# A Generalization of Uniformly Extremely Convex Banach Spaces

Authors: Suyalatu Wulede; Wurichaihu Bai; Wurina Bao
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We discuss a new class of Banach spaces which are the generalization of uniformly extremely convex spaces introduced by Wulede and Ha. We prove that the new class of Banach spaces lies strictly between either the classes ofk-uniformly rotund spaces and k-strongly convex spaces or classes of fully k-convex spaces and k-strongly convex spaces and has no inclusive relation with the class of locally k-uniformly convex spaces. We obtain in addition some characterizations and properties of this new class of Banach spaces. In particular, our results contain the main results of Wulede and Ha.


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## Body

## 1. Introduction

Different uniformly convex spaces have been defined between the uniformly convex spaces [1] and the reflexivity of the Banach spaces [2–6]. In the previous paper [7] we introduce a new class of this type, namely, uniformly extremely convex spaces. This new class of Banach spaces lies strictly between either the classes of uniformly convex spaces and strongly convex spaces or the classes of fully k-convex spaces and strongly convex spaces.Here we consider another new class of this type, namely,k-uniformly extremely convex spaces, as a generalization of uniformly extremely convex spaces and discuss its relation to the drop property, the k-uniformly rotund spaces, the full k-convex spaces, the k-strongly convex spaces, the nearly uniformly convex spaces, and k-nearly uniformly convex spaces. We also give some characterizations of k-uniformly extremely convex spaces and find that this new class of Banach spaces has the following features:(1)
1-uniformly extremely convex spaces (indeed lower case) coincide with uniformly extremely convex spaces;(2)
k-uniformly extremely convex spaces possess the drop property;(3)
k-uniformly extremely convex spaces are (k+1)-uniformly extremely convex spaces, but the converse implication is not true.Throughout this paperX denotes an infinite-dimensional real Banach space with the norm ·. The symbol X∗ denotes the dual of the space X. U(X) and S(X) denote the closed unit ball and the unit sphere of X, respectively. The symbol S(X∗) denotes the unit sphere of X∗. The symbol σ(X,X∗) denotes the weak topology of X.Letx1,x2,…,xk+1 be norm-1 elements in Banach spaces X. The k-dimensional volume enclosed by x1,x2,…,xk+1 is given by (1)Vx1,x2,…,xk+1=sup⁡11⋯1f1x1f1x2⋯f1xk+1⋮⋮⋱⋮fkx1fkx2⋯fkxk+1:f1,…,fk∈SX∗.Here, and throughout the sequel, the symbol · denotes the determinant.Sullivan [6] has introduced the k-uniformly rotund (kUR) spaces and locally k-uniformly rotund (LkUR) spaces. Fan and Glicksberg [2] have introduced the fully k-convex (kR) Banach spaces. It is well known that kUR and kR spaces imply reflexivity. About kUR and kR spaces, we have the following chain of implications [2, 6, 8]: (2)UR=1UR⟹⋯⟹kUR⟹k+1UR;2R⟹⋯⟹kR⟹k+1R;LUR=L1UR⟹⋯⟹LkUR⟹Lk+1UR.A Banach spaceX is said to be a kUR space (k≥1) [6] if, for any ϵ>0, there exists a δ(ϵ)>0 such that, for all norm-1 elements x1,x2,…,xk+1 and x1+x2+⋯+xk+1>(k+1)-δ, we have V(x1,x2,…,xk+1)<ϵ.A Banach spaceX is said to be a kR space (k≥2) [2] if, for any sequence {xn} in X such that limn1,…,nk→∞⁡1/kxn1+xn2+⋯+xnk=1, then {xn} is a Cauchy sequence in X.A pointx0∈S(X) is said to be a denting point of U(X) [8] if x0∉co¯(M(x0,ϵ)) for all ϵ>0, where M(x0,ϵ)={y:y∈U(X),y-x0≥ϵ}.Huff [3] has introduced the nearly uniformly convex (NUC) spaces as a generalization of uniformly convex Banach spaces