Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 5, Article 87, 2003
SOME COMPANIONS OF THE GRÜSS INEQUALITY IN INNER PRODUCT
SPACES
S.S. DRAGOMIR
S CHOOL OF C OMPUTER S CIENCE AND M ATHEMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
PO B OX 14428
M ELBOURNE C ITY MC 8001
V ICTORIA , AUSTRALIA .
sever.dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 24 March, 2003; accepted 31 July, 2003
Communicated by J. Sándor
A BSTRACT. Some companions of Grüss inequality in inner product spaces and applications for
integrals are given.
Key words and phrases: Grüss inequality, Inner products, Integral inequality, Inequalities for sums.
2000 Mathematics Subject Classification. Primary 26D15; Secondary 46C05.
1. I NTRODUCTION
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]),
1
1
γ
+
Γ
φ
+
Φ
x −
≤ |Φ − φ| and y −
≤ |Γ − γ|
(1.2)
e
e
2
2
2 2
holds, then we have the inequality
1
(1.3)
|hx, yi − hx, ei he, yi| ≤ |Φ − φ| |Γ − γ| .
4
1
The constant 4 is best possible in the sense that it cannot be replaced by a smaller constant.
Remark 1.2. The case for K = R for the above theorem has been published by the author in
[2].
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
039-03
2
S.S. D RAGOMIR
The following particular instances for integrals and means are useful in applications.
Corollary 1.3. 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
h
i
(1.4)
Re (Φ − f (x)) f (x) − φ ≥ 0, Re (Γ − g (x)) g (x) − γ ≥ 0
for a.e. x ∈ [a, b] , or, equivalently
1
1
γ
+
Γ
φ
+
Φ
f (x) −
≤ |Φ − φ| and g (x) −
≤ |Γ − γ|
(1.5)
2 2
2 2
for a.e. x ∈ [a, b] .
Then we have the inequality
Z b
Z b
Z b
1
1
1
(1.6)
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 1.4. 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,
for each i ∈ {1, . . . , n} , or, equivalently,
γ + Γ 1
φ + Φ 1
(1.8)
xi − 2 ≤ 2 |Φ − φ| and yi − 2 ≤ 2 |Γ − γ| ,
for each i ∈ {1, . . . , n} .
Then we have the inequality
n
n
n
1 X
1
X
X
1
1
(1.9)
xi yi −
xi ·
yi ≤ |Φ − φ| |Γ − γ| .
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.
2. A G RÜSS T YPE I NEQUALITY
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
J. Inequal. Pure and Appl. Math., 4(5) Art. 87, 2003
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G RÜSS T YPE I NEQUALITIES
3
or, equivalently (see [3]),
1
1
a
+
A
b
+
B
x −
≤ |A − a| and y −
≤ |B − b| ,
(2.2)
e
e
2
2
2
2
then we have the inequality
1
|A − a| |B − b|
(2.3)
|hx, yi − hx, ei he, yi| ≤ · q
|hx, ei he, yi| .
4
Re (āA) Re b̄B
The constant
1
4
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 .
Now, assume that u, v ∈ H, and c, C ∈ K with Re (c̄C) > 0 and Re hCv − u, u − cvi ≥ 0.
This last inequality is equivalent to
h
i
(2.5)
kuk2 + Re (c̄C) kvk2 ≤ Re Chu, vi + c̄ hu, vi
= Re C̄ + c̄ hu, vi ,
since
h
i
Re Chu, vi = Re C̄ hu, vi .
1
Dividing this inequality by [Re (Cc̄)] 2 > 0, we deduce
(2.6)
1
1
2
1
2
kuk2 + [Re (c̄C)] kvk2 ≤
Re
C̄ + c̄ hu, vi
1
.
[Re (c̄C)]
[Re (c̄C)] 2
On the other hand, by the elementary inequality
1
αp2 + q 2 ≥ 2pq, α > 0, p, q ≥ 0,
α
we deduce
1
1
2
2
2
(2.7)
2 kuk kvk ≤
1 kuk + [Re (c̄C)] kvk .
[Re (c̄C)] 2
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 |hu, vi| .
