Journal of Inequalities in Pure and
Applied Mathematics
SOME COMPANIONS OF THE GRÜSS INEQUALITY
IN INNER PRODUCT SPACES
volume 4, issue 5, article 87,
2003.
S.S. DRAGOMIR
School of Computer Science and Mathematics,
Victoria University of Technology,
PO Box 14428,
Melbourne City MC
8001, Victoria, Australia.
EMail: sever@csm.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 24 March, 2003;
accepted 31 July, 2003.
Communicated by: J. Sándor
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c
2000
Victoria University
ISSN (electronic): 1443-5756
039-03
Quit
Abstract
Some companions of Grüss inequality in inner product spaces and applications
for integrals are given.
2000 Mathematics Subject Classification: Primary 26D15; Secondary 46C05.
Key words: Grüss inequality, Inner products, Integral inequality, Inequalities for
sums.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
A Grüss Type Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3
Some Related Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4
Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
References
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
1.
Introduction
The following inequality of Grüss type in real or complex linear spaces is known
(see [1]).
Theorem 1.1. Let (H; h·, ·i) be an inner product space over K (K = C, R) and
e ∈ H, kek = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors
in H such that the condition
(1.1)
Re hΦe − x, x − φei ≥ 0 and Re hΓe − y, y − γei ≥ 0
or, equivalently (see [3]),
φ+Φ 1
γ+Γ 1
(1.2)
x − 2 e ≤ 2 |Φ − φ| and y − 2 e ≤ 2 |Γ − γ|
holds, then we have the inequality
(1.3)
|hx, yi − hx, ei he, yi| ≤
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
1
|Φ − φ| |Γ − γ| .
4
JJ
J
II
I
The constant 14 is best possible in the sense that it cannot be replaced by a
smaller constant.
Go Back
Remark 1.1. The case for K = R for the above theorem has been published by
the author in [2].
Quit
Close
Page 3 of 22
The following particular instances for integrals and means are useful in applications.
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Corollary 1.2. Let f, g : [a, b] → K (K = C, R) be Lebesgue measurable and
such that there exists the constants φ, γ, Φ, Γ ∈ K with the property
h
i
(1.4)
Re (Φ − f (x)) f (x) − φ ≥ 0,
h
i
Re (Γ − g (x)) g (x) − γ ≥ 0
for a.e. x ∈ [a, b] , or, equivalently
γ + Γ 1
φ + Φ 1
(1.5)
f (x) − 2 ≤ 2 |Φ − φ| and g (x) − 2 ≤ 2 |Γ − γ|
for a.e. x ∈ [a, b] .
Then we have the inequality
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
(1.6)
Z b
Z b
Z b
1
1
1
f (x) g (x)dx −
f (x) dx ·
g (x)dx
b − a
b−a a
b−a a
a
1
≤ |Φ − φ| |Γ − γ| .
4
The constant
1
4
is best possible.
The discrete case is incorporated in
Corollary 1.3. Let x, y ∈ Kn , with x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) and
φ, γ, Φ, Γ ∈ K be such that
(1.7)
Re (Φ − xi ) xi − φ ≥ 0 and Re [(Γ − yi ) (yi − γ)] ≥ 0,
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 4 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
for each i ∈ {1, . . . , n} , or, equivalently,
1
φ
+
Φ
xi −
≤ |Φ − φ| and
(1.8)
2 2
1
γ
+
Γ
y i −
≤ |Γ − γ| ,
2 2
for each i ∈ {1, . . . , n} .
Then we have the inequality
n
n
n
1
1 X
X
X
1
1
yi ≤ |Φ − φ| |Γ − γ| .
xi ·
xi yi −
(1.9)
n
n i=1
n i=1 4
i=1
The constant
1
4
is best possible in (1.9).
For some recent results related to Grüss type inequalities in inner product
spaces, see [3]. More applications of Theorem 1.1 for integral and discrete
inequalities may be found in [4].
The main aim of this paper is to provide other inequalities of Grüss type in
the general setting of inner product spaces over the real or complex number field
K. Applications for Lebesgue integrals are pointed out as well.
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 5 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
2.
A Grüss Type Inequality
The following Grüss type inequality in inner product spaces holds.
