On Caputo fractional derivatives via exponential (s,m)-convex functions

Author(s): Saad Ihsan Butt1, Mehroz Nadeem2, Ghulam Farid3
1Department of Mathematics, COMSATS University of Islamabad, Lahore Campus, Pakistan.
2Department of Mathematics, COMSATS University of Islamabad, Lahore Campus, Pakistan.;
3Department of Mathematics COMSATS University of Islamabad, Attock Campus, Pakistan.
Copyright © Saad Ihsan Butt, Mehroz Nadeem, Ghulam Farid. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we establish several integral inequalities including Caputo fractional derivatives for exponential (s,m)-convex functions. By using convexity for exponential (s,m)-convex functions of any positive integer order differentiable function some novel results are obtained.

Keywords: Convex function, exponential (s,m)-convex functions, Caputo-fractional derivatives.

1. Introduction

Convexity plays an important role in many features of mathematical programming including, for example, sufficient optimality conditions and duality theorems. The topic of convex functions has been treated extensively in the classical book by Hardy, Littlewood and Polya [1]. The study of fractional order derivatives and integrals is called fractional calculus. Fractional calculus have important applications in all fields of applied sciences. Fractional integration and fractional differentiation appear as basic part in the subject of partial differential equations [2, 3]. Many types of fractional integral as well as differential operators have been defined in literature. Classical Caputo-fractional derivatives were introduced by Michele Caputo in [4] in 1967. Toader [5] defined the m-convexity as follows:

Definition 1. The function Ψ:[u,v]R, is said to be convex, if we have% Ψ(τz1+(1τ)z2)τΨ(z1)+(1τ)Ψ(z2) for all z1,z2[u,v] and τ[0,1].

Definition 2.(see[6]) The function Ψ:I is exponential-convex, if Ψ(τz1+(1τ)z2)τeαz1Ψ(z1)+(1τ)eαz2Ψ(z2) for allτ[0,1]and z1,z2I and α.

Definition 3.(see[7]) The function Ψ:I[0,) is s-convex in second sense with s[0,1], if  Ψ(τz1+(1τ)z2)τsΨ(z1)+(1τ)sΨ(z2) for allτ[0,1)and z1,z2I and α.

Definition 4.(see[8]) The function Ψ:I[0,) is exponential s-convex in second sense with s[0,1], if  Ψ(τz1+(1τ)z2)τseβz1Ψ(z1)+(1τ)seβz2Ψ(z2)  for\ allτ[0,1]and\ z1,z2I and β.

Definition 5.(see[9]) The function Ψ:K is (s,m)-convex in second sense with s[0,1], and K[0,] be an interval, if  Ψ(τz1+(1τ)z2)τsΨ(z1)+(1τ)smΨ(z2) for allτ[0,1]and z1,z2[0,].

Definition 6. The function Ψ:K is exponential (s,m)-convex in second sense with s[0,1], and K[0,] be an interval, if Ψ(τz1+(1τ)z2)τseβz1Ψ(z1)+(1τ)seβz2mΨ(z2)  for allτ[0,1]and\ z1,z2[0,] and β.

The previous era of fractional calculus is as old as the history of differential calculus. They generalize the differential operators and ordinary integral. However, the fractional derivatives have some basic properties than the corresponding classical ones. On the other hand, besides the smooth requirement, Caputo derivative does not coincide with the classical derivative [10]. We give the following definition of Caputo fractional derivatives, see [2, 11, 12, 13].

Definition 7. let ΨACn[u,v] be a space of functions having nth derivatives absolutely continuous with λ>0 and λ{1,2,3,...}, n=[λ]+1. The right sided Caputo fractional derivative is as follows:

(CDu+λΨ)(z)=1Γ(nλ)uzΨ(n)(τ)(zτ)λn+1dτ,z>u.
(1)
The left sided caputo fractional derivative is as follows:
(CDvλΨ)(z)=(1)nΓ(nλ)zvΨ(n)(τ)(τz)λn+1dτ,z<v.
(2)
The Caputo fractional derivative (CDu+nΨ)(z) coincides with Ψ(n)(z) whereas (CDvnΨ)(z) coincides with Ψ(n)(z) with exactness to a constant multiplier (1)n, if Λ=n{1,2,3,...} and usual derivative Ψ(n)(z) of order n exists. In particular. we have
(CDu+0Ψ)(z)=(CDv0Ψ)(z)=Ψ(z)
(3)
where n=1 and λ=0.

