Fuzzy double controlled metric spaces and related results

Abstract

In this paper, we introduce the concept of fuzzy double controlled metric space that can be regarded as the generalization of fuzzy b-metric space, extended fuzzy b-metric space and controlled fuzzy metric space. We use two non-comparable functions α and β in the triangular inequality as: $M_{q} (x, z, t α (x, y) + s β (y, z)) \geq M_{q} (x, y, t) * M_{q} (y, z, s) .$ We prove Banach contraction principle in fuzzy double controlled metric space and generalize the Banach contraction principle in aforementioned spaces. We give some examples to support our main results. An application to existence and uniqueness of solution for an integral equation is also presented in this work.

Keywords

controlled fuzzy metric space fixed point

1 Introduction and preliminaries

The concept of metric space was given by Fréchet [8] in his dissertation in 1906. Later on, in 1922, Banach [3] proved the Banach contraction principle in his doctorate dissertation. Since then, many researchers have used this principle in various directions ([4 , 18]). Zadeh [24] introduced the concept of fuzzy set, back in 1965, as an extension of an ordinary set where each element of a set has some membership values between [0, 1] . Many mathematical structures have been shifted on fuzzy sets, see ([6 , 19]). The concept of fuzzy metric space was introduced by Kramosil and Michálek [14] which is sort of a reformulation of the statistical metric space defined by Menger [16]. Banach contraction principle in the framework of fuzzy metric space was proved by Grabiec [10]. He also gave definitions of a convergent sequence and a Cauchy sequence in the context of fuzzy metric space. In 1994, George and Veeramani [9] investigated the Hausdorff topology in fuzzy metric spaces and generalized the definition of a fuzzy metric space given by Kramosil and Michálek [14].

Bakhtin [2] gave the concept of b-metric space which generalizes the metric space. In 1993, Czerwik [7] investigated some contractive mappings in b-metric space. In 2000, Branciari [5] established the definition of a rectangular metric space and proved Banach-Caccippoli type fixed point result in this new class of metric space. In 2016, Roshan et al. [21] generalized the notion of rectangular metric space by introducing b-Branciari metric space. In 2016, Nǎdǎban [17] introduced the idea of fuzzy b-metric space which generalizes fuzzy metric space. Kamran et al. [13] introduced the generalization of a b-metric space by replacing the constant b with a function θ and proved Banach fixed point theorem on this space. In 2017, Mehmood et al. [15] introduced the concept of extended fuzzy b-metric space and proved the contraction principle. Abdeljawad et al. [1] introduced the concept of double controlled metric type space by using two non-comparable functions while Sezen [23] gave the concept of controlled fuzzy metric space.

In this paper, we introduce the notion of fuzzy double controlled metric space which generalizes the spaces of fuzzy b-metric, extended fuzzy b-metric and controlled fuzzy metric. We examine some topological properties of fuzzy double controlled metric space and extend the Banach contraction principle to this space. We follow the George and Veeramani’s sense of fuzzy metric space. Also, some examples and an application to existence and uniqueness of solution for an integral equation are ensured to support our main results.

Definition 1.1. ([22]) 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 1.2. ([17]) Let X be a nonempty set, b ≥ 1 a given real number and ∗ a continuous t-norm. A fuzzy set M on X × X × (0, ∞) is called fuzzy b-metric on X, if for all x, y, z ∈ X, the following conditions hold:

M (x, y, 0) =0 ;

M (x, y, t) =1 for all t > 0, iff x = y;

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

M (x, z, b (t + s)) ≥ M (x, y, t) ∗ M (y, z, s) for all s, t > 0;

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

The quadruple (X, M, ∗ , b) is called a fuzzy b-metric space.

Definition 1.3. ([15]) Let X be a nonempty set, α : X × X → [1, ∞) a given function and ∗ a continuous t-norm. A fuzzy set M_α on X × X × (0, ∞) is called extended fuzzy b-metric on X, if for all x, y, z ∈ X, the following conditions hold:

M_α (x, y, 0) =0 ;

M_α (x, y, t) =1 for all t > 0, iff x = y;

M_α (x, y, t) = M_α (y, x, t);

M_α (x, z, α (x, z) (t + s)) ≥ M_α (x, y, t) ∗ M_α (y, z, s) for all s, t > 0;

M_α (x, y, ·) : (0, ∞) → [0, 1] is left continuous.

