Received 29/04/15
A NEW GENERALIZATION OF OSTROWSKI TYPE
INEQUALITY FOR MAPPINGS OF BOUNDED VARIATION
HUSEYIN BUDAK AND MEHMET ZEKI SARIKAYA
Abstract. In this paper, a new generalization of Ostrowski type integral
inequality for mappings of bounded variation is obtained and the quadrature
formula is also provided.
1. Introduction
Let f : [a; b] ! R be a di¤erentiable mapping on (a; b) whoose derivative f 0 :
(a; b) ! R is bounded on (a; b) ; i.e. kf 0 k1 := sup jf 0 (t)j < 1: Then, we have
t2(a;b)
the inequality
(1.1)
f (x)
1
b
a
Zb
f (t)dt
a
"
x
1
+
4
(b
#
a+b 2
2
(b
2
a)
a) kf 0 k1 ;
for all x 2 [a; b][16]. The constant 41 is the best possible. This inequality is well
known in the literature as the Ostrowski inequality.
De…nition 1. Let P : a = x0 < x1 < ::: < xn = b be any partition of [a; b] and let
f (xi ) = f (xi+1 ) f (xi ): Then f (x) is said to be of bounded variation if the sum
m
X
i=1
j f (xi )j
is bounded for all such partitions.
Let f be of bounded variation on [a; b], and
P
(P ) denotes the sum
corresponding to the partition P of [a; b]. The number
b
_
a
(f ) := sup
nX
n
X
i=1
j f (xi )j
o
(P ) : P 2 P ([a; b]) ;
is called the total variation of f on [a; b] : Here P ([a; b]) denotes the family of partitions of [a; b] :
In [9], Dragomir proved following Ostrowski type inequalities for functions of
bounded variation:
2000 Mathematics Subject Classi…cation. 26D15, 26A45, 26D10, 41A55.
Key words and phrases. Bounded Variation, Ostrowski type inequalities, Riemann-Stieltjes
integral.
1
2
HUSEYIN BUDAK AND M EHM ET ZEKI SARIKAYA
Theorem 1. Let f : [a; b] ! R be a mapping of bounded variation on [a; b] : Then
(1.2)
Zb
f (t)dt
(b
1
(b
2
a) f (x)
a) + x
a+b
2
a
holds for all x 2 [a; b] : The constant
1
2
b
_
(f )
a
is the best possible.
We introduce the notation In : a = x0 < x1 < ::: < xn = b for a division of the
interval [a; b] with hi := xi+1 xi and v(h) = max fhi : i = 0; 1; :::; n 1g and let
intermediate points i 2 [xi ; xi+1 ] (i = 0; 1; :::; n 1). Then we have
Zb
(1.3)
f (t)dt = A(f; In ; ) + R(f; In ; )
a
where
(1.4)
A(f; In ; ) :=
n
X
f ( i )hi
i=0
and the remainder term satis…es
(1.5)
1
v(h) +
max
2
i2f0;1;:::;n
jR(f; In ; )j
v(h)
b
_
1g
i
xi + xi+1
2
b
_
a
:
a
In [7], Dragomir obtained following Ostrowski type inequality for functions of
bounded variation:
Theorem 2. Let Ik : a = x0 < x1 < ::: < xk = b be a division of the interval
[a; b] and i (i = 0; 1; :::; k + 1) be k + 2 points so that 0 = a; i 2 [xi 1 ; xi ]
(i = 1; :::; k) ; k+1 = b: If f : [a; b] ! R is of bounded variation on [a; b] ; then we
have the inequality:
(1.6)
Zb
f (x)dx
k
X
1
(h) + max
2
(h)
(
i+1
i ) f (xi )
i=0
a
b
_
i+1
xi + xi+1
; i = 0; 1; :::; k
2
1
b
_
(f )
a
(f )
a
where (h) := max f hi j i = 0; :::; n 1g ; hi := xi+1
b
_
(f ) is the total variation of f on the interval [a; b] :
xi (i = 0; 1; :::; k
1) and
a
For recent results concerning the above Ostrowski’s inequality and other related
results see [1]-[21].
A NEW GENERALIZATION OF OSTROW SKI TYPE INEQUALITIES
3
The aim of this paper is to obtain a new generalization of Ostrowski type integral
inequalities for functions of bounded variation. And we give some applications for
our results.
