Research article

Chebyshev type inequalities involving extended generalized fractional integral operators

  • Received: 26 November 2019 Accepted: 27 March 2020 Published: 13 April 2020
  • MSC : 26A33, 26D10, 33B20

  • In this paper, mainly by using the extended generalized fractional integral operator that involve a further extension of Mittag-Leffler function in the kernel, we obtain several fractional Chebyshev type integral inequalities. So, results of Dahmani et al. from [4] are generalized. Also, it is point out that new results are obtained for different fractional integral operators with the help of special selection of parameters.

    Citation: Erhan Set, M. Emin Özdemir, Sevdenur Demirbaş. Chebyshev type inequalities involving extended generalized fractional integral operators[J]. AIMS Mathematics, 2020, 5(4): 3573-3583. doi: 10.3934/math.2020232

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  • In this paper, mainly by using the extended generalized fractional integral operator that involve a further extension of Mittag-Leffler function in the kernel, we obtain several fractional Chebyshev type integral inequalities. So, results of Dahmani et al. from [4] are generalized. Also, it is point out that new results are obtained for different fractional integral operators with the help of special selection of parameters.


    In 1882, Chebyshev proved on interesting and useful integral inequality as follows:

    1babaf(x)g(x)dx(1babaf(x)dx)(1babag(x)dx) (1.1)

    where f and g are two integrable and synchronous functions on [a,b]. Here two functions f and g are called synchronous on [a,b], if

    (f(x)f(y))(g(x)g(y))0(x,y[a,b]).

    The inequality (1.1) that is well known as Chebyshev inequality has many applications in diverse research subjects such as numerical quadrature, transform theory, probability, existence of solutions of differential equations and statistical problems. Therefore, many researchers have given considerable attention to this inequality, (see [3,10,11,12,18,19,20,21,22].

    On the other hand, one of the methods used to generalize inequalities is fractional calculus. In this context, firstly, in 2009, Chebyshev inequality involving Riemann-Liouville fractional integrals is presented as the following:

    Theorem 1.1. ([2]) Let f and g be two synchronous function on [0,). Then for all t>0,α>0, we have:

    Jα(f,g)(t)Γ(α+1)tαJαf(t)Jαg(t) (1.2)

    where Jαf(t) denotes Riemann-Liouville fractional integral operator of a function f(t) and Γ is the Gamma function such that these are defined as follows (see e.g. [17]).

    Let fL[a,b]. The Riemann-Liouville fractional integrals Jαa+f and Jαbf of order α>0 are defined by

    Jαa+f(x)=1Γ(α)xa(xt)α1f(t)dt,x>0Jαbf(x)=1Γ(α)bx(tx)α1f(t)dt,x<b

    respectively. Also the gamma function is defined by

    Γ(x)=0ettα1dt.

    Here is J0a+f(x)=J0Lf(x)=f(x). In the case of α=1, the this fractional integral reduces to the classical integral.

    After this study of Belarbi and Dahmani, generalizations of Chebyshev inequality were obtained for the different types of fractional integral operator with similar technique. (see, e.g. [2,5,6,7,14,23,24,25,28].

    Recently, the new generalizations of the Riemann-Liouville fractional integral operator, have been described with the help of various extensions of the Mittag-Leffler function. Now lets give some of these operators which we will need in the second section.

    Definition 1.1. ([13]) Let α,β,ρ,λC, Re(α)>0 and Re(β)>0. Let fL[a,b] and x[a,b]. Then the fractional integral operator ϵ(α,β,ρ,λ) defined by Prabhakar is as the following:

    ϵ(α,β,ρ,λ)f(x)=xa(xt)β1Eρα,βλ(xt)αf(t)dt

    where

    Eρα,β=n=0(ρ)nznΓ(αn+β)n!

    and Γ is the Gamma function.

    Definition 1.2. ([26]) Let z,β,γ,ωC, Re(α)>max{0,Re(κ)1}, min{Re(β),Re(κ)}>0. Let fL[a,b] and x[a,b]. Then the fractional integral operator ϵω;γ,κa+;α,βφ defined by Srivastava and Saxena is as the following:

    (ϵω;γ,κa+;α,βφ)(x)=xa(xt)β1Eγ,κα,β[ω(xt)α]φ(t)dt (x>a)

    where

    Eγ,κα,β(z)=n=0(γ)κnΓ(αn+β)znn!

    and Γ is the Gamma function.

