Internat. J. Math. & Math. Sci. S0161171200001964 © Hindawi Publishing Corp.

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Internat. J. Math. & Math. Sci.
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 sequence θ = {kr } such that kr −
kr −1 → ∞ as r → ∞. A sequence x is called sθ -convergent to L provided that for each
ε > 0, limr (1/(kr − kr −1 )){the number of kr −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 sequence θ = {kr } is called a lacunary sequence if it satisfies that k0 = 0 and hr := kr − kr −1 → ∞ as r → ∞. Throughout this
paper, we follow the notation used in [7]. A sequence x is called Sθ -convergent to L provided that for each ε > 0. The intervals determined by θ is denoted by Ir := (kr −1 , kr ]
and the ratio kr /kr −1 is abbreviated by qr . For a finite set E, |E| indicate the number
of elements in E.
Let θ be a lacunary sequence, a complex number sequence x is said to be Sθ convergent to L provided that for each ε > 0.
lim
r
1 the number of kr −1 < k ≤ kr : |xk − L| ≥ ε = 0.
hr
(1.1)
In this case we write Sθ − lim x = L or xk → L(Sθ ), and we define
Sθ = x : Sθ − lim x = L, for some L .
(1.2)
In [4] it is defined that the lacunary sequence β = {lr } is called a lacunary refinement
of the lacunary sequence θ = {kr } if {kr } ⊆ {lr }. In [7], after the inclusion relationship
between Sθ and Sβ is studied for a special case if β = {lr } is a lacunary refinement
of the lacunary sequence θ = {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 sequence θ = {kr }, it is shown by [7, Thm. 7] that xk → L(Sβ ) implies xk → L(Sθ ), i.e.,
Sβ ⊆ Sθ . We begin this section by giving the converse of [7, Thm. 7].
176
JINLU LI
Theorem 1. Suppose β = {lr } is a lacunary refinement of the lacunary sequence
θ = {kr }. Let Ir := (kr −1 , kr ] and Jr := (lr −1 , lr ], r = 1, 2, 3, . . . . If there exists δ > 0 such
that
|Jj |
≥δ
|Ii |
for every Jj ⊆ Ii .
(2.1)
Then xk → L(Sθ ) implies xk → L(Sβ ), i.e., Sθ ⊆ Sβ .
Proof. For any ε > 0, and every Jj , we can find Ii such that Jj ⊆ Ii ; then we have
|I | 1 1 i
k ∈ Jj : |xk − L| ≥ ε =
k ∈ Jj : |xk − L| ≥ ε |Jj |
|Jj |
|Ii |
1 |Ii |
k ∈ Ii : |xk − L| ≥ ε ≤
|Jj |
|Ii |
1 1
k ∈ Ii : |xk − L| ≥ ε ,
≤
δ
|Ii |
(2.2)
and the proof completes immediately.
Theorem 2. Suppose β = {lr }, θ = {kr } are two lacunary sequences. Let Ir := (kr −1 ,
kr ], Jr := (lr −1 , lr ], r = 1, 2, 3, . . . , and Iij = Ii ∩Jj , i, j = 1, 2, 3, . . . . If there exists δ > 0,
such that
|Iij |
≥δ
|Ii |
for every i, j = 1, 2, 3, . . . , provided Iij ≠ ∅.
(2.3)
Then xk → L(Sθ ) implies xk → L(Sβ ), i.e., Sθ ⊆ Sβ .
Proof. Let α = β ∪ θ. Then α is a lacunary refinement of the lacunary sequence
β, also θ. The interval sequence of α is Iij = Ii ∩ Jj : Iij ≠ ∅ . From Theorem 1,
the condition in Theorem 2: |Iij |/|Ii | ≥ δ, for every i, j = 1, 2, 3, . . . , provided Iij ≠ ∅
yields that xk → L(Sθ ) implies xk → L(Sα ). Since α is also a lacunary refinement of the
lacunary sequence β, from [4, Thm. 7] we have that xk → L(Sα ) implies xk → L(Sβ ).
The proof follows immediately.
In Theorem 2, if the condition is replaced by |Iij |/|Jj | ≥ δ, for every i, j = 1, 2, 3, . . . ,
provided Iij ≠ ∅, it can be seen that xk → L(Sβ ) implies xk → L(Sθ ). Combining this
remark and Theorem 2 we get the following theorem.
Theorem 3. Suppose β = {lr }, θ = {kr } are two lacunary sequences. Let Ir := (kr −1 ,
kr ], Jr := (lr −1 , lr ], r = 1, 2, 3, . . . , and Iij = Ii ∩ Jj , i, j = 1, 2, 3, . . . . If there exists δ > 0,
such that
|Iij |
≥ δ,
|Ii | + |Jj |
for every i, j = 1, 2, 3, . . . , provided Iij ≠ ∅.
(2.4)
Then xk → L(Sθ ) if and only if xk → L(Sβ ), i.e., Sθ = Sβ .
Theorem 2 provides a sufficient condition for lacunary sequence β = {lr }, θ = {kr }
to yield the inclusion relation Sθ ⊆ Sβ . In Theorem 4 we give some necessary and
sufficient conditions for two lacunary sequences to have the relationship.
LACUNARY STATISTICAL CONVERGENCE . . .
177
Theorem 4. 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 x and a number L
such that Sβ − lim x = L and Sθ − lim x ≠ L, if and only if there exist {Si }, {ti } ⊆ N and
δ > 0 which 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 sequence x = {xk } and a number L such that Sβ −
lim x = L and Sθ −lim x ≠ L, then there exist a subsequence {ti } ⊆ N, ε > 0, and δ > 0,
such that
1 k ∈ It : |xk − L| ≥ ε ≥ δ, i = 1, 2, 3, . . . .
(2.5)
i
|Iti |
For each ti , there exist a positive integer si and a whole number ri such that
Jsi +j ∩ Iti ≠ φ for every j = 0, 1, 2, . . . , i = 1, 2, 3, . . . .
(2.6)
Then, we can write
1 k ∈ It : |xk − L| ≥ ε δ≤
i
|Iti |





