Some Extended Results of Common Fixed Point Theorems via Enhanced Categories of Contractive Mappings in Dd* - Symmetric Spaces

Definition 2.1:

²⁶ A; ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric on $\mathcal{X}\ne \varnothing$ is a map ${\mathcal{D}}_{\mathcal{d}}^{\ast }:\mathcal{X}\times \mathcal{X}\times \mathcal{X}\to [0,\infty )$ (s. t) $\forall {\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast }\in \mathcal{X}$ , the next axioms are satisfied:

In this case, ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ is called ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric and $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ is called symmetric space.

Example 2.2:

²⁶ Presume that $\mathcal{X}=[0,1]$ equipped with ${\mathcal{D}}^{\ast }$ -Symmetric map described through ${\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast })={({\mathcal{x}}^{\ast }-{\mathcal{y}}^{\ast })}^2+{({\mathcal{y}}^{\ast }-{\mathcal{z}}^{\ast })}^2+{({\mathcal{z}}^{\ast }-{\mathcal{x}}^{\ast })}^2,\forall {\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast }\in \mathcal{X}$ . Here, $({\mathcal{D}}_{\mathcal{d}}^{\ast },\mathcal{X})$ are ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric space.

Definition 2.3:

²⁶ Assume that $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ is a ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric space, therefore a sequence $\left\{{\mathcal{x}}_{\mathrm{s}}^{\ast }\right\}$ is called ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -converges to ${\mathcal{x}}^{\ast }\in \mathcal{X}$ iff ${\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}_{\mathrm{s}}^{\ast },{\mathcal{x}}_{\mathrm{s}}^{\ast },{\mathcal{x}}^{\ast })={\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}^{\ast },{\mathcal{x}}^{\ast },{\mathcal{x}}_{\mathrm{s}}^{\ast })⟶0\mathrm{as}\mathrm{s}⟶\infty .$

Remark 2.4:

Equivalent to Wilson axiom’s²⁷ in ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric space as

$\left({\mathrm{W}}_1\right)$ Given $\left\{{\mathcal{x}}_{\mathrm{s}}^{\ast }\right\}$ , ${\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast }\in \mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}_{\mathrm{s}}^{\ast },{\mathcal{x}}^{\ast },{\mathcal{x}}^{\ast })⟶0$ with ${\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathrm{x}}_{\mathrm{s}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{y}}^{\ast })⟶0⟹{\mathcal{x}}^{\ast }={\mathcal{y}}^{\ast }.$

$\left({\mathrm{W}}_2\right)$ Given $\left\{{\mathcal{x}}_{\mathrm{s}}^{\ast }\right\}$ & $\left\{{\mathcal{y}}_{\mathrm{s}}^{\ast }\right\}$ , such that ${\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast }\in \mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathrm{x}}_{\mathrm{s}}^{\ast },{\mathcal{x}}^{\ast },{\mathcal{x}}^{\ast })⟶0$ with ${\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathrm{x}}_{\mathrm{s}}^{\ast },{\mathcal{y}}_{\mathrm{s}}^{\ast },{\mathcal{y}}_{\mathrm{s}}^{\ast })⟶0⟹$ ${\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{y}}_{\mathrm{s}}^{\ast },{\mathcal{x}}^{\ast },{\mathcal{x}}^{\ast })⟶0$ .

$\left({\mathrm{W}}_3\right)$ Assume that $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ are complete, ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric space. For an arbitrary sequence $\left\{{\mathcal{x}}_{\mathrm{s}}^{\ast }\right\}\text{in}\mathcal{X}$ , we have $\underset{\mathrm{s},\mathcal{r}\to \infty }{lim}{\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}_{\mathrm{s}}^{\ast },{\mathcal{x}}_{\mathcal{r}}^{\ast },{\mathcal{x}}_{\mathcal{r}}^{\ast })=0$ , iff $\underset{\mathrm{s}\to \infty }{lim}{\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}_{\mathrm{s}}^{\ast },{\mathcal{x}}_{\mathrm{s}+1}^{\ast },{\mathcal{x}}_{\mathrm{s}+1}^{\ast })=0$ .

