This paper analyzes the existence of Nash equilibrium in a discrete-time Markov stopping game with two players. At each decision point, Player Ⅱ is faced with the choice of either ending the game and thus granting Player Ⅰ a final reward or letting the game continue. In the latter case, Player Ⅰ performs an action that affects transitions and receives a running reward from Player Ⅱ. We assume that Player Ⅰ has a constant and non-zero risk sensitivity coefficient, while Player Ⅱ strives to minimize the utility of Player Ⅰ. The effectiveness of decision strategies was measured by the risk-sensitive expected total reward of Player Ⅰ. Exploiting mild continuity-compactness conditions and communication-ergodicity properties, we found that the value function of the game is described as a single fixed point of the equilibrium operator, determining a Nash equilibrium. In addition, we provide an illustrative example in which our assumptions hold.
Citation: Jaicer López-Rivero, Hugo Cruz-Suárez, Carlos Camilo-Garay. Nash equilibria in risk-sensitive Markov stopping games under communication conditions[J]. AIMS Mathematics, 2024, 9(9): 23997-24017. doi: 10.3934/math.20241167
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This paper analyzes the existence of Nash equilibrium in a discrete-time Markov stopping game with two players. At each decision point, Player Ⅱ is faced with the choice of either ending the game and thus granting Player Ⅰ a final reward or letting the game continue. In the latter case, Player Ⅰ performs an action that affects transitions and receives a running reward from Player Ⅱ. We assume that Player Ⅰ has a constant and non-zero risk sensitivity coefficient, while Player Ⅱ strives to minimize the utility of Player Ⅰ. The effectiveness of decision strategies was measured by the risk-sensitive expected total reward of Player Ⅰ. Exploiting mild continuity-compactness conditions and communication-ergodicity properties, we found that the value function of the game is described as a single fixed point of the equilibrium operator, determining a Nash equilibrium. In addition, we provide an illustrative example in which our assumptions hold.
Fractional calculus is one of the renowned fields in recent research due to its inherent applications in various areas such as mathematical physics, fluid dynamics, mathematical biology etc. [1,2,3,4,5,6]. On the other hand, the fractional integral inequalities with the fractional operators are developed by many researchers because these inequalities are used to verify various results of applied problems [7,8]. In particular, the researchers [30,31,32,33] have recently studied many remarkable fractional integral inequalities and their applications. In [39], Mehmood et al. discussed the Hermite-Hadamard-Fejér inequality for fractional integrals involving preinvex functions. Mehreen and Anwar [40] estimated he Hermite-Hadamard and Hermite-Hadamard-Fejér type inequalities for p-convex functions by utilizing conformable fractional integrals. In [41], Almutairi and Adem Kılıçman discussed new integral inequalities of Hermite-Hadamard type involving s-convexity and studied their properties. Budak [42] establish Hermite-Hadamard-Fejér type inequalities for convex function involving fractional integrals with respect to another function. The Hermite-Hadamard inequality is defined can be found in [9] for convex function by
Ψ(m+n2)≤1n−m∫nmΨ(x)dx≤Ψ(m)+Ψ(n)2 |
Ψ:I→R,m,n∈I,m<n,m,n∈R,I⊆R and is playing a significant role in the field of inequalities and are widely used by the researchers [10].
Fejér type integral inequalities can be found in [27,28,29] by
Ψ(m+n2)∫nmΦ(x)dx≤∫nmΨ(x)Φ(x)dx≤Ψ(m)+Ψ(n)2∫nmΦ(x)dx | (1.1) |
for convex function Ψ:[m,n]→R and Φ:[m,n]→R+,m,n∈R where the function Φ is integrable and is symmetric about x=m+n2. Note that the Hermite- Hadamard inequality is obtained if Φ=1 in Fejér inequality (1.1).
The (η1,η2)-convex function has been presented [11,12,13] by obtaining the generalization of η-convex function [14,15,16,17] and preinvex function [18,19,20].
Sarikaya [25] discussed the Hermite and trapezoid inequalities related to the Hermite-Hadamard inequality, and Rostamian Delavar [13] discussed Fejér, midpoint and trapezoid type inequalities related to the Hermite-Hadamard inequalities.
Definition 1.1. [26] The convex function Ψ:I→R is defined for t∈[0,1],∀u,v∈I as follows:
Ψ[tu+(1−t)v]≤tΨ(u)+(1−t)Ψ(v). |
Definition 1.2. [13] An invex set I⊆R with respect to a real bifunction θ:I×I→R, is defined for m,n∈I,λ∈[0,1] as follows
n+λθ(m,n)∈I. |
Definition 1.3. [13] The preinvex function Ψ:I→R is defined for m,n∈I and λ∈[0,1] as follows
Ψ(n+λθ(m,n))≤λΨ(m)+(1−λ)Ψ(n), |
where I is an invex set with respect to θ.
Definition 1.4. [13] A function Ψ:I→R is said to be convex with respect to η i.e (η−convex) if it satisfies
Ψ(λm+(1−λ)n)≤Ψ(n)+λη(Ψ(m),Ψ(n)) |
for all m,n∈I and λ∈[0,1] and I⊆R is a convex function and η:Ψ(I)×Ψ(I)→R is a bifunction.
