Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 1, Issue 2, Article 12, 2000
A GRÜSS TYPE INEQUALITY FOR SEQUENCES OF VECTORS IN INNER
PRODUCT SPACES AND APPLICATIONS
S.S. DRAGOMIR
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS , V ICTORIA U NIVERSITY OF T ECHNOLOGY,
PO B OX 14428, M ELBOURNE C ITY MC 8001, V ICTORIA , AUSTRALIA
sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 5 February, 2000; accepted 29 February, 2000
Communicated by B. Mond
A BSTRACT. A Grüss type inequality for sequences of vectors in inner product spaces which
complement a recent result from [6] and applications for differentiable convex functions defined
on inner product spaces and applications for Fourier and Mellin transforms, are given.
Key words and phrases: Grüss’ Inequality, Inner Product Spaces.
2000 Mathematics Subject Classification. 26D15, 26D99, 46Cxx.
1. I NTRODUCTION
In 1935, G. Grüss proved the following integral inequality (see [11] or [12])
Z b
Z b
Z b
1
1
1
(1.1) f (x) g (x) dx −
f (x) dx ·
g (x) dx
b−a a
b−a a
b−a a
1
≤ (Φ − φ) (Γ − γ) ,
4
provided that f and g are two integrable functions on [a, b] and satisfy the condition
φ ≤ f (x) ≤ Φ and γ ≤ g (x) ≤ Γ for all x ∈ [a, b] .
(1.2)
The constant
1
4
is the best possible and is achieved for
a+b
f (x) = g (x) = sgn x −
.
2
The discrete version of (1.1) states that:
If a ≤ ai ≤ A, b ≤ bi ≤ B (i = 1, ..., n) where a, A, ai , b, B, bi are real numbers, then
n
n
n
1 X
1
X
X
1
1
(1.3)
ai b i −
ai
bi ≤ (A − a) (B − b)
n
4
n
n
i=1
i=1
i=1
ISSN (electronic): 1443-5756
c 2000 Victoria University. All rights reserved.
002-00
2
S.S. D RAGOMIR
and the constant 14 is the best possible.
In the recent paper [2], the author proved the following generalisation in inner product spaces.
Theorem 1.1. Let (X; h·, ·i) be an inner product space over K, K = C, R, and e ∈ X, kek = 1.
If φ, Φ, γ, Γ ∈ K and x, y ∈ X such that
Re hΦe − x, x − φei ≥ 0 and Re hΓe − y, y − γei ≥ 0
(1.4)
holds, then we have the inequality
|hx, yi − hx, ei he, yi| ≤
(1.5)
The constant
1
4
1
|Φ − φ| |Γ − γ| .
4
is the best possible.
It has been shown in [1] that the above theorem, for the real case, contains the usual integral
and discrete Grüss inequality and also some Grüss type inequalities for mappings defined on
infinite intervals.
R∞
Namely, if ρ : (−∞, +∞) → (−∞, +∞) is a probability density function, i.e., −∞ ρ (t) dt =
1
1
1, then ρ 2 ∈ L2 (−∞, ∞) and obviously kρ 2 k2 = 1. Consequently, if we assume that
f, g ∈ L2 (−∞, ∞) and
1
1
1
1
αρ 2 ≤ f ≤ ψρ 2 , βρ 2 ≤ g ≤ θρ 2 a.e. on (0, ∞) ,
(1.6)
then we have the inequality
Z ∞
Z
(1.7) f (t) g (t) dt −
−∞
∞
1
2
Z
∞
f (t) ρ (t) dt
−∞
In a similar way, if e = (ei )i∈N ∈ l2 (R) with
−∞
P
i∈N
g (t) ρ (t) dt ≤
1
2
1
(ψ − α) (θ − β) .
