Contractions in fuzzy rectangular b -metric spaces with application

Abstract

In this paper, by introducing the concept of a fuzzy rectangular-b-metric space, the notion of a fuzzy metric space and a fuzzy b-metric space are generalized. The well known metric fixed point theorems are established in the setting of fuzzy rectangular b-metric spaces and illustrated by examples. To show the significance of our result an application is presented to establish the existence of a solution of integral equation. Our results generalize many existing theorems in the literature.

Keywords

Fixed points Contractions

1. Introduction and Preliminaries

In 1965, Zadeh [27] introduced the concept of a fuzzy set. A fuzzy set A in X is a function with domain X and values in [0, 1]. Since then a substantial literature has been developed to investigate fuzzy sets and their applications. The notion of fuzzy maps was introduced by Heilpern [12] where some fixed point theorems for fuzzy maps are also established. For the study on fuzzy maps we refer to the work of Kamran [16] and the references therein.

A triangular norm (t-norm) is a commutative and associative binary operation *: [0, 1] × [0, 1] → [0, 1] such that a * 1 = a and ∀ a, b, c, d ∈ [0, 1] if a ≤ c and b ≤ d then a * b ≤ c * d. Some well-known examples of continuous t-norms are a ∧ b = min {a, b} , a · b = ab and a * _Lb = max {a + b - 1, 0}. For details see [23].

In 1975, Kramosil and Michalek [18] initiated the idea of a fuzzy distance between two elements of a nonempty set by using the concepts of a fuzzy set and a t-norm. This work proved to be the foundation for developing fixed point theory for fuzzy metric spaces. In 1988, by defining a Cauchy sequence, Grabiec [9] introduced a weaker form of completeness of fuzzy metric spaces which is now known as G-completeness in literature and the Cauchy sequence in such spaces is called a G-Cauchy sequence. George and Veeramani [6] generalized the concept of fuzzy metric spaces introduced by Kramosil and Michalek [18]. Given a non empty set X,and a contnuous t-norm *, the 3-tuple (X, M, *) is said to be a fuzzy metric space [18] if M is a fuzzy set on X × X × [0, ∞) satisfying the conditions M (x, y, t) >0 and M (x, y, t) = M (y, x, t) =1 iff x = y and M (x, z, t + s) ≥ M (x, y, t) * M (y, z, s). Also M (x, y, ·) is left continuous function from [0, ∞) → [0, 1] and $lim_{t \to \infty} M (x, y, t) = 1$ . Later on many authors have further studied and developed the fixed point theory for fuzzy metric spaces. For instance, see the work of Gopal et al. [8], C. Vetro [26], Roldán-López-de-Hierro [13], Turkoglu et al. [24] and the references therein.

In 1989, Bakhtin [2] introduced the concept of b-metric spaces which was further investigated by Czerwik [5] and Khamsi and Hussain [17]. For a non empty set X and a real number b ≥ 1, a function d_b : X × X → [0, ∞) is called a b-metric on X if for all x, y, z ∈ X, we have d_b (x, y) = d_b (y, x) =0 if and only if x = y and d_b (x, z) ≤ b [d_b (x, y) + d_b (y, z)]. Moreover the pair (X, d_b) is called a b-metric space. It is observed that the b-metric space is the generalization of metric space i.e. if we take b = 1 in a b-metric space then it becomes a metric space. In 2015, Hussain et al. [14] investigated the parametric b-metric and fuzzy b-metric and proved some results. In 2016, Roshan et al. [22] proved some fixed point results in b- rectangular metric spaces. They showed by an example that a sequence in b- rectangular metric space may have two limits. However there are some special situations where this is not possible. Recently,Nădăban [20] also proved some results in fuzzy b-metric spaces.

Definition 1.1. [20] Let (X, M, *) be a fuzzy b-metric space and {x_n} be a sequence in X. The sequence {x_n} is said to be convergent if there exits x ∈ X such that $lim_{n \to \infty} M (x_{n}, x, t) = 1 \forall t > 0 .$ In this case, x is called the limit of the sequence {x_n} and we write it as $lim_{n \to \infty} x_{n} = x$ or x_n → x.

Definition 1.2. [20] Let (X, M, *) be a fuzzy b-metric space and {x_n} be a sequence in X. The sequence {x_n} is said to be a Cauchy sequence if for all r ∈ (0, 1) and t > 0, there exists n_∘ ∈ N such that M (x_n, x_m, t) >1 - r, ∀ n, m ≥ n_∘ .

On the other hand, following Grabiec [9], we can extend the notion of a G-Cauchy sequence in fuzzy b-metric spaces as follows:

Definition 1.3. A sequence {x_n} in a fuzzy b-metric space (X, M, *) is said to be a G-Cauchy sequence if $lim_{n \to \infty} M (x_{n}, x_{n + q}, t) = 1$ for t > 0 and q > 0.

Therefore, in Grabiec sense, a fuzzy b-metric space in which every G-Cauchy sequence is convergent is called G-complete fuzzy b-metric space. Note that, as in fuzzy metric spaces, Definition 1.2 and Definition 1.3 are not equivalent in general. See [25] for details. In 2000, Branciari [3] introduced the concept of a rectangular metric space. Recently, George et al. [7] have introduced the concept of a rectangular b-metric space and established some interesting fixed point results for these spaces. Chugh and Kumar [4] defined the notion of a fuzzy rectangular metric space as follows:

Definition 1.4. A triplet (X, M, *) is said to be a fuzzy rectangular metric space if X is an arbitrary nonempty set, * is a continuous t-norm and M is a fuzzy set on X × X × [0, ∞) satisfying the following conditions for all x, y, z ∈ X and t, s, w > 0,

M (x, y, 0) =0

M (x, y, t) =1 if and only if x = y

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

M (x, z, t + s + w) ≥ M (x, y, t) * M (y, u, s) * M (u, z, w) for all distinct y, u ∈ X \ {x, z}.