Then (X, M_α, ∗ , α) is called an extended fuzzy b-metric space.

Definition 1.4. ([23]) Let X be a nonempty set, α : X × X → [1, ∞) a given function and ∗ a continuous t-norm. A fuzzy set M_α on X × X × (0, ∞) is called controlled fuzzy metric on X, if for all x, y, z ∈ X, the following conditions hold:

M_α (x, y, 0) =0 ;

M_α (x, y, t) =1 for all t > 0, iff x = y;

M_α (x, y, t) = M_α (y, x, t);

$M_{α} (x, z, t + s) \geq M_{α} (x, y, \frac{t}{α (x, y)}) * M_{α} (y, z, \frac{s}{α (y, z)})$ for all s, t > 0;

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

Then (X, M_α, ∗ , α) is called a controlled fuzzy metric space.

Definition 1.5. ([1]) Given non-comparable functions α, β : X × X → [1, ∞) . If q : X × X → [0, ∞) satisfies

q (x, y) =0 iff x = y,

q (x, y) = q (y, x),

q (x, y) ≤ α (x, z) q (x, z) + β (z, y) q (z, y),

for all x, y, z ∈ X . Then q is called a double controlled metric and (X, q) is called a double controlled metric space.

2 Main results

Firstly, we introduce the notion of fuzzy double controlled metric space.

Definition 2.1. Let X be a nonempty set and α, β : X × X → [1, ∞) given non-comparable functions, and ∗ a continuous t-norm. A fuzzy set M_q on X × X × (0, ∞) is called fuzzy double controlled metric on X, if for all x, y, z ∈ X, the following conditions hold:

M_q (x, y, t) >0 ;

M_q (x, y, t) =1 for all t > 0, iff x = y;

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

$M_{q} (x, z, t + s) \geq M_{q} (x, y, \frac{t}{α (x, y)}) * M_{q} (y, z, \frac{s}{β (y, z)})$ for all s, t > 0;

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

Then (X, M_q, ∗) is called a fuzzy double controlled metric space.

Remark 2.1. (i) If we take α (x, y) = β (x, y), then a fuzzy double controlled metric space becomes a controlled fuzzy metric space.

(ii) If we take α (x, y) = β (y, z) = α (x, z), then a fuzzy double controlled metric space becomes an extended fuzzy b-metric space.

(iii) If we take α (x, y) = β (x, y) = b ≥ 1, then a fuzzy double controlled metric space becomes a fuzzy b-metric space.

Example 2.1. Let X = {1, 2, 3} and α, β : X × X → [1, ∞) be two non-comparable functions given by α (x, y) = x + y + 1 and β (x, y) = x² + y² - 1. Define M_q : X × X × (0, ∞) → [0, 1] as $M_{q} (x, y, t) = \frac{min {x, y} + t}{\max {x, y} + t} .$ Then (X, M_q, ∗) is fuzzy double controlled metric space with product t-norm.

Now, α (1, 2) = α (2, 1) =4, α (1, 3) = α (3, 1) =5, α (2, 3) = α (3, 2) =6, α (1, 1) =3, α (2, 2) =5, α (3, 3) =7, and β (1, 2) = β (2, 1) =4, β (1, 3) = β (3, 1) =9, β (2, 3) = β (3, 2) =12, β (1, 1) =1, β (2, 2) =7, β (3, 3) =17 .

Axioms (M_q1) to (M_q3) and (M_q5) are easy to verify, we only prove (M_q4).

