# Hermite–Jensen–Mercer Type Inequalities for Caputo Fractional Derivatives

Authors: Jinchao Zhao; Saad Ihsan Butt; Jamshed Nasir; Zhaobo Wang; Iskander Tlili
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this article, certain Hermite–Jensen–Mercer type inequalities are proved via Caputo fractional derivatives. We established some new inequalities involving Caputo fractional derivatives, such as Hermite–Jensen–Mercer type inequalities, for differentiable mapping hn whose derivatives in the absolute values are convex.


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## Body

## 1. Introduction

In recent years, inequality theory attracts many researchers due to its applications in our daily life and within the mathematics [1–9]. Let 0<x1≤x2≤⋯≤xn and let μ=μ1,μ2,…,μn be nonnegative weights such that ∑r=1nμr=1. The Jensen inequality [10] states that h is a convex function on the interval u,v; then,(1)h∑r=1nμrxr=∑r=1nμrhxr,where ∀xr∈u,v and all μr∈0,1,r=1,2,…,n.The Hermite–Hadamard inequality asserts that if a mappingh:J⊆R⟶R is a convex function on J with u,v∈J, u<v, then(2)hu+v2≤1v−u=∫uvhτdτ≤hu+hv2.The reverse direction in the above inequality holds whenh is concave.Theorem 1 (see [11]).
Ifh is a convex function on u,v, then(3)hu+v−∑r=1nμrxr=hu+hv−∑r=1nμrhxr,(4)∀xr∈u,vand allμr∈0,1,r=1,2,…,n.Inequality (3) is known as the Jensen–Mercer inequality. Recently, inequality (3) has been generalized, see ([12–15]). For more recent and related results connected with Jensen–Mercer inequality, see ([11, 16–18]).The previous era of fractional calculus is as old as the history of differential calculus. Several fractional operators are introduced that generalize ordinary integrals. However, the fractional derivatives have some basic properties than the corresponding classical ones. On the contrary, besides the smooth requirement, the Caputo derivative does not coincide with the classical derivative [19]. Caputo fractional derivatives are introduced by the Italian mathematician Caputo in 1967. Since then, a lot of research involves Caputo fractional derivatives [20–22].The Caputo fractional derivatives are defined as in [23–26].Definition 1.
Supposeα>0 and α∉1,2,3,…,n=α+1,andh∈Cnu,v. The Caputo fractional derivatives of order α are defined as follows:(5)Du+αchz=1Γn−α∫uzhnτz−τα−n+1dτ;z>u,Dv−αchz=−1nΓn−C∫zvhnττ−zα−n+1dτ;z<v.
Ifα=n∈1,2,3,… and usual derivatives of h of order n exist, then the Caputo fractional derivatives Dcu+αhz coincide with hnz.
Specifically, we get(6)Dcu+0hz=Dcv−0hz=hz,where n=1andα=0.
In this article, by using the Jensen–Mercer inequality, we proved Hermite–Hadamard’s inequalities for fractional integrals and established some Hermite–Hadamard type inequalities for differentiable mappings whose derivatives in absolute value are convex.


