We describe the local convergence analysis of the algorithm (4) in this segment. Let the notations for the closed and open balls with center c and radius ρ>0 be \(\bar {B}(c, \rho)\) and B(c,ρ) respectively. BL(Y,X) is the notation for the set of all bounded linear operators from Y to X. k0, k1 be two positive parameters with k0≤k1. In Theorems 1 and 2, we provide the local convergence analysis of the scheme (4).
Local convergence analysis of the method (4) under Lipschitz continuity condition
We define the function Φ1 on the interval \(\left [0, \frac {1}{k_{0}}\right)\) by
$$ \Phi_{1}(u)=\frac{k_{1}u}{2(1-k_{0} u)} $$
(5)
and the parameter
$$\eta_{1}=\frac{2}{2k_{0} + k_{1}} < \frac{1}{k_{0}}. $$
It is easy to observe that Φ1(η1)=1. Again, we define the functions Φ2 and Ψ2 on \(\left [0, \frac {1}{k_{0}}\right)\) by
$$ \Phi_{2}(u)= \frac{1}{3}\left(1+2\Phi_{1}(u)\right) $$
(6)
and
$$\Psi_{2}(u)=\Phi_{2}(u)-1. $$
Now, \(\Psi _{2}(0)=-\frac {2}{3}<0\) and \({ \underset {u \to \left (\frac {1}{k_{0}}\right)^{-}}{\lim } \Psi _{2}(u)=+\infty }\). The intermediate value theorem confirms the existence of the zeros of the function Ψ2(u) in \(\left (0,\frac {1}{k_{0}}\right)\). We denote the smallest zero of Ψ2(u) in \(\left (0, \frac {1}{k_{0}}\right)\) as η2. Again, we define Φ3 and Ψ3 on \(\left [0, \frac {1}{k_{0}}\right)\) by
$$ \Phi_{3}(u)= \frac{k_{0}}{4}(1+3\Phi_{2}(u))u $$
(7)
and
$$\Psi_{3}(u)=\Phi_{3}(u)-1. $$
Now, Ψ3(0)=−1<0 and \({ \underset {u \to \left (\frac {1}{k_{0}}\right)^{-}}{\lim } \Psi _{3}(u)=+\infty }\). So, The zeros of the function Ψ3(u) lies in \((0, \frac {1}{k_{0}})\). We denote the smallest zero of Ψ3(u) in \(\left (0, \frac {1}{k_{0}}\right)\) as η3. Lastly, we define Φ4 and Ψ4 on [0,η3) by
$$ \Phi_{4}(u)= \frac{k_{1}\left[1+\frac{3}{2}\Phi_{2}(u)\right]u}{2(1-\Phi_{3}(u))} $$
(8)
and
$$\Psi_{4}(u)=\Phi_{4}(u)-1. $$
Now, Ψ4(0)=−1<0 and \({ \underset {u \to \eta _{3}^{-}}{\lim } \Psi _{4}(u)=+\infty }\). Let η4 be the notation for the smallest zero of Ψ4(u) in (0,η3). The existence of η4 is guaranteed by the intermediate value theorem. We choose
$$ R'=min\{\eta_{1}, \eta_{2}, \eta_{4}\} $$
(9)
to confirm the followings.
$$\begin{array}{*{20}l} &0\leq \Phi_{1}(u) <1, \end{array} $$
(10)
$$\begin{array}{*{20}l} &0\leq \Phi_{2}(u) <1, \end{array} $$
(11)
$$\begin{array}{*{20}l} &0\leq \Phi_{3}(u) <1 \end{array} $$
(12)
and
$$ 0\leq \Phi_{4}(u) <1, $$
(13)
for each u∈[0,R′). Also, we use the following assumptions on the Fréchet differentiable operator G:Ω⊆X→Y.
$$\begin{array}{*{20}l} G(s^{*})=0,\ G'(s^{*})^{-1} \in BL(Y, X), \end{array} $$
(14)
$$\begin{array}{*{20}l} ||G'\left(s^{*}\right)^{-1} (G'(s)-G'\left(s^{*}\right))|| \leq k_{0} ||s-s^{*}||,\ \forall s \in \Omega \end{array} $$
(15)
and
$$ ||G'\left(s^{*}\right)^{-1} \left(G'(s)-G'(t)\right)|| \leq k_{1} ||s-t||,\ \forall s, t \in \Omega. $$
(16)
Now, we discuss the local convergence analysis of the algorithm (4) in Theorem 1.
