MATEMATIQKI VESNIK
originalni nauqni rad
research paper
62, 1 (2010), 85–93
March 2010
ON A CLASS OF SEQUENCES RELATED TO THE lp SPACE
DEFINED BY A SEQUENCE OF ORLICZ FUNCTIONS
Ayhan Esi
Abstract. In this article we introduce the space m(Ω, φ, q) on generalizing the sequence
space m(φ) using the sequence of Orlicz functions. We study its different properties and obtain
some inclusion results involving the space m(Ω, φ, q).
1. Introduction
Let l∞ be the set of all real or complex sequences x = (xk ) with the norm
kxk = supk |xk | < ∞. A linear functional L on l∞ is said to be a Banach limit [1] if
it has the properties: (a) L(x) ≥ 0 if x ≥ 0 (i.e., xk ≥ 0 for all k ∈ N), (b) L(e) = 1,
where e = (1, 1, 1, . . . ), (c) L(Sx) = L(x), where the shift operator S is defined on
l∞ by (Sx)k = xk+1 .
Let B be the set of all Banach limits on l∞ . A sequence x is said to be almost
convergent to a number l if L(x) = l for all L ∈ B. Lorentz [5] has shown that x is
almost convergent to l if and only if
tmn = tmn (x) =
xn + xn+1 + · · · + xn+m
→ l as m → ∞, uniformly in n.
m+1
Throughout the article w(X), l∞ (X) and lp (X) denote respectively the spaces of
all, bounded and p-absolutely summable sequences with the elements in X, where
(X, q) is a seminormed space. By θ = (0, 0, . . . ), we denote the zero element in X.
The sequence space m(φ) was introduced by Sargent [10]. He studied some
of its properties and obtained its relationship with the space lp . Later on it was
investigated from sequence space point of view and related with summability theory by Rath and Tripathy [10], Tripathy [13],Tripathy and Sen [11], Tripathy and
Mahanta [12], Esi [4] and others.
An Orlicz function is a function M : (0, ∞] → (0, ∞] which is continuous,
non-decreasing and convex with M (0) = 0, M (x) > 0 for x > 0 and M (x) → ∞ as
x → ∞.
2010 AMS Subject Classification: 40A05, 46A45, 46E30.
Keywords and phrases: Seminorm; Orlicz function.
85
86
Ayhan Esi
An Orlicz function is said to satisfy ∆2 -condition for all values of u, if there
exists a constant K > 0, such that M (2u) ≤ KM (u), u ≥ 0.
Remark. An Orlicz function satisfies the inequality M (λx) ≤ λM (x) for all
λ with 0 < λ ≤ 1.
2. Definitions and background
Throughout the article Ps denotes the class of subsets of N, the natural numbers, those do not contain more than s elements. Further (φs ) will denote nondecreasing sequence of positive real numbers such that nφn+1 ≤ (n + 1)φn for all
n ∈ N. The class of all the sequences (φs ) satisfying this property is denoted by Φ.
The sequence space m(φ) introduced and studied by Sargent [10] which is
defined by:
o
n
1 P
|xk | < ∞ .
m(φ) = x = (xk ) : kxk = sup
s≥1,σ∈Ps φs k∈σ
Lindenstrauss and Tzafriri [6] used the idea of Orlicz function to construct the
sequence space
n
o
P ³ |xk | ´
lM = (xk ) :
M
< ∞, for some ρ > 0 .
ρ
k
The space lM becomes a Banach space with the norm
n
o
P ³ |xk | ´
M
kxk = inf ρ > 0 :
≤1
ρ
k
which is called an Orlicz sequence space. The space lM is closely related to the
space lp which is an Orlicz sequence space with M (x) = xp , 1 ≤ p < ∞.
A generalization of Orlicz sequence space is due to Woo [14]. Let Ω = (Mk )
be a sequence of Orlicz functions. Define the sequence space
n
³ |x | ´
o
P
k
l(Mk ) = x = (xk ) :
Mk
< ∞, for some ρ > 0
ρ
k
and equip this space with the norm
³ |x | ´
n
o
P
k
kxk = inf ρ > 0 :
Mk
≤1 .
ρ
k
The space l(Mk ) is a Banach space and is called a modular sequence space. The
space l(Mk ) also generalizes the concept of modulared sequence space earlier by
Nakano [7], who considered the space l(Mk ) when Mk (x) = xαk , where 1 ≤ αk < ∞
for k ≥ 1.
In the later stage different Orlicz sequence spaces were introduced and studied
by Parashar and Choudhary [8], Esi and Et [3], Esi [2] and many others.
In this article we introduce the following sequence spaces.
