On Generalization of Some Inequalities of Chebyshevs Functional Using Generalized Katugampola Fractional Integral
aa r X i v : . [ m a t h . G M ] S e p ON GENERALIZATION OF SOME INEQUALITIES OF CHEBYSHEV’SFUNCTIONAL USING GENERALIZED KATUGAMPOLA FRACTIONALINTEGRALTariq A. Aljaaidi, Deepak B. Pachpatte
Abstract
In this paper we obtain a generalization of some integral inequalities re-lated to Chebyshev‘s functional by using a generalized Katugampola fractionalintegral.
Keywords : Chebyshev’s functional; generalized fractional integral; generalizedfractional derivative. INTRODUCTION:
In (1882), Chebyshev [6] has given the following functional T ( ϕ, ψ ) := 1 b − a Z ba ϕ ( γ ) ψ ( γ ) dγ − b − a (cid:18)Z ba ϕ ( γ ) dγ (cid:19) (cid:18)Z ba ψ ( γ ) dγ (cid:19) (1.1)and its extention is obtained by Mitrinovic as (see [12]) T ( ϕ, ψ, g, h ) : = Z ba h ( γ ) dγ Z ba ϕ ( γ ) ψ ( γ ) g ( γ ) dγ + Z ba g ( γ ) dγ Z ba ϕ ( γ ) ψ ( γ ) h ( γ ) dγ − (cid:18)Z ba ϕ ( γ ) h ( γ ) dγ (cid:19) (cid:18)Z ba ψ ( γ ) g ( γ ) dγ (cid:19) (1.2) − (cid:18)Z ba ϕ ( γ ) g ( γ ) dx (cid:19) (cid:18)Z ba ψ ( γ ) h ( γ ) dγ (cid:19) , where ϕ and ψ are two integrable functions on [ a, b ] . Many researchers have givenconsiderable attention to the both functionals and a number of inequalities and anumber of extensions, generalizations and variants have appeared in the literature,1or more details see ([7], [8], [9], [10], [11]). The functional (1.2) is a Chebyshevextension of the functional (1.1) If ϕ and ψ satisfies the following condition( ϕ ( τ ) − ϕ ( γ )) ( ψ ( τ ) − ψ ( γ )) ≥ , for any τ , γ ∈ [ a, b ] , then ϕ and ψ are synchronous on [ a, b ], moreover, T ( ϕ, ψ, g, h ) ≥ ϕ and ψ are asynchronous on [ a, b ] , namely ( ϕ ( τ ) − ϕ ( γ )) ( ψ ( τ ) − ψ ( γ )) ≤ , forany τ , γ ∈ [ a, b ] . One of the most important things to note in this work is the celebrated Chebyshevfunctional [6] T ( ϕ, ψ, g ) := Z ba g ( γ ) dγ Z ba ϕ ( γ ) ψ ( γ ) g ( γ ) dγ + Z ba g ( γ ) ϕ ( γ ) dγ Z ba ψ ( γ ) g ( γ ) dγ (1.3)where ϕ and ψ are two differentiable functions on [ a, b ] , and p is positive and inte-grable functions on [ a, b ] , this functional has many applications in probability andstatistics and other fields.In [13], (see also [14]) Dragomir proved that, if ϕ and ψ are two differentiablefunctions such that ϕ ′ ∈ L s ( a, b ) , ψ ′ ∈ L v ( a, b ) , r > , s + v = 1 , then2 | T ( ϕ, ψ, g ) | ≤ k ϕ ′ k s k ψ ′ k v (cid:20)Z ba | τ − γ | g ( τ ) g ( γ ) dτ dγ (cid:21) . (1.4)For inequality (1.4), Dahmani in [15], (see also [16], [17]) proved fractional versionof the inequality | I α g ( x ) I α ( gϕψ ) ( x ) − I α ( gϕ ) ( x ) I α ( gψ ) ( x ) |≤ k ϕ ′ k s k ψ ′ k v Γ ( α ) Z ba Z ba ( x − τ ) α − ( x − γ ) α − | τ − γ | g ( τ ) g ( γ ) dτ dγ ≤ k ϕ ′ k s k ψ ′ k v x ( I α g ( x )) , (1.5)for all α > , x > . Motivated by ([15], [16], [17]), in this paper we establish some new fractionalinequalities for Chebyshev functional involving generalized Katugampola fractionalintegral. Our results in this paper are organized in three sections, first and secondsections related to Chebyshev functional in case of synchronous functions and thethird section related to Chebyshev functional in case of differentiable functions whosederivatives belong to L p ([0 , ∞ ]). 2 PRELIMINARIES:
Now, in this section we give the necessary notation and basic definitions used in oursubsequent discussion. For more details see ( [1], [2], [3] ).
