Note on fractional difference Gronwall inequalities

Note on fractional difference Gronwall inequalities

Michal Feˇckan

and Michal Pospíšil

1Department of Mathematical Analysis and Numerical Mathematics, Comenius University, Mlynská dolina, 842 48 Bratislava, Slovakia

2Mathematical Institute of Slovak Academy of Sciences, Štefánikova 49, 814 73 Bratislava, Slovakia

3Centre for Research and Utilization of Renewable Energy, Faculty of Electrical Engineering and Communication, Brno University of Technology, Technická 3058/10, 616 00 Brno, Czech Republic

Received 19 March 2014, appeared 15 September 2014 Communicated by László Hatvani

Abstract. This note is a reaction on a bunch of fractional inequalities that appeared in the last few years, and that are all based on what claims to be a fractional discrete Gronwall inequality. However, we show by a counterexample that this inequality is not correct. Stimulated by this, the main aim of this note is to propose new inequalities and illustrate the results on examples. Asymptotic properties of a solution of a linear equation are studied as well. Moreover, a brief discussion of other related results is given.

Keywords: fractional difference, backward difference, difference inequality.

2010 Mathematics Subject Classification: 39A70, 39A10, 26A33.

1 Introduction

Recently, Deekshitulu and Mohan published a set of papers on fractional difference inequal- ities. All the main results in [12,13,15] are proved using a fractional discrete version of Gronwall inequality given in [11]. In this short note, we show that the proof of this inequality is not correct, the inequality does not hold, and hence the validity of the implied results is questionable. Stimulated by these results, the main aim of this paper is to prove new inequal- ities of Gronwall type for fractional difference inequalities with linear right-hand side and constant or variable coefficients. Asymptotic properties of a solution of a linear homogeneous fractional difference equation with constant coefficient are also studied. In our results, we use a convenient Green function. In Section 4, we illustrate our results on a linear example from [14] which is also corrected in this note.

Throughout the present paper, ∇ denotes the backward difference operator defined as

∇u(_n) =_u(_n)−u(_n−1), and having the next properties.

BCorresponding author. Email: Michal.Feckan@fmph.uniba.sk

Lemma 1.1. Let∇kF(_k,j):=_F(_k,j)−F(_k−^1,j). The following holds true

∇ [ k

j

∑

F(_k,j) ]

=_F(_k−^1,k) +

∑

∇kF(_k,j),

∑

A(_k−j)∇f(_j) =

k−1 j

∑

A(_j)∇kf(_k−j),

∑

f(_j)∇g(_j) = [_f(_j)_g(_j)]^b_j=a−1− ^b

∑

j=a−1

∇f(_j+1)_g(_j).

So we will consider fractional differences corresponding to backward difference operator.

Of course, other related achievements are already done than the above-mentioned: we refer the reader to similar interesting results in [1,8,16]. We only note here that in [1], inequality of the form

∇^µ∗u(_k)≤ a(_k)_u(_k), k∈N0

was investigated (cf. equation (4.1)); in [8], the authors used Riemann–Liouville type fractional difference; and in [16], the author focused on fractional difference corresponding to forward difference operator. Hence our results are not covered in these papers.

We denote byNa = {a,a+1, . . .}the shifted set of positive integers, for simplicity N= N1, and −Na = {. . . ,a−1,a}. We assume the property of empty sum and empty product, i.e.,∑^bj=a f(_j) =0,∏^bj=a f(_j) =1 whenevera,b∈Zare such thata>_b.

2 Caputo like fractional difference

In this section, we recall some definitions of∇-based fractional operators, and we show that fractional difference considered in [9] is of Caputo type.

Definition 2.1(see [18]). Let α∈ C, p=max{^0,p0}^, p0 ∈Zbe such that 0<Re(_α+_p0)≤ ^1, and function f be defined onNa−p. We define theα-th fractional sum as

Σ^αf(_k):= ∇^p Γ(_p+_α)

∑

(_k−ρ(_j))^p+α−1_f(_j) (2.1) fork ∈Na, whereρ(_j) =_j−1 and

(_α)^β =













Γ(_α+_β)

Γ(_α) , α,α+_β∈ {/ ^{. . . ,}−^{1, 0}}^,

1, α=_β=0,

0, α=0, β∈ {/ ^{. . . ,}−^{1, 0}}^, undefined, otherwise

forα,β∈C.

