Some Geometric Properties of Lacunary Sequence Spaces ...downloads.hindawi.com/journals/aaa/2011/903736.pdfportant because Banach spaces with this property have the weak ﬁxed point

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2011, Article ID 903736, 13 pagesdoi:10.1155/2011/903736

Research ArticleSome Geometric Properties of Lacunary SequenceSpaces Related to Fixed Point Property

Chirasak Mongkolkeha and Poom Kumam

Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi(KMUTT), Bangmod, Thrungkru, Bangkok 10140, Thailand

Correspondence should be addressed to Poom Kumam, [email protected]

Received 18 June 2011; Accepted 2 November 2011

Academic Editor: Ondrej Dosly

Copyright q 2011 C. Mongkolkeha and P. Kumam. This is an open access article distributed underthe Creative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

The main purpose of this paper is considering the lacunary sequence spaces defined by Karakaya(2007), by proving the property (β) and Uniform Opial property.

1. Introduction

Let (X, ‖ · ‖) be a real Banach space and let B(X) (resp., S(X)) be a closed unit ball (resp.,the unit sphere) of X. For any subset A of X, we denote by conv(A) the convex hull of A.The Banach space X is uniformly convex (UC), if for each ε > 0 there exists δ > 0 suchthat for x, y ∈ S(X) the inequality ‖x − y‖ > ε implies ‖(x + y)/2‖ < 1 − δ (see [1]). ABanach space X has the property (β) if for each ε > 0 there exists δ > 0 such that 1 < ‖x‖ <1 + δ implies α (conv(B(X) ∪ {x}) \ B(X)) < ε, where α(A) denotes the Kuratowski measurenoncompactness of a subset A of X defined as the infimum of all ε > 0 such that A can becovered by a finite union of sets of diameter less than ε. The following characterization ofthe property (β) is very useful (see [2]): A Banach space X has the property (β) if and onlyif for each ε > 0 there exists δ > 0 such that for each element x ∈ B(X) and each sequence(xn) in B(X) with sep(xn) ≥ ε there is an index k for which ‖(x + xk)/2‖ < 1 − δ wheresep(xn) = inf{‖xn − xm‖ : n/=m} > ε. A Banach space X is nearly uniformly convex (NUC) iffor each ε > 0 and every sequence (xn) in B(X) with sep(xn) ≥ ε, there exists δ ∈ (0, 1) suchthat conv(xn) ∩ (1 − δ)B(X)/= ∅. Define for any x /∈ B(X) the drop D(x, B(X)) determinedby x by D(x, B(X)) = conv(B(X) ∪ {x}). A Banach space X has the drop property (write(D)) if for every closed set C disjoint with B(X) there exists an element x ∈ C such thatD(x, B(X)) ∩ C = {x}. A point x ∈ S(X) is an H-point of B(X) if for any sequence (xn) in Xsuch that ‖xn‖ → 1 as n → ∞, the week convergence of (xn) to x implies that ‖xn−x‖ → 0 as

2 Abstract and Applied Analysis

n → ∞. If every point in S(X) is anH-point of B(X), then X is said to have the property (H).A Banach space is said to have the uniform Kadec-Klee property (abbreviated as (UKK)) iffor every ε > 0 there exists δ > 0 such that for every sequence (xn) in S(X) with sep(xn) ≥ ε

and xnw−→ x as n → ∞, we have ‖x‖ < 1−δ. Every (UKK) Banach space hasH-property (see

[3]). The following implications are true in any Banach spaces,

(D) =⇒ (Rfx)⇑

(UC) =⇒ property(β) =⇒ (NUC) =⇒ (UKK) =⇒ property (H),(1.1)

where (Rfx) denotes the property of reflexivity (see [3–6]). A Banach space X is said to havethe Opial property (see [7]) if every sequence (xn)weakly convergent to x0 satisfies

limn→∞

inf ‖xn − x0‖ ≤ limn→∞

inf ‖xn − x‖, (1.2)

for every x ∈ X. Opial proved in [7] that the sequence space lp(1 < p < ∞) have this propertybut Lp[0, π] (p /= 2, 1 < p < ∞) do not have it. A Banach space X is said to have the uniformOpial property (see [8]), if for each ε > 0 there exists τ > 0 such that for any weakly nullsequence (xn) in S(X) and x ∈ X with ‖x‖ ≥ ε there holds

1 + τ ≤ limn→∞

inf ‖xn + x‖. (1.3)

For example, the space in [9, 10] has the uniform Opial property. The Opial property is im-portant because Banach spaces with this property have the weak fixed point property (see[11]) and the geometric property involving fixed point theory can be found, for example, in[9, 12–14].