Definition 2.1
Consider the space X pc ( a, b ) ( c ∈ R , ≤ p ≤ ∞ ) , of those complexvalued Lebesgue measurable functions ϕ on ( a, b ) for which the norm k ϕ k X pc < ∞ ,such that k ϕ k X pc = b Z x | x c ϕ | p dxx p (1 ≤ p < ∞ ) and k ϕ k X ∞ c = sup ess x ∈ ( a,b ) [ x c | ϕ | ] . In particular, when c = 1 /p, the space X pc ( a, b ) coincides with the space L p ( a, b ) . Definition 2.2
The left and right-sided fractional integrals of a function ϕ where ϕ ∈ X pc ( a, b ) , α > , and β, ρ, η, k ∈ R , are defined respectively by ρ I α,βa + ,η,k ϕ ( x ) = ρ − β x k Γ ( α ) x Z a τ ρ ( η +1) − ( x ρ − τ ρ ) − α ϕ ( τ ) dτ , ≤ a < x < b ≤ ∞ (2.1) and ρ I α,βb − ,η,k ϕ ( x ) = ρ − β x ρη Γ ( α ) b Z x τ k + ρ − ( τ ρ − x ρ ) − α ϕ ( τ ) dτ , ≤ a < x < b ≤ ∞ , (2.2) if the integral exist. In this paper we will use the left-sided fractional integrals (2.1), to present anddiscuss our new results, also we consider a = 0 , in (2.1), to obtain ρ I α,βη,k ϕ ( x ) = ρ − β x k Γ ( α ) x Z τ ρ ( η +1) − ( x ρ − τ ρ ) − α ϕ ( τ ) dτ . Now we define the following function as in [3]: let x > , α > , ρ, k, β, η ∈ R ,then Λ ρ,βx,k ( α, η ) = Γ ( η + 1)Γ ( η + α + 1) ρ − β x k + ρ ( η + α ) . η = 0 , a = 0 , k = 0 , and taking the limit ρ → , [[4], p. 79].- Weyl fractional integral, if η = 0 , a = −∞ , k = 0 , and taking the limit ρ → , [[5], p. 50].-Riemann-Liouville fractional integral, if η = 0 , k = 0 , and taking the limit ρ → , [[4], p. 69].-Katugampola fractional integral, if β = α, k = 0 , η = 0 , [[1]].-Erd´elyi-Kober fractional integral, if β = 0 , k = − ρ ( α + η ) , [ [4], p.105].-Hadamard fractional integral, if β = α, k = 0 , η = 0 + , and taking the limit ρ → , [[4], p. 110]. Generalized fractional inequality for Chebyshev’s functional:
In this section we establish inequality for Chebyshev functional [6], deals withsame parameters
Theorem 3.1
Let ϕ and ψ be two integrable and synchronous functions on [0 , ∞ ) . Then for all x > , α > , ρ > , k, β, η ∈ R we have: ρ I α,βη,k ( ϕψ ) ( x ) ≥ ρ,βx,k ( α, η ) ρ I α,βη,k ϕ ( x ) ρ I α,βη,k ψ ( x ) . (3.1) Proof.
For a synchronous functions ϕ, ψ on [0 , ∞ ) , we have for all τ ≥ , γ ≥ ϕ ( τ ) − ϕ ( γ )) ( ψ ( τ ) − ψ ( γ )) ≥ . Therefore ϕ ( τ ) ψ ( τ ) + ϕ ( γ ) ψ ( γ ) ≥ ϕ ( τ ) ψ ( γ ) + ϕ ( γ ) ψ ( τ ) . (3.2)Multiplying both sides of (3.2) by ρ − β x k Γ( α ) τ ρ ( η +1) − ( x ρ − τ ρ ) − α , where τ ∈ (0 , x ) , and integratingthe resulting inequality over (0 , x ) , with respect to the variable τ , we obtain: ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α ϕ ( τ ) ψ ( τ ) dτ + ϕ ( γ ) ψ ( γ ) ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α dτ ≥ ψ ( γ ) ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α ϕ ( τ ) dτ (3.3)+ ϕ ( γ ) ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α ψ ( τ ) dτ .
4o we have ρ I α,βη,k ( ϕψ ) ( x ) + Λ ρ,βx,k ( α, η ) ϕ ( γ ) ψ ( γ ) ≥ ψ ( γ ) ρ I α,βη,k ϕ ( x ) + ϕ ( γ ) ρ I α,βη,k ψ ( x ) . (3.4)Now multiplying both sides of (3.4) by ρ − β x k Γ( α ) γ ρ ( η +1) − ( x ρ − γ ρ ) − α , where γ ∈ (0 , x ) , over (0 , x ) , Then integrating the resulting inequality over (0 , x ) , with respect to the variable γ, we get: ρ I α,βη,k ( ϕψ ) ( x ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α dγ +Λ ρ,βx,k ( α, η ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α ϕ ( γ ) ψ ( γ ) dγ ≥ ρ I α,βη,k ϕ ( x ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α ψ ( γ ) dγ + ρ I α,βη,k ψ ( x ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α ϕ ( γ ) dγ. So we have ρ I α,βη,k ( ϕψ ) ( x ) ρ I α,βη,k (1) + Λ ρ,βx,k ( α, η ) ρ I α,βη,k ( ϕψ ) ( x ) ≥ ρ I α,βη,k ϕ ( x ) ρ I α,βη,k ψ ( x ) + ρ I α,βη,k ψ ( x ) ρ I α,βη,k ϕ ( x ) . Hence ρ I α,βη,k ( ϕψ ) ( x ) ≥ ρ,βx,k ( α, η ) ρ I α,βη,k ϕ ( x ) ρ I α,βη,k ψ ( x ) . The result has proved.Our next theorem on Chebyshev functional deals with different fractional pa-rameters:
Theorem 3.2
Let ϕ and ψ be two integrable and synchronous functions on [0 , ∞ ) . Thenfor all x > , α > , δ > , ρ > , k, β, λ, η ∈ R , we have: Λ ρ,λx,k ( δ, η ) ρ I α,βη,k ( ϕψ ) ( x ) + Λ ρ,βx,k ( α, η ) ρ I δ,λη,k ( ϕψ ) ( x ) ≥ ρ I α,βη,k ϕ ( x ) ρ I δ,λη,k ψ ( x ) + ρ I α,βη,k ψ ( x ) ρ I δ,λη,k ϕ ( x ) . Proof.
