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Article

On Some Important Class of Dynamic Hilbert’s-Type Inequalities on Time Scales

by
Hassan M. El-Owaidy
1,
Ahmed A. El-Deeb
1,*,
Samer D. Makharesh
1,
Dumitru Baleanu
2,3,4 and
Clemente Cesarano
5,*
1
Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City 11884, Cairo, Egypt
2
Institute of Space Science, 077125 Bucharest, Romania
3
Department of Mathematics, Cankaya University, 06530 Ankara, Turkey
4
Department of Medical Research, China Medical University, Taichung 40447, Taiwan
5
Section of Mathematics, International Telematic University Uninettuno, Corso Vittorio Emanuele II, 39, 00186 Rome, Italy
*
Authors to whom correspondence should be addressed.
Symmetry 2022, 14(7), 1395; https://doi.org/10.3390/sym14071395
Submission received: 8 June 2022 / Revised: 29 June 2022 / Accepted: 29 June 2022 / Published: 7 July 2022

Abstract

:
In this important work, we discuss some novel Hilbert-type dynamic inequalities on time scales. The inequalities investigated here generalize several known dynamic inequalities on time scales and unify and extend some continuous inequalities and their corresponding discrete analogues. Our results will be proved by using some algebraic inequalities, Hölder inequality, and Jensen’s inequality on time scales.

1. Introduction

The celebrated Hardy-Hilbert’s integral inequality [1] is
0 + 0 + ϝ ( ϑ ) g ( ς ) ϑ + ς d ϑ d ς π sin π p 0 + ϝ p ( ϑ ) d ϑ 1 p 0 + g q ( ς ) d ς 1 q ,
where p > 1 , q = p / p 1 . Putting p = q = 2 , we get:
0 + 0 + ϝ ( ϑ ) g ( ς ) ϑ + ς d ϑ d ς π 0 + ϝ 2 ( ϑ ) d ϑ 1 2 0 + g 2 ( ς ) d ς 1 2 ,
where the constants π and π sin π p are the best possible.
Over the past decade, a great number of dynamic Hilbert-type inequalities on time scales have been established by many researchers who were motivated by some applications; see the papers [2,3,4,5,6,7,8,9,10,11,12,13]. For more details on time scales calculus, see [14].
In this paper, we extend some generalizations of the integral Hardy-Hilbert inequality to a general time scale. As special cases of our results, we will recover some dynamic integral and discrete inequalities known in the literature.
A time scale T is an arbitrary non-empty closed subset of the real number. We define forward and backward jump operators σ : T T and ρ : T : T respectively by
σ ( ι ) : = inf { s T : s > ι } , ι T ,
ρ ( ι ) : = sup { s T : s < ι } , ι T .
We will need the following important relations between calculus on time scales T and either continuous calculus on R or discrete calculus on Z . Note that:
(i) 
If T = R , then
σ ( ι ) = ι , μ ( ι ) = 0 , ϝ Δ ( ι ) = ϝ ( ι ) , a b ϝ ( ι ) Δ ι = a b ϝ ( ι ) d ι .
(ii) 
If T = Z , then
σ ( ι ) = ι + 1 , μ ( ι ) = 1 , ϝ Δ ( ι ) = ϝ ( ι + 1 ) ϝ ( ι ) , a b ϝ ( ι ) Δ ι = ι = a b 1 ϝ ( ι ) .
Next, we write Hölder’s inequality and Jensen’s inequality on time scales.
Lemma 1
([3]). Let u ,   v T with u < v . Assume ϝ * g * C C r 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 | ϝ * ( r * , ι * ) g * ( r * , ι * ) | Δ r * Δ ι * u v u v | ϝ * ( r * , ι * ) | p Δ r * Δ ι * 1 p × u v u v | g * ( r * , ι * ) | q Δ r * Δ ι * 1 q .
This inequality is reversed if 0 < p < 1 and if p < 0 or q < 0 .
Lemma 2
([15]). Let r * , ι * R and + m * , n * + . If ϝ * C C r d 1 ( R , ( m * , n * ) ) , and ϕ : ( m * , n * ) R is convex, then
ϕ u v ω s ϝ * ( r * , ι * ) Δ 1 r * Δ 2 ι * u v ω s Δ 1 r * Δ 2 ι * u v ω s ϕ ( ϝ * ( r * , ι * ) ) Δ 1 r * Δ 2 ι * u v ω s Δ 1 r * Δ 2 ι * .
This inequality is reversed if ϕ C r d ( c , d ) , R is concave.
Theorem 1
(Chain rule on time scales [14]). Assume g : R R is continuous, g : T R is Δ -differentiable on T κ , and ϝ : R R is continuously differentiable. Then there exists c [ ι , σ ( ι ) ] R with
( ϝ g ) Δ ( ι ) = ϝ ( g ( c ) ) ( g ) Δ ( ι ) .
Definition 1.
Φ is called a supermultiplicative function on [ 0 , + ) if
Φ ( ϑ ς ) Φ ( ϑ ) Φ ( ς ) , for all ϑ , ς 0 .
Next, we write Fubini’s theorem on time scales.
Lemma 3
(Fubini’s Thoerem, see [16]). Assume that ( ϑ , Σ 1 , μ Δ ) and ( ς , Σ 2 , ν Δ ) are two finite-dimensional time scales measure spaces. Moreover, suppose that ϝ : ϑ × ς R is a delta integrable function and define the functions
ϕ ( ς ) = ϑ ϝ ( ϑ , ς ) d μ Δ ( ϑ ) , ς ς ,
and
ψ ( ϑ ) = ς ϝ ( ϑ , ς ) d ν Δ ( ς ) , ϑ ϑ .
Then ϕ is delta integrable on ς, and ψ is delta integrable on ϑ and
ϑ d μ Δ ( ϑ ) ς ϝ ( ϑ , ς ) d ν Δ ( ς ) = ς d ν Δ ( ς ) ϑ ϝ ( ϑ , ς ) d μ Δ ( ϑ ) .
Now we are ready to state and prove our main results.

