volume 4, issue 4, article 66, 2003.
Received 07 April, 2003;
accepted 30 April, 2003.
Communicated by:L. Leindler
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Journal of Inequalities in Pure and Applied Mathematics
INCLUSION THEOREMS FOR ABSOLUTE SUMMABILITY METHODS
H. S. ÖZARSLAN
Department of Mathematics, Erciyes University,
38039 Kayseri, Turkey.
EMail:seyhan@erciyes.edu.tr
c
2000Victoria University ISSN (electronic): 1443-5756 046-03
Inclusion Theorems for Absolute Summability Methods
H. S. Özarslan
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Abstract
In this paper we have proved two theorems concerning an inclusion between two absolute summability methods by using any absolute summability factor.
2000 Mathematics Subject Classification:40D15, 40F05, 40G99 Key words: Absolute summability, Summability factors, Infinite series.
Contents
1 Introduction. . . 3 2 The Main Result . . . 6
References
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1. Introduction
Let P
an be a given infinite series with partial sums(sn), andrn = nan. By un and tn we denote the n-th (C,1) means of the sequences (sn) and (rn), respectively. The seriesP
anis said to be summable|C,1|k, k ≥1, if (see [4])
(1.1)
∞
X
n=1
nk−1|un−un−1|k <∞.
But since tn = n(un−un−1)(see [7]), the condition (1.1) can also be written as
(1.2)
∞
X
n=1
1
n|tn|k <∞.
The seriesP
an is said to be summable|C,1;δ|k k ≥ 1and δ ≥ 0, if (see [5])
(1.3)
∞
X
n=1
nδk−1|tn|k <∞.
If we takeδ = 0, then|C,1;δ|ksummability is the same as|C,1|ksummability.
Let(pn)be a sequence of positive numbers such that (1.4) Pn =
n
X
v=0
pv → ∞ as n→ ∞, (P−i =p−i = 0, i≥1).
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The sequence-to-sequence transformation
(1.5) Tn = 1
Pn
n
X
v=0
pvsv
defines the sequence(Tn)of the( ¯N , pn)mean of the sequence(sn), generated by the sequence of coefficients(pn)(see [6]).
The seriesP
anis said to be summable N , p¯ n
k, k ≥1,if (see [1]) (1.6)
∞
X
n=1
Pn pn
k−1
|∆Tn−1|k <∞
and it is said to be summable
N , p¯ n;δ
k,k ≥1andδ ≥0,if (see [3]) (1.7)
∞
X
n=1
Pn pn
δk+k−1
|∆Tn−1|k <∞,
where
(1.8) ∆Tn−1 =− pn
PnPn−1 n
X
v=1
Pv−1av, n≥1.
In the special case when δ = 0 (resp. pn = 1for all values of n)
N , p¯ n;δ k
summability is the same as N , p¯ n
k(resp. |C,1;δ|k) summability.
Concerning inclusion relations between|C,1|k and N , p¯ n
ksummabilities, the following theorems are known.
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Theorem 1.1. ([1]).Letk ≥ 1and let(pn)be a sequence of positive numbers such that asn → ∞
(1.9) (i) Pn =O(npn), (ii) npn =O(Pn).
If the seriesP
anis summable|C,1|k,then it is also summable N , p¯ n
k. Theorem 1.2. ([2]). Let k ≥ 1 and let (pn) be a sequence of positive num- bers such that condition (1.9) of Theorem1.1is satisfied. If the series P
anis summable
N , p¯ n
k,then it is also summable|C,1|k.
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2. The Main Result
The aim of this paper is to generalize the above theorems for |C,1;δ|k and
N , p¯ n;δ
ksummabilities, by using a summability factors. Now, we shall prove the following theorems.
Theorem 2.1. Let k ≥ 1and0 ≤ δk < 1. Let(pn)be a sequence of positive numbers such thatPn =O(npn)and
(2.1)
∞
X
n=v+1
Pn pn
δk−1
1 Pn−1
=O (
Pv pv
δk
1 Pv
) . Let P
an be summable |C,1;δ|k. Then P
anλn is summable
N , p¯ n;δ k, if (λn)satisfies the following conditions:
(2.2) n∆λn =O
Pn npn
1−δkk ,
(2.3) λn=O
Pn npn
1k .
