Journal of Inequalities in Pure and
Applied Mathematics
HERMITE-HADAMARD TYPE INEQUALITIES FOR
INCREASING RADIANT FUNCTIONS
volume 4, issue 2, article 47,
2003.
E.V. SHARIKOV
Tver State University,
Tver, Russia.
E-Mail: a001102@tversu.ru
Received 20 November, 2002;
accepted 16 May, 2003.
Communicated by: S.S. Dragomir
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
129-02
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Abstract
We study Hermite-Hadamard type inequalities for increasing radiant functions
and give some simple examples of such inequalities.
2000 Mathematics Subject Classification: 11N05, 11N37, 26D15.
Key words: Increasing radiant functions, Abstract convexity, Hermite-Hadamard type
inequalities.
The author is very grateful to A. M. Rubinov for formulation of the tasks and very
useful discussions.
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
Contents
1
2
3
4
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Hermite-Hadamard Type Inequalities . . . . . . . . . . . . . . . . . . . . 7
Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
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1.
Introduction
In this paper we consider one generalization of Hermite-Hadamard inequalities
for the class InR of increasing radiant functions defined on the cone Rn++ =
{x ∈ Rn : xi > 0 (i = 1, . . . , n)}.
Recall that for a function f : [a, b] → R, which is convex on [a, b], we have
the following:
Z b
a+b
1
1
(1.1)
f
≤
f (x) dx ≤ (f (a) + f (b)).
2
b−a a
2
These inequalities are well known as the Hermite-Hadamard inequalities. There
are many generalizations of these inequalities for classes of non-convex functions. For more information see ([2], Section 6.5), [1] and references therein.
In this paper we consider generalizations of the inequalities from both sides of
(1.1). Some techniques and notions, which are used here, can be found in [1].
In Section 2 of this paper we give a definition of InR functions and recall some results related to these functions. In Section 3 we consider HermiteHadamard type inequalities for the class InR. Some examples of such inequalities for functions defined on R++ and R2++ are given in Section 4.
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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2.
Preliminaries
We assume that the cone Rn++ is equipped with coordinate-wise order relation.
Recall that a function f : Rn++ → R̄+ = [0, +∞] is said to be increasing
radiant (InR) if:
1. f is increasing: x ≥ y =⇒ f (x) ≥ f (y);
2. f is radiant: f (λx) ≤ λf (x) for all λ ∈ (0, 1) and x ∈ Rn++ .
For example, any function f of the following form belongs to the class InR:
X
f (x) =
ck xk11 · · · xknn ,
|k|≥1
where k = (k1 , . . . , kn ), |k| = k1 + · · · + kn , ki ≥ 0, ck ≥ 0.
For each f ∈ InR its conjugate function ([4])
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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1
f ∗ (x) =
,
f (1/x)
where 1/x = (1/x1 , . . . , 1/xn ), is also increasing and radiant. Hence any function
1
f (x) = P
−k1
n
· · · x−k
n
|k|≥1 ck x1
is InR. In the more general case we have the following InR functions:
!t
P
k1
kn
|k|≥u ck x1 · · · xn
f (x) = P
,
−k1
n
· · · x−k
n
|k|≥v dk x1
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where u, v > 0, t ≥ 1/(u + v). Indeed, these functions are increasing and for
any λ ∈ (0, 1)
k1
|k|
kn
|k|≥u λ ck x1 · · · xn
P
f (λx) =
!t
1
n
λ−|k| dk x−k
· · · x−k
1
n
!t
P
λu |k|≥u ck xk11 · · · xknn
P
1
n
λ−v |k|≥v dk x−k
· · · x−k
1
n
P
|k|≥v
≤
= λ(u+v)t f (x) ≤ λf (x).
Consider the coupling function ϕ defined on Rn++ × Rn++ :
0,
if hh, xi < 1,
(2.1)
ϕ(h, x) =
hh, xi, if hh, xi ≥ 1,
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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where
hh, xi = min{hi xi : i = 1, . . . , n}
is the so-called min-type function.
Denote by ϕh the function defined on Rn++ by the formula: ϕh (x) = ϕ(h, x).
It is known (see [4]) that the set
1
n
H=
ϕh : h ∈ R++ , c ∈ (0, +∞]
c
is the supremal generator of the class InR of all increasing radiant functions
defined on Rn++ .
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It is known also that for any InR function f
1
(2.2)
f (h)ϕ
, x ≤ f (x) for all x, h ∈ Rn++ .
h
Note that for c = +∞ we set cϕh (x) = supl>0 (lϕh (x)).
Formula (2.2) implies the following statement.
Proposition 2.1. Let f be an InR function defined on Rn++ and ∆ ⊂ Rn++ .
Then the function
1
f∆ (x) = sup f (h)ϕ
,x
h
h∈∆
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
is InR, and it possesses the properties:
1) f∆ (x) ≤ f (x) for all x ∈ Rn++ ,
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2) f∆ (x) = f (x) for all x ∈ ∆.
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3.
