# Some Characterizations ofk-Strictly Convex and k-Smooth Banach Spaces

Authors: Suyalatu Wulede
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

In this paper, some characterizations ofk-strictly convex (k-smooth, respectively) spaces are given. As special cases for k=1, some characterizations of strictly convex (smooth, respectively) spaces are given.


---

## Body

## 1. Introduction

In 1936, Clarkson [1] introduced the concept of uniformly convex Banach spaces. Consequently, some methods were found to investigate the property of Banach space from the geometric structure of the unit sphere in Banach space. This initiated the study of convexity of Banach space. Since convexity has a striking intuitive geometric meaning, many mathematicians were attracted to this field of study. Smoothness, later introduced as a dual notion of convexity, is closely related to the various properties of differentiability of norm. This prompted further in-depth study of smoothness and the further development of the smooth theory of Banach space. As two important properties of geometry in Banach space, both convexity and smoothness play a very significant role in geometry of Banach space. They not only have promoted the development of geometric theories of Banach space, but also are also widely applied to such fields as control theory, operator theory, optimal approximation theory, and fixed-point theory.In particular, the concept of a strictly convex Banach space is very interesting and has numerous applications (cf. [2–25], for example). In 1960, this concept has been generalized by Singer. He defined the so-called k-strictly convex Banach space. The k-strict convexity has important applications in approximation theory (cf. [12, 19, 26–28], for example). It is well known that every k-strictly convex space is (k+1)-strictly convex. Also, it was shown that 1-strict convexity coincides with strict convexity. In order to describe spaces dual to those being k-strictly convex space Nan and Wang introduced the notion of the so-called k-smooth space. It is well known that every k-smooth space is (k+1)-smooth space. Also, it was shown that 1-smoothness coincide with smoothness.The purpose of this paper is to give some necessary and sufficient conditions fork-strict convexity (k-smoothness, respectively).Throughout the sequel, the symbolX denotes a real Banach space and X∗ denotes its dual space. S(X)={x:x∈X,x=1}, S(X∗)={x∗:x∗∈X∗,x∗=1} denote the unit sphere in X and the unit sphere in X∗, respectively.For eachx∈S(X), set Sx={x∗:x∗∈S(X∗),x∗(x)=1}.For eachx∗∈S(X∗), set Ax∗={x:x∈S(X),x∗(x)=1}.For eachx∗∈X∗, set J(x∗)={x:x∈X,x∗(x)=x∗x}.For eachx∈X, set J∗(x)={x∗:x∗∈X∗,x∗(x)=x∗x}.Forx1,x2,…,xk+1∈SX, the k-dimensional volume enclosed by x1,x2,…,xk+1 is given by(1)Vx1,x2,…,xk+1=sup11⋯1x1∗x1x1∗x2⋯x1∗xk+1⋯⋯⋯⋯xk∗x1xk∗x2⋯xk∗xk+1:xi∗∈SX∗,i=1,…,k.Definition 1 (see [19]).
A Banach spaceX is said to be k-strictly convex space provided that, for any k+1 norm-one elements x1,…,xk+1 in X, if ∑i=1k+1‍xi=k+1, then x1,…,xk+1 are linearly dependent.Definition 2 (see [27]).
A Banach spaceX is said to be k-smooth space provided that, for any norm-one element x in X, one has dim⁡Sx≤k.Lemma 3 (see [27]).
IfX∗ is k-strictly convex, then X is k-smooth. If X∗ is k-smooth, then X is k-strictly convex.Lemma 4 (see [29]).
LetX be a Banach space. Suppose that x1,x2,…,xk+1 are norm-one elements in X and [x2,…,xk+1] denotes the affine subspace generated by {x2,…,xk+1}. Set (2)d1=distx1,x2,…,xk+1,d2=distx2,x3,…,xk+1,…,dk=xk-xk+1.Then(3)d1d2⋯dk≤Vx1,x2,…,xk+1≤kk/2d1d2⋯dk.


