Khayyam J. Math. 1 (2015), no. 1, 115–124
SOME INTEGRAL INEQUALITIES FOR
α−, m−, (α, m) −LOGARITHMICALLY CONVEX FUNCTIONS
MEVLÜT TUNÇ1∗ , EBRU YÜKSEL2
Communicated by S. Hejazian
Abstract. In this paper, the authors establish some Hermite-Hadamard type
inequalities by using elementary inequalities for functions whose first derivative
absolute values are α-, m-, (α, m)-logarithmically convex.
1. Introduction and preliminaries
In this section, we will present definitions and some results used in this paper.
Let f : I ⊆ R → R be a convex mapping defined on the interval I of real
numbers and a, b ∈ I, with a < b. The following double inequalities:
Z b
a+b
f (a) + f (b)
1
f
f (x) dx ≤
≤
2
b−a a
2
hold. This double inequality is known in the literature as the Hermite-Hadamard
inequality for convex functions (see [1]-[8]).
Definition 1.1. Let I be an interval in R. Then f : I → R, ∅ =
6 I ⊆ R is said to
be convex if
f (tx + (1 − t) y) ≤ tf (x) + (1 − t) f (y) .
(1.1)
for all x, y ∈ I and t ∈ [0, 1].
The concepts of α-, m- and (α, m)-logarithmically convex functions were introduced as follows.
Date: Received: 12 November 2014; Revised: 15 December 2014; Accepted: 23 December
2014.
∗
Corresponding author.
2010 Mathematics Subject Classification. Primary 26A15; Secondary 26A51, 26D10.
Key words and phrases. α−, m−, (α, m) −logarithmically convex, Hadamard’s inequality,
Hölder’s inequality, power mean inequality, Cauchy’s inequality.
115
116
M. TUNÇ, E. YÜKSEL
Definition 1.2. [1] A function f : [0, b] → (0, ∞) is said to be m-logarithmically
convex if the inequality
f (tx + m (1 − t) y) ≤ [f (x)]t [f (y)]m(1−t)
(1.2)
holds for all x, y ∈ [0, b], m ∈ (0, 1], and t ∈ [0, 1].
Obviously, if putting m = 1 in Definition 1.2, then f is just the ordinary
logarithmically convex on [0, b].
Definition 1.3. [8] A function f : [0, b] → (0, ∞) is said to be α-logarithmically
convex if
α
α)
f (tx + (1 − t) y) ≤ [f (x)]t [f (y)](1−t
(1.3)
holds for all x, y ∈ [0, b], α ∈ (0, 1] and t ∈ [0, 1].
Clearly, when taking α = 1 in Definition 1.3, then f becomes the ordinary
logarithmically convex on [0, b].
Definition 1.4. [1] A function f : [0, b] → (0, ∞) is said to be (α, m)-logarithmically
convex if
α
α)
f (tx + m (1 − t) y) ≤ [f (x)]t [f (y)]m(1−t
(1.4)
holds for all x, y ∈ [0, b], (α, m) ∈ (0, 1] × (0, 1] , and t ∈ [0, 1].
Clearly, when taking α = 1 in Definition 1.4, then f becomes the standard mlogarithmically convex function on [0, b], and, when taking m = 1 in Definition
1.4, then f becomes the α-logarithmically convex function on [0, b].
In [3], the following theorem which was obtained by Dragomir and Agarwal
contains the Hermite-Hadamard type integral inequality.
Theorem 1.5. [3, Theorem 2.2] Let f : I ⊆ R → R be a differentiable mapping
on I ◦ , the interior of I, a, b ∈ I ◦ with a < b. If |f 0 (x) | is convex on [a, b], then
Z b
f (a) + f (b)
(b − a) (|f 0 (a)| + |f 0 (b)|)
1
≤
−
f
(x)
dx
.
(1.5)
2
b−a a
8
Theorem 1.6. [3, Theorem 2.3] Let f : I ⊆ R → R be a differentiable mapping
on I ◦ , a, b ∈ I ◦ with a < b, and p > 1. If the new mapping |f 0 (x) |p/p−1 is convex
on [a, b], then
Z b
f (a) + f (b)
1
−
f
(x)
dx
(1.6)
2
b−a a
"
#(p−1)/p
b−a