Hermite-Hadamard Type Inequalities
Let D ⊂ Rn++ be a closed domain (in topology of Rn++ ), i.e. D is a bounded set
such that cl int D = D. Denote by Q(D) the set of all points x̄ ∈ D such that
Z
1
1
(3.1)
ϕ
, x dx = 1,
A(D) D
x̄
R
where A(D) = D dx, dx = dx1 · · · dxn .
Proposition 3.1. Let f be an InR function defined on Rn++ . If the set Q(D) is
nonempty and f is integrable on D then
Z
1
(3.2)
sup f (x̄) ≤
f (x)dx.
A(D) D
x̄∈Q(D)
Proof. First, let x̄ ∈ Q(D) and f (x̄) < +∞. Then f (x̄)ϕ(1/x̄, x) ≤ f (x) for
all x ∈ D ⊂ Rn++ (see (2.2)). By (3.1), we get
Z
1
1
f (x̄) = f (x̄)
ϕ
, x dx
A(D) D
x̄
Z
1
1
=
f (x̄)ϕ
, x dx
A(D) D
x̄
Z
1
≤
f (x)dx.
A(D) D
Now, suppose that f (x̄) = +∞.
R Then for all l > 0 function lϕ1/x̄ (x) is mi1
norant of f . Hence l ≤ A(D) D f (x) dx ∀ l > 0, that implies that function
f is not integrable on D. This contradiction shows that f (x̄) < +∞ for any
x̄ ∈ Q(D).
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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As it was done in [1], we may introduce the set Qm (D) of all maximal
elements of Q(D). It means that a point x̄ ∈ Q(D) belongs to Qm (D) if and
only if for any ȳ ∈ Q(D) : (ȳ ≥ x̄) =⇒ (ȳ = x̄). Suppose that the set Q(D)
is nonempty. It is easy to see that Q(D) is a closed set in the topology of Rn++ .
Hence, using the Zorn Lemma we conclude that Qm (D) is a nonempty closed
set and for any x̄ ∈ Q(D) there exists ȳ ∈ Qm (D), for which x̄ ≤ ȳ.
So, in assumptions of Proposition 3.1 we have the following estimate:
Z
1
(3.3)
sup f (x̄) ≤
f (x)dx.
A(D) D
x̄∈Qm (D)
Since f is an increasing function then this inequality implies inequality (3.2).
Remark 3.1. Let D ⊂ Rn++ be a closed domain and the set Q(D) be nonempty.
Then for every x̄ ∈ Q(D) inequality
Z
1
f (x̄) ≤
f (x)dx
A(D) D
is sharp. For example, if we set f = ϕ1/x̄ then (see (3.1))
Z
Z
1
1
1
1
, x̄ = 1 =
ϕ
, x dx =
f (x)dx.
f (x̄) = ϕ
x̄
A(D) D
x̄
A(D) D
Note that here we used only the values of function f on a set D. Therefore
we need the following definition.
Definition 3.1. Let D ⊂ Rn++ . A function f : D → [0, +∞] is said to be
increasing radiant on D if there exists an InR function F defined on Rn++ such
that F |D = f , that is F (x) = f (x) for all x ∈ D.
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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We assume here, as above, that for c = +∞ : cϕh (x) = supl>0 (lϕh (x)).
Proposition 3.2. Let f : D → [0, +∞] be a function defined on D ⊂ Rn++ .
Then the following assertions are equivalent:
1) f is increasing radiant on D,
2) f (h)ϕ(1/h, x) ≤ f (x) for all h, x ∈ D,
3) f is abstract convex with respect to the set of functions (1/c)ϕ(1/h) : D →
[0, +∞] with h ∈ D, c ∈ (0, +∞].
Proof. 1)=⇒2). By Definition 3.1, there exists an InR function F : Rn++ →
[0, +∞] such that F (x) = f (x) for all x ∈ D. Then Proposition 2.1 implies
that the function
1
FD (x) = sup F (h)ϕ
,x
h
h∈D
interpolates F in all points x ∈ D. Hence
1
, x = f (x) for all x ∈ D,
sup f (h)ϕ
h
h∈D
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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that implies the assertion 2)
2)=⇒3). Consider the function fD defined on D
1
fD (x) = sup f (h)ϕ
,x .
h
h∈D
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First, it is clear that fD is abstract convex with respect to the set of functions
defined on D : {(1/c)ϕ(1/h) : h ∈ D, c ∈ (0, +∞]}. Further, using 2) we get
for all x ∈ D
1
1
fD (x) ≤ f (x) = f (x)ϕ
, x ≤ sup f (h)ϕ
, x = fD (x).
x
h
h∈D
So, fD (x) = f (x) for all x ∈ D and we have the desired statement 3).
3)=⇒1). It is obvious since any function (1/c)ϕh defined on D can be considered as an elementary function (1/c)ϕh ∈ H defined on Rn++ .
Remark 3.2. We may require in Proposition 3.1, formula (3.3) and Remark 3.1
only that function f is increasing radiant and integrable on D.