    Definition 1.3. ([16]) Let α,β,γ,δC, min{Re(α),Re(β),Re(γ),Re(δ)}>0, p,q>0 and qRe(α)+p. Let fL[a,b] and x[a,b]. Then the fractional integral operator ϵγ,δ,qα,β,p,ω,a+ defined by Salim and Faraj is as the following:

    ϵγ,δ,qα,β,p,ω,a+(x)=xa(xt)β1Eγ,δ,qα,β,p(ω(xt)α)φ(t)dt

    where

    Eγ,δ,qα,β,p(z)=n=0γqnΓ(αn+β)zn(δ)pn

    and Γ is the Gamma function.

    Definition 1.4 ([15]) Let p0, q>0, ω,δ,λ,σ,c,ρC, Re(c)>0, Re(ρ)>0 and Re(σ)>0. Let fL[a,b] and x[a,b]. Then the fractional integral operator (ϵω,δ,q,ca+,ρ,σf) defined by Rahman et al. is as the following:

    (ϵω,δ,q,ca+,ρ,σf)(x)=xa(xτ)σ1Eδ,q,cp,σ(ω(xτ)ρ;p)f(τ)dτ

    where

    Eδ,q,cρ,σ(z;p)=n=0Bp(δ+nq,cδ)B(δ,cδ)(c)nqΓ(ρn+σ)znn!

    and Bp(x,y) is an extension of Beta function defined in [15]

    Bp(x,y)=10tx1(1t)y1ept(1t)dt    x,y,p>0 (1.3)

    where Re(p)>0, Re(x)>0 and Re(y)>0. Also, here B is familiar Beta function as follows:

    B(a,b)=Γ(a)Γ(b)Γ(a+b)=10ta1(1t)b1dt,   a,b>0. (1.4)

    Definition 1.5. ([1]) Let ω,α,β,σ,δ,cC, Re(α),Re(β),Re(σ),Re(δ),Re(c)>0 with p0, q>0 and 0<rq+Re(α). Let fL1[a,b] and x[a,b]. Then the generalized fractional integral operator ϵω,δ,q,r,ca+,α,β,σf is defined by

    (ϵω,δ,q,r,ca+,α,β,σf)(x;p)=xa(xt)β1Eδ,q,r,cα,β,σ(ω(xt)α;p)f(t)dt (1.5)

    where

    Eδ,q,r,cα,β,σ(z;p)=n=0Bp(δ+nq,cδ)B(δ,cδ)(c)nqΓ(αn+β)zn(σ)nr

    and Bp and B is as (1.3) and (1.4) respectively. For further information about this operator, (see [1,8,9,27]).

    Theorem 2.1. Let t be a positive function on [0,] and let f and g be two differentiable functions on [0,]. If fLr ([0,]), gLs ([0,]), r>1, r1+s1=1, then for all x>0, α>0, β>0, we have

    2|(ϵω,δ,q,r,c0+,α,β,σtfg)(x;p)(ϵω,δ,q,r,c0+,α,β,σt)(x;p)(ϵω,δ,q,r,c0+,α,β,σtf)(x;p)(ϵω,δ,q,r,c0+,α,β,σtg)(x;p)|||f||r||g||sx0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cα,β,σ(ω(xρ)α;p)dτdρ ||f||r||g||sx(ϵω,δ,q,r,c0+,α,β,σt(x;p))2. (2.1)

    Proof. Let f and g be two functions satisfying the conditions of Theorem 2.1 and let t be a positive function on [0,], Define

    H(τ,ρ):=(f(τ)f(ρ))(g(τ)g(ρ));τ,ρ(0,x),x>0. (2.2)

    Multiplying (2.2) by

    (xτ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)t(τ);τ(0,x)

    and integrating the resulting identity with respect to τ from 0 x, we can state that

    x0(xτ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)t(τ)H(τ,ρ)dτ=(ϵω,δ,q,r,c0+,α,β,σtfg)(x;p)f(ρ)(ϵω,δ,q,r,c0+,α,β,σtg)(x;p)g(ρ)(ϵω,δ,q,r,c0+,α,β,σtf)(x;p)+f(ρ)g(ρ)(ϵω,δ,q,r,c0+,α,β,σt)(x;p) (2.3)

    Now, multiplying (2.3) by

    (xρ)(β1)Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(ρ);ρ(0,x)

    and integrating the resulting identity with respect to ρ over (0,x), we can write

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)H(τ,ρ)dτdρ=(ϵω,δ,q,r,c0+,α,β,σtfg)(x;p)(ϵω,δ,q,r,c0+,α,β,σt)(x;p)(ϵω,δ,q,r,c0+,α,β,σtf)(x;p)(ϵω,δ,q,r,c0+,α,β,σtg)(x;p)(ϵω,δ,q,r,c0+,α,β,σtg)(x;p)(ϵω,δ,q,r,c0+,α,β,σtf)(x;p)+(ϵω,δ,q,r,c0+,α,β,σtfg)(x;p)(ϵω,δ,q,r,c0+,α,β,σt)(x;p)