|f 0 (a)|p/(p−1) + |f 0 (b)|p/(p−1)
.
≤
2
2 (p + 1)1/p
The aim of this paper is to establish some integral inequalities of HermiteHadamard type for α-, m-, (α, m)-logarithmically convex functions.
INEQUALITIES FOR α−, m−, (α, m) −LOGARITHMICALLY CONVEX FUNCTIONS
117
2. Hadamard Type Inequalities
In order to prove our main theorems, we need the following lemma [7].
Lemma 2.1. [7] Let f : I ⊂ R → R be a differentiable mapping on I ◦ , a, b ∈ I ◦
with a < b. If f 0 ∈ L [a, b] , then the following equality holds:
Z b
f (a) + f (b)
1
−
f (x) dx
(2.1)
2
b−a a
Z Z
b−a 1 1 0
=
[f (ta + (1 − t) b) − f 0 (sa + (1 − s) b)] (s − t) dtds.
2
0
0
A simple proof of this equality can be also done integrating by parts in the
right hand side (see [7]).
The next theorems gives a new result of the upper Hermite-Hadamard inequality for α-, m-, (α, m)-logarithmically convex functions.
Theorem 2.2. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
0
0
differentiable function on I such that
f b ∈ L (a, b) for 0 ≤ a2< b < ∞. If |f (x)|
is (α, m)-logarithmically convex on 0, m for (α, m) ∈ (0, 1] , then
Z b
f (a) + f (b)
1
f
(x)
dx
(2.2)
−
2
b−a a
(
(b−a) 0 b m
f m
,
η=1
3
≤
(b−a) 0 b m −α2 ln2 η−2α ln η+2η α −2
f m
, η<1
2
α3 ln3 η
m
where η = |f 0 (a)| / f 0 mb .
Proof. By Lemma 2.1 and since |f 0 | is an (α, m)-logarithmically convex on 0, mb ,
then we have
Z b
f (a) + f (b)
1
f
(x)
dx
−
2
b−a a
Z 1Z 1
b−a
|(f 0 (ta + (1 − t) b)) − (f 0 (sa + (1 − s) b))| |s − t| dtds
≤
2
"0Z 0Z
#
m(1−tα )
1
1
α
b−a
b
t
≤
|s − t| |f 0 (a)| f 0
dtds
2
m
0
0
"Z Z
#
m(1−sα )
1
1
α
b
b−a
s
+
|s − t| |f 0 (a)| f 0
dtds
2
m 0
0
Let 0 < k ≤ 1, 0 ≤ m ≤ 1, and 0 < n ≤ 1. Then
n
k m ≤ k nm .
When η = 1, by (2.3), we get
Z b
f (a) + f (b)
1
−
f (x) dx
2
b−a a
(2.3)
118
M. TUNÇ, E. YÜKSEL
m Z 1 Z 1
Z 1Z 1
b − a 0 b |s − t| dtds +
|s − t| dtds
≤
f
2 m 0
0
0
0
m
b − a 0 b =
f
3 m When 0 < η < 1, by (2.3), we get
Z b
f (a) + f (b)
1
−
f (x) dx
2
b−a
m Z 1 Za 1
Z 1Z 1
b − a 0 b αt
αs
≤
f
|s − t| η dtds +
|s − t| η dtds
2 m 0
0
0
0
m b − a 0 b −α2 ln2 η − 2α ln η + 4η α + α2 η α ln2 η − 2αη α ln η − 4
f
=
2 m 2α3 ln3 η
−α ln η + 2η α − αη α ln η − 2
+
2α2 ln2 η
which completes the proof.
Corollary 2.3. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
0
0
differentiable function on I such
that
f ∈ L (a, b) for 0 ≤ a < b < ∞. If |f (x)|
b
is m-logarithmically convex on 0, m for m ∈ (0, 1], then
( (b−a) 0 b m
Z b
f (a) + f (b)
f m
,
η=1
1
3
−
f (x) dx ≤
(b−a) 0 b m − ln2 η−2 ln η+2η−2
f m
, η<1
2
b−a a
2
ln3 η
where η is same as Theorem 2.2.
Corollary 2.4. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
differentiable function on I such that f 0 ∈ L (a, b) for 0 ≤ a < b < ∞. If |f 0 (x)|is
α-logarithmically convex on [0, b] for α ∈ (0, 1] , then
( (b−a) 0
Z b
f (a) + f (b)
|f (b)| ,
η=1
1
3
≤
−
f
(x)
dx
(b−a)
4η α −4α ln η−2α2 ln2 η−4
0
|f (b)|
,
η<1
2
b−a a
2
2α3 ln3 η
where η = |f 0 (a)| / |f 0 (b)| .
Theorem 2.5. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
differentiable function on I such that f 0 ∈ L (a,
b) for 0 ≤ a < b < ∞. If |f 0 (x)|q
is an (α, m)-logarithmically convex on 0, mb for (α, m) ∈ (0, 1]2 and p, q > 1
with p1 + 1q = 1, then
Z b
f (a) + f (b)
1
f
(x)
dx
(2.4)
−
2
b−a a

