Volume 9 (2008), Issue 4, Article 96, 12 pp.
HADAMARD TYPE INEQUALITIES FOR m-CONVEX AND (α, m)-CONVEX
FUNCTIONS
M. KLARIČIĆ BAKULA, M. E. ÖZDEMIR, AND J. PEČARIĆ
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE
U NIVERSITY OF S PLIT
T ESLINA 12, 21000 S PLIT
C ROATIA
milica@pmfst.hr
ATATÜRK U NIVERSITY
K. K. E DUCATION FACULTY
D EPARTMENT OF M ATHEMATICS
25240 K AMPÜS , E RZURUM
T URKEY
emos@atauni.edu.tr
FACULTY OF T EXTILE T ECHNOLOGY
U NIVERSITY OF Z AGREB
P IEROTTIJEVA 6, 10000 Z AGREB
C ROATIA
pecaric@hazu.hr
Received 03 March, 2008; accepted 31 July, 2008
Communicated by E. Neuman
A BSTRACT. In this paper we establish several Hadamard type inequalities for differentiable mconvex and (α, m)-convex functions. We also establish Hadamard type inequalities for products
of two m-convex or (α, m)-convex functions. Our results generalize some results of B.G. Pachpatte as well as some results of C.E.M. Pearce and J. Pečarić.
Key words and phrases: m-convex functions, (α, m)-convex functions, Hadamard’s inequalities.
2000 Mathematics Subject Classification. 26D15, 26A51.
1. I NTRODUCTION
The following definitions are well known in literature.
Let [0, b] , where b is greater than 0, be an interval of the real line R, and let K (b) denote the
class of all functions f : [0, b] → R which are continuous and nonnegative on [0, b] and such
that f (0) = 0.
We say that the function f is convex on [0, b] if
f (tx + (1 − t) y) ≤ tf (x) + (1 − t) f (y)
062-08
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M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR ,
AND
J. P E ČARI Ć
holds for all x, y ∈ [0, b] and t ∈ [0, 1] . Let KC (b) denote the class of all functions f ∈ K (b)
convex on [0, b] , and let KF (b) be the class of all functions f ∈ K (b) convex in mean on [0, b] ,
that is, the class of all functions f ∈ K (b) for which F ∈ KC (b) , where the mean function F
of the function f ∈ K (b) is defined by
( 1 Rx
f (t) dt, x ∈ (0, b] ;
x 0
F (x) =
0,
x = 0.
Let KS (b) denote the class of all functions f ∈ K (b) which are starshaped with respect to
the origin on [0, b] , that is, the class of all functions f with the property that
f (tx) ≤ tf (x)
holds for all x ∈ [0, b] and t ∈ [0, 1] . In [1] Bruckner and Ostrow, among others, proved that
KC (b) ⊂ KF (b) ⊂ KS (b) .
In [9] G. Toader defined m-convexity: another intermediate between the usual convexity and
starshaped convexity.
Definition 1.1. The function f : [0, b] → R, b > 0, is said to be m-convex, where m ∈ [0, 1] ,
if we have
f (tx + m (1 − t) y) ≤ tf (x) + m (1 − t) f (y)
for all x, y ∈ [0, b] and t ∈ [0, 1] . We say that f is m-concave if −f is m-convex.
Denote by Km (b) the class of all m-convex functions on [0, b] for which f (0) ≤ 0.
Obviously, for m = 1 Definition 1.1 recaptures the concept of standard convex functions on
[0, b] , and for m = 0 the concept of starshaped functions.
The following lemmas hold (see [10]).
Lemma A. If f is in the class Km (b) , then it is starshaped.
Lemma B. If f is in the class Km (b) and 0 < n < m ≤ 1, then f is in the class Kn (b).
From Lemma A and Lemma B it follows that
K1 (b) ⊂ Km (b) ⊂ K0 (b) ,
whenever m ∈ (0, 1) . Note that in the class K1 (b) are only convex functions f : [0, b] → R for
which f (0) ≤ 0, that is, K1 (b) is a proper subclass of the class of convex functions on [0, b] .
It is interesting to point out that for any m ∈ (0, 1) there are continuous and differentiable
functions which are m-convex, but which are not convex in the standard sense (see [11]).
