\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 74, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/74\hfil Generalized Riemann derivative]
{Generalized Riemann derivative}

\author[ S. R\u adulescu, P. Alexandrescu,  D.-O. Alexandrescu 
\hfil EJDE-2013/74\hfilneg]
{Sorin R\u adulescu, Petru\c s Alexandrescu, Diana-Olimpia Alexandrescu}
 

\address{Sorin R\u adulescu \newline
  Institute of Mathematical Statistics and Applied Mathematics,
  Calea 13 Septembrie, no. 13, Bucharest 5, RO-050711, Romania}
  \email{xsradulescu@gmail.com}

\address{Petru\c s Alexandrescu \newline
  Institute of Sociology, Casa Academiei Rom\^ane,
  Calea 13 Septembrie, no. 13, Bucharest 5, RO-050711, Romania}
 \email{alexandrescu\_petrus@yahoo.com}

\address{Diana-Olimpia Alexandrescu \newline
 Department of Mathematics, University of Craiova,
 200585 Craiova, Romania}
\email{alexandrescudiana@yahoo.com}

\thanks{Submitted January 10, 2013. Published March 18, 2013.}
\subjclass[2000]{26A24, 28A15}
\keywords{Riemann generalized derivative; symmetric derivative;
\hfill\break\indent Schwarz derivative; $(\sigma,\tau)$ differentiable function}

\begin{abstract}
 Initiated by Marshall Ash in 1966, the study of generalized Riemann
 derivative draw significant attention of the mathematical community
 and numerous studies where carried out since then.
 One of the major areas that benefits from these developments is
 the numerical analysis, as the use of generalized Riemann derivatives
 leads to solving a wider class of problems that are not solvable
 with the classical tools. This article studies the generalized Riemann
 derivative and its properties and establishes relationships between
 Riemann generalized derivative and the classical one.
 The existence of classical derivative implies the existence of the
 Riemann generalized derivative, and we study conditions necessary
 for the generalized Riemann derivative to imply the existence of
 the classical derivative. Furthermore, we provide conditions on
 the generalized Riemann derivative that are sufficient for the existence
 of the classical derivative.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

Marshall Ash initiated  the study of
generalized Riemann derivative in his thesis \cite{ref2} in In 1966.
Urged by Zygmund and starting
from his papers \cite{ref19,ref20,ref29}, Ash begun his studies with
symmetric derivative and Schwarz derivative of second order:
Given an interval $I$ of real numbers, $x\in \operatorname{Int}(I)$ and
$f:I\to\mathbb{R}$ a function, then for all $h\in \mathbb{R}\setminus \{0\}$
such that $x-h \in I$ and $x+h \in I$ we define the following ratios:
$$
R_1 f(x,h)=\frac{f\big( x+\frac{h}{2}\big)
-f\big( x-\frac{h}{2}\big)}{h},\quad
 R_2 f(x,h)=\frac{f(x+h)-2f(x)+f(x-h)}{h^2}.
$$
If the limit $ \lim_{h\to 0}R_1 f(x,h)=R_1 f(x)$ exists and is finite,
then $R_1 f(x)$ is known as the symmetric derivative of $f$ at $x$, cf.
\cite{ref7,ref10}.
If the limit $ \lim_{h\to 0} R_2 f(x,h)=R_2 f(x)$ exists and is finite,
then $R_2 f(x)$ is known as the Schwarz derivative of $f$ at $x$, cf.
\cite{ref21,ref22}.
Both derivatives have important applications in trigonometrical series
theory and in numerical analysis, see \cite{ref18,ref25,ref26,ref25,ref28}.
A natural generalization of these two derivatives is the generalized Riemann
derivative of order $r$ of a function $f$ at a point $x$.
We consider $a_i$, $b_i$, $i=1,2,\dots ,n$ real numbers and suppose
 that the following conditions of consistency  are satisfied:
\begin{equation} \label{e1}
 \sum_{i=1}^n a_i b_i^k=
\begin{cases}
0, & k=0,1,\dots ,r-1 \\
1, & k=r.
\end{cases}
\end{equation}
If we have  $k=0$ and $b_i=0$ in relation \eqref{e1}, then we denote
 by $b_i^k=1$.
Further we consider the  ratio
\[
R_r^{a,b}f(x,h)= \frac{\sum_{i=1}^n a_i f(x+b_i h)}{h^r}.
\]
In the case when
\[
 \lim_{h\to 0} R_r^{a,b}f(x,h)=R_r^{a,b}f(x) 
\]
exists and is finite, 
we say that the function $f$ is generalized Riemann differentiable of order
$r$ at $x$.

We denote by $D_r^{a,b}f(x)=r!\cdot R_r^{a,b}f(x)$ and we say that
$D_r^{a,b}f(x)$ is the generalized Riemann derivative of order $r$
of function $f$ at $x$.
One can easily observe that if $f$ is a function differentiable of order
$r$ in classical sense then $f$ is generalized Riemann differentiable of
 order $r$ and the two derivatives are equal. The converse does not hold.
More generally if $f$ is Peano differentiable of order $r$, then $f$ is
generalized Riemann differentiable of order $r$. The converse does not hold.

Among the most important contributions to the generalized Riemann derivative
are those of Humke, Laczkovich and  Mukhopadhyay in
\cite{ref16,ref17,ref Muk}.
In \cite{ref2}--\cite{ref6} and \cite{ref8}, Ash gives a number of problems
 linked to generalized Riemann derivative. These papers deals with the
applications of generalized Riemann derivative to some uniqueness theorems
in trigonometric series theory. It can be noted that by replacing the
classical derivative with the generalized Riemann derivative in the
process of solving ordinary differential equations, the resulting
solutions are no longer differentiable in the classical sense.
Such solutions are known in ordinary differential equations theory
as weak solutions. Therefore, it is necessary to study the system of
parameters $a_i, b_i$, $i=1,\dots ,n$ that satisfy conditions of consistency
and for which the generalized Riemann derivative coincides with the
classical derivative.

It can also be noticed that the speed of convergence of numerical scheme
associated to differential equation depends essentially on the type
of generalized Riemann derivative and consequently on the parameters
$a_i$, $b_i$, $i=1,\dots ,n$.
In the following section we study the links between generalized
Riemann derivative and classical derivative. We give conditions in
 which the existence of generalized Riemann derivative implies
the existence of classical derivative. We give sufficient  conditions
on the system of vectors $(a,b)$ that define generalized Riemann derivative,
such that any function which is generalized Riemann differentiable is
also classical differentiable.

\section{$(\sigma, \tau)$-Riemann differentiable functions}

We study further a new class of generalized differentiable functions
- the functions $(\sigma,\tau)$-Riemann differentiable.
The motivation for this definition is that we mark out
the system $(\sigma,\tau)$ in $\mathbb{K}^n\times \mathbb{K}^n$
(where $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$), on which we give
conditions such that a function $f$ is generalized Riemann differentiable
and such that a series of theorems hold.

Denote by $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let
$G\subset \mathbb{K}$ an open subset and the function $f:G\to \mathbb{K}$.
For $p, n\in \mathbb{N}^*, p\le n$, define the set:
\begin{align*}
L(p,n,\mathbb{K})
:=\Big\{&(\sigma,\tau)\in\mathbb{K}^{*n}\times\mathbb{K}^n :
  \sum_{k=1}^n\sigma_k\tau_k^j=0, \text{ for }
j\in\{0,1,\dots ,p-1\}, \\
& \sum_{k=1}^n\sigma_k \tau_k^p=p! \text{ and $\tau$ has all the components
distinct}\Big\}.
\end{align*}

 \begin{definition} \label{def2.1}\rm
 The function $f$ is $(\sigma,\tau)$-differentiable of order $p$
 at $x$ ($x$ in $G$) if $(\sigma,\tau)\in L(p,n,\mathbb{K})$ and
if the following limit exists and belongs to $\mathbb{K}$:
\[
\lim_{h\to 0}\frac{1}{h^p} \sum_{k=1}^n\sigma_k f(x+\tau_k h).
\]
If the limit above exists, we denote it by $D_p(\sigma,\tau)f(x)$.
\end{definition}

 \begin{remark} \label{rmk2.2}\rm
If $G\subset \mathbb{K}$ is an open set, $(\sigma,\tau)\in L(p,n,\mathbb{K})$,
$f:G\to \mathbb{K}$ is $(\sigma, \tau)$-differentiable at $x$, ($x\in G$),
then for all $\lambda\in K$ the function $\lambda f$ is
$(\sigma,\tau)$-differentiable at $x$ and the following relation holds:
\[
 D_p(\sigma,\tau)(\lambda f)(x)=\lambda D_p (\sigma,\tau)f(x).
\]
\end{remark}

