On generalization of different type inequalities for harmonically quasi-convex functions via fractional integrals

0.16 MB·

by  İmdat İscan

ON GENERALIZATION OF DIFFERENT TYPE INEQUALITIES
FOR HARMONICALLY QUASI-CONVEX FUNCTIONS VIA
FRACTIONAL INTEGRALS
4
1
0 I˙MDATI˙S¸CAN
2

n Abstract. In this paper, we obtained somenew estimates ongeneralization
a of Hadamard, Ostrowski and Simpson-like type inequalities for harmonically
J quasi-convexfunctions viaRiemannLiouvillefractionalintegral.

7

]
A
1. Introduction
C
. In this section we will present definitions and some results used in this paper.
h
t A function f :I ⊆R→R is said to be a convex function if
a
m f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y)
[ holds for all x,y ∈I and λ∈[0,1].

1 The notion of quasi-convexfunctions generalizesthe notion of convexfunctions.
v More precisely, a function f :[u,v]→R is said quasi-convex on [u,v] if
6
f(λx+(1−λ)y)≤sup{f(x),f(y)},
4
9
for any x,y ∈ [u,v] and λ ∈ [0,1]. Clearly, any convex function is a quasi-convex
2
function. Furthermore, there exist quasi-convex functions which are not convex
.
1 (see [4]).
0
Following inequalities are well known in the literature as Hermite-Hadamard
4
inequality, Ostrowskiinequality and Simpson inequality respectively:
1
:
v Theorem 1. Let f : I ⊆ R→R be a convex function defined on the interval I of
i real numbers and u,v ∈I with u<v. The following double inequality holds
X
v
r
u+v 1 f(u)+f(v)
a
(1.1) f ≤ f(x)dx≤ .
2 v−u 2
(cid:18) (cid:19) Z
u
Theorem 2. Let f : I ⊆ R→R be a mapping differentiable in I◦, the interior of
I, and let u,v ∈I◦ with u<v. If |f′(x)|≤M for all x∈[u,v], then the following
inequality holds
v
1 M (x−u)2+(v−x)2
f(x)− f(t)dt ≤
(cid:12) v−u (cid:12) v−u 2
(cid:12) Z (cid:12) " #
(cid:12) u (cid:12)
(cid:12) (cid:12)
for all x∈[u,(cid:12)v]. (cid:12)
(cid:12) (cid:12)
2000 Mathematics Subject Classification. 26A33,26A51,26D15.
Keywordsandphrases. Hermite–Hadamardinequality,Riemann–Liouvillefractionalintegral,
Ostrowskiinequality,Simpsontypeinequalities,harmonicallyquasi-convexfunction.
1
2 I˙MDATI˙S¸CAN
Theorem 3. Let f :[u,v]→R be a four times continuously differentiable mapping
on (u,v) and f(4) = sup f(4)(x) <∞. Then the following inequality holds:
∞
x∈(u,v)
(cid:13) (cid:13) (cid:12) (cid:12)
(cid:13) (cid:13) (cid:12) (cid:12) v
1 f(u)+f(v) u+v 1 1
+2f − f(x)dx ≤ f(4) (v−u)4.
(cid:12)3 2 2 v−u (cid:12) 2880 ∞
(cid:12)(cid:12) (cid:20) (cid:18) (cid:19)(cid:21) Zu (cid:12)(cid:12) (cid:13)(cid:13) (cid:13)(cid:13)
(cid:12) (cid:12) (cid:13) (cid:13)
(cid:12)In [5], Iscan defined harmonically convex functio(cid:12)ns and gave some Hermite-
(cid:12) (cid:12)
Hadamard type inequalities for this class of functions
Definition 1. Let I ⊂ R\{0} be a real interval. A function f : I → R is said to
be harmonically convex, if
xy
(1.2) f ≤λf(y)+(1−λ)f(x)
λx+(1−λ)y
(cid:18) (cid:19)
for all x,y ∈ I and λ ∈ [0,1]. If the inequality in (1.2) is reversed, then f is said
to be harmonically concave.
