Some Minkowski-type inequalities using generalized proportional Hadamard fractional integral operators
aa r X i v : . [ m a t h . G M ] J u l Some Minkowski-type inequalities using generalizedproportional Hadamard fractional integral operators
Asha B.N ale , Satish K.P anchal , V aijanath L.Chinchane , Department of Mathematics,Dr. Babasaheb Ambedkar MarathwadaUniversity, Aurangabad-431 004, INDIA. Department of Mathematics,Deogiri Institute of Engineering and ManagementStudies Aurangabad-431005, [email protected]/[email protected]/[email protected]
Abstract
The main objective of present investigation to obtain some Minkowski-type fractional integral inequalities using generalised proportional Hadamardfractional integral operators which is introduced by Rahman et al inthe paper (certain inequalities via generalized proportional Hadamardfractional integral operators), Advances in Differential Equations, (2019),2019:454. In addition, we establish some other fractional integral in-equalities for positive and continuous functions.
Keywords :
Minkowski-type fractional integral inequality, generalised pro-portional Hadamard fractional integral operator.
Mathematics Subject Classification :
Fractional calculus is generalization of traditional calculus into non-integerdifferential and integral order. Fractional calculus is very important due toit’s various application in field of science and technology. Fractional integralinequalities play big role in obtaining uniqueness of solution of fractionalordinary differential equations, fractional partial differential equations andfractional Boundary value problems. Recently, a number of researchers inthe fields of fractional integral inequalities have established different integralinequalities about fractional integral operators such as Riemann-Liouville,Hadamard, Saigo, generalized Katugamapola, Erd´ e lyi-Kober, Riemann-Liouvillek-fractional, Hadamard k-fractional, (k,s)-Riemann-Liouville and k-generalized(in terms hypergeometric function) fractional integral operators, see [1, 4, 5,6, 7, 10, 11, 12, 13, 14, 16, 17, 19, 20, 21, 22, 28, 30, 31]. In [9], Dahmaniinvestigated reverse Minkowski fractional integral inequality by employing1iemann-Liouville fractional integral . Ahmed Anber and et al [2] presentedsome fractional integral inequalities which is simillar to Minkowski fractionalintegral inequality using Riemann-Liouville fractional integral. V. L. Chin-chane et al [3] proposed fractional inequalities similar to Minkowski typevia Saigo fractional integral operator. In [8], V. L. Chinchane studied thereverse Minkowski fractional integral inequality by considering generalizedK-fractional integral operator is in terms of the Gauss herpergeometric func-tion. S. Mubeen and et al [18] have introduced Minkowski inequality involv-ing generalized k-fractional conformable integrals.G. Rahman et al. [23, 24, 25] established Minkowski inequality and someother fractional inequalities for convex functions by employing fractional pro-portional integral operators. In [15, 26, 27], F. Jarad et al and G.Rahman pre-sented concepts of non-local fractional proportional and generalized Hadamardproportional integrals involving exponential functions in their kernels.Motivated from [24, 25, 26, 27], our purpose in this paper is to proposesome new results using generalized Hadamard proportional integrals. Thepaper has been organized as follows, in Section 2, we recall basic definitions,remarks and lemma related to generalized Hadamard proportional integrals.In Section 3, we obtain reverse Minkowski fractional integral inequality usinggeneralized Hadamard proportional integrals, In Section 4, we present someother inequalities using generalized Hadamard proportional integrals. Here, we present some important definition, remarks and lemma of gener-alised proportional Hadamard fractional integral operator which will be usedthroughout this paper.
Definition 2.1
The left and right sided generalized proportional fractionalintegrals are respectively defined by ( a J α,β z )( x ) = 1 β α Γ( α ) Z xa e [ β − β ( x − t )] ( x − t ) α − z ( t ) dt, a < x, (2.1) and ( J α,βb z )( x ) = 1 β α Γ( α ) Z bx e [ β − β ( t − x )] ( t − x ) α − z ( t ) dt, x < b, (2.2) where the proportionality index β ∈ (0 , and α ∈ C with R ( α ) > . emark 2.1 If we consider β = 1 in (2.1) and (2.2), then we get the wellknown left and right Riemann-Liouville integrals which are respectively de-fined by ( a J α z )( x ) = 1Γ( α ) Z xa ( x − t ) α − z ( t ) dt, a < x, (2.3) and ( J αb z )( x ) = 1Γ( α ) Z bx ( t − x ) α − z ( t ) dt, x < b, (2.4) where α ∈ C with R ( α ) > . Recently, Rahman et al.[27] proposed the following generalized Hadamardproportional fractional integrals.
