The Relation Between Nabla Fractional Differences and Nabla Integer Differences

Abstract

In this paper we obtain two interrelated results.  The first result is the following inequality:\\

{\bf{Theorem.}} Assume that $f:\N_a\rightarrow\R$ satisfies $\nabla_a^\nu
f(t)\geq 0$, for each $t\in\N_{a+1}$, $\nu>0,
 \nu\notin\N_1$, and choose $N\in \N_1$, such that $N-1<\nu<N$. Then for each $k\in\N_{a+N}$, we have
\begin{equation}
\begin{split}
\nabla^{N-1}f(a+k)&\geq -\sum^{N-2}_{i=0}H_{-\nu+i}(a+k,a+i)\nabla^{i}f(a+i+1)\\
&-\sum^{k-1}_{i=N}H_{-\nu+N-2}(a+k,a+i-1)\nabla^{N-1}f(a+i),\notag
\end{split}
\end{equation}
where
$$H_{-\nu+N-2}(a+k,a+i-1)=\frac{(k-i+1)^{\overline{-\nu+N-2}}}{\Gamma(-\nu+N-1)}<0.$$
As an application of the above inequality we prove the following result:\\

{\bf{Theorem.}} Assume that $f:\N_a\rightarrow\R$ satisfies $\nabla^\nu_af(t)\geq 0$, for each $t\in\N_{a+1}$, where $5<\nu<6$.
Then $\nabla^5f(t)\geq 0$, for $t\in \N_{a+6}$.\\

\noindent This demonstrates that, in some sense, the positivity of the $\nu$-th
order fractional difference has a strong connection to the positivity of
an integer-order difference of the function $f$.