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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)=1b1−a1b1∫a1f1(ϰ)g1(ϰ)dϰ−1b1−a1b1∫a1f1(ϰ)dϰ1b1−a1b1∫a1g1(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(Φ−ϕ)(Υ−γ), | (1.6) |
where f1 and g1 are two integrable functions on [a1,b1] satisfying the assumptions
ϕ≤f1(ϰ)≤Φ,γ≤g1(ϰ)≤Υ,ϕ,Φ,γ,Υ∈R,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(Φ−ϕ)(Υ−γ)(b1∫a1P1(ϰ)dϰ)2, | (1.8) |
where
S(f1,g1,P)=12T(f1,g1,P)=b1∫a1P(ϰ)dϰb1∫a1P(ϰ)f1(ϰ)g1(ϰ)dϰ−b1∫a1P(ϰ)f1(ϰ)dϰb1∫a1P(ϰ)g1(ϰ)dϰ | (1.9) |
and
T(f1,g1,P,Q)=b1∫a1Q(ϰ)dϰb1∫a1P(ϰ)f1(ϰ)g1(ϰ)dϰ+b1∫a1P(ϰ)dϰb1∫a1Q(ϰ)f1(ϰ)g1(ϰ)dϰ−b1∫a1Q(ϰ)f1(ϰ)dϰb1∫a1P(ϰ)g1(ϰ)dϰ−b1∫a1P(ϰ)f1(ϰ)dϰb1∫a1Q(ϰ)g1(ϰ)dϰ. | (1.10) |
In [37], Dragomir established the inequality:
If f′1,g′1∈L∞(a1,b1), then
|S(f1,g1,P)|≤‖f′1‖∞‖g′1‖∞(b1∫a1P(ϰ)dϰb1∫a1ϰ2P(ϰ)dϰ−(b1∫a1ϰP(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(b1∫a1P(ϰ)dϰb1∫a1g21(ϰ)P(ϰ)dϰ−(b1∫a1g1(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(b1∫a1P(ϰ)dϰb1∫a1ϰ2P(ϰ)dϰ−(b1∫a1ϰP(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 ω≠0 be a mapping defined on [a1,b1], g1 is a differentiable strictly increasing function on [a1,b1]. The space χpω(a1,b1),1≤p<∞ is the space of all Lebesgue measurable functions f1 defined on [a1,b1] for which ‖f1‖χpω, where
‖f1‖χpω=(b1∫a1|ω(ϰ)f1(ϰ)|pg′1(ϰ)dϰ)1p,1<p<∞ | (2.1) |
and
‖f1‖χpω=esssupa1≤ϰ≤b1|ω(ϰ)f1(ϰ)|<∞. | (2.2) |
Remark 1. Clearly we see that f1∈χpω(a1,b1) ⟹ ω(ϰ)f1(ϰ)(g−11(ϰ))1/p∈Lp(a1,b1) for 1≤p<∞ and f1∈χpω(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 f1∈χpω(a1,b1) and ω≠0 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)=ω−1(ϰ)ραΓ(α)ϰ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))](Ψ(ϰ)−Ψ(μ))1−αf1(μ)ω(μ)Ψ′(μ)dμ,a1<ϰ | (2.3) |
and
ΨωΩρ;αb1f1(ϰ)=ω−1(ϰ)ραΓ(α)b1∫ϰexp[ρ−1ρ(Ψ(μ)−Ψ(ϰ))](Ψ(μ)−Ψ(ϰ))1−αf1(μ)ω(μ)Ψ′(μ)dμ,ϰ<b1, | (2.4) |
where ρ∈(0,1] is the proportionality index, α∈C,ℜ(α)>0 and Γ(ϰ)=∫∞0μϰ−1e−μdμ 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)=ω−1(ϰ)ραΓ(α)ϰ∫a1exp[ρ−1ρ(ϰ−μ)](ϰ−μ)1−αf1(μ)ω(μ)dμ,a1<ϰ | (2.5) |
and
ωΩρ;αb1f1(ϰ)=ω−1(ϰ)ραΓ(α)b1∫ϰexp[ρ−1ρ(μ−ϰ)](μ−ϰ)1−αf1(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)=ω−1(ϰ)Γ(α)ϰ∫a1f1(μ)ω(μ)dμ(ϰ−μ)1−α,a1<ϰ | (2.7) |
and
