# Some useful inequalities for nabla tempered fractional calculus

Yiheng Wei  School of Mathematics, Southeast University, Nanjing 210096, China
<EMAIL_ADDRESS>,  Linlin Zhao  School of Business, Nanjing Audit
University, Nanjing 211815, China<EMAIL_ADDRESS>,  Kai Cao  School of
Mathematics, Southeast University, Nanjing 210096, China<EMAIL_ADDRESS>and
Jinde Cao  School of Mathematics, Southeast University, Nanjing 210096, China;
Yonsei Frontier Lab, Yonsei University, Seoul 03722, South Korea
<EMAIL_ADDRESS>
###### Abstract.

This paper gives particular emphasis to the nabla tempered fractional
calculus, which involves the multiplication of the rising function kernel with
tempered functions, and provides a more flexible alternative with considerable
promise for practical applications. Some remarkable inequalities for such
nabla fractional calculus are developed and analyzed, which greatly enrich the
mathematical theory of nabla tempered fractional calculus. Numerical results
confirm the validity of the developed properties once again, which also
reveals that the introduction of tempered function provides high value and
huge potential.

###### 2020 Mathematics Subject Classification:

Primary 26A33, Secondary 39A70, 34A08, 97H30, 35B51

The work described in this paper was fully supported by the National Natural
Science Foundation of China (62273092) and the National Key R$\&$D Project of
China (2020YFA0714300).

##  1\. Introduction

Fractional calculus, as a useful analytical toolbox, has attracted an
increasing attention from scholars [15]. Due to its nonlocal property,
fractional calculus play a key role in diverse application of science and
engineering, such as diffusion model [25], compartment model [6], automatic
control [33], neural network [34], etc. Especially, tempered fractional
calculus, which introduces an extra tempered function, has many merits, and
henceforth many researchers have deployed themselves to explore valuables
results for such class of fractional calculus. A huge work has been done for
this subject [35, 3, 17, 30], which makes a positive and profound impact.

For the continuous time case, many diffusion models involving tempered
fractional calculus were established and the solvability of the resulting
equation becomes important and difficult. To achieve this aim, a lot of work
has been carried out. A Chebyshev pseudospectral scheme was developed to
discretize the space-time tempered fractional diffusion equation in [24]. A
time discretization method was established for approximating the mild solution
of the tempered fractional Feynman–Kac equation in [13]. An efficient and
stable finite difference scheme was proposed for solving space tempered
fractional diffusion equations in [23]. The tempered Lévy flights were
introduced to process the anomalous diffusion problem in [14]. The equivalence
between the tempered fractional derivative operator and the Hadamard finite-
part integral was first proved and then the fractional linear multistep method
was extended to the tempered fractional integral and derivative operators in
[22]. A class of tempered fractional neural networks was proposed and the
conditions for attractivity and Mittag–Leffler stability were provided in
[21].

For the discrete time case, the research is just in its infancy and some
properties has been explored preliminarily. The memory effect of delta
tempered fractional calculus was investigated and applied to image processing
[1]. The tempered fractional derivative on an isolated time scale was defined
and a new method was presented based on the time scale theory for numerical
discretization in [19]. A general definition for nabla discrete time tempered
fractional calculus was presented in [18]. The tempered function was chosen as
the nonzero case instead of the discrete exponential function, which greatly
enrich the potential of the tempered fractional calculus. Compared with the
continuous time case, the discrete time case performs better in computing,
storage, transport, etc. and it has greater potential in the digital era.
Though the study on discrete time tempered fractional calculus is still in
sufficient, a proliferation of results reported on discrete time fractional
calculus [12, 32, 20] could give us a lot of helpful inspiration and
reference.

The basic arithmetic and equivalence relations of fractional difference and
fractional sum were discussed in [12, 32, 20, 37]. The monotonicity of
fractional difference was explored in [16, 20, 2]. The comparison principles
were developed in [39, 9, 38, 27, 28, 29], which are dominant in the stability
analysis of discrete time fractional order systems and derive fascinating
consequences. The fractional difference inequalities were developed in [9, 38,
36], and the research was not comprehensive and some new inequalities were
still expected, including the non-convex case, the Cauchy inequality, the
Jensen inequality and the Hölder inequality, etc. Firstly, some similar
properties like the classical case can be checked for the tempered case.
Secondly, efforts can be made for the unsolved problems of the classical case.
Thirdly, this work could try to produce some new remarkable results.

Since there is still a gap in the literature concerning nabla tempered
fractional calculus, the main purpose of this study is to investigate the
nabla tempered fractional calculus further and derive some fundamental
properties which are also the main contributions of this work, including i)
the monotonicity of nabla tempered fractional difference is explored; ii)
three comparison principles for nabla tempered fractional order systems are
developed; iii) a series of tempered fractional difference inequalities are
derived; and iv) many tempered fractional sum inequalities are built. Notably,
it is not an easy task to generalize the existing properties to the new field,
since the introduction of the tempered function brings some unexpected
difficulty and damage some accustomed properties. Furthermore, it is more
difficult to discover some innovative valuable properties.

The remainder of this paper is organized as follows. Section 2 presents some
preliminaries on the classical nabla fractional calculus and the tempered
case. Section 3 is devoted to deriving the main results. Section 4 provides
three numerical examples to verify the elaborated theoretical results.
Finally, this paper ends in Section 5 with some concluding remarks.

##  2\. Preliminaries

In this section, the basic definitions for nabla fractional calculus and nabla
tempered fractional calculus are presented.

For $x:\mathbb{N}_{a+1-n}\to\mathbb{R}$, its $n$-th nabla difference is
defined by

(2.1)
${\textstyle\nabla^{n}x(k):=\sum_{i=0}^{n}(-1)^{i}\big{(}\begin{smallmatrix}n\\\
i\end{smallmatrix}\big{)}x(k-i),}$

where $n\in\mathbb{Z}_{+}$, $k\in\mathbb{N}_{a+1}:=\\{a+1,a+2,\cdots\\}$,
$a\in\mathbb{R}$, $\left(\begin{smallmatrix}p\\\
q\end{smallmatrix}\right):=\frac{\Gamma(p+1)}{\Gamma(q+1)\Gamma(p-q+1)}$ is
the generalized binomial coefficient and $\Gamma(\cdot)$ is the Gamma
function.

For $x:\mathbb{N}_{a+1}\to\mathbb{R}$, its $\alpha$-th Grünwald–Letnikov
difference/sum is defined by [32, 7]

(2.2) ${\textstyle{}_{a}^{\rm
G}\nabla_{k}^{\alpha}x(k):=\sum_{i=0}^{k-a-1}(-1)^{i}\big{(}\begin{smallmatrix}\alpha\\\
i\end{smallmatrix}\big{)}x(k-i),}$

where $\alpha\in\mathbb{R}$, $k\in\mathbb{N}_{a+1}$ and $a\in\mathbb{R}$. When
$\alpha>0$, ${}_{a}^{\rm G}\nabla_{k}^{\alpha}x(k)$ represents the difference
operation. When $\alpha<0$, ${}_{a}^{\rm G}\nabla_{k}^{\alpha}x(k)$ represents
the sum operation including the fractional order case and the integer order
case. Specially, ${}_{a}^{\rm G}\nabla_{k}^{0}x(k)=x(k)$. Even though
$\alpha=n\in\mathbb{Z}_{+}$, ${}_{a}^{\rm
G}\nabla_{k}^{\alpha}x(k)\not\equiv\nabla^{n}x(k)$ for all
$k\in\mathbb{N}_{a+1}$.

Defining the rising function
$p\overline{{}^{q}}:=\frac{\Gamma(p+q)}{\Gamma(p)}$, $p\in\mathbb{Z}_{+}$,
$q\in\mathbb{R}$, (2.2) can rewritten as

(2.3)
${\textstyle\begin{array}[]{rl}{}_{a}^{\mathrm{G}}\nabla_{k}^{\alpha}x(k)=&\sum_{i=0}^{k-a-1}(-1)^{i}\frac{\Gamma(1+\alpha)}{\Gamma(i+1)\Gamma(1+\alpha-i)}x(k-i)\\\
=&\sum_{i=0}^{k-a-1}\frac{\Gamma(i-\alpha)}{\Gamma(i+1)\Gamma(-\alpha)}x(k-i)\\\
=&\sum_{i=0}^{k-a-1}\frac{(i+1)^{\overline{-\alpha-1}}}{\Gamma(-\alpha)}x(k-i)\\\
=&\sum_{i=a+1}^{k}\frac{(k-i+1)^{\overline{-\alpha-1}}}{\Gamma(-\alpha)}x(i),\end{array}}$

where $\Gamma(\theta)\Gamma(1-\theta)=\frac{\pi}{\sin(\pi\theta)}$,
$\theta\in\mathbb{R}$ is adopted.

From the previous definitions, the $\alpha$-th Riemann–Liouville fractional
difference and Caputo fractional difference for
$x:\mathbb{N}_{a+1-n}\to\mathbb{R}$, $\alpha\in(n-1,n)$, $n\in\mathbb{Z}_{+}$,
$k\in\mathbb{N}_{a+1}$ and $a\in\mathbb{R}$ are defined by [20]

(2.4)
${\textstyle{}_{a}^{\mathrm{R}}\nabla_{k}^{\alpha}x(k):=\nabla^{n}{}_{a}^{\rm
G}\nabla_{k}^{\alpha-n}x(k),}$ (2.5)
${\textstyle{}_{a}^{\mathrm{C}}\nabla_{k}^{\alpha}x(k):={}_{a}^{\rm
G}\nabla_{k}^{\alpha-n}\nabla^{n}x(k).}$

On this basis, the following properties hold.

######  Lemma 2.1.

[20, 37] For any function $x:\mathbb{N}_{a+1-n}\to\mathbb{R}$,
$n\in\mathbb{Z}_{+}$, $a\in\mathbb{R}$, one has

(2.6) ${\textstyle{}_{a}^{\rm R}\nabla_{k}^{\alpha}{}_{a}^{\rm
G}\nabla_{k}^{-\alpha}x(k)={}_{a}^{\rm C}\nabla_{k}^{\alpha}{}_{a}^{\rm
G}\nabla_{k}^{-\alpha}x(k)=x(k),}$ (2.7) ${\textstyle{}_{a}^{\rm
G}\nabla_{k}^{-\alpha}{}_{a}^{\rm
R}\nabla_{k}^{\alpha}x(k)=x(k)-\sum_{i=0}^{n-1}\frac{(k-a)\overline{{}^{\alpha-i-1}}}{\Gamma(\alpha-i)}[{}_{a}^{\rm
R}\nabla_{k}^{\alpha-i-1}x(k)]_{k=a},}$ (2.8) ${\textstyle{}_{a}^{\rm
G}\nabla_{k}^{-\alpha}{}_{a}^{\rm
C}\nabla_{k}^{\alpha}x(k)=x(k)-\sum_{i=0}^{n-1}\frac{(k-a)\overline{{}^{i}}}{i!}[\nabla^{i}x(k)]_{k=a},}$

where $k\in\mathbb{N}_{a+1}$, $\alpha\in(n-1,n)$.

By introducing a tempered function
$w:\mathbb{N}_{a+1}\to\mathbb{R}\backslash\\{0\\}$, the concept of nabla
fractional calculus can be extended further.

For $x:\mathbb{N}_{a+1}\to\mathbb{R}$, its $\alpha$-th Grünwald–Letnikov
tempered difference/sum is defined by [18]

(2.9) ${\textstyle{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}x(k):=w^{-1}(k)_{a}^{\rm
G}\nabla_{k}^{\alpha}[w(k)x(k)],}$

where $\alpha\in\mathbb{R},k\in\mathbb{N}_{a+1},a\in\mathbb{R}$ and
$w:\mathbb{N}_{a+1}\to\mathbb{R}\backslash\\{0\\}$.

