R E S E A R C H
Open Access
Some further extensions of absolute Cesàro
summability for double series
Hamdullah ¸Sevli
*and Ekrem Sava¸s
*Correspondence:
[email protected]; [email protected] Department of Mathematics, Faculty of Arts and Sciences, ˙Istanbul Commerce University, Üsküdar, ˙Istanbul, Turkey
Abstract
In a recent paper [Sava¸s and Rhoades in Appl. Math. Lett. 22:1462-1466, 2009], the authors extended the result of Flett [Proc. Lond. Math. Soc. 7:113-141, 1957] to double summability. In this paper, we consider some further extensions of absolute Cesàro summability for double series.
MSC: 40F05; 40G05
Keywords: absolute summability; bounded operator; Cesáro matrix; double series; Hausdorff matrix
Let∞m=∞n=amnbe an infinite double series with real or complex numbers, with partial
sums
smn= m
i= n
j=
aij.
For any double sequence (xmn) we shall define
xmn=xmn–xm+,n–xm,n++xm+,n+.
Denote byAkthe sequence space defined by
A k=
(smn)∞m,n=: ∞
m= ∞
n=
(mn)k–|amn|k<∞;amn=sm–,n–
fork≥.
A four-dimensional matrix T = (tmnij :m,n,i,j= , , . . .) is said to be absolutely kth
power conservative fork≥, ifT∈B(Ak),i.e., if
∞
m= ∞
n=
(mn)k–|sm–,n–|k<∞,
then
∞
m= ∞
n=
(mn)k–|tm–,n–|k<∞,
where
tmn= ∞
i= ∞
j=
tmnijsij (m,n= , , . . .).
A double infinite Cesáro matrixC(α,β) is a double infinite Hausdorff matrix with entries
hmnij=
Eα– m–iE
β– n–j
Eα
mE
β
n
,
where
Eα
m=
m+α
m
= (α+m+ ) (m+ )(α+ )≈
mα (α+ ).
The series amnis said to be summable|C(α,β)|k,k≥,α,β> –, if (see [])
∞
m= ∞
n=
(mn)k–σm–,n–αβ k
<∞, ()
whereσmnαβdenotes themn-term of theC(α,β) transform of a sequence (smn);i.e.,
σmnαβ= Eα
mE
β
n m
i= n
j=
Eα– m–iE
β–
n–jsij. ()
Quite recently, Savaş and Rhoades [] extended the result of Flett [] to double summa-bility. Their theorem is as follows.
Theorem Letα≥γ > –,β≥δ> –,andmnamnbe a double series with partial
sums smn.If
m
namnis|C(γ,δ)|k-summable,then it is also|C(α,β)|k-summable,k≥.
It then follows that if one setsγ=δ= , thenC(α,β)∈B(A
k) for eachα,β≥. In this
paper, we consider some further extensions of absolute Cesàro summability for double series.
We shall use the following lemmas.
Lemma Ifθ> –,φ> –,θ–ϕ> andφ–ψ> ,then
∞
m=i ∞
n=j
Em–iϕ En–jψ mnEθ
mE
φ
n
=
ijEθi–ϕ–Eφj–ψ–. ()
Proof Forα> –,n≥ since
Eα
n
=
( –x)αxn–dx
and
( –x)–α=
∞
n=
we obtain
∞
m=i ∞
n=j
Em–iϕ En–jψ mnEθ
mE
φ
n
=
∞
m=i
Eϕm–i mEθ
m ∞
n=
Enψ
(n+j)En+jφ
=
∞
m=i
Eϕm–i mEθ
m ∞
n=
Eψn
( –x)φxn+j–dx
=
∞
m=i
Eϕm–i mEθ
m
( –x)φxj– ∞
n=
Eψnxn
dx
=
( –x)φ–ψ–xj–dx ∞
m=
Eϕm
(m+i)Eθ
m+i
= jEφj–ψ–
∞
m=
Eϕm
( –x)θxm+i–dx
= jEφj–ψ–
( –x)θ–ϕ–xi–dx
=
ijEθi–ϕ–Eφj–ψ–.
For single series, Lemma due to Chow [].
