Unified Semi-Local Convergence for k—Step Iterative Methods with Flexible and Frozen Linear Operator

Abstract

The aim of this article is to present a unified semi-local convergence analysis for a k-step iterative method containing the inverse of a flexible and frozen linear operator for Banach space valued operators. Special choices of the linear operator reduce the method to the Newton-type, Newton’s, or Stirling’s, or Steffensen’s, or other methods. The analysis is based on center, as well as Lipschitz conditions and our idea of the restricted convergence region. This idea defines an at least as small region containing the iterates as before and consequently also a tighter convergence analysis.

1. Introduction

Let $\mathcal{X},\mathcal{Y}$ be Banach spaces and $D\subset \mathcal{X}$ be a nonempty and open set. By $\mathcal{L}(\mathcal{X},\mathcal{Y}),$ we denote the space of bounded linear operators from $\mathcal{X}$ into $\mathcal{Y}.$ Let also $U(w,d)$ stand for an open set centered at $w\in \mathcal{X}$ and of radius $d>0$ and $\overline{U}(w,d)$ stand for its closure.

There is a plethora of problems from diverse disciplines, such as mathematics [1,2,3,4,5,6,7,8,9,10,11,12,13], optimization [3,4,5,6,7,8], mathematical programming [7,8], chemistry [7], biology [1,2,12], physics [9,13], economics [8], statistics [13], engineering [1,2,9,10,11,12,13] and other disciplines, that can be reduced to finding a solution $x^{\ast }$ of the equation:

$$F(x)=0,$$

(1)

where $F:D⟶\mathcal{Y}$ is a continuous operator. The solution $x^{\ast }$ of Equation (1) should be unique in a neighborhood about it and in closed form. However, the latter can be achieved only in special cases. This problem leads researchers to the construction of iterative methods that generate a sequence converging to $x^{\ast }.$

The most widely-used iterative method is Newton’s, defined for each $n=0,1,2,\dots ,$ by:

$$x_0\in D,x_{n+1}=x_n-F^{\prime }{(x_n)}^{-1}F(x_n).$$

(2)

Newton’s method is a special case of one-point iterative methods without memory defined for each $n=0,1,2,3\dots ,$ by:

$$x_0\in D{x}_{n+1}=R(x_n),$$

(3)

where $R:\mathcal{X}⟶\mathcal{X}$ has some properties. The order of convergence $p\in N$ depends explicitly on the first $p-1$ derivatives of the functions appearing in the method. Moreover, the computational cost increases in general especially when the convergence order increases, since successive derivatives must be computed [1,2,3,4,5,6,7,8,9,10,11,12,13].

That is why researchers and practitioners have developed iterative methods that on the one hand avoid the computation of derivatives and on the other hand achieve a high order of convergence. In particular, we unify the study of such methods by considering k-step iterative methods with a frozen linear operator defined for each $n=0,1,2,\dots ,$ by:

$$\begin{array}{ccc}\hfill x_0\in D,x_n^{(1)}& =& x_n^{(0)}-A_n^{-1}F(x_n^{(0)})\hfill \\ \hfill x_n^{(2)}& =& x_n^{(1)}-A_n^{-1}F(x_n^{(1)})\hfill \\ \hfill & \cdots \\ \hfill x_n^{(k-1)}& =& x_n^{(k-2)}-A_n^{-1}F(x_n^{(k-2)})\hfill \\ \hfill x_n^{(k)}& =& x_n^{(k-1)}-A_n^{-1}F(x_n^{(k-1)}),\hfill \end{array}$$

(4)

where $A_n=A(x_n),A:D⟶\mathcal{L}(\mathcal{X},\mathcal{Y}),x_n=x_n^{(0)}$ and $x_{n+1}=x_n^{(k)}$ for each $k=0,1,2,\dots .$ Special choices of operator A lead to well-known methods. If $k=1$ and $A(x)=F^{\prime }(x)$ for each $x\in D,$ we obtain Newton’s method (2), whereas, if $k=1,2,\dots$ and $A(x)=F^{\prime }(x)$ for each $x\in D,$ we obtain a method whose semi-local convergence was given in [12]. If $A(x)=[g_1(x),g_2(x);F]$ for each $x\in D,$ $k=1$ or $k=1,2,\dots ,$ where $g_1:X⟶X$ and $g_2:X⟶X,$ we obtain Steffensen-type methods. Stirling’s and other one-point methods are also special cases of method (4). Based on the above, it is important to study the semi-local convergence analysis of method (4). It is well known that as the convergence order increases, the convergence region decreases in general. To avoid this problem as well, we introduce a center-Lipschitz-type condition that helps us determine an at least as small region as before containing the iterates $\left\{x_n\right\}.$ This way, the resulting Lipschitz constants are at least as small. A tighter convergence analysis is obtained this way.