Theorem 2.1. Let x, y, e ∈ H with kek = 1, and the scalars a, A, b, B ∈ K
(K = C, R) such that Re (āA) > 0 and Re b̄B > 0. If
(2.1)
Re hAe − x, x − aei ≥ 0 and Re hBe − y, y − bei ≥ 0
or, equivalently (see [3]),
1
1
b
+
B
a
+
A
x −
≤ |A − a| and y −
≤ |B − b| ,
(2.2)
e
e
2
2
2
2
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
then we have the inequality
Title Page
(2.3)
1
|A − a| |B − b|
|hx, yi − hx, ei he, yi| ≤ · q
|hx, ei he, yi| .
4
Re (āA) Re b̄B
The constant 41 is best possible in the sense that it cannot be replaced by a
smaller constant.
Proof. Apply Schwartz’s inequality in (H; h·, ·i) for the vectors x − hx, ei e and
y − hy, ei e, to get (see also [1])
(2.4)
|hx, yi − hx, ei he, yi|2 ≤ kxk2 − |hx, ei|2 kyk2 − |hy, ei|2 .
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 6 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Now, assume that u, v ∈ H, and c, C ∈ K with Re (c̄C) > 0 and RehCv − u,
u − cvi ≥ 0. This last inequality is equivalent to
h
i
2
2
kuk + Re (c̄C) kvk ≤ Re Chu, vi + c̄ hu, vi
(2.5)
= Re C̄ + c̄ hu, vi ,
since
h
i
Re Chu, vi = Re C̄ hu, vi .
1
2
Dividing this inequality by [Re (Cc̄)] > 0, we deduce
1
Re C̄ + c̄ hu, vi
1
2
2
2
(2.6)
.
1 kuk + [Re (c̄C)] kvk ≤
1
[Re (c̄C)] 2
[Re (c̄C)] 2
1
αp + q 2 ≥ 2pq,
α
Contents
α > 0, p, q ≥ 0,
we deduce
(2.7)
2 kuk kvk ≤
1
[Re (c̄C)]
1
1
2
S.S. Dragomir
Title Page
On the other hand, by the elementary inequality
2
Some Companions of the Grüss
Inequality in Inner Product
Spaces
kuk2 + [Re (c̄C)] 2 kvk2 .
Making use of (2.6) and (2.7) and the fact that for any z ∈ C, Re (z) ≤ |z| , we
get
Re C̄ + c̄ hu, vi
|C + c|
kuk kvk ≤
≤
1
1 |hu, vi| .
2 [Re (c̄C)] 2
2 [Re (c̄C)] 2
JJ
J
II
I
Go Back
Close
Quit
Page 7 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Consequently
#
2
|C
+
c|
kuk2 kvk2 − |hu, vi|2 ≤
− 1 |hu, vi|2
4 [Re (c̄C)]
"
(2.8)
1 |C − c|2
= ·
|hu, vi|2 .
4 Re (c̄C)
Now, if we write (2.8) for the choices u = x, v = e and u = y, v = e
respectively and use (2.4), we deduce the desired result (2.2). The sharpness of
the constant will be proved in the case where H is a real inner product space.
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
The following corollary which provides a simpler Grüss type inequality for
real constants (and in particular, for real inner product spaces) holds.
Corollary 2.2. With the assumptions of Theorem 2.1 and if a, b, A, B ∈ R are
such that A > a > 0, B > b > 0 and
1
1
a
+
A
b
+
B
x −
(2.9)
e
≤ (A − a) and y−
e
≤ 2 (B − b) ,
2
2
2
then we have the inequality
(2.10)
|hx, yi − hx, ei he, yi| ≤
The constant
1
4
is best possible.
Title Page
Contents
JJ
J
II
I
Go Back
Close
1 (A − a) (B − b)
√
·
|hx, ei he, yi| .
4
abAB
Quit
Page 8 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Proof. The prove the sharpness of the constant 14 assume that the inequality
(2.10) holds in real inner product spaces with x = y and for a constant k > 0,
i.e.,
(A − a)2
(2.11)
kxk − |hx, ei| ≤ k ·
|hx, ei|2 (A > a > 0) ,
aA
e ≤ 12 (A − a) , or equivalently, hAe − x, x − aei ≥ 0.
provided x − a+A
2
1 √1
2
2
√
We choose H = R , x = (x1 , x2 ) ∈ R , e =
, 2 . Then we have
2
2
2
2
kxk2 − |hx, ei|2 = x21 + x22 −
2
(x1 + x2 )
(x1 − x2 )
=
,
2
2
(x1 + x2 )2
,
|hx, ei| =
2
and by (2.11) we get
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
2
(2.12)
Title Page
Contents
(x1 − x2 )2
(A − a)2 (x1 + x2 )2
≤k·
·
.
