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Article

Dynamic Hilbert-Type Inequalities with Fenchel-Legendre Transform

by
Ahmed A. El-Deeb
1,*,
Samer D. Makharesh
1 and
Dumitru Baleanu
2,3,4
1
Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City, Cairo 11884, Egypt
2
Department of Mathematics, Cankaya University, Ankara 06530, Turkey
3
Institute of Space Science, 07650 Magurele-Bucharest, Romania
4
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(4), 582; https://doi.org/10.3390/sym12040582
Submission received: 12 March 2020 / Revised: 22 March 2020 / Accepted: 26 March 2020 / Published: 7 April 2020
(This article belongs to the Special Issue Multibody Systems with Flexible Elements)

Abstract

:
Our work is based on the multiple inequalities illustrated in 2020 by Hamiaz and Abuelela. With the help of a Fenchel-Legendre transform, which is used in various problems involving symmetry, we generalize a number of those inequalities to a general time scale. Besides that, in order to get new results as special cases, we will extend our results to continuous and discrete calculus.
AMS Subject Classifications:
26D10; 26D15; 26E70; 34A40

1. Introduction

In 2020, Hamiaz and Abuelela [1] have studied the following discrete inequalities:
Theorem 1.
Suppose q , p 1 , α β 1 2 and ( b m ) m 0 ( a n ) n 0 are sequences of real numbers. Define A n = s = 1 n a s , B m = t = 1 m b t . Then
n = 1 k m = 1 r A n 2 p B m 2 q h ( n ) + h * ( m ) C 1 * ( p , q ) n = 1 k ( k n + 1 ) ( a n A n p 1 ) 2 × m = 1 r ( r m + 1 ) ( b m B m q 1 ) 2
and
n = 1 k m = 1 r A n p B m q | h ( n ) | 1 2 β + | h * ( m ) | 1 2 β α n = 1 k m = 1 r A n p B m q h ( n ) + h * ( m ) C 2 * ( p , q , k , r ) n = 1 k ( k n + 1 ) ( a n A n p 1 ) 2 1 2 × m = 1 r ( r m + 1 ) ( b m B m q 1 ) 2 1 2
unless ( a n ) or ( b m ) is null, where
C 1 * ( p , q ) = ( p q ) 2 and C 2 * ( p , q , r , k ) = p q k r .
Hilger [2] suggested time scales theory to unify discrete and continuous analysis. More Hilbert-type inequalities and other types can be seen in [1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36], see also [37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53]. For more details on time scales calculus see [54].
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 ( t ) = t , μ ( t ) = 0 , f Δ ( t ) = f ( t ) , a b f ( t ) Δ t = a b f ( t ) d t .
(ii) 
If T = Z , then
σ ( t ) = t + 1 , μ ( t ) = 1 , f Δ ( t ) = f ( t + 1 ) f ( t ) , a b f ( t ) Δ t = t = a b 1 f ( t ) .
Next is Hölder’s and Jensen’s inequality:
Lemma 1
([19]). 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 ( t ) g ( t ) Δ t a b f p ( t ) Δ t 1 p a b g q ( t ) Δ t 1 q .
Lemma 2
( [19]). 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 nonnegative with a b r ( t ) Δ t > 0 . If Φ ˘ C r d ( c ˘ , d ˘ ) , R be a convex function, then
Φ ˘ a b g ( t ) r ( t ) Δ t a b r ( t ) Δ t a b r ( t ) Φ ˘ ( g ( t ) ) Δ t a b r ( t ) Δ t .
Now, we present the Fenchel-Legendre transform and refer, for example, to [11,12,13], for more details.
Definition 1.
Assuming h : R n R { + } is a function: h + i.e., Dom ( h ) = { x R n , | h ( x ) < } . Then the Fenchel-Legendre transform is defined as:
h * : R n R { + } , y h * ( y ) = sup { < y , x > h ( x ) , x Dom ( h ) }
where < . , . > is the scalar product on R n . The mapping h h * is often be called the conjugate operation.
The domain of h * is the set of slopes of all the affine functions minorizing the function h over R n . An equivalent formula for (3) is introduced as follows:
Corollary 1.
Assuming h : R n R is differentiable, strictly convex and 1-coercive function. Then
h * ( y ) = < y , ( h ) 1 ( y ) > h ( ( h ) 1 ( y ) ) ,
y Dom ( h * ) , where < . , . > denotes the scalar product on R n .
Lemma 3
( [13]). Let h be a function and h * its Fenchel-Legendre transform. Then
< x , y > h ( x ) + h * ( y ) ,
for all x Dom ( h ) , and y Dom ( h * ) .
In addition, we will use the following definition and lemma as we will see in the proof of our results:
Definition 2.
The function Φ ˘ is said to be a submultiplicative on [ 0 , ) if
Φ ˘ ( x y ) Φ ˘ ( x ) Φ ˘ ( y ) , for all x , y 0 .
Lemma 4
([20]). Assuming T is a time scale with x , a T such that x a . If f 0 and α ˜ 1 , then
a σ ( x ) f ( τ ˇ ) Δ τ ˇ α ˜ α ˜ a σ ( x ) f ( η ) a σ ( η ) f ( τ ˇ ) Δ τ ˇ α ˜ 1 Δ η .
Next, we write Fubini’s theorem on time scales.
Lemma 5
(Fubini’s Theorem, see [55]). Assume that ( X , 1 , μ Δ ) and ( Y , 2 , ν Δ ) are two finite-dimensional time scales measure spaces. Moreover, suppose that f : X × Y R is a delta integrable function and define the functions
π ^ 1 ( y ) = X f ( x , y ) d μ Δ ( x ) , y Y ,
and
π ^ 2 ( x ) = X f ( x , y ) d ν Δ ( y ) , x X .
Then π ^ 1 is delta integrable on Y and π ^ 2 is delta integrable on X and
X d μ Δ ( x ) Y f ( x , y ) d ν Δ ( y ) = Y d ν Δ ( y ) X f ( x , y ) d μ Δ ( x ) .
In this manuscript, by using Fubini’s theorem and the Fenchel-Legendre transform, which is used in various problems involving symmetry, we extend the discrete results proved in [1] on time scales. We start from the inequalities treated in the Theorem 1. Our results can be applied to give more general forms of some previously proved inequalities through substituting h and h * by suitable functions as we will see in the following two sections.
The following section contains our main results.