In[17],Zhangetal. definedtheharmonicallyquasi-convexfunctionandsupplied
severalproperties of this kind of functions.
Definition 2. A function f : I ⊆ (0,∞) → [0,∞) is said to be harmonically
convex, if
xy
f ≤sup{f(x),f(y)}
λx+(1−λ)y
(cid:18) (cid:19)
for all x,y ∈I and λ∈[0,1].
We wouldliketopointoutthatanyharmonicallyconvexfunctiononI ⊆(0,∞)
is a harmonically quasi-convex function, but not conversely. For example, the
function
1, x∈(0,1];
f(x)=
((x−2)2, x∈[1,4].
is harmonically quasi-convex on (0,4], but it is not harmonically convex on (0,4].
Let us recall the following special functions:
(1) The Beta function:
1
β(x,y)= Γ(x)Γ(y) = tx−1(1−t)y−1dt, x,y >0,
Γ(x+y)
Z
0
(2) The hypergeometric function:
1
F (a,b;c;z)= 1 tb−1(1−t)c−b−1(1−zt)−adt, c>b>0, |z|<1 (see [11]).
2 1
β(b,c−b)
Z
0
We give some necessarydefinitions andmathematicalpreliminaries of fractional
calculus theory which are used throughout this paper.
Definition 3. Let f ∈ L[u,v]. The Riemann-Liouville integrals Jα f and Jα f
u+ v−
of oder α>0 with u≥0 are defined by
ON GENERALIZATION OF DIFFERENT TYPE INEQUALITIES 3
b
Jα f(b)= 1 (b−t)α−1f(t)dt, b>u
u+ Γ(α)
Z
u
and
v
Jα f(a)= 1 (t−a)α−1f(t)dt, a<v
v− Γ(α)
Z
a
∞
respectively, where Γ(α) is the Gamma function defined by Γ(α) = e−ttα−1dt
Z
0
and J0 f(x)=J0 f(x)=f(x).
u+ v−
In the case of α = 1, the fractional integral reduces to the classical integral. In
recent years, many athors have studied errors estimations for Hermite-Hadamard,
Ostrowski and Simpson inequalities; for refinements, counterparts, generalization
see [1, 2, 3, 7, 8, 9, 10, 12, 13, 14, 15, 16].
In [6], Iscan gave the following new identity for differentiable functions and ob-
tainedsomenew generalintegralinequalities forharmonicallyconvexfunctions via
fractional integrals.
Lemma 1. Let f : I ⊆(0,∞)→R be a differentiable function on I◦ and a,b∈I
with a<b. If f′ ∈L[a,b] then for all x∈[a,b] , λ∈[0,1] and α>0 we have:
1
(x−a)α+1 tα−λ ax
′
I (x,λ,α,a,b)= f dt
f,g (ax)α−1 A2(a,x) A (a,x)
Z t (cid:18) t (cid:19)
0
1
(b−x)α+1 tα−λ bx
′
− f dt,
(bx)α−1 A2(b,x) A (b,x)
Z t (cid:18) t (cid:19)
0
where A (u,x)=tu+(1−t)x and
t
(1.3)I (x,λ,α,a,b)
f,g
α α α α
x−a b−x x−a b−x
= (1−λ) + f(x)+λ f(a)+ f(b)
ax bx ax bx
(cid:20)(cid:18) (cid:19) (cid:18) (cid:19) (cid:21) (cid:20)(cid:18) (cid:19) (cid:18) (cid:19) (cid:21)
−Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b) , g(u)=1/u.
1/x+ 1/x−
h i
In this paper, by using of Lemma 1, we obtained a generalizationof Hadamard,
Ostrowski and Simpson type inequalities for harmonically quasi-convex functions
via Riemann Liouville fractional integral.
2. Main Results
Let f : I ⊆ (0,∞) → R be a differentiable function on I◦, the interior of
I, throughout this section we will take the notation I (x,λ,α,a,b) as in (1.3),
f,g
where a,b ∈I with a < b, x ∈ [a,b] , λ∈ [0,1], g(u)= 1/u, α> 0 and Γ is Euler
Gammafunction. Inordertoproveourmainresultsweneedthefollowingidentity.
