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J. Nonlinear Sci. Appl. 9 (2016), 139–148
Research Article
Some inequalities of Hermite–Hadamard type for
functions whose second derivatives are
(α, m)-convex
Ye Shuanga , Feng Qib,∗, Yan Wanga
a
College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City, Inner Mongolia Autonomous Region, China.
b
Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300160, China.
Communicated by C. Park
Abstract
In the paper, the authors establish a new integral identity and by this identity with the Hölder integral
inequality, discover some new Hermite–Hadamard type integral inequalities for functions whose second
c
derivatives are (α, m)-convex. 2016
All rights reserved.
Keywords: Hermite–Hadamard type inequality, (α, m)-convex function, second derivative, Hölder integral
inequality.
2010 MSC: 26D15, 26A51, 41A55.
1. Introduction
It is common knowledge in mathematical analysis that a function f : I ⊆ R → R is said to be convex on
an interval I if
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y)
for all x, y ∈ I and λ ∈ [0, 1].
Suppose that f : I ⊆ R → R is a convex function on an interval I such that a, b ∈ I and a < b. Then
the well-known Hermite–Hadamard integral inequality reads that
Z b
f (a) + f (b)
a+b
1
f
≤
f (x) d x ≤
.
2
b−a a
2
∗
Corresponding author
Email addresses: shuangye152300@sina.com (Ye Shuang), qifeng618@gmail.com, qifeng618@hotmail.com (Feng Qi),
sella110@vip.qq.com (Yan Wang)
Received 2015-06-25
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
140
Definition 1.1 ([9]). For f : [0, b] → R and m ∈ (0, 1], if
f (λx + m(1 − λ)y) ≤ λf (x) + m(1 − λ)f (y)
is valid for all x, y ∈ [0, b] and λ ∈ [0, 1], then we say that f (x) is an m-convex function on [0, b].
Definition 1.2 ([5]). For f : [0, b] → R and (α, m) ∈ (0, 1]2 , if
f (λx + m(1 − λ)y) ≤ λα f (x) + m 1 − λα f (y)
is valid for all x, y ∈ [0, b] and λ ∈ [0, 1], then we say that f (x) is an (α, m)-convex function on [0, b].
For the above convex functions, we recite the following theorems.
Theorem 1.3 ([2, Theorem 2.2]). Let f : I ◦ ⊆ R → R be a differentiable mapping and a, b ∈ I ◦ with a < b.
If |f 0 (x)| is convex on [a, b], then
Z b
f (a) + f (b)
(b − a) 0
1
f (x) d x ≤
−
|f (a)| + |f 0 (b)| .
2
b−a a
8
Theorem 1.4 ([1, Theorem 2.2]). Let I ⊇ R0 = [0, ∞) be an open interval and let f : I → R be a
differentiable function such that f 0 ∈ L[a, b] for 0 ≤ a < b < ∞. If |f 0 (x)|q is m-convex on [a, b] for some
m ∈ (0, 1] and q ≥ 1, then
0
Z b
a+b
b−a
1
|f (a)|q + m|f 0 (b/m)|q 1/q m|f 0 (a/m)|q + |f 0 (b)|q 1/q
f
−
f (x) d x≤
min
,
.
2
b−a a
4
2
2
Theorem 1.5 ([1, Theorem 3.1]). Let I ⊇ R0 be an open real interval and let f : I → R be a differentiable
function on I such that f 0 ∈ L([a, b]) for 0 ≤ a < b < ∞. If |f 0 |q is (α, m)-convex on [a, b] for some given
numbers m, α ∈ (0, 1] and q ≥ 1, then
q 1/q
Z b
f (a) + f (b)
b − a 1 1−1/q
1
b 0
q
−
f (x) d x ≤
× min v1 |f (a)| + v2 mf 0
,
2
b−a a
2
2
m q
1/q a 0
q
v2 mf 0
+
v
|f
(b)|
,
1
m where
1
1
α+ α
v1 =
(α + 1)(α + 2)
2
and
2
1
α +α+2
1
v2 =
− α .
(α + 1)(α + 2)
2
2
In recent years, some other kinds of Hermite–Hadamard type inequalities were generated in, for example,
[4, 7, 8, 10, 11, 12]. For more systematic information, please refer to monographs [3, 6] and related references
therein.
In this paper, we will establish a new integral identity and discover some new Hermite–Hadamard type
integral inequalities for functions whose second derivatives are (α, m)-convex.