In [3] S.S. Dragomir and G. Toader proved the following Hadamard type inequality for mconvex functions.
Theorem A. Let f : [0, ∞) → R be an m-convex function with m ∈ (0, 1] . If 0 ≤ a < b < ∞
and f ∈ L1 ([a, b]) then
(
)
Z b
b
a
f
(a)
+
mf
f
(b)
+
mf
1
m
m
(1.1)
f (x)dx ≤ min
,
.
b−a a
2
2
Some generalizations of this result can be found in [4].
The notion of m-convexity has been further generalized in [5] as it is stated in the following
definition:
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
I NEQUALITIES FOR m- CONVEX AND (α, m)- CONVEX
3
FUNCTIONS
Definition 1.2. The function f : [0, b] → R, b > 0, is said to be (α, m)-convex, where (α, m) ∈
[0, 1]2 , if we have
f (tx + m (1 − t) y) ≤ tα f (x) + m (1 − tα ) f (y)
for all x, y ∈ [0, b] and t ∈ [0, 1] .
α
Denote by Km
(b) the class of all (α, m)-convex functions on [0, b] for which f (0) ≤ 0.
It can be easily seen that for (α, m) ∈ {(0, 0) , (α, 0) , (1, 0) , (1, m) , (1, 1) , (α, 1)} one obtains the following classes of functions: increasing, α-starshaped, starshaped, m-convex, convex and α-convex functions respectively. Note that in the class K11 (b) are only convex functions
f : [0, b] → R for which f (0) ≤ 0, that is K11 (b) is a proper subclass of the class of all convex
functions on [0, b] . The interested reader can find more about partial ordering of convexity in
[8, p. 8, 280].
In [2] in order to prove some inequalities related to Hadamard’s inequality S. S. Dragomir
and R. P. Agarwal used the following lemma.
Lemma C. Let f : I → R, I ⊂ R, be a differentiable mapping on I̊, and a, b ∈ I, where a < b.
If f 0 ∈ L1 ([a, b]) , then
(1.2)
1
f (a) + f (b)
−
2
b−a
Z
a
b
b−a
f (x)dx =
2
Z
1
(1 − 2t) f 0 (ta + (1 − t) b)dt.
0
Here I̊ denotes the interior of I.
In [7], using the same Lemma C, C.E.M. Pearce and J. Pečarić proved the following theorem.
Theorem B. Let f : I → R, I ⊂ R, be a differentiable mapping on I 0 , and a, b ∈ I, where
a < b. If |f 0 |q is convex on [a, b] for some q ≥ 1, then
1
Z b
f (a) + f (b)
b − a |f 0 (a)|q + |f 0 (b)|q q
1
−
f (x)dx ≤
2
b−a a
4
2
and
1
Z b
b − a |f 0 (a)|q + |f 0 (b)|q q
a
+
b
1
f
.
−
f (x)dx ≤
2
b−a a
4
2
In [6] B. G. Pachpatte established two new Hadamard type inequalities for products of convex
functions. They are given in the next theorem.
Theorem C. Let f, g : [a, b] → [0, ∞) be convex functions on [a, b] ⊂ R, where a < b. Then
(1.3)
1
b−a
Z
a
b
1
1
f (x) g (x) dx ≤ M (a, b) + N (a, b) ,
3
6
where M (a, b) = f (a) g (a) + f (b) g (b) and N (a, b) = f (a) g (b) + f (b) g (a).
The main purpose of this paper is to establish new inequalities like those given in Theorems
A, B and C, but now for the classes of m-convex functions (Section 2) and (α, m)-convex
functions (Section 3).
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
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M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR ,
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J. P E ČARI Ć
2. I NEQUALITIES FOR m-C ONVEX F UNCTIONS
Theorem 2.1. Let I be an open real interval such that [0, ∞) ⊂ I. Let f : I → R be a
differentiable function on I such that f 0 ∈ L1 ([a, b]) , where 0 ≤ a < b < ∞. If |f 0 |q is
m-convex on [a, b] for some fixed m ∈ (0, 1] and q ∈ [1, ∞) , then
Z b
f (a) + f (b)
1
−
f
(x)dx
2
b−a a

 |f 0 (a)|q + m f 0
b−a
≤
min

2
4
b
m
q ! 1q
,
m f 0
a
m
!1 
q
+ |f 0 (b)|q q 
2
.

