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Research article Special Issues

The number of rational points on a class of hypersurfaces in quadratic extensions of finite fields

  • Received: 11 March 2023 Revised: 21 April 2023 Accepted: 08 May 2023 Published: 05 June 2023
  • Let q be an even prime power and let Fq be the finite field of q elements. Let f be a nonzero polynomial over Fq2 of the form f=a1xm11++asxmss+y1y2++yn1yn+y2n2t1++y2n3+y2n1 +bty2n2t+ +b1y2n2+b0y2n, where ai,bjFq2, mi1, (mi,mk)=1, ik, mi|(q+1), miZ+, 2|n, n>2, 0tn22, TrFq2/F2(bj)=1 for i,k=1,,s and j=0,1,,t. For each bFq2, let Nq2(f=b) denote the number of Fq2-rational points on the affine hypersurface f=b. In this paper, we obtain the formula of Nq2(f=b) by using the Jacobi sums, Gauss sums and the results of quadratic form in finite fields.

    Citation: Qinlong Chen, Wei Cao. The number of rational points on a class of hypersurfaces in quadratic extensions of finite fields[J]. Electronic Research Archive, 2023, 31(7): 4303-4312. doi: 10.3934/era.2023219

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  • Let q be an even prime power and let Fq be the finite field of q elements. Let f be a nonzero polynomial over Fq2 of the form f=a1xm11++asxmss+y1y2++yn1yn+y2n2t1++y2n3+y2n1 +bty2n2t+ +b1y2n2+b0y2n, where ai,bjFq2, mi1, (mi,mk)=1, ik, mi|(q+1), miZ+, 2|n, n>2, 0tn22, TrFq2/F2(bj)=1 for i,k=1,,s and j=0,1,,t. For each bFq2, let Nq2(f=b) denote the number of Fq2-rational points on the affine hypersurface f=b. In this paper, we obtain the formula of Nq2(f=b) by using the Jacobi sums, Gauss sums and the results of quadratic form in finite fields.



    In recent years, a useful extension has been proposed from the classical calculus by permitting derivatives and integrals of arbitrary orders is known as fractional calculus. It emerged from a celebrated logical conversation between Leibniz and L'Hopital in 1695 and was enhanced by different scientists like Laplace, Abel, Euler, Riemann, and Liouville [1]. Fractional calculus has gained popularity on the account of diverse applications in various areas of science and technology [2,3,4]. The concept of this new calculus was applied in several distinguished areas previously with excellent developments in the frame of novel approaches and posted scholarly papers, see [5,6,7,8,9,10,11,12,13,14,15,16,17,18]. Various notable generalized fractional integral operators such as the Riemann-Liouville, Hadamard, Caputo, Marichev-Saigo-Maeda, Riez, the Gaussian hypergeometric operators and so on, their attempts helpful for researchers to recognize the real world phenomena. Therefore, the Caputo and Riemann-Liouville was the most used fractional operators having singular kernels. It is remarkable that all the above mentioned operators are the particular cases of the operators investigated by Jarad et al. [19]. The utilities to weighted generalized fractional operators are undertaking now.

    Adopting the excellency of the above work, we introduce a new weighted framework of generalized proportional fractional integral operator with respect to monotone function Ψ. Also, some new characteristics of the aforesaid operator are apprehended to explore new ideas to amplify the fractional operators and acquire fractional integral inequalities via generalized fractional operators (see Remark 2 and 3 below).

    Recently, by employing the fractional integral operators, several researchers have established a bulk of fractional integral inequalities and their variant forms with fertile applications. These sorts of speculations have noteworthy applications in fractional differential/difference equations and fractional Schrödinger equations [20,21]. By the use of Riemann-Liouville fractional integral operator, Belarbi and Dahmani [22] contemplated the subsequent integral inequalities as follows:

    If f1 and g1 are two synchronous functions on [0,), then

    Ωα(f1g1)(ϰ)Γ(α+1)ϰαΩα(f1)(ϰ)Ωα(g1)(ϰ) (1.1)

    and

    ϰαΓ(α+1)Ωβ(f1g1)(ϰ)+ϰβΓ(β+1)Ωα(f1g1)(ϰ)Ωα(f1)(ϰ)Ωβ(g1)(ϰ)+Ωβ(f1)(ϰ)Ωα(g1)(ϰ), (1.2)

    for all ϰ>0,α,β>0. Butt et al. [23], Rashid et al. [24] and Set et al. [25] established the fractional integral inequalities via generalized fractional integral operator having Raina's function, generalized K-fractional integral and Katugampola fractional integral inequalities similar to the variants (1.1) and (1.2), respectively. Here we should emphasize that, inequalities (1.1) and (1.2) are a remarkable instrument for reconnoitering plentiful scientific regions of investigation encompassing probability theory, statistical analysis, physics, meteorology, chaos and henceforth.

    More general version of inequalities (1.1) and (1.2) proposed by Dahmani [26] by employing Riemann-Liouville fractional integral operator.

    Let f1 and g1 be two synchronous functions on [0,) and let r,s:[0,)[0,). Then

    ΩαP(ϰ)Ωα(Qf1g1)(ϰ)+ΩαQ(ϰ)Ωα(Pf1g1)(ϰ)Ωα(Qf1)(ϰ)Ωα(Pg1)(ϰ)+Ωα(Pf1)(ϰ)Ωα(Qg1)(ϰ) (1.3)

    and

    ΩαP(ϰ)Ωβ(Qf1g1)(ϰ)+ΩβQ(ϰ)Ωα(Pf1g1)(ϰ)Ωα(Qf1)(ϰ)Ωβ(Pg1)(ϰ)+Ωβ(Pf1)(ϰ)Ωα(Qg1)(ϰ) (1.4)

    for all ϰ>0,α,β>0. Chinchane and Pachpatte [27], Brahim and Taf [28] and Shen et al. [29] explored the Hadamard fractional integral inequalities, the fractional version of integral inequalities in two variable quantum deformation and the Riemann-Liouville fractional integral operator on time scale analysis coincide to variants (1.3) and (1.4), respectively.

