# OnP- and p-Convexity of Banach Spaces

Authors: Omar Muñiz-Pérez
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We show that everyU-space and every Banach space X satisfying δX(1)>0  are P(3)-convex, and we study the nonuniform version of P-convexity, which we call p-convexity.


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## Body

## 1. Introduction

Kottman introduced in 1970 the concept ofP-convexity in [1]. He proved that every P-convex space is reflexive and also that P-convexity follows from uniform convexity, as well as from uniform smoothness. In this paper we study conditions which guarantee the P-convexity of a Banach space and generalize the result of Kottman concerning uniform convexity in two different ways: every U-space and every Banach space X satisfying δX(1)>0 are P(3)-convex. There are many convexity conditions of Banach spaces which have a uniform and also a nonuniform version, for example, strictly convexity is the nonuniform version of uniform convexity, smoothness is the nonuniform version of uniform smoothness, and a u-space is the nonuniform version of a U-space, among others. We also define the concept of p-convexity, which is the nonuniform version of P-convexity and obtain some interesting results.


## 2.P-Convex Banach Spaces

Throughout this paper we adopt the following notation.(X,∥·∥) will be a Banach space and when there is no possible confusion, we simply write X. The unit ball {x∈X:∥x∥≤1} and the unit sphere {x∈X:∥x∥=1} are denoted, respectively, by BX and SX. B(y,r) will denote the closed ball with center y and radius r. The topological dual space of X is denoted by X*.


### 2.1.P-Convexity

The next concept was given by Kottman in [1].Definition 2.1.
LetX be a Banach space. For each n∈ℕ let
(2.1)P(n,X)=sup{r>0:thereexistndisjointballsofradiusrinBX}.
It is easy to see that P(n,X)≤1/2 for n≥2.Definition 2.2.
X is said to be P-convex if P(n,X)<1/2 for some n∈ℕ.The following lemma was proved in [1].Lemma 2.3.
LetX be a Banach space and n∈ℕ. Then P(n,X)<1/2 if and only if there exists ε>0 such that for any x1,x2,…,xn∈SX(2.2)min{∥xi-xj∥:1≤i,j≤n,i≠j}≤2-ε.
That is, X is P -convex if and only if X satisfies condition (2.2) for some n∈ℕ and some ε>0.Definition 2.4.
Givenn∈ℕ and ε>0 we say that X is P(ε,n)-convex if X satisfies (2.2). For each n∈ℕ, X is said to be P(n)-convex if it is P(ε,n)-convex for some ε>0.


### 2.2.P-Convexity and the Coefficient of Convexity

In [1], Kottman proved that if X is a Banach space satisfying the condition δX(2/3)>0, then X is P(3)-convex, where δX is the modulus of convexity. In this section we give a result which improves this condition, and we show that this assumption is sharp.We recall the following concepts introduced by  J. A. Clarkson in 1936.Definition 2.5.
The modulus of convexity of a Banach spaceX is the function δX:[0,2]→[0,1] defined by
(2.3)δX(ε)=inf{1-∥x+y2∥:x,y∈BX,∥x-y∥≥ε}.
The coefficient of convexity of a Banach space X is the number ε0(X) defined as
(2.4)ε0(X)=sup{ε∈[0,2]:δX(ε)=0}.
We also need the following definition given by R. C. James in 1964.Definition 2.6.
X is said to be uniformly nonsquare if there exists α>0 such that for all ξ,η∈SX(2.5)min{∥ξ-η∥,∥ξ+η∥}≤2-α.
In order to prove our theorem we need two known results which can be found in [2].Lemma 2.7 (Goebel-Kirk).
LetX be a Banach space. For each ε∈[ε0(X),2], one has the equality δX(2-2δX(ε))=1-ε/2.Lemma 2.8 (Ullán).
LetX be a Banach space. For each 0≤ε2≤ε1<2 the following inequality holds: δX(ε1)-δX(ε2)≤(ε1-ε2)/(2-ε1).Using these lemmas we obtain:Theorem 2.9.
LetX be a Banach space which satisfies δX(1)>0, that is, ε0(X)<1. Then X is P(3)-convex. Moreover, there exists a Banach space X with ε0(X)=1 which is not P(3)-convex.Proof.
Lett0=2-2-ε0(X). Clearly ε0(X)<t0<1. Let x,y,z∈SX, and suppose that ∥x-y∥>2-2δX(t0) and ∥x-z∥>2-2δX(t0). By Lemma 2.7, we have
(2.6)∥x+y2∥≤1-δ(2-2δX(t0))=1-(1-t02)=t02.
Similarly ∥(x+z)/2∥≤t0/2. Hence we get
(2.7)∥z-y∥≤∥z+x∥+∥x+y∥≤2t0.
Finally, from Lemma 2.8 it follows that
(2.8)δX(t0)=δX(t0)-δX(ε0(X))≤t0-ε0(X)2-t0=2-ε0(X)-1=1-t0.
Then ∥y-z∥≤2t0≤2-2δX(t0), and thus X is P(3)-convex.
Now consider for each1<p<∞ the space lp,∞ defined as follows. Each element x={xi}i∈lp may be represented as x=x+-x-, where the respective ith components of x+ and x- are given by (x+)i=max{xi,0} and (x-)i=max{-xi,0}. Set ∥x∥p,∞=max{∥x+∥p,∥x-∥p} where ∥·∥p stands for the lp-norm. The space lp,∞=(lp,∥·∥p,∞) satisfies ε0(lp,∞)=1 (see [3]). On the other hand let x1=e1-e3, x2=-e1+e2, x3=-e2+e3∈Slp,∞, where {ei}i is the canonical basis in lp. These points satisfy that ∥xi-xj∥p,∞=2, i≠j. Thus lp,∞ is not P(3.2)-convex.It is known that if a Banach spaceX satisfies ε0(X)<1, then X has normal structure as well as P(3)-convexity. The space X=lp,∞ is an example of a Banach space with ε0(X)=1 which does not have normal structure (see [3]) and is not P(3)-convex.Kottman also proved in [1] that every uniformly smooth space is a P-convex space. We obtain a generalization of this fact. Before we show this result we recall the next concept.Definition 2.10.
The modulus of smoothness of a Banach spaceX is the function ρX:[0,∞)→[0,∞) defined by
(2.9)ρX(t)=sup{12(∥x+ty∥+∥x-ty∥-2):x,y∈SX}
for each t≥0. X is called uniformly smooth if limt→0ρX(t)/t=0.
The proofs of the following lemmas can be found in [4, 5].Lemma 2.11.
For every Banach spaceX, one has limt→0ρX(t)/t=(1/2)ε0(X*).Lemma 2.12.
LetX be a Banach space. X is P(3)-convex if and only if X* is P(3)-convex.By Theorem2.9 and by the previous lemmas we deduce the next result.Corollary 2.13.
IfX is a Banach space satisfying limt→0ρX(t)/t<1/2, then X is P(3)-convex.With respect toP(4)-convex spaces we have this result, which is easy to prove.Proposition 2.14.
IfX is a Banach space P(ε,4)-convex, then ε0(X)≤2-ε, and hence X is uniformly nonsquare.In fact, in bidimensional normed spaces,P(4)-convexity and uniform nonsquareness coincide. The proof of this involves many calculations and can be seen in [6].Another technical proof (see [6]) shows that if X is a bidimensional normed space, then X is always P(1,5)-convex. Hence the space X=(ℝ2,∥·∥∞) is P(1,5)-convex and ε0(X)=2, and thus P(5)-convexity does not imply uniform squareness.


