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<title>On approximating the solutions of nonlinear equations by a method of Aitken-Steffensen type: On approximating the solutions of nonlinear equations by a method of Aitken-Steffensen type</title>
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<h1>On approximating the solutions of nonlinear equations by a method of Aitken-Steffensen type</h1>
<p class="authors">
<span class="author">Adrian Diaconu\(^\ast \)</span>
</p>
<p class="date">April 1st, 2014.</p>
</div>
<p>\(^\ast \)“Babeş-Bolyai" University, Faculty of Mathematics and Computer Science, st. M. Kogălniceanu Nr. 1, 3400 Cluj-Napoca, Romania, e-mail: <span class="tt">adiaconu@math.ubbcluj.ro</span>. </p>

<div class="abstract"><p> This paper completes the results that are presented in <span class="cite">
	[
	<a href="#d8" >14</a>
	]
</span>. Using as a starting point  the abstract method of the chord, in the mentioned paper we have presented an iterative method of approximation for the solutions of an equation. This method uses auxiliary sequences, and aims to improve the convergence order. The used method generalizes the method of Aitken-Steffensen. In the paper <span class="cite">
	[
	<a href="#d8" >14</a>
	]
</span> we have given the statement of the main theorem and the statement and the proof of an auxiliary proposition concerning the convergence of some recurrence sequences of real numbers. In the present paper we give the proof of the main result and at the same time we discuss an interesting special case. </p>
<p><b class="bf">MSC.</b> 65J15 </p>
<p><b class="bf">Keywords.</b> Convergence of the method of the Aitken-Steffensen type. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p> &#8195; We will take again the main elements that are presented in detail in the papers <span class="cite">
	[
	<a href="#d7" >13</a>
	]
</span>-<span class="cite">
	[
	<a href="#d8" >14</a>
	]
</span> and that constitute the basis of our result. </p>
<p>Let us consider \(X,Y\)&#160;two linear normed spaces. We note by \(\left\Vert \cdot \right\Vert _{X}:X\rightarrow \mathbb {R}\) and \(\left\Vert \cdot \right\Vert _{Y}:Y\rightarrow \mathbb {R}\) their norms respectively, and by \(\theta _{X}\) and \(\theta _{Y}\) their null elements. Through \(\left( X,Y\right) ^{\ast }\) we note the set of the linear and continuous mappings defined from \(X\) to \(Y.\) The set \(\left( X,Y\right) ^{\ast }\) is a linear normed space as well, if we define the norm \(\left\Vert \cdot \right\Vert :\left( X,Y\right) ^{\ast }\rightarrow \left[ 0,+\infty \right[ ,\) for any \(U\in \left( X,Y\right) ^{\ast }\) having \(\left\Vert U\right\Vert =\sup \big\{  \left\Vert U\left( h\right) \right\Vert : h\in X,\  \left\Vert h\right\Vert _{X}=1\big\}  .\) For the case of \(Y=\mathbb {R}\) we denote by \(X^{\ast }\) the set \(\left( X,\mathbb {R}\right) ^{\ast },\) this set representing the space of real, linear and continuous functionals defined on the linear normed space \(X.\) </p>
<p>Let us consider now a set \(D\subseteq X\) and a nonlinear mapping \(f:D\rightarrow Y.\) Using this mapping we have the equation:</p>
<div class="equation" id="f1">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) =\theta _{Y}. \label{f1}\end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> We will study the approximation of its solutions. </p>
<p>In order to clarify the aforementioned notions we have the following de- finition: </p>
<p><div class="definition_thmwrapper " id="d1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">1</span>
  </div>
  <div class="definition_thmcontent">
  <p>If we fix the nonlinear mapping \(f:D\rightarrow Y\) together with the points \(x,y\in D,\) \(x\neq y,\) any mapping \(\Gamma _{f;x,y}\in \left( X,Y\right) ^{\ast }\) that verifies the equality:</p>
<div class="equation" id="f2">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;x,y}\left( x-y\right) =f\left( x\right) -f\left( y\right) \label{f2}\end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> is called <b class="bfseries">generalized abstract divided difference of the function</b><br />\(f:D\rightarrow Y\)<b class="bfseries"> at the points</b> \(x,y.\) </p>

  </div>
</div> </p>
<p>In connection with the previous definition we have the following remark: </p>
<p><div class="remark_thmwrapper " id="r1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2</span>
  </div>
  <div class="remark_thmcontent">
  <p><b class="bfseries">a)</b> We consider the theorem according to which in every linear normed space \(\left( X,\left\Vert \cdot \right\Vert _{X}\right) ,\) for any \(a\in X \setminus \left\{  \theta _{X}\right\}  \) there exists a linear and continuous functional \(u\in X^{\ast }\) such that \(\left\Vert u\right\Vert =1\) and \(u\left( a\right) =\left\Vert a\right\Vert _{X}.\) Then, for any \(x,y\in X\) with \(x\neq y\) there exists the functional \(U_{xy}\in X^{\ast }\) such that \(\left\Vert U_{xy}\right\Vert =1\) and \(U_{xy}\left( x-y\right) =\left\Vert x-y\right\Vert _{X}.\) At the same time there exists the functional \(U_{yx}\in X^{\ast }\) such that \(\left\Vert U_{yx}\right\Vert =1\) and \(U_{yx}\left( x-y\right) =\left\Vert y-x\right\Vert _{X}\) as well. In the paper <span class="cite">
	[
	<a href="#d7" >13</a>
	]
</span> there appears the mapping \(\left[ x,y;f\right] \in \left( X,Y\right) ^{\ast },\) defined by the equality:</p>
<div class="equation" id="f3">
<p>
  <div class="equation_content">
    \begin{equation}  \left[ x,y;f\right] h=\tfrac {U_{xy}\left( h\right) f\left( x\right) +U_{yx}\left( h\right) f\left( y\right) }{\left\Vert x-y\right\Vert _{X}} \label{f3}\end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> for any \(h\in X.\) </p>
<p>This mapping verifies the equality (<a href="#f2">2</a>) and it is called <b class="bfseries">abstract divided difference</b> of the nonlinear mapping \(f:D\rightarrow Y\) at the points \(x,y\in D\) with \(x\neq y.\) This mapping is a special case of generalized abstract divided difference. Therefore we can choose \(\Gamma _{f;x,y}=\left[ x,y;f\right] .\) </p>
<p><b class="bfseries">b)</b> Let us suppose now that the space \(X\) is a space with a scalar product \(\left\langle \cdot |\cdot \right\rangle :X\times X\rightarrow \mathbb {R}.\) Defining \(\left\Vert \cdot \right\Vert _{X}:X\rightarrow \mathbb {R}\) by \(\left\Vert x\right\Vert _{X}=\sqrt{\left\langle x|x\right\rangle },\) the space \(\left( X,\left\Vert \cdot \right\Vert _{X}\right) \) is a linear normed space. </p>
<p>For any \(x,y\in X\) with \(x\neq y\) the functional \(U_{xy}\in X^{\ast }\) from <b class="bfseries">a)</b> will be defined by:</p>
<div class="displaymath" id="a0000000003">
  \[  U_{xy}\left( h\right) =\tfrac {\left\langle h|x-y\right\rangle }{\left\Vert x-y\right\Vert _{X}} \]
</div>
<p> for any \(h\in X.\) So, for the same elements \(x,y\in X\) with \(x\neq y\) we have that the abstract divided difference \(\left[ x,y;f\right] \in \left( X,Y\right) ^{\ast }\) is defined by:</p>
<div class="displaymath" id="a0000000004">
  \[  \left[ x,y;f\right] h=\tfrac {\left\langle x-y|h\right\rangle \left( f\left( x\right) -f\left( y\right) \right) }{\left\Vert x-y\right\Vert _{X}^{2}} \]
</div>
<p> for any \(h\in X.\)<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>Let us consider a initial element \(x_{0}\in D.\) Besides the main sequence<br /> \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\subseteq D\) we also use two auxiliary sequences \(\left( y_{n}\right) _{n\in \mathbb {N}^{\ast }},\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}\subseteq D.\) </p>
<p>For these auxiliary sequences we request the existence of the numbers<br /> \( K_{1},K_{2},p,q{\gt}0\) such that for any \(n\in \mathbb {N}^{\ast }\) the following inequalities are verified:</p>
<div class="displaymath" id="f44a">
  \begin{align}  \left\Vert f\left( y_{n}\right) \right\Vert _{Y}& \leq K_{1}\left\Vert f\left( x_{n}\right) \right\Vert _{Y}^{p}, \label{f44a}\\ \left\Vert f\left( z_{n}\right) \right\Vert _{Y}& \leq K_{2}\left\Vert f\left( x_{n}\right) \right\Vert _{Y}^{q}\nonumber .\end{align}
</div>
<p>Then, if for a number \(n\in \mathbb {N}^{\ast }\) we have available the elements \(y_{n},z_{n}\in D\) starting from \(x_{n}\in D,\) we will build the new iterate \(x_{n+1}\in D\) for the verification of the equality:</p>
<div class="equation" id="f45">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;y_{n},z_{n}}\left( x_{n+1}-y_{n}\right) +f\left( y_{n}\right) =\theta _{Y}. \label{f45}\end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>On account of the property of definition of the mapping \(\Gamma _{f;y_{n},z_{n}}\in \left( X,Y\right) ^{\ast }\) the equality (<a href="#f45">5</a>) is equivalent to:</p>
<div class="equation" id="f46">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;y_{n},z_{n}}\left( x_{n+1}-z_{n}\right) +f\left( z_{n}\right) =\theta _{Y}. \label{f46}\end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>If for any \(n\in \mathbb {N}^{\ast }\) there exists the mapping \(\Gamma _{f;y_{n},z_{n}}^{-1}\in \left( Y,X\right) ^{\ast }\) we have:</p>
<div class="equation" id="f47">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=y_{n}-\Gamma _{f;y_{n},z_{n}}^{-1}f\left( y_{n}\right) =z_{n}-\Gamma _{f;y_{n},z_{n}}^{-1}f\left( z_{n}\right) . \label{f47}\end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p>In connection with the main sequence \(\left( x_{n}\right) _{n\in \mathbb {N}}\subseteq D\) and also with the auxiliary sequences \(\left( y_{n}\right) _{n\in \mathbb {N}},\  \left( z_{n}\right) _{n\in \mathbb {N}}\subseteq D,\) one can see the remarks 2.1– 2.2 from the paper <span class="cite">
	[
	<a href="#d8" >14</a>
	]
</span>. </p>
<p><div class="remark_thmwrapper " id="r4">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3</span>
