On Strictly Convex and Strictly Convex According to an Index Semi-Normed Vector Spaces

Home , Convex set

Gen. Math. Notes, Vol. 4, No. 2, June 2011, pp.10-22 ISSN 2219-7184; Copyright c ICSRS Publication, 2011 www.i-csrs.org Available free online at http://www.geman.in

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 space X. 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 convexity, 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 on X, if to any x, 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 x 6= 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 with kxk = On Strictly Convex and Strictly Convex... 11

[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 point u of a convex set C in a vector space is called an extreme point of C if u = t · v + (1 − t) · w with 0 < t < 1 and v, w ∈ C, implies that u = v = w. A normed linear space X is 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 space X is strictly convex if and only if:

kx + yk = kxk + kyk , where x 6= 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 x 6= 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, 12 Artur Stringa

(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 on X (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}. Since p is a semi–norm on X, we have that M is a closed subspace of the vector space X. We note that the relation x ∼ y ⇔ (x − y) ∈ M, is a equivalent relation on X. Let us denote by X1 the X/M . For any two points x1 and x2 from the class of equivalence, xb, 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(xb) = p (x) , for some x from the equivalence class xb, is a norm on X1. Then, by [3], there exists a s.i.p. on X1 × X1:

1 h·, ·i : X1 × X1 → R, so that pb(xb) = hx,b xbi 2 , ∀xb ∈ X1. Let us construct the function:

[·, ·]: X × X → R, such that [x, y] = hx,b ybi , for x, y ∈ X. The above function is a s.p.i.p. function. So,

(i)[ x, x] = hx,b xbi ≥ 0, for x ∈ X. D E (ii)[ λx, y] = λx,c yb = hλx,b ybi = λ hx,b ybi = λ [x, y], for λ ∈ R and x, y ∈ X, D E (iii)[ x1 + x2, y] = x\1 + x2, yb = hxb1 + xb2, ybi = hxb1, ybi + hxb1, ybi = [x1, y] + + [x2, y], for x1, x2 and y ∈ X

2 2 (iv)[ x, y] = hx,b ybi ≤ hx,b xbi · hy,b ybi = [x, x] · [y, y], for x, y ∈ X. We conclude by noting that:

2 2 [x, x] = hx,b xbi = [pb(xb)] = p (x), ∀x ∈ X. On Strictly Convex and Strictly Convex... 13

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:

2 [x, y] = [x, y]1 + [x, y]2 , ∀ (x, y) ∈ X , 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.

2 2 2 2 [x, y] = ([x, y]1 + [x, y]2) = ([x, y]1) + 2 [x, y]1 · [x, y]2 + ([x, y]2) ≤ 1 2 ≤ [x, x]1 · [y, y]1 + 2 ([x, x]1 · [y, y]1 · [x, x]2 · [y, y]2) + [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, {pα}α∈A), where {pα}α∈A is a family of semi-norms and A 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-norms pα, α ∈ A. Actually the family of semi-norms {pα}α∈A can be replaced with the family of semi-norms {pA}A⊂A, A = {α1, α2, . . . , αs}, s ∈ N, where:

1 2 2 2 2 pA(x) = pα1 (x) + pα2 (x) + ··· + pαs (x) , for x ∈ X. It is clear that this function satisﬁes the ﬁrst two conditions of a semi-norm, and for the third condition we have that: v u s 1 uX p (x + y) = p2 (x + y) + ··· + p2 (x + y) 2 = p2 (x + y) ≤ A α1 αs t αi i=1 v v v u s u s u s X 2 X X ≤ u (p (x) + p (y)) ≤ u p2 (x) + u p2 (y) t αi αi t αi t αi 6 i=1 i=1 i=1

≤ pA (x) + pA (y) , for x, y ∈ X.

