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Article

Dynamic Hardy–Copson-Type Inequalities via (γ,a)-Nabla-Conformable Derivatives on Time Scales

1
Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City 11884, Cairo, Egypt
2
Department of Automation, Biomechanics and Mechatronics, Lodz University of Technology, 1/15 Stefanowski St., 90-924 Lodz, Poland
3
Department of Mathematics, Texas A&M University-Kingsville, Kingsville, TX 78363-8202, USA
*
Authors to whom correspondence should be addressed.
Submission received: 2 August 2022 / Revised: 31 August 2022 / Accepted: 2 September 2022 / Published: 5 September 2022
(This article belongs to the Section Mathematics)

Abstract

:
We prove new Hardy–Copson-type ( γ , a ) -nabla fractional dynamic inequalities on time scales. Our results are proven by using Keller’s chain rule, the integration by parts formula, and the dynamic Hölder inequality on time scales. When γ = 1 , then we obtain some well-known time-scale inequalities due to Hardy. As special cases, we obtain new continuous and discrete inequalities. Symmetry plays an essential role in determining the correct methods to solve dynamic inequalities.

1. Introduction

A great number of dynamic Hardy type inequalities on time scales have been established by many researchers who were motivated by some applications [1,2,3]. Additionally, over several decades, many generalizations, extensions, and refinements of other types of dynamic inequalities have been studied; we refer the reader to the papers [4,5,6,7,8].
In 1920, Hardy [9] established the following renowned discrete inequality.
Theorem 1.
If { a ( n ) } n = 0 is a non-negative real sequence and p > 1 , then
n = 1 1 n = 1 n a ( ) p p p 1 p n = 1 a p ( n ) .
In 1925, using the calculus of variations, Hardy himself in [10] gave the integral analogous of inequality (1) as follows:
Theorem 2.
If f is a non-negative continuous function on [ 0 , ) and p > 1 , then
0 1 ϑ 0 ϑ f ( ϱ ) d ϱ p d ϑ p p 1 p 0 f p ( ϑ ) d ϑ .
Copson [11] improved inequality (1) by replacing the arithmetic mean of a sequence by a weighted arithmetic mean in the following manner: Let b ( ) 0 and ω ( ) 0 for all . If p > 1 ,   c > 1 , then
= 1 ω ( ) [ A ¯ ( ) ] c j = 1 b ( j ) ω ( j ) p p c 1 p = 1 ω ( ) [ A ¯ ( ) ] p c b p ( ) ,
where A ¯ ( ) = j = 1 ω ( j ) , and if 0 c < 1 < p , then
= 1 ω ( ) [ A ¯ ( ) ] c j = 1 b ( j ) ω ( j ) p p 1 c p = 1 ω ( ) [ A ¯ ( ) ] p c b p ( ) .
Hardy–Copson inequalities have been developed by Bennett as follows: Let b ( ) 0 and ω ( ) 0 for all and j = 1 ω ( j ) < . If 0 c < 1 < p , then
= 1 ω ( ) [ A ¯ ( ) ] c j = 1 b ( j ) ω ( j ) p p 1 c p = 1 ω ( ) [ A ¯ ( ) ] p c b p ( ) ,
and
= 1 ω ( ) [ A ( ) ] c j = 1 b ( j ) ω ( j ) p p 1 c p = 1 ω ( ) [ A ( ) ] p c b p ( ) ,
where A ¯ ( ) = j = 1 ω ( j ) , and A ( ) = j = ω ( j ) .
If 1 < c p , then
= 1 ω ( ) [ A ¯ ( ) ] c j = 1 b ( j ) ω ( j ) p p 1 c p = 1 ω ( ) [ A ¯ ( ) ] p c b p ( ) ,
and
= 1 ω ( ) [ A ( ) ] c j = 1 b ( j ) ω ( j ) p p 1 c p = 1 ω ( ) [ A ( ) ] p c b p ( ) .
Then, the continuous version of the inequalities (3) (or (7)) and (6) (or (8)) established by Copson [12], respectively, are as follows: Let ω ( ς ) and f ( ς ) be non-negative functions and A ¯ ( ς ) = 0 ς ω ( ϱ ) d ϱ , B ( ς ) = 0 ς ω ( ϱ ) f ( ϱ ) d ϱ , B ¯ ( ς ) = ς ω ( ϱ ) f ( ϱ ) d ϱ . If 1 < c , 1 p , 1 < b , then
0 b ω ( ς ) [ A ¯ ( ς ) ] c [ B ( ς ) ] p d t p c 1 p 0 b ω ( ς ) [ A ¯ ( ς ) ] p c f p ( ς ) d t ,
and if c < 1 p ,   a > 0 , then
a ω ( ς ) [ A ¯ ( ς ) ] c [ B ¯ ( ς ) ] p d t p 1 c p a ω ( ς ) [ A ¯ ( ς ) ] p c f p ( ς ) d t .
Other refinements of continuous Hardy–Copson inequalities, which are generalizations of (9) and (10), respectively, have been introduced by Pacarić and Hanjs as follows: Let ω ( ς ) and f ( ς ) be non-negative functions and A ¯ ( ς ) = 0 ς ω ( ϱ ) d ϱ , B ( ς ) = 0 ς ω ( ϱ ) f ( ϱ ) d ϱ , B ¯ ( ς ) = ς ω ( ϱ ) f ( ϱ ) d ϱ .
If ξ 0 ,   p > 1 ,   ξ + c > 1 then
0 ω ( ς ) [ A ¯ ( ς ) ] c [ B ( ς ) ] p + ξ d t p + ξ c + ξ 1 p 0 ω ( ς ) f p ( ς ) [ B ( ς ) ] ξ [ A ¯ ( ς ) ] ξ + c p d t ,
and if ξ 0 ,   p > 1 ,   ξ + c < 1 then
0 ω ( ς ) [ A ¯ ( ς ) ] c [ B ¯ ( ς ) ] p d t p + ξ 1 c ξ p 0 ω ( ς ) f p ( ς ) [ B ¯ ( ς ) ] ξ [ A ¯ ( ς ) ] c + ξ p d t .
A time scale T is an arbitrary non-empty closed subset of the real numbers R . We assume throughout that T has the topology that it inherits from the standard topology on the real number R . For more details on time-scales calculus, see [13,14]. We define the forward jump operator σ : T T by
σ ( ς ) : = inf { ϱ T : ϱ > ς } , ς T ,
and the backward jump operator ρ : T : T is defined by
ρ ( ς ) : = sup { ϱ T : ϱ < ς } , ς T .
In the previous two definitions, we set inf = sup T (i.e., if ς is the minimum of T , then σ ( ς ) = ς ) and sup = inf T (i.e., if ς is the maximum of ς , then ρ ( ς ) = ς ), where is the empty set.
We introduce the nabla derivative of a function f : T R at a point t T κ , as follows:
Definition 1.
Let f : T R be a function, and let t T κ . We define f ( t ) as the real number (provided it exists) with the property that for any ϵ > 0 , there exists a neighborhood N of t (i.e., N = ( t δ , t + δ ) T for some δ > 0 ), such that
| [ f ρ ( t ) f ( s ) ] f ( t ) [ ρ ( t ) s ] | ϵ | ρ ( t ) s | f o r e v e r y s N .
We say that f ( t ) is the nabla derivative of f at t.
Definition 2.
We say that a function F : T R is a nabla antiderivative of f : T R if F ( t ) = f ( t ) for all t T κ . In this case, the nabla integral of f is defined by
a t f ( t ) τ = F ( t ) F ( a ) f o r a l l t T κ .
Now, we introduce the set of all ld-continuous functions in order to find a class of functions that have nabla antiderivatives.
Definition 3
(Ld-Continuous Function). We say that the function f : T R is ld-continuous if it is continuous at all left-dense points of T and its right-sided limits exist (finite) at all right-dense points of T .
Theorem 3
(Existence of Nabla Antiderivatives). Every ld-continuous function possess a nabla antiderivative.
In 2021, Kayar et al. [15], established the time-scale version unification of discrete and continuous Bennett–Leindler inequalities (10) and (11), as in the following theorem.
