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Estimation type results associated to k-fractional integral inequalities with applications
Accepted Manuscript Estimation type results associated to k-fractional integral inequalities with applications Tingsong Du, Hao Wang, Muhammad Amer Latif, Yao Zhang PII: DOI: Reference:
S1018-3647(18)31243-6 https://doi.org/10.1016/j.jksus.2018.09.010 JKSUS 733
To appear in:
Journal of King Saud University - Science
Please cite this article as: T. Du, H. Wang, M.A. Latif, Y. Zhang, Estimation type results associated to k-fractional integral inequalities with applications, Journal of King Saud University - Science (2018), doi: https://doi.org/ 10.1016/j.jksus.2018.09.010
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Estimation type results associated to k-fractional integral inequalities with applications
Abstract This paper aims to establish k-fractional integral inequalities with multiple parameters involving the first differentiable mappings. Applications of our results to random variables are also given. 2010 Mathematics Subject Classification: 26A33; 26D15; 26E60; 41A55. Key words: invex set; preinvex functions; k-fractional integrals.
1
Introduction
If h : I ⊆ R → R is a convex mapping and e1 , e2 ∈ I with e1 < e2 , then one has Z e2 e1 + e2 1 h(e1 ) + h(e2 ) h ≤ h(t)dt ≤ . 2 e 2 − e 1 e1 2 The subsequent inequality is an improvement of the inequality (1.1). e + e 1 3e + e e + 3e 1 2 1 2 1 2 h ≤ h +h 2 2 4 4 Z e2 1 ≤ h(t)dt e2 − e1 e1 1 e1 + e2 h(e1 ) + h(e2 ) h(e1 ) + h(e2 ) ≤ h + ≤ . 2 2 2 2
(1.1)
(1.2)
Many different generalizations, new extensions and improvements related to the inequality (1.1) ˙ scan [6], Khan et al. can be found in Awan et al. [1], Chu et al. [2, 3], Du et al. [4], Iqbal et al. [5], I¸ [7–12], Sarikaya and Karaca [17], Set et al. [18], Xi et al. [19]. Let us consider an invex set K. A set K ⊆ Rn is named invex set with respect to the mapping δ : K × K → Rn , if ν + λδ(µ, ν) ∈ K holds, for all µ, ν ∈ K and λ ∈ [0, 1]. A mapping h : K → R is called preinvex respecting δ, if the following inequality: h ν + λδ(µ, ν) ≤ (1 − λ)h(ν) + λh(µ) (1.3) grips, for every µ, ν ∈ K and λ ∈ [0, 1]. The preinvex function is an important substantive generalization of the convex function. Example 1.1 The function h(x) = 12 − x − 12 , (x ∈ R) with respect to the following 2(y − x), x ≥ 12 , y ≥ 12 , y ≥ x, 0, x > 12 , y ≥ 12 , y < x, x < 12 , y ≤ 12 , y > x, 0, y − x, x ≤ 12 , y ≤ 12 , y ≤ x, δ(y, x) = 1 − x − y, x < 12 , y > 12 , x + y ≥ 1, 0, x < 12 , y > 12 , x + y ≤ 1, 0, x > 12 , y < 12 , x + y ≥ 1, 1 − x − y, x > 12 , y < 12 , x + y ≤ 1 is preinvex on R while it is not convex on R. 1
(1.4)
This paper aims to obtain estimation type results related to k-fractional integral operators. For this, we suppose that the absolute value of the derivative of the considered mapping is preinvex. Next, we substitute this hypothesis with the boundedness of the derivative and with a Lipschitzian condition for the derivative of the considered mapping to derive integral inequalities with new bounds. Applications of our results to random variables are also provided. We end this section by reciting a well-known k-fractional integral operators in the literature. Definition 1.1 (Mubeen and Habibullah [15]) Let h ∈ L1 [µ, ν], the k-fractional integrals k Jµτ+ h(x) and k Jντ− h(x) of order τ > 0 are defined by Z x τ 1 τ J h(x) = (x − λ) k −1 h(λ)dλ, (0 ≤ µ < x < ν) + k µ kΓk (τ ) µ and τ k Jν − h(x)
1 = kΓk (τ )
Z
ν
τ
(λ − x) k −1 h(λ)dλ, (0 ≤ µ < x < ν),
x
respectively, where k > 0, and Γk is the k-gamma function defined as Γk (x) := Re(x) > 0, with the properties Γk (x + k) = xΓk (x) and Γk (k) = 1.
2
R∞ 0
λx−1 e−
λk k
dλ,
Main Results
Throughout this work, let K ⊆ R be an open invex subset respecting δ : K × K → R \ {0} and r1 , r2 ∈ K with r1 < r2 , and let h : K → R be a differentiable mapping such that h0 is integrable on the δ-path Pxv : v = x + δ(y, x) with x, y ∈ [r1 , r2 ]. Before stating the results, we use notations below. τ τ δ k (x, r1 ) h r1 + δ(x, r1 ) + h(r1 ) + δ k (r2 , x) h(x) + h x + δ(r2 , x) Lδ (τ, k, λ; x) :=(1 − λ) δ(r2 , r1 ) τ τ 2λ δ k (x, r1 )h r1 + 12 δ(x, r1 ) + δ k (r2 , x)h x + 12 δ(r2 , x) + δ(r2 , r1 ) τ k 2 Γk (τ + k) 1 1 τ τ − k J[r1 +δ(x,r1 )]− h r1 + δ(x, r1 ) + k Jr + h r1 + δ(x, r1 ) 1 δ(r2 , r1 ) 2 2 1 1 τ + k Jxτ+ h x + δ(r2 , x) + k J[x+δ(r δ(r2 , x) . −h x + 2 ,x)] 2 2 (2.1) If δ(µ, ν) = µ − ν with µ, ν ∈ [r1 , r2 ], then we have τ
τ
(x − r1 ) k [h(r1 ) + h(x)] + (r2 − x) k [h(x) + h(r2 )] L(τ, k, λ; x) := (1 − λ) r2 − r1 τ τ τ r +x 1 2 2λ (x − r1 ) k h 2 + (r2 − x) k h x+r 2 k Γk (τ + k) 2 (2.2) + − r2 − r1 r2 − r1 r + x r + x x + r x + r 1 2 2 1 × k Jrτ+ h + k Jxτ− h + k Jxτ+ h + k Jrτ− h . 1 2 2 2 2 2 (a) Choosing λ = 0 and τ = k = 1, the equation (2.2) reduces to L(1, 1, 0; x) := h(x) +
(r2 − x)h(r2 ) + (x − r1 )h(r1 ) 2 − r2 − r1 r2 − r1
Z
r2
h(u)du. r1
Specially, Z r2 r + r r1 + r2 h(r1 ) + h(r2 ) 2 1 2 L 1, 1, 0; := +h − h(u)du. 2 2 2 r2 − r1 r1
2
(b) Choosing λ = 1 and τ = k = 1, the equation (2.2) degenerates into Z r2 2 2 (x − r1 )h r12+x + (r2 − x)h x+r 2 2 L(1, 1, 1; x) := − h(u)du. r2 − r1 r2 − r1 r1 Specially, Z r2 3r + r r + 3r r1 + r2 2 1 2 1 2 L 1, 1, 1; := h +h − h(u)du. 2 4 4 r2 − r1 r1 We need the following lemma for our results. Lemma 2.1 One has the subsequent identity Lδ (τ, k, λ; x) ) Z 1 τ 1 + t 1 − t t − λ h0 r1 + δ(x, r1 ) dt − t k − λ h0 r1 + δ(x, r1 ) dt 2 2 0 0 (Z ) Z 1 τ 1 τ τ δ k +1 (r2 , x) 1−t 1+t 0 0 − tk − λ h x + δ(r2 , x) dt − tk − λ h x + δ(r2 , x) dt 2δ(r2 , r1 ) 2 2 0 0 τ