and showed that the NUC spaces are equivalent to reflexive spaces possessing the uniform Kadec-Klee property. The local version of NUC spaces, namely, locally nearly uniformly convex (LNUC), was studied by Kutzarova and Lin [9]. Kutzarova [4] introduced the k-nearly uniformly convex (kNUC) spaces as a generalization of nearly uniformly convex Banach spaces. In [4, 9], it is pointed out that NUC ⇒ LNUC and kNUC ⇒ NUC for every k≥2.A Banach spaceX is said to be a NUC [3] space if, for any ϵ>0, there exists a δ(ϵ)>0 such that, for any sequence {xn}⊂U(X), sep(xn)>ϵ, we have co({xn})∩(1-δ)U(X)≠∅, where sep(xn)=inf⁡{xn-xm:n≠m} and co({xn}) means the convex hull of {xn}.A Banach spaceX is said to be a LNUC [9] space if, for any norm-1 element x and ϵ>0, there exists a δ=δ(ϵ,x)>0 such that, for any sequence {xn}⊂U(X), sep(xn)>ϵ, we have co({x}∪{xn})∩(1-δ)U(X)≠∅, where co({x}∪{xn}) means the convex hull of {x} and {xn}.A Banach spaceX is said to be a kNUC [4] space, if for any ϵ>0 there exists a 0<δ(ϵ)<1 such that, for any sequence {xn}⊂U(X), sep(xn)>ϵ, there are indices {ni} and scalars λi≥0, i=1,2,…,k, with ∑i=1kλi=1 so that ∑i=1kλixni≤1-δ.Singer [10] has introduced the k-strictly convex spaces. It is well known that k-strictly convex spaces are (k+1)-strictly convex spaces; 1-strictly convex spaces (indeed lower case) coincide with strictly convex spaces; kR spaces are k-strictly convex spaces and have the drop property. Wu and Li [11] have introduced the strongly convex spaces. Wulede and Wu [12] introduced the k-strongly convex spaces as a generalization of strongly convex Banach spaces and gave an equivalent definition of k-strongly convex spaces (see Theorem  5 in [13]). It is well known that k-strongly convex spaces are k-strictly convex spaces; 1-strongly convex spaces (indeed lower case) coincide with strongly convex spaces; k-strongly convex spaces are (k+1)-strongly convex spaces, but the converse implication is not true.A Banach spaceX is said to be a k-strictly convex space [10] if, for all norm-1 elements x1,x2,…,xk+1 such that ∑i=1k+1xi=k+1, then x1,x2,…,xk+1 are linearly dependent.A Banach spaceX is said to be a strongly convex space [11] if, for any x∈S(X), {xn}⊂S(X) and for a certain functional f∈Sx such that f(xn)→1(n→∞), then xn-x→0(n→∞), where Sx={f:f∈S(X∗), f(x)=1}.A Banach spaceX is said to be a k-strongly convex space [12] if, for any norm-1 element x, ϵ>0 and for any functional f∈Sx, there is a δ(x,f,ϵ)>0 such that, for all norm-1 elements x1,…,xk and f(x+x1+⋯+xk)>(k+1)-δ, we have V(x,x1,…,xk)<ϵ.Rolewicz [14] has defined the norm · to have the drop property, if for every closed set C⊂X disjoint from U(X) there exists x∈C such that D(x,U(X))∩C={x}, where the set D(x,U(X)), the convex hull of x and U(X), is called the drop generated by x∉U(X).Lemma 1 (Kadec-Klee property).
If anyx∈S(X), {xn}⊂S(X) such that xn→wx, n→∞, and xn→x, n→∞, then xn-x→0, n→∞, where xn→wx, n→∞, means that f(xn)→f(x), n→∞, for all f∈X∗.Lemma 2 (Montesinos [15]).
LetX be a Banach space. Then X has the drop property if and only if X is reflexive and has the Kadec-Klee property.Lemma 3 (Nan and Wang [16]).
X is k-strictly convex space if and only if, for any f∈S(X∗), one has dim⁡Af≤k, where Af={x:x∈S(X), f(x)=1}.Lemma 4 (Wulede and Wu [12], Zhang and Fang [17]).
LetX be a Banach space.(i)
IfX is k-strongly convex, then X is k-strictly convex and has the Kadec-Klee property.(ii)
IfX is reflexive, k-strictly convex and has the Kadec-Klee property, then X is k-strongly convex.(iii)
IfX is k-strongly convex, {xn}⊂U(X), f∈S(X∗), and f(xn)→1, n→∞, then dist⁡(xn,Af)→0, n→∞.Lemma 5 (Zhang and Fang [17]).