Theorem 4.
Assume that T is a time scale with a [ 0 , ) . Moreover, suppose that ϕ and g are non-negative, ld-continuous, ∇-differentiable, and locally nabla integrable functions. Set
ϖ ( ς ) = r ς ϕ ( ϱ ) ϱ Ψ ( ς ) = r ς ϕ ( ϱ ) g ( ϱ ) ϱ .
Assume that Ψ ( ) < and r ϕ ( ς ) [ ϖ ( ς ) ] c + θ < . Suppose that there exists M > 0 , such that ϖ ¯ ( ς ) ϖ ¯ ρ ( ς ) M for ς ( a , ) T . If c 0 ,   p > 1 , and c + θ > 1 are real constants, then
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p ϖ c + θ ( ς ) ς c + p c + θ 1 r ϕ ( ς ) g ( ς ) Ψ c + p 1 ( ς ) [ ϖ ρ ( ς ) ] c + θ 1 ς .
(ii)
r ϕ ( ς ) Ψ c + p ( ς ) ϖ ( ς ) c + θ ς c + p c + θ 1 p r ϕ ( ς ) g p ( ς ) Ψ c ( ς ) [ ϖ ( ς ) ] ( c + θ ) ( p 1 ) [ ϖ ρ ( ς ) ] ( c + θ 1 ) p ς .
(iii)
r ϕ ( ς ) Ψ c + p ( ς ) ϖ ( ς ) c + θ ς c + p c + θ 1 p M ( c + θ ) ( p 1 ) r ϕ ( ς ) g p ( ς ) [ Ψ c ( ς ) ] [ ϖ ρ ( ς ) ] ( c + θ p ) ς .
(iv)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p ϖ c + θ ( ς ) ς c + p c + θ 1 M c + θ 1 p r ϕ ( ς ) g p ( ς ) [ Ψ c ( ς ) ] [ ϖ ( ς ) ] c + θ p ς .
(v)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p [ ϖ ρ ( ς ) ] c + θ ς c + p c + θ 1 M c + θ r ϕ ( ς ) g ( ς ) Ψ c + p 1 ( ς ) [ ϖ ρ ( ς ) ] c + θ 1 ς .
(vi)
r ϕ ( ς ) Ψ c + p ( ς ) ϖ ρ ( ς ) c + θ ς M c + θ c + p c + θ 1 p r ϕ ( ς ) g p ( ς ) [ Ψ c ( ς ) ] [ ϖ ρ ( ς ) ] c + θ p ς .
Rahmat introduced a new nabla conformable calculus in his recent paper [16]. He presents the following basic definitions and concepts.
Definition 4.
Let [ ϱ , ς ] T and ϱ < ς . The generalized time-scale power function G ^ n : T × T R + for n N 0 is defined by
G ^ n ( ς , ϱ ) = ( ς ϱ ) n , if [ ς , ϱ ] dense ; j = 0 n 1 ( ς ρ j ( ϱ ) ) , if [ ς , ϱ ] isolated ;
and its inverse function G ^ n : T × T R + is then given by
G ^ n ( ς , ϱ ) = ( ς ϱ ) n , if [ ς , ϱ ] dense ; 1 j = 0 n 1 ( ρ n ( ς ) ρ j ( ϱ ) ) , if [ ς , ϱ ] isolated .
We use the convention G ^ 0 ( ς , ϱ ) = 1 for all ϱ , ς T .
Definition 5.
Given a function f : T R and a T , f is ( γ , a ) -nabla differentiable at ς > a , if it is nabla differentiable at ς, and its ( γ , a ) -nabla derivative is defined by
a γ f ( ς ) = G ^ 1 γ ( ς , a ) f ( ς ) ς > a ,
where the function G ^ 1 γ ( ς , a ) , as defined in (15). If a γ [ f ( ς ) ] exists in some interval ( a , a + ϵ ) T , ϵ > 0 , then we define
a γ [ f ( a ) ] = lim ς a + a γ [ f ( ς ) ]
if the lim ς a + a γ [ f ( ς ) ] exists. Moreover, we call f as ( γ , a ) -nabla differentiable on T k   ( a T k ) , provided a γ [ f ( ς ) ] exists for all ς T k . The function a γ : T k R is then called the ( γ , a ) -nabla derivative of f on T k .
Next, we provide the ( γ , a ) -nabla derivatives of sums, products, and quotients of ( γ , a ) -nabla differentiable functions.
Theorem 5.
Assume f , g : T R are ( γ , a ) -nabla differentiable at t T k ,   t > a . Then:
(i)
The sum f + g : T R is ( γ , a ) -nabla differentiable at t, with
a γ ( f + g ) ( t ) = a γ f ( t ) + a γ g ( t ) .
(ii)
For all k R , then k f : T R is ( γ , a ) -nabla differentiable at t with
a γ ( k f ) ( t ) = k a γ f ( t ) .
(iii)
The product f g : T R is ( γ , a ) -nabla differentiable at t, with
a γ ( f g ) = [ a γ f ( t ) ] g ( t ) + f ρ ( t ) [ a γ g ( t ) ] .
(iv)
If g ( t ) g ρ ( t ) 0 , then f / g is ( γ , a ) -nabla differentiable at t, with
a γ f g ( t ) = [ a γ f ( t ) ] g ( t ) f ( t ) [ a γ g ( t ) ] g ( t ) g ρ ( t ) .
Lemma 1.
Let g C l d ( T ) and assume that f : R R is a continuously differentiable function. Then, ( f g ) : T R is ( γ , a ) -nabla differentiable and satisfies
a γ ( f g ) ( ς ) = 0 1 f g ( ς ) h ν ( ς ) g ( ς ) d h a γ g ( ς ) .
Lemma 2.
Let γ ( 0 , 1 ] . Assume ξ : T R is continuous and ( γ , a ) -nabla differentiable of order γ at ς T k , where ς > a and ϖ : R R is continuously differentiable. Then, there is c in the real interval [ ρ ( ς ) , ς ] , such that
a γ ( ϖ ξ ) ( ς ) = ϖ ( ξ ( c ) ) a γ ( ξ ( ς ) ) .
Lemma 3.
Suppose that d ,   b T , where b > d . If η , ξ are conformable ( γ , a ) -nabla fractional differentiable and γ ( 0 , 1 ] , then
d b η ( t ) a γ ξ ( t ) a γ t = η ( t ) ξ ( t ) d b d b a γ η ( t ) ξ ρ ( t ) a γ t .
Definition 6.
Assume that 0 < γ 1 ,   a ,   ς 1 ,   ς 2 T ,   a ς 1 ς 2 , and f C l d ( T ) , then we say that f is ( γ , a ) -nabla integrable on interval [ ς 1 , ς 2 ] if the following integral:
a γ f ( ς ) = ς 1 ς 2 f ( τ ) a γ τ = ς 1 ς 2 f ( τ ) G ^ γ 1 ( σ γ 1 ( τ ) , a ) τ .
exists and is finite.
For the case T = R , we have the classical conformable integral as defined in [17], namely
a ς f ( τ ) a γ τ = a ς f ( τ ) ( τ a ) γ 1 d τ .
For T = h Z ,   h > 0 , we have a new conformable fractional h-sum given by
a ς f ( τ ) a γ τ = τ ( a , ς ] h f ( τ ) ρ γ 1 ( τ ) a h ( γ 1 ) .
For T = q N 0 , we have a new conformable fractional q-sum given by
a ς f ( τ ) a γ τ = τ ( a , ς ] τ ( 1 q ˜ ) f ( τ ) ρ γ 1 ( τ ) a q ˜ ( γ 1 ) .
Theorem 6.
Let γ ( 0 , 1 ] and a T . Then, for any ld-continuous function f : T R , there exists a function F : T R , such that
a γ F ( ς ) = f ( ς ) f o r a l l ς T k .
The function F is called an ( γ , a ) -nabla antiderivative of f.
Lemma 4
([18]) (Hölder inequality for the γ -nabla derivative). Let d , b T , where b > d . If γ ( 0 , 1 ] and ϖ , ξ : T R , then
d b | ϖ ( ς ) ξ ( ς ) | a γ ς d b | ϖ ( ς ) | p a γ ς 1 / p d b | ξ ( ς ) | q a γ ς 1 / q ,
where p , q > 1 and 1 / p + 1 / q = 1 . This inequality is reversed if 0 < p < 1 and if p < 0 , or q < 0 .
Here, we prove the new Hardy–Copson-type dynamic via the ( γ , a ) -nabla-conformable calculus on time scales. Our inequalities have a completely new form and may be considered as extensions of [15]. As special cases, we obtain some new continuous and discrete inequalities of the Hardy-type, generalizing those obtained in the literature.
The paper is organized as follows. The original results are given and proven in Section 2. In Section 3, we stated the discussion of the results. We end with Section 4, the conclusion.
Now, we are ready to state and to prove our main results. Throughout the following, we assumed that the time-scale T is unbounded above.