2. Main Results

We start by establishing the following useful inequality:
Lemma 6.
Assume x and y R such that x + y 1 , then for γ > 0 , and α β 1 2 , we get
( x + y ) 1 γ | x | 1 2 β + | y | 1 2 β 2 α γ .
Proof. 
For x + y 1 and α β 1 , we have
( x + y ) 1 2 ( x + y ) 1 2 α β = ( x + y ) 1 2 β α ( | x | + | y | ) 1 2 β α .
From | x | + | y | 1 n | x | 1 n + | y | 1 n , for all n 1 . Thus, from (9), and since 2 β 1 , we obtain:
( x + y ) 1 2 ( | x | + | y | ) 1 2 β α | x | 1 2 β + | y | 1 2 β α .
Now, since γ > 0 , by taking the power 1 γ for both sides of (10), we get:
( x + y ) 1 γ | x | 1 2 β + | y | 1 2 β 2 α γ .
This proves our claim. ☐
In the next theorems, we will let p > 1 , q > 1 and 1 q + 1 p = 1 .
Theorem 2.
Let T be a time scale with L 1 , K 1 and s , t , t 0 , x , y T . Assume a ( τ ˇ ) 0 and b ( τ ˇ ) 0 are right-dense continuous functions on the time scales intervals [ t 0 , x ] T and [ t 0 , y ] T respectively and define
A ( s ) : = t 0 s a ( τ ˇ ) Δ τ ˇ , a n d B ( t ) : = t 0 t b ( τ ˇ ) Δ τ ˇ ,
then for σ ( s ) [ t 0 , x ] T and σ ( t ) [ t 0 , y ] T , we have that
t 0 x t 0 y A q K ( σ ( s ) ) B q L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p Δ s Δ t C 1 ( L , K , q ) t 0 x ( σ ( x ) σ ( s ) ) a ( s ) A K 1 ( σ ( s ) q Δ s × t 0 y ( σ ( y ) σ ( t ) ) b ( t ) B L 1 ( σ ( t ) q Δ t
and
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t C 2 ( L , K , p ) t 0 x ( σ ( x ) σ ( s ) ) A K 1 ( σ ( s ) ) a ( s ) q Δ s 1 q × ( t 0 y ( σ ( y ) σ ( t ) ) B L 1 ( σ ( t ) ) b ( t ) q Δ t ) 1 q
where C 1 ( L , K , q ) = ( K L ) q and C 2 ( L , K , p ) = K L ( x t 0 ) 1 p ( y t 0 ) 1 p .
Proof. 
By using the inequality (7), we obtain
A K ( σ ( s ) ) K t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) Δ η ,
B L ( σ ( t ) ) L t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) ) Δ η .
We use Lemma 1. Then from (13), we get
A K ( σ ( s ) ) K ( σ ( s ) t 0 ) 1 p t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η 1 q .
We use Lemma 1. Then from (14), we also have
B L ( σ ( t ) ) L ( σ ( t ) t 0 ) 1 p t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η 1 q .
From (15) and (16), we get
A K ( σ ( s ) ) B L ( σ ( t ) ) K L ( σ ( s ) t 0 ) 1 p ( σ ( t ) t 0 ) 1 p × t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η 1 q × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η 1 q .
From inequality (17), we have
A q K ( σ ( s ) ) B q L ( σ ( t ) ) ( K L ) q ( σ ( s ) t 0 ) q p ( σ ( t ) t 0 ) q p × t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η .
Using Lemma 3 in (17) and (18) gives
A K ( σ ( s ) ) B L ( σ ( t ) ) K L h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) 1 p × t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η 1 q × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η 1 q ,
A q K ( σ ( s ) ) B q L ( σ ( t ) ) ( K L ) q h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) q p × t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η .
Using Lemma 6 in (19) and (20) gives
A K ( σ ( s ) ) B L ( σ ( t ) ) K L | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p × t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η 1 q × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η 1 q ,
A q K ( σ ( s ) ) B q L ( σ ( t ) ) ( K L ) q | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p × t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η .
Dividing both sides of (21) and (22) by | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p and | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p respectively, we get that