X is k-strongly convex if and only if, for any x∈S(X) and f∈Sx, ϵ>0, there exist δ>0 and a compact set C⊂X with dim⁡C≤k such that F(f,δ)⊂{x:x∈X, d(x,C)<ϵ}, where the set F(f,δ)={x:x∈U(X), f(x)≥1-δ} is the slice generated by f and δ.


## 2.k-Uniformly Extremely Convex Spaces and Drop Property

Definition 6 (see [7]).
A Banach spaceX is said to be a uniformly extremely convex space if, for any sequences {xn},{yn} consisting of elements of norm-1 and for a certain functional f of norm-1, limn→∞⁡f(xn)=limn→∞⁡f(yn)=1 holds; then limn→∞⁡xn-yn=0.On the base of uniformly extremely convex spaces, now we introduce the notion ofk-uniformly extremely convex spaces as a generalization of uniformly extremely convex spaces.Definition 7.
A Banach spaceX is said to be a k-uniformly extremely convex space if, for any sequences {xn(1)},…,{xn(k+1)} consisting of elements of norm-1 and for a certain functional f of norm-1, limn→∞⁡f(xn(1))=⋯=limn→∞⁡f(xn(k+1))=1 holds; then limn→∞⁡V(xn(1),…,xn(k+1))=0.We give first a simple result which shows that the notion ofk-uniformly extremely convex space is “coherent.”Theorem 8.
IfX is k-uniformly extremely convex space, then X is (k+1)-uniformly extremely convex space.Proof.
If, for any sequences{xn(1)},…,{xn(k+2)} consisting of elements of norm-1 and for a certain functional f of norm-1, limn→∞⁡f(xn(1))=⋯=limn→∞⁡f(xn(k+2))=1 holds, then, for all 1≤j≤k+2, we have limn→∞⁡V(xn(1),…,xn(j-1),xn(j+1),…,xn(k+2))=0 by the assumption that X is k-uniformly extremely convex space. Furthermore, by the properties of determinant we have(3)limn→∞⁡Vxn1,…,xnk+2≤limn→∞⁡∑j=2k+1Vxn1,…,xnj-1,xnj+1,…,xnk+2=0;this shows that X is (k+1)-uniformly extremely convex space.Now we give a simple but useful lemma. By using this lemma we can prove that anyk-uniformly extremely convex space has the drop property. And the fact that k-uniformly extremely convex spaces include k-strongly convex spaces can be easily found. We also show that 1-uniformly extremely convex spaces coincide with uniformly extremely convex spaces by using this lemma.Lemma 9.
X is k-uniformly extremely convex if and only if, for any ϵ>0,f∈S(X∗), there exists a δ(ϵ)>0 such that, for all norm-1 elements x1,…,xk+1 and f(∑i=1k+1xi)>(k+1)-δ, one has V(x1,…,xk+1)<ϵ.Proof.
Necessity. Suppose the contrary. Then there exist ϵ0>0,f0∈S(X∗) and xii=1k+1⊂S(X) such that, for any δ=1/n,n∈N, we have f0(∑i=1k+1xi)>(k+1)-1/n, but V(x1,…,xk+1)≥ϵ0. Take xn(i)=xi(i=1,…,k+1); then {xn(i)}i=1k+1⊂S(X) and k+1-1/n<f0(∑i=1k+1xn(i))≤k+1. It follows that limn→∞⁡f0(xn(i))=1. On the other hand, by the definition of the k-uniformly extremely convex space, we have V(x1,…,xk+1)→0; this contradicts the statement that V(x1,…,xk+1)≥ϵ0.
Sufficiency. If, for any sequences {xn(1)},…,{xn(k+1)} consisting of elements of norm-1 and for a certain functional f of norm-1, limn→∞⁡f(xn(1))=⋯=limn→∞⁡f(xn(k+1))=1 holds, then limn→∞⁡f(∑i=1k+1xn(i))=k+1. Therefore, for any δ>0, there exists an integer N0∈N such that, for all n≥N0, inequality f(∑i=1k+1xn(i))>(k+1)-δ holds. For any ϵ>0, by the conditions given in Lemma 9, we have V(xn(1),…,xn(k+1))<ϵ; this means that limn→∞⁡V(xn(1),…,xn(k+1))=0.Remark 10.
1-uniformly extremely convex space (indeed lower case) coincides with uniformly extremely convex space.
