Gen. Math. Notes, Vol. 4, No. 2, June 2011, pp.10-22
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
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: aastringa@yahoo.it
(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)
(2)
(3)
(i) [x + y, z] = [x, z] + [y, z]
(ii) [λx, y] = λ [x, y] , for x, y, z ∈ X and λ ∈ R,
[x, x] > 0, for x 6= 0,
[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] ,
(4)0 [x, y]2 ≤ [x, x] · [y, y] ,
x, y and z ∈ X,
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,
x
b, inequality 0 ≤ |p (x1 ) − p (x2 )| ≤ p (x1 − x2 ), implies p(x1 ) = p(x2 ) which
gives that the function:
pb : X1 → R+ , such that ∀b
x ∈ X1 , pb (b
x) = p (x) ,
for some x from the equivalence class x
b, is a norm on X1 . Then, by [3], there
exists a s.i.p. on X1 × X1 :
1
x ∈ X1 .
h·, ·i : X1 × X1 → R, so that pb (b
x) = hb
x, x
bi 2 , ∀b
Let us construct the function:
[·, ·] : X × X → R, such that [x, y] = hb
x, ybi , for x, y ∈ X.
The above function is a s.p.i.p. function. So,
(i) [x, x] = hb
x, x
bi ≥ 0, for x ∈ X.
D
E
c
(ii) [λx, y] = λx, yb = hλb
x, ybi = λ hb
x, ybi = λ [x, y], for λ ∈ R and x, y ∈ X,
D
E
(iii) [x1 + x2 , y] = x\
b = hxb1 + xb2 , ybi = hxb1 , ybi + hxb1 , ybi = [x1 , y] +
1 + x2 , y
+ [x2 , y], for x1 , x2 and y ∈ X
(iv) [x, y]2 = hb
x, ybi2 ≤ hb
x, x
bi · hb
y , ybi = [x, x] · [y, y], for x, y ∈ X.
We conclude by noting that:
[x, x] = hb
x, x
bi = [b
p (b
x)]2 = p2 (x), ∀x ∈ X.
13
On Strictly Convex and Strictly Convex...
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) ∈ X 2 ,
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 ≤
1
≤ [x, x]1 · [y, y]1 + 2 ([x, x]1 · [y, y]1 · [x, x]2 · [y, y]2 ) 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 considered 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
pA (x) = p2α1 (x) + p2α2 (x) + · · · + p2αs (x) 2 ,
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
2 uX
2
2
pA (x + y) = pα1 (x + y) + · · · + pαs (x + y) = t
p2αi (x + y) ≤
i=1
v
v
v
u s
u s
u s
X
uX
u
uX
2
≤ t
(pαi (x) + pαi (y)) ≤ t
p2αi (x) + t
p2αi (y) 6
i=1
≤ pA (x) + pA (y) ,
i=1
i=1
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]A = [x, y]α1 + [x, y]α2 + · · · + [x, y]αs ,
for x, y ∈ X.
14
Artur Stringa
If we denote by BA∗ (0, ε) =
s
T
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 − λα x) = 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.
Let α ∈ A and denote by Xα the X/p−1
α (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 x
b and yb be two elements from
Xα such that:
pbα (b
x + yb) = pbα (b
x) + pbα (b
y)
By the structure of pbα it derives that if x ∈ x
b and y ∈ yb then:
pα (x + y) = pα (x) + pα (y).
Thus, there exists a λα > 0, such that:
pα (y − λα x) = 0 ⇔ pbα (y \
− λα x) = 0 ⇔ y \
− λα x = b
0 ⇔ yb − λα x
b=b
0⇔
yb = λα x
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 x
b and yb be
the equivalence classes of the points x and y. The following equality holds:
pbα (b
x + yb) = pbα (b
x) + pbα (b
y)
15
On Strictly Convex and Strictly Convex...
By the condition, ∃λα > 0, such that yb = λα x
b. We have:
pα (y − λα x) = pbα (y \
− λα x) = pbα (b
y − λd
y − λα x
b) = pbα (b
0) = 0. α x) = pbα (b
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 − λα x) = 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 x
b and yb,
hb
x, ybiα = pbα (b
x) · pbα (b
y ).
Theorem 1.3 implies that there exists a λα > 0, such that
yb = λα x
b ⇔ pα (y − λα x) = 0.
Conversely, suppose that x and y are two points in X, such that:
pα (x + y) = pα (x) + pα (y).
Since pbα (b
x + yb) = pbα (b
x) + pbα (b
y ) we conclude that:
hb
x, x
b + ybiα + hb
y, x
b + ybiα = hb
x + yb, x
b + ybiα = [pbα (b
x + yb)]2 =
= pbα (b
x + yb) · pbα (b
x + yb) =
= pbα (b
x + yb) (pbα (b
x) + pbα (b
y ))
therefore,
(pbα (b
x) · pbα (b
x + yb) − hb
x, x
b + ybiα ) + (pbα (b
y ) · pbα (b
x + yb) − hb
y, x
b + 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 (b
x) · pbα (b
x + yb) = hb
x, x
b + ybiα
and pbα (b
y ) · pbα (b
x + yb) = hb
y, x
b + ybiα
(1)
16
Artur Stringa
Since pbα (b
x) · pbα (b
x + yb) = hb
x, x
b + ybiα , it derives that there exists some ηα > 0,
such that x
b + yb = ηα x
b ⇔ yb = (ηα − 1)b
x. Denoting by λα = ηα − 1 we have
that:
pbα (b
x + yb) = pbα (b
x) + pbα (b
y ) ⇔ pbα (ηα x
b) = pbα (b
x) + pbα (b
y) ⇔
⇔ ηα · pbα (b
x) = pbα (b
x) + pbα (b
y ) ⇔ pbα (b
y ) = (ηα − 1)pbα (b
x)
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 = λα x
b ⇔ pα (y − λα x) = 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 x
b and yb we have the followings:
pbα (b
x) 6= 0, hb
x, ybiα = 0 and pbα (b
x + yb) = pbα (b
x).