ωΩαb1f1(ϰ)=ω−1(ϰ)Γ(α)b1∫ϰf1(μ)ω(μ)dμ(μ−ϰ)1−α,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)=ω−1(ϰ)ραΓ(α)ϰ∫a1exp[ρ−1ρ(lnϰμ)](lnϰμ)1−αf1(μ)ω(μ)μdμ,a1<ϰ | (2.9) |
and
ωΩρ;αb1f1(ϰ)=ω−1(ϰ)ραΓ(α)b1∫ϰexp[ρ−1ρ(lnμϰ)](lnμϰ)1−αf1(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)=ω−1(ϰ)Γ(α)ϰ∫a1f1(μ)ω(μ)dμμ(lnϰμ)1−α,a1<ϰ | (2.11) |
and
ωΩαb1f1(ϰ)=ω−1(ϰ)Γ(α)b1∫ϰf1(μ)ω(μ)dμμ(lnμϰ)1−α,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)=ω−1(ϰ)Γ(α)ϰ∫a1(ϰτ−μττ)α−1f1(μ)ω(μ)dμμ1−τ,a1<ϰ | (2.13) |
and
ωΩαb1f1(ϰ)=ω−1(ϰ)Γ(α)b1∫ϰ(μτ−ϰττ)α−1f1(μ)ω(μ)dμμ1−τ,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],s∈R, then we get the Definition 2 defined by Khan and Khan et al [50].
(8) Letting ω(ϰ)=ρ=1 and Ψ(ϰ)=(ϰ−a1)ττ, and Ψ(ϰ)=−(b1−ϰ)ττ,(τ>0), then we get the operator defined by Jarad et al. [51].
Theorem 2.3. For α>0,ρ∈(0,1],1≤p≤∞ and f1∈χpω(a1,b1). Then ΨωΩρ;αa1 is bounded in χpω(a1,b1) and
‖ΨωΩρ;αa1f1‖χpω≤(Ψ(b1)−Ψ(a1))α‖f1‖χpωραΓ(α+1). |
Proof. For 1≤p≤∞, we have
‖ΨωΩρ;αa1f1‖χpω=1ραΓ(α)(b1∫a1|ϰ∫a1exp[ρ−1ρΨ(ϰ)−Ψ(μ)](Ψ(ϰ)−Ψ(μ))1−αω(μ)f1(μ)Ψ′(μ)dμ|pΨ′(ϰ)dϰ)1/p=1ραΓ(α)(∫Ψ(b1)Ψ(a1)|t2∫Ψ(a1)exp[ρ−1ρ(t2−t1)](t2−t1)1−αω(Ψ−1(t1))f1(Ψ−1(t1))|pdt2)1/p. |
Using the fact that |exp[ρ−1ρ(t2−t1)]|<1. Taking into account the generalized Minkowski inequality [5], we can write
‖ΨωΩρ;αa1f1‖χpω≤1ραΓ(α)∫Ψ(b1)Ψ(a1)(|ω(Ψ−1(t1))f1(Ψ−1(t1))|pΨ(b1)∫t1(t2−t1)p(α−1)dt2)1/pdt1=1ραΓ(α)Ψ(b1)∫Ψ(a1)(|ω(Ψ−1(t1))f1(Ψ−1(t1))|((Ψ(b1)−t1)p(α−1)+1p(α−1)+1)1/pdt1. |
By employing the well-known Hölder inequality satisfying p−1+q−1=1, we obtain
‖ΨωΩρ;αa1f1‖χpω≤1ραΓ(α)(∫Ψ(b1)Ψ(a1)|ω(Ψ−1(t1))f1(Ψ−1(t1))|pdt1)1/p(∫Ψ(b1)Ψ(a1)((Ψ(b1)−t1)p(α−1)+1p(α−1)+1)q/pdt1)1/q≤1ραΓ(α)(∫b1a1|ω(ϰ)f1(ϰ)|pΨ′(ϰ)dϰ)1/p(∫Ψ(b1)Ψ(a1)((Ψ(b1)−t1)p(α−1)+1p(α−1)+1)q/pdt1)1/q≤(Ψ(b1)−Ψ(a1))α‖f1‖χpωραΓ(α+1). |
Now, for p=∞, we have
|ω(ϰ)ΨωΩρ;αa1f1(ϰ)|=1ραΓ(α)ϰ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))](Ψ(ϰ)−Ψ(μ))1−αf1(μ)ω(μ)Ψ′(μ)dμ≤1ραΓ(α)ϰ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))](Ψ(ϰ)−Ψ(μ))1−α|f1(μ)ω(μ)|Ψ′(μ)dμ,Since(|exp[ρ−1ρ(t2−t1)]|<1)≤‖f1‖χ∞ωραΓ(α)ϰ∫a1(Ψ(ϰ)−Ψ(μ))α−1dμ≤(Ψ(ϰ)−Ψ(a1))α‖f1‖χ∞ωραΓ(α+1)=(Ψ(b1)−Ψ(a1))α‖f1‖χ∞ωραΓ(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 1≤p≤∞ and let f1∈χpω(a1,b1). Then
(ΨωΩρ;αa1ΨωΩρ;βa1)f1=(ΨωΩρ;α+βa1)f1. | (2.15) |
Proof.
