Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 705–716
Research Article
Several complementary inequalities to inequalities of
Hermite-Hadamard type for s-convex functions
Feixiang Chena , Shanhe Wub,∗
a
Key Laboratory for Nonlinear Science and System Structure, Chongqing Three Georges University, Wanzhou, 404100, P. R. China.
b
Department of Mathematics and Computer Science, Longyan University, Longyan, Fujian, 364012, P. R. China.
Communicated by Yonghong Yao
Abstract
In this paper, we establish some new Hermite-Hadamard inequalities for s-convex functions via fractional
integrals. Some Hermite-Hadamard type inequalities for products of two convex and s-convex functions via
c
Riemann-Liouville integrals are also established. 2016
All rights reserved.
Keywords: Hermite-Hadamard inequality, s-convex function, Riemann-Liouville fractional integrals.
2010 MSC: 26D15, 26A51.
1. Introduction and preliminaries
If f : I → R is a convex function on the interval I, then for any a, b ∈ I with a 6= b we have the following
double inequality
Z b
f (a) + f (b)
a+b
1
≤
f (t)dt ≤
.
(1.1)
f
2
b−a a
2
This remarkable result is well known in the literature as the Hermite-Hadamard inequality.
Definition 1.1 ([7]). f : I ⊂ [0, ∞) → [0, ∞) is said to be s-convex in the second sense, or that f belongs
to the class Ks2 , if the inequality
f (αx + (1 − α)y) ≤ αs f (x) + (1 − α)s f (y)
holds for all x, y ∈ I, α ∈ [0, 1] and for some fixed s ∈ (0, 1].
∗
Corresponding author
Email addresses: cfx2002@126.com (Feixiang Chen), shanhewu@gmail.com (Shanhe Wu)
Received 2015-08-10
(1.2)
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
706
It can be easily seen that for s = 1, s-convexity reduces to ordinary convexity of functions defined on
[0, ∞).
In [6], Dragomir and Fitzpatrick proved a variant of Hermite-Hadamard inequality which holds for the
s-convex functions.
Theorem 1.2 ([6]). Suppose that f : [0, ∞) → [0, ∞) is an s-convex function in the second sense, where
s ∈ (0, 1] and let a, b ∈ [0, ∞), a < b. If f ∈ L[a, b], then the following inequality holds
Z b
a+b
1
f (a) + f (b)
s−1
2 f
≤
f (t)dt ≤
.
(1.3)
2
b−a a
s+1
In [8], İşcan gave definition of harmonically convexity as follows:
Definition 1.3 ([8]). Let I ⊆ R\{0} be a real interval. A function f : I → R is said to be harmonically
convex, if
xy
f
≤ tf (y) + (1 − t)f (x)
(1.4)
tx + (1 − t)y
for all x, y ∈ I and t ∈ [0, 1]. If the inequality in (1.4) is reversed, then f is said to be harmonically concave.
The following result of the Hermite-Hadamard type holds.
Theorem 1.4 ([8]). Let f : I ⊆ R\{0} → R be a harmonically convex function and a, b ∈ I with a < b. If
f ∈ L(a, b) then we have
Z b
2ab ab
f (x)
f (a) + f (b)
≤
dx ≤
.
(1.5)
f
a+b
b − a a x2
2
The definition of harmonically s-convex functions is proposed by İşcan in [9].
Definition 1.5 ([9]). Let I ⊆ R\{0} be a real interval. A function f : I ⊆ (0, ∞) → R is said to be
harmonically s-convex, if
xy
f
≤ ts f (y) + (1 − t)s f (x)
(1.6)
tx + (1 − t)y
for all x, y ∈ I, t ∈ [0, 1] and for some fixed s ∈ (0, 1]. If the inequality in (1.6) is reversed, then f is said to
be harmonically s-concave.
The following Hermite-Hadamard inequality for harmonically s-convex functions holds.
Theorem 1.6 ([9]). Let f : I ⊆ (0, ∞) → R be a harmonically s-convex function and a, b ∈ I with a < b.
If f ∈ L(a, b) then we have
Z b
2ab ab
f (x)
f (a) + f (b)
2s−1 f
≤
dx ≤
.
