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Hermite-Hadamard type inequalities based on the Erdélyi-Kober fractional integrals

  • In the paper, based on Erdélyi-Kober fractional integrals ρKαχ+f and ρKαχf for any χ[a,b] with fXpc(a,b), authors establish some new Hermite-Hadamard type inequalities for convex function. The obtained inequalities generalize the corresponding results for Riemann-Liouville fractional integrals by taking limits when a parameter ρ1. As applications, the error estimations of Hermite-Hadamard type inequality are also provided.

    Citation: XuRan Hai, ShuHong Wang. Hermite-Hadamard type inequalities based on the Erdélyi-Kober fractional integrals[J]. AIMS Mathematics, 2021, 6(10): 11494-11507. doi: 10.3934/math.2021666

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  • In the paper, based on Erdélyi-Kober fractional integrals ρKαχ+f and ρKαχf for any χ[a,b] with fXpc(a,b), authors establish some new Hermite-Hadamard type inequalities for convex function. The obtained inequalities generalize the corresponding results for Riemann-Liouville fractional integrals by taking limits when a parameter ρ1. As applications, the error estimations of Hermite-Hadamard type inequality are also provided.



    Fractional calculus is a field of applied mathematics and deals with derivatives and integrals of arbitrary orders (including complex orders). Although the definitions for fractional integrals are inconsistent and work in some cases but not in others, there are almost practical applications and profound impact in science, engineering, mathematics, economics, and other fields.

    Suppose that (a,b) is a finite or infinite interval of the real line R, where a<b and a,b[,+], and α is a complex number with Re(α)>0. Let Γ() be the Euler's gamma function given by

    Γ(χ)=0τχ1eτdτ.

    In [20], Podlubny introduced the left-side and right-side Riemann-Liouville fractional integrals of order α of a function f as follows:

    Rαa+f(χ)=1Γ(α)χa(χτ)α1f(τ)dτ (1.1)

    and

    Rαbf(χ)=1Γ(α)bχ(τχ)α1f(τ)dτ, (1.2)

    respectively, where f is a function on the interval [a,b] such that (χτ)α1f(τ)L[a,b] for any χ[a,b].

    In [22], Samko introduced the left-side and right-side Hadamard fractional integrals of order α of a function f as follows

    Hαa+f(χ)=1Γ(α)χa(lnχlnτ)α1f(τ)dττ (1.3)

    and

    Hαbf(χ)=1Γ(α)bχ(lnτlnχ)α1f(τ)dττ, (1.4)

    respectively, where f is a function on the interval [a,b] such that (lnχlnτ)α1f(τ)τL[a,b] for any χ[a,b].

    Suppose that Xpc(a,b) is the space of the complex-valued Lebesgue measurable functions f on [a,b] with fXpc<, that is

    Xpc(a,b)={f:[a,b]C|fXpc<},

    where the norm fXpc is

    fXpc=(ba|τcf(τ)|pdττ)1/pfor1p<andcR

    and

    fXc=esssupaτb[τc|f(τ)|]forp=andcR.

    In the sense of the above function space, Katugampola in [16] introduced the left-side and right-side fractional integrals of order α of a function fXpc(a,b) defined by

    ρKαa+f(χ)=1Γ(α)χa(χρτρρ)α1f(τ)dττ1ρ(ρ>0) (1.5)

    and

    ρKαbf(χ)=1Γ(α)bχ(τρχρρ)α1f(τ)dττ1ρ(ρ>0), (1.6)

    respectively.

    The above fractional operators are known as Erdélyi-Kober fractional integrals in [18], or Katugampola fractional integrals in [16] or ρRiemann-Liouville fractional integrals in [6], which generalized fractional integrals of Riemann-Liouville and Hadamard, respectively[17]:

    limρ1[ρKαa+f(χ)]=limρ11Γ(α)χa(χρτρρ)α1f(τ)dττ1ρ=1Γ(α)χa(χτ)α1f(τ)dτ=Rαa+f(χ) (1.7)

    and

    limρ0[ρKαa+f(χ)]=limρ01Γ(α)χa(χρτρρ)α1f(τ)dττ1ρ=1Γ(α)χa(lnχlnτ)α1f(τ)dττ=Hαa+f(χ). (1.8)

    The similar results for right-sided fractionals integral also hold.

