A fuzzy fixed point problem for constraint inequalities

Abstract

In this work, we prove a new fixed point theorem in the setting fuzzy metric spaces. The fuzzy metric space considered here is assumed to have two partial orders defined on it. We introduce a new approach to the existence of a fixed point of a function satisfying the two constraint inequalities. An example is included which illustrates new results of this paper. Moreover, an application of our result to the study of integral equations is provided.

Keywords

Common fixed points constraint inequalities G-complete fuzzy metric spaces partial order

1 Introduction

Zadeh [20] was the first to introduce the fuzzy sets. Kramosil and Michalek [8] employed this concept in the metric spaces to introduce fuzzy metric spaces. George and Veeramani [4] modified the concept of fuzzy metric spaces given by [8]. Gregori and Sapena [6] extended the Banach fixed point theorem to fuzzy contractive mappings on complete fuzzy metric spaces. Later on many authors have extended the theory of fixed point for fuzzy metric spaces and many of its generalizations. For instance, see [2 , 17–19].

In 2016, Jleli and Samet [7] established an existence result for the following problem: ${\begin{matrix} Find x \in X such that \\ x = Tx, \\ \begin{matrix} Ax ⪯_{1} Bx, \\ Cx ⪯_{2} Dx . \end{matrix} \end{matrix}$ (1) where T : X → X defined on a complete metric space equipped with two partial orders ⪯₁ and ⪯₂ and A, B, C, D : X → X are self-maps.

In this paper, our aim is to introduce a new approach for solution to problem (1) in G-complete fuzzy metric spaces. So fisrtly, a new definition is presented in this work. Later, some fixed point theorems are proved in fuzzy metric spaces in the sense of George and Veeramani having a partial order defined on it. Then, using this theorems, it is shown that there is a solution to the problem (1) in G-complete fuzzy metric spaces. Also, an example and an application to integral equations are given to illustrate the usability of our theory.

2 Preliminaries

In this section, we recall some basic definitions and notations which are helpful for understanding of this paper.

Definition 2.1. [16] A binary operation ∗ : [0, 1] × [0, 1] → [0, 1] is called a continuous triangular norm (in short, continuous t-norm) if it satisfies the following conditions:

∗ is commutative and associative;

∗ is continuous;

∗(a, 1) = a for every a∈ [0, 1] ;

∗(a, b) ≤ ∗(c, d) whenever a ≤ c, b ≤ d and a, b, c, d ∈ [0, 1] .

Definition 2.2. [4] A fuzzy metric space is an ordered triple (X, M, ∗) such that X is a nonempty set, ∗ is a continuous t-norm and M is a fuzzy set on X² ×(0, ∞) , satisfying the following conditions, for all x, y, z ∈ X, s, t > 0 :

M(x, y, t)>0 ;

M(x, y, t) =1 iff x = y ;

M(x, y, t) = M(y, x, t) ;

∗( M(x, y, t) , M(y, z, s))≤ M(x, z, t + s) ;

M(x, y, ·) :(0, ∞) →(0, 1] is continuous.

Definition 2.3. Let (X, M, ∗) be a fuzzy metric space. Then

A sequence {x_n} in X is said to be convergent to x in X, denoted by x_n → x, if and only if $lim_{n \to \infty} M (x_{n}, x, t) = 1$ for all t > 0, i.e. for each r ∈(0, 1) and t > 0, there exists $n_{0} \in ℕ$ such that M(x_n, x, t) >1 - r for all n ≥ n₀, [8, 15].

A sequence {x_n} is a M-Cauchy sequence if and only if for all ɛ ∈(0, 1) and t > 0, there exists $n_{0} \in ℕ$ such that M(x_n, x_m, t) ≥1 - ɛ for all m > n ≥ n₀ [4, 15]. A sequence {x_n} is a G-Cauchy sequence if and only if $lim_{n \to \infty} M (x_{n}, x_{n + p}, t) = 1$ for any p > 0 and t > 0, [5 , 18].

The fuzzy metric space (X, M, ∗) is called M-complete (G-complete) if every M-Cauchy (G-Cauchy) sequence is convergent.