## 2. Hermite–Hadamard–Mercer Type Inequalities for Caputo Fractional Derivatives

By using Jensen–Mercer inequalities, Hermite–Hadamard type inequalities can be expressed in Caputo fractional derivatives as follows.Theorem 2.
Suppose that a positive functionh:u,v⟶R with 0≤u<v and h∈Cnu,v. If hn is a convex function on u,v, then the following inequalities for Caputo fractional derivatives hold:(7)hnu+v−x+y2≤hnu+hnv−Γn−α+12y−xn−αDcx+αhy+−1nDcy−αhx≤hnu+hnv−hnx+y2,∀x,y∈u,v,α>0, and Γ⋅ is the gamma function.Proof.
Using the Jensen–Mercer inequality, we get(8)hnu+v−u1+v12≤hnu+hnv−hnu1+hnv12,∀u1,v1∈a,b.
Now, by changing of variablesu1=τx+1−τy and v1=1−τx+τy, ∀x,y∈u,v, and τ∈0,1 in (8), we get(9)hnu+v−x+y2≤hnu+hnv−hnτx+1−τy+hn1−τx+τy2.
Multiplying both sides byτn−α−1 and then integrating with respect to τ over 0,1, we get(10)1n−αhnu+v−x+y2≤1n−αhnu+hnv−12∫01τn−α−1hnτx+1−τy+hn1−τx+τydτ.
After simplification, we get(11)hnu+v−x+y2≤hnu+hnv−Γn−α+12y−xn−αDcx+αhy+−1nDcy−αhx,and so the first inequality of (2.1) is proved.
Now, for the proof of the second inequality of (2.1), we first note that ifhn is a convex function, then for τ∈0,1, it gives(12)hnx+y2=hnτx+1−τy+1−τx+τy2≤hnτx+1−τy+hn1−τx+τy2.
Multiplying both sides byτn−α−1 and then integrating the resulting inequality with respect to τ over 0,1, we get(13)1n−αhnx+y2≤12∫01τn−α−1hnτx+1−τy+hn1−τx+τyhnx+y2≤Γn−α+12y−xn−αDcx+αhy+−1nDcy−αhx.
Multiplying by−1 and adding hnu+hnv both sides in (13), we get the second inequality of (2.1), which completes the proof.Theorem 3.
Suppose that a positive functionh:u,v⟶R with 0≤u<v and h∈Cnu,v. If hn is a convex function on u,v, then the following inequalities for Caputo fractional derivatives hold:(14)hnu+v−x+y2≤2n−α−1Γn−α+1y−xn−α×Dcu+v−x+y/2+αhu+v−x+−1nDcu+v−x+y/2−αhu+v−y≤hnu+hnv−hnx+hny2,where ∀x,y∈u,v,α>0, and Γ⋅ is the gamma function.Proof.
To prove the first part of inequality, we use convexity ofhn to get(15)hnu+v−u1+v12=hnu+v−u1+u+v−v12≤hnu+v−u1+hnu+v−v12,for all u1,v1∈u,v. Now by writing variables u1=τ/2x+2−τ/2y and v1=2−τ/2x+τ/2y for x,y∈u,v and τ∈0,1, we get(16)2hnu+v−x+y2≤hnu+v−τ2x+2−τ2y+hnu+v−2−τ2x+τ2y.
Multiplying both sides of (16) by τn−α−1 and then integrating with respect to τ over 0,1, we get(17)2hnu+v−x+y2∫01τn−α−1dτ≤∫01τn−α−1hnu+v−τ2x+2−τ2y+hnu+v−2−τ2x+τ2ydτ.
Further simplifying gives(18)hnu+v−x+y2≤2n−α−1Γn−α+1y−xn−αDcu+v−x+y/2+αhu+v−x+−1nDcu+v−x+y/2−αhu+v−y,and so the first inequality of (14) is proved.
