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<title>Semi-local convergence of iterative methods and Banach space valued functions in abstract fractional calculus: Semi-local convergence of iterative methods and Banach space valued functions in abstract fractional calculus</title>
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<h1>Semi-local convergence of iterative methods and Banach space valued functions in abstract fractional calculus</h1>
<p class="authors">
<span class="author">George A. Anastassiou\(^\ast \) Ioannis K. Argyros\(^\bullet \)</span>
</p>
<p class="date">May 1st, 2017. Accepted: February 5, 2018. Published online: August 6, 2018.</p>
</div>
<p>\(^\ast \)Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A., e-mail: <span class="ttfamily">ganastss@memphis.edu</span>. </p>
<p>\(^\bullet \)Department of Mathematics Sciences, Cameron University, Lawton, OK 73505, USA, e-mail: <span class="ttfamily">iargyros@cameron.edu</span>. </p>

<div class="abstract"><p> We present a semi-local convergence analysis for a class of iterative methods under generalized conditions. Some applications are suggested including Banach space valued functions of fractional calculus, where all integrals are of Bochner-type. </p>
<p><b class="bf">MSC.</b> 65G99, 65H10, 26A33, 46B25, 47J25, 47J05. </p>
<p><b class="bf">Keywords.</b> iterative method, Banach space, semi-local convergence, fractional calculus, Bochner-type integral. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>Let \(B_{1},B_{2}\) stand for Banach space and let \(\Omega \) stand for an open subset of \(B_{1}\). Let also \(U\left( z,\rho \right) :=\left\{  u\in B_{1}:\left\|  u-z\right\|  {\lt}\rho \right\}  \) and let \(\overline{U}\left( z,\rho \right) \) stand for the closure of \(U\left( z,\rho \right) \). </p>
<p>Many problems in Computational Sciences, Engineering, Mathematical Chemistry, Mathematical Physics, Mathematical Economics and other disciplines can be brought in a form like </p>
<div class="equation" id="1.1">
<p>
  <div class="equation_content">
    \begin{equation}  F\left( x\right) =0 \tag {1.1} \label{1.1} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p>using Mathematical Modeling <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>–<span class="cite">
	[
	<a href="#17" >17</a>
	]
</span>, where \(F:\Omega \rightarrow B_{2}\) is a continuous operator. The solution \(x^{\ast }\) of equation (<a href="#1.1">1</a>) is sought in closed form, but this is attainable only in special cases. That explains why most solution methods for such equations are usually iterative. There is a plethora of iterative methods for solving equation (<a href="#1.1">1</a>). We can divide these methods in two categories. </p>
<p><i class="itshape">Explicit Methods</i> <span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>, <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, <span class="cite">
	[
	<a href="#15" >15</a>
	]
</span>, <span class="cite">
	[
	<a href="#16" >16</a>
	]
</span>: Newton’s method </p>
<div class="equation" id="1.2">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-F^{\prime }\left( x_{n}\right) ^{-1}F\left( x_{n}\right) . \tag {1.2} \label{1.2} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p>Secant method: </p>
<div class="equation" id="1.3">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-\left[ x_{n-1},x_{n};F\right] ^{-1}F\left( x_{n}\right) , \tag {1.3} \label{1.3} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p>where \(\left[ \cdot ,\cdot ;F\right] \) denotes a divided difference of order one on \(\Omega \times \Omega \) <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#15" >15</a>
	]
</span>, <span class="cite">
	[
	<a href="#16" >16</a>
	]
</span>. </p>
<p>Newton-like method: </p>
<div class="equation" id="1.4">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-E_{n}^{-1}F\left( x_{n}\right) , \tag {1.4} \label{1.4} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p>where \(E_{n}=E\left( F\right) \left( x_{n}\right) \) and \(E:\Omega \rightarrow \mathcal{L}\left( B_{1},B_{2}\right) \) the space of bounded linear operators from \(B_{1}\) into \(B_{2}\). Other explicit methods can be found in <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, <span class="cite">
	[
	<a href="#15" >15</a>
	]
</span>, <span class="cite">
	[
	<a href="#16" >16</a>
	]
</span> and the references there in. </p>
<p>Implicit Methods <span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>, <span class="cite">
	[
	<a href="#9" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, <span class="cite">
	[
	<a href="#16" >16</a>
	]
</span>: </p>
<div class="equation" id="1.5">
<p>
  <div class="equation_content">
    \begin{equation}  F\left( x_{n}\right) +A_{n}\left( x_{n+1}-x_{n}\right) =0 \tag {1.5} \label{1.5} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<div class="equation" id="1.6">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-A_{n}^{-1}F\left( x_{n}\right) , \tag {1.6} \label{1.6} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>where \(A_{n}=A\left( x_{n+1},x_{n}\right) =A\left( F\right) \left( x_{n+1},x_{n}\right) \) and \(A:\Omega \times \Omega \rightarrow \mathcal{L}\left( B_{1},B_{2}\right) .\) </p>
<p>There is a plethora on local as well as semi-local convergence results for explicit methods <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>–<span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#10" >10</a>
	]
</span>–<span class="cite">
	[
	<a href="#16" >16</a>
	]
</span>. However, the research on the convergence of implicit methods has received little attention. Authors, usually consider the fixed point problem </p>
<div class="equation" id="1.7">
<p>
  <div class="equation_content">
    \begin{equation}  P_{z}\left( x\right) =x, \tag {1.7} \label{1.7} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p>where </p>
<div class="equation" id="1.8">
<p>
  <div class="equation_content">
    \begin{equation}  P_{z}\left( x\right) =x+F\left( z\right) +A\left( x,z\right) \left( x-z\right) \tag {1.8} \label{1.8} \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>or </p>
<div class="equation" id="1.9">
<p>
  <div class="equation_content">
    \begin{equation}  P_{z}\left( x\right) =z-A\left( x,z\right) ^{-1}F\left( z\right) \tag {1.9} \label{1.9} \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>for methods (<a href="#1.5">5</a>) and (<a href="#1.6">6</a>), respectivelly, where \(z\in \Omega \) is given. If \(P\) is a contraction operator mapping a closed set into itself, then according to the contraction mapping principle <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, <span class="cite">
	[
	<a href="#15" >15</a>
	]
</span>, <span class="cite">
	[
	<a href="#16" >16</a>
	]
</span>, \(P_{z}\) has a fixed point \(x_{z}^{\ast }\) which can be found using the method of succesive substitutions or Picard’s method <span class="cite">
	[
	<a href="#16" >16</a>
	]
</span> defined for each fixed \(n\) by </p>
<div class="equation" id="1.10">
<p>
  <div class="equation_content">
    \begin{equation}  y_{k+1,n}=P_{x_{n}}\left( y_{k,n}\right) ,\text{ \  \  }y_{0,n}=x_{n}\text{, \   }x_{n+1}=\underset {k\rightarrow +\infty }{\lim }y_{k,n}. \tag {1.10} \label{1.10} \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p>Let us also consider the analogous explicit methods </p>
<div class="equation" id="1.11">
<p>
  <div class="equation_content">
    \begin{equation}  F\left( x_{n}\right) +A\left( x_{n},x_{n}\right) \left( x_{n+1}-x_{n}\right) =0 \tag {1.11} \label{1.11} \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<div class="equation" id="1.12">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-A\left( x_{n},x_{n}\right) ^{-1}F\left( x_{n}\right) \tag {1.12} \label{1.12} \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<div class="equation" id="1.13">
<p>
  <div class="equation_content">
    \begin{equation}  F\left( x_{n}\right) +A\left( x_{n},x_{n-1}\right) \left( x_{n+1}-x_{n}\right) =0 \tag {1.13} \label{1.13} \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p>and </p>
<div class="equation" id="1.14">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-A\left( x_{n},x_{n-1}\right) ^{-1}F\left( x_{n}\right) . \tag {1.14} \label{1.14} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p>In the present paper in Section 2, we present the semi-local convergence of method (<a href="#1.5">5</a>) and method (<a href="#1.6">6</a>). Section 3 contains the semi-local convergence of method (<a href="#1.11">11</a>), method (<a href="#1.12">12</a>), method (<a href="#1.13">13</a>) and method (<a href="#1.14">14</a>). Some applications to Abstract Fractional Calculus are suggested in Section 4 on a certain Banach space valued functions, where all the integrals are of Bochner-type <span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>. </p>
