# Strongh-Convexity and Separation Theorems

Authors: Teodoro Lara; Nelson Merentes; Kazimierz Nikodem
Journal: International Journal of Analysis
 (2016)

---

## Abstract

Jensen inequality for stronglyh-convex functions and a characterization of pairs of functions that can be separated by a strongly h-convex function are presented. As a consequence, a stability result of the Hyers-Ulam type is obtained.


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## Body

## 1. Introduction

LetX be a normed space, let D be a convex subset of X, and let c
>
0
. A function f
:
D
→
R is calledstrongly convex with modulus
c (see, e.g., [1, 2]) if (1)
f
t
x
+
1
-
t
y
≤
t
f
x
+
1
-
t
f
y
-
c
t
1
-
t
x
-
y
2for all x
,
y
∈
D and t
∈
[
0,1
]
. Recall also that the usual notion of convex functions corresponds to the case c
=
0
. Strongly convex functions, introduced by Polyak [3], play an important role in optimization theory and mathematical economics. Many properties and applications of them can be found in the literature (see, e.g., [2, 4–7] and the references therein).In [8] Varošanec introduced the notion of h-convexity. Let h
:
[
0,1
]
→
R
+ be a given function. A function f
:
D
→
R is said to be h
-convex if(2)
f
t
x
+
1
-
t
y
≤
h
t
f
x
+
h
1
-
t
f
yfor all x
,
y
∈
D and t
∈
[
0,1
]. This notion unifies and generalizes the known classes of convex functions, s-convex functions, Godunova-Levin functions, and P-functions, which are obtained by putting in (2) h
(
t
)
=
t, h
(
t
)
=
t
s (where s
∈
(
0,1
)), h
(
t
)
=
1
/
t (with h
(
0
)
=
0), and h
(
t
)
=
1, respectively. Some properties of them can be found, for example, in [8–13].Combining the above two ideas we say that a functionf
:
D
→
R isstrongly
h
-convex with modulus
c (cf. [14]) if(3)
f
t
x
+
1
-
t
y
≤
h
t
f
x
+
h
1
-
t
f
y
-
c
t
1
-
t
x
-
y
2for all x
,
y
∈
D and t
∈
[
0,1
]
.In this note we present a Jensen-type inequality for such functions and give a characterization of pairs of functions that can be separated by a stronglyh-convex one. Separation (or sandwich) theorems, that is, theorems providing conditions under which two given functions can be separated by a function from some special class, play an important role in many fields of mathematics and have various applications. In the literature one can find numerous results of this type (see, e.g., [12, 15–24]).


