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doi: 10.17951/a.2022.76.1.61-71
ANNALES
UNIVERSITATIS MARIAE CURIE-SKŁODOWSKA
L U B L I N – P O L O N I A
VOL. LXXVI, NO. 1, 2022 SECTIO A 61–71
NERMIN OKIˇ
CIĆ, AMRA REKIĆ-VUKOVIĆ and VEDAD PAˇ
SIĆ
A new characterization of strict convexity
on normed linear spaces
Abstract. We consider relations between the distance of a set Aand the
distance of its translated set A+xfrom 0, for x∈A, in a normed linear space.
If the relation d(0, A +x)< d(0, A)+ kxkholds for exactly determined vectors
x∈A, where Ais a convex, closed set with positive distance from 0, which
we call (TP) property, then this property is equivalent to strict convexity of
the space. We show that in uniformly convex spaces the considered property
holds.
1. Introduction. Translation of the set as a simple transformation was
not often considered in normed spaces. The distance of the set and of its
translated set from 0 in the space show some regularities. Properties of
the considered space determine the behavior of the mentioned distances for
convex and closed sets in the given normed space. In the main part of this
paper we will show that using this property, we obtain a characterization
of strict convexity of the normed space, but also that uniform convexity
determines the relation between the distance of the set and its translated
set from 0 in the considered normed space.
The metric space in which we have a vector structure is called a linear
metric space. If the metric is obtained from the norm, we call such space
a normed linear space. In what follows we denote by Xthe normed linear
space. BX={x∈X|d(0, x)≤1}and SX={x∈X|d(0, x)=1}denote
2010 Mathematics Subject Classification. 46B20, 52A27.
Key words and phrases. Translation, uniform convexity, strict convexity.
62 N. Okiˇcić, A. Rekić-Vuković and V. Paˇsić
the unit ball and the unit sphere in the given space, where the metric is
induced by the norm of the space, d(x, y) = kx−yk. A set Ain Xis said to
be convex if, for all xand yin Aand all λ∈[0,1], the point λx + (1 −λ)y
also belongs to A. By conv Aand convAwe denote the convex hull and the
closure of the convex hull of the set Arespectively.
2. Some translation properties. The translation of the set A⊂Xby
the vector x∈Xis the set
A+x={y+x|y∈A}.
Let Abe a nonempty subset of a normed space X. For every x∈X, the
distance between the point xand the set Ais denoted by d(x, A)and is
defined by the following formula
d(x, A) = inf{d(x, y)|y∈A}.
Lemma 2.1. Let Xbe a normed linear space and A⊂Xbe a convex,
closed set with d(0, A)>0. Then, for all x∈Awe have d(0, A +x)>0.
Proof. Suppose that there exists x0∈Awith d(0, A +x0) = 0. Let (xn+
x0)n∈Nbe the minimizing sequence such that
d(0, xn+x0) = kxn+x0k → 0, n → ∞.
We conclude that xn+x0→0when n→ ∞, i.e.,
xn→ −x0, n → ∞ .
Since (xn)n∈N⊂Aand Ais closed, we see that −x0∈A. So, we have
x0,−x0∈Aand because of convexity of the set, we conclude that 0∈A.
But this contradicts the assumption that d(0, A)>0.
Lemma 2.2. Let Xbe a normed linear space and let A⊂Xbe a convex
set with d(0, A)>0. Then for an arbitrary x∈A,
d(0, A +x)> d(0, A).
Proof. Suppose that for some x0∈Awe have
d(0, A +x0)≤d(0, A).
One can choose minimizing sequences (xn)n∈N⊂Aand (yn)n∈N⊂A+x0,
that is
(1) kxnk → d(0, A), n → ∞ ,
(2) kynk → d(0, A +x0), n → ∞,
where yn=zn+x0for some zn∈A(n∈N) and such that the inequality
(3) kzn+x0k≤kxnk
A new characterization of strict convexity... 63
holds for almost all n∈N. Let ε < d(0,A)
2be arbitrary. From (1) and (2)
we get
(∃n1∈N) (∀n∈N) (n≥n1⇒ kxnk< d(0, A) + ε),
(∃n2∈N) (∀n∈N) (n≥n2⇒ kzn+x0k< d(0, A +x0) + ε).