2. A lemma
For establishing some new integral inequalities of Hermite–Hadamard type for functions whose second
derivatives are (α, m)-convex, we need the following lemma.
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
141
Lemma 2.1. Let f : I ⊆ R → R be differentiable on I ◦ and a, b ∈ I with a < b. If f 00 ∈ L([a, b]), then
Z b
f (a) + f (b)
1
f (x) d x
−
2
b−a a
Z 1
2a + b
(b − a)2
2 00
=
(2 − t − t )f ta + (1 − t)
dt
54
3
0
Z 1
2a + b
a + 2b
2 00
(2 + t − t )f t
+
dt
+ (1 − t)
3
3
0
Z 1
a + 2b
2 00
(3t − t )f t
+
+ (1 − t)b d t .
3
0
Proof. Integrating by part and changing the variable of definite integral yield
Z 1
2a + b
2 00
(2 − t − t )f ta + (1 − t)
dt
3
0
Z 1
2a + b
6
2a + b
3
=−
+
(1 + 2t)f 0 ta + (1 − t)
dt
f0
a−b
3
a−b 0
3
6
27
9
2a + b
0 2a + b
=−
f
+
f (a) −
f
a−b
3
(a − b)2
(a − b)2
3
Z 1
Z 1 18
2a + b
2a + b
a + 2b
2 00
−
f ta + (1 − t)
dt
(2 + t − t )f t
+ (1 − t)
dt
(a − b)2 0
3
3
3
0
6
2a + b
6
a + 2b
=
f0
−
f0
a−b
3
a−b
3
Z 1
3
2a + b
a + 2b
0
−
(1 − 2t)f t
+ (1 − t)
dt
a−b 0
3
3
9
2a + b
a + 2b
6
0 2a + b
0 a + 2b
f
−f
+
f
+f
=
a−b
3
3
(a − b)2
3
3
Z 1 18
2a + b
a + 2b
−
f t
+ (1 − t)
dt
(a − b)2 0
3
3
and
Z
0
1
a + 2b
(3t − t )f t
+ (1 − t)b d t
3
Z 1
3
a + 2b
6
0 a + 2b
0
(3 − 2t)f t
=
f
−
+ (1 − t)b d t
a−b
3
a−b 0
3
6
a + 2b
9
a + 2b
27
=
f0
−
f
+
f (b)
2
a−b
3
(a − b)
3
(a − b)2
Z 1 a + 2b
18
f t
−
+ (1 − t)b d t.
(a − b)2 0
3
2
00
Lemma 2.1 is thus proved.
3. Hermite–Hadamard type inequalities for (α, m)-convex functions
Now we start out to establish some new Hermite–Hadamard type integral inequalities for functions whose
second derivatives are (α, m)-convex.
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
142
Theorem 3.1. Let f : R0 → R
be twice differentiable and f 00 ∈ L([a, b]) for 0 ≤ a < b. If |f 00 |q is an
b
(α, m)-convex function on 0, m
for (α, m) ∈ (0, 1]2 and q ≥ 1, then
Z b
f (a) + f (b)
1
f (x) d x
−
2
b−a a
1/q
1
(b − a)2
≤
324
(α + 1)(α + 2)(α + 3)
(
00 q
00 2a + b q 1/q
2
1−1/q
× 7
6(3α + 7) f (a) + mα 7α + 42α + 59 f
3m 00 2a + b q
00 a + 2b q 1/q
2
2
1−1/q
6 2α + 11α + 13 f
+ 13
+ mα 13α + 66α + 77 f
3
3m )
00 a + 2b q
00 b q 1/q
1−1/q
2
2
+7
6 2α + 9α + 7 f
.