The next Lemma is analogue of Lemma²⁷ in $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ :

Lemma 2.5:

Each ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ , describes a symmetric ${\mathcal{d}}_{{\mathcal{D}}_{\mathcal{d}}^{\ast }}$ on $\mathcal{X}$ via:

${\mathcal{d}}_{{\mathcal{D}}_{\mathcal{d}}^{\ast }}({\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast })={\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{y}}^{\ast })+{\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{y}}^{\ast },{\mathcal{x}}^{\ast },{\mathcal{x}}^{\ast }),\forall {\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast }\in \mathcal{X}$ , that is:

${\mathcal{d}}_{{\mathcal{D}}_{\mathcal{d}}^{\ast }}({\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast })=2{\mathcal{D}}_{\mathcal{d}}^{\ast }({\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{y}}^{\ast }),\forall {\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast }\in \mathcal{X}.$

Definition 2.6:

²⁸ Assume that $\mathfrak{F}\mathfrak{,}\mathcal{G}$ : $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })⟶(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ are single-valued self-maps If $\mathfrak{F}\left({\mathcal{x}}^{\ast }\right)=\mathcal{G}\left({\mathcal{x}}^{\ast }\right)={\mathcal{w}}^{\ast }$ for ${\mathcal{x}}^{\ast }\in \mathcal{X}$ , thus ${\mathcal{x}}^{\ast }$ is said to be a coincidence point of $\mathfrak{F}$ & $\mathcal{G}$ , and ${\mathcal{w}}^{\ast }$ is the point of coincidence of $\mathfrak{F}$ and $\mathcal{G}$ .

Definition 2.7:

²⁹ Assume that $\mathfrak{F}\mathfrak{,}\mathcal{G}:(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })\to (\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ are single-valued self-maps. If $\mathfrak{F}\left({\mathcal{x}}^{\ast }\right)=\mathcal{G}\left({\mathrm{x}}^{\ast }\right)={\mathrm{x}}^{\ast }$ for ${\mathcal{x}}^{\ast }\in \mathcal{X},$ consequently ${\mathcal{x}}^{\ast }$ is called the common fixed point (C.F.P) of $\mathfrak{F}$ and $\mathrm{G}$ .

Definition 2.8:

²⁶ A pair $(\mathcal{F},\mathcal{G})$ of self-maps of ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ are called weakly compatible maps (Concisely, W.C.M) if they are commute at their coincidence points.

Definition 2.9:

⁶ Self-maps $\mathcal{F}$ and $\mathcal{G}$ of $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ are called occasionally weakly compatible iff $\exists \mathcal{x}\in \mathcal{X}$ which is coincidence point of $\mathcal{F}$ and $\mathcal{G}$ which $\mathcal{F}$ and $\mathcal{G}$ are commute.

Remark 2.10:

It is clear that each pair of weakly compatible maps is (O.W.C) map, other than the opposite, and not generally correct.

Definition 2.11:

³⁰ $\mathrm{\mu }:[0,\infty )\to [0,\infty )$ is said to be an altering distance if $\mathrm{\mu }$ is continuous and nondecreasing with $\mathrm{\mu }\left(\mathcal{t}\right)=0$ iff $\mathcal{t}=0$ .

Definition 2.12:

¹⁰ Two self-maps $\mathrm{\eta }$ and $\mathrm{\xi }$ of $(\mathcal{X},\mathrm{d})$ satisfy the property of the common limit in the range of $\mathrm{\xi }$ , indicated via $\left({\mathrm{C}\mathbb{L}\mathrm{R}}_{\mathrm{\xi }}\right)$ , if $\exists \left\{{\mathcal{x}}_{\mathrm{s}}^{\ast }\right\}$ in $\mathcal{X}$ (s. t) $\underset{s\to \infty }{\mathit{lim}}\eta {\mathcal{x}}_s^{\ast }=\underset{s\to \infty }{\mathit{lim}}\xi {\mathcal{x}}_s^{\ast }=\xi \mathcal{p}$ for $\mathcal{p}\in \mathcal{X}$ .