M (x, y, .) : (0, ∞) → [0, 1] is left continuous, and $lim_{t \to \infty} M (x, y, t) = 1$ .

Example 1.1. Consider a rectangular metric space (X, d_r), Define M:X × X × [0, ∞) → [0, 1] by $M (x, y, t) = \frac{t}{t + d_{r} (x, y)}$ for all x, y ∈ X and t > 0 and let * be a t-norm on X. Then it is easy to see that (X, M, *) is a fuzzy rectangular metric space.

In this paper we aim to extend the concept of a rectangular b-metric space by defining the notion of a fuzzy rectangular b-metric space. We first establish the Banach contraction principle and then the fixed point theorem of Hicks and Rhoads [11] in the setting of fuzzy rectangular b-metric spaces.

2. Main results

Following the notion of Nădăban [20] of fuzzy b-metric space, we, now generalize Definition 1.4 by introducing the idea of fuzzy rectangular b-metric space as follows.

Definition 2.1. Let X be a nonempty set, * a continuous t-norm and M_b be a fuzzy set on X × X × [0, ∞). A triplet (X, M_b, *) is said to be a fuzzy rectangular b- metric space if there is b ≥ 1 and M_b satisfies the following conditions for all x, y, z ∈ X and t, s, w > 0,

M_b (x, y, 0) =0

M_b (x, y, t) =1 if and only if x = y

M_b (x, y, t) = M_b (y, x, t)

M_b (x, z, b (t+ s + w)) ≥ M_b (x, y, t) * M_b (y, u, s) * M_b (u, z, w) ∀ distinct y, u ∈X \ {x, z}.

M_b (x, y, .) : (0, ∞) → [0, 1] is left continuous, and $lim_{t \to \infty} M_{b} (x, y, t) = 1$ .

Example 2.1. Let (X, d_b) be a rectangular b-metric space and define M_b:X × X × [0, ∞) → [0, 1] by $M_{b} (x, y, t) = {\begin{matrix} \frac{t}{t + d_{b} (x, y)} & if t > 0 \\ 0 & if t = 0 \end{matrix}$

Then (X, M_b, *) is a fuzzy rectangular b-metric space with the t-norm a * b = min {a, b}.In fact, to prove Property FRBM-4 of Definition 2.1, let x, y, z ∈ X and t, s, w > 0. Without restraining the generality, we assume that $\begin{matrix} M_{b} (x, y, t) \leq M_{b} (y, u, s) and \\ M_{b} (x, y, t) \leq M_{b} (u, z, w) \end{matrix}$

Which implies that: $\frac{t}{t + d_{b} (x, y)} \leq \frac{s}{s + d_{b} (y, u)}$ and $\frac{t}{t + d_{b} (x, y)} \leq \frac{w}{w + d_{b} (u, z)}$ and hence, we have ${td}_{b} (y, u) \leq s d_{b} (x, y) and t d_{b} (u, z) \leq {wd}_{b} (x, y)$ $\Rightarrow (s + w) d_{b} (x, y) \geq t (d_{b} (y, u) + d_{b} (u, z)) .$ (1)

Now, note that $\begin{matrix} M_{b} (x, z, b (t + s + w)) \geq M_{b} (x, y, t) \\ \Leftrightarrow \frac{b (t + s + w)}{b (t + s + w) + d_{b} (x, z)} \geq \frac{t}{t + d_{b} (x, y)} \\ \Leftrightarrow \frac{b (t + s + w)}{b (t + s + w) + b [d_{b} (x, y) + d_{b} (y, u) + d_{b} (u, z)]} \\ \geq \frac{t}{t + d_{b} (x, y)} \\ \Leftrightarrow \frac{t + s + w}{t + s + w + d_{b} (x, y) + d_{b} (y, u) + d_{b} (u, z)} \\ \geq \frac{t}{t + d_{b} (x, y)} \\ \Leftrightarrow (s + w) d_{b} (x, y) \geq t (d_{b} (y, u) + d_{b} (u, z)) \end{matrix}$

Which is the same as (1). Hence it follows that: $\begin{matrix} M_{b} (x, z, b (t + s + w)) \\ \geq M_{b} (x, y, t) * M_{b} (y, u, s) * M_{b} (u, z, w) \end{matrix}$

Therefore, (X, M_b, *) is a fuzzy rectangular b-metric space.

The following example shows that a fuzzy rectangular b-metric space need not be a fuzzy rectangular metric space.

Example 2.2. Let $X = ℕ$ and * be a continuoust-norm. Let M_b be a fuzzy set on X × X × [0, ∞) defined by $M_{b} (x, y, t) = e^{- \frac{(x - y)^{2}}{t}}$ for all x, y ∈ X and t > 0. Then (X, M_b, *) is a fuzzy rectangular b-metric space with coefficient b = 3. But it is not a fuzzy rectangular metric space.

