Abstract
The Hermite–Hadamard inequality is the first principal result for convex functions defined on a interval of real numbers with a natural geometrical interpretation and a loose number of applications for particular inequalities. In this paper we proposed the Hermite–Hadamard and midpoint type inequalities for functions whose first and second derivatives in absolute value are s-convex through the instrument of generalized fractional integral operator and a considerable amount of results for special means which can naturally be deduced.
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1 Introduction
The Hermite–Hadamard inequality, which is the first fundamental result for convex mappings with a natural geometrical interpretation and many applications, has drawn attention much interest in elementary mathematics. A number of mathematicians have devoted their efforts to generalise, refine, counterpart and extend it for different classes of functions such as using convex mappings.
The inequalities discovered by Hermite and Hadamard for convex functions are considerable significant in the literature (see, e.g., [6, 8, 13, p.137]). These inequalities state that if \( f:I\rightarrow {\mathbb {R}}\) is a convex function on the interval I of real numbers and \(a,b\in I\) with \(a<b\), then
Both inequalities hold in the reversed direction if f is concave. We note that Hadamard’s inequality may be regarded as a refinement of the concept of convexity and it follows easily from Jensen’s inequality. Hadamard’s inequality for convex functions has received renewed attention in recent years and a remarkable variety of refinements and generalizations have been found (see, for example, [2, 3, 5, 7, 10, 12, 15, 17,18,19,20, 22]) and the references cited therein.
The overall structure of the study takes the form of five sections including introduction. The remainder of this work is organized as follows: In Sect. 2, the generalised version of fractional integral operator are summarised, along with the very first results. In Sect. 3 the Hermite–Hadamard type inequalities for generalized fractional integral operators are introduced while in Sects. 4 and 5 midpoint type inequalities for functions whose first and second derivatives in absolute value are s-convex with generalized fractional integral operators are presented and we also provide some corollary for theorems. Some conclusions and further directions of research are discussed in Sect. 6.
2 Preliminaries
Now we reviewed some definitions and theorems which will be used in the proof of our main cumulative results.
Definition 1
[4] Let \(s\in (0,1].\) A function \(f:[0,\infty )\mathbb { \rightarrow }[0,\infty )\) is said to be s-convex (in the second sense), or that f belongs to the class \(K_{s}^{2},\) if
for all \(x,y\in [0,\infty )\) and \(\lambda \in [0,1].\)
An s-convex function was introduced in Breckner’s paper [4] and a number of properties and connections with s-convexity in the first sense were discussed in paper [9].
In addition to this, Raina [14] defined the following results connected with the general class of fractional integral operators.
where the coefficients \(\sigma \left( k\right) \)\(\left( k\in {\mathbb {N}}_{0}= \mathbb {N\cup }\left\{ 0\right\} \right) \) is a bounded sequence of positive real numbers and \(\mathcal {R}\) is the set of real numbers. With the help of (2.1), Raina [14] and Agarwal et al. [1] defined the following left-sided and right-sided fractional integral operators, respectively, as follows:
where \(\lambda ,\rho >0,\omega \in {\mathbb {R}}\), and \(f\left( t\right) \) is such that the integrals on the right side exists.
It is easy to verify that \(\mathcal {J}_{\rho ,\lambda ,a+;\omega }^{\alpha }f(x)\) and \(\mathcal {J}_{\rho ,\lambda ,b-;\omega }^{\alpha }f(x)\) are bounded integral operators on \(L\left( a,b\right) \), if
In fact, for \(f\in L\left( a,b\right) \), we have
and
where
The importance of these operators stems indeed from their generality. Many useful fractional integral operators can be obtained by specializing the coefficient \(\sigma \left( k\right) \). Here, we just point out that the classical Riemann–Liouville fractional integrals \(I_{a^{+}}^{\alpha }\) and \( I_{b^{-}}^{\alpha }\) of order \(\alpha \) defined by (see [11])
and
follow easily by setting
in (2.2) and (2.3), and the boundedness of (2.7) and (2.8) on \(L\left( a,b\right) \) is also inherited from (2.5) and (2.6) (see [1]).
Yaldiz and Sarikaya [21] gave the following Hermite–Hadamard inequality for the generalized fractional integral operators:
Theorem 1
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be a convex function on \(\left[ a,b\right] \) with \(a<b\), then the following inequalities for fractional integral operators hold
with \(\lambda >0.\)
The main purpose of this paper is to introduce new type Hermite Hadamard and midpoint integral inequalities with the aid of generalized fractional integral operators and establish some results connected with the them.
3 Hermite–Hadamard type inequalities for generalized fractional integral operators
In this section, we will present a theorem for Hermite–Hadamard type inequalities with generalized fractional integral operators which is the generalization of previous work.