A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p K L t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η 1 q × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η 1 q ,
A q K ( σ ( s ) ) B q L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p ( K L ) q t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η × t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η .
From (23) by using Lemma 1 we obtain
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t K L ( x t 0 ) 1 p ( y t 0 ) 1 p t 0 x t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η Δ s ) 1 q × t 0 y t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η Δ t ) 1 q .
From (24), we get
t 0 x t 0 y A q K ( σ ( s ) ) B q L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p Δ s Δ t ( K L ) q t 0 x t 0 σ ( s ) a ( η ) A K 1 ( σ ( η ) q Δ η Δ s ) × t 0 y t 0 σ ( t ) b ( η ) B L 1 ( σ ( η ) q Δ η Δ t ) .
Applying Fubini’s Theorem on (25) and (26) gives
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t K L ( x t 0 ) 1 p ( y t 0 ) 1 p t 0 x ( x σ ( s ) ) a ( s ) A K 1 ( σ ( s ) q Δ s 1 q × t 0 y ( y σ ( t ) ) b ( t ) B L 1 ( σ ( t ) q Δ t 1 q ,
t 0 x t 0 y A q K ( σ ( s ) ) B q L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p Δ s Δ t ( K L ) q t 0 x ( x σ ( s ) ) a ( s ) A K 1 ( σ ( s ) q Δ t ) × t 0 y ( y σ ( t ) ) b ( t ) B L 1 ( σ ( t ) q Δ t .
Using the facts σ ( x ) x , σ ( y ) y yields
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t C 2 ( L , K , p ) t 0 x ( σ ( x ) σ ( s ) ) a ( s ) A K 1 ( σ ( s ) q Δ s 1 q × t 0 y ( σ ( y ) σ ( t ) ) b ( t ) B L 1 ( σ ( t ) q Δ t 1 q ,
t 0 x t 0 y A q K ( σ ( s ) ) B q L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p Δ s Δ t C 1 ( L , K , q ) t 0 x ( σ ( x ) σ ( s ) ) a ( s ) A K 1 ( σ ( s ) 2 Δ s × t 0 y ( σ ( y ) σ ( t ) ) b ( t ) B L 1 ( σ ( t ) q Δ t .
This completes the proof. ☐
Theorem 3.
Let a ( τ ˇ ) , b ( η ) , A ( s ) and B ( t ) be defined as in Theorem 2, thus
t 0 x t 0 y A q ( σ ( s ) ) B q ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p Δ s Δ t t 0 x ( σ ( x ) σ ( s ) ) a q ( s ) Δ s t 0 y ( σ ( y ) σ ( t ) ) b q ( t ) Δ t
and
t 0 x t 0 y A ( σ ( s ) ) B ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t ( x t 0 ) 1 p ( y t 0 ) 1 p t 0 x ( σ ( x ) σ ( s ) ) a q ( s ) Δ s 1 q t 0 y ( σ ( y ) σ ( t ) ) b q ( t ) Δ t 1 q .
Proof. 
Put K = L = 1 in (11) and (12). This completes the proof. ☐
In Theorem 2, if we choose T = R , then we have relation (1) and the next results:
Corollary 2.
If a ( s ) 0 , b ( t ) 0 . Define A ( s ) : = 0 s a ( η ) d η and B ( t ) : = 0 t b ( η ) d η , then
0 x 0 y A q K ( s ) B q L ( t ) | h ( s ) | 1 2 β + | h * ( t ) | 1 2 β 2 q α p d s d t C 1 ( L , K , q ) 0 x ( x s ) a ( s ) A K 1 ( s ) q d s × 0 y ( y t ) b ( t ) B L 1 ( t ) q d t .
and
0 x 0 y A K ( s ) B L ( t ) | h ( s ) | 1 2 β + | h * ( t ) | 1 2 β 2 α p d s d t C 3 ( L , K , p ) 0 x ( x s ) A K 1 ( s ) a ( s ) q d s 1 q × ( 0 y ( y t ) B L 1 ( t ) b ( t ) q d t ) 1 q
where
C 3 ( L , K , p ) = K L ( x y ) 1 p .
In Theorem 2, if we chose T = Z , then we get (2), and the next result:
Corollary 3.
If a ( n ) and b ( m ) 0 . Define
A ( n ) = s = 0 n a ( s ) , B ( m ) = k = 0 m b ( k ) .
Then
n = 1 N m = 1 M A q L ( n ) B q K ( m ) | h ( n + 1 ) | 1 2 β + | h * ( m + 1 ) | 1 2 β 2 q α p C 1 ( K , L , q ) n = 1 N ( N + 1 ( n + 1 ) ) ( a ( n ) A L 1 ( n ) ) q × m = 1 M ( M + 1 ( m + 1 ) ) ( b ( m ) B L 1 ( m ) ) q
and
n = 1 N m = 1 M A L ( n ) B K ( m ) | h ( n + 1 ) | 1 2 β + | h * ( m + 1 ) | 1 2 β 2 α p C 4 ( K , L , p ) n = 1 N ( N + 1 ( n + 1 ) ) ( a ( n ) A L 1 ( n ) ) q 1 q × m = 1 M ( M + 1 ( m + 1 ) ) ( b ( m ) B L 1 ( m ) ) q 1 q
where
C 4 ( K , L , p ) = K L ( N M ) 1 p .
