Best Proximity Point Results in Non-Archimedean Modular Metric Space

Abstract

In this paper, we introduce the new notion of Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )$-contractive mapping and investigate the existence and uniqueness of the best proximity point for such mappings in non-Archimedean modular metric space using the weak $P_{\lambda }$-property. Meanwhile, we present an illustrative example to emphasize the realized improvements. These obtained results extend and improve certain well-known results in the literature.

1. Introduction and Preliminaries

Modular metric spaces are a natural and interesting generalization of classical modulars over linear spaces, like Lebesgue, Orlicz, Musielak–Orlicz, Lorentz, Orlicz–Lorentz, Calderon–Lozanovskii spaces and others. The concept of modular metric spaces was introduced in [1,2]. Here, we look at modular metric spaces as the nonlinear version of the classical one introduced by Nakano [3] on vector spaces and modular function spaces introduced by Musielak [4] and Orlicz [5].

Recently, many authors studied the behavior of the electrorheological fluids, sometimes referred to as “smart fluids” (e.g., lithium polymethacrylate). A perfect model for these fluids is obtained by using Lebesgue and Sobolev spaces, $L^p$ and $W^{1,p}$, in the case that p is a function [6].

Let X be a nonempty set and $\omega :(0,+\infty )\times X\times X\to [0,+\infty ]$ be a function; for simplicity, we will write:

$${\omega }_{\lambda }(x,y)=\omega (\lambda ,x,y),$$

for all $\lambda >0$ and $x,y\in X$.

Definition 1.

[1,2] A function $\omega :(0,+\infty )\times X\times X\to [0,+\infty ]$ is called a modular metric on X if the following axioms hold:

(i) 

$x=y$ if and only if ${\omega }_{\lambda }(x,y)=0$ for all $\lambda >0$;

(ii) 

${\omega }_{\lambda }(x,y)={\omega }_{\lambda }(y,x)$ for all $\lambda >0$ and $x,y\in X$;

(iii) 

${\omega }_{\lambda +\mu }(x,y)\le {\omega }_{\lambda }(x,z)+{\omega }_{\mu }(z,y)$ for all $\lambda ,\mu >0$ and $x,y,z\in X$.

If in the above definition, we utilize the condition:

(i’)

${\omega }_{\lambda }(x,x)=0$ for all $\lambda >0$ and $x\in X$;

instead of (i), then $\omega$ is said to be a pseudomodular metric on X. A modular metric $\omega$ on X is called regular if the following weaker version of (i) is satisfied:

$$x=y\text{if and only if}{\omega }_{\lambda }(x,y)=0\text{for some}\lambda >0.$$

Again, $\omega$ is called convex if for $\lambda ,\mu >0$ and $x,y,z\in X$, the inequality holds:

$${\omega }_{\lambda +\mu }(x,y)\le \frac{\lambda }{\lambda +\mu }{\omega }_{\lambda }(x,z)+\frac{\mu }{\lambda +\mu }{\omega }_{\mu }(z,y).$$

Remark 1.

Note that if ω is a pseudomodular metric on a set X, then the function $\lambda \to {\omega }_{\lambda }(x,y)$ is decreasing on $(0,+\infty )$ for all $x,y\in X$. That is, if $0<\mu <\lambda ,$ then:

$${\omega }_{\lambda }(x,y)\le {\omega }_{\lambda -\mu }(x,x)+{\omega }_{\mu }(x,y)={\omega }_{\mu }(x,y).$$

Definition 2.

References [1,2] suppose that ω be a pseudomodular on X and $x_0\in X$ and fixed. Therefore, the two sets:

$$X_{\omega }=X_{\omega }\left(x_0\right)=\{x\in X:{\omega }_{\lambda }(x,x_0)\to 0as\lambda \to +\infty \}$$

and:

$$X_{\omega }^{*}=X_{\omega }^{*}\left(x_0\right)=\{x\in X:\exists \lambda =\lambda \left(x\right)>0suchthat{\omega }_{\lambda }(x,x_0)<+\infty \}.$$

$X_{\omega }$ and $X_{\omega }^{*}$ are called modular spaces (around $x_0$).

It is evident that $X_{\omega }\subset X_{\omega }^{*}$, but this inclusion may be proper in general. Assume that $\omega$ is a modular on X; from [1,2], we derive that the modular space $X_{\omega }$ can be equipped with a (nontrivial) metric, induced by $\omega$ and given by:

$$d_{\omega }(x,y)=\inf \{\lambda >0:{\omega }_{\lambda }(x,y)\le \lambda \}\text{for all}x,y\in X_{\omega }.$$

Note that if $\omega$ is a convex modular on X, then according to [1,2], the two modular spaces coincide, i.e., $X_{\omega }^{*}=X_{\omega }$, and this common set can be endowed with the metric $d_{\omega }^{*}$ given by:

$$d_{\omega }^{*}(x,y)=\inf \{\lambda >0:{\omega }_{\lambda }(x,y)\le 1\}\text{for all}x,y\in X_{\omega }.$$

Such distances are called Luxemburg distances.

Example 2.1 presented by Abdou and Khamsi [7] is an important motivation for developing the modular metric spaces theory. Other examples may be found in [1,2].

Definition 3.

Reference [8] assume $X_{\omega }$ to be a modular metric space, M a subset of $X_{\omega }$ and ${\left(x_n\right)}_{n\in \mathbb{N}}$ be a sequence in $X_{\omega }$. Therefore:

(1) 

${\left(x_n\right)}_{n\in \mathbb{N}}$ is called ω-convergent to $x\in X_{\omega }$ if and only if ${\omega }_{\lambda }(x_n,x)\to 0$, as $n\to +\infty$ for all $\lambda >0$. x will be called the ω-limit of $\left(x_n\right)$.

(2) 

${\left(x_n\right)}_{n\in N}$ is called ω-Cauchy if ${\omega }_{\lambda }(x_m,x_n)\to 0$, as $m,n\to +\infty$ for all $\lambda >0$.

(3) 

M is called ω-closed if the ω-limit of a ω-convergent sequence of M always belong to M.

(4) 

M is called ω-complete if any ω-Cauchy sequence in M is ω-convergent to a point of $M.$

(5) 

M is called ω-bounded if for all $\lambda >0$, we have ${\delta }_{\omega }\left(M\right)=\sup \{{\omega }_{\lambda }(x,y);x,y\in M\}<+\infty .$

Recently Paknazar et al. [9] introduced the following concept.

Definition 4.

If in Definition 1, we replace (iii) by:

$$\left(iv\right){\omega }_{\max \{\lambda ,\mu \}}(x,y)\le {\omega }_{\lambda }(x,z)+{\omega }_{\mu }(z,y)$$

for all $\lambda ,\mu >0$ and $x,y,z\in X$

Then, $X_{\omega }$ is called the non-Archimedean modular metric space. Since (iv) implies (iii), every non-Archimedean modular metric space is a modular metric space.

One of the most important generalizations of Banach contraction mappings was given by Geraghty [10] in the following form.

Theorem 1

(Geraghty [10]). Suppose that $(X,d)$ is a complete metric space and $T:X\to X$ is self-mapping. Suppose that there exists $\beta :[0,+\infty )\to [0,1)$ satisfying the condition:

$$\beta \left(t_n\right)\to 1\mathit{implies}{t}_n\to 0,asn\to +\infty .$$

If T satisfies the following inequality:

$$d(Tx,Ty)\le \beta \left(d\right(x,y\left)\right)d(x,y),\mathit{for}\mathit{all}x,y\in X,$$

(1)

hence T has a unique fixed point.

Moreover, Kirk [11] explored some significant generalizations of the Banach contraction principle to the case of non-self mappings. Let A and B be nonempty subsets of a metric space $(X,d)$. A mapping $T:A\to B$ is called a k-contraction if there exists $k\in [0,1)$, such that $d(Tx,Ty)\le kd(x,y)$, for all $x,y\in A$. Evidently, k-contraction coincides with Banach contraction mapping if we take $A=B$.

Furthermore, a non-self contractive mapping may not have a fixed point. In this case, we try to find an element x such that $d(x,Tx)$ is minimum, i.e., x and $Tx$ are in close proximity to each other. It is clear that $d(x,Tx)$ is at least $d(A,B)=\inf \left\{d\right(x,y):x\in A,y\in B\}$. We are interested in investigating the existence of an element x such that $d(x,Tx)=d(A,B)$. In this case, x is a best proximity point of the non-self-mapping T. Evidently, a best proximity point reduces to a fixed point T as a self-mapping.

The reader can refer to [12,13,14,15,16]. Note that best proximity point theorems furnish an approximate solution to the equation $Tx=x$, when there are not any fixed points for T.

Here, we collect some notions and concepts that will be utilized throughout the rest of this work. We denote by $A_0$ and $B_0$ the following sets:

$$\begin{array}{c}\hfill A_0=\{x\in A:d(x,y)=d(A,B)\mathrm{for}\mathrm{some}y\in B\},\\ \hfill B_0=\{y\in B:d(x,y)=d(A,B)\mathrm{for}\mathrm{some}x\in A\}.\end{array}$$

(2)

In 2003, Kirk et al. [12] established sufficient conditions for determining when the sets $A_0$ and $B_0$ are nonempty.