The family {pA}A⊂A is ﬁltered since for A1 ⊂ A2, we have pA1 (x) ≤ pA2 (x), for all x ∈ X and furthermore this family is related to the s.p.i.–products family

{[x, y]A}A⊂A by the equation [x, y] = [x, y] + [x, y] + ··· + [x, y] , for x, y ∈ X. A α1 α2 αs 14 Artur Stringa

s ∗ T If we denote by BA (0, ε) = Bαi (0, ε) the neighborhoods of the origin of i=1 the X space for the topology of the semi-normed family {pα}α∈A and with BA(0, ε) = {x ∈ X/pA(x) < ε} the neighborhoods of the origin of the X space for the topology of the semi-normed family {pA}A⊂A the inclusions BA(0, ε) ⊂ ∗ √ BA(0, ε) ⊂ sBA(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, {pα}α∈A) is called strictly convex according to the index α ∈ A,if for every two points x and y in X, such that:

pα(x) 6= 0, pα(y) 6= 0 and pα(x + y) = pα(x) + pα(y), there exists a λα > 0, such that pα(y − λαx) = 0.

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

−1 Let α ∈ A and denote by Xα the X/pα (0).

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

Proof. Suppose that the semi-normed vector space (X, {pα}α∈A) is strictly convex according to the index α ∈ A and let xb and yb be two elements from Xα such that: pbα (xb + yb) = pbα(xb) + pbα(yb) By the structure of pbα it derives that if x ∈ xb and y ∈ yb then:

pα(x + y) = pα(x) + pα(y).

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

pα(y − λαx) = 0 ⇔ pbα(y\− λαx) = 0 ⇔ y\− λαx = b0 ⇔ yb − λαxb = b0 ⇔ yb = λαx.b

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

pbα (xb + yb) = pbα(xb) + pbα(yb) On Strictly Convex and Strictly Convex... 15

By the condition, ∃λα > 0, such that yb = λαxb. We have:

pα(y − λαx) = pbα(y\− λαx) = pbα(yb − λdαx) = pbα(yb − λαxb) = pbα(b0) = 0.

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

Theorem 2.4 The semi-normed vector space (X, {pα}α∈A) is strictly convex according to the index α ∈ A if and only if for every two points x and y in

X, such that pα(x) 6= 0, pα(y) 6= 0 and [x, y]α = pα(x) · pα(y), there is a λα > 0 such that pα(y − λαx) = 0.

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

pα(x) 6= 0, pα(y) 6= 0 and [x, y]α = pα(x) · pα(y).

We note that for the equivalence classes xb and yb,

hx,b ybiα = pbα(xb) · pbα(yb).

Theorem 1.3 implies that there exists a λα > 0, such that

yb = λαxb ⇔ pα(y − λαx) = 0. Conversely, suppose that x and y are two points in X, such that:

pα(x + y) = pα(x) + pα(y).

Since pbα (xb + yb) = pbα(xb) + pbα(yb) we conclude that: 2 hx,b xb + ybiα + hy,b xb + ybiα = hxb + y,b xb + ybiα = [pbα(xb + yb)] = = pbα(xb + yb) · pbα(xb + yb) = = pbα(xb + yb)(pbα(xb) + pbα(yb)) therefore,

(pbα(xb) · pbα(xb + yb) − hx,b xb + ybiα) + (pbα(yb) · pbα(xb + yb) − hy,b xb + ybiα) = 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:

pbα(xb) · pbα(xb + yb) = hx,b xb + ybiα and pbα(yb) · pbα(xb + yb) = hy,b xb + ybiα (1) 16 Artur Stringa

Since pbα(xb) · pbα(xb + yb) = hx,b xb + ybiα, it derives that there exists some ηα > 0, such that xb + yb = ηαxb ⇔ yb = (ηα − 1)xb. Denoting by λα = ηα − 1 we have that:

pbα(xb + yb) = pbα(xb) + pbα(yb) ⇔ pbα(ηαxb) = pbα(xb) + pbα(yb) ⇔ ⇔ ηα · pbα(xb) = pbα(xb) + pbα(yb) ⇔ pbα(yb) = (ηα − 1)pbα(xb) that imply ηα − 1 > 0 (we used the fact that pα(x) 6= 0 and pα(y) 6= 0). Therefore, there exists a λα > 0, such that yb = λαxb ⇔ pα(y − λαx) = 0. Theorem 2.5 The semi—normed vector space X, {pα}α∈A is strictly convex according to the index α ∈ A if and only if for every two points x and y in X, such that pα(x) 6= 0, [x, y]α = 0 and pα(x + y) = pα(x), we have that pα(y) = 0.