Remark 1.
Taking p = q = 2 in Corollary 3 gives the result due to Hamiaz and Abuelela ([1], Theorem 3).
Corollary 4.
With the hypotheses of Theorem 2 we have:
t 0 x t 0 y A q K ( σ ( s ) ) B q L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 q α p Δ s Δ t C 1 ( L , K , q ) { h t 0 x ( σ ( x ) σ ( s ) ) a ( s ) A K 1 ( σ ( s ) q Δ s + h * t 0 y ( σ ( y ) σ ( t ) ) b ( t ) B L 1 ( σ ( t ) q Δ t }
and
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t C 2 ( L , K , p ) { h t 0 x ( σ ( x ) σ ( s ) ) A K 1 ( σ ( s ) ) a ( s ) q Δ s + h * t 0 y ( σ ( y ) σ ( t ) ) B L 1 ( σ ( t ) ) b ( t ) q Δ t ) 1 q .
Proof. 
Using the Fenchel-Young inequality (5) in (11) and (12). This proves the claim. ☐
Theorem 4.
Assuming the time scale T with s , t , t 0 , x , y T , A ( s ) and B ( t ) are defined as in Theorem 2. Suppose f ( τ ˇ ) 0 and g ( η ) 0 are right-dense continuous functions on [ t 0 , x ] T and [ t 0 , y ] T respectively. Suppose that Φ ˘ 0 and Ψ ˘ 0 are convex, and submultiplicative functions on [ 0 , ) . Furthermore assume that
F ( s ) : = t 0 s f ( τ ˇ ) Δ τ ˇ , a n d G ( t ) : = t 0 t g ( η ) Δ η ,
then for σ ( s ) [ t 0 , x ] T and σ ( t ) [ t 0 , y ] T , we have that
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t M 1 ( p ) t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) q Δ s 1 q × ( t 0 y ( σ ( y ) σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) q Δ t ) 1 q
where
M 1 ( p ) = t 0 x Φ ˘ ( F σ ( s ) ) F σ ( s ) p Δ s 1 p t 0 y Ψ ˘ ( G σ ( t ) ) G σ ( t ) p Δ t 1 p .
Proof. 
From the properties of Φ ˘ and using (2), we obtain
Φ ˘ ( A σ ( s ) ) = Φ ˘ F ( σ ( s ) ) t 0 σ ( s ) f ( τ ˇ ) a ( τ ˇ ) f ( τ ˇ ) Δ τ ˇ t 0 σ ( s ) f ( τ ˇ ) Δ τ ˇ Φ ˘ ( F ( σ ( s ) ) Φ ˘ t 0 σ ( s ) f ( τ ˇ ) a ( τ ˇ ) f ( τ ˇ ) Δ τ ˇ t 0 σ ( s ) f ( τ ˇ ) Δ τ ˇ Φ ˘ ( F ( σ ( s ) ) F ( σ ( s ) ) t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) Δ τ ˇ .
Using (1) in (29), we see that
Φ ˘ ( A σ ( s ) ) Φ ˘ ( F σ ( s ) ) F σ ( s ) ( σ ( s ) t 0 ) 1 p t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) q Δ τ ˇ 1 q .
In addition, from the convexity and submultiplicative property of Ψ ˘ , we get by using (2) and (1):
Ψ ˘ ( B σ ( t ) ) Ψ ˘ ( G σ ( t ) ) G σ ( t ) ( σ ( t ) t 0 ) 1 p t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) q Δ η 1 q .
From (30) and (31), we have
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) ( σ ( s ) t 0 ) 1 p ( σ ( t ) t 0 ) 1 p ( Φ ˘ ( F σ ( s ) ) F σ ( s ) t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) q Δ τ ˇ 1 q × ( Ψ ˘ ( G σ ( t ) ) G σ ( t ) t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) q Δ η 1 q
Using (5) on ( σ ( s ) t 0 ) 1 p ( σ ( t ) t 0 ) 1 p gives:
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) 1 p ( Φ ˘ ( F σ ( s ) ) F σ ( s ) t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) q Δ τ ˇ 1 q × ( Ψ ˘ ( G σ ( t ) ) G σ ( t ) t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) q Δ η 1 q .
Applying Lemma 6 on the right hand side of (33), we see that
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p × ( Φ ˘ ( F σ ( s ) ) F σ ( s ) t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) q Δ τ ˇ 1 q × ( Ψ ˘ ( G σ ( t ) ) G σ ( t ) t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) q Δ η 1 q .
From (34), we have
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p ( Φ ˘ ( F σ ( s ) ) F σ ( s ) t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) q Δ τ ˇ 1 q × ( Ψ ˘ ( G σ ( t ) ) G σ ( t ) t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) q Δ η 1 q .
From (35), we obtain