(1.7)
a+b
b − a a x2
s+1
In [4], Chen and Wu discussed Fejér and Hermite-Hadamard type inequalities for Harmonically convex
functions and presented the following inequality:
Theorem 1.7 ([4]). Let f : I ⊆ R\{0} → R be a harmonically convex function and a, b ∈ I with a < b. If
f ∈ L(a, b) then we have
Z b
Z
2ab Z b p(x)
f (x)
f (a) + f (b) b p(x)
f
dx ≤
p(x)dx ≤
dx,
(1.8)
a + b a x2
x2
2
x2
a
a
where p : [a, b] → R is nonnegative, integrable and satisfies
p
ab x
=p
ab
.
a+b−x
(1.9)
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
707
Some refinements of the Hermite-Hadamard inequality for convex functions have been extensively investigated by a number of authors (e.g., [1], [2], [3], [4] and [5]).
In [11], Pachpatte established two new Hermite-Hadamard type inequalities for products of convex
functions as follows.
Theorem 1.8. Let f and g be real-valued, nonnegative and convex functions on [a, b]. Then
Z b
1
f (x)g(x)dx ≤
b−a a
a + b a + b
2f
g
≤
2
2
1
1
M (a, b) + N (a, b),
3
6
Z b
1
1
1
f (x)g(x)dx + M (a, b) + N (a, b),
b−a a
6
3
(1.10)
(1.11)
where M (a, b) = f (a)g(a) + f (b)g(b) and N (a, b) = f (a)g(b) + f (b)g(a).
Some Hermite-Hadamard type inequalities for products of two convex and s-convex functions are proposed by Kirmaci et al. in [10].
Theorem 1.9 ([10]). Let f, g : [a, b] → R, a, b ∈ [0, ∞), a < b, be functions such that g, f g ∈ L[a, b]. If f
is convex and nonnegative on [a, b] and g is s-convex on [a, b] for some fixed s ∈ (0, 1], then
1
b−a
Z
b
f (x)g(x)dx ≤
a
1
1
M (a, b) +
N (a, b),
s+2
(s + 1)(s + 2)
(1.12)
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (b)g(a).
Theorem 1.10 ([10]). Let f, g : [a, b] → R, a, b ∈ [0, ∞), a < b, be functions such that f , g and f g ∈ L[a, b].
If f is s1 -convex and g is s2 -convex on [a, b] for some fixed s1 , s2 ∈ (0, 1], then
b
Z
1
b−a
f (x)g(x)dx ≤
a
1
M (a, b) + β(s2 + 1, s1 + 1)N (a, b),
s1 + s2 + 1
(1.13)
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (b)g(a).
Theorem 1.11 ([10]). Let f, g : [a, b] → R, a, b ∈ [0, ∞), a < b, be functions such that f g ∈ L[a, b]. If f is
convex and nonnegative on [a, b] and g is s-convex on [a, b] for some fixed s ∈ (0, 1], then
Z b
a + b a + b
1
1
1
g
≤
f (x)g(x)dx +
M (a, b) +
N (a, b),
2 f
2
2
b−a a
(s + 1)(s + 2)
s+2
s
(1.14)
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (a)g(b).
In [5], Chen and Wu discussed Hermite-Hadamard type inequalities for Harmonically s-convex functions
and obtained the following result:
Theorem 1.12 ([5]). Let f, g : [a, b] → [0, ∞), a, b ∈ (0, ∞), a < b, be functions such that f , g, f g ∈ L[a, b].
If f is harmonically s1 -convex and g is harmonically s2 -convex on [a, b] for some fixed s1 , s2 ∈ (0, 1], then
ab
b−a
Z
a
b
f (x)g(x)
1
Γ(1 + s1 )Γ(s2 + 1)
dx ≤
M (a, b) +
N (a, b),
2
x
1 + s1 + s2
Γ(s1 + s2 + 2)
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (b)g(a).
Theorem 1.13 ([5]). Let f, g : [a, b] → [0, ∞), a, b ∈ (0, ∞), a < b, be functions such that f , g, f g ∈ L[a, b].
