# Monotonicity Analysis of Fractional Proportional Differences

Authors: Iyad Suwan; Shahd Owies; Muayad Abussa; Thabet Abdeljawad
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this work, the nabla discrete new Riemann–Liouville and Caputo fractional proportional differences of order0<ε<1 on the time scale ℤ are formulated. The differences and summations of discrete fractional proportional are detected on ℤ, and the fractional proportional sums associated to ∇cRχε,ρz with order 0<ε<1 are defined. The relation between nabla Riemann–Liouville and Caputo fractional proportional differences is derived. The monotonicity results for the nabla Caputo fractional proportional difference are proved; specifically, if ∇c−1Rχε,ρz>0 then χz is ερ−increasing, and if χz is strictly increasing on ℕc and χc>0, then ∇c−1Rχε,ρz>0. As an application of our findings, a new version of the fractional proportional difference of the mean value theorem (MVT) on ℤ is proved.


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## Body

## 1. Introduction

Many problems in science, engineering, and media can be formulated using continuous and discrete fractional calculus [1–14]. The fractional sums and differences and their monotonicity properties are deeply studied in [15–25]. In [26], Atangana and Baleanu solved the fractional heat transfer model using new fractional derivatives with exponential kernels, and they presented many applications of the new notations of fractional derivatives. Applications of discrete fractional calculus are successfully discussed by many researchers in the last decade, for example, in [27–29]. Recently, studying the monotonicity for fractional difference operators with nonsingular discrete kernels is under focus [30, 31]. Monotonicity results for fractional difference operators with discrete exponential kernels were studied in [32] when the time step h=1. In [3], deep monotonicity analysis is done for nabla h−discrete fractional differences with a discrete Mittag–Leffler kernel in the time scale hℤ with 0<ε<1 and 0<h≤1. The results of the research generalized those obtained in [22] where 0<ε<0.5 and h=1. After that, monotonicity analysis of fractional proportional differences is studied and then the results are prettified by formulating a new version of mean value theorem as an application. In [33], the nabla fractional sums and differences of order 0<ε<1 on the time scale hℤ where 0<h≤1 are formulated, and the monotonicity results for the nabla h−Caputo fractional difference operator were concluded. In this paper, the authors formulated the nabla discrete new Riemann–Liouville (RL) and Caputo fractional proportional differences of order 0<ε<1 on the time scale ℤ. They also proved a new version of the fractional proportional difference of the mean value theorem (MVT) on ℤ.The article is organized as follows: Section2 presents the main definitions and needed preliminaries. In Section 3, the monotonicity results for fractional proportional differences are classified. In Section 4, we formulate a new version of the mean value theorem as an application. Finally, we provide the conclusions in Section 5.


