Erschienen in:

Open Access 01.12.2012 | Research

On the refinements of the Hermite-Hadamard inequality

verfasst von: Sadia Khalid, Josip Pečarić

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2012

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Abstract

In this paper, we present some refinements of the classical Hermite-Hadamard integral inequality for convex functions. Further, we give the concept of n-exponential convexity and log-convexity of the functions associated with the linear functionals defined by these inequalities and prove monotonicity property of the generalized Cauchy means obtained via these functionals. Finally, we give several examples of the families of functions for which the results can be applied.

MSC: 26D15.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Both authors worked jointly on the results and they read and approved the final manuscript.

1 Introduction

One of the most well-known inequalities in mathematics for convex functions is so called Hermite-Hadamard integral inequality (see [[1], p.137])

f (\frac{a + b}{2}) \leq \frac{1}{b - a} \int_{a}^{b} f (t) d t \leq \frac{f (a) + f (b)}{2},

(1)

provided that for an interval

[a, b] \subseteq R

f : [a, b] \to R

is a convex function. If the function f is concave, then (1) holds in the reverse direction. It gives an estimate from below and above of the mean value of a convex function. These inequalities for convex functions play an important role in nonlinear analysis. In recent years there have been many extensions, generalizations and similar type results of the inequalities in (1) (see [2‐4]). These classical inequalities have been improved and generalized in many ways and applied for special means including Stolarsky-type means, logarithmic and p-logarithmic means. Also, many interesting applications of Hermite-Hadamard inequality can be found in [1].

In this paper, we present some refinements of the first Hermite-Hadamard integral inequality. Further, we study the n-exponential convexity and log-convexity of the functions associated with the linear functionals defined as differences of the left-hand and the right-hand sides of these inequalities. We also prove monotonicity property of the generalized Cauchy means obtained via these functionals. Finally, we give several examples of the families of functions for which the obtained results can be applied.

2 Main results

We shall start with the following refinement of the first Hermite-Hadamard inequality for differentiable convex functions.

Theorem 2.1 Let

a, b \in R

with

a < b

and

f : [a, b] \to R

be a differentiable convex function. Then the function

H (x) = \frac{1}{b - a} [\int_{a}^{x} f (t) d t + (b - x) f (x)] - f (\frac{2 b x - x^{2} - a^{2}}{2 (b - a)})

(2)

is increasing on

[a, b]

and for all

x, y \in [a, b]

such that

x \leq y

, we have

0 \leq H (x) \leq H (y) \leq \frac{1}{b - a} \int_{a}^{b} f (t) d t - f (\frac{a + b}{2}) .

(3)

Proof We have

H^{'} (x) = \frac{b - x}{b - a} [f^{'} (x) - f^{'} (\frac{2 b x - x^{2} - a^{2}}{2 (b - a)})],

where

\frac{b - x}{b - a} \geq 0

, as

a < b

and

x \in [a, b]

. If we prove that

f^{'} (x) - f^{'} (\frac{2 b x - x^{2} - a^{2}}{2 (b - a)}) \geq 0,

(4)

then

H (x)

will be increasing on

[a, b]

For

x \geq a

, we have

x - \frac{2 b x - x^{2} - a^{2}}{2 (b - a)} \geq 0

. Since f is a differentiable convex function defined on

[a, b]

f^{'}

is increasing on

[a, b]

, and so (4) holds, which in turn implies that

H^{'} (x) \geq 0

, showing that

H (x)

is increasing on

[a, b]

Now, as

H (x)

is increasing on

[a, b]

, for any

x, y \in [a, b]

such that

a \leq x \leq y \leq b

, we have

H (a) \leq H (x) \leq H (y) \leq H (b) .

(5)

x = a

and at

x = b

, (2) gives

H (a) = 0

and

H (b) = \frac{1}{b - a} \int_{a}^{b} f (t) d t - f (\frac{a + b}{2})

respectively. By using these values of

H (a)

and

H (b)

in (5), we have (3). □

Remark 2.2 If f is strictly convex, then

H (x)

is strictly increasing on

[a, b)

and strict inequalities hold in (3).