If f is harmonically s1 -convex and g is harmonically s2 -convex on [a, b] for some fixed s1 , s2 ∈ (0, 1], then
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
2
s1 +s2 −1
708
Z b
2ab 2ab f (x)g(x)
ab
Γ(1 + s1 )Γ(s2 + 1)
f
g
≤
dx + M (a, b)
2
a+b
a+b
b−a a
x
Γ(s1 + s2 + 2)
1
N (a, b),
+
s2 + s1 + 1
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (a)g(b).
Sarikaya et al. [12] proved the following interesting inequalities of Hermite-Hadamard type involving
Riemann-Liouville fractional integrals.
Theorem 1.14 ([12]). Let f : [a, b] → R be a positive function with a < b and f ∈ L[a, b]. If f is a convex
function on [a, b], then the following inequalities for fractional integrals hold:
f(
f (a) + f (b)
a+b
Γ(α + 1) α
)≤
[Ja+ f (b) + Jbα− f (a)] ≤
,
α
2
2(b − a)
2
(1.15)
with α > 0.
We remark that the symbol Jaα+ and Jbα− f denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order α ≥ 0 with a ≥ 0 which are defined by
Z x
1
α
Ja+ f (x) =
(x − t)α−1 f (t)dt,
x > a,
Γ(α) a
and
Jbα− f (x)
1
=
Γ(α)
Z
b
(t − x)α−1 f (t)dt,
x < b,
x
R∞
respectively. Here, Γ(α) is the Gamma function defined by Γ(α) = 0 e−t tα−1 dt.
Chen and Wu [3] investigated the Hermite-Hadamard type inequalities for products of two h-convex
functions and established the following inequality:
Theorem 1.15. Let f ∈ SX(h1 , I), g ∈ SX(h2 , I), a, b ∈ I, a < b, be functions such that f g ∈ L[a, b], and
h1 h2 ∈ L1 [0, 1], then the following inequality for fractional integrals holds:
Z 1
Γ(α)
α
α
[J + f (b)g(b) + Jb− f (a)g(a)] ≤ M (a, b)
tα−1 [h1 (t)h2 (t) + h1 (1 − t)h2 (1 − t)]dt
(b − a)α a
0
Z 1
+N (a, b)
tα−1 [h1 (t)h2 (1 − t) + h1 (1 − t)h2 (t)]dt,
0
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (b)g(a).
Theorem 1.16. Let f ∈ SX(h1 , I), g ∈ SX(h2 , I), a, b ∈ I, a < b, be functions such that f g ∈ L1 [a, b],
and h1 h2 ∈ L[0, 1], then the following inequality for fractional integrals holds:
a + b a + b
1
Γ(α)
f
g
≤
[J α+ f (b)g(b) + Jbα− f (a)g(a)]
1
1
2
2
(b − a)α a
αh1 ( 2 )h2 ( 2 )
Z 1
+ M (a, b)
tα−1 [h1 (t)h2 (1 − t) + h1 (1 − t)h2 (t)]dt
0
Z 1
+ N (a, b)
tα−1 [h1 (t)h2 (t) + h1 (1 − t)h2 (1 − t)]dt,
0
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (b)g(a).
In [13], Set et al. established the following Hermite-Hadamard inequalities for s-convex functions in the
second sense via fractional integrals.
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
709
Theorem 1.17 ([13]). Let f : [a, b] → R be a positive function with 0 ≤ a < b and f ∈ L[a, b]. If f is
s-convex in the second sense on [a, b] for some fixed s ∈ (0, 1], then the following inequalities for fractional
integrals with α > 0 hold:
a + b
h 1
i f (a) + f (b)
Γ(α + 1) α
s−1
α
2 f
≤
[J + f (b) + Jb− f (a)] ≤
+ β(α, s + 1)
,
(1.16)
2
2(b − a)α a
α+s
2
where β is Euler Beta function.
In this paper, we obtain some new Hermite-Hadamard type inequalities for s-convex functions via
Riemann-Liouville fractional integrals. Several Hermite-Hadamard type inequalities for products of two
convex and s-convex functions are also established.