The $n$-th nabla tempered difference, the $\alpha$-th Riemann–Liouville
tempered fractional difference and Caputo tempered fractional difference of
$x:\mathbb{N}_{a+1-n}\to\mathbb{R}$ can be defined by

(2.10) ${\textstyle\nabla^{n,w(k)}x(k):=w^{-1}(k)\nabla^{n}[w(k)x(k)],}$
(2.11)
${\textstyle{}_{a}^{\mathrm{R}}\nabla_{k}^{\alpha,w(k)}x(k):=w^{-1}(k){}_{a}^{\mathrm{R}}\nabla_{k}^{\alpha}[w(k)x(k)],}$
(2.12) ${\textstyle{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k):=w^{-1}(k){}_{a}^{\rm
C}\nabla_{k}^{\alpha}[w(k)x(k)],}$

respectively, where
$\alpha\in(n-1,n),n\in\mathbb{Z}_{+},k\in\mathbb{N}_{a+1},a\in\mathbb{R}$ and
$w:\mathbb{N}_{a+1}\to\mathbb{R}\backslash\\{0\\}$. On this basis, the
following relationships hold

(2.13) ${\textstyle{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k)=\nabla^{n,w(k)}{}_{a}^{\rm
G}\nabla_{k}^{\alpha-n,w(k)}x(k),}$

(2.14) ${\textstyle{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)={}_{a}^{\rm
G}\nabla_{k}^{\alpha-n,w(k)}\nabla^{n,w(k)}x(k).}$

The equivalent condition of $w:\mathbb{N}_{a+1}\to\mathbb{R}\backslash\\{0\\}$
is finite nonzero. In this work, when $w(k)=(1-\lambda)^{k-a}$,
$\lambda\in\mathbb{R}\backslash\\{1\\}$, the operations
$\nabla^{n,w(k)}x(k),{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k)$, ${}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k),{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)$
could be abbreviate as $\nabla^{n,\lambda}x(k),{}_{a}^{\rm
G}\nabla_{k}^{\alpha,\lambda}x(k),{}_{a}^{\rm
R}\nabla_{k}^{\alpha,\lambda}x(k)$, ${}_{a}^{\rm
C}\nabla_{k}^{\alpha,\lambda}x(k)$, respectively. Notably, this special case
is different from the one in [18], which facilitates the use and analysis.
Compared to existing results, the tempered function $w(k)$ is no longer
limited to the exponential function, which makes this work more general and
practical.

By using the linearity, the following lemma can be derived immediately, which
is simple while useful for understanding such fractional calculus.

######  Lemma 2.2.

For any function $x:\mathbb{N}_{a+1-n}\to\mathbb{R}$, $n\in\mathbb{Z}_{+}$,
$a\in\mathbb{R}$, finite nonzero $w(k)$, $\alpha\in(n-1,n)$,
$k\in\mathbb{N}_{a+1}$, $\lambda\in\mathbb{R}\backslash\\{0\\}$, one has

(2.15)
${\textstyle\left\\{\begin{array}[]{rl}{\nabla^{n,w(k)}}x(k)=&{\nabla^{n,\rho
w(k)}}x(k),\\\
{}_{a}^{\rm{G}}\nabla_{k}^{\alpha,w(k)}x(k)=&{}_{a}^{\rm{G}}\nabla_{k}^{\alpha,\lambda
w(k)}x(k),\\\
{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}x(k)=&{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,\lambda
w(k)}x(k),\\\
{}_{a}^{\rm{R}}\nabla_{k}^{\alpha,w(k)}x(k)=&{}_{a}^{\rm{R}}\nabla_{k}^{\alpha,\lambda
w(k)}x(k),\\\
{}_{a}^{\rm{C}}\nabla_{k}^{\alpha,w(k)}x(k)=&{}_{a}^{\rm{C}}\nabla_{k}^{\alpha,\lambda
w(k)}x(k).\end{array}\right.}$

Note that Lemma 2.2 is indeed the scale invariance. When $\lambda=-1$, the
sign of $w(k)$ is just reversed to $\lambda w(k)$. From this, one is ready to
claim that if a property on tempered calculus holds for $w(k)>0$, it also
holds for $w(k)<0$.

##  3\. Main Results

In this section, a series of inequalities concerning nabla tempered fractional
calculus will be developed.

######  Theorem 3.1.

For any $\alpha\in(0,1)$, $w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$
with $\frac{w(a)}{w(k)}>0$, $k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has

(3.1) ${\textstyle{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k)\leq{}_{a-1}^{\hskip
9.0pt\rm{R}}\nabla_{k}^{\alpha,w(k)}x(k)\leq{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k),x(a)\geq 0,}$ (3.2) ${\textstyle{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k)\geq{}_{a-1}^{\hskip
9.0pt\rm{R}}\nabla_{k}^{\alpha,w(k)}x(k)\geq{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k),x(a)\leq 0.}$

###### Proof.

Let $z(k):=w(k)x(k)$. By using (2.3) and (2.4), one has

(3.3) $\begin{array}[]{rl}{}_{a-1}^{\hskip
9.0pt\rm{R}}\nabla_{k}^{\alpha}z(k)=&\nabla_{a-1}^{\hskip
9.0pt\rm{G}}\nabla_{k}^{\alpha-1}z(k)\\\
=&\nabla\sum\nolimits_{j=a}^{k}{\frac{{({k-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}z(j)}\\\
=&{}_{a}^{\rm
R}\nabla_{k}^{\alpha}z(k)+\nabla\frac{{({k-a+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}z(a)\\\
=&{}_{a}^{\rm
R}\nabla_{k}^{\alpha}z(k)+\frac{{({k-a+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}z(a).\end{array}$

Combining (3.3) and (2.11) yields

(3.4) ${\textstyle{}_{a-1}^{\hskip
9.0pt\rm{R}}\nabla_{k}^{\alpha,w(k)}x(k)={}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k)+\frac{{({k-a+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}\frac{w(a)}{w(k)}x(a).}$

From the definition, one has

(3.5)
${\textstyle\begin{array}[]{l}{}_{a}^{\rm{R}}\nabla_{k}^{\alpha,w(k)}x(k)\\\
={\nabla^{1,w(k)}}{}_{a}^{\rm{G}}\nabla_{k}^{\alpha-1,w(k)}x(k)\\\
={w^{-1}}(k)\nabla\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}}w(j)x(j)\\\
={w^{-1}}(k)\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\nabla}[w(j)x(j)]+\frac{{(k-a)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\frac{{w(a)}}{{w(k)}}x(a)\\\
={w^{-1}}(k)\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}w(j){\nabla^{1,w(j)}}}x(j)+\frac{{(k-a)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\frac{{w(a)}}{{w(k)}}x(a)\\\
={w^{-1}}(k){}_{a}^{\rm{G}}\nabla_{k}^{\alpha-1,w(k)}{\nabla^{1,w(k)}}x(k)+\frac{{(k-a)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\frac{{w(a)}}{{w(k)}}x(a)\\\
={}_{a}^{\rm{C}}\nabla_{k}^{\alpha,w(k)}x(k)+\frac{{(k-a)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\frac{{w(a)}}{{w(k)}}x(a).\end{array}}$

Under given conditions, it is not difficult to obtain that
$\frac{w(a)}{w(k)}>0$,
$\frac{{({k-a+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}=\frac{\Gamma({k-a-\alpha})}{\Gamma({-\alpha})\Gamma({k-a+1})}<0$,
$\frac{{({k-a})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}=\frac{\Gamma({k-a-\alpha})}{\Gamma({1-\alpha})\Gamma({k-a})}>0$,
$k\in\mathbb{N}_{a+1}$. Besides, the following relationship holds

(3.6)
${\textstyle\frac{{({k-a+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}+\frac{{({k-a})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}=\frac{{\Gamma({k-a-\alpha+1})}}{{\Gamma({1-\alpha})\Gamma({k-a+1})}}>0.}$

From this, the magnitude of ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)$,
${}_{a-1}^{\hskip 9.0pt\rm{R}}\nabla_{k}^{\alpha,w(k)}x(k)$ and ${}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k)$ mainly depend on the sign of $u(a)$. Thus,
(3.1) and (3.2) establish immediately. ∎

When $\frac{w(a)}{w(k)}<0$ is adopted, the sign of inequality should reverse.
Additionally, if $w(k)=1$, $k\in\mathbb{N}_{a+1}$, (3.1) reduces to [16, Lemma
2.2]. The condition $w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with
$\frac{w(a)}{w(k)}>0$ is equivalent to $w(k)>0$ or $w(k)<0$, $\forall
k\in\mathbb{N}_{a}$.

######  Theorem 3.2.

If $\nabla^{1,w(k)}x(k)$ is nonnegative, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has ${}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k)$ is nonnegative. Besides, if $x(a)\geq 0$, then
${}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}x(k)$ is also nonnegative.

###### Proof.

From the definition, one has

(3.7)
${\textstyle\begin{array}[]{rl}{}_{a}^{\rm{C}}\nabla_{k}^{\alpha,w(k)}x(k)=&{}_{a}^{\rm{G}}\nabla_{k}^{\alpha-1,w(k)}{\nabla^{1,w(k)}}x(k)\\\
=&\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}}\frac{w(j)}{w(k)}{\nabla^{1,w(j)}}x(j).\end{array}}$

Since $\frac{w(j)}{w(k)}>0$, $\nabla^{1,w(k)}x(k)\geq 0$,
$k\in\mathbb{N}_{a+1}$, and
$\frac{{(k-j+1)\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}=\frac{{\Gamma(k-j+1-\alpha)}}{{\Gamma(k-j+1)\Gamma({1-\alpha})}}>0$,
for any $\alpha\in(0,1)$, $j\in\mathbb{N}_{a+1}^{k}$, one has
${}_{a}^{\rm{C}}\nabla_{k}^{\alpha,w(k)}x(k)\geq 0$.

Under the given conditions, it is not difficult to obtain
${}_{a}^{\rm{C}}\nabla_{k}^{\alpha,w(k)}x(k)\geq 0$,
$\frac{{({k-a})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}$
$=\frac{{\Gamma(k-a-\alpha)}}{{\Gamma(k-a)\Gamma({1-\alpha})}}>0$,
$\frac{w(a)}{w(k)}>0$, $\forall\alpha\in(0,1)$, $j\in\mathbb{N}_{a+1}^{k}$. If
$x(a)\geq 0$, the relationship ${}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k)={}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k)+\frac{{({k-a})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\frac{w(a)}{w(k)}x(a)$
will lead to the desired result
${}_{a}^{\rm{R}}\nabla_{k}^{\alpha,w(k)}x(k)\geq 0$ immediately. ∎

Theorem 3.2 can be regarded as the generalization of subsection 3.18 of [20]
and Theorem 21 of [2]. Actually, the range of the order can be wider. If
$\nabla^{n,w(k)}x(k)$ is nonnegative, then for any $\alpha\in(n-1,n)$,
$n\in\mathbb{Z}_{+}$, $w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with
$\frac{w(a)}{w(k)}>0$, $k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has
${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)$ is nonnegative. It will be more
interesting to construct the sufficient condition for $\nabla^{1,w(k)}x(k)\geq
0$ with the assumption ${}_{a}^{\mathrm{C}}\nabla_{k}^{\alpha,w(k)}x(k)\geq 0$
or ${}_{a}^{\mathrm{R}}\nabla_{k}^{\alpha,w(k)}x(k)\geq 0$

######  Theorem 3.3.

If ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)\geq{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}y(k)$ with $x(a)=y(a)$, then for any
$\alpha\in(0,1)$, $w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with
$\frac{w(a)}{w(k)}>0$, $k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has
$x(k)\geq y(k)$.