Furthermore, in [14], the authors proved that any pair $(A,B)$ of nonempty closed convex subsets of a real Hilbert space satisfies the P-property. Clearly for any nonempty subset A of $(X,d)$, the pair $(A,A)$ has the P-property.

Recently, Zhang et al. [16] introduced the following notion and showed that it is weaker than the P-property.

Definition 5.

Let $(A,B)$ be a pair of nonempty subsets of a metric space $(X,d)$ with $A_0\ne \mathrm{\varnothing }.$ Then, the pair $(A,B)$ is said to have the weak P-property if and only if for any $x_1,x_2\in A_0$ and $y_1,y_2\in B_0:$

$$d(x_1,y_1)=d(A,B)\mathrm{and}d(x_2,y_2)=d(A,B)\Rightarrow d(x_1,x_2)\le d(y_1,y_2).$$

(3)

Finally, we recall the following result of Caballero et al. [17].

Theorem 2.

Assume that $(A,B)$ is a pair of nonempty closed subsets of a complete metric space $(X,d)$, such that $A_0$ is nonempty. Let $T:A\to B$ be a Geraghty-contraction satisfying $T\left(A_0\right)\subseteq B_0.$ Assume that the pair $(A,B)$ has the P-property. Then, there exists a unique $x^{*}\in A$ such that $d(x^{*},T{x}^{*})=d(A,B).$

Recently, Kumam et al. [18] introduced the useful notion of triangular $\alpha$-proximal admissible mapping as follows. See also [19]:

Definition 6

(Reference [18]). Let A and B be two nonempty subsets ofa metric space $(X,d)$ and $\alpha :A\times A\to [0,+\infty )$ be a function. We say that a non-self-mapping $T:A\to B$ is triangular α-proximal admissible if, for all $x,y,z,x_1,x_2,u_1,u_2\in A$:

$$\begin{gathered} (T1)\left\{\begin{array}{c}\alpha (x_1,x_2)\ge 1\hfill \\ d(u_1,T{x}_1)=d(A,B)\hfill \\ d(u_2,T{x}_2)=d(A,B)\hfill \end{array}\right.⟹\alpha (u_1,u_2)\ge 1, \\ (T2)\left\{\begin{array}{c}\alpha (x,z)\ge 1\hfill \\ \alpha (z,y)\ge 1\hfill \end{array}\right.⟹\alpha (x,y)\ge 1. \end{gathered}$$

Let $\mathrm{\Theta }$ denote the set of all functions $\theta :R^{{+}^4}\to R^{+}$ satisfying:

• (${\mathrm{\Theta }}_1$) $\theta$ is continuous and increasing in all of its variables;
• (${\mathrm{\Theta }}_2$) $\theta (t_1,t_2,t_3,t_4)=0$ iff $t_1.t_2.t_3.t_4=0$.

For more details on $\mathrm{\Theta }$, see [20].

Let $\mathcal{F}$ denote the set of all functions $\beta :[0,+\infty )\to [0,1)$ satisfying the condition:

$$\beta \left(t_n\right)\to 1\mathrm{implies}{t}_n\to 0,\mathrm{as}n\to +\infty .$$

2. Best Proximity Point Results

At first, we introduce the following concept, which will be suitable for our main Theorem.

Definition 7.

Suppose that $(A,B)$ is a pair of nonempty subsets of a modular metric space $X_{\omega }$ with $A_0^{\lambda }\ne \mathrm{\varnothing }$ for all $\lambda >0.$ We say the pair $(A,B)$ has the weak $P_{\lambda }$-property if and only if for any $x_1,x_2\in A_0$, $y_1,y_2\in B_0$ and $\lambda >0:$

$${\omega }_{\lambda }(x_1,y_1)={\omega }_{\lambda }(A,B)\mathrm{and}{\omega }_{\lambda }(x_2,y_2)=d(A,B)\Rightarrow {\omega }_{\lambda }(x_1,x_2)\le {\omega }_{\lambda }(y_1,y_2),$$

(4)

where:

$${\omega }_{\lambda }(A,B)=:\inf \left\{{\omega }_{\lambda }(x,y)\right|x\in A\mathrm{and}y\in B\},$$

$$A_0^{\lambda }=:\{x\in A:{\omega }_{\lambda }(x,y)={\omega }_{\lambda }(A,B)\mathrm{for}\mathrm{some}y\in B\}.$$

Now, let us introduce the concept of Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )$-contractive mapping.

Definition 8.

Let A and B be two nonempty subsets of a modular metric space $X_{\omega }$ where $A_0^{\lambda }\ne \mathrm{\varnothing }$ for all $\lambda >0$ and $\alpha :X_{\omega }\times X_{\omega }\to [0,\infty )$ is a function. A mapping $T:A\to B$ is said to be a Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )-$contractive mapping if there exists $\beta \in \mathcal{F}$ and $\theta \in \mathrm{\Theta }$, such that for all $x,y\in A$ and $\lambda >0$ with $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $\alpha (x,y)\ge 1$, one has:

$${\omega }_{\lambda }(Tx,Ty)\le \beta \left(M(x,y)\right)M(x,y)+\gamma \left(N(x,y,\theta )\right)N(x,y,\theta )$$

(5)

where $\gamma :[0,\infty )\to [0,\infty )$ is a bounded function, ${\omega }_{\lambda }^{*}(x,y)={\omega }_{\lambda }(x,y)-{\omega }_{\lambda }(A,B),$

$$\begin{array}{c}\hfill M(x,y)=\max \{{\omega }_{\lambda }(x,y),\frac{{\omega }_{\lambda }(x,Tx)+{\omega }_{\lambda }(y,Ty)}{2}-{\omega }_{\lambda }(A,B),\\ \hfill \frac{{\omega }_{\lambda }(x,Ty)+{\omega }_{\lambda }(y,Tx)}{2}-{\omega }_{\lambda }(A,B)\}\end{array}$$

and:

$$\begin{array}{c}\hfill N(x,y,\theta )=\theta ({\omega }_{\lambda }(x,Tx)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(y,Ty)-{\omega }_{\lambda }(A,B),\\ \hfill {\omega }_{\lambda }(x,Ty)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(y,Tx)-{\omega }_{\lambda }(A,B)).\end{array}$$

Now, we are ready to prove our main result.

Theorem 3.

Let A and B be two nonempty subsets of a non-Archimedean modular metric space $X_{\omega }$ with ω regular, such that A is $\omega -$complete and $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Assume that T is a Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )$-contractive mapping satisfying the following assertions:

(i) 

$T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$, and the pair $(A,B)$ satisfies the weak $P_{\lambda }$-property,

(ii) 

T is a triangular α-proximal admissible mapping,

(iii) 

there exist elements $x_0$ and $x_1$ in $A_0^{\lambda }$ for all $\lambda >0$, such that:

$${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)\mathit{and}\alpha (x_0,x_1)\ge 1$$

(iv) 

if $\left\{x_n\right\}$ is a sequence in A, such that $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ with $x_n\to x\in A$ as $n\to \infty$, then $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}.$

Then, there exists an $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$ Further, the best proximity point is unique if, for every $x,y\in A$, such that ${\omega }_{\lambda }(x,Tx)={\omega }_{\lambda }(A,B)={\omega }_{\lambda }(y,Ty)$, we have $\alpha (x,y)\ge 1$.

Proof. 

By (iii), there exist elements $x_0$ and $x_1$ in $A_0^{\lambda }$ for all $\lambda >0$, such that:

$${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)\mathrm{and}\alpha (x_0,x_1)\ge 1.$$

On the other hand, $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$. Therefore, there exists $x_2\in A_0$, such that:

$${\omega }_{\lambda }(x_2,T{x}_1)={\omega }_{\lambda }(A,B).$$

Now, since T is triangular $\alpha$-proximal admissible, we have $\alpha (x_1,x_2)\ge 1.$ That is:

$${\omega }_{\lambda }(x_2,T{x}_1)={\omega }_{\lambda }(A,B)\mathrm{and}\alpha (x_1,x_2)\ge 1.$$

Again, since $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$, there exists $x_3\in A_0^{\lambda }$, such that:

$${\omega }_{\lambda }(x_3,T{x}_2)={\omega }_{\lambda }(A,B).$$

Thus, we have:

$${\omega }_{\lambda }(x_2,T{x}_1)={\omega }_{\lambda }(A,B)\mathrm{and}{\omega }_{\lambda }(x_3,T{x}_2)={\omega }_{\lambda }(A,B)\mathrm{and}\alpha (x_1,x_2)\ge 1.$$

Again, since T is triangular $\alpha$-proximal admissible, $\alpha (x_2,x_3)\ge 1.$ Hence:

$${\omega }_{\lambda }(x_3,T{x}_2)={\omega }_{\lambda }(A,B)\mathrm{and}\alpha (x_2,x_3)\ge 1.$$

Continuing this process, we get:

$${\omega }_{\lambda }(x_{n+1},T{x}_n)={\omega }_{\lambda }(A,B)\mathrm{and}\alpha (x_n,x_{n+1})\ge 1\mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

