# Convex \((\alpha,\beta)\)-Generalized Contraction and Its Applications in Matrix Equations
## 2 Preliminaries
: _Let \(X\) be a complete Busemann space, and \(\Phi:X\to X\) be a convex \((\alpha,\beta)\)-generalized contraction mapping. Then, \(\Phi\) has a unique fixed point in \(X\)._  
Proof. : Let \(\xi_{0}\in X\) and the following sequence be defined as follows:  
\[\xi_{n+1}=W(\xi_{n},\Phi(\xi_{n}),\alpha)\quad\text{ for all }n\in\mathbb{N} \cup\{0\} \tag{4}\]  
From the condition on space, we have  
\[\Omega(\xi_{n},\xi_{n+1})=\alpha\Omega(\xi_{n},\Phi(\xi_{n})). \tag{5}\]  
From (8) \(\lim_{n\to\infty}\Omega(\xi_{n},\Phi(\xi^{\dagger}))=0\). Therefore, \(\xi^{\dagger}\) is a fixed point of \(\Phi\). To prove the uniqueness, let \(q\) be another fixed point of \(\Phi\). Then,  
\[0<\Omega(\xi^{\dagger},q) = \Omega(\Phi(W(\xi^{\dagger},\Phi(\xi^{\dagger}),\alpha)\xi^{ \dagger},\Phi(q))\leq(1-\alpha)\beta\Omega(\xi^{\dagger},q)+\alpha\beta\Omega (\Phi(\xi^{\dagger}),q)\] \[= \beta\Omega(\xi^{\dagger},q)<\Omega(\xi^{\dagger},q)\]  
a contradiction unless \(\xi^{\dagger}=q\). Therefore, \(\Phi\) has a unique fixed point. Corollary 1: _Let \(X\) be a Banach space, and \(\Phi:X\to X\) be a convex \((\alpha,\beta)\)-generalized contraction mapping. Then, \(\Phi\) has a unique fixed point in \(X\)._  
**Example 1**. : _Let \(X=[0,4]\subset\mathbb{R}\) with the usual norm. Define \(\Psi:X\to X\) by_  
\[\Psi(\xi)= \begin{cases}0,&\text{if }\xi\
eq\ 4\\ 1,&\text{if }\xi=4.\end{cases} \]  
_First, we show that \(\Psi\) is convex \((\alpha,\beta)\)-generalized contraction mapping for \(\alpha=\dfrac{1}{2}\) and \(\beta=\dfrac{2}{3}\). We consider the following cases:_  
_(1)_ _If_ \(\xi\
eq 4\) _and_ \(\varrho\
eq 4\)_, then the condition is trivially satisfied._  
_(2)_ _If_ \(\xi\
eq 4\)_, and_ \(\varrho=4\)_, then_  
\[\|\Psi((1-\alpha)\xi+\alpha\Psi(\xi))-\Psi(\varrho)\| = \left\|\Psi\bigg{(}\bigg{(}1-\dfrac{1}{2}\bigg{)}\times\xi+\dfrac {1}{2}\Psi(\xi)\bigg{)}-\Psi(\varrho)\right\|\] \[= (1-\alpha)\beta\|4-\xi\|+4\alpha\beta\] \[= (1-\alpha)\beta\|\xi-\varrho\|+\alpha\beta\|\Psi(\xi)-\varrho\|.\]  
_(3)_ _If_ \(\xi=4\)_, and_ \(\varrho\
eq 4\)_, then_  
\[\|\Psi((1-\alpha)\xi+\alpha\Psi(\xi))-\Psi(\varrho)\| = \left\|\Psi\bigg{(}\bigg{(}1-\dfrac{1}{2}\bigg{)}\times 4+\dfrac {1}{2}\Psi(4)\bigg{)}-\Psi(\varrho)\right\|\] \[= \left\|\Psi\bigg{(}\dfrac{5}{2}\bigg{)}\right\|=0<(1-\alpha)\beta \|\xi-\varrho\|+\alpha\beta\|\Psi(\xi)-\varrho\|.\]  
_(4)_ _If_ \(\xi=4\)_, and_ \(\varrho=4\)_, then_  
\[\|\Psi((1-\alpha)\xi+\alpha\Psi(\xi))-\Psi(\varrho)\| = \left\|\Psi\bigg{(}\bigg{(}1-\dfrac{1}{2}\bigg{)}\times 4+ \dfrac{1}{2}\Psi(4)\bigg{)}-\Psi(4)\right\|\] \[= \left\|\Psi\bigg{(}\dfrac{5}{2}\bigg{)}-1\right\|=1\] \[= (1-\alpha)\beta\|\xi-\varrho\|+\alpha\beta\|\Psi(\xi)-\varrho\|.\]  
_On the other hand, \(\Psi\) is not continuous mapping, thus \(\Psi\) is not contraction mapping (even nonexpansive mapping)._  
**Example 2** ([20], Example 1.3.1). : _Suppose \(X=\mathbb{R}\) and \(\Phi:X\to X\) is defined by_  
\[\Psi(\xi)= \begin{cases}0,&\text{if }\xi\in(-\infty,2]\\ -\frac{1}{3},&\text{if }\xi\in(2,+\infty).\end{cases} \]  
_Although mapping \(\Phi\) is discontinuous and does not satisfy the contraction mapping property, mapping \(\Phi^{2}\) can be considered a contraction mapping._If the Picard-Banach contraction mapping principle cannot be used, we may find the following fixed-point theorem to be a useful option. This theorem is discussed in many places. Theorem 2 ([20], Theorem 1.3.2): _Let \(X\) be a complete metric space and \(\Phi:X\to X\) a mapping. If \(\exists\) a \(N\in\mathbb{N}\) such that \(\Phi^{N}\) is a contraction, then \(F(\Phi)=\{\xi^{*}\}\)._  
Theorem 3: _Let \(X\) be a complete Busemann space, \(G:X\to X\) be a mapping and \(\exists\) a \(N\in\mathbb{N}\) such that \(G^{N}\) is a convex \((\alpha,\beta)\)-generalized contraction mapping.