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Article

Some New Generalizations of Reverse Hilbert-Type Inequalities on Time Scales

1
Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City 11884, Egypt
2
Department of Mathematical Sciences, College of Science, Princess Nourah Bint Abdulrahman University, P.O. Box 84428, Riyadh 11671, Saudi Arabia
3
Department of Mathematics, Faculty of Science, Benha University, Benha 13518, Egypt
4
Section of Mathematics, International Telematic University Uninettuno, Corso Vittorio Emanuele II, 39, 00186 Rome, Italy
5
Department of Mathematics, College of Science, King Khalid University, P.O. Box 9004, Abha 61413, Saudi Arabia
6
Department of Mathematics, Faculty of Science, Al-Azhar University, Assiut 71524, Egypt
*
Authors to whom correspondence should be addressed.
Symmetry 2022, 14(4), 750; https://doi.org/10.3390/sym14040750
Submission received: 2 March 2022 / Revised: 20 March 2022 / Accepted: 22 March 2022 / Published: 6 April 2022

Abstract

:
This manuscript develops the study of reverse Hilbert-type inequalities by applying reverse Hölder inequalities on T . We generalize the reverse inequality of Hilbert-type with power two by replacing the power with a new power β , β > 1 . The main results are proved by using Specht’s ratio, chain rule and Jensen’s inequality. Our results (when T = N ) are essentially new. Symmetrical properties play an essential role in determining the correct methods to solve inequalities.

