Open AccessArticle

Hermite-Hadamard Type Inequalities for Interval (h₁, h₂)-Convex Functions

Yanrong An

¹,

Guoju Ye

^2,*,

Dafang Zhao

³ and

Wei Liu

School of Business, Nanjing University, Nanjing 210098, China

College of Science, Hohai University, Nanjing 210098, China

School of Mathematics and Statistics, Hubei Normal University, Huangshi 435002, China

Author to whom correspondence should be addressed.

Mathematics 2019, 7(5), 436; https://doi.org/10.3390/math7050436

Submission received: 18 April 2019 / Revised: 13 May 2019 / Accepted: 14 May 2019 / Published: 17 May 2019

(This article belongs to the Special Issue Inequalities)

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Abstract

We introduce the concept of interval

(h_{1}, h_{2})

-convex functions. Under the new concept, we establish some new interval Hermite-Hadamard type inequalities, which generalize those in the literature. Also, we give some interesting examples.

Keywords:

Hermite-Hadamard inequality; interval-valued functions; (h₁, h₂)-convex

1. Introduction

Interval analysis was introduced in numerical analysis by Moore in the celebrated book [1]. Over the past 50 years, it has attracted considerable interest and has been applied in various fields, such as interval differential equations [2], aeroelasticity [3], aerodynamic load analysis [4], and so on. For more profound results and applications, see [5,6,7,8,9].

It is known that inequalities play an important role in almost all branches of mathematics as well as in other areas of science. Among the many types of inequalities, those carrying the names of Jensen, Hermite-Hadamard, Hardy, Ostrowski, Minkowski and Opial et al. have a deep significance and have made a great impact in substantial fields of research. Recently, some of these inequalities have been extended to interval-valued functions by Chalco-Cano et al.; see, e.g., [10,11,12,13,14,15,16]. Surprisingly enough, interval Hermite-Hadamard type inequalities has perhaps not received enough attention [17]. For convenience, we recall the classical Hermite-Hadamard inequality. Let f be convex, then

f (\frac{u + v}{2}) \leq \frac{1}{v - u} \int_{u}^{v} f (t) d t \leq \frac{f (u) + f (v)}{2} .

This inequality has been developed for different classes of convexity [18,19,20,21,22,23,24,25,26]. Especially, since the h-convex concept was proposed by Varosanec in 2007 [27], a number of authors have already studied more refined Hermite-Hadamard inequalities involving h-convex functions [28,29,30,31,32,33].

In 2018, Awan et al. introduced

(h_{1}, h_{2})

-convex functions and proved the following inequality [34]:

Theorem 1.

Let

f : [u, v] \to R

. If f is

(h_{1}, h_{2})

-convex, and

h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2}) \neq 0

. Then

\frac{1}{2 h_{1} (\frac{1}{2}) h_{2} (\frac{1}{2})} f (\frac{u + v}{2}) \leq \frac{1}{v - u} \int_{u}^{v} f (t) d t \leq [f (u) + f (v)] \int_{0}^{1} h_{1} (x) h_{2} (1 - x) d x .

Motivated by Awan et al., our main objective is to generalize the results above by constructing interval Hermite-Hadamard type inequalities for

(h_{1}, h_{2})

-convex functions. Also, we present some examples to illustrate our theorems. Our results generalize some known inequalities presented in [17,32,34,35]. Furthermore, the present results can be considered as tools for further research in interval convex analysis, interval nonlinear programming, inequalities for fuzzy-interval-valued functions, among others.

We give preliminaries in Section 2. In Section 3, we introduce interval

(h_{1}, h_{2})

-convex concept, and obtain some interval Hermite-Hadamard type inequalities. Moreover, some interesting examples are given. In Section 4, we give conclusions and future work.