2. Main Results

First, we enlist the following assumptions for the proofs of our main results:
( S 1 )
T be time scales with ι 0 , υ , ς , s , ι T , ( = 1 , , n ) .
( S 2 )
λ ( s , ι ) are nonnegative, delta integrable functions defined on [ ι 0 , υ ) T × [ ι 0 , ς ) T ( = 1 , , n ) .
( S 3 )
λ ( s , ι ) have a partial Δ - derivatives λ Δ 1 ( s , ι ) and λ Δ 2 ( s , ι ) with respect s and ι respectively.
( S 4 )
All functions used in this section are integrable according to Δ sense.
( S 5 )
λ ( s , ι ) C r d 2 [ ι 0 , υ ) T × [ ι 0 , ς ) T , [ 0 , ) ( = 1 , , n ) .
( S 6 )
p ( ξ , τ ) are n positive delta integrable functions defined for ξ ( ι 0 , s ) T , τ ( ι 0 , ι ) T .
( S 7 )
p ( ξ ) and q ( τ ) are positive delta integrable functions defined for ξ ( ι 0 , s ) T , τ ( ι 0 , ι ) T .
( S 8 )
Φ ( = 1 , , n ) are n real-valued nonnegative concave and supermultiplicative functions defined on ( 0 , ) .
( S 9 )
υ and ς are positive real numbers.
( S 10 )
s [ ι 0 , υ ) T and ι [ ι 0 , ς ) T .
( S 11 )
λ ( ι 0 , ι ) = λ ( s , ι 0 ) = 0 , ( = 1 , , n ) .
( S 12 )
λ Δ 1 Δ 2 ( s , ι ) = λ Δ 2 Δ 1 ( s , ι ) .
( S 13 )
P ( s , ι ) = ι 0 ι ι 0 s p ( ξ ) q ( τ ) Δ ξ Δ τ .
( S 14 )
Λ ( s , ι ) = ι 0 s ι 0 ι λ ( ξ , τ ) Δ ξ Δ τ .
( S 15 )
P ( s , ι ) = ι 0 s ι 0 ι p ( ξ , τ ) Δ ξ Δ τ .
( S 16 )
Λ ( s , ι ) = 1 P ( ξ , τ ) ι 0 s ι 0 ι p ( ξ , τ ) λ ( ξ , τ ) Δ ξ Δ τ .
( S 17 )
γ ( 1 , ) , γ = 1 γ , γ = = 1 n γ , and γ = = 1 n γ = n γ , ( = 1 , , n ) .
( S 18 )
0 < β < 1 .
( S 19 )
h 2 .
( S 20 )
= 1 n 1 γ = 1 γ .
( S 21 )
h 1 .
( S 22 )
λ ( ξ ) C r d 1 [ ι 0 , υ ] T , ( = 1 , , n ) .
( S 23 )
υ is positive real number.
( S 24 )
Λ ( s ) = ι 0 s λ ( ξ ) Δ ξ .
( S 25 )
s [ ι 0 , υ ) T .
( S 26 )
p ( ξ ) are n positive functions.
( S 27 )
P ( s ) = ι 0 s p ( ξ ) Δ ξ .
( S 28 )
Λ ( s ) = 1 P ( s ) ι 0 s p ( ξ ) λ ( ξ ) Δ ξ .
( S 29 )
λ ( ι 0 ) = 0 .
Now, we are ready to state and prove the main results that extend several results in the literature.
Theorem 2.
Let S 1 , 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
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( λ ( s , ) ) 1 γ = 1 n γ ( s 0 ) ( 0 ) γ Δ s n Δ n Δ s 1 Δ 1 G ( υ 1 ς 1 , , υ n ς n ) × = 1 n ι 0 υ ι 0 ς ( ρ ( υ ) s ) ( ρ ( ς ) ι ) p ( s ) q ( ι ) Φ λ Δ 2 Δ 1 ( s , ι ) p ( s ) q ( ι ) 1 γ Δ s Δ ι γ
where
G ( υ 1 ς 1 , , υ n ς n ) = = 1 n ι 0 υ ι 0 ς Φ ( P ( s , ι ) ) P ( s , ι ) 1 γ Δ s Δ ι γ .
Proof. 
From the hypotheses of Theorem 2, we obtain
λ ( s , ι ) = ι 0 s ι 0 λ Δ 2 Δ 1 ( ξ , τ ) Δ ξ Δ τ .
From (10) and S 8 , it is easy to observe that
Φ ( λ ( s , ι ) ) = Φ P ( s , ι ) ι 0 s 0 p ( ξ ) q ( τ ) λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) Δ ξ Δ τ ι 0 s ι 0 ι p ( ξ ) q ( τ ) Δ ξ Δ τ Φ ( P ( s , ι ) ) Φ ι 0 s ι 0 ι p ( ξ ) q ( τ ) λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) Δ ξ Δ τ ι 0 s ι 0 ι p ( ξ ) q ( τ ) Δ ξ Δ τ .
By using inverse Jensen’s dynamic inequality, we get
Φ ( λ ( s , ι ) ) Φ ( P ( s , ι ) ) P ( s , ι ) ι 0 s ι 0 ι p ( ξ ) q ( τ ) Φ λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) Δ ξ Δ τ .