## 2. Definitions and Preliminary Results

Definition 1.
The discrete proportional difference of order0<ρ≤1 for the function χ is defined by(1)∇ρχz=1−ρχz+ρ∇χz,z∈ℕc+1=c+1,c+2,c+3,…,and c≥0 is an integer.Definition 2.
Letz∈ℕc, 0<ρ≤1, and p=ρ−1/ρ, then e^pz,c=ρz−c.Definition 3.
For any real numberα, the α rising function is zα¯=Γz+α/Γz, such that z∈ℂ\…,−2,−1,0, 0α¯=0, where Γz is the gamma function.Definition 4.
(nabla fractional proportional sums).For a functionχ:ℕc⟶ℝ, ρ>0, and ε∈ℂ, 0<Reε<1, the nabla left fractional proportional sum of χ starting at c is defined by(2)∇−ε,ρcχz=1Γε∫cze^pz−s,0z−ςsε−1¯χs∇s=1Γε∑ι=c+1ze^pz−ι,0z−ςιε−1¯χι,z∈ℕc.For the functionχ:dℕ=d,d−1,d−2,…⟶ℝ, the nabla right fractional proportional sum ending at d is defined by(3)∇d−ε,ρχz=1Γε∫zde^ps−z,0s−ςzε−1¯χsΔs=1Γε∑ι=zd−1e^pι−z,0ι−ςzε−1¯χι,z∈dℕ.We notice that by settingρ=1, the given definitions of the fractional sums are generalizations of the Riemann fractional sums.Lemma 1.
Letχ:ℕc⟶ℝ be a function, p=ρ−1/ρ, 0<ε<1, and 0<ρ≤1, then(4)∇−ε,ρc∇χz=∇c∇−ε,ρχz−z−cε−1¯Γεe^pz−1,cχc.Proof.
(5)
∇
c
∇
−
ε
,
ρ
χ
z
=
1
Γ
ε
∑
ι
=
c
+
1
z
e
^
p
z
−
ι
,
0
z
−
ς
ι
ε
−
1
¯
∇
χ
ι
=
1
Γ
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
ε
−
1
¯
χ
ι
−
χ
ι
−
1
=
1
Γ
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
ε
−
1
¯
χ
ι
−
1
Γ
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
ε
−
1
¯
χ
ι
−
1
=
1
Γ
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
ε
−
1
¯
χ
ι
−
1
Γ
ε
∑
ι
=
c
z
−
1
ρ
z
−
ι
−
1
z
−
ς
ι
+
1
ε
−
1
¯
χ
ι
=
1
Γ
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
ε
−
1
¯
χ
ι
−
∑
ι
=
c
z
−
1
ρ
z
−
1
−
ι
z
−
1
−
ς
ι
ε
−
1
¯
χ
ι
=
1
Γ
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
ε
−
1
¯
χ
ι
−
∑
ι
=
c
+
1
z
−
1
ρ
z
−
1
−
ι
z
−
1
−
ς
ι
ε
−
1
¯
χ
ι
=
1
Γ
ε
ρ
z
−
1
−
c
z
−
1
−
ς
c
ε
−
1
¯
χ
c
=
1
Γ
ε
∇
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
ε
−
1
¯
χ
ι
−
1
Γ
ε
ρ
z
−
1
−
c
z
−
1
−
c
+
1
ε
−
1
¯
χ
c
=
∇
c
∇
−
ε
,
ρ
χ
z
−
z
−
c
ε
−
1
¯
Γ
ε
e
^
p
z
−
1
,
c
χ
c
.Lemma 2.
Letχ:ℕc⟶ℝ, p=ρ−1/ρ, 0<ε<1, and 0<ρ≤1, then(6)∇c∇−ε,ρχρz=∇ρ∇cχ−ε,ρz−ρΓεz−cε−1¯e^pz−1,cχc.Proof.
(7)
∇
ρ
χ
z
=
1
−
ρ
χ
z
+
ρ
∇
χ
z
,hence,(8)∇c∇−ε,ρχρz=∇c1−ρχz+ρ∇χz−ε,ρ=1−ρ∇cχ−ε,ρz+ρ∇c∇−ε,ρχzusing Lemma1=1−ρ∇cχ−ε,ρz+ρ∇ρ∇cχ−ε,ρz−z−cε−1¯Γεe^pz−1,cχcc=1−ρ∇cχ−ε,ρz+ρ∇ρ∇cχ−ε,ρz−ρΓεz−cε−1¯e^pz−1,cχc=∇ρ∇cχ−ε,ρz−ρΓεz−cε−1¯e^pz−1,cχc.
Note that ifρ=1, we get(9)∇c∇−εχz=∇c∇−εχz−1Γεz−cε−1¯χc.Definition 5.
