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Article

Some New Inverse Hilbert Inequalities on Time Scales

by
Ahmed A. El-Deeb
1,*,
Samer D. Makharesh
1 and
Barakah Almarri
2
1
Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City 11884, Cairo, Egypt
2
Department of Mathematical Sciences, College of Sciences, Princess Nourah Bint Abdulrahman University, P.O. Box 84428, Riyadh 11671, Saudi Arabia
*
Author to whom correspondence should be addressed.
Symmetry 2022, 14(11), 2234; https://doi.org/10.3390/sym14112234
Submission received: 7 September 2022 / Revised: 29 September 2022 / Accepted: 4 October 2022 / Published: 25 October 2022
(This article belongs to the Section Mathematics)

Abstract

:
Several inverse integral inequalities were proved in 2004 by Yong. It is our aim in this paper to extend these inequalities to time scales. Furthermore, we also apply our inequalities to discrete and continuous calculus to obtain some new inequalities as special cases. Our results are proved using some algebraic inequalities, inverse Hölder’s inequality and inverse Jensen’s inequality on time scales. Symmetry plays an essential role in determining the correct methods to solve dynamic inequalities.
MSC:
26D10; 26D15; 26E70; 34A40

1. Introduction

The form of the established classical discrete Hardy–Hilbert double series inequality [1] is given as follows: If { a m } 0 , { b n } 0 , 0 < n = 1 a n p < and 0 < 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 ,
where p > 1 , q = p p 1 .
The continuous versions of inequality (1) is given by:
0 0 f ( x ) g ( y ) x + y d x d y π sin π p 0 f p ( x ) d x 1 p 0 g p ( x ) d x 1 q ,
unless f 0 or g 0 , where f and g are measurable non-negative functions such that 0 f p ( x ) d x < and 0 g p ( x ) d x < . The constant π sin π p , in (1) and (2), is the best possible.
In [2], Pachpatte proved that if f C 1 [ [ 0 , x ] , R + ] , g C 1 [ [ 0 , y ] , R + ] with f ( 0 ) = g ( 0 ) = 0 and p , q are two positive functions defined for t [ 0 , x ) and τ [ 0 , y ) , with P ( t ) = 0 t p ( τ ) d τ and Q ( t ) = 0 t q ( τ ) d τ for s [ 0 , x ) and t [ 0 , y ) , where x , y are positive real numbers. Let Φ and Ψ be two real-valued non-negative, convex and sub-multiplicative functions defined on [ 0 , ) . Then,
0 x 0 y Φ ( f ( s ) ) Ψ ( g ( t ) ) s + t d s d t L ( x , y ) ( 0 x ( x s ) p ( s ) Φ f ( s ) p ( s ) 2 d s 1 2 × ( 0 y ( y t ) q ( t ) Ψ g ( t ) q ( t ) 2 d t 1 2 ,
where
L ( x , y ) = 1 2 0 x Φ ( P ( s ) ) P ( s ) 2 d s 1 2 0 y Ψ ( Q ( t ) ) Q ( t ) 2 d t 1 2 .
In 2004, Yong [3] studied the following integral inequality:
Theorem 1.
Let l , m 1 r 0 and f ( σ ) 0 , g ( τ ) 0 for σ ( 0 , ξ ) , τ ( 0 , ζ ) , where ξ , ζ are positive real numbers and define
Θ ( ) : = 0 f ( σ ) d σ , a n d Ξ ( ς ) : = 0 ς g ( τ ) d τ ,
for ( 0 , ξ ) and ς ( 0 , ζ ) . Then, for p 1 + q 1 = 1 , p < 0 or 0 < p < 1
0 ξ 0 ζ Θ l ( ) Ξ m ( ς ) ( r + ς r ) 2 2 r p d d ς l m ( ξ ζ ) 1 p 0 ξ ( ξ ) Θ f ( ) q d 1 q × 0 ζ ( ζ ς ) Ξ g ( ς ) q d ς 1 q
unless f 0 or g 0 , where Θ f ( ) = Θ l 1 ( ) f ( ) , Ξ g ( ς ) = Ξ m 1 ( ς ) g ( ς ) .
In 2009, Yang [4] studied the following integral inequality:
Theorem 2.
Let p , q 0 , α > 1 , γ > 1 and f ( σ ) 0 , g ( τ ) 0 for σ ( 0 , ξ ) , τ ( 0 , ζ ) , where ξ , ζ are positive real numbers and define
Θ ( ) : = 0 f ( σ ) d σ , a n d Ξ ( ς ) : = 0 ς g ( τ ) d τ ,
for ( 0 , ξ ) and ς ( 0 , ζ ) . Then,
0 ξ 0 ζ Θ p ( ) Ξ q ( ς ) γ ( α 1 ) ( α + γ ) α γ + α ς ( α 1 ) ( α + γ ) α γ d d ς D ( p , q , ξ , ζ , α , γ ) 0 ξ ( ξ ) Θ p 1 ( ) f ( ) α d 1 α × 0 ζ ( ζ ς ) Ξ q 1 ( ς ) g ( ς ) γ d ς 1 γ
unless f 0 or g 0 , where D ( p , q , ξ , ζ , α , γ ) = p q α + γ ξ α 1 α ζ γ 1 γ .
In this paper, we prove some new dynamic inequalities of Hilbert type and their converses on time scales. From these inequalities, as special cases, we formulate some special integral and discrete inequalities. Symmetry plays an essential role in determining the correct methods to solve dynamic inequalities.
Now, we present some fundamental concepts and effects on time scales which are beneficial for deducing our main results. In 1988, S. Hilger [5] presented time scale theory to unify continuous and discrete analysis. For some Hilbert-type integral, dynamic inequalities and other types of inequalities on time scales, see the papers [2,3,6,7,8,9,10,11,12,13,14,15,16]. For more details on time scale calculus see [17].
We 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 ( ς ) = ς , μ ( ς ) = 0 , f Δ ( ς ) = f ( ς ) , a b f ( ς ) Δ ς = a b f ( ς ) d ς .
( i i )
If T = Z , then
σ ( ς ) = ς + 1 , μ ( ς ) = 1 , f Δ ( ς ) = f ( ς + 1 ) f ( ς ) , a b f ( ς ) Δ ς = ς = a b 1 f ( ς ) .
Next, we write Hölder’s inequality and Jensen’s inequality on time scales.
Lemma 1
(Dynamic Hölder’s Inequality [18]). Let a , b T and f, g C r d ( [ a , b ] T , [ 0 , ) ) . If p, q > 1 with 1 p + 1 q = 1 , then
a b f ( ς ) g ( ς ) Δ ς a b f p ( ς ) Δ ς 1 p a b g q ( ς ) Δ ς 1 q .
This inequality is reversed if 0 < p < 1 and if p < 0 or q < 0 .
Lemma 2
(Dynamic Jensen’s Inequality [18]). Let a, b T and c, d R . Assume that g C r d [ a , b ] T , [ c , d ] and r C r d [ a , b ] T , R are non-negative with a b r ( ς ) Δ ς > 0 . If ϕ C r d ( c , d ) , R is a convex function, then
ϕ a b g ( ς ) r ( ς ) Δ ς a b r ( ς ) Δ ς a b r ( ς ) ϕ ( g ( ς ) ) Δ ς a b r ( ς ) Δ ς .
This inequality is reversed if ϕ C r d ( c , d ) , R is concave.
Moreover, we use the following definition and lemma as we see in the proof of our results:
Definition 1.
Λ is called a supermultiplicative function on [ 0 , ) if
Λ ( x y ) Λ ( ξ ) Λ ( ζ ) , for all ξ , ζ 0 .
Lemma 3
([19]). Let T be a time scale with ξ , a T such that ξ a . If f 0 and α 1 , then
a ξ f ( τ ) Δ τ α α a ξ f ( η ) a η f ( τ ) Δ τ α 1 Δ η .
Now, we present the formula that reduces double integrals to single integrals, which is desired in [9].
Lemma 4.
Let χ : T R and u , , θ T . Then,
u u θ χ ( τ ) Δ τ Δ θ = u ( σ ( θ ) ) χ ( θ ) Δ θ f o r T ,
assuming the integrals considered exist.
The following section contains our main results:

2. Main Results

In the next theorems, we assume that p < 0 and 1 p + 1 q = 1 .
Theorem 3.
Let T be time scale with L , K 1 and , ς , ς 0 , ξ , ζ T . Assume a ( τ ) and b ( τ ) are two non-negative and right-dense continuous functions on [ ς 0 , ξ ] and [ ς 0 , ζ ] , respectively, and define
ψ ( ) : = ς 0 a ( τ ) Δ τ a n d φ ( ς ) : = ς 0 ς b ( τ ) Δ τ ,
then, for [ ς 0 , ξ ] and ς [ ς 0 , ζ ] , we have that
ς 0 ξ ς 0 ζ ψ K ( ) φ L ( ς ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς C 1 ( K , L , p , p * ) ς 0 ξ ( ρ ( ξ ) σ ( ) ) a ( ) ψ K 1 ( ) p Δ 1 p × ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) b ( ς ) φ L 1 ( ς ) p * Δ ς 1 p * ,
where
C 1 ( K , L , p , p * ) = K L ( p + p * ) p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * .
Proof. 
By using the inequality (11), we obtain
ψ K ( ) K ς 0 a ( η ) ψ K 1 ( η ) Δ η ,
φ L ( ς ) L ς 0 ς b ( η ) φ L 1 ( η ) Δ η .
Applying inverse Hölder’s inequality on the right hand side of (14) with indices p and p p 1 , we have
ψ K ( ) K ( ς 0 ) p 1 p ς 0 a ( η ) ψ K 1 ( η ) p Δ η 1 p .
Applying inverse Hölder’s inequality on the right hand side of (15) with indices p * and p * p * 1 , we also have that
φ L ( ς ) L ( ς ς 0 ) p * 1 p * ς 0 ς b ( η ) φ L 1 ( η ) p * Δ η 1 p * .
From (16) and (17), we obtain
ψ K ( ) φ L ( ς ) K L ( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * × ς 0 a ( η ) ψ K 1 ( η ) p Δ η 1 p × ς 0 ς b ( η ) φ L 1 ( η ) p * Δ η 1 p * .
Using the following inequality
λ 1 α 1 λ 2 α 2 1 α α 1 λ 1 + α 2 λ 2 α ,
where α 1 , α 2 < 0 and λ 1 , λ 2 > 0 . Now, by setting λ 1 = ( ς 0 ) ( p 1 ) , λ 2 = ( ς ς 0 ) ( p * 1 ) , α 1 = 1 p , α 2 = 1 p * and α = α 1 + α 2 . we obtain that
( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * p p * p + p * ( ς 0 ) p 1 p + ( ς ς 0 ) p * 1 p * p + p * p p * .
Substituting (20) in (18), yields
ψ K ( ) φ L ( ς ) K L ( p + p * ) p + p * p p * p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * × ς 0 a ( η ) ψ K 1 ( η ) p Δ η 1 p × ς 0 ς b ( η ) φ L 1 ( η ) p * Δ η 1 p * .
Dividing both side of (21) by p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * , we obtain that
ψ K ( ) φ L ( ς ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * K L ( p + p * ) p + p * p p * ς 0 a ( η ) ψ K 1 ( η ) p Δ η 1 p × ς 0 ς b ( η ) φ L 1 ( η ) p * Δ η 1 p * .
Integrating both sides of (22) from ς 0 to ξ and from ς 0 to ζ , and applying inverse Hölder’s inequality with indices p ,   p p 1 and p * , p * p * 1 , we obtain
ς 0 ξ ς 0 ζ ψ K ( ) φ L ( ς ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς K L ( p + p * ) p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * ς 0 ξ ς 0 a ( η ) ψ K 1 ( η ) p Δ η Δ 1 p × ς 0 ζ ς 0 ς b ( η ) φ L 1 ( η ) p * Δ η Δ ς 1 p * .
Applying Lemma 4 on the right hand side of (23), we have
ς 0 ξ ς 0 ζ ψ K ( ) φ L ( ς ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς K L ( p + p * ) p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * ς 0 ξ ( ξ σ ( ) ) a ( ) ψ K 1 ( ) p Δ 1 p × ς 0 ζ ( ζ σ ( ς ) ) b ( ς ) φ L 1 ( ς ) p * Δ ς 1 p * = C 1 ( K , L , p , p * ) ς 0 ξ ( ξ σ ( ) ) a ( ) ψ K 1 ( ) p Δ 1 p × ς 0 ζ ( ζ σ ( ς ) ) b ( ς ) φ L 1 ( ς ) p * Δ ς 1 p * .
By using the facts ξ ρ ( ξ ) and ζ ρ ( ζ ) , we obtain