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t t 0 x ( Φ ˘ ( F σ ( s ) ) F σ ( s ) t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) q Δ τ ˇ 1 q Δ s × t 0 y ( Ψ ˘ ( G σ ( t ) ) G σ ( t ) t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) q Δ η 1 q Δ t .
From (36), by using (1), we have
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t t 0 x Φ ˘ ( F σ ( s ) ) F σ ( s ) p Δ s 1 p t 0 x t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) q Δ τ ˇ Δ s 1 q × t 0 y Ψ ˘ ( G σ ( t ) ) G σ ( t ) p Δ t 1 p t 0 y t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) q Δ η Δ t 1 q .
From (37), by using (5), we obtain
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t M 1 ( p ) t 0 x ( x σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) q Δ s 1 q × t 0 y ( y σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) q Δ t 1 q .
By using the facts σ ( x ) x and σ ( y ) y , we obtain
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t M 1 ( p ) t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) q Δ s 1 q × t 0 y ( σ ( y ) σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) q Δ t 1 q
where
M 1 ( p ) = t 0 x Φ ˘ ( F σ ( s ) ) F σ ( s ) p Δ s 1 p t 0 y Ψ ˘ ( G σ ( t ) ) G σ ( t ) p Δ t 1 p .
This completes the proof. ☐
In Theorem 4, taking T = R , we have (1) and the result:
Corollary 5.
Assume that a ( s ) 0 , b ( t ) 0 , f ( τ ˇ ) 0 and g ( η ) 0 , we define
A ( s ) : = 0 s a ( η ) d η , B ( t ) : = 0 t b ( η ) d η , F ( s ) : = 0 s f ( τ ˇ ) d τ ˇ , a n d G ( t ) : = 0 t g ( η ) d η .
Then
0 x 0 y Φ ˘ ( A ( s ) Ψ ˘ ( B ( t ) ) | h ( s ) | 1 2 β + | h * ( t ) | 1 2 β 2 α p d s d t M 2 ( p ) 0 x ( x s ) f ( s ) Φ ˘ a ( s ) f ( s ) q d s 1 q × ( 0 y ( y t ) g ( t ) Ψ ˘ b ( t ) g ( t ) q d t ) 1 q
where
M 2 ( p ) = 0 x Φ ˘ ( F ( s ) ) F ( s ) p d s 1 p 0 y Ψ ˘ ( G ( t ) ) G ( t ) p d t 1 p .
In Theorem 4, taking T = Z , gives (2) and the result:
Corollary 6.
Assume that a ( n ) 0 , b ( m ) 0 , f ( n ) 0 , g ( m ) 0 are sequences of real numbers. Define
A ( n ) = s = 0 n a ( s ) , B ( m ) = k = 0 m b ( k ) , F ( n ) = s = 0 n f ( s ) and G ( m ) = k = 0 m g ( k ) .
Then
n = 1 N m = 1 M Φ ˘ ( A ( n ) ) Ψ ˘ ( B ( m ) ) | h ( n + 1 ) | 1 2 β + | h * ( m + 1 ) | 1 2 β 2 α p M 3 ( p ) n = 1 N ( N + 1 ( n + 1 ) ) f ( n ) Φ ˘ a ( n ) f ( n ) q 1 q × m = 1 M ( M + 1 ( m + 1 ) ) g ( m ) Ψ ˘ b ( m ) g ( m ) q 1 q
where
M 3 ( p ) = n = 1 N Φ ˘ ( F ( n ) F ( n ) p 1 p m = 1 M Ψ ˘ ( G ( m ) G ( m ) p 1 p
Remark 2.
In Corollary 6, if p = q = 2 we get the result due to Hamiaz and Abuelela ([1], Theorem 5).
Corollary 7.
Under the hypotheses of Theorem 4 the following inequality hold:
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t M 1 ( p ) [ h t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) q Δ s + h * t 0 y ( σ ( y ) σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) q Δ t ) 1 q .
Proof. 
Using (5) in (28). This proves our claim. ☐
Lemma 7.
With hypotheses of Theorem 4, we get:
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) 2 Ψ ˘ ( B σ ( t ) ) 2 h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) Δ s Δ t M 4 t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) 4 Δ s 1 2 t 0 y ( σ ( t ) σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) 4 Δ t 1 2
where
M 4 = t 0 x Φ ˘ ( F σ ( s ) ) 4 ( F σ ( s ) ) 4 ( σ ( s ) t 0 ) Δ s 1 2 t 0 y Ψ ˘ ( G σ ( t ) ) 4 ( G σ ( t ) ) 4 ( σ ( t ) t 0 ) Δ t 1 2 .
Proof. 
From (30) and (31) and by using Fenchel-Young inequality with p = q = 2 we have
Φ ˘ ( A σ ( s ) ) 2 Ψ ˘ ( B σ ( t ) ) 2 h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) ( Φ ˘ ( F σ ( s ) ) 2 ( F σ ( s ) ) 2 t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) 2 Δ τ ˇ × ( Ψ ˘ ( G σ ( t ) ) 2 ( G σ ( t ) ) 2 t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) 2 Δ η .