Since ϕ and ψ are synchronous functions on [0 , ∞ ) , by similar argumentsas in the proof of Theorem (3.1) we can write ρ I α,βη,k ( ϕψ ) ( x ) + Λ ρ,βx,k ( α, η ) ϕ ( γ ) ψ ( γ ) ≥ ψ ( γ ) ρ I α,βη,k ϕ ( x ) + ϕ ( γ ) ρ I α,βη,k ψ ( x ) . (3.5)5ow multiplying both sides of (3.5) by ρ − λ x k Γ( δ ) γ ρ ( η +1) − ( x ρ − γ ρ ) − δ , where γ ∈ (0 , x ) , thenintegrating the resulting inequality over (0 , x ) , with respect to the variable γ, weobtain: ρ I α,βη,k ( ϕψ ) ( x ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ dγ +Λ ρ,βx,k ( α, η ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ ϕ ( γ ) ψ ( γ ) dγ ≥ ρ I α,βη,k ϕ ( x ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ ψ ( γ ) dγ + ρ I α,βη,k ψ ( x ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ ϕ ( γ ) dγ. (3.6)So we have Λ ρ,λx,k ( δ, η ) ρ I α,βη,k ( ϕψ ) ( x ) + Λ ρ,βx,k ( α, η ) ρ I δ,λη,k ( ϕψ ) ( x ) ≥ ρ I α,βη,k ϕ ( x ) ρ I δ,λη,k ψ ( x ) + ρ I α,βη,k ψ ( x ) ρ I δ,λη,k ϕ ( x ) . Hence the proof. Generalized fractional inequality for extended Chebyshev’s functional:
In this section, we consider the extended Chebyshev functional in case of syn-chronous functions (1.2). To prove our theorem in this section, we prove the followinglemma:
Lemma 4.1
Let ϕ and ψ be two integrable and synchronous functions on [0 , ∞ ) , suppose s, v : [0 , ∞ ) → [0 , ∞ ) , Then for all x > , α > , ρ > , k, β, η ∈ R we have: ρ I α,βη,k ( sϕψ ) ( x ) ρ I α,βη,k v ( x ) + ρ I α,βη,k s ( x ) ρ I α,βη,k ( vϕψ ) ( x ) ≥ ρ I α,βη,k ( sϕ ) ( x ) ρ I α,βη,k ( vψ ) ( x ) + ρ I α,βη,k ( sψ ) ( x ) ρ I α,βη,k ( vϕ ) ( x ) . Proof.
For a synchronous functions ϕ, ψ on [0 , ∞ ) , we have for all τ ≥ , γ ≥ ϕ ( τ ) − ϕ ( γ )) ( ψ ( τ ) − ψ ( γ )) ≥ . Therefore ϕ ( τ ) ψ ( τ ) + ϕ ( γ ) ψ ( γ ) ≥ ϕ ( τ ) ψ ( γ ) + ϕ ( γ ) ψ ( τ ) . (4.1)6ow multiplying both sides of (4.1) by ρ − β x k Γ( α ) τ ρ ( η +1) − s ( τ )( x ρ − τ ρ ) − α , where τ ∈ (0 , x ) , thenintegrating the resulting inequality over (0 , x ) , with respect to the variable τ , weget: ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α s ( τ ) ϕ ( τ ) ψ ( τ ) dτ + ϕ ( γ ) ψ ( γ ) ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α s ( τ ) dτ ≥ ψ ( γ ) ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α s ( τ ) ϕ ( τ ) dτ + ϕ ( γ ) ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α s ( τ ) ψ ( τ ) dτ . So we have ρ I α,βη,k ( sϕψ ) ( x ) + ϕ ( γ ) ψ ( γ ) ρ I α,βη,k s ( x ) ≥ ψ ( γ ) ρ I α,βη,k ( sϕ ) ( x ) + ϕ ( γ ) ρ I α,βη,k ( sψ ) ( x ) . (4.2)Now multiplying both sides of (4.2) by ρ − β x k Γ( α ) γ ρ ( η +1) − v ( γ )( x ρ − γ ρ ) − α , where γ ∈ (0 , x ) , andintegrating the resulting inequality over (0 , x ) , with respect to the variable γ, weobtain: ρ I α,βη,k ( sϕψ ) ( x ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α v ( γ ) dγ + ρ I α,βη,k s ( x ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α v ( γ ) ϕ ( γ ) ψ ( γ ) dγ ≥ ρ I α,βη,k ( sϕ ) ( x ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α v ( γ ) ψ ( γ ) dγ + ρ I α,βη,k ( sψ ) ( x ) ρ − β x k Γ ( α ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − α v ( γ ) ϕ ( γ ) dγ. Therefore ρ I α,βη,k ( sϕψ ) ( x ) ρ I α,βη,k v ( x ) + ρ I α,βη,k s ( x ) ρ I α,βη,k ( vϕψ ) ( x ) ≥ ρ I α,βη,k ( sϕ ) ( x ) ρ I α,βη,k ( vψ ) ( x ) + ρ I α,βη,k ( sψ ) ( x ) ρ I α,βη,k ( vϕ ) ( x ) . Hence the proof. 7 heorem 4.1
