On Convexity, Monotonicity and Positivity Analysis for Discrete Fractional Operators Defined Using Exponential Kernels

Abstract

This article deals with analysing the positivity, monotonicity and convexity of the discrete nabla fractional operators with exponential kernels from the sense of Riemann and Caputo operators. These operators are called discrete nabla Caputo–Fabrizio–Riemann and Caputo–Fabrizio–Caputo fractional operators. Further, some of our results concern sequential nabla Caputo–Fabrizio–Riemann and Caputo–Fabrizio–Caputo fractional differences, such as $\left({{}_a{}^{CFR}\nabla _{}^{}}^{\mu }{{}_b{}^{CFC}\nabla _{}^{}}^{\upsilon }h\right)\left(x\right)$, for various values of start points a and b, and for orders $\upsilon$ and $\mu$ in different ranges. Three illustrative examples of the main lemmas in the case of Riemann–Liouville are given at the end of the article.

1. Introduction

The discrete fractional calculus ($\mathtt{DFC}$) is a field of mathematical analysis and it is a new branch of continuous fractional calculus that is responsible for studying the discrete operators of the sum and difference of fractional order on domains of discrete functions. The definitions of $\mathtt{DFC}$ models with singular and nonsingular kernels have been studied by researchers in recent years (see [1,2,3,4,5,6]). Currently, many definitions of nabla (or backward) fractional differences, ${\nabla }^{\upsilon }$, have been adopted in order to modify and generalize the concept of ordinary nabla difference, so that for $\upsilon =1$, the first-order nabla difference can be recovered, which is usually defined for any function $\mathtt{h}$ as $\nabla \mathtt{h}\left(\mathtt{x}\right)=\mathtt{h}\left(\mathtt{x}\right)-\mathtt{h}(\mathtt{x}-1)$. The important goal of these discrete operators is to convert the integral equations to system of sum equations, so their consideration provides more information regarding the nature of the development being modelled. Due to the association of $\mathtt{DFC}$ to practical ventures that are extensively employed to nanotechnology, optics, infectious diseases, physics, chaos theory, economics, medicine, among other areas, many researchers are devoting their attention to generalize the continuous problems to discrete fractional dynamics (see [1,7,8,9]).

For decades, monotonicity, positivity and convexity analyses of discrete fractional operators have been perhaps one of the most fundamental studies in the context of discrete fractional calculus due to a wide variety of representations, forms and applications. Due to these new challenges and trends of discrete fractional calculus, many scholars are devoting their attention to establishing new monotonicity results for discrete generalized nabla fractional operators with discrete singular and non-singular kernels on both time scales $\mathcal{Z}$ and $h\mathcal{Z}$.

Moreover, there has been a considerable increase in examining monotonicity analysis. We can list several current remarkable research studies; for instance, Abdeljawad and Baleanu [10] for discrete nabla Attangana-Baleanue fractional operators on the time scale $\mathcal{Z}$. Meantime, Suwan et al. [11] on the time scale $h\mathcal{Z}$. In 2012, Abdeljawad and Abdallaa [12] for discrete nabla and delta Riemann–Liouville and Caputo fractional operators using dual identities on the time scale $\mathcal{Z}$. Recently, Goodrich et al. [13] for discrete Caputo–Fabrizio fractional operators. Mohammed et al. [14] for discrete generalized nabla Attangana-Baleanue fractional operators with discrete generalized Mittag–Leffler kernels on the time scale $\mathcal{Z}$, and the references therein.

On the other hand, it was found that plenty of the researchers and scholars devoted their studies to establishing conditions which ensure that the analysis results to the aforementioned discrete fractional operators are positive, $\upsilon$-increasing or $\upsilon$-decreasing, we refer here to the published papers [15,16,17].

To our best knowledge, the contribution of the present article is to obtain and examine monotonicity, positivity and convexity analyses of discrete nabla Caputo–Fabrizio ($\mathtt{CF}$) fractional operators in the Riemann–Liouville and Caputo settings via the existing relationship between them. By using this relation we can establish some new monotonicity, positivity and convexity results, which will mix with discrete nabla Caputo–Fabrizio–Caputo ($\mathtt{CFC}$) and Caputo–Fabrizio–Riemann ($\mathtt{CFR}$) fractional differences, and these have never been done before.

The remaining sections of this article are organized as follows: Section 2 is dedicated to recalling the basic concept of discrete $\mathtt{CFC}$ fractional operators and some related properties including the relationship between discrete nabla $\mathtt{CFC}$ and $\mathtt{CFR}$ fractional differences; Section 3 deals with the analysis of the results of the discrete nabla $\mathtt{CF}$ fractional operators: Section 3.1 is dedicated to the analysis of positivity and monotonicity results; Section 3.2 is dedicated to the analysis of monotonicity and $\upsilon$-convexity results; Section 3.3 is dedicated to the analysis of convexity results in some specific domains; Section 4 includes the discussion of results by means of presenting some fractional difference initial value problems; finally, we present a brief conclusion and expectations in the last section.

2. Basic Formulations

At first, we consider the definitions of the left discrete nabla $\mathtt{CF}$ fractional operators on ${\mathrm{N}}_a:=\{a,a+1,\dots \}$.

Definition 1 (see [18]). Let $\upsilon \in [0,1)$ and $a\in \mathsf{R}$. Then, for any function $\mathtt{h}$ defined on ${\mathrm{N}}_a$, the left discrete nabla $\mathtt{CFC}$ fractional difference is given by:

$$\begin{array}{cc}\hfill \left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)& =\frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\rho \left(\mathtt{s}\right)}\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\hfill \end{array}$$

(1)

and the $\mathtt{CFR}$ fractional difference is given by:

$$\begin{array}{cc}\hfill \left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)& =\frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }{\nabla }_{\mathtt{x}}\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\rho \left(\mathtt{s}\right)}\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right){\nabla }_{\mathtt{x}}\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\hfill \end{array}$$

(2)

respectively, where $\mathcal{B}\left(\upsilon \right)$ is a normalizing positive constant.