p1

2
 (b − a) f 0 b m
,
η=1
m
(p+1)(p+2)
≤
1
1
m
p
q

2

(b − a) f 0 mb × η(αq,αq)−1
,
η<1
(p+1)(p+2)
ln η(αq,αq)
where η (α, α) is same as Theorem 2.2.
INEQUALITIES FOR α−, m−, (α, m) −LOGARITHMICALLY CONVEX FUNCTIONS
119
Proof. Since |f 0 |q is an (α, m)-logarithmically convex on 0, mb , from Lemma 2.1
and the well known Hölder inequality, we have
(2.5)
Z b
f (a) + f (b)
1
−
f
(x)
dx
2
b−a a
Z 1Z 1
b−a
≤
|(f 0 (ta + (1 − t) b)) − (f 0 (sa + (1 − s) b))| |s − t| dtds
2
0
0
m(1−tα )
Z 1Z 1
b−a
b tα 0
0
dtds
≤
|s − t| |f (a)| f
2
m 0
0
m(1−sα )
Z Z
b−a 1 1
b sα 0
0
+
dtds
|s − t| |f (a)| f
2
m 0
0
m Z 1 Z 1
p1
b − a 0 b p
≤
f
|s − t| dtds
2 m 0
0
"Z Z
1 #
1 Z Z
1
1
0
1
q
α
η qt dtds
×
1
q
α
η qs dtds
+
0
0
0
If η = 1, by (2.3), we obtain
Z b
f (a) + f (b)
1
f (x) dx
−
2
b−a a
m Z 1 Z 1
p1
0 b p
|s − t| dtds
≤ (b − a) f
m 0
0
m p1
0 b 2
= (b − a) f
m (p + 1) (p + 2)
If η < 1, by (2.3), we obtain
Z b
f (a) + f (b)
1
−
f (x) dx
2
b−a a
m Z 1 Z 1
p1
b − a 0 b p
≤
f
|s − t| dtds
2 m 0
0
"Z Z
1 Z Z
1
1
×
η
0
qtα
0
1
q
dtds
1
+
η
0
qsα
(2.6)
1q #
dtds
0
m p1 1q
0 b 2
η
(αq,
αq)
−
1
×
= (b − a) f
m (p + 1) (p + 2)
ln η (αq, αq)
which completes the proof.
Corollary 2.6. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
differentiable function on I such that f 0 ∈ L (a, b) for 0 ≤ a < b < ∞. If |f 0 (x)|q is
120
M. TUNÇ, E. YÜKSEL
an m-logarithmically convex on 0, mb for m ∈ (0, 1] and p = q = 2, then
Z b
f (a) + f (b)
1
−
f (x) dx
2
b−a a
(
m r
1,
η=1
0 b 1
12
≤ (b − a) f
×
η(2,2)−1
m 6
,
η<1
ln η(2,2)
Corollary 2.7. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
differentiable function on I such that f 0 ∈ L (a, b) for 0 ≤ a < b < ∞. If |f 0 (x)|is
α-logarithmically convex on [0, b] for α ∈ (0, 1] , then
Z b
f (a) + f (b)
1
−
f
(x)
dx
2
b−a a
p1 ( 1,
η=1
2
1q
×
≤ (b − a) |f 0 (b)|
η(αq,αq)−1
(p + 1) (p + 2)
,
η<1
ln η(αq,αq)
where η = |f 0 (a)| / |f 0 (b)| .
Theorem 2.8. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
q
0
0
differentiable function on I such that
fb∈
L (a, b) for 0 ≤ a 2< b < ∞. If |f (x)|
