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On Some Important Dynamic Inequalities of Hardy–Hilbert-Type on Timescales
 
 
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Article

(γ,a)-Nabla Reverse Hardy–Hilbert-Type Inequalities on Time Scales

by
Ahmed A. El-Deeb
1,*,
Dumitru Baleanu
2,3 and
Jan Awrejcewicz
4,*
1
Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City, Cairo 11884, Egypt
2
Institute of Space Science, 077125 Bucharest, Romania
3
Department of Mathematics, Cankaya University, Ankara 06530, Turkey
4
Department of Automation, Biomechanics and Mechatronics, Lodz University of Technology, 1/15 Stefanowski St., 90-924 Lodz, Poland
*
Authors to whom correspondence should be addressed.
Symmetry 2022, 14(8), 1714; https://doi.org/10.3390/sym14081714
Submission received: 2 July 2022 / Revised: 30 July 2022 / Accepted: 8 August 2022 / Published: 17 August 2022

Abstract

:
In this article, using a ( γ ,a)-nabla conformable integral on time scales, we study several novel Hilbert-type dynamic inequalities via nabla time scales calculus. Our results generalize various inequalities on time scales, unifying and extending several discrete inequalities and their corresponding continuous analogues. We say that symmetry plays an essential role in determining the correct methods with which to solve dynamic inequalities.

1. Introduction

Hardy [1] proved the classical discrete double series inequality of the Hilbert type: If { a m } , { b n } 0 and n = 1 a n p < and n = 1 b n q < , then we have
n = 1 m = 1 a n b m m + n π sin π p n = 1 a n p 1 p m = 1 b m q 1 q ,
Unless all the sequence { a m } or { b n } is null. Hardy also established the following integral analogous of the inequality (1):
0 0 ϑ ( ξ ) g ( ζ ) ξ + ζ d x d y π sin π p 0 ϑ p ( ξ ) d x 1 p 0 g p ( ξ ) d x 1 p ,
Unless ϑ 0 or g 0 , where ϑ and g are measurable non-negative functions, such that 0 ϑ p ( ξ ) d x < and 0 g p ( ξ ) d x < and p > 1 , p = p ( p 1 ) , with the best constant Π sin π p , in (1) and (2).
In [2], Pachpatte established the following Hilbert-type integral inequalities under the following conditions: If h 1 , l 1 , and f ( ς ) 0 , g ( ς ) 0 , for ς ( 0 , ξ ) and τ ( 0 , ζ ) , where ξ and ζ are positive real numbers and define ϑ ( s ) = 0 s f ( ς ) d ς and G ( ς ) = 0 ς g ( τ ) d τ , for s ( 0 , ξ ) and ς ( 0 , ζ ) , and Φ , Ψ are two real-valued non-negative, convex, and submultiplicative functions defined on ( 0 , ] , then
0 ξ 0 ζ ϑ h ( s ) G l ( χ ) s + χ d s d χ 1 2 h l ( ξ ζ ) 1 2 0 ξ ( ξ s ) ϑ h 1 ( s ) ϑ ( s ) 2 d s 1 2 × 0 ζ ( ζ χ ) G l 1 g ( χ ) 2 d χ 1 2 ,
and
0 ξ 0 ζ Φ ( ϑ ( s ) ) Ψ ( G ( χ ) ) s + χ d s d χ L ( ξ , ζ ) 0 ξ ( ξ s ) p ( s ) Φ ϑ ( s ) p ( s ) 2 d s 1 2 × 0 ζ ( ζ χ ) q ( χ ) Ψ g ( χ ) q ( χ ) 2 d χ 1 2
where
L ( ξ , ζ ) = 1 2 0 ξ Φ ( P ( s ) ) P ( s ) 2 d s 1 2 0 ζ Ψ ( Q ( χ ) ) Q ( χ ) 2 d χ 1 2 ,
and
0 ξ 0 ζ P ( s ) Q ( χ ) Φ ( ϑ ( s ) ) Ψ ( G ( χ ) ) s + χ d s d χ 1 2 ( ξ ζ ) 1 2 0 ξ ( ξ s ) p ( s ) Φ ϑ ( s ) 2 d s 1 2 × 0 ζ ( ζ χ ) q ( χ ) Ψ g ( χ ) 2 d χ 1 2 .
For more results on Hilbert-type inequalities and other, please see [3,4,5,6]. See also [7,8,9,10,11,12,13]. As we know that time scale T is an arbitrary, non-empty closed subset of the set of real numbers R , and the jump operators, forward and backward, are defined by σ ( ς ) : = inf { ι T : ι > ς } , ρ ( ς ) : = sup { ι T : ι < ς } . For more details, see [7]. The following relations are used.
(i)
For T = R , then
σ ( ς ) = ρ ( ς ) = ς , μ ( ς ) = ν ( ς ) = 0 , ϑ Δ ( ς ) = ϑ ( ς ) = ϑ ( ς ) , a b ϑ ( ς ) Δ ς = a b ϑ ( ς ) ς = a b ϑ ( ς ) d ς .
(ii)
For T = Z , then
σ ( ς ) = ς + 1 , ρ ( ς ) = ς 1 , μ ( ς ) = ν ( ς ) = 1 , ϑ Δ ( ς ) = Δ ϑ ( ς ) , ϑ ( ς ) = ϑ ( ς ) , a b ϑ ( ς ) Δ ς = ς = a b 1 ϑ ( ς ) , a b ϑ ( ς ) ς = ς = a + 1 b ϑ ( ς ) .
For more details on the conformable nabla calculus, see [3].
We suppose that C C l d denotes the set of all l d -continuous functions ϑ ( ξ , ζ ) in ξ and ζ and C C l d 1 is the set of all functions in C C l d for which both the first partial derivative a 1 γ and the first partial derivative a 2 γ exist in C C l d . Similarly, we can define C C l d 2 .
In order to obtain our result in this paper, we need the following lemmas.
Lemma 1
(Reversed Dynamic Hölder’s Inequality). Suppose u , v T with u < v . Assume ϑ , g C C l d 1 ( [ u , v ] T × [ u , v ] T , R ) be integrable functions and 1 p + 1 q = 1 with p < 1 , then
u v u v | ϑ ( r , χ ) g ( r , χ ) | a γ r a γ χ u v u v | ϑ ( r , χ ) | p a γ r a γ χ 1 p × u v u v | g ( r , χ ) | q a γ r a γ χ 1 q .
Proof. 
This lemma is a direct extension of the (Lemma 9, [14]). □
Lemma 2
(Reversed Dynamic Jensen’s inequality). Let r, χ R and m , n . If ϑ C C l d 1 ( R , ( m , n ) ) and Φ : ( m , n ) R is concave, then
ϕ u v ω s ϑ ( r , χ ) a 1 γ r a 2 γ χ u v ω s a 1 γ r a 2 γ χ u v ω s ϕ ( ϑ ( r , χ ) ) a 1 γ r a 2 γ χ u v ω s a 1 γ r a 2 γ χ .
Proof. 
This lemma is a direct extension of the (Lemma 10, [14]). □
Definition 1.
Φ is called a supermultiplicative function on [ 0 , ) if
Φ ( ξ ζ ) Φ ( ξ ) Φ ( ζ ) , for all ξ , ζ 0 .
In this paper, we establish a ( γ ,a)-nabla conformable integral inequality of Hardy–Hilbert type on a time scale. In special cases, we will recover some dynamic continuous and discrete inequalities known in the literature. Symmetry plays an essential role in determining the correct methods with which to solve dynamic inequalities.
Now, our main results will be presented.