Let ϕ and ψ be two integrable and synchronous functions on [0 , ∞ ) , and suppose f, g, h : [0 , ∞ ) → [0 , ∞ ) . Then for all x > , α > , ρ > , k, β, η ∈ R we have: ρ I α,βη,k h ( x ) [ ρ I α,βη,k ( f ϕψ ) ( x ) ρ I α,βη,k g ( x ) + 2 ρ I α,βη,k f ( x ) ρ I α,βη,k ( gϕψ ) ( x )+ ρ I α,βη,k g ( x ) ρ I α,βη,k ( f ϕψ ) ( x )] + 2 ρ I α,βη,k f ( x ) ρ I α,βη,k ( hϕψ ) ( x ) ρ I α,βη,k g ( x ) ≥ ρ I α,βη,k h ( x ) h ρ I α,βη,k ( f ϕ ) ( x ) ρ I α,βη,k ( gψ ) ( x ) + ρ I α,βη,k ( f ψ ) ( x ) ρ I α,βη,k ( gϕ ) ( x ) i + ρ I α,βη,k f ( x ) h ρ I α,βη,k ( hϕ ) ( x ) ρ I α,βη,k ( gψ ) ( x ) + ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( gϕ ) ( x ) i + ρ I α,βη,k g ( x ) h ρ I α,βη,k ( hϕ ) ( x ) ρ I α,βη,k ( f ψ ) ( x ) + ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( f ϕ ) ( x ) i . (4.3) Proof.
In lemma (4.1) putting s = f, v = g, then multiplying both sides of theresulting inequality by ρ I α,βη,k h ( x ) , we get: ρ I α,βη,k h ( x ) ρ I α,βη,k ( f ϕψ ) ( x ) ρ I α,βη,k g ( x )+ ρ I α,βη,k h ( x ) ρ I α,βη,k f ( x ) ρ I α,βη,k ( gϕψ ) ( x ) ≥ ρ I α,βη,k h ( x ) ρ I α,βη,k ( f ϕ ) ( x ) ρ I α,βη,k ( gψ ) ( x ) (4.4)+ ρ I α,βη,k h ( x ) ρ I α,βη,k ( f ψ ) ( x ) ρ I α,βη,k ( gϕ ) ( x ) . Now putting s = h, v = g, in lemma (4.1) and multiplying both sides of the resultinginequality by ρ I α,βη,k f ( x ) , we obtain: ρ I α,βη,k f ( x ) ρ I α,βη,k ( hϕψ ) ( x ) ρ I α,βη,k g ( x )+ ρ I α,βη,k f ( x ) ρ I α,βη,k h ( x ) ρ I α,βη,k ( gϕψ ) ( x ) ≥ ρ I α,βη,k f ( x ) ρ I α,βη,k ( hϕ ) ( x ) ρ I α,βη,k ( gψ ) ( x ) (4.5)+ ρ I α,βη,k f ( x ) ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( gϕ ) ( x ) . Now putting s = h, v = f, in lemma (4.1),then multiplying both sides of the resultinginequality by ρ I α,βη,k g ( x ) , we get: ρ I α,βη,k g ( x ) ρ I α,βη,k ( hϕψ ) ( x ) ρ I α,βη,k f ( x )+ ρ I α,βη,k g ( x ) ρ I α,βη,k h ( x ) ρ I α,βη,k ( f ϕψ ) ( x ) ≥ ρ I α,βη,k g ( x ) ρ I α,βη,k ( hϕ ) ( x ) ρ I α,βη,k ( f ψ ) ( x ) (4.6)+ ρ I α,βη,k g ( x ) ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( f ϕ ) ( x ) . By adding the inequalities(4.4, 4.5, 4.6) we get the inequality (4.3).Now we give the lemma required for proving our next theorem for differentparameter. 8 emma 4.2
Let ϕ and ψ be two integrable and synchronous functions on [0 , ∞ ) , suppose that s, v : [0 , ∞ ) → [0 , ∞ ) , then for all x > , α > , δ > , ρ > , k, β, λ, η ∈ R , we have: ρ I α,βη,k ( sϕψ ) ( x ) ρ I δ,λη,k v ( x ) + ρ I α,βη,k s ( x ) ρ I δ,λη,k ( vϕψ ) ( x ) ≥ ρ I α,βη,k ( sϕ ) ( x ) ρ I δ,λη,k ( vψ ) ( x ) + ρ I α,βη,k ( sψ ) ( x ) ρ I δ,λη,k ( vϕ ) ( x ) . Proof.
