# A New Characterization ofk-Uniformly Rotund Banach Space

Authors: Suyalatu Wulede; Tingting Li; Xuan Qin
Journal: Journal of Function Spaces
 (2015)

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## Abstract

A new characterization ofk-uniformly rotund Banach space with
1<P<+∞ is given. Moreover, a corresponding result in the
locally k-uniformly rotund Banach space with 1<P<+∞ 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. The concept of a uniformly rotund (or uniformly convex) Banach space was first introduced by Clarkson [1] in 1936 and this class of Banach space is very interesting and has numerous applications (cf. [2–8]). In 1979, Sullivan [9] introduced the k-uniformly rotund spaces as a generalization of uniformly rotund Banach spaces. Indeed, the 1-uniformly rotund Banach spaces coincide with usually uniformly rotund Banach spaces.The purpose of this paper is to give a character inequality ofk-uniformly rotund Banach space (or locally k-uniformly rotund Banach space) with 1<p<+∞. Throughout the sequel, the symbol X denotes a real Banach space and X* denotes its dual space. B(X)={x:x∈X,x≤1} and S(X)={x:x∈X,x=1} denote, respectively, the unit ball and the unit sphere in X. For x1,x2,…,xk+1∈SX, the k-dimensional volume enclosed by x1,x2,…,xk+1 is given by
(1)A(x1,x2,…,xk+1)=sup⁡11⋯1f1(x1)f1(x2)⋯f1(xk+1)⋯⋯⋯⋯fk(x1)fk(x2)⋯fk(xk+1);11⋯1f1(x1)f1(x2)⋯f1(xk+1)⋯⋯⋯⋯fk(x1)fk(x2)⋯fk(xk+1)fi∈S(X*),i=1,…,k.A Banach spaceX is said to be k-uniformly rotund [9] if, for any ϵ>0, there is a δ(ϵ)>0, such that, for x1,x2,…,xk+1∈S(X), if x1+x2+⋯+xk+1>(k+1)-δ then Ax1,x2,…,xk+1<ϵ.A Banach spaceX is said to be locally k-uniformly rotund [9] if, for ∀ϵ>0, x∈S(X), there is a δ(ϵ,x)>0, such that, for x1,…,xk∈S(X), if x+x1+⋯+xk>(k+1)-δ then Ax,x1,…,xk<ϵ.


## 2. A New Characterization ofk-Uniformly Rotund Banach Spaces

Theorem 1.
Let1<p<+∞, let X be a Banach space, and let M be an arbitrary bounded subset of X. Then, X is k-uniformly rotund space if and only if, for any ϵ>0, there exists 0<δ(ϵ,p)<1, such that the inequality
(2)x+x1+⋯+xkk+1p≤(1-δ(ϵ,p))xp+x1p+⋯+xkpk+1
holds for all x∈M and x1,…,xk∈X with Ax,x1,…,xk≥ϵ.In order to prove Theorem1, we give three lemmas.Lemma 2 (Yu [10]).
X is k-uniformly rotund space if and only if for any k+1 sequences
(3)x1nn=1∞,x2nn=1∞,…,xk+1nn=1∞⊂X,
if xin→a(i=1,2,…,k+1), x1n+x2n+⋯+xk+1n→(k+1)a, n→∞, then
(4)Ax1n,x2n,…,xk+1n⟶0,n⟶∞.For the sake of completeness of this paper, here we present the proof of Lemma2.The sufficiency of Lemma2 is clear.The Proof of Necessity. Without loss of generality, we may assume that a=1.Suppose thatxinn=1∞⊂X(i=1,2,…,k+1), satisfying the conditions given in Lemma 2. Then, for each δ>0, by the assumption that x1n+x2n+⋯+xk+1n→k+1n→∞, there exists N0>0, such that the inequality
(5)x1n+x2n+⋯+xk+1n>k+1-δ2
holds for all n>N0.On the other hand, sincexin→1(i=1,2,…,k+1, n→∞), so there exists N1≥N0, such that the inequality
(6)xin-1<δ2(k+1),(i=1,2,…,k+1)
holds for all n>N1.Therefore, by lettingyin=xin/xin, i=1,2,…,k+1, n=1,2,…, we can deduce that
(7)y1n+y2n+⋯+yk+1n=x1n+⋯+xk+1n+y1n-x1n+⋯+yk+1n-xk+1n≥x1n+⋯+xk+1n-∑i=1k+1xinxin-xin>k+1-δ2-δk+1δk+2=k+1-δ.