Definition 2.13:

¹¹ The pairs $(\mathrm{\xi },\mathrm{\eta })$ and $(\mathcal{F},\mathcal{G})$ of self-maps in Symmetric space $(\mathcal{X},\mathrm{d})$ are called gratify property of common limit range related to the maps $\mathrm{\eta }$ and $\mathcal{G}$ , indicated via $\left({\mathrm{C}\mathbb{L}\mathrm{R}}_{(\mathrm{\eta },\mathcal{G})}\right)$ , if $\exists \left\{{\mathcal{x}}_{\mathrm{s}}^{\ast }\right\}$ and $\left\{{\mathcal{y}}_{\mathrm{s}}^{\ast }\right\}$ (s. t) $\underset{s\to \infty }{\mathit{lim}}\xi {\mathcal{x}}_s^{\ast }=\underset{s\to \infty }{\mathit{lim}}\eta {\mathcal{x}}_s^{\ast }=\underset{s\to \infty }{\mathit{lim}}\mathcal{F}{\mathcal{y}}_s^{\ast }=\underset{s\to \infty }{\mathit{lim}}\mathcal{G}{\mathcal{y}}_s^{\ast }=\mathcal{t}$ with $\mathcal{t}=\mathrm{\eta }\mathcal{p}=\mathcal{Gq}$ for some $\mathcal{t},\mathcal{p},\mathcal{q}\in \mathcal{X}$ .

Remark 2.14:

¹⁰ If $\mathrm{\xi }=\mathcal{F}$ and $\mathrm{\eta }=\mathcal{G}$ , in that case Definition-2.12 reduces to $\left({\mathrm{C}\mathbb{L}\mathrm{R}}_{\mathrm{\eta }}\right)$ property.

Remark 3.1:

Throughout this manuscript, assume that the next properties are hold:

(i) Let $ℂ=\left\{\begin{array}{c}\psi /\psi :[0,\infty )\to [0,\infty )\text{is lower semi continuous and}\\ \text{nondecreasing}\mathrm{map}\text{where}\psi \left(\mathcal{t}\right)=0⟺\mathcal{t}=0\end{array}\right\}$

(ii) Assume that; $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}$ and $\mathcal{J}$ are three pairs of self-maps of a ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ , where

Wherever, $\mu$ is altering distance map, $\psi \in ℂ$ with

Theorem 3.2:

Let $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}\text{and}\mathcal{J}$ are three pairs of self-maps of a ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric space $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ satisfy inequality (1) and the following conditions:

(i) If $(S,\mathfrak{F}\mathcal{F})$ and $(\mathbb{Q},\mathcal{GJ})$ gratify ${\mathrm{C}\mathbb{L}\mathrm{R}}_{(\mathfrak{F}\mathcal{F},\mathcal{GJ})}$ property.

(ii) If $(S,\mathfrak{F}\mathcal{F})$ and $(\mathbb{Q},\mathcal{GJ})$ are weakly compatible maps.

In this case, the maps $S,\mathbb{Q},\mathfrak{F}\mathcal{F}$ and $\mathcal{GJ}$ have a unique (C. F. P) in $\mathcal{X}$ . Additionally, $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}$ and $\mathcal{J}$ contain a unique(C. F. P) provided that the pairs of maps $(\mathfrak{F},\mathcal{F}),(S,\mathfrak{F}),(S,\mathcal{F}),(\mathcal{G},\mathcal{J}),(\mathbb{Q},\mathcal{G})$ and $(\mathbb{Q},\mathcal{J})$ are commuting.