The second main result is another refinement of the first Hermite-Hadamard inequality for differentiable convex functions.

Theorem 2.3 Let

a, b \in R

with

a < b

and

f : [a, b] \to R

be a differentiable convex function. Then the function

\bar{H} (x) = \frac{1}{b - a} [\int_{x}^{b} f (t) d t + (x - a) f (x)] - f (\frac{x^{2} + b^{2} - 2 a x}{2 (b - a)})

(6)

is decreasing on

[a, b]

and for any

x, y \in [a, b]

such that

x \leq y

, we have

0 \leq \bar{H} (y) \leq \bar{H} (x) \leq \frac{1}{b - a} \int_{a}^{b} f (t) d t - f (\frac{a + b}{2}) .

(7)

Proof We have

{\bar{H}}^{'} (x) = \frac{x - a}{b - a} [f^{'} (x) - f^{'} (\frac{x^{2} + b^{2} - 2 a x}{2 (b - a)})],

where

\frac{x - a}{b - a} \geq 0

, as

a < b

and

x \in [a, b]

. If we prove that

f^{'} (x) - f^{'} (\frac{x^{2} + b^{2} - 2 a x}{2 (b - a)}) \leq 0,

(8)

then

\bar{H} (x)

will be decreasing on

[a, b]

For

x \leq b

, we have

\frac{x^{2} + b^{2} - 2 a x}{2 (b - a)} - x \geq 0

. Since f is a differentiable convex function defined on

[a, b]

f^{'}

is increasing on

[a, b]

, and so (8) holds, which in turn implies that

{\bar{H}}^{'} (x) \leq 0

, showing that

\bar{H} (x)

is decreasing on

[a, b]

Now, as

\bar{H} (x)

is decreasing on

[a, b]

, for any

x, y \in [a, b]

such that

a \leq x \leq y \leq b

, we have

\bar{H} (b) \leq \bar{H} (y) \leq \bar{H} (x) \leq \bar{H} (a) .

(9)

x = a

and at

x = b

, (6) gives

\bar{H} (a) = \frac{1}{b - a} \int_{a}^{b} f (t) d t - f (\frac{a + b}{2})

and

\bar{H} (b) = 0

respectively. By using these values of

\bar{H} (a)

and

\bar{H} (b)

in (9), we have (7). □

Remark 2.4 If f is strictly convex, then

\bar{H} (x)

is strictly decreasing on

(a, b]

and strict inequalities hold in (7).

Let us observe the inequalities (3) and (7). Motivated by them, we define two functionals

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-155/MediaObjects/13660_2012_Article_310_Equ10_HTML.gif

(10)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-155/MediaObjects/13660_2012_Article_310_Equ11_HTML.gif

(11)

where

x, y \in [a, b]

and the functions H and

\bar{H}

are as in (2) and (6) respectively. If f is a differentiable convex function defined on

[a, b]

, then Theorems 2.1 and 2.3 imply that

Φ_{i} (x, y; f) \geq 0

i = 1, 2

. Now, we give mean value theorems for the functionals

Φ_{i}

i = 1, 2

Theorem 2.5 Let

f \in C^{2} [a, b]

and

x, y \in [a, b]

be such that

x \leq y

. Let

Φ_{1}

and

Φ_{2}

be linear functionals defined as in (10) and (11). Then there exists

ξ_{1}, ξ_{2} \in [a, b]

such that

Φ_{i} (x, y; f) = \frac{f^{″} (ξ_{i})}{2} Φ_{i} (x, y; f_{0}), i = 1, 2,

(12)

where

f_{0} (z) = z^{2}

Proof Analogous to the proof of Theorem 2.4 in [5]. □

Theorem 2.6 Let

f, g \in C^{2} [a, b]

and

x, y \in [a, b]

be such that

x \leq y

, where

g^{″} (x) \neq 0

for every

x \in [a, b]

. Let

Φ_{1}

and

Φ_{2}

be linear functionals defined as in (10) and (11). If

Φ_{1}

and

Φ_{2}

are positive, then there exists

ξ_{1}, ξ_{2} \in [a, b]

such that

\frac{Φ_{i} (x, y; f)}{Φ_{i} (x, y; g)} = \frac{f^{″} (ξ_{i})}{g^{″} (ξ_{i})}, i = 1, 2 .

