

International Journal of Analysis and Applications
ISSN 2291-8639
Volume 12, Number 2 (2016), 180-187
http://www.etamaths.com

ON CHEBYSHEV FUNCTIONAL AND OSTROWSKI-GRÜS TYPE

INEQUALITIES FOR TWO COORDINATES

ATIQ UR REHMAN∗ AND GHULAM FARID

Abstract. In this paper, we construct Chebyshev functional and Grüss inequality on two coordi-

nates. Also we establish Ostrowski-Grüss type inequality on two coordinates. Related mean value

theorems of Lagrange and Cauchy type are also given.

1. Introduction

Let f,g : [a,b] → R be integrable functions. We consider

(1) T(f,g) :=
1

b−a

∫ b
a

f(x)dx
1

b−a

∫ b
a

g(x)dx−
1

b−a

∫ b
a

f(x)g(x)dx.

If f and g are monotonic in same direction on [a,b], then

(2) T(f,g) ≥ 0.

If f and g are monotonic in opposite directions on interval [a,b], then the reverse of the inequality (2)
is valid.

The Chebyshev functional (1) has a long history and an extensive repertoire of applications in
many fields including numerical quadrature, transform theory, probability and statistical problems
and special functions. It is worthwhile noting that a number of identities relating to the Chebyshev
functional already exist. In [11, Chapter IX and X], one can see lots of results related to the Chebyshev
functional. One of them is famous as Korkine’s identity (see [11, p. 243 ]) given by

(3) T(f,g) =
1

2(b−a)2

∫ b
a

∫ b
a

(f(x) −f(y)) (g(x) −g(y)) dxdy.

This identity is often used to prove an inequality due to Grüss for functions bounded above and below
(see in [6]). In literature it is known as Grüss inequality.

Theorem 1.1. Let f,g : [a,b] → R be integrable functions such that φ ≤ f(x) ≤ ϕ and γ ≤ g(x) ≤ Γ
for all x ∈ [a,b], where φ, ϕ, γ and Γ are real constants. Then

(4) |T(f,g)| ≤
1

4
(ϕ−φ)(Γ −γ).

An other celebrated inequality in this respect is by Ostrowski sated in the following theorem (see
[12]).

Theorem 1.2. Let f : I → R, where I is an interval in R, be a mapping differentiable in Io the
interior of I and a, b ∈ Io, a < b, If

∣∣∣f′ (t)∣∣∣ ≤ M, for all t ∈ [a,b], then we have
(5)

∣∣∣∣∣f(x) − 1b−a
∫ b
a

f(t)dt

∣∣∣∣∣ ≤
[

1

4
+

(x− a+b
2

)2

(b−a)2

]
(b−a)M,

for all x ∈ [a,b].

2000 Mathematics Subject Classification. 26B15, 26A51, 34L15.
Key words and phrases. Chebyshev inequality; Chebyshev functional; Grüss inequality; Ostrowski-Grüss inequality;

mean value theorems.

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.

180


ON CHEBYSHEV FUNCTIONAL AND OSTROWSKI-GRÜS... 181

Inequality in (5) is well known as Ostrowski inequality and has interesting consequences in numerical
integration (see [3]). It has been improved by Dragomir and Wang in [4] using Grüss inequality in
terms of the lower and upper bounds of the first derivative. That Ostrowski-Grüss type inequality is
stated in the following theorem.

Theorem 1.3. Let f : [a,b] → R be continuous on [a,b] and differentiable on (a,b) and its derivative
satisfy the condition γ ≤ f′(x) ≤ Γ for all x ∈ [a,b], then we have the inequality∣∣∣∣∣f(x) − 1b−a

∫ b
a

f(t)dt−
(
f(b) −f(a)

b−a

)(
x−

a + b

2

)∣∣∣∣∣ ≤ 14 (b−a)(Γ −γ),(6)
for all x ∈ [a,b].

In [2], Barnett et al. pointed out a similar result to the above for twice differentiable mappings in
terms of the upper and lower bounds of the second derivative.

Theorem 1.4. Let f : [a,b] → R be continuous on [a,b] and twice differentiable on (a,b) and assume
that the second derivative f′′ : (a,b) → R satisfies the condition γ ≤ f′′(x) ≤ Γ for all x ∈ [a,b]. Then,
for all x ∈ [a,b], we have inequality∣∣∣∣∣f(x) −

(
x−

a + b

2

)
f′(x) +

[
(b−a)2

24
+

1

2

(
x−

a + b

2

)2](
f′(b) −f′(a)

b−a

)

−
1

b−a

∫ b
a

f(t)dt

∣∣∣∣∣ ≤ 18 (Γ −γ)
[

1

2
(b−a) +

∣∣∣∣x− a + b2
∣∣∣∣
]2
.(7)

Many authors considered different generalization of Chebyshev functional on two coordinates and
found the bounds of these functional, for example see [1, 14] and references there in.

