Research article

The Minkowski’s inequality by means of a generalized fractional integral

  • Received: 19 December 2017 Accepted: 23 February 2018 Published: 07 March 2018
  • We use the definition of a fractional integral, recently proposed by Katugampola, to establish a generalization of the reverse Minkowski's inequality. We show two new theorems associated with this inequality, as well as state and show other inequalities related to this fractional operator.

    Citation: J. Vanterler da C. Sousa, E. Capelas de Oliveira. The Minkowski’s inequality by means of a generalized fractional integral[J]. AIMS Mathematics, 2018, 3(1): 131-147. doi: 10.3934/Math.2018.1.131

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  • We use the definition of a fractional integral, recently proposed by Katugampola, to establish a generalization of the reverse Minkowski's inequality. We show two new theorems associated with this inequality, as well as state and show other inequalities related to this fractional operator.



    Studies involving integral inequalities are important in several areas of science: mathematics, physics, engineering, among others, in particular we mention: initial value problem, linear transformation stability, integral-differential equations, and impulse equations [1,2].

    The space of p-integrable functions Lp(a,b) plays a relevant role in the study of inequalities involving integrals and sums. Further, it is possible to extend this space of p-integrable functions, to the space of the measurable Lebesgue functions, denoted by Xpc(a,b), in which the space Lp(a,b) is contained [3]. Thus, new results involving integral inequalities have been possible and consequently, some applications have been made [1,2]. We mention few of them, the inequalities of: Minkowski, Hölder, Hardy, Hermite-Hadamard, Jensen, among others [4,5,6,7,12,13,14].

    On the other hand, when we consider the non-integer order calculus or fractional calculus, as widely known, we are able in some cases adapt the theoretical model to the experimental data, in addition, it is used to generalize integrals and derivatives, in integrating inequalities. There are many definitions of fractional integrals, for example: Riemann-Liouville, Hadamard, Liouville, Weyl, Erdélyi-Kober and Katugampola [3,15,16,17,18]. Recently, Khalil et al. [19] and Adeljawad [20], introduced the local conformable fractional integrals and derivatives. From such fractional integrals, one obtains generalizations of the inequalities: Hadamard, Hermite-Hadamard, Opial, Grüss, Ostrowski, Gronwall among others [21,22,23,24,25,26,27,28,29,30,31,32].

    Recently, Katugampola [33] proposed a fractional integral unifying other well known ones: Riemann-Liouville, Hadamard, Weyl, Liouville and Erdélyi-Kober. Motivated by this formulation, we present a generalization of the reverse Minkowski's inequality [34,35,36], using the fractional integral introduced by Katugampola. We point out that studies in this direction, involving fractional integrals, are growing in several branches of mathematics [23,37,38].

    The work is organized as follows: In section 1, we present the definition of the fractional integral, as well as its particular cases. We present the main theorems involving the reverse Minkowski's inequality, as well as the suitable spaces for such definitions. In section 2, our main result, we propose the reverse Minkowski's inequality using the fractional integral. In section 3, we discuss other inequalities involving this fractional integral. Concluding remarks close the article.


    1. Prelimiaries

    In this section, we present the reverse Minkowski's inequality theorem associated with the classical Riemann integral and its respective generalization via Riemann-Liouville and Hadamard fractional integrals. In addition, we present the fractional integral introduced by Katugampola, and we conclude with a theorem in order to recover particular cases.

    Erhan et al. [5] address the inequalities of Hermite-Hadamard and reverse Minkowski for two functions f and g by means of the classical Riemann integral. On the other hand, Lazhar [7] also proposed a work related to the inequality involving integrals, that is, Hardy's inequality and the reverse Minkowski's inequality. Two theorems below have been motivation for the works performed so far, via the Riemann-Liouville and Hadamard integrals, involving the reverse Minkowski's inequality.

    Definition 1. [3] The space Xpc(a,b)(cR,1p) consists of those complex-valued Lebesgue measurable functions f on (a,b), for which fXpc< with

    fXpc=(ba|xcf(x)|pdxx)1/p(1p<)

    and

    fXc=supessx(a,b)[xc|f(x)|].

    In particular, when c=1/p the space Xpc(a,b) coincides with the space Lp(a,b). Note that, the regularity of the function, can be obtained by the inequality of the norms below,

    fXpc(a,b)CfLp(a,b).