On a class of sequences related to the lp space
87
Let Ω = (Mk ) be a sequence of Orlicz functions. Then
n
³ ³ t (x) ´´
o
mn
b
l∞ (Ω, q) = x = (xk ) : sup Mm q
< ∞, for some ρ > 0 ,
ρ
m,n
m(Ω, φ, q) =
n
³ ³ t (x) ´´
o
1 P
mn
x = (xk ) : sup sup
Mm q
< ∞, for some ρ > 0 ,
ρ
m,n s≥1,σ∈Ps φs m∈σ
where
tmn (x) =
xn + xn+1 + · · · + xn+m
.
m+1
Let x = (xk ) be a sequence, then S(X) denotes the set of all permutations of
the elements of x = (xk ) i.e.,
S(X) = {(xπ(k) ) : π(k) is a permutation on N }.
A sequence space E is said to be symmetric if S(X) ⊂ E for all x ∈ E.
A sequence space E is said to be solid (or normal) if (αk xk ) ∈ E, whenever
x = (xk ) ∈ E and for all sequences of scalars (αk ) with |αk | ≤ 1 for all k ∈ B.
A sequence space E is said to be monotone, if it contains the canonical preimages of its step spaces.
Lemma. A sequence space E is monotone if E is solid.
3. Main results
In this section we prove some results involving the sequence spaces m(Ω, φ, q)
and b
l∞ (Ω, q).
Theorem 1.
field C.
m(Ω, φ, q) and b
l∞ (Ω, q) are linear spaces over the complex
Proof. Let x = (xk ), y = (yk ) ∈ m(Ω, φ, q) and α, β ∈ C. Then there exist
positive numbers ρ1 and ρ2 such that
³ ³ t (x) ´´
1 P
mn
sup sup
Mm q
<∞
ρ1
m,n s≥1,σ∈Ps φs m∈σ
and
sup
sup
m,n s≥1,σ∈Ps
³ ³ t (y) ´´
1 P
mn
Mm q
< ∞.
φs m∈σ
ρ
Let ρ3 = max(2|α|ρ1 , 2|β|ρ2 ). Since Mk is non-decreasing convex function for all k
and q is a seminorm, we have
³ ³ t (αx) ´
³ ³ t (αx + βy) ´´
³ t (βy) ´´
P
P
mn
mn
mn
Mm q
Mm q
≤
+q
ρ3
ρ3
ρ3
m∈σ
m∈σ
88
Ayhan Esi
³ ³ t (x) ´´
³ ³
´´
P
mn
(y)
Mm q
+
Mm q tmn
.
ρ2
ρ1
m∈σ
m∈σ
³ ³ t (αx + βy) ´´
1 P
mn
Mm q
⇒ sup sup
ρ3
m,n s≥1,σ∈Ps φs m∈σ
³ ³ t (x) ´´
³ ³ t (y) ´´
1 P
1 P
mn
mn
Mm q
+sup sup
Mm q
.
φs m∈σ
ρ1
ρ2
m,n s≥1,σ∈Ps φs m∈σ
≤
≤ sup
sup
m,n s≥1,σ∈Ps
P
Hence m(Ω, φ, q) is a linear space.
The proof for the case b
l∞ (Ω, q) is a routine work in view of the above proof.
Theorem 2. The space m(Ω, φ, q) is a seminormed space, seminormed by
o
n
³ ³ t (x) ´´
1 P
mn
≤1 .
g(x) = inf ρ > 0 : sup sup
Mm q
ρ
m,n s≥1,σ∈Ps φs m∈σ
Proof. Clearly g(x) ≥ 0 for all x ∈ m(Ω, φ, q) and g(θ) = 0. Let ρ1 > 0 and
ρ2 > 0 be such that
³ ³ t (x) ´´
1 P
mn
sup sup
Mm q
≤1
ρ1
m,n s≥1,σ∈Ps φs m∈σ
and
³ ³ t (y) ´´
1 P
mn
Mm q
≤ 1.
φ
ρ2
m,n s≥1,σ∈Ps s m∈σ
Let ρ = ρ1 + ρ2 . Then we have
³ ³ t (x + y) ´´
1 P
mn
Mm q
sup sup
ρ
m,n s≥1,σ∈Ps φs m∈σ
´´
³ ³ t (x)
1 P
mn
(y)
≤ sup sup
Mm q
+ tmn
ρ
2
ρ1
m,n s≥1,σ∈Ps φs m∈σ
n
³
³
´´
³ ³
´´o
1 P
ρ1
tmn (x)
ρ2
(y)
≤ sup sup
Mm q
+
Mm q tmn
ρ2
ρ1
ρ1 + ρ2
m,n s≥1,σ∈Ps φs m∈σ ρ1 + ρ2
³
³
´´
tmn (x)
ρ1
1 P
Mm q
≤
sup sup
+
ρ1 + ρ2 m,n s≥1,σ∈Ps φs m∈σ
ρ1
³ ³ t (y) ´´
1 P
ρ2
mn
sup sup
Mm q
≤ 1.