+mα 7α + 30α + 23 f
3
m b
Proof. Because |f 00 |q is (α, m)-convex on 0, m
, by Lemma 2.1 and the Hölder integral inequality, we have
Z b
f (a) + f (b)
1
f (x) d x
−
2
b−a a
Z 1
2a + b (b − a)2
2 00
(2 − t − t ) f ta + (1 − t)
≤
dt
54
3
0
Z 1
a + 2b 2a + b
2 00
+ (1 − t)
+
(2 + t − t ) f t
dt
3
3
0
Z 1
a + 2b
2 00
+ (1 − t)b d t
+
(3t − t ) f t
3
0
Z 1
1−1/q
2
(b − a)
2
≤
(2 − t − t ) d t
54
0
Z 1
1/q
2a + b q
2 00
(2 − t − t ) f ta + (1 − t)
×
dt
3
0
1/q
Z 1
1−1/q Z 1
a + 2b q
2a + b
2
2 00
+ (1 − t)
+
(2 + t − t ) d t
(2 + t − t ) f t
dt
3
3
0
0
q 1/q )
Z 1
1−1/q Z 1
a
+
2b
+ (1 − t)b d t
+
(3t − t2 ) d t
(3t − t2 ) f 00 t
3
0
0
q 1/q
Z
1
q
(b − a)2
7 1−1/q
2
α 00
α 00 2a + b ≤
(2 − t − t ) t f (a) + m(1 − t ) f
dt
54
6
3m 0
1−1/q Z 1
q 1/q
13
2a + b q
α 00 a + 2b f
+
(2 + t − t2 ) tα f 00
+
m(1
−
t
)
dt
6
3
3m 0
1−1/q Z 1
q
q 1/q )
7
a
+
2b
b + m(1 − tα )f 00
+
(3t − t2 ) tα f 00
dt
6
3
m 0
1/q
(b − a)2
1
=
324
(α + 1)(α + 2)(α + 3)
(
00 q
00 2a + b q 1/q
1−1/q
2
× 7
6(3α + 7) f (a) + mα 7α + 42α + 59 f
3m Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
143
00 2a + b q
00 a + 2b q 1/q
2
+ mα 13α + 66α + 77 f
6 2α + 11α + 13 f
+ 13
3
3m q 1/q )
q
a
+
2b
+mα 7α2 + 30α + 23 f 00 b + 71−1/q 6 2α2 + 9α + 7 f 00
.
3
m 1−1/q
2
The proof of Theorem 3.1 is completed.
Corollary 3.2. Under the conditions of Theorem 3.1, if α = m = 1, we have
Z b
f (a) + f (b)
1
f (x) d x
−
2
b−a a
"
"
q #1/q
q 00
+ f
13 f 00 2a+b
7(b − a)2  5|f 00 (a)|q + 9 f 00 2a+b
3
3
+
≤
324 
14
7
2

" #
q
+ 5 |f 00 (b)|q 1/q 
9 f 00 a+2b
3
+
.

14
Corollary 3.3. Under the conditions of Theorem 3.1, if α = m = q = 1, we have
"
Z b
+ 22 f 00
f (a) + f (b)
(b − a)2 5 |f 00 (a)| + 22 f 00 2a+b
1
3
f (x) d x ≤
−
2
b−a a
12
54
a+2b
3
a+2b
3
q #1/q
#
+ 5 |f 00 (b)|
.
Theorem 3.4. Let
differentiable and f 00 ∈ L([a, b]) for 0 ≤ a < b. If |f 00 |q is
fb : R0 → R be twice
2
(α, m)-convex on 0, m for (α, m) ∈ (0, 1] and q > 1, then
Z b
f (a) + f (b)
1
−
f (x) d x
2
b−a a
1/p 1/q
1
1
(b − a)2
≤
54
(p + 1)(p + 2)
2(α + 1)(α + 2)
00 2a + b q 1/q
1/p 00 q
p+2
p+1
× 3
− 2 (p + 4)
2 f (a) + mα(α + 3) f
3m 00 2a + b q
00 a + 2b q 1/q
1/p
p+1
+ 2 (p + 4) − 2p − 5
2(2α + 3) f
+ mα(3α + 5) f
3
3m q 1/q )
00 a + 2b q
1/p
p+2
p+1
+ mα(α + 1) f 00 b + 3
− 2 (p + 4)
2(α + 1) f
,
3
m (3.1)
where
1
p
+
1
q
= 1.