Definition 2.2
The left sided generalized Hadamard proportional fractionalintegral of order α > and proportional index β ∈ (0 , is defined by ( a H α,β z )( x ) = 1 β α Γ( α ) Z xa e [ β − β ( lnx − lnt )] ( lnx − lnt ) α − z ( t ) t dt, a < x. (2.5) Definition 2.3
The right sided generalized Hadamard proportional fractionalintegral of order α > and proportional index β ∈ (0 , is defined by ( H α,βb z )( x ) = 1 β α Γ( α ) Z bx e [ β − β ( lnt − lnx )] ( lnt − lnx ) α − z ( t ) t dt, x < b. (2.6) Definition 2.4
The one sided generalized Hadamard proportional fractionalintegral of order α > and proportional index β ∈ (0 , is defined by ( H α,β ,x z )( x ) = 1 β α Γ( α ) Z x e [ β − β ( lnx − lnt )] ( lnx − lnt ) α − z ( t ) t dt, t > , (2.7) where Γ( α ) is the classical well known gamma function. Remark 2.2
If we consider β = 1 , then (2.5)-(2.7) will led to the followingwell known Hadamard fractional integrals ( a H α z )( x ) = 1Γ( α ) Z xa ( lnx − lnt ) α − z ( t ) t dt, a < x, (2.8)( H αb z )( x ) = 1Γ( α ) Z bx ( lnt − lnx ) α − z ( t ) t dt, x < b, (2.9) and ( H α ,x z )( x ) = 1Γ( α ) Z x ( lnx − lnt ) α − z ( t ) t dt, x > . (2.10)3ne can easily prove the following results: Lemma 2.1 ( H α,β ,x e [ β − β ( lnx )] ( lnx ) λ − )( x ) = Γ( λ ) β α Γ( α + λ ) e [ β − β ( lnx )] ( lnx ) α + λ − , (2.11) and the semigroup property ( H α,β ,x )( H λ,β ,x ) z ( x ) = ( H α + λ,β ,x ) z ( x ) . (2.12) Remark 2.3 If β = 1 , then (2.11) will reduce to the result of [29] as definedby ( H α ,x ( lnx ) λ − )( x ) = Γ( λ )Γ( α + λ ) ( lnx ) α + λ − . (2.13) In this section, we establish reverse Minkowski fractional integral inequalityinvolving generalized proportional Hadamard fractional integral operators.
Theorem 3.1
Let p ≥ and let f , g be two positive function on [1 , ∞ ) , suchthat for all x > , H α,β ,x [ f p ( x )] < ∞ , H α,β ,x [ g q ( x )] < ∞ . If < m ≤ f ( τ ) g ( τ ) ≤ M , τ ∈ (1 , x ) we have h H α,β ,x [ f p ( x )] i p + h H α,β ,x [ g q ( x )] i p ≤ M ( m + 2)( m + 1)( M + 1) h H α,β ,x [( f + g ) p ( x )] i p , (3.1) where α > , β ∈ (0 , α ∈ C and R ( α ) > . Proof : Using the condition f ( τ ) g ( τ ) ≤ M , τ ∈ (1 , x ), x >
1, we can write( M + 1) p f ( τ ) ≤ M p ( f + g ) p ( τ ) . (3.2)Consider ψ ( x, τ ) = 1 β α Γ( α ) τ e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − . (3.3)Clearly, we can say that the function ψ ( x, τ ) remain positive because for all τ ∈ (1 , x ) , ( x > , α, β >