2. Main Results

Theorem 7.
Assume that T is a time scale with 0 r T and ς a . Moreover, suppose that g and ϕ are non-negative ( γ , a ) -nabla fractional differentiable and locally integrable functions on [ r , ) T . Set ϖ ( ς ) = r ς ϕ ( ϱ ) a γ ϱ and Ψ ( ς ) = r ς ϕ ( ϱ ) g ( ϱ ) a γ ϱ , and assume that Ψ ( ) < and r ϕ ( ς ) ϖ ( ς ) a γ ς < . Suppose that there exists Λ > 0 , such that ϖ ( ς ) ϖ ρ ( ς ) Λ for ς [ r , ) T . If c 0 ,   p > 1 ,   c + θ > 1 , and 0 < γ 1 are real constants, then
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ a γ ς .
(ii)
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 a γ ς 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ρ ( ς ) ] ( c + θ γ ) p a γ ς .
(iii)
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 a γ ς 1 + c + p γ c + θ γ p Λ ( c + θ γ + 1 ) ( p 1 ) r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] ( c + θ p γ + 1 ) a γ ς .
(iv)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ Λ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] 1 + c γ [ ϖ ( ς ) ] c + θ p γ + 1 a γ ς .
(v)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ Λ c + θ γ + 1 r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ a γ ς .
(vi)
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ρ ( ς ) c + θ γ + 1 a γ ς Λ c + θ γ + 1 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] c + θ p γ + 1 a γ ς .
Proof. 
(i)
By using the integration by parts to the left-hand side of inequality (25), we have a γ u ( ς ) = ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 and v ( ς ) = [ Ψ ( ς ) ] c + p γ + 1 ; it becomes
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς = u ( ς ) Ψ c + p γ + 1 ( ς ) | r + r u ρ ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς
where u ( ς ) = ς ϕ ( ϱ ) [ ϖ ( ϱ ) ] c + θ γ + 1 a γ ϱ . By using Ψ ( r ) = 0 and u ( ) = 0 , one can have
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς = r u ρ ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς .
By chain rule, and using a γ Ψ ( ς ) = ϕ ( ς ) g ( ς ) 0 , we obtain
a γ Ψ c + p γ + 1 ( ς ) = ( c + p γ + 1 ) Ψ c + p γ ( d ) a γ Ψ ( ς ) = ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) Ψ c + p γ ( d ) ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) .
Since a γ ϖ ( ς ) = ϕ ( ς ) 0 and c + θ > 1 , by chain rule, we get
a γ [ ϖ γ c θ ( ς ) ] = 0 1 ( γ c θ ) a γ ϖ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h = 0 1 ( c + θ γ ) ϕ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h ( c + θ γ ) 0 1 ϕ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ( ς ) ] c + θ γ + 1 d h = ( c + θ γ ) ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 a γ [ ϖ γ c θ ( ς ) ] ( c + θ γ ) .
Therefore,
u ρ ( ς ) = ρ ( ς ) ϕ ( ϱ ) [ ϖ ( ϱ ) ] c + θ γ + 1 a γ ϱ ρ ( ς ) a γ [ ϖ γ c θ ( ϱ ) ] ( c + θ γ ) a γ ϱ = ϖ γ c θ ( ) + [ ϖ ρ ( ς ) ] γ c θ c + θ γ [ ϖ ρ ( ς ) ] γ c θ c + θ γ .
Substituting (33) and (35) into (32) leads to
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ a γ ς ,
which is the desired result (25).
(ii)
Since
r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) ϖ ρ ( ς ) c + θ γ a γ ς = r ϕ 1 p ( ς ) g ( ς ) [ Ψ ( ς ) ] c γ + 1 p [ ϖ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) p [ ϖ ρ ( ς ) ] c + θ γ × ϕ ( ς ) Ψ 1 + c + p γ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 p 1 p a γ ς .
Applying the Hölder inequality with the constants p and p / p 1 to the right-hand side of above equation leads to the inequality
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 a γ ς ( 1 + c + p γ ) ( c + θ γ ) r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ρ ( ς ) ] ( c + θ γ ) p a γ ς 1 p × r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 a γ ς p 1 p
therefore,
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 a γ ς 1 + c + p γ c + θ γ p × r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ρ ( ς ) ] ( c + θ γ ) p a γ ς ,
which is the desired inequality (26).
(iii)
To obtain inequality (27), we use
[ ϖ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ρ ( ς ) ] ( c + θ γ ) p = ϖ ( ς ) ϖ ρ ( ς ) ( c + θ γ + 1 ) ( p 1 ) [ ϖ ρ ( ς ) ] c θ [ ϖ ρ ( ς ) ] p γ + 1 = Λ ( c + θ γ + 1 ) ( p 1 ) [ ϖ ρ ( ς ) ] ( c + θ p γ + 1 )
in inequality (26), and then inequality (26) becomes
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 a γ ς 1 + c + p γ c + θ γ p Λ ( c + θ γ + 1 ) ( p 1 ) r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] ( c + θ p γ + 1 ) a γ ς .
(iv)
In order to obtain inequality (28), we use inequality (25) and the constant Λ as follows.
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ a γ ς c + p γ + 1 c + θ γ Λ c + θ γ r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ( ς ) ] c + θ γ a γ ς .
Now, by using the Hölder inequality with the constants p ,   p / p 1 , we get that
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ Λ c + θ γ r ϕ 1 p g ( ς ) [ Ψ ( ς ) ] 1 + c γ p ϖ c + θ p γ + 1 p ( ς ) ϕ ( ς ) [ Ψ ( ς ) ] 1 + c + p γ [ ϖ ( ς ) ] c + θ γ + 1 p 1 p a γ ς
thus,
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ Λ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] 1 + c γ [ ϖ ( ς ) ] c + θ p γ + 1 a γ ς .
(v)
By using the integration by parts to the left-hand side of inequality (29), we have a γ u ( ς ) = ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 and v ( ς ) = [ Ψ ( ς ) ] c + p γ + 1 , which becomes
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς = u ( ς ) Ψ c + p γ + 1 ( ς ) | r + r u ρ ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς
where u ( ς ) = ς ϕ ( ϱ ) [ ϖ ρ ( ϱ ) ] c + θ γ + 1 a γ ϱ . By using Ψ ( r ) = 0 and u ( ) = 0 , one can have
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς = r u ρ ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς .
By following the procedure in the proof of ( i ) , we arrive at inequality (34) as