Definition 1.5. [13] Let Ψ:I→R, η1:I×I⟶R, and η2:Ψ(I)×Ψ(I)→R, then Ψ is called (η1,η2)-convex function if
Ψ(x+λη1(y,x))≤Ψ(x)+λη2(Ψ(y),Ψ(x)) |
holds for all x,y∈I and λ∈[0,1].
Example 1.1. [12] Let Ψ be the function such that Ψ:R+→R+ defined by
Ψ(x)={x,for0≤x≤1;1,forx>1.}. |
Let the two bifunctions η1:R+×R+→R and η2:R+×R+→R defined by
η1(x,y)={−y,for0≤y≤1;x+y,fory>1.} |
η2(x,y)={x+y,forx≤y;2(x+y),forx>y.} |
Then Ψ is (η1,η2)-convex function.
Definition 1.6. [43] The Pochhammer's symbol is defined for t∈N as
(ℑ)t={1,fort=0,ℑ≠0,ℑ(ℑ+1)⋯(ℑ+t−1),fort≥1.} |
(ℑ)n=Γ(ℑ+n)Γ(ℑ)(ℑ)kn=Γ(ℑ+kn)Γ(ℑ) |
for ℑ∈C, where Γ being the gamma function.
Definition 1.7. [43] The integral representation of gamma function is defined as
Γ(t)=∫∞0xt−1e−xdx |
for ,ℜ(t)>0.
Definition 1.8. [35,36,37] The Classical beta function is defined for ℜ(m)>0 and ℜ(n)>0,
B(m,n)=∫10tm−1(1−t)n−1dt,=Γ(m)Γ(n)Γ(m+n). |
Definition 1.9. [38] Extended beta functions is defined for ℜ(m)>0,ℜ(n)>0,ℜ(p)>0 is
Bp(m,n)=∫10tm−1(1−t)n−1exp(−pt(1−t))dt. |
Definition 1.10. [34] Ali et al. defined and investigated the generalized Bessel-Maitland function (eight-parameters) with a new fractional integral operator and discussed its properties and relations with Mittag-Leffler functions. The function of generalized Bessel-Maitland as follows:
Jν,η,ρ,γμ,ξ,m,σ(y)=∞∑p=0(θ)ξp(ϑ)σp(−y)pΓ(ϕp+ψ+1)(δ)mp, |
where ϕ,ψ,θ,δ,ϑ∈C, ℜ(ϕ)>0, ℜ(ν)≥−1, ℜ(θ)>0, ℜ(δ)>0, ℜ(ϑ)>0; ξ,m,σ≥0 and m,ξ>ℜ(ϕ)+σ.
Definition 1.11. The extended generalized Bessel-Maitland function is defined for μ,ν,η,ρ,γ,c∈C, ℜ(μ)>0, ℜ(ν)≥−1, ℜ(η)>0, ℜ(ρ)>0, ℜ(γ)>0; ξ,m,σ≥0 and m,ξ>ℜ(μ)+σ by
Jμ,ξ,m,σ,cv,η,ρ,γ(ω;p)=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v+1)(ρ)mn(−ω)n. |
In the recent era of research, the field of fractional calculus has gained more recognition due to its wide range of applications in different sciences [44,45]. Such new developments in fractional calculus motivate the researchers to establish some new innovative ideas to unify the fractional operators and propose new inequalities involving new fractional operators.
The Hermite-Hadamard integral inequalities and their extensions have been widely studied for a different type of convexities [46,47,48,49,50]. Here, we defined the following generalized fractional integral operators containing generalized Bessel-Maitland function as its kernel defined, which are the generalization of many well-known fractional integrals:
Definition 1.12. The generalized fractional integral operators with extended generalized Bessel-Maitland function acting as kernel, is defined for μ,ν,η,ρ,γ,c∈C, ℜ(μ)>0, ℜ(ν)≥−1, ℜ(η)>0, ℜ(ρ)>0, ℜ(γ)>0; ξ,m,σ≥0 and m,ξ>ℜ(μ)+σ as follows
(Tμ,ξ,m,σ,cv,η,ρ,γ;p+f)(x,r)=∫xp(x−t)vJμ,ξ,m,σ,cv,η,ρ,γ(ω(x−t)μ;r)f(t)dt |
and
(Tμ,ξ,m,σ,cv,η,ρ,γ;q−f)(x,r)=∫qx(t−x)vJμ,ξ,m,σ,cv,η,ρ,γ(ω(t−x)μ;r)f(t)dt. |
Remark 1.1. 1. If we put r=0, w=0 and replacing v by v−1 in definition 1.12, we get the Riemann-Liouville fractional integral operators [21].
2. If we put σ=0 and replace v by v−1 in definition 1.12, ω by −ω, we get generalized fractional integral operator containing extended generalized Mittag-Leffler function as their kernels defined by Andric et al.[22].
3. If we put r=0, ξ=0 and replacing v by v−1, ω by −ω in definition 1.12, we get generalized fractional integral operator containing generalized Mittag-Leffler function as their kernels defined in [23].
4. If we put r=0, ξ=0, σ=0, ρ=m=1 in definition 1.12, we get the Srivastava fractional integral operator [24].
This paper aims to obtain Hermite Hadamard and Fejér inequalities using generalized fractional integral having extended generalized Bessel-Maitland function as its kernel.
The structure of the paper follows: In section 2, we present Hermite-Hadamard inequalities for convex function using generalized fractional operator. Section 3 is devoted to Trapezoid type inequalities related to Hermite-Hadamard inequalities. Fejér type inequalities for (η1,η2)- convex function using the generalized fractional operator are presented in section 4.
