On General Summability Factor Theorems

Volume 2007, Article ID 10892,10pages doi:10.1155/2007/10892

Research Article

On General Summability Factor Theorems

Ekrem Savas¸

Received 17 August 2006; Revised 6 December 2006; Accepted 2 January 2007

Recommended by Ram Mohapatra

The goal of this paper is to obtain sufficient and (different) necessary conditions for a seriesan, which is absolutely summable of orderkby a triangular matrix methodA,

1< k≤s <∞, to be such thatanλnis absolutely summable of ordersby a triangular

matrix methodB. As corollaries, we obtain two inclusion theorems.

Copyright © 2007 Ekrem Savas¸. This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In the recent papers [1,2], the author obtained necessary and sufficient conditions for a seriesanwhich is absolutely summable of orderkby a weighted mean method,

1< k≤s <∞, to be such thatanλnis absolutely summable of ordersby a triangular

matrix method. In this paper, we obtain sufficient and (different) necessary conditions for a seriesanwhich is absolutely summable|A|kto imply the seriesanλnwhich is

absolutely summable|B|s.

LetTbe a lower triangular matrix,{sn}a sequence.

Then

Tn:= n

ν=0

tnνsν. (1)

A seriesanis said to be summable|T|k,k≥1 if

∞

n=1

nk−1_T

We may associate withTtwo lower triangular matricesTandTas follows:

tnν= n

r=νtnr, n,ν=0, 1, 2,...,

tnν=tnν−tn−1,ν, n=1, 2, 3,....

(3)

Withsn:=ni=0aiλi,

yn:= n

i=0

tnisi= n

i=0

tni i

ν=0

aνλν= n

ν=0

aνλν n

i=νtni= n

ν=0

tnνaνλν,

Yn:=yn−yn−1= n

ν=0

(tnν−tn−1)λνaν= n

ν=0

tnνλνaν.

(4)

We will callT as a triangle ifT is lower triangular andtnn=0 for eachn. The

nota-tionΔνanνmeansanν−an,ν+1. The notationλ∈(|A|k,|B|s) will be used to represent the

statement that ifanis summable|A|k, thenanλnis summable|B|s.

Theorem 1. Let 1< k≤s <∞. Let{λn}be a sequence of constants, A andB triangles satisfying

(i)|bnnλn|/|ann| =O(ν1/s−1/k),

(ii) (n|Xn|)s−k=O(1),

(iii)|ann−an+1,n| =O(|annan+1,n+1|),

(iv)n−_ν=01|Δν(bnνλν)| =O(|bnnλn|),

(v)∞_n=ν+1(n|bnnλn|)s−1|Δν(bnνλν)| =O(νs−1|bννλν|s),

(vi)n−_ν=01|bννλν+1||bn,ν+1λν+1| =O(|bnnλn+1|),

(vii)∞_n=ν+1(n|bnnλn+1|)s−1|bn,ν+1λν+1| =O((ν|bννλν+1|)s−1),

(viii)∞_n=1ns−1| n

ν=2bnνλν ν−2

i=0aνiXi|s=O(1), thenλ∈(|A|k,|B|s).

Proof. Ifyndenotes thenth term of theB-transform of a sequence{sn}, then

yn= n

i=0

bnisi= n

i=0

bni i

ν=0

λνaν= n

ν=0

λνaν n

i=νbni= n

ν=0

bnνλνaν,

yn−1= n−1

ν=0

bn−1,νλνaν,

(5)

Yn:=yn−yn−1= n

ν=0

bnνλνaν, (6)

wheresn=ni=0λiai.