From (40), by using (1) with p = q = 2 , we obtain
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) 2 Ψ ˘ ( B σ ( t ) ) 2 h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) Δ s Δ t t 0 x ( Φ ˘ ( F σ ( s ) ) 2 ( F σ ( s ) ) 2 t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) 2 Δ τ ˇ Δ s × t 0 y Ψ ˘ ( G σ ( t ) ) 2 ( G σ ( t ) ) 2 t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) 2 Δ η Δ t t 0 x Φ ˘ ( F σ ( s ) ) 2 ( F σ ( s ) ) 2 ( σ ( s ) t 0 ) 1 2 t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) 4 Δ τ ˇ 1 2 ) Δ s × t 0 y Ψ ˘ ( G σ ( t ) ) 2 ( G σ ( t ) ) 2 ( σ ( t ) t 0 ) 1 2 t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) 4 Δ η 1 2 Δ t t 0 x Φ ˘ ( F σ ( s ) ) 4 ( F σ ( s ) ) 4 ( σ ( s ) t 0 ) Δ s 1 2 t 0 x t 0 σ ( s ) f ( τ ˇ ) Φ ˘ a ( τ ˇ ) f ( τ ˇ ) 4 Δ τ ˇ Δ s 1 2 × t 0 y Ψ ˘ ( G σ ( t ) ) 4 ( G σ ( t ) ) 4 ( σ ( t ) t 0 ) Δ t 1 2 t 0 y t 0 σ ( t ) g ( η ) Ψ ˘ b ( η ) g ( η ) 4 Δ η Δ t 1 2 .
Applying (5) on (41), we obtain
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) 2 Ψ ˘ ( B σ ( t ) ) 2 h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) Δ s Δ t t 0 x Φ ˘ ( F σ ( s ) ) 4 ( F σ ( s ) ) 4 ( σ ( s ) t 0 ) Δ s 1 2 t 0 x ( x σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) 4 Δ s 1 2 × t 0 y Ψ ˘ ( G σ ( t ) ) 4 ( G σ ( t ) ) 4 ( σ ( t ) t 0 ) Δ t 1 2 t 0 y ( t σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) 4 Δ t 1 2 = M 4 t 0 x ( x σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) 4 Δ s 1 2 t 0 y ( t σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) 4 Δ t 1 2 .
Since σ ( x ) x and σ ( y ) y , from the last inequality above, we have
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) 2 Ψ ˘ ( B σ ( t ) ) 2 h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) Δ s Δ t M 4 t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ a ( s ) f ( s ) 4 Δ s 1 2 t 0 y ( σ ( t ) σ ( t ) ) g ( t ) Ψ ˘ b ( t ) g ( t ) 4 Δ t 1 2
where M 4 defined as in (39). This proves our claim. ☐
Theorem 5.
Assume the time scale T with t , s , x 0 , t 0 , y T . Suppose that b ( τ ˇ ) 0 and a ( τ ˇ ) 0 are right-dense continuous functions on [ t 0 , y ] T and [ t 0 , x ] T . Let G , F , g , f , Ψ ˘ and Φ ˘ be as assumed in Theorem 4. Furthermore assume that
A ( s ) : = 1 F ( s ) t 0 s a ( τ ˇ ) f ( τ ˇ ) Δ τ ˇ , a n d B ( t ) : = 1 G ( t ) t 0 t b ( η ) g ( η ) Δ η ,
then for σ ( s ) [ t 0 , x ] T and σ ( t ) [ t 0 , y ] T , we have that
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) F σ ( s ) G σ ( t ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t M 5 ( p ) t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ a ( s ) q Δ s 1 q × ( t 0 y ( σ ( y ) σ ( t ) ) g ( t ) Ψ ˘ b ( t ) q Δ t ) 1 q
where
M 5 ( p ) = ( x t 0 ) 1 p ( y t 0 ) 1 p .
Proof. 
From (42), we see that
Φ ˘ ( A σ ( s ) ) = Φ ˘ 1 F σ ( s ) t 0 σ ( s ) f ( τ ˇ ) a ( τ ˇ ) Δ τ ˇ .
Applying (1) on (45), we obtain
Φ ˘ ( A σ ( s ) ) ( σ ( s ) t 0 ) 1 p F σ ( s ) t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ 1 q .
From (46), we get
Φ ˘ ( A σ ( s ) ) F σ ( s ) ( σ ( s ) t 0 ) 1 p t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ 1 q .
Similarly, we obtain
Ψ ˘ ( B σ ( t ) ) G σ ( t ) ( σ ( t ) t 0 ) 1 p t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η 1 q .
From (47) and (48), we observe that
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) ( σ ( s ) t 0 ) 1 p ( σ ( t ) t 0 ) 1 p × t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ 1 q t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η 1 q .