+
ρ1 + ρ2 m,n s≥1,σ∈Ps φs m∈σ
ρ2
sup
sup
Since the ρ0 s are non-negative, we have
n
³ ³
´´
o
1 P
g(x + y) = inf ρ > 0 : sup sup
Mm q tmn (x+y)
≤
1
ρ
m,n s≥1,σ∈Ps φs m∈σ
n
³
³
´´
o
1 P
(x)
≤ inf ρ1 > 0 : sup sup
Mm q tmn
≤
1
ρ1
m,n s≥1,σ∈Ps φs m∈σ
n
o
³
³
1 P
tmn (y) ´´
+ inf ρ2 > 0 : sup sup
≤1 .
Mm q
ρ2
m,n s≥1,σ∈Ps φs m∈σ
⇒ g(x + y) ≤ g(x) + g(y).
89
On a class of sequences related to the lp space
Next for λ ∈ C, without loss of generality, let λ 6= 0, then
n
´´
o
³ ³
1 P
≤1
g(λx) = inf ρ > 0 : sup sup
Mm q tmnρ(λx)
m,n s≥1,σ∈Ps φs m∈σ
n
³ ³
´´
o
1 P
ρ
= inf |λ|r > 0 : sup sup
Mm q tmnr(x)
≤ 1 , where r = |λ|
m,n s≥1,σ∈Ps φs m∈σ
n
³ ³
´´
o
1 P
= |λ| inf r > 0 : sup sup
Mm q tmnr(x)
≤ 1 = |λ|g(x).
m,n s≥1,σ∈Ps φs m∈σ
This completes the proof.
Proposition 3. The space b
l∞ (Ω, q) is a seminormed space, seminormed by
n
³ ³ t (x) ´´
o
mn
h(x) = inf ρ > 0 : sup Mm q
≤1 .
ρ
m,n
The proof of Proposition 3 is a consequence of the above theorem.
Theorem 4. m(Ω, φ, q) ⊂ m(Ω, ψ, q) if and only if sups≥1
Proof. Let sups≥1
sup
sup
m,n s≥1,σ∈Ps
φs
ψs
φs
ψs
< ∞.
< ∞ and x = (xk ) ∈ m(Ω, φ, q). Then
³ ³ t (x) ´´
1 P
mn
Mm q
< ∞ for some ρ > 0.
φs m∈σ
ρ
³ ³ t (x) ´´
1 P
mn
Mm q
⇒ sup sup
ρ
m,n s≥1,σ∈Ps ψs m∈σ
³ ³ t (x) ´´
³
´
φs
1 P
mn
Mm q
≤ sup
sup sup
< ∞.
ψ
φ
ρ
s≥1 s m,n s≥1,σ∈Ps s m∈σ
φs
ψs =
φ
limi→∞ ψssi
i
Then x = (xk ) ∈ m(Ω, ψ, q). Conversely, suppose that sups≥1
∞. Then
there exists a sequence of natural numbers (si ) such that
x = (xk ) ∈ m(Ω, φ, q), then there exists ρ > 0 such that
= ∞. Let
sup
sup
m,n s≥1,σ∈Ps
³ ³ t (x) ´´
1 P
mn
Mm q
< ∞.
φs m∈σ
ρ
Now we have
sup
sup
m,n s≥1,σ∈Ps
³ ³ t (x) ´´
1 P
mn
Mm q
ψs m∈σ
ρ
³
´
³ ³ t (x) ´´
φs
1 P
mn
≥ sup i sup sup
Mm q
= ∞.
ψ
φ
ρ
i≥1 si m,n i≥1,σ∈Psi si m∈σ
Therefore x = (xk ) ∈
/ m(Ω, ψ, q) which is a contradiction. Hence sups≥1
φs
ψs
< ∞.
90
Ayhan Esi
The following result is a consequence of Theorem 4.
Corollary 5. Let Ω = (Mm ) be a sequence of Orlicz functions. Then
m(Ω, φ, q) = m(Ω, ψ, q) if and only if sups≥1 ξs < ∞ and sups≥1 ξs−1 < ∞, where
φs
for all s = 1, 2, 3, . . . .