Lemma Let≤k≤r<∞andα,β> –.For i,j≥,let
Aij=Aij(α,β) = ∞
m=i ∞
n=j |Eα–
m–i|r/k|E
β– n–j|r/k
mn(Eα
m)r/k(E
β
n)r/k
. ()
Then,if k=r,
Aij=
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
O(i–α–j–β–), ifα,β< ,
O(i–α–j–), ifα< ,β≥,
O(i–j–β–), ifα≥,β< ,
O(i–j–), ifα,β≥.
()
If k<r<∞,then
Aij=
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
O(i–αkr–j–
βr
k–), ifα< –k/r,β< –k/r, O(i–αkr–j–kr logj), ifα< –k/r,β= –k/r, O(i–αkr–j–kr), ifα< –k/r,β> –k/r, O(i–αkr–j–
βr
k–logi), ifα= –k/r,β< –k/r, O(i–rkj–
r
k logilogj), ifα= –k/r,β= –k/r, O(i–αkr–j–
r
klogi), ifα= –k/r,β> –k/r, O(i–rkj–
βr
k–), ifα> –k/r,β< –k/r, O(i–rkj–
βr
k–logj), ifα> –k/r,β= –k/r, O(i–rkj–
r
k), ifα> –k/r,β> –k/r.
Proof Since|Eα
n| ≤K(α)nαfor allα, andEnα≥M(α)nαforα> –, whereK(α) andM(α)
are positive constants depending only onα, ifk<r<∞, then
Aij=O() ∞
m=i ∞
n=j
m–αkr–n–
βr
k–(m–i+ )
(α–)r
k (n–j+ )
(β–)r k
=O() i–
m=i
m–αkr–(m–i+ )
(α–)r k +
∞
m=i
m–αkr–(m–i+ )
(α–)r k
×
j–
n=j
n–βkr–(n–j+ )
(β–)r k +
∞
n=j
n–βkr–(n–j+ )
(β–)r k
=O()
i–αkr–
i–
m=i
(m–i+ )(α–)k r +
∞
m=i
m–kr–
×
j–βkr–
j–
n=j
(n–j+ )(β–)k r +
∞
n=j
n–rk–
=O()i–αkr–j–
βr k–
i
m= j
n=
m(α–)k rn
(β–)r
k +O()i–αkr–j–r/k
i
m=
m(α–)k r
+O()i–r/kj–βkr–
j
n=
n(β–)k r +O()(ij)–r/k.
According as (α–)rk and (β–)rk = –, < – or > –, we have (). The case k=ris proved
similarly.
For single series, Lemma due to Mehdi []. We now prove the following theorem.
Theorem Let r≥k≥.
(i) It holds thatC(α,β)∈(A
k,Ar)for eachα,β> –k/r.
(ii) Ifα,β= –k/rand the condition∞m=∞n=(mn)k–logmlogn|a
mn|k=O()is satisfied thenC(α,β)∈(A
k,Ar).
(iii) If the condition∞m=∞n=mk+(–α)rk–nk+(–β)kr–|a
mn|k=O()is satisfied then
C(α,β)∈(Ak,Ar)for each–k/r<α,β< –k/r.
(iv) Ifβ= –k/rand the condition∞m=∞n=mk+(–α)rk–nk–logn|amn|k=O()is
satisfied thenC(α,β)∈(Ak,A
r)for each–k/r<α< –k/r.
(v) If the condition∞m=∞n=mk+(–α)rk–nk–|amn|k=O()is satisfied then C(α,β)∈(A
k,Ar)for each–k/r<α< –k/randβ> –k/r.
(vi) Ifα= –k/rand the condition∞m=∞n=mk–nk+(–β)kr–logm|a
mn|k=O()is satisfied thenC(α,β)∈(A
k,Ar)for each–k/r<β< –k/r.
(vii) If the condition∞m=∞n=mk–nk–logm|a
mn|k=O()is satisfied then
C(α,β)∈(A
k,Ar)for eachα> –k/r,β< –k/r.
(viii) If the condition∞m=∞n=mk–nk+(–β)rk–|a
mn|k=O()is satisfied then
C(α,β)∈(Ak,Ar)for eachα> –k/rand–k/r<β< –k/r.