In this section, we obtain the Hermite-Hadamard inequalities for convex function using generalized fractional operator as follows:
Theorem 2.1. Let Ψ:[u,v]→R be a convex function where 0<u<v and Ψ∈L1[u,v]. If Ψ is an increasing function on [u,v], then for the generalized fractional integrals defined in definition 1.12, we have
Ψ(u+v2)(Tμ,ξ,m,σ,cv′,η,ρ,γ;v−1)(u,p)≤12[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(v)−Ψ)(u,p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Ψ)(v,p)]≤Ψ(u)+Ψ(v)2(Tμ,ξ,m,σ,cv′,η,ρ,γ;u+1)(v,p). |
Proof. By the convexity of Ψ on the interval [u,v], let x,y∈[u,v] with t=12, we have
Ψ(x+y2)≤Ψ(x)+Ψ(y)2, |
where if we takes
x=tu+(1−t)v,y=(1−t)u+tv |
leads to
2Ψ(u+v2)≤Ψ(tu+(1−t)v)+Ψ((1−t)u+tv). |
Multiplying both sides by (1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p) and integrating the resulting inequality on [0,1] with respect to t, we have
2Ψ(u+v2)∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)dt≤∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ(tu+(1−t)v)dt+∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ((1−t)u+tv)dt |
2Ψ(u+v2)∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫10(1−t)v′+μndt≤∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×[∫10(1−t)v′+μnΨ(tu+(1−t)v)dt+∫10(1−t)v′+μnΨ((1−t)u+tv)dt]. | (2.1) |
By making suitable substitutions in inequality (2.1), we obtain
2Ψ(u+v2)(Tμ,ξ,m,σ,cv′,η,ρ,γ;v−1)(u,p)≤[∫vu(z−uv−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(z−uv−u)μ;p)Ψ(z)dz+∫vu(v−zv−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(v−zv−u)μ;p)Ψ(z)dz]orΨ(u+v2)(Tμ,ξ,m,σ,cv′,η,ρ,γ;v−1)(u,p)≤12[(Tμ,ξ,m,σ,cv′,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;v−Ψ)(u;p)]. | (2.2) |
For second part of inequality, again using the convexity of Ψ,
Ψ(tu+(1−t)v)≤tΨ(u)+(1−t)ψ(v) |
and
Ψ((1−t)u+tv)≤(1−t)Ψ(u)+tψ(v). |
Addition of these inequalities, gives
Ψ(tu+(1−t)v)+Ψ((1−t)u+tv)≤(Ψ(u)+ψ(v)). |
Multiplying both sides by (1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p) and integrating the resulting inequality on [0,1] with respect to t, we get
∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ(tu+(1−t)v)dt+∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ((1−t)u+tv)dt≤(Ψ(u)+Ψ(v))∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)dt. |
Making substitution in the integrals involved leads to
12[(Tμ,ξ,m,σ,cv′,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;v−Ψ)(u;p)]≤Ψ(u)+Ψ(v)2(Tμ,ξ,m,σ,cv′,η,ρ,γ;u+1)(v,p), | (2.3) |
combining (2.2) and (2.3), we get the desired result.
The Trapezoid type inequalities related to the Hermite-Hadamard inequalities are presented in this section.
Lemma 3.1. Let a function Ψ:I→R with I=[u,v]⊆R, Ψ′∈L1[u,v] be a differentiable function on (u,v). Then for the generalized fractional integrals defined in definition (1.12), we have
Ψ(u)+Ψ(v)2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]=v−u2IwhereI=∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ′(ut+(1−t)v)dt+∫10−tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(ut+(1−t)v)dt. |
Proof. If we consider the integral
I=∫10(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ′(ut+(1−t)v)dt+∫10−tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(ut+(1−t)v)dt. |
Let
I=I1+I2. |
Firstly, we consider I1
I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫10(1−t)v′+μnΨ′(ut+(1−t)v)dt. |
Integrating by parts, we have
I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[(1−t)v′+μnΨ(ut+(1−t)vu−v|10+v′+μnu−v∫10(1−t)v′+μn−1Ψ(ut+(1−t)v)dt]I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[Ψ(v)v−u−v′+μn(v−u)2∫vu(x−uv−u)v′+μn−1Ψ(x)dx]I1=Ψ(v)v−uJμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−1(v−u)2(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u)+Ψ)(v;p). |
On the same lines, we get
I2=Ψ(u)v−uJμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−1(v−u)2(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)impliesI=Ψ(u)+Ψ(v)v−uJμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−1(v−u)2×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)+Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u)+Ψ](v;p)]. |
Multiplying by v−u2, we have the required result.
By using Lemma 3.1, we present the following theorem.
Theorem 3.1. Let a function Ψ:I=[u,v]→R with I∈R be a differentiable function on (u,v). Also, suppose that |Ψ′| is a convex function on I, then for the generalized fractional integrals in definition 1.12, we have
|Ψ(u)+Ψ(v)2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]|≤v−u2|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(1)μ;p)−12v′Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[|Ψ′(u)|+|Ψ′(v)|], |
where v′≥0.