    Let us define the most distinguished Chebyshev functional [30]:

    T(f1,g1)=1b1a1b1a1f1(ϰ)g1(ϰ)dϰ1b1a1b1a1f1(ϰ)dϰ1b1a1b1a1g1(ϰ)dϰ, (1.5)

    where f1 and g1 are two integrable functions on [a1,b1]. In [31], Grüss proposed the well-known generalization:

    |T(f1,g1)|14(Φϕ)(Υγ), (1.6)

    where f1 and g1 are two integrable functions on [a1,b1] satisfying the assumptions

    ϕf1(ϰ)Φ,γg1(ϰ)Υ,ϕ,Φ,γ,ΥR,ϰ[a1,b1]. (1.7)

    The inequality (1.6) is known to be Grüss inequality. In recent years, the Grüss type integral inequality has been the subject of very active research. Mathematicians and scientists can see them in research papers, monographs, and textbooks devoted to the theory of inequalities [32,33,34,35] such as, Dragomir [36] demonstrated certain variants with the supposition of vectors and continuous mappings of selfadjoint operators in Hilbert space similar to (1.6). In this context, f1 and g1 are holding the assumptions (1.7), Dragomir [37] derived several functionals in two and three variable sense as follows:

    |S(f1,g1,P)|14(Φϕ)(Υγ)(b1a1P1(ϰ)dϰ)2, (1.8)

    where

    S(f1,g1,P)=12T(f1,g1,P)=b1a1P(ϰ)dϰb1a1P(ϰ)f1(ϰ)g1(ϰ)dϰb1a1P(ϰ)f1(ϰ)dϰb1a1P(ϰ)g1(ϰ)dϰ (1.9)

    and

    T(f1,g1,P,Q)=b1a1Q(ϰ)dϰb1a1P(ϰ)f1(ϰ)g1(ϰ)dϰ+b1a1P(ϰ)dϰb1a1Q(ϰ)f1(ϰ)g1(ϰ)dϰb1a1Q(ϰ)f1(ϰ)dϰb1a1P(ϰ)g1(ϰ)dϰb1a1P(ϰ)f1(ϰ)dϰb1a1Q(ϰ)g1(ϰ)dϰ. (1.10)

    In [37], Dragomir established the inequality:

    If f1,g1L(a1,b1), then

    |S(f1,g1,P)|f1g1(b1a1P(ϰ)dϰb1a1ϰ2P(ϰ)dϰ(b1a1ϰP(ϰ)dϰ)2). (1.11)

    Moreover, author [37] proved numerous variants for Lipschitzian functions as follows:

    If f1 is L-g1-Lipschitzian on [a1,b1], that is

    |f1(μ)fν|L|g1(μ)g1(ν)|,L>0,μ,ν[a1,b1]. (1.12)

    and

    |S(f1,g1,P)|L(b1a1P(ϰ)dϰb1a1g21(ϰ)P(ϰ)dϰ(b1a1g1(ϰ)P(ϰ)dϰ)2). (1.13)

    Furthermore, if f1 and g1 are L1 and L2-Lipschitzian functions on [a1,b1], then

    |S(f1,g1,P)|L1L2(b1a1P(ϰ)dϰb1a1ϰ2P(ϰ)dϰ(b1a1ϰP(ϰ)dϰ)2). (1.14)

    Owing to the above tendency, Dhamani et al. [38] proposed the fractional integral inequalities in the Riemann-Liouville parallel to variant (1.6) with the suppositions (1.7). Additionally, Dahamani and Benzidane [39] introduced weighted Grüss type inequality via (α,β)-fractional q-integral inequality resemble to (1.8) under the hypothesis of (1.5). Author [40,41] derived the extended functional of (1.10) by employing Riemann-Liouville integral corresponds to variants (1.11), (1.13) and (1.14), respectively. In this flow, Set et al. [42] contemplated the Grüss type inequalities considering the generalized K-fractional integral. Chen et al. [43] obtained the novel refinements of Hermite-Hadamard type inequalities for n-polynomial p-convex functions within the generalized fractional integral operators. Abdeljawad et al. [44] derived the Simpson's type inequalities for generalized p-convex functions involving fractal set. Jarad et al. [45] investigated the properties of the more general form of generalized proportional fractional operators in Laplace transforms.

    The motivation of this paper is twofold. First, we propose a novel framework named weighted generalized proportional fractional integral operator based on characteristics, as well as considering the boundedness and semi-group property and able to be widely applied to many scientific results. Second, the current operator employed to the extended weighted Chebyshev and Grüss type inequalities for exploring the analogous versions of (1.5) and (1.6). Some special cases are pictured with new fractional operators which are not computed yet. Interestingly, particular cases are designed for Riemann-Liouville fractional integral, generalized Riemann-Liouville fractional integral and generalized proportional fractional integral inequalities. It is worth mentioning that these operators have the ability to recapture several generalizations in the literature by considering suitable assumptions of Ψ,ω and ρ.

    In this section, we demonstrate the space where the weighted fractional integrals are bounded and also, provide certain specific features of these operators.

    Definition 2.1 ([19])Let ω0 be a mapping defined on [a1,b1], g1 is a differentiable strictly increasing function on [a1,b1]. The space χpω(a1,b1),1p< is the space of all Lebesgue measurable functions f1 defined on [a1,b1] for which f1χpω, where

    f1χpω=(b1a1|ω(ϰ)f1(ϰ)|pg1(ϰ)dϰ)1p,1<p< (2.1)

    and

    f1χpω=esssupa1ϰb1|ω(ϰ)f1(ϰ)|<. (2.2)

    Remark 1. Clearly we see that f1χpω(a1,b1) ω(ϰ)f1(ϰ)(g11(ϰ))1/pLp(a1,b1) for 1p< and f1χpω(a1,b1) ω(ϰ)f1(ϰ)L(a1,b1).

    Now, we show a novel fractional integral operator which is known as the weighted generalized proportional fractional integral operator with respect to monotone function Ψ.

    Definition 2.2. Let f1χpω(a1,b1) and ω0 be a function on [a1,b1]. Also, assume that Ψ is a continuously differentiable function on [a1,b1] with ψ>0 on [a1,b1]. Then the left and right-sided weighted generalized proportional fractional integral operator with respect to another function Ψ of order α>0 are described as:

    ΨωΩρ;αa1f1(ϰ)=ω1(ϰ)ραΓ(α)ϰa1exp[ρ1ρ(Ψ(ϰ)Ψ(μ))](Ψ(ϰ)Ψ(μ))1αf1(μ)ω(μ)Ψ(μ)dμ,a1<ϰ (2.3)

    and

    ΨωΩρ;αb1f1(ϰ)=ω1(ϰ)ραΓ(α)b1ϰexp[ρ1ρ(Ψ(μ)Ψ(ϰ))](Ψ(μ)Ψ(ϰ))1αf1(μ)ω(μ)Ψ(μ)dμ,ϰ<b1, (2.4)

    where ρ(0,1] is the proportionality index, αC,(α)>0 and Γ(ϰ)=0μϰ1eμdμ is the Gamma function.