Let x = 1, y = 2, and z = 3. Then $\begin{matrix} M_{q} (1, 3, t + s) & = \frac{min {1, 3} + t + s}{\max {1, 3} + t + s} \\ = \frac{1 + t + s}{3 + t + s} . \end{matrix}$ Now, $\begin{matrix} M_{q} (1, 2, \frac{t}{α (1, 2)}) & = \frac{min {1, 2} + \frac{t}{α (1, 2)}}{\max {1, 2} + \frac{t}{α (1, 2)}} \\ = \frac{1 + \frac{t}{4}}{2 + \frac{t}{4}} \\ = \frac{4 + t}{8 + t}, \end{matrix}$ and $\begin{matrix} M_{q} (2, 3, \frac{s}{β (2, 3)}) & = \frac{min {2, 3} + \frac{s}{β (2, 3)}}{\max {2, 3} + \frac{s}{β (2, 3)}} \\ = \frac{2 + \frac{s}{12}}{3 + \frac{s}{12}} \\ = \frac{24 + s}{36 + s} . \end{matrix}$ Clearly, $\begin{matrix} \frac{1 + t + s}{3 + t + s} \geq \frac{4 + t}{8 + t} \cdot \frac{24 + s}{36 + s}, for all t, s > 0 . \end{matrix}$ So, $\begin{matrix} M_{q} (1, 3, t + s) & \geq M_{q} (1, 2, \frac{t}{α (1, 2)}) \\ * M_{q} (2, 3, \frac{s}{β (2, 3)}) . \end{matrix}$ On the same steps, one can prove the remaining cases. Hence (X, M_q, ∗) is a fuzzy double controlled metric space.

Remark 2.2. If we take minimum t-norm instead of product t-norm in Example 2.1, then X, M_q, ∗) is not a fuzzy double controlled metric space. For instance, let x = 1, y = 2, z = 3, and t = 0.04, s = 0.05 with α (x, y) = x + y + 1, β (x, y) = x² + y² - 1, then $M_{q} (1, 3, 0.04 + 0.05) = \frac{1 + 0.09}{3 + 0.09} = 0.3528,$ and $\begin{matrix} M_{q} (1, 2, \frac{0.04}{α (1, 2)}) & = \frac{1 + \frac{0.04}{4}}{2 + \frac{0.04}{4}} = \frac{1 + 0.01}{2 + 0.01} \\ = 0.5025, \end{matrix}$ and $\begin{matrix} M_{q} (2, 3, \frac{0.05}{β (2, 3)}) & = \frac{2 + \frac{0.05}{12}}{3 + \frac{0.05}{12}} = \frac{2 + 0.0042}{3 + 0.0042} \\ = 0.6671, \end{matrix}$ Clearly, $\begin{matrix} M_{q} (1, 3, 0.04 + 0.05) & ≱ M_{q} (1, 2, \frac{0.04}{α (1, 2)}) \\ * M_{q} (2, 3, \frac{0.05}{β (2, 3)}) . \end{matrix}$ Hence (X, M_q, ∗) is not a fuzzy double controlled metric space with minimum t-norm.

Definition 2.2. Let (X, M_q, ∗) be a fuzzy double controlled metric space, then we define an open ball B (x, r, t) with centre x, radius r, r ∈]0, 1 [, and t > 0 as follows: $\begin{matrix} B (x, r, t) = {y \in X : M_{q} (x, y, t) > 1 - r} . \end{matrix}$

We show in the next example that (X, M_q, ∗) is not Hausdorff.

Example 2.2. Let X = {1, 2, 3} and α, β : X × X → [1, ∞) be two non-comparable functions given by α (x, y) = x + y + 1 and β (x, y) = x² + y² - 1. Define M_q : X × X × (0, ∞) → [0, 1] as $M_{q} (x, y, t) = \frac{min {x, y} + t}{\max {x, y} + t} .$ Then (X, M_q, ∗) is fuzzy double controlled metric space with product t-norm. Consider the open ball B (1, 0.4, 5) with centered at 1, radius r = 0.4 and t = 5. Then $\begin{matrix} B (1, 0.4, 5) = {y \in X : M_{q} (1, y, 5) > 0.6} . \end{matrix}$ Now, $\begin{matrix} M_{q} (1, 2, 5) = \frac{1 + 5}{2 + 5} = \frac{6}{7} = 0.8571, \\ M_{q} (1, 3, 5) = \frac{1 + 5}{3 + 5} = \frac{6}{8} = 0.6666 . \end{matrix}$ Thus B (1, 0.4, 5) = {2, 3}. Now consider the open ball B (2, 0.6, 10) with centered at 2, radius r = 0.6 and t = 10. Then, $\begin{matrix} B (2, 0.6, 10) = {y \in X : M_{q} (2, y, 10) > 0.4} . \end{matrix}$ Now, $\begin{matrix} M_{q} (2, 1, 10) = \frac{1 + 10}{2 + 10} = \frac{11}{12} = 0.9166, \\ M_{q} (2, 3, 10) = \frac{2 + 10}{3 + 10} = \frac{12}{13} = 0.9230 . \end{matrix}$ So, B (2, 0.6, 10) = {1, 3}. Now, $\begin{matrix} B (1, 0.4, 5) \cap B (2, 0.6, 10) & = {2, 3} \cap {1, 3} \\ = {3} \neq \emptyset . \end{matrix}$ Hence, a fuzzy double controlled metric space is not Hausdorff.

