• Sonuç bulunamadı

An inclusion theorem for double Cesaro matrices over the space of absolutely k-convergent double series

N/A
N/A
Protected

Academic year: 2023

Share "An inclusion theorem for double Cesaro matrices over the space of absolutely k-convergent double series"

Copied!
5
0
0

Yükleniyor.... (view fulltext now)

Tam metin

(1)

Contents lists available atScienceDirect

Applied Mathematics Letters

journal homepage:www.elsevier.com/locate/aml

An inclusion theorem for double Cesàro matrices over the space of absolutely k-convergent double series

Ekrem Savaş

a,

, B.E. Rhoades

b

aDepartment of Mathematics, Istanbul Ticaret University, Üsküdar, Istanbul, Turkey

bDepartment of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA

a r t i c l e i n f o

Article history:

Received 15 July 2008

Received in revised form 5 February 2009 Accepted 30 March 2009

Keywords:

Absolute summability factors Double triangular summability

a b s t r a c t

In this work it is shown that if C(α, β)and C(γ , δ)are double Cesàro matrices with α ≥ γ > −1, β ≥ δ > −1, then any double series that is| C(γ , δ) |k-summable is also|C(α, β) |k-summable, k≥1.

© 2009 Published by Elsevier Ltd

A doubly infinite Cesàro matrix C

(α, β)

is a doubly infinite Hausdorff matrix with entries hmnij

=



m+α−i1 mi

 

n+β−j1 nj



m+α α

 

n+β β

 , α, β ≥

0

.

A series

P P

amnwith partial sums smnis said to be absolutely C

(α, β)

-summable, of order k

1, if

X

m=1

X

n=1

(

mn

)

k1

|

11

σ

m(α,β)1,n1

|

k

< ∞,

(1)

where

σ

mn(α,β)

:=

1

m+α α

 

n+β β



m

X

i=0 n

X

j=0



m

+ α −

i

1 m

i

 

n

+ β −

j

1 n

j



sij

and where, for any double sequence

{

umn

}

, and for any fourfold sequence

{

amnij

}

, we define

11umn

=

umn

um+1,n

um,n+1

+

um+1,n+1

,

11amnij

=

amnij

am+1,n,i,j

am,n+1,i,j

+

am+1,n+1,i,j

,

ijamnij

=

amnij

am,n,i+1,j

am,n,i,j+1

+

am,n,i+1,j+1

,

i 0amnij

=

amnij

am,n,i+1,j

,

and

0jamnij

=

amnij

am,n,i,j+1

.

The one-dimensional version of(1)appears in [1].

Corresponding author.

E-mail addresses:ekremsavas@yahoo.com,esavas@iticu.edu(E. Savaş),rhoades@indiana.edu(B.E. Rhoades).

0893-9659/$ – see front matter©2009 Published by Elsevier Ltd doi:10.1016/j.aml.2009.03.015

(2)

For brevity, let H

=

C

(α, β)

. Associated with H are two matrices H andH defined by

ˆ

h

¯

mnij

=

m

X

µ=i n

X

ν=j

hmnµν

,

0

i

m

,

0

j

n

,

m

,

n

=

0

,

1

, . . . ,

and

h

ˆ

mnij

=

11h

¯

m1,n1,i,j

,

0

i

m

,

0

j

n

,

m

,

n

=

1

,

2

, . . . .

It is easily verified thath

ˆ

0000

= ¯

h0000

=

h0000. In [2] it was shown that, if A is a double triangular matrix with the row sums in each direction constant (i.e.,

P

m

µ=0amnµν

= P

m1

µ=0am1,nµν

=

a

(

n

, ν)

, and

P

n

ν=0amnµν

= P

n1

ν=0am,n1,µ,ν

=

b

(

m

, µ)

), then

ˆ

amnij

=

i1

X

µ=0 j1

X

ν=0

11am1,n1,µ,ν

.

(2)

Using the standard interpretation that the right hand side of(2)is zero whenever i or j is zero, formula(2)remains valid for these choices. (Under the hypotheses that the row sums are constant in each direction, a direct calculation verifies that a

ˆ

mn0j

= ˆ

amni0

= ˆ

amn00

=

0.) Eq.(2)is certainly true for double Cesàro matrices, since the row sums in each direction are 1.

