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On generalization of different type inequalities for harmonically quasi-convex functions via fractional integrals
Applied Mathematics and Computation 275 (2016) 287–298
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
On generalization of different type inequalities for harmonically quasi-convex functions via fractional integrals ˙ Imdat I˙ scan ¸ Department of Mathematics, Faculty of Sciences and Arts, Giresun University, Giresun, Turkey
a r t i c l e
i n f o
a b s t r a c t
MSC: 26A33 26A51 26D15
In this paper, a new general identity for fractional integrals have been defined. By using of this identity, new estimates on generalization of Hadamard, Ostrowski and Simpson like type inequalities for harmonically quasi-convex functions via the Riemann–Liouville fractional integral have been obtained.
Keywords: Hermite–Hadamard inequality Riemann–Liouville fractional integral Ostrowski inequality Simpson inequalityHarmonically quasi-convex function
© 2015 Elsevier Inc. All rights reserved.
1. Introduction In this section we will present definitions and some results used in this paper. A function f : I ⊆ R → R is said to be a convex function if
f (tx + (1 − t )y ) ≤ t f (x ) + (1 − t ) f (y ) holds for all x, y ∈ I and t ∈ [0, 1]. The notion of quasi-convex functions generalizes the notion of convex functions. More precisely, a function f : [a, b]→ R is said quasi-convex on [a, b] if
f (α x + (1 − α )y ) ≤ sup { f (x ), f (y )}, for any x, y ∈ [a, b] and α ∈ [0, 1]. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex (see [3]). Following inequalities are well known in the literature as Hermite–Hadamard’s inequality, Ostrowski’s inequality and Simpson’s inequality, respectively. Theorem 1. Let f : I ⊆ R → R be a convex function defined on the interval I of real numbers and a, b ∈ I with a < b. The following double inequality holds
f
a+b 2
≤
1 b−a
b a
f (x )dx ≤
f (a ) + f (b ) . 2
E-mail address: [email protected], [email protected] http://dx.doi.org/10.1016/j.amc.2015.11.074 0096-3003/© 2015 Elsevier Inc. All rights reserved.
(1.1)
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Theorem 2. Let f : I ⊆ R → R be a mapping differentiable in I°, the interior of I, and let a, b ∈ I° with a < b. If |f (x)| ≤ M for all x ∈ [a, b], then the following inequality holds
b 1 M (x − a )2 + (b − x )2 f (x ) − b − a f (t )dt ≤ b − a 2 a
for all x ∈ [a, b]. Theorem 3. Let f : [a, b]→ R be a four times continuously differentiable mapping on (a, b) and f (4 ) ∞ = supx ∈ (a, b)| f (4 ) (x )| < ∞. Then the following inequality holds:
b 1 1 a+b 1 f (a ) + f (b ) f ( 4 ) ( b − a )4 . + 2f − f (x )dx ≤ 3 ∞ 2 2 b−a 2880 a
In [4], Iscan defined harmonically convex functions and gave some Hermite–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
f
xy tx + (1 − t )y
≤ t f (y ) + (1 − t ) f (x )
for all x, y ∈ I and t ∈ [0, 1]. If the inequality in (2.1) is reversed, then f is said to be harmonically concave. In [16], Zhang et al. defined the harmonically quasi-convex function and supplied several properties of this kind of functions. Definition 2. A function f: I⊆(0, ∞) → [0, ∞) is said to be harmonically quasi-convex, if
f
xy tx + (1 − t )y
≤ sup { f (x ), f (y )}
for all x, y ∈ I and t ∈ [0, 1]. We would like to point out that any harmonically convex function on I ⊆ (0, ∞ ) is a harmonically quasi-convex function, but not conversely. For example, the function
f (x ) =
1, x ∈ (0, 1]; (x − 2 )2 ,x ∈ [1, 4].
is harmonically quasi-convex on (0, 4], but it is not harmonically convex on (0, 4]. For some results concerning harmonically convex and harmonically quasi-convex functions we refer the reader to papers [4,6–8,10–13,16] and references therein. Let us recall the following special functions: (1) The beta function:
β (x, y ) =
(x )(y ) = (x + y )
1 0
t x−1 (1 − t )y−1 dt, x, y > 0,
(2) The hypergeometric function: 2 F1
(a, b; c; z ) =
1 β (b, c − b)
1 0
t b−1 (1 − t )c−b−1 (1 − zt )−a dt, c > b > 0, |z| < 1
(see [9]). We give some necessary definitions and mathematical preliminaries of fractional calculus theory which are used throughout this paper. Definition 3. Let f ∈ L[a, b]. The Riemann–Liouville integrals Jaα+ f and Jbα− f of order α > 0 with a ≥ 0 are defined by α f (x ) = Ja+
1
(α )
x a
(x − t )α−1 f (t )dt, x > a
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289
and
α f (x ) = Jb−
b
1
(α )
(t − x )α−1 f (t )dt, x < b
x
respectively, where (α ) is the gamma function defined by (α ) =
∞ 0
e−t t α −1 dt and Ja0+ f (x ) = Jb0− f (x ) = f (x ).
In the case of α = 1, the fractional integral reduces to the classical integral. In recent years, many authors have studied errors estimations for Hermite–Hadamard’s, Ostrowski’s and Simpson’s inequalities; for new results and generalizations see [1,2,5–8,14,15]. In this paper, new identity for fractional integrals have been defined. By using of this identity, we obtained a generalization of Hadamard’s, Ostrowski’s and Simpson’s type inequalities for harmonically quasi-convex functions via the 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
α
b − x α
x−a α x−a α b−x S f (x, λ, α , a, b) = (1 − λ ) + f (x ) + λ f (a ) + f (b ) ax bx ax bx
α α −(α + 1 ) J1/x+ ( f ◦ g)(1/a ) + J1/x− ( f ◦ g)(1/b) where a, b ∈ I with a < b, x ∈ [a, b] , λ ∈ [0, 1], g(u ) = 1/u, α > 0 and is Euler’s gamma function. In order to prove our main results we need the following identity. 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:
S f (x, λ, α , a, b) =
(x − a )α+1 (ax )α−1
1 0
α +1 1
tα − λ ax (b − x ) f dt − At (a, x ) (bx )α−1 At2 (a, x )
0
tα − λ f At2 (b, x )
bx dt, At (b, x )
where At (u, x ) = tu + (1 − t )x. Proof. By integration by parts and twice changing the variable, for x = a, we can state that
ax(x − a )
0
1
tα − λ ax f dt 2 A ( At (a, x ) t a, x )
ax
(t α − λ )df At (a, x ) 0 ax 1 1 ax
= (t α − λ ) f αt α−1 f dt − At (a, x ) 0 At (a, x ) 0 ax α 1/a 1
α−1 1 −u = (1 − λ ) f (x ) + λ f (a ) − α f ( )du
=
1
x−a
= (1 − λ ) f (x ) + λ f (a ) −
ax α x−a
a
u
1/x
α (α + 1 )J1/x+ ( f ◦ g)(1/a )
(2.2)
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and similarly for x = b we get
bx(x − b) =
0
1
1 0
tα − λ f At2 (a, x )
(t α − λ )df
bx dt At (b, x )
bx At (b, x )
1 1 bx α −1 f − α t dt At (b, x ) 0 0 α 1/x
