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On double Hausdorff summability method

Journal of Inequalities and Applications20142014:240

https://doi.org/10.1186/1029-242X-2014-240

  • Received: 9 December 2013
  • Accepted: 6 May 2014
  • Published:

Abstract

Das (Proc. Camb. Philos. Soc. 67:321-326, 1970) proved that every conservative Hausdorff matrix is absolutely k th power conservative. Savaş and Rhoades (Anal. Math. 35:249-256, 2009) proved the result of Das for double Hausdorff summability. In this paper we will consider the double Endl-Jakimovski (E-J) generalization and we will prove the corresponding result of Savaş and Şevli (J. Comput. Anal. Appl. 11:702-710, 2009) for double E-J generalized Hausdorff matrices.

MSC:40F05, 40G05.

Keywords

  • absolute summability
  • conservative matrix
  • double series
  • Hausdorff matrices

Introduction and background

The basic theory of Hausdorff transformations for double sequences was developed by Adams [1] in 1933. Later a few authors studied double Hausdorff matrices; see e.g. Ramanujan [2] and Ustina [3].

Several generalizations of Hausdorff matrices have been made. One of them is the Endl-Jakimovski, or E-J generalization defined independently by Endl [4] and Jakimovski [5] as follows.

Let β be a real number, let ( μ n ) be a real sequence, and let Δ be the forward difference operator defined by Δ μ k = μ k μ k + 1 , Δ n ( μ k ) = Δ ( Δ n 1 μ k ) . Then the infinite matrix ( H ( β ) , μ n ( β ) ) = ( H β , μ ) = ( h n k ( β ) ) is defined by
h n k ( β ) = { ( n + β n k ) Δ n k μ k ( β ) , 0 k n , 0 , k > n ,
and the associated matrix method is called a generalized Hausdorff matrix and generalized Hausdorff method, respectively. The moment sequence μ n ( β ) is given by
μ n ( β ) = 0 1 t n + β d χ ( t ) ,

where χ ( t ) B V [ 0 , 1 ] . The case β = 0 corresponds to ordinary Hausdorff summability.

In a recent paper [6], the first author jointly with Savaş has extended the result of Das [7] to the E-J matrices; i.e., all conservative E-J matrices are absolutely k th power conservative for k 1 . Thereafter, Savaş and Rhoades [8] proved the result of Das [7] for double Hausdorff summability. In this paper we will consider double E-J generalization and we will prove the corresponding result of [6] for double E-J generalized Hausdorff matrices.

Let m = 0 n = 0 a m n be an infinite double series with real or complex numbers, with partial sums
s m n = i = 0 m j = 0 n a i j .
For any double sequence ( u m n ) we shall define
Δ 11 u m n = u m n u m + 1 , n u m , n + 1 + u m + 1 , n + 1 .
Denote by A k 2 the sequence space defined by
A k 2 = { ( s m n ) m , n = 0 : m = 1 n = 1 ( m n ) k 1 | a m n | k < ; a m n = Δ 11 s m 1 , n 1 }

for k 1 .

A four-dimensional matrix T = ( t m n i j : m , n , i , j = 0 , 1 , ) is said to be absolutely k th power conservative for k 1 , if T B ( A k 2 ) ; i.e., if
m = 1 n = 1 ( m n ) k 1 | Δ 11 s m 1 , n 1 | k < ,
then
m = 1 n = 1 ( m n ) k 1 | Δ 11 t m 1 , n 1 | k < ,
where
t m n = i = 0 j = 0 t m n i j s i j ( m , n = 0 , 1 , ) ,

see e.g. [9, 10] and the references contained therein.

A double Hausdorff matrix has entries
h m n i j = ( m i ) ( n j ) Δ 1 m i Δ 2 n j μ i j ,
where { μ i j } is any real or complex sequence and
Δ 1 m i Δ 2 n j μ i j = s = 0 m i t = 0 n j ( 1 ) i + j ( m i s ) ( n j t ) μ i + s j + t .
For double Hausdorff matrices, the necessary and sufficient condition for H to be conservative is the existence of a function χ ( s , t ) B V [ 0 , 1 ] × [ 0 , 1 ] such that
0 1 0 1 | d χ ( s , t ) | < ,
and
μ m n = 0 1 0 1 s m t n d χ ( s , t ) .