  </div>
  <div class="remark_thmcontent">
  <p>It is clear that if the first of the inequalities (<a href="#f44a">4</a>) is verified for any \(n\in \mathbb {N}\) with a certain \(K_{1}{\gt}0,\) this inequality is verified with any number \(K\geq \max \left\{  1,K_{1}\right\}  .\) The situation is identical regarding the second inequality from (<a href="#f44a">4</a>). In conclusion we can suppose that in these relations we have \(K_{1}=K_{2}=K\geq 1.\)</p>
<p>Identically, we can suppose that in the inequalities:</p>
<div class="displaymath" id="a0000000005">
  \begin{align*}  \left\Vert y_{n}-x_{n}\right\Vert _{X}& \leq a\left\Vert f\left( x_{n}\right) \right\Vert _{Y},\\ \left\Vert z_{n}-x_{n}\right\Vert _{X}& \leq b\left\Vert f\left( x_{n}\right) \right\Vert _{Y}, \end{align*}
</div>
<p> that are true for any \(n\in \mathbb {N},\) we can have \(b=a\geq 1.\) </p>
<p>In conclusion, for the main sequence \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\subseteq D\) together with the auxiliary sequences \(\left( y_{n}\right) _{n\in \mathbb {N}^{\ast }},\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}\subseteq D\) we can suppose that for any \(n\in \mathbb {N}^{\ast }\) we have \(y_{n}\neq z_{n}\) and there exist the numbers \(K,a\geq 1\) such that for any \(n\in \mathbb {N}^{\ast }\) the following inequalities are verified:</p>
<div class="displaymath" id="f53">
  \begin{align}  \left\Vert f\left( y_{n}\right) \right\Vert _{Y}\leq &  K\left\Vert f\left( x_{n}\right) \right\Vert _{Y}^{p}, \label{f53}\\ \left\Vert f\left( z_{n}\right) \right\Vert _{Y}\leq &  K\left\Vert f\left( x_{n}\right) \right\Vert _{Y}^{q}, \nonumber \\ \left\Vert y_{n}-x_{n}\right\Vert _{X}\leq & a\left\Vert f\left( x_{n}\right) \right\Vert _{Y}, \nonumber \\ \left\Vert z_{n}-x_{n}\right\Vert _{X}\leq &  a\left\Vert f\left( x_{n}\right) \right\Vert _{Y}. \nonumber \end{align}
</div>

  </div>
</div> </p>
<p>In connection with the stated problem we consider, for the real numbers \(p,q\geq 1,\) the following equation in \(x\) on the interval \(\left[ 0,+\infty \right[ :\)</p>
<div class="equation" id="f54">
<p>
  <div class="equation_content">
    \begin{equation}  x^{p+q-1}+2x^{2}+2x-1=0. \label{f54}\end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>We have the following remarks: <div class="remark_thmwrapper " id="r5">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  <p><b class="bfseries">a)</b> The equation (<a href="#f54">9</a>) has an unique root \(\alpha \in \left] 0,1\right[ .\) </p>
<p><b class="bfseries">b)</b> If \(\alpha \in \left] 0,1\right[ \) is the root of the equation (<a href="#f54">9</a>) one verifies the following inequalities as well:</p>
<div class="equation" id="f55">
<p>
  <div class="equation_content">
    \begin{equation}  \alpha ^{2}+\alpha -1<0,\qquad \alpha ^{2}+2\alpha -1<0,\qquad 2\alpha ^{2}+2\alpha -1<0 \label{f55}\end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> and these inequalities are equivalent to the following inequalities respectively:</p>
<div class="equation" id="f56">
<p>
  <div class="equation_content">
    \begin{equation}  0<\tfrac {\alpha ^{2}}{1-\alpha }<1,\qquad 0<\tfrac {\alpha }{1-\alpha -\alpha ^{2}}<1,\qquad 0<\tfrac {\alpha ^{2}}{1-2\alpha -\alpha ^{2}}<1. \label{f56}\end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>

  </div>
</div> </p>
<p>Let us consider now the numbers \(a,K,L,B_{0},R_{0}{\gt}0\) and the numbers \(p,q\geq 1\) and using these numbers we build the real number sequences \(\left( u_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( s_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( v_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( w_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( t_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( B_{n}\right) _{n\in \mathbb {N}^{\ast }}\) and \(\left( R_{n}\right) _{n\in \mathbb {N}^{\ast }}\) using the following recurrence relations:</p>
<div class="displaymath" id="f57">
  \begin{align}  u_{n}& =LKB_{n}^{2}R_{n}^{p}\label{f57},\\ s_{n}& =LKB_{n}^{2}R_{n}^{q},\nonumber \\ v_{n}& =aL^{2}K^{2}\cdot \tfrac {B_{n}^{3}R_{n}^{p+q}}{1-u_{n}},\nonumber \\ w_{n}& =\tfrac {LKB_{n}^{2}R_{n}^{q}}{\left( 1-u_{n}\right) \left( 1-v_{n}\right) },\nonumber \\ t_{n}& =\tfrac {aL^{2}K^{2}B_{n}^{3}R_{n}^{p+q}}{\left( 1-u_{n}\right) \left( 1-v_{n}\right) \left( 1-w_{n}\right) },\nonumber \\ B_{n+1}& =\tfrac {B_{n}}{\left( 1-u_{n}\right) \left( 1-v_{n}\right) \left( 1-w_{n}\right) \left( 1-t_{n}\right) },\nonumber \\ R_{n+1}& =LK^{2}B_{n}^{2}R_{n}^{p+q}.\nonumber \end{align}
</div>
<p>It is obvious that this construction has a meaning if for any \(n\in \mathbb {N}^{\ast }\) we have that \(u_{n},\  v_{n},\  w_{n},\  t_{n}\in \mathbb {R\diagdown }\left\{  1\right\}  \) and \(B_{n},R_{n}{\gt}0.\) </p>
<p>It is clear that for any \(n\in \mathbb {N}^{\ast }\) we have:</p>
<div class="displaymath" id="f58">
  \begin{align}  v_{n}& =\tfrac {a}{B_{n}}\cdot \tfrac {u_{n}s_{n}}{1-u_{n}},\label{f58}\\ w_{n}& =\tfrac {s_{n}}{\left( 1-u_{n}\right) \left( 1-v_{n}\right) },\nonumber \\ t_{n}& =\tfrac {v_{n}}{\left( 1-v_{n}\right) \left( 1-w_{n}\right) },\nonumber \\ R_{n+1}& =\tfrac {u_{n}s_{n}}{LB_{n}^{2}},\nonumber \end{align}
</div>
<p> as well. </p>
<p>Referring to the sequences that are defined by the relations (<a href="#f57">12</a>) we have the following proposition: </p>
<p><div class="proposition_thmwrapper " id="p1">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">5</span>
  </div>
  <div class="proposition_thmcontent">
  <p>If the following inequalities are verified:</p>
<div class="equation" id="f59">
<p>
  <div class="equation_content">
    \begin{equation}  a\leq B_{0}\leq \tfrac {1}{\sqrt{L}}\cdot \min \left\{  K^{\frac{p-q+1}{2\left( q-1\right) }},K^{\frac{q-p+1}{2\left( p-1\right) }}\right\}  \label{f59}\end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p> (with the specification that for \(q=1\) the expression that has \(q-1\) in its denominator is \(+\infty ,\) and the same for the expression that has \(p-1\) in its denominator) and:</p>
<div class="equation" id="f60">
<p>
  <div class="equation_content">
    \begin{equation}  d=\tfrac {LKB_{0}^{2}}{\alpha ^{2}}\cdot \max \nolimits ^{\frac{1}{p+q-1}}\left\{  \tfrac {R_{0}^{p\left( p+q-1\right) }K^{p-q+1}}{\left( LB_{0}^{2}\right) ^{q-1}},\tfrac {R_{0}^{q\left( p+q-1\right) }K^{q-p+1}}{\left( LB_{0}^{2}\right) ^{p-1}}\right\}  <1 \label{f60}\end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p> where \(\alpha \in \left] 0,1\right[ \) is the unique root of the equation <span class="rm">(<a href="#f54">9</a>)</span>, then for any \(n\in \mathbb {N}^{\ast }\) we have the following inequalities:</p>
<div class="displaymath" id="f61">
  \begin{align}  u_{n}& \leq \alpha d^{\left( p+q\right) ^{n}}{\lt}\alpha {\lt}1\label{f61},\\ s_{n}& \leq \alpha d^{\left( p+q\right) ^{n}}{\lt}\alpha {\lt}1,\nonumber \\ v_{n}& \leq \tfrac {\alpha ^{2}}{\alpha -1}\cdot d^{2\left( p+q\right) ^{n}}{\lt}\tfrac {\alpha ^{2}}{\alpha -1}{\lt}1,\nonumber \\ w_{n}& \leq \tfrac {\alpha }{1-\alpha -\alpha ^{2}}\cdot d^{\left( p+q\right) ^{n}}{\lt}\tfrac {\alpha }{1-\alpha -\alpha ^{2}}{\lt}1,\nonumber \\ t_{n}& \leq \tfrac {\alpha ^{2}}{1-2\alpha -\alpha ^{2}}d^{2\left( p+q\right) ^{n}}{\lt}\tfrac {\alpha ^{2}}{1-2\alpha -\alpha ^{2}}{\lt}1,\nonumber \\ B_{n+1}& \leq \tfrac {B_{n}}{1-2\alpha -2\alpha ^{2}},\nonumber \\ R_{n+1}& \leq \tfrac {\alpha ^{2}}{LB_{0}^{2}}d^{2\left( p+q\right) ^{n}}.\nonumber \end{align}
</div>

  </div>
</div> </p>
<p>The proof of this proposition has been given in the paper <p><span class="cite">
	[
	<a href="#d8" >14</a>
	]
</span>. </p>
</p>
<h1 id="a0000000006">2 The main result</h1>
<p>In this section we present the statement and the proof of the main theorem regarding the convergence of the sequences \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }},\  \left( y_{n}\right) _{n\in \mathbb {N}^{\ast }},\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}\) \(\subseteq D\subseteq X.\) </p>
<p><div class="theorem_thmwrapper " id="t2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p>We suppose that the following conditions hold: </p>
<ul class="itemize">
  <li><p>The linear normed space \(\left( X,\left\Vert \cdot \right\Vert _{X}\right) \) is a Banach space; </p>
</li>
  <li><p>The mapping \(f:D\rightarrow Y\) admits for any \(x,y\in D\) with \(x\neq y\) a generalized abstract divided difference \(\Gamma _{f;x,y}\in \left( X,Y\right) ^{\ast }\) and there exists a number \(L{\gt}0\) such that for any \(x,y,z\in D\) with \(x\neq y,\  y\neq z\) we have the inequality:</p>
<div class="displaymath" id="a0000000007">
  \[  \left\Vert \Gamma _{f;x,y}-\Gamma _{f;y,z}\right\Vert \leq L\left\Vert x-z\right\Vert _{X};  \]
</div>
</li>
  <li><p>The main approximant sequence \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\) together with the secondary sequences \(\left( y_{n}\right) _{n\in \mathbb {N}^{\ast }}\) and \(\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}\) are such that for any \(n\in \mathbb {N}^{\ast }\) the following equality is fulfilled:</p>
<div class="equation" id="f63p">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;y_{n},z_{n}}\left( x_{n+1}-y_{n}\right) +f\left( y_{n}\right) =\theta _{Y} \label{f63p}\end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> together with the inequalities (<a href="#f53">8</a>) with the constants \(a,K{\gt}1\) and \(p,q\geq 1.\) We also have that \(f\left( y_{n}\right) ,f\left( z_{n}\right) \in Y\diagdown \left\{  \theta _{Y}\right\}  ,\) \(y_{n}\neq z_{n}\) and we are in one of the following situations: </p>
<ul class="itemize">
  <li><p>\(x_{n}\neq y_{n}\) and \(y_{n+1}\neq z_{n},\) </p>
</li>
</ul>
<p>or </p>
<ul class="itemize">