For a bounded subset A ⊂ X, the set measure of noncompactness was defined in [15]by

α(A) = inf{ε > 0 : A can be covered by finitely many sets of diameter ≤ ε

}. (1.4)

The ball measure of noncompactness was defined in [16, 17] by

β(A) = inf{ε > 0 : A can be covered by finitely many balls of diameter ≤ ε

}. (1.5)

The functions α and β are called the Kuratowski measure of noncompactness and the Haus-dorffmeasure of noncompactness inX, respectively.We can associate these functions with thenotions of the set-contraction and ball contraction (see [18]). These notions are very usefultools to study nonlinear operator propblems (see [8, 18]). For each ε > 0 define that Δ(ε) =inf{1 − inf [‖x‖ : x ∈ A]:A is closed convex subset of B(X) with β(A) ≥ ε}. The function Δis called the modulus of noncompact convexity (see [16]). A Banach space X is said to haveproperty (L) if lim ε→ 1−Δ(ε) = 1. It has been proved in [8] that property (L) is a useful tool in

Abstract and Applied Analysis 3

the fixed point theory and that a Banach space X has property (L) if and only if it is reflexiveand has the uniform Opial property.

For a real vector space X, a function ρ : X → [0,∞] is called amodular if it satisfies thefollowing conditions:

(i) ρ(x) = 0 if and only if x = 0;

(ii) ρ(αx) = ρ(x) for all scalar α with |α| = 1;

(iii) ρ(αx + βy) ≤ ρ(x) + ρ(y), for all x, y ∈ X and all α, β ≥ 0 with α + β = 1;

the modular ρ is called convex if

(iv) ρ(αx + βy) ≤ αρ(x) + βρ(y), for all x, y ∈ X and all α, β ≥ 0 with α + β = 1.

For modular ρ on X, the space

Xρ ={x ∈ X : ρ(λx) −→ 0 as λ −→ 0+

}(1.6)

is called the modular space.A sequence (xn) in Xρ is called modular convergent to x ∈ Xρ if there exists a λ > 0 such

that ρ(λ(xn − x)) → 0 as n → ∞.A modular ρ is said to satisfy the Δ2-condition (ρ ∈ Δ2) if for any ε > 0 there exist

constants K ≥ 2 and a > 0 such that

ρ(2u) ≤ Kρ(u) + ε (1.7)

for all u ∈ Xρ with ρ(u) ≤ a.If ρ satisfies the Δ2-condition for any a > 0 with K ≥ 2 dependent on a, we say that ρ

satisfies the strong Δ2-condition (ρ ∈ Δs2).

By a lacunary sequence θ = (kr), where k0 = 0, we will mean an increasing sequenceof nonnegative integers with kr −kr−1 → ∞ as r → ∞. The intervals determined by θ will bedenote by Ir = (kr−1, kr]. We write hr = kr − kr−1 and the ratio kr/kr−1, will be denoted by qr .The space of lacunary strongly convergent sequence Nθ was defined by Freedman et al. [19]as

Nθ =

{

x = (xk) : limr→∞

1hr

∑

k∈Ir|xk − l| = 0, for some l

}

. (1.8)

It is well known that there is very closed connection between the space of lacunary strong-ly convergent sequence and the space of strongly Cesaro summability sequences. This con-nection can be found in [18–23], because a lot of these connection, a lot of geometric propertyof Cesaro sequence spaces can generalize the lacunary sequence spaces.

