Erschienen in:

Open Access 01.12.2014 | Research

On the $(p, h)$ -convex function and some integral inequalities

verfasst von: Zhong Bo Fang, Renjie Shi

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2014

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Abstract

In this paper, we introduce a new class of

(p, h)

-convex functions which generalize P-functions and convex,

h, p, s

-convex, Godunova-Levin functions, and we give some properties of the functions. Moreover, we establish the corresponding Schur, Jensen, and Hadamard types of inequalities.

MSC:35K65, 35B33, 35B40.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript.

1 Introduction

Let I and J be intervals in R. To motivate our work, let us recall the definitions of some special classes of functions.

Definition 1 [1]

A function

f : I \to R

is said to be a Godunova-Levin function or belongs to the class

Q (I)

if f is non-negative and

f (α x + (1 - α) y) \leq \frac{f (x)}{α} + \frac{f (y)}{1 - α}

for all

x, y \in I

and

α \in (0, 1)

The class

Q (I)

was firstly described in [1] by Godunova and Levin. Some further properties of it are given in [2, 3]. It has been known that non-negative convex and monotone functions belong to this class of functions.

Definition 2 [4]

Let

s \in (0, 1)

be a fixed real number. A function

f : [0, \infty) \to [0, \infty)

is said to be an s-convex function (in the second sense) or belongs to the class

K_{s}^{2}

, if

f (α x + (1 - α) y) \leq α^{s} f (x) + {(1 - α)}^{s} f (y)

for all

x, y \in I

and

α \in [0, 1]

An s-convex function was introduced by Breckner [4] and a number of properties and connections with s-convexity (in the first sense) were discussed in [5]. Of course, s-convexity means just convexity when

s = 1

Definition 3 [2]

A function

f : I \to R

is said to be a P-function or belongs to the class

P (I)

, if f is non-negative and

f (α x + (1 - α) y) \leq f (x) + f (y)

for all

x, y \in I

and

α \in [0, 1]

For some results on the class

P (I)

, see [6, 7].

Definition 4 [8]

Let I be a p-convex set. A function

f : I \to R

is said to be a p-convex function or belongs to the class

P C (I)

, if

f ({[α x^{p} + (1 - α) y^{p}]}^{\frac{1}{p}}) \leq α f (x) + (1 - α) f (y)

for all

x, y \in I

and

α \in [0, 1]

Remark 1 [8]

An interval I is said to be a p-convex set if

{[α x^{p} + (1 - α) y^{p}]}^{\frac{1}{p}} \in I

for all

x, y \in I

and

α \in [0, 1]

, where

p = 2 k + 1

p = \frac{n}{m}

n = 2 r + 1

m = 2 t + 1

, and

k, r, t \in N

Definition 5 [9]

Let

h : J \to R

be a non-negative and non-zero function. We say that

f : I \to R

is an h-convex function or that f belongs to the class

S X (I)

, if f is non-negative and

f (α x + (1 - α) y) \leq h (α) f (x) + h (1 - α) f (y)

for all

x, y \in I

and

α \in (0, 1)

The h- and p-convex functions were introduced by Varšanec, Zhang and Wan, and a number of properties and Jensen’s inequalities of the functions were established (cf. [8]). As one can see, the definitions of the P-function, convex,

h, p, s

-convex, Godunova-Levin functions have similar forms. This observation leads us to generalize these varieties of convexity.

2 Definitions and basic results

In this section, we give new definitions and properties of the

(p, h)

-convex function. Throughout this paper, we assume that

(0, 1) \subseteq J

, f and h are real non-negative functions defined on I and J, respectively, and the set I is p-convex when

f \in g h x (p, h, I)

f \in g h v (p, h, I)

. We first give a definition of the new class of convex functions.

Definition 6 Let

h : J \to R

be a non-negative and non-zero function. We say that

f : I \to R

is a

(p, h)

-convex function or that f belongs to the class

g h x (h, p, I)

, if f is non-negative and

f ({[α x^{p} + (1 - α) y^{p}]}^{\frac{1}{p}}) \leq h (α) f (x) + h (1 - α) f (y)

(2.1)

for all

x, y \in I

and

α \in (0, 1)

. Similarly, if the inequality sign in (2.1) is reversed, then f is said to be a

(p, h)

-concave function or belong to the class

g h v (h, p, I)

Remark 2 It can be obviously seen that if

h (α) = α

, then all non-negative p-convex and p-concave functions belong to

g h x (h, p, I)

and

g h v (h, p, I)

, respectively; if

h (α) = α

and

p = 1

, then all non-negative convex functions belong to

g h x (h, p, I)

; if

h (α) = \frac{1}{α}

and

p = 1

, then

Q (I) = g h x (h, p, I)

; if

h (α) = α^{s}

s \in (0, 1)

, and

p = 1

, then

K_{s}^{2} \subseteq g h x (h, p, I)

; if

h (α) = 1

and

p = 1

, then

P (I) \subseteq g h x (h, p, I)

, and if

p = 1

, then

S X (I) \subseteq g h x (h, p, I)

Example 1 Let

h_{k} (α) = α^{k}

, where

k \leq 1

and

α > 0

. If f is a function defined as

f (x) = x^{p}

, where p is an odd number and

x \geq 0

, we then have

f ({[α x^{p} + (1 - α) y^{p}]}^{\frac{1}{p}}) \leq α f (x) + (1 - α) f (y) \leq h_{k} (α) f (x) + h_{k} (1 - α) f (y),

and hence, f belongs to

g h x (h_{k}, p, I)

Next, we discuss some interesting properties of

(p, h)

-convex (concave) functions, which include linearity, product, composition properties, and an ordered property of h and p. In addition, we give some interesting properties of the

(p, h)

-convex function, when h is a super(sub)-multiplicative function.