1 α −1 1 bx = (1 − λ ) f (x ) + λ f (b ) − α u− f ( )du
= (t α − λ ) f
bx At (b, x )
= (1 − λ ) f (x ) + λ f (b ) −
b−x
bx b−x
α
1/b
b
u
α (α + 1 )J1/x− ( f ◦ g)(1/b)
(2.3)
b−x α α Multiplying both sides of (2.2) and (2.3) by ( x−a ax ) and ( bx ) , respectively, and adding the resulting identities we obtain the desired result. For x = a and for x = b, the identities
S f (a, λ, α , a, b) = −
S f (b, λ, α , a, b) =
(b − a )α+1 (ba )α−1
(b − a )α+1 (ab)α−1
1 0
1 0
tα − λ f At2 (b, a )
tα − λ f At2 (a, b)
ab dt, At (b, a )
ab dt At (a, b)
can be proved respectively easily by performing an integration by parts in the integrals from the right-hand side and changing the variable. 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
α +1 1/q 1/q a S f (x, λ, α , a, b) ≤ C 1−1/q (α , λ ) × (x − a ) f (x )q , f (a )q sup C , λ , q, α 1 2 x (ax )α−1 x2
q q 1/q 1/q x (b − x )α+1 + sup f (x ) , f (b) C3 α , λ, q, , α −1 2 b (bx ) b where
2αλ1+ α + 1 − λ, α+1 1
C1 (α , λ ) =
a 1 a a α , λ, q, = − λ.2 F1 2q; 1; 2, 1 − .2 F1 2q; α + 1; α + 2, 1 − x α+1 x x
a 1 a 1 .2 F1 2q; α + 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − , x α+1 x
C2
x 1 x x α , λ, q, .2 F1 2q; 1; α + 2, 1 − = − λ.2 F1 2q; 1; 2, 1 − b α+1 b b
x 1 x 1 .2 F1 2q; 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − . b α+1 b
C3
(2.4)
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Proof. Since |f |q is harmonically quasi-convex on [a, b], from Lemma 1, property of the modulus and using the power-mean inequality we have
1
α +1 1 α ax bx |t − λ| (b − x )α+1 |t α − λ| dt S f (x, λ, α , a, b) ≤ (x − a ) f f A (a, x ) dt + 2 α −1 2 A ( b, x ) (ax )α−1 ( bx ) At (a, x ) At (b, x ) t t ≤
(x − a )α+1 (ax )α−1
0
1
1− 1q |t α − λ|dt
0
(b − x )α+1 + (bx )α−1
1 0
1
1− 1q |t α − λ|dt
0
q ax |t α − λ| f dt At (a, x ) At2q (a, x )
1 0
1q
0
q 1q bx |t α − λ| dt f 2q A ( b, x ) t At (b, x )
⎧ ⎨
1 1/q q q 1/q |t α − λ| (x − a )α+1 ≤ sup f (x ) , f (a ) dt (α , λ ) ⎩ (ax )α−1 At2q (a, x ) 0 1 1/q ⎫ ⎬ α +1 α 1/q |t − λ| (b − x ) f (x )q , f (b)q + sup dt (bx )α−1 ⎭ At2q (b, x ) C11−1/q
(2.5)
0
where by simple computation we obtain
C1 (α , λ ) =
1
|t α − λ|dt
0
2αλ1+ α + 1 − λ, α+1 1
=
1 0
1
a a |t α − λ| .2 F1 2q; α + 1; α + 2, 1 − dt = x−2q − λ.2 F1 2q; 1; 2, 1 − 2q α+1 x x At (a, x )
a 1 a 1 .2 F1 2q; α + 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − , x α+1 x
(2.6)
(2.7)
and
1 0
1
x x |t α − λ| .2 F1 2q; 1; α + 2, 1 − dt = b−2q − λ.2 F1 2q; 1; 2, 1 − 2q α+1 b b At (b, x )
x 1 x 1 .2 F1 2q; 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − , b α+1 b
Hence, if we use (2.6)–(2.8) in (2.5 ), we obtain the desired result. This completes the proof. Corollary 1. Under the assumptions of Theorem 4 with q = 1, the inequality (2.4) reduced to the following inequality
α +1 f (x ), f (a ) C2 α , λ, 1, a S f (x, λ, α , a, b) ≤ (x − a ) sup x (ax )α−1 x2
α +1 1/q x (b − x ) + sup f (x ), f (b) C3 α , λ, 1, , b (bx )α−1 b2
(2.8)
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Corollary 2. Under the assumptions of Theorem 4 with α = 1, the inequality (2.4) reduced to the following inequality
b ab b ( x − a ) f ( a ) + a ( b − x ) f ( b ) f u ( ) S f (x, λ, α , a, b) = (1 − λ ) f (x ) + λ − du x (b − a ) b−a u2 a 2 1− 1q
1/q 1/q a ab 2λ − 2λ + 1 ( x − a )2 f (x )q , f (a )q ≤ sup C 1, λ , q, 2 b−a 2 x x2
2 q q 1/q 1/q x (b − x ) + sup f (x ) , f (b) C3 1, λ, q, , 2
ab b−a
b
b
specially for x = H = 2ab/(a + b), we get
1− 1q b ab f (a ) + f (b ) f (u ) b − a 2λ2 − 2λ + 1 − du ≤ (1 − λ ) f (H ) + λ 2 b−a 2 u2 4ab a
× a2C21/q 1, λ, q,
a+b 2b
sup
1 1 f (H )q , f (a )q q + H 2C 1/q 1, λ, q, 2a f (H )q , f (b)q q (2.9) sup . 3 a+b
Corollary 3. In Theorem 4, (1) If we take x = H = 2ab/(a + b), λ =
1 3,
then we get the following Simpson type inequality for fractional integrals
α 1
α α −1 α [ f (a ) + 4 f (H ) + f (b)] − ab 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 6 b−a 1
q q 1/q 1/q 1 a b − a 1−1/q α, C1 ≤ a2 sup f (H ) , f (a ) C2 α , , q, 4ab 3 3 H
q q 1/q 1/q H 1 +H 2 sup f (H ) , f (a ) C3 α , , q, , 3
b
specially for α = 1, we get
b 1 ab f (u ) b − a 5 1− q 1 du ≤ 6 [ f ( a ) + 4 f ( H ) + f ( b )] − b − a u2 4ab 18 a 1
1 q q 1q a+b 2a 1 × a2C21/q 1, , q, sup f (H ), f (a ) q + H 2C31/q 1, , q, sup f (H ) , f (b) . 3
2b
3
a+b
(2) If we take x = H = 2ab/(a + b), λ = 0, then we get the following midpoint type inequality for fractional integrals
α
α −1 (α + 1 J α α f (H ) − ab 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− b−a 1 1− 1q q q 1q b−a a+b ≤ a2C21/q α , 0, q, sup f (H ) , f (a ) 4ab α + 1 2b
q q 1q 2a 2 1/q +H C3 α , 0, q, sup f (H ) , f (b) , a+b
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specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 1− q 2 1/q f (H )q , f (a )q q du ≤ a C 1, 0, q, sup f (H ) − b − a 2 2b u2 4ab 2 a 1
q q q 2a +H 2C 1/q 1, 0, q, sup f (H ) , f (b) . a+b
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 )
α ab α −1 α − 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 2 b−a α 1− 1q q q 1q b−a a+b ≤ a2C21/q α , 1, q, sup f (H ) , f (a ) 4ab α + 1 2b
q 1q 2a q +H 2C 1/q α , 1, q, sup f (H ) , f (b) , a+b
3
specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 1− q 2 1/q f (a ) + f (b ) f (H )q , f (a )q q − du ≤ a C 1, 1, q, sup 2 2 b−a 2b u2 4ab 2 a 1
q q q 2a +H 2C31/q 1, 1, q, sup f (H ) , f (b) . a+b
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 (2.4) integrals
α α x − a α
b−x ab α −1 (α + 1 J α α + f ( x ) − 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− ax bx b−a
M a x (x − a )α+1 1/q (b − x )α+1 1/q ≤ C2 α , 0, q, + C3 α , 0, q, . 1 α −1 2 α −1 2 x b (bx ) b (α + 1 )1− q (ax ) x
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
α +1 q q 1/q S f (x, λ, α , a, b) ≤ C2 α , λ, 1, a (x − a ) sup f (x ) , f (a ) α −1 2 x (ax ) x
α +1 1/q x (b − x ) q f (x ) , f (b)q +C3 α , λ, 1, sup b (bx )α −1 b2
(2.10)
where C2 and C3 are defined as in Theorem 4. Proof. Since |f |q is harmonically quasi-convex on [a, b], from Lemma 1,using property of the modulus and the power-mean inequality we have
1 1− 1q 1 1q
α q (x − a )α+1 |t α − λ| ax |t − λ| S f (x, λ, α , a, b) ≤ dt f A (a, x ) dt (ax )α−1 At2 (a, x ) At2 (a, x ) t +
(b − x )α+1 (bx )α−1
≤ C2
α , λ, 1,
0
1 0
|t α − λ| dt At2 (b, x )
1− 1q
0
a (x − a )α +1 sup x (ax )α −1 x2
which completes the proof.