### 2.3. Relation betweenU-Spaces and P-Convex Spaces

In this section we show thatP-convexity follows from U-convexity. The following concept was introduced by Lau in 1978 [7].Definition 2.15.
A Banach spaceX is called a U-space if for any ε>0 there exists δ>0 such that
(2.10)x,y∈SX,f(x-y)>ε,forsomef∈∇(x)⇒∥x+y2∥≤1-δ,
where for each x∈X(2.11)∇(x)={f∈SX*:f(x)=∥x∥}.
The modulus of this type of convexity was introduced by Gao in [8] and further studied by Mazcuñán-Navarro [9] and Saejung [10]. The following result is proved in [8].Lemma 2.16.
LetX be a Banach space. If X is U-space, then X is uniformly nonsquare,From the above we obtain the next theorem which is a generalization of Kottman's result, who showed in [1] that P(3)-convexity follows from uniform convexity.Theorem 2.17.
IfX is a U-space, then  X is P( 3)-convex.Proof.
By Lemma2.16 we have that there exists α>0 such that for all ξ,η∈SX(2.12)min{∥ξ-η∥,∥ξ+η∥}≤2-α.
Since X is a U-space, for ε=α/2 there exists δ>0 such that
(2.13)x,y∈SX,f(x-y)≥α2,forsomef∈∇(x)⇒∥x+y2∥≤1-δ.
We claim that X is P(β,3)-convex, where β=min{α,δ}. Indeed, proceeding by contradiction, assume that there exist x,y,z∈SX such that
(2.14)min{∥x-y∥,∥x-z∥,∥y-z∥}>2-β.
Define w=-y and u=-z, and let f∈∇(w). If f(w-x)≥α/2, then
(2.15)∥w+x2∥<1-δ.
Therefore 2-δ≤2-β<∥x-y∥<2-2δ, which is not possible. Hence f(w-x)<α/2. Similarly we prove f(w+u)<α/2. Also ∥x+u∥=∥x-z∥>2-β≥2-α, and hence, by (2.12) we have f(x-u)≤∥x-u∥≤2-α. By the above we have
(2.16)2=2f(w)=f(w-x)+f(x-u)+f(u+w)<α2+2-α+α2=2
which is a contradiction.