Let n0= max{n1, n2}. Then for all n∈N,
(4) n≥n0⇒ kxnk< d(0, A) + ε∧ kzn+x0k< d(0, A +x0) + ε .
Since x0and znbelong to the set Awhich is convex, for an arbitrary
λ∈[0,1] we have λx0+ (1 −λ)zn∈A. In particular, for λ=1
2, because of
(3), the following inequality holds
1
2x0+1
2zn
≤1
2kxnk.
Using (4) and the choice of the ε, we get
1
2x0+1
2zn
<1
2(d(0, A) + ε)<3
4d(0, A).
So, x∗=1
2x0+1
2zn∈Aand
kx∗k<3
4d(0, A)< d(0, A) = inf{kxk | x∈A},
which is an obvious contradiction.
In the general case, if the distance of the set Afrom 0 is obtained by the
sequence (xn)n∈N⊂A, i.e.,
kxnk → d(0, A), n → ∞ ,
then for x0∈Athe sequence (xn+x0)n∈Ndoes not have to be minimizing
for the distance of the set A+x0from 0. Indeed, consider the space R2with
the Euclidean metric and A⊂R2given by A={(x, y)∈R2|x= 1 , y ∈
[−1,1]}. The set Ais clearly convex and closed and equality d(0, A)=1=
d((0,0),(1,0)) holds. Consider then x0= (1,1) ∈Aand translation of the
set Aby this vector, i.e.,
A+x0={(x, y)∈R2|x= 2 , y ∈[0,2]}.
Then
d(0, A +x0) = inf {d((0,0),(x, y) + (1,1)) |(x, y)∈A}
= inf{d((0,0),(2, y + 1)) |y∈[−1,1]}
= inf np4+(y+ 1)2|y∈[−1,1]o
=2=d((0,0),(2,0)) .
Since d(0, A) = d(0, x∗), where x∗= (1,0) ∈Aand since
d(0, x∗+x0) = d((0,0),(2,1)) = √5,
64 N. Okiˇcić, A. Rekić-Vuković and V. Paˇsić
we have
d(0, A +x0)=2<√5 = d(0, x∗+x0).
Lemma 2.3. In the general case of a normed linear space X, for A⊂X
and α∈Rwe have
1. d(x, αA) = |α|d(α−1x, A)for an arbitrary x∈X.
2. d(x, A) = d(x+y, A +y)for arbitrary x, y ∈X.
3. d(x, x +z) = d(y, y +z)for arbitrary x, y, z ∈X.
Lemma 2.4. Let Xbe a normed linear space and A⊂Xbe such that
d(0, A)>0. For an arbitrary x∈Ait holds
d(0, A +x)≤d(0, A) + kxk.
Proof. Let x∈Abe arbitrary and fixed. Then
A+x={a+x|a∈A}.
For an arbitrary y∈Awe have
d(0, A +x) = inf {ka+xk | a∈A}≤ky+xk ≤ kyk+kxk.
Since the left hand side of the above inequality does not depend on y, we
see that
d(0, A +x)≤inf {kyk+kxk | y∈A}= inf{kyk | y∈A}+kxk
=d(0, A) + kxk.
Lemma 2.5. Let Xbe a normed linear space and let A⊂Xbe a convex and
closed set with d(0, A)>0. If there exists x∗∈Asuch that d(0, A) = kx∗k,
then for all t∈[1,+∞)such that tx∗∈A, we have
d(0, A +tx∗) = d(0, A) + tkx∗k.
Proof. Let t∈[1,+∞)be such that tx∗∈A. Lemma 2.4 implies
d(0, A +tx∗)≤d(0, A) + ktx∗k=d(0, A) + tkx∗k.
Suppose that d(0, A +tx∗)< d(0, A) + tkx∗k. Then there exists y∈Asuch
that
ky+tx∗k< d(0, A) + tkx∗k= (1 + t)kx∗k.
