Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 2, Article 27, 2002
ON MULTIDIMENSIONAL GRÜSS TYPE INEQUALITIES
B.G. PACHPATTE
57, S HRI N IKETEN C OLONY
AURANGABAD - 431 001,
(M AHARASHTRA ) I NDIA .
bgpachpatte@hotmail.com
Received 14 August, 2001; accepted 25 January, 2002.
Communicated by R.N. Mohapatra
A BSTRACT. In this paper we establish some new multidimensional integral inequalities of the
Grüss type by using a fairly elementary analysis.
Key words and phrases: Multidimensional, Grüss type inequalities, Partial Derivatives, Continuous Functions, Bounded,
Identities.
2000 Mathematics Subject Classification. 26D15, 26D20.
1. I NTRODUCTION
The following inequality is well known in the literature as the Grüss inequality [2] (see also
[4, p. 296]):
1
b−a
Z
b
f (x) g (x) dx −
a
1
b−a
Z
b
f (x) dx
a
1
b−a
Z
b
g (x) dx
a
≤ (M − m) (N − n) ,
provided that f, g : [a, b] → R are integrable on [a, b] and
m ≤ f (x) ≤ M, n ≤ g (x) ≤ N,
for all x ∈ [a, b] , where m, M, n, N are given constants.
Since the appearance of the above inequality in 1935, it has evoked considerable interest and
many variants, generalizations and extensions have appeared, see [1, 4] and the references cited
therein. The main purpose of the present paper is to establish some new integral inequalities of
the Grüss type involving functions of several independent variables. The analysis used in the
proofs is elementary and our results provide new estimates on inequalities of this type.
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
063-01
2
B.G. PACHPATTE
2. S TATEMENT OF R ESULTS
In what follows, R denotes the set of real numbers. Let ∆ = [a, b]×[c, d], ∆0 = (a, b)×(c, d),
Ω = [a, k]×[b, m]×[c, n], Ω0 = (a, k)×(b, m)×(c, n) for a, b, c, d, k, m, n in R. For functions α
2 α(x,y)
3 β(x,y,z)
and β defined respectively on ∆ and Ω, the partial derivatives ∂ ∂y∂x
and ∂ ∂z∂y∂x
are denoted
by D2 D1 α (x, y) and D3 D2 D1 β (x, y, z) . Let
D = {x = (x1 , . . . , xn ) : ai < xi < bi (i = 1, 2, . . . , n)}
and D̄ be the closure of D. For a function e defined on D̄ the partial derivatives
∂e(x)
(i = 1, 2, . . . , n) are denoted by Di e (x) .
∂xi
First, we give the following notations used to simplify the details of presentation.
A (D2 D1 f (x, y)) = A [a, c; x, y; b, d; D2 D1 f (s, t)]
Z xZ y
Z
=
D2 D1 f (s, t) dtds −
a
c
x
Z
D2 D1 f (s, t) dtds
a
Z bZ
−
d
y
y
Z bZ
D2 D1 f (s, t) dtds +
x
c
d
D2 D1 f (s, t) dtds,
x
y
B(D3 D2 D1 f (r, s, t))
= B [a, b, c; r, s, t; k, m, n; D3 D2 D1 f (u, v, w)]
Z rZ sZ t
Z rZ sZ n
=
D3 D2 D1 f (u, v, w) dwdvdu −
D3 D2 D1 f (u, v, w) dwdvdu
a
b
c
a
b
t
Z rZ mZ t
Z kZ sZ t
−
D3 D2 D1 f (u, v, w) dwdvdu −
D3 D2 D1 f (u, v, w) dwdvdu
a
s
c
r
b
c
Z rZ mZ n
Z kZ mZ t
+
D3 D2 D1 f (u, v, w) dwdvdu +
D3 D2 D1 f (u, v, w) dwdvdu
a
s
t
r
s
c
Z kZ sZ n
Z kZ mZ n
+
D3 D2 D1 f (u, v, w) dwdvdu −
D3 D2 D1 f (u, v, w) dwdvdu,
r
b
t
r
s
t
E (f (x, y)) = E [a, c; x, y; b, d; f ]
1
= [f (x, c) + f (x, d) + f (a, y) + f (b, y)]
2
1
− [f (a, c) + f (a, d) + f (b, c) + f (b, d)] ,
4
L (f (r, s, t)) = L [a, b, c; r, s, t; k, m, n; f ]
1
= [f (a, b, c) + f (k, m, n)]
8
1
− [f (r, b, c) + f (r, m, n) + f (r, m, c) + f (r, b, n)]
4
1
− [f (a, s, c) + f (k, s, n) + f (a, s, n) + f (k, s, c)]
4
J. Inequal. Pure and Appl. Math., 3(2) Art. 27, 2002
http://jipam.vu.edu.au/
M ULTIDIMENSIONAL G RÜSS T YPE I NEQUALITIES
3
1
[f (a, b, t) + f (k, m, t) + f (k, b, t) + f (a, m, t)]
4
1
1
+ [f (a, s, t) + f (k, s, t)] + [f (r, b, t) + f (r, m, t)]
2
2
1
+ [f (r, s, c) + f (r, s, n)] .
