≤kr:|x_k−L| ≥ε} =0.In this paper, we study the general description of inclusion between two arbitrary lacunary sequences convergent.

Keywords and phrases. Lacunary sequence, statistical convergence.

2000 Mathematics Subject Classification. Primary 40A05, 40C05, 40D05; Secondary 11B05.

1. Introduction. An increasing integer sequenceθ= {kr}is called a lacunary se- quence if it satisfies thatk0=0 andhr:=kr−kr−1→ ∞asr→ ∞. Throughout this paper, we follow the notation used in [7]. A sequencexis calledSθ-convergent toLpro- vided that for eachε >0. The intervals determined byθis denoted byIr:=(kr−1,kr] and the ratiokr/kr−1is abbreviated byqr. For a finite setE,|E|indicate the number of elements inE.

Let θ be a lacunary sequence, a complex number sequence x is said to be Sθ- convergent toLprovided that for eachε >0.

limr

1 hr

the number ofkr−1< k≤kr:|xk−L| ≥ε

=0. (1.1)

In this case we writeSθ−limx=Lorxk→L(Sθ), and we define Sθ=

x:Sθ−limx=L,for someL

. (1.2)

In [4] it is defined that the lacunary sequenceβ= {lr}is called a lacunary refinement of the lacunary sequenceθ= {kr}if{kr} ⊆ {lr}. In [7], after the inclusion relationship betweenSθ and Sβ is studied for a special case ifβ= {lr}is a lacunary refinement of the lacunary sequenceθ= {kr}, an open problem is raised: what is the general description of the inclusion between two arbitrary lacunary methods. In this paper, we study the inclusion properties of different lacunary methods and solve this open problem.

2. Inclusion properties. If β= {lr}is a lacunary refinement of the lacunary se- quenceθ= {kr}, it is shown by [7, Thm. 7] thatxk→L(Sβ)impliesxk→L(Sθ), i.e., Sβ⊆Sθ. We begin this section by giving the converse of [7, Thm. 7].

Theorem1. Supposeβ= {lr}is a lacunary refinement of the lacunary sequence θ= {kr}. LetIr:=(kr−1,kr]andJr:=(lr−1,lr],r=1,2,3,... .If there existsδ >0such that

|Jj|

|Ii| ≥δ for everyJj⊆Ii. (2.1) Thenxk→L(Sθ)impliesxk→L(Sβ), i.e.,Sθ⊆Sβ.

Proof. For anyε >0, and everyJj, we can findIisuch thatJj⊆Ii; then we have 1

|Jj|

k∈Jj:|xk−L| ≥ε= |Ii|

|Jj| 1

|Ii|

k∈Jj:|xk−L| ≥ε

≤ |Ii|

|Jj| 1

|Ii|

k∈Ii:|xk−L| ≥ε

≤ 1

δ 1

|Ii|

k∈Ii:|xk−L| ≥ε,

(2.2)

and the proof completes immediately.

Theorem2. Supposeβ= {lr},θ= {kr}are two lacunary sequences. LetIr:=(kr−1, kr],Jr:=(lr−1,lr],r=1,2,3,..., andIij=Ii∩Jj,i,j=1,2,3,... .If there existsδ >0, such that

|Iij|

|Ii| ≥δ for everyi,j=1,2,3,...,providedIij≠∅. (2.3) Thenxk→L(Sθ)impliesxk→L(Sβ), i.e.,Sθ⊆Sβ.

Proof. Letα=β∪θ. Thenαis a lacunary refinement of the lacunary sequence β, alsoθ. The interval sequence of α is

Iij =Ii∩Jj :Iij ≠∅

. From Theorem 1, the condition in Theorem 2:|Iij|/|Ii| ≥δ, for everyi,j=1,2,3,..., providedIij≠∅ yields thatxk→L(Sθ)impliesxk→L(Sα). Sinceαis also a lacunary refinement of the lacunary sequenceβ, from [4, Thm. 7] we have thatxk→L(Sα)impliesxk→L(Sβ).

The proof follows immediately.

In Theorem 2, if the condition is replaced by|Iij|/|Jj| ≥δ, for everyi,j=1,2,3,..., providedIij≠∅, it can be seen thatxk→L(Sβ)impliesxk→L(Sθ). Combining this remark and Theorem 2 we get the following theorem.

Theorem3. Supposeβ= {lr},θ= {kr}are two lacunary sequences. LetIr:=(kr−1, kr],Jr:=(lr−1,lr],r=1,2,3,..., andIij=Ii∩Jj,i,j=1,2,3,... .If there existsδ >0, such that

|Iij|

|Ii|+|Jj| ≥δ, for everyi,j=1,2,3,...,providedIij≠∅. (2.4)

Thenxk→L(Sθ)if and only ifxk→L(Sβ), i.e.,Sθ=Sβ.