2. Preliminaries

For the basic notations and definitions on interval analysis, see [17]. The family of all intervals and positive intervals of

R

are denoted by

R_{I}

and

R_{I}^{+}

, respectively. For interval

[\underset{̲}{u}, \bar{u}]

and

[\underset{̲}{v}, \bar{v}]

, the Hausdorff distance is defined by

d ([\underset{̲}{u}, \bar{u}], [\underset{̲}{v}, \bar{v}]) = max \{| \underset{̲}{u} - \underset{̲}{v} |, | \bar{u} - \bar{v} |\} .

Then,

(R_{I}, d)

is complete.

A set of numbers

{t_{i - 1}, ξ_{i}, t_{i}}_{i = 1}^{m}

is said to be a tagged partition P of

[u, v]

u = t_{0} < t_{1} < \dots < t_{m} = v

and if

t_{i - 1} \leq ξ_{i} \leq t_{i}

for all

i = 1, 2, \dots, m

. Moreover, if we let

Δ t_{i} = t_{i} - t_{i - 1}

, then the partition is called

δ

-fine if

Δ t_{i} < δ

for each i. We denote by

P (δ, [u, v])

the family of all

δ

-fine partitions of

[u, v]

. Given

P \in P (δ, [u, v])

, we define a integral sum of

f : [u, v] \to R_{I}

as follows:

S (f, P, δ, [u, v]) = \sum_{i = 1}^{m} f (ξ_{i}) (t_{i} - t_{i - 1}) .

Definition 1.

Let

f : [u, v] \to R_{I}

. f is called

I R

-integrable on

[u, v]

with

I R

-integral

A = (I R) \int_{u}^{v} f (t) d t

, if there exists an

A \in R_{I}

such that for any

ϵ > 0

there exists a

δ > 0

such that

d (S (f, P, δ, [u, v]), A) < ϵ

for each

P \in P (δ, [u, v])

. Let

{IR}_{([u, v])}

denote the set of all

I R

-integrable functions on

[u, v]

Definition 2.

Let

h_{1}, h_{2} : [0, 1] \subseteq J \to R^{+}

such that

h_{1}, h_{2} ≢ 0

(Awan et al. [34]).

f : J \to R^{+}

is called

(h_{1}, h_{2})

-convex, or that

f \in S X ((h_{1}, h_{2}), J, R)

, if for any

s, t \in J

and

x \in (0, 1)

one has

f (x s + (1 - x) t) \leq h_{1} (x) h_{2} (1 - x) f (s) + h_{1} (1 - x) h_{2} (x) f (t) .

(1)

Remark 1.

h_{2} \equiv 1

, then Definition 2 reduces to h-convex in [27].

h_{1} = h_{2} \equiv 1

, then Definition 2 reduces to P-function in [18].

h_{1} (t) = t^{s}

h_{2} \equiv 1

, then Definition 2 reduces to s-convex in [36].

We end this section of preliminaries by introducing the new concept of interval

(h_{1}, h_{2})

-convexity. This idea is inspired by Costa [12]. Note that for interval

[\underset{̲}{u}, \bar{u}]

and

[\underset{̲}{v}, \bar{v}]

, the inclusion “

\subseteq

” is defined by

[\underset{̲}{u}, \bar{u}] \subseteq [\underset{̲}{v}, \bar{v}] ⟺ \underset{̲}{v} \leq \underset{̲}{u}, \bar{u} \leq \bar{v} .

Definition 3.

Let

h_{1}, h_{2} : [0, 1] \subseteq J \to R^{+}

such that

h_{1}, h_{2} ≢ 0

f : J \to R_{I}^{+}

is called interval

(h_{1}, h_{2})

-convex, if for all

s, t \in J

and

x \in (0, 1)

one has

h_{1} (x) h_{2} (1 - x) f (s) + h_{1} (1 - x) h_{2} (x) f (t) \subseteq f (x s + (1 - x) t) .

(2)

The set of all interval

(h_{1}, h_{2})

-convex function is denoted by

S X ((h_{1}, h_{2}), J, R_{I}^{+})

3. Interval Hermite-Hadamard Type Inequality

In what follows, let

H (x, y) = h_{1} (x) h_{2} (y)

for

x, y \in [0, 1]

Theorem 2.

