Volume 2009, Article ID 729045, 13 pagesdoi:10.1155/2009/729045 Research Article On Generalized Paranormed Statistically Convergent Sequence Spaces Defined by Orlicz Function Metin Bas¸a
Trang 1Volume 2009, Article ID 729045, 13 pages
doi:10.1155/2009/729045
Research Article
On Generalized Paranormed
Statistically Convergent Sequence Spaces
Defined by Orlicz Function
Metin Bas¸arir and Selma Altunda ˘g
Department of Mathematics, Faculty of Science and Arts, Sakarya University, 54187 Sakarya, Turkey
Correspondence should be addressed to Metin Bas¸arir,basarir@sakarya.edu.tr
Received 8 May 2009; Revised 3 August 2009; Accepted 26 August 2009
Recommended by Andrei Volodin
We define generalized paranormed sequence spaces c σ, M, p, q, s, c0σ, M, p, q, s, mσ, M, p, q, s, and m0σ, M, p, q, s defined over a seminormed sequence space X, q We establish some
inclusion relations between these spaces under some conditions
Copyrightq 2009 M Bas¸arir and S Altunda˘g This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
1 Introduction
wX, cX, c0X, cX, c0X, l∞X, mX, m0X will represent the spaces of all,
con-vergent, null, statistically concon-vergent, statistically null, bounded, bounded statistically
convergent, and bounded statistically null X-valued sequence spaces throughout the paper,
where X, q is a seminormed space, seminormed by q For X C, the space of complex numbers, these spaces represent the w, c, c0, c, c0, l∞, m, m0 which are the spaces of all, convergent, null, statistically convergent, statistically null, bounded, bounded statistically convergent, and bounded statistically null sequences, respectively The zero sequence is
denoted by θ θ, θ, θ, , where θ is the zero element of X.
The idea of statistical convergence was introduced by Fast1 and studied by various authors see 2 4 The notion depends on the density of subsets of the set N of natural
numbers A subset E of N is said to have density δE if
δE lim n → ∞ n1 n
k1
where χ E is the characteristic function of E.
Trang 2A sequence x x k is said to be statistically convergent to the number L i.e., x k ∈
c if for every ε > 0
In this case, we write x k stat→ L or stat − lim x L.
Let σ be a mapping of the set of positive integers into itself A continuous linear functional φ on l∞, the space of real bounded sequences, is said to be an invariant mean
or σ-mean if and only if
1 φx ≥ 0 when the sequence x x n has x n ≥ 0 for all n ∈ N,
2 φe 1, where e 1, 1, ,
3 φx σn φx for all x ∈ l∞.
The mappings σ are one to one and such that σ k n / n for all positive integers n and k, where σ k n denotes the kth iterate of the mapping σ at n Thus φ extends the limit functional
on c, the space of convergent sequences, in the sense that φx lim x for all x ∈ c In that case σ is translation mapping n → n 1, a σ-mean is often called a Banach limit, and V σ, the set of bounded sequences all of whose invariant means are equal, is the set of almost convergent sequences5
If x x n , set Tx Tx n x σn It can be shown 6 that
V σ
x x n : limm t mn x L e uniformly in n, L σ − lim x
1.3
where t mn x x n Tx n T m x n /m 1.
Several authors including Schaefer 7, Mursaleen 6, Savas 8, and others have studied invariant convergent sequences
An Orlicz function is a function M : 0, ∞ → 0, ∞, which is continuous, nondecreasing, and convex with M0 0, Mx > 0 for x > 0 and Mx → ∞ as x → ∞ If the convexity of an Orlicz function M is replaced by
x y≤ Mx My
then this function is called modulus function, introduced and investigated by Nakano9 and followed by Ruckle10, Maddox 11, and many others
Lindenstrauss and Tzafriri 12 used the idea of Orlicz function to construct the sequence space
l M
x ∈ w :∞
k1
M
|x
k|
ρ
which is called an Orlicz sequence space
Trang 3The space l Mbecomes a Banach space with the norm
x inf
ρ > 0 :∞
k1
M
|x
k|
ρ
The space l M is closely related to the space l pwhich is an Orlicz sequence space with
and Choudhary13, Bhardwaj and Singh 14, and many others
It is well known that since M is a convex function and M0 0 then Mtx ≤ tMx for all t with 0 < t < 1.