Note that, for all x, y, z, u ∈ X and s, w > 0, we have $(x - z)^{2} = (x - y + y - u + u - z)^{2}$ $\Rightarrow (x - z)^{2} \leq 3 {(x - y)^{2} + (y - u)^{2} + (u - z)^{2}}$

It, therefore, follows that $\begin{matrix} M_{b} (x, z, t) \\ = e^{- \frac{(x - z)^{2}}{t + s + w}} \\ \geq e^{- \frac{3 {(x - y)^{2} + (y - u)^{2} + (u - z)^{2}}}{t + s + w}} \\ = e^{- \frac{3 {(x - y)^{2}}}{t + s + w}} . e^{- \frac{3 {(y - u)^{2}}}{t + s + w}} . e^{- \frac{3 {(u - z)^{2}}}{t + s + w}} \\ \geq e^{- \frac{3 {(x - y)^{2}}}{t}} . e^{- \frac{3 {(y - u)^{2}}}{s}} . e^{- \frac{3 {(u - z)^{2}}}{w}} \\ = e^{- \frac{{(x - y)^{2}}}{\frac{t}{3}}} . e^{- \frac{3 {(y - u)^{2}}}{\frac{s}{3}}} . e^{- \frac{3 {(u - z)^{2}}}{\frac{w}{3}}} \\ = M_{b} (x, y, \frac{t}{3}) * M_{b} (y, u, \frac{s}{3}) * M_{b} (u, z, \frac{w}{3}) \end{matrix}$

Hence (X, M_b, *) is a fuzzy rectangular b-metric space and not a fuzzy rectangular metric space.

Following Grabiec, we now introduce the concept of a G-convergent sequence, and a G-Cauchy sequence in fuzzy rectangular b-metric spaces as follows:

Definition 2.2. Let (X, M_b, *) be a fuzzy rectangular b-metric space and let {x_n} be a sequence in X. Then

{x_n} is said to be a G-convergent sequence if there exits x ∈ X such that $M_{b} (x_{n}, x, t) = 1, \forall t > 0 .$

{x_n} in X is said to be a G-Cauchy sequence if $lim_{n \to \infty} M_{b} (x_{n}, x_{n + q}, t) = 1 .$

If every G-Cauchy sequence is convergent then (X, M_b, *) is called a G-complete fuzzy rectangular b-metric space.

Example 2.3. Let $X = ℝ \forall x, y \in X$ define $M_{b} (x, y, t) = {\begin{matrix} \frac{t}{t + (x - y)^{2}} & if t > 0 \\ 0 & if t = 0 \end{matrix}$ then M_b is a fuzzy rectangular b-metric.

Let ${x_{n}} = \frac{1}{n^{2}} \forall n \in ℕ$ be a sequence in X, then {x_n} converges to 0. Now $lim_{n \to \infty} M_{b} (x_{n}, 0, t) = lim_{n \to \infty} \frac{t}{t + \frac{1}{n^{2}}} = 1 .$

Hence {x_n} is a G-convergent sequence

Example 2.4. Consider again the fuzzy rectangular b-metric space given in Example 2.3. Let the sequence {x_n} be given by $x_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \forall n \in ℕ .$ Now for all $q \in ℕ$ , we have $\begin{matrix} M_{b} (x_{n + q}, x_{n}, t) & = \frac{t}{t + \frac{1}{(n + 1)^{2}} + \dots + \frac{1}{(n + q)^{2}}} \\ \to 1 as n \to \infty . \end{matrix}$

Hence {x_n} is a G-Cauchy sequence.

One can easily prove the following lemma from [10] in the setting fuzzy rectangular b-metric spaces.

Lemma 2.1. Let (X, M, *) be a complete fuzzy metric space and M (x, y, kt) ≥ M (x, y, t) for all x, y ∈ X, k ∈ (0, 1) and t > 0 then x = y.

We now state and prove the Banach Contraction Theorem in the setting of fuzzy rectangular b-metric spaces.

Theorem 2.1. (Banach Contraction Theorem in fuzzy rectangular b-metric spaces) Let (X, M_b, *) be a G-complete fuzzy rectangular b-metric space with b ≥ 1 such that $lim_{t \to \infty} M_{b} (x, y, t) = 1 \forall x, y \in X .$ (2)

Let T:X → X be a mapping satisfying $M_{b} (Tx, Ty, kt) \geq M_{b} (x, y, t)$ (3) for all $x, y \in X, k \in [0, \frac{1}{b}) .$ Then T has a unique fixed point.

Proof. Fix an arbitrary point a₀ ∈ X and for n = 0, 1, 2, ⋯ , start an iterative process a_n+1 = Ta_n. Successively applying Inequality (3), we get for all n, t > 0, $M_{b} (a_{n}, a_{n + 1}, t) \geq M_{b} (a_{0}, a_{1}, \frac{t}{k^{n}}),$ (4)

Since (X, M_b, *) is a fuzzy rectangular b-metric space, for the sequence {a_n}, writing $t = \frac{t}{3} + \frac{t}{3} + \frac{t}{3}$ and using the rectangular inequality given in FRBM-4, on M_b (a_n, a_n+p, t) in the following two cases.

Case-1: If p is odd say p = 2m + 1 where $m \in ℕ$ , we have $\begin{matrix} M_{b} & (a_{n}, a_{n + 2 m + 1}, t) \geq M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) * \\ M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) * M_{b} (a_{n + 2}, a_{n + 2 m + 1}, \frac{t}{3 b}) \\ \geq & M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) * M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) * \\ M_{b} (a_{n + 2}, a_{n + 3}, \frac{t}{(3 b)^{2}}) * M_{b} (a_{n + 3}, a_{n + 4}, \frac{t}{(3 b)^{2}}) * \\ M_{b} (a_{n + 4}, a_{n + 2 m + 1}, \frac{t}{(3 b)^{2}}) \\ \geq & M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) * M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) * \\ M_{b} (a_{n + 2}, a_{n + 3}, \frac{t}{(3 b)^{2}}) * M_{b} (a_{n + 3}, a_{n + 4}, \frac{t}{(3 b)^{2}}) * \\ M_{b} & (a_{n + 4}, a_{n + 5}, \frac{t}{(3 b)^{3}}) * \dots * M_{b} (a_{n + 2 m}, a_{n + 2 m + 1}, \frac{t}{(3 b)^{m}}) \end{matrix}$