(13)

Proof Analogous to the proof of Theorem 2.6 in [5]. □

Remark 2.7 If the inverse of the function

\frac{f^{″}}{g^{″}}

exists, then (13) gives

ξ_{i} = {(\frac{f^{″}}{g^{″}})}^{- 1} (\frac{Φ_{i} (x, y; f)}{Φ_{i} (x, y; g)}), i = 1, 2 .

(14)

3 n-exponential convexity and log-convexity of the Hermite-Hadamard differences

We begin this section by recollecting the definitions and properties which are going to be explored here and also some useful characterizations of these properties. Throughout the paper, I is an open interval in

R

Definition 1 A function

h : I \to R

is n-exponentially convex in the Jensen sense on I if

\sum_{i, j = 1}^{n} α_{i} α_{j} h (\frac{x_{i} + x_{j}}{2}) \geq 0

holds for every

α_{i} \in R

and

x_{i} \in I

i = 1, \dots, n

(see [5]).

Definition 2 A function

h : I \to R

is n-exponentially convex on I if it is n-exponentially convex in the Jensen sense and continuous on I.

Remark 3.1 From the above definition it is clear that 1-exponentially convex functions in the Jensen sense are nonnegative functions. Also, n-exponentially convex functions in the Jensen sense are k-exponentially convex functions in the Jensen sense for all

k \in N

k \leq n

By definition of positive semi-definite matrices and some basic linear algebra, we have the following proposition.

Proposition 3.2 If h is n-exponentially convex in the Jensen sense, then the matrix

{[h (\frac{x_{i} + x_{j}}{2})]}_{i, j = 1}^{k}

is a positive semi-definite matrix for all

k \in N

k \leq n

. Particularly,

det {[h (\frac{x_{i} + x_{j}}{2})]}_{i, j = 1}^{k} \geq 0 for every k \in N, k \leq n, x_{i} \in I, i = 1, \dots, n .

Definition 3 A function

h : I \to R

is exponentially convex in the Jensen sense if it is n-exponentially convex in the Jensen sense for all

n \in N

Definition 4 A function

h : I \to R

is exponentially convex if it is exponentially convex in the Jensen sense and continuous.

Lemma 3.3 A function

h : I \to (0, \infty)

is log-convex in the Jensen sense, that is, for every

x, y \in I

h^{2} (\frac{x + y}{2}) \leq h (x) h (y)

holds if and only if the relation

α^{2} h (x) + 2 α β h (\frac{x + y}{2}) + β^{2} h (y) \geq 0

holds for every

α, β \in R

and

x, y \in I

Remark 3.4 It follows that a function is log-convex in the Jensen sense if and only if it is 2-exponentially convex in the Jensen sense. Also, by using the basic convexity theory, a function is log-convex if and only if it is 2-exponentially convex.

The following result will be useful further (see [[1], p.2]).

Lemma 3.5 If f is a convex function defined on an interval I and

x_{1} \leq y_{1}

x_{2} \leq y_{2}

x_{1} \neq x_{2}

y_{1} \neq y_{2}

, then the following inequality is valid

\frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} \leq \frac{f (y_{2}) - f (y_{1})}{y_{2} - y_{1}} .

If the function f is concave, the inequality reverses.

Definition 5 The second order divided difference of a function

f : [a, b] \to R

at mutually distinct points

y_{0}, y_{1}, y_{2} \in [a, b]

is defined recursively by

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-155/MediaObjects/13660_2012_Article_310_Equ15_HTML.gif

(15)

Remark 3.6 The value

[y_{0}, y_{1}, y_{2}; f]

is independent of the order of the points

y_{0}

y_{1}

and

y_{2}

. This definition may be extended to include the case in which some or all the points coincide (see [[1], p.16]). Namely, taking the limit

y_{1} \to y_{0}

in (15), we get

lim_{y_{1} \to y_{0}} [y_{0}, y_{1}, y_{2}; f] = [y_{0}, y_{0}, y_{2}; f] = \frac{f (y_{2}) - f (y_{0}) - f^{'} (y_{0}) (y_{2} - y_{0})}{{(y_{2} - y_{0})}^{2}}, y_{2} \neq y_{0},

provided that

f^{'}

exists; and furthermore, taking the limits

y_{i} \to y_{0}

i = 1, 2

, in (15), we get

lim_{y_{2} \to y_{0}} lim_{y_{1} \to y_{0}} [y_{0}, y_{1}, y_{2}; f] = [y_{0}, y_{0}, y_{0}; f] = \frac{f^{″} (y_{0})}{2},

provided that

f^{″}

exists.