\begin{remark} \label{rmk2.3} \rm
If $f$, $g:G\to\mathbb{K}$ are two functions $(\sigma,\tau)$ differentiable
at $x$, then the function $f+g$ is $(\sigma, \tau)$ differentiable at
 $x$ and the following relation holds:
\[
 D_p(\sigma,\tau)(f+g)(x)=D_p(\sigma,\tau)f(x)+D_p(\sigma,\tau)g(x).
\]
 The proof is straightforward and we let the reader to complete it, if needed.
\end{remark}

\begin{remark} \label{rmk2.4}\rm
 Let $G\subset \mathbb{R}$ with $\operatorname{Int} G\neq\emptyset$ and
 $x\in \operatorname{Int} G$, $f:G\to \mathbb{R}$ and
$(\sigma,\tau)\in L(p,n,\mathbb{R})$. If $f$ is differentiable of order
$p$ at $x$ in the classic sense, then $D_p(\sigma,\tau)f(x)$ exists and
these two derivatives are equal:
\[
D_p(\sigma,\tau)f(x)=f^{(p)}(x).
\]
\end{remark}

 \begin{theorem} \label{thm2.5}
Let $G\subset\mathbb{K}$ such that $\operatorname{Int} G\neq\emptyset$,
$x\in \operatorname{Int} G$, $f:G\to \mathbb{K}$ and
$(\sigma',\tau')\in L(p,m,\mathbb{K})$,
$(\sigma'',\tau'')\in L(p,n,\mathbb{K})$. If $D_p(\sigma',\tau')f(x)$
and $D_p(\sigma'',\tau'')f(x)$ exist and belong to $\mathbb{K}$,
then they are equal.
\end{theorem}

\begin{proof} Using Definition \ref{def2.1}, we obtain
\begin{align*}
D_p(\sigma',\tau')f(x)
&=\frac{1}{p!}p!D_p(\sigma',\tau')f(x)\\
&=\frac{1}{p!}\sum_{j=1}^n\sigma_j''\tau_j''^pD_p(\sigma',\tau')f(x)\\
&=\frac{1}{p!}\sum_{j=1}^n\sigma_j''\tau_j''^p \lim_{j\to 0}
 \frac{1}{h^p}\sum_{k=1}^m \sigma_k'f(x+\tau_k'h)\\
&=\frac{1}{p!}\sum_{j=1}^n\sigma_j''\tau_j''^p \lim_{h\to 0} 
 \frac{1}{(\tau_j''h)^p}\sum_{k=1}^m\sigma_k'f(x+\tau_k'\tau_k''h)\\
&=\frac{1}{p!}\lim_{h\to 0} \frac{1}{h^p} \sum_{j=1}^n \sum_{k=1}^m 
 \sigma_j''\tau_j''^p \sigma_k' \frac{1}{\tau_j''^p}f(x+\tau_k'\tau_j''h)\\
&=\frac{1}{p!}\lim_{h\to 0} \frac{1}{h^p} \sum_{j=1}^n \sum_{k=1}^m 
 \sigma_j''\sigma_k' f(x+\tau_k'\tau_j''h)\\
&=\frac{1}{p!}  \lim_{h\to 0} \sum_{k=1}^m \sigma_k' \sum_{j=1}^n 
 \sigma_j'' f(x+\tau_k'\tau_j''h)\\
&=\frac{1}{p!} \sum_{k=1}^m\sigma_k'\tau_k'^p \lim_{h\to 0} 
 \frac{1}{(h\tau_k')^p}\sum_{j=1}^n \sigma_j'' f(x+\tau_j''(\tau_k'h))\\
&=\frac{1}{p!} \sum_{k=1}^m \sigma_k'\tau_k'^p D_p (\sigma'',\tau'')f(x)\\
&=\frac{1}{p!} p! D_p(\sigma'',\tau'')f(x)=D_p(\sigma'',\tau'')f(x).
\end{align*}
\end{proof}

This theorem shows that if a function is differentiable in the generalized 
sense of order $p$ in relation with two systems $(\sigma', \tau')$ 
and $(\sigma'',\tau'')$, then these two generalized derivatives of order 
$p$ are equal.
We can formulate with the aid of notion of divided difference, the 
$(\sigma,\tau)$ derivative. 

\begin{remark} \label{rmk2.6} 
Let $f:G\to\mathbb{K}$, $G\subset\mathbb{K}$ open set. 

1. Then $f$ is $(\sigma,\tau)$ differentiable if and only if 
the following limit exists:
\[
 \lim_{h\to 0}\sum_{j=1}^n \sigma_j \tau_j [x+\tau_j h,x;f]=D_1(\sigma,\tau)f(x).
\]
with $(\sigma,\tau)\in L(1,n,\mathbb{K})$.

2. If $p=n-1$ and $(\sigma,\tau)\in L(p,n,\mathbb{K})$ and $G\subset K$
 open set and $f:G\to\mathbb{K}$, then
\[
 D_p(\sigma,\tau)f(x)=(n-1)!\lim_{h\to 0}[x+\tau_1 h, x+\tau_2 h,
\dots ,x+\tau_n h;f]. 
\]
\end{remark}
The following theorem gives conditions on a
 $(\sigma, \tau)$-differentiable  function at a point $x$ and on the system 
of vectors $(\sigma, \tau)$ for such  function to become 
differentiable in the classical sense and such that the two derivatives 
to be equal.

\begin{theorem} \label{thm2.7} 
Let $G\subset \mathbb{R}$, $(\sigma,\tau)\in L(1,n,\mathbb{R})$,
 $x\in \operatorname{Int} G\neq\emptyset$ and $f:G\to \mathbb{R}$. 
Suppose that the following conditions hold:
\begin{itemize}
\item[(i)] $f$ is $(\sigma,\tau)$-differentiable at $x$;
\item[(ii)] there exist left and right derivatives $f_l'(x)$ and 
 $f_r'(x)$ and are finite;
\item[(iii)] $ \sum_{\tau_k<0}\sigma_k\tau_k\neq \sum_{\tau_k>0}\sigma_k\tau_k$.
\end{itemize}
Under these conditions $f$ is differentiable at $x$ and 
$D_1(\sigma,\tau)f(x)=f'(x)$.
\end{theorem}

\begin{proof}
On the one hand,  by the definition of $(\sigma,\tau)$-derivative of
 $f$ we obtain:
\begin{equation} \label{e2.1}
\begin{aligned}
&D_1(\sigma,\tau)f(x)\\
&=\lim_{h\to 0, h>0}\frac{1}{h}\sum_{k=1}^n\sigma_k f(x+\tau_k h)\\
&=\lim_{h\to 0, h>0}\frac{1}{h}\sum_{k=1}^n\sigma_k [f(x+\tau_k h)-f(x)]\\
&= \lim_{h\to 0, h>0,\tau_k\neq 0} \sum_{k=1}^n \sigma_k\tau_k
 \frac{f(x+\tau_k h)-f(x)}{\tau_k h}\\
&= \lim_{h\to 0, h>0} \Big\{ \sum_{\tau_k>0}^n\sigma_k \tau_k
 \frac{f(x+\tau_k h)-f(x)}{\tau_k h}
 +\sum_{\tau_k<0} \sigma_k \tau_k\frac{f(x+\tau_k h)-f(x)}{\tau_k h} \Big\} \\
&= \Big( \sum_{\tau_k>0}\sigma_k \tau_k\Big)f'_r(x)
 +\Big( \sum_{\tau_k<0}\sigma_k\tau_k\Big)f'_l(x).
\end{aligned}
\end{equation}
On the other hand, we have
\begin{equation}
\begin{aligned}
& D_1(\sigma,\tau)f(x)\\
&=\lim_{h\to0, h<0}\frac{1}{h}\sum_{k=1}^n\sigma_kf(x+\tau_k h)\\
&=\lim_{h'\to 0, h'>0}\frac{1}{-h'}\sum_{k=1}^n\sigma_k f(x-\tau_k h')\\
&=\lim_{h'\to 0, h'>0}\sum_{k=1}^n \sigma_k \tau_k
 \frac{f(x-\tau_k h')-f(x)}{-\tau_k h'}\\
&=\lim_{h'\to 0, h'>0}\Big\{ \sum_{\tau_k>0}\sigma_k \tau_k
 \frac{f(x-\tau_k h')-f(x)}{-\tau_k h'}+\sum_{\tau_k<0}\sigma_k\tau_k
  \frac{f(x-\tau_k h')-f(x)}{-\tau_k h'}\Big\}\\
&= f_l'(x)\sum_{\tau_k>0}\sigma_k\tau_k+f'_r(x)\sum_{\tau_k<0}\sigma_k \tau_k.
\end{aligned} \label{e2.2}
\end{equation}
From \eqref{e2.1} and \eqref{e2.2} we obtain
\[
\Big( \sum_{\tau_k>0}\sigma_k \tau_k-\sum_{\tau_k<0}
\sigma_k \tau_k\Big)(f_r'(x)-f_l'(x))=0.
\]
Taking into account (iii) it follows that:
 $f_l'(x)=f_r'(x)=f'(x)$.
 This lead us to conclude that $f$ is differentiable at $x$.
In addition we have
\[
 D_1 (\sigma, \tau)f(x)=f'(x)\Big[ \sum_{\tau_k>0}\sigma_k\tau_k
+\sum_{\tau_k<0}\sigma_k\tau_k\Big]=f'(x)\sum_{k=1}^n\sigma_k\tau_k=f'(x).
\]
\end{proof}