### 2.4. The Dual Concept ofP-Convexity

In [1], Kottman introduces a property which turns out to be the dual concept of P-convexity. In this section we characterize the dual of a P-convex space in an easier way. We begin by showing Kottman's characterization.Definition 2.18.
LetX be a Banach space and ε>0. A convex subset A of BX is said to be ε-flat if A⋂(1-ε)BX=∅. A collection 𝔇 of ε-flats is called complemented if for each pair of ε-flatsA and B in 𝔇 we have that A⋃B has a pair of antipodal points. For any n∈ℕ we define
(2.17)F(n,X)=inf{ε>0:BXhasacomplementedcollection𝔇ofε-flatssuchthatCard(𝔇)=n}.Theorem 2.19 (Kottman).
LetX be a Banach space and  n∈ℕ. Then (a)
F(n,X*)=0⇔P(n,X)=1/2.(b)
P(n,X*)=1/2⇔F(n,X)=0.Now we defineP-smoothness and prove that it turns out to be the dual concept of P-convexity. The advantage of this characterization is that it uses only simple concepts, and one does not need ε-flats. Besides in the proof of the duality we do not need Helly's theorem nor the theorem of Hahn-Banach, as Kottman does in Theorem 2.19.Definition 2.20.
LetX be a Banach space and δ>0. For each f,g∈X* set S(f,g,δ)={x∈BX:f(x)≥1-δ,g(x)≥1-δ}. Given δ>0 and n∈ℕ, X is said to be P(δ,n)-smooth if for each f1,f2,…,fn∈SX* there exist 1≤i,j≤n,i≠j, such that S(fi,-fj,δ)=∅. X is said to be P(n)-smooth if it is P(δ,n)-smooth for some δ>0, and X is said to be P-smooth if it is P(δ,n)-smooth for some δ>0 and some n∈ℕ.Proposition 2.21.
LetX be a Banach space. Then
(a)
X is P(n)-convex if and only if X* is P(n)-smooth.(b)
X is P(n)-smooth if and only if X* is P(n)-convex.Proof.
(a) LetX be a P(ε,n)-convex space. Let x1**,…,xn**∈SX**. We will show that there exist 1≤i,j≤n,i≠j, such that S(xi**,-xj**,ε/4)=∅. Since X is P-convex, it is also reflexive. Therefore x1**=𝚥(x1),…,xn**=𝚥𝚥(xn) for some x1,…,xn∈SX, where 𝚥 is the canonical injection from X to X**. By hypothesis, there exist 1≤i,j≤n, i≠j, such that ∥xi-xj∥≤2-ε. Therefore it is enough to prove that
(2.18){f∈BX*:f(xi)≥1-ε4,-f(xj)≥1-ε4}=∅.
We proceed by contradiction supposing that there exists f∈BX* such that f(xi)≥1-ε/4 and -f(xj)≥1-ε/4. Then
(2.19)2-ε≥∥xi-xj∥≥f(xi-xj)≥2-ε2,
which is not possible; consequently X* is P(ε/4,n)-smooth.
Now letX be a Banach space such that X* is P(ε,n)-smooth. Let x1,…,xn∈SX. By hypothesis, there exist 1≤i,j≤n, i≠j, such that S(𝚥(xi),-𝚥(xj),ε)=∅, that is, for each f∈BX* we have f(xi)<1-ε or -f(xj)<1-ε. We will see that ∥xi-xj∥≤2-ε. We again proceed by contradiction supposing that ∥xi-xj∥=∥𝚥(xi-xj)∥>2-ε. There exists f∈SX* such that 𝚥(xi-xj)(f)=f(xi)-f(xj)>2-ε. If f(xi)<1-ε, then(2.20)1=∥f∥∥xj∥≥-f(xj)>2-ε-f(xi)>1
which is not possible. Similarly if -f(xj)<1-ε, we obtain a contradiction. Thus ∥xi-xj∥≤2-ε, and consequently X is P(ε,n)-convex. The proof of (b) is analogous to the proof of (a).Therefore the conditionsX is P(n)-smooth and F(n,X)>0 must be equivalent.


## 2.1.P-Convexity

The next concept was given by Kottman in [1].Definition 2.1.
LetX be a Banach space. For each n∈ℕ let
(2.1)P(n,X)=sup{r>0:thereexistndisjointballsofradiusrinBX}.
It is easy to see that P(n,X)≤1/2 for n≥2.Definition 2.2.
X is said to be P-convex if P(n,X)<1/2 for some n∈ℕ.The following lemma was proved in [1].Lemma 2.3.
LetX be a Banach space and n∈ℕ. Then P(n,X)<1/2 if and only if there exists ε>0 such that for any x1,x2,…,xn∈SX(2.2)min{∥xi-xj∥:1≤i,j≤n,i≠j}≤2-ε.
That is, X is P -convex if and only if X satisfies condition (2.2) for some n∈ℕ and some ε>0.Definition 2.4.
Givenn∈ℕ and ε>0 we say that X is P(ε,n)-convex if X satisfies (2.2). For each n∈ℕ, X is said to be P(n)-convex if it is P(ε,n)-convex for some ε>0.