Example 2.1. [18] Let $X = ℝ,$ the set of all real numbers. For x, y ∈ X and t ≥ 0, define $M (x, y, t) = {\begin{matrix} \frac{t}{t + | x - y |}, & x, y \in X, t > 0, \\ 0, & x, y \in X, t = 0 . \end{matrix}$ Then M is a fuzzy metric on $ℝ .$ Let $u_{n} = 1 + \frac{1}{2} + \dots + \frac{1}{n},$ for $n \in ℕ .$ Then, $\begin{matrix} M ( & u_{n}, u_{n + p}, t) \\ = \frac{t}{t + 1 / (n + 1) + \dots + 1 / (n + p)} \to 1, \end{matrix}$ as n→ ∞ for all p > 0 . Hence {u_n} is a G-Cauchy sequence. But obviously {u_n} is not a M-Cauchy sequence. In fact, if {u_n} is a M-Cauchy sequence, then it is a Cauchy sequence in the standard metric space $ℝ$ which gives a contradiction.

Definition 2.4. [14] Let $γ : [0, 1) \to ℝ$ be a strictly increasing, continuous mapping and for each sequence of positive numbers ${a_{n}}_{n \in ℕ}$ , $lim_{n \to \infty} a_{n} = 1$ if and only if $lim_{n \to \infty} γ (a_{n}) = + \infty .$ Let Γ be the family of all γ functions.

A mapping T : X → X is said to be a γ-contraction if there exists δ ∈(0, 1) such that $\begin{matrix} M ( & Tx, Ty, t) < 1 \Rightarrow \\ γ (M (Tx, Ty, t)) \geq γ (M (x, y, t)) + δ, \end{matrix}$ (2) for all x, y ∈ X and γ ∈ Γ.

Example 2.2. [14] Let $γ_{i} : [0, 1) \to ℝ$ where i ∈ {1, 2, 3, 4} defined by

γ₁(t) =1/(1 - t);

γ₂(t) =1/(1 - t) + t;

γ₃(t) =1/(1 - t²);

$γ_{4} (t) = 1 / \sqrt{1 - t}$ .

Then γ_i ∈ Γ for all i .

Remark 2.1. [14] Since γ is strictly increasing and by (2), it is easy to conclude that every γ-contraction T is a contractive mapping, that is, $M (Tx, Ty, t) > M (x, y, t),$ (3) for all x, y ∈ X with Tx ≠ Ty . Thus every γ-contraction is a continuous mapping.

Definition 2.5. [7] Let (X, d) be a metric space and ⪯ be a partial order on X . Then, the partial order ⪯ is d-regular if the following condition is satisfied: For every sequences {a_n}, {b_n} ⊂ X such that ${\begin{matrix} lim_{n \to \infty} d (a_{n}, a) = lim_{n \to \infty} d (b_{n}, b) = 0 \\ a_{n} ⪯ b_{n} forall n \end{matrix} \Rightarrow a ⪯ b$ where (a, b) ∈ X × X .

Definition 2.6. [7] Let X be a nonempty set endowed with two partial orders ⪯₁ and ⪯₂. Let T, A, B, C, D : X → X be given operators. Then, the operator T is (A, B, C, D, ⪯ ₁, ⪯ ₂) -stable, if the following condition is satisfied: $x \in X, Ax ⪯_{1} Bx \Rightarrow CTx ⪯_{2} DTx .$

3 The results

In this section, we present a definition and one of the our main results. Also, we consider γ-contraction given by [14] in G-complete fuzzy metric spaces.

Definition 3.1. Let (X, M, ∗) be a fuzzy metric space and ⪯ be a partial order on X . Then, the partial order ⪯ is M-regular, if the following condition is satisfied: For every sequences {x_n},{y_n} ⊂ X such that

${\begin{matrix} lim_{n \to \infty} M (x_{n}, x, t) = lim_{n \to \infty} M (y_{n}, y, t) = 1 \\ x_{n} ⪯ y_{n} forall n \end{matrix} \Rightarrow x ⪯ y$

where (x, y) ∈ X × X .

