1 Introduction

It is common knowledge in mathematical analysis that a function f:IRR is said to be convex on an interval I if the inequality

f ( λ x + ( 1 λ ) y ) λf(x)+(1λ)f(y)
(1.1)

is valid for all x,yI and λ[0,1]. If f:IRR is a convex function on I and a,bI with a<b, then the double inequality

f ( a + b 2 ) 1 b a a b f(x)dx f ( a ) + f ( b ) 2
(1.2)

holds. This double inequality is known in the literature as Hermite-Hadamard’s integral inequality for convex functions. The definition of convex functions and Hermite-Hadamard’s integral inequality (1.2) have been generalized, refined, and extended by many mathematicians in a lot of references. Some of them may be recited as follows.

Theorem 1.1 ([[1], Theorems 2.2 and 2.3])

Let f: I RR be differentiable on I and a,b I with a<b.

  1. (1)

    If | f (x)| is a convex function on [a,b], then

    | f ( a ) + f ( b ) 2 1 b a a b f ( x ) d x | b a 8 ( | f ( a ) | + | f ( b ) | ) .
    (1.3)
  2. (2)

    If | f ( x ) | p / ( p 1 ) for p>1 is a convex function on [a,b], then

    | f ( a ) + f ( b ) 2 1 b a a b f ( x ) d x | b a 2 ( p + 1 ) 1 / p [ | f ( a ) | p / ( p 1 ) + | f ( b ) | p / ( p 1 ) 2 ] ( p 1 ) / p .
    (1.4)

Theorem 1.2 ([[2], Theorems 2.2 and 2.3])

Let f: I RR be a differentiable mapping on I and a,b I with a<b. If | f | p / ( p 1 ) for p>1 is convex on [a,b], then

| f ( a + b 2 ) 1 b a a b f ( x ) d x | b a 16 ( 4 p + 1 ) 1 / p [ ( | f ( a ) | p / ( p 1 ) + 3 | f ( b ) | p / ( p 1 ) ) ( p 1 ) / p + ( 3 | f ( a ) | p / ( p 1 ) + | f ( b ) | p / ( p 1 ) ) ( p 1 ) / p ]
(1.5)

and

| f ( a + b 2 ) 1 b a a b f ( x ) d x | b a 4 ( 4 p + 1 ) 1 / p [ | f ( a ) | + | f ( b ) | ] .
(1.6)

Definition 1.1 ([3])

A function f:IR[0,) is said to be quasi-convex if

f ( λ x + ( 1 λ ) y ) sup { f ( x ) , f ( y ) }
(1.7)

holds for all x,yI and λ[0,1].

Theorem 1.3 ([[4], Theorem 2])

Let f:IRR be differentiable on I such that f L([a,b]) and a,b I with a<b. If | f | is quasi-convex on [a,b], then

| a b f ( x ) d x b a 6 [ f ( a ) + 4 f ( a + b 2 ) + f ( b ) ] | ( b a ) 4 1152 [ max { | f ( a ) | , | f ( a + b 2 ) | } + max { | f ( a + b 2 ) | , | f ( b ) | } ] .
(1.8)

Definition 1.2 ([5])

Let s(0,1]. A function f: R 0 R 0 is said to be s-convex in the second sense if

f ( λ x + ( 1 λ ) y ) λ s f(x)+ ( 1 λ ) s f(y)
(1.9)

for all x,yI and λ[0,1].

Theorem 1.4 ([[6], Theorem 3.1])

Let f:I R 0 R be differentiable on I , a,b I with a<b, and f L([a,b]). If q1 and | f | is s-convex in the second sense on [a,b] for s(0,1], then

| f ( a ) + f ( b ) 2 1 b a a b f ( x ) d x b a 12 [ f ( b ) f ( a ) ] | ( b a ) 3 192 [ 2 2 s ( s + 6 + 2 s + 2 s ) ( s + 2 ) ( s + 3 ) ( s + 4 ) ] 1 / q [ | f ( a ) | q + | f ( b ) | q ] 1 / q .
(1.10)

For more information on Hermite-Hadamard type inequalities, please refer to [719], for example, and to monographs [20, 21] and related references therein.

In this paper, we will create some new integral inequalities of Hermite-Hadamard type for functions whose third derivatives are convex.

2 Lemma

For establishing some new integral inequalities of Hermite-Hadamard type for functions whose third derivatives are convex, we need an integral identity below.