Using (4) on each factor on the right hand side of the above inequality we get $\begin{matrix} M_{b} & (a_{n}, a_{n + 2 m + 1}, t) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{t}{3 {bk}^{n}}) * M_{b} (a_{0}, a_{1}, \frac{t}{3 {bk}^{n + 1}}) * \\ M_{b} (a_{0}, a_{1}, \frac{t}{(3 b)^{2} k^{n + 2}}) * M_{b} (a_{0}, a_{1}, \frac{t}{(3 b)^{2} k^{n + 3}}) \\ * \dots * M_{b} (a_{0}, a_{1}, \frac{t}{(3 b)^{m} k^{n + 2 m}}) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{t}{(3 b) k^{n}}) * M_{b} (a_{0}, a_{1}, \frac{t}{(3 bk) k^{n}}) * \\ M_{b} (a_{0}, a_{1}, \frac{t}{(3 bk)^{2} k^{n}}) * M_{b} (a_{0}, a_{1}, \frac{t}{(3 bk)^{2} k^{n + 1}}) \\ * \dots * M_{b} (a_{0}, a_{1}, \frac{t}{(3 bk)^{m} k^{n + m}}) \end{matrix}$

Case-2: If p is even say $p = 2 m; m \in ℕ$ , then we have $\begin{matrix} M_{b} (a_{n}, a_{n + 2 m}, t) \\ \geq & M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) * M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) * \\ M_{b} (a_{n + 2}, a_{n + 2 m}, \frac{t}{3 b}) \\ \geq & M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) * M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) \\ * & M_{b} (a_{n + 2}, a_{n + 3}, \frac{t}{(3 b)^{2}}) * M_{b} (a_{n + 3}, a_{n + 4}, \frac{t}{(3 b)^{2}}) \\ * & M_{b} (a_{n + 4}, a_{n + 2 m}, \frac{t}{(3 b)^{2}}) \\ \geq & M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) * M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) \\ * & M_{b} (a_{n + 2}, a_{n + 3}, \frac{t}{(3 b)^{2}}) * M_{b} (a_{n + 3}, a_{n + 4}, \frac{t}{(3 b)^{2}}) \\ * & M_{b} (a_{n + 4}, a_{n + 5}, \frac{t}{(3 b)^{3}}) * \dots * \\ M_{b} (a_{n + 2 m - 4}, a_{n + 2 m - 3}, \frac{t}{(3 b)^{m - 1}}) * \\ M_{b} (a_{n + 2 m - 3}, a_{n + 2 m - 2}, \frac{t}{(3 b)^{m - 1}}) * \\ M_{b} (a_{n + 2 m - 2}, a_{n + 2 m}, \frac{t}{(3 b)^{m - 1}}) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{t}{3 {bk}^{n}}) * M_{b} (a_{0}, a_{1}, \frac{t}{3 {bk}^{n + 1}}) * \\ M_{b} (a_{0}, a_{1}, \frac{t}{(3 b)^{2} k^{n + 2}}) * M_{b} (a_{0}, a_{1}, \frac{t}{(3 b)^{2} k^{n + 3}}) \\ * \dots * M_{b} (a_{0}, a_{2}, \frac{t}{(3 b)^{m - 1} k^{n + 2 m - 2}}) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{t}{(3 b) k^{n}}) * M_{b} (a_{0}, a_{1}, \frac{t}{(3 bk) k^{n}}) * \\ M_{b} (a_{0}, a_{1}, \frac{t}{(3 bk)^{2} k^{n}}) * M_{b} (a_{0}, a_{1}, \frac{t}{(3 bk)^{2} k^{n + 1}}) \\ * \dots * M_{b} (a_{0}, a_{2}, \frac{t}{(3 bk)^{m - 1} k^{n + m - 1}}) \end{matrix}$

Therefore, from Case 1 and Case 2, together with (2) it follows that for all $p \in ℕ$ we have $lim_{n \to \infty} M_{b} (a_{n}, a_{n + p}, t) \geq 1 * 1 * \dots * 1 = 1$

Hence {a_n} is G-Cauchy sequence. Since (X, M_b, *) is a G-complete fuzzy rectangular b- metric space so there exists u ∈ X such that $lim_{n \to \infty} a_{n} = u .$

We now show that u is fixed point of T. $\begin{matrix} M_{b} (u, Tu, t) & \geq & M_{b} (u, a_{n}, \frac{t}{3 b}) * M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) \\ * & M_{b} (a_{n + 1}, Tu, \frac{t}{3 b}) \\ \geq & M_{b} (u, a_{n}, \frac{t}{3 b}) * M_{b} ({Ta}_{n - 1}, {Ta}_{n}, \frac{t}{3 b}) \\ * & M_{b} ({Ta}_{n}, Tu, \frac{t}{3 b}) \\ \geq & M_{b} (u, a_{n}, \frac{t}{3 b}) * M_{b} (a_{n - 1}, a_{n}, \frac{t}{3 bk}) \\ * & M_{b} (a_{n}, u, \frac{t}{3 bk}) \\ ⟶ 1 * 1 * 1 = 1 as n \to \infty . \end{matrix}$

Which shows that Tu = u is a fixed point.

Uniqueness:

Assume Tv = v for some v ∈ X, then $\begin{matrix} M_{b} & (v, u, t) = M_{b} (Tv, Tu, t) \\ \geq & M_{b} (v, u, \frac{t}{k}) = M_{b} (Tv, Tu, \frac{t}{k}) \\ \geq & M_{b} (v, u, \frac{t}{k^{2}}) \geq \dots \geq M_{b} (v, u, \frac{t}{k^{n}}) \\ ⟶ 1 as n ⟶ \infty \end{matrix}$

Thus u = v . Hence the fixed point is unique.