(ΨωΩρ;αa1ΨωΩρ;βa1f1)(ϰ)=ω−1(ϰ)ραΓ(α)ϰ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))](Ψ(ϰ)−Ψ(μ))1−αω(μ)(ΨωΩρ;βa1f1)(μ)Ψ′(μ)dμ=ω−1(ϰ)ρα+βΓ(α)Γ(β)ϰ∫a1μ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))](Ψ(ϰ)−Ψ(μ))1−αexp[ρ−1ρ(Ψ(μ)−Ψ(ν))](Ψ(μ)−Ψ(ν))1−β×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)=ω−1(ϰ)ρα+βΓ(α)Γ(β)1∫0θβ−1(1−θ)α−1dθϰ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))](Ψ(ϰ)−Ψ(ν))1−α−βω(ν)f1(ν)Ψ′(ν)dν=ω−1(ϰ)ρα+βΓ(α)Γ(β)Γ(α)Γ(β)Γ(α+β)ϰ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))](Ψ(ϰ)−Ψ(ν))1−α−βω(ν)f1(ν)Ψ′(ν)dν=(ΨωΩρ;α+βa1f1)(ϰ), |
where B(α,β)=Γ(α)Γ(β)Γ(α+β)=1∫0θβ−1(1−θ)α−1dθ 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],α∈C with ℜ(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[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]Q(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−α and integrating the resulting inequality with respect to μ from 0 to ϰ, we get
1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]Q(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−αf1(μ)g1(μ)dμ+f1(ν)g1(ν)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]Q(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−αdμ≥f1(ν)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]Q(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−αg1(ν)dν+g1(ν)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]Q(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−αf1(μ)dμ. | (3.4) |
Taking product both sides of the above equation by ω−1(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[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]P(ν)ω(ν)Ψ′(ν)ραΓ(α)(Ψ(ϰ)−Ψ(ν))1−α and integrating the resulting inequality with respect to ν from 0 to ϰ. Then, multiplying by ω−1(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],α∈C with ℜ(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,β∈C with ℜ(α),ℜ(β)>0.
Proof. If we multiply both sides of (3.2) by exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]Q(ν)ω(ν)Ψ′(ν)ρβΓ(β)(Ψ(ϰ)−Ψ(ν))1−β and integrating the resulting inequality with respect to ν from 0 to ϰ, we have
f1(μ)g1(μ)ρβΓ(β)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]Q(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βdν+f1(ν)g1(ν)ρβΓ(β)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]Q(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βdν≥g1(μ)ρβΓ(β)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]Q(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βf1(ν)dν+f1(μ)ρβΓ(β)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]Q(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βg1(ν)dν. | (3.13) |
Taking product both sides of the above equation by ω−1(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[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]P(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−α and integrating the resulting inequality with respect to ν from 0 to ϰ, we have
ΨωΩρ;β0+Q(ϰ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]P(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αf1(μ)g1(μ)dμ+ΨωΩρ;β0+(Qf1g1)(ϰ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]P(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αdμ≥ΨωΩρ;β0+(Qg1)(ϰ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]P(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αf1(μ)dμ+ΨωΩρ;β0+(Qf1)(ϰ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]P(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αg1(μ)dμ. | (3.15) |
Taking product both sides of the above equation by ω−1(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,β∈C with ℜ(α),ℜ(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)(Φ−f1(ϰ))(f1(ϰ)−ϕ)) | (3.22) |
holds for all ρ∈(0,1],α∈C with ℜ(α)>0.