  <li><p>\(x_{n}\neq z_{n}\) and \(z_{n+1}\neq y_{n}.\) </p>
</li>
</ul>
</li>
  <li><p>The mapping \(\Gamma _{f;y_{0},z_{0}}\in \left( X,Y\right) ^{\ast }\) is invertible and \(\Gamma _{f;y_{0},z_{0}}^{-1}\in \left( Y,X\right) ^{\ast }.\) </p>
</li>
  <li><p>If we note:</p>
<div class="displaymath" id="a0000000008">
  \begin{align*}  B_{0}& =\max \left\{  a,\Vert \Gamma _{f;y_{0},z_{0}}^{-1}\Vert \right\} ,\\ R_{0}& =\left\Vert f\left( x_{0}\right) \right\Vert _{Y},\\ \overline{K}& =\max \big\{  K,\  \big( B_{0}\sqrt{L}\big) ^{\frac{2\left( q-1\right) }{p-q+1}},\big( B_{0}\sqrt{L}\big) ^{\frac{2\left( p-1\right) }{q-p+1}}\big\} ,\\ d& =\tfrac {L\overline{K}B_{0}^{2}}{\alpha ^{2}}\cdot \max \left\{  \tfrac {R_{0}^{p\left( p+q-1\right) }\overline{K}^{p-q+1}}{\left( LB_{0}^{2}\right) ^{q-1}},\tfrac {R_{0}^{q\left( p+q-1\right) }\overline{K}^{q-p+1}}{\left( LB_{0}^{2}\right) ^{p-1}}\right\}  ,\\ \delta & =2aR_{0}+\tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot \tfrac {d^{2}}{1-d^{2\left( p+q-1\right) }}+\tfrac {2\alpha }{L\overline{K}B_{0}}\cdot \tfrac {d}{1-d^{p+q-1}}, \end{align*}
</div>
<p> where \(\alpha \in \left] 0,1\right[ \) is the unique root of the equation <span class="rm">(<a href="#f54">9</a>)</span>, the conditions \(d{\lt}1\) and \(S\left( x_{0},\delta \right) =\left\{  \left. x\in X\right/ \left\Vert x-x_{0}\right\Vert _{X}\leq \delta \right\}  \) are fulfilled. </p>
</li>
</ul>
<p>If the previous hypotheses  are true, then the following conclusions are true as well: </p>
<ul class="itemize">
  <li><p>for any \(n\in \mathbb {N}^{\ast }\) we have that \(x_{n},y_{n},z_{n}\in S\left( x_{0},\delta \right) ,\) there exists the mapping \(\Gamma _{f;y_{n},z_{n}}^{-1}\in \left( Y,X\right) ^{\ast }\) and:</p>
<div class="equation" id="K">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=y_{n}-\Gamma _{f;y_{n},z_{n}}^{-1}f\left( y_{n}\right) =z_{n}-\Gamma _{f;y_{n},z_{n}}^{-1}f\left( z_{n}\right) ; \label{K}\end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
</li>
  <li><p>the sequences \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }},\left( y_{n}\right) _{n\in \mathbb {N}^{\ast }},\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}\subseteq X\) are convergent to the limit \(\overline{x}\in S\left( x_{0},\delta \right) \) and \(f\left( \overline{x}\right) =\theta _{Y};\) </p>
</li>
  <li><p>for any \(n\in \mathbb {N}^{\ast }\) the following inequalities are fulfilled:</p>
<div class="displaymath" id="K1">
  \begin{align} & \left\Vert x_{n+1}-x_{n}\right\Vert _{X}\leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot d^{2\left( p+q\right) ^{n-1}}+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot d^{\left( p+q\right) ^{n}}; \label{K1} \\ & \left\Vert x_{n}-\overline{x}\right\Vert _{X}\leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot \tfrac {d^{2\left( p+q\right) ^{n-1}}}{1-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }} +\tfrac {\alpha }{L\overline{K}B_{0}}\cdot \tfrac {d^{\left( p+q\right) ^{n}}}{1-d^{\left( p+q\right) ^{n}\left( p+q-1\right) }};\label{K2} \\ & \max \left\{  \left\Vert y_{n}-\overline{x}\right\Vert _{X},\left\Vert z_{n}-\overline{x}\right\Vert _{X}\right\}  \leq \label{K3}\\ & \leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot d^{2\left( p+q\right) ^{n-1}}\cdot \tfrac {2-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }}{1-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }} +\tfrac {\alpha }{L\overline{K}B_{0}}\cdot \tfrac {d^{\left( p+q\right) ^{n}}}{1-d^{\left( p+q\right) ^{n}\left( p+q-1\right) }}. \nonumber \end{align}
</div>
</li>
</ul>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000009">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>For more clarity in the case of the hypothesis <b class="bfseries">iii)</b> we suppose that the situation <b class="bfseries">iii</b>\(_{1}\)<b class="bfseries">)</b> is fulfilled, namely for any \(n\in \mathbb {N}^{\ast }\) we will suppose that we have:</p>
<div class="displaymath" id="a0000000010">
  \[  f\left( y_{n}\right) ,f\left( z_{n}\right) \in Y\diagdown \left\{  \theta _{Y}\right\}  ,\  y_{n}\neq z_{n},\  x_{n}\neq y_{n},\  y_{n+1}\neq z_{n},  \]
</div>
<p> while the equality (<a href="#f63p">17</a>) is verified. </p>
<p>Using the constants \(L,B_{0},\overline{K},R_{0}{\gt}0\) we generate the real number sequences \(\left( u_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( s_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( v_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( w_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( t_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( B_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( R_{n}\right) _{n\in \mathbb {N}^{\ast }}\) on the basis of the relations (<a href="#f57">12</a>) in which the constant \(K{\gt}0\) is replaced by \(\overline{K}{\gt}0.\) </p>
<p>From the expression of \(\overline{K}\) we immediately deduce that:</p>
<div class="displaymath" id="a0000000011">
  \[  \overline{K}^{\tfrac {p-q+1}{2\left( q-1\right) }}\geq B_{0}\sqrt{L},\  \overline{K}^{\tfrac {q-p+1}{2\left( p-1\right) }}\geq B_{0}\sqrt{L},  \]
</div>
<p> therefore:</p>
<div class="displaymath" id="a0000000012">
  \[  B_{0}\leq \tfrac {1}{\sqrt{L}}\cdot \min \big\{  \overline{K}^{\tfrac {p-q+1}{2\left( q-1\right) }},\overline{K}^{\tfrac {q-p+1}{2\left( p-1\right) }}\big\}  ,  \]
</div>
<p> so the double inequality (<a href="#f59">14</a>) is true. The condition imposed to \(d,\) expressed by the inequality (<a href="#f60">15</a>), from the hypothesis of the Proposition <a href="#p1">5</a> is fulfilled on the basis of the hypotheses of the theorem. </p>
<p>On account of the Proposition <a href="#p1">5</a> we deduce that for any \(n\in \mathbb {N}^{\ast }\) the relations (<a href="#f61">16</a>) are true. </p>
<p>We now prove that for any \(n\in \mathbb {N}^{\ast }\) the following relations are true: </p>
<ul class="itemize">
  <li><p>\(x_{n},y_{n},z_{n}\in S\left( x_{0},\delta \right) ;\) </p>
</li>
  <li><p>\(\left\Vert f\left( x_{n}\right) \right\Vert _{Y}\leq R_{n};\) </p>
</li>
  <li><p>there exists the mapping \(\Gamma _{f;y_{n},z_{n}}^{-1}\in \left( Y,X\right) ^{\ast },\) \(\Vert \Gamma _{f;y_{n},z_{n}}^{-1}\Vert \leq B_{n}\) and:</p>
<div class="equation" id="63pp">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=y_{n}-\Gamma _{f;y_{n},z_{n}}^{-1}f\left( y_{n}\right) =z_{n}-\Gamma _{f;y_{n},z_{n}}^{-1}f\left( z_{n}\right) . \label{63pp}\end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
</li>
</ul>
<p>In order to prove these relations we will use the method of the mathematical induction. </p>
<p>For \(n=0\) we notice the following statements: </p>
<ul class="itemize">
  <li><p>Evidently \(x_{0}\in S\left( x_{0},\delta \right) \) and:</p>
<div class="displaymath" id="a0000000013">
  \[  \left\Vert y_{0}-x_{0}\right\Vert _{X}\leq a\left\Vert f\left( x_{0}\right) \right\Vert _{Y}=aR_{0}\leq \delta ,  \]
</div>
<p> therefore \(y_{0}\in S\left( x_{0},\delta \right) .\) </p>
<p>Similarly:</p>
<div class="displaymath" id="a0000000014">
  \[  \left\Vert z_{0}-x_{0}\right\Vert _{X}\leq a\left\Vert f\left( x_{0}\right) \right\Vert _{Y}=aR_{0}\leq \delta  \]
</div>
<p> so we have that \(z_{0}\in S\left( x_{0},\delta \right) \) as well. </p>
</li>
  <li><p>Evidently \(\left\Vert f\left( x_{0}\right) \right\Vert _{Y}\leq R_{0};\) </p>
</li>
  <li><p>The existence of the mapping \(\Gamma _{f;y_{0},z_{0}}^{-1}\in \left( Y,X\right) ^{\ast }\) is assured from the hypotheses of the theorem and the inequality \(\Vert \Gamma _{f;y_{0},z_{0}}^{-1}\Vert \leq B_{0}\) is assured  from the definition of the number \(B_{0}.\) </p>
</li>
</ul>
<p>We suppose that the inequalities <b class="bfseries">a) - c)</b> are true for any number \(n\in \mathbb {N}^{\ast },\) \(n\leq k\) and we prove them for \(n=k+1.\)</p>
<p><b class="bfseries">a)</b> For any \(j\in \{ 0,1,...,k\} \) we have that:</p>
<div class="displaymath" id="a0000000015">
  \begin{align*}  \left\Vert x_{j+1}-x_{j}\right\Vert _{X}& \leq \left\Vert y_{j}-x_{j}\right\Vert _{X}+\Vert \Gamma _{f;y_{j},z_{j}}^{-1}\Vert \cdot \left\Vert f\left( y_{j}\right) \right\Vert _{Y}\\ & \leq a\left\Vert f\left( x_{j}\right) \right\Vert _{Y}+B_{j}\left\Vert f\left( x_{j}\right) \right\Vert _{Y}^{p}\\ & =aR_{j}+B_{j}R_{j}^{p}\\ & =aR_{j}+\tfrac {B_{j}^{2}R_{j}^{p}}{B_{j}}\\ & \leq aR_{j}+\tfrac {u_{j}}{L\overline{K}B_{0}}. \end{align*}
</div>
<p>For \(j=0\) we obtain:</p>
<div class="displaymath" id="a0000000016">
  \[  \left\Vert x_{1}-x_{0}\right\Vert _{X}\leq aR_{0}+\tfrac {u_{0}}{L\overline{K}B_{0}}=aR_{0}+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot d.  \]
</div>
<p>For \(j\in \left\{  1,2,...,k\right\}  \) we obtain:</p>
<div class="equation" id="f63ppp">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert x_{j+1}-x_{j}\right\Vert _{X}\leq a\cdot \tfrac {\alpha ^{2}}{LB_{0}^{2}}\cdot d^{2\left( p+q\right) ^{j-1}}+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot d^{\left( p+q\right) ^{j}}. \label{f63ppp}\end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p>Therefore:</p>
<div class="displaymath" id="a0000000017">
  \begin{align*}  \left\Vert x_{k+1}-x_{0}\right\Vert _{X}& \leq \Bigg\Vert {\displaystyle \sum \limits _{j=0}^{k}} \left( x_{j+1}-x_{j}\right) \Bigg\Vert _{X}\\ & \leq {\displaystyle \sum \limits _{j=0}^{k}} \left\Vert x_{j+1}-x_{j}\right\Vert _{X}\medskip \\ & \leq aR_{0}+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot d+\tfrac {a\alpha ^{2}}{LB_{0}^{2}}{\displaystyle \sum \limits _{j=0}^{k-1}} d^{2\left( p+q\right) ^{j}}+\tfrac {\alpha }{L\overline{K}B_{0}}{\displaystyle \sum \limits _{j=0}^{k}} d^{\left( p+q\right) ^{j}}. \end{align*}
</div>
<p>In the last expression we have:</p>
<div class="displaymath" id="a0000000018">
