Hermite–Hadamard-Type Inequalities for Coordinated Convex Functions Using Fuzzy Integrals

Abstract

In this paper, some estimates of third and fourth inequalities in Hermite–Hadamard-type inequalities for coordinated convex functions are proved using the non-additivity of the integrals and Fubini’s theorem for fuzzy integrals. That is, the results are obtained in the fuzzy context and using the Lebesgue measure. Several examples are provided on how to evaluate these estimates in order to illustrate the obtained results.

1. Introduction

Sugeno investigated the theory of fuzzy measures and fuzzy integrals as a tool for modeling non-deterministic issues. The fuzzy integral (also known as the Sugeno integral) has profoundly amazing mathematical characteristics that have been noted by several authors. Sugeno pioneered the study of the theory of nonadditive measures and integrals, frequently referred to as fuzzy measures and fuzzy integrals [1]. Ralescu and Adams [2] investigated numerous equivalent definitions of fuzzy integral, whereas Wang and Klir [3,4,5] offered an outline of fuzzy measure theory and generalized fuzzy measure theory, respectively. Fuzzy measures and the Sugeno integral have also been successfully employed in a variety of domains, including decision-making [6] and artificial intelligence [7]. Integral inequalities are useful tools in a variety of theoretical and practical applications. The investigation of inequalities for the Sugeno integral was started by Román-Flores et al. in [8,9,10,11,12,13], and was subsequently expanded upon by Ouyang et al. in [14,15,16].

Many researchers studied celebrated inequalities using the Sugeno integral; for example, Hu [17] proved the following Chebyshev-type inequalities for a Sugeno-like integral by using a binary operation called g-seminorm.

Theorem 1 

([17]). Let $H:B^n\to B$ be a left continuous and non-decreasing n-place function. Let $u:B\to B$ be any strictly monotone increasing bijection, and suppose that $f_1,\dots ,f_n:X\to B$ is any comonotone system. If the g-seminorm G satisfies

$$G\left(\phi \left(H\left(x_1,\dots ,x_n\right)\right),c\right)\ge \left\{\begin{array}{c}H\left(G\left(\phi \left(x_1\right),c\right),\phi \left(x_2\right),\dots ,\phi \left(x_n\right)\right)\\ \\ H\left(\phi \left(x_1\right),G\left(\phi \left(x_2\right),c\right),\dots ,\phi \left(x_n\right)\right)\\ \\ \cdots \\ H\left(\phi \left(x_1\right),\phi \left(x_2\right),\dots ,G\left(\phi \left(x_n\right),c\right)\right)\end{array}\right\}$$

for all $\left(x_1,\dots ,x_n\right)\in B$, then for any $A\subseteq \Sigma$ we have

$${\int }_{GA}\phi \left(H\left(f_1,\dots ,f_n\right)\right)d\mu \ge H\left({\int }_{GA}\phi \left(f_1\right)d\mu ,\dots ,{\int }_{GA}\phi \left(f_n\right)d\mu \right).$$

Caballero and Sadarangani [18] developed a Cauchy–Schwarz-type inequality for the Sugeno integral.

Theorem 2 

([18]). Let $\left(X,\Sigma ,\mu \right)$ be a fuzzy measure space, $f,g\in {\mathcal{F}}_{+}\left(X\right)$, (${\mathcal{F}}_{+}\left(X\right)$ is the set of all non-negative measurable functions with respect to Σ) and f and g comonotone functions, $A\in \Sigma$ with ${\int }_{A}f·gd\mu \le 1$. Then,

$${\left({\int }_{A}f·gd\mu \right)}^2\le \left({\int }_A{f}^{2}d\mu \vee {\int }_A{g}^{2}d\mu \right).$$

Agahi et al. [19] shown a generalization of the Stolarsky inequality for a Sugeno integral as follows:

Theorem 3 

([19]). Let $f:\left[0,1\right]\to \left[0,1\right]$ be a nonincreasing function, $\left(\left[0,1\right],\mathcal{B}\left(\left[0,1\right]\right),m\right)$ a fuzzy measure space, and define $h:\left[0,1\right]\to \left[0,1\right]$ by $h\left(a\right)=m\left(\left[0,m\right]\right)$ for $a\in \left[0,1\right]$. Let $\beta ,\gamma$ be automorphisms on $\left[0,1\right]$ (i.e., $\beta ,\gamma :\left[0,1\right]\to \left[0,1\right]$ are increasing bijections) and $\alpha ={\left({\beta }^{-1}★{\alpha }^{-1}\right)}^{-1}$ is a continuous aggregation function that is jointly strictly increasing and bounded from above by min, and which is dominated by h, i.e., for all $x,y$ from $\left[0,1\right]$, it holds

$$h\left(x★y\right)\ge h\left(x\right)★h\left(y\right),$$

then

$$\left(S\right){\int }_0^{1}f\left(\alpha \right)dm\ge \left(S\right){\int }_0^{1}f\left(\beta \right)dm★\left(S\right){\int }_0^{1}f\left(\gamma \right)dm,$$

where $f\left(\alpha \right)$ means the composite function defined on $\left[0,1\right]$ and given by $f\left(\alpha \right)\left(x\right)=f\left(\alpha \left(x\right)\right)$.

We also refer the reader to Ouyang et al. [16], which contains the study of the Minkowski type for the Sugeno integral on abstract spaces.

Theorem 4 

([16]). Let μ be an arbitrary fuzzy measure on $\left[0,a\right]$ and $f,g:\left[0,a\right]\to \mathbb{R}$ be two real-valued measurable functions such that $\left(S\right){\int }_0^{a}fd\mu \le 1$ and $\left(S\right){\int }_0^{a}gd\mu \le 1$. If $f,g$ are both non-decreasing, then the inequality

$$\left(S\right){\int }_0^{a}fgd\mu \ge \left(\left(S\right){\int }_0^{a}fd\mu \right)\left(\left(S\right){\int }_0^{a}gd\mu \right)$$

holds.

Caballero and Sadarangani [20] have shown that the classical Hermite–Hadamard inequalities [21,22] do not hold true for fuzzy integrals in general and established some Hermite–Hadamard-type inequalities for the Sugeno integral with peculiar examples to validate their results.

The main results from [20] are stated in the following theorems.

Theorem 5 

([20]). Let $f:\left[a,b\right]\to \left[0,\infty \right)$ be a convex function and μ the Lebesgue measure on $\mathbb{R}$.

(a) 

If $f\left(a\right)<f\left(b\right)$, then

$${\int }_a^{b}fd\mu \le min\left\{\frac{\left(b-a\right)f\left(b\right)}{f\left(b\right)-f\left(a\right)+b-a},b-a\right\}.$$

(b) 

If $f\left(a\right)=f\left(b\right)$, then

$${\int }_a^{b}fd\mu \le min\left\{f\left(a\right),b-a\right\}.$$

(c) 

If $f\left(a\right)>f\left(b\right)$, then

$${\int }_a^{b}fd\mu \le min\left\{\frac{\left(b-a\right)f\left(a\right)}{f\left(a\right)-f\left(b\right)+b-a},b-a\right\}.$$

Recently, the integral inequalities for Sugeno integrals using different kinds of convexities and other results on several other types of inequalities based on Sugeno integrals are a thought-provoking topic to many authors in the field of fuzzy integrals, see for instance [20,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37] and the references cited therein.

The integral inequalities for coordinated convex functions in the context of Sugeno integrals have not been investigated in any study so far. Hence, motivated by the ongoing research about the integral inequalities for the Sugeno integrals involving the different kinds of convex functions, it is the main objective of this paper to obtain Hermite–Hadamard-type inequalities for Sugeno double integrals by using coordinated convex functions.

2. Main Results

In order to proceed to our results, we first give some basic notations and properties of Sugeno integral.

Definition 1 

([20]). Suppose that Σ is σ-algebra of subsets of $\mathbb{R}$ and that $\mu :\Sigma \to \left[0,\infty \right)$ is non-negative extended real valued set function, then μ is said to be fuzzy measure if and only if:

1. 

$\mu (\varnothing )=0$,

2. 

E, $F\in \Sigma$ and $E\subseteq F$ imply that $\mu \left(E\right)\le \mu \left(F\right)$ (monotonicity),

3. 

$\left\{E_n\right\}\subseteq \Sigma$, $E_1\subseteq E_2\subseteq \dots$, imply ${\displaystyle \underset{n\to \infty }{lim}}\mu \left(E_n\right)=\mu \left(\underset{n=1}{\stackrel{\infty }{\cup }}{E}_n\right)$ (continuity from below),

4. 

$\left\{E_n\right\}\subseteq \Sigma$, $E_1\subseteq E_2\subseteq \dots$, $\mu \left(E_1\right)<\infty$, imply ${\displaystyle \underset{n\to \infty }{lim}}\mu \left(E_n\right)=\mu \left(\underset{n=1}{\stackrel{\infty }{\cap }}{E}_n\right)$ (continuity from above).

If f is a non-negative real-valued function defined on $\mathbb{R}$, we will denote by $L_{\alpha }f=\left\{x\in \mathbb{R}:f\left(x\right)\ge \alpha \right\}=\left\{f\ge \alpha \right\}$ the $\alpha$-level of f, for $\alpha >0$ and $L_{0}f=\overline{\{x\in \mathbb{R}:f\left(x\right)>0\}}$=supp$f,$ the support of f. It may be noted that if $\alpha \le \beta$, then $\left\{f\le \alpha \right\}\subseteq \left\{f\le \beta \right\}$. If $\mu$ is fuzzy measure on $\left(\mathbb{R},\Sigma \right)$, by ${\mathcal{F}}^{\mu }\left(\mathbb{R}\right)$, we mean all $\mu$-measurable functions from $\mathbb{R}$ to $\left[0,\infty \right)$.

Definition 2 

([20]). Suppose that μ is a fuzzy measure on $\left(\mathbb{R},\Sigma \right)$. If $f\in$${\mathcal{F}}^{\mu }\left(\mathbb{R}\right)$ and $A\in \Sigma$ then the Sugeno integral (or fuzzy integral) of f on A with respect to the fuzzy measure μ is defined as:

$${\int }_{A}fd\mu =\underset{\alpha \ge 0}{\bigvee }\left[\alpha \wedge \mu \left(A\cap \left\{f\ge \alpha \right\}\right)\right],$$

where ∨ and ∧ denote the supremum and infimum on $\left[0,\infty \right)$, respectively. In particular, if $A=X$, then

$${\int }_{X}fd\mu =\int fd\mu =\underset{\alpha \ge 0}{\bigvee }\left[\alpha \wedge \mu \left(\left\{f\ge \alpha \right\}\right)\right].$$

The following properties of the Sugeno integral are well known and can be found in [5].