(Riemann–Liouville (RL) fractional proportional differences)For0<ρ≤1, ε∈ℂ, 0<Reε<1, and χ be a function defined on ℕc or on ℕd, then the left Riemann–Liouville fractional proportional difference starting at c is defined by(10)∇cRχε,ρz=∇ρ∇cχ−1−ε,ρz=∇ρΓ1−ε∫cze^pz−s,0z−ςs−ε¯χs∇s=∇ρΓ1−ε∑ι=c+1ze^pz−ι,0z−ςι−ε¯χι=∇ρΓ1−ε∑ι=c+1zρz−ιz−ςι−ε¯χι,and the right Riemann–Liouville fractional proportional difference ending at d is defined by(11)∇Rχdε,ρz=−Δρ∇d−1−ε,ρχz=−ΔρΓ1−ε∑ι=zd−1ρι−zι−ςz−ε¯χι.We notice that by settingρ=1, the given definitions of the fractional differences are generalizations of the Riemann fractional differences.Definition 6.
(Caputo fractional proportional differences)For0<ρ≤1, ε∈ℂ, 0<Reε<1, and χ be a function defined on ℕc or on ℕd, then the left Caputo fractional proportional difference starting at c is defined by(12)∇cCχε,ρz=∇c∇−1−ε,ρχρz=1Γ1−ε∫cze^pz−s,0z−ςs−ε¯∇ρχs∇s=1Γ1−ε∑ι=c+1zρz−ιz−ςι−ε¯∇ρχι,and the right Caputo fractional proportional difference ending at d is defined by(13)∇Cχdε,ρz=∇d−1−ε,ρ−Δρχz=1Γ1−ε∑ι=zd−1ρι−zι−ςz−ε¯−Δρχι.We notice that by settingρ=1, the given definitions of the fractional differences are generalizations of the Caputo fractional differences.Proposition 1.
(the relation between nabla RL and Caputo fractional proportional differences)For anyε∈ℂ, 0<Reε<1, and 0<ρ≤1, the relation between nabla RL and Caputo fractional proportional differences is given as follows:(i)
∇cCχε,ρz=∇cRχε,ρz−z−c−ε¯/Γ1−εe^pz,cχc.(ii)
∇cCχε,ρz=∇cRχε,ρz−d−z−ε¯/Γ1−εe^pd,zχd.Proof.
(14)
∇
c
C
χ
ε
,
ρ
z
=
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
∇
ρ
χ
ι
=
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
1
−
ρ
χ
ι
+
ρ
∇
χ
ι
=
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
1
−
ρ
χ
ι
+
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
ρ
∇
χ
ι
=
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
1
−
ρ
χ
ι
+
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
ρ
χ
ι
−
χ
ι
−
1
=
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
1
−
ρ
χ
ι
+
1
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
ρ
χ
ι
−
1
Γ
1
−
ε
∑
ι
=
c
z
−
1
ρ
z
−
ι
−
1
z
−
ς
ι
+
1
−
ε
¯
ρ
χ
k
=
1
−
ρ
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
χ
ι
+
ρ
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
χ
ι
−
ρ
Γ
1
−
ε
∑
ι
=
c
+
1
z
−
1
ρ
z
−
1
−
ι
z
−
1
−
ς
ι
−
ε
¯
χ
ι
−
ρ
Γ
1
−
ε
ρ
z
−
1
−
c
z
−
1
−
ς
c
−
ε
¯
χ
c
=
1
−
ρ
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
χ
ι
+
ρ
∇
Γ
1
−
ε
∑
ι
=
c
+
1
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
χ
ι
−
ρ
Γ
1
−
ε
ρ
z
−
1
−
c
z
−
1
−
c
+
1
−
ε
¯
χ
c
=
∇
ρ
Γ
1
−
ε
∑
ι
=
c
+
1
z
e
^
p
z
−
ι
,
0
z
−
ς
ι
−
ε
¯
χ
ι
−
z
−
c
−
ε
¯
Γ
1
−
ε
ρ
z
−
c
χ
c
=
∇
c
C
χ
ε
,
ρ
z
−
z
−
c
−
ε
¯
Γ
1
−
ε
e
^
p
z
,
c
χ
c
.
(15)
∇
C
χ
d
ε
,
ρ
z
=
