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Some new Riemann Liouville fractional integral inequalities Jessada Tariboona 1 Sotiris K Ntouyasb 2and Weerawat Sudsutada a Department of Mathematics Faculty of Applied Science King Mongkut s University of Technology North Bangkok Bangkok Thailand E mail jessadat kmutnb ac th wrw sst b Department of Mathematics University of Ioannina 451 10 Ioannina Greece E mail sntouyas uoi gr Abstract In this paper some new fractional integral inequalities are established Keywords fractional integral fractional integral inequalities Gr uss inequality 2010 Mathematics Subject Classifi cations 26D10 26A33 1Introduction In 2 see also 3 the Gr uss inequality is defi ned as the integral inequality that establishes a connection between the integral of the product of two functions and the product of the integrals The inequality is as follows If f and g are two continuous functions on a b satisfying m f t M and p g t P for all t a b m M p P R then fl fl fl fl 1 b a Z b a f t g t dt 1 b a 2 Z b a f t dt Z b a g t dt fl fl fl fl 1 4 M m P p The literature on Gr uss type inequalities is now vast and many extensions of the classical inequality were intensively studied by many authors In the past sev eral years by using the Riemann Liouville fractional integrals the fractional integral inequalities and applications have been addressed extensively by several researchers For example we refer the reader to 4 5 6 7 8 9 10 and the references 1Corresponding author 2Member of Nonlinear Analysis and Applied Mathematics NAAM Research Group at King Abdulaziz University Jeddah Saudi Arabia 2J Tariboon S K Ntouyas and W Sudsutad cited therein Dahmani et al 1 gave the following fractional integral inequalities by using the Riemann Liouville fractional integrals Let f and g be two integrable functions on 0 satisfying the following conditions m f t M p g t P m M p P R t 0 For all t 0 0 0 then fl fl fl fl t 1 J fg t J f t J g t fl fl fl fl t 1 2 M m P p and t 1 J fg t t 1 J fg t J f t J g t J f t J g t 2 M t 1 J f t J f t m t 1 J f t m t 1 M t 1 J f t P t 1 J g t J g t p t 1 J g t p t 1 P t 1 J g t In this paper we use the Riemann Liouville fractional integrals to establish some new fractional integral inequalities of Gr uss type We replace the constants appeared as bounds of the functions f and g by four integrable functions From our results the above inequalities of 1 and the classical Gr uss inequalities can be deduced as some special cases In Section 2 we briefl y review the necessary defi nitions Our results are given in Section 3 The proof technique is close to that presented in 1 But the obtained results are new and also can be applied to unbounded functions as shown in examples 2Preliminaries Defi nition 2 1 The Riemann Liouville fractional integral of order 0 of a func tion g L1 0 R is defi ned by J f t Z t 0 t s 1 f s ds Some new Riemann Liouville fractional integral inequalities3 and J0f t f t where is the Gamma function For the convenience of establishing our results we give the semigroup property J J f t J f t 0 0 which implies the commutative property J J f t J J f t From Defi nition 2 1 if f t t then we have J t 1 1 t 0 1 t 0 3Main Results Theorem 3 1 Let f be an integrable function on 0 Assume that H1 There exist two integrable functions 1 2on 