Proposition 1 

([5]). If μ is a fuzzy measure on $\left(\mathbb{R},\Sigma \right)$, $A\in \Sigma$ and f, $g\in$${\mathcal{F}}^{\mu }\left(\mathbb{R}\right),$ then

1. 

${\int }_{A}fd\mu \le \mu \left(A\right)$.

2. 

${\int }_{A}kd\mu =k\wedge \mu \left(A\right)$.

3. 

If $g\le f$ on A, then ${\int }_{A}gd\mu \le {\int }_{A}fd\mu$.

4. 

$\mu \left(A\cap \left\{f\ge \alpha \right\}\right)\ge \alpha \Rightarrow {\int }_{A}fd\mu \ge \alpha$.

5. 

$\mu \left(A\cap \left\{f\ge \alpha \right\}\right)\le \alpha \Rightarrow {\int }_{A}fd\mu \le \alpha$.

6. 

${\int }_{A}fd\mu <\alpha$ if and only if there exists $\gamma <\alpha$ such that $\mu \left(A\cap \left\{f\ge \gamma \right\}\right)<\alpha$.

7. 

${\int }_{A}fd\mu >\alpha$ if and only if there exists $\gamma >\alpha$ such that $\mu \left(A\cap \left\{f\ge \gamma \right\}\right)>\alpha$.

Definition 3 

([20]). If μ is a fuzzy measure on $\left(\mathbb{R},\Sigma \right)$, $A\in \Sigma$ and $f\in$${\mathcal{F}}^{\mu }\left(\mathbb{R}\right),$ then the survival function F associated to f on A is defined by

$$F\left(\alpha \right)=\mu \left(A\cap \left\{f\ge \alpha \right\}\right),$$

where $\alpha \ge 0$.

Remark 1 

([20]). Consider the survival function F associated to f on A, that is, $F\left(\alpha \right)=\mu \left(A\cap \left\{f\ge \alpha \right\}\right)$. If $\mu \left(A\cap \left\{f\ge \alpha \right\}\right)\ge \alpha$ and $\mu \left(A\cap \left\{f\ge \alpha \right\}\right)\le \alpha$, from (4) and (5) of Proposition 1, we obtain

$$F\left(\alpha \right)=\alpha \Rightarrow {\int }_{A}fd\mu =\alpha .$$

The above equation implies that any fuzzy integral can be calculated by solving the equation $F\left(\alpha \right)=\alpha$.

Theorem 6 

([38]). Let $(X,\mathcal{X},\mu )$ and $(Y,\mathcal{Y},\nu )$ be two fuzzy measure spaces and f be a $\mathcal{X}\times \mathcal{Y}$-measurable function. Then, there exists a fuzzy measure m on $X\times Y$ such that

$${\int }_{\mathcal{Y}}\left({\int }_{\mathcal{X}}f\left(x,y\right)d\mu \right)d\nu ={\int }_{\mathcal{X}}\left({\int }_{\mathcal{Y}}f\left(x,y\right)d\nu \right)d\mu ={\int }_{\mathcal{X}\times \mathcal{Y}}f\left(x,y\right)dm$$

Considering the characteristic function of $A\times B\in \mathcal{X}\times \mathcal{Y}$, we have $m=\mu \wedge \nu$.

Remark 2 

([13]). In the sequel, μ will denote the Lebesgue measure on $\mathbb{R}$ and $\mu \times \mu$ will denote the Lebesgue measure on $\mathbb{R}\times \mathbb{R}$. We recall that if $A,B$ are two μ-measurable subsets of $\mathbb{R}$, then

$$\left(\mu \times \mu \right)(A\times B)=\mu \left(A\right)\mu \left(B\right).$$

For more details on Sugeno integral we refer the interested readers to [1,5].

The classical Hermite–Hadamard inequalities provide estimates of the mean value of a convex function $f:[a,b]\to \mathbb{R}$

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le \frac{f\left(a\right)+f\left(b\right)}{2}.$$

(1)

The inequalities (1) were first discovered by C. Hermite [22] in 1893 and J. Hadamard proved it independently again in [21]. The inequalities (1) have numerous generalizations and extensions, we the refer the reader to [20,35,37,39,40,41,42,43,44,45,46,47] and the references cited therein.

We recall some definitions and Hermite–Hadamard-type integral inequalities for coordinated convex functions on $[a,b]\times [c,d]$ in ${\mathbb{R}}^2$.

Definition 4 

([39]). Let $\Delta =:[a,b]\times [c,d]$ in ${\mathbb{R}}^2$ with $a<b$ and $c<d$ be a bidimensional interval. A mapping $f:\Delta \to \mathbb{R}$ is said to be convex on Δ if the inequality

$$f(\lambda x+(1-\lambda )z,\lambda y+(1-\lambda \left)w\right)\le \lambda f(x,y)+(1-\lambda )f(z,w)$$

holds for all $(x,y),(z,w)\in \Delta$ and $\lambda \in$$[0,1]$.

Dragomir [39] modified Definition 4 of convex functions on $\Delta$, known as coordinated convex functions as follows.

Definition 5 

([39]). A function $f:\Delta \to \mathbb{R}$ is said to be convex on the coordinates on Δ if the partial mappings $f_y:[a,b]\to \mathbb{R}$, $f_y\left(u\right)=f(u,y)$ and $f_x:[c,d]\to \mathbb{R}$, $f_x\left(v\right)=f(x,v)$ are convex where defined for all $x\in [a,b]$, $y\in [c,d]$.

Remark 3 

([41]). It is clear that if a function $f:\Delta \to \mathbb{R}$ is convex on the coordinates on Δ, then

$$\begin{array}{cc}\hfill & f(tx+(1-t)z,sy+(1-s\left)w\right)\hfill \\ \hfill & \le tsf(x,y)+t(1-s)f(x,w)+s(1-t)f(z,y)+(1-t)(1-s)f(z,w),\hfill \end{array}$$

holds for all $\left(t,s\right)\in [0,1]\times \left[0,1\right]$ and $x,z\in [a,b],y,w\in [c,d]$.

The following Hermite–Hadamard-type for coordinated convex functions on the rectangle from the plane ${\mathbb{R}}^2$ were proved in ([39], Theorem 1, page 778):

Theorem 7 

([39]). If $f:\Delta \to \mathbb{R}$ is coordinated convex on Δ, then

$$\begin{array}{c}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\hfill \\ \le \frac{1}{2}\left[\frac{1}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)dx+\frac{1}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)dy\right]\\ \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dydx\\ \le \frac{1}{4}\left[\frac{1}{b-a}{\int }_a^b\left[f\left(x,c\right)+f\left(x,d\right)\right]dx+\frac{1}{d-c}{\int }_c^d\left[f\left(a,y\right)+f\left(b,y\right)\right]dy\right]\\ \hfill \le \frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}.\end{array}$$

(2)

The above inequalities are sharp.

We will see that this inequality does not hold true for fuzzy integrals in general.

Example 1. 

Take $X\times Y=\left[0,1\right]\times \left[0,1\right]$ and let μ be the Lebesgue measure on X and Y. Then, $\mu \times \mu$ will denote the Lebesgue measure on $X\times Y$. Let $f\left(x,y\right):X\times Y\to [0,\infty )$ be defined as $f\left(x,y\right)=\frac{1}{4}xy$, then the function is convex on the coordinates on $X\times Y$. Now, we calculate the Sugeno integral ${\int }_{{\left[0,1\right]}^2}\frac{1}{4}xyd\left(\mu \times \mu \right)$. By using the Fubini theorem for fuzzy integrals, we observe that

$${\int }_{{\left[0,1\right]}^2}\frac{1}{4}xyd\left(\mu \times \mu \right)={\int }_{\left[0,1\right]}\left({\int }_{\left[0,1\right]}\frac{1}{4}xyd\mu \right)d\mu$$

Let F be the survival function associated to $f\left(x,y\right)=\frac{1}{4}xy$ on $\left[0,1\right]$, then

$$\alpha =F\left(\alpha \right)=\mu \left(\left[0,1\right]\cap \left\{\frac{1}{4}xy\ge \alpha \right\}\right)=\mu \left(\left[0,1\right]\cap \left\{y\ge \frac{4\alpha }{x}\right\}\right)=1-\frac{4\alpha }{x}.$$

Thus,

$$\alpha =\frac{x}{x+4}$$

According to Remark 1, we obtain

$${\int }_{\left[0,1\right]}\frac{1}{4}xyd\mu =\frac{x}{x+4}.$$

Hence

$${\int }_{{\left[0,1\right]}^2}\frac{1}{4}xyd\left(\mu \times \mu \right)={\int }_{\left[0,1\right]}\frac{x}{x+4}d\mu =3-2\sqrt{2}\approx 0.1716.$$

$$\frac{1}{4}\left[{\int }_{\left[0,1\right]}\frac{1}{4}xd\mu +{\int }_{\left[0,1\right]}\frac{1}{4}yd\mu \right]=\frac{1}{4}\left[\frac{1}{5}+\frac{1}{5}\right]=0.1.$$

Finally,

$$\frac{f\left(0,0\right)+f\left(0,1\right)+f\left(1,0\right)+f\left(1,1\right)}{4}=0.0625.$$

We can see that the third and the fourth inequalities in (2) are not satisfied in the fuzzy context.

Example 2. 