Proof. Suppose that q = 1. From Lemma C we have
(2.1)
Z b
Z
b−a 1
f (a) + f (b)
1
−
f (x)dx ≤
|1 − 2t| |f 0 (ta + (1 − t) b)| dt.
2
b−a a
2
0
Since |f 0 | is m-convex on [a, b] we know that for any t ∈ [0, 1]
0
b |f (ta + (1 − t) b)| = f ta + m (1 − t)
m b 0
,
≤ t |f (a)| + m (1 − t) f 0
m 0
hence
Z b
f (a) + f (b)
1
−
f
(x)dx
2
b−a a
Z 1
b−a
b 0
≤
|1 − 2t| t |f (a)| + m (1 − t) f 0
dt
2
m 0
Z b−a 1
b 0
=
t |1 − 2t| |f (a)| + m (1 − t) |1 − 2t| f 0
dt
2
m 0
(Z 1 2
b−a
b 0
=
t (1 − 2t) |f (a)| + m (1 − t) (1 − 2t) f 0
dt
2
m 0
)
Z 1
b +
t (2t − 1) |f 0 (a)| + m (1 − t) (2t − 1) f 0
dt
1
m 2
b b−a
0
=
|f (a)| + m f 0
.
8
m Analogously we obtain
Z b
f (a) + f (b)
b − a 0 a 1
0
≤
−
f
(x)dx
m
f
+
|f
(b)|
,
2
b−a a
8
m
which completes the proof for this case.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
I NEQUALITIES FOR m- CONVEX AND (α, m)- CONVEX
5
FUNCTIONS
Suppose now that q > 1. Using the well known Hölder inequality for q and p = q/ (q − 1)
we obtain
Z 1
|1 − 2t| |f 0 (ta + (1 − t) b)| dt
0
Z 1
1
1
=
|1 − 2t|1− q |1 − 2t| q |f 0 (ta + (1 − t) b)| dt
0
Z
≤
(2.2)
1
q−1
Z
q
|1 − 2t| dt
1
1q
|1 − 2t| |f 0 (ta + (1 − t) b)| dt .
q
0
0
0 q
Since |f | is m-convex on [a, b] we know that for every t ∈ [0, 1]
q
b q
q
0
0
(2.3)
|f (ta + (1 − t) b)| ≤ t |f (a)| + m (1 − t) f 0
,
m hence from (2.1) , (2.2) and (2.3) we have
Z b
f (a) + f (b)
1
−
f
(x)dx
2
b−a a
Z 1
q−1
q 1q
Z 1
q
0
b−a
b
dt
≤
|1 − 2t| dt
|1 − 2t| f ta + m (1 − t)
2
m 0
0
q 1
Z 1
q−1
q
q
b−a
b 1
q
|f 0 (a)| + m f 0
≤
|1 − 2t| dt
2
4
m 0
q ! 1q
q
b − a m |f 0 (a)| + m f 0 mb =
2
4
and analogously
Z b
f (a) + f (b)
b−a
1
≤
−
f
(x)dx
2
b−a a
4
m f 0
a
m
!1
q
+ |f 0 (b)|q q
2
,
which completes the proof.
Theorem 2.2. Suppose that all the assumptions of Theorem 2.1 are satisfied. Then
Z b
f a + b − 1
f
(x)dx
2
b−a a

! 1q 
0 b q ! 1q
0 a q
q
q


0
0
|f (a)| + m f m
m f m + |f (b)|
b−a
,
.
≤
min


4
2
2
Proof. Our starting point here is the identity (see [7, Theorem 2])
Z b
Z b
a+b
1
1
f
−
f (x)dx =
S (x) f 0 (x)dx,
2
b−a a
b−a a
where

 x − a, x ∈ a, a+b
;
2
S (x) =
 x − b, x ∈ a+b , b .