Definition 2 (see [19]). The nth order nabla difference operator is defined by $\left({\nabla }^n\mathtt{h}\right)\left(\mathtt{x}\right)=\left(\nabla \left({\nabla }^{n-1}\mathtt{h}\right)\right)\left(\mathtt{x}\right)$ for each $\mathtt{x}\in {\mathrm{N}}_{a+n}$. Also, for $\mathtt{h}:{\mathrm{N}}_{a-n}\to \mathsf{R}$ and $n<\upsilon \le n+1$, the left discrete nabla $\mathtt{CFC}$ and $\mathtt{CFR}$ fractional differences can be expressed as follows:

$$\begin{array}{cc}\hfill \left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)& =\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon -n}{\nabla }^n\mathtt{h}\right)\left(\mathtt{x}\right)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)& =\left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon -n}{\nabla }^n\mathtt{h}\right)\left(\mathtt{x}\right)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\hfill \end{array}$$

respectively.

It is of interest to recall the relation between discrete nabla $\mathtt{CFC}$ and $\mathtt{CFR}$ fractional differences.

Proposition 1.

For any $\upsilon \in (0,1)$, we have:

$$\begin{array}{c}\hfill \left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)=\left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)-\frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\mathtt{h}\left(a\right){(1-\upsilon )}^{\mathtt{x}-a}\left(\forall \mathtt{x}\in {\mathrm{N}}_a\right).\end{array}$$

3. Results and Analysis

In this section, we focus on implementing positivity, monotonicity and $\upsilon$-convexity analyses for the discrete nabla $\mathtt{CF}$ fractional operators discussed in the previous section.

Definition 3 (see [11,12,15,16]). Let a function $\mathtt{h}$ be defined on ${\mathrm{N}}_a$ with $\mathtt{h}\left(a\right)\ge 0$ and $0<\upsilon \le 1$. Then, $\mathtt{h}$ is an υ-monotone increasing on ${\mathrm{N}}_a$ if

$$\begin{array}{c}\hfill \mathtt{h}(\mathtt{x}+1)\ge \upsilon \mathtt{h}\left(\mathtt{x}\right),\end{array}$$

and υ-monotone decreasing on ${\mathrm{N}}_a$ if

$$\begin{array}{c}\hfill \mathtt{h}(\mathtt{x}+1)\le \upsilon \mathtt{h}\left(\mathtt{x}\right),\end{array}$$

for each $\mathtt{x}$ in ${\mathrm{N}}_a$.

Remark 1.Observe that:

• The υ-monotone strictly increasing can be obtained from the Definition 3 by replacing “≥” sign with “>” sign;
• The υ-monotone strictly decreasing can be obtained from the Definition 3 by replacing “≤” sign with “<” sign.

Definition 4 (see [20]). Let $1\le \upsilon \le 2$ and $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$. Then, $\mathtt{h}$ is called an υ-convex function on ${\mathrm{N}}_a$, if

$$\begin{array}{c}\hfill \mathtt{h}\left(\mathtt{x}\right)-\upsilon \mathtt{h}(\mathtt{x}-1)+(\upsilon -1)\mathtt{h}(\mathtt{x}-2)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right).\end{array}$$

In other words, we say that $\mathtt{h}$ is an υ-convex function if and only if $(\nabla \mathtt{h})$ is an $(\upsilon -1)$-monotone increasing.

3.1. Positivity and Monotonicity Results

Lemma 1.

If a function $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ satisfies

$$\begin{array}{c}\hfill \left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\mathtt{h}\left(a\right){(1-\upsilon )}^{\mathtt{x}-a}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\end{array}$$

and $\upsilon \in \left(0,1\right)$, then $\mathtt{h}\left(\mathtt{x}\right)$ is positive and υ–monotone increasing on ${\mathrm{N}}_a$.

Proof. 

Considering $\frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\mathtt{h}\left(a\right){(1-\upsilon )}^{\mathtt{x}-a}\ge 0$, we see that $\mathtt{h}\left(a\right)\ge 0$ since $\frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }>0$ and ${(1-\upsilon )}^{\mathtt{x}-a}>0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$. Also, having $\left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\mathtt{h}\left(a\right){(1-\upsilon )}^{\mathtt{x}-a}$ gives that $\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ by Proposition 1. Following Lemma 3.1 in [13], we find that $\mathtt{h}\left(\mathtt{x}\right)$ is positive and $\upsilon$–monotone increasing on ${\mathrm{N}}_a$ as required. □

Theorem 1.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ and $0<\upsilon ,\mu <1$ with $0<\upsilon +\mu \le 1$. If $\mathtt{h}$ satisfies $\mathtt{h}\left(a\right)\ge 0$ and

$$\begin{array}{c}\hfill \left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)\mathcal{B}\left(\mu \right)}{1-\mu }\left(\nabla \mathtt{h}\right)(a+1){(1-\mu )}^{\mathtt{x}-a-1}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\end{array}$$

then $\mathtt{h}\left(\mathtt{x}\right)$ is positive and $\upsilon +\mu$–monotone increasing on ${\mathrm{N}}_a$.

Proof. 

Set $\mathtt{g}\left(\mathtt{x}\right):=\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)$, it follows from Definition 1 that:

$$\begin{array}{cc}\hfill \mathtt{g}(a+1)& =\mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{a+1}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{a+1-\mathtt{s}}=\mathcal{B}\left(\upsilon \right)\left(\nabla \mathtt{h}\right)(a+1).\hfill \end{array}$$

(3)

From the assumption, we have:

$$\begin{array}{c}\hfill \left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\mu }\mathtt{g}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\mu \right)}{1-\mu }\mathtt{g}(a+1){(1-\mu )}^{\mathtt{x}-a-1}\ge 0.\end{array}$$

(4)

Thus, by making use of Lemma 1, we see that $\mathtt{g}\left(\mathtt{x}\right)$ is positive and $\mu$–monotone increasing on ${\mathrm{N}}_{a+1}$. This means that:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \mathtt{g}\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\hfill \\ \hfill & \mathtt{g}\left(\mathtt{x}\right)\ge \mu \mathtt{g}(\mathtt{x}-1)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right).\hfill \end{array}\end{array}$$

