Journal of Inequalities in Pure and
Applied Mathematics
SOME GRÜSS TYPE INEQUALITIES IN INNER PRODUCT SPACES
S.S. DRAGOMIR
School of Computer Science & Mathematics
Victoria University,
PO Box 14428, MCMC
Melbourne, Victoria 8001,
Australia.
E-Mail: sever.dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
volume 4, issue 2, article 42,
2003.
Received 12 March, 2003;
accepted 12 April, 2003.
Communicated by: B. Mond
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
032-03
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Abstract
Some new Grüss type inequalities in inner product spaces and applications for
integrals are given.
2000 Mathematics Subject Classification: Primary 26D15, 46D05 Secondary 46C99
Key words: Grüss’ Inequality, Inner products, Integral inequalities, Discrete Inequalities
Contents
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
1
2
3
4
5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
An Equivalent Assumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
A Refinement of the Grüss Inequality . . . . . . . . . . . . . . . . . . . . 9
Some Companion Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
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1.
Introduction
In [1], the author has proved the following Grüss type inequality in real or complex inner product spaces.
Theorem 1.1. Let (H, h·, ·i) be an inner product space over K (K = R,C) 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 conditions
(1.1)
Re hΦe − x, x − ϕei ≥ 0 and Re hΓe − y, y − γei ≥ 0
hold, then we have the inequality
(1.2)
|hx, yi − hx, ei he, yi| ≤
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
1
|Φ − ϕ| · |Γ − γ| .
4
The constant 41 is best possible in the sense that it cannot be replaced by a
smaller constant.
Some particular cases of interest for integrable functions with real or complex values and the corresponding discrete versions are listed below.
Corollary 1.2. Let f, g : [a, b] → K (K = R,C) be Lebesgue integrable and
such that
h
i
h
i
(1.3) Re (Φ − f (x)) f (x) − ϕ ≥ 0, Re (Γ − g (x)) g (x) − γ ≥ 0
for a.e. x ∈ [a, b] , where ϕ, γ, Φ, Γ are real or complex numbers and z̄ denotes
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the complex conjugate of z. Then we have the inequality
(1.4)
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.
Some Grüss’ Type Inequalities
in Inner Product Spaces
The discrete case is embodied in
Corollary 1.3. Let x, y ∈Kn and ϕ, γ, Φ, Γ are real or complex numbers such
that
(1.5)
Re [(Φ − xi ) (xi − ϕ)] ≥ 0,
Re [(Γ − yi ) (yi − γ)] ≥ 0
for each i ∈ {1, . . . , n} . Then we have the inequality
n
n
n
1 X
1 X 1
1X
xi ·
yi ≤ |Φ − ϕ| · |Γ − γ| .
(1.6)
xi yi −
n
n i=1
n i=1 4
i=1
The constant
1
4
is best possible.
For other applications of Theorem 1.1, see the recent paper [2].
In the present paper we show that the condition (1.1) may be replaced by
an equivalent but simpler assumption and a new proof of Theorem 1.1 is produced. A refinement of the Grüss type inequality (1.2) , some companions and
applications for integrals are pointed out as well.
S.S. Dragomir
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2.
An Equivalent Assumption
The following lemma holds.
Lemma 2.1. Let a, x, A be vectors in the inner product space (H, h·, ·i) over
K (K = R,C) with a 6= A. Then
Re hA − x, x − ai ≥ 0
if and only if
x − a + A ≤ 1 kA − ak .
2 2
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
Proof. Define
2
a
+
A
1
2
.
I1 := Re hA − x, x − ai , I2 := kA − ak − x
−
4
2 A simple calculation shows that
I1 = I2 = Re [hx, ai + hA, xi] − Re hA, ai − kxk2
and thus, obviously, I1 ≥ 0 iff I2 ≥ 0, showing the required equivalence.
The following corollary is obvious
Corollary 2.2. Let x, e ∈ H with kek = 1 and δ, ∆ ∈ K with δ 6= ∆. Then
Re h∆e − x, x − δei ≥ 0
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if and only if
1
δ
+
∆
x −
· e
≤ 2 |∆ − δ| .
