International Journal of Analysis and Applications
ISSN 2291-8639
Volume 11, Number 2 (2016), 137-145
http://www.etamaths.com

HERMITE-HADAMARD TYPE INEQUALITIES FOR p-CONVEX FUNCTIONS

İMDAT İŞCAN∗

Abstract. In this paper, the author establishes some new Hermite-Hadamard type inequalities for

p-convex functions. Some natural applications to special means of real numbers are also given.

1. Introduction

Let f : I ⊂ R → R be a convex function defined on the interval I of real numbers and a,b ∈ I with
a < b. The following inequality

(1.1) f

(
a + b

2

)
≤

1

b−a

b∫
a

f(x)dx ≤
f(a) + f(b)

2

holds. This double inequality is known in the literature as Hermite-Hadamard integral inequality
for convex functions. Note that some of the classical inequalities for means can be derived from (1.1)
for appropriate particular selections of the mapping f. Both inequalities hold in the reversed direction
if f is concave. For some results which generalize, improve and extend the inequalities (1.1) we refer
the reader to the recent papers (see [2, 3, 5, 6, 8, 9, 12]).

In [3], Dragomir gave the following Lemma:

Lemma 1. Let f : I◦ ⊂ R → R be a differentiable mapping on I◦ and a,b ∈ I◦ with a < b. If
f′ ∈ L[a,b], then the following equality holds:

(1.2)
f(a) + f(b)

2
−

1

b−a

b∫
a

f(x)dx =
b−a

2

1∫
0

(1 − 2t) f′ (ta + (1 − t)b) dt.

By using this Lemma, Dragomir obtained the following Hermite-Hadamard type inequalities for
convex functions:

Theorem 1. Let f : I◦ ⊂ R → R be a differentiable mapping on I◦ and a,b ∈ I◦ with a < b. If |f′|
is convex on [a,b] , then the following inequality holds:

(1.3)

∣∣∣∣∣∣f(a) + f(b)2 − 1b−a
b∫

a

f(x)dx

∣∣∣∣∣∣ ≤ (b−a) (|f
′(a)| + |f′(b)|)

8
.

Theorem 2. Let f : I◦ ⊂ R → R be a differentiable mapping on I◦ and a,b ∈ I◦ with a < b and
p > 1. If the new mapping |f′|q is convex on [a,b] , then the following inequality holds:

(1.4)

∣∣∣∣∣∣f(a) + f(b)2 − 1b−a
b∫

a

f(x)dx

∣∣∣∣∣∣ ≤ (b−a)2(p + 1)1/p
[
|f′(a)|q + |f′(b)|q

2

]1/q
,

where 1/p + 1/q = 1.

2010 Mathematics Subject Classification. 26D15; 26A51.
Key words and phrases. p-convex function; Hermite-Hadamard type inequality; hypergeometric function.

c©2016 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

137



138 İŞCAN

Let A (a,b; t) = ta + (1 − t)b, G (a,b; t) = atb1−t, H (a,b; t) = ab/(ta + (1 − t)b) and Mp (a,b; t) =
(tap + (1 − t)bp)1/p ,p ∈ R\{0} , be the weighted arithmetic, geometric, harmonic , power of order
p means of two positive real numbers a and b with a 6= b for t ∈ [0, 1] , respectively. Mp (a,b; t) is
continuous and strictly increasing with respect to t ∈ R for fixed p ∈ R\{0} and a,b > 0 with a > b.
See [13, 7] for some kinds of convexity obtained by using weighted means.

In [7], the author, gave definition Harmonically convex and concave functions as follow.

Definition 1. Let I ⊂ R\{0} be a real interval. A function f : I → R is said to be harmonically
convex, if

(1.5) f

(
xy

tx + (1 − t)y

)
≤ tf(y) + (1 − t)f(x)

for all x,y ∈ I and t ∈ [0, 1]. If the inequality in (1.5) is reversed, then f is said to be harmonically
concave.

The following result of the Hermite-Hadamard type holds for harmonically convex functions.

Theorem 3 ([7]). Let f : I ⊂ R\{0}→ R be a harmonically convex function and a,b ∈ I with a < b.
If f ∈ L[a,b] then the following inequalities hold

(1.6) f

(
2ab

a + b

)
≤

ab

b−a

b∫
a

f(x)

x2
dx ≤

f(a) + f(b)

2
.

The above inequalities are sharp.