Quite recently, Savaş and Rhoades [8] extended the result of Das [7] to double Hausdorff summability. Their theorem is as follows.

Theorem 1 [8]

Let H be a conservative double Hausdorff matrix. Then H B ( A k 2 ) .

Our purpose is to achieve the result established in [7] for double E-J Hausdorff summability.

Main results

The matrix δ ( α , β ) = ( δ m n i j ( α , β ) ) , whose elements are defined by
δ m n i j ( α , β ) = { ( 1 ) i + j ( m + α m i ) ( n + β n j ) , i m , j n , 0 , otherwise ,

is called a difference matrix, where α and β are real numbers.

Theorem 2 The difference matrix δ ( α , β ) = ( δ m n i j ( α , β ) ) is its own inverse.

Proof Let
a m n k l = i = 0 m j = 0 n δ m n i j ( α , β ) δ i j k l ( α , β ) ,
thus A = δ ( α , β ) δ ( α , β ) . For any double sequence ( u r s )
r = 0 m s = 0 n a m n r s u r s = r = 0 m s = 0 n i = 0 m j = 0 n δ m n i j ( α , β ) δ i j r s ( α , β ) u r s = r = 0 m s = 0 n ( 1 ) r + s u r s i = r m j = s n ( 1 ) i + j ( m + α m i ) ( n + β n j ) ( i + α i r ) ( j + β j s ) = r = 0 m s = 0 n ( 1 ) r + s u r s ( m + α m r ) ( n + β n s ) i = r m j = s n ( 1 ) i + j ( m r m i ) ( n s n j ) = u r s ,
since
i = r m j = s n ( 1 ) i + j ( m r m i ) ( n s n j ) = { ( 1 ) r + s , m = r , n = s , 0 , otherwise .

 □

Let ( μ m n ( α , β ) ) be a given sequence and μ ( α , β ) = ( μ m n i j ( α , β ) ) be a diagonal matrix whose only non-zero entries are μ m n ( α , β ) = μ m n m n ( α , β ) . The transformation matrix
H ( α , β ) = δ ( α , β ) μ ( α , β ) δ ( α , β )

is called a double E-J generalized Hausdorff matrix corresponding to the sequence ( μ m n ( α , β ) ) .

Theorem 3 A matrix H ( α , β ) = ( h m n i j ( α , β ) ) is a double E-J generalized Hausdorff matrix corresponding to the sequence ( μ m n ( α , β ) ) if and only if its elements have the form
h m n i j ( α , β ) = ( m + α m i ) ( n + β n j ) Δ 1 m i Δ 2 n j μ i j ( α , β ) ,
where
Δ 1 m i Δ 2 n j μ i j ( α , β ) : = r = 0 m i s = 0 n j ( 1 ) r + s ( m i r ) ( n j s ) μ i + r , j + s ( α , β ) .
Proof Let H ( α , β ) = δ ( α , β ) μ ( α , β ) δ ( α , β ) be a double E-J Hausdorff matrix. Applying this to a double sequence ( s m n ) we have
t m n = i = 0 m j = 0 n h m n i j ( α , β ) s i j = i = 0 m j = 0 n r = 0 m s = 0 n δ m n r s ( α , β ) μ r s ( α , β ) δ r s i j ( α , β ) s i j = i = 0 m j = 0 n r = 0 m s = 0 n ( 1 ) r + s ( m + α m r ) ( n + β n s ) μ r s ( α , β ) ( 1 ) i + j ( r + α r i ) ( s + β s j ) s i j = i = 0 m j = 0 n ( 1 ) i + j ( m + α m i ) ( n + β n j ) r = i m s = j n ( 1 ) r + s ( m i m r ) ( n j n s ) μ r s ( α , β ) s i j = i = 0 m j = 0 n ( m + α m i ) ( n + β n j ) r = 0 m i s = 0 n j ( 1 ) r + s ( m i r ) ( n j s ) μ i + r , j + s ( α , β ) s i j .
Hence
h m n i j ( α , β ) = ( m + α m i ) ( n + β n j ) r = 0 m i s = 0 n j ( 1 ) r + s ( m i r ) ( n j s ) μ i + r , j + s ( α , β ) .