(5)

From $\mathtt{h}\left(a\right)\ge 0$, (5) and Proposition 1, we have:

$$\begin{array}{c}\hfill \left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\mathtt{h}\left(a\right){(1-\upsilon )}^{\mathtt{x}-a}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right).\end{array}$$

Again, from Lemma 1, we see that $\mathtt{h}\left(\mathtt{x}\right)$ is positive and $\upsilon$–monotone increasing on ${\mathrm{N}}_a$. This gives:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \mathtt{h}\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_a\right),\hfill \\ \hfill & \mathtt{h}\left(\mathtt{x}\right)\ge \mu \mathtt{h}(\mathtt{x}-1)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right).\hfill \end{array}\end{array}$$

(6)

Thus, the positivity of $\mathtt{h}$ is proved. It remains to show that $\mathtt{h}$ is $\upsilon +\mu$–monotone increasing on ${\mathrm{N}}_a$. That is, we shall show that:

$$\begin{array}{c}\hfill \mathtt{h}\left(\mathtt{x}\right)\ge (\upsilon +\mu )\mathtt{h}(\mathtt{x}-1)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\end{array}$$

(7)

by using induction. From (3) and (5), we get $\mathtt{g}(a+1)=\mathcal{B}\left(\upsilon \right)\left(\nabla \mathtt{h}\right)(a+1)\ge 0$ and it gives $\mathtt{h}(a+1)\ge \mathtt{h}\left(a\right)$. By making use of this, (5) and (6), we get:

$$\begin{array}{c}\hfill \underset{\mathrm{since}\upsilon +\mu \le 1\mathrm{by}\mathrm{assumption}}{\underbrace{\mathtt{h}(a+1)\ge (\upsilon +\mu )\mathtt{h}(a+1)}}\ge (\upsilon +\mu )\mathtt{h}\left(a\right).\end{array}$$

This implies that the rule is true for $\mathtt{x}=a+1$. In view of (5), we have for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$:

$$\begin{array}{cc}\hfill \mathtt{g}\left(\mathtt{x}\right)-\mu \mathtt{g}(\mathtt{x}-1)& =\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)-\mu \left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)(\mathtt{x}-1)\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}-\mu \mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}-1}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-1-\mathtt{s}}\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\left(\mathtt{h}\left(\mathtt{s}\right)-\mathtt{h}(\mathtt{s}-1)\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\hfill \\ \hfill & -\mu \mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}-1}\left(\mathtt{h}\left(\mathtt{s}\right)-\mathtt{h}(\mathtt{s}-1)\right){(1-\upsilon )}^{\mathtt{x}-1-\mathtt{s}}\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =\mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}-\mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\mathtt{h}(\mathtt{s}-1){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\hfill \\ \hfill & -\mu \mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}-1}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-1-\mathtt{s}}+\mu \mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}-1}\mathtt{h}(\mathtt{s}-1){(1-\upsilon )}^{\mathtt{x}-1-\mathtt{s}}\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\left[\mathtt{h}\left(\mathtt{x}\right)-{(1-\upsilon )}^{\mathtt{x}-a-1}\mathtt{h}\left(a\right)-\frac{\upsilon }{1-\upsilon }\sum _{\mathtt{s}=a+1}^{\mathtt{x}-1}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]\hfill \\ \hfill & -\mu \mathcal{B}\left(\upsilon \right)\left[\mathtt{h}(\mathtt{x}-1)-{(1-\upsilon )}^{\mathtt{x}-a-2}\mathtt{h}\left(a\right)-\frac{\upsilon }{1-\upsilon }\sum _{\mathtt{s}=a+1}^{\mathtt{x}-2}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}-1}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =\mathcal{B}\left(\upsilon \right)\left[\mathtt{h}\left(\mathtt{x}\right)-{(1-\upsilon )}^{\mathtt{x}-a-1}\mathtt{h}\left(a\right)-\upsilon \mathtt{h}(\mathtt{x}-1)-\frac{\upsilon }{1-\upsilon }\sum _{\mathtt{s}=a+1}^{\mathtt{x}-2}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]\hfill \\ \hfill & -\mu \mathcal{B}\left(\upsilon \right)\left[\mathtt{h}(\mathtt{x}-1)-{(1-\upsilon )}^{\mathtt{x}-a-2}\mathtt{h}\left(a\right)-\frac{\upsilon }{{(1-\upsilon )}^2}\sum _{\mathtt{s}=a+1}^{\mathtt{x}-2}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\{\left[\mathtt{h}\left(\mathtt{x}\right)-\upsilon \mathtt{h}(\mathtt{x}-1)-\mu \mathtt{h}(\mathtt{x}-1)\right]-[1-\upsilon -\mu ]{(1-\upsilon )}^{\mathtt{x}-a-2}\mathtt{h}\left(a\right)\hfill \\ \hfill & -\frac{\upsilon [1-\upsilon -\mu ]}{{(1-\upsilon )}^2}\sum _{\mathtt{s}=a+1}^{\mathtt{x}-2}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\}\ge 0.\hfill \end{array}$$

Since $\mathcal{B}\left(\upsilon \right)>0$, the last inequality implies that:

$$\begin{array}{cc}\hfill \mathtt{h}\left(\mathtt{x}\right)-\upsilon \mathtt{h}(\mathtt{x}-1)-\mu \mathtt{h}(\mathtt{x}-1)& \ge [1-\upsilon -\mu ]{(1-\upsilon )}^{\mathtt{x}-a-2}\mathtt{h}\left(a\right)\hfill \\ \hfill & +\frac{\upsilon [1-\upsilon -\mu ]}{{(1-\upsilon )}^2}\sum _{\mathtt{s}=a+1}^{\mathtt{x}-2}\mathtt{h}\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\hfill \end{array}$$

since $0<\upsilon <1$, $0<\upsilon +\mu \le 1$, and $\mathtt{h}\left(\mathtt{x}\right)\ge 0$ by (7) for every $\mathtt{x}\in {\mathrm{N}}_a$. Hence the result. □

Corollary 1.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ and $0<\upsilon ,\mu <1$. If $\mathtt{h}$ satisfies $\mathtt{h}\left(a\right)\ge 0$ and

$$\begin{array}{c}\hfill \left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)\mathcal{B}\left(\mu \right)}{1-\mu }\left(\nabla \mathtt{h}\right)(a+1){(1-\mu )}^{\mathtt{x}-a-1}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\end{array}$$

then $\mathtt{h}\left(\mathtt{x}\right)$ is positive and υ–monotone increasing on ${\mathrm{N}}_a$.

