International Journal of Analysis and Applications
ISSN 2291-8639
Volume 12, Number 2 (2016), 142-156
http://www.etamaths.com

ON OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS OF TWO

VARIABLES WITH BOUNDED VARIATION

HÜSEYIN BUDAK∗ AND MEHMET ZEKI SARIKAYA

Abstract. In this paper, we establish a new generalization of Ostrowski type inequalities for func-

tions of two independent variables with bounded variation and apply it for qubature formulae. Some

connections with the rectangle, the midpoint and Simpson’s rule are also given.

1. Introduction

Let f : [a,b] → R be a differentiable mapping on (a,b) whoose derivative f′ : (a,b) → R is bounded
on (a,b) , i.e. ‖f′‖∞ := sup

t∈(a,b)
|f′(t)| < ∞. Then we have the inequality

(1.1)

∣∣∣∣∣∣f(x) − 1b−a
b∫

a

f(t)dt

∣∣∣∣∣∣ ≤
[

1

4
+

(
x− a+b

2

)2
(b−a)2

]
(b−a)‖f′‖∞ ,

for all x ∈ [a,b][19]. The constant 1
4

is the best possible. This inequality is well known in the literature
as the Ostrowski inequality.

In [11], Dragomir proved following Ostrowski type inequalities related functions of bounded varia-
tion:

Theorem 1. Let f : [a,b] → R be a mapping of bounded variation on [a,b] . Then∣∣∣∣∣∣
b∫

a

f(t)dt− (b−a) f(x)

∣∣∣∣∣∣ ≤
[

1

2
(b−a) +

∣∣∣∣x− a + b2
∣∣∣∣
] b∨

a

(f)

holds for all x ∈ [a,b] . The constant 1
2

is the best possible.

2. Preliminaries and Lemmas

In 1910, Fréchet [16] has given the following characterization for the double Riemann-Stieltjes
integral. Assume that f(x,y) and α(x,y) are defined over the rectangle Q = [a,b]× [c,d]; let R be the
divided into rectangular subdivisions, or cells, by the net of straight lines x = xi, y = yi,

a = x0 < x1 < ... < xn = b, and c = y0 < y1 < ... < ym = d;

let ξi,ηj be any numbers satisfying ξi ∈ [xi−1,xi] , ηj ∈ [yj−1,yj] , (i = 1, 2, ...,n; j = 1, 2, ...,m); and
for all i,j let

∆11α(xi,yj) = α(xi−1,yj−1) −α(xi−1,yj) −α(xi,yj−1) + α(xi,yj).

Then if the sum

S =

n∑
i=1

m∑
j=1

f (ξi,ηj) ∆11α(xi,yj)

2010 Mathematics Subject Classification. 26D15, 26B30, 26D10, 41A55.

Key words and phrases. bounded variation; Ostrowski type inequalities; Riemann-Stieltjes integral.

c©2016 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

142



OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS... 143

tends to a finite limit as the norm of the subdivisions approaches zero, the integral of f with respect
to α is said to exist. We call this limit the restricted integral, and designate it by the symbol

(2.1)

b∫
a

d∫
c

f(x,y)dydxα(x,y).

If in the above formulation S is replaced by the sum

S∗ =

n∑
i=1

m∑
j=1

f (ξij,ηij) ∆11α(xi,yj),

where ξij,ηij are numbers satisfying ξij ∈ [xi−1,xi] , ηij ∈ [yj−1,yj] , we call the limit, when it exist,
the unrestricted integral, and designate it by the symbol

(2.2)

b∫
a

d∫
c

f(x,y)dydxα(x,y).

Clearly, the existence of (2.2) implies both the existence of (2.1) and its equality (2.2). On the other
hand, Clarkson ([8]) has shown that the existence of (2.1) does not imply the existence of (2.2).

In [7], Clarkson and Adams gave the following definitions of bounded variation for functions of two
variables:

2.1. Definitions. The function f(x,y) is assumed to be defined in rectangle R(a ≤ x ≤ b, c ≤ y ≤ d).
By the term net we shall, unless otherwise specified mean a set of parallels to the axes:

x = xi(i = 0, 1, 2, ...,m), a = x0 < x1 < ... < xm = b;

y = yj(j = 0, 1, 2, ...,n), c = y0 < y1 < ... < yn = d.

Each of the smaller rectangles into which R is devided by a net will be called a cell. We employ the
notation

∆11f(xi,yj) = f(xi+1,yj+1) −f(xi+1,yj) −f(xi,yj+1) + f(xi,yj),
∆f(xi,yj) = f(xi+1,yj+1) −f(xi,yj).

The total variation function, φ(x) [ψ(y)] , is defined as the total variation of f(x,y) [f(x,y)] considered
as a function of y [x] alone in interval (c,d) [(a,b)], or as +∞ if f(x,y) [f(x,y)] is of unbounded
variation.