2. Main Results

First, we suppose the following assumptions:
( S 1 )
T be time scales with χ 0 , ξ , ζ , s , χ T , ( = 1 , , n ) .
( S 2 )
ϑ ( s , χ ) are non-negative, nabla integrable functions defined as [ χ 0 , ξ ) T × [ χ 0 , ζ ) T ( = 1 , , n ) .
( S 3 )
ϑ ( s , χ ) have partial a γ - derivatives ϑ a 1 γ ( s , χ ) and ϑ a 2 γ ( s , χ ) with respect to s and χ , respectively.
( S 4 )
All functions used in this section are integrable according to a γ sense.
( S 5 )
ϑ ( s , χ ) C l d 2 [ χ 0 , ξ ) T × [ χ 0 , ζ ) T , [ 0 , ) ( = 1 , , n ) .
( S 6 )
p ( ς , τ ) are n positive nabla-integrable functions defined for ς ( χ 0 , s ) T , τ ( χ 0 , χ ) T .
( S 7 )
p ( ς ) and q ( τ ) are positive nabla-integrable functions defined for ς ( χ 0 , s ) T , τ ( χ 0 , χ ) T .
( S 8 )
Φ and Ψ , ( = 1 , , n ) are n real-valued, non-negative concave and supermultiplicative functions defined on ( 0 , ) .
( S 9 )
ξ and ζ are positive real numbers.
( S 10 )
s [ χ 0 , ξ ) T and χ [ χ 0 , ζ ) T .
( S 11 )
ϑ ( χ 0 , χ ) = ϑ ( s , χ 0 ) = 0 , ( = 1 , , n ) .
( S 12 )
ϑ a 1 γ a 2 γ ( s , χ ) = ϑ a 2 γ a 1 γ ( s , χ ) .
( S 13 )
P ( s , χ ) = χ 0 χ χ 0 s p ( ς ) q ( τ ) a γ ς a γ τ .
( S 14 )
Θ ( s , χ ) = χ 0 s χ 0 χ ϑ ( ς , τ ) a γ ς a γ τ .
( S 15 )
P ( s , χ ) = χ 0 s χ 0 χ p ( ς , τ ) a γ ς a γ τ .
( S 16 )
Θ ( s , χ ) = 1 P ( ς , τ ) χ 0 s χ 0 χ p ( ς , τ ) ϑ ( ς , τ ) a γ ς a γ τ .
( S 17 )
α ( 1 , ) , α = 1 α , α = = 1 n α , and α = = 1 n α = n α , ( = 1 , , n ) .
( S 18 )
0 < β < 1 .
( S 19 )
h 2 .
( S 20 )
= 1 n 1 α = 1 α .
( S 21 )
h 1 .
( S 22 )
ϑ ( ς ) C l d 1 [ χ 0 , ξ ] T , ( = 1 , , n ) .
( S 23 )
ξ is a positive real number.
( S 24 )
Θ ( s ) = χ 0 s ϑ ( ς ) a γ ς .
( S 25 )
s [ χ 0 , ξ ) T .
( S 26 )
p ( ς ) are n positive functions.
( S 27 )
P ( s ) = χ 0 s p ( ς ) a γ ς .
( S 28 )
Θ ( s ) = 1 P ( s ) χ 0 s p ( ς ) ϑ ( ς ) a γ ς .
( S 29 )
ϑ ( χ 0 ) = 0 .
The first important inequality is stated in the following theorem:
Theorem 1.
Let S 1 , S 4 , S 5 , S 14 , S 6 , S 15 and S 8 be satisfied. Then, for S 10 , S 18 and S 20 , we find that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 L ( ξ 1 ζ 1 , , ξ n y n ) × = 1 n χ 0 ξ χ 0 ζ ( ρ ( ξ ) s ) ( ρ ( ζ ) χ ) p ( s , χ ) Φ ϑ ( s , χ ) p ( s , χ ) β a γ s a γ χ 1 β
where
L ( ξ 1 ζ 1 , , ξ n y n ) = = 1 n χ 0 ξ χ 0 ζ Φ ( P ( s , χ ) ) P ( s , χ ) α a γ s a γ χ 1 α .
Proof. 
From the hypotheses of Theorem 1, S 14 , S 15 , and S 8 , it is easy to observe that
Φ ( Θ ( s , χ ) ) = Φ P ( s , χ ) χ 0 s χ 0 χ p ( ς , τ ) ϑ ( ς , τ ) p ( ς , τ ) a γ ς a γ τ χ 0 s χ 0 χ p ( ς , τ ) a γ ς a γ τ Φ ( P ( s , χ ) ) Φ χ 0 s χ 0 χ p ( ς , τ ) ϑ ( ς , τ ) p ( ς , τ ) a γ ς a γ τ χ 0 s χ 0 χ p ( ς , τ ) a γ ς a γ τ .
Using inverse Jensen dynamic inequality, we obtain that
Φ ( Θ ( s , χ ) ) Φ ( P ( s , χ ) ) P ( s , χ ) χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) p ( ς , τ ) a γ ς a γ τ .
Applying inverse Hölder’s inequality on the right-hand side of (13) with indices α and β , it is easy to observe that
Φ ( Θ ( s , χ ) ) Φ ( P ( s , χ ) ) P ( s , χ ) ( s χ 0 ) ( χ χ 0 ) 1 α χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) p ( ς , τ ) β a γ ς a γ τ 1 β .