 □

For double E-J Hausdorff matrices, the necessary and sufficient condition for H ( α , β ) to be conservative is the existence of a function χ ( s , t ) B V [ 0 , 1 ] × [ 0 , 1 ] such that
0 1 0 1 | d χ ( s , t ) | < ,
and
μ m n ( α , β ) = 0 1 0 1 s m + α t n + β d χ ( s , t ) .
Theorem 4 Given a function χ ( s , t ) B V [ 0 , 1 ] × [ 0 , 1 ] , a bounded variation in the unit square, the corresponding double E-J Hausdorff transformation ( t m n ) , of a sequence ( s m n ) , may be defined by
t m n = i = 0 m j = 0 n ( m + α m i ) ( n + β n j ) s i j 0 1 0 1 s i + α ( 1 s ) m i t j + β ( 1 t ) n j d χ ( s , t ) .
Proof For i m and j n ,
h m n i j ( α , β ) = k = i m l = j n δ m n k l ( α , β ) μ k l ( α , β ) δ k l i j ( α , β ) = k = i m l = j n δ m n k l ( α , β ) 0 1 0 1 s k + α t l + β d χ ( s , t ) δ k l i j ( α , β ) = 0 1 0 1 k = i m l = j n ( 1 ) k + l ( m + α m k ) ( n + β n l ) ( 1 ) i + j ( k + α k i ) ( l + β l j ) s k + α t l + β d χ ( s , t ) = 0 1 0 1 k = i m l = j n ( 1 ) k + l + i + j ( m + α m i ) ( n + β n j ) ( m i m k ) ( n j n l ) s k + α t l + β d χ ( s , t ) = 0 1 0 1 ( m + α m i ) ( n + β n j ) k = 0 m i l = 0 n j ( 1 ) k + l ( m i k ) ( n j l ) s k + i + α t l + j + β d χ ( s , t ) = 0 1 0 1 ( m + α m i ) ( n + β n j ) s i + α t j + β ( k = 0 m i l = 0 n j ( 1 ) k + l ( m i k ) ( n j l ) s k t l ) d χ ( s , t ) = 0 1 0 1 ( m + α m i ) ( n + β n j ) s i + α t j + β ( 1 s ) m i ( 1 t ) n j d χ ( s , t ) .

 □

Theorem 5 Let H ( α , β ) be a conservative double E-J Hausdorff matrix. Then H ( α , β ) B ( A k 2 ) , α , β 0 .

As tools to prove our result, we need to the following lemmas.

Lemma 1 [6]

Let k 1 , n v and α 0 . Then
E m + α k 1 E m μ μ + α 1 E μ + α k 1 E m μ μ + α + k 2 .

The following lemma is a double version of [11].