Take $X\times Y=\left[0,1\right]\times \left[0,1\right]$ and let μ be the Lebesgue measure on X and Y. Then, $\mu \times \mu$ will denote the Lebesgue measure on $X\times Y$. Let $f\left(x,y\right):X\times Y\to [0,\infty )$ be defined as $f\left(x,y\right)=4xy$, then the function is convex on the coordinates on $X\times Y$. Now, we calculate the Sugeno integral ${\int }_{{\left[0,1\right]}^2}4xyd\left(\mu \times \mu \right)$. By using the Fubini theorem for fuzzy integrals and Remark 1, we observe that $f\left(\frac{1}{2},\frac{1}{2}\right)=1$ and

$${\int }_{{\left[0,1\right]}^2}4xyd\left(\mu \times \mu \right)={\int }_{\left[0,1\right]}\left({\int }_{\left[0,1\right]}4xyd\mu \right)d\mu .$$

Suppose that F is the survival function associated to $f\left(x,y\right)=4xy$ on $\left[0,1\right]$. Then,

$$\alpha =F\left(\alpha \right)=\mu \left(\left[0,1\right]\cap \left\{4xy\ge \alpha \right\}\right)=\mu \left(\left[0,1\right]\cap \left\{y\ge \frac{\alpha }{4x}\right\}\right)=1-\frac{\alpha }{4x}.$$

Thus,

$$\alpha =\frac{4x}{4x+1}$$

According to Remark 1, we obtain

$${\int }_{\left[0,1\right]}4xyd\mu =\frac{4x}{4x+1}.$$

Hence,

$${\int }_{{\left[0,1\right]}^2}4xyd\left(\mu \times \mu \right)={\int }_{\left[0,1\right]}\frac{4x}{4x+1}d\mu =\frac{9-\sqrt{17}}{8}\approx 0.6096.$$

Lastly,

$$\frac{1}{2}\left[{\int }_{\left[0,1\right]}f\left(x,\frac{1}{2}\right)d\mu +{\int }_{\left[0,1\right]}f\left(\frac{1}{2},y\right)d\mu \right]=\frac{1}{2}\left[{\int }_{\left[0,1\right]}2xd\mu +{\int }_{\left[0,1\right]}2yd\mu \right]=\frac{1}{3}\approx 0.3333,$$

which shows that the first and the second inequalities in (2) are also not satisfied in the fuzzy framework.

Now we prove estimates for the third and the fourth inequalities in (2) but for the Sugeno integral. In order to obtain our main results, we will use a non-additivity assumption of integrals together with the Lebesgue measure, that is, we use the fuzzy context to prove the results.

Our first result gives estimates of the integrals involved in the fourth inequality in (2) for fuzzy integrals over the interval $\left[0,1\right]$.

Theorem 8. 

Let $g:\left[0,1\right]\times \left[0,1\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[0,1\right]\times \left[0,1\right]$.

1. 

If $g\left(1,1\right)+g\left(1,0\right)>g\left(0,1\right)+g\left(0,0\right)$, then

$$\begin{array}{c}\hfill {\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \le min\left\{1,\frac{g\left(1,1\right)+g\left(1,0\right)}{1+g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)}\right\}.\end{array}$$

(3)

2. 

If $g\left(1,1\right)+g\left(0,1\right)>g\left(1,0\right)+g\left(0,0\right)$, then

$$\begin{array}{c}\hfill {\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \le min\left\{1,\frac{g\left(1,1\right)+g\left(0,1\right)}{1+g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)}\right\}.\end{array}$$

(4)

Proof. 

By the coordinated convexity of g on $\left[0,1\right]\times \left[0,1\right]$ that

$$\begin{array}{cc}\hfill g(x,0)+g(x,1)& =g\left(\left(1-x\right)·0+x·1,1\right)+g\left(\left(1-x\right)·0+x·1,0\right)\hfill \\ \hfill & \le \left(1-x\right)g\left(0,1\right)+xg\left(1,1\right)+\left(1-x\right)g\left(0,0\right)+xg\left(1,0\right)\hfill \\ \hfill & =g\left(0,1\right)+g\left(0,0\right)+x\left[g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)\right]\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{cc}\hfill & {\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \hfill \\ \hfill & \le {\int }_{\left[0,1\right]}\left\{g\left(0,1\right)+g\left(0,0\right)+x\left[g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)\right]\right\}d\mu \hfill \\ \hfill & ={\int }_{\left[0,1\right]}{h}_1\left(x\right)d\mu .\hfill \end{array}$$

If we consider the survival function F together with the assumption $g\left(1,1\right)+g\left(1,0\right)>g\left(0,1\right)+g\left(0,0\right)$. Then, according to Remark 1, we obtain

$$\begin{array}{c}\alpha =\mu \left(\left[0,1\right]\cap \left\{\begin{array}{c}g\left(0,1\right)+g\left(0,0\right)\\ +x\left[g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)\right]\ge \alpha \end{array}\right\}\right)\hfill \\ \hfill =1-\frac{\alpha -g\left(0,1\right)-g\left(0,0\right)}{g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)}.\end{array}$$

(5)

The solution of the Equation (5) is $\alpha =\frac{g\left(1,1\right)+g\left(1,0\right)}{1+g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \le {\int }_{\left[0,1\right]}{h}_1\left(x\right)d\mu \le \mu \left(\left[0,1\right]\right)=1.$$

(6)

The solution $\alpha =\frac{g\left(1,1\right)+g\left(1,0\right)}{1+g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)}$ of the Equation (5) together with (6) prove the inequality (3).

Since g is convex on the coordinates on $\left[0,1\right]\times \left[0,1\right]$, we find that

$$\begin{array}{cc}\hfill g(0,y)+g(1,y)& =g\left(0,\left(1-y\right)·0+y·1\right)+g\left(1,\left(1-y\right)·0+y·1\right)\hfill \\ \hfill & \le \left(1-y\right)g\left(0,0\right)+yg\left(0,1\right)+\left(1-y\right)g\left(1,0\right)+yg\left(1,1\right)\hfill \\ \hfill & =g\left(1,0\right)+g\left(0,0\right)+y\left[g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)\right]\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{cc}\hfill & {\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \hfill \\ \hfill & \le {\int }_{\left[0,1\right]}\left\{g\left(1,0\right)+g\left(0,0\right)+y\left[g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)\right]\right\}d\mu \hfill \\ \hfill & ={\int }_{\left[0,1\right]}{h}_2\left(y\right)d\mu .\hfill \end{array}$$

Suppose F is the survival function and $g\left(1,1\right)+g\left(0,1\right)>g\left(1,0\right)+g\left(0,0\right)$, then according to Remark 1, we obtain

$$\begin{array}{c}\beta =\mu \left(\left[0,1\right]\cap \left\{\begin{array}{c}g\left(1,0\right)+g\left(0,0\right)\\ +y\left[g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)\right]\ge \beta \end{array}\right\}\right)\hfill \\ \hfill =1-\frac{\beta -g\left(1,0\right)-g\left(0,0\right)}{g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)}.\end{array}$$

(7)

The solution of the Equation (7) is $\beta =\frac{g\left(1,1\right)+g\left(0,1\right)}{1+g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \le {\int }_{\left[0,1\right]}{h}_2\left(y\right)d\mu \le \mu \left(\left[0,1\right]\right)=1.$$

(8)

The solution $\beta =\frac{g\left(1,1\right)+g\left(0,1\right)}{1+g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)}$ of the Equation (7) together with (8) prove the inequality (4). □

Remark 4. 

If $g\left(1,1\right)+g\left(1,0\right)=g\left(0,1\right)+g\left(0,0\right)$, then

$${\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \le 1\wedge \left[g\left(1,1\right)+g\left(1,0\right)\right].$$

(9)

If $g\left(1,1\right)+g\left(0,1\right)=g\left(1,0\right)+g\left(0,0\right)$, then

$${\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \le 1\wedge \left[g\left(1,1\right)+g\left(0,1\right)\right].$$

(10)

The second result provides an estimate of the first integral of the third inequality in (2) for fuzzy integrals over the interval $\left[0,1\right]$.

Theorem 9. 

Let $g:\left[0,1\right]\times \left[0,1\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$ such that $g(0,1)>g(0,0)$ and $g(0,0)+g(1,1)>g(1,0)+g(0,1)$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[0,1\right]\times \left[0,1\right]$, then

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le min\left\{1,\alpha \right\},$$

(11)

where α is a positive solution of the equation

$$\begin{array}{cc}\left(g(0,0)-g(1,0)-g(0,1)+g(1,1)\right){\alpha }^2\hfill & \\ & \hfill +\left(g(1,0)+g(0,1)-2g(1,1)-1\right)\alpha +g(1,1)=0.\end{array}$$

(12)

Proof. 

Since g is a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$. Therefore, for $\left(x,y\right)\in \left[0,1\right]\times \left[0,1\right]$, we obtain

$$\begin{array}{c}g(x,y)=g\left(\left(1-x\right)·0+x·1,\left(1-y\right)·0+y·1\right)\hfill \\ \le \left(1-x\right)\left(1-y\right)g(0,0)+\left(1-x\right)yg(0,1)\\ \hfill +x\left(1-y\right)g(1,0)+xyg(1,1)=h(x,y).\end{array}$$

Suppose F is the survival function with respect to the variable x together with $g(0,1)>g(0,0)$ and $g(0,0)+g(1,1)>g(1,0)+g(0,1)$. By 3. of Proposition 1 and by using the Fubini theorem for fuzzy integrals and Remark 1, we have

$$\begin{array}{c}{\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{{\left[0,1\right]}^2}\left[\left(1-x\right)\left(1-y\right)g(0,0)+\left(1-x\right)yg(0,1)\right.\hfill \\ \left.+x\left(1-y\right)g(1,0)+xyg(1,1)\right]d\left(\mu \times \mu \right)={\int }_{\left[0,1\right]}\left({\int }_{\left[0,1\right]}h(x,y)d\mu \right)d\mu .\\ \hfill ={\int }_{\left[0,1\right]}\frac{g(0,1)+\left[g(1,1)-g(0,1)\right]x}{1+g(0,1)-g(0,0)+\left[g(0,0)+g(1,1)-g(1,0)-g(0,1)\right]x}d\mu =\alpha ,\end{array}$$

(13)

where $\alpha$ is a positive solution of the equation

$$\begin{array}{cc}\left(g(0,0)-g(1,0)-g(0,1)+g(1,1)\right){\alpha }^2\hfill & \\ & \hfill +\left(g(1,0)+g(0,1)-2g(1,1)-1\right)\alpha +g(1,1)=0.\end{array}$$

(14)

However, according to 1. of Proposition 1, we obtain

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le \mu \times \mu \left(\left[0,1\right]\times \left[0,1\right]\right)=\mu \left(\left[0,1\right]\right)\mu \left(\left[0,1\right]\right)=1.$$

(15)

A positive solution of (14) and (15), we obtain (11). □

Remark 5. 

If $g(0,0)+g(1,1)=g(1,0)+g(0,1)$, $g(1,1)=g(0,1)$ and $g(0,1)>g(0,0)$ in Theorem 9, then from (13) we obtain

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le 1\wedge \frac{g(0,1)}{1+g(0,1)-g(0,0)}.$$

(16)

Another estimate of the first integral of the third inequality in (2) for fuzzy integrals over the interval $\left[0,1\right]$ can be determined as given in the following remark.