2
aA
2
Now, if we let x1 =
√a ,
2
x2 =
A
√
2
(A > a > 0) , then obviously
JJ
J
II
I
Go Back
2 X
A
a
√ − xi
hAe − x, x − aei =
xi − √
= 0,
2
2
i=1
which shows that the condition (2.9) is fulfilled, and by (2.12) we get
2
2
Close
Quit
Page 9 of 22
2
(A − a)
(A − a) (a + A)
≤k·
·
4
aA
4
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
for any A > a > 0. This implies
(A + a)2 k ≥ aA
(2.13)
for any A > a > 0.
Finally, let a = 1 − q, A = 1 + q, q ∈ (0, 1) . Then from (2.13) we get
4k ≥ 1 − q 2 for any q ∈ (0, 1) which produces k ≥ 41 .
Remark 2.1. If hx, ei , hy, ei are assumed not to be zero, then the inequality
(2.3) is equivalent to
hx, yi
1
|A − a| |B − b|
(2.14)
,
hx, ei he, yi − 1 ≤ 4 · q
Re (āA) Re b̄B
while the inequality (2.10) is equivalent to
hx, yi
1 (A − a) (B − b)
≤ ·
√
(2.15)
−
1
.
4
hx, ei he, yi
abAB
The constant
1
4
is best possible in both inequalities.
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 10 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
3.
Some Related Results
The following result holds.
Theorem 3.1. Let (H; h·, ·i) be an inner product space over K (K = C, R) . If
γ, Γ ∈ K, e, x, y ∈ H with kek = 1 and λ ∈ (0, 1) are such that
(3.1)
Re hΓe − (λx + (1 − λ) y) , (λx + (1 − λ) y) − γei ≥ 0,
or, equivalently,
(3.2)
λx + (1 − λ) y − γ + Γ e ≤ 1 |Γ − γ| ,
2 2
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
then we have the inequality
(3.3)
Re [hx, yi − hx, ei he, yi] ≤
1
1
·
|Γ − γ|2 .
16 λ (1 − λ)
1
The constant 16
is the best possible constant in (3.3) in the sense that it cannot
be replaced by a smaller one.
Proof. We know that for any z, u ∈ H one has
Re hz, ui ≤
1
kz + uk2 .
4
Then for any a, b ∈ H and λ ∈ (0, 1) one has
(3.4)
Re ha, bi ≤
1
kλa + (1 − λ) bk2 .
4λ (1 − λ)
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 11 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Since
hx, yi − hx, ei he, yi = hx − hx, ei e, y − hy, ei ei
(as kek = 1),
using (3.4), we have
(3.5)
Re [hx, yi − hx, ei he, yi]
= Re [hx − hx, ei e, y − hy, ei ei]
1
≤
kλ (x − hx, ei e) + (1 − λ) (y − hy, ei e)k2
4λ (1 − λ)
1
=
kλx + (1 − λ) y − hλx + (1 − λ) y, ei ek2 .
4λ (1 − λ)
Since, for m, e ∈ H with kek = 1, one has the equality
(3.6)
2
2
S.S. Dragomir
Title Page
2
km − hm, ei ek = kmk − |hm, ei| ,
Contents
JJ
J
then by (3.5) we deduce the inequality
(3.7)
Some Companions of the Grüss
Inequality in Inner Product
Spaces
Re [hx, yi − hx, ei he, yi]
1
≤
kλx + (1 − λ) yk2 − |hλx + (1 − λ) y, ei|2 .
4λ (1 − λ)
II
I
Go Back
Close
Quit
Now, if we apply Grüss’ inequality
Page 12 of 22
1
0 ≤ kak − |ha, ei| ≤ |Γ − γ|2
4
2
2
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
provided
Re hΓe − a, a − γei ≥ 0,
for a = λx + (1 − λ) y, we have
(3.8)
kλx + (1 − λ) yk2 − |hλx + (1 − λ) y, ei|2 ≤
1
|Γ − γ|2 .