Using the following inequality on the term ( s χ 0 ) ( χ χ 0 ) 1 α ,
= 1 n m 1 α α = 1 n 1 α m 1 α ,
we get that
= 1 n Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α = 1 n Φ ( P ( s , χ ) ) P ( s , χ ) χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) p ( ς , τ ) 1 β a γ ς a γ τ 1 β .
Integrating both sides of (16) over s , χ from χ 0 to ξ , ζ ( = 1 , , n ) , we obtain that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1
= 1 n χ 0 ξ χ 0 ζ Φ ( P ( s , χ ) ) P ( s , χ ) χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) p ( ς , τ ) β a γ ς a γ τ 1 β a γ s a γ χ .
Applying inverse Hölder’s inequality on the right-hand side of (17) with indices α and β , it is easy to observe that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1
= 1 n χ 0 ξ χ 0 ζ Φ ( P ( s , χ ) ) P ( s , χ ) α a γ s a γ χ 1 α
× = 1 n χ 0 ξ χ 0 ζ χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) p ( ς , τ ) β a γ ς a γ τ a γ s a γ χ 1 β .
Using Fubini’s theorem, we observe that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 L ( ξ 1 ζ 1 , , ξ n y n ) × = 1 n χ 0 ξ χ 0 ζ ( ξ s ) ( ζ χ ) p ( s , χ ) Φ ϑ ( s , χ ) p ( s , χ ) β a γ s a γ χ 1 β .
Using the fact ξ ρ ( ξ ) , and ζ ρ ( ζ ) , , we get that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 L ( ξ 1 ζ 1 , , ξ n y n ) × = 1 n χ 0 ξ χ 0 ζ ( ρ ( ξ ) s ) ( ρ ( ζ ) χ ) p ( s , χ ) Φ ϑ ( s , χ ) p ( s , χ ) β a γ s a γ χ 1 β .
This completes the proof. □
Remark 1.
In Theorem 1, if T = R , γ = 1 , we get the result due to Zhao et al. [8] (Theorem 2).
As a special case of Theorem 1, when T = Z , γ = 1 , we have ρ ( n ) = n 1 , and we get the following result.
Corollary 1.
Let { a s , χ , m s , m χ } and { p s , χ , m s , m χ } , ( = 1 , , n ) be n sequences of non-negative numbers defined for m s = 1 , , k s , and m χ = 1 , , k χ , and define
A s , χ , m s , m χ = m ς m s m η m χ a s , χ , m ς , m η P s , χ , m s , m χ = m ς m s m η m χ p s , χ , m ς , m η .
Then,
m s 1 k s 1 m χ 1 k χ 1 m s n k s n m χ n k χ n = 1 n Φ ( A s , χ , m s , m χ ) α = 1 n 1 α ( m s m χ ) 1 α C ( k s 1 k χ 1 , , k s n k χ n ) × = 1 n m s k s m χ k χ ( k s ( m s 1 ) ) ( k χ ( m χ 1 ) ) P s , χ , m s , m χ Φ a s , χ , m s , m χ P s , χ , m s , m χ β 1 β
where
C ( k s 1 k χ 1 , , k s n k χ n ) = = 1 n m s k s m χ k χ Φ ( P s , χ , m s , m χ ) P s , χ , m s , m χ β ) 1 β .
Remark 2.
Let ϑ ( ς , τ ) , p ( ς , τ ) , P ( ς , τ ) , and ϑ ( ς , τ ) change to ϑ ( ς ) , p ( ς ) , P ( s ) and ϑ ( s ) , respectively, and with suitable changes, we have the following new corollary:
Corollary 2.
Let S 22 , S 23 , S 24 , S 26 , S 27 be satisfied. Then, for S 18 , S 20 and S 25 we find that
χ 0 ξ 1 χ 0 ξ n = 1 n Φ ( Θ ( s ) α = 1 n 1 α ( s χ 0 ) 1 α a γ s 1 a γ s n L * ( ξ 1 , , ξ n ) = 1 n χ 0 ξ ( ρ ( ξ ) s ) p ( s ) Φ ϑ ( s ) p ( s ) β a γ s 1 β
where
L * ( ξ 1 , , ξ n ) = = 1 n χ 0 ξ Φ ( P ( s ) ) P ( s ) α a γ s 1 α .
Remark 3.
In Corollary 2, if we take n = 2 , β = 1 2 , then the Inequality (21) changes to
χ 0 ξ 1 χ 0 ξ 1 Φ 1 ( Θ 1 ( s 1 ) ) Φ 2 ( Θ 2 ( s 2 ) ) ( s 1 χ 0 ) + ( s 2 χ 0 ) 2 a γ s 1 a γ s 2 L * * ( ξ 1 , ξ 2 ) ( χ 0 ξ 1 ( ρ ( ξ 1 ) s 1 ) p 1 ( s 1 ) Φ ϑ 1 ( s 1 ) p 1 ( s 1 ) 2 a γ s 1 ) 1 2 × χ 0 ξ 2 ( ρ ( ξ 2 ) s 2 ) p 2 ( s 2 ) Ψ ϑ 2 ( s 2 ) p 2 ( s 2 ) 2 a γ s 2 1 2