4 I˙MDATI˙S¸CAN
Theorem 4. Let f : I ⊆ (0,∞) → R be a differentiable function on I◦ such that
f′ ∈ L[a,b], where a,b ∈ I◦ with a < b. If |f′|q is harmonically quasi-convex on
[a,b] for some fixed q ≥ 1, then for x ∈ [a,b], λ ∈ [0,1] and α > 0 the following
inequality holds
|I (x,λ,α,a,b)|≤C1−1/q(α,λ)
f,g 1
(2.1) × (x−a)α+1 sup |f′(x)|q,|f′(a)|q 1/qC1/q α,λ,q, a
(ax)α−1x2q 2 x
(
(cid:16) (cid:17)
(cid:2) (cid:8) (cid:9)(cid:3)
+(b−x)α+1 sup |f′(x)|q,|f′(a)|q 1/qC1/q α,λ,q,x ,
(bx)α−1b2q 3 b
)
(cid:16) (cid:17)
(cid:2) (cid:8) (cid:9)(cid:3)
where
2αλ1+α1 +1
C (α,λ)= −λ,
1
α+1
a
C α,λ,q,
2
x
1(cid:16) (cid:17) a a
= . F (2q;α+1;α+2,1− )−λ. F (2q;1;2,1− )
2 1 2 1
α+1 x x
+2λ1+α1. 2F1(2q;1;2,λ1/α 1− a )− 1 .2F1(2q;α+1;α+2,λ1/α 1− a ) ,
x α+1 x
(cid:20) (cid:16) (cid:17) (cid:16) (cid:17) (cid:21)
x
C α,λ,q,
3
b
1(cid:16) (cid:17) x x
= . F (2q;1;α+2,1− )−λ. F (2q;1;2,1− )
2 1 2 1
α+1 b b
+2λ1+α1. 2F1(2q;1;2,λ1/α 1− x )− 1 .2F1(2q;1;α+2,λ1/α 1− x ) .
b α+1 b
(cid:20) (cid:16) (cid:17) (cid:16) (cid:17) (cid:21)
Proof. Since|f′|q isharmonicallyquasi-convexon[a,b],fromLemma1,propertyof
the modulus and using the power-mean inequality we have
1
(x−a)α+1 |tα−λ| ax
′
|I (x,λ,α,a,b)|≤ f dt
f,g (ax)α−1 A2(a,x) A (a,x)
Z t (cid:12) (cid:18) t (cid:19)(cid:12)
0 (cid:12) (cid:12)
1 (cid:12) (cid:12)
(b−x)α+1 |tα−λ| bx (cid:12) (cid:12)
′
+ f dt
(bx)α−1 A2(b,x) A (b,x)
Z t (cid:12) (cid:18) t (cid:19)(cid:12)
0 (cid:12) (cid:12)
(x−a)α+1 1 (cid:12)(cid:12) 1−1q 1 |t(cid:12)(cid:12)α−λ| ax q q1
≤ |tα−λ|dt f′ dt
(ax)α−1 Z  Z A2tq(a,x)(cid:12) (cid:18)At(a,x)(cid:19)(cid:12) 
0 0 (cid:12) (cid:12)
(b−x)α+1 1  1−q1 1 |tα−λ|(cid:12)(cid:12) bx (cid:12)(cid:12) q  q1
+ |tα−λ|dt f′ dt
(bx)α−1 Z  Z A2tq(b,x)(cid:12) (cid:18)At(b,x)(cid:19)(cid:12) 
0 0 (cid:12) (cid:12)
   (cid:12) (cid:12) 
(cid:12) (cid:12)
ON GENERALIZATION OF DIFFERENT TYPE INEQUALITIES 5
1 1/q
1−1/q (x−a)α+1 ′ q ′ q 1/q |tα−λ|
≤ C (α,λ) sup |f (x)| ,|f (a)| dt
1  (ax)α−1  A2q(a,x) 
 (cid:2) (cid:8) (cid:9)(cid:3) Z0 t
 
(b−x)α+1 ′ q ′ q 1/q 1 |tα−λ| 1/q
(2.2) + sup |f (x)| ,|f (a)| dt
(bx)α−1  A2q(a,x)  