<h1 id="a0000000003">2 Semi-local Convergence for Implicit methods</h1>
<p>We present the semi-local convergence analysis of method (<a href="#1.6">6</a>) using conditions \(\left( S\right) \): </p>
<p>\((s_{1})\)  \(F:\Omega \subset B_{1}\rightarrow B_{2}\) is continuous and \(A\left( x,y\right) \in \mathcal{L}\left( B_{1},B_{2}\right) \) for each \(\left( x,y\right) \in \Omega \times \Omega .\) </p>
<p>\((s_{2})\) There exist \(\beta {\gt}0\) and \(\Omega _{0}\subset B_{1}\) such that \(A\left( x,y\right) ^{-1}\in \mathcal{L}\left( B_{2},B_{1}\right) \) for each \(\left( x,y\right) \in \Omega _{0}\times \Omega _{0}\  \)and </p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  \|  A\left( x,y\right) ^{-1}\|  \leq \beta ^{-1}. \end{equation*}
</div>
<p>Set \(\Omega _{1}=\Omega \cap \Omega _{0}\). </p>
<p>\((s_{3})\) There exists a continuous and nondecreasing function \(\psi :[0,+\infty )^{3}\rightarrow \lbrack 0,+\infty )\) such that for each \(x,y\in \Omega _{1}\) </p>
<div class="displaymath" id="a0000000005">
  \begin{align*} & \left\|  F\left( x\right) -F\left( y\right) -A\left( x,y\right) \left( x-y\right) \right\| \leq \\ \nonumber & \leq \beta \psi \left( \left\|  x-y\right\|  ,\left\|  x-x_{0}\right\|  ,\left\|  y-x_{0}\right\|  \right) \left\|  x-y\right\|  . \end{align*}
</div>
<p>\((s_{4})\) For each \(x\in \Omega _{0}\) there exists \(y\in \Omega _{0}\) such that </p>
<div class="displaymath" id="a0000000006">
  \begin{equation*}  y=x-A\left( y,x\right) ^{-1}F\left( x\right) . \end{equation*}
</div>
<p>\((s_{5})\) For \(x_{0}\in \Omega _{0}\) and \(x_{1}\in \Omega _{0}\) satisfying \((s_{4})\) there exists \(\eta \geq 0\) such that </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*}  \left\|  A\left( x_{1},x_{0}\right) ^{-1}F\left( x_{0}\right) \right\|  \leq \eta . \end{equation*}
</div>
<p>\((s_{6})\) Define \(q\left( t\right) :=\psi \left( \eta ,t,t\right) \) for each \(t\in \lbrack 0,+\infty )\). Equation </p>
<div class="displaymath" id="a0000000008">
  \begin{equation*}  t\left( 1-q\left( t\right) \right) -\eta =0 \end{equation*}
</div>
<p>has positive solutions. Denote by \(s\) the smallest such solution. </p>
<p>\((s_{7})\)  \(\overline{U}\left( x_{0},s^{{}}\right) \subset \Omega \), where </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  s^{{}}=\tfrac {\eta }{1-q_{0}^{{}}}\text{ \  and \  }q_{0}^{{}}=\psi \left( \eta ,s,s\right) . \end{equation*}
</div>
<p>Next, we present the semi-local convergence analysis for method (<a href="#1.6">6</a>) using the conditions \(\left( S\right) \) and the preceding notation. </p>
<p><div class="theorem_thmwrapper " id="t2.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.1</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Assume that the conditions \((S)\) hold. Then, sequence \(\left\{  x_{n}\right\}  \) generated by method <a href="#1.6" class="eqref">6</a> starting at \(x_{0}\in \Omega \) is well defined in \(U\left( x_{0},s^{{}}\right) \), remains in \(U\left( x_{0},s^{{}}\right) \) for each \(n=0,1,2,...\) and converges to a solution \(x^{\ast }\in \overline{U}\left( x_{0},s^{{}}\right) \) of equation \(F\left( x\right) =0\). Moreover, suppose that there exists a continuous and nondecreasing function \( \psi _{1}:[0,+\infty )^{4}\rightarrow \lbrack 0,+\infty )\) such that for each \(x,y,z\in \Omega _{1}\)</p>
<div class="displaymath" id="a0000000010">
  \begin{align*} & \left\Vert F\left( x\right) -F\left( y\right) -A\left( z,y\right) \left( x-y\right) \right\Vert \leq \\ & \leq \beta \psi _{1}\left( \left\Vert x-y\right\Vert ,\left\Vert x-x_{0}\right\Vert ,\left\Vert y-x_{0}\right\Vert ,\left\Vert z-x_{0}\right\Vert \right) \left\Vert x-y\right\Vert \end{align*}
</div>
<p>and \(q_{1}=\psi _{1}\left( \eta ,s^{{}},s^{{}},s^{{}}\right) {\lt}1.\) </p>
<p>Then, \(x^{\ast }\) is the unique solution of equation \(F\left( x\right) =0\) in \(\overline{U}\left( x_{0},s^{{}}\right) .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000011">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By the definition of \(s^{{}}\) and \(\left( s_{5}\right) \), we have \(x_{1}\in U\left( x_{0},s^{{}}\right) \). The proof is based on mathematical induction on \(k\). Suppose that \(\left\Vert x_{k}-x_{k-1}\right\Vert \leq q_{0}^{k-1}\eta \) and \(\left\Vert x_{k}-x_{0}\right\Vert \leq s.\) </p>
<p>We get by (<a href="#1.6">6</a>), \(\left( s_{2}\right)\)–\(\left( s_{5}\right) \) in turn that </p>
<div class="displaymath" id="a0000000012">
  \begin{align}  \big\Vert x_{k+1}-x_{k}\big\Vert & =\big\Vert A_{k}^{-1}F\left( x_{k}\right) \big\Vert \nonumber \\ & =\big\Vert A_{k}^{-1}\left( F\left( x_{k}\right) -F\left( x_{k-1}\right) -A_{k-1}\left( x_{k}-x_{k-1}\right) \right) \big\Vert \nonumber \\ & \leq \big\Vert A_{k}^{-1}\big\Vert \left\Vert F\left( x_{k}\right) -F\left( x_{k-1}\right) -A_{k-1}\left( x_{k}-x_{k-1}\right) \right\Vert \nonumber \\ & \leq \beta ^{-1}\beta \psi \left( \left\Vert x_{k}-x_{k-1}\right\Vert ,\left\Vert x_{k-1}-x_{0}\right\Vert ,\left\Vert y_{k}-x_{0}\right\Vert \right) \left\Vert x_{k}-x_{k-1}\right\Vert \nonumber \\ & \leq \psi \left( \eta ,s,s\right) \left\Vert x_{k}-x_{k-1}\right\Vert =q_{0}\left\Vert x_{k}-x_{k-1}\right\Vert \nonumber \leq \\ &  \leq q_{0}^{k}\left\Vert x_{1}-x_{0}\right\Vert \leq q_{0}^{k}\eta \tag {2.1} \label{2.1} \end{align}
</div>
<p>and </p>
<div class="displaymath" id="a0000000013">
  \begin{align*}  \left\Vert x_{k+1}-x_{0}\right\Vert & \leq \left\Vert x_{k+1}-x_{k}\right\Vert +...+\left\Vert x_{1}-x_{0}\right\Vert \\ & \leq q_{0}^{k}\eta +...+\eta =\tfrac {1-q_{0}^{k+1}}{1-q_{0}}\eta {\lt}\tfrac {\eta }{1-q_{0}}=s. \end{align*}
</div>
<p>The induction is completed. Moreover, we have by (<a href="#2.1">14</a>) that for \(m=0,1,2,...\) </p>
<div class="displaymath" id="a0000000014">
  \begin{equation*}  \left\Vert x_{k+m}-x_{k}\right\Vert \leq \tfrac {1-q_{0}^{m}}{1-q_{0}}q_{0}^{k}\eta . \end{equation*}
</div>
<p>It follows from the preceding inequation that sequence \(\left\{  x_{k}\right\}  \) is complete in a Banach space \(B_{1}\) and as such it converges to some \(x^{\ast }\in \overline{U}\left( x_{0},s\right) \) (since \(\overline{U}\left( x_{0},s\right) \) is a closed ball). By letting \(k\rightarrow +\infty \) in (<a href="#2.1">14</a>) we get \(F\left( x^{\ast }\right) =0\). To show the uniqueness part, let \(x^{\ast \ast }\in U\left( x_{0},s\right) \) be a solution of equation \(F\left( x\right) =0.\  \)By using (<a href="#1.6">6</a>) and the hypothesis on \(\psi _{1}\), we obtain in turn that </p>
<div class="displaymath" id="a0000000015">
  \begin{align*} & \left\Vert x^{\ast \ast }-x_{k+1}\right\Vert =\\ \nonumber & =\big\Vert x^{\ast \ast }-x_{k}+A_{k}^{-1}F\left( x_{k}\right) -A_{k}^{-1}F\left( x^{\ast \ast }\right) \big\Vert \\ \nonumber & \leq \big\Vert A_{k}^{-1}\big\Vert \left\Vert F\left( x^{\ast \ast }\right) -F\left( x_{k}\right) -A_{k}\left( x^{\ast \ast }-x_{k}\right) \right\Vert \\ \nonumber & \leq \beta ^{-1}\beta \psi _{1}\left( \left\Vert x^{\ast \ast }-x_{k}\right\Vert ,\left\Vert x_{k-1}-x_{0}\right\Vert ,\left\Vert x_{k}-x_{0}\right\Vert ,\left\Vert x^{\ast \ast }-x_{0}\right\Vert \right) \left\Vert x^{\ast \ast }-x_{k}\right\Vert \\ \nonumber & \leq q_{1}\left\Vert x^{\ast \ast }-x_{k}\right\Vert \leq q_{1}^{k+1}\left\Vert x^{\ast \ast }-x_{0}\right\Vert , \end{align*}
</div>
<p>so \(\underset {k\rightarrow +\infty }{\lim }x_{k}=x^{\ast \ast }\). We have shown that \(\underset {k\rightarrow +\infty }{\lim }x_{k}=x^{\ast }\), so \(x^{\ast }=x^{\ast \ast }.\) <div class="proof_wrapper" id="a0000000016">
  <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><div class="remark_thmwrapper " id="r2.2">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2.2</span>
  </div>
  <div class="remark_thmcontent">
  <p>(1) The equation in \(\left( s_{6}\right) \) is used to determine the smallness of \(\eta \). It can be replaced by a stronger condition as follows. Choose \(\mu \in \left( 0,1\right) \). Denote by \(s_{0}\) the smallest positive solution of equation \(q\left( t\right) =\mu \). Notice that if function \(q\) is strictly increasing, we can set \(s_{0}=q^{-1}\left( \mu \right) \). Then, we can suppose instead of \(\left( s_{6}\right) :\) </p>
<p>\(\left( s_{6}^{\prime }\right) \)  \(\eta \leq \left( 1-\mu \right) s_{0}\) </p>
<p>which is a stronger condition than \(\left( s_{6}\right) \). </p>
<p>However, we wanted to leave the equation in \(\left( s_{6}\right) \) as uncluttered and as weak as possible. </p>
<p>(2) Condition \(\left( s_{2}\right) \) can become part of condition \(\left( s_{3}\right) \) by considering </p>
<p>\(\left( s_{3}\right) ^{\prime }\) There exists a continuous and nondecreasing function \(\varphi :[0,+\infty )^{3}\rightarrow \lbrack 0,+\infty )\) such that for each \(x,y\in \Omega _{1}\) </p>
<div class="displaymath" id="a0000000017">