Definition 1. (Vitali-Lebesque-Fréchet-de la Vallée Poussin). The function f(x,y) is said tobe of
bounded variation if the sum

m−1 , n−1∑
i=0 , j=0

|∆11f(xi,yj)|

is bounded for all nets.

Definition 2. (Fréchet). The function f(x,y) is said tobe of bounded variation if the sum

m−1 , n−1∑
i=0 , j=0

�i�j |∆11f(xi,yj)|

is bounded for all nets and all possible choices of �i = ±1 and �j = ±1.

Definition 3. (Hardy-Krause). The function f(x,y) is said tobe of bounded variation if it satisfies
the condition of Definition 1 and if in addition f(x,y) is of bounded variation in y (i.e. φ(x) is finite)
for at least one x and f(x,y) is of bounded variation in y (i.e. ψ(y) is finite) for at least one y.

Definition 4. (Arzelà). Let (xi,yi) (i = 0, 1, 2, ...,m) be any set of points satisfiying the conditions

a = x0 < x1 < ... < xm = b;

c = y0 < y1 < ... < ym = d.



144 BUDAK AND SARIKAYA

Then f(x,y) is said tobe of bounded variation if the sum
m∑
i=1

|∆f(xi,yi)|

is bounded for all such sets of points.

Therefore, one can define the consept of total variation of a function of two variables, as follows:

Let f be of bounded variation on Q = [a,b]×[c,d], and let
∑

(P) denote the sum
n∑

i=1

m∑
j=1

|∆11f(xi,yj)|

corresponding to the partition P of Q. The number∨
Q

(f) :=

d∨
c

b∨
a

(f) := sup
{∑

(P) : P ∈ P (Q)
}
,

is called the total variation of f on Q. Here P ([a,b]) denotes the family of partitions of [a,b] .
In [17], authors proved foolowing Lemmas related double Riemann-Stieltjes integral:

Lemma 1. (Integrating by parts) If f ∈ RS(α) on Q, then α ∈ RS(f) on Q, and we have
d∫

c

b∫
a

f(t,s)dtdsα(t,s) +

d∫
c

b∫
a

α(t,s)dtdsf(t,s)(2.3)

= f(b,d)α(b,d) −f(b,c)α(b,c) −f(a,d)α(a,d) + f(a,c)α(a,c).

Lemma 2. Assume that g ∈ RS(α) on Q and α is of bounded variation on Q, then

(2.4)

∣∣∣∣∣∣
d∫

c

b∫
a

g(x,y)dxdyα(x,y)

∣∣∣∣∣∣ ≤ sup(x,y)∈Q |g(x,y)|
∨
Q

(α) .

In [17], Jawarneh and Noorani obtained following Ostrowski type inequality for functions of two
variables with bounded variation:

Theorem 2. Let f : Q →→ R be mapping of bounded variation on Q. Then for all (x,y) ∈ Q, we
have inequality ∣∣∣∣∣∣(b−a) (d− c) f(x,y) −

d∫
c

b∫
a

f(t,s)dtds

∣∣∣∣∣∣(2.5)
≤

[
1

2
(b−a) +

∣∣∣∣x− a + b2
∣∣∣∣
][

1

2
(d− c) +

∣∣∣∣y − c + d2
∣∣∣∣
]∨

Q

(f)

where
∨
Q

(f) denotes the total (double) variation of f on Q.

For more information and recent developments on inequalities for mappings of bounded variation,
please refer to([1]-[6],[9]-[15],[17],[18],[20]-[24]).

The aim of this paper is to establish a new generzlization of Ostrowski type inequalities for func-
tions of two independent variables with bounded variation and apply it for qubature formulae. Some
connections with the rectangle, the midpoint and Simpson’s rule are also given.

3. Main Results

First, we give the following notations used in main our Theorem;
Let

∆n,m := {(x0,y0) , (x0,y1) , ..., (x0,ym) , (x1,y0) , ..., (x1,ym) , ..., (xn,y0) , (xn,y1) , ..., (xn,ym)}
is a partition of Q = [a,b]×[c,d] satisfaying a = x0, b = xn, y0 = c, ym = d with α0 = a, αi ∈ [xi−1,xi]
(i = 1, ...,n) , αn+1 = b and β0 = c, βj ∈ [yj−1,yj] (j = 1, ...,m) , βm+1 = d.

υ(h) := max{hi| i = 0, ...,n− 1} , hi := xi+1 −xi,



OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS... 145

υ(l) := max{lj| j = 0, ...,m− 1} , lj := yj+1 −yj.

