Research Article
ON THE GENERALIZED INTEGRAL INEQUALITIES FOR TWICE DIFFERENTIABLE MAPPINGS
Mehmet Zeki SARIKAYA1, Samet ERDEN2, Hüseyin BUDAK*3
1Department of Mathematics, Düzce University, DÜZCE; ORCID: 0000-0002-6165-9242 2Department of Mathematics, Bartın University, BARTIN; ORCID: 0000-0001-8430-7533 3Department of Mathematics, Düzce University, DÜZCE; ORCID: 0000-0001-8843-955X Received: 14.11.2018 Revised: 27.12.2018 Accepted: 29.12.2018
ABSTRACT
In this paper, an important integral equality is derived. Then, we establish several new inequalities for some twice differentiable mappings that are connected with the celebrated Hermite-Hadamard type and Ostrowski type integral inequalities. Some of the new inequalities are obtained by using Grüss inequality and Chebyshev inequality. The results presented here would provide extensions of those given in earlier works
Keywords: Hermite-Hadamard inequality, Ostrowski inequality, Grüss inequality, Čebyšev inequality, Hölder inequality.
1. INTRODUCTION
In 1938, Ostrowski established the integral inequality which is one of the fundamental inequalities of mathematics as follows (see, [19]):
Let
f
:
[
a
,
b
]
R
be a differentiable mapping on( b
a
,
)
whose derivativeR
)
,
(
:
a
b
f
is bounded on( b
a
,
),
i.e.,sup
(
)
.
) , (
f
t
f
b a t Then, the inequality holds:
b
a
f
a
b
x
dt
t
f
a
b
x
f
b a b a)
(
)
(
)
(
4
1
)
(
1
)
(
2 2 2 (1.1)for all x in
[ b
a
,
].
The constant 14 is the best possible.
Inequality (1.1) has wide applications in numerical analysis and in the theory of some special means; estimating error bounds for some special means, some mid-point, trapezoid and Simpson rules and quadrature rules, etc. Hence, inequality (1.1) has attracted considerable attention and interest from mathematicians and researchers. Due to this, over the years, the interested reader is also refered to ([1]-[4], [9], [12], [13], [15], [20]-[28]) for integral inequalities in several
* Corresponding Author: e-mail: huseyinbudak@duzce.edu.tr, tel: (380) 541 24 04 / 3939
independent variables. In addition, the current approach of obtaining the bounds, for a particular quadrature rule, have depended on the use of Peano kernel. The general approach in the past has involved the assumption of bounded derivatives of degree greater than one.
Definition 1. The function
f
:
[
a
,
b
]
R
R
, is said to be convex if the following inequality holds)
(
)
1
(
)
(
)
)
1
(
(
x
y
f
x
f
y
f
for all
x
,
y
[
a
,
b
]
and
0
,
1
.
We say thatf
is concave if( f
)
is convex. The following inequality is well known in the literature as the Hermite-Hadamard integral inequality (see, [8]):(1.2)
where
f :
I
R
R
is a convex function on the intervalI
of real numbers andI
b
a
,
witha
b
.A largely applied inequality for convex functions, due to its geometrical significance, is Hadamard's inequality, (see [7], [29]-[33]) which has generated a wide range of directions for extension and a rich mathematical literature.
Integration with weight functions is used in countless mathematical problems such as approximation theory, spectral analysis, statistical analysis and the theory of distributions. Grüss developed an integral inequality [14 in 1935. The integral inequality that establishes a connection between the integral of the product of two functions and the product of the integrals is known in the literature as the Grüss inequality. The Grüss inequality is as follows.
Theorem 1. Let
f
,
g
:
a
,
b
R
be integrable functions such that
f
x
and
g
x
for allx
[ b
a
,
],
where
,
,
,
are constants. Then
,
4
1
1
1
1
a
f
x
g
x
dx
b
a
f
x
dx
b
a
g
x
dx
b
b a b a b awhere the constant
4
1 is sharp (see, [14]).