    Consequently,

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)H(τ,ρ)dτdρ=2((ϵω,δ,q,r,c0+,α,β,σtfg)(x;p)(ϵω,δ,q,r,c0+,α,β,σt)(x;p)(ϵω,δ,q,r,c0+,α,β,σtf)(x;p)(ϵω,δ,q,r,c0+,α,β,σtg)(x;p)) (2.4)

    On the other and, we have

    H(τ,ρ):=ρτρτf(y)g(z)dydz. (2.5)

    Using Hölder inequality for double integral, we can write

    |H(τ,ρ)||ρτρτ|f(y)|rdydz|r1|ρτρτ|g(z)|sdydz|s1 (2.6)

    Since,

    |ρτρτ|f(y)|rdydz|r1=|τρ|r1|ρτ|f(y)|rdy|r1 (2.7)

    and

    |ρτρτ|g(z)|sdydz|s1=|τρ|s1|ρτ|g(z)|sdz|s1 (2.8)

    then, we can estimate H as follows:

    |H(τ,ρ)||τρ||ρτ|f(y)|rdy|r1|ρτ|g(z)|sdz|s1 (2.9)

    On the other hand, we have

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)|H(τ,ρ)|dτdρx0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)×|ρτ|f(y)|rdy|r1|ρτ|g(z)|sdz|s1dτdρ (2.10)

    Applying again Hölder inequality to the right-hand side of (2.10), we can state that

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)|H(τ,ρ)|dτdρ[x0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)|ρτ|f(y)|rdy|dτdρ]r1×[x0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)|ρτ|g(z)|sdz|dτdρ]s1.

    Now, using the fact the

    |ρτ|f(y)|rdy|||f||rr,,|ρτ|g(z)|sdz|||g||ss, (2.11)

    we obtain

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)|H(τ,ρ)|dτdρ[||f||rr,x0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cα,β,σ(ω(xρ)α;p)dτdρ]r1×[||g||ss,x0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cα,β,σ(ω(xρ)α;p)dτdρ]s1. (2.12)

    From (2.12), we get

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)|H(τ,ρ)|dτdρ||f||r||g||s[x0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cα,β,σ(ω(xρ)α;p)dτdρ]r1×[x0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cα,β,σ(ω(xρ)α;p)dτdρ]s1. (2.13)

    Since r1+s1=1, then we have

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)|H(τ,ρ)|dτdρ||f||r||g||s[x0x0(xτ)(β1)(xρ)(β1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cα,β,σ(ω(xρ)α;p)dτdρ]. (2.14)

    By the relations (2.4) and (2.14) and using the properties of the modulus, we get the first inequality in (2.1). We have

    0τx,0ρx.

    Hence,

    0|τρ|x.

    Therefore, we have

    x0x0(xτ)(β1)(xρ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)|H(τ,ρ)|dτdρ||f||r||g||sx[x0x0(xτ)(β1)(xρ)(β1)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cα,β,σ(ω(xρ)α;p)t(τ)t(ρ)dτdρ]=||f||r||g||s x(ϵω,δ,q,r,c0+,α,β,σt(x))2. (2.15)

    Theorem (2.1) is thus proved.

    In Theorem 2.1, if we set t(x)=1, we arrive at the following corollary :

    Corollary 2.1. Let f and g be two differentiable functions on [0,]. If fLr([0,]), gLs([0,]), r>1, r1+s1=1, then for all x>0, α>0, β>0, we have:

    |(ϵω,δ,q,r,c0+,α,β,σfg)(x;p)1(ϵω,δ,q,r,c0+,α,β,σ)(1)(ϵω,δ,q,r,c0+,α,β,σf)(x;p)(ϵω,δ,q,r,c0+,α,β,σg)(x;p)|12(||f||r||g||sx(ϵω,δ,q,r,c0+,α,β,σ)(1)). (2.16)

    Corollary 2.2. For different choices of parameters in (2.1) we can establish the corresponding fractional integral inequalities such as

    (i) setting p=0, we get Chebyshev inequality for the Salim-Faraj fractional integral operator, defined in [16],

    (ii) setting σ=r=1, we get Chebyshev inequality for the fractional integral operator defined by Rahman et al. in [15],

    (iii) setting p=0 and σ=r=1, we get Chebyshev inequality for the Srivastava-Tomovski fractional integral operator defined in [26],

    (iv) setting p=0 and σ=r=q=1, we get Chebyshev inequality for the Prabhakar fractional integral operator defined in [13].

    Remark 2.1. In (2.1) setting p=ω=0, we get the inequality (3.1) in [4].