Letxndenote thenth term of theA-transform of a seriesan, then as in (6),

Xn:=xn−xn−1= n

ν=0

SinceAis a triangle, it has a unique two-sided inverse, which we will denote byA. Thus we may solve (7) foranto obtain

an= n

ν=0

a

nνXν. (8)

Substituting (8) into (6) yields

Yn= n

ν=0

bnνλνaν= n

ν=0

bnνλν _ν

i=0

a νiXi

=

n

ν=0

bnνλν _ν−2

i=0

a

νiXi+aν,ν−1Xν−1+a_ννXν

= n

ν=0

bnνλνaννXν+ n

ν=1

bnνλνaν,ν−1Xν−1+ n

ν=2

bnνλν ν−2

i=0

a νiXi

=bnnλnannXn+

n−1

ν=0

bnνλνaννXν+

n−1

ν=0

bn,ν+1λν+1aν+1,νXν+

n

ν=2

bnνλν ν−2

i=0

a νiXi

=b_ann nnλnXn+

n−1

ν=0

_bnνλνa

νν+bn,ν+1λν+1a_ν+1,ν Xν+ n

ν=2

bnνλν ν−2

i=0

a νiXi

=b_ann nnλnXn+

n−1

ν=0

_bnνλνa

νν+bn,ν+1λν+1a_νν−bn,ν+1λν+1a_νν+bn,ν+1λν+1a_ν+1,ν Xν

+

n

ν=2

bnνλν ν−2

i=0

aνiXi

=bnn annλnXn+

n−1

ν=0

Δνbnνλν

aνν Xν+ n−1

ν=0

bn,ν+1λν+1aνν+aν+1,ν Xν+ n

ν=2

bnνλν ν−2

i=0

a νiXi.

(9)

Using the fact that

a

νν+aν+1,ν=_a1 νν

_a

νν−aν+1,ν

aν+1,ν+1

, (10)

and substituting (10) into (9), we have the following:

Yn=b_ann nnλnXn+

n−1

ν=0

Δνbnνλν

aνν Xν+ n−1

ν=0

bn,ν+1λν+1 _a

νν−aν+1,ν

aννaν+1,ν+1

Xν+ n

ν=2

bnνλν ν−2

i=0

a νiXi

=Tn1+Tn2+Tn3+Tn4, say.

By Minkowski’s inequality, it is sufficient to show that

∞

n=1

ns−1_T

nis<∞, i=1, 2, 3, 4. (12)

Using (i),

J1:= ∞

n=1

ns−1_T n1s=

∞

n=1

ns−1bnnλn

ann Xn s

=O(1)

∞

n=1

ns−1_n1/s−1/k s_Xns

=O(1)

∞

n=1

nk−1_X

nkns−s/k−k+1Xns−k .

(13)

Butns−s/k−k+1_|X

n|s−k=(n1−1/k_|X

n|)s−k=O((n|Xn|)s−k)=O(1), from (ii) ofTheorem 1. Sinceanis summable|A|k,J1=O(1).

Using (i), (iv), (v), (ii), and H¨older’s inequality,

J2:= ∞

n=1

ns−1_T n2s=

∞

n=1

ns−1

n−1

ν=0

Δνbnνλν

aνν Xν s

≤ ∞

n=1

ns−1 _n−1

ν=0

ν1/s−1/k_b

ννλν−1Δνbnνλν Xν s

=O(1)

∞

n=1

ns−1 _n−1

ν=0

ν1−s/k_b

ννλν−sΔνbnνλν Xνs

× _n−1

ν=0

_Δνbnνλν s−1

=O(1)

∞

n=1

nbnnλn s−1 n−1

ν=0

ν1−s/k_b

ννλν−sΔνbnνλν Xνs

=O(1)

∞

ν=1

ν1−s/k_b

ννλν−sXνs ∞

n=ν+1

nbnnλn s−1Δνbnνλν

=O(1)

∞

ν=1

ν1−s/k_b

ννλν−sXνsνs−1bννλνs=O(1) ∞

ν=1

νs−s/k_X νs

=O(1)

∞

ν=1

νk−1_X

νkνs−s/k−k+1Xνs−k =O(1) ∞

ν=1

νk−1_X

νk=O(1).

