## 1 Introduction and Preliminaries
Now, to show the uniqueness, we assume there exists \(y\
eq\mu\) (\(M_{d}(\mu,y)>0\)) in \(\mathcal{P}\) such that the mapping \(T\) is an \(\alpha\)-\(\theta\)-Geraghty-Pata-Suzuki-type proximal contraction, so  
\[\frac{1}{1+\mu+\gamma}M_{d}^{*}(\mu,T\mu)\leq M_{d}^{*}(\mu,T\mu)=M_{d}(\mu,T \mu)-M_{d}(\mathcal{P},\mathcal{Q})=0<M_{d}(\mu,y).\]  
Thus, the above inequality can be written as,  
\[\frac{1}{1+\mu+\gamma}M_{d}^{*}(\mu,T\mu)\leq M_{d}(\mu,y),\]  
which implies  
\[M_{d}(\mu,y) \leq \frac{\alpha(\mu,y)}{\theta(\mu,y)}M_{d}(\mu,y) \tag{12}\] \[\leq \frac{(1-\epsilon)}{2}\beta(m(\mu,y))m(\mu,y)+\Lambda(\epsilon)^{ \mu}\psi(\epsilon)[1+\|\mu\|+\|y\|+\|y\|]^{\gamma}\]  
Taking the limit \(\epsilon\to 0\), then inequality (12) becomes  
\[M_{d}(\mu,y) \leq \frac{\beta(m(\mu,y))}{2}m(\mu,y) \tag{13}\] \[< \beta(m(\mu,y))m(\mu,y)\] (14) \[\leq m(\mu,y), \tag{14}\]  
where  
\[m(\mu,y) = \max\{M_{d}(\mu,y),M_{d}(\mu,T\mu)-M_{d}(\mathcal{P},\mathcal{Q} ),M_{d}(y,T\eta)-M_{d}(\mathcal{P},\mathcal{Q}),\] \[\frac{M_{d}(\mu,T\mu)+M_{d}(y,Ty)}{2}-M_{d}(\mathcal{P},\mathcal{ Q}),M_{d}(y,\mu)+M_{d}(\mu,T\mu)-M_{d}(\mathcal{P},\mathcal{Q}),\] \[\frac{1}{1+\mu+\gamma}M_{d}^{*}(\mu,T\mu)\leq M_{d}^{*}(\mu,T\mu)=M_{d}(\mu,T \mu)-M_{d}(\mathcal{P},\mathcal{Q})=0<M_{d}(\mu,y).\]0 \[\frac{1}{1+\mu+\gamma}M_{d}^{*}(\mu,T\mu)\leq M_{d}^{*}(\mu,T\mu)=M_{d}(\mu,T \mu)-M_{d}(\mathcal{P},\mathcal{Q})=0<M_{d}(\mu,y).\]1  
Then, inequality (14) becomes  
\[\frac{1}{1+\mu+\gamma}M_{d}^{*}(\mu,T\mu)\leq M_{d}^{*}(\mu,T\mu)=M_{d}(\mu,T \mu)-M_{d}(\mathcal{P},\mathcal{Q})=0<M_{d}(\mu,y).\]2  
which is a contradiction; therefore, \(\mu=y\), which shows that \(\mu\) is the unique best proximity point of mapping \(T\). **Example 1**. : _Consider the usual metric space \(\Omega=\mathbb{R}\). Let_  
\[\frac{1}{1+\mu+\gamma}M_{d}^{*}(\mu,T\mu)\leq M_{d}^{*}(\mu,T\mu)=M_{d}(\mu,T \mu)-M_{d}(\mathcal{P},\mathcal{Q})=0<M_{d}(\mu,y).\]3  
_be two closed subsets of \(\Omega\), where_  
\[\frac{1}{1+\mu+\gamma}M_{d}^{*}(\mu,T\mu)\leq M_{d}^{*}(\mu,T\mu)=M_{d}(\mu,T \mu)-M_{d}(\mathcal{P},\mathcal{Q})=0<M_{d}(\mu,y).\]4Define a mapping \(T:\mathcal{P}\rightarrow\mathcal{Q}\) as  
\[T\mu=\left\{ \begin{array}{l}\frac{2}{3};\text{ when }\mu\in\{\frac{1}{3},\frac{7}{3}\},\\ \frac{8}{3};\text{ when }\mu\in\{\frac{3}{3},\frac{5}{3}\}.\end{array} \right.\]  
Obviously, \(T(\mathcal{P}_{0})\subseteq\mathcal{Q}_{0}\). Further, suppose that \(\alpha(\mu,y)=2M_{d}(\mu,y)\) and \(\theta(\mu,y)=M_{d}(\mu,y)\). If we take \(\mu=\frac{1}{3}\), \(y=\frac{3}{3}\in\mathcal{P}_{0}\), then \(\alpha(\mu,y)=2(\frac{2}{3})>\frac{2}{3}=\theta(\mu,y)\), and we have  
\[\alpha(\mu,y)>\theta(\mu,y)\] \[M_{d}(\frac{3}{3},T_{3}^{1})=M_{d}(\mathcal{P},\mathcal{Q})=M_{d}( \frac{7}{3},T_{3}^{3})\quad\bigg{\}}\text{ implies }\alpha(\frac{3}{3},\frac{7}{3})>\theta(\frac{3}{3},\frac{7}{3}).\]  
Define \(\psi(s)=se^{s}\) for all \(s\geq 0\). If \(\mu=2\) and \(\gamma\in[0,2]\), assuming that \(\lambda=\gamma\), taking \(\epsilon=\frac{1}{10}\) and verifying that \(T\) satisfies the conditions of an \(\alpha\)-\(\theta\)-Geraghty-Pata-Suzuki-type proximal contraction, it is evident the mapping \(T\) satisfies the following condition:  
\[\frac{\alpha(u,v)}{\theta(u,v)}M_{d}(u,v)\leq\frac{(1-\epsilon)}{2}\beta(m( \mu,y))m(\mu,y)+\Lambda(\epsilon)^{\mu}\psi(\epsilon)[1+\|\mu\|+\|y\|+\|v\|]^ {\gamma} \tag{15}\]  
for all \(u,v,\mu\) and \(y\in\mathcal{P}_{0}\). For this setting and the following simple calculation, we have  
\[m(\frac{1}{3},\frac{3}{3}) = \max\{M_{d}(\frac{1}{3},\frac{3}{3}),M_{d}(\frac{1}{3},\frac{2}{3 })-\frac{1}{3},M_{d}(\frac{3}{3},\frac{8}{3})-\frac{1}{3},\frac{M_{d}(\frac{ 1}{3},\frac{2}{3})+M_{d}(\frac{3}{3},\frac{8}{3})}{2}\] \[-\frac{1}{3},M_{d}(\frac{3}{3},\frac{2}{3})-\frac{1}{3},\frac{M_{ d}(\frac{1}{3},\frac{8}{3})+M_{d}(\frac{3}{3},\frac{2}{3})}{2}-\frac{1}{3}\}= \frac{4}{3}.\]  
Define  
\[\beta(u)=\left\{ \begin{array}{l}1,\text{ if }u=0,\\ \frac{\ln(1+u)+1}{u},\text{ if }u>0.\end{array}