4
|ei |2 = 1 and x = (xi )i∈N , y = (yi )i∈N ∈ l2 (R)
are such that
(1.8)
αei ≤ xi ≤ ψei , βei ≤ yi ≤ θei
for all i ∈ N, then we have
X
1
X
X
xi yi −
xi ei
yi ei ≤ (ψ − α) (θ − β) .
(1.9)
4
i∈N
i∈N
i∈N
In the recent paper [6], the author also proved the following discrete inequality in inner
product spaces:
n
n
n
X
1
X
X
(1.10)
p i ai x i −
p i ai
pi xi ≤ |A − a| kX − xk
4
i=1
i=1
i=1
provided xi ∈ H, ai ∈ K (K = C, R) and a, A ∈ K, x, X ∈ H are such that
(1.11)
Re [(A − ai ) (āi − ā)] ≥ 0 and
Re hX − xi , xi − xi ≥ 0 for all i ∈ {1, ..., n} .
The constant 41 is sharp.
For other recent developments of the Grüss inequality, see the papers [1]-[6], [10] and the
website http://rgmia.vu.edu.au/Gruss.html
In this paper we point out some other Grüss type inequalities in inner product spaces which
will complement the above result (1.10).
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
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A G RÜSS T YPE I NEQUALITY
3
2. P RELIMINARY R ESULTS
The following lemma is of interest in itself (see also [6]).
Lemma 2.1. Let (H; h·, ·i) be an inner product space over the real or complex number field K,
n
P
xi ∈ H and pi ≥ 0 (i = 1, ..., n) such that
pi = 1 (n ≥ 2).
i=1
If x, X ∈ H are such that
Re hX − xi , xi − xi ≥ 0 for all i ∈ {1, ..., n} ,
(2.1)
then we have the inequality
0≤
(2.2)
n
X
i=1
The constant
1
4
2
n
X
1
2
pi kxi k − pi xi ≤ kX − xk2 .
4
i=1
is sharp.
Proof. Define
*
I1 :=
X−
n
X
pi xi ,
i=1
and
I2 :=
n
X
n
X
+
pi xi − x
i=1
pi hX − xi , xi − xi .
i=1
Then
I1 =
n
X
i=1
and
I2 =
n
X
2
n
n
X
X
pi hX, xi i − hX, xi − pi xi +
pi hxi , xi
i=1
pi hX, xi i − hX, xi −
i=1
n
X
i=1
2
pi kxi k +
i=1
n
X
pi hxi , xi .
i=1
Consequently
I1 − I2 =
(2.3)
n
X
i=1
2
n
X
2
pi kxi k − pi xi .
i=1
Taking the real value in (2.3) we can state
2
n
n
X
X
(2.4)
pi kxi k2 − pi xi i=1
i=1
*
+
n
n
n
X
X
X
= Re X −
pi xi ,
pi xi − x −
pi Re hX − xi , xi − xi ,
i=1
i=1
i=1
which is an identity of interest in itself.
Using the assumption (2.1), we can conclude, by (2.4), that
2
+
*
n
n
n
n
X
X
X
X
(2.5)
pi kxi k2 − pi xi ≤ Re X −
pi xi ,
pi xi − x .
i=1
i=1
i=1
i=1
It is known that if y, z ∈ H, then
(2.6)
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
4 Re hz, yi ≤ kz + yk2 ,
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4
S.S. D RAGOMIR
with equality iff z = y.
Now, by (2.6), we can state that
2
*
+
n
n
n
n
X
X
X
X
1
1
Re X −
pi xi ,
pi xi − x ≤ X −
pi xi +
pi xi − x = kX − xk2 .
4
4
i=1
i=1
i=1
i=1
Using (2.5), we can easily deduce (2.2).
To prove the sharpness of the constant 14 , let us assume that the inequality (2.2) holds with a
constant c > 0, i.e.,
n
2
n
X
X
2
(2.7)
0≤
pi kxi k − pi xi ≤ c kX − xk2
i=1
i=1
for all pi , xi and x, X as in the hypothesis of Lemma 2.1.