Lemma 2.1 Let f:IRR be a three times differentiable mapping on I and a,b I with a<b. If f L([a,b]), then

f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] = ( b a ) 3 96 [ 0 1 t ( 1 t ) ( 2 t ) f ( 1 t 2 a + 1 + t 2 b ) d t 0 1 t ( 1 t ) ( 2 t ) f ( 1 + t 2 a + 1 t 2 b ) d t ] .
(2.1)

Proof Integrating by part and changing variable of definite integral yield

0 1 t ( 1 t ) ( 2 t ) f ( 1 t 2 a + 1 + t 2 b ) d t = 2 b a 0 1 ( 3 t 2 6 t + 2 ) f ( 1 t 2 a + 1 + t 2 b ) d t = 4 ( b a ) 2 [ f ( b ) + 2 f ( a + b 2 ) ] + 48 ( b a ) 3 0 1 ( t 1 ) d f ( 1 t 2 a + 1 + t 2 b ) = 4 ( b a ) 2 [ f ( b ) + 2 f ( a + b 2 ) ] + 48 ( b a ) 3 f ( a + b 2 ) 48 ( b a ) 3 0 1 f ( 1 t 2 a + 1 + t 2 b ) d t

and

0 1 t ( 1 t ) ( 2 t ) f ( 1 + t 2 a + 1 t 2 b ) d t = 2 b a 0 1 ( 3 t 2 6 t + 2 ) f ( 1 + t 2 a + 1 t 2 b ) d t = 4 ( b a ) 2 [ f ( a ) + 2 f ( a + b 2 ) ] 48 ( b a ) 3 0 1 ( t 1 ) d f ( 1 + t 2 a + 1 t 2 b ) = 4 ( b a ) 2 [ f ( a ) + 2 f ( a + b 2 ) ] 48 ( b a ) 3 f ( a + b 2 ) + 48 ( b a ) 3 0 1 f ( 1 + t 2 a + 1 t 2 b ) d t .

Lemma 2.1 is thus proved. □

3 Hermite-Hadamard type inequalities for convex functions

Basing on Lemma 2.1, we now start out to establish some new integral inequalities of Hermite-Hadamard type for functions whose third derivatives are convex.

Theorem 3.1 Let f:IRR be three times differentiable on I and f L([a,b]) for a,b I with a<b. If | f | q for q1 is convex on [a,b], then

| f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 384 { [ 4 | f ( a ) | q + 11 | f ( b ) | q 15 ] 1 / q + [ 11 | f ( a ) | q + 4 | f ( b ) | q 15 ] 1 / q } .
(3.1)

Proof Since | f | q is convex on [a,b], by Lemma 2.1 and Hölder’s inequality, we have

| f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 { 0 1 t ( 1 t ) ( 2 t ) | f ( 1 t 2 a + 1 + t 2 b ) | d t + 0 1 t ( 1 t ) ( 2 t ) | f ( 1 + t 2 a + 1 t 2 b ) | d t } ( b a ) 3 96 [ 0 1 t ( 1 t ) ( 2 t ) d t ] 1 1 / q × { [ 0 1 t ( 1 t ) ( 2 t ) | f ( 1 t 2 a + 1 + t 2 b ) | q d t ] 1 / q + [ 0 1 t ( 1 t ) ( 2 t ) | f ( 1 + t 2 a + 1 t 2 b ) | q d t ] 1 / q } ( b a ) 3 96 ( 1 4 ) 1 1 / q { [ 1 2 0 1 t ( 1 t ) 2 ( 2 t ) | f ( a ) | q d t + 1 2 0 1 t ( 1 t 2 ) ( 2 t ) | f ( b ) | q d t ] 1 / q + [ 1 2 0 1 t ( 1 t 2 ) ( 2 t ) | f ( a ) | q d t + 1 2 0 1 t ( 1 t ) 2 ( 2 t ) | f ( b ) | q d t ] 1 / q } = ( b a ) 3 384 { [ 4 | f ( a ) | q + 11 | f ( b ) | q 15 ] 1 / q + [ 11 | f ( a ) | q + 4 | f ( b ) | q 15 ] 1 / q } .