2
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
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M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR ,
AND
J. P E ČARI Ć
We have
Z b
f a + b − 1
f (x)dx
2
b−a a
"Z a+b
#
Z b
2
1
(x − a) |f 0 (x)| dx +
(b − x) |f 0 (x)| dx
≤
a+b
b−a a
2
#
"Z 1
Z 1
2
t |f 0 (ta + (1 − t) b)| dt +
(1 − t) |f 0 (ta + (1 − t) b)| dt
= (b − a)
1
2
0
"Z
≤ (b − a)
1
2
0
b t t |f (a)| + m (1 − t) f 0
dt
m 0
#
b +
dt
(1 − t) t |f (a)| + m (1 − t) f 0
1
m 2
b−a
b 0
=
|f (a)| + m f 0
,
8
m Z
1
0
and analogously
Z b
b − a 0 a a
+
b
1
0
f
≤
−
f
(x)dx
m
f
+
|f
(b)|
.
2
b−a a
8
m
This completes the proof for the case q = 1.
An argument similar to the one used in the proof of Theorem 2.1 gives the proof for the case
q ∈ (1, ∞) .
As a special case of Theorem 2.1 for m = 1, that is for |f 0 |q convex on [a, b] , we obtain the
first inequality in Theorem B. Similarly, as a special case of Theorem 2.2 we obtain the second
inequality in Theorem B.
Theorem 2.3. Let I be an open real interval such that [0, ∞) ⊂ I. Let f : I → R be a
differentiable function on I such that f 0 ∈ L1 ([a, b]) , where 0 ≤ a < b < ∞. If |f 0 |q is
m-convex on [a, b] for some fixed m ∈ (0, 1] and q ∈ (1, ∞) , then
(2.4)
q−1
1
Z b
1
q
f (a) + f (b)
b−a q−1
1
q
q
−
f (x)dx ≤
µ1 + µ2
2
b−a a
4
2q − 1
1
1
b−a
q
q
≤
µ1 + µ2 ,
4
where
(
µ1 = min
(
µ2 = min
|f 0 (a)|q + m f 0
2
a+b
2m
|f 0 (b)|q + m f 0
2
a+b
2m
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
q 0
f
,
a+b
2
q 0
f
,
a+b
2
q
+ m f 0
a
m
q )
b
m
q )
2
q
+ m f 0
2
,
.
http://jipam.vu.edu.au/
I NEQUALITIES FOR m- CONVEX AND (α, m)- CONVEX
7
FUNCTIONS
Proof. If |f 0 |q is m-convex from Theorem A we have
Z 1
q
2
|f 0 (ta + (1 − t) b)| dt ≤ µ1 ,
Z
1
2
1
2
2
q
|f 0 (ta + (1 − t) b)| dt ≤ µ2 ,
0
hence
Z b
f (a) + f (b)
1
−
f
(x)dx
2
b−a a
Z 1
b−a
≤
|1 − 2t| |f 0 (ta + (1 − t))| dt
2
#
"0Z 1
Z 1
2
b−a
(2t − 1) |f 0 (ta + (1 − t) b)| dt .
=
(1 − 2t) |f 0 (ta + (1 − t) b)| dt +
1
2
0
2
Using Hölder’s inequality for q ∈ (1, ∞) and p = q/ (q − 1) we obtain
Z b
f (a) + f (b)
1
−
f
(x)dx
2
b−a a

! q−1
! 1q
Z 1
Z 1
q
2
2
q
b−a
q
≤
(1 − 2t) q−1 dt
|f 0 (ta + (1 − t) b)| dt
2
0
0
1
Z
! q−1
q
q
1
2
b−a
≤
4
since
Z
q−1
2q − 1
1
2
q
|f 0 (ta + (1 − t) b)| dt
(2t − 1) q−1 dt
+
! 1q 
1
Z
1
2
q−1
q
(1 − 2t)
1
q
1
q
µ1 + µ2
q
q−1
1
Z
0
,
q
(2t − 1) q−1 dt =
dt =
1
2

q−1
.
2 (2q − 1)
This completes the proof of the first inequality in (2.4) . The second inequality in (2.4) follows
from the fact
q−1
q
1
q−1
<
< 1,
q ∈ (1, ∞) .
