calculs&amp_nbsp_08.dvi


CUBO A Mathematical Journal

Vol.12, No¯ 03, (121–138). October 2010

Calculations in New Sequence Spaces and

Application to Statistical Convergence

BRUNO DE MALAFOSSE

LMAH Université du Havre, BP 4006 IUT Le Havre,

76610 Le Havre. France

email: bdemalaf@wanadoo.fr

AND

VLADIMIR RAKOČEVIĆ1

Department of Mathematics, University of Niš,

Videgradska 33, 18000 Niš, Serbia

email: vrakoc@bankerinter.net

ABSTRACT

In this paper we recall recent results that are direct consequences of the fact that

(w∞ (λ) , w∞ (λ)) is a Banach algebra. Then we define the set Wτ = Dτw∞ and charac-

terize the sets Wτ (A) where A is either of the operators ∆, Σ, ∆(λ), or C (λ). Afterwards

we consider the sets [A1, A2]Wτ of all sequences X such that A1 (λ)
(∣∣A2

(
µ
)

X
∣∣) ∈ Wτ where

A1 and A2 are of the form C (ξ), C
+ (ξ), ∆(ξ), or ∆+ (ξ) and it is given necessary conditions

to get
[
A1 (λ) , A2

(
µ
)]

Wτ
in the form Wξ. Finally we apply the previous results to statis-

tical convergence. So we have conditions to have xk → L (S ( A)) where A is either of the

infinite matrices D1/τC (λ) C
(
µ
)
, D1/τ∆(λ)∆

(
µ
)
, D1/τ∆(λ) C

(
µ
)
. We also give conditions to

have xk → 0 (S ( A)) where A is either of the operators D1/τC
+ (λ)∆

(
µ
)
, D1/τC

+ (λ) C
(
µ
)
,

D1/τC
+

(λ) C+
(
µ
)
, or D1/τ∆(λ) C

+
(
µ
)
.

1Supported by Grant No. 144003 of the Ministry of Science, Technology and Development, Republic of Serbia.



122 Bruno de Malafosse & Vladimir Rakočević
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12, 3 (2010)

RESUMEN

Recordamos resultados recientes que son consecuencia directa del hecho de que (w∞(λ),

w∞(λ)) es una algebra de Banach. Entonces nosotros definimos el conjunto Wτ = Dτw∞
y caracterizamos los conjuntos Wτ (A) donde A es uno de los siguientes operadores ∆, Σ,

∆(λ), o C (λ). Después consideramos los conjuntos [A1, A2 ]Wτ de todas las sucesiones X

tal que A1 (λ)
(∣∣A2

(
µ
)
X

∣∣) ∈ Wτ donde A1 y A2 son de la forma C (ξ), C+ (ξ), ∆(ξ), o ∆+ (ξ)
y son dadas condiciones necesarias para obtener

[
A1 (λ) , A2

(
µ
)]

Wτ
en la forma Wξ. Final-

mente, aplicamos los resultados previos para tener xk → L (S ( A)) donde A es una de las

matrices infinitas D1/τC (λ) C
(
µ
)
, D1/τ∆(λ)∆

(
µ
)
, D1/τ∆(λ) C

(
µ
)
. Nosotros también damos

condiciones para tener xk → 0 (S ( A)) donde A es uno de los operadores D1/τC
+ (λ)∆

(
µ
)
,

D1/τC
+ (λ) C

(
µ
)
, D1/τC

+ (λ) C+
(
µ
)
, o D1/τ∆(λ) C

+
(
µ
)
.

Key words and phrases: Banach algebra, statistical convergence, A−statistical convergence,

infinite matrix.

Math. Subj. Class.: 40C05, 40F05, 40J05, 46A15.

1 Introduction

In this paper we consider spaces generalizing the well-known sets w0 and w∞ introduced

and studied by Maddox [12, 13]. Recall that w0 and w∞ are the sets of strongly summable

and strongly bounded sequences. In [15] Malkowsky and Rakočević gave characterizations

of matrix maps between w0, w, or w∞ and w
p
∞ and between w

0, w, or w∞ and l1. In [2] de

Malafosse defined the spaces wα (λ), w
(c)
α (λ) and w

0
α (λ) of all sequences that are α−strongly

bounded, summable and summable to zero respectively. For instance recall that wα (λ) is the

set of all sequences (xn)n such that 1/λn
∑n

m=1
|xm| = αnO (1) as n tends to infinity. It was

shown that these spaces can be written in the form sξ, s
(c)
ξ

and s0
ξ

under some condition on α

and λ.

More recently in [5] it was shown that if λ is a sequence exponentially bounded then

(w∞ (λ) , w∞ (λ)) is a Banach algebra. This result led to consider bijective operators mapping

between w∞ (λ). Here we will use these results to study sets of the form Wτ = Dτw∞, Wτ (∆(λ)),

Wτ (C (λ)) and Wτ
(
C+ (λ)

)
generalizing the well-known set of strongly bounded sequences c∞ =

w∞
(
∆

(
µ
))

where µn = n for all n. These results lead to the study of statistical convergence

which was introduced by Steinhaus in 1949, see [16], and studied by several authors such as

Fast [7], Fridy, Orhan [8-11] and Connor. Here we will deal with the notion of A− statistical

convergence which generalizes the notion of statistical convergence, see [6], where A belongs

to a special class of operators.