Remark 3.3. We may consider a more general case of Hermite-Hadamard type
inequalities for InR functions. Let f be an increasing radiant function on D.
Then Proposition 3.2 implies that f (h)ϕ(1/h, x) ≤ f (x) for all h, x ∈ D. If
f (x̄) < +∞ and f is integrable on D then
Z
Z
1
(3.4)
f (x̄)
ϕ
, x dx ≤
f (x)dx.
x̄
D
D
This inequality is sharp for any x̄ ∈ D since we have the equality in (3.4) for
f = ϕ(1/x̄) .
Proposition 3.2 implies also that the class InR is broad enough.
Proposition 3.3. Let S ⊂ Rn++ be a set such that every point x ∈ S is maximal
in S. Then for any function f : S → [0, +∞] there exists an increasing radiant
function F : Rn++ → [0, +∞], for which F |S = f .
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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Proof. It is sufficient to check only that f (h)ϕ(1/h, x) ≤ f (x) for all h, x ∈
S. If h = x then ϕ(1/h, x) = 1, f (h) = f (x). If h 6= x then h1/h, xi =
mini xi /hi < 1 since h is a maximal point in S, hence ϕ(1/h, x) = 0 and
f (h)ϕ(1/h, x) = 0 ≤ f (x).
In particular, Proposition 3.3 holds if S = {x ∈ Rn++ : (x1 )p + · · · + (xn )p =
1}, where p > 0.
Now we present two assertions supported by the definition of function ϕ.
Recall that a set Ω ⊂ Rn++ is said to be normal if for each x ∈ Ω we have
(y ∈ Ω for all y ≤ x). The normal hull N (Ω) of a set Ω is defined as follows:
N (Ω) = {x ∈ Rn++ : (∃ y ∈ Ω) x ≤ y} (see, for example, [3]).
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
Proposition 3.4. Let D, Ω ⊂ Rn++ be closed domains and D ⊂ Ω. If the set
Q(Ω) is nonempty and
E.V. Sharikov
(Ω\D) ⊂ N (Q(Ω))
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(3.5)
then the set Q(D) consists of all points x̄ ∈ Ω such that
Z
1
1
ϕ
, x dx = 1.
A(D) Ω
x̄
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Proof. If D = Ω then the assertion is clear. Assume that D 6= Ω. Since D, Ω
are closed domains and D ⊂ Ω then
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A(D) < A(Ω).
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(3.6)
Let x̄ ∈ Ω and
(3.7)
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1
A(D)
Z
ϕ
Ω
1
, x dx = 1.
x̄
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We show that ϕ(1/x̄, x) = 0 for all x ∈ Ω\D. If x ∈ Ω\D then, by (3.5), there
exists a point ȳ ∈ Q(Ω) : ȳ ≥ x; hence h1/x̄, xi ≤ h1/x̄, ȳi. Suppose that
h1/x̄, ȳi ≥ 1. Then ȳ ≥ x̄ =⇒ 1/ȳ ≤ 1/x̄. Since ȳ ∈ Q(Ω) then, by (3.6) and
(3.7)
Z
1
1
1=
ϕ
, x dx
A(Ω) Ω
ȳ
Z
1
1
<
ϕ
, x dx
A(D) Ω
ȳ
Z
1
1
≤
ϕ
, x dx = 1.
A(D) Ω
x̄
So, we have the inequalities: h1/x̄, xi ≤ h1/x̄, ȳi < 1. Therefore ϕ(1/x̄, x) = 0
for all x ∈ Ω\D =⇒
Z
Z
1
1
1
1
1=
ϕ
, x dx =
ϕ
, x dx.
A(D) Ω
x̄
A(D) D
x̄
The equality (ϕ(1/x̄, ·) = 0 on Ω\D) implies also that x̄ 6= x for all x ∈ Ω\D,
hence x̄ 6∈ Ω\D =⇒ x̄ ∈ D. Thus, we have the established result: x̄ ∈ Q(D).
Conversely, let x̄ ∈ Q(D). For any x ∈ Ω\D there exists ȳ ∈ Q(Ω) such
that ȳ ≥ x =⇒ h1/x̄, xi ≤ h1/x̄, ȳi. Moreover, we may assume that ȳ is a
maximal point in Q(Ω), i.e. ȳ ∈ Qm (Ω). First, we check that
1
(3.8)
, x ≤ 1 for all x ∈ Ω\D, ȳ ∈ Qm (Ω).
ȳ
Hermite-Hadamard Type
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Indeed, if x ∈ Ω\D then for some z̄ ∈ Qm (Ω): x ≤ z̄ =⇒ h1/ȳ, xi ≤ h1/ȳ, z̄i.
But h1/ȳ, z̄i ≤ 1 since ȳ, z̄ ∈ Qm (Ω) (otherwise, if h1/ȳ, z̄i > 1 then z̄ >
ȳ =⇒ ȳ 6∈ Qm (Ω)).