## 2. Characterizations ofk- Strictly Convex Banach Spaces

Theorem 5.
A Banach spaceX is k-strictly convex if and only if, for any k+1 norm-one elements x1,…,xk+1 in X and real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, the equality ∑i=1k+1‍λixi=1 holds; then x1,…,xk+1 are linearly dependent.Proof.
Necessity. Suppose the contrary; i.e., there are k+1 linearly independent norm-one elements x1,…,xk+1 of X and real numbers 0<λi<1,i=1,…,k+1 satisfying(4)∑i=1k+1λi=1,∑i=1k+1λixi=1.Now we are going to show that ∑i=1k+1‍xi=k+1.
If the above equality does not hold, then(5)∑i=1k+1xi<k+1.This yields (6)1=∑i=1k+1λik∑i=1k+1λixi=λ1kx1+⋯+λk+1kxk+1+⋯+k!λ1λ2⋯λk+1∑i=1k+1λixi<λ1k+⋯+λk+1k+⋯+k!λ1λ2⋯λk+1k+1<λ1+λ2+⋯+λk+1kλ1+λ2+⋯+λk+1=1;this is impossible.
Combining∑i=1k+1‍xi=k+1 with the assumption that X is k-strictly convex space, we know that x1,…,xk+1 are linearly dependent; this contradicts the fact that x1,…,xk+1 are linearly independent elements of X.
Sufficiency. Suppose that, for any k+1 norm-one elements x1,…,xk+1 in X, the following equality ∑i=1k+1‍xi=k+1 holds.
Letλi=1/k+1,i=1,…,k+1; then we have (7)∑i=1k+1λi=1,∑i=1k+1λixi=10<λi<1.From the conditions given here, we know that x1,…,xk+1 are linearly dependent. This shows that X is a k-strictly convex space.Corollary 6.
A Banach spaceX is strictly convex if and only if, for any norm-one elements x,y in X and real numbers 0<λ,μ<1 with λ+μ=1, the equality λx+μy=1 holds; then x,y are linearly dependent.Theorem 7.
A Banach spaceX is k-strictly convex if and only if, for any k+1 linearly independent norm-one elements x1,…,xk+1 in X, real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, and any functional x∗∈X∗∖{0}, one has x∗(∑i=1k+1‍λixi)<x∗.Proof.
Necessity. Suppose that x1,…,xk+1 are any k+1 linearly independent norm-one elements in X and 0<λi<1,i=1,…,k+1 are arbitrary real numbers with ∑i=1k+1‍λi=1.
Clearly,∑i=1k+1‍λixi≤1 and the inequality x∗(∑i=1k+1‍λixi)≤x∗ holds for any functional x∗∈X∗∖0.
Now we are going to show that(8)x∗∑i=1k+1λixi<x∗.If the above inequality is not true, then (9)x∗∑i=1k+1λixi=x∗.Therefore,(10)x∗=x∗∑i=1k+1λixi≤x∗∑i=1k+1λixi,which leads to 1≤∑i=1k+1λixi. Hence ∑i=1k+1‍λixi=1. By the k-strict convexity of X and Theorem 5, we know that x1,…,xk+1 are linearly dependent; hence we reach a contradiction.
Sufficiency. Suppose that x1,…,xk+1 are any k+1 norm-one elements of X and ∑i=1k+1‍xi=k+1. Then, by Definition 1, it is necessary to show that x1,…,xk+1 are linearly dependent.
Ifx1,…,xk+1 are linearly independent, put λi=1/k+1, and then(11)∑i=1k+1λi=1,∑i=1k+1λixi=1,0<λi<1,i=1,…,k+1. From the conditions given here, we know that the inequality(12)x∗∑i=1k+1λixi<x∗holds for any functional x∗∈X∗∖{0}.