###### Proof.

Following from ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)\geq{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}y(k)$, there must exist a compensation sequence
$c(k)\geq 0$ such that

(3.8) ${\textstyle{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)={}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}y(k)+c(k).}$

Taking $\alpha$-th Grünwald–Letnikov tempered sum for both sides of (3.8)
yields

(3.9) ${\textstyle
x(k)-\frac{w(a)}{w(k)}x(a)=y(k)-\frac{w(a)}{w(k)}y(a)+{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}c(k).}$

Since $x(a)=y(a)$, the key item is
${}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}c(k)$. According to the
nonnegativeness of $c(k)$ and the invariance of the sign of $w(k)$, one has
${}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}c(k)\geq 0$. As a result, $x(k)\geq
y(k)$ is implied in (3.9). ∎

Notably, in Theorem 3.3, if $x(a)=y(a)$ is replaced by $x(a)\geq y(a)$, the
conclusion still holds. If $x(a)=y(a)$ is replaced by ${[{{}_{a}^{\rm
R}\nabla_{k}^{\alpha-1,w(k)}x(k)}]_{k=a}}={[{{}_{a}^{\rm
R}\nabla_{k}^{\alpha-1,w(k)}y(k)}]_{k=a}}$, the conclusion still holds. By
using ${}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}x(k)={}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k)+\frac{{({k-a})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\frac{w(a)}{w(k)}x(a)$,
one obtains that if ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)\geq{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}y(k)$ is replaced by ${}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k)\geq{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}y(k)$,
the conclusion still can be guaranteed. It is generally difficult to obtain
the relationship of two sequences’ fractional difference and therefore the
corresponding inequalities are considered.

######  Theorem 3.4.

If ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)\geq\mu x(k)+\gamma$,
${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}y(k)\leq\mu y(k)+\gamma$, where
$x(a)=y(a)$, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$\mu<0$, $\gamma\in\mathbb{R}$, $k\in{\mathbb{N}}_{a+1}$, $a\in\mathbb{R}$,
one has $x(k)\geq y(k)$.

###### Proof.

For ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}u(k)\geq\mu x(k)+\gamma$, there
must exist a compensation sequence $c_{1}(k)\geq 0$ such that

(3.10) ${\textstyle{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)=\mu
x(k)+\gamma+c_{1}(k).}$

Letting $\hat{x}(k)=w(k)x(k)$, $\hat{c}_{1}(k)=w(k)c_{1}(k)$, then one has
${}_{a}^{\rm C}\nabla_{k}^{\alpha}\hat{x}(k)=\mu\hat{x}(k)+\gamma
w(k)+\hat{c}_{1}(k)$. Therefore, by using the nabla linear system theory, it
follows

(3.11)
${\textstyle\begin{array}[]{rl}\hat{x}(k)=&\hat{x}(a){{\mathcal{F}}_{\alpha,1}}({\mu,k,a})+{\rm\gamma}w(k)\ast{{\mathcal{F}}_{\alpha,\alpha}}({\mu,k,a})\\\
&+\hat{c}_{1}(k)\ast{{\mathcal{F}}_{\alpha,\alpha}}({\mu,k,a}).\end{array}}$

In a similar way, letting $\hat{y}(k)=w(k)y(k)$, $\hat{c}_{2}(k)=w(k)c_{2}(k)$
where $c_{2}(k)$ is a nonnegative compensation sequence, then one has
$\hat{y}(a)=\hat{x}(a)$ and ${}_{a}^{\rm
C}\nabla_{k}^{\alpha}\hat{y}(k)=\mu\hat{y}(k)+\gamma w(k)-\hat{c}_{2}(k)$,
which implies

(3.12)
${\textstyle\begin{array}[]{rl}\hat{y}(k)=&\hat{y}(a){{\mathcal{F}}_{\alpha,1}}({\mu,k,a})+{\rm\gamma}w(k)\ast{{\mathcal{F}}_{\alpha,\alpha}}({\mu,k,a})\\\
&-\hat{c}_{2}(k)\ast{{\mathcal{F}}_{\alpha,\alpha}}({\mu,k,a}).\end{array}}$

Since ${{\mathcal{F}}_{\alpha,1}}({\mu,k,a})>0$,
${{\mathcal{F}}_{\alpha,\alpha}}({\mu,k,a})>0$ hold for any $\alpha\in(0,1)$,
$\mu<0$, $k\in{\mathbb{N}}_{a+1}$, $a\in\mathbb{R}$, when $w(k)>0$, one has
$\hat{c}_{1}(k)\geq 0$, $\hat{c}_{2}(k)\geq 0$ and $\hat{x}(k)\geq\hat{y}(k)$,
which implies $x(k)\geq y(k)$. When $w(k)<0$, one has $\hat{c}_{1}(k)\leq 0$,
$\hat{c}_{2}(k)\leq 0$ and $\hat{x}(k)\leq\hat{y}(k)$. In a similar way, the
desired result $x(k)\geq y(k)$ follows. The case of $w(k)<0$ can also be
proved by applying Lemma 2.2. ∎

To be more practical, the following nonlinear case is discussed.

######  Theorem 3.5.

If ${}_{a}^{\rm{C}}\nabla_{k}^{\alpha,w(k)}x(k)\geq-\gamma({x(k)})+h(k)$,
${}_{a}^{\rm{C}}\nabla_{k}^{\alpha,w(k)}y(k)\leq-\gamma({y(k)})$ $+h(k)$,
where $x(a)=y(a)$, $x(k),y(k)\geq 0$, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
locally Lipschitz class $\mathcal{K}$ function $\gamma$, bounded $h(k)$,
$k\in{\mathbb{N}}_{a+1}$, $a\in\mathbb{R}$, one has $x(k)\geq y(k)$.

###### Proof.

When $w(k)>0$, by combining (2.12) and the given conditions, one has

(3.13)
${\textstyle{}_{a}^{\rm{C}}\nabla_{k}^{\alpha}[w(k)x(k)]\geq-w(k)\gamma(x(k))+w(k)h(k),}$
(3.14)
${\textstyle{}_{a}^{\rm{C}}\nabla_{k}^{\alpha}[w(k)y(k)]\leq-w(k)\gamma(y(k))+w(k)h(k).}$

By combining (3.13) and (3.14), one has

(3.15)
${\textstyle{}_{a}^{\rm{C}}\nabla_{k}^{\alpha}[w(k)y(k)-w(k)x(k)]\leq-w(k)[\gamma(y(k))-\gamma(x(k))].}$

By substituting $k=a+1$ into (3.15), one has
$w(a+1)y(a+1)-w(a+1)x(a+1)\leq-w(a+1)[\gamma(y(a+1))-\gamma(x(a+1))]$ which
implies

(3.16) ${\textstyle y(a+1)+\gamma({y(a+1)})\leq x(a+1)+\gamma({x(a+1)}).}$

The monotonicity of $f(z)=z+\gamma(z)$, $z\geq 0$ results in $y(a+1)\leq
x(a+1)$.

Now, let us assume that there exist a constant $k_{1}\in\mathbb{N}_{a+1}$ such
that

(3.17) $\left\\{\begin{array}[]{l}y(k)\leq
x(k),k\in\mathbb{N}_{a}^{{k_{1}}-1},\\\ y(k)>x(k),k=k_{1}.\end{array}\right.$

From (3.17), the definition of Caputo fractional difference and the formula of
summation by parts give

(3.18)
$\begin{array}[]{l}\\{{}_{a}^{\rm{C}}\nabla_{k}^{\alpha}[w(k)y(k)-w(k)x(k)]\\}_{k={k_{1}}}\\\
=\sum\nolimits_{j=a+1}^{{k_{1}}}{\frac{{({{k_{1}}-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\\{{\nabla[w(j)y(j)]-\nabla[w(j)x(j)]}\\}}\\\
=\sum\nolimits_{j=a+1}^{{k_{1}}}{\frac{{({{k_{1}}-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\nabla\\{w(j)[y(j)-x(j)]\\}}\\\
=\frac{{({{k_{1}}-j})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}w(j)[{y(j)-x(j)}]\big{|}_{j=a}^{j={k_{1}}}\\\
\hskip
12.0pt+\sum\nolimits_{j=a+1}^{{k_{1}}}{\frac{{({{k_{1}}-j+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}w(j)[{y(j)-x(j)}]}\\\
=\sum\nolimits_{j=a+1}^{{k_{1}-1}}{\frac{{({{k_{1}}-j+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}w(j)[{y(j)-x(j)}]}\\\
\hskip 12.0pt+w(k_{1})[y(k_{1})-x(k_{1})]\\\ >0,\end{array}$

where $\frac{{0\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}=0$,
$x(a)=y(a)$,
$\frac{{({{k_{1}}-j+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}=\frac{\Gamma({k_{1}-j-\alpha})}{\Gamma({k_{1}-j+1})\Gamma({-\alpha})}<0$,
$i\in\mathbb{N}_{a+1}^{k_{1}-1}$ and $w(k)>0$, $k\in\mathbb{N}_{a+1}$ are
adopted.

By using (3.15) and (3.17), one has

(3.19)
$\begin{array}[]{l}\\{{}_{a}^{\rm{C}}\nabla_{k}^{\alpha}[w(k)y(k)-w(k)x(k)]\\}_{k={k_{1}}}\\\
\leq-w(k_{1})[\gamma(y(k_{1}))-\gamma(x(k_{1}))]\\\ \leq 0,\end{array}$

which contradicts (3.18). Consequently, $k_{1}$ does not exist. From this, it
is not difficult to obtain the desired result $x(k)\geq y(k)$ for all
$k\in{\mathbb{N}}_{a+1}$.

When $w(k)<0$, $\forall k\in{\mathbb{N}}_{a+1}$, the desired result $x(k)\geq
y(k)$, $\forall k\in{\mathbb{N}}_{a+1}$ can also be obtained by applying Lemma
2.2. ∎

The comparison principle in Theorem 3.6 \- Theorem 3.8 are inspired by Lemma
2.10 in [9], Lemma 3.14 in [29], Lemma 3.1 in [28], Lemma 3.4 in [38], Lemma
2.16 in [39], Lemma 3.3 in [27], etc. Compared with the existing ones, the
newly developed theorems are more practical.

######  Theorem 3.6.

If $W:\mathbb{R}^{n}\to\mathbb{R}$ is differentiable convex, $W({0})=0$,
$V(x(k)):=w^{-1}(k)W(z(k))$, $z(k):=w(k)x(k)$ and $\frac{{\rm
d}W({z(k)})}{{{\rm d}z(k)}}=\frac{{\rm d}V({x(k)})}{{{\rm d}x(k)}}$ holds
almost everywhere, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, $a\in\mathbb{R}$, one has

(3.20) ${\textstyle{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}V({x(k)})\leq\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.21)
${\textstyle{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}V({x(k)})\leq\frac{{{\rm
d}V({x(k)})}}{{{\rm d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.22) ${\textstyle{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}V({x(k)})\leq\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k).}$

###### Proof.

Similarly, only the case of $w(k)>0$ is considered here. Owing to the
differentiable convex condition, one has $W({z(j)})-W({z(k)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}[{z(j)-z(k)}]\geq 0$,
$j\in\mathbb{N}_{a}^{k-1}$, $k\in\mathbb{N}_{a+1}$.

On the basis of these conclusions, the subsequent proof of this theorem will
be divided into three parts.

$\blacktriangleright$ Part 1: the Caputo case

By using (2.3), (2.12) and the formula of summation by parts, it follows

(3.23) $\begin{array}[]{l}w(k)[{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}V({x(k)})-\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)]\\\ ={}_{a}^{\rm
C}\nabla_{k}^{\alpha}W({z(k)})-\frac{{{\rm d}W({z(k)})}}{{{\rm
d}{z^{\rm{T}}}(k)}}{}_{a}^{\rm C}\nabla_{k}^{\alpha}z(k)\\\ ={}_{a}^{\rm
G}\nabla_{k}^{\alpha-1}\big{[}{\nabla W({z(k)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}\nabla z(k)}\big{]}\\\
=\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}}\big{[}{\nabla
W({z(j)})-\frac{{{\rm d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}\nabla
z(j)}\big{]}\\\ =\sum\nolimits_{j=a+1}^{k}{f({j-1})\nabla g(j)}\\\
={f({j})g(j)}|_{j=a}^{j=k}-\sum\nolimits_{j=a+1}^{k}{\nabla
f(j)g(j)},\end{array}$

where $f(j):=\frac{{({k-j})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}$,
$g(j):=W({z(j)})-W({z(k)})-\frac{{{\rm d}W({z(k)})}}{{{\rm
d}{z^{\rm{T}}}(k)}}[{z(j)-z(k)}]$.