2. Main results
In order to prove our main theorems, we need the following fundamental integral identity including the
second order derivatives of a given function via Riemann-Liouville integrals which was established by Wang
et al. in [14].
Lemma 2.1 ([14]). Let f : [a, b] → R be a twice differentiable mapping on (a, b) with a < b. If f 00 ∈ L[a, b],
then the following equality for fractional integrals hold:
Z
Γ(α + 1) α
f (a) + f (b)
(b − a)2 1 (1 − t)α+1 + tα+1 − 1 00
α
[J + f (b) + Jb− f (a)] −
=
f (ta + (1 − t)b)dt. (2.1)
2(b − a)α a
2
2
α+1
0
Theorem 2.2. Let f : [a, b] → R be a twice differentiable mapping on (a, b) with a < b such that f 00 ∈ L[a, b].
If |f 00 | is s-convex on [a, b], then the following inequality holds:
Γ(α + 1)
f (a) + f (b) α
α
[J + f (b) + Jb− f (a)] −
2(b − a)α a
2
|f 00 (a)| + |f 00 (b)|
1
(b − a)2 1
−
− β(α + 2, s + 1)
.
≤
2
s+1 α+s+2
α+1
Proof. From lemma 2.1, we get
Γ(α + 1)
f (a) + f (b) α
α
[J
f
(b)
+
J
f
(a)]
−
+
b−
2(b − a)α a
2
(2.2)
Z
(b − a)2 1 (1 − t)α+1 + tα+1 − 1 00
≤
|f (ta + (1 − t)b)|dt.
2
α+1
0
Because (1 − t)α+1 + tα+1 ≤ 1 for any t ∈ [0, 1] and |f 00 | is s-convex on [a, b], we get
Z 1
(1 − t)α+1 + tα+1 − 1 00
|f (ta + (1 − t)b)|dt
α+1
0
Z 1
1 − (1 − t)α+1 − tα+1 s 00
≤
t |f (a)| + (1 − t)s |f 00 (b)| dt
α+1
0
1
|f 00 (a)| + |f 00 (b)|
1
=
−
− β(α + 2, s + 1)
.
s+1 α+s+2
α+1
(2.3)
Now by (2.2), (2.3), we can obtain the desired result.
Theorem 2.3. Let f : [a, b] → R be a twice differentiable mapping on (a, b) with a < b such that f 00 ∈ L[a, b].
If |f 00 |q is s-convex on [a, b] with q ≥ 1, then the following inequality holds:
Γ(α + 1)
f (a) + f (b) α
α
[J
f
(b)
+
J
f
(a)]
−
+
−
b
2(b − a)α a
2
1 1
(b − a)2 α 1− 1q 1
1
00
q
q
00
q
q
≤
×
− β(α + 2, s + 1) −
|f (a| + |f (b)| ) .
2(α + 1) α + 2
s+1
α+s+2
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
710
Proof. From lemma 2.1 and the power mean inequality for q, we have
Γ(α + 1)
f (a) + f (b) α
α
[J
f
(b)
+
J
f
(a)]
−
+
−
b
2(b − a)α a
2
Z
i1− 1
(b − a)2 h 1
q
α+1
α+1
(1 − (1 − t)
−t
)dt
≤
2(α + 1) 0
1
Z 1
q
(1 − (1 − t)α+1 − tα+1 )|f 00 (ta + (1 − t)b)|q dt
×
0
Z
i1− 1
(b − a)2 h 1
q
≤
(1 − (1 − t)α+1 − tα+1 )dt
2(α + 1) 0
1
Z 1
00
q
(1 − (1 − t)α+1 − tα+1 )(ts |f 00 (a|q + (1 − t)s |f (b)|q )dt
×
0
=
1
1 (b − a)2 α 1− 1q 1
1
00
q
q
|f 00 (a|q + |f (b)|q ) ,
− β(α + 2, s + 1) −
2(α + 1) α + 2
s+1
α+s+2
which completes the proof.
Theorem 2.4. Let f : [a, b] → R be a twice differentiable mapping on (a, b) with a < b such that f 00 ∈ L[a, b].