    For more results on the fractional integrals please see [1,4,8,13,14,15,21,24] and the references therein.

    For any convex function f:[a,b]R, the following double inequalities

    f(a+b2)1babaf(τ)dτf(a)+f(b)2 (1.9)

    are known as Hermite-Hadamard inequality[12,19].

    In [5], Chen et al. established the following Hermite-Hadamard type inequalities based on the Katugampola fractional integrals.

    Theorem 1.1. Suppose that f:[aρ,bρ]R is a positive function with ρ>0 and 0a<b, and fXpc(aρ,bρ). If f is a convex function on [a,b], then for any α>0

    f(aρ+bρ2)ραΓ(α+1)2(bρaρ)α[ρKαa+f(bρ)+ρKαbf(aρ)]f(aρ)+f(bρ)2, (1.10)

    where the fractional integrals are considered for the function f(χρ) and evaluated at a and b, respectively.

    Furthermore, Chen et al. also gave some right estimations of the Hermite-Hadamard type inequalities for the Katugampola fractional integrals in [5].

    Theorem 1.2. Suppose that f:[aρ,bρ]R is a differentiable function on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If |f| is a convex function on [aρ,bρ], then for any α>0

    |f(aρ)+f(bρ)2ραΓ(α+1)2(bρaρ)α[ρKαa+f(bρ)+ρKαbf(aρ)]|bρaρ2(α+1)[|f(aρ)|+|f(bρ)|], (1.11)

    where the fractional integrals are considered for the function f(χρ) and evaluated at a and b, respectively.

    Theorem 1.3. Suppose that f:[aρ,bρ]R is a differentiable function on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If |f| is convex on [aρ,bρ], then for any α>0

    |f(aρ)+f(bρ)2ραΓ(α+1)2(bρaρ)α[ρKαa+f(bρ)+ρKαbf(aρ)]|bρaρ2(α+1)(112α)[|f(aρ)|+|f(bρ)|], (1.12)

    where the fractional integrals are considered for the function f(χρ) and evaluated at a and b, respectively.

    For more results on the convexity and Hermite-Hadamard type inequalities please see [2,7,9,10,11,23] and the references therein.

    In the paper, based on Erdélyi-Kober fractional integrals ρKαχ+f(bρ) and ρKαχf(aρ) for any χ[a,b] with fXpc(a,b), authors establish some new Hermite-Hadamard type inequalities for convex function. The obtained inequalities generalize the corresponding results for Riemann-Liouville fractional integrals by taking limits when a parameter ρ1. As applications, the error estimations of Hermite-Hadamard type inequality are also provided.

    Firstly, we establish Hermite-Hadamard type inequality for the Erdélyi-Kober fractional integrals ρKαχ+f(bρ) and ρKαχf(aρ) for any χ[a,b] with fXpc(a,b).

    Theorem 2.1. Suppose that f:[aρ,bρ]R is a function with ρ>0 and 0a<b, and fXpc(aρ,bρ). If f is a convex function on [aρ,bρ], then for any α>0 and any χ[a,b],

    f(1α+1aρ+bρ2+αα+1χρ)ραΓ(α+1)2[ρKαχ+f(bρ)(bρχρ)α+ρKαχf(aρ)(χρaρ)α]1α+1[f(aρ)+f(bρ)2+αf(χρ)]. (2.1)

    Proof. It easy to follow that

    ραΓ(α+1)2[ρKαχ+f(bρ)(bρχρ)α+ρKαχf(aρ)(χρaρ)α]=ρα2[bχ(bρtρbρχρ)α1tρ1f(tρ)bρχρdt+χa(tρaρχρaρ)α1tρ1f(tρ)χρaρdt]. (2.2)

    Making the integral transformations tρ=τρχρ+(1τρ)bρ and tρ=(1τρ)aρ+τρχρ respectively in (2.2), it is obtained that

    ραΓ(α+1)2[ρKαχ+f(bρ)(bρχρ)α+ρKαχf(aρ)(χρaρ)α]=ρα210τρα1[f(τρχρ+(1τρ)bρ)+f((1τρ)aρ+τρχρ)]dτ. (2.3)

    Then by the convexity of f and Jensen's inequality, we obtain

    ρα210τρα1[f(τρχρ+(1τρ)bρ)+f((1τρ)aρ+τρχρ)]dτρα10τρα1f(τρχρ+(1τρ)aρ+bρ2)dτραf(10τρα1[τρχρ+(1τρ)aρ+bρ2]dτ10τρα1dτ)10τρα1dτ=f(1α+1aρ+bρ2+αα+1χρ), (2.4)

    which completes the left inequality of Theorem 2.1.