Proof. 

This is a direct consequence of Theorem 1. □

3.2. Monotonicity and $\upsilon$-Convexity Results

Lemma 2 (see Lemma 4.1 in [13]). Let $\upsilon \in \left(0,1\right)$. If a function $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ satisfies $\left(\nabla \mathtt{h}\right)(a+1)\ge 0$ and

$$\begin{array}{c}\hfill {\nabla }_t\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\end{array}$$

then $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$.

Lemma 3.

Let $0<\upsilon <1$. If a function $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ satisfies

$$\begin{array}{c}\hfill \left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\upsilon }{\nabla }_t\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\left(\nabla \mathtt{h}\right)(a+1){(1-\upsilon )}^{\mathtt{x}-a-1}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\end{array}$$

then $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$.

Proof. 

We know from $\frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\left(\nabla \mathtt{h}\right)(a+1){(1-\upsilon )}^{\mathtt{x}-a-1}\ge 0$ that $\left(\nabla \mathtt{h}\right)(a+1)\ge 0$ where we know that $\frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }>0$ and ${(1-\upsilon )}^{\mathtt{x}-a-1}>0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$. Also, considering $\left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }{\nabla }_t\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\left(\nabla \mathtt{h}\right)(a+1){(1-\upsilon )}^{\mathtt{x}-a-1}$, we have that $\left({{}_{a+1}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}{\nabla }_t\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ by Proposition 1. Then, by using Lemma 4.2 in [13], we get $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$ as required. □

Theorem 2.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$, $0<\upsilon <1$ and $1<\mu <2$. If $\mathtt{h}$ satisfies $\left(\nabla \mathtt{h}\right)(a+1)\ge 0$ and

$$\begin{array}{cc}\hfill & \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}\left(\upsilon \right)\mathcal{B}(\mu -1)}{2-\mu }\left[\left(\nabla \mathtt{h}\right)(a+2)-\upsilon \left(\nabla \mathtt{h}\right)(a+1)\right]{(2-\mu )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right),\hfill \end{array}$$

then $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$.

Proof. 

Set $\mathtt{g}\left(\mathtt{x}\right):=\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)$. Then, from Definition 1, we have:

$$\begin{array}{cc}\hfill \left(\nabla \mathtt{g}\right)(a+2)& =\left(\nabla {{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)(a+2)\hfill \\ \hfill & ={\nabla }_{\mathtt{x}}{\left[\mathcal{B}\left(\upsilon \right)\sum _{\mathtt{s}=a+1}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]}_{\mathtt{x}=a+2}\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\left[\sum _{\mathtt{s}=a+1}^{a+2}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{a+2-\mathtt{s}}-\sum _{\mathtt{s}=a+1}^{a+1}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(1-\upsilon )}^{a+1-\mathtt{s}}\right]\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\left[\left(\nabla \mathtt{h}\right)(a+1)(1-\upsilon )+\left(\nabla \mathtt{h}\right)(a+2)-\left(\nabla \mathtt{h}\right)(a+1)\right]\hfill \\ \hfill & =\mathcal{B}\left(\upsilon \right)\left[\left(\nabla \mathtt{h}\right)(a+2)-\upsilon \left(\nabla \mathtt{h}\right)(a+1)\right].\hfill \end{array}$$

(8)

From Definition 2, the assumption and (8), we have:

$$\begin{array}{cc}\hfill \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\mu -1}\nabla \mathtt{g}\right)\left(\mathtt{x}\right)& =\left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\mu -1}\nabla {{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & =\left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}\left(\upsilon \right)\mathcal{B}(\mu -1)}{2-\mu }\left[\left(\nabla \mathtt{h}\right)(a+2)-\upsilon \left(\nabla \mathtt{h}\right)(a+1)\right]{(2-\mu )}^{\mathtt{x}-a-2}\hfill \\ \hfill & =\frac{\mathcal{B}(\mu -1)}{2-\mu }\left(\nabla \mathtt{g}\right)(a+2){(2-\mu )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\hfill \end{array}$$

Moreover, we know that $0<\mu -1<1$, where $1<\mu <2$. Thus, $\left(\nabla \mathtt{g}\right)\left(\mathtt{x}\right)=\left(\nabla {{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$ by Lemma 3. Now, since $\left(\nabla \mathtt{h}\right)(a+1)\ge 0$ by assumption, therefore, $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$ by Lemma 2. Hence, the desired result is achieved. □

Theorem 3.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$, $1<\upsilon <2$ and $0<\mu <1$. If $\mathtt{h}$ satisfies $\left(\nabla \mathtt{h}\right)(a+1)\ge 0$ and

$$\begin{array}{cc}\hfill & \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_{a+1}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}(\upsilon -1)\mathcal{B}\left(\mu \right)}{1-\mu }\left({\nabla }^2\mathtt{h}\right)(a+2){(1-\mu )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right),\hfill \end{array}$$

then $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$.

Proof. 

