On Strictly Convex and Strictly Convex

ISSN 2219-7184; Copyright ICSRS Publication, 2011c www.i-csrs.org

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On Strictly Convex and Strictly Convex

According to an Index Semi-Normed Vector Spaces

Artur Stringa

Department of Mathematics, Faculty of Natural Sciences, University of Tirana, Albania

E-mail: [email protected] (Received:22-5-11 /Accepted:16-6-11)

Abstract

The purpose of this paper is giving the notion of strictly convex semi–

normed vector spaces according to an index and the notion of an extreme point of a convex set C in a vector space X, according to the semi-norm p in the spaceX. We extend to semi-normed vector spaces, via semi–pre-inner–

products, some known results on strictly convex normed vector spaces, which are characterized in terms of semi-inner-products.

Keywords: extreme point, semi–norm, semi–pre–inner–product, strict con- vexity, strict convexity according to an index.

1 Introduction

Definition 1.1 [3] Let X be a real vector space. We shall say that a real semi-inner-product is defined onX, if to anyx, y ∈X there corresponds a real number [x, y] and the following properties hold:

(1) (i) [x+y, z] = [x, z] + [y, z]

(ii) [λx, y] =λ[x, y], for x, y, z ∈X and λ∈R, (2) [x, x]>0, for x6= 0,

(3) [x, y]² ≤[x, x]·[y, y].

A vector space with a semi–inner–product (in short s.i.p.) is called a semi–

inner–product space (in short s.i.p.s.). It is a normed linear space withkxk=

[x, x]^1/2[3]. The topology on a s.i.p.s. is the one induced by this norm. It is further prove in [3] that every normed vector space can be made into s.i.p.s.

(in general, in infinitely many different ways).

A pointu of a convex setC in a vector space is called an extreme point of C ifu=t·v+ (1−t)·wwith 0< t <1 andv, w∈C, implies that u=v =w.

A normed linear spaceXis strictly convex if each point of the unit sphere is an extreme point of the unit ball [2]. The following characterizations on strictly convex spaces are available.

Theorem 1.2 [5]A normed linear spaceX is strictly convex if and only if:

kx+yk=kxk+kyk, where x6=y, implies y=λx, for some real λ >0.

Theorem 1.3 [1] A s.i.p.s. X is strictly convex if and only if:

[x, y] =kxk · kyk, where x6=y, implies y=λx, for some real λ >0.

Theorem 1.4 [2] On a s.i.p.s. X, the following conditions are equivalent:

(i) X is strictly convex ;

(ii) If ky+zk ≤ kyk and [z, y] = 0, then z = 0 ; (iii) If ky+zk=kyk and [z, y] = 0, then z = 0;

(iv) If A is a bounded linear on X, if kI + Ak ≤ 1 and if [Ax, x] = 0, for some x in X, then Ax= 0.

Aiming to generalize the condition (2) in the definition of s.i.p. function, we have introduced [7] the concept of the semi-pre-inner-product function, which is a generalization of the s.i.p. function’s concept:

Definition 1.5 [7] Let X be a real vector space. Consider a functional defined on X×X as follows:

X×X →R (x, y)7→[x, y]

If [x, y] satisfies the postulates:

(1)⁰ [x, x]≥0, x∈X,

(2)⁰ [λx, y] =λ[x, y], λ∈R and x, y ∈X,

(3)⁰ [x+y, z] = [x, z] + [y, z], x, y and z∈X, (4)⁰ [x, y]² ≤[x, x]·[y, y], x, y ∈X.

then we say that is a semi-pre-inner-product onX (in short s.p.i.p.). The pair (X,[·,·]) is called a semi-pre- inner-product space (in short s.p.i.p.s.).

The following theorem proves the existence of the s.p.i.p.:

Theorem 1.6 [7] For every semi-norm function p in the vector space X, there is a s.p.i.p. [·,·] , such that p²(x) = [x, x],∀x∈X.