    Remark 2. Some particular fractional operators are the special cases of (2.3) and (2.4).

    (1) Setting Ψ(ϰ)=ϰ, in Definition (2.2), then we get the weighted generalized proportional fractional operators stated as follows:

    ωΩρ;αa1f1(ϰ)=ω1(ϰ)ραΓ(α)ϰa1exp[ρ1ρ(ϰμ)](ϰμ)1αf1(μ)ω(μ)dμ,a1<ϰ (2.5)

    and

    ωΩρ;αb1f1(ϰ)=ω1(ϰ)ραΓ(α)b1ϰexp[ρ1ρ(μϰ)](μϰ)1αf1(μ)ω(μ)dμ,ϰ<b1. (2.6)

    (2) Setting Ψ(ϰ)=ϰ and ρ=1 in Definition (2.2), then we get the weighted Riemann-Liouville fractional operators stated as follows:

    ωΩαa1f1(ϰ)=ω1(ϰ)Γ(α)ϰa1f1(μ)ω(μ)dμ(ϰμ)1α,a1<ϰ (2.7)

    and

    ωΩαb1f1(ϰ)=ω1(ϰ)Γ(α)b1ϰf1(μ)ω(μ)dμ(μϰ)1α,ϰ<b1. (2.8)

    (3) Setting Ψ(ϰ)=lnϰ and a1>0 in Definition (2.2), we get the weighted generalized proportional Hadamard fractional operators stated as follows:

    ωΩρ;αa1f1(ϰ)=ω1(ϰ)ραΓ(α)ϰa1exp[ρ1ρ(lnϰμ)](lnϰμ)1αf1(μ)ω(μ)μdμ,a1<ϰ (2.9)

    and

    ωΩρ;αb1f1(ϰ)=ω1(ϰ)ραΓ(α)b1ϰexp[ρ1ρ(lnμϰ)](lnμϰ)1αf1(μ)ω(μ)μdμ,ϰ<b1. (2.10)

    (4) Setting Ψ(ϰ)=lnϰ and a1>0 along with ρ=1 in Definition (2.2), then we get the weighted Hadamard fractional operators stated as follows:

    ωΩαa1f1(ϰ)=ω1(ϰ)Γ(α)ϰa1f1(μ)ω(μ)dμμ(lnϰμ)1α,a1<ϰ (2.11)

    and

    ωΩαb1f1(ϰ)=ω1(ϰ)Γ(α)b1ϰf1(μ)ω(μ)dμμ(lnμϰ)1α,ϰ<b1. (2.12)

    (5) Setting Ψ(ϰ)=ϰττ(τ>0) in Definition (2.2), then we get the weighted generalized fractional operators in terms of Katugampola stated as follows:

    ωΩαa1f1(ϰ)=ω1(ϰ)Γ(α)ϰa1(ϰτμττ)α1f1(μ)ω(μ)dμμ1τ,a1<ϰ (2.13)

    and

    ωΩαb1f1(ϰ)=ω1(ϰ)Γ(α)b1ϰ(μτϰττ)α1f1(μ)ω(μ)dμμ1τ,ϰ<b1. (2.14)

    Remark 3. Several existing integral operators can be derived from Definition 2.2 as follows:

    (1) Letting ω(ϰ)=1, then we get the Definition 4 proposed by Rashid et al. [46] and Definition 3.2 introduced by Jarad et al. [47], independently.

    (2) Letting ω(ϰ)=1,Ψ(ϰ)=ϰ, then we get the Definition 3.4 defined by Jarad et al. [48].

    (3) Letting ω(ϰ)=1 and Ψ(ϰ)=lnϰ along with a1>0, then we get the Definition 2.1 defined by Rahman et al. [49].

    (4) Letting ω(ϰ)=ρ=1 and Ψ(ϰ)=lnϰ along with a1>0, then we get the operator defined by Kilbas et al. [3] and Smako et al. [5], respectively.

    (5) Letting ω(ϰ)=ρ=1 and Ψ(ϰ)=ϰ, then we get the operator defined by Kilbas et al [3].

    (6) Letting ω(ϰ)=1 and Ψ(ϰ)=ϰττ,(τ>0), then we get the operator defined by Katugampola et al. [7].

    (7) Letting ω(ϰ)=ρ=1 and Ψ(ϰ)=ϰτ+sτ+s,τ(0,1],sR, then we get the Definition 2 defined by Khan and Khan et al [50].

    (8) Letting ω(ϰ)=ρ=1 and Ψ(ϰ)=(ϰa1)ττ, and Ψ(ϰ)=(b1ϰ)ττ,(τ>0), then we get the operator defined by Jarad et al. [51].

    Theorem 2.3. For α>0,ρ(0,1],1p and f1χpω(a1,b1). Then ΨωΩρ;αa1 is bounded in χpω(a1,b1) and

    ΨωΩρ;αa1f1χpω(Ψ(b1)Ψ(a1))αf1χpωραΓ(α+1).

    Proof. For 1p, we have

    ΨωΩρ;αa1f1χpω=1ραΓ(α)(b1a1|ϰa1exp[ρ1ρΨ(ϰ)Ψ(μ)](Ψ(ϰ)Ψ(μ))1αω(μ)f1(μ)Ψ(μ)dμ|pΨ(ϰ)dϰ)1/p=1ραΓ(α)(Ψ(b1)Ψ(a1)|t2Ψ(a1)exp[ρ1ρ(t2t1)](t2t1)1αω(Ψ1(t1))f1(Ψ1(t1))|pdt2)1/p.

    Using the fact that |exp[ρ1ρ(t2t1)]|<1. Taking into account the generalized Minkowski inequality [5], we can write

    ΨωΩρ;αa1f1χpω1ραΓ(α)Ψ(b1)Ψ(a1)(|ω(Ψ1(t1))f1(Ψ1(t1))|pΨ(b1)t1(t2t1)p(α1)dt2)1/pdt1=1ραΓ(α)Ψ(b1)Ψ(a1)(|ω(Ψ1(t1))f1(Ψ1(t1))|((Ψ(b1)t1)p(α1)+1p(α1)+1)1/pdt1.