In fact, by Lemma9 we know that X is 1-uniformly extremely convex space if and only if, for any ϵ>0, f∈S(X∗), there exists a δ(ϵ,f)>0 such that, for any norm-1 elements x,y and f(x+y)>2-δ, we have (4)Vx,y=sup⁡11fxfy:f∈SX∗=supf∈SX∗⁡fx-fy=x-y<ϵ,and also if and only if X is uniformly extremely convex space.Theorem 11.
X is k-uniformly extremely convex space if and only if X is k-strictly convex space and has the drop property.Proof.
Necessity. Suppose that X is k-uniformly extremely convex space; by the definition of k-strongly convex space and a condition which characterizes k-uniformly extremely convex space in Lemma 9, it is easy to see that X is k-strongly convex space. From Lemma 4(i), we know that X is k-strictly convex space and has the Kadec-Klee property.
Now we are going to prove thatX has the drop property. In fact, from Lemma 2, it is sufficient to prove that X is reflexive. Suppose that X is not reflexive. Using the well-known James’ theorem, for each 0<ϵ<1, we can choose 0<θ<1 so that θ>1-δ(ϵ)/k+1 and θk>ϵ, and x1,…,xk+1⊂SX, {x1∗},…,{xk+1∗}⊂S(X∗) so that (5)xj∗xi=θifj≤i0ifj>i.Here δ(ϵ) is the function required in Lemma 9.
Now we have that(6)x1∗x1+x2+⋯+xk+1=k+1θ>k+1-δϵ.On the other hand it is easy to check that (7)Vx1,…,xk+1≥11⋯1x2∗x1x2∗x2⋯x2∗xk+1⋮⋮⋱⋮xk+1∗x1xk+1∗x2⋯xk+1∗xk+1=θk>ϵ,which gives the required contradiction.
Sufficiency. From the assumption that X is k-strictly convex space and has the drop property, we can deduce, by Lemma 4(ii), that X is k-strongly convex space and reflexive. Observing the definition of k-strongly convex space and a condition which characterizes k-uniformly extremely convex space in Lemma 9, by the reflexivity of X, it is easy to see that X is k-uniformly extremely convex space.Corollary 12 (see [7]).
X is uniformly extremely convex space if and only if X is strictly convex space and has the drop property.Noticing the procedure of proving Theorem11 we can deduce the following.Corollary 13.
IfX is k-uniformly extremely convex space, then X is k-strongly convex space.Now we are going to show that the converse to Corollary13 is not true. In [12], it is proved that LkUR spaces are k-strongly convex spaces. In general, LkUR spaces need not be reflexive since L1UR is just the usual definition of LUR space [18, 19]. It follows that there exists a k-strongly convex space X which is not reflexive. Hence X is not a k-uniformly extremely convex space since X is not reflexive.Corollary 14.
X is k-uniformly extremely convex space if and only if X is reflexive and, for any x∈S(X) and f∈Sx, ϵ>0, there exist δ>0 and a compact set C⊂X with dim⁡C≤k such that F(f,δ)⊂{x:x∈X,d(x,C)<ϵ}, where the set F(f,δ)={x:x∈U(X),f(x)≥1-δ} is the slice generated by f and δ.Proof.
It is immediate from Corollary13, Theorem 11, and Lemmas 2, 4, and 5.Theorem 15.
X is k-uniformly extremely convex space if and only if X is reflexive and, for any f∈S(X∗), one has dim⁡Af≤k, Af∩co¯(U(X)∖VAf)=∅, where the set VAf, which includes set Af, is arbitrary open set with regard to norm topology (X,·).Proof.
Necessity. Suppose that X is k-uniformly extremely convex space; then by Theorem 11 we know that X is k-strictly convex space and reflexive. For any f∈S(X∗), by the reflexivity of X, there exists x∈S(X) such that f(x)=1; hence x∈Af. Combining the fact that X is k-strictly convex space with Lemma 3 we can deduce that dim⁡Af≤k.
Now we are going to prove the equalityAf∩co¯(U(X)∖VAf)=∅.
Firstly, we will prove that, for anyz∉VAf and every open set VAf (where VAf⊃Af) with regard to norm topology (X,·), there exists a scalar r>0 such that dist(z,Af)≥r.