Assume that n = 2, p1 = p2 = 12 , x1 = x and x2 = X with x, X ∈ H and x 6= X. Then,
obviously,
hX − x1 , x1 − xi = hX − x2 , x2 − xi = 0,
which shows that the condition (2.1) holds.
If we replace n, p1 , p2 , x1 , x2 in (2.7), we obtain
2
2 !
2
2
2
X
X
x
+
X
1
= kX − xk ≤ c kX − xk2 ,
pi kxi k2 −
pi xi =
kxk2 + kXk2 − 2 2
4
i=1
i=1
from where we deduce c ≥ 14 , which proves the sharpness of the constant factor 14 .
Remark 2.2. The assumption (2.1) can be replaced by the more general condition
n
X
(2.8)
pi Re hX − xi , xi − xi ≥ 0
i=1
and the conclusion (2.2) will still remain valid.
The following corollary is natural.
Corollary 2.3. Let ai ∈ K, pi ≥ 0 (i = 1, ..., n) (n ≥ 2) with
n
P
pi = 1. If a, A ∈ K are such
i=1
that
(2.9)
Re [(A − ai ) (āi − ā)] ≥ 0 for all i ∈ {1, ..., n} ,
then we have the inequality
(2.10)
0≤
n
X
i=1
2
n
X
1
pi |ai | − pi ai ≤ |A − a|2 .
4
i=1
2
The constant 14 is sharp.
The proof follows by the above Lemma 2.1 by choosing H = K, hx, yi := xȳ, xi = ai ,
x = a, X = A. We omit the details.
Remark 2.4. The condition (2.9) can be replaced by the more general assumption
(2.11)
n
X
pi Re [(A − ai ) (āi − ā)] ≥ 0.
i=1
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
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A G RÜSS T YPE I NEQUALITY
5
Remark 2.5. If we assume that K = R, then (2.8) is equivalent with
a ≤ ai ≤ A for all i ∈ {1, ..., n}
(2.12)
and then, with the assumption (2.12), we get the discrete Grüss type inequality
!2
n
n
X
X
1
(2.13)
0≤
pi a2i −
p i ai
≤ (A − a)2
4
i=1
i=1
and the constant
1
4
is sharp.
3. A D ISCRETE I NEQUALITY OF G RÜSS T YPE
The following Grüss type inequality holds.
Theorem 3.1. Let (H; h·, ·i) be an inner product space over K; K = C, R, xi , yi ∈ H, pi ≥ 0
n
P
pi = 1. If x, X, y, Y ∈ H are such that
(i = 0, ..., n) (n ≥ 2) with
i=1
(3.1)
Re hX − xi , xi − xi ≥ 0 and Re hY − yi , yi − yi ≥ 0 for all i ∈ {1, ..., n} ,
then we have the inequality
* n
+
n
n
X
1
X
X
(3.2)
pi hxi , yi i −
pi xi ,
pi yi ≤ kX − xk kY − yk .
4
i=1
i=1
i=1
The constant
1
4
is sharp.
Proof. A simple calculation shows that
* n
+
n
n
n
X
X
X
1X
(3.3)
pi hxi , yi i −
pi xi ,
pi yi =
pi pj hxi − xj , yi − yj i .
2
i=1
i=1
i=1
i,j=1
Taking the modulus in both parts of (3.3), and using the generalized triangle inequality, we
obtain
* n
+
n
n
n
X
1X
X
X
(3.4)
pi hxi , yi i −
pi xi ,
pi yi ≤
pi pj |hxi − xj , yi − yj i| .
2
i=1
i=1
i=1
i,j=1
By Schwartz’s inequality in inner product spaces we have
(3.5)
|hxi − xj , yi − yj i| ≤ kxi − xj k kyi − yj k
for all i, j ∈ {1, ..., n} , and therefore
* n
+
n
n
n
X
1X
X
X
(3.6)
pi hxi , yi i −
pi xi ,
pi yi ≤
pi pj kxi − xj k kyi − yj k .