CUBO
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Calculations in New Sequence Spaces ... 123

The paper is organized as follows. In Section 2 among other things we recall a recent

result on the operators ∆ρ and ∆
T
ρ considered as map from w∞ (λ) to itself. In Sections 3

and 4 our aim is to give necessary conditions to have Wτ (A) in the form Wξ when A is either

one of the matrices ∆(λ), C (λ) or C+ (λ). Then we consider spaces generalizing the well-

known set of all strongly bounded sequences [C,∆] = c∞ defined and studied by Maddox.

Then we will define the sets [A1, A2]Wτ of all sequences X with A1 (λ)
(∣∣A2

(
µ
)

X
∣∣) ∈ Wτ where

A1 and A2 are of the form C (ξ), C
+ (ξ), ∆(ξ), or ∆+ (ξ) and we will give necessary conditions

to get
[
A1 (λ) , A2

(
µ
)]

in the form Wτ. In Section 5 we apply these results to A− statistical

convergence, where A is equal to D1/τ A1 A2 and A1, A2 are of the form C (ξ), ∆(ξ), ∆
(
µ
)
, or

C+ (ξ).

2 Well Known Results

For a given infinite matrix A = (anm)n,m≥1 we define the operators An for any integer n ≥ 1,

by

An (X ) =
∞∑

m=1

anm xm (1)

where X = (xn)n≥1, the series intervening in the second member being convergent. So we are

led to the study of the infinite linear system

An (X ) = bn n = 1, 2, ... (2)

where B = (bn)n≥1 is a one-column matrix and X the unknown, see [2-5]. The equations (2)

can be written in the form A X = B, where A X = (An (X ))n≥1. In this paper we shall also

consider A as an operator from a sequence space into another sequence space.

We will write s for the set of all complex sequences and ℓ∞ for the set of all bounded

sequences.

Let E and F be any subsets of s. When A maps E into F we write that A ∈ (E, F).

So for every X ∈ E, A X ∈ F, (A X ∈ F means that for each n ≥ 1 the series defined by yn =∑
∞
m=1

anm xm is convergent and ( yn)n≥1 ∈ F).

Body Math For any subset E of s, we put

Body Math

AE = {Y ∈ s : Y = A X for some X ∈ E} . (3)

If F is a subset of s, we shall denote

F (A) = FA = {X ∈ s : Y = A X ∈ F} . (4)



124 Bruno de Malafosse & Vladimir Rakočević
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12, 3 (2010)

In all what follows we will use the set

U
+
=

{
(un )n≥1 ∈ s : un > 0 for all n

}

and the notation e = (1, ..., 1, ...). So for λ = (λn)n≥1 ∈ U
+ we will consider the sets of strongly

bounded and strongly summable sequences, respectively, that is

w∞ (λ) =

{
X = (xn)n≥1 ∈ s : sup

n

1

λn

n∑

m=1

|xm| < ∞

}
,

w
0

(λ) =

{
X = (xn)n≥1 ∈ s : lim

n→∞

1

λn

n∑

m=1

|xm| = 0

}

and

w (λ) =
{
X = (xn)n≥1 ∈ s : X − l e ∈ w

0
(λ) for some l ∈ C

}

were studied by Malkowsky, with the concept of exponentially bounded sequences, see [3].

Recall that Maddox [12, 13], defined and studied the sets w∞ (λ) = w∞, w0 (λ) = w
0 and w (λ) =

w where λn = n for all n.

A Banach space E of complex sequences with the norm ‖‖E is a BK space if each projection

Pn : X 7→ Pn X = xn is continuous. A BK space E is said to have AK if every sequence X =

(xn)n≥1 ∈ E has a unique representation X =
∑

∞
n=1

xn e n where e n is the sequence with 1 in the

n-th position and 0 otherwise.

Recall that a nondecreasing sequence λ = (λn)n≥1 ∈ U
+ is exponentially bounded if there

is an integer m ≥ 2 such that for all non-negative integers ν there is at least one term

λn ∈ I
(ν)
m =

[
mν, mν+1 − 1

]
. It was shown (cf. [14, Lemma 1]) that a non-decreasing sequence

λ = (λn)n≥1 is exponentially bounded if and only if there are reals s ≤ t such that for some

subsequence
(
λni

)
i≥1

0 < s ≤
λni

λni+1
≤ t < 1 for all i = 1, 2, ...;

such a sequence is called an associated subsequence. Consider now the norm

‖X‖λ = sup
n

(
1

λn

n∑

m=1

|xm|

)
.

In [5] it was shown that if λ = (λn)n≥1 ∈ U
+ is exponentially bounded the class (w∞ (λ) , w∞ (λ))

is a Banach algebra with the norm

‖A‖(w∞(λ),w∞(λ)) = sup
X 6=0

(
‖A X‖λ

‖X‖λ

)
. (5)



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12, 3 (2010)

Calculations in New Sequence Spaces ... 125

For ρ =
(
ρn

)
n≥1

consider now the following matrices

∆
+
ρ =




1 −ρ1

. .

1 −ρn

0 . .




and ∆ρ =




1

−ρ1 1 0

. .

−ρn−1 1 .

. .




.

It can easily be shown that if ρ =
(
ρn

)
n≥1

and (λn+1/λn)n≥1 ∈ ℓ∞ then ∆
+
ρ ∈ (w∞ (λ) , w∞ (λ)).

We also see that ∆ρ ∈ (w∞ (λ) , w∞ (λ)) for ρ, (λn−1/λn)n≥2 ∈ ℓ∞. Recall the next result which is

a direct consequence of [5, Theorem 5.1 and Theorem 5.12].

Lemma 2.1. Let λ ∈ U+ be a sequence exponentially bounded.