By using the property of rising function and taking first order difference
with respect to the variable $j$, one has $\nabla
f(j)=-\frac{{({k-j+1})\overline{{}^{-\alpha-1}}}}{{\Gamma({-\alpha})}}$.
According to the sign of Gamma function, one has $\nabla f(j)\geq 0$,
$j\in\mathbb{N}_{a+1}^{k-1}$, $[\nabla f(j)]_{j=k}=-1$, $f(a)>0$ and $f(k)=0$.
Due to the differentiable convex property of $W(z(k))$ with regard to $z(k)$
(see [40]), one obtains $g(j)\geq 0$, $j\in\mathbb{N}_{a}^{k-1}$ and $g(k)=0$.
Then the equation in (3.26) can be rewritten as

(3.24) ${\textstyle\begin{array}[]{l}w(k)[{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}V({x(k)})-\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k)]\\\
=-f(a)g(a)-\sum\nolimits_{j=a+1}^{k-1}{\nabla f(j)g(j)}\\\ \leq
0.\end{array}}$

Because of the positivity of $w(k)$, (3.24) implies (3.20) firmly.

$\blacktriangleright$ Part 2: the Riemann–Liouville case

By using the derived result in (3.24) and the property in (2.11), one has

(3.25) $\begin{array}[]{l}w(k)[{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}V({x(k)})-\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}x(k)]\\\ ={}_{a}^{\rm
R}\nabla_{k}^{\alpha}W({z(k)})-\frac{{{\rm d}W({z(k)})}}{{{\rm
d}{z^{\rm{T}}}(k)}}{}_{a}^{\rm R}\nabla_{k}^{\alpha}z(k)\\\ ={}_{a}^{\rm
C}\nabla_{k}^{\alpha}W({z(k)})-\frac{{{\rm d}W({z(k)})}}{{{\rm
d}{z^{\rm{T}}}(k)}}{}_{a}^{\rm C}\nabla_{k}^{\alpha}z(k)\\\ \hskip
12.0pt+\frac{{({k-a})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}W({z(a)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm
d}{z^{\rm{T}}}(k)}}\frac{{({k-a})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}z(a)\\\
=-\sum\nolimits_{j=a+1}^{k-1}{\nabla f(j)g(j)}+f(a)[W({z(k)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}z(k)].\end{array}$

Letting $z(j)=0$ in $g(j)$, $W({0})=0$ gives that $W({z(k)})-\frac{{{\rm
d}W({x(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}z(k)\leq 0$. By applying (3.24), (3.25)
and $w(k)>0$, it follows

(3.26) ${\textstyle{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}V({x(k)})-\frac{{{\rm
d}V({x(k)})}}{{{\rm d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k)\leq 0.}$

$\blacktriangleright$ Part 3: the Grünwald–Letnikov case

Based on (2.3), one has

(3.27) $\textstyle{\begin{array}[]{l}w(k)[{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}V({x(k)})-\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k)]\\\ ={}_{a}^{\rm
G}\nabla_{k}^{\alpha}W({z(k)})-\frac{{{\rm d}W({z(k)})}}{{{\rm
d}{z^{\rm{T}}}(k)}}{}_{a}^{\rm G}\nabla_{k}^{\alpha}z(k)\\\
=\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha-1}}}}{{\Gamma(-\alpha)}}\big{[}{W({z(j)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}z(j)}\big{]}.}\end{array}}$

Since it is difficult to judge the sign of ${W({z(j)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}z(j)}$, an effective way is to use the
skills of magnifying and shrinking of inequality, transforming the unknown
unfamiliar case into the known familiar case. As such, $W({0})=0$ and
$W({z(k)})-\frac{{{\rm d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}z(k)\leq 0$ can
be guaranteed. Since
$\nabla\frac{{(k-j)\overline{{}^{-\alpha}}}}{{\Gamma(1-\alpha)}}=-\frac{{(k-j+1)\overline{{}^{-\alpha-1}}}}{{\Gamma(-\alpha)}}$,
it becomes
${\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha-1}}}}{{\Gamma(-\alpha)}}}=\frac{{(k-a)\overline{{}^{-\alpha}}}}{{\Gamma(1-\alpha)}}\geq
0}$. Applying those properties in (3.23), the following desired results can be
obtained

(3.28) $\begin{array}[]{l}w(k)[{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}V({x(k)})-\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k)]\\\
\leq\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha-1}}}}{{\Gamma(-\alpha)}}\big{[}{W({z(j)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}z(j)}\big{]}}\\\ \hskip
12.0pt-\frac{{(k-a)\overline{{}^{-\alpha}}}}{{\Gamma(1-\alpha)}}\big{[}{W({z(k)})-\frac{{{\rm
d}W({z(k)})}}{{{\rm d}{z^{\rm{T}}}(k)}}z(k)}\big{]}\\\
=\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{-\alpha-1}}}}{{\Gamma(-\alpha)}}g(j)}\\\
=-\sum\nolimits_{j=a+1}^{k-1}{\nabla f(j)g(j)}-g(k)\\\ \leq 0,\end{array}$

which implies (3.22). Till now, the proof has been completed. ∎

Theorem 3.6 is inspired by [11, Theorem 1] and [8, Lemma 1]. Note that the
condition $W({0})=0$ is not necessary for the Caputo case. In Theorem 3.6, the
considered function $W(z(k))$ is general and it can be Volterra function,
power function, logarithmic function, integral upper limit function and
tangent function, etc. For multivariate composite functions, the following
corollary follows.

######  Corollary 1.

If $W:\mathbb{R}^{p}\times\mathbb{R}^{q}\to\mathbb{R}$ is differentiable
convex, $W({0,0})=0$, $V(x(k)):=w^{-1}(k)W(\hat{u}(k),\hat{v}(k))$,
$\hat{u}(k):=w(k)u(k)$, $\hat{v}(k):=w(k)v(k)$, $\frac{{\partial
W({\hat{u}(k),\hat{v}(k)})}}{{\partial\hat{u}(k)}}$ $=\frac{{\partial
V({u(k),v(k)})}}{{\partial u(k)}}$ and $\frac{{\partial
W({\hat{u}(k),\hat{v}(k)})}}{{\partial\hat{u}(k)}}=\frac{{\partial
V({u(k),v(k)})}}{{\partial v(k)}}$ hold almost everywhere, then for any
$\alpha\in(0,1)$, $w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with
$\frac{w(a)}{w(k)}>0$, $k\in{\mathbb{N}_{a+1}}$, $a\in\mathbb{R}$, one has

(3.29) $\textstyle{\begin{array}[]{rl}{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}V({u(k),v(k)})\leq&\frac{{\partial
V({u(k),v(k)})}}{{\partial u^{\rm T}(k)}}{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}u(k)\\\ &+\frac{{\partial V({u(k),v(k)})}}{{\partial
v^{\rm T}(k)}}{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}v(k),\end{array}}$ (3.30)
$\textstyle{\begin{array}[]{rl}{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}V({u(k),v(k)})\leq&\frac{{\partial
V({u(k),v(k)})}}{{\partial u^{\rm T}(k)}}{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}u(k)\\\ &+\frac{{\partial V({u(k),v(k)})}}{{\partial
v^{\rm T}(k)}}{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}v(k),\end{array}}$ (3.31)
$\textstyle{\begin{array}[]{rl}{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}V({u(k),v(k)})\leq&\frac{{\partial
V({u(k),v(k)})}}{{\partial u^{\rm T}(k)}}{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}u(k)\\\ &+\frac{{\partial V({u(k),v(k)})}}{{\partial
v^{\rm T}(k)}}{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}v(k).\end{array}}$

In Theorem 3.6, $W:\mathbb{R}^{n}\to\mathbb{R}$ is assume to be differentiable
convex. If it is convex but not differentiable, the following theorem can be
developed.

######  Theorem 3.7.

If $W:\mathbb{R}^{n}\to\mathbb{R}$ is convex, $W({0})=0$, $z(k):=w(k)x(k)$,
$V(x(k)):=w^{-1}(k)W(w(k)x(k))$ and $\zeta(z(k))=\zeta(x(k))$ holds almost
everywhere, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, $a\in\mathbb{R}$, one has

(3.32) ${\textstyle{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}V({x(k)})\leq\zeta^{\rm T}(x(k)){}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.33) ${\textstyle{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}V({x(k)})\leq\zeta^{\rm T}(x(k)){}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.34) ${\textstyle{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}V({x(k)})\leq\zeta^{\rm T}(x(k)){}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}x(k),}$

where $\zeta(z(k))\in\partial W(z(k))$ and $\zeta(x(k))\in\partial V(x(k))$.

###### Proof.

The main idea of this proof is also to construct non-negative terms by using
the properties of convex functions. Different from Theorem 3.6, letting
$g(j):=W({z(j)})-W({z(k)})-\zeta^{\rm T}(z(k))[{z(j)-z(k)}]$. By using the
basic property of subgradient [40], one finds $g(j)\geq 0$,
$j\in{\mathbb{N}_{a}^{k}}$ and $g(k)=0$ when $w(k)>0$, which coincide with the
property of $g(j)$ in Theorem 3.6. Along this way, the proof could be
completed similarly. For simplity, it is omitted here. ∎

Theorem 3.7 is inspired by [41, Theorem 2], [31, Lemma 10] and [36, Theorem 3,
Theorem 7]. When $W({z(k)})$ is differentiable, the subgradient $\zeta(z(k))$
is strengthened as the true gradient $\frac{{{\rm d}W({z(k)})}}{{{\rm
d}z(k)}}$. Therefore, Theorem 3.6 is a special case of Theorem 3.7. When
$n=1$, $w(k)=1$, $V({x(k)})=|x(k)|$, it follows ${\mathop{\rm
sgn}}({x(k)})\in\partial V(x(k))$. At this point, the following corollary can
be obtained.

######  Corollary 2.

For any $x:\mathbb{N}_{a+1}\to\mathbb{R}$, $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, $a\in\mathbb{R}$, one has

(3.35) ${\textstyle{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}|x(k)|\leq{\rm
sgn}(x(k)){}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.36)
${\textstyle{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}|x(k)|\leq{\rm
sgn}(x(k)){}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.37)
${\textstyle{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}|x(k)|\leq{\rm
sgn}(x(k)){}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k).}$

Notably, the complex condition of $W(z(k))$ in Theorem 3.6 and Theorem 3.7 has
been removed in Corollary 2. It would be more practical to reduce the
complexity of the given condition for previous theorems. Also, the scalar case
can be extended to the vector case. Along this way, the fractional difference
inequalities on $\|x(k)\|_{1}$ or $\|x(k)\|_{\infty}$ can be developed.

If it is differentiable but not convex, three cases will be discussed, i.e.,
the concave function, the monotone function and the synchronous function. The
latter two cases are neither convex nor concave.

######  Theorem 3.8.

If $W:\mathbb{R}^{n}\to\mathbb{R}$ is differentiable concave, $W({0})=0$,
$V(x(k)):=w^{-1}(k)W(z(k))$, $z(k):=w(k)x(k)$ and $\frac{{\rm
d}W({z(k)})}{{{\rm d}z(k)}}=\frac{{\rm d}V({x(k)})}{{{\rm d}x(k)}}$ holds
almost everywhere, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, $a\in\mathbb{R}$, one has

(3.38) ${\textstyle{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}V({x(k)})\geq\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.39)
${\textstyle{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}V({x(k)})\geq\frac{{{\rm
d}V({x(k)})}}{{{\rm d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}x(k),}$ (3.40) ${\textstyle{}_{a}^{\rm
G}\nabla_{k}^{\alpha,w(k)}V({x(k)})\geq\frac{{{\rm d}V({x(k)})}}{{{\rm
d}{x^{\rm{T}}}(k)}}{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k).}$

###### Proof.

Define $f(j)$, $W(z(k))$ and $z(k)$ like the proof of Theorem 3.6. Let
$g(j):=W({z(j)})-W({z(k)})-\frac{{{\rm d}W({z(k)})}}{{{\rm
d}{z^{\rm{T}}}(k)}}[{z(j)-z(k)}]$. When the convex condition is replaced by
the concave condition, $g(j)\leq 0$, $j\in{\mathbb{N}_{a}^{k}}$ and $g(k)=0$.
In a similar way, the inequalities in (3.38) - (3.40) can be derived. ∎

Notably, to remove the coupling and prettify the conclusion, the complicated
conditions are introduced in Theorem 3.6 \- Theorem 3.8. In future study, a
plain condition on $V(x(k))$ is expected.