2
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Remark 2.1. If H = C, then
Re [(A − x) (x̄ − ā)] ≥ 0
if and only if
1
a
+
A
≤ |A − a| ,
x −
2 2
where
a, x,
A ∈ C. If H = R, and A > a then a ≤ x ≤ A if and only if
x − a+A ≤ 1 |A − a| .
2
2
The following lemma also holds.
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
Lemma 2.3. Let x, e ∈ H with kek = 1. Then one has the following representation
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(2.1)
2
2
2
0 ≤ kxk − |hx, ei| = inf kx − λek .
Contents
λ∈K
Proof. Observe, for any λ ∈ K, that
hx − λe, x − hx, ei ei = kxk2 − |hx, ei|2 − λ he, xi − he, xi kek2
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= kxk2 − |hx, ei|2 .
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Using Schwarz’s inequality, we have
2
kxk2 − |hx, ei|2 = |hx − λe, x − hx, ei ei|2
2
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2
≤ kx − λek kx − hx, ei ek
= kx − λek2 kxk2 − |hx, ei|2
J. Ineq. Pure and Appl. Math. 4(2) Art. 42, 2003
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giving the bound
kxk2 − |hx, ei|2 ≤ kx − λek2 , λ ∈ K.
(2.2)
Taking the infimum in (2.2) over λ ∈ K, we deduce
kxk2 − |hx, ei|2 ≤ inf kx − λek2 .
λ∈K
Since, for λ0 = hx, ei , we get kx − λ0 ek2 = kxk2 − |hx, ei|2 , then the representation (2.1) is proved.
We are able now to provide a different proof for the Grüss type inequality
in inner product spaces mentioned in the Introduction, than the one from paper
[1].
Theorem 2.4. Let (H, h·, ·i) be an inner product space over K (K = R,C) 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 conditions (1.1) hold, or, equivalently, the following
assumptions
ϕ+Φ 1
γ+Γ 1
(2.3)
x − 2 · e ≤ 2 |Φ − ϕ| ,
y − 2 · e ≤ 2 |Γ − γ|
are valid, then one has the inequality
|hx, yi − hx, ei he, yi| ≤
(2.4)
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
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1
|Φ − ϕ| · |Γ − γ| .
4
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J. Ineq. Pure and Appl. Math. 4(2) Art. 42, 2003
The constant
1
4
is best possible.
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Proof. It can be easily shown (see for example the proof of Theorem 1 from
[1]) that
(2.5)
1 1
|hx, yi − hx, ei he, yi| ≤ kxk2 − |hx, ei|2 2 kyk2 − |hy, ei|2 2 ,
for any x, y ∈ H and e ∈ H, kek = 1. Using Lemma 2.3 and the conditions
(2.3) we obviously have that
1
1
ϕ
+
Φ
2
2 2
≤ |Φ − ϕ|
kxk − |hx, ei|
= inf kx − λek ≤ x
−
·
e
2
λ∈K
2
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
and
1
1
γ
+
Γ
2
2 2
kyk − |hy, ei|
= inf ky − λek ≤ y−
· e
≤ 2 |Γ − γ|
λ∈K
2
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and by (2.5) the desired inequality (2.4) is obtained.
The fact that 14 is the best possible constant, has been shown in [1] and we
omit the details.
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3.
A Refinement of the Grüss Inequality
The following result improving (1.1) holds
Theorem 3.1. Let (H, h·, ·i) be an inner product space over K (K = R,C) 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 conditions (1.1), or, equivalently, (2.3) hold, then we have
the inequality
(3.1)
1
|Φ − ϕ| · |Γ − γ|
4
1
1
− [Re hΦe − x, x − ϕei] 2 [Re hΓe − y, y − γei] 2 .
|hx, yi − hx, ei he, yi| ≤
Proof. As in [1], we have
(3.2)
(3.3)
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
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|hx, yi − hx, ei he, yi|2 ≤ kxk2 − |hx, ei|2 kyk2 − |hy, ei|2 ,
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kxk2 − |hx, ei|2
h
= Re (Φ − hx, ei) hx, ei − ϕ
i
− Re hΦe − x, x − ϕei
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and
(3.4)
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kyk2 − |hy, ei|2
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h
= Re (Γ − hy, ei) hy, ei − γ
i
− Re hΓe − x, x − γei .