0
q 1q 1 α bx |t − λ| dt f 2 At (b, x ) At (b, x )
α +1 1/q 1/q x (b − x ) f (x )q , f (a )q f (x )q , f (b)q + C3 α , λ, 1, sup b (bx )α −1 b2
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Corollary 5. Under the assumptions of Theorem 5 with q = 1, the inequality (2.10) reduced to the following inequality
α +1 S f (x, λ, α , a, b) ≤ C2 α , λ, 1, a (x − a ) sup f (x ), f (a ) x (ax )α −1 x2
α +1 x (b − x ) +C3 α , λ, 1, sup f (x ), f (b) α −1 2 b (bx ) b
Corollary 6. Under the assumptions of Theorem 5 with α = 1, the inequality (2.10) reduced to the following inequality
b ab b ( x − a ) f ( a ) + a ( b − x ) f ( b ) f u ( ) S f (x, λ, α , a, b) = (1 − λ ) f (x ) + λ − du 2 x (b − a ) b−a u a
q q 1/q a ab ( x − a )2 ≤ sup f (x ) , f (a ) C2 1, λ, 1, 2 b−a x x
2 1/q q q x (b − x ) + sup f (x ) , f (b) C3 1, λ, 1, , 2
ab b−a
b
b
specially for x = H = 2ab/(a + b), we get
b ab f (a ) + f (b ) f (u ) b − a 1 − λ f H + λ − du ( ) ( ) ≤ 4ab 2 b−a u2 a q q 1q a+b sup f (H ) , f (a ) × a2C2 1, λ, 1, 2b
q q 1q 2a + H 2C3 1, λ, 1, sup f (H ) , f (b) . a+b
(2.11)
Corollary 7. In Theorem 5, (1) If we take x = H = 2ab/(a + b), λ =
1 3,
then we get the following Simpson type inequality for fractional integrals
α 1
α α −1 α [ f (a ) + 4 f (H ) + f (b)] − ab 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 6 b−a
q q 1/q 1 a b−a 2 ≤ a sup f (H ) , f (a ) C2 α , , 1, 4ab
3
q q 1/q 1 x 2 +H sup f (H ) , f (b) C3 α , , 1, ,
x
3
b
specially for α = 1, we get
b q q 1q ab a+b 1 f (u ) b − a 1 2 × a C2 1, , 1, du ≤ sup f (H ) , f (a ) 6 [ f ( a ) + 4 f ( H ) + f ( b )] − b − a 2 3 2b u 4ab a 1
q q 1q 2a +H 2C3 1, , 1, sup f (H ) , f (b) . 3
a+b
(2) If we take x = H = 2ab/(a + b), λ = 0, then we get the following midpoint type inequality for fractional integrals
α
α −1 (α + 1 J α α f (H ) − ab 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− b−a 1 q q q b−a 2 a+b ≤ a C2 α , 0, 1, sup f (H ) , f (a ) 4ab 2b
q q 1q 2a +H 2C3 α , 0, 1, sup f (H ) , f (b) , a+b
specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 1− q 2 1/q f (H )q , f (a )q q du ≤ a C 1, 0, q, sup f (H ) − b − a 2 2b u2 4ab 2 a
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+H 2C31/q 1, 0, q,
2a a+b
sup
f (H )q , f (b)q
1 q
295
.
(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 )
α ab α −1 α − 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 2 b−a q q 1q b−a 2 a+b ≤ a C2 α , 1, 1, sup f (H ) , f (a ) 4ab 2b
q q 1q 2a +H 2C3 α , 1, 1, sup f (H ) , f (b) , a+b
specially for α = 1, we get
b 1 ab a+b f (u ) b − a 2 1/q f (a ) + f (b ) f (H )q , f (a )q q − du ≤ a C 1, 1, 1, sup 2 2 b−a 2b u2 4ab a 1
q q q 2a +H 2C 1/q 1, 1, 1, sup f (H ) , f (b) . a+b
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 (2.10) integrals
α α x − a α
b−x ab α −1 (α + 1 J α α + f ( x ) − 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− ax bx b−a
a x (x − a )α+1 (b − x )α+1 ≤M C2 α , 0, 1, + C3 α , 0, 1, . α −1 2 α −1 2 x b (ax ) x (bx ) b Theorem 6. 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
α +1 1/q S f (x, λ, α , a, b) ≤ C 1/q (α , λ ) C 1/p α , λ, p, a (x − a ) f (x )q , f (a )q sup 1 2 x (ax )α −1 x2
α +1 q q 1/q x (b − x ) f (x ) , f (b) +C31/p α , λ, p, sup b (bx )α −1 b2
(2.12)
where C1 , C2 and C3 are defined as in Theorem 4. Proof. Since |f |q is harmonically quasi-convex on [a, b], from Lemma 1, using property of the modulus and the Hölder inequality we have
1 1p 1 1q
q (x − a )α+1 |t α − λ| ax S f (x, λ, α , a, b) ≤ dt |t α − λ| f dt At (a, x ) (ax )α−1 At2p (a, x ) α +1
(b − x ) + (bx )α−1
0
1 0
|t α − λ| dt At2p (b, x )
1p
0
1 α |t − λ| f 0
bx At (b, x )
q 1q dt
1/q a (x − a )α +1 f (x )q , f (a )q sup (α , λ ) C21/p α , λ, p, x (ax )α −1 x2
α +1 q q 1/q x (b − x ) 1/p α , λ, p, +C3 sup f (x ) , f (b) b (bx )α −1 b2 ≤
C11/q
which completes the proof.