    By employing the well-known Hölder inequality satisfying p1+q1=1, we obtain

    ΨωΩρ;αa1f1χpω1ραΓ(α)(Ψ(b1)Ψ(a1)|ω(Ψ1(t1))f1(Ψ1(t1))|pdt1)1/p(Ψ(b1)Ψ(a1)((Ψ(b1)t1)p(α1)+1p(α1)+1)q/pdt1)1/q1ραΓ(α)(b1a1|ω(ϰ)f1(ϰ)|pΨ(ϰ)dϰ)1/p(Ψ(b1)Ψ(a1)((Ψ(b1)t1)p(α1)+1p(α1)+1)q/pdt1)1/q(Ψ(b1)Ψ(a1))αf1χpωραΓ(α+1).

    Now, for p=, we have

    |ω(ϰ)ΨωΩρ;αa1f1(ϰ)|=1ραΓ(α)ϰa1exp[ρ1ρ(Ψ(ϰ)Ψ(μ))](Ψ(ϰ)Ψ(μ))1αf1(μ)ω(μ)Ψ(μ)dμ1ραΓ(α)ϰa1exp[ρ1ρ(Ψ(ϰ)Ψ(μ))](Ψ(ϰ)Ψ(μ))1α|f1(μ)ω(μ)|Ψ(μ)dμ,Since(|exp[ρ1ρ(t2t1)]|<1)f1χωραΓ(α)ϰa1(Ψ(ϰ)Ψ(μ))α1dμ(Ψ(ϰ)Ψ(a1))αf1χωραΓ(α+1)=(Ψ(b1)Ψ(a1))αf1χωραΓ(α+1).

    This ends the proof.

    Our next result is the semi group property for weighted generalized proportional fractional integral operator with respect to monotone function.

    Theorem 2.4. For α,β>0,ρ(0,1] with 1p and let f1χpω(a1,b1). Then

    (ΨωΩρ;αa1ΨωΩρ;βa1)f1=(ΨωΩρ;α+βa1)f1. (2.15)

    Proof.

    (ΨωΩρ;αa1ΨωΩρ;βa1f1)(ϰ)=ω1(ϰ)ραΓ(α)ϰa1exp[ρ1ρ(Ψ(ϰ)Ψ(μ))](Ψ(ϰ)Ψ(μ))1αω(μ)(ΨωΩρ;βa1f1)(μ)Ψ(μ)dμ=ω1(ϰ)ρα+βΓ(α)Γ(β)ϰa1μa1exp[ρ1ρ(Ψ(ϰ)Ψ(μ))](Ψ(ϰ)Ψ(μ))1αexp[ρ1ρ(Ψ(μ)Ψ(ν))](Ψ(μ)Ψ(ν))1β×ω(ν)f1(ν)Ψ(ν)Ψ(μ)dμdν.

    By making change of variable technique θ=Ψ(μ)Ψ(a1)Ψ(ϰ)Ψ(a1), we can write

    (ΨωΩρ;αa1ΨωΩρ;βa1f1)(ϰ)=ω1(ϰ)ρα+βΓ(α)Γ(β)10θβ1(1θ)α1dθϰa1exp[ρ1ρ(Ψ(ϰ)Ψ(ν))](Ψ(ϰ)Ψ(ν))1αβω(ν)f1(ν)Ψ(ν)dν=ω1(ϰ)ρα+βΓ(α)Γ(β)Γ(α)Γ(β)Γ(α+β)ϰa1exp[ρ1ρ(Ψ(ϰ)Ψ(ν))](Ψ(ϰ)Ψ(ν))1αβω(ν)f1(ν)Ψ(ν)dν=(ΨωΩρ;α+βa1f1)(ϰ),

    where B(α,β)=Γ(α)Γ(β)Γ(α+β)=10θβ1(1θ)α1dθ is known to be Euler Beta function.

    This section contains some significant generalizations for weighted integral inequalities by employing weighted generalized proportional fractional integral operator, for the consequences relating to (1.1) and (1.2), it is suppose that all mappings are integrable in the Riemann sense.

    Throughout this investigation, we use the following assumptions:

    I. Let f1 and g1 be two synchronous functions on [0,).

    II. Let Ψ:[0,)(0,) is an increasing function with continuous derivative Ψ on the interval (0,).

    Lemma 3.1. If the supposition I and II are satisfied and let Q and P be two non-negative continuous mappings on [0,). Then the inequality

    ΨωΩρ;α0+(P)(ϰ)ΨωΩρ;α0+(Qf1g1)(ϰ)+ΨωΩρ;α0+(Pf1g1)(ϰ)ΨωΩρ;α0+(Q)(ϰ)ΨωΩρ;α0+(Pg1)(ϰ)ΨωΩρ;α0+(Qf1)(ϰ)+ΨωΩρ;α0+(Pf1)(ϰ)ΨωΩρ;α0+(Qg1)(ϰ), (3.1)

    holds for all ρ(0,1],αC with (α)>0.

    Proof. Since f1 and g1 are two synchronous functions on [0,), then for all μ>0 and ν>0, we have

    (f1(μ)f1(ν))(g1(μ)g1(ν))0. (3.2)

    By (3.2), we write

    f1(μ)g1(μ)+f1(ν)g1(ν)g1(μ)f1(ν)+g1(ν)f1(μ). (3.3)

    If we multiply both sides of (3.3) by exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]Q(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1α and integrating the resulting inequality with respect to μ from 0 to ϰ, we get

    1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]Q(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1αf1(μ)g1(μ)dμ+f1(ν)g1(ν)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]Q(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1αdμf1(ν)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]Q(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1αg1(ν)dν+g1(ν)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]Q(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1αf1(μ)dμ. (3.4)

    Taking product both sides of the above equation by ω1(ϰ) and in view of Definition (2.2), we have

    ΨωΩρ;α0+(Qf1g1)(ϰ)+f1(ν)g1(ν)ΨωΩρ;α0+(Q)(ϰ)g1(ν)ΨωΩρ;α0+(Qf1)(ϰ)+f1(ν)ΨωΩρ;α0+(Qg1)(ϰ). (3.5)

    Further, if we multiply both sides of (3.5) by exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]P(ν)ω(ν)Ψ(ν)ραΓ(α)(Ψ(ϰ)Ψ(ν))1α and integrating the resulting inequality with respect to ν from 0 to ϰ. Then, multiplying by ω1(ϰ) and in view of Definition 2.2, we obtain

    ΨωΩρ;α0+(P)(ϰ)ΨωΩρ;α0+(Qf1g1)(ϰ)+ΨωΩρ;α0+(Pf1g1)(ϰ)ΨωΩρ;α0+(Q)(ϰ)ΨωΩρ;α0+(Pg1)(ϰ)ΨωΩρ;α0+(Qf1)(ϰ)+ΨωΩρ;α0+(Pf1)(ϰ)ΨωΩρ;α0+(Qg1)(ϰ), (3.6)

    which implies (3.1).