Proof. By the given hypothesis and utilizing (1.7). For any μ,ν∈[0,∞), we have
(Φ−f1(ν))(f1(μ)−ϕ)+(Φ−f1(μ))(f1(ν)−ϕ)−(Φ−f1(μ))(f1(μ)−ϕ)−(Φ−f1(ν))(f1(ν)−ϕ)≤f21(μ)+f21(ν)−2f1(μ)f1(ν). | (3.23) |
Multiplying both sides of (3.23) by exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)ραΓ(α)(Ψ(ϰ)−Ψ(ν))1−α and integrating the resulting inequality with respect to ν from 0 to ϰ, we have
(f1(μ)−ϕ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−α(Φ−f1(ν))dν+(Φ−f1(μ))ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−α(f1(ν)−ϕ)dν−(Φ−f1(μ))(f1(μ)−ϕ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−αdν−1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−α(Φ−f1(ν))(f1(ν)−ϕ)dν≤f21(μ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−αdν+1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−αf21(ν)dν−2f1(μ)ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−αf1(ν)dν. | (3.24) |
Taking product both sides of the above equation by ω−1(ϰ) and in view of Definition (2.2), we obtain
(ΦΨωΩρ;α0+r(ϰ)−ΨωΩρ;α0+(rf1)(ϰ))(f1(μ)−ϕ)+(Φ−f1(μ))(ΨωΩρ;α0+(rf1)(ϰ)−ϕΨωΩρ;α0+r(ϰ))−(Φ−f1(μ))(f1(μ)−ϕ)ΨωΩρ;α0+r(ϰ)−ΨωΩρ;α0+(r(ϰ)(Φ−f1(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[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−α 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)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α(f1(μ)−ϕ)dμ+(ΨωΩρ;α0+(rf1)(ϰ)−ϕΨωΩρ;α0+r(ϰ))1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α(Φ−f1(μ))dμ−(1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α(Φ−f1(μ))(f1(μ)−ϕ)dμ)ΨωΩρ;α0+r(ϰ)−ΨωΩρ;α0+(r(ϰ)(Φ−f1(ν))(f1(ν)−ϕ)1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αdν≤(1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αf21(μ)dμ)ΨωΩρ;α0+r(ϰ)+(1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αdμ)ΨωΩρ;α0+(rf21)(ϰ)−2(1ραΓ(α)ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−αf1(μ)dμ)ΨωΩρ;α0+(rf1)(ϰ). | (3.26) |
Taking product both sides of the above equation by ω−1(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)(Φ−f1(ϰ))(f1(ϰ)−ϕ))ΨωΩρ;α0+r(ϰ)−ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Φ−f1(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)|≤(Φ−ϕ)(Υ−γ)4(ΨωΩρ;α0+r(ϰ))2 | (3.28) |
holds for all ρ∈(0,1],α∈C with ℜ(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[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−αexp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)ραΓ(α)(Ψ(ϰ)−Ψ(ν))1−α and integrating the resulting inequality with respect to μ and ν from 0 to ϰ, we can state that
1ρ2αΓ2(α)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−αT(μ,ν)dμdν=1ρ2αΓ2(α)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−α×(f1(μ)−f1(ν))(g1(μ)−g1(ν))dμdν. | (3.30) |
Taking product both sides of the above equation by ω−1(ϰ) and in view of Definition (2.2), we obtain
ω−2(ϰ)ρ2αΓ2(α)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−αT(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
(ω−2(ϰ)ρ2αΓ2(α)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−αT(μ,ν)dμdν)2≤(ω−2(ϰ)ρ2αΓ2(α)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−α(f1(μ)−f1(ν))dμdν)(ω−2(ϰ)ρ2αΓ2(α)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]r(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−α(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 (Φ−f1(μ))(f1(μ)−ϕ)≥0 and (Υ−g1(μ))(g1(μ)−γ)≥0, we have
ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Φ−f1(μ))(f1(μ)−ϕ))≥0, | (3.33) |
and
ΨωΩρ;α0+r(ϰ)ΨωΩρ;α0+(r(ϰ)(Υ−g1(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,a2∈R, 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)(Φ−ϕ))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)(Υ−γ))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,β∈C with ℜ(α),ℜ(β)>0.