######  Theorem 3.9.

If the function $\phi:{\mathbb{N}_{a+1}}\to\mathbb{R}$ is monotonically
decreasing and $x:{\mathbb{N}_{a+1}}\to\mathbb{R}^{n}$ is non-negative, then
for any $\alpha\in(0,1)$, $w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$
with $\frac{w(a)}{w(k)}>0$, $k\in{\mathbb{N}_{a+1}}$, $a\in\mathbb{R}$, one
has

(3.41) $\textstyle{{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}[\phi^{\rm
T}(k)x(k)]\leq\phi^{\rm T}(k){}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}x(k),}$
(3.42) $\textstyle{{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}[\phi^{\rm
T}(k)x(k)]\leq\phi^{\rm T}(k){}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}x(k),}$
(3.43) $\textstyle{{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}[\phi^{\rm
T}(k)x(k)]\leq\phi^{\rm T}(k){}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k).}$

###### Proof.

Setting $w(k)>0$ and defining
$f(j):=\frac{{({k-j})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}$ as
previous, it follows $\nabla f(j)\geq 0$, $j\in\mathbb{N}_{a+1}^{k-1}$,
$[\nabla f(j)]_{j=k}=-1$, $f(a)>0$ and $f(k)=0$. Letting $z(k):=w(k)x(k)$,
$g(j):=[{\phi(j)-\phi(k)}]^{\rm T}z(j)$, one has $g(j)\geq 0$,
$j\in\mathbb{N}_{a}^{k}$ and $g(k)=0$. Then the proof can be proceed as
follows

(3.44) $\begin{array}[]{l}w(k)\\{{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}[\phi^{\rm T}(k)x(k)]-\phi^{\rm T}(k){}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}x(k)\\}\\\ ={}_{a}^{\rm
C}\nabla_{k}^{\alpha}[\phi^{\rm T}(k)z(k)]-\phi^{\rm T}(k){}_{a}^{\rm
C}\nabla_{k}^{\alpha}z(k)\\\
=\sum\nolimits_{j=a+1}^{k}{\frac{{{{({k-j+1})}^{\overline{-\alpha}}}}}{{\Gamma({-\alpha+1})}}\nabla[\phi^{\rm
T}(j)z(j)]}\\\ \hskip 12.0pt-\phi^{\rm
T}(k)\sum\nolimits_{j=a+1}^{k}{\frac{{{{({k-j+1})}^{\overline{-\alpha}}}}}{{\Gamma({-\alpha+1})}}\nabla
z(j)}\\\ =\sum\nolimits_{j=a+1}^{k}{f(j-1)\nabla g(j)}\\\
=f(j)g(j)|_{j=a}^{j=k}-\sum\nolimits_{j=a+1}^{k}{\nabla f(j)g(j)}\\\
=-f(a)g(a)-\sum\nolimits_{j=a+1}^{k-1}{\nabla f(j)g(j)}\\\ \leq 0,\end{array}$

which implies the correctness of (3.41). Along this way, the remainder proof
can be completed smoothly. ∎

Theorem 3.9 is inspired by [26, Section 3]. In the proof of Theorem 3.9,
$g(j)\geq 0$, $j\in\mathbb{N}_{a}^{k}$ and $g(k)=0$ hold since $\phi$ is
monotonically decreasing. If $\phi$ is monotonically increasing, one has
$g(j)\leq 0$, $j\in\mathbb{N}_{a}^{k}$ and $g(k)=0$. On this basis, the sign
of inequality should reverse.

Before introducing the third case, a key definition is provided here.

######  Definition 3.10.

Let $u,v:\mathbb{N}_{a}\to\mathbb{R}^{n}$, $a\in\mathbb{R}$,
$n\in\mathbb{Z}_{+}$. $u$ and $v$ are said synchronous if $[{u(j)-u(k)}]^{\rm
T}[{v(j)-v(k)}]\geq 0$ holds for any $j,k\in\mathbb{N}_{a}$. $u$ and $v$ are
said asynchronous, if $[{u(j)-u(k)}]^{\rm T}[{v(j)-v(k)}]$ $\leq 0$ holds for
any $j,k\in\mathbb{N}_{a}$.

######  Theorem 3.11.

If $u,v:\mathbb{N}_{a}\to\mathbb{R}^{n}$ are synchronous, $a\in\mathbb{R}$,
$n\in\mathbb{Z}_{+}$, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$,one has

(3.45) $\begin{array}[]{l}{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}[{u^{\rm
T}(k)v(k)}]+[{u^{\rm T}(k)v(k)}]{}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}1\\\
\leq u^{\rm T}(k){}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}{v(k)}+v^{\rm
T}(k){}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}{u(k)},\end{array}$ (3.46)
$\begin{array}[]{l}{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}[{u^{\rm
T}(k)v(k)}]+[{u^{\rm T}(k)v(k)}]{}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}1\\\
\leq u^{\rm T}(k){}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}{v(k)}+v^{\rm
T}(k){}_{a}^{\rm R}\nabla_{k}^{\alpha,w(k)}{u(k)},\end{array}$ (3.47)
$\begin{array}[]{l}{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}[{u^{\rm
T}(k)v(k)}]+[{u^{\rm T}(k)v(k)}]{}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}1\\\
\leq u^{\rm T}(k){}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}{v(k)}+v^{\rm
T}(k){}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}{u(k)}.\end{array}$

###### Proof.

Define $f(j):=\frac{{({k-j})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}$,
$g(j)=w(j)[{u(j)-u(k)}]^{\rm T}[{v(j)-v(k)}]$. For the case of $w(k)>0$, one
has $g(j)\geq 0$, $j\in\mathbb{N}_{a}^{k}$ and $g(k)=0$. On this basis, one
obtains

(3.48) $\begin{array}[]{l}w(k)\\{{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}[{u(k)v(k)}]+[{u(k)v(k)}]{}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}1\\}\\\ -w(k)[u(k){}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}{v(k)}+v(k){}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}{u(k)}]\\\
=\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\nabla{[w(j){u(j)v(j)}]}}\\\
\hskip
12.0pt+[{u(k)v(k)}]\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\nabla{w(j)}}\\\
\hskip
12.0pt-u(k)\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}{{\nabla[w(j)v(j)]}}}\\\
\hskip
12.0pt-v(k)\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}{{\nabla[w(j)u(j)]}}}\\\
=\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{-\alpha}}}}{{\Gamma({1-\alpha})}}\nabla\\{{w(j)[{u(j)-u(k)}][{v(j)-v(k)}]}\\}}\\\
=\sum\nolimits_{j=a+1}^{k}{f(j-1)\nabla g(j)}\\\
=f(j)g(j)|_{j=a}^{j=k}-\sum\nolimits_{j=a+1}^{k}{\nabla f(j)g(j)}\\\
=-f(a)g(a)-\sum\nolimits_{j=a+1}^{k-1}{\nabla f(j)g(j)}\\\ \leq 0,\end{array}$

which leads to the desired result in (3.45). For the case of $w(k)<0$, the
desired result in (3.45) can be derived by using Lemma 2.2. Similarly, the
results in (3.46), (3.47) can be established successfully. ∎

From the synchronous concept, one has $u(k)v(k)\geq 0$. When $w(k)$ is
constant or increasing, one has ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}1\geq
0$. By using Theorem 3.1 further, one obtains ${}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}1\geq 0$ and
${}_{a}^{\rm{G}}\nabla_{k}^{\alpha,w(k)}1\geq 0$. To make it more practical,
the following corollary can be developed.

######  Corollary 3.

If $u,v:\mathbb{N}_{a}\to\mathbb{R}^{n}$ are synchronous, $a\in\mathbb{R}$,
$n\in\mathbb{Z}_{+}$, then for any $\alpha\in(0,1)$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$
and ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}1\geq 0$, $k\in{\mathbb{N}_{a+1}}$,
one has

(3.49) ${}_{a}^{\rm C}\nabla_{k}^{\alpha,w(k)}[{u(k)v(k)}]\leq u(k){}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}{v(k)}+v(k){}_{a}^{\rm
C}\nabla_{k}^{\alpha,w(k)}{u(k)},$ (3.50) ${}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}[{u(k)v(k)}]\leq u(k){}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}{v(k)}+v(k){}_{a}^{\rm
R}\nabla_{k}^{\alpha,w(k)}{u(k)},$ (3.51)
${}_{a}^{\rm{G}}\nabla_{k}^{\alpha,w(k)}[{u(k)v(k)}]\leq
u(k){}_{a}^{\rm{G}}\nabla_{k}^{\alpha,w(k)}{v(k)}+v(k){}_{a}^{\rm{G}}\nabla_{k}^{\alpha,w(k)}{u(k)}.$

Corollary 3 is inspired by [4, Lemma 1], [5, Theorem 3.10] and [36, Theorem
1]. When $v(k)=u(k)$ or $v(k)={\mathop{\rm sgn}}(u(k))$, some special cases of
Theorem 3.11 can also be developed. Besides, when the synchronous condition is
replaced by the asynchronous case, the sign of inequality should be flipped.

Besides the fractional difference inequalities, the following fractional sum
inequalities can be derived.

######  Theorem 3.12.

If $u,v:\mathbb{N}_{a}\to\mathbb{R}^{n}$ are synchronous,
$n\in\mathbb{Z}_{+}$, then for any $\alpha>0$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has

(3.52) ${\textstyle{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[{u^{\rm
T}(k)v(k)}]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\geq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u^{\rm
T}(k)}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{v(k)}.}$

If $u,v:\mathbb{N}_{a}\to\mathbb{R}^{n}$ are asynchronous,
$n\in\mathbb{Z}_{+}$, then for any $\alpha>0$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has

(3.53) ${\textstyle{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[{u^{\rm
T}(k)v(k)}]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\leq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u^{\rm
T}(k)}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{v(k)}.}$

###### Proof.

With the help of Lemma 2.2, if Theorem 3.12 holds for $w(k)>0$, then Theorem
3.12 holds for $w(k)>0$. As a result, only the case of $w(k)>0$ is considered.
From the definition of synchronous functions, one has $u^{\rm T}(j)v(j)+u^{\rm
T}(i)v(i)$ $\geq u^{\rm T}(j)v(i)+u^{\rm T}(i)v(j)$ for any
$i,j\in\mathbb{N}_{a+1}$. Multiplying this inequality by the positive factor
$\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)$ on both
left and right hand sides, and then summing both sides with respect to $j$
over the interval $(a+1,k)$ yield

(3.54)
$\begin{array}[]{l}\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)[{u^{\rm
T}(j)v(j)+u^{\rm T}(i)v(i)}]}\\\
=w(k){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm T}(k)v(k)]+w(k)u^{\rm
T}(i)v(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\\\
\geq\sum\nolimits_{j=a+1}^{k}{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)[{u^{\rm
T}(j)v(i)+u^{\rm T}(i)v(j)}]}\\\ =w(k)v^{\rm
T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u(k)+w(k)u^{\rm
T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v(k).\end{array}$

Since $w(k)>0$, one has

(3.55) $\begin{array}[]{l}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]+u^{\rm T}(i)v(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\\\ \geq
v^{\rm T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u(k)+u^{\rm
T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v(k).\end{array}$

Multiplying the inequality in (3.55) by the positive factor
$\frac{{({k-i+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(i)$ on both
left and right hand sides, and then summing both sides with respect to $i$
over the interval $(a+1,k)$, one has

(3.56)
$\begin{array}[]{l}\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]}\\\
+\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(i)u^{\rm
T}(i)v(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1}\\\
=2w(k){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\\\
\geq\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(i)v^{\rm
T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u(k)}\\\ \hskip
12.0pt+\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(i)u^{\rm
T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v(k)}\\\
=2w(k){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u^{\rm
T}(k){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v(k),\end{array}$

which implies (3.52). In a similar way, if $u(k)$ and $v(k)$ are asynchronous,
$u^{\rm T}(j)v(j)+u^{\rm T}(i)v(i)\leq u^{\rm T}(j)v(i)+u^{\rm T}(i)v(j)$ and
then (3.53) can be derived. ∎

The inequalities in Theorem 3.12 can be further generalized from the viewpoint
of order number.