In [18, p.40], Čebyšev's inequality is given by the following Theorem:
Theorem 2. Let
f
,
g
:
a
,
b
R
be integrable and monotone functions on
a,
b
and letp
be a positive and integrable function on the same interval. Then
x
g
x
dx
p
x
dx
p
x
g
x
dx
p
x
g
x
dx
f
x
p
b a b a b a b a
(
)
(
)
with equality if and only if one of the functions
f ,
g
reduces to a constantf
andg
are monotone in the opposite sense, the reverse inequality holds.In [5], Bullen proved the following inequality which is known as Bullen's inequality for convex function.
Let
f :
I
R
R
be a convex function on the intervalI
of real numbers andI
b
a
,
witha
b
.
The inequality.
)
(
)
(
1
)
(
1
bf
x
dx
f
a
b
f
a
f
b
2
)
(
)
(
)
(
1
2
b
f
a
f
dx
x
f
a
b
b
a
f
b a
In this study, using twice differentiable functions, we give new inequalities that are connected with the celebrated Hermite-Hadamard type and Ostrowski type integral inequalities. The results presented here would provide extensions of those given in earlier works.
2. MAIN RESULTS
In order to prove our main results we need the following lemma:
Lemma 1. Let
f
:
I
R
R
be twice differentiable function onI
(the interior of the intervalI
) such thatf
L
a
,
b
wherea
,
b
I
witha
b
. Then the following identity holds:
f
S
dt
t
f
a
b
b
f
a
f
h
x
f
h
x
f
b
a
x
h
a
b
a
x
x
b
dt
t
f
t
x
K
a
b
h b a h b a:
1
2
2
2
1
2
2
1
,
2
1
2 2
(2.1) for
b
t
x
h
b
t
t
b
x
t
a
h
a
t
t
a
t
x
K
a b a b h,
,
:
,
2 2 whereh
0
,
1
anda
h
b2a
x
b
h
b2a.
Proof . Firstly, we arrange the operations:
.
2
2
2
2
,
2 2dt
t
f
b
t
a
b
h
dt
t
f
b
t
dt
t
f
a
t
a
b
h
dt
t
f
a
t
dt
t
f
a
b
h
b
t
t
b
dt
t
f
a
b
h
a
t
t
a
dt
t
f
t
x
K
b x b x x a x a b x x a h b a
By integration by parts twice, we have
.
2
2
2
2
2
2
,
2 2x
f
x
b
x
f
b
f
a
b
h
dt
t
f
x
f
x
b
x
f
x
b
x
f
a
f
x
f
a
x
a
b
h
dt
t
f
x
f
a
x
x
f
a
x
dt
t
f
t
x
K
b x x a h b a
(2.2)If we arrange the equality (2.2), then we obtain desired equality (2.1). Hence, the proof is completed.
Remark 1. In Lemma 1, let
h
1
andx
a2b.
Then, we have the equality
b
a
f
t
dt
b
f
a
f
b
a
f
dt
t
f
t
b
a
K
a
b
b a b a
1
2
2
2
1
,
2
2
1
1which is proved by Kirmaci and Dikici in [16] and Minculete et. al in [17].
Theorem 3. Let
f
:
I
R
R
be twice differentiable function onI
, the interior of the intervalI
,
wherea
,
b
I
witha
b
. Iff
:
a
,
b
R
is bounded on
a,
b
, denote (
)
,
sup
,
f
t
f
b a tthen we have the inequality
2 22 2 3 3 2 3)
(
)
(
4
1
4
)
(
6
24
b
a
x
a
b
h
a
b
x
b
a
x
a
b
h
f
f
S
b a hfor all
h
0
,
1
and.
2 2 a b a b
x
b
h
h
a
.