  \[ {\displaystyle \sum \limits _{j=0}^{k-1}} d^{2\left( p+q\right) ^{j}}=d^{2}+d^{2\left( p+q\right) }+...+d^{2\left( p+q\right) ^{k-1}}=d^{2}{\displaystyle \sum \limits _{j=0}^{k-1}} d^{2\left( p+q\right) ^{j}-2}.  \]
</div>
<p>As \(p,q\geq 1\) we have that:</p>
<div class="displaymath" id="a0000000019">
  \begin{align*}  2\left( p+q\right) ^{j}-2& =2\left( p+q-1\right) \left[ 1+\left( p+q\right) +...+\left( p+q\right) ^{j-1}\right] \\ & \geq 2j\left( p+q-1\right), \end{align*}
</div>
<p> and as \(d{\lt}1\) we deduce that \(d^{2\left( p+q\right) ^{j}-2}\leq \left[ d^{2\left( p+q-1\right) }\right] ^{j}\), therefore:</p>
<div class="displaymath" id="a0000000020">
  \[ {\displaystyle \sum \limits _{j=0}^{k-1}} d^{2\left( p+q\right) ^{j}}\leq d^{2}\sum _{j=0}^{\infty }\left[ d^{2\left( p+q-1\right) }\right] ^{j}=\tfrac {d^{2}}{1-d^{2\left( p+q-1\right) }}.  \]
</div>
<p>Similarly</p>
<div class="displaymath" id="a0000000021">
  \[ {\displaystyle \sum \limits _{j=0}^{k}} d^{\left( p+q\right) ^{j}}\leq \tfrac {d}{1-d^{p+q-1}}.  \]
</div>
<p>Therefore</p>
<div class="displaymath" id="a0000000022">
  \begin{align*}  \left\Vert x_{k+1}-x_{0}\right\Vert _{X}& \leq aR_{0}+\tfrac {a\alpha ^{2}}{LB_{0}^{2}}\left( d^{2}+\tfrac {d^{2}}{1-d^{2\left( p+q-1\right) }}\right)+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot \tfrac {d}{1-d^{p+q-1}}\\ & \leq \delta , \end{align*}
</div>
<p> so \(x_{k+1}\in S\left( x_{0},\delta \right) .\) </p>
<p>From here it is clear that:</p>
<div class="displaymath" id="a0000000023">
  \begin{align*}  \left\Vert y_{k+1}-x_{0}\right\Vert _{X}& \leq \left\Vert y_{k+1}-x_{k+1}\right\Vert _{X}+\left\Vert x_{k+1}-x_{0}\right\Vert _{X}\\ & \leq a\left\Vert f\left( x_{k+1}\right) \right\Vert _{Y}+\left\Vert x_{k+1}-x_{0}\right\Vert _{X} \\ & \leq aR_{k+1}+\left\Vert x_{k+1}-x_{0}\right\Vert _{X}\\ & \leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot d^{2\left( p+q\right) ^{k}}+\left\Vert x_{k+1}-x_{0}\right\Vert _{X}. \end{align*}
</div>
<p>As \(d{\lt}1\) and  \(\left( p+q\right) ^{k}\geq 1\) it is clear that \(d^{2\left( p+q\right) ^{K}}\leq d^{2},\) therefore:</p>
<div class="displaymath" id="a0000000024">
  \begin{align*}  \left\Vert y_{k+1}-x_{0}\right\Vert _{X}& \leq 2aR_{0}+\tfrac {a\alpha ^{2}}{LB_{0}^{2}}\left( d^{2}+\tfrac {d^{2}}{1-d^{2\left( p+q-1\right) }}\right)+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot \tfrac {d}{1-d^{p+q-1}}\\ & =\delta , \end{align*}
</div>
<p> therefore \(y_{k+1}\in S\left( x_{0},\delta \right).\) </p>
<p>As</p>
<div class="displaymath" id="a0000000025">
  \begin{align*}  \left\Vert z_{k+1}-x_{0}\right\Vert _{X}& \leq \left\Vert z_{k+1}-x_{k+1}\right\Vert _{X}+\left\Vert x_{k+1}-x_{0}\right\Vert _{X}\\ & \leq a\left\Vert f\left( x_{k+1}\right) \right\Vert _{Y}+\left\Vert x_{k+1}-x_{0}\right\Vert _{X}, \end{align*}
</div>
<p> we obtain for \(\left\Vert z_{k+1}-x_{0}\right\Vert _{X}\) the same delimitation as for \(\left\Vert y_{k+1}-x_{0}\right\Vert _{X}\) and the expression of this delimitation is \(\delta .\) Therefore we have that \(z_{k+1}\in S\left( x_{0},\delta \right) .\) </p>
<p><b class="bfseries">b)</b> We know that:</p>
<div class="displaymath" id="a0000000026">
  \begin{align*}  \left\Vert f\left( x_{k+1}\right) \right\Vert _{Y}& =\left\Vert f\left( x_{k+1}\right) -\theta _{Y}\right\Vert _{Y}\\ & =\left\Vert f\left( x_{k+1}\right) -f\left( y_{k}\right) -\Gamma _{f;y_{k},z_{k}}\left( x_{k+1}-y_{k}\right) \right\Vert _{Y}\\ & \leq \left\Vert \Gamma _{f;y_{k},x_{k+1}}-\Gamma _{f;y_{k},z_{k}}\right\Vert \cdot \left\Vert x_{k+1}-y_{k}\right\Vert _{X}\\ & \leq L\left\Vert x_{k+1}-y_{k}\right\Vert _{X}\cdot \left\Vert x_{k+1}-z_{k}\right\Vert _{X}. \end{align*}
</div>
<p>On account of the equalities (<a href="#f47">7</a>) that are true for \(n=k\) we have that:</p>
<div class="displaymath" id="a0000000027">
  \[  \left\Vert f\left( x_{k+1}\right) \right\Vert _{Y}\leq L\Vert \Gamma _{f;y_{k},z_{k}}^{-1}\Vert ^{2}\cdot \left\Vert f\left( y_{k}\right) \right\Vert _{Y}\cdot \left\Vert f\left( z_{k}\right) \right\Vert _{Y}.  \]
</div>
<p>Using the hypothesis of the induction and the hypothesis of the verification of the inequalities (<a href="#f53">8</a>) we obtain:</p>
<div class="displaymath" id="a0000000028">
  \[  \left\Vert f\left( x_{k+1}\right) \right\Vert _{Y}\leq LK^{2}B_{k}^{2}\left\Vert f\left( x_{k}\right) \right\Vert _{Y}^{p+q}\leq L\overline{K}^{2}B_{k}^{2}R_{k}^{p+q}=R_{k+1}.  \]
</div>
<p><b class="bfseries">c)</b> From  the existence of the mapping \(\Gamma _{f;y_{k},z_{k}}^{-1}\in \left( Y,X\right) ^{\ast },\) of the fact that \(f\left( y_{k}\right) \neq \theta _{Y},\) \(f\left( z_{k}\right) \neq \theta _{Y}\) and using the equalities:</p>
<div class="displaymath" id="a0000000029">
  \[  x_{k+1}=y_{k}-\Gamma _{f;y_{k},z_{k}}^{-1}f\left( y_{k}\right) =z_{k}-\Gamma _{f;y_{k},z_{k}}^{-1}f\left( z_{k}\right)  \]
</div>
<p> we deduce that \(x_{k+1}\neq y_{k}\) and \(z_{k+1}\neq z_{k}.\) </p>
<p>So, the following mapping has a meaning:</p>
<div class="displaymath" id="a0000000030">
  \[  U_{k}=\Gamma _{f;y_{k},z_{k}}^{-1}\left( \Gamma _{f;y_{k},z_{k}}-\Gamma _{f;x_{k+1},z_{k}}\right) \in \left( X,X\right) ^{\ast }.  \]
</div>
<p>From here it is clear that:</p>
<div class="equation" id="f64">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;x_{k+1},z_{k}}=\Gamma _{f;y_{k},z_{k}}\left( \mathbf{I}_{X}-U_{k}\right) . \label{f64}\end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p>Considering the fact that \(y_{k},z_{k}\in S\left( x_{0},\delta \right) \) and taking into account what we have proved at <b class="bfseries">a)</b> and as we have \(x_{k+1}\in S\left( x_{0},\delta \right) ,\) we deduce that:</p>
<div class="displaymath" id="a0000000031">
  \begin{align*}  \left\Vert U_{k}\right\Vert & \leq \Vert \Gamma _{f;y_{k},z_{k}}^{-1}\Vert \cdot \left\Vert \Gamma _{f;y_{k},z_{k}}-\Gamma _{f;x_{k+1},z_{k}}\right\Vert \\ & \leq B_{k}L\left\Vert x_{k+1}-y_{k}\right\Vert _{X}\leq \end{align*}
</div>
<div class="displaymath" id="a0000000032">
  \begin{align*} & \leq LB_{k}\Vert \Gamma _{f;y_{k},z_{k}}^{-1}\Vert \cdot \left\Vert f\left( y_{k}\right) \right\Vert _{Y}\\ & \leq LB_{k}^{2}K\left\Vert f\left( x_{k}\right) \right\Vert _{Y}^{p}\\ & =LKB_{k}^{2}R_{k}^{p}\\ & =u_{k}\leq \alpha {\lt}1. \end{align*}
</div>
<p>From here, using the well known theorem of Banach (on account of the fact that \(\left( X,\left\Vert \cdot \right\Vert _{X}\right) \) is a Banach space), we deduce that there exists the mapping \(\left( \mathbf{I}_{X}-U_{k}\right) ^{-1}\in \left( X,X\right) ^{\ast }\) and</p>
<div class="displaymath" id="a0000000033">
  \[  \Vert \left( \mathbf{I}_{X}-U_{k}\right) ^{-1}\Vert \leq \tfrac {1}{1-\left\Vert U_{k}\right\Vert }\leq \tfrac {1}{1-u_{k}}.  \]
</div>
<p>From the existence of the mappings \(\Gamma _{f;y_{k},z_{k}}^{-1}\in \left( Y,X\right) ^{\ast }\) and \(\left( \mathbf{I}_{X}-U_{k}\right) ^{-1}\) \( \in \left( X,X\right) ^{\ast }\) using the equality (<a href="#f64">24</a>) we deduce the existence of the mapping \(\Gamma _{f;x_{k+1},z_{k}}^{-1}\in \left( Y,X\right) ^{\ast }\) and:</p>
<div class="equation" id="f65">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;x_{k+1},z_{k}}^{-1}=\left( \mathbf{I}_{X}-U_{k}\right) ^{-1}\Gamma _{f;y_{k},z_{k}}^{-1}. \label{f65}\end{equation}
  </div>
  <span class="equation_label">25</span>
</p>
</div>
<p>From the equality (<a href="#f65">25</a>) we have the inequality:</p>
<div class="displaymath" id="a0000000034">
  \[  \Vert \Gamma _{f;x_{k+1},z_{k}}^{-1}\Vert \leq \Vert \left( \mathbf{I}_{X}-U_{k}\right) ^{-1}\Vert \cdot \Vert \Gamma _{f;y_{k},z_{k}}^{-1}\Vert \leq \tfrac {B_{k}}{1-u_{k}} \]
</div>
<p> as well. </p>
<p>Also the mapping</p>
<div class="displaymath" id="a0000000035">
  \[  S_{k}=\Gamma _{f;y_{k},z_{k}}^{-1}\left( \Gamma _{f;y_{k},z_{k}}-\Gamma _{f;y_{k},x_{k+1}}\right) \in \left( X,X\right) ^{\ast } \]
</div>
<p> has a meaning. We can write that:</p>
<div class="equation" id="f66">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;y_{k},x_{k+1}}=\Gamma _{f;y_{k},z_{k}}\left( \mathbf{I}_{X}-S_{k}\right) . \label{f66}\end{equation}
  </div>
  <span class="equation_label">26</span>
</p>
</div>
<p>In the same way as in the case of the mapping \(U_{k}\) we have that:</p>
<div class="displaymath" id="a0000000036">
  \[  \left\Vert S_{k}\right\Vert \leq LKB_{k}^{2}R_{k}^{q}=s_{k}\leq \alpha {\lt}1,  \]
</div>
<p> therefore there exists the mapping \(\left( \mathbf{I}_{X}-S_{k}\right) ^{-1}\in \left( X,X\right) ^{\ast }\), so there exists the mapping \(\Gamma _{f;y_{k},x_{k+1}}^{-1}\in \left( Y,X\right) ^{\ast }\) as well. </p>
<p>As \(\Gamma _{f;y_{k},x_{k+1}}^{-1}=\left( \mathbf{I}_{X}-S_{k}\right) ^{-1}\Gamma _{f;y_{k},z_{k}}^{-1}\) we have that:</p>
<div class="displaymath" id="a0000000037">
  \[  \Vert \Gamma _{f;y_{k},x_{k+1}}^{-1}\Vert \leq \Vert \left( \mathbf{I}_{X}-S_{k}\right) ^{-1}\Vert \cdot \Vert \Gamma _{f;y_{k},z_{k}}^{-1}\Vert \leq \tfrac {B_{k}}{1-s_{k}}.  \]
</div>
<p>As \(y_{k+1}\neq z_{k}\) the following mapping has a meaning</p>
<div class="displaymath" id="a0000000038">
  \[  V_{k}=\Gamma _{f;x_{k+1},z_{k}}^{-1}\left( \Gamma _{f;x_{k+1},z_{k}}-\Gamma _{f;y_{k+1},z_{k}}\right) \in \left( X,X\right) ^{\ast },  \]
</div>
<p> from where evidently</p>
<div class="equation" id="f67">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;y_{k+1},z_{k}}=\Gamma _{f;x_{k+1},z_{k}}\left( \mathbf{I}_{X}-V_{K}\right) . \label{f67}\end{equation}
  </div>
  <span class="equation_label">27</span>
</p>
</div>
<p>As \(x_{k+1},z_{k},y_{k+1}\in S\left( x_{0},\delta \right) \subseteq D\) we deduce that:</p>
<div class="displaymath" id="a0000000039">