The proof of Theorem 3.1 is complete. □

Corollary 3.1 Under conditions of Theorem  3.1, if q=1, we have

| f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 384 [ | f ( a ) | + | f ( b ) | ] .
(3.2)

Theorem 3.2 Let f:IRR be three times differentiable on I and f L([a,b]) for a,b I with a<b. If | f | q for q>1 is convex on [a,b] and if qr and s0, then

| f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 [ 2 B ( 2 q r 1 q 1 , 2 q s 1 q 1 ) B ( 3 q r 2 q 1 , 2 q s 1 q 1 ) ] 1 1 / q × ( 1 2 ) 1 / q { ( [ 2 B ( r + 1 , s + 2 ) B ( r + 2 , s + 2 ) ] | f ( a ) | q + [ 2 B ( r + 1 , s + 1 ) + B ( r + 2 , s + 1 ) B ( r + 3 , s + 1 ) ] | f ( b ) | q ) 1 / q + ( [ 2 B ( r + 1 , s + 1 ) + B ( r + 2 , s + 1 ) B ( r + 3 , s + 1 ) ] | f ( a ) | q + [ 2 B ( r + 1 , s + 2 ) B ( r + 2 , s + 2 ) ] | f ( b ) | q ) 1 / q } ,

where B(x,y) is the classical Beta function, which may be defined for Re(x)>0 and Re(y)>0 by

B(x,y)= 0 1 t x 1 ( 1 t ) y 1 dt.
(3.3)

Proof By Lemma 2.1, Hölder’s inequality, and the convexity of | f | q on [a,b], we have

| f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 { 0 1 t ( 1 t ) ( 2 t ) | f ( 1 t 2 a + 1 + t 2 b ) | d t + 0 1 t ( 1 t ) ( 2 t ) | f ( 1 + t 2 a + 1 t 2 b ) | d t } ( b a ) 3 96 ( 0 1 t ( q r ) / ( q 1 ) ( 1 t ) ( q s ) / ( q 1 ) ( 2 t ) d t ) 1 1 / q × { ( 0 1 t r ( 1 t ) s ( 2 t ) | f ( 1 t 2 a + 1 + t 2 b ) | q d t ) 1 / q + ( 0 1 t r ( 1 t ) s ( 2 t ) | f ( 1 + t 2 a + 1 t 2 b ) | q d t ) 1 / q } ( b a ) 3 96 [ 2 B ( 2 q r 1 q 1 , 2 q s 1 q 1 ) B ( 3 q r 2 q 1 , 2 q s 1 q 1 ) ] 1 1 / q × { [ 1 2 0 1 t r ( 1 t ) s + 1 ( 2 t ) | f ( a ) | q d t + 1 2 0 1 t r ( 1 + t ) ( 1 t ) s ( 2 t ) | f ( b ) | q d t ] 1 / q + [ 1 2 0 1 t r ( 1 + t ) ( 1 t ) s ( 2 t ) | f ( a ) | q d t + 1 2 0 1 t r ( 1 t ) s + 1 ( 2 t ) | f ( b ) | q d t ] 1 / q } = ( b a ) 3 96 [ 2 B ( 2 q r 1 q 1 , 2 q s 1 q 1 ) B ( 3 q r 2 q 1 , 2 q s 1 q 1 ) ] 1 1 / q × ( 1 2 ) 1 / q { [ [ 2 B ( r + 1 , s + 2 ) B ( r + 2 , s + 2 ) ] | f ( a ) | q + [ 2 B ( r + 1 , s + 1 ) + B ( r + 2 , s + 1 ) B ( r + 3 , s + 1 ) ] | f ( b ) | q ] 1 / q + [ [ 2 B ( r + 1 , s + 1 ) + B ( r + 2 , s + 1 ) B ( r + 3 , s + 1 ) ] | f ( a ) | q + [ 2 B ( r + 1 , s + 2 ) B ( r + 2 , s + 2 ) ] | f ( b ) | q ] 1 / q } .

The proof of Theorem 3.2 is completed. □

Corollary 3.2 Under conditions of Theorem  3.2,

  1. (1)

    if r=0, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 [ 2 B ( 2 q 1 q 1 , 2 q s 1 q 1 ) B ( 3 q 2 q 1 , 2 q s 1 q 1 ) ] 1 1 / q ( 1 2 ( s + 1 ) ( s + 2 ) ( s + 3 ) ) 1 / q × { [ ( s + 1 ) ( 2 s + 5 ) | f ( a ) | q + ( 2 s 2 + 11 s + 13 ) | f ( b ) | q ] 1 / q + [ ( 2 s 2 + 11 s + 13 ) | f ( a ) | q + ( s + 1 ) ( 2 s + 5 ) | f ( b ) | q ] 1 / q } ;
  2. (2)