1. Introduction

In [1], Hardy established that
i = 1 m = 1 Ξ i ϝ m i + m π sin π l i = 1 Ξ i l 1 l m = 1 ϝ m q 1 q ,
where Ξ i , ϝ m 0 with 0 < i = 1 Ξ i l < , 0 < m = 1 ϝ m q < and l > 1 , 1 / l + 1 / q = 1 . The continuous form (see [2]) of (1) is
0 0 φ ( ϑ ) ψ ( y ) ϑ + y d ϑ d y π sin π l 0 φ l ( ϑ ) d ϑ 1 l 0 ψ q ( y ) d y 1 q ,
where φ , ψ 0 are measurable functions such that 0 < 0 φ l ( ϑ ) d ϑ < and 0 < 0 ψ q ( y ) d y < . The constant π / sin ( π / l ) in both (1) and (2) is sharp. In [2], Hardy showed that if d > 1 , q > 1 , 1 / d + 1 / q 1 and 0 < λ = 2 ( 1 / d + 1 / q ) 1 , then
i = 1 n = 1 Ξ i ϝ n ( i + n ) λ K ( d , q ) i = 1 Ξ i d 1 d n = 1 ϝ n q 1 q .
In [3], Hölder proved that
k = 1 n ζ k y k k = 1 n ζ k α 1 α k = 1 n y k β 1 β ,
where ζ k and y k are positive sequences and α , β > 1 such that 1 / α + 1 / β = 1 . The continuous form of (3) is
ϱ b ψ ( τ ) ϖ ( τ ) d τ ϱ b ψ α ( τ ) d τ 1 α ϱ b ϖ β ( τ ) d τ 1 β ,
where α , β > 1 such that 1 / α + 1 / β = 1 and ψ , ϖ C ( ϱ , b ) , R + .
In [4], Zhao and Cheung proved that if ψ ( ζ ) , ϖ ( ζ ) 0 are continuous functions and ψ 1 / α ( ζ ) ϖ 1 / β ( ζ ) is integrable on [ ϱ , c ] , then
ϱ c ψ α ( ζ ) d ζ 1 α ϱ c ϖ β ( ζ ) d ζ 1 β ϱ c S Y ψ α ( ζ ) X ϖ β ( ζ ) ψ ( ζ ) ϖ ( ζ ) d ζ ,
with
X = ϱ c ψ α ( ζ ) d ζ , Y = ϱ c ϖ β ( ζ ) d ζ , α > 1 and 1 α + 1 β = 1 ,
where S ( . ) is Specht’s ratio function (see [5]) and defined as
S ( u ) = u 1 / ( u 1 ) e log u 1 / ( u 1 ) , u 1 and S ( 1 ) = 1 .
In [4], the authors proved that if ψ , ϖ C ( ϱ , c ) , R + and m > 0 , then
ϱ c ψ m + 1 ( ζ ) ϖ m ( ζ ) d ζ ϱ c S G ψ m + 1 ( ζ ) F ϖ m + 1 ( ζ ) ψ ( ζ ) d ζ m + 1 ϱ c ϖ ( ζ ) d ζ m ,
where
G = ϱ c ϖ ( ζ ) d ζ and F = ϱ c ψ m + 1 ( ζ ) ϖ m ( ζ ) d ζ .
In addition, they proved the discrete case of (4) and established that
i = 1 ϱ i m + 1 b i m i = 1 S B ϱ i m + 1 A b i m + 1 ϱ i i = 1 b i m ,
where B = i = 1 b i and A = i = 1 ϱ i m + 1 / b i m .
In 2019, Zhao and Cheung [6] studied the reverse Hilbert inequalities and proved that if 0 d , q 1 and { λ i } 1 k , { ψ n } 1 r are nonnegative and decreasing sequences of real numbers with k , r N , then
i = 1 k n = 1 r S d , q , k , r , i , n s = 1 i λ s d t = 1 n ψ t q i n 1 2 2 C d , q , k , r i = 1 k λ i s = 1 i λ s d 1 2 k i + 1 1 2 × n = 1 r ψ n t = 1 n ψ t q 1 2 r n + 1 1 2 ,
where
C d , q , r , s = 1 2 d q k r 1 2 ,
and
S d , q , k , r , i , n = S k s = 1 i λ s τ = 1 s λ τ d 1 2 s = 1 k k s + 1 λ s τ = 1 s λ τ d 1 2 × S i λ u τ = 1 u λ τ d 1 2 s = 1 i λ s τ = 1 s λ τ d 1 2 × S r t = 1 n ψ t τ = 1 t ψ τ q 1 2 t = 1 r r t + 1 ψ t τ = 1 t ψ τ q 1 2 × S n ψ v τ = 1 v ψ τ q 1 2 t = 1 n ψ t τ = 1 t ψ τ q 1 2 ,
where
S i λ u τ = 1 u λ τ d 1 2 s = 1 i λ s τ = 1 s λ τ d 1 2 = max S i λ 1 τ = 1 1 λ τ d 1 2 s = 1 i λ s τ = 1 s λ τ d 1 2 ; S i λ i τ = 1 i λ τ d 1 2 s = 1 i λ s τ = 1 s λ τ d 1 2 ,
and
S n ψ v τ = 1 v ψ τ q 1 2 t = 1 n ψ t τ = 1 t ψ τ q 1 2 = max S n ψ 1 τ = 1 1 ψ τ q 1 2 t = 1 n ψ t τ = 1 t ψ τ q 1 2 ; S n ψ n τ = 1 n ψ τ q 1 2 t = 1 n ψ t τ = 1 t ψ τ q 1 2 .
In addition, they proved that if { λ i } 1 k , { ϖ n } 1 r are nonnegative sequences and { d i } 1 k , { q n } 1 r are positive sequences with k , r N , then
i = 1 k n = 1 r S k , r , i , n ϕ ( Λ i ) ψ ( Ω n ) i n 1 2 2 N k , r s = 1 k d s ϕ λ s d s 2 k s + 1 1 2 × t = 1 r q t ψ ϖ t q t 2 r t + 1 1 2 ,
with
N k , r = 1 2 i = 1 k ϕ ( D i ) D i 2 1 2 n = 1 r ψ ( Q n ) Q n 2 1 2 ,
S k , r , i , n = S s = 1 k d s ϕ λ s d s 2 k s + 1 ϕ ( D i ) D i 2 i = 1 k ϕ ( D i ) D i 2 s = 1 i d s ϕ λ s d s 2 × S t = 1 r q t ψ ϖ t q t 2 r t + 1 ψ ( Q n ) Q n 2 n = 1 r ψ ( Q n ) Q n 2 t = 1 n q t ψ ϖ t q t 2 ,
