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<h1>Semilocal convergence conditions for the secant method, using recurrent functions</h1>
<p class="authors">
<span class="author">Ioannis K. Argyros\(^\ast \) Saïd Hilout\(^\S \)</span>
</p>
<p class="date">October 1, 2010</p>
</div>
<p>\(^\ast \)Cameron University, Department of Mathematics Sciences, Lawton, OK 73505, USA, e-mail: <span class="tt">iargyros@cameron.edu</span>. </p>
<p>\(^\S \)Poitiers University, Laboratoire de Mathématiques et Applications, Bd. Pierre et Marie Curie, Téléport 2, B.P. 30179, 86962 Futuroscope Chasseneuil Cedex, France, e-mail: <span class="tt">said.hilout@math.univ-poitiers.fr</span>. </p>

<div class="abstract"><p> Using our new concept of recurrent functions, we present new sufficient convergence conditions for the secant method to a locally unique solution of a nonlinear equation in a Banach space. We combine Lipschitz and center–Lipschitz conditions on the divided difference operator to obtain the semilocal convergence analysis of the secant method. Our error bounds are tighter than earlier ones. Moreover, under our convergence hypotheses, we can expand the applicability of the secant method in cases not covered before <span class="cite">
	[
	<a href="#623-6" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-7" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-10" >12</a>
	]
</span>–<span class="cite">
	[
	<a href="#potra-01" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-11" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-12" >19</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-13" >21</a>
	]
</span>. Application and examples are also provided in this study. </p>
<p><b class="bf">MSC.</b> 65H10, 65B05, 65G99, 65N30, 47H17, 49M15. </p>
<p><b class="bf">Keywords.</b> Recurrent functions, semilocal convergence, secant method, Banach space, majorizing sequence, divided difference, Fréchet derivative. </p>
</div>
<h1 id="a0000000002">1  Introduction</h1>
<p> In this study we are concerned with the problem of approximating a locally unique solution \(x^\star \) of equation </p>
<div class="equation" id="1.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{1.1} F(x) =0, \end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p> where \(F\) is a Fréchet–differentiable operator defined on a convex subset \(\mathcal{D}\) of a Banach space \(\mathcal{X}\) with values in a Banach space \(\mathcal{Y}\). </p>
<p>The field of computational sciences has seen a considerable development in mathematics, engineering sciences, and economic equilibrium theory. For example, dynamic systems are mathematically modeled by difference or differential equations, and their solutions usually represent the states of the systems. For the sake of simplicity, assume that a time–invariant system is driven by the equation \(\dot{x} =T(x)\), for some suitable operator \(T\), where \(x\) is the state. Then the equilibrium states are determined by solving equation (<a href="#1.1">1.1</a>). Similar equations are used in the case of discrete systems. The unknowns of engineering equations can be functions (difference, differential, and integral equations), vectors (systems of linear or nonlinear algebraic equations), or real or complex numbers (single algebraic equations with single unknowns). Except in special cases, the most commonly used solution methods are iterative–when starting from one or several initial approximations a sequence is constructed that converges to a solution of the equation. Iteration methods are also applied for solving optimization problems. In such cases, the iteration sequences converge to an optimal solution of the problem at hand. Since all of these methods have the same recursive structure, they can be introduced and discussed in a general framework. We note that in computational sciences, the practice of numerical analysis for finding such solutions is essentially connected to variants of Newton’s method. </p>
<p>We consider the secant method (SM) in the form </p>
<div class="equation" id="1.2">
<p>
  <div class="equation_content">
    \begin{equation} \label{1.2} x_{n +1}= x_n -\delta F(x_{n -1} ,x_n)^{-1} \, \,  F(x_n) \quad (n\geq 0), \quad (x_{-1}, x_0 \in \mathcal{D}) \end{equation}
  </div>
  <span class="equation_label">1.2</span>
</p>
</div>
<p> where, \(\delta F(x,y) \in \mathcal{L}(\mathcal{X},\mathcal{Y})\) \((x,y\in \mathcal{D})\) is a consistent approximation of the Fréchet–derivative of \(F\) <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >11</a>
	]
</span>. \( \mathcal{L}(\mathcal{X}, \mathcal{Y})\) denotes the space of bounded linear operators from \(\mathcal{X}\) into \(\mathcal{Y}\). (SM) is an alternative method of Newton’s method (NM) </p>
<div class="displaymath" id="a0000000003">
  \[  x_{n+1}= x_n -F' (x_n) ^{-1} \, \, F(x_n) , \quad (n \geq 0), \quad (x_0 \in \mathcal{D}) .  \]
</div>
<p> Dennis <span class="cite">
	[
	<a href="#623-6" >8</a>
	]
</span>, Potra <span class="cite">
	[
	<a href="#potra-01" >14</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-11" >16</a>
	]
</span>, Bosarge and Falb <span class="cite">
	[
	<a href="#4" >7</a>
	]
</span>, Hernández, M.J. Rubio and J.A. Ezquerro <span class="cite">
	[
	<a href="#623-7" >9</a>
	]
</span>, Argyros <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, and others <span class="cite">
	[
	<a href="#8" >10</a>
	]
</span>, <span class="cite">
	[
	<a href="#10" >13</a>
	]
</span>, <span class="cite">
	[
	<a href="#12" >20</a>
	]
</span>, have provided sufficient convergence conditions for (SM) based on Lipschitz–type conditions on \(\delta F\) (see, also relevant works in <span class="cite">
	[
	<a href="#623-2" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-3" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#arg-hil-polimetrica1" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-10" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#669-P1" >17</a>
	]
</span>, <span class="cite">
	[
	<a href="#669-P2" >18</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-12" >19</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-13" >21</a>
	]
</span>). </p>
<p>In the previous mentioned references, the conditions usually associated with the semilocal convergence of secant method (<a href="#1.2">1.2</a>) are: </p>
<ol class="enumerate">
  <li><p>\(F\) is a nonlinear operator defined on a convex subset \(\mathcal{D}\) of a Banach space \(\mathcal{X}\) with values in a Banach space \(\mathcal{Y}\); </p>