Property 1 If

f, g \in g h x (h, p, I)

and

λ > 0

, then

f + g, λ f \in g h x (h, p, I)

. Similarly, if

f, g \in g h v (h, p, I)

and

λ > 0

, then

f + g, λ f \in g h v (h, p, I)

Proof The proof immediately follows from the definitions of the classes

g h x (h, p, I)

and

g h v (h, p, I)

. □

Property 2 Let

h_{1}

and

h_{2}

be non-negative functions defined on an interval J with

h_{2} \leq h_{1}

(0, 1)

. If

f \in g h x (h_{2}, p, I)

, then

f \in g h x (h_{1}, p, I)

. Similarly, if

f \in g h v (h_{1}, p, I)

, then

f \in g h v (h_{2}, p, I)

Proof If

f \in g h x (h_{2}, p, I)

, then for any

x, y \in I

and

α \in (0, 1)

we have

\begin{array}{rcl} f ({[α x^{p} + (1 - α) y^{p}]}^{\frac{1}{p}}) & \leq & h_{2} (α) f (x) + h_{2} (1 - α) f (y) \\ \leq & h_{1} (α) f (x) + h_{1} (1 - α) f (y), \end{array}

and hence,

f \in g h x (h_{1}, p, I)

. □

Property 3 Let

f \in g h x (h, p_{1}, I)

(a)

For

I \subseteq (0, 1]

, if f is monotone increasing (monotone decreasing), and

p_{2} \geq p_{1} > 0

p_{2} \leq p_{1} < 0

, and (

p_{1} \geq p_{2} > 0

p_{1} \leq p_{2} < 0

), then

f \in g h x (h, p_{2}, I)

(b)

For

I \subseteq [1, \infty)

, if f is monotone increasing (monotone decreasing), and

p_{1} \geq p_{2} > 0

p_{1} \leq p_{2} < 0

, and (

p_{2} \geq p_{1} > 0

p_{2} \leq p_{1} < 0

), then

f \in g h x (h, p_{2}, I)

Let

f \in g h v (h, p_{1}, I)

(c)

For

I \subseteq (0, 1]

, if f is monotone increasing (monotone decreasing), and

p_{1} \geq p_{2} > 0

p_{1} \leq p_{2} < 0

, and (

p_{2} \geq p_{1} > 0

p_{2} \leq p_{1} < 0

), then

f \in g h v (h, p_{2}, I)

(d)

For

I \subseteq [1, \infty)

, if f is monotone increasing (monotone decreasing), and

p_{2} \geq p_{1} > 0

p_{2} \leq p_{1} < 0

, and (

p_{1} \geq p_{2} > 0

p_{1} \leq p_{2} < 0

), then

f \in g h v (h, p_{2}, I)

Proof (a) Setting

g (p) = {(α x^{p} + (1 - α) y^{p})}^{\frac{1}{p}}

, we have

g^{'} (p) = \frac{1}{p} {(α x^{p} + (1 - α) y^{p})}^{\frac{1}{p} - 1} (α x^{p} ln (x) + (1 - α) y^{p} ln (y)) .

When

p > 0

and

x, y \in (0, 1]

, we have

g^{'} (p) < 0

, and so

g (p_{2}) \leq g (p_{1})

. We then obtain

f (g (p_{2})) \leq f (g (p_{1})) \leq h (α) f (x) + (1 - α) f (y),

since f is monotone increasing and

f \in g h x (h, p_{1}, I)

. Therefore, we get

f \in g h x (h, p_{2}, I)

The results of (b), (c), and (d) follow by similar arguments as above. □

Property 4 Let f and g be similarly ordered functions on I, i.e.,

(f (x) - f (y)) (g (x) - g (y)) \geq 0

(2.2)

for all

x, y \in I

. If

f \in g h x (h_{1}, p, I)

g \in g h x (h_{2}, p, I)

, and

h (α) + h (1 - α) \leq c

for all

α \in (0, 1)

, where

h (t) = max (h_{1} (t), h_{2} (t))

and c is a fixed positive number, then the product fg belongs to

g h x (c h, p, I)

. Similarly, let f and g be oppositely ordered, i.e.,

(f (x) - f (y)) (g (x) - g (y)) \leq 0

for all

x, y \in I

. If

f \in g h v (h_{1}, p, I)

g \in g h v (h_{2}, p, I)

, and

h (α) + h (1 - α) \geq c

for all

α \in (0, 1)

, where

h (t) = min (h_{1} (t), h_{2} (t))

and c is a fixed positive number, then the product fg belongs to

g h v (c h, p, I)

Proof We only give a proof for the first part, since the result of the second part of this theorem follows by a similar argument. By (2.2), we have

f (x) g (x) + f (y) g (y) \geq f (x) g (y) + f (y) g (x) .

Let α and β be positive numbers such that

α + β = 1

. We then obtain

\begin{array}{rcl} f g ({[α x^{p} + β y^{p}]}^{\frac{1}{p}}) & \leq & (h_{1} (α) f (x) + h_{1} (β) f (y)) (h_{2} (α) g (x) + h_{2} (β) g (y)) \\ \leq & h^{2} (α) f g (x) + h (α) h (β) f (x) g (y) + h (α) h (β) f (y) g (x) + h^{2} (β) f g (y) \\ \leq & h^{2} (α) f g (x) + h (α) h (β) f (x) g (x) + h (α) h (β) f (y) g (y) + h^{2} (β) f g (y) \\ = & (h (α) + h (β)) (h (α) f g (x) + h (β) f g (y)) \\ \leq & c h (α) f g (x) + c h (β) f g (y), \end{array}

which completes the proof. □

Definition 7 [9]

A function

h : I \to R

is called a super-multiplicative function if

h (x y) \geq h (x) h (y)

(2.3)

for all

x, y \in J

If the inequality sign in (2.3) is reversed, then h is said to be a sub-multiplicative function, and if the equality holds in (2.3), then h is called a multiplicative function.

Example 2 Let

h (x) = c e^{x}

. If

c = 1

, then h is a multiplicative function. If

c > 1

, then h is a sub-multiplicative function, and if

0 < c < 1

, then h is a super-multiplicative function.

Property 5 Let I be an interval such that

0 \in I

. We then have the following.