ς 0 ξ ς 0 ζ ψ K ( ) φ L ( ς ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς C 1 ( K , L , p , p * ) ς 0 ξ ( ρ ( ξ ) σ ( ) ) a ( ) ψ K 1 ( ) p Δ 1 p × ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) b ( ς ) φ L 1 ( ς ) p * Δ ς 1 p * .
This completes the proof. □
Theorem 4.
Let a ( τ ) , b ( η ) , ψ ( ) and φ ( ς ) be defined as Theorem 3. Then, we have
ς 0 ξ ς 0 ζ ψ ( ) φ ( ς ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * ( p + p * ) p + p * p p * ς 0 ξ ( ρ ( ξ ) σ ( ) ) a ( ) p Δ 1 p × ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) b ( ς ) p * Δ ς 1 p * .
Proof. 
Put K = L = 1 in (13). This completes the proof. □
As a special case of Theorem 3, when T = R , we have ρ ( ξ ) = ξ , ρ ( ζ ) = ζ , σ ( ) = , σ ( ς ) = ς , and we obtain the following result:
Corollary 1.
Assume that a ( ) and b ( ς ) are non-negative functions and define ψ ( ) : = 0 a ( η ) d η and φ ( ς ) : = 0 ς b ( η ) d η . Then,
0 ξ 0 ζ ψ K ( ) φ L ( ς ) p * p 1 + p ς p * 1 p + p * p p * d d ς C 3 ( K , L , p , p * ) 0 ξ ( ξ ) a ( ) ψ K 1 ( ) p d 1 p × 0 ζ ( ζ ς ) b ( ς ) φ L 1 ( ς ) p * d ς 1 p * ,
where
C 3 ( L , K , p , p * ) = K L ξ p 1 p ζ p * 1 p * ( p + p * ) p + p * p p * .
As special case of Theorem 3, when T = Z , we have ρ ( ξ ) = ξ 1 , ρ ( ζ ) = ζ 1 , σ ( ) = + 1 , σ ( ς ) = ς + 1 , and we obtain the following result:
Corollary 2.
Assume that a ( n ) and b ( m ) are non-negative sequences and define
ψ ( n ) = = 0 n a ( ) and φ ( m ) = k = 0 m b ( k ) .
Then,
n = 1 N m = 1 M ψ L ( n ) φ K ( m ) p * n p 1 + p m p * 1 p + p * p p * C 4 ( K , L , p , p * ) n = 1 N ( ( N 1 ) ( n + 1 ) ) ( a ( n ) ψ L 1 ( n ) ) p 1 p × m = 1 M ( ( M 1 ) ( m + 1 ) ) ( b ( m ) φ L 1 ( m ) ) p * 1 p * ,
where
C 4 ( K , L , p ) = K L N p 1 p M p * 1 p * ( p + p * ) p + p * p p * .
Theorem 5.
Let T be a time scale with , ς , ς 0 , ξ , ζ T , ψ ( ) and φ ( ς ) be as defined in Theorem 3. Let f ( τ ) and g ( η ) be two non-negative and right-dense continuous functions on [ ς 0 , ξ ] and [ ς 0 , ζ ] , respectively. Suppose that Λ and Υ are non-negative, concave and supermultiplicative functions defined on [ 0 , ) . Furthermore, assume that
Θ ( ) : = ς 0 f ( τ ) Δ τ a n d Ξ ( ς ) : = ς 0 ς g ( η ) Δ η ,
then, for [ ς 0 , ξ ] and ς [ ς 0 , ζ ] , we have that
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς M 1 ( p ) ς 0 ξ ( ρ ( ξ ) σ ( ) ) f ( ) Λ a ( ) f ( ) p Δ 1 p × ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) g ( ς ) Υ b ( ς ) g ( ς ) p * Δ ς 1 p * ,
where
M 1 ( p ) = 1 p + p * p + p * p p * ς 0 ξ Λ ( Θ ( ) Θ ( ) p p 1 Δ p 1 p ς 0 ζ Υ ( Ξ ( ς ) ) Ξ ( ς ) p * p * 1 Δ ς p * 1 p * .
Proof. 
Since Λ is a concave and supermultiplicative function, we obtain by applying inverse Jensen’s inequality that
Λ ( ψ ( ) ) = Λ Θ ( ) ) ς 0 f ( τ ) a ( τ ) f ( τ ) Δ τ ς 0 f ( τ ) Δ τ Λ ( Θ ( ) ) Λ ς 0 f ( τ ) a ( τ ) f ( τ ) Δ τ ς 0 f ( τ ) Δ τ Λ ( Θ ( ) ) Θ ( ) ς 0 f ( τ ) Λ a ( τ ) f ( τ ) Δ τ .
Applying inverse Hölder’s inequality with indices p and p p 1 on the right hand side of (26), we see that
Λ ( ψ ( ) ) Λ ( Θ ( ) Θ ( ) ( ς 0 ) p 1 p ς 0 f ( τ ) Λ a ( τ ) f ( τ ) p Δ τ 1 p .
Moreover, since Υ is a concave and supermultiplicative function, we obtain by applying inverse Jensen’s inequality and inverse Hölder’s inequality with indices p * and p * p * 1 that we have