    Theorem 3.2. Under the assumption of I, II and let r, s and t be three non-negative continuous functions on [0,). Then the inequality

    2ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+s(ϰ)ΨωΩρ;α0+(tf1g1)(ϰ)+ΨωΩρ;α0+(sf1g1)(ϰ)ΨωΩρ;α0+t(ϰ))+2ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+s(ϰ)ΨωΩρ;α0+t(ϰ)ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;α0+(tf1)(ϰ)+ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(tg1)(ϰ))+ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;α0+(tf1)(ϰ)+ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;α0+(tg1)(ϰ))+ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)+ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ)) (3.7)

    holds for all ρ(0,1],αC with (α)>0.

    Proof. By means of Lemma 3.1 and setting P=r,Q=s, we can write

    ΨωΩρ;α0+s(ϰ)ΨωΩρ;α0+(tf1g1)(ϰ)+ΨωΩρ;α0+(sf1g1)(ϰ)ΨωΩρ;α0+t(ϰ)ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;α0+(tf1)(ϰ)+ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(tg1)(ϰ). (3.8)

    Conducting product both sides of (3.8) by ΨωΩρ;α0+r(ϰ), we obtain

    ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+s(ϰ)ΨωΩρ;α0+(tf1g1)(ϰ)+ΨωΩρ;α0+(sf1g1)(ϰ)ΨωΩρ;α0+t(ϰ))ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;α0+(tf1)(ϰ)+ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(tg1)(ϰ)). (3.9)

    By means of Lemma 3.1 and setting P=r,Q=t, we can write

    ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(tf1g1)(ϰ)+ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+t(ϰ)ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;α0+(tf1)(ϰ)+ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;α0+(tg1)(ϰ). (3.10)

    Conducting product of (3.10) by ΨωΩρ;α0+s(ϰ), we obtain

    ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(tf1g1)(ϰ)+ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+t(ϰ))ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;α0+(tf1)(ϰ)+ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;α0+(tg1)(ϰ)). (3.11)

    By similar argument as we did before, yields

    ΨωΩρ;α0+t(ϰ)(ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(sf1g1)(ϰ)+ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+t(ϰ))ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)+ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ)). (3.12)

    Adding (3.9), (3.11) and (3.12), we get the desired inequality (3.8).

    Lemma 3.3. Under the assumption of I, II and let Q and P be two non-negative continuous functions on [0,). Then the inequality

    ΨωΩρ;α0+(Pf1g1)(ϰ)ΨωΩρ;β0+Q(ϰ)+ΨωΩρ;α0+P(ϰ)ΨωΩρ;β0+(Qf1g1)(ϰ)ΨωΩρ;α0+(Pf1)(ϰ)ΨωΩρ;β0+(Qg1)(ϰ)+ΨωΩρ;α0+(Pg1)(ϰ)ΨωΩρ;β0+(Qf1)(ϰ),

    holds for all ρ(0,1],α,βC with (α),(β)>0.

    Proof. If we multiply both sides of (3.2) by exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]Q(ν)ω(ν)Ψ(ν)ρβΓ(β)(Ψ(ϰ)Ψ(ν))1β and integrating the resulting inequality with respect to ν from 0 to ϰ, we have

    f1(μ)g1(μ)ρβΓ(β)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]Q(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1βdν+f1(ν)g1(ν)ρβΓ(β)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]Q(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1βdνg1(μ)ρβΓ(β)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]Q(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1βf1(ν)dν+f1(μ)ρβΓ(β)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]Q(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1βg1(ν)dν. (3.13)

    Taking product both sides of the above equation by ω1(ϰ) and in view of Definition (2.2), we have

    f1(μ)g1(μ)ΨωΩρ;β0+Q(ϰ)+ΨωΩρ;β0+(Qf1g1)(ϰ)f1(μ)ΨωΩρ;β0+(Qg1)(ϰ)+g1(μ)ΨωΩρ;β0+(Qf1)(ϰ). (3.14)

    Again, multiplying both sides of (3.14) by exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]P(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1α and integrating the resulting inequality with respect to ν from 0 to ϰ, we have

    ΨωΩρ;β0+Q(ϰ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]P(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αf1(μ)g1(μ)dμ+ΨωΩρ;β0+(Qf1g1)(ϰ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]P(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αdμΨωΩρ;β0+(Qg1)(ϰ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]P(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αf1(μ)dμ+ΨωΩρ;β0+(Qf1)(ϰ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]P(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αg1(μ)dμ. (3.15)

    Taking product both sides of the above equation by ω1(ϰ) and in view of Definition (2.2), we obtain

    ΨωΩρ;α0+(Pf1g1)(ϰ)ΨωΩρ;β0+Q(ϰ)+ΨωΩρ;α0+P(ϰ)ΨωΩρ;β0+(Qf1g1)(ϰ)ΨωΩρ;α0+(Pf1)(ϰ)ΨωΩρ;β0+(Qg1)(ϰ)+ΨωΩρ;α0+(Pg1)(ϰ)ΨωΩρ;β0+(Qf1)(ϰ),

    which implies (3.13).

    Theorem 3.4. Under the assumptions I, II and let r, s and t be three non-negative continuous functions on [0,). Then the inequality

    ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+(sf1g1)(ϰ)ΨωΩρ;β0+t(ϰ)+2ΨωΩρ;α0+s(ϰ)ΨωΩρ;β0+(tf1g1)(ϰ)+ΨωΩρ;β0+t(ϰ)ΨωΩρ;α0+(sf1g1)(ϰ))+(ΨωΩρ;β0+t(ϰ)ΨωΩρ;α0+s(ϰ)+ΨωΩρ;α0+t(ϰ)ΨωΩρ;β0+s(ϰ))ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;β0+(tg1)(ϰ)+ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;β0+(tf1)(ϰ))+ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(tg1)(ϰ)+ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(tf1)(ϰ))+ΨωΩρ;α0+t(ϰ)(ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)+ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(sf1)(ϰ)) (3.16)

    holds for all ρ(0,1],α,βC with (α),(β)>0.