In this paper, we establish several new integral inequalities including Caputo fractional derivatives for exponential (s,m)-convex functions. By using convexity for exponential (s,m)-convex functions of any positive integer order differentiable function some novel results are given. The purpose of this paper is to introduce some fractional inequalities for the Caputo-fractional derivatives via (s,m)-convex functions which have derivatives of any integer order.

2. Main Results

First we give the following estimate of the sum of left and right handed Caputo fractional derivatives.

Theorem 1. Let f:IR be a real valued n-time differentiable function where n is a positive integer. If f(n) is a positive (s,m)-convex function, then for u,vI;u<v and λ1,λ21, the following inequality for Caputo fractional derivatives holds:

Γ(nλ1+1)(CDu+λ11f)(u)+Γ(nλ2+1)(CDvλ21f)(u)(zu)nλ1+1eβuf(n)(u)+(vz)nλ2+1eβvf(n)(v)s+1+(m)eβzf(n)(z)[(zu)nλ1+1+(vz)nλ2+1s+1].
(4)

Proof. Let us consider the function f on the interval [u,z],z[u,v] and n is a positive integer. For τ[u,z] and n>α, the following inequality holds:

(zτ)nλ1(zu)nλ1.
(5)
Since f(n) is exponential (s,m)-convex therefore for τ[u,z], we have
f(n)(τ)(zτzu)seβuf(n)(u)+m(τuzu)seβzf(n)(z).
(6)
Multiplying inequalities (5) and (6), then integrating with respect to τ over [u,z], we have uz(zτ)nλ1f(n)(τ)dτ(zu)nλ1(zu)s[eβuf(n)(u)uz(zτ)sdτ+meβzf(n)(z)uz(τu)sdτ].
Γ(nλ1+1)(CDu+λ11f)(z)(zu)nλ1+1s+1[eβuf(n)(u)+meβzf(n)(z)].
(7)
Now we consider function f on the interval [z,v],z[u,v]. For τ[z,v], the following inequality holds:
(τz)nλ2(vz)nλ2.
(8)
Since f(n) is exponential (s,m)-convex on [u,v], therefore for τ[z,v], we have
f(n)(τ)(τzvz)seβvf(n)(v)+m(vτvz)seβzf(n)(z).
(9)
Multiplying inequalities (8) and (9), then integrating with respect to τ over [z,v], we have zv(τz)nλ2f(n)(τ)dτ(vz)nλ2(vz)s[eβvf(n)(v)zv(τz)sdτ+meβzf(n)(z)zv(vτ)sdτ]
Γ(nλ2+1)(CDvλ21f)(z)(vz)nλ2+1s+1[eβvf(n)(v)+meβzf(n)(z)].
(10)
Adding (7) and (10) we get the required inequality in (4).

Corollary 1. By setting λ1=λ2 in (4) we get the following fractional integral inequality:

Γ(nλ1+1)((CDu+λ11f)(z)+(CDvλ11f)(z))(zu)nλ1+1eβuf(n)(u)+(vz)nλ1+1eβvf(n)(v)s+1+meβzf(n)(z)[(zu)nλ1+1+(vz)nλ1+1s+1].
(11)

Remark 1. By setting s=1 the inequality will be of the form:

Γ(nλ1+1)((CDu+λ11f)(z)+(CDvλ11f)(z))(zu)nλ1+1eβuf(n)(u)+(vz)nλ1+1eβvf(n)(v)2+meβzf(n)(z)[(zu)nλ1+1+(vz)nλ1+12].
(12)

Remark 2. By setting λ1=λ2, β=0, s=1 and m=1, we will get Corollary 2.1 of [14].

Now, we give the next result stated in the following theorem.

Theorem 2. Let f:IR be a real valued n-time differentiable function where n is a positive integer. If |f(n+1)| is exponential (s,m)-convex function, then for u,vI;u0, the following inequality for Caputo fractional derivatives holds

|Γ(nλ1+1)(CDu+λ1f)(z)+Γ(nλ2+1)(CDvλ2f)(z)((zu)nλ1f(n)(u)+(vz)nλ2f(n)(v))|(zu)λ1+1eβu|f(n+1)(u)|+(vz)λ2+1eβv|f(n+1)(v)|s+1+meβz|f(n+1)(z)|((zu)λ1+1+(vz)λ2+1)s+1.
(13)

Proof. Since |f(n+1)| is exponential (s,m)-convex function and n is a positive integer, therefore for τ[u,z] and n>α, we have |f(n+1)(τ)|(zτzu)seβu|f(n+1)(u)|+m(τuzu)seβz|f(n+1)(z)| from which we can write