Now, let’s prove our first result.

Theorem 3.1. Let (X, M, ∗) be a G-complete fuzzy metric space endowed with two partial orders ⪯₁ and ⪯ ₂. Let T, A, B, C, D : X → X be given operators. Suppose that the following conditions are satisfied:

⪯_i is M-regular, i=1,2;

A, B, C and D are continuous;

there exists x₀ ∈ X such that ${Ax}_{0} ⪯_{1} {Bx}_{0};$

T is (A, B, C, D, ⪯ ₁, ⪯ ₂)-stable;

T is (C, D, A, B, ⪯ ₂, ⪯ ₁)-stable;

there exists γ ∈ Γ such that Ax ⪯ ₁Bx and Cy ⪯ ₂Dy ⇒ T is a γ-contraction.

Then

the sequence {Tⁿx₀} converges to some z ∈ X satisfying $Az ⪯_{1} Bz and Cz ⪯_{2} Dz;$

the point z ∈ X is a solution to (1).

Proof. (a) Let x₀ be an arbitrary element of X such that ${Ax}_{0} ⪯_{1} {Bx}_{0} .$ Such an element exists from (H₃). Let us consider the sequence {x_n} ⊂ X defined by $x_{n} = T^{n} x_{0}, n = 0, 1, 2, . . .$ Since T is (A, B, C, D, ⪯ ₁, ⪯ ₂)-stable, we have ${Ax}_{0} ⪯_{1} {Bx}_{0} \Rightarrow {CTx}_{0} ⪯_{2} {DTx}_{0},$ that is, ${Cx}_{1} ⪯_{2} {Dx}_{1} .$ Then, we have ${Ax}_{0} ⪯_{1} {Bx}_{0} and {Cx}_{1} ⪯_{2} {Dx}_{1} .$ Since T is (C, D, A, B, ⪯ ₁, ⪯ ₂)-stable, we get ${Cx}_{1} ⪯_{2} {Dx}_{1} \Rightarrow {ATx}_{1} ⪯_{1} {BTx}_{1},$ that is, ${Ax}_{2} ⪯_{1} {Bx}_{2 .}$ Again, Since T is (A, B, C, D, ⪯ ₁, ⪯ ₂)-stable, we have ${Ax}_{2} ⪯_{1} {Bx}_{2} \Rightarrow {CTx}_{2} ⪯_{2} {DTx}_{2},$ that is, ${Cx}_{3} ⪯_{2} {Dx}_{3} .$ By induction, we get ${Ax}_{2 n} ⪯_{1} {Bx}_{2 n} and {Cx}_{2 n + 1} ⪯_{2} {Dx}_{2 n + 1},$ (4)n = 0, 1, 2, . . .. Using (4) and (H₆), by symmetry, we have $\begin{matrix} γ ( & M ({Tx}_{n - 1}, {Tx}_{n}, t)) \\ = γ (M (x_{n}, x_{n + 1}, t)) \\ \geq γ (M (x_{n - 1}, x_{n}, t)) + δ . \end{matrix}$ Repeating this process, we get $\begin{matrix} γ ( & M ({Tx}_{n - 1}, {Tx}_{n}, t)) \\ \geq γ (M (x_{n - 1}, x_{n}, t)) + δ \\ = γ (M ({Tx}_{n - 2}, {Tx}_{n - 1}, t)) + δ \\ \geq γ (M (x_{n - 2}, x_{n - 1}, t)) + 2 δ \\ . . . \\ \geq γ (M (x_{0}, x_{1}, t)) + n δ . \end{matrix}$ (5) Letting n→ ∞, from (5) we get $lim_{n \to \infty} γ (M ({Tx}_{n - 1}, {Tx}_{n}, t)) = + \infty .$ (6) Then, we have $lim_{n \to \infty} M ({Tx}_{n - 1}, {Tx}_{n}, t) = 1 .$ (7) Now, we want to show that {x_n} is a Cauchy sequence. $\begin{matrix} γ ( & M (x_{n + p}, x_{n}, t)) \\ \geq γ (M (x_{n + p - 1}, x_{n - 1}, t)) + δ \\ = γ (M ({Tx}_{n + p - 2}, {Tx}_{n - 2}, t)) + δ \\ \geq γ (M (x_{n + p - 2}, x_{n - 2}, t)) + 2 δ \\ . . . \\ \geq γ (M (x_{p}, x_{0}, t)) + n δ . \end{matrix}$ (8) Since for some fixed p, γ(M(x_p, x₀, t)) is fixed, hence on letting limit n→ ∞ in (8), we get $lim_{n \to \infty} γ (M (x_{n + p}, x_{n}, t)) = + \infty .$ (9) Then, we have $lim_{n \to \infty} M (x_{n + p}, x_{n}, t) = 1 .$ (10) Thus, {x_n} is a Cauchy sequence in X. From the completeness of (X, M, ∗) there exists z ∈ X such that $lim_{n \to \infty} x_{n} = z .$ That is, $lim_{n \to \infty} M (x_{n}, z, t) = 1 .$ (11) On the other hand, from (4), we have ${Ax}_{2 n} ⪯_{1} {Bx}_{2 n}, n = 0, 1, 2, . . .$ Using the continuity of A and B, it follows from (11) that $lim_{n \to \infty} M ({Ax}_{2 n}, Az, t) = lim_{n \to \infty} M ({Bx}_{2 n}, Bz, t) = 1 .$ Since ⪯₁ is a M-regular, we get $Az ⪯_{1} Bz .$ (12) Similarly, from (4), we have ${Ax}_{2 n + 1} ⪯_{2} {Bx}_{2 n + 1}, n = 0, 1, 2, . . .$ Using the continuity of C and D, it follows from (11) that $\begin{matrix} lim_{n \to \infty} M ({Cx}_{2 n + 1}, Cz, t) \\ = lim_{n \to \infty} M ({Dx}_{2 n + 1}, Dz, t) = 1 . \end{matrix}$ Since ⪯₂ is a M-regular, we get $Cz ⪯_{2} Dz .$ (13) The proof of (a) is completed. (b) Now, we show that z ∈ X is a solution to problem (1). The continuity of T yields $\begin{matrix} M (Tz, z, t) & = lim_{n \to \infty} M ({Tx}_{n}, x_{n}, t) \\ = lim_{n \to \infty} M (x_{n + 1}, x_{n}, t) = 1, \end{matrix}$ and thus $M (Tz, z, t) = 1 .$ Therefore z ∈ X is a solution to (1).