    Proof. By means of Lemma 3.3 and setting P=s,Q=t, we can write

    ΨωΩρ;α0+(sf1g1)(ϰ)ΨωΩρ;β0+t(ϰ)+ΨωΩρ;α0+s(ϰ)ΨωΩρ;β0+(tf1g1)(ϰ)ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;β0+(tg1)(ϰ)+ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;β0+(tf1)(ϰ). (3.17)

    Conducting product both sides of (3.17) by ΨωΩρ;α0+r(ϰ), we obtain

    ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+(sf1g1)(ϰ)ΨωΩρ;β0+t(ϰ)+ΨωΩρ;α0+s(ϰ)ΨωΩρ;β0+(tf1g1)(ϰ))ΨωΩρ;α0+r(ϰ)(ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;β0+(tg1)(ϰ)+ΨωΩρ;α0+(sg1)(ϰ)ΨωΩρ;β0+(tf1)(ϰ)). (3.18)

    Again, by means of Lemma 3.3 and setting P=r,Q=t, we can write

    ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;β0+t(ϰ)+ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(tf1g1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(tg1)(ϰ)+ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(tf1)(ϰ). (3.19)

    Conducting product both sides of (3.19) by ΨωΩρ;α0+s(ϰ), we obtain

    ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;β0+t(ϰ)+ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(tf1g1)(ϰ))ΨωΩρ;α0+s(ϰ)(ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(tg1)(ϰ)+ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(tf1)(ϰ)). (3.20)

    By similar arguments as we did before, yields

    ΨωΩρ;α0+t(ϰ)(ΨωΩρ;α0+(sf1g1)(ϰ)ΨωΩρ;β0+r(ϰ)+ΨωΩρ;α0+s(ϰ)ΨωΩρ;β0+(rf1g1)(ϰ))ΨωΩρ;α0+t(ϰ)(ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)+ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(sf1)(ϰ)). (3.21)

    Adding (3.18), (3.20) and (3.21), we get the desired inequality (3.16).

    Remark 4. Theorem 3.2 and Theorem 3.4 lead to the following conclusions:

    (1) Let f1 and g1 are the asynchronous functions on [0,), then (3.8) and (3.16) are reversed.

    (2) Let r,s and t are negative on [0,), then (3.8) and (3.16) are reversed.

    (3) Let r,s are positive t is negative on [0,), then (3.8) and (3.16) are reversed.

    In the next, we derive certain novel Grüss-type integral inequalities via weighted generalized proportional fractional integral operators.

    Lemma 3.5. Suppose an integrable function f1 defined on [0,) satisfying the assertions I,II and (1.7) on [0,) and let a continuous function r defined on [0,). Then the inequality

    ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf21)(ϰ)(ΨωΩρ;α0+(rf1)(ϰ))2(ΦΨωΩρ;α0+x(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ)) (3.22)

    holds for all ρ(0,1],αC with (α)>0.

    Proof. By the given hypothesis and utilizing (1.7). For any μ,ν[0,), we have

    (Φf1(ν))(f1(μ)ϕ)+(Φf1(μ))(f1(ν)ϕ)(Φf1(μ))(f1(μ)ϕ)(Φf1(ν))(f1(ν)ϕ)f21(μ)+f21(ν)2f1(μ)f1(ν). (3.23)

    Multiplying both sides of (3.23) by exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)ραΓ(α)(Ψ(ϰ)Ψ(ν))1α and integrating the resulting inequality with respect to ν from 0 to ϰ, we have

    (f1(μ)ϕ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1α(Φf1(ν))dν+(Φf1(μ))ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1α(f1(ν)ϕ)dν(Φf1(μ))(f1(μ)ϕ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1αdν1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1α(Φf1(ν))(f1(ν)ϕ)dνf21(μ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1αdν+1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1αf21(ν)dν2f1(μ)ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1αf1(ν)dν. (3.24)

    Taking product both sides of the above equation by ω1(ϰ) and in view of Definition (2.2), we obtain

    (ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(f1(μ)ϕ)+(Φf1(μ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))(Φf1(μ))(f1(μ)ϕ)ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ))f21(μ)ΨωΩρ;α0+r(ϰ)+ΨωΩρ;α0+(rf21)(ϰ)2f1(μ)ΨωΩρ;α0+(rf1)(ϰ). (3.25)

    Multiplying both sides of (3.25) by exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1α and integrating the resulting inequality with respect to μ from 0 to ϰ, we have

    (ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ν))1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α(f1(μ)ϕ)dμ+(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α(Φf1(μ))dμ(1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α(Φf1(μ))(f1(μ)ϕ)dμ)ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Φf1(ν))(f1(ν)ϕ)1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αdν(1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αf21(μ)dμ)ΨωΩρ;α0+r(ϰ)+(1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αdμ)ΨωΩρ;α0+(rf21)(ϰ)2(1ραΓ(α)ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1αf1(μ)dμ)ΨωΩρ;α0+(rf1)(ϰ). (3.26)

    Taking product both sides of the above equation by ω1(ϰ) and in view of Definition (2.2), we obtain

    (ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))+(ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))ΨωΩρ;α0+(r(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ))ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ))ΨωΩρ;α0+(rf21)(ϰ)ΨωΩρ;α0+r(ϰ)+ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf21)(ϰ)2ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ), (3.27)

    which gives (3.22) and proves the lemma.

    Theorem 3.6. Suppose two integrable functions f1 and g1 defined on [0,) satisfying the assertions I,II and (1.7) on [0,) and let a continuous function r defined on [0,). Then the inequality

    |ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ)|(Φϕ)(Υγ)4(ΨωΩρ;α0+r(ϰ))2 (3.28)

    holds for all ρ(0,1],αC with (α)>0.