Let us denote $\left(\nabla \mathtt{g}\right)\left(\mathtt{x}\right):=\left({{}_{a+1}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)$. Then, from Definitions 1 and 2, we have:

$$\begin{array}{cc}\hfill \left(\nabla \mathtt{g}\right)(a+2)& =\left({{}_{a+1}{}^{CFC}\nabla _{}^{}}^{\upsilon -1}\nabla \mathtt{h}\right)(a+2)\hfill \\ \hfill & ={\left[\mathcal{B}(\upsilon -1)\sum _{\mathtt{s}=a+2}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}^2\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]}_{\mathtt{x}=a+2}\hfill \\ \hfill & =\mathcal{B}(\upsilon -1)\sum _{\mathtt{s}=a+2}^{a+2}\left({\nabla }_{\mathtt{s}}^2\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{a+2-\mathtt{s}}\hfill \\ \hfill & =\mathcal{B}(\upsilon -1)\left({\nabla }^2\mathtt{h}\right)(a+2).\hfill \end{array}$$

(9)

By making use of Definition 2, the assumption and (9), we have:

$$\begin{array}{cc}\hfill \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\mu }\nabla \mathtt{g}\right)\left(\mathtt{x}\right)& =\left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_{a+1}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}(\upsilon -1)\mathcal{B}\left(\mu \right)}{1-\mu }\left({\nabla }^2\mathtt{h}\right)(a+2){(1-\mu )}^{\mathtt{x}-a-2}\hfill \\ \hfill & =\frac{\mathcal{B}\left(\mu \right)}{1-\mu }\left(\nabla \mathtt{g}\right)(a+2){(1-\mu )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\hfill \end{array}$$

Thus, $\left(\nabla \mathtt{g}\right)\left(\mathtt{x}\right)=\left({{}_{a+1}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)=\left({{}_{a+1}{}^{CFC}\nabla _{}^{}}^{\upsilon -1}\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$ by Lemma 3. Now, since $\left(\nabla \mathtt{h}\right)(a+1)\ge 0$ by assumption and $0<\upsilon -1<1$ where $1<\upsilon <2$. Therefore, $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$ by Lemma 2, and thus the proof is completed. □

Our last result in this subsection is about $\upsilon$-convexity.

Lemma 4.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ and $1<\upsilon <2$. If $\mathtt{h}$ satisfies

$$\begin{array}{cc}\hfill & \left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}(\upsilon -1)}{2-\upsilon }\left(\nabla \mathtt{h}\right)(a+1){(2-\upsilon )}^{\mathtt{x}-a-1}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\hfill \end{array}$$

then $\mathtt{h}$ is positive and monotone increasing on ${\mathrm{N}}_a$. In addition, $\mathtt{h}$ is υ-convex on ${\mathrm{N}}_a$.

Proof. 

Since from the assumption and Definition 2:

$$\begin{array}{cc}\hfill \left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\upsilon -1}\nabla \mathtt{h}\right)\left(\mathtt{x}\right)& =\left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}(\upsilon -1)}{2-\upsilon }\left(\nabla \mathtt{h}\right)(a+1){(2-\upsilon )}^{\mathtt{x}-a-1}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\hfill \end{array}$$

and since $0<\upsilon -1<1$ where $1<\upsilon <2$, we have $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$ by Lemma 3. Therefore, $\mathtt{h}$ is monotone increasing and positive on ${\mathrm{N}}_a$.

Now, we shall show that $\mathtt{h}$ is $\upsilon$-convex on ${\mathrm{N}}_a$. From Definition 1 and the assumption, we have for any $\mathtt{x}\in {\mathrm{N}}_{a+2}$:

$$\begin{array}{cc}\hfill & \left({{}_{a+1}{}^{CFR}\nabla _{}^{}}^{\upsilon -1}\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & =\mathcal{B}(\upsilon -1)\left[{\nabla }_{\mathtt{x}}\sum _{\mathtt{s}=a+2}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]\hfill \\ \hfill & =\mathcal{B}(\upsilon -1)\left[\sum _{\mathtt{s}=a+2}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}-\sum _{\mathtt{s}=a+2}^{\mathtt{x}-1}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-1-\mathtt{s}}\right]\hfill \\ \hfill & =\mathcal{B}(\upsilon -1)\left[\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)+\sum _{\mathtt{s}=a+2}^{\mathtt{x}-1}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}-\frac{1}{2-\upsilon }\sum _{\mathtt{s}=a+2}^{\mathtt{x}-1}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =\mathcal{B}(\upsilon -1)\left[\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)+\frac{1-\upsilon }{2-\upsilon }\sum _{\mathtt{s}=a+2}^{\mathtt{x}-1}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]\hfill \\ \hfill & =\mathcal{B}(\upsilon -1)\left[\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)+(1-\upsilon )\left(\nabla \mathtt{h}\right)(\mathtt{x}-1)+\frac{1-\upsilon }{2-\upsilon }\sum _{\mathtt{s}=a+2}^{\mathtt{x}-2}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right]\hfill \\ \hfill & \ge 0.\hfill \end{array}$$

(10)

Since $1<\upsilon <2,\mathcal{B}(\upsilon -1)>0$, $\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ and ${(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}>0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$ and $\mathtt{s}=a+2,a+3,\dots ,\mathtt{x}-2$, it follows from (10) that

$$\begin{array}{cc}\hfill \left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)& \ge (\upsilon -1)\left(\nabla \mathtt{h}\right)(\mathtt{x}-1)+\frac{\upsilon -1}{2-\upsilon }\sum _{\mathtt{s}=a+2}^{\mathtt{x}-2}\left({\nabla }_{\mathtt{s}}\mathtt{h}\right)\left(\mathtt{s}\right){(2-\upsilon )}^{\mathtt{x}-\mathtt{s}}]\hfill \\ \hfill & \ge (\upsilon -1)\left(\nabla \mathtt{h}\right)(\mathtt{x}-1)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\hfill \end{array}$$

which implies that $\mathtt{h}$ is $\upsilon$-convex on ${\mathrm{N}}_a$. Hence the proof. □

3.3. Convexity Results

Lemma 5.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ and $2<\upsilon <3$. If $\mathtt{h}$ satisfies

$$\begin{array}{cc}\hfill & \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left({\nabla }^2\mathtt{h}\right)(a+2){(3-\upsilon )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right),\hfill \end{array}$$

then $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$.

Proof. 