where
L * * ( ξ 1 , ξ 2 ) = 4 χ 0 ξ 1 Φ 1 ( P 1 ( s 1 ) ) P 1 ( s 1 ) 1 a γ s 1 1 χ 0 ξ 2 Φ 2 ( P 2 ( s 2 ) ) P 2 ( s 2 ) 1 a γ s 2 1
.
Remark 4.
In Corollary 3, if we take T = R , γ = 1 , then the inequality (22) changes to
0 ξ 1 0 ξ 1 Φ 1 ( Θ 1 ( s 1 ) ) Φ 2 ( Θ 2 ( s 2 ) ) ( s 1 + s 2 ) 2 d s 1 d s 2 L * * ( ξ 1 , ξ 2 ) 0 ξ 1 ( ξ 1 s 1 ) p 1 ( s 1 ) Φ ϑ 1 ( s 1 ) p 1 ( s 1 ) 2 d s 1 1 2 × 0 ξ 2 ( ξ 2 s 2 ) p 2 ( s 2 ) Ψ ϑ 2 ( s 2 ) p 2 ( s 2 ) 2 d s 2 1 2
where
L * * ( ξ 1 , ξ 2 ) = 4 0 ξ 1 Φ 1 ( P 1 ( s 1 ) ) P 1 ( s 1 ) 1 d s 1 1 0 ξ 2 Φ 2 ( P 2 ( s 2 ) ) P 2 ( s 2 ) 1 d s 2 1 .
This is an inverse of the Inequality (4), which was proved by Pachpatte [15].
Corollary 3.
In Corollary 2, if we take β = n 1 n the Inequality (21) becomes
χ 0 ξ 1 χ 0 ξ n = 1 n Φ ( Θ ( s ) = 1 n ( s χ 0 ) n n 1 a γ s n a γ s 1 L * ( ξ 1 , , ξ n ) = 1 n χ 0 ξ ( ρ ( ξ ) s ) p ( s ) Φ ϑ ( s ) p ( s ) n 1 n a γ s n n 1
where
L * ( ξ 1 , , ξ n ) = n n n 1 = 1 n χ 0 ξ Φ ( P ( s ) ) P ( s ) ( n 1 ) a γ s 1 n 1 .
Theorem 2.
Let S 1 , S 4 , S 5 , S 6 , S 9 , S 15 , and S 16 be satisfied. Then for S 10 , S 18 and S 20 , we have that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n P ( s , χ ) Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 = 1 n ( ξ χ 0 ) ( ζ χ 0 ) 1 α χ 0 ξ χ 0 ζ ( ρ ( ξ ) s ) ( ρ ( ζ ) χ ) p ( s , χ ) Φ ϑ ( s , χ ) β a γ s a γ χ 1 β .
Proof. 
From the hypotheses of Theorem 2, and using inverse Jensen dynamic inequality, we have
Φ ( ϑ ( s , χ ) ) = Φ 1 P ( s , χ ) χ 0 s χ 0 χ p ( ς , τ ) Θ ( ς , τ ) a γ ς a γ τ 1 P ( s , χ ) χ 0 s χ 0 χ p ( σ , τ ) Φ ϑ ( ς , τ ) a γ ς a γ τ .
Applying inverse Hölder’s inequality on the right-hand side of (24) with indices α and β , , it is easy to observe that
Φ ( Θ ( s , χ ) ) 1 P ( s , χ ) ( s χ 0 ) ( χ χ 0 ) 1 α ( χ 0 s χ 0 χ ( p ( ς , τ ) Φ ϑ ( ς , τ ) ) β a γ ς a γ τ ) 1 β .
Using the Inequality (15), on the term ( s χ 0 ) ( χ χ 0 ) 1 α , we get that
P ( s , χ ) Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) β a γ ς a γ τ 1 β
Integrating both sides of (25) over s , χ from χ 0 to ξ , ζ ( = 1 , , n ) , we get that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n P ( s , χ ) Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 = 1 n χ 0 ξ χ 0 ζ χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) β a γ σ a γ τ 1 β .
Applying inverse Hölder’s inequality on the right hand side of (17) with indices α and β , it is easy to observe that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n P ( s , χ ) Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 = 1 n ( ξ χ 0 ) ( ζ χ 0 ) 1 α ( χ 0 ξ χ 0 ζ χ 0 s χ 0 χ p ( ς , τ ) Φ ϑ ( ς , τ ) β a γ ς a γ τ a γ s a γ χ ) 1 β .
Using Fubini’s theorem, we observe that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n P ( s , χ ) Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 = 1 n ( ξ χ 0 ) ( ζ χ 0 ) 1 α ( χ 0 ξ χ 0 ζ ( ξ s ) ( ζ χ ) p ( s , χ ) Φ ϑ ( s , χ ) β a γ s a γ χ ) 1 β .