</li>
  <li><p>\(x_{-1}\) and \(x_0\) are two points belonging to the interior \(\mathcal{D}^0\) of \(\mathcal{D}\) and satisfying the inequality </p>
<div class="displaymath" id="a0000000004">
  \[  \parallel x_0 -x_{-1} \parallel \leq c;  \]
</div>
</li>
  <li><p>\(F\) is Fréchet–differentiable on \(\mathcal{D}^0\), and there exists an operator <br />\(\delta F \,  \colon \,  \mathcal{D}^0 \times \mathcal{D}^0 \to \mathcal{L}(\mathcal{X},\mathcal{Y})\), such that the linear operator \(A \! =\! \delta F(x_{-1} ,x_0)\) is invertible, its inverse \(A^{-1}\) is bounded, and: </p>
<div class="displaymath" id="a0000000005">
  \[  \parallel A^{-1} \,  F(x_0) \parallel \leq \eta ;  \]
</div>
<div class="displaymath" id="a0000000006">
  \[  \parallel A^{-1} \,  ( \delta F(x,y) -F'(z)) \parallel \leq \ell \,  (\parallel x-z\parallel +\parallel y -z\parallel ), \quad {\rm for \, \,  all} \quad x,y,z\in \mathcal{D};  \]
</div>
<div class="displaymath" id="a0000000007">
  \[  \overline{U} (x_0,r)= \{  x\in \mathcal{X}\,  : \,  \parallel x- x_0 \parallel \leq r \}  \subseteq \mathcal{D}^0 ,  \]
</div>
<p> for some \(r{\gt}0\) depending on \(\ell \), \(c\), and \(\eta \); and </p>
<div class="equation" id="1.3">
<p>
  <div class="equation_content">
    \begin{equation} \label{1.3} \ell \, \,  c + 2 \, \,  \sqrt{\ell \, \,  \eta } \leq 1 . \end{equation}
  </div>
  <span class="equation_label">1.3</span>
</p>
</div>
</li>
</ol>
<p>The sufficient convergence condition (<a href="#1.3">1.3</a>) is easily violated. Indeed, let \(\ell =1\), \(\eta = .18\), and \(c=.185\). Then, (<a href="#1.3">1.3</a>) does not holds, since </p>
<div class="displaymath" id="a0000000008">
  \[ \ell \, \,  c + 2 \, \,  \sqrt{\ell \, \,  \eta }=1.033528137 .  \]
</div>
<p> Moreover, our recently found corresponding conditions are also violated <span class="cite">
	[
	<a href="#ar-hi-24" >6</a>
	]
</span> (see, Remark <a href="#R.2.4">4</a>(c)). Hence, there is not guarantee that equation (<a href="#1.1">1.1</a>) under the information (\(\ell , c, \eta \)) has a solution that can be found using (SM). In this study we are motivated by optimization considerations, and the above observation. </p>
<p>Here, using a combination of Lipschitz and center–Lipschitz conditions, we provide a semilocal convergence analysis for (SM). Our error bounds are tighter, and our convergence conditions hold in cases where the corresponding hypotheses in earlier references <span class="cite">
	[
	<a href="#623-6" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-7" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-10" >12</a>
	]
</span>–<span class="cite">
	[
	<a href="#potra-01" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-11" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-12" >19</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-13" >21</a>
	]
</span> are violated. Applications and examples are also provided in this study. </p>
<h1 id="a0000000009">2 Semilocal convergence analysis of (SM)</h1>
<p> We need the following result on majorizing sequences for (SM). <div class="lemma_thmwrapper " id="L.2.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(\ell _0 {\gt} 0\), \(\ell {\gt} 0\), \(c \geq 0\), and \( \eta \in (0,c]\) be given constants. Assume: </p>
<div class="equation" id="2.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.1} (\ell + 2 \, \,  \ell _0 ) \, \,  \eta + \ell _0 \, \,  c < 1 ; \end{equation}
  </div>
  <span class="equation_label">2.1</span>
</p>
</div>
<p> for </p>
<div class="displaymath" id="a0000000010">
  \[  \delta _0= \displaystyle \tfrac {\ell \, \,  (c+\eta )} {1- \ell _0 \, \,  (c+\eta )},  \]
</div>
<div class="displaymath" id="a0000000011">
  \[  \delta = \displaystyle \tfrac {2 \,  (1- ((\ell + 2\,  \ell _0) \,  \eta + \ell _0 \,  c))} { (\ell + 2\,  \ell _0) \,  \eta + \sqrt{ ((\ell + 2\,  \ell _0) \,  \eta )^2+ 4 \,  \ell _0 \,  \eta \,  (1- ((\ell + 2\,  \ell _0) \,  \eta + \ell _0 \,  c) }) } ,  \]
</div>
<p> \(\delta _1\) the unique positive root of polynomial \(f\) in \((0,1)\) </p>
<div class="equation" id="2.2">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.2} f(t)= \ell _0 \, \,  t^3 + (\ell _0+\ell ) \, \,  t^2 -\ell , \end{equation}
  </div>
  <span class="equation_label">2.2</span>
</p>
</div>
<p> given by </p>
<div class="equation" id="2.2-ab">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.2-ab} \delta _1 = \displaystyle \tfrac {1}{2} \, \,  \displaystyle \tfrac {-\ell + \sqrt{\ell ^2 + 4 \,  \ell \,  \ell _0}}{\ell _0}, \end{equation}
  </div>
  <span class="equation_label">2.3</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.3">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.3} \delta _0 \leq \delta \leq \delta _1 . \end{equation}
  </div>
  <span class="equation_label">2.4</span>
</p>
</div>
<p>Then, scalar sequence \(\{ t_n \} \) (\(n\geq -1\)) given by </p>
<div class="equation" id="2.4">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.4} t_{-1}=0, \, \,  t_0 =c, \, \,  t_1 =c+\eta ,\, \,  t_{n +2}= t_{n +1}+ \displaystyle \tfrac {\ell \, \,  (t_{n +1} -t_{n -1}) \, \,  (t_{n +1}- t_n) }{1 -\ell _0 \, \,  (t_{n +1}- t_0 +t_n)} \end{equation}
  </div>
  <span class="equation_label">2.5</span>
</p>
</div>
<p> is non–decreasing, bounded from above by </p>
<div class="equation" id="2.5">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.5} t^{\star \star } = \displaystyle \tfrac {\eta }{1 -\delta } +c , \end{equation}
  </div>
  <span class="equation_label">2.6</span>
</p>
</div>
<p> and converges to its unique least upper bound \(t^\star \) such that </p>
<div class="equation" id="2.6">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.6} 0 \leq t^\star \leq t^{\star \star } . \end{equation}
  </div>
  <span class="equation_label">2.7</span>
</p>
</div>
<p> Moreover, the following estimates hold for all \(n\geq 0\): </p>
<div class="equation" id="2.7">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.7} 0\leq t_{n +2} -t_{n +1} \leq \delta \, \,  (t_{n +1}- t_n) \leq \delta ^{n +1} \, \,  \eta . \end{equation}
  </div>
  <span class="equation_label">2.8</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000012">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> In view of (<a href="#2.1">2.1</a>), we have \(\delta _0 \in [0,1)\), and \(\delta {\gt}0\). </p>
<p>We shall show using mathematical induction on \(k\geq 0\) </p>