Let us denote $\mathtt{g}\left(\mathtt{x}\right):=\left(\nabla \mathtt{h}\right)\left(\mathtt{x}\right)$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$. Then, from Definitions 1 and 2 and the assumption, we have:

$$\begin{array}{cc}\hfill \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\upsilon }\nabla \mathtt{g}\right)\left(\mathtt{x}\right)& =\left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\upsilon -2}{\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left({\nabla }^2\mathtt{h}\right)(a+2){(3-\upsilon )}^{\mathtt{x}-a-2}\hfill \\ \hfill & =\frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left(\nabla \mathtt{g}\right)(a+2){(3-\upsilon )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\hfill \end{array}$$

We also know from $2<\upsilon <3$ that $0<\upsilon -2<1$. Therefore, $\left(\nabla \mathtt{g}\right)\left(\mathtt{x}\right)=\left({\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$ for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$ by Lemma 3. This means that $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$, and hence the proof is completed. □

Lemma 6 (see Lemma 5.2 in [13]). Let $0<\upsilon <1$. If a function $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$ satisfies $\left(\nabla \mathtt{h}\right)(a+2)\ge \left(\nabla \mathtt{h}\right)(a+1)\ge 0$ and

$$\begin{array}{c}\hfill {\nabla }_t^2\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right),\end{array}$$

then $\left({\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$.

Now, we will state and prove for sequential operators our first result on convexity in the following theorem.

Theorem 4.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$, $2<\upsilon <3$ and $0<\mu <1$. If $\mathtt{h}$ satisfies $\left({\nabla }^2\mathtt{h}\right)(a+2)\ge 0$ and

$$\begin{array}{cc}\hfill & \left({{}_{a+3}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}(\upsilon -2)\mathcal{B}\left(\mu \right)}{1-\mu }\left({\nabla }^3\mathtt{h}\right)(a+3){(1-\mu )}^{\mathtt{x}-a-3}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+4}\right),\hfill \end{array}$$

then $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$.

Proof. 

First, we denote $\mathtt{g}\left(\mathtt{x}\right):=\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)$ for each $\mathtt{x}\in {\mathrm{N}}_{a+3}$. Considering,

$$\begin{array}{cc}\hfill \mathtt{g}(a+3)& =\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon -2}{\nabla }^2\mathtt{h}\right)(a+3)\hfill \\ \hfill & =\mathcal{B}(\upsilon -2)\sum _{\mathtt{s}=a+3}^{a+3}\left({\nabla }_{\mathtt{s}}^3\mathtt{h}\right)\left(\mathtt{s}\right){(3-\upsilon )}^{a+3-\mathtt{s}}\hfill \\ \hfill & =\mathcal{B}(\upsilon -2)\left({\nabla }^3\mathtt{h}\right)(a+3).\hfill \end{array}$$

(11)

By making use of (11) into the assumption, we obtain:

$$\begin{array}{cc}\hfill \left({{}_{a+3}{}^{CFR}\nabla _{}^{}}^{\mu }\mathtt{g}\right)\left(\mathtt{x}\right)& =\left({{}_{a+3}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}(\upsilon -2)\mathcal{B}\left(\mu \right)}{1-\mu }\left({\nabla }^3\mathtt{h}\right)(a+3){(1-\mu )}^{\mathtt{x}-a-3}\hfill \\ \hfill & =\frac{\mathcal{B}\left(\mu \right)}{1-\mu }\mathtt{g}(a+3){(1-\mu )}^{\mathtt{x}-a-3}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+4}\right).\hfill \end{array}$$

Therefore, $\mathtt{g}$ is positive and $\upsilon$-monotone increasing on ${\mathrm{N}}_{a+3}$ by Lemma 1. That is,

$$\begin{array}{c}\hfill \mathtt{g}\left(\mathtt{x}\right)=\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\end{array}$$

From the relationship in Proposition 1 and Definition 1, it follows that:

$$\begin{array}{c}\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)=\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon -2}{\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ =\left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\upsilon -2}{\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\\ \hfill -\frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left({\nabla }^2\mathtt{h}\right)(a+2){(3-\upsilon )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\end{array}$$

Since $3-\upsilon >0,\mathcal{B}(\upsilon -2)>0$, ${(3-\upsilon )}^{\mathtt{x}-a-2}>0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+3}$, and $\left({\nabla }^2\mathtt{h}\right)(a+2)\ge 0$, it follows that:

$$\begin{array}{c}\hfill \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left({\nabla }^2\mathtt{h}\right)(a+2){(3-\upsilon )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right),\end{array}$$

which implies that $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$ by Lemma 5. Thus the proof is completed. □

The following results in Theorems 5 and 6 can provide 2 variations on Theorem 4 regarding where the pair of parameters $(\upsilon ,\mu )$ lives.

Theorem 5.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$, $2<\upsilon <3$ and $1<\mu <2$. If $\mathtt{h}$ satisfies $\left({\nabla }^2\mathtt{h}\right)(a+2)\ge 0$ and

$$\begin{array}{c}\left({{}_{a+4}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}(\upsilon -2)\mathcal{B}(\mu -1)}{2-\mu }\hfill \\ \hfill \times \left[\left({\nabla }^3\mathtt{h}\right)(a+4)-(\upsilon -2)\left({\nabla }^3\mathtt{h}\right)(a+3)\right]{(2-\mu )}^{\mathtt{x}-a-4}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+5}\right),\end{array}$$

then $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$.

Proof. 

Denoting $\mathtt{g}\left(\mathtt{x}\right):=\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)$ for each $\mathtt{x}\in {\mathrm{N}}_{a+3}$. For $\mathtt{x}\in {\mathrm{N}}_{a+4}$, we have

$$\begin{array}{cc}\hfill \left(\nabla \mathtt{g}\right)\left(\mathtt{x}\right)& =\nabla \left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon -2}{\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & =\mathcal{B}(\upsilon -2)\nabla \sum _{\mathtt{s}=a+3}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}^3\mathtt{h}\right)\left(\mathtt{s}\right){(3-\upsilon )}^{\mathtt{x}-\mathtt{s}}\hfill \\ \hfill & =\mathcal{B}(\upsilon -2)\left[\sum _{\mathtt{s}=a+3}^{\mathtt{x}}\left({\nabla }_{\mathtt{s}}^3\mathtt{h}\right)\left(\mathtt{s}\right){(3-\upsilon )}^{\mathtt{x}-\mathtt{s}}-\sum _{\mathtt{s}=a+3}^{\mathtt{x}-1}\left({\nabla }_{\mathtt{s}}^3\mathtt{h}\right)\left(\mathtt{s}\right){(3-\upsilon )}^{\mathtt{x}-1-\mathtt{s}}\right]\hfill \\ \hfill & =\mathcal{B}(\upsilon -2)\left[\left({\nabla }_{\mathtt{s}}^3\mathtt{h}\right)\left(\mathtt{x}\right)-\frac{\upsilon -2}{3-\upsilon }\sum _{\mathtt{s}=a+3}^{\mathtt{x}-1}\left({\nabla }_{\mathtt{s}}^3\mathtt{h}\right)\left(\mathtt{s}\right){(3-\upsilon )}^{\mathtt{x}-\mathtt{s}}\right].\hfill \end{array}$$