a γ [ ϖ γ c θ ( ς ) ] ( c + θ γ ) ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 ( c + θ γ ) ϕ ( ς ) Λ c + θ γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1
which implies ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 Λ c + θ γ + 1 a γ [ ϖ γ c θ ( ς ) ] c + θ γ . Then, we obtain that
u ρ ( ς ) = ρ ( ς ) ϕ ( ϱ ) [ ϖ ρ ( ϱ ) ] c + θ γ + 1 a γ ϱ ρ ( ς ) Λ c + θ γ + 1 a γ [ ϖ γ c θ ( ϱ ) ] ( c + θ γ ) a γ ϱ = Λ c + θ γ + 1 ϖ γ c θ ( ) + [ ϖ ρ ( ς ) ] γ c θ c + θ γ Λ c + θ γ + 1 [ ϖ ρ ( ς ) ] γ c θ c + θ γ .
Substituting (33) and (40) into (39) leads to
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ Λ c + θ γ + 1 r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ a γ ς .
which is the desired result (29).
(vi)
Since
r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) ϖ ρ ( ς ) c + θ γ a γ ς = r ϕ 1 p ( ς ) g ( ς ) [ Ψ ( ς ) ] c γ + 1 p [ ϖ ρ ( ς ) ] c + θ p γ + 1 p ϕ ( ς ) Ψ 1 + c + p γ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 p 1 p a γ ς .
Applying the Hölder inequality with the constants p and p / p 1 to the right-hand side of the above equation leads to the inequality
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 a γ ς Λ c + θ γ + 1 ( 1 + c + p γ ) ( c + θ γ ) r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] c + θ p γ + 1 a γ ς 1 p
× r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς p 1 p
therefore,
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ρ ( ς ) c + θ γ + 1 a γ ς Λ c + θ γ + 1 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] c + θ p γ + 1 a γ ς ,
which is the desired inequality (30).
Remark 1.
In Theorem 7, if we take γ = 1 , then we get Theorem 4.
Corollary 1.
In Theorem 7, if we take T = R and Λ = 1 , then we get
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ς a ) γ 1 d t c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ( ς ) ] c + θ γ ( ς a ) γ 1 d t .
(ii)
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 ( ς a ) γ 1 d t 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 ( ς a ) γ 1 d t .
where
ϖ ( ς ) = r ς ϕ ( ϱ ) ( ϱ a ) 1 γ d ϱ , and Ψ ( ς ) = r ς ϕ ( ϱ ) g ( ϱ ) ( ϱ a ) 1 γ d ϱ .
Remark 2.
In inequality (45), if we take r = 0 ,   γ = 1 , and c = 0 , we get inequality (9).
Remark 3.
In inequality (45), if we take r = 0 , and γ = 1 , we get inequality (11).
Corollary 2.
In Theorem 7, if we take T = h Z , then we get
(i)
ς = r h ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ ς = r h ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ( h t h ) ] c + θ γ ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(ii)
ς = r h ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) 1 + c + p γ c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( h t h ) ] c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(iii)
ς = r h ϕ ( ς ) Ψ 1 + c + p ( ς ) ϖ ( ς ) c + θ ( ρ γ 1 ( ς ) a ) h ( γ 1 ) 1 + c + p γ c + θ γ p Λ ( c + θ γ + 1 ) ( p 1 ) ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( h t h ) ] ( c + θ p γ + 1 ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(iv)
ς = r h ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ Λ c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] 1 + c γ [ ϖ ( ς ) ] c + θ p γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(v)
ς = r h ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ( h t h ) ] c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ Λ c + θ γ + 1 ς = r h ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ( h t h ) ] c + θ γ ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(vi)
ς = r h ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( h t h ) c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) Λ c + θ γ + 1 1 + c + p γ c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( h t h ) ] c + θ p γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
where
ϖ ( ς ) = h ς = r h ς h 1 ϕ ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) , and Ψ ( ς ) = h ς = r h ς h 1 ϕ ( ϱ ) g ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
Theorem 8.
Assume that T is a time scale with 0 r T and ς a . Moreover, suppose that g and ϕ are non-negative ( γ , a ) -nabla fractional differentiable and locally integrable functions on [ r , ) T . Set ϖ ( ς ) = ς ϕ ( ϱ ) a γ ϱ and Ψ ( ς ) = ς ϕ ( ϱ ) g ( ϱ ) a γ ϱ , and assume that Ψ ( r ) < and r ϕ ( ς ) ϖ ρ ( ς ) a γ ς < . Suppose that there exists Λ > 0 , such that ϖ ρ ( ς ) ϖ ( ς ) Λ for ς [ r , ) T . If c 0 ,   p > 1 ,   c + θ > 1 , and 0 < γ 1 are real constants, then
(i)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ a γ ς .
(ii)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ ϖ ρ ( ς ) c + θ γ + 1 a γ ς 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( ς ) ] ( c + θ γ ) p a γ ς .
(iii)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ ϖ ρ ( ς ) c + θ γ + 1 a γ ς 1 + c + p γ c + θ γ p Λ ( c + θ γ + 1 ) ( p 1 ) r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 a γ ς .
(iv)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ Λ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] 1 + c γ [ ϖ ρ ( ς ) ] c + θ p γ + 1 a γ ς .
(v)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ Λ c + θ γ + 1 r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ a γ ς .
(vi)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ ϖ ( ς ) c + θ γ + 1 a γ ς Λ c + θ γ + 1 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 a γ ς .
Proof. 
(i)
By using the integration by parts to the left-hand side of inequality (46), we have a γ u ( ς ) = ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 and v ( ς ) = [ Ψ ( ς ) ] c + p γ + 1 ; it becomes
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς = u ( ς ) Ψ c + p γ + 1 ( ς ) | r + r u ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς
where u ( ς ) = r ς ϕ ( ϱ ) [ ϖ ρ ( ϱ ) ] c + θ γ + 1 a γ ϱ . By using Ψ ( ) = 0 and u ( r ) = 0 , one can have