Theorem 3. If f : Q → R is of bounded variatin on Q, then we have the inequality

∣∣∣∣∣∣
n∑

i=0

m∑
j=0

(αi+1 −αi) (βj+1 −βj) f(xi,yj) −
b∫

a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣(3.1)
≤

[
1

2
υ(h) + max

i∈{0,...,n−1}

∣∣∣∣αi+1 − xi + xi+12
∣∣∣∣
]

×
[

1

2
υ(l) + max

j∈{0,...,m−1}

∣∣∣∣βj+1 − yj + yj+12
∣∣∣∣
] b∨

a

d∨
c

(f)

≤ υ(h)υ(l)
b∨
a

d∨
c

(f)

where
b∨
a

d∨
c

(f) is the total variation of f on Q.

Proof. Let us consider the mappings K and L given by

K(t) =




t−α1, t ∈ [a,x1)
t−α2, t ∈ [x1,x2)
...
t−αn−1, t ∈ [xn−2,xn−1)
t−αn, t ∈ [xn−1,b]

, L(s) =




s−β1, s ∈ [c,y1)
s−β2, s ∈ [y1,y2)
...
s−βm−1, s ∈ [ym−2,ym−1)
s−βm, s ∈ [ym−1,d] .

Integrating by parts using Lemma 1, we obtain

b∫
a

d∫
c

K(t)L(s)dsdtf(t,s) =

n−1∑
i=0

m−1∑
j=0




xi+1∫
xi

yj+1∫
yj

K(t)L(s)dsdtf(t,s)


(3.2)

=

n−1∑
i=0

m−1∑
j=0




xi+1∫
xi

yj+1∫
yj

(t−αi+1) (s−βj+1) dsdtf(t,s)




=

n−1∑
i=0

m−1∑
j=0

[(xi+1 −αi+1) (yj+1 −βj+1) f(xi+1,yj+1)

−(xi+1 −αi+1) (yj −βj+1) f(xi+1,yj)

−(xi −αi+1) (yj+1 −βj+1) f(xi,yj+1)

+ (xi −αi+1) (yj −βj+1) f(xi,yj) −
xi+1∫
xi

yj+1∫
yj

f(t,s)dsdt






146 BUDAK AND SARIKAYA

=

n∑
i=1

m∑
j=1

(xi −αi) (yj −βj) f(xi,yj)

−
n∑

i=1

m−1∑
j=0

(xi −αi) (yj −βj+1) f(xi,yj)

−
n−1∑
i=0

m∑
j=1

(xi −αi+1) (yj −βj) f(xi,yj)

+

n−1∑
i=0

m−1∑
j=0

(xi −αi+1) (yj −βj+1) f(xi,yj) −
b∫

a

d∫
c

f(t,s)dsdt.

In last equality, we have

n∑
i=1

m∑
j=1

(xi −αi) (yj −βj) f(xi,yj)(3.3)

= (b−αn) (d−βm) f(b,d) + (b−αn)
m−1∑
j=1

(yj −βj) f(b,yj)

+ (d−βm)
n−1∑
i=1

(xi −αi) f(xi,d) +
n−1∑
i=1

m−1∑
j=1

(xi −αi) (yj −βj) f(xi,yj).

Similarly, we have

n∑
i=1

m−1∑
j=0

(xi −αi) (yj −βj+1) f(xi,yj)(3.4)

= (b−αn) (c−β1) f(b,c) + (b−αn)
m−1∑
j=1

(yj −βj+1) f(b,yj)

+ (c−β1)
n−1∑
i=1

(xi −αi) f(xi,c) +
n−1∑
i=1

m−1∑
j=1

(xi −αi) (yj −βj+1) f(xi,yj),

n−1∑
i=0

m∑
j=1

(xi −αi+1) (yj −βj) f(xi,yj)(3.5)

= (a−α1) (d−βm) f (a,d) + (a−α1)
m−1∑
j=1

(yj −βj) f(a,yj)

+ (d−βm)
n−1∑
i=1

(xi −αi+1) f(xi,d) +
n−1∑
i=1

m−1∑
j=1

(xi −αi+1) (yj −βj) f(xi,yj)

and

n−1∑
i=0

m−1∑
j=0

(xi −αi+1) (yj −βj+1) f(xi,yj)(3.6)

= (a−α1) (c−β1) f (a,c) + (a−α1)
m−1∑
j=1

(yj −βj+1) f(a,yj)

+ (c−β1)
n−1∑
i=1

(xi −αi+1) f(xi,c) +
n−1∑
i=1

m−1∑
j=1

(xi −αi+1) (yj −βj+1) f(xi,yj).



OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS... 147

Adding (3.3)-(3.6) in last equality of (3.2), we obtain

b∫
a

d∫
c

K(t)L(s)dsdtf(t,s)

= (b−αn) (d−βm) f(b,d) + (b−αn) (β1 − c) f(b,c)

+ (α1 −a) (d−βm) f (a,d) + (α1 −a) (β1 − c) f (a,c)

+(b−αn)
m−1∑
j=1

(βj+1 −βj) f(b,yj) + (α1 −a)
m−1∑
j=1

(βj+1 −βj) f(a,yj)

+ (d−βm)
n−1∑
i=1

(αi+1 −αi) f(xi,d) + (β1 − c)
n−1∑
i=1

(αi+1 −αi) f(xi,c)

n−1∑
i=1

m−1∑
j=1

(αi+1 −αi) (βj+1 −βj) f(xi,yj) −
b∫

a

d∫
c

f(t,s)dsdt

=

n∑
i=0

m∑
j=0

(αi+1 −αi) (βj+1 −βj) f(xi,yj) −
b∫

a

d∫
c

f(t,s)dsdt.

On the other hand, we have

∣∣∣∣∣∣
b∫

a

d∫
c

K(t)L(s)dsdtf(t,s)

∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
n−1∑
i=0

m−1∑
j=0




xi+1∫
xi

yj+1∫
yj

K(t)L(s)dsdtf(t,s)



∣∣∣∣∣∣∣(3.7)

≤
n−1∑
i=0

m−1∑
j=0

∣∣∣∣∣∣∣
xi+1∫
xi

yj+1∫
yj

(t−αi+1) (s−βj+1) dsdtf(t,s)

∣∣∣∣∣∣∣ .
Using Lemma 2 in the last part of the (3.7), we have

∣∣∣∣∣∣∣
xi+1∫
xi

yj+1∫
yj

(t−αi+1) (s−βj+1) dsdtf(t,s)

∣∣∣∣∣∣∣(3.8)
≤ sup

t∈[xi,xi+1]
s∈[yj,yj+1]

[|t−αi+1| |s−βj+1|]
xi+1∨
xi

yj+1∨
yj

(f)

= max{αi+1 −xi,xi+1 −αi+1}max{βj+1 −yj,yj+1 −βj+1}
xi+1∨
xi

yj+1∨
yj

(f)

=

[
1

2
(xi+1 −xi) +

∣∣∣∣αi+1 − xi + xi+12
∣∣∣∣
]

×
[

1

2
(yj+1 −yj) +

∣∣∣∣βj+1 − yj + yj+12
∣∣∣∣
]xi+1∨

xi

yj+1∨
yj

(f).



148 BUDAK AND SARIKAYA

Putting (3.8) in (3.7), we obtain

∣∣∣∣∣∣
b∫

a

d∫
c

K(t)L(s)dsdtf(t,s)

∣∣∣∣∣∣(3.9)
≤

n−1∑
i=0

m−1∑
j=0

[
1

2
(xi+1 −xi) +

∣∣∣∣αi+1 − xi + xi+12
∣∣∣∣
]

×
[

1

2
(yj+1 −yj) +

∣∣∣∣βj+1 − yj + yj+12
∣∣∣∣
]xi+1∨

xi

yj+1∨
yj

(f)

≤ max
i∈[0,...,n−1]

[
1

2
(xi+1 −xi) +

∣∣∣∣αi+1 − xi + xi+12
∣∣∣∣
]

× max
j∈[0,...,m−1]

[
1

2
(yj+1 −yj) +

∣∣∣∣βj+1 − yj + yj+12
∣∣∣∣
]n−1∑

i=0

m−1∑
j=0


xi+1∨

xi

yj+1∨
yj

(f)




≤
[

1

2
υ(h) + max

i∈[0,...,n−1]

∣∣∣∣αi+1 − xi + xi+12
∣∣∣∣
]

×
[

1

2
υ(l) + max

j∈[0,...,m−1]

∣∣∣∣βj+1 − yj + yj+12
∣∣∣∣
] b∨

a

d∨
c

(f)

which completes the proof of first inequality in (3.1).
In last inequality in (3.9), we get

(3.10)

∣∣∣∣αi+1 − xi + xi+12
∣∣∣∣ ≤ 12hi and maxi∈[0,...,n−1]

∣∣∣∣αi+1 − xi + xi+12
∣∣∣∣ ≤ 12υ(h),

and similarly,

(3.11) max
j∈[0,...,m−1]

∣∣∣∣βj+1 − yj + yj+12
∣∣∣∣ ≤ 12υ(l).