By using the fact ξ ρ ( ξ ) , and ζ ρ ( ζ ) , we get that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n P ( s , χ ) Φ ( Θ ( s , χ ) ) α = 1 n 1 α ( s χ 0 ) ( χ χ 0 ) 1 α a γ s n a γ χ n a γ s 1 a γ χ 1 = 1 n ( ξ χ 0 ) ( ζ χ 0 ) 1 α ( χ 0 ξ χ 0 ζ ( ρ ( ξ ) s ) ( ρ ( ζ ) χ ) p ( s , χ ) Φ ϑ ( s , χ ) β a γ s a γ χ ) 1 β
This completes the proof. □
Remark 5.
In Theorem 2, if T = R , we get the result due to Zhao et al. [8] (Theorem 3).
As a special case of Theorem 2, when T = Z , γ = 1 , we have ρ ( n ) = n 1 , we get the following result.
Corollary 4.
Let { a s , χ , m s , m χ } and { p s , χ , m s , m χ } , ( = 1 , , n ) be n sequences of non-negative numbers defined for m s = 1 , , k s , and m χ = 1 , , k χ , and define
A s , χ , m s , m χ = 1 P s , χ , m s , m χ m ς m s m η m χ a s , χ , m ς , m η p s , χ , m ς , m η , P s , χ , m s , m χ = m ς m s m η m χ p s , χ , m ς , m η .
Then,
m s 1 k s 1 m χ 1 k χ 1 m s n k s n m χ n k χ n = 1 n P s , χ , m s , m χ Φ ( A s , χ , m s , m χ ) α = 1 n 1 α ( m s m χ ) 1 α = 1 n ( k s k χ ) 1 α ( m s k s m χ k χ ( k s ( m s 1 ) ) ( k χ ( m χ 1 ) ) p s , χ , m s , m χ Φ a s , χ , m s , m χ β ) 1 β .
Remark 6.
Let ϑ ( ς , τ ) , p ( ς , τ ) , P ( ς , τ ) be defined as above and
Θ ( s , χ ) = 1 P ( s , χ ) χ 0 s 0 χ p ( ς , τ ) ϑ ( ς , τ ) a γ ς a γ τ
changes to ϑ ( ς ) , p ( ς ) , P ( s ) , and
Θ ( s ) = 1 P ( s ) χ 0 s p ( ς ) ϑ ( ς ) a γ ς .
respectively, and with suitable changes, we have the following new corollary:
Corollary 5.
Let S 22 , S 23 , S 26 , S 27 and S 28 be satisfied. Then, for S 18 , S 20 and S 25 , we find that
χ 0 ξ 1 χ 0 ξ n = 1 n P ( s ) Φ ( Θ ( s ) α = 1 n 1 α ( s χ 0 ) 1 α a γ s n a γ s 1 = 1 n ( ξ χ 0 ) 1 α χ 0 ξ ( ρ ( ξ ) s ) p ( s ) Φ ϑ ( s ) β a γ s 1 β .
Corollary 6.
In Corollary 5, if we take n = 2 , β = 1 2 , then the inequality (21) changes to
χ 0 ξ 1 χ 0 ξ 1 P 1 ( s 1 ) P 2 ( s 2 ) Φ 1 ( Θ 1 ( s 1 ) ) Φ 2 ( Θ 2 ( s 2 ) ) ( s 1 χ 0 ) + ( s 2 χ 0 ) 2 a γ s 1 a γ s 2 4 ( ξ 1 χ 0 ) ( ξ 2 χ 0 ) 1 × χ 0 ξ 1 ( ρ ( ξ 1 ) s 1 ) p 1 ( s 1 ) Φ 1 ϑ 1 ( s 1 ) 2 a γ s 1 1 2 ( χ 0 ξ 2 ( ρ ( ξ 2 ) s 2 ) ( p 2 ( s 2 ) Φ 2 ϑ 2 ( s 2 ) ) 2 a γ s 2 ) 1 2 .
Remark 7.
In Corollary 6, if we take T = R , γ = 1 , then Inequality (28) changes to
0 ξ 1 0 ξ 1 P 1 ( s 1 ) P 2 ( s 2 ) Φ 1 ( Θ 1 ( s 1 ) ) Φ 2 ( Θ 2 ( s 2 ) ) ( s 1 + s 2 ) 2 d s 1 d s 2 4 ξ 1 ξ 2 1 × 0 ξ 1 ( ξ 1 s 1 ) p 1 ( s 1 ) Φ 1 ϑ 1 ( s 1 ) 2 d s 1 1 2 0 ξ 2 ( ξ 2 s 2 ) p 2 ( s 2 ) Φ 2 ϑ 2 ( s 2 ) 2 d s 2 1 2 .
This is an inverse of Inequality (5), which was proved by Pachpatte [15].
Corollary 7.
In Corollary 6, let p 1 ( s 1 ) = p 2 ( s 2 ) = 1 , then P 1 ( s 1 ) = s 1 , P 2 ( s 2 ) = s 2 . Therefore, Inequality (28) changes to
χ 0 ξ 1 χ 0 ξ 1 Φ 1 ( Θ 1 ( s 1 ) ) Φ 2 ( Θ 2 ( s 2 ) ) ( s 1 s 2 ) 1 ( s 1 χ 0 ) + ( s 2 χ 0 ) 2 a γ s 1 a γ s 2 4 ( ξ 1 χ 0 ) ( ξ 2 χ 0 ) 1 × χ 0 ξ 1 ( ρ ( ξ 1 ) s 1 ) Φ 1 ϑ 1 ( s 1 ) 2 a γ s 1 1 2 χ 0 ξ 2 ( ρ ( ξ 2 ) s 2 ) Φ 2 ϑ 2 ( s 2 ) 2 a γ s 2 1 2 .
Remark 8.
In Corollary 7, if we take T = R , then, the Inequality (30) changes to
0 ξ 1 0 ξ 1 Φ 1 ( Θ 1 ( s 1 ) ) Φ 2 ( Θ 2 ( s 2 ) ) ( s 1 s 2 ) 1 s 1 + s 2 2 d s 1 d s 2 4 ξ 1 ξ 2 1 × 0 ξ 1 ( ξ 1 s 1 ) Φ 1 ϑ 1 ( s 1 ) 2 d s 1 1 2 0 ξ 2 ( ξ 2 s 2 ) Φ 2 ϑ 2 ( s 2 ) 2 d s 2 1 2 .