(a)

f \in g h x (h, p, I)

f (0) = 0

, and h is super-multiplicative, then the inequality

f ({[α x^{p} + β y^{p}]}^{\frac{1}{p}}) \leq h (α) f (x) + h (β) f (y)

(2.4)

holds for all

x, y \in I

and all

α, β > 0

such that

α + β \leq 1

(b)

Let h be a non-negative function with

h (α) < \frac{1}{2}

for some

α \in (0, \frac{1}{2})

. If f is a non-negative function satisfying (2.4) for all

x, y \in I

and all

α, β > 0

with

α + β \leq 1

, then

f (0) = 0

(c)

f \in g h v (h, p, I)

f (0) = 0

, and h is sub-multiplicative, then the inequality

f ({[α x^{p} + β y^{p}]}^{\frac{1}{p}}) \geq h (α) f (x) + h (β) f (y)

(2.5)

holds for all

x, y \in I

and all

α, β > 0

such that

α + β \leq 1

(d)

Let h be a non-negative function with

h (α) > \frac{1}{2}

for some

α \in (0, \frac{1}{2})

. If f is a non-negative function satisfying (2.5) for all

x, y \in I

and all

α, β > 0

with

α + β \leq 1

, then

f (0) = 0

Proof (a) Let

α, β > 0

α + β = γ < 1

, and let a and b be numbers such that

a = \frac{α}{γ}

and

b = \frac{β}{γ}

. We then have

a + b = 1

and

\begin{array}{rcl} f ({[α x^{p} + β y^{p}]}^{\frac{1}{p}}) & = & f ({[a γ x^{p} + b γ y^{p}]}^{\frac{1}{p}}) \\ \leq & h (a) f (γ^{\frac{1}{p}} x) + h (b) f (γ^{\frac{1}{p}} y) \\ = & h (a) f ({[γ x^{p} + (1 - γ) 0^{p}]}^{\frac{1}{p}}) + h (b) f ({[γ y^{p} + (1 - γ) 0^{p}]}^{\frac{1}{p}}) \\ \leq & h (a) h (γ) f (x) + h (a) h (1 - γ) f (0) \\ + h (b) h (γ) f (y) + h (b) h (1 - γ) f (0) \\ = & h (a) h (γ) f (x) + h (b) h (γ) f (y) \\ \leq & h (a γ) f (x) + h (b γ) f (y) = h (α) f (x) + h (β) f (y) . \end{array}

(b) If

f (0) \neq 0

, then

f (0) > 0

. Setting

x = y = 0

in (2.4), we get

f (0) \leq h (α) f (0) + h (β) f (0) .

By setting

α = β

, where

α \in (0, \frac{1}{2})

, and dividing both sides of the inequality above by

f (0)

, we obtain

2 h (α) \geq 1

for all

α \in (0, \frac{1}{2})

, which is a contradiction to the assumption

h (α) < \frac{1}{2}

for some

α \in (0, \frac{1}{2})

, and so

f (0) = 0

The results of (c) and (d) follow by using similar arguments as above, and so we omit the proofs here. □

Corollary 1 Let

h_{s} (x) = x^{s}

, where

s, x > 0

, and let

0 \in I

. For all

f \in g h x (h_{s}, p, I)

, inequality (2.4) holds for all

α, β > 0

with

α + β \leq 1

if and only if

f (0) = 0

. For all

f \in g h v (h_{s}, p, I)

, inequality (2.5) holds for all

α, β > 0

with

α + β \leq 1

if and only if

f (0) = 0

Proof Let

α, β > 0

α + β = γ < 1

, and let a and b be positive numbers such that

a = \frac{α}{γ}

and

b = \frac{β}{γ}

. We then have

a + b = 1

and

\begin{array}{rcl} f ({[α x^{p} + β y^{p}]}^{\frac{1}{p}}) & = & f ({[a γ x^{p} + b γ y^{p}]}^{\frac{1}{p}}) \\ \leq & a^{s} f (γ^{\frac{1}{p}} x) + b^{s} f (γ^{\frac{1}{p}} y) \\ = & a^{s} f ({[γ x^{p} + (1 - γ) 0^{p}]}^{\frac{1}{p}}) + b^{s} f ({[γ y^{p} + (1 - γ) 0^{p}]}^{\frac{1}{p}}) \\ \leq & a^{s} γ^{s} f (x) + a^{s} {(1 - γ)}^{s} f (0) + b^{s} γ^{s} f (y) + b^{s} {(1 - γ)}^{s} f (0) \\ = & a^{s} γ^{s} f (x) + b^{s} γ^{s} f (y) \\ = & α^{s} f (x) + β^{s} f (y) . \end{array}

Setting

x = y = α = β = 0

in (2.4), we get

f (0) \leq 0

, while

f (0) \geq 0

by the definition of the

(p, h)

-convex function, and hence

f (0) = 0

. □

Property 6 Suppose that

h_{i} : J_{i} \to (0, \infty)

i = 1, 2

, are functions such that

h_{2} (J_{2}) \subseteq J_{1}

and

h_{2} (α) + h_{2} (1 - α) \leq 1

for all

α \in (0, 1)

, and that

f : I_{1} \to [0, \infty)

and

g : I_{2} \to [0, \infty)

are functions with

g (I_{2}) \subseteq I_{1}

0 \in I_{1}

, and

f (0) = 0

h_{1}

is a super-multiplicative function,

f \in S X (h_{1}, I_{1})

, and f is increasing (decreasing) and

g \in g h x (h_{2}, p, I_{2})

(

g \in g h v (h_{2}, p, I_{2})

), then the composite function

f \circ g

belongs to

g h x (h_{1} \circ h_{2}, p, I_{2})

. If

h_{1}

is a sub-multiplicative function,

f \in S V (h_{1}, I_{1})

, and f is increasing (decreasing) and

g \in g h v (h_{2}, p, I_{2})

(

g \in g h x (h_{2}, p, I_{2})

), then the composite function

f \circ g

belongs to

g h v (h_{1} \circ h_{2}, p, I_{2})

Proof If

g \in g h x (h_{2}, p, I_{2})

and f is an increasing function, then we have

(f \circ g) ({[α x^{p} + (1 - α) y^{p}]}^{\frac{1}{p}}) \leq f (h_{2} (α) g (x) + h_{2} (1 - α) g (y))

for all

x, y \in I_{2}

and

α \in (0, 1)

. Using Property 5(a) with

p = 1

, we obtain

f (h_{2} (α) g (x) + h_{2} (1 - α) g (y)) \leq h_{1} (h_{2} (α)) f (g (x)) + h_{1} (h_{2} (1 - α)) f (g (y)),

which implies that

f \circ g

belongs to

g h x (h_{1} \circ h_{2}, p, I_{2})

. □

If f is a convex or concave function, then we may give a similar statement on the composite function of f and g.

