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Commun.Fac.Sci.Univ.Ank.Ser. A1 Math. Stat. Volume 70, Number 1, Pages 269–278 (2021) DOI: 10.31801/cfsuasmas.771172 ISSN 1303–5991 E-ISSN 2618–6470

Received by the editors: July 18, 2020; Accepted: December 20, 2020

SEVERAL INTEGRAL INEQUALITIES FOR GENERALIZED RIEMANN-LIOUVILLE FRACTIONAL OPERATORS

Juan Gabriel GALEANO DELGADO1, Juan E. NÁPOLES VALDÉS2, and Edgardo PÉREZ REYES1 1Facultad de Ciencias e Ingeniería, Universidad del Sinú Elías Bechara Zainúm, Montería, COLOMBIA 2 UNNE, FaCENA Ave. Libertad 5450, Corrientes 3400, ARGENTINA 2UTN-FRRE, French 414, Resistencia, Chaco 3500, ARGENTINA

Abstract. In this paper, using a generalized integral operator, of the Riemann- Liouville type, de…ned and studied in a previous work by the authors, we obtain various integral inequalities for positive functions, which contains several re- ported in the literature. Various remarks carried out throughout the work and pointed out in the Conclusions, show the scope and strength of our results, in particular, it is shown that under particular cases of the considered kernel, several known fractional integral operators are obtained.

1. INTRODUCTION Calculus, using di¤erent notions of derivatives and integrals of arbitrary order, has become in recent years one of the centers of attention of mathematical re- searchers, both pure and applied. By other hand, one of the most developed mathematical areas in the last 20 years is that of Integral Inequalities, associated with di¤erent functional notions: convex, synchronous functions within the framework of Riemann, fractional and generalized integral operators. Throughout the work we use the functions (see [8,9,10,11])and k (cf. de…ned by [2]):

2020 Mathematics Subject Classi…cation. Primary 26A33; Secondary 26D10, 47A63. Keywords and phrases. Generalized fractional operator, Riemann-Liouville integral, integral inequalities. [email protected]; [email protected] author; edgardo- [email protected] 0000-0002-3042-933X; 0000-0003-2470-1090; 0000-0002-7666-1636.

c 2021 Ankara University Communications Faculty of Sciences University of Ankara-Series A1 Mathematics and Statistics 269 270 J.G.G.DELGADO,J.E.N.VALDES,E.P.REYES

1 z 1  (z) =  e d; (z) > 0; (1) < Z0 1 z 1  k=k k(z) =  e d; k > 0: (2) 0 Z z k 1 z It is clear that if k 1 we have k(z) (z), k(z) = (k) and ! ! k k(z + k) = zk(z). As well, we de…ne the k-beta function as follows
1 1 u 1 v 1 Bk(u; v) =  k (1 ) k d; k Z0 1 u v k(u)k(v) notice that Bk(u; v) = B( ; ) and Bk(u; v) = . k k k k(u+v) In [3] the following fractional integral operator of the Riemann-Liouville type is de…ned and its main properties are studied. De…nition 1. The k-generalized fractional Riemann-Liouville integral of order with R, and s = 1 of an integrable function (u) on [0; ), are given as follows:2 6 1

u s k 1 F (; s) ()d J (u) = ; (3) F;a1 1 k ( ) k k a1 [ F(u; )] u Z with F (; 0) = 1 and F(u; ) =  F (; s)d. In this work, it was shown thatR many of the known integral operators can be obtained as particular cases of them. The main purpose of this paper, using the generalized fractional integral operator of the Riemann-Liouville type, from De…nition 1, is to establish several integral inequalities, which contain as particular cases, several of those reported in the literature.

2. MAIN RESULTS Below we present several integral inequalities, in the framework of the operators of De…nition 1, the …rst of them is the following.