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Using the elementary inequality
4 Re ab ≤ |a + b|2 ; a, b ∈ K (K = R,C)
we may state that
(3.5)
h
i 1
Re (Φ − hx, ei) hx, ei − ϕ ≤ |Φ − ϕ|2
4
and
(3.6)
h
Re (Γ − hy, ei) hy, ei − γ
i
≤
1
|Γ − γ|2 .
4
Consequently, by (3.2) − (3.6) we may state that
(3.7)
|hx, yi − hx, ei he, yi|2
1 2
1
2
2
|Φ − ϕ| − [Re hΦe − x, x − ϕei]
≤
4
1 2
1
2
2
×
|Γ − γ| − [Re hΓe − y, y − γei]
.
4
Finally, using the elementary inequality for positive real numbers
m2 − n2 p2 − q 2 ≤ (mp − nq)2
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
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we have
1 2
1
2
|Φ − ϕ| − [Re hΦe − x, x − ϕei] 2
4
1 2
1
2
×
|Γ − γ| − [Re hΓe − y, y − γei] 2
4
2
1
1
1
≤
|Φ − ϕ| · |Γ − γ| − [Re hΦe − x, x − ϕei] 2 [Re hΓe − y, y − γei] 2 ,
4
giving the desired inequality (3.1) .
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
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4.
Some Companion Inequalities
The following companion of the Grüss inequality in inner product spaces holds.
Theorem 4.1. Let (H, h·, ·i) be an inner product space over K (K = R,C) and
e ∈ H, kek = 1. If γ, Γ ∈ K and x, y ∈ H are such that
x+y x+y
(4.1)
Re Γe −
,
− γe ≥ 0
2
2
Some Grüss’ Type Inequalities
in Inner Product Spaces
or, equivalently,
(4.2)
x + y γ + Γ 1
2 − 2 · e ≤ 2 |Γ − γ| ,
then we have the inequality
S.S. Dragomir
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(4.3)
1
Re [hx, yi − hx, ei he, yi] ≤ |Γ − γ|2 .
4
The constant 14 is best possible in the sense that it cannot be replaced by a
smaller constant.
Proof. Start with the well known inequality
(4.4)
1
Re hz, ui ≤ kz + uk2 ; z, u ∈ H.
4
Since
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hx, yi − hx, ei he, yi = hx − hx, ei e, y − hy, ei ei
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then using (4.4) we may write
(4.5)
Re [hx, yi − hx, ei he, yi] = Re [hx − hx, ei e, y − hy, ei ei]
1
≤ kx − hx, ei e + y − hy, ei ek2
4
2
x + y
x
+
y
=
−
,
e
·
e
2
2
2 2
x + y x+y
=
, e .
2 −
2
If we apply Grüss’ inequality in inner product spaces for, say, a = b =
get
2
x + y 2 x + y
1
(4.6)
, e ≤ |Γ − γ|2 .
2 −
2
4
Some Grüss’ Type Inequalities
in Inner Product Spaces
x+y
,
2
we
Making use of (4.5) and (4.6) we deduce (4.3) .
The fact that 14 is the best possible constant in (4.3) follows by the fact that if
in (4.1) we choose x = y, then it becomes Re hΓe − x, x − γei ≥ 0, implying
0 ≤ kxk2 − |hx, ei|2 ≤ 41 |Γ − γ|2 , for which, by Grüss’ inequality in inner
product spaces, we know that the constant 14 is best possible.
The following corollary might be of interest if one wanted to evaluate the
absolute value of
Re [hx, yi − hx, ei he, yi] .
S.S. Dragomir
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Corollary 4.2. Let (H, h·, ·i) be an inner product space over K (K = R,C) and
e ∈ H, kek = 1. If γ, Γ ∈ K and x, y ∈ H are such that
x±y x±y
(4.7)
Re Γe −
,
− γe ≥ 0
2
2
or, equivalently,
(4.8)
x ± y γ + Γ 1
2 − 2 · e ≤ 2 |Γ − γ| ,
then we have the inequality
(4.9)
1
|Re [hx, yi − hx, ei he, yi]| ≤ |Γ − γ|2 .