We conclude that
1
1 + ty+t
1 + tx∗
<kx∗k
and since the set Ais convex, 1
1+ty+t
1+tx∗∈Aholds. Hence, the last
inequality gives a contradiction with the assumption that d(0, A) = kx∗k.
Therefore
d(0, A +tx∗) = d(0, A) + tkx∗k.
As a special case of Lemma 2.5, for t= 1, there exists x∗∈Asuch that
d(0, A) = kx∗kand the following equality holds
d(0, A +x∗)=2kx∗k= 2d(0, A) = d(0, A) + kx∗k.
A new characterization of strict convexity... 65
3. Main results. We recall some notions.
The normed space Xis called uniformly convex, for short (UC), if for
every ε,0< ε ≤2, there exists δ=δ(ε)>0such that for every x, y ∈BX,
x+y
2
>1−δ⇒ kx−yk< ε ,
or equivalently
kx−yk ≥ ε⇒
x+y
2
≤1−δ .
The normed space Xis called strictly convex, for short (SC), if its unit
sphere SXdoes not contain nontrivial segments, that is, for every x, y ∈SX,
x6=y,[x, y]6⊂ SX. This means that for all t∈(0,1), we have k(1 −t)x+
tyk<1, or equivalently, if the equality k(1 −t)x+tyk= 1 holds for some
x, y ∈SXand some t∈(0,1), then x=y. For more about uniform and
strict convexity see [1] and [3].
The following proposition contains some equivalent conditions for strict
convexity (see [2]).
Proposition 3.1. For a normed space X, the following conditions are
equivalent:
(1) Xis strictly convex;
(2) for every x, y ∈SXwith x6=y,kx+yk<2;
(3) for every x, y ∈X\{0}, the equality kx+yk=kxk+kykimplies
y=αx for some α > 0.
Evidently, every (UC) space is a (SC) space.
Lemma 2.5 holds for the spaces that have the property that for a convex
and closed set A, with d(0, A)>0, there exists x∗∈Asuch that d(0, A) =
kx∗k. For example, Hilbert and reflexive spaces have this property.
Let Xbe a normed space, ∅ 6=Z⊂Xand x∈X. By PZ(x)we denote
the set
PZ(x) = {z∈Z|d(x, Z) = kx−zk}.
In [2], it was shown that if Zis a convex and closed set and Xis a strictly
convex space, then the following holds
(∀x∈X\Z) card(PZ(x)) ≤1.
In particular, if Xis a uniformly convex space, then
(∀x∈X\Z) card(PZ(x)) = 1 .
It is clear that Lemma 2.5 holds in (UC) spaces, but it is also clear that
this assertion can not be applied to (SC) spaces because we do not know if
there exists x∗∈Zsuch that d(0, Z) = kx∗k.
Let us prove a few auxiliary claims.
66 N. Okiˇcić, A. Rekić-Vuković and V. Paˇsić
Lemma 3.2. Let Xbe a normed linear space and x, y ∈Xbe nonzero
vectors such that kx+yk=kxk+kyk. Then for vectors x0=x
kxk,y0=
y
kyk∈SXthe equality kx0+y0k= 2 holds.
Proof. Let x∈Xbe an arbitrary nonzero vector and x0=x
kxk. Consider
the set A={λx |λ∈(0,+∞)}. Suppose that A∩SX={x1, x2}. Because
x1, x2∈A, there exist λ1, λ2∈(0,+∞)such that x1=λ1xand x2=λ2x.
Because x1, x2∈SX, we have kλ1xk=kλ2xk= 1, so we conclude that
λ1=λ2, that is x1=x2. Obviously, for λ=1
kxk,x0∈Aand kx0k= 1,
that is x0∈SX. Therefore, A∩SX={x0}.