Theorem 2.2. Let k ≥ 1and0 ≤ δk < 1. Let(pn)be a sequence of positive numbers such that npn = O(Pn) and satisfies the condition (2.1). Let P
an be summable
N , p¯ n;δ
k.ThenP
anλnis summable|C,1;δ|k,if(λn)satisfies the following conditions:
(2.4) n∆λn =O
npn Pn
1−δkk
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(2.5) λn=O
npn Pn
1k .
Remark 2.1. It may be noted that, if we takeλn = 1andδ= 0in Theorem2.1 and Theorem 2.2, then we get Theorem 1.1 and Theorem1.2, respectively. In this case condition (2.1) reduces to
m+1
X
n=v+1
pn PnPn−1
=
m+1
X
n=v+1
1 Pn−1
− 1 Pn
=O 1
Pv
as m → ∞,
which always holds.
Proof of Theorem2.1. Since
tn= 1 n+ 1
n
X
v=1
vav
we have that
an= n+ 1
n tn−tn−1. Let(Tn)denote the( ¯N , pn)mean of the seriesP
anλn. Then, by definition and changing the order of summation, we have
Tn= 1 Pn
n
X
v=0
pv
v
X
i=0
aiλi = 1 Pn
n
X
v=0
(Pn−Pv−1)avλv.
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Then, forn≥1, we have
Tn−Tn−1 = pn PnPn−1
n
X
v=1
Pv−1avλv.
By Abel’s transformation, we have Tn−Tn−1 = pn
PnPn−1 n−1
X
v=1
Pv∆λvv+ 1
v tv− pn PnPn−1
n−1
X
v=1
pvλvv+ 1 v tv
+ pn PnPn−1
n−1
X
v=1
Pv
v λv+1tv+ pn
Pnλnn+ 1 n tn
=Tn,1+Tn,2+Tn,3+Tn,4, say.
Since
|Tn,1 +Tn,2+Tn,3+Tn,4|k ≤4k(|Tn,1|k+|Tn,2|k+|Tn,3|k+|Tn,4|k),
to complete the proof of the theorem, it is enough to show that (2.6)
∞
X
n=1
Pn pn
δk+k−1
|Tn,r|k <∞ for r = 1,2,3,4.
Now, when k > 1, applying Hölder’s inequality with indicesk andk0, where
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k +k10 = 1, we get
m+1
X
n=2
Pn pn
δk+k−1
|Tn,1|k
≤
m+1
X
n=2
Pn pn
δk−1
1 Pn−1k
n−1
X
v=1
Pvv+ 1
v |tv| |∆λv|
!k
=O(1)
m+1
X
n=2
Pn pn
δk−1 1 Pn−1k
n−1
X
v=1
|tv| |v∆λv| Pv vpvpv
!k
=O(1)
m+1
X
n=2
Pn pn
δk−1
1 Pn−1k
n−1
X
v=1
|tv| |v∆λv|pv
!k
=O(1)
m+1
X
n=2
Pn pn
δk−1
1 Pn−1
n−1
X
v=1
|tv|k|v∆λv|kpv 1 Pn−1
n−1
X
v=1
pv
!k−1
=O(1)
m
X
v=1
|tv|k|v∆λv|kpv
m+1
X
n=v+1
Pn pn
δk−1 1 Pn−1
=O(1)
m
X
v=1
|tv|k|v∆λv|k Pv
pv δk−1
=O(1)
m
X
v=1
vδk−1|tv|k =O(1) as m → ∞.
by virtue of the hypotheses of Theorem2.1.
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Again using Hölder’s inequality,
m+1
X
n=2
Pn pn
δk+k−1
|Tn,2|k
=O(1)
m+1
X
n=2
Pn pn
δk−1 1 Pn−1
n−1
X
v=1
pv|tv|k|λv|k 1 Pn−1
n−1
X
v=1
pv
!k−1
=O(1)
m
X
v=1
pv|tv|k|λv|k
m+1
X
n=v+1
Pn pn
δk−1
1 Pn−1
=O(1)
m
X
v=1
Pv
pv δk−1
|tv|k|λv|k
=O(1)
m
X
v=1
vδk−1|tv|k|λv|k
=O(1)
m
X
v=1
vδk−1|tv|k=O(1) as m → ∞, by virtue of the hypotheses of Theorem2.1.