(i) If

lim
n→∞

(
λn+1

λn

)
< ∞ and lim

n→∞

∣∣ρn
∣∣ <

1

lim
n→∞

(
λn+1
λn

) , (6)

for given B ∈ w∞ (λ) the equation ∆
+
ρ X = B has a unique solution in w∞ (λ).

(ii) If

lim
n→∞

∣∣ρn
∣∣ <

1

lim
n→∞

(
λn−1
λn

) , (7)

then for any given B ∈ w∞ (λ) the equation ∆ρ X = B has a unique solution in w∞ (λ).

When λ is a strictly increasing sequence tending to infinity we obtain similar results on

the Banach algebra
(
w0 (λ) , w0 (λ)

)
with the norm ‖A‖(w∞(λ),w∞(λ)).

3 On the Sets Wτ (A) Where A is Either ∆(λ), C (λ) or C
+ (λ)

In the following we will use the operators represented by C (λ) and ∆(λ). Let U be the set of

all sequences (un )n≥1 with un 6= 0 for all n. We define C (λ) for λ = (λn)n≥1 ∈ U, by

[C (λ)]nm =





1

λn
if m ≤ n,

0 otherwise.

We will write C (λ)T = C+ (λ), C (e) = Σ, Σ+ = ΣT , and for λn = n, the matrix C1 = C ((n)n) is

called the Cesaro operator. If It can be proved that the matrix ∆(λ) with

[∆(λ)]nm =





λn if m = n,

−λn−1 if m = n − 1 and n ≥ 2,

0 otherwise,



126 Bruno de Malafosse & Vladimir Rakočević
CUBO

12, 3 (2010)

is the inverse of C (λ), see [2, 3]. We will use the following sets

Γ =

{
X ∈ U

+
: lim

n→∞

(
xn−1

xn

)
< 1

}
,

Γ
+

=

{
X ∈ U

+
: lim

n→∞

(
xn+1

xn

)
< 1

}
.

Note that X ∈ Γ+ if and only if 1/X ∈ Γ.

For given sequence τ = (τn)n≥1 ∈ U
+, we write Dτ for the diagonal matrix defined by

[Dτ]nn = τn for all n. For any subset E of s, we write

DτE =

{
X = (xn)n≥1 ∈ s :

(
xn

τn

)

n

∈ E

}
.

We put Wτ = Dτw∞ for τ ∈ U
+, that is

Wτ =

{
X : ‖X‖Wτ = sup

n

(
1

n

∞∑

m=1

|xm|

τm

)
< ∞

}
.

It can easily be seen that Wτ = w∞ (D1/τ) is a BK space with norm ‖‖Wτ , (cf. [17, Theorem

4.3.6, p. 52]). In all that follows we will use the convention that the entries with subscripts

strictly less than 1 are equal to zero. Then we are interested in the study of the following sets

where λ, τ ∈ U+.

Wτ (∆(λ)) =

{
X : sup

n

(
1

n

n∑

m=1

1

τm
|λm xm −λm−1 xm−1|

)
< ∞

}
,

Wτ (C (λ)) =

{
X : sup

n

1

n

n∑

m=1

(
1

λmτm

m∑

k=1

|xk|

)
< ∞

}
,

Wτ
(
C

+
(λ)

)
=

{
X : sup

n

1

n

n∑

m=1

(
1

τm

∞∑

k=m

|xk|

λk

)
< ∞

}
.

Note that for λn = n and τ = e, Wτ (∆(λ)) is the well known set of all strongly and bounded

sequences c∞. We obtain the following result that is a direct consequence of Lemma 2.1.

Proposition 3.1. (i) If τ ∈ Γ then the operators ∆ and Σ are bijective from Wτ into itself and

Wτ (∆) = Wτ, Wτ (Σ) = Wτ.

(ii) a) If λτ ∈ Γ then

Wτ (C (λ)) = Wλτ.

b) If τ ∈ Γ then

Wτ (∆(λ)) = Wτ/λ.



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Calculations in New Sequence Spaces ... 127

(iii) Let τ ∈ Γ+. Then

a) the operators ∆+ and Σ+ are bijective from Wτ into itself and

Wτ
(
Σ
+
)
= Wτ.

b) the operator C+ (λ) is bijective from Wλτ into Wτ and

Wτ
(
C

+
(λ)

)
= Wλτ.

Proof. (i) By Lemma 2.1 where ρn = τn−1/τn and λn = n for all n, we easily see that if

lim
n→∞

τn−1

τn
<

1

limn→∞
(

n−1
n

) = 1,

that is τ ∈ Γ, then D1/τ∆Dτ is bijective from w∞ to itself. This means that ∆ is bijective from

Dτw∞ to itself. Since Σ is also bijective from Dτw∞ to itself, this shows Wτ (∆) = Wτ and

Wτ (Σ) = Wτ.

(ii) We have X ∈ Wτ (C (λ)) if and only if ΣX ∈ Dλτw∞ = Wλτ. This means that X ∈ Wλτ (Σ)

and by (i) the condition λτ ∈ Γ implies Wλτ (Σ) = Wλτ. Then Wτ (C (λ)) = Wλτ and C (λ) is

bijective from Wλτ to Wτ. Since ∆(λ) = C (λ)
−1 we conclude ∆(λ) bijective from Wτ to Wλτ and

Wλτ (∆(λ)) = Wτ. We deduce that for τ ∈ Γ, Wτ (∆(λ)) = Wτ/λ.