## 2. Jensen-Type Inequality

In the whole paper we assume thatX is a real inner product space (i.e., the norm · in X is induced by an inner product: x
=
〈
x
∣
x
〉). D is a convex nonempty subset of X and c is a positive constant.A functionh
:
[
0,1
]
→
R is said to be multiplicative if (4)
h
s
t
=
h
s
h
t
,
s
,
t
∈
0,1
.Note that if h is multiplicative, then it is nonnegative and either h
=
0 or h
(
1
)
=
1
. In what follows we assume that h
≠
0.The following result is a counterpart of the classical Jensen inequality for stronglyh-convex functions. It generalizes the Jensen-type inequality for strongly convex functions obtained in [4]. Similar results for h-convex functions are proved in [8, 12].Theorem 1.
Leth
:
[
0,1
]
→
R be a multiplicative function such that h
(
t
)
≥
t for all t
∈
[
0,1
]
. If a function f
:
D
→
R is strongly h-convex with modulus c, then(5)
f
∑
i
=
1
n
t
i
x
i
≤
∑
i
=
1
n
h
t
i
f
x
i
-
c
∑
i
=
1
n
t
i
x
i
-
x
¯
2
,for all n
∈
N, x
1
,
…
,
x
n
∈
D, and t
1
,
…
,
t
n
>
0 with t
1
+
⋯
+
t
n
=
1 and x
¯
=
t
1
x
1
+
⋯
+
t
n
x
n
.Proof.
Forn
=
1 inequality (5) is trivial and for n
=
2 it follows from the definition of strong h-convexity (note that t
1
x
1
-
x
¯
2
+
t
2
x
2
-
x
¯
2
=
t
1
t
2
x
1
-
x
2
2). Now, assuming (2) holds for some n, we will prove it for n
+
1. By the definition of strong h-convexity we get(6)
f
∑
i
=
1
n
+
1
t
i
x
i
=
f
1
-
t
n
+
1
∑
i
=
1
n
t
i
1
-
t
n
+
1
x
i
+
t
n
+
1
x
n
+
1
≤
h
1
-
t
n
+
1
f
∑
i
=
1
n
t
i
1
-
t
n
+
1
x
i
+
h
t
n
+
1
f
x
n
+
1
-
c
1
-
t
n
+
1
s
-
x
¯
2
+
t
n
+
1
x
n
+
1
-
x
¯
2
,where(7)
s
=
∑
i
=
1
n
t
i
1
-
t
n
+
1
x
i
,
x
¯
=
1
-
t
n
+
1
s
+
t
n
+
1
x
n
+
1
=
∑
i
=
1
n
+
1
t
i
x
i
.By the inductive assumption we have (8)
f
∑
i
=
1
n
t
i
1
-
t
n
+
1
x
i
≤
∑
i
=
1
n
h
t
i
1
-
t
n
+
1
f
x
i
-
c
∑
i
=
1
n
t
i
1
-
t
n
+
1
x
i
-
s
2
.Now, using the above inequalities, the multiplicativity of h, and the assumption h
(
t
)
≥
t, we obtain (9)
f
∑
i
=
1
n
+
1
t
i
x
i
≤
∑
i
=
1
n
+
1
h
t
i
f
x
i
-
c
h
1
-
t
n
+
1
∑
i
=
1
n
t
i
1
-
t
n
+
1
x
i
-
s
2
+
1
-
t
n
+
1
s
-
x
¯
2
+
t
n
+
1
x
n
+
1
-
x
¯
2
≤
∑
i
=
1
n
+
1
h
t
i
f
x
i
-
c
∑
i
=
1
n
t
i
x
i
-
s
2
+
1
-
t
n
+
1
s
-
x
¯
2
+
t
n
+
1
x
n
+
1
-
x
¯
2
.To finish the proof it is enough to show that (10)
∑
i
=
1
n
t
i
x
i
-
s
2
+
1
-
t
n
+
1
s
-
x
¯
2
+
t
n
+
1
x
n
+
1
-
x
¯
2
=
∑
i
=
1
n
+
1
t
i
x
i
-
x
¯
2
,or, equivalently,(11)
L
≔
∑
i
=
1
n
t
i
x
i
-
s
2
-
x
i
-
x
¯
2
=
-
1
-
t
n
+
1
s
-
x
¯
2
.Since(12)
x
i
-
s
2
-
x
i
-
x
¯
2
=
s
-
x
¯
∣
s
+
x
¯
-
2
x
i
,
x
¯
-
t
n
+
1
x
n
+
1
=
1
-
t
n
+
1
s
,we have(13)
L
=
∑
i
=
1
n
s
-
x
¯
∣
s
+
x
¯
-
2
x
i
=
s
-
x
¯
∣
∑
i
=
1
n
t
i
s
+
x
¯
-
2
x
i
=
s
-
x
¯
∣
1
-
t
n
+
1
s
+
x
¯
-
2
x
¯
-
t
n
+
1
x
n
+
1
=
s
-
x
¯
∣
1
-
t
n
+
1
s
+
x
¯
-
2
1
-
t
n
+
1
s
=
s
-
x
¯
∣
1
-
t
n
+
1
x
¯
-
s
=
-
1
-
t
n
+
1
s
-
x
¯
2
,which finishes the proof.