Let w be the space of all real sequences. Let p = (pr) be a bounded sequence of thepositive real numbers. In 2007, Karakaya [24] introduced the new sequence spaces l(p, θ)involving lacunary sequence as follows:

l(p, θ)=

{

x = (x(i)) :∞∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

< ∞}

(1.9)


and paranorm on l(p, θ) is given by

‖x‖l(p,θ) =( ∞∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr)1/M

, (1.10)

whereM = suprpr . If pr = p for all r ∈ N, we will use the notation lp(θ) in place of l(p, θ). Thenorm on lp(θ) is given by

‖x‖lp(θ) =( ∞∑

r=1

(1hr

∑

i∈Ir|x(i)|

)p)1/p

. (1.11)

By using the properties of lacunary sequence in the space l(p, θ), we get the following se-quences. If θ = (2r), then l(p, θ) = ces(p). If θ = (2r) and pr = p for all r ∈ N, then l(p, θ) = cesp.For x ∈ l(p, θ) defined the modular on l(p, θ) by

�(x) =∞∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

. (1.12)

It is easy to see that if limr→∞ sup pr < ∞ then � ∈ Δs2. The Luxembourg norm on l(p, θ) is

defined by

‖x‖ = inf{ε > 0 : �

(x

ε

)≤ 1}. (1.13)

The Luxembourg norm on lp(θ) can be reduced to a usual norm on lp(θ) [24], that is,

‖x‖ = ‖x‖lp(θ). (1.14)

Throughout this paper, we assume that limr→∞ inf pr > 1 and limr→∞ sup pr < ∞ andfor x ∈ w, i ∈ N, we denote

ei =

⎛

⎜⎝

i−1 times︷︸︸︷0, 0, . . . , 0, 1, 0, 0, 0, . . .

⎞

⎟⎠,

x|i = (x(1), x(2), x(3), . . . , x(i), 0, 0, 0, . . .),

x|N−i = (0, 0, 0, . . . , x(i + 1), x(i + 2), . . .).

(1.15)

The following results are very important for our consideration.


Lemma 1.1 (see [25, Lemma 2.1]). If ρ ∈ Δs2, then for any L > 0 and ε > 0, there exists δ =

δ(L, ε) > 0 such that

∣∣ρ(u + v) − ρ(u)

∣∣ < ε, (1.16)

whenever u, v ∈ Xρ with ρ(u) ≤ L, and ρ(v) ≤ δ.

Lemma 1.2 (see [25, Lemma 2.3]). Convergence in norm and in modular are equivalent in Xρ ifρ ∈ Δ2.

Lemma 1.3 (see [25, Corollary 2.2, Lemma 2.3]). If ρ ∈ Δ2, then for any sequence (xn) in Xρ,‖xn‖ → 0 if and only if ρ(xn) → 0 as n → ∞.

Lemma 1.4 (see [25, Lemma 2.4]). If ρ ∈ Δs2, then for any ε > 0 there exists δ = δ(ε) > 0 such

that ‖x‖ ≥ 1 + δ whenever ρ(x) ≥ 1 + ε.

Lemma 1.5 (see [24, Lemma 2.3]). The functional � is a convex modular on l(p, θ).

Lemma 1.6 (see [24, Lemma 2.5]). (i) For any x ∈ l(p, θ), if ‖x‖ < 1, then �(x) ≤ ‖x‖.(ii) For any x ∈ l(p, θ), ‖x‖ = 1 if and only if �(x) = 1.

2. The Main Results

In this section, we prove the property (β) and uniform Opial property in lacunary sequenceand connect to the fixed point property. First we shall give some results which are veryimportant for our consideration.

Lemma 2.1. For any x ∈ l(p, θ), there exists k0 ∈ N and λ ∈ (0, 1) such that �(xk/2) ≤ ((1 −λ)/2)�(xk) for all k ∈ N with k ≥ k0, where

xk =

⎛

⎜⎝

k−1︷︸︸︷0, 0, . . . , 0, x(k), x(k + 1), x(k + 2), . . .

⎞

⎟⎠. (2.1)

Proof. Let k ∈ N be fixed. So there exist rk ∈ N such that k ∈ Irk . Let α be a real number suchthat 1 < α ≤ limr→∞ inf pr , then there exists k0 ∈ N such that α < prk for all k ≥ k0. Chooseλ ∈ (0, 1) as a real such that (1/2)α ≤ (1 − λ)/2. Then for each x ∈ l(p, θ) and k ≥ k0, we have

�

(xk

2

)

=∞∑

r=rk

(1hr

∑

i∈Ir

∣∣∣∣x(i)2

∣∣∣∣

)pr

=∞∑

r=rk

(12

)pr(

1hr

∑

i∈Ir|x(i)|

)pr


≤(12

)α ∞∑

r=rk

(1hr

∑

i∈Ir|x(i)|

)pr

≤ 1 − λ

2�(xk).