If p ∈ N,α ∈ −/ N0,k ∈ Na and f is defined on Na−p, formula (2.1) can be simplified to the casep=0 (cf. equation (2.2) in [18]) using

∇^p Γ(_p+_α)

∑

(_k−ρ(_j))^p+α−1_f(_j) = ¹ Γ(_α)

∑

(_k−ρ(_j))^α−1_f(_j), (2.2)

i.e.,

Σ^αf(_k) = ¹ Γ(_α)

∑

(_k−ρ(_j))^α−1_f(_j) (2.3) fork ∈Na, f defined onNa−p. We note that the latter equality is valid wheneverα∈R\−N0. Σ^α is sometimes denoted as∇^−α.

Fractional sum (2.3) is a discrete analogue to Riemann–Liouville fractional integral (cf.

[21]).

Analogically to forward difference based Riemann–Liouville [5] and Caputo [3,4] like fractional difference, Deekshitulu and Mohan proposed in [9] the next definition of a∇^-based fractional difference.

Definition 2.2. Let µ ∈ (0, 1) and f be defined on N0. Then we define the µ-th fractional difference of a function f as

∇^µf(_k) =

k−1

∑

(j−µ j

)

∇kf(_k−j) (2.4)

fork∈ N, where(^b_n)withb∈R,n∈Zis a generalized binomial coefficient given by (b

n )

=









Γ(b+1)

Γ(b−n+1)_Γ(n+1), n>0,

1, n=0,

0, n<0.

Here we used the lower indexk to denote the variable affected by operator∇^.

From now on, we assumeµ ∈ (0, 1). By [9, Remark 3.2],Σ^−µu(_n) =∇^µu(_n)onN ifu is defined on N0. However, this equivalence is obtained because of an incorrect application of operator∇. We provide a simple counterexample.

Example 2.3. Consideru(_k) =_b^k fork ∈N0andb>0. Then [∇^µu(_k)]_k=1=

[k−1

∑

(j−µ j

)

∇ku(_k−j) ]

k=1

= (−µ

0 )

[∇u(_k)]_k=1 =_b−^1.

On the other side, 0<1−µ<1. Thus p=1 in (2.1). Moreover, sinceuis defined onN0, then a=1. Therefore,

[Σ^−µu(_k)^]

k=1=

[ ∇ Γ(1−µ)

∑

(_k−ρ(_j))^−µ_u(_j) ]

k=1

= [

1 Γ(−µ)

∑

(_k−ρ(_j))^−µ−1_u(_j) ]

k=1

= [

1 Γ(−µ)

∑

Γ(_k−j−µ) Γ(_k−j+1)^u(_j)

]

k=1

=_b.

Here we applied the identity (2.2).

In reality, Σ^−µ is closely related to Riemann–Liouville like ∇-based fractional difference discussed in [6], while the following lemma states that ∇^µ is of Caputo type.

Lemma 2.4. Letµ∈ (0, 1)_,ν=1−µand function f be defined onN0. Then∇^µf(_k) =_Σ^ν(∇f(_k)) onN.

Proof. Fork ∈Nwe expand the left-hand side by (2.4), and apply Lemma1.1to get

∇^µf(_k) =

k−1

∑

(j−µ j

)

∇kf(_k−j) = ¹ Γ(_ν)

k−1 j

∑

Γ(_j+_ν)

Γ(_j+1)∇kf(_k−j)

= ¹ Γ(_ν)

∑

Γ(_k−j+_ν)

Γ(_k−j+1)∇f(_j) = ¹ Γ(_ν)

∑

(_k−ρ(_j))^ν−1∇f(_j) which, by (2.1) with a=1, p=0, is exactly what has to be proved.