2
Our main results are given in the following theorems.
−
Theorem 2.1. Let f, g : ∆ → R be continuous functions on ∆; D2 D1 f (x, y),
D2 D1 g (x, y) exist on ∆0 and are bounded, i.e.
kD2 D1 f k∞ =
kD2 D1 gk∞ =
sup |D2 D1 f (x, y)| < ∞,
(x,y)∈∆0
sup |D2 D1 g (x, y)| < ∞.
(x,y)∈∆0
Then
Z bZ
d
f (x, y) g (x, y) dydx
(2.1)
a
c
1
−
2
Z bZ
d
[E (f (x, y)) g (x, y) + E (g (x, y)) f (x, y)] dydx
a
c
1
≤ (b − a) (d − c)
8
Z bZ
d
(|g (x, y)| kD2 D1 f k∞ + |f (x, y)| kD2 D1 gk∞ ) dydx.
a
c
Theorem 2.2. Let p, q : Ω → R be continuous functions on Ω; D3 D2 D1 p (r, s, t),
D3 D2 D1 q (r, s, t) exist on Ω0 and are bounded, i.e.
kD3 D2 D1 pk∞ =
kD3 D2 D1 qk∞ =
sup
|D3 D2 D1 p (r, s, t)| < ∞,
(r,s,t)∈Ω0
sup
|D3 D2 D1 q (r, s, t)| < ∞.
(r,s,t)∈Ω0
Then
Z
k
Z
m
n
Z
p (r, s, t) q (r, s, t) dtdsdr
(2.2)
a
b
c
1
−
2
≤
Z
k
Z
m
Z
n
[L (p (r, s, t)) q (r, s, t) + L (q (r, s, t)) p (r, s, t)] dtdsdr
a
b
c
1
(k − a) (m − b) (n − c)
16
Z kZ mZ n
×
(|q (r, s, t)| kD3 D2 D1 pk∞ + |p (r, s, t)| kD3 D2 D1 qk∞ ) dtdsdr.
a
b
c
n
Theorem 2.3. Let f, g : R → R be continuous functions on D̄ and differentiable on D whose
derivatives Di f (x), Di g (x) are bounded, i.e.,
kDi f k∞ = sup |Di f (x)| < ∞,
x∈D
kDi gk∞ = sup |Di g (x)| < ∞.
x∈D
J. Inequal. Pure and Appl. Math., 3(2) Art. 27, 2002
http://jipam.vu.edu.au/
4
B.G. PACHPATTE
Then
(2.3)
1
M
Z
f (x) g (x) dx −
D
1
M
Z
1
≤
2M 2
f (x) dx
D
Z X
n
1
M
Z
g (x) dx
D
(|g (x)| kDi f k∞ + |f (x)| kDi gk∞ ) Ei (x) dx,
D i=1
where
M =
n
Y
Z
|xi − yi | dy,
(bi − ai ) , Ei (x) =
D
i=1
dx = dx1 · · · dxn ,
dy = dy1 · · · dyn .
3. P ROOF OF T HEOREM 2.1
From the hypotheses we have the following identities (see [6]):
1
f (x, y) = E (f (x, y)) + A (D2 D1 f (x, y)) ,
4
1
g (x, y) = E (g (x, y)) + A (D2 D1 g (x, y)) ,
4
(3.1)
(3.2)
for (x, y) ∈ ∆. Multiplying (3.1) by g (x, y) and (3.2) by f (x, y) and adding the resulting
identities, then integrating on ∆ and rewriting we have
Z bZ
d
(3.3)
a
c
f (x, y) g (x, y) dydx
Z Z
1 b d
=
(E (f (x, y)) g (x, y) + E (g (x, y)) f (x, y)) dydx
2 a c
Z Z
1 b d
+
(A (D2 D1 f (x, y)) g (x, y) + A (D2 D1 g (x, y)) f (x, y)) dydx.
8 a c
From the properties of modulus and integrals, it is easy to see that
Z bZ d
(3.4)
|A (D2 D1 f (x, y))| ≤
|D2 D1 f (s, t)| dtds,
a
c
Z bZ
|A (D2 D1 g (x, y))| ≤
(3.5)
d
|D2 D1 g (s, t)| dtds.
a
c
From (3.3) – (3.5) we observe that
Z bZ
d
f (x, y) g (x, y) dydx
a
c
Z Z
1 b d
−
(E (f (x, y)) g (x, y) + E (g (x, y)) f (x, y)) dydx
2 a c
Z Z
1 b d
≤
(|g (x, y)| |A (D2 D1 f (x, y))| + |f (x, y)| |A (D2 D1 g (x, y))|) dydx
8 a c
J. Inequal. Pure and Appl. Math., 3(2) Art. 27, 2002
http://jipam.vu.edu.au/
M ULTIDIMENSIONAL G RÜSS T YPE I NEQUALITIES
1
≤
8
Z bZ
d
Z bZ
d
|g (x, y)|
a
|D2 D1 f (s, t)| dtds
c
a
c
Z bZ
+ |f (x, y)|
1
(b − a) (d − c)
≤
8
5
d
|D2 D1 g (s, t)| dtds dydx
Z bZ
a
d
c
(|g (x, y)| kD2 D1 f k∞ + |f (x, y)| kD2 D1 gk∞ ) dydx,
a
c
which is the required inequality in (2.1). The proof is complete.