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

pα(x) 6= 0, [x, y]α = 0 and pα(x + y) = pα(x)

For the equivalence classes of xb and yb we have the followings:

pbα(xb) 6= 0, hx,b ybiα = 0 and pbα(xb + yb) = pbα(xb).

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

pbα(yb) = 0 ⇔ pα(y) = 0. Conversely, suppose that x 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 + yb) = pbα(xb) + pbα(yb). Let’s

pbα(yb) λα = . pbα(xb)

Therefore, λα > 0. Considering the points ub = λαxb− yb and vb = xb+ yb, we have that: pbα(yb) xb · (pbα(xb) + pbα(yb)) pbα(ub + vb) = pbα(xb + λαxb) = pbα xb + xb = pbα = pbα(xb) pbα(xb)

pbα(xb) + pbα(yb) = pbα(xb) = pbα(xb) + pbα(yb) = pbα(xb + yb) = pbα(vb), pbα(xb) On Strictly Convex and Strictly Convex... 17 furthermore, pbα(vb) 6= 0. Due to the equalities (1) in Theorem 2.4, we can conclude that:

hu,b vbiα = hλαxb − y,b xb + ybiα = λα hx,b xb + ybiα − hy,b xb + ybiα = pbα(yb) = · pbα(xb) · pbα(xb + yb) − pbα(yb) · pbα(xb + yb) = 0. pbα(xb)

Finally, for the points ub and vb the following equalities hold:

pbα(ub + vb) = pbα(vb) and hu,b vbiα = 0.

Let u1 ∈ ub and v1 ∈ vb. The following equalities are true:

pα(u1 + v1) = pα(v1), [u1, v1]α = 0 and pα(v1) = pα(vb) 6= 0. From the assumption, it derives that:

pα(u1) = 0 ⇒ pbα(ub) = 0 ⇒ pα(y−λαx) = 0.

Let p 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 2 to the semi-norm p, if x0 = tu + (1 − t)v with 0 < t < 1 and (u, v) ∈ C , implies that p(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 p is extreme point for C, since if p(x0 − u) = p(x0 − v) = p(u − v) = 0, we have that x0 = u = v. Theorem 2.7 The semi-normed vector space X, {pα}α∈A is strictly convex according to the index α ∈ A if and only if every point x0 ∈ Sα(0, 1) = ∗ {x ∈ X/pα(x) = 1} is an extreme point of the set Bα(0, 1) = {x ∈ X/pα(x) ≤ 1} according to the semi-norm pα. Proof. Let’s denote by x0 = tu + (1 − t)v with 0 < t < 1, pα(u) ≤ 1, pα(v) ≤ 1 and pα(x0) = 1

From the deﬁnition of the sum and multiplication by scalars of the equivalence classes in Xα we have that xb0 = tub + (1 − t)vb. Since Xα is strictly convex, from [2] we have that xb0 = ub = vb. Therefore, xb0 − ub = xb0 − vb = ub − vb = b0 ⇒ pα (x0 − u) = pα(x0 − v) = pα(u − v). Conversely, to prove that the semi--normed vector space X, {pα}α∈A is strictly convex according to the index α ∈ A its suﬃcient to prove that Xα is strictly convex. 18 Artur Stringa

Let xb0 and yb0 be two distinct points from zero, such that hxb0, yb0iα = pbα(xb0)· xb0 yb0 pbα(yb0). The points xb1 = and yb1 = are in the the closed ball: pbα(xb0) pbα(yb0)

∗ Bbα(0, 1) = {xb ∈ Xα/pbα(xb) ≤ 1} .