In the sense of the above lemma, we add the lower index∗as done in [3,4] to denote the Caputo nature of the difference, i.e., ∇^µ∗ := ∇^µ in the rest of the paper. Next, in [9, Remark 3.2] the properties ofΣ^µ were translated to∇^µ∗. The above discussion implies that this is also a mistake, and∇^µ∗ does not have to possess such properties. Nevertheless, we do not go into details, as we do not need the properties in the present paper.

3 Linear fractional difference equation

In this section, we derive a solution of a nonhomogeneous linear fractional difference equa- tion in terms of a Green function. Particular case of a constant coefficient at linear term is investigated in details.

First, we provide a lemma transforming a fractional difference equation to a corresponding fractional sum equation and a direct corollary.

Lemma 3.1. Letµ∈(0, 1), u, f be real functions defined onN0, andN0×R, respectively. For any n∈N, if

∇^µ∗u(_k+1) = _f(_k,u(_k)), ∀k=0, 1, . . . ,n−^{1, (3.1)} then

u(_k) =_u(0) +

k−1 j

∑

A_µ(_k−^1,j)_f(_j,u(_j)), ∀k ∈0, 1, . . . ,n (3.2) with A_µ(_k,j) =^{(k−j+µ−1}

k−j )

for0≤j≤k.

Proof. We show that applying Σ^µ to equation (3.1) results in equation (3.2). Letk ∈ {0, 1, . . . , n−¹}be arbitrary and fixed. By Lemma 2.4, Σ^µ(∇^µ∗u(_k+1)) = _Σ^µ(_Σ^ν(∇u(_k+1)))for k ∈ N0, where ν= 1−µ. Property 2.(ii) in [18] says that ifµ∈ Candν∈/ N, thenΣ^µΣ^ν =_Σ^µ+ν. Hence

Σ^µ(

∇^µ_∗u(_k+1)⁾= _Σ¹(∇u(_k+1)) =

∑

∇u(_j+1) =_u(_k+1)−u(0)

= _Σ^µ_f(_k,u(_k)) = ¹ Γ(_µ)

∑

(_k−ρ(_j))^µ−1_f(_j,u(_j))

=

∑

(k−j+_µ−¹ k−j

)

f(j,u(j)) =

∑

A_µ(k,j)f(j,u(j)). This completes the proof.

Corollary 3.2. Letµ ∈ (0, 1), u, f be real functions defined on N0, and N0×R, respectively. For any n ∈N, if

∇^µ∗u(_k+1)≤ f(_k,u(_k)), ∀k=0, 1, . . . ,n−^1, then

u(k)≤u(0) +

k−1

∑

A_µ(k−1,j)f(j,u(j)), ∀k=0, 1, . . . ,n.

Proof. Since A_µ(_k,j) > 0 for each 0 ≤ j ≤ k < n, the statement immediately follows from definition ofΣ^µ.

Note that the above corollary holds true with≥instead of≤.

Next, we derive a solution of a linear initial value problem. Let h(k)denote the solution of the problem

∇^µ∗h(_k+1) =_a(_k)_h(_k), k∈N0

h(0) =1, (3.3)

and define a Green function{gj(_k)}k∈N0,k,j∈N0as

∇^µ∗gj(_k+1) =_a(_k)_gj(_k) +_δj(_k), k∈N0

gj(0) =0, (3.4)

whereδj(_k) =0 forj̸=_k andδj(_j) =1. Thenv(_k) =_u0h(_k),k ∈N0solves

∇^µ∗v(k+1) =a(k)v(k), k∈N0

v(0) =_u0, (3.5)

and

u(_k) =

k−1 j

∑

gj(_k)_b(_j), k ∈N0 (3.6) solves the equation

∇^µ_∗u(_k+1) =_a(_k)_u(_k) +_b(_k), k∈N0

u(0) =0. (3.7)

Here we note thatgj(k) =0 for 0≤ k≤ jandgj(j+1) =1. So setting e

gj(k):=gj(k+1), k=−1, 0, . . . , formula (3.6) becomes

u(_k) =

k−1

∑

e

gj(_k−¹)_b(_j), k ∈N0, (3.8) while (3.4) gives

∇^µ∗egj(_k+1) =_a(_k+1)_gej(_k), k ≥j e

gj(_j) =1, gej(_k) =0, −¹≤k <_j ^(3.9) for j ∈ N0. Using ∇^µ_∗gej(_j) = 1, we can directly verify that (3.8) solves (3.4). Thus there are two different ways, (3.6) and (3.8), how to define a solution u of (3.7). The following lemma uses (3.6), and concludes the above arguments.