Lemma 2 ([7]). Let f : I ⊂ R\{0}→ R be a differentiable function on I◦ and a,b ∈ I with a < b. If
f′ ∈ L[a,b] then

f(a) + f(b)

2
−

ab

b−a

b∫
a

f(x)

x2
dx

=
ab (b−a)

2

1∫
0

1 − 2t
(tb + (1 − t)a)2

f′
(

ab

tb + (1 − t)a

)
dt.(1.7)

Using this Lemma, the following inequalities hold.

Theorem 4 ([7]). Let f : I ⊂ (0,∞) → R be a differentiable function on I◦, a,b ∈ I with a < b, and
f′ ∈ L[a,b]. If |f′|q is harmonically convex on [a,b] for q ≥ 1, then∣∣∣∣∣∣f(a) + f(b)2 − abb−a

b∫
a

f(x)

x2
dx

∣∣∣∣∣∣(1.8)
≤

ab (b−a)
2

λ
1−1

q

1

[
λ2 |f′ (a)|

q
+ λ3 |f′ (b)|

q]1q ,
where

λ1 =
1

ab
−

2

(b−a)2
ln

(
(a + b)

2

4ab

)
,

λ2 =
−1

b (b−a)
+

3a + b

(b−a)3
ln

(
(a + b)

2

4ab

)
,

λ3 =
1

a (b−a)
−

3b + a

(b−a)3
ln

(
(a + b)

2

4ab

)
= λ1 −λ2.



HERMITE-HADAMARD TYPE INEQUALITIES 139

Theorem 5 ([7]). Let f : I ⊂ (0,∞) → R be a differentiable function on I◦, a,b ∈ I with a < b, and
f′ ∈ L[a,b]. If |f′|q is harmonically convex on [a,b] for q > 1, 1

p
+ 1

q
= 1, then∣∣∣∣∣∣f(a) + f(b)2 − abb−a

b∫
a

f(x)

x2
dx

∣∣∣∣∣∣(1.9)
≤

ab (b−a)
2

(
1

p + 1

)1
p (
µ1 |f′ (a)|

q
+ µ2 |f′ (b)|

q)1q ,
where

µ1 =

[
a2−2q + b1−2q [(b−a) (1 − 2q) −a]

]
2 (b−a)2 (1 −q) (1 − 2q)

,

µ2 =

[
b2−2q −a1−2q [(b−a) (1 − 2q) + b]

]
2 (b−a)2 (1 −q) (1 − 2q)

.

In [16], Zhang and Wan gave definition of p-convex function as follow:

Definition 2. Let I be a p-convex set. A function f : I → R is said to be a p-convex function or
belongs to the class PC(I), if

f
(

[αxp + (1 −α)yp]1/p
)
≤ αf(x) + (1 −α)f(y)

for all x,y ∈ I and α ∈ [0, 1].

Remark 1 ([16]). An interval I is said to be a p-convex set if [αxp + (1 −α)yp]1/p ∈ I for all x,y ∈ I
and α ∈ [0, 1], where p = 2k + 1 or p = n/m, n = 2r + 1, m = 2t + 1 and k,r,t ∈ N.

Remark 2 ([10]). If I ⊂ (0,∞) be a real interval and p ∈ R\{0}, then
[αxp + (1 −α)yp]1/p ∈ I for all x,y ∈ I and α ∈ [0, 1].

According to Remark 2, we can give a different version of the definition of p-convex function as
follow:

Definition 3 ([10]). Let I ⊂ (0,∞) be a real interval and p ∈ R\{0} . A function f : I → R is said
to be a p-convex function, if

(1.10) f
(

[αxp + (1 −α)yp]1/p
)
≤ αf(x) + (1 −α)f(y)

for all x,y ∈ I and α ∈ [0, 1]. If the inequality in (1.10) is reversed, then f is said to be p-concave.

According to Definition 3, It can be easily seen that for p = 1 and p = −1, p-convexity reduces to
ordinary convexity and harmonically convexity of functions defined on I ⊂ (0,∞), respectively.

Example 1. Let f : (0,∞) → R, f(x) = xp,p 6= 0, and g : (0,∞) → R, g(x) = c, c ∈ R, then f and
g are both p-convex and p-concave functions.

In [4, Theorem 5], if we take I ⊂ (0,∞), p ∈ R\{0} and h(t) = t , then we have the following
Theorem.