Proof. If we consider the following integral expression
|Ψ(u)+Ψ(v)2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]|=|v−u2I|≤v−u2∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×∫10|(1−t)v′+μn−tv′+μn||Ψ′(ut+(1−t)v)|dt≤v−u2∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×∫10|(1−t)v′+μn−tv′+μn|[|Ψ′(u)|t+(1−t)|Ψ′(v)|]dt]=v−u2∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×[∫120((1−t)v′+μn−tv′+μn)[|Ψ′(u)|t+(1−t)|Ψ′(v)|]dt+∫112(tv′+μn−(1−t)v′+μn)[|Ψ′(u)|t+(1−t)|Ψ′(v)|]dt]. |
Solving the integrals involved by using integrating by parts method, we obtain
|Ψ(u)+Ψ(v)2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]|≤v−u2|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(1)μ;p)−12v′Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[|Ψ′(u)|+|Ψ′(v)|]. |
Here, we present Fejér type inequalities for (η1,η2)-convex function by using the generalized fractional operator in definition 1.12.
Theorem 4.1. Let Ψ:I→R, be an (η1,η2)-convex functions such that η2 is an integrable bi-function on Ψ(I)×Ψ(I) and for any u,v∈I,η1(v,u)>0 with Ψ∈L1[u,u+η1(v,u)] and the function Φ:[u,u+η1(v,u)]→R+ is integrable and symmetric to u+12η1(v,u) i.e., Φ(2u+η1(v,u)−x)=Φ(x), where I⊆R be an invex set with respect to η1 such that
η1(v+t2η1(u,v),v+t1η1(u,v))=(t2−t1)η1(u,v), | (4.1) |
for all t1,t2∈[0,1]. Then for the generalized fractional integrals defined in definition (1.12), the following Fejér type inequality holds:
Ψ(2u+η1(v,u)2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Φ)(u,p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(u+η1(v,u),p)]−12∫u+η1(v,u)u[(u+η1(v,u)−x)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(u+η1(v,u)−x)μ;p)+(x−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(x−u)μ;p)]×η2(Ψ(x),Ψ(2u+η1(v,u)−x)).Φ(x)dx≤(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u,p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u),p). |
Proof. By the (η1,η2)-convexity of the function Ψ and using (4.1), we get
Ψ(2u+η1(v,u)2)=Ψ(2u+(1+t)η1(v,u)2−tη1(v,u)2)=Ψ(2u+(1+t)η1(v,u)2+12η1(u+(1−t)2η1(v,u),u+(1+t)2η1(v,u))), |
or
Ψ(2u+η1(v,u)2)≤Ψ(2u+(1+t)η1(v,u)2+12η2(Ψ(2u+(1−t)η1(v,u)2),Ψ(2u+(1+t)η1(v,u)2))). | (4.2) |
Now by adapting the same procedure as above, we obtain
Ψ(2u+η1(v,u)2)≤Ψ(2u+(1−t)η1(v,u)2+12η2(Ψ(2u+(1+t)η1(v,u)2),Ψ(2u+(1−t)η1(v,u)2))). | (4.3) |
Using the generalized fractional integral operators defined in 12, we have
I1=(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u;p)=∫u+η1(v,u)u(x−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(x−u)μ;p)Ψ(x)Φ(x)dx=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫u+η1(v,u)u(x−u)v′+μnΨ(x)Φ(x)dx |
I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v+1)(ρ)mn(−ω)n[∫u+12η1(v,u)u(x−u)v′+μnΨ(x)Φ(x)dx+∫u+η1(v,u)u+12η1(v,u)(x−u)v′+μnΨ(x)Φ(x)dx]. | (4.4) |
By making the substitution x=2u+(1−t)η1(v,u)2 and x=2u+(1+t)η1(v,u)2 respectively in the integrals appearing in (4.4), we have
I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n(η1(v,u)2)v′+μn+1[∫10(1−t)v′+μnΨ(2u+(1−t)η1(v,u)2)Φ(2u+(1−t)η1(v,u)2)dt+∫10(1+t)v′+μnΨ(2u+(1+t)η1(v,u)2)Φ(2u+(1+t)η1(v,u)2dt)]. |
By using the inequalities (4.2) and (4.3), we proceed
I1≥∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n(η1(v,u)2)v′+μn+1[∫10(1−t)v′+μnΨ(2u+η1(v,u)2)Φ(2u+(1−t)η1(v,u)2)dt−12∫10(1−t)v′+μnη2(Ψ(2u+(1+t)η1(v,u)2),Ψ(2u+(1−t)η1(v,u)2))Φ(2u+(1−t)η1(v,u)2)dt+∫10(1+t)v′+μnΨ(2u+η1(v,u)2)Φ(2u+(1+t)η1(v,u)2)dt−12∫10(1+t)v′+μnη2(Ψ(2u+(1−t)η1(v,u)2),Ψ(2u+(1+t)η1(v,u)2))Φ(2u+(1+t)η1(v,u)2)dt], |
or
I1≥Ψ(2u+η1(v,u)2)(η1(v,u)2)v′+μn+1∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[∫10(1−t)v′+μnΦ(2u+(1−t)η1(v,u)2)dt−12∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×(η1(v,u)2)v′+μn+1∫10(1−t)v′+μnη2(Ψ(2u+(1+t)η1(v,u)2),Ψ(2u+(1−t)η1(v,u)2))×Φ(2u+(1−t)η1(v,u)2)dt+Ψ(2u+η1(v,u)2)(η1(v,u)2)v′+μn+1∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×∫10(1+t)v′+μnΦ(2u+(1+t)η1(v,u)2)dt−12(η1(v,u)2)v′+μn+1∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫10(1+t)v′+μnη2(Ψ(2u+(1−t)η1(v,u)2),Ψ(2u+(1+t)η1(v,u)2))Φ(2u+(1+t)η1(v,u)2)dt]. |
Again by simplification and using the above mentioned substitution as well as the symmetry of Φ to u+12η1(v,u) leads to the following
I1≥Ψ(2u+η1(v,u)2)∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[∫u+η1(v,u)u(x−u)v′+μnΦ(x)dx−12∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×∫u+η1(v,u)u(u+η1(v,u)−x)v′+μnη2(Ψ(x),Ψ(2u+η1(v,u)−x))Φ(x)dx. |
It follows that
I1≥Ψ(2u+η1(v,u)2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Φ)(u,p)−12∫u+η1(v,u)u(u+η1(v,u)−x)v′×Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(u+η1(v,u)−x)μ;p)×η2(Ψ(x),Ψ(2u+η1(v,u)−x)).Φ(x)dx. |
Now, consider
I2=(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u),p)=∫u+η1(v,u)u(u+η1(v,u)−x)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(u+η1(v,u)−x)μ;p)Ψ(x)Φ(x)dx. |
Solving on the same pattern as used above, we get