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
−
Δ
ρ
χ
ι
=
−
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
1
−
ρ
χ
ι
+
ρ
Δ
χ
ι
=
−
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
1
−
ρ
χ
ι
−
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
ρ
χ
ι
+
1
−
χ
ι
=
−
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
1
−
ρ
χ
ι
−
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
ρ
χ
ι
+
1
+
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
ρ
χ
ι
=
−
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
1
−
ρ
χ
ι
−
1
Γ
1
−
ε
∑
ι
=
z
+
1
d
ρ
ι
−
1
−
z
ι
−
1
−
ς
z
−
ε
¯
ρ
χ
ι
+
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
ρ
χ
ι
=
−
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
1
−
ρ
χ
ι
−
1
Γ
1
−
ε
ρ
d
−
1
−
z
d
−
1
−
ς
z
−
ε
¯
ρ
χ
d
−
1
Γ
1
−
ε
∑
ι
=
z
+
1
d
−
1
ρ
ι
−
z
+
1
ι
−
ς
z
+
1
−
ε
¯
ρ
χ
ι
+
1
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
ρ
χ
ι
=
−
1
−
ρ
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
χ
ι
−
ρ
Δ
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
ρ
χ
ι
−
1
Γ
1
−
ε
ρ
d
−
z
d
−
1
−
z
+
1
−
ε
¯
χ
d
=
−
Δ
ρ
Γ
1
−
ε
∑
ι
=
z
d
−
1
ρ
ι
−
z
ι
−
ς
z
−
ε
¯
χ
ι
−
1
Γ
1
−
ε
ρ
d
−
z
d
−
z
−
ε
¯
χ
d
=
∇
R
χ
d
ε
,
ρ
z
−
d
−
z
−
ε
¯
Γ
1
−
ε
e
^
p
d
,
z
χ
d
.
Numerical calculations have been done in order to verify the first equation in Proposition1. The values used are c=2.5, ρ=0.7, and ε=0.3. The results are illustrated in Figure 1.
In addition to that, the data are presented in Table1.Figure 1
The relation between nabla Riemann and Caputo fractional proportional differences.Table 1
The relation between nabla Riemann and Caputo fractional proportional differences.
z
Caputo
Riemann
Difference
3.5
2.181077
−0.415443
2.596520
4.5
5.786530
3.190010
2.596520
5.5
10.787558
8.649248
2.138310
6.5
17.274639
15.611509
1.663130
7.5
25.339691
24.081105
1.258585
8.5
35.058768
34.121524
0.937244
9.5
46.490945
45.800344
0.690601
10.5
59.680789
59.175722
0.505066
11.5
74.661373
74.294052
0.367321
12.5
91.456990
91.190999
0.265991
13.5
110.085354
109.893402
0.191952
14.5
130.559312
130.421178
0.138134
15.5
152.888147
152.788974
0.099173
16.5
177.078543
177.007482
0.071061
17.5
203.135301
203.084469
0.050832Lemma 3.
Let0<ε<1, z,ι∈ℕc, and ι<z, then ∇z−ςι−ε¯=−εz−ςι−ε−1¯.Proof.
(16)
∇
z
−
ς
ι
−
ε
¯
=
z
−
ς
ι
−
ε
¯
−
z
−
1
−
ς
ι
−
ε
¯
=
z
−
ι
+
1
−
ε
¯
−
z
−
ι
−
ε
¯
=
Γ
z
−
ι
−
ε
+
1
Γ
z
−
ι
+
1
−
Γ
z
−
ι
−
ε
Γ
z
−
ι
=
Γ
z
−
ι
−
ε
Γ
z
−
ι
z
−
ι
−
ε
z
−
ι
−
1
=
Γ
z
−
ι
+
1
−
ε
−
1
Γ
z
−
ι
−
ε
z
−
ι
=
−
ε
Γ
z
−
ι
+
1
−
ε
−
1
Γ
z
−
ι
+
1
=
−
ε
z
−
ι
+
1
−
ε
−
1
¯
=
−
ε
z
−
ς
ι
−
ε
−
1
¯
.


