Bol. Soc. Mat. Mex. DOI 10.1007/s40590-016-0087-9 ORIGINAL ARTICLE

On some fractional Hermite–Hadamard inequalities via s-convex and s-Godunova–Levin functions and their applications Zhuoyan Gao1 · Mengmeng Li2 · JinRong Wang2

Received: 9 August 2015 / Accepted: 28 January 2016 © Sociedad Matemática Mexicana 2016

Abstract In this paper, we establish two fractional integral equalities involving once and twice differential functions. Then, we apply such equalities to give some fractional Hermite–Hadamard inequalities via s-convex and s-Godunova–Levin functions. Some applications to special means of positive real numbers are also given. Keywords Fractional Hermite–Hadamard inequalities · s-Convex function · s-Godunova–Levin function Mathematics Subject Classification

26A33 · 26A51 · 26D15

1 Introduction Throughout this paper we will use the following notations and conventions. Let I = [a, b] ⊆ R + ∪ {0}. Let f : I → R be a convex function on I . Then f satisfies the following well-known Hermite–Hadamard inequality

This work is supported by National Natural Science Foundation of China (11201091).

B

JinRong Wang [email protected] Zhuoyan Gao [email protected] Mengmeng Li [email protected]

1

College of Applied Mathematics, Shanxi University of Finance and Economics, Taiyuan 030031, Shanxi, People’s Republic of China

2

Department of Mathematics, Guizhou University, Guiyang 550025, Guizhou, People’s Republic of China

Z. Gao et al.

 f

a+b 2

 ≤

1 b−a



b

f (x) dx ≤

a

f (a) + f (b) . 2

For recent results concerning fractional Hermite–Hadamard inequalities and concepts of all kinds of convexity see [1–3,6,8,10,13,15,16,18] and references therein. Very recently, Tunç [17] established the following equality. Lemma 1.1 (See [17, Lemma 1]) Let f : I → R be a differentiable function and f  ∈ L[u, v]. Then (v − x)(v f (v) − u f (x)) + (x − u)(v f (x) − u f (u)) 1 − 2 (v − u) v−u  (v − x)2 1 (tu + (1 − t)v) f  (t x + (1 − t)v) dt = (v − u)2 0  (x − u)2 1 + (tv + (1 − t)u) f  (t x + (1 − t)u) dt (v − u)2 0



v

f (μ) dμ

u

for each t ∈ [0, 1] and x ∈ [u, v]. On the basis of the above equality, the author established some inequalities of Hadamard-like via convex and s-convex functions. In this paper, we extend Lemma 1.1 to fractional case. That is, we establish the following lemmas. Lemma 1.2 Let f : [a, b] → R be a differentiable function and f  ∈ L[a, b]. For any 0 < α ≤ 1 and a < x < b, the following equality for fractional integrals holds: (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)] (b − a)2   a (α + 1) b α α − R L Jx − f (a) + R L Jx + f (b) (b − a)2 (x − a)α−1 (b − x)α−1  (x − a)2 1 α (at + x) f  (t x + (1 − t)a) dt = (b − a)2 0  (b − x)2 1 − (b(1 − t)α + x) f  (tb + (1 − t)x) dt, (b − a)2 0 where the symbols R L Jaα+ f and R L Jbα− f denote the left-sided and right-sided Riemann–Liouville fractional integrals of the order α ∈ R + (see [9]) defined by ( R L Jaα+ ( R L Jbα−

 x 1 f )(x) = (x − t)α−1 f (t) dt, (0 ≤ a < x ≤ b), (α) a  b 1 f )(x) = (t − x)α−1 f (t) dt, (0 ≤ a ≤ x < b), (α) x

respectively, and (·) is the Gamma function.

On some fractional Hermite–Hadamard inequalities via s-convex…

Lemma 1.3 Let f : [a, b] → R be twice differentiable function and f  ∈ L[a, b]. For any 0 < α ≤ 1 and a < x < b, the following equality for fractional integrals holds:   (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a) 1 + (b − a)2 b−x x −a   x (α + 1)(α + 1) b α α − J f (b) + J f (a) + − R L R L x x (b − a)2 (b − x)α+1 (x − a)α+1  1 (x − b) = (t α+1 x + tb) f  (t x + (1 − t)b) dt (b − a)2 0  (a − x) 1 + ((1 − t)α+1 b + (1 − t)x) f  (ta + (1 − t)x) dt. (b − a)2 0 In the present paper, we show some new fractional Hermite–Hadamard type inequalities involving left-sided and right-sided Riemann–Liouville fractional integrals using our equalities in Lemmas 1.2 and 1.3 via Definitions 1.4 and 1.5. Finally, some applications to special means of real numbers are given as well. To end this section, we collect the following definitions. Definition 1.4 (See [4,5]) The function f : I → R is said to be s-convex, where s ∈ (0, 1], if for every x, y ∈ I and λ ∈ [0, 1], we have f (λx + (1 − λ)y) ≤ λs f (x) + (1 − λ)s f (y). Definition 1.5 (See [11, Definition 3]) The function f : I → R is said to be s-Godunova–Levin function, where s ∈ (0, 1], if for every x, y ∈ I and λ ∈ (0, 1), we have f (λx + (1 − λ)y) ≤