(6)

Since $(A,B)$ has the weak $P_{\lambda }$-property, we derive that:

$${\omega }_{\lambda }(x_n,x_{n+1})\le {\omega }_{\lambda }(T{x}_{n-1},T{x}_n)\mathrm{for}\mathrm{any}n\in \mathbb{N}.$$

(7)

Now, by (6), we get:

$${\omega }_{\lambda }(x_{n-1},T{x}_{n-1})\le {\omega }_{\lambda }(x_{n-1},x_n)+{\omega }_{\lambda }(x_n,T{x}_{n-1})={\omega }_{\lambda }(x_{n-1},x_n)+{\omega }_{\lambda }(A,B).$$

(8)

Clearly, if there exists $n_0\in \mathbb{N}$, such that ${\omega }_{\lambda }(x_{n_0},x_{n_0+1})=0$, then we have nothing to prove. In fact:

$$0={\omega }_{\lambda }(x_{n_0},x_{n_0+1})={\omega }_{\lambda }(T{x}_{n_0-1},T{x}_{n_0}).$$

Since $\omega$ is regular, we get, $T{x}_{n_0-1}=T{x}_{n_0}.$ Thus, we conclude that:

$${\omega }_{\lambda }(A,B)={\omega }_{\lambda }(x_{n_0},T{x}_{n_0-1})={\omega }_{\lambda }(x_{n_0},T{x}_{n_0}).$$

For the rest of the proof, we suppose that ${\omega }_{\lambda }(x_n,x_{n+1})>0$ for any $n\in \mathbb{N}.$ Now, from (8), we deduce that:

$$\frac{1}{2}{\omega }_{\lambda }^{*}(x_{n-1},T{x}_{n-1})\le {\omega }_{\lambda }^{*}(x_{n-1},T{x}_{n-1})\le {\omega }_{\lambda }(x_n,x_{n-1}).$$

(9)

Applying (6) and (7), we obtain:

$$\begin{array}{cc}\hfill M(x_{n-1},x_n)& =\max \{{\omega }_{\lambda }(x_{n-1},x_n),\frac{{\omega }_{\lambda }(x_{n-1},T{x}_{n-1})+{\omega }_{\lambda }(x_n,T{x}_n)}{2}-{\omega }_{\lambda }(A,B),\hfill \\ & \frac{{\omega }_{\lambda }(x_{n-1},T{x}_n)+{\omega }_{\lambda }(x_n,T{x}_{n-1})}{2}-{\omega }_{\lambda }(A,B)\}\hfill \\ & \le \max \left\{{\omega }_{\lambda }(x_{n-1},x_n),\right.\hfill \\ & \frac{{\omega }_{\lambda }(x_{n-1},x_n)+{\omega }_{\lambda }(x_n,T{x}_{n-1})+{\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)}{2}-{\omega }_{\lambda }(A,B),\hfill \\ & \frac{{\omega }_{\lambda }(x_{n-1},x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)+{\omega }_{\lambda }(x_n,T{x}_{n-1})}{2}-{\omega }_{\lambda }(A,B)\}\hfill \\ & =\max \left\{{\omega }_{\lambda }(x_{n-1},x_n),\right.\hfill \\ & \frac{{\omega }_{\lambda }(x_{n-1},x_n)+{\omega }_{\lambda }(A,B)+{\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(A,B)}{2}-{\omega }_{\lambda }(A,B),\hfill \\ & \frac{{\omega }_{\lambda }(x_{n-1},x_{n+1})+{\omega }_{\lambda }(A,B)+{\omega }_{\lambda }(A,B)}{2}-{\omega }_{\lambda }(A,B)\}\hfill \\ & =\max \left\{{\omega }_{\lambda }(x_{n-1},x_n),\frac{{\omega }_{\lambda }(x_{n-1},x_n)+{\omega }_{\lambda }(x_n,x_{n+1})}{2},\frac{{\omega }_{\lambda }(x_{n-1},x_{n+1})}{2}\right\}\hfill \\ & \le \max \{{\omega }_{\lambda }(x_{n-1},x_n),\frac{{\omega }_{\lambda }(x_{n-1},x_n)+{\omega }_{\lambda }(x_n,x_{n+1})}{2}\}\hfill \\ & \le \max \{{\omega }_{\lambda }(x_{n-1},x_n),{\omega }_{\lambda }(x_n,x_{n+1})\}.\hfill \end{array}$$

Thus:

$$M(x_{n-1},x_n)\le \max \{{\omega }_{\lambda }(x_{n-1},x_n),{\omega }_{\lambda }(x_n,x_{n+1})\}.$$

(10)

Furthermore:

$$\begin{array}{cc}\hfill N(x_{n-1},x_n,\theta )=\theta & ({\omega }_{\lambda }(x_{n-1},T{x}_{n-1})-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ \hfill & {\omega }_{\lambda }(x_{n-1},T{x}_n)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_n,T{x}_{n-1})-{\omega }_{\lambda }(A,B))\hfill \\ \hfill =\theta & ({\omega }_{\lambda }(x_{n-1},T{x}_{n-1})-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ \hfill & {\omega }_{\lambda }(x_{n-1},T{x}_n)-{\omega }_{\lambda }(A,B),0)=0.\hfill \end{array}$$

(11)

Since T is a Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )$-contractive mapping, we have:

$$\begin{array}{cc}\hfill {\omega }_{\lambda }(x_n,x_{n+1})& \le {\omega }_{\lambda }(T{x}_{n-1},T{x}_n)\hfill \\ & \le \beta \left(M(x_{n-1},x_n)\right)M(x_{n-1},x_n)+\gamma \left(N(x_{n-1},x_n,\theta )\right)N(x_{n-1},x_n,\theta )\hfill \\ & <M(x_{n-1},x_n)+\gamma \left(N(x_{n-1},x_n,\theta )\right)N(x_{n-1},x_n,\theta ).\hfill \end{array}$$

(12)

From (10) to (12), we deduce:

$${\omega }_{\lambda }(x_n,x_{n+1})<\max \{{\omega }_{\lambda }(x_{n-1},x_n),{\omega }_{\lambda }(x_n,x_{n+1})\}.$$

Now if, $\max \{{\omega }_{\lambda }(x_{n-1},x_n),{\omega }_{\lambda }(x_n,x_{n+1})\}={\omega }_{\lambda }(x_n,x_{n+1})$ then,

$${\omega }_{\lambda }(x_n,x_{n+1})<{\omega }_{\lambda }(x_n,x_{n+1}),$$

which is a contradiction. Hence:

$$\begin{array}{c}\hfill \begin{array}{c}{\omega }_{\lambda }(x_{n-1},x_n)\le M(x_{n-1},x_n)\le \max \{{\omega }_{\lambda }(x_{n-1},x_n),{\omega }_{\lambda }(x_n,x_{n+1})\}={\omega }_{\lambda }(x_{n-1},x_n),\hfill \end{array}\end{array}$$

and so:

$$M(x_{n-1},x_n)={\omega }_{\lambda }(x_{n-1},x_n),$$

(13)

for all $n\in \mathbb{N}$. Now, by (12), we get:

$$\begin{array}{cc}\hfill {\omega }_{\lambda }(x_n,x_{n+1})& ={\omega }_{\lambda }(T{x}_{n-1},T{x}_n)\hfill \\ & \le \beta \left({\omega }_{\lambda }(x_{n-1},x_n)\right){\omega }_{\lambda }(x_{n-1},x_n)\hfill \\ & <{\omega }_{\lambda }(x_{n-1},x_n),\hfill \end{array}$$

(14)

for all $n\in \mathbb{N}.$ Consequently, $\left\{{\omega }_{\lambda }(x_n,x_{n+1})\right\}$ is a non-increasing sequence, which is bounded from below, and so, $\underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,x_{n+1}):=L$ exists. Let $L>0$. Then, from (14), we have:

$$\frac{{\omega }_{\lambda }(x_n,x_{n+1})}{{\omega }_{\lambda }(x_{n-1},x_n)}\le \beta \left({\omega }_{\lambda }(x_{n-1},x_n)\right)\le 1,$$

for each $n\ge 1,$ which implies:

$$\underset{n\to \infty }{\lim }\beta \left({\omega }_{\lambda }(x_n,x_{n+1})\right)=1.$$

On the other hand, since $\beta \in \mathcal{F}$, we conclude:

$$L=\underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,x_{n+1})=0.$$

(15)