Using the idea of George and Veeramani [9], we now give definitions of a convergent sequence and a Cauchy sequence in a fuzzy double controlled metric space.

Definition 2.3. Let (X, M_q, ∗) be a fuzzy double controlled metric space and {x_n} be a sequence in X. Then {x_n} is said to be:

a convergent sequence, if there exists x ∈ X such that $lim_{n \to \infty} M_{q} (x_{n}, x, t) = 1, for all t > 0,$

a Cauchy sequence if and only if for each ε > 0, t > 0, there exists $n_{0} \in ℕ$ such that $M_{q} (x_{n}, x_{m}, t) \geq 1 - ε, for all m, n \geq n_{0} .$

If every Cauchy sequence converges in X, then (X, M_q, ∗) is called complete fuzzy double controlled metric space.

Following lemmas are needed in the proof of our main results.

Lemma 2.1. Let {x_n} be a Cauchy sequence in fuzzy double controlled metric space (X, M_q, ∗) such that x_m ≠ x_n whenever $m, n \in ℕ$ with m ≠ n . Then the sequence {x_n} can converge to at most one limit point.

Proof. Suppose on the contrary that x_n → x and x_n → y, for x ≠ y. Then $lim_{n \to \infty} M_{q} (x_{n}, x, t) = 1$ and $lim_{n \to \infty} M_{q} (x_{n}, y, t) = 1$ for all t > 0. Consider $\begin{matrix} M_{q} (x, y, t) & \geq M_{q} (x, x_{n}, \frac{\frac{t}{2}}{α (x, x_{n})}) \\ * M_{q} (x_{n}, y, \frac{\frac{t}{2}}{β (x_{n}, y)}) \\ \to 1 * 1, \end{matrix}$ as n → ∞ . So M_q (x, y, t) ≥1 ∗ 1 =1. Hence x = y, that is the sequence {x_n} converges to at most one limit point.

Lemma 2.2. Let (X, M_q, ∗) be a fuzzy double controlled metric space. If for some k ∈ (0, 1) and for any x, y ∈ X, t > 0, $M_{q} (x, y, t) \geq M_{q} (x, y, \frac{t}{k})$ , then x = y.

Proof. The proof is similar in [20]. Hence, we omit it.

Now we prove Banach contraction principle in the setting of fuzzy double controlled metric space.

Theorem 2.1. Let $α, β : X \times X \to [1, \frac{1}{k})$ (k ∈ (0, 1)) be two non-comparable functions and (X, M_q, ∗) be a complete fuzzy double controlled metric space such that $lim_{t \to \infty} M_{q} (x, y, t) = 1 .$ (1) Let T : X → X be a mapping satisfying $\begin{matrix} M_{q} (Tx, Ty, kt) \geq \\ min {M_{q} (x, y, t), M_{q} (x, Tx, t), M_{q} (y, Ty, t)}, \end{matrix}$ (2)for all x, y ∈ X . Then T has a unique fixed point.