is (α, m)-logarithmically convex on 0, m for (α, m) ∈ (0, 1] , and then
Z b
b − a 0 b m
f (a) + f (b)
1
f
f (x) dx ≤
−
2
b−a a
3 m 
η=1
 1,
1q 1q 1
×
1−
ϕ+1
2ϕ−2
ϕ−1
q
− [ln
− 21−ϕ
− 2ϕ+1
+ [ln
η<1
 32 31
ln ϕ
ln ϕ
[ln ϕ]3
ϕ]2
ϕ]2
where η (α, α) is same as Theorem 2.2, and ϕ = η (αq, αq).
Proof. Since |f 0 |q is an (α, m)-logarithmically convex on 0, mb , for q ≥ 1, from
Lemma 2.1 and the well known power mean integral inequality, we get
Z b
f (a) + f (b)
1
−
f
(x)
dx
2
b−a a
Z 1Z 1
b−a
≤
|(f 0 (ta + (1 − t) b)) − (f 0 (sa + (1 − s) b))| |s − t| dtds
2
0
0
Z 1 Z 1
1− 1q Z 1 Z 1
1q
b−a
q
0
|s − t| |f (ta + (1 − t) b)| dtds
≤
|s − t| dtds
2
0
0
0
0
1q
Z 1 Z 1
1− 1q Z 1 Z 1
b−a
q
+
|s − t| dtds
|s − t| |f 0 (sa + (1 − s) b)| dtds
2
0
0
0
0
m Z 1 Z 1
1− 1q Z 1 Z 1
1q
b − a 0 b qtα
≤
f
|s − t| dtds
|s − t| η dtds
2 m 0
0
0
0
m Z 1 Z 1
1− 1q Z 1 Z 1
1q
b − a 0 b α
+
f
|s − t| dtds
|s − t| η qs dtds
2 m 0
0
0
0
INEQUALITIES FOR α−, m−, (α, m) −LOGARITHMICALLY CONVEX FUNCTIONS
121
When η = 1, by (2.3), we obtain
Z b
f (a) + f (b)
1
−
f
(x)
dx
2
b−a a
1q
1− 1q m Z 1 Z 1
0 b b−a 1
|s − t| dtds
≤
f m 2
3
0
0
1− 1q m Z 1 Z 1
1q
0 b b−a 1
f
|s − t| dtds
+
2
3
m 0
0
m
b − a 0 b =
f
3 m When η < 1, by (2.3), we obtain
Z b
f (a) + f (b)
1
−
f
(x)
dx
2
b−a a
1q
1− 1q m Z 1 Z 1
0 b b−a 1
αqt
f
|s − t| η dtds
≤
2
3
m 0
0
1q
1− 1q m Z 1 Z 1
0 b b−a 1
f
|s − t| η αqs dtds
+
2
3
m 0
0
1− 1q m
0 b b−a 1
f
=
2
3
m (
1
2η (αq, αq) − 2
η (αq, αq) + 1
1 − η (αq, αq) q
×
−
−
[ln (η (αq, αq))]3 [ln (η (αq, αq))]2 2 ln (η (αq, αq))
1 )
η (αq, αq) − 1
η (αq, αq) + 1 q
+
−
,
[ln (η (αq, αq))]2 2 ln (η (αq, αq))
which completes the proof.
Corollary 2.9. Let I ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
q
0
0
differentiable function on I such
f ∈ L (a, b) for 0 ≤ a < b < ∞. If |f (x)| is
that
b
m-logarithmically convex on 0, m for m ∈ (0, 1], then
Z b
f (a) + f (b)
1
f
(x)
dx
−
2
b−a a

b−a 0 b m
,
η=1
f m

3


h
i 1q

1  (b−a) 1 1− q 0 b m
2η(q,q)−2
η(q,q)+1
f m
− [ln
− 21−η(q,q)
2
3
ln η(q,q)
[ln η(q,q)]3
η(q,q)]2
≤
, η<1

h
i 1q 

η(q,q)−1
η(q,q)+1

+ [ln η(q,q)]2 − 2 ln η(q,q)

