Let δX(ϵ) and R(1,X) be the modulus of convexity and the Domínguez-Benavides coefficient, respectively. According to these two geometric parameters, we obtain a sufficient condition for normal structure, that is, a Banach space
X has normal structure if
2δX(1+ϵ)>max{(R(1,x)-1)ϵ,1-(1-ϵ/R(1,X)-1)} for some ϵ∈[0,1] which generalizes the known result by
Gao and Prus.

1. Introduction

Let X be a Banach space. Throughout the paper, denote by SX,BX the unit sphere and unit ball of X, respectively. Recall
that a Banach space X is said to have normal structure (resp., weak,
normal, structure) if for every closed bounded (resp., weakly compact) convex
subset C of X with diam C>0, there exists x∈C such that sup{∥x−y∥:y∈C}<diamC, where diam C=sup{∥x−y∥:x,y∈C}. For a reflexive Banach space, the normal
structure and weak normal structure are the same.
Recently a good deal of investigations have focused on finding the sufficient
conditions with various geometrical constants for a Banach space to have normal
structure (see, e.g., [1–5]). The geometric condition sufficient for normal
structure in terms of the modulus of convexity is given by Goebel [6], who proved that X has normal structure provided that δX(1)>0. Here the function δX(ϵ):[0,2]→[0,1], defined by Clarkson [7] asδX(ϵ)=inf{1−∥x+y∥2:x,y∈BX,∥x−y∥≥ϵ},is called
the modulus of
convexity of X. Later Gao and Prus generalized the above
results as the following (see [2, 8]).

Theorem 1.1.

A Banach space X has normal structure provided that δX(1+ϵ)>ϵ/2 for some ϵ∈[0,1].

In this paper, we obtain a class of Banach spaces with
normal structure, which involves the coefficient R(1,X). This coefficient is defined by
Domínguez Benavides [9] asR(1,X)=sup{lim infn→∞∥x+xn∥},where the supremum is taken over
all x∈X with ∥x∥≤1 and all weakly null sequence (xn) in BX such thatD[(xn)]:=lim supn→∞(lim supm→∞∥xn−xm∥)≤1.Obviously, 1≤R(1,X)≤2.

2. Main Results

Let us begin this section with a sufficient condition
for a Banach space X having weak normal structure and the idea in
the following proof is due to [5, Lemma 5].

Lemma 2.1.

Let X be a Banach space for which BX* is w*-sequentially compact. If X does not have weak normal structure, then for
any η>0,
there exist x1,x2∈SX and f1,f2∈SX*, such that

|∥x1−x2∥−1|<η;

|fi(xj)|<η for i≠j and fi(xi)=1,i,j=1,2;

∥x1+x2∥≤R(1,X)(1+η).

Proof.

Assume
that X does not have weak normal structure. It is
well known that (see, e.g., [10]) there exists a sequence {xn} in X satisfying

xn is weakly convergent to 0;

diam ({xn}n=1∞)=1=limn∥xn−x∥ for all x∈clco{xn}n=1∞.

Since BX* is w*-sequentially compact, we can find {fn} in SX* satisfying

fn(xn)=∥xn∥ for all n;

fn→w*f for some f∈BX*.

Let η∈(0,1) sufficiently small and ϵ=η/3. Then, by the properties of the sequence (xn),
we can choose n1∈ℕ such that|f(xn1)|<ϵ2,1−ϵ≤∥xn1∥≤1.Note that the sequence {xn} is weakly null and verifies D[{xn}]=1. It follows from the definition of R(1,X) thatliminfn∥xn+xn1∥≤R(1,X).The rest of
the proof is similar to that of [5, Lemma 5].

Theorem 2.2.

A Banach space X has normal structure provided that δX(1+ϵ)>f(ϵ) for some ϵ∈[0,1], where the function f(ϵ) is defined asf(ϵ):={(R(1,X)−1)ϵ2,0≤ϵ≤1R(1,X),12(1−1−ϵR(1,X)−1),1R(1,X)<ϵ≤1.

Proof.

Observe that X is uniformly nonsquare [11] and then reflexive. Therefore
normal structure and weak normal structure coincide.