(cid:2) (cid:8) (cid:9)(cid:3) Z0 t 
 
where by simple computation we obtain 
1
C (α,λ) = |tα−λ|dt
1
Z
0
2αλ1+α1 +1
(2.3) = −λ,
α+1
1
|tα−λ|
(2.4) dt
A2q(a,x)
Z t
0
1 a a
= x−2q . F (2q;α+1;α+2,1− )−λ. F (2q;1;2,1− )
2 1 2 1
α+1 x x
(cid:26)
+2λ1+α1. 2F1(2q;1;2,λ1/α 1− a )− 1 .2F1(2q;α+1;α+2,λ1/α 1− a ) ,
x α+1 x
(cid:20) (cid:16) (cid:17) (cid:16) (cid:17) (cid:21)
and
1
|tα−λ|
(2.5) dt
A2q(b,x)
Z t
0
1 x x
= b−2q . F (2q;1;α+2,1− )−λ. F (2q;1;2,1− )
2 1 2 1
α+1 b b
(cid:26)
+2λ1+α1. 2F1(2q;1;2,λ1/α 1− x )− 1 .2F1(2q;1;α+2,λ1/α 1− x ) ,
b α+1 b
(cid:20) (cid:16) (cid:17) (cid:16) (cid:17) (cid:21)
Hence, If we use (2.3), (2.4) and (2.5) in (2.2), we obtain the desired result. This
completes the proof. (cid:3)
Corollary 1. Under the assumptions of Theorem 4 with q =1, the inequality (2.1)
reduced to the following inequality
|I (x,λ,α,a,b)|
f,g
(x−a)α+1 a
′ ′
≤ [sup{|f (x)|,|f (a)|}]C α,λ,1,
(ax)α−1x2 2 x
(
(cid:16) (cid:17)
(b−x)α+1 x
′ ′ 1/q
+ [sup{|f (x)|,|f (a)|}] C α,λ,1, ,
(bx)α−1b2 3 b
)
(cid:16) (cid:17)
6 I˙MDATI˙S¸CAN
Corollary 2. Under theassumptions of Theorem 4withα=1, theinequality(2.1)
reduced to the following inequality
ab
|I (x,λ,α,a,b)|
f,g
b−a
(cid:18) (cid:19)
b
b(x−a)f(a)+a(b−x)f(b) ab f(u)
= (1−λ)f(x)+λ − du
(cid:12) x(b−a) b−a u2 (cid:12)
(cid:12) (cid:20) (cid:21) Z (cid:12)
(cid:12) a (cid:12)
≤ (cid:12)(cid:12)(cid:12) ab 2λ2−2λ+1 1−1q (x−a)2 sup |f′(x)|q,|f′(a)|q 1(cid:12)(cid:12)(cid:12)/qC1/q 1,λ,q, a
b−a 2 x2q 2 x
(cid:18) (cid:19)(cid:18) (cid:19) ( (cid:2) (cid:8) (cid:9)(cid:3) (cid:16) (cid:17)
+(b−x)2 sup |f′(x)|q,|f′(a)|q 1/qC1/q 1,λ,q,x ,
b2q 3 b
)
(cid:16) (cid:17)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for x=H =2ab/(a+b), we get
f(a)+f(b) ab b f(u) b−a 2λ2−2λ+1 1−q1
(1−λ)f(H)+λ − du ≤
(cid:12) 2 b−a u2 (cid:12) 4ab 2
(cid:12) (cid:18) (cid:19) Z (cid:12) (cid:18) (cid:19)
(cid:12) a (cid:12)
×(cid:12)(cid:12)(cid:12) aH2H2q2C21/q 1,λ,q,a2+bb sup |f′(H)|q,|f′(a)(cid:12)(cid:12)(cid:12)|q 1q
(cid:26) (cid:18) (cid:19)
+ b2H2C1/q 1,λ,q, 2a s(cid:0)up |(cid:8)f′(H)|q,|f′(b)|q (cid:9)q1(cid:1) .