n

1


k∈
Jsi +j ∩ Iti : |xk − L| ≥ ε =

|It | 
i
=
j=0
1
|Iti |
ri
k ∈ Js
j=0
i +j
∩ Iti : |xk − L| ≥ ε 1
k ∈ Js +j ∩ It : |xk − L| ≥ ε =
i
i
|Iti | j=0,r
i
1
k ∈ Js +j ∩ It : |xk − L| ≥ ε +
i
i
|Iti | 0<j<r
i
1
k ∈ Js +j ∩ It : |xk − L| ≥ ε =
i
i
|Iti | j=0,r
i
1
k ∈ Js +j : |xk − L| ≥ ε +
i
|Iti | 0<j<r
i
1
=
k ∈ Jsi +j ∩ Iti : |xk − L| ≥ ε |Iti | j=0,r
i
|Js +j | 1 i
k ∈ Js +j : |xk − L| ≥ ε .
+
i
|I
|
|J
|
si +j
ti
0<j<r
(2.7)
i
Since Sβ − lim x = L, if si + j is sufficiently large, we have
1
|Jsi +j |
k ∈ Js
0<j<ri
i +j
δ
: |xk − L| ≥ ε < .
2
(2.8)
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JINLU LI
It can be seen that
o<j<ri (|Jsi +j |/|Iti |)
≤ 1. Hence, we get
1 k ∈ It : |xk − L| ≥ ε i
|Iti |
1
k ∈ Js +j ∩ It : |xk − L| ≥ ε + δ .
≤
i
i
|Iti | j=o,r
2
δ≤
(2.9)
i
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 k ∈ Js ∩ It : |xk − L| ≥ ε ≥ δ ,
(2.10)
i
i
|Iti |
4
or
1 k ∈ Js +r ∩ It : |xk − L| ≥ ε ≥ δ .
(2.11)
i
i
i
|Iti |
4
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 such si , ti chosen in the proof of (iii), from (2.10), we have
δ
1 k ∈ Js ∩ It : |xk − L| ≥ ε ≤
i
i
4
|Iti |
(2.12)
|Jsi |
1 k ∈ Js ∩ It : |xk − L| ≥ ε .
≤
i
i
|Iti |
|Jsi |
Since (1/|Jsi |)|{k ∈ Jsi ∩ Iti : |xk − L| ≥ ε}| → 0, as i → ∞, (2.12) implies |Jsi |/|Iti | → ∞,
as i → ∞, which proves (ii). It is clear that the intervals Jsi , Iti chosen in the proof of
(iii) satisfy (i).
(ii) Sufficiency. Suppose that for the two lacunary sequences β = {lr }, θ = {kr }
there exist sequences {si }, {ti } ⊆ N and δ > 0 which satisfy the conditions (i), (ii), and
(iii) in the theorem. Define