Let x, y ∈Xbe nonzero vectors such that kx+yk=kxk+kyk. Without
lost of generality let kxk>1and kyk>1. Suppose that for vectors x0=x
kxk
and y0=y
kykthe inequality kx0+y0k<2holds. Now we have
kx+yk=|k|xkx0+kyky0||
=k(x0+y0)+(kxk − 1)x0+ (kyk − 1)y0k
≤ kx0+y0k+ (kxk − 1)kx0k+ (kyk − 1)ky0k
<2 + kxk − 1 + kyk − 1 = kxk+kyk.
This contradicts the initial assumption kx+yk=kxk+kyk. As it is true
that kx0+y0k ≤ 2, we conclude that it must be kx0+y0k= 2.
Lemma 3.3. Let Xbe a normed linear space and x, y ∈SX. It holds
[x, y]⊂SXif and only if kx+yk= 2.
Proof. Let x, y ∈SXbe such that kx+yk= 2. For an arbitrary λ∈[0,1]
it holds
k(1 −λ)x+λyk ≤ 1.
Suppose that there exists λ0∈(0,1) such that k(1 −λ0)x+λ0yk<1. Then
we have
2 = kx+yk=k(1 −λ0)x+λ0y+λ0x+ (1 −λ0)yk
≤ k(1 −λ0)x+λ0yk+kλ0x+ (1 −λ0)yk
<1 + λ0+ (1 −λ0)=2,
which is a contradiction. Therefore, for all λ∈[0,1] it holds (1 −λ)x+λy ∈
SX, that is [x, y]⊂SX.
Conversely, assume that [x, y]⊂SX(x6=y). Then x+y
2∈SX, that is
kx+yk= 2.
Lemma 3.4. Let Xbe a normed linear space and x, y ∈Xbe such that the
equality kx+yk=kxk+kykholds. Then, for arbitrary α, β ∈R+it holds
kαx +βyk=αkxk+βkyk.
A new characterization of strict convexity... 67
Proof. Let x, y ∈Xbe such that
(5) kx+yk=kxk+kyk.
For arbitrary α, β ∈R+then it holds
(6) kαx +βyk ≤ αkxk+βkyk.
If the vectors xand yare collinear, that is, x=λy,λ∈R+, then we have
kαx +βyk=kαλy +βyk
=|αλ +β|kyk=αkλyk+βkyk
=αkxk+βkyk.
Now, suppose that xand yare not collinear. Based on Proposition 3.1, this
means that Xis not an (SC) space. Let α, β ∈R+be such that
(7) kαx +βyk< αkxk+βkyk.
Let λ1, λ2∈R+be such that x=λ1x0and y=λ2y0, where x0, y0∈SX.
Based on the Lemma 3.3 and Lemma 3.2, we have [x0, y0]⊂SX. The
inequality (7) becomes
kαx +βyk< αλ1kx0k+βλ2ky0k=αλ1+βλ2.
Dividing this inequality by αλ1+βλ26= 0 we get
(8)
αλ1
αλ1+βλ2
x0+βλ2
αλ1+βλ2
y0
<1.
Because the vector x=αλ1
αλ1+βλ2x0+βλ2
αλ1+βλ2y0is a convex combination of
vectors x0and y0, this means x∈SX, which contradicts (8). So, using (6),
we get
kαx +βyk=αkxk+βkyk.
Theorem 3.5. Let Xbe a uniformly convex space and let A⊂Xbe a
convex, closed set with d(0, A)>0. For x∗∈Asuch that d(0, A) = kx∗k,
we denote
Ax∗={z∈A|z=λx∗for some 1≤λ < +∞}.
Then for all x∈A\Ax∗we have
d(0, A +x)< d(0, A) + kxk.
Proof. For arbitrary x∈A, using Lemma 2.4, we have
d(0, A +x)≤d(0, A) + kxk
and if x∈Ax∗, then
d(0, A +x) = d(0, A) + kxk.
So, let x∈A\Ax∗be arbitrary and let us suppose that
d(0, A +x) = d(0, A) + kxk.
68 N. Okiˇcić, A. Rekić-Vuković and V. Paˇsić
Since A+x⊂Xis convex and closed, because of uniform convexity of the
space, there exists y∗=x0+x∈A+xsuch that
(9) ky∗k=d(0, A +x) = kx0+xk=d(0, A) + kxk=kx∗k+kxk.