Corollary 2.10. LetI ⊃ [0, ∞) be an open interval and let f : I → (0, ∞) be a
differentiable function on I such that f 0 ∈ L (a, b) for 0 ≤ a < b < ∞. If |f 0 (x)|
122
M. TUNÇ, E. YÜKSEL
is α-logarithmically convex on [0, b] for α ∈ (0, 1] , then
Z b
b−a 0
f (a) + f (b)
1
f (x) dx ≤
−
|f (b)|
2
b−a a
3

1, η=1



i 1q
h
1

 3 1 1− q
η(αq,αq)+1
1−η(αq,αq)
2η(αq,αq)−2
− [ln(η(αq,αq))]
2 − 2 ln(η(αq,αq))
2 3
[ln(η(αq,αq))]3
≤
, η<1

i 1q h


η(αq,αq)−1
η(αq,αq)+1

+ [ln(η(αq,αq))]2 − 2 ln(η(αq,αq))

where η = |f 0 (a)| / |f 0 (b)| .
Theorem 2.11. Let f : I ⊂ R+ → R+ be differentiable on I ◦ , a, b ∈ I, with
a < b and f 0 ∈ L ([a, b]) . If |f 0 | is an (α, m)-logarithmically convex 0, mb for
(α, m) ∈ (0, 1]2 and µ1 , µ2 , τ1 , τ2 > 0 with µ1 + τ1 = 1 and µ2 + τ2 = 1, then
Z b
f (a) + f (b)
(b − a) 0 b m
1
f
−
f (x) dx ≤
(2.7)
2
b−a a
2 m 
2µ31
2µ32

+
+ τ1 + τ2 ,
η=1
 (2µ1 +1)(µ
(2µ2 +1)(µ2 +1)
1 +1)
α α
α α
×
3
3
η
η
,
−1
,
−1
2µ1
2µ2

+ (2µ2 +1)(µ
+ τ1 τ1 ατ1 α + τ2 τ2 ατ2 α , η < 1
 (2µ1 +1)(µ
1 +1)
2 +1)
ln η
,
ln η
τ1 τ1
,
τ2 τ2
where η (α, α) is same as Theorem 2.2.
Proof. Since |f 0 |q is an (α, m)-logarithmically convex on 0, mb , from Lemma 2.1,
we have
Z b
f (a) + f (b)
1
f
(x)
dx
(2.8)
−
2
b−a a
Z Z
(b − a) 1 1 0
≤
|(f (ta + (1 − t) b)) − (f 0 (sa + (1 − s) b))| |s − t| dtds
2
"0Z 0Z
#
m(1−tα )
1
1
α
(b − a)
b
t
≤
|s − t| |f 0 (a)| f 0
dtds
2
m 0
0
"Z Z
#
m(1−sα )
1
1
α
b
(b − a)
s
|s − t| |f 0 (a)| f 0
dtds
+
2
m 0
0
m Z 1 Z 1
Z 1Z 1
(b − a) 0 b tα
sα
=
f
|s − t| η dtds +
|s − t| η dtds
2 m 0
0
0
0
1
1
for all t ∈ [0, 1] . Using the well known inequality rt ≤ µr µ + τ t τ , on the right
side of (2.8), we get
Z 1Z 1
Z 1Z 1
α
tα
|s − t| η dtds +
|s − t| η s dtds
(2.9)
0
0
0
0
Z 1Z 1 α
Z 1Z 1
1
t
µ1
≤ µ1
|s − t| dtds + τ1
η τ1 dtds
0
0
0
0
INEQUALITIES FOR α−, m−, (α, m) −LOGARITHMICALLY CONVEX FUNCTIONS
Z
1
Z
1
|s − t|
+µ2
0
1
µ2
Z
≤
Z
1
dtds + τ2
0
1
α
|s − t| η s dtds
(2.10)
0
0
0
0
1
Z
sα
η τ2 dtds
0
When η = 1, by (2.3), we get
Z
Z 1Z 1
tα
|s − t| η dtds +
0
1
123
2µ31
2µ32
+
+ τ1 + τ2
(2µ1 + 1) (µ1 + 1) (2µ2 + 1) (µ2 + 1)
When η < 1, by (2.3), we get
(2.11)
Z
1
Z
1
α
Z
1
Z
1
α
|s − t| η t dtds +
|s − t| η s dtds
0
0
0
0
Z 1Z 1
Z 1Z 1 α
1
t
|s − t| µ1 dtds + τ1
≤ µ1
η τ1 dtds
0
0
0
0
Z 1Z 1 α
Z 1Z 1
s
1
η τ2 dtds
+µ2
|s − t| µ2 dtds + τ2
0
0
0
0
Z 1Z 1
Z 1Z 1
1
1
|s − t| µ2 dtds
≤ µ1
|s − t| µ1 dtds + µ2
0
0
0
0
Z 1Z 1
Z 1Z 1
αt
αs
+τ1
η τ1 dtds + τ2
η τ2 dtds
0
0
0
0
2µ31
2µ32
=
+
(2µ1 + 1) (µ1 + 1) (2µ2 + 1) (µ2 + 1)
α α
α α
η τ1 , τ1 − 1
η τ2 , τ2 − 1
+ τ2
+τ1
α α
α α
ln η τ1 , τ1
ln η τ2 , τ2
from (2.8)-(2.11), which completes the proof.
Corollary 2.12. Under the assumptions of Theorem 2.11, and µ = µ1 = µ2 > 0,
τ = τ1 = τ2 > 0 with µ + τ = 1, then we have
Z b
f (a) + f (b)
1
−
f
(x)
dx
2
b−a a