Since ϕ and ψ are synchronous functions on [0 , ∞ ) , by similar argumentsas in the proof of lemma (4.1), we can write ρ I α,βη,k ( sϕψ ) ( x ) + ϕ ( γ ) ψ ( γ ) ρ I α,βη,k s ( x ) ≥ ψ ( γ ) ρ I α,βη,k ( sϕ ) ( x ) + ϕ ( γ ) ρ I α,βη,k ( sψ ) ( x ) . (4.7)Multiplying both sides of (4.7) by ρ − λ x k Γ( δ ) γ ρ ( η +1) − v ( γ )( x ρ − γ ρ ) − δ , where γ ∈ (0 , x ) , then integrat-ing resulting inequality over (0 , x ) , with respect to the variable γ, we obtain: ρ I α,βη,k ( sϕψ ) ( x ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ v ( γ ) dγ + ρ I α,βη,k s ( x ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ v ( γ ) ϕ ( γ ) ψ ( γ ) dγ ≥ ρ I α,βη,k ( sϕ ) ( x ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ v ( γ ) ψ ( γ ) dγ + ρ I α,βη,k ( sψ ) ( x ) ρ − λ x k Γ ( δ ) Z x γ ρ ( η +1) − ( x ρ − γ ρ ) − δ v ( γ ) ϕ ( γ ) dγ. Therefore ρ I α,βη,k ( sϕψ ) ( x ) ρ I δ,λη,k v ( x ) + ρ I α,βη,k s ( x ) ρ I δ,λη,k ( vϕψ ) ( x ) ≥ ρ I α,βη,k ( sϕ ) ( x ) ρ I δ,λη,k ( vψ ) ( x ) + ρ I α,βη,k ( sψ ) ( x ) ρ I δ,λη,k ( vϕ ) ( x ) . Hence the proof.
Theorem 4.2
Let ϕ and ψ be two integrable and synchronous functions on [0 , ∞ ) , and suppose f, g, h : [0 , ∞ ) → [0 , ∞ ) . Then for all x > , α > , δ > , ρ > , k, β, λ, η ∈ R , we have: ρ I α,βη,k h ( x ) [ ρ I α,βη,k ( f ϕψ ) ( x ) ρ I δ,λη,k g ( x ) + 2 ρ I α,βη,k f ( x ) ρ I δ,λη,k ( gϕψ ) ( x )+ ρ I α,βη,k g ( x ) ρ I δ,λη,k ( f ϕψ ) ( x )]+ h ρ I α,βη,k f ( x ) ρ I δ,λη,k g ( x ) + ρ I α,βη,k g ( x ) ρ I δ,λη,k f ( x ) i ρ I α,βη,k ( hϕψ ) ( x ) ≥ ρ I α,βη,k h ( x ) h ρ I α,βη,k ( f ϕ ) ( x ) ρ I δ,λη,k ( gψ ) ( x ) + ρ I α,βη,k ( f ψ ) ( x ) ρ I δ,λη,k ( gϕ ) ( x ) i + ρ I α,βη,k f ( x ) h ρ I α,βη,k ( hϕ ) ( x ) ρ I δ,λη,k ( gψ ) ( x ) + ρ I α,βη,k ( hψ ) ( x ) ρ I δ,λη,k ( gϕ ) ( x ) i + ρ I α,βη,k g ( x ) h ρ I α,βη,k ( hϕ ) ( x ) ρ I δ,λη,k ( f ψ ) ( x ) + ρ I α,βη,k ( hψ ) ( x ) ρ I δ,λη,k ( f ϕ ) ( x ) i . (4.8) Proof.
In lemma (4.2), putting s = f, v = g, we can write: ρ I α,βη,k ( f ϕψ ) ( x ) ρ I δ,λη,k g ( x ) + ρ I α,βη,k f ( x ) ρ I δ,λη,k ( gϕψ ) ( x ) ≥ ρ I α,βη,k ( f ϕ ) ( x ) ρ I δ,λη,k ( gψ ) ( x ) + ρ I α,βη,k ( f ψ ) ( x ) ρ I δ,λη,k ( gϕ ) ( x ) . (4.9)Multiplying both sides of (4.9) by ρ I α,βη,k h ( x ) , we get ρ I α,βη,k h ( x ) ρ I α,βη,k ( f ϕψ ) ( x ) ρ I δ,λη,k g ( x )+ ρ I α,βη,k h ( x ) ρ I α,βη,k f ( x ) ρ I δ,λη,k ( gϕψ ) ( x ) ≥ ρ I α,βη,k h ( x ) ρ I α,βη,k ( f ϕ ) ( x ) ρ I δ,λη,k ( gψ ) ( x )+ ρ I α,βη,k h ( x ) ρ I α,βη,k ( f ψ ) ( x ) ρ I δ,λη,k ( gϕ ) ( x ) . (4.10)Now putting s = h, v = g, in lemma (4.2), then multiplying both sides of theresulting inequality by ρ I α,βη,k f ( x ) , we obtain: ρ I α,βη,k f ( x ) ρ I α,βη,k ( hϕψ ) ( x ) ρ I δ,λη,k g ( x )+ ρ I α,βη,k f ( x ) ρ I α,βη,k h ( x ) ρ I δ,λη,k ( gϕψ ) ( x ) ≥ ρ I α,βη,k f ( x ) ρ I α,βη,k ( hϕ ) ( x ) ρ I δ,λη,k ( gψ ) ( x )+ ρ I α,βη,k f ( x ) ρ I α,βη,k ( hψ ) ( x ) ρ I δ,λη,k ( gϕ ) ( x ) . (4.11)Now putting s = h, v = f, in lemma (4.2) and multiplying both sides of the resultinginequality by ρ I α,βη,k g ( x ) , we get: ρ I α,βη,k g ( x ) ρ I α,βη,k ( hϕψ ) ( x ) ρ I δ,λη,k f ( x )+ ρ I α,βη,k g ( x ) ρ I α,βη,k h ( x ) ρ I δ,λη,k ( f ϕψ ) ( x ) ≥ ρ I α,βη,k g ( x ) ρ I α,βη,k ( hϕ ) ( x ) ρ I δ,λη,k ( f ψ ) ( x )+ ρ I α,βη,k g ( x ) ρ I α,βη,k ( hψ ) ( x ) ρ I δ,λη,k ( f ϕ ) ( x ) . (4.12)10y adding the inequalities(4.10), (4.11), (4.12) we get the required inequality (4.8). Generalized fractional inequality for Chebyshev functional for differen-tiable functions : Our results in present section are for Chebyshev functional in case of differentiablefunctions whose derivatives belong to L p ([0 , ∞ ]) [6].First we give the following lemma: Lemma 5.1
Let h be a positive function on [0 , ∞ ) , and let ϕ, ψ, be two differen-tiable functions on [0 , ∞ ) . Then for all x > , α > , ρ > , k, β, η ∈ R we have: ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) H ( τ , γ ) dτ dγ = h ρ I α,βη,k ( hϕψ ) ( x ) ρ I α,βη,k h ( x ) − ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( hϕ ) ( x ) i . (5.1) Proof.