2
i=1
i=1
i,j=1
i=1
Using the Cauchy-Buniakowsky-Schwartz inequality for double sums, we can state that
n
1X
(3.7)
pi pj kxi − xj k kyi − yj k
2 i,j=1
≤
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
n
1X
pi pj kxi − xj k2
2 i,j=1
! 12
×
n
1X
pi pj kyi − yj k2
2 i,j=1
! 12
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S.S. D RAGOMIR
and, a simple calculation shows that,
2
n
n
n
X
X
1X
pi pj kxi − xj k2 =
pi kxi k2 − pi xi 2 i,j=1
i=1
i=1
and
2
n
n
n
X
X
1X
2
2
pi pj kyi − yj k =
pi kyi k − pi yi .
2 i,j=1
i=1
i=1
We obtain
+
* n
n
n
X
X
X
(3.8) pi hxi , yi i −
pi xi ,
pi yi i=1
i=1
i=1

2  12 
2  12
n
n
n
n
X
X
X

X
2
2


≤
pi kxi k − pi xi ×
pi kyi k − pi yi  .
i=1
i=1
i=1
i=1
Using Lemma 2.1, we know that

2  12
n
n
X
X
1

pi kxi k2 − pi xi  ≤ kX − xk
2
i=1
i=1
and

2  21
n
n
X
X
1
2

pi kyi k − pi yi  ≤ kY − yk .
2
i=1
i=1
Therefore, by (3.8) we may deduce the desired inequality (3.3).
To prove the sharpness of the constant 41 , let us assume that (3.2) holds with a constant c > 0,
i.e.,
* n
+
n
n
X
X
X
(3.9)
pi hxi , yi i −
pi xi ,
pi yi ≤ c kX − xk kY − yk
i=1
i=1
i=1
under the above assumptions for pi , xi , yi , x, X, y, Y and n ≥ 2.
If we choose n = 2, x1 = x, x2 = X, y1 = y, y2 = Y (x 6= X, y 6= Y ) and p1 = p2 = 12 ,
then
* 2
+
2
2
2
X
X
X
1X
pi hxi , yi i −
pi xi ,
pi yi
=
pi pj hxi − xj , yi − yj i
2 i,j=1
i=1
i=1
i=1
X
=
pi pj hxi − xj , yi − yj i
1≤i<j≤2
1
hx − X, y − Y i
4
* 2
+
2
2
X
1
X
X
pi hxi , yi i −
pi xi ,
pi yi = |hx − X, y − Y i| .
4
i=1
i=1
i=1
=
and then
Choose X − x = z, Y − y = z, z 6= 0. Then using (3.9), we derive
1
kzk2 ≤ c kzk2 , z 6= 0
4
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
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A G RÜSS T YPE I NEQUALITY
7
which implies that c ≥ 14 , and the theorem is proved.
Remark 3.2. The condition (3.1) can be replaced by the more general assumption
(3.10)
n
X
pi Re hX − xi , xi − xi ≥ 0,
i=1
n
X
pi Re hY − yi , yi − yi ≥ 0
i=1
and the conclusion (3.2) still remains valid.
The following corollary for real or complex numbers holds.
Corollary 3.3. Let ai , bi ∈ K (K = C, R) , pi ≥ 0 (i = 1, ..., n) with
n
P
pi = 1. If a, A, b, B ∈
i=1
K are such that
Re [(A − ai ) (āi − ā)] ≥ 0 and
(3.11)
Re (B − bi ) b̄i − b̄ ≥ 0,
then we have the inequality
n
n
n
X
1
X
X
(3.12)
pi ai b̄i −
p i ai
pi b̄i ≤ |A − a| |B − b|
4
i=1
i=1
i=1
and the constant
1
4
is sharp.