On the other hand, by Hahn-Banach theorem, we know that there is a norm-one functionalx0∗∈X∗ such that (13)x0∗∑i=1k+1λixi=x0∗;hence we reach a contradiction.Corollary 8.
A Banach spaceX is strictly convex if and only if, for any linearly independent norm-one elements x,y in X, real numbers 0<λ,μ<1 with λ+μ=1, and any functional x∗∈X∗∖{0}, one has x∗(λx+μy)<x∗.Theorem 9.
LetX be a Banach space and x be any norm-one element of X. Then X is k-strictly convex if and only if, for any k norm-one elements x1,…,xk in X linearly independent with x, real numbers 0<λ,λi<1,i=1,…,k with λ+∑i=1k‍λi=1, and any functional x∗∈Sx, one has x∗(∑i=1k‍λixi)<1-λ.Proof.
Necessity. Suppose that x1,…,xk are any k norm-one elements x1,…,xk of X linearly independent with any norm-one element x of X and 0<λ,λi<1,i=1,…,k are arbitrary real numbers with λ+∑i=1k‍λi=1. By the k-strict convexity of X and Theorem 7, we know that(14)x∗λx+∑i=1kλixi<x∗=1.Hence, taking into account the fact that x∗∈Sx, we derive(15)x∗∑i=1kλixi<1-λ.Sufficiency. If X is not k-strictly convex, then there are k+1 linearly independent norm-one elements x,x1,…,xk of X such that x+∑i=1k‍xi=k+1. Let us take x0=x+x1+⋯+xk/k+1. Then, by Hahn-Banach theorem we know that there is a norm-one functional x0∗∈X∗ such that x0∗(x0)=x0∗(x+x1+⋯+xk/k+1)=1. This leads to x0∗∈Sx. Let λ=λi=1/k+1,i=1,…,k, and then we have λ+∑i=1k‍λi=1, but x0∗(∑i=1k‍λixi)=1-λ; hence we reach a contradiction.Corollary 10.
LetX be a Banach space and x be any norm-one element of X. Then X is strictly convex if and only if, for any norm-one element y in X linearly independent with x∈S(X), real numbers 0<λ,μ<1 with λ+μ=1, and any functional x∗∈Sx, one has x∗(μy)<1-λ.Theorem 11.
A Banach spaceX is k-strictly convex if and only if, for any k+1 norm-one elements x1,…,xk+1 in X and real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, there is some functional x∗∈X∗∖{0} such that x∗∈J∗(∑i=1k+1‍λixi) and J∗(∑i=1k+1‍λixi)=∩J∗(λixi); then x1,…,xk+1 are linearly dependent.Proof.
Necessity. Let x1,…,xk+1 be any k+1 norm-one elements of X and 0<λi<1,i=1,…,k+1 are arbitrary real numbers with ∑i=1k+1‍λi=1. Suppose that there is some functional x∗∈X∗∖{0} such that x∗∈J∗(∑i=1k+1‍λixi) and J∗(∑i=1k+1‍λixi)=∩J∗(λixi).
Now we are going to prove thatx1,…,xk+1 are linearly dependent.
Ifx1,…,xk+1 are linearly independent, then ∑i=1k+1‍λixi≠0. From x∗∈J∗(∑i=1k+1‍λixi)=∩J∗(λixi), we have(16)x∗∑i=1k+1λixi=x∗∑i=1k+1λixi,x∗λixi=x∗λixi=λix∗;it follows that(17)x∗∑i=1k+1λixi=∑i=1k+1λix∗=x∗.This yields ∑i=1k+1‍λixi=1 because of x∗≠0. Hence, in view of the above real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, we know that x1,…,xk+1 are linearly dependent from Theorem 5, a contradiction.
Sufficiency. Suppose that x1,…,xk+1 are any k+1 norm-one elements of X and 0<λi<1,i=1,…,k+1 are arbitrary real numbers satisfying(18)∑i=1k+1λi=1,∑i=1k+1λixi=1.