Corollary 2.1. (Banach Contraction Theorem in fuzzy rectangular metric spaces)

Let (X, M, *) be a G-complete fuzzy rectangular metric space such that $lim_{t \to \infty} M (x, y, t) = 1 \forall x, y \in X$

Let T:X → X be a mapping satisfying $M (Tx, Ty, kt) \geq M (x, y, t)$ for all x, y ∈ X, 0 < k < 1 Then T has a unique fixed point.

Proof. Taking b = 1, the proof follows immediately from Theorem 2.1.

Now we introduce the notion of T-orbitally upper semi continuous function in fuzzy metric spaces and then prove the fixed point theorem of Hicks and Rhoades [11] in the setting of fuzzy rectangular b-metric space.

Definition 2.3. Let T:X → X be a self mapping and for $a_{0} \in X, O (a_{0}) = {a_{0}, {Ta}_{0}, T^{2} a_{0}, \dots}$ be the orbit of a₀ . A function F:X → [0, 1] is said to be T-orbitally upper semi continuous at u ∈ X if ${a_{n}} \subset O (a_{0})$ and $a_{n} \to u, \Rightarrow F (u) \geq \underset{n \to \infty}{lim sup} F (a_{n})$

The following example illustrates the above definition.

Example 2.5. Consider the set X = [0, 2] and let the self map T defined on X be $Tx = \frac{1}{2} x^{2} .$ Choose an element $a_{0} = \frac{1}{2}$ in X, then we have $O (a_{0}) = O (\frac{1}{2}) = {\frac{1}{2}, \frac{1}{2^{3}}, \frac{1}{2^{7}}, \dots} .$

Clearly, for any sequence ${a_{n}} \subset O (\frac{1}{2})$ we have a_n → 0. Consider a function F : X → [0, 1] given by $F (x) = {\begin{matrix} 1 & if x = 0 \\ \sqrt{2 x - x^{2}} & if 0 < x \leq 2 \end{matrix} .$ Now F (0) =1 and a_n → u = 0 implies that $F (0) = 1 > 0 = \underset{n \to \infty}{lim sup} F (a_{n})$ $= \underset{n \to \infty}{lim sup} \sqrt{2 a_{n} - a_{n}^{2}} .$

It follows that F is T-orbitally upper semi-continuous at u = 0.

Theorem 2.2. Let (X, M_b, *) be a G-complete fuzzy rectangular b-metric space. Let T:X → X be a mapping and there exists a₀ ∈ X such that $M_{b} (Tx, T^{2} x, kt) \geq M_{b} (x, Tx, t) for each x \in O (a_{0})$ (5)

where $k \in (0, \frac{1}{3 b})$ . Then Tⁿa₀ → a ∈ X.

Furthermore a is fixed point of T if and only if F (x) = M_b (x, Tx, t) is T-orbitally upper semi continuous at a₀.

Proof. For a₀ ∈ X, we define an iterative scheme {a_n} by a_n = Tⁿa₀. With x = Ta₀ and successive application of (5) we get $\begin{matrix} M_{b} (T^{n} a_{0}, T^{n + 1} a_{0}, kt) & = M_{b} (a_{n}, a_{n + 1}, kt) \\ \geq M_{b} (a_{0}, a_{1}, \frac{t}{k^{n}}) \end{matrix}$ (6)

For any $p \in ℕ$ , we have $\begin{matrix} M_{b} (T^{n} a_{0}, T^{n + p} a_{0}, t) & = M_{b} (a_{n}, a_{n + p}, t) . \end{matrix}$

As in the proof of Theorem 2.1, starting with M_b (a_n, a_n+p, t) together with (6) we get for all $p \in ℕ$ $lim_{n \to \infty} M_{b} (T^{n} a_{0}, T^{n + p} a_{0}, t) \geq 1 * 1 * \dots * 1 = 1$

Hence {Tⁿa₀} is G-Cauchy sequence. Since X is G-complete, so there is a point a ∈ X such that a_n = Tⁿa₀ → a ∈ X.

Suppose that that F is upper semi continuous at a ∈ X then $\begin{matrix} M_{b} (a, Ta, t) & \geq lim_{n \to \infty} sup M_{b} (T^{n} a_{0}, T^{n + 1} a_{0}, t) \\ \geq lim_{n \to \infty} sup M_{b} (a_{0}, a_{1}, \frac{t}{k^{n}}) \to 1 . \end{matrix}$

So, we have a = Ta.

Conversely, suppose a = Ta and $x \in O (x)$ with a_n → a, then $\begin{matrix} F (a) = M_{b} (a, Ta, t) & = 1 \geq lim_{n \to \infty} sup F (a_{n}) \\ = M_{b} (T^{n} a_{0}, T^{n + 1} a_{0}, t) . \end{matrix}$

The the following corollary becomes an immediate consequence of Theorem 2.2 by setting b = 1.

Corollary 2.2. Let (X, M, *) be a G-complete fuzzy rectangular metric space. Let T:X → X be a mapping satisfying $M (Tx, T^{2} x, kt) \geq M (x, Tx, t)$ (7)for each $x \in O (a_{0}) for a_{0} \in X and 0 < k < 1$ .

Here $a_{n} = T^{n} a_{0} (n \in ℕ)$ , then Tⁿa₀ → a ∈ X. Furthermore a is fixed point of T if and only if F (x) = M (x, Tx, t) is T-orbitally upper semi continuous at a.

We now furnish an example to illustrate Theorem 2.1.

