Dual Spaces of Generalized Cesaro Sequence Space and Related Matrix Mapping

International Journal of Mathematics and Statistics Invention (IJMSI)
E-ISSN: 2321 – 4767 P-ISSN: 2321 - 4759
www.ijmsi.org Volume 4 Issue 4 || April. 2016 || PP-44-50
www.ijmsi.org 44 | Page
Dual Spaces of Generalized Cesaro Sequence Space and Related
Matrix Mapping
Md. Fazlur Rahman1
, A B M Rezaul Karim*2
Department of Mathematics, Eden University College, Dhaka, Bangladesh.
ABTRACT: In this paper we define the generalized Cesaro sequence spaces 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠). We prove the space
𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) is a complete paranorm space. In section-2 we determine its Kothe-Toeplitz dual. In section-3 we
establish necessary and sufficient conditions for a matrix A to map 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 to 𝑙∞ and 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) to c,
where 𝑙∞ is the space of all bounded sequences and c is the space of all convergent sequences. We also get some
known and unknown results as remarks.
KEYWORDS: Sequence space, Kothe-Toeplitz dual, Matrix transformation.
I. INTRODUCTION
Let 𝜔 be the space of all (real or complex) sequences and let 𝑙∞, 𝑐 𝑎𝑛𝑑 𝑐0 are respectively the Banach spaces
of bounded sequences, convergent sequences and null sequences. Let 𝑝 = 𝑝 𝑘 be a bounded sequence of
strictly positive real numbers. Then 𝑙(𝑝) was defined by Maddox [7] as
𝑙 𝑝 = 𝑥 = 𝑥 𝑘 ∈ 𝜔: |𝑥 𝑘 | 𝑝 𝑘 < ∞
∞
𝑘=1
with 0 < 𝑝 𝑘 ≤
𝑠𝑢𝑝
𝑘
𝑝 𝑘 = 𝐻 < ∞.
In [9] Shiue introduce the Cesaro sequence space 𝑐𝑒𝑠 𝑝 as
𝑐𝑒𝑠𝑝 = 𝑥 = 𝑥 𝑘 ∈ 𝜔:
1
𝑛
𝑥 𝑘
𝑛
𝑘=1
𝑝
< ∞
∞
𝑛=1
𝑓𝑜𝑟 1 < 𝑝 < ∞
𝑎𝑛𝑑 𝑐𝑒𝑠∞ = 𝑥 = 𝑥 𝑘 ∈ 𝜔:
𝑠𝑢𝑝
𝑛 ≥ 1
1
𝑛
|𝑥 𝑘 |
𝑛
𝑘=1
𝑓𝑜𝑟 𝑝 = ∞.
𝐼𝑛 [5] Leibowitz studied some properties of this space and showed that it is a Banach space. Lim [10] defined
this space in a different norm as
𝑐𝑒𝑠𝑝 = 𝑥 = 𝑥 𝑘 ∈ 𝜔:
1
2 𝑟
𝑥 𝑘
𝑟
𝑝
< ∞
∞
𝑟=0
𝑓𝑜𝑟 1 < 𝑝 < ∞
𝑎𝑛𝑑 𝑐𝑒𝑠∞ = 𝑥 = 𝑥 𝑘 ∈ 𝜔:
𝑠𝑢𝑝
𝑟 ≥ 0
1
2 𝑟
𝑥 𝑘 < ∞ 𝑓𝑜𝑟 𝑝 = ∞
where 𝑑𝑒𝑛𝑜𝑡𝑒𝑠𝑟 a sum over the ranges [2 𝑟
, 2 𝑟+1
), determined its dual spaces and characterize some matrix
classes. Later in [11] Lim extended this space 𝑐𝑒𝑠𝑝 𝑡𝑜 𝑐𝑒𝑠(𝑝) for the sequence 𝑝 = (𝑝𝑟) with inf 𝑝𝑟 > 0 and
defined as
authoringCorrespond*
𝑐𝑒𝑠 𝑝 = 𝑥 = 𝑥 𝑘 ∈ 𝜔:
1
2 𝑟
|𝑥 𝑘 |
𝑟
𝑝 𝑟
< ∞
∞
𝑟=0
.