<div class="equation" id="2.8">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.8} 0\leq t_{k +2} -t_{k +1} \leq \delta \, \,  (t_{k +1}- t_k) . \end{equation}
  </div>
  <span class="equation_label">2.9</span>
</p>
</div>
<p> By (<a href="#2.4">2.5</a>) for \(k=0\), we must show </p>
<div class="displaymath" id="a0000000013">
  \[  0\leq \displaystyle \tfrac {\ell \, \,  (t_1 - t_{-1})} {1 - \ell _0 \, \,  t_1} \leq \delta \quad {\rm or } \quad 0\leq \displaystyle \tfrac {\ell \, \,  (c + \eta )} {1 - \ell _0 \, \,  (c+ \eta )} \leq \delta ,  \]
</div>
<p> which is true from (<a href="#2.1">2.1</a>), and the choice of \(\delta \geq \delta _0\). </p>
<p>Let assume that (<a href="#2.8">2.9</a>) holds for \(k\leq n+1\). The induction hypothesis gives </p>
<div class="equation" id="2.9">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.9} \begin{array}{lll} t_{k +2} &  \leq &  t_{k +1} +\delta \, \,  (t_{k +1}-t_k) \\ & \leq &  t_k +\delta \, \,  (t_k -t_{k -1}) +\delta \, \,  (t_{k +1}- t_k)\\ &  \leq &  t_1+\delta \, \,  (t_1-t_0) +\cdots +\delta \, \,  (t_{k +1}-t_k)\\ & \leq &  c +\eta +\delta \, \,  \eta +\cdots +\delta ^{k +1} \, \,  \eta \\ & =&  c+ \displaystyle \tfrac {1- \delta ^{k +2}}{1 -\delta } \, \,  \eta < \displaystyle \tfrac {\eta }{1 -\delta } +c= t^{\star \star } . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.10</span>
</p>
</div>
<p> Moreover, we have: </p>
<div class="equation" id="2.10">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.10} \begin{array}{l} \ell \, \,  (t_{k +2} - t_{k } ) + \delta \, \,  \ell _0 \, \,  (t_{k +2} -t_0 + t_{k +1} )\\ \leq \ell \, \,  {\bigg(} (t_{k +2} - t_{k +1})+ (t_{k +1} -t_{k }) {\bigg)} + \delta \, \,  \ell _0 \, \,  {\bigg(} \displaystyle \tfrac {1- \delta ^{k +2}}{1 -\delta } + \displaystyle \tfrac {1- \delta ^{k +1}}{1 -\delta } {\bigg)}\, \,  \eta + \delta \, \,  \ell _0\, \,  c\\ \leq \ell \, \,  ( \delta ^k + \delta ^{k+1} ) \, \,  \eta + \displaystyle \tfrac {\delta \, \,  \ell _0}{1-\delta }\, \,  (2- \delta ^{k+1} - \delta ^{k+2} )\, \,  \eta + \delta \, \,  \ell _0 \, \,  c . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.11</span>
</p>
</div>
<p> We prove now (<a href="#2.8">2.9</a>). By (<a href="#2.4">2.5</a>), we can have estimate </p>
<div class="equation" id="2.11">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.11} \ell \, \,  ( \delta ^k + \delta ^{k+1} ) \, \,  \eta + \displaystyle \tfrac {\delta \, \,  \ell _0}{1-\delta }\, \,  (2- \delta ^{k+1} - \delta ^{k+2} )\, \,  \eta + \delta \, \,  \ell _0 \, \,  c \leq \delta \end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>
<p> or </p>
<div class="equation" id="2.12">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.12} \ell \, \,  ( \delta ^{k-1} + \delta ^{k} ) \, \,  \eta + \ell _0 \, \,  {\bigg(} (1+\delta + \cdots + \delta ^k) + (1+\delta + \cdots + \delta ^{k+1}) {\bigg)} \, \,  \eta + \ell _0 \, \,  c - 1 \leq 0. \end{equation}
  </div>
  <span class="equation_label">2.13</span>
</p>
</div>
<p> In view of (<a href="#2.12">2.13</a>), we are motivated to define (for \(\delta =s\)) the functions for \(k\geq 1\) </p>
<div class="equation" id="2.13">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.13} f_k (s) = \ell \, \,  ( s ^{k-1} + s ^{k} ) \, \,  \eta + \ell _0 \, \,  {\bigg(} 2\, \,  (1+s + \cdots + s ^k) + s ^{k+1} {\bigg)} \, \,  \eta + \ell _0 \, \,  c - 1 . \end{equation}
  </div>
  <span class="equation_label">2.14</span>
</p>
</div>
<p> We need the relationship between two consecutive functions \(f_k\). Using (<a href="#2.13">2.14</a>), we obtain </p>
<div class="equation" id="2.14">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.14} \begin{array}{lll} f_{k+1} (s) & = &  \ell \, \,  ( s ^{k}\!  +\!  s ^{k+1} ) \, \,  \eta \! +\!  \ell _0 \, \,  {\bigg(} 2\, \,  (1\! +\! s \! + \! \cdots \!  + \! s ^{k+1})\!  +\!  s ^{k+2} {\bigg)} \, \,  \eta + \ell _0 \, \,  c - 1 \\ & =&  f_k(s) + \ell \, \,  ( s ^{k+1} - s^{k-1} ) \, \,  \eta + \ell _0 \, \,  ( s ^{k+1} + s^{k+2} )\, \,  \eta \\ & =&  f(s)\, \,  s^{k-1} \, \,  \eta + f_k(s). \end{array} \end{equation}
  </div>
  <span class="equation_label">2.15</span>
</p>
</div>
<p> Note that \(\delta \) is the unique positive root of polynomial \(f_1\). </p>
<p>Instead of (<a href="#2.12">2.13</a>), we shall show </p>
<div class="equation" id="2.15">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.15} f_k (\delta ) \leq 0 \qquad (k \geq 0). \end{equation}
  </div>
  <span class="equation_label">2.16</span>
</p>
</div>
<p> Estimate (<a href="#2.15">2.16</a>) holds for \(k=1\), as equality. Using (<a href="#2.14">2.15</a>), we get in turn </p>
<div class="displaymath" id="a0000000014">
  \[  f_{2} (\delta ) = f_{1} (\delta ) + f(\delta ) \, \delta \,  \eta \leq 0 ,  \]
</div>
<p> since, </p>
<div class="displaymath" id="a0000000015">
  \[  f_1(\delta ) =0 , \quad f(\delta ) \leq 0 \quad ({\rm by} \, \,  (\ref{2.2}), \, \,  {\rm and} \, \,  \delta _1 \geq \delta ).  \]
</div>
<p> Assume (<a href="#2.15">2.16</a>) holds for \(m \leq k\). Then, again by (<a href="#2.14">2.15</a>), we have: </p>
<div class="displaymath" id="a0000000016">
  \[  f_{k+1} (\delta ) = f_{k} (\delta ) + f(\delta ) \, \delta ^{k-1} \,  \eta \leq 0,  \]
</div>
<p> which completes the induction for (<a href="#2.15">2.16</a>). </p>
<p>Moreover, define function \(f_{\infty }\) on \([0,1)\) by </p>
<div class="equation" id="2.16">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.16} f_{\infty } (s) = \displaystyle \lim _{k\longrightarrow \infty } f_k (s) . \end{equation}
  </div>
  <span class="equation_label">2.17</span>
</p>
</div>
<p> Using (<a href="#2.15">2.16</a>), and (<a href="#2.16">2.17</a>), we obtain </p>
<div class="equation" id="2.17">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.17} f_{\infty } (\delta ) = \displaystyle \lim _{k\longrightarrow \infty } f_k (\delta ) \leq 0 . \end{equation}
  </div>
  <span class="equation_label">2.18</span>
</p>
</div>
<p>Hence, we showed sequence \(\{ t_n\} \) (\(n\geq -1\)) is non–decreasing and bounded above from by \(t^{\star \star } \), so that (<a href="#2.7">2.8</a>) holds. It follows that there exists \(t^\star \in [0, t^{\star \star } ]\), so that \(\displaystyle \lim _{n\longrightarrow \infty } t_n= t^\star \). </p>