    if s=0, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 [ 2 B ( 2 q r 1 q 1 , 2 q 1 q 1 ) B ( 3 q r 2 q 1 , 2 q 1 q 1 ) ] 1 1 / q [ 1 2 ( r + 1 ) ( r + 2 ) ( r + 3 ) ] 1 / q × { [ ( r + 5 ) | f ( a ) | q + ( 2 r 2 + 11 r + 13 ) | f ( b ) | q ] 1 / q + [ ( 2 r 2 + 11 r + 13 ) | f ( a ) | q + ( r + 5 ) | f ( b ) | q ] 1 / q } ;
  3. (3)

    if r=s=0, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 [ 2 B ( 2 q 1 q 1 , 2 q 1 q 1 ) B ( 3 q 2 q 1 , 2 q 1 q 1 ) ] 1 1 / q ( 3 2 ) 1 / q × { [ 5 | f ( a ) | q + 13 | f ( b ) | q 18 ] 1 / q + [ 13 | f ( a ) | q + 5 | f ( b ) | q 18 ] 1 / q } ;
  4. (4)

    if r=q, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 [ ( 5 q 2 s 3 ) ( q 1 ) ( q s ) 2 + ( 5 q 3 s 2 ) ( q 1 ) ] 1 1 / q ( 1 2 ) 1 / q × { [ [ 2 B ( q + 1 , s + 2 ) B ( q + 2 , s + 2 ) ] | f ( a ) | q + [ 2 B ( q + 1 , s + 1 ) + B ( q + 2 , s + 1 ) B ( q + 3 , s + 1 ) ] | f ( b ) | q ] 1 / q + [ [ 2 B ( q + 1 , s + 1 ) + B ( q + 2 , s + 1 ) B ( q + 3 , s + 1 ) ] | f ( a ) | q + [ 2 B ( q + 1 , s + 2 ) B ( q + 2 , s + 2 ) ] | f ( b ) | q ] 1 / q } ;
  5. (5)

    if s=q, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 ( ( q 1 ) ( 4 q r 3 ) ( 2 q r 1 ) ( 3 q r 2 ) ) 1 1 / q ( 1 2 ) 1 / q × { [ [ 2 B ( r + 1 , q + 2 ) B ( r + 2 , q + 2 ) ] | f ( a ) | q + [ 2 B ( r + 1 , q + 1 ) + B ( r + 2 , q + 1 ) B ( r + 3 , q + 1 ) ] | f ( b ) | q ] 1 / q + [ [ 2 B ( r + 1 , q + 1 ) + B ( r + 2 , q + 1 ) B ( r + 3 , q + 1 ) ] | f ( a ) | q + [ 2 B ( r + 1 , q + 2 ) B ( r + 2 , q + 2 ) ] | f ( b ) | q ] 1 / q } ;
  6. (6)

    if r=s=q, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | 3 1 / q ( b a ) 3 64 { [ [ 2 B ( q + 1 , q + 2 ) B ( q + 2 , q + 2 ) ] | f ( a ) | q + [ 2 B ( q + 1 , q + 1 ) + B ( q + 2 , q + 1 ) B ( q + 3 , q + 1 ) ] | f ( b ) | q ] 1 / q + [ [ 2 B ( q + 1 , q + 1 ) + B ( q + 2 , q + 1 ) B ( q + 3 , q + 1 ) ] | f ( a ) | q + [ 2 B ( q + 1 , q + 2 ) B ( q + 2 , q + 2 ) ] | f ( b ) | q ] 1 / q } .

Theorem 3.3 Let f:IRR be three times differentiable on I and f L([a,b]) for a,b I with a<b. If | f | q is convex on [a,b] for q>1 and q0, then

| f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 [ ( 2 ξ + 2 + 1 ) ξ 2 ξ + 2 + 5 ξ 3 + 6 ξ 2 + 11 ξ + 6 ] 1 1 / q [ 1 ( + 1 ) ( + 2 ) ( + 3 ) ( + 4 ) ] 1 / q × { [ ( 2 + 1 2 ( 2 + 1 + 1 ) + 2 + 3 7 ) | f ( a ) | q + ( ( 2 + 1 + 1 ) 2 + 2 ( 7 × 2 + 5 ) 3 × 2 + 3 + 27 ) | f ( b ) | q ] 1 / q + [ ( ( 2 + 1 + 1 ) 2 + 2 ( 7 × 2 + 5 ) 3 × 2 + 3 + 27 ) | f ( a ) | q + ( 2 + 1 2 ( 2 + 1 + 1 ) + 2 + 3 7 ) | f ( b ) | q ] 1 / q } ,

where ξ= q q 1 .