δ k +1 (x, r1 ) = 2δ(r2 , r1 )
(Z
1
τ k
(2.3) for k-fractional integrals with x ∈ (r1 , r2 ), τ > 0, k > 0 and λ ∈ [0, 1]. Proof Using integration by parts and appropriate substitutions, such as µ = r1 + r1 + 1−t 2 δ(x, r1 ) · · · , we can obtain the identity (2.3). Example 2.1 If we take r1 = 12 , r2 = 1, τ = 1 = k, h(x) = 12 − x − 12 , and ( 2(µ − ν), µ ≥ ν, δ(µ, ν) = 0, µ < ν,
1+t 2 δ(x, r1 ),
ν =
(2.4)
then all the assumptions in Lemma 2.1 are satisfied. Clearly, the left of the equation (2.3) is 3 1 2 x − 21 2 − 2x − 2 + 2(1− x)[(1 − x) + (x − 1)] Lδ (1, 1, λ; x) = (1 − λ) 2 1 − 12 2 x − 12 [1 − x] + 2(1 − x)[1 − 1] + 2λ 2 1 − 12 Z x Z 2x− 12 Z 1 Z 2−x (1 − t)dt + (1 − t)dt + (1 − t)dt + (1 − t)dt = 0 −2 1 2
x
x
1
and the right of the equation (2.3) is Z 1 Z 1 Z 1 Z 1 1 2 4 x− −(t − λ)dt + (t − λ)dt − 4(1 − x)2 −(t − λ)dt + (t − λ)dt = 0. 2 0 0 0 0 Corollary 2.1 If δ(µ, ν) = µ − ν with µ, ν ∈ [r1 , r2 ] in Lemma 2.1, then one has L(τ, k, λ; x) τ
Z 1 1 + t 1 − t τ 1−t 1+t 0 0 k t −λ h x+ r1 dt − t −λ h x+ r1 dt 2 2 2 2 0 0 Z 1 Z 1 τ 1 + t 1 − t τ τ (r2 − x) k +1 1−t 1+t − t k − λ h0 x+ r2 dt − t k − λ h0 x+ r2 dt , 2(r2 − r1 ) 2 2 2 2 0 0 (2.5)
(x − r1 ) k +1 = 2(r2 − r1 )
Z
1
τ k
which is a new form of Lemma 2.1. Remark 2.1 (a) In Lemma 2.1, putting λ = 0 and k = 1, one has Lemma 2.8 in Noor et al. [16]. (b) In Corollary 2.1, taking λ = 0 and k = 1, one has Lemma 1 in Mihai and Mitroi [14]. Further, choosing τ = 1, one has Lemma 1 in Latif [13] . 3
Theorem 2.1 If |h0 |q for q > 1 is preinvex on K, then the coming inequality for k-fractional integrals with x ∈ (r1 , r2 ), τ > 0, k > 0, p−1 + q −1 = 1 and λ ∈ [0, 1] grips: Lδ (τ, k, λ; x) (" τ +1 1 0 1 ) δ k (x, r1 ) 1 |h0 (r1 )|q + 3|h0 (x)|q q 3|h (r1 )|q + |h0 (x)|q q p ≤ ρ (τ, k, λ, p) + 2|δ(r2 , r1 )| 4 4 (2.6) (" τ +1 1 1) δ k (r2 , x) 1 3|h0 (x)|q + |h0 (r2 )|q q |h0 (x)|q + 3|h0 (r2 )|q q + ρ p (τ, k, λ, p) + , 2|δ(r2 , r1 )| 4 4 where k , τ p+k τ p+k τ p+k k 2k p ρ(τ, k, λ, p) = τ p+k − τ p+k λ τ − λ + 2λ τ , k τ +1) Γ(p+1)Γ( , k
λ = 0, 0 < λ < 1, λ = 1.
Γ( τ +1+p)
Proof According to Lemma 2.1, the Holder inequality and preinvexity of |h0 |q , one gets Z 1 τ p p1 q q k +1 (x, r )| τk 1 Lδ (τ, k, λ; x) ≤ |δ χ1 + χ2 t − λ dt 2|δ(r2 , r1 )| 0 τ +1 p p1 δ k (r2 , x) Z 1 τ q q k + χ3 + χ4 , t − λ dt 2|δ(r2 , r1 )| 0 where Z
1
q 1+t 0 δ(x, r1 ) dt h r1 + 2 0 Z 1 1−t 0 1+t 0 |h0 (r1 )|q + 3|h0 (x)|q q q ≤ |h (r1 )| + |h (x)| dt = , 2 2 4 0 Z 1 q 1−t 3|h0 (r1 )|q + |h0 (x)|q 0 χ2 = δ(x, r1 ) dt ≤ , h r1 + 2 4 0 Z 1 q 1−t 3|h0 (x)|q + |h0 (r2 )|q 0 χ3 = δ(r2 , x) dt ≤ , h x + 2 4 0 Z 1 q 1+t |h0 (x)|q + 3|h0 (r2 )|q 0 χ4 = δ(r2 , x) dt ≤ . h x + 2 4 0 When λ = 0, we have Z 1 p k τk , t − λ dt = τp + k 0 χ1 =
when λ = 1, we have Z 0
1
τ p Γ(p + 1)Γ( τk + 1) k , t − λ dt = Γ τk + 1 + p
when 0 < λ < 1, we have Z 0
1
k
Z τ p k t − λ dt =
λτ
τ
λ − tk
p
Z dt +
k
0
Z ≤ =
1
τ p t k − λ dt
λτ 1
k λτ
k
t
τp k
−λ
p
Z dt +
λτ
λp − t
τp k
dt
0
τ p+k τ p+k k 2k − λ τ − λp + 2λ τ . τp + k τp + k
The above inequality is obtained by using the fact that (α − β)κ ≤ ακ − β κ , α ≥ β ≥ 0, κ ≥ 1. The proof of Theorem 2.1 is completed. 4
(2.7)
Corollary 2.2 Choosing λ = 1, k = 1 and δ(µ, ν) = µ − ν for µ, ν ∈ [r1 , r2 ] in Theorem 2.1, one obtains L(τ, 1, 1; x) 2 (x − r )τ h r1 +x ) + (r − x)τ h x+r2 2τ Γ(τ + 1) 1 2 2 2 := − r2 − r1 r2 − r1 r + x r + x x + r x + r 1 1 2 2 τ τ τ τ × Jr+ h + Jx− h + Jx+ h + Jr− h 1 2 2 2 2 2 ( 1 q2 +1 Γ(p + 1)Γ( 1 + 1) (x − r )τ +1 1 1 1 0 q 0 q q 0 q 0 q q τ ≤ |h (r1 )| + 3|h (x)| + 3|h (r1 )| + |h (x)| 2 r2 − r1 Γ τ1 + 1 + p ) 1 1 (r2 − x)τ +1 0 q 0 q q 0 q 0 q q + |h (x)| + 3|h (r2 )| + 3|h (x)| + |h (r2 )| . r2 − r1 (2.8) 2 Specially, putting τ = 1 and x = r1 +r 2 , one has Z r2 r + 3r 3r1 + r2 2 1 2 h h(u)du + h − 4 4 r2 − r1 r1 ( r + r q q1 r + r q q1 1 p1 1 q2 +1 r − r 2 1 1 2 1 2 ≤ |h0 (r1 )|q + 3 h0 + 3|h0 (r1 )|q + h0 p+1 2 4 2 2 q1 q1 ) 0 r1 + r2 q 0 r1 + r2 q 0 q 0 q + h + 3 h + 3|h (r2 )| + |h (r2 )| 2 2 1 p1 1 q3 +1 r − r h 1 i 1 1 1 2 1 ≤ 1 q + 3 q + 5 q + 7 q |h0 (r1 )| + |h0 (r2 )| . p+1 2 4 (2.9)
The second inequality is obtained by applying the convexity of |h0 |q and the succeeding fact that n n n X X X (%i + σi )s ≤ (%i )s + (σi )s , %i , σi > 0, 0 ≤ s < 1. i=1
i=1
i=1
Theorem 2.2 If |h0 |q for q ≥ 1 is preinvex on K, then the coming inequality for k-fractional integrals with x ∈ (r1 , r2 ), τ > 0, k > 0 and λ ∈ [0, 1] grips: ( τ 1 q k +1 (x, r1 ) 1− q1 0 0 q q 1 Lδ (τ, k, λ; x) ≤ δ ∆ ∆1 − ∆2 h (r1 ) + ∆1 + ∆2 h (x) 2|δ(r2 , r1 )| 1 2 1 ) q 0 0 q q 1 + ∆1 + ∆2 h (r1 ) + ∆1 − ∆2 h (x) 2 ( τ +1 1 (2.10) δ k (r2 , x) 1− 1 1 q q 0 q 0 q + ∆ ∆1 + ∆2 h (x) + ∆1 − ∆2 h (r2 ) 2|δ(r2 , r1 )| 1 2 1 ) q q 0 q 0 1 + ∆1 − ∆2 h (x) + ∆1 + ∆2 h (r2 ) , 2 where 1
Z ∆1 = 0
τ k t − λ dt =
τ +k k τ − λ + 2λ τ τ +k τ +k
(2.11)
and Z ∆2 = 0
1
τ t t k − λ dt =
τ +2k k λ τ − +λ τ . τ + 2k 2 τ + 2k
5
(2.12)
Proof Using Lemma 2.1 and the power mean inequality, one has τ Z 1 1− q1 1 k +1 (x, r1 ) 1 τk Lδ (τ, k, λ; x) ≤ δ I1 q + I2 q t − λ dt 2|δ(r2 , r1 )| 0 τ +1 1− q1 1 δ k (r2 , x) Z 1 τ 1 k + I3 q + I4 q , t − λ dt 2|δ(r2 , r1 )| 0 where Z
