# New Characterizations ofk-Uniformly Extremely Convex Banach Spaces

Authors: Suyalatu Wulede
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Two characterizations ofk-uniformly extremely convex spaces are given in this paper. As special cases for k=1, two necessary and sufficient conditions of uniform extreme convexity are presented and a characteristic inequality of uniformly extremely convex spaces is given.


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## Body

## 1. Introduction

In the geometric theory of Banach spaces, the concept of uniform convexity plays a very significant role and is frequently used in functional analysis. In 2012, this concept has been generalized by Wulede and Ha [1]. They defined the so-called uniform extreme convexity and showed many interesting connections between this concept and several others (see [1–8]). Recently, Wulede et al. [9], generalizing the uniform extreme convexity, introduced the so-called k-uniform extreme convexity. Every k-uniformly extremely convex space is (k+1)-uniformly extremely convex, but the converse implication is not true (see [9]). It is shown that 1-uniform extreme convexity coincides with uniform extreme convexity (see [9]).In 1960, Singer [10] introduced k-strictly convex spaces. It is well known that every k-strictly convex space is (k+1)-strictly convex space. Also, it was shown that 1-strict convexity coincides with strict convexity.Throughout the sequel, the symbolX denotes a real Banach space and X∗ denotes its dual space. B(X)={x:x∈X,x≤1}, S(X)={x:x∈X,x=1} denote the closed unit ball and the unit sphere in X, respectively. For x1,x2,…,xk+1∈SX, the k-dimensional volume enclosed by x1,x2,…,xk+1 is given by(1)Vx1,x2,…,xk+1=sup11⋯1f1x1f1x2⋯f1xk+1⋯⋯⋯⋯fkx1fkx2⋯fkxk+1:fi∈SX∗,i=1,…,k.In 1972, Daneš [11] proved the so-called “Drop Theorem”. In 1987, modifying the assumption of Daneš drop theorem, Rolewicz [6] began the study of drop property for the closed unit ball. He defined the norm · to have the drop property, if for every closed set C⊂X disjoint from closed unit ball B(X) there exists a x∈C such that D(x,B(X))∩C={x}, where the set D(x,B(X)), the convex hull of x and B(X), is called the drop generated by x∉B(X).Definition 1 (see [1]).
A Banach spaceX is said to be uniformly extremely convex space provided that for any sequences {xn},{yn} of norm-one elements of X, if there is a norm-one functional f∈X∗ such that limn→∞⁡f(xn)=limn→∞⁡f(yn)=1, then limn→∞⁡xn-yn=0.Definition 2 (see [9]).
A Banach spaceX is said to be k-uniformly extremely convex space provided that for any sequences {xn(i)}i=1k+1 of norm-one elements of X, if there is a norm-one functional f∈X∗ such that limn→∞⁡f(xn(i))=1 for i=1,…,k+1, then limn→∞⁡V(xn(1),xn(2),…,xn(k+1))=0.Lemma 3 (see [9]).
A Banach space X isk-uniformly extremely convex if and only if for any ϵ>0 there exists δ(ϵ)>0 such that for each norm-one functional f∈X∗ and for all norm-one elements x1,…,xk+1 of X with f(x1+x2+⋯+xk+1)>(k+1)-δ, one has V(x1,x2,…,xk+1)<ϵ.Lemma 4 (Singer [10], Wulede [9]).
Every strictly convex (uniformly extremely convex, respectively) space isk-strictly convex (k-uniformly extremely convex, respectively).Lemma 5 (Wulede [1, 9]).
LetX be a Banach space; let k be an integer with k≥1. Then, X is k-uniformly extremely convex if and only if X is k-strictly convex and has the drop property.Lemma 6 (see [12]).
Let1<p<+∞; then we have(2)1+t1+⋯+tkk+1p≤1+t1p+⋯+tkpk+1,where t1,t2,…,tk≥0, and the sign of equality holds if and only if t1=t2=⋯=tk=1.