Lemma 3.3. The initial value problem

∇^µ∗u(_k+1) =_a(_k)_u(_k) +_b(_k), k∈ N0

u(0) =_u0 has a solution

u(_k) =_u0h(_k) +

k−1

∑

gj(_k)_b(_j), k∈N0.

Proof. The considered problem is decomposed to a homogeneous equation with a nontriv- ial initial condition, of the form (3.5), and a nonhomogeneous equation with a zero initial condition, of the form (3.7). Consequently, the superposition principle is applied.

The rest of this section is devoted to the case of constant functiona(k). Proposition 3.4. Letµ∈(0, 1)_{, a,u0} ∈Rand u fulfill

∇^µ∗u(_k+1) =_au(_k), k ∈N0

u(0) =_u0. (3.10)

Then u has the form

u(_k) =_u0





1+

∑

j=1

∑

i1,i2,...,ij≥0

∑_l^j=1i_l≤k−j

∏

aΓ(_il+_µ) Γ(_µ)_Γ(_il+1)





, k∈N0. (3.11)

Proof. First, we apply Lemma3.1to get a corresponding fractional sum equation u(_k) =_u0+

k−1

∑

A_µ(_k−^1,j)_au(_j), k ∈N0. (3.12)

Note that A_µ(_k,j) = _Aµ(_k+_α,j+_α) = _Γ(^Γ_µ⁽₎^k_Γ⁻_(k^j₋^+µ_j+⁾₁₎ for any α ∈ [−j,∞), 0 ≤ j ≤ k. For simplicity, we denoteB_µ(k−j) =aA_µ(k,j)for 0≤ j≤k, i.e.,

B_µ(_n) = ^aΓ(_n+_µ)

Γ(_µ)_Γ(_n+1)^{, n} ∈N0. Consequently,

u(_k) =_u0+

k−1 j

∑

B_µ(_k−¹−j)_u(_j), k∈N0. (3.13) We claim that then

u(_k) =_u0





1+

∑

j=1

∑

i1,i2,...,ij≥0

∑^j_l=1i_l≤k−j

∏

B_µ(_il)





, k∈N0 (3.14)

what is the statement of the proposition.

We prove the claim by induction with respect tok. Ifk = 0, then due to the empty sum propertyu(0) =_u0in both (3.13) and (3.14). Now, let us assume that (3.14) holds for 0, 1, . . . ,k, and we show that it is true also fork+1. Using (3.13) and the inductive hypothesis, we have

u(_k+1) =_u0+

∑

B_µ(_k−j)_u(_j)

=_u0+

∑

B_µ(_k−j)_u0 [

1+

∑

q=1

∑

i1,i2,...,iq≥0

∑^ql=1il≤j−q

∏

B_µ(_il) ]

=_u0 [

1+

∑

B_µ(_k−j)

| {z }

=:S₁

+

∑

B_µ(_k−j)

∑

q=1

∑

i1,i2,...,iq≥0

∑^ql=1i_l≤j−q

∏

B_µ(_il)

| {z }

=:S2

] .

(3.15)

Changing k−j+1→j, we get S1 =

k+1 j

∑

B_µ(_j−¹) =

∑

j=1

∑

0≤i_j≤k

∏

B_µ(_ij) =

∑

j=1

∑

i1,i2,...,ij≥0

∑^j_l=1i_l≤k+1−j

∏

B_µ(_il).

Similarly inS2:

S2=

k+1 j

∑

B_µ(_j−¹)

k+1−j

q

∑

∑

i1,i2,...,iq≥0

∑^ql=1i_l≤k+1−j−q

∏

B_µ(_il)

=

k+1 j

∑

k+1−j q

∑

B_µ(_j−¹)

∑

i1,i2,...,iq≥0

∑^q_l=1i_l+j−1≤k−q

∏

B_µ(_il).