1
τ q 1+t k δ(x, r1 ) dt t − λ h0 r1 + 2 0 Z 1 1 − t q 1 + t τk 0 0 q ≤ t − λ h (r1 ) + h (x) dt 2 2 0 Z 1 Z 1 Z 1 q Z 1 τ τ τ 1 t τk − λ dt − t k − λ dt + = t t k − λ dt h0 (r1 ) + t t k − λ dt h0 (x) , 2 0 0 0 0 Z 1 q 1−t τk I2 = δ(x, r1 ) dt t − λ h0 r1 + 2 0 Z 1 Z 1 Z 1 q Z 1 τ τ τ τ 1 0 0 k k k k ≤ t − λ dt + t t − λ dt h (r1 ) + t − λ dt − t t − λ dt h (x) , 2 0 0 0 0 Z 1 q 1−t τk I3 = δ(r2 , x) dt t − λ h0 x + 2 0 Z 1 Z 1 Z 1 Z 1 τ τ τ τ 1 0 q 0 k k k k ≤ t − λ dt + t t − λ dt h (x) + t − λ dt − t t − λ dt h (r2 ) , 2 0 0 0 0 Z 1 q 1+t τk I4 = δ(r2 , x) dt t − λ h0 x + 2 0 Z 1 Z 1 Z 1 q Z 1 τ τ τ τ 1 t k − λ dt − t k − λ dt + ≤ t t k − λ dt h0 (x) + t t k − λ dt h0 (r2 ) . 2 0 0 0 0 I1 =
These last four inequalities clasp due to the preinvexity of |h0 |q . This ends the proof. Corollary 2.3 Choosing λ = 1, k = 1 and δ(µ, ν) = µ − ν with µ, ν ∈ [r1 , r2 ] in Theorem 2.2, one obtains ( 1 q τ +1 τ (x − r ) τ + 3 0 3τ + 5 0 q q 1 L(τ, 1, 1; x) ≤ h (r1 ) + h (x) 2(τ + 1)(r2 − r1 ) 4(τ + 2) 4(τ + 2) 1) q 3τ + 5 0 τ + 3 0 q q + h (r1 ) + h (x) 4(τ + 2) 4(τ + 2) (2.13) ( q q1 τ (r2 − x)τ +1 τ + 3 0 q 3τ + 5 0 + h (x) + h (r2 ) 2(τ + 1)(r2 − r1 ) 4(τ + 2) 4(τ + 2) ) q q1 3τ + 5 0 q τ + 3 0 + . h (x) + h (r2 ) 4(τ + 2) 4(τ + 2) 2 Specially, taking τ = 1, x = r1 +r and utilizing similar arguments as in Corollary 2.2, one gets 2 Z r2 r + 3r 3r1 + r2 2 1 2 h +h − h(u)du 4 4 r2 − r1 r1 ( q 1 r + r q q1 1 2 (r2 − r1 ) 1 0 q 2 0 r1 + r2 q q 0 1 2 ≤ + h (r1 ) + h0 h (r1 ) + h 16 3 3 2 3 3 2 (2.14) ) q q1 2 r + r q 1 q q1 1 0 r1 + r2 q 2 0 0 1 0 2 + h + h + h (r2 ) + h (r2 ) 3 2 3 3 2 3 1h i 1 1 1 (r2 − r1 ) 1 q q1 ≤ 1 + 2 q + 4 q + 5 q |h0 (r1 )| + |h0 (r2 )| . 16 2
6
Remark 2.2 Theorem 2.1 and 2.2 will be reduced to Theorem 2 and 3 in Latif [13], respectively, if we choose λ = 0, k = 1 = τ and δ(µ, ν) = µ − ν for µ, ν ∈ [r1 , r2 ].
3
New estimation results
For obtaining new estimation type results, we deal with the boundedness and the Lipschitzian condition of h0 , respectively. Theorem 3.1 Assume that there exists constants m < M such that −∞ < m ≤ h0 (y) ≤ M < ∞ for all y ∈ K, then the inequality τ τ τ +k k +1 (x, r )| + |δ k +1 (r , x)| 2τ λ τ + k 1 2 Lδ (τ, k, λ; x) ≤ (M − m) |δ (3.1) −λ 2|δ(r2 , r1 )| τ +k holds with τ > 0, k > 0, λ ∈ [0, 1] and x ∈ (r1 , r2 ). Proof From Lemma 2.1, we have (Z τ 1 0 m + M τ δ k +1 (x, r1 ) 1+t k Lδ (τ, k, λ; x) = t − λ h r1 + δ x, r1 − dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 m + M τ 1−t − t k − λ h r1 + δ x, r1 − dt 2 2 0 (Z τ 1 0 m + M τ δ k +1 (r2 , x) 1−t k − t −λ h x+ δ r2 , x − dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 m + M τ 1+t − tk − λ h x + δ r2 , x − dt . 2 2 0 ≤ h0 (y) − m+M ≤ M − m+M 2 2 , one obtains τ ( Z 1 k +1 (x, r1 ) m + M 1+t τk 0 Lδ (τ, k, λ; x) ≤ δ dt δ x, r1 − t − λ h r1 + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 m + M 1−t τk 0 dt + δ x, r1 − t − λ h r1 + 2 2 0 τ +1 ( δ k (r2 , x) Z 1 τ m + M 1−t 0 dt k + δ r2 , x − t − λ h x + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 m + M 1+t τk 0 dt + δ r2 , x − t − λ h x + 2 2 0 τ +1 τ +1 δ k (x, r1 ) M − m Z 1 τ δ k (r2 , x) M − m Z 1 τ ≤ t k − λ dt + t k − λ dt |δ(r2 , r1 )| 2 |δ(r , r )| 2 2 1 0 0 τ τ τ +k (M − m) |δ k +1 (x, r1 )| + |δ k +1 (r2 , x)| 2τ λ τ + k = −λ . 2|δ(r2 , r1 )| τ +k
Using the fact that m −
m+M 2
(3.2)
The proof is completed. Corollary 3.1 In Theorem 3.1, choosing δ(µ, ν) = µ − ν for µ, ν ∈ [r1 , r2 ], and τ = k = 1, one gets (x − r1 )[h(r1 ) + h(x)] + (r2 − x)[h(x) + h(r2 )] |L(1, 1, λ; x)| = (1 − λ) r2 − r1 Z r2 2 (x − r1 )h r12+x + (r2 − x)h x+r 2 2 + 2λ − h(u)du (3.3) r2 − r1 r2 − r1 r1 (M − m) (x − r1 )2 + (r2 − x)2 2λ2 − 2λ + 1 ≤ . 2(r2 − r1 ) 2 7
2 Remark 3.1 Taking x = r1 +r in Corollary 3.1, one can see the following. 2 (a) For λ = 0, we have Z r2 h(r ) + h(r ) (M − m)(r − r ) r + r 2 1 2 1 2 2 1 +h − h(u)du ≤ . 2 2 r2 − r1 r1 8
(b) For λ = 12 , we have h(r ) + h(r ) 1 r + r h 1 2 1 2 + h + 4 2 2
3r1 +r2 4
+h 2
r1 +3r2 4
2 − r2 − r1
Z
r2
r1
(3.4)
(M − m)(r − r ) 2 1 h(u) ≤ . 16 (3.5)
(c) For λ = 1, we have Z r2 r + 3r 3r1 + r2 (M − m)(r2 − r1 ) 2 1 2 h +h − h(u)du ≤ . 4 4 r2 − r1 r1 8
(3.6)
Theorem 3.2 Assume that h0 satisfies Lipschitz condition on K for some L > 0, then the inequality τ τ 2k k +2 (x, r )| + |δ k +2 (r , x)| 2τ λ τ +1 + 2k 1 2 Lδ (τ, k, λ; x) ≤ L |δ −λ 4|δ(r2 , r1 )| τ + 2k
(3.7)
grips for all x ∈ (r1 , r2 ), µ > 0, k > 0 and λ ∈ [0, 1]. Proof From Lemma 2.1, we have (Z τ 1 0 τ δ k +1 (x, r1 ) 1+t 1 0 k Lδ (τ, k, λ; x) = t − λ h r1 + δ x, r1 − h r1 + δ x, r1 dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 τ 1−t 1 0 − t k − λ h r1 + δ x, r1 − h r1 + δ x, r1 dt 2 2 0 (Z τ 1 0 τ δ k +1 (r2 , x) 1−t 1 0 k − t −λ h x+ δ r2 , x − h x + δ r2 , x dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 τ 1+t 1 0 − tk − λ h x + δ r2 , x − h x + δ r2 , x dt . 2 2 0 Utilizing the fact that h0 satisfies Lipschitz condition on K for L > 0, we have τ ( Z 1 δ k +1 (x, r1 ) 1+t 1 τk Lδ (τ, k; λ, x) ≤ δ x, r1 − h0 r1 + δ(x, r1 ) dt t − λ h0 r1 + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 0 1 − t 1 τk + δ x, r1 − h0 r1 + δ(x, r1 ) dt t − λ h r1 + 2 2 0 τ +1 ( Z 1 δ k (r2 , x) 1−t 1 τk + δ r2 , x − h0 x + δ(r2 , x) dt t − λ h0 x + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 1+t 1 τk + δ r2 , x − h0 x + δ(r2 , x) dt , t − λ h0 x + 2 2 0 where 0 h r1 + 1 + t δ(x, r1 ) − h0 r1 + 1 δ(x, r1 ) ≤ L r1 + 1 + t δ(x, r1 ) − r1 + 1 δ(x, r1 ) = tL δ(x, r1 ) , 2 2 2 2 2 0 h r1 + 1 − t δ(x, r1 ) − h0 r1 + 1 δ(x, r1 ) ≤ tL δ(x, r1 ) , 2 2 2 8