For $\mathtt{x}=a+4$, it follows that:

$$\begin{array}{cc}\hfill \left(\nabla \mathtt{g}\right)(a+4)& =\mathcal{B}(\upsilon -2)\left[\left({\nabla }^3\mathtt{h}\right)(a+4)-(\upsilon -2)\left({\nabla }^3\mathtt{h}\right)(a+3)\right].\hfill \end{array}$$

(12)

Using (12) in the assumption to obtain

$$\begin{array}{cc}\hfill \left({{}_{a+4}{}^{CFR}\nabla _{}^{}}^{\mu }\mathtt{g}\right)\left(\mathtt{x}\right)& =\left({{}_{a+3}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}(\upsilon -2)\mathcal{B}(\mu -1)}{2-\mu }\left[\left({\nabla }^3\mathtt{h}\right)(a+4)-(\upsilon -2)\left({\nabla }^3\mathtt{h}\right)(a+3)\right]{(2-\mu )}^{\mathtt{x}-a-4}\hfill \\ \hfill & =\frac{\mathcal{B}(\mu -1)}{2-\mu }\left(\nabla \mathtt{g}\right)(a+4){(2-\mu )}^{\mathtt{x}-a-4}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+5}\right).\hfill \end{array}$$

Therefore, $\mathtt{g}$ is positive and monotone increasing on ${\mathrm{N}}_{a+3}$ by Lemma 4. That is,

$$\begin{array}{c}\hfill \mathtt{g}\left(\mathtt{x}\right)=\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\end{array}$$

Again, from the relationship in Proposition 1 and Definition 1, it follows that:

$$\begin{array}{cc}\hfill \left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)& =\left({{}_{a+2}{}^{CFC}\nabla _{}^{}}^{\upsilon -2}{\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & =\left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\upsilon -2}{\nabla }^2\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & -\frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left({\nabla }^2\mathtt{h}\right)(a+2){(3-\upsilon )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\hfill \end{array}$$

Considering $3-\upsilon >0,\mathcal{B}(\upsilon -2)>0$, ${(3-\upsilon )}^{\mathtt{x}-a-2}>0$ for each $\mathtt{x}\in {\mathrm{N}}_{a+3}$, and $\left({\nabla }^2\mathtt{h}\right)(a+2)\ge 0$, it follows that:

$$\begin{array}{c}\hfill \left({{}_{a+2}{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left({\nabla }^2\mathtt{h}\right)(a+2){(3-\upsilon )}^{\mathtt{x}-a-2}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right)\end{array}$$

and which implies that $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$ by Lemma 5. Thus, the proof is completed. □

Theorem 6.

Let $\mathtt{h}:{\mathrm{N}}_a\to \mathsf{R}$, $0<\upsilon <1$ and $2<\mu <3$. If $\mathtt{h}$ satisfies $\left(\nabla \mathtt{h}\right)(a+3)\ge (\upsilon +1)\left(\nabla \mathtt{h}\right)(a+2)\ge 0$ and

$$\begin{array}{c}\left({{}_{a+3}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge \frac{\mathcal{B}\left(\upsilon \right)\mathcal{B}(\mu -2)}{3-\mu }[\left(\nabla \mathtt{h}\right)(a+3)\hfill \\ \hfill -(1+\upsilon )\left(\nabla \mathtt{h}\right)(a+2)+{\upsilon }^2\left(\nabla \mathtt{h}\right)(a+1)]{(3-\mu )}^{\mathtt{x}-a-3}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+4}\right),\end{array}$$

then $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$.

Proof. 

Denoting $\mathtt{g}\left(\mathtt{x}\right):=\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)$ for each $\mathtt{x}\in {\mathrm{N}}_{a+1}$. Then, from the proof of Theorem 5.5 in [13], we have:

$$\begin{array}{cc}\hfill \left({\nabla }^2\mathtt{g}\right)(a+3)& =\mathcal{B}\left(\upsilon \right)\left[\left(\nabla \mathtt{h}\right)(a+3)-(1+\upsilon )\left(\nabla \mathtt{h}\right)(a+2)+{\upsilon }^2\left(\nabla \mathtt{h}\right)(a+1)\right].\hfill \end{array}$$

(13)

Using (13) into the assumption to obtain:

$$\begin{array}{cc}\hfill \left({{}_{a+3}{}^{CFR}\nabla _{}^{}}^{\mu }\mathtt{g}\right)\left(\mathtt{x}\right)& =\left({{}_{a+3}{}^{CFR}\nabla _{}^{}}^{\mu }{{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\hfill \\ \hfill & \ge \frac{\mathcal{B}\left(\upsilon \right)\mathcal{B}(\mu -2)}{3-\mu }[\left(\nabla \mathtt{h}\right)(a+3)-(1+\upsilon )\left(\nabla \mathtt{h}\right)(a+2)\hfill \\ \hfill & +{\upsilon }^2\left(\nabla \mathtt{h}\right)(a+1)]{(3-\mu )}^{\mathtt{x}-a-3}\hfill \\ \hfill & =\frac{\mathcal{B}(\mu -2)}{3-\mu }\left({\nabla }^2\mathtt{g}\right)(a+3){(3-\mu )}^{\mathtt{x}-a-3}\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+4}\right).\hfill \end{array}$$

Therefore, $\mathtt{g}$ is convex on ${\mathrm{N}}_{a+3}$ by Lemma 5. That is,

$$\begin{array}{c}\hfill \left({\nabla }^2\mathtt{g}\right)\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\end{array}$$

That is,

$$\begin{array}{c}\hfill {\nabla }^2\left({{}_a{}^{CFC}\nabla _{}^{}}^{\upsilon }\mathtt{h}\right)\left(\mathtt{x}\right)\ge 0\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+3}\right).\end{array}$$

Since by assumption $\left(\nabla \mathtt{h}\right)(a+2)\ge \left(\nabla \mathtt{h}\right)(a+1)\ge 0$, we can deduce that $\mathtt{h}$ is convex on ${\mathrm{N}}_{a+2}$ by Lemma 6. This completes the proof. □

4. Applications

In this section, we provide some numerical investigation tests of discrete nabla $\mathtt{CFR}$ fractional differences equations with various initial data.