b2q 3 a+b
(cid:18) (cid:19) (cid:27)
(cid:0) (cid:8) (cid:9)(cid:1)
Corollary 3. In Theorem 4,
(1) If we take x = H = 2ab/(a+b), λ = 1, then we get the following Simpson
3
type inequality for fractional integrals
α
1 ab
[f(a)+4f(H)+f(b)]− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
6 b−a 1/H+ 1/H−
≤ (cid:12)(cid:12)(cid:12)(cid:12)b4−abaC11−1/q α,31 aH2H2q2(cid:18)sup (cid:19)|f′(H)|q,|f′(a)|qh 1/qC21/q α,31,q,Ha i(cid:12)(cid:12)(cid:12)(cid:12)
(cid:18) (cid:19)(cid:26) (cid:18) (cid:19)
+b2H2 sup |f′(H)|q,|f′(a(cid:2))|q (cid:8)1/qC1/q α,1,q,H(cid:9)(cid:3) ,
b2q 3 3 b
(cid:18) (cid:19)(cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for α=1, we get
1 ab b f(u) b−a 5 1−q1
[f(a)+4f(H)+f(b)]− du ≤
(cid:12)6 b−a u2 (cid:12) 4ab 18
(cid:12) Z (cid:12) (cid:18) (cid:19)
(cid:12) a (cid:12)
×(cid:12)(cid:12)(cid:12) aH2H2q2C21/q 1,13,q,a2+bb [sup{|f′(H)|(cid:12)(cid:12)(cid:12),|f′(a)|}]q1
(cid:26) (cid:18) (cid:19)
+b2H2C1/q 1,1,q, 2a sup |f′(H)|q,|f′(b)|q q1 .
b2q 3 3 a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
ON GENERALIZATION OF DIFFERENT TYPE INEQUALITIES 7
(2) If we take x=H =2ab/(a+b), λ=0, then we get the following midpoint type
inequality for fractional integrals
ab α
f(H)− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
b−a 1/H+ 1/H−
(cid:12) (cid:18) (cid:19) (cid:12)
≤ (cid:12)(cid:12)(cid:12)b−a 1 1−q1 a2H2C1/q αh,0,q,a+b sup |f′(H)|q,|f′(a)|q q1i(cid:12)(cid:12)(cid:12)
4ab α+1 H2q 2 2b
(cid:18) (cid:19) (cid:26) (cid:18) (cid:19)
+b2H2C1/q α,0,q, 2a sup |f′(H)|q,|f′((cid:2)b)|q (cid:8)q1 , (cid:9)(cid:3)
b2q 3 a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for α=1, we get
b
ab f(u)
f(H)− du
(cid:12) b−a u2 (cid:12)
(cid:12) Z (cid:12)
(cid:12) a (cid:12)
≤ (cid:12)(cid:12)(cid:12)b−a 1 1−q1 a2H2C1(cid:12)(cid:12)(cid:12)/q 1,0,q,a+b sup |f′(H)|q,|f′(a)|q q1
4ab 2 H2q 2 2b
(cid:18) (cid:19) (cid:26) (cid:18) (cid:19)
+b2H2C1/q 1,0,q, 2a sup |f′(H)|q,(cid:2)|f′((cid:8)b)|q q1 . (cid:9)(cid:3)
b2q 3 a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
(3) If we take x=H =2ab/(a+b), λ=1, then we get the following trapezoid type
inequality for fractional integrals
α
f(a)+f(b) ab
− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
2 b−a 1/H+ 1/H−
≤ (cid:12)(cid:12)(cid:12)(cid:12)b−a α 1(cid:18)−q1 a2(cid:19)H2C1/q α,1,q,ha+b sup |f′(H)|q,|f′(a)|q q1 i(cid:12)(cid:12)(cid:12)(cid:12)