4
If the inner product space H is real, then (for m, M ∈ R, M > m)
x±y x±y
(4.10)
Me −
,
− me ≥ 0
2
2
or, equivalently,
(4.11)
x ± y m + M 1
· e
≤ 2 (M − m) ,
2 −
2
S.S. Dragomir
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implies
(4.12)
Some Grüss’ Type Inequalities
in Inner Product Spaces
|hx, yi − hx, ei he, yi| ≤
1
(M − m)2 .
4
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In both inequalities (4.9) and (4.12) , the constant
1
4
is best possible.
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Proof. We only remark that, if
x−y x−y
Re Γe −
,
− γe ≥ 0
2
2
holds, then by Theorem 4.1, we get
Re [− hx, yi + hx, ei he, yi] ≤
1
|Γ − γ|2 ,
4
Some Grüss’ Type Inequalities
in Inner Product Spaces
showing that
(4.13)
1
Re [hx, yi − hx, ei he, yi] ≥ − |Γ − γ|2 .
4
Making use of (4.3) and (4.13) we deduce the desired result (4.9) .
Finally, we may state and prove the following dual result as well
Proposition 4.3. Let (H, h·, ·i) be an inner product space over K (K = R,C)
and e ∈ H, kek = 1. If ϕ, Φ ∈ K and x, y ∈ H are such that
h
i
(4.14)
Re (Φ − hx, ei) hx, ei − ϕ ≤ 0,
S.S. Dragomir
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then we have the inequalities
(4.15)
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1
kx − hx, ei ek ≤ [Re hx − Φe, x − ϕei] 2
√
1
2
≤
kx − Φek2 + kx − ϕek2 2 .
2
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Proof. We know that the following identity holds true (see (3.3))
(4.16)
kxk2 − |hx, ei|2
i
h
= Re (Φ − hx, ei) hx, ei − ϕ + Re hx − Φe, x − ϕei .
Using the assumption (4.14) and the fact that
kxk2 − |hx, ei|2 = kx − hx, ei ek2 ,
by (4.16) we deduce the first inequality in (4.15) .
The second inequality in (4.15) follows by the fact that for any v, w ∈ H
one has
1
Re hw, vi ≤
kwk2 + kvk2 .
2
The proposition is thus proved.
Some Grüss’ Type Inequalities
in Inner Product Spaces
S.S. Dragomir
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5.
Integral Inequalities
Let (Ω, Σ, µ) be a measure space consisting of a set Ω, a σ−algebra of parts Σ
and a countably additive and positive measure µ 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
5.1. If f, g, h ∈ L2 (Ω, K) and ϕ, Φ, γ, Γ ∈ K, are such that
R
2
|h (s)| dµ (s) = 1 and
Ω
Z
h
i
Re (Φh (s) − f (s)) f (s) − ϕh (s) dµ (s) ≥ 0,
(5.1)
ZΩ
h
i
Re (Γh (s) − g (s)) g (s) − γh (s) dµ (s) ≥ 0
Ω
or, equivalently
(5.2)
Some Grüss’ Type Inequalities
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S.S. Dragomir
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! 12
2
Z Φ
+
ϕ
1
f (s) −
h (s) dµ (s)
≤ |Φ − ϕ| ,
2
2
Ω
! 12
2
Z Γ
+
γ
1
g (s) −
h (s) dµ (s)
≤ |Γ − γ| ,
2
2
Ω
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then we have the following refinement of the Grüss integral inequality
(5.3)
Z
Z
Z
f (s) g (s)dµ (s) − f (s) h (s)dµ (s) h (s) g (s)dµ (s)
Ω
Ω
Ω
1
≤ |Φ − ϕ| · |Γ − γ|
4 Z
h
i
−
Re (Φh (s) − f (s)) f (s) − ϕh (s) dµ (s)
Ω
Z
×
12
i
Re (Γh (s) − g (s)) g (s) − γh (s) dµ (s) .
h
Ω
The constant
1
4
S.S. Dragomir
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
We omit the details.
Remark 5.1. It is obvious that a sufficient condition for (5.1) to hold is
h
i
Re (Φh (s) − f (s)) f (s) − ϕh (s) ≥ 0,
and
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h
i
Re (Γh (s) − g (s)) g (s) − γh (s) ≥ 0,
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for µ−a.e. s ∈ Ω, or equivalently,
Φ
+
ϕ
f (s) −
≤
h
(s)
2
g (s) − Γ + γ h (s) ≤
2
1
|Φ − ϕ| |h (s)| and
2
1
|Γ − γ| |h (s)| ,
2
for µ−a.e. s ∈ Ω.