If |f 00 |q is s-convex on [a, b] with q > 1, then the following inequality holds:
Γ(α + 1)
1 |f 00 (a|q + |f 00 (b)|q 1
f (a) + f (b) (b − a)2 2
p
q
α
α
,
[Ja+ f (b) + Jb− f (a)] −
1−
≤
α
2(b − a)
2
2(α + 1)
p(α + 1) + 1
s+1
where
1 1
+ = 1.
p q
Proof. From lemma 2.1 and using the Hölder inequality, we have
Γ(α + 1)
f (a) + f (b) α
α
[J
f
(b)
+
J
f
(a)]
−
+
−
b
2(b − a)α a
2
Z
(b − a)2 1 1 − (1 − t)α+1 − tα+1 00
≤
|f (ta + (1 − t)b)|dt
2
α+1
0
Z
p 1 Z 1
1
(b − a)2 1 p
q
α+1
α+1
00
q
≤
1 − (1 − t)
−t
dt
|f (ta + (1 − t)b)| dt
2(α + 1) 0
0
Z 1
1Z 1
1
2
(b − a)
p
q
≤
1 − (1 − t)p(α+1) − tp(α+1) dt
|f 00 (ta + (1 − t)b)|q dt
2(α + 1) 0
0
1 Z 1
1
(b − a)2 2
p
q
=
1−
|f 00 (ta + (1 − t)b)|q dt .
2(α + 1)
p(α + 1) + 1
0
(2.4)
Because |f 00 |q is s-convex, we have
Z
0
1
00
|f 00 (a|q + |f (b)|q
|f (ta + (1 − t)b)| dt ≤
.
s+1
00
q
Here we use
(1 − (1 − t)α+1 − tα+1 )p ≤ 1 − (1 − t)p(α+1) − tp(α+1)
for any t ∈ [0, 1], which follows from
(A − B)p ≤ Ap − B p
for any A > B ≥ 0 and p ≥ 1. From (2.4) and (2.5), we complete the proof.
(2.5)
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
711
We note that the Beta and Gamma functions are defined, respectively, as follows
Z ∞
Γ(x) =
e−t tx−1 dt, x > 0,
0
Z 1
(1 − t)y−1 tx−1 dt, x > 0, y > 0,
β(x, y) =
0
and
β(x, y) =
Γ(x)Γ(y)
.
Γ(x + y)
Theorem 2.5. Let f, g : [a, b] → R, a, b ∈ [0, ∞), a < b, be functions such that g, f g ∈ L[a, b]. If f is
convex and nonnegative and g is s-convex on [a, b] for some fixed s ∈ (0, 1], then the following inequality for
fractional integrals holds:
Γ(α)
[J α+ f (b)g(b) + Jbα− f (a)g(a)]
(b − a)α a
1
1
≤
+ β(α, s + 2) M (a, b) + β(α + 1, s + 1) +
N (a, b),
α+s+1
(α + s)(α + s + 1)
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (b)g(a).
Proof. Since f is convex and g is s-convex on [a, b], then for t ∈ [0, 1] we get
f (ta + (1 − t)b) ≤ tf (a) + (1 − t)f (b),
(2.6)
g(ta + (1 − t)b) ≤ ts g(a) + (1 − t)s g(b).
(2.7)
and
From (2.6) and (2.7), we get
f (ta + (1 − t)b)g(ta + (1 − t)b) ≤ ts+1 f (a)g(a) + (1 − t)s+1 f (b)g(b) + t(1 − t)s f (a)g(b) + (1 − t)ts f (b)g(a).
Similarly, we have
f ((1 − t)a + tb)g((1 − t)a + tb) ≤ (1 − t)s+1 f (a)g(a) + ts+1 f (b)g(b) + (1 − t)ts f (a)g(b) + t(1 − t)s f (b)g(a).
So
f (ta + (1 − t)b)g(ta + (1 − t)b) + f ((1 − t)a + tb)g((1 − t)a + tb)
≤ (ts+1 + (1 − t)s+1 )[f (a)g(a) + f (b)g(b)]
+ (t(1 − t)s + (1 − t)ts )[f (a)g(b) + f (b)g(a)].
