On new inequalities via Riemann-Liouville fractional integration
aa r X i v : . [ m a t h . C A ] M a y ON NEW INEQUALITIES VIA RIEMANN-LIOUVILLEFRACTIONAL INTEGRATION
MEHMET ZEKI SARIKAYA ⋆ ♣ AND HASAN OGUNMEZ (cid:7)
Abstract.
In this paper, we extend the Montogomery identities for the Riemann-Liouville fractional integrals. We also use this Montogomery identities to es-tablish some new integral inequalities for convex functions. Introduction
The inequality of Ostrowski [16] gives us an estimate for the deviation of thevalues of a smooth function from its mean value. More precisely, if f : [ a, b ] → R isa differentiable function with bounded derivative, then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( x ) − b − a b Z a f ( t ) dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ "
14 + ( x − a + b ) ( b − a ) ( b − a ) k f ′ k ∞ for every x ∈ [ a, b ]. Moreover the constant 1 / f : [ a, b ] → R is differentiable on [ a, b ] with the first derivative f ′ integrableon [ a, b ] , then Montgomery identity holds: f ( x ) = 1 b − a b Z a f ( t ) dt + b Z a P ( x, t ) f ′ ( t ) dt, where P ( x, t ) is the Peano kernel defined by P ( x, t ) := t − ab − a , a ≤ t < xt − bb − a , x ≤ t ≤ b. Recently, several generalizations of the Ostrowski integral inequality are consid-ered by many authors; for instance covering the following concepts: functions ofbounded variation, Lipschitzian, monotonic, absolutely continuous and n -times dif-ferentiable mappings with error estimates with some special means together withsome numerical quadrature rules. For recent results and generalizations concerning Mathematics Subject Classification.
Key words and phrases.
Riemann-Liouville fractional integral, convex function, Ostrowskiinequality. ⋆ corresponding author. ⋆ ♣ AND HASAN OGUNMEZ (cid:7)
Ostrowski’s inequality, we refer the reader to the recent papers [3], [6], [9]-[11],[13]-[15].In this article, we use the Riemann-Liouville fractional integrals to establish somenew integral inequalities of Ostrowski’s type. From our results, the weighted andthe classical Ostrowski’s inequalities can be deduced as some special cases.2.
Fractional Calculus
Firstly, we give some necessary definitions and mathematical preliminaries offractional calculus theory which are used further in this paper. More details, onecan consult [7], [12].
Definition 1.
The Riemann-Liouville fractional integral operator of order α ≥ with a ≥ is defined as J αa f ( x ) = 1Γ( α ) x Z a ( x − t ) α − f ( t ) dt,J a f ( x ) = f ( x ) . Recently, many authors have studied a number of inequalities by used the Riemann-Liouville fractional integrals, see ([1], [2], [4], [5]) and the references cited therein.3.
Main Results
In order to prove our main results, we need the following identities, which correctsthe result proved in [1]:
Lemma 1.
Let f : I ⊂ R → R be differentiable function on I ◦ with a, b ∈ I ( a < b )and f ′ ∈ L [ a, b ] , then (3.1) f ( x ) = Γ( α ) b − a ( b − x ) − α J αa f ( b ) − J α − a ( P ( x, b ) f ( b )) + J αa ( P ( x, b ) f ′ ( b )) , α ≥ , where P ( x, t ) is the fractional Peano kernel defined by (3.2) P ( x, t ) := t − ab − a ( b − x ) − α Γ( α ) , a ≤ t < xt − bb − a ( b − x ) − α Γ( α ) , x ≤ t ≤ b. N NEW INEQUALITIES 3
Proof.