Theorem 1
Let s∗∈Ω. Suppose the Fréchet differentiable operator G:Ω⊆X→Y obeys (14)–(16) and
$$ \bar{B}\left(s^{*}, R'\right) \subseteq \Omega, $$
(17)
where R′ is given in (9). Starting from s0∈B(s∗,R′) the scheme (4) produces the sequence {sk} which is well defined, {sk}k≥0∈B(s∗,R′) and converges to s∗. Also, the followings hold ∀k≥0
$$ ||t_{k}-s^{*}|| \leq \Phi_{1}\left(||s_{k}-s^{*}||\right)||s_{k}-s^{*}||<||s_{k}-s^{*}||<R', $$
(18)
$$ \left|\left|\left[G'(s_{k}) + 3G'\left(\frac{1}{3}(s_{k}+2t_{k})\right)\right]^{-1}G'(s^{*})\right|\right| \leq \frac{1}{4\left(1-\Phi_{3}\left(||s_{k}-s^{*}||\right)\right)} $$
(19)
and
$$ ||s_{k+1}-s^{*}|| \leq \Phi_{4}(||s_{k}-s^{*}||)||s_{k}-s^{*}||<||s_{k}-s^{*}||<R', $$
(20)
where the functions Φ1, Φ3, and Φ4 are provided in (5), (7), and (8) respectively. For \(\delta \in [R', \frac {2}{k_{0}})\), the equation G(s)=0 has only one solution s∗ in \(\bar {B}\left (s^{*}, \delta \right) \cap \Omega \).
Proof
It follows from (9), (15) and the assumption s0∈B(s∗,R′) that
$$ ||G'\left(s^{*}\right)^{-1} \left(G'(s_{0})-G'\left(s^{*}\right)\right)|| \leq k_{0} ||s_{0}-s^{*}||< k_{0}R'<1. $$
(21)
Now, Banach Lemma on invertible operators [17–21] ensures that G′(s0)−1∈BL(Y,X) and
$$ ||G'\left(s_{0}\right)^{-1}G'\left(s^{*}\right)|| \leq \frac{1}{1-k_{0} ||s_{0}-s^{*}||} < \frac{1}{1-k_{0}R'}. $$
(22)
Therefore, t0 is well defined. Again,
$$ {}\begin{aligned} t_{0}-s^{*} &=s_{0}-s^{*}-G'\left(s_{0}\right)^{-1}G\left(s_{0}\right) \\ &=-\left[G'(s_{0})^{-1}G'(s^{*})\right]\! \left[\int_{0}^{1} G'\left(s^{*}\right)^{-1} \left(G'\left(s^{*}+\theta\left(s_{0}-s^{*}\right)\right)\,-\,G'\left(s_{0}\right)\right) \left(s_{0}\!-s^{*}\right)\ d\theta \right]. \end{aligned} $$
(23)
Using (5), (9), (10), (16), (22) and (23), we find
$$ {}{\begin{aligned} ||t_{0}-s^{*}|| & \leq \left[ ||G'(s_{0})^{-1}G'\left(s^{*}\right)|| \right] \left[ \left|\left|\int_{0}^{1} G'\left(s^{*}\right)^{-1} \left(G'\left(s^{*}+\theta\left(s_{0}-s^{*}\right)\right)-G'\left(s_{0}\right)\right) \left(s_{0}-s^{*}\right)\ d\theta \right|\right| \right] \\ & \leq \frac{k_{1} ||s_{0}-s^{*}||}{2\left(1-k_{0}||s_{0}-s^{*}||\right)}||s_{0}-s^{*}|| \\ & = \Phi_{1}\left(||s_{0}-s^{*}||\right) ||s_{0}-s^{*}|| < ||s_{0}-s^{*}|| < R'. \end{aligned}} $$
(24)
So, (18) holds for k=0. Now,
$$ \begin{aligned} \left|\left|\frac{1}{3}(s_{0}+2t_{0})-s^{*}\right|\right| & \leq \frac{1}{3} \left|\left|(s_{0}+2t_{0})-3s^{*}\right|\right| \\ & \leq \frac{1}{3} \left(||s_{0}-s^{*}|| + 2||t_{0}-s^{*}||\right) \\ & \leq \frac{1}{3} \left(||s_{0}-s^{*}|| + 2\Phi_{1}\left(||s_{0}-s^{*}||\right) ||s_{0}-s^{*}||\right) \\ & = \frac{1}{3} \left(1 + 2\Phi_{1}\left(||s_{0}-s^{*}||\right)\right) ||s_{0}-s^{*}|| \\ & = \Phi_{2}\left(||s_{0}-s^{*}||\right) ||s_{0}-s^{*}|| < ||s_{0}-s^{*}|| < R'. \end{aligned} $$