  \begin{align*}  \left\Vert V_{k}\right\Vert &  \leq \Vert \Gamma _{f;x_{k+1},z_{k}}^{-1}\Vert \cdot \left\Vert \Gamma _{f;x_{k+1},z_{k}}-\Gamma _{f;y_{k+1},z_{k}}\right\Vert _{X}\\ & \leq \Vert \Gamma _{f;x_{k+1},z_{k}}^{-1}\Vert L\left\Vert x_{k+1}-y_{k+1}\right\Vert \\ & \leq aL\cdot \tfrac {B_{k}}{1-u_{k}}\cdot \left\Vert f\left( x_{k+1}\right) \right\Vert _{Y}\\ & =\tfrac {aLB_{k}}{1-u_{k}}\cdot L\overline{K}^{2}B_{k}^{2}R_{k}^{p+q}\\ & =\tfrac {aL^{2}\overline{K}^{2}B_{k}^{3}R_{k}^{p+q}}{1-u_{k}}\\ & =v_{k}{\lt}\tfrac {\alpha ^{2}}{1-\alpha }{\lt}1. \end{align*}
</div>
<p>From here we deduce that there exists the mapping \(\left( \mathbf{I}_{X}-V_{k}\right) ^{-1}\in \left( X,X\right) ^{\ast }\) and \(\Vert \left( \mathbf{I}_{X}-V_{k}\right) ^{-1}\Vert \leq \tfrac {1}{1-v_{k}}.\) </p>
<p>Taking into account of the existence of the mappings \(\Gamma _{f;x_{k+1},z_{k}}^{-1}\in \left( Y,X\right) ^{\ast }\) and \(\left( \mathbf{I}_{X}-V_{k}\right) ^{-1}\in \left( X,X\right) ^{\ast }\) we deduce the existence of the mapping:</p>
<div class="displaymath" id="a0000000040">
  \[  \Gamma _{f;y_{k+1},z_{k}}^{-1}=\left( \mathbf{I}_{X}-V_{k}\right) ^{-1}\Gamma _{f;x_{k+1},z_{k}}^{-1}\in \left( Y,X\right) ^{\ast } \]
</div>
<p> and the relations:</p>
<div class="equation" id="f68">
<p>
  <div class="equation_content">
    \begin{equation}  \Vert \Gamma _{f;y_{k+1},z_{k}}^{-1}\Vert \leq \Vert \left( \mathbf{I}_{X}-V_{k}\right) ^{-1}\Vert \cdot \Vert \Gamma _{f;x_{k+1},z_{k}}^{-1}\Vert \leq \tfrac {B_{k}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) }. \label{f68}\end{equation}
  </div>
  <span class="equation_label">28</span>
</p>
</div>
<p>At the same time from the hypotheses we have \(y_{k+1}\neq x_{k+1},\) so the mapping:</p>
<div class="equation" id="f69">
<p>
  <div class="equation_content">
    \begin{equation}  W_{k}=\Gamma _{f;y_{k+1},z_{k}}^{-1}\left( \Gamma _{f;y_{k+1},z_{k}}-\Gamma _{f;y_{k+1},x_{k+1}}\right) \in \left( X,X\right) ^{\ast } \label{f69}\end{equation}
  </div>
  <span class="equation_label">29</span>
</p>
</div>
<p> has a meaning. </p>
<p>Obviously we have:</p>
<div class="equation" id="f70">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;y_{k+1},x_{k+1}}=\Gamma _{f;y_{k+1},z_{k}}\left( \mathbf{I}_{X}-W_{k}\right) . \label{f70}\end{equation}
  </div>
  <span class="equation_label">30</span>
</p>
</div>
<p>As \(z_{k},y_{k+1},x_{k+1}\in S\left( x_{0},\delta \right) \subseteq D\) we deduce that</p>
<div class="displaymath" id="a0000000041">
  \begin{align*}  \left\Vert W_{k}\right\Vert & \leq \Vert \Gamma _{f;y_{k+1},z_{k}}^{-1}\Vert \cdot \left\Vert \Gamma _{f;y_{k+1},z_{k}}-\Gamma _{f;y_{k+1},x_{k+1}}\right\Vert \\ & \leq \Vert \Gamma _{f;y_{k+1},z_{k}}^{-1}\Vert \cdot L\left\Vert x_{k+1}-z_{k}\right\Vert _{X}\\ & \leq L\cdot \tfrac {B_{k}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) }\cdot \Vert \Gamma _{f;y_{k+1},z_{k}}^{-1}\Vert \cdot \left\Vert f\left( z_{k}\right) \right\Vert _{Y}^{q}\\ & =\tfrac {LB_{k}^{2}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) }\cdot \overline{K}\left\Vert f\left( x_{k}\right) \right\Vert _{Y}^{q}\\ & =L\overline{K}\cdot \tfrac {B_{k}^{2}R_{k}^{q}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) }\\ & =w_{k}{\lt}\tfrac {\alpha }{1-\alpha -\alpha ^{2}}{\lt}1, \end{align*}
</div>
<p> therefore there exists the mapping \(\left( \mathbf{I}_{X}-W_{k}\right) ^{-1}\in \left( X,X\right) ^{\ast }\) and we have the inequality \(\Vert \left( \mathbf{I}_{X}-W_{k}\right) ^{-1}\Vert \leq \tfrac {1}{1-\left\Vert W_{k}\right\Vert }\leq \tfrac {1}{1-w_{k}}.\) </p>
<p>Considering the existence of the mapping \(\Gamma _{f;y_{k+1},z_{k}}^{-1}\in \left( Y,X\right) ^{\ast }\) from the previous evaluation we deduce on the basis of the equality (<a href="#f70">30</a>) the existence of the mapping \(\Gamma _{f;y_{k+1},x_{k+1}}^{-1}\in \left( Y,X\right) ^{\ast }\) such that</p>
<div class="displaymath" id="a0000000042">
  \[  \Gamma _{f;y_{k+1},x_{k+1}}^{-1}=\left( \mathbf{I}_{X}-W_{k}\right) ^{-1}\Gamma _{f;y_{k+1},z_{k}}^{-1} \]
</div>
<p> and</p>
<div class="displaymath" id="a0000000043">
  \begin{align*}  \Vert \Gamma _{f;y_{k+1},x_{k+1}}^{-1}\Vert & \leq \Vert \left( \mathbf{I}_{X}-W_{k}\right) ^{-1}\Vert \cdot \Vert \Gamma _{f;y_{k+1},z_{k}}^{-1}\Vert \\ & \leq \tfrac {B_{k}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) \left( 1-w_{k}\right) }. \end{align*}
</div>
<p>Evidently \(y_{k+1}\neq z_{k+1}\) therefore the mapping:</p>
<div class="equation" id="f71">
<p>
  <div class="equation_content">
    \begin{equation}  T_{k}=\Gamma _{f;y_{k+1},x_{k+1}}^{-1}\left( \Gamma _{f;y_{k+1},x_{k+1}}-\Gamma _{f;y_{k+1},z_{k+1}}\right) \in \left( X,X\right) ^{\ast } \label{f71}\end{equation}
  </div>
  <span class="equation_label">31</span>
</p>
</div>
<p> has a meaning. </p>
<p>From (<a href="#f71">31</a>) we obviously have that:</p>
<div class="equation" id="f72">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;y_{k+1},z_{k+1}}=\Gamma _{f;y_{k+1},x_{k+1}}\left( \mathbf{I}_{X}-T_{k}\right) . \label{f72}\end{equation}
  </div>
  <span class="equation_label">32</span>
</p>
</div>
<p>As \(y_{k+1},x_{k+1},z_{k+1}\in S\left( x_{0},\delta \right) \subseteq D\) we deduce that</p>
<div class="displaymath" id="a0000000044">
  \begin{align*}  \left\Vert T_{k}\right\Vert & \leq \Vert \Gamma _{f;y_{k+1},x_{k+1}}^{-1}\Vert \cdot \left\Vert \Gamma _{f;y_{k+1},x_{k+1}}-\Gamma _{f;y_{k+1},z_{k+1}}\right\Vert \\ & \leq \Vert \Gamma _{f;y_{k+1},x_{k+1}}^{-1}\Vert \cdot L\left\Vert x_{k+1}-z_{k+1}\right\Vert \\ & \leq \tfrac {aLB_{k}\left\Vert f\left( x_{k+1}\right) \right\Vert _{Y}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) \left( 1-w_{k}\right) }\\ & =\tfrac {aLB_{k}R_{k+1}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) \left( 1-w_{k}\right) }\\ & =\tfrac {aLB_{k}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) \left( 1-w_{k}\right) }L\overline{K}^{2}B_{k}^{2}R_{k}^{p+q}\\ & =\tfrac {aL^{2}\overline{K}^{2}B_{k}^{3}R_{k}^{p+q}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) \left( 1-w_{k}\right) }\\ & =t_{k}{\lt}\tfrac {\alpha ^{2}}{1-2\alpha -\alpha ^{2}}{\lt}1. \end{align*}
</div>
<p>Therefore, based on the same theorem of Banach, there exists the mapping \(\left( \mathbf{I}_{X}-T_{k}\right) ^{-1}\in \left( X,X\right) ^{\ast }\) and \(\Vert \left( \mathbf{I}_{X}-T_{k}\right) ^{-1}\Vert \leq \tfrac {1}{1-\left\Vert T_{k}\right\Vert }\leq \tfrac {1}{1-t_{k}}.\) </p>
<p>Adding to the last information the fact that there exists the mapping \(\Gamma _{f;y_{k+1},x_{k+1}}^{-1}\in \left( Y,X\right) ^{\ast },\) we deduce on the basis of the equality (<a href="#f72">32</a>) the existence of the mapping:</p>
<div class="displaymath" id="a0000000045">
  \[  \Gamma _{f;y_{k+1},z_{k+1}}^{-1}=\left( \mathbf{I}_{X}-T_{k}\right) ^{-1}\Gamma _{f;y_{k+1},x_{k+1}}^{-1}\in \left( Y,X\right) ^{\ast } \]
</div>
<p> together with the inequality:</p>
<div class="displaymath" id="a0000000046">
  \begin{align*}  \Vert \Gamma _{f;y_{k+1},z_{k+1}}^{-1}\Vert & \leq \Vert \left( \mathbf{I}_{X}-T_{k}\right) ^{-1}\Vert \cdot \Vert \Gamma _{f;y_{k+1},x_{k+1}}^{-1}\Vert \\ & \leq \tfrac {B_{k}}{\left( 1-u_{k}\right) \left( 1-v_{k}\right) \left( 1-w_{k}\right) \left( 1-t_{k}\right) }=B_{k+1}. \end{align*}
</div>
<p>The existence of the mapping \(\Gamma _{f;y_{k+1},z_{k+1}}^{-1}\in \left( Y,X\right) ^{\ast }\) and the inequality \(\Vert \Gamma _{f;y_{k+1},z_{k+1}}^{-1}\Vert \leq B_{k+1}\) express the fact that the proposition <b class="bfseries">c)</b> is true for \(n=k+1.\) </p>
<p>Therefore the propositions <b class="bfseries">a) - c)</b> are true for \(n=k+1.\) </p>
<p>On account of the principle of mathematical induction these relations are true for any number \(n\in \mathbb {N}.\) </p>
<p>We will now show that the sequence \(\left( x_{n}\right) _{n\in \mathbb {N}}\) is a Cauchy sequence. </p>
<p>Taking into account of the inequality (<a href="#f63ppp">23</a>) we deduce that for any \(n,p\in \mathbb {N}\) we have the following inequality:</p>
<div class="displaymath" id="a0000000047">
  \begin{align*}  \left\Vert x_{n+p}-x_{n}\right\Vert _{X}& =\Big\Vert {\displaystyle \sum \limits _{j=n}^{n+p-1}} \left( x_{j+1}-x_{j}\right) \Big\Vert _{X}\\ & \leq {\displaystyle \sum \limits _{j=n}^{n+p-1}} \left\Vert x_{j+1}-x_{j}\right\Vert _{X}\\ & \leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}{\displaystyle \sum \limits _{j=n}^{n+p-1}} d^{2\left( p+q\right) ^{j-1}}+\tfrac {\alpha }{L\overline{K}B_{0}}{\displaystyle \sum \limits _{j=n}^{n+p-1}} d^{\left( p+q\right) ^{j}}. \end{align*}
</div>
<p>But, as \(p,q\geq 1,\) we deduce that for any \(s\in \mathbb {N\cup }\left\{  0\right\} \)</p>
<div class="displaymath" id="a0000000048">
  \begin{align*}  2\left( p+q\right) ^{n-1+s}-2\left( p+q\right) ^{n-1}& =2\left( p+q\right) ^{n-1}\left[ \left( p+q\right) ^{s}-1\right] \\ & {\gt}2s\left( p+q\right) ^{n-1}\left( p+q-1\right) , \end{align*}
</div>
<p> therefore as \(d{\lt}1\) we deduce that:</p>
<div class="displaymath" id="a0000000049">
  \begin{align*}  {\displaystyle \sum \limits _{j=n}^{n+p-1}} d^{2\left( p+q\right) ^{j-1}}& {\lt}d^{2\left( p+q\right) ^{n-1}}{\displaystyle \sum \limits _{s=0}^{p-1}} \left[ d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }\right] ^{s}\\ & {\lt}\tfrac {d^{2\left( p+q\right) ^{n-1}}}{1-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }}. \end{align*}
</div>
<p>Similarly</p>
<div class="displaymath" id="a0000000050">
  \[ {\displaystyle \sum \limits _{j=n}^{n+p-1}} d^{\left( p+q\right) ^{j}}{\lt}\tfrac {d^{\left( p+q\right) ^{n}}}{1-d^{\left( p+q\right) ^{n}\left( p+q-1\right) }},  \]
</div>
<p> therefore</p>
<div class="displaymath" id="f73">
  \begin{align} \left\Vert x_{n+p}-x_{n}\right\Vert _{X}& \leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot \tfrac {d^{2\left( p+q\right) ^{n-1}}}{1-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }}+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot \tfrac {d^{\left( p+q\right) ^{n}}}{1-d^{\left( p+q\right) ^{n}\left( p+q-1\right) }}. \label{f73}\end{align}