((zτzu)seβu|f(n+1)(u)|+m(τuzu)seβz|f(n+1)(z)|)f(n+1)(τ)(zτzu)seβu|f(n+1)(u)|+m(τuzu)seβz|f(n+1)(z)|.
(14)
We consider the second inequality of inequality (14)
f(n+1)(τ)(zτzu)seβu|f(n+1)(u)|+m(τuzu)seβz|f(n+1)(z)|.
(15)
Now for λ1>0, we have
(zτ)nλ1(zu)nλ1,τ[u,z].
(16)
The product of last two inequalities give (zτ)nλ1f(n+1)(τ)(zu)nλ1s((zτ)seβu|f(n+1)(u)|+m(τu)seβz|f(n+1)(z)|). Integrating with respect to τ over [u,z], we have
uz(zτ)nλ1f(n+1)(τ)dτ(zu)nλ1s[eβu|f(n+1)(u)|uz(ztτ)sdτ+meβz|f(n+1)(z)|uz(τu)sdτ]=(zu)nλ1+1[eβu|f(n+1)(u)|+meβz|f(n+1)(z)|s+1],
(17)
and uz(zτ)nλ1f(n+1)(τ)dτ=f(n)(τ)(zτ)nλ1|uz+(nλ1)uz(zτ)nλ11f(n)(τ)dτ=f(n)(u)(zu)nλ1+Γ(nλ1+1)(CDu+λ1f)(z). Therefore (17) takes the form:
Γ(nλ1+1)(CDu+λ1f)(z)f(n)(u)(zu)nλ1(zu)nλ1+1[eβu|f(n+1)(u)|+meβz|f(n+1)(z)|s+1].
(18)
If one consider from (14) the first inequality and proceed as we did for the second inequality, then following inequality can be obtained:
f(n)(u)(zu)nλ1Γ(nλ1+1)(CDu+λ1f)(z)(zu)nλ1+1[eβu|f(n+1)(u)|+meβz|f(n+1)(z)|s+1].
(19)
From (18) and (19), we get
|Γ(nλ1+1)(CDu+λ1f)(z)f(n)(u)(zu)nλ1|(zu)nλ1+1[eβu|f(n+1)(u)|+meβz|f(n+1)(z)|s+1].
(20)
On the other hand, for τ[z,v], using convexity of |f(n+1)| as a exponential (s,m)-convex function, we have
|f(n+1)(τ)|(τzvz)seβv|f(n+1)(v)|+m(vτvz)seβz|f(n+1)(z)|.
(21)
Also for τ[z,v] and β>0, we have
(τz)nλ2(vz)nλ2.
(22)
By adopting the same treatment as we have done for (14) and (16) one can obtain from (21) and (22) the following inequality:
|Γ(nλ2+1)(CDvλ2f)(z)f(n)(v)(vz)nλ2|(vz)nλ2+1[eβv|f(n+1)(v)|+meβz|f(n+1)(z)|s+1].
(23)
By combining the inequalities (20) and (23) via triangular inequality we get the required inequality.

It is interesting to see the following inequalities as a special case.

Corollary 2. By setting λ1=λ2 in (13), we get the following fractional integral inequality: |Γ(nλ1+1)[(CDu+λ1f)(z)+(CDvλ1f)(z)]((zu)nλ1f(n)(u)+(vz)nλ1f(n)(v))|(zu)nλ1+1eβu|f(n+1)(u)|+(vz)nλ1+1eβv|f(n+1)(v)|s+1+meβz|f(n+1)(z)|[(zu)nλ1+1+(vz)nλ1+1]s+1.

Remark 3. By setting s=1 the inequality will be of the form, |Γ(nλ1+1)[(CDu+λ1f)(z)+(CDvλ1f)(z)]((zu)nλ1f(n)(u)+(vz)nλ1f(n)(v))|(zu)nλ1+1eβu|f(n+1)(u)|+(vz)nλ1+1eβv|f(n+1)(v)|2+meβz|f(n+1)(z)|[(zu)nλ1+1+(vz)nλ1+1]2.

Remark 4. By setting λ1=λ2, β=0, s=1 and m=1, we will get Corollary 2.2 of [14].

Before going to the next theorem we observe the following result.