Property 7 Let

f : I_{1} \to [0, \infty)

and

g : I_{2} \to [0, \infty)

be functions with

g (I_{2}) \subseteq I_{1}

. If the function f is convex and increasing (decreasing), and

g \in g h x (h, p, I_{2})

(

g \in g h v (h, p, I_{2})

) with

h (α) + h (1 - α) = 1

for

α \in (0, 1)

, then

f \circ g

belongs to

g h x (h, p, I_{2})

. If the function f is concave and increasing (decreasing), and

g \in g h v (h, p, I_{2})

(

g \in g h x (h, p, I_{2})

) with

h (α) + h (1 - α) = 1

for

α \in (0, 1)

, then

f \circ g

belongs to

g h v (h, p, I_{2})

Proof If

g \in g h x (h, p, I_{2})

and f is an increasing function, we then have

(f \circ g) ({[α x^{p} + (1 - α) y^{p}]}^{\frac{1}{p}}) \leq f (h (α) g (x) + h (1 - α) g (y))

for all

x, y \in I_{2}

and

α \in (0, 1)

. Since

h (α) + h (1 - α) = 1

and f is convex, we obtain

f (h (α) g (x) + h (1 - α) g (y)) \leq h (α) f (g (x)) + h (1 - α) f (g (y)),

which implies that

f \circ g

belongs to

g h x (h, p, I_{2})

. □

3 Schur-type inequalities

In this section, we establish Schur-type inequalities of

(p, h)

-convex functions.

Theorem 1 Let

h : J \to R

be a non-negative super-multiplicative function and let

f : I \to R

be a function such that

f \in g h x (h, p, I)

. Then for all

x_{1}, x_{2}, x_{3} \in I

such that

x_{1} < x_{2} < x_{3}

and

x_{3}^{p} - x_{1}^{p}, x_{3}^{p} - x_{2}^{p}, x_{2}^{p} - x_{1}^{p} \in J

, the following inequality holds:

h (x_{3}^{p} - x_{2}^{p}) f (x_{1}) - h (x_{3}^{p} - x_{1}^{p}) f (x_{2}) + h (x_{2}^{p} - x_{1}^{p}) f (x_{3}) \geq 0 .

(3.1)

If the function h is sub-multiplicative and

f \in g h v (h, p, I)

, then the inequality sign in (3.1) is reversed.

Proof Let

f \in g h x (h, p, I)

and let

x_{1}, x_{2}, x_{3} \in I

be the numbers stated in this theorem. Then one can easily see that

\frac{x_{3}^{p} - x_{2}^{p}}{x_{3}^{p} - x_{1}^{p}}, \frac{x_{2}^{p} - x_{1}^{p}}{x_{3}^{p} - x_{1}^{p}} \in (0, 1) \subseteq J and \frac{x_{3}^{p} - x_{2}^{p}}{x_{3}^{p} - x_{1}^{p}} + \frac{x_{2}^{p} - x_{1}^{p}}{x_{3}^{p} - x_{1}^{p}} = 1 .

We also have

h (x_{3}^{p} - x_{2}^{p}) = h (\frac{x_{3}^{p} - x_{2}^{p}}{x_{3}^{p} - x_{1}^{p}} (x_{3}^{p} - x_{1}^{p})) \geq h (\frac{x_{3}^{p} - x_{2}^{p}}{x_{3}^{p} - x_{1}^{p}}) h (x_{3}^{p} - x_{1}^{p})

and

h (x_{2}^{p} - x_{1}^{p}) \geq h (\frac{x_{2}^{p} - x_{1}^{p}}{x_{3}^{p} - x_{1}^{p}}) h (x_{3}^{p} - x_{1}^{p}) .

Setting

α = \frac{x_{3}^{p} - x_{2}^{p}}{x_{3}^{p} - x_{1}^{p}}

x = x_{1}

, and

y = x_{3}

in (2.1), we have

x_{2}^{p} = α x^{p} + (1 - α) y^{p}

and

\begin{aligned} f (x_{2}) & \leq h (\frac{x_{3}^{p} - x_{2}^{p}}{x_{3}^{p} - x_{1}^{p}}) f (x_{1}) + h (\frac{x_{2}^{p} - x_{1}^{p}}{x_{3}^{p} - x_{1}^{p}}) f (x_{3}) \\ \leq \frac{h (x_{3}^{p} - x_{2}^{p})}{h (x_{3}^{p} - x_{1}^{p})} f (x_{1}) + \frac{h (x_{2}^{p} - x_{1}^{p})}{h (x_{3}^{p} - x_{1}^{p})} f (x_{3}) . \end{aligned}

(3.2)

Assuming

h (x_{3}^{p} - x_{1}^{p}) > 0

and multiplying both sides of the inequality above by

h (x_{3}^{p} - x_{1}^{p})

, we obtain inequality (3.1). □

Remark 3 In fact, if

f (x) = x^{λ}

λ \in R

h (x) = h_{- 1} (x) = \frac{1}{x}

p = 1

, and

x_{1}, x_{2}, x_{3} \in I = (0, 1)

, then inequality (3.1) gives the Schur inequality, see [[10], p.177].

The following corollary gives a Schur-type inequality for the

(p, h)

-convex function.

Corollary 2 If

f : I = (0, 1) \to I

belongs to the class

g h x (h_{- k}, p, I)

and

h_{- k} = \frac{1}{x^{k}}

, then we have the inequality

\begin{aligned} f (x_{1}) {(x_{3}^{p} - x_{1}^{p})}^{k} {(x_{2}^{p} - x_{1}^{p})}^{k} - f (x_{2}) {(x_{3}^{p} - x_{2}^{p})}^{k} {(x_{2}^{p} - x_{1}^{p})}^{k} \\ + f (x_{3}) {(x_{3}^{p} - x_{1}^{p})}^{k} {(x_{3}^{p} - x_{2}^{p})}^{k} \geq 0 \end{aligned}

(3.3)

for all

x_{1}, x_{2}, x_{3} \in I

with

x_{1} < x_{2} < x_{3}

. If

f \in g h v (h_{- k}, p, I)

, then the inequality sign in (3.3) is reversed. If

k = 1

p = 1

, and

f (x) = x^{λ}

λ \in R

, then

f \in g h x (h_{- 1}, 1, I)

and inequality (3.3) gives the Schur inequality.