Using (iii), (vi), (vii), (ii), and H¨older’s inequality,

J3:= ∞

n=1

ns−1_T n3s=

∞

n=1

ns−1

n−1

ν=0

bn,ν+1λν+1 _a

νν−aν+1,ν

aννaν+1,ν+1 Xν s ≤∞ n=1

ns−1 _n−1

ν=0

_{bn,ν+1λν+1}aνν−aν+1,ν

aννaν+1,ν+1 Xν

s

=O(1)

∞

n=1

ns−1 _n−1

ν=0

_{bn,ν+1λν+1Xν}s

=O(1)

∞

n=1

ns−1 _n−1

ν=0

bννλν+1 _bννλν+1

_{bn,ν+1λν+1Xν}s

=O(1)

∞

n=1

ns−1 _n−1

ν=0

_bννλν+11−s_b

n,ν+1λν+1Xνs

× _n−1

ν=0

_{bννλν+1bn,ν+1λν+1}s−1

=O(1)

∞

n=1

nbnnλn+1 s−1 n−1

ν=0

_bννλν+11−s

bn,ν+1λν+1Xνs

=O(1)

∞

ν=0

_bννλν+11−s_X νs

∞

n=ν+1

nbnnλn+1 s− 1

bn,ν+1λν+1

=O(1)

∞

ν=0

_bννλn+11−s

Xνsνs−1bννλν+1s−1=O(1) ∞

ν=0

νs−1_X νs

=O(1)

∞

ν=0

νk−1_X

νkνXν s−k=O(1) ∞

ν=0

νk−1_X

νk=O(1).

(15)

From (viii),

∞

n=1

ns−1_T n4s=

∞

n=1

ns−1

n

ν=2

bnνλν ν−2

i=0

a νiXi

s

=O(1). (16)

We now state sufficient conditions, whenk=s.

Corollary1. Let{λn}be a sequence of constants,AandBtriangles satisfying

(i)|bnn|/|ann| =O(1/|λn|),

(ii)|ann−an+1,n| =O(|annan+1,n+1|),

(iii)n−_ν=01|Δν(bnνλν)| =O(|bnnλn|),

(iv)∞_n=ν+1(n|bnnλn|)k−1|Δν(bnνλν)| =O(νk−1|bννλν|k),

(v)n−_ν=01|bννλν+1||bn,ν+1λν+1| =O(|bnnλn+1|),

(vi)∞_n=ν+1(n|bnnλn+1|)k−1|bn,ν+1λn+1| =O((ν|bννλν+1|)k−1),

(vii)∞_n=1nk−1| _n

ν=2bnνλν _ν−2

A weighted mean matrix is a lower triangular matrix with entries pk/Pn, 0≤k≤n,

wherePn=nk=0pk.

Corollary2. Let1< k≤s <∞. Let{λn}be a sequence of constants, letBbe a triangle such thatB, and let{pn}satisfy

(i)bννλν=O((pν/Pν)ν1/s−1/k),

(ii) (n|Xn|)s−k=O(1),

(iii)n−_ν=11|Δν(bnνλν)| =O(|bnnλn|),

(iv)∞_n=ν+1(n|bnnλn|)s−1|Δν(bnνλν)| =O(νs−1|bννλν|s),

(v)n−_ν=11|bννλν||bn,νλν| =O(|bnnλn|),

(vi)∞_n=ν+1(n|bnnλn|)s−1|bn,νλν| =O((ν|bν,νλν|)s−1), thenλ∈(|N,pn|k,|B|s).

Proof. Conditions (i), (ii), (iii)–(vii) ofTheorem 1reduce to conditions (i)–(vi), respec-tively, ofCorollary 1.

WithA=(N,pn),

ann−an+1,n=_Ppn n−

pn

Pn+1=

pnpn+1

PnPn+1 =annan+1,n+1, (17)

and condition (ii) ofTheorem 1is automatically satisfied.