Proof:

Because $(S,\mathfrak{F}\mathcal{F})$ and $(\mathbb{Q},\mathcal{GJ})$ satisfy $\left({\mathrm{C}\mathbb{L}\mathrm{R}}_{(\mathfrak{F}\mathcal{F},\mathcal{GJ})}\right)$ property, we can discover two sequences $\left\{{\mathcal{x}}_{\mathrm{s}}^{\ast }\right\}$ and $\left\{{\mathcal{y}}_{\mathrm{s}}^{\ast }\right\}$ in $\mathcal{X}$ (s. t), $\underset{s\to \infty }{\mathit{lim}}S{\mathcal{x}}_s^{\ast }=\underset{s\to \infty }{\mathit{lim}}\mathfrak{F}\mathcal{F}{\mathcal{x}}_s^{\ast }=\underset{s\to \infty }{\mathit{lim}}\mathbb{Q}{\mathcal{y}}_s^{\ast }=\underset{s\to \infty }{\mathit{lim}}\mathcal{GJ}{\mathcal{y}}_s^{\ast }=\mathcal{t}$ with $\mathcal{t}=\mathfrak{F}\mathcal{Fp}=\mathcal{GJq}$ for some $\mathcal{t},\mathcal{p},\mathcal{q}\in \mathcal{X}$ .

Firstly verify $\mathfrak{F}\mathcal{Fp}=S\mathcal{p}$ . From inequality (1) and ${\mathcal{x}}^{\ast }=\mathcal{p},{\mathcal{y}}^{\ast }={\mathcal{z}}^{\ast }={\mathcal{y}}_{\mathrm{s}}^{\ast }$ , we get

Such that

Consequently

Let $s\to \infty$ in inequality (2), $\mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathcal{t})\right)\le \mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathcal{t})\right)-\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathcal{t})\right)$ which implies $\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathcal{t})\right)=0$ , as a result ${\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathcal{t})=0$ , that is, $S\mathcal{p}=\mathcal{t}=\mathfrak{F}\mathcal{Fp}$ , illustrating $\mathcal{p}$ is the coincidence point of $S$ and $\mathfrak{F}\mathcal{F}$ .

Because $(S,\mathfrak{F}\mathcal{F})$ is weakly compatible, we have $S\left(\mathfrak{F}\mathcal{F}\right)\mathcal{p}=\left(\mathfrak{F}\mathcal{F}\right)S\mathcal{p}$ , therefore $S\mathcal{t}=\mathfrak{F}\mathcal{\text{Ft}}$ .

Secondly verify: $\mathbb{Q}\mathcal{q}=\mathcal{GJq}$ . From inequality (1) (s. t), ${\mathcal{x}}^{\ast }={\mathcal{x}}_{\mathrm{s}}^{\ast },{\mathcal{y}}^{\ast }={\mathcal{z}}^{\ast }=\mathcal{q}$ , get that

Where

Therefore,

Selecting limit as $s\to \infty$ in inequality (3),

This implies that $\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathbb{Q}\mathcal{q})\right)=0$ , as a result ${\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathbb{Q}\mathcal{q})=0$ ,that is, $\mathbb{Q}\mathcal{q}=\mathcal{t}=\mathcal{GJq}$ , explaining $\mathcal{q}$ is the coincidence point of $\mathbb{Q}$ and $\mathcal{GJ}$ . As $(\mathbb{Q},\mathcal{GJ})$ is (W.C.M), we obtain $\mathbb{Q}\left(\mathcal{GJ}\right)\mathcal{q}=\left(\mathcal{GJ}\right)\mathbb{Q}\mathcal{q}$ , and consequently $\mathbb{Q}\mathcal{t}=\mathcal{GJt}$ .