<p>That completes the proof of Lemma <a href="#L.2.1">1</a>. <div class="proof_wrapper" id="a0000000017">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We shall study (SM) for triplets \((F, x_{-1},x_0)\) belonging to the class <br />\(\mathcal{C}(\ell ,\ell _0 ,\eta ,c, \delta )\) defined as follows: <div class="definition_thmwrapper " id="D.2.2">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2</span>
  </div>
  <div class="definition_thmcontent">
  <p> Let \(\ell \), \(\ell _0\), \(\eta \), \(c\), \(\delta \) be non–negative constants satisfying the hypotheses of Lemma <span class="rm"><a href="#L.2.1">1</a></span>. A triplet \((F, x_{-1} ,x_0)\) belongs to the class \(\mathcal{C}(\ell ,\ell _0 ,\eta ,c, \delta )\) if: </p>
<ol class="enumerate">
  <li><p>\(F\) is a nonlinear operator defined on a convex subset \(\mathcal{D}\) of a Banach space \(\mathcal{X}\) with values in a Banach space \(\mathcal{Y}\); </p>
</li>
  <li><p>\(x_{-1}\) and \(x_0\) are two points belonging to the interior \(\mathcal{D}^0\) of \(\mathcal{D}\) and satisfying the inequality </p>
<div class="displaymath" id="a0000000018">
  \[  \parallel x_0 -x_{-1} \parallel \leq c;  \]
</div>
</li>
  <li><p>\(F\) is Fréchet–differentiable on \(\mathcal{D}^0\), and there exists an operator <br />\(\delta F \,  \colon \,  \mathcal{D}^0 \times \mathcal{D}^0 \to \mathcal{L}(\mathcal{X},\mathcal{Y})\), such that, \(A^{-1} =\delta F(x_{-1} ,x_0) ^{-1}\in \mathcal{L}(\mathcal{Y}, \mathcal{X})\), and for all \(x,y,z\in \mathcal{D}\), the following hold </p>
<div class="displaymath" id="a0000000019">
  \[  \parallel A^{-1} \,  F(x_0) \parallel \leq \eta ,  \]
</div>
<div class="displaymath" id="a0000000020">
  \[  \parallel A^{-1 }\,  ( \delta F(x,y) -F'(z) ) \parallel \leq \ell \, \,  (\parallel x-z\parallel +\parallel y -z\parallel ),  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000021">
  \[  \parallel A^{-1 }\,  ( \delta F(x,y) -F'(x_0) ) \parallel \leq \ell _0 \,  (\parallel x-x_0\parallel +\parallel y -x_0\parallel ) ;  \]
</div>
</li>
  <li><p>\(\overline{U} (x_0 ,t^\star ) \subseteq \mathcal{D}_c =\{  x\in \mathcal{D}\,  : \,  F\, \,  {\rm is \, \,  continuous \, \,  at} \, \,  x \}  \subseteq \mathcal{D},\) where, \(t^\star \) is given in Lemma <a href="#L.2.1">1</a>; </p>
</li>
  <li><p>For </p>
<div class="displaymath" id="a0000000022">
  \[  \delta _2 = \displaystyle \tfrac {2\,  (1- \ell _0 \,  (c+ \eta ))}{1- \ell _0 \,  c},  \]
</div>
<p> we have </p>
<div class="displaymath" id="a0000000023">
  \[  \left\{  \begin{array}{lll} \delta \leq \delta _1 & {\rm if}&  \delta _2 {\lt} \delta _1 \\ \delta {\lt} \delta _2 & {\rm if}&  \delta _1 \leq \delta _2 . \end{array} \right.  \]
</div>
</li>
</ol>

  </div>
</div> </p>
<p>The semilocal convergence theorem for (SM) is as follows. <div class="theorem_thmwrapper " id="T.2.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> If \((F, x_{-1} ,x_0) \in \mathcal{C}(\ell ,\ell _0 ,\eta ,c, \delta )\), then, the sequence \(\{  x_n\} \) \((n\geq -1)\) generated by (SM) is well defined, remains in \(\overline{U} (x_0 ,t^\star )\) for all \(n\geq 0\), and converges to a unique solution \(x^\star \in \overline{U} (x_0 ,t^\star )\) of <span class="rm">(<a href="#1.1">1.1</a>)</span>. </p>
<p>Moreover the following estimates hold for all \(n\geq 0\) </p>
<div class="equation" id="2.18">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.18} \parallel x_{n }- x_{n - 1} \parallel \leq t_{n }- t_{n -1} , \end{equation}
  </div>
  <span class="equation_label">2.19</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.19">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.19} \parallel x_n -x^\star \parallel \leq t^\star - t_n \end{equation}
  </div>
  <span class="equation_label">2.20</span>
</p>
</div>
<p> where, \(\{  t_n\} \), \((n\geq 0)\) is given by <span class="rm">(<a href="#2.4">2.5</a>)</span>. </p>
<p>Furthermore, if there exists \(R{\gt}0\), such that </p>
<div class="equation" id="2.20">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.20} R \geq t^\star - t_0 \quad {\rm and} \quad \ell _0 \,  {\bigg(} \displaystyle \tfrac {\eta }{1- \delta } + c +R {\bigg)} \leq 1 , \end{equation}
  </div>
  <span class="equation_label">2.21</span>
</p>
</div>
<p> then, the solution \(x^\star \) is unique in \( {U} (x_0, R) \). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> First, we show that \(L =\delta F(x_k,x_{k+1})\) is invertible for \(x_k , x_{k+1} \in \overline{U} (x_0, t^\star )\). By (<a href="#2.4">2.5</a>), (<a href="#2.5">2.6</a>), <b class="bf">(\(\mathcal{A}_2\))</b> and <b class="bf">(\(\mathcal{A}_3\))</b>, we have </p>
<div class="displaymath" id="2.23">
  \begin{align} \label{2.23} \parallel I -A^{-1} \,  L\parallel &  = \parallel A^{-1} \,  (L-A)\parallel \\ &  \leq \parallel A^{-1} (L -F’(x_0))\parallel + \parallel A^{-1} (F’(x_0) -A)\parallel \nonumber \\ &  \leq \ell _0 \,  (\parallel x_k -x_0\parallel + \parallel x_{k+1} -x_0\parallel + \parallel x_0 -x_{-1} \parallel )\nonumber \\ & \leq \ell _0 \,  (t_k - t_0 + t_{k+1} - t_0 + c) \nonumber \\ & \leq \ell _0 \,  (t^\star - t_0 + t^\star - t_0 + c) \nonumber \\ & \leq \ell _0 \,  { \bigg(} 2 \,  {\bigg(}\displaystyle \tfrac {\eta }{1- \delta } +c{\bigg)} -c {\bigg)} {\lt} 1 ,\nonumber \end{align}
</div>
<p> by <b class="bf">(\(\mathcal{A}_5\))</b>. </p>
<p>Using the Banach Lemma on invertible operators <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >11</a>
	]
</span>, and (<a href="#2.23">2.22</a>), \(L\) is invertible and </p>
<div class="equation" id="2.24">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.24} \parallel L^{-1} \,  A\parallel \leq {\bigg(}1 -\ell _0 \,  (\parallel x_k -x_0\parallel + \parallel x_{k+1}-x_0\parallel +c){\bigg)}^{-1} . \end{equation}
  </div>
  <span class="equation_label">2.23</span>
</p>
</div>
<p> By <b class="bf">(\(\mathcal{A}_3\))</b>, we have </p>
<div class="equation" id="2.25">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.25} \parallel A^{-1} \,  (F'(u) -F'(v)) \parallel \leq 2 \,  \ell \,  \parallel u-v\parallel , \quad u,v\in \mathcal{D}^0. \end{equation}
  </div>
  <span class="equation_label">2.24</span>
</p>
</div>
<p> We can write the identity </p>
<div class="equation" id="2.26">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.26} F(x) -F(y) =\displaystyle \int ^1_0 F'(y +t (x -y)) \,  {\rm d}t \,  (x-y) \end{equation}
  </div>
  <span class="equation_label">2.25</span>
</p>
</div>
<p> then, for all \(x,y,u,v\in \mathcal{D}^0\), we obtain </p>