Theorem 6. Let f : I ⊂ (0,∞) → R be a p-convex function, p ∈ R\{0}, and a,b ∈ I with a < b. If
f ∈ L[a,b] then we have

(1.11) f

([
ap + bp

2

]1/p)
≤

p

bp −ap

b∫
a

f(x)

x1−p
dx ≤

f(a) + f(b)

2
.

For some results related to p-convex functions and its generalizations, we refer the reader to see
[4, 10, 15, 14, 16].

In [15, Lemma 2.4], if we take I ⊂ (0,∞) and p ∈ R\{0} , then we have the following Lemma.



140 İŞCAN

Lemma 3. Let f : I ⊂ (0,∞) → R be a differentiable function on I◦ and a,b ∈ I with a < b and
p ∈ R\{0}. If f′ ∈ L[a,b] then

f(a) + f(b)

2
−

p

bp −ap

b∫
a

f(x)

x1−p
dx

=
bp −ap

2p

1∫
0

1 − 2t
[tap + (1 − t)bp]1−1/p

f′
(

[tap + (1 − t)bp]1/p
)
dt.(1.12)

Remark 3. In Lemma 3,
(i) If we take p = 1, then we have inequality (1.2) in Lemma 1.
(ii) If we take p = −1, then we have inequality (1.7) in Lemma 2.

For finding some new inequalities of Hermite-Hadamard type for functions whose derivatives are
p-convex, we need Lemma 3.

We recall the following special functions
(1) The Beta function:

β (x,y) =
Γ(x)Γ(y)

Γ(x + y)
=

1∫
0

tx−1 (1 − t)y−1 dt, x,y > 0,

(2) The hypergeometric function:

2F1 (a,b; c; z) =
1

β (b,c− b)

1∫
0

tb−1 (1 − t)c−b−1 (1 −zt)−a dt, c > b > 0, |z| < 1 (see [11]).

The main purpose of this paper is to establish some new results connected with the right-hand side of
the inequalities (1.11) for p-convex functions.

2. Main Results

We obtain the another version of [15, Theorem 3.2] as follow:

Theorem 7. Let f : I ⊂ (0,∞) → R be a differentiable function on I◦, a,b ∈ I◦ with a < b, p ∈ R\{0}
and f′ ∈ L[a,b]. If |f′|q is p-convex on [a,b] for q ≥ 1, then∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap

b∫
a

f(x)

x1−p
dx

∣∣∣∣∣∣(2.1)
≤

bp −ap

2p
C

1−1
q

1

[
C2 |f′ (a)|

q
+ C3 |f′ (b)|

q]1q ,
where

C1 = C1(a,b; p) =
1

4

(
ap + bp

2

)1
p
−1

×
[

2F1

(
1 −

1

p
, 2; 3;

ap − bp

ap + bp

)
+2 F1

(
1 −

1

p
, 2; 3;

bp −ap

ap + bp

)]
,

C2 = C2(a,b; p) =
1

24

(
ap + bp

2

)1
p
−1 [

2F1

(
1 −

1

p
, 2; 4;

ap − bp

ap + bp

)
+6.2F1

(
1 −

1

p
, 2; 3;

bp −ap

ap + bp

)
+2 F1

(
1 −

1

p
, 2; 4;

bp −ap

ap + bp

)]
,

C3 = C3(a,b; p) = C1 −C2,



HERMITE-HADAMARD TYPE INEQUALITIES 141

Proof. From Lemma 3 and using the Power mean integral inequality, we have∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap
b∫

a

f(x)

x1−p
dx

∣∣∣∣∣∣
≤

bp −ap

2p

1∫
0

∣∣∣∣∣ 1 − 2t[tap + (1 − t)bp]1−1/p
∣∣∣∣∣
∣∣∣f′([tap + (1 − t)bp]1/p)∣∣∣dt

≤
bp −ap

2p


 1∫

0

|1 − 2t|
[tap + (1 − t)bp]1−1/p

dt


1−

1
q

×


 1∫

0

|1 − 2t|
[tap + (1 − t)bp]1−1/p

∣∣∣f′([tap + (1 − t)bp]1/p)∣∣∣q dt



1
q

.