    On the another hand, by the convexity of f again, we have

    ρα210τρα1[f(τρχρ+(1τρ)bρ)+f((1τρ)aρ+τρχρ)]dτρα210τρα1[τρf(χρ)+(1τρ)f(bρ)+(1τρ)f(aρ)+τρf(χρ)]dτ=ρα10τρα1[τρf(χρ)+(1τρ)f(aρ)+f(bρ)2])dτ=1α+1[f(aρ)+f(bρ)2+αf(χρ)], (2.5)

    which completes the right inequality of Theorem 2.1.

    Corollary 2.1.1. With the assumptions of Theorem 2.1 and taking χρ=aρ+bρ2, it reduces that

    f(aρ+bρ2)2α1ραΓ(α+1)(bρaρ)α[ρKαρaρ+bρ2+f(bρ)+ρKαρaρ+bρ2f(aρ)]1α+1[f(aρ)+f(bρ)2+αf(aρ+bρ2)]f(aρ)+f(bρ)2. (2.6)

    In particular, taking limits when χa and χb respectively in the inequality (2.1), and using the L'Hospital rule, we have the following result.

    Corollary 2.1.2. With the assumptions of Theorem 2.1 and taking limits when χa and χb respectively, it reduces that

    f(aρ+bρ2)12[f((2α+1)aρ+bρ2(α+1))+f(aρ+(2α+1)bρ2(α+1))]f(aρ)+f(bρ)4+ραΓ(α+1)4(bρaρ)α[ρKαa+f(bρ)+ρKαbf(aρ)]f(aρ)+f(bρ)2. (2.7)

    Now we give a interest equality.

    Lemma 3.1. Suppose that f:[aρ,bρ]R is differentiable on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If the generalized fractional integrals exist, then for any α>0 and any χ[a,b], the equality

    f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]=ρ210(1τρα)τρ1[(bρχρ)f(τρχρ+(1τρ)bρ)(χρaρ)f((1τρ)aρ+τρχρ)]dτ (3.1)

    holds, where the fractional integrals are considered for the function f(χρ) and evaluated at a and b, respectively.

    Proof. Using integration by parts, it easy to follow that

    ρ210(1τρα)τρ1[(bρχρ)f(τρχρ+(1τρ)bρ)(χρaρ)f((1τρ)aρ+τρχρ)]dτ=f(bρ)2ρα210τρα1f(τρχρ+(1τρ)bρ)dτ+f(aρ)2ρα210τρα1f((1τρ)aρ+τρχρ)dτ=f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α], (3.2)

    which completes the proof of Lemma 3.1.

    In particular, taking limits when χa and χb respectively in the identity (3.1), and summing the obtained identities, we obtain

    f(aρ)+f(bρ)2ραΓ(α+1)2(bρaρ)α[ρKαa+f(bρ)+ρKαbf(aρ)]=ρ(bρaρ)210[(1τρ)ατρα]τρ1f(τρaρ+(1τρ)bρ)dτ, (3.3)

    which is Lemma 2.4 in [5].

    Next, we establish some integral inequalities by the differentiability, the convexity and Lemma 3.1.

    If the function |f| is convex, then the following integral inequality holds.

    Theorem 3.1. Suppose that f:[aρ,bρ]R is differentiable on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If |f| is convex on [aρ,bρ], then the inequality

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|α4(α+1)(α+2)[(α+3)(χρaρ)|f(aρ)|+(α+1)(bρaρ)|f(χρ)|+(α+3)(bρχρ)|f(bρ)|] (3.4)

    holds for any α>0 and any χ[a,b].