Define H ( τ , γ ) = ( ϕ ( τ ) − ϕ ( γ )) ( ψ ( τ ) − ψ ( γ )) ; τ , γ ∈ (0 , x ) , x > . (5.2)Multiplying both sides of (5.2) by ρ − β x k Γ( α ) τ ρ ( η +1) − h ( τ )( x ρ − τ ρ ) − α , where τ ∈ (0 , x ) , we get ρ − β x k Γ ( α ) τ ρ ( η +1) − ( x ρ − τ ρ ) − α h ( τ ) H ( τ , γ )= ρ − β x k Γ ( α ) τ ρ ( η +1) − ( x ρ − τ ρ ) − α h ( τ ) ϕ ( τ ) ψ ( τ ) − ρ − β x k Γ ( α ) τ ρ ( η +1) − ( x ρ − τ ρ ) − α h ( τ ) ϕ ( τ ) ψ ( γ ) − ρ − β x k Γ ( α ) τ ρ ( η +1) − ( x ρ − τ ρ ) − α h ( τ ) ϕ ( γ ) ψ ( τ ) + ρ − β x k Γ ( α ) τ ρ ( η +1) − ( x ρ − τ ρ ) − α h ( τ ) ϕ ( γ ) ψ ( γ ) . (5.3)Now integrating (5.3) over (0 , x ) , with respect to the variable τ , we obtain: ρ − β x k Γ ( α ) Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α h ( τ ) H ( τ , γ ) dτ = ρ I α,βη,k ( hϕψ ) ( x ) − ψ ( γ ) ρ I α,βη,k ( hϕ ) ( x ) − ϕ ( γ ) ρ I α,βη,k ( hψ ) ( x ) + ϕ ( γ ) ψ ( γ ) ρ I α,βη,k h ( x ) . (5.4)11ow multiplying both sides of (5.4) by ρ − β x k Γ( α ) γ ρ ( η +1) − h ( γ )( x ρ − γ ρ ) − α , where γ ∈ (0 , x ) , andintegrating the resulting identity with respect to γ from 0 to x , we get ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) H ( τ , γ ) dτ dγ = 2 h ρ I α,βη,k ( hϕψ ) ( x ) ρ I α,βη,k h ( x ) − ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( hϕ ) ( x ) i . Which is (5.1) .
Theorem 5.1
Let h be a positive function on [0 , ∞ ) , and let ϕ, ψ, be two differ-entiable functions on [0 , ∞ ) . Suppose that ϕ ′ ∈ L s ([0 , ∞ )) , ψ ′ ∈ L v ([0 , ∞ )) , s > , s + v = 1 . Then for all x > , α > , ρ > , k, β, η ∈ R we have: (cid:12)(cid:12)(cid:12) ρ I α,βη,k ( hϕψ ) ( x ) ρ I α,βη,k h ( x ) − ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( hϕ ) ( x ) (cid:12)(cid:12)(cid:12) ≤ k ϕ ′ k s k ψ ′ k v ρ − β ) x k Γ ( α ) × (cid:20)Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | dτ dγ (cid:21) ≤ k ϕ ′ k s k ψ ′ k v x (cid:16) ρ I α,βη,k h ( x ) (cid:17) . (5.5) Proof.