f (y) f (x) + . λs (1 − λ)s

2 Proofs of Lemmas 1.2 and 1.3 2.1 Proofs of Lemma 1.2 Proof It suffices to note that  (x − a)2 1 α (at + x) f  (t x + (1 − t)a) dt (b − a)2 0  (b − x)2 1 − (b(1 − t)α + x) f  (tb + (1 − t)x) dt (b − a)2 0 := I1 − I2 .

I =

Integrating by parts

(1)

Z. Gao et al.

 (x − a)2 1 α (at + x) f  (t x + (1 − t)a) dt (b − a)2 0  (x − a)[(a + x) f (x) − x f (a)] aα(x − a) 1 α−1 = − t f (t x + (1 − t)a) dt (b − a)2 (b − a)2 0    x (x − a)[(a + x) f (x) − x f (a)] τ − a α−1 aα = − f (τ ) dτ (b − a)2 (b − a)2 a x −a (x − a)[(a + x) f (x) − x f (a)] a(α + 1) α = − (2) R L Jx − f (a), 2 (b − a) (b − a)2 (x − a)α−1

I1 :=

and similarly we get I2 :=

(b − x)2 (b − a)2



1

(b(1 − t)α + x) f  (tb + (1 − t)x) dt

0

  b b − τ α−1 αb (b − x)[x f (b) − (x + b) f (x)] + f (τ ) dτ = (b − a)2 (b − a)2 x b−x (b − x)[x f (b) − (x + b) f (x)] b(α + 1) α = + (3) R L Jx + f (b). (b − a)2 (b − a)2 (b − x)α−1 Submitting (2) and (3) to (1), we have I :=

(x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)] (b − a)2   a (α + 1) b α α − R L Jx − f (a) + R L Jx + f (b) . (b − a)2 (x − a)α−1 (b − x)α−1 

2.2 Proofs of Lemma 1.3 Proof It suffices to note that  (x − b) 1 α+1 I = (t x + tb) f  (t x + (1 − t)b) dt (b − a)2 0  (a − x) 1 + ((1 − t)α+1 b + (1 − t)x) f  (ta + (1 − t)x) dt (b − a)2 0 := I3 + I4 . Integrating by parts  (x − b) 1 α+1 (t x + tb) f  (t x + (1 − t)b) dt (b − a)2 0  1 (x + b) f  (x) 1 = − ((α + 1)xt α + b) f  (t x + (1 − t)b) dt (b − a)2 (b − a)2 0

I3 :=

(4)

On some fractional Hermite–Hadamard inequalities via s-convex…

(x + b) f  (x) (x(α + 1) + b) f (x) − b f (b) − (b − a)2 (x − b)(b − a)2   x τ − b α−1 α(α + 1)x 1 dτ + f (τ ) (x − b)(b − a)2 b x −b x −b (x + b) f  (x) (x(α + 1) + b) f (x) − b f (b) = + (b − a)2 (b − x)(b − a)2 x(α + 1)(α + 1) α − R L Jx + f (b), (b − x)α+1 (b − a)2 =

(5)

and similarly we get I4 := = =

=

 (a − x) 1 ((1 − t)α+1 b + (1 − t)x) f  (ta + (1 − t)x) dt (b − a)2 0  1 −(x + b) f  (x) 1 + ((α + 1)b(1 − t)α + x) f  (ta + (1 − t)x) dt (b − a)2 (b − a)2 0 −(x + b) f  (x) x f (a) − (b(α + 1) + x) f (x) + (b − a)2 (a − x)(b − a)2   a a − τ α−1 bα(α + 1) 1 dτ + f (τ ) 2 (a − x)(b − a) x a−x a−x −(x + b) f  (x) x f (a) − (b(α + 1) + x) f (x) + (b − a)2 (a − x)(b − a)2 b(α + 1)(α + 1) α − (6) R L Jx − f (a). (x − a)α+1 (b − a)2

Submitting (5) and (6) to (4), we have I :=

  (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a) 1 + (b − a)2 b−x x −a   x (α + 1)(α + 1) b α α − R L Jx + f (b) + R L Jx − f (a) . (b − a)2 (b − x)α+1 (x − a)α+1 

3 Some new fractional Hermite–Hadamard inequalities The following elementary inequality will be used in this section. Lemma 3.1 (See [18,19]) For A > 0, B > 0, when θ ≥ 1 it holds Aθ + B θ ≤ (A + B)θ ≤ 2θ−1 (Aθ + B θ ). Now we are ready to present the main results in this paper.