Noticing thatAf is compact set in X, for any z∉VAf, we can deduce that there exists x∈Af such that x-z=dist(z,Af)=rz. Now we claim that there exists minimum value of rz denoted by r, such that dist(z,Af)≥r for any z∉VAf. In fact, if rz does not have minimum value, then 1/n is impossible to be minimum value for any integer n. Hence, there exist zn∉VAf and xn∈Af such that xn-zn=dist(zn,Af)<1/n. Since Af is compact, the above sequence {xn} has the convergent subsequence; without loss of generality and letting the convergent subsequence be {xn} itself, then xn→x0,x0∈Af. Noticing that xn-zn=dist(zn,Af)<1/n, we can deduce that zn→x0,x0∈Af⊂VAf.
On the other hand, combining the fact thatX∖VAf is closed set with zn∈X∖VAf,zn→x0, we can deduce that x0∈X∖VAf. This contradicts x0⊂VAf.
Secondly, we will prove that forVAf there exists a scalar m>0 such that the inequality f(x)>f(y)+m holds for all x∈Af and y∈U(X)∖VAf.
If the above inequality is not true, then there existsyn∈U(X)∖VAf such that f(yn)→f(x)=1,n→∞. By the condition given in Theorem 15, Corollary 13, and Lemma 4(iii), we have dist(yn,Af)→0,n→∞. On the other hand, by yn∈U(X)∖VAf, we can deduce that dist(yn,Af)↛0,n→∞; this contradicts the statement that dist(yn,Af)→0,n→∞. Hence we have (8)fx-m≥sup⁡fy:y∈UX∖VAf=sup⁡fy:y∈co¯UX∖VAf;this shows that x∉co¯(U(X)∖VAf). By the arbitrary of x∈Af, we can deduce that Af∩co¯(U(X)∖VAf)=∅.
Sufficiency. By Lemmas 2 and 3, Theorem 11, and the condition given in Theorem 15, only we need to prove that X has the Kadec-Klee property. Let x∈S(X), {xn}n=1∞⊂S(X), and xn→wx,n→∞. By the well-known James’ theorem, there exists f∈S(X∗) such that f(x)=1; it follows that x∈Af.
Case  1. If {xn}n=1∞∩Af=∅, then {xn}n=1∞ is relatively compact. Otherwise, every point of Af is not accumulation point of {xn}n=1∞. Hence, for any x∈Af there exists ϵ0>0 such that {y∈X:y-x<ϵ0} does not contain any point of {xn}n=1∞. We construct an open set VAf=⋃x∈Af{y∈X:y-x<ϵ0} with regard to norm topology (X,·); then Af⊂VAf and VAf∩{xn}n=1∞=∅. Since co¯(U(X)∖VAf) is bounded closed convex set with regard to norm topology (X,·), co¯w(U(X)∖VAf)=co¯(U(X)∖VAf). Noticing that co¯w(U(X)∖VAf) is bounded set with regard to weak topology σ(X,X∗), we know that co¯w(U(X)∖VAf) is compact set with regard to weak topology σ(X,X∗). Hence there is a function g∈X∗ which separates Af and co¯(U(X)∖VAf); that is, there is a scalar l>0 such that g(Af)-l>sup⁡g(co¯(U(X)∖VAf)). Evidently, {xn}n=1∞⊂co¯(U(X)∖VAf); it follows that g(Af)-g({xn}n=1∞)>l. This contradicts the assumption that xn→wx,n→∞.
Case  2. If {xn}n=1∞∩Af≠∅, then ({xn}n=1∞∖Af)∩Af=∅. According to Case  1 we know that {xn}n=1∞∖Af is relatively compact set. Hence {xn}n=1∞∩Af is compact set since Af is bounded closed convex set in certain finite dimensional subspace of X. On the other hand, it is obvious that {xn}n=1∞=({xn}n=1∞∖Af)∪({xn}n=1∞∩Af); hence {xn}n=1∞¯=({xn}n=1∞∖Af)¯∪({xn}n=1∞∩Af)¯. This shows that {xn}n=1∞ is relatively compact.
Consequently, in Cases  1 and 2, we always conclude that{xn}n=1∞ is relatively compact. Furthermore, by the assumption that xn→wx,n→∞, we can deduce that xn-x→0,n→∞. This completes the proof that X has the Kadec-Klee property.In particular, considering the special case of Theorem15 when k=1, we obtained Theorem  2.5 in [7] as a corollary.Corollary 16.
