1 Introduction

Let f:[a,b]R be a convex function, then the inequality

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

is known as the Hermite-Hadamard inequality (see [1] for more information). Since then, some refinements of the Hermite-Hadamard inequality on convex functions have been extensively investigated by a number of authors (e.g., [2, 3] and [4]). In [5], the first author obtained a new refinement of the Hermite-Hadamard inequality for convex functions. The Hermite-Hadamard inequality was generalized in [6] to an r-convex positive function which is defined on an interval [a,b]. A positive function f is called r-convex on [a,b], if for each x,y[a,b] and t[0,1],

f ( t x + ( 1 t ) y ) { , 1 r r 0 , [ f ( y ) ] 1 t t , r = 0 .

It is obvious 0-convex functions are simply log-convex functions and 1-convex functions are ordinary convex functions. One should note that if f is r-convex in [a,b], then f r is a convex function (r>0).

Some refinements of the Hadamard inequality for r-convex functions could be found in [7] and [8]. In [9], Bessenyei studied Hermite-Hadamard-type inequalities for generalized 3-convex functions. In [7], the authors showed that if f is r-convex in [a,b] and 0<r1, then

1 b a a b f(x)dx r r + 1 [ f r ( a ) + f r ( b ) ] 1 r .
(2)

In this paper, first we show that if f is r-convex in [a,b] and r1, then

1 b a a b f(x)dx [ 1 2 ( f r ( a ) + f r ( b ) ) ] 1 r .
(3)

In Theorem 2.3, we prove the following inequality for r-convex functions:

1 b a a b f(x)dx r r + 1 f r + 1 ( b ) f r + 1 ( a ) f r ( b ) f r ( a ) (r>0).
(4)

The inequality (4) is an extension and refinement of (2) and (3). In Theorem 2.4, we show that r-convexity implies s-convexity (0rs). We employ this result in Theorem 2.6 and Corollary 2.7 to refine the Hermite-Hadamard inequality by r-convexity (0r1). Finally, we generalize some results in [7] without using Minkowski’s inequality. Indeed, we obtain refinements for the product of an r-convex function f and an s-convex function g (r,s0).

2 Main results

Theorem 2.1 Let f:[a,b](0,) be r-convex and r1. Then the following inequality holds:

1 b a a b f(x)dx 1 2 [ ( f r ( a ) + f r ( b ) ) ] 1 r .

Proof Since r1, by Jensen’s inequality, we have

( 1 b a a b f ( x ) d x ) r 1 b a a b f r (x)dx.

By convexity of f r and the right side of (1), we obtain

1 b a a b f r (x)dx 1 2 ( f r ( a ) + f r ( b ) ) .

Thus,

1 b a a b f(x)dx [ 1 2 ( f r ( a ) + f r ( b ) ) ] r .

 □

Corollary 2.2 Let f:[a,b](0,) be a 1-convex function. Then

1 b a a b f(x)dx 1 2 ( f ( b ) + f ( a ) ) .
(5)

Theorem 2.3 Let f:[a,b](0,) be r-convex and r0. Then the following inequalities hold:

1 b a a b f(x)dx{ r r + 1 ( f r + 1 ( b ) f r + 1 ( a ) f r ( b ) f r ( a ) ) , r 0 , ln f ( b ) f ( a ) , r = 0 .

Proof First, let r>0. Since f is r-convex, for all t[0,1], we have

f ( t a + ( 1 t ) b ) [ t f r ( a ) + ( 1 t ) f r ( b ) ] 1 r .

It is easy to observe that

1 b a a b f ( x ) d x = 0 1 f ( t a + ( 1 t ) b ) d t 0 1 [ t f r ( a ) + ( 1 t ) f r ( b ) ] 1 r d t = 0 1 [ t ( f r ( a ) f r ( b ) ) + f r ( b ) ] 1 r d t .

By substitution t( f r (a) f r (b))+ f r (b)=z, we obtain

1 b a a b f ( x ) d x 1 f r ( b ) f r ( a ) f r ( a ) f r ( b ) z 1 r d z = 1 f r ( b ) f r ( a ) 1 1 + 1 r [ z 1 + 1 r ] f r ( a ) f r ( b ) = r r + 1 ( f r + 1 ( b ) f r + 1 ( a ) f r ( b ) f r ( a ) ) .

For r=0, we have

f ( t x + ( 1 t ) y ) [ f ( x ) ] t [ f ( y ) ] 1 t .

So,

1 b a a b f ( x ) d x = 0 1 f ( t a + ( 1 t ) b ) d t 0 1 [ f ( a ) ] t [ f ( b ) ] 1 t d t = f ( b ) 0 1 [ f ( a ) f ( b ) ] t d t = f ( b ) [ f ( a ) f ( b ) ] t ln f ( a ) f ( b ) | 0 1 = [ f ( b ) f ( a ) ] ln f ( b ) f ( a ) .

The proof is completed. □

With the hypotheses of Theorem 2.3, if f(a)=f(b), its proving process shows that 1 b a a b f(x)dx can be dominated by f(a) where r0.