Multiplying both sides of above inequality by tα−1 , then integrating the resulting inequality with respect to
t over [0, 1], we obtain
Z 1
Z 1
α−1
t
f (ta + (1 − t)b)g(ta + (1 − t)b)dt +
tα−1 f ((1 − t)a + tb)g((1 − t)a + tb)dt
0
0
Z b
Z a
du
v − a α−1
dv
b − u α−1
f (u)g(u)
+
f (v)g(v)
=
b
−
a
a
−
b
b
−
a
b
−a
a
b
Γ(α)
=
[J α+ f (b)g(b) + Jbα− f (a)g(a)]
(b − a)α a
Z 1
≤ [f (a)g(a) + f (b)g(b)]
tα−1 (ts+1 + (1 − t)s+1 )dt
0
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
Z
712
1
tα−1 (t(1 − t)s + (1 − t)ts )dt
+ [f (a)g(b) + f (b)g(a)]
0
1
=
+ β(α, s + 2) [f (a)g(a) + f (b)g(b)]
α+s+1
1
[f (a)g(b) + f (b)g(a)]
+ β(α + 1, s + 1) +
(α + s)(α + s + 1)
1
1
=
+ β(α, s + 2) M (a, b) + β(α + 1, s + 1) +
N (a, b).
α+s+1
(α + s)(α + s + 1)
So
Γ(α)
[J α+ f (b)g(b) + Jbα− f (a)g(a)]
(b − a)α a
1
1
≤
+ β(α, s + 2) M (a, b) + β(α + 1, s + 1) +
N (a, b),
α+s+1
(α + s)(α + s + 1)
which completes the proof.
Remark 2.6. Taking α = 1 in Theorem 2.5, we have
1
b−a
Z
b
1
M (a, b) N (a, b)
1
+ β(1, s + 2)
+ β(2, s + 1) +
s+2
2
(1 + s)(s + 2)
2
1
1
=
M (a, b) +
N (a, b)
s+2
(1 + s)(s + 2)
f (x)g(x)dx ≤
a
which is the result of (1.12).
Remark 2.7. Choosing f (x) = 1 for all x ∈ [a, b] in Theorem 2.5 gives
Γ(α)
1
α
α
[J
g(b)
+
J
g(a)]
≤
+ β(α, s + 2) + β(α + 1, s + 1)
+
−
a
b
α
(b − a)
α+s+1
1
[g(a) + g(b)]
+
(α + s)(α + s + 1)
1
=
+ β(α, s + 1) [g(a) + g(b)],
α+s
which is the right hand side of (1.16).
Theorem 2.8. Let f, g : [a, b] → R, a, b ∈ [0, ∞), a < b, be functions such that f , g, f g ∈ L[a, b]. If f is
s1 -convex and g is s2 -convex on [a, b] for some fixed s1 , s2 ∈ (0, 1], then the following inequality for fractional
integrals holds:
Γ(α)
1
α
α
+
β(α,
s
+
s
+
1)
M (a, b)
[J
f
(b)g(b)
+
J
f
(a)g(a)]
≤
+
−
1
2
b
(b − a)α a
α + s1 + s2
+ β(α + s1 , s2 + 1) + β(α + s2 , s1 + 1) N (a, b),
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (b)g(a).
Proof. Since f is s1 -convex and g is s2 -convex on [a, b], then for t ∈ [0, 1] we get
f (ta + (1 − t)b) ≤ ts1 f (a) + (1 − t)s1 f (b),
(2.8)
g(ta + (1 − t)b) ≤ ts2 g(a) + (1 − t)s2 g(b).
(2.9)
and
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
713
From (2.8) and (2.9), we get
f (ta+(1−t)b)g(ta+(1−t)b) ≤ ts1 +s2 f (a)g(a)+(1−t)s1 +s2 f (b)g(b)+ts1 (1−t)s2 f (a)g(b)+(1−t)s1 ts2 f (b)g(a).
Similarly, we have
f ((1−t)a+tb)g((1−t)a+tb) ≤ (1−t)s1 +s2 f (a)g(a)+ts1 +s2 f (b)g(b)+(1−t)s1 ts2 f (a)g(b)+ts1 (1−t)s2 f (b)g(a).