4 Jensen-type inequalities

In this section, we introduce some Jensen-type inequalities of

(p, h)

-convex functions.

Theorem 2 Let

w_{1}, \dots, w_{n}

be positive real numbers with

n \geq 2

. If h is a non-negative super-multiplicative function and if

f \in g h x (h, p, I)

and

x_{1}, \dots, x_{n} \in I

, then we have the inequality

f ({[\frac{1}{W_{n}} \sum_{i = 1}^{n} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) \leq \sum_{i = 1}^{n} h (\frac{w_{i}}{W_{n}}) f (x_{i}), where W_{n} = \sum_{i = 1}^{n} w_{i} .

(4.1)

If h is sub-multiplicative and

f \in g h v (h, p, I)

, then the inequality sign in (4.1) is reversed.

Proof When

n = 2

, inequality (4.1) holds by (2.1) with

α = \frac{w_{1}}{W_{2}}

. Assuming inequality (4.1) holds for

n - 1

, we obtain

\begin{array}{rcl} f ({[\frac{1}{W_{n}} \sum_{i = 1}^{n} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) & = & f ({[\frac{w_{n}}{W_{n}} x_{n}^{p} + \sum_{i = 1}^{n - 1} \frac{w_{i}}{W_{n}} x_{i}^{p}]}^{\frac{1}{p}}) \\ = & f ({[\frac{w_{n}}{W_{n}} x_{n}^{p} + \frac{W_{n - 1}}{W_{n}} \sum_{i = 1}^{n - 1} \frac{w_{i}}{W_{n - 1}} x_{i}^{p}]}^{\frac{1}{p}}) \\ \leq & h (\frac{w_{n}}{W_{n}}) f (x_{n}) + h (\frac{W_{n - 1}}{W_{n}}) f ({[\sum_{i = 1}^{n - 1} \frac{w_{i}}{W_{n - 1}} x_{i}^{p}]}^{\frac{1}{p}}) \\ \leq & h (\frac{w_{n}}{W_{n}}) f (x_{n}) + h (\frac{W_{n - 1}}{W_{n}}) \sum_{i = 1}^{n - 1} h (\frac{w_{i}}{W_{n - 1}}) f (x_{i}) \\ \leq & \sum_{i = 1}^{n} h (\frac{w_{i}}{W_{n}}) f (x_{i}), \end{array}

and, hence, the result follows by mathematical induction. □

Remark 4 For

h (α) = α

and

p = 1

, inequality (4.1) becomes the classical Jensen inequality.

Theorem 3 Let

w_{1}, \dots, w_{n}

be positive real numbers and let

(m, M)

be an interval in I. If

h : (0, \infty) \to R

is a non-negative super-multiplicative function and

f \in g h x (h, p, I)

, then for all

x_{1}, \dots, x_{n} \in (m, M)

we have the inequality

\begin{aligned} \sum_{i = 1}^{n} h (\frac{w_{i}}{W_{n}}) f (x_{i}) \leq & f (m) \sum_{i = 1}^{n} h (\frac{w_{i}}{W_{n}}) h (\frac{M^{p} - x_{i}^{p}}{M^{p} - m^{p}}) \\ + f (M) \sum_{i = 1}^{n} h (\frac{w_{i}}{W_{n}}) h (\frac{x_{i}^{p} - m^{p}}{M^{p} - m^{p}}) . \end{aligned}

(4.2)

If h is a non-negative sub-multiplicative function and

f \in g h v (h, p, I)

, then the inequality sign in (4.2) is reversed.

Proof Setting

x_{1} = m

x_{2} = x_{i}

, and

x_{3} = M

in (3.2), we get the inequalities

f (x_{i}) \leq h (\frac{M^{p} - x_{i}^{p}}{M^{p} - m^{p}}) f (m) + h (\frac{x_{i}^{p} - m^{p}}{M^{p} - m^{p}}) f (M), i = 1, \dots, n .

Multiplying both sides of the above inequality with

h (\frac{w_{i}}{W_{n}})

and adding all inequalities side by side for

i = 1, \dots, n

, we obtain (4.2). □

Let K be a finite nonempty set of positive integers and let F be an index set function defined by

F (K) = h (W_{K}) f ({[\frac{1}{W_{K}} \sum_{i \in K} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) - \sum_{i \in K} h (w_{i}) f (x_{i}), where W_{K} = \sum_{i \in K} w_{i} .

Theorem 4 Let

h : (0, \infty) \to R

be a non-negative function, and let M and K be finite nonempty sets of positive integers such that

M \cap K = \emptyset

. If h is super-multiplicative and

f : I \to R

belongs to the class

g h x (h, p, I)

, then for

w_{i} > 0

x_{i} \in I

i \in M \cup K

we have the inequality

F (M \cup K) \leq F (M) + F (K) .

(4.3)

If h is sub-multiplicative and

f \in g h v (h, p, I)

, then the inequality sign in (4.3) is reversed.