    On the other hand, we can also note that space Xpc(a,b), implicitly involves the fractional Sobolev space W1,pX [8,9,10,11].

    Theorem 1. [5] Let f,gLp(a,b) be two positive functions, with 1p. If 0<mf(t)g(t)M, for m,MR+ and t[a,b], then

    (bafp(t)dt)1p+(bagp(t)dt)1pc1(ba(fp+gp)(t)dt)1p, (1)

    with c1=M(m+1)+(M+1)(m+1)(M+1).

    Theorem 2. [5] Let f,gLp(a,b) be two positive functions, with 1p. If 0<mf(t)g(t)M, for m,MR+ and t[a,b], then

    (bafp(t)dt)2p+(bagp(t)dt)2pc2(bafp(t)dt)1p(bagp(t)dt)1p, (2)

    with c2=(M+1)(m+1)M2.

    We present the definitions of the fractional integrals that will be useful in the development of the article: Riemann-Liouville fractional integral, Hadamard integral, Erdélyi-Kober integral, Katugampola integral, Weyl integral and Liouville integral.

    Definition 2. [3,16] Let [a,b](<a<b<) be a finite interval on the real-axis R. The Riemann-Liouville fractional integrals (leftsidedandrightsided) of order αC, Re(α)>0 of a real function fLp(a,b), are defined by

    Jαa+f(x):=1Γ(α)xaf(t)(xt)1αdt, a<x<b (3)

    and

    Jαbf(x):=1Γ(α)bxf(t)(tx)1αdt, a<x<b, (4)

    respectively.

    Definition 3. [3,16] Let (a,b)(0a<b<) be a finite or infinite interval on the half-axis R+. The Hadamard fractional integrals (leftsidedandrightsided) of order αC, Re(α)>0 of a real function fLp(a,b) are defined by

    Hαa+f(x):=1Γ(α)xa(logxt)α1f(t)tdt,  a<x<b (5)

    and

    Hαbf(x):=1Γ(α)bx(logtx)α1f(t)tdt, a<x<b (6)

    respectively.

    Definition 4. [3,16] Let (a,b)(a<b) be a finite or infinite interval or half-axis R+. Also let Re(α)>0, σ>0 and ηC. The Erdélyi-Kober fractional integrals (leftsidedandrightsided) of order αC of a real function fLp(a,b) are defined by

    Iαa+,σ,ηf(x):=σxσ(α+η)Γ(α)xatσ(η+1)1(xσtσ)1αf(t)dt, a<x<b (7)

    and

    Iαb,σ,ηf(x):=σxσηΓ(α)bxtσ(1ηα)1(tσxσ)1αf(t)dt, a<x<b, (8)

    respectively.

    Definition 5. [17] Let [a,b]R be a finite interval. The Katugampola fractional integrals (leftsidedandrightsided) of order αC, ρ>0, Re(α)>0 of a real function fXpc(a,b) are defined by

    ρIαa+f(x):=ρ1αΓ(α)xatρ1(xρtρ)1αf(t)dt, x>a (9)

    and

    ρIαbf(x):=ρ1αΓ(α)bxtρ1(tρxρ)1αf(t)dt, x<b, (10)

    respectively.

    Definition 6. [39] The Weyl fractional integrals of order αC, Re(α)>0 of a real function f locally integrable into (,) being x are defined by

    xWα=xIαf(x):=1Γ(α)xf(t)(xt)1αdt (11)

    and

    Wαx=Iαxf(x):=1Γ(α)xf(t)(tx)1αdt, (12)

    respectively.

    Definition 7. [3,16] Let a continuous function by parts in R=(,). The Liouville fractional integrals (leftsidedandrightsided) of order αC, Re(α)>0 of a real function f, are defined by

    Iα+f(x):=1Γ(α)xf(t)(xt)1αdt (13)

    and

    Iαf(x):=1Γ(α)xf(t)(tx)1αdt, (14)

    respectively.

    Zoubir [35] established the reverse Minkowski's inequality and another result that refers to the inequality via Riemann-Liouville fractional integral according to the following two theorems.