  \begin{align*} & \big\Vert A\left( x,y\right) ^{-1}\left[ F\left( x\right) -F\left( y\right) -A\left( x,y\right) \left( x,y\right) \right] \big\Vert \leq \\ & \leq \varphi \left( \left\Vert x-y\right\Vert ,\left\Vert x-x_{0}\right\Vert ,\left\Vert y-x_{0}\right\Vert \right) \left\Vert x-y\right\Vert . \end{align*}
</div>
<p>Notice that </p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  \varphi \left( u_{1},u_{2},u_{3}\right) \leq \psi \left( u_{1},u_{2},u_{3}\right) \end{equation*}
</div>
<p>for each \(u_{1}\geq 0\), \(u_{2}\geq 0\) and \(u_{3}\geq 0\). Similarly, a function \(\varphi _{1}\) can replace \(\psi _{1}\) for the uniqueness of the solution part. These replacements are of Mysovskii-type <span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>, <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, <span class="cite">
	[
	<a href="#15" >15</a>
	]
</span> and influence the weaking of the convergence criterion in \(\left( s_{6}\right) \), error bounds and the precision of \(s\). </p>
<p>(3) Suppose that there exist \(\beta {\gt}0\), \(\beta _{1}{\gt}0\) and \(L\in \mathcal{L}\left( B_{1},B_{2}\right) \) with \(L^{-1}\in \mathcal{L}\left( B_{2},B_{1}\right) \) such that </p>
<div class="displaymath" id="a0000000019">
  \begin{align*}  \|  L^{-1}\|  & \leq \beta ^{-1}\\ \left\|  A\left( x,y\right) -L\right\| &  \leq \beta _{1} \end{align*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  \beta _{2}:=\beta ^{-1}\beta _{1}{\lt}1. \end{equation*}
</div>
<p>Then, it follows from the Banach lemma on invertible operators <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, and </p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  \|  L^{-1}\|  \left\|  A\left( x,y\right) -L\right\|  \leq \beta ^{-1}\beta _{1}=\beta _{2}{\lt}1 \end{equation*}
</div>
<p>that \(A\left( x,y\right) ^{-1}\in \mathcal{L}\left( B_{2},B_{1}\right) \). Let \(\beta =\tfrac {\beta ^{-1}}{1-\beta _{2}}\). Then, under these replacements, condition \((s_{2})\) is implied, therefore it can be dropped from the conditions \(\left( S\right) \). </p>
<p>(4) Clearly method (<a href="#1.5">5</a>) converges under the conditions \(\left( S\right) \), since (<a href="#1.6">6</a>) implies (<a href="#1.5">5</a>). </p>
<p>(5) We wanted to leave condition \(\left( s_{4}\right) \) as uncluttered as possible, since in practice equations (<a href="#1.6">6</a>) (or (<a href="#1.5">5</a>)) may be solvable in a way avoiding the already mentioned conditions of the contraction mapping principle. However, in what follows we examine the solvability of method (<a href="#1.5">5</a>) under a stronger version of the contraction mapping principle using the conditions \(\left( V\right) :\) </p>
<p>\(\left( v_{1}\right) =\left( s_{1}\right) .\) </p>
<p>\(\left( v_{2}\right) \) There exist functions \(w_{1}:[0,+\infty )^{4}\rightarrow \lbrack 0,+\infty )\), \(w_{2}:[0,+\infty )^{4}\rightarrow \lbrack 0,+\infty )\) continuous and nondecreasing such that for each \(x,y,z\in \Omega \) </p>
<div class="displaymath" id="a0000000022">
  \begin{align*}  \left\|  I+A\left( x,z\right) -A\left( y,z\right) \right\|  &  \leq w_{1}\left( \left\|  x-y\right\|  ,\left\|  x-x_{0}\right\|  ,\left\|  y-x_{0}\right\|  ,\left\|  z-x_{0}\right\|  \right) \\ \left\|  A\left( x,z\right) -A\left( y,z\right) \right\|  &  \leq w_{2}\left( \left\|  x-y\right\|  ,\left\|  x-x_{0}\right\|  ,\left\|  y-x_{0}\right\|  ,\left\|  z-x_{0}\right\|  \right) ||x-y|| \end{align*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000023">
  \begin{equation*}  w_{1}\left( 0,0,0,0\right) =w_{2}\left( 0,0,0,0\right) =0. \end{equation*}
</div>
<p>Set </p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  h\left( t,t,t,t\right) = \begin{cases}  w_{1}\left( 2t,t,t,t\right) +w_{2}\left( 2t,t,t,t\right) \left( t+\left\|  x_{0}\right\|  \right) \text{, \  }& z\neq x_{0} \\[2mm] w_{1}\left( 2t,t,t,0\right) +w_{2}\left( 2t,t,t,0\right) \left\|  x_{0}\right\|  \text{, \  }&  z=x_{0}.\end{cases}\end{equation*}
</div>
<p>\(\left( v_{3}\right) \) There exists \(\tau {\gt}0\) satisfying  </p>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  h\left( t,t,t,t\right) {\lt}1 \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  h\left( t,t,0,t\right) t+\left\|  F\left( x_{0}\right) \right\|  \leq t \end{equation*}
</div>
<p>\(\left( v_{4}\right) \)  \(\overline{U}\left( x_{0},\tau \right) \subseteq D\). <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="t2.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.3</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Suppose that the conditions \(\left( V\right) \) are satisfied. Then, equation <a href="#1.5" class="eqref">5</a> is uniquely solvable for each \(n=0,1,2,...\). Moreover, if \(A_{n}^{-1}\in \mathcal{L}\left( B_{2},B_{1}\right) \), the equation <a href="#1.6" class="eqref">6</a> is also uniquely solvable for each \(n=0,1,2,...\) </p>

  </div>
</div> </p>
<p><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> The result is an application of the contraction mapping principle. Let \(x,y,z\in U\left( x_{0},\tau \right) \). By the definition of operator \(P_{z}\), \(\left( v_{2}\right) \) and \(\left( v_{3}\right) \), we get in turn that </p>
<div class="displaymath" id="a0000000028">
  \begin{align*} & \left\|  P_{z}\left( x\right) -P_{z}\left( y\right) \right\|  =\\ \nonumber & =\left\|  \left( I+A\left( x,z\right) -A\left( y,z\right) \right) \left( x-y\right) -\left( A\left( x,z\right) -A\left( y,z\right) \right) z\right\|  \\ \nonumber & \leq \left\|  I+A\left( x,z\right) -A\left( y,z\right) \right\|  \|  x-y\|  +\left\|  A\left( x,z\right) -A\left( y,z\right) \right\|  \|  z\|  \\ \nonumber & \leq \Big[ w_{1}(\big\|  x-y\|  ,\|  x-x_{0}\|  ,\|  y-x_{0}\|  ,\|  z-x_{0}\|  \big) +\\ \nonumber & \qquad + w_{2}\big( \left\|  x-y\right\|  ,\left\|  x-x_{0}\right\|  ,\left\|  y-x_{0}\right\|  ,\left\|  z-x_{0}\right\|  \big) \left( \left\|  z-x_{0}\right\|  + \left\|  x_{0}\right\|  \right) \Big] \left\|  x-y\right\|  \\ \nonumber & \leq h\left( \tau ,\tau ,\tau ,\tau \right) \left\|  x-y\right\|  \end{align*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000029">
  \begin{align*}  \left\|  P_{z}\left( x\right) -x_{0}\right\| & \leq \left\|  P_{z}\left( x\right) -P_{z}\left( x_{0}\right) \right\|  +\left\|  P_{z}\left( x_{0}\right) -x_{0}\right\| \\ \nonumber & \leq h\left( \left\|  x-x_{0}\right\|  ,\left\|  x-x_{0}\right\|  ,0,\left\|  z-x_{0}\right\|  \right) \left\|  x-x_{0}\right\|  +\left\|  F\left( x_{0}\right) \right\| \\ \nonumber & \leq h\left( \tau ,\tau ,0,\tau \right) \tau +\left\|  F\left( x_{0}\right) \right\|  \leq \tau . \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000030">
  <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><div class="remark_thmwrapper " id="r2.4">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2.4</span>
  </div>
  <div class="remark_thmcontent">
  <p>Section 2 and Section 3 have an interest independent of Section 4. It is worth noticing that the results especially of Theorem <a href="#t2.1">2.1</a> can apply in Abstract Fractional Calculus as illustrated in Section 4. By specializing function \(\psi \), we can apply the results of say Theorem <a href="#t2.1">2.1</a> in the examples suggested in Section 4. In particular for (<a href="#4.1">-3</a>), we choose \(\psi \left( u_{1},u_{2},u_{3}\right) =\frac{c_{1}u_{1}^{p}}{\left( p+1\right) \beta }\) for \(u_{1}\geq 0\), \(u_{2}\geq 0,\) \(u_{3}\geq 0\) and \(c_{1}\), \(p\) are given in Section 4. Similar choices for the other examples of Section 4. It is also worth noticing that estimate (<a href="#4.2">-2</a>) derived in Section 4 is of independent interest but not needed in Theorem <a href="#t2.1">2.1</a>.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000031">3 Semi-local convergence for explicit methods</h1>
<p>A specialization of Theorem <a href="#t2.1">2.1</a> can be utilized to study the semi-local convergence of the explicit methods given in the introduction of this study. In particular, for the study of method (<a href="#1.12">12</a>) (and consequently of method (<a href="#1.11">11</a>)), we use the conditions \(\left( S^{\prime }\right) :\) </p>
<p>\(\left( s_{1}^{\prime }\right) \)  \(F:\Omega \subset B_{1}\rightarrow B_{2}\) is continuous and \(A\left( x,x\right) \in \mathcal{L}\left( B_{1},B_{2}\right) \) for each \(x\in \Omega .\) </p>
<p>\((s_{2}^{\prime })\) There exist \(\beta {\gt}0\) and \(\Omega _{0}\subset B_{1}\) such that \(A\left( x,x\right) ^{-1}\in \mathcal{L}\left( B_{2},B_{1}\right) \) for each \(x\in \Omega _{0}\) and </p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  \|  A\left( x,x\right) ^{-1}\|  \leq \beta ^{-1}. \end{equation*}
</div>
<p>Set \(\Omega _{1}=\Omega \cap \Omega _{0}\). </p>
<p>\((s_{3}^{\prime })\) There exist continuous and nondecreasing functions \(\psi _{0}:[0,+\infty )^{3}\rightarrow \lbrack 0,+\infty )\), \(\psi _{2}:[0,+\infty )^{3}\rightarrow \lbrack 0,+\infty )\) with \(\psi _{0}\left( 0,0,0\right) =\psi _{2}\left( 0,0,0\right) =0\) such that for each \(x,y\in \Omega _{1}\) </p>