Theorem 2
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be a function with \(0\le a<b\) and \(f\in L_{1}\left[ a,b\right] .\) If f is a s-convex function on \(\left[ a,b\right] ,\) then we have the following inequalities for generalized fractional integral operators:
where \(\sigma _{0,s}(k)=\frac{\sigma (k)}{\rho k+s+\lambda },\)\(k=0,1,2,...\) and
Proof
Since f is s-convex function on \(\left[ a,b\right] ,\) we have for \(x,y\in \left[ a,b\right] \)
For \(x=\frac{t}{2}a+\frac{2-t}{2}b\) and \(y=\frac{2-t}{2}a+\frac{t}{2}b,\) we obtain
Multiplying both sides of (3.2) by \(t^{\lambda -1}\mathcal {F}_{\rho ,\lambda }^{\sigma }\left[ \omega \left( \frac{b-a}{2}\right) ^{\rho }t^{\rho }\right] ,\) then integrating the resulting inequality with respect to t over \(\left[ 0,1\right] ,\) we get
For \(u=\frac{t}{2}a+\frac{2-t}{2}b\) and \(v=\frac{2-t}{2}a+\frac{t}{2}b,\) we obtain
and the first inequality is proved.
For the proof of the second inequality (3.1), we first note that if f is a s-convex function, it yields
and
By adding these inequalities together, one has the following inequality:
Then multiplying both sides of (3.3) by \(t^{\lambda -1}\mathcal {F} _{\rho ,\lambda }^{\sigma }\left[ \omega \left( \frac{b-a}{2}\right) ^{\rho }t^{\rho }\right] \) and integrating the resulting inequality with respect to t over \(\left[ 0,1\right] ,\) we obtain
That is,
Hence, the proof is completed. \(\square \)
Corollary 1
If we take \(s=1\) in Theorem 2, then we have the following inequality
Corollary 2
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Theorem 2, then we have the following inequality for Riemann–Lioville fractional integral operators
where
Remark 1
Choosing \(s=1\) in Corollary 2, then we have the following inequality for Riemann–Lioville fractional integral operators
which was given by Sarikaya and Yıldırım [16].
4 Midpoint type inequalities for differentiable functions with generalized fractional integral operators
In this section, firstly we need to give a lemma for differentiable functions which will help us to prove our main theorems. Then, we present some theorems which are the generalization of those given in earlier works.
Lemma 1
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( f^{\prime }\in L\left[ a,b\right] ,\) then we have the following identity for generalized fractional integral operators:
Proof
Integrating by parts gives
and similarly we get
By subtracting equation (4.3) from (4.2), we have
By re-arranging the last equality above, we get the desired result. \(\square \)
Theorem 3
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( \left| f^{\prime }\right| \) is s-convex function in the second sense, then we have the following inequality for generalized fractional integral operators:
where \(\sigma _{1,s}(k)=\frac{\sigma (k)}{\rho k+s+\lambda +1},\)\(k=0,1,2,\ldots \) and \(A_{1}(\lambda ,s)\) is defined as in Theorem 2.
Proof
Taking modulus of (4.1) and using s-convexity of \(\left| f^{\prime }\right| \), we have
where \(\sigma _{1,s}(k)\) and \(A_1(\lambda ,s)\) are defined above. Thus, the proof is completed. \(\square \)
Corollary 3
If we take \(s=1\) in Theorem 3, then we have
Corollary 4
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Theorem 3, then we have the following inequality for Riemann–Lioville fractional integral operators
where \(B_{1}(\alpha ,s)\) is defined as in Corollary 2.
Remark 2
Chosing \(s=1\) in Corollary 4, we obtain following inequality
which was given by Sarikaya and Yıldırım in [16].
Theorem 4
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( \left| f^{\prime }\right| ^{q},\)\(q>1,\) is s-convex function in the second sense, then we have the following inequality for generalized fractional integral operators:
where \(\frac{1}{p}+\frac{1}{q}=1\) and
Proof
Taking modulus of (4.1) and using well-known Hölder inequality, we obtain
Since \(\left| f^{\prime }\right| ^{q},\)\(q>1,\) is s-convex, we have
and similarly
By substituting inequalities (4.8) and (4.9) into (4.7), we get the desired result (4.6). \(\square \)
Corollary 5
If we take \(s=1\) in Theorem 4, then we get
Proof
The proof of the first inequality in (4.10) is obvious. For the proof of second inequality, let \(a_{1}=3\left| f^{\prime }\left( a\right) \right| ^{q},\)\(b_{1}=\left| f^{\prime }\left( b\right) \right| ^{q},\)\(a_{2}=\left| f^{\prime }\left( a\right) \right| ^{q}\) and \( b_{2}=3\left| f^{\prime }\left( b\right) \right| ^{q}.\) Using the fact that,
the desired result can be obtained straightforwardly. \(\square \)
Corollary 6
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Theorem 4, then we have the following inequality for Riemann–Liouville fractional integral operators
Remark 3
Choosing \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Corollary 5, we have the following inequality
which is the same result given by Sarikaya and Yıldırım [16].
Theorem 5
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( \left| f^{\prime }\right| ^{q},\)\(q\ge 1,\) is s-convex function in the second sense, then we have the following inequality for generalized fractional integral operators:
where \(\sigma _{1,s}(k),\)\(k=0,1,2,\ldots \) and \(A_1(\lambda ,s)\) are defined as in Theorem 3.