The following definition of a real valued convex function is characterized by second order divided difference (see [[1], p.15]).

Definition 6 A function

f : [a, b] \to R

is said to be convex if and only if for all choices of three distinct points

y_{0}, y_{1}, y_{2} \in [a, b]

[y_{0}, y_{1}, y_{2}; f] \geq 0

Next, we study the n-exponential convexity and log-convexity of the functions associated with the linear functionals

Φ_{i}

(

i = 1, 2

) defined in (10) and (11).

Theorem 3.7 Let

Ω = {f_{s} : s \in I \subseteq R}

be a family of differentiable functions defined on

[a, b]

such that the function

s \mapsto [y_{0}, y_{1}, y_{2}; f_{s}]

is n-exponentially convex in the Jensen sense on I for every three mutually distinct points

y_{0}, y_{1}, y_{2} \in [a, b]

. Let

Φ_{i}

(

i = 1, 2

) be linear functionals defined as in (10) and (11). Then the following statements hold.

(i) The function

s \mapsto Φ_{i} (x, y; f_{s})

is n-exponentially convex in the Jensen sense on I.

(ii) If the function

s \mapsto Φ_{i} (x, y; f_{s})

is continuous on I, then it is n-exponentially convex on I.

Proof The idea of the proof is the same as in [5].

(i) Let

α_{j} \in R

(

j = 1, \dots, n

) and consider the function

g (y) = \sum_{j, k = 1}^{n} α_{j} α_{k} f_{\frac{s_{j} + s_{k}}{2}} (y),

where

s_{j} \in I

and

f_{\frac{s_{j} + s_{k}}{2}} \in Ω

. Then

[y_{0}, y_{1}, y_{2}; g] = \sum_{j, k = 1}^{n} α_{j} α_{k} [y_{0}, y_{1}, y_{2}; f_{\frac{s_{j} + s_{k}}{2}}]

and since

[y_{0}, y_{1}, y_{2}; f_{\frac{s_{j} + s_{k}}{2}}]

is n-exponentially convex in the Jensen sense on I by assumption, it follows that

[y_{0}, y_{1}, y_{2}; g] = \sum_{j, k = 1}^{n} α_{j} α_{k} [y_{0}, y_{1}, y_{2}; f_{\frac{s_{j} + s_{k}}{2}}] \geq 0 .

And so by using Definition 6, we conclude that g is a convex function. Hence

Φ_{i} (x, y; g) \geq 0, i = 1, 2,

which is equivalent to

\sum_{j, k = 1}^{n} α_{j} α_{k} Φ_{i} (x, y; f_{\frac{s_{j} + s_{k}}{2}}) \geq 0, i = 1, 2,

and so we conclude that the function

s \mapsto Φ_{i} (x, y; f_{s})

is n-exponentially convex in the Jensen sense on I.

(ii) If the function

s \mapsto Φ_{i} (x, y; f_{s})

is continuous on I, then from (i) and by Definition 2 it follows that it is n-exponentially convex on I.

□

Corollary 3.8 Let

Ω = {f_{s} : s \in I \subseteq R}

be a family of differentiable functions defined on

[a, b]

such that the function

s \mapsto [y_{0}, y_{1}, y_{2}; f_{s}]

is exponentially convex in the Jensen sense on I for every three mutually distinct points

y_{0}, y_{1}, y_{2} \in [a, b]

. Let

Φ_{i}

(

i = 1, 2

) be linear functionals defined as in (10) and (11). Then the following statements hold.