Proof. Taking product (3.30) by exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)ραΓ(α)(Ψ(ϰ)−Ψ(μ))1−αexp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)ρβΓ(β)(Ψ(ϰ)−Ψ(ν))1−β and integrating the resulting inequality with respect to μ and ν from 0 to ϰ, we can state that
1ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βT(μ,ν)dμdν=1ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−β×(f1(μ)−f1(ν))(g1(μ)−g1(ν))dμdν. | (3.41) |
Taking product both sides of the above equation by ω−2(ϰ) and utilizing Definition (2.2), we have
ω−2(ϰ)ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βT(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)(Φ−f1(ϰ))(f1(ϰ)−ϕ))ΨωΩρ;α0+r(ϰ)−ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(r(ϰ)(Φ−f1(ϰ))(f1(ϰ)−ϕ)) | (3.43) |
holds for all ρ∈(0,1],α,β∈C with ℜ(α),ℜ(β)>0.
Proof. Multiplying both sides of (3.25) by exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)ρβΓ(β)(Ψ(ϰ)−Ψ(μ))1−β and integrating the resulting inequality with respect to μ from 0 to ϰ. Then, by multiplying with ω−1(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)(Φ−f1(ϰ))(f1(ϰ)−ϕ))ΨωΩρ;α0+r(ϰ)−ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(r(ϰ)(Φ−f1(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 ρ∈(0,1],α,β∈C with ℜ(α),ℜ(β)>0.
Proof. Since (Φ−f1(μ))(f1(μ)−ϕ)≥0 and (Υ−g1(μ))(g1(μ)−γ)≥0, we have
−ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(s(ϰ)(Φ−f1(ϰ))(f1(ϰ)−ϕ))−ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(r(ϰ)(Φ−f1(ϰ))(f1(ϰ)−ϕ))≤0 | (3.46) |
and
−ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(s(ϰ)(Υ−g1(ϰ))(g1(ϰ)−γ))−ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(r(ϰ)(Υ−g1(ϰ))(g1(ϰ)−γ))≤0. | (3.47) |
Utilizing Lemma 3.8 to f1 and g1, and utilizing Lemma 3.7 and the inequalities (3.46) and (3.47), yields (3.45).
Theorem 3.10. 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)|≤ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+s(ϰ)(Φ−ϕ)(Υ−γ) | (3.48) |
holds for all ρ∈(0,1],α,β∈C with ℜ(α),ℜ(β)>0.
Proof. Taking into consideration the assumption (1.7), we have
|f1(μ)−f1(ν)|≤Φ−ϕ,|g1(μ)−g1(ν)|≤Υ−γ,μ,ν∈[0,∞), | (3.49) |
which implies that
|T(μ,ν)|=|f1(μ)−f1(ν)||g1(μ)−g1(ν)|≤(Φ−ϕ)(Υ−γ). | (3.50) |
From (3.42) and (3.50), we obtain that
|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)|≤ω−2(ϰ)ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βT(μ,ν)dμdν≤ω−2(ϰ)ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−β((Φ−ϕ)(Υ−γ))dμdν=ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+s(ϰ)(Φ−ϕ)(Υ−γ). | (3.51) |
This ends the proof.
Theorem 3.11. 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)|≤L(ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(sg21)(ϰ)+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(rg21)(ϰ)−2ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)) | (3.52) |
holds for all ρ∈(0,1],α,β∈C with ℜ(α),ℜ(β)>0.
Proof. Taking into consideration the assumption (1.12), we have
|f1(μ)−f1(ν)|≤L|g1(μ)−g1(ν)|μ,ν∈[0,∞), | (3.53) |
which implies that
|T(μ,ν)|=|f1(μ)−f1(ν)||g1(μ)−g1(ν)|≤L(g1(μ)−g1(ν))2. | (3.54) |
From (3.42) and (3.54), we obtain that
|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)|≤ω−2(ϰ)ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βT(μ,ν)dμdν≤Lω−2(ϰ)ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−β(g1(μ)−g1(ν))2dμdν=L(ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(sg21)(ϰ)+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(rg21)(ϰ)−2ΨωΩρ;α0+(rg1)(ϰ)ΨωΩρ;β0+(sg1)(ϰ)). | (3.55) |
This ends the proof.