Z. Gao et al.

Theorem 3.2 Let f : [a, b] → R be a differentiable mapping on (a, b) with a < b. If | f  | ∈ L[a, b] and | f  | is s-convex function. Then for any 0 < α ≤ 1 and a < x < b, the following inequality for fractional integrals holds:   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − J f (a) + J f (b) RL x− RL x+  2 α−1 α−1 (b − a) (x − a) (b − x)   (x − a)2 a x ≤ + | f  (x)| (b − a)2 α+s+1 s+1    x a(α + 1)(s + 1) + | f  (a)| + (α + s + 2) s+1   2 b(α + 1)(s + 1) x (b − x) + | f  (b)| + (b − a)2 (α + s + 2) s+1    x b  + | f (x)| . + α+s+1 s+1 Proof Using Lemma 1.2 via f  ∈ L[a, b] and | f  | is s-convex function, we have   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − J f (a) + J f (b) RL x− RL x+  2 α−1 α−1 (b − a) (x − a) (b − x)  (x − a)2 1 α ≤ (at + x)| f  (t x + (1 − t)a)| dt (b − a)2 0  (b − x)2 1 + (b(1 − t)α + x)| f  (tb + (1 − t)x)| dt (b − a)2 0  (x − a)2 1 α ≤ (at + x)(t s | f  (x)| + (1 − t)s | f  (a)|) dt (b − a)2 0  (b − x)2 1 + (b(1 − t)α + x)(t s | f  (b)| + (1 − t)s | f  (x)|) dt (b − a)2 0   a x (x − a)2 + | f  (x)| = (b − a)2 α+s+1 s+1    a(α + 1)(s + 1) x | f  (a)| + + (α + s + 2) s+1   2 b(α + 1)(s + 1) x (b − x) + | f  (b)| + (b − a)2 (α + s + 2) s+1    x b + | f  (x)| . + α+s+1 s+1 This completes the proof.



On some fractional Hermite–Hadamard inequalities via s-convex…

Theorem 3.3 Let f : [a, b] → R be a differentiable mapping on (a, b) with a < b. If | f  | ∈ L[a, b] and | f  | is s-Godunova–Levin function, then for any 0 < α ≤ 1 and a < x < b, the following inequality for fractional integrals holds:   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − R L Jx − f (a) + R L Jx + f (b)  2 α−1 α−1 (b − a) (x − a) (b − x)   (x − a)2 a x ≤ + | f  (x)| (b − a)2 α−s+1 1−s    x a(α + 1)(1 − s) + | f  (a)| + (α − s + 2) 1−s   2 x b(α + 1)(1 − s) (b − x) + | f  (b)| + (b − a)2 (α − s + 2) 1−s    x b + | f  (x)| . + α−s+1 1−s Proof Using Lemma 1.2 via f  ∈ L[a, b] and | f  | is s-Godunova–Levin function, we have   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − J f (a) + J f (b) − + RL x RL x  2 α−1 α−1 (b − a) (x − a) (b − x)  (x − a)2 1 α ≤ (at + x)| f  (t x + (1 − t)a)| dt (b − a)2 0  (b − x)2 1 + (b(1 − t)α + x)| f  (tb + (1 − t)x)| dt (b − a)2 0  (x − a)2 1 α ≤ (at + x)(t −s | f  (x)| + (1 − t)−s | f  (a)|) dt (b − a)2 0  (b − x)2 1 + (b(1 − t)α + x)(t −s | f  (b)| + (1 − t)−s | f  (x)|) dt (b − a)2 0   a x (x − a)2 + | f  (x)| = (b − a)2 α−s+1 1−s    x a(α + 1)(1 − s) + | f  (a)| + (α − s + 2) 1−s   2 b(α + 1)(1 − s) x (b − x) + | f  (b)| + (b − a)2 (α − s + 2) 1−s    b x | f  (x)| . + + α−s+1 1−s This complete the proof.



Z. Gao et al.

Corollary 3.4 In Theorem 3.2, if we choose x =

a+b 2 ,

we have

  (a + b)(4 f ( a+b ) − f (a) − f (b))  2   4(b − a)  2α−1 (α + 1) a R L J(αa+b )− f (a) + b − (b − a)α+1 2 ≤

α R L J( a+b )+ 2

   f (b)  

1 1 M1 + M2 , 4 4

where      a+b  a+b a   f M1 = +  α + s + 1 2(s + 1)  2   a(α + 1)(s + 1) a+b + + | f  (a)|, (α + s + 2) 2(s + 1)   b(α + 1)(s + 1) a+b | f  (b)| M2 = + (α + s + 2) 2(s + 1)       a+b  a+b b  . f + +  α + s + 1 2(s + 1)  2 

Corollary 3.5 In Theorem 3.3, if we choose x =

a+b 2 ,

we have

  (a + b)(4 f ( a+b ) − f (a) − f (b))  2   4(b − a)  2α−1 (α + 1) − a R L J(αa+b )− f (a) + b (b − a)α+1 2 ≤

α R L J( a+b )+ 2

   f (b)  