(25)
So, \(\frac {1}{3}\left (s_{0}+2t_{0}\right) \in B\left (s^{*}, R'\right)\). Then, our claim is \(\left [G'(s_{0}) + 3G'\left (\frac {1}{3}(s_{0}+2t_{0})\right)\right ]^{-1} \in BL(Y, X)\). The equations (7), (9), (12), (15), (24), and (25) are used to deduce
$${}\begin{aligned} & \left|\left|(4G'(s^{*}))^{-1} \left[G'(s_{0}) + 3G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)-4G'\left(s^{*}\right)\right]\right|\right| \\ & \leq \frac{1}{4}\left[||G'\left(s^{*}\right)^{-1}\left(G'(s_{0})-G'\left(s^{*}\right)\right)||+ 3\left|\left|G'\left(s^{*}\right)^{-1}\left(G'\left(\frac{1}{3}\left(s_{0}+2t_{0}\right)\right)-G'(s^{*})\right)\right|\right|\right] \\ & \leq \frac{k_{0}}{4}\left[||s_{0}-s^{*}||+3\left|\left|\frac{1}{3}(s_{0}+2t_{0})-s^{*}\right|\right|\right] \\ & \leq \frac{k_{0}}{4}\left[||s_{0}-s^{*}||+3\Phi_{2}\left(||s_{0}-s^{*}||\right)||s_{0}-s^{*}||\right] \\ & = \frac{k_{0}}{4}\left[1+3\Phi_{2}\left(||s_{0}-s^{*}||\right)\right] ||s_{0}-s^{*}||\\ & = \Phi_{3}\left(||s_{0}-s^{*}||\right)<\Phi_{3}\left(R'\right)<1. \end{aligned} $$
Now, we obtain \(\left [G'(s_{0}) + 3G'\left (\frac {1}{3}(s_{0}+2t_{0})\right)\right ]^{-1} \in BL(Y, X)\) using Banach Lemma on invertible operators. Also,
$$ \left|\left|\left[G'(s_{0}) + 3G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)\right]^{-1}G'(s^{*})\right|\right| \leq \frac{1}{4(1-\Phi_{3}(||s_{0}-s^{*}||))}. $$
(26)
Hence, s1 is well defined. We use (8) (9), (13), (16), (25), and (26) to derive
$$ {}\begin{aligned} ||s_{1}-s^{*}|| & \leq \left(\left|\left|\left[G'(s_{0}) + 3G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)\right]^{-1}G'\left(s^{*}\right)\right|\right| \right) \\ & \left(\left|\left|\int_{0}^{1} G'\left(s^{*}\right)^{-1} \left(G'(s_{0})-G'\left(s^{*}+\theta\left(s_{0}-s^{*}\right)\right)\right) \left(s_{0}-s^{*}\right)\ d\theta \right|\right| \right. \\ & \left. + 3\left|\left|\int_{0}^{1} G'\left(s^{*}\right)^{-1} \left(G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)-G'\left(s^{*}+\theta\left(s_{0}-s^{*}\right)\right)\right) \left(s_{0}-s^{*}\right)\ d\theta \right|\right| \right) \\ & \leq \frac{\frac{k_{1}}{2} ||s_{0}-s^{*}||^{2} + 3k_{1} \int_{0}^{1}\left(||\frac{1}{3}(s_{0}+2t_{0})-s^{*}-\theta\left(s_{0}-s^{*}\right)||\right) d\theta ||s_{0}-s^{*}||}{4\left(1-\Phi_{3}(||s_{0}-s^{*}||)\right)} \\ & \leq \frac{\frac{k_{1}}{2} ||s_{0}-s^{*}||^{2} + 3k_{1} \left(||\frac{1}{3}(s_{0}+2t_{0})-s^{*}||+\frac{||s_{0}-s^{*}||}{2}\right) ||s_{0}-s^{*}||}{4(1-\Phi_{3}(||s_{0}-s^{*}||))} \\ & \leq \frac{\frac{k_{1}}{2} ||s_{0}-s^{*}||^{2} + 3k_{1} \left[\Phi_{2}\left(||s_{0}-s^{*}||\right) ||s_{0}-s^{*}||+\frac{||s_{0}-s^{*}||}{2}\right] ||s_{0}-s^{*}||}{4(1-\Phi_{3}(||s_{0}-s^{*}||))} \\ & = \frac{(2k_{1} ||s_{0}-s^{*}|| + 3k_{1}\Phi_{2}\left(||s_{0}-s^{*}||\right) ||s_{0}-s^{*}||) ||s_{0}-s^{*}||}{4(1-\Phi_{3}(||s_{0}-s^{*}||))} \\ & = \frac{k_{1}\left[\left(1 + \frac{3}{2}\Phi_{2}\left(||s_{0}-s^{*}||\right)\right)||s_{0}-s^{*}||\right] ||s_{0}-s^{*}||}{2(1-\Phi_{3}(||s_{0}-s^{*}||))} \\ & = \Phi_{4}(||s_{0}-s^{*}||)||s_{0}-s^{*}|| < ||s_{0}-s^{*}||<R'. \end{aligned} $$
(27)
Therefore, we prove that (20) is true for k=0. We find the estimates (18)-(20) by substituting sk, tk and sk+1 in place of s0, t0 and s1 respectively in the earlier estimations. From the inequality ||sk+1−s∗||≤Φ4(R′)||sk−s∗||<R′, we have sk+1∈B(s∗,R′) and \({{\lim }_{k \to \infty } s_{k}=s^{*}}\). Now, we have to show the uniqueness part. Let t∗(≠s∗)∈B(s∗,δ) be such that 0=G(t∗). Consider \(B=\int _{0}^{1} G'\left (t^{*}+\theta \left (s^{*}-t^{*}\right)\right)\ d\theta \). From Eq. (15), we get
$$\begin{aligned} ||G'(s^{*})^{-1} (B-G'(s^{*}))|| & \leq \int_{0}^{1} k_{0} ||t^{*}+\theta(s^{*}-t^{*})-s^{*}||\ d\theta \\ & \leq \frac{k_{0}}{2}||s^{*}-t^{*}|| \\ & \leq \frac{k_{0} \delta}{2} <1. \end{aligned} $$
Applying Banach Lemma, we confirm that B−1∈BL(Y,X). The identity 0=G(s∗)−G(t∗)=B(s∗−t∗) implies that s∗=t∗. □
Local convergence analysis of the method (4) under Hölder continuity condition
There are numerous problems for which the technique based on Lipschitz condition fails to solve without using higher-order derivatives. As an illustration, we consider the following problem given in [6].
$$G(s)(x)=s(x)-3 \int_{0}^{1} G_{1}(x, y)\ s(y)^{\frac{5}{4}}\ dy, $$
where s(x)∈C[0,1] and G1(x,y) is Green’s function defined on [0,1]×[0,1] by
$$G_{1}(x, y)= \left\{\begin{array}{cc} (1-x)y, & \text{if~~ \(y\leq x\)} \\ x(1-y), & \text{if~~ \(x\leq y\)} \end{array}\right. $$
Then,
$$||G'(s)-G'(t)|| \leq \frac{15}{32} ||s-t||^{\frac{1}{4}}. $$
It is important to note that G′ does not obey Lipschitz condition. However, G′ is Hölder continuous. So, we discuss the local convergence of the method (4) with Hölder continuous first-order Fréchet derivative. This analysis also generalizes the local convergence analysis presented in the previous section.