Theorem 3.12. Suppose two integrable functions f1 and g1 defined on [0,∞) satisfying the assertions I,II and the lipschitzian condition with the constants M1 and M2 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)|≤M1M2(ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(ϰ2s(ϰ))+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(ϰ2r(ϰ))−2ΨωΩρ;α0+(ϰr(ϰ))ΨωΩρ;β0+(ϰs(ϰ))) | (3.56) |
holds for all ρ∈(0,1],α,β∈C with ℜ(α),ℜ(β)>0.
Proof. By the given hypothesis, we have
|f1(μ)−f1(ν)|≤M1|μ−ν||g1(μ)−g1(ν)|≤M2|μ−ν|μ,ν∈[0,∞), | (3.57) |
which implies that
|T(μ,ν)|=|f1(μ)−f1(ν)||g1(μ)−g1(ν)|≤M1M2(μ−ν)2. | (3.58) |
From (3.42) and (3.58), we obtain that
|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)|≤ω−2(ϰ)ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−βT(μ,ν)dμdν≤Lω−2(ϰ)ραΓ(α)ρβΓ(β)ϰ∫0ϰ∫0exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))]r(μ)ω(μ)Ψ′(μ)(Ψ(ϰ)−Ψ(μ))1−α×exp[ρ−1ρ(Ψ(ϰ)−Ψ(ν))]s(ν)ω(ν)Ψ′(ν)(Ψ(ϰ)−Ψ(ν))1−β(μ−ν)2dμdν=M1M2(ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(ϰ2s(ϰ))+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(ϰ2r(ϰ))−2ΨωΩρ;α0+(ϰr(ϰ))ΨωΩρ;β0+(ϰs(ϰ))). | (3.59) |
This ends the proof.
Corollary 1. Let f1 and g1 be two differentiable functions on [0,∞) and let r and s be two non-negative continuous functions 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)|≤‖f′1‖∞‖g′1‖∞(ΨωΩρ;α0+r(ϰ)ΨωΩρ;β0+(ϰ2s(ϰ))+ΨωΩρ;β0+s(ϰ)ΨωΩρ;α0+(ϰ2r(ϰ))−2ΨωΩρ;α0+(ϰr(ϰ))ΨωΩρ;β0+(ϰs(ϰ))) | (3.60) |
holds for all ρ∈(0,1],α,β∈C with ℜ(α),ℜ(β)>0.
Proof. We have f1(μ)−f1(ν)=μ∫νf′1(ϰ)dϰ and g1(μ)−g1(ν)=μ∫νg′1(ϰ)dϰ. That is, |f1(μ)−f1(ν)|≤‖f′1‖∞|μ−ν|, |g1(μ)−g1(ν)|≤‖g′1‖∞|μ−ν|,μ,ν∈[0,∞), and the immediate consequence follows from Theorem 3.12. This completes the proof.
Example 3.13. Let ρ,α>0,q1,q2>1 with q−11+q−12=1, and ω≠0 be a function on [0,∞). Let f1 be an integrable function defined on [0,∞) and ΨωΩρ;αa+1f1 be the weighted generalized proportional fractional integral operator satisfying assumption II. Then we have
|(ΨωΩρ;αa+1f1)(ϰ)|≤Θ‖(f1∘ω)(μ)‖L1(a1,ϰ), |
where
Θ=ω−1(ϰ)(−1)α−1Γ(α){(ρq1(ρ−1))α−1+1/q1}1/q1Φ1/q1(q1(α−1)+1,q1(ρ−1)ρ(Ψ(ϰ)−Ψ(a1))) |
and
Φ(α,ϰ)=ϰ∫0e−vvα−1dv |
is the incomplete gamma function [52,53].
Proof. It follows from Definition 2.2 and the modulus property that
|(ΨωΩρ;αa+1f1)(ϰ)|≤ω−1(ϰ)ραΓ(ρ)ϰ∫a1exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))](Ψ(ϰ)−Ψ(μ))1−αΨ′(μ)|f1(μ)ω(μ)|dμ |
for ϰ>a1.