(i) The function

s \mapsto Φ_{i} (x, y; f_{s})

is exponentially convex in the Jensen sense on I.

(ii) If the function

s \mapsto Φ_{i} (x, y; f_{s})

is continuous on I, then it is exponentially convex on I.

Corollary 3.9 Let

Ω = {f_{s} : s \in I \subseteq R}

be a family of differentiable functions defined on

[a, b]

such that the function

s \mapsto [y_{0}, y_{1}, y_{2}; f_{s}]

is 2-exponentially convex in the Jensen sense on I for every three mutually distinct points

y_{0}, y_{1}, y_{2} \in [a, b]

. Let

Φ_{i}

(

i = 1, 2

) be linear functionals defined as in (10) and (11). Further, assume

Φ_{i} (x, y; f_{s})

(

i = 1, 2

) is strictly positive for

f_{s} \in Ω

. Then the following statements hold:

(i) If the function

s \mapsto Φ_{i} (x, y; f_{s})

is continuous on I, then it is 2-exponentially convex on I and so, it is log-convex.

(ii) If the function

s \mapsto Φ_{i} (x, y; f_{s})

is differentiable on I, then for every

s, q, u, v \in I

such that

s \leq u

and

q \leq v

, we have

μ_{s, q} (x, y, Φ_{i}, Ω) \leq μ_{u, v} (x, y, Φ_{i}, Ω), i = 1, 2,

(16)

where

μ_{s, q} (x, y, Φ_{i}, Ω) = {\begin{matrix} {(\frac{Φ_{i} (x, y; f_{s})}{Φ_{i} (x, y; f_{q})})}^{\frac{1}{s - q}}, & s \neq q, \\ exp (\frac{\frac{d}{d s} Φ_{i} (x, y; f_{s})}{Φ_{i} (x, y; f_{s})}), & s = q, \end{matrix}

(17)

for

f_{s}, f_{q} \in Ω

Proof The idea of the proof is the same as in [5].

(i) The claim is an immediate consequence of Theorem 3.7 and Remark 3.4.

(ii) Since by (i) the function

s \mapsto Φ_{i} (x, y; f_{s})

is log-convex on I, that is, the function

s \mapsto log Φ_{i} (x, y; f_{s})

is convex on I. Applying Lemma 3.5 with setting

f (z) = log Φ_{i} (x, y; f_{z})

(

i = 1, 2

), we get

\frac{log Φ_{i} (x, y; f_{s}) - log Φ_{i} (x, y; f_{q})}{s - q} \leq \frac{log Φ_{i} (x, y; f_{u}) - log Φ_{i} (x, y; f_{v})}{u - v}

(18)

for

s \leq u

q \leq v

s \neq q, u \neq v

; and therefore conclude that

μ_{s, q} (x, y, Φ_{i}, Ω) \leq μ_{u, v} (x, y, Φ_{i}, Ω), i = 1, 2 .

s = q

, we consider the limit when

q \to s

in (18) and conclude that

μ_{s, s} (x, y, Φ_{i}, Ω) \leq μ_{u, v} (x, y, Φ_{i}, Ω), i = 1, 2 .

The case

u = v

can be treated similarly.

□

Remark 3.10 Note that the results from Theorem 3.7, Corollary 3.8 and Corollary 3.9 still hold when two of the points

y_{0}, y_{1}, y_{2} \in [a, b]

coincide, say

y_{1} = y_{0}

, for a family of differentiable functions

f_{s}

such that the function

s \mapsto [y_{0}, y_{1}, y_{2}; f_{s}]

is n-exponentially convex in the Jensen sense (exponentially convex in the Jensen sense, log-convex in the Jensen sense on I); and furthermore, they still hold when all three points coincide for a family of twice differentiable functions with the same property. The proofs are obtained by recalling Remark 3.6 and by using suitable characterizations of convexity.

4 Examples

In this section, we present several families of functions which fulfill the conditions of Theorem 3.7, Corollary 3.8 and Corollary 3.9 (Remark 3.10). In this way, we can construct large families of functions which are exponentially convex.