Proof. [7] Let M =p⁻¹(0) = {x∈X/p(x) = 0}. Sincep is a semi–norm on X, we have thatM is a closed subspace of the vector space X. We note that the relationx∼y⇔(x−y)∈M, is a equivalent relation onX. Let us denote byX1 the X/M. For any two points x1 and x2 from the class of equivalence, bx, inequality 0 ≤ |p(x₁)−p(x₂)| ≤ p(x₁−x₂), implies p(x₁) = p(x₂) which gives that the function:

pb:X₁ →R⁺, such that∀xb∈X₁,pb(x) =b p(x),

for some x from the equivalence classx, is a norm onb X₁. Then, by [3], there exists a s.i.p. onX₁×X₁:

h·,·i:X₁×X₁ →R, so thatpb(bx) =hx,b xib ¹² ,∀bx∈X₁. Let us construct the function:

[·,·] :X×X →R, such that [x, y] =hx,b yib , for x, y ∈X.

The above function is a s.p.i.p. function. So, (i) [x, x] =hbx,xi ≥b 0, for x∈X.

(ii) [λx, y] =D λx,c ybE

=hλx,b yib =λhbx,byi=λ[x, y], forλ∈R and x, y ∈X, (iii) [x₁+x₂, y] = D

x\₁+x₂,byE

= hxb₁+xb₂,yib = hxb₁,yib +hxb₁,yib = [x₁, y] + + [x₂, y], for x₁, x₂ and y∈X

(iv) [x, y]² =hbx,byi² ≤ hbx,bxi · hby,byi= [x, x]·[y, y], for x, y ∈X.

We conclude by noting that:

[x, x] =hbx,xib = [pb(x)]b ² =p²(x),∀x∈X.

2 Main Results

Theorem 1.6 allows us to extend some results related to the semi-inner-products on semi-normed spaces.

At first, we note that, if [·,·]₁ and [·,·]₂ are two semi-pre-inner–products on X, then the function such that:

[x, y] = [x, y]₁+ [x, y]₂,∀(x, y)∈X²,

is a s.p.i.p.. It is obvious that the function [·,·] satisfies the conditions (1)⁰, (2)⁰, (3)⁰, of Definition 1.5. Let us prove that it satisfies also the condition (4)⁰. [x, y]² = ([x, y]₁+ [x, y]₂)² = ([x, y]₁)²+ 2 [x, y]₁·[x, y]₂+ ([x, y]₂)² ≤

≤ [x, x]₁·[y, y]₁+ 2 ([x, x]₁·[y, y]₁·[x, x]₂·[y, y]₂)¹² + [x, x]₂·[y, y]₂ ≤

≤ [x, x]₁·[y, y]₁+ [x, x]₁·[y, y]₂+ [x, x]₂·[y, y]₁+ [x, x]₂·[y, y]₂ =

= ([x, x]₁+ [x, x]₂)·([y, y]₁+ [y, y]₂) = [x, x]·[y, y].

By induction one can prove that every finite sum of s.p.i.–products is also a s.p.i.p. This fact makes it possible that every semi-normed space (X,{p_α}α∈A), where{p_α}α∈A is a family of semi-norms andA is an index set, can be consid- ered filtered and the filtration concept can be related with the filtration of the s.p.i.products [x, y]_α, corresponding to the semi-normsp_α,α∈ A. Actually the family of semi-norms {p_α}α∈A can be replaced with the family of semi-norms {pA}A⊂A, A={α1, α2, . . . , αs}, s∈N, where:

pA(x) = p²_α1(x) +p²_α2(x) +· · ·+p²_αs(x)¹₂

, forx∈X.

It is clear that this function satisfies the first two conditions of a semi-norm, and for the third condition we have that:

p_A(x+y) = p²_α

1(x+y) +· · ·+p²_αs(x+y)¹₂

= v u u t

s

X

i=1

p²_α

i(x+y)≤

≤ v u u t

s

X

i=1

(pαi(x) +pαi(y))² ≤ v u u t

s

X

i=1

p²_αi(x) + v u u t

s

X

i=1

p²_αi(y)6

≤ p_A(x) +p_A(y), for x, y ∈X.