\subsection*{Remarks}
(1) If $G$ is an interval, $f:G\to \mathbb{R}$ and conditions 
(i) and (ii) from Theorem \ref{thm2.7} hold, and 
\begin{itemize}
\item[(iii')] $ \sum_{\tau_k<0}\sigma_k\tau_k=\sum_{\tau_k>0}\sigma_k \tau_k$,
\end{itemize}
then
\[
 D_1(\sigma,\tau)f(x)=\frac{1}{2}[f'_l(x)+f'_r(x)].
\]

(2) Condition (iii') is very important because if the system $(\sigma, \tau)$ 
satisfies it, then a large class of non-differentiable functions at a 
point or on a finite set becomes $(\sigma, \tau)$-differentiable.
Let us consider the  function $f:\mathbb{R}\to\mathbb{R}$,
\[
f(x)=\begin{cases}
\alpha x, & x<0\\
\beta x, & x\ge 0
\end{cases}
\]
for $\alpha, \beta\in\mathbb{R}^*$, $\alpha\neq\beta$. 
This function is not differentiable at $x=0$. However we have
\begin{align*}
(D_1(\sigma,\tau)f)_r(x)
&=\lim_{h\to 0, h>0} \frac{1}{h}\sum_{k=1}^n\sigma_k f(\tau_k h)\\
&=\lim_{h\to 0,h>0}\frac{1}{h} \Big[ \sum_{\tau_k>0}\sigma_k \beta \tau_k h
+\sum_{\tau_k<0}\sigma_k\alpha\tau_k h\Big]\\
&=\beta\sum_{\tau_k>0} \sigma_k \tau_k +\alpha \sum_{\tau_k<0}\sigma_k\tau_k.
\end{align*}
For
\[
\sum_{\tau_k>0}\sigma_k\tau_k =\sum_{\tau_k<0}\sigma_k \tau_k=\frac{1}{2},
\]
 we obtain
\[
 (D_1(\sigma,\tau)f)_r(x)=\frac{\alpha+\beta}{2}.
\]
Similarly we can prove that
\[
(D_1(\sigma,\tau)f)_l(x)=\frac{\alpha+\beta}{2}.
\]
We conclude that function $f$ is $(\sigma,\tau)$-differentiable at 
$x=0$ and moreover, is $(\sigma,\tau)$-differentiable on $\mathbb{R}$.

(3) As we can easily observe, condition (iii) from Theorem \ref{thm2.7}
 holds for  the classical derivative. 

The $(\sigma, \tau)$-derivative has a lot of interesting properties 
that shall be further studied. 
We shall give a theorem for the Riemann generalized derivative of 
the product of two functions.

\begin{theorem} \label{thm2.8} 
Let $G\subset \mathbb{K}$ open set, $f$, $g:G\to \mathbb{K}$ and 
$(\sigma,\tau)\in L(1,n,\mathbb{K})$. If the following conditions hold:
\begin{itemize}
\item[(1)] $f$ is Lipschitz and $(\sigma,\tau)$ differentiable;

\item[(2)] $g$ is continuous and $(\sigma,\tau)$ differentiable,
\end{itemize}
then $f\cdot g$ is $(\sigma,\tau)$ differentiable and we have the 
following relation
\begin{equation}
D_1(\sigma,\tau)(f\cdot g)(x)=f(x)\cdot D_1(\sigma,\tau)g(x)
+g(x)\cdot D_1(\sigma,\tau)f(x),\quad x\in G \label{e2.3}
\end{equation}
\end{theorem}

\begin{proof}
 It is easy to show that
\begin{equation}
\begin{aligned}
&\frac{1}{h}\sum_{j=1}^n \sigma_j[f(x+\tau_jh)-f(x)][g(x+\tau_jh)-g(x)]\\
&= \frac{1}{h}\sum_{j=1}^n \sigma_j f(x+\tau_j h)\cdot g(x+\tau_j h)\\
&\quad -f(x)\cdot\frac{1}{h}\sum_{j=1}^n\sigma_j g(x+\tau_jh)-g(x)\cdot
 \frac{1}{h}\sum_{j=1}^n \sigma_j f(x+\tau_j h)\\
&\quad +f(x)\cdot g(x)\cdot \frac{1}{h} \sum_{j=1}^n \sigma_j.
\end{aligned} \label{e2.4}
\end{equation}
for $x\in G$, $h\neq0$, $|h|$ small enough.
From (1)  there exists $L\ge 0$ such that $|f(x)-f(y)|\le L|x-y|$,
for all $x$, $y\in G$.
By applying the modulus,  the left side of  \eqref{e2.4} becomes
\begin{align*}
&\big|\sum_{j=1}^n \sigma_j \frac{f(x+\tau_j h)-f(x)}{h}(g(x+\tau_j h)
-g(x))\big|\\
&\le  \sum_{j=1}^n| \sigma_j|\cdot \big| \frac{f(x+\tau_j h)
-f(x)}{h}\big|\cdot |g(x+\tau_j h)-g(x)|\\
&\le  L\sum_{j=1}^n|\sigma_j|\cdot|\tau_j|\cdot |g(x+\tau_j h)-g(x)|
\end{align*}
 for all $x\in G$, $|h|$ small enough.
Using that $g$ is continuous, the left side of \eqref{e2.4} tends
 to 0 when $h\to 0$. So we obtain
\[
0=D_1(\sigma,\tau)(f\cdot g)(x)-f(x)\cdot D_1(\sigma,\tau)g(x)-g(x)
\cdot D_1(\sigma,\tau)f(x)\quad\text{for }x\in G
\]
which proves completes the proof.
\end{proof}

\begin{corollary} \label{coro2.9} 
Let $G\subset \mathbb{K}$ open set, $f$, $g:G\to \mathbb{K}$ and 
$(\sigma,\tau)\in L(1,n,\mathbb{K})$. If the following conditions hold:\
\begin{itemize}
\item[(1)] $f$ is locally Lipschitz,
\item[(2)] $g$ is continuous and $(\sigma,\tau)$ differentiable,
\end{itemize}
then $f\cdot g$ is $(\sigma,\tau)$ differentiable almost everywhere and
 relation \eqref{e2.3} holds almost everywhere.
\end{corollary}

\begin{proof} We will use that any Lipschitz function is almost everywhere 
differentiable.
 Many properties of the classical derivative correspond to $(\sigma,\tau)$ 
derivative and when these properties does not hold on general case, 
we shall find conditions on $f$ or on the system $(\sigma,\tau)$ such 
as these properties remain available. Such a situation shall be reveal 
in a theorem bellow.
\end{proof}

\subsection*{Notation} 
Let $G\subset\mathbb{K}$ such that $\operatorname{Int} G\neq\emptyset$ , 
$f:G\to \mathbb{K}$, $x\in \operatorname{Int} G$ and $\lambda\in \mathbb{K}$, 
$\lambda\neq -1$. We define:
 \[
 R(\lambda)f(x):=\lim_{h\to 0} \frac{f(x+h)-f(x-\lambda h)}{(1+\lambda)h}
 \]
in the hypotheses in which this limit belongs to $\mathbb{K}$, and we 
say that $f$ is $R(\lambda)$ differentiable at $x$.
 We define:
 \[
  \sigma=\Big( \frac{1}{1+\lambda},-\frac{1}{1+\lambda}\Big), \quad 
\tau=\Big( 1, -\lambda\Big).
 \]
 We notice that $(\sigma,\tau)\in L(1,2,\mathbb{K})$ and 
$R(\lambda)f(x)=D_1(\sigma,\tau)f(x)$.

 \begin{theorem} \label{thm2.11} 
Let $G\subset K$ open set, $x\in G$ and $f,g:G\to \mathbb{K}$, 
$\lambda \in \mathbb{K}$, $\lambda\neq -1$, such that:
\begin{itemize}
\item[(1)] $f,g$ are continuous at $x$;
\item[(2)] $f,g$ are $R(\lambda)$-differentiable at $x$.
\end{itemize}
Under these conditions the function $h=fg$ is $R(\lambda)$ differentiable 
at $x$ and we have the  formula
 \[
 R(\lambda)(fg)(x)=g(x)R(\lambda)f(x)+f(x)R(\lambda)g(x), \quad x\in G.
 \]
\end{theorem}

\begin{proof} The proof  is based on
\begin{align*}
&R(\lambda)(fg)(x)\\
&= \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x-\lambda h)g(x-\lambda h)}{(1+\lambda)h}\\
&= \lim_{h\to 0}\frac{f(x+h)[g(x+h)-g(x-\lambda h)]
 +g(x-\lambda h)[f(x+h)-f(x-\lambda h)]}{(1+\lambda)h}\\
&=f(x)R(\lambda)g(x)+g(x)R(\lambda)f(x).
\end{align*}
\end{proof}

 \subsection*{Remarks}
(1)  For $ p=1$, $n=2$, $\sigma_1 =\frac{1}{2}$, $\sigma_2=-\frac{1}{2}$,
 $\tau_1=1$, $\tau_2=-1$ we obtain the symmetric Riemann derivative.