Let

f : [u, v] \to R_{I}^{+}

h_{1}, h_{2} : [0, 1] \to R^{+}

and

H (\frac{1}{2}, \frac{1}{2}) \neq 0

. If

f \in S X ((h_{1}, h_{2}), [u, v], R_{I}^{+})

and

f \in {IR}_{([u, v])}

, then

\frac{1}{2 H (\frac{1}{2}, \frac{1}{2})} f (\frac{u + v}{2}) \supseteq \frac{1}{v - u} \int_{u}^{v} f (t) d t \supseteq [f (u) + f (v)] \int_{0}^{1} H (x, 1 - x) d x .

(3)

Proof.

By hypothesis, we have

H (\frac{1}{2}, \frac{1}{2}) f (x u + (1 - x) v) + H (\frac{1}{2}, \frac{1}{2}) f ((1 - x) u + x v) \subseteq f (\frac{u + v}{2}) .

Then

\begin{matrix} \int_{0}^{1} \underset{̲}{f} (x u + (1 - x) v) d x + \int_{0}^{1} \underset{̲}{f} ((1 - x) u + x v) d x \geq \frac{1}{H (\frac{1}{2}, \frac{1}{2})} \int_{0}^{1} \underset{̲}{f} (\frac{u + v}{2}) d x, \\ \int_{0}^{1} \bar{f} (x u + (1 - x) v) d x + \int_{0}^{1} \bar{f} ((1 - x) u + x v) d x \leq \frac{1}{H (\frac{1}{2}, \frac{1}{2})} \int_{0}^{1} \bar{f} (\frac{u + v}{2}) d x . \end{matrix}

It follows that

\begin{matrix} \frac{2}{v - u} \int_{u}^{v} \underset{̲}{f} (t) d t \geq \frac{1}{H (\frac{1}{2}, \frac{1}{2})} \int_{0}^{1} \underset{̲}{f} (\frac{u + v}{2}) d x = \frac{1}{H (\frac{1}{2}, \frac{1}{2})} \underset{̲}{f} (\frac{u + v}{2}), \\ \frac{2}{v - u} \int_{u}^{v} \bar{f} (t) d t \leq \frac{1}{H (\frac{1}{2}, \frac{1}{2})} \int_{0}^{1} \bar{f} (\frac{u + v}{2}) d x = \frac{1}{H (\frac{1}{2}, \frac{1}{2})} \bar{f} (\frac{u + v}{2}) . \end{matrix}

This implies

\frac{1}{H (\frac{1}{2}, \frac{1}{2})} [\underset{̲}{f} (\frac{u + v}{2}), \bar{f} (\frac{u + v}{2})] \supseteq \frac{2}{v - u} [\int_{u}^{v} \underset{̲}{f} (t) d t, \int_{u}^{v} \bar{f} (t) d t] .

Thus,

\frac{1}{2 H (\frac{1}{2}, \frac{1}{2})} f (\frac{u + v}{2}) \supseteq \frac{1}{v - u} \int_{u}^{v} f (t) d t .

In the same way as above, we have

\frac{1}{v - u} \int_{u}^{v} f (t) d t \supseteq [f (u) + f (v)] \int_{0}^{1} H (x, 1 - x) d x,

and the result follows. □

Remark 2.

H (x, y) \equiv h_{1} (x)

, then Theorem 2 reduces to ([17], Theorem 4.1).

h_{1} (x) = x^{s}

h_{2} \equiv 1

, then Theorem 2 reduces to ([37], Theorem 4).

h_{1} = h_{2} \equiv 1

, then inequality (3) in Theorem 2 reduces to inequality for P-function.

\underset{̲}{f} = \bar{f}

, then Theorem 2 reduces to ([34], Theorem 1). Furthermore, If

h_{2} \equiv 1

, then we get ([32], Theorem 6).

Example 1.