0. Multiplying both side of (3.2) by ψ ( x, τ ), then4ntegrating resulting identity with respect to τ from 1 to x , we get( M + 1) p β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − f p ( τ ) dττ ≤ M p β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − ( f + g ) p ( τ ) dττ , (3.4)which is equivalent to H α,β ,x [ f p ( x )] ≤ M p ( M + 1) p h H α,β ,x [( f + g ) p ( x )] i , (3.5)hence, we can write h H α,β ,x [ f p ( x )] i p ≤ M ( M + 1) h H α,β ,x [( f + g ) p ( x )] i p . (3.6)On other hand, using condition m ≤ f ( τ ) g ( τ ) , we obtain(1 + 1 m ) g ( τ ) ≤ m ( f ( τ ) + g ( τ )) , (3.7)therefore, (1 + 1 m ) p g p ( τ ) ≤ ( 1 m ) p ( f ( τ ) + g ( τ )) p . (3.8)Now, multiplying both side of (3.8) by ψ ( x, τ ), ( τ ∈ (1 , x ), x > G ( x, τ ) is defined by (3.3). Then integrating resulting identity with respectto τ from 1 to x , we have h H α,β ,x [ g p ( x )] i p ≤ m + 1) h H α,β ,x [( f + g ) p ( x )] i p . (3.9)The inequalities (3.1) follows on adding the inequalities (3.6) and (3.9).Our second result is as follows. Theorem 3.2
Let p ≥ and f , g be two positive function on [1 , ∞ ) , suchthat for all x > , H α,β ,x [ f p ( x )] < ∞ , H α,β ,x [ g q ( x )] < ∞ . If < m ≤ f ( τ ) g ( τ ) ≤ M ,we have h H α,β ,x [ f p ( x )] i p + h H α,β ,x [ g q ( x )] i p ≥ ( ( M + 1)( m + 1) M − h H α,β ,x [ f p ( x )] i p + h I α,β,η ,x [ g q ( x )] i p , (3.10) where α > , β ∈ (0 , α ∈ C and R ( α ) > . roof : Multiplying the inequalities (3.6) and (3.9), we obtain( M + 1)( m + 1) M h H α,β ,x [ f p ( x )] i p h H α,β ,x [ g q ( x )] i p ≤ (cid:16) [ H α,β ,x [( f ( x ) + g ( x )) p ]] p (cid:17) . (3.11)Applying Minkowski inequalities to the right hand side of (3.11), we have( h H α,β ,x [( f ( x ) + g ( x )) p ] i p ) ≤ ( h H α,β ,x [ f p ( x )] i p + h H α,β ,x [ g q ( x )] i p ) , (3.12)which implies that h H α,β ,x [( f ( x ) + g ( x )) p ] i p ≤ h H α,β ,x [ f p ( x )] i p + h H α,β ,x [ g q ( x )] i p + 2 h H α,β ,x [ f p ( x )] i p h H α,β ,x [ g q ( x )] i p , (3.13)using (3.11) and (3.13) we obtain (3.10). Theorem 3.2 is thus proved. Here, we establish some new integral inequalities using generalized propor-tional Hadamard fractional integral operators.