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς = r u ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς .
By chain rule, and by using a γ Ψ ( ς ) = ϕ ( ς ) g ( ς ) 0 , we obtain
a γ Ψ c + p γ + 1 ( ς ) = ( c + p γ + 1 ) Ψ c + p γ ( d ) a γ Ψ ( ς ) = ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) Ψ c + p γ ( d ) ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ .
Since a γ ϖ ( ς ) = ϕ ( ς ) 0 and c + θ > 1 , by chain rule, we get
a γ [ ϖ γ c θ ( ς ) ] = 0 1 ( γ c θ ) a γ ϖ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h = 0 1 ( c + θ γ ) ϕ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h ( c + θ γ ) 0 1 ϕ ( ς ) [ ( 1 h ) ϖ ρ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h = ( c + θ γ ) ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ [ ϖ ( ς ) ] γ c θ c + θ γ .
Therefore,
u ( ς ) = r ς ϕ ( ϱ ) [ ϖ ρ ( ϱ ) ] c + θ γ + 1 a γ ϱ r ς a γ [ ϖ γ c θ ( ϱ ) ] ( c + θ γ ) a γ ϱ = ϖ γ c θ ( r ) + [ ϖ ( ς ) ] γ c θ c + θ γ [ ϖ ( ς ) ] γ c θ c + θ γ .
Substituting (54) and (56) into (53) leads to
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ a γ ς ,
which is the desired result (46).
(ii)
Since
r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ ϖ ( ς ) c + θ γ a γ ς = r ϕ 1 p ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 p [ ϖ ρ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) p [ ϖ ( ς ) ] c + θ γ ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ [ ϖ ρ ( ς ) ] c + θ γ + 1 p 1 p a γ ς .
Applying the Hölder inequality with the constants p and p / p 1 to the right-hand side of the above equation leads to inequality
r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ ϖ ρ ( ς ) c + θ γ + 1 a γ ς ( 1 + c + p γ ) ( c + θ γ ) r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( ς ) ] ( c + θ γ ) p a γ ς 1 p
× r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς p 1 p
therefore,
r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ ϖ ρ ( ς ) c + θ γ + 1 a γ ς 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( ς ) ] ( c + θ γ ) p a γ ς ,
which is the desired inequality (47).
(iii)
To obtain inequality (48), we use
[ ϖ ρ ( ς ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( ς ) ] ( c + θ γ ) p = ϖ ρ ( ς ) ϖ ( ς ) ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( ς ) ] c θ [ ϖ ( ς ) ] p γ + 1 = Λ ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( ς ) ] ( c + θ p γ + 1 )
in inequality (47). Then, the desired inequality (48) can be proven directly.
(iv)
In order to obtain inequality (49), we use inequality (46) and the constant Λ as follows.
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ a γ ς c + p γ + 1 c + θ γ Λ c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ρ ( ς ) ] c + θ γ a γ ς .
Now, by using the Hölder inequality with the constants p ,   p / p 1 , we get that
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ Λ c + θ γ r ϕ 1 p g ( ς ) [ Ψ ρ ( ς ) ] 1 + c γ p [ ϖ ρ ( ς ) ] c + θ p γ + 1 p ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ [ ϖ ρ ( ς ) ] c + θ γ + 1 p 1 p a γ ς
thus,
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ Λ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] 1 + c γ [ ϖ ρ ( ς ) ] c + θ p γ + 1 a γ ς .
(v)
By using the integration by parts to the left-hand side of inequality (50), we have a γ u ( ς ) = ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 and v ( ς ) = [ Ψ ( ς ) ] c + p γ + 1 ; it becomes
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ( ς ) ] c + θ γ + 1 a γ ς = u ( ς ) Ψ c + p γ + 1 ( ς ) | r + r u ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς
where u ( ς ) = r ς ϕ ( ϱ ) [ ϖ ( ϱ ) ] c + θ γ + 1 a γ ϱ . By using Ψ ( ) = 0 and u ( r ) = 0 , one can have
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ( ς ) ] c + θ γ + 1 a γ ς = r u ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς .
By following the procedure in the proof of ( i ) , we arrive at inequality (55) as
a γ [ ϖ γ c θ ( ς ) ] ( c + θ γ ) ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 ( c + θ γ ) ϕ ( ς ) Λ c + θ γ + 1 [ ϖ ( ς ) ] c + θ γ + 1
which implies ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 Λ c + θ γ + 1 a γ [ ϖ γ c θ ( ς ) ] c + θ γ . Then, we obtain that
u ( ς ) = r ς ϕ ( ϱ ) [ ϖ ( ϱ ) ] c + θ γ + 1 a γ ϱ r ς Λ c + θ γ + 1 a γ [ ϖ γ c θ ( ϱ ) ] ( c + θ γ ) a γ ϱ = Λ c + θ γ + 1 ϖ γ c θ ( r ) + [ ϖ ( ς ) ] γ c θ c + θ γ Λ c + θ γ + 1 [ ϖ ( ς ) ] γ c θ c + θ γ .
Substituting (54) and (62) into (61) leads to
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 [ ϖ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ Λ c + θ γ + 1 r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ a γ ς .
which is the desired result (50).
(vi)
Since,
r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ ϖ ( ς ) c + θ γ a γ ς = r ϕ 1 p ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 p [ ϖ ( ς ) ] c + θ p γ + 1 p ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ [ ϖ ( ς ) ] c + θ γ + 1 p 1 p a γ ς .
Applying the Hölder inequality with the constants p and p / p 1 to the right-hand side of the above equation leads to the inequality
r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ ϖ ( ς ) c + θ γ + 1 a γ ς Λ c + θ γ + 1 ( 1 + c + p γ ) ( c + θ γ ) r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 a γ ς 1 p
× r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ [ ϖ ( ς ) ] c + θ γ + 1 a γ ς p 1 p
therefore,
r ϕ ( ς ) [ Ψ ρ ( ς ) ] 1 + c + p γ ϖ ( ς ) c + θ γ + 1 a γ ς Λ c + θ γ + 1 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 a γ ς ,
which is the desired inequality (51).
Corollary 3.
In Theorem 8, if we take γ = 1 , then we get
(i)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p [ ϖ ρ ( ς ) ] c + θ ς c + p c + θ 1 r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p 1 [ ϖ ( ς ) ] c + θ 1 ς .