Applying inverse Hölder’s inequality on the right hand side of (12) with indices 1 / γ and 1 / γ , we obtain
Φ ( λ ( s , ι ) ) Φ ( P ( s , ι ) ) P ( s , ι ) ( s ι 0 ) ( ι ι 0 ) γ × ι 0 s ι 0 ι p ( ξ ) q ( τ ) Φ λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) 1 γ Δ ξ Δ τ γ .
Using the following inequality on the term ( s ι 0 ) ( 0 ) γ , where γ < 0 and ϱ > 0
= 1 n ϱ γ 1 γ = 1 n γ ϱ γ ,
we obtain that
= 1 n Φ ( λ ( s , ι ) ) = 1 n Φ ( P ( s , ι ) ) P ( s , ι ) 1 γ = 1 n γ ( s ι 0 ) ( ι ι 0 ) γ × ι 0 s ι 0 ι p ( ξ ) q ( τ ) Φ λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) 1 γ Δ ξ Δ τ γ .
From (15), we obtain that
= 1 n Φ ( λ ( s , ι ) ) 1 γ = 1 n γ ( s ι 0 ) ( ι ι 0 ) γ = 1 n Φ ( P ( s , ι ) ) P ( s , ι ) ι 0 s ι 0 ι p ( ξ ) q ( τ ) Φ λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) 1 γ Δ ξ Δ τ γ .
Integrating both sides of (16) over s , ι from 0 to υ , ς ( = 1 , , n ) , we get
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( λ ( s , ) ) 1 γ = 1 n γ ( s 0 ) ( 0 ) γ Δ s n Δ n Δ s 1 Δ 1 = 1 n ι 0 υ ι 0 ς Φ ( P ( s , ι ) ) P ( s , ι ) ι 0 s ι 0 ι p ( ξ ) q ( τ ) Φ λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) 1 γ Δ ξ Δ τ γ Δ s Δ ι .
Applying inverse Hölder’s inequality on the right hand side of (17) with indices 1 / γ and 1 / γ , we obtain
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( λ ( s , ) ) 1 γ = 1 n γ ( s 0 ) ( 0 ) γ Δ s 1 Δ 1 Δ s n Δ n = 1 n ι 0 υ ι 0 ς Φ ( P ( s , ι ) ) P ( s , ι ) 1 γ Δ s Δ ι γ × = 1 n ι 0 υ ι 0 ς ι 0 s ι 0 ι p ( ξ ) q ( τ ) Φ λ Δ 2 Δ 1 ( ξ , τ ) p ( ξ ) q ( τ ) 1 γ Δ ξ Δ τ Δ s Δ ι γ .
By using Fubini’s theorem, we observe that
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( λ ( s , ) ) 1 γ = 1 n γ ( s 0 ) ( 0 ) γ Δ s n Δ n Δ s 1 Δ 1 G ( υ 1 ς 1 , , υ n ς n ) × = 1 n ι 0 υ ι 0 ς ( υ s ) ( ς ι ) p ( s ) q ( ι ) Φ λ Δ 2 Δ 1 ( s , ι ) p ( s ) q ( ι ) 1 γ Δ s Δ ι γ .
By using the facts υ ρ ( υ ) and ς ρ ( ς ) , we get
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( λ ( s , ) ) 1 γ = 1 n γ ( s 0 ) ( 0 ) γ Δ s n Δ n Δ s 1 Δ 1 G ( υ 1 ς 1 , , υ n ς n ) × = 1 n ι 0 υ ι 0 ς ( ρ ( υ ) s ) ( ρ ( ς ) ι ) p ( s ) q ( ι ) Φ λ Δ 2 Δ 1 ( s , ι ) p ( s ) q ( ι ) 1 γ Δ s Δ ι γ .
This completes the proof. □
Remark 1.
In Theorem 2, if T = Z , we get the result due to Zhao et al. ([17], Theorem 1.5).
Remark 2.
In Theorem 2, if we take T = R , we get inequality due to Zhao et al. [17].
Remark 3.
Let S 1 , S 2 , S 9 , S 11 , S 7 , S 13 , S 3 and S 12 be satisfied and let Φ , γ , γ , γ, and γ be as in Theorem 2. Similar to proof of Theorem 2, we have
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( λ ( s , ) ) 1 γ = 1 n γ ( s 0 ) ( 0 ) γ Δ s n Δ n Δ s 1 Δ 1 G * ( υ 1 ς 1 , , υ n ς n ) × = 1 n ι 0 υ ι 0 ς ( σ ( υ ) s ) ( σ ( ς ) ι ) p ( s ) q ( ι ) Φ λ Δ 2 Δ 1 ( s , ι ) p ( s ) q ( ι ) 1 γ Δ s Δ ι γ .
where
G * ( υ 1 ς 1 , , υ n ς n ) = 1 ( γ ) γ = 1 n ι 0 υ ι 0 ς Φ ( P ( s , ι ) ) P ( s , ι ) 1 γ Δ s Δ ι γ .