Note that if we put r=1 in Theorem 2.3, we can obtain again the inequality (5).

Theorem 2.4 Let f:[a,b](0,) be r-convex on [a,b] and 0rs. Then f is s-convex. In particular, if f is r-convex and 0r1, then f is convex.

In order to prove the above theorem, we need the following lemma.

Lemma 2.5 If 0α1 and 0rs, then the following inequalities hold for every pair of non-negative real numbers x and y:

x α y 1 α ( α x r + ( 1 α ) y r ) 1 r ( α x s + ( 1 α ) y s ) 1 s .
(6)

Proof The left side of the inequality is clear by Young’s inequality. The right side is obvious if either x or y equals zero. So, let x>0 and y>0. Consider f:[0,)R defined by

f(t)= ( α t r + 1 α ) s ( α t s + 1 α ) r .

Then f (t)=rsα[ t r 1 ( α t r + 1 α ) s 1 t s 1 ( α t s + 1 α ) r 1 ]. So, t=1 is a critical point of f. By an easy calculation, we see that f (1)=rsα(1α)(rs)0. It follows that f attains its maximum at t=1. Thus, f(t)f(1)=0. This shows that

( α t r + 1 α ) s ( α t s + 1 α ) r .

Now, if we put t= x y in the above inequality, we get

( α x r + ( 1 α ) y r ) s ( α x s + ( 1 α ) y s ) r .

Therefore, we can deduce the right side of (6) by taking rs th root. □

Proof of Theorem 2.4 Since f is r-convex, by Lemma 2.5 for all x,y[a,b] and t[0,1], we have

f ( t x + ( 1 t ) y ) { t [ f r ( x ) + ( 1 t ) f r ( y ) ] 1 r [ t f s ( x ) + ( 1 t ) f s ( y ) ] 1 s , 0 < r s , [ f ( y ) ] 1 t t [ t f s ( x ) + ( 1 t ) f s ( y ) ] 1 s , 0 = r s .

Hence, f is s-convex. □

Theorem 2.6 Let f:[a,b](0,) be r-convex on [a,b] and 0rs. Then the following inequalities hold:

1 b a a b f(x)dx{ r r + 1 f r + 1 ( b ) f r + 1 ( a ) f r ( b ) f r ( a ) s s + 1 ( f s + 1 ( b ) f s + 1 ( a ) f s ( b ) f s ( a ) ) , 0 < r s , ln f ( b ) f ( a ) s s + 1 ( f s + 1 ( b ) f s + 1 ( a ) f s ( b ) f s ( a ) ) , 0 = r < s .

Proof The left side of the inequalities is clear by Theorem 2.3. For the right side, by the inequality in (6), we have

[ t f r ( x ) + ( 1 t ) f r ( b ) ] 1 r [ t f s ( a ) + ( 1 t ) f s ( b ) ] 1 s .

By integrating it on [0,1], we obtain

0 1 [ t f r ( a ) + ( 1 t ) f r ( b ) ] dt 0 1 [ t f s ( a ) + ( 1 t ) f s ( b ) ] 1 t dt.

Thus,

r r + 1 ( f r + 1 ( b ) f r + 1 ( a ) f ( b ) f ( a ) ) s s + 1 ( f s + 1 ( b ) f s + 1 ( a ) f s ( b ) f s ( a ) ) .

Also, another inequality can be deduced by integrating the inequalities in (6) if we replace x and y by f(x) and f(y), respectively. □

Corollary 2.7 Let f:[a,b](0,) be r-convex and 0r1. Then

f ( a + b 2 ) 1 b a a b f ( x ) d x [ f ( b ) f ( a ) ] ln f ( b ) f ( a ) r r + 1 ( f r + 1 ( b ) f r + 1 ( a ) f r ( b ) f r ( a ) ) f ( a ) + f ( b ) 2 .

In other words, when f is r-convex and 0r1, we can refine the Hermite-Hadamard inequalities through Theorem  2.6.

Theorem 2.8 Let f,g:[a,b](0,) be r-convex and s-convex functions respectively on [a,b] and r,s>0. Then the following inequality holds:

1 b a a b f ( x ) g ( x ) d x 1 2 ( r r + 2 ) ( f r + 2 ( b ) f r + 2 ( a ) f r ( b ) f r ( a ) ) + 1 2 ( s s + 2 ) ( g s + 2 ( b ) g s + 2 ( a ) g s ( b ) g s ( a ) ) ( f ( a ) f ( b ) , g ( a ) g ( b ) ) .

Proof Since f is r-convex and g is s-convex, for all t[0,1], we have

f ( t a + ( 1 t ) b ) [ t f r ( a ) + ( 1 t ) f r ( b ) ] 1 r , g ( t a + ( 1 t ) b ) [ t g s ( a ) + ( 1 t ) g s ( b ) ] 1 s .