Thereforeh

ˆ

mni0

= ˆ

hmn0j

=

0.

σ

mn(α,β)

=

m

X

i=0 n

X

j=0

hmnijsij

=

m

X

i=0 n

X

j=0 i

X

µ=0 j

X

ν=0

hmnijaµν

=

m

X

µ=0 n

X

ν=0 m

X

i=µ n

X

j=ν hmnijaµν

=

m

X

µ=0 n

X

ν=0

h

¯

mnµνaµν

,

and a direct calculation verifies that

Xmn

:=

11

σ

m(α,β)1,n1

=

m

X

i=1 n

X

j=1

h

ˆ

mnijaij

,

(3)

since

h

¯

m1,n1,m

=

hm1,n1,µ,n

= ˆ

hm,n1,µ,n

= ˆ

hm1,n,m,n

=

0

.

Theorem 1. Let

α ≥ γ > −

1

, β ≥ δ > −

1, and

P

m

P

namnbe a double series with partial sums smn. If

P

m

P

namnis

|

C

(γ , δ)|

k-summable, then it is also

|

C

(α, β)|

k-summable.

We will require the following lemmas in the proof of the theorem.

Lemma 1.

τ

mn(α,β)

=

mn11

σ

m(α,β)1,n1, where

τ

mn(α,β)

:=

1

m+α α

 

n+β β



m

X

i=0 n

X

j=0



m

+ α −

i

1 m

i

 

n

+ β −

j

1 n

j



ijaij

Proof. By analogy to formula (2.1.3) of [1] we may write 1



n+β β



n

X

j=0



n

+ β −

j

1 n

j



jaij

=

n

in

− σ

i,n1

),

where

σ

in

=

1



n+β β



n

X

j=0



n

+ β −

j

1 n

j



j

X

ν=0

aiν

.

Define bij

= P

j

ν=0aνj. Then we may write 1

m+α α



m

X

j=0



m

+ α −

j

1 m

j



jbij

=

m

( ¯σ

mj

− ¯ σ

m1,j

),

(3)

where

σ ¯

mj

=

1

m+α α



m

X

i=0



m

+ α −

i

1 m

i



i

X

µ=0

bµj

=

1

m+α α



m

X

i=0



m

+ α −

i

1 m

i



sij

.

Thus

τ

mn(α,β)

=

n

m+α α



m

X

i=0



m

+ α −

i

1 m

i



i

in

− σ

i,n1

)

=

n

1



n+β β



n

X

j=0



n

+ β −

j

1 n

j



1

m+α α



m

X

i=0



m

+ α −

i

1 m

i



ibij

− 

1

n+β−1 β



n1

X

j=0



n

+ β −

j

2 n

j

1



1

m+α α



m

X

i=0



m

+ α −

i

1 m

i



ibij

=

mn

1



n+β β



n

X

j=0



n

+ β −

j

1 n

j

 (

1

m+α α



m

X

i=0



m

+ α −

i

1 m

i



sij

− 

1

m+α−1 α



m1

X

i=0



m

+ α −

i

2 m

i

1



sij

− 

1

n+β−1 β



n1

X

j=0



n

+ β −

j

2 n

j

1

 (

1

m+α α



m

X

i=0



m

+ α −

i

1 m

i



sij

1



m+α−1 α



m1

X

i=0



m

+ α −

i

2 m

i

1



sij

 =

mn11

σ

m(α,β)1,n1

.



Lemma 2. For any

, η ≥

0

, α, β > −

1,

τ

mn(α+,β+η)

=

1

m+α+

α+

 

n+β+η β+η



m

X

µ=1 n

X

ν=1



m

+  − µ −

1 m

− µ

 

n

+ η − ν −

1 n

− ν

  µ + α α

  ν + β β

 τ

µν(α,β)

.