(2). For $ p=2$, $n=3$, $\sigma_1=1$, $\sigma_2=-2$, $\sigma_3=1$, $\tau_1=1$,
 $\tau_2=0$, $\tau_3=-1$ we find the Schwarz derivative.

(3) It is also easy to notice that $R(\lambda)$ extend the symmetric
 Riemann derivative, which is $R(1)$ and the classical derivative,
 which is $R(0)$.

 In conclusion, the set of $(\sigma,\tau)$-differentiable functions 
is larger than the set of classical differentiable functions.

If $(\sigma,\tau)\in L(p,n,\mathbb{K})$ and $G$ an open set, 
then we denote 
\[
\mathfrak{T}_p(\sigma,\tau)(G,\mathbb{K})=\{f:G\to \mathbb{K}| f
\text{ is $(\sigma,\tau)$-differentiable everywhere on $G$}\}.
\]

\begin{theorem} \label{thm2.12}
 If $(\sigma,\tau)\in L(p,n,\mathbb{K})$ and if any function from
 $\mathfrak{T}_p(\sigma,\tau)(G,\mathbb{K})$ is $p$ times differentiable 
in classical sense, then there exists $j\in \{1,2,\dots ,n\}$ such that 
$\tau_j=0$.
\end{theorem}

\begin{proof} 
Indeed, if $\tau_j\neq0$, for all $j\in \{1,2,\dots ,n\}$ we prove that 
there exist a function in $\mathfrak{T}_p(\sigma,\tau)(G,\mathbb{K})$ 
which is $(\sigma, \tau)$ differentiable and does not have the property 
in the theorem. This function is $f:G\to\mathbb{K}$,
\[ f(x) = \begin{cases}
      1 & \text{if }x=a \\
      0 & \text{if }x\in G\setminus \{a\}. 
   \end{cases} 
\]
This function is discontinuous at $a$ and consequently is not differentiable 
in the classical sense at $a$. However
$D(\sigma, \tau)f(x)=0$
and this leads us to a contradiction.
\end{proof}

More properties of $(\sigma,\tau)$-Riemann differentiable functions were 
studied  in \cite{ref1, ref14,ref27}.

\section{Main Results}

If we replace the classical derivative of the function  with the Riemann 
generalized derivative in the process of solving numerical ordinary 
differential equations, we shall obtain new solutions that are not 
differentiable in the classical sense (see cite{ref9,ref12}). 
This kind of solutions are well known in ordinary differential equations
 theory as weak solutions. Therefore is needed to study the system of 
parameters $(\sigma,\tau)$ that satisfy the consistency conditions 
and for which, the Riemann generalized derivative which is defined by 
this system of parameters, is equal to classical derivative. 

The speed of convergence of the numerical scheme associated to the ordinary
 differential equation, depends on the type of derivative, consequently 
on the system of parameters $(\sigma,\tau)$.
We shall further give some conditions in which a $(\sigma,\tau)$-differentiable 
function is classical differentiable.
 
 \begin{lemma} \label{lem3.1} 
Let $a$, $b\in\mathbb{K}$ such that $|a| \le |b|<1$ and the function 
$\varphi :\mathbb{K}\to \mathbb{K}$. Suppose that the conditions below hold:
\begin{itemize}
\item[(i)] $ \lim_{x\to 0} x\varphi(x)=0$;

\item[(ii)] $\lim_{x\to 0}[\varphi(x)-a\varphi(bx)]=0$, $x\in \mathbb{K}$.
\end{itemize}
 Then  $ \lim_{x\to 0}\varphi (x)=0$.
\end{lemma}

\begin{proof}
 If $b=0$ results that $a=0$ and the statement is available in this case.
 For $b$ in $\mathbb{K}^*$, we denote 
 \begin{equation}
  g(x):=\varphi(x)-a\varphi(bx), \quad x\in \mathbb{K} \label{e3.1}.
 \end{equation}
This condition is equivalent to $\varphi(x)=g(x)+a\varphi(bx)$,
$ x\in \mathbb{K}$.
 Repeating the transformation: $x\to bx$, we obtain
\begin{gather*}
\varphi(bx)=g(bx)+a\varphi(b^2x), \quad x\in \mathbb{K}\\
\varphi(b^2x)=g(b^2x)+a\varphi(b^3x), \quad x\in \mathbb{K}\\
\dots \\
\varphi(b^{n-1}x)=g(b^{n-1}x)+a\varphi(b^nx), \quad x\in \mathbb{K}, \;
n\in \mathbb{N}^*.
\end{gather*}
It follows that
\[
\varphi(x)=g(x)+ag(bx)+a^2g(b^2x)+\dots +a^{n-1}g(b^{n-1}x)+a^{n}\varphi(b^{n}x),
\]
for $ x\in \mathbb{K}$, $n\in\mathbb{N}^*$.
From (1) and (ii), we have
$ \lim_{x\to 0}g(x)=0$.
For $0<|x|<r$, $n\in\mathbb{N}^*$ we obtain
\begin{align*}
|\varphi(x)|
&\le  \sup_{0<|y|\le r}|g(y)|+|a|\sup_{0<|y|\le r}|g(by)|+\dots\\
&\quad +|a|^{n-1}\sup_{0<|y|\le r}|g(b^{n-1}y)|+|\frac{a}{b}|^n\cdot
  \frac{1}{|x|}\cdot |(b^nx)\varphi(b^nx)|.
\end{align*}
For $\epsilon >0$ we consider $V_{\epsilon}$ - the disc of radius
 $\epsilon$ centered in $0$ such that for any $x\in V_r\setminus\{0\}$ we have
\[
|\varphi(x)|\le \frac{1}{1-|a|}\cdot \sup_{0<|y|\le r} |g(y)|
+\frac{1}{|x|}\cdot \overline{\lim_{n\to\infty}}|(b^n x)\varphi(b^nx)|.
\]
As a consequence we obtain
\[
|\varphi(x)|\le \frac{1}{1-|a|}\cdot \sup_{0<|y|\le r}|g(y)|,\quad
\text{for $x$ in }V_r\setminus\{0\}
\]
and because  $ \lim_{r\to 0}\Big( \sup_{0<|y|\le r}|g(y)|\Big)=0$
we conclude that
$ \lim_{x\to 0}\varphi (x)=0$.
\end{proof}

We remark that Lemma \ref{lem3.1} remains true if we have the condition below
 instead of condition (ii):
\begin{itemize}
\item[(ii')] $  \lim_{x\to 0}[\varphi(x)+a\varphi(bx)]=0$, $x\in \mathbb{K}$.
\end{itemize}
In this chase the auxiliary function is
$g(x):=\varphi(x)+a\varphi(bx)$, $\quad x\in \mathbb{K}$,
and 
\[
\varphi (x)=g(x)-ag(bx)+a^2g(b^2x)-a^3g(b^3x)+\dots
 +(-a)^{n-1}g(b^{n-1}x)+(-1)^n a^n\varphi (b^nx),
\]
for $x\in \mathbb{K}$ and $n\in \mathbb{N}^*$.