Proof. Let a₀ ∈ X be an arbitrary point. If Ta₀ = a₀, then a₀ is the required fixed point. If Ta₀ ≠ a₀, then there exists a₁ ∈ X such that Ta₀ = a₁. Continuing in this way, we have $a_{2} = {Ta}_{1} = T ({Ta}_{0}) = T^{2} a_{0},$

and $a_{3} = {Ta}_{2} = T ({Ta}_{1}) = T^{2} a_{1} = T^{2} ({Ta}_{0}) = T^{3} a_{0},$ and, we have ${Ta}_{n} = T ({Ta}_{n - 1}) = \dots = T^{n + 1} a_{0} = a_{n + 1} .$ Hence, we get the iterative sequence a_n = Ta_n-1 = Tⁿa₀. Now, by (2) $\begin{matrix} M_{q} (a_{n}, a_{n + 1}, t) \\ = M_{q} ({Ta}_{n - 1}, {Ta}_{n}, t) \\ \geq min {M_{q} (a_{n - 1}, a_{n}, \frac{t}{k}), M (a_{n - 1}, {Ta}_{n - 1}, \frac{t}{k}), \\ M_{q} (a_{n}, {Ta}_{n}, \frac{t}{k})} \\ = min {M_{q} (a_{n}, a_{n - 1}, \frac{t}{k}), M_{q} (a_{n}, a_{n + 1}, \frac{t}{k})} . \end{matrix}$ If $M_{q} (a_{n}, a_{n + 1}, t) \geq M_{q} (a_{n}, a_{n + 1}, \frac{t}{k})$ , then by Lemma 2.2 a_n = a_n+1 for all $n \in ℕ$ and a_n is a fixed point of T. So, we deduce $\begin{matrix} M_{q} (a_{n}, a_{n + 1}, t) & \geq M_{q} (a_{n - 1}, a_{n}, \frac{t}{k}) \\ \geq M_{q} (a_{n - 2}, a_{n - 1}, \frac{t}{k^{2}}) \\ ⋮ \\ \geq M_{q} (a_{0}, a_{1}, \frac{t}{k^{n}}) . \end{matrix}$ (3) Let $n, m \in ℕ$ such that n < m, then $\begin{matrix} M_{q} (a_{n}, a_{m}, t) \\ \geq M_{q} (a_{n}, a_{n + 1}, \frac{\frac{t}{2}}{α (a_{n}, a_{n + 1})}) * M_{q} (a_{n + 1}, a_{m}, \frac{\frac{t}{2}}{β (a_{n + 1}, a_{m})}) \\ \geq M_{q} (a_{n}, a_{n + 1}, \frac{\frac{t}{2}}{α (a_{n}, a_{n + 1})}) * M_{q} (a_{n + 1}, a_{n + 2}, \frac{\frac{t}{2^{2}}}{α (a_{n + 1}, a_{n + 2}) β (a_{n + 1}, a_{m})}) \\ * M_{q} (a_{n + 2}, a_{m}, \frac{\frac{t}{2^{2}}}{β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m})}) \\ \geq M_{q} (a_{n}, a_{n + 1}, \frac{\frac{t}{2}}{α (a_{n}, a_{n + 1})}) * M_{q} (a_{n + 1}, a_{n + 2}, \frac{\frac{t}{2^{2}}}{α (a_{n + 1}, a_{n + 2}) β (a_{n + 1}, a_{m})}) \\ * M_{q} (a_{n + 2}, a_{n + 3}, \frac{\frac{t}{2^{3}}}{α (a_{n + 2}, a_{n + 3}) β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m})}) \\ * M_{q} (a_{n + 3}, a_{m}, \frac{\frac{t}{2^{3}}}{β (a_{n + 3}, a_{m}) β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m})}) \end{matrix}$ $\begin{matrix} \geq M_{q} (a_{n}, a_{n + 1}, \frac{\frac{t}{2}}{α (a_{n}, a_{n + 1})}) * M_{q} (a_{n + 1}, a_{n + 2}, \frac{\frac{t}{2^{2}}}{α (a_{n + 1}, a_{n + 2}) β (a_{n + 1}, a_{m})}) \\ * M_{q} (a_{n + 2}, a_{n + 3}, \frac{\frac{t}{2^{3}}}{α (a_{n + 2}, a_{n + 3}) β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m})}) \\ * M_{q} (a_{n + 3}, a_{n + 4}, \frac{\frac{t}{2^{4}}}{α (a_{n + 3}, a_{n + 4}) β (a_{n + 