For positive sequence of real numbers 𝑝 𝑛 , 𝑞 𝑛 𝑎𝑛𝑑 𝑄 𝑛 = 𝑞1 + 𝑞2 + ⋯ … . +𝑞 𝑛 , Johnson and Mohapatra [14]
defined the Cesaro sequence space 𝑐𝑒𝑠 𝑝, 𝑞 𝑎𝑠
𝑐𝑒𝑠(𝑝, 𝑞) = 𝑥 = 𝑥 𝑘 ∈ 𝜔:
1
𝑄 𝑛
𝑞 𝑘
𝑛
𝑘=1
𝑥 𝑘
𝑝 𝑟
< ∞
∞
𝑛=1
and studied some inclusion relations. What amounts to the same thing defined by Khan and Rahman [4] as

Dual Spaces of Generalized Cesaro Sequence Space and Related Matrix Mapping
𝑐𝑒𝑠 𝑝, 𝑞 = 𝑥 = 𝑥 𝑘 ∈ 𝜔:
1
𝑄2 𝑟
𝑞 𝑘|𝑥 𝑘 |
𝑟
𝑝 𝑟
< ∞
∞
𝑟=0
for 𝑝 = 𝑝𝑟 with inf 𝑝𝑟 > 0, 𝑄2 𝑟 = 𝑞2 𝑟+ 𝑞2 𝑟+1 + ⋯ … … . . … + 𝑞2 𝑟+1−1 and denotes𝑟 a sum over the
ranges [2 𝑟
, 2 𝑟+1
). They determined it’s Kothe –Toeplitz dual and characterized some matrix classes.
The main purpose of this paper is to define the generalized Cesaro sequence space 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠). We determine
the Kothe-Toeplitz dual of 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) and then consider the matrix mapping
𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 𝑡𝑜 𝑙∞ 𝑎𝑛𝑑 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 𝑡𝑜 𝑐.
𝐼𝑛 [2] Bulut and Cakar defined and studied the sequence space 𝑙 𝑝, 𝑠 , in [3] Khan and Khan defined and
investigated the Cesaro sequence space 𝑐𝑒𝑠 𝑝, 𝑠 , in [12] we defined and studied the Riesz sequence space
𝑟 𝑞
(𝑢, 𝑝, 𝑠) of non-absolute type and in [13] we defined and studied the generalized weighted Cesaro sequence
space 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 . In the same vein we define generalized Cesaro sequence space 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 in the following
way.
DEFINITION. For 𝑠 ≥ 0 we define
𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 = 𝑥 = 𝑥 𝑘 ∈ 𝜔: (𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 |𝑥 𝑘 |
𝑟
𝑝 𝑟
< ∞
∞
𝑟=0
where (𝑞 𝑘 ) is a bounded sequence of real numbers, 𝑝 = 𝑝𝑟 with inf 𝑝𝑟 > 0, 𝑄2 𝑟 = 𝑞2 𝑟 + 𝑞2 𝑟+1 +
⋯ … … … … … . . +𝑞2 𝑟+1−1 𝑎𝑛𝑑 denotes𝑟 a sum over the range 2 𝑟
≤ 𝑘 < 2 𝑟+1
. With regard notation, the dual
space of 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , that is, the space of all continuous linear functional on 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 will be denoted by
𝑐𝑒𝑠∗
(𝑝, 𝑞, 𝑠). We write
𝐴 𝑟 𝑛 =
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑛,𝑘
where for each n the maximum with respect to k in [2 𝑟
, 2 𝑟+1
).
Throughout the paper the following well-known inequality (see [7] or [8]) will be frequently used. For any
positive integer 𝐸 > 1 and any two complex numbers a and b we have
|𝑎𝑏| ≤ 𝐸 |𝑎|𝑡
𝐸−𝑡
+ |𝑏|𝑡
(1)
where 𝑝 > 1 𝑎𝑛𝑑
1
𝑝
+
1
𝑞
= 1.
To begin with, we show that the space 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 is a paranorm space paranormed by
𝑔 𝑥 = (𝑄2 𝑟 )−𝑠 1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘𝑟
𝑝 𝑟
∞
𝑟=0
1/𝑀
(2)
provided 𝐻 =
𝑠𝑢𝑝
𝑟
𝑝𝑟 < ∞ 𝑎𝑛𝑑 𝑀 = max 1, 𝐻 .
Clearly
𝑔 𝜃 = 0
𝑔 −𝑥 = 𝑔(𝑥),
where 𝜃 = (0, 0, 0, … … … … … … … … … . . )
Since 𝑝𝑟 ≤ 𝑀, 𝑀 ≥ 1 𝑠𝑜 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑥, 𝑦 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 𝑤𝑒 have by Minkowski’s inequality
(𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘 + 𝑦 𝑘
𝑟
𝑝 𝑟∞
𝑟=0
1/𝑀
≤ (𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
(𝑞 𝑘 𝑥 𝑘
𝑟
+ 𝑞 𝑘 |𝑦 𝑘 |)
𝑝 𝑟∞
𝑟=0
1/𝑀
≤ (𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟∞
𝑟=0
1/𝑀
+ (𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 |𝑦 𝑘
𝑟
|
𝑝 𝑟∞
𝑟=0
1/𝑀
which shows that g is subadditive.