If we add (3.10) and (3.11) in (3.9), the proof of theorem is completed. �

Now, using the result of the Theorem 3, we give some applications as follows:

Corollary 1. With the assumptions of Theorem 3. If we choose

α0 = a, α1 =
a + x1

2
, α2 =

x1 + x2
2

, ...,αn−1 =
xn−2 + xn−1

2
, αn =

xn−1 + b

2
, αn+1 = b

and

β0 = c, β1 =
c + y1

2
, β2 =

y1 + y2
2

, ...,βn−1 =
ym−2 + ym−1

2
, βn =

ym−1 + d

2
, βm+1 = d



OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS... 149

in Theorem 3, then we have the inequality∣∣∣∣14 [(b−xn−1) (d−ym−1) f(b,d) + (b−xn−1) (y1 − c) f(b,c)
+ (x1 −a) (d−ym−1) f (a,d) + (x1 −a) (y1 − c) f (a,c)

+(b−xn−1)
m−1∑
j=1

(yj+1 −yj−1) f(b,yj) + (x1 −a)
m−1∑
j=1

(yj+1 −yj−1) f(a,yj)

+ (d−ym−1)
n−1∑
i=1

(xi+1 −xi−1) f(xi,d) + (y1 − c)
n−1∑
i=1

(xi+1 −xi−1) f(xi,c)

+

n−1∑
i=1

m−1∑
j=1

(xi+1 −xi−1) (xi+1 −xi−1) f(xi,yj)


− b∫

a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣
≤

1

4
υ(h)υ(l)

b∨
a

d∨
c

(f).

Corollary 2. In Corollary 1, if we take xi := a + (b − a) in (i = 0, 1, ...,n) and yj := c + (d − c)
j
m

(j = 0, 1, ...,m), then we have the inequality∣∣∣∣∣∣(b−a) (d− c)4nm

f(b,d) + f(b,c) + f (a,d) + f (a,c) + 2


m−1∑

j=1

f

(
b,

(m− j) c + jd
m

)

+

m−1∑
j=1

f

(
a,

(m− j) c + jd
m

)
+

n−1∑
i=1

f

(
(n− i) a + ib

n
,d

)
+

n−1∑
i=1

f

(
(n− i) a + ib

n
,c

)

+4

n−1∑
i=1

m−1∑
j=1

f

(
(n− i) a + ib

n
,

(m− j) c + jd
m

)− b∫
a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣
≤

(b−a) (d− c)
4nm

b∨
a

d∨
c

(f).

Corollary 3. Under assumption Theorem 3, choosing x0 = a, x1 = b, α0 = a, α1 = α, α2 = b, y0 = c,
y1 = d, β0 = c, β1 = β and β2 = d, we obtain the inequality

|(α−a) (β − c) f(a,c) + (α−a) (d−β) f(a,d)(3.12)

+ (b−α) (β − c) f(b,c) + (b−α) (d−β) f(b,d) −
b∫

a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣
≤

[
1

2
(b−a) +

∣∣∣∣α− a + b2
∣∣∣∣
][

1

2
(d− c) +

∣∣∣∣β − c + d2
∣∣∣∣
] b∨

a

d∨
c

(f).

Remark 1. a) If we put α = b and β = d in (3.12), then we have the ”left rectangle inequality”∣∣∣∣∣∣(b−a) (d− c) f(a,c) −
b∫

a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣ ≤ (b−a) (d− c)
b∨
a

d∨
c

(f),



150 BUDAK AND SARIKAYA

b) If take α = a and β = c in (3.12), then we have the ”right rectangle inequality”

∣∣∣∣∣∣(b−a) (d− c) f(b,d) −
b∫

a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣ ≤ (b−a) (d− c)
b∨
a

d∨
c

(f),

c) Similarliy, if we put α = a+b
2

and β = c+d
2

in (3.12), then we get the ”trapezoid inequality”

∣∣∣∣∣∣f(b,d) + f(b,c) + f(a,d) + f(a,c)4 − 1(b−a) (d− c)
b∫

a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣
≤

1

4

b∨
a

d∨
c

(f).

Corollary 4. Under assumption Theorem 3, taking a ≤ x ≤ b, a ≤ α1 ≤ x ≤ α2 ≤ b, c ≤ y ≤ d,
c ≤ β1 ≤ y ≤ β2 ≤ d we obtain the inequality

|(α1 −a) (β1 − c) f(a,c) + (α1 −a) (β2 −β1) f(a,y)(3.13)

+ (α1 −a) (d−β2) f(a,d) + (α2 −α1) (β1 − c) f(x,c)

+ (α2 −α1) (β2 −β1) f(x,y) + (α2 −α1) (d−β2) f(x,d)

+ (b−α2) (β1 − c) f(b,c) + (b−α2) (β2 −β1) f(b,y)

+ (b−α2) (d−β2) f(b,d) −
b∫

a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣
≤

1

4

[
1

2
(b−a) +

∣∣∣∣x− a + b2
∣∣∣∣ +
∣∣∣∣α1 − a + x2

∣∣∣∣ +
∣∣∣∣α2 − x + b2

∣∣∣∣
+

∣∣∣∣
∣∣∣∣α1 − a + x2

∣∣∣∣−
∣∣∣∣α2 − x + b2

∣∣∣∣
∣∣∣∣
]