In this paper we give the Chebyshev functional and Grüss inequality on two coordinates and establish
the Ostrowski-Grüss type inequality on two coordinates in terms of lower and upper bounds of first
and second order partial derivatives. Also we give Lagrange and Cauchy type mean value theorems
for the Chebyshev functional, as given in [5].

2. Main Results

Let ∆ = [a,b]×[c,d] be a bi-dimensional interval in R2 and f : ∆ → R be a mapping. If x = (x1,x2)
and y = (y1,y2), then we say x ≤ y if x1 ≤ y1 and x2 ≤ y2. Also we say f is monotonically increasing
on ∆ if for all x,y ∈ ∆

f(x) ≤ f(y)
when x ≤ y.
If we take F(x) =

∫ b
a
f(x,t)dt, provided that the integral exists, then one can note that

F(x) =

∫ d
c

f(x,t)dt ≤
∫ d
c

f(y,t)dt = F(y)

for x < y, that is F is monotonically increasing on [a,b].
In the following theorem, we introduce Chebyshev functional on two coordinates and generalize the

Chebyshev inequality on a rectangle from the plane.

Theorem 2.1. Let f,g : ∆ → R be integrable functions. We consider

(8) A(f; ∆) =
1

(b−a)(d− c)

∫ b
a

∫ d
c

f(x,y)dydx,

and

T(f,g; ∆) = A(f; ∆)A(g; ∆) −A(fg; ∆)(9)

If f and g are monotonic in same direction on ∆, then

(10) T(f,g; ∆) ≥ 0.


182 REHMAN AND FARID

Proof. Considering the monotonicity of f and g on second coordinate and using (2), we get

1

d− c

∫ d
c

f(x,y)g(x,y)dy ≤
1

(d− c)2

∫ d
c

f(x,y)dy

∫ d
c

g(x,y)dy.

Integrating above inequality over [a,b], we have

1

d− c

∫ b
a

∫ d
c

f(x,y)g(x,y)dydx ≤
1

(d− c)2

∫ b
a

(∫ d
c

f(x,y)dy

∫ d
c

g(x,y)dy

)
dx.(11)

Now if we take F(x) =
∫d
c
f(x,y)dy, then F is monotonic on [a,b] by considering monotonicity of f

on first coordinate. Similarly, we take G(x) =
∫d
c
g(x,y)dy, then G is monotonic on [a,b]. If f and g

are monotone in same direction so are F and G, then using the Chebyshev inequality, one has

1

(b−a)

∫ b
a

F(x)G(x)dx ≤
1

(b−a)2

∫ b
a

F(x)dx

∫ b
a

G(x)dx.(12)

Using the above inequality in (11), we get

A(fg; ∆) ≤ A(f; ∆)A(g; ∆),
which is equivalent to required result. �

It is easy to find that

T (f,g; ∆) =

1

2(b−a)2(d− c)2

∫ b
a

∫ d
c

∫ b
a

∫ d
c

(f(x,y) −f(u,v))(g(x,y) −g(u,v))dxdydudv.

This identity can be considered as Korkine’s identity in two coordinates. Using this identity one can
prove the following result similar to the proof of Theorem 1.1 (see also [11, p. 296])

Theorem 2.2. Let f,g : ∆ → R be integrable functions such that ϕ ≤ f(x,y) ≤ φ and γ ≤ g(x,y) ≤ Γ,
for all x,y ∈ ∆ where φ, ϕ, γ and Γ are constants. Then

(13) |T(f,g : ∆)| ≤
1

4
(φ−ϕ)(Γ −γ).

In [9], an important result related to Grüss inequality is given, we can have a similar result related
to Grüss inequality on two coordinates.

If f,g : ∆ → R be integrable functions, then
T (f,f; ∆) ≥ 0

and a following inequality holds:

T 2(f,g; ∆) ≤ T (f,f; ∆)T (g,g; ∆).(14)
By the combination of inequalities (13) and (14), we obtain the following result.

Theorem 2.3. Let f,g : ∆ → R be two integrable functions. If ϕ ≤ f(x,y) ≤ φ, for all x ∈ [a,b] and
y ∈ [c,d], where φ and ϕ are some constants, then

(15) |T(f,g; ∆)| ≤
1

2
(φ−ϕ)

√
T(g,g; ∆).