<div class="displaymath" id="a0000000033">
  \begin{align*} & \left\|  F\left( x\right) -F\left( y\right) -A\left( y,y\right) \left( x-y\right) \right\| \leq \\ \nonumber & \leq \beta \psi _{0}\left( \left\|  x-y\right\|  ,\left\|  x-x_{0}\right\|  ,\left\|  y-x_{0}\right\|  \right) \left\|  x-y\right\|  \end{align*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  \left\|  A\left( x,y\right) -A\left( y,y\right) \right\|  \leq \beta \psi _{2}\left( \left\|  x-y\right\|  ,\left\|  x-x_{0}\right\|  ,\left\|  y-x_{0}\right\|  \right) . \end{equation*}
</div>
<p>Set \(\psi =\psi _{0}+\psi _{2}.\) </p>
<p>\((s_{4}^{\prime })\) There exist \(x_{0}\in \Omega _{0}\) and \(\eta \geq 0\) such that \(A\left( x_{0},x_{0}\right) ^{-1}\in \mathcal{L}\left( B_{2},B_{1}\right) \) and \(\  \)</p>
<div class="displaymath" id="a0000000035">
  \begin{equation*}  \|  A\left( x_{0},x_{0}\right) ^{-1}F\left( x_{0}\right)\|  \leq \eta . \end{equation*}
</div>
<p>\((s_{5}^{\prime })=\left( s_{6}\right) \) </p>
<p>\((s_{6}^{\prime })=\left( s_{7}\right) \). </p>
<p>Next, we present the following semi-local convergence analysis of method (<a href="#1.12">12</a>) using the \(\left( S^{\prime }\right) \) conditions and the preceding notation. </p>
<p><div class="proposition_thmwrapper " id="p3.1">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">3.1</span>
  </div>
  <div class="proposition_thmcontent">
  <p>Suppose that the conditions \((S^{\prime })\) are satisfied. Then, sequence \(\left\{  x_{n}\right\}  \) generated by method <a href="#1.12" class="eqref">12</a> starting at \(x_{0}\in \Omega \) is well defined in \(U\left( x_{0},s\right) \), remains in \(U\left( x_{0},s\right) \) for each \(n=0,1,2,\ldots \) and converges to a unique solution \(x^{\ast }\in \overline{U}\left( x_{0},s\right) \) of equation \(F\left( x\right) =0\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000036">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We follow the proof of Theorem <a href="#t2.1">2.1</a> but use instead the analogous estimate </p>
<div class="displaymath" id="a0000000037">
  \begin{align*}  \left\Vert F\left( x_{k}\right) \right\Vert & =\left\Vert F\left( x_{k}\right) -F\left( x_{k-1}\right) -A\left( x_{k-1},x_{k-1}\right) \left( x_{k}-x_{k-1}\right) \right\Vert \\ \nonumber & \leq \left\Vert F\left( x_{k}\right) -F\left( x_{k-1}\right) -A\left( x_{k},x_{k-1}\right) \left( x_{k}-x_{k-1}\right) \right\Vert \\ \nonumber & \quad +\left\Vert \left( A\left( x_{k},x_{k-1}\right) -A\left( x_{k-1},x_{k-1}\right) \right) \left( x_{k}-x_{k-1}\right) \right\Vert \\ \nonumber & \leq \left[ \psi _{0}\left( \left\Vert x_{k}-x_{k-1}\right\Vert ,\left\Vert x_{k-1}-x_{0}\right\Vert ,\left\Vert x_{k}-x_{0}\right\Vert \right)\right.\\ \nonumber & \quad +\left. \psi _{2}\left( \left\Vert x_{k}-x_{k-1}\right\Vert ,\left\Vert x_{k-1}-x_{0}\right\Vert ,\left\Vert x_{k}-x_{0}\right\Vert \right) \right] \left\Vert x_{k}-x_{k-1}\right\Vert \\ \nonumber & =\psi \left( \left\Vert x_{k}-x_{k-1}\right\Vert ,\left\Vert x_{k-1}-x_{0}\right\Vert ,\left\Vert x_{k}-x_{0}\right\Vert \right) \left\Vert x_{k}-x_{k-1}\right\Vert . \end{align*}
</div>
<p>The rest of the proof is identical to the one in Theorem <a href="#t2.1">2.1</a> until the uniqueness part for which we have the corresponding estimate </p>
<div class="displaymath" id="a0000000038">
  \begin{align*}  \left\Vert x^{\ast \ast }-x_{k+1}\right\Vert &  = \big\Vert x^{\ast \ast }-x_{k}+A_{k}^{-1}F\left( x_{k}\right) -A_{k}^{-1}F\left( x^{\ast \ast }\right)\big\Vert \\ \nonumber & \leq \big\Vert A_{k}^{-1}\big\Vert \left\Vert F\left( x^{\ast \ast }\right) -F\left( x_{k}\right) -A_{k}\left( x^{\ast \ast }-x_{k}\right) \right\Vert \\ \nonumber & \leq \beta ^{-1}\beta \psi _{0}\left( \left\Vert x^{\ast \ast }-x_{k}\right\Vert ,\left\Vert x_{k-1}-x_{0}\right\Vert ,\left\Vert x_{k}-x_{0}\right\Vert \right)\\ \nonumber & \leq q\left\Vert x^{\ast \ast }-x_{k}\right\Vert \leq q^{k+1}\left\Vert x^{\ast \ast }-x_{0}\right\Vert . \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000039">
  <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><div class="remark_thmwrapper " id="r3.2">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.2</span>
  </div>
  <div class="remark_thmcontent">
  <p>Comments similar to the ones given in Section 2 can follows but for method (<a href="#1.13">13</a>) and method (<a href="#1.14">14</a>) instead of method (<a href="#1.5">5</a>) and method (<a href="#1.6">6</a>), respectively.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000040">4 Applications to Fractional Calculus</h1>
<p>Here we deal with Banach space \(\left( X,\left\Vert \cdot \right\Vert \right) \) valued functions \(f\) of real domain \(\left[ a,b\right] \). All integrals here are of Bochner-type, see <span class="cite">
	[
	<a href="#14" >14</a>
	]
</span>. The derivatives of \(f\) are defined similarly to numerical ones, see <span class="cite">
	[
	<a href="#17" >17</a>
	, 
	pp.
	
	83–86
	
	and
	
	p.
	
	93
	]
</span>. </p>
<p>In this section we apply the earlier numerical methods to \(X\)-valued fractional calculus for solving \(f\left( x\right) =0\). </p>
<p>Here we would like to establish for \(\left[ a,b\right] \subseteq \mathbb {R}\), \(a{\lt}b\), \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \), \(p\in \mathbb {N}\), that </p>
<div class="equation" id="4.1">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f\left( y\right) -f\left( x\right) -A\left( x,y\right) \left( y-x\right) \right\|  \leq c_{1}\tfrac {\left| x-y\right| ^{p+1}}{p+1}, \tag {4.1} \label{4.1} \end{equation}
  </div>
  <span class="equation_label">-3</span>
</p>
</div>
<p>\(\forall \) \(x,y\in \left[ a,b\right] \), where \(c_{1}{\gt}0\), and </p>
<div class="equation" id="4.2">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  A\left( x,x\right) -A\left( y,y\right) \right\|  \leq c_{2}\left| x-y\right| , \tag {4.2} \label{4.2} \end{equation}
  </div>
  <span class="equation_label">-2</span>
</p>
</div>
<p>with \(c_{2}{\gt}0,\) \(\forall \) \(x,y\in \left[ a,b\right] \). </p>
<p>Above \(A\) stands for a \(X\)-valued differential operator to be defined and presented per case in the next, it will be denoted as \(A_{+}\left( f\right) \), \(A_{-}\left( f\right) \) in the \(X\)-valued fractional cases, and \(A_{0}\left( f\right) \) in the \(X\)-valued ordinary case. </p>
<p>We examine the following cases: </p>
<p>I) Here see <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span>. </p>
<p>Let \(x,y\in \left[ a,b\right] \) such that \(x\geq y\), \(\nu {\gt}0\), \(\nu \notin \mathbb {N}\), such that \(p=\left[ v\right] \), \(\left[ \cdot \right] \) the integral part, \(\alpha =\nu -p\) (\(0{\lt}\alpha {\lt}1\)). </p>
<p>Let \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \) and define </p>
<div class="equation" id="4.3.">
<p>
  <div class="equation_content">
    \begin{equation}  \left( J_{\nu }^{y}f\right) \left( x\right) :=\tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{\nu -1}f\left( t\right) dt,\text{ \   }y\leq x\leq b, \tag {4.3} \label{4.3.} \end{equation}
  </div>
  <span class="equation_label">-1</span>
</p>
</div>
<p><i class="it">the \(X\)-valued left generalized Riemann-Liouville fractional integral</i>. </p>
<p>Here \(\Gamma \) stands for the gamma function. </p>
<p>Clearly here it holds \(\left( J_{\nu }^{y}f\right) \left( y\right) =0\). We define \(\left( J_{\nu }^{y}f\right) \left( x\right) =0\) for \(x{\lt}y\). By <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span> \(\left( J_{\nu }^{y}f\right) \left( x\right) \) is a continuous function in \(x\), for a fixed \(y\). </p>
<p>We define the subspace \(C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \) of \(C^{p}\left( \left[ a,b\right] ,X\right) :\) </p>
<div class="equation" id="4.4">
<p>
  <div class="equation_content">
    \begin{equation}  C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) :=\left\{  f\in C^{p}\left( \left[ a,b\right] ,X\right) :J_{1-\alpha }^{y}f^{\left( p\right) }\in C^{1}\left( \left[ y,b\right] ,X\right) \right\}  . \tag {4.4} \label{4.4} \end{equation}
  </div>
  <span class="equation_label">0</span>
</p>
</div>
<p>So let \(f\in C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \), we define the <i class="it">\(X\)-valued generalized \(\nu \)-fractional derivative of</i> \(f\) over \(\left[ y,b\right] \) as </p>
<div class="equation" id="4.5">
<p>
  <div class="equation_content">
    \begin{equation}  D_{y}^{\nu }f=\left( J_{1-\alpha }^{y}f^{\left( p\right) }\right) ^{\prime }, \tag {4.5} \label{4.5} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p>that is, </p>
<div class="equation" id="4.6.">
<p>
  <div class="equation_content">