4
Utilising (3.7) and (3.8) we deduce the desired inequality (3.3). To prove the
1
sharpness of the constant 16
, assume that (3.3) holds with a constant C > 0,
provided (3.1) is valid, i.e.,
(3.9)
Re [hx, yi − hx, ei he, yi] ≤ C ·
1
|Γ − γ|2 .
λ (1 − λ)
If in (3.9) we choose x = y, provided (3.1) holds with x = y and λ ∈ (0, 1) ,
then
(3.10)
kxk2 − |hx, ei|2 ≤ C ·
1
|Γ − γ|2 ,
λ (1 − λ)
provided
Re hΓe − x, x − γei ≥ 0.
Since we know, in Grüss’ inequality, the constant 41 is best possible, then by
(3.10), one has
1
C
≤
for λ ∈ (0, 1) ,
4
λ (1 − λ)
1
giving, for λ = 12 , C ≥ 16
.
The theorem is completely proved.
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 13 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
The following corollary is a natural consequence of the above result.
Corollary 3.2. Assume that γ, Γ, e, x, y and λ are as in Theorem 3.1. If
Re hΓe − (λx ± (1 − λ) y) , (λx ± (1 − λ) y) − γei ≥ 0,
(3.11)
or, equivalently,
λx ± (1 − λ) y − γ + Γ e ≤ 1 |Γ − γ|2 ,
2 2
(3.12)
then we have the inequality
(3.13)
|Re [hx, yi − hx, ei he, yi]| ≤
The constant
1
16
1
1
·
|Γ − γ|2 .
16 λ (1 − λ)
is best possible in (3.13).
Re hΓe − (λx − (1 − λ) y) , (λx − (1 − λ) y) − γei ≥ 0,
which implies that
1
1
·
|Γ − γ|2
16 λ (1 − λ)
giving
(3.14)
S.S. Dragomir
Title Page
Proof. Using Theorem 3.1 for (−y) instead of y, we have that
Re [− hx, yi + hx, ei he, yi] ≤
Some Companions of the Grüss
Inequality in Inner Product
Spaces
Contents
JJ
J
II
I
Go Back
Close
Quit
1
1
Re [hx, yi − hx, ei he, yi] ≥ − ·
|Γ − γ|2 .
16 λ (1 − λ)
Consequently, by (3.3) and (3.14) we deduce the desired inequality (3.13).
Page 14 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Remark 3.1. If M, m ∈ R with M > m and, for λ ∈ (0, 1) ,
1
M
+
m
λx + (1 − λ) y −
(3.15)
e
≤ 2 (M − m)
2
then
1
1
·
(M − m)2 .
16 λ (1 − λ)
If (3.15) holds with “±” instead of “+” , then
hx, yi − hx, ei he, yi ≤
(3.16)
|hx, yi − hx, ei he, yi| ≤
Remark 3.2. If λ =
(3.17)
1
1
·
(M − m)2 .
16 λ (1 − λ)
1
2
in (3.1) or (3.2), then we obtain the result from [3], i.e.,
x+y x+y
Re Γe −
,
− γe ≥ 0
2
2
x + y γ + Γ 1
2 − 2 e ≤ 2 |Γ − γ|
implies
Title Page
JJ
J
II
I
Go Back
Re [hx, yi − hx, ei he, yi] ≤
(3.19)
The constant
S.S. Dragomir
Contents
or, equivalently
(3.18)
Some Companions of the Grüss
Inequality in Inner Product
Spaces
1
4
1
,
2
1
|Γ − γ|2 .
4
is best possible in (3.19).
For λ =
Corollary 3.2 and Remark 3.1 will produce the corresponding
results obtained in [3]. We omit the details.
Close
Quit
Page 15 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
4.
Integral Inequalities
Let (Ω, Σ, µ) be a measure space consisting of a set Ω, Σ a σ−algebra of parts
and µ a countably additive and positive measure on Σ with values in R∪ {∞} .
Denote by L2 (Ω, K) the Hilbert space of all real or complex valued functions f
defined on Ω and 2−integrable on Ω, i.e.,
Z
|f (s)|2 dµ (s) < ∞.