Example 2.6. Let X = [0, 1] and define M_b:X × X × [0, ∞) → [0, 1] by $M_{b} (x, y, t) = {\begin{matrix} e^{- \frac{(x - y)^{2}}{t}} & if t > 0 \\ 0 & if t = 0 \end{matrix} .$

It is easy to verify that (X, M_b, *) is a G-complete fuzzy rectangular b- metric space with the coefficient b = 3 . Let k ∈ (0, 1) and define T : X → X by $T (x) = \sqrt{k} x$ . Now for all t > 0 we have $\begin{matrix} M_{b} (Tx, Ty, kt) & = M_{b} (\sqrt{k} x, \sqrt{k} y, kt) \\ = e^{- \frac{(\sqrt{k} x - \sqrt{k} y)^{2}}{kt}} \\ = e^{- \frac{(x - y)^{2}}{t}} \\ = M_{b} (x, y, t) \end{matrix}$

Hence all the conditions of Theorem 2.1 are satisfied and x = 0 ∈ [0, 1] is a unique fixed point of T.

Following [22], for a real number b > 1, let F_b denotes the class of all functions $β [0, \infty) \to [0, \frac{1}{b})$ satisfying the following condition: $\underset{n \to \infty}{lim sup} β (t_{n}) = \frac{1}{b} implies lim_{n \to \infty} t_{n} = 0$

We now establish a fixed point result, analogue to [22, Theorem 1], in the setting of G-complete fuzzy rectangular b-metric spaces, as follows:

Theorem 2.3. Let (X, M_b, *) be a G-complete fuzzy rectangular b-metric space with b ≥ 1 such that $lim_{t \to \infty} M_{b} (x, y, t) = 1 \forall x, y \in X$ (8)

Let T:X → X be a mapping satisfying $M_{b} (Tx, Ty, β (M_{b} (x, y, t)) t) \geq δ (x, y, t)$ (9)

∀ x, y ∈ X and for some β ∈ F_b . Where $\begin{matrix} δ (x, y, t) & = & min {\frac{M_{b} (x, Tx, t) [1 + M_{b} (y, Ty, t)]}{1 + M_{b} (Tx, Ty, t)}, \\ \frac{M_{b} (y, Ty, t) [1 + M_{b} (x, Tx, t)]}{1 + M_{b} (x, y, t)}, \\ \frac{M_{b} (x, Tx, t) [2 + M_{b} (x, Ty, t)]}{1 + M_{b} (x, Ty, t) + M_{b} (y, Tx, t)}, M_{b} (x, y, t)} . \end{matrix}$

Then T has a unique fixed point.

Proof. For any arbitrary point a₀ ∈ X, we choose a sequence {a_n} in X and start with iterative process a_n+1 = Ta_n . For all n, t > 0, we have $\begin{matrix} M_{b} & (a_{n}, a_{n + 1}, t) = M_{b} ({Ta}_{n - 1}, {Ta}_{n}, t) \\ \geq δ (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) \end{matrix}$ (10)□

Now, $\begin{matrix} δ (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) \\ = & min {\frac{M_{b} (a_{n - 1}, {Ta}_{n - 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [1 + M_{b} (a_{n}, {Ta}_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} ({Ta}_{n - 1}, {Ta}_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, \\ \frac{M_{b} (a_{n}, {Ta}_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [1 + M_{b} (a_{n - 1}, {Ta}_{n - 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, \\ \frac{M_{b} (a_{n - 1}, {Ta}_{n - 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [2 + M_{b} (a_{n - 1}, {Ta}_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n - 1}, {Ta}_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) + M_{b} (a_{n}, {Ta}_{n - 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})} \\ = & min {\frac{M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [1 + M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, \\ \frac{M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [1 + M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, \\ \frac{M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [2 + M_{b} (a_{n - 1}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n - 1}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) + M_{b} (a_{n}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})} \\ = & min {\frac{M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [1 + M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, \\ \frac{M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [1 + M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})}, \\ \frac{M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) [2 + M_{b} (a_{n - 1}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})]}{1 + M_{b} (a_{n - 1}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}) + 1}, M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})} \\ = & min {M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}), M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})} \end{matrix}$ If $\begin{matrix} min {M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}), \\ M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n}, a_{n + 1}, t))}} \\ = M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}), \end{matrix}$ then from (10), $M_{b} (a_{n}, a_{n + 1}, t) \geq M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})$

Since $β \in [0, \frac{1}{b})$ , so by Lemma 1.2, there is nothing to prove.

If $\begin{matrix} min {M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}), \\ M_{b} (a_{n}, a_{n + 1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})} \\ = M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))}), \end{matrix}$ then from (10), $M_{b} (a_{n}, a_{n + 1}, t) \geq M_{b} (a_{n - 1}, a_{n}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t))})$

Continuing in this way, we have $\begin{matrix} M_{b} (a_{n}, a_{n + 1}, t) \geq \\ M_{b} {a_{0}, a_{1}, \frac{t}{β (M_{b} (a_{n - 1}, a_{n}, t) . \dots β (M_{b} (a_{0}, a_{1}, t)}} \end{matrix}$ (11)

Case-1: If p is odd say p = 2m + 1 where $m \in ℕ$ , we have $\begin{matrix} M_{b} & (a_{n}, a_{n + 2 m + 1}, t) \geq M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) \\ M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) * M_{b} (a_{n + 2}, a_{n + 3}, \frac{t}{(3 b)^{2}}) \\ * & M_{b} (a_{n + 3}, a_{n + 4}, \frac{t}{(3 b)^{2}}) * M_{b} (a_{n + 4}, a_{n + 5}, \frac{t}{(3 b)^{3}}) \\ \dots * M_{b} (a_{n + 2 m}, a_{n + 2 m + 1}, \frac{t}{(3 b)^{m}}) \end{matrix}$