2. Main Results
Theorem 3. Let f : [a; b] ! R be a mapping of bounded variation on [a; b] : Then,
for all x 2 [a; b] ; we have
(2.1)
(b
a) 1
(x
f (x) +
2
a) f (a) + (b
2
x) f (b)
Zb
f (t)dt
a
b
1
where
2
a
2
2 [0; 1] and
d
_
b
_
a+b
2
+ x
(f )
a
(f ) denotes the total variation of f on [c; d] :
c
Proof. De…ne the mapping K (x; t) by,
8
a+
< t
K (x; t) =
:
t
b
x a
2
;
b x
2
a
t
x
x<t
b:
Integrating by parts, we get
Zb
(2.2)
K (x; t)df (t)
a
=
Zx
t
x
a+
a
df (t) +
2
t
+
=
x
a
t
(x
+
t
b
b
x
df (t)
2
x
a
=
Zb
a) 1
b
x
2
x
a
f (t)
2
b+
a
b
x
2
2
Zx
a
Zb
b
f (t)
x
f (x) +
f (b) + (b
f (t)dt
f (t)dt
x
x
a
2
x) 1
2
f (a)
f (x)
Zb
f (t)dt
a
=
(b
a) 1
2
f (x) +
(x
a) f (a) + (b
2
x) f (b)
Zb
a
f (t)dt:
4
HUSEYIN BUDAK AND M EHM ET ZEKI SARIKAYA
It is well known that if g; f : [a; b] ! R are such that g is continuous on [a; b] and
Rb
f is of bounded variation on [a; b] ; then g(t)df (t) exist and
a
Rb
(2.3)
sup jg(t)j
g(t)df (t)
t2[a;b]
a
On the other hand, using (2.3), we get
Zb
b
_
(f ):
a
K (x; t)df (t)
a
Zx
t
x
a+
a
Zb
df (t) +
2
a
1
2
=
1
a) 1
2
b
2
(f ) + sup t
x
df (t)
2
(f ) + (b
x) 1
2
a
a
2
a; b
xg
b
_
b
2
b
_
b
x _
(f )
x
(f )
x
(f )
a
a+b
2
+ x
This completes the proof.
Remark 1. If we choose
nequality (1.2).
b+
t2[x;b]
a
x
_
max fx
2
x
a _
x
a
t2[a;x]
(x
b
b
x
sup t
=
t
b
_
(f ):
a
= 0 in Theorem 3, then the inequality (2.1) reduces the
Corollary 1. Under the assumption of Theorem 3 with
following inequality
= 1, then we have the
(2.4)
1
(b
2
a) f (x) +
(x
a) f (a) + (b
2
x) f (b)
Zb
f (t)dt
a
1 b a
+ x
2
2
Remark 2. If we take x =
b
a
2
f
a+b
2
a+b
2
a+b
2
b
_
(f ):
a
in Corollary 1, then we have the inequality
f (a) + f (b)
+
2
Zb
f (t)dt
a
which was given by Alomari in [3]. The constant
1
4
1
(b
4
a)
b
_
(f )
a
is the best possible.
A NEW GENERALIZATION OF OSTROW SKI TYPE INEQUALITIES
Corollary 2. Under the assumption of Theorem 3 with
inequality
=
2
3,
5
then we get the
(2.5)
2
(b
3
a) f (x) +
(x
a) f (a) + (b
3
x) f (b)
Zb
f (t)dt
a
2 b a
+ x
3
2
Remark 3. If we take x =
equality
b
b
_
a+b
2
a+b
2
(f ):
a
in Corollary 2, then we have the Simpson’s in-
a f (a) + f (b)
+ 2f
3
2
Zb
a+b
2
f (t)dt
1
(b
3
a
a)
b
_
(f )
a
which was given by Dragomir in [7].