Proof. Setting g = f in (13), we get

(16) T (f,f; ∆) = |T (f,f; ∆)| ≤
1

4
(φ−ϕ)2.

Combining (16) with (14) we get

T 2(f,g; ∆) ≤
1

4
(φ−ϕ)2T (g,g; ∆),

this is equivalent to required result (15). �

In the following result we construct Ostrowski-Grüss type inequality on two coordinates in terms of
the lower and upper bounds of the first order partial derivatives.


ON CHEBYSHEV FUNCTIONAL AND OSTROWSKI-GRÜS... 183

Theorem 2.4. Let f : ∆ → R be continuous on ∆ and its partial derivative satisfy the condition
γ1 ≤ ∂f∂x ≤ Γ1 and γ2 ≤

∂f
∂y
≤ Γ2 on ∆. Then we have

∣∣∣∣∣
∫ b
a

f(x,c) + f(x,d)

2
dx +

∫ d
c

f(a,y) + f(b,y)

2
dy −

(
1

b−a
+

1

d− c

)
∫ b
a

∫ d
c

f(x,y)dxdy

∣∣∣∣∣ ≤ (b−a)(d− c)4 [(Γ2 −γ2) + (Γ1 −γ1)] .(17)
Proof. For all (x,y) ∈ ∆, consider two mappings fy : [a,b] → R and fx : [c,d] → R defined by
fy(t) = f(t,y) and fx(t) = f(x,t) respectively.

Applying (6) for mapping fy at x = b, we have∣∣∣∣∣f(b,y) − 1b−a
∫ b
a

f(t,y)dt−
f(b,y) −f(a,y)

2

∣∣∣∣∣ ≤ 14 (b−a)(Γ1 −γ1).
On integrating over [c,d], we have∣∣∣∣∣

∫ d
c

f(b,y)dy −
1

b−a

∫ b
a

∫ d
c

f(x,y)dydx−
(d− c)(f(b,y) −f(a,y))

2

∣∣∣∣∣
≤

1

4
(b−a)(d− c)(Γ1 −γ1).(18)

Applying (6) for mapping fy at x = a and then integrating over [c,d], we get∣∣∣∣∣
∫ d
c

f(a,y)dy −
1

b−a

∫ b
a

∫ d
c

f(x,y)dydx +
(d− c)(f(b,y) −f(a,y))

2

∣∣∣∣∣
≤

1

4
(b−a)(d− c)(Γ1 −γ1).(19)

Addition of (18) and (19) lead us to∣∣∣∣∣
∫ d
c

f(a,y) + f(b,y)

2
dy −

1

b−a

∫ b
a

∫ d
c

f(x,y)dydx

∣∣∣∣∣
≤

1

4
(b−a)(d− c)(Γ1 −γ1).(20)

Similarly using inequalities getting after applying (6) for mapping fx first at y = c then at y = d and
integrating over [a,b], we can have∣∣∣∣∣

∫ b
a

f(x,c) + f(x,d)

2
dx−

1

d− c

∫ b
a

∫ d
c

f(x,y)dydx

∣∣∣∣∣ ≤ 14 (b−a)(d− c)(Γ2 −γ2).(21)
Using (20) and (21), we have (17). �

In the following we establish the similar result to the Theorem 2.4 for twice differentiable mappings
in terms of the lower and upper bounds of the second order partial derivative.

Theorem 2.5. Let f : ∆2 → R be continuous on ∆2 and differentiable for all x ∈ (a,b) and y ∈ (c,d)
and assume that the second order partial derivative satisfies the condition γ2 ≤ ∂

2f
∂x2
≤ Γ2 for all


184 REHMAN AND FARID

x ∈ [a,b] and γ1 ≤ ∂
2f
∂y2
≤ Γ1 for all y ∈ [c,d], then we have

(22)

∣∣∣∣∣12
[∫ b

a

(f(x,c) + f(x,d)) dx +

∫ d
c

(f(a,y) + f(b,y)) dy

]
+

1

12[
(b−a)

∫ d
c

(
∂f(a,y)

∂x
−
∂f(b,y)

∂x

)
dy + (d− c)

∫ b
a

(
∂f(x,c)

∂y
−
∂f(x,d)

∂y

)
dx

]
−

(
1

b−a
+

1

d− c

)∫ b
a

∫ d
c

f(x,y)dydx

∣∣∣∣∣ ≤ 18 (d− c)(b−a) ((Γ1 −γ1)(b−a)
+(Γ2 −γ2)(d− c)) .