Λ i = s = 1 i S i d s ϕ λ s d s 2 s = 1 i d s ϕ λ s d s 2 λ s ,
Ω n = t = 1 n S n q t ψ ϖ t q t 2 t = 1 n q t ψ ϖ t q t 2 ϖ t ;
D i = s = 1 i S i d s ϕ λ s d s 2 s = 1 i d s ϕ λ s d s 2 d s ;
and
Q n = t = 1 n S n q t ψ ϖ t q t 2 t = 1 n q t ψ ϖ t q t 2 q t ,
where ϕ , ψ are nonnegative, concave and supermultiplicative functions.
In [6], the authors proved that if { λ i } 1 k , { ϖ n } 1 r are nonnegative sequenceswith k , r N , then
i = 1 k n = 1 r S k , r , i , n Λ i Ω n i n 1 2 k r 1 2 i = 1 k λ i 2 k i + 1 1 2 n = 1 r ϖ n 2 r n + 1 1 2 ,
with
S k , r , i , n = S s = 1 k λ s 2 k s + 1 k s = 1 i λ s 2 S t = 1 r ϖ t 2 r t + 1 r t = 1 n ϖ t 2 ,
Λ i = s = 1 i S i λ s 2 s = 1 i λ s 2 λ s and Ω n = t = 1 n S n ϖ t 2 t = 1 n ϖ t 2 ϖ t .
Furthermore, many authors studied the inequalities of Hilbert-type, see [7,8,9,10,11,12,13,14,15].
In the last decades, the time scale theory was discovered which is a unification of the continuous calculus and discrete calculus. A time scale T is an arbitrary nonempty closed subset of the real numbers R . Many authors established some dynamic inequalities of Hilbert-type on time scales. For example, in 2021, AlNemer et al. [16] studied some reversed dynamic inequalities of Hilbert-type and proved that if a T , 0 α , β 1 and λ , ψ are nonnegative and decreasing functions, then the inequality
a σ ( s ) a σ ( r ) S α , β , t , ξ , r , s a σ ( t ) λ ( τ ) Δ τ α a σ ( ξ ) ψ ( τ ) Δ τ β σ ( t ) a 1 2 σ ( ξ ) a 1 2 Δ t Δ ξ 2 C α , β , r , s a σ ( r ) λ ( t ) a σ ( t ) λ ( τ ) Δ τ α 1 2 σ ( r ) t Δ t 1 2 × a σ ( s ) ψ ( ξ ) a σ ( ξ ) ψ ( τ ) Δ τ β 1 2 σ ( s ) ξ Δ ξ 1 2 ,
holds for all r , s [ a , ] T ,with
C α , β , r , s = 1 2 α β σ ( r ) a 1 2 σ ( s ) a 1 2 ,
and
S α , β , t , ξ , r , s = S σ ( t ) a λ ( ζ ) a σ ( ζ ) λ ( τ ) Δ τ α 1 2 a σ ( t ) λ ( ϰ ) a σ ( ϰ ) λ ( τ ) Δ τ α 1 2 Δ ϰ × S σ ( ξ ) a ψ ( η ) a σ ( η ) ψ ( τ ) Δ τ β 1 2 a σ ( ξ ) ψ ( z ) a σ ( z ) ψ ( τ ) Δ τ β 1 2 Δ z × S σ ( r ) a a σ ( t ) λ ( ζ ) a σ ( ζ ) λ ( τ ) Δ τ α 1 2 a σ ( r ) λ ( ϰ ) a σ ( ϰ ) λ ( τ ) Δ τ α 1 2 σ ( r ) ϰ Δ ϰ × S σ ( s ) a a σ ( ξ ) ψ ( η ) a σ ( η ) ψ ( τ ) Δ τ β 1 2 a σ ( s ) ψ ( z ) a σ ( z ) ψ ( τ ) Δ τ β 1 2 σ ( s ) z Δ z .
Such that
S σ ( t ) a λ ( ζ ) a σ ( ζ ) λ ( τ ) Δ τ α 1 2 a σ ( t ) λ ( ϰ ) a σ ( ϰ ) λ ( τ ) Δ τ α 1 2 Δ ϰ = max S σ ( t ) a λ ( a ) a σ ( a ) λ ( τ ) Δ τ α 1 2 a σ ( t ) λ ( ϰ ) a σ ( ϰ ) λ ( τ ) Δ τ α 1 2 Δ ϰ ; S σ ( t ) a λ ( t ) a σ ( t ) λ ( τ ) Δ τ α 1 2 a σ ( t ) λ ( ϰ ) a σ ( ϰ ) λ ( τ ) Δ τ α 1 2 Δ ϰ ,
and
S σ ( ξ ) a ψ ( η ) a σ ( η ) ψ ( τ ) Δ τ β 1 2 a σ ( ξ ) ψ ( z ) a σ ( z ) ψ ( τ ) Δ τ β 1 2 Δ z = max S σ ( ξ ) a ψ ( a ) a σ ( a ) ψ ( τ ) Δ τ β 1 2 a σ ( ξ ) ψ ( z ) a σ ( z ) ψ ( τ ) Δ τ β 1 2 Δ z ; S σ ( ξ ) a ψ ( ξ ) a σ ( ξ ) ψ ( τ ) Δ τ β 1 2 a σ ( ξ ) ψ ( z ) a σ ( z ) ψ ( τ ) Δ τ β 1 2 Δ z ,
where the function S ( . ) is the Specht ratio (see [5]) which is defined as follows:
S ( h ) = h 1 / ( h 1 ) e log h 1 / ( h 1 ) , h 1 , S ( 1 ) = 1 .
The aim of this manuscript is to use reverse Hölder inequalities with Specht’s ratio on time scales T to establish some new generalizations of reverse Hilbert-type inequalities. In particular, we generalize the inequality (8) by replacing the power 2 with a new power β , β > 1 .
The following is a breakdown of the paper’s structure. In Section 2, we cover some fundamentals of time scale theory as well as several time scale lemmas that will be useful in Section 3, where we prove our findings. As specific examples (when T = N ), our major results yield (5)–(7) proven by Zhao and Cheung [6]. In addition, we obtain the inequality (8) proved by AlNemer et al. [16].