    Proof. By the given hypothesis stated in Theorem 3.6. Also, assume that μ,ν be defined by

    T(μ,ν)=(f1(μ)f1(ν))(g1(μ)g1(ν)),μ,ν[0,ϰ],ϰ>0. (3.29)

    Multiplying both sides of (3.30) by exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1αexp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)ραΓ(α)(Ψ(ϰ)Ψ(ν))1α and integrating the resulting inequality with respect to μ and ν from 0 to ϰ, we can state that

    1ρ2αΓ2(α)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1αT(μ,ν)dμdν=1ρ2αΓ2(α)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1α×(f1(μ)f1(ν))(g1(μ)g1(ν))dμdν. (3.30)

    Taking product both sides of the above equation by ω1(ϰ) and in view of Definition (2.2), we obtain

    ω2(ϰ)ρ2αΓ2(α)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1αT(μ,ν)dμdν=2ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1g1)(ϰ)2ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ). (3.31)

    Thanks to the weighted Cauchy-Schwartz integral inequality for double integrals, we can write that

    (ω2(ϰ)ρ2αΓ2(α)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1αT(μ,ν)dμdν)2(ω2(ϰ)ρ2αΓ2(α)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1α(f1(μ)f1(ν))dμdν)(ω2(ϰ)ρ2αΓ2(α)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]r(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1α(g1(μ)g1(ν))dμdν)=4(ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf21)(ϰ)(ΨωΩρ;α0+(rf1)(ϰ))2)×(ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rg21)(ϰ)(ΨωΩρ;α0+(rg1)(ϰ))2). (3.32)

    Since (Φf1(μ))(f1(μ)ϕ)0 and (Υg1(μ))(g1(μ)γ)0, we have

    ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Φf1(μ))(f1(μ)ϕ))0, (3.33)

    and

    ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Υg1(μ))(g1(μ)γ))0. (3.34)

    Therefore, from (3.33), (3.34) and Lemma 3.5, we get

    ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf21)(ϰ)(ΨωΩρ;α0+(rf1)(ϰ))2(ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ)) (3.35)

    and

    ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rg21)(ϰ)(ΨωΩρ;α0+(rg1)(ϰ))2(ΥΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rg1)(ϰ))(ΨωΩρ;α0+(rg1)(ϰ)γΨωΩρ;α0+r(ϰ)). (3.36)

    Combining (3.30), (3.31), (3.35) and (3.36), we deduce that

    (ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(xf1g1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ))2(ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;α0+(rf)(ϰ)ϕΨωΩρ;α0+r(ϰ))×(ΥΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rg1)(ϰ))(ΨωΩρ;α0+(rg1)(ϰ)γΨωΩρ;α0+r(ϰ)). (3.37)

    Taking into consideration the elementary inequality 4a1a2(a1+a2)2,a1,a2R, we can state that

    4(ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))(ΨωΩρ;α0+r(ϰ)(Φϕ))2 (3.38)

    and

    4(ΥΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rg1)(ϰ))(ΨωΩρ;α0+(rg1)(ϰ)γΨωΩρ;α0+r(ϰ))(ΨωΩρ;α0+r(ϰ)(Υγ))2. (3.39)

    From (3.37)-(3.39), we obtain (3.28). This completes the proof of Theorem 3.6.

    Lemma 3.7. Suppose two integrable functions f1 and g1 defined on [0,) satisfying the assertions I,II and (1.7) on [0,) and let two continuous function r and s defined on [0,). Then the inequality

    (ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(sf1g1)(ϰ)+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ))2(ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(sf21)(ϰ)+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(rf21)(ϰ)2ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(sf1)(ϰ))×(ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(sg21)(ϰ)+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(rg21)(ϰ)2ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)) (3.40)

    holds for all ρ(0,1],α,βC with (α),(β)>0.

    Proof. Taking product (3.30) by exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)ραΓ(α)(Ψ(ϰ)Ψ(μ))1αexp[ρ1ρ(Ψ(ϰ)Ψ(ν))]s(ν)ω(ν)Ψ(ν)ρβΓ(β)(Ψ(ϰ)Ψ(ν))1β and integrating the resulting inequality with respect to μ and ν from 0 to ϰ, we can state that

    1ραΓ(α)ρβΓ(β)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]s(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1βT(μ,ν)dμdν=1ραΓ(α)ρβΓ(β)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]s(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1β×(f1(μ)f1(ν))(g1(μ)g1(ν))dμdν. (3.41)

    Taking product both sides of the above equation by ω2(ϰ) and utilizing Definition (2.2), we have

    ω2(ϰ)ραΓ(α)ρβΓ(β)ϰ0ϰ0exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)(Ψ(ϰ)Ψ(μ))1α×exp[ρ1ρ(Ψ(ϰ)Ψ(ν))]s(ν)ω(ν)Ψ(ν)(Ψ(ϰ)Ψ(ν))1βT(μ,ν)dμdν=ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(sf1g1)(ϰ)+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ). (3.42)

    Then, thanks to the weighted Cauchy-Schwartz integral inequality for double integrals, we conclude (3.40).

    Lemma 3.8. Suppose an integrable function f1 defined on [0,) satisfying the assertions I and II on [0,) and let two continuous function r and s defined on [0,). Then the inequality

    ΨωΩρ;β0+(sf21)(ϰ)ΨωΩρ;α0+r(ϰ)+ΨωΩρ;α0+(rf21)(ϰ)ΨωΩρ;β0+s(ϰ)2ΨωΩρ;β0+(sf1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)(ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;β0+(sf1)(ϰ)ϕΨωΩρ;β0+s(ϰ))+(ΦΨωΩρ;β0+s(ϰ)ΨωΩρ;β0+(sf1)(ϰ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))ΨωΩρ;β0+(s(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ))ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(r(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ)) (3.43)

    holds for all ρ(0,1],α,βC with (α),(β)>0.

    Proof. Multiplying both sides of (3.25) by exp[ρ1ρ(Ψ(ϰ)Ψ(μ))]r(μ)ω(μ)Ψ(μ)ρβΓ(β)(Ψ(ϰ)Ψ(μ))1β and integrating the resulting inequality with respect to μ from 0 to ϰ. Then, by multiplying with ω1(ϰ) and in view of Definition 2.2, concludes

    (ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;β0+(sf1)(ϰ)ϕΨωΩρ;β0+s(ϰ))+(ΦΨωΩρ;β0+s(ϰ)ΨωΩρ;β0+(sf1)(ϰ))(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))ΨωΩρ;β0+(s(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ))ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(r(ϰ)(Φf1(ϰ))(f1(ϰ)ϕ))ΨωΩρ;β0+(sf21)(ϰ)ΨωΩρ;α0+r(ϰ)+ΨωΩρ;α0+(rf21)(ϰ)ΨωΩρ;β0+s(ϰ)2ΨωΩρ;β0+(sf1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ), (3.44)

    which gives (3.43) and proves the lemma.