Since, ${\omega }_{\lambda }(x_n,T{x}_{n-1})={\omega }_{\lambda }(A,B)$ holds for all $n\in \mathbb{N}$ and $(A,B)$ satisfies the weak $P_{\lambda }$-property, so for all $m,n\in \mathbb{N}$ with $n<m$, we obtain, ${\omega }_{\lambda }(x_m,x_n)\le {\omega }_{\lambda }(T{x}_{m-1},T{x}_{n-1}).$ Note that:

$$\begin{array}{cc}\hfill M(x_m,x_n)& {\displaystyle =\max \{{\omega }_{\lambda }(x_m,x_n),\frac{{\omega }_{\lambda }(x_m,T{x}_m)+{\omega }_{\lambda }(x_n,T{x}_n)}{2}-{\omega }_{\lambda }(A,B),}\hfill \\ & {\displaystyle \frac{{\omega }_{\lambda }(x_m,T{x}_n)+{\omega }_{\lambda }(x_n,T{x}_m)}{2}-{\omega }_{\lambda }(A,B)\}}\hfill \\ & \le \max \{{\omega }_{\lambda }(x_m,x_n),\hfill \\ & {\displaystyle \frac{{\omega }_{\lambda }(x_m,x_{m+1})+{\omega }_{\lambda }(x_{m+1},T{x}_m)+{\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)}{2}-{\omega }_{\lambda }(A,B),}\hfill \\ & {\displaystyle \frac{{\omega }_{\lambda }(x_m,x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)+{\omega }_{\lambda }(x_n,x_{m+1})+{\omega }_{\lambda }(x_{m+1},T{x}_m)}{2}-{\omega }_{\lambda }(A,B)\}}\hfill \\ & {\displaystyle =\max \left\{{\omega }_{\lambda }(x_m,x_n),\frac{{\omega }_{\lambda }(x_m,x_{m+1})+{\omega }_{\lambda }(x_n,x_{n+1})}{2},{\omega }_{\lambda }(x_m,x_{n+1})\right\}}\hfill \\ & {\displaystyle \le \max \{{\omega }_{\lambda }(x_m,x_n),\frac{{\omega }_{\lambda }(x_m,x_{m+1})+{\omega }_{\lambda }(x_n,x_{n+1})}{2},}\hfill \\ & {\omega }_{\lambda }(x_m,x_n)+{\omega }_{\lambda }(x_n,x_{n+1})\}.\hfill \end{array}$$

As ${\displaystyle \underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,x_{n+1})=0,}$ we have:

$$\begin{array}{c}\underset{m,n\to \infty }{\lim }{\omega }_{\lambda }(x_m,x_n)\le \underset{m,n\to \infty }{\lim }M(x_m,x_n)\le \underset{m,n\to \infty }{\lim }{\omega }_{\lambda }(x_m,x_n),\hfill \end{array}$$

that is:

$$\begin{array}{c}\underset{m,n\to \infty }{\lim }M(x_m,x_n)=\underset{m,n\to \infty }{\lim }{\omega }_{\lambda }(x_m,x_n).\hfill \end{array}$$

(16)

Furthermore:

$$\begin{array}{cc}\hfill 0& \le N(x_m,x_n,\theta )\hfill \\ & =\theta ({\omega }_{\lambda }(x_m,T{x}_m)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x_m,T{x}_n)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_n,T{x}_m)-{\omega }_{\lambda }(A,B))\hfill \\ & \le \theta ({\omega }_{\lambda }(x_m,x_{m+1})+{\omega }_{\lambda }(A,B)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x_m,T{x}_n)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_n,T{x}_m)-{\omega }_{\lambda }(A,B))\hfill \\ & \le \theta ({\omega }_{\lambda }(x_m,x_{m+1}),{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_m,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x_n,T{x}_m)-{\omega }_{\lambda }(A,B)).\hfill \end{array}$$

Again, by ${\displaystyle \underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,x_{n+1})=0,}$ we have:

$$\begin{array}{cc}\hfill 0& {\displaystyle \le \underset{m,n\to \infty }{\lim }N(x_m,x_n,\theta )}\hfill \\ & \le \underset{m,n\to \infty }{\lim }\theta ({\omega }_{\lambda }(x_m,x_{m+1}),{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_m,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x_n,T{x}_m)-{\omega }_{\lambda }(A,B))\hfill \\ & \le \underset{m,n\to \infty }{\lim }\theta (0,{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(x_m,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x_n,T{x}_m)-{\omega }_{\lambda }(A,B))=0.\hfill \end{array}$$

That is:

$$\underset{m,n\to \infty }{\lim }N(x_m,x_n,\theta )=0.$$

(17)

Now, we show that $\left\{x_n\right\}$ is a Cauchy sequence. On the contrary, assume that:

$$\epsilon =\underset{m,n\to \infty }{\text{lim sup }}{\omega }_{\lambda }(x_n,x_m)>0.$$

(18)

Now, since ${\displaystyle \underset{n\to +\infty }{\lim }{\omega }_{\lambda }(x_n,x_{n+1})=0,}$ then:

$$\begin{array}{ccc}\hfill {\omega }_{\lambda }(A,B)& \le & \underset{m\to +\infty }{\lim }{\omega }_{\lambda }(x_m,T{x}_m)\hfill \\ & \le & \underset{m\to +\infty }{\lim }[{\omega }_{\lambda }(x_m,x_{m+1})+{\omega }_{\lambda }(x_{m+1},T{x}_m)]\hfill \\ & =& \underset{m\to +\infty }{\lim }[{\omega }_{\lambda }(x_m,x_{m+1})+{\omega }_{\lambda }(A,B)]={\omega }_{\lambda }(A,B),\hfill \end{array}$$

which implies that $\underset{m\to +\infty }{\lim }{\omega }_{\lambda }(x_m,T{x}_m)={\omega }_{\lambda }(A,B)$, that is:

$$\underset{m\to +\infty }{\lim }\frac{1}{2}{\omega }_{\lambda }^{*}(x_m,T{x}_m)=\underset{m\to +\infty }{\lim }\frac{1}{2}[{\omega }_{\lambda }(x_m,T{x}_m)-{\omega }_{\lambda }(A,B)]=0.$$

On the other hand, from (18), it is follows that there exists $N\in \mathbb{N}$, such that, for all $m,n\ge N$, we have:

$$\frac{1}{2}{\omega }_{\lambda }^{*}(x_m,T{x}_m)\le {\omega }_{\lambda }(x_n,x_m).$$

Furthermore, we can show that:

$$\alpha (x_m,x_n)\ge 1,\mathrm{where}n>m.$$

(19)

Indeed, since T is a triangular $\alpha$-proximal admissible mapping and:

$$\left\{\begin{array}{c}\alpha (x_m,x_{m+1})\ge 1\hfill \\ \alpha (x_{m+1},x_{m+2})\ge 1\hfill \end{array}\right.,$$

from Condition (T2) of Definition 6, we have:

$$\alpha (x_m,x_{m+2})\ge 1.$$

Again, since T is a triangular $\alpha$-proximal admissible mapping and:

$$\left\{\begin{array}{c}\alpha (x_m,x_{m+2})\ge 1\hfill \\ \alpha (x_{m+2},x_{m+3})\ge 1\hfill \end{array}\right.,$$

from Condition (T2) of Definition 6, we have:

$$\alpha (x_m,x_{m+3})\ge 1.$$

Continuing this process, we get (19).

Now, using the triangle inequality, we have:

$${\omega }_{\lambda }(x_n,x_m)\le {\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(x_{n+1},x_{m+1})+{\omega }_{\lambda }(x_{m+1},x_m).$$

(20)

From (5) and (20) we have:

$$\begin{array}{cc}\hfill {\omega }_{\lambda }(x_n,x_m)& \\ \hfill \le & {\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(T{x}_n,T{x}_m)+{\omega }_{\lambda }(x_{m+1},x_m)\hfill \\ \hfill \le & {\omega }_{\lambda }(x_n,x_{n+1})+\beta \left(M(x_n,x_m)\right)M(x_n,x_m)+\gamma \left(N(x_n,x_m,\theta )\right)N(x_n,x_m,\theta )\hfill \\ & +{\omega }_{\lambda }(x_{m+1},x_m).\hfill \end{array}$$

(21)

Now, (16), (17), (21) and: ${\displaystyle \underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,x_{n+1})=0,}$ imply:

$$\begin{array}{cc}\hfill \underset{m,n\to \infty }{\lim }{\omega }_{\lambda }(x_n,x_m)\le & \underset{m,n\to \infty }{\lim }\beta \left(M(x_n,x_m)\right)\underset{m,n\to \infty }{\lim }M(x_m,x_n)\hfill \\ & +\underset{m,n\to \infty }{\lim }\gamma \left(N(x_n,x_m,\theta )\right)\underset{m,n\to \infty }{\lim }N(x_m,x_n,\theta )\hfill \\ \hfill =& \underset{m,n\to \infty }{\lim }\beta \left(M(x_n,x_m)\right)\underset{m,n\to \infty }{\lim }{\omega }_{\lambda }(x_m,x_n).\hfill \end{array}$$

By (18), we get:

$$1\le \underset{m,n\to \infty }{\lim }\beta \left(M(x_n,x_m)\right).$$

Therefore, $\underset{m,n\to \infty }{\lim }\beta \left(M(x_n,x_m)\right)=1$, so $\underset{m,n\to \infty }{\lim }M(x_n,x_m)=0$. This implies:

$$\underset{m,n\to \infty }{\lim }{\omega }_{\lambda }(x_n,x_m)=0,$$

which is a contradiction. Therefore, $\left\{x_n\right\}$ is a Cauchy sequence. Since $\left(x_n\right)\subset A$ and $(A,d)$ is a complete metric space, we can find $x^{*}\in A$, such that $x_n\to x^{*}$ as $n\to \infty .$ From (iv), we know that, $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}.$ Next, using (14), we have:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\omega }_{\lambda }^{*}(x_n,T{x}_n)& ={\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B)\hfill \\ & \le {\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)-{\omega }_{\lambda }(A,B)\hfill \\ & ={\omega }_{\lambda }(x_n,x_{n+1}),\hfill \end{array}\end{array}$$