The rest of the article is organized as follows: Section 2 contains the conditions to be used in the semi-local convergence that follows in Section 3. Final remarks are given in the concluding Section 4.

2. Convergence Conditions

We shall assume that $U(x_0,r\eta )\subseteq D$ for some $r>1$ and $\eta >0.$ The semi-local convergence analysis of Method (4) is based on Condition (A) (see also the Conclusion Section 4):

(a1)

$F:D⟶\mathcal{Y}$ is a differentiable operator in the sense of Fréchet, $A(x)\in \mathcal{L}(\mathcal{X},\mathcal{Y})$, and there exists $x_0\in D,\beta >0,\eta >0$ such that $A{(x_0)}^{-1}\in \mathcal{L}(\mathcal{Y},\mathcal{X}),$

$$\parallel A{(x_0)}^{-1}\parallel \le \beta \mathrm{and}\parallel A{(x_0)}^{-1}F(x_0)\parallel \le \eta .$$

(a2)

There exist $L>0,\ell \in (0,\frac{1}{\beta })$ such that for each $x,y\in D$:

$$\parallel A(x)-A(x_0)\parallel \le L\parallel x-x_0\parallel +\ell .$$

Set $D_0=D\cap U(x_0,\frac{1-\beta \ell }{\beta L\eta }).$

(a3)

There exist $K>0,M>0,\mu >0$ such that for each $x,y\in D_0$:

$$\parallel F^{\prime }(y)-F^{\prime }(x)\parallel \le K\parallel x-y\parallel ,$$

$$\parallel A(x)-F^{\prime }(x)\parallel \le M\parallel x-x_0\parallel +\mu .$$

(a4)

There exist $\overline{\beta }>0$ and $K_0>0$ such that $F^{\prime }{(x_0)}^{-1}\in \mathcal{L}(\mathcal{Y},\mathcal{X}),$ $\parallel F^{\prime }{(x_0)}^{-1}\parallel \le \overline{\beta }$ and for each $x\in D_0$:

$$\parallel F^{\prime }(x)-F^{\prime }(x_0)\parallel \le K_0\parallel x-x_0\parallel .$$

(a5)

There exists $r^{\ast }\ge r$ such that $r<\frac{2}{\overline{\beta }{K}_0\eta }=r_0$. Set $D_1=D\cap \overline{U}(x_0,r^{\ast }\eta ).$

From now on, we assume Condition (A).

3. Semi-Local Convergence

We need some auxiliary results to show the semi-local convergence of Method (4).

Lemma 1.

Suppose that there exists $r>1$ such that $x_n^{(i)}\in U(x_0,r\eta )\subset D$ for each $i=1,2,\dots k,k\ge 1,n\in \mathbb{N}$ and for:

$$\beta \ell <1,r<\frac{1-\beta \ell }{\beta L\eta }.$$

(5)

Then, method (4) is well defined.

Proof. 

We have that $F(x_n^{(i)})$ are well defined for each $i=1,2,\dots ,k,k\ge 1.$ Using (5), (a1) and (a2), we have in turn that:

$$\begin{array}{ccc}\hfill \parallel A_0^{-1}(A_n-A_0)\parallel & \le & \parallel A_0^{-1}\parallel \parallel A_n-A_0\parallel \hfill \\ \hfill & \le & \beta (L\parallel x_n-x_{)}\parallel +\ell )\hfill \\ \hfill & \le & \beta (Lr\eta +\ell )<1.\hfill \end{array}$$

(6)

By (6) and the Banach lemma on invertible operators [3,4,5,6,7,11], we deduce $A_n^{-1}\in \mathcal{L}(\mathcal{Y},\mathcal{X})$ and:

$$\parallel A_n^{-1}\parallel \le \frac{\beta }{1-\beta (Lr\eta +\ell )}.$$

(7)