4ab α+1 H2q 2 2b
(cid:18) (cid:19) (cid:26) (cid:18) (cid:19)
+b2H2C1/q α,1,q, 2a sup |f′(H)|q,|f′((cid:2)b)|q (cid:8)q1 , (cid:9)(cid:3)
b2q 3 a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for α=1, we get
b
f(a)+f(b) ab f(u)
− du
(cid:12) 2 b−a u2 (cid:12)
(cid:12) Z (cid:12)
(cid:12) a (cid:12)
≤ (cid:12)(cid:12)(cid:12)b−a 1 1−q1 a2H2C1/q 1,(cid:12)(cid:12)(cid:12)1,q,a+b sup |f′(H)|q,|f′(a)|q q1
4ab 2 H2q 2 2b
(cid:18) (cid:19) (cid:26) (cid:18) (cid:19)
+b2H2C1/q 1,1,q, 2a sup |f′(H)|q,(cid:2)|f′((cid:8)b)|q q1 . (cid:9)(cid:3)
b2q 3 a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
Corollary 4. Let the assumptions of Theorem 4 hold. If |f′(x)|≤M for all x∈
[a,b] and λ = 0, then we get the following Ostrowski type inequality for fractional
from the inequality (2.1) integrals
x−a α b−x α ab α
+ f(x)− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
ax bx b−a 1/x+ 1/x−
(cid:12)(cid:12)(cid:12)(cid:20)(cid:18) M (cid:19) (x(cid:18)−a)α+(cid:19)1 (cid:21) (cid:18) a (cid:19) (b−x)α+1 h x i(cid:12)(cid:12)(cid:12)
≤ (cid:12) C1/q α,0,q, + C1/q α,0,q, . (cid:12)
(α+1)1−1q "(ax)α−1x2q 2 x (bx)α−1b2q 3 b #
(cid:16) (cid:17) (cid:16) (cid:17)
8 I˙MDATI˙S¸CAN
Theorem 5. Let f : I ⊆ (0,∞) → R be a differentiable function on I◦ such that
f′ ∈ L[a,b], where a,b ∈ I◦ with a < b. If |f′|q is harmonically quasi-convex on
[a,b] for some fixed q ≥ 1, then for x ∈ [a,b], λ ∈ [0,1] and α > 0 the following
inequality holds
a (x−a)α+1 ′ q ′ q 1/q
(2|.I6) (x,λ,α,a,b)| ≤ C α,λ,1, sup |f (x)| ,|f (a)|
f,g 2 x (ax)α−1x2q
(cid:16) (cid:17)x (b−x)α+1(cid:2) (cid:8) ′ q ′ (cid:9)q(cid:3) 1/q
+C α,λ,1, sup |f (x)| ,|f (b)|
3 b (bx)α−1b2q
(cid:16) (cid:17)
(cid:2) (cid:8) (cid:9)(cid:3)
where C and C are defined as in Theorem 4.
2 3
Proof. Since|f′|q is harmonically quasi-convexon [a,b], from Lemma 1,using prop-
erty of the modulus and the power-mean inequality we have
|I (x,λ,α,a,b)|
f,g
(x−a)α+1 1 |tα−λ| 1−1q 1 |tα−λ| ax q q1
′
≤ dt f dt
(ax)α−1  A2(a,x)   A2(a,x) A (a,x) 
Z t Z t (cid:12) (cid:18) t (cid:19)(cid:12)
0 0 (cid:12) (cid:12)
(b−x)α+1 1 |tα−λ|  1−1q 1 |tα−λ|(cid:12)(cid:12) bx (cid:12)(cid:12) q  1q
′
+ dt f dt
(bx)α−1  A2(b,x)   A2(b,x) A (b,x) 
Z t Z t (cid:12) (cid:18) t (cid:19)(cid:12)
0 0 (cid:12) (cid:12)
   (cid:12) (cid:12) 
(cid:12) (cid:12)
a (x−a)α+1 ′ q ′ q 1/q
≤ C α,λ,1, sup |f (x)| ,|f (a)|
2 x (ax)α−1x2q