So
f (ta + (1 − t)b)g(ta + (1 − t)b) + f ((1 − t)a + tb)g((1 − t)a + tb)
≤ (ts1 +s2 + (1 − t)s1 +s2 )[f (a)g(a) + f (b)g(b)]
+ (ts1 (1 − t)s2 + (1 − t)s1 ts2 )[f (a)g(b) + f (b)g(a)].
Multiplying both sides of above inequality by tα−1 , then integrating the resulting inequality with respect to
t over [0, 1], we obtain
Z 1
Z 1
α−1
tα−1 f ((1 − t)a + tb)g((1 − t)a + tb)dt
t
f (ta + (1 − t)b)g(ta + (1 − t)b)dt +
0
0
Z a
Z b
b − u α−1
v − a α−1
du
dv
=
+
f (u)g(u)
f (v)g(v)
b−a
a−b
b−a
b−a
b
a
Γ(α)
[J α+ f (b)g(b) + Jbα− f (a)g(a)]
=
(b − a)α a
Z 1
≤ [f (a)g(a) + f (b)g(b)]
tα−1 (ts1 +s2 + (1 − t)s1 +s2 )dt
0
Z 1
+ [f (a)g(b) + f (b)g(a)]
tα−1 (ts1 (1 − t)s2 + (1 − t)s1 ts2 )dt
0
1
+ β(α, s1 + s2 + 1) [f (a)g(a) + f (b)g(b)]
=
α + s1 + s2
+ β(α + s1 , s2 + 1) + β(α + s2 , s1 + 1) [f (a)g(b) + f (b)g(a)]
1
+ β(α, s1 + s2 + 1) M (a, b) + β(α + s1 , s2 + 1) + β(α + s2 , s1 + 1) N (a, b).
=
α + s1 + s2
So
Γ(α)
1
α
α
[J
f
(b)g(b)
+
J
f
(a)g(a)]
≤
+
β(α,
s
+
s
+
1)
M (a, b)
+
−
1
2
b
(b − a)α a
α + s1 + s2
+ β(α + s1 , s2 + 1) + β(α + s2 , s1 + 1) N (a, b),
which completes the proof.
Remark 2.9. Putting α = 1 in Theorem 2.8 leads to
Z b
M (a, b)
1
1
f (x)g(x)dx ≤
+ β(1, s1 + s2 + 1)
b−a a
s1 + s2 + 1
2
N (a, b)
+ β(1 + s1 , s2 + 1) + β(1 + s2 , s1 + 1)
2
1
=
M (a, b) + β(s2 + 1, s1 + 1)N (a, b),
s1 + s2 + 1
which is the result of (1.13).
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
714
Theorem 2.10. Let f, g : [a, b] → R, a, b ∈ [0, ∞), a < b, be functions such that f g ∈ L[a, b]. If f is convex
and nonnegative on [a, b] and g is s-convex on [a, b] for some fixed s ∈ (0, 1], then
a + b a + b
Γ(α + 1) α
2s f
[J + f (b)g(b) + Jbα− f (a)g(a)]
g
≤
2
2
2(b − a)α a
1
1
+ M (a, b) β(α + 1, s + 1) +
2
(α + s)(α + s + 1)
1
1
+ N (a, b) β(α, s + 2) +
,
2
α+s+1
where M (a, b) = f (a)g(a) + f (b)g(b), N (a, b) = f (a)g(b) + f (a)g(b).