An Orlicz funtion M is said to satisfyΔ2-condition for all values of u, if there exists constant K > 0, such that M2u ≤ KMu u ≥ 0 The Δ2-condition is equivalent to the
inequality MLu ≤ KLMu for all values of u and for L > 1 being satisfied 15
The notion of paranormed space was introduced by Nakano16 and Simons 17 Later on it was investigated by Maddox 18, Lascarides 19, Rath and Tripathy 20, Tripathy and Sen21, Tripathy 22, and many others
The following inequality will be used throughout this paper Let p p k be a sequence
of positive real numbers with 0 < p k ≤ sup p k G and let D max1, 2 G−1 Then for
a k , b k ∈ C, the set of complex numbers for all k ∈ N, we have 23
|a k b k|p k ≤ D |a k|p k |b k|p k
2 Definitions and Notations
A sequence space E is said to be solid or normal if α k x k ∈ E, whenever x k ∈ E and for
all sequencesα k of scalars with |α k | ≤ 1 for all k ∈ N.
A sequence space E is said to be symmetric if x k ∈ E implies x πk ∈ E, where πk
is a permutation ofN
A sequence space E is said to be monotone if it contains the canonical preimages of its
step spaces
Throughout the paper p p k will represent a sequence of positive real numbers and
X, q a seminormed space over the field C of complex numbers with the seminorm q We
define the following sequence spaces:
c
σ, M, p, q, s
x k ∈ l∞X : k −s M
q x σ k n − L
ρ
p k
stat
−→ 0,
as k −→ ∞, uniformly in n, s ≥ 0, for some ρ > 0, L ∈ X ,
c0
σ, M, p, q, s
x k ∈ l∞X : k −s M
q x σ k n
ρ
p k
stat
−→ 0,
as k −→ ∞, uniformly in n, s ≥ 0, for some ρ > 0
,
Trang 4σ, M, p, q, s
x k ∈ l∞X : sup
k,n k −s M
q x σ k n
ρ
p k
< ∞,
s ≥ 0, for some ρ > 0
,
σ, M, p, q, s
x k ∈ l∞X : lim
j
1
j
j
k1
q x σ k n − L
ρ
p k
0,
uniformly in n, s ≥ 0, for some ρ > 0
2.1
We write
m
σ, M, p, q, s
cσ, M, p, q, s
∩ l∞σ, M, p, q, s
,
m0
σ, M, p, q, s
c0
σ, M, p, q, s
∩ l∞
σ, M, p, q, s
If Mx x, qx |x|, s 0, σn n1 for each n and k 0 then these spaces reduce
to the spaces
c
p
x k ∈ w : |x k − L| p k −→ 0, as k −→ ∞, L ∈ Xstat ,
c0
p
x k ∈ w : |x k|p k −→ 0, as k −→ ∞stat ,
l∞
p
x k ∈ w : sup
k |x k|p k < ∞ ,
m
p
cp
∩ l∞p
,
m0
p
c0
p
∩ l∞p
,
2.3
defined by Tripathy and Sen21
Firstly, we give some results; those will help in establishing the results of this paper
Lemma 2.1 21 For two sequences p k and t k one has m0p ⊇ m0t if and only if
lim infk∈K p k /t k > 0, where K ⊆ N such that δK 1.
Lemma 2.2 21 Let h inf p k and G sup p k , then the followings are equivalent:
i G < ∞ and h > 0,
ii mp m.
Lemma 2.3 21 Let K {n1, n2, n3, } be an infinite subset of N such that δK 0 Let
Then T is uncountable.
Trang 5Lemma 2.4 24 If a sequence space E is solid then E is monotone.