    Theorem 2.2. Let t be a positive function on [0,] and let f and g be two differentiable functions on [0,]. If fLr([0,]), gLs([0,]), r>1, r1+s1=1, then for all x>0, α>0, β>0, λ>0, θ>0, we have

    |(ϵω,δ,q,r,c0+,α,β,σt)(x;p)(ϵω,δ,q,r,c0+,λ,θ,ptfg)(x;p)+(ϵω,δ,q,r,c0+,λ,θ,pt)(x;p)(ϵω,δ,q,r,c0+,α,β,σtfg)(x;p)(ϵω,δ,q,r,c0+,α,β,σtf)(x;p)(ϵω,δ,q,r,c0+,λ,θ,ptg)(x;p)(ϵω,δ,q,r,c0+,λ,θ,ptf)(x;p)(ϵω,δ,q,r,c0+,α,β,σtg)(x;p)|||f||r||g||sx0x0(xτ)(β1)(xρ)(θ1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)dτdρ||f||r||g||sx(ϵω,δ,q,r,c0+,α,β,σt)(x;p)(ϵω,δ,q,r,c0+,λ,θ,pt)(x;p). (2.17)

    Proof. Using the identity (2.3), we can state that

    x0x0(xτ)(β1)(xρ)(θ1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)t(τ)t(ρ)H(τ,ρ)dτdρ=(ϵω,δ,q,r,c0+,α,β,σt)(x;p)(ϵω,δ,q,r,c0+,λ,θ,ptfg)(x;p)+(ϵω,δ,q,r,c0+,λ,θ,pt)(x;p)(ϵω,δ,q,r,c0+,α,β,σtfg)(x;p)(ϵω,δ,q,r,c0+,α,β,σtf)(x;p)(ϵω,δ,q,r,c0+,λ,θ,ptg)(x;p)(ϵω,δ,q,r,c0+,λ,θ,ptf)(x;p)(ϵω,δ,q,r,c0+,α,β,σtg)(x;p). (2.18)

    From the relation (2.9), we can obtain the following estimation

    x0(xτ)(β1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)t(τ)|H(τ,ρ)|dτx0(xτ)(β1)|τρ|Eδ,q,r,cα,β,σ(ω(xτ)α;p)t(τ)×|ρτ|f(y)|rdy|r1|ρτ|g(z)|sdz|s1dτ (2.19)

    Therefore, we have

    x0x0(xτ)(β1)(xρ)(θ1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)t(τ)t(ρ)|H(τ,ρ)|dτdρx0x0(xτ)(β1)(xρ)(θ1)|τρ|Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)t(τ)t(ρ)|ρτ|f(y)|rdy|r1|ρτ|g(z)|sdz|s1dτdρ. (2.20)

    Applying Hölder inequality for double integral to the right-hand side of (2.20), yields

    x0x0(xτ)(β1)(xρ)(θ1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)t(τ)t(ρ)|H(τ,ρ)|dτdρ[x0x0(xτ)(β1)(xρ)(θ1)|τρ|Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)t(τ)t(ρ)|ρτ|f(y)|rdy|dτdρ]r1×[x0x0(xτ)(β1)(xρ)(θ1)|τρ|Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)t(τ)t(ρ)|ρτ|g(z)|sdz|dτdρ]s1. (2.21)

    By (2.11) and (2.21), we get

    x0x0(xτ)(β1)(xρ)(θ1)Eδ,q,r,cα,β,σ(ω(xτ)α;p)×Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)t(τ)t(ρ)|H(τ,ρ)|dτdρ||f||r||g||sx0x0(xτ)(β1)(xρ)(θ1)|τρ|t(τ)t(ρ)×Eδ,q,r,cα,β,σ(ω(xτ)α;p)Eδ,q,r,cλ,θ,p(ω(xρ)λ;p)dτdρ. (2.22)

    Using (2.18) and (2.22) and the properties of modulus, we get the first inequality in (2.17).

    Corollary 2.3. For different choices of parameters in (2.17) we can establish the corresponding fractional integral inequalities such as

    (i) setting p=0, we get Chebyshev inequality for the Salim-Faraj fractional integral operator, defined in [16],

    (ii) setting σ=r=1, we get Chebyshev inequality for the fractional integral operator defined by Rahman et al. in [15],

    (iii) setting p=0 and σ=r=1, we get Chebyshev inequality for the Srivastava-Tomovski fractional integral operator defined in [26],

    (iv) setting p=0 and σ=r=q=1, we get Chebyshev inequality for the Prabhakar fractional integral operator defined in [13].

    Remark 2.2. Applying Theorem 2.2 for β=θ, α=λ, we obtain theorem 2.1.

    Remark 2.3. In (2.17) setting p=ω=0, we get the inequality (3.17) in [4].

    The authors declare there is no conflicts of interest in this paper.



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