4 Some consequences

In this section, we present some corollaries and an example.

4.1 A fixed point problem under one constraint equality

Here, we are concerned with the following problem: Find x ∈ X such that ${\begin{matrix} x = Tx \\ Ax = Bx \end{matrix}$ (14) where T, A, B : X → X are given operators and (X, M, ∗) be a G-complete fuzzy metric space endowed with a certain partial orders ⪯ . Observe that (1) is equivalent to (14) with $⪯_{1} = ⪯_{2} = ⪯, C = B and D = A .$ Then from Theorem 3.1, we obtain the following result.

Corollary 4.1. Let (X, M, ∗) be a G-complete fuzzy metric space endowed with a partial order ⪯. Let T, A, B : X → X be given operators. Suppose that the following conditions are satisfied:

⪯ is M-regular;

A and B are continuous;

there exists x₀ ∈ X such that ${Ax}_{0} ⪯ {Bx}_{0};$

for all x ∈ X, we have $Ax ⪯ Bx \Rightarrow BTx ⪯ ATx;$

for all x ∈ X, we have $Bx ⪯ Ax \Rightarrow ATx ⪯ BTx;$

there exists γ ∈ Γ such that Ax ⪯ Bx and By ⪯ Ay ⇒ T is a γ-contraction.

Then

the sequence {Tⁿx₀} converges to some z ∈ X satisfying Az = Bz ;

the point z ∈ X is a solution to (14).