1, if k ∈ Js ∩ It ,
i
i
xk =
(2.13)
0, otherwise.
For any 1 < ε < 0, if j ≠ si for any i = 1, 2, 3, . . . , (1/|Jj |){k ∈ Jj : |xk − 0| ≥ ε} =
∅|/|Jj = 0; if j = si , for some i, (1/|Js |)|{k ∈ Js ∩ : |xk − 0| ≥ ε}| = |Js ∩ It |/|Js | ≤
i
i
i
i
i
|Iti |/|Jsi | → 0. Hence Sβ − lim x = 0. But (1/|Iti |)|{k ∈ Iti : |xk − 0| ≥ ε}| = |Jsi ∩
Iti |/|Iti | ≥ δ, for i = 1, 2, 3, . . . . This implies that Sθ − lim x ≠ 0.
Corollary 1. Suppose β = {lr }, θ = {kr } are two lacunary sequences. Let Jr :=
(lr −1 , lr ], Ir := (kr −1 , kr ], r = 1, 2, 3, . . . . If there exist {si }, {ti } ⊆ N such that Jsi ⊇ Iti , i =
1, 2, 3, . . . , and limi |Jsi |/|Iti | = ∞, then there exists a sequence x and a number L such
that Sβ − lim x = L and Sθ − lim x ≠ L.
Proof. The conditions of the Corollary 1 imply all the conditions of the Theorem 4.
So the corollary follow Theorem 4 immediately.
LACUNARY STATISTICAL CONVERGENCE . . .
179
3. The Space Xθ . Let θ = {kr } be a lacunary sequence. A Sθ -convergent sequence
x is not necessary to be bounded. The superior norm in l∞ does not apply to the set
Sθ , where l∞ 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 of l∞ , Sθ ∩ l∞ ,
we get a closed subset of l∞ .
Theorem 5. Sθ ∩ l∞ is a closed subset of l∞ with the superior norm.
Proof. Suppose {x n } ⊆ Sθ ∩ l∞ is a convergent sequence and it converges to x ∈
l . We need to prove that x ∈ Sθ ∩ l. Assume x n → Ln (Sθ ), n = 1, 2, 3, . . . . Take a
positive decreasing convergent sequence {εn }, then for every n = 1, 2, 3, . . . , there is
a positive integer mn , such that if n ≥ mn , then x − x n ∞ ≤ εn /4. Without loss of
generality, we just assume mn = n. For the fixed n, there is krn such that
1 k ∈ Ir : x n − Ln ≥ εn < 1 ,
n
k
hrn
4 4
(3.1)
1 k ∈ Ir : x n+1 − Ln+1 ≥ εn+1 < 1 .
n
k
hr n
4
4
∞
Then
εn
εn+1
∩ k ∈ Irn : xkn+1 − Ln+1 <
≠ ∅.
k ∈ Irn : xkn − Ln <
4
4
Pick a member k in this intersection, we can write
Ln − Ln+1 ≤ x n − Ln + x n+1 − Ln+1 + x n − x n+1 k
k
k
k
≤ xkn − Ln + xkn+1 − Ln+1 + x − x n ∞ + x − x n+1 ∞
(3.2)
(3.3)
εn εn+1 εn εn+1
+
+
+
≤ εn .
≤
4
4
4
4
This implies that {Ln } is a Cauchy sequence in C, and there is a complex number L
such that Ln → L, as n → ∞. We need to prove that x → L(Sθ ). For any ε > 0, there is
n, such that εn < ε/4, |x − x n ∞ < ε/4, |Ln − L| < ε/4, then
1 k ∈Ir : |x − L| ≥ ε hr
1 k ∈ Ir : |x n − Ln | + x − x n ∞ + |Ln − L| ≥ ε ≤
hr
(3.4)
1 k ∈ Ir : |x n − Ln | + ε + ε ≥ ε ≤
hr
4 4
1 k ∈ Ir : |x n − Ln | ≥ ε → 0, as r → ∞.
=
hr
2 This gives that x → L(Sθ ), which completes the proof.
Let θ = {kr } be a lacunary sequence. for every L ∈ C, define
XL := x ∈ Sθ : x → L(Sθ )
and
Xθ := XL : L ∈ C .
(3.5)
If x → L(Sθ ), y → K(Sθ ), and a ∈ C, it can be seen that
x + y → (L + K)(Sθ )
and
ax → aL(Sθ ),
(3.6)
180
JINLU LI
where x + y = {xn + yn } and ax = {axn }. Then the following definitions on Xθ well
defined
XL + XK := XL+K
and
aXL = XaL .
(3.7)
For every XL ∈ Xθ , we define a norm of it by XL := |L|.
Theorem 6. With the addition, scalar multiplication and norm defined on Xθ as
above, Xθ is a Banach space.
Proof. To prove that Xθ is complete norm space, we just notice that {XLn } is a
Cauchy sequence of Xθ , if and only if {Ln } is a Cauchy sequence of C, and for any
complex number L, the sequence with constant terms L should converge to L(Sθ ).
Rest of the proof is clear.
Acknowledgement. The author wish to thank Prof. J. A. Fridy for his communication and useful suggestions.
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Li: Department of Mathematics, Shawnee State University, Portsmouth, OH 45662,
USA
E-mail address: [email protected]
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