## 2.2.P-Convexity and the Coefficient of Convexity

In [1], Kottman proved that if X is a Banach space satisfying the condition δX(2/3)>0, then X is P(3)-convex, where δX is the modulus of convexity. In this section we give a result which improves this condition, and we show that this assumption is sharp.We recall the following concepts introduced by  J. A. Clarkson in 1936.Definition 2.5.
The modulus of convexity of a Banach spaceX is the function δX:[0,2]→[0,1] defined by
(2.3)δX(ε)=inf{1-∥x+y2∥:x,y∈BX,∥x-y∥≥ε}.
The coefficient of convexity of a Banach space X is the number ε0(X) defined as
(2.4)ε0(X)=sup{ε∈[0,2]:δX(ε)=0}.
We also need the following definition given by R. C. James in 1964.Definition 2.6.
X is said to be uniformly nonsquare if there exists α>0 such that for all ξ,η∈SX(2.5)min{∥ξ-η∥,∥ξ+η∥}≤2-α.
In order to prove our theorem we need two known results which can be found in [2].Lemma 2.7 (Goebel-Kirk).
LetX be a Banach space. For each ε∈[ε0(X),2], one has the equality δX(2-2δX(ε))=1-ε/2.Lemma 2.8 (Ullán).
LetX be a Banach space. For each 0≤ε2≤ε1<2 the following inequality holds: δX(ε1)-δX(ε2)≤(ε1-ε2)/(2-ε1).Using these lemmas we obtain:Theorem 2.9.
LetX be a Banach space which satisfies δX(1)>0, that is, ε0(X)<1. Then X is P(3)-convex. Moreover, there exists a Banach space X with ε0(X)=1 which is not P(3)-convex.Proof.
Lett0=2-2-ε0(X). Clearly ε0(X)<t0<1. Let x,y,z∈SX, and suppose that ∥x-y∥>2-2δX(t0) and ∥x-z∥>2-2δX(t0). By Lemma 2.7, we have
(2.6)∥x+y2∥≤1-δ(2-2δX(t0))=1-(1-t02)=t02.
Similarly ∥(x+z)/2∥≤t0/2. Hence we get
(2.7)∥z-y∥≤∥z+x∥+∥x+y∥≤2t0.
Finally, from Lemma 2.8 it follows that
(2.8)δX(t0)=δX(t0)-δX(ε0(X))≤t0-ε0(X)2-t0=2-ε0(X)-1=1-t0.
Then ∥y-z∥≤2t0≤2-2δX(t0), and thus X is P(3)-convex.
Now consider for each1<p<∞ the space lp,∞ defined as follows. Each element x={xi}i∈lp may be represented as x=x+-x-, where the respective ith components of x+ and x- are given by (x+)i=max{xi,0} and (x-)i=max{-xi,0}. Set ∥x∥p,∞=max{∥x+∥p,∥x-∥p} where ∥·∥p stands for the lp-norm. The space lp,∞=(lp,∥·∥p,∞) satisfies ε0(lp,∞)=1 (see [3]). On the other hand let x1=e1-e3, x2=-e1+e2, x3=-e2+e3∈Slp,∞, where {ei}i is the canonical basis in lp. These points satisfy that ∥xi-xj∥p,∞=2, i≠j. Thus lp,∞ is not P(3.2)-convex.It is known that if a Banach spaceX satisfies ε0(X)<1, then X has normal structure as well as P(3)-convexity. The space X=lp,∞ is an example of a Banach space with ε0(X)=1 which does not have normal structure (see [3]) and is not P(3)-convex.Kottman also proved in [1] that every uniformly smooth space is a P-convex space. We obtain a generalization of this fact. Before we show this result we recall the next concept.Definition 2.10.
The modulus of smoothness of a Banach spaceX is the function ρX:[0,∞)→[0,∞) defined by
(2.9)ρX(t)=sup{12(∥x+ty∥+∥x-ty∥-2):x,y∈SX}
for each t≥0. X is called uniformly smooth if limt→0ρX(t)/t=0.
The proofs of the following lemmas can be found in [4, 5].Lemma 2.11.
For every Banach spaceX, one has limt→0ρX(t)/t=(1/2)ε0(X*).Lemma 2.12.
LetX be a Banach space. X is P(3)-convex if and only if X* is P(3)-convex.By Theorem2.9 and by the previous lemmas we deduce the next result.Corollary 2.13.
IfX is a Banach space satisfying limt→0ρX(t)/t<1/2, then X is P(3)-convex.With respect toP(4)-convex spaces we have this result, which is easy to prove.Proposition 2.14.
IfX is a Banach space P(ε,4)-convex, then ε0(X)≤2-ε, and hence X is uniformly nonsquare.In fact, in bidimensional normed spaces,P(4)-convexity and uniform nonsquareness coincide. The proof of this involves many calculations and can be seen in [6].Another technical proof (see [6]) shows that if X is a bidimensional normed space, then X is always P(1,5)-convex. Hence the space X=(ℝ2,∥·∥∞) is P(1,5)-convex and ε0(X)=2, and thus P(5)-convexity does not imply uniform squareness.


## 2.3. Relation betweenU-Spaces and P-Convex Spaces

In this section we show thatP-convexity follows from U-convexity. The following concept was introduced by Lau in 1978 [7].Definition 2.15.
A Banach spaceX is called a U-space if for any ε>0 there exists δ>0 such that
(2.10)x,y∈SX,f(x-y)>ε,forsomef∈∇(x)⇒∥x+y2∥≤1-δ,
where for each x∈X(2.11)∇(x)={f∈SX*:f(x)=∥x∥}.
The modulus of this type of convexity was introduced by Gao in [8] and further studied by Mazcuñán-Navarro [9] and Saejung [10]. The following result is proved in [8].Lemma 2.16.
LetX be a Banach space. If X is U-space, then X is uniformly nonsquare,From the above we obtain the next theorem which is a generalization of Kottman's result, who showed in [1] that P(3)-convexity follows from uniform convexity.Theorem 2.17.
IfX is a U-space, then  X is P( 3)-convex.Proof.
By Lemma2.16 we have that there exists α>0 such that for all ξ,η∈SX(2.12)min{∥ξ-η∥,∥ξ+η∥}≤2-α.
Since X is a U-space, for ε=α/2 there exists δ>0 such that
(2.13)x,y∈SX,f(x-y)≥α2,forsomef∈∇(x)⇒∥x+y2∥≤1-δ.
We claim that X is P(β,3)-convex, where β=min{α,δ}. Indeed, proceeding by contradiction, assume that there exist x,y,z∈SX such that
(2.14)min{∥x-y∥,∥x-z∥,∥y-z∥}>2-β.
Define w=-y and u=-z, and let f∈∇(w). If f(w-x)≥α/2, then
(2.15)∥w+x2∥<1-δ.
Therefore 2-δ≤2-β<∥x-y∥<2-2δ, which is not possible. Hence f(w-x)<α/2. Similarly we prove f(w+u)<α/2. Also ∥x+u∥=∥x-z∥>2-β≥2-α, and hence, by (2.12) we have f(x-u)≤∥x-u∥≤2-α. By the above we have
(2.16)2=2f(w)=f(w-x)+f(x-u)+f(u+w)<α2+2-α+α2=2
which is a contradiction.