## 3. Monotonicity Results

The following two monotonicity definitions are given in [18].Definition 7.
Lety:ℕa⟶ℝ be a function satisfying ya≥0, 0<α<1. Then, yt is called an α−increasing function on ℕa if yt+1≥αyt∀t∈ℕa.Definition 8.
Lety:ℕa⟶ℝ be a function satisfying ya≤0, 0<α<1. Then, yt is called an α−decreasing function on ℕa if yt+1≤αyt∀t∈ℕa.
In the following, we report the new proportional monotonicity main results.Theorem 1.
Letχ:ℕc⟶ℝ be a function, and suppose that (∇ε,ρc−1Rχ) z≥0 for 0<ε<1 and 0<ρ≤1,z∈ℕc−1. Then, χzisερ−increasing.Proof.
(17)
∇
c
−
1
R
χ
ε
,
ρ
z
=
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ρ
Γ
1
−
ε
∑
ι
=
c
z
e
^
p
z
−
ι
,
0
z
−
ς
ι
−
ε
¯
χ
ι
=
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ρ
Γ
1
−
ε
∑
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c
z
ρ
z
−
ι
z
−
ς
ι
−
ε
¯
χ
ι
.
Let(18)Sz=∑ι=czρz−ιz−ςι−ε¯χι.
Then,(19)∇c−1Rχε,ρz=∇ρΓ1−εSz.
Hence, from the assumption, we have∇ρSz≥0. That is,(20)∇ρSz=1−ρSz+ρ∇Sz=1−ρSz+ρSz−Sz−1=Sz−ρSz+ρSz−ρSz−1=Sz−ρSz−1=∑ι=czρz−ιz−ςι−ε¯χι−ρ∑ι=cz−1ρz−1−ιz−1−ςι−ε¯χι=z−ςz−ε¯χz+∑ι=cz−1ρz−ιz−ςι−ε¯χι−∑ι=cz−1ρz−1−ιz−1−ςι−ε¯χι=z−z+1−ε¯χz+∑ι=cz−1ρz−ιχιz−ςι−ε¯−z−1−ςι−ε¯=1−ε¯χz+∑ι=cz−1ρz−ιχι∇z−ςι−ε¯=Γ1−εΓ1χz+∑ι=cz−1ρz−ιχι−εz−ςι−ε−1¯=Γ1−εχz−ε∑ι=cz−1ρz−ιχιz−ςι−ε−1¯≥0.
Therefore,(21)∇c−1Rχε,ρz=∇ρΓ1−εSz=1Γ1−εΓ1−εχz−ε∑ι=cz−1ρz−ιχιz−ςι−ε−1¯=χz−εΓ1−ε∑ι=cz−1ρz−ιχιz−ςι−ε−1¯=0.
Hence,(22)χz≥εΓ1−ε∑ι=cz−1ρz−ιχιz−ςι−ε−1¯.
Clearly,χc−1=0. So, we can start the induction from the next step. When z=c, we get χc≥0; also, when z=c+1, we have(23)χc+1≥εΓ1−ε∑ι=ccρc+1−ιχιc+1−ςι−ε−1¯=εΓ1−ερc+1−cχcc+1−ςc−ε−1¯=εΓ1−ερχcc+1−c+1−ε−1¯=ερΓ1−εχcΓ1−εΓ2=ερχc.
Now forz+1, replace z by z+1, then we get(24)χz+1≥εΓ1−ε∑ι=czρz+1−ιχιz+1−ςι−ε−1¯≥εΓ1−ερz+1−zχzz+1−ςz−ε−1¯=εΓ1−ερχzz+1−z+1−ε−1¯=ερΓ1−εχz2−ε−1¯=ερχz.
Hence,χz is ερ−increasing which completes the proof.
Using Theorem1 and Proposition 1 we can state the following Caputo fractional proportional difference monotonicity result.Corollary 1.
Letχ:ℕc−1⟶ℝ be a function, and suppose that for 0<ε<1 and 0<ρ≤1. Suppose that(25)∇c−1Cχε,ρz≥−e^pz,c−1Γ1−εz−c+1−ε¯χc−1,z∈ℕc−1,then χzisερ−increasing.Proof.
(26)
∇
c
−
1
C
χ
ε
,
ρ
z
=
∇
c
−
1
R
χ
ε
,
ρ
z
−
e
^
p
z
,
c
−
1
Γ
1
−
ε
z
−
c
+
1
−
ε
¯
χ
c
−
1
,
∀
z
∈
ℕ
c
−
1
,now, from the assumption we have(27)∇ε,ρc−1Cχz≥−e^pz,c−1Γ1−εz−c+1−ε¯χc−1,z∈ℕc−1,z∈ℕc−1,hence,(28)∇c−1Cχε,ρz+e^pz,c−1Γ1−εz−c+1−ε¯χc−1≥0,z∈ℕc−1,z∈ℕc−1,which means that ∇c−1Rχε,ρz≥0.
Now, from Theorem1, we conclude that χz is ερ−increasing.Theorem 2.
Letχ:ℕc−1⟶ℝ be a function which satisfies χc≥0, and suppose that for 0<ε<1 and 0<ρ≤1. If χz is increasing on ℕc, then we have(29)∇c−1Rχε,ρz≥0,∀z∈ℕc−1.Proof.