Suppose that

h_{1} (x) = x

h_{2} (x) \equiv 1

for

x \in [0, 1]

[u, v] = [- 1, 1]

, and

f : [u, v] \to R_{I}^{+}

be defined by

f (t) = [t^{2}, 4 - e^{t}]

. Then

\begin{matrix} \frac{1}{2 H (\frac{1}{2}, \frac{1}{2})} f (\frac{u + v}{2}) = f (0) = [0, 3], \\ \frac{1}{v - u} \int_{u}^{v} f (t) d t = \frac{1}{2} [\int_{- 1}^{1} t^{2} d t, \int_{- 1}^{1} (6 - e^{t}) d t] = [\frac{1}{3}, 4 - \frac{e - e^{- 1}}{2}], \\ [f (u) + f (v)] \int_{0}^{1} H (x, 1 - x) d x = [1, 4 - \frac{e + e^{- 1}}{2}] . \end{matrix}

Then, we obtain that

[0, 3] \supseteq [\frac{1}{3}, 4 - \frac{e - e^{- 1}}{2}] \supseteq [\frac{1}{3}, 4 - \frac{e + e^{- 1}}{2}] .

Consequently, Theorem 2 is verified.

The next result generalizes Theorem 3.1 of [35] and Theorem 4.3 of [17].

Theorem 3.

Let

f : [u, v] \to R_{I}^{+}

h_{1}, h_{2} : [0, 1] \to R^{+}

and

H (\frac{1}{2}, \frac{1}{2}) \neq 0

. If

f \in S X ((h_{1}, h_{2}), [u, v], R_{I}^{+})

and

f \in {IR}_{([u, v])}

, then

\begin{matrix} \frac{1}{4 H^{2} (\frac{1}{2}, \frac{1}{2})} f (\frac{u + v}{2}) \supseteq Δ_{1} \supseteq \frac{1}{v - u} \int_{u}^{v} & f (t) d t \\ \supseteq Δ_{2} \supseteq [f (u) + f (v)] [\frac{1}{2} + H (\frac{1}{2}, \frac{1}{2})] \int_{0}^{1} H (x, 1 - x) d x, \end{matrix}

where

\begin{matrix} Δ_{1} = \frac{1}{4 H (\frac{1}{2}, \frac{1}{2})} [f (\frac{3 u + v}{4}) + f (\frac{u + 3 v}{4})], \\ Δ_{2} = [\frac{f (u) + f (v)}{2} + f (\frac{u + v}{2})] \int_{0}^{1} H (x, 1 - x) d x . \end{matrix}

Proof.

For

[u, \frac{u + v}{2}]

, one has

\begin{matrix} H (\frac{1}{2}, \frac{1}{2}) f (x u + (1 - x) \frac{u + v}{2}) + H & (\frac{1}{2}, \frac{1}{2}) f ((1 - x) u + x \frac{u + v}{2}) \\ \subseteq f (\frac{x u + (1 - x) \frac{u + v}{2}}{2} + \frac{(1 - x) u + x \frac{u + v}{2}}{2}) = f (\frac{3 u + v}{4}) . \end{matrix}

Consequently, we get

\frac{1}{4 H (\frac{1}{2}, \frac{1}{2})} f (\frac{3 u + v}{4}) \supseteq \frac{1}{v - u} \int_{u}^{\frac{u + v}{2}} f (t) d t .

In the same way as above, for

[\frac{u + v}{2}, v]

, we have

\frac{1}{4 H (\frac{1}{2}, \frac{1}{2})} f (\frac{u + 3 v}{4}) \supseteq \frac{1}{v - u} \int_{\frac{u + v}{2}}^{v} f (t) d t .

Hence,

Δ_{1} = \frac{1}{4 H (\frac{1}{2}, \frac{1}{2})} [f (\frac{3 u + v}{4}) + f (\frac{u + 3 v}{4})] \supseteq \frac{1}{v - u} \int_{u}^{v} f (t) d t .