## 3. Separation by Stronglyh-Convex Functions

It is proved in [15] that two functions f
,
g
:
D
→
R defined on a convex subset D of a vector space can be separated by a convex function if and only if (14)
f
∑
i
=
1
n
t
i
x
i
≤
∑
i
=
1
n
t
i
g
x
i
,for all n
∈
N, x
1
,
…
,
x
n
∈
D, and t
1
,
…
,
t
n
>
0 with t
1
+
⋯
+
t
n
=
1.In this section we present counterparts of that result related to strongh-convexity.Theorem 2.
Letf
,
g
:
D
→
R be given functions and h
:
[
0,1
]
→
R be a multiplicative function such that h
(
t
)
≥
t for all t
∈
[
0,1
]
. If there exists a function φ
:
D
→
R strongly h-convex with modulus c such that(15)
f
≤
φ
≤
g
o
n
D
,then(16)
f
∑
i
=
1
n
t
i
x
i
≤
∑
i
=
1
n
h
t
i
g
x
i
-
c
∑
i
=
1
n
t
i
x
i
-
x
¯
2
,for all n
∈
N, x
1
,
…
,
x
n
∈
D, and t
1
,
…
,
t
n
>
0 with t
1
+
⋯
+
t
n
=
1 and x
¯
=
t
1
x
1
+
⋯
+
t
n
x
n
.Proof.
By the Jensen inequality for stronglyh-convex functions (Theorem 1) we have(17)
f
∑
i
=
1
n
t
i
x
i
≤
φ
∑
i
=
1
n
t
i
x
i
≤
∑
i
=
1
n
h
t
i
φ
x
i
-
c
∑
i
=
1
n
t
i
x
i
-
x
¯
2
≤
∑
i
=
1
n
h
t
i
g
x
i
-
c
∑
i
=
1
n
t
i
x
i
-
x
¯
2
.Theorem 3.
Letf
,
g
:
D
→
R be given functions and h
:
[
0,1
]
→
R be a multiplicative function such that h
(
t
)
≤
t for all t
∈
[
0,1
]
. If(18)
f
∑
i
=
1
n
t
i
x
i
≤
∑
i
=
1
n
h
t
i
g
x
i
-
c
∑
i
=
1
n
t
i
x
i
-
x
¯
2
,for all n
∈
N, x
1
,
…
,
x
n
∈
D, and t
1
,
…
,
t
n
>
0 with t
1
+
⋯
+
t
n
=
1 and x
¯
=
t
1
x
1
+
⋯
+
t
n
x
n, then there exists a function φ
:
D
→
R strongly h-convex with modulus c such that (19)
f
≤
φ
≤
g
o
n
D
.Proof.
Fixx
∈
D and define a function φ
:
D
→
R by(20)
φ
x
=
inf
⁡
∑
i
=
1
n
h
t
i
g
x
i
-
∑
i
=
1
n
t
i
x
i
-
x
2
:
n
∈
N
,
x
1
,
…
,
x
n
∈
D
,
t
1
,
…
,
t
n
∈
0,1
such  that
t
1
+
⋯
+
t
n
=
1
,
x
=
t
1
x
1
+
⋯
+
t
n
x
n
.By (18) the definition is correct and f
(
x
)
≤
φ
(
x
) for all x
∈
D. On the other hand, taking n
=
1 in the above definition (and, consequently, t
1
=
1, x
=
x
1) and using the fact that h
(
1
)
=
1, we get φ
(
x
)
≤
g
(
x
) for all x
∈
D.
To prove thatφ is strongly h-convex with modulus c, fix x
,
y
∈
D and t
∈
[
0,1
]. Take arbitrary u
1
,
…
,
u
n
∈
D, α
1
,
…
,
α
n
∈
[
0,1
] and v
1
,
…
,
v
m
∈
D, β
1
,
…
,
β
m
∈
[
0,1
] such that α
1
+
⋯
+
α
n
=
1, β
1
+
⋯
+
β
m
=
1 and x
=
α
1
u
1
+
⋯
+
α
n
u
n, y
=
β
1
v
1
+
⋯
+
β
m
v
m. Since ∑
i
=
1
n
‍
t
α
i
+
∑
j
=
1
m
‍
(
1
-
t
)
β
j
=
1, the point t
x
+
(
1
-
t
)
y is a convex combination of u
1
,
…
,
u
n, v
1
,
…
,
v
m, and(21)
t
x
+
1
-
t
y
=
∑
i
=
1
n
t
α
i
u
i
+
∑
j
=
1
m
1
-
t
β
j
v
j
.Therefore, by the definition of φ we have(22)
φ
t
x
+
1
-
t
y
≤
∑
i
=
1
n
h
t
α
i
g
u
i
+
∑
j
=
1
m
h
1
-
t
β
j
g
v
j
-
c
t
∑
i
=
1
n
α
i
u
i
-
m
2
+
1
-
t
∑
j
=
1
m
β
j
v
j
-
m
2
,where m
=
t
x
+
(
1
-
t
)