I2≥Ψ(2u+η1(v,u)2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;u+Φ)(u+η1(v,u),p)×−12∫u+η1(v,u)u(x−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(x−u)μ;p)×η2(Ψ(x),Ψ(2u+η1(v,u)−x)).Φ(x)dx. |
By adding I1 and I2, we get the required inequality.
Lemma 4.1. Let I be an invex subset of R with respect to η1:I×I⟶R. Let u,v∈I satisfying η1(v,u)>0 and Φ:[u,u+η1(v,u)]⟶R be integrable and symmetric about u+12η1(v,u). Then for the integrals defined in definition (1.12), the following holds;
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Ψ)(u+η1(v,u);p)=(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u)−Ψ)(u;p)=12[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]. |
Proof. If we consider
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Ψ)(u+η1(v,u);r)=∫u+η1(v,u)u(u+η1(v,u)−x)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(u+η1(v,u)−x)μ;p)Ψ(x)dx. | (4.5) |
By substituting x=2u+η1(v,u)−t in (4.5), we get
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Ψ)(u+η1(v,u);r)=∫u+η1(v,u)u(t−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t−u)μ;p)Ψ(2u+η1(v,u)−t)dt |
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Ψ)(u+η1(v,u);p)=(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p). | (4.6) |
Addition of (Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p) in Eq (4.6) on both sides, leads to the required result.
Theorem 4.2. Let Ψ:I→R, be an (η1,η2)-convex functions such that η2 is an integrable bifunction on Ψ(I)×Ψ(I) and for any u,v∈I, η1(v,u)>0 with Ψ∈L1[u,u+η1(v,u)] and the function Φ:[u,u+η1(v,u)]→R+ is integrable and symmetric to u+12η1(v,u) i.e Φ(2u+η1(v,u)−x)=Φ(x), where I⊆R be an invex set with respect to η1. Then for the generalized fractional integrals defined in definition (1.12), the following Fejér type inequality holds:
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u,p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u),p)≤(2Ψ(u)+η2(Ψ(v),Ψ(u))2)×[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u,p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u),p)]. |
Proof. If we consider the integral
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u;p)=∫u+η1(v,u)u(x−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(x−u)μ;p)Ψ(x)Φ(x)dx=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫u+η1(v,u)u(x−u)v′+μnΨ(x)Φ(x)dx. |
Making substitution x=u+tη1(v,u) leads to following integral,
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u;p)=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×(η1(v,u))v′+μn+1∫10(t)v′+μnΨ(u+tη1(v,u))Φ(u+tη1(v,u))dt. |
Using the (η1,η2)-convexity of Ψ, we get
≤∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×(η1(v,u))v′+μn+1∫10(t)v′+μn(Ψ(u)+tη2(Ψ(v),Ψ(u)))Φ(u+tη1(v,u))dt |
=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n(η1(v,u))v′+μn+1[Ψ(u)∫10(t)v′+μnΦ(u+tη1(v,u))dt+η2(Ψ(v),Ψ(u))∫10(t)v′+μn+1Φ(u+tη1(v,u))dt]. | (4.7) |
Now, consider
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u);p)=∫u+η1(v,u)u(u+η1(v,u)−x)v′×Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(u+η1(v,u)−x)μ;p)Ψ(x)Φ(x)dx=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫u+η1(v,u)u(u+η1(v,u)−x)v′+μnΨ(x)Φ(x)dx |
≤∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×(η1(v,u))v′+μn+1∫10(t)v′+μn(Ψ(u)+(1−t)η2(Ψ(v),Ψ(u)))Φ(u+(1−t)η1(v,u))dt |
=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n(η1(v,u))v′+μn+1[Ψ(u)∫10(t)v′+μnΦ(u+(1−t)η1(v,u))dt+η2(Ψ(v),Ψ(u))×∫10(t)v′+μn(1−t)Φ(u+(1−t)η1(v,u))dt]. | (4.8) |
Adding Eqs (4.7) and (4.8) and using the symmetry of Φ about u+12η1(v,u), we have
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ(u;p))≤∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n(η1(v,u))v′+μn+1×[2Ψ(u)∫10(t)v′+μnΦ(u+tη1(v,u))dt+η2(Ψ(v),Ψ(u))∫10(t)v′+μnΦ(u+tη1(v,u))dt]=(2Ψ(u)+η2(Ψ(v),Ψ(u)))∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n(η1(v,u))v′+μn+1×∫10(t)v′+μnΦ(u+(t)η1(v,u))dt.=(2Ψ(u)+η2(Ψ(v),Ψ(u))2)(2Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Φ)(u;p). |
By using lemma 4.1, we have
(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u);p)≤(2Ψ(u)+η2(Ψ(v),Ψ(u))2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Φ)(u;p)]. |
Corollary 4.1. In Fejér type inequalities defined in Theorems 4.1 and 4.2, if we take η1(u,v)=u−v, ∀u,v∈I,
Ψ(u+v2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(v;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(v)−Φ)(u;p)]−12∫vu[(v−x)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(v−x)μ;p)+(x−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(x−u)μ;p)]×η2(Ψ(x),Ψ(u+v−x)).Φ(x)dx≤(Tμ,ξ,m,σ,cv′,η,ρ,γ;(v)−ΨΦ)(u;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(v;p)≤(2Ψ(u)+η2(Ψ(v),Ψ(u))2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(v;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(v)−Φ)(u;p)], |
which is Fejér inequality for generalized fractional integral can be obtained by considering the function Ψ to be η-convex.