2
2
2
2
,
2
,
2
1
,
2
1
L
a
b
f
dt
a
b
h
b
t
t
b
dt
a
b
h
a
t
t
a
a
b
f
dt
t
x
K
a
b
f
dt
t
f
t
x
K
a
b
dt
t
f
t
x
K
a
b
f
S
b x x a h b a h b a h a h
Now, let us observe that
2
3
3
2 3 3p
r
p
q
p
r
p
q
dt
q
t
p
t
dt
t
q
p
t
dt
q
t
p
t
r q q p r p
for all
r
,
p
,
q
such thatp
q
r
.
Then we get that
4
3
24
2
2 3 3 3a
x
a
b
h
a
x
a
b
h
dt
a
b
h
a
t
a
t
x a
and
.
4
3
24
2
2 3 3 3x
b
a
b
h
x
b
a
b
h
dt
a
b
h
b
t
b
t
b x
Then, we have
2
2
3 3 3 34
3
12
x
a
b
x
a
b
h
x
b
a
x
a
b
h
L
and the proof is thus completed. Remark 2. If we choose
h
1
with2
b a
x
in Theorem 3, then we have the following inequality
f
t
dt
b
a
f
a
b
b
f
a
f
b
a
f
b a48
1
2
2
2
1
2Remark 3. If we choose
h
0
and2
b a
x
in Theorem 3, then we have the following midpoint inequality
f
t
dt
b
a
f
a
b
b
a
f
b a24
1
2
2which is given by Cerone et al. in [6].
Theorem 4. Let
f
:
I
R
R
be twice differentiable function onI
, the interior of the intervalI
,
wherea
,
b
I
witha
b
. Iff
is belong toL
p
a
,b
,
p
1
and,
1
1 1
q
p then we have the inequality
q x b a b h M a x a b h M q q B a b h a b f f S q p h 1 , 2 , 2 1 , 1 2 2 ) ( 2 1 2 (2.3)for all
h
0
,
1
and,
2 2 a b a b
x
b
h
h
a
and whereB
p
,
q
is Euler's Beta function.Proof . We take absolute value of (2.1). Using Hölder's inequality, we find that
.
)
(
2
,
)
(
2
,
)
(
2
1
,
)
(
2
1
1 1 1 1 q q q pK
a
b
f
dt
t
x
K
a
b
f
dt
t
x
K
dt
t
f
a
b
dt
t
f
t
x
K
a
b
f
S
p q h b a p q h b a p b a h b a h
(2.4)We need to calculate the integral
K
to prove the theorem.
dt
a
b
h
b
t
t
b
dt
a
b
h
a
t
t
a
dt
t
x
K
K
q q b x q q x a q h b a2
2
,
(2.5)
4 3 2 1 2 2 2 22
2
2
2
I
I
I
I
dt
a
b
h
b
t
t
b
dt
t
a
b
h
b
t
b
dt
a
b
h
a
t
a
t
dt
t
a
b
h
a
a
t
q q b a b h b q q a b h b x q q x a b h a q q a b h a a
Now, we calculate four integrals above . For integral
I
1, using the change of variableu
h
a
t
b a 2
and fromdt
h
b2adu
,
we write
1
,
1
2
1
2
2
1 2 1 0 1 2 2 1
q
q
B
a
b
h
du
u
u
a
b
h
dt
t
a
b
h
a
a
t
I
q q q q q q a b h a a (2.6)and, for integral
I
4, using the change of variablet
b
h
b2a
h
b2au
and from,
2du
h
dt
ba we write
1
,
1
2
1
2
2
1 2 1 0 1 2 2 4
q
q
B
a
b
h
du
u
u
a
b
h
dt
a
b
h
b
t
t
b
I
q q q q q q b a b h b (2.7) where
,
1
1
1,
1 0du
u
u
q
p
B
p
q
p
,
q
0
.
Before calculating the other integrals, let us define that
c
,
d
u
u
c
du
.