Ω
The following proposition holds
Proposition 4.1.R If f, g, h ∈ L2 (Ω, K) and ϕ, Φ, γ, Γ ∈ K, are so that Re (Φϕ) >
0, Re (Γγ) > 0, Ω |h (s)|2 dµ (s) = 1 and
Z
h
i
(4.1)
Re (Φh (s) − f (s)) f (s) − ϕh (s) dµ (s) ≥ 0
ZΩ
h
i
Re (Γh (s) − g (s)) g (s) − γh (s) dµ (s) ≥ 0
Ω
or, equivalently
(4.2)
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
JJ
J
II
I
Go Back
! 12
2
Z Φ
+
ϕ
1
f (s) −
h (s) dµ (s)
≤ |Φ − ϕ| ,
2
2
Ω
! 12
2
Z 1
Γ
+
γ
g (s) −
h (s) dµ (s)
≤ |Γ − γ| ,
2
2
Ω
Close
Quit
Page 16 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
then we have the following Grüss type integral inequality
Z
Z
Z
(4.3) f (s) g (s)dµ (s) − f (s) h (s)dµ (s) h (s) g (s)dµ (s)
Ω
Ω
Ω
Z
Z
|Φ − ϕ| |Γ − γ| 1
.
≤ ·p
f
(s)
h
(s)dµ
(s)
h
(s)
g
(s)dµ
(s)
4
Re (Φϕ̄) Re (Γγ̄) Ω
Ω
The constant
1
4
is best possible.
The proof follows by Theorem 3.1 on choosing H = L2 (Ω, K) with the
inner product
Z
hf, gi :=
f (s) g (s)dµ (s) .
Ω
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
We omit the details.
Remark 4.1. It is obvious that a sufficient condition for (4.1) to hold is
h
i
Re (Φh (s) − f (s)) f (s) − ϕh (s) ≥ 0,
and
h
i
Re (Γh (s) − g (s)) g (s) − γh (s) ≥ 0,
for µ−a.e. s ∈ Ω, or equivalently,
f (s) − Φ + ϕ h (s) ≤
2
Γ
+
γ
g (s) −
h (s) ≤
2
for µ−a.e. s ∈ Ω.
Title Page
Contents
JJ
J
II
I
Go Back
1
|Φ − ϕ| |h (s)| and
2
1
|Γ − γ| |h (s)| ,
2
Close
Quit
Page 17 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
The following result may be stated as well.
Corollary 4.2. If z, Z, t, T ∈ K, with Re (z̄Z) , Re (t̄T ) > 0, µ (Ω) < ∞ and
f, g ∈ L2 (Ω, K) are such that:
h
i
Re (Z − f (s)) f (s) − z̄ ≥ 0,
(4.4)
h
i
Re (T − g (s)) g (s) − t̄ ≥ 0 for a.e. s ∈ Ω
Some Companions of the Grüss
Inequality in Inner Product
Spaces
or, equivalently
(4.5)
z
+
Z
f (s) −
≤
2 t
+
T
g (s) −
≤
2 1
|Z − z| ,
2
1
|T − t| for a.e. s ∈ Ω;
2
S.S. Dragomir
Title Page
Contents
then we have the inequality
(4.6)
Z
1
f (s) g (s)dµ (s)
µ (Ω)
Ω
Z
Z
1
1
f (s) dµ (s) ·
g (s)dµ (s)
−
µ (Ω) Ω
µ (Ω) Ω
1
|Z − z| |T − t|
≤ ·p
4
Re (z̄Z) Re (t̄T )
Z
Z
1
1
× f (s) dµ (s) ·
g (s)dµ (s) .
µ (Ω)
µ (Ω)
Ω
Ω
JJ
J
II
I
Go Back
Close
Quit
Page 18 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Remark 4.2. The case of real functions incorporates the following interesting
inequality
R
µ (Ω) Ω f (s) g (s) dµ (s)
1 (Z − z) (T − t)
√
(4.7)
R f (s) dµ (s) R g (s) dµ (s) − 1 ≤ 4 ·
ztZT
Ω
Ω
provided µ (Ω) < ∞,
z ≤ f (s) ≤ Z, t ≤ g (s) ≤ T
for µ − a.e. s ∈ Ω, where z, t, Z, T are real numbers and the integrals at the
denominator are not zero. Here the constant 41 is best possible in the sense
mentioned above.
Using Theorem 3.1 we may state the following result as well.