######  Theorem 3.13.

If $u,v:\mathbb{N}_{a}\to\mathbb{R}^{n}$ are synchronous,
$n\in\mathbb{Z}_{+}$, then for any $\alpha,\beta>0$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has

(3.57) $\begin{array}[]{l}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}1+{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}[u^{\rm
T}(k)v(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\\\
\geq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u^{\rm
T}(k)}{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}{v(k)}+{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}{u^{\rm
T}(k)}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{v(k)}.\end{array}$

If $u,v:\mathbb{N}_{a}\to\mathbb{R}^{n}$ are asynchronous,
$n\in\mathbb{Z}_{+}$, then for any $\alpha,\beta>0$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has

(3.58) $\begin{array}[]{l}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}1+{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}[u^{\rm
T}(k)v(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\\\
\leq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u^{\rm
T}(k)}{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}{v(k)}+{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}{u^{\rm
T}(k)}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{v(k)}.\end{array}$

###### Proof.

Similarly, only the case of $w(k)>0$ is considered. For the synchronous case,
recalling the proof of Theorem 3.12 yields the inequality (3.55). Multiplying
this inequality by the positive factor
$\frac{{({k-i+1})\overline{{}^{\beta-1}}}}{{\Gamma(\beta)}}w(i)$ on both left
and right hand sides, and then summing both sides with respect to $i$ over the
interval $(a+1,k)$, one has

(3.59)
$\begin{array}[]{l}\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\beta-1}}}}{{\Gamma(\beta)}}w(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]}\\\
+\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\beta-1}}}}{{\Gamma(\beta)}}w(i)u^{\rm
T}(i)v(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1}\\\
=w(k){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}1\\\ \hskip
12.0pt+w(k){}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}[u^{\rm
T}(k)v(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1\\\
\geq\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\beta-1}}}}{{\Gamma(\beta)}}w(i)v^{\rm
T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u(k)}\\\ \hskip
12.0pt+\sum\nolimits_{i=a+1}^{k}{\frac{{({k-i+1})\overline{{}^{\beta-1}}}}{{\Gamma(\beta)}}w(i)u^{\rm
T}(i){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v(k)}\\\
=w(k){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u^{\rm
T}(k){}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}v(k)\\\ \hskip
12.0pt+w(k){}_{a}^{\rm{G}}\nabla_{k}^{-\beta,w(k)}u^{\rm
T}(k){}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v(k).\end{array}$

Considering $w(k)>0$, the inequality (3.57) can be derived. Likewise, if
$u(k)$ and $v(k)$ are asynchronous, (3.58) can be obtained directly. ∎

The inequalities in Theorem 3.12 can be further generalized from the viewpoint
of sequence number.

######  Theorem 3.14.

If $u_{i}:\mathbb{N}_{a+1}\to\mathbb{R}$, $i=1,2,\cdots,n$,
$n\in\mathbb{Z}_{+}$, are positive increasing functions, then for any
$\alpha>0$, $w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with
$\frac{w(a)}{w(k)}>0$, $k\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$, one has

(3.60)
${\textstyle{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}\prod\nolimits_{i=1}^{n}{{u_{i}}(k)}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{n-1}\geq\prod\nolimits_{i=1}^{n}{{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{i}}(k)}.}$

###### Proof.

The proof will be proven by induction.

$\blacktriangleright$ Part 1: when $n=1$, (3.60) reduces to

(3.61)
${\textstyle{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{1}}(k)\geq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{1}}(k),}$

which holds naturally.

$\blacktriangleright$ Part 2: assume that when $n=m\in\mathbb{Z}_{+}$, (3.60)
holds, i.e.,

(3.62)
${\textstyle{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}\prod\nolimits_{i=1}^{m}{{u_{i}}(k)}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{m-1}\geq\prod\nolimits_{i=1}^{m}{{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{i}}(k)}.}$

Defining $v_{m}(k):=\prod\nolimits_{i=1}^{m}{{u_{i}}(k)}$ and using Theorem
3.12, one has

(3.63)
$\begin{array}[]{l}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}\prod\nolimits_{i=1}^{m+1}{{u_{i}}(k)}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{m}\\\
={}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[{u_{m+1}}(k){v_{m}}(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,\lambda}1[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{m-1}\\\
\geq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{m+1}(k)}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{v_{m}(k)}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{m-1}\\\
\geq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{m+1}(k)}\prod\nolimits_{i=1}^{m}{{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{i}}(k)}\\\
=\prod\nolimits_{i=1}^{m+1}{{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{u_{i}}(k)},\end{array}$

where ${}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1>0$ is adopted. (3.63) means
that (3.60) holds for $n=m+1$. All of these end the proof. ∎

Theorem 3.12 \- Theorem 3.14 are generalized from the Chebyshev inequalities
from [10, 5, 17, 36].

######  Theorem 3.15.

If $x:\mathbb{N}_{a+1}^{b}\to[m,M]$, $m,M\in\mathbb{R}$, then for any
$\alpha>0$, $k,b\in\mathbb{N}_{a+1}$, $a\in\mathbb{R}$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
one has

(3.64) ${\textstyle m{}_{a}^{\rm G}\nabla_{k}^{-\alpha,w(k)}1\leq{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}x(k)\leq M{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}1.}$

###### Proof.

By using the definition and the condition, $w(k)>0$,
$\frac{{(k-j+1)\overline{{}^{\alpha-1}}}}{{\Gamma\left(\alpha\right)}}>0$,
$x(k)\in[m,M]$, one has

(3.65)
${\textstyle\begin{array}[]{rl}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{x}(k)=&{w^{-1}}(k)\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{\alpha-1}}}}{{\Gamma\left(\alpha\right)}}w(j){x}(j)}\\\
\geq&m{w^{-1}}(k)\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{\alpha-1}}}}{{\Gamma\left(\alpha\right)}}w(j)}\\\
=&m{}_{a}^{\rm G}\nabla_{k}^{-\alpha,w(k)}1,\end{array}}$ (3.66)
${\textstyle\begin{array}[]{rl}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}{x}(k)=&{w^{-1}}(k)\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{\alpha-1}}}}{{\Gamma\left(\alpha\right)}}w(j){x}(j)}\\\
\leq&M{w^{-1}}(k)\sum\nolimits_{j=a+1}^{k}{\frac{{(k-j+1)\overline{{}^{\alpha-1}}}}{{\Gamma\left(\alpha\right)}}w(j)}\\\
=&M{}_{a}^{\rm G}\nabla_{k}^{-\alpha,w(k)}1.\end{array}}$

The proof is thus completed. ∎

Theorem 3.15 is the valuation theorem of nabla tempered fractional sum.

######  Theorem 3.16.

For any $\alpha>0$, $u,v:\mathbb{N}_{a+1}\to\mathbb{R}^{n}$,
$n\in\mathbb{Z}_{+}$, $a\in\mathbb{R}$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, one has

(3.67) ${\textstyle\\{{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]\\}^{2}\leq{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)u(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[v^{\rm T}(k)v(k)].}$

###### Proof.

Similarly, the $w(k)>0$ case is equivalent of the $w(k)<0$ case. Consequently,
only the case of $w(k)>0$ is considered. Let
$\hat{u}(j):=\sqrt{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)}u(j)$,
$\hat{v}(j):=\sqrt{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)}v(j)$.
The following quadratic function is constructed

(3.68)
${\textstyle\begin{array}[]{rl}f(x)=&\big{[}\sum\nolimits_{j=a+1}^{k}{\hat{u}^{\rm
T}(j)\hat{u}(j)}\big{]}x^{2}+2\big{[}\sum\nolimits_{j=a+1}^{k}{\hat{u}^{\rm
T}(j)\hat{v}(j)}\big{]}x\\\ &+\sum\nolimits_{j=a+1}^{k}{\hat{v}^{\rm
T}(j)\hat{v}(j)},\end{array}}$

where $x\in\mathbb{R}$.

By using basic mathematical derivation, one has

(3.69) ${\textstyle
f(x)=\sum\nolimits_{j=a+1}^{k}{[\hat{u}(j)x+\hat{v}(j)]^{\rm
T}[\hat{u}(j)x+\hat{v}(j)]},}$

which means that its discriminant is nonnegative, i.e.,

(3.70)
${\textstyle\begin{array}[]{rl}\Delta=&4\big{[}\sum\nolimits_{j=a+1}^{k}{\hat{u}^{\rm
T}(j)\hat{v}(j)}\big{]}^{2}-4\big{[}\sum\nolimits_{j=a+1}^{k}{\hat{u}^{\rm
T}(j)\hat{u}(j)}\big{]}\big{[}\sum\nolimits_{j=a+1}^{k}{\hat{v}^{\rm
T}(j)\hat{v}(j)}\big{]}\\\ \leq&0.\end{array}}$

Multiplying this inequality by the positive factor $\frac{1}{4w^{2}(k)}$ on
both left and right hand sides and substituting $\hat{u}(j)$, $\hat{v}(j)$
into (3.70), one has

(3.71)
${\textstyle\begin{array}[]{l}\\{{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)v(k)]\\}^{2}-{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u^{\rm
T}(k)u(k)]{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[v^{\rm T}(k)v(k)]\\\
=\frac{1}{4w^{2}(k)}\Delta\leq 0,\end{array}}$

which completes the proof. ∎

######  Theorem 3.17.

If $f:\mathbb{D}\to\mathbb{R}$ is convex, then for any $\alpha>0$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, $a\in\mathbb{R}$, one has

(3.72) ${\textstyle
f\big{(}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{-1}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}x(k)\big{)}\leq[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{-1}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}f(x(k)),}$

where $\mathbb{D}\subseteq\mathbb{R}^{n}$.

###### Proof.

Letting
$\lambda_{j}:=[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{-1}\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}\frac{w(j)}{w(k)}$,
it is not difficult to obtain $\sum\nolimits_{j=a+1}^{k}{\lambda_{j}}=1$ and
$\lambda_{j}>0$, $j\in\mathbb{N}_{a+1}^{k}$, $k\in\mathbb{N}_{a+1}$. By using
(2.9) and the newly defined $\lambda_{j}$, one has

(3.73) ${\textstyle
f\big{(}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{-1}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}x(k)\big{)}=f\big{(}\sum\nolimits_{j=a+1}^{k}{\lambda_{j}x(j)}\big{)},}$
(3.74)
${\textstyle[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}1]^{-1}{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}f(x(k))=\sum\nolimits_{j=a+1}^{k}{\lambda_{j}f(x(j))}.}$

Now, the problem becomes

(3.75) ${\textstyle
f\big{(}\sum\nolimits_{j=a+1}^{k}{\lambda_{j}x(j)}\big{)}\leq\sum\nolimits_{j=a+1}^{k}{\lambda_{j}f(x(j))}.}$

The mathematical induction will be adopted once again.

$\blacktriangleright$ Part 1: when $k=a+1$, $\lambda_{a+1}=1$, (3.75) holds
naturally.

$\blacktriangleright$ Part 2: when $k=a+2$, $\lambda_{a+1}=\frac{\alpha
w(a+1)}{\alpha w(a+1)+w(a+2)}$, $\lambda_{a+2}=\frac{w(a+2)}{\alpha
w(a+1)+w(a+2)}$, (3.75) becomes

(3.76)
${\textstyle\begin{array}[]{l}f(\lambda_{a+1}x(a+1)+\lambda_{a+2}x(a+2))\\\
\leq\lambda_{a+1}f(x(a+1))+\lambda_{a+2}f(x(a+2)).\end{array}}$

By using the convexity of $f$, (3.72) can be obtained.