 □

Let ${\mu }_1=max\{\mu ,\ell \},K_1=max\{K,L\},K_2=K_1+M$ ${\rho }_n(t)=K_2{\eta }_{n}t+{\mu }_1$ and ${\rho }_n={\rho }_n(r).$ We assume from now on that the previous hypotheses are satisfied. Let $n=0$ and $i=1.$ Then, we have:

$$\parallel x_0^{(i)}-x_0^{(0)}\parallel =\parallel A_0^{-1}F(x_0^{(0)})\parallel \le \eta .$$

Set:

$${\eta }_0=\eta ,{\beta }_0=\beta ,{\rho }_0={\rho }_0(r),h_0=h_0(r)={\beta }_0{\rho }_0.$$

By the first step in Method (4), we can write:

$$\begin{array}{ccc}\hfill F(x_0^{(1)})& =& F(x_0^{(1)})-F(x_0^{(0)})-F^{\prime }(x_0^{(0)})(x_0^{(1)}-x_0^{(0)})\hfill \\ & \hfill & +(F^{\prime }(x_0^{(0)})-A(x_0^{(0)}))(x_0^{(1)}-x_0^{(0)}),\hfill \end{array}$$

so:

$$\begin{array}{ccc}\hfill \parallel F(x_0^{(1)})\parallel & \le & \parallel {\int }_0^1[F^{\prime }(x_0^{(0)}+\theta (x_0^{(1)}-x_0^{(0)}))-F^{\prime }(x_0^{(0)})]d\theta (x_0^{(1)}-x_0^{(0)})\parallel \hfill \\ & & +\parallel F^{\prime }(x_0^{(0)})-A(x_0^{(0)})\parallel \parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \\ & \le & \frac{1}{2}{K}_0{\eta }_0\parallel x_0^{(1)}-x_0^{(0)}\parallel +(M\parallel x_0^{(1)}-x_0^{(0)}\parallel +\mu )\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \\ & =& (\frac{1}{2}{K}_0{\eta }_0+\mu )\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \\ & \le & {\rho }_0\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \\ \hfill \parallel x_0^{(2)}-x_0^{(1)}\parallel & =& \parallel A_0^{-1}F(x_0^{(1)})\parallel \hfill \\ & \le & \parallel A_0^{-1}\parallel \parallel F(x_0^{(1)})\parallel \hfill \\ & \le & {\beta }_0{\rho }_0\parallel x_0^{(1)}-x_0^{(0)}\parallel =h_0\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \end{array}$$

and:

$$\begin{array}{ccc}\hfill \parallel x_0^{(2)}-x_0^{(0)}\parallel & \le & \parallel x_0^{(2)}-x_0^{(1)}\parallel +\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \\ & \le & h_0\parallel x_0^{(1)}-x_0^{(0)}\parallel +\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \\ & =& (1+h_0){\eta }_0.\hfill \end{array}$$

Similarly, we can write:

$$\begin{array}{ccc}\hfill F(x_0^{(2)})& =& F(x_0^{(2)})-F(x_0^{(1)})-F^{\prime }(x_0^{(0)})(x_0^{(2)}-x_0^{(1)})\hfill \\ & \hfill & +(F^{\prime }(x_0^{(0)})-A(x_0^{(0)}))(x_0^{(2)}-x_0^{(1)}),\hfill \end{array}$$

so:

$$\begin{array}{ccc}\hfill \parallel F(x_0^{(2)})\parallel & \le & \parallel {\int }_0^1[F^{\prime }(x_0^{(1)}+\theta (x_0^{(2)}-x_0^{(1)}))-F^{\prime }(x_0^{(0)})]d\theta (x_0^{(2)}-x_0^{(1)})\parallel \hfill \\ & & +\parallel F^{\prime }(x_0^{(0)})-A(x_0^{(0)})\parallel \parallel x_0^{(2)}-x_0^{(1)}\parallel \hfill \\ & \le & K_0{\int }_0^1[(1-\theta )\parallel x_0^{(1)}-x_0^{(0)}\parallel +\theta \parallel x_0^{(2)}-x_0^{(0)}\parallel ]d\theta \parallel x_0^{(2)}-x_0^{(1)}\parallel \hfill \\ & & (M\parallel x_0^{(1)}-x_0^{(0)}\parallel +\mu )\parallel x_0^{(2)}-x_0^{(1)}\parallel \hfill \\ & =& (K_{0}r{\eta }_0+\mu )\parallel x_0^{(2)}-x_0^{(1)}\parallel \hfill \\ & \le & {\rho }_0\parallel x_0^{(2)}-x_0^{(1)}\parallel \hfill \\ \hfill \parallel x_0^{(3)}-x_0^{(2)}\parallel & =& \parallel A_0^{-1}F(x_0^{(2)})\parallel \hfill \\ & \le & {\beta }_0{\rho }_0\parallel x_0^{(2)}-x_0^{(1)}\parallel =h_0\parallel x_0^{(2)}-x_0^{(1)}\parallel \hfill \\ & \le & h_0^2\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \end{array}$$

