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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 163, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/163\hfil Caputo fractional difference]
{Some relations between the Caputo fractional difference operators and
integer-order differences}

\author[B. Jia, L. Erbe, A. Peterson \hfil EJDE-2015/163\hfilneg]
{Baoguo Jia, Lynn Erbe,  Allan Peterson}

\address{Baoguo Jia  \newline
School of Mathematics and Computer Science\\
Sun Yat-Sen University\\
Guangzhou 510275, China.\newline
Department of Mathematics\\
University of Nebraska-Lincoln\\
Lincoln, NE 68588-0130, USA}
\email{mcsjbg@mail.sysu.edu.cn}

\address{Lynn Erbe \newline
Department of Mathematics\\
University of Nebraska-Lincoln\\
Lincoln, NE 68588-0130, USA}
\email{lerbe2@math.unl.edu}

\address{Allan Peterson \newline
Department of Mathematics\\
University of Nebraska-Lincoln\\
Lincoln, NE 68588-0130, USA}
\email{apeterson1@math.unl.edu}

\thanks{Submitted May 30, 2015. Published June 17, 2015.}
\subjclass[2010]{39A12, 39A70}
\keywords{Caputo fractional difference; monotonicity; Taylor monomial}

\begin{abstract}
 In this article, we are concerned with the relationships between
 the sign of Caputo fractional differences and integer nabla differences.
 In particular, we show that if
 $N-1<\nu<N$, $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$,
 $\nabla^\nu_{a^*}f(t)\geq 0$, for  $t\in\mathbb{N}_{a+1}$ and
 $\nabla^{N-1}f(a)\geq 0$, then $\nabla^{N-1}f(t)\geq 0$ for $t\in\mathbb{N}_a$.
 Conversely, if $N-1<\nu<N$, $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$, and
 $\nabla^{N}f(t)\geq 0$ for $t\in\mathbb{N}_{a+1}$, then
 $\nabla^{\nu}_{a^*}f(t)\geq 0$, for each $t\in\mathbb{N}_{a+1}$.
 As applications of these two results, we get that
 if $1<\nu<2$, $f:\mathbb{N}_{a-1}\to\mathbb{R}$, $\nabla^\nu_{a^*}f(t)\geq 0$
 for $t\in\mathbb{N}_{a+1}$ and $f(a)\geq f(a-1)$, then
 $f(t)$ is an increasing function for  $t\in \mathbb{N}_{a-1}$.
 Conversely if $0<\nu<1$, $f:\mathbb{N}_{a-1}\to\mathbb{R}$ and  $f$ is an
 increasing function for $t\in\mathbb{N}_{a}$, then
 $\nabla^\nu_{a^*}f(t)\geq 0$, for each $t\in\mathbb{N}_{a+1}$.
 We also give a counterexample to show that the above assumption
 $f(a)\geq  f(a-1)$ in the last result is essential.
 These results demonstrate that, in some sense, the positivity of the
 $\nu$-th order Caputo fractional difference has a strong connection to the
 monotonicity of $f(t)$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction} 

If $a$ is a real number, then we use the notation
$$
\mathbb{N}_a:=\{a,a+1,a+2,\dots\}.
$$
If $f:\mathbb{N}_a\to\mathbb{R}$, then we define the nabla 
(backwards difference) operator by
$$
\nabla f(t)=f(t)-f(t-1), \quad t\in\mathbb{N}_{a+1}.
$$
Discrete fractional calculus has generated a lot of interest in
recent years. Some of the work has employed the delta (forward)  or nabla
(forward difference) operator. We refer the readers to 
\cite{ae,bp1,a1, gp,ft1}, for example. It
seems, however, that more work has been developed for the backward
or nabla difference operator and we refer the readers to 
\cite{bp2,ft}. There has also been some work to develop relations
between the forward and backward fractional operators,
$\Delta_a^\nu$ and $\nabla_a^\nu$ \cite{ae1} (see also \cite{gp})
and fractional calculus on time scales \cite{bp1}. Anastassiou
\cite{a2} has introduced the study of nabla fractional calculus in
the case of Caputo fractional difference.