3}, a_{m}) β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m})}) \\ ⋮ \\ * M_{q} (a_{m - 2}, a_{m - 1}, \frac{\frac{t}{2^{m - 1}}}{α (a_{m - 2}, a_{m - 1}) β (a_{m - 2}, a_{m}) β (a_{m - 3}, a_{m}) \dots β (a_{n + 1}, a_{m})}) \\ * M_{q} (a_{m - 1}, a_{m}, \frac{\frac{t}{2^{m - 1}}}{β (a_{m - 1}, a_{m}) β (a_{m - 2}, a_{m}) \dots β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m})}) . \end{matrix}$ Now applying (3) on each term of right hand side of above inequality, we deduce $\begin{matrix} M_{q} (a_{n}, a_{m}, t) \\ \geq M_{q} (a_{0}, a_{1}, \frac{\frac{t}{2}}{α (a_{n}, a_{n + 1}) k^{n}}) * M_{q} (a_{0}, a_{1}, \frac{\frac{t}{2^{2}}}{α (a_{n + 1}, a_{n + 2}) β (a_{n + 1}, a_{m}) k^{n + 1}}) \\ * M_{q} (a_{0}, a_{1}, \frac{\frac{t}{2^{3}}}{α (a_{n + 2}, a_{n + 3}) β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m}) k^{n + 2}}) \\ * M_{q} (a_{0}, a_{1}, \frac{\frac{t}{2^{4}}}{α (a_{n + 3}, a_{n + 4}) β (a_{n + 3}, a_{m}) β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m}) k^{n + 3}}) \\ ⋮ \\ * M_{q} (a_{0}, a_{1}, \frac{\frac{t}{2^{m - 1}}}{α (a_{m - 2}, a_{m - 1}) β (a_{m - 2}, a_{m}) β (a_{m - 3}, a_{m}) \dots β (a_{n + 1}, a_{m}) k^{n + m - 2}}) \\ * M_{q} (a_{0}, a_{1}, \frac{\frac{t}{2^{m - 1}}}{β (a_{m - 1}, a_{m}) β (a_{m - 2}, a_{m}) \dots β (a_{n + 2}, a_{m}) β (a_{n + 1}, a_{m}) k^{n + m - 1}}) . \end{matrix}$ Since k ∈ (0, 1) , by taking limit as n→ ∞ on the above inequality and together with (1), we get $lim_{n \to \infty} M_{q} (a_{n}, a_{m}, t) = 1 * 1 * \dots * 1 = 1,$ which shows that the sequence {a_n} is a Cauchy sequence in X. Since X is complete fuzzy double controlled metric space, so there exists some a ∈ X such that $lim_{n \to \infty} M_{q} (a_{n}, a, t) = 1 .$ (4) From (2), we have $\begin{matrix} M_{q} (a_{n + 1}, Ta, t) \\ = M_{q} ({Ta}_{n}, Ta, t) \\ \geq min {M (a_{n}, a, \frac{t}{k}), M_{q} (a_{n}, {Ta}_{n}, \frac{t}{k}), \\ M_{q} (a, Ta, \frac{t}{k})} \\ = min {M_{q} (a_{n}, a, \frac{t}{k}), M_{q} (a_{n}, a_{n + 1}, \frac{t}{k}), \\ M_{q} (a, Ta, \frac{t}{k})} . \end{matrix}$ Taking limit as n→ ∞ and using (1)1, we get $\begin{matrix} M_{q} (a, Ta, t) \geq M_{q} (a, Ta, \frac{t}{k}), t > 0, \end{matrix}$ i.e., a is a fixed point of T. We now prove that a is unique. Suppose on contrary that b is also a fixed point of T. Then, $\begin{matrix} M_{q} (a, b, t) = M_{q} (Ta, Tb, t) \\ \geq min {M_{q} (a, b, \frac{t}{k}), M_{q} (a, Ta, \frac{t}{k}), M_{q} (b, Tb, \frac{t}{k})} \\ \geq min {M_{q} (a, b, \frac{t}{k}), M_{q} (a, a, \frac{t}{k}), M_{q} (b, b, \frac{t}{k})} \\ \geq M_{q} (a, b, \frac{t}{k}), \end{matrix}$ and so a = b. This completes the proof.

