## 3 Optimal Coincidence Point Results
Furthermore, since \(T\mu_{1}\in T(\mathcal{P}_{0})\subseteq\mathcal{Q}_{0}\) and \(g\mu_{2}\in\mathcal{P}_{0}\subseteq g(\mathcal{P}_{0})\), it follows that there exists an element \(\mu_{2}\in\mathcal{P}_{0}\) such that \(M_{d}(g\mu_{2},T\mu_{1})=M_{d}(\mathcal{P},\mathcal{Q})\). As the mapping \(T\) is \(\alpha\)-\(\theta\)-proximal admissible, then \(\alpha(g\mu_{1},g\mu_{2})\geq\theta(g\mu_{1},g\mu_{2})\). Further, the mapping \(g\) satisfies the \((\alpha,\theta)_{R}\)-property, which implies \(\alpha(\mu_{1},\mu_{2})\geq\theta(\mu_{1},\mu_{2})\), where \(g\mu_{1},g\mu_{2}\in\mathcal{P}_{0}\). Since \(\lambda\geq 1\), and \(\gamma\in[0,  
which is a contradiction; hence,  
\[m_{g}(\mu_{n+m},\mu_{n})\leq\max\{M_{d}(g\mu_{n+m},g\mu_{n}),M_{d}(g\mu_{n+1},g \mu_{n+m+1})\}=M_{d}(g\mu_{n+m},g\mu_{n}).\]  
Choose  
\[\epsilon=1-(\frac{n}{n+1})^{\mu}\leq\frac{\mu}{n+1}.\]  
Then, from (23),  
\[P_{n+1} \leq (n+1)^{\mu}((\frac{n}{n+1})^{\mu})\beta(M_{d}(g\mu_{n+m},g\mu_{n} ))M_{d}(g\mu_{n+m},g\mu_{n})+\] \[(n+1)^{\mu}\Lambda(\frac{\mu}{n+1})^{\mu}\psi(\frac{\mu}{n+1})[1 +3c_{n}]^{\gamma}\] \[\leq n^{\mu}M_{d}(g\mu_{n+m},g\mu_{n})+\Lambda(\mu)^{\mu}\psi(\frac{ \mu}{n+1})[1+3c_{n}]^{\gamma},\]  
and we get  
\[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]0 \[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]1 \[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]2 \[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]3  
Since \(P_{0}=0\), this gives  
\[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]4  
After division by \(n^{\mu}\), we obtain that  
\[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]5  
On taking the limit as \(n\to+\infty\), we can show that \(\{g\mu_{n}\}\) is a Cauchy sequence. For each \(\epsilon\in[0,1]\), from inequality (18), we have  
\[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]6 \[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]7 \[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]8  
Since  
\[\frac{\alpha(g\mu,g\mu)}{\theta(g\mu,gy)}M_{d}(gu,gv)\leq\frac{(1-e)}{2}\beta (m_{g}(\mu,y))m_{g}(\mu,y)+\Lambda(\epsilon)^{\lambda}\psi(\epsilon)[1+\|\mu \|+\|y\|+\|v\|]^{\gamma},\]9  
and the mapping \(g\) satisfies the \((\alpha,\theta)_{R}\)-property, then we have  
\[m_{g}(\mu,y) = \max\{M_{d}(g\mu,gy),M_{d}^{*}(g\mu,T\mu),M_{d}^{*}(gy,Ty),\frac{ M_{d}^{*}(g\mu,T\mu)+M_{d}^{*}(gy,Ty)}{2},\]0  
for each \(\epsilon\in[0,1]\); hence, \(\{g\mu_{n}\}\subseteq\mathcal{P}_{0}\) is a Cauchy sequence in a complete metric space \((\Omega,M_{d})\), and every Cauchy sequence in a complete metric space is convergent. Since \(\mathcal{P}_{0}\) is closed, there exists some \(g\mu\in\mathcal{P}_{0}\subseteq g(\mathcal{P}_{0})\) such that \(g\mu_{n}\to g\mu\) in \(\mathcal{P}_{0}\).