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Corollary 9. Under the assumptions of Theorem 6 with α = 1, the inequality (2.12) reduced to the following inequality
b f (u ) S f (x, λ, α , a, b) = (1 − λ ) f (x ) + λ b(x − a ) f (a ) + a(b − x ) f (b) − ab du x (b − a ) b−a u2 a 2 1q
1/q 1/p a ab 2λ − 2λ + 1 ( x − a )2 f (x )q , f (a )q ≤ sup C 1, λ , p, 2 b−a 2 x x2
2 q q 1/q 1/p x (b − x ) + sup f (x ) , f (a ) C3 1, λ, p, , 2
ab b−a
b
b
specially for x = H = 2ab/(a + b), we get
1 b ab f (a ) + f (b ) f (u ) b − a 2λ2 − 2λ + 1 q − du ≤ (1 − λ ) f (H ) + λ 2 b−a 2 u2 4ab a
× a2C21/p 1, λ, p,
a+b 2b
sup
1 1 f (H )q , f (a )q q + H 2C 1/p 1, λ, p, 2a f (H )q , f (b)q q (2.13) sup . 3 a+b
Corollary 10. In Theorem 6, (1) If we take x = H = 2ab/(a + b), λ =
1 3,
then we get the following Simpson type inequality for fractional integrals
1/q 1/p a+b 1 S f (x, λ, α , a, b) ≤ b − a C 1−1/q α , 1 a2 sup f (H )q , f (a )q , p, C , α 2 4ab 1 3 3 2b
1/q 2a 1 q q +H 2 sup f (H ) , f (a ) C31/p α , , p, ,
ab b−a
α
3
a+b
specially for α = 1, we get
b 1 1 ab a+b 1 f (u ) b − a 5 q 1 2 1/p , p, du ≤ × a C 1, sup f (H ), f (a ) q 6 [ f ( a ) + 4 f ( H ) + f ( b )] − b − a 2 2 4ab 18 3 2b u a 1
q q 1q 2a 2 1/p +H C 1, , p, sup f (H ) , f (b) . 3
3
a+b
(2) If we take x = H = 2ab/(a + b), λ = 0, then we get the following midpoint type inequality for fractional integrals
α
α α −1 α f (H ) − ab 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) b−a 1 1q q q 1q b−a a+b ≤ a2C21/p α , 0, p, sup f (H ) , f (a ) 4ab α + 1 2b
q q 1q 2a +H 2C 1/p α , 0, p, sup f (H ) , f (b) , 3
a+b
specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 q 2 1/p f (H )q , f (a )q q du ≤ a C 1, 0, p, sup f (H ) − b − a 2 2b u2 4ab 2 a 1
q q q 2a +H 2C 1/p 1, 0, p, sup f (H ) , f (b) . 3
a+b
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(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 )
α ab α −1 α − 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 2 b−a α 1q q q 1q b−a a+b 2 1/p ≤ a C2 sup f (H ) , f (a ) α , 1, p, 4ab α + 1 2b
q q 1q 2a +H 2C 1/p α , 1, p, sup f (H ) , f (b) , a+b
3
specially for α = 1, we get
b 1 q q 1q ab a+b f (u ) b − a 1 q 2 1/p f (a ) + f (b ) − du ≤ a C2 1, 1, p, sup f (H ) , f (a ) 2 2 b−a 2b u 4ab 2 a
q q 1q 2a +H 2C 1/p 1, 1, p, sup f (H ) , f (b) . a+b
3
Corollary 11. Let the assumptions of Theorem 6 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 (2.12) integrals
α α x − a α
α b−x ab α −1 α + f (x ) − 2 (α + 1 ) J1/x+ ( f ◦ g)(1/a ) + J1/x− ( f ◦ g)(1/b) ax bx b−a
M a x (x − a )α+1 1/p (b − x )α+1 1/p α α ≤ C , 0, p, + C , 0, p, . 1 α −1 2 2 x b (bx )α−1 b2 3 (α + 1 ) q (ax ) x
3. Application to special means Let us recall the following special means of two nonnegative number a, b with b > a: (1) The arithmetic mean
A = A(a, b) :=
a+b . 2
(2) The geometric mean
G = G(a, b) :=
a+b . 2
(3) The harmonic mean
H = H (a, b) :=
2ab . a+b
(4) The n-logarithmic mean
Ln = Ln (a, b) := Let f (x ) =
xn+1 n+1 ,
bn+1 − an+1 (n + 1 )(b − a )
1n
, n ∈ R\{−1, 0}.
x > 0, n ≥ 1 and q ≥ 1. Then, the function | f (x )|q = xnq is harmonically quasi-convex on (0, ∞).
Proposition 1. For b > a > 0, n > 1, λ ∈ [0, 1] and q ≥ 1, we have
1− 1q b − a 2λ2 − 2λ + 1 (1 − λ )H n+1 + λA an+1 , bn+1 − G2 Ln−1 ≤ n + 1 ( ) n−1
× a2 H nC21/q 1, λ, q,
A b
+ H 2 bnC31/q 1, λ, q,
a A
4G2
2
,
specially for n = 1 we get
2 1− 1q
A a 1/q 1/q 2 2 (1 − λ )H 2 + λA a2 , b2 − G2 ≤ b − a 2λ − 2λ + 1 × a HC 1, λ , q, + H bC 1, λ , q, , 2 3 2 2G
where C2 and C3 are defined as in Theorem 4.
2
b
A
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xn+1 n+1 ,
Proof. The assertion follows from inequality (2.9) in Corollary 2, for f (x ) =
x > 0.
Proposition 2. For b > a > 0, n > 1, λ ∈ [0, 1] and q ≥ 1, we have
A a b−a 2 n 2 n (1 − λ )H n+1 + λA an+1 , bn+1 − G2 Ln−1 ≤ n + 1 × a H C 1, λ , 1, + H b C 1, λ , 1, , ( ) 2 3 n−1 2 b
4G
A
specially for n = 1 we get
(1 − λ )H 2 + λA a2 , b2 − G2 ≤ b − a × a2 HC2 1, λ, 1, A + H 2 bC3 1, λ, 1, a , 2 b
2G
A
where C2 and C3 are defined as in Theorem 4. Proof. The assertion follows from inequality (2.11) in Corollary 6, for f (x ) =
xn+1 n+1 ,
x > 0.
Proposition 3. For b > a > 0, n > 1, λ ∈ [0, 1] and q ≥ 1, we have
2 1q n+1 n+1 b − a 2 λ − 2 λ + 1 n+1 2 n−1 (1 − λ )H + λA a , b − G Ln−1 ≤ (n + 1 )
× a2 H nC21/p 1, λ, p,
A b
+ H 2 bnC31/p 1, λ, p,
a A
4G2
2
,
specially for n = 1 we get
2 1q
A a 1/p 1/p 2 2 (1 − λ )H 2 + λA a2 , b2 − G2 ≤ b − a 2λ − 2λ + 1 × a HC 1, λ , p, + H bC 1, λ , p, , 2 3 2 2G
2
b
A
where C2 and C3 are defined as in Theorem 4. Proof. The assertion follows from inequality (2.13) in Corollary 9, for f (x ) =
xn+1 n+1 ,
x > 0.