(22)

and:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\omega }_{\lambda }^{*}(x_{n+1},T{x}_{n+1})& ={\omega }_{\lambda }(x_{n+1},T{x}_{n+1})-{\omega }_{\lambda }(A,B)\hfill \\ & \le {\omega }_{\lambda }(T{x}_n,T{x}_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)-{\omega }_{\lambda }(A,B)\hfill \\ & ={\omega }_{\lambda }(T{x}_n,T{x}_{n+1})\hfill \\ & ={\omega }_{\lambda }(x_{n+1},x_{n+2})\hfill \\ & \le {\omega }_{\lambda }(x_n,x_{n+1}).\hfill \end{array}\end{array}$$

(23)

Therefore, (22) and (23) imply that:

$$\begin{array}{c}\hfill \frac{1}{2}[{\omega }_{\lambda }^{*}(x_n,T{x}_n)+{\omega }_{\lambda }^{*}(x_{n+1},T{x}_{n+1})]\le {\omega }_{\lambda }(x_n,x_{n+1}).\end{array}$$

(24)

Now, suppose that:

$$\frac{1}{2}{\omega }_{\lambda }^{*}(x_n,T{x}_n)>{\omega }_{\lambda }(x_n,x^{*})\mathrm{and}\frac{1}{2}{\omega }_{\lambda }^{*}(x_{n+1},T{x}_{n+1})>{\omega }_{\lambda }(x_{n+1},x^{*}),$$

for some $n\in \mathbb{N}$. Hence, using (24), we can write:

$$\begin{array}{cc}\hfill {\omega }_{\lambda }(x_n,x_{n+1})& \le {\omega }_{\lambda }(x_n,x^{*})+{\omega }_{\lambda }(x_{n+1},x^{*})\hfill \\ & <\frac{1}{2}[{\omega }_{\lambda }^{*}(x_n,T{x}_n)+{\omega }_{\lambda }^{*}(x_{n+1},T{x}_{n+1})]\hfill \\ & \le {\omega }_{\lambda }(x_n,x_{n+1}),\hfill \end{array}$$

which is a contradiction. Then, for any $n\in \mathbb{N}$, either:

$$\frac{1}{2}{\omega }_{\lambda }^{*}(x_n,T{x}_n)\le {\omega }_{\lambda }(x_n,x^{*})\mathrm{or}\frac{1}{2}{\omega }_{\lambda }^{*}(x_{n+1},T{x}_{n+1})\le {\omega }_{\lambda }(x_{n+1},x^{*})$$

holds.

We shall show that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$. Suppose, to the contrary, that:

$${\omega }_{\lambda }(x^{*},T{x}^{*})\ne {\omega }_{\lambda }(A,B).$$

From (5) with $x=x_n$ and $y=x^{*}$, we get:

$${\omega }_{\lambda }(T{x}_n,T{x}^{*})\le \beta \left(M(x_n,x^{*})\right)M(x_n,x^{*})+\gamma \left(N(x_n,x^{*},\theta )\right)N(x_n,x^{*},\theta ).$$

(25)

On the other hand:

$$\begin{array}{cc}\hfill M(x_n,& x^{*})\hfill \\ & {\displaystyle =\max \{{\omega }_{\lambda }(x_n,x^{*}),\frac{{\omega }_{\lambda }(x_n,T{x}_n)+{\omega }_{\lambda }(x^{*},T{x}^{*})}{2}-{\omega }_{\lambda }(A,B),}\hfill \\ & {\displaystyle \frac{{\omega }_{\lambda }(x_n,T{x}^{*})+{\omega }_{\lambda }(x^{*},T{x}_n)}{2}-{\omega }_{\lambda }(A,B)\}}\hfill \\ & {\displaystyle \le \max \{{\omega }_{\lambda }(x_n,x^{*}),\frac{{\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)+{\omega }_{\lambda }(x^{*},T{x}^{*})}{2}-{\omega }_{\lambda }(A,B),}\hfill \\ & {\displaystyle \frac{{\omega }_{\lambda }(x_n,x^{*})+{\omega }_{\lambda }(x^{*},T{x}^{*})+{\omega }_{\lambda }(x^{*},x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)}{2}-{\omega }_{\lambda }(A,B)\}}\hfill \\ & {\displaystyle =\max \{{\omega }_{\lambda }(x_n,x^{*}),\frac{{\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(A,B)+{\omega }_{\lambda }(x^{*},T{x}^{*})}{2}-{\omega }_{\lambda }(A,B),}\hfill \\ & {\displaystyle \frac{{\omega }_{\lambda }(x_n,x^{*})+{\omega }_{\lambda }(x^{*},T{x}^{*})+{\omega }_{\lambda }(x^{*},x_{n+1})+{\omega }_{\lambda }(A,B)}{2}-{\omega }_{\lambda }(A,B)\},}\hfill \end{array}$$

and so:

$$\begin{array}{c}\hfill \underset{k\to \infty }{\lim }M(x_n,x^{*})\le \frac{{\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B)}{2}.\end{array}$$

(26)

Furthermore, we have:

$$\begin{array}{cc}\hfill {\omega }_{\lambda }(x^{*},T{x}^{*})& \le {\omega }_{\lambda }(x^{*},T{x}_n)+{\omega }_{\lambda }(T{x}_n,T{x}^{*})\hfill \\ & \le {\omega }_{\lambda }(x^{*},x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)+{\omega }_{\lambda }(T{x}_n,T{x}^{*})\hfill \\ & \le {\omega }_{\lambda }(x^{*},x_{n+1})+{\omega }_{\lambda }(A,B)+{\omega }_{\lambda }(T{x}_n,T{x}^{*}).\hfill \end{array}$$

Taking limit as $n\to \infty$ in the above inequality, we have:

$$\begin{array}{c}\hfill {\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B)\le \underset{n\to \infty }{\lim }{\omega }_{\lambda }(T{x}_n,T{x}^{*}).\end{array}$$

(27)

Further, we get:

$${\omega }_{\lambda }(x_n,T{x}_n)\le {\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(x_{n+1},T{x}_n)={\omega }_{\lambda }(x_n,x_{n+1})+{\omega }_{\lambda }(A,B).$$

Taking the limit as $n\to \infty$ in the above inequality, we get:

$$\underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,T{x}_n)\le {\omega }_{\lambda }(A,B),$$

and so, $\underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,T{x}_n)={\omega }_{\lambda }(A,B)$. Now, we have:

$$\begin{array}{cc}\hfill \underset{n\to \infty }{\lim }N(x_n,x^{*},\theta )& \\ & =\theta (\underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,T{x}_n)-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B),\underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,T{x}^{*})-{\omega }_{\lambda }(A,B),\hfill \\ & \underset{n\to \infty }{\lim }{\omega }_{\lambda }(x^{*},T{x}_n)-{\omega }_{\lambda }(A,B))\hfill \\ & =\theta (0,{\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B),\hfill \\ & \underset{n\to \infty }{\lim }{\omega }_{\lambda }(x_n,T{x}^{*})-{\omega }_{\lambda }(A,B),\underset{n\to \infty }{\lim }{\omega }_{\lambda }(x^{*},T{x}_n)-{\omega }_{\lambda }(A,B))=0,\hfill \end{array}$$

that is:

$$\begin{array}{c}\hfill {\displaystyle \underset{n\to \infty }{\lim }N(x_n,x^{*},\theta )=0.}\end{array}$$

(28)

From (25) to (28), we deduce that:

$$\begin{array}{cc}\hfill {\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B)& {\displaystyle \le \underset{n\to \infty }{\lim }{\omega }_{\lambda }(T{x}_n,T{x}^{*})}\hfill \\ & {\displaystyle \le \underset{n\to \infty }{\lim }\beta \left(M(x_n,x^{*})\right)\underset{n\to \infty }{\lim }M(x_n,x^{*})}\hfill \\ & {\displaystyle +\underset{n\to \infty }{\lim }\gamma \left(N(x_n,x^{*},\theta )\right)\underset{n\to \infty }{\lim }N(x_n,x^{*},\theta )}\hfill \\ & {\displaystyle =\underset{n\to \infty }{\lim }\beta \left(M(x_n,x^{*})\right)(\frac{{\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B)}{2})}\hfill \\ & <{\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B),\hfill \end{array}$$

which is a contradiction. Therefore, ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$, and $x^{*}$ is a best proximity point of T. We now show the uniqueness of the best proximity point of $T.$ Suppose that $x^{*}$ and $y^{*}$ are two distinct best proximity points of T. This implies:

$${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)={\omega }_{\lambda }(y^{*},T{y}^{*}).$$

(29)