4.2 bfA common fixed point problem

Here, we are concerned with the following problem: Find x ∈ X such that ${\begin{matrix} x = Tx \\ x = Bx \end{matrix}$ (15) where T, B : X → X are given operators and (X, M, ∗) be a G-complete fuzzy metric space endowed with a certain partial order ⪯ . Observe that (15) is equivalent to (14) with A = I_X, the identity mapping on X . So, take A = I_X in Corollary 4.1, we obtain the following result.

Corollary 4.2. Let (X, M, ∗) be a G-complete fuzzy metric space endowed with a partial order ⪯. Let T, B : X → X be given operators. Suppose that the following conditions are satisfied:

⪯ is M-regular;

B is continuous;

there exists x₀ ∈ X such that $x_{0} ⪯ {Bx}_{0};$

for all x ∈ X, we have $x ⪯ Bx \Rightarrow BTx ⪯ Tx;$

for all x ∈ X, we have $Bx ⪯ x \Rightarrow Tx ⪯ BTx;$

there exists γ ∈ Γ such that x ⪯ Bx and By ⪯ y ⇒ T is a γ -contraction.

Then

the sequence {Tⁿx₀} converges to some z ∈ X satisfying z = Bz ;

the point z ∈ X is a solution to (15).

Example 4.1. Let $X = ℝ \times$ [0, 1] . Define t-norm ∗ : [0, 1] × [0, 1] → [0, 1] by p ∗ q = min {p, q} and define fuzzy metric M by $M (u, v, t) = \frac{t}{t + d (u, v)},$ for all u, v ∈ X and t > 0, where u =(x, y) , v =(z, w) and $d (u, v) = d ((x, y), (z, w)) = \sqrt{(x - z)^{2} + (y - w)^{2}} .$ Then (X, M, ∗) is a G-complete fuzzy metric space. Let ⪯ be the partial order on X defined by $(x, y) ⪯ (z, w) \Leftrightarrow x ⪯ z, y ⪯ w,$ for all (x, y) , (z, w) ∈ X . Let $γ : [0, 1] \to ℝ$ such that $γ (a) = \frac{1}{1 - a^{2}}$ for all a ∈ [0, 1]. Now, define T, B : X → X by $T (x, y) = (- \frac{x}{2}, 1) and B (x, y) = (- x, 1) .$ Observe that $u ⪯ Bu \Leftrightarrow u \in {(x, y) : x \in ℝ^{-} \cup {0}, y \in [0, 1]},$ for all u ∈ X . However, $Bv ⪯ v \Leftrightarrow v \in {(z, w) : z \in ℝ^{+} \cup {0}, w = 1},$ for all v ∈ X . Let u ∈ X be such that u ⪯ Bu . Then, we have $\begin{matrix} BTu & = BT (u) = B (- \frac{x}{2}, 1) \\ = (\frac{x}{2}, 1) ⪯ (- \frac{x}{2}, 1) = Tu, \end{matrix}$ and let v ∈ X be such that Bv ⪯ v . Then, we have $Tv = (- \frac{x}{2}, 1) ⪯ (\frac{x}{2}, 1) = B (- \frac{x}{2}, 1) = BTv .$ Now, let (u, v) ∈ X × X be such that $u ⪯ Bu and Bv ⪯ v .$ Now, putting u =(x, y) , v =(z, w) in (H₆), we have $\begin{matrix} γ (M (Tu, Tv, t)) \\ = γ (M ((- \frac{x}{2}, 1), (- \frac{z}{2}, 1), t)) = γ (\frac{t}{t + \frac{| x - z |}{2}}) \\ = \frac{1}{1 - (\frac{t}{t + \frac{| x - z |}{2}})^{2}} > \frac{1}{1 - (\frac{t}{t + | x - z |})^{2}} \\ = γ (\frac{t}{t + | x - z |}) \\ = γ (M (u, v, t)) . \end{matrix}$ That is, we obtain $γ (M (Tu, Tv, t)) > γ (M (u, v, t)) .$ Therefore, there exists a δ ∈(0, 1) such that $γ (M (u_{1}, u_{2}, t)) \geq γ (M (x_{1}, x_{2}, t)) + δ .$ Then it is easy to see that T is a γ-contraction. It now follows from Corollary 4.2 that (0, 1) is a common fixed point of T and B.