0 h x + 1 − t δ(r2 , x) − h0 a + 1 δ(r2 , x) ≤ tL δ(r2 , x) , 2 2 2 0 h x + n + t δ(r2 , x) − h0 a + 1 δ(r2 , x) ≤ tL δ(r2 , x) . n+1 2 2 After a simple calculation, we obtain the desired result. Corollary 3.2 In Theorem 3.2, choosing δ(µ, ν) = µ − ν for µ, ν ∈ [a, b], and τ = k = 1, one has L (x − r1 )3 + (r2 − x)3 2λ3 + 2 (3.8) |L(1, 1, λ; x)| ≤ −λ . 4(r2 − r1 ) 3 2 Remark 3.2 Putting x = r1 +r in Corollary 3.2, one can see the following. 2 (a) For λ = 0, we have Z r2 h(r ) + h(r ) L(r − r )2 r + r 2 1 2 1 2 2 1 +h − h(u)du ≤ . 2 2 r2 − r1 r1 24
(b) For λ = 12 , we h(r ) + h(r ) 1 2 + 4
have 1 r1 + r2 h h + 2 2
3r1 +r2 4
+h 2
r1 +3r2 4
2 − r2 − r1
Z
r2
r1
L(r − r )2 2 1 h(u) ≤ . (3.10) 64
(c) For λ = 1, we have Z r2 r + 3r 3r1 + r2 L(r2 − r1 )2 2 1 2 h ≤ h(u)du . + h − 4 4 r2 − r1 r1 48
4
(3.9)
(3.11)
Applications for random variables
Let X be a random variable in [r1 , r2 ], with the probability density mapping p : [r1 , r2 ] → [0, 1], and with the cumulative distribution mapping Z x F (x) = P(X ≤ x) = p(λ)dλ. a
Using the fact that E(X) =
R r2 r1
λdF (λ) = r2 −
R r2 r1
F (λ)dλ, one has the following results.
2 Proposition 4.1 In Theorem 2.2, taking λ = 1, k = 1 = τ , x = r1 +r and δ(u, v) = u − v with 2 u, v ∈ [r1 , r2 ], one has 2 P X ≤ 3r1 + r2 + P X ≤ r1 + 3r2 − (r2 − E(X)) 4 4 r2 − r1 (4.1) i 1 1 1 (r2 − r1 ) 1 q1 h q1 ≤ 1 + 2 q + 4 q + 5 q |p(r1 )| + |p(r2 )| . 16 2 2 Proposition 4.2 In Theorem 2.1, putting λ = 1, k = 1 = τ , x = r1 +r and δ(u, v) = u − v with 2 u, v ∈ [r1 , r2 ], one gets 2 P X ≤ 3r1 + r2 + P X ≤ r1 + 3r2 − (r − E(X)) 2 4 4 r2 − r1 (4.2) 1 3 1 p 1 q +1 r − r h 1 i 1 1 1 2 1 ≤ 1 q + 3 q + 5 q + 7 q |p(r1 )| + |p(r2 )| . p+1 2 4
Remark 4.1 Applications can be given based on the obtained results to special means, and we omit the details.
5
Conclusion
Four main results of the trapezium-like inequalities involving the mappings δ are hereby obtained. More new results can be derived by choosing different mappings δ and the special parameter values for λ, k and τ . 9
Competing interests The authors declare that there are no competing interests.
Author’s Contributions All authors read and approved the final manuscript.
Acknowledgments This work was partially supported by the National Natural Science foundation of China under Grant (No. 61374028), and sponsored by Research Fund for Excellent Dissertation of China Three Gorges University (No. 2018SSPY132 and No. 2018SSPY134).
References [1] Awan, M. U., Noor, M. A., Mihai, M. V., Noor, K. I., Fractional Hermite-Hadamard inequalities for differentiable s-Godunova-Levin functions, Filomat, 30 (12), (2016), 3235-3241. [2] Chu, Y. M., Khan, M. A., Ali, T., Dragomir, S. S., Inequalities for α-fractional differentiable functions, J. Inequal. Appl., Article Number 93, (2017), 12 pages. [3] Chu, Y. M., Khan, M. A., Khan T. U., Ali, T., Generalizations of Hermite-Hadamard type inequalities for M T -convex functions, J. Nonlinear Sci. Appl., 9, (2016), 4305-4316. [4] Du, T. S., Li, Y. J., Yang, Z. Q., A generalization of Simpson’s inequality via differentiable mapping using extended (s, m)-convex functions, Appl. Math. Comput., 293, (2017), 358-369. [5] Iqbal, A., Khan, M. A., Ullah, S., Chu, Y. M., Kashuri, A., Hermite-Hadamard type inequalities pertaining conformable fractional integrals and their applications, AIP advances, 8, Article Number 075101, (2018), 1-18. ˙ scan, I., ˙ Hermite-Hadamard type inequalities for harmonically (α, m)-convex functions, Hacet. J. [6] I¸ Math. Stat. , 45 (2), (2016), 381-390. [7] Khan, M. A., Ali, T., Dragomir, S. S., Sarikaya, M. Z., Hermite-Hadamard type inequalities for conformable fractional integrals, Rev. R. Acad. Cienc. Exactas F´ıs. Nat. Ser. A Math. RACSAM, (2017), doi:10.1007/s13398-017-0408-5 [8] Khan, M. A., Ali, T., Khan, T. U., Hermite-Hadamard type inequalities with applications, Fasciculi Mathematici, 59, (2017), 57-74. [9] Khan, M. A., Begum, S., Khurshid, Y., Chu, Y. M., Ostrowski type inequalities involving conformable fractional integrals, J. Inequal. Appl., Article Number 70, (2018), 14 pages. [10] Khan, M. A., Chu, Y. M., Kashuri, A., Liko, R., Ali, G., Conformable fractional integrals versions of Hermite-Hadamard inequalities and their generalizations, J. Funct. Spaces, 2018, Article ID 6928130, 9 pages. [11] Khan, M. A., Chu, Y. M., Khan, T. U., Khan, J., Some new inequalities of Hermite–Hadamard type for s-convex functions with applications, Open Math., 15, (2017), 1414-1430. [12] Khan, M. A., Iqbal, A., Suleman, M., Chu, Y. M., Hermite-Hadamard type inequalities for fractional integrals via green’s function, J. Inequal. Appl., Article Number 161, (2018), 15 pages. [13] Latif, M. A., Inequalities of Hermite-Hadamard type for functions whose derivatives in absolute value are convex with applications, Arab. J. Math. Sci., 21 (1), (2015), 84-97. [14] Mihai, M. V., Mitroi, F. C., Hermite-Hadamard type inequalities obtained via Riemann-Liouville fractional calculus, Acta Math. Univ. Comenianae, Vol. LXXXIII, 2, (2014), 209-215.