M
q q d c
If we use the change of variable
t
a
u
forI
2 andb
t
u
forI
3,
then we get
a
x
a
b
h
M
du
a
b
h
u
u
dt
a
b
h
a
t
a
t
I
q q a x a b h q q x a b h a,
2
2
2
2 2 2 (2.8) and
.
,
2
2
2
2 2 3
x
b
a
b
h
M
du
a
b
h
u
u
dt
a
b
h
t
b
t
b
I
q q x b a b h q q a b h b x (2.9)Substituting (2.6- (2.9) in (2.5), we obtain the equality
.
,
2
,
2
1
,
1
2
2
,
1 2
x
b
a
b
h
M
a
x
a
b
h
M
q
q
B
a
b
h
dt
t
x
K
q q h b a (2.10)If we substitute the equality (2.10) in (2.4), then we easily deduce the required inequality (2.3) which completes the proof.
Corollary 1. Let
f
be as in Theorem 4. If we use the equality
.
1
!
!
1
!
,
1 1 0
u
u
r
du
q
q
k
r
k
d
q
k
c
d
c
M
k q k q k q k q k q q d c
.
1
2
!
!
1
!
1
,
1
2
2
)
(
2
1 1 2 1 1 2 0 1 2 qk
q
x
b
a
x
k
k
q
q
q
q
B
a
b
h
a
b
f
f
S
k q a b h k q k q k q a b h k q k q p h
Corollary 2. Under the same assumptions of Theorem 4 with
2
b a
x
andh
1
,
then the following inequality holds:
.
8
1
,
1
1
2
2
2
1
11 1 p b af
q
q
B
a
b
dt
t
f
a
b
b
f
a
f
b
a
f
q q
Remark 4. If we choose
x
a2b andh
0
in Theorem 4, then we have the following midpoint inequality
p b af
q
a
b
dt
t
f
a
b
b
a
f
q q
1 11
2
8
1
2
1which is proved by Dragomir et al. in [11].
Theorem 5. Let
f
:
I
R
R
be twice differentiable function onI
(the interior of the intervalI
) such thatf
L
a
,
b
wherea
,
b
I
witha
b
. If the mappings
b
x
t
t
f
h
b
t
t
b
x
a
t
t
f
h
a
t
t
a
t
a b a b,
,
,
,
2 2
is convex on
a
,b
,
then we have the inequality
b
x
x
b
h
b
a
f
x
a
b
x
F
f
S
a
b
x
F
h
8
2
1
2
(2.11) where
.
2
2
2
2
2
2
2
2
2 2
x
a
a
x
h
b
a
f
x
a
b
x
x
b
h
b
a
f
x
b
x
F
for all
h
0
,
1
anda
h
b2a
x
b
h
b2a.
Proof. If we use left hand side of Hermite-Hadamard's inequality for the mappings
, then we get
2
2
2
2
2
1
x
a
f
a
b
h
x
a
a
x
a
x
dt
t
a
x
x a
(2.12)and
2
2
2
2
2
1
x
b
f
a
b
h
b
x
x
b
b
x
dt
t
x
b
b x
(2.13)The inequality (2.12) and (2.13) are multiplied by
x
a
andb
x
,
respectively. Then, these two inequalities are added, we have
x
t
dt
b
a
S
f
F
h b a
2
2
(2.14) Applying the Bullen's inequality for the mappings
, we get
b
x
x
b
h
b
a
f
x
a
x
f
a
b
h
x
a
a
x
x
a
x
a
dt
t
a
x
x a
2
2
1
2
2
2
2
2
2
2
(2.15) and
.
2
2
1
2
2
2
2
2
2
2
x
f
a
b
h
b
x
x
b
b
x
f
a
b
h
b
x
b
x
x
b
b
x
dt
t
x
b
b x
(2.16)The inequality (2.15) and (2.16) are multiplied by
x
a
/
2
and
b
x
/
2
,
respectively. Then, these two inequalities are added, we have
t
dt
F
x
b
a
b
x
x
b
h
b
a
f
x
b a
4
1
2
(2.17) Because of (2.14) and (2.17), we easily the deduce required inequality (2.11) which completes the proof.Remark 5. In Theorem 5, let
h
1
and.