1 1 N1 + N2 , 4 4

where      a+b  a+b a f  +  α − s + 1 2(1 − s)  2   a(α + 1)(1 − s) a+b + + | f  (a)|, (α − s + 2) 2(1 − s)   a+b b(α + 1)(1 − s) + | f  (b)| N2 = (α − s + 2) 2(1 − s)       a+b  b a+b f . + +  α − s + 1 2(1 − s)  2 

N1 =

On some fractional Hermite–Hadamard inequalities via s-convex…

From Corollaries 3.4 and 3.5, we have   (a + b)(4 f ( a+b ) − f (a) − f (b))  2   4(b − a)  2α−1 (α + 1) a R L J(αa+b )− f (a) + b − (b − a)α+1 2 ≤

α R L J( a+b )+ 2

   f (b)  

1 min{M1 + M2 , N1 + N2 }. 4

Theorem 3.6 Let f : [a, b] → R be a differentiable. If | f  |q ∈ L[a, b] and | f  |q (q > 1) is s-convex function. Then for any 0 < α ≤ 1, a < x < b, the following inequality for fractional integrals holds:   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]    (b − a)2    a (α + 1) b  α α − R L Jx − f (a) + R L Jx + f (b)  2 α−1 α−1  (b − a) (x − a) (b − x)  1p   1 2 p−1 a p | f (a)|q + | f  (x)|q q p−1 p +2 x pα + 1 s+1 1    1 p | f  (b)|q + | f  (x)|q q (b − x)2 2 p−1 b p p−1 p +2 + x , (b − a)2 pα + 1 s+1

(x − a)2 ≤ (b − a)2

where

1 p

+

1 q



= 1.

Proof Using Lemmas 1.2 and 3.1, Hölder’s inequality via | f  |q ∈ L[a, b] and | f  |q is s-convex function, we have   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]    (b − a)2    a (α + 1) b  α α − R L Jx − f (a) + R L Jx + f (b)  2 α−1 α−1  (b − a) (x − a) (b − x)  (x − a)2 1 α ≤ (at + x)| f  (t x + (1 − t)a)| dt (b − a)2 0  (b − x)2 1 + (b(1 − t)α + x)| f  (tb + (1 − t)x)| dt (b − a)2 0  1  1p  1  q1 (x − a)2 α p  q (at + x) dt | f (t x + (1 − t)a)| dt ≤ (b − a)2 0 0  1  1p  1  q1 (b − x)2 α p  q + (b(1 − t) + x) dt | f (tb + (1 − t)x)| dt (b − a)2 0 0

Z. Gao et al.

≤

(x − a)2 (b − a)2



1

2 p−1 (a p t αp + x p ) dt

 1p 

0

| f  (t x + (1 − t)a)|q dt

 q1

0

 1p  1  q1 (b − p−1 p αp p  q + 2 (b (1 − t) + x ) dt | f (tb+(1−t)x)| dt (b − a)2 0 0   1p   1 | f (a)|q + | f  (x)|q q (x − a)2 2 p−1 a p p−1 p +2 = x (b − a)2 pα + 1 s+1 1    1 p | f  (b)|q + | f  (x)|q q (b − x)2 2 p−1 b p p−1 p + 2 + x , (b − a)2 pα + 1 s+1 x)2



1

1

where we use the following fact due to [17, Theorem 2], 

| f  (a)|q + | f  (x)|q , s+1 0  1 | f  (b)|q + | f  (x)|q . | f  (tb + (1 − t)x)|q dt ≤ s+1 0 1

| f  (t x + (1 − t)a)|q dt ≤



This completes the proof. Corollary 3.7 In Theorem 3.6, if we choose x =

a+b 2 ,

we have

  (a + b)(4 f ( a+b ) − f (a) − f (b))  2   4(b − a)    2α−1 (α + 1)  α α − a R L J( a+b )− f (a) + b R L J( a+b )+ f (b)   (b − a)α+1 2 2 1  1  q q 1 2 p−1 a p (a + b) p p | f  (a)|q + | f  ( a+b 2 )| ≤ + 4 pα + 1 2 s+1  1 1   q q (a + b) p p | f  (b)|q + | f  ( a+b 1 2 p−1 b p 2 )| + + . 4 pα + 1 2 s+1 Theorem 3.8 Let f : [a, b] → R be a differentiable on (a, b) with a < b, | f  |q ∈ L[a, b] and | f  |q (q > 1) is s-convex function, then for any 0 < α ≤ 1, a < x < b we have   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − J f (a) + J f (b) RL x− RL x+  2 α−1 α−1 (b − a) (x − a) (b − x)   1− 1 1− 1 q q (x − a)2 a b (b − x)2 ≤ +x +x ψ(α, s, x)+ ω(α, s, x), 2 2 (b − a) α+1 (b − a) α+1