Now, for the proof of the second inequality of (14), we first note that if hn is a convex function, then by employing Jensen–Mercer inequality (3) for τ∈0,1 gives(19)hnu+v−τ2x+2−τ2y≤hnu+hnv−τ2hnx+2−τ2hny,(20)hnu+v−2−τ2x+τ2y≤hnu+hnv−2−τ2hnx+τ2hny.
By adding the inequalities of (19) and (20), we get(21)hnu+v−τ2x+2−τ2y+hnu+v−2−τ2x+τ2y≤2hnu+hnv−hnx+hny.
Multiplying both sides byτn−α−1 and then integrating over τ∈0,1, we get(22)∫01τn−α−1hnu+v−τ2x+2−τ2y+hnu+v−2−τ2x+τ2ydτ≤2hnu+hnv−hnx+hny∫01τn−α−1dτ.
After simplification, we get(23)2n−α−1Γn−α+1y−xn−α×Dcu+v−x+y/2+αhu+v−x+−1nDcu+v−x+y/2−αhu+v−y≤hnu+hnv−hnx+hny2.
Now, concatenating (18) and (23), we get (14).
Now, we introduce some new lemmas involving Caputo fractional derivatives.Lemma 1.
Suppose thath:u,v⟶R is a differentiable mapping on u,v with u<v and h∈Cn+1u,v, then the following equality for Caputo fractional derivatives holds:(24)hnu+v−x+hnu+v−y2−Γn−α+12y−xn−α×Dcu+v−y+αhu+v−x+−1nDcu+v−x−αhu+v−y=y−x2∫01τn−α−1−τn−αhn+1u+v−τx+1−τydτ,where ∀x,y∈u,v,α>0,τ∈0,1, and Γ⋅ is the gamma function.Proof.
It suffices to note that(25)I=y−x2I1−I2,where(26)I1=∫01τn−αhn+1u+v−τx+1−τydτ=hnu+v−xy−x−n−αy−x∫01τn−α−1hnu+v−τx+1−τydτ=hnu+v−xy−x−Γn−α+1y−xn−α+1−1nDcu+v−x−αhu+v−yI2=∫011−τn−αfn+1u+v−τx+1−τydτ=−hnu+v−yy−x+n−αy−x∫01∫011−τn−α−1hnu+v−τx+1−τydτ=−hnu+v−yy−x+Γn−α+1y−xn−α+1Dcu+v−y+αhu+v−x,and replacing the values of I1 and I2 in (25), we get (24).Remark 1.
If we takex=u and y=v in Lemma 1, then it reduces to Remark 2.5 in [25].Lemma 2.
Suppose thath:u,v⟶R is a differentiable mapping on u,v with u<v and h∈Cn+1u,v, then the following equality for Caputo fractional derivatives holds:(27)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αDcu+v−x+y/2+αhu+v−x+−1nDcu+v−x+y/2−αhu+v−y=y−x4∫01τn−αhn+1u+v−τ2x+2−τ2ydτ−∫01τn−αhn+1u+v−2−τ2x+τ2ydτ.Proof.
It suffices to note that(28)I=y−x4I1−I2,(29)I1=∫01tn−αhn+1u+v−τ2x+2−τ2ydτ=2y−xhnu+v−x+y2−2n−αy−x∫01τn−α−1hnu+v−τ2x+2−τ2ydτ=2y−xhnu+v−x+y2−−1n2n−α+1Γn−α+1y−xn−α+1Dcu+v−x+y/2−αhu+v−y,(30)I2=∫01τn−αhn+1u+v−2−τ2x+τ2ydτ=−2y−xhnu+v−x+y2+2n−αy−x∫01τn−α−1hnu+v−2−τ2x+τ2ydτ=−2y−xhnu+v−x+y2+2n−α+1Γn−α+1y−xn−α+1Dcu+v−x+y/2+αhu+v−x, and combining (29) and (30) with (28), we get (27).Remark 2.
If we takex=u and y=v in Lemma 2, then it reduces to Lemma 2 in [24].Theorem 4.
Suppose thath:u,v⟶R is a differentiable mapping on u,v with u<v and h∈Cn+1u,v. If hn+1 is convex function on u,v, then the following inequality for Caputo fractional derivatives holds:(31)hnu+v−x+hnu+v−y2−Γn−α+12y−xn−αcDu+v−y+αfu+v−x+−1ncDu+v−x−αhu+v−y≤y−xn−α+11−12n−αhn+1u+hn+1v−hn+1x+hn+1y2,where ∀x,y∈u,v,α>0,τ∈0,1, and Γ. is the gamma function.Proof.