This work is motivated by the paper by  Dahal and Goodrich \cite{dg}. They
obtained some interesting
monotonicity results for the delta fractional difference operator.
In this paper, we prove the following corresponding results for Caputo
fractional differences.

\begin{theorem} \label{thmA}
 Assume that $N-1<\nu<N$, $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$,
 $\nabla^\nu_{a^*}f(t)\geq 0$
for  $t\in\mathbb{N}_{a+1}$ and $\nabla^{N-1}f(a)\geq 0$. 
Then $\nabla^{N-1}f(t)\geq 0$ for $t\in\mathbb{N}_a$
\end{theorem}

\begin{theorem} \label{thmB} 
Assume $N-1<\nu<N$, $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$, and
$\nabla^{N}f(t)\geq 0$ for $t\in\mathbb{N}_{a+1}$.
   Then $\nabla^{\nu}_{a^*}f(t)\geq 0$, for each $t\in\mathbb{N}_{a+1}$.
\end{theorem}

As applications, we have:

\begin{corollary}\label{alp} 
Assume that $1<\nu<2$, $f:\mathbb{N}_{a-1}\to\mathbb{R}$,
 $\nabla^\nu_{a^*}f(t)\geq 0$
for $t\in\mathbb{N}_{a+1}$ and $f(a)\geq f(a-1)$, then
$f(t)$ is an increasing function for  $t\in \mathbb{N}_{a-1}$.
\end{corollary}


\begin{corollary} 
Assume $0<\nu<1$, $f:\mathbb{N}_{a-1}\to\mathbb{R}$ and  $f$ is an increasing
function for $t\in\mathbb{N}_{a}$. Then
 $\nabla^\nu_{a^*}f(t)\geq 0$,
for each $t\in\mathbb{N}_{a+1}$.
\end{corollary}

 We also give a counterexample to show that the above assumption $f(a)\geq
 f(a-1)$ in Corollary \ref{alp} is essential.


\section{Caputo fractional difference}

The following definitions appear in \cite[Chapter 3]{gp}. 
First we define the nabla fractional sum of 
$f:\mathbb{N}_{a+1}\to\mathbb{R}$ based at $a$.

\begin{definition}\label{dnfs} Let
$f:\mathbb{N}_{a+1}\to\mathbb{R}$ be given and $\nu>0$, then the
\emph{nabla fractional sum} of $f$ based at $a$ is defined by
\begin{equation}\label{al}
\nabla^{-\nu}_{a}f(t)=\int^t_aH_{\nu-1}(t,\rho(s)) f(s)\nabla s,
\end{equation}
for $t\in\mathbb{N}_a$, where by convention $\nabla_a^{-\nu}f(a)=0$.
\end{definition}

Next we define the Caputo fractional difference in terms of the nabla
fractional sum.

\begin{definition}\label{dnfd}
 Assume $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$ and $\mu>0$.
Then the $\mu$-th Caputo nabla fractional difference of $f$ based at $a$ is
defined by
$$
\nabla^\mu_{a^*}f(t)=\nabla_a^{-(N-\mu)}\nabla^Nf(t)
$$
for $t\in\mathbb{N}_{a+1}$, where $N=\lceil\mu\rceil$, $\lceil\cdot\rceil$
the ceiling of number, $m\in\mathbb{N}$.
\end{definition}

 Let $\Gamma$ denote the gamma function, then the \emph{rising function}
$t^{\overline{r}}$ is defined by
$$
t^{\overline{r}}=\frac{\Gamma(t+r)}{\Gamma (t)},
$$
for those values of $t$ and $r$ such that
the right hand side of this last equation makes sense. We also use the
convention that if the numerator is well defined and the denominator is not
well defined, then $t^{\overline{r}}:=0$.
We define the $\mu$-th degree Taylor monomial based at $a$ by
$$
H_{\mu}(t,a):=\frac{(t-a)^{\overline{\mu}}}{\Gamma(\mu+1)}.
$$
We will use the following power rule (see \cite[Chapter 3]{gp}):
\begin{align}\label{prn}
 \nabla H_{\mu}(t_0,t)=-H_{\mu-1}(t_0,\rho(t)),
\end{align}
where $t_0\in\mathbb{N}_a$.

Then (see \cite[Chapter 3]{gp}) if
$f:\mathbb{N}_{a+1}\to\mathbb{R}$ and $N-1<\mu< N$, $N\in\mathbb{N}_1$,
then the $\mu$-th nabla fractional difference is given by
\begin{equation}\label{al2}
\nabla^{\mu}_{a}f(t)=\int^t_aH_{-\mu-1}(t,\rho(s)) f(s)\nabla s,
\end{equation}
for $t\in\mathbb{N}_a$, where by convention $\nabla_a^{\mu}f(a)=0$.