2. Definitions and Basic Lemmas

A time scale T is defined as an arbitrary nonempty closed subset of the real numbers R and the forward jump operator is defined by: σ ( τ ) : = inf { r T : r > τ } . The set of all such rd-continuous functions is ushered by C r d ( T , R ) and for any function U : T R , the notation U σ ( τ ) denotes U ( σ ( τ ) ) .
The derivatives of U ϖ and U / ϖ (where ϖ ϖ σ 0 ) are given by
( U ϖ ) Δ = U Δ ϖ + U σ ϖ Δ = U ϖ Δ + U Δ ϖ σ , U ϖ Δ = U Δ ϖ U ϖ Δ ϖ ϖ σ .
The integration by parts formula on T is
υ 0 υ λ ( τ ) φ Δ ( τ ) Δ τ = λ ( τ ) φ ( τ ) υ 0 υ υ 0 υ λ Δ ( τ ) φ σ ( τ ) Δ τ .
The time scales chain rule is
( ϖ φ ) Δ ( τ ) = ϖ φ ϰ φ Δ τ , where ϰ τ , σ τ ,
where it is supposed that ϖ : R R is continuously differentiable and φ : T R is Δ -differentiable. For further information on the time scale calculus, see [17,18].
Definition 1
([19]). A function G : J R + is supermultiplicative if
G ϰ s G ϰ G s , ϰ , s J R .
Inequality (10) holds with equality if G is the identity map ( i . e . , G ( ϰ ) = ϰ ). G is said to be a submultiplicative function if the last inequality has the opposite sign.
Lemma 1.
If ϱ T , λ is a nonnegative rd-continuous function and 0 < γ 1 , then
ϱ σ ( s ) λ ( τ ) Δ τ γ γ ϱ σ ( s ) ϱ σ ( ϑ ) λ ( τ ) Δ τ γ 1 λ ( ϑ ) Δ ϑ .
Proof. 
Using the time scales chain rule on the term ϱ ϑ λ ( τ ) Δ τ , we obtain
ϱ ϑ λ ( τ ) Δ τ γ Δ = γ ϱ ζ λ ( τ ) Δ τ γ 1 λ ( ϑ ) , ζ [ ϑ , σ ( ϑ ) ] .
Since ζ σ ( ϑ ) , then we have (note 0 < γ 1 ) that
ϱ ζ λ ( τ ) Δ τ γ 1 ϱ σ ( ϑ ) λ ( τ ) Δ τ γ 1 ,
Substituting (13) into (12), we see
ϱ ϑ λ ( τ ) Δ τ γ Δ γ ϱ σ ( ϑ ) λ ( τ ) Δ τ γ 1 λ ( ϑ ) .
Integrating (14) over ϑ from ϱ to σ ( s ) , we have
ϱ σ ( s ) ϱ ϑ λ ( τ ) Δ τ γ Δ Δ ϑ γ ϱ σ ( s ) ϱ σ ( ϑ ) λ ( τ ) Δ τ γ 1 λ ( ϑ ) Δ ϑ .
This means that
ϱ σ ( s ) λ ( τ ) Δ τ γ γ ϱ σ ( s ) ϱ σ ( ϑ ) λ ( τ ) Δ τ γ 1 λ ( ϑ ) Δ ϑ ,
which is (11). □
Lemma 2
(Specht’s ratio [5]). Let α , β be positive numbers, d > 1 and 1 / d + 1 / q = 1 . Then,
S α β α 1 / d β 1 / q α d + β q ,
where
S ( u ) = u 1 / ( u 1 ) e log u 1 / ( u 1 ) , u 1 .
Lemma 3
([5]). Let S ( . ) be as defined in Lemma 2. Then, S ( l ) is strictly decreasing for 0 < l < 1 and strictly increasing for l > 1 . In addition, the following equations are true
S ( 1 ) = 1 and S ( l ) = S ( 1 l ) l > 0 .
Lemma 4
([20], when α = 1 ). If f , g C [ ϱ , c ] T , R + such that f γ , g ν are Δ-integrable on [ ϱ , c ] T and let β > 1 and 1 / β + 1 / ν = 1 , then
ϱ c S Y f β ( ζ ) X g ν ( ζ ) f ( ζ ) g ( ζ ) Δ ζ ϱ c f β ( ζ ) Δ ζ 1 β ϱ c g ν ( ζ ) Δ ζ 1 ν ,
where X = ϱ c f β ( ζ ) Δ ζ and Y = ϱ c g ν ( ζ ) Δ ζ .
Lemma 5
(Jensen’s inequality). Let ζ 0 , ζ T and r 0 , d R . If λ C r d ( [ ζ 0 , ζ ] T , R ) , φ: [ ζ 0 , ζ ] T ( r 0 , d ) is rd-continuous and Ψ: ( r 0 , d ) R is continuous and convex, then
Ψ 1 ζ 0 ζ λ ( τ ) Δ τ ζ 0 ζ λ ( τ ) φ ( τ ) Δ τ 1 ζ 0 ζ λ ( τ ) Δ τ ζ 0 ζ λ ( τ ) Ψ ( φ ( τ ) ) Δ τ .
Lemma 6.
Let ϱ T , λ , ψ 0 be decreasing functions and 0 < d , q 1 , β > 1 . Then,
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ = max S σ ( t ) ϱ λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ; S σ ( t ) ϱ λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ,
and
S σ ( ξ ) ϱ ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y = max S σ ( ξ ) ϱ ψ ( ϱ ) ϱ σ ( ϱ ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y ; S σ ( ξ ) ϱ ψ ( ξ ) ϱ σ ( ξ ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y .
Proof. 
We have for ϑ y that
ϱ σ ( ϑ ) λ ( τ ) Δ τ ϱ σ ( y ) λ ( τ ) Δ τ ,
and then (where 0 < d 1 ),
ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 ϱ σ ( y ) λ ( τ ) Δ τ d 1 .
Since λ is decreasing, we have
λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β λ ( y ) ϱ σ ( y ) λ ( τ ) Δ τ d 1 β ,
thus the function λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β is decreasing. Therefore, we have for ϱ ϑ that
λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β .
Integrating (20) over ϑ from ϱ to σ ( t ) , we obtain
σ ( t ) ϱ λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ,
and then,
σ ( t ) ϱ λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ 1 .
Since the function λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β is decreasing, we obtain that
λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β .
Integrating the last inequality over ϑ from ϱ to σ ( t ) , we have
ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ϱ σ ( t ) λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β Δ ϑ = σ ( t ) ϱ λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β ,
and then,
σ ( t ) ϱ λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ 1 .
From (21) and (22), we observe
σ ( t ) ϱ λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ . . . 1 . . . σ ( t ) ϱ λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ .
Since S ( . ) is decreasing on 0 , 1 and increasing on 1 , , we find that one of
S σ ( t ) ϱ λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ,
and
S σ ( t ) ϱ λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ,
is maximum (where S ( 1 ) = 1 ), and it is in the form
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ = max S σ ( t ) ϱ λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ; S σ ( t ) ϱ λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ,
which is (18). Similarly, with respect to the decreasing function ψ when 0 < q 1 , we have
S σ ( ξ ) ϱ ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y = max S σ ( ξ ) ϱ ψ ( ϱ ) ϱ σ ( ϱ ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y ; S σ ( ξ ) ϱ ψ ( ξ ) ϱ σ ( ξ ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y ,
which is (19). □