This is an inverse inequality of the following inequality, which was proved by Pachpatte [8].
0 ξ 0 ζ Φ ( Θ ( s ) ) Ψ ( G ( χ ) ) ( s χ ) 1 s + χ d s d χ 1 2 ξ ζ 1 2 × 0 ξ ( ξ s 1 ) Φ ϑ ( s ) 2 d s 1 2 0 ζ ( ζ χ ) Ψ g ( χ ) 2 d χ 1 2 .
Corollary 8.
In Corollary 5, if we take β = n 1 n ( = 1 , , n ) Inequality (27)
χ 0 ξ 1 χ 0 ξ n = 1 n P ( s ) Φ ( Θ ( s ) = 1 n ( s χ 0 ) n n 1 a γ s n a γ s 1 n n n 1 = 1 n ( ξ χ 0 ) 1 n 1 χ 0 ξ ( ρ ( ξ ) s ) p ( s ) Φ ϑ ( s ) n 1 n a γ s n n 1 .
Theorem 3.
Let S 1 , S 4 , S 2 , S 9 , S 11 , S 7 , S 13 , S 3 , S 12 , S 8 and S 17 be satisfied. Then, for S 10 we have that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( ϑ ( s , χ ) ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α a γ s n a γ χ n a γ s 1 a γ χ 1 G ( ξ 1 ζ 1 , , ξ n y n ) × = 1 n ( χ 0 ξ χ 0 ζ ( ρ ( ξ ) s ) ( ρ ( ζ ) χ ) p ( s ) q ( χ ) Φ ϑ a 2 γ a 1 γ ( s , χ ) p ( s ) q ( χ ) 1 α a γ s a γ χ ) α ,
where
G ( ξ 1 ζ 1 , , ξ n y n ) = = 1 n χ 0 ξ χ 0 ζ Φ ( P ( s , χ ) ) P ( s , χ ) 1 α a γ s a γ χ α .
Proof. 
From the hypotheses of Theorem 3, we obtain
ϑ ( s , χ ) = χ 0 s χ 0 χ ϑ a 2 γ a 1 γ ( ς , τ ) a γ ς a γ τ .
From (32) and S 8 , it is easy to observe that
Φ ( ϑ ( s , χ ) ) = Φ P ( s , χ ) χ 0 s χ 0 χ p ( ς ) q ( τ ) ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) a γ ς a γ τ χ 0 s χ 0 χ p ( ς ) q ( τ ) a γ ς a γ τ
Φ ( P ( s , χ ) ) Φ χ 0 s χ 0 χ p ( ς ) q ( τ ) ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) a γ ς a γ τ χ 0 s χ 0 χ p ( ς ) q ( τ ) a γ ς a γ τ .
Using inverse Jensen’s dynamic inequality, we get that
Φ ( ϑ ( s , χ ) ) Φ ( P ( s , χ ) ) P ( s , χ ) χ 0 s χ 0 χ p ( ς ) q ( τ ) Φ ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) a γ ς a γ τ .
Applying inverse Hölder’s inequality on the right-hand side of (34) with indices 1 / α and 1 / α , we obtain
Φ ( ϑ ( s , χ ) ) Φ ( P ( s , χ ) ) P ( s , χ ) ( s χ 0 ) ( χ χ 0 ) α
× χ 0 s χ 0 χ p ( ς ) q ( τ ) Φ ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) 1 α a γ ς a γ τ α .
Using the following inequality on the term ( s χ 0 ) ( χ χ 0 ) α , where α < 0 and λ > 0 .
= 1 n λ α 1 α = 1 n α λ α .
We obtain that
= 1 n Φ ( ϑ ( s , χ ) ) = 1 n Φ ( P ( s , χ ) ) P ( s , χ ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α
× χ 0 s χ 0 χ p ( ς ) q ( τ ) Φ ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) 1 α a γ ς a γ τ α .
From (37), we find that
= 1 n Φ ( ϑ ( s , χ ) ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α
= 1 n Φ ( P ( s , χ ) ) P ( s , χ ) χ 0 s χ 0 χ p ( ς ) q ( τ ) Φ ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) 1 α a γ ς a γ τ α .
Integrating both sides of (38) over s , χ from χ 0 to ξ , ζ ( = 1 , , n ) , we find that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( ϑ ( s , χ ) ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α a γ s n a γ χ n a γ s 1 a γ χ 1
= 1 n χ 0 ξ χ 0 ζ Φ ( P ( s , χ ) ) P ( s , χ ) χ 0 s χ 0 χ p ( ς ) q ( τ ) Φ ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) 1 α a γ ς a γ τ α a γ s a γ χ
Applying inverse Hölder’s inequality on the right-hand side of (39) with indices 1 / α and 1 / α , , we obtain