 \begin{theorem}\label{al10}
 Assume that $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$, and $\nabla^\mu_{a^*}f(t)\geq 0$, 
for each  $t\in\mathbb{N}_{a+1}$, with $N-1<\mu <N$. Then
\begin{equation} \label{lynn}
\begin{aligned}
\nabla^{N-1}f(a+k)
&\geq \sum^{k-1}_{i=1}\Big[\frac{(k-i+1)^{\overline{N-\mu-2}}}{\Gamma(N-\mu-1)}\Big]
\nabla^ { N-1 }f(a+i-1)\\
&\quad + H_{N-\mu-1}(a+k,a)\nabla^{N-1}f(a),
\end{aligned}
\end{equation}
for $k\in\mathbb{N}_1$ (note by our convention on sums the first term on the right
hand side is zero when $k=1$).
\end{theorem}

\begin{proof} 
If $t=a+1$ we have 
\begin{align*}
 0\le\nabla^{\mu}_{a^*}f(a+1)
&=\nabla_{a}^{-(N-\mu)}\nabla^Nf(t) \\
&=\int^{a+1}_aH_{N-\mu-1}(a+1,\rho(s))\nabla^{N} f(s)\nabla s\\
&=H_{N-\mu-1}(a+1,a)\nabla^{N}f(a+1)\\
&=\nabla^Nf(a+1)=\nabla^{N-1} f(a+1)-\nabla^{N-1}f(a),
\end{align*}
where we used $H_{N-\mu-1}(a+1,a)=1$.
Solving for $\nabla^{N-1} f(a+1)$ we get the inequality
$$
\nabla^{N-1}f(a+1)\ge \nabla^{N-1}f(a)
$$
which gives us the inequality \eqref{lynn} for $t=a+1$.
Hence the inequality \eqref{lynn} holds for $t=a+1$.

Next consider the case $t=a+k$ for $k\ge 2$.
Taking $t=a+k$, $k\ge 2$ we have from \eqref{al} we have
\begin{align*} 
0&\leq\nabla^{\mu}_{a^*}f(t)\\
&=\nabla_a^{-(N-\mu)}\nabla^Nf(t)\\
&=\int^t_aH_{N-\mu-1}(t,\rho(s))\nabla^N f(s)\nabla s\\
&=\int^{a+k}_aH_{N-\mu-1}(a+k,\rho(s))\nabla^N f(s)\nabla s\\
&=\sum^k_{i=1}H_{N-\mu-1}(a+k,a+i-1)\nabla^N f(a+i)\\
&=\sum_{i=1}^kH_{N-\mu-1}(a+k,a+i-1)\left[\nabla^{N-1}f(a+i)-\nabla^{N-1}
f(a+i-1) \right] \\
&=\sum^{k}_{i=1}H_{N-\mu-1}(a+k,a+i-1)\nabla^{N-1}f(a+i)\\
&\quad-\sum^{k}_{i=1}H_{N-\mu-1}(a+k,a+i-1)\nabla^{N-1}f(a+i-1)\\
&=\nabla^{N-1}f(a+k)+\sum^{k-1}_{i=1}H_{N-\mu-1}(a+k,a+i-1)\nabla^{N-1}f(a+i)\\
&-H_{N-\mu-1}(a+k,a)\nabla^{N-1}f(a)\\
&\quad-\sum^{k}_{i=2}H_{N-\mu-1}
(a+k,a+i-1)\nabla^{N-1} f(a+i-1),
\end{align*}
where we used $H_{N-\mu-1}(a+k,a+k-1)=1$. 
It follows that
\begin{align*}
&0\le\nabla^{N-1}f(a+k)+\sum^{k-1}_{i=1}H_{N-\mu-1}(a+k,a+i-1)\nabla^{N-1}
f(a+i)\\
&-H_{N-\mu-1}(a+k,a)\nabla^{N-1}f(a)-\sum^{k-1}_{i=1}H_{N-\mu-1}
(a+k,a+i)\nabla^{N-1} f(a+i)\\
&=\nabla^{N-1}f(a+k)-H_{N-\mu-1}(a+k,a)\nabla^{N-1}f(a+i)\\
&-\sum^{k-1}_{i=1}\Big[H_{N-\mu-1}(a+k,a+i)\\
&\quad\quad\quad\quad\quad\quad-H_{N-\mu-1}(a+k,
a+i-1)\Big ] \nabla^{ N-1 }f(a+i).
\end{align*}
It follows that
\begin{align*}
0&\le\nabla^{N-1}f(a+k)-H_{N-\mu-1}(a+k,a)\nabla^{N-1 } f(a)\\
&\quad -\sum^{k-1}_{i=1}\nabla_s H_{N-\mu-1}(a+k,s)|_{s=a+i}\nabla^{ N-1
}f(a+i)\\
&\overset{\eqref{prn}}=\nabla^{N-1}f(a+k)-H_{N-\mu-1}(a+k,a)\nabla^{
N-1}
f(a)\\
&\quad +\sum^{k-1}_{i=1} H_{N-\mu-2}(a+k,a+i-1)\nabla^{ N-1
}f(a+i)\\
&=\nabla^{N-1}f(a+k)-H_{N-\mu-1}(a+k,a)\nabla^{N-1}
f(a)\\
&\quad +\sum^{k-1}_{i=1}\Big[\frac{(k-i+1)^{\overline{N-\mu-2}}}{\Gamma(N-\mu-1)}
\Big]\nabla^{N-1}f(a+i).
\end{align*}