Finally we have to check the continuity of scalar multiplication. From the definition of 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠), we have
inf 𝑝𝑟 > 0. So, we may assume that inf 𝑝𝑟 ≡ 𝜌 > 0. Now for any complex 𝜆 with |𝜆 | < 1, we have

𝑔 𝜆𝑥 = (𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘|𝜆𝑥 𝑘
𝑟
|
𝑝 𝑟∞
𝑟=0
1/𝑀
= 𝜆
𝑝 𝑟
𝑀 𝑄2 𝑟
−𝑠
1
𝑄2 𝑟
𝑞 𝑘 |𝑥 𝑘
𝑟
|
𝑝 𝑟∞
𝑟=0
1
𝑀
≤
𝑠𝑢𝑝
𝑟
𝜆
𝑝 𝑟
𝑀 𝑔(𝑥)
≤ 𝜆
𝜌
𝑀 𝑔 𝑥 → 0 𝑎𝑠 𝜆 → 0
above. It is quite routine to show that 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) is a metric space with the metric 𝑑(𝑥, 𝑦) = 𝑔(𝑥 − 𝑦)
provided that 𝑥, 𝑦 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , where g is defined by (2). And using a similar method to that in
([3],[4],[13])one can show that 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) is complete under the metric mentioned.
II. KOTHE-TOEPLITZ DUALS
If X is a sequence space we define ([1], [6])
𝑋|+|
= 𝑋 𝛼
= 𝑎 = 𝑎 𝑘 ∈ 𝜔: 𝑎 𝑘 𝑥 𝑘 < ∞,
𝑘
𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑥 ∈ 𝑋
𝑋+
= 𝑋 𝛽
= 𝑎 = 𝑎 𝑘 ∈ 𝜔: 𝑎 𝑘 𝑥 𝑘 𝑖𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑥 ∈ 𝑋
𝑘
Now we are going to give the following theorem by which the generalized Kothe-Toeplitz dual 𝑐𝑒𝑠+
(𝑝, 𝑞, 𝑠)
will be determined.
Theorem 1: If 1 < 𝑝𝑟 ≤
𝑠𝑢𝑝
𝑟
𝑝𝑟 < ∞ 𝑎𝑛𝑑
1
𝑝 𝑟
+
1
𝑡 𝑟
= 1, 𝑓𝑜𝑟 𝑟 = 0, 1, 2, … … ., then
𝑐𝑒𝑠+
𝑝, 𝑞, 𝑠 = [𝑐𝑒𝑠(𝑝, 𝑞, 𝑠)] 𝛽
= 𝑎 = 𝑎 𝑘 : (𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
𝑄2 𝑟
𝑚𝑎𝑥
𝑟
(𝑞 𝑘
−1
𝑎 𝑘 )
𝑡 𝑟
𝐸−𝑡 𝑟 < ∞, 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝐸 > 1∞
𝑟=0 .
Proof : Let 1 < 𝑝𝑟 ≤
𝑠𝑢𝑝
𝑟
𝑝𝑟 < ∞ 𝑎𝑛𝑑
1
𝑝 𝑟
+
1
𝑡 𝑟
= 1, 𝑓𝑜𝑟 𝑟 = 0,1,2, … …. . Define
𝜇 𝑡, 𝑠 =
𝑎 = 𝑎 𝑘 : (𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘
𝑡 𝑟
𝐸−𝑡 𝑟 < ∞, 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝐸 > 1∞
𝑟=0 . (3)
We want to show that 𝑐𝑒𝑠+
𝑝, 𝑞, 𝑠 = 𝜇 𝑡, 𝑠 . Let 𝑥 ∈ 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) and 𝑎 ∈ 𝜇 𝑡, 𝑠 . Then using inequality (1)
we get
𝑎 𝑘 𝑥 𝑘 = 𝑎 𝑘 𝑥 𝑘
𝑟
∞
𝑟=0
∞
𝑘=1
= 𝑞 𝑘
−1
𝑎 𝑘 𝑞 𝑘 𝑥 𝑘
𝑟
∞
𝑟=0
≤
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘 𝑞 𝑘 𝑥 𝑘
𝑟
∞
𝑟=0
= 𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘 (𝑄2 𝑟 )
𝑠
𝑝 𝑟
1
𝑄2 𝑟
(𝑄2 𝑟 )
−
𝑠
𝑝 𝑟 𝑞 𝑘 𝑥 𝑘
𝑟
∞
𝑟=0
≤ 𝐸 𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘
𝑡 𝑟
(𝑄2 𝑟 )
𝑠 𝑡 𝑟
𝑝 𝑟 𝐸−𝑡 𝑟 + (𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟∞
𝑟=0
= 𝐸 𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘
𝑡 𝑟
(𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
𝐸−𝑡 𝑟 + (𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟∞
𝑟=0
∞
𝑟=0
< ∞
which implies that the series 𝑎 𝑘 𝑥 𝑘
∞
𝑘=1 convergent.