and:

$$\begin{array}{ccc}\hfill \parallel x_0^{(3)}-x_0^{(0)}\parallel & \le & \parallel x_0^{(3)}-x_0^{(2)}\parallel +\parallel x_0^{(2)}-x_0^{(1)}\parallel +\parallel x_0^{(1)}-x_0^{(0)}\parallel \hfill \\ & \le & (1+h_0+h_0^2){\eta }_0.\hfill \end{array}$$

Hence, we arrive at:

Lemma 2.

The following assertions hold for $n=0,i=1,2,3,\dots ,k-1:$

$$\begin{array}{ccc}\hfill \parallel F(x_0^{(i)})& \le & \rho \parallel x_0^{(i)}-x_0^{(i-1)}\parallel ,\hfill \end{array}$$

(8)

$$\begin{array}{ccc}\hfill \parallel x_0^{(i)}-x_0^{(i-1)}\parallel & \le & h_0\parallel x_0^{(i-1)}-x_0^{(i-2)}\parallel ,\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill \parallel x_0^{(i)}-x_0^{(0)}\parallel & \le & (1+h_0+\dots +h_0^{j-1}){\eta }_0\hfill \end{array}$$

(10)

and:

$$\parallel x_0^{(k)}-x_0^{(0)}\parallel \le (1+h_0+\dots +h_0^{k-1}){\eta }_0.$$

(11)

Proof. 

We have that for each $\theta \in [0,1],$ $x_0^{(i-1)}+\theta (x_0^{(i)}-x_0^{(i-1)})\in U(x_0,r\eta ),$ since $x_0^{(i)},x_0^{(i-1)}\in U(x_0,r\eta )$ and:

$$\begin{array}{cc}& \parallel x_0^{(i-1)}-x_0^{(0)}+\theta (x_0^{(i)}-x_0^{(i-1)})\parallel \hfill \\ \le & (1-\theta )\parallel x_0^{(i-1)}-x_0^{(0)}\parallel +\theta \parallel x_0^{(i)}-x_0^{(0)}\parallel \hfill \\ \le & (1-\theta )r\eta +\theta r\eta =rn.\hfill \end{array}$$

Then, as previously, using Method (4), we can write:

$$\begin{array}{ccc}\hfill F(x_0^{(i)})& =& {\int }_0^1[F(x_0^{(i-1)}+\theta (x_0^{(i)}-x_0^{(i-1)}))-F^{\prime }(x_0^{(0)})]d\theta (x_0^{(i)}-x_0^{(i-1)})\hfill \\ & \hfill & +(F^{\prime }(x_0^{(0)})-A(x_0^{(0)}))(x_0^{(i)}-x_0^{(i-1)}),\hfill \end{array}$$

so:

$$\begin{array}{ccc}\hfill \parallel F(x_0^{(i)})\parallel & \le & K_0{\int }_0^1[(1-\theta )\parallel x_0^{(i-1)}-x_0^{(0)}\parallel +\theta \parallel x_0^{(i)}-x_0^{(0)}\parallel ]d\theta \parallel x_0^{(i)}-x_0^{(i-1)}\parallel \hfill \\ & & +\parallel F^{\prime }(x_0^{(0)})-A(x_0^{(0)})\parallel \parallel x_0^{(i)}-x_0^{(i-1)}\parallel \hfill \\ & =& (K_{0}r\eta +\mu )\parallel x_0^{(i)}-x_0^{i-1)}\parallel \hfill \\ & \le & \rho \parallel x_0^{(i)}-x_0^{(i-1)}\parallel ,\hfill \\ \hfill \parallel x_0^{(i+1)}-x_0^{(i)}\parallel & =& \parallel A_0^{-1}F(x_0^{(i)})\parallel \hfill \\ & \le & \beta \rho \parallel x_0^{(i)}-x_0^{(i-1)}\parallel =h_0\parallel x_0^{(i)}-x_0^{(i-1)}\parallel \hfill \end{array}$$