Assume first that X fails to have weak normal structure. Fix η>0 sufficiently small and ϵ∈[0,1].
It follows that there exist x1,x2∈SX and f1,f2∈SX*, satisfying the condition in Lemma 2.1. Next,
denote by R:=R(1,X) and consider two cases for ϵ∈[0,1].

Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M83"><mml:mrow><mml:mi>ϵ</mml:mi><mml:mo>∈</mml:mo><mml:mo stretchy="false">[</mml:mo><mml:mn>0</mml:mn><mml:mo>,</mml:mo><mml:mn>1</mml:mn><mml:mo>/</mml:mo><mml:mi>R</mml:mi><mml:mo stretchy="false">]</mml:mo></mml:mrow></mml:math></inline-formula>).

Now let us putx=x1−x21+η,y=(1−(R−1)ϵ)x1+ϵx21+η,and so x∈BX,∥y∥=∥ϵ1+η(x1+x2)+1−Rϵ1+ηx1∥≤Rϵ+(1−Rϵ)=1,and also that∥x−y∥=∥(R−1)ϵ1+ηx1−1+ϵ1+ηx2∥≥1+ϵ1+ηf2(x2)−(R−1)ϵ1+ηf2(x1)≥1+ϵ−η1+η,∥x+y∥=∥(2−(R−1))ϵ1+ηx1−1−ϵ1+ηx2∥≥(2−(R−1))ϵ1+ηf1(x1)−1−ϵ1+ηf1(x2)≥(1−2η1+η)(2−(R−1)ϵ).By the definition of modulus of
convexity,(1−2η1+η)(2−(R−1)ϵ)≤∥x+y∥≤2(1−δX(∥x−y∥))≤2(1−δX(1+ϵ−η1+η)),or equivalently,(2−(R−1)ϵ)≤2(1−δX(1+ϵ−η1+η))(1+2η1−η).Letting η→0, we have2δX(1+ϵ)≤(R−1)ϵ,which contradicts our
hypothesis.

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M92"><mml:mrow><mml:mi>ϵ</mml:mi><mml:mo>∈</mml:mo><mml:mo stretchy="false">(</mml:mo><mml:mn>1</mml:mn><mml:mo>/</mml:mo><mml:mi>R</mml:mi><mml:mn>,1</mml:mn><mml:mo stretchy="false">]</mml:mo></mml:mrow></mml:math></inline-formula>).

In this
case R>1, otherwise ϵ>1. Letx′=x2−x11+η,y′=(1−(R−1)ϵ′)x1+ϵ′x21+η,where ϵ′=1−(R−1)ϵ∈[0,1/R).
It follows from Case 1 that x,y∈BX,∥x−y∥≥(1−2η1+η)(2−(R−1)ϵ′),∥x+y∥≥1+ϵ′−η1+η.This implies
thatδX(2−(R−1)ϵ′)≤12(1−ϵ′),which is equivalent toδX(1+ϵ)≤12(1−1−ϵR−1).This is a contradiction.

Remark 2.3.

(1) It
is readily seen that f(ϵ)≤ϵ/2 for any ϵ∈[0,1] and
Theorem 2.2 is therefore a generalization of
Theorem 1.1. Moreover this generalization is strict whenever X is the space with R(1,X)<2.

(2) Consider the space X=ℝ2 with the norm ∥(x,y)∥:=max(|x|,|y|,|x−y|). It is known that δX(ϵ)=max{0,(ϵ−1)/2} [8] and R(1,X)=1, then X has normal structure from Theorem 2.2, but lies
out of the scope of Theorem 1.1.

Corollary 2.4.

Let X be a Banach space with R(1,X)=1 and δX(1+ϵ)>0forsomeϵ∈[0,1], then X has normal structure.

Corollary 2.5.

If X is a Banach space withδX(1+1R(1,X))>12(1−1R(1,X)),then X has normal structure.

Remark 2.6.

Corollary 2.5 is equivalent to
[4, Corollary 24].

Acknowledgments

The authors
would like to express their sincere thanks to the referees for their valuable
suggestions on this paper. This work is supported by NSF of Education Department
of Henan Province (2008A110012).

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