×
[

1

2
(d− c) +

∣∣∣∣y − c + d2
∣∣∣∣ +
∣∣∣∣β1 − c + y2

∣∣∣∣ +
∣∣∣∣β2 − y + d2

∣∣∣∣
+

∣∣∣∣
∣∣∣∣β1 − c + y2

∣∣∣∣−
∣∣∣∣β2 − y + d2

∣∣∣∣
∣∣∣∣
] b∨

a

d∨
c

(f)

≤
[

1

2
(b−a) +

∣∣∣∣x− a + b2
∣∣∣∣
][

1

2
(d− c) +

∣∣∣∣y − c + d2
∣∣∣∣
] b∨

a

d∨
c

(f)

≤ (b−a) (d− c)
b∨
a

d∨
c

(f).

Remark 2. If we put α1 = a, α2 = b and β1 = c, β2 = d in (3.13), then the inequality (3.13) reduces
the inequality (2.5).



OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS... 151

Remark 3. If we choose α1 =
5a+b
6
, α2 =

a+5b
6
, x ∈

[
5a+b
6
, a+5b

6

]
, β1 =

5c+d
6
, β2 =

c+5d
6

and

y ∈
[
5c+d
6
, c+5d

6

]
in (3.13), then we have the ”Simpson’s rule inequality”

∣∣∣∣(b−a) (d− c)
[
f(b,d) + f(b,c) + f(a,d) + f(a,c)

36

+
f
(
a, c+d

2

)
+ f

(
a+b
2
,c
)

+ f
(
b, c+d

2

)
+ f

(
a+b
2
,d
)

9

+
4

9
f

(
a + b

2
,
c + d

2

)]
−

b∫
a

d∫
c

f(t,s)dsdt

∣∣∣∣∣∣
≤

(b−a) (d− c)
9

b∨
a

d∨
c

(f)

which is proved by Jawarneh and Noorani in [17].

4. Some Composite Qubature Formula

Let us consider the arbitrary division In : a = x0 < x1 < ... < xn = b, and Jm : c = y0 < y1 < ... <
ym = d, hi := xi+1 −xi, and lj := yj+1 −yj,

υ(h) := max{hi| i = 0, ...,n− 1} ,

υ(l) := max{lj| j = 0, ...,m− 1} .

Then, the following theorem holds.

Theorem 4. Let f : Q → R is of bounded variatin on Q and ξi ∈ [xi,xi+1] (i = 0, ...,n− 1) , ηj ∈
[yj,yj+1] (j = 0, ...,m− 1) . Then we have the qubature formula:

b∫
a

d∫
c

f(t,s)dsdt(4.1)

=

n−1∑
i=0

m−1∑
j=0

(ξi −xi) (ηj −yj) f(xi,yj)

+

n−1∑
i=0

m−1∑
j=0

(ξi −xi) (yj+1 −ηj) f(xi,yj+1)

+

n−1∑
i=0

m−1∑
j=0

(xi+1 − ξi) (ηj −yj) f(xi+1,yj)

+

n−1∑
i=0

m−1∑
j=0

(xi+1 − ξi) (yj+1 −ηj) f(xi+1,yj+1) + R(ξ,η,In,Jm,f).



152 BUDAK AND SARIKAYA

The remainder R(ξ,η,In,Jm,f) satisfies

|R(ξ,η,In,Jm,f)|(4.2)

≤
[

1

2
υ(h) + max

i∈{0,..,n−1}

{∣∣∣∣ξi − xi + xi+12
∣∣∣∣
}]

×
[

1

2
υ(l) + max

j∈{0,..,m−1}

{∣∣∣∣ηj − yj + yj+12
∣∣∣∣
}] b∨

a

d∨
c

(f)

≤ υ(h)υ(l)
b∨
a

d∨
c

(f)

for all ξi ∈ [xi,xi+1] (i = 0, ...,n− 1) and ηj ∈ [yj,yj+1] (j = 0, ...,m− 1) .

Proof. Aplying Corollary 3 on the bidimentional interval [xi,xi+1] × [yj,yj+1] , we get
|(ξi −xi) (ηj −yj) f(xi,yj)(4.3)

+ (ξi −xi) (yj+1 −ηj) f(xi,yj+1) + (xi+1 − ξi) (ηj −yj) f(xi+1,yj)

+ (xi+1 − ξi) (yj+1 −ηj) f(xi+1,yj+1) −
xi+1∫
xi

yj+1∫
yj

f(t,s)dsdt

∣∣∣∣∣∣∣
≤

[
1

2
hi +

∣∣∣∣ξi − xi + xi+12
∣∣∣∣
][

1

2
lj +

∣∣∣∣ηj − yj + yj+12
∣∣∣∣
]xi+1∨

xi

yj+1∨
yj

(f).

