Semi-local convergence of iterative methods and Banach space valued functions in abstract fractional calculus
George A. Anastassiou\(^\ast \) Ioannis K. Argyros\(^\bullet \)
May 1st, 2017. Accepted: February 5, 2018. Published online: August 6, 2018.
\(^\ast \)Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A., e-mail: ganastss@memphis.edu.
\(^\bullet \)Department of Mathematics Sciences, Cameron University, Lawton, OK 73505, USA, e-mail: iargyros@cameron.edu.
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.
MSC. 65G99, 65H10, 26A33, 46B25, 47J25, 47J05.
Keywords. iterative method, Banach space, semi-local convergence, fractional calculus, Bochner-type integral.
1 Introduction
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 \(U\left( z,\rho \right) \).
Many problems in Computational Sciences, Engineering, Mathematical Chemistry, Mathematical Physics, Mathematical Economics and other disciplines can be brought in a form like
\begin{equation} F\left( x\right) =0 \tag {1.1} \label{1.1} \end{equation}
1
using Mathematical Modeling
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–
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, where \(F:\Omega \rightarrow B_{2}\) is a continuous operator. The solution \(x^{\ast }\) of equation (1) 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 (1). We can divide these methods in two categories.
Explicit Methods
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,
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: Newton’s method
\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}
2
Secant method:
\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}
3
where \(\left[ \cdot ,\cdot ;F\right] \) denotes a divided difference of order one on \(\Omega \times \Omega \)
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.
Newton-like method:
\begin{equation} x_{n+1}=x_{n}-E_{n}^{-1}F\left( x_{n}\right) , \tag {1.4} \label{1.4} \end{equation}
4
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
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and the references there in.
Implicit Methods
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:
\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}
5
\begin{equation} x_{n+1}=x_{n}-A_{n}^{-1}F\left( x_{n}\right) , \tag {1.6} \label{1.6} \end{equation}
6
where \(A_{n}=A\left( x_{n+1},x_{n}\right) =A\left( F\right) \left( x_{n+1},x_{n}\right) \) and \(A:\Omega \times \Omega \rightarrow \mathcal{L}\left( B_{1},B_{2}\right) .\)
There is a plethora on local as well as semi-local convergence results for explicit methods
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. However, the research on the convergence of implicit methods has received little attention. Authors, usually consider the fixed point problem
\begin{equation} P_{z}\left( x\right) =x, \tag {1.7} \label{1.7} \end{equation}
7
where
\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}
8
or
\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}
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for methods (5) and (6), 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
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, \(P_{z}\) has a fixed point \(x_{z}^{\ast }\) which can be found using the method of succesive substitutions or Picard’s method
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defined for each fixed \(n\) by
\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}
10
Let us also consider the analogous explicit methods
\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}
11
\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}
12
\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}
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and
\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}
14
In the present paper in Section 2, we present the semi-local convergence of method (5) and method (6). Section 3 contains the semi-local convergence of method (11), method (12), method (13) and method (14). 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
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2 Semi-local Convergence for Implicit methods
We present the semi-local convergence analysis of method (6) using conditions \(\left( S\right) \):
\((s_{1})\) \(F:\Omega \subset B_{1}\rightarrow B_{2}\) is continuous and \(A\left( x,y\right) \in \mathcal{L}\left( B_{1},B_{2}\right) \) for each \(\left( x,y\right) \in \Omega \times \Omega .\)
\((s_{2})\) 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
\begin{equation*} \| A\left( x,y\right) ^{-1}\| \leq \beta ^{-1}. \end{equation*}
Set \(\Omega _{1}=\Omega \cap \Omega _{0}\).