This is an inverse form of the inequality (9).
Corollary 1.
Let S 22 , S 23 , S 25 , S 26 , S 27 , S 29 , S 17 , and S 8 be satisfied. Then we have
ι 0 υ 1 ι 0 υ n = 1 n Φ ( λ ( s ) ) 1 γ = 1 n γ ( s ι 0 ) γ Δ s n Δ s 1 G * * ( υ 1 , , υ n ) × = 1 n ι 0 υ ( ρ ( υ ) s ) p ( s ) Φ λ Δ ( s ) p ( s ) 1 γ Δ s γ ,
where
G * * ( υ 1 , , υ n ) = = 1 n ι 0 υ Φ ( P ( s ) ) P ( s ) 1 γ Δ s γ .
Remark 4.
In Corollary 1, if we take T = Z , we get an inverse form of inequality due to Handley [18].
Remark 5.
In Corollary 1, if we take T = R , we get an inverse form of inequality due to Handley [18].
Remark 6.
In inequality (20) taking n = 2 , γ 1 = γ 2 = 2 , then γ 1 = γ 2 = 1 , we have
ι 0 υ 1 ι 0 υ 2 = 1 n Φ 1 ( λ 1 ( s 1 ) ) Φ 1 ( λ 2 ( s 2 ) ) ( s 1 ι 0 ) + ( s 2 ι 0 ) 2 Δ s 1 Δ s 2 D ( υ 1 , υ 2 ) ι 0 υ 1 ( ρ ( υ 1 ) s 1 ) p 1 ( s 1 ) Φ 1 λ 1 Δ ( s 1 ) p 1 ( s 1 ) 1 2 Δ s 1 2 × ι 0 υ 2 ( ρ ( υ 2 ) s 2 ) p 2 ( s 2 ) Φ 2 λ 2 Δ ( s 2 ) p 2 ( s 2 ) 1 2 Δ s 2 2 .
where
D ( υ 1 , υ 2 ) = 4 ι 0 υ 1 Φ 1 ( P 1 ( s 1 ) ) P 2 ( s 1 ) 1 Δ s 1 1 ι 0 υ 2 Φ 2 ( P 2 ( s 2 ) ) P 2 ( s 2 ) 1 Δ s 2 1 .
Remark 7.
If we take T = Z , the inequality (21) is an inverse of inequality due to Pachpatte [19].
Remark 8.
If we take T = R , the inequality (21) is an inverse of inequality due to Pachpatte [19].
Theorem 3.
Let S 1 , S 5 , S 14 , S 6 , S 15 , and S 8 be satisfied. Then for S 10 , S 18 and S 20 , we have
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 L ( υ 1 ς 1 , , υ n ς n ) × = 1 n ι 0 υ ι 0 ς ( ρ ( υ ) s ) ( ρ ( ς ) ι ) p ( s , ι ) Φ λ ( s , ι ) p ( s , ι ) β Δ s Δ ι 1 β
where
L ( υ 1 ς 1 , , υ n ς n ) = = 1 n ι 0 υ ι 0 ς Φ ( P ( s , ι ) ) P ( s , ι ) γ Δ s Δ ι 1 γ .
Proof. 
From the hypotheses of Theorem 3, S 14 , S 15 , and S 8 , it is easy to observe that
Φ ( Λ ( s , ι ) ) = Φ P ( s , ι ) ι 0 s ι 0 ι p ( ξ , τ ) λ ( ξ , τ ) p ( ξ , τ ) Δ ξ Δ τ ι 0 s ι 0 p ( ξ , τ ) Δ ξ Δ τ Φ ( P ( s , ι ) ) Φ ι 0 s ι 0 ι p ( ξ , τ ) λ ( ξ , τ ) p ( ξ , τ ) Δ ξ Δ τ ι 0 s ι 0 p ( ξ , τ ) Δ ξ Δ τ .
By using inverse Jensen dynamic inequality, we obtain that
Φ ( Λ ( s , ι ) ) Φ ( P ( s , ι ) ) P ( s , ι ) ι 0 s 0 p ( ξ , τ ) Φ λ ( ξ , τ ) p ( ξ , τ ) Δ ξ Δ τ .
Applying inverse Hölder’s inequality on the right hand side of (24) with indices γ and β , it is easy to observe that
Φ ( Λ ( s , ι ) ) Φ ( P ( s , ι ) ) P ( s , ι ) ( s ι 0 ) ( ι ι 0 ) 1 γ ι 0 s ι 0 ι p ( ξ , τ ) Φ λ ( ξ , τ ) p ( ξ , τ ) β Δ ξ Δ τ 1 β .
By using inequality the following inequality on the term ( s ι 0 ) ( ι ι 0 ) 1 γ , we get
= 1 n m 1 α α = 1 n 1 α m 1 α ,
= 1 n Φ ( Λ ( s , ι ) ) γ = 1 n 1 γ ( s ι 0 ) ( ι ι 0 ) 1 γ = 1 n Φ ( P ( s , ι ) ) P ( s , ι ) ι 0 s ι 0 ι p ( ξ , τ ) Φ λ ( ξ , τ ) p ( ξ , τ ) 1 β Δ ξ Δ τ 1 β .
Integrating both sides of (27) over s , ι from 0 to υ , ς ( = 1 , , n ) , we obtain that
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 = 1 n ι 0 υ ι 0 ς Φ ( P ( s , ι ) ) P ( s , ι ) ι 0 s ι 0 ι p ( ξ , τ ) Φ λ ( ξ , τ ) p ( ξ , τ ) β Δ ξ Δ τ 1 β Δ s Δ ι .