<div class="equation" id="2.27">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.27} \begin{array}{l} \parallel A_0^{-1}\,  (F(x) -F(y) -F’(u) (x-y)) \parallel \\ \leq \ell \, \,  (\parallel x-u\parallel + \parallel y-u\parallel ) \,  \parallel x-y \parallel \end{array} \end{equation}
  </div>
  <span class="equation_label">2.26</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.28">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.28} \begin{array}{l} \parallel A_0^{-1} \,  (F(x) -F(y) -\delta F(u,v) \,  (x-y)) \parallel \\ \leq \ell \,  (\parallel x-v \parallel + \parallel y -v \parallel + \parallel u -v\parallel ) \parallel x-y \parallel . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.27</span>
</p>
</div>
<p> By a continuity argument (<a href="#2.25">2.24</a>)–(<a href="#2.28">2.27</a>) remain valid if \(x\) and/or \(y\) belong to \(\mathcal{D}_c\). </p>
<p>Now we show (<a href="#2.18">2.19</a>). If (<a href="#2.18">2.19</a>) holds for all \(n \leq k\) and if \(\{  x_n\} \) \((n\geq 0)\) is well defined for \(n = 0,1,2,\cdots ,k\), then </p>
<div class="equation" id="2.29">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.29} \parallel x_n - x_0 \parallel \leq t_n -t_0 < t^\star - t_0 ,\quad n\leq k. \end{equation}
  </div>
  <span class="equation_label">2.28</span>
</p>
</div>
<p> That is (<a href="#1.2">1.2</a>) is well defined for \(n =k+1\). For \(n =-1\), and \(n=0\), (<a href="#2.18">2.19</a>) reduces to \(\parallel x_{-1} -x_0\parallel \leq c\), and \(\parallel x_0 -x_1\parallel \leq \eta \). Suppose (<a href="#2.18">2.19</a>) holds for \(n = -1,0, 1,\cdots ,k\) \((k\geq 0)\). By (<a href="#2.24">2.23</a>), (<a href="#2.28">2.27</a>), and </p>
<div class="equation" id="2.30">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.30} F(x_{k +1}) =F(x_{k +1}) -F(x_k) - \delta F(x_{k -1} ,x_k) \, \,  (x_{k +1}- x_k) \end{equation}
  </div>
  <span class="equation_label">2.29</span>
</p>
</div>
<p> we obtain the following estimate </p>
<div class="equation" id="2.31">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.31} \begin{array}{lll} \parallel x_{k +2}- x_{k +1}\parallel &  =&  \parallel \delta F(x_k ,x_{k +1})^{-1} \,  F(x_{k +1}) \parallel \\ &  \leq &  \parallel \delta F(x_k ,x_{k +1})^{-1} \,  A\parallel \,  \parallel A^{-1} \,  F(x_{k +1}) \parallel \\ &  \leq &  \displaystyle \tfrac {\ell \,  (\parallel x_{k +1}- x_k\parallel + \parallel x_k -x_{k -1} \parallel )}{1-\ell _0 \,  (\parallel x_{k +1}-x_0\parallel + \parallel x_k -x_0 \parallel +c)} \parallel x_{k +1}- x_k\parallel \\ &  \leq &  \displaystyle \tfrac {\ell \,  (t_{k+1} -t_{k } +t_{k } -t_{k-1})} {1 -\ell _0 \,  ( t_{k+1} -t_{0}+ t_k -t_0 +t_0 -t_{-1})} \, \,  (t_{k+1} -t_{k }) \\ & =&  t_{k +2}- t_{k +1} , \end{array} \end{equation}
  </div>
  <span class="equation_label">2.30</span>
</p>
</div>
<p> and the induction for (<a href="#2.18">2.19</a>) is completed. It follows from (<a href="#2.18">2.19</a>), and Lemma <a href="#L.2.1">1</a> that\(\{  x_n\} \) \((n\geq -1)\) is Cauchy in a Banach space \(\mathcal{X}\), and as such it converges to some \(x^\star \in \overline{U} (x_0,t^\star )\) (since \(\overline{U} (x_0 ,t^\star )\) is a closed set). By letting \(k\to \infty \) in (<a href="#2.31">2.30</a>), we obtain \(F(x^\star ) =0\). Estimate (<a href="#2.19">2.20</a>) follows from (<a href="#2.18">2.19</a>) by using standard majoration techniques <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#arg-hil-polimetrica1" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >11</a>
	]
</span>. </p>
<p>Finally, for showing the uniqueness in \( \overline{U} (x_0, t^\star ) \), let \(y^\star \in \overline{U} (x_0, t^\star ) \) be a solution (<a href="#1.1">1.1</a>). Set </p>
<div class="displaymath" id="a0000000025">
  \[  \mathcal{M} = \displaystyle \int _0^1 F'(y^\star + t\, \,  (y^\star - x^\star ) ) \, \,  {\rm d}t.  \]
</div>
<p> It then follows by <b class="bf">(\(\mathcal{A}_3\))</b>, and <b class="bf">(\(\mathcal{A}_5\))</b>: </p>
<div class="equation" id="2.32">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.32} \begin{array}{lll} \parallel A^{-1} \,  (A- \mathcal{M})\parallel &  =&  \ell _0 \,  (\parallel y^\star - x_0\parallel + \parallel x^\star - x_0 \parallel + \parallel x_0 - x_{-1} \parallel )\\ & \leq &  \ell _0 \,  ((t^\star - t_0) + (t^\star - t_0) + t_0)\\ &  \leq &  \ell _0 \,  {\bigg(} 2 \,  {\bigg(} \displaystyle \tfrac {\eta }{1- \delta } + c {\bigg)} - c {\bigg)} = \ell _0 \,  {\bigg(} \displaystyle \tfrac { 2 \,  \eta }{1- \delta } + c {\bigg)} < 1 . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.31</span>
</p>
</div>
<p> It follows from (<a href="#2.32">2.31</a>), and the Banach lemma on invertible operators that \(\mathcal{M}^{-1}\) exists on \({U} (x_0, t^\star )\). </p>
<p>Using the identity: </p>
<div class="equation" id="2.33">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.33} F(x^\star ) - F (y^\star ) = \mathcal{M}\, \,  (x^\star - y^\star ) \end{equation}
  </div>
  <span class="equation_label">2.32</span>
</p>
</div>
<p> we deduce \(x^\star = y^\star \). Finally, we shall show uniqueness in \({U} (x_0, R)\). As in (<a href="#2.32">2.31</a>), we arrive at </p>
<div class="displaymath" id="a0000000026">
  \[  \parallel A^{-1} \, \,  (A - \mathcal{M})\parallel {\lt} \ell _0 \, \,  {\bigg(} \displaystyle \tfrac { \eta }{1- \delta } + c + R {\bigg)} \leq 1,  \]
</div>
<p> by (<a href="#2.20">2.21</a>)–(<a href="#2.24">2.23</a>). </p>
<p>That completes the proof of Theorem <a href="#T.2.3">3</a>. <div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <div class="remark_thmwrapper " id="R.2.4">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  
<ol class="enumerate">
  <li><p>The point \(t^{\star \star } \) given in closed form by (<a href="#2.5">2.6</a>) can replace \(t^\star \) in Theorem <a href="#T.2.3">3</a>. </p>
</li>
  <li><p>If we impose the condition \((5 \,  \ell _0 + 2 \,  \ell ) \,  \eta + \ell _0 \,  c {\gt}1 ,\) hence \(\delta \in (0,1)\). Moreover, if </p>
<div class="displaymath" id="a0000000028">
  \[  f(\delta ) \leq 0,  \]
</div>
<p> then, </p>
<div class="displaymath" id="a0000000029">
  \[  \delta \leq \delta _1  \]
</div>
<p> (see also part (c) that follows). </p>
</li>
  <li><p>Returning back to the example given in the introduction, say \(\ell _0 =.9\), we obtain \(\delta _0 = .543559196\), \(\delta = .554824435\), \(\delta _1 = .6359783661\), whereas (<a href="#2.1">2.1</a>) holds, since \(.6705 {\lt} 1\). That is our results can apply, whereas the ones using (<a href="#1.3">1.3</a>) cannot. </p>