    \begin{equation}  \left( D_{y}^{\nu }f\right) \left( x\right) =\tfrac {1}{\Gamma \left( 1-\alpha \right) }\tfrac {d}{dx}\int _{y}^{x}\left( x-t\right) ^{-\alpha }f^{\left( p\right) }\left( t\right) dt, \tag {4.6} \label{4.6.} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p>which exists for \(f\in C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \), for \(a\leq y\leq x\leq b.\) </p>
<p>Here we consider \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \) such that \(f\in C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \), for every \(y\in \left[ a,b\right] \), which means also that \(f\in C_{x+}^{\nu }\left( \left[ a,b\right] ,X\right) \), for every \(x\in \left[ a,b\right] \) (<i class="it">i.e.</i> exchange roles of \(x\) and \(y\)), we write that as \(f\in C_{+}^{\nu }\left( \left[ a,b\right] ,X\right) .\) </p>
<p>That is </p>
<div class="equation" id="4.7">
<p>
  <div class="equation_content">
    \begin{equation}  \left( D_{x}^{\nu }f\right) \left( y\right) =\tfrac {1}{\Gamma \left( 1-\alpha \right) }\tfrac {d}{dy}\int _{x}^{y}\left( y-t\right) ^{-\alpha }f^{\left( p\right) }\left( t\right) dt \tag {4.7} \label{4.7} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p>exists for \(f\in C_{x+}^{\nu }\left( \left[ a,b\right] ,X\right) \), for \(a\leq x\leq y\leq b.\) </p>
<p>We mention the following left generalized \(X\)-valued fractional Taylor formula (\(f\in C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \), \(\nu {\gt}1\)), see <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span>. </p>
<p>It holds </p>
<div class="equation" id="4.8">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) -f\left( y\right) =\sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{\nu -1}\left( D_{y}^{\nu }f\right) \left( t\right) dt, \tag {4.8} \label{4.8} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p>all \(x,y\in \left[ a,b\right] \) with \(x\geq y\). </p>
<p>Similarly for \(f\in C_{x+}^{\nu }\left( \left[ a,b\right] ,X\right) \) we have </p>
<div class="equation" id="4.9">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( y\right) -f\left( x\right) =\sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{\nu -1}\left( D_{x}^{\nu }f\right) \left( t\right) dt, \tag {4.9} \label{4.9} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>all \(x,y\in \left[ a,b\right] \) with \(y\geq x\). </p>
<p>So here we work with \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \), such that \(f\in C_{+}^{\nu }\left( \left[ a,b\right] ,X\right) .\) </p>
<p>We define the \(X\)-valued left linear fractional operator </p>
<div class="equation" id="4.10">
<p>
  <div class="equation_content">
    \begin{equation}  \left( A_{+}\left( f\right) \right) \left( x,y\right) :=\left\{  \begin{array}{l} \sum \limits _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k-1}+\left( D_{y}^{\nu }f\right) \left( x\right) \tfrac {\left( x-y\right) ^{\nu -1}}{\Gamma \left( \nu +1\right) }\text{, \  }x>y, \\[4mm] \sum \limits _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k-1}+\left( D_{x}^{\nu }f\right) \left( y\right) \tfrac {\left( y-x\right) ^{\nu -1}}{\Gamma \left( \nu +1\right) }\text{, \  }y>x, \\[4mm] f^{\left( p-1\right) }\left( x\right) \text{, \  }x=y.\end{array}\right. \tag {4.10} \label{4.10} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>Notice that (see <span class="cite">
	[
	<a href="#12" >12</a>
	, 
	p. 3
	]
</span>) </p>
<div class="displaymath" id="4.11">
  \begin{align}  \left\Vert \left( A_{+}\left( f\right) \right) \left( x,x\right) -\left( A_{+}\left( f\right) \right) \left( y,y\right) \right\Vert & =\big\Vert f^{\left( p-1\right) }\left( x\right) -f^{\left( p-1\right) }\left( y\right) \big\Vert \tag {4.11} \label{4.11} \\ & \leq \big\Vert f^{\left( p\right) }\big\Vert _{\infty }\left\vert x-y\right\vert \text{, \  }\forall \text{ }x,y\in \left[ a,b\right] ,\nonumber \end{align}
</div>
<p>so that condition (<a href="#4.2">-2</a>) is fulfilled. </p>
<p>Next we will prove condition (<a href="#4.1">-3</a>). It is trivially true if \(x=y\). So we examine the case of \(x\neq y\). </p>
<p>We distinguish the subcases: </p>
<p>1) \(x{\gt}y:\) We observe that </p>
<div class="displaymath" id="a0000000041">
  \begin{align*} & \left\Vert f\left( y\right) -f\left( x\right) -A_{+}\left( f\right) \left( x,y\right) \left( y-x\right) \right\Vert =\\ \nonumber & =\left\Vert f\left( x\right) -f\left( y\right) -A_{+}\left( f\right) \left( x,y\right) \left( x-y\right) \right\Vert \overset {\text{(by (\ref{4.8}), (\ref{4.10}))}}{=}\\ \nonumber & =\left\Vert \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{\nu -1}\left( D_{y}^{\nu }f\right) \left( t\right) dt-\right. \tag {4.12} \label{4.12.}\\ & \quad -\left. \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}-\left( D_{y}^{\nu }f\right) \left( x\right) \tfrac {\left( x-y\right) ^{\nu }}{\Gamma \left( \nu +1\right) }\right\Vert \nonumber \\ & =\left\Vert \tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{\nu -1}\left( D_{y}^{\nu }f\right) \left( t\right) dt-\tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{\nu -1}\left( D_{y}^{\nu }f\right) \left( x\right) dt\right\Vert = \end{align*}
</div>
<p>by <span class="cite">
	[
	<a href="#1" >1</a>
	, 
	p.
	
	426,
	
	Th.
	
	11.43
	]
</span>, </p>
<div class="equation" id="4.13">
<p>
  <div class="equation_content">
    \begin{equation}  =\tfrac {1}{\Gamma \left( \nu \right) }\left\Vert \int _{y}^{x}\left( x-t\right) ^{\nu -1}\left( \left( D_{y}^{\nu }f\right) \left( t\right) -\left( D_{y}^{\nu }f\right) \left( x\right) \right) dt\right\Vert \tag {4.13} \label{4.13} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>(by <span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>)</p>
<div class="displaymath" id="a0000000042">
  \begin{equation*}  \leq \tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{\nu -1}\left\Vert \left( D_{y}^{\nu }f\right) \left( t\right) -\left( D_{y}^{\nu }f\right) \left( x\right) \right\Vert dt \leq \end{equation*}
</div>
<p>\(\Big(\)we assume that </p>
<div class="equation" id="4.14">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert \left( D_{y}^{\nu }f\right) \left( t\right) -\left( D_{y}^{\nu }f\right) \left( x\right) \right\Vert \leq \lambda _{1}\left( y\right) \left\vert t-x\right\vert ^{p+1-\nu }, \tag {4.14} \label{4.14} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>for all \(x,y,t\in \left[ a,b\right] \) with \(x\geq t\geq y\), with \(\lambda _{1}\left( y\right) {\gt}0\) and \(\underset {y\in \left[ a,b\right] }{\sup }\lambda _{1}\left( y\right) =:\lambda _{1}{\lt}\infty \), also it is \(0{\lt}p+1-\nu {\lt}1 \Big)\) </p>
<div class="equation" id="4.15.">
<p>
  <div class="equation_content">
    \begin{equation}  \leq \tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{\nu -1}\left( x-t\right) ^{p+1-\nu }dt \tag {4.15} \label{4.15.} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<div class="displaymath" id="a0000000043">
  \begin{equation*}  =\tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( x-t\right) ^{p}dt=\tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }\tfrac {\left( x-y\right) ^{p+1}}{\left( p+1\right) }. \end{equation*}
</div>
<p>We have proved condition (<a href="#4.1">-3</a>) </p>
<div class="equation" id="4.16">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert f\left( y\right) -f\left( x\right) -A_{+}\left( f\right) \left( x,y\right) \left( y-x\right) \right\Vert \leq \tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }\tfrac {\left( x-y\right) ^{p+1}}{\left( p+1\right) }\text{, \quad for }x>y. \tag {4.16} \label{4.16} \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>2) \(x{\lt}y:\) We observe that </p>
<div class="displaymath" id="a0000000044">
  \begin{align} & \left\Vert f\left( y\right) -f\left( x\right) -\left( A_{+}\left( f\right) \right) \left( x,y\right) \left( y-x\right) \right\Vert \overset {\text{(by (\ref{4.9}), (\ref{4.10}))}}{=} \nonumber \\ & =\left\Vert \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{\nu -1}\left( D_{x}^{\nu }f\right) \left( t\right) dt-\right. \tag {4.17} \label{4.17} \\ & \qquad -\left. \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k}-\left( D_{x}^{\nu }f\right) \left( y\right) \tfrac {\left( y-x\right) ^{\nu }}{\Gamma \left( \nu +1\right) }\right\Vert \nonumber \\ & =\left\Vert \tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{\nu -1}\left( D_{x}^{\nu }f\right) \left( t\right) dt-\left( D_{x}^{\nu }f\right) \left( y\right) \tfrac {\left( y-x\right) ^{\nu }}{\Gamma \left( \nu +1\right) }\right\Vert \nonumber \\ & =\left\Vert \tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{\nu -1}\left( D_{x}^{\nu }f\right) \left( t\right) dt-\tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{\nu -1}\left( D_{x}^{\nu }f\right) \left( y\right) dt\right\Vert \tag {4.18} \label{4.18} \\ & =\tfrac {1}{\Gamma \left( \nu \right) }\left\Vert \int _{x}^{y}\left( y-t\right) ^{\nu -1}\left( \left( D_{x}^{\nu }f\right) \left( t\right) -\left( D_{x}^{\nu }f\right) \left( y\right) \right) dt\right\Vert \nonumber \\ & \leq \tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{\nu -1}\left\Vert \left( D_{x}^{\nu }f\right) \left( t\right) -\left( D_{x}^{\nu }f\right) \left( y\right) \right\Vert dt\nonumber \leq \end{align}