By using Lemma1 and the Jensen–Mercer inequality, we get(32)hnu+v−x+hnu+v−y2−Γn−α+12y−xn−αcDu+v−y+αhu+v−x+−1ncDu+v−x−αhu+v−y≤y−x2∫01τn−α−1−τn−αhn+1u+v−τx+1−τydτ≤y−x2∫01τn−α−1−τn−αhn+1u+hn+1v−τhn+1x+1−τhn+1ydτ≤y−x2I1+I2,where(33)I1=∫0121−τn−α−τn−αhn+1u+hn+1v−τhn+1x+1−τhn+1y=hn+1u+hn+1v1n−α+1−2−n+αn−α+1−hn+1x1n−α+1n−α+2−2−n+α−2n−α+2+hn+1y1n−α+2−2−n+α−1n−α+1I2=∫121τn−α−1−τn−αhn+1u+hn+1v−τhn+1x+1−τhn+1y=hn+1u+hn+1v1n−α+1−2−n+αn−α+1−hn+1x1n−α+2−2−n+α−1n−α+1+hn+1y1n−α+1n−α+2−2−n+α−1n−α+1,and putting the values of I1 and I2 in (32), we get (31).Remark 3.
If we takex=u and y=v in Theorem 4, then it reduces to Corollary 2.7 in [25].Theorem 5.
Suppose thath:u,v⟶R is a differentiable mapping on u,v with u<v and h∈Cn+1u,v. If hn+1 is a convex function on u,v, then the following inequality for Caputo fractional derivatives holds:(34)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x2n−α+1hn+1u+hn+1b−hn+1x+hn+1y2,where ∀x,y∈u,v,α>0,τ∈0,1, and Γ⋅ is the gamma function.Proof.
By using Lemma2 and the Jensen–Mercer inequality, we get(35)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x4∫01τn−αhn+1u+v−τ2x+2−τ2ydτ+∫01τn−αhn+1u+v−2−τ2x+τ2ydτ≤y−x4∫01τn−αhn+1u+hn+1v−τ2hn+1x+2−τ2hn+1ydτ+∫01τn−αhn+1u+hn+1v−2−τ2hn+1x+τ2hn+1ydτ≤y−x2n−α+1hn+1u+hn+1v−hn+1x+hn+1y2,which completes the proof.Theorem 6.
Suppose thath:u,v⟶R is a differentiable mapping on u,v with u<v and h∈Cn+1u,v. If hn+1q is a convex function on u,vand q>1, then the following inequality for Caputo fractional derivatives holds:(36)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x164np−αp+11/p41/q⋅2hn+1u+hn+1v−31/q+1hn+1x+hn+1y,where ∀x,y∈u,v,α>0,τ∈0,1, and Γ⋅ is the gamma function.Proof.
By using Lemma2 and applying the famous Hölder integral inequality, we get(37)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x4∫01τn−αpdτ1/p.∫01hn+1u+v−τ2x+2−τ2yqdτ1/q+∫01τn−αpdτ1/p.∫01τn+1u+v−2−τ2x+τ2yqdτ1/q≤y−x41np−αp+11/p×∫01hn+1uq+hn+1vq−τ2hn+1xq+2−τ2hn+1yqdτ1/q+∫01hn+1uq+hn+1vq−2−τ2hn+1xq+τ2hn+1yqdτ1/q≤y−x41np−αp+11/phn+1uq+hn+1vq−hn+1x|q+3hn+1y|q41/q+hn+1uq+hn+1vq−3hn+1xq+hn+1yq41/q.
By applying Minkowski’s inequality, we get(38)y−x164np−αp+11/p4hn+1uq+hn+1vq−hn+1xq+3hn+1yq1/q+4hn+1uq+hn+1vq−3hn+1xq+hn+1yq1/q≤y−x164np−αp+11/p41/q.2hn+1u+hn+1v−31/q+1hn+1x+hn+1y,which completes the proof.