Third verify: $S\mathcal{t}=\mathfrak{F}\mathcal{\text{Ft}}=\mathcal{t}$ . From inequality (1) (s. t) ${\mathcal{x}}^{\ast }=\mathcal{t},{\mathcal{y}}^{\ast }={\mathcal{z}}^{\ast }=\mathcal{q}$ , obtain

Such that

Consequently

Thus, $\mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{t},S\mathcal{t},\mathcal{t})\right)\le \mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{t},S\mathcal{t},\mathcal{t})\right)-\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{t},S\mathcal{t},\mathcal{t})\right)⟹$ $\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{t},S\mathcal{t},\mathcal{t})\right)=0$ , as a result ${\mathcal{D}}^{\ast }(S\mathcal{t},S\mathcal{t},\mathcal{t})=0$ , that is, $S\mathcal{t}=\mathcal{t}$ . Consequently, $S\mathcal{t}=\mathfrak{F}\mathcal{\text{Ft}}=\mathcal{t}$ .

Finally verify: $\mathbb{Q}\mathcal{t}=\mathcal{GJt}=\mathcal{t}.$ From inequality-(1) (s. t) ${\mathcal{x}}^{\ast }=\mathcal{p},{\mathcal{y}}^{\ast }={\mathcal{z}}^{\ast }=\mathcal{t}$ , we obtain

Such that

Consequently, $\mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathbb{Q}\mathcal{t})\right)\le \mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathbb{Q}\mathcal{t})\right)-\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathbb{Q}\mathcal{t})\right)⟹\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathbb{Q}\mathcal{t})\right)=0.$ Therefore ${\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathbb{Q}\mathcal{t})=0$ , that is, $\mathbb{Q}\mathcal{t}=\mathcal{t}$ . Consequently, $\mathbb{Q}\mathcal{t}=\mathcal{GJt}=\mathcal{t}$ , thus $S\mathcal{t}=\mathfrak{F}\mathcal{\text{Ft}}=\mathbb{Q}\mathcal{t}=\mathcal{GJt}=\mathcal{t}$ , demonstrating $\mathcal{t}$ is (C. F. P) of $S,\mathbb{Q},\mathfrak{F}\mathcal{F}$ and $\mathcal{JG}$ .

Uniqueness: Assume that $(\mathcal{n}\ne \mathcal{t})$ is different (C. F. P) for $S,\mathbb{Q},\mathfrak{F}\mathcal{F}$ and $\mathcal{JG}$ . Consequently, we have $S\mathcal{n}=\mathfrak{F}\mathcal{Fn}=\mathbb{Q}\mathcal{n}=\mathcal{GJn}=\mathcal{n}$ . Now, from inequality (1) with ${\mathcal{x}}^{\ast }=\mathcal{t},{\mathcal{y}}^{\ast }={\mathcal{z}}^{\ast }=\mathcal{n}$ , obtain

Such that

Therefore,

Consequently, ${\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},\mathcal{n})=0$ , that is, $\mathcal{t}=\mathcal{n}$ . Thus, $S,\mathfrak{F}\mathcal{F},\mathbb{Q},\mathcal{JG}$ have unique common fixed point in $\mathcal{X}$ .

Next, we shall verify $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}$ and $\mathcal{J}$ have unique common fixed point.

Because, $(\mathfrak{F},\mathcal{F}),(S,\mathfrak{F}),(S,\mathcal{F})$ are commuting, so $\mathfrak{F}\mathcal{t}=\mathfrak{F}\left(\mathfrak{F}\mathcal{\text{Ft}}\right)=\left(\mathfrak{F}\mathcal{F}\right)\mathfrak{F}\mathcal{t}$ with $\mathfrak{F}\mathcal{t}=\mathfrak{F}\left(S\mathcal{t}\right)=S\left(\mathfrak{F}\mathcal{t}\right)$ .

In addition, $\mathcal{\text{Ft}}=\mathcal{F}\left(\mathfrak{F}\mathcal{\text{Ft}}\right)=\left(\mathfrak{F}\mathcal{F}\right)\mathcal{\text{Ft}}$ with $\mathcal{\text{Ft}}=\mathcal{F}\left(S\mathcal{t}\right)=S\left(\mathcal{\text{Ft}}\right)$ .

