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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 kr−1→ ∞asr → ∞. A sequence xis called sθ-convergent toLprovided that for each ε >0, limr(1/(kr−kr−1)){the number ofkr−1< k≤kr:|xk−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 writeSθlimx=Lorxk→L(Sθ), and we define Sθ=

x:Sθlimx=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].

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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.

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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 thatSβlimx=LandSθlimx≠L, if and only if there exist{Si},{ti} ⊆Nand δ >0which satisfies the following conditions:

(i) Jsi∩Iti∅;

(ii) limi|Jsi|/|Iti| = ∞;

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

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

1

|Iti|

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∩Itiφ for everyj=0,1,2,..., i=1,2,3,... . (2.6) Then, we can write

δ≤ 1

|Iti|

k∈Iti:|xk−L| ≥ε

= 1

|Iti|



k∈

n

j=0

Jsi+j

∩Iti:|xk−L| ≥ε



= 1

|Iti| ri

j=0

k∈Jsi+j∩Iti:|xk−L| ≥ε

= 1

|Iti| j=0,r

i

k∈Jsi+j∩Iti:|xk−L| ≥ε +

1

|Iti| 0<j<ri

k∈Jsi+j∩Iti:|xk−L| ≥ε

= 1

|Iti| j=0,r

i

k∈Jsi+j∩Iti:|xk−L| ≥ε +

1

|Iti| 0<j<ri

k∈Jsi+j:|xk−L| ≥ε

= 1

|Iti| j=0,r

i

k∈Jsi+j∩Iti:|xk−L| ≥ε

+

0<j<ri

|Jsi+j|

|Iti| 1

|Jsi+j|

k∈Jsi+j:|xk−L| ≥ε.

(2.7)

SinceSβlimx=L, ifsi+jis sufficiently large, we have 1

|Jsi+j| 0<j<ri

k∈Jsi+j:|xk−L| ≥ε<δ

2. (2.8)

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It can be seen that

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

1

|Iti|

k∈Iti:|xk−L| ≥ε

1

|Iti| j=o,r

i

k∈Jsi+j∩Iti:|xk−L| ≥ε+δ

2. (2.9)

This implies that(1/|Iti|)

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

1

|Iti|

k∈Jsi∩Iti:|xk−L| ≥ε≥δ

4, (2.10)

or

1

|Iti|

k∈Jsi+ri∩Iti:|xk−L| ≥ε≥δ

4. (2.11)

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

|Jsi|

|Iti| 1

|Jsi|

k∈Jsi∩Iti:|xk−L| ≥ε.

(2.12)

Since(1/|Jsi|)|{k∈Jsi∩Iti:|xk−L| ≥ε}| →0, asi→ ∞, (2.12) implies|Jsi|/|Iti| → ∞, 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∈Jsi∩Iti,

0, otherwise. (2.13)

For any 1< ε <0, ifjsi for anyi=1,2,3,...,(1/|Jj|){k∈Jj:|xk0| ≥ε}= ∅|/|Jj=0; ifj=si, for somei,(1/|Jsi|)|{k∈Jsi:|xk0| ≥ε}| = |Jsi∩Iti|/|Jsi| ≤

|Iti|/|Jsi| → 0. Hence Sβlimx =0. But (1/|Iti|)|{k∈ Iti : |xk0| ≥ ε}| = |Jsi Iti|/|Iti| ≥δ, fori=1,2,3,... .This implies thatSθlimx≠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 thatJsi⊇Iti,i= 1,2,3,..., andlimi|Jsi|/|Iti| = ∞, then there exists a sequencexand a numberLsuch thatSβlimx=LandSθlimx≠L.

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

So the corollary follow Theorem 4immediately.

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3. The SpaceXθ. Letθ= {kr}be a lacunary sequence. ASθ-convergent sequence xis not necessary to be bounded. The superior norm inldoes not apply to the set Sθ, wherelis 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∩lis a closed subset oflwith the superior norm.

Proof. Suppose{xn} ⊆Sθ∩lis 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 sequencen}, 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:xkn−Ln≥εn

4 <1

4, 1

hrn

k∈Irn:xn+1k −Ln+1≥εn+1

4

<1

4. (3.1)

Then

k∈Irn:xnk−Lnn

4

k∈Irn:xkn+1−Ln+1n+1

4

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

Ln−Ln+1≤xkn−Ln+xkn+1−Ln+1+xnk−xkn+1

≤xkn−Ln+xkn+1−Ln+1+x−xn+x−xn+1

≤εn

4 n+1

4 n

4 n+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

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:xL(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 axaL(Sθ), (3.6)

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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.

References

[1] R. C. Buck,Generalized asymptotic density, Amer. J. Math.75(1953), 335–346. MR 14,854f.

Zbl 050.05901.

[2] J. S. Connor,The statistical and strongp-Cesaro convergence of sequences, Analysis8 (1988), no. 1–2, 47–63. MR 89k:40013. Zbl 653.40001.

[3] H. Fast, Sur la convergence statistique, Colloq. Math. 2 (1951), 241–244. MR 14,29c.

Zbl 044.33605.

[4] A. R. Freedman, J. J. Sember, and M. Raphael,Some Cesaro-type summability spaces, Proc.

London Math. Soc.37(1978), no. 3, 508–520. MR 80c:40007. Zbl 424.40008.

[5] J. A. Fridy,On statistical convergence, Analysis5(1985), no. 4, 301–313. MR 87b:40001.

Zbl 588.40001.

[6] J. A. Fridy and H. I. Miller,A matrix characterization of statistical convergence, Analysis 11(1991), no. 1, 59–66. MR 92e:40001. Zbl 727.40001.

[7] J. A. Fridy and C. Orhan,Lacunary statistical convergence, Pacific J. Math.160(1993), no. 1, 43–51. MR 94j:40014. Zbl 794.60012.

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

E-mail address:[email protected]

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