Proof By Lemma 2.1, Hölder’s inequality, and the convexity of | f | q on [a,b], we have

| f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 { 0 1 t ( 1 t ) ( 2 t ) | f ( 1 t 2 a + 1 + t 2 b ) | d t + 0 1 t ( 1 t ) ( 2 t ) | f ( 1 + t 2 a + 1 t 2 b ) | d t } ( b a ) 3 96 ( 0 1 t ( 1 t ) ( 2 t ) ( q ) / ( q 1 ) d t ) 1 1 / q × { ( 0 1 t ( 1 t ) ( 2 t ) | f ( 1 t 2 a + 1 + t 2 b ) | q d t ) 1 / q + ( 0 1 t ( 1 t ) ( 2 t ) | f ( 1 + t 2 a + 1 t 2 b ) | q d t ) 1 / q } ( b a ) 3 96 ( 0 1 t ( 1 t ) ( 2 t ) ( q ) / ( q 1 ) d t ) 1 1 / q × { [ 1 2 0 1 t ( 1 t ) ( 2 t ) [ ( 1 t ) | f ( a ) | q + ( 1 + t ) | f ( b ) | q ] d t ] 1 / q + [ 1 2 0 1 t ( 1 t ) ( 2 t ) [ ( 1 + t ) | f ( a ) | q + ( 1 t ) | f ( b ) | q ] d t ] 1 / q } = ( b a ) 3 96 [ ( 2 ξ + 2 + 1 ) ξ 2 ξ + 2 + 5 ξ 3 + 6 ξ 2 + 11 ξ + 6 ] 1 1 / q [ 1 ( + 1 ) ( + 2 ) ( + 3 ) ( + 4 ) ] 1 / q × { [ ( 2 + 1 2 ( 2 + 1 + 1 ) + 2 + 3 7 ) | f ( a ) | q + ( ( 2 + 1 + 1 ) 2 + 2 ( 7 × 2 + 5 ) 3 × 2 + 3 + 27 ) | f ( b ) | q ] 1 / q + [ ( ( 2 + 1 + 1 ) 2 + 2 ( 7 × 2 + 5 ) 3 × 2 + 3 + 27 ) | f ( a ) | q + ( 2 + 1 2 ( 2 + 1 + 1 ) + 2 + 3 7 ) | f ( b ) | q ] 1 / q } .

The proof of Theorem 3.3 is complete. □

Corollary 3.3 Under conditions of Theorem  3.3.

  1. (1)

    if =0, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 ( 1 6 ) 1 / q [ ( q 1 ) 2 ( 6 q + 2 ( 3 q 2 ) / ( q 1 ) 5 ) ( 2 q 1 ) ( 3 q 2 ) ( 4 q 3 ) ] 1 1 / q × { [ | f ( a ) | q + 3 | f ( b ) | q 4 ] 1 / q + [ 3 | f ( a ) | q + | f ( b ) | q 4 ] 1 / q } ,
    (3.4)
  2. (2)

    if =q, we have

    | f ( a + b 2 ) 1 b a a b f ( x ) d x + b a 24 [ f ( b ) f ( a ) ] | ( b a ) 3 96 ( 1 6 ) 1 1 / q [ 1 ( q + 1 ) ( q + 2 ) ( q + 3 ) ( q + 4 ) ] 1 / q × { [ ( 2 q + 1 q 2 ( 2 q + 1 + 1 ) q + 2 q + 3 7 ) | f ( a ) | q + ( ( 2 q + 1 + 1 ) q 2 + 2 ( 7 × 2 q + 5 ) q 3 × 2 q + 3 + 27 ) | f ( b ) | q ] 1 / q + [ ( ( 2 q + 1 + 1 ) q 2 + 2 ( 7 × 2 q + 5 ) q 3 × 2 q + 3 + 27 ) | f ( a ) | q + ( 2 q + 1 q 2 ( 2 q + 1 + 1 ) q + 2 q + 3 7 ) | f ( b ) | q ] 1 / q } .