\((s_{3})\) There exists a continuous and nondecreasing function \(\psi :[0,+\infty )^{3}\rightarrow \lbrack 0,+\infty )\) such that for each \(x,y\in \Omega _{1}\)
\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*}
\((s_{4})\) For each \(x\in \Omega _{0}\) there exists \(y\in \Omega _{0}\) such that
\begin{equation*} y=x-A\left( y,x\right) ^{-1}F\left( x\right) . \end{equation*}
\((s_{5})\) For \(x_{0}\in \Omega _{0}\) and \(x_{1}\in \Omega _{0}\) satisfying \((s_{4})\) there exists \(\eta \geq 0\) such that
\begin{equation*} \left\| A\left( x_{1},x_{0}\right) ^{-1}F\left( x_{0}\right) \right\| \leq \eta . \end{equation*}
\((s_{6})\) Define \(q\left( t\right) :=\psi \left( \eta ,t,t\right) \) for each \(t\in \lbrack 0,+\infty )\). Equation
\begin{equation*} t\left( 1-q\left( t\right) \right) -\eta =0 \end{equation*}
has positive solutions. Denote by \(s\) the smallest such solution.
\((s_{7})\) \(\overline{U}\left( x_{0},s^{{}}\right) \subset \Omega \), where
\begin{equation*} s^{{}}=\tfrac {\eta }{1-q_{0}^{{}}}\text{ \ and \ }q_{0}^{{}}=\psi \left( \eta ,s,s\right) . \end{equation*}
Next, we present the semi-local convergence analysis for method (6) using the conditions \(\left( S\right) \) and the preceding notation.
Theorem
2.1
Assume that the conditions \((S)\) hold. Then, sequence \(\left\{ x_{n}\right\} \) generated by method 6 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 \(x,y,z\in \Omega _{1}\)
\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*}
and \(q_{1}=\psi _{1}\left( \eta ,s^{{}},s^{{}},s^{{}}\right) {\lt}1.\)
Then, \(x^{\ast }\) is the unique solution of equation \(F\left( x\right) =0\) in \(\overline{U}\left( x_{0},s^{{}}\right) .\)
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 \(\left\Vert x_{k}-x_{0}\right\Vert \leq s.\)
We get by (6), \(\left( s_{2}\right)\)–\(\left( s_{5}\right) \) in turn that
\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}
and
\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*}
The induction is completed. Moreover, we have by (14) that for \(m=0,1,2,...\)
\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*}
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 (14) 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 (6) and the hypothesis on \(\psi _{1}\), we obtain in turn that
\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*}
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 \(x^{\ast }=x^{\ast \ast }.\)
Theorem
2.3
Suppose that the conditions \(\left( V\right) \) are satisfied. Then, equation 5 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 6 is also uniquely solvable for each \(n=0,1,2,...\)
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
\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*}
and
\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*}
3 Semi-local convergence for explicit methods
A specialization of Theorem 2.1 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 (12) (and consequently of method (11)), we use the conditions \(\left( S^{\prime }\right) :\)
\(\left( s_{1}^{\prime }\right) \) \(F:\Omega \subset B_{1}\rightarrow B_{2}\) is continuous and \(A\left( x,x\right) \in \mathcal{L}\left( B_{1},B_{2}\right) \) for each \(x\in \Omega .\)
\((s_{2}^{\prime })\) 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
\begin{equation*} \| A\left( x,x\right) ^{-1}\| \leq \beta ^{-1}. \end{equation*}
Set \(\Omega _{1}=\Omega \cap \Omega _{0}\).
\((s_{3}^{\prime })\) 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 \(x,y\in \Omega _{1}\)
\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*}
and
\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*}
Set \(\psi =\psi _{0}+\psi _{2}.\)
\((s_{4}^{\prime })\) 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 \(\ \)
\begin{equation*} \| A\left( x_{0},x_{0}\right) ^{-1}F\left( x_{0}\right)\| \leq \eta . \end{equation*}
\((s_{5}^{\prime })=\left( s_{6}\right) \)
\((s_{6}^{\prime })=\left( s_{7}\right) \).
Next, we present the following semi-local convergence analysis of method (12) using the \(\left( S^{\prime }\right) \) conditions and the preceding notation.
Proposition
3.1
Suppose that the conditions \((S^{\prime })\) are satisfied. Then, sequence \(\left\{ x_{n}\right\} \) generated by method 12 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 \(F\left( x\right) =0\).