By the assumption that X is k-uniformly rotund space, we may take δ=δ(ϵ/2) for any ϵ>0. Therefore, by the above proof, there exists an N1 corresponding to δ=δ(ϵ/2) such that the inequality
(8)Ay1n,y2n,…,yk+1n<ϵ2
holds for all n>N1.Furthermore, by using inequality (8), we easily obtain the desired result that
(9)Ax1n,x2n,…,xk+1n⟶0,n⟶∞.Lemma 3.
Let1<p<+∞, then one has
(10)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.Proof.
(1∘) When k=1, we construct a function ft1=1+t1/2p-1+t1p/2; then
(11)f′(t1)=p21+t12p-1-t1p-1.
Obviously,
(12)f′t1=0ift1=1f′t1>0ift1<1f′t1<0ift1>1.
It is easy to see that the function f(t1) attains its maximum value at point t1=1 and f(1)=0. Hence f(t1)≤f(1)=0; that is,
(13)1+t12p≤1+t1p2.
And the sign of equality holds if and only if t1=1.
(2∘) Suppose the conclusion of Lemma 3 is true when k=n-1; that is, the inequality
(14)1+t1+⋯+tn-1np≤1+t1p+⋯+tn-1pn
holds and the sign of equality holds if and only if t1=t2=⋯=tn-1=1.
(3∘) When k=n, we construct a multivariate function
(15)ft1,t2,…,tn=1+t1+⋯+tnn+1p-1+t1p+⋯+tnpn+1;
then
(16)∂f∂tn=pn+11+t1+⋯+tnn+1p-1-tnp-1.
Now, let us fix variables t1,t2,…,tn-1. Then the function f(t1,t2,…,tn) attains its maximum value at point tn=(1+t1+⋯+tn-1)/n. Hence
(17)f(t1,t2,…,tn)≤1+t1+⋯+tn-1+1+t1+⋯+tn-1/nn+1p-1+t1p+⋯+tn-1p+1+t1+⋯+tn-1/npn+1=nn+11+t1+⋯+tn-1np-1+t1p+⋯+tn-1pn≤0.
This shows that the inequality
(18)1+t1+⋯+tnn+1p≤1+t1p+⋯+tnpn+1
holds and the sign of equality holds if and only if t1=t2=⋯=tn=1.
Combining (1∘), (2∘), and (3∘), we have
(19)1+t1+⋯+tkk+1p≤1+t1p+⋯+tkpk+1,
and the sign of equality holds if and only if t1=t2=⋯=tk=1.Lemma 4.
Let1<p<+∞, then one has
(20)1+t1+⋯+tk/k+1p1+t1p+⋯+tkp/k+1≤1k+1p-1k-1+1+t1p1+t1p1/(p-1)p-1,
where t1,t2,…,tk≥0.Proof.
(1∘) When k=1, the conclusion of Lemma 4 is obvious. When k=2, we construct a function
(21)ft1,t2=1+t1+t2/3p1+t1p+t2p/3=13p-11+t1+t2p1+t1p+t2p;
then
(22)∂f∂t2=13p-1p1+t1+t2p-11+t1p-t2p-1-t2p-1t11+t1p+t2p2.