Using the weak $P_1$-property, we have:

$${\omega }_{\lambda }(x^{*},y^{*})\le {\omega }_{\lambda }(T{x}^{*},T{y}^{*}).$$

(30)

Since:

$$\begin{array}{cc}\hfill M(x^{*},y^{*})& \\ \hfill =\max \{& {\omega }_{\lambda }(x^{*},y^{*}),\frac{{\omega }_{\lambda }(x^{*},T{x}^{*})+{\omega }_{\lambda }(y^{*},T{y}^{*})}{2}-{\omega }_{\lambda }(A,B),\hfill \\ & \frac{{\omega }_{\lambda }(x^{*},T{y}^{*})+{\omega }_{\lambda }(y^{*},T{x}^{*})}{2}-{\omega }_{\lambda }(A,B)\}\hfill \\ \hfill =\max \{& {\omega }_{\lambda }(x^{*},y^{*}),0,\frac{{\omega }_{\lambda }(x^{*},T{y}^{*})+{\omega }_{\lambda }(y^{*},T{x}^{*})}{2}-{\omega }_{\lambda }(A,B)\}\hfill \\ \hfill \le \max \{& {\omega }_{\lambda }(x^{*},y^{*}),0,\hfill \\ & \frac{{\omega }_{\lambda }(x^{*},T{x}^{*})+{\omega }_{\lambda }(T{x}^{*},T{y}^{*})+{\omega }_{\lambda }(y^{*},T{y}^{*})+{\omega }_{\lambda }(T{y}^{*},T{x}^{*})}{2}-{\omega }_{\lambda }(A,B)\}\hfill \\ \hfill \le \max \{& {\omega }_{\lambda }(x^{*},y^{*}),0,\hfill \\ & \frac{{\omega }_{\lambda }(A,B)+{\omega }_{\lambda }(x^{*},y^{*})+{\omega }_{\lambda }(A,B)+{\omega }_{\lambda }(y^{*},x^{*})}{2}-{\omega }_{\lambda }(A,B)\}\hfill \\ \hfill ={\omega }_{\lambda }(x^{*}& ,y^{*}).\hfill \end{array}$$

Furthermore:

$$\begin{array}{cc}\hfill N(x^{*},y^{*}& ,\theta )\hfill \\ & =\theta ({\omega }_{\lambda }(x^{*},T{x}^{*})-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(y^{*},T{y}^{*})-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x^{*},T{y}^{*})-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(y^{*},T{x}^{*})-{\omega }_{\lambda }(A,B))\hfill \\ & =\theta ({\omega }_{\lambda }(A,B)-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(A,B)-{\omega }_{\lambda }(A,B),\hfill \\ & {\omega }_{\lambda }(x^{*},T{y}^{*})-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(y^{*},T{x}^{*})-{\omega }_{\lambda }(A,B))\hfill \\ & =\theta \left(0,0,{\omega }_{\lambda }(x^{*},T{y}^{*})-{\omega }_{\lambda }(A,B),{\omega }_{\lambda }(y^{*},T{x}^{*})-{\omega }_{\lambda }(A,B)\right)=0.\hfill \end{array}$$

As T is a Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )$-contractive mapping and $\frac{1}{2}{\omega }_{\lambda }^{*}(x^{*},T{x}^{*})=0\le {\omega }_{\lambda }(x^{*},y^{*})$ and $\alpha (x^{*},y^{*})\ge 1$, then, we obtain:

$$\begin{array}{cc}\hfill {\omega }_{\lambda }(x^{*},y^{*})& \le {\omega }_{\lambda }(T{x}^{*},T{y}^{*})\hfill \\ & \le \beta \left(M(x^{*},y^{*})\right)M(x^{*},y^{*})+\gamma \left(N(x^{*},y^{*},\theta )\right)N(x^{*},y^{*},\theta )\hfill \\ & =\beta \left({\omega }_{\lambda }(x^{*},y^{*})\right){\omega }_{\lambda }(x^{*},y^{*})\hfill \\ & <{\omega }_{\lambda }(x^{*},y^{*}),\hfill \end{array}$$

which is a contradiction. This completes the proof of the theorem. ☐

If in Theorem 3, we take $\beta \left(t\right)=r$ where $r\in [0,1)$ and $\gamma \left(t\right)=L$ where $L\ge 0$, then we obtain the following best proximity point result.

Corollary 1.

Let $(A,B)$ be a pair of nonempty subsets of a non-Archimedean modular metric space $X_{\omega }$ with ω regular, such that A is complete and $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Let $T:A\to B$ be a non-self mapping, such that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$ and for all $x,y\in A$ with $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $\alpha (x,y)\ge 1$; one has:

$${\omega }_{\lambda }(Tx,Ty)\le rM(x,y)+LN(x,y,\theta )$$

where $r\in [0,1)$, $L\ge 0$ and $\theta \in \mathrm{\Theta }.$ Suppose that the pair $(A,B)$ has the weak $P_1$-property and the following assertions hold:

(i) 

T is a triangular α-proximal admissible mapping,

(ii) 

there exist elements $x_0$ and $x_1$ in $A_0^{\lambda }$ for all $\lambda >0$, such that:

$${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)\mathit{and}\alpha (x_0,x_1)\ge 1.$$

(iii) 

if $\left\{x_n\right\}$ is a sequence in A, such that $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}$ with $x_n\to x\in A$ as $n\to \infty$, then $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}.$

Then, there exists an $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$ Further, the best proximity point is unique if, for every $x,y\in A$, such that ${\omega }_{\lambda }(x,Tx)={\omega }_{\lambda }(A,B)={\omega }_{\lambda }(y,Ty)$, we have: $\alpha (x,y)\ge 1$.

If in Corollary 1 we take, $\theta (t_1,t_2,t_3,t_4)=\min \{t_1,t_2,t_3,t_4\}$, we obtain the following best proximity result.

Corollary 2.

Let $(A,B)$ be a pair of nonempty subsets of a non-Archimedean modular metric space $X_{\omega }$ with ω regular, such that A is complete and $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Let $T:A\to B$ be a non-self mapping, such that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$ and for all $x,y\in A$ with $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $\alpha (x,y)\ge 1$; we have:

$${\omega }_{\lambda }(Tx,Ty)\le rM(x,y)+LN(x,y)$$

where $r\in [0,1)$, $L\ge 0$,

$$\begin{array}{cc}\hfill M(x,y)=& \max \{{\omega }_{\lambda }(x,y),\frac{{\omega }_{\lambda }(x,Tx)+{\omega }_{\lambda }(y,Ty)}{2}-{\omega }_{\lambda }(A,B),\hfill \\ & \frac{{\omega }_{\lambda }(x,Ty)+{\omega }_{\lambda }(y,Tx)}{2}-{\omega }_{\lambda }(A,B)\}\hfill \end{array}$$

and:

$$N(x,y)=\min \left\{{\omega }_{\lambda }(x,Tx),{\omega }_{\lambda }(y,Ty),{\omega }_{\lambda }(x,Ty),{\omega }_{\lambda }(y,Tx)\right\}-{\omega }_{\lambda }(A,B).$$

Suppose that the pair $(A,B)$ has the weak $P_{\lambda }$-property and the following assertions hold:

(i) 

T is a triangular α-proximal admissible mapping,

(ii) 

there exist elements $x_0$ and $x_1$ in $A_0^{\lambda }$ for all $\lambda >0$, such that:

$${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)\mathit{and}\alpha (x_0,x_1)\ge 1.$$

(iii) 

if $\left\{x_n\right\}$ is a sequence in A, such that $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}$ with $x_n\to x\in A$ as $n\to \infty$, then $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}.$

Then, there exists an $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$ Further, the best proximity point is unique if, for every $x,y\in A$, such that ${\omega }_{\lambda }(x,Tx)={\omega }_{\lambda }(A,B)={\omega }_{\lambda }(y,Ty)$, we have $\alpha (x,y)\ge 1$.

The following example illustrates our results.

Example 1.

Consider the space $X={\mathbb{R}}^2$ endowed with the non-Archimedean modular metric $\omega :X\times X\to (0,+\infty )$ given by:

$${\omega }_{\lambda }((x_1,x_2),(y_1,y_2))=\frac{1}{\lambda }(|x_1-y_1|+|x_2-y_2|),$$

for all $(x_1,x_2),(y_1,y_2)\in X$. Define the sets:

$$A=\left\{\right(1,0),(4,5),(5,4\left)\right\}\cup (-\infty ,-1]\times (-\infty ,-1]$$

and:

$$B=\left\{\right(0,0),(0,4),(4,0\left)\right\}\cup [10,\infty )\times [10,\infty )$$

so that ${\omega }_{\lambda }(A,B)=\frac{1}{\lambda }$, $A_0^{\lambda }=\left\{(1,0)\right\}$, $B_0^{\lambda }=\left\{(0,0)\right\}$ for all $\lambda >0$, and the pair $(A,B)$ has the weak $P_{\lambda }$-property. Furthermore, let $T:A\to B$ be defined by:

$$T(x_1,x_2)=\left\{\begin{array}{cc}(10x_1^2,15x_2^4)\hfill & \mathrm{if}{x}_1,x_2\in (-\infty ,-1],\hfill \\ (x_1,0)\hfill & \mathrm{if}{x}_1,x_2\notin (-\infty ,-1]\mathrm{with}{x}_1\le x_2,\hfill \\ (0,x_2)\hfill & \mathrm{if}{x}_1,x_2\notin (-\infty ,-1]\mathrm{with}{x}_1>x_2.\hfill \end{array}\right.$$

Notice that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$.