Applying inverse Hölder’s inequality on the right hand side of (28) with indices γ and β , it is easy to observe that
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 = 1 n ι 0 υ ι 0 ς Φ ( P ( s , ι ) ) P ( s , ι ) γ Δ s Δ ι 1 γ × = 1 n ι 0 υ ι 0 ς ι 0 s ι 0 ι p ( ξ , τ ) Φ λ ( ξ , τ ) p ( ξ , τ ) β Δ ξ Δ τ Δ s Δ ι 1 β .
Using Fubini’s theorem, we observe that
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 L ( υ 1 ς 1 , , υ n ς n ) × = 1 n ι 0 υ ι 0 ς ( υ s ) ( ς ι ) p ( s , ι ) Φ λ ( s , ι ) p ( s , ι ) β Δ s Δ ι 1 β .
By using the facts υ ρ ( υ ) and ς ρ ( ς ) , we get
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 L ( υ 1 ς 1 , , υ n ς n ) × = 1 n ι 0 υ ι 0 ς ( ρ ( υ ) s ) ( ρ ( ς ) ι ) p ( s , ι ) Φ λ ( s , ι ) p ( s , ι ) β Δ s Δ ι 1 β .
This completes the proof. □
Remark 9.
In Theorem 3, if T = R , we get the result due to Zhao et al. [20] (Theorem 2).
As a special case of Theorem 3, when T = Z , we have ρ ( n ) = n 1 , and we get the following result.
Corollary 2.
Let { a s , ι , m s , m ι } and { p s , ι , m s , m ι } , ( = 1 , , n ) be n sequences of non-negative numbers defined for m s = 1 , , k s , and m ι = 1 , , k , and define
A s , ι , m s , m ι = m ξ m s m η m ι a s , ι , m ξ , m η P s , ι , m s , m ι = m ξ m s m η m ι p s , ι , m ξ , m η .
Then
m s 1 k s 1 m ι 1 k ι 1 m s n k s n m ι n k ι n = 1 n Φ ( A s , ι , m s , m ι ) γ = 1 n 1 γ ( m s m ι ) 1 γ C ( k s 1 k ι 1 , , k s n k ι n ) × = 1 n m s k s m ι k ι ( k s ( m s 1 ) ) ( k ι ( m ι 1 ) ) P s , ι , m s , m ι Φ a s , ι , m s , m ι P s , ι , m s , m ι β 1 β
where
C ( k s 1 k ι 1 , , k s n k ι n ) = = 1 n m s k s m ι k ι Φ ( P s , ι , m s , m ι ) P s , ι , m s , m ι β ) 1 β .
Remark 10.
Let λ ( ξ , τ ) , p ( ξ , τ ) , P ( ξ , τ ) , and λ ( ξ , τ ) change to λ ( ξ ) , p ( ξ ) , P ( s ) , and λ ( s ) , respectively, and with suitable changes, and we have the following result:
Corollary 3.
Let S 22 , S 23 , S 24 , S 26 , S 27 and S 8 be satisfied. Then for S 18 ,   S 20 and S 25 we have that
ι 0 υ 1 ι 0 υ n = 1 n Φ ( Λ ( s ) γ = 1 n 1 γ ( s ι 0 ) 1 γ Δ s n Δ s 1 L * ( υ 1 , , υ n ) = 1 n ι 0 υ ( ρ ( υ ) s ) p ( s ) Φ λ ( s ) p ( s ) β Δ s 1 β
where
L * ( υ 1 , , υ n ) = = 1 n ι 0 υ Φ ( P ( s ) ) P ( s ) γ Δ s 1 γ .
Corollary 4.
In Corollary 3, if we take n = 2 , β = 1 2 , then the inequality (31) changes to
ι 0 υ 1 ι 0 υ 2 Φ 1 ( Λ 1 ( s 1 ) ) Φ 2 ( Λ 2 ( s 2 ) ) ( s 1 ι 0 ) + ( s 2 ι 0 ) 2 Δ s 1 Δ s 2 L * * ( υ 1 , υ 2 ) ι 0 υ 1 ( ρ ( υ 1 ) s 1 ) p 1 ( s 1 ) Φ λ 1 ( s 1 ) p 1 ( s 1 ) 2 Δ s 1 1 2 × ι 0 υ 2 ( ρ ( υ 2 ) s 2 ) p 2 ( s 2 ) Ψ λ 2 ( s 2 ) p 2 ( s 2 ) 2 Δ s 2 1 2 .
where
L * * ( υ 1 , υ 2 ) = 4 ι 0 υ 1 Φ 1 ( P 1 ( s 1 ) ) P 1 ( s 1 ) 1 Δ s 1 1 ι 0 υ 2 Φ 2 ( P 2 ( s 2 ) ) P 2 ( s 2 ) 1 Δ s 2 1 .
Remark 11.
In Corollary 4, if we take T = R , then the inequality (32) changes to
0 υ 1 0 υ 1 Φ 1 ( Λ 1 ( s 1 ) ) Φ 2 ( Λ 2 ( s 2 ) ) ( s 1 + s 2 ) 2 d s 1 d s 2 L * * ( υ 1 , υ 2 ) 0 υ 1 ( υ 1 s 1 ) p 1 ( s 1 ) Φ λ 1 ( s 1 ) p 1 ( s 1 ) 2 d s 1 1 2 × 0 υ 2 ( υ 2 s 2 ) p 2 ( s 2 ) Ψ λ 2 ( s 2 ) p 2 ( s 2 ) 2 d s 2 1 2 ,