Now we are going to prove that there is some functionalx∗∈X∗∖{0} such that(19)x∗∈J∗∑i=1k+1λixi,J∗∑i=1k+1λixi=∩J∗λixi.
Firstly, we will prove that there is some functionalx∗∈X∗∖{0} such that x∗∈J∗(∑i=1k+1‍λixi).
For∑i=1k+1‍λixi with ∑i=1k+1‍λixi=1, by Hahn-Banach theorem we choose a norm-one functional x∗∈X∗ such that x∗(∑i=1k+1‍λixi)=1; i.e., x∗(∑i=1k+1‍λixi)=x∗∑i=1k+1‍λixi. Hence we have x∗∈J∗(∑i=1k+1‍λixi).
Secondly, we will prove thatJ∗(∑i=1k+1‍λixi)=∩J∗(λixi).
(i) Ify∗∈J∗(∑i=1k+1‍λixi), then y∗(∑i=1k+1‍λixi)=y∗∑i=1k+1‍λixi=y∗. Hence, in view of the inequality y∗(λixi)≤λiy∗xi=λiy∗ and ∑i=1k+1‍λi=1,0<λi<1,i=1,…,k+1, we have(20)y∗λixi=λiy∗xi=y∗λixi,i=1,…,k+1.This shows that y∗∈∩J∗(λixi).
(ii) Ify∗∈∩J∗(λixi), then y∗(λixi)=y∗λixi=λiy∗xi,i=1,…,k+1; it follows that(21)y∗∑i=1k+1λixi=∑i=1k+1λiy∗xi=∑i=1k+1λiy∗=y∗=y∗∑i=1k+1λixi.This shows that y∗∈J∗(∑i=1k+1‍λixi).
Combining (i) and (ii) we haveJ∗(∑i=1k+1‍λixi)=∩J∗(λixi).
Hence, from the conditions given here, we know thatx1,…,xk+1 are linearly dependent. So X is k-strictly convex space which follows from Theorem 5.Corollary 12.
A Banach spaceX is k-strictly convex if and only if, for any norm-one elements x,y in X and real numbers 0<λ,μ<1 with λ+μ=1, there is some functional x∗∈X∗∖{0} such that x∗∈J∗(λx+μy) and J∗(λx+μy)=J∗(λx)∩J∗(μy); then x,y are linearly dependent.


## 3. Characterizations ofk- Smooth Banach Spaces

Theorem 13.
A Banach spaceX is k-smooth if and only if, for any k+1 linearly independent norm-one functionals x1∗,…,xk+1∗∈X∗, real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, and any element x∈X∖{0}, one has ∑i=1k+1‍λixi∗(x)<x.Proof.
Necessity. Suppose that X is a k-smooth space. Then, by Definition 2, we have dim⁡Sx≤k for any norm-one element x of X. Suppose that x1∗,…,xk+1∗ are any k+1 linearly independent norm-one functionals of X∗ and 0<λi<1,i=1,…,k+1 are arbitrary real numbers with ∑i=1k+1λi=1.
Clearly,∑i=1k+1λixi∗≤1 and the inequality ∑i=1k+1λixi∗(x)≤x holds for any element x∈X∖{0}.
Now we are going to show that∑i=1k+1‍λixi∗(x)<x.
If the above inequality is not true, then we have(22)∑i=1k+1λixi∗xx=1.
Clearly,xi∗x/x≤1 for every i=1,…,k+1. Combining the above equality with 0<λi<1,i=1,…,k+1, we have xi∗(x/x)=1.
Lety=x/x, and then y is a norm-one element of X and xi∗∈Sy,i=1,…,k+1. This leads to dim⁡Sy≥k+1, a contradiction.
Sufficiency. Suppose that for any k+1 linearly independent norm-one functions x1∗,…,xk+1∗∈X∗, real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, and any element x∈X∖{0} we have ∑i=1k+1‍λixi∗(x)<x.