Thanks to Theorem 2, one has

\begin{matrix} \frac{1}{4 {[H (\frac{1}{2}, \frac{1}{2})]}^{2}} f (\frac{u + v}{2}) \\ = \frac{1}{4 {[H (\frac{1}{2}, \frac{1}{2})]}^{2}} f (\frac{1}{2} \cdot \frac{3 u + v}{4} + \frac{1}{2} \cdot \frac{u + 3 v}{4}) \\ \supseteq \frac{1}{4 {[H (\frac{1}{2}, \frac{1}{2})]}^{2}} [H (\frac{1}{2}, \frac{1}{2}) f (\frac{3 u + v}{4}) + H (\frac{1}{2}, \frac{1}{2}) f (\frac{u + 3 v}{4})] \\ \supseteq Δ_{1} \\ \supseteq \frac{1}{v - u} \int_{u}^{v} f (t) d t \\ \supseteq \frac{1}{2} [f (u) + f (v) + 2 f (\frac{u + v}{2})] \int_{0}^{1} H (x, 1 - x) d x \\ = Δ_{2} \\ \supseteq [\frac{f (u) + f (v)}{2} + H (\frac{1}{2}, \frac{1}{2}) (f (u) + f (v))] \int_{0}^{1} H (x, 1 - x) d x \\ \supseteq [f (u) + f (v)] [\frac{1}{2} + H (\frac{1}{2}, \frac{1}{2})] \int_{0}^{1} H (x, 1 - x) d x, \end{matrix}

and the result follows. □

Example 2.

Furthermore, by Example 1, we have

\begin{matrix} Δ_{1} = \frac{1}{2} [f (- \frac{1}{2}), f (\frac{1}{2})] = [\frac{1}{4}, 4 - \frac{e^{\frac{1}{2}} + e^{- \frac{1}{2}}}{2}], \\ Δ_{2} = \frac{1}{2} ([1, 4 - \frac{e + e^{- 1}}{2}] + [0, 3]) = [\frac{1}{2}, \frac{7}{2} - \frac{e + e^{- 1}}{4}], \\ [f (u) + f (v)] [\frac{1}{2} + H (\frac{1}{2}, \frac{1}{2})] \int_{0}^{1} H (x, 1 - x) d x = [1, 4 - \frac{e + e^{- 1}}{2}] . \end{matrix}

Then, we obtain that

[0, 3] \supseteq [\frac{1}{4}, 4 - \frac{e^{\frac{1}{2}} + e^{- \frac{1}{2}}}{2}] \supseteq [\frac{1}{3}, 4 - \frac{e - e^{- 1}}{2}] \supseteq [\frac{1}{2}, \frac{7}{2} - \frac{e + e^{- 1}}{4}] \supseteq [1, 4 - \frac{e + e^{- 1}}{2}] .

Consequently, Theorem 3 is verified.

Similarly, we get the following result, which generalizes Theorem 3 of [34] and Theorem 4.5 of [17].

Theorem 4.

Let

f, g : [u, v] \to R_{I}^{+}

h_{1}, h_{2} : [0, 1] \to R^{+}

and

H (\frac{1}{2}, \frac{1}{2}) \neq 0

. If

f, g \in S X ((h_{1}, h_{2}), [u, v], R_{I}^{+})

and

f g \in {IR}_{([u, v])}

, then

\frac{1}{v - u} \int_{u}^{v} f (t) g (t) d t \supseteq M (u, v) \int_{0}^{1} H^{2} (x, 1 - x) d x + N (u, v) \int_{0}^{1} H (x, x) H (1 - x, 1 - x) d x,

where

M (u, v) = f (u) g (u) + f (v) g (v), N (u, v) = f (u) g (v) + f (v) g (u) .

Example 3.

Suppose that

h_{1} (x) = x

h_{2} (x) \equiv 1

[u, v] = [0, 1]

and

f (t) = [t^{2}, 4 - e^{t}], g (t) = [t, 3 - t^{2}] .