10
[15] Mubeen, S., Habibullah, G. M., k-Fractional integrals and application, Int. J. Contemp. Math. Sciences, 7 (2), (2012), 89-94. [16] Noor, M. H., Noor, K. I., Mihai, M. V., Awan, M. U., Fractional Hermite-Hadamard inequalities for some classes of differentiable perinvex functions, U.P.B. Sci. Bull. Series A., 78 (3), (2016), 163-174. [17] Sarikaya, M. Z., Karaca, A., On the k-Riemann-Liouville fractional integral and applications, Int. J. Stat. Math. 1 (3), (2014), 33-43. [18] Set, E., Tomar, M., Sarikaya, M. Z., On generalized Gr¨ uss type inequalities for k-fractional integrals, Appl. Math. Comput., 269, (2016), 29-34. [19] Xi, B-Y. Qi, F., Zhang, T-Y., Some inequalities of Hermite-Hadamard type for m-harmonicarithmetically convex functions, ScienceAsia, 41, (2015), 357-361.
11
S1018-3647(18)31243-6 https://doi.org/10.1016/j.jksus.2018.09.010 JKSUS 733
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Journal of King Saud University - Science
Please cite this article as: T. Du, H. Wang, M.A. Latif, Y. Zhang, Estimation type results associated to k-fractional integral inequalities with applications, Journal of King Saud University - Science (2018), doi: https://doi.org/ 10.1016/j.jksus.2018.09.010
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Estimation type results associated to k-fractional integral inequalities with applications
Abstract This paper aims to establish k-fractional integral inequalities with multiple parameters involving the first differentiable mappings. Applications of our results to random variables are also given. 2010 Mathematics Subject Classification: 26A33; 26D15; 26E60; 41A55. Key words: invex set; preinvex functions; k-fractional integrals.
1
Introduction
If h : I ⊆ R → R is a convex mapping and e1 , e2 ∈ I with e1 < e2 , then one has Z e2 e1 + e2 1 h(e1 ) + h(e2 ) h ≤ h(t)dt ≤ . 2 e 2 − e 1 e1 2 The subsequent inequality is an improvement of the inequality (1.1). e + e 1 3e + e e + 3e 1 2 1 2 1 2 h ≤ h +h 2 2 4 4 Z e2 1 ≤ h(t)dt e2 − e1 e1 1 e1 + e2 h(e1 ) + h(e2 ) h(e1 ) + h(e2 ) ≤ h + ≤ . 2 2 2 2
(1.1)
(1.2)
Many different generalizations, new extensions and improvements related to the inequality (1.1) ˙ scan [6], Khan et al. can be found in Awan et al. [1], Chu et al. [2, 3], Du et al. [4], Iqbal et al. [5], I¸ [7–12], Sarikaya and Karaca [17], Set et al. [18], Xi et al. [19]. Let us consider an invex set K. A set K ⊆ Rn is named invex set with respect to the mapping δ : K × K → Rn , if ν + λδ(µ, ν) ∈ K holds, for all µ, ν ∈ K and λ ∈ [0, 1]. A mapping h : K → R is called preinvex respecting δ, if the following inequality: h ν + λδ(µ, ν) ≤ (1 − λ)h(ν) + λh(µ) (1.3) grips, for every µ, ν ∈ K and λ ∈ [0, 1]. The preinvex function is an important substantive generalization of the convex function. Example 1.1 The function h(x) = 12 − x − 12 , (x ∈ R) with respect to the following 2(y − x), x ≥ 12 , y ≥ 12 , y ≥ x, 0, x > 12 , y ≥ 12 , y < x, x < 12 , y ≤ 12 , y > x, 0, y − x, x ≤ 12 , y ≤ 12 , y ≤ x, δ(y, x) = 1 − x − y, x < 12 , y > 12 , x + y ≥ 1, 0, x < 12 , y > 12 , x + y ≤ 1, 0, x > 12 , y < 12 , x + y ≥ 1, 1 − x − y, x > 12 , y < 12 , x + y ≤ 1 is preinvex on R while it is not convex on R. 1
(1.4)
This paper aims to obtain estimation type results related to k-fractional integral operators. For this, we suppose that the absolute value of the derivative of the considered mapping is preinvex. Next, we substitute this hypothesis with the boundedness of the derivative and with a Lipschitzian condition for the derivative of the considered mapping to derive integral inequalities with new bounds. Applications of our results to random variables are also provided. We end this section by reciting a well-known k-fractional integral operators in the literature. Definition 1.1 (Mubeen and Habibullah [15]) Let h ∈ L1 [µ, ν], the k-fractional integrals k Jµτ+ h(x) and k Jντ− h(x) of order τ > 0 are defined by Z x τ 1 τ J h(x) = (x − λ) k −1 h(λ)dλ, (0 ≤ µ < x < ν) + k µ kΓk (τ ) µ and τ k Jν − h(x)
1 = kΓk (τ )
Z
ν
τ
(λ − x) k −1 h(λ)dλ, (0 ≤ µ < x < ν),
x
respectively, where k > 0, and Γk is the k-gamma function defined as Γk (x) := Re(x) > 0, with the properties Γk (x + k) = xΓk (x) and Γk (k) = 1.