Proof. By Lemma 2.1 and the Hölder integral inequality, we have
Z b
f (a) + f (b)
1
−
f (x) d x
2
b−a a
Z 1
00
(b − a)2
2a + b ≤
(1 − t) (2 + t) f ta + (1 − t)
dt
54
3
0
Z 1
00 2a + b
a + 2b +
(1 + t)(2 − t) f t
+ (1 − t)
dt
3
3
0
Z 1
00 a + 2b
+
t(3 − t) f t
+ (1 − t)b d t
3
0
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
Z 1
1/p Z 1
1/q
00
(b − a)2
2a + b q
p
≤
(1 − t)(2 + t) d t
(1 − t) f ta + (1 − t)
dt
54
3
0
0
1/q
Z 1
1/p Z 1
00 2a + b
a + 2b q
p
+ (1 − t)
+
(1 + t)(2 − t) d t
(1 + t) f t
dt
3
3
0
0
Z 1
1/p Z 1 q 1/q )
00 a + 2b
p
+
t(3 − t) d t
t f t
+ (1 − t)b d t
,
3
0
0
144
(3.2)
where
Z 1
2p+1 (p + 4) − 2p − 5
3p+2 − 2p+1 (p + 4)
(1 + t)(2 − t)p d t =
,
,
(p + 1)(p + 2)
(p + 1)(p + 2)
0
0
(3.3)
Z 1
3p+2 − 2p+1 (p + 4)
p
t(3 − t) d t =
.
(p + 1)(p + 2)
0
b
Using the (α, m)-convexity of |f 00 |q on 0, m
, we obtain
q
q Z 1
Z 1
00
q
00
2a
+
b
α
α
dt ≤
f (a) + m(1 − t ) f 00 2a + b d t
(1 − t) f ta + (1 − t)
(1
−
t)
t
3
3m 0
0
00 q
00 2a + b q
1
,
2 f (a) + mα(α + 3) f
=
2(α + 1)(α + 2)
3m Z
Z
1
0
1
(1 − t)(2 + t)p d t =
q
Z
00 a + 2b
t f t
+ (1 − t)b d t ≤
3
q q
b α 00 a + 2b α 00
t t f
+ m(1 − t ) f
dt
3
m 0
q q
00 a + 2b 1
b 2(α + 1) f
+ mα(α + 1) f 00
,
=
2(α + 1)(α + 2)
3
m 1
and
Z
0
1
00 2a + b
a + 2b q
+ (1 − t)
(1 + t) f t
dt
3
3
q
Z 1
00 a + 2b q
α 00 2a + b α
dt
(1 + t) t f
≤
+ m(1 − t ) × f
3
3m 0
q 00 2a + b q
1
+ mα(3α + 5) f 00 a + 2b ;
2(2α + 3) f
=
2(α + 1)(α + 2)
3
3m Substituting the equalities in (3.3) into the above inequalities and the inequality (3.2) yields the inequality (3.1). The proof of Theorem 3.4 is complete.
Corollary 3.5. Under the conditions of Theorem 3.4, if α = m = 1, we have
1/p 1/q
Z b
f (a) + f (b)
(b − a)2
1
1
1
−
f (x) d x ≤
2
b−a a
54
(p + 1)(p + 2)
6
(
00 2a + b q 1/q
1/p 00 q
p+2
p+1
f (a) + 2 f
× 3
− 2 (p + 4)
3
+ 2
p+1
(p + 4) − 2p − 5
+ 3p+2 − 2p+1 (p + 4)
q 1/q
00 2a + b q
+ 4 f 00 a + 2b 5 f
3
3
)
00 a + 2b q 00 q 1/q
2 f
.
+ f (b)
3
1/p
1/p
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
145
Theorem 3.6. Let f : R0 → R
be twice differentiable and f 00 ∈ L([a, b]) for 0 ≤ a < b. If |f 00 |q is an
b
(α, m)-convex function on 0, m
for (α, m) ∈ (0, 1]2 , q > 1, and 0 < r ≤ q, then
Z b
f (a) + f (b)
1
−
f
(x)
d
x
2
b−a a
(
1−1/q q
(q − 1)(7q − 2r − 5)
(b − a)2
3α + 2r + 5
≤
×
B(α + 1, r + 1) f 00 (a)
54
(2q − r − 1)(3q − r − 2)
α+r+2
2r + 5
3α + 2r + 5
+m
−
B(α + 1, r + 1)
(r + 1)(r + 2)
α+r+2
1−1/q p+1
00 2a + b q 1/q
00 2a + b q
α+3
2 (p + 4) − 2p − 5
× f
f
+
3m (p + 1)(p + 2)
(α + 1)(α + 2) 3
q 1/q 1−1/q
00 a + 2b α(3α + 7)
(q − 1)(7q − 2r − 5)
f
+m
+
2(α + 1)(α + 2)
3m
(2q − r − 1)(3q − r − 2)
q
2α + 2r + 5
f 00 a + 2b ×
(α + r + 1)(α + r + 2) 3
q 1/q )
00 b 2α + 2r + 5
2r + 5
f
−
,
+ m
(r + 1)(r + 2) (α + r + 1)(α + r + 2) m where
1
p
+
1
q
(3.4)
= 1 and B(α, β) denotes the well-known Beta function which may be defined by
Z
B(α, β) =
1
tα−1 (1 − t)β−1 d t,
α, β > 0.