(ix) Ifβ= –k/rand the condition∞m=∞n=mk–nk–logn|amn|k=O()is satisfied thenC(α,β)∈(Ak,A
Proof We shall show that (σmnαβ)∈Ar,i.e., ∞ m= ∞ n=
(mn)r–σ
αβ
m–,n– r
<∞.
Letτmnαβ denote themn-term of theC(α,β)-transform in terms of (mnamn),i.e.,
τmnαβ= Eα mE β n m i= n j=
Eαm–i–Eβn–j–ijaij.
Forα,β> –, since
τmnαβ=mnσmnαβ–σm,n–αβ –σm–,nαβ +σm–,n–αβ ,
to prove the theorem, it will be sufficient to show that
∞ m= ∞ n=
(mn)–τmnαβr<∞. ()
Using Hölder’s inequality, we have
∞ m= ∞ n= mnτ
αβ mn r = ∞ m= ∞ n= mn Eα mE β n m i= n j=
Eα– m–iE
β– n–jijaij
r ≤ ∞ m= ∞ n= mn(Eα
m)r(E
β
n)r
m i= n j=
Eαm–i–Eβn–j–ikjk|aij|k
r/k × m i= n j=
Em–iα–Eβn–j– (k–)r/k . Since m i= n j=
Eαm–i–Eβn–j–=
Eα–+
m–
i=
Eαm–i–
Eβ–+
n–
j=
Eβn–j–
=
Eα–
+ m– i=
Eα– m–i
Eβ–+
n– j=
Eβn–j– =
Eα–+ m i=
Eαm–i––Emα––Eα– × Eβ–+
n j=
En–jβ––Eβ– n –E
β–
=Eα–
+Eαm––Eα–E
β– +E
β
n––E
β– ,
and using the fact that
Em–α
Eα
m
=O() and E β
n–
Enβ
we obtain
∞
m= ∞
n=
mnτ
αβ
mn r
≤
∞
m= ∞
n=
(Eα
m)(k–)r/k(E
β
n)(k–)r/k
mn(Eα
m)r(E
β
n)r
×
m
i= n
j=
Eα– m–iE
β– n–ji
kjk|a ij|k
r/k
≤
∞
m= ∞
n=
(Eα
m)–r/k(E
β
n)–r/k
mn
×
m
i= n
j=
Eα– m–iE
β– n–j(ij)
–k/r+k/r
|aij|k
/r
(ij)–(r–k)+k(r–k)/r|aij|k(r–k)/r
r/k
.
Applying Hölder’s inequality with indicesr/k,r/(r–k), we deduce that
∞
m= ∞
n=
mnτ
αβ
mn r
≤
∞
m= ∞
n=
(Eα
m)–r/k(E
β
n)–r/k
mn
m
i= n
j=
Eα– m–i
r/k
Eβn–j–r/k(ij)k–+r/k|aij|k
×
m
i= n
j=
(ij)k–|aij|k
(r–k)/k
.
Since (smn)∈Ak, we have
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|a ij|k(ij)r/k
∞
m=i ∞
n=j |Eα–
m–i|r/k|E
β– n–j|r/k
mn(Eα
m)r/k(E
β
n)r/k
.
(i) From Lemma , ifα,β> –k/r, then
∞
m=i ∞
n=j
(Eα– m–i)r/k(E
β– n–j)r/k
mn(Eα
m)r/k(E
β
n)r/k
= (ij)–
E r k–
i E
r k–
j
=O(ij)–r/k.
Therefore, for the caseα,β> –k/r, we have
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|aij|k(ij)r/k(ij)–r/k
=O()
∞
i= ∞
j=
(ij)k–|aij|k=O(),
(ii) Ifα,β= –k/r, from Lemma , then
∞
m=i ∞
n=j |Eα–
m–i|r/k|E
β– n–j|r/k
mn(Eα
m)r/k(E
β
n)r/k
=Oi–krj–krlogilogj.
Hence,
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–logilogj|aij|k=O().
(iii) If –k/r<α,β< –k/r, from Lemma , then
∞
m=i ∞
n=j |Eα–
m–i|r/k|E
β– n–j|r/k
mn(Eα
m)r/k(E
β
n)r/k
=Oi–αkr–j–
βr k–,
and then
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k+r/k–i–αkr–j–
βr k–|aij|k
=O()
∞
i= ∞
j=
ik+(–α)rk–jk+(–β)rk–|aij|k=O().