(ii)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p ϖ ρ ( ς ) c + θ ς c + p c + θ 1 p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c [ ϖ ρ ( ς ) ] ( c + θ ) ( p 1 ) [ ϖ ( ς ) ] ( c + θ 1 ) p ς .
(iii)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p ϖ ρ ( ς ) c + θ ς c + p c + θ 1 p Λ ( c + θ ) ( p 1 ) r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c [ ϖ ( ς ) ] ( c + θ p ) ς .
(iv)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p [ ϖ ρ ( ς ) ] c + θ ς c + p c + θ 1 Λ c + θ p r ϕ ( ς ) g p ( ς ) [ Ψ c ( ς ) ] [ ϖ ρ ( ς ) ] c + θ p ς .
(v)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p [ ϖ ( ς ) ] c + θ ς c + p c + θ 1 Λ c + θ r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p [ ϖ ( ς ) ] c + θ 1 ς .
(vi)
r ϕ ( ς ) Ψ c + p ( ς ) ϖ ( ς ) c + θ ς Λ c + θ c + p c + θ 1 p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c [ ϖ ( ς ) ] c + θ p ς .
where
ϖ ( ς ) = ς ϕ ( ϱ ) ϱ , and Ψ ( ς ) = ς ϕ ( ϱ ) g ( ϱ ) ϱ .
which is Theorem 3.8 in [15].
Corollary 4.
In Theorem 8, if we take T = R and Λ = 1 , then we get
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ς a ) γ 1 d t c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) Ψ c + p γ ( ς ) [ ϖ ( ς ) ] c + θ γ ( ς a ) γ 1 d t .
(ii)
r ϕ ( ς ) Ψ 1 + c + p γ ( ς ) ϖ ( ς ) c + θ γ + 1 ( ς a ) γ 1 d t 1 + c + p γ c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 ( ς a ) γ 1 d t .
where
ϖ ( ς ) = r ς ϕ ( ϱ ) ( ϱ a ) 1 γ d ϱ , and Ψ ( ς ) = r ς ϕ ( ϱ ) g ( ϱ ) ( ϱ a ) 1 γ d ϱ .
Corollary 5.
In Theorem 8, if we take T = h Z , then we get
(i)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] c + p γ + 1 [ ϖ ( h t h ) ] c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ ς = r h ϕ ( ς ) g ( ς ) [ Ψ ( h t h ) ] c + p γ [ ϖ ( ς ) ] c + θ γ ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(ii)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] 1 + c + p γ ϖ ( h t h ) c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) 1 + c + p γ c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( h t h ) ] c γ + 1 [ ϖ ( h t h ) ] ( c + θ γ + 1 ) ( p 1 ) [ ϖ ( ς ) ] ( c + θ γ ) p ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(iii)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] 1 + c + p γ ϖ ( h t h ) c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) 1 + c + p γ c + θ γ p Λ ( c + θ γ + 1 ) ( p 1 ) ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( h t h ) ] c γ + 1 [ ϖ ( ς ) ] ( c + θ p γ + 1 ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(iv)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] c + p γ + 1 [ ϖ ( h t h ) ] c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ Λ c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] 1 + c γ [ ϖ ( h t h ) ] c + θ p γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(v)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] c + p γ + 1 [ ϖ ( h t h ) ] c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ Λ c + θ γ + 1 ς = r h ϕ ( ς ) g ( ς ) [ Ψ ( h t h ) ] c + p γ [ ϖ ( ς ) ] c + θ γ ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(vi)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] 1 + c + p γ ϖ ( ς ) c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) Λ c + θ γ + 1 1 + c + p γ c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( h t h ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
where
ϖ ( ς ) = h ϱ = ς h ϕ ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) , and Ψ ( ς ) = h ϱ = ς h ϕ ( ϱ ) g ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
Theorem 9.
Assume that T is a time scale with 0 r T and ς a . Moreover, suppose that g and ϕ are non-negative ( γ , a ) -nabla fractional differentiable and locally integrable functions on [ r , ) T . Set ϖ ( ς ) = r ς ϕ ( ϱ ) a γ ϱ and Ψ ( ς ) = ς ϕ ( ϱ ) g ( ϱ ) a γ ϱ , and assume that Ψ ( r ) < and r ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς < . If c 0 ,   p > 1 ,   0 < c + θ < γ and 0 < γ 1 are real constants, then
(i)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ a γ ς .
(ii)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 a γ ς .
Proof. 
(i)
By using the integration by parts to the left-hand side of inequality (67), we have a γ u ( ς ) = ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 and v ( ς ) = [ Ψ ( ς ) ] c + p γ + 1 , and it becomes
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς = u ( ς ) Ψ c + p γ + 1 ( ς ) | r + r u ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς
where u ( ς ) = r ς ϕ ( ϱ ) [ ϖ ( ϱ ) ] c + θ γ + 1 a γ ϱ . By using Ψ ( ) = 0 and u ( r ) = 0 , one can have
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς = r u ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς .
By chain rule, and by using a γ Ψ ( ς ) = ϕ ( ς ) g ( ς ) 0 , we obtain
a γ Ψ c + p γ + 1 ( ς ) = ( c + p γ + 1 ) Ψ c + p γ ( d ) a γ Ψ ( ς ) = ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) Ψ c + p γ ( d ) ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ .
Since a γ ϖ ( ς ) = ϕ ( ς ) 0 and 0 < c + θ < γ , by chain rule, we get
a γ [ ϖ γ c θ ( ς ) ] = 0 1 ( γ c θ ) a γ ϖ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h = 0 1 ( γ c θ ) ϕ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h
( γ c θ ) 0 1 ϕ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ( ς ) ] c + θ γ + 1 d h = ( γ c θ ) ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1
ϕ ( ς ) [ ϖ ( ς ) ] c + θ γ + 1 a γ [ ϖ γ c θ ( ς ) ] ( γ c θ ) .
Therefore,
u ( ς ) = r ς ϕ ( ϱ ) [ ϖ ( ϱ ) ] c + θ γ + 1 a γ ϱ r ς a γ [ ϖ γ c θ ( ϱ ) ] γ c θ a γ ϱ = ϖ γ c θ ( r ) + [ ϖ ( ς ) ] γ c θ γ c θ [ ϖ ( ς ) ] γ c θ γ c θ .
Substituting (71) and (74) into (70) leads to
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ a γ ς ,
which is the desired result (67).
(ii)