References [1] M.W. Alomari, M. Darus, U.S. Kirmaci, Refinements of Hadamard-type inequalities for quasi-convex functions with applications to trapezoidal formula and to special means, Comput. Math. Appl. 59 (2010) 225–232. [2] M. Alomari, S. Hussain, Two inequalities of Simpson type for quasi-convex functions and applications, Appl. Math. E-Notes 11 (2011) 110–117. [3] D.A. Ion, Some estimates on the Hermite–Hadamard inequality through quasi-convex functions, Ann. Univ. Craiova Math. Comput. Sci. Ser. 34 (2007) 82–87. [4] I.˙ I˙ scan, ¸ Hermite–Hadamard type inequalities for harmonically convex functions, Hacet. J. Math. Stat. 43 (6) (2014) 935–942. [5] I.˙ I˙ scan, ¸ On new general integral inequalities for quasi-convex functions and their applications, Palest. J. Math. 4 (1) (2015) 21–29. [6] I.˙ I˙ scan, ¸ Hermite–Hadamard and Simpson-like type inequalities for differentiable harmonically convex functions, J. Math. 2014 (2014) 10. Article ID 346305. [7] I.˙ I˙ scan, ¸ S. Wu, Hermite–Hadamard type inequalities for harmonically convex functions via fractional integrals, Appl. Math. Comput. 238 (2014) 237–244. [8] I.˙ I˙ scan, ¸ S. Numan, Ostrowski type inequalities for harmonically quasi-convex functions, Electron. J. Math. Anal. Appl. 2 (2) (2014) 189–198. [9] A.A. Kilbas, H.M. Srivastava, J.J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier, Amsterdam, 2006. [10] J. Park, Hermite–Hadamard and Simpson-like type inequalities for differentiable harmonically quasi-convex functions, Int. J. Math. Anal. 8 (33) (2014) 1629–1645. [11] J. Park, Some generalized inequalities for differentiable harmonically quasi-convex functions, Int. J. Math. Anal. 8 (38) (2014) 1893–1906. [12] J. Park, Remarks on the Hermite–Hadamard type inequalities for harmonically quasi-convex functions, Appl. Math. Sci. 8 (105) (2014) 5213–5224. [13] J. Park, Generalized Hermite–Hadamard-like type inequalities for differentiable harmonically quasi-convex functions, Appl. Math. Sci. 8 (105) (2014) 5225– 5238. [14] E. Set, M.E. Ozdemir, M.Z.S. kaya, On new inequalities of Simpson’s type for quasi-convex functions with applications, Tamkang J. Math. 43 (3) (2012) 357–364. [15] M.Z.S. kaya, E. Set, H.Y. z, N. Basak, ¸ Hermite–Hadamard’s inequalities for fractional integrals and related fractional inequalities, Math. Comput. Model. 57 (9-10) (2013) 2403–2407. [16] T.-Y. Zhang, A.-P. Ji, F. Qi, Integral inequalities of Hermite–Hadamard type for harmonically quasi-convex functions, Proc. Jangjeon Math. Soc. 16 (3) (2013) 399–407.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
On generalization of different type inequalities for harmonically quasi-convex functions via fractional integrals ˙ Imdat I˙ scan ¸ Department of Mathematics, Faculty of Sciences and Arts, Giresun University, Giresun, Turkey
a r t i c l e
i n f o
a b s t r a c t
MSC: 26A33 26A51 26D15
In this paper, a new general identity for fractional integrals have been defined. By using of this identity, new estimates on generalization of Hadamard, Ostrowski and Simpson like type inequalities for harmonically quasi-convex functions via the Riemann–Liouville fractional integral have been obtained.
Keywords: Hermite–Hadamard inequality Riemann–Liouville fractional integral Ostrowski inequality Simpson inequalityHarmonically quasi-convex function
© 2015 Elsevier Inc. All rights reserved.
1. Introduction In this section we will present definitions and some results used in this paper. A function f : I ⊆ R → R is said to be a convex function if
f (tx + (1 − t )y ) ≤ t f (x ) + (1 − t ) f (y ) holds for all x, y ∈ I and t ∈ [0, 1]. The notion of quasi-convex functions generalizes the notion of convex functions. More precisely, a function f : [a, b]→ R is said quasi-convex on [a, b] if
f (α x + (1 − α )y ) ≤ sup { f (x ), f (y )}, for any x, y ∈ [a, b] and α ∈ [0, 1]. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex (see [3]). Following inequalities are well known in the literature as Hermite–Hadamard’s inequality, Ostrowski’s inequality and Simpson’s inequality, respectively. Theorem 1. Let f : I ⊆ R → R be a convex function defined on the interval I of real numbers and a, b ∈ I with a < b. The following double inequality holds
f
a+b 2
≤
1 b−a
b a
f (x )dx ≤
f (a ) + f (b ) . 2
E-mail address: [email protected], [email protected] http://dx.doi.org/10.1016/j.amc.2015.11.074 0096-3003/© 2015 Elsevier Inc. All rights reserved.
(1.1)
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Theorem 2. Let f : I ⊆ R → R be a mapping differentiable in I°, the interior of I, and let a, b ∈ I° with a < b. If |f (x)| ≤ M for all x ∈ [a, b], then the following inequality holds
b 1 M (x − a )2 + (b − x )2 f (x ) − b − a f (t )dt ≤ b − a 2 a
for all x ∈ [a, b]. Theorem 3. Let f : [a, b]→ R be a four times continuously differentiable mapping on (a, b) and f (4 ) ∞ = supx ∈ (a, b)| f (4 ) (x )| < ∞. Then the following inequality holds:
b 1 1 a+b 1 f (a ) + f (b ) f ( 4 ) ( b − a )4 . + 2f − f (x )dx ≤ 3 ∞ 2 2 b−a 2880 a
In [4], Iscan defined harmonically convex functions and gave some Hermite–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
f
xy tx + (1 − t )y
≤ t f (y ) + (1 − t ) f (x )
for all x, y ∈ I and t ∈ [0, 1]. If the inequality in (2.1) is reversed, then f is said to be harmonically concave. In [16], Zhang et al. defined the harmonically quasi-convex function and supplied several properties of this kind of functions. Definition 2. A function f: I⊆(0, ∞) → [0, ∞) is said to be harmonically quasi-convex, if
f
xy tx + (1 − t )y
≤ sup { f (x ), f (y )}
for all x, y ∈ I and t ∈ [0, 1]. We would like to point out that any harmonically convex function on I ⊆ (0, ∞ ) is a harmonically quasi-convex function, but not conversely. For example, the function
f (x ) =
1, x ∈ (0, 1]; (x − 2 )2 ,x ∈ [1, 4].
is harmonically quasi-convex on (0, 4], but it is not harmonically convex on (0, 4]. For some results concerning harmonically convex and harmonically quasi-convex functions we refer the reader to papers [4,6–8,10–13,16] and references therein. Let us recall the following special functions: (1) The beta function:
β (x, y ) =
(x )(y ) = (x + y )
1 0
t x−1 (1 − t )y−1 dt, x, y > 0,
(2) The hypergeometric function: 2 F1
(a, b; c; z ) =
1 β (b, c − b)
1 0
t b−1 (1 − t )c−b−1 (1 − zt )−a dt, c > b > 0, |z| < 1
(see [9]). We give some necessary definitions and mathematical preliminaries of fractional calculus theory which are used throughout this paper. Definition 3. Let f ∈ L[a, b]. The Riemann–Liouville integrals Jaα+ f and Jbα− f of order α > 0 with a ≥ 0 are defined by α f (x ) = Ja+
1
(α )
x a
(x − t )α−1 f (t )dt, x > a
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and
α f (x ) = Jb−
b
1
(α )
(t − x )α−1 f (t )dt, x < b
x
respectively, where (α ) is the gamma function defined by (α ) =
∞ 0
e−t t α −1 dt and Ja0+ f (x ) = Jb0− f (x ) = f (x ).