Theorem 2 provides a sufficient condition for lacunary sequenceβ= {lr},θ= {kr} to yield the inclusion relation Sθ⊆Sβ. In Theorem 4we give some necessary and sufficient conditions for two lacunary sequences to have the relationship.

Theorem4. Suppose β = {lr},θ= {kr} are two lacunary sequences. Let Jr :=

(lr−1,lr],Ir :=(kr−1,kr],r =1,2,3,... .There exists a sequence xand a numberL such thatSβ−limx=LandSθ−limx≠L, if and only if there exist{Si},{ti} ⊆Nand δ >0which satisfies the following conditions:

(i) Jsi∩Iti≠∅;

(ii) limi|Js_i|/|It_i| = ∞;

(iii) |Js_i∩It_i|/|It_i| ≥δ,i=1,2,3,... . Proof

(i) Necessity. If there exists a sequencex= {xk}and a numberLsuch thatSβ− limx=LandSθ−limx≠L, then there exist a subsequence{ti} ⊆N,ε >0, andδ >0, such that

1

|It_i|

k∈Iti:|xk−L| ≥ε≥δ, i=1,2,3,... . (2.5)

For eachti, there exist a positive integersiand a whole numberrisuch that

Jsi+j∩Iti≠φ for everyj=0,1,2,..., i=1,2,3,... . (2.6) Then, we can write

δ≤ 1

|It_i|

k∈It_i:|xk−L| ≥ε

= 1

|Iti|



k∈



ⁿ

j=0

Js_i+j



∩It_i:|xk−L| ≥ε





= 1

|Iti| ri

j=0

k∈Js_i+j∩It_i:|xk−L| ≥ε

= 1

|It_i| _j=0,r

i

k∈Jsi+j∩Iti:|xk−L| ≥ε +

1

|Iti| 0<j<ri

k∈Js_i+j∩It_i:|xk−L| ≥ε

= 1

|Iti| _j=0,r

i

k∈Js_i+j∩It_i:|xk−L| ≥ε +

1

|It_i| 0<j<r_i

k∈Js_i+j:|xk−L| ≥ε

= 1

|It_i| _j=0,r

i

k∈Jsi+j∩Iti:|xk−L| ≥ε

+

0<j<r_i

|Js_i+j|

|It_i| 1

|Js_i+j|

k∈Js_i+j:|xk−L| ≥ε.

(2.7)

SinceSβ−limx=L, ifsi+jis sufficiently large, we have 1

|Js_i+j| 0<j<r_i

k∈Jsi+j:|xk−L| ≥ε<δ

2. (2.8)

It can be seen that

o<j<ri(|Js_i+j|/|It_i|)≤1. Hence, we get δ≤

1

|It_i|

k∈Iti:|xk−L| ≥ε

≤ 1

|It_i| _j=o,r

i

k∈Jsi+j∩Iti:|xk−L| ≥ε+δ

2. (2.9)

This implies that(1/|It_i|)

j=o,ri|{k∈Js_i+j∩It_i:|xk−L| ≥ε}| ≥δ/2, which ensures that at least one of the following inequalities hold:

1

|Iti|

k∈Js_i∩It_i:|xk−L| ≥ε≥δ

4, (2.10)

or

1

|Iti|

k∈Js_i+r_i∩It_i:|xk−L| ≥ε≥δ

4. (2.11)

Suppose the first one holds, i.e.,(1/|It_i|){k∈Js_i∩It_i :|xk−L| ≥ε}≥δ/4. From this inequality and since{k∈Jsi∩Iti :|xk−L| ≥ε}≤ |Jsi∩Iti|, we conclude that δ/4≤ |Js_i∩It_i|/|It_i|which proves (iii).

For suchsi,tichosen in the proof of (iii), from (2.10), we have δ

4≤ 1

|Iti|

k∈Jsi∩Iti:|xk−L| ≥ε

≤ |Js_i|

|Iti| 1

|Jsi|

k∈Js_i∩It_i:|xk−L| ≥ε.

(2.12)

Since(1/|Js_i|)|{k∈Js_i∩It_i:|xk−L| ≥ε}| →0, asi→ ∞, (2.12) implies|Js_i|/|It_i| → ∞, asi→ ∞, which proves (ii). It is clear that the intervalsJsi,Iti chosen in the proof of (iii) satisfy (i).

(ii) Sufficiency. Suppose that for the two lacunary sequencesβ= {lr},θ= {kr} there exist sequences{si},{ti} ⊆Nandδ >0 which satisfy the conditions (i), (ii), and (iii) in the theorem. Define

xk=





1, ifk∈Js_i∩It_i,

0, otherwise. (2.13)

For any 1< ε <0, ifj≠si for anyi=1,2,3,...,(1/|Jj|){k∈Jj:|xk−0| ≥ε}= ∅|/|Jj=0; ifj=si, for somei,(1/|Js_i|)|{k∈Js_i∩:|xk−0| ≥ε}| = |Js_i∩It_i|/|Js_i| ≤

|Iti|/|Jsi| → 0. Hence Sβ−limx =0. But (1/|Iti|)|{k∈ Iti : |xk−0| ≥ ε}| = |Jsi∩ It_i|/|It_i| ≥δ, fori=1,2,3,... .This implies thatSθ−limx≠0.