Solving the above inequality for $\nabla^{N-1}f(a+k)$, we obtain the
desired inequality \eqref{lynn}. Next we consider for $1\le i\le k-1$,
\begin{align*}
&\frac{(k-i+1)^{\overline{N-\mu-2}}}{
\Gamma(N-\mu-1)}=\frac{\Gamma(N-\mu+k-i-1)}{\Gamma(k-i+1)\Gamma(N-\mu-1)}\\
&=\frac{(N-\mu+k-i-2)\dots (N-\mu-1)}{(k-i)!}<0
\end{align*}
since $N<\mu+1$. Also
\begin{align*}
H_{N-\mu-1}(a+k,a)
&=\frac{k^{\overline{N-\mu-1}}}{\Gamma(N-\mu)}\\
&=\frac{\Gamma(N-\mu+k-1)}{\Gamma(k)\Gamma(N-\mu)}\\
&=\frac{(N-\mu+k-2)\dots(N-\mu)}{(k-1)!}>0.\\
\end{align*}
\end{proof}

From Theorem \ref{al10}, we have the following result.

\begin{theorem}\label{jia6}
Assume that $N-1<\nu<N$, $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$,
 $\nabla^\nu_{a^*}f(t)\geq 0$
for $t\in\mathbb{N}_{a+1}$ and $\nabla^{N-1}f(a)\geq 0$. Then 
$\nabla^{N-1}f(t)\geq 0$ for $t\in\mathbb{N}_{a}$.
\end{theorem}

\begin{proof}
 By using the principle of strong induction, we
prove that the conclusion of the theorem is correct.

By assumption, the result holds for $t=a$.
Suppose that $\nabla^{N-1}f(t)\geq 0$, for $t=a,a+1,\dots,a+k-1$.
From Theorem \ref{al10} and \eqref{lynn}, we have
$\nabla^{N-1}f(a+k)\geq 0$ and the proof is complete.
\end{proof}


Taking $N=2$ and $N=3$, we can get the following corollaries.

\begin{corollary}\label{2.3} 
Assume that $1<\nu<2$, $f:N_{a-1}\to\mathbb{R}$,
 $\nabla^\nu_{a^*}f(t)\geq 0$
for $t\in\mathbb{N}_{a+1}$  and $f(a)\geq f(a-1)$, then
$f(t)$ is increasing for $t\in\mathbb{N}_{a-1}$.
\end{corollary}

\begin{corollary} 
Assume that $2<\nu<3$, $f:N_{a-2}\to\mathbb{R}$,
 $\nabla^\nu_{a^*}f(t)\geq 0$
for $t\in\mathbb{N}_{a+1}$  and $\nabla^2f(a)\geq 0$, then
$\nabla f(t)$ is increasing for $t\in\mathbb{N}_{a}$.
\end{corollary}

In the following, we give the inverse proposition of Theorem
\ref{jia6}.


\begin{theorem} 
Assume that $N-1<\nu<N$, $f:\mathbb{N}_{a-N+1}\to\mathbb{R}$,
$\nabla^{N}f(t)\geq 0$ for $t\in \mathbb{N}_{a+1}$,
   then $\nabla^\nu_{a^*}f(t)\geq 0$, for each $t\in\mathbb{N}_{a+1}$.
\end{theorem}