Since every (UC) space is also an (SC) space, and since vectors x∗and x
are not collinear it holds
(10) kx∗+xk<kx∗k+kxk.
Using (9) and (10), we conclude that
kx∗+xk<kx0+xk=ky∗k=d(0, A +x).
Since x∗+x∈A+x, this is an obvious contradiction, so we have
d(0, A +x)< d(0, A) + kxk.
For simplicity we use the following notation. We say that a normed space
Xhas the translation property, Xis a (TP) space for short, if every convex,
closed set Awith positive distance from 0, such that there exists x∗∈A
with d(0, A) = kx∗k, satisfies the condition d(0, A +x)< d(0, A) + kxkfor
all x∈A\Ax∗. Now, Theorem 3.5 tells us that every (UC) space is a (TP)
space.
Consider the space l1with the norm
kxk=kxkl1+kxkl2=
∞
X
n=1 |xn|+ ∞
X
n=1 |xn|2!
1
2
.
It is clear that for arbitrary x∈l1we have
kxkl1≤ kxk ≤ 2kxkl1,
so these two norms are equivalent. The standard space l1is not reflexive,
therefore l1is nonreflexive after renorming with the equivalent norm k · k.
Consequently, the renormed space l1is not a (UC) space, since uniform
convexity implies reflexivity ([1]).
We will show that the renormed space l1is a (TP) space. Let A⊂
l1be a convex, closed set for which exists x∗= (ξn)n∈N∈Asuch that
d(0, A) = kx∗k>0. Suppose there exists z= (zn)n∈N∈A\Ax∗with
d(0, A +z) = d(0, A) + kzk. Then,
d(0, A +z) = d(0, A) + kzk=kx∗k+kzk≥kx∗+zk ≥ d(0, A +z).
We conclude that kx∗+zk=kx∗k+kzk. Hence,
∞
X
n=1 |ξn+zn|+ ∞
X
n=1 |ξn+zn|2!
1
2
=
∞
X
n=1 |ξn|+
∞
X
n=1 |zn|+ ∞
X
n=1 |ξn|2!
1
2
+ ∞
X
n=1 |zn|2!
1
2
.
A new characterization of strict convexity... 69
Equality in the general Minkowski inequality holds if and only if ξi·zi≥0
(i∈N), for p= 1, that is |zi|p=c|ξi|p(i∈N), for p > 1. In our case this
means that for all n∈Nwe have zn=cξnfor some c > 0, i.e., z=cx∗.
This is a contradiction with the choice of the element z∈A\Ax∗. Thus,
for an arbitrary z∈A\Ax∗,
d(0, A +z)< d(0, A) + kzk
holds and because of arbitrariness of the set A, this establishes the trans-
lation property of the renormed space l1. This proves that there exists a
(TP) space that is not a (UC) space.
Theorem 3.6. Let Xbe a normed linear space and let A⊂Xbe a convex,
closed set with d(0, A)>0such that there exists x∗∈Awith kx∗k=d(0, A).
The inequality
d(0, A +x)< d(0, A) + kxk,
holds for all x∈A\Ax∗if and only if the space Xis strictly convex.
Proof. Suppose that the space Xis not a (TP) space. This means that
there exists a closed and convex set A⊂Xfor which there exists x∗∈A
such that d(0, A) = kx∗k>0and there exists x∈A\Ax∗such that
d(0, A +x) = d(0, A) + kxk.
Therefore,
d(0, A +x) = kx∗k+kxk ≥ kx∗+xk ≥ d(0, A +x).
We conclude that
kx∗k+kxk=kx∗+xk.
Since x∗and xare not collinear, we conclude that the space Xis not strictly
convex. Using contraposition, we get that if Xis (SC), then Xis (TP).
Suppose now that space Xis not strictly convex. This means that there
exist x, y ∈SX,x6=yand [x, y]⊂SXor equivalently
(∀λ∈[0,1]) λx + (1 −λ)y∈SX.