Example 1.

In this example, a positivity analysis result is achieved. For $\upsilon \in (0,1)$ and $c\ge 0$, we consider the fractional difference initial value problem:

$$\begin{array}{cc}\hfill \left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{y}\right)(\mathtt{x}+1)& =\mathtt{h}\left(\mathtt{x},\mathtt{y}\left(\mathtt{x}\right)\right)\left(\forall \mathtt{x}\in {\mathrm{N}}_a\right),\hfill \\ \hfill \mathtt{y}\left(a\right)& =c.\hfill \end{array}$$

The chosen $\mathtt{h}$ should satisfy

$$\begin{array}{c}\hfill \mathtt{h}\left(\mathtt{x},\mathtt{y}\left(\mathtt{x}\right)\right)\ge \frac{\mathcal{B}\left(\upsilon \right)}{1-\upsilon }\mathtt{y}\left(a\right){(1-\upsilon )}^{\mathtt{x}+1-a}\ge 0\left(\forall (\mathtt{x},\mathtt{y})\in {\mathrm{N}}_a\times [0,\infty )\right).\end{array}$$

Therefore, the solution function $\mathtt{h}\left(\mathtt{x}\right)$ is positive and υ–monotone increasing for each $\mathtt{x}\in {\mathrm{N}}_a$ by Lemma 1.

Example 2.

This example is dedicated to a monotonicity analysis result. For $\upsilon \in (1,2)$ and $c\ge 0$, we consider the fractional difference initial value problem:

$$\begin{array}{cc}\hfill \left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{y}\right)(\mathtt{x}+1)& =\mathtt{h}\left(\mathtt{x},\mathtt{y}\left(\mathtt{x}\right)\right)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+1}\right),\hfill \\ \hfill \left(\nabla \mathtt{y}\right)\left(a\right)& =c.\hfill \end{array}$$

As before, the chosen $\mathtt{h}$ should satisfy

$$\begin{array}{c}\hfill \mathtt{h}\left(\mathtt{x},\mathtt{y}\left(\mathtt{x}\right)\right)\ge \frac{\mathcal{B}(\upsilon -1)}{2-\upsilon }\left(\nabla \mathtt{y}\right)\left(a\right){(2-\upsilon )}^{\mathtt{x}-a}\ge 0\left(\forall (\mathtt{x},\mathtt{y})\in {\mathrm{N}}_{a+1}\times [0,\infty )\right).\end{array}$$

Therefore, the solution function $\mathtt{h}\left(\mathtt{x}\right)$ is positive and monotone increasing, and thus $\mathtt{h}\left(\mathtt{x}\right)$ is υ-convex for each $\mathtt{x}\in {\mathrm{N}}_a$ by Lemma 4.

Example 3.

Our last example is dedicated to a convexity analysis result. Let $\upsilon \in (2,3)$ and $c\ge 0$. Consider the fractional difference initial value problem:

$$\begin{array}{cc}\hfill \left({{}_a{}^{CFR}\nabla _{}^{}}^{\upsilon }\mathtt{y}\right)(\mathtt{x}+1)& =\mathtt{h}\left(\mathtt{x},\mathtt{y}\left(\mathtt{x}\right)\right)\left(\forall \mathtt{x}\in {\mathrm{N}}_{a+2}\right),\hfill \\ \hfill \left({\nabla }^2\mathtt{y}\right)\left(a\right)& =c.\hfill \end{array}$$

The chosen $\mathtt{h}$ should satisfy

$$\begin{array}{c}\hfill \mathtt{h}\left(\mathtt{x},\mathtt{y}\left(\mathtt{x}\right)\right)\ge \frac{\mathcal{B}(\upsilon -2)}{3-\upsilon }\left({\nabla }^2\mathtt{y}\right)\left(a\right){(3-\upsilon )}^{\mathtt{x}-a}\ge 0\left(\forall (\mathtt{x},\mathtt{y})\in {\mathrm{N}}_{a+1}\times [0,\infty )\right).\end{array}$$

Therefore, the solution function $\mathtt{h}\left(\mathtt{x}\right)$ is convex for each $\mathtt{x}\in {\mathrm{N}}_{a+2}$ by Lemma 5.

5. Conclusions

It is of interest to notice the great development in the study of monotonicity and positivity analyses of the discrete fractional operators in recent times. Extended by these investigations, the aim of this study is to consider analysing the positivity, monotonicity and convexity of the discrete nabla $\mathtt{CF}$ fractional operators for Riemann and Caputo operators. Their correlations to the non-negativity of both non-sequential and sequential $\mathtt{CFR}$-type and $\mathtt{CFC}$-type fractional differences are also considered in different ranges. The correctness and validity of the theoretical analysis in Lemmas 1, 4 and 5 are verified through Examples 1–3.

It is worth mentioning, as a main conclusion to observe, that the action of $\mathcal{Q}$–operator (see [21]) can confirm that the results of this article are correct to the right discrete nabla $\mathtt{CF}$ fractional operators on ${}_b\mathrm{N}:=\{\dots ,b-1,b\}$, which is defined by $\left(\mathcal{Q}\mathtt{h}\right)\left(\mathtt{x}\right)=\mathtt{h}(a+b-\mathtt{x})$.