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volume 3, issue 2, article 27, 2002.

Received 14 August, 2001.;

accepted 25 January, 2002.

Communicated by:R.N. Mohapatra

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Journal of Inequalities in Pure and Applied Mathematics

ON MULTIDIMENSIONAL GRÜSS TYPE INEQUALITIES

B.G. PACHPATTE

57, Shri Niketen Colony Aurangabad - 431 001, (Maharashtra) India.

EMail:bgpachpatte@hotmail.com

c

2000Victoria University ISSN (electronic): 1443-5756 063-01

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On Multidimensional Grüss Type Inequalities

B.G. Pachpatte

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J. Ineq. Pure and Appl. Math. 3(2) Art. 27, 2002

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Abstract

In this paper we establish some new multidimensional integral inequalities of the Grüss type by using a fairly elementary analysis.

2000 Mathematics Subject Classification:26D15, 26D20.

Key words: Multidimensional, Grüss type inequalities, Partial Derivatives, Continu- ous Functions, Bounded, Identities.

Contents

1 Introduction. . . 3

2 Statement of Results. . . 4

3 Proof of Theorem 2.1 . . . 9

4 Proof of Theorem 2.2 . . . 11

5 Proof of Theorem 2.3 . . . 13 References

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On Multidimensional Grüss Type Inequalities

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1. Introduction

The following inequality is well known in the literature as the Grüss inequality [2] (see also [4, p. 296]):

1 b−a

Z b

a

f(x)g(x)dx− 1

b−a Z b

a

f(x)dx 1 b−a

Z b

a

g(x)dx

≤(M−m) (N −n), provided thatf, g: [a, b]→Rare integrable on[a, b]and

m≤f(x)≤M, n≤g(x)≤N, for allx∈[a, b],wherem, M, n, N are given constants.

Since the appearance of the above inequality in 1935, it has evoked consider- able 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.

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On Multidimensional Grüss Type Inequalities

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2. Statement of Results

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) fora, b, c, d, k, m, ninR. For functionsαandβ defined respectively on∆and Ω,the partial derivatives2∂y∂xα(x,y) and 3∂z∂y∂xβ(x,y,z) are denoted byD2D1α(x, y)and D3D2D1β(x, y, z).Let

D={x= (x1, . . . , xn) :ai < xi < bi (i= 1,2, . . . , n)}

andD¯ be the closure ofD.For a functionedefined onD¯ the partial derivatives

∂e(x)

∂xi (i= 1,2, . . . , n)are denoted byDie(x).

First, we give the following notations used to simplify the details of presen- tation.

A(D2D1f(x, y)) = A[a, c;x, y;b, d;D2D1f(s, t)]

= Z x

a

Z y

c

D2D1f(s, t)dtds− Z x

a

Z d

y

D2D1f(s, t)dtds

− Z b

x

Z y

c

D2D1f(s, t)dtds+ Z b

x

Z d

y

D2D1f(s, t)dtds, B(D3D2D1f(r, s, t))

=B[a, b, c;r, s, t;k, m, n;D3D2D1f(u, v, w)]

= Z r

a

Z s

b

Z t

c

D3D2D1f(u, v, w)dwdvdu

− Z r

a

Z s

b

Z n

t

D3D2D1f(u, v, w)dwdvdu

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On Multidimensional Grüss Type Inequalities

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− Z r

a

Z m

s

Z t

c

D3D2D1f(u, v, w)dwdvdu

− Z k

r

Z s

b

Z t

c

D3D2D1f(u, v, w)dwdvdu +

Z r

a

Z m

s

Z n

t

D3D2D1f(u, v, w)dwdvdu +

Z k

r

Z m

s

Z t

c

D3D2D1f(u, v, w)dwdvdu +

Z k

r

Z s

b

Z n

t

D3D2D1f(u, v, w)dwdvdu

− Z k

r

Z m

s

Z n

t

D3D2D1f(u, v, w)dwdvdu,

E(f(x, y)) = E[a, c;x, y;b, d;f]

= 1

2[f(x, c) +f(x, d) +f(a, y) +f(b, y)]

− 1

4[f(a, c) +f(a, d) +f(b, c) +f(b, d)], L(f(r, s, t)) =L[a, b, c;r, s, t;k, m, n;f]

= 1

8[f(a, b, c) +f(k, m, n)]

− 1

4[f(r, b, c) +f(r, m, n) +f(r, m, c) +f(r, b, n)]

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On Multidimensional Grüss Type Inequalities

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− 1

4[f(a, s, c) +f(k, s, n) +f(a, s, n) +f(k, s, c)]

− 1

4[f(a, b, t) +f(k, m, t) +f(k, b, t) +f(a, m, t)]

+ 1

2[f(a, s, t) +f(k, s, t)] + 1

2[f(r, b, t) +f(r, m, t)]

+ 1

2[f(r, s, c) +f(r, s, n)]. Our main results are given in the following theorems.

