# An Improved Convergence Analysis of a Multi-Step Method with High-Efficiency Indices
## 2 Semi-Local Analysis  
We will define scalar majorizing sequences for our semi-local analysis [4]. For \(\alpha_{0}=0\), \(\beta_{0}\geq 0\) and \(L_{0},L_{1}\geq 0\), define the scalar sequences \(\{\alpha_{n}\}\) and \(\{\beta_{n}\}\) such that  
\[\alpha_{n+1} = \beta_{n}+\frac{L_{1}}{1-L_{0}\alpha_{n}}\bigg{[}\frac{1}{1-L_{0} \alpha_{n}}\bigg{(}2+L_{0}\beta_{n}+\frac{L_{0}L_{1}(\beta_{n}-\alpha_{n})^{2} }{4(1-L_{0}\alpha_{n})}\bigg{)}\] \[+\frac{3}{2(1-L_{0}\alpha_{n})}\bigg{(}L_{1}(\beta_{n}-\alpha_{n} )+\frac{L_{1}^{2}(\beta_{n}-\alpha_{n})^{2}}{4(1-L_{0}\alpha_{n})}\bigg{)}+ \frac{3}{2}\bigg{]}(\beta_{n}-\alpha_{n})^{2},\] \[\delta_{n+1} = \frac{L_{1}}{2}(\alpha_{n+1}-\alpha_{n})^{2}+(1+L_{0}\alpha_{n}) (\alpha_{n+1}-\beta_{n})\] and \[\beta_{n+1} = \alpha_{n+1}+\frac{\delta_{n+1}}{1-L_{0}\alpha_{n+1}}.\]  
**Lemma 1**. : _Assume there exists \(\mu\geq 0\) such that_  
\[L_{0}\alpha_{n}<1\text{ and }\alpha_{n}\leq\mu\ \forall n\in\mathbb{N}\cup\{0\}. \tag{5}\]  
_Then, the sequences \(\{\alpha_{n}\}\) and \(\{\beta_{n}\}\) defined by (5) are convergent to some \(\lambda\in[\beta_{0},\mu]\) and \(0\leq\alpha_{n}\leq\beta_{n}\leq\alpha_{n+1}\leq\lambda\)._  
Proof. : The scalar sequences \(\{\alpha_{n}\}\) and \(\{\beta_{n}\}\) are non-decreasing and bounded above \(\mu\). Hence, \(\exists\ \lambda\in[\beta_{0},\mu]\) such that \(\{\alpha_{n}\}\) and \(\{\beta_{n}\}\) converges to \(\lambda\). Let \(U(s,r)\) be the ball centered at \(s\) with radius \(r\) and \(\bar{U}(s,r)\) be its closure. For the convergence analysis, we use the following assumptions. * \(\exists\) an initial point \(s_{0}\in\Omega\) such that \(\|\mathcal{E}^{\prime}(s_{0})^{-1}\mathcal{E}(s_{0})\|<\beta_{0}\). * There exist an operator \(\mathcal{G}\in\mathcal{B}(X,Y)\) (the set of all bounded linear operators from \(X\) to \(Y\)) and constant \(L_{0}>0\) with \[\|\mathcal{G}^{-1}(\mathcal{E}^{\prime}(s)-\mathcal{G})\|\leq L_{0}\|s-s_{0}\|, \ \forall s\in\Omega.\] Set \(\Omega_{1}=\Omega\cap U(s_{0},\frac{1}{L_{0}})\). * There exists a constant \(L_{1}>0\) with \[\|\mathcal{G}^{-1}(\mathcal{E}^{\prime}(s)-\mathcal{E}^{\prime}(t))\|\leq L_ {1}\|s-t\|,\ \ \forall s,t\in\Omega_{1}.\] \((a_{4})\): \(\bar{U}(s_{0},\lambda)\subset\Omega\). From assumption \((a_{2})\), we have \(\forall\ s\in\Omega_{1}\)  
\[\|\mathcal{G}^{-1}(\mathcal{E}^{\prime}(s)-\mathcal{G})\| \leq L_{0}\|s-s_{0}\|<1 \tag{6}\]  
and hence by Banach Lemma (BL) on invertible operators [4], we get \(\mathcal{E}^{\prime}(s)^{-1}\in\mathcal{B}(Y,X)\) and  
\[\|\mathcal{E}^{\prime}(s)^{-1}\mathcal{G}\|\leq\frac{1}{1-L_{0}\|s-s_{0}\|}, \ \forall\ s\in\Omega_{1}. \tag{7}\]  
Further, we will be using the following mean value theorem (MVT) [4]:\[\mathcal{AE}(\
u_{1})-\mathcal{AE}(\
u_{2})=\int_{0}^{1}\mathcal{AE}^{\prime}(\
u_{ 2}+\tau(\
u_{1}-\
u_{2}))d\tau(\
u_{1}-\
u_{2}),\ \forall\
u_{1},\
u_{2}\in\Omega. \tag{8}\]  
Next, the main semi-local result uses the conditions \((a_{1})-(a_{4})\). Theorem 1: _Under the assumptions \((a_{1})-(a_{4})\), the sequence \(\{s_{n}\}\) defined by method (3) with \(s_{0}\) satisfies_  
\[+\frac{3}{2(1-L_{0}\alpha_{n})}\bigg{(}L_{1}(\beta_{n}-\alpha_{n} )+\frac{L_{1}^{2}(\beta_{n}-\alpha_{n})^{2}}{4(1-L_{0}\alpha_{n})}\bigg{)}+ \frac{3}{2}\bigg{]}(\beta_{n}-\alpha_{n})^{2},\]0  
_and_  
\[+\frac{3}{2(1-L_{0}\alpha_{n})}\bigg{(}L_{1}(\beta_{n}-\alpha_{n} )+\frac{L_{1}^{2}(\beta_{n}-\alpha_{n})^{2}}{4(1-L_{0}\alpha_{n})}\bigg{)}+ \frac{3}{2}\bigg{]}(\beta_{n}-\alpha_{n})^{2},\]1  
_Moreover, the sequence \(s_{n}\) converges to some \(s^{*}\in\bar{U}(s_{0},\lambda)\) with \(\mathcal{AE}(s^{*})=0\)._  
Proof: We will be using mathematical induction to prove the result. Using \((a_{1})\) and first step of (3),  
\[+\frac{3}{2(1-L_{0}\alpha_{n})}\bigg{(}L_{1}(\beta_{n}-\alpha_{n} )+\frac{L_{1}^{2}(\beta_{n}-\alpha_{n})^{2}}{4(1-L_{0}\alpha_{n})}\bigg{)}+ \frac{3}{2}\bigg{]}(\beta_{n}-\alpha_{n})^{2},\]2  
Thus, \(t_{0}\in U(s_{0},\lambda)\) and (9) is true for \(n=0\).