## 3 Semi-Local Convergence of [PERSON]'s Method
### Auxiliary Scalar Sequences  
From Conditions (A1)-(A2), we define \(\gamma=\varepsilon\mathpzc{f}\), \(b_{0}=\mathpzc{h}\) and \(\delta=\frac{\gamma}{(1+p)(1-\gamma)}\), where \(\varepsilon=\frac{K_{0}}{K}\in[0,1]\). Now, we define \(b_{1}=\frac{\delta^{p}}{1-\gamma}b_{0}\) and:  
\[a_{n}=\frac{b_{n}}{(1+p)(1-b_{n})},\quad n\geq 1, \tag{4}\]  
\[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]  
Observe that we consider the case \(b_{0}>0\), since if \(b_{0}=0\), a trivial problem results, as the solution of the equation \(F(x)=0\) is \(x_{0}\). Next, we prove the following four recurrence relations for Sequences (1), (4) and (5):  
\[\|\Gamma_{1}\|=\|[F^{\prime}(x_{1})]^{-1}\|\leq\frac{\|\Gamma_{0}\|}{1-\gamma}, \tag{6}\]  
\[\|x_{2}-x_{1}\|\leq\delta\|x_{1}-x_{0}\|, \tag{7}\]  
\[K\|\Gamma_{1}\|\|x_{2}-x_{1}\|^{p}\leq b_{1}, \tag{8}\]  
\[\|x_{2}-x_{0}\|\leq(1+\delta)\|x_{1}-x_{0}\|, \tag{9}\]  
provided that:  
\[x_{1}\in\Omega\qquad\text{and}\qquad\gamma<1. \tag{10}\]  
If \(x_{1}\in\Omega\), then:  
\[\|I-\Gamma_{0}F^{\prime}(x_{1})\|\leq\|\Gamma_{0}\|\|F^{\prime}(x_{0})-F^{ \prime}(x_{1})\|\leq K\beta\eta^{p}=\mathpzc{h}<1.\]Then, by the [PERSON] lemma on invertible operators, it follows that the operator \(\Gamma_{1}\) exists and:  
\[\|\Gamma_{1}\|\leq\frac{\|\Gamma_{0}\|}{1-\|I-\Gamma_{0}F^{\prime}(x_{1})\|}\leq \frac{\|\Gamma_{0}\|}{1-\gamma}.\]  
After that, from [PERSON]'s series and Sequence (1), we have:  
\[\|F(x_{1})\|=\left\|\int_{0}^{1}(F^{\prime}(x_{0}+\tau(x_{1}-x_{0}))-F^{\prime }(x_{0}))(x_{1}-x_{0})\,d\tau\right\|\leq\frac{K_{0}\eta^{p}}{1+p}\|x_{1}-x_{0}\|.\]  
Thus,  
\[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]0 \[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]1 \[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]2  
provided that \(\delta\leq 1\) and \(a_{1}<1\). Now, we prove in the same way as above the following four recurrence relations for Sequences (1), (4) and (5):  
\[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]3 \[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]4 (12) \[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]5 (13) \[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]6  
provided that:  
\[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]7  
In addition, we generalize the last recurrence relations to every point of Sequence (1), so that we can guarantee that (1) is a Cauchy sequence from them. For this, we analyze the scalar sequences defined in (4) and (5) in order to prove later the semi-local convergence of Sequence (1). For this, it suffices to see that (1) is a Cauchy sequence and (10) and (15) are true for all \(x_{n}\) and \(b_{n-1}\) with \(n\geq 3\). We begin by presenting a technical lemma. **Lemma 2**. : _If \(\gamma\leq\frac{1+p}{2+p}\) and \(b_{1}\) is such that_  
\[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]8  
_then:_  
* _the sequences_ \(\{b_{n}\}\) _and_ \(\{a_{n}\}\) _are strictly decreasing,_
* \(a_{n}<1\) _and_ \(b_{n}<1\)_, for all_ \(n\geq 1\)_._  
_If \(b_{1}=1-a_{1}^{p}<\frac{1+p}{2+p}\), then \(a_{n}=a_{1}<1\) and \(b_{n}=b_{1}<1\) for all \(n\geq 2\)._  
**Proof. ** We first consider the case in which \(b_{1}\) satisfies (16). Item (a) is proven by mathematical induction on \(n\). As \(b_{1}+a_{1}^{p}<1\), then \(b_{2}<b_{1}\) and \(a_{2}<a_{1}\). If we now suppose that \(b_{j}<b_{j-1}\) and \(a_{j}<a_{j-1}\), for all \(i=2,3,\ldots,n\), then:  
\[b_{n}=\frac{b_{n-1}a_{n-1}^{p}}{1-b_{n-1}},\quad n\geq 2. \tag{5}\]9  
As a result, the sequences \(\{a_{n}\}\) and \(\{b_{n}\}\) are strictly decreasing for all \(n\geq 2\). To see Item (b), we have \(a_{n}<a_{1}<1\) and \(b_{n}<b_{1}<1\), for all \(n\geq 2\), by Item (a) and the conditions given in (16). Second, if \(b_{1}=1-a_{1}^{p}\), then \(b_{n}=b_{1}=1-a_{1}^{p}<1\), for all \(n\geq 2\). Moreover, if \(b_{1}<\frac{1+p}{2+p}\), then we have \(a_{n}=a_{1}<1\), for all \(n\geq 2\).