Summing the inequality (4.3) over i from 0 to n− 1 and j from 0 to m− 1, we get
(4.4)

|R(ξ,η,In,Jm,f)|

≤
n−1∑
i=0

m−1∑
j=0

[
1

2
hi +

∣∣∣∣ξi − xi + xi+12
∣∣∣∣
][

1

2
lj +

∣∣∣∣ηj − yj + yj+12
∣∣∣∣
]xi+1∨

xi

yj+1∨
yj

(f)

≤ max
i∈{0,..,n−1}

{
1

2
hi +

∣∣∣∣ξi − xi + xi+12
∣∣∣∣
}

× max
j∈{0,..,m−1}

{
1

2
lj +

∣∣∣∣ηj − yj + yj+12
∣∣∣∣
}n−1∑

i=0

m−1∑
j=0

xi+1∨
xi

yj+1∨
yj

(f)

≤
[

1

2
υ(h) + max

i∈{0,..,n−1}

{∣∣∣∣ξi − xi + xi+12
∣∣∣∣
}]

×
[

1

2
υ(l) + max

j∈{0,..,m−1}

{∣∣∣∣ηj − yj + yj+12
∣∣∣∣
}] b∨

a

d∨
c

(f)

which copletes the proof of first inequality in (4.2).
In last inequality

(4.5)

∣∣∣∣ξi − xi + xi+12
∣∣∣∣ ≤ 12hi and maxi∈[0,...,n−1]

∣∣∣∣ξi − xi + xi+12
∣∣∣∣ ≤ 12υ(h),

and similarly,

(4.6) max
j∈[0,...,m−1]

∣∣∣∣ηj − yj + yj+12
∣∣∣∣ ≤ 12υ(l).



OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS... 153

If we add (4.5) and (4.6) in (4.4), we obtain the required result. �

Corollary 5. Let f, In and Jm be as above.
1) If we choose ξi = xi+1 and ηj = yj+1 in (4.1), then we have the ”left rectangle rule”

b∫
a

d∫
c

f(t,s)dsdt =

n−1∑
i=0

m−1∑
j=0

f(xi,yj)hilj + RL(In,Jm,f).

2) If we choose ξi = xi and ηj = yj in (4.1), then we have the ”right rectangle rule”

b∫
a

d∫
c

f(t,s)dsdt =

n−1∑
i=0

m−1∑
j=0

f(xi+1,yj+1)hilj + RR(In,Jm,f).

3) Finally, if we choose ξi =
xi+xi+1

2
and ηj =

xi+xi+1
2

in (4.1), then we have the ”trapezoid rule”

b∫
a

d∫
c

f(t,s)dsdt

=
1

4

n−1∑
i=0

m−1∑
j=0

[f(xi,yj) + f(xi,yj+1) + f(xi+1,yj) + f(xi+1,yj+1)] hilj + RT (In,Jm,f).

Theorem 5. Let f, In and Jm be as above and xi ≤ α
(1)
i ≤ ξi ≤ α

(2)
i ≤ xi+1, yj ≤ β

(1)
j ≤ ηj ≤ β

(2)
j ≤

yj+1. Then we have the qubature formula

b∫
a

d∫
c

f(t,s)dsdt(4.7)

=

n−1∑
i=0

m−1∑
j=0

(
α
(1)
i −xi

)(
β
(1)
j −yj

)
f(xi,yj)

+

n−1∑
i=0

m−1∑
j=0

(
α
(1)
i −xi

)(
β
(2)
j −β

(1)
j

)
f(xi,ηj)

+

n−1∑
i=0

m−1∑
j=0

(
α
(1)
i −xi

)(
yj+1 −β

(2)
j

)
f(xi,yj+1)

+

n−1∑
i=0

m−1∑
j=0

(
α
(2)
i −α

(1)
i

)(
β
(1)
j −yj

)
f(ξi,yj)

+

n−1∑
i=0

m−1∑
j=0

(
α
(2)
i −α

(1)
i

)(
β
(2)
j −β

(1)
j

)
f(ξi,ηj)

+

n−1∑
i=0

m−1∑
j=0

(
α
(2)
i −α

(1)
i

)(
yj+1 −β

(2)
j

)
f(ξi,yj+1)

+

n−1∑
i=0

m−1∑
j=0

(
xi+1 −α

(2)
i

)(
β
(1)
j −yj

)
f(xi+1,yj)

+

n−1∑
i=0

m−1∑
j=0

(
xi+1 −α

(2)
i

)(
β
(2)
j −β

(1)
j

)
f(xi+1,ηj)

+

n−1∑
i=0

m−1∑
j=0

(
xi+1 −α

(2)
i

)(
yj+1 −β

(2)
j

)
f(xi+1,yj+1)

+R(ξ,η,α
(1)
i ,α

(2)
i ,β

(1)
j ,β

(2)
j In,Jm,f).