where
L * * ( υ 1 , υ 2 ) = 4 0 υ 1 Φ 1 ( P 1 ( s 1 ) ) P 1 ( s 1 ) 1 d s 1 1 0 υ 2 Φ 2 ( P 2 ( s 2 ) ) P 2 ( s 2 ) 1 d s 2 1 .
This is an inverse of the inequality due to Pachpatte [21].
Corollary 5.
In Corollary 3, if we take β = n 1 n , the inequality (31) becomes
ι 0 υ 1 ι 0 υ n = 1 n Φ ( λ ( s ) = 1 n ( s ι 0 ) n n 1 Δ s 1 Δ s n L * ( υ 1 , , υ n ) = 1 n ι 0 υ ( ρ ( υ ) s ) p ( s ) Φ λ ( s ) p ( s ) n 1 n Δ s n n 1
where
L * ( υ 1 , , υ n ) = n n n 1 = 1 n ι 0 υ Φ ( P ( s ) ) P ( s ) ( n 1 ) Δ s 1 n 1 .
Theorem 4.
Let S 1 , S 5 , S 6 , S 9 , S 15 , and S 16 be satisfied. Then for S 10 , S 18 , and S 20 , we have
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n P ( s , ) Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 = 1 n ( υ ι 0 ) ( ς ι 0 ) 1 γ ι 0 υ ι 0 ς ( ρ ( υ ) s ) ( ρ ( ς ) ι ) p ( s , ι ) Φ λ ( s , ι ) β Δ s Δ ι 1 β .
Proof. 
From the hypotheses of Theorem 4, and by using inverse Jensen dynamic inequality, we have
Φ ( Λ ( s , ι ) ) = Φ 1 P ( s , ) 0 s 0 p ( ξ , τ ) λ ( ξ , τ ) Δ ξ Δ τ 1 P ( s , ι ) ι 0 s ι 0 ι p ( σ , τ ) Φ λ ( ξ , τ ) Δ ξ Δ τ .
Applying inverse Hölder’s inequality on the right hand side of (35) with indices γ and β , it is easy to observe that
Φ ( Λ ( s , ι ) ) 1 P ( s , ) ( s ι 0 ) ( ι ι 0 ) 1 γ 0 s 0 p ( ξ , τ ) Φ λ ( ξ , τ ) β Δ ξ Δ τ 1 β .
By using the inequality (26) on the term ( s ι 0 ) ( ι ι 0 ) 1 γ , we get
P ( s , ι ) Φ ( Λ ( s , ι ) ) γ = 1 n 1 γ ( s ι 0 ) ( ι ι 0 ) 1 γ ι 0 s ι 0 ι p ( ξ , τ ) Φ λ ( ξ , τ ) β Δ ξ Δ τ 1 β .
Integrating both sides of (36) over s , ι from ι 0 to υ , ς ( = 1 , , n ) , we get
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n P ( s , ) Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 = 1 n ι 0 υ 0 ς 0 s 0 p ( ξ , τ ) Φ λ ( ξ , τ ) β Δ σ Δ τ 1 β .
Applying inverse Hölder’s inequality on the right hand side of (37) with indices γ and β , it is easy to observe that
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n P ( s , ) Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s 1 Δ 1 Δ s n Δ n = 1 n ( υ ι 0 ) ( ς ι 0 ) 1 γ ι 0 υ 0 ς 0 s 0 p ( ξ , τ ) Φ λ ( ξ , τ ) β Δ ξ Δ τ Δ s Δ 1 β .
By using Fubini’s theorem, we observe that
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n P ( s , ) Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 = 1 n ( υ ι 0 ) ( ς ι 0 ) 1 γ ι 0 υ ι 0 ς ( υ s ) ( ς ) p ( s , ) Φ λ ( s , ) β Δ s s Δ 1 β .
By using the facts υ ρ ( υ ) and ς ρ ( ς ) , we get
ι 0 υ 1 ι 0 ς 1 0 υ n 0 ς n = 1 n P ( s , ) Φ ( Λ ( s , ) ) γ = 1 n 1 γ ( s 0 ) ( 0 ) 1 γ Δ s n Δ n Δ s 1 Δ 1 = 1 n ( υ ι 0 ) ( ς ι 0 ) 1 γ ι 0 υ ι 0 ς ( ρ ( υ ) s ) ( ρ ( ς ) ι ) p ( s , ι ) Φ λ ( s , ι ) β Δ s Δ ι 1 β .
This completes the proof. □
Remark 12.
In Theorem 4, if T = R , we get the result due to Zhao et al. [20] (Theorem 3).
As a special case of Theorem 4, when T = Z , we have ρ ( n ) = n 1 , and we get the following result.
Corollary 6.