2
R∞ 0
λx−1 e−
λk k
dλ,
Main Results
Throughout this work, let K ⊆ R be an open invex subset respecting δ : K × K → R \ {0} and r1 , r2 ∈ K with r1 < r2 , and let h : K → R be a differentiable mapping such that h0 is integrable on the δ-path Pxv : v = x + δ(y, x) with x, y ∈ [r1 , r2 ]. Before stating the results, we use notations below. τ τ δ k (x, r1 ) h r1 + δ(x, r1 ) + h(r1 ) + δ k (r2 , x) h(x) + h x + δ(r2 , x) Lδ (τ, k, λ; x) :=(1 − λ) δ(r2 , r1 ) τ τ 2λ δ k (x, r1 )h r1 + 12 δ(x, r1 ) + δ k (r2 , x)h x + 12 δ(r2 , x) + δ(r2 , r1 ) τ k 2 Γk (τ + k) 1 1 τ τ − k J[r1 +δ(x,r1 )]− h r1 + δ(x, r1 ) + k Jr + h r1 + δ(x, r1 ) 1 δ(r2 , r1 ) 2 2 1 1 τ + k Jxτ+ h x + δ(r2 , x) + k J[x+δ(r δ(r2 , x) . −h x + 2 ,x)] 2 2 (2.1) If δ(µ, ν) = µ − ν with µ, ν ∈ [r1 , r2 ], then we have τ
τ
(x − r1 ) k [h(r1 ) + h(x)] + (r2 − x) k [h(x) + h(r2 )] L(τ, k, λ; x) := (1 − λ) r2 − r1 τ τ τ r +x 1 2 2λ (x − r1 ) k h 2 + (r2 − x) k h x+r 2 k Γk (τ + k) 2 (2.2) + − r2 − r1 r2 − r1 r + x r + x x + r x + r 1 2 2 1 × k Jrτ+ h + k Jxτ− h + k Jxτ+ h + k Jrτ− h . 1 2 2 2 2 2 (a) Choosing λ = 0 and τ = k = 1, the equation (2.2) reduces to L(1, 1, 0; x) := h(x) +
(r2 − x)h(r2 ) + (x − r1 )h(r1 ) 2 − r2 − r1 r2 − r1
Z
r2
h(u)du. r1
Specially, Z r2 r + r r1 + r2 h(r1 ) + h(r2 ) 2 1 2 L 1, 1, 0; := +h − h(u)du. 2 2 2 r2 − r1 r1
2
(b) Choosing λ = 1 and τ = k = 1, the equation (2.2) degenerates into Z r2 2 2 (x − r1 )h r12+x + (r2 − x)h x+r 2 2 L(1, 1, 1; x) := − h(u)du. r2 − r1 r2 − r1 r1 Specially, Z r2 3r + r r + 3r r1 + r2 2 1 2 1 2 L 1, 1, 1; := h +h − h(u)du. 2 4 4 r2 − r1 r1 We need the following lemma for our results. Lemma 2.1 One has the subsequent identity Lδ (τ, k, λ; x) ) Z 1 τ 1 + t 1 − t t − λ h0 r1 + δ(x, r1 ) dt − t k − λ h0 r1 + δ(x, r1 ) dt 2 2 0 0 (Z ) Z 1 τ 1 τ τ δ k +1 (r2 , x) 1−t 1+t 0 0 − tk − λ h x + δ(r2 , x) dt − tk − λ h x + δ(r2 , x) dt 2δ(r2 , r1 ) 2 2 0 0 τ
δ k +1 (x, r1 ) = 2δ(r2 , r1 )
(Z
1
τ k
(2.3) for k-fractional integrals with x ∈ (r1 , r2 ), τ > 0, k > 0 and λ ∈ [0, 1]. Proof Using integration by parts and appropriate substitutions, such as µ = r1 + r1 + 1−t 2 δ(x, r1 ) · · · , we can obtain the identity (2.3). Example 2.1 If we take r1 = 12 , r2 = 1, τ = 1 = k, h(x) = 12 − x − 12 , and ( 2(µ − ν), µ ≥ ν, δ(µ, ν) = 0, µ < ν,
1+t 2 δ(x, r1 ),
ν =
(2.4)
then all the assumptions in Lemma 2.1 are satisfied. Clearly, the left of the equation (2.3) is 3 1 2 x − 21 2 − 2x − 2 + 2(1− x)[(1 − x) + (x − 1)] Lδ (1, 1, λ; x) = (1 − λ) 2 1 − 12 2 x − 12 [1 − x] + 2(1 − x)[1 − 1] + 2λ 2 1 − 12 Z x Z 2x− 12 Z 1 Z 2−x (1 − t)dt + (1 − t)dt + (1 − t)dt + (1 − t)dt = 0 −2 1 2
x
x
1
and the right of the equation (2.3) is Z 1 Z 1 Z 1 Z 1 1 2 4 x− −(t − λ)dt + (t − λ)dt − 4(1 − x)2 −(t − λ)dt + (t − λ)dt = 0. 2 0 0 0 0 Corollary 2.1 If δ(µ, ν) = µ − ν with µ, ν ∈ [r1 , r2 ] in Lemma 2.1, then one has L(τ, k, λ; x) τ
Z 1 1 + t 1 − t τ 1−t 1+t 0 0 k t −λ h x+ r1 dt − t −λ h x+ r1 dt 2 2 2 2 0 0 Z 1 Z 1 τ 1 + t 1 − t τ τ (r2 − x) k +1 1−t 1+t − t k − λ h0 x+ r2 dt − t k − λ h0 x+ r2 dt , 2(r2 − r1 ) 2 2 2 2 0 0 (2.5)
(x − r1 ) k +1 = 2(r2 − r1 )
Z
1
τ k
which is a new form of Lemma 2.1. Remark 2.1 (a) In Lemma 2.1, putting λ = 0 and k = 1, one has Lemma 2.8 in Noor et al. [16]. (b) In Corollary 2.1, taking λ = 0 and k = 1, one has Lemma 1 in Mihai and Mitroi [14]. Further, choosing τ = 1, one has Lemma 1 in Latif [13] . 3
Theorem 2.1 If |h0 |q for q > 1 is preinvex on K, then the coming inequality for k-fractional integrals with x ∈ (r1 , r2 ), τ > 0, k > 0, p−1 + q −1 = 1 and λ ∈ [0, 1] grips: Lδ (τ, k, λ; x) (" τ +1 1 0 1 ) δ k (x, r1 ) 1 |h0 (r1 )|q + 3|h0 (x)|q q 3|h (r1 )|q + |h0 (x)|q q p ≤ ρ (τ, k, λ, p) + 2|δ(r2 , r1 )| 4 4 (2.6) (" τ +1 1 1) δ k (r2 , x) 1 3|h0 (x)|q + |h0 (r2 )|q q |h0 (x)|q + 3|h0 (r2 )|q q + ρ p (τ, k, λ, p) + , 2|δ(r2 , r1 )| 4 4 where k , τ p+k τ p+k τ p+k k 2k p ρ(τ, k, λ, p) = τ p+k − τ p+k λ τ − λ + 2λ τ , k τ +1) Γ(p+1)Γ( , k
λ = 0, 0 < λ < 1, λ = 1.
Γ( τ +1+p)
Proof According to Lemma 2.1, the Holder inequality and preinvexity of |h0 |q , one gets Z 1 τ p p1 q q k +1 (x, r )| τk 1 Lδ (τ, k, λ; x) ≤ |δ χ1 + χ2 t − λ dt 2|δ(r2 , r1 )| 0 τ +1 p p1 δ k (r2 , x) Z 1 τ q q k + χ3 + χ4 , t − λ dt 2|δ(r2 , r1 )| 0 where Z
1
q 1+t 0 δ(x, r1 ) dt h r1 + 2 0 Z 1 1−t 0 1+t 0 |h0 (r1 )|q + 3|h0 (x)|q q q ≤ |h (r1 )| + |h (x)| dt = , 2 2 4 0 Z 1 q 1−t 3|h0 (r1 )|q + |h0 (x)|q 0 χ2 = δ(x, r1 ) dt ≤ , h r1 + 2 4 0 Z 1 q 1−t 3|h0 (x)|q + |h0 (r2 )|q 0 χ3 = δ(r2 , x) dt ≤ , h x + 2 4 0 Z 1 q 1+t |h0 (x)|q + 3|h0 (r2 )|q 0 χ4 = δ(r2 , x) dt ≤ . h x + 2 4 0 When λ = 0, we have Z 1 p k τk , t − λ dt = τp + k 0 χ1 =