Proof. The summations on the right hand side can be written in the form L1

:=

m

X

µ=1



m

+  − µ −

1 m

− µ



n

X

ν=1



n

+ η − ν −

1 n

− ν



µ

X

i=1

X

ν

j=1

 µ + α −

i

1

µ −

i

  ν + β −

j

1

ν −

j



ijaij

=

m

X

i=1 n

X

j=1

ijaij

m

X

µ=i



m

+  − µ −

1 m

− µ

  µ + α −

i

1

µ −

i



n

X

ν=j



n

+ η − ν −

1 n

− ν

  ν + β −

j

1

ν −

j

 .

Using identity (2.1.1) from [1],

m

X

µ=i



m

+  − µ −

1 m

− µ

  µ + α −

i

1

µ −

i



=

mi

X

t=0



m

+  −

t

i

1 m

t

i

 

t

+ α −

1 t



=



m

+ α +  −

i

1 m

i

 .

Similarly,

n

X

ν=j



n

+ η − ν −

1 n

− ν

  ν + β −

j

1

ν −

j



=



n

+ β + η −

j

1 n

j



.

(4)

Thus L1

m+α+

α+

 

n+β+η β+η

 =

1

m+α+

α+

 

n+β+η β+η



m

X

i=1 n

X

j=1



m

+ α +  −

i

1 m

i

 

n

+ β + η −

j

1 n

j



ijaij

= τ

mn(α+,β+η)

.



To prove the theorem, we shall first prove that

(

X

m=1

X

n=1

(

mn

)

1

| τ

mn(α+,β+η)

|

k

)

1/k

A

X

m=1

X

n=1

(

mn

)

1

| τ

mn(α,β)

|

k

!

1/k

,

(4)

where A is a positive constant depending only on k

, α, β, , η

, and where

, η ≥

0

, α, β > −

1.

Let us have

λ >

1 to be chosen later, and let

λ

0denote the conjugate index of

λ

. Using the identity inLemma 2and Hölder’s inequality we have

| τ

mn(α+,β+η)

| ≤

1

m+α+

α+

 

n+β+η β+η



m

X

µ=1 n

X

ν=1



m

+  − µ −

1 m

− µ

 

n

+ η − ν −

1 n

− ν



1

×

 µ + α α

  ν + β β



1/k

| τ

µν(α,β)

|

×



m

+  − µ −

1 m

− µ

 

n

+ η − ν −

1 n

− ν



10

 µ + α α

  ν + β β



1/k0

1

m+α+

α+

 

n+β+η β+η





m

X

µ=1 n

X

ν=1



m

+  − µ −

1 m

− µ



×



n

+ η − ν −

1 n

− ν



k

 µ + α α

  ν + β β



| τ

µν(α,β)

|

k



1/k

×



m

X

µ=1 n

X

ν=1



m

+  − µ −

1 m

− µ

 

n

+ η − ν −

1 n

− ν



k00

 µ + α α

  ν + β β



1/k0

.

Now choose

λ

so that k0

( −

1

)/λ

0and k0

(η −

1

)/λ

0

> −

1. Then

m

X

µ=1 n

X

ν=1



m

+  − µ −

1 m

− µ

 

n

+ η − ν −

1 n

− ν



k00

 µ + α α

  ν + β β



Bmk0(−1)/λ0+α+1nk0(η−1)/λ0+β+1

.

Hence

| τ

mn(α+,β+η)

|

k

Bmσnφ

m

X

µ=1 n

X

ν=1



m

+  − µ −

1 m

− µ

 

n

+ η − ν −

1 n

− ν



k

 µ + α α

  ν + β β



| τ

µν(α,β)

|

k

,

where

σ = −

k

( −

1

)

λ

0

k

(α +

1

)

k0

+

k

(α + ) = α +

1

+

k

( −

1

) λ ,

and

φ = β +

1

+

k

(η −

1

) λ .

Hence

1

mn

| τ

mn(α+,β+η)

|

k

Bm1σn1φ

m

X

µ=1 n

X

ν=1



m

+  − µ −

1 m

− µ

 

n

+ η − ν −

1 n

− ν



k(−1)/λ

× µ

α+1

µ

1

ν

β+1

ν

1

| τ

µν(α,β)

|

k

.