Remark 6. 

Since g is a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$, we obtain

$$g(x,y)=g\left(\left(1-x\right)·0+x·1,y\right)\le \left(1-x\right)g(0,y)+xg(1,y)\le g(0,y)+g(1,y)$$

and

$$g(x,y)=g\left(x,\left(1-y\right)·0+y·1\right)\le \left(1-y\right)g(x,0)+yg(x,1)\le g(x,0)+g(x,1).$$

Hence, by 1., 3. of Proposition 1 and Fubini theorem for fuzzy integrals, we obtain

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu$$

(17)

and

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu .$$

(18)

Thus, from (3), (4), (17) and (18), we find that

$$\begin{array}{c}{\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le min\left\{1,\frac{g\left(1,1\right)+g\left(1,0\right)}{1+g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)},\right.\hfill \\ \hfill \left.\frac{g\left(1,1\right)+g\left(0,1\right)}{1+g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)}\right\}.\end{array}$$

(19)

It is clear that from (11) and (19) that the inequality

$$\begin{array}{c}{\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le min\left\{1,\alpha ,\frac{g\left(1,1\right)+g\left(1,0\right)}{1+g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)},\right.\hfill \\ \hfill \left.\frac{g\left(1,1\right)+g\left(0,1\right)}{1+g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)}\right\}\end{array}$$

(20)

holds, where α is a positive solution of the Equation (12).

The following example illustrates how to obtain an estimate for the first integral of the third inequality in (2) using fuzzy integrals.

Example 3. 

Take $X=\left[0,1\right]$, $Y=\left[0,1\right]$ and let μ be the Lebesgue measure on X and Y. Suppose that $g:\left[0,1\right]\times \left[0,1\right]\to [0,\infty )$ is defined as $g(x,y)=x^2{e}^y$ and $\mu \times \mu$ is the Lebesgue measure on $X\times Y$.

Since $g\left(1,1\right)+g\left(1,0\right)>g\left(0,1\right)+g\left(0,0\right)$, hence by 1. of Theorem 8, we have

$${\int }_{\left[0,1\right]}\left(1+e\right)x^{2}d\mu \le min\left\{1,\frac{e+1}{2+e}\right\}=\frac{e+1}{2+e}\approx 0.78806.$$

We observe that $g\left(1,1\right)+g\left(0,1\right)>g\left(1,0\right)+g\left(0,0\right)$, hence by 2. of Theorem 8, we obtain

$${\int }_{\left[0,1\right]}{e}^{y}d\mu \le 1.$$

Finally, (20) gives

$${\int }_{{\left[0,1\right]}^2}{x}^2{e}^{y}d\left(\mu \times \mu \right)\le min\left\{1,\alpha ,\frac{e+1}{2+e}\right\},$$

where α is a positive root of the equation

$$\left(e-1\right){\alpha }^2-2e\alpha +e=0.$$

The solution of this equation is

$${\alpha }_1=\frac{e+e^{\frac{1}{2}}}{e-1}\approx 2.5415$$

and

$${\alpha }_2=\frac{e-e^{\frac{1}{2}}}{e-1}\approx 0.62246.$$

Thus,

$${\int }_{{\left[0,1\right]}^2}{x}^2{e}^{y}d\left(\mu \times \mu \right)\le \frac{e-e^{\frac{1}{2}}}{e-1}\approx 0.62246.$$

The next result gives different estimates of the integral involved in the fourth inequality in (2) for fuzzy integrals over the interval $\left[0,1\right]$.

Theorem 10. 

Let $g:\left[0,1\right]\times \left[0,1\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[0,1\right]\times \left[0,1\right]$.

1. 

If $g\left(0,0\right)+g\left(0,1\right)<g\left(1,0\right)+g\left(1,1\right)$, then

$$\begin{array}{c}\hfill {\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \le min\left\{1,\frac{g\left(0,0\right)+g\left(0,1\right)}{1+g\left(0,0\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(1,1\right)}\right\}.\end{array}$$

(21)

2. 

If $g\left(1,0\right)+g\left(0,0\right)<g\left(1,1\right)+g\left(0,1\right)$, then

$$\begin{array}{c}\hfill {\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \le min\left\{1,\frac{g\left(0,0\right)+g\left(1,0\right)}{1+g\left(1,0\right)+g\left(0,0\right)-g\left(1,1\right)-g\left(0,1\right)}\right\}.\end{array}$$

(22)

Proof. 

By the coordinated convexity of g on $\left[0,1\right]\times \left[0,1\right]$, we find that

$$\begin{array}{cc}\hfill g(x,0)+g(x,1)& =g\left(\left(1-x\right)·0+x·1,1\right)+g\left(\left(1-x\right)·0+x·1,0\right)\hfill \\ \hfill & \le \left(1-x\right)g\left(0,1\right)+xg\left(1,1\right)+\left(1-x\right)g\left(0,0\right)+xg\left(1,0\right)\hfill \\ \hfill & =g\left(0,1\right)+g\left(0,0\right)+x\left[g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)\right]\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{cc}\hfill & {\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \hfill \\ \hfill & \le {\int }_{\left[0,1\right]}\left\{g\left(0,1\right)+g\left(0,0\right)+x\left[g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)\right]\right\}d\mu \hfill \\ \hfill & ={\int }_{\left[0,1\right]}{h}_1\left(x\right)d\mu .\hfill \end{array}$$

If we consider the survival function F together with the condition $g\left(0,0\right)+g\left(0,1\right)<g\left(1,0\right)+g\left(1,1\right)$, then according to Remark 1, we obtain

$$\begin{array}{c}\alpha =\mu \left(\left[0,1\right]\cap \left\{\begin{array}{c}g\left(0,1\right)+g\left(0,0\right)\\ +x\left[g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)\right]\ge \alpha \end{array}\right\}\right)\hfill \\ \hfill =\frac{\alpha -g\left(0,1\right)-g\left(0,0\right)}{g\left(1,1\right)+g\left(1,0\right)-g\left(0,1\right)-g\left(0,0\right)}.\end{array}$$

(23)

The solution of the Equation (23) is $\alpha =\frac{g\left(0,0\right)+g\left(0,1\right)}{1+g\left(0,0\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(1,1\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \le {\int }_{\left[0,1\right]}{h}_1\left(x\right)d\mu \le \mu \left(\left[0,1\right]\right)=1.$$

(24)

The solution of Equation (23) together with (24) give us the required inequality (21).

Since g is convex on the coordinates on $\left[0,1\right]\times \left[0,1\right]$, we find that

$$\begin{array}{cc}\hfill g(0,y)+g(1,y)& =g\left(0,\left(1-y\right)·0+y·1\right)+g\left(1,\left(1-y\right)·0+y·1\right)\hfill \\ \hfill & \le \left(1-y\right)g\left(0,0\right)+yg\left(0,1\right)+\left(1-y\right)g\left(1,0\right)+yg\left(1,1\right)\hfill \\ \hfill & =g\left(1,0\right)+g\left(0,0\right)+y\left[g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)\right]\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{cc}\hfill & {\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \hfill \\ \hfill & \le {\int }_{\left[0,1\right]}\left\{g\left(1,0\right)+g\left(0,0\right)+y\left[g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)\right]\right\}d\mu \hfill \\ \hfill & ={\int }_{\left[0,1\right]}{h}_2\left(y\right)d\mu .\hfill \end{array}$$

Suppose F is the survival function and $g\left(1,0\right)+g\left(0,0\right)<g\left(1,1\right)+g\left(0,1\right)$, then according to Remark 1, we obtain

$$\begin{array}{c}\beta =\mu \left(\left[0,1\right]\cap \left\{\begin{array}{c}g\left(1,0\right)+g\left(0,0\right)\\ +y\left[g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)\right]\ge \beta \end{array}\right\}\right)\hfill \\ \hfill =\frac{\beta -g\left(1,0\right)-g\left(0,0\right)}{g\left(1,1\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(0,0\right)}.\end{array}$$

(25)

The solution of Equation (25) is $\beta =\frac{g\left(0,0\right)+g\left(1,0\right)}{1+g\left(1,0\right)+g\left(0,0\right)-g\left(1,1\right)-g\left(0,1\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \le {\int }_{\left[0,1\right]}{h}_2\left(y\right)d\mu \le \mu \left(\left[0,1\right]\right)=1.$$

(26)

The solutions of Equation (25) and inequality (26) give us inequality (22). □

Remark 7. 

If $g\left(1,1\right)+g\left(1,0\right)=g\left(0,1\right)+g\left(0,0\right)$ in Theorem 8, then $g\left(0,0\right)+g\left(0,1\right)=g\left(1,0\right)+g\left(1,1\right)$

$${\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu \le 1\wedge \left[g\left(0,1\right)+g\left(0,0\right)\right]$$

(27)

and if $g\left(1,1\right)+g\left(0,1\right)=g\left(1,0\right)+g\left(0,0\right)$, then

$${\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu \le 1\wedge \left[g\left(1,0\right)+g\left(0,0\right)\right].$$

(28)

We can obtain some different estimates of the integral involved in the fourth inequality in (2) for fuzzy integrals over the interval $\left[0,1\right]$ if we replace “>” with “<” in the assumptions of Theorem 10.

Theorem 11. 

Let $g:\left[0,1\right]\times \left[0,1\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$ such that $g(0,0)<g(0,1)$ and $g(0,0)+g(1,1)<g(1,0)+g(0,1)$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[0,1\right]\times \left[0,1\right]$, then

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le min\left\{1,\alpha \right\},$$

(29)

where α is a positive solution of the equation

$$\begin{array}{c}\left(g(0,0)-g(1,0)-g(0,1)+g(1,1)\right){\alpha }^2\hfill \\ \hfill +\left(g(1,0)+g(0,1)-2g(0,0)-1\right)\alpha +g(0,0)=0.\end{array}$$

(30)

Proof. 