0
Proof. From Lemma 2.1 and the Hölder integral inequality, we have
Z b
1
f (a) + f (b)
−
f (x) d x
|
2
b−a a
Z 1
2
00
(b − a)
2a + b ≤
(1 − t) (2 + t) f ta + (1 − t)
dt
54
3
0
Z 1
00 2a + b
a + 2b + (1 − t)
+
(1 + t)(2 − t) f t
dt
3
3
0
Z 1
00 a + 2b
+
t(3 − t) f t
+ (1 − t)b d t
(3.5)
3
0
Z 1
1/p Z 1
1/q
q−r
(b − a)2
2a + b q
≤
(1 − t) q−1 (2 + t) d t
d
t
(1 − t)r (2 + t) f 00 ta + (1 − t)
54
3
0
0
Z 1
1/p Z 1
1/q
00 2a + b
a + 2b q
p
(1 + t) (2 − t) d t
(2 − t) f t
+
+ (1 − t)
dt
3
3
0
0
Z 1
1/p Z 1
q 1/q )
00 a + 2b
q−r
r
+
t q−1 (3 − t) d t
.
t (3 − t) f t
+ (1 − t)b d t
3
0
0
A straightforward computation gives
Z 1
Z 1
q−r
q−r
(q − 1)(7q − 2r − 5)
q−1
(1 − t) (2 + t) d t =
t q−1 (3 − t) d t =
,
(2q − r − 1)(3q − r − 2)
0
0
Z 1
2p+1 (p + 4) − 2p − 5
(1 + t)p (2 − t) d t =
.
(p + 1)(p + 2)
0
(3.6)
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
146
b
By the (α, m)-convexity of |f 00 |q on 0, m
, we have
1
Z
00
2a + b q
(1 − t) (2 + t) f ta + (1 − t)
dt
3
Z 1
q
2a + b q
(1 − t)r (2 + t) tα f 00 (a) +m(1 − tα ) f 00
≤
dt
3m 0
q
3α + 2r + 5
B(α + 1, r + 1) f 00 (a)
=
α+r+2
00 2a + b q
2r + 5
3α + 2r + 5
;
+m
−
B(α + 1, r + 1) f
(r + 1)(r + 2)
α+r+2
3m r
0
1
Z
0
Z
0
00 2a + b
a + 2b q
(2 − t) f t
+(1 − t)
dt
3
3
q
q Z 1
α 00 2a + b α 00 a + 2b ≤
+
m(1
−
t
)
dt
f
(2 − t) t f
3
3m 0
q
00 2a + b 00 a + 2b q
α+3
α(3α + 7)
;
f
=
+ m 2(α + 1)(α + 2) f
(α + 1)(α + 2) 3
3m 1
q
00 a + 2b
t (3 − t) f t
+(1 − t)b) d t
3
q q
Z 1
b α 00
r
α 00 a + 2b + m(1 − t ) f
dt
≤
t (3 − t) t f
3
m 0
00 a + 2b q
2α + 2r + 5
f
=
(α + r + 1)(α + r + 2) 3
q
00 b 2r + 5
2α + 2r + 5
f
.
+m
−
(r + 1)(r + 2) (α + r + 1)(α + r + 2) m r
Substituting the equalities in (3.6) to the above inequalities and the inequality (3.5) yields the inequality (3.4). The proof of Theorem 3.6 is complete.
Corollary 3.7. Under the conditions of Theorem 3.6, if α = m = 1, then
(
1/p
Z b
f (a) + f (b)
(b − a)2
1
(q − 1)(7q − 2r − 5)
−
f (x) d x ≤
2
b−a a
54
(2q − r − 1)(3q − r − 2)
00 q
00 2a + b q 1/q
2(r + 4)
2r + 7
f
×
f (a) +
(r + 1)(r + 2)(r + 3)
(r + 2)(r + 3) 3
p+1
1/p 1/q
2 (p + 4) − 2p − 5
2 00 2a + b q 5 00 a + 2b q
+
+
f
f
(p + 1)(p + 2)
3
3
6
3
1/p 00 a + 2b q
(q − 1)(7q − 2r − 5)
2r + 7
+
f
(2q − r − 1)(3q − r − 2)
(r + 2)(r + 3) 3
1/q )
00 q
2(r + 4)
f (b)
+
.
