# Local Convergence of an Efficient Multipoint Iterative Method in Banach Space
## 2 Local Convergence Analysis  
Certain real functions and parameters appearing in the local convergence analysis of Equation (2) are introduced. Set \(S=[0,\infty]\), let \(w_{0}:S\times S\to S\), \(w_{1}:S\to S\) be continuous and increasing functions with \(w_{0}(0,0)=0\). Suppose that equation  
\[w_{0}(w_{1}(x)x,x)=1 \tag{3}\]  
has at least one positive solution. Let \(\rho_{0}\) be the smallest such solution. Set \(S_{0}=[0,\rho_{0})\). Let also \(w:S_{0}\times S_{0}\to S\), \(v:S_{0}\to S\) and \(w_{2}:S_{0}\to S\) be continuous and increasing functions with \(w(0,0)=0\). Define functions \(p_{1}\) and \(\tilde{p}_{1}\) in the interval \(S_{0}\) by  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]  
and  
\[\tilde{p}_{1}(x)=p_{1}(x)-1,\]  
where \(a(x)=1-w_{0}(w_{1}(x)x,x)\). We have \(\bar{p}_{1}(0)=-1<0\) and \(\bar{p}_{1}(x)\to\infty\) as \(x\to\rho_{0}^{-}\). The intermediate value theorem implies that equation \(\bar{p}_{1}(x)=0\) has at least one solution in \((0,\rho_{0})\). Let \(r_{1}\) be the smallest such solution. Further assume that equation  
\[w_{0}\big{(}w_{1}(p_{1}(x)x)p_{1}(x)x,p_{1}(x)x\big{)}=1 \tag{4}\]  
possesses at least one positive solution. Denote by \(\rho_{1}\) the smallest such solution. Set \(\rho_{2}=\min\{\rho_{0},\rho_{1}\}\) and \(S_{1}=[0,\rho_{2})\). Define functions \(p_{2}\) and \(\bar{p}_{2}\) on the interval \(S_{1}\) by  
\[p_{2}(x)=\bigg{(}p_{1}(p_{1}(x)x)+\frac{d(x)}{a(x)b(x)}v(p_{1}(x)x)+\frac{1}{4} \big{[}4h(x)+5h^{2}(x)\big{]}\frac{v(p_{1}(x)x)}{a(x)}\bigg{)}p_{1}(x)\]  
and  
\[\bar{p}_{2}(x)=p_{2}(x)-1,\]  
where  
\[b(x) =1-w_{0}(w_{1}(p_{1}(x)x)p_{1}(x)x,p_{1}(x)x),\] \[d(x) =w_{0}(w_{1}(x)x,x)+w_{0}(w_{1}(p_{1}(x)x),p_{1}(x)x)\]  
and  
\[h(x)=\frac{d(x)}{a(x)}.\]  
Then, we also get \(\bar{p}_{2}(0)=-1\) and \(\bar{p}_{2}(x)\to\infty\) as \(x\to\rho_{2}^{-}\). Denote by \(r_{2}\) the smallest solution of equation \(\bar{p}_{2}(x)=0\) in \((0,\rho_{2})\). Assume that equation  
\[w_{0}\big{(}w_{1}(p_{2}(x)x)p_{2}(x)x,p_{2}(x)x\big{)}=1 \tag{5}\]  
possesses at least one positive solution. Denote by \(\rho_{3}\) the smallest such solution. Set \(\rho=\min\{\rho_{2},\rho_{3}\}\) and \(S_{2}=[0,\rho)\). Define functions \(p_{3}\) and \(\bar{p}_{3}\) on the interval \(S_{2}\) by  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]0  
and  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]1  
where  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]2  
and  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]3  
We obtain again \(\bar{p}_{3}(0)=-1\) and \(\bar{p}_{3}(x)\to\infty\) as \(x\to\rho^{-}\). Denote by \(r_{3}\) the smallest solution of equation \(\bar{p}_{3}(x)=0\) in \((0,\rho)\). Define parameter \(r\) by  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]4  
Next, we present the local convergence of Equation (2) using the conditions (A) and the notations mentioned above. **Theorem 1**. : _Suppose that the conditions (A) hold. Then, sequence \(\{u_{n}\}\) generated for \(u_{0}\in U(u_{*},r)-\{u_{*}\}\) is well defined, remains in \(U(u_{*},r)\) for each \(n=0,1,2......\) and converges to \(u_{*}\), so that_  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]5 \[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]6  
_and_  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]7  
_where the functions \(p_{i}\) are given previously and \(r\) is defined in Equation (6)._  
Proof. : We shall show estimates for Equations (17)-(19) using mathematical induction. Let \(u\in U(u_{*},r)-\{u_{*}\}.\) By \((a_{1})\), \((a_{2})\), \((a_{4})\) and Equation (6), we obtain in turn  
\[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]8 \[p_{1}(x)=\frac{w(w_{1}(x)x,x)}{a(x)}\]9 \[\tilde{p}_{1}(x)=p_{1}(x)-1,\]0 \[\tilde{p}_{1}(x)=p_{1}(x)-1,\]1  
where we also used that  
\[\tilde{p}_{1}(x)=p_{1}(x)-1,\]2  
so \(u+Q(u)-u_{*}\in U(u_{*},\bar{r})\). It follows from Equation (20) and the Banach perturbation lemma on invertible operators [8] that \([u+Q(u),u;F]^{-1}\in L(B,B)\) and  
\[\tilde{p}_{1}(x)=p_{1}(x)-1,\]3  
and \(y_{0}\), \(z_{0}\), \(u_{1}\) are well defined by Equation (2) for \(n=0\).