\begin{lemma} \label{lem3.2} 
Let $a$, $b\in \mathbb{K}$ such that $|a|\ge|b|>1$ and 
$\varphi :\mathbb{K}\to \mathbb{K}$ for which the following conditions hold:
\begin{itemize}
\item[(i)] $ \lim_{x\to 0} x\varphi(x)=0$,
\item[(ii)] $ \lim_{x\to 0} [\varphi (x)+a \varphi (bx)]=0$.
\end{itemize}
Then $ \lim_{x\to 0}\varphi(x)=0$.
\end{lemma}

\begin{proof} To prove this we shall use Lemma \ref{lem3.1} with transformations:
 $ x=y/b=b'y$, $b\in \mathbb{K}^*$, $ b'=1/b$, 
$ a'=1/a$, $a'\in \mathbb{K}^*$ with $1>|b'|\ge|a'|$.
For the case $b=0$ results that $a=0$ and condition (ii) validate the statement.
\end{proof}

\begin{lemma} \label{lem3.3}
 Let $a=(a_1,a_2,\dots ,a_n)\in \mathbb{K}^n$, 
$b=(b_1,b_2,\dots ,b_n)\in \mathbb{K}^n$ with $|a_j|\le|b_j|<1$, 
for all $j\in \{1,2,\dots ,n\}$ and the function 
$\varphi:\mathbb{K}\to\mathbb{K}$ with the properties:
\begin{itemize}
\item[(i)] $ \lim_{x\to 0}x\varphi(x)=0$;
\item[(ii)] 
\begin{align*}
&\lim_{x\to 0}\Big\{ \varphi(x)+\sum_{j=1}^n a_j \varphi(b_j x)
+\sum_{j<k}a_j a_k \varphi(b_jb_k x)+\dots \\
&+a_1a_2\dots a_n \varphi (b_1\dots b_n x)\Big\}=0.
\end{align*}
\end{itemize}
Then $ \lim_{x\to 0}\varphi(x)=0$.
\end{lemma}

\begin{proof} 
We shall repeatedly use Lemma \ref{lem3.1} with condition (ii'). 
For $1\le j\le n$ we define
\[
(L_j\varphi)(x):=\varphi(x)+a_j\varphi(b_j x), \quad x\in \mathbb{K}
\]
and hence
$ \lim_{x\to 0}(L_j\varphi)(x)=0$ which implies 
$ \lim_{x\to 0}\varphi(x)=0$, for $j\in \{1,\dots,n\}$.
Using the  relation
\[
(L_j L_k\varphi)(x)=\varphi (x)+[a_k\varphi (b_kx)
+a_j\varphi(b_j x)]+a_j a_k\varphi(b_jb_kx),
\]
for  $1\le j$, $k\le n$, $x\in \mathbb{K}$.
By induction we obtain
\begin{align*}
(L_1 L_2\dots L_n\varphi)(x)
&=\varphi (x)+ \sum_{j=1}^n a_j\varphi (b_j x)
+\sum_{j<k}a_ja_k\varphi(b_jb_kx)+\dots\\
 &\quad  +a_1a_2\dots a_n\varphi(b_1\dots b_nx),
\end{align*}
for  $x\in \mathbb{K}$.
Therefore, the proof has the following logical scheme:
\[
\lim_{x\to 0}(L_jL_k\varphi)(x)=\lim_{x\to 0}L_j (L_k\varphi(x))=0
\]
implies
\[
\lim_{x\to 0}(L_k\varphi)(x)=0\Rightarrow 
\lim_{x\to 0}\varphi(x)=0,\quad j,k\in \{1,\dots ,n\}.
\]
This completes the proof.
\end{proof}

\begin{lemma} \label{lem3.4} Let the polynomial function
\[
P(x)=\alpha_0+\alpha_1 x+\dots +\alpha _n x^n\in \mathbb{K}[x]
\]
with roots $(x_j)_{1\le j\le n}$, that have the property $|x_j|<1$, 
for all $j\in \{1,\dots ,n\}$.
Let $\varphi :\mathbb{K}\to\mathbb{K}$ satisfy the following conditions:
\begin{itemize}
\item[(1)] $ \lim_{x\to 0}x\varphi(x)=0$;
\item[(2)] $ \lim_{x\to 0}\sum_{k=0}^n(-1)^k\alpha_k\varphi(b^k x)=0$ 
where $ \max_{1 \le j\le n}|x_j|\le |b|<1$.
\end{itemize}
Then $ \lim_{x\to 0}\varphi(x)=0$.
\end{lemma}

For the proof of the above lemma, it is sufficient to consider 
$b_1=b_2=\dots =b_n=b$, $a_j=x_j$ 
in Lemma \ref{lem3.3} and to take into account Vi\`ete relations.


\begin{lemma} \label{lem3.5} 
Let $a_k$, $b_k\in \mathbb{K}^*$, $k\in \{1,2,\dots ,n\}$ and the 
function $\varphi :\mathbb{K}\to\mathbb{K}$ with the following properties:
\begin{itemize}
\item[(1)] $\varphi$ is bounded on a neighborhood of the origin;
\item[(2)] $ \sum_{j=1}^n |a_j|<1 $;
\item[(3)] $b= \max_{1\le j\le n}|b_j|\le 1$;
\item[(4)] $ \lim_{x\to 0}[\varphi(x)-\sum_{j=1}^na_j\varphi(b_j x)]=0$.
\end{itemize}
Then $ \lim_{x\to 0}\varphi(x)=0$.
\end{lemma}

\begin{proof} We define the function
\[
g(x):=\varphi(x)- \sum_{j=1}^na_j\varphi(b_j x), \quad\text{for }x\in \mathbb{K}.
\]
Note that $ \lim_{x\to 0}g(x)=0$.
 We define the operator $L:\mathbb{K}^{\mathbb{K}}\to \mathbb{K}^{\mathbb{K}}$,
\[
L\psi(x):= \sum_{j=1}^n a_j\psi(b_j x), \quad 
\psi\in \mathbb{K}^{\mathbb{K}}, x\in \mathbb{K}.
\]
Note that the operator $L$ is linear.
We can write: $\varphi -L\varphi =g$ as $\varphi=L \varphi +g$ and iterate to
obtain
$L\varphi=L^2\varphi+Lg$, but $L\varphi=\varphi-g$, which implies
\[
\varphi=L^2\varphi+Lg+g.
\]
By induction we obtain
\[
\varphi=L^k\varphi+L^{k-1}g+\dots +Lg+g, \quad k\ge 1.
\]
As $L\varphi(x)= \sum_{j=1}^n a_j\varphi(b_j x)$, $x\in \mathbb{K}$,  
we deduce that
\[
|L\varphi(x)|\le \sum_{j=1}^n|a_j|\cdot |\varphi(b_j x)|, \quad
 x\in \mathbb{K}.
\]
Denote 
\[
\widetilde{\varphi}(r):= \sup_{0<|x|\le r}|\varphi(x)|, \quad r\in (0,\infty).
\]
Then we have
\[
\widetilde {L\varphi}(r)\le  \Big( \sum_{j=1}^n |a_j|\Big)\widetilde{\varphi}(br), 
\quad r\in  (0,\infty).
\]
In the same manner we obtain
 \[
 \big(\widetilde{L^2\varphi}\big)(r)\le \Big(\sum_{j=1}^n |a_j|\Big)
\widetilde{L\varphi}(br)
\le \Big(\sum_{j=1}^n|a_j|\Big)^2\widetilde{\varphi}(b^2r),\quad 
r\in (0,\infty)
\]
and similarly,
\[
\Big( \widetilde{L^k \varphi}\Big)(r)
\le\Big(  \sum_{j=1}^n|a_j|^k\Big)\widetilde{\varphi}(b^k r), \quad
 r\in (0,\infty),\; k\in \mathbb{N}^*.
\]
If we denote  $a:=\sum_{j=1}^n|a_j|$, then
\[
\widetilde{L^k\varphi}(r)\le a^k\widetilde{\varphi}(b^k r), 
\quad r\in (0,\infty), \; k\in\mathbb{N}^*.
\]
If follows that
\begin{align*}
\widetilde{\varphi}(r)
&\le\widetilde{L^k\varphi}(r)
 + \widetilde{L^{k-1} g}(r)+\dots +\widetilde{Lg}(r)+\widetilde{g}(r)\\
&\le  a^k\widetilde{\varphi}(r)+a^{k-1}\widetilde{g}(r)+\dots 
 +a\widetilde{g}(r)+\widetilde{g}(r)\\
&\le a^k\widetilde{\varphi}(r) +\frac{1}{1-a}\widetilde{g}(r).
\end{align*}
The above relation holds for all $k\in\mathbb{N}^*$ and under hypotheses
 (1) and (2), that is  $\varphi$ is bounded on a neighborhood of 
the origin and $a\in (0,1)$. 
This implies that
\[
\widetilde{\varphi}(r)\le \frac{1}{1-a}\widetilde{g}(r),\quad r\in (0,\infty).
\]
As $ \lim_{r\to 0} \widetilde{g}(r)=0$, it follows that 
$\lim_{r\to 0}\widetilde{\varphi}(r)=0$ and in conclusion we obtain
$ \lim_{x\to 0}\varphi(x)=0$.
\end{proof}