Proof Setting

x = {[\frac{1}{W_{M}} \sum_{i \in M} w_{i} x_{i}^{p}]}^{\frac{1}{p}}

y = {[\frac{1}{W_{K}} \sum_{i \in K} w_{i} x_{i}^{p}]}^{\frac{1}{p}}

, and

α = \frac{W_{M}}{W_{M \cup K}}

in (2.1), we obtain the inequality

\begin{aligned} f ({[\frac{1}{W_{M \cup K}} \sum_{i \in M \cup K} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) \\ \leq h (\frac{W_{M}}{W_{M \cup K}}) f ({[\frac{1}{W_{M}} \sum_{i \in M} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) + h (\frac{W_{K}}{W_{M \cup K}}) f ({[\frac{1}{W_{K}} \sum_{i \in K} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) . \end{aligned}

Multiplying both sides of the above inequality with

h (W_{M \cup K})

, we get the inequality

\begin{aligned} h (W_{M \cup K}) f ({[\frac{1}{W_{M \cup K}} \sum_{i \in M \cup K} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) \\ \leq h (W_{M}) f ({[\frac{1}{W_{M}} \sum_{i \in M} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) + h (W_{K}) f ({[\frac{1}{W_{K}} \sum_{i \in K} w_{i} x_{i}^{p}]}^{\frac{1}{p}}) . \end{aligned}

Subtracting

\sum_{i \in M \cup K} h (w_{i}) f (x_{i})

from both sides of the inequality above and using the identity

\sum_{i \in M \cup K} h (w_{i}) f (x_{i}) = \sum_{i \in M} h (w_{i}) f (x_{i}) + \sum_{i \in K} h (w_{i}) f (x_{i})

, we obtain (4.3). □

A simple consequence of Theorem 4 is stated in the following corollary without proof.

Corollary 3 Let

h : (0, \infty) \to R

be a non-negative super-multiplicative function. If

w_{i} > 0

i = 1, \dots, n

, and

M_{k} = {1, \dots, K}

, then for

f \in g h x (h, p, I)

we have

F (M_{n}) \leq F (M_{n - 1}) \leq \dots \leq F (M_{2}) \leq 0

(4.4)

and

F (M_{n}) \leq min_{1 \leq i < j \leq n} {h (w_{i} + w_{j}) f ({[\frac{w_{i} x_{i}^{p} + w_{j} x_{j}^{p}}{w_{i} + w_{j}}]}^{\frac{1}{p}}) - h (w_{i}) f (x_{i}) - h (w_{j}) f (x_{j})} .

(4.5)

If h is sub-multiplicative and

f \in g h v (h, p, I)

, then the inequality signs in (4.4) and (4.5) are reversed, and min is replaced with max.

Remark 5 Some results obtained from Theorem 4 and Corollary 3 are given in [[11], p.7], when

h (α) = α

p = 1

, and h is a convex or concave function.

5 Hadamard-type inequalities

In this section, we give some Hadamard-type inequalities of

(p, h)

-convex functions.

Theorem 5 If

f \in g h x (h, p, I) \cap L_{1} ([a, b])

for

a, b \in I

with

a < b

, then we have

\frac{1}{2 h (\frac{1}{2})} f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \leq \frac{p}{b^{p} - a^{p}} \int_{a}^{b} x^{p - 1} f (x) d x \leq (f (a) + f (b)) \int_{0}^{1} h (t) d t .

(5.1)

Proof Setting

x^{p} = \frac{y - a}{b - a} b^{p} + \frac{b - y}{b - a} a^{p}

, we get

\frac{p}{b^{p} - a^{p}} \int_{a}^{b} x^{p - 1} f (x) d x = \frac{1}{b - a} \int_{a}^{b} f ({[\frac{y - a}{b - a} b^{p} + \frac{b - y}{b - a} a^{p}]}^{\frac{1}{p}}) d y .

By using inequality (2.1) we obtain

f ({[\frac{y - a}{b - a} b^{p} + \frac{b - y}{b - a} a^{p}]}^{\frac{1}{p}}) \leq h (\frac{y - a}{b - a}) f (b) + h (\frac{b - y}{b - a}) f (a),

and hence, by integrating the above inequality over

[a, b]

, we have

\begin{array}{rcl} \int_{a}^{b} f ({[\frac{y - a}{b - a} b^{p} + \frac{b - y}{b - a} a^{p}]}^{\frac{1}{p}}) d y & \leq & f (b) \int_{a}^{b} h (\frac{y - a}{b - a}) d y + f (a) \int_{a}^{b} h (\frac{b - y}{b - a}) d y \\ = & (b - a) (f (a) + f (b)) \int_{0}^{1} h (t) d t, \end{array}

which gives the second inequality.

Setting

y = \frac{1}{2} (a + b) + t

, we obtain

\begin{aligned} \int_{- \frac{1}{2} (b - a)}^{\frac{1}{2} (b - a)} f ({[\frac{1}{2} (a^{p} + b^{p}) + \frac{b^{p} - a^{p}}{b - a} t]}^{\frac{1}{p}}) d t \\ = \int_{0}^{\frac{1}{2} (b - a)} f ({[\frac{1}{2} (a^{p} + b^{p}) + \frac{b^{p} - a^{p}}{b - a} t]}^{\frac{1}{p}}) d t \\ + \int_{0}^{\frac{1}{2} (b - a)} f ({[\frac{1}{2} (a^{p} + b^{p}) - \frac{b^{p} - a^{p}}{b - a} t]}^{\frac{1}{p}}) d t \\ \geq \frac{1}{h (1 / 2)} \int_{0}^{\frac{1}{2} (b - a)} f ({[\frac{1}{2} (a^{p} + b^{p})]}^{\frac{1}{p}}) d t = \frac{b - a}{2 h (1 / 2)} f ({[\frac{1}{2} (a^{p} + b^{p})]}^{\frac{1}{p}}), \end{aligned}

and, hence, the first inequality follows. □

Remark 6 If

h (α) = α

and

p = 1

, then inequality (5.1) gives the classical Hadamard inequality.

Theorem 6 Suppose that f and g are functions such that

f \in g h x (h_{1}, p, I)

g \in g h x (h_{2}, p, I)

f g \in L_{1} ([a, b])

, and

h_{1} h_{2} \in L_{1} ([0, 1])

with

a, b \in I

and

a < b

. We then have

\begin{aligned} \frac{p}{b^{p} - a^{p}} \int_{a}^{b} x^{p - 1} f (x) g (x) d x \leq & M (a, b) \int_{0}^{1} h_{1} (t) h_{2} (t) d t \\ + N (a, b) \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t, \end{aligned}

(5.2)

where

M (a, b) = f (a) g (a) + f (b) g (b)

and

N (a, b) = f (a) g (b) + f (b) g (a)

Proof Since

f \in g h x (h_{1}, p, I)

and

g \in g h x (h_{2}, p, I)

, we have

\begin{matrix} f ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) \leq h_{1} (t) f (a) + h_{1} (1 - t) f (b), \\ g ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) \leq h_{2} (t) g (a) + h_{2} (1 - t) g (b) \end{matrix}

for all

t \in [0, 1]