In the case of α = 1, the fractional integral reduces to the classical integral. In recent years, many authors have studied errors estimations for Hermite–Hadamard’s, Ostrowski’s and Simpson’s inequalities; for new results and generalizations see [1,2,5–8,14,15]. In this paper, new identity for fractional integrals have been defined. By using of this identity, we obtained a generalization of Hadamard’s, Ostrowski’s and Simpson’s type inequalities for harmonically quasi-convex functions via the 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
α
b − x α
x−a α x−a α b−x S f (x, λ, α , a, b) = (1 − λ ) + f (x ) + λ f (a ) + f (b ) ax bx ax bx
α α −(α + 1 ) J1/x+ ( f ◦ g)(1/a ) + J1/x− ( f ◦ g)(1/b) where a, b ∈ I with a < b, x ∈ [a, b] , λ ∈ [0, 1], g(u ) = 1/u, α > 0 and is Euler’s gamma function. In order to prove our main results we need the following identity. 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:
S f (x, λ, α , a, b) =
(x − a )α+1 (ax )α−1
1 0
α +1 1
tα − λ ax (b − x ) f dt − At (a, x ) (bx )α−1 At2 (a, x )
0
tα − λ f At2 (b, x )
bx dt, At (b, x )
where At (u, x ) = tu + (1 − t )x. Proof. By integration by parts and twice changing the variable, for x = a, we can state that
ax(x − a )
0
1
tα − λ ax f dt 2 A ( At (a, x ) t a, x )
ax
(t α − λ )df At (a, x ) 0 ax 1 1 ax
= (t α − λ ) f αt α−1 f dt − At (a, x ) 0 At (a, x ) 0 ax α 1/a 1
α−1 1 −u = (1 − λ ) f (x ) + λ f (a ) − α f ( )du
=
1
x−a
= (1 − λ ) f (x ) + λ f (a ) −
ax α x−a
a
u
1/x
α (α + 1 )J1/x+ ( f ◦ g)(1/a )
(2.2)
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and similarly for x = b we get
bx(x − b) =
0
1
1 0
tα − λ f At2 (a, x )
(t α − λ )df
bx dt At (b, x )
bx At (b, x )
1 1 bx α −1 f − α t dt At (b, x ) 0 0 α 1/x
1 α −1 1 bx = (1 − λ ) f (x ) + λ f (b ) − α u− f ( )du
= (t α − λ ) f
bx At (b, x )
= (1 − λ ) f (x ) + λ f (b ) −
b−x
bx b−x
α
1/b
b
u
α (α + 1 )J1/x− ( f ◦ g)(1/b)
(2.3)
b−x α α Multiplying both sides of (2.2) and (2.3) by ( x−a ax ) and ( bx ) , respectively, and adding the resulting identities we obtain the desired result. For x = a and for x = b, the identities
S f (a, λ, α , a, b) = −
S f (b, λ, α , a, b) =
(b − a )α+1 (ba )α−1
(b − a )α+1 (ab)α−1
1 0
1 0
tα − λ f At2 (b, a )
tα − λ f At2 (a, b)
ab dt, At (b, a )
ab dt At (a, b)
can be proved respectively easily by performing an integration by parts in the integrals from the right-hand side and changing the variable. 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
α +1 1/q 1/q a S f (x, λ, α , a, b) ≤ C 1−1/q (α , λ ) × (x − a ) f (x )q , f (a )q sup C , λ , q, α 1 2 x (ax )α−1 x2
q q 1/q 1/q x (b − x )α+1 + sup f (x ) , f (b) C3 α , λ, q, , α −1 2 b (bx ) b where
2αλ1+ α + 1 − λ, α+1 1
C1 (α , λ ) =
a 1 a a α , λ, q, = − λ.2 F1 2q; 1; 2, 1 − .2 F1 2q; α + 1; α + 2, 1 − x α+1 x x
a 1 a 1 .2 F1 2q; α + 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − , x α+1 x
C2
x 1 x x α , λ, q, .2 F1 2q; 1; α + 2, 1 − = − λ.2 F1 2q; 1; 2, 1 − b α+1 b b
x 1 x 1 .2 F1 2q; 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − . b α+1 b
C3
(2.4)
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Proof. Since |f |q is harmonically quasi-convex on [a, b], from Lemma 1, property of the modulus and using the power-mean inequality we have
1
α +1 1 α ax bx |t − λ| (b − x )α+1 |t α − λ| dt S f (x, λ, α , a, b) ≤ (x − a ) f f A (a, x ) dt + 2 α −1 2 A ( b, x ) (ax )α−1 ( bx ) At (a, x ) At (b, x ) t t ≤
(x − a )α+1 (ax )α−1
0
1
1− 1q |t α − λ|dt
0
(b − x )α+1 + (bx )α−1
1 0
1
1− 1q |t α − λ|dt
0
q ax |t α − λ| f dt At (a, x ) At2q (a, x )
1 0
1q
0
q 1q bx |t α − λ| dt f 2q A ( b, x ) t At (b, x )
⎧ ⎨
1 1/q q q 1/q |t α − λ| (x − a )α+1 ≤ sup f (x ) , f (a ) dt (α , λ ) ⎩ (ax )α−1 At2q (a, x ) 0 1 1/q ⎫ ⎬ α +1 α 1/q |t − λ| (b − x ) f (x )q , f (b)q + sup dt (bx )α−1 ⎭ At2q (b, x ) C11−1/q
(2.5)
0
where by simple computation we obtain
C1 (α , λ ) =
1
|t α − λ|dt
0
2αλ1+ α + 1 − λ, α+1 1
=
1 0
1
a a |t α − λ| .2 F1 2q; α + 1; α + 2, 1 − dt = x−2q − λ.2 F1 2q; 1; 2, 1 − 2q α+1 x x At (a, x )
a 1 a 1 .2 F1 2q; α + 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − , x α+1 x
(2.6)
(2.7)
and
1 0
1
x x |t α − λ| .2 F1 2q; 1; α + 2, 1 − dt = b−2q − λ.2 F1 2q; 1; 2, 1 − 2q α+1 b b At (b, x )
x 1 x 1 .2 F1 2q; 1; α + 2, λ1/α 1 − +2λ1+ α . 2 F1 2q; 1; 2, λ1/α 1 − − , b α+1 b
Hence, if we use (2.6)–(2.8) in (2.5 ), we obtain the desired result. This completes the proof. Corollary 1. Under the assumptions of Theorem 4 with q = 1, the inequality (2.4) reduced to the following inequality
α +1 f (x ), f (a ) C2 α , λ, 1, a S f (x, λ, α , a, b) ≤ (x − a ) sup x (ax )α−1 x2
α +1 1/q x (b − x ) + sup f (x ), f (b) C3 α , λ, 1, , b (bx )α−1 b2
(2.8)
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Corollary 2. Under the assumptions of Theorem 4 with α = 1, the inequality (2.4) reduced to the following inequality
b ab b ( x − a ) f ( a ) + a ( b − x ) f ( b ) f u ( ) S f (x, λ, α , a, b) = (1 − λ ) f (x ) + λ − du x (b − a ) b−a u2 a 2 1− 1q
1/q 1/q a ab 2λ − 2λ + 1 ( x − a )2 f (x )q , f (a )q ≤ sup C 1, λ , q, 2 b−a 2 x x2
2 q q 1/q 1/q x (b − x ) + sup f (x ) , f (b) C3 1, λ, q, , 2
ab b−a
b
b
specially for x = H = 2ab/(a + b), we get
1− 1q b ab f (a ) + f (b ) f (u ) b − a 2λ2 − 2λ + 1 − du ≤ (1 − λ ) f (H ) + λ 2 b−a 2 u2 4ab a
× a2C21/q 1, λ, q,
a+b 2b
sup
1 1 f (H )q , f (a )q q + H 2C 1/q 1, λ, q, 2a f (H )q , f (b)q q (2.9) sup . 3 a+b
Corollary 3. In Theorem 4, (1) If we take x = H = 2ab/(a + b), λ =
1 3,
then we get the following Simpson type inequality for fractional integrals
α 1
α α −1 α [ f (a ) + 4 f (H ) + f (b)] − ab 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 6 b−a 1
q q 1/q 1/q 1 a b − a 1−1/q α, C1 ≤ a2 sup f (H ) , f (a ) C2 α , , q, 4ab 3 3 H
q q 1/q 1/q H 1 +H 2 sup f (H ) , f (a ) C3 α , , q, , 3
b
specially for α = 1, we get
b 1 ab f (u ) b − a 5 1− q 1 du ≤ 6 [ f ( a ) + 4 f ( H ) + f ( b )] − b − a u2 4ab 18 a 1