Therefore,
𝑎 ∈ 𝑑𝑢𝑎𝑙 𝑜𝑓 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 = 𝑐𝑒𝑠+
(𝑝, 𝑞, 𝑠). This shows, 𝜇(𝑡, 𝑠) ⊂ 𝑐𝑒𝑠+
(𝑝, 𝑞, 𝑠)
Conversely, suppose that 𝑎 𝑘 𝑥 𝑘 is convergent for all 𝑥 ∈ 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) but 𝑎 ∉ 𝜇(𝑡, 𝑠). Then

(𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘
𝑡 𝑟
𝐸−𝑡 𝑟 = ∞
∞
𝑟=0
for every integer 𝐸 > 1.
So, we can define a sequence 0 = 𝑛 0 < 𝑛 1 < 𝑛 2 < ⋯ … … … . …, such that 𝛾 = 0, 1, 2, … … … …. , we
have
𝑀𝛾 = (𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘
𝑡 𝑟
𝛾 + 2 −
𝑡 𝑟
𝑝 𝑟
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
> 1
Now we define a sequence 𝑥 = 𝑥 𝑘 in the following way:
𝑥 𝑁(𝑟) = 𝑄2 𝑟
𝑡 𝑟
𝑎 𝑁(𝑟)
𝑡 𝑟−1
𝑠𝑔𝑛 𝑎 𝑁(𝑟)(𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
(𝛾 + 2)−𝑡 𝑟 𝑀𝛾
−1
for 𝑛 𝛾 ≤ 𝑟 ≤ 𝑛 𝛾 + 1 − 1, 𝛾 = 0, 1, 2, … … … … … …, and 𝑥 𝑘 = 0 for 𝑘 ≠ 𝑁 𝑟 , where 𝑁 𝑟 is such
that
𝑎 𝑁(𝑟) =
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘 , the maximum is taken with respect to k in 2 𝑟
, 2 𝑟+1
.
Therefore .
𝑎 𝑘 𝑥 𝑘
2 𝑛 𝛾+1 −1
𝑘=2 𝑛(𝛾)
= 𝑄2 𝑟 𝑎 𝑁(𝑟)
𝑡 𝑟
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
(𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
(𝛾 + 2)−𝑡 𝑟 𝑀𝛾
−1
= 𝑀𝛾
−1
(𝛾 + 2)−1
𝑄2 𝑟 𝑎 𝑁(𝑟)
𝑡 𝑟
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
(𝑄2 𝑟 ) 𝑠(𝑡 𝑟−1)
(𝛾 + 2)−𝑡 𝑟 /𝑝 𝑟
= 𝑀𝛾
−1
𝑀𝛾 (𝛾 + 2)−1
= (𝛾 + 2)−1
It follows that
𝑎 𝑘 𝑥 𝑘
∞
𝑘=1
= (𝛾 + 2)−1
∞
𝛾=0
diverges.
Moreover
(𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
= (𝑄2 𝑟 )−𝑠
𝑄2 𝑟
𝑠 𝑡 𝑟−1
𝑄2 𝑟
𝑡 𝑟−1
𝑎 𝑁(𝑟)
𝑡 𝑟−1
(𝛾 + 2)−𝑡 𝑟 𝑀𝛾
−1
𝑝 𝑟
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
= (𝑄2 𝑟 )−𝑠
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
𝑄2 𝑟
(𝑠+1) 𝑡 𝑟−1 𝑝 𝑟
𝑎 𝑁(𝑟)
𝑡 𝑟−1 𝑝 𝑟
(𝛾 + 2)−𝑡 𝑟 𝑝 𝑟 𝑀𝛾
−𝑝 𝑟
= (𝑄2 𝑟 )−𝑠
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
𝑄2 𝑟
(𝑠+1)𝑡 𝑟
𝑎 𝑁(𝑟)
𝑡 𝑟
(𝛾 + 2)−𝑡 𝑟 𝑝 𝑟 𝑀𝛾
−𝑝 𝑟
= (𝛾 + 2)−2
𝑀𝛾
−1
𝑄2 𝑟
𝑠 𝑡 𝑟−1
(𝑄2 𝑟 𝑎 𝑁 𝑟 )𝑡 𝑟 𝛾 + 2 2−𝑡 𝑟−𝑝 𝑟 𝑀𝛾
1−𝑝 𝑟
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
= (𝛾 + 2)−2
𝑀𝛾
−1
𝑄2 𝑟
𝑠 𝑡 𝑟−1
(𝑄2 𝑟 𝑎 𝑁 𝑟 )𝑡 𝑟 𝛾 + 2 2−𝑡 𝑟/𝑝 𝑟 𝑀𝛾
1−𝑝 𝑟
(𝛾 + 2)2−𝑡 𝑟− 𝑝 𝑟 +𝑡 𝑟/𝑝 𝑟
𝑛 𝛾+1 −1
𝑟=𝑛(𝛾)
= (𝛾 + 2)−2
𝑀𝛾
−1
𝑀𝛾 𝑀𝛾
1−𝑝 𝑟
𝛾 + 2 1−𝑝 𝑟
= (𝛾 + 2)−2
𝑀𝛾
−𝑝 𝑟/𝑡 𝑟
𝛾 + 2 −𝑝 𝑟/𝑡 𝑟
=
(𝛾 + 2)−2
𝑀𝛾
𝑝 𝑟/𝑡 𝑟
𝛾 + 2 𝑝 𝑟/𝑡 𝑟
< (𝛾 + 2)−2
< ∞.