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( ϑ ( s , χ ) ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α a γ s n a γ χ n a γ s 1 a γ χ 1 = 1 n χ 0 ξ χ 0 ζ Φ ( P ( s , χ ) ) P ( s , χ ) 1 α a γ s a γ χ α × = 1 n χ 0 ξ χ 0 ζ χ 0 s χ 0 χ p ( ς ) q ( τ ) Φ ϑ a 2 γ a 1 γ ( ς , τ ) p ( ς ) q ( τ ) 1 α a γ ς a γ τ a γ s a γ χ α
By using Fubini’s theorem, we observe that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( ϑ ( s , χ ) ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α a γ s n a γ χ n a γ s 1 a γ χ 1 G ( ξ 1 ζ 1 , , ξ n y n ) × = 1 n χ 0 ξ χ 0 ζ ( ξ s ) ( ζ χ ) p ( s ) q ( χ ) Φ ϑ a 2 γ a 1 γ ( s , χ ) p ( s ) q ( χ ) 1 α a γ s a γ χ α .
Using the fact ξ ρ ( ξ ) , and ζ ρ ( ζ ) , we get that
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( ϑ ( s , χ ) ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α a γ s n a γ χ n a γ s 1 a γ χ 1 G ( ξ 1 ζ 1 , , ξ n y n ) × = 1 n χ 0 ξ χ 0 ζ ( ρ ( ξ ) s ) ( ρ ( ζ ) χ ) p ( s ) q ( χ ) Φ ϑ a 2 γ a 1 γ ( s , χ ) p ( s ) q ( χ ) 1 α a γ s a γ χ α .
This completes the proof. □
Remark 9.
In Theorem 3, if T = Z , γ = 1 , we get the result due to Zhao et al. [9] (Theorem 1.5).
Remark 10.
In Theorem 3, if we take T = R , γ = 1 , we get the result due to Zhao et al. [9] (Theorem 1.6).
Remark 11.
Let S 1 , S 2 , S 9 , S 11 , S 7 , S 13 , S 3 and S 12 be satisfied and let Φ , α , α ,   α , and α be the same as Theorem 3. Similar to proof of Theorem 3, we have
χ 0 ξ 1 χ 0 ζ 1 χ 0 ξ n χ 0 ζ n = 1 n Φ ( ϑ ( s , χ ) ) 1 α = 1 n α ( s χ 0 ) ( χ χ 0 ) α a γ s n a γ χ n a γ s 1 a γ χ 1 G * ( ξ 1 ζ 1 , , ξ n y n ) × = 1 n χ 0 ξ χ 0 ζ ( σ ( ξ ) s ) ( σ ( ζ ) χ ) p ( s ) q ( χ ) Φ ϑ a 2 γ a 1 γ ( s , χ ) p ( s ) q ( χ ) 1 α a γ s a γ χ α .
where
G * ( ξ 1 ζ 1 , , ξ n y n ) = 1 ( α ) α = 1 n χ 0 ξ χ 0 ζ Φ ( P ( s , χ ) ) P ( s , χ ) 1 α a γ s a γ χ α .
This is an inverse form of Inequality (31).
Corollary 9.
Let S 22 , S 23 , S 25 , S 26 , S 27 , S 29 , S 17 and S 8 be satisfied. Then, we have that
χ 0 ξ 1 χ 0 ξ n = 1 n Φ ( ϑ ( s ) ) 1 α = 1 n α ( s χ 0 ) α a γ s n a γ s 1 G * * ( ξ 1 , , ξ n ) × = 1 n χ 0 ξ ( ρ ( ξ ) s ) p ( s ) Φ ϑ a γ ( s ) p ( s ) 1 α a γ s α .
where
G * * ( ξ 1 , , ξ n ) = = 1 n χ 0 ξ Φ ( P ( s ) ) P ( s ) 1 α a γ s α .
Remark 12.
In Corollary 9, if we take T = Z , γ = 1 , we get an inverse form of inequality due to Handley et al. [16].
Remark 13.
In Corollary 9, if we take T = R , γ = 1 , we get an inverse form of inequality due to Handley et al. [16].
Remark 14.
In Inequality (42), taking n = 2 , α 1 = α 2 = 2 , then α 1 = α 2 = 1 , we have
χ 0 ξ 1 χ 0 ξ 2 = 1 n Φ 1 ( ϑ 1 ( s 1 ) ) Φ 1 ( ϑ 2 ( s 2 ) ) ( s 1 χ 0 ) + ( s 2 χ 0 ) 2 a γ s 1 a γ s 2 D ( ξ 1 , ξ 2 ) χ 0 ξ 1 ( ρ ( ξ 1 ) s 1 ) p 1 ( s 1 ) Φ 1 ϑ 1 a γ ( s 1 ) p 1 ( s 1 ) 1 2 a γ s 1 2 × χ 0 ξ 2 ( ρ ( ξ 2 ) s 2 ) p 2 ( s 2 ) Φ 2 ϑ 2 a γ ( s 2 ) p 2 ( s 2 ) 1 2 a γ s 2 2 .
where
D ( ξ 1 , ξ 2 ) = 4 χ 0 ξ 1 Φ 1 ( P 1 ( s 1 ) ) P 2 ( s 1 ) 1 a γ s 1 1 χ 0 ξ 2 Φ 2 ( P 2 ( s 2 ) ) P 2 ( s 2 ) 1 a γ s 2 1 .
Remark 15.
If we take T = Z , γ = 1 , Inequality (43) is an inverse of inequality due to Pachpatte [2].
Remark 16.
If we take T = R , γ = 1 , Inequality (43) is an inverse of inequality due to Pachpatte [2].