Proof. We can write
a+b
(1 − t)a + tb ta + (1 − t)b
=
+
,
2
2
2
so
f
a + b a + b
g
2
2
ta + (1 − t)b (1 − t)a + tb ta + (1 − t)b (1 − t)a + tb +
g
+
=f
2
2
2
2
h
ih
i
1
≤ s+1 f (ta + (1 − t)b) + f ((1 − t)a + tb) g(ta + (1 − t)b) + g((1 − t)a + tb)
2
1 h
= s+1 f (ta + (1 − t)b)g(ta + (1 − t)b) + f ((1 − t)a + tb)g((1 − t)a + tb)
2
i
+ f (ta + (1 − t)b)g((1 − t)a + tb) + f ((1 − t)a + tb)g(ta + (1 − t)b)
i
1 h
≤ s+1 f (ta + (1 − t)b)g(ta + (1 − t)b) + f ((1 − t)a + tb)g((1 − t)a + tb)
2
(
h
ih
i
1
+ s+1 tf (a) + (1 − t)f (b) (1 − t)s g(a) + ts g(b)
2
)
h
ih
i
s
s
+ (1 − t)f (a) + tf (b) t g(a) + (1 − t) g(b)
=
1 h
i
f
(ta
+
(1
−
t)b)g(ta
+
(1
−
t)b)
+
f
((1
−
t)a
+
tb)g((1
−
t)a
+
tb)
2s+1 (
)
i
h
i
h
1
+ s+1 t(1 − t)s + (1 − t)ts M (a, b) + (1 − t)s+1 + ts+1 N (a, b) .
2
Multiplying both sides of above inequality by tα−1 , then integrating the resulting inequality with respect to
t over [0, 1], we get
Z
a + b a + b Z 1
1 h 1 α−1
α−1
f
g
t
dt ≤ s+1
t
f (ta + (1 − t)b)g(ta + (1 − t)b)dt
2
2
2
0
0
Z 1
i
+
tα−1 f ((1 − t)a + tb)g((1 − t)a + tb)dt
0
(
Z 1
1
+ s+1 M (a, b)
tα−1 [t(1 − t)s + (1 − t)ts ]dt
2
0
)
Z 1
+ N (a, b)
tα−1 [(1 − t)s+1 + ts+1 ]dt .
0
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
715
That is
i
1 a + b a + b
1 h Γ(α)
α
α
[J
f
(b)g(b)
+
J
f
(a)g(a)]
f
g
≤ s+1
+
b−
α
2
2
2
(b − a)α a
(
Z 1
1
+ s+1 M (a, b)
tα−1 [t(1 − t)s + (1 − t)ts ]dt
2
0
)
Z 1
tα−1 [(1 − t)s+1 + ts+1 ]dt .
+ N (a, b)
0
From
1
Z
tα−1 [t(1 − t)s + (1 − t)ts ]dt = β(α + 1, s + 1) +
0
and
1
Z
tα−1 [(1 − t)s+1 + ts+1 ]dt = β(α, s + 2) +
0
1
(α + s)(α + s + 1)
1
,
α+s+1
we get
2s f
a + b a + b
Γ(α + 1) α
g
≤
[J + f (b)g(b) + Jbα− f (a)g(a)]
2
2
2(b − a)α a
1
1
+ M (a, b) β(α + 1, s + 1) +
2
(α + s)(α + s + 1)
1
1
+ N (a, b) β(α, s + 2) +
,
2
α+s+1
which completes the proof.
Remark 2.11. Setting α = 1 in Theorem 2.10, then
2s f
Z b
a + b a + b
1
g
≤
f (x)g(x)dx
2
2
b−a a
1
1
+ M (a, b) β(2, s + 1) +
2
(s + 1)(s + 2)
1
1 + N (a, b) β(1, s + 2) +
2
s+2
Z b
1
1
1
=
f (x)g(x)dx +
M (a, b) +
N (a, b),
b−a a
(s + 1)(s + 2)
s+2
which is the result of (1.14).
Remark 2.12. Choosing f (x) = 1 for all x ∈ [a, b] in Theorem 2.10, we have
2s g
a + b
2
≤
Γ(α + 1) α
1 g(a) + g(b)
α
[J
g(b)
+
J
g(a)]
+
β(α,
s
+
1)
+
.
+
−
b
2(b − a)α a
α+s
2
Acknowledgements:
The authors would like to thank the reviewers and the editors for their valuable suggestions and comments.
The present investigation was supported, in part, by the Natural Science Foundation of Chongqing
Municipal Education Commission (No.KJ1501009), and, in part, by the Foundation of Scientific Research
Project of Fujian Province Education Department of China (No.JK2013051).
F. Chen, S. Wu, J. Nonlinear Sci. Appl. 9 (2016), 705–716
716
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