Following is the Banach contraction principle in the setting of fuzzy double controlled metric space.

Corollary 2.1. Let (X, M_q, ∗) be a complete fuzzy double controlled metric space such that $lim_{t \to \infty} M_{q} (x, y, t) = 1$ and $α, β : X \times X \to [1, \frac{1}{k})$ be given functions where k ∈ (0, 1). Let T : X → X be a self mapping satisfying $M_{q} (Tx, Ty, kt) \geq M_{q} (x, y, t), t > 0,$ for all x, y ∈ X . Then T has a unique fixed point.

Proof. If we take min {M_q (x, y, t) , M_q (x, Tx, t) , M_q (y, Ty, t)} = M_q (x, y, t) in (2), then the result follows from Theorem 2.1 on the same lines.

Now we give an example to support our main result.

Example 2.3. Let X = [0, 1] and $α, β : X \times X \to [1, \frac{1}{k})$ be defined as α (x, y) =2 (x + y) and β (x, y) =2 (x² + y² + 1). Define $M_{q} (x, y, t) = e^{- \frac{(x - y)^{2}}{t}}, for all x, y \in X, t > 0 .$ Then (X, M_q, ∗) is a complete fuzzy double controlled metric space with product t-norm. Define T : X → X by $T (x) = \frac{1 - 2^{- x}}{3} .$ Then, $\begin{matrix} M_{q} (Tx, Ty, kt) \\ = M_{q} (\frac{1 - 2^{- x}}{3}, \frac{1 - 2^{- y}}{3}, kt) \\ = e^{- \frac{(\frac{1 - 2^{- x}}{3} - \frac{1 - 2^{- y}}{3})^{2}}{kt}} \\ = e^{- \frac{(2^{- x} - 2^{- y})^{2}}{9 kt}} \\ \geq e^{- \frac{(x - y)^{2}}{9 kt}} \\ \geq e^{- \frac{(x - y)^{2}}{t}} \\ = M_{q} (x, y, t) \\ \geq min {M_{q} (x, y, t), M_{q} (x, Tx, t), \\ M_{q} (y, Ty, t)}, \end{matrix}$ $for all x, y \in X, where k \in [\frac{1}{9}, 1) .$ Thus, all the conditions of Theorem 2.1 are satisfied. Hence, T has a unique fixed point which is x = 0.

Corollary 2.1 generalizes Theorem 2 of [23] as follows.

Corollary 2.2. Let (X, M_q, ∗) be a complete controlled fuzzy metric space such that $lim_{t \to \infty} M_{q} (x, y, t) = 1,$ and $α : X \times X \to [1, \frac{1}{k})$ be a function where k ∈ (0, 1). Let T : X → X be a self mapping satisfying $M_{q} (Tx, Ty, kt) \geq M_{q} (x, y, t), t > 0,$ for all x, y ∈ X . Then T has a unique fixed point.

Corollary 2.1 generalizes Theorem 2.1 of [15] as follows.

Corollary 2.3. Let (X, M_q, ∗) be a complete extended fuzzy b-metric space such that $lim_{t \to \infty} M_{q} (x, y, t) = 1,$ and $α : X \times X \to [1, \frac{1}{k})$ be a function where k ∈ (0, 1) . Let T : X → X be a self mapping satisfying $M_{q} (Tx, Ty, kt) \geq M_{q} (x, y, t), t > 0,$ for all x, y ∈ X . Then T has a unique fixed point.

3 Application

Fixed point theory is considered as a powerful tool to study the existence and uniqueness of solution of an integral equation. In this section, we give an application of fuzzy double controlled metric spaces to integral equations.

Consider the integral equation $x (s) = p (s) + \int_{0}^{s} G (s, r, x (r)) dr,$ (5) for all s ∈ [0, I] where I > 0 .

Let $X : = C ([0, I], ℝ)$ be the space of all real valued continuous functions defined on [0, I] . Observe that X is a complete metric space with respect to sup-metric $d (x, y) = sup_{s \in [0, I]} | x (s) - y (s) | .$ Also, the space (X, M_q, ∗) with $M_{q} (x, y, t) = e^{- \frac{{sup}_{s \in [0, I]} | x (s) - y (s) |}{t}},$ for all x, y ∈ X and t > 0, and a ∗ b = ab for all a, b ∈ [0, 1] , is a complete fuzzy double controlled metric space.