when λ = 1, we have Z 0
1
τ p Γ(p + 1)Γ( τk + 1) k , t − λ dt = Γ τk + 1 + p
when 0 < λ < 1, we have Z 0
1
k
Z τ p k t − λ dt =
λτ
τ
λ − tk
p
Z dt +
k
0
Z ≤ =
1
τ p t k − λ dt
λτ 1
k λτ
k
t
τp k
−λ
p
Z dt +
λτ
λp − t
τp k
dt
0
τ p+k τ p+k k 2k − λ τ − λp + 2λ τ . τp + k τp + k
The above inequality is obtained by using the fact that (α − β)κ ≤ ακ − β κ , α ≥ β ≥ 0, κ ≥ 1. The proof of Theorem 2.1 is completed. 4
(2.7)
Corollary 2.2 Choosing λ = 1, k = 1 and δ(µ, ν) = µ − ν for µ, ν ∈ [r1 , r2 ] in Theorem 2.1, one obtains L(τ, 1, 1; x) 2 (x − r )τ h r1 +x ) + (r − x)τ h x+r2 2τ Γ(τ + 1) 1 2 2 2 := − r2 − r1 r2 − r1 r + x r + x x + r x + r 1 1 2 2 τ τ τ τ × Jr+ h + Jx− h + Jx+ h + Jr− h 1 2 2 2 2 2 ( 1 q2 +1 Γ(p + 1)Γ( 1 + 1) (x − r )τ +1 1 1 1 0 q 0 q q 0 q 0 q q τ ≤ |h (r1 )| + 3|h (x)| + 3|h (r1 )| + |h (x)| 2 r2 − r1 Γ τ1 + 1 + p ) 1 1 (r2 − x)τ +1 0 q 0 q q 0 q 0 q q + |h (x)| + 3|h (r2 )| + 3|h (x)| + |h (r2 )| . r2 − r1 (2.8) 2 Specially, putting τ = 1 and x = r1 +r 2 , one has Z r2 r + 3r 3r1 + r2 2 1 2 h h(u)du + h − 4 4 r2 − r1 r1 ( r + r q q1 r + r q q1 1 p1 1 q2 +1 r − r 2 1 1 2 1 2 ≤ |h0 (r1 )|q + 3 h0 + 3|h0 (r1 )|q + h0 p+1 2 4 2 2 q1 q1 ) 0 r1 + r2 q 0 r1 + r2 q 0 q 0 q + h + 3 h + 3|h (r2 )| + |h (r2 )| 2 2 1 p1 1 q3 +1 r − r h 1 i 1 1 1 2 1 ≤ 1 q + 3 q + 5 q + 7 q |h0 (r1 )| + |h0 (r2 )| . p+1 2 4 (2.9)
The second inequality is obtained by applying the convexity of |h0 |q and the succeeding fact that n n n X X X (%i + σi )s ≤ (%i )s + (σi )s , %i , σi > 0, 0 ≤ s < 1. i=1
i=1
i=1
Theorem 2.2 If |h0 |q for q ≥ 1 is preinvex on K, then the coming inequality for k-fractional integrals with x ∈ (r1 , r2 ), τ > 0, k > 0 and λ ∈ [0, 1] grips: ( τ 1 q k +1 (x, r1 ) 1− q1 0 0 q q 1 Lδ (τ, k, λ; x) ≤ δ ∆ ∆1 − ∆2 h (r1 ) + ∆1 + ∆2 h (x) 2|δ(r2 , r1 )| 1 2 1 ) q 0 0 q q 1 + ∆1 + ∆2 h (r1 ) + ∆1 − ∆2 h (x) 2 ( τ +1 1 (2.10) δ k (r2 , x) 1− 1 1 q q 0 q 0 q + ∆ ∆1 + ∆2 h (x) + ∆1 − ∆2 h (r2 ) 2|δ(r2 , r1 )| 1 2 1 ) q q 0 q 0 1 + ∆1 − ∆2 h (x) + ∆1 + ∆2 h (r2 ) , 2 where 1
Z ∆1 = 0
τ k t − λ dt =
τ +k k τ − λ + 2λ τ τ +k τ +k
(2.11)
and Z ∆2 = 0
1
τ t t k − λ dt =
τ +2k k λ τ − +λ τ . τ + 2k 2 τ + 2k
5
(2.12)
Proof Using Lemma 2.1 and the power mean inequality, one has τ Z 1 1− q1 1 k +1 (x, r1 ) 1 τk Lδ (τ, k, λ; x) ≤ δ I1 q + I2 q t − λ dt 2|δ(r2 , r1 )| 0 τ +1 1− q1 1 δ k (r2 , x) Z 1 τ 1 k + I3 q + I4 q , t − λ dt 2|δ(r2 , r1 )| 0 where Z
1
τ q 1+t k δ(x, r1 ) dt t − λ h0 r1 + 2 0 Z 1 1 − t q 1 + t τk 0 0 q ≤ t − λ h (r1 ) + h (x) dt 2 2 0 Z 1 Z 1 Z 1 q Z 1 τ τ τ 1 t τk − λ dt − t k − λ dt + = t t k − λ dt h0 (r1 ) + t t k − λ dt h0 (x) , 2 0 0 0 0 Z 1 q 1−t τk I2 = δ(x, r1 ) dt t − λ h0 r1 + 2 0 Z 1 Z 1 Z 1 q Z 1 τ τ τ τ 1 0 0 k k k k ≤ t − λ dt + t t − λ dt h (r1 ) + t − λ dt − t t − λ dt h (x) , 2 0 0 0 0 Z 1 q 1−t τk I3 = δ(r2 , x) dt t − λ h0 x + 2 0 Z 1 Z 1 Z 1 Z 1 τ τ τ τ 1 0 q 0 k k k k ≤ t − λ dt + t t − λ dt h (x) + t − λ dt − t t − λ dt h (r2 ) , 2 0 0 0 0 Z 1 q 1+t τk I4 = δ(r2 , x) dt t − λ h0 x + 2 0 Z 1 Z 1 Z 1 q Z 1 τ τ τ τ 1 t k − λ dt − t k − λ dt + ≤ t t k − λ dt h0 (x) + t t k − λ dt h0 (r2 ) . 2 0 0 0 0 I1 =
These last four inequalities clasp due to the preinvexity of |h0 |q . This ends the proof. Corollary 2.3 Choosing λ = 1, k = 1 and δ(µ, ν) = µ − ν with µ, ν ∈ [r1 , r2 ] in Theorem 2.2, one obtains ( 1 q τ +1 τ (x − r ) τ + 3 0 3τ + 5 0 q q 1 L(τ, 1, 1; x) ≤ h (r1 ) + h (x) 2(τ + 1)(r2 − r1 ) 4(τ + 2) 4(τ + 2) 1) q 3τ + 5 0 τ + 3 0 q q + h (r1 ) + h (x) 4(τ + 2) 4(τ + 2) (2.13) ( q q1 τ (r2 − x)τ +1 τ + 3 0 q 3τ + 5 0 + h (x) + h (r2 ) 2(τ + 1)(r2 − r1 ) 4(τ + 2) 4(τ + 2) ) q q1 3τ + 5 0 q τ + 3 0 + . h (x) + h (r2 ) 4(τ + 2) 4(τ + 2) 2 Specially, taking τ = 1, x = r1 +r and utilizing similar arguments as in Corollary 2.2, one gets 2 Z r2 r + 3r 3r1 + r2 2 1 2 h +h − h(u)du 4 4 r2 − r1 r1 ( q 1 r + r q q1 1 2 (r2 − r1 ) 1 0 q 2 0 r1 + r2 q q 0 1 2 ≤ + h (r1 ) + h0 h (r1 ) + h 16 3 3 2 3 3 2 (2.14) ) q q1 2 r + r q 1 q q1 1 0 r1 + r2 q 2 0 0 1 0 2 + h + h + h (r2 ) + h (r2 ) 3 2 3 3 2 3 1h i 1 1 1 (r2 − r1 ) 1 q q1 ≤ 1 + 2 q + 4 q + 5 q |h0 (r1 )| + |h0 (r2 )| . 16 2
6
Remark 2.2 Theorem 2.1 and 2.2 will be reduced to Theorem 2 and 3 in Latif [13], respectively, if we choose λ = 0, k = 1 = τ and δ(µ, ν) = µ − ν for µ, ν ∈ [r1 , r2 ].