Since g is a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$. Hence, for $\left(x,y\right)\in \left[0,1\right]\times \left[0,1\right]$, we have the inequality

$$\begin{array}{c}g(x,y)=g\left(\left(1-x\right)·0+x·1,\left(1-y\right)·0+y·1\right)\hfill \\ \le \left(1-x\right)\left(1-y\right)g(0,0)+\left(1-x\right)yg(0,1)\\ \hfill +x\left(1-y\right)g(1,0)+xyg(1,1)=h(x,y).\end{array}$$

Suppose F is the survival function with respect to the variable x together with $g(0,0)<g(0,1)$ and G is the survival function with respect to the variable y together with $g(1,0)+g(0,1)<g(0,0)+g(1,1)$. By 3. of Proposition 1 and by using the Fubini theorem for fuzzy integrals and Remark 1, we have

$$\begin{array}{c}{\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{{\left[0,1\right]}^2}\left[\left(1-x\right)\left(1-y\right)g(0,0)+\left(1-x\right)yg(0,1)\right.\hfill \\ \left.+x\left(1-y\right)g(1,0)+xyg(1,1)\right]d\left(\mu \times \mu \right)={\int }_{\left[0,1\right]}\left({\int }_{\left[0,1\right]}h(x,y)d\mu \right)d\mu .\\ \hfill ={\int }_{\left[0,1\right]}\frac{g(0,0)+\left[g(1,0)-g(0,0)\right]x}{1+g(0,0)-g(0,1)+\left[g(1,0)+g(0,1)-g(0,0)-g(1,1)\right]x}d\mu =\alpha ,\end{array}$$

(31)

where $\alpha$ is a positive solution of the equation

$$\begin{array}{c}\left(g(0,0)-g(1,0)-g(0,1)+g(1,1)\right){\alpha }^2\hfill \\ \hfill +\left(g(1,0)+g(0,1)-2g(0,0)-1\right)\alpha +g(0,0)=0.\end{array}$$

(32)

However, according to 1. of Proposition 1, we obtain

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le \mu \times \mu \left(\left[0,1\right]\times \left[0,1\right]\right)=\mu \left(\left[0,1\right]\right)\mu \left(\left[0,1\right]\right)=1.$$

(33)

A positive solution of (32) and inequality (33) give us the desired inequality (29). □

Remark 8. 

If $g(0,0)+g(1,1)=g(1,0)+g(0,1)$, $g(0,0)=g(1,0)$ and $g(0,0)<g(0,1)$ in Theorem 11, then from (31) we obtain

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le 1\wedge \frac{g(0,0)}{1+g(0,0)-g(0,1)}.$$

(34)

A complementary estimate of the first integral of the third inequality in (2) for fuzzy integrals over the interval $\left[0,1\right]$ is given in the remark below.

Remark 9. 

Since $g$ is a convex function on the coordinates on $\left[0,1\right]\times \left[0,1\right]$, we obtain

$$g(x,y)=g\left(\left(1-x\right)·0+x·1,y\right)\le \left(1-x\right)g(0,y)+xg(1,y)\le g(0,y)+g(1,y)$$

and

$$g(x,y)=g\left(x,\left(1-y\right)·0+y·1\right)\le \left(1-y\right)g(x,0)+yg(x,1)\le g(x,0)+g(x,1).$$

Hence by 1., 3. of Proposition 1 and Fubini theorem for fuzzy integrals, we obtain

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{\left[0,1\right]}\left[g(0,y)+g(1,y)\right]d\mu$$

(35)

and

$${\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{\left[0,1\right]}\left[g(x,0)+g(x,1)\right]d\mu .$$

(36)

Thus, from (21), (22), (35) and (36), we obtain

$$\begin{array}{c}{\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le min\left\{1,\frac{g\left(0,0\right)+g\left(0,1\right)}{1+g\left(0,0\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(1,1\right)},\right.\hfill \\ \hfill \left.\frac{g\left(0,0\right)+g\left(1,0\right)}{1+g\left(1,0\right)+g\left(0,0\right)-g\left(1,1\right)-g\left(0,1\right)}\right\}.\end{array}$$

(37)

It is clear from (29) and (37) that the inequality

$$\begin{array}{c}{\int }_{{\left[0,1\right]}^2}g(x,y)d\left(\mu \times \mu \right)\le min\left\{1,\alpha ,\frac{g\left(0,0\right)+g\left(0,1\right)}{1+g\left(0,0\right)+g\left(0,1\right)-g\left(1,0\right)-g\left(1,1\right)},\right.\hfill \\ \hfill \left.\frac{g\left(0,0\right)+g\left(1,0\right)}{1+g\left(1,0\right)+g\left(0,0\right)-g\left(1,1\right)-g\left(0,1\right)}\right\}\end{array}$$

(38)

holds, where α is a positive solution of the Equation (30).

The following example illustrates the method for the calculation of the estimates of the first integral of the third inequality in (2).

Example 4. 

Take $X=\left[0,1\right]$, $Y=\left[0,1\right]$ and let μ be the Lebesgue measure on X and Y. Suppose that $g:\left[0,1\right]\times \left[0,1\right]\to [0,\infty )$ is defined as $g(x,y)=\left(1+x-ln\left(x+1\right)\right)e^{-y}$ and $\mu \times \mu$ be the Lebesgue measure on $X\times Y$.

Since $g\left(0,0\right)+g\left(0,1\right)<g\left(1,0\right)+g\left(1,1\right)$, hence by 1. of Theorem 9, we have

$${\int }_{\left[0,1\right]}\left(1+e^{-1}\right)\left(1+x-ln\left(x+1\right)\right)d\mu \le \frac{1+e^{-1}}{\left(1+e^{-1}\right)ln2-e^{-1}}\approx 2.3573.$$

We observe that $g\left(1,0\right)+g\left(0,0\right)<g\left(1,1\right)+g\left(0,1\right)$, hence by 2. of Theorem 9, we obtain

$${\int }_{\left[0,1\right]}\left(3-ln2\right)e^{-y}d\mu \le \frac{3-ln2}{\left(e^{-1}-1\right)ln2-3e^{-1}+4}\approx 0.93843.$$

Finally, (30) gives

$$\left(1-e^{-1}\right)\left(ln2-1\right){\alpha }^2+\left(e^{-1}-ln2-1\right)\alpha +1=0.$$

The solutions of this equation are

$$\begin{array}{cc}\hfill {\alpha }_1& =\frac{e^{-1}\left(e-\sqrt{5e^2-6e+e^2{ln}^{2}2+2eln2-2e^{2}ln2+1}-1+eln2\right)}{2\left(1-e^{-1}\right)\left(ln2-1\right)}\hfill \\ \hfill & \approx 0.68574\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\alpha }_2& =\frac{e^{-1}\left(e+\sqrt{5e^2-6e+e^2{ln}^{2}2+2eln2-2e^{2}ln2+1}-1+eln2\right)}{2\left(1-e^{-1}\right)\left(ln2-1\right)}\hfill \\ \hfill & \approx -7.5181\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill & {\int }_{{\left[0,1\right]}^2}\left[x+1-ln\left(1+x\right)\right]e^{-y}d\left(\mu \times \mu \right)\hfill \\ \hfill & \le \frac{e^{-1}\left(e-\sqrt{5e^2-6e+e^2{ln}^{2}2+2eln2-2e^{2}ln2+1}-1+eln2\right)}{2\left(1-e^{-1}\right)\left(ln2-1\right)}\hfill \\ \hfill & \approx 0.68574.\hfill \end{array}$$

In the sequel, we will prove the general case of Theorems 8–11 and examples related to these theorems.

Theorem 12. 

Let $g:\left[a,b\right]\times \left[c,d\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[a,b\right]\times \left[c,d\right]$.

1. 

If $g\left(b,d\right)+g\left(b,c\right)>g\left(a,d\right)+g\left(a,c\right)$, then

$$\begin{array}{c}{\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \hfill \\ \hfill \le min\left\{b-a,\frac{\left(b-a\right)\left(g\left(b,d\right)+g\left(b,c\right)\right)}{b-a+g\left(b,d\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(a,c\right)}\right\}.\end{array}$$

(39)

2. 

If $g\left(a,d\right)+g\left(b,d\right)>g\left(a,c\right)+g\left(b,c\right)$, then

$$\begin{array}{c}{\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu \hfill \\ \hfill \le min\left\{d-c,\frac{\left(d-c\right)\left(g\left(a,d\right)+g\left(b,d\right)\right)}{d-c+g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}\right\}.\end{array}$$

(40)

Proof. 

By the coordinated convexity of g on $\left[a,b\right]\times \left[c,d\right]$, we find that

$$\begin{array}{cc}\hfill & g(x,c)+g(x,d)\hfill \\ \hfill & =g\left(\left(1-\frac{x-a}{b-a}\right)·a+\frac{x-a}{b-a}·b,d\right)+g\left(\left(1-\frac{x-a}{b-a}\right)·a+\frac{x-a}{b-a}·b,c\right)\hfill \\ \hfill & \le \left(\frac{b-x}{b-a}\right)g\left(a,d\right)+\left(\frac{x-a}{b-a}\right)g\left(b,d\right)+\left(\frac{b-x}{b-a}\right)g\left(a,c\right)+\left(\frac{x-a}{b-a}\right)g\left(b,c\right).\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{c}{\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \hfill \\ \le {\int }_{\left[a,b\right]}\left\{\left(\frac{b-x}{b-a}\right)g\left(a,d\right)+\left(\frac{x-a}{b-a}\right)g\left(b,d\right)\right.\\ \hfill \left.+\left(\frac{b-x}{b-a}\right)g\left(a,c\right)+\left(\frac{x-a}{b-a}\right)g\left(b,c\right)\right\}d\mu ={\int }_{\left[a,b\right]}{h}_1\left(x\right)d\mu .\end{array}$$

If we consider the survival function F and $g\left(b,d\right)+g\left(b,c\right)>g\left(a,d\right)+g\left(a,c\right)$, then according to Remark 1, we obtain

$$\begin{array}{c}\alpha =\mu \left(\left[a,b\right]\cap \left\{\begin{array}{c}\left(\frac{b-x}{b-a}\right)g\left(a,d\right)+\left(\frac{x-a}{b-a}\right)g\left(b,d\right)\\ +\left(\frac{b-x}{b-a}\right)g\left(a,c\right)+\left(\frac{x-a}{b-a}\right)g\left(b,c\right)\ge \alpha \end{array}\right\}\right)\hfill \\ \hfill =b-\frac{\alpha \left(b-a\right)+ag\left(b,d\right)-bg\left(a,d\right)+ag\left(b,c\right)-bg\left(a,c\right)}{g\left(a,d\right)+g\left(b,c\right)-g\left(b,d\right)-g\left(a,c\right)}.\end{array}$$

(41)

The solution of Equation (41) is $\alpha =\frac{\left(b-a\right)\left(g\left(b,d\right)+g\left(b,c\right)\right)}{b-a+g\left(b,d\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(a,c\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \le {\int }_{\left[a,b\right]}{h}_1\left(x\right)d\mu \le \mu \left(\left[a,b\right]\right)=b-a.$$

(42)

The solutions of Equations (41) and (42) give us (39).