(iii) a) By Lemma 2.1 with ρn = τn+1/τn and λn = n we have ∆
+
ρ = D1/τ∆

+Dτ and ∆
+ is

bijective from Dτw∞ = Wτ into itself for τ ∈ Γ
+ and it is the same for Σ+. Now the equation

Σ
+ X = Y for Y ∈ Wτ is equivalent to

∞∑

m=n

xm = yn for all n. (8)

We deduce (8) has a unique solution X = ( yn − yn+1)n≥1 = ∆
+Y ∈ Wτ and Wτ

(
Σ
+
)
= Wτ.

b) We have

Wτ
(
C

+
(λ)

)
=

{
X : Σ

+
D1/λ X ∈ Wτ

}
= DλWτ

(
Σ
+
)
.

Now as we have seen above since τ ∈ Γ+ we get Wτ
(
Σ
+
)
= Wτ and

Wτ
(
C

+
(λ)

)
= DλWτ

(
Σ
+
)
= DλWτ = Wλτ.

This gives the conclusion.



128 Bruno de Malafosse & Vladimir Rakočević
CUBO

12, 3 (2010)

4 Calculations in New Sequence Spaces

4.1 The sets [C,∆]Wτ , [C, C]Wτ ,
[
C+,∆

]
Wτ

,
[
C+, C

]
Wτ

and
[
C+, C+

]
Wτ

.

In [4], were defined and studied the sets

[A1, A2] =
[
A1 (λ) , A2

(
µ
)]
=

{
X ∈ s : A1 (λ)

(∣∣A2
(
µ
)

X
∣∣) ∈ Dτl∞

}

where |X| = (|xn|)n≥1, A1 and A2 of the form C (ξ), C
+ (ξ), ∆(ξ), or ∆+ (ξ) for ξ ∈ U+. It was

given necessary conditions to get
[
A1 (λ) , A2

(
µ
)]

in the form sγ.

Similarly in the following we will put

[A1, A2]Wτ =
[
A1 (λ) , A2

(
µ
)]

Wτ
=

{
X ∈ s : A1 (λ)

(∣∣A2
(
µ
)

X
∣∣) ∈ Wτ

}

for λ, µ, τ ∈ U+. We can explicitly write the previous sets [A1, A2]Wτ as follows.

[C,∆]Wτ =

{
X : sup

n

(
1

n

n∑

m=1

1

λmτm

m∑

k=1

∣∣µk xk −µk−1 xk−1
∣∣
)
< ∞

}
,

[C, C]Wτ =

{
X : sup

n

(
1

n

n∑

m=1

(
1

λmτm

m∑

k=1

1

µk

∣∣∣∣∣
k∑

i=1

xi

∣∣∣∣∣

))
< ∞

}
,

[
C

+
,∆

]
Wτ

=

{
X : sup

n

(
1

n

n∑

m=1

(
1

τm

∞∑

k=m

1

λk

∣∣µk xk −µk−1 xk−1
∣∣
))

< ∞

}
,

[
C

+
, C

]
Wτ

=

{
X : sup

n

(
1

n

n∑

m=1

(
1

τm

∞∑

k=m

1

λk

1

µk

∣∣∣∣∣
k∑

i=1

xi

∣∣∣∣∣

))
< ∞

}
,

[
C

+
, C

+
]
Wτ

=

{
X : sup

n

(
1

n

n∑

m=1

(
1

τm

∞∑

k=m

1

λk

∣∣∣∣∣
∞∑

i=k

xi

µi

∣∣∣∣∣

))
< ∞

}
.

Note that if λn = µn for all n we get the well known set of sequences that are strongly

bounded [C,∆]We = c∞ (λ). We can state the following.

Theorem 4.1. Let λ, µ,τ ∈ U+.

(i) If λτ ∈ Γ then

[C,∆]Wτ = Wλτ/µ;

(ii) if λτ, λµτ ∈ Γ then

[C, C]Wτ = Wλµτ;

(iii) if τ ∈ Γ+ and λτ ∈ Γ then
[
C

+
,∆

]
Wτ

= Wλτ/µ;



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Calculations in New Sequence Spaces ... 129

(iv) if τ ∈ Γ+ and λµτ ∈ Γ then

[
C

+
, C

]
Wτ

= Wλµτ;

(v) if τ, λτ ∈ Γ+ then
[
C

+
, C

+
]
Wτ

= Wλµτ.

Proof. In the following we will use the fact that for any ξ ∈ U+ we have |X| ∈ Wξ if and only if

X ∈ Wξ.

(i) We have C (λ)
(∣∣∆

(
µ
)

X
∣∣) ∈ Wτ if and only if

∣∣∆
(
µ
)

X
∣∣ ∈ Wτ (C (λ)) and by Proposition 3.1,

since λτ ∈ Γ we get Wτ (C (λ)) = Wλτ. Then by Proposition 3.1 (ii) we have Wλτ
(
∆

(
µ
))

= Wλτ/µ

and we conclude ∆
(
µ
)
X ∈ Wλτ if and only if X ∈ Wλτ

(
∆

(
µ
))

= Wλτ/µ, that is [C,∆]Wτ = Wλτ/µ.

(ii) Here we have C (λ)
(∣∣C

(
µ
)

X
∣∣) ∈ Wτ if and only if

∣∣C
(
µ
)

X
∣∣ ∈ Wτ (C (λ)); and since λτ ∈ Γ

by Proposition 3.1 we have Wτ (C (λ)) = Wλτ. So X ∈ [C, C]Wτ if and only if C
(
µ
)

X ∈ Wλτ, that

is X ∈ Wλτ
(
C

(
µ
))

. Then by Proposition 3.1 (ii) a) λµτ ∈ Γ implies Wλτ
(
C

(
µ
))

= Wλµτ and we

have shown (ii).