Proof. For all (x,y) ∈ ∆, consider two mappings fy : [a,b] → R and fx : [c,d] → R defined by
fy(t) = f(t,y) and fx(t) = f(x,t) respectively.

Applying (7) for mapping fy at x = b, we have∣∣∣∣∣f(b,y) − (b−a)6
(
∂f(a,y)

∂x
+ 2

∂f(b,y)

∂x

)
−

1

b−a

∫ b
a

f(t,y)dt

∣∣∣∣∣
≤

1

8
(Γ1 −γ1)(b−a)2.

Integrating over [c,d], we have∣∣∣∣∣
∫ d
c

f(b,y)dy −
(b−a)

6

∫ d
c

(
∂f(a,y)

∂x
+ 2

∂f(b,y)

∂x

)
dy

−
1

b−a

∫ b
a

∫ d
c

f(x,y)dydx

∣∣∣∣∣ ≤ 18 (d− c)(Γ1 −γ1)(b−a)2.(23)
Applying (7) for mapping fy at x = a and integrating over [c,d], we get∣∣∣∣∣

∫ d
c

f(a,y)dy +
(b−a)

6

∫ d
c

(
∂f(b,y)

∂x
+ 2

∂f(a,y)

∂x

)
dy

−
1

b−a

∫ b
a

∫ d
c

f(x,y)dydx

∣∣∣∣∣ ≤ 18 (d− c)(Γ1 −γ1)(b−a)2.(24)
Using (23) and (24), we get

(25)

∣∣∣∣∣12
∫ d
c

(f(a,y)dy + f(b,y))dy +
(b−a)

12

∫ d
c

(
∂f(a,y)

∂x
−
∂f(b,y)

∂x

)
dy

−
1

b−a

∫ b
a

∫ d
c

f(x,y)dydx

∣∣∣∣∣ ≤ 18 (d− c)(Γ1 −γ1)(b−a)2.
Similarly using inequalities getting after applying (7) for mapping fx first at y = c then at y = d and
integrating over [a,b], we have

(26)

∣∣∣∣∣12
∫ b
a

(f(x,c)dy + f(x,d))dx +
(d− c)

12

∫ b
a

(
∂f(x,c)

∂y
−
∂f(x,d)

∂y

)
dx

−
1

d− c

∫ b
a

∫ d
c

f(x,y)dydx

∣∣∣∣∣ ≤ 18 (b−a)(Γ2 −γ2)(d− c)2.
Using (25) and (26), we get (22). �


ON CHEBYSHEV FUNCTIONAL AND OSTROWSKI-GRÜS... 185

3. Mean Value Theorems

In this section, we give mean value theorems of Lagrange and Cauchy type for Chebyshev functional
on two coordinates. Before presenting our main results, one can note: if a function f : ∆ → R has
non-negative first order partial derivatives on ∆, then it is increasing on ∆.

Lemma 3.1. Let f : ∆ → R be an integrable function and also monotonically increasing on coor-
dinates, such that m1 ≤

∂f(x,y)
∂x

≤ M1 and m2 ≤
∂f(x,y)
∂y

≤ M2 for all interior points (x,y) in ∆.
Consider the functions h,k : ∆ → R defined as

h(x,y) = max{M1,M2}(x + y) −f(x,y)

and

k(x,y) = f(x,y) − min{m1,m2}(x + y).
Then h and k are monotonically increasing on ∆.

Proof. Since

(27)
∂h(x,y)

∂x
= max{M1,M2}−

∂f(x,y)

∂x
≥ 0

and

(28)
∂h(x,y)

∂y
=
∂f(x,y)

∂y
− min{m1,m2}≥ 0

for all interior points (x,y) in ∆, h is monotonically increasing on coordinates.
Similarly it can also be proved that k is monotonically increasing on coordinates on ∆. �

Theorem 3.2. Let f,g : ∆ → R be functions such that f has continuous partial derivatives of first
order in ∆ and g is increasing on ∆. Then there exists (ξ1,η1) and (ξ2,η2) in the interior of ∆ such
that

(29) T(f,g; ∆) =
∂f(ξ1,η1)

∂x
T(r,g; ∆)

and

(30) T(f,g; ∆) =
∂f(ξ2,η2)

∂y
T(r,g; ∆),

where r(x,y) = x + y and T(r,g; ∆) 6= 0.