On some fractional Hermite–Hadamard inequalities via s-convex…

where 

 a x + | f  (x)|q α+s+1 s+1  1  q a(α + 1)(s + 1) x  q + , + | f (a)| (α + s + 2) s+1   x b(α + 1)(s + 1) + | f  (b)|q ω(α, s, x) = (α + s + 2) s+1 1   q x b  q + | f (x)| + . α+s+1 s+1

ψ(α, s, x) =

Proof Using Lemma 1.2 and the well-known power mean inequality and | f  |q is s-convex function, we have   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − J f (a) + J f (b) − + R L R L x x  2 α−1 α−1 (b − a) (x − a) (b − x)  (x − a)2 1 α ≤ (at + x)| f  (t x + (1 − t)a)| dt (b − a)2 0  (b − x)2 1 + (b(1 − t)α + x)| f  (tb + (1 − t)x)| dt (b − a)2 0  1 1− q1  1  q1 (x − a)2 α α  q (at + x) dt (at + x)| f (t x + (1 − t)a)| dt ≤ (b − a)2 0 0  1 1− q1 (b − x)2 α + (b(1 − t) + x) dt (b − a)2 0  1  q1 α  q × (b(1 − t) + x)| f (tb + (1 − t)x)| dt 0

≤

 1 1− q1 (x − a)2 α (at + x) dt (b − a)2 0  1  q1 α s  q s  q × (at + x)(t | f (x)| + (1 − t) | f (a)| ) dt 0

 1 1− q1 (b − x)2 α + (b(1 − t) + x) dt (b − a)2 0  1  q1 α s  q s  q × (b(1 − t) + x)(t | f (b)| + (1 − t) | f (x)| ) dt 0

Z. Gao et al.

≤

(x −a)2 (b−a)2



a +x α+1

1− 1 q

ψ(α, s, x)+

(b − x)2 (b − a)2



b +x α+1

1− 1 q

ω(α, s, x). 

This completes the proof.

Theorem 3.9 Let f : [a, b] → R be a differentiable on (a, b) with a < b, | f  |q ∈ L[a, b] and | f  |q (q > 1) is s-Godunova–Levin function, then for any 0 < α ≤ 1, a < x < b we have   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − J f (a) + J f (b) RL x− RL x+  2 α−1 α−1 (b − a) (x − a) (b − x)    1 1 p (x − a)2 | f (x)|q ap | f  (a)|q q ≤ + + x (b − a)2 pα + 1 1−s 1−s  1    1 p | f (b)|q bp | f  (x)|q q (b − x)2 + + +x , (b − a)2 pα + 1 1−s 1−s where

1 p

+

1 q

= 1.

Proof Using Lemma 1.2, Holder’s inequality and the fact that | f  |q is s-Godunova– Levin function, one has   (x − a)[(a + x) f (x) − x f (a)] + (b − x)[(x + b) f (x) − x f (b)]   (b − a)2    a (α + 1) b α α  − J f (a) + J f (b) RL x− RL x+  2 α−1 α−1 (b − a) (x − a) (b − x)  (x − a)2 1 α ≤ (at + x)| f  (t x + (1 − t)a)| dt (b − a)2 0  (b − x)2 1 + (b(1 − t)α + x)| f  (tb + (1 − t)x)| dt (b − a)2 0  1  1p  1  q1 (x − a)2 α p  q (at + x) dt | f (t x + (1 − t)a)| dt ≤ (b − a)2 0 0  1  1p  1  q1 (b − x)2 α p  q + (b(1 − t) + x) dt | f (tb + (1 − t)x)| dt (b − a)2 0 0 ⎡ ⎤  1  1p  1  1p (x − a)2 ⎣ ≤ a p t pα dt + x p dt ⎦ (b − a)2 0 0 

1

× 0

(t

−s



−s



| f (x)| + (1 − t) | f (a)| ) dt q

q

 q1

On some fractional Hermite–Hadamard inequalities via s-convex…

⎡ ⎤  1  1p  1  1p (b − x)2 ⎣ + b p (1 − t) pα dt + x p dt ⎦ (b − a)2 0 0 

1

×

(t

−s

0



−s



| f (b)| + (1 − t) | f (x)| ) dt q

q

 q1

 1 1 p | f  (x)|q ap | f  (a)|q q + +x pα + 1 1−s 1−s    1 1 p | f (b)|q bp | f  (x)|q q (b − x)2 + + + x , (b − a)2 pα + 1 1−s 1−s

(x − a)2 = (b − a)2



where we use the following fact due to Minkowski’s inequality for p > 1:  

1

α

(at + x) dt p

 1p

 ≤

0 1

α

(b(1 − t) + x) dt

0

p

1

p pα

a t  1p

0

 1p dt



1

+

1

≤

b (1 − t) p

 1p

x dt 0



p

pα

dt

 1p

,



1

+

0

p

x dt

.