We follow the proof of Theorem
2.1 but use instead the analogous estimate
\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*}
The rest of the proof is identical to the one in Theorem 2.1 until the uniqueness part for which we have the corresponding estimate
\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*}
4 Applications to Fractional Calculus
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
[
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. The derivatives of \(f\) are defined similarly to numerical ones, see
[
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,
pp.
83–86
and
p.
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.
In this section we apply the earlier numerical methods to \(X\)-valued fractional calculus for solving \(f\left( x\right) =0\).
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
\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}
-3
\(\forall \) \(x,y\in \left[ a,b\right] \), where \(c_{1}{\gt}0\), and
\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}
-2
with \(c_{2}{\gt}0,\) \(\forall \) \(x,y\in \left[ a,b\right] \).
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.
We examine the following cases:
I) Here see
[
4
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,
[
5
]
.
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, \(\alpha =\nu -p\) (\(0{\lt}\alpha {\lt}1\)).
Let \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \) and define
\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}
-1
the \(X\)-valued left generalized Riemann-Liouville fractional integral.
Here \(\Gamma \) stands for the gamma function.
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
[
4
]
\(\left( J_{\nu }^{y}f\right) \left( x\right) \) is a continuous function in \(x\), for a fixed \(y\).
We define the subspace \(C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \) of \(C^{p}\left( \left[ a,b\right] ,X\right) :\)
\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}
0
So let \(f\in C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \), we define the \(X\)-valued generalized \(\nu \)-fractional derivative of \(f\) over \(\left[ y,b\right] \) as
\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}
1
that is,
\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}
2
which exists for \(f\in C_{y+}^{\nu }\left( \left[ a,b\right] ,X\right) \), for \(a\leq y\leq x\leq b.\)
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.e. exchange roles of \(x\) and \(y\)), we write that as \(f\in C_{+}^{\nu }\left( \left[ a,b\right] ,X\right) .\)
That is
\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}
3
exists for \(f\in C_{x+}^{\nu }\left( \left[ a,b\right] ,X\right) \), for \(a\leq x\leq y\leq b.\)
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
[
5
]
.
It holds
\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}
4
all \(x,y\in \left[ a,b\right] \) with \(x\geq y\).
Similarly for \(f\in C_{x+}^{\nu }\left( \left[ a,b\right] ,X\right) \) we have
\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}
5
all \(x,y\in \left[ a,b\right] \) with \(y\geq x\).
So here we work with \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \), such that \(f\in C_{+}^{\nu }\left( \left[ a,b\right] ,X\right) .\)
We define the \(X\)-valued left linear fractional operator
\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}
6
Notice that (see
[
12
,
p. 3
]
)
\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}
so that condition (-2) is fulfilled.
Next we will prove condition (-3). It is trivially true if \(x=y\). So we examine the case of \(x\neq y\).
We distinguish the subcases:
1) \(x{\gt}y:\) We observe that
\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*}
by
[
1
,
p.
426,
Th.
11.43
]
,
\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}
5
(by
[
8
]
)
\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*}
\(\Big(\)we assume that
\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}
6
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 \(0{\lt}p+1-\nu {\lt}1 \Big)\)
\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}
7
\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*}
We have proved condition (-3)
\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}
8
2) \(x{\lt}y:\) We observe that
\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}
\(\Big(\)we assume here that
\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}
11
for all \(x,y,t\in \left[ a,b\right] \) with \(y\geq t\geq x\), \(\lambda _{2}\left( x\right) {\gt}0\) and \(\underset {x\in \left[ a,b\right] }{\sup }\lambda _{2}\left( x\right) =:\lambda _{2}{\lt}\infty \Big) \)
\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}
12
\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*}
We have proved that
\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}
13
for all \(x,y\in \left[ a,b\right] \) such that \(y{\gt}x\).
Call \(\lambda :=\max \left( \lambda _{1},\lambda _{2}\right) .\)
Conclusion. We have proved condition (-3), in detail that
\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}
14
II) Here see
[
3
]
and
[
5
]
.
Let \(x,y\in \left[ a,b\right] \) such that \(x\leq y\), \(\nu {\gt}0\), \(\nu \notin \mathbb {N}\), such that \(p=\left[ v\right] \), \(\alpha =\nu -p\) (\(0{\lt}\alpha {\lt}1\)).