\begin{proof}
From \eqref{al}, taking $t=a+k$,  we have
\begin{align}\label{lynn30}
\nabla^{-\mu}_{a^*}f(t)
 &=\nabla_a^{-(N-\mu)}\nabla^Nf(t)\\
 &=\int^t_aH_{N-\mu-1}(t,\rho(s))\nabla^N f(s)\nabla s\\
 &=\sum^k_{i=1}H_{N-\mu-1}(a+k,a+i-1)\nabla^N f(a+i).
\end{align}
Since
\begin{equation} \label{al30}
\begin{aligned}
H_{N-\mu-1}(a+k,a+i-1)
&=\frac{(k-i+1)^{\overline{N-\mu-1}}}{\Gamma(N-\mu)}\\
&=\frac{\Gamma(k+N-i-\mu)}{
\Gamma(N-\mu)\Gamma(k-i+1)}\\
 &=\frac{(-\mu+k+N-i)\dots(N-\mu+1)(N-\mu)}{(k-i)!} >0,
\end{aligned}
\end{equation}
where we used $\mu<N$, from \eqref{lynn30} and \eqref{al30} we get
that $\nabla^\nu_{a^*}f(t)\geq 0$, for each $t\in\mathbb{N}_{a+1}$,
\end{proof}

Taking $N=1$ and $N=2$, we get the following corollaries.

\begin{corollary}
Assume that $0<\nu<1$, $f:\mathbb{N}_{a-1}\to\mathbb{R}$ and $f$ 
is an increasing function for $t\in\mathbb{N}{a}$. Then
 $\nabla^{\nu}_{a^*}f(t)\geq 0$,
for $t\in\mathbb{N}_{a+1}$.
\end{corollary}

\begin{corollary}\label{2.7}
Assume that $1<\nu<2$, $f:\mathbb{N}_{a-1}\to\mathbb{R}$ and  $\nabla^2f(t)\geq 0$ for
$t\in\mathbb{N}_{a+1}$. Then  $\nabla^\nu_{a^*}f(t)\geq 0$,
for each $t\in\mathbb{N}_{a+1}$.
\end{corollary}

 In the following, we will give a counterexample to show
that the assumption in Corollary \ref{2.3} ``$f(a)\geq  f(a-1)$" is essential. 
To verify this example we will use the following simple lemma.

\begin{lemma}
Assume that $f''(t)\geq 0$ on $[a,\infty)$. Then
$\nabla^\nu_{a^*}f(t)\geq 0$, for each $t\in\mathbb{N}_{a+1}$, with
$1<\nu<2$.
\end{lemma}

\begin{proof}
By Taylor's Theorem,
\begin{gather}\label{1.2}
f(a+i+1)=f(a+i)+f^{\prime}(a+i)+\frac{f''(\xi^i)}{2},\quad
\xi^i\in[a+i,a+i+1],\\
\label{1.3}
f(a+i-1)=f(a+i)-f^{\prime}(a+i)+\frac{f''(\eta^i)}{2},\quad
\eta^i\in[a+i-1,a+i]
\end{gather}
for $i=0,1,\dots,k-1$.
Using \eqref{1.2} and \eqref{1.3}, we have
\begin{equation}\label{1.4}
\begin{aligned}
\nabla^2f(a+i+1)
&=f(a+i+1)-2f(a+i)+f(a+i-1)\\
&=\frac{f''(\xi^i)+f''(\eta^i)}{2}
\geq  0.
\end{aligned}
\end{equation}
From \eqref{1.4} and Corollary \ref{2.7}, we obtain
$\nabla^\nu_{a^*}f(t)\geq 0$, for each $t\in\mathbb{N}_{a+1}$, with
$1<\nu<2$.
\end{proof}

\begin{example} \rm
Let $f(t)=-\sqrt{t}$, $a=2$. We have $f''(t)\geq 0$,
for $t\geq 1$. By Corollary \ref{2.7}, we have
$\nabla^\nu_{a^*}f(t)\geq 0$
\end{example}

Note that $f(a-1)=f(1)=-1>f(a)=-\sqrt{2}$. Therefore $f(x)$ does not
satisfy the assumptions of Corollary \ref{2.3}. 
In fact, $f(t)$ is decreasing, for $t\geq 1$.

We conclude this note by mentioning a representative consequence of
Corollary \ref{2.3}.

\begin{corollary}
Let $h:\mathbb{N}_{a+1}\times \mathbb{R}\to\mathbb{R}$ be a nonnegative,
continuous function. Then any solution of the Caputo nabla fractional 
difference equation
\begin{equation}
\nabla^\nu_{a^*}y(t)=h(t,y(t)),\quad t\in\mathbb{N}_{a+1},\quad 1<\nu<2
\end{equation}
satisfying $\nabla y(a)= A\ge 0$ is increasing on $\mathbb{N}_{a-1}$.
\end{corollary}

\subsection*{Acknowledgments}
This work is supported by the National Natural Science Foundation
 of China (No.11271380).


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\end{document}