This illustrates $\mathfrak{F}\mathcal{t}$ and $\mathcal{\text{Ft}}$ are (C. F. Ps) of $\mathfrak{F}\mathcal{F}$ and $S$ . Therefore, via the uniqueness of common fixed point, we obtain $\mathfrak{F}\mathcal{t}=\mathcal{\text{Ft}}=\mathcal{t}$ .

Likewise, because $(\mathcal{G},\mathcal{J}),(\mathbb{Q},\mathcal{G})$ and $(\mathbb{Q},\mathcal{J})$ are commuting, obtain $\mathcal{Gt}=\mathcal{G}\left(\mathbb{Q}\mathcal{t}\right)=\left(\mathbb{Q}\mathcal{G}\right)\mathcal{t}$ with $\mathcal{Gt}=\mathcal{G}\left(\mathcal{GJt}\right)=\mathcal{GJ}\left(\mathcal{Gt}\right)$ .

Moreover, $\mathcal{Jt}=\mathcal{J}\left(\mathbb{Q}t\right)=\mathbb{Q}\left(\mathcal{Jt}\right)$ and $\mathcal{Jt}=\mathcal{J}\left(\mathcal{GJt}\right)=\mathcal{GJ}\left(\mathcal{Jt}\right)$ . This explains $\mathcal{Gt}$ and $\mathcal{Jt}$ are common fixed point of $\mathcal{GJ}$ and $\mathbb{Q}$ . Consequently, via the uniqueness of common fixed point, obtain $\mathcal{Gt}=\mathcal{Jt}=\mathcal{t}$ . As a result, $S\mathcal{t}=\mathbb{Q}\mathcal{t}=\mathfrak{F}\mathcal{t}=\mathcal{\text{Ft}}=\mathcal{Gt}=\mathcal{Jt}=\mathcal{t}$ , verifying $\mathcal{t}$ is unique common fixed point to $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}$ and $\mathcal{J}.$

Corollary 3.3:

If $S,\mathbb{Q},\mathcal{F}$ and $\mathcal{J}$ are two pairs of self-maps of ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -Symmetric $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ satisfying the following conditions:

(i) If $\mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S{\mathcal{x}}^{\ast },\mathbb{Q}{\mathcal{y}}^{\ast },\mathbb{Q}{\mathcal{z}}^{\ast })\right)\le \mu \left(\mathbb{L}({\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast })\right)-\psi \left(\mathbb{L}({\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast })\right)\forall {\mathcal{x}}^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast }\in \mathcal{X}.$ (s. t), $\mathrm{\mu }$ alters the distance map with $\psi :{ℝ}^{+}\to {ℝ}^{+}$ , is a lower semi-continuous and nondecreasing map (s. t) $\psi \left(\mathcal{t}\right)=0$ iff $\mathcal{t}=0$ with

(ii) If $(S,\mathcal{F})$ and $(\mathbb{Q},\mathcal{J})$ satisfy $\left({\mathrm{C}\mathbb{L}\mathrm{R}}_{(\mathcal{F},\mathcal{J})}\right)$ property.

(iii) If $(S,\mathcal{F})$ and $(\mathbb{Q},\mathcal{J})$ are (W. C. M). In this case, $S,\mathbb{Q},\mathcal{F}$ and $J$ contained unique (C. F. P) in $\mathcal{X}$ .

Proof:

Its follows immediately from Theorem-3.2, via putting $\mathcal{G}=\mathfrak{F}={\mathbb{I}}_{\mathcal{X}}$ (Identity map).