1 q q 1q a+b 2a 1 × a2C21/q 1, , q, sup f (H ), f (a ) q + H 2C31/q 1, , q, sup f (H ) , f (b) . 3
2b
3
a+b
(2) If we take x = H = 2ab/(a + b), λ = 0, then we get the following midpoint type inequality for fractional integrals
α
α −1 (α + 1 J α α f (H ) − ab 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− b−a 1 1− 1q q q 1q b−a a+b ≤ a2C21/q α , 0, q, sup f (H ) , f (a ) 4ab α + 1 2b
q q 1q 2a 2 1/q +H C3 α , 0, q, sup f (H ) , f (b) , a+b
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293
specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 1− q 2 1/q f (H )q , f (a )q q du ≤ a C 1, 0, q, sup f (H ) − b − a 2 2b u2 4ab 2 a 1
q q q 2a +H 2C 1/q 1, 0, q, sup f (H ) , f (b) . a+b
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 )
α ab α −1 α − 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 2 b−a α 1− 1q q q 1q b−a a+b ≤ a2C21/q α , 1, q, sup f (H ) , f (a ) 4ab α + 1 2b
q 1q 2a q +H 2C 1/q α , 1, q, sup f (H ) , f (b) , a+b
3
specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 1− q 2 1/q f (a ) + f (b ) f (H )q , f (a )q q − du ≤ a C 1, 1, q, sup 2 2 b−a 2b u2 4ab 2 a 1
q q q 2a +H 2C31/q 1, 1, q, sup f (H ) , f (b) . a+b
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 (2.4) integrals
α α x − a α
b−x ab α −1 (α + 1 J α α + f ( x ) − 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− ax bx b−a
M a x (x − a )α+1 1/q (b − x )α+1 1/q ≤ C2 α , 0, q, + C3 α , 0, q, . 1 α −1 2 α −1 2 x b (bx ) b (α + 1 )1− q (ax ) x
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
α +1 q q 1/q S f (x, λ, α , a, b) ≤ C2 α , λ, 1, a (x − a ) sup f (x ) , f (a ) α −1 2 x (ax ) x
α +1 1/q x (b − x ) q f (x ) , f (b)q +C3 α , λ, 1, sup b (bx )α −1 b2
(2.10)
where C2 and C3 are defined as in Theorem 4. Proof. Since |f |q is harmonically quasi-convex on [a, b], from Lemma 1,using property of the modulus and the power-mean inequality we have
1 1− 1q 1 1q
α q (x − a )α+1 |t α − λ| ax |t − λ| S f (x, λ, α , a, b) ≤ dt f A (a, x ) dt (ax )α−1 At2 (a, x ) At2 (a, x ) t +
(b − x )α+1 (bx )α−1
≤ C2
α , λ, 1,
0
1 0
|t α − λ| dt At2 (b, x )
1− 1q
0
a (x − a )α +1 sup x (ax )α −1 x2
which completes the proof.
0
q 1q 1 α bx |t − λ| dt f 2 At (b, x ) At (b, x )
α +1 1/q 1/q x (b − x ) f (x )q , f (a )q f (x )q , f (b)q + C3 α , λ, 1, sup b (bx )α −1 b2
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Corollary 5. Under the assumptions of Theorem 5 with q = 1, the inequality (2.10) reduced to the following inequality
α +1 S f (x, λ, α , a, b) ≤ C2 α , λ, 1, a (x − a ) sup f (x ), f (a ) x (ax )α −1 x2
α +1 x (b − x ) +C3 α , λ, 1, sup f (x ), f (b) α −1 2 b (bx ) b
Corollary 6. Under the assumptions of Theorem 5 with α = 1, the inequality (2.10) reduced to the following inequality
b ab b ( x − a ) f ( a ) + a ( b − x ) f ( b ) f u ( ) S f (x, λ, α , a, b) = (1 − λ ) f (x ) + λ − du 2 x (b − a ) b−a u a
q q 1/q a ab ( x − a )2 ≤ sup f (x ) , f (a ) C2 1, λ, 1, 2 b−a x x
2 1/q q q x (b − x ) + sup f (x ) , f (b) C3 1, λ, 1, , 2
ab b−a
b
b
specially for x = H = 2ab/(a + b), we get
b ab f (a ) + f (b ) f (u ) b − a 1 − λ f H + λ − du ( ) ( ) ≤ 4ab 2 b−a u2 a q q 1q a+b sup f (H ) , f (a ) × a2C2 1, λ, 1, 2b
q q 1q 2a + H 2C3 1, λ, 1, sup f (H ) , f (b) . a+b
(2.11)
Corollary 7. In Theorem 5, (1) If we take x = H = 2ab/(a + b), λ =
1 3,
then we get the following Simpson type inequality for fractional integrals
α 1
α α −1 α [ f (a ) + 4 f (H ) + f (b)] − ab 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 6 b−a
q q 1/q 1 a b−a 2 ≤ a sup f (H ) , f (a ) C2 α , , 1, 4ab
3
q q 1/q 1 x 2 +H sup f (H ) , f (b) C3 α , , 1, ,
x
3
b
specially for α = 1, we get
b q q 1q ab a+b 1 f (u ) b − a 1 2 × a C2 1, , 1, du ≤ sup f (H ) , f (a ) 6 [ f ( a ) + 4 f ( H ) + f ( b )] − b − a 2 3 2b u 4ab a 1
q q 1q 2a +H 2C3 1, , 1, sup f (H ) , f (b) . 3
a+b
(2) If we take x = H = 2ab/(a + b), λ = 0, then we get the following midpoint type inequality for fractional integrals
α
α −1 (α + 1 J α α f (H ) − ab 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− b−a 1 q q q b−a 2 a+b ≤ a C2 α , 0, 1, sup f (H ) , f (a ) 4ab 2b
q q 1q 2a +H 2C3 α , 0, 1, sup f (H ) , f (b) , a+b
specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 1− q 2 1/q f (H )q , f (a )q q du ≤ a C 1, 0, q, sup f (H ) − b − a 2 2b u2 4ab 2 a
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+H 2C31/q 1, 0, q,
2a a+b
sup
f (H )q , f (b)q
1 q
295
.
(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 )
α ab α −1 α − 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 2 b−a q q 1q b−a 2 a+b ≤ a C2 α , 1, 1, sup f (H ) , f (a ) 4ab 2b
q q 1q 2a +H 2C3 α , 1, 1, sup f (H ) , f (b) , a+b
specially for α = 1, we get
b 1 ab a+b f (u ) b − a 2 1/q f (a ) + f (b ) f (H )q , f (a )q q − du ≤ a C 1, 1, 1, sup 2 2 b−a 2b u2 4ab a 1
q q q 2a +H 2C 1/q 1, 1, 1, sup f (H ) , f (b) . a+b
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 (2.10) integrals
α α x − a α
b−x ab α −1 (α + 1 J α α + f ( x ) − 2 f ◦ g 1/a ) + J f ◦ g 1/b ) ) ( )( ( )( 1/H+ 1/H− ax bx b−a
a x (x − a )α+1 (b − x )α+1 ≤M C2 α , 0, 1, + C3 α , 0, 1, . α −1 2 α −1 2 x b (ax ) x (bx ) b Theorem 6. 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
α +1 1/q S f (x, λ, α , a, b) ≤ C 1/q (α , λ ) C 1/p α , λ, p, a (x − a ) f (x )q , f (a )q sup 1 2 x (ax )α −1 x2
α +1 q q 1/q x (b − x ) f (x ) , f (b) +C31/p α , λ, p, sup b (bx )α −1 b2
(2.12)
where C1 , C2 and C3 are defined as in Theorem 4. Proof. Since |f |q is harmonically quasi-convex on [a, b], from Lemma 1, using property of the modulus and the Hölder inequality we have
1 1p 1 1q
q (x − a )α+1 |t α − λ| ax S f (x, λ, α , a, b) ≤ dt |t α − λ| f dt At (a, x ) (ax )α−1 At2p (a, x ) α +1
(b − x ) + (bx )α−1
0
1 0
|t α − λ| dt At2p (b, x )
1p
0
1 α |t − λ| f 0
bx At (b, x )
q 1q dt
1/q a (x − a )α +1 f (x )q , f (a )q sup (α , λ ) C21/p α , λ, p, x (ax )α −1 x2
α +1 q q 1/q x (b − x ) 1/p α , λ, p, +C3 sup f (x ) , f (b) b (bx )α −1 b2 ≤
C11/q
which completes the proof.