3 Main Results
Theorem 3.1 cσ, M, p, q, s, c0σ, M, p, q, s, mσ, M, p, q, s, m0σ, M, p, q, s are linear spaces.
Proof Let x k , y k ∈ cσ, M, p, q, s Then there exist ρ1, ρ2 positive real numbers and K,
L ∈ X such that
q x σ k n − K
ρ1
p k
stat
−→ 0, as k −→ ∞, uniformly in n,
q y σ k n − L
ρ2
p k
stat
−→ 0, as k −→ ∞, uniformly in n.
3.1
Let α, β be scalars and let ρ3 max2|α|ρ1, 2|β|ρ2 Then by 1.7 we have
k −s
M
q
αx σ k n βy σ k n − αK βL
ρ3
p k
≤ k −s M
q x σ k n − K 2ρ1
q y σ k n − L
2ρ2
p k
≤ k −s 1
2p k M
q x σ k n − K
ρ1
M
q y σ k n − L
ρ2
p k
≤ D
q x σ k n − K
ρ1
p k
k −s M
q y σ k n − L
ρ2
p k
stat
−→ 0 as k −→ ∞, uniformly in n.
3.2
Hencecσ, M, p, q, s is a linear space.
The rest of the cases will follow similarly
Theorem 3.2 The spaces mσ, M, p, q, s and m0σ, M, p, q, s are paranormed spaces, paranormed
by
gx inf
ρ p m /H: sup
k k −s M
q x σ k n
ρ
≤ 1, uniformly in n, s ≥ 0, ρ > 0, m ∈ N , 3.3
where H max1, sup p k .
Trang 6Proof We prove the theorem for the space m0σ, M, p, q, s The proof for the other space can
be proved by the same way Clearly gx g−x for all x ∈ m0σ, M, p, q, s and gθ 0 Let x, y ∈ m0σ, M, p, q, s Then we have ρ1, ρ2> 0 such that
sup
k k −s M
q x σ k n
ρ1
≤ 1, uniformly in n,
sup
k
k −s M
q y σ k n
ρ2
≤ 1, uniformly in n.
3.4
Let ρ ρ1 ρ2 Then by the convexity of M, we have
sup
k k −s M
q x σ k n y σ k n
ρ
≤ sup
k
k −s M ρ1
ρ1 ρ2
q x σ k n
ρ1
ρ2
ρ1 ρ2
q y σ k n
ρ2
≤ ρ1
ρ1 ρ2sup
k
k −s M
q x σ k n
ρ1
ρ ρ2
1 ρ2
sup
k k −s M
q y σ k n
ρ2
≤ 1, uniformly in n.
3.5
Hence from above inequality, we have
g
x y
inf
ρ p m /H: sup
k k −s M
q x σ k n y σ k n
ρ
≤ 1, uniformly in n, ρ > 0, m∈ N
≤ inf
ρ p m /H
1 : sup
k k −s M
q x σ k n
ρ1
≤ 1, uniformly in n, ρ1> 0
inf
ρ p m /H
2 : sup
k k −s M
q y σ k n
ρ1
≤ 1, uniformly in n, ρ2> 0
gx gy
.
3.6
Trang 7For the continuity of scalar multiplication let λ / 0 be any complex number Then by
the definition of g we have
gλx inf
ρ p m /H: sup
k
k −s M
q λx σ k n
ρ
≤ 1, uniformly in n, ρ > 0
inf
r|λ| p m /H: sup
k
k −s M
q x σ k n
r
≤ 1, uniformly in n, r > 0 ,
3.7
where r ρ/|λ|.
Since|λ| p m ≤ max1, |λ| H , we have |λ| p m /H ≤ max1, |λ| H1/H Then
1, |λ| H1/Hinf
r p m /H: sup
k k −s M
q x σ k n
r
≤ 1, uniformly in n, r > 0
max1, |λ| H1/H · gx,
3.8
and therefore gλx converges to zero when gx converges to zero or λ converges to zero Hence the spaces mσ, M, p, q, s and m0σ, M, p, q, s are paranormed by g.