Therefore

(𝑄2 𝑟 )−𝑠
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟
≤ (𝛾 + 2)−2
< ∞
∞
𝑟=0
That is, 𝑥 ∈ 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) which is a contradiction to our assumption.
Hence 𝑎 ∈ 𝜇(𝑡, 𝑠). That is, 𝜇 𝑡, 𝑠 ⊃ 𝑐𝑒𝑠+
𝑝, 𝑞, 𝑠 .
Then combining the two results, we get 𝑐𝑒𝑠+
𝑝, 𝑞, 𝑠 = 𝜇(𝑡, 𝑠).
The continuous dual of 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) is determined by the following theorem.
Theorem 2: Let 1 < 𝑝𝑟 ≤
𝑠𝑢𝑝
𝑟
𝑝𝑟 < ∞. Then continuous dual 𝑐𝑒𝑠∗
(𝑝, 𝑞, 𝑠) is isomorphic to 𝜇(𝑡, 𝑠), which is
defined by (3)
Proof: It is easy to check that each 𝑥 ∈ 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) can be written in the form
𝑥 = 𝑥 𝑘 𝑒 𝑘
∞
𝑘=1
, 𝑤𝑕𝑒𝑟𝑒 𝑒 𝑘 = (0, 0, 0, … … … 0, 1, 0, … … … … … … . . )
and the 1 appears at the k-th place. Then for any 𝑓 ∈ 𝑐𝑒𝑠∗
(𝑝, 𝑞, 𝑠) we have
𝑓 𝑥 = 𝑥 𝑘 𝑓 𝑒 𝑘 = 𝑥 𝑘
∞
𝑘=1
∞
𝑘=1 𝑎 𝑘 .
(4)
where 𝑓 𝑒 𝑘 = 𝑎 𝑘. By theorem 1, the convergence of 𝑎 𝑘 𝑥 𝑘 for every x in 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) implies that 𝑎 ∈
𝜇(𝑡, 𝑠).
If 𝑥 ∈ 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠) and if we take 𝑎 ∈ 𝜇(𝑡, 𝑠), then by theorem 1, 𝑎 𝑘 𝑥 𝑘 converges and clearly defines a linear
functional on 𝑐𝑒𝑠(𝑝, 𝑞, 𝑠). Using the same kind of argument as in theorem 1, it is easy to check that
𝑎 𝑘 𝑥 𝑘 ≤ 𝐸 𝑄2 𝑟
𝑠 𝑡 𝑟−1
𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑞 𝑘
−1
𝑎 𝑘
𝑡 𝑟
𝐸−𝑡 𝑟 + 1
∞
𝑟=0
∞
𝑘=1
𝑔(𝑥)
whenever 𝑔(𝑥) ≤ 1, where 𝑔(𝑥) is defined by (2). Hence 𝑎 𝑘 𝑥 𝑘 defines an element of 𝑐𝑒𝑠∗
𝑝, 𝑞, 𝑠 .
Furthermore, it is easy to see that representation (4) is unique. Hence we can define a mapping
𝑇: 𝑐𝑒𝑠∗
𝑝, 𝑞, 𝑠 → 𝜇 𝑡, 𝑠 .
By 𝑇 𝑓 = (𝑎1, 𝑎2, … … … … … … ) where the 𝑎 𝑘 appears in representation (4). It is evident that 𝑇 is linear and
bijective. Hence 𝑐𝑒𝑠∗
𝑝, 𝑞, 𝑠 is isomorphic to 𝜇 𝑡, 𝑠 .