It is easy to see that the function f(t1,t2) attains its maximum value at point t2=(1+t1p)/(1+t1)1/(p-1). Hence
(23)ft1,t2≤13p-11+t1+1+t1p/1+t11/(p-1)p1+t1p+1+t1p/1+t1p/(p-1)=13p-11+1+t1p1+t1p1/(p-1)p-1.
(2∘) Suppose the conclusion of Lemma 4 is true when k=n-1; that is, we have
(24)1+t1+⋯+tn-1/np(1+t1p+⋯+tn-1p)/n≤1np+1n-2+1+t1p1+t1p1/(p-1)p-1.
(3∘) When k=n, we construct a multivariate function
(25)ft1,t2,…,tn=1+t1+⋯+tn/n+1p1+t1p+⋯+tnp/n+1=1n+1p-11+t1+⋯+tnp1+t1p+⋯+tnp;
then(26)∂f∂tn=1n+1p-1p1+t1+⋯+tnp-11+t1p+⋯+tn-1p-tnp1+t1+⋯+tn-11+t1p+⋯+tnp2.Now, let us fix variables t1,t2,…,tn-1. Then the function f(t1,t2,…,tn) attains its maximum value at point tn=(1+t1p+⋯+tn-1p)/(1+t1+⋯+tn-1)1/(p-1). Hence,(27)ft1,t2,…,tn≤1n+1p-11+t1+⋯+tn-1+1+t1p+⋯+tn-1p/1+t1+⋯+tn-11/(p-1)p1+t1p+⋯+tn-1p+1+t1p+⋯+tn-1p/1+t1+⋯+tn-1p/(p-1)=1n+1p-11+1+t1+⋯+tn-1p1+t1p+⋯+tn-1p1/(p-1)p-1.
By using inequality (24) we have(28)1+t1+⋯+tn/n+1p1+t1p+⋯+tnp/n+1≤1n+1p-11+1+t1+⋯+tn-1p1+t1p+⋯+tn-1p1/p-1p-1≤1n+1p-1n-1+1+t1p1+t1p1/(p-1)p-1.
Combining (1∘), (2∘), and (3∘), we have
(29)1+t1+⋯+tk/(k+1)p(1+t1p+⋯+tkp)/(k+1)≤1k+1p-1k-1+1+t1p1+t1p1/(p-1)p-1.Proof of Theorem1.
Proof of Sufficiency. Suppose that, for ∀ϵ>0, there is a 0<δ1=δ1(ϵ,p)<1, such that the inequality
(30)x1+x2+⋯+xk+1k+1p≤1-δ1
holds for all x1,x2,…,xk+1∈S(X) with Ax1,x2,…,xk+1≥ϵ.
Letδ2=1-1-δ11/p>0 and δ=(k+1)δ2; then x1+x2+⋯+xk+1≤1-δ.
By the definition ofk-uniformly rotund space, we know that X is k-uniformly rotund space.
Proof of Necessity. Suppose inequality (2) is not true. Then there exist b∈R+, ϵ0>0, such that, for ∀1/n>0, there exist x1nn=1∞,x2nn=1∞,…,xk+1nn=1∞⊂X, satisfying x1n≤b. When Ax1n,x2n,…,xk+1n≥ϵ0, we have
(31)x1n+x2n+⋯+xk+1nk+1p>1-1nx1np+x2np+⋯+xk+1npk+1.
Take u1n=x1n/x1n, u2n=x2n/x1n,…,uk+1n=xk+1n/x1n. Then
(32)u1n=1,Au1n,u2n,…,uk+1n=1x1nkAx1n,x2n,…,xk+1n≥ϵ0bk.
By Lemma 3 we know that
(33)1-1n<u1n+u2n+⋯+uk+1n/k+1pu1np+u2np+⋯+uk+1np/(k+1)≤1+u2n+⋯+uk+1n/k+1p1+u2np+⋯+uk+1np/(k+1)≤1.