Since g is convex on the coordinates on $\left[a,b\right]\times \left[c,d\right]$, we find that

$$\begin{array}{cc}\hfill & g(a,y)+g(b,y)\hfill \\ \hfill & =g\left(a,\left(1-\frac{y-c}{d-c}\right)·c+\left(\frac{y-c}{d-c}\right)·d\right)+g\left(b,\left(1-\frac{y-c}{d-c}\right)·c+\frac{y-c}{d-c}·d\right)\hfill \\ \hfill & \le \left(\frac{d-y}{d-c}\right)g\left(a,c\right)+\left(\frac{y-c}{d-c}\right)g\left(a,d\right)+\left(\frac{d-y}{d-c}\right)g\left(b,c\right)+\left(\frac{y-c}{d-c}\right)g\left(b,d\right).\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{c}{\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu \le {\int }_{\left[c,d\right]}\left\{\left(\frac{d-y}{d-c}\right)g\left(a,c\right)+\left(\frac{y-c}{d-c}\right)g\left(a,d\right)\right.\hfill \\ \hfill \left.+\left(\frac{d-y}{d-c}\right)g\left(b,c\right)+\left(\frac{y-c}{d-c}\right)g\left(b,d\right)\right\}d\mu ={\int }_{\left[c,d\right]}{h}_2\left(y\right)d\mu .\end{array}$$

Suppose F is the survival function and $g\left(a,d\right)+g\left(b,d\right)>g\left(a,c\right)+g\left(b,c\right)$, then according to Remark 1, we obtain

$$\begin{array}{c}\beta =\mu \left(\left[c,d\right]\cap \left\{\begin{array}{c}\left(\frac{d-y}{d-c}\right)g\left(a,c\right)+\left(\frac{y-c}{d-c}\right)g\left(a,d\right)\\ +\left(\frac{d-y}{d-c}\right)g\left(b,c\right)+\left(\frac{y-c}{d-c}\right)g\left(b,d\right)\ge \beta \end{array}\right\}\right)\hfill \\ \hfill =d-\frac{\left(d-c\right)\beta +cg\left(a,d\right)+cg\left(b,d\right)-dg\left(a,c\right)-dg\left(b,c\right)}{g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}.\end{array}$$

(43)

The solution of Equation (43) is $\beta =\frac{\left(d-c\right)\left(g\left(a,d\right)+g\left(b,d\right)\right)}{d-c+g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[c,d\right]}\left[g(c,y)+g(d,y)\right]d\mu \le {\int }_{\left[c,d\right]}{h}_2\left(y\right)d\mu \le \mu \left(\left[c,d\right]\right)=d-c.$$

(44)

The solution of Equation (43) and the inequality (44) give us the inequality (40). □

Remark 10. 

If $g\left(b,d\right)+g\left(b,c\right)=g\left(a,d\right)+g\left(a,c\right)$ in Theorem 12, then

$${\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \le \left(b-a\right)\wedge \left(g\left(b,d\right)+g\left(b,c\right)\right)$$

(45)

and if $g\left(a,d\right)+g\left(b,d\right)=g\left(a,c\right)+g\left(b,c\right)$, then

$${\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu \le \left(d-c\right)\wedge \left(g\left(a,d\right)+g\left(b,d\right)\right).$$

(46)

Theorem 13. 

Let $g:\left[a,b\right]\times \left[c,d\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$ such that $bg(a,d)+ag\left(b,c\right)>bg(a,c)+ag\left(b,d\right)$ and $g(b,d)+g(a,c)>g(a,d)+g(b,c)$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[a,b\right]\times \left[c,d\right]$, then

$${\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le min\left\{\left(b-a\right)\left(d-c\right),\alpha \right\},$$

(47)

where α is a positive solution of the equation

$$\begin{array}{c}\left(g(b,d)-g(a,d)-g(b,c)+g(a,c)\right){\alpha }^2+\left(bc+ad-ac-bd\right.\hfill \\ \left.+\left(d-c\right)g(a,d)+\left(a-b+c-d\right)g\left(b,d\right)+\left(b-a\right)g(b,c)\right)\alpha \\ \hfill +\left(ac-bc-ad+bd\right)g(b,d)=0.\end{array}$$

(48)

Proof. 

Since g is a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$. Therefore, for $\left(x,y\right)\in \left[a,b\right]\times \left[c,d\right]$, we obtain

$$\begin{array}{c}g(x,y)=g\left(\left(1-\frac{x-a}{b-a}\right)·a+\left(\frac{x-a}{b-a}\right)·b,\left(1-\frac{y-c}{d-c}\right)·c+\left(\frac{y-c}{d-c}\right)·d\right)\hfill \\ \le \left(\frac{b-x}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(a,c)+\left(\frac{b-x}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(a,d)\\ \hfill +\left(\frac{x-a}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(b,c)+\left(\frac{x-a}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(b,d)=h(x,y).\end{array}$$

Suppose F is the survival function with respect to the variable x together with $bg(a,d)+ag\left(b,c\right)>bg(a,c)+ag\left(b,d\right)$ and $g(b,d)+g(a,c)>g(a,d)+g(b,c)$. By 3. of Proposition 1 and by using the Fubini theorem for fuzzy integrals and Remark 1, we have

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{\left[a,b\right]\times \left[c,d\right]}\left[\left(\frac{b-x}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(a,c)\right.\hfill \\ +\left(\frac{x-a}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(b,c)+\left(\frac{b-x}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(a,d)\\ \left.+\left(\frac{x-a}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(b,d)\right]d\left(\mu \times \mu \right)={\int }_{\left[a,b\right]}\left({\int }_{\left[c,d\right]}h(x,y)d\mu \right)d\mu .\\ \hfill ={\int }_{\left[a,b\right]}\frac{\left(d-c\right)\left(bg(a,d)-ag\left(b,d\right)+\left(g(b,d)-g(a,d)\right)x\right)}{\begin{array}{c}\left(b-a\right)\left(d-c\right)+b\left(g(a,d)-g(a,c)\right)+a\left(g\left(b,c\right)-g\left(b,d\right)\right)\\ +\left(g(b,d)+g(a,c)-g(a,d)-g(b,c)\right)x\end{array}}d\mu =\alpha ,\end{array}$$

(49)

where $\alpha$ is a positive solution of the equation

$$\begin{array}{c}\left(g(b,d)-g(a,d)-g(b,c)+g(a,c)\right){\alpha }^2+\left(bc+ad-ac-bd\right.\hfill \\ \left.+\left(d-c\right)g(a,d)+\left(a-b+c-d\right)g\left(b,d\right)+\left(b-a\right)g(b,c)\right)\alpha \\ \hfill +\left(ac-bc-ad+bd\right)g(b,d)=0.\end{array}$$

(50)

However, according to 1. of Proposition 1, we obtain

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le \mu \times \mu \left(\left[a,b\right]\times \left[c,d\right]\right)\hfill \\ \hfill =\mu \left(\left[a,b\right]\right)\mu \left(\left[c,d\right]\right)=\left(b-a\right)\left(d-c\right).\end{array}$$

(51)

A positive solution of (50) and the inequality (51) give us the desired inequality (47). □

Remark 11. 

If in Theorem 13 the conditions $bg(a,d)+ag\left(b,c\right)>bg(a,c)+ag\left(b,d\right)$, $g(b,d)+g(a,c)=g(a,d)+g(b,c)$ and $g(b,d)=g(a,d)$ hold, then

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\hfill \\ \hfill \le \frac{\left(b-a\right)\left(d-c\right)\left(bg(a,d)-ag\left(b,d\right)\right)}{\left(b-a\right)\left(d-c\right)+b\left(g(a,d)-g(a,c)\right)+a\left(g\left(b,c\right)-g\left(b,d\right)\right)}.\end{array}$$

(52)

Remark 12. 

Since g is a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$, we obtain

$$\begin{array}{cc}\hfill g(x,y)& =g\left(\left(1-\frac{x-a}{b-a}\right)·a+\left(\frac{x-a}{b-a}\right)·b,\frac{y-c}{d-c}\right)\hfill \\ \hfill & \le \left(1-\frac{x-a}{b-a}\right)g(a,y)+\left(\frac{x-a}{b-a}\right)g(b,y)\le g(a,y)+g(b,y)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill g(x,y)& =g\left(x,\left(1-\frac{y-c}{d-c}\right)·c+y·d\right)\hfill \\ \hfill & \le \left(1-\frac{y-c}{d-c}\right)g(x,c)+\frac{y-c}{d-c}g(x,d)\le g(x,c)+g(x,d).\hfill \end{array}$$

Hence, by 1., and 3. of Proposition 1 and the Fubini theorem for fuzzy integrals, we obtain

$${\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le \left(b-a\right)\wedge {\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu$$

(53)

and

$${\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le \left(d-c\right)\wedge {\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu .$$

(54)

Thus, from (39), (40), (53) and (54), we observe that the inequality

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le min\left\{b-a,d-c,\right.\hfill \\ \hfill \left.\frac{\left(b-a\right)\left(g\left(b,d\right)+g\left(b,c\right)\right)}{b-a+g\left(b,d\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(a,c\right)},\frac{\left(d-c\right)\left(g\left(a,d\right)+g\left(b,d\right)\right)}{d-c+g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}\right\}\end{array}$$

(55)

holds.

It is clear from (54) and (55) that the inequality

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le min\left\{\left(b-a\right)\left(d-c\right),b-a,d-c,\alpha ,\right.\hfill \\ \hfill \frac{\left(b-a\right)\left(g\left(b,d\right)+g\left(b,c\right)\right)}{b-a+g\left(b,d\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(a,c\right)},\frac{\left(d-c\right)\left(g\left(a,d\right)+g\left(b,d\right)\right)}{d-c+g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}\end{array}$$

(56)

holds, where α is a positive solution of the Equation (48).

Theorem 14. 

Let $g:\left[a,b\right]\times \left[c,d\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[a,b\right]\times \left[c,d\right]$.

1. 