Let { a s , ι , m s , m ι } and { p s , ι , m s , m ι } , ( = 1 , , n ) be n sequences of non-negative numbers defined for m s = 1 , , k s and m ι = 1 , , k ι , and define
A s , ι , m s , m ι = 1 P s , ι , m s , m ι m ξ m s m η m ι a s , ι , m ξ , m η p s , ι , m ξ , m η , P s , ι , m s , m ι = m ξ m s m η m ι p s , ι , m ξ , m η .
Then
m s 1 k s 1 m ι 1 k ι 1 m s n k s n m ι n k ι n = 1 n P s , ι , m s , m ι Φ ( A s , ι , m s , m ι ) γ = 1 n 1 γ ( m s m ι ) 1 γ = 1 n ( k s k ι ) 1 γ m s k s m ι k ι ( k s ( m s 1 ) ) ( k ι ( m ι 1 ) ) p s , ι , m s , m ι Φ a s , ι , m s , m ι β 1 β .
Remark 13.
Let λ ( ξ , τ ) , p ( ξ , τ ) , P ( ξ , τ ) and
λ ( s , ι ) = 1 P ( s , ι ) ι 0 s 0 ι p ( ξ , τ ) Λ ( ξ , τ ) Δ ξ Δ τ
changes to λ ( ξ ) , p ( ξ ) , P ( s ) and
λ ( s ) = 1 P ( s ) ι 0 s p ( ξ ) λ ( ξ ) Δ ξ .
respectively, and with suitable changes, we have the following result:
Corollary 7.
Let S 22 , S 23 , S 26 , S 27 , and S 28 be satisfied. Then for S 18 , S 20 , and S 25 , we have
ι 0 υ 1 ι 0 υ n = 1 n P ( s ) Φ ( Λ ( s ) γ = 1 n 1 γ ( s ι 0 ) 1 γ Δ s n Δ s 1 = 1 n ( υ ι 0 ) 1 γ ι 0 υ ( ρ ( υ ) s ) p ( s ) Φ λ ( s ) β Δ s 1 β .
Corollary 8.
In Corollary 7, if we take n = 2 , β = 1 2 , then the inequality (39) changes to
ι 0 υ 1 ι 0 υ 2 P 1 ( s 1 ) P 2 ( s 2 ) Φ 1 ( Λ 1 ( s 1 ) ) Φ 2 ( Λ 2 ( s 2 ) ) ( s 1 ι 0 ) + ( s 2 ι 0 ) 2 Δ s 1 Δ s 2 4 ( υ 1 ι 0 ) ( υ 2 ι 0 ) 1 × ι 0 υ 1 ( ρ ( υ 1 ) s 1 ) p 1 ( s 1 ) Φ 1 λ 1 ( s 1 ) 2 Δ s 1 1 2 ι 0 υ 2 ( ρ ( υ 2 ) s 2 ) p 2 ( s 2 ) Φ 2 λ 2 ( s 2 ) 2 Δ s 2 1 2 .
Remark 14.
In Corollary 8, if we take T = R , then the inequality (40) changes to
0 υ 1 0 υ 1 P 1 ( s 1 ) P 2 ( s 2 ) Φ 1 ( Λ 1 ( s 1 ) ) Φ 2 ( Λ 2 ( s 2 ) ) ( s 1 + s 2 ) 2 d s 1 d s 2 4 υ 1 υ 2 1 × 0 υ 1 ( υ 1 s 1 ) p 1 ( s 1 ) Φ 1 λ 1 ( s 1 ) 2 d s 1 1 2 0 υ 2 ( υ 2 s 2 ) p 2 ( s 2 ) Φ 2 λ 2 ( s 2 ) 2 d s 2 1 2 .
This is an inverse of the inequality due to Pachpatte [21].
Corollary 9.
In Corollary 8, let p 1 ( s 1 ) = p 2 ( s 2 ) = 1 , then P 1 ( s 1 ) = s 1 , P 2 ( s 2 ) = s 2 . Therefore, the inequality (40) changes to
ι 0 υ 1 0 υ 2 Φ 1 ( Λ 1 ( s 1 ) ) Φ 2 ( Λ 2 ( s 2 ) ) ( s 1 s 2 ) 1 ( s 1 0 ) + ( s 2 0 ) 2 Δ s 1 Δ s 2 4 ( υ 1 0 ) ( υ 2 0 ) 1 × ι 0 υ 1 ( ρ ( υ 1 ) s 1 ) Φ 1 λ 1 ( s 1 ) 2 Δ s 1 1 2 ι 0 υ 2 ( ρ ( υ 2 ) s 2 ) Φ 2 λ 2 ( s 2 ) 2 Δ s 2 1 2 .
Remark 15.
In Corollary 9, if we take T = R , then the inequality (42) changes to
0 υ 1 0 υ 1 Φ 1 ( Λ 1 ( s 1 ) ) Φ 2 ( Λ 2 ( s 2 ) ) ( s 1 s 2 ) 1 s 1 + s 2 2 d s 1 d s 2 4 υ 1 υ 2 1 × 0 υ 1 ( υ 1 s 1 ) Φ 1 λ 1 ( s 1 ) 2 d s 1 1 2 0 υ 2 ( υ 2 s 2 ) Φ 2 λ 2 ( s 2 ) 2 d s 2 1 2 .
This is an inverse inequality of the following inequality, which was proved by Pachpatte [20].
0 υ 0 ς Φ ( Λ ( s ) ) Ψ ( G ( ι ) ) ( s ι ) 1 s + ι d s d 1 2 υ ς 1 2 × 0 υ ( υ s 1 ) Φ λ ( s ) 2 d s 1 2 0 ς ( ς ι ) Ψ g ( ) 2 d 1 2 .
Corollary 10.
In Corollary 7, if we take β = n 1 n ( = 1 , , n ) , the inequality (39)
ι 0 υ 1 ι 0 υ n = 1 n P ( s ) Φ ( Λ ( s ) = 1 n ( s ι 0 ) n n 1 Δ s 1 Δ s n n n n 1 = 1 n ( υ ι 0 ) 1 n 1 ι 0 υ ( ρ ( υ ) s ) p ( s ) Φ λ ( s ) n 1 n Δ s n n 1 .