The proof is obvious by Theorem 3.1 applied for the inner product space (C, h·, ·i) where
hx, yi = x · ȳ. We omit the details.
Remark 3.4. The condition (3.11) can be replaced by the more general condition
(3.13)
n
X
pi Re [(A − ai ) (āi − ā)] ≥ 0,
i=1
n
X
pi Re (B − bi ) b̄i − b̄ ≥ 0
i=1
and the conclusion of the above corollary will still remain valid.
Remark 3.5. If we assume that ai , bi , a, b, A, B are real numbers, then (3.11) is equivalent to
a ≤ ai ≤ A, b ≤ bi ≤ B for all i ∈ {1, ..., n}
(3.14)
and (3.12) becomes
(3.15)
n
n
n
X
1
X
X
0≤
p i ai b i −
p i ai
pi bi ≤ (A − a) (B − b) ,
4
i=1
i=1
i=1
which is the classical Grüss inequality for sequences of real numbers.
4. A PPLICATIONS FOR C ONVEX F UNCTIONS
Let (H; h·, ·i) be a real inner product space and F : H → R a Fréchet differentiable convex
mapping on H. Then we have the “gradient inequality”
(4.1)
F (x) − F (y) ≥ h∇F (y) , x − yi
for all x, y ∈ H, where ∇F : H → H is the gradient operator associated to the differentiable
convex function F .
The following theorem holds.
Theorem 4.1. Let F : H → R be as above and xi ∈ H (i = 1, ..., n). Suppose that there exists
the vectors γ, φ ∈ H such that hxi − γ, φ − xi i ≥ 0 for all i ∈ {1, ..., m} and m, M ∈ H
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8
S.S. D RAGOMIR
such that h∇F (xi ) − m, M − ∇F (xi )i ≥ 0 for all i ∈ {1, ..., m}. Then for all pi ≥ 0
m
P
(i = 1, ..., m) with Pm :=
pi > 0, we have the inequality
i=1
1
0≤
Pm
(4.2)
m
X
pi F (xi ) − F
i=1
Proof. Choose in (4.1), x =
F
(4.3)
1
PM
m
1 X
pi xi
Pm i=1
m
P
m
1 X
pi xi
Pm i=1
!
≤
1
kφ − γk kM − mk .
4
pi xi and y = xj to obtain
i=1
!
*
− F (xj ) ≥
m
1 X
∇F (xj ) ,
pi xi − xj
Pm i=1
+
for all j ∈ {1, ..., n}.
If we multiply (4.3) by pj ≥ 0 and sum over j from 1 to m, we have
! m
* m
+ m
m
m
X
X
X
1 X
1 X
Pm F
pi xi −
pj F (xj ) ≥
pj ∇F (xj ) ,
pi xi −
h∇F (xi ) , xi i .
Pm i=1
Pm j=1
j=1
i=1
i=1
Dividing by Pm > 0, we obtain the inequality
(4.4)
!
m
1 X
pi xi
Pm i=1
*
+
m
m
m
1 X
1 X
1 X
≤
pi h∇F (xi ) , xi i −
pi ∇F (xi ) ,
pi xi
Pm i=1
Pm i=1
Pm i=1
m
1 X
0 ≤
pi F (xi ) − F
Pm i=1
which is a generalisation for the case of inner product spaces of the result by Dragomir and Goh
established in 1996 for the case of differentiable mappings defined on Rn [9].
Applying Theorem 3.1 for real inner product spaces, X = φ, x = γ, yi = ∇F (xi ), y = m,
Y = M and n = m, we easily deduce
*
+
m
m
m
1 X
1 X
1 X
1
(4.5)
pi hxi , ∇F (xi )i −
pi xi ,
pi ∇F (xi ) ≤ kΦ − φk kM − mk
Pm i=1
Pm i=1
Pm i=1
4
and then, by (4.4) and (4.5) we can conclude that the desired inequality (4.2) holds.