</div>
<p>From the fact that \(d{\lt}1\) we deduce that:</p>
<div class="displaymath" id="a0000000051">
  \[  \lim _{n\rightarrow \infty }\tfrac {d^{2\left( p+q\right) ^{n-1}}}{1-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }}=\lim _{n\rightarrow \infty }\tfrac {d^{\left( p+q\right) ^{n}}}{1-d^{\left( p+q\right) ^{n}\left( p+q-1\right) }}=0,  \]
</div>
<p> therefore \(\lim \limits _{n\rightarrow \infty }\underset {p\in \mathbb {N}}{\rm unif}\left\Vert x_{n+p}-x_{n}\right\Vert _{X}=0\) and this equality expresses the fact that \(\left( x_{n}\right) _{n\in \mathbb {N}}\subseteq D\subseteq X\) is a Cauchy sequence. </p>
<p>From the quality of the linear normed space \(\left( X,\left\Vert \cdot \right\Vert _{X}\right) \) of being a Banach space we deduce that the sequence \(\left( x_{n}\right) _{n\in \mathbb {N}}\subseteq X\) is a convergent sequence, therefore there exists an element \(\overline{x}\in X\) such that \(\overline{x}=\lim \limits _{n\rightarrow \infty }x_{n}.\) </p>
<p>If in the inequality (<a href="#f73">33</a>) we tend to the limit as \(p\rightarrow \infty ,\) we obtain the inequality (<a href="#K2">20</a>) form the conclusion of the theorem. </p>
<p>From the inequality (<a href="#K2">20</a>) with \(n=0\)  we deduce that:</p>
<div class="displaymath" id="a0000000052">
  \begin{align*}  \left\Vert \overline{x}-x_{0}\right\Vert _{X}& \leq \left\Vert \overline{x}-x_{1}\right\Vert _{X}+\left\Vert x_{1}-x_{0}\right\Vert _{X}\\ & \leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot \tfrac {d^{2}}{1-d^{2\left( p+q-1\right) }}+\tfrac {\alpha }{L\overline{K}B_{0}}\cdot \tfrac {d^{p+q}}{1-d^{\left( p+q\right) \left( p+q-1\right) }} +aR_{0}+\tfrac {\alpha d}{L\overline{K}B_{0}}\\ & \leq \delta , \end{align*}
</div>
<p> and this inequality expresses the fact that \(\overline{x}\in S\left( x_{0},\delta \right) .\) </p>
<p>The existence of the mapping \(\Gamma _{f;y_{n},z_{n}}^{-1}\in \left( Y,X\right) ^{\ast }\) allows for the expressions of the element \(x_{n+1}\) under the forms (<a href="#K">18</a>). </p>
<p>Obviously, for any \(n\in \mathbb {N}\) we have the inequalities:</p>
<div class="displaymath" id="a0000000053">
  \begin{align} \left\Vert y_{n}-\overline{x}\right\Vert _{X}& \leq \left\Vert y_{n}-x_{n}\right\Vert _{X}+\left\Vert x_{n}-\overline{x}\right\Vert _{X}\nonumber \\ & \leq a\left\Vert f\left( x_{n}\right) \right\Vert _{Y}+\left\Vert x_{n}-\overline{x}\right\Vert _{X}\nonumber \\ & \leq aR_{n}+\left\Vert x_{n}-\overline{x}\right\Vert _{X}\nonumber \\ & \leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot d^{2\left( p+q\right) ^{n-1}}+\left\Vert x_{n}-\overline{x}\right\Vert _{X}.\label{f74}\end{align}
</div>
<p>We have a similar estimate for \(\left\Vert z_{n}-\overline{x}\right\Vert _{X}\) as well. We obtain in this way the estimates (<a href="#K3">21</a>). The inequality (<a href="#K3">21</a>) expresses in fact that:</p>
<div class="displaymath" id="a0000000054">
  \[  \lim _{n\rightarrow \infty }\left\Vert y_{n}-\overline{x}\right\Vert _{X}=\lim _{n\rightarrow \infty }\left\Vert z_{n}-\overline{x}\right\Vert _{X}=0.  \]
</div>
<p>We still need to show that \(\overline{x}\in D\) is the solution of the equation \(f\left( x\right) =\theta _{Y}.\) </p>
<p>We will first remark that from the fact that for any \(n\in \mathbb {N}\) we have that \(\left\Vert f\left( x_{n}\right) \right\Vert _{Y}\leq R_{n}\leq \tfrac {\alpha ^{2}}{LB_{0}^{2}}\cdot d^{2\left( p+q\right) ^{n-1}}\) and from \(d{\lt}1\) we deduce that<br /> \(\lim \limits _{n\rightarrow \infty }\left\Vert f\left( x_{n}\right) \right\Vert _{Y}=0.\) </p>
<p>Let us consider a number \(n\in \mathbb {N}\) arbitrarily. As \(f\left( y_{n}\right) \neq \theta _{Y}\) we deduce that \(y_{n}\neq \overline{x},\) therefore as \(x_{n}\neq y_{n}\) we will have that:</p>
<div class="displaymath" id="a0000000055">
  \begin{align*}  0& \leq \left\Vert f\left( \overline{x}\right) \right\Vert _{Y}\\ & \leq \left\Vert f\left( y_{n}\right) -f\left( \overline{x}\right) \right\Vert _{Y}+\left\Vert f\left( y_{n}\right) \right\Vert _{Y}\\ & \leq \left\Vert \Gamma _{f;\overline{x},y_{n}}\right\Vert \cdot \left\Vert y_{n}-\overline{x}\right\Vert _{X}+K\left\Vert f\left( x_{n}\right) \right\Vert _{Y}^{p}. \end{align*}
</div>
<p>Here, we have:</p>
<div class="displaymath" id="a0000000056">
  \begin{align*}  \left\Vert \Gamma _{f;\overline{x},y_{n}}\right\Vert & \leq \left\Vert \Gamma _{f;\overline{x},y_{n}}-\Gamma _{f;\overline{x},y_{0}}\right\Vert +\left\Vert \Gamma _{f;\overline{x},y_{0}}\right\Vert \medskip \\ & \leq L\left\Vert y_{n}-y_{0}\right\Vert _{X}+\left\Vert \Gamma _{f;\overline{x},y_{0}}\right\Vert \\ &  \leq L\left( \left\Vert y_{n}-x_{0}\right\Vert _{X}+\left\Vert y_{0}-x_{0}\right\Vert _{X}\right) +\left\Vert \Gamma _{f;\overline{x},y_{0}}\right\Vert \\ &  \leq L\left( \delta +\left\Vert y_{0}-x_{0}\right\Vert _{X}\right) +\left\Vert \Gamma _{f;\overline{x},y_{0}}\right\Vert , \end{align*}
</div>
<p> therefore</p>
<div class="equation" id="f75">
<p>
  <div class="equation_content">
    \begin{equation} 0\leq \left\Vert f\left( \overline{x}\right) \right\Vert _{Y}\leq \left[ L\left( \delta +\left\Vert y_{0}-x_{0}\right\Vert _{X}\right) +\left\Vert \Gamma _{f;\overline{x},y_{0}}\right\Vert \right] \cdot \left\Vert y_{n}-\overline{x}\right\Vert _{X}+K\left\Vert f\left( x_{n}\right) \right\Vert _{Y}^{p}. \label{f75}\end{equation}
  </div>
  <span class="equation_label">35</span>
</p>
</div>
<p>As \(\lim \limits _{n\rightarrow \infty }\left\Vert y_{n}-\overline{x}\right\Vert _{X}=0=\lim \limits _{n\rightarrow \infty }\left\Vert f\left( x_{n}\right) \right\Vert _{Y}\) and \(p{\gt}1\) we deduce that:</p>
<div class="displaymath" id="a0000000057">
  \[  \lim _{n\rightarrow \infty }\left\{  \left[ L\left( \delta +\left\Vert y_{0}-x_{0}\right\Vert _{X}\right) +\left\Vert \Gamma _{f;\overline{x},y_{0}}\right\Vert \right] \cdot \left\Vert y_{n}-\overline{x}\right\Vert _{X}+K\left\Vert f\left( x_{n}\right) \right\Vert _{Y}^{p}\right\}  =0,  \]
</div>
<p> from where on account of the relation (<a href="#f75">35</a>) we deduce that \(\left\Vert f\left( \overline{x}\right) \right\Vert _{Y}=0,\) namely \(f\left( \overline{x}\right) =\theta _{Y}.\) </p>
<p>With that the theorem is proven.</p>
<h1 id="a0000000058">3 Remarks on the margin of the main result</h1>
<p><div class="remark_thmwrapper " id="r6">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">7</span>
  </div>
  <div class="remark_thmcontent">
  <p>With the  hypotheses of the Theorem <a href="#t2">6</a> the sequence:</p>
<div class="displaymath" id="a0000000059">
  \[  ( \Vert \Gamma _{f;y_{n},z_{n}}^{-1}\Vert ) _{n\in \mathbb {N}\text{ }^{\ast }} \]
</div>
<p> is bounded, and for any \(n\in \mathbb {N}^{\ast }\) the following inequality takes place:</p>
<div class="equation" id="f75p">
<p>
  <div class="equation_content">
    \begin{equation}  \Vert \Gamma _{f;y_{n},z_{n}}^{-1}\Vert \leq B_{0}\mathbf{e}^{G\left( d,\alpha \right) } \label{f75p}\end{equation}
  </div>
  <span class="equation_label">36</span>
</p>
</div>
<p> where:</p>
<div class="displaymath" id="a0000000060">
  \begin{align} G\left( d,\alpha \right) & =\tfrac {\alpha \left( 2-3\alpha -\alpha ^{2}\right) }{\left( 1-\alpha \right) \left( 1-2\alpha -\alpha ^{2}\right) }\cdot \tfrac {d}{1-d^{p+q-1}}\nonumber \\ & \quad +\tfrac {\alpha ^{2}\left( 2-3\alpha -3\alpha ^{2}\right) }{\left( 1-\alpha -\alpha ^{2}\right) \left( 1-2\alpha -2\alpha ^{2}\right) }\cdot \tfrac {d^{2}}{1-d^{2\left( p+q-1\right) }}. \label{f76}\end{align}
</div>
<p>Also the sequence \(\left( \left\Vert \Gamma _{f;y_{n},z_{n}}\right\Vert \right) _{n\in \mathbb {N}\text{ }^{\ast }}\) is bounded and for any \(n\in \mathbb {N}^{\ast }\) we have the inequality:</p>
<div class="equation" id="f76a">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert \Gamma _{f;y_{n},z_{n}}\right\Vert \leq \left\Vert \Gamma _{f;y_{0},z_{0}}\right\Vert +2\alpha \left( K+\tfrac {a\alpha }{B_{0}^{2}}\right) \cdot \tfrac {d}{1-d^{p+q-1}}. \label{f76a}\hfil \qed \end{equation}
  </div>
  <span class="equation_label">38</span>
</p>
</div>

  </div>
</div> </p>
<p>Indeed, from the recurrence relation of the sequence \(\left( B_{n}\right) _{n\in \mathbb {N}^{\ast }}\) we deduce that:</p>
<div class="displaymath" id="a0000000061">
  \[  B_{n}\leq \tfrac {B_{0}}{{\displaystyle \prod \limits _{j=0}^{n-1}} \left[ \left( 1-u_{j}\right) \left( 1-v_{j}\right) \left( 1-w_{j}\right) \left( 1-t_{j}\right) \right] }.  \]
</div>
<p>Obviously</p>
<div class="displaymath" id="a0000000062">
  \begin{align*} & \tfrac {1}{{\displaystyle \prod \limits _{j=0}^{n-1}} \left[ \left( 1-u_{j}\right) \left( 1-v_{j}\right) \left( 1-w_{j}\right) \left( 1-t_{j}\right) \right] }\leq \\ & \leq \left[ \tfrac {1}{4n}{\displaystyle \sum \limits _{j=0}^{n-1}} \left( \tfrac {1}{1-u_{j}}+\tfrac {1}{1-v_{j}}+\tfrac {1}{1-w_{j}}+\tfrac {1}{1-t_{j}}\right) \right] ^{4n}\medskip \\ & =\left[ 1+\tfrac {1}{4n}{\displaystyle \sum \limits _{j=0}^{n-1}} \left( \tfrac {u_{j}}{1-u_{j}}+\tfrac {v_{j}}{1-v_{j}}+\tfrac {w_{j}}{1-w_{j}}+\tfrac {t_{j}}{1-t_{j}}\right) \right] ^{4n}\\ & \leq \left[ 1\! +\! \tfrac {1}{4n}\left( Q_{1}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} u_{j}+Q_{2}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} v_{j}+Q_{3}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} w_{j}+Q_{4}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} t_{j}\right) \right] ^{4n}, \end{align*}
</div>
<p> where</p>
<div class="displaymath" id="a0000000063">