Corollary1. Supposeβ= {lr}, θ= {kr}are two lacunary sequences. LetJr :=

(lr−1,lr],Ir:=(kr−1,kr],r=1,2,3,... .If there exist{si},{ti} ⊆Nsuch thatJs_i⊇It_i,i= 1,2,3,..., andlimi|Jsi|/|Iti| = ∞, then there exists a sequencexand a numberLsuch thatSβ−limx=LandSθ−limx≠L.

Proof. The conditions of the Corollary 1 imply all the conditions of the Theorem 4.

So the corollary follow Theorem 4immediately.

3. The SpaceXθ. Letθ= {kr}be a lacunary sequence. ASθ-convergent sequence xis not necessary to be bounded. The superior norm inl^∞does not apply to the set Sθ, wherel^∞is the set of bounded complex numbers. Hence we need to define a new norm on a suitable space. If we apply the superior norm in the subset ofl^∞,Sθ∩l^∞, we get a closed subset ofl^∞.

Theorem5. Sθ∩l^∞is a closed subset ofl^∞with the superior norm.

Proof. Suppose{xⁿ} ⊆Sθ∩l^∞is a convergent sequence and it converges tox∈ l^∞. We need to prove that x∈Sθ∩l. Assume xⁿ→Ln(Sθ),n=1,2,3,... .Take a positive decreasing convergent sequence{εn}, then for everyn=1,2,3,..., there is a positive integermn, such that if n≥mn, thenx−xⁿ∞≤εn/4. Without loss of generality, we just assumemn=n. For the fixedn, there iskrnsuch that

1 hrn

k∈Irn:x_kⁿ−Ln≥εn

4 <1

4, 1

hrn

k∈Irn:xⁿ⁺¹_k −Ln+1≥εn+1

4

<1

4. (3.1)

Then

k∈Irn:xⁿ_k−Ln<εn

4

∩

k∈Irn:x_kⁿ⁺¹−Ln+1<εn+1

4

≠∅. (3.2) Pick a memberkin this intersection, we can write

Ln−Ln+1≤x_kⁿ−Ln+x_kⁿ⁺¹−Ln+1+xⁿ_k−x_kⁿ⁺¹

≤x_kⁿ−Ln+x_kⁿ⁺¹−Ln+1+x−xⁿ_∞+x−xⁿ⁺¹_∞

≤εn

4 +εn+1

4 +εn

4 +εn+1

4 ≤εn.

(3.3)

This implies that{Ln}is a Cauchy sequence inC, and there is a complex numberL such thatLn→L, asn→ ∞. We need to prove thatx→L(Sθ). For anyε >0, there is n, such thatεn< ε/4,|x−xⁿ∞< ε/4,|Ln−L|< ε/4, then

1 hr

k∈Ir:|x−L| ≥ε

≤ 1

hr

k∈Ir:|xⁿ−Ln|+x−xⁿ∞+|Ln−L| ≥ε

≤ 1

hr

k∈Ir:|xⁿ−Ln|+ε 4+ε

4≥ε

= 1

hr

k∈Ir:|xⁿ−Ln| ≥ε 2

→0, asr → ∞.

(3.4)

This gives thatx→L(Sθ), which completes the proof.

Letθ= {kr}be a lacunary sequence. for everyL∈C, define XL:=

x∈Sθ:x →L(Sθ)

and Xθ:=

XL:L∈C

. (3.5)

Ifx→L(Sθ),y→K(Sθ), anda∈C, it can be seen that

x+y →(L+K)(Sθ) and ax →aL(Sθ), (3.6)

wherex+y= {xn+yn}andax= {axn}. Then the following definitions onXθwell defined

XL+XK:=XL+K and aXL=XaL. (3.7) For everyXL∈Xθ, we define a norm of it byXL:= |L|.

Theorem6. With the addition, scalar multiplication and norm defined on Xθ as above,Xθis a Banach space.

Proof. To prove thatXθ is complete norm space, we just notice that{XLn}is a Cauchy sequence ofXθ, if and only if{Ln}is a Cauchy sequence ofC, and for any complex number L, the sequence with constant termsL should converge toL(Sθ).

Rest of the proof is clear.

Acknowledgement. The author wish to thank Prof. J. A. Fridy for his communi- cation and useful suggestions.

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Li: Department of Mathematics, Shawnee State University, Portsmouth, OH45662, USA

E-mail address:[email protected]

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