Making use of the well-known Hölder inequality, we obtain
|(ΨωΩρ;αa+1f1)(ϰ)|≤ω−1(ϰ)ραΓ(ρ)(ϰ∫a1q1exp[ρ−1ρ(Ψ(ϰ)−Ψ(μ))](Ψ(ϰ)−Ψ(μ))q1(1−α)Ψ′(μ)dμ)1/q1‖f1∘ω(μ)‖L1(a1,ϰ). |
Let θ=Ψ(ϰ)−Ψ(μ). Then elaborated computations lead to
|(ΨωΩρ;αa+1f1)(ϰ)|≤(−1)α−1ω−1(ϰ)ραΓ(α){(ρq1(ρ−1))α−1+1/q1}1/q1×Φ1/q1(q1(α−1)+1,q1(ρ−1)ρ(Ψ(ϰ)−Ψ(a1)))‖f1∘ω(μ)‖L1(a1,ϰ). |
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 f1 and g1 be two synchronous functions on [0,∞). Assume that Q and P be two non-negative continuous mappings on [0,∞). Then the inequality
ωΩρ;α0+(P)(ϰ)ωΩρ;α0+(Qf1g1)(ϰ)+ωΩΩρ;α0+(Pf1g1)(ϰ)ΨωΩρ;α0+(Q)(ϰ)≥ωΩρ;α0+(Pg1)(ϰ)ωΩρ;α0+(Qf1)(ϰ)+ωΩρ;α0+(Pf1)(ϰ)ωΩρ;α0+(Qg1)(ϰ), |
holds for all ρ∈(0,1],α∈C with ℜ(α)>0.
Proof. Letting Ψ(ϰ)=ϰ and Lemma 3.1 yields the proof of Lemma 4.1.
Lemma 4.2. Let f1 and g1 be two synchronous functions on [0,∞). Assume that Q and P be two non-negative continuous mappings on [0,∞). Then the inequality
ωΩρ;α0+(P)(ϰ)ωΩρ;α0+(Qf1g1)(ϰ)+ωΩΩρ;α0+(Pf1g1)(ϰ)ΨωΩρ;α0+(Q)(ϰ)≥ωΩρ;α0+(Pg1)(ϰ)ωΩρ;α0+(Qf1)(ϰ)+ωΩρ;α0+(Pf1)(ϰ)ωΩρ;α0+(Qg1)(ϰ), |
holds for all ρ∈(0,1],α∈C with ℜ(α)>0.
Proof. Letting Ψ(ϰ)=ϰ and Lemma 3.1 yields the proof of Lemma 4.2.
Lemma 4.3. Under the assumption of Lemma 3.1, then the inequality
ΨωΩα0+(P)(ϰ)ΨωΩα0+(Qf1g1)(ϰ)+ΨωΩα0+(Pf1g1)(ϰ)ΨωΩρ;α0+(Q)(ϰ)≥ΨωΩα0+(Pg1)(ϰ)ΨωΩα0+(Qf1)(ϰ)+ΨωΩρ;α0+(Pf1)(ϰ)ΨωΩα0+(Qg1)(ϰ), |
holds for all α∈C with ℜ(α)>0.
Proof. Letting ρ=1 and Lemma 3.1 yields the proof of Lemma 4.3.
Lemma 4.4. Under the assumption of Lemma 4.2, then the inequality
ωΩα0+(P)(ϰ)ωΩα0+(Qf1g1)(ϰ)+ωΩα0+(Pf1g1)(ϰ)ωΩρ;α0+(Q)(ϰ)≥ωΩα0+(Pg1)(ϰ)ωΩα0+(Qf1)(ϰ)+ωΩρ;α0+(Pf1)(ϰ)ωΩα0+(Qg1)(ϰ), |
holds for all α∈C with ℜ(α)>0.
Proof. Letting ρ=1,Ψ(ϰ)=ϰ and Lemma 3.1 yields the proof of Lemma 4.4.
