# On double Hausdorff summability method

## 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.

## Introduction and background

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

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

$μ n ( β ) = ∫ 0 1 t n + β dχ(t),$

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

In a recent paper , the first author jointly with Savaş has extended the result of Das  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  proved the result of Das  for double Hausdorff summability. In this paper we will consider double E-J generalization and we will prove the corresponding result of  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 $χ(s,t)∈BV[0,1]×[0,1]$ such that

$∫ 0 1 ∫ 0 1 | d χ ( s , t ) | <∞,$

and

$μ m n = ∫ 0 1 ∫ 0 1 s m t n dχ(s,t).$

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

Theorem 1 

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

Our purpose is to achieve the result established in  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 $χ(s,t)∈BV[0,1]×[0,1]$ such that

$∫ 0 1 ∫ 0 1 | d χ ( s , t ) | <∞,$

and

$μ m n ( α , β ) = ∫ 0 1 ∫ 0 1 s m + α t n + β dχ(s,t).$

Theorem 4 Given a function $χ(s,t)∈BV[0,1]×[0,1]$, 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

$t m n = ∑ i = 0 m ∑ j = 0 n ( m + α m − i ) ( n + β n − j ) s i j ∫ 0 1 ∫ 0 1 s i + α ( 1 − s ) m − i t j + β ( 1 − t ) n − j dχ(s,t).$

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 

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 .

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 , if $( t m n )$ and $( τ m n )$ are the $(H, μ m n )$ transformation of $( s m n )$ and $(mn a m n )$, respectively, then

$τ m n =mn Δ 11 t m − 1 , n − 1 .$

A similar consequence can be proved for $( H ( α , β ) , μ ( α , β ) )$, see ; 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,

$μ m n ( α , β ) = ∫ 0 1 ∫ 0 1 s m + α t n + β dχ(s,t),$

and

$Δ 1 m − i Δ 2 n − j μ i j ( α , β ) = ∫ 0 1 ∫ 0 1 s i + α ( 1 − s ) m − i t j + β ( 1 − t ) n − j dχ(s,t)$

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

$μ m n ( α , β ) = ∫ 0 1 ∫ 0 1 u m + α v n + β γδ ( 1 − u ) γ − 1 ( 1 − v ) δ − 1 dudv,$

where

$χ(u,v)=γδ ∫ 0 u ∫ 0 v ( 1 − s ) γ − 1 ( 1 − t ) δ − 1 dsdt.$

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 $γ,δ=1$,

is a double E-J Hausdorff matrix.

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## Acknowledgements

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

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Correspondence to Hamdullah Şevli.

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The authors contributed equally and significantly in writing this paper. Both authors read and approved the final manuscript.

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