By definition of P ( x, t ), we haveΓ( α ) J αa ( P ( x, b ) f ′ ( b )) = b Z a ( b − t ) α − P ( x, t ) f ′ ( t ) dt = x Z a ( b − t ) α − (cid:18) t − ab − a (cid:19) f ′ ( t ) dt + b Z x ( b − t ) α − (cid:18) t − bb − a (cid:19) f ′ ( t ) dt = 1 b − a x Z a ( b − t ) α − ( t − a ) f ′ ( t ) dt − b Z x ( b − t ) α f ′ ( t ) dt = 1 b − a ( I + I ) . Integrating by parts, we can state: I = ( b − t ) α − ( t − a ) f ( t ) (cid:12)(cid:12) xa − x Z a [ − ( α − b − t ) α − ( t − a ) + ( b − t ) α − ] f ( t ) dt (3.3)= ( b − x ) α − ( x − a ) f ( x ) + ( α − x Z a ( b − t ) α − ( t − a ) f ( t ) dt − x Z a ( b − t ) α − f ( t ) dt and similary,(3.4) I = − ( b − t ) α f ( t ) | bx − α b Z x ( b − t ) α − f ( t ) dt = ( b − x ) α f ( x ) − α b Z x ( b − t ) α − f ( t ) dt Adding (3.3) and (3.4), we getΓ( α ) J αa ( P ( x, b ) f ′ ( b )) = 1 b − a ( b − a )( b − x ) α − f ( x ) + ( α − x Z a ( b − t ) α − ( t − a ) f ( t ) dt − α b Z x ( b − t ) α − f ( t ) dt − x Z a ( b − t ) α − f ( t ) dt . MEHMET ZEKI SARIKAYA ⋆ ♣ AND HASAN OGUNMEZ (cid:7)
If we add and subtract the integral ( α − b R x ( b − t ) α − ( t − b ) f ( t ) dt to the right handside of the equation above, then we haveΓ( α ) J αa ( P ( x, b ) f ′ ( b )) = 1 b − a (cid:8) ( b − a )( b − x ) α − f ( x )+( b − a )( α − b Z a ( b − t ) α − P ( x, t ) f ( t ) dt − b Z a ( b − t ) α − f ( t ) dt = ( b − x ) α − f ( x ) + ( α − b Z a ( b − t ) α − P ( x, t ) f ( t ) dt − b − a b Z a ( b − t ) α − f ( t ) dt = ( b − x ) α − f ( x ) − Γ( α ) b − a J αa f ( b ) + Γ( α ) J α − a ( P ( x, b ) f ( b )) . Multiplying the both sides by ( b − x ) − α , we obtain J αa ( P ( x, b ) f ′ ( b )) = f ( x ) − Γ( α ) b − a ( b − x ) − α J αa f ( b ) + J α − a ( P ( x, b ) f ( b ))and so f ( x ) = Γ( α ) b − a ( b − x ) − α J αa f ( b ) − J α − a ( P ( x, b ) f ( b )) + J αa ( P ( x, b ) f ′ ( b )) . This completes the proof. (cid:3)
Remark 1.
If we choose α = 1 , the formula (3.1) reduces to the classical Mont-gomery Identity with P ( x, b ) f ( b ) = 0 , x ≤ t ≤ b . However, because of the pacesof the < and ≤ signs in [1] , the correspınding result therein does not reduce to theclassical Montgomery Identity. Theorem 1.
Let f : [ a, b ] → R be a convex function on [ a, b ] . Then for any x ∈ ( a, b ) , the following inequality holds: α ( α + 1) (cid:20) α ( b − x ) b − a f ′ + ( x ) − (cid:18) ( b − a ) α ( b − x ) − α + α ( b − x ) b − a − ( α + 1)( b − x ) (cid:19) f ′− ( x ) (cid:21) (3.5) ≤ Γ( α ) b − a ( b − x ) − α J αa f ( b ) − J α − a ( P ( x, b ) f ( b )) − f ( x ) , α ≥ . Proof.
From Lemma 1, we have f ( x ) − Γ( α ) b − a ( b − x ) − α J αa f ( b ) + J α − a ( P ( x, b ) f ( b ))(3.6) = ( b − x ) − α b − a x Z a ( b − t ) α − ( t − a ) f ′ ( t ) dt − b Z x ( b − t ) α f ′ ( t ) dt . Since f is convex, then for any x ∈ ( a, b ) we have the following inequalities(3.7) f ′ ( t ) ≤ f ′− ( x ) for a.e. t ∈ [ a, x ](3.8) f ′ ( t ) ≥ f ′ + ( x ) for a.e. t ∈ [ x, b ] . N NEW INEQUALITIES 5
If we multiply (3.7) by ( b − t ) α − ( t − a ) ≥ , t ∈ [ a, x ] , α ≥ a, x ] , we get x Z a ( b − t ) α − ( t − a ) f ′ ( t ) dt ≤ x Z a ( b − t ) α − ( t − a ) f ′ − ( t ) dt (3.9) = 1 α ( α + 1) (cid:2) ( b − a ) α +1 + ( b − x ) α [ α ( b − x ) − ( α + 1) ( b − a )] (cid:3) f ′− ( x )and if we multiply (3.8) by ( b − t ) α ≥ , t ∈ [ x, b ] , α ≥ x, b ] , we also get(3.10) b Z x ( b − t ) α f ′ ( t ) dt ≥ b Z x ( b − t ) α f ′ + ( t ) dt = ( b − x ) α +1 α + 1 f ′ + ( x ) . Finally, if we subtract (3.10) from (3.9) and use the representtation (3.6) we deducethe desired inequality (3.5). (cid:3)
Corollary 1.