Theorem 2.1. Letf, g : ∆ →Rbe continuous functions on∆;D2D1f(x, y), D2D1g(x, y)exist on0 and are bounded, i.e.

kD2D1fk = sup

(x,y)∈∆0

|D2D1f(x, y)|<∞, kD2D1gk = sup

(x,y)∈∆0

|D2D1g(x, y)|<∞.

Then (2.1)

Z b

a

Z d

c

f(x, y)g(x, y)dydx

− 1 2

Z b

a

Z d

c

[E(f(x, y))g(x, y) +E(g(x, y))f(x, y)]dydx

≤ 1

8(b−a) (d−c) Z b

a

Z d

c

(|g(x, y)| kD2D1fk

+|f(x, y)| kD2D1gk)dydx.

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On Multidimensional Grüss Type Inequalities

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Theorem 2.2. Letp, q : Ω→Rbe continuous functions onΩ;D3D2D1p(r, s, t), D3D2D1q(r, s, t)exist on0and are bounded, i.e.

kD3D2D1pk = sup

(r,s,t)∈Ω0

|D3D2D1p(r, s, t)|<∞, kD3D2D1qk = sup

(r,s,t)∈Ω0

|D3D2D1q(r, s, t)|<∞.

Then

Z k

a

Z m

b

Z n

c

p(r, s, t)q(r, s, t)dtdsdr (2.2)

− 1 2

Z k

a

Z m

b

Z n

c

[L(p(r, s, t))q(r, s, t) +L(q(r, s, t))p(r, s, t)]dtdsdr

≤ 1

16(k−a) (m−b) (n−c)

× Z k

a

Z m

b

Z n

c

(|q(r, s, t)| kD3D2D1pk +|p(r, s, t)| kD3D2D1qk)dtdsdr.

Theorem 2.3. Let f, g : Rn → Rbe continuous functions onand differen- tiable onDwhose derivativesDif(x),Dig(x)are bounded, i.e.,

kDifk = sup

x∈D

|Dif(x)|<∞, kDigk = sup

x∈D

|Dig(x)|<∞.

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On Multidimensional Grüss Type Inequalities

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Then (2.3)

1 M

Z

D

f(x)g(x)dx− 1

M Z

D

f(x)dx 1 M

Z

D

g(x)dx

≤ 1 2M2

Z

D n

X

i=1

(|g(x)| kDifk+|f(x)| kDigk)Ei(x)dx, where

M =

n

Y

i=1

(bi−ai), Ei(x) = Z

D

|xi−yi|dy, dx = dx1· · ·dxn, dy=dy1· · ·dyn.

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3. Proof of Theorem 2.1

From the hypotheses we have the following identities (see [6]):

f(x, y) = E(f(x, y)) + 1

4A(D2D1f(x, y)), (3.1)

g(x, y) = E(g(x, y)) + 1

4A(D2D1g(x, y)), (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 b

a

Z d

c

f(x, y)g(x, y)dydx (3.3)

= 1 2

Z b

a

Z d

c

(E(f(x, y))g(x, y) +E(g(x, y))f(x, y))dydx + 1

8 Z b

a

Z d

c

(A(D2D1f(x, y))g(x, y) + A(D2D1g(x, y))f(x, y))dydx.

From the properties of modulus and integrals, it is easy to see that (3.4) |A(D2D1f(x, y))| ≤

Z b

a

Z d

c

|D2D1f(s, t)|dtds,

(3.5) |A(D2D1g(x, y))| ≤ Z b

a

Z d

c

|D2D1g(s, t)|dtds.