Let’s ub = txb1+(1−t)yb1, 0 < t < 1. We have pbα(ub) ≤ tpbα(xb1)+(1−t)pbα(yb1) ≤ 1. Furthermore,

hu,b yb0iα = t · hxb1, yb0iα + (1 − t) · hyb1, yb0iα = t · p (x ) · p (y ) (1 − t) · p (y ) · p (y ) = α b0 α b0 + α b0 α b0 = pα(xb0) pbα(yb0) = pbα(yb0)[t + (1 − t)] = pbα(yb0).

So, pbα(yb0) = hu,b yb0iα = pbα(ub) · pbα(yb0) ⇒ pbα(ub) = 1, which means that:

ub ∈ Scα(0, 1) = {xb ∈ Xα/pbα(xb) = 1} .

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

pα(x1 − u) = pα(y1 − u) = pα(x1 − y1) = 0 ⇒ ub = xb1 = yb1 ⇔ yb0 = λα · xb0 ⇔ pbα(yb0 − λα · xb0) = 0,

pbα(yb0) where λα = > 0. From Theorem 1.3, we conclude that Xα is strictly pbα(xb0) 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 cα such that pα(T x) ≤ cαpα(x), for x ∈ X. Theorem 2.9 Let X, {pα}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:

pα(I + T ) ≤ 1 and [T x, x]α = 0,

where pα(I + T ) = sup {pα(x + T x)/pα(x) ≤ 1}, then pα(T x) = 0; On Strictly Convex and Strictly Convex... 19

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

[z, y]α = 0 and pα(z + y) = pα(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, pα(z + y) = pα(y) and pα(z) 6= 0 Let us consider the operator T : X → X, deﬁned by: 1 T x = 2 · [x, y]α · (z + y) − x, for x ∈ X. (pα(y))

This operator is in L0, since:

1 |[x, y]α| pα(T x) ≤ 2 · |[x, y]α| · pα(y + z) + pα(x) = + pα(x) ≤ (pα(y)) pα(y)

pα(x)pα(y) ≤ + pα(x) = (1 + 1)pα(x) = cαpα(x), for x ∈ X. pα(y) We note that:

pα(I + T ) = sup {pα(x + T x/pα(x) ≤ 1} = 1 = sup pα 2 · [x, y]α · (z + y) /pα(x) ≤ 1 = (pα(y)) [x, y]α pα(x)pα(y) = sup /pα(x) ≤ 1 ≤ sup /pα(x) ≤ 1 = pα(y) pα(y)

= sup {pα(x)/pα(x) ≤ 1} = 1. On the other hand [y, y]α [T y, y]α = 2 (y + z) − y, y = [z, y]α = 0, pα(y) α while T y = z and pα(T y) = pα(z) 6= 0. This fact contradicts the proposition

(1) of the theorem, since although pα(I + T ) ≤ 1 and [T y, y]α = 0, we have that pα(T y) 6= 0. Theorem 2.10 Let us suppose that the semi–normed vector space X, {pα}α∈A is strictly convex according to the α ∈ A. Then, if z and y are two points in X such that:

pα(y) 6= 0, [z, y]α = 0 and pα(z + y) ≤ pα(y) then pα(z) = 0. 20 Artur Stringa

Proof. Let us construct the corresponding classes of equivalence zb and yb of the points z and y. Since pbα(zb + yb) = pα(z + y) ≤ pα(y) = pbα(yb) and hz,b ybiα = [z, y]α = 0, then from Theorem 1.4/(iii), we conclude that zb = b0, therefore pα(z) = pbα(zb) = 0. Theorem 2.11 Let X, {pα}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: ,

pα(y) 6= 0, [z, y]α = 0 and pα(y + z) ≤ pα(y),

then, pα(z) = 0;

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

pα(I + T ) ≤ 1 and [T x, x]α = 0

then pα(T x) = 0.