Then

\begin{matrix} \frac{1}{v - u} \int_{u}^{v} f (t) g (t) d t = \int_{0}^{1} [t^{3}, (4 - e^{t}) (3 - t^{2})] d t = [\frac{1}{4}, \frac{35}{3} - 2 e], \\ M (u, v) \int_{0}^{1} H^{2} (x, 1 - x) d x = M (0, 1) \int_{0}^{1} x^{2} d x = [\frac{1}{3}, \frac{17}{3} - \frac{2}{3} e], \\ N (u, v) \int_{0}^{1} H (x, x) H (1 - x, 1 - x) d x = N (0, 1) \int_{0}^{1} x^{2} d x = [0, 3 - \frac{e}{2}] . \end{matrix}

It follows that

[\frac{1}{4}, \frac{35}{3} - 2 e] \supseteq [\frac{1}{3}, \frac{17}{3} - \frac{2}{3} e] + [0, 3 - \frac{e}{2}] = [\frac{1}{3}, \frac{26}{3} - \frac{7}{6} e] .

Consequently, Theorem 4 is verified.

The next result generalizes Theorem 2 of [34] and Theorem 4.6 of [17].

Theorem 5.

Let

f, g : [u, v] \to R_{I}^{+}

h_{1}, h_{2} : [0, 1] \to R^{+}

, and

H (\frac{1}{2}, \frac{1}{2}) \neq 0

. If

f, g \in S X ((h_{1}, h_{2}), [u, v], R_{I}^{+})

and

f g \in {IR}_{([u, v])}

, then

\begin{matrix} \frac{1}{2 H^{2} (\frac{1}{2}, \frac{1}{2})} f (\frac{u + v}{2}) g (\frac{u + v}{2}) & \supseteq \frac{1}{v - u} \int_{u}^{v} f (t) g (t) d t + N (u, v) \int_{0}^{1} H^{2} (x, 1 - x) d x \\ + M (u, v) \int_{0}^{1} H (x, x) H (1 - x, 1 - x) d x . \end{matrix}

Proof.

By hypothesis, one has

\begin{matrix} f (\frac{u + v}{2}) g (\frac{u + v}{2}) \\ \supseteq H^{2} (\frac{1}{2}, \frac{1}{2}) [\underset{̲}{f} (x u + (1 - x) v) \underset{̲}{g} (x u + (1 - x) v), \bar{f} (x u + (1 - x) v) \bar{g} (x u + (1 - x) v)] \\ + H^{2} (\frac{1}{2}, \frac{1}{2}) [\underset{̲}{f} (x u + (1 - x) v) \underset{̲}{g} ((1 - x) u + x v), \bar{f} (x u + (1 - x) v) \bar{g} ((1 - x) u + x v)] \\ + H^{2} (\frac{1}{2}, \frac{1}{2}) [\underset{̲}{f} ((1 - x) u + x v) \underset{̲}{g} (x u + (1 - x) v), \bar{f} ((1 - x) u + x v) \bar{g} (x u + (1 - x) v)] \\ + H^{2} (\frac{1}{2}, \frac{1}{2}) [\underset{̲}{f} ((1 - x) u + x v) \underset{̲}{g} ((1 - x) u + x v), \bar{f} ((1 - x) u + x v) \bar{g} ((1 - x) u + x v)] \\ \supseteq H^{2} (\frac{1}{2}, \frac{1}{2}) [f (x u + (1 - x) v) g (x u + (1 - x) v) + f ((1 - x) u + x v) g ((1 - x) u + x v)] \\ + H^{2} (\frac{1}{2}, \frac{1}{2}) [(H (x, 1 - x) f (u) + H (1 - x, x) f (v)) (H (1 - x, x) g (u) + H (x, 1 - x) g (v)) \\ + (H (1 - x, x) f (u) + H (x, 1 - x) f (v)) (H (x, 1 - x) g (u) + H (1 - x, x) g (v))] \\ = H^{2} (\frac{1}{2}, \frac{1}{2}) [f (x u + (1 - x) v) g (x u + (1 - x) v) + f ((1 - x) u + x v) g ((1 - x) u + x v)] \\ + 2 H^{2} (\frac{1}{2}, \frac{1}{2}) [H (x, x) H (1 - x, 1 - x) M (u, v) + H^{2} (x, 1 - x) N (u, v)] \end{matrix}

Integrating over

[0, 1]

, and the result follows. □

Example 4.