Theorem 2. Let ' be a positive non-decreasing continuous function on [a1; a2] + and let h :[a1; a2] R be a positive continuous function. Then for a1 <  a2;  > 0;  > 0; > 0!and s = 1, we have  6 s k   s k s k  s k  J [( a1) h()' ()] J h() J [( a1) h()] J [h()' ()]: F;a1 F;a1  F;a1 F;a1 (4)

Proof. Since the function ' is positive continuous and non-decreasing on [a1; a2] then, for all  > 0,  > 0, u; v [a1; ], with v u, and a1 <  a2, we have 2       ((u a1) (v a1) )(' (u) ' (v)) 0: (5)  SEVERALINTEGRALINEQUALITIES 271

So, from (5) we get         (u a1) ' (u) + (v a1) ' (v) (u a1) ' (v) + (v a1) ' (u): (6)  F (u;s)h(u) Multiplying both sides of (6) by 1 , then we integrate the resulting kk( )[F(;u)] k inequality with respect to u over (a1; ), we get

s J k [( a )h()'()]+(v a )'(v) s J k h() '(v) s J k [( a )h()] F;a1 1 1 F;a1 F;a1 1   s k  + (v a1) J [h()' ()]: (7) F;a1 F (v;s)h(v) Now, multiplying both sides of (7) by 1 , then we integrate the re- kk( )[F(;v)] k sulting inequality with respect to v over (a1; ), it holds that

s J k h() s J k [( a )h()'()]+ s J k [( a )h()'()] s J k h() F;a1 F;a1 1 F;a1 1 F;a1  s k  s k  s k  s k  J [( a1) h()] J [h()' ()]+ J [h()' ()] J [( a1) h()]; F;a1 F;a1 F;a1 F;a1 (8) which implies (4).  Remark 3. If we take the kernel F (; s) =  s, with k > 0 and s R, s = 1, the previous theorem becomes the Theorem 6 of [5]. 2 6 In the following result, two functional parameters are used, extending the previ- ous theorem.

Theorem 4. Let ' be a positive non-decreasing continuous function on [a1; a2] + and let h :[a1; a2] R be a positive continuous function. Then for a1 <  a2;  > 0;  > 0; > 0!, > 0 and s = 1, we have  6

s k   s k s k   s k J [( a1) h()' ()] J h() + J [( a1) h()' ()] J h() F;a1 F;a1 F;a1 F;a1

s k  s k  J [h()' ()] J [( a1) h()]+  F;a1 F;a1

s k  s k  J [( a1) h()] J [h()' ()]: (9) F;a1 F;a1 F (v;s)h(v) Proof. Multiplying both sides of (7) by , then we integrate the 1 kk( )[F(;v)] k resulting inequality with respect to v over (a1; ), we obtain (9).  Remark 5. If in Theorem 2 we take = , we obtain Theorem 4. Remark 6. Under the same conditions as the previous Remark, this result covers Theorem 7 of [5]. The following result generalizes the Theorem 2, by including an appropriate h function. 272 J.G.G.DELGADO,J.E.N.VALDES,E.P.REYES

Theorem 7. Let ' and be two positive continuous function on [a1; a2] such that + ' is non-decreasing, is non-increasing, and let h :[a1; a2] R be a positive ! continuous function. Then for a1 <  a2;  > 0;  > 0; > 0 and s = 1, we have  6

s J k [h() ()] s J k [h()'()] s J k h() s J k [h()'() ()] F;a1 F;a1  F;a1 F;a1 (10)

Proof. Since the functions ' and are positive continuous on [a1; a2] with ' is non-decreasing and is non-increasing, then for all  > 0,  > 0, u; v [a1; ], 2 with v u, and a1 <  a2, we have   '(u) (v) + '(v) (u) '(u) (u) + '(v) (v): (11)  F (u;s)h(u) Multiplying (11) by 1 , then we integrate the resulting inequality kk( )[F(;u)] k with respect to u over (a1; ), it holds that

(v) sJ k [h()'()] + '(v) sJ k [h() ()] sJ k [h()'() ()] F;a1 F;a1 F;a1  + '(v) (v) sJ k h(): (12) F;a1 F (v;s)h(v) Now, we multiple (12) by 1 , then we integrate the resulting inequality kk( )[F(;v)] k with respect to v over (a1; ), we get

s J k [h() ()] s J k [h()'()] + s J k [h()'()] s J k [h() ()] F;a1 F;a1 F;a1 F;a1  s J k h() s J k [h()'() ()] + s J k [h()'() ()] s J k h(); F;a1 F;a1 F;a1 F;a1 which implies (10).  Remark 8. As before, if we have F (; s) =  s, with k > 0 and s R, s = 1, this theorem reduces to Theorem 9 of [5]. 2 6