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From (3.3) – (3.5) we observe that

Z b

a

Z d

c

f(x, y)g(x, y)dydx

− 1 2

Z b

a

Z d

c

(E(f(x, y))g(x, y) +E(g(x, y))f(x, y))dydx

≤ 1 8

Z b

a

Z d

c

(|g(x, y)| |A(D2D1f(x, y))|

+|f(x, y)| |A(D2D1g(x, y))|)dydx

≤ 1 8

Z b

a

Z d

c

|g(x, y)|

Z b

a

Z d

c

|D2D1f(s, t)|dtds +|f(x, y)|

Z b

a

Z d

c

|D2D1g(s, t)|dtds

dydx

≤ 1

8(b−a) (d−c) Z b

a

Z d

c

(|g(x, y)| kD2D1fk +|f(x, y)| kD2D1gk)dydx,

which is the required inequality in (2.1). The proof is complete.

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4. Proof of Theorem 2.2

From the hypotheses we have the following identities (see [5]):

p(r, s, t) = L(p(r, s, t)) + 1

8B(D3D2D1p(r, s, t)), (4.1)

q(r, s, t) = L(q(r, s, t)) + 1

8B(D3D2D1q(r, s, t)), (4.2)

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 onΩand rewriting we have

Z k

a

Z m

b

Z n

c

p(r, s, t)q(r, s, t)dtdsdr (4.3)

= 1 2

Z k

a

Z m

b

Z n

c

[L(p(r, s, t))q(r, s, t) +L(q(r, s, t))p(r, s, t)]dtdsdr + 1

16 Z k

a

Z m

b

Z n

c

(B(D3D2D1p(r, s, t))q(r, s, t) +B(D3D2D1q(r, s, t))p(r, s, t))dtdsdr.

From the properties of modulus and integrals, we observe that

|B(D3D2D1p(r, s, t))| ≤ Z k

a

Z m

b

Z n

c

|D3D2D1p(u, v, w)|dwdvdu, (4.4)

|B(D3D2D1q(r, s, t))| ≤ Z k

a

Z m

b

Z n

c

|D3D2D1q(u, v, w)|dwdvdu.

(4.5)

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Now, from (4.3) – (4.5) and following the same arguments as in the proof of Theorem2.1with suitable changes, we get the required inequality in (2.2).

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5. Proof of Theorem 2.3

Letx ∈ D, y¯ ∈ D. From then−dimensional version of the mean value theo- rem, we have (see [3]):

(5.1) f(x)−f(y) =

n

X

i=1

Dif(c) (xi−yi), wherec= (y1+δ(x1−y1), . . . , yn+δ(xn−yn)),0< δ <1.

Integrating (5.1) with respect toy,we obtain (5.2) f(x) mesD=

Z

D

f(y)dy+

n

X

i=1

Z

D

Dif(c) (xi−yi)dy, wheremesD=Qn

i=1(bi−ai) =M.Similarly, we obtain (5.3) g(x) mesD=

Z

D

g(y)dy+

n

X

i=1

Z

D

Dig(c) (xi−yi)dy.

Multiplying (5.2) byg(x)and (5.3) byf(x)and adding the resulting identities, then integrating onDand noting thatmes>0, we have

Z

D

f(x)g(x)dx (5.4)

= 1 M

Z

D

f(x)dx Z

D

g(x)dx

+ 1 2M

"

Z

D

g(x)

n

X

i=1

Z

D

Dif(c) (xi−yi)dy

! dx

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+ Z

D

f(x)

n

X

i=1

Z

D

Dig(c) (xi−yi)dy

! dx

# .

From (5.4) we observe that

1 M

Z

D

f(x)g(x)dx− 1

M Z

D

f(x)dx 1 M

Z

D

g(x)dx

≤ 1 2M2

Z

D n

X

i=1

(|g(x)| kDifk+|f(x)| kDigk)Ei(x)dx.

This is the required inequality in (2.3). The proof is complete.

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References

[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.

[2] G. GRÜSS, Über das maximum des absoluten Betrages von

1 b−a

Rb

a f(x)g(x)dx − (b−a)1 2

Rb

a f(x)dxRb

a g(x)dx, Math. Z., 39 (1935), 215–226.

[3] G.V. MILOVANOVI ´C, On some integral inequalities, Univ. Beograd Publ.

Elek. Fak. Ser. Mat. Fiz., No498–No541 (1975), 112–124.

[4] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C 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.

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