2
2q − 1
Theorem 2.4. Let f, g : [0, ∞) → [0, ∞) be such that f g is in L1 ([a, b]), where 0 ≤ a < b <
∞. If f is m1 -convex and g is m2 -convex on [a, b] for some fixed m1 , m2 ∈ (0, 1] , then
Z b
1
f (x)g (x) dx ≤ min {M1 , M2 } ,
b−a a
where
1
b
b
M1 =
f (a) g (a) + m1 m2 f
g
3
m1
m2
1
b
b
+
m2 f (a) g
+ m1 f
g (a) ,
6
m2
m1
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
8
M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR ,
AND
J. P E ČARI Ć
1
a
a
M2 =
g
f (b) g (b) + m1 m2 f
3
m1
m2
1
a
a
+
m2 f (b) g
+ m1 f
g (b) .
6
m2
m1
Proof. We have
b
b
f ta + m1 (1 − t)
≤ tf (a) + m1 (1 − t) f
,
m1
m1
b
b
g ta + m2 (1 − t)
≤ tg(a) + m2 (1 − t) g
,
m2
m2
for all t ∈ [0, 1] . f and g are nonnegative, hence
b
b
f ta + m1 (1 − t)
g ta + m2 (1 − t)
m1
m2
b
b
2
≤ t f (a) g(a) + m2 t (1 − t) f (a) g
+ m1 t (1 − t) f
g(a)
m2
m1
b
b
2
+ m1 m2 (1 − t) f
g
.
m1
m2
Integrating both sides of the above inequality over [0, 1] we obtain
Z 1
f (ta + (1 − t) b)g (ta + (1 − t) b) dt
0
Z b
1
f (x)g (x) dx
=
b−a a
1
b
b
≤
f (a) g (a) + m1 m2 f
g
3
m1
m2
1
b
b
+
m2 f (a) g
+ m1 f
g (a) .
6
m2
m1
Analogously we obtain
Z b
1
1
a
a
f (x)g (x) dx ≤
f (b) g (b) + m1 m2 f
g
b−a a
3
m1
m2
1
a
a
+
m2 f (b) g
+ m1 f
g (b) ,
6
m2
m1
hence
1
b−a
Z
b
f (x)g (x) dx ≤ min {M1 , M2 } .
a
Remark 1. If in Theorem 2.4 we choose a 1-convex (convex) function g : [0, ∞) → [0, ∞)
defined by g (x) = 1 for all x ∈ [0, ∞), we obtain
(
)
Z b
a
f (a) + mf mb f (b) + mf m
1
f (x) dx ≤ min
,
,
b−a a
2
2
which is (1.1) . If the functions f and g are 1-convex we obtain (1.3) .
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
I NEQUALITIES FOR m- CONVEX AND (α, m)- CONVEX
9
FUNCTIONS
3. I NEQUALITIES FOR (α, m)-C ONVEX F UNCTIONS
In this section on two examples we ilustrate how the same inequalities as in Section 2 can be
obtained for the class of (α, m)-convex functions.
Theorem 3.1. Let I be an open real interval such that [0, ∞) ⊂ I. Let f : I → R be a
differentiable function on I such that f 0 ∈ L1 ([a, b]) , where 0 ≤ a < b < ∞. If |f 0 |q is
(α, m)-convex on [a, b] for some fixed α, m ∈ (0, 1] and q ∈ [1, ∞) , then
Z b
f (a) + f (b)
1
−
f
(x)dx
2
b−a a
(
q 1
q−1
b−a 1 q
b q
q
0
,
· min
ν1 |f (a)| + ν2 m f 0
≤
m 2
2
a q 1 q
q
0
ν1 |f (b)| + ν2 m f 0
,
m
where
α 1
1
α+
,
ν1 =
(α + 1) (α + 2)
2
2
α α +α+2
1
1
ν2 =
−
.
(α + 1) (α + 2)
2
2
Proof. Suppose that q = 1. From Lemma A we have
Z b
Z
f (a) + f (b)
b−a 1
1
(3.1)
−
f (x)dx ≤
|1 − 2t| |f 0 (ta + (1 − t) b)| dt.