    Theorem 3. [35] Let α>0, p1 and f,g two positive functions in [0,), such that x>0, Jαfp(x)< and Jαgp(x)<. If 0<mf(t)g(t)M, for m,MR+ and t[0,x], then

    (Jαfp(x))1p+(Jαgp(x))1pc1(Jα(f+g)p(x))1p, (15)

    where c1=M(m+1)+(M+1)(m+1)(M+1).

    Theorem 4. [35] Let α>0, p1 and f,g two positive functions in [0,), such that x>0, Jαfp(x)< and Jαgp(x)<. If 0<mf(t)g(t)M, for m,MR+ and t[0,x], then

    (Jαfp(x))2p+(Jαgp(x))2pc2(Jαfp(x))1p(Jαgp(x))1p, (16)

    where c2=(M+1)(m+1)M2.

    In 2014, Chinchane et al. [36] and Taf et al. [40] also established the reverse Minkowski's inequality via Hadamard fractional integral as in two theorems below.

    Theorem 5. [36,40] Let α>0, p1 and f,g two positive functions in [0,), such that x>0, Hα1fp(x)< and Hα1gp(x)<. If 0<mf(t)g(t)M, for m,MR+ and t[0,x], then

    (Hα1fp(x))1p+(Hα1gp(x))1pc1(Hα1(f+g)p(x))1p, (17)

    where c1=M(m+1)+(M+1)(m+1)(M+1).

    Theorem 6. [36,40] Let α>0, p1 and f,g two positive functions in [0,), such that x>0, Hα1fp(x)< and Hα1gp(x)<. If 0<mf(t)g(t)M, for m,MR+ and t[0,x], then

    (Hα1fp(x))2p+(Hα1gp(x))2pc2(Hα1fp(x))1p(Hα1gp(x))1p (18)

    where c2=(M+1)(m+1)M2.

    In 2014 Chinchane et al. [41] and recently Chinchane [42], established the reverse Minkowski's inequality via fractional integral of Saigo and the k-fractional integral, respectively.

    In 2017, Katugampola [33] introduced a fractional integral that unifies the six fractional integrals above mentioned. Finally, we introduce this integral and with a theorem we study their respective particular cases.

    Definition 8. [33] Let φXpc(a,b), α>0 and β,ρ,η,κR. Then, the fractional integrals of a function f, left and right, are given by

    ρIα,βa+,η,κφ(x):=ρ1βxκΓ(α)xaτρ(η+1)1(xρτρ)1αφ(τ)dτ, 0a<x<b (19)

    and

    ρIα,βb,η,κφ(x):=ρ1βxρηΓ(α)bxτκ+ρ1(τρxρ)1αφ(τ)dτ,0a<x<b (20)

    respectively, if integrals exist.

    From now on, let's work only with the left integral, Eq.(19), because with the right integral we have a similar treatment.

    Remark 1. Let α>0 and β,ρ,η,κR. Then for fXpc(a,b), with a<x<b, we have [33]:

    1. For κ=0, η=0 and the limit ρ1, in Eq.(19), we get the Riemann-Liouville fractional integral, i.e; Eq.(3).

    2. With β=α, κ=0, η=0, we take the limit ρ0+ and using the 'Hospital role, in Eq.(19), we get the Hadamard fractional integral, i.e; Eq.(5).

    3. In the case β=0 and κ=ρ(α+η), in Eq.(19), we get the Erdélyi-Kober fractional integral, i.e; Eq.(7).

    4. For β=α, κ=0 and η=0, in Eq.(19), we get Katugampola fractional integral, i.e; Eq.(9).

    5. With κ=0, η=0, a= and take the limit ρ1, in Eq.(19), we get Weyl fractional integral, i.e; Eq.(11).

    6. With κ=0, η=0, a=0 and take the limit ρ1, in Eq.(19), we get Liouville fractional integral, i.e; Eq.(13).


    2. Reverse Minkowski fractional integral inequality

    In this section, our main contribution, we establish and prove the reverse Minkowski's inequality via generalized fractional integral Eq.(19) and a theorem that refers to the reverse Minkowski's inequality.

    Theorem 7. Let α>0, ρ,η,κ,βR and p1. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0<mf(t)g(t)M, for m,MR+ and t[a,x], then

    (ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1pc1(ρIα,βa+,η,κ(f+g)p(x))1p, (21)

    with c1=M(m+1)+(M+1)(m+1)(M+1).