Since(30)∇c−1Rχε,ρz=χz−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χι,z∈ℕc−1,when z=c, we have from the assumption ∇c−1Rχε,ρc=χc≥0.
Clearly,χc−1=0. So, we can start the induction from the next step.
Assume that∇c−1Rχε,ρi≥0,∀i<z. We shall show that ∇c−1Rχε,ρz≥0.
Since, from assumption,χz is increasing, it follows that χz≥χz−1≥χc≥0,∀z∈ℕc:(31)∇c−1Rχε,ρz=χz−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χι=χz−εΓ1−ερz−z+1z−ςz−1−ε−1¯χz−1−εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χι=χz−ερχz−1−εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χι+εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χz−1−εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χz−1=χz−ερχz−1+εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χz−1−χι−εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χz−1≥χz−ερχz−1−εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χz−1=χz−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χz−1=χz−χz−1+χz−1−εΓ1−ε∑ι=cz−2ρz−ιz−ςι−ε−1¯χz−1≥χz−1−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χz−1=χz−11−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯≥0.Theorem 3.
Letχ:ℕc−1⟶ℝ be a function which satisfies χc>0 and be strictly increasing on ℕc, where 0<ε<1 and 0<ρ≤1. Then,(32)∇c−1Rχε,ρ>0,z∈ℕc−1.Proof.
Since(33)∇c−1Rχε,ρz=χz−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χι,z∈ℕc−1,when z=c, we have ∇c−1Rχε,ρc=χc>0. Clearly, χc−1=0, and so we can start the induction from the next step.
Assume that∇c−1Rχε,ρi>0,∀i<z. We shall show that ∇c−1Rχε,ρz>0.
Since, from assumption,χz is increasing it follows that χz>χz−1>χc>0,∀z∈ℕc:(34)∇c−1Rχε,ρz=χz−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χι>χz−χz−1+χz−1−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χz−1>χz−1−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯χz−1=χz−11−εΓ1−ε∑ι=cz−1ρz−ιz−ςι−ε−1¯>0.Theorem 4.
Letχ:ℕc−1⟶ℝ be a function, and suppose that (∇c−1Rχε,ρ) z≤0 for 0<ε<1 and 0<ρ≤1,z∈ℕc−1. Then, χzisερ−decreasing.Proof.
Letθ:ℕc−1⟶ℝ be a function such that θz=−χz; hence,(35)∇c−1Rθε,ρz=∇c−1R−χε,ρz=−∇c−1Rχε,ρz≥0.
Now by Theorem1, we conclude that θz is ερ−increasing.
Hence,(36)θz+1≥ερθz,which is(37)−χz+1≥ερ−χz,χz+1≤ερχz,that is to say, χz is ερ−decreasing.Theorem 5.
Let a functionχ:ℕc−1⟶ℝ be decreasing on ℕc such that χc≤0. Then, for 0<ε<1 and 0<ρ≤1, we have(38)∇c−1Rχε,ρz≤0,∀z∈ℕc−1.Proof.
The proof follows by applying Theorem2 to θz=−χz.
Using Theorem4.3 in [4] we can state the following.Theorem 6.
(see [4]). For any 0<ε<1, 0<ρ≤1, p=ρ−1/ρ, and χ:ℕc+1⟶ℝ, the following equality holds:(39)∇cR−ε,ρ∇c−1Rχε,ρz=χz−e^pz,cΓεz−c+1ε−1¯χc.