2
b−a a
2
0
Since |f 0 | is (α, m)-convex on [a, b] we know that for any t ∈ [0, 1]
0
b
α
0
α
≤ t |f (a)| + m (1 − t ) f 0 b ,
f ta + m (1 − t)
m m thus we have
Z b
f (a) + f (b)
1
−
f (x)dx
2
b−a a
Z 1
b−a
b α
0
α 0
≤
|1 − 2t| t |f (a)| + m (1 − t ) f
dt
2
m 0
Z b−a 1 α
b 0
α
=
t |1 − 2t| |f (a)| + m (1 − t ) |1 − 2t| f 0
dt.
2
m 0
We have
α 1
1
t |1 − 2t| dt =
α+
= ν1 ,
(α + 1) (α + 2)
2
0
2
α Z 1
1
α +α+2
1
α
(1 − t ) |1 − 2t| dt =
−
= ν2 ,
(α + 1) (α + 2)
2
2
0
Z
1
α
hence
Z b
f (a) + f (b)
b−a
0 b 1
0
f
.
−
f
(x)dx
≤
ν
|f
(a)|
+
ν
m
1
2
2
b−a a
2
m J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
10
M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR , AND J. P E ČARI Ć
Analogously we obtain
Z b
a f (a) + f (b)
b−a
1
0
−
ν1 |f (b)| + ν2 m f 0
f (x)dx ≤
,
2
b−a a
2
m
which completes the proof for this case.
Suppose now that q ∈ (1, ∞) . Similarly to Theorem 2.1 we have
Z 1
(3.2)
|1 − 2t| |f 0 (ta + (1 − t) b)| dt
0
Z
≤
1
Z
q−1
q
|1 − 2t| dt
0
1
1q
|1 − 2t| |f 0 (ta + (1 − t) b)| dt .
q
0
0 q
Since |f | is (α, m)-convex on [a, b] we know that for every t ∈ [0, 1]
q
q
0
b
q
α
0
α
≤ t |f (a)| + m (1 − t ) f 0 b ,
f ta + m (1 − t)
(3.3)
m
m hence from (3.1) , (3.2) and (3.3) we obtain
Z b
f (a) + f (b)
1
−
f (x)dx
2
b−a a
Z 1
q−1
Z 1
q 1q
q
0
b−a
b
dt
≤
|1 − 2t| dt
|1 − 2t| f ta + m (1 − t)
2
m
0
0
q 1
q−1 b−a 1 q
b q
q
0
≤
ν1 |f (a)| + ν2 m f 0
2
2
m and analogously
q−1
Z b
a q 1
f (a) + f (b)
b−a 1 q 1
0
q
q
0
≤
−
f
(x)dx
ν
|f
(b)|
+
ν
m
,
f
1
2
2
b−a a
2
2
m
which completes the proof.
Observe that if in Theorem 3.1 we have α = 1 the statement of Theorem 3.1 becomes the
statement of Theorem 2.1.
Theorem 3.2. Let f, g : [0, ∞) → [0, ∞) be such that f g is in L1 ([a, b]), where 0 ≤ a < b <
∞. If f is (α1 , m1 )-convex and g is (α2 , m2 )-convex on [a, b] for some fixed α1 , m1 , α2 , m2 ∈
(0, 1] , then
Z b
1
f (x)g (x) dx ≤ min {N1 , N2 } ,
b−a a
where
f (a) g (a)
1
1
b
N1 =
+ m2
−
f (a) g
α1 + α2 + 1
α1 + 1 α1 + α2 + 1
m2
1
1
b
+ m1
−
f
g(a)
α2 + 1 α1 + α2 + 1
m1
1
1
1
b
b
+ m1 m2 1 −
−
+
f
g
,
α1 + 1 α2 + 1 α1 + α2 + 1
m1
m2
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
I NEQUALITIES FOR m- CONVEX AND (α, m)- CONVEX FUNCTIONS
11
and
f (b) g (b)
1
1
a
N2 =
+ m2
−
f (b) g
α1 + α2 + 1
α1 + 1 α1 + α2 + 1
m2
1
a
1
−
f
g(b)
+ m1
α2 + 1 α1 + α2 + 1
m1
1
1
1
a
a
+ m1 m2 1 −
−
+
f
g
.