Theorem 9. Let ' and be two positive continuous function on [a1; a2] such that + ' is non-decreasing, is non-increasing, and let h :[a1; a2] R be a positive ! continuous function. Then for a1 <  a2;  > 0;  > 0; > 0; > 0 and s = 1, we have  6

s J k [h() ()] s J k [h()'()] + s J k [h()'()] s J k [h() ()] F;a1 F;a1 F;a1 F;a1 

s J k h() s J k [h()'() ()] + s J k [h()'() ()] s J k h(): F;a1 F;a1 F;a1 F;a1 (13)

F (v;s)h(v) Proof. If we multiple (12) by , then we integrate the resulting in- 1 kk( )[F(;v)] k equality with respect to v over (a1; ), we obtain (13).  SEVERALINTEGRALINEQUALITIES 273

Remark 10. If in Theorem 9 we put = we obtain Theorem 7. Remark 11. Theorem 10 of [5] is obtained from Theorema 9, considering, as before, the kernel F (; s) =  s, with k > 0 and s R, s = 1. 2 6 Theorem 12. Let ' be a positive decreasing continuous function on [a1; a2] and + let h :[a1; a2] R be a positive continuous function. Then for a1 <  a2;   > 0;  > 0!; > 0 and s = 1, we have   6

sJ k [( a )h()'()] sJ k [h()'()] F;a1 1 F;a1  s k   s k  J [( a1) h()' () J [h()' ()]: (14) F;a1 F;a1 Proof. Since the function ' is positive continuous and decreasing on [a1; a2] then, for all   0,  > 0, u; v [a1; ], with v u, and a1 <  a2, we have  2  

            (u a1) ' (v) + (v a1) ' (u) (u a1) ' (u) + (v a1) ' (v): (15)  F (u;s)h(u)' (u) Multiplying both sides of (15) by 1 , then we integrate the resulting kk( )[F(;u)] k inequality with respect to u over (a1; ), we get

' (v) s J k [( a )h()'()] + (v a ) s J k [h()'()] F;a1 1 1 F;a1  s k      s k  J [( a1) h()' ()] + (v a1) ' (v) J [h()' ()]: (16) F;a1 F;a1 F (v;s)h(v)' (v) Now, multiplying both sides of (16) by 1 , then we integrate the kk( )[F(;v)] k resulting inequality with respect to v over (a1; ), it holds that

sJ k [h()'()] sJ k [( a )h()'()]+ F;a1 F;a1 1 sJ k [( a )h()'()] sJ k [h()'()] F;a1 1 F;a1  sJ k [h()'()] sJ k [( a )h()'()]+ F;a1 F;a1 1 s k   s k  J [( a1) h()' ()] J [h()' ()]; (17) F;a1 F;a1 which implies (14).  Remark 13. This Theorem contains as a particular case Theorem 12 of [5], under the same conditions as the previous Remarks. The latest results of our work is the extension of the previous Theorems consid- ering the product of a family of appropriate functions.

Theorem 14. Let 'j, j = 1; 2; : : : ; n are n positive continuous and decreasing + functions on [a1; a2] and let h :[a1; a2] R be a positive continuous function. ! Then for a1 <  a2; > 0, s = 1,  > 0 and  r > 0 with r 1; 2; : : : ; n , we have  6  2 f g 274 J.G.G.DELGADO,J.E.N.VALDES,E.P.REYES

n n

s k  j s k  j J h()' () ' () J ( a1) h() ' () F;a1 2 r j 3 F;a1 2 j 3  j=r j=1 Y6 Y 4 n 5 4 n 5

s k j s k   j J h() ' () J ( a1) h()' () ' () : (18) F;a1 2 j 3 F;a1 2 r j 3 j=1 j=r Y Y6 4 5 4 5 Proof. Since the function 'r is positive continuous and decreasing on [a1; a2] then, for all  > 0, a1 <  a2, u; v [a1; ], with v u, and  r > 0, for any …xed r 1; 2; : : : ; n , we get 2   2 f g    r  r ((u a1) (v a1) )(' (v) ' (u)) 0: (19) r r  So, from (19), we deduce that