</div>
<p>\(\Big(\)we assume here that </p>
<div class="equation" id="4.19">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert \left( D_{x}^{\nu }f\right) \left( t\right) -\left( D_{x}^{\nu }f\right) \left( y\right) \right\Vert \leq \lambda _{2}\left( x\right) \left\vert t-y\right\vert ^{p+1-\nu }, \tag {4.19} \label{4.19} \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p>for all \(x,y,t\in \left[ a,b\right] \) with \(y\geq t\geq x\), \(\lambda _{2}\left( x\right) {\gt}0\) and \(\underset {x\in \left[ a,b\right] }{\sup }\lambda _{2}\left( x\right) =:\lambda _{2}{\lt}\infty \Big) \) </p>
<div class="equation" id="4.20">
<p>
  <div class="equation_content">
    \begin{equation}  \leq \tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{\nu -1}\left( y-t\right) ^{p+1-\nu }dt \tag {4.20} \label{4.20} \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<div class="displaymath" id="a0000000045">
  \begin{equation*}  =\tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( y-t\right) ^{p}dt=\tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }\tfrac {\left( y-x\right) ^{p+1}}{\left( p+1\right) }. \end{equation*}
</div>
<p>We have proved that </p>
<div class="equation" id="4.21">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert f\left( y\right) -f\left( x\right) -\left( A_{+}\left( f\right) \right) \left( x,y\right) \left( y-x\right) \right\Vert \leq \tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }\tfrac {\left( y-x\right) ^{p+1}}{\left( p+1\right) }\text{, } \tag {4.21} \label{4.21} \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p>for all \(x,y\in \left[ a,b\right] \) such that \(y{\gt}x\). </p>
<p>Call \(\lambda :=\max \left( \lambda _{1},\lambda _{2}\right) .\) </p>
<p><i class="itshape">Conclusion.</i> We have proved condition (<a href="#4.1">-3</a>), in detail that </p>
<div class="equation" id="4.22.">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f\left( y\right) -f\left( x\right) -\left( A_{+}\left( f\right) \right) \left( x,y\right) \left( y-x\right) \right\|  \leq \tfrac {\lambda }{\Gamma \left( \nu \right) }\tfrac {\left| x-y\right| ^{p+1}}{\left( p+1\right) }\text{, \  }\forall \text{ }x,y\in \left[ a,b\right] . \tag {4.22} \label{4.22.} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p>II) Here see <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span> and <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span>. </p>
<p>Let \(x,y\in \left[ a,b\right] \) such that \(x\leq y\), \(\nu {\gt}0\), \(\nu \notin \mathbb {N}\), such that \(p=\left[ v\right] \), \(\alpha =\nu -p\) (\(0{\lt}\alpha {\lt}1\)). </p>
<p>Let \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \) and define </p>
<div class="equation" id="4.23">
<p>
  <div class="equation_content">
    \begin{equation}  \left( J_{y-}^{\nu }f\right) \left( x\right) :=\tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}f\left( z\right) dz,\text{ \   }a\leq x\leq y, \tag {4.23} \label{4.23} \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p><i class="it">the \(X\)-valued right generalized Riemann-Liouville fractional integral</i>. </p>
<p>Define the subspace of functions </p>
<div class="equation" id="4.24">
<p>
  <div class="equation_content">
    \begin{equation}  C_{y-}^{\nu }\left( \left[ a,b\right] ,X\right) :=\left\{  f\in C^{p}\left( \left[ a,b\right] ,X\right) :J_{y-}^{1-\alpha }f^{\left( p\right) }\in C^{1}\left( \left[ a,y\right] ,X\right) \right\}  . \tag {4.24} \label{4.24} \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p>Define the <i class="it">\(X\)-valued right generalized \(\nu \)-fractional derivative of </i>\(f\) over \(\left[ a,y\right] \) as </p>
<div class="equation" id="4.25">
<p>
  <div class="equation_content">
    \begin{equation}  D_{y-}^{\nu }f:=\left( -1\right) ^{p-1}\left( J_{y-}^{1-\alpha }f^{\left( p\right) }\right) ^{\prime }. \tag {4.25} \label{4.25} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p>Notice that </p>
<div class="equation" id="4.26">
<p>
  <div class="equation_content">
    \begin{equation}  J_{y-}^{1-\alpha }f^{\left( p\right) }\left( x\right) =\tfrac {1}{\Gamma \left( 1-\alpha \right) }\int _{x}^{y}\left( z-x\right) ^{-\alpha }f^{\left( p\right) }\left( z\right) dz, \tag {4.26} \label{4.26} \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p>exists for \(f\in C_{y-}^{\nu }\left( \left[ a,b\right] ,X\right) \), and</p>
<div class="equation" id="4.27">
<p>
  <div class="equation_content">
    \begin{equation}  \left( D_{y-}^{\nu }f\right) \left( x\right) =\tfrac {\left( -1\right) ^{p-1}}{\Gamma \left( 1-\alpha \right) }\tfrac {d}{dx}\int _{x}^{y}\left( z-x\right) ^{-\alpha }f^{\left( p\right) }\left( z\right) dz, \tag {4.27} \label{4.27} \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p><i class="it">i.e.</i>, </p>
<div class="equation" id="4.28">
<p>
  <div class="equation_content">
    \begin{equation}  \left( D_{y-}^{\nu }f\right) \left( x\right) =\tfrac {\left( -1\right) ^{p-1}}{\Gamma \left( p-\nu +1\right) }\tfrac {d}{dx}\int _{x}^{y}\left( z-x\right) ^{p-\nu }f^{\left( p\right) }\left( z\right) dz. \tag {4.28} \label{4.28} \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p>Here we consider \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \) such that \(f\in C_{y-}^{\nu }\left( \left[ a,b\right] ,X\right) \), for every \(y\in \left[ a,b\right] \), which means also that \(f\in C_{x-}^{\nu }\left( \left[ a,b\right] ,X\right) \), for every \(x\in \left[ a,b\right] \) (<i class="it">i.e.</i> exchange roles of \(x\) and \(y\)), we write that as \(f\in C_{-}^{\nu }\left( \left[ a,b\right] ,X\right) .\) </p>
<p>That is, </p>
<div class="equation" id="4.29">
<p>
  <div class="equation_content">
    \begin{equation}  \left( D_{x-}^{\nu }f\right) \left( y\right) =\tfrac {\left( -1\right) ^{p-1}}{\Gamma \left( p-\nu +1\right) }\tfrac {d}{dy}\int _{y}^{x}\left( z-y\right) ^{p-\nu }f^{\left( p\right) }\left( z\right) dz \tag {4.29} \label{4.29} \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p>exists for \(f\in C_{x-}^{\nu }\left( \left[ a,b\right] ,X\right) \), for \(a\leq y\leq x\leq b.\) </p>
<p>We mention the following \(X\)-valued right generalized fractional Taylor formula (\(f\in C_{y-}^{\nu }\left( \left[ a,b\right] ,X\right) \), \(\nu {\gt}1\)), see <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span>. </p>
<p>It holds </p>
<div class="equation" id="4.30">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) -f\left( y\right) =\sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}\left( D_{y-}^{\nu }f\right) \left( z\right) dz, \tag {4.30} \label{4.30} \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p>all \(x,y\in \left[ a,b\right] \) with \(x\leq y\). </p>
<p>Similarly for \(f\in C_{x-}^{\nu }\left( \left[ a,b\right] ,X\right) \) we have </p>
<div class="equation" id="4.31">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( y\right) -f\left( x\right) =\sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( D_{x-}^{\nu }f\right) \left( z\right) dz, \tag {4.31} \label{4.31} \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p>all \(x,y\in \left[ a,b\right] \) with \(x\geq y\). </p>
<p>So here we work with \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \), such that \(f\in C_{-}^{\nu }\left( \left[ a,b\right] ,X\right) .\) </p>
<p>We define the \(X\)-valued right linear fractional operator </p>
<div class="equation" id="4.32.">
<p>
  <div class="equation_content">
    \begin{equation}  A_{-}\left( f\right) \left( x,y\right) :=\left\{  \begin{array}{l} \sum \limits _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k-1}-\left( D_{x-}^{\nu }f\right) \left( y\right) \tfrac {\left( x-y\right) ^{\nu -1}}{\Gamma \left( \nu +1\right) }\text{, \  }x>y, \\[4mm] \sum \limits _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k-1}-\left( D_{y-}^{\nu }f\right) \left( x\right) \tfrac {\left( y-x\right) ^{\nu -1}}{\Gamma \left( \nu +1\right) }\text{, \  }y>x, \\[4mm] f^{\left( p-1\right) }\left( x\right) \text{, \  }x=y.\end{array}\right. \tag {4.32} \label{4.32.} \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p>Condition (<a href="#4.2">-2</a>) is fulfilled, the same as in (<a href="#4.11">7</a>), now for \(A_{-}\left( f\right) \left( x,x\right) .\) </p>
<p>We would like to prove that </p>
<div class="equation" id="4.33">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f\left( x\right) -f\left( y\right) -\left( A_{-}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  \leq c\cdot \tfrac {\left| x-y\right| ^{p+1}}{p+1}\text{, } \tag {4.33} \label{4.33} \end{equation}
  </div>
  <span class="equation_label">25</span>