ξs = ψ
s
Theorem 6. Let M, M1 , M2 ∈ Ω be Orlicz functions satisfying ∆2 -condition.
Then
(a) m(M1 , φ, q) ⊂ m(M oM1 , φ, q),
(b) m(M1 , φ, q) ∩ m(M2 , φ, q) ⊂ m(M1 + M2 , φ, q).
Proof. (a) Let x = (xk ) ∈ m(M1 , φ, q), then there exists ρ > 0 such that
³ ³
´´
1 P
sup sup
M1 q tmnρ(x)
< ∞.
m,n s≥1,σ∈Ps φs m∈σ
Let 0 < ε < 1 and δ with 0 < δ < 1 such that M (t) < ε for 0 ≤ t ≤ δ. Write
ym,n = M1 (q( tmnρ(x) )) and for any σ ∈ Ps , consider
P
P
P
M (ym,n ) =
M (ym,n ) + M (ym,n )
m∈σ
1
2
where the first summation is over ym,n ≤ δ and the second is over ym,n > δ. By
the Remark, we have
P
P
P
M (ym,n ) ≤ M (1) ym,n ≤ M (2) ym,n
(3.1)
1
1
ym,n
δ
1
ym,n
δ ,
≤ 1+
since M is non-decreasing and
For ym,n > δ, we have ym,n <
convex, so
³
ym,n ´ 1
1 ³ 2ym,n ´
M (ym,n ) < M 1 +
< M (2) + M
.
δ
2
2
δ
Since M satisfies ∆2 -condition, so
M (ym,n ) <
Hence
P
2
1 ym,n
1 ym,n
ym,n
K
M (2) + K
M (2) = K
M (2).
2
δ
2
δ
δ
M (ym,n ) ≤ max(1, Kδ −1 M (2))
P
2
ym,n
(3.2)
By (3.1) and (3.2), we have x = (xk ) ∈ m(M oM1 , φ, q).
(b) The proof is trivial.
Taking M1 (x) = x in Theorem 6 (a), we have the following result.
Corollary 7. Let M ∈ Ω be Orlicz function satisfying ∆2 -condition. Then
m(φ, q) ⊂ m(M, φ, q).
From Theorem 4 and Corollary 7, we have:
On a class of sequences related to the lp space
91
Corollary 8. Let M ∈ Ω be Orlicz function satisfying ∆2 -condition. Then
φs
m(φ, q) ⊂ m(M, Ψ, q) if and only if sups≥1 Ψ
< ∞.
s
Theorem 9. Let Ω = (Mm ) be a sequence of Orlicz functions. Then the
sequence space m(Ω, φ, q) is solid and symmetric.
Proof. Let x = (xk ) ∈ m(Ω, φ, q). Then there exists ρ > 0 such that
³ ³ t (x) ´´
1 P
mn
sup sup
Mm q
< ∞.
φ
ρ
m,n s≥1,σ∈Ps s m∈σ
(3.3)
Let (λm ) be a sequence of scalars with |λm | ≤ 1 for all m ∈ N. Then the result
follows from (3.3) and the following inequality
³ ³ t (λx) ´´
³ ³
´´
P
P
mn
Mm q
≤
|λm |Mm q tmnρ(x)
(by Remark)
ρ
m∈σ
m∈σ
´´
³ ³
P
≤
Mm q tmnρ(x) .
m∈σ
The symmetricity of the space follows from the definition of the space m(Ω, φ, q)
and symmetric sequence space.
The following result follows from Theorem 9 and Lemma.
Corollary 10. Let Ω = (Mm ) be a sequence of Orlicz functions. Then the
sequence space m(Ω, φ, q) is monotone.
The proof of the following result is a routine verification.
Proposition 11. Let Ω = (Mm ) be a sequence of Orlicz functions. Then the
sequence space b
l∞ (Ω, q) is solid and as such is monotone.
Theorem 12. Let Ω = (Mm ) be a sequence of Orlicz functions.
m(Ω, φ, q) ⊂ b
l∞ (Ω, q).
Then
Proof. Let x = (xk ) ∈ m(Ω, φ, q). Then there exists ρ > 0 such that
³ ³ t (x) ´´
³ ³
´´
1 P
mn
Mm q
< ∞ ⇒ supm,n φ11 Mm q tmnρ(x)
<∞
sup sup
ρ
m,n s≥1,σ∈Ps φs m∈σ
for some ρ > 0, (on taking cardinality of σ to be 1). This implies that x = (xk ) ∈
b
l∞ (Ω, q). This completes the proof.
Theorem 13. Let Ω = (Mm ) be a sequence of Orlicz functions.
m(Ω, φ, q) = b
l∞ (Ω, q) if and only if sups≥1 φss < ∞.