154 BUDAK AND SARIKAYA

The remainder R(ξ,η,α
(1)
i ,α

(2)
i ,β

(1)
j ,β

(2)
j ,In,Jm,f) satisfies∣∣∣R(ξ,η,α(1)i ,α(2)i ,β(1)j ,β(2)j ,In,Jm,f)∣∣∣(4.8)

≤
[

1

2
υ(h) + max

i∈{0,..,n−1}

{∣∣∣∣ξi − xi + xi+12
∣∣∣∣
}]

×
[

1

2
υ(l) + max

j∈{0,..,m−1}

{∣∣∣∣ηj − yj + yj+12
∣∣∣∣
}] b∨

a

d∨
c

(f)

≤ υ(h)υ(l)
b∨
a

d∨
c

(f).

Proof. Aplying Corollary 4 on the bidimentional interval [xi,xi+1] × [yj,yj+1] , we have
(4.9) ∣∣∣(α(1)i −xi)(β(1)j −yj)f(xi,yj) + (α(1)i −xi)(β(2)j −β(1)j )f(xi,ηj)

+
(
α
(1)
i −xi

)(
yj+1 −β

(2)
j

)
f(xi,yj+1) +

(
α
(2)
i −α

(1)
i

)(
β
(1)
j −yj

)
f(ξi,yj)

+
(
α
(2)
i −α

(1)
i

)(
β
(2)
j −β

(1)
j

)
f(ξi,ηj) +

(
α
(2)
i −α

(1)
i

)(
yj+1 −β

(2)
j

)
f(ξi,yj+1)

+
(
xi+1 −α

(2)
i

)(
β
(1)
j −yj

)
f(xi+1,yj) +

(
xi+1 −α

(2)
i

)(
β
(2)
j −β

(1)
j

)
f(xi+1,ηj)

+
(
xi+1 −α

(2)
i

)(
yj+1 −β

(2)
j

)
f(xi+1,yj+1) −

b∫
a

d∫
c

f(t,s)dsdt.

∣∣∣∣∣∣
≤

[
1

2
hi +

∣∣∣∣ξi − xi + xi+12
∣∣∣∣
][

1

2
lj +

∣∣∣∣ηj − yj + yj+12
∣∣∣∣
]xi+1∨

xi

yj+1∨
yj

(f).

Summing the inequality (4.9) over i from 0 to n− 1 and j from 0 to m− 1, then we get∣∣∣R(ξ,η,α(1)i ,α(2)i ,β(1)j ,β(2)j ,In,Jm,f)∣∣∣
≤

n−1∑
i=0

m−1∑
j=0

[
1

2
hi +

∣∣∣∣ξi − xi + xi+12
∣∣∣∣
][

1

2
lj +

∣∣∣∣ηj − yj + yj+12
∣∣∣∣
]xi+1∨

xi

yj+1∨
yj

(f)

≤ max
i∈{0,..,n−1}

{
1

2
hi +

∣∣∣∣ξi − xi + xi+12
∣∣∣∣
}

× max
j∈{0,..,m−1}

{
1

2
lj +

∣∣∣∣ηj − yj + yj+12
∣∣∣∣
}n−1∑

i=0

m−1∑
j=0

xi+1∨
xi

yj+1∨
yj

(f)

≤
[

1

2
υ(h) + max

i∈{0,..,n−1}

{∣∣∣∣ξi − xi + xi+12
∣∣∣∣
}]

×
[

1

2
υ(l) + max

j∈{0,..,m−1}

{∣∣∣∣ηj − yj + yj+12
∣∣∣∣
}] b∨

a

d∨
c

(f)

≤ υ(h)υ(l)
b∨
a

d∨
c

(f).



OSTROWSKI TYPE INEQUALITIES FOR FUNCTIONS... 155

This completes the proof of Theorem. �

Corollary 6. Under assumption of Theorem 5 with α
(1)
i = xi, α

(2)
i = xi+1, ξi =

xi+xi+1
2

, β
(1)
j = yj

β
(2)
j = yj+1 and ηj =

yj+yj+1
2

then we have the ”midpoint rule”

b∫
a

d∫
c

f(t,s)dsdt =

n−1∑
i=0

m−1∑
j=0

f

(
xi + xi+1

2
,
yj + yj+1

2

)
hilj + RM (In,Jm,f)

where the remainder satisfies

|RM (In,Jm,f)| ≤
1

4
υ(h)υ(l)

b∨
a

d∨
c

(f).

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156 BUDAK AND SARIKAYA

Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce-Turkey

∗Corresponding author: hsyn.budak@gmail.com