</p>
</div>
<p>for any \(x,y\in \left[ a,b\right] \), where \(c{\gt}0.\) </p>
<p>When \(x=y\) the last condition (<a href="#4.33">25</a>) is trivial. We assume \(x\neq y\). </p>
<p>We distinguish the subcases: </p>
<p>1) \(x{\gt}y:\) We observe that </p>
<div class="displaymath" id="4.34">
  \begin{align} & \left\|  \left( f\left( x\right) -f\left( y\right) \right) -\left( A_{-}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  = \tag {4.34} \label{4.34} \\ & =\left\|  \left( f\left( y\right) -f\left( x\right) \right) -\left( A_{-}\left( f\right) \right) \left( x,y\right) \left( y-x\right) \right\|  \nonumber \\ & =\left\|  \left( \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( D_{x-}^{\nu }f\right) \left( z\right) dz\right) -\right.\nonumber \\ & \qquad -\left. \left( \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( x\right) }{k!}\left( y-x\right) ^{k-1}-\left( D_{x-}^{\nu }f\right) \left( y\right) \tfrac {\left( x-y\right) ^{\nu -1}}{\Gamma \left( \nu +1\right) }\right) \left( y-x\right) \right\|  \tag {4.35} \label{4.35.} \\ & =\left\|  \tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( D_{x-}^{\nu }f\right) \left( z\right) dz+\left( D_{x-}^{\nu }f\right) \left( y\right) \tfrac {\left( x-y\right) ^{\nu -1}}{\Gamma \left( \nu +1\right) }\left( y-x\right) \right\|  \nonumber \\ & =\left\|  \tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( D_{x-}^{\nu }f\right) \left( z\right) dz-\left( D_{x-}^{\nu }f\right) \left( y\right) \tfrac {\left( x-y\right) ^{\nu }}{\Gamma \left( \nu +1\right) }\right\|  \nonumber \\ & =\tfrac {1}{\Gamma \left( \nu \right) }\left\|  \int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( D_{x-}^{\nu }f\right) \left( z\right) dz-\int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( D_{x-}^{\nu }f\right) \left( y\right) dz\right\|  \nonumber \\ & =\tfrac {1}{\Gamma \left( \nu \right) }\left\|  \int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( \left( D_{x-}^{\nu }f\right) \left( z\right) -\left( D_{x-}^{\nu }f\right) \left( y\right) \right) dz\right\|  \tag {4.36} \label{4.36} \\ \nonumber & \leq \tfrac {1}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( z-y\right) ^{\nu -1}\left\|  \left( D_{x-}^{\nu }f\right) \left( z\right) -\left( D_{x-}^{\nu }f\right) \left( y\right) \right\|  dz \leq \end{align}
</div>
<p>\(\big(\)we assume that </p>
<div class="equation" id="4.37">
<p>
  <div class="equation_content">
    \begin{equation}  \leq \left\|  \left( D_{x-}^{\nu }f\right) \left( z\right) -\left( D_{x-}^{\nu }f\right) \left( y\right) \right\|  \leq \lambda _{1}\left| z-y\right| ^{p+1-\nu }, \tag {4.37} \label{4.37} \end{equation}
  </div>
  <span class="equation_label">29</span>
</p>
</div>
<p>\(\lambda _{1}{\gt}0\), for all \(x,z,y\in \left[ a,b\right] \) with \(x\geq z\geq y\big)\)</p>
<div class="equation" id="4.38.">
<p>
  <div class="equation_content">
    \begin{equation}  \leq \tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( z-y\right) ^{\nu -1}\left( z-y\right) ^{p+1-\nu }dz= \tag {4.38} \label{4.38.} \end{equation}
  </div>
  <span class="equation_label">30</span>
</p>
</div>
<div class="displaymath" id="a0000000046">
  \begin{equation*}  =\tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }\int _{y}^{x}\left( z-y\right) ^{p}dz=\tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }\tfrac {\left( x-y\right) ^{p+1}}{p+1}=\rho _{1}\tfrac {\left( x-y\right) ^{p+1}}{p+1}, \end{equation*}
</div>
<p>where \(\rho _{1}:=\tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }{\gt}0.\) </p>
<p>We have proved, when \(x{\gt}y\), that </p>
<div class="equation" id="4.39">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f\left( x\right) -f\left( y\right) -\left( A_{-}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  \leq \rho _{1}\tfrac {\left( x-y\right) ^{p+1}}{p+1}. \tag {4.39} \label{4.39} \end{equation}
  </div>
  <span class="equation_label">31</span>
</p>
</div>
<p>2) \(y{\gt}x:\) We observe that </p>
<div class="displaymath" id="a0000000047">
  \begin{align} & \left\|  f\left( x\right) -f\left( y\right) -\left( A_{-}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  =\nonumber \\ & =\left\|  \Bigg( \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}+\tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}\left( D_{y-}^{\nu }f\right) \left( z\right) dz\Bigg) \right. \nonumber \\ & \qquad -\left. \Bigg( \sum _{k=1}^{p-1}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k-1}-\left( D_{y-}^{\nu }f\right) \left( x\right) \tfrac {\left( y-x\right) ^{\nu -1}}{\Gamma \left( \nu +1\right) }\Bigg) \left( x-y\right) \right\|  \tag {4.40} \label{4.40} \\ & =\left\|  \tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}\left( D_{y-}^{\nu }f\right) \left( z\right) dz-\left( D_{y-}^{\nu }f\right) \left( x\right) \tfrac {\left( y-x\right) ^{\nu }}{\Gamma \left( \nu +1\right) }\right\|  \nonumber \\ & =\left\|  \tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}\left( D_{y-}^{\nu }f\right) \left( z\right) dz-\tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}\left( D_{y-}^{\nu }f\right) \left( x\right) dz\right\|  \tag {4.41} \label{4.41} \\ & =\tfrac {1}{\Gamma \left( \nu \right) }\left\|  \int _{x}^{y}\left( z-x\right) ^{\nu -1}\left( \left( D_{y-}^{\nu }f\right) \left( z\right) -\left( D_{y-}^{\nu }f\right) \left( x\right) \right) dz\right\|  \tag {4.42} \label{4.42} \\ & \leq \tfrac {1}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}\left\|  \left( D_{y-}^{\nu }f\right) \left( z\right) -\left( D_{y-}^{\nu }f\right) \left( x\right) \right\|  dz\nonumber \end{align}
</div>
<p>(we assume that </p>
<div class="equation" id="4.43">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  \left( D_{y-}^{\nu }f\right) \left( z\right) -\left( D_{y-}^{\nu }f\right) \left( x\right) \right\|  \leq \lambda _{2}\left| z-x\right| ^{p+1-\nu }, \tag {4.43} \label{4.43} \end{equation}
  </div>
  <span class="equation_label">35</span>
</p>
</div>
<p>\(\lambda _{2}{\gt}0,\) for all \(y,z,x\in \left[ a,b\right] \) with \(y\geq z\geq x\)) </p>
<div class="displaymath" id="4.44">
  \begin{align} & \leq \tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{\nu -1}\left( z-x\right) ^{p+1-\nu }dz= \tag {4.44} \label{4.44} \\ & =\tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }\int _{x}^{y}\left( z-x\right) ^{p}dz=\tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }\tfrac {\left( y-x\right) ^{p+1}}{p+1}. \end{align}
</div>
<p>We have proved, for \(y{\gt}x\), that </p>
<div class="equation" id="4.45">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f\left( x\right) -f\left( y\right) -\left( A_{-}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  \leq \rho _{2}\tfrac {\left( y-x\right) ^{p+1}}{p+1}\text{, } \tag {4.45} \label{4.45} \end{equation}
  </div>
  <span class="equation_label">38</span>
</p>
</div>
<p>where \(\rho _{2}:=\tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }{\gt}0.\) </p>
<p>Set \(\lambda :=\max \left( \lambda _{1},\lambda _{2}\right) \) and \(\rho :=\tfrac {\lambda }{\Gamma \left( \nu \right) }{\gt}0.\) </p>
<p><i class="itshape">Conclusion.</i> We have proved (<a href="#4.1">-3</a>) that </p>
<div class="equation" id="4.46">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f\left( x\right) -f\left( y\right) -\left( A_{-}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  \leq \rho \tfrac {\left| x-y\right| ^{p+1}}{p+1}\text{, \  } \tag {4.46} \label{4.46} \end{equation}
  </div>
  <span class="equation_label">39</span>
</p>
</div>
<p>for any \(x,y\in \left[ a,b\right] .\) </p>
<p>III) Let again \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \), \(p\in \mathbb {N}\), \(x,y\in \left[ a,b\right] \). </p>
<p>By vector \(X\)-valued Taylor’s formula we have, see <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#17" >17</a>
	]
</span>, </p>
<div class="equation" id="4.47">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) -f\left( y\right) =\sum _{k=1}^{p}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}+\tfrac {1}{\left( p-1\right) !}\int _{y}^{x}\left( x-t\right) ^{p-1}\left( f^{\left( p\right) }\left( t\right) -f^{\left( p\right) }\left( y\right) \right) dt, \tag {4.47} \label{4.47} \end{equation}
  </div>
  <span class="equation_label">40</span>
</p>
</div>
<p>\(\forall \) \(x,y\in \left[ a,b\right] .\) </p>
<p>We define the \(X\)-valued function </p>
<div class="equation" id="4.48">
<p>
  <div class="equation_content">
    \begin{equation}  \left( A_{0}\left( f\right) \right) \left( x,y\right) :=\left\{  \begin{array}{l} \sum \limits _{k=1}^{p}\frac{f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k-1}\text{, \  }x\neq y, \\[3mm] f^{\left( p-1\right) }\left( x\right) \text{, \  }x=y.\end{array}\right. \tag {4.48} \label{4.48} \end{equation}
  </div>
  <span class="equation_label">41</span>
</p>
</div>
<p>Then it holds, by <span class="cite">
	[
	<a href="#12" >12</a>
	, 
	p.