Now we switch the sums ∑^k_j=⁺1¹∑^k_q⁺=¹1^−j =_∑^k_q⁺₌¹_1∑^k_j=⁺₁^1−q. Note that the second sum is empty for q=_k+1. Hence

S2=

∑

k+1−q j

∑

B_µ(_j−1)

∑

i₁,i₂,...,iq≥0

∑^ql=1i_l+j−1≤k−q

∏

B_µ(_il).

Note that j−1 takes values 0, 1, . . . ,k−q. So we can denoteiq+1 = _j−1 and merge the last two sums to obtain

S2=

∑

q=1

∑

i₁,i2,...,i_q+1≥0

∑^q_l=⁺1¹i_l≤k−q q+1

∏

B_µ(_il).

Finally, we changeq+1→q

S2 =

k+1

q

∑

∑

i1,i2,...,iq≥0

∑^ql=1il≤k+1−q

∏

B_µ(i_l),

and after summingS1andS2, (3.14) is obtained fork+1.

Now we present an alternative proof without using induction principle.

Alternative proof of Proposition3.4. If a =0, then by (3.12), the statement is proved. From now on, we assume thata̸=0. By using the formula [19, Problem 7, p. 15]

nlim→∞

Γ(_n+_µ) Γ(_n+1)ⁿ

1−µ =1, (3.16)

it is clear that

1 lim sup

n→∞

n

√|B_µ(_n)| =1.

So the power series

∆µ(_x) =

∑

B_µ(_n)_xⁿ has the radius of convergence 1. Furthermore, (3.16) gives that

1 a

∑

B_µ(_n) = +_∞,

and using ^Bµ_a⁽ⁿ⁾ >0 we see that 1 a lim

x→1⁻∆_µ(_x) = +_∞. (3.17)

Next, we know that the sequence{|B_µ(_n)|}^∞_n=0is decreasing (see the proof of Lemma4.6) with lim_n→∞B_µ(_n) =0 (see (3.16)). So the Leibnitz criterion implies the convergence of the series

∑

B_µ(_n)(−¹)ⁿ, and the Abel theorem [23, p. 9] implies

x→−lim1⁺∆_µ(_x) = lim

x→1⁻

∑

B_µ(_n)(−1)ⁿ_xⁿ=

∑

B_µ(_n)(−1)ⁿ=_∆µ(−1). Next, Lemma4.6 implies

|u(_k)| ≤ |u0|(1+|a|)^k, k∈N0. Hence

1 lim sup

n→∞

n

√|u(_n)| ≥ ¹

1+|a|^{, (3.18)}

thus the power series

U(_x) =

∑

u(_n)_xⁿ

has the radius of convergence RU greater than or equal to _1+|¹_a|. Consequently, we start with 0̸=|x|< ₁₊¹_|a|. Then using (3.13), we derive

U(_x)_∆µ(_x) = (∞

i

∑

u(_i)_xⁱ ) (∞

j

∑

B_µ(_j)_x^j )

=

∑

k=1

∑

i+j=k−1

u(_i)_Bµ(_j)_x^k−1=

∑

k−1 j

∑

u(_j)_Bµ(_k−¹−j)_x^k−1

=

∑

(u(k)−u0)x^k−1 = ^∑

∞k=0u(_k)_x^k−u(0)

x − ^u0

1−x

= ^U(_x)

x − ^u0

x(1−x)^.

(3.19)

Solving (3.19) we obtain

U(_x) = ^u0

(1−x)(1−x∆_µ(_x))^{. (3.20)} From (3.20) we obtain

∑

u(n)xⁿ =u0

(∞ j

∑

x^j ) 



∑

i=0

xⁱ ( ∞

k

∑

B_µ(k)x^k )_i

. (3.21)

By expanding the right-hand side of (3.21) and comparing the powers of x, we immediately get (3.14).