Also
m+1
X
n=2
Pn
pn
δk+k−1
|Tn,3|k
≤
m+1
X
n=2
Pn pn
δk−1
1 Pn−1k
n−1
X
v=1
Pv
|λv+1| v |tv|
!k
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=O(1)
m+1
X
n=2
Pn pn
δk−1
1 Pn−1k
n−1
X
v=1
vpv|λv+1| v |tv|
!k
=O(1)
m+1
X
n=2
Pn
pn δk−1
1 Pn−1
n−1
X
v=1
pv|λv+1|k|tv|k 1 Pn−1
n
X
v=1
pv
!k−1
=O(1)
m
X
v=1
pv|λv+1|k|tv|k
m+1
X
n=v+1
Pn pn
δk−1 1 Pn−1
=O(1)
m
X
v=1
|tv|k Pv
pv
δk−1
|λv+1|k
=O(1)
m
X
v=1
vδk−1|tv|k=O(1) as m→ ∞, by virtue of the hypotheses of Theorem2.1.
Lastly
m
X
n=1
Pn pn
δk+k−1
|Tn,4|k =O(1)
m
X
n=1
Pn pn
δk−1
|λn|k|tn|k
=O(1)
m
X
n=1
Pn npn
δk−1
|λn|k|tn|knδk−1
=O(1)
m
X
n=1
nδk−1|tn|k =O(1) as m → ∞, by virtue of the hypotheses of Theorem2.1.
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This completes the proof of Theorem2.1.
Proof of Theorem2.2. Let (Tn) denotes the( ¯N , pn) mean of the series P an. We have
Tn= 1 Pn
n
X
v=0
pvsv = 1 Pn
n
X
v=0
(Pn−Pv−1)av. Since
Tn−Tn−1 = pn
PnPn−1 n
X
v=0
Pv−1av,
we have that
(2.7) an = −Pn
pn ∆Tn−1+ Pn−2
pn−1∆Tn−2. Let
tn= 1 n+ 1
n
X
v=1
vavλv.
By using (2.7) we get tn= 1
n+ 1
n
X
v=1
v
−Pv
pv ∆Tv−1+ Pv−2 pv−1
∆Tv−2
λv
= 1 n+ 1
n−1
X
v=1
(−v)Pv
pv∆Tv−1λv− nPnλn
(n+ 1)pn∆Tn−1
+ 1 n+ 1
n
X
v=1
vPv−2 pv−1
∆Tv−2λv
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= 1 n+ 1
n−1
X
v=1
(−v)Pv
pv∆Tv−1λv
+ 1 n+ 1
n−1
X
v=1
(v+ 1)Pv−1
pv ∆Tv−1λv+1− nPnλn
(n+ 1)pn∆Tn−1
= 1 n+ 1
n−1
X
v=1
∆Tv−1
pv {−vλvPv + (v+ 1)λv+1Pv−1} − nPnλn
(n+ 1)pn∆Tn−1
= 1 n+ 1
n−1
X
v=1
∆Tv−1
pv {−vλvPv + (v+ 1)λv+1(Pv −pv)} − nPnλn
(n+ 1)pn∆Tn−1
= 1 n+ 1
n−1
X
v=1
∆Tv−1
pv {−vλvPv + (v+ 1)λv+1Pv−(v+ 1)λv+1pv}
− nPnλn
(n+ 1)pn∆Tn−1
= 1 n+ 1
n−1
X
v=1
∆Tv−1
pv {−(∆vλv)Pv−(v+ 1)λv+1pv} − nPnλn
(n+ 1)pn∆Tn−1
= 1 n+ 1
n−1
X
v=1
−Pv
pv ∆Tv−1{v∆λv−λv+1} − 1 n+ 1
n−1
X
v=1
∆Tv−1(v+ 1)λv+1
− nPnλn
(n+ 1)pn∆Tn−1
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=− 1 n+ 1
n−1
X
v=1
vPv
pv ∆λv∆Tv−1+ 1 n+ 1
n−1
X
v=1
Pv
pv∆Tv−1λv+1
− 1
n+ 1
n−1
X
v=1
(v+ 1)λv+1∆Tv−1− nPnλn
(n+ 1)pn∆Tn−1
=tn,1+tn,2+tn,3+tn,4, say.
Since
|tn,1+tn,2 +tn,3+tn,4|k ≤4k(|tn,1|k+|tn,2|k+| |tn,3|k+|t4|k), to complete the proof of Theorem2.2, it is enough to show that
∞
X
n=1
nδk−1|tn,r|k <∞ f or r= 1,2,3,4.