    Theorem 3.9. Suppose two integrable functions f1 and g1 defined on [0,) satisfying the assertions I,II and (1.7) on [0,) and let two continuous function r and s defined on [0,). Then the inequality

    (ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(sf1g1)(ϰ)+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(rf1g1)(ϰ)ΨωΩρ;α0+(rf1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)ΨωΩρ;α0+(sf1)(ϰ)ΨωΩρ;α0+(rg1)(ϰ))2{(ΦΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rf1)(ϰ))(ΨωΩρ;β0+(sf1)(ϰ)ϕΨωΩρ;β0+s(ϰ))+(ΨωΩρ;α0+(rf1)(ϰ)ϕΨωΩρ;α0+r(ϰ))(ΦΨωΩρ;β0+s(ϰ)ΨωΩρ;β0+(sf1)(ϰ))}×{(ΥΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(rg1)(ϰ))(ΨωΩρ;β0+(sg1)(ϰ)γΨωΩρ;β0+s(ϰ))+(ΨωΩρ;α0+(rg1)(ϰ)γΨωΩρ;α0+r(ϰ))(ΥΨωΩρ;β0+s(ϰ)ΨωΩρ;β0+(sg1)(ϰ))} (3.45)

    holds for all with

    Proof. Since and we have

    (3.46)

    and

    (3.47)

    Utilizing Lemma 3.8 to and and utilizing Lemma 3.7 and the inequalities (3.46) and (3.47), yields (3.45).

    Theorem 3.10. Suppose two integrable functions and defined on satisfying the assertions and on and let two continuous function and defined on . Then the inequality

    (3.48)

    holds for all with

    Proof. Taking into consideration the assumption (1.7), we have

    (3.49)

    which implies that

    (3.50)

    From (3.42) and (3.50), we obtain that

    (3.51)

    This ends the proof.

    Theorem 3.11. Suppose two integrable functions and defined on satisfying the assertions and on and let two continuous function and defined on . Then the inequality

    (3.52)

    holds for all with

    Proof. Taking into consideration the assumption (1.12), we have

    (3.53)

    which implies that

    (3.54)

    From (3.42) and (3.54), we obtain that

    (3.55)

    This ends the proof.

    Theorem 3.12. Suppose two integrable functions and defined on satisfying the assertions and the lipschitzian condition with the constants and and let two continuous function and defined on . Then the inequality

    (3.56)

    holds for all with

    Proof. By the given hypothesis, we have

    (3.57)

    which implies that

    (3.58)

    From (3.42) and (3.58), we obtain that

    (3.59)

    This ends the proof.

    Corollary 1. Let and be two differentiable functions on and let and be two non-negative continuous functions on Then the inequality

    (3.60)

    holds for all with

    Proof. We have and That is, and the immediate consequence follows from Theorem 3.12. This completes the proof.

    Example 3.13. Let with and be a function on Let be an integrable function defined on and be the weighted generalized proportional fractional integral operator satisfying assumption Then we have

    where

    and

    is the incomplete gamma function [52,53].

    Proof. It follows from Definition 2.2 and the modulus property that

    for

    Making use of the well-known Hölder inequality, we obtain

    Let Then elaborated computations lead to

    Here, we aim at present some new generalizations via weighted generalized proportional fractional, weighted generalized Riemann-Liouville and weighted Riemann-Liouville fractional integral operators, which are the new estimates of the main consequences.

    Lemma 4.1. Let and be two synchronous functions on Assume that and be two non-negative continuous mappings on Then the inequality

    holds for all with

    Proof. Letting and Lemma 3.1 yields the proof of Lemma 4.1.

    Lemma 4.2. Let and be two synchronous functions on Assume that and be two non-negative continuous mappings on Then the inequality

    holds for all with

    Proof. Letting and Lemma 3.1 yields the proof of Lemma 4.2.

    Lemma 4.3. Under the assumption of Lemma 3.1, then the inequality

    holds for all with

    Proof. Letting and Lemma 3.1 yields the proof of Lemma 4.3.

    Lemma 4.4. Under the assumption of Lemma 4.2, then the inequality

    holds for all with

    Proof. Letting and Lemma 3.1 yields the proof of Lemma 4.4.

    Theorem 4.5. Let and be two synchronous functions on Assume that and be three non-negative continuous functions on Then the inequality

    holds for all with

    Proof. Letting and Theorem 3.2 yields the proof of Theorem 4.5.

    Theorem 4.6. Under the assumption of and let and be three non-negative continuous functions on Then the inequality

    holds for all with

    Proof. Letting and Theorem 3.2 yields the proof of Theorem 4.6.

    Theorem 4.7. Under the assumption of Theorem 4.5, then the inequality

    holds for all with

    Proof. Letting and Theorem 3.2 yields the proof of Theorem 4.7.

    Remark 5. The computed results lead to the following conclusion:

    (1) Setting and and using the relation (2.7), (2.8) and the assumption , then Theorem 3.6 and Theorem 3.9 reduces to the known results due to Dahmani et al. [38].

    (2) Setting and using the relation (2.7), (2.8) and the assumption , then Theorem 3.10–3.12, and Corollary 1 reduces to the known results due to Dahmani et al. [38] and Dahmani [40], respectively.

    A new generalized fractional integral operator is proposed in this paper. The novel investigation is used to generate novel weighted fractional operators in the Riemann-Liouville, generalized Riemann-Liouville, Hadamard, Katugampola, Generalized proportional fractional, generalized Hadamard proportional fractional and henceforth, which effectively alleviates the adverse effect of another function and proportionality index Utilizing the weighted generalized proportional fractional operator technique, we derived the analogous versions of the extended Chebyshev and Grüss type inequalities that improve the accuracy and efficiency of the proposed technique. Contemplating the Remark 2 and 3, several existing results can be identified in the literature. Some innovative particular cases constructed by this method are tested and analyzed for statistical theory, fractional Schrödinger equation [20,21]. The results show that the method proposed in this paper can stably and efficiently generate integral inequalities for convexity with better operators performance, thus providing a reliable guarantee for its application in control theory [54].

    The authors declare that they have no competing interests.

    The authors would like to express their sincere thanks to referees for improving the article and also thanks to Natural Science Foundation of China (Grant Nos. 61673169) for providing financial assistance to support this research. The authors would like to express their sincere thanks to the support of Taif University Researchers Supporting Project Number (TURSP-2020/217), Taif University, Taif, Saudi Arabia.



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