4µ3
m 
+ 2τ,
η=1
(2µ+1)(µ+1)
(b − a) 0 b α α
≤
f
×
3
η
,
−1
)
(
4µ
 (2µ+1)(µ+1)
2 m + 2τ ln ητ ατ , α , η < 1
(τ τ )
References
1. R.F. Bai, F. Qi, B.Y. Xi, Hermite-Hadamard type inequalities for the m- and (α, m)logarithmically convex functions. Filomat, 27 (2013), 1–7.
2. S.S. Dragomir, C.E.M. Pearce, Selected Topics on Hermite-Hadamard Inequalities and Applications, RGMIA monographs, Victoria University, 2000. [Online:
http://www.staff.vu.edu.au/RGMIA/monographs/hermite-hadamard.html].
124
M. TUNÇ, E. YÜKSEL
3. S.S. Dragomir, R.P. Agarwal, Two inequalities for differantiable mappings and applications
to special means of real numbers and trapezoidal formula, Appl. Math. Lett., 11(5) (1998),
91-95.
4. J. Hadamard, Étude sur les propriétés des fonctions entières et en particulier d’une fonction
considerée par Riemann, J. Math Pures Appl., 58, (1893) 171–215.
5. D.S. Mitrinović, J. Pečarić, A.M. Fink, Classical and New Inequalities in Analysis, KluwerAcademic, Dordrecht, 1993.
6. J.E. Pečarić, F. Proschan, Y.L. Tong, Convex Functions, Partial Orderings, and Statistical
Applications, Academic Press Inc., 1992.
7. M.Z. Sarıkaya, E. Set, M.E. Özdemir, New inequalities of Hermite-Hadamard Type, Volume
12, Issue 4, 2009, Art.11, RGMIA Online: http://rgmia.org/papers/v12n4/set2.pdf
8. M. Tunç, E. Yüksel, İ. Karabayır, On some inequalities for functions whose second derivetives absolute values are α-, m-, (α, m)-logarithmically convex, Georgian Mathematical
Journal, Accepted.
1
Department of Mathematics, Faculty of Science and Arts, Mustafa Kemal
University, Hatay, 31000, Turkey.
E-mail address: mevluttttunc@gmail.com
2
Department of Mathematics, Faculty of Science and Arts, Ağrı İbrahim
Çeçen University, Ağrı, 04000, Turkey.
E-mail address: yuksel.ebru90@hotmail.com