## 2.4. The Dual Concept ofP-Convexity

In [1], Kottman introduces a property which turns out to be the dual concept of P-convexity. In this section we characterize the dual of a P-convex space in an easier way. We begin by showing Kottman's characterization.Definition 2.18.
LetX be a Banach space and ε>0. A convex subset A of BX is said to be ε-flat if A⋂(1-ε)BX=∅. A collection 𝔇 of ε-flats is called complemented if for each pair of ε-flatsA and B in 𝔇 we have that A⋃B has a pair of antipodal points. For any n∈ℕ we define
(2.17)F(n,X)=inf{ε>0:BXhasacomplementedcollection𝔇ofε-flatssuchthatCard(𝔇)=n}.Theorem 2.19 (Kottman).
LetX be a Banach space and  n∈ℕ. Then (a)
F(n,X*)=0⇔P(n,X)=1/2.(b)
P(n,X*)=1/2⇔F(n,X)=0.Now we defineP-smoothness and prove that it turns out to be the dual concept of P-convexity. The advantage of this characterization is that it uses only simple concepts, and one does not need ε-flats. Besides in the proof of the duality we do not need Helly's theorem nor the theorem of Hahn-Banach, as Kottman does in Theorem 2.19.Definition 2.20.
LetX be a Banach space and δ>0. For each f,g∈X* set S(f,g,δ)={x∈BX:f(x)≥1-δ,g(x)≥1-δ}. Given δ>0 and n∈ℕ, X is said to be P(δ,n)-smooth if for each f1,f2,…,fn∈SX* there exist 1≤i,j≤n,i≠j, such that S(fi,-fj,δ)=∅. X is said to be P(n)-smooth if it is P(δ,n)-smooth for some δ>0, and X is said to be P-smooth if it is P(δ,n)-smooth for some δ>0 and some n∈ℕ.Proposition 2.21.
LetX be a Banach space. Then
(a)
X is P(n)-convex if and only if X* is P(n)-smooth.(b)
X is P(n)-smooth if and only if X* is P(n)-convex.Proof.
(a) LetX be a P(ε,n)-convex space. Let x1**,…,xn**∈SX**. We will show that there exist 1≤i,j≤n,i≠j, such that S(xi**,-xj**,ε/4)=∅. Since X is P-convex, it is also reflexive. Therefore x1**=𝚥(x1),…,xn**=𝚥𝚥(xn) for some x1,…,xn∈SX, where 𝚥 is the canonical injection from X to X**. By hypothesis, there exist 1≤i,j≤n, i≠j, such that ∥xi-xj∥≤2-ε. Therefore it is enough to prove that
(2.18){f∈BX*:f(xi)≥1-ε4,-f(xj)≥1-ε4}=∅.
We proceed by contradiction supposing that there exists f∈BX* such that f(xi)≥1-ε/4 and -f(xj)≥1-ε/4. Then
(2.19)2-ε≥∥xi-xj∥≥f(xi-xj)≥2-ε2,
which is not possible; consequently X* is P(ε/4,n)-smooth.
Now letX be a Banach space such that X* is P(ε,n)-smooth. Let x1,…,xn∈SX. By hypothesis, there exist 1≤i,j≤n, i≠j, such that S(𝚥(xi),-𝚥(xj),ε)=∅, that is, for each f∈BX* we have f(xi)<1-ε or -f(xj)<1-ε. We will see that ∥xi-xj∥≤2-ε. We again proceed by contradiction supposing that ∥xi-xj∥=∥𝚥(xi-xj)∥>2-ε. There exists f∈SX* such that 𝚥(xi-xj)(f)=f(xi)-f(xj)>2-ε. If f(xi)<1-ε, then(2.20)1=∥f∥∥xj∥≥-f(xj)>2-ε-f(xi)>1
which is not possible. Similarly if -f(xj)<1-ε, we obtain a contradiction. Thus ∥xi-xj∥≤2-ε, and consequently X is P(ε,n)-convex. The proof of (b) is analogous to the proof of (a).Therefore the conditionsX is P(n)-smooth and F(n,X)>0 must be equivalent.