Theorem 3.3 Let X, q be complete seminormed space, then the spaces mσ, M, p, q, s and
m0σ, M, p, q, s are complete.
Proof We prove it for the case m0σ, M, p, q, s and the other case can be established similarly Let x s xs
σ k n be a Cauchy sequence in m0σ, M, p, q, s for all k, n ∈ N Then gx i −x j → 0,
as i, j → ∞ For a given ε > 0, let r > 0 and δ > 0 to be such that ε/rδ > 0 Then there exists
a positive integer N such that
g
x i − x j
< ε
Using definition of paranorm we get
inf
⎧
⎨
⎩ρ p k /H: supk k −s M
⎛
⎝q
⎛
⎝x i σ k n − x j σ k n
ρ
⎞
⎠
⎞
⎠ ≤ 1, uniformly in n, ρ > 0
⎫
⎬
⎭<
ε
rδ .
3.10
Hence x iis a Cauchy sequence inX, q Therefore for each ε 0 < ε < 1 there exists a positive integer N such that
q
x i − x j
Trang 8Using continuity of M, we find that
sup
k k −s M
q
x i− limj x j
ρ
Thus
sup
k k −s M
q
x i − x
ρ
Taking infimum of such ρ s we get
inf
ρ p k /H : sup
k
k −s M
q
x i − x
ρ
for all i ≥ N and j → ∞ Since x i ∈ m0σ, M, p, q, s and M is continuous, it follows that
x ∈ m0σ, M, p, q, s This completes the proof of the theorem.
Theorem 3.4 Let M1and M2be two Orlicz functions satisfyingΔ2-condition Then
i Zσ, M1, p, q, s ⊆ Zσ, M2◦ M1, p, q, s,
ii Zσ, M1, p, q, s ∩ Zσ, M2, p, q, s ⊆ Zσ, M1 M2, p, q, s,
where Z c, m, c0, and m0.
x k ∈ c0σ, M1, p, q, s Then for a given 0 < ε < 1, there exists ρ > 0 such that there exists a
subset K of N with δK 1, where
K
k ∈ N : k −s M1
q x σ k n
ρ
p k
< ε B
,
B max
1, sup
1
k −s1/p k
p k
.
3.15
If we take a k k −s1/p k M1qx σ k n /ρ then a p k
k < ε/B < 1 implies that a k < 1.
Hence we have by convexity of M,
M2◦ M1
q x σ k n
ρ
M2
a k
k −s1/p k
≤ a k M2
1
k −s1/p k
. 3.16
Thus
k −s M2a kp k ≤ k −s
a k
k −s1/p k
p k
≤ k −s Ba kp k ≤ Ba kp k < ε. 3.17
Trang 9Hence by3.15 it follows that for a given ε > 0, there exists ρ > 0 such that
δ
k ∈ N : k −s M2
q x σ k n
ρ
p k
< ε
Thereforex k ∈ c0σ, M2◦ M1, p, q, s.
ii We prove this part for the case Z c0and the other cases will follow similarly Letx k ∈ c0σ, M1, p, q, s ∩ c0σ, M2, p, q, s Then by using 1.7 it can be shown that
x k ∈ c0σ, M1 M2, p, q, s Hence
c0
σ, M1, p, q, s
∩ c0
σ, M2, p, q, s
⊆ c0
σ, M1 M2, p, q, s
This completes the proof
Theorem 3.5 For any sequence p p k of positive real numbers and for any two seminorms q1and
q2on X one has
Z
σ, M, p, q1, s
∩ Zσ, M, p, q2, s
where Z c, m, c0, and m0.