Lemma 2 For 0 s 1 , 0 t 1 , α 0 and β 0
i = 0 m j = 0 n ( m + α i ) ( n + β j ) ( 1 s ) m ( 1 t ) n s m + α i t n + β j 1 .
Proof of Theorem 5 Let ( t m n ) be the double E-J transform of a double sequence ( s m n ) ; i.e.,
t m n = μ = 0 m v = 0 n h m n μ v ( α , β ) s μ v .
We will demonstrate that
m = 1 n = 1 ( m n ) k 1 | a m n | k < m = 1 n = 1 ( m n ) k 1 | Δ 11 t m 1 , n 1 | k < .
(1)
Write
t m n = μ = 0 m v = 0 n b μ v .
Then b m n = Δ 11 t m 1 , n 1 . For k 1
E m k 1 = ( m + k 1 m ) = ( m + k 1 k 1 ) = ( m + k 1 ) ! m ! ( k 1 ) ! = Γ ( m + k ) Γ ( m + 1 ) Γ ( k ) .
Then
E m k 1 m k 1 Γ ( k ) m k 1 , m k 1 E m k 1 E m + α k 1 .
Due to this (1) is equivalent to
m = 1 n = 1 E m + α k 1 E n + β k 1 | a m n | k < m = 1 n = 1 E m + α k 1 E n + β k 1 | b m n | k < .
(2)
For s [ 0 , 1 ] and t [ 0 , 1 ] define
ϕ m n ( s , t ) = μ = 1 m v = 1 n E m μ μ + α 1 E n v v + β 1 s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v a μ v .
(3)
It follows from the Hölder inequality that
| ϕ m n ( s , t ) | k = | μ = 1 m v = 1 n E m μ μ + α 1 E n v v + β 1 s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v a μ v | k μ = 1 m v = 1 n E m μ μ + α 1 E n v v + β 1 s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v | a μ v | k × { μ = 1 m v = 1 n E m μ μ + α 1 E n v v + β 1 s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v } k 1 .
From Lemma 2
μ = 1 m v = 1 n E m μ μ + α 1 E n v v + β 1 s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v = μ = 1 m v = 1 n ( m + α 1 m μ ) ( n + β 1 n v ) s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v = μ = 0 m 1 v = 0 n 1 ( m + α 1 m μ 1 ) ( n + β 1 n v 1 ) s μ + α + 1 t v + β + 1 ( 1 s ) m μ 1 ( 1 t ) n v 1 = s t μ = 0 m 1 v = 0 n 1 ( m + α 1 m μ 1 ) ( n + β 1 n v 1 ) s μ + α t v + β ( 1 s ) m μ 1 ( 1 t ) n v 1 = O ( s t ) .
Hence
| ϕ m n ( s , t ) | k = O ( 1 ) ( s t ) k 1 μ = 1 m v = 1 n E m μ μ + α 1 E n v v + β 1 s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v | a μ v | k
and from Lemma 1
m = 1 n = 1 E m + α k 1 E n + β k 1 | ϕ m n ( s , t ) | k = O ( 1 ) m = 1 n = 1 E m + α k 1 E n + β k 1 ( s t ) k 1 × μ = 1 m v = 1 n E m μ μ + α 1 E n v v + β 1 s μ + α t v + β ( 1 s ) m μ ( 1 t ) n v | a μ v | k = O ( 1 ) ( s t ) k 1 μ = 1 v = 1 s μ + α t v + β | a μ v | k × m = μ n = v E m + α k 1 E n + β k 1 E m μ μ + α 1 E n v v + β 1 ( 1 s ) m μ ( 1 t ) n v = O ( 1 ) ( s t ) k 1 μ = 1 v = 1 s μ + α t v + β | a μ v | k E μ + α k 1 E v + β k 1 × m = μ n = v E m μ μ + α + k 2 E n v v + β + k 2 ( 1 s ) m μ ( 1 t ) n v = O ( 1 ) ( s t ) k 1 μ = 1 v = 1 s μ + α t v + β | a μ v | k E μ + α k 1 E v + β k 1 s μ α k + 1 t v β k + 1 = O ( 1 ) μ = 1 v = 1 E μ + α k 1 E v + β k 1 | a μ v | k .
From Lemma of [8], if ( t m n ) and ( τ m n ) are the ( H , μ m n ) transformation of ( s m n ) and ( m n a m n ) , respectively, then
τ m n = m n Δ 11 t m 1 , n 1 .
A similar consequence can be proved for ( H ( α , β ) , μ ( α , β ) ) , see [6]; i.e.,
τ m n = ( m + α ) ( n + β ) Δ 11 t m 1 , n 1 .
Hence
b m n = 1 ( m + α ) ( n + β ) τ m n = 1 ( m + α ) ( n + β ) i = 0 m j = 0 n ( m + α m i ) ( n + β n j ) Δ 1 m i Δ 2 n j μ i j ( α , β ) ( i + α ) ( j + β ) a i j = i = 0 m j = 0 n ( m + α 1 m i ) ( n + β 1 n j ) Δ 1 m i Δ 2 n j μ i j ( α , β ) a i j = i = 0 m j = 0 n E m i i + α 1 E n j j + β 1 Δ 1 m i Δ 2 n j μ i j ( α , β ) a i j .
Since H ( α , β ) is conservative, μ n ( α , β ) is a moment sequence,
μ m n ( α , β ) = 0 1 0 1 s m + α t n + β d χ ( s , t ) ,
and
Δ 1 m i Δ 2 n j μ i j ( α , β ) = 0 1 0 1 s i + α ( 1 s ) m i t j + β ( 1 t ) n j d χ ( s , t )
from Theorem 4. In view of (3) we can deduce that
b m n = i = 0 m j = 0 n E m i i + α 1 E n j j + β 1 0 1 0 1 s i + α ( 1 s ) m i t j + β ( 1 t ) n j d χ ( s , t ) a i j = 0 1 0 1 ( i = 0 m j = 0 n E m i i + α 1 E n j j + β 1 s i + α t j + β ( 1 s ) m i ( 1 t ) n j a i j ) d χ ( s , t ) = 0 1 0 1 ϕ m n ( s , t ) d χ ( s , t ) .
Using Minkowski’s inequality we get
{ m = 1 n = 1 E m + α k 1 E n + β k 1 | b m n | k } 1 / k = { m = 1 n = 1 E m + α k 1 E n + β k 1 | 0 1 0 1 ϕ m n ( s , t ) d χ ( s , t ) | k } 1 / k 0 1 0 1 | d χ ( s , t ) | { m = 1 n = 1 E m + α k 1 E n + β k 1 | ϕ m n ( s , t ) | k } 1 / k = O ( 1 ) 0 1 0 1 | d χ ( s , t ) | { μ = 1 v = 1 E μ + α k 1 E v + β k 1 | a μ v | k } 1 / k .