4. P ROOF OF T HEOREM 2.2
From the hypotheses we have the following identities (see [5]):
1
(4.1)
p (r, s, t) = L (p (r, s, t)) + B (D3 D2 D1 p (r, s, t)) ,
8
1
(4.2)
q (r, s, t) = L (q (r, s, t)) + B (D3 D2 D1 q (r, s, t)) ,
8
for (r, s, t) ∈ Ω. Multiplying (4.1) by q (r, s, t) and (4.2) by p (r, s, t) and adding the resulting
identities, then integrating on Ω and rewriting we have
Z kZ mZ n
(4.3)
p (r, s, t) q (r, s, t) dtdsdr
a
b
c
Z Z Z
1 k m n
=
[L (p (r, s, t)) q (r, s, t) + L (q (r, s, t)) p (r, s, t)] dtdsdr
2 a b
c
Z kZ mZ n
1
+
(B (D3 D2 D1 p (r, s, t)) q (r, s, t)
16 a b
c
+ B (D3 D2 D1 q (r, s, t)) p (r, s, t)) dtdsdr.
From the properties of modulus and integrals, we observe that
Z kZ mZ n
(4.4)
|B (D3 D2 D1 p (r, s, t))| ≤
|D3 D2 D1 p (u, v, w)| dwdvdu,
a
b
c
Z kZ mZ n
(4.5)
|B (D3 D2 D1 q (r, s, t))| ≤
|D3 D2 D1 q (u, v, w)| dwdvdu.
a
b
c
Now, from (4.3) – (4.5) and following the same arguments as in the proof of Theorem 2.1 with
suitable changes, we get the required inequality in (2.2).
5. P ROOF OF T HEOREM 2.3
Let x ∈ D̄, y ∈ D. From the n−dimensional version of the mean value theorem, we have
(see [3]):
n
X
(5.1)
f (x) − f (y) =
Di f (c) (xi − yi ) ,
i=1
where c = (y1 + δ (x1 − y1 ) , . . . , yn + δ (xn − yn )) , 0 < δ < 1.
Integrating (5.1) with respect to y, we obtain
Z
n Z
X
(5.2)
f (x) mesD =
f (y) dy +
Di f (c) (xi − yi ) dy,
D
J. Inequal. Pure and Appl. Math., 3(2) Art. 27, 2002
i=1
D
http://jipam.vu.edu.au/
6
where mesD =
(5.3)
B.G. PACHPATTE
Qn
(bi − ai ) = M. Similarly, we obtain
Z
n Z
X
g (x) mesD =
g (y) dy +
Di g (c) (xi − yi ) dy.
i=1
D
i=1
D
Multiplying (5.2) by g (x) and (5.3) by f (x) and adding the resulting identities, then integrating
on D and noting that mesD > 0, we have
Z
Z
Z
1
(5.4)
f (x) g (x) dx =
f (x) dx
g (x) dx
M
D
D
D
"Z
!
n Z
X
1
+
g (x)
Di f (c) (xi − yi ) dy dx
2M
D
i=1 D
! #
Z
n Z
X
+
f (x)
Di g (c) (xi − yi ) dy dx .
D
i=1
D
From (5.4) we observe that
Z
Z
Z
1
1
1
f (x) g (x) dx −
f (x) dx
g (x) dx
M D
M D
M D
Z X
n
1
≤
(|g (x)| kDi f k∞ + |f (x)| kDi gk∞ ) Ei (x) dx.
2M 2 D i=1
This is the required inequality in (2.3). The proof is complete.
R EFERENCES
[1] S.S. DRAGOMIR, Some integral inequalities of Grüss type, Indian J. Pure & Appl. Math., 31(4)
(2000), 379–415. ONLINE: RGMIA Research Report Collection, 1(2) (1998), 97–113.
Rb
1
[2] G. GRÜSS, Über das maximum des absoluten Betrages von b−a
a f (x) g (x) dx −
Rb
Rb
1
f (x) dx a g (x) dx, Math. Z., 39 (1935), 215–226.
(b−a)2 a
[3] G.V. MILOVANOVIĆ, On some integral inequalities, Univ. Beograd Publ. Elek. Fak. Ser. Mat. Fiz.,
No498–No541 (1975), 112–124.
[4] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and new Inequalities in Analysis,
Kluwer Academic Publishers, Dordrecht, 1993.
[5] B.G. PACHPATTE, On an inequality of Ostrowski type in three independent variables, J. Math. Anal.
Appl., 249 (2000), 583–591.
[6] B.G. PACHPATTE, On a new Ostrowski type inequality in two independent variables, Tamkang J.
Math., 32 (2001), 45–49.
J. Inequal. Pure and Appl. Math., 3(2) Art. 27, 2002
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