Under the assumptions Theorem 1 with α = 1 , we have (cid:2) ( b − x ) f ′ + ( x ) − ( a − x ) f ′− ( x ) (cid:3) ≤ b Z a f ( t ) dt − ( b − a ) f ( x ) . The proof of the Corrollary 1 is proved by Dragomir in [10]. Hence, our resultsin Theorem 1 are generalizations of the corresponding results of Dragomir [10].
Remark 2.
If we take x = a + b in Corollary 1, we get ≤ b − a (cid:20) f ′ + ( a + b − f ′− ( a + b (cid:21) ≤ b − a b Z a f ( t ) dt − f ( a + b . Theorem 2.
Let f : [ a, b ] → R be a convex function on [ a, b ] . Then for any x ∈ [ a, b ] , the following inequality holds: Γ( α ) b − a ( b − x ) − α J αa f ( b ) − J α − a ( P ( x, b ) f ( b )) − f ( x )(3.11) ≤ α ( α + 1) (cid:20) α ( b − x ) b − a f ′− ( b ) − (cid:18) ( b − a ) α ( b − x ) − α + α ( b − x ) b − a − ( α + 1)( b − x ) (cid:19) f ′ + ( a ) (cid:21) , α ≥ . Proof.
Assume that f ′ + ( a ) and f ′− ( b ) are finite. Since f is convex on [ a, b ], then wehave the following inequalities(3.12) f ′ ( t ) ≥ f ′ + ( a ) for a.e. t ∈ [ a, x ](3.13) f ′ ( t ) ≤ f ′− ( b ) for a.e. t ∈ [ x, b ] . MEHMET ZEKI SARIKAYA ⋆ ♣ AND HASAN OGUNMEZ (cid:7)
If we multiply (3.12) by ( b − t ) α − ( t − a ) ≥ , t ∈ [ a, x ] , α ≥ a, x ] , we have x Z a ( b − t ) α − ( t − a ) f ′ ( t ) dt ≥ x Z a ( b − t ) α − ( t − a ) f ′ + ( a ) dt (3.14) = 1 α ( α + 1) (cid:2) ( b − a ) α +1 + ( b − x ) α [ α ( b − x ) − ( α + 1) ( b − a )] (cid:3) f ′ + ( a )and if we multiply (3.13) by ( b − t ) α ≥ , t ∈ [ x, b ] , α ≥ x, b ] , we also have(3.15) b Z x ( b − t ) α f ′ ( t ) dt ≤ b Z x ( b − t ) α f ′ + ( t ) dt = ( b − x ) α +1 α + 1 f ′− ( b ) . Finally, if we subtract (3.14) from (3.15) and use the representtation (3.6) we deducethe desired inequality (3.11). (cid:3)
Corollary 2.
Under the assumptions Theorem 2 with α = 1 , we have b Z a f ( t ) dt − ( b − a ) f ( x ) ≤ (cid:2) ( b − x ) f ′− ( b ) − ( a − x ) f ′ + ( a ) (cid:3) . The proof of the Corrollary 2 is proved by Dragomir in [10]. So, our results inTheorem 2 are generalizations of the corresponding results of Dragomir [10].
Remark 3.
If we take x = a + b in Corollary 2, we get ≤ b − a b Z a f ( t ) dt − f ( a + b ≤ b − a (cid:2) f ′− ( b ) − f ′ + ( a ) (cid:3) . Now, we extend the Lemma 1 as follows:
Theorem 3.
Let f : I ⊂ R → R be twice differentiable function on I ◦ with f ′ ∈ L [ a, b ] , then the following identity holds: (3.16)(1 − λ ) f ( x ) = Γ( α ) b − a ( b − x ) − α J αa f ( b ) − λ (cid:16) b − ab − x (cid:17) α − f ( a ) − J α − a ( P ( x, b ) f ( b )) + J αa ( P ( x, b ) f ′ ( b )) , α ≥ , where P ( x, t ) is the fractional Peano kernel defined by (3.17) P ( x, t ) := t − (1 − λ ) a − λbb − a ( b − x ) − α Γ( α ) , a ≤ t < xt − (1 − λ ) b − λab − a ( b − x ) − α Γ( α ) , x ≤ t ≤ b for ≤ λ ≤ . N NEW INEQUALITIES 7
Proof.