(cid:16) (cid:17)x (b−x)α+1(cid:2) (cid:8) ′ q ′ q(cid:9)(cid:3) 1/q
+C α,λ,1, sup |f (x)| ,|f (b)|
3 b (bx)α−1b2q
(cid:16) (cid:17)
(cid:2) (cid:8) (cid:9)(cid:3)
which completes the proof. (cid:3)
Corollary 5. Under the assumptions of Theorem 5 with q =1, the inequality (2.6)
reduced to the following inequality
a (x−a)α+1
′ ′
|I (x,λ,α,a,b)| ≤ C α,λ,1, [sup{|f (x)|,|f (a)|}]
f,g 2 x (ax)α−1x2
(cid:16) (cid:17)
x (b−x)α+1
′ ′
+C α,λ,1, [sup{|f (x)|,|f (b)|}]
3 b (bx)α−1b2
(cid:16) (cid:17)
ON GENERALIZATION OF DIFFERENT TYPE INEQUALITIES 9
Corollary 6. Under theassumptions of Theorem 5withα=1, theinequality(2.6)
reduced to the following inequality
ab
|I (x,λ,α,a,b)|
f,g
b−a
(cid:18) (cid:19)
b
b(x−a)f(a)+a(b−x)f(b) ab f(u)
= (1−λ)f(x)+λ − du
(cid:12) x(b−a) b−a u2 (cid:12)
(cid:12) (cid:20) (cid:21) Z (cid:12)
(cid:12) a (cid:12)
≤ (cid:12)(cid:12)(cid:12) ab (x−a)2 sup |f′(x)|q,|f′(a)|q 1/qC 1,λ,1, a (cid:12)(cid:12)(cid:12)
b−a x2 2 x
(cid:18) (cid:19)( (cid:2) (cid:8) (cid:9)(cid:3) (cid:16) (cid:17)
(b−x)2 ′ q ′ q 1/q x
+ sup |f (x)| ,|f (a)| C 1,λ,1, ,
b2 3 b
)
(cid:16) (cid:17)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for x=H =2ab/(a+b), we get
b
f(a)+f(b) ab f(u) b−a
(1−λ)f(H)+λ − du ≤
(cid:12) 2 b−a u2 (cid:12) 4ab
(cid:12) (cid:18) (cid:19) Z (cid:12)
(cid:12) a (cid:12)
(cid:12)(cid:12)(cid:12)× a2C2 1,λ,1,a+b sup |f′(H)|q,|f′(a)|q 1q(cid:12)(cid:12)(cid:12)
2b
(cid:26) (cid:18) (cid:19)
+H2C 1,λ,1, 2a (cid:0)sup (cid:8)|f′(H)|q,|f′(b)|q (cid:9)(cid:1)q1 .
3
a+b
(cid:18) (cid:19) (cid:27)
(cid:0) (cid:8) (cid:9)(cid:1)
Corollary 7. In Theorem 5,
(1) If we take x = H = 2ab/(a+b), λ = 1, then we get the following Simpson
3
type inequality for fractional integrals
α
1 ab
[f(a)+4f(H)+f(b)]− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
6 b−a 1/H+ 1/H−
≤ (cid:12)(cid:12)(cid:12)(cid:12)b−a a2 sup |f′(H)|q,|f(cid:18)′(a)|q (cid:19)1/qC2 α,1,1, ah i(cid:12)(cid:12)(cid:12)(cid:12)
4ab 3 x
(cid:26) (cid:18) (cid:19)
+H2 sup(cid:2)|f′((cid:8)H)|q,|f′(a)|q 1/q(cid:9)C(cid:3) α,1,1,x ,
3
3 b
(cid:18) (cid:19)(cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for α=1, we get
b
1 ab f(u) b−a
[f(a)+4f(H)+f(b)]− du ≤
(cid:12)6 b−a u2 (cid:12) 4ab
(cid:12) Z (cid:12)
(cid:12) a (cid:12)
(cid:12)(cid:12)(cid:12)× a2C2 1,1,1,a+b [sup{|f′(H)|,|f′((cid:12)(cid:12)(cid:12)a)|}]1q
3 2b
(cid:26) (cid:18) (cid:19)
+H2C 1,1,1, 2a sup |f′(H)|q,|f′(b)|q 1q .