Theorem 4.1
Let p > , p + q = 1 and f , g be two positive function on [1 , ∞ ) , such that H α,β ,x [ f ( x )] < ∞ , H α,β ,x [ g ( x )] < ∞ . If < m ≤ f ( τ ) g ( τ ) ≤ M < ∞ , τ ∈ (1 , x ) we have h H α,β ,x [ f ( x )] i p h H α,β ,x [ g ( x )] i q ≤ ( Mm ) pq h H α,β ,x [[ f ( x )] p [ g ( t )] q ] i , (4.1) hold, where α > , β ∈ (0 , α ∈ C and R ( α ) > . Proof:-
Since f ( τ ) g ( τ ) ≤ M , τ ∈ [1 , x ] x >
1, therefore[ g ( τ )] p ≥ M − q [ f ( τ )] q , (4.2)and also, [ f ( τ )] p [ g ( τ )] q ≥ M − q [ f ( τ )] q [ f ( τ )] p ≥ M − q [ f ( τ )] q + q ≥ M − q [ f ( τ )] . (4.3)6ultiplying both side of (4.3) by ψ ( x, τ ), ( τ ∈ (1 , x ), x > ψ ( x, τ )is defined by (3.3). Then integrating resulting identity with respect to τ from1 to x , we have1 β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − f ( τ ) p g ( τ ) q ( τ ) dττ ≤ M − q β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − f ( τ )) dττ , (4.4)which implies that, H α,β ,x h [ f ( x )] p [ g ( x )] q i ≤ M − q h H α,β ,x f ( x ) i . (4.5)Consequently, (cid:16) H α,β ,x h [ f ( x )] p [ g ( x )] q i(cid:17) p ≤ M − pq h H α,β ,x f ( x ) i p , (4.6)on other hand, since mg ( τ ) ≤ f ( τ ), τ ∈ [1 , x ), x >
1, then we have[ f ( τ )] p ≥ m p [ g ( τ )] p , (4.7)multiplying equation (4.7) by [ g ( τ )] q , we have,[ f ( τ )] p [ g ( τ )] q ≥ m p [ g ( τ )] q [ g ( τ )] p = m p [ g ( τ )] . (4.8)Multiplying both side of (4.8) by ψ ( x, τ ), ( τ ∈ (1 , x ), x > ψ ( x, τ )is defined by (3.3). Then integrating resulting identity with respect to τ from1 to x , we have1 β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − f ( τ ) p g ( τ ) q ( τ ) dττ ≤ m p β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [ g ( τ )] dττ , (4.9)that is, H α,β ,x h [ f ( x )] p [ g ( x )] q i ≤ m p h H α,β ,x g ( x ) i . (4.10)Hence we can write, (cid:16) H α,β ,x h [ f ( x )] p [ g ( x )] q i(cid:17) q ≤ m pq h H α,β ,x f ( x ) i q , (4.11)multiplying equation (4.6) and (4.11) we get the result (4.1).7 heorem 4.2 Let f and g be two positive function on [1 , ∞ [ , such that H α,β ,x [ f p ( x )] < ∞ , H α,β ,x [ g q ( x )] < ∞ . x > , If < m ≤ f ( τ ) p g ( τ ) q ≤ M < ∞ , τ ∈ [1 , x ] . Then we have h H α,β ,x f p ( x ) i p h H α,β ,x g q ( x ) i q ≤ ( Mm ) pq h H α,β ,x ( f ( x ) g ( x )) i hold, (4.12) where p > , p + q = 1 , α > , β ∈ (0 , α ∈ C and R ( α ) > . Proof:-
Replacing f ( τ ) and g ( τ ) by f ( τ ) p and g ( τ ) q , τ ∈ [1 , x ], x > Theorem 4.3 let f and g be two integrable functions on [1 , ∞ ] such that p + q = 1 , p > , and < m < f ( τ ) g ( τ ) < M, τ ∈ (1 , x ) , x > . Then we have H α,β ,x { f g } ( x ) ≤ p − M p p ( M + 1) p (cid:16) H α,β ,x [ f p + g p ]( x ) (cid:17) + 2 q − q ( m + 1) q (cid:16) H α,β ,x [ f q + g q ]( x ) (cid:17) , (4.13) where α > , β ∈ (0 , α ∈ C and R ( α ) > . Proof:-
Since, f ( τ ) g ( τ ) < M, τ ∈ (1 , x ) , x > , we have( M + 1) f ( τ ) ≤ M ( f + g )( τ ) . (4.14)Taking p th power on both side of (4.14)and multiplying resulting identityby ψ ( x, τ ), which is positive because τ ∈ [1 , x ], x >