If $g\left(b,d\right)+g\left(b,c\right)<g\left(a,d\right)+g\left(a,c\right)$, then

$$\begin{array}{c}{\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \hfill \\ \hfill \le min\left\{b-a,\frac{\left(b-a\right)\left(g\left(a,d\right)+bg\left(a,c\right)\right)}{b-a+g\left(a,d\right)+g\left(a,c\right)-g\left(b,c\right)-g\left(b,d\right)}\right\}.\end{array}$$

(57)

2. 

If $g\left(a,d\right)+g\left(b,d\right)<g\left(a,c\right)+g\left(b,c\right)$, then

$$\begin{array}{c}{\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu \hfill \\ \hfill \le min\left\{d-c,\frac{\left(d-c\right)\left(g\left(a,c\right)+g\left(b,c\right)\right)}{d-c+g\left(a,c\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(b,d\right)}\right\}.\end{array}$$

(58)

Proof. 

By the coordinated convexity of g on $\left[a,b\right]\times \left[c,d\right]$ that

$$\begin{array}{cc}\hfill & g(x,c)+g(x,d)\hfill \\ \hfill & =g\left(\left(1-\frac{x-a}{b-a}\right)·a+\frac{x-a}{b-a}·b,d\right)+g\left(\left(1-\frac{x-a}{b-a}\right)·a+\frac{x-a}{b-a}·b,c\right)\hfill \\ \hfill & \le \left(\frac{b-x}{b-a}\right)g\left(a,d\right)+\left(\frac{x-a}{b-a}\right)g\left(b,d\right)+\left(\frac{b-x}{b-a}\right)g\left(a,c\right)+\left(\frac{x-a}{b-a}\right)g\left(b,c\right).\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{c}{\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \hfill \\ \le {\int }_{\left[a,b\right]}\left\{\left(\frac{b-x}{b-a}\right)g\left(a,d\right)+\left(\frac{x-a}{b-a}\right)g\left(b,d\right)\right.\\ \hfill \left.+\left(\frac{b-x}{b-a}\right)g\left(a,c\right)+\left(\frac{x-a}{b-a}\right)g\left(b,c\right)\right\}d\mu ={\int }_{\left[a,b\right]}{h}_1\left(x\right)d\mu .\end{array}$$

If we consider the survival function F and $g\left(b,d\right)+g\left(b,c\right)<g\left(a,d\right)+g\left(a,c\right)$, then according to Remark 1, we obtain that

$$\begin{array}{c}\alpha =\mu \left(\left[a,b\right]\cap \left\{\begin{array}{c}\left(\frac{b-x}{b-a}\right)g\left(a,d\right)+\left(\frac{x-a}{b-a}\right)g\left(b,d\right)\\ +\left(\frac{b-x}{b-a}\right)g\left(a,c\right)+\left(\frac{x-a}{b-a}\right)g\left(b,c\right)\ge \alpha \end{array}\right\}\right)\hfill \\ \hfill =\frac{\left(b-a\right)\alpha +ag\left(b,d\right)-bg\left(a,d\right)-bg\left(a,c\right)+ag\left(b,c\right)}{g\left(a,d\right)+g\left(a,c\right)-g\left(b,d\right)-g\left(b,c\right)}-a.\end{array}$$

(59)

The solution of Equation (59) is $\alpha =\frac{\left(b-a\right)\left(g\left(a,d\right)+g\left(a,c\right)\right)}{b-a+g\left(a,d\right)+g\left(a,c\right)-g\left(b,c\right)-g\left(b,d\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \le {\int }_{\left[a,b\right]}{h}_1\left(x\right)d\mu \le \mu \left(\left[a,b\right]\right)=b-a.$$

(60)

The solution of Equation (59) and the inequality (60) give the inequality (57).

Since g is convex on the coordinates on $\left[a,b\right]\times \left[c,d\right]$, we find that

$$\begin{array}{cc}\hfill & g(a,y)+g(b,y)\hfill \\ \hfill & =g\left(a,\left(1-\frac{y-c}{d-c}\right)·c+\left(\frac{y-c}{d-c}\right)·d\right)+g\left(b,\left(1-\frac{y-c}{d-c}\right)·c+\frac{y-c}{d-c}·d\right)\hfill \\ \hfill & \le \left(\frac{d-y}{d-c}\right)g\left(a,c\right)+\left(\frac{y-c}{d-c}\right)g\left(a,d\right)+\left(\frac{d-y}{d-c}\right)g\left(b,c\right)+\left(\frac{y-c}{d-c}\right)g\left(b,d\right).\hfill \end{array}$$

and hence by 3. of Proposition 1, we obtain

$$\begin{array}{c}{\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu \hfill \\ \le {\int }_{\left[c,d\right]}\left\{\left(\frac{d-y}{d-c}\right)g\left(a,c\right)+\left(\frac{y-c}{d-c}\right)g\left(a,d\right)\right.\\ \hfill \left.+\left(\frac{d-y}{d-c}\right)g\left(b,c\right)+\left(\frac{y-c}{d-c}\right)g\left(b,d\right)\right\}d\mu ={\int }_{\left[c,d\right]}{h}_2\left(y\right)d\mu .\end{array}$$

Suppose F is the survival function and if $g\left(a,d\right)+g\left(b,d\right)<g\left(a,c\right)+g\left(b,c\right)$, then according to Remark 1, we obtain

$$\begin{array}{c}\beta =\mu \left(\left[c,d\right]\cap \left\{\begin{array}{c}\left(\frac{d-y}{d-c}\right)g\left(a,c\right)+\left(\frac{y-c}{d-c}\right)g\left(a,d\right)\\ +\left(\frac{d-y}{d-c}\right)g\left(b,c\right)+\left(\frac{y-c}{d-c}\right)g\left(b,d\right)\ge \beta \end{array}\right\}\right)\hfill \\ \hfill =\frac{\beta \left(d-c\right)+cg\left(a,d\right)+cg\left(b,d\right)-dg\left(a,c\right)-dg\left(b,c\right)}{g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}-c.\end{array}$$

(61)

The solution of Equation (61) is $\beta =\frac{\left(d-c\right)\left(g\left(a,c\right)+g\left(b,c\right)\right)}{d-c+g\left(a,c\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(b,d\right)}$.

Applying 1. of Proposition 1, we obtain

$${\int }_{\left[c,d\right]}\left[g(c,y)+g(d,y)\right]d\mu \le {\int }_{\left[c,d\right]}{h}_2\left(y\right)d\mu \le \mu \left(\left[c,d\right]\right)=d-c.$$

(62)

The solution of Equation (61) and inequality (62) yield the inequality (58). □

Remark 13. 

If $g\left(b,d\right)+g\left(b,c\right)=g\left(a,d\right)+g\left(a,c\right)$ in Theorem 14, then

$${\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu \le \left(b-a\right)\wedge \left(g\left(a,d\right)+bg\left(a,c\right)\right)$$

(63)

and if $g\left(a,d\right)+g\left(b,d\right)=g\left(a,c\right)+g\left(b,c\right)$, then

$${\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu \le \left(d-c\right)\wedge \left(g\left(a,c\right)+g\left(b,c\right)\right).$$

(64)

Theorem 15. 

Let $g:\left[a,b\right]\times \left[c,d\right]\to [0,\infty )$ be a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$ such that $bg\left(a,c\right)+ag\left(b,d\right)<bg\left(a,d\right)+ag\left(b,c\right)$ and $g\left(a,d\right)+g\left(b,c\right)<g\left(b,d\right)+g\left(a,c\right)$. Let $\mu \times \mu$ be the Lebesgue measure on $\left[a,b\right]\times \left[c,d\right]$, then

$${\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le min\left\{\left(b-a\right)\left(d-c\right),\alpha \right\},$$

(65)

where α is a positive solution of the equation

$$\begin{array}{c}\left(g(a,d)-g(b,d)-g(a,c)+g(b,c)\right){\alpha }^2+\left(ac+bd-bc-ad\right.\hfill \\ \left.+\left(c-d\right)g(b,c)+\left(b-a+d-c\right)g\left(a,c\right)+\left(a-b\right)g(a,d)\right)\alpha \\ \hfill +\left(bc-ac-bd+ad\right)g(a,c)=0.\end{array}$$

(66)

Proof. 

Since g is a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$. Therefore, for $\left(x,y\right)\in \left[a,b\right]\times \left[c,d\right]$, we obtain

$$\begin{array}{c}g(x,y)=g\left(\left(1-\frac{x-a}{b-a}\right)·a+\left(\frac{x-a}{b-a}\right)·b,\left(1-\frac{y-c}{d-c}\right)·c+\frac{y-c}{d-c}·d\right)\hfill \\ \le \left(\frac{b-x}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(a,c)+\left(\frac{b-x}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(a,d)\\ \hfill +\left(\frac{x-a}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(b,c)+\left(\frac{x-a}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(b,d)=h(x,y).\end{array}$$

Suppose F is the survival function with respect to the variable x together with the assumptions $bg\left(a,c\right)+ag\left(b,d\right)<bg\left(a,d\right)+ag\left(b,c\right)$ and $g\left(a,d\right)+g\left(b,c\right)<g\left(b,d\right)+g\left(a,c\right)$. By 3. of Proposition 1 and by using the Fubini theorem for fuzzy integrals and Remark 1, we have

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le {\int }_{\left[a,b\right]\times \left[c,d\right]}\left[\left(\frac{b-x}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(a,c)\right.\hfill \\ +\left(\frac{b-x}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(a,d)+\left(\frac{x-a}{b-a}\right)\left(\frac{d-y}{d-c}\right)g(b,c)\\ \left.+\left(\frac{x-a}{b-a}\right)\left(\frac{y-c}{d-c}\right)g(b,d)\right]d\left(\mu \times \mu \right)={\int }_{\left[a,b\right]}\left({\int }_{\left[c,d\right]}h(x,y)d\mu \right)d\mu .\\ \hfill ={\int }_{\left[a,b\right]}\frac{(d-c)(bg\left(a,c\right)-ag\left(b,c\right)+(g\left(b,c\right)-g\left(a,c\right))x)}{\begin{array}{c}\left(b-a\right)\left(d-c\right)+b(g\left(a,c\right)-g\left(a,d\right))+a(g\left(b,d\right)\\ -g\left(b,c\right))+(g\left(a,d\right)+g\left(b,c\right)-g\left(b,d\right)-g\left(a,c\right))x\end{array}}d\mu =\alpha ,\end{array}$$

(67)

where $\alpha$ is a positive solution of the equation

$$\begin{array}{c}\left(g(a,d)-g(b,d)-g(a,c)+g(b,c)\right){\alpha }^2+\left(ac+bd-bc-ad\right.\hfill \\ \left.+\left(c-d\right)g(b,c)+\left(b-a+d-c\right)g\left(a,c\right)+\left(a-b\right)g(a,d)\right)\alpha \\ \hfill +\left(bc-ac-bd+ad\right)g(a,c)=0.\end{array}$$

(68)

However, according to 1. of Proposition 1, we obtain

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le \mu \times \mu \left(\left[a,b\right]\times \left[c,d\right]\right)\hfill \\ \hfill =\mu \left(\left[a,b\right]\right)\mu \left(\left[c,d\right]\right)=\left(b-a\right)\left(d-c\right).\end{array}$$

(69)

A positive solution of (68) and the inequality (69) prove the desired inequality (66). □

Remark 14. 