\begin{lemma} \label{lem3.6}
 Let $a_k$, $b_k\in \mathbb{K}^*$, $k\in \{1,2,\dots ,n\}$ and the 
functions $\varphi :\mathbb{K}\to\mathbb{K}$ with the following properties:
\begin{itemize}
\item[(1)] $ \lim_{x\to 0}x\varphi(x)=0$;
\item[(2)] $ \sum_{j=1}^n\big|\frac{a_j}{b_j}\big|\le 1$;
\item[(3)] $ b=\max_{j}|b_j|<1$;
\item[(4)] $ \lim_{x\to 0}\big[ \varphi(x)-\sum_{j=1}^n a_j \varphi(b_jx)\big]=0$.
\end{itemize}
Then $ \lim_{x\to 0}\varphi(x)=0$.
\end{lemma}

\begin{proof}
 We define the operator 
$L:\mathbb{K}^{\mathbb{K}}\to \mathbb{K}^{\mathbb{K}}$:
\[
L\psi(x):= \sum_{j=1}^n a_j\psi (b_j x),\quad 
\psi\in \mathbb{K}^{\mathbb{K}},\; x\in \mathbb{K}
\]
and the function $g(x):=\varphi(x)-L\varphi(x)$, $x\in \mathbb{K}$.
Iterating, we have 
\[
\varphi=L^k\varphi+L^{k-1}g+\dots +Lg+g.
\]
This leads us to define for a function $\psi:\mathbb{K}\to\mathbb{K}$ 
bounded on a neighborhood of origin
\[
\widetilde{\psi}(r):= \sup_{0<|x|\le r}|\psi(x)|, \quad r\in (0,\infty).
\]
We notice that $\widetilde{\psi}$ is increasing. Denote 
$u(x):=x\varphi(x)$, $\quad x\in \mathbb{K}$.
Then we can write
\[
|L\varphi(x)|\le \sum_{j=1}^n|a_j|\cdot |\varphi(b_j x)|,\quad x\in \mathbb{K}
\]
and further we have
\[
 |xL\varphi(x)|\le \sum_{j=1}^n \big| \frac{a_j}{b_j}\big|
\cdot |b_j x\varphi(b_j x)|=\sum_{j=1}^n \big|\frac{a_j}{b_j}\big|\cdot|u(b_j x)|, 
\quad x\in \mathbb{K}.
\]
Similarly, we obtain
\[
|xL^2 \varphi (x)|\le  \sum_{j=1}^n \big|\frac{a_j}{b_j}\big|\cdot |b_j x L\varphi (b_j x)|\le
\sum_{j,k=1}^n\big|\frac{a_j}{b_j}\big|\cdot \big|\frac{a_k}{b_k}\big|
\cdot |b_j b_k x\varphi(b_j b_k)|, \quad x\in \mathbb{K}.
\]
As
\[
|L^2\varphi(x)|\le \frac{1}{|x|}\sum_{j,k=1}^n\big|\frac{a_j}{b_j}\big| 
\cdot\big|\frac{a_k}{b_k}\big|\cdot |u(b_jb_kx)|, \quad x\in \mathbb{K}^*,
\]
we obtain
\[
|L^k\varphi(x)|\le \frac{1}{|x|} \sum_{j_1, \dots ,j_k}
\big| \frac{a_{j_1}}{b_{j_1}}\big|\cdot \dots \cdot 
\big| \frac{a_{j_k}}{b_{j_k}}\big|\cdot |u(b_{j_1}\dots b_{j_k}x)|, 
\quad x\in \mathbb{K}^*, k\in \mathbb{N}^*
\]
So we have
\[
 |L^k\varphi (x)|
\le \frac{1}{|x|} \Big ( \sum_{j=1}^n\big|\frac{a_j}{b_j}\big|\Big)^k 
|\widetilde{u}(b^k|x|)|
\le \frac{1}{|x|}\widetilde{u}(b^k|x|), \quad x\in \mathbb{K}^*
\]
where $a:= \sum_{j=1}^n |a_j|$, 
$ a=\sum_{j=1}^n\big|\frac{a_j}{b_j}\big|\cdot |b_j|\le \sum_{j=1}^n |b_j|=b<1$.
Then we have
\begin{gather*}
|Lg(x)|\le  \Big( \sum_{j=1}^n |a_j|\Big)\widetilde{g} (b|x|),\quad
 x\in \mathbb{K}^*, \\
|\widetilde{L^k g}(r)|\le \Big( \sum_{j=1}^n |a_j|\Big)^k\widetilde{g}(b^k r),
\quad r\in (0,\infty),\; k\in \mathbb{N}^*.
\end{gather*}
As
\[
|\varphi (x)|\le  \frac{1}{|x|} \cdot \widetilde{g}(b^k|x|)
+\frac{1}{1-a} \widetilde{g}(|x|), \quad x\in \mathbb{K}^*, k\in\mathbb{N}^*,
\]
 letting $k\to \infty$ we obtain
\[
|\varphi(x)|\le  \frac{1}{1-a}\widetilde{g}(|x|),\quad x\in \mathbb{K}^*.
\]
We can now conclude that $ \lim_{x\to 0}\varphi (x)=0$.
\end{proof}

From Lemma \ref{lem3.6} can be easy obtained the following lemma.

\begin{lemma} \label{lem3.7} 
Let $n\ge 2$, $\varphi:\mathbb{K}\to \mathbb{K}$, $\alpha_j$, 
$\beta_j\in \mathbb{K}^*$, $j\in \{ 1,\dots ,n\}$, such that conditions
 below hold:
\begin{itemize}
\item[(1)] $ \lim_{x\to 0} x\varphi (x)=0$;
\item[(2)] $ \max_{2\le j\le n}|\beta_j|< |\beta_1|$;
\item[(3)] $ \sum_{j=2}^n \big| \frac{\alpha_j}{\beta_j}\big|
 \le \big| \frac{\alpha_1}{\beta_1}\big|$;
\item[(4)] $ \lim_{x\to 0}\sum_{j=1}^n \alpha_j \varphi(\beta_j x)=0$.
\end{itemize}
Then $ \lim_{x\to 0}\varphi(x)=0$.
\end{lemma}

\begin{proof} For this proof, we denote
\[
\frac{\alpha_j}{\alpha_1}=-a_{j-1}, \quad 
\frac{\beta_j}{\beta_1}=b_{j-1}, \quad 
\beta_1 x=t,\quad j\in \{2,\dots ,n\} .
\]
Then
\begin{align*}
 \lim_{x\to 0}\sum_{j=1}^n \alpha _j \varphi (\beta_j x)
&=\alpha_1 \lim_{x\to 0}\sum_{j=1}^n \frac{\alpha_j}{\alpha_1} 
\varphi \Big( \frac{\beta_j}{\beta_1}\beta_1 x\Big)\\
&= \alpha_1 \lim_{t\to 0}\Big[ \varphi(t)-\sum_{j=2}^n a_{j-1}
 \varphi(b_{j-1}t)\Big]=0.
\end{align*}
 Therefore, condition (4) in Lemma \ref{lem3.6} is verified. Further we have
 \[
 \sum_{j=2}^n \big| \frac{\alpha_j}{\beta_j}\big|
=\sum_{j=2}^n \big| \frac{a_{j-1}\alpha_1}{b_{j-1}\beta_1}\big|
=\big| \frac{\alpha_1}{\beta_1}\big|\sum_{k=1}^{n-1}\big|
\frac{a_k}{b_k}\big|<\big| \frac{\alpha_1}{\beta_1}\big|
 \]
if and only if $ \sum_{k=1}^{n-1}\big|\frac{a_k}{b_k}\big|<1$. Also
  \[
 \max_{2\le j\le n}|\beta_j|=\max_{2\le j\le n} |\beta_1 b_{j-1}|<|\beta_1|.
 \]
Applying Lemma \ref{lem3.6}, it results that $ \lim_{x\to 0}\varphi(x)=0$.
\end{proof}

\begin{lemma} \label{lem3.8} Let $n\ge 2$, $\varphi:\mathbb{K}\to\mathbb{K}$, 
$\alpha_j$, $\beta_j\in \mathbb{K}^*$, $j\in \{1,2,\dots ,n\}$, 
such that the following conditions hold:
\begin{itemize}
\item[(1)] $\varphi$ is bounded on a neighborhood of the origin;
\item[(2)] $ \sum_{j=2}^n|\alpha_j|<|\alpha_1|$;
\item[(3)] $\beta= \max_{1\le j\le n}|\beta_j|<|\beta_1|$;
\item[(4)] $ \lim_{x\to 0} \sum_{j=1}^n \alpha_j \varphi(\beta_j x)=0$.
\end{itemize}
Then $ \lim_{x\to 0}\varphi(x)=0$.
\end{lemma}