(2.2)

Lemma 2.2. For any x ∈ l(p, θ) and ε ∈ (0, 1) there exists δ ∈ (0, 1) such that �(x) ≤ 1 − ε implies‖x‖ ≤ 1 − δ.

Proof. Suppose that the lemma does not hold, then there exist ε > 0 and xn ∈ l(p, θ) such that�(xn) ≤ 1 − ε and 1/2 ≤ ‖xn‖ ↗ 1. Let an = (1/‖xn‖) − 1. Then an → 0 as n → ∞. LetL = sup{�(2xn) : n ∈ N}. Since � ∈ Δs

2 there exists K ≥ 2 such that

�(2u) ≤ K�(u) + 1, (2.3)

for every u ∈ l(p, θ)with �(u) < 1. By (2.3), we have �(2xn) ≤ K�(xn)+1 ≤ K+1 for all n ∈ N.Hence 0 < L < ∞. By Lemmas 1.5 and 1.6(ii), we have

1 = �

(xn

‖xn‖)

= �(2anxn + (1 − an)xn)

≤ an�(2xn) + (1 − an)�(xn)

≤ anL + (1 − ε) −→ 1 − ε,

(2.4)

which is a contradiction.

Theorem 2.3. The space l(p, θ) is Banach spaces with respect to the Luxemburg norm.

Proof. Let (xn) = (xn(i)) be a Cauchy sequence in l(p, θ) and ε ∈ (0, 1). Thus there existsN ∈ N such that ‖xn − xm‖ < εM for all n,m ≥ N. By Lemma 1.6 (i), we have

�(xn − xm) ≤ ‖xn − xm‖ < εM ∀n,m ≥ N. (2.5)

That is,

∞∑

r=1

(1hr

∑

i∈Ir|xn(i) − xm(i)|

)pr

< εM ∀n,m ≥ N. (2.6)

For fixed r, we get that

|xn(i) − xm(i)| < ε ∀n,m ≥ N. (2.7)


Thus let (xn(i)) be a Cauchy sequence in R for all i ∈ N. Since R is complete, then there existsx(i) ∈ R such that xm(i) → x(i) as m → ∞ for all i ∈ N. Thus for fixed r, we have

|xn(i) − x(i)| < ε as m −→ ∞, ∀n ≥ N. (2.8)

This implies that, for all n ≥ N,

�(xn − xm) −→ �(xn − x) as m −→ ∞. (2.9)

This means that, for all n ≥ N,

∞∑

r=1

(1hr

∑

i∈Ir|xn(i) − xm(i)|

)pr

−→∞∑

r=1

(1hr

∑

i∈Ir|xn(i) − x(i)|

)pr

(2.10)

asm → ∞. By (2.6), we have

�(xn − x) ≤ ‖xn − xm‖ < εM ≤ ε ∀n ≥ N. (2.11)

This implies that xn → x as n → ∞. So we have xN − x ∈ l(p, θ). By the linearity of thesequence space l(p, θ), we have x = (xN − x) + xN ∈ l(p, θ). Therefore the sequence spacel(p, θ) is Banach space, with respect to the Luxemburg norm and the proof is complete.

Theorem 2.4. The space l(p, θ) has property (β).

Proof. Let ε > 0 and (xn) ⊂ B(l(p, θ)) with sep(xn) ≥ ε. For each k ∈ N, there exist rk ∈ N suchthat k is a minimal element in Irk . Let

xkn =

⎛

⎜⎝

k−1︷︸︸︷0, 0, . . . , 0, xn(k), xn(k + 1), xn(k + 2), . . .

⎞

⎟⎠. (2.12)

Since for each i ∈ N, (xn(i))∞n=1 is bounded. By using the diagonal method, we have that for

each k ∈ N we can find subsequence (xnj ) of (xn) such that (xnj (i)) converges for each i ∈ N.Therefore, for any k ∈ N there exists an increasing sequence (tk) such that sep((xk

nj)j>tk) ≥ ε.