A matrixAis said to be factorable ifank=bnckfor eachnandk.

SinceAis a weighted mean matrix,Ais a factorable triangle and it is easy to show that its inverse is bidiagonal. Therefore condition (viii) ofTheorem 1is trivially satisfied.

We now turn our attention to obtaining necessary conditions.

Theorem2. Let1< k≤s <∞, and letAandBbe two lower triangular matrices withA satisfying

∞

n=ν+1n k−1_Δνa

nνk=Oaννk . (18)

Then necessary conditions forλ∈(|A|k,|B|s)are

(i)|bννλν| =O(|aνν|ν1/s−1/k),

(ii)∞_n=ν+1ns−1_|Δνb

nνλν|s)=O(|aνν|sνs−s/k),

(iii)∞_n=ν+1ns−1_|b

n,ν+1λν+1|s=O(∞_n=ν+1nk−1|an,ν+1|k)s/k.

Proof. Define

A∗_=a i:

aiis summable|A|k

,

B∗_=b i:

biλiis summable|B|s

WithYnandXnas defined by (6) and (7), the spacesA∗andB∗are BK-spaces, with

norms given by

a 1=

_X0k

+

∞

n=1

nk−1_X nk

1/k

,

a 2=

_Y0s

+

∞

n=1

ns−1_Y ns

1/s

,

(20)

respectively.

From the hypothesis of the theorem, a 1<∞implies that a 2<∞. The inclusion

mapi:A∗→B∗_{defined byi(x)=xis continuous, sinceA}∗_andB∗_{are BK-spaces.}

Ap-plying the Banach-Steinhaus theorem, there exists a constantK >0 such that

a 2≤K a 1. (21)

Letendenote thenth coordinate vector. From (6) and (7), with{an}defined byan=

en−en+1,n=ν,an=0 otherwise, we have

Xn= ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

0, n <ν,

anν, n=ν, Δνanν, n >ν,

Yn= ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

0, n <ν,

bnνλν, n=ν, Δνbnνλν , n >ν.

(22)

From (20),

a 1=

νk−1_a ννk+

∞

n=ν+1

nk−1_Δνa nνk

1/k

,

a 2=

νs−1_b ννλνs+

∞

n=ν+1

ns−1_Δνb nνλν s

1/s

,

(23)

recalling thatbνν=bνν=bνν. From (21), using (18), we obtain

νs−1_b ννλνs+

∞

n=ν+1

ns−1_Δνb nνλν s

≤Ks

νk−1_a ννk+

∞

n=ν+1n

k−1_Δνa nνk

s/k ≤Ks

νk−1_a

ννk+O(1)aννk s/k

=Oaννkνk−1+ 1 s/k=Oνk−1aννk s/k.

The above inequality will be true if and only if each term on the left-hand side is O(νk−1_|a

νν|k)s/k. Using the first term,

νs−1_b

ννλνs=Oνk−1aννk s/k, (25)

which implies that|bννλν| =O(|aνν|ν1/s−1/k), and (i) is necessary.

Using the second term, we obtain

∞

n=ν+1

ns−1_Δνb

nνλν s=Oνk−1aννk s/k=Oνs−s/kaννs , (26)

which is condition (ii).

If we now definean=en+1forn=ν,an=0 otherwise, then from (6) and (7), we obtain

Xn= ⎧ ⎨ ⎩

0, n≤ν,

an,ν+1, n >ν,

Yn= ⎧ ⎨ ⎩

0, n≤ν,

bn,ν+1λν+1, n >ν.

(27)

The corresponding norms are

a 1= _∞

n=ν+1n k−1_a

n,ν+1k 1/k

,

a 2=

_∞

n=ν+1n s−1_b

n,ν+1λν+1s 1/s

.

(28)

Applying (21),

∞

n=ν+1n s−1_b

n,ν+1λν+1s≤Ks _∞

n=ν+1n k−1_a

n,ν+1k s/k

, (29)

which implies condition (iii).