Using (11) on each factor on the right hand side of the above inequality we get $\begin{matrix} a_{n + 2 m + 1}, t) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{t}{3 b β (M_{b} (a_{n - 1}, a_{n}, t) \dots β (M_{b} (a_{0}, a_{1}, t)}) * M_{b} (a_{0}, a_{1}, \frac{t}{3 b β (M_{b} (a_{n}, a_{n + 1}, t) \dots β (M_{b} (a_{0}, a_{1}, t)}) * \\ \dots * M_{b} (a_{0}, a_{1}, \frac{t}{(3 b)^{m} β (M_{b} (a_{n + 2 m - 1}, a_{n + 2 m}, t) \dots β (M_{b} (a_{0}, a_{1}, t)}) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{b^{n} t}{(3 b)}) * M_{b} (a_{0}, a_{1}, \frac{b^{n + 1} t}{(3 b)}) * M_{b} (a_{0}, a_{1}, \frac{b^{n + 2} t}{(3 b)^{2}}) * M_{b} (a_{0}, a_{1}, \frac{b^{n + 3} t}{(3 b)^{2}}) * \dots * M_{b} (a_{0}, a_{1}, \frac{b^{n + 2 m - 1} t}{(3 b)^{m}}) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{b^{n - 1} t}{3}) * M_{b} (a_{0}, a_{1}, \frac{b^{n} t}{3}) * M_{b} (a_{0}, a_{1}, \frac{b^{n} t}{(3)^{2}}) * M_{b} (a_{0}, a_{1}, \frac{b^{n + 1} t}{(3)^{2}}) * \dots * M_{b} (a_{0}, a_{1}, \frac{b^{n + m - 1} t}{(3)^{m}}) \end{matrix}$

Case-2: If p is even say $p = 2 m; m \in ℕ$ , then we have $\begin{matrix} M_{b} (a_{n}, a_{n + 2 m}, t) \geq M_{b} (a_{n}, a_{n + 1}, \frac{t}{3 b}) * M_{b} (a_{n + 1}, a_{n + 2}, \frac{t}{3 b}) * M_{b} (a_{n + 2}, a_{n + 3}, \frac{t}{(3 b)^{2}}) * M_{b} (a_{n + 3}, a_{n + 4}, \frac{t}{(3 b)^{2}}) * \\ M_{b} (a_{n + 4}, a_{n + 5}, \frac{t}{(3 b)^{3}}) * \dots * M_{b} (a_{n + 2 m - 4}, a_{n + 2 m - 3}, \frac{t}{(3 b)^{m - 1}}) * \\ M_{b} (a_{n + 2 m - 3}, a_{n + 2 m - 1}, \frac{t}{(3 b)^{m - 1}}) * M_{b} (a_{n + 2 m - 2}, a_{n + 2 m}, \frac{t}{(3 b)^{m - 1}}) \end{matrix}$ $\begin{matrix} \geq & M_{b} (a_{0}, a_{1}, \frac{t}{3 b β (M_{b} (a_{n - 1}, a_{n}, t) \dots β (M_{b} (a_{0}, a_{1}, t)}) * M_{b} (a_{0}, a_{1}, \frac{t}{3 b β (M_{b} (a_{n}, a_{n + 1}, t) \dots β (M_{b} (a_{0}, a_{1}, t)}) * \dots * \\ M_{b} & (a_{0}, a_{2}, \frac{t}{(3 b)^{m - 1} β (M_{b} (a_{n + 2 m - 1}, a_{n + 2 m - 3}, t) \dots β (M_{b} (a_{0}, a_{1}, t)}) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{b^{n - 1} t}{(3 b)}) * M_{b} (a_{0}, a_{1}, \frac{b^{n} t}{(3 b)}) * M_{b} (a_{0}, a_{1}, \frac{b^{n + 1} t}{(3 b)^{2}}) * \dots * M_{b} (a_{0}, a_{2}, \frac{b^{n + 2 m - 2} t}{(3 b)^{m - 1}}) \\ \geq & M_{b} (a_{0}, a_{1}, \frac{b^{n - 2} t}{3}) * M_{b} (a_{0}, a_{1}, \frac{b^{n - 1} t}{3}) * M_{b} (a_{0}, a_{1}, \frac{b^{n - 1} t}{(3)^{2}}) * \dots * M_{b} (a_{0}, a_{2}, \frac{b^{n + m - 3} t}{(3)^{m - 1}}) \end{matrix}$

Therefore, from Case-1 and Case-2 together with (8), it follows that for all $p \in ℕ$ , we have $lim_{n \to \infty} M_{b} (a_{n}, a_{n + p}, t) \geq 1 * 1 * \dots * 1 = 1$

Hence {a_n} is G-Cauchy sequence. Since (X, M_b, *) is a G-complete fuzzy rectangular b- metric space so there exists u ∈ X such that $lim_{n \to \infty} a_{n} = u .$

Using a_n = Ta_n-1, we get $\begin{matrix} M_{b} & (u, Tu, t) \geq M_{b} (u, a_{n}, \frac{t}{3 b}) * M_{b} ({Ta}_{n - 1}, {Ta}_{n}, \frac{t}{3 b}) * \\ M_{b} & ({Ta}_{n}, Tu, \frac{t}{3 b}) \geq M_{b} (u, a_{n}, \frac{t}{3 b}) * \\ M_{b} & (a_{n - 1}, a_{n}, \frac{t}{3 b β (M_{b} (a_{n - 1}, a_{n}, t)}) \\ * & M_{b} (a_{n}, u, \frac{t}{3 b β (M_{b} (a_{n}, u, t)}) \\ ⟶ & 1 * 1 * 1 = 1 as n \to \infty . \end{matrix}$

Which shows that Tu = u is a fixed point.

Uniqueness:

Assume Tv = v for some v ∈ X, then $\begin{matrix} M_{b} & (v, u, t) = M_{b} (Tv, Tu, t) \end{matrix}$ $\begin{matrix} \geq M_{b} (v, u, \frac{t}{β (M_{b} (v, u, t)}) = M_{b} (Tv, Tu, \frac{t}{k}) \\ \geq M_{b} (v, u, \frac{t}{(β (M_{b} (v, u, t)))^{2}}) \geq \dots \\ \geq M_{b} (v, u, \frac{t}{(β (M_{b} (v, u, t)))^{n}}) \\ = M_{b} (v, u, b^{n} t) ⟶ 1 as n ⟶ \infty \end{matrix}$

Thus u = v

Hence the fixed point is unique.