1, then integrate theresulting identity with respect to τ from 1 to x , we get( M + 1) p β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − f ( τ ) p dττ ≤ M p β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [( f + g ) p ( τ )] dττ , (4.15)therefore, H α,β ,x [ f p ( x )] ≤ M p ( M + 1) p H α,β ,x [( f + g ) p ( x )] , (4.16)on other hand, 0 < m < f ( τ ) g ( τ ) , τ ∈ (1 , x ) , x > , we can write( m + 1) g ( τ ) ≤ ( f + g )( τ ) , (4.17)8aking q th power on both side (4.17) and multiplying resulting identity by ψ ( x, τ ), which is positive because τ ∈ [1 , x ], x > τ from 1 to x , we get( m + 1) q β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − g q ( τ ) dττ ≤ β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [( f + g ) q ( τ )] dττ , (4.18)consequently, we have H α,β ,x [ g q ( x )] ≤ m + 1) q H α,β ,x [( f + g ) q ( x )] . (4.19)Now, using Young inequality[ f ( τ ) g ( τ )] ≤ f p ( τ ) p + g q ( τ ) q . (4.20)Multiplying both side of (4.20) by ψ ( x, τ ), which is positive because τ ∈ [1 , x ], x >
1, then integrate the resulting identity with respect to τ from 1 to x , weget H α,β ,x [ f ( x ) g ( x ))] ≤ p H α,β ,x [ f p ( x )] + 1 q H α,β ,x [ g q ( x )] , (4.21)from equation (4.16), (4.19) and (4.21) we get H α,β ,x [ f ( x ) g ( x ))] ≤ M p p ( M + 1) p H α,β ,x [( f + g ) p ( x )]+ 1 q ( m + 1) q H α,β ,x [( f + g ) q ( x )] , (4.22)now using the inequality ( a + b ) r ≤ r − ( a r + b r ) , r > , a, b ≥ , we have H α,β ,x [( f + g ) p ( x )] ≤ p − H α,β ,x [( f p + g p )( x )] , (4.23)and H α,β ,x [( f + g ) q ( x )] ≤ q − H α,β ,x [( f q + g q )( x )] . (4.24)Injecting (4.23), (2.24) in (4.22) we get required inequality (4.13). Thiscomplete the proof. Theorem 4.4
Let f , g be two positive functions defined on [1 , ∞ ) , such that g is non-decreasing. If H α,β ,x f ( x ) ≥ H α,β ,x g ( x ) , x > . (4.25) then for all α > , β ∈ (0 , α ∈ C and R ( α ) > , H α,β ,x f γ − δ ( x ) ≤ H α,β ,x f γ ( x ) g − δ ( x ) , (4.26) hold. roof:- We use arithmetic-geometric inequality, for γ > δ >
0, we have: γγ − δ f γ − δ ( τ ) − δγ − δ g γ − δ ( τ ) ≤ f γ ( τ ) g − δ ( τ ) , τ ∈ (1 , x ) , x > . (4.27)Now, multiplying both side of (4.27) by ψ ( x, τ ), which is positive because τ ∈ [1 , x ], x >
1, then integrate the resulting identity with respect to τ from1 to x , we get γγ − δβ α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − f γ − δ ( τ ) dττ − δγ − δβ α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − g γ − δ ( τ ) dττ ≤ β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − f γ ( τ ) g − δ ( τ ) dττ , (4.28)consequently, γγ − δ H α,β ,x [ f γ − δ ( x )] − δγ − δ H α,β ,x [ g γ − δ ( x )] ≤ H α,β ,x [ f γ ( x ) g − δ ( x )] , (4.29)which implies that, γγ − δ H α,β ,x [ f γ − δ ( x )] ≤ H α,β ,x [ f γ ( x ) g − δ ( x )] + δγ − δ H α,β ,x [ g γ − δ ( x )] , (4.30)that is H α,β ,x [ f γ − δ ( x )] ≤ γ − δγ H α,β ,x [ f γ ( t ) g − δ ( x )] + δγ H α,β ,x [ f γ − δ ( x )] , (4.31)thus we get the result (4.26). Theorem 4.5