In lemma (5.1), from the identity (5.2), we can write H ( τ , γ ) = Z γτ Z γτ ϕ ′ ( t ) ψ ′ ( r ) dtdr. By applying Holder inequality for double integral, we obtain: | H ( τ , γ ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z γτ Z γτ | ϕ ′ ( t ) | s dtdr (cid:12)(cid:12)(cid:12)(cid:12) s (cid:12)(cid:12)(cid:12)(cid:12)Z γτ Z γτ | ψ ′ ( r ) | v dtdr (cid:12)(cid:12)(cid:12)(cid:12) v ≤ | τ − γ | s (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ϕ ′ ( t ) | s dt (cid:12)(cid:12)(cid:12)(cid:12) s ! | τ − γ | v (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ψ ′ ( r ) | v dr (cid:12)(cid:12)(cid:12)(cid:12) v ! ≤ | τ − γ | (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ϕ ′ ( t ) | s dt (cid:12)(cid:12)(cid:12)(cid:12) s (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ψ ′ ( r ) | v dr (cid:12)(cid:12)(cid:12)(cid:12) v . (5.6)Using inequality (5.6) in left-hand side of lemma (5.1), we can write ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) × | τ − γ | (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ϕ ′ ( t ) | s dt (cid:12)(cid:12)(cid:12)(cid:12) s (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ψ ′ ( r ) | v dr (cid:12)(cid:12)(cid:12)(cid:12) v dτ dγ. (5.7)12gain applying Holder inequality to the right-hand side of inequality (5.7), we obtainfrom (5.7) that ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ (cid:20) ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ |× (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ϕ ′ ( t ) | s dt (cid:12)(cid:12)(cid:12)(cid:12) dτ dγ (cid:21) s × (cid:20) ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ψ ′ ( r ) | v dr (cid:12)(cid:12)(cid:12)(cid:12) dτ dγ (cid:21) v . (5.8)Since (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ϕ ′ ( t ) | s dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ k ϕ ′ k ss , (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ψ ′ ( r ) | v dr (cid:12)(cid:12)(cid:12)(cid:12) ≤ k ψ ′ k vv . (5.9)Then from (5.8), we have ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ (cid:20) k ϕ ′ k ss ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | dτ dγ (cid:21) s × (cid:20) k ψ ′ k vv ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | dτ dγ (cid:21) v . So we have ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ k ϕ ′ k s k ψ ′ k v ρ − β ) x k Γ ( α ) × (cid:20)Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | dτ dγ (cid:21) s × (cid:20)Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | dτ dγ (cid:21) v . ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ k ϕ ′ k s k ψ ′ k v ρ − β ) x k Γ ( α ) × (cid:20)Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | dτ dγ (cid:21) . (5.10)Using lemma (5.1), and the inequality (5.10), with the properties of the modulus,we get 2 (cid:12)(cid:12)(cid:12) ρ I α,βη,k ( hϕψ ) ( x ) ρ I α,βη,k h ( x ) − ρ I α,βη,k ( hψ ) ( x ) ρ I α,βη,k ( hϕ ) ( x ) (cid:12)(cid:12)(cid:12) ≤ k ϕ ′ k s k ψ ′ k v ρ − β ) x k Γ ( α ) × (cid:20)Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | τ − γ | dτ dγ (cid:21) . (5.11)Which proves frist part of (5.5). To prove the second inequality of (5.5), we have0 ≤ τ ≤ x, ≤ γ ≤ x. Then 0 ≤ | τ − γ | ≤ x. Hence ρ − β ) x k Γ ( α ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ k ϕ ′ k s k ψ ′ k v ρ − β ) x k Γ ( α ) × (cid:20)Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − α h ( τ ) h ( γ ) dτ dγ (cid:21) . = k ϕ ′ k s k ψ ′ k v x (cid:16) ρ I α,βη,k h ( x ) (cid:17) . (5.12)Which proves second inequality of (5.5). Hence, Theorem (5.1) is proved.Now we give the following theorem with different parameters:14 heorem 5.2 Let h be a positive function on [0 , ∞ ) , and let ϕ, ψ, be two differ-entiable functions on [0 , ∞ ) . suppose that ϕ ′ ∈ L s ([0 , ∞ )) , ψ ′ ∈ L v ([0 , ∞ )) , s > , s + v = 1 . Then for all x > , α > , δ > , ρ > , k, β, λ, η ∈ R , we have: ρ I α,βη,k ( hϕψ ) ( x ) ρ I δ,λη,k h ( x ) − ρ I δ,λη,k ( hψ ) ( x ) ρ I α,βη,k ( hϕ ) ( x ) ρ I δ,λη,k ( hϕ ) ( x ) ρ I α,βη,k ( hψ ) ( x ) + ρ I δ,λη,k ( hϕψ ) ( x ) ρ I α,βη,k h ( x ) ≤ k ϕ ′ k s k ψ ′ k v ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) × "Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( τ ) h ( γ ) | τ − γ | dτ dγ ≤ k ϕ ′ k s k ψ ′ k v x (cid:16) ρ I α,βη,k h ( x ) ρ I δ,λη,k h ( x ) (cid:17) . (5.13) Proof.