\begin{proof} We proceed analogously as in Lemma \ref{lem3.7} and we make the 
transformations:
\[
\alpha_j=-\alpha_1 a_{j-1}, \quad \beta_j=\beta_1 b_{j-1}, \quad 
\beta_1 x=t, \quad j\in {1,\dots ,n}.
\]
 Further, we notice that the conditions from Lemma \ref{lem3.5} hold. Indeed,
\[
\sum_{j=2}^n|\alpha_j|=\sum_{j=2}^n|a_{j-1}\alpha_1|
=|\alpha_1|\sum_{k=1}^{n-1}|a_k|<|\alpha_1|
\]
if and only if $ \sum_{k=1}^{n-1}|a_k|<1$.
As
\begin{align*}
0&=\lim_{x\to 0}\sum_{j=1}^n \alpha_j\varphi(\beta_jx)\\
 &=\alpha_1\lim_{x\to 0}\sum_{j=1}^n \frac{\alpha_j}{\alpha_1}\varphi
\Big( \frac{\beta_j}{\beta_1}\beta_1 x \Big)\\
&=\alpha_1  \lim_{t\to 0}\Big[ \varphi(t)-\sum_{j=2}^na_{j-1}
\varphi(b_{j-1}t)\Big]\\
&=\alpha_1 \lim_{t\to 0}\Big[ \varphi(t)-\sum_{k=1}^{n-1}a_k\varphi(b_k t)\Big],
\end{align*}
according to Lemma \ref{lem3.5}, it results that $ \lim_{x\to 0}\varphi(x)=0$.
 \end{proof}

The following theorem establishes conditions in which a 
 $(\sigma,\tau)$-differentiable function at a point is classical 
differentiable at that point.

\begin{theorem} \label{thm3.9} 
We consider the function $f:\mathbb{K}\to \mathbb{K}$, $x\in \mathbb{K}$ 
and the numbers $a_j$, $b_j\in \mathbb{K}^*$, $j\in \{ 1,2,\dots ,n\}$ 
such that the following conditions hold:
\begin{itemize}
\item[(1)] $ \sum_{j=1}^n a_j =1$;
\item[(2)] $ \max_{2\le j\le n}|b_j|<|b_1|$;
\item[(3)] $ \sum_{j=2}^n \big|\frac{a_j}{b_j}\big|\le \big|
  \frac{a_1}{b_1} \big|$;
\item[(4)] $f$ is continuous at $x$;
\item[(5)] $f$ is $(\sigma,\tau)$ differentiable at $x$, where 
$ \sigma= \Big( \frac{a_1}{b_1},\dots ,\frac{a_n}{b_n},
-\sum_{j=1}^n\frac{a_j}{b_j}\Big)$, $\tau=(b_1,\dots ,b_n,0)$.
\end{itemize}
Then $f$ is classical differentiable at the point $x$ and  
$D_1(\sigma,\tau)f(x)=f'(x)$.
\end{theorem}

\begin{proof} We shall use the Lemma \ref{lem3.7} with the following notation
\[
\varphi(h):=  \frac{f(x+h)-f(x)-\ell h}{h}, \quad h \in \mathbb{K}^*,
\]
where $\ell :=D_1(\sigma,\tau)f(x)$.
We notice that $f$ is continuous at $x$, which is equivalent to:
\[
0= \lim_{h\to 0}[f(x+h)-f(x)]=\lim_{h\to 0}
\big[ h\cdot \frac{f(x+h)-f(x)}{h}-\ell h \big]=\lim_{h\to 0}h\varphi(h);
\]
that is, condition (1) from Lemma \ref{lem3.7} holds.
 On the one hand we have
\[
\ell=\lim_{h\to 0}\frac{1}{h} \sum_{k=1}^{n+1}\sigma_k f(x+\tau_k h)
=\lim_{h\to 0}\frac{1}{h}\Big[ \sum_{k=1}^n \frac{a_k}{b_k} (f(x+b_kh)-f(x))\Big].
\]
On the other hand side we have
\begin{align*}
\lim_{h\to 0} \sum_{k=1}^n a_k \varphi(b_k h)
&=\lim_{h\to 0}\sum_{k=1}^n a_k \frac{f(x+b_k h)-f(x)- \ell b_k h}{b_k h}\\
&= \lim_{h\to 0} \frac{1}{h} \sum_{k=1}^n \frac{a_k}{b_k}(f(x+b_k h)-f(x))-\ell\\
&=\ell-\ell=0.
\end{align*}
The conditions for Lemma \ref{lem3.7} being satisfied, it results that
$\lim_{x\to 0}\varphi(x)=0$. This is equivalent with $f$ differentiable 
at $x$ and $f'(x)=\ell$.
\end{proof}

\begin{definition} \label{def3.10} \rm
 Let $V\subset \mathbb{K}$, a neighborhood of $0$, 
$a=(a_1,\dots ,a_n)\in \mathbb{K}^n$, $b=(b_1,\dots ,b_n)\in \mathbb{K}^n$ 
and the function $\varphi: V\to \mathbb{K}$. We say that the system $(a,b)$ 
satisfies  \textit{condition $(C_1)$} if the following conditions are satisfied:
\begin{itemize}
\item[(i)] $ \lim_{x\to 0}x\varphi(x)=0$,
\item[(ii)] $ \lim_{x\to 0}\sum_{k=1}^n a_k \varphi(b_k x)=0$
\end{itemize}
 imply $ \lim_{x\to 0}\varphi(x)=0$.
\end{definition}

 \begin{proposition} \label{prop3.11}
 Let $a$, $b\in \mathbb{K}^n$, $c$, $d\in \mathbb{K}^m$. If the systems 
$(a,b)$ and $(c,d)$ satisfy condition $(C_1)$, then the system 
$((a_i,c_j),(b_k,d_l))$, $i$, $k\in \{ 1,\dots ,n\}$, $j$, 
$l\in \{ 1,\dots ,m \}$ also satisfy condition $(C_1)$.
\end{proposition}

 \begin{proof} Indeed, it is sufficient to consider the function
 \[
 g(x)= \sum_{j=1}^m c_j \varphi(d_jx), \quad x\in \mathbb{K}.
 \]
 Then
 \[
  \sum_{k=1}^n a_k g(b_k x)=\sum_{k=1}^n a_k \sum_{j=1}^m c_j \varphi(b_k d_j x)
= \sum_{k=1}^n\sum_{j=1}^m a_k c_j \varphi (b_k d_j x),
\]
for $x\in \mathbb{K}$.
\end{proof}

\begin{definition} \label{def3.12} \rm
The system $(a,b)$ with $a\in \mathbb{K}^n$, $b\in \mathbb{K}^n$ satisfies
 \textit{condition $(C_2)$} if for all functions 
$\varphi:\mathbb{K}\to \mathbb{K}$ with the properties:
\begin{itemize}
\item[(i)] $\varphi$ is bounded on a neighborhood of origin;
\item[(ii)] $ \lim_{x\to 0} \sum_{k=1}^n a_k \varphi (b_k x)=0$;
results that $ \lim_{x\to 0}\varphi(x)=0$.
\end{itemize}
\end{definition}

\subsection*{Remarks}
(1) It is easy to observe that for Definition \ref{def3.12},
we can state a result similar to Proposition \ref{prop3.11}.