For q∈(0,1], we define Θ1 on \(\left [0, \left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)\) by
$$ \Theta_{1}(w)=\frac{k_{1}w^{q}}{(q+1)(1-k_{0} w^{q})} $$
(28)
and the parameter
$$\xi_{1}=\left(\frac{(q+1)}{(q+1)k_{0} + k_{1}}\right)^{\frac{1}{q}} < \left(\frac{1}{k_{0}}\right)^{\frac{1}{q}}. $$
Observe that Θ1(ξ1)=1. Again, we define functions Θ2 and Γ2 on \(\left [0, \left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)\) by
$$ \Theta_{2}(w)= \frac{1}{3}(1+2\Theta_{1}(w)) $$
(29)
and
$$\Gamma_{2}(w)=\Theta_{2}(w)-1. $$
Now, \(\Gamma _{2}(0)=-\frac {2}{3}<0\) and \({ \underset {w \to \left (\left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)^{-}}{\lim }\Gamma _{2}(w)=+\infty }\). The intermediate value theorem confirms the existence of the zeros of the function Γ2(w) in \(\left (0, \left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)\). We denote the smallest zero of Γ2(w) in \(\left (0, \left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)\) as ξ2. Again, we define Θ3 and Γ3 on \(\left [0, \left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)\) by
$$ \Theta_{3}(w)= \frac{k_{0}}{4}\left[1+3\Theta_{2}(w)^{q}\right] w^{q} $$
(30)
and
$$\Gamma_{3}(w)=\Theta_{3}(w)-1. $$
Now, Γ3(0)=−1<0 and \({\underset {w \to \left (\left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)^{-}} {\lim } \Gamma _{3}(w)=+\infty }\). So, the zeros of the function Γ3(w) lies in \(\left (0, \left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)\). We denote the smallest zero of Γ3(w) in \(\left (0, \left (\frac {1}{k_{0}}\right)^{\frac {1}{q}}\right)\) as ξ3. Lastly, we define Θ4 and Γ4 on [0,ξ3) by
$$ \Theta_{4}(w)= \frac{ k_{1} \left(\frac{4}{q+1} + 3 \Theta_{2}(w)^{q} \right) w^{q}}{4(1-\Theta_{3}(w))} $$
(31)
and
$$\Gamma_{4}(w)=\Theta_{4}(w)-1. $$
Now, Γ4(0)=−1<0 and \({\underset {w \to \xi _{3}^{-}}{\lim } \Gamma _{4}(w)=+\infty }\). Let ξ4 be the notation for the smallest zero of Γ4(w) in (0,ξ3). The existence of ξ4 is guaranteed by the intermediate value theorem. We choose
$$ R=min\left\{\xi_{1}, \xi_{2}, \xi_{4}\right\} $$
(32)
to confirm the followings.
$$\begin{array}{*{20}l} &0\leq \Theta_{1}(w) <1, \end{array} $$
(33)
$$\begin{array}{*{20}l} &0\leq \Theta_{2}(w) <1, \end{array} $$
(34)
$$\begin{array}{*{20}l} &0\leq \Theta_{3}(w) <1 \end{array} $$
(35)
and
$$ 0\leq \Theta_{4}(w) <1, $$
(36)
for each w∈[0,R). Also, we use the following assumptions on the Fréchet differentiable operator G:Ω⊆X→Y.