and:

$$\begin{array}{ccc}\hfill \parallel x_0^{(i+1)}-x_0^{(0)}\parallel & \le & \parallel x_0^{(i+1)}-x_0^{(i)}\parallel +\parallel x_0^{(i)}-x_0^{(1)}\parallel \hfill \\ & \le & (1+h_0+\dots +h_0^i){\eta }_0,\hfill \end{array}$$

which show Estimates (8)–(10), respectively. Estimate (11) follows from (10) for $i=k.$ □

It follows that $x_0^{(i)}$ for $i=1,2,\dots ,k-1,$ $x_0^{(i)}=x_1$ belong in $U(x_0,r\eta ).$ Define:

$$T_0(t)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}k=1\hfill \\ 1+h_0+\dots +h_0^{k-1},\hfill & \mathrm{if}k=2,3,\dots .\hfill \end{array}\right.$$

(12)

Next, we study Method (4) for $n=1$ in an analogous way to $n=0.$ It follows from Lemma 1 that $A{(x_1)}^{-1}\in \mathcal{L}(\mathcal{Y},\mathcal{X})$ and:

$${\parallel A(x)}^{-1}\parallel \le \frac{\beta }{1-h_0}:={\beta }_1.$$

(13)

Hence, $x_1^{(1)}=x_1^{(0)}-A{(x_1)}^{-1}F(x_1^{(0)}),$ with $x_1^{(0)}=x_0^{(k)}=x_1$, is well defined,

$$\parallel F(x_1^{(0)})\parallel \le {\rho }_1\parallel x_0^{(k)}-x_0^{(k-1)}\parallel ,$$

so:

$$\begin{array}{ccc}\hfill \parallel x_1^{(1)}-x_0^{(0)}\parallel & =& \parallel A{(x_1)}^{-1}F(x_1^{(0)})\parallel \hfill \\ \hfill & \le & \parallel A{(x_1)}^{-1}\parallel \parallel F(x_1^{(0)})\parallel \hfill \\ \hfill & \le & {\beta }_1{\rho }_1\parallel x_0^{(k)}-x_0^{(k-1)}\parallel \hfill \\ \hfill & \le & \frac{\beta {\rho }_1}{1-h_0}{h}_0^{k-1}{\eta }_0\hfill \\ \hfill & \le & \frac{h_0^k}{1-h_0}{\eta }_0\hfill \\ \hfill & =& h_1{\eta }_0={\eta }_1,\hfill \end{array}$$

(14)

where:

$$h_1=\frac{h_0^k}{1-h_0}.$$

(15)

Define as previously,

$$T_1(t)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}k=1\hfill \\ 1+h_1+\dots +h_1^{k-1},\hfill & \mathrm{if}k=2,3,\dots .\hfill \end{array}\right.$$

Then, we have again that:

$$\begin{array}{ccc}\hfill \parallel A{(x_1)}^{-1}\parallel & \le & \parallel F(x_1^{(i)})\parallel \hfill \\ & \le & {\rho }_1\parallel x_1^{(i)}-x_1^{(i-1)}\parallel \hfill \\ \hfill \parallel x_1^{(i+1)}-x_1^{(i)}\parallel & \le & h_1\parallel x_1^{(i)}-x_1^{(i-1)}\parallel ,\hfill \\ \hfill \parallel F(x_1^{(k)})\parallel & \le & {\rho }_1\parallel x_1^{(k)}-x_1^{(k-1)}\parallel ,\hfill \\ \hfill \parallel x_1^{(k)}-x_1^{(k-1)}\parallel & =& \parallel x_2-x_1^{(k-1)}\parallel \le h_1\parallel x_1^{(k-1)}-x_1^{(k-2)}\parallel \hfill \end{array}$$

and:

$$\parallel x_1^{(k)}-x_1^{(0)}\parallel =\parallel x_2-x_1\parallel \le T_1(r){\eta }_1.$$