    Proof. Using Lemma 3.1 and the convexity of the function |f|, we have

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ210(1τρα)τρ1[(bρχρ)|f(τρχρ+(1τρ)bρ)|+(χρaρ)|f((1τρ)aρ+τρχρ)|]dτρ(bρχρ)210(1τρα)τρ1[τρ|f(χρ)|+(1τρ)|f(bρ)|]dτ+ρ(χρaρ)210(1τρα)τρ1[(1τρ)|f(aρ)|+τρ|f(χρ)|]dτ. (3.5)

    By simple computation, the inequality (3.4) is obtained which completes the proof of Theorem 3.1.

    If we take χρ=aρ+bρ2 in the inequality (3.4), then we get a integral inequality.

    |f(aρ)+f(bρ)22α1ραΓ(α+1)(bρaρ)α[ρKαρaρ+bρ2f(aρ)+ρKαρaρ+bρ2+f(bρ)]|α(bρaρ)8(α+1)(α+2)[(α+3)|f(aρ)|+2(α+1)|f(aρ+bρ2)|+(α+3)|f(bρ)|]. (3.6)

    Also, making limits when ρ1 in the inequality (3.6), we immediately get the integral inequalities for Riemann-Liouville fractional integrals:

    |f(a)+f(b)22α1Γ(α+1)(ba)α[Rαa+b2f(a)+Rαa+b2+f(b)]|α(ba)8(α+1)(α+2)[(α+3)|f(a)|+2(α+1)|f(a+b2)|+(α+3)|f(b)|]. (3.7)

    If the function |f|q(q>1) is convex, then the below inequality holds.

    Theorem 3.2. Suppose that f:[aρ,bρ]R is differentiable on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If |f|q(q>1) is convex on [aρ,bρ], then the inequality

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|12α[B(2qr1q1,ρqr1ρα(q1))]11/q{(χρaρ)[B(r+1,ρ+r+1ρα)|f(χρ)|q+(B(r+1,r+1ρα)B(r+1,ρ+r+1ρα))|f(aρ)|q]1/q+(bρχρ)[B(r+1,ρ+r+1ρα)|f(χρ)|q+(B(r+1,r+1ρα)B(r+1,ρ+r+1ρα))|f(bρ)|q]1/q} (3.8)

    holds for any α>0, χ[a,b] and 0rq, where B(μ,ν) is classical beta function defined by

    B(μ,ν)=10sμ1(1s)ν1ds(μ>0,ν>0). (3.9)

    Proof. Using the identity (3.1), Hölder's inequality and the convexity of the function |f|q(q>1), we have

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ210(1τρα)τρ1[(bρχρ)|f(τρχρ+(1τρ)bρ)|+(χρaρ)|f((1τρ)aρ+τρχρ)|]dτρ(bρχρ)2(10(1τρα)qrq1τq(ρ1)rq1dτ)11/q×[10(1τρα)rτr[τρ|f(χρ)|q+(1τρ)|f(bρ)|q]dτ]1/q+ρ(χρaρ)2(10(1τρα)qrq1τq(ρ1)rq1dτ)11/q×[10(1τρα)rτr[τρ|f(aρ)|q+(1τρ)|f(χρ)|q]dτ]1/q. (3.10)

    By direct calculation, we obtain the inequality (3.8) which completes the proof of Theorem 3.2.

    In particular, making r=0, r=1 and r=q, respectively, then

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ1/q2α11/q(ρ+1)1/q[B(2q1q1,ρq1ρα(q1))]11/q×{(χρaρ)[|f(χρ)|q+ρ|f(aρ)|q]1/q+(bρχρ)[|f(χρ)|q+ρ|f(bρ)|q]1/q} (3.11)

    and

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ2α(q1)22/q21+1/q[(ρ+2)(ρ+2+ρα)(ρα+2)](1(ρq2)[ρq2+ρα(q1)])11/q×{(χρaρ)[2(2+ρα)|f(χρ)|q+ρ(ρ+2+ρα)|f(aρ)|q]1/q+(bρχρ)[2(2+ρα)|f(χρ)|q+ρ(ρ+2+ρα)|f(bρ)|q]1/q} (3.12)

    and

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ11/q2α1/q(q1ρqq1)11/q{(χρaρ)[B(q+1,ρ+q+1ρα)|f(χρ)|q+(B(q+1,q+1ρα)B(q+1,ρ+q+1ρα))|f(aρ)|q]1/q+(bρχρ)[B(q+1,ρ+q+1ρα)|f(χρ)|q+(B(q+1,q+1ρα)B(q+1,ρ+q+1ρα))|f(bρ)|q]1/q}. (3.13)

    Furthermore, utilizing Lemma 3.1 reduces to the below inequalities.