3. Conclusions

In this article, we presented some investigations of the ( γ , a ) -nabla Hilbert inequality on time scales. Some dynamic integral and discrete inequalities, known in the literature, are generalized as special cases of our results. We obtained the discrete and the continuous inequalities as special cases of our main results. In future work, I will ask if it is possible to generalize these results using a q-difference operator.

Author Contributions

A.A.E.-D., D.B. and J.A.; formal analysis, A.A.E.-D., D.B. and J.A.; investigation, A.A.E.-D., D.B. and J.A.; writing–original draft preparation, A.A.E.-D., D.B. and J.A.; writing–review and editing, A.A.E.-D., D.B. and J.A. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Conflicts of Interest

The authors declare no conflict of interest.

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El-Deeb, A.A.; Baleanu, D.; Awrejcewicz, J. (γ,a)-Nabla Reverse Hardy–Hilbert-Type Inequalities on Time Scales. Symmetry 2022, 14, 1714. https://doi.org/10.3390/sym14081714

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El-Deeb AA, Baleanu D, Awrejcewicz J. (γ,a)-Nabla Reverse Hardy–Hilbert-Type Inequalities on Time Scales. Symmetry. 2022; 14(8):1714. https://doi.org/10.3390/sym14081714

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El-Deeb, Ahmed A., Dumitru Baleanu, and Jan Awrejcewicz. 2022. "(γ,a)-Nabla Reverse Hardy–Hilbert-Type Inequalities on Time Scales" Symmetry 14, no. 8: 1714. https://doi.org/10.3390/sym14081714

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