$$\begin{array}{*{20}l} G(s^{*})=0,\ G'(s^{*})^{-1} \in BL(Y, X), \end{array} $$
(37)
$$\begin{array}{*{20}l} ||G'(s^{*})^{-1} (G'(s)-G'(s^{*}))|| \leq k_{0} ||s-s^{*}||^{q},\ \forall s \in \Omega \end{array} $$
(38)
and
$$ ||G'\left(s^{*}\right)^{-1} (G'(s)-G'(t))|| \leq k_{1} ||s-t||^{q},\ \forall s, t \in \Omega. $$
(39)
Theorem 2
Let s∗∈Ω. Suppose the Fréchet differentiable operator G:Ω⊆X→Y obeys (37)–(39) and
$$ \bar{B}(s^{*}, R) \subseteq \Omega, $$
(40)
where R is given in (32). Starting from s0∈B(s∗,R), the scheme (4) yields the sequence {sk} which is well defined, {sk}k≥0∈B(s∗,R) and converges to s∗. Also, the following holds ∀k≥0
$$\begin{array}{*{20}l} ||t_{k}-s^{*}|| \leq \Theta_{1}\left(||s_{k}-s^{*}||\right)||s_{k}-s^{*}||<||s_{k}-s^{*}||<R, \end{array} $$
(41)
$$\begin{array}{*{20}l} \left|\left|\left[G'(s_{k}) + 3G'\left(\frac{1}{3}\left(s_{k}+2t_{k}\right)\right)\right]^{-1}G'(s^{*})\right|\right| \leq \frac{1}{4(1-\Theta_{3}(||s_{k}-s^{*}||))} \end{array} $$
(42)
and
$$ ||s_{k+1}-s^{*}|| \leq \Theta_{4}\left(||s_{k}-s^{*}||\right)||s_{k}-s^{*}||<||s_{k}-s^{*}||<R, $$
(43)
where the functions Θ1, Θ3, and Θ4 are provided in (28), (30), and (31) respectively. For \(\Delta \in \left [R, \left (\frac {q+1}{k_{0}} \right)^{\frac {1}{q}} \right)\), the solution s∗ is the only solution of G(s)=0 in \(\bar {B}(s^{*}, \Delta) \cap \Omega \).
Proof
It follows from (32), (38) and the assumption s0∈B(s∗,R) that
$$ ||G'\left(s^{*}\right)^{-1} \left(G'\left(s_{0})-G'(s^{*}\right)\right)|| \leq k_{0} ||s_{0}-s^{*}||^{q}< k_{0}R^{q}<1. $$
(44)
Now, Banach Lemma on invertible operators [17–21] ensures that G′(s0)−1∈BL(Y,X) and
$$ ||G'(s_{0})^{-1}G'\left(s^{*}\right)|| \leq \frac{1}{1-k_{0} ||s_{0}-s^{*}||^{q}} < \frac{1}{1-k_{0}R^{q}}. $$
(45)
Therefore, t0 is well defined. Again,
$$\begin{array}{*{20}l} {}t_{0}-s^{*} &=s_{0}-s^{*}-G'(s_{0})^{-1}G(s_{0}) \\ &=\!-\left[G'(s_{0})^{-1}G'(s^{*})\right] \left[\int_{0}^{1} G'\left(s^{*}\right)^{-1} \left(G'\left(s^{*}+\theta\left(s_{0}-s^{*}\right)\right)-G'\left(s_{0}\right)\right) (s_{0}-s^{*})\ d\theta \right]. \end{array} $$
(46)
Using (28), (32), (33), (39), (45) and (46), we find
$$ {}{\begin{aligned} ||t_{0}-s^{*}|| & \leq \left[ ||G'(s_{0})^{-1}G'(s^{*})|| \right] \left[ \left|\left|\int_{0}^{1} G'(s^{*})^{-1} (G'(s^{*}+\theta(s_{0}-s^{*}))-G'(s_{0})) (s_{0}-s^{*})\ d\theta \right|\right| \right] \\ & \leq \frac{k_{1} ||s_{0}-s^{*}||^{q}}{(q+1)(1-k_{0}||s_{0}-s^{*}||^{q})}||s_{0}-s^{*}|| \\ & = \Theta_{1}(||s_{0}-s^{*}||) ||s_{0}-s^{*}|| < ||s_{0}-s^{*}|| < R. \end{aligned}} $$