Lemma 3.4:

Assume that; $\mathcal{X}\ne \varnothing$ , with $\xi$ and $\eta$ are (O. W. C) maps. If $\xi$ and $\eta$ include a unique point of coincidence ${\mathcal{w}}^{\ast }=\xi \mathcal{t}=\eta \mathcal{t}$ for $\mathcal{t}\in \mathcal{X}$ , then ${\mathcal{w}}^{\ast }$ is unique (C. F. P) of $\xi$ and $\eta$ .

Theorem 3.5:

If $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}$ and $J$ are three pairs of self-maps in ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -symmetric $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ satisfies the following conditions:

(i) If $\mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S{\mathcal{x}}^{\ast },\mathbb{Q}{\mathcal{y}}^{\ast },\mathbb{Q}{\mathcal{z}}^{\ast })\right)\le \mu \left(\mathcal{M}(x^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast })\right)-\psi \left(\mathcal{M}(x^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast })\right),\forall x^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast }\in \mathcal{X}$ , (s. t), $\mathsf{\mu }$ μ alters the distance map with $\psi :{ℝ}^{+}\to {ℝ}^{+}$ , is a lower semi-continuous and nondecreasing map (s. t) $\psi \left(\mathcal{t}\right)=0$ iff $\mathcal{t}=0$ with

(ii) If $(S,\mathfrak{F}\mathcal{F})$ and $(\mathbb{Q},\mathcal{GJ})$ are (O. W. C) maps. In this case, the maps $S,\mathbb{Q},\mathfrak{F}\mathcal{F}$ and $\mathcal{GJ}$ have a unique (C. F. P) in $\mathcal{X}$ . Additional $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}$ and $\mathcal{J}$ include unique (C.F.P), provided the pairs of maps $(\mathfrak{F},\mathcal{F}),(S,\mathfrak{F}),(S,\mathcal{F}),(\mathcal{G},\mathcal{J}),(\mathbb{Q},\mathcal{G})$ and $(\mathbb{Q},\mathcal{J})$ are commuting.

Proof:

Assuming that $(S,\mathfrak{F}\mathcal{F})$ and $(\mathbb{Q},\mathcal{GJ})$ are (O. W. C), we can discover $\mathcal{p}$ and $\mathcal{q}$ in $\mathcal{X}$ (s. t) $S\mathcal{p}=\mathfrak{F}\mathcal{Fp}={\mathcal{p}}^1$ and $S\left(\mathfrak{F}\mathcal{F}\right)\mathcal{p}=\left(\mathfrak{F}\mathcal{F}\right)S\mathcal{p}$ with $\mathbb{Q}\mathcal{q}=\mathcal{GJq}={\mathcal{q}}^1$ and $\mathbb{Q}\left(\mathcal{GJ}\right)\mathcal{q}=\left(\mathcal{GJ}\right)\mathbb{Q}\mathcal{q}$ .

Firstly, verify: $S\mathcal{p}=\mathbb{Q}\mathcal{q}$ . From condition (i), (s. t) ${\mathcal{x}}^{\ast }=\mathcal{p},{\mathcal{y}}^{\ast }={\mathcal{z}}^{\ast }=\mathcal{q}$ , we get

Such that

Consequently, $\mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathbb{Q}\mathcal{q})\right)\le \mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathbb{Q}\mathcal{q})\right)-\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathbb{Q}\mathcal{q})\right)$ which implies that $\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathbb{Q}\mathcal{q})\right)=0$ , as a result ${\mathcal{D}}_{\mathcal{d}}^{\ast }(S\mathcal{p},S\mathcal{p},\mathbb{Q}\mathcal{q})=0$ , that is, $S\mathcal{p}=\mathbb{Q}\mathcal{q}$ . Therefore, $S\mathcal{p}=\mathfrak{F}\mathcal{Fp}=\mathbb{Q}\mathcal{q}=\mathcal{GJq}$ .