<p>Let us define \(\delta _{\infty }\) by </p>
<div class="displaymath" id="a0000000030">
  \[  \delta _{\infty } = \displaystyle \tfrac {1- \ell _0 \,  (c + 2\,  \eta )}{1- \ell _0 \,  c}.  \]
</div>
<p> The corresponding to (<a href="#2.3">2.4</a>) condition in <span class="cite">
	[
	<a href="#ar-hi-24" >6</a>
	]
</span> is given by </p>
<div class="displaymath" id="a0000000031">
  \[  \delta _0 \leq \delta _1 \leq \delta _{\infty }.  \]
</div>
<p> But we have </p>
<div class="displaymath" id="a0000000032">
  \[  \delta _{\infty } =.611277744 {\lt} \delta _1 = .6359783661.  \]
</div>
<p> Hence, again the results in this study apply, but not the ones in our study <span class="cite">
	[
	<a href="#ar-hi-24" >6</a>
	]
</span>. </p>
<p>That is the sufficient convergence conditions in this study are different from ones in <span class="cite">
	[
	<a href="#ar-hi-24" >6</a>
	]
</span>. We conclude that as far as the convergence domains go, in practice we shall test all of them, to see which one (if any) applies. </p>
</li>
</ol>
<p><span class="qed">â–¡</span></p>

  </div>
</div> <div class="remark_thmwrapper " id="R.2.5">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">5</span>
  </div>
  <div class="remark_thmcontent">
  
<ol class="enumerate">
  <li><p>Let us define the majoring sequence \(\{  w_n\} \) used in <span class="cite">
	[
	<a href="#623-6" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-7" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-10" >12</a>
	]
</span>–<span class="cite">
	[
	<a href="#potra-01" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-11" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-12" >19</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-13" >21</a>
	]
</span> (under condition (<a href="#1.3">1.3</a>)): </p>
<div class="equation" id="2.34">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.34} w_{-1}=0, \, \,  w_0 =c, \, \,  w_1 =c+\eta ,\, \,  w_{n +2}= w_{n +1}+ \displaystyle \tfrac {\ell \,  (w_{n +1} -w_{n -1}) \,  (w_{n +1}- w_n) }{1 -\ell \,  (w_{n +1}- w_0 +w_n)} . \end{equation}
  </div>
  <span class="equation_label">2.33</span>
</p>
</div>
<p>Note that in general </p>
<div class="equation" id="2.35">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.35} \ell _0 \leq \ell \end{equation}
  </div>
  <span class="equation_label">2.34</span>
</p>
</div>
<p> holds , and \(\displaystyle \tfrac {\ell }{\ell _0}\) can be arbitrarily large <span class="cite">
	[
	<a href="#623-3" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>. In the case \(\ell _0 = \ell \), then \(t_n = w_n\) (\(n \geq -1\)). Otherwise: </p>
<div class="equation" id="2.36">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.36} t_n < w_n, \quad t_{n+1} - t_n \leq w_{n+1} - w_n , \end{equation}
  </div>
  <span class="equation_label">2.35</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.37">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.37} 0 \leq t^\star - t_n \leq w^{\star } - w_n, \quad w^{\star } = \displaystyle \lim _{n \longrightarrow \infty } w_n . \end{equation}
  </div>
  <span class="equation_label">2.36</span>
</p>
</div>
<p> Note also that strict inequality holds in (<a href="#2.36">2.35</a>) for \(n \geq 1\), if \(\ell _0 {\lt} \ell \). </p>
<p>The proof of (<a href="#2.36">2.35</a>), (<a href="#2.37">2.36</a>) can be found in <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>. Note that the only difference in the proofs is that the conditions of Lemma <a href="#L.2.1">1</a> are used here, instead of the ones in <span class="cite">
	[
	<a href="#argyros-CMJ-05" >3</a>
	]
</span>. However this makes no difference between the proofs. </p>
<p>Finally, note that (<a href="#1.3">1.3</a>) is the sufficient convergence condition for sequence (<a href="#2.34">2.33</a>). </p>
</li>
  <li><p>It turns out from the proof of Theorem <a href="#T.2.3">3</a> that \(\{  v_n\} \) given by </p>
<div class="equation" id="2.38-a">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.38-a} v_{-1}=0, \, \,  v_0 =c, \, \,  v_1 =c+\eta ,\, \,  v_{n +2}= v_{n +1}+ \displaystyle \tfrac {\ell _1 \,  (v_{n +1} -v_{n -1}) \,  (v_{n +1}- v_n) }{1 -\ell _0 \,  (v_{n +1}- v_0 +v_n)} , \end{equation}
  </div>
  <span class="equation_label">2.37</span>
</p>
</div>
<p> where, </p>
<div class="displaymath" id="a0000000033">
  \[  \ell _1 = \left\{  \begin{array}{lll} \ell _0 & {\rm if}&  n=0 \\ \ell & {\rm if}&  n{\gt}0 \end{array} \right.  \]
</div>
<p> is a finer majorizing sequence for \(\{ x_n \} \) than \(\{  t_n \} \), if \(\ell _0 {\lt} \ell \). </p>
<p>Moreover, we have </p>
<div class="equation" id="2.39-a">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.39-a} v_n < t_n, \quad v_{n+1} - v_n < t_{n+1} - t_n , \end{equation}
  </div>
  <span class="equation_label">2.38</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.40-a">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.40-a} 0\leq v^\star - v_n \leq t^\star - t_n, \quad v^{\star } = \displaystyle \lim _{n \longrightarrow \infty } v_n . \end{equation}
  </div>
  <span class="equation_label">2.39</span>
</p>
</div>
</li>
</ol>
<p><span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000034">3 Examples</h1>
<p> In this section, we present some numerical examples. <div class="example_thmwrapper " id="E.3.1">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">6</span>
  </div>
  <div class="example_thmcontent">