Now we verify that h1/x̄, xi < 1 for all x ∈ Ω\D. If x ∈ Ω\D then for
some ȳ ∈ Qm (Ω) : h1/x̄, xi ≤ h1/x̄, ȳi. Suppose that h1/x̄, ȳi ≥ 1. Then
ȳ ≥ x̄ and therefore, using inclusion x̄ ∈ Q(D), we get
Z
1
1
(3.9)
1=
ϕ
, x dx
A(D) D
x̄
Z
1
1
>
ϕ
, x dx
A(Ω) D
x̄
Z
1
1
≥
ϕ
, x dx.
A(Ω) D
ȳ
Let D1 = {x ∈ Ω\D : h1/ȳ, xi < 1}, D2 = {x ∈ Ω\D : h1/ȳ, xi = 1}. It
follows from (3.8) that Ω\D = D1 ∪ D2 (D1 ∩ D2 = ∅), hence
Z
Z
Z
1
1
1
ϕ
, x dx =
ϕ
, x dx +
ϕ
, x dx
ȳ
ȳ
ȳ
Ω\D
D1
D2
Z
Z
1
, x dx =
dx.
=
ϕ
ȳ
D2
D2
R
But the last integral D2 dx is also equal to zero, since the set D2 has no interior
points. Thus, by (3.9)
Z
Z
1
1
1
1
1>
ϕ
, x dx =
ϕ
, x dx.
A(Ω) D
ȳ
A(Ω) Ω
ȳ
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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This inequality contradicts the inclusion ȳ ∈ Qm (Ω). So, we conclude that the
inequality h1/x̄, ȳi ≥ 1 is impossible. Hence h1/x̄, xi ≤ h1/x̄, ȳi < 1 for all
x ∈ Ω\D and ȳ = ȳ(x) ∈ Qm (Ω), which implies the required equality:
Z
Z
1
1
1
1
ϕ
, x dx =
ϕ
, x dx.
1=
A(D) D
x̄
A(D) Ω
x̄
Corollary 3.5. Let D1 , D2 ⊂ Rn++ be a closed domains such that
A(D1 ) = A(D2 ).
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
If there exists a closed domain Ω ⊂ Rn++ , for which the set Q(Ω) is nonempty
and
Di ⊂ Ω, (Ω\Di ) ⊂ N (Q(Ω))
(i = 1, 2),
then
Q(D1 ) = Q(D2 ).
Proposition 3.6. Let D, Ω ⊂ Rn++ be closed domains and D ⊂ Ω. If
(3.10)
N (Ω\D) ∩ D = ∅,
then the set Q(D) consists of all points x̄ ∈ D such that
Z
1
1
ϕ
, x dx = 1.
A(D) Ω
x̄
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Proof. Formula (3.10) implies that if x̄ ∈ D then x̄ 6∈ N (Ω\D). It means that
for all
1
1
x ∈ Ω\D : x < x̄ =⇒
, x < 1 =⇒ ϕ
, x = 0.
x̄
x̄
Thus, for any x̄ ∈ D
Z
Z
1
1
1
1
ϕ
, x dx = 1 ⇐⇒
ϕ
, x dx = 1
A(D) Ω
x̄
A(D) D
x̄
⇐⇒ x̄ ∈ Q(D).
Now consider the generalization of the inequality from the right-hand side
of (1.1). Let f be an increasing radiant function defined on a closed domain
D ⊂ Rn++ , and f is integrable on D. Then f (h)ϕ(1/h, x) ≤ f (x) for all
h, x ∈ D. In particular, f (h)h1/h, xi ≤ f (x) if h1/h, xi ≥ 1. Hence for all
x≥h
+
1
f (x)
f (h) ≤
= h,
f (x),
h1/h, xi
x
where h(y) = hh, yi+ = maxi hi yi is the so-called max-type function. So, if
x̄ ∈ D and x̄ ≥ x for all x ∈ D, then f (x) ≤ hx, 1/x̄i+ f (x̄) for any x̄ ∈ D.
This reduces to the following assertion.
Proposition 3.7. Let the function f be increasing radiant and integrable on D.
If x̄ ∈ D and x̄ ≥ x for all x ∈ D, then
+
Z
Z 1
(3.11)
f (x)dx ≤ f (x̄)
x,
dx.
x̄
D
D
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Radiant Functions
E.V. Sharikov
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Inequality (3.11) is sharp since we get equality for f (x) = hx, 1/x̄i+ .
In the more general case we have the following inequalities:
f (x) ≤ hx, 1/x̄i+ sup f (y) for all x̄ ≥ x.
y∈D
Hence
)
(
+
1
: x̄ ≥ x, x̄ ∈ D
for all x ∈ D
f (x) ≤ sup f (y) inf
x,
x̄
y∈D
and therefore
)
(
+
Z
Z
1
: x̄ ≥ x, x̄ ∈ D dx.
(3.12)
f (x)dx ≤ sup f (y)
inf
x,
x̄
y∈D
D
D
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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4.
Examples
Here we describe the set Q(D) for some special domains D of the cones R++
and R2++ .