It follows that
(34)limn→∞u1n+u2n+⋯+uk+1n/(k+1)pu1np+u2np+⋯+uk+1np/(k+1)=1.
Now we prove that(35)limn→∞uin=1,i=2,3,…,k+1.
If (35) does not hold, then there exists subsequence ujni of ujn satisfying
(36)ujni≤αj-1<1orujni≥αj-1>1,j=2,…,k+1.
Hence, by Lemma 4 we have
(37)u1ni+u2ni+⋯+uk+1ni/(k+1)pu1nip+u2nip+⋯+uk+1nip/(k+1)≤1k+1p-11+u2ni+⋯+uk+1nip1+u2nip+⋯+uk+1nip≤1k+1p-1k-1+1+u2nip1+u2nip1/(p-1)p-1.
Considering a function1+tp/(1+tp) and noticing that
(38)1+tp1+tp′t=p1+tp-11+tp-ptp-11+tp1+tp2=p1+tp-11-tp-11+tp2,
we know that 1+tp/(1+tp) is strictly increasing (decreasing) function when t<1(t>1) and attains its maximum value at point t=1.
Hence(39)u1ni+u2ni+⋯+uk+1ni/(k+1)pu1nip+u2nip+⋯+uk+1nip/(k+1)≤1k+1p-1k-1+1+u2nip1+u2nip1/p-1p-1≤1k+1p-1k-1+1+α1p1+α1p1/p-1p-1<1;
this contradicts (34), so limn→∞u2n=1.
Similarly, we can deduce thatlimn→∞u3n=1,…,limn→∞uk+1n=1. It follows that
(40)limn→∞u1n+u2n+⋯+uk+1nk+1=1.
Let zn=uk+1n/uk+1n; then
(41)zn-uk+1n=uk+1nuk+1n-uk+1n⟶0,n⟶∞,1=limn→∞u1n+u2n+⋯+uk+1nk+1≤limn→∞u1n+u2n+⋯+znk+1+uk+1n-znk+1=limn→∞u1n+u2n+⋯+znk+1≤1.
This means that
(42)limn→∞u1n+u2n+⋯+znk+1=1.
But
(43)limn→∞Au1n,u2n,…,zn=limn→∞Au1n,u2n,…,uk+1n≥ϵobk.
This contradicts that X is k-uniformly rotund space from Lemma 2.Theorem 5.
Let1<p<+∞, let X be a Banach space, and let M be an arbitrary bounded subset of X. Then, X is locally k-uniformly rotund space if and only if, for any ϵ>0 and x1∈M, there exists 0<δ(ϵ,p,x1)<1, such that the inequality
(44)x1+x2+⋯+xk+1k+1p≤(1-δ(ϵ,p,x1))x1p+x2p+⋯+xk+1pk+1
holds for all x2,x3,…,xk+1∈X with Ax1,x2,…,xk+1≥ϵ.The proof of Theorem5 is greatly similar to the proof of Theorem 1.In particular, considering the special cases of Theorems1 and 5 when k=1, we give a new characterization of uniformly rotund (resp., locally uniformly rotund) Banach space; that is, we have the following two corollaries.Corollary 6.
Let1<p<+∞, let X be a Banach space, and let M be an arbitrary bounded subset of X. Then, X is uniformly rotund space if and only if, for any ϵ>0, there exists 0<δ(ϵ,p)<1, such that the inequality
(45)x+y2p≤(1-δ(ϵ,p))xp+yp2
holds for all x∈M and y∈X with x-y≥ϵ.Corollary 7.
Let1<p<+∞, let X be a Banach space, and let M be an arbitrary bounded subset of X. Then, X is locally uniformly rotund space if and only if, for any ϵ>0 and x∈M, there exists 0<δ(ϵ,p,x)<1, such that the inequality
(46)x+y2p≤(1-δ(ϵ,p,x))xp+yp2
holds for all y∈X with x-y≥ϵ.



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