III. MATRIX TRANSFORMATIONS
In the following theorems we shall characterize the matrix classes (𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑙∞) and 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑐 . Let
𝐴 = 𝑎 𝑛,𝑘 𝑛, 𝑘 = 1,2, … … .. be an infinite matrix of complex numbers and X, Y two subsets of the space of
complex sequences. We say that the matrix A defines a matrix transformation from X into Y and denote it by
𝐴 ∈ (𝑋, 𝑌) if for every sequence 𝑥 = 𝑥 𝑘 ∈ 𝑋 the sequence 𝐴 𝑥 = 𝐴 𝑛 (𝑥) is in Y, where
𝐴 𝑛 𝑥 = 𝑎 𝑛,𝑘 𝑥 𝑘
∞
𝑘=1
provided the series on the right is convergent.
Theorem 3: Let 1 < 𝑝𝑟 ≤
𝑠𝑢𝑝
𝑟
𝑝𝑟 < ∞. Then 𝐴 ∈ (𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑙∞) if and only if there exists an integer 𝐸 > 1,
such that 𝑈 𝐸, 𝑠 < ∞, where
𝑈 𝐸, 𝑠 =
𝑠𝑢𝑝
𝑛
𝑄2 𝑟 𝐴 𝑟 𝑛
𝑡 𝑟
∞
𝑟=0
𝑄2 𝑟
𝑠 𝑡 𝑟−1 𝐸−𝑡 𝑟
and
1
𝑝 𝑟
+
1
𝑡 𝑟
= 1, 𝑟 = 0, 1, 2, … … … … …
Proof: Sufficiency: Suppose there exists an integer 𝐸 > 1, such that 𝑈 𝐸, 𝑠 < ∞. Then by inequality (1), we
have
𝑎 𝑛,𝑘 𝑥 𝑘 = |𝑎 𝑛,𝑘 | 𝑥 𝑘
𝑟
∞
𝑟=0
∞
𝑘=1
=
𝑎 𝑛,𝑘
𝑞 𝑘
𝑟
𝑞 𝑘
∞
𝑟=0
𝑥 𝑘
≤
𝑚𝑎𝑥
𝑟
𝑎 𝑛,𝑘
𝑞 𝑘
∞
𝑟=0
𝑞 𝑘 𝑥 𝑘
𝑟
= (𝑄2 𝑟 )
𝑠
𝑝 𝑟 𝑄2 𝑟
𝑚𝑎𝑥
𝑟
𝑎 𝑛,𝑘
𝑞 𝑘
(𝑄2 𝑟 )
−
𝑠
𝑝 𝑟
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
∞
𝑟=0
≤ 𝐸 (𝑄2 𝑟 )
𝑠 𝑡 𝑟
𝑝 𝑟 𝑄2 𝑟 𝐴 𝑟(𝑛) 𝑡 𝑟 𝐸−𝑡 𝑟 + (𝑄2 𝑟 )
−
𝑠
𝑝 𝑟
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟∞
𝑟=0

≤ 𝐸 𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
∞
𝑟=0
𝑄2 𝑟 𝐴 𝑟(𝑛) 𝑡 𝑟 𝐸−𝑡 𝑟 + 𝑄2 𝑟
−𝑠
∞
𝑟=0
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟
< ∞.
Therefore, 𝐴 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑙∞ .
Necessity: Suppose that 𝐴 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑙∞ , but
𝑠𝑢𝑝
𝑛
𝑡 𝑟
𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
∞
𝑟=0
𝐸−𝑡 𝑟 = ∞ 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝐸 > 1.
Then 𝑎 𝑛,𝑘 𝑥 𝑘
∞
𝑘=1 converges for every 𝑛 𝑎𝑛𝑑 𝑥 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 ,
whence 𝑎 𝑛,𝑘 𝑘=1,2,……
∈ 𝑐𝑒𝑠+
(𝑝, 𝑞, 𝑠) for every n. By theorem 1, it follows that each 𝐴 𝑛 defined by
𝐴 𝑛 𝑥 = 𝑎 𝑛,𝑘 𝑥 𝑘
∞
𝑘=1
is an element of 𝑐𝑒𝑠∗
(𝑝, 𝑞, 𝑠). Since 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 is complete and since
𝑠𝑢𝑝
𝑛
𝐴 𝑛 (𝑥) < ∞ on 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , by the
uniform boundedness principle there exists a number L independent of n and a number 𝛿 < 1, such that
𝐴 𝑛 (𝑥) ≤ 𝐿
(5)
for every n and 𝑥 ∈ 𝑆[𝜃, 𝛿], where 𝑆[𝜃, 𝛿] is the closed sphere in 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 with centre at the origin 𝜃 and
radius 𝛿.