y. By the multiplicativity of h we have(23)
h
t
α
i
=
h
t
h
α
i
,
h
1
-
t
β
j
=
h
1
-
t
h
β
j
.Note also that(24)
∑
i
=
1
n
α
i
u
i
-
m
2
=
∑
i
=
1
n
α
i
u
i
2
-
2
u
i
∣
m
+
m
2
=
∑
i
=
1
n
α
i
u
i
2
-
2
x
∣
m
+
m
2
=
∑
i
=
1
n
α
i
u
i
-
x
2
+
x
-
m
2
,and, similarly,(25)
∑
j
=
1
m
β
j
v
j
-
m
2
=
∑
j
=
1
m
β
j
v
j
-
y
2
+
y
-
m
2
.Hence, using the fact that h
(
t
)
≤
t, we get(26)
t
∑
i
=
1
n
α
i
u
i
-
m
2
+
1
-
t
∑
j
=
1
m
β
j
v
j
-
m
2
≥
h
t
∑
i
=
1
n
α
i
u
i
-
x
2
+
h
1
-
t
∑
j
=
1
m
β
j
v
j
-
y
2
+
t
x
-
m
2
+
1
-
t
y
-
m
2
=
h
t
∑
i
=
1
n
α
i
u
i
-
x
2
+
h
1
-
t
∑
j
=
1
m
β
j
v
j
-
y
2
+
t
1
-
t
x
-
y
2
.Substituting (23) and (26) into (22), we obtain(27)
φ
t
x
+
1
-
t
y
≤
h
t
∑
i
=
1
n
h
α
i
g
u
i
-
c
∑
i
=
1
n
α
i
u
i
-
x
2
+
h
1
-
t
∑
j
=
1
m
h
β
j
g
v
j
-
c
∑
j
=
1
m
β
j
v
j
-
y
2
-
c
t
1
-
t
x
-
y
2
.Now, taking the infimum in the first term and next in the second term of the right hand side of (27) and using the definition of φ, we get(28)
φ
t
x
+
1
-
t
y
≤
h
t
φ
x
+
h
1
-
t
φ
y
-
c
t
1
-
t
x
-
y
2
,which shows that φ is strongly h-convex with modulus c and finishes the proof.Remark 4.
The method used in the proof of Theorem3 is similar to that in [12, Theorem3]. However, in our case we assume additionally that h
(
t
)
≤
t, t
∈
[
0,1
]. The following example shows that this assumption is essential. Let h
(
t
)
=
1, t
∈
[
0,1
] and consider f
,
g
:
[
0,1
]
→
R, f
=
-
1 and g
=
0. Then condition (18) is satisfied with c
=
1, but there is no function φ strongly h-convex (with any modulus) satisfying -
1
=
f
≤
φ
≤
g
=
0. Indeed, if φ is strongly h-convex with h
=
1, then φ is nonnegative and some of its values are positive (putting y
=
x in the definition of strong h-convexity, we get φ
≥
0, but φ
=
0 is not strongly h-convex).As a consequence of Theorem3 we obtain the following Hyers-Ulam-type stability result for strongly h-convex functions.Letε be a positive constant. We say that a function g
:
D
→
R is ε-strongly h-convex with modulus c if (29)
g
∑
i
=
1
n
t
i
x
i
≤
∑
i
=
1
n
h
t
i
g
x
i
-
c
∑
i
=
1
n
t
i
x
i
-
x
¯
2
+
ε
,for all n
∈
N, x
1
,
…
,
x
n
∈
D, and t
1
,
…
,
t
n
>
0 with t
1
+
⋯
+
t
n
=
1 and x
¯
=
t
1
x
1
+
⋯
+
t
n
x
n.Corollary 5.
Leth
:
[
0,1
]
→
R be a multiplicative function such that h
(
t
)
≤
t for all t
∈
[
0,1
]
. If a function g
:
D
→
R is ε-strongly h-convex with modulus c, then there exists a function φ
:
D
→
R strongly h-convex with modulus c such that(30)
g
x
-
ε
≤
φ
x
≤
g
x
,
x
∈
D
.Proof.
Definef
(
x
)
=
g
(
x
)
-
ε, x
∈
D. Then f and g satisfy (18). Therefore, by Theorem 3, there exists φ
:
D
→
R, strongly h-convex with modulus c, such that g
-
ε
≤
φ
≤
g, on D
.



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*Source: 7160348-2016-11-10.xml*