Now, consider the function $\beta :[0,+\infty )\to [0,1)$ given by:

$$\beta \left(t\right)=\left\{\begin{array}{cc}0\hfill & \mathrm{if}t=0,\hfill \\ \frac{\ln (1+t)}{t}\hfill & \mathrm{if}0<t\le 1,\hfill \\ \frac{8}{9}\hfill & \mathrm{if}1<t\le 10,\hfill \\ \frac{10}{11}\hfill & \mathrm{if}t>10,\hfill \end{array}\right.$$

and note that $\beta \in \mathcal{F}$. Furthermore, define $\alpha :X\times X\to [0,\infty )$ by:

$$\alpha (x,y)=\left\{\begin{array}{cc}2,\hfill & x,y\in \left\{\right(1,0),(4,5),(5,4\left)\right\}\hfill \\ \frac{1}{4},\hfill & \mathrm{otherwise}.\hfill \end{array}\right..$$

Clearly, ${\omega }_{\lambda }((1,0),T(1,0))={\omega }_{\lambda }(A,B)=\frac{1}{\lambda }$ and $\alpha \left(\right(1,0),(1,0\left)\right)\ge 1$.

Assume that $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $\alpha (x,y)\ge 1$, for some $x,y\in A$. Then:

$$\left\{\begin{array}{cc}x=(1,0),y=(4,5)\hfill & \mathrm{or}\hfill \\ x=(1,0),y=(5,4)\hfill & \mathrm{or}\hfill \\ y=(1,0),x=(4,5)\hfill & \mathrm{or}\hfill \\ y=(1,0),x=(5,4).\hfill & \end{array}\right.$$

Since ${\omega }_{\lambda }(Tx,Ty)={\omega }_{\lambda }(Ty,Tx)$ and $M(x,y)=M(y,x)$ for all $x,y\in A$, without any loss of generality, we can assume that:

$$(x,y)=\left(\right(1,0),(4,5\left)\right)\mathrm{or}(x,y)=\left(\right(1,0),(5,4\left)\right).$$

Now, we want to distinguish the following cases:

(i)

if $(x,y)=\left(\right(1,0),(4,5\left)\right)$, then:

$${\omega }_{\lambda }(T(1,0),T(4,5))=\frac{4}{\lambda }\le \frac{8}{9}·\frac{8}{\lambda }=\beta \left(M((1,0),(4,5))\right)\left[M((1,0),(4,5))\right];$$

(ii)

if $(x,y)=\left(\right(1,0),(5,4\left)\right)$, then:

$${\omega }_{\lambda }(T(1,0),T(5,4))=4\le \frac{8}{9}·\frac{8}{\lambda }=\beta \left(M((1,0),(5,4))\right)\left[M((1,0),(5,4))\right].$$

Consequently, we have:

$$\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)\mathrm{and}\alpha (x,y)\ge 1\Rightarrow {\omega }_{\lambda }(Tx,Ty)\le \beta \left(M(x,y)\right)\left[M(x,y)\right]$$

and hence, T is a Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )$-contractive mapping with $\gamma \left(t\right)=0.$ Let:

$$\left\{\begin{array}{c}\alpha (x,y)\ge 1\hfill \\ {\omega }_{\lambda }(u,Tx)={\omega }_{\lambda }(A,B)=\frac{1}{\lambda }\hfill \\ {\omega }_{\lambda }(v,Ty)={\omega }_{\lambda }(A,B)=\frac{1}{\lambda },\hfill \end{array}\right.$$

then:

$$\left\{\begin{array}{c}x,y\in \left\{\right(1,0),(4,5),(5,4\left)\right\}\hfill \\ {\omega }_{\lambda }(u,Tx)={\omega }_{\lambda }(A,B)=\frac{1}{\lambda }\hfill \\ {\omega }_{\lambda }(v,Ty)={\omega }_{\lambda }(A,B)=\frac{1}{\lambda },\hfill \end{array}\right.$$

and so, $u=v=(1,0)$. i.e., $\alpha (u,v)\ge 1.$ Furthermore, assume that $\alpha (x,y)\ge 1$ and $\alpha (y,z)\ge 1$. Then, $x,y,z\in \left\{\right(1,0),(4,5),(5,4\left)\right\}$, i.e., $\alpha (x,z)\ge 1.$ Therefore, T is a triangular $\alpha -$proximal admissible mapping. Moreover, if $\left\{x_n\right\}$ is a sequence, such that $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ and $x_n\to x$ as $n\to +\infty$, then $\left\{x_n\right\}\subseteq \{(1,0),(4,5),(5,4)\}$, and hence, $x\in \left\{\right(1,0),(4,5),(5,4\left)\right\}.$ Consequently, $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}.$ Hence, as you see, all of the conditions of Theorem 3 hold true, and T has a unique best proximity point. Here, $x=(1,0)$ is the unique best proximity point of T.

If in Theorem 3, we take $\alpha (x,y)=1$ for all $x,y\in A$, then we can deduce the following corollary.

Corollary 3.

Let $(A,B)$ be a pair of nonempty subsets of a non-Archimedean modular metric space $X_{\omega }$ with ω regular, such that A is complete and $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Let $T:A\to B$ be a non-self mapping, such that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$, and there exists $\beta \in \mathcal{F}$ and $\theta \in \mathrm{\Theta }$, such that $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ implies:

$${\omega }_{\lambda }(Tx,Ty)\le \beta \left(M(x,y)\right)M(x,y)+\gamma \left(N(x,y,\theta )\right)N(x,y,\theta ).$$

Suppose that the pair $(A,B)$ has the weak $P_{\lambda }$-property. Then, there exists a unique $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$

We investigate the Suzuki-type result of Zhang et al. [16] in the setting of non-Archimedean modular metric space as follows:

Corollary 4.

Let $(A,B)$ be a pair of nonempty and closed subsets of a complete non-Archimedean modular metric space $X_{\omega }$ with ω regular, such that $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Let $T:A\to B$ be a non-self mapping, such that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$, and there exists $r\in [0,1)$, such that $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ implies:

$${\omega }_{\lambda }(Tx,Ty)\le r{\omega }_{\lambda }(x,y)$$

for all $x,y\in A$. Suppose that the pair $(A,B)$ has the weak $P_{\lambda }$-property. Then there exists a unique point $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$

Corollary 5.

(Suzuki-type result of Suzuki [21]) Let $(A,B)$ be a pair of nonempty and closed subsets of a complete non-Archimedean modular metric space $X_{\omega }$ with ω regular, such that $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Let $T:A\to B$ be a non-self mapping, such that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$, and there exists $r\in [0,1)$, such that $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ implies:

$${\omega }_{\lambda }(Tx,Ty)\le r\left[\frac{{\omega }_{\lambda }(x,Tx)+{\omega }_{\lambda }(y,Ty)}{2}-{\omega }_{\lambda }(A,B)\right]$$

(31)

for all $x,y\in A$. Suppose that the pair $(A,B)$ has the weak $P_{\lambda }$-property. Therefore, there exists a unique point $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$

Corollary 6.

Let $(A,B)$ be a pair of nonempty subsets of a non-Archimedean modular metric space $X_{\omega }$ with ω regular, such that A is complete and $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Let $T:A\to B$ be a non-self mapping, such that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$, and there exists $r\in [0,1)$, such that $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ implies:

$${\omega }_{\lambda }(Tx,Ty)\le r\left[\frac{{\omega }_{\lambda }(x,Ty)+{\omega }_{\lambda }(y,Tx)}{2}-{\omega }_{\lambda }(A,B)\right]$$

(32)

for all $x,y\in A_0$. Suppose that the pair $(A,B)$ has the weak $P_{\lambda }$-property. Then, there exists a unique point $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$

3. Best Proximity Point Results in Metric Spaces Endowed with a Graph

Consistent with Jachymski [22], let $X_{\omega }$ be a modular metric space, and $\Delta$ denotes the diagonal of the Cartesian product $X_{\omega }\times X_{\omega }$. Assume that G is a directed graph, such that the set $V\left(G\right)$ of its vertices coincides with $X_{\omega }$ and the set $E\left(G\right)$ of its edges contains all loops, i.e., $E\left(G\right)\supseteq \Delta$. We suppose that G has no parallel edges. We identify G with the pair $\left(V\right(G),E(G\left)\right)$. Furthermore, we may handle G as a weighted graph (see [23], p. 309) by assigning to every edge the distance between its vertices. If x and y are vertices in a graph G, then a path in G from x to y of length N $(N\in \mathbb{N})$ is a sequence ${\left\{x_i\right\}}_{i=0}^N$ of $N+1$ vertices, such that $x_0=x,x_N=y$ and $(x_{i-1},x_i)\in E\left(G\right)$ for $i=1,\dots ,N.$ The foremost fixed point result in this area was given by Jachymski [22].