Applying the Lemma 3 on the term ( σ ( s ) t 0 ) 1 p ( σ ( t ) t 0 ) 1 p , gives:
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) h ( σ ( s ) t 0 ) + h * ( σ ( t ) t 0 ) 1 p t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ 1 q × t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η 1 q .
From 6 and (50), we obtain
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p × t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ 1 q t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η 1 q .
Dividing both sides of (51) by | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p , we get
Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ 1 q × t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η 1 q .
Taking delta-integral for (52), yields:
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t t 0 x t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ 1 q Δ s t 0 y t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η 1 q Δ t .
Using (1) in (53), yield:
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t ( x t 0 ) 1 p ( y t 0 ) 1 p t 0 x t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ Δ s 1 q × t 0 y t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η Δ t 1 q = M 5 ( p ) t 0 x t 0 σ ( s ) f ( τ ˇ ) Φ ˘ [ a ( τ ˇ ) ] q Δ τ ˇ Δ s 1 q × t 0 y t 0 σ ( t ) g ( η ) Ψ ˘ [ b ( η ) ] q Δ η Δ t 1 q ,
where M 5 defined as in (44). From (5) and (54), we get:
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t = M 5 ( p ) t 0 x ( x σ ( s ) ) f ( s ) Φ ˘ [ a ( s ) ] q Δ s 1 q × t 0 y ( y σ ( t ) ) g ( t ) Ψ ˘ [ b ( t ) ] q Δ t 1 q .
By using the fact σ ( x ) x and σ ( y ) y , we obtain
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) G σ ( t ) F σ ( s ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t = M 5 ( p ) t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ [ a ( s ) ] q Δ s 1 q × t 0 y ( σ ( y ) σ ( t ) ) g ( t ) Ψ ˘ [ b ( t ) ] q Δ t 1 q .
This completes the proof. ☐
Taking T = R in Theorem 5 with relation (1), we have:
Corollary 8.
Assume g ( t ) 0 , b ( t ) 0 , f ( s ) 0 , a ( s ) 0 . Define
A ( s ) : = 1 F ( s ) 0 s f ( τ ˇ ) a ( τ ˇ ) d τ ˇ and B ( t ) : = 1 G ( t ) 0 t g ( τ ˇ ) b ( τ ˇ ) d τ ˇ ,
F ( s ) : = 0 s f ( τ ˇ ) d τ ˇ and G ( t ) : = 0 t g ( τ ˇ ) d τ ˇ .
Then
0 x 0 y Φ ˘ ( A ( s ) ) Ψ ˘ ( B ( t ) ) F ( s ) G ( t ) | h ( s ) | 1 2 β + | h * ( t ) | 1 2 β 2 α p d s d t M 6 ( p ) 0 x ( x s ) f ( s ) Φ ˘ a ( s ) q d s 1 q . × ( 0 y ( y t ) g ( t ) Ψ ˘ b ( t ) q d t ) 1 q
where
M 6 ( p ) = ( x ) 1 p ( y ) 1 p
Taking T = Z in Theorem 5 with relation (2), gives:
Corollary 9.
Assume g ( n ) 0 , b ( n ) 0 , f ( n ) 0 , a ( n ) 0 . Define
A ( n ) : = 1 F ( n ) s = 0 n f ( s ) a ( s ) and B ( m ) : = 1 G ( m ) k = 0 m g ( k ) b ( k ) .
F ( n ) : = s = 0 n f ( s ) and G ( m ) : = k = 0 m g ( k ) .
Then
n = 1 N m = 1 M Φ ˘ ( A ( n ) ) Ψ ˘ ( B ( m ) ) F ( n ) G ( m ) | h ( n + 1 ) | 1 2 β + | h * ( m + 1 ) | 1 2 β 2 α p M 7 ( p ) n = 1 N ( N + 1 ( n + 1 ) ) f ( n ) Φ ˘ a ( n ) q 1 q . × ( m = 1 M ( M + 1 ( m + 1 ) ) g ( m ) Ψ ˘ b ( m ) q ) 1 q
where
M 7 ( p ) = ( N M ) 1 p .
Remark 3.
In Corollary 9, if p = q = 2 we get the result due to Hamiaz and Abuelela ([1], Theorem 7).
Corollary 10.
With the hypotheses of Theorem 5, we get:
t 0 x t 0 y Φ ˘ ( A σ ( s ) ) Ψ ˘ ( B σ ( t ) ) F σ ( s ) G σ ( t ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t M 5 ( p ) { h t 0 x ( σ ( x ) σ ( s ) ) f ( s ) Φ ˘ a ( s ) q Δ s + h * t 0 y ( σ ( y ) σ ( t ) ) g ( t ) Ψ ˘ b ( t ) q Δ t ) 1 q .
where M 5 defined as in (44).
Proof. 
We apply the Fenchel-Young inequality (5) in (43). This completes the proof. ☐