Suppose that $bg\left(a,c\right)+ag\left(b,d\right)<bg\left(a,d\right)+ag\left(b,c\right)$, $g\left(b,c\right)=g\left(a,c\right)$ and $g\left(a,d\right)+g\left(b,c\right)=g\left(b,d\right)+g\left(a,c\right)$ in Theorem 15, then

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\hfill \\ \hfill \le \frac{\left(b-a\right)(d-c)\left(bg\left(a,c\right)-ag\left(b,c\right)\right)}{\left(b-a\right)\left(d-c\right)+b\left(g\left(a,c\right)-g\left(a,d\right)\right)+a\left(g\left(b,d\right)-g\left(b,c\right)\right)}.\end{array}$$

(70)

Remark 15. 

Since $g$ is a convex function on the coordinates on $\left[a,b\right]\times \left[c,d\right]$, we obtain

$$\begin{array}{cc}\hfill g(x,y)& =g\left(\left(1-\frac{x-a}{b-a}\right)·a+\left(\frac{x-a}{b-a}\right)·b,\frac{y-c}{d-c}\right)\hfill \\ \hfill & \le \left(1-\frac{x-a}{b-a}\right)g(a,y)+\left(\frac{x-a}{b-a}\right)g(b,y)\le g(a,y)+g(b,y)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill g(x,y)& =g\left(x,\left(1-\frac{y-c}{d-c}\right)·c+y·d\right)\hfill \\ \hfill & \le \left(1-\frac{y-c}{d-c}\right)g(x,c)+\left(\frac{y-c}{d-c}\right)g(x,d)\le g(x,c)+g(x,d).\hfill \end{array}$$

Hence, by 1. and 3. of Proposition 1 and the Fubini theorem for fuzzy integrals, we obtain

$${\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le \left(b-a\right)\wedge {\int }_{\left[c,d\right]}\left[g(a,y)+g(b,y)\right]d\mu$$

(71)

and

$${\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le \left(d-c\right)\wedge {\int }_{\left[a,b\right]}\left[g(x,c)+g(x,d)\right]d\mu .$$

(72)

Thus, from (57), (58), (71) and (72), we find that

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\hfill \\ \le min\left\{b-a,d-c,\frac{\left(b-a\right)\left(g\left(b,d\right)+g\left(b,c\right)\right)}{b-a+g\left(b,d\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(a,c\right)},\right.\\ \hfill \left.\frac{\left(d-c\right)\left(g\left(a,d\right)+g\left(b,d\right)\right)}{d-c+g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}\right\}.\end{array}$$

(73)

It is clear from (72) and (73), that the inequality

$$\begin{array}{c}{\int }_{\left[a,b\right]\times \left[c,d\right]}g(x,y)d\left(\mu \times \mu \right)\le min\left\{\left(b-a\right)\left(d-c\right),b-a,d-c,\alpha ,\right.\hfill \\ \frac{\left(b-a\right)\left(g\left(b,d\right)+g\left(b,c\right)\right)}{b-a+g\left(b,d\right)+g\left(b,c\right)-g\left(a,d\right)-g\left(a,c\right)},\\ \hfill \left.\frac{\left(d-c\right)\left(g\left(a,d\right)+g\left(b,d\right)\right)}{d-c+g\left(a,d\right)+g\left(b,d\right)-g\left(a,c\right)-g\left(b,c\right)}\right\},\end{array}$$

(74)

holds, where α is a positive solution of the Equation (66).

Example 5. 

Take $X=\left[0,1\right]$, $Y=\left[0,1\right]$ and let μ be the Lebesgue measure on X and Y. Suppose that $g:\left[0,1\right]\times \left[0,1\right]\to [0,\infty )$ is defined as $g(x,y)=\left(x-arctanx\right)e^y$ and $\mu \times \mu$ be the Lebesgue measure on $X\times Y$.

Since $g\left(b,d\right)+g\left(b,c\right)>g\left(a,d\right)+g\left(a,c\right)$, hence by 1. of Theorem 12, we have

$${\int }_{\left[0,1\right]}\left(1+e\right)\left(x-arctanx\right)d\mu \le min\left\{1,\frac{4e-\pi -\pi e+4}{4e-\pi -\pi e+8}\right\}.$$

We observe that $g\left(a,d\right)+g\left(b,d\right)>g\left(a,c\right)+g\left(b,c\right)$, hence by 2. of Theorem 12, we obtain

$${\int }_{\left[0,1\right]}\left(1-\frac{\pi }{4}\right)e^{y}d\mu \le min\left\{1,\frac{\left(4-\pi \right)e}{\pi +4e-\pi e}\right\}.$$

Finally, (56) gives

$$\begin{array}{c}\hfill {\int }_{{\left[0,1\right]}^2}\left(x-arctanx\right)e^{y}d\left(\mu \times \mu \right)\le min\left\{1,\alpha ,\frac{4e-\pi -\pi e+4}{4e-\pi -\pi e+8},\frac{\left(4-\pi \right)e}{\pi +4e-\pi e}\right\},\end{array}$$

where α is a positive root of the equation

$$\left(4-\pi \right)\left(e-1\right){\alpha }^2-\left(2\left(4-\pi \right)e+\pi \right)\alpha +\left(4-\pi \right)e=0.$$

The solution of this equation is

$${\alpha }_1=\frac{\sqrt{64e-16\pi e+{\pi }^2}-8e-\pi +2\pi e}{2\left(\pi -4\right)\left(e-1\right)}\approx 0.31793$$

and

$${\alpha }_2=-\frac{\pi +8e+\sqrt{64e-16\pi e+{\pi }^2}-2\pi e}{2\left(\pi -4\right)\left(e-1\right)}\approx 4.9759.$$

Thus,

$$\begin{array}{cc}\hfill & {\int }_{{\left[0,1\right]}^2}\left(x-arctanx\right)e^{y}d\left(\mu \times \mu \right)\hfill \\ \hfill & \le \frac{\sqrt{64e-16\pi e+{\pi }^2}-8e-\pi +2\pi e}{2\left(\pi -4\right)\left(e-1\right)}\approx 0.31793.\hfill \end{array}$$

Example 6. 

Take $X=\left[0,1\right]$, $Y=\left[1,2\right]$ and let μ be the Lebesgue measure on X and Y. Suppose that $g:\left[0,1\right]\times \left[1,2\right]\to [0,\infty )$ is defined as $g(x,y)=\left(x^2-3x+\frac{5}{2}\right)e^{-\sqrt{y}}$ and $\mu \times \mu$ be the Lebesgue measure on $X\times Y$.

Since $g\left(b,d\right)+g\left(b,c\right)<g\left(a,d\right)+g\left(a,c\right)$, hence by 1. of Theorem 14, we have, then

$${\int }_{\left[0,1\right]}\left(e^{-1}+e^{-\sqrt{2}}\right)\left(x^2-3x+\frac{5}{2}\right)d\mu \le \frac{5\left(e^{-\sqrt{2}}+e^{-1}\right)}{2\left(1+2e^{-\sqrt{2}}+2e^{-1}\right)}.$$

We observe that $g\left(a,d\right)+g\left(b,d\right)<g\left(a,c\right)+g\left(b,c\right)$, hence by 2. of Theorem 14, we obtain

$${\int }_{\left[1,2\right]}2e^{-\sqrt{y}}d\mu \le \frac{2e^{-1}}{1+2e^{-1}-2e^{-\sqrt{2}}}.$$

Finally, (74) gives

$$\begin{array}{c}{\int }_{\left[0,1\right]\times \left[1,2\right]}\left(x^2-3x+\frac{5}{2}\right)e^{-\sqrt{y}}d\left(\mu \times \mu \right)\hfill \\ \hfill \le min\left\{1,\alpha ,\frac{5\left(e^{-\sqrt{2}}+e^{-1}\right)}{2\left(1+2e^{-\sqrt{2}}+2e^{-1}\right)},\frac{e^{-1}}{1+2e^{-1}-2e^{-\sqrt{2}}}\right\},\end{array}$$

where α is a positive root of the equation

$$4\left(e^{-\sqrt{2}}-e^{-1}\right){\alpha }^2+\left(2+9e^{-1}-5e^{-\sqrt{2}}\right)\alpha -5e^{-1}=0.$$

The solution of this equation is

$${\alpha }_1\approx 0.47685$$

and

$${\alpha }_2\approx 7.7294.$$

Thus,

$${\int }_{\left[0,1\right]\times \left[1,2\right]}\left(x^2-3x+\frac{5}{2}\right)e^{-\sqrt{y}}d\left(\mu \times \mu \right)\le \frac{e^{-1}}{1+2e^{-1}-2e^{-\sqrt{2}}}.$$

3. Conclusions

In this study, we discussed the theory of fuzzy integrals, also known as Sugeno integrals, and their properties. Since integral inequalities are useful tools in several theoretical and applied fields, this is why mathematicians have developed analogues of a number of integral inequalities such as Cauchy–Schwarz-type inequality, Stolarsky inequality and Minkowski inequality using fuzzy integrals. It has been shown that classical Hermite–Hadamard inequalities do not hold true for fuzzy integrals in general. In this research, we prove that the Hermite–Hadamard-type integral inequalities (2) established for coordinated convex functions do not hold for the Sugeno integral. We also prove some estimates of some inequalities involved in (2) using the Sugeno integral and Fubini theorem of fuzzy integrals and support our study by providing peculiar examples. This study could have the potential to encourage the researchers already working in this field to further explore the topic of mathematical inequalities in the fuzzy context of functions of two or more variables. The research conducted in this paper can also be extended for other generalizations of coordinated convex functions and not limited to the Hermite–Hadamard-type inequalities for coordinated convex functions, but also for coordinated Jensen’s type inequalities.