The family{p_A}A⊂A is filtered since forA₁ ⊂A₂, we havep_A1(x)≤p_A2(x), for allx ∈X and furthermore this family is related to the s.p.i.–products family {[x, y]_A}_A⊂A by the equation

[x, y]_A= [x, y]_α

1 + [x, y]_α

2 +· · ·+ [x, y]_α

s, for x, y ∈X.

If we denote by B_A^∗ (0, ε) =

s

T

i=1

B_αi(0, ε) the neighborhoods of the origin of the X space for the topology of the semi-normed family {p_α}_α∈A and with B_A(0, ε) ={x∈X/p_A(x)< ε}the neighborhoods of the origin of theX space for the topology of the semi-normed family{p_A}_A⊂A the inclusionsB_A(0, ε)⊂ B_A^∗(0, ε)⊂√

sB_A(0, ε) show that the topologies of these families coincide.

At [5] and [2] some results in the normed vector spaces, characterized in terms of s.i.–products are formulated and proved. Our aim is to extend them in the semi-normed spaces, via s.p.i.–products. Firstly, let us give the following definitions.

Definition 2.1 The semi—normed vector space(X,{p_α}α∈A)is called strict- ly convex according to the index α ∈ A,if for every two points x and y in X, such that:

p_α(x)6= 0, p_α(y)6= 0 and p_α(x+y) = p_α(x) +p_α(y), there exists aλ_α >0, such that p_α(y−λ_αx) = 0.

Definition 2.2 The semi—normed vector space(X,{p_α}α∈A)is called strict- ly convex if it is strictly convex according to the index α, for allα ∈ A.

Letα ∈ A and denote by X_α the X/p⁻¹_α (0).

Theorem 2.3 The semi—normed vector space(X,{pα}α∈A)is strictly con- vex according to the index α∈ A if and only if the corresponding space X_α is strictly convex.

Proof. Suppose that the semi-normed vector space (X,{p_α}α∈A) is strictly convex according to the index α ∈ A and let xb and ybbe two elements from X_α such that:

pb_α(bx+y) =b pb_α(x) +b pb_α(y)b

By the structure of pb_α it derives that if x∈xband y∈ybthen:

p_α(x+y) = p_α(x) +p_α(y).

Thus, there exists a λ_α >0, such that:

p_α(y−λ_αx) = 0⇔pb_α(y\−λ_αx) = 0 ⇔y\−λ_αx=b0⇔yb−λ_αxb=b0⇔ yb = λ_αx.b

Conversely, let us suppose that Xα is a strictly convex normed space and let x, y be two points in X, such that p_α(x+y) = p_α(x) +p_α(y). Let bx and by be the equivalence classes of the pointsx and y. The following equality holds:

pb_α(bx+y) =b pb_α(x) +b pb_α(y)b

By the condition,∃λ_α >0, such that yb=λ_αx. We have:b

p_α(y−λ_αx) =pb_α(y\−λ_αx) =pb_α(yb−λd_αx) = pb_α(yb−λ_αbx) =pb_α(b0) = 0.

Theorem 2.3 allows us to find some similar characteristics of being strictly convex in a semi-normed vector space (X,{p_α}α∈A).

Theorem 2.4 The semi-normed vector space (X,{pα}α∈A) is strictly con- vex according to the indexα ∈ A if and only if for every two pointsx and y in X, such thatp_α(x)6= 0, p_α(y)6= 0 and [x, y]_α =p_α(x)·p_α(y), there is a λ_α >0 such that pα(y−λαx) = 0.