Now choose an integer 𝐺 > 1, such that
𝐺𝛿 𝑀
> 𝐿.
Since
𝑠𝑢𝑝
𝑛
𝑄2 𝑟 𝐴 𝑟(𝑛) 𝑡 𝑟
∞
𝑟=0
𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
𝐺−𝑡 𝑟 = ∞
there exists an integer 𝑚0 > 1, such that
𝑅 = 𝑄2 𝑟 𝐴 𝑟(𝑛) 𝑡 𝑟 𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
𝑚0
𝑟=0
𝐺−𝑡 𝑟
> 1 (6)
Define a sequence 𝑥 = 𝑥 𝑘 as follows:
𝑥 𝑘 = 0 𝑖𝑓 𝑘 ≥ 2 𝑚0+1
𝑥 𝑁(𝑟) = 𝑄2 𝑟
𝑡 𝑟
𝛿 𝑀/𝑝 𝑟 𝑠𝑔𝑛 𝑎 𝑛,𝑁(𝑟) 𝑎 𝑛,𝑁(𝑟)
𝑡 𝑟−1
𝑅−1
𝐺−𝑡 𝑟/𝑝 𝑟 𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
and 𝑥 𝑘 = 0 𝑖𝑓 𝑘 ≠ 𝑁(𝑟) for 0 ≤ 𝑟 ≤ 𝑚0, where 𝑁(𝑟) is the smallest integer such that
𝑎 𝑛,𝑁(𝑟) =
𝑚𝑎𝑥
𝑟
|𝑎 𝑛,𝑘 |
𝑞 𝑘
Then one can easily show that 𝑔 𝑥 ≤ 𝛿 but 𝐴 𝑛 𝑥 > 𝐿, which contradicts (5). This complete the proof of
the theorem.
Theorem 4. Let 1 < 𝑝𝑟 ≤
𝑠𝑢𝑝
𝑟
𝑝𝑟 < ∞. Then 𝐴 ∈ (𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑐) if and only if
(i) 𝑎 𝑛,𝑘 → 𝛼 𝑘 𝑛 → ∞, 𝑘 𝑖𝑠 𝑓𝑖𝑥𝑒𝑑 and
(ii) there exists an integer 𝐸 > 1, such that 𝑈 𝐸, 𝑠 < ∞, where
𝑈 𝐸, 𝑠 =
𝑠𝑢𝑝
𝑛
𝑡 𝑟∞
𝑟=0 𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
𝐸−𝑡 𝑟 and
1
𝑝 𝑟
+
1
𝑡 𝑟
= 1, 𝑟 = 0, 1, 2, … … … … …
Proof: Necessity. Suppose 𝐴 ∈ (𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑐). Then 𝐴 𝑛 (𝑥)exists for each 𝑛 ≥ 1 𝑎𝑛𝑑
𝐿𝑖𝑚
𝑛 → ∞
𝐴 𝑛 (𝑥) exists
for every 𝑥 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 . Therefore by an argument similar to that in theorem 3 we have condition (ii).
Condition (i) is obtained by taking 𝑥 = 𝑒 𝑘 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , where 𝑒 𝑘 is a sequence with 1 at the k-th place and
zeros elsewhere.
Sufficiency. The conditions of the theorem imply that

𝑄2 𝑟
𝑚𝑎𝑥
𝑟
|𝛼 𝑘|
𝑞 𝑘
𝑡 𝑟
𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
𝐸−𝑡 𝑟 ≤ 𝑈 𝐸, 𝑠
∞
𝑟=0
< ∞ (7)
By (7) it is easy to check that 𝛼 𝑘𝑘 𝑥 𝑘 is absolutely convergent for each 𝑥 ∈ 𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 . For each 𝑥 ∈
𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 and 𝜀 > 0, we can choose an integer 𝑚0 > 1, such that
𝑔 𝑚0
𝑥 = 𝑄2 𝑟
− 𝑠
1
𝑄2 𝑟
𝑞 𝑘 𝑥 𝑘
𝑟
𝑝 𝑟∞
𝑟=𝑚0
< 𝜀 𝑀
Then by the proof of theorem 2 and by inequality (1), we have
𝑎 𝑛,𝑘 − 𝛼 𝑘 𝑥 𝑘
∞
𝑘=2 𝑚 0
≤ 𝐸 𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
𝑄2 𝑟 𝐵𝑟 (𝑛) 𝑡 𝑟
∞
𝑟=𝑚0
𝐸−𝑡 𝑟 + 1 𝑔 𝑚0
(𝑥)
1/𝑀
< 𝐸 2𝑈 𝐸, 𝑠 + 1 𝜀,
where 𝐵𝑟 𝑛 =
𝑚𝑎𝑥
𝑟
|𝑎 𝑛,𝑘−𝛼 𝑘|
𝑞 𝑘
and
𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
𝑄2 𝑟 𝐵𝑟 𝑛
𝑡 𝑟
𝐸−𝑡 𝑟 ≤ 2𝑈 𝐸, 𝑠 < ∞
∞
𝑟=𝑚0
It follows immediately that
𝐿𝑖𝑚
𝑛 → ∞
𝑎 𝑛,𝑘 𝑥 𝑘 = 𝛼 𝑘 𝑥 𝑘
∞
𝑘=1
∞
𝑘=1
This shows that 𝐴 ∈ (𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑐) which proved the theorem.