Example 4.1 Consider the family of functions

Ω_{1} = {g_{s} : R \to [0, \infty) : s \in R}

defined by

g_{s} (x) = {\begin{matrix} \frac{1}{s^{2}} e^{s x}, & s \neq 0, \\ \frac{1}{2} x^{2}, & s = 0 . \end{matrix}

We have

\frac{d^{2}}{d x^{2}} g_{s} (x) = e^{s x} > 0

which shows that

g_{s}

is convex on

R

for every

s \in R

and

s \mapsto \frac{d^{2}}{d x^{2}} g_{s} (x)

is exponentially convex by Example 4.1 given in [6]. From [6], we then also have that

s \mapsto [y_{0}, y_{1}, y_{2}; g_{s}]

is exponentially convex and so

s \mapsto [y_{0}, y_{1}, y_{2}; g_{s}]

is exponentially convex in Jensen sense. Now, by using Corollary 3.8, we have

s \mapsto Φ_{i} (x, y; g_{s})

(

i = 1, 2

) are exponentially convex in Jensen sense. Since these mappings are continuous (although the mapping

s \mapsto g_{s}

is not continuous for

s = 0

s \mapsto Φ_{i} (x, y; g_{s})

(

i = 1, 2

) are exponentially convex.

For this family of functions, by taking

Ω = Ω_{1}

in (17),

Ξ_{s, q; 1}^{i} : = μ_{s, q} (x, y, Φ_{i}, Ω_{1})

(

i = 1, 2

) are of the form

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-155/MediaObjects/13660_2012_Article_310_Equs_HTML.gif

where

\begin{aligned} \hat{x} & = \frac{2 b x - x^{2} - a^{2}}{2 (b - a)}, \hat{y} = \frac{2 b y - y^{2} - a^{2}}{2 (b - a)}, \\ \tilde{x} & = \frac{x^{2} + b^{2} - 2 a x}{2 (b - a)}, \tilde{y} = \frac{y^{2} + b^{2} - 2 a y}{2 (b - a)} . \end{aligned}

(19)

By using Theorem 2.6, it can be seen that

M_{s, q} (x, Φ_{i}, Ω_{1}) = log μ_{s, q} (x, Φ_{i}, Ω_{1}), i = 1, 2,

satisfy

min {a, b} \leq M_{s, q} (x, Φ_{i}, Ω_{1}) \leq max {a, b}

, showing that

M_{s, q} (x, Φ_{i}, Ω_{1})

(

i = 1, 2

) are means.

Example 4.2 Consider the family of functions

Ω_{2} = {f_{s} : (0, \infty) \to R : s \in R}

defined by

f_{s} (x) = {\begin{matrix} \frac{x^{s}}{s (s - 1)}, & s \neq 0, 1, \\ - ln x, & s = 0, \\ x ln x, & s = 1 . \end{matrix}

Here,

\frac{d^{2}}{d x^{2}} f_{s} (x) = x^{s - 2} = e^{(s - 2) ln x} > 0

which shows that

f_{s}

is convex for

x > 0

and

s \mapsto \frac{d^{2}}{d x^{2}} f_{s} (x)

is exponentially convex by Example 4.2 given in [6]. From [6], we have

s \mapsto [y_{0}, y_{1}, y_{2}; f_{s}]

is exponentially convex. Arguing as in Example 4.1, we have

s \mapsto Φ_{i} (x, y; f_{s})

(

i = 1, 2

) are exponentially convex.

By taking

Ω = Ω_{2}

in (17),

Ξ_{s, q; 2}^{i} : = μ_{s, q} (x, y, Φ_{i}, Ω_{2})

(

i = 1, 2

) for

x, y > 0

, where

x, y \in [a, b]

, are of the form

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-155/MediaObjects/13660_2012_Article_310_Equw_HTML.gif

where

\hat{x}

\hat{y}

\tilde{x}

and

\tilde{y}

are the same as in (19). If

Φ_{i}

(

i = 1, 2

) are positive, then Theorem 2.6 applied for

f = f_{s} \in Ω_{2}

and

g = f_{q} \in Ω_{2}

yields that there exists

ξ_{i} \in [a, b]

such that

ξ_{i}^{s - q} = \frac{Φ_{i} (x, y; f_{s})}{Φ_{i} (x, y; f_{q})}, i = 1, 2 .