Theorem 3.1. Let T : X → X be an integral operator defined by $Tx (s) = p (s) + \int_{0}^{s} G (s, r, x (r)) dr,$ where $p \in C ([0, I], ℝ),$ and $G \in C ([0, I] \times [0, I] \times ℝ, ℝ) .$ If there exists g : [0, I] × [0, I] → [0, ∞) such that for all s, r ∈ [0, I] , $g (s, r) \in L^{1} ([0, I], ℝ),$ and for all x, y ∈ X, we get $\begin{matrix} | G (s, r, x (r)) - G (s, r, y (r)) | \\ \leq g (s, r) | x (r) - y (r) |, \end{matrix}$ where $\int_{0}^{s} g (s, r) dr$ is bounded on [0, I] and $0 < sup_{s \in [0, I]} \int_{0}^{s} g (s, r) dr \leq k < 1 .$

Then the integral equation (5) has a unique solution in X.

Proof. Let x, y ∈ X and consider $\begin{matrix} M_{q} (Tx, Ty, kt) \\ = e^{- \frac{{sup}_{s \in [0, I]} | Tx (s) - Ty (s) |}{kt}} \\ \geq e^{- \frac{{sup}_{s \in [0, I]} \int_{0}^{s} | G (s, r, x (r)) - G (s, r, y (r)) | dr}{kt}} \\ \geq e^{- \frac{{sup}_{s \in [0, I]} \int_{0}^{s} g (s, r) | x (r) - y (r) | dr}{kt}} \\ \geq e^{- \frac{| x (r) - y (r) | {sup}_{s \in [0, I]} \int_{0}^{s} g (s, r) dr}{kt}} \\ \geq e^{- \frac{k | x (r) - y (r) |}{kt}} = e^{- \frac{| x (r) - y (r) |}{t}} \\ \geq e^{- \frac{{sup}_{r \in [0, I]} | x (r) - y (r) |}{t}} = M_{q} (x, y, t) \\ \geq min {M_{q} (x, y, t), M_{q} (x, Tx, t), \\ M_{q} (y, Ty, t)} . \end{matrix}$ Since all the conditions of Theorem 2.1 are satisfied, then T has a unique fixed point and hence the integral equation (5) has a unique solution in X.

Example 3.1. Consider the following integral equation in $X = C ([0, 1], ℝ),$ $\begin{matrix} x (s) & = sin (s) + \frac{2}{3} cos (s) - \frac{1}{3} {cos}^{2} (s) \\ + \int_{0}^{s} \frac{x (r)}{3} dr, \end{matrix}$ (6) here $p (s) = sin (s) + \frac{2}{3} cos (s) - \frac{1}{3} {cos}^{2} (s)$ and $G (s, r, x (r)) = \frac{x (r)}{3}$ which are continuous.

Note that, $\begin{matrix} | G (s, r, x (r)) - G (s, r, y (r)) | \\ = | \frac{x (r)}{3} - \frac{y (r)}{3} | \\ = \frac{1}{3} | x (r) - y (r) | \\ \leq g (s, r) | x (r) - y (r) | \end{matrix}$ where $g (s, r) = \frac{1}{2 s}$ so that $sup_{s \in [0, 1]} \int_{0}^{s} g (s, r) dr = \frac{1}{2} < 1 .$

Since all the conditions of Theorem 3.1 are satisfied, so the integral equation (6) has a unique solution.

4 Conclusion

In this present study, we have introduced the concept of a fuzzy double controlled metric space using two controlled functions which generalizes fuzzy b-metric space, extended fuzzy b-metric space and controlled fuzzy metric space. We have proved the Banach contraction principle in the context of fuzzy double controlled metric space and generalized the Banach fixed point theorem in above mentioned spaces. We have given the criteria when a Cauchy sequence in a fuzzy double controlled metric space can have at most one limit point. We also showed that a fuzzy double controlled metric space is not Hausdorff. In future, new theorems and results can be proved in fuzzy double controlled metric spaces.

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