Definition 9

(Reference [22]). Let $(X,d)$ be a modular metric space endowed with a graph G.We say that a self-mapping $T:X\to X$ is a Banach G-contraction or simply a G-contraction if T preserves the edges of G, that is:

$$forallx,y\in X,(x,y)\in E\left(G\right)⟹(Tx,Ty)\in E\left(G\right)$$

and T decreases the weights of the edges of G in the following way:

$$\exists \alpha \in (0,1)suchthatforallx,y\in X,(x,y)\in E\left(G\right)⟹d(Tx,Ty)\le \alpha d(x,y).$$

We define the following notion for modular metric spaces.

Definition 10.

Let $X_{\omega }$ be a modular metric space endowed with a graph G. We say that a self-mapping $T:X\to X$ is a Banach G-contraction or simply a G-contraction if T preserves the edges of G, that is:

$$forallx,y\in X,(x,y)\in E\left(G\right)⟹(Tx,Ty)\in E\left(G\right)$$

and T decreases the weights of the edges of G in the following way:

$$\exists \alpha \in (0,1)suchthatforallx,y\in X,(x,y)\in E\left(G\right)⟹{\omega }_{\lambda }(Tx,Ty)\le \alpha {\omega }_{\lambda }(x,y).$$

Definition 11.

Let A and B be two nonempty subsets of a non-Archimedean modular metric space $X_{\omega }$ endowed with a graph G and $A_0\ne \mathrm{\varnothing }$. A mapping $T:A\to B$ is said to be a Suzuki-type $G-(\beta ,\theta ,\gamma )$-contractive mapping if there exists $\beta \in \mathcal{F}$ and $\theta \in \mathrm{\Theta }$, such that for all $x,y\in A$ with $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $(x,y)\in E\left(G\right)$, one has:

$${\omega }_{\lambda }(Tx,Ty)\le \beta \left(M(x,y)\right)M(x,y)+\gamma \left(N(x,y,\theta )\right)N(x,y,\theta )$$

(33)

and:

$$\left\{\begin{array}{c}(x,y)\in E\left(G\right)\hfill \\ {\omega }_{\lambda }(u,Tx)={\omega }_{\lambda }(A,B)\hfill \\ {\omega }_{\lambda }(v,Ty)={\omega }_{\lambda }(A,B)\hfill \end{array}\right.⟹(u,v)\in E\left(G\right).$$

Theorem 4.

Let A and B be two nonempty subsets of a non-Archimedean modular metric space $X_{\omega }$ with ω regular endowed with a graph G, such that A is complete and $A_0^{\lambda }$ is nonempty for all $\lambda >0$. Assume that T is a Suzuki-type $G-(\beta ,\theta ,\gamma )$-contractive mapping satisfying the following assertions:

(i) 

$T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$, and the pair $(A,B)$ satisfies the weak P-property,

(ii) 

$(x,y)\in E\left(G\right)$ and $(y,z)\in E\left(G\right)$ implies $(x,z)\in E\left(G\right)$,

(iii) 

there exist elements $x_0$ and $x_1$ in $A_0^{\lambda }$ for all $\lambda >0$, such that:

$${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)\mathit{and}(x_0,x_1)\in E\left(G\right).$$

(iv) 

if $\left\{x_n\right\}$ is a sequence in A, such that $(x_n,x_{n+1})\in E\left(G\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ with $x_n\to x\in A$ as $n\to \infty$, then $(x_n,x)\in E\left(G\right)$ for all $n\in \mathbb{N}.$

Then, there exists an $x^{*}$ in A, such that ${\omega }_{\lambda }(x^{*},T{x}^{*})={\omega }_{\lambda }(A,B)$ for all $\lambda >0.$

Proof. 

Define $\alpha :X\times X\to [0,+\infty )$ with:

$$\alpha (x,y)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}(x,y)\in E\left(G\right)\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

At first, we show that T is a triangular $\alpha$-proximal admissible mapping. For this goal, assume:

$$\left\{\begin{array}{c}\alpha (x,y)\ge 1\hfill \\ {\omega }_{\lambda }(u,Tx)={\omega }_{\lambda }(A,B)\hfill \\ {\omega }_{\lambda }(v,Ty)={\omega }_{\lambda }(A,B).\hfill \end{array}\right.$$

Therefore, we have:

$$\left\{\begin{array}{c}(x,y)\in E\left(G\right)\hfill \\ {\omega }_{\lambda }(u,Tx)={\omega }_{\lambda }(A,B)\hfill \\ {\omega }_{\lambda }(v,Ty)={\omega }_{\lambda }(A,B).\hfill \end{array}\right.$$

Since T is a Suzuki-type $G-(\beta ,\theta ,\gamma )$-contractive mapping, we get $(u,v)\in E\left(G\right)$, that is $\alpha (u,v)\ge 1$. Furthermore, let $\alpha (x,z)\ge 1$ and $\alpha (z,y)\ge 1$, then $(x,z)\in E\left(G\right)$ and $(z,y)\in E\left(G\right)$. Consequently, from (iii), we deduce that $(x,y)\in E\left(G\right)$, that is, $\alpha (x,y)\ge 1.$ Thus, T is a triangular $\alpha$-proximal admissible mapping with $T\left(A_0\right)\subseteq B_0$. Now, assume that, $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $\alpha (x,y)\ge 1$. Then, $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $(x,y)\in E\left(G\right)$. As T is a Suzuki-type $G-(\beta ,\theta ,\gamma )$-contraction, then we get:

$${\omega }_{\lambda }(Tx,Ty)\le \beta \left(M(x,y)\right)M(x,y)+\gamma \left(N(x,y,\theta )\right)N(x,y,\theta ),$$

and so, T is a Suzuki-type $(\alpha ,\beta ,\theta ,\gamma )$-contractive mapping. From (iii), there exist $x_0,x_1\in A_0$, such that ${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)$ and $(x_0,x_1)\in E\left(G\right)$, that is ${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)$ and $\alpha (x_0,x_1)\ge 1.$ Hence, all of the conditions of Theorem 3 are satisfied, and so, T has a best proximity point.  ☐

4. Best Proximity Point Results in Partially-Ordered Metric Spaces

The existence of best proximity points in partially-ordered metric spaces has been investigated in recent years by many authors (see, [24] and the references therein). In this section, we introduce a new notion of Suzuki-type ordered $(\beta ,\theta ,\gamma )$-contractive mapping and investigate the existence of the best proximity points for such mappings in partially-ordered non-Archimedean modular metric spaces by using the weak $P_{\lambda }$-property.

Definition 12.

Let $X_{\omega }$ be a partially-ordered modular metric space. We say that a non-self-mapping $T:A\to B$ is proximally ordered-preserving if and only if, for all $x_1,x_2,u_1,u_2\in A$:

$$\begin{array}{c}\hfill \begin{array}{c}\left\{\begin{array}{c}{x}_1⪯x_2\hfill \\ {\omega }_{\lambda }(u_1,T{x}_1)={\omega }_{\lambda }(A,B)\hfill \\ {\omega }_{\lambda }(u_2,T{x}_2)={\omega }_{\lambda }(A,B)\hfill \end{array}\right.⟹u_1⪯u_2.\hfill \end{array}\end{array}$$

Definition 13.

Let A and B be two nonempty closed subsets of a partially-ordered modular metric space $X_{\omega }$ and $A_0\ne \mathrm{\varnothing }$. A mapping $T:A\to B$ is said to be a Suzuki-type ordered $(\beta ,\theta ,\gamma )$-contractive mapping if there exists $\beta \in \mathcal{F}$ and $\theta \in \mathrm{\Theta }$, such that for all $x,y\in A$ with $\frac{1}{2}{\omega }_{\lambda }^{*}(x,Tx)\le {\omega }_{\lambda }(x,y)$ and $x⪯y$, we have:

$${\omega }_{\lambda }(Tx,Ty)\le \beta \left(M(x,y)\right)M(x,y)+\gamma \left(N(x,y,\theta )\right)N(x,y,\theta ).$$

Theorem 5.

Let A and B be two nonempty closed subsets of a partially-ordered non-Archimedean modular metric space with ω regular, such that A is complete, $A_0^{\lambda }$ is nonempty for all $\lambda >0$ and the pair $(A,B)$ has the weak $P_{\lambda }$-property. Assume that $T:A\to B$ satisfies the following conditions:

(i) 

T is proximally ordered-preserving, such that $T\left(A_0^{\lambda }\right)\subseteq B_0^{\lambda }$ for all $\lambda >0$,

(ii) 

there exist elements $x_0,x_1\in A_0$, such that:

$${\omega }_{\lambda }(x_1,T{x}_0)={\omega }_{\lambda }(A,B)\mathit{and}{x}_0⪯x_1,$$

(iii) 

T is a Suzuki-type ordered $(\beta ,\theta ,\gamma )$-contractive mapping,

(iv) 

if $\left\{x_n\right\}$ is an increasing sequence in A converging to $x\in A,$ then $x_n⪯x$ for all $n\in \mathbb{N}.$

Then, T has a best proximity point.