Let a, b ∈ R be numbers such that 0 ≤ a < b. We denote by [a, b] the
segment {x ∈ R++ : a ≤ x ≤ b}.
Example 4.1. Let D = [a, b] ⊂ R++ , where 0 ≤ a < b. By definition, the set
Q(D) consists of all points x̄ ∈ D, for which
Z
Z b 1
1
1
1
ϕ
, x dx =
ϕ
, x dx = 1.
A(D) D
x̄
b−a a
x̄
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
We have:
ϕ
1
,x
x̄
(
=
0,
x
,
x̄
if x < x̄,
if x ≥ x̄.
Hence, if x̄ ∈ D = [a, b] then
Z b
Z b x
1 2
1
(4.1)
ϕ
, x dx =
dx =
(b − x̄2 ).
x̄
2x̄
a
x̄ x̄
So, a point x̄ ∈ [a, b] belongs to Q(D) if and only if
1
(b2 − x̄2 ) = 1 ⇐⇒ x̄2 + 2(b − a)x̄ − b2 = 0.
2(b − a)x̄
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We get
(4.2)
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x̄ =
p
(b − a)2 + b2 − (b − a).
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Show that for the point (4.2)
a < x̄ <
(4.3)
Since b > a ≥ 0 then x̄ =
Further,
x̄ <
p
a+b
.
2
(b − a)2 + b2 − (b − a) >
√
b2 − (b − a) = a.
p
a+b
a+b
3b − a
⇐⇒ (b − a)2 + b2 < (b − a) +
=
2
2
2
2
2
2
⇐⇒ 4(b − a) + 4b < (3b − a)
⇐⇒ 0 < b2 + 2ab − 3a2 .
The last inequality follows
np from the same conditions
o b > a ≥ 0.
Thus, Q([a, b]) =
(b − a)2 + b2 − (b − a) . Remark 3.1 implies that for
every InR function f ∈ L1 [a, b]
f
p
(b −
a)2
+
b2
− (b − a) ≤
1
b−a
Z
b
f (x)dx
a
and this inequality is sharp. (Compare it with the corresponding estimate for
convex functions (1.1), see also (4.3)).
Remark 3.3 and formula (4.1) imply the following inequalities
Z b
2u
(4.4)
f (u) ≤ 2
f (x)dx,
b − u2 a
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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which are sharp in the class of all InR functions f ∈ L1 [a, b] and hold for any
u ∈ [a, b). In particular, we get for u = (a + b)/2
Z b
a+b
4(a + b)
f
≤
f (x)dx.
2
(a + 3b)(b − a) a
Note that here
1
4(a + b)
>
.
(a + 3b)(b − a)
b−a
Further, Proposition 3.7 implies that
Z b
Z b
b 2 − a2
x
f (x)dx ≤ f (b)
dx =
f (b),
2b
a
a b
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
hence
Z b
1
a+b
f (x)dx ≤
f (b)
b−a a
2b
for every InR function f ∈ L1 [a, b].
Let D ⊂ R2++ , x̄ = (x̄1 , x̄2 ) ∈ D. We denote by D(x̄) the set {x ∈ D :
x1 ≥ x̄1 , x2 ≥ x̄2 }. It is clear that
Z
Z
Z
1
1
x1 x2
ϕ
, x dx =
, x dx =
min
,
dx1 dx2 .
x̄
x̄
x̄1 x̄2
D
D(x̄)
D(x̄)
In order to calculate such integrals we represent the set D(x̄) as a union D1 (x̄)∪
D2 (x̄), where
x2
x1
x1
x2
D1 (x̄) = x ∈ D(x̄) :
≤
, D2 (x̄) = x ∈ D(x̄) :
≤
.
x̄2
x̄1
x̄1
x̄2
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Then
Z
ϕ
D
Z
Z
1
1
1
, x dx =
, x dx +
, x dx
x̄
x̄
x̄
D1 (x̄)
D2 (x̄)
Z
Z
1
1
=
x2 dx1 dx2 +
x1 dx1 dx2 .
x̄2 D1 (x̄)
x̄1 D2 (x̄)
In the next examples we will use the number k, which possesses the properties:
(4.5)
2k 3 − 3k 2 − 3k + 1 = 0,
0 < k < 1.
Let g(k) = 2k 3 − 3k 2 − 3k + 1. We have: g(0) > 0, g(1) < 0, g 0 (k) =
6k 2 − 6k − 3 < 6k − 6k − 3 < 0 for all k ∈ (0, 1). So, there exists a unique
solution of the equation (4.5), which belongs to the interval (0, 1). We denote
this solution by the same symbol k.