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

pα(x + T x) ≤ pα(I + T ) · pα(x), for x ∈ X

If x ∈ X and pα(x) = 0, then equality holds. If x ∈ X and pα(x) 6= 0, we note that: pα(x + T x) x x x = pα + T = pα(x1 + T x1), where x1 = . pα(x) pα(x) pα(x) pα(x)

pα(x + T x) Since pα(x1) = 1, pα(x1 + T x1) ≤ pα(I + T ), which gives ≤ pα(x) pα(I +T ), so we conclude that pα(x+T x) ≤ pα(I +T )·pα(x) for x ∈ X. From the condition we have that pα(I+T ) ≤ 1, therefore we have pα(x+T x) ≤ pα(x), for x ∈ X. Finally, pα(x + T x) ≤ pα(x), for x ∈ X and [T x, x]α = 0, for x ∈ 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, {pα}α∈A be a semi–normed vector space. The following propositions are equivalent: 1. The space X, {pα}α∈A is strictly convex according to the α ∈ A; On Strictly Convex and Strictly Convex... 21

2. If y and z are two points in X that satisfying the conditions :

pα(y) 6= 0, [z, y]α = 0 and pα(y + z) ≤ pα(y),

then pα(z) = 0;

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

pα(I + T ) ≤ 1 and [T x, x]α = 0

then pα(T x) = 0. Proof. From Theorem 2.10 we have that (1) ⇒ (2). From Theorem 2.3 and from the fact that the condition pα(y + z) = pα(y) is weaker than the following one pα(y + z) ≤ pα(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, {pα}α∈A a semi-normed vector space. We denote with:

\ ∗ \ S(0, 1) = Sα(0, 1) and B (0, 1) = Bα(0, 1). α∈A α∈A Let us suppose that the family of the s.p.i. products, corresponding to the family of semi—norms, is separated, i.e.:

2 ∀x ∈ X, ∃α ∈ A, such that [x, x]α = pα(x) 6= 0. Theorem 2.13 The Haussdorf vector space X, {pα}α∈A is strictly convex if and only if every point x0 ∈ S(0, 1) (S(0, 1) 6= Φ) , is a extreme point for the set B∗(0, 1).

∗ Proof. Suppose that x0 = tu+(1−t)v, where x0 ∈ S(0, 1) and u, v ∈ B (0, 1). Since X, {pα}α∈A is a convex structure for all α ∈ A Theorem 2.7 implies that: pα(x0 − u) = pα(x0 − v) = pα(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 that pα(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—norm pα. Using Theorem 2.7 on the set ∗ B (0, 1) we have that the space X, {pα} is strictly convex according to α∈A the index α. Since α ∈ A is an arbitrary index, the space X, {pα}α∈A is strictly convex. 22 Artur Stringa

References

[1] E. Berkson, Some types of Banach algebras, Hermitian and Bade func- tionals, Trans. Amer. Math. Soc., 116(1965), 376.

[2] E. Torrance, Stricly convex spaces via semi–inner–product spaces ortogo- nality, Proc. Amer. Math. Soc., 26(1970), 108-110.

[3] G. Lumer, Semi–inner–product spaces, Trans. Amer. Math. Soc., 100(1) (1961), 29–43.

[4] J.R. Gilles, Classes of semi–inner–product spaces, Trans. Amer. Math. Soc., 129(3) (1967), 436–446.

[5] K.R. Unni and C. Puttamadaiah, Some remarks on strictly convex semi– inner–product spaces, Bull. Calc. Math. Soc. India, 77(1980), 261-265.

[6] M.M. Pavle, Sur la transervalit dans des espaces norm´e, Mathematica Balcanica, 1(1971), 171–176.

[7] X.H. Teliti and A. Stringa, The semi–pre–inner–product functions, 3–rd International Conference On Algebra and Functional Analysis, Elbasan, Albania, 15–16 May, (2010).