Corollary 4.2. In Theorems 4.1 and 4.2, if we put η2(u,v)=u−v, forallu,v∈Ψ(I), then
Ψ(2u+η1(v,u)2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Φ)(u;p)]≤(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−ΨΦ)(u;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(u+η1(v,u);p)≤(Ψ(u)+Ψ(v)2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Φ)(u;p)], |
which is Fejér inequality for generalized fractional integral can be obtained by considering the function Ψ to be preinvex convex.
Corollary 4.3. In Fejér type inequalities 4.1 and 4.2, if we put η1(u,v)=u−v,∀u,v∈I and η2(x,y)=x−y,∀x,y∈Ψ(I),
Ψ(u+v2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(v;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(v)−Φ)(u;p)]≤(Tμ,ξ,m,σ,cv′,η,ρ,γ;(v)−ΨΦ)(u;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+ΨΦ)(v;p)≤(Ψ(u)+Ψ(v)2)[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Φ)(v;p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(v)−Φ)(u;p)], |
which is Fejér type inequality for generalized fractional integral can be obtained by considering the function Ψ to be convex.
Corollary 4.4. In corollary 4.3, if we take Φ=1, we get Hermite-Hadamard type inequality for convex function discussed in 2.1.
Corollary 4.5. In Fejér type inequalities defined in 4.1 and 4.2, if we take Φ=1 then can obtain the Hermite-Hadamard type inequality for (η1,η2)-convex function as
Ψ(2u+η1(v,u)2)(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−1)(u,p)−14∫u+η1(v,u)u[(u+η1(v,u)−x)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(u+η1(v,u)−x)μ;p)+(x−u)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(x−u)μ;p)]×η2(Ψ(x),Ψ(2u+η1(v,u)−x))dx≤12[(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+Ψ)(u+η1(v,u),p)+(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u+η1(v,u))−Ψ)(u,p)]≤(2Ψ(u)+η2(Ψ(v),Ψ(u))2)(Tμ,ξ,m,σ,cv′,η,ρ,γ;(u)+1)(u+η1(v,u),p). |
In the section, we discuss the midpoint and trapezoid type inequalities connected to Hermite-Hadamard inequalities for the function whose absolute value of the derivative is (η1,η2)-convex function. The following lemma will help us in the next result.
Lemma 5.1. Let a function Ψ:I→R with I∈R, Ψ∈L1[u,u+η1(v,u)] be a differentiable function where I is taken to be an open invex set with respect to η1:I×I→R with η1(v,u)>0, for u,v∈I. Then for the generalized fractional integrals defined in definition 1.12, we have
η1(v,u)24∑k=1Ik=Ψ(2u+η1(v,u)2)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]whereI1=∫120tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(u+tη1(v,u))dtI2=∫120−(t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(u+(1−t)η1(v,u))dtI3=∫112tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(u+tη1(v,u))dt−∫112Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)Ψ′(u+tη1(v,u))dtI4=∫112Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)Ψ′(u+(1−t)η1(v,u))dt−∫112tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(u+(1−t)η1(v,u))dt. |
Proof. If we consider
I1=∫120tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(u+tη1(v,u))dtI1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫120(t)v′+μnΨ′(u+tη1(v,u))dt. |
Solving the integrals by using integrating by parts method leads to
I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[tv′+μnΨ(u+tη1(v,u))η1(v,u)|120−v′+μnη1(v,u)∫120(t)v′+μn−1Ψ(u+tη1(v,u))dt]=∞∑n=0βp(η+ξη)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[(2)−(v′+μn)η1(v,u)Ψ(2u+η1(v,u))2−v′+μnη1(v,u)∫120(t)v′+μn−1Ψ(u+tη1(v,u))dt]. |
Similarly,
I2=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[(2)−(v′+μn)η1(v,u)Ψ(2u+η1(v,u))2−v′+μnη1(v,u)∫120(t)v′+μn−1Ψ(u+(1−t)η1(v,u))dt]. |
Now
I3=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫112(tv′+μn−1)Ψ′(u+tη1(v,u))dtI3=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[(tv′+μn−1)Ψ(u+tη1(v,u))η1(v,u)|112−v′+μnη1(v,u)∫112(t)v′+μn−1Ψ(u+tη1(v,u))dt]=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[1−2−(v′+μn)η1(v,u)Ψ((2u+η1(v,u))2)−v′+μnη1(v,u)∫112(t)v′+μn−1Ψ(u+tη1(v,u))dt]. |
Similarly,
I4=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[1−2−(v′+μn)η1(v,u)Ψ((2u+η1(v,u))2)−v′+μnη1(v,u)∫112(t)v′+μn−1Ψ(u+(1−t)η1(v,u))dt]. |
Adding I1,I2,I3 and I4, we proceed to the desired result as,
4∑k=1Ik=2η1(v,u)Ψ(2u+η1(v,u)2)∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n−∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)nv′+μnη1(v,u)×[∫10tv′+μn−1Ψ(u+tη1(v,u))dt+∫10tv′+μn−1Ψ(u+(1−t)η1(v,u))dt]orη1(v,u)24∑k=1Ik=Ψ(2u+η1(v,u)2)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]. |