(47)
So, (41) holds for k=0. Now,
$$ \begin{aligned} \left|\left|\frac{1}{3}(s_{0}+2t_{0})-s^{*}\right|\right| & \leq \frac{1}{3} \left|\left|(s_{0}+2t_{0})-3s^{*}\right|\right| \\ & \leq \frac{1}{3} \left(||s_{0}-s^{*}|| + 2||t_{0}-s^{*}||\right) \\ & \leq \frac{1}{3} \left(||s_{0}-s^{*}|| + 2\Theta_{1}\left(||s_{0}-s^{*}||\right) ||s_{0}-s^{*}||\right) \\ & = \frac{1}{3} \left(1 + 2\Theta_{1}\left(||s_{0}-s^{*}||\right)\right) ||s_{0}-s^{*}|| \\ & = \Theta_{2}\left(||s_{0}-s^{*}||\right) ||s_{0}-s^{*}|| < ||s_{0}-s^{*}|| < R. \end{aligned} $$
(48)
So, \(\frac {1}{3}(s_{0}+2t_{0}) \in B(s^{*}, R)\). Then, our claim is \(\left [G'(s_{0}) + 3G'\left (\frac {1}{3}(s_{0}+2t_{0})\right)\right ]^{-1} \in BL(Y, X)\). The Eqs. (30), (32), (35), (38), (47), and (48) are used to deduce
$$\begin{aligned} & \left|\left|(4G'(s^{*}))^{-1} \left[G'(s_{0}) + 3G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)-4G'(s^{*})\right]\right|\right| \\ & \leq \frac{1}{4}\left[||G'(s^{*})^{-1}(G'(s_{0})-G'(s^{*}))||+ 3\left|\left|G'(s^{*})^{-1}\left(G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)-G'(s^{*})\right)\right|\right|\right] \\ & \leq \frac{k_{0}}{4}\left[||s_{0}-s^{*}||^{q}+3\left|\left|\frac{1}{3}(s_{0}+2t_{0})-s^{*}\right|\right|^{q}\right] \\ & \leq \frac{k_{0}}{4}\left[||s_{0}-s^{*}||^{q}+3\Theta_{2}(||s_{0}-s^{*}||)^{q}||s_{0}-s^{*}||^{q}\right] \\ & = \frac{k_{0}}{4}[1+3\Theta_{2}(||s_{0}-s^{*}||)^{q}] ||s_{0}-s^{*}||^{q}\\ & = \Theta_{3}(||s_{0}-s^{*}||)<\Theta_{3}(R)<1. \end{aligned} $$
Now, we obtain \(\left [G'(s_{0}) + 3G'\left (\frac {1}{3}(s_{0}+2t_{0})\right)\right ]^{-1} \in BL(Y, X)\) using Banach Lemma on invertible operators. Also,
$$ \left|\left|\left[G'(s_{0}) + 3G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)\right]^{-1}G'(s^{*})\right|\right| \leq \frac{1}{4(1-\Theta_{3}(||s_{0}-s^{*}||))}. $$
(49)
Hence, s1 is well defined. We use (31) (32), (36), (39), (48), and (49) to derive