$\blacktriangleright$ Part 3: assume that when $k=m\in\mathbb{N}_{a+2}$,
(3.75) holds, i.e.,
$f\big{(}\sum\nolimits_{j=a+1}^{m}{\lambda_{j}x(j)}\big{)}$
$\leq\sum\nolimits_{j=a+1}^{m}{\lambda_{j}f(x(j))}$. Since
$\sum\nolimits_{j=a+1}^{m+1}{\lambda_{j}}=1$, it follows

(3.77) ${\textstyle\sum\nolimits_{j=a+1}^{m}{\lambda_{j}}=1-\lambda_{m+1}.}$

Defining $\eta_{j}:=\frac{\lambda_{j}}{1-\lambda_{m+1}}$,
$j\in\mathbb{N}_{a+1}^{m}$, then one has

(3.78)
${\textstyle\sum\nolimits_{j=a+1}^{m}{\eta_{j}}=\sum\nolimits_{j=a+1}^{m}{\frac{\lambda_{j}}{1-\lambda_{m+1}}}=1,}$

and $\eta_{j}>0$, $j\in\mathbb{N}_{a+1}^{m}$.

On this basis, one has

(3.79)
${\textstyle\begin{array}[]{l}f\big{(}\sum\nolimits_{j=a+1}^{m+1}{\lambda_{j}x(j)}\big{)}\\\
=f\big{(}\lambda_{m+1}x(m+1)+\sum\nolimits_{j=a+1}^{m}{\lambda_{j}x(j)}\big{)}\\\
=f\big{(}\lambda_{m+1}x(m+1)+(1-\lambda_{m+1})\sum\nolimits_{j=a+1}^{m}{\eta_{j}x(j)}\big{)}\\\
\leq\lambda_{m+1}f(x(m+1))+(1-\lambda_{m+1})f\big{(}\sum\nolimits_{j=a+1}^{m}{\eta_{j}x(j)}\big{)}\\\
\leq\lambda_{m+1}f(x(m+1))+(1-\lambda_{m+1})\sum\nolimits_{j=a+1}^{m}{\eta_{j}f(x(j))}\\\
=\lambda_{m+1}f(x(m+1))+\sum\nolimits_{j=a+1}^{m}{\lambda_{j}f(x(j))}\\\
\leq\sum\nolimits_{j=a+1}^{m+1}{\lambda_{j}f(x(j))},\end{array}}$

which means that (3.75) holds for $k=m+1$.

To sum up, the proof is done. ∎

######  Theorem 3.18.

For any $\alpha>0$, $u,v:\mathbb{N}_{a+1}\to\mathbb{R}_{+}$, $a\in\mathbb{R}$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, $\frac{1}{p}+\frac{1}{q}=1$, if $p>1$, one has

(3.80)
${\textstyle{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u(k)v(k)]\leq[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u^{p}(k)]^{\frac{1}{p}}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v^{q}(k)]^{\frac{1}{q}}},$

and if $p\in(0,1)$, one has

(3.81)
${\textstyle{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}[u(k)v(k)]\geq[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}u^{p}(k)]^{\frac{1}{p}}[{}_{a}^{\rm{G}}\nabla_{k}^{-\alpha,w(k)}v^{q}(k)]^{\frac{1}{q}}}.$

###### Proof.

Similarly, consider $w(k)>0$, $k\in{\mathbb{N}_{a+1}}$. Let
$a_{j}:=\sqrt[p]{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)}u(j)$,
$b_{j}:=\sqrt[q]{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)}v(j)$.
It follows $a_{j},b_{j}\geq 0$, $\forall j\in\mathbb{N}_{a+1}^{k}$. Bearing
this in mind, (3.80) can be expressed as

(3.82)
${\textstyle\sum\nolimits_{j=a+1}^{k}{a_{j}b_{j}}\leq[\sum\nolimits_{j=a+1}^{k}{a_{j}^{p}}]^{\frac{1}{p}}[\sum\nolimits_{j=a+1}^{k}{b_{j}^{q}}}]^{\frac{1}{q}},$

which is actually the discrete time Hölder inequality. (3.81) can be expressed
as

(3.83)
${\textstyle\sum\nolimits_{j=a+1}^{k}{a_{j}b_{j}}\geq[\sum\nolimits_{j=a+1}^{k}{a_{j}^{p}}]^{\frac{1}{p}}[\sum\nolimits_{j=a+1}^{k}{b_{j}^{q}}}]^{\frac{1}{q}}.$

When $p\in(0,1)$, one has $q=\frac{p}{p-1}$. Let $p^{\prime}:=\frac{1}{p}$,
$q^{\prime}:=\frac{1}{1-p}$, it is not difficult to obtain
$p^{\prime},q^{\prime}>1$ and $\frac{1}{p^{\prime}}+\frac{1}{q^{\prime}}=1$.
By applying the discrete time Hölder inequality once again, it yields

(3.84)
${\textstyle\begin{array}[]{rl}\sum\nolimits_{j=a+1}^{k}{a_{j}^{p}}=&\sum\nolimits_{j=a+1}^{k}{({a_{j}b_{j}})^{p}b_{j}^{-p}}\\\
\leq&[\sum\nolimits_{j=a+1}^{k}{({a_{j}b_{j}})^{pp^{\prime}}}]^{\frac{1}{p^{\prime}}}[\sum\nolimits_{j=a+1}^{k}{b_{j}^{-pq^{\prime}}}]^{\frac{1}{q^{\prime}}}\\\
\leq&(\sum\nolimits_{j=a+1}^{k}{{a_{j}b_{j}}})^{p}[\sum\nolimits_{j=a+1}^{k}{b_{j}^{q}}]^{1-p}.\end{array}}$

Calculating the $\frac{1}{p}$ power on both sides of (3.84) gives

(3.85)
${\textstyle\begin{array}[]{rl}[\sum\nolimits_{j=a+1}^{k}{a_{j}^{p}}]^{\frac{1}{p}}\leq&\sum\nolimits_{j=a+1}^{k}{{a_{j}b_{j}}}[\sum\nolimits_{j=a+1}^{k}{b_{j}^{q}}]^{\frac{1-p}{p}}\\\
=&\sum\nolimits_{j=a+1}^{k}{{a_{j}b_{j}}}[\sum\nolimits_{j=a+1}^{k}{b_{j}^{q}}]^{-\frac{1}{q}},\end{array}}$

which implies (3.83). ∎

######  Theorem 3.19.

For any $\alpha>0$, $u,v:\mathbb{N}_{a+1}\to\mathbb{R}_{+}$, $a\in\mathbb{R}$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, if $p\geq 1$, one has

(3.86) ${\textstyle\\{{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[u(k)+v(k)]^{p}\\}^{\frac{1}{p}}\leq[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}u^{p}(k)]^{\frac{1}{p}}+[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha}v^{p}(k)]^{\frac{1}{p}}},$

and if $p\in(0,1)$, one has

(3.87) ${\textstyle\\{{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[u(k)+v(k)]^{p}\\}^{\frac{1}{p}}\geq[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}u^{p}(k)]^{\frac{1}{p}}+[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha}v^{p}(k)]^{\frac{1}{p}}}.$

###### Proof.

Similarly, consider $w(k)>0$, $k\in{\mathbb{N}_{a+1}}$. Let
$a_{j}:=\sqrt[p]{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)}u(j)$,
$b_{j}:=\sqrt[q]{\frac{{({k-j+1})\overline{{}^{\alpha-1}}}}{{\Gamma(\alpha)}}w(j)}v(j)$.
It follows $a_{j},b_{j}>0$, $\forall j\in\mathbb{N}_{a+1}^{k}$. Along this
way, (3.86) can be expressed as

(3.88)
${\textstyle[\sum\nolimits_{j=a+1}^{k}{(a_{j}+b_{j})^{p}}]^{\frac{1}{p}}\leq(\sum\nolimits_{j=a+1}^{k}{a_{j}^{p}})^{\frac{1}{p}}+(\sum\nolimits_{j=a+1}^{k}{b_{j}^{q}}})^{\frac{1}{q}},$

which is just the Minkonski inequality in discrete time domain. Similarly,
(3.87) can be expressed as

(3.89)
${\textstyle[\sum\nolimits_{j=a+1}^{k}{(a_{j}+b_{j})^{p}}]^{\frac{1}{p}}\geq(\sum\nolimits_{j=a+1}^{k}{a_{j}^{p}})^{\frac{1}{p}}+(\sum\nolimits_{j=a+1}^{k}{b_{j}^{q}}})^{\frac{1}{q}}.$

The proof completes immediately. ∎

In a similar way, the number of sequences could be extended to finite.

######  Corollary 4.

For any $\alpha>0$, $u_{i}:\mathbb{N}_{a+1}\to\mathbb{R}_{+}$,
$i=1,2,\cdots,n$, $n\in\mathbb{Z}_{+}$, $a\in\mathbb{R}$,
$w:\mathbb{N}_{a}\to\mathbb{R}\backslash\\{0\\}$ with $\frac{w(a)}{w(k)}>0$,
$k\in{\mathbb{N}_{a+1}}$, if $p\geq 1$, one has

(3.90) ${\textstyle\\{{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[\sum\nolimits_{i=1}^{n}u_{i}(k)]^{p}\\}^{\frac{1}{p}}\leq\sum\nolimits_{i=1}^{n}[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}u_{i}^{p}(k)]^{\frac{1}{p}}},$

and if $p\in(0,1)$, one has

(3.91) ${\textstyle\\{{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[\sum\nolimits_{i=1}^{n}u_{i}(k)]^{p}\\}^{\frac{1}{p}}\geq\sum\nolimits_{i=1}^{n}[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}u_{i}^{p}(k)]^{\frac{1}{p}}}.$

Theorem 3.16 is the generalization of Cauchy inequality or Carlson inequality.
Theorem 3.17 is the generalization of Jensen inequality. Theorem 3.18 is the
generalization of Hölder inequality. Theorem 3.19 is the generalization of
Minkonski inequality. In a similar way, some integral like inequalities could
be further developed for the nabla tempered fractional sum.

##  4\. Simulation study

In this section, six examples are provided to test the relation between
theoretical and the simulated results.

######  Example 4.1.

To examine Theorem 3.6 and Theorem 3.7, a new variable is introduced as
$e(k)={}_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}V(x(k))-\zeta^{\rm
T}(x(k))_{a}^{\rm G}\nabla_{k}^{\alpha,w(k)}x(k)$. When $V(x(k))$ is
differentiable, $\zeta(x(k))$ denotes the true gradient $\frac{{\rm
d}V(x(k))}{{\rm d}x(k)}$. Setting $a=0$, $k\in\mathbb{N}_{a+1}^{100}$,
$\alpha=0.01,0.02,\cdots,1$, the following four cases are considered.

$\left\\{\begin{array}[]{ll}{\rm{case}}\;1:\;V(x(k))={x^{2}}(k)\,,&x(k)=\sin(10k);\\\
{\rm{case}}\;2:\;V(x(k))={x^{2}}(k)\,,&x(k)=\cos(10k);\\\
{\rm{case}}\;3:\;V(x(k))=|x(k)|\,,&x(k)=\sin(10k);\\\
{\rm{case}}\;4:\;V(x(k))=|x(k)|\,,&x(k)=\cos(10k).\end{array}\right.$

Since $[{}_{a}^{\mathrm{G}}\nabla_{k}^{\alpha}x(k)]_{k=a+1}=x(a+1)$,
$e(a+1)=-x^{2}(a+1)$ for case 1, case 2 and $e(a+1)=0$ for case 3, case 4.
Consequently, for each $\alpha$ the maximum of $e(k)$ is calculated with
$k\in\mathbb{N}_{a+2}$ instead of $k\in\mathbb{N}_{a+1}$. Letting the tempered
function be positive and linearly increasing as $w(k)=0.5(k-a)$, the simulated
results are shown in Figure 1(a). Letting the tempered function be negative
and linearly decreasing i.e., $w(k)=0.5(a-k)$, the corresponding simulated
results are shown in Figure 1(b). It is shown that $e(k)\leq 0$ holds for all
the elaborated conditions, which confirms the developed inequalities. To
demonstrate more details, the relationship between $W(z(k))$ and $z(k)$ is
shown in Figure 2. It can be found that when $w(k)>0$, $W(z(k))$ is the convex
function of $z(k)$ on the whole for case 1, case 2 and $W(z(k))$ is the convex
function of $z(k)$ for case 3, case 4. When $w(k)<0$, $W(z(k))$ is the convex
function of $z(k)$ on the whole for case 1, case 2 and $W(z(k))$ is the convex
function of $z(k)$ for case 3, case 4. All of these coincide with the
theoretical analysis.