Example 4.2. Let D ⊂ R2++ be the triangle with vertices (0, 0), (a, 0) and
(0, b), that is
n
o
x1 x2
D = x ∈ R2++ :
+
≤1 .
a
b
If x̄ ∈ D then we get
abx̄2
x̄1
a
2
D1 (x̄) = x ∈ R++ : x̄2 ≤ x2 ≤
,
x2 ≤ x1 ≤ a − x2 ,
ax̄2 + bx̄1 x̄2
b
abx̄1
x̄2
b
2
D2 (x̄) = x ∈ R++ : x̄1 ≤ x1 ≤
,
x1 ≤ x2 ≤ b − x1 .
ax̄2 + bx̄1 x̄1
a
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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Therefore
Z
D1 (x̄)
Z
Z a−(a/b)x2
1
1 (abx̄2 )/(ax̄2 +bx̄1 )
, x dx =
dx2
x2 dx1 .
x̄
x̄2 x̄2
(x̄1 /x̄2 )x2
This reduces to
Z
1
x̄2 /b
ab x̄2 ab x̄2 x̄1 x̄2 ab
, x dx =
−
·
+
·
+
.
x̄
6 (x̄1 /a + x̄2 /b)2
2 b
3 b a
b
D1 (x̄)
By analogy,
Z
1
ab
x̄1 /a
ab x̄1 ab x̄1 x̄1 x̄2 , x dx =
·
−
·
+
·
+
.
x̄
6 (x̄1 /a + x̄2 /b)2
2 a
3 a a
b
D2 (x̄)
Thus, the sum of these quantities is
Z
1
(4.6)
ϕ
, x dx
x̄
D
ab
1
ab x̄1 x̄2 ab x̄1 x̄2 2
=
·
−
+
+
+
.
6 (x̄1 /a + x̄2 /b)
2 a
b
3 a
b
Since A(D) = (ab)/2 then for x̄ ∈ D
x̄
1
1
x̄2 2 x̄1 x̄2 2
1
−
+
+
+
=1
3 (x̄1 /a + x̄2 /b)
a
b
3 a
b
x̄
x̄
x̄
x̄2 3
x̄2 2
x̄2 1
1
1
⇐⇒ 2
+
−3
+
−3
+
+ 1 = 0.
a
b
a
b
a
b
x̄ ∈ Q(D) ⇐⇒
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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Using inequalities 0 < (x̄1 /a + x̄2 /b) ≤ 1 for x̄ ∈ D we get
n
o
x̄1 x̄2
2
Q(D) = x̄ ∈ R++ :
+
=k ,
a
b
where k is the solution of (4.5).
In the more general case we have inequality (see (3.4) and (4.6))
Z
6u
f (x̄1 , x̄2 ) ≤
f (x)dx,
ab(1 − 3u2 + 2u3 ) D
where u = u(x̄1 , x̄2 ) = x̄1 /a + x̄2 /b < 1, function f is increasing radiant and
integrable on D.
Consider now inequality (3.12) for our triangle D. We show that
)
(
+
x
1
x2 1
: x̄ ≥ x, x̄ ∈ D =
+
.
inf
x,
x̄
a
b
Let x̄ = (x̄1 , x̄2 ) = (x1 /(x1 /a + x2 /b), x2 /(x1 /a + x2 /b)). Then x̄ ≥ x and
x̄ ∈ D since x̄1 /a + x̄2 /b = 1. Hence
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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(
)
+
1
inf
x,
: x̄ ≥ x, x̄ ∈ D
x̄
(
≤ max x1
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x1
a
+
x1
x2
b
, x2
x1
a
+
x2
x2
b
)
=
x1 x2
+ .
a
b
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Suppose that the converse inequality does not hold, then hx, 1/x̄i+ < x1 /a +
x2 /b for some x̄ ≥ x, x̄ ∈ D, hence x/(x1 /a + x2 /b) < x̄. But this implies that
x̄ 6∈ D.
Thus, it follows from (3.12) that
Z
Z x1 x2 f (x)dx ≤ sup f (y)
+
dx.
a
b
y∈D
D
D
Calculation gives the quantity
Z D
ab
x1 x2 +
dx = .
a
b
3
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
Since A(D) = ab/2 then the final result is
Z
1
2
f (x)dx ≤ sup f (y).
A(D) D
3 y∈D
Example 4.3. Now let Ω be the triangle from Example 4.2:
n
o
x1 x2
2
Ω = x ∈ R++ :
+
≤1 .
a
b
Denote by D the subset of Ω such that
k
x1 k
x2 x1 x2
Ω\D = x ∈ Ω : < , < ,
+
<k .
3
a 3
b a
b
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R2++
Then (Ω\D) ⊂ N (Q(Ω)) = {x ∈
: x1 /a + x2 /b ≤ k}. Note that
A(Ω\D) = (1/18)k 2 ab, hence A(D) = (ab)/2 − (1/18)k 2 ab = ab(1/2 −
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k 2 /18). It follows from Proposition 3.4 and formula (4.6) (with Ω instead of D)
that a point x̄ ∈ Ω belongs to Q(D) if and only if
1
ab
1
ab x̄1 x̄2 ab x̄1 x̄2 2
−
+
+
+
=1
ab(1/2 − k 2 /18) 6 (x̄1 /a + x̄2 /b)
2 a
b
3 a
b
x̄
x̄
x̄2 3
x̄2 2
k 2 x̄1 x̄2 1
1
+
−3
+
− 3−
+
+ 1 = 0.