0



This completes the proof. Corollary 3.10 In Theorem 3.9, if we choose x =

 1p

a+b 2 ,

we have

  (a + b)(4 f ( a+b ) − f (a) − f (b))  2   4(b − a)    2α−1 (α + 1)  α α a R L J( a+b )− f (a) + b R L J( a+b )+ f (b)  −  (b − a)α+1 2 2 1   1 q  q q p | f  ( a+b 1 ap a+b 2 )| + | f (a)| ≤ + 4 pα + 1 2 1−s   1 1 q  q q p | f  ( a+b bp 1 a+b 2 )| + | f (b)| + + . 4 pα + 1 2 1−s Theorem 3.11 Let f : [a, b] → R be twice differentiable. If f  ∈ L[a, b] and | f  | is s-convex function, then for any 0 < α ≤ 1 and a < x < b we have     1 (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a)  +  (b − a)2 b−x x −a    x (α + 1)(α + 1) b α α  − J f (b) + J f (a) RL x+ RL x−  2 α+1 α+1 (b − a) (b − x) (x − a)   b−x x b ≤ + | f  (x)| (b − a)2 α+s+2 s+2

Z. Gao et al.

  b x(α + 2)(s + 1) + | f  (b)| (α + s + 3) (s + 1)(s + 2)   b(α + 2)(s + 1) x x −a + | f  (a)| + (b − a)2 (α + s + 3) (s + 1)(s + 2)    b x | f  (x)| . + + α+s+2 s+2 

+

Proof Using Lemma 1.3 via | f  | is s-convex function, we have     1 (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a)  +  (b − a)2 b−x x −a    x (α + 1)(α + 1) b α α  − J f (b) + J f (a) RL x+ RL x−  2 α+1 α+1 (b − a) (b − x) (x − a)  1 (b − x) ≤ (t α+1 x + tb)| f  (t x + (1 − t)b)| dt (b − a)2 0  (x − a) 1 + ((1 − t)α+1 b + (1 − t)x)| f  (ta + (1 − t)x)| dt (b − a)2 0  (b − x) 1 α+1 ≤ (t x + tb)(t s | f  (x)| + (1 − t)s | f  (b)|) dt (b − a)2 0  (x − a) 1 + ((1 − t)α+1 b + (1 − t)x)(t s | f  (a)| + (1 − t)s | f  (x)|) dt (b − a)2 0   x b b−x + | f  (x)| = (b − a)2 α+s+2 s+2    b x(α + 2)(s + 1)  + | f (b)| + (α + s + 3) (s + 1)(s + 2)   b(α + 2)(s + 1) x x −a + | f  (a)| + (b − a)2 (α + s + 3) (s + 1)(s + 2)    x b  + | f (x)| . + α+s+2 s+2 This completes the proof.



Theorem 3.12 Let f : [a, b] → R be twice differentiable. If f  ∈ L[a, b] and | f  | is s-Godunova–Levin function, then for any 0 < α ≤ 1 and a < x < b we have     1 (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a)  +  (b − a)2 b−x x −a    x (α + 1)(α + 1) b α α  − J f (b) + J f (a) RL x+ RL x−  2 α+1 α+1 (b − a) (b − x) (x − a)   b−x x b ≤ + | f  (x)| (b − a)2 α−s+2 2−s

On some fractional Hermite–Hadamard inequalities via s-convex…

  b x(α + 2)(1 − s) + | f  (b)| (α − s + 3) (1 − s)(2 − s)   b(α + 2)(1 − s) x x −a + | f  (a)| + (b − a)2 (α − s + 3) (1 − s)(2 − s)    b x | f  (x)| . + + α−s+2 2−s 

+

Proof Using Lemma 1.3 via | f  | is s-Godunova–Levin function, we have     1 (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a)  +  (b − a)2 b−x x −a    x b (α + 1)(α + 1) α α  J f (b) + J f (a) − + − R L R L x x  (b − a)2 (b − x)α+1 (x − a)α+1  (b − x) 1 α+1 ≤ (t x + tb)| f  (t x + (1 − t)b)| dt (b − a)2 0  (x − a) 1 ((1 − t)α+1 b + (1 − t)x)| f  (ta + (1 − t)x)| dt + (b − a)2 0  (b − x) 1 α+1 (t x + tb)(t −s | f  (x)| + (1 − t)−s | f  (b)|) dt ≤ (b − a)2 0  (x − a) 1 ((1 − t)α+1 b + (1 − t)x)(t −s | f  (a)| + (1 − t)−s | f  (x)|) + (b − a)2 0   b−x x b = + | f  (x)| (b − a)2 α−s+2 2−s    b x(α + 2)(1 − s) + | f  (b)| + (α − s + 3) (1 − s)(2 − s)   b(α + 2)(1 − s) x x −a + | f  (a)| + (b − a)2 (α − s + 3) (1 − s)(2 − s)    b x + + | f  (x)| . α−s+2 2−s This completes the proof.



