Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 2, Article 47, 2003
HERMITE-HADAMARD TYPE INEQUALITIES FOR INCREASING RADIANT
FUNCTIONS
E.V. SHARIKOV
T VER S TATE U NIVERSITY,
T VER , RUSSIA
a001102@tversu.ru
Received 20 November, 2002; accepted 16 May, 2003
Communicated by S.S. Dragomir
A BSTRACT. We study Hermite-Hadamard type inequalities for increasing radiant functions and
give some simple examples of such inequalities.
Key words and phrases: Increasing radiant functions, Abstract convexity, Hermite-Hadamard type inequalities.
2000 Mathematics Subject Classification. 11N05, 11N37, 26D15.
1. I NTRODUCTION
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 Hermite-Hadamard type inequalities for the class
InR. Some examples of such inequalities for functions defined on R++ and R2++ are given in
Section 4.
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
The author is very grateful to A. M. Rubinov for formulation of the tasks and very useful discussions.
129-02
2
E.V. S HARIKOV
2. P RELIMINARIES
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])
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
where u, v > 0, t ≥ 1/(u + v). Indeed, these functions are increasing and for any λ ∈ (0, 1)
!t
P
k1
|k|
kn
λ
c
x
·
·
·
x
k
1
n
|k|≥u
f (λx) = P
−k1
−|k|
n
dk x1 · · · x−k
n
|k|≥v λ
!t
P
λu |k|≥u ck xk11 · · · xknn
≤
P
1
n
λ−v |k|≥v dk x−k
· · · x−k
1
n
= 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,
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++ .
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)).
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H ERMITE -H ADAMARD T YPE I NEQUALITIES
3
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∈∆
is InR, and it possesses the properties:
1) f∆ (x) ≤ f (x) for all x ∈ Rn++ ,
2) f∆ (x) = f (x) for all x ∈ ∆.
3. H ERMITE -H ADAMARD T YPE I NEQUALITIES
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
Z
Z
1
1
1
1
1
f (x̄) = f (x̄)
ϕ
, x dx =
f (x̄)ϕ
, x dx ≤
f (x)dx.
A(D) D
x̄
A(D) D
x̄
A(D) D
Now, suppose
that f (x̄) = +∞. Then for all l > 0 function lϕ1/x̄ (x) is minorant of f . Hence
R
1
l ≤ A(D)
f
(x)
dx ∀ l > 0, that implies that function f is not integrable on D. This contraD
diction shows that f (x̄) < +∞ for any x̄ ∈ Q(D).
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 3.2. 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
f (x̄) = ϕ
, x̄ = 1 =
ϕ
, x dx =
f (x)dx.
x̄
A(D) D
x̄
A(D) D
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4
E.V. S HARIKOV
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.
We assume here, as above, that for c = +∞ : cϕh (x) = supl>0 (lϕh (x)).
Proposition 3.3. 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
,x
FD (x) = sup F (h)ϕ
h
h∈D
interpolates F in all points x ∈ D. Hence
1
sup f (h)ϕ
, x = f (x) for all x ∈ D,
h
h∈D
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
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 3.4. We may require in Proposition 3.1, formula (3.3) and Remark 3.2 only that function f is increasing radiant and integrable on D.
Remark 3.5. 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 3.3 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 3.3 implies also that the class InR is broad enough.
Proposition 3.6. 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 .
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H ERMITE -H ADAMARD T YPE I NEQUALITIES
5
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 3.6 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]).
Proposition 3.7. Let D, Ω ⊂ Rn++ be closed domains and D ⊂ Ω. If the set Q(Ω) is nonempty
and
(Ω\D) ⊂ N (Q(Ω))
(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
Proof. If D = Ω then the assertion is clear. Assume that D 6= Ω. Since D, Ω are closed
domains and D ⊂ Ω then
A(D) < A(Ω).
(3.6)
Let x̄ ∈ Ω and
(3.7)
1
A(D)
Z
ϕ
Ω
1
, x dx = 1.
x̄
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
Z
Z
1
1
1
1
1
1
ϕ
, x dx <
ϕ
, x dx ≤
ϕ
, x dx = 1.
1=
A(Ω) Ω
ȳ
A(D) Ω
ȳ
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
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
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6
E.V. S HARIKOV
x̄ ∈ Q(D), we get
(3.9)
1
1=
A(D)
Z
ϕ
D
Z
Z
1
1
1
1
1
, x dx >
ϕ
, x dx ≥
ϕ
, x dx.
x̄
A(Ω) D
x̄
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
Z
Z
1
1
1
1
ϕ
, x dx =
ϕ
, x dx +
ϕ
, x dx =
ϕ
, x dx =
dx.
ȳ
ȳ
ȳ
ȳ
D1
D2
D2
D2
Ω\D
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
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 3.8. Let D1 , D2 ⊂ Rn++ be a closed domains such that
A(D1 ) = A(D2 ).