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Corollary 9. Under the assumptions of Theorem 6 with α = 1, the inequality (2.12) reduced to the following inequality
b f (u ) S f (x, λ, α , a, b) = (1 − λ ) f (x ) + λ b(x − a ) f (a ) + a(b − x ) f (b) − ab du x (b − a ) b−a u2 a 2 1q
1/q 1/p a ab 2λ − 2λ + 1 ( x − a )2 f (x )q , f (a )q ≤ sup C 1, λ , p, 2 b−a 2 x x2
2 q q 1/q 1/p x (b − x ) + sup f (x ) , f (a ) C3 1, λ, p, , 2
ab b−a
b
b
specially for x = H = 2ab/(a + b), we get
1 b ab f (a ) + f (b ) f (u ) b − a 2λ2 − 2λ + 1 q − du ≤ (1 − λ ) f (H ) + λ 2 b−a 2 u2 4ab a
× a2C21/p 1, λ, p,
a+b 2b
sup
1 1 f (H )q , f (a )q q + H 2C 1/p 1, λ, p, 2a f (H )q , f (b)q q (2.13) sup . 3 a+b
Corollary 10. In Theorem 6, (1) If we take x = H = 2ab/(a + b), λ =
1 3,
then we get the following Simpson type inequality for fractional integrals
1/q 1/p a+b 1 S f (x, λ, α , a, b) ≤ b − a C 1−1/q α , 1 a2 sup f (H )q , f (a )q , p, C , α 2 4ab 1 3 3 2b
1/q 2a 1 q q +H 2 sup f (H ) , f (a ) C31/p α , , p, ,
ab b−a
α
3
a+b
specially for α = 1, we get
b 1 1 ab a+b 1 f (u ) b − a 5 q 1 2 1/p , p, du ≤ × a C 1, sup f (H ), f (a ) q 6 [ f ( a ) + 4 f ( H ) + f ( b )] − b − a 2 2 4ab 18 3 2b u a 1
q q 1q 2a 2 1/p +H C 1, , p, sup f (H ) , f (b) . 3
3
a+b
(2) If we take x = H = 2ab/(a + b), λ = 0, then we get the following midpoint type inequality for fractional integrals
α
α α −1 α f (H ) − ab 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) b−a 1 1q q q 1q b−a a+b ≤ a2C21/p α , 0, p, sup f (H ) , f (a ) 4ab α + 1 2b
q q 1q 2a +H 2C 1/p α , 0, p, sup f (H ) , f (b) , 3
a+b
specially for α = 1, we get
b 1 1 ab a+b f (u ) b − a 1 q 2 1/p f (H )q , f (a )q q du ≤ a C 1, 0, p, sup f (H ) − b − a 2 2b u2 4ab 2 a 1
q q q 2a +H 2C 1/p 1, 0, p, sup f (H ) , f (b) . 3
a+b
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297
(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 )
α ab α −1 α − 2 (α + 1 ) J1/H+ ( f ◦ g)(1/a ) + J1/H− ( f ◦ g)(1/b) 2 b−a α 1q q q 1q b−a a+b 2 1/p ≤ a C2 sup f (H ) , f (a ) α , 1, p, 4ab α + 1 2b
q q 1q 2a +H 2C 1/p α , 1, p, sup f (H ) , f (b) , a+b
3
specially for α = 1, we get
b 1 q q 1q ab a+b f (u ) b − a 1 q 2 1/p f (a ) + f (b ) − du ≤ a C2 1, 1, p, sup f (H ) , f (a ) 2 2 b−a 2b u 4ab 2 a
q q 1q 2a +H 2C 1/p 1, 1, p, sup f (H ) , f (b) . a+b
3
Corollary 11. Let the assumptions of Theorem 6 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 (2.12) integrals
α α x − a α
α b−x ab α −1 α + f (x ) − 2 (α + 1 ) J1/x+ ( f ◦ g)(1/a ) + J1/x− ( f ◦ g)(1/b) ax bx b−a
M a x (x − a )α+1 1/p (b − x )α+1 1/p α α ≤ C , 0, p, + C , 0, p, . 1 α −1 2 2 x b (bx )α−1 b2 3 (α + 1 ) q (ax ) x
3. Application to special means Let us recall the following special means of two nonnegative number a, b with b > a: (1) The arithmetic mean
A = A(a, b) :=
a+b . 2
(2) The geometric mean
G = G(a, b) :=
a+b . 2
(3) The harmonic mean
H = H (a, b) :=
2ab . a+b
(4) The n-logarithmic mean
Ln = Ln (a, b) := Let f (x ) =
xn+1 n+1 ,
bn+1 − an+1 (n + 1 )(b − a )
1n
, n ∈ R\{−1, 0}.
x > 0, n ≥ 1 and q ≥ 1. Then, the function | f (x )|q = xnq is harmonically quasi-convex on (0, ∞).
Proposition 1. For b > a > 0, n > 1, λ ∈ [0, 1] and q ≥ 1, we have
1− 1q b − a 2λ2 − 2λ + 1 (1 − λ )H n+1 + λA an+1 , bn+1 − G2 Ln−1 ≤ n + 1 ( ) n−1
× a2 H nC21/q 1, λ, q,
A b
+ H 2 bnC31/q 1, λ, q,
a A
4G2
2
,
specially for n = 1 we get
2 1− 1q
A a 1/q 1/q 2 2 (1 − λ )H 2 + λA a2 , b2 − G2 ≤ b − a 2λ − 2λ + 1 × a HC 1, λ , q, + H bC 1, λ , q, , 2 3 2 2G
where C2 and C3 are defined as in Theorem 4.
2
b
A
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298
xn+1 n+1 ,
Proof. The assertion follows from inequality (2.9) in Corollary 2, for f (x ) =
x > 0.
Proposition 2. For b > a > 0, n > 1, λ ∈ [0, 1] and q ≥ 1, we have
A a b−a 2 n 2 n (1 − λ )H n+1 + λA an+1 , bn+1 − G2 Ln−1 ≤ n + 1 × a H C 1, λ , 1, + H b C 1, λ , 1, , ( ) 2 3 n−1 2 b
4G
A
specially for n = 1 we get
(1 − λ )H 2 + λA a2 , b2 − G2 ≤ b − a × a2 HC2 1, λ, 1, A + H 2 bC3 1, λ, 1, a , 2 b
2G
A
where C2 and C3 are defined as in Theorem 4. Proof. The assertion follows from inequality (2.11) in Corollary 6, for f (x ) =
xn+1 n+1 ,
x > 0.
Proposition 3. For b > a > 0, n > 1, λ ∈ [0, 1] and q ≥ 1, we have
2 1q n+1 n+1 b − a 2 λ − 2 λ + 1 n+1 2 n−1 (1 − λ )H + λA a , b − G Ln−1 ≤ (n + 1 )
× a2 H nC21/p 1, λ, p,
A b
+ H 2 bnC31/p 1, λ, p,
a A
4G2
2
,
specially for n = 1 we get
2 1q
A a 1/p 1/p 2 2 (1 − λ )H 2 + λA a2 , b2 − G2 ≤ b − a 2λ − 2λ + 1 × a HC 1, λ , p, + H bC 1, λ , p, , 2 3 2 2G
2
b
A
where C2 and C3 are defined as in Theorem 4. Proof. The assertion follows from inequality (2.13) in Corollary 9, for f (x ) =
xn+1 n+1 ,
x > 0.
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