Proof. Suppose that the semi-normed vector space (X,{p_α}α∈A) is strictly convex according to the index α ∈ A. Theorem 2.3 implies that the normed vector spaceX_α is strictly convex. Let xand y be two points in X such that:

p_α(x)6= 0, p_α(y)6= 0 and [x, y]_α =p_α(x)·p_α(y).

We note that for the equivalence classesxband y,b hx,b yib _α=pbα(bx)·pbα(by).

Theorem 1.3 implies that there exists a λ_α >0, such that by=λ_αbx⇔p_α(y−λ_αx) = 0.

Conversely, suppose thatx and y are two points in X, such that:

p_α(x+y) = p_α(x) +p_α(y).

Sincepbα(bx+by) =pbα(bx) +pbα(by) we conclude that:

hx,b bx+byi_α+hy,b xb+yib_α = hbx+y,b xb+byi_α = [pb_α(bx+by)]² =

= pb_α(xb+by)·pb_α(xb+by) =

= pb_α(xb+by) (pb_α(x) +b pb_α(y))b therefore,

(pb_α(x)b ·pb_α(bx+y)b − hx,b bx+byi_α) + (pb_α(y)b ·pb_α(xb+by)− hby,bx+yib_α) = 0.

From the condition (4)⁰ of the Definition 1.5, each of the brackets in the above equality is nonnegative and since their sum is zero, we have that:

pb_α(bx)·pb_α(xb+by) =hx,b xb+byi_α and pb_α(by)·pb_α(xb+by) = hy,b bx+yib_α (1)

Sincepb_α(bx)·pb_α(xb+y) =b hbx,bx+yib_α, it derives that there exists some η_α >0, such that bx+by = η_αbx ⇔ by = (η_α−1)bx. Denoting by λ_α = η_α−1 we have that:

pbα(xb+y)b = pbα(x) +b pbα(y)b ⇔pbα(ηαbx) = pbα(x) +b pbα(y)b ⇔

⇔ η_α·pb_α(x) =b pb_α(bx) +pb_α(y)b ⇔pb_α(y) = (ηb _α−1)pb_α(bx) that imply η_α−1>0 (we used the fact that p_α(x)6= 0 and p_α(y)6= 0).

Therefore, there exists a λ_α >0, such that yb=λ_αxb⇔p_α(y−λ_αx) = 0.

Theorem 2.5 The semi—normed vector space X,{p_α}_α∈A

is strictly con- vex according to the index α ∈ A if and only if for every two points x and y in X, such that p_α(x) 6= 0, [x, y]_α = 0 and p_α(x+y) = p_α(x), we have that p_α(y) = 0.

Proof. Suppose that x and y are two points inX, such that:

p_α(x)6= 0,[x, y]_α= 0 and p_α(x+y) = p_α(x) For the equivalence classes of bx and by we have the followings:

pb_α(bx)6= 0,hbx,byi_α = 0 andpb_α(bx+y) =b pb_α(x).b

Theorem 2.3 implies that the space X_α is strictly convex. From Theorem 1.4/(iii) we conclude that:

pb_α(y) = 0b ⇔p_α(y) = 0.

Conversely, suppose thatx and y are two points in X, such that:

pα(x)6= 0, pα(y)6= 0 and pα(x+y) =pα(x) +pα(x).

We note that pbα(xb+y) =b pbα(x) +b pbα(y). Let’sb λα = pb_α(y)b

pb_α(bx).

Therefore,λ_α>0. Considering the pointsbu=λ_αbx−byandbv =xb+by, we have that:

pb_α(bu+bv) = pb_α(xb+λ_αx) =b pb_α

xb+ pb_α(by) pb_α(x)b bx

=pb_α

xb·(pb_α(bx) +pb_α(by)) pb_α(x)b

=

= pb_α(bx) +pb_α(by)

pb_α(bx) pb_α(bx) =pb_α(bx) +pb_α(by) = pb_α(xb+by) =pb_α(bv),

furthermore, pb_α(vb) 6= 0. Due to the equalities (1) in Theorem 2.4, we can conclude that:

hbu,bvi_α = hλ_αxb−y,b xb+byi_α =λ_αhbx,bx+yib_α− hy,b xb+byi_α =

= pb_α(y)b

pbα(bx)·pb_α(bx)·pb_α(bx+by)−pb_α(y)b ·pb_α(bx+y) = 0.b Finally, for the points buand bv the following equalities hold:

pb_α(bu+vb) =pb_α(bv) and hbu,bvi_α = 0.