(iii) For any given X ∈
[
C+,∆

]
Wτ

we have ∆
(
µ
)

X ∈ Wτ
(
C+ (λ)

)
and for τ ∈ Γ+ we have

Wτ
(
C+ (λ)

)
= Wλτ. Now the condition λτ ∈ Γ implies X ∈

[
C+,∆

]
Wτ

if and only if X ∈ Wλτ
(
∆

(
µ
))
=

Wλτ/µ and we have shown (iii).

(iv) Let X ∈
[
C+, C

]
Wτ

. We have τ ∈ Γ+ implies Wτ
(
C+ (λ)

)
= Wλτ and so X ∈

[
C+, C

]
Wτ

if and only if C
(
µ
)

X ∈ Wλτ. Now since λµτ ∈ Γ we have Wλτ
(
C

(
µ
))

= Wλµτ and we conclude[
C+, C

]
Wτ

= Wλµτ.

(v) As above X ∈
[
C+, C+

]
Wτ

if and only if C+
(
µ
)

X ∈ Wτ
(
C+ (λ)

)
and the condition τ ∈ Γ+

implies Wτ
(
C+ (λ)

)
= Wλτ. Since λτ ∈ Γ

+ we conclude Wλτ
(
C+

(
µ
))
= Wλµτ that is

[
C+, C+

]
Wτ

=

Wλµτ.

Now we are led to study sets of the form [∆, A2 ]Wτ for A2 ∈
{
∆,∆, C+

}
.

4.2 The sets [∆,∆]Wτ , [∆, C]Wτ and
[
∆, C+

]
Wτ

Using the convention µ0 = 0, and the notation ∆
(
µ
)
xm = µm xm − µm−1 xm−1 for m ≥ 1 we

explicitly have

[∆,∆]Wτ =

{
X : sup

n

(
1

n

n∑

m=1

1

τm

∣∣λm
∣∣∆

(
µ
)
xm

∣∣ −λm−1
∣∣∆

(
µ
)
xm−1

∣∣∣∣
)
< ∞

}
,

[∆, C]Wτ =

{
X : sup

n

(
1

n

n∑

m=1

1

τm

∣∣∣∣∣λm

∣∣∣∣∣
1

µm

m∑

k=1

xk

∣∣∣∣∣ −λm−1

∣∣∣∣∣
1

µm−1

m−1∑

k=1

xk

∣∣∣∣∣

∣∣∣∣∣

)
< ∞

}
,



130 Bruno de Malafosse & Vladimir Rakočević
CUBO

12, 3 (2010)

[
∆, C

+
]
Wτ

=

{
X : sup

n

(
1

n

n∑

m=1

1

τm

∣∣∣∣∣λm

∣∣∣∣∣
∞∑

k=m

xk

µk

∣∣∣∣∣ −λm−1

∣∣∣∣∣
∞∑

k=m−1

xk

µk

∣∣∣∣∣

∣∣∣∣∣

)
< ∞

}
.

As a direct consequence of Proposition 3.1 we also obtain the following results.

Theorem 4.2. Let λ, µ, τ ∈ U+. Then

(i) If τ, τ/λ ∈ Γ then

[∆,∆]Wτ = Wτ/λµ.

(ii) If τ, τµ/λ ∈ Γ then

[∆, C]Wτ = Wτµ/λ.

(iii) If τ, τ/λ ∈ Γ+ then
[
∆, C

+
]
Wτ

= Wτµ/λ.

Proof. (i) Let X ∈ [∆,∆]Wτ . Since τ ∈ Γ we have Wτ (∆(λ)) = Wτ/λ and ∆(λ)
∣∣∆

(
µ
)

X
∣∣ ∈ Wτ means

∆
(
µ
)

X ∈ Wτ/λ. We conclude Wτ/λ
(
∆

(
µ
))
= Wτ/λµ for τ/λ ∈ Γ.

(ii) Reasoning as above since τ ∈ Γ we have X ∈ [∆, C]Wτ if and only if C
(
µ
)

X ∈ Wτ/λ. We

conclude since the condition τµ/λ ∈ Γ implies Wτ/λ
(
C

(
µ
))
= Wτµ/λ.

(iii) Here under the conditions τ, τ/λ ∈ Γ+, we have X ∈
[
∆, C+

]
Wτ

if and only if X ∈

Wτ/λ
(
C+

(
µ
))
= Wτµ/λ.

The previous results can be applied to the case when w∞ is replaced by w
0.

4.3 The sets [A1, A2]W 0τ

Using the Banach algebra
(
w0 (λ) , w0 (λ)

)
we get similar results to those given above replacing

w∞ (λ) by w
0 (λ) and Wτ by W

0
τ = Dτw

0. Note that X ∈ W 0τ if and only if

1

n

n∑

m=1

|xm|

τm
→ 0 (n → ∞) .

By [17, Theorem 4.3.6, p. 52] the set W 0τ is a BK space with AK normed by ‖‖Wτ . So we can

state the following.

Proposition 4.3. Let λ, µ ∈ U+.

