# Journal of Inequalities in Pure and Applied Mathematics ```Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 2, Article 27, 2002
ON MULTIDIMENSIONAL GR&Uuml;SS TYPE INEQUALITIES
B.G. PACHPATTE
57, S HRI N IKETEN C OLONY
(M AHARASHTRA ) I NDIA .
[email protected]
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
Gr&uuml;ss type by using a fairly elementary analysis.
Key words and phrases: Multidimensional, Gr&uuml;ss 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 Gr&uuml;ss inequality  (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 Gr&uuml;ss 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
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]&times;[c, d], ∆0 = (a, b)&times;(c, d),
Ω = [a, k]&times;[b, m]&times;[c, n], Ω0 = (a, k)&times;(b, m)&times;(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 &lt; xi &lt; 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 R&Uuml;SS 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)| &lt; ∞,
(x,y)∈∆0
sup |D2 D1 g (x, y)| &lt; ∞.
(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)| &lt; ∞,
(r,s,t)∈Ω0
sup
|D3 D2 D1 q (r, s, t)| &lt; ∞.
(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
&times;
(|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)| &lt; ∞,
x∈D
kDi gk∞ = sup |Di g (x)| &lt; ∞.
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 &middot; &middot; &middot; dxn ,
dy = dy1 &middot; &middot; &middot; dyn .
3. P ROOF OF T HEOREM 2.1
From the hypotheses we have the following identities (see ):
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 R&Uuml;SS 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 ):
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 ):
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 &lt; δ &lt; 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 &gt; 0, we have
Z
Z
Z
1
(5.4)
f (x) g (x) dx =
f (x) dx
g (x) dx
M
D
D
D
&quot;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.
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1
 G. GR&Uuml;SS, &Uuml;ber 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
 G.V. MILOVANOVIĆ, On some integral inequalities, Univ. Beograd Publ. Elek. Fak. Ser. Mat. Fiz.,
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