Letu1 ∈bu and v1 ∈vb. The following equalities are true:

p_α(u₁+v₁) = p_α(v₁),[u₁, v₁]_α = 0 and p_α(v₁) =p_α(bv)6= 0.

From the assumption, it derives that:

p_α(u₁) = 0⇒pb_α(u) = 0b ⇒p_α(y−λ_αx) = 0.

Letp be a semi-norm in the space X and C a convex subspace of X.

Definition 2.6 A point x₀ ∈C, is called an extreme point of C, according to the semi-norm p, if x₀ = tu+ (1 −t)v with 0 < t < 1 and (u, v) ∈ C², implies thatp(x₀−u) = p(x₀−v) = p(u−v) = 0.

If the semi-norm p is a norm, then the extreme point according to the semi- norm pis extreme point for C, since if p(x0−u) =p(x0−v) =p(u−v) = 0, we have thatx₀ =u=v.

Theorem 2.7 The semi-normed vector space X,{p_α}_α∈A

is strictly con- vex according to the index α ∈ A if and only if every point x0 ∈ Sα(0,1) = {x∈X/p_α(x) = 1}is an extreme point of the setB_α^∗(0,1) ={x∈X/p_α(x)≤1}

according to the semi-norm p_α. Proof. Let’s denote by

x₀ =tu+ (1−t)v with 0< t <1, p_α(u)≤1, p_α(v)≤1 and p_α(x₀) = 1 From the definition of the sum and multiplication by scalars of the equivalence classes in X_α we have that xb₀ = tbu+ (1−t)bv. Since X_α is strictly convex, from [2] we have thatxb₀ =ub=vb. Therefore,xb₀−ub=xb₀−bv =bu−bv =b0⇒ p_α(x₀−u) =p_α(x₀−v) = p_α(u−v).

Conversely, to prove that the semi--normed vector space X,{p_α}_α∈A is strictly convex according to the indexα∈ A its sufficient to prove that X_α is strictly convex.

Letxb₀andyb₀be two distinct points from zero, such thathxb₀,yb₀i_α =pb_α(xb₀)·

pb_α(yb₀). The pointsxb₁ = xb₀

pb_α(xb₀) and yb₁ = yb₀

pb_α(yb₀) are in the the closed ball:

Bb_α^∗(0,1) = {bx∈X_α/pb_α(x)b ≤1}.

Let’sub=txb₁+(1−t)yb₁,0< t <1. We havepb_α(bu)≤tpb_α(xb₁)+(1−t)pb_α(yb₁)≤1.

Furthermore,

hbu,yb₀i_α = t· hxb₁,yb₀i_α+ (1−t)· hyb₁,yb₀i_α =

= t·p_α(xb₀)·p_α(yb₀)

p_α(xb₀) +(1−t)·p_α(yb₀)·p_α(yb₀) pb_α(yb₀) =

= pb_α(yb₀) [t+ (1−t)] = pb_α(yb₀).

So, pb_α(yb₀) =hu,b yb₀i_α =pb_α(bu)·pb_α(yb₀)⇒pb_α(u) = 1, which means that:b

ub∈Sc_α(0,1) ={xb∈X_α/pb_α(x) = 1}b .