In lemma (5.1), multiplying both sides of (5.4) by ρ − λ x k Γ( δ ) γ ρ ( η +1) − h ( γ )( x ρ − γ ρ ) − δ , where γ ∈ (0 , x ) , and integrating the resulting identity with respect to γ from 0 to x , wecan write ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( γ ) h ( τ ) H ( τ , γ ) dτ = ρ I α,βη,k ( hϕψ ) ( x ) ρ I δ,λη,k h ( x ) − ρ I δ,λη,k ( hψ ) ( x ) ρ I α,βη,k ( hϕ ) ( x ) − ρ I δ,λη,k ( hϕ ) ( x ) ρ I α,βη,k ( hψ ) ( x ) + ρ I δ,λη,k ( hϕψ ) ( x ) ρ I α,βη,k h ( x ) . (5.14)Using inequality (5.6) in left-hand side of (5.14), we can write ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( γ ) h ( τ ) | H ( τ , γ ) | dτ ≤ ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( γ ) h ( τ ) × | τ − γ | (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ϕ ′ ( t ) | s dt (cid:12)(cid:12)(cid:12)(cid:12) s (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ψ ′ ( r ) | v dr (cid:12)(cid:12)(cid:12)(cid:12) v dτ dγ. (5.15)15y applying Holder inequality for double integral to above inequality, we obtain: ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ " ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( τ ) h ( γ ) | τ − γ |× (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ϕ ′ ( t ) | s dt (cid:12)(cid:12)(cid:12)(cid:12) dτ dγ (cid:21) s × " ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( τ ) h ( γ ) | τ − γ | (cid:12)(cid:12)(cid:12)(cid:12)Z γτ | ψ ′ ( r ) | v dr (cid:12)(cid:12)(cid:12)(cid:12) dτ dγ (cid:21) v . (5.16)Using (5.9) and (5.16), we can write ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( τ ) h ( γ ) | H ( τ , γ ) | dτ dγ ≤ k ϕ ′ k s k ψ ′ k v ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) × "Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( τ ) h ( γ ) | τ − γ | dτ dγ . (5.17)Using (5.14) and (5.17), with the properties of the modulus, we get ρ I α,βη,k ( hϕψ ) ( x ) ρ I δ,λη,k h ( x ) − ρ I δ,λη,k ( hψ ) ( x ) ρ I α,βη,k ( hϕ ) ( x ) ρ I δ,λη,k ( hϕ ) ( x ) ρ I α,βη,k ( hψ ) ( x ) + ρ I δ,λη,k ( hϕψ ) ( x ) ρ I α,βη,k h ( x ) ≤ k ϕ ′ k s k ψ ′ k v ρ − ( β − λ ) x k Γ ( α ) Γ ( δ ) × "Z x Z x τ ρ ( η +1) − ( x ρ − τ ρ ) − α γ ρ ( η +1) − ( x ρ − γ ρ ) − δ h ( τ ) h ( γ ) | τ − γ | dτ dγ . (5.18)Which proves first inequality of (4.2). Second inequality can be proved similarly. Remark 5.3 If k = 0 , η = 0 and taking ρ → in the theorems (5.1) and (5.2), weget |I α h ( x ) I α ( hϕψ ) ( x ) − I α ( hϕ ) ( x ) I α ( hψ ) ( x ) |≤ k ϕ ′ k s k ψ ′ k v Γ ( α ) Z ba Z ba ( x − τ ) α − ( x − γ ) α − | τ − γ | h ( τ ) h ( γ ) dτ dγ ≤ k ϕ ′ k s k ψ ′ k v x ( I α h ( x )) nd I α ( hϕψ ) ( x ) I δ h ( x ) − I δ ( hψ ) ( x ) I α ( hϕ ) ( x ) I δ ( hϕ ) ( x ) I α ( hψ ) ( x ) + I δ ( hϕψ ) ( x ) I α h ( x ) ≤ k ϕ ′ k s k ψ ′ k v Γ ( α ) Γ ( δ ) Z x Z x ( x − τ ) α − ( x − γ ) δ − h ( τ ) h ( γ ) | τ − γ | dτ dγ ≤ k ϕ ′ k s k ψ ′ k v x (cid:0) I α h ( x ) I δ h ( x ) (cid:1) . respectively, as in (see [15]). Similarly we can get the remaining five cases of gen-eralized fractional integral mentioned at the preliminaries. References [1] Katugampola, U.N., A new approach to generalized fractional derivatives, Bull.Math. Anal. Appl. 6 (4) (2014), 1–15.[2] Katugampola, U.N., New fractional integral unifying six existing fractional in-tegrals, pp. 6. (2016), arXiv:1612.08596 (eprint).[3] Sousa J., Oliveira D. S., Capelas de Oliveira E., Gr¨uss-Type Inequalities byMeans of Generalized Fractional Integrals, Bull Braz Math Soc,2 (4) (2019).[4] Kilbas, A.A., Srivastava, H.M., Trujillo, J.J., Theory and Applications of theFractional Differential Equations, vol. 204. Elsevier, Amsterdam (2006).[5] Camargo, R.F., Oliveira, E.C., Fractional Calculus (in Portuguese), EditoraLivraria da F´ısica, S˜ao Paulo (2015).[6] Chebyshev P.L. , Sur les expressions approximatives des integrales definies parles autres prises entre les mˆemes limites, Proc. Math. Soc. Charkov, 2 (1882),93–98.[7] Malamud S.M. , Some complements to the Jenson and Chebyshev inequalitiesand a problem of W. Walter, Proc. Amer. Math. Soc., 129 (9) (2001), 2671–2678.[8] Marinkovic S., Rajkovic P. and Stankovic M., The inequalities for some typesqintegrals, Comput. Math. Appl., 56 (2008), 2490–2498.[9] Pachpatte B.G., A note on Chebyshev-Gr¨uss type inequalities for differentialfunctions, Tamsui Oxf. J. Manag. Sci., 22 (1) (2006), 29–36.1710] Gavrea I., On Chebyshev type inequalities involving functions whose derivativesbelog to L pp