Next, we continue for $n=2.$ By Lemma 1, $A{(x_2)}^{-1}\in \mathcal{L}(\mathcal{Y},\mathcal{X})$ and:

$$\parallel A{(x_2)}^{-1}\parallel \le \frac{\beta }{1-h_0}:={\beta }_2.$$

(16)

Notice that ${\beta }_2={\beta }_1.$ Then, for $i=1$ and since $x_2=x_2^{(0)},$ we get as in (14):

$$\begin{array}{ccc}\hfill \parallel x_2^{(1)}-x_2^{(0)}\parallel & =& \parallel A{(x_2)}^{-1}F(x_2^{(0)})\parallel \hfill \\ \hfill & \le & \parallel A{(x_2)}^{-1}\parallel \parallel F(x_2^{(0)})\parallel \hfill \\ \hfill & \le & {\beta }_2{\rho }_2\parallel x_1^{(k)}-x_1^{(k-1)}\parallel \hfill \\ \hfill & =& h_2{\eta }_1={\eta }_2,\hfill \end{array}$$

(17)

where $h_2=\frac{h_1^k}{1-h_0}.$ Then, as before, we can write:

$$\parallel A{(x_2)}^{-1}\parallel \le {\beta }_2,$$

$$\parallel A{(x_2)}^{-1}F(x_2)\parallel \le {\eta }_2,$$

$$T_2(t)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}k=1\hfill \\ 1+h_2+\dots +h_2^{k-1},\hfill & \mathrm{if}k=2,3,\dots .\hfill \end{array}\right.$$

so $x_2^{(i)},x_3\in U(x_2,T_2(r){\eta }_2)$ for $i=1,2,\dots ,k-1.$ We are motivated by the preceding items to define recurrent relations:

$${\beta }_n={\beta }_1,$$

$${\eta }_n=\frac{1}{1-h_0}{h}_{n-1}^k{\eta }_{n-1},$$

$$h_n={\beta }_n{\rho }_n{\eta }_n,$$

$$T_n(t)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}k=1\hfill \\ 1+h_n+\dots +h_n^{k-1},\hfill & \mathrm{if}k=2,3,\dots .\hfill \end{array}\right.$$

(18)

Hence, we arrive at:

Lemma 3.

Suppose that the hypotheses of Lemma 1 hold. Then, $x_n^{(i)},x_{n+1}\in U(x_n,T_n(r){\eta }_n)$ for each $i=1,2,\dots ,k-1.$

Proof. 

As in the cases $n=1,2,$ we get for each $n=1,2,3,\dots :$

$$F(x_n^{(i)})\parallel \le {\rho }_n(r)\parallel x_n^{(i)}-x_n^{(i-1)}\parallel ,i=1,2,\dots ,k$$

and for $i=1,2,\dots ,k-1,$

$$\parallel x_n^{(i+1)}-x_n^{(i)}\parallel \le h_n\parallel x_n^{(i)}-x_n^{(i-1)}\parallel \le h_n^i\parallel x_n^{(i)}-x_n^{(i-1)}\parallel ,$$

$$\parallel x_n^{(i+1)}-x_n^{(0)}\parallel \le (1+h_n+h_n^2+\dots +h_n^i){\eta }_n.$$

That is, we obtain:

$$\parallel F(x_n^{(k-1)})\parallel \le {\rho }_n(r)\parallel x_n^{(k-1)}-x_n^{(k-2)}\parallel ,$$

$$\parallel x_n^{(k)}-x_n^{(k-1)}\parallel =\parallel x_{n+1}-x_n^{(k-1)}\parallel \le h_n\parallel x_n^{(k-1)}-x_n^{(k-2)}\parallel \le h_n^{k-1}\parallel x_n^{(1)}-x_n^{(0)}\parallel$$

and:

$$\parallel x_n^{(k)}-x_n^{(0)}\parallel =\parallel x_{n+1}-x_n^{(0)}\parallel \le (1+h_n+h_n^2+\dots h_n^{k-1}){\eta }_n.$$

 □

Define function $\phi$ on the interval $[0,1]$ by:

$$\phi (t)=t^{k-1}+t-1,k=1,2,\dots .$$

(19)

We have that $\phi (0)=-1$ and $\phi (1)=1>0.$ It then follows from the intermediate value theorem that equation $\phi (t)=0$ has at least one solution in $(0,1).$ Denote by s the smallest such solution. Notice that for:

$$r<\frac{s-{\mu }_1}{K_2\eta },\eta \ne 0and{\mu }_1<s,$$

(20)

a simple inductive argument shows that:

$$h_{n+1}<h_0\le sforeachn=1,2,\dots .$$

(21)

and:

$$T_n(r)<T_{n-1}(r).$$

(22)

Hence, we arrive at:

Lemma 4.