Ifu0=0 thenU(_x) =0. So we suppose thatu0 ̸=0. Next, using (3.16) we see that B_µ(_n)∼ _Γ ^a

(_µ)_n^1−µ asn→∞. So by results of [23, pp. 224–225], we have

(1−x)_∆µ(_x)∼ (1−x)

∑

a Γ(_µ)_n^1−µx

n∼a(1−x)^1−µ as x→1⁻, so

xlim→1⁻(1−x)_∆µ(_x) =0, while we recall (3.17). Then clearly

xlim→1⁻|U(_x)|= +_∞

due to (3.20), thus the radius of convergenceRU ofU(x)is less than or equal to 1, and we get 1

1+|a| ≤ RU ≤^{1. (3.22)}

On the other hand, we know

|∆_µ(_x)| ≤ |a| [

1+ ¹ Γ(_µ)

∑

Γ(_n+_µ) Γ(_n+1)|x|ⁿ

]

≤ |a| [

1+ ¹ Γ(_µ)

∑

|x|ⁿ n^1−µ

]

=|a| [

1+ ¹

Γ(_µ)^Li1−µ(|x|) ]

where we applied the estimation of the ratio of gamma functions Γ(_n+_µ)

Γ(_n+1) ≤ ¹

n^1−µ, n∈N, µ∈ (0, 1) from [17], and used the notation

Li_ν(_x) =

∑

xⁿ n^ν = ¹

Γ(_ν)

∫ _∞

0

xt^ν−1

e^t−xdt, x,ν ∈(0, 1) for the polylogarithm function [20, Section 7.12]. Consequently,

|¹−x∆_µ(_x)| ≥¹− |x||∆_µ(_x)| ≥¹−F(|x|) (3.23) for

F(x) =x|a| [

1+ ¹

Γ(_µ)^Li1−µ(x) ]

.

We note that F is increasing on(0, 1). The right-hand side of (3.23) is positive if and only if

|x|<_F⁻¹(1). Note thatF⁻¹(1)> ¹

1+|a|. To see this, we estimate Li_1−µ

( 1 1+|a|

)

<

∑

( 1 1+|a|

)_n

= ¹

|a| < ^Γ(_µ)

|a| ^, i.e.,

|a| 1+|a|

[

1+ ¹

Γ(_µ)^Li1−µ ( 1

1+|a| )]

= _F ( 1

1+|a| )

<1.

This improves (3.22) to

F⁻¹(1)≤ RU ≤^{1. (3.24)}

Summarizing the above arguments, we obtain the next result.

Proposition 3.5. The solution u(_n)of initial value problem(3.10)with u0̸=0fulfills lim sup

n→∞

n

√|u(_n)|= ¹ RU

(3.25) for RU satisfying(3.24).

To get a better result, we note ifa >0 thenx∆µ(_x)is increasing on[0, 1)from 0 to+_∞. So for anya>0, there is a unique 1>_ra,µ> _F⁻¹(1)solving equation

1=_ra,µ∆µ(_ra,µ).

Then|1−x∆_µ(_x)|>0 for any|x|<_r|a|,µ. Hence (3.24) is improved to

r_|a|,µ ≤ RU ≤1. (3.26)

Note that estimation (3.24) as well as (3.26) gives better estimate on the asymptotic property (3.25) than (3.18) derived from Lemma4.6. Moreover, Proposition3.5yields the next corollary.

Corollary 3.6. If the solution u(_n)_of (3.10)with u0̸=0satisfies u(_n)→⁰as n →∞, then the rate of convergence is slower than any exponential one, i.e., there are no constants c1 > 0andϖ ∈ (0, 1) so that|u(_n)| ≤c1ϖⁿfor any n∈N0.

Proof. In contrary, ifc1 >0,ϖ∈ (0, 1)are such that|u(_n)| ≤c1ϖⁿ∀n∈ N0, then

RU = ¹

lim sup

n→∞

n

√|u(_n)| ≥ _ϖ¹ >1

what contradicts (3.26).

On the other hand, ifa>0,u0 ̸=0 then 1 u0

lim

x→r⁻_a,µU(_x) = +_∞,

so RU = _ra,µ, and thus {u(_n)}n∈N0 is unbounded satisfying (3.25). A related result is derived in [7, Theorem 5.1].