3
New estimation results
For obtaining new estimation type results, we deal with the boundedness and the Lipschitzian condition of h0 , respectively. Theorem 3.1 Assume that there exists constants m < M such that −∞ < m ≤ h0 (y) ≤ M < ∞ for all y ∈ K, then the inequality τ τ τ +k k +1 (x, r )| + |δ k +1 (r , x)| 2τ λ τ + k 1 2 Lδ (τ, k, λ; x) ≤ (M − m) |δ (3.1) −λ 2|δ(r2 , r1 )| τ +k holds with τ > 0, k > 0, λ ∈ [0, 1] and x ∈ (r1 , r2 ). Proof From Lemma 2.1, we have (Z τ 1 0 m + M τ δ k +1 (x, r1 ) 1+t k Lδ (τ, k, λ; x) = t − λ h r1 + δ x, r1 − dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 m + M τ 1−t − t k − λ h r1 + δ x, r1 − dt 2 2 0 (Z τ 1 0 m + M τ δ k +1 (r2 , x) 1−t k − t −λ h x+ δ r2 , x − dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 m + M τ 1+t − tk − λ h x + δ r2 , x − dt . 2 2 0 ≤ h0 (y) − m+M ≤ M − m+M 2 2 , one obtains τ ( Z 1 k +1 (x, r1 ) m + M 1+t τk 0 Lδ (τ, k, λ; x) ≤ δ dt δ x, r1 − t − λ h r1 + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 m + M 1−t τk 0 dt + δ x, r1 − t − λ h r1 + 2 2 0 τ +1 ( δ k (r2 , x) Z 1 τ m + M 1−t 0 dt k + δ r2 , x − t − λ h x + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 m + M 1+t τk 0 dt + δ r2 , x − t − λ h x + 2 2 0 τ +1 τ +1 δ k (x, r1 ) M − m Z 1 τ δ k (r2 , x) M − m Z 1 τ ≤ t k − λ dt + t k − λ dt |δ(r2 , r1 )| 2 |δ(r , r )| 2 2 1 0 0 τ τ τ +k (M − m) |δ k +1 (x, r1 )| + |δ k +1 (r2 , x)| 2τ λ τ + k = −λ . 2|δ(r2 , r1 )| τ +k
Using the fact that m −
m+M 2
(3.2)
The proof is completed. Corollary 3.1 In Theorem 3.1, choosing δ(µ, ν) = µ − ν for µ, ν ∈ [r1 , r2 ], and τ = k = 1, one gets (x − r1 )[h(r1 ) + h(x)] + (r2 − x)[h(x) + h(r2 )] |L(1, 1, λ; x)| = (1 − λ) r2 − r1 Z r2 2 (x − r1 )h r12+x + (r2 − x)h x+r 2 2 + 2λ − h(u)du (3.3) r2 − r1 r2 − r1 r1 (M − m) (x − r1 )2 + (r2 − x)2 2λ2 − 2λ + 1 ≤ . 2(r2 − r1 ) 2 7
2 Remark 3.1 Taking x = r1 +r in Corollary 3.1, one can see the following. 2 (a) For λ = 0, we have Z r2 h(r ) + h(r ) (M − m)(r − r ) r + r 2 1 2 1 2 2 1 +h − h(u)du ≤ . 2 2 r2 − r1 r1 8
(b) For λ = 12 , we have h(r ) + h(r ) 1 r + r h 1 2 1 2 + h + 4 2 2
3r1 +r2 4
+h 2
r1 +3r2 4
2 − r2 − r1
Z
r2
r1
(3.4)
(M − m)(r − r ) 2 1 h(u) ≤ . 16 (3.5)
(c) For λ = 1, we have Z r2 r + 3r 3r1 + r2 (M − m)(r2 − r1 ) 2 1 2 h +h − h(u)du ≤ . 4 4 r2 − r1 r1 8
(3.6)
Theorem 3.2 Assume that h0 satisfies Lipschitz condition on K for some L > 0, then the inequality τ τ 2k k +2 (x, r )| + |δ k +2 (r , x)| 2τ λ τ +1 + 2k 1 2 Lδ (τ, k, λ; x) ≤ L |δ −λ 4|δ(r2 , r1 )| τ + 2k
(3.7)
grips for all x ∈ (r1 , r2 ), µ > 0, k > 0 and λ ∈ [0, 1]. Proof From Lemma 2.1, we have (Z τ 1 0 τ δ k +1 (x, r1 ) 1+t 1 0 k Lδ (τ, k, λ; x) = t − λ h r1 + δ x, r1 − h r1 + δ x, r1 dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 τ 1−t 1 0 − t k − λ h r1 + δ x, r1 − h r1 + δ x, r1 dt 2 2 0 (Z τ 1 0 τ δ k +1 (r2 , x) 1−t 1 0 k − t −λ h x+ δ r2 , x − h x + δ r2 , x dt 2δ(r2 , r1 ) 2 2 0 ) Z 1 0 τ 1+t 1 0 − tk − λ h x + δ r2 , x − h x + δ r2 , x dt . 2 2 0 Utilizing the fact that h0 satisfies Lipschitz condition on K for L > 0, we have τ ( Z 1 δ k +1 (x, r1 ) 1+t 1 τk Lδ (τ, k; λ, x) ≤ δ x, r1 − h0 r1 + δ(x, r1 ) dt t − λ h0 r1 + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 0 1 − t 1 τk + δ x, r1 − h0 r1 + δ(x, r1 ) dt t − λ h r1 + 2 2 0 τ +1 ( Z 1 δ k (r2 , x) 1−t 1 τk + δ r2 , x − h0 x + δ(r2 , x) dt t − λ h0 x + 2|δ(r2 , r1 )| 2 2 0 ) Z 1 1+t 1 τk + δ r2 , x − h0 x + δ(r2 , x) dt , t − λ h0 x + 2 2 0 where 0 h r1 + 1 + t δ(x, r1 ) − h0 r1 + 1 δ(x, r1 ) ≤ L r1 + 1 + t δ(x, r1 ) − r1 + 1 δ(x, r1 ) = tL δ(x, r1 ) , 2 2 2 2 2 0 h r1 + 1 − t δ(x, r1 ) − h0 r1 + 1 δ(x, r1 ) ≤ tL δ(x, r1 ) , 2 2 2 8
0 h x + 1 − t δ(r2 , x) − h0 a + 1 δ(r2 , x) ≤ tL δ(r2 , x) , 2 2 2 0 h x + n + t δ(r2 , x) − h0 a + 1 δ(r2 , x) ≤ tL δ(r2 , x) . n+1 2 2 After a simple calculation, we obtain the desired result. Corollary 3.2 In Theorem 3.2, choosing δ(µ, ν) = µ − ν for µ, ν ∈ [a, b], and τ = k = 1, one has L (x − r1 )3 + (r2 − x)3 2λ3 + 2 (3.8) |L(1, 1, λ; x)| ≤ −λ . 4(r2 − r1 ) 3 2 Remark 3.2 Putting x = r1 +r in Corollary 3.2, one can see the following. 2 (a) For λ = 0, we have Z r2 h(r ) + h(r ) L(r − r )2 r + r 2 1 2 1 2 2 1 +h − h(u)du ≤ . 2 2 r2 − r1 r1 24
(b) For λ = 12 , we h(r ) + h(r ) 1 2 + 4
have 1 r1 + r2 h h + 2 2
3r1 +r2 4
+h 2
r1 +3r2 4
2 − r2 − r1
Z
r2
r1
L(r − r )2 2 1 h(u) ≤ . (3.10) 64
(c) For λ = 1, we have Z r2 r + 3r 3r1 + r2 L(r2 − r1 )2 2 1 2 h ≤ h(u)du . + h − 4 4 r2 − r1 r1 48
4
(3.9)
(3.11)
Applications for random variables
Let X be a random variable in [r1 , r2 ], with the probability density mapping p : [r1 , r2 ] → [0, 1], and with the cumulative distribution mapping Z x F (x) = P(X ≤ x) = p(λ)dλ. a
Using the fact that E(X) =
R r2 r1
λdF (λ) = r2 −
R r2 r1
F (λ)dλ, one has the following results.
2 Proposition 4.1 In Theorem 2.2, taking λ = 1, k = 1 = τ , x = r1 +r and δ(u, v) = u − v with 2 u, v ∈ [r1 , r2 ], one has 2 P X ≤ 3r1 + r2 + P X ≤ r1 + 3r2 − (r2 − E(X)) 4 4 r2 − r1 (4.1) i 1 1 1 (r2 − r1 ) 1 q1 h q1 ≤ 1 + 2 q + 4 q + 5 q |p(r1 )| + |p(r2 )| . 16 2 2 Proposition 4.2 In Theorem 2.1, putting λ = 1, k = 1 = τ , x = r1 +r and δ(u, v) = u − v with 2 u, v ∈ [r1 , r2 ], one gets 2 P X ≤ 3r1 + r2 + P X ≤ r1 + 3r2 − (r − E(X)) 2 4 4 r2 − r1 (4.2) 1 3 1 p 1 q +1 r − r h 1 i 1 1 1 2 1 ≤ 1 q + 3 q + 5 q + 7 q |p(r1 )| + |p(r2 )| . p+1 2 4
Remark 4.1 Applications can be given based on the obtained results to special means, and we omit the details.
5
Conclusion
Four main results of the trapezium-like inequalities involving the mappings δ are hereby obtained. More new results can be derived by choosing different mappings δ and the special parameter values for λ, k and τ . 9
Competing interests The authors declare that there are no competing interests.
Author’s Contributions All authors read and approved the final manuscript.
Acknowledgments This work was partially supported by the National Natural Science foundation of China under Grant (No. 61374028), and sponsored by Research Fund for Excellent Dissertation of China Three Gorges University (No. 2018SSPY132 and No. 2018SSPY134).
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