(r + 1)(r + 2)(r + 3)
Theorem 3.8. Let f : R0 → R
be twice differentiable and f 00 ∈ L([a, b]) for 0 ≤ a < b. If |f 00 |q is an
b
(α, m)-convex function on 0, m
for (α, m) ∈ (0, 1]2 and q > 1, then
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
147
Z b
f (a) + f (b)
1
f (x) d x
−
2
b−a a
1/p 1/p
(b − a)2
1
≤
3p+1 − 2p+1 (p + 2)
54
(p + 1)(p + 2)
00 q
00 2a + b q 1/q
1
− B (α + 1, q + 1) f
× B (α + 1, q + 1) f (a) + m
q+1
3m 00 2a + b q
1/p
2α + 3
p+1
+ 2 (p + 4) − 2p − 5
f
(α + 1)(α + 2) 3
00 a + 2b q 1/q
1/p
3α2 + 5α
p+1
p+1
+m
f
+
3
−
2
(p
+
2)
2(α + 1)(α + 2) 3m q 1/q )
q
00 a + 2b 00 b α
1
f
+m
f
,
×
α+q+1
3
(α + q + 1)(q + 1) m where
1
p
+
1
q
= 1 and B(α, β) is the Beta function.
b
, by Lemma 2.1 and the Hölder integral inequality, we have
Proof. Since |f 00 |q is (α, m)-convex on 0, m
Z b
f (a) + f (b)
1
−
f (x) d x
2
b−a a
(Z
1/p Z 1
1/q
1
2a + b q
(b − a)2
p
q 00
(2 + t) d t
×
(1 − t) f ta + (1 − t)
≤
d
t
54
3
0
0
Z 1
1/p Z 1
1/q
00 2a + b
a + 2b q
p
+
(1 + t)(2 − t) d t
(1 + t) f t
+ (1 − t)
dt
3
3
0
0
Z 1
1/p Z 1 q 1/q )
a + 2b
p
q 00
+
(3 − t) d t
t f t
+ (1 − t)b d t
3
0
0
1/p
1/p
1
(b − a)2
3p+1 − 2p+1 (p + 2)
≤
54
(p + 1)(p + 2)
00 q
00 2a + b q 1/q
1
× B (α + 1, q + 1) f (a) + m
− B (α + 1, q + 1) f
q+1
3m 00 2a + b q
00 a + 2b q 1/q
1/p
2α + 3
3α2 + 5α
p+1
+ 2 (p + 4) − 2p − 5
f
+ m 2(α + 1)(α + 2) f
(α + 1)(α + 2) 3
3m q 1/q )
00 a + 2b q
00 b 1/p
α
1
p+1
p+1
f
+m
f
+ 3
−2
(p + 2)
.
α+q+1
3
(α + q + 1)(q + 1) m The proof of Theorem 3.8 is complete.
Corollary 3.9. Under the conditions of Theorem 3.8, if α = m = 1, then
Z b
f (a) + f (b)
1
−
f (x) d x
2
b−a a
1/p (
1/p
(b − a)2
1
≤
×
3p+1 − 2p+1 (p + 2)
54
(p + 1)(p + 2)
1
1 00 2a + b q 1/q
00
q
×
|f (a)| +
f
(q + 1)(q + 2)
q + 2
3
Y. Shuang, F. Qi, Y. Wang, J. Nonlinear Sci. Appl. 9 (2016), 139–148
148
1/p 5 00 2a + b q 2 00 a + 2b q 1/q
f
+ f
+ 2 (p + 4) − 2p − 5
6
3
3
3
q
1/q )
1
1
a
+
2b
1/p
f 00
+
+ 3p+1 − 2p+1 (p + 2)
|f 00 (b)|q
.
q + 2
3
(q + 1)(q + 2)
p+1
Acknowledgments
This work was partially supported by the National Natural Science Foundation of China under Grant
No. 11361038.
The authors thank anonymous referees for their valuable comments on and careful corrections to the
original version of this paper.
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