Remark 4.2. The conditions
hxi − γ, φ − xi i ≥ 0, h∇F (xi ) − m, M − ∇F (xi )i ≥ 0,
(4.6)
for all i ∈ {1, ..., m} can be replaced by the more general conditions
m
m
X
X
(4.7)
pi hxi − γ, φ − xi i ≥ 0 and
pi h∇F (xi ) − m, M − ∇F (xi )i ≥ 0
i=1
i=1
and the conclusion (4.2) will still be valid.
Remark 4.3. Even if the inequality (4.2) is not as sharp as (4.4), it can be more useful in
practice when only some bounds of the gradient operator ∇F and of the vectors xi (i = 1, ..., n)
are known. In other words, it provides the opportunity to estimate the difference
!
m
m
1 X
1 X
∆ (F, x, p) :=
pi F (xi ) − F
pi xi ,
Pm i=1
Pm i=1
where the differences kφ − γk and kM − mk are known.
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
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A G RÜSS T YPE I NEQUALITY
9
Remark 4.4. For example, if we know that h∇F (xi ) − m, M − ∇F (xi )i ≥ 0 for all i ∈
{1, ..., m} and the vectors xi (i = 1, ..., n) are not too far from each other in the sense that
hxi − γ, φ − xi i ≥ 0 for all i ∈ {1, ..., m} and kφ − γk ≤ kM4ε
(ε > 0), then by (4.2), we
−mk
can conclude that
0 ≤ ∆ (F, x, p) ≤ ε.
5. A PPLICATIONS FOR S OME D ISCRETE T RANSFORMS
Let (H; h·, ·i) be an inner product space over K, K = C, R and x̄ = (x1 , ..., xn ) be a sequence
of vectors in H.
For a given m ∈ K, define the discrete Fourier transform
(5.1)
Fw (x̄) (m) =
n
X
exp (2wimk) × xk , m = 1, ..., n.
k=1
The complex number
n
P
exp (2wimk) hxk , yk i is actually the usual Fourier transform of the
k=1
vector (hx1 , y1 i , ..., hxn , yn i) ∈ Kn and will be denoted by
(5.2)
Fw (x̄ · ȳ) (m) =
n
X
exp (2wimk) hxk , yk i , m = 1, ..., n.
k=1
The following result holds.
Theorem 5.1. Let x̄, ȳ ∈ H n be sequences of vectors such that there exists the vectors c, C, y, Y ∈
H with the properties
(5.3)
Re hC − exp (2wimk) xk , exp (2wimk) xk − ci ≥ 0, k, m = 1, ..., n
and
(5.4)
Re hY − yk , yk − yi ≥ 0, k = 1, ..., n.
Then we have the inequality
*
+
n
n
X
1
(5.5)
yk ≤ kC − ck kY − yk ,
Fw (x̄ · ȳ) (m) − Fw (x̄) (m) ,
4
n k=1
for all m ∈ {1, ..., n}.
The proof follows by Theorem 3.1 applied for pk = n1 and for the sequences xk → ck =
exp (2wimk) xk and yk (k = 1, ..., n). We omit the details.
We can also consider the Mellin transform
(5.6)
M (x̄) (m) :=
n
X
k m−1 xk , m = 1, ..., n,
k=1
n
of the sequence x̄ = (x1 , ..., xn ) ∈ H .
We remark that the complex number
n
P
k m−1 hxk , yk i is actually the Mellin transform of the
k=1
vector (hx1 , y1 i , ..., hxn , yn i) ∈ Kn and will be denoted by
(5.7)
M (x̄ · ȳ) (m) :=
n
X
k m−1 hxk , yk i .
k=1
The following theorem holds.