Suppose that (20) holds. Then, sequences $\left\{h_n\right\}$ and $\left\{T_n(r)\right\}$ are decreasing.

Proof. 

It follows immediately from (19). □

Taking into account $x_n^{(i)}\in U(x_0,r{\eta }_n)$ and (20)–(22), we can obtain in turn the estimate:

$$\begin{array}{ccc}\hfill \parallel x_n^{(i)}-x_0\parallel & \le & \parallel x_n^{(i)}-x_n^{(0)}\parallel +\sum _{i=0}^{n-1}\parallel x_{n-i}-x_{n-i-1}\parallel \hfill \\ \hfill & \le & \sum _{i=0}^n{T}_i(r){\eta }_i\le T_0(r)\sum _{i=0}^n{\eta }_i\hfill \\ \hfill & =& T_0(r)[\eta +{\eta }_1+{\eta }_2+\dots ]\hfill \\ \hfill & \le & T_0(r)[1+\frac{1}{1-h_0}{h}_0^k+{(\frac{1}{1-h_0})}^2{h}_0^{2k}+\dots +{(\frac{1}{1-h_0})}^n{h}_0^{nk}+\dots ]\eta \hfill \\ \hfill & =& T_0(r)[1+\frac{1}{1-h_0}{h}_0^k(1+(\frac{1}{1-h_0})h_0^k+\dots +{(\frac{1}{1-h_0})}^{n-1}{h}_0^{(n-1)k}+\dots )]\eta \hfill \\ \hfill & =& T_0(r)[1+\frac{h_0^k}{1-h_0}\frac{1}{1-\frac{h_0^k}{1-h_0}}]\eta \hfill \\ \hfill & =& T_0(r)\frac{1-h_0}{1-h_0-h_0^k}\eta ,\hfill \end{array}$$

(23)

where we also used that:

$$\begin{array}{ccc}\hfill {\eta }_n& =& \frac{1}{1-h_0}{h}_{n-1}^k{\eta }_{n-1}\hfill \\ \hfill & \le & \frac{1}{1-h_0}{h}_0^k\frac{1}{1-h_0}{h}_0^k{\eta }_{n-2}\hfill \\ \hfill & \cdots \\ \hfill & \le & {(\frac{1}{1-h_0})}^n{h}_0^{nk}\eta .\hfill \end{array}$$

(24)

Then, we can show:

Theorem 1.

Suppose Condition (A) is satisfied and for each fixed number of steps $k,$ equation:

$$T_0(t)[\frac{1-h_0}{1-h_0-h_0^k}]\eta =t$$

(25)

has at least one positive solution. Denote by r the smallest such solution. Moreover, suppose that (20) is satisfied and $U(x_0,r\eta )\subseteq D.$ Then, sequence $\left\{x_n\right\}$ generated by Method (4) is well defined, remains in $\overline{U}(x_0,r\eta )$ for each $n=0,1,2,\dots ,i=1,2,\dots ,k$ and converges to a solution $x^{\ast }\in \overline{U}(x_0,r\eta )$ of equation $F(x)=0.$ The solution $x^{\ast }$ is unique in $D_1.$

Proof. 

It follows from the previous results that $x_n^{(i)}$ and $x_n^{(k)}=x_n$ belong in $U(x_0,r\eta ).$ We must show that sequence $\left\{x_n\right\}$ is complete:

$$\begin{array}{ccc}\hfill \parallel x_{n+j}-x_n\parallel & \le & \sum _{i=1}^j\parallel x_{n+i}-x_{n_i-1}\parallel \hfill \\ \hfill & \le & \sum _{i=1}^j{T}_{n+i-1}(r){\eta }_{n+i-1}\hfill \\ \hfill & \le & T_0(r)\sum _{i=1}^j{\eta }_{n+i-1}\le T_0(r)\sum _{i=0}^{j-1}{\eta }_{n+i}\hfill \\ \hfill & \le & T_0(r)\sum _{i=0}^{j-1}{(\frac{h_0^k}{1-h_0})}^{n+i}\eta \hfill \\ \hfill & \le & T_0(r)\frac{(\frac{h_0^k}{1-h_0})-{(\frac{h_0^k}{1-h_0})}^{n+j}}{1-\frac{h_0^k}{1-h_0}}\eta ,\hfill \end{array}$$