⇐⇒ 2
a
b
a
b
3
a
b
It is easy to check that there exists a unique solution s of the equation:
2s3 − 3s2 − (3 − k 2 /3)s + 1 = 0,
0 < s ≤ 1.
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
Hence
o
n
x̄1 x̄2
Q(D) = x̄ ∈ R2++ :
+
=s .
a
b
We may establish also that s > k.
0
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0
Remark 4.1. For any other closed domain D such that (Ω\D ) ⊂ N (Q(Ω)) =
{x ∈ R2++ : x1 /a + x2 /b ≤ k} the set Q(D0 ) has the same form, i.e. it is
intersection of R2++ and a line (x̄1 /a + x̄2 /b) = s0 with some s0 : k < s0 < 1.
Example 4.4. Let Ω be the same triangle: Ω = {x ∈ R2++ : (x1 /a + x2 /b) ≤
1}. Let D ⊂ Ω and
a
b
Ω\D = x ∈ Ω : x1 < , x2 <
.
2
2
Then Ω\D is the normal set, hence N (Ω\D) ∩ D = (Ω\D) ∩ D is the empty
set. Since A(Ω\D) = ab/4 then A(D) = ab/2 − ab/4 = ab/4. By Proposition
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3.6, we have for x̄ ∈ D
ab
1
1
ab x̄1 x̄2 ab x̄1 x̄2 2
x̄ ∈ Q(D) ⇐⇒
−
+
+
+
=1
ab/4 6 (x̄1 /a+ x̄2 /b) 2 a
b
3 a
b
x̄ x̄ 3
x̄
x̄2 2 3 x̄1 x̄2 1
2
1
+
−3
+
−
+
+ 1 = 0.
⇐⇒ 2
a
b
a
b
2 a
b
So,
o
n
x̄1 x̄2
Q(D) = D ∩ x̄ ∈ R2++ :
+
=p
a
b
o
n
a x̄1 x̄2
2
+
=p
= x̄ ∈ R++ : x̄1 ≥ ,
2 a
b
b x̄1 x̄2
2
∪ x̄ ∈ R++ : x̄2 ≥ ,
+
=p ,
2 a
b
3
2
where 2p − 3p − (3/2)p + 1 = 0, 0 < p ≤ 1.
E.V. Sharikov
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The following two examples were considered in [1] for ICAR functions defined on R2+ . Note that the coefficient k plays here the same role as the number
(1/3) in [1].
Example 4.5. Consider the triangle D with vertices (0, 0), (a, 0) and (a, va):
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If x̄ ∈ D then
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D1 (x̄) =
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
x ∈ R2++ : x̄1 ≤ x1 ≤ a, x̄2 ≤ x2 ≤
x̄2
x1 ,
x̄1
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x̄2
2
D2 (x̄) = x ∈ R++ : x̄1 ≤ x1 ≤ a,
x1 ≤ x2 ≤ vx1 .
x̄1
Calculation gives the following quantities
Z
Z
Z (x̄2 /x̄1 )x1
1
1 a
x2 dx1 dx2 =
dx1
x2 dx2
x̄2 D1 (x̄)
x̄2 x̄1
x̄2
3
a x̄1
a
= x̄2
− +
,
6x̄21 2
3
1
x̄1
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
Z
Z vx1
1 a
x1 dx1 dx2 =
dx1
x1 dx2
x̄1 x̄1
D2 (x̄)
(x̄2 /x̄1 )x1
3
3
vx̄21
a
va
x̄1
=
−
− x̄2
−
.
3x̄1
3
3x̄21
3
Z
E.V. Sharikov
Title Page
Contents
Further,
Z
ϕ
D
1
,x
x̄
dx =
va3 vx̄21
−
3x̄1
3
+ x̄2
2x̄1 a
a3
− − 2
3
2 6x̄1
.
Since A(D) = va2 /2 then a point x̄ ∈ D belongs to Q(D) if and only if
2
2 a
2 x̄1
−
3 x̄1 3 a2
+
x̄2
va
2
4 x̄1
1a
−1−
=1
3 a
3 x̄21
x̄31
x̄1
x̄31
x̄21
⇐⇒ x̄2 1 + 3 2 − 4 3 = vx̄1 2 − 3 − 2 3 .
a
a
a
a
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In particular, if x̄2 = vx̄1 then we get the equation 2(x̄1 /a)3 − 3(x̄1 /a)2 −
3(x̄1 /a) + 1 = 0, hence (x̄1 /a) = k. So, the point (ka, vka) belongs to Q(D).
This implies that for each InR function f , which is integrable on D:
Z
1
f (x)dx.
f (ka, vka) ≤
A(D) D
2
If x̄2 = vx̄1 /2 then then
√ equation has the form
√ (x̄1 /a) +
√2(x̄1 /a) −1 = 0. This
shows that (x̄1 /a) = 2 − 1, therefore ( 2 − 1)a, v( 2 − 1)a/2 ∈ Q(D).