By similar way in proof of the Lemma 1, we haveΓ( α ) J αa ( P ( x, b ) f ′ ( b )) = b Z a ( b − t ) α − P ( x, t ) f ′ ( t ) dt = Γ( α )( b − x ) − α b − a x Z a ( b − t ) α − ( t − (1 − λ ) a − λb ) f ′ ( t ) dt (3.18) + b Z x ( b − t ) α − ( t − (1 − λ ) b − λa ) f ′ ( t ) dt = Γ( α )( b − x ) − α b − a ( J + J ) . Integrating by parts, we can state: J = ( b − x ) α − ( x − (1 − λ ) a − λb ) f ( x ) + ( b − a ) α f ( a )+( α − x Z a ( b − t ) α − ( t − (1 − λ ) a − λb ) f ( t ) dt − x Z a ( b − t ) α − f ( t ) dt and similary, J = − ( b − x ) α ( x − (1 − λ ) b − λa ) f ( x )+( α − b Z x ( b − t ) α − ( t − (1 − λ ) a − λb ) f ( t ) dt − b Z x ( b − t ) α − f ( t ) dt. Thus, using J and J in (3.18) we get (3.16) which completes the proof. (cid:3) Remark 4.
We note that in the special cases, if we take λ = 0 in Theorem 3, thenwe get (3.1) with the kernel P ( x, t ) . Theorem 4.
Let f : [ a, b ] → R be differentiable on ( a, b ) such that f ′ ∈ L [ a, b ] , where a < b. If (cid:12)(cid:12)(cid:12) f ′ ( x ) (cid:12)(cid:12)(cid:12) ≤ M for every x ∈ [ a, b ] and α ≥ , then the followinginequalty holds: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 − λ ) f ( x ) − Γ( α ) b − a ( b − x ) − α J αa f ( b ) + λ (cid:18) b − ab − x (cid:19) α − f ( a ) + J α − a ( P ( x, b ) f ( b )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Mα ( α + 1) (cid:8) ( b − a ) α ( b − x ) − α (cid:2) λ α +1 + 2(1 − λ ) α +1 + λ ( b − a ) − (cid:3) (3.19)+( b − x ) (cid:20) α b − xb − a − ( α + 1) (cid:21)(cid:27) . MEHMET ZEKI SARIKAYA ⋆ ♣ AND HASAN OGUNMEZ (cid:7)
Proof.
From Theorem 3, we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 − λ ) f ( x ) = Γ( α ) b − a ( b − x ) − α J αa f ( b ) − λ (cid:18) b − ab − x (cid:19) α − f ( a ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ α ) b Z a ( b − t ) α − | P ( x, t ) | (cid:12)(cid:12)(cid:12) f ′ ( t ) (cid:12)(cid:12)(cid:12) dt = ( b − x ) − α b − a x Z a ( b − t ) α − | t − (1 − λ ) a − λb | (cid:12)(cid:12)(cid:12) f ′ ( t ) (cid:12)(cid:12)(cid:12) dt (3.20)+ b Z x ( b − t ) α − | t − (1 − λ ) b − λa | (cid:12)(cid:12)(cid:12) f ′ ( t ) (cid:12)(cid:12)(cid:12) dt ≤ M ( b − x ) − α b − a x Z a ( b − t ) α − | t − (1 − λ ) a − λb | dt + b Z x ( b − t ) α − | t − (1 − λ ) b − λa | dt = M ( b − x ) − α b − a { J + J } . By simple computation, we obtain J = x Z a ( b − t ) α − | t − (1 − λ ) a − λb | dt = λb +(1 − λ ) a Z a ( b − t ) α − ( λb + (1 − λ ) a − t ) dt + x Z λb +(1 − λ ) a ( b − t ) α − ( t − λb − (1 − λ ) a ) dt = ( b − a ) α +1 α ( α + 1) (cid:2) − λ ) α +1 + λ ( b − a ) − (cid:3) + ( b − x ) α α ( α + 1) [ α ( b − x ) − (1 − λ )( b − a ) ( α + 1)]and similarly J = b Z x ( b − t ) α − | t − (1 − λ ) b − λa | dt = λa +(1 − λ ) b Z x ( b − t ) α − ( λa + (1 − λ ) b − t ) dt + b Z λa +(1 − λ ) b ( b − t ) α − ( t − λa − (1 − λ ) b ) dt = 2 λ α +1 ( b − a ) α +1 α ( α + 1) + ( b − x ) α α ( α + 1) [ α ( b − x ) − λ ( b − a ) ( α + 1)] . Using J and J in (3.20), we obtain (3.19). (cid:3) N NEW INEQUALITIES 9
Remark 5.
We note that in the special cases, if we take λ = 0 in Theorem 4,then it reduces Theorem 4.1 proved by Anastassiou et. al. [1] . So, our results aregeneralizations of the corresponding results of Anastassiou et. al. [1] . References [1] G. Anastassiou, M.R. Hooshmandasl, A. Ghasemi and F. Moftakharzadeh,
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E-mail address : [email protected] (cid:7) Department of Mathematics, Faculty of Science and Arts, Afyon Kocatepe Uni-versity, Afyon-TURKEY
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