Υ ( φ ( ς ) ) Υ ( Ξ ( ς ) ) Ξ ( ς ) ( ς ς 0 ) p * 1 p * ς 0 ς g ( η ) Υ b ( η ) g ( η ) p * Δ η 1 p * .
From (27) and (28), we have
Λ ( ψ ( ) ) Υ ( φ ( ς ) ) ( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * Λ ( Θ ( ) Θ ( ) ς 0 f ( τ ) Λ a ( τ ) f ( τ ) p Δ τ 1 p × Υ ( Ξ ( ς ) ) Ξ ( ς ) ς 0 ς g ( η ) Υ b ( η ) g ( η ) p * Δ η 1 p * .
By using inequality (19), we obtain that
( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * p p * p + p * ( ς 0 ) p 1 p + ( ς ς 0 ) p * 1 p * p + p * p p * .
From (29) and (30), we have that
Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p p * p + p * ( ς 0 ) p 1 p + ( ς ς 0 ) p * 1 p * p + p * p p * × Λ ( Θ ( ) Θ ( ) ς 0 f ( τ ) Λ a ( τ ) f ( τ ) p Δ τ 1 p Υ ( Ξ ( ς ) ) Ξ ( ς ) ς 0 ς g ( η ) Υ b ( η ) g ( η ) p * Δ η 1 p * .
Then,
Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * 1 p + p * p + p * p p * Λ ( Θ ( ) Θ ( ) ς 0 f ( τ ) Λ a ( τ ) f ( τ ) p Δ τ 1 p × Υ ( Ξ ( ς ) ) Ξ ( ς ) ς 0 ς g ( η ) Υ b ( η ) g ( η ) p * Δ η 1 p * .
Integrating both sides of (32) from ς 0 to ξ and from ς 0 and ζ , we obtain
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς 1 p + p * p + p * p p * ς 0 ξ Λ ( Θ ( ) Θ ( ) ς 0 f ( τ ) Λ a ( τ ) f ( τ ) p Δ τ 1 p Δ × ς 0 ζ Υ ( Ξ ( ς ) ) Ξ ( ς ) ς 0 ς g ( η ) Υ b ( η ) g ( η ) p * Δ η 1 p * Δ ς .
Applying inverse Hölder’s inequality with indices p , p p 1 and p * , p * 1 p * on the right hand of side (33), we have
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς 1 p + p * p + p * p p * ς 0 ξ Λ ( Θ ( ) Θ ( ) p p 1 Δ p 1 p ς 0 ξ ς 0 f ( τ ) Λ a ( τ ) f ( τ ) p Δ τ Δ 1 p × ς 0 ζ Υ ( Ξ ( ς ) ) Ξ ( ς ) p * p * 1 Δ ς p * 1 p * ς 0 ζ ς 0 ς g ( η ) Υ b ( η ) g ( η ) p * Δ η Δ ς 1 p * .
Applying Lemma 4 on the right hand side of (34), we obtain
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς M 1 ( p ) ς 0 ξ ( ξ σ ( ) ) f ( ) Λ a ( ) f ( ) p Δ 1 p × ς 0 ζ ( ζ σ ( ς ) ) g ( ς ) Υ b ( ς ) g ( ς ) p * Δ ς 1 p * .
By using the facts ξ ρ ( ξ ) and ζ ρ ( ζ ) , we obtain
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς M 1 ( p ) ς 0 ξ ( ρ ( ξ ) σ ( ) ) f ( ) Λ a ( ) f ( ) p Δ 1 p × ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) g ( ς ) Υ b ( ς ) g ( ς ) p * Δ ς 1 p * ,
where
M 1 ( p ) = 1 p + p * p + p * p p * ς 0 ξ Λ ( Θ ( ) Θ ( ) p p 1 Δ p 1 p ς 0 ζ Υ ( Ξ ( ς ) ) Ξ ( ς ) p * p * 1 Δ ς p * 1 p * .
This completes the proof. □
As a special case of Theorem 5, when T = R , we have ρ ( ξ ) = ξ ρ ( ζ ) = ζ , σ ( ) = , σ ( ς ) = ς , and we obtain the following result:
Corollary 3.
Assume that a ( ) , b ( ς ) , f ( τ ) and g ( η ) are non-negative functions and define
ψ ( ) : = 0 a ( η ) d η , φ ( ς ) : = 0 ς b ( η ) d η , Θ ( ) : = 0 f ( τ ) d τ , a n d Ξ ( ς ) : = 0 ς g ( η ) d η .
Then,
0 ξ 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) p * p 1 + p ς p * 1 p + p * p p * d d ς M 2 ( p ) 0 ξ ( ξ ) f ( ) Λ a ( ) f ( ) p d 1 p × 0 ζ ( ζ ς ) g ( ς ) Υ b ( ς ) g ( ς ) p * d ς 1 p * ,
where
M 2 ( p ) = 1 p + p * p + p * p p * 0 ξ Λ ( Θ ( ) Θ ( ) p p 1 d p 1 p 0 ζ Υ ( Ξ ( ς ) ) Ξ ( ς ) p * p * 1 d ς p * 1 p * .
As a special case of Theorem 5, when T = Z , we have ρ ( ξ ) = ξ 1 , ρ ( ζ ) = ζ 1 , σ ( ) = + 1 , σ ( ς ) = ς + 1 , and we obtain the following result.