  \begin{align*}  Q_{1}\left( \alpha \right) & =\tfrac {1}{1-\alpha },\\ Q_{2}\left( \alpha \right) & =\tfrac {1}{1-\tfrac {\alpha ^{2}}{1-\alpha }}=\tfrac {1-\alpha }{1-\alpha -\alpha ^{2}},\\ Q_{3}\left( \alpha \right) & =\tfrac {1}{1-\tfrac {\alpha }{1-\alpha -\alpha ^{2}}}=\tfrac {1-\alpha -\alpha ^{2}}{1-2\alpha -\alpha ^{2}},\\ Q_{4}\left( \alpha \right)&  =\tfrac {1}{1-\tfrac {\alpha ^{2}}{1-2\alpha -\alpha ^{2}}}=\tfrac {1-2\alpha -\alpha ^{2}}{1-2\alpha -2\alpha ^{2}}. \end{align*}
</div>
<p>It is obvious that</p>
<div class="displaymath" id="a0000000064">
  \begin{align*}  {\displaystyle \sum \limits _{j=0}^{n-1}} u_{j}& \leq \alpha {\displaystyle \sum \limits _{j=0}^{n-1}} d^{\left( p+q\right) ^{j}}\leq \tfrac {\alpha d}{1-d^{p+q-1}},\\ {\displaystyle \sum \limits _{j=0}^{n-1}} v_{j}& \leq \tfrac {\alpha ^{2}}{1-\alpha }{\displaystyle \sum \limits _{j=0}^{n-1}} d^{2\left( p+q\right) ^{j}}\leq \tfrac {\alpha ^{2}d^{2}}{\left( 1-\alpha \right) \left( 1-d^{2\left( p+q-1\right) }\right) },\\ {\displaystyle \sum \limits _{j=0}^{n-1}} w_{j}& \leq \tfrac {\alpha }{1-\alpha -\alpha ^{2}}{\displaystyle \sum \limits _{j=0}^{n-1}} d^{\left( p+q\right) ^{j}}\leq \tfrac {\alpha d}{\left( 1-\alpha -\alpha ^{2}\right) \left( 1-d^{p+q-1}\right) },\\ {\displaystyle \sum \limits _{j=0}^{n-1}} t_{j}& \leq \tfrac {\alpha ^{2}}{1-2\alpha -\alpha ^{2}}{\displaystyle \sum \limits _{j=0}^{n-1}} d^{2\left( p+q\right) ^{j}}\leq \tfrac {\alpha ^{2}d^{2}}{\left( 1-2\alpha -\alpha ^{2}\right) \left( 1-d^{2\left( p+q-1\right) }\right) }. \end{align*}
</div>
<p>Therefore</p>
<div class="displaymath" id="a0000000065">
  \begin{align*} & Q_{1}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} u_{j}+Q_{2}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} v_{j}+Q_{3}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} w_{j}+Q_{4}\left( \alpha \right) {\displaystyle \sum \limits _{j=0}^{n-1}} t_{j}\leq \\ & \leq \tfrac {\alpha d}{\left( 1-\alpha \right) \left( 1-d^{p+q-1}\right) }+\tfrac {\alpha ^{2}d^{2}}{\left( 1-\alpha -\alpha ^{2}\right) \left( 1-d^{2\left( p+q-1\right) }\right) }\\ & \quad +\tfrac {\alpha d}{\left( 1-2\alpha -\alpha ^{2}\right) \left( 1-d^{p+q-1}\right) }+\tfrac {\alpha ^{2}d^{2}}{\left( 1-2\alpha -2\alpha ^{2}\right) \left( 1-d^{2\left( p+q-1\right) }\right) }\\ & =G\left( d,\alpha \right) . \end{align*}
</div>
<p>As</p>
<div class="displaymath" id="a0000000066">
  \[  B_{n}\leq B_{0}\left[ 1+\tfrac {1}{4n}G\left( d,\alpha \right) \right] ^{4n}\leq B_{0}\mathbf{e}^{G\left( d,\alpha \right) },  \]
</div>
<p> the inequality (<a href="#f75p">36</a>) is proven. </p>
<p>In order to establish the inequality (<a href="#f76a">38</a>), we note that for any \(n\in \mathbb {N}\) we have:</p>
<div class="displaymath" id="a0000000067">
  \[  \Gamma _{f;y_{n},z_{n}}=\Gamma _{f;y_{0},z_{0}}+\sum _{j=0}^{n-1}\left( \Gamma _{f;y_{j+1},z_{j+1}}-\Gamma _{f;y_{j},z_{j}}\right) ,  \]
</div>
<p> from where:</p>
<div class="displaymath" id="a0000000068">
  \[  \left\Vert \Gamma _{f;y_{n},z_{n}}\right\Vert \leq \left\Vert \Gamma _{f;y_{0},z_{0}}\right\Vert +\sum _{j=0}^{n-1}\left\Vert \Gamma _{f;y_{j+1},z_{j+1}}-\Gamma _{f;y_{j},z_{j}}\right\Vert .  \]
</div>
<p>For any \(j\in \mathbb {N}\) we have</p>
<div class="displaymath" id="a0000000069">
  \begin{align*}  \left\Vert \Gamma _{f;y_{j+1},z_{j+1}}\! -\! \Gamma _{f;y_{j},z_{j}}\right\Vert & \leq \left\Vert \Gamma _{f;y_{j+1},z_{j+1}}\! -\! \Gamma _{f;y_{j+1},z_{j}}\right\Vert  +\left\Vert \Gamma _{f;y_{j+1},z_{j}}\! -\! \Gamma _{f;y_{j},z_{j}}\right\Vert \\ & \leq L\left( \left\Vert y_{j+1}-y_{j}\right\Vert _{X}+\left\Vert z_{j+1}-z_{j}\right\Vert _{X}\right) . \end{align*}
</div>
<p>But</p>
<div class="displaymath" id="a0000000070">
  \begin{align*}  \left\Vert y_{j+1}-y_{j}\right\Vert _{X}& \leq \left\Vert y_{j+1}-x_{j+1}\right\Vert _{X}+\left\Vert x_{j+1}-y_{j}\right\Vert _{X}\\ & \leq a\left\Vert f\left( x_{j+1}\right) \right\Vert _{Y}+\left\Vert \Gamma _{f;y_{j},z_{j}}^{-1}\right\Vert \cdot \left\Vert f\left( x_{j}\right) \right\Vert _{Y}^{p}\\ & =aR_{j+1}+KB_{j}R_{j}^{p}\\ & \leq \tfrac {a\alpha ^{2}}{LB_{0}^{2}}\cdot d^{2\left( p+q\right) ^{j}}+\tfrac {K\alpha }{B_{0}}\cdot d^{\left( p+q\right) ^{j}}\\ & {\lt}\tfrac {\alpha \left( KB_{0}^{2}+a\alpha \right) }{LB_{0}^{2}}d^{\left( p+q\right) ^{j}}. \end{align*}
</div>
<p>For \(\left\Vert z_{j+1}-z_{j}\right\Vert _{X}\) we obtain the same estimate. </p>
<p>Therefore, for any \(n\in \mathbb {N}\) we obtain:</p>
<div class="displaymath" id="a0000000071">
  \[  \left\Vert \Gamma _{f;y_{n},z_{n}}\right\Vert \leq \left\Vert \Gamma _{f;y_{0},z_{0}}\right\Vert +\tfrac {2\alpha \left( KB_{0}^{2}+a\alpha \right) }{LB_{0}^{2}}\sum _{j=0}^{n-1}d^{\left( p+q\right) ^{j}},  \]
</div>
<p> from where the inequality (<a href="#f76a">38</a>) derives directly. </p>
<p><div class="remark_thmwrapper " id="r7">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">8</span>
  </div>
  <div class="remark_thmcontent">
  <p>For any \(n\in \mathbb {N}\) we have the following inequality:</p>
<div class="equation" id="f77">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert x_{n}-\overline{x}\right\Vert _{X}\leq \tfrac {\alpha }{LB_{0}\left[ 1-d^{2\left( p+q-1\right) }\right] }\cdot \left( \tfrac {\alpha }{B_{0}}+\tfrac {1}{K}\right) d^{2\left( p+q\right) ^{n-1}}, \label{f77}\end{equation}
  </div>
  <span class="equation_label">39</span>
</p>
</div>
<p> and this inequality show that the convergence order of the studied iterative method is \(p+q.\) </p>
<p>Indeed, from the fact that \(p,q\geq 1\) and \(d{\lt}1,\) we deduce that \(\left( p+q\right) ^{n}\geq \)<br />\(\geq 2\left( p+q\right) ^{n-1},\) therefore \(d^{\left( p+q\right) ^{n}}\leq d^{2\left( p+q\right) ^{n-1}}\) and </p>
<div class="displaymath" id="a0000000072">
  \[  d^{\left( p+q\right) ^{n}\left( p+q-1\right) }\leq d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }.  \]
</div>
<p>So</p>
<div class="displaymath" id="a0000000073">
  \[  \tfrac {d^{\left( p+q\right) ^{n}}}{1-d^{\left( p+q\right) ^{n}\left( p+q-1\right) }}\leq \tfrac {d^{2\left( p+q\right) ^{n-1}}}{1-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }},  \]
</div>
<p> therefore:</p>
<div class="displaymath" id="a0000000074">
  \[  \left\Vert x_{n}-\overline{x}\right\Vert _{X}\leq \tfrac {\alpha }{LB_{0}}\cdot \left( \tfrac {\alpha }{B_{0}}+\tfrac {1}{K}\right) \cdot \tfrac {d^{2\left( p+q\right) ^{n-1}}}{1-d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }}.  \]
</div>
<p>But \(\overline{K}\geq K,\) and as \(n\geq 1\) we have that \(d^{2\left( p+q\right) ^{n-1}\left( p+q-1\right) }\leq d^{2\left( p+q-1\right) }\) and in this way we obtain the inequality (<a href="#f77">39</a>). <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000075">4 Noticeable special cases</h1>
<p><div class="remark_thmwrapper " id="r8">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p>An important special case is that in which for any \(n\in \mathbb {N}\) we have \(x_{n}=z_{n}.\) This case is admitted from the hypotheses, and it is necessary for the following conditions to hold:</p>
<div class="displaymath" id="a0000000076">
  \[  f\left( x_{n}\right) ,f\left( y_{n}\right) \in Y\diagdown \left\{  \theta _{Y}\right\}  ,\  x_{n}\neq y_{n},\  x_{n}\neq y_{n+1} \]
</div>
<p> and:</p>
<div class="displaymath" id="a0000000077">
  \[  \Gamma _{f;x_{n},y_{n}}\left( x_{n+1}-x_{n}\right) +f\left( x_{n}\right) =\theta _{Y},  \]
</div>
<p> using for this aim the property that is expressed by the equality (<a href="#f46">6</a>). </p>
<p>For any \(n\in \mathbb {N}\) there exists the mapping \(\Gamma _{f;x_{n},y_{n}}^{-1}\in \left( Y,X\right) ^{\ast },\) therefore the recurrence relation of the sequence \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\) will be:</p>
<div class="displaymath" id="a0000000078">
  \[  x_{n+1}=x_{n}-\Gamma _{f;x_{n},y_{n}}^{-1}f\left( x_{n}\right) .  \]
</div>
<p> the inequality (<a href="#f77">39</a>) in this case becomes:</p>
<div class="equation" id="f79p">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert x_{n}-\overline{x}\right\Vert _{X}\leq \tfrac {\alpha }{LB_{0}\left( 1-d^{2p}\right) }\cdot \left( \tfrac {\alpha }{B_{0}}+\tfrac {1}{K}\right) \cdot d^{2\left( p+1\right) ^{n-1}}, \label{f79p}\end{equation}
  </div>
  <span class="equation_label">40</span>
</p>
</div>
<p> therefore the convergence order of the method is \(p+1\).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>One verifies the hypotheses of the Theorem <a href="#t2">6</a> in the case of \(q=1,\) therefore we have the conclusions of this theorem in this case. </p>
<p><div class="remark_thmwrapper " id="r9">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">10</span>
  </div>
  <div class="remark_thmcontent">
  <p>We will now consider an even more special case. Let be a mapping \(U\in \left( X,Y\right) ^{\ast }\diagdown \left\{  \Theta \right\}  \) (\(\Theta \) being the null mapping) and we will choose the sequence \(\left( y_{n}\right) _{n\in \mathbb {N}^{\ast }}\) by the relation \(y_{n}=x_{n}-Uf\left( x_{n}\right) \) for any \(n\in \mathbb {N}^{\ast }.\) At the same time we will choose \(z_{n}=x_{n}.\) This case comes in the framework of the previous more general case. </p>
<p>Therefore the sequence \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\) will be chosen such that for any \(n\in \mathbb {N}^{\ast }\) the following equality is verified:</p>
<div class="equation" id="f80">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;x_{n},x_{n}-Uf\left( x_{n}\right) }\left( x_{n+1}-x_{n}\right) +f\left( x_{n}\right) =\theta _{Y}, \label{f80}\end{equation}
  </div>
  <span class="equation_label">41</span>
</p>
</div>
<p> or if for any \(n\in \mathbb {N}^{\ast }\) there exists the mapping \(\Gamma _{f;x_{n},x_{n}-Uf\left( x_{n}\right) }^{-1}\in \left( Y,X\right) ^{\ast }\) we have the equality:</p>
<div class="equation" id="f81">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-\Gamma _{f;x_{n},x_{n}-Uf\left( x_{n}\right) }^{-1}f\left( x_{n}\right) . \label{f81}\hfil \qed \end{equation}