Remark 3.7. Taking limitµ→¹⁻in (3.10) gives

∇u(k+1) =au(k), k∈N0

u(0) =_u0

which has solutionu(_k) =_u0(1+_a)^k. On the other hand, from (3.11), we get

µlim→1⁻u(k)c=u0





1+

∑

a^j

∑

i1,i2,...,ij≥0

∑l^j=1i_l≤k−j

1







=_u0





1+

∑

a^j

k−j

m

∑

∑

i1,i2,...,ij≥0

∑_l^j=1i_l=m

1





=_u0 [

1+

∑

a^j

k−j m

∑

(m+_j−¹ j−1

)]

=_u0 [

1+

∑

(k j )

a^j ]

=_u0(1+_a)^k.

So we call the bracket in (3.11), the generalized binomial. Note that linear Riemann–Liouville fractional difference equations are solved in [7] leading to discrete Mittag-Leffler functions.

Next, we derive a formula for a Green function satisfying (3.4) with constant a.

Proposition 3.8. Let µ ∈ (0, 1)_{, a} ∈ R. Then for any i ∈ N0, the Green function {g^c_i(_k)}k∈N0

satisfying

∇^µ∗g^c_i(_k+1) =_agi^c(_k) +_δi(_k), k∈N0,

g^c_i(0) =0 (3.27)

has the form

g^c_i(_k) =

k−i j

∑

a^j−1

∑

i₁,i₂,...,i_j≥0

∑l^j=1i_l=k−i−j

∏

Γ(_il+_µ)

Γ(_µ)_Γ(_il+1)^{, k} ∈N0. (3.28)

Proof. Leti∈ N0 be arbitrary and fixed. As in the proof of Proposition3.4, applying Lemma 3.1, one can see that

g^c_i(_k) =

k−1 j

∑

(B_µ(_k−¹−j)_g^c_i(_j) +_Aµ(_k−^1,j)_δi(_j)⁾, k∈ N0. (3.29)

In particular,g^c_i(0) =0 which agrees with (3.28).

Let (3.28) be valid at 0, 1, . . . ,k. Then by (3.29), g^c_i(_k+1) =

∑

B_µ(_k−j)

j−i q

∑

a^q−1

∑

i₁,i₂,...,iq≥0

∑^q_l=1i_l=j−i−q

∏

C_µ(_il)

| {z }

=:S

+

∑

C_µ(_k−j)_δi(_j) (3.30)

where C_µ(_k−j) = _Aµ(_k,j). There are three possible cases. If i ∈ {/ 0, 1, . . . ,k}^, g^c_i(_k+1) = 0 due to the empty sum property. The same is true for (3.28).

Ifi= _{k, thenS}=0 and (3.30) implies g_i^c(_k+1) =

∑

C_µ(_k−j)_δi(_j) =_Cµ(0) =1.

The same holds for (3.28).

Finally, ifi∈ {0, 1, . . . ,k−1}, then S=

∑

j−i q

∑

a^qC_µ(_k−j)

∑

i1,i2,...,iq≥0

∑^ql=1i_l=j−i−q

∏

C_µ(_il)

=

k−i q

∑

∑

a^qC_µ(_k−j)

∑

i1,i2,...,iq≥0

∑^ql=1i_l=j−i−q

∏

C_µ(_il).

Note thatk−jtakes the values 0, 1, . . . ,k−q−i. So we denoteiq+1 =_k−jand write S=

k−i q

∑

a^q

∑

i₁,i2,...,i_q+1≥0

∑^q_l=⁺1¹i_l=k−i−q q+1

∏

C_µ(_il).

On substituteq+1→q, we get S=

k+1−i q

∑

a^q−1

∑

i1,i2,...,iq≥0

∑^ql=1il=k+1−i−q

∏

C_µ(_il)

and (3.30) becomes (3.28).

4 Fractional difference inequalities

In this section, we explain where the problem lies of the fractional Gronwall inequality es- tablished in [11]. Then we propose our alternative fractional difference inequalities of the