Theorem 2.3 implies that the space Xα is strictly convex. From Theorem
1.4/(iii) we conclude that:
pbα (b
y ) = 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 (b
x + yb) = pbα (b
x) + pbα (b
y ). Let’s
λα =
pbα (b
y)
.
pbα (b
x)
Therefore, λα > 0. Considering the points u
b = λα x
b − yb and vb = x
b + yb, we have
that:
pbα (b
y)
x
b · (pbα (b
x) + pbα (b
y ))
pbα (b
u + vb) = pbα (b
x + λα x
b) = pbα x
b+
x
b = pbα
=
pbα (b
x)
pbα (b
x)
=
pbα (b
x) + pbα (b
y)
pbα (b
x) = pbα (b
x) + pbα (b
y ) = pbα (b
x + yb) = pbα (b
v ),
pbα (b
x)
17
On Strictly Convex and Strictly Convex...
furthermore, pbα (b
v ) 6= 0. Due to the equalities (1) in Theorem 2.4, we can
conclude that:
hb
u, vbiα = hλα x
b − yb, x
b + ybiα = λα hb
x, x
b + ybiα − hb
y, x
b + ybiα =
pbα (b
y)
=
· pbα (b
x) · pbα (b
x + yb) − pbα (b
y ) · pbα (b
x + yb) = 0.
pbα (b
x)
Finally, for the points u
b and vb the following equalities hold:
pbα (b
u + vb) = pbα (b
v ) and
hb
u, vbiα = 0.
Let u1 ∈ u
b and v1 ∈ vb. The following equalities are true:
pα (u1 + v1 ) = pα (v1 ), [u1 , v1 ]α = 0 and pα (v1 ) = pα (b
v ) 6= 0.
From the assumption, it derives that:
pα (u1 ) = 0 ⇒ pbα (b
u) = 0 ⇒ pα (y−λα x) = 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
to the semi-norm p, if x0 = tu + (1 − t)v with 0 < t < 1 and (u, v) ∈ C 2 ,
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 seminorm 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 = tb
u + (1 − t)b
v . Since Xα is strictly convex,
from [2] we have that xb0 = u
b = vb. Therefore, xb0 − u
b = xb0 − vb = u
b − vb = b
0⇒
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 , yb0 iα = pbα (xb0 )·
xb0
yb0
pbα (yb0 ). The points xb1 =
and yb1 =
are in the the closed ball:
pbα (xb0 )
pbα (yb0 )
b ∗ (0, 1) = {b
B
x ∈ Xα /pbα (b
x) ≤ 1} .
α
Let’s u
b = txb1 +(1−t)yb1 , 0 < t < 1. We have pbα (b
u) ≤ tpbα (xb1 )+(1−t)pbα (yb1 ) ≤ 1.
Furthermore,
hb
u, yb0 iα = t · hxb1 , yb0 iα + (1 − t) · hyb1 , yb0 iα =
t · pα (xb0 ) · pα (yb0 ) (1 − t) · pα (yb0 ) · pα (yb0 )
+
=
=
pα (xb0 )
pbα (yb0 )
= pbα (yb0 ) [t + (1 − t)] = pbα (yb0 ).
So, pbα (yb0 ) = hb
u, yb0 iα = pbα (b
u) · pbα (yb0 ) ⇒ pbα (b
u) = 1, which means that:
cα (0, 1) = {b
u
b∈S
x ∈ Xα /pbα (b
x) = 1} .
Let us take the points x1 ∈ xb1 , y1 ∈ yb1 and u ∈ u
b. 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 ⇒ u
b = xb1 = yb1 ⇔ yb0 = λα · xb0
⇔ pbα (yb0 − λα · xb0 ) = 0,
where λα =
convex.
pbα (yb0 )
> 0. From Theorem 1.3, we conclude that Xα is strictly
pbα (xb0 )
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) ≤
cα pα (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;
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On Strictly Convex and Strictly Convex...
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:
Tx =
1
· [x, y]α · (z + y) − x,
(pα (y))2
for x ∈ X.
This operator is in L0 , since:
|[x, y]α |
1
+ pα (x) ≤
2 · |[x, y]α | · pα (y + z) + pα (x) =
pα (y)
(pα (y))
pα (x)pα (y)
≤
+ pα (x) = (1 + 1)pα (x) = cα pα (x), for x ∈ X.
pα (y)
pα (T x) ≤
We note that:
pα (I + T ) = sup {pα (x + T x/pα (x) ≤ 1} =
1
= sup pα
· [x, y]α · (z + y) /pα (x) ≤ 1 =
(pα (y))2
[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
(y + z) − y, y
[T y, y]α =
p2α (y)
= [z, y]α = 0,
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 (b
z + yb) = pα (z + y) ≤ pα (y) = pbα (b
y ) and
hb
z , ybiα = [z, y]α = 0, then from Theorem 1.4/(iii), we conclude that zb = b
0,
therefore pα (z) = pbα (b
z ) = 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)
≤
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.
Since pα (x1 ) = 1, pα (x1 + T x1 ) ≤ pα (I + T ), which gives
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;
21
On Strictly Convex and Strictly Convex...
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:
\
\
Bα (0, 1).
Sα (0, 1) and B ∗ (0, 1) =
S(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.:
∀x ∈ X, ∃α ∈ A, such that [x, x]α = p2α (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 }α∈A is strictly convex according to
the index α. Since α ∈ A is an arbitrary index, the space X, {pα }α∈A is
strictly convex.
22
Artur Stringa
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