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
http://jipam.vu.edu.au/
10
S.S. D RAGOMIR
Theorem 5.2. Let x̄, ȳ ∈ H n be sequences of vectors such that there exist the vectors d, D, y, Y ∈
H with the properties
(5.8)
Re D − k m−1 xk , k m−1 xk − d ≥ 0
for all k, m ∈ {1, ..., n} , and (5.4) is fulfilled.
Then we have the inequality
+
*
n
n
X
1
y
kD − dk kY − yk
(5.9)
M
(x̄
·
ȳ)
(m)
−
M
(x̄)
(m)
,
k ≤
4
n k=1
for all m ∈ {1, ..., n}.
The proof follows by Theorem 3.1 applied for pk = n1 and for the sequences xk → dk = kxk
and yk (k = 1, ..., n). We omit the details.
Another result which connects the Fourier transforms for different parameters w also holds.
Theorem 5.3. Let x̄, ȳ ∈ H n and w, z ∈ K. If there exists the vectors e, E, f, F ∈ H such that
Re hE − exp (2wimk) xk , exp (2wimk) xk − ei ≥ 0, k, m = 1, ..., n
and
Re hF − exp (2zimk) yk , exp (2zimk) yk − f i ≥ 0, k, m = 1, ..., n
then we have the inequality:
1
1
1
1
≤ kE − ek kF − f k ,
Fw+z (x̄ · ȳ) (m) −
F
(x̄)
(m)
,
F
(ȳ)
(m)
w
z
n
4
n
n
for all m ∈ {1, ..., n}.
The proof follows by Theorem 3.1 for the sequences exp (2wimk) xk , exp (2zimk) yk (k = 1, ..., n).
We omit the details.
R EFERENCES
[1] S.S. DRAGOMIR, Grüss inequality in inner product spaces, Austral. Math. Soc. Gazette, 26(2)
(1999), 66–70.
[2] S.S. DRAGOMIR, A generalization of Grüss’ inequality in inner product spaces and applications,
J. Math. Anal. Appl., 237 (1999), 74–82.
[3] S.S. DRAGOMIR, A Grüss type integral inequality for mappings of r-Hölder’s type and applications for trapezoid formula, Tamkang J. of Math., 31(1) (2000), 43–47.
[4] S.S. DRAGOMIR, Some discrete inequalities of Grüss type and applications in guessing theory,
Honam Math. J., 21(1) (1999), 115–126.
[5] S.S. DRAGOMIR, Some integral inequalities of Grüss type, Indian J. of Pure and Appl. Math.,
31(4) (2000), 397–415.
[6] S.S. DRAGOMIR, A Grüss type discrete inequality in inner product spaces and applications, J.
Math. Anal. Appl., (in press).
[7] S.S. DRAGOMIR AND G.L. BOOTH, On a Grüss-Lupaş type inequality and its applications for
the estimation of p-moments of guessing mappings, Math. Comm., (in press).
[8] S.S. DRAGOMIR AND I. FEDOTOV, An inequality of Grüss’ type for Riemann-Stieltjes integral
and applications for special means, Tamkang J. of Math., 29(4) (1998), 286–292.
[9] S.S. DRAGOMIR AND C.J. GOH, A counterpart of Jensen’s discrete inequality for differentiable
convex mappings and applications in information theory, Mathl. Comput. Modelling, 24(2) (1996),
1–11.
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
http://jipam.vu.edu.au/
A G RÜSS T YPE I NEQUALITY
11
[10] A.M. FINK, A treatise on Grüss’ inequality, submitted.
[11] G. GRÜSS, Über das Maximum des absoluten Betrages von
Rb
Rb
1
2 a f (x)dx a g(x)dx, Math. Z., 39 (1935), 215–226.
(b−a)
1
b−a
Rb
a
f (x)g(x)dx −
[12] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis,
Kluwer Academic Publishers, Dordrecht, 1993.
J. Inequal. Pure and Appl. Math., 1(2) Art. 12, 2000
http://jipam.vu.edu.au/