Next, we present mid-point type inequalities related to Hermite-Hadamard inequalities:
Theorem 5.1. Consider a function Ψ:I→R with I∈R, Ψ′∈L1[u,u+η1(v,u)] be a differentiable function where I is taken to be an open invex set with respect to η1:I×I→R with η1(v,u)>0 for u,v∈I. Then for the generalized fractional integrals defined in definition 1.12, we have
|Ψ(2u+η1(v,u)2)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]|≤η1(v,u)2[|Ψ′(u)|+|Ψ′(v)|+12η2(|Ψ′(u)|,|Ψ′(v)|)+12.η2(|Ψ′(v)|,|Ψ′(u)|)], |
for 0<v′+μn≤1.
Proof. By using lemma 5.1 and using the property of absolute function for addition, leads to
|Ψ(2u+η1(v,u)2)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u) | (5.1) |
[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]| | (5.2) |
≤η1(v,u)24∑k=1|Ik|. | (5.3) |
To solve |Ik|,k=1,2,3,4, we further move by using (η1,η2)-convexity of |Ψ′|
|I1|≤∫120tv′|Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)||Ψ′(u+tη1(v,u))|dt≤∫120tv′|Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)||Ψ′(u)|dt+∫120(t)v′+1|Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)|η2(|Ψ′(v)|,|Ψ′(u)|)dt≤∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|[|Ψ′(u)|12v′+μn+1(v′+μn+1)+η2(|Ψ′(v)|,|Ψ′(u)|)12v′+μn+2(v′+μn+2)]Analogously|I2|≤∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|[|Ψ′(v)|12v′+μn+1(v′+μn+1)+η2(|Ψ′(u)|,|Ψ′(v)|)12v′+μn+2(v′+μn+2)]. |
For |Ik|,k=3,4. We will use the fact that for all j∈(0,1] and u1,u2∈[0,1]. Therefore, we have
|uj1−uj2|≤|u1−u2|j |
|I3|≤∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×[|Ψ′(u)|12v′+μn+1(v′+μn+1)+η2(|Ψ′(v)|,|Ψ′(u)|)v′+μn+32v′+μn+2(v′+μn+1)(v′+μn+2)] |
and
|I4|≤∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×[|Ψ′(v)|12v′+μn+1(v′+μn+1)+η2(|Ψ′(u)|,|Ψ′(v)|)v′+μn+32v′+μn+2(v′+μn+1)(v′+μn+2)]. |
Using the above evaluated absolute values in (5.1), we have
|Ψ(2u+η1(v,u)2)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]|≤∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|η1(v,u)2v′+μn+1(v′+μn+1)[|Ψ′(u)|+|Ψ′(v)|+12.η2(|Ψ′(v)|,|Ψ′(u)|)+12η2(|Ψ′(u)|,|Ψ′(v)|)]=η1(v,u)2v+1|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[|Ψ′(u)|+|Ψ′(v)|+12.η2(|Ψ′(v)|,|Ψ′(u)|)+12η2(|Ψ′(u)|,|Ψ′(v)|)]. |
Corollary 5.1. In Theorem 5.1, if we take η1(u,v)=u−v, ∀u,v∈I, then
|Ψ(u+v2)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]|≤v−u2v+1|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[|Ψ′(u)|+|Ψ′(v)|+12.η2(|Ψ′(v)|,|Ψ′(u)|)+12η2(|Ψ′(u)|,|Ψ′(v)|)]. |
Corollary 5.2. In Theorem 5.1, if we take η1(u,v)=u−v, ∀u,v∈I and η2(x,y)=x−y, ∀x,y∈Ψ(I), then
|Ψ(u+v2)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]|≤v−u2v′+1|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[|Ψ′(u)|+|Ψ′(v)|]. |
Lemma 5.2. If we consider a function Ψ:I→R with I∈R, Ψ∈L1[u,u+η1(v,u)] be a differentiable function where I is taken to be an open invex set with respect to η1:I×I→R with η1(v,u)>0 for u,v∈I. Then for the generalized fractional integrals defined in definition (1.12), we have
Ψ(u)+Ψ(u+η1(v,u))2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]=η1(v,u)2IwhereI=∫10tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(u+tη1(v,u))dt+∫10−(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ′(u+tη1(v,u))dt. |
Proof. We consider the fractional integral
I=∫10tv′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(t)μ;p)Ψ′(u+tη1(v,u))dt+∫10−(1−t)v′Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1−t)μ;p)Ψ′(u+tη1(v,u))dt. |
Let
I=I1+I2. |
First, we consider I1
I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n∫10tv′+μnΨ′(u+tη1(v,u))dt. |
Integrating by parts, we have
I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n[tv′+μnΨ(u+tη1(v,u))η1(v,u)|10−v′+μnη1(v,u)∫10tv′+μn−1Ψ(u+tη1(v,u))dt]I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×[Ψ(u+η1(v,u))η1(v,u)−v+μnη1(v,u)∫10tv′+μn−1Ψ(u+tη1(v,u))dt]I1=∞∑n=0βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n×[Ψ(u+η1(v,u))η1(v,u)−v′+μn(η1(v,u))2∫u+η1(v,u)u(x−uu+η1(v,u)−u)v′+μn−1Ψ(x)dx]I1=Ψ(u+η1(v,u))η1(v,u)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−1(η1(v,u))2(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p). |
On the same lines, we get
I2=Ψ(u)η1(v,u)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−1(η1(v,u))2(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u)+Ψ)(u+η1(v,u);p)I=Ψ(u)+Ψ(u+η1(v,u))η1(v,u)Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−1(η1(v,u))2×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)+Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u)+Ψ)(u+η1(v,u);p)]. |
Multiplying by η1(v,u)2, we get the required result.