Corollary 4.
Assume that a ( n ) , b ( m ) , f ( n ) and g ( m ) are non-negative sequences and define
ψ ( n ) = = 0 n a ( ) , φ ( m ) = k = 0 m b ( k ) , Θ ( n ) = = 0 n f ( ) and Ξ ( m ) = k = 0 m g ( k ) .
Then,
n = 1 N m = 1 M Λ ( ψ ( n ) ) Υ ( φ ( m ) ) p * n p 1 + p m p * 1 p + p * p p * M 3 ( p ) n = 1 N ( ( N 1 ) ( n + 1 ) ) f ( n ) Λ a ( n ) f ( n ) p 1 p × m = 1 M ( ( M 1 ) ( m + 1 ) ) g ( m ) Υ b ( m ) g ( m ) p * 1 p * ,
where
M 3 ( p ) = 1 p + p * p + p * p p * n = 1 N Λ ( Θ ( n ) Θ ( n ) p p 1 p 1 p m = 1 M Υ ( Ξ ( m ) ) Ξ ( m ) p * p * 1 p * 1 p * .
Theorem 6.
Let T be a time scale with , ς , ς 0 , ξ , ζ T . Let f and g be two non-negative and right-dense continuous functions on [ ς 0 , ξ ] and [ ς 0 , ζ ] , respectively. Suppose that Λ and Υ are non-negative, concave and supermultiplicative functions defined on [ 0 , ) and define
Θ ( ) : = 1 ς 0 ς 0 f ( τ ) Δ τ a n d Ξ ( ς ) : = 1 ς ς 0 ς 0 ς g ( τ ) Δ τ ,
then, for [ ς 0 , ξ ] and ς [ ς 0 , ζ ] , we have that
ς 0 ξ ς 0 ζ Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς E ( p , p * ) ς 0 ξ ( ρ ( ξ ) σ ( ) ) Λ f ( ) p Δ 1 p ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) Υ g ( ς ) p * Δ ς 1 p * ,
where
E ( p , p * ) = 1 p + p * p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * .
Proof. 
By assumption and by using the inverse Jensen inequality, we see that
Λ ( Θ ( ) ) = Λ 1 ς 0 ς 0 f ( τ ) Δ τ 1 ς 0 ς 0 Λ f ( τ ) Δ τ .
By applying inverse Hölder’s inequality on (37) with indices, p , p p 1 , we have
Λ ( Θ ( ) ) 1 ς 0 ( ς 0 ) p 1 p ς 0 Λ f ( τ ) p Δ τ 1 p .
This implies that
Λ ( Θ ( ) ) ( ς 0 ) ( ς 0 ) p 1 p ς 0 Λ f ( τ ) p Δ τ 1 p .
Analogously,
Υ ( Ξ ( ς ) ) ( ς ς 0 ) ( ς ς 0 ) p * 1 p * ς 0 ς Υ g ( τ ) p * Δ τ 1 p * .
From (39) and (40), we obtain
Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) ( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * ς 0 Λ f ( τ ) p Δ τ 1 p × ς 0 ς Υ g ( τ ) p * Δ τ 1 p * .
By using inequality (19), we obtain that
( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * p p * p + p * ( ς 0 ) p 1 p + ( ς ς 0 ) p * 1 p * p + p * p p * .
Then,
Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) p p * p + p * ( ς 0 ) p 1 p + ( ς ς 0 ) p * 1 p * p + p * p p * ς 0 Λ f ( τ ) p Δ τ 1 p ς 0 ς Υ g ( τ ) p * Δ τ 1 p * .
From (43), we have
Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * 1 p + p * p + p * p p * ς 0 Λ f ( τ ) p Δ τ 1 p × ς 0 ς Υ g ( τ ) p * Δ τ 1 p * .
Taking delta integrating on both sides of (44), first over from ς 0 to ξ and then over ς from ς 0 to ζ , we find that
ς 0 ξ ς 0 ζ Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς 1 p + p * p + p * p p * ς 0 ξ ς 0 Λ f ( τ ) p Δ τ 1 p Δ × ς 0 ζ ς 0 ς Υ g ( τ ) p * Δ τ 1 p * Δ ς .
By applying inverse Hölder’sinequality on (45) with indices p ,   p p 1 and p * , p * p * 1 , we get
ς 0 ξ ς 0 ζ Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς 1 p + p * p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * × ς 0 ξ ς 0 Λ f ( τ ) p Δ τ Δ 1 p ς 0 ζ ς 0 ς Υ g ( τ ) p * Δ τ Δ ς 1 p * .
Applying Lemma 4 on (46), we fined that