  <p> In the following table, we validate our Remark <a href="#R.2.4">4</a>(c) and <a href="#R.2.5">5</a>(b). <span class="rm"><div class="centered"> <b class="bf">Comparison table.</b> <br /><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> (<a href="#2.4">2.5</a>) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> (<a href="#2.34">2.33</a>) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> (<a href="#2.38-a">2.37</a>) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(n\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(t_{n+1} - t_n \) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(w_{n+1} - w_n \) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(v _{n+1} - v_n \) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>1</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0978406552 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> .1034645669 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> .0880565897 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>2</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0645024008 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> .0834298221 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> .0548615926 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>3</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0380319604</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> .0947064307</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> .0259951222 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>4 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0213029402</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>-1.250121362 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> .0091844655</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>5 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0097492108</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0578542134 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0016385499</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>6 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0029765775</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0165794721</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0000946072 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>7 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0004197050</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -.1376680823</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>8.821 \(\times 10^{-7}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>8 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>.0000163476</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>-.5014996853</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>5 \(\times 10^{-10} \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>9 </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 8.21\(\times 10^{-8}\)</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> .5290112928</p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
  </tr>
</table> </div> </span> The table shows that our error bounds \(v_{n+1}-v_n\), and \(t_{n+1}-t_n\) are finer than \(w_{n+1} -w_n\) given in <span class="cite">
	[
	<a href="#623-6" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-7" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-10" >12</a>
	]
</span>–<span class="cite">
	[
	<a href="#potra-01" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-11" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-12" >19</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-13" >21</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> <div class="example_thmwrapper " id="E.3.4">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">7</span>
  </div>
  <div class="example_thmcontent">
  <p> Define the scalar function \(F\) by \(F(x)= c_0 \,  x + c_1 + c_2 \,  \sin e^{c_3\,  x}\), \(x_0=0\), where \(c_i\), \(i=0,1,2,3\) are given parameters. Define linear operator \(\delta F(x,y)\) by </p>
<div class="displaymath" id="a0000000035">
  \[  \delta F(x,y) = \displaystyle \int _0^1 F'(y+t \,  (x-y)) \,  {\rm d}t = c_0 + c_2 \,  \displaystyle \tfrac {\sin e^{c_3\,  x} - \sin e^{c_3\,  y}}{x- y }.  \]
</div>
<p> Then it can easily be seen that for \(c_3\) large and \(c_2\) sufficiently small, \(\displaystyle \tfrac {\ell }{\ell _0}\) can be arbitrarily large. That is (<a href="#2.3">2.4</a>) may be satisfied but not (<a href="#1.3">1.3</a>).<span class="qed">â–¡</span></p>

  </div>
</div> <div class="example_thmwrapper " id="E.3.6-ab">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">8</span>
  </div>
  <div class="example_thmcontent">
  <p><span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span> <b class="bf">(Newton’s method case)</b> Let \(\mathcal{X}=\mathcal{Y}= {\mathcal C} [0,1] \) be the space of real–valued continuous functions defined on the interval \([0,1]\), equipped with the max–norm \(\parallel . \parallel \). Let \(\theta \in [0,1]\) be a given parameter. Consider the "Cubic" Chandrasekhar integral equation </p>
<div class="equation" id="4.6">
<p>
  <div class="equation_content">
    \begin{equation}  \label{4.6} u(s) = u^3(s)+ \lambda \,  u(s) \,  \displaystyle \int _0 ^1 q(s,t) \,  u(t) \,  {\rm d}t + y(s) - \theta . \end{equation}
  </div>
  <span class="equation_label">3.1</span>
</p>
</div>
<p> Here the kernel \(q(s,t)\) is a continuous function of two variables defined on \( [0,1] \times [0,1] \). The parameter \(\lambda \) in (<a href="#4.6">3.1</a>) is a real number called the "albedo" for scattering, and \(y(s)\) is a given continuous function defined on \([0,1]\) and \(x(s)\) is the unknown function sought in \({\mathcal C} [0,1] \). For simplicity, we choose \(u_0(s)=y(s)=1\), and \(q(s,t)=\displaystyle \tfrac {s}{s+t}\), for all \(s \in [0,1]\), and \(t \in [0,1]\), with \(s+t \neq 0\). If we let \(\mathcal{D}= U(u_0, 1-\theta )\), and define the operator \(F\) on \(\mathcal{D}\) by </p>
<div class="equation" id="4.7">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.7} F(x)(s) = x^3(s)- x(s) + \lambda \,  x(s) \,  \displaystyle \int _0 ^1 q(s,t) \,  x(t) \,  {\rm d}t + y(s) - \theta , \end{equation}
  </div>
  <span class="equation_label">3.2</span>
</p>
</div>
<p> for all \(s\in [0,1]\), then every zero of \(F\) satisfies equation (<a href="#4.6">3.1</a>). </p>
<p>We have the estimates </p>
<div class="displaymath" id="a0000000036">
  \[  \displaystyle \max _{0 \leq s \leq 1} | \displaystyle \int _0^1 \displaystyle \tfrac {s}{s+t} \,  {\rm d}t | = \ln 2 .  \]
</div>
<p>Therefore, if we set \(\xi = \parallel F'(u_0)^{-1}\parallel \), then the hypotheses of Theorem <a href="#T.2.3">3</a> (see <b class="bf">(\(\mathcal{A}_3\))</b>) correspond to the usual Lipschitz and center–Lipschitz conditions for (NM) (see <span class="cite">
	[
	<a href="#ar-hi-24" >6</a>
	, 
	Theorem 3.4
	]
</span>), such that </p>
<div class="displaymath" id="a0000000037">
  \[  \eta = \xi \,  ( |\lambda |\,  \ln 2 + 1 - \theta ), \]
</div>
<div class="displaymath" id="a0000000038">
  \[  \ell = 2\,  \,  \xi \, \,  ( |\lambda |\,  \ln 2 + 3\,  (2 - \theta ) )\quad {\rm and} \quad \ell _0 = \xi \,  \,  ( 2\,  |\lambda |\,  \ln 2 + 3\,  (3 - \theta ) ).  \]
</div>
<p> It follows from an equivalent Theorem for (NM) to Theorem <a href="#T.2.3">3</a> that if condition </p>
<div class="displaymath" id="a0000000039">