Taking B = T in Corollary 4.2, we obtain the following result.

Corollary 4.3. Let (X, M, ∗) be a G-complete fuzzy metric space endowed with a partial order ⪯. Let T : X → X be a given operator. Suppose that the following conditions are satisfied:

⪯ is M-regular;

T is continuous;

there exists x₀ ∈ X such that $x_{0} ⪯ {Tx}_{0};$

for all x ∈ X, we have $x ⪯ Tx \Rightarrow T^{2} x ⪯ Tx;$

for all x ∈ X, we have $Tx ⪯ x \Rightarrow Tx ⪯ T^{2} x;$

there exists γ ∈ Γ such that x ⪯ Tx and Ty ⪯ y ⇒ T is a γ -contraction.

Then, the sequence {Tⁿx₀} converges to some z ∈ X.

5 An application

Inspired by [1] and [11], we apply our theory to examine the existence of solution for the following system of integral equations: ${\begin{matrix} u (s) = p (s) + \int_{0}^{s} G (s, r, u (r)) dr \\ u (s) = p (s) + \int_{0}^{s} H (s, r, u (r)) dr \end{matrix}$ (16) for all s ∈ [0, J] where J > 0 .

Let $X : = C ([0, J], ℝ)$ be the space of all real valued continuous functions defined on [0, J] . Observe that X is a complete metric space with respect to sup-metric $d (u, v) = sup_{s \in [0, J]} | u (s) - v (s) | .$ Also, the space (X, M, ∗) with $M (u, v, t) = \frac{t}{t + d (u, v)}, \forall u, v \in X and t > 0$ and a ∗ b = ab for all a, b ∈ [0, 1] , is a G-complete fuzzy metric space.

Consider define two operator equations T, B : X → X as follows: ${\begin{matrix} Tu (s) = p (s) + \int_{0}^{s} G (s, r, u (r)) dr \\ Bu (s) = p (s) + \int_{0}^{s} H (s, r, u (r)) dr \end{matrix}$ (17) where $p \in C ([0, J], ℝ)$ and $G, H \in C ([0, J] \times [0, J] \times ℝ, ℝ) .$ Then the existence of a solution for the system (16) is equivalent to the existence of a common fixed point of T and B.

We analyze the Equation (16) under the following assumptions:

H(s, r, u) is decreasing related to the third variable;

there exists u₀ ∈ X such that, for all s ∈ [0, J], we have $u_{0} (s) \leq p (s) + \int_{0}^{s} H (s, r, u_{0} (r)) dr;$

(i) for all u ∈ X, we have $\begin{matrix} u (s) \leq p (s) + \int_{0}^{s} H (s, r, u (r)) dr \\ \Rightarrow H (s, r, u (r)) \leq G (s, r, u (r)), \end{matrix}$ (ii) for all u ∈ X, we have $\begin{matrix} p (s) + \int_{0}^{s} H (s, r, u (r)) dr \leq u (s) \\ \Rightarrow G (s, r, u (r)) \leq H (s, r, u (r)); \end{matrix}$

there exists g : [0, J] × [0, J] → [0, + ∞) such that for all r, s ∈ [0, J] , $g (s, r) \in L^{1} ([0, J], ℝ),$ and for all u, v ∈ X with u ≤ Bu and Bv ≤ v, we get $\begin{matrix} | G (s, r, u (r)) - G (s, r, v (r)) | \\ \leq g (s, r) | u (r) - v (r) |, \end{matrix}$ where $\int_{0}^{s} g (s, r) dr$ is bounded on [0, J] and $0 < sup_{s \in [0, J]} \int_{0}^{s} g (s, r) dr \leq λ < 1 .$

Theorem 5.1. Assume that the conditions (A₁)-(A₄) are satisfied. Then the system of integral Equations (16) has a solution in X.