Here, we are able to give trapezoid-type inequalities related to Hermite-Hadamard inequalities:
Theorem 5.2. If we consider a function Ψ:I→R with I∈R, Ψ′∈L1[u,u+η1(v,u)] be a differentiable function where I is taken to be an open invex set with respect to η1:I×I→R with η1(v,u)>0 for u,v∈I. Suppose also that |Ψ′| is an (η1,η2)-convex function on I. Then for the generalized fractional integrals defined in definition 1.12, we have
|Ψ(u)+Ψ(u+η1(v,u))2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]|≤η1(v,u)2|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(1)μ;p)−12v′Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[2|Ψ′(u)|+η2(|Ψ′(v)|,|Ψ′(u)|)], |
where v′≥0.
Proof.
|Ψ(u)+Ψ(u+η1(v,u))2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]|=|η1(v,u)2I|≤η1(v,u)2∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×∫10|tv′+μn−(1−t)v′+μn||Ψ′(u+tη1(v,u))|dt≤η1(v,u)2∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×∫10|tv′+μn−(1−t)v′+μn|[|Ψ′(u)|+tη2(|Ψ′(v)|,|Ψ′(u)|)]dt=η1(v,u)2∞∑n=0|βp(η+ξn,c−η)(c)ξn(γ)σnβ(η,c−η)Γ(μn+v′+1)(ρ)mn(−ω)n|×[∫120((1−t)v′+μn−tv′+μn)[|Ψ′(u)|+tη2(|Ψ′(v)|,|Ψ′(u)|)dt]+∫112(tv′+μn−(1−t)v′+μn)[|Ψ′(u)|+tη2(|Ψ′(v)|,|Ψ′(u)|)dt]. |
Solving the integrals involved by using integrating by parts method, we obtain the desired result
|Ψ(u)+Ψ(u+η1(v,u))2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12η1(v,u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(u+η1(v,u);p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(u+η1(v,u))−Ψ)(u;p)]|≤η1(v,u)2|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(1)μ;p)−12v′Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[2|Ψ′(u)|+η2(|Ψ′(v)|,|Ψ′(u)|)]. |
Corollary 5.3. In theorem 5.2, η1(u,v)=u−v, ∀u,v∈I gives
|Ψ(u)+Ψ(v)2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]≤v−u2|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(1)μ;p)−12v′Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[2|Ψ′(u)|+η2(|Ψ′(v)|,|Ψ′(u)|)]. |
Corollary 5.4. In theorem (5.2), η1(u,v)=u−v, ∀u,v∈I, and η2(u,v)=u−v, ∀u,v∈Ψ(I) gives
|Ψ(u)+Ψ(v)2Jμ,ξ,m,σ,cv′,η,ρ,γ(ω(1)μ;p)−12(v−u)×[(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;u+Ψ)(v;p)+(Tμ,ξ,m,σ,cv′−1,η,ρ,γ;(v)−Ψ)(u;p)]|≤v−u2|Jμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(1)μ;p)−12vJμ,ξ,m,σ,cv′+1,η,ρ,γ(ω(12)μ;p)|[|Ψ′(u)|+|Ψ′(v)|]. |
Various researchers have studied integral inequalities due to their wide applications in both pure and applied mathematics. This paper discussed the new version of integral inequalities such as Hermite-Hadamard type and trapezoid type inequalities for the convex function by utilizing generalized fractional integrals concerning the extended Wright generalized Bessel function as a kernel. Also, we established new mid-point type and trapezoidal type integral inequalities for (η1,η2)-convex function related to Hermite-Hadamard and Fejér type inequalities. All the inequalities presented in this paper are more general than the inequalities available in the literature, which can easily observe from the corollaries.
The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University for funding this work through Research Group Program under Grant No. RGP. 2/51/42.
The authors declare that they have no competing interest.
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