ς 0 ξ ς 0 ζ Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς 1 p + p * p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * × ς 0 ξ ( ξ σ ( ) ) Λ f ( ) p Δ 1 p ς 0 ζ ( ζ σ ( ς ) ) Υ g ( ς ) p * Δ ς 1 p * = E ( p , p * ) ς 0 ξ ( ξ σ ( ) ) ϕ f ( ) p Δ 1 p ς 0 ζ ( ζ σ ( ς ) ) Υ g ( ς ) p * Δ ς 1 p * .
By using the facts ξ ρ ( ξ ) and ζ ρ ( ζ ) , we obtain
ς 0 ξ ς 0 ζ Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ( ς 0 ) ( ς ς 0 ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς E ( p , p * ) ς 0 ξ ( ρ ( ξ ) σ ( ) ) Λ f ( ) p Δ 1 p ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) Υ g ( ς ) p * Δ ς 1 p * ,
where
E ( p , p * ) = 1 p + p * p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * .
This completes the proof. □
As a special case of Theorem 6, when T = R , we obtain the following conclusion.
Corollary 5.
Assume that f and g are non-negative functions and define
Θ ( ) : = 1 0 f ( τ ) d τ a n d Ξ ( ς ) : = 1 ς 0 ς g ( τ ) d τ ,
then, for [ 0 , ξ ] and ς [ 0 , ζ ] , we have that
0 ξ 0 ζ Λ ( Θ ( ) ) Υ ( Ξ ( ς ) ) ς p * p 1 + p ς p * 1 p + p * p p * d d ς E 1 ( p , p * ) 0 ξ ( ξ ) Λ f ( ) p d 1 p × 0 ζ ( ζ ς ) Υ g ( ς ) p * d ς 1 p * ,
where
E 1 ( p , p * ) = 1 p + p * p + p * p p * ξ p 1 p ζ p * 1 p * .
As a special case of Theorem 6, when T = Z , we obtain the following conclusion.
Corollary 6.
Assume that f ( n ) and g ( m ) are two non-negative sequences of real numbers and define
Θ ( n ) : = 1 n = 1 n f ( ) a n d Ξ ( m ) : = 1 m k = 1 m g ( k ) ,
then,
n = 1 N m = 1 M Λ ( Θ ( n ) ) Υ ( Ξ ( m ) ) n m p * n p 1 + p m p * 1 p + p * p p * E 2 ( p , p * ) n = 1 N ( ( N 1 ) ( n + 1 ) ) Λ f ( n ) p 1 p × m = 1 M ( ( M 1 ) ( m + 1 ) ) Υ g ( m ) p * 1 p * ,
where
E 2 ( p , p * ) = 1 p + p * p + p * p p * N p 1 p M p * 1 p * .
Theorem 7.
Let T be a time scale with , ς , ς 0 , ξ , ζ T . Let a ( τ ) and b ( τ ) be two non-negative and right-dense continuous functions on [ ς 0 , ξ ] and [ ς 0 , ζ ] , respectively. Let Θ , Ξ , f , g , Λ and Υ be as assumed in Theorem 5. Furthermore, assume that
ψ ( ) : = 1 Θ ( ) ς 0 a ( τ ) f ( τ ) Δ τ a n d φ ( ς ) : = 1 Ξ ( ς ) ς 0 ς b ( η ) g ( η ) Δ η ,
then, for [ ς 0 , ξ ] and ς [ ς 0 , ζ ] , we have that
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 ) p + p * p p * Δ Δ ς D 1 ( p ) ς 0 ξ ( ξ σ ( ) ) f ( ) Λ [ a ( ) ] p Δ 1 p ς 0 ζ ( ζ σ ( ς ) ) g ( ς ) Υ [ b ( ς ) ] p * Δ ς 1 p * ,
where
D 1 ( p ) = ( 1 p + p * ) p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * .
Proof. 
From (47), we see that
Λ ( ψ ( ) ) = Λ 1 Θ ( ) ς 0 f ( τ ) a ( τ ) Δ τ .
Applying inverse Hölder’s inequality with indices p and p p 1 on the right hand side of (48), we obtain
Λ ( ψ ( ) ) ( ς 0 ) p 1 p Θ ( ) ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ 1 p .
From (49), we obtain that
Λ ( ψ ( ) ) Θ ( ) ( ς 0 ) p 1 p ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ 1 p .
Similarly, we obtain
Υ ( φ ( ς ) ) Ξ ( ς ) ( ς ς 0 ) p * 1 p * ς 0 ς g ( η ) Υ [ b ( η ) ] p * Δ η 1 p * .
From (50) and (51), we observe that
Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) ( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * × ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ 1 p ς 0 ς g ( η ) Υ [ b ( η ) ] p * Δ η 1 p * .
Applying the inequality (19) on the term ( ς 0 ) p 1 p ( ς ς 0 ) p * 1 p * , we obtain the following inequality
Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) p p * p + p * ( ς 0 ) p 1 p + ( ς ς 0 ) p * 1 p * p + p * p p * × ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ 1 p ς 0 ς g ( η ) Υ [ b ( η ) ] p * Δ η 1 p * .
Dividing both sides of (53) by p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * , we obtain that
Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * ( 1 p + p * ) p + p * p p * ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ 1 p × ς 0 ς g ( η ) Υ [ b ( η ) ] p * Δ η 1 p * .
Integrating both sides of (54) from ς 0 to ξ and ς 0 to ζ , we obtain
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς ( 1 p + p * ) p + p * p p * ς 0 ξ ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ 1 p Δ ς 0 ζ ς 0 ς g ( η ) Υ [ b ( η ) ] p * Δ η 1 p * Δ ς .
Applying inverse Hölder’s inequality again with indices p , p p 1 and p * , p * p * 1 on the right hand side of (55), we have
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς ( 1 p + p * ) p + p * p p * ( ξ ς 0 ) p 1 p ( ζ ς 0 ) p * 1 p * ς 0 ξ ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ Δ 1 p × ς 0 ζ ς 0 ς g ( η ) Υ [ b ( η ) ] p * Δ η Δ ς 1 p * = D 1 ( p ) ς 0 ξ ς 0 f ( τ ) Λ [ a ( τ ) ] p Δ τ Δ 1 p ς 0 ζ ς 0 ς g ( η ) Υ [ b ( η ) ] p * Δ η Δ ς 1 p * .
Applying Lemma 4 on the right hand side of (56), we obtain that
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς D 1 ( p ) ς 0 ξ ( ξ σ ( ) ) f ( ) Λ [ a ( ) ] p Δ 1 p ς 0 ζ ( ζ σ ( ς ) ) g ( ς ) Υ [ b ( ς ) ] p * Δ ς 1 p * .
By using the facts ξ ρ ( ξ ) and ζ ρ ( ζ ) , we obtain
ς 0 ξ ς 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Ξ ( ς ) Θ ( ) p * ( ς 0 ) p 1 + p ( ς ς 0 ) p * 1 p + p * p p * Δ Δ ς D 1 ( p ) ς 0 ξ ( ρ ( ξ ) σ ( ) ) f ( ) Λ [ a ( ) ] p Δ 1 p ς 0 ζ ( ρ ( ζ ) σ ( ς ) ) g ( ς ) Υ [ b ( ς ) ] p * Δ ς 1 p * .
This completes the proof. □
As a special case of Theorem 7, when T = R , we have ρ ( ξ ) = ξ , ρ ( ζ ) = ζ , σ ( ) = , σ ( ς ) = ς , and we obtain the following result:
Corollary 7.
Assume that a ( ) , b ( ς ) , f ( ) and g ( ς ) are non-negative functions and define
ψ ( ) : = 1 Θ ( ) 0 f ( τ ) a ( τ ) d τ and φ ( ς ) : = 1 Ξ ( ς ) 0 ς g ( τ ) b ( τ ) d τ ,
Θ ( ) : = 0 f ( τ ) d τ and Ξ ( ς ) : = 0 ς g ( τ ) d τ .
Then,
0 ξ 0 ζ Λ ( ψ ( ) ) Υ ( φ ( ς ) ) Θ ( ) Ξ ( ς ) p * p 1 + p ς p * 1 p + p * p p * d d ς D 2 ( p ) 0 ξ ( ξ ) f ( ) Λ a ( ) p d 1 p × ( 0 ζ ( ζ ς ) g ( ς ) Υ b ( ς ) p * d ς ) 1 p * ,
where
D 2 ( p ) = ( 1 p + p * ) p + p * p p * ( ξ ) p 1 p ( ζ ) p * 1 p * .
As a special case of Theorem 7, when T = Z , we have ρ ( ξ ) = ξ 1 , ρ ( ζ ) = ζ 1 , σ ( ) = + 1 , σ ( ς ) = ς + 1 , and we obtain the following result:
Corollary 8.
Assume that a ( n ) , b ( m ) , f ( n ) and g ( m ) are non-negative sequences and define
ψ ( n ) : = 1 Θ ( n ) = 0 n f ( ) a ( ) and φ ( m ) : = 1 Ξ ( m ) k = 0 m g ( k ) b ( k ) ,
Θ ( n ) : = = 0 n f ( ) and Ξ ( m ) : = k = 0 m g ( k ) .
Then
n = 1 N m = 1 M Λ ( ψ ( n ) ) Υ ( φ ( m ) ) Θ ( n ) Ξ ( m ) p * n p 1 + p m p * 1 p + p * p p * D 3 ( p ) n = 1 N ( ( N 1 ) ( n + 1 ) ) f ( n ) Λ a ( n ) p 1 p . × ( m = 1 M ( ( M 1 ) ( m + 1 ) ) g ( m ) Υ b ( m ) p * ) 1 p * ,
where
D 3 ( p ) = ( 1 p + p * ) p + p * p p * ( N ) p 1 p ( M ) p * 1 p * .