    Proof 1. Using the condition f(t)g(t)M, t[a,x], we can write

    f(t)M(f(t)+g(t))Mf(t),

    which implies,

    (M+1)pfp(t)Mp(f(t)+g(t))p. (22)

    Multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(22) and integrating with respect to the variable t, we have

    (M+1)pρ1βxκΓ(α)xatρ(η+1)1(xρtρ)1αfp(t)dtMpρ1βxκΓ(α)xatρ(η+1)1(xρtρ)1α(f+g)p(t)dt. (23)

    Consequently, we can write

    (ρIα,βa+,η,κfp(x))1pMM+1(ρIα,βa+,η,κ(f+g)p(x))1p. (24)

    On the other hand, as mg(t)f(t), follows

    (1+1m)pgp(t)(1m)p(f(t)+g(t))p. (25)

    Further, multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(25) and integrating with respect to the variable t, we have

    (ρIα,βa+,η,κgp(t))1p1m+1(ρIα,βa+,η,κ(f+g)p(t))1p. (26)

    From Eq.(24) and Eq.(26), the result follows.

    Eq.(21) is the so-called reverse Minkowski's inequality associated with the Katugampola fractional integral.

    Theorem 8. Let α>0, ρ,η,κ,βR and p1. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0<mf(t)g(t)M, for m,MR+ and t[a,x], then

    (ρIα,βa+,η,κfp(x))2p+(ρIα,βa+,η,κgp(x))2pc2(ρIα,βa+,η,κfp(x))1p(ρIα,βa+,η,κgp(x))1p (27)

    with c2=(M+1)(m+1)M2.

    Proof 2. Carrying out the product between (24) and Eq.(26), we have

    (M+1)(m+1)M(ρIα,βa+,η,κfp(x))1p(ρIα,βa+,η,κgp(x))1p(ρIα,βa+,η,κ(f+g)p(x))2p. (28)

    Using the Minkowski's inequality, on the right side of Eq.(28), we have

    (M+1)(m+1)M(ρIα,βa+,η,κfp(x))1p(ρIα,βa+,η,κgp(x))1p((ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1p)2. (29)

    So, from Eq.(29), we conclude that

    ((M+1)(m+1)M2)(ρIα,βa+,η,κfp(x))1p(ρIα,βa+,η,κgp(x))1p(ρIα,βa+,η,κfp(x))2p+(ρIα,βa+,η,κgp(x))2p.

    Note that, if β=α, κ=0, η=0 and the limit ρ1, in Eq.(19), we recover Riemann-Liouville fractional integral, Eq.(3). In this sense, choosing a+=0, and substituting in Theorem 8 and Theorem 9, we obtain, as particular cases, the respective Theorem 3 and Theorem 4, which correspond to the inequality via Riemann-Liouville fractional integral. On the other hand, if β=α, κ=0, η=0, and the limit ρ0+ and using the L'Hôpital rule, in Eq.(19), we obtain the Hadamard fractional integral, Eq.(5). Similarly, choosing a=1 and substituting in Theorem 8 and Theorem 9, we obtain, as particular cases, the Theorem 5 and Theorem 6, respectively.


    3. Other fractional integral inequalities

    In this section we generalize the results discussed by Chinchane [42], Sulaiman [43] and Sroysang [44] on the reverse Minkowski's inequality via Riemann integral, using the fractional integral proposed by Katugampola [33].

    Theorem 9. Let α>0, ρ,η,κ,βR, p1 and 1p+1q=1. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0<mf(t)g(t)M, for m,MR+ and t[a,x], then

    (ρIα,βa+,η,κf(x))1p(ρIα,βa+,η,κg(x))1q(Mm)1pq(ρIα,βa+,η,κf1p(x)g1q(x)). (30)

    Proof 3. Using the condition f(t)g(t)M, t[a,x] with x>a, we have

    f(t)Mg(t)g1q(t)M1qf1q(t). (31)

    Multiplying by f1p(t) both sides of Eq.(30), we can rewrite it as follows

    f1p(t)g1q(t)M1qf(t). (32)