## 4. Application: Mean Value Theorem (MVT)

First, for the sake of simplification, depending on Theorem6, we shall write(40)∇cR−ε,ρ∇c−1Rχε,ρz=χz−Sz,cχc,where(41)Sz,c=e^pz,cΓεz−c+1ε−1¯.Theorem 7.
(the fractional proportional difference MVT)LetΘ and θ be functions defined on ℕc∩dℕ=c,c+1,c+2,…,d−2,d−1,d, where c≡dmod1. Assume that θ is strictly increasing, θc>0, and 0<ε<1 and 0<ρ≤1. Then, there exist s1,s2∈ℕc∩d,ℕ such that(42)∇c−1RΘε,ρs1∇c−1Rθε,ρs1≤Θd−Sd,cΘcθd−Sd,cθc≤∇c−1RΘε,ρs2∇c−1Rθε,ρs2.Proof.
First we need to show thatθd−Sd,cθc>0. Since θ is strictly increasing, then by Theorem 3 we have(43)∇c−1Rθε,ρz>0,∀z,∈ℕc∩d,ℕ.
Applying the fractional sum operator associated to∇cRθε,ρz on both sides of the inequality, by means of (40), we get(44)∇cR∇c−1Rθε,ρ−ε,ρz>∇c−1R0−ε,ρ∀z,∈ℕc∩d,ℕ,or we have(45)θz−Sz,cθc>0.∀z∈ℕc∩dℕ.
Forz=d, we get(46)θd−Sd,cθc>0.
To prove the theorem, we use the proof by contradiction. Assume (42) is not true, then either(47)Θd−Sd,cΘcθd−Sd,cθc<∇c−1RΘε,ρz∇c−1Rθε,ρz,∀z∈ℕc∩dℕ,or(48)Θd−Sd,cΘcθd−Sd,cθc>∇c−1RΘε,ρz∇c−1Rθε,ρz,∀z∈ℕc∩dℕ.
Again, sinceθ is strictly increasing, then by Theorem 3 we conclude that(49)∇c−1RΘε,ρz,∀z∈ℕc∩dℕ.
Hence, (47) becomes(50)Θd−Sd,cΘcθd−Sd,cθc∇c−1Rθε,ρz<∇c−1RΘε,ρz,∀z∈ℕc∩dℕ.
Applying the fractional sum operator on both sides of the inequality atz=d and by making use of (43), we see that(51)Θd−Sd,cΘcθd−Sd,cθcθd−Sd,cθc<Θd−Sd,cΘc,and hence, Θd<Θd, which is a contradiction. In a similar way, (48) leads to contradiction.


## 5. Conclusions

The conclusions of this article are summarized as follows:(1)
The summation and difference of a discrete fractional proportional have been detected.(2)
The nabla discrete new Riemann–Liouville and Caputo fractional proportional differences of order0<ε<1 on the time scale ℤ have been formulated.(3)
The fractional proportional sums associated to∇cRχε,ρz with order 0<ε<1 have been defined.(4)
The relation between nabla Riemann–Liouville and Caputo fractional proportional differences has been detected.(5)
The monotonicity results for the nabla Caputo fractional proportional difference which are if∇c−1Rχε,ρz>0, then χz is ερ−increasing; if χz is strictly increasing on ℕc and χc>0, then ∇c−1Rχε,ρz>0 has been proved as well.(6)
A new version of the fractional proportional difference of the mean value theorem onℤ has been proved as an application.



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