Then
Proof. We have m(Ω, ψ, q) = b
l∞ (Ω, q) if ψs = s for all s ∈ N. By Theorem 4
and Theorem 9, it follows that m(Ω, φ, q) = b
l∞ (Ω, q) if and only if sups≥1 φss < ∞.
This completes the proof.
92
Ayhan Esi
Since the proof of the following proposition is not hard, we give it without
proof.
Proposition 14. Let Ω = (Mm ) be a sequence of Orlicz functions and q1 and
q2 be seminorms. Then
(a) m(Ω, φ, q1 ) ∩ m(Ω, φ, q2 ) ⊂ m(Ω, φ, q1 + q2 ) and b
l∞ (Ω, q1 ) ∩ b
l∞ (Ω, q2 ) ⊂
b
l∞ (Ω, q1 + q2 ),
(b) If q1 is stronger than q2 , then m(Ω, φ, q1 ) ⊂ m(Ω, φ, q2 ) and b
l∞ (Ω, q1 ) ⊂
b
l∞ (Ω, q2 ).
Theorem 15. Let Ω = (Mm ) be a sequence of Orlicz functions and (X, q) be
complete. Then {m(Ω, φ, q), g} is also complete.
Let (xi ) be a Cauchy sequence in m(Ω, φ, q),where xi = (xik ) =
for i ∈ N. Let r > 0 and x0 > 0 be fixed. Then
a positive integer n0 such that
Proof.
(xi1 , xi2 , xi3 , . . . ) ∈ m(Ω, φ, q)
for each rxε 0 > 0, there exists
ε
, for all i, j ≥ n0 .
rx0
n
³ ³ t (xi − xj ) ´´
o
1 P
mn
⇒ inf ρ > 0 : sup sup
Mm q
≤ 1 < ε,
ρ
m,n s≥1,σ∈Ps φs m∈σ
g(xi − xj ) <
(3.4)
for all i, j ≥ n0 . We have for all for all i, j ≥ n0
³ ³ t (xi − xj ) ´´
1 P
mn
Mm q
≤ 1, for some ρ > 0.
φ
g(xi − xj )
m,n s≥1,σ∈Ps s m∈σ
³ ³ |t (xi − xj )| ´´
1
mn
⇒ sup Mm q
≤1
g(xi − xj )
m,n φ1
³ ³ t (xi − xj ) ´´
mn
⇒ Mm q
≤ φ1 , for all i, j ≥ n0 and m, n ∈ N.
g(xi − xj )
⇒ sup
sup
We can find r > 0 such that
with Mm for all m, such that
rx0
x0
2 ηm ( 2 )
≥ φ1 , where ηm is the kernel associated
³ ³ |t (xi − xj )| ´´ rx
x0
mn
0
Mm q
≤
ηm ( )
i
j
g(x − x )
2
2
⇒ q(tmn (xi − xj )) <
rx0 ε
ε
.
= , for each m, n ∈ N.
2 rx0
2
In particular
q(t0n (xi − xj )) = q(xin − xjn ) <
rx0
ε
ε
·
= , for each fixed n.
2 rx0
2
Hence (xi ) is a Cauchy sequence in (X, q), which is complete. Therefore for each
n ∈ N, there exists xn ∈ X such that q(xin − xjn ) → 0 as i → ∞. Using the
On a class of sequences related to the lp space
93
continuity of Mm for all m and q is a seminorm, so we have
³ ³ t (xi − lim
j ´´
1 P
mn
j→∞ x )
Mm q
≤1
ρ
m,n s≥1,σ∈Ps φs m∈σ
³ ³ t (xi − x) ´´
1 P
mn
⇒ sup sup
Mm q
≤ 1, for some ρ > 0.
φ
ρ
m,n s≥1,σ∈Ps s m∈σ
sup
sup
Now taking the infimum of such ρ0 s, by (3.4), we get
n
inf ρ > 0 : sup
sup
m,n s≥1,σ∈Ps
³ ³ t (xi − x) ´´
o
1 P
mn
Mm q
≤ 1 < ε, for all i ≥ n0 .
φs m∈σ
ρ
Since m(Ω, φ, q) is linear space and (xi ) and (xi − x) are in m(Ω, φ, q), so it follows
that
(x) = (x − xi ) + (xi ) ∈ m(Ω, φ, q).
Hence m(Ω, φ, q) is complete.
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(received 24.03.2009, in revised form 31.07.2009)
Adiyaman University, Science and Art Faculty, Department of Mathematics, 02040, Adiyaman,
Turkey
E-mail: aesi23@hotmail.com