(i) If λτ ∈ Γ then [C,∆]W0τ
= W 0

λτ/µ
;

(ii) if λτ, λµτ ∈ Γ then [C, C]W0τ
= W 0

λµτ
;

(iii) if τ ∈ Γ+ and λτ ∈ Γ then
[
C+,∆

]
W0τ

= W 0
λτ/µ

;



CUBO
12, 3 (2010)

Calculations in New Sequence Spaces ... 131

(iv) if τ ∈ Γ+ and λµτ ∈ Γ then
[
C+, C

]
W0τ

= W 0
λµτ

;

(v) if τ, λτ ∈ Γ+ then
[
C+, C+

]
W0τ

= W 0
λµτ

;

(vi) if τ, τ/λ ∈ Γ then [∆,∆]W0τ
= W 0

τ/λµ
;

(vii) if τ, τµ/λ ∈ Γ then [∆, C]W0τ
= W 0

τµ/λ
;

(viii) if τ, τ/λ ∈ Γ+ then
[
∆, C+

]
W0τ

= W 0
τµ/λ

.

We immediatly get the next remark.

Remark 4.4. It can easily be seen that in Proposition 4.3 each of the sets [A1, A2]W0τ
is equal

to W 0τ (A1 A2). This result is a direct consequence of the previous proofs and of the fact that W
0
τ

is of absolute type, that is |X| ∈ W 0τ if and only if X ∈ W
0
τ .

These results can be applied to statistical convergence.

5 Application to A−Statistical Convergence

In this section we will give conditions to have xk → L (S (A)) where A is either of the infinite

matrices D1/τC (λ) C
(
µ
)
, D1/τ∆(λ)∆

(
µ
)
, or D1/τ∆(λ) C

(
µ
)
. Then we give conditions to have

xk → 0 (S (A)) where A is either of the operators D1/τC
+ (λ)∆

(
µ
)
, D1/τC

+ (λ) C
(
µ
)
, D1/τC

+ (λ)

C+
(
µ
)

and D1/τ∆(λ) C
+

(
µ
)
.

The sequence X = (xn)n≥1 is said to be statiscally convergent to the number L if

lim
n→∞

1

n
|{k ≤ n : |xk − L| ≥ ε}| = 0 for all ε > 0,

where the vertical bars indicate the number of elements in the enclosed set. In this case we

will write xk → L (S) or st − lim X = L.

Let A ∈ (E, F) for given L ∈ C and for every ε > 0 we will use the notation

Iε (A) = {k ≤ n : |[A X ]k − L| ≥ ε} ,

(where we assume that every series [A X ]k = Ak (X ) =
∑

∞
m=1

akm xm for k ≥ 1 is convergent).

We will say that X = (xn)n≥1 is A− statistically convergent to L if for every ε > 0,

lim
n→∞

1

n
|Iε (A)| = 0.

Then we will write xk → L (S ( A)) and for A = I, xk → L (S (I)) means that st − lim X = L, (cf.

[6]).

Now we require a lemma where we will put T−1 e = l̃ = (l n)n≥1 for given triangle T, that

is T = (tnm)n,m≥1 with tnn 6= 0 and tnm = 0 if m > n for all n, m.



132 Bruno de Malafosse & Vladimir Rakočević
CUBO

12, 3 (2010)

We can state the following.

Lemma 5.1. If X − Ll̃ ∈ w0 (T) then xk is T− statistically convergent to L.

Proof. The condition X − Ll̃ ∈ w0 (T) means that T
(
X − Ll̃

)
∈ w0. Since

T X − Le = T
(
X − LT

−1
e
)
= T

(
X − Ll̃

)

for any ε > 0 we have

yn =
1

n

n∑

k=1

|[T X ]k − L| =
1

n

n∑

k=1

∣∣[T
(
X − Ll̃

)]
k

∣∣

≥
1

n

∑

k∈Iε(T)

∣∣[T
(
X − Ll̃

)]
k

∣∣

≥
1

n

∑

k∈Iε(T)

ε

≥
ε

n
|{k ≤ n : |[T X ]k − L| ≥ ε}|.

We conclude that X − Ll̃ ∈ w0 (T) implies yn → 0 (n → ∞) and xk → L (S (T)).

We are led to state the next results.

Theorem 5.2. (i) Let λτ, λτµ ∈ Γ. If

lim
n→∞

1

n

n∑

k=1

∣∣xk − L
[
λkµkτk +

(
µk−1 +µk

)
λk−1τk−1 −λk−2µk−2τk−2

]∣∣
λkµkτk

= 0 (9)

then xk → L
(
S

(
D1/τC (λ) C

(
µ
)))

, that is for every ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣
1

λkτk

k∑

i=1

1

µi

(
i∑

j=1

x j

)
− L

∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0.

(ii) Let τ, τ/λ ∈ Γ. If

lim
n→∞

1

n

n∑

k=1

λkµk

τk

∣∣∣∣∣xk − L
(

1

µk

k∑

i=1

1

λi

i∑

j=1

τj

)∣∣∣∣∣ = 0

then xk → L
(
S

(
D1/τ∆(λ)∆

(
µ
)))

, that is for every ε > 0

lim
n→∞

1

n

∣∣∣∣
{

k ≤ n :

∣∣∣∣
1

τk

[
λk∆

(
µ
)
xk −λk−1∆

(
µ
)
xk−1

]
− L

∣∣∣∣ ≥ ε
}∣∣∣∣ = 0.

(iii) Let τ, τµ/λ ∈ Γ. If

lim
n→∞

1

n

n∑

k=1

λk

µkτk

∣∣∣∣∣xk − L
[(

µk

λk
−

µk−1

λk−1

)
k−1∑

i=1

τi +
µk

λk
τk

]∣∣∣∣∣ = 0



CUBO
12, 3 (2010)

Calculations in New Sequence Spaces ... 133

then xk → L
(
S

(
D1/τ∆(λ) C

(
µ
)))

, that is for every ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣
1

τk

[(
λk

µk
−

λk−1

µk−1

)
k−1∑

i=1

xi +
λk

µk
xk

]
− L

∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0.