Proof The proof follows from the fact that the zero sequence belongs to each of the classes the
sequence spaces involved in the intersection
The proof of the following result is easy, so omitted
Proposition 3.6 Let M be an Orlicz function which satisfies Δ2−condition, and let q1and q2be two
i c0σ, M, p, q1, s ⊆ cσ, M, p, q1, s,
ii m0σ, M, p, q1, s ⊆ mσ, M, p, q1, s,
iii Zσ, M, p, q1, s ∩ Zσ, M, p, q2, s ⊆ Zσ, M, p, q1 q2, s where Z c, m, c0, and m0,
iv if q1is stronger than q2, then
Z
σ, M, p, q1, s
⊆ Zσ, M, p, q2, s
where Z c, m, c0, and m0.
Theorem 3.7 The spaces Zσ, M, p, q, s are not solid, where Z c and m.
Proof To show that the spaces are not solid in general, consider the following example Let Mx x p 1 ≤ p < ∞, p k 1/p for all k, qx sup i |x i |, where x x i ∈ l∞ and
sequencex k , where x k x i
k ∈ l∞is defined by x i
k k, k, k, , k i2, i ∈ N and
x i
k 2, 2, 2, , k / i2, i ∈ N for each fixed k ∈ N Hence x k ∈ Zσ, M, p, q, s for Z c and m Let α k 1, 1, 1, if k is odd and α k θ, otherwise Then α k x k /∈ Zσ, M, p, q, s for
Z c and m Thus Zσ, M, p, q, s is not solid for Z c and m.
Trang 10The proof of the following result is obvious in view ofLemma 2.4.
Proposition 3.8 The space Zσ, M, p, q, s is solid as well as monotone for Z c0and m0.
Theorem 3.9 The spaces Zσ, M, p, q, s are not symmetric, where Z c, m, c0, and m0.
Proof To show that the spaces are not symmetric, consider the following examples Let Mx x p 1 ≤ p < ∞, p k 1/p for all k, qx sup i |x i |, where x x i ∈ l∞ and
sequencex k defined by x k 1, 1, 1, if k i2, i ∈ N, and x k θ, otherwise Then
x k ∈ Zσ, M, p, q, s for Z c0and m0 Lety k be a rearrangement of x k, which is defined
as y k 1, 1, 1, if k is odd and y k θ, otherwise Then y k /∈ Zσ, M, p, q, s for Z c0
and m0
To show for Z c and m, let p k 1 for all k odd and p k 2−1for all k even Let X R3
and qx max{|x1|, |x2|, |x3|}, where x x1, x2, x3 ∈ R3 Let Mx x4and σ n n 1 for all n ∈ N Then we have σ k n n k for all k, n ∈ N We consider
x k
⎧
⎨
⎩
1, 1, 1, i2≤ k < i2 2i − 1, i ∈ N,
Thenx k ∈ Zσ, M, p, q, s for Z c and m We consider the rearrengement y k of x k as
y k
⎧
⎨
⎩
1, 1, 1, k is odd,
Theny k /∈ Zσ, M, p, q, s for Z c and m Thus the spaces Zσ, M, p, q, s are not symmetric
in general, where Z c, m, c0and m0
Proposition 3.10 For two sequences p k and t k one has m0σ, M, p, q, s ⊇ m0σ, M, t, q, s if
and only if lim inf k∈K p k /t k > 0, where K ⊆ N such that δK 1.
The following result is a consequence of the above result
Corollary 3.11 For two sequences p k and t k one has m0σ, M, p, q, s m0σ, M, t, q, s if and
only if lim inf k∈K p k /t k > 0 and lim inf k∈K t k /p k > 0, where K ⊆ N such that δK 1.
The following result is obvious in view ofLemma 2.2
Proposition 3.12 Let h inf p k and G sup p k , then the followings are equivalent:
i G < ∞ and h > 0,
ii mσ, M, p, q, s mσ, M, q, s.
... class="text_page_counter">Trang 7For the continuity of scalar multiplication let λ / be any complex number Then by< /i>
the definition...
Trang 5Lemma 2.4 24 If a sequence space E is solid then E is monotone.
3...
and therefore gλx converges to zero when gx converges to zero or λ converges to zero Hence the spaces mσ, M, p, q, s and m0σ, M, p, q, s are paranormed by g.