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 3.9. 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
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 ⇐⇒ x̄ ∈ Q(D).
A(D) Ω
x̄
A(D) D
x̄
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H ERMITE -H ADAMARD T YPE I NEQUALITIES
7
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
+
f (x)
1
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 3.10. 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
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
(
f (x) ≤ sup f (y) inf
y∈D
1
x,
x̄
)
+
: x̄ ≥ x, x̄ ∈ D
for all x ∈ D
and therefore
(
)
+
1
f (x)dx ≤ sup f (y)
inf
x,
: x̄ ≥ x, x̄ ∈ D dx.
x̄
y∈D
D
D
Z
(3.12)
Z
4. E XAMPLES
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
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
1
x
1 2
(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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8
E.V. S HARIKOV
We get
x̄ =
(4.2)
p
(b − a)2 + b2 − (b − a).
Show that for the point (4.2)
a+b
.
2
√
p
Since b > a ≥ 0 then x̄ = (b − a)2 + b2 − (b − a) > b2 − (b − a) = a. Further,
p
a+b
a+b
3b − a
x̄ <
⇐⇒ (b − a)2 + b2 < (b − a) +
=
2
2
2
2
2
2
⇐⇒ 4(b − a) + 4b < (3b − a)
a < x̄ <
(4.3)
⇐⇒ 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 3.2 implies that for every InR
function f ∈ L1 [a, b]
Z b
p
1
2
2
(b − a) + b − (b − a) ≤
f (x)dx
f
b−a a
and this inequality is sharp. (Compare it with the corresponding estimate for convex functions
(1.1), see also (4.3)).
Remark 3.5 and formula (4.1) imply the following inequalities
Z b
2u
(4.4)
f (u) ≤ 2
f (x)dx,
b − u2 a
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
4(a + b)
a+b
≤
f (x)dx.
f
2
(a + 3b)(b − a) a
Note that here
4(a + b)
1
>
.
(a + 3b)(b − a)
b−a
Further, Proposition 3.10 implies that
Z b
Z b
x
b 2 − a2
f (x)dx ≤ f (b)
dx =
f (b),
2b
a
a b
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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H ERMITE -H ADAMARD T YPE I NEQUALITIES
9
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:
2k 3 − 3k 2 − 3k + 1 = 0,
(4.5)
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
x̄2
b
abx̄1
2
,
x1 ≤ x2 ≤ b − x1 .
D2 (x̄) = x ∈ R++ : x̄1 ≤ x1 ≤
ax̄2 + bx̄1 x̄1
a
Therefore
Z
Z
Z a−(a/b)x2
1
1 (abx̄2 )/(ax̄2 +bx̄1 )
, x dx =
dx2
x2 dx1 .
x̄
x̄2 x̄2
D1 (x̄)
(x̄1 /x̄2 )x2
This reduces to
Z
D1 (x̄)
By analogy,
Z
D2 (x̄)
1
ab
x̄2 /b
ab x̄2 ab x̄2 x̄1 x̄2 , x dx =
−
·
+
·
+
.
x̄
6 (x̄1 /a + x̄2 /b)2
2 b
3 b a
b
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
Thus, the sum of these quantities is
Z
1
ab
1
ab x̄1 x̄2 ab x̄1 x̄2 2
(4.6)
ϕ
, x dx =
·
−
+
+
+
.
x̄
6 (x̄1 /a + x̄2 /b)
2 a
b
3 a
b
D
Since A(D) = (ab)/2 then for x̄ ∈ D
x̄
1
1
x̄2 2 x̄1 x̄2 2
1
x̄ ∈ Q(D) ⇐⇒
−
+
+
+
=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
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).
J. Inequal. Pure and Appl. Math., 4(2) Art. 47, 2003
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10
E.V. S HARIKOV
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
inf
x,
: x̄ ≥ x, x̄ ∈ D =
+
.
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
(
)
(
)
+
x1
x2
x1
x2
+
+
1
x1 x2
b
b
inf
x,
: x̄ ≥ x, x̄ ∈ D ≤ max x1 a
, x2 a
=
+ .
x̄
x1
x2
a
b
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 x1 x2 ab
+
dx = .
a
b
3
D
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
Ω = x ∈ R2++ :
+
≤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
Then (Ω\D) ⊂ N (Q(Ω)) = {x ∈ R2++ : 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 − k 2 /18). It follows from Proposition 3.7 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
⇐⇒ 2
+
−3
+
− 3−
+
+ 1 = 0.
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,
Hence
0 < s ≤ 1.
n
o
x̄1 x̄2
2
Q(D) = x̄ ∈ R++ :
+
=s .
a
b
J. Inequal. Pure and Appl. Math., 4(2) Art. 47, 2003
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H ERMITE -H ADAMARD T YPE I NEQUALITIES
11
We may establish also that s > k.