    Now, multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(31) and integrating with respect to the variable t, we have

    xaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1αM1qf(t)dtxaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1αf1p(t)g1q(t)dt. (33)

    So, the inequality follows

    M1pq(ρIα,βa+,η,κf(x))1p(ρIα,βa+,η,κf1p(x)g1q(x))1p. (34)

    On the order hand, we have

    m1pg1p(t)f1p(t), x>a. (35)

    Multiplying by g1q(t) both sides of Eq.(35) and using the relation 1p+1q=1, we have

    m1pg(t)f1p(t)g1q(t). (36)

    Multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(36) and integrating with respect to the variable t, we have

    m1pq(ρIα,βa+,η,κg(x))1q(ρIα,βa+,η,κf1p(x)g1q(x))1q. (37)

    Evaluating the product between Eq.(34) and Eq.(37) and using the relation 1p+1q=1, we conclude that

    (ρIα,βa+,η,κf(x))1p(ρIα,βa+,η,κg(x))1q(Mm)1pq(ρIα,βa+,η,κf1p(x)g1q(x))1p.

    Theorem 10. Let α>0, ρ,η,κ,βR, p1 and 1p+1q=1. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0<mf(t)g(t)M, for m,MR+ and t[a,x], then

    ρIα,βa+,η,κf(x)g(x)c3(ρIα,βa+,η,κ(fp+gp)(x))+c4(ρIα,βa+,η,κ(fq+gq)(x)), (38)

    with c3=2p1Mpp(M+1)p and c4=2p1q(m+1)q.

    Proof 4. Using the hypothesis, we have the following identity

    (M+1)pfp(t)Mp(f+g)p(t). (39)

    Multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(39) and integrating with respect to the variable t, we get

    xaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α(M+1)pfp(t)dtxaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1αMp(f+g)p(t)dt.

    In this way, we have

    ρIα,βa+,η,κfp(x)Mp(M+1)pρIα,βa+,η,κ(f+g)p(x). (40)

    On the other hand, as 0<m<f(t)g(t), t(a,x), we have

    (m+1)qgq(t)(f+g)q(t). (41)

    Again, multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(41) and integrating with respect to the variable t, we get

    ρIα,βa+,η,κgq(x)1(m+1)qρIα,βa+,η,κ(f+g)q(x). (42)

    Considering Young's inequality, [45]

    f(t)g(t)fp(t)p+gq(t)q, (43)

    multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(43) and integrating with respect to the variable t, we have

    ρIα,βa+,η,κ(fg)(x)1p(ρIα,βa+,η,κfp(x))+1q(ρIα,βa+,η,κgq(x)). (44)

    Thus, using Eq.(40), Eq.(42) and Eq.(44), we get

    ρIα,βa+,η,κ(fg)(x)1p(ρIα,βa+,η,κfp(x))+1q(ρIα,βa+,η,κgq(x))(ρIα,βa+,η,κfp(x))+(ρIα,βa+,η,κgq(x))Mpp(M+1)p(ρIα,βa+,η,κ(f+g)p(x))+1q(m+1)q(ρIα,βa+,η,κ(f+g)q(x)). (45)

    Using the following inequality, (a+b)r2p1(ar+br), r>1, a,b0, we get

    ρIα,βa+,η,κ(f+g)p(x)2p1ρIα,βa+,η,κ(fp+gp)(x) (46)

    and

    ρIa+,η,κα,β(f+g)q(x)2q1ρIα,βa+,η,κ(fq+gq)(x). (47)

    Thus, replacing Eq.(46) and Eq.(47) at Eq.(45), we conclude that

    ρIα,βa+,η,κ(fg)(x)2p1Mpp(M+1)p(ρIα,βa+,η,κ(fp+gp)(x))+2q1q(m+1)q(ρIα,βa+,η,κ(fq+gq)(x)).

    Theorem 11. Let α>0, ρ,η,κ,βR and p1. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0<mf(t)g(t)M, for m,MR+ and t[a,x], then

    M+1Mc(ρIα,βa+,η,κ(f(x)cg(x)))1p(ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1pm+1mc(ρIα,βa+,η,κ(f(x)cg(x)))1p (48)

    Proof 5. Considering 0<c<mM, we can write

    mcMcmc+mmc+MMc+M(M+1)(mc)(m+1)(Mc).