Suppose that f , g and h be positive and continuous functionson [1 , ∞ ) , such that ( g ( τ ) − g ( σ )) (cid:18) f ( σ ) h ( σ ) − f ( τ ) h ( τ ) (cid:19) ≥ τ, σ ∈ [1 , x ) x > , (4.32) then for all α > , β ∈ (0 , α ∈ C and R ( α ) > , H α,β ,x [ f ( x )] H α,β ,x [ h ( x )] ≥ H α,β ,x [( gf )( x )] H α,β ,x [( gh )( x )] . (4.33) Hold. roof : Since f , g and h be three positive and continuous functions on [1 , ∞ [by (4.32), we can write g ( τ ) f ( σ ) h ( σ ) + g ( σ ) f ( τ ) h ( τ ) − g ( σ ) f ( σ ) h ( σ ) − g ( τ ) f ( τ ) h ( τ ) ≥ τ, σ ∈ (1 , x ) , x > . (4.34)Now, multiplying equation (4.34) by h ( σ ) h ( τ ), on both side, we have, g ( τ ) f ( σ ) h ( τ ) − g ( τ ) f ( τ ) h ( σ ) − g ( σ ) f ( σ ) h ( τ ) + g ( σ ) f ( τ ) h ( σ ) ≥ . (4.35)Now multiplying equation (4.35) by ψ ( x, τ ), which is positive because τ ∈ [1 , x ], x >
1, then integrate the resulting identity with respect to τ from 1 to x , we get f ( σ ) β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [ g ( τ ) h ( τ )] dττ − h ( σ ) β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [ f ( τ ) g ( τ )] dττ + f ( σ ) g ( σ ) β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [ h ( τ )] dττg ( σ ) h ( σ ) β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [ f ( τ )] dττ ≥ , (4.36)we get f ( σ ) H α,β ,x [( gh )( x )] + g ( σ ) h ( σ ) H α,β ,x [ f ( x )] − g ( σ ) f ( ρ ) H α,β ,x [ h ( x )] − h ( σ ) H α,β ,x [( gf )( x )] ≥ . (4.37)Again multiplying (4.37) by ψ ( x, σ ),in view of equation (3.3)and which ispositive because σ ∈ [1 , x ], x >
1, then integrate the resulting identity withrespect to σ from 1 to x , we get H α,β ,x [ f ( x )] H α,β ,x [( gh )( x )] − H α,β ,x [ h ( x )] H α,β ,x [( gf )( x )] − H α,β ,x [( gf )( x )] H α,β ,x [ h ( x )]+ H α,β ,x [( gh )( x )] H α,β ,x [ f ( x )] ≥ , (4.38)which implies that, H α,β ,x [ f ( x )] H α,β ,x [( gh )( x )] ≥ H α,β ,x [ h ( x )] H α,β ,x [( gf )( x )] , (4.39)we get H α,β ,x [ f ( x )] H α,β ,x [ h ( x )] ≥ H α,β ,x [( gf )( x )] H α,β ,x [( gh )( x )] . (4.40)This completes the proof. 11 heorem 4.6 Suppose that f , g and h be positive and continuous functionson [1 , ∞ ) , such that ( g ( τ ) − g ( σ )) (cid:18) f ( σ ) h ( σ ) − f ( τ ) h ( τ ) (cid:19) ≥ , τ, σ ∈ (1 , x ) x > , (4.41) then for all α > , β ∈ (0 , α ∈ C and R ( α ) > , H α,β ,x [ f ( x )] H φ,ϕ ,x [( gh )( x )] + H φ,ϕ ,x [ f ( x )] H α,β ,x [( gh )( x )] H α,β ,x [ h ( x )] H φ,ϕ ,x [( gf )( x )] + H φ,ϕ ,x [ h ( x )] H α,β ,x [( gf )( x )] ≥ , (4.42) hold. Proof :Multiplying equation (4.37) by ϕ φ Γ( φ ) σ e [ ϕ − ϕ ( lnx − lnσ )] ( lnx − lnσ ) φ − ,which remain positive because σ ∈ (1 , x ) , ( x > , φ, ϕ >