Further, we may set in (3.11) x̄ = (a, va):
Z
Z
nx x o
2
1
f (x)dx ≤ f (a, va)
max
,
dx1 dx2
a
va
D
D
Z
x1
= f (a, va)
dx1 dx2
D a
Z
Z vx1
f (a, va) a
=
dx1
x1 dx2
a
0
0
va2
=
f (a, va).
3
Thus,
1
A(D)
Z
2
f (x)dx ≤ f (a, va).
3
D
Example 4.6. Let D be the square:
D = {x ∈ R2++ : x1 ≤ 1, x2 ≤ 1}.
We consider two possible cases for x̄ ∈ D : (x̄2 /x̄1 ) ≤ 1 and (x̄2 /x̄1 ) ≥ 1.
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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a) If (x̄2 /x̄1 ) ≤ 1 then we have
Z
Z
Z (x̄2 /x̄1 )x1
1
1 1
x2 dx1 dx2 =
dx1
x2 dx2
x̄2 D1 (x̄)
x̄2 x̄1
x̄2
x̄2
1
2x̄1
=
−1+
,
2 3x̄21
3
1
x̄1
Z
Z 1
1 1
x1 dx1 dx2 =
dx1
x1 dx2
x̄1 x̄1
D2 (x̄)
(x̄2 /x̄1 )x1
1 1
x̄2
1
=
− x̄1 +
x̄1 − 2 .
2 x̄1
3
x̄1
Z
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
Hence
Title Page
Z
ϕ
D
1
1
, x dx =
x̄
2
1
− x̄1
x̄1
+
x̄2
6
4x̄1 − 3 −
1
x̄21
.
Since A(D) = 1 then we get the equation for x̄ ∈ Q(D)
1
2
1
− x̄1
x̄1
x̄2
+
6
1
4x̄1 − 3 − 2 = 1
x̄1
⇐⇒ x̄2 1 + 3x̄21 − 4x̄31 = 3x̄1 1 − 2x̄1 − x̄21 .
b) If (x̄2 /x̄1 ) ≥ 1 then we get the symmetric equation
x̄1 1 + 3x̄22 − 4x̄32 = 3x̄2 1 − 2x̄2 − x̄22 .
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Thus, the set Q(D) can be represented as the union of two sets:
x̄ ∈ R2++ : x̄2 ≤ x̄1 ≤ 1, x̄2 1 + 3x̄21 − 4x̄31 = 3x̄1 1 − 2x̄1 − x̄21
and
x̄ ∈ R2++ : x̄1 ≤ x̄2 ≤ 1, x̄1 1 + 3x̄22 − 4x̄32 = 3x̄2 1 − 2x̄2 − x̄22 .
In particular, if x̄1 = x̄2 then
x̄ ∈ Q(D) ⇐⇒ 0 < x̄1 ≤ 1,
1+
2x̄31
⇐⇒ 0 < x̄1 ≤ 1,
3x̄21
−
−
3x̄21
4x̄31
= 3 1 − 2x̄1 −
− 3x̄1 + 1 = 0 .
x̄21
This implies that (k, k) ∈ Q(D).
At last we investigate inequality (3.11) with x̄ = (1, 1) for the square D:
Z
Z
f (x)dx ≤ f (1, 1)
max{x1 , x2 }dx1 dx2 .
D
D
Since A(D) = 1 and
Z
Z
max{x1 , x2 }dx1 dx2 =
D
1
Z
x1
dx1
x1 dx2 +
0
Z 1
1
(1 − x21 )
= +
dx1
3
2
0
1 1 1
2
= + − =
3 2 6
3
0
Z
1
Z
E.V. Sharikov
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1
dx1
0
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
x2 dx2
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J. Ineq. Pure and Appl. Math. 4(2) Art. 47, 2003
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then
1
A(D)
Z
2
f (x)dx ≤ f (1, 1),
3
D
and this estimate holds for every increasing radiant and integrable on D function f .
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
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J. Ineq. Pure and Appl. Math. 4(2) Art. 47, 2003
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References
[1] S.S. DRAGOMIR, J. DUTTA AND A.M. RUBINOV, HermiteHadamard-type inequalities for increasing convex-along-rays functions, RGMIA Res. Rep. Coll., 4(4) (2001), Article 4. [ONLINE
http://rgmia.vu.edu.au/v4n4.html]
[2] A.M. RUBINOV, Abstract convexity and global optimization. Kluwer Academic Publishers, Boston-Dordrecht-London, (2000).
[3] A.M. RUBINOV AND B.M. GLOVER, Duality for increasing positively
homogeneous functions and normal sets, RAIRO-Operations Research, 32
(1998), 105–123.
Hermite-Hadamard Type
Inequalities for Increasing
Radiant Functions
E.V. Sharikov
[4] E.V. SHARIKOV, Increasing radiant functions, (submitted).
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J. Ineq. Pure and Appl. Math. 4(2) Art. 47, 2003
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