  r   r   r   r (u a1) ' (v)+(v a1) ' (u) (u a1) ' (u)+(v a1) ' (v): (20) r r  r r n j F (u;s)h(u) j=1 'j (u) Multiplying both sides of (20) by 1 , then we integrate the re- kk( )[FQ(;u)] k sulting inequality with respect to u over (a1; ), we get

n

 r s k  j ' (v) J ( a1) h() ' () + r F;a1 2 j 3 j=1 Y 4 5 n

 s k  j (v a1) J h()' () ' () F;a1 2 r j 3  j=r Y6 4 n 5

s k   j J ( a1) h()' () ' () + F;a1 2 r j 3 j=r Y6 4 5 n

  r s k j (v a1) ' (v) J h() ' () : (21) r F;a1 2 j 3 j=1 Y n 4j 5 F (v;s)h(v) j=1 'j (v) Lastly, multiplying both sides of (21) by 1 , then we integrate the kk( )[FQ(;v)] k resulting inequality with respect to v over (a1; ), which concludes the inequality (18). 

Theorem 15. Let 'j, j = 1; 2; : : : ; n are n positive continuous and decreasing + functions on [a1; a2] and let h :[a1; a2] R be a positive continuous function. ! Then for a1 <  a2; > 0, > 0, s = 1,  > 0 and  r > 0 with r 1; 2; : : : ; n , we have 6  2 f g n n

s k  j s k  j J ( a1) h() ' () J h()' () ' () F;a1 2 j 3 F;a1 2 r j 3 j=1 j=r Y Y6 4 5 4 5 SEVERALINTEGRALINEQUALITIES 275

n n

s k  j s k  j + J h()' () ' () J ( a1) h() ' () F;a1 2 r j 3 F;a1 2 j 3  j=r j=1 Y6 Y 4 5 n 4 n 5

s k   j s k j J ( a1) h()' () ' () J h() ' () F;a1 2 r j 3 F;a1 2 j 3 j=r j=1 Y6 Y 4 n 5 4 n 5

s k j s k   j + J h() ' () J ( a1) h()' () ' () : (22) F;a1 2 j 3 F;a1 2 r j 3 j=1 j=r Y Y6 4 5 4 n j 5 F (v;s)h(v) j=1 'j (v) Proof. If we multiple both sides of (21) by , then we integrate 1 kk( )[FQ(;v)] k the resulting inequality with respect to v over (a1; ), with which we deduce the inequality (22).  Remark 16. If in Theorem 15 we put = we obtain Theorem 14.

Theorem 17. Let 'j, j = 1; 2; : : : ; n and are positive continuous functions on [a1; a2], such that is increasing and 'j, j = 1; 2; : : : ; n are decreasing on [a1; a2] + and let h :[a1; a2] R be a positive continuous function. Then for a1 <  a2; !  > 0, s = 1,  > 0 and  r > 0 with r 1; 2; : : : ; n , we have 6  2 f g n n

s J k h()' () ' j () s J k ()h() ' j () F;a1 2 r j 3 F;a1 2 j 3  j=r j=1 Y6 Y 4 n 5 4 n5

s J k h() ' j () s J k ()h()' () ' j () : (23) F;a1 2 j 3 F;a1 2 r j 3 j=1 j=r Y Y6 4 5 4 5 Proof. Since the functions 'r and are positive continuous on [a1; a2] with 'r decreasing and increasing; then, for all  > 0, a1 <  a2, u; v [a1; ], with  2 v u, and  r > 0, for any …xed r 1; 2; : : : ; n , we get   2 f g   r   r   r   r (u)' (v) + (v)' (u) (u)' (u) + (v)' (v): (24) r r  r r n j F (u;s)h(u) j=1 'j (u) Multiplying both sides of (24) by 1 , then we integrate the re- kk( )[FQ(;u)] k sulting inequality with respect to u over (a1; ), we deduce

n n

 r s k  j  s k  j ' (v) J ()h() ' () + (v) J h()' () ' () r F;a1 2 j 3 F;a1 2 r j 3  j=1 j=r Y Y6 4 n 5 4 n 5 s k   j   r s k j J ()h()' () ' () + (v)' (v) J h() ' () : F;a1 2 r j 3 r F;a1 2 j 3 j=r j=1 Y6 Y 4 5 4 5(25) 276 J.G.G.DELGADO,J.E.N.VALDES,E.P.REYES

n j F (v;s)h(v) j=1 'j (v) Lastly, multiplying both sides of (25) by 1 , then we integrate the kk( )[FQ(;v)] k resulting inequality with respect to v over (a1; ), which concludes the inequality (23). 