Proof
Taking modulus of (4.1), using well-known power mean inequality and s -convexity of \(\left| f^{\prime }\right| ^{q}\), we obtain
which completes the proof. \(\square \)
Corollary 7
If we take \(s=1\) in Theorem 5, then we get
Corollary 8
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Theorem 5, then we have the following inequality for Riemann–Liouville fractional integral operators
where \(B_1(\alpha ,s)\) is defined as in Corollary 4.
Remark 4
Choosing \(s=1\) in Corollary 8, we have the following inequality
which is the same result given by Sarikaya and Yıldırım [16].
5 Midpoint type inequalities for twice differentiable functions with generalized fractional integral operators
In Sect. 5, firstly we need to give a lemma for twice differentiable functions which will help us to prove our main theorems. Then, we present some theorems which are the generalization of those given in earlier works.
Lemma 2
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be twice differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( f^{\prime \prime }\in L\left[ a,b\right] ,\) then we have the following identity for generalized fractional integral operators:
Proof
Integrating by parts we have
Similarly, we have
Combining of (5.2), (5.3) and (5.4), we obtain
Mutiplying both sides of (5.5) by \(\frac{(b-a)^{2}}{8\mathcal {F}_{\rho ,\lambda +1}^{\sigma }\left[ w\left( b-a\right) ^{\rho }\right] }\) completes the proof. \(\square \)
Theorem 6
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be twice differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( \left| f^{\prime \prime }\right| \) is s-convex function in the second sense, then we have the following inequality for generalized fractional integral operators:
where \(\sigma _{2,s}(k)=\frac{\sigma (k)}{\rho k+\lambda +s+2},\)\( k=0,1,2,... \) and \(A_2(\lambda ,s)\) is defined by
Proof
Taking modulus both sides of (3.1) and using s-convexity of \( \left| f^{\prime \prime }\right| \), we obtain
Thus, the proof is completed. \(\square \)
Corollary 9
If we choose \(s=1\) Theorem 6, then we have the following inequality
Corollary 10
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Theorem 6, then we have the following inequality for Riemann–Liouville fractional integral operators
where
which is the same result in [12]
Theorem 7
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be twice differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( \left| f^{\prime \prime }\right| ^{q},\)\(q>1,\) is s-convex function in the second sense, then we have the following inequality for generalized fractional integral operators:
where \(\frac{1}{p}+\frac{1}{q}=1\) and \(C_{2}(\lambda ,p )\) is defined by
Proof
Taking modulus both sides of (3.1) and using well-known Hölder inequality, we have
Since\(\left| f^{\prime \prime }\right| ^{q}\) is s-convex function in the second sense, we get
and similarly
If we substitute the inequalities (5.11) and (5.12) in (5.10), we obtain desired result (5.8).\(\quad \square \)
Corollary 11
If we choose \(s=1\) Theorem 7, then \(\left| f^{\prime \prime }\right| ^{q},\)\(q>1,\) is convex and we have the following inequality
\(\frac{1}{p}+\frac{1}{q}=1\) and \(C_{2}(p,\lambda )\) is defined as in (5.9).
Proof
The proof can be done by following the similar steps with Corollary 5. \(\square \)
Corollary 12
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Theorem 7, then we have the following inequality for Riemann–Liouville fractional integral operators
which is the same result in [12].
Corollary 13
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Corollary 11, we have
Theorem 8
Let \(f:\left[ a,b\right] \rightarrow \mathbb {R} \) be twice differentiable function on \(\left( a,b\right) \) with \(a<b.\) If \( \left| f^{\prime \prime }\right| ^{q},\)\(q\ge 1,\) is s-convex function in the second sense, then we have the following inequality for generalized fractional integral operators:
where \(\sigma _{2,s}(k),\)\(k=0,1,2,\ldots \) and \(A_2(\lambda ,s)\) are defined as in Theorem 6.
Proof
Taking modulus both sides of (3.1) and using well-known power mean inequality, we have
By a simple computation, we obtain
Using the equality (5.14) and s-convexity of \(\left| f^{\prime \prime }\right| ^{q},\) we have
which completes the proof. \(\square \)
Corollary 14
If we choose \(s=1\) Theorem 5, then \(\left| f^{\prime \prime }\right| ^{q},\)\(q\ge 1,\) is convex and we have the following inequality
Corollary 15
If we take \(\lambda =\alpha ,\ \sigma (0)=1,\ w=0\) in Theorem 7, then we have the following inequality for Riemann–Liouville fractional integral operators
which is the same result in [12].
6 Concluding remarks
In this study, we consider the Hermite–Hadamard and midpoint type inequalities for functions whose first and second derivatives in absolute value are s-convex and related results to establish new type inequalities involving generalised fractional integral operator. The results presented in this study would provide generalizations of those given in earlier works.
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Budak, H., Usta, F., Sarikaya, M.Z. et al. On generalization of midpoint type inequalities with generalized fractional integral operators. RACSAM 113, 769–790 (2019). https://doi.org/10.1007/s13398-018-0514-z
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DOI: https://doi.org/10.1007/s13398-018-0514-z