0 such that 1 t f t 2 t for allt 0 Then for t 0 0 we have J 1 t J f t J 2 t J f t J 2 t J 1 t J f t J f t 3 1 Proof From H1 for all 0 0 we have 2 f f 1 0 Therefore 2 f 1 f 1 2 f f 3 2 Multiplying both sides of 3 2 by t 1 0 t we get f t 1 2 1 t 1 f 1 t 1 2 f t 1 f 3 3 4J Tariboon S K Ntouyas and W Sudsutad Integrating both sides of 3 3 with respect to on 0 t we obtain f Z t 0 t 1 2 d 1 Z t 0 t 1 f d 1 Z t 0 t 1 2 d f Z t 0 t 1 f d which yields f J 2 t 1 J f t 1 J 2 t f J f t 3 4 Multiplying both sides of 3 4 by t 1 0 t we have J 2 t t 1 f J f t t 1 1 J 2 t t 1 1 J f t t 1 f 3 5 Integrating both sides of 3 5 with respect to on 0 t we get J 2 t Z t 0 t 1 f d J f t Z t 0 t 1 1 d J 2 t Z t 0 t 1 1 d J f t Z t 0 t 1 f d Hence we deduce inequality 3 1 as requested This completes the proof As a special case of Theorems 3 1 we obtain the following result Corollary 3 1 Let f be an integrable function on 0 satisfying m f t M for all t 0 and m M R Then for t 0 and 0 we have m t 1 J f t M t 1 J f t mM t 1 1 J f t J f t Example 3 1 Let f be a function satisfying t f t t 1 for t 0 Then for t 0 and 0 we have 2t 1 2 t 1 J f t t 1 2 t 1 t 1 2 J f t 2 Theorem 3 2 Let f and g be two integrable functions on 0 Suppose that H1 holds and moreover we assume that H2 There exist 1and 2integrable functions on 0 such that 1 t g t 2 t for allt 0 Some new Riemann Liouville fractional integral inequalities5 Then for t 0 0 the following inequalities hold a J 1 t J f t J 2 t J g t J 1 t J 2 t J f t J g t b J 1 t J g t J 2 t J f t J 1 t J 2 t J f t J g t c J 2 t J 2 t J f t J g t J 2 t J g t J 2 t J f t d J 1 t J 1 t J f t J g t J 1 t J g t J 1 t J f t Proof To prove a from H1 and H2 we have for t 0 that 2 f g 1 0 Therefore 2 g 1 f 1 2 f g 3 6 Multiplying both sides of 3 6 by t 1 0 t we get g t 1 2 1 t 1 f 1 t 1 2 g t 1 f 3 7 Integrating both sides of 3 7 with respect to on 0 t we obtain g Z t 0 t 1 2 d 1 Z t 0 t 1 f d 1 Z t 0 t 1 2 d g Z t 0 t 1 f d Then we have g J 2 t 1 J f t 1 J 2 t g J f t 3 8 Multiplying both sides of 3 8 by t 1 0 t we have J 2 t t 1 g J f t t 1 1 J 2 t t 1 1 J f t t 1 g 3 9 Integrating both sides of 3 9 with respect to on 0 t we get the desired inequality a To prove b d we use the following inequalities b 2 g f 1 0 c 2 f g 2 0 d 1 f g 1 0 As a special case of Theorem 3 2 we have the following Corollary 6J Tariboon S K Ntouyas and W Sudsutad Corollary 3 2 Let f and g be two integrable functions on 0 Assume that H3 There exist real constants m M n N such that m f t Mandn g t Nfor allt 0 Then for t 0 0 we have a1 nt 1 J f t Mt 1 J g t nMt 1 1 J f t J g t b1 mt 1 J g t Nt 1 J f t mNt 1 1 J f t J g t c1 MNt 1 1 J f t J g t Mt 1 J g t Nt 1 J f t d1 mnt 1 1 J f t J g t mt 1 J g t nt 1 J f t Lemma 3 1 Let f be an integrable function on 0 and 1 2are two integrable functions on 0 Assume that the condition H1 holds Then for t 0 0 we have t 1 J f2 t J f t 2 J 2 t J f t J f t J 1 t t 1 J 2 t f t f t 1 t t 1 J 1f t J 1 t J f t t 1 J 2f t J 2 t J f t J 1 t J 2 t t 1 J 1 2 t 3 10 Proof For any 0 and 0 we have 2 f f 1 2 f f 1 2 f f 1 2 f f 1 f2 f2 2f f 2 f 1 f 1 2 2 f 1 f 1 2 2 f 1 2 1 f 2 f 1 2 1 f 3 11 Multiplying 3 11 by t 1 0 t t 0 and integrating the resulting identity with respect to from 0 to t we get 2 f J f t J 1 t J 2 t J f t f 1 Some new Riemann Liouville fractional integral inequalities7 