    Theorem 3.3. Suppose that f:[aρ,bρ]R is differentiable on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If |f|q(q>1) is convex on [aρ,bρ], then the inequality

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|12α[B(2qr1q1,ρ(qr)+r1ρα(q1))]11/q{(χρaρ)[B(r+1,ρ(r+1)r+1ρα)|f(χρ)|q+(B(r+1,ρrr+1ρα)B(r+1,ρ(r+1)r+1ρα))|f(aρ)|q]1/q+(bρχρ)[B(r+1,ρ(r+1)r+1ρα)|f(χρ)|q+(B(r+1,ρrr+1ρα)B(r+1,ρ(r+1)r+1ρα))|f(bρ)|q]1/q} (3.14)

    holds for any α>0, χ[a,b] and 0rq, where B(μ,ν) is classical beta function defined in (3.9).

    Proof. Using the identity (3.1), Hölder's inequality and the convexity of the function |f|q(q>1), we have

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ210(1τρα)τρ1[(bρχρ)|f(τρχρ+(1τρ)bρ)|+(χρaρ)|f((1τρ)aρ+τρχρ)|]dτρ(bρχρ)2(10(1τρα)qrq1τ(ρ1)(qr)q1dτ)11/q×[10(1τρα)rτ(ρ1)r[τρ|f(χρ)|q+(1τρ)|f(bρ)|q]dτ]1/q+ρ(χρaρ)2(10(1τρα)qrq1τ(ρ1)(qr)q1dτ)11/q×[10(1τρα)rτ(ρ1)r[τρ|f(aρ)|q+(1τρ)|f(χρ)|q]dτ]1/q. (3.15)

    By direct calculation, we obtain the inequality (3.14) which completes the proof of Theorem 3.3.

    In particular, making r=1 and r=q, respectively, then

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|α211/q(α+1)(α+2)1/q{(χρaρ)[(α+1)|f(χρ)|q+(α+3)|f(aρ)|q]1/q+(bρχρ)[(α+1)|f(χρ)|q+(α+3)|f(bρ)|q]1/q} (3.16)

    and

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ11/q2α1/q{(χρaρ)[B(q+1,ρ(q+1)q+1ρα)|f(χρ)|q+(B(q+1,ρqq+1ρα)B(q+1,ρ(q+1)q+1ρα))|f(aρ)|q]1/q+(bρχρ)[B(q+1,ρ(q+1)q+1ρα)|f(χρ)|q+(B(q+1,ρqq+1ρα)B(q+1,ρ(q+1)q+1ρα))|f(bρ)|q]1/q}. (3.17)

    Theorem 3.4. Suppose that f:[aρ,bρ]R is differentiable on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If |f|q(q>1) is convex on [aρ,bρ], then the inequality

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ1/q2α11/q[(ρrr+1)(ρ(r+1)r+1)]1/q[B(2q1q1,(ρ1)(qr)+q1ρα(q1))]11/q×{(χρaρ)[(ρrr+1)|f(χρ)|q+ρ|f(aρ)|q]1/q+(bρχρ)[(ρrr+1)|f(χρ)|q+ρ|f(bρ)|q]1/q} (3.18)

    holds for any α>0, χ[a,b] and 0rq, where B(μ,ν) is classical beta function defined in (3.9).

    Proof. Using the identity (3.1), Hölder's inequality and the convexity of the function |f|q(q>1), we have

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ210(1τρα)τρ1[(bρχρ)|f(τρχρ+(1τρ)bρ)|+(χρaρ)|f((1τρ)aρ+τρχρ)|]dτρ(bρχρ)2(10(1τρα)qq1τ(ρ1)(qr)q1dτ)11/q×[10τ(ρ1)r[τρ|f(χρ)|q+(1τρ)|f(bρ)|q]dτ]1/q+ρ(χρaρ)2(10(1τρα)qq1τ(ρ1)(qr)q1dτ)11/q×[10τ(ρ1)r[τρ|f(aρ)|q+(1τρ)|f(χρ)|q]dτ]1/q. (3.19)

    By direct calculation, we obtain the inequality (3.18) which completes the proof of Theorem 3.4.