Employing the same procedure of the proof of ( i i ) of Theorem 7, we obtain the desired result (68).
Corollary 6.
In Theorem 9, if we take γ = 1 , we get that
(i)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p ϖ c + θ ( ς ) ς c + p c + θ 1 r ϕ ( ς ) g ( ς ) [ Ψ ρ ( ς ) ] c + p 1 [ ϖ ( ς ) ] c + θ 1 ς .
(ii)
r ϕ ( ς ) [ Ψ ρ ( ς ) ] c + p ϖ c + θ ( ς ) ς c + p c + θ 1 p r ϕ ( ς ) g p ( ς ) [ Ψ ρ ( ς ) ] c [ ϖ ( ς ) ] c + θ p ς .
where
ϖ ( ς ) = r ς ϕ ( ϱ ) ϱ and Ψ ( ς ) = ς ϕ ( ϱ ) g ( ϱ ) ϱ
which is Theorem 3.13 in [15].
Corollary 7.
In Theorem 9, if we take T = R , we get that
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ς a ) γ 1 d t c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ ( ς a ) γ 1 d t .
(ii)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ς a ) γ 1 d t c + p γ + 1 c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 ( ς a ) γ 1 d t .
where
ϖ ( ς ) = r ς ϕ ( ϱ ) ( ϱ a ) γ 1 d t and Ψ ( ς ) = ς ϕ ( ϱ ) g ( ϱ ) ( ϱ a ) γ 1 d ϱ
Corollary 8.
In Theorem 9, if we take T = h Z , we get that
(i)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ ς = r h ϕ ( ς ) g ( ς ) [ Ψ ( h t h ) ] c + p γ [ ϖ ( ς ) ] c + θ γ ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(ii)
ς = r h ϕ ( ς ) [ Ψ ( h t h ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( h t h ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
where
ϖ ( ς ) = h ϱ = r h ς h 1 ϕ ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) and Ψ ( ς ) = h ϱ = r h ϕ ( ϱ ) g ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
Theorem 10.
Assume that T is a time scale with 0 r T and ς a . Moreover, suppose that g and ϕ are non-negative ( γ , a ) -nabla fractional differentiable and locally integrable functions on [ r , ) T . Set ϖ ( ς ) = ς ϕ ( ϱ ) a γ ϱ and Ψ ( ς ) = r ς ϕ ( ϱ ) g ( ϱ ) a γ ϱ , and assume that Ψ ( ) < and r ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς < . If c 0 ,   p > 1 ,   0 < c + θ < γ and 0 < γ 1 are real constants, then
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ( ς ) ] c + p γ [ ϖ ρ ( ς ) ] c + θ γ a γ ς .
(ii)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ρ ( ς ) ] c + θ p γ + 1 a γ ς .
Proof. 
(i)
By using the integration by parts to the left-hand side of inequality (75), we have a γ u ( ς ) = ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 and v ( ς ) = [ Ψ ( ς ) ] c + p γ + 1 ; it becomes
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς = u ( ς ) Ψ c + p γ + 1 ( ς ) | r + r u ρ ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς
where u ( ς ) = ς ϕ ( ϱ ) [ ϖ ρ ( ϱ ) ] c + θ γ + 1 a γ ϱ . By using Ψ ( r ) = 0 and u ( ) = 0 , one can have
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς = r u ρ ( ς ) a γ Ψ c + p γ + 1 ( ς ) a γ ς .
By chain rule, and using a γ Ψ ( ς ) = ϕ ( ς ) g ( ς ) 0 , we obtain
a γ Ψ c + p γ + 1 ( ς ) = ( c + p γ + 1 ) Ψ c + p γ ( d ) a γ Ψ ( ς ) = ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) Ψ c + p γ ( d ) ( c + p γ + 1 ) ϕ ( ς ) g ( ς ) [ Ψ ( ς ) ] c + p γ .
Since a γ ϖ ( ς ) = ϕ ( ς ) 0 and 0 < c + θ < γ , by chain rule, we get
a γ [ ϖ γ c θ ( ς ) ] = 0 1 ( γ c θ ) a γ ϖ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h = 0 1 ( γ c θ ) ϕ ( ς ) [ ( 1 h ) ϖ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h ( γ c θ ) 0 1 ϕ ( ς ) [ ( 1 h ) ϖ ρ ( ς ) + h ϖ ρ ( ς ) ] c + θ γ + 1 d h = ( γ c θ ) ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 ϕ ( ς ) [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ [ ϖ γ c θ ( ς ) ] ( γ c θ ) .
Therefore,
u ρ ( ς ) = ρ ( ς ) ϕ ( ϱ ) [ ϖ ρ ( ϱ ) ] c + θ γ + 1 a γ ϱ ρ ( ς ) a γ [ ϖ γ c θ ( ϱ ) ] ( γ c θ ) a γ ϱ = ϖ γ c θ ( ) + [ ϖ ρ ( ς ) ] γ c θ γ c θ [ ϖ ρ ( ς ) ] γ c θ γ c θ .
Substituting (79) and (81) into (78) leads to
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ρ ( ς ) ] c + θ γ + 1 a γ ς c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ( ς ) ] c + p γ [ ϖ ρ ( ς ) ] c + θ γ a γ ς .
which is the desired result (75).
(ii)
Employing the same procedure of the proof of ( i i ) of Theorem 7, we obtain the desired result (76).
Corollary 9.
In Theorem 10, if we take γ = 1 , we get that
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p [ ϖ ρ ( ς ) ] c + θ ς c + p c + θ 1 r ϕ ( ς ) g ( ς ) [ Ψ ( ς ) ] c + p 1 [ ϖ ρ ( ς ) ] c + θ 1 ς .
(ii)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p [ ϖ ρ ( ς ) ] c + θ ς c + p c + θ 1 p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c [ ϖ ρ ( ς ) ] c + θ p ς .
where
ϖ ( ς ) = r ς ϕ ( ϱ ) ϱ and Ψ ( ς ) = ς ϕ ( ϱ ) g ( ϱ ) ϱ
which is Theorem 3.19 in [15].
Corollary 10.
In Theorem 10, if we take T = R , we get that
(i)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ς a ) γ 1 d t c + p γ + 1 c + θ γ r ϕ ( ς ) g ( ς ) [ Ψ ( ς ) ] c + p γ [ ϖ ( ς ) ] c + θ γ ( ς a ) γ 1 d t .
(ii)
r ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 ϖ c + θ γ + 1 ( ς ) ( ς a ) γ 1 d t c + p γ + 1 c + θ γ p r ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( ς ) ] c + θ p γ + 1 ( ς a ) γ 1 d t .
where
ϖ ( ς ) = ς ϕ ( ϱ ) ( ϱ a ) γ 1 d t and Ψ ( ς ) = r ς ϕ ( ϱ ) g ( ϱ ) ( ϱ a ) γ 1 d ϱ
Corollary 11.
In Theorem 10, if we take T = h Z , we get that
(i)
ς = r h ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ( ς ) ] c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ ς = r h ϕ ( ς ) g ( ς ) [ Ψ ( ς ) ] c + p γ [ ϖ ( h t h ) ] c + θ γ ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
(ii)
ς = r h ϕ ( ς ) [ Ψ ( ς ) ] c + p γ + 1 [ ϖ ( h t h ) ] c + θ γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) c + p γ + 1 c + θ γ p ς = r h ϕ ( ς ) g p ( ς ) [ Ψ ( ς ) ] c γ + 1 [ ϖ ( h t h ) ] c + θ p γ + 1 ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .
where
ϖ ( ς ) = h ϱ = ς h ϕ ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) and Ψ ( ς ) = h ϱ = ς h ς h 1 ϕ ( ϱ ) g ( ϱ ) ( ρ γ 1 ( ς ) a ) h ( γ 1 ) .