## 3.p-Convex Banach Spaces

In this section we introduce the nonuniform version ofP-convexity and we call it p-convexity.Definition 3.1.
LetX be a Banach space and n∈ℕ. X is said to be p(n)-convex if for any x1,…,xn∈SX, there exist 1≤i,j≤n, i≠j, such that ∥xi-xj∥<2. X is said to be p-convex if is p(n)-convex for some n∈ℕ.Kottman defined the concept ofP-convexity in terms of the intersection of balls. We will do something similar to give an equivalent definition of p-convexity. It is easy to see that in a normed space any two closed balls of radius 1/2 contained in the unit ball have non empty intersection. If the radius is less than 1/2, for example, in l1 for every n and for every r<1/2, then there exist n closed balls of radius r so that no two of them intersect. In fact let {ei}i=1∞ be the canonical basis of l1. Then the closed balls of radius r<1/2 centered at the points (1/2)ei, i∈ℕ are disjoint and contained in the unit ball. However, if X is p(n)-convex, we will see that for any n points in the unit ball there exists r<1/2 so that if the n closed balls centered at these n points are contained in the unit ball, there are two different balls with non empty intersection. To prove this we need the following lemma, which was shown in [11].Lemma 3.2.
LetX be a Banach space and x,y∈X, x,y≠0. Then
(3.1)∥x∥x∥-y∥y∥∥≥1min{∥x∥,∥y∥}(∥x-y∥-|∥x∥-∥y∥|).Lemma 3.3.
X is a p(n)-convex space if and only if for any y1,…,yn∈BX there exists r∈(0,1/2) such that, if B(yi,r)⊂BX for all i=1,…,n, then there are 1≤i,j≤n, i≠j, so that
(3.2)B(yi,r)∩B(yj,r)≠∅.Proof.
Assume thatX satisfies condition (3.2), and let x1,…,xn∈SX. Let r∈(0,1/2) be the number which satisfies condition (3.2) for x1/2,…,xn/2. It is easy to see that B(xi/2,r)⊂BX for each i=1,…,n. Therefore there exist 1≤i,j≤n, i≠j, such that
(3.3)B(xi2,r)∩B(xj2,r)≠∅.
Let
(3.4)y∈B(xi2,r)∩B(xj2,r).
We have
(3.5)∥xi-xj2∥≤∥xi2-y∥+∥xj2-y∥<2r<1,
and thus X is p(n)-convex. Now we suppose that there exist y1,…,yn∈BX such that for any ρ∈(0,1/2) we have
(3.6)B(yi,12-ρ)⊂BX
for all i=1,…,n, and
(3.7)B(yi,12-ρ)∩B(yj,12-ρ)=∅,
for all i,j=1,…,n, i≠j. We verify that X is not p(n)-convex in four steps.(a)
Take∥yi-yj∥>1-2ρ for any i,j=1,…,n, i≠j.(b)
Take1/2-3ρ<∥yi∥≤1/2+ρ, for all i=1,…,n. To verify this claim we note that ∥yi/∥yi∥-yi∥≥1/2-ρ for all i, because if ∥yi/∥yi∥-yi∥<1/2-ρ for some i, then yi/∥yi∥∈intB(yi,1/2-ρ)⊂intBX, which is not possible. Hence, as ∥yi/∥yi∥-yi∥=1-∥yi∥, it follows that ∥yi∥=1-∥yi/∥yi∥-yi∥≤1/2+ρ, for each i=1,…,n. Now, if ∥yi∥≤1/2-3ρ for some i, we have by (a) that for any j≠i, 1-2ρ<∥yi-yj∥≤∥yi∥+∥yj∥≤(1/2-3ρ)+(1/2+ρ)=1-2ρ which is not possible.(c)
Take|∥yi∥-∥yj∥|<4ρ, for any i,j=1,…,n, i≠j. Indeed, by (b) we get -4ρ=(1/2-3ρ)-(1/2+ρ)<∥yi∥-∥yj∥<(1/2+ρ)-(1/2-3ρ)=4ρ.(d)
From (a), (b), (c), and by Lemma3.2, we have(3.8)∥yi∥yi∥-yj∥yj∥∥≥1∥yi∥(∥yi-yj∥-|∥yi∥-∥yj∥|)>2-16ρ1+2ρ
for any i,j=1,…,n, i≠j. Since ρ>0 is arbitrary, as ρ→0, we obtain ∥yi/∥yi∥-yj/∥yj∥∥=2, for all i,j=1,…,n, i≠j, and thus X is not p(n)-convex.Next we give some examples of spaces which are notp-convex. The first is not reflexive and the last one is superreflexive.Example 3.4.
c0,and consequently,C[0,1] and l∞ are not p-convex spaces. Indeed, let {ei}i=1∞ be the canonical basis in c0.For each n∈ℕ we define ui=∑j=1nλi,jej, where λi,j=1 if j≠i,λi,i=-1, and i=1,…,n.Clearly u1,…,un∈Sc0, and for each i≠j we have ∥ui-uj∥∞=2.Example 3.5.
LetX denote the space obtained by renorming l2 as follows.For x=(xi)i∈ℕ∈l2 set
(3.9)∥|x|∥=max{supi,j|xi-xj|,(∑i=1∞xi2)1/2}.
Then ∥x∥≤∥|x|∥≤2∥x∥, where ∥·∥ stands for the l2-norm and X is superreflexive. On the other hand, the canonical basis {en}n in l2 satisfies ∥ei-ej∥∞=2 for each i≠j. Thus X is not p-convex.Now we will mention several properties that implyp-convexity.Recall the following concepts. LetX be a Banach space. X is said to be a u-space if it satisfies the following implication:
(3.10)x,y∈SX,∥x+y2∥=1⇒∇(x)=∇(y).X is said to be smooth if for any x∈SX, there exists a unique f∈SX* such that f(x)=1. That is, for each x∈SX, ∇(x) contains a single point. X is called strictly convex if the following implication holds:
(3.11)∀x,y∈BX:x≠y⇒∥x+y2∥<1.Proposition 3.6.
Every smooth space, every strictly convex space and every u-space are p(3)-convex space.Proof.
Every smooth space and every strictly convex space areu-space. It suffices to show that p(3)-convexity follows from being u-space. If X is a u-space, then for any x,y∈SX the following inequality holds: min{∥x-y∥,∥x+y∥}<2. Indeed, if we suppose that there exist x,y∈SX such that ∥x+y∥=∥x-y∥=2, then ∇(x)=∇(y) and ∇(x)=∇(-y), which is not possible. Suppose that X is not p(3)-convex, and there exist x,y,z∈SX so that ∥x-y∥=∥y-z∥=∥z-x∥=2. Since (1/2)∥x-y∥=(1/2)∥y-z∥=1,we have ∇(x)=∇(-y)=∇(z). Let f∈∇(-y); then f(x+z)≤∥x+z∥<2, and
(3.12)2=f(x)+f(-y)=f(x+z)-f(z)+f(-y)=f(x+z)<2.
Thus X is p(3)-convex.ObviouslyP-convexity implies p-convexity; however, a p-convex space is not necessarily P-convex, even if the space is reflexive as the following example shows.Example 3.7.
Let{rk}k=1∞ be a sequence of real numbers such that rk>1 for each k∈ℕ and rk↓1,when k→∞.Consider the space X=∑k=1∞⊕2lrk.It is known that this space is strictly convex, hence it is alsop(3)-convex.It is also known that X is reflexive.However X is notP-convex.Indeed, let ε>0.We choose k∈ℕ such that 2-ε<21/rk. If{ei}i=1∞ is the canonical basis of lrk, we have that ∥ei-ej∥rk=21/rk>2-ε for all i,j∈ℕ,i≠j, and hence X is not aP-convex space.We have obtained a result which shows a strong relation betweenP-convexity and p-convexity with respect to the ultrapower of Banach spaces. We recall the definition and some results regarding ultrapowers which can be found in [4].A filter𝔘 on I is called an ultrafilter on I if 𝔘 is a maximal element from 𝒫 with respect to the set inclusion. 𝔘 is an ultrafilter on I if and only if for all A⊂I either A∈𝔘 or I∖A∈𝔘. Let {Xi}i∈I be a family of Banach spaces, and let(3.13)l∞(Xi)={{xi}i∈I∈∏i∈IXi:sup{∥xi∥Xi:i∈I}<∞}.
If we define ∥{xi}i∈I∥∞=sup{∥xi∥Xi:i∈I} for each {xi}i∈I∈l∞(Xi), then ∥·∥∞ defines a norm in ł∞(Xi), and (l∞(Xi),∥·∥∞) is a Banach space. If 𝔘 is a free ultrafilter on I, then for each {xi}i∈I∈l∞(Xi) we have lim𝔘xi always exists and is unique. Let 𝔘 be an ultrafilter on I, and define (3.14)𝒩𝔘={{xi}∈l∞(Xi):lim𝔘∥xi∥=0}.𝒩𝔘 is a closed subspace of l∞(Xi). Theultraproduct of {Xi}i∈I with respect to the ultrafilter 𝔘 on I is the quotient space l∞(Xi)/𝒩𝔘 equipped with the quotient norm, which is denoted by {Xi}𝔘 and its elements by {xi}𝔘. If Xi=X for all i∈I, then {X}𝔘={Xi}𝔘 is called the ultrapower of X. The quotient norm in {Xi}𝔘, (3.15)∥{xi}𝔘∥=inf{∥{xi+yi}i∥∞:{yi}i∈𝒩𝔘},
satisfies the equality (3.16)∥{xi}𝔘∥=lim𝔘∥xi∥Xi,foreach{xi}𝔘∈{Xi}𝔘.
If 𝔘 is nontrivial, then X can be embedded into {X}𝔘 isometrically. We will write X̃i instead of {Xi}𝔘 and x̃ instead of {xi}𝔘 unless we need to specify the ultrafilter we are talking about.It is known thatX is uniformly convex if and only if X̃ is strictly convex, X is uniformly smooth if and only if X̃ is smooth, and X is a U-space if and only if X̃ is a u-space (see [12]). Similarly we obtain the following result.Theorem 3.8.
LetX be a Banach space and  m∈ℕ. The following are equivalent:(a)
X̃ is P(m)-convex.(b)
X is P(m)-convex,(c)
X̃ is p(m)-convex,Proof.
(a)⇒(b). Let {xi(n)}n∈x̃i, x̃i∈SX̃, i=1,…,m. Since lim𝔘∥xi(n)∥X=∥x̃i∥X̃=1 for all i, there exists a subsequence {xi(nk)}k of {xi(n)}n such that limk→∞∥xi(nk)∥X=1 and ∥xi(nk)∥X>0, for all k∈ℕ. Define
(3.17)yi(nk)=xi(nk)∥xi(nk)∥X,Γi,j={k∈ℕ:∥yi(nk)-yj(nk)∥X≤2-ε},
for each i,j=1,…,m, i≠j. We verify that there exist 1≤i,j≤m, i≠j, such that Γi,j∈𝔘. We proceed by contradiction assuming that, Γi,j∉𝔘 for all i≠j. Hence ℕ∖Γi,j∈𝔘 for all i≠j, and consequently ℕ∖(⋃i≠jΓi,j)≠∅, therefore there exists k0∈ℕ∖(⋃i≠jΓi,j). Thus we have ∥yi(nk0)-yj(nk0)∥>2-ε for each i≠j, and X is not P(m)-convex, which is a contradiction. Therefore there exist 1≤i,j≤m, i≠j, such that Γi,j∈𝔘, and hence lim𝔘∥yi(nk)-yj(nk)∥X≤2-ε. Finally, note that
(3.18)∥xi(nk)-xj(nk)∥X≤∥xi(nk)-yi(nk)∥X+∥xj(nk)-yj(nk)∥X+∥yi(nk)-yj(nk)∥X=|1-∥xi(nk)∥X|+|1-∥xi(nk)∥X|+∥yi(nk)-yj(nk)∥X,∥x̃i-x̃j∥X̃=lim𝔘∥xi(n)-xj(n)∥X=lim𝔘∥xi(nk)-xj(nk)∥X≤lim𝔘|1-∥xi(nk)∥X|+lim𝔘|1-∥xi(nk)∥X|+lim𝔘∥yi(nk)-yj(nk)∥X≤2-ε.
Therefore X̃ is P(m)-convex.
(b)⇒(c) is obvious.