Proof. (i) First by Proposition 4.3 (ii) and Remark 4.4, we easily see that for λτ, λτµ ∈ Γ we

have W 0τ
(
C (λ) C

(
µ
))
= W 0

λµτ
. Then putting T = D1/τC (λ) C

(
µ
)

we get

w
0

(T) = W
0
τ

(
C (λ) C

(
µ
))
= W

0
λµτ

. (10)

Then l̃ = T−1 e = ∆
(
µ
)
∆(λ) Dτ e for each n with

l n =
[
∆

(
µ
)
∆(λ) Dτ e

]
n
= λnµnτn +

(
µn−1 +µn

)
λn−1τn−1 −λn−2µn−2τn−2 (11)

Using (10) and (11) we see that condition (9) is equivalent X − Ll̃ ∈ w0 (T). We conclude by

Lemma 5.1 that xk → L (S (T)). This completes the proof of (i).

(ii) By Proposition 4.3 (vi) and Remark 4.4, since τ, τ/λ ∈ Γ we have W 0τ
(
∆(λ)∆

(
µ
))

= W 0
τ/λµ

.

Then putting T′ = D1/τ∆(λ)∆
(
µ
)

we get

w
0
(
T

′
)
= W

0
τ

(
∆(λ)∆

(
µ
))
= W

0
τ/λµ

. (12)

Since l̃′ = T′−1 e = C
(
µ
)
C (λ) Dτ e we have

l
′
n =

[
C

(
µ
)
C (λ) Dτ e

]
n
=

1

µn

n∑

i=1

1

λi

(
i∑

j=1

τj

)
for all n.

By Lemma 5.1 we conclude xk → L
(
S

(
D1/τ∆(λ)∆

(
µ
)))

for all X with

lim
n→∞

1

n

n∑

k=1

∣∣xk − Ll′k
∣∣ λkµk

τk
= 0

This shows (ii).

(iii) Again by Proposition 4.3 (vii) and Remark 4.4, since τ, τµ/λ ∈ Γ we have W 0τ
(
∆(λ) C

(
µ
))
=

W 0
τµ/λ

. Then putting T
′′

= D1/τ∆(λ) C
(
µ
)

we get

w
0
(
T

′′
)
= W

0
τ

(
∆(λ) C

(
µ
))
= W

0
τµ/λ

. (13)

Writing l̃′′ = T
′′
−1 e = ∆

(
µ
)
C (λ) Dτ e we successively get

Dτ e = (τn)n≥1 , C (λ) Dτ e =

((
n∑

i=1

τi

)
/λn

)

n≥1

and

∆
(
µ
)
C (λ) Dτ e =

(
µn

λn

n∑

i=1

τi −
µn−1

λn−1

n−1∑

i=1

τi

)

n≥1

.



134 Bruno de Malafosse & Vladimir Rakočević
CUBO

12, 3 (2010)

So for each n we have

l
′′
n =

[
∆

(
µ
)
C (λ) Dτ e

]
n
=

(
µn

λn
−

µn−1

λn−1

)
n−1∑

i=1

τi +
µn

λn
xk.

We conclude that for every X with

lim
n→∞

1

n

n∑

k=1

∣∣xk − Ll′′k
∣∣ λk
µkτk

= 0

then xk → L
(
S

(
T′′

))
. Finally we easily get

[
T

′′
X

]
n

=
1

τn

(
λn

µn

n∑

i=1

xi −
λn−1

µn−1

n−1∑

i=1

xi

)

=
1

τn

[(
λn

µn
−

λn−1

µn−1

)
n−1∑

i=1

xi +
λn

µn
xn

]
.

This shows (iii).

We are led to illustrate the previous results with some examples where we must have in

mind that the condition xk/τk → 0 (k → ∞) implies X ∈ W
0
τ .

Example 5.3. The condition

lim
n→∞

1

n

n∑

k=1

∣∣∣∣
xk

2k
−

7

4
L

∣∣∣∣ = 0

for given L ∈ C implies xk → L
(
S

(
D(n/2n )n C1Σ

))
, that is, for each ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣
1

2k

k∑

i=1

i∑

j=1

x j − L

∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0. (14)

Indeed it is enough to apply Theorem 5.2 (i) with λk = k, τk = 2
k/k and µk = 1 for all k. Note

that if xk/2
k
→ 7L/4 (k → ∞) then xk → L

(
S

(
D(n/2n )n C1Σ

))
.

We can also state the next application.

Example 5.4. If limn→∞ (1/n)
∑n

k=1
|xk|/k2

k
= 0 then xk → L

(
S

(
D(2−n )n ∆C1

))
, that is for each

ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣
1

2k

(
1

k
−

1

k − 1

)
k−1∑

i=1

xi +
1

k
xk

∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0.

This result is a direct consequence of Theorem 5.2 (iii) with λk = 1, τk = 2
k and µk = k for all

k. Again note that we have xk → L
(
S

(
D(2−n )n ∆C1

))
if xk/k2

k
→ 0 (k → ∞).



CUBO
12, 3 (2010)

Calculations in New Sequence Spaces ... 135

In the following we will use the previous Proposition 4.3 and the expressions of

W 0τ
(
C+ (λ)∆

(
µ
))

=
[
C+,∆

]
W0τ

, W 0τ
(
C+ (λ) C

(
µ
))

=
[
C+, C

]
W0τ

, W 0τ
(
C+ (λ) C+

(
µ
))

=
[
C+, C+

]
W0τ

and W 0τ
(
∆(λ) C+

(
µ
))

=
[
∆, C+

]
W0τ

. We now require a lemma which is a direct consequence

of Lemma 5.1.