    Thus, we conclude that

    M+1Mcm+1mc.

    Also, we have

    mcf(t)cg(t)g(t)Mc

    which implies,

    (f(t)cg(t))p(Mc)pgp(t)(f(t)cg(t))p(mc)p. (49)

    Again, we have

    1Mg(t)f(t)1mmccmf(t)cg(t)cf(t)MccM,

    which implies,

    (MMc)p(f(t)cg(t))pfp(t)(mmc)p(f(t)cg(t))p. (50)

    Multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(49) and integrating with respect to the variable t, we have

    xaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α(f(t)cg(t))p(Mc)pdtxaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1αgp(t)dtxaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α(f(t)cg(t))p(mc)pdt.

    In this way, we obtain

    1Mc(ρIα,βa+,η,κ(f(x)cg(x))p)1p(ρIα,βa+,η,κgp(x))1p1mc(ρIα,βa+,η,κ(f(x)cg(x))p)1p. (51)

    Realizing the same procedure as in Eq.(50), we have

    MMc(ρIα,βa+,η,κ(f(x)cg(x))p)1p(ρIα,βa+,η,κfp(x))1pmmc(ρIα,βa+,η,κ(f(x)cg(x))p)1p. (52)

    Adding Eq.(51) and Eq.(52), we conclude that

    M+1Mc(ρIα,βa+,η,κ(f(x)cg(x))p)1p(ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1pm+1mc(ρIα,βa+,η,κ(f(x)cg(x))p)1p.

    Theorem 12. Let α>0, ρ,η,κ,βR and p1. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0af(t)A and 0bg(t)B, t[a,x], then

    (ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1pc5(ρIα,βa+,η,κ(f+g)p(x))1p, (53)

    with c5=A(a+B)+B(A+b)(A+b)(a+B).

    Proof 6. By hypothesis, it follows that

    1B1g(t)1b. (54)

    Realizing the product between Eq.(54) and 0<af(t)A, we have

    aBf(t)g(t)Ab. (55)

    From Eq.(55), we get

    gp(t)(Ba+B)p(f(t)+g(t))p (56)

    and

    fp(t)(Ab+A)p(f(t)+g(t))p. (57)

    Multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(56) and integrating with respect to the variable t, we have

    xaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1αgp(t)dtxaρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α(Ba+B)p(f(t)+g(t))pdt.

    Thus, it follows that

    (ρIα,βa+,η,κgp(x))1pBa+B(ρIα,βa+,η,κ(f+g)p(x))1p. (58)

    Similarly, we perform the calculations for Eq.(57), we get

    (ρIα,βa+,η,κfp(x))1pAb+A(ρIα,βa+,η,κ(f+g)p(x))1p. (59)

    Adding Eq.(58) and Eq.(59), we conclude that

    (ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1pA(a+B)+B(b+A)(a+B)(b+A)(ρIα,βa+,η,κ(f+g)p(x))1p.

    Theorem 13. Let α>0 and ρ,η,κ,βR. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0<mf(t)g(t)M, for m,MR+ and t[a,x], then

    1M(ρIα,βa+,η,κf(x)g(x))1(m+1)(M+1)(ρIα,βa+,η,κ(f+g)2(x))1m(ρIα,βa+,η,κf(x)g(x)). (60)

    Proof 7. Taking 0<mf(t)g(t)M, t[a,x], we have

    g(t)(m+1)g(t)+f(t)g(t)(M+1). (61)

    Also, it follows that 1Mg(t)f(t)1m, which implies,

    g(t)(M+1M)g(t)+f(t)g(t)(m+1m). (62)

    Evaluating the product between Eq.(61) and Eq.(62), we have

    f(t)g(t)M(g(t)+f(t))2(m+1)(M+1)f(t)g(t)m. (63)

    Multiplying by ρ1βxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(63) and integrating with respect to the variable t, we have

    ρ1βxκMΓ(α)xatρ(η+1)1(xρtρ)1αf(t)g(t)dtc6ρ1βxκΓ(α)xatρ(η+1)1(xρtρ)1α(g(t)+f(t))2dtρ1βxκmΓ(α)xatρ(η+1)1(xρtρ)1αf(t)g(t)dt,

    with c6=1(m+1)(M+1).