3. Conclusions

In this article, we gave some generalizations of the Hardy-Hilbert inequality on a general time scale, and some dynamic integral and discrete inequalities known in the literature were extended as special cases of our results.

Author Contributions

Conceptualization, H.M.E.-O., A.A.E.-D., S.D.M., D.B. and C.C.; formal analysis, H.M.E.-O., A.A.E.-D., S.D.M., D.B. and C.C.; investigation, H.M.E.-O., A.A.E.-D., S.D.M., D.B. and C.C.; writing—original draft preparation, H.M.E.-O., A.A.E.-D., S.D.M., D.B. and C.C.; writing—review and editing, H.M.E.-O., A.A.E.-D., S.D.M., D.B. and C.C. 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-Owaidy, H.M.; El-Deeb, A.A.; Makharesh, S.D.; Baleanu, D.; Cesarano, C. On Some Important Class of Dynamic Hilbert’s-Type Inequalities on Time Scales. Symmetry 2022, 14, 1395. https://doi.org/10.3390/sym14071395

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El-Owaidy HM, El-Deeb AA, Makharesh SD, Baleanu D, Cesarano C. On Some Important Class of Dynamic Hilbert’s-Type Inequalities on Time Scales. Symmetry. 2022; 14(7):1395. https://doi.org/10.3390/sym14071395

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El-Owaidy, Hassan M., Ahmed A. El-Deeb, Samer D. Makharesh, Dumitru Baleanu, and Clemente Cesarano. 2022. "On Some Important Class of Dynamic Hilbert’s-Type Inequalities on Time Scales" Symmetry 14, no. 7: 1395. https://doi.org/10.3390/sym14071395

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