Proof. Let X be equipped with the partial order ⪯ given by $u, v \in X, u ⪯ v \Leftrightarrow u (s) \leq v (s), s \in [0, J] .$ Then, from definition, ≤ is M-regular. Also, B is obviously continuous and from assumption (A₂) , u₀ ≤ Bu₀. Now, we prove the following three steps to complete the proof.

Step 1: Prove that for all s ∈ [0, J] and u ∈ X, $u (s) \leq Bu (s) \Rightarrow BTu (s) \leq Tu (s) .$ Let s ∈ [0, J] , u ∈ X and u(s) ≤ Bu(s) . Applying the part (i) of (A₃), we have H(s, r, u(r)) ≤ G(s, r, u(r)) . By using (17), we deduce that Bu(s) ≤ Tu(s) . Since u(s) ≤ Bu(s) ≤ Tu(s) , the part (i) of (A₃) and (A₁) imply that $\begin{matrix} H (s, r, Tu (r)) & \leq H (s, r, Bu (r)) \\ \leq H (s, r, u (r)) \\ \leq G (s, r, u (r)) . \end{matrix}$ Therefore H(s, r, Tu(r)) ≤ G(s, r, u(r)) . By using (17), we get BTu(s) ≤ Tu(s) .

Step 2: Show that for all s ∈ [0, J] and u ∈ X, $Bu (s) \leq u (s) \Rightarrow Tu (s) \leq BTu (s) .$

Let s ∈ [0, J] , u ∈ X and Bu(s) ≤ u(s) . Applying the part (ii) of (A₃), we have G(s, r, u(r)) ≤ H(s, r, u(r)) . By using (17), we deduce that Tu(s) ≤ Bu(s) . Since Tu(s) ≤ Bu(s) ≤ u(s) , the part (ii) of (A₃) and (A₁) imply that $\begin{matrix} G (s, r, u (r)) & \leq H (s, r, u (r)) \\ \leq H (s, r, Bu (r)) \\ \leq H (s, r, Tu (r)) . \end{matrix}$ Therefore G(s, r, u(r)) ≤ H(s, r, Tu(r)) . By using (17), we get Tu(s) ≤ BTu(s) . Step 3: Finally, for all u, v ∈ X such that u ≤ Bu and Bv ≤ v, we prove that T is a γ-contraction. By assumption (A₄), we have $\begin{matrix} | Tu (s) - Tv (s) | \\ \leq \int_{0}^{s} | G (s, r, u (r)) - G (s, r, v (r)) | dr \\ \leq \int_{0}^{s} g (s, r) | u (r) - v (r) | dr \\ \leq d (u, v) \int_{0}^{s} g (s, r) dr \\ \leq λ d (u, v) . \end{matrix}$ It follows that $\begin{matrix} d (Tu, Tv) \leq λ d (u, v) . \end{matrix}$ Since 0 < λ < 1, we deduce $\begin{matrix} \frac{1}{1 - M (Tu, Tv, t)} & = 1 + \frac{t}{d (Tu, Tv)} \\ \geq 1 + \frac{t}{λ d (u, v)} \\ = 1 + \frac{1}{λ} \frac{t}{d (u, v)} \\ \geq 1 + \frac{t}{d (u, v)} + λ \\ = \frac{1}{1 - M (u, v, t)} + λ . \end{matrix}$ Taking γ(x) =1/(1 - x) and δ = λ by the last inequality, we infer that $γ (M (Tu, Tv, t)) \geq γ (M (u, v, t)) + δ .$ Finally, applying Corollary 4.2, T and B have a solution in X which is a solution of the system of integral Equations (16).

6 Conclusion

Fixed point theory for fuzzy metric spaces is recognized to be one of the basic approach for solving various mathematical problems. Here, we have obtained sufficient conditions for the existence of a fixed point of a certain operator under two constraint inequalities with respect to two partial orders in G-complete fuzzy metric spaces. An interesting question is the existence of a best proximity point of a certain operator under constraint inequalities in fuzzy metric spaces. Such a question will be studied in a future work.

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