  \[  h _A = \displaystyle \tfrac {1}{8} \,  {\bigg(} \ell + 4 \,  \ell _0 + \sqrt{\ell ^2 + 8 \,  \ell \,  \ell _0} {\bigg)} \,  \eta \leq \displaystyle \tfrac {1}{2} ,  \]
</div>
<p> holds, then problem (<a href="#4.6">3.1</a>) has a unique solution near \(u_0\). This assumption is weaker than the one given before using the Newton–Kantorovich hypothesis. Note also that \(\ell _0 {\lt} \ell \) for all \(\theta \in [0,1]\).<span class="qed">â–¡</span></p>

  </div>
</div> <div class="example_thmwrapper " id="E.3.6'">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">9</span>
  </div>
  <div class="example_thmcontent">
  <p> <b class="bf">(secant method case)</b> Let \(\mathcal{X}=\mathcal{Y}= {\mathcal C} [0,1] \), equipped with the norm \( \parallel x \parallel = \displaystyle \max _{0 \leq s \leq 1} |x(s)| \). Consider the following nonlinear boundary value problem <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span> </p>
<div class="displaymath" id="a0000000040">
  \[  \left\{  \begin{array}[c]{c}u^{\prime \prime }= - u^3 - \gamma \, \,  u^2 \\ u(0)=0,\quad u(1)=1. \end{array} \right.  \]
</div>
<p> It is well known that this problem can be formulated as the integral equation </p>
<div class="equation" id="ref-4-1">
<p>
  <div class="equation_content">
    \begin{equation} \label{ref-4-1} u(s)=s+\displaystyle \int _{0}^{1}Q(s,t) \, \,  (u^3(t) + \gamma \, \,  u^2(t) ) \, \,  {\rm d}t \end{equation}
  </div>
  <span class="equation_label">3.3</span>
</p>
</div>
<p> where, \(Q\) is the Green function: </p>
<div class="displaymath" id="a0000000041">
  \[  Q(s,t)= \left\{  \begin{array}[c]{c} t\, \,  (1- s), \quad t\leq s\\ s\, \,  (1-t), \quad s{\lt} t. \end{array} \right.  \]
</div>
<p> We observe that </p>
<div class="displaymath" id="a0000000042">
  \[  \displaystyle \max _{0\leq s \leq 1} \, \,  \displaystyle \int _0^1 |Q(s,t)|\,  {\rm d}t = \displaystyle \tfrac {1}{8}.  \]
</div>
<p> Then problem (<a href="#ref-4-1">3.3</a>) is in the form (<a href="#1.1">1.1</a>), where, \(F\,  : \,  \mathcal{D}\longrightarrow \mathcal{Y}\) is defined as </p>
<div class="displaymath" id="a0000000043">
  \[  [F(x)] \,  (s)=x(s) - s - \displaystyle \int _{0}^{1}Q(s,t) \, \,  (x^3(t) + \gamma \, \,  x^2(t) ) \, \,  {\rm d}t .  \]
</div>
<p> It is easy to verify that the Fréchet derivative of \(F\) is defined in the form </p>
<div class="displaymath" id="a0000000044">
  \[  [F^{\prime }(x)v] \,  (s)=v(s)- \displaystyle \int _{0}^{1} Q(s,t) \, \,  (3 \, \,  x^2(t) + 2\, \,  \gamma \, \,  x(t) ) \, \,  v(t) \, \,  {\rm d}t .  \]
</div>
<p> Let </p>
<div class="displaymath" id="a0000000045">
  \[  \delta F(x,y)= \displaystyle \int _0^1 F'(y+t \,  (x-y)) \,  {\rm d}t .  \]
</div>
<p> If we set \(u_0 (s) = s\), and \(\mathcal{D}= U (u_0, R)\), then since \(\parallel u_0 \parallel =1\), it is easy to verify that \(U(u_0, R) \subset U(0, R+1)\). It follows that \(2 \, \,  \gamma {\lt} 5\), then (see <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>) </p>
<div class="displaymath" id="a0000000046">
  \[  \parallel I - F'(u_0) \parallel \leq \displaystyle \tfrac {3 + 2\, \,  \gamma }{8} , \quad \parallel F'(u_0)^{-1} \parallel \leq \displaystyle \tfrac {8 }{5 - 2 \, \,  \gamma } ,  \]
</div>
<div class="displaymath" id="a0000000047">
  \[  \parallel F(u_0) \parallel \leq \displaystyle \tfrac {1+\gamma }{8} , \quad \parallel F(u_0)^{-1} \, \,  F(u_0) \parallel \leq \displaystyle \tfrac {1+\gamma }{5 - 2 \, \,  \gamma } .  \]
</div>
<p> On the other hand, for \(x, y \in \mathcal{D}\), we have </p>
<div class="displaymath" id="a0000000048">
  \[  [(F^{\prime }(x) - F^{\prime } (y))v] \,  (s)=- \displaystyle \int _{0}^{1} Q(s,t) \,  (3 \,  x^2(t) - 3 \,  y^2(t) + 2\,  \gamma \,  (x(t)- y(t)) ) \,  v(t) \,  {\rm d}t .  \]
</div>
<p> Consequently (see <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>), </p>
<div class="displaymath" id="a0000000049">
  \[  \parallel F'(x) - F'(y) \parallel \leq \displaystyle \tfrac { \gamma + 6 \,  R + 3}{4} \,  \parallel x - y \parallel ,  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000050">
  \[  \parallel F'(x) - F'(u_0) \parallel \leq \displaystyle \tfrac {2 \,  \gamma + 3 \,  R + 6}{8} \,  \parallel x - u_0 \parallel .  \]
</div>
<p>Define linear operator \(\delta F(x,y)\) by </p>
<div class="displaymath" id="a0000000051">
  \[  \delta F(x,y) = \displaystyle \int _0^1 F'(y+t \,  (x-y)) \,  {\rm d}t .  \]
</div>
<p> Then, conditions of Theorem <a href="#T.2.3">3</a> hold with </p>
<div class="displaymath" id="a0000000052">
  \[  \eta = \displaystyle \tfrac {1 + \gamma }{ 5 - 2 \,  \gamma }, \quad \ell = \displaystyle \tfrac { \gamma + 6 \,  R + 3 }{ 8 }, \quad \ell _0= \displaystyle \tfrac {2 \,  \gamma + 3 \,  R + 6 }{ 16 }.  \]
</div>
<p>Note also that \(\ell _0 {\lt} \ell \).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000053">Conclusion</h1>
<p> We provided new sufficient convergence conditions for (SM) to a locally unique solution of a nonlinear equation in a Banach space. Using our new concept of recurrent functions, and combining Lipschitz and center–Lipschitz conditions on the divided difference operator, we obtained the semilocal convergence analysis of (SM). Our error bounds are more precise than earlier ones, and under our convergence hypotheses we can cover cases where earlier conditions are violated <span class="cite">
	[
	<a href="#623-6" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-7" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-10" >12</a>
	]
</span>–<span class="cite">
	[
	<a href="#potra-01" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-11" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-12" >19</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-13" >21</a>
	]
</span>. Applications, and numerical examples are also provided in this study. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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</dd>
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</dd>
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  <dd><p><i class="sc">J.M. Ortega</i> and <i class="sc">W.C. Rheinboldt</i>, <i class="it">Iterative Solution of Nonlinear Equations in Several Variables</i>, Academic Press, New York, 1970. </p>
</dd>
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</dd>
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</dd>
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</dd>
</dl>


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