The following result may be stated as well.
Corollary 5.2. If z, Z, t, T ∈ K, µ (Ω) < ∞ and f, g ∈ L2 (Ω, K) are such
that:
h
i
(5.4)
Re (Z − f (s)) f (s) − z̄ ≥ 0,
h
i
Re (T − g (s)) g (s) − t̄ ≥ 0 for a.e. s ∈ Ω
S.S. Dragomir
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or, equivalently
(5.5)
Some Grüss’ Type Inequalities
in Inner Product Spaces
f (s) − z + Z ≤
2 g (s) − t + T ≤
2 1
|Z − z| ,
2
1
|T − t| for a.e. s ∈ Ω
2
then we have the inequality
Z
1
f (s) g (s)dµ (s)
(5.6) µ (Ω) Ω
Z
Z
1
1
−
f (s) dµ (s) ·
g (s)dµ (s)
µ (Ω) Ω
µ (Ω) Ω
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1
1
≤ |Z − z| |T − t| −
4
µ (Ω)
Z
h
i
Re (Z − f (s)) f (s) − z̄ dµ (s)
Ω
Z
×
12
i
Re (T − g (s)) g (s) − t̄ dµ (s) .
h
Ω
Using Theorem 4.1 we may state the following result as well.
R
Proposition 5.3. If f, g, h ∈ L2 (Ω, K) and γ, Γ ∈ K are such that Ω |h (s)|2 dµ (s)
= 1 and
Z
f (s) + g (s)
(5.7)
Re Γh (s) −
2
Ω
"
#)
f (s) + g (s)
×
− γ̄ h̄ (s)
dµ (s) ≥ 0
2
or, equivalently,
! 12
2
Z f (s) + g (s) γ + Γ
1
(5.8)
−
h (s) dµ (s)
≤ |Γ − γ| ,
2
2
2
Ω
then we have the inequality
Z
i
h
(5.9)
I :=
Re f (s) g (s) dµ (s)
Ω
Z
Z
− Re
f (s) h (s)dµ (s) · h (s) g (s)dµ (s)
Ω
≤
1
|Γ − γ|2 .
4
Some Grüss’ Type Inequalities
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Ω
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If (5.7) and (5.8) hold with “ ± ” instead of “ + ”, then
(5.10)
|I| ≤
1
|Γ − γ|2 .
4
Remark 5.2. It is obvious that a sufficient condition for (5.7) to hold is
#)
(
"
f (s) + g (s)
f (s) + g (s)
·
− γ̄ h̄ (s)
≥0
(5.11)
Re Γh (s) −
2
2
for a.e. s ∈ Ω, or equivalently
1
f (s) + g (s) γ + Γ
≤ |Γ − γ| |h (s)| for a.e. s ∈ Ω.
(5.12)
−
h
(s)
2
2
2
Finally, the following corollary holds.
Corollary 5.4. If Z, z ∈ K, µ (Ω) < ∞ and f, g ∈ L2 (Ω, K) are such that
"
!#
f (s) + g (s)
f (s) + g (s)
(5.13) Re Z −
−z
≥ 0 for a.e. s ∈ Ω
2
2
or, equivalently
f (s) + g (s) z + Z 1
≤ |Z − z| for a.e. s ∈ Ω,
(5.14)
−
2
2 2
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then we have the inequality
Z
h
i
1
J :=
Re f (s) g (s) dµ (s)
µ (Ω) Ω
Z
Z
1
1
g (s)dµ (s)
− Re
f (s) dµ (s) ·
µ (Ω) Ω
µ (Ω) Ω
1
≤ |Z − z|2 .
4
If (5.13) and (5.14) hold with “ ± ” instead of “ + ”, then
(5.15)
|J| ≤
1
|Z − z|2 .
4
Remark 5.3. 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.
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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 AND I. GOMM, Some integral and discrete
versions of the Grüss inequality for real and complex functions
and sequences, RGMIA Res. Rep. Coll., 5(3) (2003), Article
[http://rgmia.vu.edu.au/v5n3.html]
Some Grüss’ Type Inequalities
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S.S. Dragomir
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