X is uniformly extremely convex space if and only if X is reflexive and for any f∈S(X∗), one has dim⁡Af=1, Af∩co¯(U(X)∖VAf)=∅, where the set VAf, which includes set Af, is arbitrary open set with regard to norm topology (X,·). In other words, X is uniformly extremely convex space if and only if X is reflexive and every point of S(X) is a denting point of U(X).To show that the converse to Theorem8 is not true, we consider the following example.Example 17.
There exists ak-uniformly extremely convex space X which is not a (k-1)-uniformly extremely convex space.Letk≥2 be an integer, and let i1<i2<⋯<ik. For each x=(a1,a2,…)∈l2, define (9)xi1,…,ik2=∑j=1kaij2+∑i≠i1,…,ikai2.From [20] we know that Xi1,…,ik=(l2,·i1,…,ik) is a kUR space. It is easy to see that kUR space is k-uniformly extremely convex space from the definition of kUR space and a condition which characterizes k-uniformly extremely convex space in Lemma 9. Hence Xi1,…,ik is a k-uniformly extremely convex space. It follows from Theorem 11 that Xi1,…,ik is a k-strictly convex space but is not a (k-1)-strictly convex space that follows from [16]. Hence Xi1,…,ik is not a (k-1)-uniformly extremely convex space.


## 3. The Relations betweenk-Uniformly Extremely Convex Space and Various Other Types of Convex Space

Now we give a list of examples to distinguishk-uniformly extremely convex spaces from kR, kUR, kNUC, and NUC spaces.(i)
We are ready now to distinguishk-uniformly extremely convex and kR spaces.SincekR spaces are k-strictly convex spaces and have the drop property, it follows from Theorem 11 that kR spaces are k-uniformly extremely convex, but the converse is not true.Example 18.
There exists ak-uniformly extremely convex space X which is not a kR space for every k≥2.Letk≥2 be an integer, and let i1<i2<⋯<ik. For each x=(a1,a2,…)∈l2, define (10)xi1,…,ik2=∑j=1kaij2+∑i≠i1,…,ikai2,and let Xi1,…,ik=(l2,·i1,…,ik). For x∈l2, let xk=supi1<i2<⋯<ik⁡xi1,…,ik, Xk=(l2,xk). It follows from [20] that Xk is a kUR space but is not a kR space. Hence Xk is a k-uniformly extremely convex space since X is a kUR space. (ii)
We are ready now to distinguishk-uniformly extremely convex and kUR spaces.Example 19.
For allk≥1, there exists a k-uniformly extremely convex space X which is not a kUR space.LetE=(l2,·); for x=(a1,a2,…)∈E, define (11)x2=a1+a22+a32+⋯1/22+a222+⋯+ann2+⋯2.It follows from [2] that X=(∑‍⨁E)l2 is a 2R space; furthermore, X=(∑‍⨁E)l2 is a k-uniformly extremely convex space but is not a kUR space [20]. (iii)
We are ready now to distinguishk-uniformly extremely convex and LkUR spaces.We consider a nonreflexive LkUR space X. Then X is not a k-uniformly extremely convex space since X is not reflexive. On the other hand, we consider Example 19; then X=(∑‍⨁E)l2 is a 2R space and it follows that X is a k-uniformly extremely convex space for all k≥1. But X is not a LkUR space that follows from [21].(iv)
We are ready now to distinguishk-uniformly extremely convex spaces and NUC or kNUC spaces.Example 20.
For allk≥1, there exists a k-uniformly extremely convex space X which is neither a NUC nor a kNUC space for all k≥2.Let(X,·) be the l2-sum of {ln,n≥2}; then (X,·) is a 2R space with normalized basis {en}. Define, ∀x=∑n=1∞anen∈X, (12)x=a1+∑n=2∞anen2+∑n=2∞ann21/2.By Theorem  4 in [9], we know that (X,·) is a 2R space but is not a LNUC space. It follows that X is a k-uniformly extremely convex space for all k≥1 but is neither a NUC nor a kNUC space for all k≥2.Remark 21.
(i) The class ofk-uniformly extremely convex spaces lies strictly between the classes of kUR spaces and the k-strongly convex spaces.
(ii) The class ofk-uniformly extremely convex spaces lies strictly between the classes of kR spaces and the class of k-strongly convex spaces.
(iii) The class ofk-uniformly extremely convex spaces has no inclusive relation with the class of LkUR spaces.In particular, considering the special case of Remark21 when k=1, we obtained the main conclusions of [7], that is, Remarks 3.5  and  3.7 in [7].



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