. Because f and g are non-negative, we get the inequality

\begin{aligned} f ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) g ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) \\ \leq h_{1} (t) h_{2} (t) f (a) g (a) + h_{1} (1 - t) h_{2} (t) f (b) g (a) + h_{1} (t) h_{2} (1 - t) f (a) g (b) \\ + h_{1} (1 - t) h_{2} (1 - t) f (b) g (b) . \end{aligned}

Integrating both sides of the above inequality over

(0, 1)

, we obtain the inequality

\begin{aligned} \int_{0}^{1} f ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) g ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) d t \\ \leq f (a) g (a) \int_{0}^{1} h_{1} (t) h_{2} (t) d t + f (b) g (a) \int_{0}^{1} h_{1} (1 - t) h_{2} (t) d t \\ + f (a) g (b) \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t + f (b) g (b) \int_{0}^{1} h_{1} (1 - t) h_{2} (1 - t) d t . \end{aligned}

Setting

x = {[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}

, we get

\frac{p}{b^{p} - a^{p}} \int_{a}^{b} x^{p - 1} f (x) g (x) d x \leq M (a, b) \int_{0}^{1} h_{1} (t) h_{2} (t) d t + N (a, b) \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t .

□

Theorem 7 Let

f \in g h x (h_{1}, p, I)

g \in g h x (h_{2}, p, I)

be functions such that

f g \in L_{1} ([a, b])

and

h_{1} h_{2} \in L_{1} ([0, 1])

, and let

a, b \in I

with

a < b

. We then have

\begin{aligned} \frac{1}{2 h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2})} f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) - \frac{p}{b^{p} - a^{p}} \int_{a}^{b} x^{p - 1} f (x) g (x) d x \\ \leq M (a, b) \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t + N (a, b) \int_{0}^{1} h_{1} (t) h_{2} (t) d t . \end{aligned}

(5.3)

Proof Since

\frac{a^{p} + b^{p}}{2} = \frac{t a^{p} + (1 - t) b^{p}}{2} + \frac{(1 - t) a^{p} + t b^{p}}{2}

, we have

\begin{aligned} f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \\ = f ({[\frac{t a^{p} + (1 - t) b^{p}}{2} + \frac{(1 - t) a^{p} + t b^{p}}{2}]}^{\frac{1}{p}}) g ({[\frac{t a^{p} + (1 - t) b^{p}}{2} + \frac{(1 - t) a^{p} + t b^{p}}{2}]}^{\frac{1}{p}}) \\ = h_{1} (\frac{1}{2}) [f ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) + f ({[(1 - t) a^{p} + t b^{p}]}^{\frac{1}{p}})] \\ \times h_{2} (\frac{1}{2}) [g ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) + g ({[(1 - t) a^{p} + t b^{p}]}^{\frac{1}{p}})] \\ \leq h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [f ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) g ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}})] \\ + h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [f ({[(1 - t) a^{p} + t b^{p}]}^{\frac{1}{p}}) g ({[(1 - t) a^{p} + t b^{p}]}^{\frac{1}{p}})] \\ + h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [h_{1} (t) f (a) + h_{1} (1 - t) f (b)] [h_{2} (1 - t) g (a) + h_{2} (t) g (b)] \\ + h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [h_{1} (1 - t) f (a) + h_{1} (t) f (b)] [h_{2} (t) g (a) + h_{2} (1 - t) g (b)] \\ = h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [f ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}}) g ({[t a^{p} + (1 - t) b^{p}]}^{\frac{1}{p}})] \\ + h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [f ({[(1 - t) a^{p} + t b^{p}]}^{\frac{1}{p}}) g ({[(1 - t) a^{p} + t b^{p}]}^{\frac{1}{p}})] \\ + h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [(h_{1} (t) h_{2} (1 - t) + h_{1} (1 - t) h_{2} (t)) M (a, b)] \\ + h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) [(h_{1} (t) h_{2} (t) + h_{1} (1 - t) h_{2} (1 - t)) N (a, b)] . \end{aligned}

Integrating the above inequality over

[0, 1]

, we obtain

\begin{aligned} \frac{1}{2 h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2})} f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) - \frac{p}{b^{p} - a^{p}} \int_{a}^{b} x^{p - 1} f (x) g (x) d x \\ \leq M (a, b) \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t + N (a, b) \int_{0}^{1} h_{1} (t) h_{2} (t) d t . \end{aligned}

□

Theorem 8 Let

f \in g h x (h_{1}, p, I)

and

g \in g h x (h_{2}, p, I)

be functions such that

f g \in L_{1} ([a, b])

h_{1} h_{2} \in L_{1} ([0, 1])

, and let

a, b \in I

with

a < b

. We then have the inequality

\begin{aligned} \frac{p^{2}}{2 {(b^{p} - a^{p})}^{2}} \int_{a}^{b} \int_{a}^{b} \int_{0}^{1} x^{p - 1} y^{p - 1} f ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) d x d y d t \\ \leq \frac{p}{b^{p} - a^{p}} \int_{0}^{1} h_{1} (t) h_{2} (t) d t \int_{a}^{b} x^{p - 1} f (x) g (x) d x \\ + \int_{0}^{1} h_{1} (t) d t \int_{0}^{1} h_{2} (t) d t \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t [M (a, b) + N (a, b)] . \end{aligned}

(5.4)

Proof Since

f \in g h x (h_{1}, p, I)

and

g \in g h x (h_{2}, p, I)

, we have

\begin{matrix} f ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) \leq h_{1} (t) f (x) + h_{1} (1 - t) f (y), \\ g ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) \leq h_{2} (t) g (x) + h_{2} (1 - t) g (y) \end{matrix}

for all

t \in [0, 1]

. Because f and g are non-negative, we get the inequality

\begin{aligned} f ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) \\ \leq h_{1} (t) h_{2} (t) f (x) g (x) + h_{1} (1 - t) h_{2} (t) f (y) g (x) + h_{1} (t) h_{2} (1 - t) f (x) g (y) \\ + h_{1} (1 - t) h_{2} (1 - t) f (y) g (y) . \end{aligned}