3. Main Results

Theorem 1.
Let ϱ T , 0 d , q 1 and λ , ψ be nonnegative and decreasing functions. If β > 1 , ν > 1 with 1 / β + 1 / ν = 1 , then
ϱ σ ( s ) ϱ σ ( r ) S d , q , t , ξ , r , s , β ϱ σ ( t ) λ ( τ ) Δ τ d ϱ σ ( ξ ) ψ ( τ ) Δ τ q σ ( t ) ϱ 1 ν σ ( ξ ) ϱ 1 ν Δ t Δ ξ ν C d , q , r , s ϱ σ ( r ) λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β σ ( r ) t Δ t 1 β × ϱ σ ( s ) ψ ( ξ ) ϱ σ ( ξ ) ψ ( τ ) Δ τ q 1 β σ ( s ) ξ Δ ξ 1 β ,
where
C d , q , r , s , ν = 1 ν d q σ ( r ) ϱ 1 ν σ ( s ) ϱ 1 ν ,
and
S d , q , t , ξ , r , s , β = S σ ( r ) ϱ ϱ σ ( t ) λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( r ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β σ ( r ) ϑ Δ ϑ × S σ ( s ) ϱ ϱ σ ( ξ ) ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( s ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β σ ( s ) y Δ y × S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ × S σ ( ξ ) ϱ ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y ,
such that
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ = max S σ ( t ) ϱ λ ( ϱ ) ϱ σ ( ϱ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ; S σ ( t ) ϱ λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ,
and
S σ ( ξ ) ϱ ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y = max S σ ( ξ ) ϱ ψ ( ϱ ) ϱ σ ( ϱ ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y ; S σ ( ξ ) ϱ ψ ( ξ ) ϱ σ ( ξ ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y .
Proof. 
Applying (11) with γ = d , we obtain
ϱ σ ( t ) λ ( τ ) Δ τ d d ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 Δ ϑ .
Multiplying the last inequality by
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ,
we obtain
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ϱ σ ( t ) λ ( τ ) Δ τ d d ϱ σ ( t ) S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ × λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 Δ ϑ .
From Lemma 6, the last inequality becomes
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ ϱ σ ( t ) λ ( τ ) Δ τ d d ϱ σ ( t ) S σ ( t ) ϱ λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ × λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 Δ ϑ .
Similarly, we have for ψ and 0 < q 1 that
S σ ( ξ ) ϱ ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y ϱ σ ( ξ ) ψ ( τ ) Δ τ q q ϱ σ ( ξ ) S σ ( ξ ) ϱ ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y × ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 Δ y .
From (25) and (26), we see that
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ × S σ ( ξ ) ϱ ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y × ϱ σ ( t ) λ ( τ ) Δ τ d ϱ σ ( ξ ) ψ ( τ ) Δ τ q d q ϱ σ ( t ) S σ ( t ) ϱ λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ × λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 × 1 Δ ϑ × ϱ σ ( ξ ) S σ ( ξ ) ϱ ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y × ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 × 1 Δ y .
Applying (16) on the right hand side of (27), we have
S σ ( t ) ϱ λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ × S σ ( ξ ) ϱ ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y × ϱ σ ( t ) λ ( τ ) Δ τ d ϱ σ ( ξ ) ψ ( τ ) Δ τ q d q σ ( t ) ϱ 1 ν ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ 1 β × σ ( ξ ) ϱ 1 ν ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y 1 β .
Multiplying (28) by
S σ ( r ) ϱ ϱ σ ( t ) λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( r ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β σ ( r ) ϑ Δ ϑ × S σ ( s ) ϱ ϱ σ ( ξ ) ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( s ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β σ ( s ) y Δ y ,
we obtain
S d , q , t , ξ , r , s , β ϱ σ ( t ) λ ( τ ) Δ τ d ϱ σ ( ξ ) ψ ( τ ) Δ τ q d q σ ( t ) ϱ 1 ν S σ ( r ) ϱ ϱ σ ( t ) λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( r ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β σ ( r ) ϑ Δ ϑ × ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ 1 β × S σ ( s ) ϱ ϱ σ ( ξ ) ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( s ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β σ ( s ) y Δ y × σ ( ξ ) ϱ 1 ν ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y 1 β .
Dividing the two sides of (29) by σ ( t ) ϱ 1 ν σ ( ξ ) ϱ 1 ν and then taking the integration over t from ϱ to σ ( r ) and the integration over ξ from ϱ to σ ( s ) , we have
ϱ σ ( s ) ϱ σ ( r ) S d , q , t , ξ , r , s , β ϱ σ ( t ) λ ( τ ) Δ τ d ϱ σ ( ξ ) ψ ( τ ) Δ τ q σ ( t ) ϱ 1 ν σ ( ξ ) ϱ 1 ν Δ t Δ ξ d q ϱ σ ( r ) S σ ( r ) ϱ ϱ σ ( t ) λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β ϱ σ ( r ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β σ ( r ) ϑ Δ ϑ × ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ 1 β Δ t × ϱ σ ( s ) S σ ( s ) ϱ ϱ σ ( ξ ) ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β ϱ σ ( s ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β σ ( s ) y Δ y × ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y 1 β Δ ξ .
Applying (9) on the term
ϱ σ ( r ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β σ ( r ) ϑ Δ ϑ ,