$$ {}\begin{aligned} ||s_{1}-s^{*}|| & \leq \left(\left|\left|\left[G'(s_{0}) + 3G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)\right]^{-1}G'(s^{*})\right|\right| \right) \\ & \left(\left|\left|\int_{0}^{1} G'\left(s^{*}\right)^{-1} \left(G'\left(s_{0}\right)-G'\left(s^{*}+\theta\left(s_{0}-s^{*}\right)\right)\right) \left(s_{0}-s^{*}\right)\ d\theta \right|\right| \right. \\ &\left. + 3\left|\left|\int_{0}^{1} G'\left(s^{*}\right)^{-1} \left(G'\left(\frac{1}{3}(s_{0}+2t_{0})\right)-G'\left(s^{*}+\theta\left(s_{0}-s^{*}\right)\right)\right) (s_{0}-s^{*})\ d\theta \right|\right| \right) \\ & \leq \frac{\frac{k_{1}}{q+1} ||s_{0}-s^{*}||^{q+1} + 3k_{1} \int_{0}^{1}\left(||\frac{1}{3}\left(s_{0}+2t_{0}\right)-s^{*}-\theta\left(s_{0}-s^{*}\right)||^{q}\right)\ d\theta\ ||s_{0}-s^{*}||}{4(1-\Theta_{3}(||s_{0}-s^{*}||))} \\ & \leq \frac{\frac{k_{1}}{q+1} ||s_{0}-s^{*}||^{q+1} + 3k_{1} \left(||\frac{1}{3}(s_{0}+2t_{0})-s^{*}||^{q}+\frac{||s_{0}-s^{*}||^{q}}{q+1}\right) ||s_{0}-s^{*}||}{4(1-\Theta_{3}(||s_{0}-s^{*}||))} \\ & \leq \frac{\frac{k_{1}}{q+1} ||s_{0}-s^{*}||^{q+1} + 3k_{1} \left[\Theta_{2}(||s_{0}-s^{*}||)^{q} ||s_{0}-s^{*}||^{q}+\frac{||s_{0}-s^{*}||^{q}}{q+1}\right] ||s_{0}-s^{*}||}{4(1-\Theta_{3}(||s_{0}-s^{*}||))} \\ & = \frac{ \left(\frac{k_{1}}{q+1} ||s_{0}-s^{*}||^{q} + 3k_{1} \left[\Theta_{2}(||s_{0}-s^{*}||)^{q} ||s_{0}-s^{*}||^{q}+\frac{||s_{0}-s^{*}||^{q}}{q+1}\right] \right) ||s_{0}-s^{*}||}{4(1-\Theta_{3}(||s_{0}-s^{*}||))} \\ & = \frac{ \left(\frac{4k_{1}}{q+1} ||s_{0}-s^{*}||^{q} + 3k_{1} \left[\Theta_{2}(||s_{0}-s^{*}||)^{q} ||s_{0}-s^{*}||^{q}\right] \right) ||s_{0}-s^{*}||}{4(1-\Theta_{3}(||s_{0}-s^{*}||))} \\ & = \frac{ k_{1} \left(\frac{4}{q+1} ||s_{0}-s^{*}||^{q} + 3 \left[\Theta_{2}(||s_{0}-s^{*}||)^{q} ||s_{0}-s^{*}||^{q}\right] \right) ||s_{0}-s^{*}||}{4(1-\Theta_{3}(||s_{0}-s^{*}||))} \\ & = \Theta_{4}(||s_{0}-s^{*}||)||s_{0}-s^{*}|| < ||s_{0}-s^{*}||<R. \end{aligned} $$
(50)
Thus, we prove that (43) holds for k=0. We find the estimates (41)-(43) by substituting sk, tk, and sk+1 in place of s0, t0, and s1 respectively in the preceding estimations. From the inequality ||sk+1−s∗||≤Θ4(R)||sk−s∗||<R, we have sk+1∈B(s∗,R) and \({{\lim }_{k \to \infty } s_{k}=s^{*}}\). Now, we have to show the uniqueness part. Let t∗(≠s∗)∈B(s∗,Δ) be such that 0=G(t∗). Consider \(A=\int _{0}^{1} G'(\theta s^{*}+t^{*}(1-\theta))\ d\theta \). From Eq. (38), we get
$$\begin{aligned} ||G'(s^{*})^{-1} (A-G'(s^{*}))|| & \leq \int_{0}^{1} k_{0} ||t^{*}+\theta\left(s^{*}-t^{*}\right)-s^{*}||^{q}\ d\theta \\ & \leq \frac{k_{0}}{q+1}||s^{*}-t^{*}||^{q} \\ & \leq \frac{k_{0} \Delta^{q}}{q+1} <1. \end{aligned} $$
Applying Banach Lemma, we confirm that A−1∈BL(Y,X). The identity 0=G(s∗)−G(t∗)=A(s∗−t∗) implies that s∗=t∗. □