(iv) Ifβ= –k/rand –k/r<α< –k/r, from Lemma , then
∞
m=i ∞
n=j |Eα–
m–i|r/k|E
β– n–j|r/k
mn(Eα
m)r/k(E
β
n)r/k
=Oi–αkr–j–kr logj,
therefore, we have
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|aij|k(ij)r/ki– αr
k–j–krlogj
=O()
∞
i= ∞
j=
ik+(–α)kr–jk–logj|aij|k=O().
(v) If –k/r<α< –k/randβ> –k/r, then
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|aij|k(ij)r/ki– αr
k–j–kr
=O()
∞
i= ∞
j=
ik+(–α)rk–jk–|aij|k=O(),
(vi) Ifα= –k/rand –k/r<β< –k/r, then
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|aij|k(ij)r/ki–
r kj–
βr k–logi
=O()
∞
i= ∞
j=
ik–jk+(–β)kr–logi|aij|k=O(),
by using Lemma .
(vii) Ifα= –k/randβ> –k/r, then
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|aij|k(ij)r/ki–
r kj–krlogi
=O()
∞
i= ∞
j=
ik–jk–logi|aij|k=O(),
by using Lemma .
(viii) Ifα> –k/rand –k/r<β< –k/r, then
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|aij|k(ij)r/ki–
r kj–
βr k–
=O()
∞
i= ∞
j=
ik–jk+(–β)kr–|a
ij|k=O(),
by using Lemma .
(ix) Ifα> –k/randβ= –k/r, then
∞
m= ∞
n=
mnτ
αβ
mn r
=O()
∞
i= ∞
j=
(ij)k–|aij|k(ij)r/ki–
r kj–
βr k–logj
=O()
∞
i= ∞
j=
ik–jk–logj|aij|k=O(),
by using Lemma .
The one-dimensional version of Theorem appears in []. By (), Theorem includes the following theorem with the special caser=k.
Theorem Let k≥.
(i) It holds thatC(α,β)∈B(Ak)for eachα,β≥. (ii) If the condition∞m=∞n=mk–α–nk–β–|a
mn|k=O()is satisfied then
C(α,β)∈B(A
k)for each– <α< and– <β< .
(iii) If the condition∞m=∞n=mk–α–nk–|a
mn|k=O()is satisfied then
C(α,β)∈B(A
k)for each– <α< andβ≥.
(iv) If the condition∞m=∞n=mk–nk–β–|a
mn|k=O()is satisfied then
Remark Theorem moderates Theorem of []. Since Holder’s inequality is valid if each of the terms is nonnegative, it should be added the absolute values of the binomial coefficients in the proof of Theorem of [], when – <α< and/or – <β< . Therefore, if we replace the binomial coefficients with their absolute values, then the inequality () of [] will be true. So, we should add the conditions, given above in (ii), (iii) and (iv) of Theorem in the statement of Theorem of [], for the cases – <α< and/or – <β< .
Corollary Letθmnα =Eα
m m
i=Em–iα–sin=C(α, )(smn).
(i) It holds thatC(α, )∈B(A
k)for eachα≥.
(ii) If the condition∞m=∞n=mk–α–nk–|a
mn|k=O()is satisfied thenC(α, )∈B(Ak) for each– <α< .
Corollary Letθmnβ = Eβn
n j=E
β–
n–jsmj= (C, ,β)(smn).
(i) It holds thatC(,β)∈B(A
k)for eachβ≥.
(ii) If the condition∞m=∞n=mk–nk–β–|a
mn|k=O()is satisfied thenC(,β)∈B(Ak) for each– <β< .
Corollary Letσmn=(m+)(n+)
m i=
n
j=sij= (C, , )(smn).Then C(, )∈B(Ak).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors contributed equally and significantly in writing this paper. Both the authors read and approved the final manuscript.
Acknowledgements
Dedicated to Professor Hari M. Srivastava.
Received: 14 December 2012 Accepted: 20 March 2013 Published: 2 April 2013
References
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doi:10.1186/1029-242X-2013-144
Cite this article as:¸Sevli and Sava¸s:Some further extensions of absolute Cesàro summability for double series.