IfX is not k-smooth, then there is some norm-one element x0 of X such that dim⁡Sx0>k. This means that there are k+1 linearly independent norm-one functionals x1∗,…,xk+1∗ such that xi∗∈Sx0,i=1,…,k+1. Put λi=1/k+1; then 0<λi<1,i=1,…,k+1, ∑i=1k+1‍λi=1 and ∑i=1k+1‍λixi∗(x0)=x0, a contradiction.Corollary 14.
A Banach spaceX is smooth if and only if for any linearly independent norm-one functionals x∗,y∗ of X∗, real numbers 0<λ,μ<1 with λ+μ=1, and any element x∈X∖{0} one has (λx∗+μy∗)(x)<x.Theorem 15.
LetX be a Banach space and x∗ be any norm-one functionals of X∗ which attains its norm on S(X). Then X is k-smooth if and only if, for any k norm-one functionals x1∗,…,xk∗∈X∗ linearly independent with x∗, real numbers 0<λ,λi<1,i=1,…,k with λ+∑i=1k‍λi=1, and any element x∈Ax∗, one has ∑i=1k‍λixi∗(x)<1-λ.Proof.
Necessity. Let x∗∈S(X∗),x∈Ax∗. Suppose that x1∗,…,xk∗∈X∗ are any k norm-one functionals of X∗ linearly independent with x∗ and 0<λ,λi<1,i=1,…,k are arbitrary real numbers with λ+∑i=1k‍λi=1. By the k-smoothness of X and Theorem 13, we know that(23)λx∗x+∑i=1kλixi∗x<x=1.Hence, taking into account the fact that x∈Ax∗, we derive(24)∑i=1kλixi∗x<1-λ.Sufficiency. If X is not k-smooth, then there is some norm-one elements x0 of X such that dim⁡Sx0>k. This means that there are k+1 linearly independent norm-one functionals x0∗,x1∗,…,xk∗∈X∗ such that xi∗∈Sx0,i=0,1,…,k; it follows that x0∈Ax0∗.
Letλ=λi=1/k+1,i=1,…,k, and then λ+∑i=1k‍λi=1, but ∑i=1k‍λixi∗(x0)=1-λ; hence we reach a contradiction.Corollary 16.
LetX be a Banach space and x∗ be any norm-one functionals of X∗ which attains its norm on S(X). Then X is smooth if and only if, for any norm-one functional y∗∈X∗ linearly independent with x∗, real numbers 0<λ,μ<1 with λ+μ=1, and any element x∈Ax∗, one has μy∗(x)<1-λ.Theorem 17.
A Banach spaceX is k-smooth if and only if, for any k+1 norm-one functionals x1∗,…,xk+1∗∈X∗ and real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, there is some element x∈X∖{0} such that x∈J(∑i=1k+1‍λixi∗) and J(∑i=1k+1‍λixi∗)=∩J(λixi∗); then x1∗,…,xk+1∗ are linearly dependent.Proof.
Necessity. Let x1∗,…,xk+1∗ be any k+1 norm-one functionals in X∗ and 0<λi<1,i=1,…,k+1 are any real numbers with ∑i=1k+1‍λi=1. Suppose that there is some element x∈X∖{0} such that (25)x∈J∑i=1k+1λixi∗,J∑i=1k+1λixi∗=∩Jλixi∗. Then(26)∑i=1k+1λixi∗x=∑i=1k+1λixi∗x,λixi∗x=λixi∗x,i=0,1,…,k+1.
For∑i=1k+1‍λixi∗, we have two possibilities: either ∑i=1k+1‍λixi∗=0, or ∑i=1k+1‍λixi∗≠0.
Case (I). If ∑i=1k+1λixi∗=0, then ∑i=1k+1λixi∗=0. This means that x1∗,…,xk+1∗ are linearly dependent, so we obtain the desired result.
Case (II). If ∑i=1k+1‍λixi∗≠0, we may assume that xi∗≠0 for i=1,…,k+1. Indeed, if some xi∗=0, we also obtain the desired result.