Corollary 1. Let 1 < 𝑝𝑟 ≤
𝑠𝑢𝑝
𝑟
𝑝𝑟 < ∞. Then 𝐴 ∈ (𝑐𝑒𝑠 𝑝, 𝑞, 𝑠 , 𝑐0) if and only if
(i) 𝑎 𝑛,𝑘 → 0 𝑛 → ∞, 𝑘 𝑖𝑠 𝑓𝑖𝑥𝑒𝑑
(ii) there exists an integer E >1 such that 𝑈 𝐸, 𝑠 < ∞, where
𝑈 𝐸, 𝑠 =
𝑠𝑢𝑝
𝑛
(𝑄2 𝑟 𝐴 𝑟(𝑛))𝑡 𝑟
∞
𝑟=0
𝑄2 𝑟
𝑠 (𝑡 𝑟−1)
𝐸−𝑡 𝑟 𝑎𝑛𝑑
1
𝑝𝑟
+
1
𝑡 𝑟
= 1, 𝑟 = 0, 1, 2, … … ….
Remarks:
(1) If 𝑠 = 0 then we get the results of Khan and Rahman [4]
(2) If 𝑠 = 0, 𝑞 𝑛 = 1 for every n then we get the results of Lim [11]
(3) When 𝑠 = 0, 𝑞 𝑛 = 1 𝑎𝑛𝑑 𝑝 𝑛 = 𝑝 for all n then the results of Lim [10] follows.
(4) If 𝑠 ≥ 1 then specializing the sequences (𝑝 𝑛 ) and (𝑞 𝑛 ) we get many unknown results.
(5)
REFERENCES
[1] [1] B. Choudhury and S. K. Mishra, On Kothe-Toeplitz duals of certain sequence spaces and their matrix transformations, Indian
J. pure appl. Math, 24(15), 291-301, May 1993.
[2] [2] E. Bulut and O Cakar, The sequence space 𝑙(𝑝, 𝑠) and related matrix transformation, communication de la faculatedes science d
L’universite D’Ankara Tome, 28(1979), 33-44.
[3] [3] F.M. KHAN and M.A. KHAN, The sequence space 𝑐𝑒𝑠(𝑝, 𝑠) and related matrix transformations, Research Ser.Mat.,21(1991);
95-104.
[4] [4] F.M. KHAN and M.F. RAHMAN, Infinite matrices and Cesaro sequence spaces, Analysis Mathematica, 23(1997), 3-11.
[5] [5] G. M. Leibowitz, A note on the Cesaro sequence spaces, Tamkang J. of Math.,2(1971),151-157
[6] [6] H. Kizmaz, Canadian Math. Bull. 24(2)(1981),169-176.
[7] [7] I,J. MADDOX, continuous and Köthe-Toeplitz dual of certain sequence spaces, Proc. Camb. phil. Soc., 65(1969),431-435.
[8] [8] I,J. MADDOX, Elements of Functional Analysis, Cambridge University Press Cambridge, second edition, 1988.
[9] [9] J.S. shiue, On the Cesaro sequence spaces, Tamkang J. of Math. 1(1970), 19-25.
[10] [10] K.P. LIM, Matrix transformation in the Cesaro sequence spaces, KyungpookMath. J. , 14(1974),221-227
[11] [11] K.P. LIM, Matrix transformation on certain sequence space, Tamkang J. of Math. 8(1977), 213-220.
[12] [12] M. F. Rahman and A.B.M. Rezaul Karim, Generalized Riesz sequence space of Non-absolute Type and Some Matrix
Mapping. Pure and Applied Mathematics Journal.(2015); 4(3): 90-95.
[13] [13] M.F. Rahman and A.B.M. Rezaul Karim, Dual spaces of Generalized Weighted Cesaro sequence space and related Matrix
Mapping. Bulletin of Mathematics and Statistics Research ,vol.4.Issue.1.2016(January-March).
[14] [14] P.D. Johnson Jr. and R.N. Mohapatra, density of finitely non-zero sequences in some sequence spaces, Math. Japonica 24, No.
3(1979), 253-262.

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