Corollary3. LetAandBbe two lower triangular matrices withAsatisfying

∞

n=ν+1n k−1_Δνa

nνk=Oaννk . (30)

Then necessary conditions forλ∈(|A|k,|B|k)are

(i)|bννλν| =O(|aνν|),

(ii) (∞_n=ν+1nk−1_|Δνb

nνλν|k)1/k=O(|aνν|ν1−1/k),

(iii)∞_n=ν+1nk−1_|b

Corollary4. Let1< k≤s <∞. LetBbe a lower triangular matrix,{pn}is a sequence satisfying

∞

n=ν+1

nk−1 pn

PnPn−1 k

=O_P1_k

ν

. (31)

Then necessary conditions forλ∈(|N,pn|k,|B|s)are

(i)Pν|bννλν|/pν=O(ν1/s−1/k),

(ii)∞_n=ν+1ns−1_|Δν(b

nνλν)|s=O(νs−s/k(pν/Pν)s),

(iii)∞_n=ν+1ns−1_|b

n,ν+1λν+1|s=O(1).

Proof. WithA=(N,pn), (18) becomes (31), and conditions (i)–(iii) ofTheorem 2

be-come conditions (i)–(iii) ofCorollary 4, respectively.

Every summability factor theorem becomes an inclusion theorem by setting each λn=1.

Corollary5 (see [3]). Let1< k≤s <∞. LetAandBbe triangles satisfying

(i)|bnn|/|ann| =O(ν1/s−1/k),

(ii) (n|Xn|)s−k=O(1),

(iii)|ann−an+1,n| =O(|annan+1,n+1|),

(iv)n−_ν=01|Δν(bnν)| =O(|bnn|),

(v)∞_n=ν+1(n|bnn|)s−1|Δν(bnν)| =O(νs−1|bνν|s),

(vi)n−_ν=01|bνν||bn,ν+1| =O(|bnn|),

(vii)∞_n=ν+1(n|bnn|)s−1|bn,ν+1| =O((ν|bνν|)s−1),

(viii)∞_n=1ns−1| n

ν=2bnν ν−2

i=0aνiXi|s=O(1). Then ifanis summable|A|k, it is summable|B|s.

Corollary6 (see [4]). Let be{pn}a sequence of positive constants,Bis a triangle satisfy-ing

(i)Pn|bnn| =O((pn)ν1/s−1/k_),

(ii) (n|Xn|)s−k=O(1),

(iii)n−_ν=01|Δν(bnν)| =O(|bnn|),

(iv)∞_n=ν+1(n|bnn|)s−1|Δν(bnν)| =O(νs−1|bνν|s),

(v)n−_ν=01|bννbn,ν+1| =O(|bnn|),

(vi)∞_n=ν+1(n|bnn|)s−1|bn,ν+1| =O((ν|bνν|)s−1). Then ifanis summable|N,pn|k, it is summable|B|s.

Acknowledgment

References

[1] E. Savas¸, “On absolute summability factors,” The Rocky Mountain Journal of Mathematics, vol. 33, no. 4, pp. 1479–1485, 2003.

[2] E. Savas¸, “Necessary conditions for inclusion relations for absolute summability,”Applied Math-ematics and Computation, vol. 151, no. 2, pp. 523–531, 2004.

[3] E. Savas¸, “General inclusion relations for absolute summability,”Mathematical Inequalities & Applications, vol. 8, no. 3, pp. 521–527, 2005.

[4] B. E. Rhoades and E. Savas¸, “On inclusion relations for absolute summability,” International Journal of Mathematics and Mathematical Sciences, vol. 32, no. 3, pp. 129–138, 2002.

Ekrem Savas¸: Department of Mathematics, Faculty of Science and Arts, Istanbul Ticaret University, Uskudar, 34672 Istanbul, Turkey