Therefore

M

X

m=1 N

X

n=1

(

mn

)

1

| τ

mn(α+,β+η)

|

k

B

M

X

µ=1 N

X

ν=1

λ

µν

(µν)

1

| τ

µν(α,β)

|

k

,

(5)

where

λ

µν

= µ

α+1

ν

β+1

X

M

m=µ N

X

n=ν

m1σn1φ



m

+

k

( −

1

)/λ − µ

m

− µ

 

n

+

k

(η −

1

)/λ − ν

n

− ν



≤ µ

α+1

ν

β+1

X

m=µ

X

n=ν

m1σn1φ



m

+

k

( −

1

)/λ − µ

m

− µ

 

n

+

k

(η −

1

)/λ − ν

n

− ν



≤ µ

α+1

ν

β+1

X

m=µ

X

n=ν



m

+

k

( −

1

)/λ − µ

m

− µ

 

n

+

k

(η −

1

)/λ − ν

n

− ν

 Z

1 0

(

1

x

)

σxmdx

Z

1 0

(

1

y

)

φyndy

= µ

α+1

ν

β+1

Z

1

0

(

1

x

)

σxµ−1

(

X

m=µ



m

+

k

( −

1

)/λ − µ

m

− µ



xmµ

)

dx

× Z

1

0

(

1

y

)

φyν−1

(

X

n=ν



n

+

k

(η −

1

)/λ − ν

n

− ν



ynν

)

dy

≤ µ

α+1

ν

β+1

Z

1

0

(

1

x

)

σ −k(−1)/λ−1xµ−1dx

Z

1

0

(

1

y

)

φ−k(η−1)/λ−1yν−1dy

B

µ

α+1

µ

k(−1)/λ−σ

ν

β+1

ν

k(η−1)/λ−φ

=

B

,

since

α +

1

+

k

( −

1

)/λ − σ =

0 and

β +

1

+

k

(η −

1

)/λ − φ =

0.

To finish the proof, in(4)set

 = γ − α, η = δ − β

. Define

A

= (

{

smn

} :

X

m=1

X

n=1

(

mn

)

k1

|

amn

|

k

< ∞ )

.

If a doubly infinite matrix A sums every double sequence inA, then A is said to be absolutely kth-power conservative.

Corollary 1. For

α, β ≥

0, C

(α, β)

is absolutely kth-power conservative.

Proof. InTheorem 1set

γ = δ =

0. 

The one-dimensional version ofCorollary 1appears in [1].

Acknowledgements

We wish to thank the referees for their careful reading of the manuscript and for their helpful suggestions. The second author acknowledges support from the Scientific and Technical Research Council of Turkey for the preparation of this work.

References

[1] T.M. Flett, On an extension of absolute summability and some theorems of Littlewood and Paley, Proc. London Math. Soc. 7 (1957) 113–141.

[2] Ekrem Savaş, B.E. Rhoades, Double absolute summability factor theorems and applications, Nonlinear Anal. 69 (2008) 189–200.

Referanslar

Benzer Belgeler

Evaluation management of the capacity of the sport club managers who participated in first degree championship and Iraqi football team in 2000 – 2001, the

It appears to be a case of involuntary manslaughter and the penalty should not be beheading by any means.’ ‘Californiadesi’ would, however, point to the one aspect of the case

In the framework of the current Cooperation Agreement in Science and Technology, The Scientific and Technological Research Council of Turkey (TÜBİTAK) and The

However, Turkey has always been careful on this issue. The 1960 coup G¶HWDW opened many new pages and brought a change in Turkey in many respects. First of all, the new

Therefore, the created optimal models of spatial connections for technical and vocational colleges based on integration approaches of work and education places can create

PRFSs and publication support systems based on bibliometric measures generally use the number of papers published in refereed journals and their impact in terms of

PRFSs and publication support systems based on bibliometric measures generally use the number of papers published in refereed journals and their impact in terms of citations as

• Additional numbers of papers published due to support program in these years were negligible (564, 651 and 826 papers, respectively). • So, the support program had no