2
S.S. Dragomir
Title Page
R
2
Proposition 4.3. If f, g, h ∈ L (Ω, K) and γ, Γ ∈ K are such that Ω |h (s)| dµ (s) =
1 and
Z
(4.8)
Re [Γh (s) − (λf (s) + (1 − λ) g (s))]
Ω
h
io
× λf (s) + (1 − λ) g (s) − γ̄ h̄ (s) dµ (s) ≥ 0
or, equivalently,
(4.9)
Some Companions of the Grüss
Inequality in Inner Product
Spaces
! 12
2
Z 1
λf (s) + (1 − λ) g (s) − γ + Γ h (s) dµ (s)
≤ |Γ − γ| ,
2
2
Ω
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 19 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
then we have the inequality
Z
h
i
(4.10) I :=
Re f (s) g (s) dµ (s)
Ω
Z
Z
f (s) h (s)dµ (s) · h (s) g (s)dµ (s)
− Re
Ω
Ω
1
1
·
|Γ − γ|2 .
≤
16 λ (1 − λ)
1
The constant 16
is best possible.
If (4.8) and (4.9) hold with “ ± ” instead of “ + ” (see Corollary 3.2), then
1
1
|I| ≤
·
|Γ − γ|2 .
16 λ (1 − λ)
(4.11)
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Remark 4.3. It is obvious that a sufficient condition for (4.8) to hold is
(4.12)
Re [Γh (s) − (λf (s) + (1 − λ) g (s))]
h
io
× λf (s) + (1 − λ) g (s) − γ̄ h̄ (s) ≥ 0
for a.e. s ∈ Ω, or equivalently
1
γ
+
Γ
λf (s) + (1 − λ) g (s) −
≤ |Γ − γ| |h (s)|
(4.13)
h
(s)
2
2
for a.e. s ∈ Ω.
Finally, the following corollary holds.
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 20 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
Corollary 4.4. If Z, z ∈ K, µ (Ω) < ∞ and f, g ∈ L2 (Ω, K) are such that
(4.14)
Re (Z − (λf (s) + (1 − λ) g (s)))
i
× λf (s) + (1 − λ) g (s) − z ≥ 0
for a.e. s ∈ Ω, or, equivalently
1
z
+
Z
λf (s) + (1 − λ) g (s) −
≤ |Z − z| ,
(4.15)
2 2
for a.e. s ∈ Ω, then we have the inequality
Z
h
i
1
J :=
Re f (s) g (s) dµ (s)
µ (Ω) Ω
Z
Z
1
1
f (s) dµ (s) ·
g (s)dµ (s)
− Re
µ (Ω) Ω
µ (Ω) Ω
1
1
≤
·
|Z − z|2 .
16 λ (1 − λ)
If (4.14) and (4.15) hold with “ ± ” instead of “ + ”, then
(4.16)
1
1
|J| ≤
·
|Z − z|2 .
16 λ (1 − λ)
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Remark 4.4. It is obvious that if one chooses the discrete measure above, then
all the inequalities in this section may be written for sequences of real or complex numbers. We omit the details.
Page 21 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au
References
[1] S.S. DRAGOMIR, A generalization of Grüss’ inequality in inner product
spaces and applications, J. Math. Anal. Appl., 237 (1999), 74–82.
[2] S.S. DRAGOMIR, Grüss inequality in inner product spaces, Australian
Math. Soc. Gazette, 29(2) (1999), 66–70.
[3] S.S. DRAGOMIR, Some Grüss’ type inequalities in inner product spaces,
J. Ineq. Pure & Appl. Math., 4(2) (2003), Article 42. [ONLINE: http:
//jipam.vu.edu.au/v4n2/032_03.html].
[4] S.S. DRAGOMIR AND I. GOMM, Some integral and discrete versions of
the Grüss inequality for real and complex functions and sequences, Tamsui
Oxford Journal of Math. Sci., 19(1) (2003), 66–77, Preprint: RGMIA Res.
Rep. Coll., 5(3) (2003), Article 9 [ONLINE http://rgmia.vu.edu.
au/v5n3.html].
Some Companions of the Grüss
Inequality in Inner Product
Spaces
S.S. Dragomir
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 22 of 22
J. Ineq. Pure and Appl. Math. 4(5) Art. 87, 2003
http://jipam.vu.edu.au