    In particular, making r=1 and r=q, respectively, then

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|121+1/qα11/q[B(2q1q1,1α)]11/q{(χρaρ)[|f(χρ)|q+|f(aρ)|q]1/q+(bρχρ)[|f(χρ)|q+|f(bρ)|q]1/q} (3.20)

    and

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ1/q2α11/q[(ρqq+1)(ρ(q+1)q+1)][B(2q1q1,1ρα)]11/q×{(χρaρ)[(ρqq+1)|f(χρ)|q+ρ|f(aρ)|q]1/q+(bρχρ)[(ρqq+1)|f(χρ)|q+ρ|f(bρ)|q]1/q}. (3.21)

    Theorem 3.5. Suppose that f:[aρ,bρ]R is differentiable on (aρ,bρ) with ρ>0 and 0a<b, and fXpc(aρ,bρ). If |f|q(q>1) is convex on [aρ,bρ], then the inequality

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|12α[B(2qr1q1,ρq1ρα(q1))]11/q{(χρaρ)[B(r+1,ρ+1ρα)|f(χρ)|q+(B(r+1,1ρα)B(r+1,ρ+1ρα))|f(aρ)|q]1/q+(bρχρ)[B(r+1,ρ+1ρα)|f(χρ)|q+(B(r+1,1ρα)B(r+1,ρ+1ρα))|f(bρ)|q]1/q} (3.22)

    holds for any α>0, χ[a,b] and 0rq, where B(μ,ν) is classical beta function defined in (3.9).

    Proof. Using the identity (3.1), Hölder's inequality and the convexity of the function |f|q(q>1), we have

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ210(1τρα)τρ1[(bρχρ)|f(τρχρ+(1τρ)bρ)|+(χρaρ)|f((1τρ)aρ+τρχρ)|]dτρ(bρχρ)2(10(1τρα)qrq1τq(ρ1)q1dτ)11/q×[10(1τρα)r[τρ|f(χρ)|q+(1τρ)|f(bρ)|q]dτ]1/q+ρ(χρaρ)2(10(1τρα)qrq1τq(ρ1)q1dτ)11/q×[10(1τρα)r[τρ|f(aρ)|q+(1τρ)|f(χρ)|q]dτ]1/q. (3.23)

    By direct calculation, we obtain the inequality (3.22) which completes the proof of Theorem 3.5.

    In particular, making r=1 and r=q, respectively, then

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ2α(q1)22/q2[(ρ+1)(ρα+1)ρα+ρ+1)]1/q[1(ρq1)[ρq1+ρα(q1)]]11/q×{(χρaρ)[(1+ρα)|f(χρ)|q+ρ(ρα+ρ+2)|f(aρ)|q]1/q+(bρχρ)[(1+ρα)|f(χρ)|q+ρ(ρα+ρ+2)|f(bρ)|q]1/q} (3.24)

    and

    |f(aρ)+f(bρ)2ραΓ(α+1)2[ρKαχf(aρ)(χρaρ)α+ρKαχ+f(bρ)(bρχρ)α]|ρ11/q2α1/q(q1ρq1)11/q{(χρaρ)[B(q+1,ρ+1ρα)|f(χρ)|q+(B(q+1,1ρα)B(q+1,ρ+1ρα))|f(aρ)|q]1/q+(bρχρ)[B(q+1,ρ+1ρα)|f(χρ)|q+(B(q+1,1ρα)B(q+1,ρ+1ρα))|f(bρ)|q]1/q}. (3.25)

    In this article, we firstly establish Hermite-Hadamard type inequality for the Erdélyi-Kober fractional integrals (2.1). Taking limits when χa and χb respectively in the inequality (2.1), it is the generalization and the refinement of Theorem 2.1 in the reference [5]. In addition, as an application, the error estimations of Hermite-Hadamard type inequality have been investigated. The derived inequalities can be seen as a generalization for Riemann-Liouville fractional integrals when ρ1 and the Corollaries presented in the paper show that the results of this paper generalize and extend many existing results.

    This work was supported PhD Research Foundation of Inner Mongolia Minzu University (No. BS402) and Foundation of the Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region (No. NJZY20119) and Natural Science Foundation of Inner Mongolia (No. 2019MS01007), China.

    The authors declare that they have no conflict of interest.



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