Let us take the points x1 ∈ xb1, y1 ∈ yb1 and u ∈ u. One can see that:b x1 ∈ B_α^∗(0,1), y₁ ∈B_α^∗(0,1) and u∈B_α^∗(0,1). From the assumption it derives that:

p_α(x₁−u) = p_α(y₁−u) =p_α(x₁−y₁) = 0 ⇒bu=xb₁ =yb₁ ⇔yb₀ =λ_α·xb₀

⇔ pb_α(yb₀−λ_α·xb₀) = 0, where λ_α = pb_α(yb₀)

pb_α(xb₀) > 0. From Theorem 1.3, we conclude that X_α is strictly

convex.

Let us try to extend some results related to the strict convexity of a normed space. Firstly we give the following definition:

Definition 2.8 The operator T : X,{pα}_α∈A

→ X,{pα}_α∈A

belongs to the class L₀ if for every α ∈ A there exists a constant c_α such that p_α(T x)≤ c_αp_α(x), for x∈X.

Theorem 2.9 Let X,{p_α}_α∈A

be a semi–normed vector space. For every α ∈ A the proposition 1. implies the proposition 2., where 1. and 2. are the following propositions:

1. If the operator T ∈ L₀ satisfies the conditions:

p_α(I+T)≤1 and [T x, x]_α = 0,

where p_α(I+T) = sup{p_α(x+T x)/p_α(x)≤1}, then p_α(T x) = 0;

2. If z and y (p_α(y)6= 0) satisfy the conditions:

[z, y]_α = 0 and p_α(z+y) = p_α(y) then pα(z) = 0.

Proof. On the contrary, let us suppose that the points z and y are such that:

[z, y]_α = 0, p_α(z+y) = p_α(y) and p_α(z)6= 0 Let us consider the operator T :X →X, defined by:

T x= 1

(p_α(y))² ·[x, y]_α·(z+y)−x, forx∈X.

This operator is inL₀, since:

p_α(T x) ≤ 1

(pα(y))² · |[x, y]_α| ·p_α(y+z) +p_α(x) = |[x, y]_α|

p_α(y) +p_α(x)≤

≤ p_α(x)p_α(y)

p_α(y) +pα(x) = (1 + 1)pα(x) = cαpα(x), for x∈X.

We note that:

p_α(I+T) = sup{p_α(x+T x/p_α(x)≤1}=

= sup

p_α

1

(p_α(y))² ·[x, y]_α·(z+y)

/p_α(x)≤1

=

= sup

[x, y]_α

p_α(y)/p_α(x)≤1

≤sup

p_α(x)p_α(y)

p_α(y) /p_α(x)≤1

=

= sup{p_α(x)/p_α(x)≤1}= 1.

On the other hand [T y, y]_α =

[y, y]_α

p²_α(y)(y+z)−y, y

α

= [z, y]_α = 0,

while T y =z and p_α(T y) = p_α(z)6= 0. This fact contradicts the proposition (1) of the theorem, since although p_α(I +T) ≤ 1 and [T y, y]_α = 0, we have

that p_α(T y)6= 0.

Theorem 2.10 Let us suppose that the semi–normed vector space X,{p_α}_α∈A is strictly convex according to the α ∈ A. Then, if z and y are two points in X such that:

p_α(y)6= 0,[z, y]_α = 0 and p_α(z+y)≤p_α(y) then p_α(z) = 0.

Proof. Let us construct the corresponding classes of equivalence bz and by of the points z and y. Since pb_α(zb+y) =b p_α(z +y) ≤ p_α(y) = pb_α(by) and hz,b yib_α = [z, y]_α = 0, then from Theorem 1.4/(iii), we conclude that zb= b0,

thereforep_α(z) =pb_α(bz) = 0.

Theorem 2.11 Let X,{p_α}_α∈A

be a semi–normed vector space. For all α∈ A the following propositions are equivalent:

1. If y and z are two points in X that satisfying the conditions: , p_α(y)6= 0,[z, y]_α = 0 and p_α(y+z)≤p_α(y), then, p_α(z) = 0;

2. If the operator T ∈ L₀ satisfies the conditions:

p_α(I+T)≤1 and [T x, x]_α = 0 then p_α(T x) = 0.