Hence for each k ∈ N there exists sequence of positive integers (sk)∞k=1 with s1 < s2 < s3 < · · ·

such that ‖xksk‖ ≥ ε/2, and since � ∈ Δs

2, by Lemma 1.3 we may assume that there exists η > 0such that �(xk

sk) ≥ η for all k ∈ N, that is,

∞∑

r=rk

(1hr

∑

i∈Ir

∣∣∣xksk(i)

∣∣∣

)pr

≥ η (2.13)


for all k ∈ N. On the other hand by Lemma 2.1, there exist k0 ∈ N and λ ∈ (0, 1) such that

�

(uk

2

)

≤ 1 − λ

2�(uk)

(2.14)

for all u ∈ l(p, θ) and k ≥ k0. From Lemma 2.2, there exist δ ∈ (0, 1) such that for any y ∈l(p, θ),

�(y) ≤ 1 − λη

4=⇒ ∥∥y∥∥ ≤ 1 − δ. (2.15)

Since again � ∈ Δs2, by Lemma 1.1, there exists δ0 such that

∣∣�(u + v) − �(u)∣∣ <

λη

4, (2.16)

whenever �(u) ≤ 1 and �(v) ≤ δ0. Since x ∈ B(l(p, θ)), we have that �(x) ≤ 1. Then there exitsk ≥ k0 such that �(xk) ≤ δ0. We put u = xk

sk and v = xk,

�(u2

)=

∞∑

r=rk

(1hr

∑

i∈Ir

∣∣∣∣xsk(i)2

∣∣∣∣

)pr

< 1, �(v2

)=

∞∑

r=rk

(1hr

∑

i∈Ir

∣∣∣∣x(i)2

∣∣∣∣

)pr

< δ0. (2.17)

From (2.14) and (2.16), we have

∞∑

r=rk

(1hr

∑

i∈Ir

∣∣∣∣x(i) + xsk(i)

2

∣∣∣∣

)pr

= �(u + v

2

)≤ �(u2

)+λη

4≤ 1 − λ

2(�(u)

)+λη

4. (2.18)

By (2.13), (2.16), (2.18), and convexity of function f(t) = |t|pr , for all r ∈ N, we have

�

(x + xsk

2

)=

∞∑

r=1

(1hr

∑

i∈Ir

∣∣∣∣x(i) + xsk(i)

2

∣∣∣∣

)pr

=rk−1∑

r=1

(1hr

∑

i∈Ir

∣∣∣∣x(i) + xsk(i)

2

∣∣∣∣

)pr

+∞∑

r=rk

(1hr

∑

i∈Ir

∣∣∣∣x(i) + xsk(i)

2

∣∣∣∣

)pr

≤ 12

(rk−1∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

+rk−1∑

r=1

(1hr

∑

i∈Ir|xsk(i)|

)pr)


+∞∑

r=rk

(1hr

∑

i∈Ir

∣∣∣∣xsk(i)2

∣∣∣∣

)pr

+λη

4

≤ 12

(rk−1∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

+rk−1∑

r=1

(1hr

∑

i∈Ir|xsk(i)|

)pr)

+1 − λ

2

∞∑

r=rk

(1hr

∑

i∈Ir|xsk(i)|

)pr

+λη

4

=12

rk−1∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

+12

rk−1∑

r=1

(1hr

∑

i∈Ir|xsk(i)|

)pr

+1 − λ

2

∞∑

r=rk

(1hr

∑

i∈Ir|xsk(i)|

)pr

+λη

4

=12

rk−1∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

+12

∞∑

r=1

(1hr

∑

i∈Ir|xsk(i)|

)pr

− λ

2

∞∑

r=rk

(1hr

∑

i∈Ir|xsk(i)|

)pr

+λη

4

≤ 12+12− λη

2+λη

4

= 1 − λη

4.

(2.19)

So it follows from (2.15) that

∥∥∥∥x + xsk

2

∥∥∥∥ ≤ 1 − δ. (2.20)

Therefore, the space l(p, θ) has property (β).

By the facts presented in the introduction, following results are obtained directly fromTheorem 2.4.

Corollary 2.5. The space lp(θ) has property (β).

Corollary 2.6. The space l(p, θ) is nearly uniform convexity, has drop property, and is reflexive.

Corollary 2.7. The space l(p, θ) has property (UKK).

Corollary 2.8 (see [24, Theorem 2.9]). The space l(p, θ) has property (H).