    Thus, we conclude that

    1M(ρIα,βa+,η,κf(x)g(x))1(m+1)(M+1)(ρIα,βa+,η,κ(g(x)+f(x))2)1m(ρIα,βa+,η,κf(x)g(x)).

    Theorem 14. Let α>0, ρ,η,κ,βR and p1. Let f,gXpc(a,x) be two positive functions in [0,), x>a. If 0<mf(t)g(t)M, for m,MR+ and t[a,x], then

    (ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1p2(ρIα,βa+,η,κhp(f(x),g(x)))1p,

    with h(f(x),g(x))=max{M[(Mm+1)f(x)Mg(x)],(m+M)g(x)f(x)m}.

    Proof 8. From the hypothesis, 0<mf(t)g(t)M, t[a,x], we have

    0<mM+mf(t)g(t) (64)

    and

    M+mf(t)g(t)M. (65)

    Thus, using Eq.(64) and Eq.(65), we get

    g(t)<(M+m)g(t)f(t)mh(f(t),g(t)), (66)

    where h(f(t),g(t))=max{M[(Mm+1)f(t)Mg(t)],(M+m)g(t)f(t)m}.

    Using the hypothesis, it follows that 0<1Mg(t)f(t)1m. In this way, we obtain

    1M1M+1mg(t)f(t) (67)

    and

    1M+1mg(t)f(t)1m. (68)

    Then, from Eq.(67) and Eq.(68), we have

    1M(1m+1M)f(t)g(t)f(t)1m,

    which can be rewritten as

    f(t)M(1m+1M)f(t)Mg(t)=M(M+m)f(t)M2mg(t)Mm=(Mm+1)f(t)Mg(t)M[(Mm+1)f(t)Mg(t)]h(f(t),g(t)). (69)

    Thus, using Eq.(66) and Eq.(69), we can write

    fp(t)hp(f(t),g(t)) (70)

    and

    gp(t)hp(f(t),g(t)). (71)

    Multiplying by ρ1αxκtρ(η+1)1Γ(α)(xρtρ)1α both sides of Eq.(70) and integrating with respect to the variable t, we have

    xaρ1αxκtρ(η+1)1Γ(α)(xρtρ)1αfp(t)dtxaρ1αxκtρ(η+1)1Γ(α)(xρtρ)1αhp(f(t),g(t))dt.

    In this way, we obtain

    (ρIα,βa+,η,κfp(x))1p(ρIα,βa+,η,κhp(f(x),g(x)))1p. (72)

    Using the same procedure as above, for Eq.(71), we have

    (ρIα,βa+,η,κgp(x))1p(ρIα,βa+,η,κhp(f(x),g(x)))1p. (73)

    Thus, using Eq.(72) and Eq.(73), we conclude that

    (ρIα,βa+,η,κfp(x))1p+(ρIα,βa+,η,κgp(x))1p2(ρIα,βa+,η,κhp(f(x),g(x)))1p.

    Using Eq.(19) and Theorem 7 with the convenient conditions for each respective fractional integral, we have the previous theorems, that is, Theorem 10 to Theorem 15 introduced and demonstrated above, contain as particular cases, each result involving the following fractional integrals: Riemann-Liouville, Hadamard, Liouville, Weyl, Erdélyi-Kober, and Katugampola.


    4. Concluding remarks

    After a brief introduction to the fractional integral, proposed by Katugampola and fractional integrals in the sense of Riemann-Liouville and Hadamard, we generalize the reverse Minkowski's inequality obtaining, as a particular case, the inequality involving the fractional integral in the Riemann-Liouville sense and Hadamard sense [33]. We also show other inequalities using the Katugampola fractional integral. The application of this fractional integral can be used to generalize several inequalities, among them, we mention the Grüss-type inequality, recently introduced and proved [46]. A continuation of this work, with this formulation of fractional integral, consists in generalize the inequalities of Hermite-Hadamard and Hermite-Hadamard-Féjer [27,28,29,30]. Moreover, we will discuss inequalities via M-fractional integral according to [47].


    Acknowledgment

    We also thank anonymous referees by the suggestions that improved the paper.


    Conflict of Interest

    All authors declare no conflicts of interest in this paper.


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