Proof. Since p_α(I +T) = sup{p_α(x+T x)/p_α(x)≤1} ≤ 1, it derives that the following is true:

p_α(x+T x)≤p_α(I+T)·p_α(x), for x∈X

Ifx∈X and p_α(x) = 0, then equality holds. If x∈X and p_α(x)6= 0, we note that:

p_α(x+T x) p_α(x) =p_α

x p_α(x)+T

x p_α(x)

=p_α(x₁+T x₁), where x₁ = x p_α(x). Since p_α(x₁) = 1, p_α(x₁ + T x₁) ≤ p_α(I +T), which gives p_α(x+T x)

p_α(x) ≤ pα(I+T), so we conclude thatpα(x+T x)≤pα(I+T)·pα(x) forx∈X. From the condition we have thatp_α(I+T)≤1, therefore we havep_α(x+T x)≤p_α(x), forx∈X.Finally, p_α(x+T x)≤p_α(x), for x∈X and [T x, x]_α = 0, forx∈X

imply that pα(T x) = 0.

As a corollary of Theorems 2.9, 2.10 and 2.11 we conclude the following, which is a generalization of Theorem 1.4.

Theorem 2.12 Let X,{p_α}_α∈A

be a semi–normed vector space. The fo- llowing propositions are equivalent:

1. The space X,{p_α}_α∈A

is strictly convex according to the α∈ A;

2. If y and z are two points in X that satisfying the conditions : p_α(y)6= 0,[z, y]_α = 0 and p_α(y+z)≤p_α(y), then p_α(z) = 0;

3. If the operator T from L₀ , satisfies the conditions:

p_α(I +T)≤1 and [T x, x]_α = 0 then p_α(T x) = 0.

Proof. From Theorem 2.10 we have that (1)⇒(2).

From Theorem 2.3 and from the fact that the condition p_α(y+z) =p_α(y) is weaker than the following one p_α(y+z) ≤ p_α(y) we have that (2) ⇒ (1).

From Theorems 2.3 and 2.9 we have that (3)⇒(1).

From theorem 2.11 we have that (2) ⇒ (3). Finally, we have that (1) ⇔

(2)⇔(3).

Lel’s X,{p_α}_α∈A

a semi-normed vector space. We denote with:

S(0,1) = \

α∈A

S_α(0,1) and B^∗(0,1) = \

α∈A

B_α(0,1).

Let us suppose that the family of the s.p.i. products, corresponding to the family of semi—norms, is separated, i.e.:

∀x∈X,∃α∈A, such that [x, x]_α =p²_α(x)6= 0.

Theorem 2.13 The Haussdorf vector space X,{p_α}_α∈A

is strictly convex if and only if every pointx₀ ∈S(0,1) (S(0,1)6= Φ), is a extreme point for the set B^∗(0,1).

Proof. Suppose thatx₀ =tu+(1−t)v, wherex₀ ∈S(0,1) andu, v ∈B^∗(0,1).

Since X,{p_α}_α∈A

is a convex structure for all α ∈ A Theorem 2.7 implies that:

p_α(x₀−u) =p_α(x₀−v) =p_α(u−v) = 0

Since the above equalities hold for all α ∈ A than x₀ =u =v. Really, if we suppose the contrary, for instance x₀ 6= u then there exists an index α ∈ A such thatp_α(x₀−u)6= 0.

Conversely, let us consider an index α ∈ A. Since the point x₀ ∈ S(0,1) is an extreme point of B^∗(0,1), we conclude that the point x₀ ∈ S(0,1) is a extreme point according to the semi—normp_α. Using Theorem 2.7 on the set B^∗(0,1) we have that the space X,{p_α}_α∈A

is strictly convex according to the index α. Since α ∈ A is an arbitrary index, the space X,{p_α}_α∈A

is

strictly convex.

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