Hence, by p-convexity of |f′|q on [a,b], we have∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap
b∫

a

f(x)

x1−p
dx

∣∣∣∣∣∣
≤

bp −ap

2p


 1∫

0

|1 − 2t|
[tap + (1 − t)bp]1−1/p

dt


1−

1
q

×


 1∫

0

||1 − 2t||
[
t |f′ (a)|q + (1 − t) |f′ (b)|q

]
[tap + (1 − t)bp]1−1/p

dt




1
q

≤
bp −ap

2p
C

1−1
q

1

[
C2 |f′ (a)|

q
+ C3 |f′ (b)|

q]1q .
It is easily check that

1∫
0

|1 − 2t|
[tap + (1 − t)bp]1−1/p

dt = C1(a,b; p),

1∫
0

|1 − 2t|t
[tap + (1 − t)bp]1−1/p

dt = C2(a,b; p),

1∫
0

|1 − 2t|(1 − t)
[tap + (1 − t)bp]1−1/p

dt = C1(a,b; p) −C2(a,b; p).

�

Remark 4. If we take p = −1 in Theorem 7, then we have inequality (1.8) in Theorem 4.

If we take q = 1 in Theorem 7, then we have the following corollary.

Corollary 1. Let f : I ⊂ (0,∞) → R be a differentiable function on I◦, a,b ∈ I◦ with a < b,
p ∈ R\{0} and f′ ∈ L[a,b]. If |f′| is p-convex on [a,b], then∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap

b∫
a

f(x)

x1−p
dx

∣∣∣∣∣∣
≤

bp −ap

2p
[C2 |f′ (a)| + C3 |f′ (b)|] ,

where C2 and C3 are defined as in Theorem 7.



142 İŞCAN

Remark 5. If we take p = 1 in Corollary 1, then we have inequality (1.3) in Theorem 1.

Theorem 8. Let f : I ⊂ (0,∞) → R be a differentiable function on I◦, a,b ∈ I with a < b, p ∈ R\{0}
and f′ ∈ L[a,b]. If |f′|q is p-convex on [a,b] for q > 1, 1

r
+ 1

q
= 1, then

∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap
b∫

a

f(x)

x1−p
dx

∣∣∣∣∣∣(2.2)
≤

bp −ap

2p

(
1

r + 1

)1
r (
C4 |f′ (a)|

q
+ C5 |f′ (b)|

q)1q ,
where

C4 = C4(a,b; p; q)

=




1
2aqp−q

.2F1

(
q − q

p
, 1; 3; 1 −

(
b
a

)p)
, p < 0

1
2bqp−q

.2F1

(
q − q

p
, 2; 3; 1 −

(
a
b

)p)
, p > 0

,

C5 = C5(a,b; p; q)

=




1
2aqp−q

.2F1

(
q − q

p
, 2; 3; 1 −

(
b
a

)p)
, p < 0

1
2bqp−q

.2F1

(
q − q

p
, 1; 3; 1 −

(
a
b

)p)
, p > 0

.

Proof. From Lemma 3, Hölder’s inequality and the p-convexity of |f′|q on [a,b],we have,

∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap
b∫

a

f(x)

x1−p
dx

∣∣∣∣∣∣
≤

bp −ap

2p


 1∫

0

|1 − 2t|r dt




1
r

×


 1∫

0

1

[tap + (1 − t)bp]q−q/p
∣∣∣f′([tap + (1 − t)bp]1/p)∣∣∣q dt




1
q

≤
bp −ap

2p

(
1

r + 1

)1
r

×


 1∫

0

t |f′ (a)|q + (1 − t) |f′ (b)|q

[tap + (1 − t)bp]q−q/p
dt




1
q

,

where an easy calculation gives

1∫
0

t

[tap + (1 − t)bp]q−q/p
dt(2.3)

=




1
2aqp−q

.2F1

(
q − q

p
, 1; 3; 1 −

(
b
a

)p)
, p < 0

1
2bqp−q

.2F1

(
q − q

p
, 2; 3; 1 −

(
a
b

)p)
, p > 0



HERMITE-HADAMARD TYPE INEQUALITIES 143

and

1∫
0

1 − t
[tap + (1 − t)bp]q−q/p

dt(2.4)

=




1
2aqp−q

.2F1

(
q − q

p
, 2; 3; 1 −

(
b
a

)p)
, p < 0

1
2bqp−q

.2F1

(
q − q

p
, 1; 3; 1 −

(
a
b

)p)
, p > 0

.

Substituting equations (2.3) and (2.4) into the above inequality results in the inequality (2.2), which
completes the proof. �

Remark 6. In Theorem 8,
(i) If we take p = 1, then we have inequality (1.4) in Theorem 2.
(ii) If we take p = −1, then we have the inequality (1.9) in Theorem 5.