Lety=x/x,yi∗=xi∗/xi∗,i=1,…,k+1, and then yi∗ is a norm-one functional of X∗ for i=1,…,k+1 and ∑i=1k+1‍λiyi∗(y)=1. By Theorem 13, we know that y1∗,…,yk+1∗ are linearly dependent. Hence, x1∗,…,xk+1∗ are linearly dependent also, so we obtain the desired result.
Sufficiency. Suppose that x is any norm-one element of X and x1∗,…,xk+1∗ are any k+1 norm-one functionals of Sx. Then, for any real numbers 0<λi<1 with ∑i=1k+1‍λi=1, we have(27)1=∑i=1k+1λixi∗x≤∑i=1k+1λixi∗x≤∑i=1k+1λixi∗x=1;it follows that (28)∑i=1k+1λixi∗=1,∑i=1k+1λixi∗x=∑i=1k+1λixi∗x.Hence we have x∈J(∑i=1k+1‍λixi∗).
(i) Ify∈J(∑i=1k+1‍λixi∗), then(29)∑i=1k+1λixi∗y=∑i=1k+1λixi∗y=∑i=1k+1λixi∗y=y.Hence, in view of the inequality (λixi∗)(y)≤λiy and ∑i=1k+1‍λi=1,0<λi<1,i=1,…,k+1, we have(30)λixi∗y=λiy=λixi∗y=λixi∗y,i=1,…,k+1.This shows that y∈∩J(λixi∗).
(ii) Ify∈∩J(λixi∗), then (λixi∗)(y)=λixi∗y=λixi∗y,i=1,…,k+1; it follows that(31)∑i=1k+1λixi∗y=∑i=1k+1λixi∗y=∑i=1k+1λixi∗y=∑i=1k+1λiy=y=∑i=1k+1λixi∗y.This shows that y∈J(∑i=1k+1‍λixi∗).
Combining (i) and (ii) we haveJ(∑i=1k+1‍λixi∗)=∩J(λixi∗).
Hence, from the conditions given here, we know thatx1∗,…,xk+1∗ are linearly dependent. This means that dim⁡Sx≤k. It follows from Definition 2 that X is k-smooth space.Corollary 18.
A Banach spaceX is smooth if and only if, for all norm-one functionals x∗,y∗∈X∗ and real numbers 0<λ,μ<1 with λ+μ=1, there is some element x∈X∖{0} such that x∈J(λx∗+μy∗) and J(λx∗+μy∗)=J(λx∗)∩J(μy∗); then x∗,y∗ are linearly dependent.Theorem 19.
A finite-dimensional Banach spaceX is k-smooth if and only if for any k+1 norm-one functionals x1∗,…,xk+1∗∈X and real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, the equality ∑i=1k+1‍λixi∗=1 holds; then x1∗,…,xk+1∗ are linearly dependent.Proof.
Necessity. Suppose that x1∗,…,xk+1∗ are any k+1 norm-one functionals of X∗ and 0<λi<1,i=1,…,k+1 are arbitrary real numbers satisfying(32)∑i=1k+1λi=1,∑i=1k+1λixi∗=1.Since X is a finite-dimensional Banach space, so X is reflexive. Therefore, there is a norm-one element x in X such that(33)∑i=1k+1λixi∗x=1;it follows from 0<λi<1,i=1,…,k+1,∑i=1k+1‍λi=1 that(34)xi∗x=1,i=1,…,k+1This means that xi∗∈Sx,i=1,…,k+1. By Definition 2 we know that x1∗,…,xk+1∗ are linearly dependent.
Sufficiency. It is immediate from Theorem 5 and Lemma 3.Corollary 20.
A finite-dimensional BanachX is smooth if and only if, for any norm-one functionals x∗,y∗∈X and real numbers 0<λ,μ<1 with λ+μ=1, the equality λx∗+μy∗=1 holds; then x∗,y∗ are linearly dependent.Theorem 21.