Consider the set A= conv x, y, x+y
4. It is obvious that Ais a convex and
closed set. Suppose now that there exists a∈Awith kak<1
2. This means
that there are µ1, µ2, µ3≥0such that
a=µ1x+µ2y+µ3x∗, µ1+µ2+µ3= 1 ,
where x∗=x+y
4. Then we have
kak=
µ1x+µ2y+µ3
x+y
4
=
µ1+µ3
4x+µ2+µ3
4y
.
Based on Lemma 3.4, we now have
kak=µ1+µ2+µ3
2= 1 −µ3
2<1
2,
70 N. Okiˇcić, A. Rekić-Vuković and V. Paˇsić
which means that µ3>1. This is of course impossible due to the conditions
on the coefficients in the decomposition of the vector a. So for every a∈A
we have kak ≥ 1
2. Because x∗∈Aand
kx∗k=
1
2x+y
2
=1
2,
since x+y
2∈SX, we conclude that d(0, A) = 1
2.
Since xand x∗are not collinear, we get x∈A\Ax∗. Let us make the
translation of the set Aby the vector x. We have
x
2
=1
2=kx∗kand
x
2+x
=3
2,kx+x∗k ≤ 3
2.
From Lemma 3.4 we have
(11) kx+x∗k=
5
4x+1
4y
=5
4+1
4=3
2.
Suppose there exists a∈Asuch that ka+xk<kx∗+xk. Then a=
µ1x+µ2y+µ3x∗, where µ1+µ2+µ3= 1 and µ1, µ2, µ3≥0. Now again
based on Lemma 3.4, we have
ka+xk=
µ1+µ3
4+ 1x+µ2+µ3
4y
=µ1+µ2+µ3
2+ 1 = 2 −µ3
2.
Due to the assumption ka+xk<3
2we now have 2−µ3
2<3
2, which
is equivalent to µ3>1. This is a contradiction with the choice of the
coefficient µ3.
Therefore, for any a∈Awe have ka+xk ≥ kx∗+xk=3
2. Because
x∗∈A, we conclude that
(12) d(0, A +x) = inf
a∈Aka+xk=kx∗+xk.
Using (11) and (12), we conclude that
d(0, A +x) = d(0, A) + kxk.
Therefore, if the space is not an (SC) space, then it is possible to construct
a convex, closed set Asuch that d(0, A) = kx∗k>0for some x∗∈A, so
that there exists x∈A\Ax∗for which d(0, A +x) = d(0, A) + kxk. This
means that the space is not a (TP) space. Using contraposition, we get the
result.
Theorem 3.6 tells us that the translation property is equivalent to strict
convexity, which gives a new characterization of strict convexity.
A new characterization of strict convexity... 71
References
[1] Ayerbe Toledano, J. M., Dom´ınguez Benavides, T., López Acedo, G., Measures of
Noncompactness in Metric Fixed Point Theory, Birkh¨auser Verlag, Basel, 1997.
[2] Cobzas, S., Geometric properties of Banach spaces and the existence of nearest and
farthest points, Abstr. Appl. Anal. 2005(3) (2005), 259–285.
[3] Istr˘at¸escu, V. I., Strict Convexity and Complex Strict Convexity: Theory and Appli-
cations, Marcel Dekker, Inc., New York, 1984.
Nermin Okiˇcić Amra Rekić-Vuković
Department of Mathematics Department of Mathematics
Faculty of Natural Sciences Faculty of Natural Sciences
and Mathematics and Mathematics
University of Tuzla University of Tuzla
Univerzitetska 4, 75000 Tuzla Univerzitetska 4, 75000 Tuzla
Bosnia and Herzegovina Bosnia and Herzegovina
e-mail: nermin.okicic@untz.ba e-mail: amra.rekic@untz.ba
Vedad Paˇsić
Department of Mathematics
Faculty of Natural Sciences
and Mathematics
University of Tuzla
Univerzitetska 4, 75000 Tuzla
Bosnia and Herzegovina
e-mail: vedad.pasic@untz.ba
Received February 23, 2022