(c)⇒(a). Suppose that X is not P(m)-convex. Hence for any n∈ℕ there exist x1(n),…,xm(n)∈SX such that ∥xi(n)-xj(n)∥X>2-1/n for all i,j=1,…,m, i≠j. Define x̃i={xi(n)}𝔘 for each i=1,…,m. Clearly x̃i∈SX̃ for all i, because ∥x̃i∥X̃=lim𝔘∥xi(n)∥X=1, and also,(3.19)∥x̃i-x̃j∥X̃=lim𝔘∥xi(n)-xj(n)∥X=limn→∞∥xi(n)-xj(n)∥X=2,
for each i≠j. Hence X̃ is not p(m)-convex.By the above theorem we can deduce the following known result.Corollary 3.9.
IfX is P -convex, then X is superreflexive.Proof.
IfX is P-convex, then X̃ is P-convex and therefore is reflexive. However in ultrapower reflexivity and superreflexivity are equivalent, hence X̃ is superreflexive, and consequently X is superreflexive.Now we turn our attention to some results regarding thep-convexity and the P-convexity of quotient spaces. To prove them we need the following concept.Definition 3.10.
A subspaceY of a normed space X is said to be proximinal if for all x∈X there exists y∈Y such that d(x,Y)=∥x-y∥.It is easy to see that every proximinal subspaceY of a Banach space X is closed.Proposition 3.11.
IfX is p(n)-convex and Y is a proximinal subspace of X, then X/Y is p (n)-convex.Proof.
Letq:X→X/Y be the quotient function. By the proximinality of Y we have q(BX)=BX/Y. Let x̃1,…,x̃n∈SX/Y and x1,…,xn∈SX such that x̃i=q(xi). Since X is p(n)-convex, there exist 1≤i,j≤n, i≠j, such that ∥xi-xj∥<2, and consequently ∥x̃i-x̃j∥<2.Corollary 3.12.
LetX be p(n)-convex and reflexive. If Y is a closed subspace of X, then X/Y is p(n)-convex.Proof.
It is shown in [13] that a Banach space X is reflexive if and only if each closed subspace of X is proximinal, and thus the corollary is a consequence of Proposition 3.11.Similarly we can prove that ifX is P(ε,n)-convex and Y is a closed subspace of X, then X/Y is P(ε,n)-convex.We obtain two results involvingψ-direct sums of p-convex spaces. Next we will define these sums as in [14] by Saito, et al.Definition 3.13.
SetΨ={ψ:[0,1]→ℝ∣ψisacontinuousconvexfunction,max{1-t,t}≤ψ(t)≤1,forall0≤t≤1.}
Let(X,∥·∥X) and (Y,∥·∥Y) be Banach spaces. For each ψ∈Ψ, one defines the norm ∥·∥ψ in X⊕Y as ∥(0,0)∥ψ=0 and for each (x,y)≠(0,0)(3.20)∥(x,y)∥ψ=(∥x∥X+∥y∥Y)ψ(∥y∥Y∥x∥X+∥y∥Y).In [15] it is shown that (X⊕Y,∥·∥ψ) is a Banach space, denoted by X⊕ψY called the ψ-direct and sum of X and Y.The proof of the following theorem is similar to the proof of Theorem3.5 in [16], which shows the corresponding result for P-convex spaces.Theorem 3.14.
LetX and  Y be Banach spaces and  ψ∈Ψ. Then  X⊕ψY is p-convex if and only if  X and  Y are p-convex.In [17] there is a theorem stating several equivalent conditions for strict convexity. We prove a similar result for p-convexity.Lemma 3.15.
LetX be a Banach space. The next assertions are equivalent.(a)
X is p(n)-convex.(b)
For anyq∈(1,∞) and for any x1,…,xn∈X, not all zero, there exist 1≤i,j≤n, i≠j, such that ∥xi-xj∥<2(q-1)/q(∥xi∥q+∥xj∥q)1/q.(c)
For someq∈(1,∞) and for any x1,…,xn∈X, not all zero, there exist 1≤i,j≤n, i≠j, such that ∥xi-xj∥<2(q-1)/q(∥xi∥q+∥xj∥q)1/q.Proof.
The implications(b)⇒(c)⇒(a) are immediate. We verify (a)⇒(b). Let q∈(1,∞) and x1,…,xn∈X, not all zero. If xj=0 and xi≠0 for some 1≤i,j≤n, then it is clear that ∥xi-xj∥<2(q-1)/q(∥xi∥q+∥xj∥q)1/q. Suppose that x1,…,xn∈X∖{0}. There exist 1≤i,j≤n, i≠j, such that
(3.21)∥xi∥xi∥-xj∥xj∥∥<2.
If ∥xj∥≤∥xi∥ by Lemma 3.2 we get
(3.22)∥xi-xj∥≤∥xj∥∥xi∥xi∥-xj∥xj∥∥+∥xi∥+∥xj∥<∥xi∥+∥xj∥.
As the function t↦tq is convex we obtain that
(3.23)∥xi-xj2∥q<(∥xi∥+∥xj∥2)q≤12(∥xi∥q+∥xj∥q).
Thus ∥xi-xj∥<2(q-1)/q(∥xi∥q+∥xj∥q)1/q.Proposition 3.16.
Let{Xi}i∈I be a family of p(n)-convex spaces, where the index set I≠∅ has any cardinality. Then the space X=lq(Xi) (1<q<∞) is p(n)-convex.Proof.
Letx(k)={xi(k)}i∈I∈X, 1≤k≤n, not all zero. Let i0∈I be such that xi0(k)≠0, for some k∈{1,…,n}. As Xi0 is a p(n)-convex space, we have by the preceding lemma that there exist 1≤l,m≤n such that
(3.24)∥xi0(l)-xi0(m)∥q<2q-1(∥xi0(l)∥q+∥xi0(m)∥q).
By the above we obtain
(3.25)∥x(l)-x(m)∥qq=∑i∈I∥xi(l)-xi(m)∥q<∑i∈I2q-1(∥xi(l)∥q+∥xi(m)∥q)=2q-1(∥x(l)∥qq+∥x(m)∥qq).
Therefore, by the previous lemma, X is p(n)-convex.



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