(a) $w(k)=0.5(k-a)$.

(b) $w(k)=0.5(a-k)$.

Figure 1. The evolution of the error with respect to the order.

(a) $w(k)=0.5(k-a)$.

(b) $w(k)=0.5(a-k)$.

Figure 2. The evolution of $W(z(k))$ with respect to $z(k)$.

Letting $w(k)=0.4^{k-a}+0.5$ and $w(k)=0.4^{a-k}+0.5$, the corresponding
results are displayed in Figure 2 and Figure 3, respectively. Note that
$w(k)=0.4^{k-a}+0.5$ is positive and exponentially decreasing.
$w(k)=0.4^{a-k}+0.5$ is positive and exponentially increasing. Figure 3 shows
that the maximum of $e(k)$ in the mentioned conditions are non-positive, which
confirms the correctness of the developed fractional difference inequalities
firmly. Figure 4(a) shows that $W(z(k))$ is the convex function of $z(k)$
almost everywhere for case 1, case 2 and $W(z(k))$ is the convex function of
$z(k)$ for case 3, case 4. Though the points are uneven dispersion in Figure
4(b), the convexity of $W(z(k))$ can be found from the trend closely.

(a) $w(k)=0.4^{k-a}+0.5$.

(b) $w(k)=0.4^{a-k}+0.5$.

Figure 3. The evolution of the error with respect to the order.

(a) $w(k)=0.4^{k-a}+0.5$.

(b) $w(k)=0.4^{a-k}+0.5$.

Figure 4. The evolution of $W(z(k))$ with respect to $z(k)$.

######  Example 4.2.

To examine the Chebyshev fractional sum inequality, define $e(k):={}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[{u^{\rm T}(k)v(k)}]{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}1-{}_{a}^{\rm G}\nabla_{k}^{-\alpha,w(k)}{u^{\rm
T}(k)}{}_{a}^{\rm G}\nabla_{k}^{-\alpha,w(k)}{v(k)}$. Setting $a=0$,
$\alpha=0.01,0.02,\cdots,4$, considering different $w(k)$ and the following
four cases

$\left\\{\begin{array}[]{l}{\rm{case}}\;1:\;u(k)=\sin(k-a),v(k)={\rm
sgn}(\sin(k-a));\\\ {\rm{case}}\;2:\;u(k)=\sin(k-a),v(k)=2\sin(k-a)+1;\\\
{\rm{case}}\;3:\;u(k)=k-a,v(k)=(k-a)^{2};\\\
{\rm{case}}\;4:\;u(k)=\frac{1}{k-a},v(k)=\frac{1}{(k-a)^{2}},\end{array}\right.$

the simulated results are shown in Figure 5 and Figure 6.

Figure 5. The diagram of $u(k)$ and $v(k)$.

(a) $w(k)={(-1)^{k-a}}+2$

(b) $w(k)={(-1)^{k-a}}-2$

(c) $w(k)=\sin(k-a)+2$

(d) $w(k)=\sin(k-a)-2$

Figure 6. The evolution of the error with respect to the order.

The synchronous property of $u(k)$ and $v(k)$ can be confirmed by Figure 5. In
particular, $u(k),v(k)$ are continuously oscillating in case 1 and case 2.
$u(k),v(k)$ are monotonically increasing in case 3. $u(k),v(k)$ are
monotonically decreasing in case 4. In Figure 6(a) and Figure 6(c), the
tempered functions $w(k)$ are positive. In Figure 6(b) and Figure 6(d), the
tempered functions $w(k)$ are negative. From the quantitative analysis, it can
be observed that the minimum of $e(k)$ in all four cases are $0$, which
verifies the validity of Theorem 3.12 firmly.

######  Example 4.3.

To examine the Cauchy fractional sum inequality, define
$e(k):=\big{\\{}{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[u(k)v(k)]\big{\\}}^{2}-{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[u^{2}(k)]{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[v^{2}(k)]$. Setting $a=0$,
$\alpha=0.01,0.02,\cdots,4$ and considering the following four cases

$\left\\{\begin{array}[]{l}{\rm{case}}\;1:\;u(k)=\sin(10k)+2,v(k)={\cos(10k)}-2;\\\
{\rm{case}}\;2:\;u(k)=\sin(10k){(1-\lambda)^{a-k}},v(k)={\cos(10k)}{(1-\lambda)^{k-a}},\lambda=2;\\\
{\rm{case}}\;3:\;u(k)=\sin(10k)+\nu_{1},v(k)=\cos(10k)+\nu_{2},\nu_{1},\nu_{2}=\texttt{randn}(\texttt{size}(k));\\\
{\rm{case}}\;4:\;u(k)=\nu_{1}\sin(10k),v(k)=\nu_{2}\cos(10k),\nu_{1},\nu_{2}=\texttt{randn}(\texttt{size}(k)),\end{array}\right.$

the simulated results are shown in Figure 7 and Figure 8.

Figure 7. The diagram of $u(k)$ and $v(k)$.

(a) $w(k)={(-1)^{k-a}}+2$

(b) $w(k)={(-1)^{k-a}}-2$

(c) $w(k)=\sin(k-a)+2$

(d) $w(k)=\sin(k-a)-2$

Figure 8. The evolution of the error with respect to the order.

Notably, $u(k),v(k)$ are continuously oscillating in case 1 and case 2.
$u(k),v(k)$ are continuously oscillating functions combined with the additive
white noise in case 3. $u(k),v(k)$ are continuously oscillating functions
combined with the multiplicative white noise in case 4. The tempered functions
$w(k)$ are selected as Example 4.2. It can be found that the maximum of $e(k)$
in all the mentioned cases are negative, which confirms the correctness of
Theorem 3.16 firmly. Since $\mathop{\lim}\limits_{\alpha\to
0}{}_{a}^{\mathrm{G}}\nabla_{k}^{\alpha}x(k)=x(k)$, it follows
$\mathop{\lim}\limits_{\alpha\to 0}e(k)=0$ which coincides with the obtained
simulated results. Besides, for given $u(k),v(k),w(k)$, with the increase of
$\alpha$, the value of function $f(x)$ in (3.68) increases gradually. As a
result, the discriminant $\Delta$ in (3.70) decreases gradually, which just
matches the variation tendency of the plotted curves.

######  Example 4.4.

To examine the Jensen fractional sum inequality, define
$e(k):=f\big{(}[{}_{a}^{\rm G}\nabla_{k}^{-\alpha,w(k)}1]^{-1}{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}x(k)\big{)}-[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}1]^{-1}{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}f(x(k))$. Setting $a=0$,
$\alpha=0.01,0.02,\cdots,4$, $x(k)=-1:0.01:1$ and considering the following
four cases

$\left\\{\begin{array}[]{l}{\rm{case}}\;1:\;f(x(k))=|x(k)|^{1.5};\\\
{\rm{case}}\;2:\;f(x(k))=x^{2}(k);\\\ {\rm{case}}\;3:\;f(x(k))={\rm
e}^{0.5x(k)};\\\ {\rm{case}}\;4:\;f(x(k))=\max\\{|x(k)|^{1.5},x^{2}(k),{\rm
e}^{0.5x(k)}\\},\end{array}\right.$

the simulated results are shown in Figure 9 and Figure 10.

Figure 9. $f(x(k))$ with regard to $x(k)$.

(a) $w(k)={(-1)^{k-a}}+2$

(b) $w(k)={(-1)^{k-a}}-2$

(c) $w(k)=\sin(k-a)+2$

(d) $w(k)=\sin(k-a)-2$

Figure 10. The evolution of the error with respect to the order.

From Figure 9, it can be found that all the mentioned four functions are all
convex. The maximum of $e(k)$ in all the mentioned cases are negative, which
coincides with the theoretical results in Theorem 3.17 firmly.

######  Example 4.5.

To examine the Hölder fractional sum inequality, define $e(k):={}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[u(k)v(k)]-[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}u^{p}(k)]^{\frac{1}{p}}[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}v^{q}(k)]^{\frac{1}{q}}$. Setting $a=0$,
$\alpha=0.01,0.02,\cdots,4$ and considering the following four cases

$\left\\{\begin{array}[]{l}{\rm{case}}\;1:\;u(k)=1+0.5\sin(10k),v(k)=1+0.5\cos(10k);\\\
{\rm{case}}\;2:\;u(k)=2+(-1)^{k-a},v(k)=2-(-1)^{k-a};\\\
{\rm{case}}\;3:\;u(k)=|\sin(10k)|+0.01,v(k)=|\cos(10k)|+0.01;\\\
{\rm{case}}\;4:\;u(k)=|\texttt{randn}(\texttt{size}(k))|+0.01,v(k)=|\texttt{randn}(\texttt{size}(k))|+0.01,\end{array}\right.$

the simulated results are shown in Figure 11-Figure 13.

Figure 11. The diagram of $u(k)$ and $v(k)$

(a) $w(k)={(-1)^{k-a}}+2$

(b) $w(k)={(-1)^{k-a}}-2$

(c) $w(k)=\sin(k-a)+2$

(d) $w(k)=\sin(k-a)-2$

Figure 12. The evolution of the error with respect to the order ($p=1.5$).

(a) $w(k)={(-1)^{k-a}}+2$

(b) $w(k)={(-1)^{k-a}}-2$

(c) $w(k)=\sin(k-a)+2$

(d) $w(k)=\sin(k-a)-2$

Figure 13. The evolution of the error with respect to the order ($p=0.5$).

From the diagram in Figure 12, it can be observed that $u(k),v(k)$ are
positive, which satisfies the needed conditions. When $p=1.5$, the maximum of
$e(k)$ is negative. With the increase of $\alpha$, $e(k)$ gets smaller and
smaller. When $p=0.5$, the minimum of $e(k)$ is positive. With the increase of
$\alpha$, $e(k)$ gets bigger and bigger. All of these illustrate the
correctness of Theorem 3.18.

######  Example 4.6.

To examine the Minkonski fractional sum inequality, define
$e(k):=\\{{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}[u(k)+v(k)]^{p}\\}^{\frac{1}{p}}-[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}u^{p}(k)]^{\frac{1}{p}}-[{}_{a}^{\rm
G}\nabla_{k}^{-\alpha,w(k)}v^{p}(k)]^{\frac{1}{p}}$. Setting $a=0$,
$\alpha=0.01,0.02,\cdots,4$ and considering the mentioned four cases in
Example 4.5, the simulated results are shown in Figure 14 and Figure 15.

(a) $w(k)={(-1)^{k-a}}+2$

(b) $w(k)={(-1)^{k-a}}-2$

(c) $w(k)=\sin(k-a)+2$

(d) $w(k)=\sin(k-a)-2$

Figure 14. The evolution of the error with respect to the order ($p=1.5$).

(a) $w(k)={(-1)^{k-a}}+2$

(b) $w(k)={(-1)^{k-a}}-2$

(c) $w(k)=\sin(k-a)+2$

(d) $w(k)=\sin(k-a)-2$

Figure 15. The evolution of the error with respect to the order ($p=0.5$).

It can be also found that when $p=1.5$, the maximum of $e(k)$ is negative.
When $p=0.5$, the minimum of $e(k)$ is positive. All of obtained simulated
results illustrate the correctness of Theorem 3.19.

##  5\. Concluding remarks

In this paper, the nabla tempered fractional calculus has been investigated
systemically. A series of interesting and promising inequalities are developed
rigorously including the monotonicity, the comparison principle, the
difference inequality, the sum inequality. These properties have huge
potentials in stability analysis and controller design. The proposed
properties could greatly enrich the comprehension of nabla tempered fractional
calculus and facilitate its applications. In addition, it is hoped that the
techniques and consequences of this work could inspire the researchers to
explore more interesting sequel in this area.

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