Proof. Since f has continuous partial derivatives of first order in ∆, there exist real numbers m1, m2,

M1 and M2, such that m1 ≤
∂f(x,y)
∂x

≤ M1 and m2 ≤
∂f(x,y)
∂y

≤ M2 for all (x,y) ∈ ∆.
Now consider function h defined in Lemma 3.1. As h is increasing on coordinates in ∆, therefore

T (h,g; ∆) ≥ 0,

that is

T (max{M1,M2}r −f,g; ∆) ≥ 0.
This gives us

(31) T (f,g; ∆) ≤ max{M1,M2}T (r,g; ∆).

On the other hand for the function k defined in Lemma 3.1, one has

(32) min{m1,m2}T (r,g; ∆) ≤ T (f,g; ∆).

As T (r,g; ∆) 6= 0, combining above inequalities (31) and (32), we get

min{m1,m2}≤
T (f,g; ∆)

T (r,g; ∆)
≤ max{M1,M2}.

Then there exist (ξ1,η1) and (ξ2,η2) in the interior of ∆, such that

T (f,g; ∆)

T (r,g; ∆)
=
∂f(ξ1,η1)

∂x


186 REHMAN AND FARID

and
T (f,g; ∆)

T (r,g; ∆)
=
∂f(ξ2,η2)

∂y
.

Hence the required results are proved. �

Theorem 3.3. Let f,g : ∆ → R be functions having partial derivatives in ∆ and g is increasing on
∆. Then there exists (ξi,ηi), i = 1, 2, 3, 4 in the interior of ∆ such that

(33) T(f,g; ∆) =
∂f(ξ1,η1)

∂x

∂g(ξ3,η3)

∂x
T(r,r; ∆)

and

(34) T(f,g; ∆) =
∂f(ξ2,η2)

∂y

∂g(ξ4,η4)

∂y
T(r,r; ∆),

where r(x,y) = x + y.

Proof. Since T(r,g; ∆) = T(g,r; ∆) and r(x,y) = x+y is increasing on ∆, by Theorem 3.2 there exists
(ξ3,η3) in the interior of ∆ such that

(35) T (r,g; ∆) =
∂g(ξ3,η3)

∂x
T (r,r; ∆)

Using above expression in (29) gives us (33).
In a similar way, one can deduce (34). �

In [15], Pečarić gave many interesting result related to Chebyshev functional. A similar result is
also valid for Chebyshev functional on two coordinates. Namely, the following corollary.

Corollary 3.4. Let f,g : ∆ → R be functions, such that g is increasing on ∆ and f has partial
derivatives of first order in ∆ with

∣∣∣∂f∂x∣∣∣ ≤ M1, ∣∣∣∂f∂y∣∣∣ ≤ M2, ∣∣∣∂g∂x∣∣∣ ≤ N1 and ∣∣∣∂f∂x∣∣∣ ≤ N2. Then one has
(36) T(f,g; ∆) ≤ MiNiT(r,r; ∆), i = 1, 2,

where r(x,y) = x + y.

Theorem 3.5. Let f1,f2,g : ∆ → R be functions, such that f has partial derivatives of first order in
∆ and g is increasing on ∆. Then there exists (ξ1,η1) and (ξ2,η2) in the interior of ∆ such that

T(f1,g; ∆)

T(f2,g; ∆)
=

∂f1(ξ1,η1)
∂x

∂f2(ξ1,η1)
∂x

and

T(f1,g; ∆)

T(f2,g; ∆)
=

∂f1(ξ1,η1)
∂y

∂f2(ξ1,η1)
∂y

.

Proof. We define the function h : ∆ → R, such that

h = c1f1 − c2f2,

where c1 = T (f2,g; ∆) and c2 = T (f1,g; ∆).
Now, using Theorem 3.2 with f = h, we have

0 =

(
c1
∂f1(ξ1,η1)

∂x
− c2

∂f2(ξ1,η1)

∂x

)
T (r,g; ∆)

and

0 =

(
c1
∂f1(ξ2,η2)

∂y
− c2

∂f2(ξ2,η2)

∂y

)
T (r,g; ∆).

Since T (r,g; ∆) 6= 0, we have
c2
c1

=
∂f1(ξ1,η1)

∂x
∂f2(ξ1,η1)

∂x


ON CHEBYSHEV FUNCTIONAL AND OSTROWSKI-GRÜS... 187

and

c2
c1

=

∂f1(ξ1,η1)
∂y

∂f2(ξ1,η1)
∂y

,

this complete the proof. �

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COMSATS Institute of Information Technology, Attock Campus, Pakistan

∗Corresponding author: atiq@mathcity.org