Theorem 3.13 Let f : [a, b] → R be twice differentiable. If | f  |q ∈ L[a, b] and | f  |q (q > 1) is s-convex function, then for any 0 < α ≤ 1 and a < x < b, the following inequality for fractional integrals holds:     1 (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a)  +  (b − a)2 b−x x −a    x (α + 1)(α + 1) b α α  − J f (b) + J f (a) RL x+ RL x−  2 α+1 α+1 (b − a) (b − x) (x − a)

Z. Gao et al.

1  1 2 p−1 x p 2 p−1 b p p | f  (x)|q | f  (b)|q q + + p(α + 1) + 1 p+1 s+1 s+1 1    1 2 p−1 b p 2 p−1 x p p | f  (a)|q | f  (x)|q q x −a + + + , (b − a)2 p(α + 1) + 1 p+1 s+1 s+1

≤

where

b−x (b − a)2

1 p

+

1 q



= 1.

Proof Using Lemmas 3.1 and 1.3, Holder’s inequality via and | f  |q is s-convex function, we have     1 (x(α + 1) + b) f (x) − b f (b) (b(α + 1) + x) f (x) − x f (a)  +  (b − a)2 b−x x −a    x (α + 1)(α + 1) b α α  − J f (b) + J f (a) RL x+ RL x−  2 α+1 α+1 (b − a) (b − x) (x − a)  1 (b − x) ≤ (t α+1 x + tb)| f  (t x + (1 − t)b)| dt (b − a)2 0  (x − a) 1 + ((1 − t)α+1 b + (1 − t)x)| f  (ta + (1 − t)x)| dt (b − a)2 0  1p  1  q1  1 b−x α+1 p  q (t x + tb) dt | f (t x + (1 − t)b)| dt ≤ (b − a)2 0 0  1  1p x −a α+1 p + ((1 − t) b + (1 − t)x) dt (b − a)2 0  1  q1  q × | f (ta + (1 − t)x)| dt 0

 1  1p b−x p−1 p p(α+1) p p ≤ 2 (x t + t b ) dt (b − a)2 0  1  q1 s  q s  q × (t | f (x)| + (1 − t) | f (b)| ) dt 0

 1  1p x −a p−1 p(α+1 p p p + 2 ((1 − t) )b + (1 − t) x ) dt (b − a)2 0  1  q1 s  q s  q × (t | f (a)| + (1 − t) | f (x)| ) dt 0

1  1 2 p−1 x p 2 p−1 b p p | f  (x)|q | f  (b)|q q + + p(α + 1) + 1 p+1 s+1 s+1 1    1 2 p−1 b p 2 p−1 x p p | f  (a)|q | f  (x)|q q x −a + + + . (b − a)2 p(α + 1) + 1 p+1 s+1 s+1

b−x = (b − a)2



On some fractional Hermite–Hadamard inequalities via s-convex…



This complete the proof. Corollary 3.14 In Theorem 3.13, if we choose x =

a+b 2 ,

then we have

  ((a +3b)(α+1)+(a +3b)) f ( a+b ) − 2b f (b) − (a +b) f (a) (α+1)(α+1)  2 −   (b − a)3 (b − a)2  α   2 (a + b) b2α+1  α α × R L J a+b + f (b) + R L J a+b − f (a)  ( 2 ) ( 2 )  (b − a)α+1 (b − a)α+1 1  1  q  q q 1 (a + b) p 2 p−1 b p p | f  ( a+b 2 )| + | f (b)| ≤ + 2(b − a) 2( p(α + 1) + 1) p+1 s+1  1 1   q  q q 2 p−1 b p (a + b) p p | f  ( a+b 1 2 )| + | f (a)| + + . 2(b − a) p(α + 1) + 1 2( p + 1) s+1 Remark 3.15 According to the similar methods of the references [7,12], one can prove that if s ∈ (0, 1] then the inequalities in Theorems 3.2, 3.3, 3.6, 3.8, 3.9, 3.11, 3.12, and 3.13 are sharp if f ∈ { f : I → R : f (x) = b1 x + b2 , b1 , b2 ∈ R}.

4 Applications to some special means Consider the following special means (see Pearce and Pe˘cari´c [14]) for arbitrary real numbers α, β, α = β as follows: H (α, β) =

2 1 α

+

1 β

, α, β ∈ R \ {0},

α+β , α, β ∈ R, 2 β −α ¯ , |α| = |β|, αβ = 0, L(α, β) = ln |β| − ln |α| 1  n+1 − α n+1 n ¯L n (α, β) = β , n ∈ Z \ {−1, 0}, α, β ∈ R, α = β. (n + 1)(β − α) A(α, β) =

Now, using the theory results in Sect. 3 we give some applications to special means of real numbers. Theorem 4.1 For some s ∈ (0, 1], n ∈ Z \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds:   (4( a+b )n − a n − bn )  2 A(a, b)   2(b − a)       a a + b a + b b  − ,b + L¯ n a, L¯ n   2(b − a) n 2 2(b − a) n 2