3. Conclusions and Discussion

In this article, with the help of the inverse Hölder’s inequality and inverse Jensen’s inequality on time scales, we discussed and proved several new generalizations of the integral retarded inequalities given in [3]. Moreover, we generalized a number of other inequalities to a general time scale. Finally, as a special case, we studied the discrete and continuous inequalities. As a future work, we intend to generalize these inequalities by using alpha-conformable fractional derivatives on time scales. Furthermore, we will extend these results to diamond alpha calculus.

Author Contributions

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

Funding

Princess Nourah bint Abdulrahman University Researchers Supporting Project number PNURSP2022R216, Princess Nourah bint Abdulrahman University, Riyadh, Saudi Arabia.

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.; Almarri, B. Some New Inverse Hilbert Inequalities on Time Scales. Symmetry 2022, 14, 2234. https://doi.org/10.3390/sym14112234

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El-Deeb AA, Makharesh SD, Almarri B. Some New Inverse Hilbert Inequalities on Time Scales. Symmetry. 2022; 14(11):2234. https://doi.org/10.3390/sym14112234

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El-Deeb, Ahmed A., Samer D. Makharesh, and Barakah Almarri. 2022. "Some New Inverse Hilbert Inequalities on Time Scales" Symmetry 14, no. 11: 2234. https://doi.org/10.3390/sym14112234

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El-Deeb, A. A., Makharesh, S. D., & Almarri, B. (2022). Some New Inverse Hilbert Inequalities on Time Scales. Symmetry, 14(11), 2234. https://doi.org/10.3390/sym14112234

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