Suppose that ${\mathcal{p}}^1$ is another point where ${S\mathcal{p}}^1=\mathfrak{F}\mathcal{F}{\mathcal{p}}^1$ . In this case, obtain from condition (i), that ${S\mathcal{p}}^1=\mathfrak{F}\mathcal{F}{\mathcal{p}}^1=\mathbb{Q}\mathcal{q}=\mathcal{GJq}$ . Thus $,S\mathcal{p}={S\mathcal{p}}^1$ , that is, $\mathcal{p}={\mathcal{p}}^1$ , demonstrating $S$ and $\mathfrak{F}\mathcal{F}$ include a unique point of coincidence. Consequently, via Lemma-3.4, get $S$ and $\mathfrak{F}\mathcal{F}$ include unique (C.F.P), namely, $\mathcal{t}$ . Likewise, it can be verified $\mathbb{Q}$ and $\mathcal{GJ}$ include unique (C.F.P), say ${\mathcal{t}}^1$ .

Secondly verify, $\mathcal{t}={\mathcal{t}}^1$ . From condition (i), (s. t) ${\mathcal{x}}^{\ast }=\mathcal{t},{\mathcal{y}}^{\ast }={\mathcal{z}}^{\ast }={\mathcal{t}}^1$ , get

Where

Therefore, $\mu \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},{\mathcal{t}}^1)\right)\le \mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},{\mathcal{t}}^1)\right)-\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},{\mathcal{t}}^1)\right)⟹\psi \left({\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},{\mathcal{t}}^1)\right)=0$ , as a result ${\mathcal{D}}_{\mathcal{d}}^{\ast }(\mathcal{t},\mathcal{t},{\mathcal{t}}^1)=0$ , that is, $\mathcal{t}={\mathcal{t}}^1$ . Consequently, $S,\mathbb{Q},\mathfrak{F}\mathcal{F}$ and $\mathcal{GJ}$ are unique (C. F. P). The rest of the evidence is similar to of Theorem-3.2, for this reason, obtain $S\mathcal{t}=\mathbb{Q}\mathcal{t}=\mathfrak{F}\mathcal{t}=\mathcal{\text{Ft}}=\mathcal{Gt}=\mathcal{Jt}=\mathcal{t}$ , verifying $\mathcal{t}$ is unique(C.F.P) to $S,\mathbb{Q},\mathfrak{F},\mathcal{F},\mathcal{G}$ and $\mathcal{J}$ .

Corollary 3.6:

If $S,\mathbb{Q},\mathcal{F}$ and $\mathcal{J}$ are two pairs of self-maps of a ${\mathcal{D}}_{\mathcal{d}}^{\ast }$ -symmetric $(\mathcal{X},{\mathcal{D}}_{\mathcal{d}}^{\ast })$ satisfying the next conditions:

(i) If $\mathrm{\mu }\left({\mathcal{D}}_{\mathcal{d}}^{\ast }(S{\mathcal{x}}^{\ast },\mathbb{Q}{\mathcal{y}}^{\ast },\mathbb{Q}{\mathcal{z}}^{\ast })\right)\le \mu \left(\mathcal{M}(x^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast })\right)-\psi \left(\mathcal{M}(x^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast })\right),\forall x^{\ast },{\mathcal{y}}^{\ast },{\mathcal{z}}^{\ast }\in \mathcal{X}$ , (s. t), $\mathsf{\mu }$ μ altering distance map with $\psi :{ℝ}^{+}⟶{ℝ}^{+}$ , is lower semi-continuous and nondecreasing map (s. t) $\psi \left(\mathcal{t}\right)=0$ , iff $\mathcal{t}=0$ , with

(ii) If $(S,\mathcal{F})$ and $(\mathbb{Q},\mathcal{J})$ are (O. W. C) maps. In this case, maps $S,\mathbb{Q},\mathcal{F}$ and $\mathcal{J}$ have a unique common fixed point in $\mathcal{X}$ .

Proof:

This follows directly from Theorem-3.5, by putting $\mathfrak{F}=\mathcal{G}={\mathbb{I}}_{\mathcal{X}}$ (Identity map).