## 2. Characterizations ofk-Uniformly Extremely Convex Spaces

Theorem 7.
A Banach spaceX is k-uniformly extremely convex if and only if for any k+1 sequences {xn(i)}i=1k+1⊂X, if limn→∞⁡xn(i)=τ,i=1,…,k+1 and for some norm-one functional f∈X∗, limn→∞⁡f(xn(1)+xn(2)+⋯+xn(k+1))=(k+1)τ, then limn→∞⁡V(xn(1),xn(2),…,xn(k+1))=0.Proof.
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Necessity. Without loss of generality, we may assume that τ≠0 and suppose that for any k+1 sequences {xn(i)}i=1k+1⊂X with limn→∞⁡xn(i)=τ,i=1,…,k+1, and for some norm-one functional f∈X∗, limn→∞⁡fxn1+xn2+⋯+xnk+1=k+1τ.
Letyn(i)=xn(i)/xn(i),i=1,…,k+1, and then we have yn(i)=1,i=1,…,k+1, and(3)limn→∞fyn1+yn2+⋯+ynk+1=limn→∞fxn1xn1+xn2xn2+⋯+xnk+1xnk+1=k+1.It follows that limn→∞⁡f(yn(i))=1,i=1,…,k+1. By the assumption that X is a k-uniformly extremely convex space, we have limn→∞⁡V(yn(1),yn(2),…,yn(k+1))=0. Hence,(4)limn→∞Vyn1,yn2,…,ynk+1=limn→∞Vxn1xn1,xn2xn2,…,xnk+1xnk+1=1τklimn→∞Vxn1,xn2,…,xnk+1=0.It follows that limn→∞V(xn(1),xn(2),…,xn(k+1))=0.
Sufficiency. Suppose that {xn(i)}i=1k+1 are sequences of norm-one elements of X such that for some norm-one functional f∈X∗ the equality limn→∞f(xn(i))=1 holds for i=1,…,k+1. It follows that limn→∞f(xn(1)+xn(2)+⋯+xn(k+1))=k+1. From the conditions given here, we have(5)limn→∞Vxn1,xn2,…,xnk+1=0.This shows that X is k-uniformly extremely convex space.Theorem 8.
Let1<p<+∞, X be a Banach space, and M be an arbitrary bounded subset of X. Then X is k-uniformly extremely convex if and only if for any ϵ>0 there exists 0<δ(ϵ,p)<1 such that the inequality(6)fx1+x2+⋯+xk+1k+1p≤1-δϵ,px1p+x2p+⋯+xk+1pk+1holds for any norm-one functional f∈X∗ and all elements x1,x2,…,xk+1∈M with(7)Vx1,x2,…,xk+1≥ϵ.Proof.
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Sufficiency. Suppose that for every ϵ>0 there is a 0<δ1=δ1(ϵ,p)<1 such that the inequality(8)fx1+x2+⋯+xk+1k+1p≤1-δ1holds for any norm-one functional f∈X∗ and all norm-one elements x1,x2,…,xk+1 of X with Vx1,x2,…,xk+1≥ϵ.
Let us denoteδ2=1-1-δ11/p>0 and take δ(ϵ)=(k+1)δ2. Then we have(9)fx1+x2+⋯+xk+1≤k+1-δ.By Lemma 3, we know that X is a k-uniformly extremely convex space.
Necessity. Suppose the inequality (6) is not true. Then there exist r>0 and ϵ0>0 such that for each n∈ N there exist {xn(i)}i=1k+1⊂X satisfying xn(1)≤r for i=1,…,k+1, V(xn(1),xn(2),…,xn(k+1))≥ϵ0 and(10)fxn1+xn2+⋯+xnk+1k+1p>1-1nxn1p+xn2p+⋯+xnk+1pk+1.
Putrn=max1≤i≤k+1⁡xn(i) and take yn(1)=xn(1)/rn,yn(2)=xn(2)/rn,…,yn(k+1)=xn(k+1)/rn,ϵ1=ϵ0/rk. Rearranging the vectors, we can assume that yn(1)=1. We have rn≤r for every n and(11)Vyn1,yn2,…,ynk+1=1rnkVxn1,xn2,…,xnk+1≥ϵ0rk=ϵ1.
Combining inequality (10) with Lemma 6 we derive that(12)1-1n<fyn1+yn2+⋯+ynk+1/k+1pyn1p+yn2p+⋯+ynk+1p/k+1≤1+yn2+⋯+ynk+1/k+1p1+yn2p+⋯+ynk+1p/k+1≤1.It follows that(13)limn→∞fyn1+yn2+⋯+ynk+1/k+1pyn1p+yn2p+⋯+ynk+1p/k+1=1=limn→∞1+yn2+⋯+ynk+1/k+1p1+yn2p+⋯+ynk+1p/k+1.
We shall show that(14)limn→∞⁡yni=1for every i=1,…,k+1. To this end, it is enough to show that every subsequence of the sequence yin has a further subsequence converging to 1. Since every bounded sequence has a convergent subsequence, without loss of generality, we may assume that the sequence yin converges to some ti. The right equality in (13) shows that(15)1k+1∑i=1k+1tip=1k+1∑i=1k+1tiwhere t1=1. This is the equality in Ho¨lder’s inequality, so it holds only if(16)1k+1∑m=1k+1tmp=tipfor every i=1,…,k+1. Consequently, ti=t1=1 for every i=1,…,k+1 and the left equality in (13) shows that(17)limn→∞fyn1+yn2+⋯+ynk+1=k+1.
Combining equality (17) with the assumption that X is a k-uniformly extremely convex space, by Theorem 7 we have(18)limn→∞Vyn1,yn2,…,ynk+1=0;this contradicts (11).In particular, considering the special cases of Theorems7 and 8 when k = 1, we obtain a necessary and sufficient condition for uniform extreme convexity and a characteristic inequality of uniformly extremely convex spaces.Corollary 9.
A Banach spaceX is uniformly extremely convex if and only if for any sequences {xn},{yn} in X, if  limn→∞⁡xn=limn→∞⁡yn=τ and for some norm-one functional f∈X∗, limn→∞⁡f(xn+yn)=2τ, then limn→∞⁡xn-yn=0.Corollary 10.
Let1<p<+∞, X be a Banach space, and M be an arbitrary bounded subset of X. Then X is uniformly extremely convex if and only if for any ϵ>0, there exists 0<δ(ϵ,p)<1 such that the inequality(19)fx+yp2p≤1-δϵ,pxp+yp2holds for any norm-one functional f∈X∗ and all elements x,y∈M with x-y≥ϵ.In addition to two characterizations of uniformly extremely convex spaces presented in Lemmas4 and 5, we can also obtain another feature of uniformly extremely convex spaces as follows.Corollary 11.
A Banach spaceX is uniformly extremely convex if and only if X is strictly convex and k-uniformly extremely convex.



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