Since the functions

ξ_{i} \mapsto ξ_{i}^{s - q}

(

i = 1, 2

) are invertible for

s \neq q

, we then have

min {a, b} \leq {(\frac{Φ_{i} (x, y; f_{s})}{Φ_{i} (x, y; f_{q})})}^{\frac{1}{s - q}} \leq max {a, b}, i = 1, 2,

(20)

which, together with the fact that

μ_{s, q} (x, y, Φ_{i}, Ω_{2})

(

i = 1, 2

) are continuous, symmetric and monotonous (by (16)), shows that

μ_{s, q} (x, y, Φ_{i}, Ω_{2})

(

i = 1, 2

) are means.

Now, by the substitutions

x \to x^{t}

y \to y^{t}

s \to \frac{s}{t}

q \to \frac{q}{t}

(

t \neq 0

s \neq q

), where

x, y \in [a, b]

, from (20) we get

min {a^{t}, b^{t}} \leq {(\frac{Φ_{i} (x^{t}, y^{t}; f_{s / t}}{Φ_{i} (x^{t}, y^{t}; f_{q / t}})}^{\frac{t}{s - q}} \leq max {a^{t}, b^{t}} .

We define a new mean (for

i = 1, 2

) as follows:

μ_{s, q; t} (x, y, Φ_{i}, Ω_{2}) = {\begin{matrix} {(μ_{\frac{s}{t}, \frac{q}{t}} (x^{t}, y^{t}, Φ_{i}, Ω_{2}))}^{\frac{1}{t}}, & t \neq 0, \\ μ_{s, q} (ln x, ln y, Φ_{i}, Ω_{1}), & t = 0 . \end{matrix}

These new means are also monotonous. More precisely, for

s, q, u, v \in R

such that

s \leq u

q \leq v

s \neq q

u \neq v

, we have

μ_{s, q; t} (x, y, Φ_{i}, Ω_{2}) \leq μ_{u, v; t} (x, y, Φ_{i}, Ω_{2}), i = 1, 2 .

We know that

μ_{\frac{s}{t}, \frac{q}{t}} (x^{t}, y^{t}, Φ_{i}, Ω_{2}) \leq μ_{\frac{u}{t}, \frac{v}{t}} (x^{t}, y^{t}, Φ_{i}, Ω_{2}), i = 1, 2,

equivalently

{(\frac{Φ_{i} (x^{t}, y^{t}; f_{s / t})}{Φ_{i} (x^{t}, y^{t}; f_{q / t})})}^{\frac{t}{s - q}} \leq {(\frac{Φ_{i} (x^{t}, y^{t}; f_{u / t})}{Φ_{i} (x^{t}, y^{t}; f_{v / t})})}^{\frac{t}{u - v}}, i = 1, 2,

for

s, q, u, v \in I

such that

s / t \leq u / t

q / t \leq v / t

and

t \neq 0

, since

μ_{s, q} (x, y, Φ_{i}, Ω_{2})

(

i = 1, 2

) are monotonous in both parameters, the claim follows. For

t = 0

, we obtain the required result by taking the limit

t \to 0

Remark 4.3 If we make the substitutions

x \to a

y \to b

s \to s - 1

and

t \to t - 1

in our means

μ_{s, q} (x, y, Φ_{i}, Ω_{2})

and

μ_{s, q; t} (x, y, Φ_{i}, Ω_{2})

(

i = 1, 2

), then the results for the

E_{1} (r, t)

means and the generalized

E_{1}

means given in [7] are recaptured. In this way our results for means are the generalizations of the above mentioned means.

Example 4.4 Consider the family of functions

Ω_{3} = {h_{s} : (0, \infty) \to (0, \infty) : s \in (0, \infty)}

defined by

h_{s} (x) = {\begin{matrix} \frac{s^{- x}}{{ln}^{2} s}, & s \neq 1, \\ \frac{x^{2}}{2}, & s = 1 . \end{matrix}

We have

\frac{d^{2}}{d x^{2}} h_{s} (x) = s^{- x} > 0

which shows that

h_{s}

is convex for all

s > 0

. Exponential convexity of

s \mapsto \frac{d^{2}}{d x^{2}} h_{s} (x) = s^{- x}

(0, \infty)

is given by Example 4.3 in [6]. Arguing as in Example 4.1, we have

s \mapsto Φ_{i} (x, y; h_{s})

(

i = 1, 2

) are exponentially convex.