with u ( ϑ ) = σ ( r ) ϑ and v Δ ( ϑ ) = λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β , we obtain
ϱ σ ( r ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β σ ( r ) ϑ Δ ϑ = σ ( r ) ϑ v ( ϑ ) ϱ σ ( r ) + ϱ σ ( r ) v σ ( ϑ ) Δ ϑ ,
where v ( ϑ ) = ϱ ϑ λ ( θ ) ϱ σ ( θ ) λ ( τ ) Δ τ d 1 β Δ θ , and then (where v ( ϱ ) = 0 ),
ϱ σ ( r ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β σ ( r ) ϑ Δ ϑ = ϱ σ ( r ) ϱ σ ( ϑ ) λ ( θ ) ϱ σ ( θ ) λ ( τ ) Δ τ d 1 β Δ θ Δ ϑ .
Similarly, we see that
ϱ σ ( s ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β σ ( s ) y Δ y = ϱ σ ( s ) ϱ σ ( y ) ψ ( θ ) ϱ σ ( θ ) ψ ( τ ) Δ τ q 1 β Δ θ Δ y .
Substituting (31) and (32) into (30) and, then, by applying (16), we observe that
ϱ σ ( s ) ϱ σ ( r ) S d , q , t , ξ , r , s , β ϱ σ ( t ) λ ( τ ) Δ τ d ϱ σ ( ξ ) ψ ( τ ) Δ τ q σ ( t ) ϱ 1 ν σ ( ξ ) ϱ 1 ν Δ t Δ ξ d q ϱ σ ( r ) S σ ( r ) ϱ ϱ σ ( t ) λ ( ζ ) ϱ σ ( ζ ) λ ( τ ) Δ τ d 1 β Δ ζ ϱ σ ( r ) ϱ σ ( ϑ ) λ ( θ ) ϱ σ ( θ ) λ ( τ ) Δ τ d 1 β Δ θ Δ ϑ × ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ 1 β × 1 Δ t × ϱ σ ( s ) S σ ( s ) ϱ ϱ σ ( ξ ) ψ ( η ) ϱ σ ( η ) ψ ( τ ) Δ τ q 1 β Δ η ϱ σ ( s ) ϱ σ ( y ) ψ ( θ ) ϱ σ ( θ ) ψ ( τ ) Δ τ q 1 β Δ θ Δ y × ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y 1 β × 1 Δ ξ d q σ ( r ) ϱ 1 ν ϱ σ ( r ) ϱ σ ( t ) λ ( ϑ ) ϱ σ ( ϑ ) λ ( τ ) Δ τ d 1 β Δ ϑ Δ t 1 β × σ ( s ) ϱ 1 ν ϱ σ ( s ) ϱ σ ( ξ ) ψ ( y ) ϱ σ ( y ) ψ ( τ ) Δ τ q 1 β Δ y Δ ξ 1 β .
From (31)–(33), the last inequality becomes
ϱ σ ( s ) ϱ σ ( r ) S d , q , t , ξ , r , s , β ϱ σ ( t ) λ ( τ ) Δ τ d ϱ σ ( ξ ) ψ ( τ ) Δ τ q σ ( t ) ϱ 1 ν σ ( ξ ) ϱ 1 ν Δ t Δ ξ d q σ ( r ) ϱ 1 ν ϱ σ ( r ) λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β σ ( r ) t Δ t 1 β × σ ( s ) ϱ 1 ν ϱ σ ( s ) ψ ( ξ ) ϱ σ ( ξ ) ψ ( τ ) Δ τ q 1 β σ ( s ) ξ Δ ξ 1 β = ν C d , q , r , s ϱ σ ( r ) λ ( t ) ϱ σ ( t ) λ ( τ ) Δ τ d 1 β σ ( r ) t Δ t 1 β × ϱ σ ( s ) ψ ( ξ ) ϱ σ ( ξ ) ψ ( τ ) Δ τ q 1 β σ ( s ) ξ Δ ξ 1 β ,
which is (23). □
Remark 1.
If ν = β = 2 , we obtain (8) proved by AlNemer et al. [16].
Remark 2.
When T = N , ϱ = 1 and ν = β = 2 , in Theorem 1, we obtain (5) as demonstrated in [6].
Remark 3.
As a special case of Theorem 1 (when T = R ), we have that if 0 d , q 1 and λ , ψ are nonnegative and decreasing functions and assume that β > 1 , ν > 1 with 1 / β + 1 / ν = 1 , then
0 s 0 r S d , q , t , ξ , r , s , β 0 t λ ( τ ) d τ d 0 ξ ψ ( τ ) d τ q t 1 ν ξ 1 ν d t d ξ ν C d , q , r , s 0 r λ ( t ) 0 t λ ( τ ) d τ d 1 β r t d t 1 β × 0 s ψ ( ξ ) 0 ξ ψ ( τ ) d τ q 1 β s ξ d ξ 1 β ,
where
C d , q , r , s , ν = 1 ν d q r 1 ν s 1 ν ,
and
S d , q , t , ξ , r , s , β = S r 0 t λ ( ζ ) 0 ζ λ ( τ ) d τ d 1 β d ζ 0 r λ ( ϑ ) 0 ϑ λ ( τ ) d τ d 1 β r ϑ d ϑ × S s 0 ξ ψ ( η ) 0 η ψ ( τ ) d τ q 1 β d η 0 s ψ ( y ) 0 y ψ ( τ ) d τ q 1 β s y d y × S t λ ( ζ ) 0 ζ λ ( τ ) d τ d 1 β 0 t λ ( ϑ ) 0 ϑ λ ( τ ) d τ d 1 β d ϑ × S ξ ψ ( η ) 0 η ψ ( τ ) d τ q 1 β 0 ξ ψ ( y ) 0 y ψ ( τ ) d τ q 1 β d y ,
Theorem 2.
Let ϱ T , λ , ϖ be nonnegative and d , q be positive functions. If ϕ , ψ 0 are concave and supermultiplicative functions and β > 1 , ν > 1 with 1 / β + 1 / ν = 1 , then
ϱ σ ( s ) ϱ σ ( r ) S t , r , s , ζ , ν , β ϕ ( Λ ( t ) ) ψ ( Ω ( ζ ) ) σ ( t ) ϱ 1 ν σ ( ζ ) ϱ 1 ν Δ t Δ ζ ν M r , s , ν ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ 1 β × ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y 1 β ,
holds for all r , s [ ϱ , ] T , with
M r , s , ν = 1 ν ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t 1 ν ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ 1 ν ,
S t , r , s , ζ , ν , β = S ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ ϕ ( D ( t ) ) D ( t ) ν ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ × S ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y ψ ( Q ( ζ ) ) Q ( ζ ) ν ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y ,
Λ ( t ) = ϱ σ ( t ) S σ ( t ) ϱ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ λ ( ϑ ) Δ ϑ ,
Ω ( ζ ) = ϱ σ ( ζ ) S σ ( ζ ) ϱ q ( y ) ψ ϖ ( y ) q ( y ) β ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y ϖ ( y ) Δ y ,
D ( t ) = ϱ σ ( t ) S σ ( t ) ϱ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ d ( ϑ ) Δ ϑ ,
and
Q ( ζ ) = ϱ σ ( ζ ) S σ ( ζ ) ϱ q ( y ) ψ ϖ ( y ) q ( y ) β ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y q ( y ) Δ y .
Proof. 
Using the fact that ϕ is a supermultiplicative function, applying Jensen’s inequality and then applying (16), we find
ϕ ( Λ ( t ) ) = ϕ D ( t ) ϱ σ ( t ) S σ ( t ) ϱ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ d ( ϑ ) λ ( ϑ ) / d ( ϑ ) Δ ϑ ϱ σ ( t ) S σ ( t ) ϱ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ d ( ϑ ) Δ ϑ ϕ ( D ( t ) ) ϕ ϱ σ ( t ) S σ ( t ) ϱ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ d ( ϑ ) λ ( ϑ ) d ( ϑ ) Δ ϑ ϱ σ ( t ) S σ ( t ) ϱ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ d ( ϑ ) Δ ϑ ϕ ( D ( t ) ) D ( t ) ϱ σ ( t ) S σ ( t ) ϱ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) × 1 Δ ϑ ϕ ( D ( t ) ) D ( t ) σ ( t ) ϱ 1 ν ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ 1 β .