Z. Gao et al.

     a a +b a +b a +b n−1 a(2)(s +1) n−1 + + a + 2+s 2(s +1) 2 (3+s) 2(s +1)       b(2)(s +1) a + b n−1 a +b a+b b n n−1 + + b + + . 4 (3+s) 2(s +1) 2+s 2(s + 1) 2

n ≤ 4



Proof Applying Theorem 3.2 for f (t) = t n , α = 1 and x = result immediately.

a+b 2

one can obtain the 

Theorem 4.2 For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds:   (4( a+b )n − a n − bn )  2 A(a, b)   2(b − a)       a a + b a + b b  − ,b + L¯ nn a, L¯ nn   2(b − a) 2 2(b − a) 2       n a a + b n−1 a +b a +b a(2)(1 − s) n−1 ≤ + + a + 4 2 − s 2(1 − s) 2 (3 − s) 2(1−s)       b(2)(1 − s) a +b n−1 a +b a +b b n n−1 + + b + + . 4 (3 − s) 2(1 − s) 2 − s 2(1−s) 2 Proof Applying Theorem 3.3 for f (t) = t n , α = 1 and x = result immediately.

a+b 2

one can obtain the 

Theorem 4.3 For some s ∈ (0, 1], n ∈ Z \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds:   (4( a+b )n − a n − bn )  2 A(a, b)   2(b − a)       a + b a + b a b  n n − ,b + L¯ a, L¯   2(b − a) n 2 2(b − a) n 2 1  1  q(n−1) q n 2 p−1 a p (a + b) p p a q(n−1) + ( a+b 2 ) ≤ + 4 p+1 2 s+1 n + 4 where

1 p

+

1 q



2 p−1 b p (a + b) p + p+1 2

 1p 

q(n−1) q1  bq(n−1) + a+b 2 , s+1

= 1, 1 < q < ∞.

Proof Applying Theorem 3.6 for f (t) = t n , α = 1 and x = result immediately.

a+b 2

one can obtain the 

On some fractional Hermite–Hadamard inequalities via s-convex…

Theorem 4.4 For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds:  ( 1 − 1 − 1)  2(a+b) a b A(a, b)   2(b − a)    1 1 b a  + −  a+b ¯ ¯ a+b , b)  2(b − a) L(a, 2(b − a) ) L( 2 2   1 1 p q 1 ap 4q a+b 1 ≤ + + 2q 4 p+1 2 (1 − s)(a + b)2q a (1 − s)   1 1 p q bp 4q 1 a+b 1 + + + , 4 p+1 2 (1 − s)(a + b)2q b2q (1 − s) where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof Applying Theorem 3.9 for f (t) = 1t , α = 1 and x = result immediately.

a+b 2

one can obtain the 

Theorem 4.5 For some s ∈ (0, 1], n ∈ Z \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds:    3 a+b n (a + 3b) − 2bn+1 − (a + b)a n  2   (b − a)3     a+b 4b 2(a + b) ¯ n a + b  n ¯ , b) + − L ( L a,   (b − a)3 n 2 (b − a)3 n 2  1 1   n(n − 1) (a + b) p 2 p−1 b p p (a + b)(n−2)q b(n−2)q q ≤ + + 2(b − a) 2(2 p + 1) p+1 s+1 2(n−2)q (s + 1) 1  1  (a + b) p p (a + b)(n−2)q a (n−2)q q n(n − 1) 2 p−1 b p + + + , 2(b − a) 2 p + 1 2( p + 1) s+1 2(n−2)q (s + 1) where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof Applying Theorem 3.13 for f (t) = t n , α = 1 and x = result immediately.

a+b 2

one can obtain the 

Theorem 4.6 For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds:

Z. Gao et al.

  6(a + 3b) a+b 2  − −  3 3  (a + b)(b − a) a(b − a) (b − a)3    2(a + b) 1 1 4b  + −  a+b ¯ a+b , b) (b − a)3 L(a, ¯  (b − a)3 L( ) 2 2

1  1 q 8q (a + b) p 2 p−1 b p p 1 + + 3q 3q 2(2 p + 1) p+1 (a + b) (s + 1) b s + 1 1   p−1 p  1 q 8q 2 (a + b) p p 1 b 1 + + + , 2(b − a) 2 p + 1 2( p + 1) (a + b)3q (s + 1) a 3q s + 1

1 ≤ 2(b − a)

where

1 p

+

1 q



= 1, 1 < q < ∞.

Proof Applying Theorem 3.13 for f (t) = 1t , α = 1 and x = result immediately.

a+b 2

one can obtain the 

Acknowledgments The authors are grateful to the referees for their careful reading of the manuscript and valuable comments. The authors thank the help from the editor too. Compliance with ethical standards Conflict of interest

The authors declare that they have no competing interests.

Authors’s contributions This work was carried out in collaboration between all authors. GZY, LMM and WJR proved the theorems, interpreted the results and wrote the article. All authors defined the research theme, read and approved the manuscript.

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