	
	3
	]
</span>, </p>
<div class="displaymath" id="4.49">
  \begin{align}  \left\Vert \left( A_{0}\left( f\right) \right) \left( x,x\right) -\left( A_{0}\left( f\right) \right) \left( y,y\right) \right\Vert & = \Vert f^{\left( p-1\right) }\left( x\right) -f^{\left( p-1\right) }\left( y\right)\Vert \tag {4.49} \label{4.49} \\ \nonumber & \leq \Vert f^{\left( p\right) }\Vert _{\infty }\left\vert x-y\right\vert \end{align}
</div>
<p>\(\forall \text{ }x,y\in \left[ a,b\right] ,\) so that condition (<a href="#4.2">-2</a>) is fulfilled. </p>
<p>Next we observe that </p>
<div class="displaymath" id="a0000000048">
  \begin{align} & \left\|  f\left( x\right) -f\left( y\right) -\left( A_{0}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  =\nonumber \\ & =\left\|  \sum _{k=1}^{p}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}+\tfrac {1}{\left( p-1\right) !}\int _{y}^{x}\left( x-t\right) ^{p-1}\left( f^{\left( p\right) }\left( t\right) -f^{\left( p\right) }\left( y\right) \right) dt\right. \tag {4.50} \label{4.50} \\ & \left. \qquad -\sum _{k=1}^{p}\tfrac {f^{\left( k\right) }\left( y\right) }{k!}\left( x-y\right) ^{k}\right\|  \nonumber \\ & =\tfrac {1}{\left( p-1\right) !}\left\|  \int _{y}^{x}\left( x-t\right) ^{p-1}\left( f^{\left( p\right) }\left( t\right) -f^{\left( p\right) }\left( y\right) \right) dt\right\|  =:\left( \xi \right) . \tag {4.51} \label{4.51} \end{align}
</div>
<p>Here we assume that </p>
<div class="equation" id="4.52">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f^{\left( p\right) }\left( t\right) -f^{\left( p\right) }\left( y\right) \right\|  \leq c\left| t-y\right| ,\text{ \  }\forall \text{ }t,y\in \left[ a,b\right] \text{, \  }c>0. \tag {4.52} \label{4.52} \end{equation}
  </div>
  <span class="equation_label">45</span>
</p>
</div>
<p>1) Subcase of \(x{\gt}y:\) We have that (by <span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>) </p>
<div class="displaymath" id="a0000000049">
  \begin{align}  \left( \xi \right) & \leq \tfrac {1}{\left( p-1\right) !}\int _{y}^{x}\left( x-t\right) ^{p-1}\left\Vert f^{\left( p\right) }\left( t\right) -f^{\left( p\right) }\left( y\right) \right\Vert dt\nonumber \\ & \leq \tfrac {c}{\left( p-1\right) !}\int _{y}^{x}\left( x-t\right) ^{p-1}\left( t-y\right) ^{2-1}dt \tag {4.53} \label{4.53} \\ & =c\tfrac {\Gamma \left( p\right) \Gamma \left( 2\right) }{\left( p-1\right) !\Gamma \left( p+2\right) }\left( x-y\right) ^{p+1}=c\tfrac {\left( p-1\right) !}{\left( p-1\right) !\left( p+1\right) !}\left( x-y\right) ^{p+1}\nonumber \\ & =\tfrac {c\left( x-y\right) ^{p+1}}{\left( p+1\right) !}. \nonumber \end{align}
</div>
<p>Hence </p>
<div class="equation" id="4.54">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \xi \right) \leq c\tfrac {\left( x-y\right) ^{p+1}}{\left( p+1\right) !}\text{, \  }x>y. \tag {4.54} \label{4.54} \end{equation}
  </div>
  <span class="equation_label">47</span>
</p>
</div>
<p>2) Subcase of \(y{\gt}x.\) </p>
<p>We have that </p>
<div class="displaymath" id="4.55">
  \begin{align}  \left( \xi \right) & =\tfrac {1}{\left( p-1\right) !}\left\|  \int _{x}^{y}\left( t-x\right) ^{p-1}\left( f^{\left( p\right) }\left( y\right) -f^{\left( p\right) }\left( t\right) \right) dt\right\|  \tag {4.55} \label{4.55} \\ & \leq \tfrac {1}{\left( p-1\right) !}\int _{x}^{y}\left( t-x\right) ^{p-1}\left\|  f^{\left( p\right) }\left( y\right) -f^{\left( p\right) }\left( t\right) \right\|  dt\nonumber \\ & \leq \tfrac {c}{\left( p-1\right) !}\int _{x}^{y}\left( t-x\right) ^{p-1}\left( y-t\right) dt\nonumber \\ & =\tfrac {c}{\left( p-1\right) !}\int _{x}^{y}\left( y-t\right) ^{2-1}\left( t-x\right) ^{p-1}dt \tag {4.56} \label{4.56} \\ & =\tfrac {c}{\left( p-1\right) !}\tfrac {\Gamma \left( 2\right) \Gamma \left( p\right) }{\Gamma \left( p+2\right) }\left( y-x\right) ^{p+1}=\tfrac {c}{\left( p-1\right) !}\tfrac {\left( p-1\right) !}{\left( p+1\right) !}\left( y-x\right) ^{p+1}\nonumber \\ & =c\tfrac {\left( y-x\right) ^{p+1}}{\left( p+1\right) !}.\nonumber \end{align}
</div>
<p>That is </p>
<div class="equation" id="4.57">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \xi \right) \leq c\tfrac {\left( y-x\right) ^{p+1}}{\left( p+1\right) !}\text{, \  }y>x. \tag {4.57} \label{4.57} \end{equation}
  </div>
  <span class="equation_label">50</span>
</p>
</div>
<p>Therefore it holds </p>
<div class="equation" id="4.58">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \xi \right) \leq c\tfrac {\left| x-y\right| ^{p+1}}{\left( p+1\right) !}\text{, \  all }x,y\in \left[ a,b\right] \text{ such that }x\neq y. \tag {4.58} \label{4.58} \end{equation}
  </div>
  <span class="equation_label">51</span>
</p>
</div>
<p>We have found that </p>
<div class="equation" id="4.59">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f\left( x\right) -f\left( y\right) -\left( A_{0}\left( f\right) \right) \left( x,y\right) \left( x-y\right) \right\|  \leq c\tfrac {\left| x-y\right| ^{p+1}}{\left( p+1\right) !}\text{, \  }c>0, \tag {4.59} \label{4.59} \end{equation}
  </div>
  <span class="equation_label">52</span>
</p>
</div>
<p>for all \(x\neq y\). </p>
<p>When \(x=y\) inequality (<a href="#4.59">52</a>) holds trivially, so (<a href="#4.1">-3</a>) is true for any \(x,y\in \left[ a,b\right] .\) </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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<dl class="bibliography">
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  <dd><p><i class="sc">C.D. Aliprantis, K.C. Border</i>, <i class="itshape">Infinite Dimensional Analysis</i>, Springer, New York, 2006. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.jmaa.2010.01.047 "> <i class="sc">S. Amat, S. Busquier, S. Plaza</i>, <i class="itshape">Chaotic dynamics of a third-order Newton-type method</i>, J. Math. Anal. Appl., <b class="bf">366</b> (2010) 1, 24–32. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><a href ="https://doi.org/10.4067/s0716-09172017000100009"> <i class="sc">G.A. Anastassiou</i>, <i class="itshape">Strong Right Fractional Calculus for Banach space valued functions</i>, Revista Proyecciones, <b class="bf">36</b> (2017) 1, 149–186. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">G.A. Anastassiou</i>, <i class="itshape">A strong Fractional Calculus Theory for Banach space valued functions</i>, Nonlinear Functional Analysis and Applications (Korea), accepted for publication, 2017. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">G.A. Anastassiou</i>, <i class="itshape">Strong mixed and generalized fractional calculus for Banach space valued functions</i>, Mat. Vesnik, <b class="bf">69</b> (2017) no. 3, pp. 176–191. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.jmaa.2004.04.008"> <i class="sc">I.K. Argyros</i>, <i class="itshape">A unifying local-semilocal convergence analysis and applications for two-point Newton-like methods in Banach space</i>, J. Math. Anal. Appl., <b class="bf">298</b> (2004), 374–397. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="7">7</a></dt>
  <dd><p><i class="sc">I.K. Argyros, A. Magrenan</i>, <i class="itshape">Iterative methods and their dynamics with applications</i>, CRC Press, New York, 2017. </p>
</dd>
  <dt><a name="8">8</a></dt>
  <dd><p><i class="it">Bochner integral</i>. Encyclopedia of Mathematics. URL: http://www.encyclopedia ofmath.org/index.php?title=Bochner_integral&amp; oldid=38659. </p>
</dd>
  <dt><a name="9">9</a></dt>
  <dd><p><a href ="https://doi.org/10.1112/jlms/s1-37.1.74"> <i class="sc">M. Edelstein</i>, <i class="itshape">On fixed and periodic points under contractive mappings</i>, J. London Math. Soc., <b class="bf">37</b> (1962), 74–79. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="10">10</a></dt>
  <dd><p><i class="sc">J.A. Ezquerro, J.M. Gutierrez, M.A. Hernandez, N. Romero, M.J. Rubio</i>, <i class="itshape">The Newton method: From Newton to Kantorovich</i> (Spanish), Gac. R. Soc. Mat. Esp., <b class="bf">13</b> (2010), 53–76. </p>
</dd>
  <dt><a name="11">11</a></dt>
  <dd><p><i class="sc">L.V. Kantorovich, G.P. Akilov</i>, <i class="itshape">Functional Analysis in Normed Spaces</i>, Pergamon Press, New York, 1982. </p>
</dd>
  <dt><a name="12">12</a></dt>
  <dd><p><i class="sc">G.E. Ladas, V. Lakshmikantham</i>, <i class="itshape">Differential equations in abstract spaces,</i> Academic Press, New York, London, 1972. </p>
</dd>
  <dt><a name="13">13</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.amc.2014.09.061"> <i class="sc">A. Alberto Magrenan</i>, <i class="itshape">A new tool to study real dynamics: The convergence plane</i>, Appl. Math. Comput., <b class="bf">248</b> (2014), 215–224. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="14">14</a></dt>
  <dd><p><i class="sc">J. Mikusinski</i>, <i class="itshape">The Bochner integral</i>, Academic Press, New York, 1978. </p>
</dd>
  <dt><a name="15">15</a></dt>
  <dd><p><i class="sc">F.A. Potra, V. Ptak</i>, <i class="itshape">Nondiscrete induction and iterative processes</i>, Pitman Publ., London, 1984. </p>
</dd>
  <dt><a name="16">16</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.jco.2009.05.001"> <i class="sc">P.D. Proinov</i>, <i class="itshape">New general convergence theory for iterative processes and its applications to Newton-Kantorovich type theorems</i>, J. Complexity, <b class="bf">26</b> (2010), 3–42. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="17">17</a></dt>
  <dd><p><i class="sc">G.E. Shilov</i>, <i class="itshape">Elementary Functional Analysis</i>, Dover Publications, Inc., New York, 1996. </p>
</dd>
</dl>


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