Similarly, we can obtain
ψ ( Ω ( ζ ) ) ψ ( Q ( ζ ) ) Q ( ζ ) σ ( ζ ) ϱ 1 ν ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y 1 β .
Multiplying both sides of (35) and (36), respectively, by
S ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ ϕ ( D ( t ) ) D ( t ) ν ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ ,
and
S ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y ψ ( Q ( ζ ) ) Q ( ζ ) ν ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y ,
and then multiplying these inequalities, we obtain
S ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ ϕ ( D ( t ) ) D ( t ) ν ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ ϕ ( Λ ( t ) ) × S ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y ψ ( Q ( ζ ) ) Q ( ζ ) ν ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y ψ ( Ω ( ζ ) ) ϕ ( D ( t ) ) D ( t ) σ ( t ) ϱ 1 ν ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ 1 β × S ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ ϕ ( D ( t ) ) D ( t ) ν ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ × ψ ( Q ( ζ ) ) Q ( ζ ) σ ( ζ ) ϱ 1 ν ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y 1 β × S ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y ψ ( Q ( ζ ) ) Q ( ζ ) ν ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y .
By dividing the two sides of (37) on σ ( t ) ϱ 1 ν σ ( ζ ) ϱ 1 ν and then taking the integration over ζ from ϱ to σ ( s ) and, then, the integration over t from ϱ to σ ( r ) , we obtain
ϱ σ ( s ) ϱ σ ( r ) S t , r , s , ζ , ν , β ϕ ( Λ ( t ) ) ψ ( Ω ( ζ ) ) σ ( t ) ϱ 1 ν σ ( ζ ) ϱ 1 ν Δ t Δ ζ ϱ σ ( r ) S ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ ϕ ( D ( t ) ) D ( t ) ν ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ × ϕ ( D ( t ) ) D ( t ) ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ 1 β Δ t × ϱ σ ( s ) S ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y ψ ( Q ( ζ ) ) Q ( ζ ) ν ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y × ψ ( Q ( ζ ) ) Q ( ζ ) ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y 1 β Δ ζ .
By using the integration by parts, we can see that
ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ = ϱ σ ( r ) ϱ σ ( ϑ ) d ( θ ) ϕ λ ( θ ) d ( θ ) β Δ θ Δ ϑ .
In addition, we can obtain that
ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y = ϱ σ ( s ) ϱ σ ( y ) q ( θ ) ψ ϖ ( θ ) q ( θ ) β Δ θ Δ y .
Substituting (39) and (40) into (38), we have
ϱ σ ( s ) ϱ σ ( r ) S t , r , s , ζ , ν , β ϕ ( Λ ( t ) ) ψ ( Ω ( ζ ) ) σ ( t ) ϱ 1 ν σ ( ζ ) ϱ 1 ν Δ t Δ ζ ϱ σ ( r ) S ϱ σ ( r ) ϱ σ ( ϑ ) d ( θ ) ϕ λ ( θ ) d ( θ ) β Δ θ Δ ϑ ϕ ( D ( t ) ) D ( t ) ν ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ × ϕ ( D ( t ) ) D ( t ) ϱ σ ( t ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β Δ ϑ 1 β Δ t × ϱ σ ( s ) S ϱ σ ( s ) ϱ σ ( y ) q ( θ ) ψ ϖ ( θ ) q ( θ ) β Δ θ Δ y ψ ( Q ( ζ ) ) Q ( ζ ) ν ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y × ψ ( Q ( ζ ) ) Q ( ζ ) ϱ σ ( ζ ) q ( y ) ψ ϖ ( y ) q ( y ) β Δ y 1 β Δ ζ .
Applying (16) with γ = ν = ν on the R.H.S. of (41), we have
ϱ σ ( s ) ϱ σ ( r ) S t , r , s , ζ , ν , β ϕ ( Λ ( t ) ) ψ ( Ω ( ζ ) ) σ ( t ) ϱ 1 ν σ ( ζ ) ϱ 1 ν Δ t Δ ζ ϱ σ ( r ) ϕ ( D ( t ) ) D ( t ) ν Δ t 1 ν ϱ σ ( r ) ϱ σ ( ϑ ) d ( θ ) ϕ λ ( θ ) d ( θ ) β Δ θ Δ ϑ 1 β × ϱ σ ( s ) ψ ( Q ( ζ ) ) Q ( ζ ) ν Δ ζ 1 ν ϱ σ ( s ) ϱ σ ( y ) q ( θ ) ψ ϖ ( θ ) q ( θ ) β Δ θ Δ y 1 β = ν M r , s , ν ϱ σ ( r ) ϱ σ ( ϑ ) d ( θ ) ϕ λ ( θ ) d ( θ ) β Δ θ Δ ϑ 1 β × ϱ σ ( s ) ϱ σ ( y ) q ( θ ) ψ ϖ ( θ ) q ( θ ) β Δ θ Δ y 1 β .
From (39) and (40), the Inequality (42) becomes
ϱ σ ( s ) ϱ σ ( r ) S t , r , s , ζ , ν , β ϕ ( Λ ( t ) ) ψ ( Ω ( ζ ) ) σ ( t ) ϱ 1 ν σ ( ζ ) ϱ 1 ν Δ t Δ ζ ν M r , s , ν ϱ σ ( r ) d ( ϑ ) ϕ λ ( ϑ ) d ( ϑ ) β σ ( r ) ϑ Δ ϑ 1 β × ϱ σ ( s ) q ( y ) ψ ϖ ( y ) q ( y ) β σ ( s ) y Δ y 1 β ,
which is (34). □
Remark 4.
If T = N , ϱ = 1 and ν = β = 2 , in Theorem 2, then we obtain (6) as demonstrated in [6].
By putting ϕ ( ϑ ) = ϑ and ψ ( y ) = y in Theorem 2, we have the following theorem.
Theorem 3.
Assume that ϱ T and λ , ϖ are nonnegative functions and β > 1 , ν > 1 with 1 / β + 1 / ν = 1 . Then, for all r , s [ ϱ , ] T , we have
ϱ σ ( s ) ϱ σ ( r ) S t , r , s , ζ , ν , β Λ ( t ) Ω ( ζ ) σ ( t ) ϱ 1 ν σ ( ζ ) ϱ 1 ν Δ t Δ ζ ν M r , s , ν ϱ σ ( r ) λ β ( ϑ ) σ ( r ) ϑ Δ ϑ 1 β × ϱ σ ( s ) ϖ β ( y ) σ ( s ) y Δ y 1 β ,
where
M r , s , ν = 1 ν σ ( r ) ϱ 1 ν σ ( s ) ϱ 1 ν ,
S t , r , s , ζ , ν , β = S ϱ σ ( r ) λ β ( ϑ ) σ ( r ) ϑ Δ ϑ σ ( r ) ϱ ϱ σ ( t ) λ β ( ϑ ) Δ ϑ S ϱ σ ( s ) ϖ β ( y ) σ ( s ) y Δ y σ ( s ) ϱ ϱ σ ( ζ ) ϖ β ( y ) Δ y ,
Λ ( t ) = ϱ σ ( t ) S σ ( t ) ϱ λ β ( ϑ ) ϱ σ ( t ) λ β ( ϑ ) Δ ϑ λ ( ϑ ) Δ ϑ ,
Ω ( ζ ) = ϱ σ ( ζ ) S σ ( ζ ) ϱ ϖ β ( y ) ϱ σ ( ζ ) ϖ β ( y ) Δ y ϖ ( y ) Δ y ,
D ( t ) = ϱ σ ( t ) S σ ( t ) ϱ λ β ( ϑ ) ϱ σ ( t ) λ β ( ϑ ) Δ ϑ d ( ϑ ) Δ ϑ ,
and
Q ( ζ ) = ϱ σ ( ζ ) S σ ( ζ ) ϱ ϖ β ( y ) ϱ σ ( ζ ) ϖ β ( y ) Δ y q ( y ) Δ y .
Remark 5.
As a special case of Theorem 3, when T = N , ϱ = 1 and ν = β = 2 , we obtain (7) as was proved by Zhao and Cheung [6].