3. Discussion

In this work, firstly, we began with the Introduction section, which contains a brief recall of the necessary results of the Hardy inequalities and the notions of time-scale calculus. Then, the original results are then given and proven in Section 2, since in the first theorem we established seven new inequalities that may be considered as a generalization of Theorem 4. Additionally, we obtain Inequalities (9) and (11) as a special case of our main results. After that, in the second theorem, we proved six new inequalities that may be considered as a generalization of Theorem 3.8 in [15]. In the last theorem, we investigated two main results, which generalize Theorem 3.19 in [15] After each theorem, we discussed the cases of time scales T = R , T = Z , and T = h Z .

4. Conclusions

In this manuscript, by employing the ( γ , a ) -nabla-conformable fractional calculus on the time scales of Rahmat et al. [16], several new Hardy-type inequalities were proven. The results extend several dynamic inequalities known in the literature, being new, even in the discrete and continuous settings.

Author Contributions

Conceptualization, A.A.E.-D., S.D.M., R.P.A. and J.A.; formal analysis, A.A.E.-D., S.D.M., R.P.A. and J.A.; investigation, A.A.E.-D., S.D.M., R.P.A. and J.A.; writing—original draft preparation, A.A.E.-D., S.D.M., R.P.A. and J.A.; writing—review and editing, A.A.E.-D., S.D.M., R.P.A. and J.A. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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El-Deeb, A.A.; Makharesh, S.D.; Awrejcewicz, J.; Agarwal, R.P. Dynamic Hardy–Copson-Type Inequalities via (γ,a)-Nabla-Conformable Derivatives on Time Scales. Symmetry 2022, 14, 1847. https://0-doi-org.brum.beds.ac.uk/10.3390/sym14091847

AMA Style

El-Deeb AA, Makharesh SD, Awrejcewicz J, Agarwal RP. Dynamic Hardy–Copson-Type Inequalities via (γ,a)-Nabla-Conformable Derivatives on Time Scales. Symmetry. 2022; 14(9):1847. https://0-doi-org.brum.beds.ac.uk/10.3390/sym14091847

Chicago/Turabian Style

El-Deeb, Ahmed A., Samer D. Makharesh, Jan Awrejcewicz, and Ravi P. Agarwal. 2022. "Dynamic Hardy–Copson-Type Inequalities via (γ,a)-Nabla-Conformable Derivatives on Time Scales" Symmetry 14, no. 9: 1847. https://0-doi-org.brum.beds.ac.uk/10.3390/sym14091847

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