0. Then integratingresulting identity with respect to σ from 1 to x , we have, H φ,ϕ ,x [ f ( x )] H α,β ,x [( gh )( x )] − H φ,ϕ ,x [ h ( x )] H α,β ,x [( gf )( x )] − H φ,ϕ ,x [( gf )( x )] H α,β ,x [ h ( x )]+ H φ,ϕ ,x [( gh )( x )] H α,β ,x [ f ( x )] ≥ , (4.43)we get, H φ,ϕ ,x [ f ( x )] H α,β ,x [( gh )( x )] + H φ,ϕ ,x [( gh )( x )] H α,β ,x [ f ( x )] ≥ H φ,ϕ ,x [ h ( x )] H α,β ,x [( gf )( x )] + H φ,ϕ ,x [( gf )( x )] H α,β ,x [ h ( x )] , (4.44)this gives the required inequality (4.42). Theorem 4.7
Suppose that f and h are two positive continuous functionsuch that f ≤ h on [0 , ∞ ) . If fh is decreasing and f is increasing on [1 , ∞ ) ,then for any p ≥ , For all x > , α, β > we have H α,β ,x [ f ( x )] H α,β ,x [ h ( x )] ≥ H α,β ,x [ f p ( x )] H α,β ,x [ h p ( x )] . (4.45) Proof : We take g = f p − in theorem 4.5. H α,β ,x [ f ( x )] H α,β ,x [ h ( x )] ≥ H α,β ,x [( f f p − )( x )] H α,β ,x [( hf p − )( x )] . (4.46)Since f ≤ h on [1 , ∞ ), then we can write, hf p − ≤ h p . (4.47)12ultiplying equation (4.47) by ψ ( x, τ ), which is positive because τ ∈ [1 , x ], x >
1, then integrate the resulting identity with respect to τ from 1 to x , weget 1 β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [ f p − h ( τ )] dττ ≤ β α Γ( α ) Z x e [ β − β ( lnx − lnτ )] ( lnx − lnτ ) α − [ h p ( τ )] dττ , (4.48)implies that H α,β ,x [ hf p − ( x )] ≤ H α,β ,x [ h p ( x )] , (4.49)and so we have, H α,β ,x [( f f p − )( x )] H α,β ,x [( hf p − )( x )] ≥ H α,β ,x [ f p ( x )] H α,β ,x [ h p ( x )] , (4.50)then from equation (4.46) and (4.50), we obtain (4.45). Theorem 4.8
Suppose that f and h are two positive continuous functionsuch that f ≤ h on [1 , ∞ ) . If fh is decreasing and f is increasing on [1 , ∞ )[ ,then for any p ≥ , for all x > , σ, β, φ, ϕ > H α,β ,x [ f ( x )] H φ,ϕ ,x h p [( x )] + H φ,ϕ ,x [ f ( x )] H α,β ,x [ h p ( x )] H α,β ,x [ h ( x )] H φ,ϕ ,x [ f p ( x )] + H φ,ϕ ,x [ h ( x )] H α,β ,x [ f p ( x )] ≥ . (4.51) Proof : We take g = f p − in theorem 4.6, then we obtain H α,β ,x [ f ( x )] H φ,ϕ ,x [ hf p − ( x )] + H φ,ϕ ,x [ f ( x )] H α,β ,x [ hf p − ( x )] H α,β ,x [ h ( x )] H φ,ϕ ,x [ f p ( x )] + H φ,ϕ ,x [ h ( x )] H α,β ,x [ f p ( x )] ≥ , (4.52)then by hypothesis, f ≤ h on [1 , ∞ ), which implies that hf p − ≤ h p . (4.53)Now, multiplying both side of (4.53) by ϕ φ Γ( φ ) σ e [ ϕ − ϕ ( lnx − lnσ )] ( lnx − lnσ ) φ − ,which remain positive because σ ∈ (1 , x ) , ( x > , φ, ϕ >
0. Then integratingresulting identity with respect to σ from 1 to x , we have, H φ,ϕ ,x [ hf p − ( x )] ≤ H φ,ϕ ,x [ h p ( x )] , (4.54)multiplying on both side of (4.50) by H α,β ,x [ f ( x )], we obtain H α,β ,x [ f ( x )] H φ,ϕ ,x [ hf p − ( x )] ≤ H α,β ,x [ f ( x )] H φ,ϕ ,x [ h p ( x )] , (4.55)hence by (4.49) and (4.55), we obtain H α,β ,x [ f ( x )] H φ,ϕ ,x [ hf p − ( x )] + H φ,ϕ ,x [ f ( x )] H α,β ,x [ hf p − ( x )] ≤ H α,β ,x [ f ( x )] H φ,ϕ ,x [ h p ( x )] + H φ,ϕ ,x [ f ( x )] H α,β ,x [ h p ( x )] . (4.56)By (4.52) and (4.56), we complete the proof of this theorem.13 eferences [1] G. A. Anastassiou, Fractional Differentiation Inequalities,
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