α1 + 1 α2 + 1 α1 + α2 + 1
m1
m2
Proof. Since f is (α1 , m1 )-convex and g is (α2 , m2 )-convex on [a, b] we have
b
b
α1
α1
f ta + m1 (1 − t)
≤ t f (a) + m1 (1 − t ) f
,
m1
m1
b
b
α2
α2
g ta + m2 (1 − t)
≤ t g(a) + m2 (1 − t ) g
,
m2
m2
for all t ∈ [0, 1] . The functions f and g are nonnegative, hence
f (ta + (1 − t) b)g(ta + (1 − t) b) ≤ tα1 +α2 f (a) g(a)
b
b
α1
α2
α2
α1
+ m2 t (1 − t ) f (a) g
+ m1 t (1 − t ) f
g(a)
m2
m1
b
b
α1
α2
+ m1 m2 (1 − t ) (1 − t ) f
g
.
m1
m2
Integrating both sides of the above inequality over [0, 1] we obtain
Z 1
f (ta + (1 − t) b)g (ta + (1 − t) b) dt
0
Z b
1
=
f (x)g (x) dx
b−a a
f (a) g (a)
1
1
b
+ m2
−
f (a) g
≤
α1 + α2 + 1
α1 + 1 α1 + α2 + 1
m2
1
1
b
+ m1
−
f
g(a)
α2 + 1 α1 + α2 + 1
m1
1
1
1
b
b
+ m1 m2 1 −
−
+
f
g
.
α1 + 1 α2 + 1 α1 + α2 + 1
m1
m2
Analogously we have
Z b
1
f (x)g (x) dx
b−a a
f (b) g (b)
1
1
a
≤
+ m2
−
f (b) g
α1 + α2 + 1
α1 + 1 α1 + α2 + 1
m2
1
1
a
+ m1
−
f
g(b)
α2 + 1 α1 + α2 + 1
m1
1
1
1
a
a
+ m1 m2 1 −
−
+
f
g
,
α1 + 1 α2 + 1 α1 + α2 + 1
m1
m2
which completes the proof.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/
12
M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR , AND J. P E ČARI Ć
If in Theorem 3.2 we have α1 = α2 = 1, the statement of Theorem 3.2 becomes the statement
of Theorem 2.4.
R EFERENCES
[1] A.M. BRUCKNER AND E. OSTROW, Some function classes related to the class of convex functions, Pacific J. Math., 12 (1962), 1203–1215.
[2] S.S. DRAGOMIR AND R.P. AGARWAL, Two inequalities for differentiable mappings and applications to special means of real numbers and trapezoidal formula, Appl. Math. Lett., 11(5) (1998),
91–95.
[3] S.S. DRAGOMIR AND G. TOADER, Some inequalities for m-convex functions, Studia Univ.
Babeş-Bolyai Math., 38(1) (1993), 21–28.
[4] M. KLARIČIĆ BAKULA, J. PEČARIĆ AND M. RIBIČIĆ, Companion inequalities to Jensen’s
inequality for m-convex and (α, m)-convex functions, J. Inequal. Pure & Appl. Math., 7(5) (2006),
Art. 194. [ONLINE: http://jipam.vu.edu.au/article.php?sid=811].
[5] V.G. MIHEŞAN, A generalization of the convexity, Seminar on Functional Equations, Approx. and
Convex., Cluj-Napoca (Romania) (1993).
[6] B.G. PACHPATTE, On some inequalities for convex functions, RGMIA Res. Rep.Coll., 6(E) (2003),
[ONLINE: http://rgmia.vu.edu.au/v6(E).html].
[7] C.E.M. PEARCE AND J. PEČARIĆ, Inequalities for differentiable mappings with application to
special means and quadrature formulae, Appl. Math. Lett., 13(2) (2000), 51–55.
[8] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, Inc. (1992).
[9] G. TOADER, Some generalizations of the convexity, Proceedings of the Colloquium on Approximation and Optimization, Univ. Cluj-Napoca, Cluj-Napoca, 1985, 329-338.
[10] G. TOADER, On a generalization of the convexity, Mathematica, 30 (53) (1988), 83-87.
[11] S. TOADER, The order of a star-convex function, Bull. Applied & Comp. Math., 85-B (1998),
BAM-1473, 347–350.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 96, 12 pp.
http://jipam.vu.edu.au/