(2) Condition $(C_1)$ and condition $(C_2)$ are related as follows:
 $(C_2) \Rightarrow (C_1)$. 
Indeed, let $\varphi$ bounded on an arbitrary neighborhood of origin 
which satisfies the condition
\[
 \lim_{x\to 0} \sum_{k=1}^n a_k \varphi(b_k x)=0.
\]
As $\varphi$ is bounded on a neighborhood of origin results that
 $ \lim_{x\to 0} x\varphi(x)=0$.

\begin{theorem} \label{thm3.13}
 Let $f:\mathbb{K}\to \mathbb{K}$, $x_0\in \mathbb{K}$ fixed, 
$a_j\in \mathbb{K}$, $b_j\in \mathbb{K}^*$ $j\in {1,\dots ,n}$ 
such that the following condition hold:
\begin{itemize}
\item[(1)] $ \sum_{j=1}^n a_j=1$;
\item[(2)] $f$ is continuous at $x_0$;
\item[(3)] $f$ is $(\sigma, \tau)$- differentiable at $x_0$, where
\[
\sigma=\Big( \frac{a_1}{b_1},\dots ,\frac{a_n}{b_n},
-\sum_{k=1}^n \frac{a_k}{b_k}\Big), \quad\tau=(b_1,\dots ,b_n,0);
\]
\item[(4)] the system $(a,b)$ satisfies condition $(C_1)$.
\end{itemize}
Then $f$ is classical differentiable at $x_0$ and  
$D_1 (\sigma, \tau)f(x_0) =f'(x_0)$.
\end{theorem}

\begin{proof} Let $\varphi:\mathbb{K}^*\to \mathbb{K}$,
\[
 \varphi(h):= \frac{f(x_0+h)-f(x_0)-\ell h}{h}
\]
where $\ell:=D_1 (\sigma,\tau)f(x_0).$\\
On the right hand side, because $f$ is $(\sigma,\tau)$-differentiable 
at $x_0$, $(\sigma,\tau)\in L(1,n+1,\mathbb{K})$ then
\begin{align*}
D_1(\sigma,\tau)f(x_0)
&= \lim_{h\to 0} \frac{1}{h} \sum_{k=1}^{n+1} \sigma_k f(x_0+\tau_k h) \\
&=\lim_{h\to 0} \frac{1}{h} \Big[ \sum_{k=1}^n\frac{a_k}{b_k}f(x_0 +b_k h)
 -\sum_{k=1}^n\frac{a_k}{b_k}f(x_0)\Big]\\
&=\lim_{h\to 0} \sum_{k=1}^n a_k \frac{f(x_0+b_kh)-f(x_0)}{b_kh}=\ell.
\end{align*}
On the left hand side, because $f$ is continuous at $x_0$ we have
\[
 \lim_{h\to 0} h\varphi(h)=\lim_{h\to 0}[f(x_0+h)-f(x_0)-\ell h]=0
\]
and as $f$ is $(\sigma,\tau)$-differentiable at $x_0$ we obtain
\begin{align*}
\lim_{h\to 0} \sum_{k=1}^n a_k \varphi (b_k h)
&=\lim_{h\to 0}\sum_{k=1}^n a_k \frac{f(x_0+b_k h)-f(x_0)-\ell b_k h}{b_k h}\\
&=\lim_{h\to 0}\sum_{k=1}^n a_k \Big( \frac{f(x_0+b_k h)-f(x_0)}{b_k h}-\ell \Big)\\
&=\lim_{h\to 0}\sum_{k=1}^n a_k \frac{f(x_0+b_k h)-f(x_0)}{b_k h}
-\ell \sum_{k=1}^n a_k=\ell-\ell=0.
\end{align*}
Therefore, conditions (1) and (2) from Definition \ref{def3.10} are satisfied;
 that is, the system $(\sigma,\tau)$ satisfies condition $(C_1)$.
 As a consequence we have:
$ \lim_{h\to 0} \varphi (h)=0$
which is equivalent to $f'(x_0)=\ell$.
\end{proof}

\begin{theorem} \label{thm3.14} 
Let $A\subset \mathbb{K}$ such that $\operatorname{Int} A\neq \emptyset$,
 $x\in \operatorname{Int} A$, $a_j$, $b_j\in \mathbb{K}^*$,
 $1\le j\le n$, $n\ge 2$ and $f: A\to \mathbb{K}$ with the following properties:
\begin{itemize}
\item[(1)] $f$ is Lipschitz on a neighborhood of $x$;
\item[(2)] $ \max_{1\le j\le n} |b_j|\le |b_1|$;
\item[(3)] $ \sum_{j=2}^n|a_j|\le |a_1|$;
\item[(4)] $ \sum_{j=1}^n a_j =1$;
\item[(5)] $f$ is $(\sigma, \tau)$ differentiable at $x$, where
\[
\sigma =\Big( \frac{a_1}{b_1},\dots , \frac{a_n}{b_n}, 
-\sum_{k=1}^n \frac{a_k}{b_k}\Big), \quad\tau=(b_1,\dots ,b_n, 0).
\]
\end{itemize}
Then $f$ is differentiable at $x$ and
$D_1 (\sigma,\tau)f(x)=f'(x)$.
\end{theorem}

\begin{proof} 
We shall use Lemma \ref{lem3.8}. First we observe the equivalence between the 
following two statements:
``$\varphi$ bounded on a neighborhood of origin'' if and only if
``there exists $M>0$ such that $|\varphi(h)|\le M$, for all 
$h\in V_{\epsilon}(0)$, where $V_{\epsilon}(0)$ is a symmetric
 neighborhood of origin of length $2\epsilon$, $(\epsilon>0)$''
if and only if
\[
 \big| \frac{f(x+h)-f(x)-\ell h}{h}\big|\le M,\quad \forall 
h\in V_{\epsilon}(0),
\]
 where we denoted $\ell:=D_1 (\sigma,\tau)f(x)$. 
This is further equivalent to
\[
|f(x+h)-f(x)|\le (M+|\ell|)|h|,\quad\forall h\in V_{\epsilon}(0)
\]
if and only if
\[
|f(y)-f(x)|\le (M+|\ell|)|y-x|, \quad\forall y\in V_{\epsilon}(0)\subset A, 
\]
where $y=x+h$,
which implies that $f$ is Lipschitz at $x$.
Similarly, we have
\[
0=\lim_{h\to 0} \sum_{j=1}^n a_j \varphi (b_j h)
=\lim_{h\to 0}\sum_{j=1}^n a_j \frac{f(x+b_j h)-f(x)-\ell b_j h}{b_j h}
\]
if and only if
\[
\lim_{h\to 0} \sum_{j=1}^n a_j \frac{f(x+b_j h)-f(x)}{b_j h}
=\ell \sum_{j=1}^n a_j=\ell\,.
\]
However,
\begin{align*}
\ell
&=D_1 (\sigma,\tau)f(x)
=\lim_{h\to 0}\frac{1}{h}\sum_{k=1}^{n+1}\sigma_k f(x+\tau_k h)\\
&=\lim_{h\to 0} \frac{1}{h}\sum_{k=1}^n \frac{a_k}{b_k}[f(x+b_k h)-f(x)]. 
\end{align*}
As the conditions for Lemma \ref{lem3.8} are satisfied, it results that
\[
\lim_{h\to 0}\varphi(h)=0
\Leftrightarrow \frac{f(x+h)-f(x)}{h}=\ell 
\Leftrightarrow f'(x)=D_1 (\sigma,\tau)f(x).
\]
\end{proof}

\begin{theorem} \label{thm3.15} 
Let $G\subset \mathbb{K}$ such that $\operatorname{Int} G\neq 0$, 
$x\in \operatorname{Int} G$, $a_j\in \mathbb{K}$, $b_j\in \mathbb{K}^*$,
$j\in \{1,\dots ,n \}$ and $f:G\to \mathbb{K}$ with properties:
\begin{itemize}
\item[(1)] $f$ is Lipschitz on a neighborhood of $x$;
\item[(2)] $ \sum_{j=1}^n a_j=1$;
\item[(3)] $f$ is $(\sigma,\tau)$-differentiable at $x$, where
\[
\sigma=\Big( \frac{a_1}{b_1},\dots ,\frac{a_n}{b_n}, 
-\sum_{k=1}^n\frac{a_k}{b_k}\Big), \tau=(b_1,\dots ,b_n,0);
\]
\item[(4)] the system $(a,b)$ satisfy condition $(C_2)$.
\end{itemize}
Then $f$ is classical differentiable at $x$ and
$D_1 (\sigma,\tau)f(x)=f'(x)$.
\end{theorem}

\begin{proof} 
The condition $f$-Lipschitz at $x$ is equivalent to $\varphi$ bounded 
on a neighborhood of origin, where 
$ \varphi(h)=\frac{1}{h}[f(x+h)-f(x)-\ell h]$ as we could observe 
in Theorem \ref{thm3.15}. Moreover, condition (ii) from Definition 
\ref{def3.12} of 
condition $(C_2)$ is also satisfied. From here results that 
$ \lim_{h\to 0} \varphi(h)=0$ which is equivalent to
\[
\ell=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
\]
that is $D_1(\sigma, \tau)f(x)=f'(x)$.
\end{proof}

\subsection*{Conclusion} 
Theorems \ref{thm3.13}--\ref{thm3.15} are general criteria which state that 
if we find systems $(a,b)$, ($(a,b)\in \mathbb{K}^n\times \mathbb{K}^{*n}$) 
that satisfy conditions $(C_1)$ and $(C_2)$, then any
 $(\sigma,\tau)$-differentiable function at a point, satisfying the 
conditions from these theorems, is classical differentiable at that 
point and the two derivatives are equal. 


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\end{document}