1
2 [Re (c̄C)] 2
2 [Re (c̄C)] 2
Consequently
"
#
2
|C
+
c|
(2.8)
kuk2 kvk2 − |hu, vi|2 ≤
− 1 |hu, vi|2
4 [Re (c̄C)]
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.
The following corollary which provides a simpler Grüss type inequality for real constants
(and in particular, for real inner product spaces) holds.
J. Inequal. Pure and Appl. Math., 4(5) Art. 87, 2003
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S.S. D RAGOMIR
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
a+A 1
b+B 1
(2.9)
x − 2 e ≤ 2 (A − a) and y − 2 e ≤ 2 (B − b) ,
then we have the inequality
1 (A − a) (B − b)
√
|hx, ei he, yi| .
(2.10)
|hx, yi − hx, ei he, yi| ≤ ·
4
abAB
The constant 41 is best possible.
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
1
a+A provided x − 2 e ≤ 2 (A − a) , or equivalently,
hAe− x, x − aei ≥ 0.
2
2
We choose H = R2 , x = (x1 , x2 ) ∈ R2 , e =
√1 , √1
2
2
. Then we have
(x1 + x2 )2
(x1 − x2 )2
kxk − |hx, ei| = + −
=
,
2
2
(x1 + x2 )2
2
|hx, ei| =
,
2
and by (2.11) we get
2
2
x21
x22
(x1 − x2 )2
(A − a)2 (x1 + x2 )2
≤k·
·
.
2
aA
2
Now, if we let x1 = √a2 , x2 = √A2 (A > a > 0) , then obviously
2 X
A
a
√ − xi
hAe − x, x − aei =
xi − √
= 0,
2
2
i=1
(2.12)
which shows that the condition (2.9) is fulfilled, and by (2.12) we get
(A − a)2
(A − a)2 (a + A)2
≤k·
·
4
aA
4
for any A > a > 0. This implies
(2.13)
(A + a)2 k ≥ aA
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 2.3. If hx, ei , hy, ei are assumed not to be zero, then the inequality (2.3) is equivalent
to
hx, yi
1
≤ · q|A − a| |B − b| ,
(2.14)
−
1
hx, ei he, yi
4
Re (āA) Re b̄B
while the inequality (2.10) is equivalent to
hx, yi
1 (A − a) (B − b)
≤ ·
√
(2.15)
−
1
.
hx, ei he, yi
4
abAB
J. Inequal. Pure and Appl. Math., 4(5) Art. 87, 2003
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G RÜSS T YPE I NEQUALITIES
The constant
1
4
5
is best possible in both inequalities.
3. S OME R ELATED R ESULTS
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
Re hΓe − (λx + (1 − λ) y) , (λx + (1 − λ) y) − γei ≥ 0,
(3.1)
or, equivalently,
(3.2)
λx + (1 − λ) y − γ + Γ e ≤ 1 |Γ − γ| ,
2 2
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
1
(3.4)
Re ha, bi ≤
kλa + (1 − λ) bk2 .
4λ (1 − λ)
Since
hx, yi − hx, ei he, yi = hx − hx, ei e, y − hy, ei ei
using (3.4), we have
(3.5)
(as kek = 1),
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)
km − hm, ei ek2 = kmk2 − |hm, ei|2 ,
then by (3.5) we deduce the inequality
(3.7)
Re [hx, yi − hx, ei he, yi]
1
≤
kλx + (1 − λ) yk2 − |hλx + (1 − λ) y, ei|2 .
4λ (1 − λ)
Now, if we apply Grüss’ inequality
0 ≤ kak2 − |ha, ei|2 ≤
J. Inequal. Pure and Appl. Math., 4(5) Art. 87, 2003
1
|Γ − γ|2
4
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S.S. D RAGOMIR
provided
Re hΓe − a, a − γei ≥ 0,
for a = λx + (1 − λ) y, we have
1
|Γ − γ|2 .