(26)

$(\frac{h_0^k}{1-h_0}<1),$ so $\left\{x_n\right\}$ is complete in a Banach space X, and as such, it converges to some $x^{\ast }\in \overline{U}(x_0,r\eta ),$ since $\overline{U}(x_0,r\eta )$ is a closed set. Moreover, we have:

$$\begin{array}{ccc}\hfill \parallel F(x_n)\parallel & =& \parallel F(x_{n-1}^{(k)})\parallel \le {\rho }_{n-1}(r)\parallel x_{n-1}^{(k)}-x_{n-1}^{(k-1)}\parallel \hfill \\ & \le & {\rho }_{n-1}(r)h_{n-1}^{k-1}\parallel x_n^{(1)}-x_n^{(0)}\parallel \hfill \\ \hfill & \le & h_n^{k-1}\parallel x_n^{(1)}-x_n^{(0)}\parallel \le h_n^{k-1}{\eta }_n\hfill \\ \hfill & \le & h_n^{k-1}{(\frac{h_0^k}{1-h_0})}^n\le h_0^{k-1}{(\frac{h_0^k}{1-h_0})}^n\hfill \\ \hfill & \le & {(\frac{h_0^k}{1-h_0})}^n⟶0asn⟶\infty ,\hfill \end{array}$$

so $F(x^{\ast })=0.$ Furthermore, to show the uniqueness part, let $y^{\ast }\in D_1$ with $F(y^{\ast })=0.$ Set $Q={\int }_0^1{F}^{\prime }(x^{\ast }+\theta (y^{\ast }-x^{\ast }))d\theta .$ By (a4) and (a6), we get in turn that:

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{(x_0)}^{-1}\parallel \parallel Q-F(x_0)\parallel & \le & \overline{\beta }{\int }_0^1[K_0\parallel x^{\ast }+\theta (y^{\ast }-x^{\ast })-x_0\parallel ]d\theta \hfill \\ \hfill & \le & \beta [\frac{\parallel x^{\ast }-x_0\parallel +\parallel y^{\ast }-x_0\parallel }{2}]\hfill \\ \hfill & \le & \beta [\frac{K_0(r+r^{\ast })\eta }{2}]<1\hfill \end{array}$$

(27)

by (27), so $Q^{-1}\in \mathcal{L}(\mathcal{Y},\mathcal{X}).$ Then, from the identity $0=F(y^{\ast })-F(x^{\ast })=Q(y^{\ast }-x^{\ast }),$ we conclude that $x^{\ast }=y^{\ast }.$

Remark 1.

As noted in the Introduction, even if specialized to $A(x)=F^{\prime }(x),$ Theorem 1 can give better results, since $K_0\le K.$ As an example, consider the uniqueness result in [12], where:

$$r<\frac{2}{\overline{\beta }K\eta }=r_1,$$

but $r_0<r_1$ for $K_0<K.$

4. Conclusions

We presented a semi-local convergence analysis for a k-steps iterative method with a flexible and frozen linear operator. The results obtained in this article reduce to the ones given in [1,2,12], if we choose $A(x)=F^{\prime }(x)$ for each $x\in D.$ On top of that, in the special case, our results have the following advantages over these works:

(1)

Larger convergence region, leading to more initial points;

(2)

Tighter upper bound estimates on $\parallel x_{n+1}-x_n\parallel$, as well as $\parallel x_n-x^{\ast }\parallel$, which means that fewer iterations are needed to arrive at a desired error tolerance.

(3)

The information on the location of the solution is at least as precise.

These advantages are obtained, since we locate a ball inside the old ball containing the iterates. Then, the Lipschitz constants depend on the smallest ball and that is why these constants are at least as small as the old ones. It is also worth noticing that these advantages are attained, because the new constants are special cases of the old ones. That is no additional effort is required to compute the new constants. A plethora of numerical examples where the new constants are strictly smaller than the old ones can be found in [3,4,5,6,7,8]. Finally, other choices of operator A lead to methods not studied before.