3
3 a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
10 I˙MDATI˙S¸CAN
(2) If we take x=H =2ab/(a+b), λ=0, then we get the following midpoint type
inequality for fractional integrals
α
ab
f(H)− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
b−a 1/H+ 1/H−
≤ (cid:12)(cid:12)(cid:12)(cid:12)b−a a2(cid:18)C2 α,(cid:19)0,1,a+b suph|f′(H)|q,|f′(a)|q q1 i(cid:12)(cid:12)(cid:12)(cid:12)
4ab 2b
(cid:26) (cid:18) (cid:19)
+H2C α,0,1, 2a sup(cid:2)|f′((cid:8)H)|q,|f′(b)|q q1 (cid:9),(cid:3)
3
a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for α=1, we get
b
ab f(u)
f(H)− du
(cid:12) b−a u2 (cid:12)
(cid:12) Z (cid:12)
(cid:12) a (cid:12)
≤ (cid:12)(cid:12)(cid:12)b−a 1 1−q1 a2H2C1(cid:12)(cid:12)(cid:12)/q 1,0,q,a+b sup f′ 2ab q,|f′(a)|q q1
4ab 2 H2q 2 2b a+b
(cid:18) (cid:19) ( (cid:18) (cid:19)(cid:20) (cid:26)(cid:12) (cid:18) (cid:19)(cid:12) (cid:27)(cid:21)
(cid:12) (cid:12)
+b2H2C1/q 1,0,q, 2a sup f′ 2ab q,(cid:12)(cid:12)|f′(b)|q q1(cid:12)(cid:12) .
b2q 3 a+b a+b
(cid:18) (cid:19)(cid:20) (cid:26)(cid:12) (cid:18) (cid:19)(cid:12) (cid:27)(cid:21) )
(cid:12) (cid:12)
(3) If we take x=H =2ab/(a+b), λ=1(cid:12), then we ge(cid:12)t the following trapezoid type
(cid:12) (cid:12)
inequality for fractional integrals
α
f(a)+f(b) ab
− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
2 b−a 1/H+ 1/H−
≤ (cid:12)(cid:12)(cid:12)(cid:12)b−a a2C2 α(cid:18),1,1,a(cid:19)+b sup |f′(Hh)|q,|f′(a)|q q1 i(cid:12)(cid:12)(cid:12)(cid:12)
4ab 2b
(cid:26) (cid:18) (cid:19)
+H2C α,1,1, 2a sup(cid:2)|f′((cid:8)H)|q,|f′(b)|q q1 (cid:9),(cid:3)
3
a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
specially for α=1, we get
b
f(a)+f(b) ab f(u)
− du
(cid:12) 2 b−a u2 (cid:12)
(cid:12) Z (cid:12)
(cid:12) a (cid:12)
≤ (cid:12)(cid:12)(cid:12)b4−aba a2C21/q 1,1,1,a2+bb (cid:12)(cid:12)(cid:12)sup |f′(H)|q,|f′(a)|q q1
(cid:26) (cid:18) (cid:19)
+H2C1/q 1,1,1, 2a sup(cid:2)|f′((cid:8)H)|q,|f′(b)|q q1 (cid:9).(cid:3)
3 a+b
(cid:18) (cid:19) (cid:27)
(cid:2) (cid:8) (cid:9)(cid:3)
Corollary 8. Let the assumptions of Theorem 5 hold. If |f′(x)|≤M for all x∈
[a,b] and λ = 0, then we get the following Ostrowski type inequality for fractional
from the inequality (2.6) integrals
x−a α b−x α ab α
+ f(x)− 2α−1Γ(α+1) Jα (f ◦g)(1/a)+Jα (f ◦g)(1/b)
ax bx b−a 1/x+ 1/x−
(cid:12)(cid:12)(cid:12)(cid:20)(cid:18) (x−a(cid:19))α+1(cid:18) (cid:19) (cid:21)a (b(cid:18)−x)α+(cid:19)1 x h i(cid:12)(cid:12)(cid:12)
≤ M(cid:12) C α,0,1, + C α,0,1, . (cid:12)
(ax)α−1x2 2 x (bx)α−1b2 3 b
" #
(cid:16) (cid:17) (cid:16) (cid:17)