In this case by taking

Ω = Ω_{3}

in (17),

Ξ_{s, q; 3}^{i} : = μ_{s, q} (x, y, Φ_{i}, Ω_{3})

(

i = 1, 2

) for

x, y > 0

, where

x, y \in [a, b]

, are of the form

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-155/MediaObjects/13660_2012_Article_310_Equaf_HTML.gif

where

\hat{x}

\hat{y}

\tilde{x}

and

\tilde{y}

are the same as in (19). By using Theorem 2.6, it follows that

M_{s, q} (x, Φ_{i}, Ω_{3}) = - L (s, q) log μ_{s, q} (x, Φ_{i}, Ω_{3}), i = 1, 2,

satisfy

min {a, b} \leq M_{s, q} (x, Φ_{i}, Ω_{3}) \leq max {a, b}

and so

M_{s, q} (x, Φ_{i}, Ω_{3})

(

i = 1, 2

) are means, where

L (s, q)

is a logarithmic mean defined by

L (s, q) = \frac{s - q}{log s - log q}

s \neq q

L (s, s) = s

Example 4.5 Consider the family of functions

Ω_{4} = {k_{s} : (0, \infty) \to (0, \infty) : s \in (0, \infty)}

defined by

k_{s} (x) = \frac{e^{- x \sqrt{s}}}{s} .

Here,

\frac{d^{2}}{d x^{2}} k_{s} (x) = e^{- x \sqrt{s}} > 0

which shows that

k_{s}

is convex for all

s > 0

. Exponential convexity of

s \mapsto \frac{d^{2}}{d x^{2}} k_{s} (x) = e^{- x \sqrt{s}}

(0, \infty)

is given by Example 4.4 in [6]. Arguing as in Example 4.1, we have

s \mapsto Φ_{i} (x, y; k_{s})

(

i = 1, 2

) are exponentially convex.

In this case by taking

Ω = Ω_{4}

in (17),

Ξ_{s, q; 4}^{i} : = μ_{s, q} (x, y, Φ_{i}, Ω_{4})

(

i = 1, 2

) for

x, y > 0

, where

x, y \in [a, b]

, are of the form

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-155/MediaObjects/13660_2012_Article_310_Equaj_HTML.gif

where

\hat{x}

\hat{y}

\tilde{x}

and

\tilde{y}

are the same as in (19). By using Theorem 2.6, it is easy to see that

M_{s, q} (x, Φ_{i}, Ω_{4}) = - (\sqrt{s} + \sqrt{q}) log μ_{s, q} (x, Φ_{i}, Ω_{4}), i = 1, 2,

satisfy

min {a, b} \leq M_{s, q} (x, Φ_{i}, Ω_{4}) \leq max {a, b}

, showing that

M_{s, q} (x, Φ_{i}, Ω_{4})

(

i = 1, 2

) are means.

Remark 4.6 From (16), it is clear that

μ_{s, q} (x, y, Φ_{i}, Ω)

(

i = 1, 2

) for

Ω = Ω_{1}, Ω_{2}, Ω_{3} and Ω_{4}

are monotonous functions in parameters s and q.

Acknowledgements

The second author’s research was supported by the Croatian Ministry of Science, Education and Sports, under the Research Grant 117-1170889-0888.

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

Both authors worked jointly on the results and they read and approved the final manuscript.

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On the refinements of the Hermite-Hadamard inequality

Abstract

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Authors’ contributions

1 Introduction

2 Main results

3 n-exponential convexity and log-convexity of the Hermite-Hadamard differences

4 Examples

Acknowledgements

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Abstract

Competing interests

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1 Introduction

2 Main results

3 n-exponential convexity and log-convexity of the Hermite-Hadamard differences

4 Examples

Acknowledgements

Competing interests

Authors’ contributions

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