Now, when k > 1 applying Hölder’s inequality with indicesk and k0, where
1
k +k10 = 1, we have that
m+1
X
n=2
nδk−1|tn,1|k ≤
m+1
X
n=2
nδk−1 1 nk
n−1
X
v=1
Pv
pvv|∆λv| |∆Tv−1|
!k
≤
m+1
X
n=2
1 n2−δk
n−1
X
v=1
Pv pv
k
|∆λv|kvk|∆Tv−1|k 1 n
n−1
X
v=1
1
!k−1
≤
m+1
X
n=2
1 n2−δk
n−1
X
v=1
Pv pv
k
|∆λv|kvk|∆Tv−1|k
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=O(1)
m
X
v=1
Pv pv
k
|∆λv|kvk|∆Tv−1|k
m+1
X
n=v+1
1 n2−δk
=O(1)
m
X
v=1
Pv pv
k
|∆λv|kvk|∆Tv−1|k 1 v1−δk
=O(1)
m
X
v=1
Pv pv
δk+k−1
|∆Tv−1|k=O(1) as m→ ∞, by virtue of the hypotheses of Theorem2.2.
Again using Hölder’s inequality,
m+1
X
n=2
nδk−1|tn,2|k ≤
m+1
X
n=2
nδk−1−k
n−1
X
v=1
|λv+1|Pv pv
|∆Tv−1|
!k
≤
m+1
X
n=2
1 n2−δk
n−1
X
v=1
Pv pv
k
|λv+1|k|∆Tv−1|k 1 n
n−1
X
v=1
1
!k−1
=O(1)
m
X
v=1
Pv pv
k
|λv+1|k|∆Tv−1|k
m+1
X
n=v+1
1 n2−δk
=O(1)
m
X
v=1
Pv
pv k
|λv+1|k|∆Tv−1|k 1 v1−δk
=O(1)
m
X
v=1
Pv pv
)δk+k−1|∆Tv−1|k
=O(1) as m→ ∞,
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Also, we have that
m+1
X
n=2
nδk−1|tn,3|k ≤
m+1
X
n=2
nδk−1−k
n−1
X
v=1
(v+ 1)|λv+1| |∆Tv−1|
!k
=O(1)
m+1
X
n=2
nδk−1−k
n−1
X
v=1
v|λv+1| |∆Tv−1|
!k
=O(1)
m+1
X
n=2
1 n2−δk
n−1
X
v=1
vk|λv+1|k|∆Tv−1|k 1 n
n−1
X
v=1
1
!k−1
=O(1)
m+1
X
n=2
1 n2−δk
n−1
X
v=1
vk|λv+1|k|∆Tv−1|k
=O(1)
m
X
v=1
vk|λv+1|k|∆Tv−1|k
m+1
X
n=v+1
1 n2−δk
=O(1)
m
X
v=1
vδk+k−1|∆Tv−1|k
=O(1)
m
X
v=1
Pv
pv
δk+k−1
|∆Tv−1|k
=O(1) as m→ ∞, by virtue of the hypotheses of Theorem2.2.
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Finally, we have that
m
X
n=1
nδk−1|tn,4|k =O(1)
m
X
n=1
Pn
pn k
nδk−1|λn|k|∆Tn−1|k
=O(1)
m
X
n=1
Pn
pn
δk+k−1
|∆Tn−1|k
=O(1) as m→ ∞, by virtue of the hypotheses of Theorem2.2.
Therefore, we get that
m
X
n=1
nδk−1|tn,r|k =O(1) as m→ ∞, for r= 1,2,3,4.
This completes the proof of Theorem2.2.
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References
[1] H. BOR, On two summability methods, Math. Proc. Cambridge Phil. Soc., 97 (1985), 147–149.
[2] H. BOR, A note on two summability methods, Proc. Amer. Math. Soc., 98 (1986), 81–84.
[3] H. BOR, On the local property of
N , p¯ n;δ
k summability of factored Fourier series, J. Math. Anal. Appl., 179 (1993), 646–649.
[4] T.M. FLETT, On an extension of absolute summability and some theorems of Littlewood and Paley, Proc. London Math. Soc., 7 (1957), 113–141.
[5] T.M. FLETT, Some more theorems concerning the absolute summability of Fourier series, Proc. London Math. Soc., 8 (1958), 357–387.
[6] G.H. HARDY, Divergent Series, Oxford University Press., Oxford, (1949).
[7] E. KOGBETLIANTZ, Sur les series absolument sommables par la methode des moyannes aritmetiques, Bull. Sci. Math., 49(2) (1925), 234–256.