Corollary 2.9. The space lp(θ) is nearly uniform convexity, has drop property, and is reflexive.


Corollary 2.10. The space lp(θ) has property (UKK) and (H).

Theorem 2.11. The space l(p, θ) has uniform Opial property.

Proof. Take any ε > 0 and x ∈ l(p, θ) with ‖x‖ ≥ ε. Let (xn) be weakly null sequence inS(l(p, θ)). By limr→∞ sup pr < ∞, that is, � ∈ Δs

2, hence by Lemma 1.2 there exists δ ∈ (0, 1)independent of x such that �(x) > δ. Also, by � ∈ Δs

2, Lemma 1.1 asserts that there existsδ1 ∈ (0, δ) such that

∣∣�(y + z

) − �(y)∣∣ <

δ

4, (2.21)

whenever �(y) ≤ 1 and �(z) ≤ δ1. Choose r0 ∈ N such that

∞∑

r=r0+1

(1hr

∑

i∈Ir|x(i)|

)pr

<δ14. (2.22)

So, we have

δ <r0∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

+∞∑

r=r0+1

(1hr

∑

i∈Ir|x(i)|

)pr

≤r0∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

+δ14,

(2.23)

which implies that

r0∑

r=1

(1hr

∑

i∈Ir|x(i)|

)pr

> δ − δ14

> δ − δ

4=

3δ4. (2.24)

Since xnw−→ 0, then there exists n0 ∈ N such that

3δ4

≤r0∑

r=1

(1hr

∑

i∈Ir|xn(i) + x(i)|

)pr

(2.25)

for all n > n0, since weak convergence implies coordinatewise convergence. Again, by xnw−→

0, then there exists n1 ∈ N such that

∥∥∥xn|k0

∥∥∥ < 1 −(1 − δ

4

)1/M(2.26)


for all n > n1 where k0 is a minimal element in Ir0+1 and M ∈ N with pr ≤ M for all r ∈ N.Hence, by the triangle inequality of the norm, we get

∥∥∥xn|

N−k0

∥∥∥ >

(1 − δ

4

)1/M

. (2.27)

It follows by the definition of ‖ · ‖ that we have

1 < �

(xn|

N−k0

(1 − (δ/4))1/M

)

=∞∑

r=r0+1

((1/hr)

∑i∈Ir |xn(i)|

(1 − (δ/4))1/M

)pr

≤(

1

(1 − (δ/4))1/M

)M ∞∑

r=r0+1

(1hr

∑

i∈Ir|xn(i)|

)pr

(2.28)

Which implies that

∞∑

r=r0+1

(1hr

∑

i∈Ir|xn(i)|

)pr

> 1 − δ

4(2.29)

for all n > n1. By inequality (2.21), (2.25), and (2.29), it yields for any n > n1 that

�(xn + x) =r0∑

r=1

(1hr

∑


)pr

+∞∑

r=r0+1

(1hr

∑


)pr

≥ 3δ4

+∞∑

r=r0+1

(1hr

∑

i∈Ir|xn(i)|

)pr

− δ

4

≥ 3δ4

+(1 − δ

4

)− δ

4

≥ 1 +δ

4.

(2.30)

Since � ∈ Δs2 and by Lemma 1.4 there exists τ depending on δ only such that ‖xn + x‖ ≥ 1 + τ ,

which implies that limn→∞ inf ‖xn + x‖ ≥ 1 + τ , hence the prove is complete.

By the facts presented in the introduction and the reflexivity of l(p, θ), we get thefollowing results.

Corollary 2.12. The space lp(θ) has uniform Opial property.

Corollary 2.13. The space l(p, θ) has property (L) and the fixed point property.

Corollary 2.14. The space lp(θ) has property (L) and the fixed point property.


Acknowledgments

The authors are grateful to the referees for their valuable comments. They also wish tothank the Higher Education Research Promotion and National Research University Project ofThailand, Office of the Higher Education Commission for financial support (under the Projectno. 54000267). Mr. Chirasak Mongkolkeha was supported by the Thailand Research Fundthrough the Royal Golden Jubilee Program under Grant PHD/0029/2553 for Ph.D. programat KMUTT, Thailand. The second author was supported by the Commission on HigherEducation, the Thailand Research Fund, and the King Mongkuts University of TechnologyThonburi (Grant no. MRG5380044).

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