Corollary 3. Under the assumption of Theorem 3. Suppose that f 2 C 1 [a; b] ;
then we have
Zb
(x a) f (a) + (b x) f (b)
(b a) 1
f (x) +
f (t)dt
2
2
a
1
b
2
a
2
a+b
2
+ x
kf 0 k1
for all x 2 [a; b] : Here as subsequently k:k1 is the L1 norm
0
kf k1 :=
Zb
f 0 (t)dt:
a
Corollary 4. Under the assumption of Theorem 3. Let f : [a; b] ! R be a Lipschitzian with the constant L > 0: Then
Zb
(x a) f (a) + (b x) f (b)
(b a) 1
f (x) +
f (t)dt
2
2
a
1
b
2
a
2
+ x
a+b
2
(b
a) L
for all x 2 [a; b] :
Corollary 5. Under the assumption of Theorem 3. Let f : [a; b] ! R be a
monotone mapping on [a; b] : Then
(b
a) 1
2
f (x) +
(x
a) f (a) + (b
2
x) f (b)
Zb
a
1
b
2
a
2
+ x
a+b
2
jf (b)
f (a)j
f (t)dt
6
HUSEYIN BUDAK AND M EHM ET ZEKI SARIKAYA
for all x 2 [a; b] :
3. Application to Quadrature Formula
We now introduce the intermediate points i 2 [xi ; xi+1 ] (i = 0; 1; :::; n 1) in
the division In : a = x0 < x1 < ::: < xn = b. Let hi := xi+1 xi and v(h) =
max fhi : i = 0; 1; :::; n 1g and de…ne the sum
(3.1)
A(f; In ; )
:
=
n
X
1
i=0
(
f ( i )hi +
2
xi ) f (xi ) + (xi+1
2
i
i ) f (xi+1 )
:
Then the following Theorem holds:
Theorem 4. Let f be as Theorem 3. Then
Zb
(3.2)
f (t)dt = A(f; In ; ) + R(f; In ; )
a
where A(f; In ; ) is de…ned as above and the remainder term R(f; In ; ) satis…es
(3.3)jR(f; In ; )j
1
1
v(h) +
max
2
i2f0;1;:::;n
2
1
v(h)
2
b
_
i
1g
b
_
xi + xi+1
2
(f )
a
(f ):
a
Proof. Application of Theorem 3 to the interval [xi ; xi+1 ] (i = 0; 1; :::; n
1) gives
(3.4)
1
2
f ( i )hi +
(
i
xi ) f (xi ) + (xi+1
2
x
Zi+1
i ) f (xi+1 )
f (t)dt
xi
1
2
hi
+
2
xi + xi+1
2
i
xi+1
_
(f )
xi
for all i 2 f0; 1; :::; n 1g :
Summing the inequality (3.4) over i from 0 to n
triangle inequality, we have
jR(f; In ; )j
1
1
1
2
2
2
n
X
hi
+
2
i=0
max
i
i2f0;1;:::;n 1g
1 and using the generalized
xi + xi+1
2
hi
+
2
1
v(h) +
max
2
i2f0;1;:::;n
i
1g
xi+1
_
(f )
xi
xi + xi+1
2
i
which completes the proof of the …rst inequality in (3.3)
n x_
i+1
X
(f )
i=0 xi
xi + xi+1
2
b
_
a
(f )
A NEW GENERALIZATION OF OSTROW SKI TYPE INEQUALITIES
7
For the second inequality in (3.3), we show that
xi + xi+1
2
i
hi
i 2 f0; 1; :::; n
2
and
max
xi + xi+1
2
i
i2f0;1;:::;n 1g
1g
1
v(h)
2
which completes the proof.
Remark 4. If we choose
= 0; we get (1.3) with (1.4) and (1.5).
Remark 5. If we choose
=
Zb
2
3
and
i
=
xi +xi+1
,
2
then we have
f (t)dt = AS (f; In ) + RS (f; In )
a
where
n
AS (f; In ) =
n
2X
1X
[f (xi ) + f (xi+1 )] hi +
f
6 i=0
3 i=0
xi + xi+1
2
hi
and the remainder term RS (f; In ) satis…es
b
_
1
v(h) (f )
3
a
jRS (f; In )j
which were given by Dragomir in [7].
Corollary 6. Choosing
= 1 gives
Zb
f (t)dt = A(f; In ; ) + R(f; In ; )
a
where
A(f; In ; ) =
n
X
1
(
f ( i )hi +
2
i=0
i
xi ) f (xi ) + (xi+1
2
i ) f (xi+1 )
and the remainder term R(f; In ; ) satis…es
1 1
v(h) +
max
2 2
i2f0;1;:::;n
jR(f; In ; )j
1g
i
xi + xi+1
2
b
_
b
_
1
v(h) (f ):
2
a
Particularly, if we take
i
=
n
A(f; In ) =
and
xi +xi+1
,then
2
1X
f
2 i=0
we have
xi + xi+1
2
+
f (xi ) + f (xi+1 )
hi
2
b
jR(f; In ; )j
_
1
v(h) (f ):
4
a
a
(f )
8
HUSEYIN BUDAK AND M EHM ET ZEKI SARIKAYA
Corollary 7. Let f : [a; b] ! R be a Lipschitzian with the constant L > 0: Then
we have (3.1) and (3.2) and the remainder term satis…es
jR(f; In ; )j
L 1
L 1
2
2
1
v(h) +
max
2
i2f0;1;:::;n
v(h) (b
1g
i
xi + xi+1
2
(b
a)
a) :
Corollary 8. Let f : [a; b] ! R be a monotone mapping on [a; b] : Then we get
(3.1) and (3.2) and the remainder term satis…es
jR(f; In ; )j
1
1
2
2
1
v(h) +
max
2
i2f0;1;:::;n
v(h) jf (b)
1g
i
xi + xi+1
2
jf (b)
f (a)j
f (a)j :
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Department of Mathematics, Faculty of Science and Arts, Düzce University, DüzceTURKEY
E-mail address : [email protected]
E-mail address : [email protected]