Furthermore, by Example 3, we get

\begin{matrix} \frac{1}{2 H^{2} (\frac{1}{2}, \frac{1}{2})} f (\frac{u + v}{2}) g (\frac{u + v}{2}) = 2 f (\frac{1}{2}) g (\frac{1}{2}) = [\frac{1}{4}, 22 - \frac{11}{2} \sqrt{e}], \\ N (u, v) \int_{0}^{1} H^{2} (x, 1 - x) d x = N (0, 1) \int_{0}^{1} x^{2} d x = [0, 6 - e], \\ M (u, v) \int_{0}^{1} H (x, x) H (1 - x, 1 - x) d x = M (0, 1) \int_{0}^{1} (x - x^{2}) d x = [\frac{1}{6}, \frac{17}{6} - \frac{e}{3}] . \end{matrix}

It follows that

[\frac{1}{4}, 22 - \frac{11}{2} \sqrt{e}] \supseteq [0, 6 - e] + [\frac{1}{6}, \frac{17}{6} - \frac{e}{3}] + [\frac{1}{4}, \frac{35}{3} - 2 e] = [\frac{5}{12}, \frac{123}{6} - \frac{10}{3} e] .

Consequently, Theorem 5 is verified.

4. Conclusions

We introduced interval

(h_{1}, h_{2})

-convex and presented some new interval Hermite-Hadamard type inequalities. Our results generalize some known Hermite-Hadamard type inequalities and will be useful in developing the theory of interval differential (or integral) inequalities and interval convex analysis. As a future research direction, we intend to investigate inequalities for fuzzy-interval-valued functions, and some applications in interval nonlinear programming.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Funding

This research is supported by the National Key Research and Development Program of China (2018YFC1508106), the Fundamental Research Funds for the Central Universities (2017B19714 and 2017B07414) and Natural Science Foundation of Jiangsu Province (BK20180500).

Conflicts of Interest

The authors declare no conflict of interest.

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An, Y.; Ye, G.; Zhao, D.; Liu, W. Hermite-Hadamard Type Inequalities for Interval (h₁, h₂)-Convex Functions. Mathematics 2019, 7, 436. https://doi.org/10.3390/math7050436

AMA Style

An Y, Ye G, Zhao D, Liu W. Hermite-Hadamard Type Inequalities for Interval (h₁, h₂)-Convex Functions. Mathematics. 2019; 7(5):436. https://doi.org/10.3390/math7050436

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An, Yanrong, Guoju Ye, Dafang Zhao, and Wei Liu. 2019. "Hermite-Hadamard Type Inequalities for Interval (h₁, h₂)-Convex Functions" Mathematics 7, no. 5: 436. https://doi.org/10.3390/math7050436

APA Style

An, Y., Ye, G., Zhao, D., & Liu, W. (2019). Hermite-Hadamard Type Inequalities for Interval (h₁, h₂)-Convex Functions. Mathematics, 7(5), 436. https://doi.org/10.3390/math7050436

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Hermite-Hadamard Type Inequalities for Interval (h₁, h₂)-Convex Functions

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1. Introduction

2. Preliminaries

3. Interval Hermite-Hadamard Type Inequality

4. Conclusions

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Hermite-Hadamard Type Inequalities for Interval (h1, h2)-Convex Functions

Abstract

1. Introduction

2. Preliminaries

3. Interval Hermite-Hadamard Type Inequality

4. Conclusions

Author Contributions

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Hermite-Hadamard Type Inequalities for Interval (h₁, h₂)-Convex Functions