  </div>
  <span class="equation_label">42</span>
</p>
</div>

  </div>
</div> </p>
<p>For the convergence of the iterative method generated by the relation (<a href="#f80">41</a>) we have the following corollary: </p>
<p><div class="corollary_thmwrapper " id="c1">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">11</span>
  </div>
  <div class="corollary_thmcontent">
  <p>If the following hypotheses are fulfilled: </p>
<ul class="itemize">
  <li><p>\(\left( X,\left\Vert \cdot \right\Vert _{X}\right) \) is a Banach space; </p>
</li>
  <li><p>the mapping \(f:D\rightarrow Y\) verifies the hypothesis <b class="bfseries">ii)</b> of the Theorem <span class="rm"><a href="#t2">6</a></span> with a constant \(L{\gt}0\) and there exists a number \(\lambda {\gt}0\) such that for any \(x,y\in D\) the following inequality is true:</p>
<div class="displaymath" id="a0000000079">
  \[  \left\Vert \Gamma _{f;x,y}\right\Vert \leq \lambda ;  \]
</div>
</li>
  <li><p>the sequence \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\subseteq D\) verifies for any \(n\in \mathbb {N}^{\ast }\) the equality <span class="rm">(<a href="#f80">41</a>)</span> with a mapping \(U\in \left( X,Y\right) ^{\ast }\diagdown \left\{  \Theta \right\}  ;\) </p>
</li>
  <li><p>the mapping \(\Gamma _{f;x_{0},x_{0}-Uf\left( x_{0}\right) }\in \left( X,Y\right) ^{\ast }\) is invertible and \(\Gamma _{f;x_{0},x_{0}-Uf\left( x_{0}\right)\! \! \! \! \!  }^{-1}\in \) \(\in \left( Y,X\right) ^{\ast };\) </p>
</li>
  <li><p>denoting:</p>
<div class="displaymath" id="a0000000080">
  \begin{align*}  B_{0}& =\max \left\{  \left\Vert U\right\Vert ,\big\Vert \Gamma _{f;x_{0},x_{0}-Uf\left( x_{0}\right) }^{-1}\big\Vert \right\}  ,\\ R_{0}& =\left\Vert f\left( x_{0}\right) \right\Vert _{Y},\medskip \\ d& =\tfrac {LB_{0}^{2}R_{0}\left( 1+\lambda \left\Vert U\right\Vert \right) ^{2}}{\alpha },\\ \delta & =2\left\Vert U\right\Vert R_{0}+\tfrac {\alpha }{LB_{0}}\cdot \left( \alpha +\tfrac {2}{1+\lambda \left\Vert U\right\Vert }\right) \cdot \tfrac {d}{1-d}\end{align*}
</div>
<p> with \(\alpha =\tfrac {\sqrt{17}-3}{4},\) the relations \(d{\lt}1\) and \(S\left( x_{0},\delta \right) \subseteq D\) are true. </p>
<p>Then we have the following conclusions: </p>
</li>
  <li><p>for any \(n\in \mathbb {N}^{\ast }\) we have that \(x_{n},x_{n}-Uf\left( x_{n}\right) \in S\left( x_{0},\delta \right) ,\) there exists \(\Gamma _{f;x_{n},x_{n}-Uf\left( x_{n}\right) }^{-1}\in \left( Y,X\right) ^{\ast }\) and the equality <span class="rm">(<a href="#f81">42</a>)</span> is true; </p>
</li>
  <li><p>the sequences \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( x_{n}-Uf\left( x_{n}\right) \right) _{n\in \mathbb {N}^{\ast }}\) are convergent to the same limit \(\overline{x}\in S\left( x_{0},\delta \right) \) for which \(f\left( \overline{x}\right) =\theta _{Y};\) </p>
</li>
  <li><p>for any \(n\in \mathbb {N}^{\ast }\) the following inequalities are true:</p>
<div class="equation" id="C1">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert x_{n+1}-x_{n}\right\Vert _{X}\leq \tfrac {\alpha }{LB_{0}}\left( \tfrac {a}{B_{0}}+\tfrac {\alpha }{1+\lambda \left\Vert U\right\Vert }\right) d^{2^{n}}, \label{C1}\end{equation}
  </div>
  <span class="equation_label">43</span>
</p>
</div>
<div class="equation" id="C2">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert x_{n}-\overline{x}\right\Vert _{X}\leq \tfrac {\alpha }{LB_{0}}\left( \tfrac {a}{B_{0}}+\tfrac {\alpha }{1+\lambda \left\Vert U\right\Vert }\right) \tfrac {d^{2^{n}}}{1-d^{2^{n}}}. \label{C2}\end{equation}
  </div>
  <span class="equation_label">44</span>
</p>
</div>
</li>
</ul>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000081">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>For any \(n\in \mathbb {N}^{\ast }\) we will note \(y_{n}=x_{n}-Uf\left( x_{n}\right) ,\) \(z_{n}=x_{n}.\) </p>
<p>For the verification of the hypothesis <b class="bfseries">iii)</b> of the Theorem <a href="#t2">6</a> we have the following relations:</p>
<div class="displaymath" id="a0000000082">
  \begin{align*}  \left\Vert y_{n}-x_{n}\right\Vert _{X}\leq & \left\Vert U\right\Vert \cdot \left\Vert f\left( y_{n}\right) \right\Vert _{Y}, \\ \left\Vert f\left( y_{n}\right) \right\Vert _{Y}\leq & \left\Vert f\left( x_{n}\right) \right\Vert _{Y}+\left\Vert f\left( y_{n}\right) -f\left( x_{n}\right) \right\Vert _{Y} \\ \leq & \left\Vert f\left( x_{n}\right) \right\Vert _{Y}+\left\Vert \Gamma _{f;x_{n},y_{n}}\right\Vert \cdot \left\Vert y_{n}-x_{n}\right\Vert \\ \leq & \left( 1+\lambda \left\Vert U\right\Vert \right) \left\Vert f\left( x_{n}\right) \right\Vert _{Y}, \end{align*}
</div>
<p> therefore the inequalities (<a href="#f53">8</a>) will be verified with \(a=\left\Vert U\right\Vert ,\) \(K=1+\lambda \left\Vert U\right\Vert ,\) if we take into account the conditions on the sequence \(\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}=\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\) as well. </p>
<p>Obviously, from \(p=q=1\) we deduce that \(\overline{K}=K\) and the others constants from the statement of Theorem <a href="#t2">6</a>, have the values from the present corollary. </p>
<p>By the application in this case of Theorem <a href="#t2">6</a> we can deduce the conclusions j)-jjj) from the previous statement. </p>
<p>An other special case is the case in which we obtain the sequences \(\left( y_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}\subseteq D\) with the help of the iterative operators \(Q_{1},Q_{2}:X\rightarrow X\) that verify \(Q_{i}\left( D\right) \subseteq D\) for any \(i\in \left\{  1,2\right\}  .\) In this case we can choose \(y_{n}=Q_{1}\left( x_{n}\right) \) and \(z_{n}=Q_{2}\left( x_{n}\right) ,\) obtaining the sequence \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\) by the verification for any \(n\in \mathbb {N}^{\ast }\) of the equality:</p>
<div class="equation" id="f82">
<p>
  <div class="equation_content">
    \begin{equation}  \Gamma _{f;Q_{1}\left( x_{n}\right) ,Q_{2}\left( x_{n}\right) }\left( x_{n+1}-Q_{1}\left( x_{n}\right) \right) +f\left( Q_{1}\left( x_{n}\right) \right) =\theta _{Y}. \label{f82}\end{equation}
  </div>
  <span class="equation_label">45</span>
</p>
</div>
<p>In this equality the roles of the operators \(Q_{1}\) and \(Q_{2}\) can be inverted. </p>
<p>If for any \(n\in \mathbb {N}^{\ast }\) there exists the mapping \(\Gamma _{f;Q_{1}\left( x_{n}\right) ,Q_{2}\left( x_{n}\right) }^{-1}\in \left( Y,X\right) ^{\ast }\) the relation (<a href="#f82">45</a>) is equivalent to:</p>
<div class="equation" id="f83">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=Q_{1}\left( x_{n}\right) -\Gamma _{f;Q_{1}\left( x_{n}\right) ,Q_{2}\left( x_{n}\right) }^{-1}f\left( Q_{1}\left( x_{n}\right) \right) . \label{f83}\end{equation}
  </div>
  <span class="equation_label">46</span>
</p>
</div>
<p>If for any \(x\in D\) and \(i\in \left\{  1,2\right\}  \) we have the following relations:</p>
<div class="equation" id="f84">
<p>
  <div class="equation_content">
    \begin{equation}  \left\{  \begin{array}[c]{c}\left\Vert f\left( Q_{i}\left( x\right) \right) \right\Vert _{Y}\leq K\left\Vert f\left( x\right) \right\Vert _{Y}^{p_{i}},\medskip \\ \left\Vert Q_{i}\left( x\right) -x\right\Vert _{X}\leq a\left\Vert f\left( x\right) \right\Vert _{Y}, \end{array} \right. \label{f84}\end{equation}
  </div>
  <span class="equation_label">47</span>
</p>
</div>
<p> for the sequences that are chosen in the manner showed the hypotheses of the Theorem <a href="#t2">6</a> are fulfilled, therefore we have the conclusions of this theorem. One obtains the convergence order \(p_{1}+p_{2}.\) </p>
<p>The convergence order can be increased if we replace the mapping \(Q_{2}\) with the mapping \(Q_{2}\circ Q_{1}:X\rightarrow X.\) </p>
<p>In this case:</p>
<div class="displaymath" id="a0000000083">
  \begin{align*}  \left\Vert f\left( Q_{2}\circ Q_{1}\right) \left( x\right) \right\Vert _{Y}& =\left\Vert f\left( Q_{2}\left( Q_{1}\left( x\right) \right) \right) \right\Vert _{Y}\\ & \leq K\left\Vert f\left( Q_{1}\left( x\right) \right) \right\Vert _{Y}^{p_{2}}\\ & \leq K^{1+p_{2}}\left\Vert f\left( x\right) \right\Vert _{Y}^{p_{1}+p_{2}},\\ \left\Vert \left( Q_{2}\circ Q_{1}\right) \left( x\right) -x\right\Vert _{X}& \leq \left\Vert Q_{2}\left( Q_{1}\left( x\right) \right) -Q_{1}\left( x\right) \right\Vert _{X}+\left\Vert Q_{1}\left( x\right) -x\right\Vert _{X}\\ & \leq a\left\Vert f\left( Q_{1}\left( x\right) \right) \right\Vert _{Y}+a\left\Vert f\left( x\right) \right\Vert _{Y}\\ & \leq a\left( K\left\Vert f\left( x\right) \right\Vert _{Y}^{p}+\left\Vert f\left( x\right) \right\Vert _{Y}\right). \end{align*}
</div>
<p>Usually we are interested in the set of points from around the solutions, for which \(\left\Vert f\left( x\right) \right\Vert _{Y}\leq 1,\) therefore as \(p\geq 1\) we have \(\left\Vert f\left( x\right) \right\Vert _{Y}^{p}\leq \left\Vert f\left( x\right) \right\Vert _{Y}\) and so:</p>
<div class="displaymath" id="a0000000084">
  \[  \left\Vert \left( Q_{2}\circ Q_{1}\right) \left( x\right) -x\right\Vert _{X}\leq a\left( 1+K\right) \left\Vert f\left( x\right) \right\Vert _{Y}.  \]
</div>
<p>Therefore using the sequences \(\left( y_{n}\right) _{n\in \mathbb {N}^{\ast }},\) \(\left( z_{n}\right) _{n\in \mathbb {N}^{\ast }}\) that are defined by \(y_{n}=Q_{1}\left( x_{n}\right) ,\) \(z_{n}=\left( Q_{2}\circ Q_{1}\right) \left( x_{n}\right) \) for the main sequence \(\left( x_{n}\right) _{n\in \mathbb {N}^{\ast }}\) we obtain the convergence order \(p_{1}+p_{1}p_{2}.\) </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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</dd>
</dl>


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