## 3. New Hölder’s and Improved Iscan Inequalities

Theorem 7.
Suppose thath:u,v⟶R is a differentiable mapping on u,v with u<v and h∈Cn+1u,v. If hn+1q is a convex function on u,vand q>1, then the following inequality for Caputo fractional derivatives holds:(39)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x41n−αp+1n−αp+21/p⋅12hn+1uq+hn+1vq−112hn+1xq+512hn+1yq1/q+1n−αp+21/p12hn+1uq+hn+1vq−16hn+1xq+13hn+1yq1/q+1n−αp+1n−αp+21/p12hn+1uq+hn+1vq−512hn+1xq+112hn+1yq1/q+1n−αp+21/p×12hn+1uq+hn+1vq−13hn+1xq+16%hn+1yq1/q,where ∀x,y∈u,v,α>0,τ∈0,1, and Γ⋅ is the gamma function.Proof.
By using Lemma2 with Jensen–Mercer inequality and applying the Hölder–Iscan integral inequality (Theorem 1.4 [27]), we get(40)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x4∫011−ττnp−αpdτ1/p∫011−τhn+1u+v−τ2x+2−τ2yqdτ1/q+∫01τnp−αp+1dτ1/p∫01τhn+1u+v−τ2x+2−τ2yqdτ1/q+∫011−ττnp−αpdτ1/p∫011−τfn+1u+v−2−τ2x+τ2yqdτ1/q+∫01τnp−αp+1dτ1/p∫01τhn+1u+v−2−τ2x+τ2yqdτ1/q.
By the convexity ofhn+1|q(41)hn+1u+v−τ2x+2−τ2yq≤hn+1uq+hn+1vq−τ2hn+1xq+2−τ2hn+1yq.
By using calculus tools, one can have required result.Theorem 8.
Suppose thath:u,v⟶R is a differentiable mapping on u,v with u<v and h∈Cn+1u,v. If hn+1q is a convex function on u,v, p>1, and q=p/p−1, then the following inequality for Caputo fractional derivatives holds:(42)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x41n−α+1n−α+21−1/q×hn+1uq+hn+1vqn−α+1n−α+2−hn+1xq2n−α+2n−α+3+hn+1yq2n−α+1n−α+2n−α+3+1n−α+21−1/q×hn+1uq+hn+1vqn−α+2−hn+1xq2n−α+3+hn+1yq2n−α+2n−α+31/q+1n−α+1n−α+21−1/q×hn+1aq+hn+1vqn−α+1n−α+2−n−α+5hn+1xq2n−α+1n−α+2n−α+3+hn+1yq2n−α+2n−α+31/q+1n−α+21−1/q×hn+1aq+hn+1vqn−α+2−n−α+4hn+1xq2n−α+2n−α+3+hn+1yq2n−α+31/q,where ∀x,y∈u,v,α>0,τ∈0,1,andΓ⋅is the gamma function.Proof.
By using Lemma2 with Jensen–Mercer inequality and applying the improved power-mean integral inequality (Theorem 1.5 [27]), we get(43)hnu+v−x+y2−2n−α−1Γn−α+1y−xn−αcDu+v−x+y/2+αhu+v−x+−1ncDu+v−x+y/2−αhu+v−y≤y−x41n−α+1n−α+21−1/q×hn+1uq+hn+1vqn−α+1n−α+2−hn+1xq2n−α+2n−α+3+hn+1yq2n−α+1n−α+2n−α+31/q+1n−α+21−1/qhn+1uq+hn+1vqn−α+2−hn+1xq2n−α+3+hn+1yq2n−α+2n−α+31/q+1n−α+1n−α+21−1/qhn+1uq+hn+1vqn−α+1n−α+2−n−α+5hn+1xq2n−α+1n−α+2n−α+3+hn+1yq2n−α+2n−α+31/q+1n−α+21−1/q×hn+1uq+hn+1vqn−α+2−n−α+4hn+1xq2n−α+2n−α+3+hn+1yq2n−α+31/q.
After some simplifications, we get required result.


## 4. Conclusion

In this paper, we presented Hermite–Hadamard–Mercer inequalities for convex functions via Caputo fractional derivatives. We also developed some new bounds using Hölder–Iscan and improved power-mean integral inequalities. Our results will attract attentions of many researchers working in the field of inequalities and enable them to think further for other generalized convex functions. One may think to extend these results for higher order convex functions. The works above can also build up for a convex function of two variables. For further directions, we refer to [28–33].



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