Theorem 9. Let f : I ⊂ (0,∞) → R be a differentiable function on I◦, a,b ∈ I with a < b, p ∈ R\{0}
and f′ ∈ L[a,b]. If |f′|q is p-convex on [a,b] for q > 1, 1

r
+ 1

q
= 1, then∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap

b∫
a

f(x)

x1−p
dx

∣∣∣∣∣∣(2.5)
≤

bp −ap

2p
C

1
r
6

(
1

q + 1

)1
q
(
|f′ (a)|q + |f′ (b)|q

2

)1
q

,

where

C6 = C6(a,b; p; r)

=




1
apr−r

.2F1

(
r − r

p
, 1; 2; 1 −

(
b
a

)p)
, p < 0

1
bpr−r

.2F1

(
r − r

p
, 1; 2; 1 −

(
a
b

)p)
, p > 0

,

Proof. From Lemma 3, Hölder’s inequality and the p-convexity of |f′|q on [a,b],we have,∣∣∣∣∣∣f(a) + f(b)2 − pbp −ap
b∫

a

f(x)

x1−p
dx

∣∣∣∣∣∣
≤

bp −ap

2p


 1∫

0

1

[tap + (1 − t)bp]r−r/p
dt




1
r

×


 1∫

0

|1 − 2t|q
∣∣∣f′([tap + (1 − t)bp]1/p)∣∣∣q dt




1
q

≤
bp −ap

2p
C

1
r
6 (a,b; p; r)

(
1

q + 1

)1
q
(
|f′ (a)|q + |f′ (b)|q

2

)1
q

,

where an easy calculation gives

C6(a,b; p; r) =

1∫
0

1

[tap + (1 − t)bp]r−r/p
dt(2.6)

=




1
apr−r

.2F1

(
r − r

p
, 1; 2; 1 −

(
b
a

)p)
, p < 0

1
bpr−r

.2F1

(
r − r

p
, 1; 2; 1 −

(
a
b

)p)
, p > 0



144 İŞCAN

and

(2.7)

1∫
0

|1 − 2t|q tdt =
1∫

0

|1 − 2t|q (1 − t) dt =
1

2(q + 1)
.

Substituting equations (2.6) and (2.7) into the above inequality results in the inequality (2.5), which
completes the proof. �

3. Some applications for special means

Let us recall the following special means of two nonnegative number a,b with b > a :

(1) The arithmetic mean

A = A (a,b) :=
a + b

2
.

(2) The geometric mean

G = G (a,b) :=
√
ab.

(3) The harmonic mean

H = H (a,b) :=
2ab

a + b
.

(4) The power mean

Mr = Mr (a,b) :=

(
ar + br

2

)1/r
, r 6= 0.

(5) The Logarithmic mean

L = L (a,b) :=
b−a

ln b− ln a
.

(6) The p-Logarithmic mean

Lp = Lp (a,b) :=

(
bp+1 −ap+1

(p + 1)(b−a)

)1
p

, p ∈ R\{−1, 0} .

(7) the Identric mean

I = I (a,b) =
1

e

(
bb

aa

) 1
b−a

.

These means are often used in numerical approximation and in other areas. However, the following
simple relationships are known in the literature:

H ≤ G ≤ L ≤ I ≤ A.

It is also known that Lp is monotonically increasing over p ∈ R, denoting L0 = I and L−1 = L.

Proposition 1. Let 0 < a < b and p ∈ (−∞, 1)\{−1} . Then we have the following inequality

Mp.L
p−1
p−1 ≤ L

p
p ≤ A.L

p−1
p−1

Proof. The assertion follows from the inequality (1.11) in Theorem 6, for f : (0,∞) → R, f(x) = x. �

Proposition 2. Let 0 < a < b and p > 1. Then we have the following inequality

H(ap,bp) ≤ Lp−1p−1.L ≤ A(a
p,bp).

Proof. The assertion follows from the inequality (1.11) in Theorem 6, for f : (0,∞) → R, f(x) =
x−p. �

Proposition 3. Let 0 < a < b. Then we have the following inequality

LppH ≤ L
p−1
p−1G

2+2p ≤ LppMp.

Proof. The assertion follows from the inequality (1.11) in Theorem 6, for f : (0,∞) → R, f(x) =
1/x. �



HERMITE-HADAMARD TYPE INEQUALITIES 145

4. Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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Department of Mathematics, Faculty of Arts and Sciences,, Giresun University, 28100, Giresun, Turkey

∗Corresponding author: imdat.iscan@giresun.edu.tr