Theorem 4.5. Let f1 and g1 be two synchronous functions on [0,∞). Assume that r, s and t be three non-negative continuous functions on [0,∞). Then the inequality
2ωΩρ;α0+r(ϰ)(ωΩρ;α0+s(ϰ)ωΩρ;α0+(tf1g1)(ϰ)+ωΩρ;α0+(sf1g1)(ϰ)ωΩρ;α0+t(ϰ))+2ωΩρ;α0+(rf1g1)(ϰ)ωΩρ;α0+s(ϰ)ωΩρ;α0+t(ϰ)≥ωΩρ;α0+r(ϰ)(ωΩρ;α0+(sg1)(ϰ)ωΩρ;α0+(tf1)(ϰ)+ωΩρ;α0+(sf1)(ϰ)ωΩρ;α0+(tg1)(ϰ))+ωΩρ;α0+s(ϰ)(ωΩρ;α0+(rg1)(ϰ)ωΩρ;α0+(tf1)(ϰ)+ωΩρ;α0+(rf1)(ϰ)ωΩρ;α0+(tg1)(ϰ))+ωΩρ;α0+s(ϰ)(ωΩρ;α0+(sg1)(ϰ)ωΩρ;α0+(rf1)(ϰ)+ωΩρ;α0+(sf1)(ϰ)ωΩρ;α0+(rg1)(ϰ)) |
holds for all ρ∈(0,1],α∈C with ℜ(α)>0.
Proof. Letting Ψ(ϰ)=ϰ and Theorem 3.2 yields the proof of Theorem 4.5.
Theorem 4.6. Under the assumption of I, II and let r, s and t be three non-negative continuous functions on [0,∞). Then the inequality
2ΨωΩα0+r(ϰ)(ΨωΩα0+s(ϰ)ΨωΩα0+(tf1g1)(ϰ)+ΨωΩα0+(sf1g1)(ϰ)ΨωΩα0+t(ϰ))+2ΨωΩα0+(rf1g1)(ϰ)ΨωΩα0+s(ϰ)ΨωΩα0+t(ϰ)≥ΨωΩα0+r(ϰ)(ΨωΩα0+(sg1)(ϰ)ΨωΩα0+(tf1)(ϰ)+ΨωΩα0+(sf1)(ϰ)ΨωΩα0+(tg1)(ϰ))+ΨωΩα0+s(ϰ)(ΨωΩα0+(rg1)(ϰ)ΨωΩα0+(tf1)(ϰ)+ΨωΩα0+(rf1)(ϰ)ΨωΩα0+(tg1)(ϰ))+ΨωΩα0+s(ϰ)(ΨωΩα0+(sg1)(ϰ)ΨωΩα0+(rf1)(ϰ)+ΨωΩα0+(sf1)(ϰ)ΨωΩα0+(rg1)(ϰ)) |
holds for all α∈C with ℜ(α)>0.
Proof. Letting ρ=1 and Theorem 3.2 yields the proof of Theorem 4.6.
Theorem 4.7. Under the assumption of Theorem 4.5, then the inequality
2ωΩα0+r(ϰ)(ωΩα0+s(ϰ)ωΩα0+(tf1g1)(ϰ)+ωΩα0+(sf1g1)(ϰ)ωΩα0+t(ϰ))+2ωΩα0+(rf1g1)(ϰ)ωΩα0+s(ϰ)ωΩα0+t(ϰ)≥ωΩα0+r(ϰ)(ωΩα0+(sg1)(ϰ)ωΩα0+(tf1)(ϰ)+ωΩα0+(sf1)(ϰ)ωΩα0+(tg1)(ϰ))+ωΩα0+s(ϰ)(ωΩα0+(rg1)(ϰ)ωΩα0+(tf1)(ϰ)+ωΩα0+(rf1)(ϰ)ωΩα0+(tg1)(ϰ))+ωΩα0+s(ϰ)(ωΩα0+(sg1)(ϰ)ωΩα0+(rf1)(ϰ)+ωΩα0+(sf1)(ϰ)ωΩα0+(rg1)(ϰ)) |
holds for all α∈C with ℜ(α)>0.
Proof. Letting ρ=1,Ψ(ϰ)=ϰ and Theorem 3.2 yields the proof of Theorem 4.7.
Remark 5. The computed results lead to the following conclusion:
(1) Setting ρ=1,Ψ(ϰ)=ϰ and r(ϰ)=s(ϰ)=1, and using the relation (2.7), (2.8) and the assumption ω(ϰ)=1, then Theorem 3.6 and Theorem 3.9 reduces to the known results due to Dahmani et al. [38].
(2) Setting ρ=1,Ψ(ϰ)=ϰ and using the relation (2.7), (2.8) and the assumption ω(ϰ)=1, 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 Ψ and proportionality index ρ. 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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