Lemma 5.5. Let A be an infinite matrix. If X ∈ w0 (A) then

xk → 0 (S (A)) .

we deduce the next results.

Theorem 5.6. (i) Let τ ∈ Γ+ and λτ ∈ Γ. If

lim
n→∞

1

n

n∑

k=1

|xk|

λkτk
µk = 0 (15)

then xk → 0
(
S

(
D1/τC

+ (λ)∆
(
µ
)))

, that is for every ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣
1

τk

∞∑

i=k

µi xi −µi−1 xi−1

λi

∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0. (16)

(ii) Let τ ∈ Γ+ and λµτ ∈ Γ. If

lim
n→∞

1

n

n∑

k=1

|xk|

λkµkτk
= 0 (17)

then xk → 0
(
S

(
D1/τC

+ (λ) C
(
µ
)))

, that is for every ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣
1

τk

∞∑

i=k

1

λi

(
1

µi

i∑

j=1

x j

)∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0. (18)

(iii) Let τ, λτ ∈ Γ+. If

lim
n→∞

1

n

n∑

k=1

|xk|

λkµkτk
= 0 (19)

then xk → 0
(
S

(
D1/τC

+ (λ) C+
(
µ
)))

, that is for every ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣
1

τk

∞∑

i=k

1

λi

(
∞∑

j=i

x j

µj

)∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0. (20)

(iv) Let τ, τ/λ ∈ Γ+. If

lim
n→∞

1

n

n∑

k=1

λk |xk|

µkτk
= 0

then xk → 0
(
S

(
D1/τ∆(λ) C

+
(
µ
)))

, that is for every ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

1

τk

∣∣∣∣∣(λk −λk−1)
∞∑

i=k−1

xi

µi
+

λk

µk
xk

∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0. (21)



136 Bruno de Malafosse & Vladimir Rakočević
CUBO

12, 3 (2010)

Proof. (i) Condition (15) implies X ∈ W 0
λτ/µ

and by Proposition 4.3 and Remark 4.4 since τ ∈ Γ+

and λτ ∈ Γ we have W 0
λτ/µ

= W 0τ
(
C+ (λ)∆

(
µ
))

and X ∈ W 0τ
(
C+ (λ) ∆

(
µ
))

. Now it can be easily

seen that
[
D1/τC

+
(λ)∆

(
µ
)]

n
=

1

τn

∞∑

i=n

µi xi −µi−1 xi−1

λi
,

so by Lemma 5.5 with A = D1/τC
+ (λ)∆

(
µ
)

we conclude xk → 0
(
S

(
D1/τC

+ (λ)∆
(
µ
)))

. This

shows (i).

(ii) Here condition (17) means X ∈ W 0
λµτ

and by Proposition 4.3 and Remark 4.4 since

τ ∈ Γ+ and λµτ ∈ Γ we have W 0
λµτ

= W 0τ
(
C+ (λ) C

(
µ
))

and X ∈ W 0τ
(
C+ (λ) C

(
µ
))

. Now since

[
D1/τC

+
(λ) C

(
µ
)]

n
=

1

τn

∞∑

i=n

1

λi

(
1

µi

i∑

j=1

x j

)
,

by Lemma 5.5 where A′ = D1/τC
+ (λ) C

(
µ
)
, we conclude xk → 0

(
S

(
D1/τC

+ (λ) C
(
µ
)))

. So we

have shown (ii).

(iii) can be obtained reasoning as above with A′′ = D1/τC
+ (λ) C+

(
µ
)

and so xk → 0(
S

(
D1/τC

+ (λ) C+
(
µ
)))

.

(iv) can also be obtained similarly. It is enough to put A′′′ = D1/τ∆(λ) C
+

(
µ
)
. An elemen-

tary calculation gives

[
A
′′′

X
]

k
=

1

τk

[
(λk −λk−1)

∞∑

i=k−1

xi

µi
+

λk

µk
xk

]

and we conclude that xk → 0
(
S

(
D1/τ∆(λ) C

+
(
µ
)))

, that is (21).

We can state the next example

Example 5.7. for each ε > 0 and for every X ∈ W 0
3/2

we have xk → 0
(
S

(
D(2n )n Σ

+C ((3n)n)
))

,

that is

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣2
k

∞∑

i=1

1

3i

(
i∑

j=1

x j

)∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0. (22)

It is enough to apply Theorem 5.6 (ii) with τk = 2
−k, µk = 3

k and λk = 1 for all k. So if

(2/3)
k xk → 0 (k → ∞) then (22) holds.

We also have the next example.

Example 5.8. From Theorem 5.6 (iii) with λk = µk = k and τk = 2
−k the condition

lim
n→∞

1

n

n∑

k=1

2
k |xk|

k2
= 0



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12, 3 (2010)

Calculations in New Sequence Spaces ... 137

implies xk → 0
(
S

(
D(2n )n C1C

+
1

))
that is, for each ε > 0

lim
n→∞

1

n

∣∣∣∣∣

{
k ≤ n :

∣∣∣∣∣2
k

∞∑

i=k

1

i

(
∞∑

j=i

x j

j

)∣∣∣∣∣ ≥ ε
}∣∣∣∣∣ = 0. (23)

As in the previous cases (23) holds if 2k xk/k
2
→ 0 (k → ∞).

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138 Bruno de Malafosse & Vladimir Rakočević
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