Corollary 2.3. Let (X, M, *) be a G-complete fuzzy rectangular metric space such that $lim_{t \to \infty} M (x, y, t) = 1 \forall x, y \in X$ (12)

Let T: X → X be a mapping satisfying $M (Tx, Ty, β (M (x, y, t)) t) \geq δ (x, y, t)$ (13) ∀ x, y ∈ X and for some β ∈ F_b . Where $\begin{matrix} δ (x, y, t) & = min {\frac{M (x, Tx, t) [1 + M (y, Ty, t)]}{1 + M (Tx, Ty, t)}, \\ \frac{M (y, Ty, t) [1 + M (x, Tx, t)]}{1 + M (x, y, t)}, \\ \frac{M (x, Tx, t) [2 + M (x, Ty, t)]}{1 + M (x, Ty, t) + M (y, Tx, t)}, M (x, y, t)} . \end{matrix}$

Then T has a unique fixed point.

Proof. Taking b = 1, the proof follows immediately from Theorem 2.3 □

3. Application

Fixed point theorems for operators in (ordered) metric spaces are widely investigated and have found various applications in differential and integral equations (see [1, 21] and references therein). Inspired by Mishra et al. [19], we now present an application of our main fixed point result stated in Theorem 2.1. In particular, we show the existence of the solution of an integral equation of the form $x (s) = g (s) + \int_{0}^{s} F (s, r, x (r)) dr,$ (14) for all s ∈ [0, a] where a > 0. Let $C ([0, a], ℝ)$ be the space of all continuous functions defined on [0, a] equipped with the product t-norm, i.e, a * b = ab for all a, b ∈ [0, 1] and define the G-complete fuzzy rectangular b-metric on $C ([0, a], ℝ)$ by $M_{b} (x, y, t) = e^{- \frac{{sup}_{s \in [0, a]} | x (s) - y (s) |^{2}}{t}}$ for all t > 0 and $x, y \in C ([0, a], ℝ) .$

The following theorem proves the existence of a solution of the integral equation (14).

Theorem 3.1. Let $T : C ([0, a], ℝ) \to C ([0, a], ℝ)$ be the integral operator given by $T (x (s)) = g (s) + \int_{0}^{s} F (s, r, x (r)) dr, g \in C ([0, a], ℝ)$ where $F \in C ([0, a] \times [0, a] \times ℝ, ℝ)$ satisfies the following condition:

there exists f : [0, a] × [0, a] → [0, + ∞] such that for all $r, s \in [0, a], f (s, r) \in L^{1} ([0, a], ℝ)$ and $\forall x, y \in C ([0, a], ℝ)$ , we have $| F (s, r, x (r)) - F (s, r, y (r)) |^{2} \leq f^{2} (s, r) | x (r) - y (r) |^{2}$ where $sup_{s \in [0, a]} \int_{0}^{s} f^{2} (s, r) dr \leq k < 1 .$ Then the integral equation has the solution $x^{*} \in C ([0, a], ℝ) .$

Proof. For all $x, y \in C ([0, a], ℝ)$ , we have $\begin{matrix} M_{b} (T (x (s)), T (y (s)), kt) \\ = e^{- \frac{{sup}_{s \in [0, a]} | T (x (s)) - T (y (s)) |^{2}}{kt}} \\ \geq e^{- \frac{{sup}_{s \in [0, a]} \int_{0}^{s} | F (s, r, x (r)) - F (s, r, y (r)) |^{2} dr}{kt}} \\ \geq e^{- \frac{{sup}_{s \in [0, a]} \int_{0}^{s} f^{2} (s, r) | x (r) - y (r) |^{2} dr}{kt}} \\ \geq e^{- \frac{| x (r) - y (r) |^{2} {sup}_{s \in [0, a]} \int_{0}^{s} f^{2} (s, r) dr}{kt}} \\ \geq e^{- \frac{k | x (r) - y (r) |^{2}}{kt}} = e^{- \frac{| x (r) - y (r) |^{2}}{t}} \\ \geq e^{- \frac{{sup}_{r \in [0, a]} | x (r) - y (r) |^{2}}{t}} = M_{b} (x, y, t) \end{matrix}$ Hence $x^{*} \in C ([0, a], ℝ)$ is a fixed point of T, which is the solution of integral equation (14). □

4. Conclusion

In the present study, we have proved the famous Banach fixed point theorem for fuzzy rectangular b-metric spaces and furnished an example to illustrate our theorem. In this way, we have generalized the main result of Grabeic [9]. Moreover, by restricting the contraction mapping to the elements in the orbit of a point in fuzzy rectangular b-metric space, we have also proved an analogue of the fixed point theorem of Hicks and Rhoads [11] in the setting of fuzzy rectangular b-metric spaces. Thus our results are more general than the existing results in the fuzzy set theory.

Fuzzy set theory has been found to be useful not only in decision making problems arising in physical and social sciences but also has application in multi-attribute decision making. For a combined study of fuzzy set theory and rough set theory and its application in making an optimal decision see the recent work presented in [15 , 28–30] and the references therein. The structure of a fuzzy metric space might also be useful to solve fixed point problems related to some sort of distance between the programs to measure, for instance, the complexity of an algorithm.

Competing interests

The authors declare that they have no competing interests.

Author’s contributions

The authors contributed equally to this work.

Footnotes

Acknowledgments

The authors are thankful to the editor and referees for their comments and suggestions for the improvement of the overall presentation of this work.

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