J 2 t f t f t 1 t 2 f f 1 t 1 J f2 t f2 t 1 2f J f t 2 J f t f J 1 t 2 J 1 t f J 2 t 1 J f t 1 J 2 t J 2f t J 1 2 t J 1f t 2 f t 1 1 2 t 1 1 f t 1 3 12 Multiplying 3 12 by t 1 0 t t 0 and integrating the resulting identity with respect to from 0 to t we have J 2 t J f t J f t J 1 t J 2 t J f t J f t J 1 t J 2 t f t f t 1 t t 1 J 2 t f t f t 1 t t 1 t 1 J f2 t t 1 J f2 t 2J f t J f t J 2 t J f t J 1 t J f t J 1 t J 2 t J 2 t J f t J 1 t J f t J 1 t J 2 t t 1 J 2f t t 1 J 1 2 t t 1 J 1f t t 1 J 2f t t 1 J 1 2 t t 1 J 1f t 3 13 which implies 3 10 If 1 t m and 2 t M m M R for all t 0 then inequality 3 10 reduces to the following corollary 1 Lemma 3 2 Corollary 3 3 Let f be an integrable function on 0 satisfying m f t M for all t 0 The for all t 0 0 we have t 1 J f2 t J f t 2 M t 1 J f t J f t m t 1 t 1 J M f t f t m 3 14 8J Tariboon S K Ntouyas and W Sudsutad Theorem 3 3 Let f and g be two integrable functions on 0 and 1 2 1and 2are four integrable functions on 0 satisfying the conditions H1 and H2 on 0 Then for all t 0 0 we have fl fl fl fl t 1 J fg t J f t J g t fl fl fl fl p T f 1 2 T g 1 2 3 15 where T u v w is defi ned by T u v w J w t J u t J u t J v t t 1 J vu t J v t J u t t 1 J wu t J w t J u t J v t J w t t 1 J vw t 3 16 Proof Let f and g be two integrable functions defi ned on 0 satisfying H1 and H2 Defi ne H f f g g 0 t t 0 3 17 Multiplying both sides of 3 17 by t 1 t 1 2 0 t and integrating the resulting identity with respect to and from 0 to t we can state that 1 2 2 Z t 0 Z t 0 t 1 t 1H d d t 1 J fg t J f t J g t 3 18 Applying the Cauchy Schwarz inequality to 3 18 we have t 1 J fg t J f t J g t 2 t 1 J f2 t J f t 2 t 1 J g2 t J g t 2 3 19 Since 2 t f t f t 1 t 0 and 2 t g t g t 1 t 0 for t 0 we have t 1 J 2 t f t f t 1 t 0 and t 1 J 2 t g t g t 1 t 0 Some new Riemann Liouville fractional integral inequalities9 Thus from Lemma 3 1 we get t 1 J f2 t J f t 2 J 2 t J f t J f t J 1 t t 1 J 1f t J 1 t J f t t 1 J 2f t J 2 t J f t J 1 t J 2 t t 1 J 1 2 t T f 1 2 3 20 and t 1 J g2 t J g t 2 J 2 t J g t J g t J 1 t t 1 J 1g t J 1 t J g t t 1 J 2g t J 2 t J g t J 1 t J 2 t t 1 J 1 2 t T g 1 2 3 21 From 3 19 3 20 and 3 21 we obtain 3 15 Remark 3 1 If T f 1 2 T f m M and T g 1 2 T g p P m M p P R then inequality 3 15 reduces to fl fl fl fl t 1 J fg t J f t J g t fl fl fl fl t 2 1 2 M m P p 3 22 see 1 Theorem 3 1 Example 3 2 Let f and g be two functions satisfying t f t t 1 and t 1 g t t for t 0 Then for t 0 and 0 we have fl fl fl fl t 1 J fg t J f t J g t fl fl fl fl p T f t t 1 T g t 1 t where T f t t 1 t 1 2 t 1 J f t J f t t 1 2 t 1 J tf t t 1 2 J f t 10J Tariboon S K Ntouyas and W Sudsutad t 1 J t 1 f t t 1 2 t 1 J f t t 1 2 t 1 2 t 1 t 1 2t 2 3 t 1 2 and T g t 1 t t 1 2 J g t J g t t 1 2 t 1 t 1 J t 1 g t t 1 2 t 1 J g t t 1 J tg t t 1 2 J g t t 1 2 t 1 t 1 2 t 1 2t 2 3 t 1 2 Confl ict of Interests The authors declare that there is no confl ict of interests regarding the publication of this article Acknowledgement This research was funded by King Mongkut s University of Technology North Bangkok Thailand Project Code KMUTNB GRAD 56 02 References 1 Z Dahmani L Tabharit S Taf New generalisations of Gruss inequality using Riemann Liouville fractional integrals Bull Math Anal Appl 2 2010 93 99 2 G Gr uss Uber das Maximum des absoluten Betrages von 1 b a R b a f t g t