Thus,

1 b a a b f ( x ) g ( x ) d x = 0 1 f ( t a + ( 1 t ) b ) g ( t a + ( 1 t ) b ) d t 0 1 [ t f r ( a ) + ( 1 t ) f r ( b ) ] 1 r [ t g s ( a ) + ( 1 t ) g s ( b ) ] 1 s d t = 0 1 [ t ( f r ( a ) f r ( b ) ) + f r ( b ) ] 1 r [ t ( g s ( a ) g s ( b ) ) + g s ( b ) ] 1 s d t .

Applying Cauchy’s inequality, we get

(7)

Similar to the proof of Theorem 2.3 and by substitution t( f r (a) f r (b))+ f r (b)=z, we obtain

0 1 [ t ( f r ( a ) f r ( b ) ) + f r ( b ) ] 2 r dt= r r + 2 ( f r + 2 ( b ) f r + 2 ( a ) f r ( b ) f r ( a ) ) .
(8)

Similarly,

0 1 [ t ( g s ( a ) g s ( b ) ) + g s ( b ) ] 1 s dt= s s + 2 ( g s + 2 ( b ) g s + 2 ( a ) g s ( b ) g s ( a ) ) .
(9)

Using (7), (8) and (9), we can obtain the desired result. □

Remark 2.9 If the conditions of Theorem 2.8 hold, and rs, by Theorem 2.4, f is s-convex. Thus, the result of Theorem 2.8 could be as follows:

1 b a a b f(x)g(x)dx 1 2 ( s s + 2 ) [ f s + 2 ( b ) f s + 2 ( a ) f s ( b ) f s ( a ) + g s + 2 ( b ) g s + 2 ( a ) g s ( b ) g s ( a ) ] .

If f=g, we have

1 b a a b f 2 (x)dx s s + 2 ( f s + 2 ( b ) f s + 2 ( a ) f s ( b ) f s ( a ) ) .

Now, if f=g and r=s=2 in Theorem 2.8, we have

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

which is the same result as in [[7], Corollary 2.5]. This shows that Theorem 2.8 is a generalization of [[7], Theorem 2.3]. In fact, the condition r,s2 is redundant.

Theorem 2.10 Let f,g:[a,b](0,) be 0-convex on [a,b]. Then the following inequality holds:

1 b a a b f(x)g(x)dx [ f ( b ) g ( b ) f ( a ) g ( a ) ] ln f ( b ) g ( b ) f ( a ) g ( a ) .

Proof Since f and g are 0-convex, for all t[0,1], we have

f ( t x + ( 1 t ) y ) [ f ( x ) ] t [ f ( y ) ] 1 t , g ( t x + ( 1 t ) y ) [ g ( x ) ] t [ g ( y ) ] 1 t .

For all x,y[0,1], and thus

1 b a a b f ( x ) g ( x ) d x = 0 1 [ f ( t a + ( 1 t ) b ) ] [ g ( t x + ( 1 t ) y ) ] d t 0 1 [ f ( a ) ] t [ f ( b ) ] 1 t [ g ( a ) ] t [ g ( b ) ] 1 t d t = f ( b ) g ( b ) 0 1 [ f ( a ) g ( a ) f ( b ) g ( b ) ] t d t = [ f ( b ) g ( b ) f ( a ) g ( a ) ] ln f ( b ) g ( b ) f ( a ) g ( a ) .

 □

Corollary 2.11 With the hypotheses of the above theorem and f=g, we have

1 b a a b f 2 (x)dx [ f 2 ( b ) f 2 ( a ) ] ln f 2 ( b ) f 2 ( a ) .

Theorem 2.12 Let f,g:[a,b](0,) be r-convex and 0-convex functions respectively on [a,b] and r>0. Then the following inequality holds:

Proof Since f is r-convex and g is 0-convex, for all t[0,1], we have

f ( t a + ( 1 t ) b ) [ t f r ( a ) + ( 1 t ) f r ( b ) ] 1 r , g ( t x + ( 1 t ) y ) [ g ( x ) ] t [ g ( y ) ] 1 t .

Thus,

1 b a a b f ( x ) g ( x ) d x = 0 1 f ( t a + ( 1 t ) b ) g ( t a + ( 1 t ) b ) d t 0 1 [ t f r ( a ) + ( 1 t ) f r ( b ) ] 1 r [ g ( a ) ] t [ g ( b ) ] 1 t d t = 0 1 [ t ( f r ( a ) f r ( b ) ) + f r ( b ) ] 1 r [ g ( a ) ] t [ g ( b ) ] 1 t d t .

Again, Cauchy’s inequality shows that

(10)

We have

0 1 [ t ( f r ( a ) f r ( b ) ) + f r ( b ) ] 2 r dt= r r + 2 ( f r + 2 ( b ) f r + 2 ( a ) f r ( b ) f r ( a ) ) .
(11)

Similar to the proof of Theorem 2.10, we can show that

0 1 [ g ( a ) ] 2 t [ g ( b ) ] 2 2 t dt= 1 2 [ g ( b ) 2 g ( a ) 2 ] ln g ( b ) g ( a ) .
(12)

Using (10), (11) and (12), we can obtain the desired result. □