4. Conclusions

In this paper, we establish some new generalizations of reverse Hilbert-type inequalities by applying reverse Hölder inequalities with the Specht ratio function on time scales. We generalize a number of those inequalities to a general time-scale measure space. In addition to this, in order to obtain some new inequalities as special cases, we also extend our inequalities to a discrete and continuous calculus. In future work, we will continue to generalize more fractional dynamic inequalities by using Specht’s ratio, Kantorovich’s ratio and n-tuple fractional integral. In particular, such inequalities can be introduced by using fractional integrals and fractional derivatives of the Riemann–Liouville-type on time scales. It will also be very interesting to introduce such inequalities in quantum calculations.

Author Contributions

Software and writing—original draft, H.M.R., G.A. and A.I.S.; writing—review and editing, H.M.R., O.B. and M.Z. All authors have read and agreed to the published version of the manuscript.

Funding

This research project was funded by Deanship of Scientific research, Princess Nourah Bint Abdulrahman University.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

This research project was funded by Deanship of Scientific Research, Princess Nourah bint Abdulrahman University, through the Program of Research Project Funding After Publication, Grant No (PRFA-P-42-14).

Conflicts of Interest

The authors declare no conflict of interest.

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Rezk, H.M.; AlNemer, G.; Saied, A.I.; Bazighifan, O.; Zakarya, M. Some New Generalizations of Reverse Hilbert-Type Inequalities on Time Scales. Symmetry 2022, 14, 750. https://doi.org/10.3390/sym14040750

AMA Style

Rezk HM, AlNemer G, Saied AI, Bazighifan O, Zakarya M. Some New Generalizations of Reverse Hilbert-Type Inequalities on Time Scales. Symmetry. 2022; 14(4):750. https://doi.org/10.3390/sym14040750

Chicago/Turabian Style

Rezk, Haytham M., Ghada AlNemer, Ahmed I. Saied, Omar Bazighifan, and Mohammed Zakarya. 2022. "Some New Generalizations of Reverse Hilbert-Type Inequalities on Time Scales" Symmetry 14, no. 4: 750. https://doi.org/10.3390/sym14040750

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