Multiplying both sides of the above inequality with

\frac{p^{2} x^{p - 1} y^{p - 1}}{{(b^{p} - a^{p})}^{2}}

and integrating the result over

[a, b]

and

[0, 1]

, we obtain the inequality

\begin{aligned} \frac{p^{2}}{{(b^{p} - a^{p})}^{2}} \int_{a}^{b} \int_{a}^{b} \int_{0}^{1} x^{p - 1} y^{p - 1} f ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) d x d y d t \\ \leq \int_{0}^{1} h_{1} (t) h_{2} (t) d t [\frac{p^{2}}{{(b^{p} - a^{p})}^{2}} (\int_{a}^{b} x^{p - 1} f (x) g (x) d x \int_{a}^{b} y^{p - 1} d y \\ + \int_{a}^{b} y^{p - 1} f (y) g (y) d y \int_{a}^{b} x^{p - 1} d x)] \\ + 2 \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t [\frac{p^{2}}{{(b^{p} - a^{p})}^{2}} \int_{a}^{b} x^{p - 1} f (x) d x \int_{a}^{b} y^{p - 1} f (y) d y] . \end{aligned}

By (5.1), we have the inequality

\begin{aligned} \frac{p^{2}}{2 {(b^{p} - a^{p})}^{2}} \int_{a}^{b} \int_{a}^{b} \int_{0}^{1} x^{p - 1} y^{p - 1} f ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) y^{p}]}^{\frac{1}{p}}) d x d y d t \\ \leq \frac{p}{b^{p} - a^{p}} \int_{0}^{1} h_{1} (t) h_{2} (t) d t \int_{a}^{b} x^{p - 1} f (x) g (x) d x \\ + \int_{0}^{1} h_{1} (t) d t \int_{0}^{1} h_{2} (t) d t \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t [M (a, b) + N (a, b)] . \end{aligned}

□

Theorem 9 Let

f \in g h x (h_{1}, p, I)

g \in g h x (h_{2}, p, I)

be functions such that

f g \in L_{1} ([a, b])

h_{1} h_{2} \in L_{1} ([0, 1])

, and let

a, b \in I

with

a < b

. We then have the inequality

\begin{aligned} \int_{a}^{b} \int_{0}^{1} x^{p - 1} f ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) d t d x \\ \leq \int_{0}^{1} h_{1} (t) h_{2} (t) d t \int_{a}^{b} x^{p - 1} f (x) g (x) d x + \frac{b^{p} - a^{p}}{p} [M (a, b) + N (a, b)] \\ \times [h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) \int_{0}^{1} h_{1} (t) h_{2} (t) d t \\ + [h_{1} (\frac{1}{2}) \int_{0}^{1} h_{2} (t) d t + h_{2} (\frac{1}{2}) \int_{0}^{1} h_{1} (t) d t] \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t] . \end{aligned}

(5.5)

Proof Since

f \in g h x (h_{1}, p, I)

and

g \in g h x (h_{2}, p, I)

, we have the inequalities

\begin{matrix} f ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \leq h_{1} (t) f (x) + h_{1} (1 - t) f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}), \\ g ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \leq h_{2} (t) g (x) + h_{2} (1 - t) g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \end{matrix}

for all

t \in [0, 1]

. Because f and g are non-negative, we get the inequality

\begin{aligned} f ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \\ \leq h_{1} (t) h_{2} (t) f (x) g (x) + h_{1} (1 - t) h_{2} (t) f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g (x) \\ + h_{1} (t) h_{2} (1 - t) f (x) g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \\ + h_{1} (1 - t) h_{2} (1 - t) f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) . \end{aligned}

Multiplying both sides of the inequality above with

x^{p - 1}

and integrating the result over

[a, b]

and

[0, 1]

, we obtain

\begin{aligned} \int_{a}^{b} \int_{0}^{1} x^{p - 1} f ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) d t d x \\ \leq \int_{0}^{1} h_{1} (t) h_{2} (t) d t [\int_{a}^{b} x^{p - 1} f (x) g (x) d x + \frac{b^{p} - a^{p}}{p} f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}})] \\ + \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t [g ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \int_{a}^{b} x^{p - 1} f (x) d x \\ + f ({[\frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) \int_{a}^{b} x^{p - 1} g (x) d x] . \end{aligned}

By inequality (5.1), we have

\begin{aligned} \int_{a}^{b} \int_{0}^{1} x^{p - 1} f ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) g ({[t x^{p} + (1 - t) \frac{a^{p} + b^{p}}{2}]}^{\frac{1}{p}}) d t d x \\ \leq \int_{0}^{1} h_{1} (t) h_{2} (t) d t [\int_{a}^{b} x^{p - 1} f (x) g (x) d x \\ + \frac{b^{p} - a^{p}}{p} h_{1} (\frac{1}{2}) (f (a) + f (b)) h_{2} (\frac{1}{2}) (g (a) + g (b))] \\ + \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t \frac{b^{p} - a^{p}}{p} h_{2} (\frac{1}{2}) (f (a) + f (b)) (g (a) + g (b)) \int_{0}^{1} h_{1} (t) d t \\ + \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t \frac{b^{p} - a^{p}}{p} h_{1} (\frac{1}{2}) (f (a) + f (b)) (g (a) + g (b)) \int_{0}^{1} h_{2} (t) d t \\ = \int_{0}^{1} h_{1} (t) h_{2} (t) d t \int_{a}^{b} x^{p - 1} f (x) g (x) d x + \frac{b^{p} - a^{p}}{p} [M (a, b) + N (a, b)] \\ \times [h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) \int_{0}^{1} h_{1} (t) h_{2} (t) d t \\ + [h_{1} (\frac{1}{2}) \int_{0}^{1} h_{2} (t) d t + h_{2} (\frac{1}{2}) \int_{0}^{1} h_{1} (t) d t] \int_{0}^{1} h_{1} (t) h_{2} (1 - t) d t] . \end{aligned}

□

Acknowledgements

This work is supported by the Natural Science Foundation of Shandong Province of China (ZR2012AM018) and the Fundamental Research Funds for the Central Universities (No. 201362032). The authors would like to deeply thank all the reviewers for their insightful and constructive comments.

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript.

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