Therefore the proof of Theorem 5 is complete. □

Specially, if we take α = 0 and β = 0 in Theorem 5, we get Theorem 1 as a corollary.

The following is an example of a double E-J Hausdorff matrix.

A doubly infinite Cesàro matrix ( C , γ , δ ) is a doubly infinite Hausdorff matrix with entries
h m n i j = ( m + γ i 1 n i ) ( n + δ j 1 n j ) ( m + γ γ ) ( n + δ δ ) , γ , δ 0 .

We use the following to denote the corresponding E-J generalizations of the ( C , γ , δ ) .

( C ( α , β ) , γ , δ ) has moment sequence
μ m n ( α , β ) = 0 1 0 1 u m + α v n + β γ δ ( 1 u ) γ 1 ( 1 v ) δ 1 d u d v ,
where
χ ( u , v ) = γ δ 0 u 0 v ( 1 s ) γ 1 ( 1 t ) δ 1 d s d t .
For i m and j n ,
h m n i j ( α , β ) = 0 1 0 1 ( m + α m i ) ( n + β n j ) u i + α v j + β ( 1 u ) m i ( 1 v ) n j d χ ( u , v ) = 0 1 0 1 ( m + α m i ) ( n + β n j ) × u i + α v j + β ( 1 u ) m i ( 1 v ) n j γ δ ( 1 u ) γ 1 ( 1 v ) δ 1 d u d v = γ δ ( m + α m i ) ( n + β n j ) 0 1 0 1 u i + α ( 1 u ) m i + γ 1 v j + β ( 1 v ) n j + δ 1 d u d v = γ δ ( m + α m i ) ( n + β n j ) B ( i + α + 1 , m i + γ ) B ( j + β + 1 , n j + δ ) = γ Γ ( m + α + 1 ) Γ ( m i + γ ) Γ ( m i + 1 ) Γ ( m + α + γ + 1 ) δ Γ ( n + β + 1 ) Γ ( n j + δ ) Γ ( n j + 1 ) Γ ( n + β + δ + 1 ) = E m i γ 1 E n j δ 1 E m + α γ E n + β δ .
For the special case γ , δ = 1 ,
( C ( α , β ) , 1 , 1 ) = { 1 ( m + α + 1 ) ( n + β + 1 ) , i m  and  j n , 0 , otherwise

is a double E-J Hausdorff matrix.

Declarations

Acknowledgements

This work is supported by Istanbul Commerce University Scientific Research Projects Coordination Unit.

Authors’ Affiliations

(1)
Department of Mathematics, İstanbul Commerce University, Sütlüce/Beyoğlu, İstanbul, Turkey

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