Lemma 1. Let f:[u,v]R, be a exponential (s,m)-convex function. If f is exponentially symmetric about u+v2, then the following inequality holds

f(u+v2)12s(eβzf(z))(1+m)z[u,v].
(24)

Proof. Since f is exponential (s,m)-convex we have

f(u+v2)12s[eβ(uτ+(1τ)v)f(uτ+(1τ)v)+meβ(u(1τ)+vτ)f(u(1τ)+vτ)].
(25)
Since f is symmetric about a+b2, therefore we get f(u+vz)=f(vτ+(1τ)u).
f(u+v2)12s[eβ(uτ+(1τ)v)f(uτ+(1τ)v)+meβ(u+vz)f(u+vz)].
(26)
By substituting z=(uτ+(1τ)v) where z[u,v], we get f(u+v2)12s(eβzf(z)+meβ(u+vz)f(u+vz)). Also f is exponentially symmetric about u+v2, therefore we have f(u+vz)=f(z) and inequality in (24) holds.

Theorem 2. Let f:IR be a real valued n-time differentiable function where n is a positive integer. If f(n) is a positive exponential (s,m)- convex and symmetric about u+v2, then for u,vI;u<v and λ1,λ21, the following inequality for Caputo fractional derivatives holds

h(β)2s2(1+m)(1nλ1+1+1nλ2+1)f(n)(u+v2)Γ(nλ2+1)(CDvλ21f)(u)2(vu)nλ2+1+Γ(nλ1+1)(CDu+λ11f)(v)2(vu)nλ1+1mf(n)(u)+f(n)(v)(s+1).
(27)
where h(β)=eβv for β<0 and h(β)=eβu for β0.

Proof. For z[u,v], we have

(za)nλ2(vu)nλ2.
(28)
Also f is exponential (s,m)-convex function, we have
f(n)(z)(zuvu)seβvf(n)(v)+(bxvu)seβumf(n)(u).
(29)
Multiplying (28) and (29) and then integrating with respect to z over [u,v], we have uv(zu)nλ2f(n)(z)dz(vu)nλ2(vu)s(uveβv(f(n)(v)(zu)s+eβumf(n)(u)(vz)s)dz). From which we have
Γ(nλ2+1)(CDvλ21f)(u)(vu)nλ2+1eβvf(n)(v)+eβumf(n)(u)s+1.
(30)
On the other hand for z[u,v] we have
(vz)nλ1(vu)nλ1.
(31)
Multiplying (29) and (31) and then integrating with respect to z over [u,v], we get uv(vz)nλ1f(n)(z)dz(vu)nλ1+1eβumf(n)(u)+eβvf(n)(v)s+1. From which we have
Γ(nλ1+1)(CDu+λ11f)(v)(vu)nλ1+1eβumf(n)(u)+eβvf(n)(v)s+1.
(32)
Adding (30) and (32) we get the second inequality. Γ(nλ2+1)(CDvλ21f)(u)2(vu)nλ2+1+Γ(nλ1+1)(CDu+λ11f)(v)2(vu)nλ1+1eβumf(n)(u)+eβvf(n)(v)s+1. Since f(n) is exponential s-convex and symmetric about u+v2 using Lemma ???, we have
f(n)(u+v2)12s(eβzfn(z)(1+m)),z[u,v].
(33)
Multiplying with (zu)nλ2 on both sides and then integrating over [u,v], we have
f(n)(u+v2)uv(zu)nλ2dz(1+m)h(β)2suv(zu)nλ2f(n)(z)dz.
(34)
By definition of Caputo fractional derivatives for exponential (s,m)-convex function, one can have
f(n)(u+v2)12(nλ2+1)(1+m)h(β)2sΓ(nλ2+1)(CDvλ21f)(u)2(vu)nλ2+1.
(35)
Multiplying (33) with (vz)nλ1, then integrating over [u,v], one can get
f(n)(u+v2)12(nλ1+1)(1+m)h(β)2sΓ(nλ1+1)(CDu+λ11f)(v)2(vu)nλ1+1.
(36)
Adding (35) and (36), we get the first inequality.

Corollary 3. If we put λ1=λ2 in (27), then we get h(β)2s(1+m)f(n)(u+v2)1(nλ1+1)Γ(nλ1+1)(2)(vu)λ1+1[(CDvλ1+1f)(u)+(CDu+λ1+1f)(v)]eβuf(n)(u)+eβvf(n)(v)s+1 \noindent where h(β)=eβv for β<0 and h(β)=eβu for β0.

Remark By setting γ=0, s=1 and s=1 in Theorem 3 we will get Theorem 2.3 of [14].

Autho Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflict of Interests

The authors declare no conflict of interest.

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