A finite-dimensional Banach spaceX is k-smooth if and only if for any k+1 norm-one functionals x1∗,…,xk+1∗ of X∗, real numbers 0<λi<1,i=1,…,k+1 with ∑i=1k+1‍λi=1, and ∑i=1k+1‍λixi∗=1, one has V(x1∗,x2∗,…,xk+1∗)=0.Proof.
Necessity. Suppose that x1∗,…,xk+1∗ are any k+1 norm-one functionals of X∗ and 0<λi<1,i=1,…,k+1 are arbitrary real numbers satisfying(35)∑i=1k+1λi=1,∑i=1k+1λixi∗=1.By Hahn-Banach theorem, we know that there is a norm-one functional x∗∗∈X∗∗ such that x∗∗(∑i=1k+1‍λixi∗)=1. By the assumption that 0<λi<1,i=1,…,k+1,∑i=1k+1‍λi=1, we have x∗∗(xi∗)=1 for every i=1,…,k+1. Since X is a finite-dimensional Banach space, so X is reflexive. Therefore, x∗∗∈X=X∗∗. Furthermore, x∗∗ may be denoted by x=x∗∗. For every i=1,…,k+1, the equality x∗∗(xi∗)=1 may be rewritten by xi∗(x)=1. This shows that x1∗,…,xk+1∗∈Sx. By the k-smoothness of X we know that x1∗,…,xk+1∗ are linearly dependent. Without loss of generality, we may assume that(36)x1∗=∑i=2k+1μixi∗.This leads to(37)x1∗x=∑i=2k+1μixi∗x. By x1∗,…,xk+1∗∈Sx, we know that ∑i=2k+1‍μi=1. Hence we have(38)sup11⋯1x1∗x1x2∗x1⋯xk+1∗x1⋯⋯⋯⋯x1∗xkx2∗xk⋯xk+1∗xk:xi∈SX,i=1,⋯,k=sup1-∑i=2k+1μi1⋯1x1∗-∑i=2k+1μixi∗x1x2∗x1⋯xk+1∗x1⋯⋯⋯⋯x1∗-∑i=2k+1μixi∗xkx2∗xk⋯xk+1∗xk:xi∈SX,i=1,…,k.=0.
On the other hand, by Goldstine-Weston Theorem [30], it is easy to see that(39)Vx1∗,x2∗,…,xk+1∗=sup11⋯1x1∗x1x2∗x1⋯xk+1∗x1⋯⋯⋯⋯x1∗xkx2∗xk⋯xk+1∗xk:xi∈SX,i=1,…,k. This leads to (40)Vx1∗,x2∗,…,xk+1∗=0.Sufficiency. Suppose that x1∗,…,xk+1∗ are any k+1 norm-one functionals of X∗ and 0<λi<1,i=1,…,k+1 are arbitrary real numbers satisfying(41)∑i=1k+1λi=1,∑i=1k+1λixi∗=1.Then we have(42)Vx1∗,x2∗,…,xk+1∗=0.Let di denote the distance from xi∗ to the affine subspace [xi+1∗,…,xk+1∗]. Then by Lemma 4 and the above equality we have(43)d1d2⋯dk≤Vx1∗,x2∗,…,xk+1∗=0. This shows at least one of d1,d2,…,di,…,dk is equal to zero. We may assume that (44)di=distxi∗,xi+1∗,…,xk+1∗=0.It shows that x1∗,…,xk+1∗ are linearly dependent. Combining Theorem 5 and Lemma 3 we know that X is k-smooth space.Corollary 22.
A finite-dimensional Banach spaceX is smooth if and only if, for all norm-one functionals x∗,y∗ of X∗, real numbers 0<λ,μ<1 with λ+μ=1, and λx∗+μy∗=1, one has x∗=y∗.



---

*Source: 9527954-2019-02-06.xml*