Remark 4.1. For any other closed domain D0 such that (Ω\D0 ) ⊂ 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
b
a
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 3.9, we have for x̄ ∈ D
ab
1
ab x̄1 x̄2 ab x̄1 x̄2 2
1
x̄ ∈ Q(D) ⇐⇒
−
+
+
+
=1
ab/4 6 (x̄1 /a + x̄2 /b)
2 a
b
3 a
b
x̄
x̄
x̄2 3
x̄2 2 3 x̄1 x̄2 1
1
⇐⇒ 2
+
−3
+
−
+
+ 1 = 0.
a
b
a
b
2 a
b
So,
n
o
x̄1 x̄2
Q(D) = D ∩ x̄ ∈ R2++ :
+
=p
a
b
n
o a x̄1 x̄2
b x̄1 x̄2
2
2
= x̄ ∈ R++ : x̄1 ≥ ,
+
= p ∪ x̄ ∈ R++ : x̄2 ≥ ,
+
=p ,
2 a
b
2 a
b
where 2p3 − 3p2 − (3/2)p + 1 = 0, 0 < p ≤ 1.
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):
D = {x ∈ R2++ : x1 ≤ a, x2 ≤ vx1 }.
If x̄ ∈ D then
x̄2
2
D1 (x̄) = x ∈ R++ : x̄1 ≤ x1 ≤ a, x̄2 ≤ x2 ≤ x1 ,
x̄1
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
a x̄1
= x̄2
− +
,
6x̄21 2
3
Z
Z
Z vx1
1
1 a
x1 dx1 dx2 =
dx1
x1 dx2
x̄1 D2 (x̄)
x̄1 x̄1
(x̄2 /x̄1 )x1
3
3
va
vx̄21
a
x̄1
=
−
− x̄2
−
.
3x̄1
3
3x̄21
3
Further,
3
Z
1
va
vx̄21
2x̄1 a
a3
ϕ
, x dx =
−
+ x̄2
− − 2 .
x̄
3x̄1
3
3
2 6x̄1
D
J. Inequal. Pure and Appl. Math., 4(2) Art. 47, 2003
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12
E.V. S HARIKOV
Since A(D) = va2 /2 then a point x̄ ∈ D belongs to Q(D) if and only if
2 a
2 x̄21
x̄2 4 x̄1
1 a2
−
+
−1−
=1
3 x̄1 3 a2
va 3 a
3 x̄21
x̄31
x̄1
x̄21
x̄31
⇐⇒ x̄2 1 + 3 2 − 4 3 = vx̄1 2 − 3 − 2 3 .
a
a
a
a
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 (ka, vka) ≤
f (x)dx.
A(D) D
2
If x̄2 = vx̄√1 /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
1
2
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,
Z
1
2
f (x)dx ≤ f (a, va).
A(D) D
3
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.
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
Hence
Z
ϕ
D
1
1 1
x̄2
1
, x dx =
− x̄1 +
4x̄1 − 3 − 2 .
x̄
2 x̄1
6
x̄1
J. Inequal. Pure and Appl. Math., 4(2) Art. 47, 2003
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H ERMITE -H ADAMARD T YPE I NEQUALITIES
1
2
13
Since A(D) = 1 then we get the equation for x̄ ∈ Q(D)
1
x̄2
1
− x̄1 +
4x̄1 − 3 − 2 = 1 ⇐⇒ x̄2 1 + 3x̄21 − 4x̄31 = 3x̄1 1 − 2x̄1 − x̄21 .
x̄1
6
x̄1
b) If (x̄2 /x̄1 ) ≥ 1 then we get the symmetric equation
x̄1 1 + 3x̄22 − 4x̄32 = 3x̄2 1 − 2x̄2 − x̄22 .
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 + 3x̄21 − 4x̄31 = 3 1 − 2x̄1 − x̄21
⇐⇒ 0 < x̄1 ≤ 1, 2x̄31 − 3x̄21 − 3x̄1 + 1 = 0 .
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
0
1
Z
dx1
x1
Z
x1 dx2 +
0
1
Z
1
dx1
0
x2 dx2
x1
Z 1
1
(1 − x21 )
= +
dx1
3
2
0
1 1 1
2
= + − =
3 2 6
3
then
Z
1
2
f (x)dx ≤ f (1, 1),
A(D) D
3
and this estimate holds for every increasing radiant and integrable on D function f .
R EFERENCES
[1] S.S. DRAGOMIR, J. DUTTA AND A.M. RUBINOV, Hermite-Hadamard-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, BostonDordrecht-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.
[4] E.V. SHARIKOV, Increasing radiant functions, (submitted).
J. Inequal. Pure and Appl. Math., 4(2) Art. 47, 2003
http://jipam.vu.edu.au/