Let \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \) and define
\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}
15
the \(X\)-valued right generalized Riemann-Liouville fractional integral.
Define the subspace of functions
\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}
16
Define the \(X\)-valued right generalized \(\nu \)-fractional derivative of \(f\) over \(\left[ a,y\right] \) as
\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}
17
Notice that
\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}
18
exists for \(f\in C_{y-}^{\nu }\left( \left[ a,b\right] ,X\right) \), and
\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}
19
i.e.,
\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}
20
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.e. exchange roles of \(x\) and \(y\)), we write that as \(f\in C_{-}^{\nu }\left( \left[ a,b\right] ,X\right) .\)
That is,
\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}
21
exists for \(f\in C_{x-}^{\nu }\left( \left[ a,b\right] ,X\right) \), for \(a\leq y\leq x\leq b.\)
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
[
5
]
.
It holds
\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}
22
all \(x,y\in \left[ a,b\right] \) with \(x\leq y\).
Similarly for \(f\in C_{x-}^{\nu }\left( \left[ a,b\right] ,X\right) \) we have
\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}
23
all \(x,y\in \left[ a,b\right] \) with \(x\geq y\).
So here we work with \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \), such that \(f\in C_{-}^{\nu }\left( \left[ a,b\right] ,X\right) .\)
We define the \(X\)-valued right linear fractional operator
\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}
24
Condition (-2) is fulfilled, the same as in (7), now for \(A_{-}\left( f\right) \left( x,x\right) .\)
We would like to prove that
\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}
25
for any \(x,y\in \left[ a,b\right] \), where \(c{\gt}0.\)
When \(x=y\) the last condition (25) is trivial. We assume \(x\neq y\).
We distinguish the subcases:
1) \(x{\gt}y:\) We observe that
\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}
\(\big(\)we assume that
\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}
29
\(\lambda _{1}{\gt}0\), for all \(x,z,y\in \left[ a,b\right] \) with \(x\geq z\geq y\big)\)
\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}
30
\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*}
where \(\rho _{1}:=\tfrac {\lambda _{1}}{\Gamma \left( \nu \right) }{\gt}0.\)
We have proved, when \(x{\gt}y\), that
\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}
31
2) \(y{\gt}x:\) We observe that
\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}
(we assume that
\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}
35
\(\lambda _{2}{\gt}0,\) for all \(y,z,x\in \left[ a,b\right] \) with \(y\geq z\geq x\))
\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}
We have proved, for \(y{\gt}x\), that
\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}
38
where \(\rho _{2}:=\tfrac {\lambda _{2}}{\Gamma \left( \nu \right) }{\gt}0.\)
Set \(\lambda :=\max \left( \lambda _{1},\lambda _{2}\right) \) and \(\rho :=\tfrac {\lambda }{\Gamma \left( \nu \right) }{\gt}0.\)
Conclusion. We have proved (-3) that
\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}
39
for any \(x,y\in \left[ a,b\right] .\)
III) Let again \(f\in C^{p}\left( \left[ a,b\right] ,X\right) \), \(p\in \mathbb {N}\), \(x,y\in \left[ a,b\right] \).
By vector \(X\)-valued Taylor’s formula we have, see
[
3
]
,
[
4
]
,
[
17
]
,
\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}
40
\(\forall \) \(x,y\in \left[ a,b\right] .\)
We define the \(X\)-valued function
\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}
41
Then it holds, by
[
12
,
p.
3
]
,
\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}
\(\forall \text{ }x,y\in \left[ a,b\right] ,\) so that condition (-2) is fulfilled.
Next we observe that
\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}
Here we assume that
\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}
45
1) Subcase of \(x{\gt}y:\) We have that (by
[
8
]
)
\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}
Hence
\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}
47
2) Subcase of \(y{\gt}x.\)
We have that
\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}
That is
\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}
50
Therefore it holds
\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}
51
We have found that
\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}
52
for all \(x\neq y\).
When \(x=y\) inequality (52) holds trivially, so (-3) is true for any \(x,y\in \left[ a,b\right] .\)
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