3. Some Applications

We can apply our inequalities to obtain different formulas of Hilbert-type inequalities by suggesting h * ( y ) and h ( x ) by some functions:
In (12), as a special case, if we take h ( x ) = x 2 2 , we have h * ( x ) = x 2 2 see [12], we get
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t = t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) ( σ ( s ) t 0 ) ) 1 β + ( σ ( t ) t 0 ) 1 β 2 α p Δ s Δ t 1 2 α p β C 2 ( L , K , p ) t 0 x ( σ ( x ) σ ( s ) ) A K 1 ( σ ( s ) ) a ( s ) q Δ s 1 q × t 0 y ( σ ( y ) σ ( t ) ) B L 1 ( σ ( t ) ) b ( t ) q Δ t 1 q ,
where C 2 ( L , K , p ) defined as in Theorem 2. Consequently, for α = β = 1 , inequality (55) produces
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) ( σ ( s ) t 0 ) ) + ( σ ( t ) t 0 ) 2 p Δ s Δ t 1 2 1 p C 2 ( L , K , p ) t 0 x ( σ ( x ) σ ( s ) ) A K 1 ( σ ( s ) ) a ( s ) q Δ s 1 q × ( t 0 y ( σ ( y ) σ ( t ) ) B L 1 ( σ ( t ) ) b ( t ) q Δ t ) 1 q .
On the other hand if we take h ( n ) = n r r , r > 1 , then h * ( m ) = m k k where 1 r + 1 k = 1 and n , m R + , then (12) gives
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) | h ( σ ( s ) t 0 ) | 1 2 β + | h * ( σ ( t ) t 0 ) | 1 2 β 2 α p Δ s Δ t = t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) ( k ( σ ( s ) t 0 ) r ) 1 2 β + ( r ( σ ( t ) t 0 ) k ) 1 2 β 2 α p Δ s Δ t 1 r k α p β C 2 ( L , K , p ) t 0 x ( σ ( x ) σ ( s ) ) A K 1 ( σ ( s ) ) a ( s ) q Δ s 1 q × ( t 0 y ( σ ( y ) σ ( t ) ) B L 1 ( σ ( t ) ) b ( t ) q Δ t ) 1 q .
Clearly, when β = 1 2 α , the inequality (57) becomes
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) ( k ( σ ( s ) t 0 ) r ) α + ( r ( σ ( t ) t 0 ) k ) α 2 α p Δ s Δ t 1 r k 2 α 2 p C 2 ( L , K , p ) t 0 x ( σ ( x ) σ ( s ) ) A K 1 ( σ ( s ) ) a ( s ) q Δ s 1 q × ( t 0 y ( σ ( y ) σ ( t ) ) B L 1 ( σ ( t ) ) b ( t ) q Δ t ) 1 q .
If β = α = 1 . From (57), we get
t 0 x t 0 y A K ( σ ( s ) ) B L ( σ ( t ) ) ( k ( σ ( s ) t 0 ) r ) 1 2 + ( r ( σ ( t ) t 0 ) k ) 1 2 2 p Δ s Δ t 1 r k 1 p C 2 ( L , K , p ) t 0 x ( σ ( x ) σ ( s ) ) A K 1 ( σ ( s ) ) a ( s ) q Δ s 1 q × ( t 0 y ( σ ( y ) σ ( t ) ) B L 1 ( σ ( t ) ) b ( t ) q Δ t ) 1 q .

4. Conclusions

In this paper, with the help of a Fenchel-Legendre transform, which is used in various problems involving symmetry, we generalized a number of Hilbert-type inequalities to a general time scale. Besides that, in order to obtain some new inequalities as special cases, we also extended our inequalities to discrete and continuous calculus. In the future, we can generalize these inequalities in a different way by using other mathematical tools.

Author Contributions

All authors have contributed equally. All authors have read and agreed to the published version of the manuscript.

Funding

The authors declare that they have received no funding from any funding body.

Conflicts of Interest

The authors declare that they have no competing interests.

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El-Deeb, A.A.; Makharesh, S.D.; Baleanu, D. Dynamic Hilbert-Type Inequalities with Fenchel-Legendre Transform. Symmetry 2020, 12, 582. https://doi.org/10.3390/sym12040582

AMA Style

El-Deeb AA, Makharesh SD, Baleanu D. Dynamic Hilbert-Type Inequalities with Fenchel-Legendre Transform. Symmetry. 2020; 12(4):582. https://doi.org/10.3390/sym12040582

Chicago/Turabian Style

El-Deeb, Ahmed A., Samer D. Makharesh, and Dumitru Baleanu. 2020. "Dynamic Hilbert-Type Inequalities with Fenchel-Legendre Transform" Symmetry 12, no. 4: 582. https://doi.org/10.3390/sym12040582

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