Theorem 18. Let 'j, j = 1; 2; : : : ; n and are positive continuous functions on [a1; a2], such that is increasing and 'j, j = 1; 2; : : : ; n are decreasing on [a1; a2] + and let h :[a1; a2] R be a positive continuous function. Then for a1 <  a2; !  > 0, > 0, s = 1,  > 0 and  r > 0 with r 1; 2; : : : ; n , we have 6  2 f g n n

s J k h()' () ' j () s J k ()h() ' j () F;a1 2 r j 3 F;a1 2 j 3 j=r j=1 Y6 Y 4 n 5 4 n 5

+ s J k h()' () ' j () s J k ()h() ' j () F;a1 2 r j 3 F;a1 2 j 3  j=r j=1 Y6 Y 4 n 5 4 n 5

s J k h() ' j () s J k ()h()' () ' j () F;a1 2 j 3 F;a1 2 r j 3 j=1 j=r Y Y6 4 n 5 4 n 5

+ s J k h() ' j () s J k ()h()' () ' j () : (26) F;a1 2 j 3 F;a1 2 r j 3 j=1 j=r Y Y6 4 5 4 n j 5 F (v;s)h(v) j=1 'j (v) Proof. If we multiple both sides of (25) by , then we integrate 1 kk( )[FQ(;v)] k the resulting inequality with respect to v over (a1; ), with which we deduce the inequality (26).  Remark 19. If in Theorem 18 we put = we obtain Theorem 17. Remark 20. Theorems 14, 15, 17 and 18, are generalizations of Theorems 18, 19, 21 and 22 of [5], respectively.

3. CONCLUSIONS In this work we have obtained various inequalities of the Hermite-Hadamard type, in the case of di¤erent notions of convexity, and using a generalized fractional operator, which allows obtaining as particular cases, several of those reported in the literature. We want to point out, in addition to the observations made throughout the work, that with di¤erent choices of the F kernel we can obtain, as particular cases, several well-known integral operators. So, for example, if 1 A) The classic Riemann integral is obtained with F (t; ) = t , = 1 and = k (with notation changed). SEVERALINTEGRALINEQUALITIES 277

1 B) If F (t; ) = t and = k we obtain the fractional Riemann-Liouville integral. C) Considering F (t; ) = ts with s = 1, we can write the right sided operator x f(t)dt k 1 as follows J f (x) = and similarly the left sided F;a+ k k( ) a 1 (x t) k integral. The k-Riemann-Liouville fractionalR integral of Mubeen and Habibullah (see [7]).

D) Katugampola fractional integral of [6] is obtained, taking F (t; ) = t (the notation is changed). s E) If we put F = t with s = 1, then we get the right sided Hadamard fractional integral of [4]. F) An integral operator with non-singular kernel can also be obtained from our 1 De…nition 1. Thus, considering F (t; ) = exp t , if = 1 we have that 1 F = 1. In this case F ( F+(x; t); ) = exp (x t) , a slight modi…cation of the   operator de…ned by Kirane and Toberek inh [1]. i

From all of the above, we can conclude that many of the integral inequalities obtained in the framework of these integral operators, can be obtained as particular cases from those obtained in this work.

Authors Contribution Statement The authors contributed equally to this work. All authors of the submitted research paper have directly participated in the plan- ning, execution, or analysis of study.

Declaration of Competing Interest The authors declare that there is no con‡ict of interest regarding the publication of this article. Besides, the contents of the manuscript have not been submitted, copyrighted or published elsewhere and the visual-graphical materials such as photograph, drawing, picture, and document within the article do not have any copyright issue. Finally, all authors of the paper have read and approved the …nal version submitted.

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