2
b a
4
3
4
3
128
1
2
2
2
1
4
3
4
3
64
2 2b
a
f
b
a
f
a
b
dt
t
f
a
b
b
f
a
f
b
a
f
b
a
f
b
a
f
a
b
b awhich is proved by Kirmaci and Dikici in [16] and Minculete et. al in [17].
Corollary 3. Under assumptions of Theorem 5 with
h
0
andx
a2b,
we have the inequality
.
4
3
4
3
64
2
1
2
4
4
3
4
3
128
2 2
b
a
f
b
a
f
a
b
b
a
f
dt
t
f
a
b
b
a
f
b
a
f
b
a
f
a
b
b aTheorem 6. Let
f
:
I
R
R
be twice differentiable function onI
such that
f
on[
a
,
x
)
and
f
on[ b
x
,
],
wherea
,
b
I
witha
b
. Iff
is integrable on
a,
b
, then we have the inequality
f
t
dt
x
b
dt
t
f
a
x
x
b
b
f
a
x
a
f
a
b
h
x
f
x
b
a
x
a
b
h
x
f
x
b
a
b x x a
2
2
2
2
4
2
2 (2.18)
2
16
2
16
4
1
3
4
1
3
4
1
2 2 2 2 3 2 2 3 2 2a
b
h
b
x
x
b
a
b
h
a
b
h
a
x
x
a
a
b
h
x
f
b
f
x
b
x
b
a
b
h
x
b
a
f
x
f
a
x
a
x
a
b
h
a
x
for all
h
0
,
1
and.
2 2 a b a b
x
b
h
h
a
Proof . Using Grüss inequality, we find that
m
n
n
m
dt
t
f
dt
a
b
h
b
t
t
b
x
b
dt
t
f
dt
a
b
h
a
t
t
a
a
x
dt
t
f
a
b
h
b
t
t
b
x
b
dt
t
f
a
b
h
a
t
t
a
a
x
b x b x x a x a b x x a4
1
2
1
2
1
2
1
2
1
2 2 (2.19)Here, we have that
x
a
f
x
f
a
f
x
a
x
a
b
h
dt
t
f
a
x
x
f
x
f
a
x
dt
t
f
a
b
h
a
t
t
a
a
x
x a x a
2
2
2
2
1
(2.20)
.
2
2
2
2
1
x
f
x
b
x
f
b
f
x
b
a
b
h
dt
t
f
x
b
x
f
x
f
x
b
dt
t
f
a
b
h
b
t
t
b
x
b
b x x
(2.21)Now, we calculate bounds
m
,
n
,
m
and
n
,
16
2
sup
2 2 ) , [a
b
h
a
b
h
a
t
t
a
m
x a t
(2.22)
,
2
2
inf
) , [
a
b
h
a
x
x
a
a
b
h
a
t
t
a
n
x a t (2.23)
,
16
2
sup
2 2 ] , [a
b
h
a
b
h
b
t
t
b
m
b x t
(2.24)
.
2
2
inf
] , [
b
a
h
b
x
x
b
a
b
h
b
t
t
b
n
b x t (2.25)On the other hand,
,
3
4
2
3 2a
x
a
x
a
b
h
dt
a
b
h
a
t
t
a
x a
(2.26)
,
3
4
2
3 2x
b
x
b
a
b
h
dt
a
b
h
b
t
t
b
b x
(2.27)
t
dt
f
x
f
a
,
f
t
dt
f
b
f
x
.
f
b x x a
(2.28) If we substitute (2.20)-(2.28) in (2.19), then we obtain the inequality (2.18) which completes the proof.Corollary 4. In Theorem 6, let
h
1
and.
2
b a