4
Utilising (3.7) and (3.8) we deduce the desired inequality (3.3). To prove the sharpness of the
1
constant 16
, assume that (3.3) holds with a constant C > 0, provided (3.1) is valid, i.e.,
kλx + (1 − λ) yk2 − |hλx + (1 − λ) y, ei|2 ≤
(3.8)
Re [hx, yi − hx, ei he, yi] ≤ C ·
(3.9)
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
kxk2 − |hx, ei|2 ≤ C ·
(3.10)
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 = 21 , C ≥ 16
.
The theorem is completely proved.
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
|Re [hx, yi − hx, ei he, yi]| ≤
(3.13)
The constant
1
16
1
1
·
|Γ − γ|2 .
16 λ (1 − λ)
is best possible in (3.13).
Proof. Using Theorem 3.1 for (−y) instead of y, we have that
Re hΓe − (λx − (1 − λ) y) , (λx − (1 − λ) y) − γei ≥ 0,
which implies that
Re [− hx, yi + hx, ei he, yi] ≤
1
1
·
|Γ − γ|2
16 λ (1 − λ)
Re [hx, yi − hx, ei he, yi] ≥ −
1
1
·
|Γ − γ|2 .
16 λ (1 − λ)
giving
(3.14)
Consequently, by (3.3) and (3.14) we deduce the desired inequality (3.13).
J. Inequal. Pure and Appl. Math., 4(5) Art. 87, 2003
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G RÜSS T YPE I NEQUALITIES
7
Remark 3.3. If M, m ∈ R with M > m and, for λ ∈ (0, 1) ,
1
M
+
m
λx + (1 − λ) y −
≤ (M − m)
(3.15)
e
2
2
then
1
1
·
(M − m)2 .
16 λ (1 − λ)
If (3.15) holds with “±” instead of “+” , then
1
1
(3.16)
|hx, yi − hx, ei he, yi| ≤
·
(M − m)2 .
16 λ (1 − λ)
hx, yi − hx, ei he, yi ≤
Remark 3.4. If λ =
(3.17)
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
or, equivalently
x + y γ + Γ 1
2 − 2 e ≤ 2 |Γ − γ|
(3.18)
implies
Re [hx, yi − hx, ei he, yi] ≤
(3.19)
The constant
1
4
1
|Γ − γ|2 .
4
is best possible in (3.19).
For λ = 12 , Corollary 3.2 and Remark 3.3 will produce the corresponding results obtained in
[3]. We omit the details.
4. I NTEGRAL I NEQUALITIES
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. If f, g, h ∈ L2 (Ω, K) and ϕ, Φ, γ, Γ ∈ K, are so that Re (Φϕ) > 0, Re (Γγ) >
R
2
0, Ω |h (s)| dµ (s) = 1 and
Z
i
h
(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)
! 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
J. Inequal. Pure and Appl. Math., 4(5) Art. 87, 2003
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8
S.S. D RAGOMIR
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
ZΩ
Z
1
|Φ − ϕ| |Γ − γ| f (s) h (s)dµ (s) h (s) g (s)dµ (s) .
≤ ·p
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
We omit the details.
Remark 4.2. 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
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 4.3. If z, Z, t, T ∈ K, with Re (z̄Z) , Re (t̄T ) > 0, µ (Ω) < ∞ and f, g ∈ L2 (Ω, K)
are such that:
i
h
(4.4)
Re (Z − f (s)) f (s) − z̄ ≥ 0,
h
i
Re (T − g (s)) g (s) − t̄ ≥ 0 for a.e. s ∈ Ω
or, equivalently
(4.5)
f (s) − z + Z ≤
2 t
+
T
g (s) −
≤
2 1
|Z − z| ,
2
1
|T − t| for a.e. s ∈ Ω;
2
then we have the inequality
Z
Z
Z
1
1
1
(4.6) f (s) g (s)dµ (s) −
f (s) dµ (s) ·
g (s)dµ (s)
µ (Ω) Ω
µ (Ω) Ω
µ (Ω) Ω
Z
Z
1
|Z − z| |T − t| 1
1
≤ ·p
f (s) dµ (s) ·
g (s)dµ (s) .
4
µ (Ω) Ω
Re (z̄Z) Re (t̄T ) µ (Ω) Ω
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G RÜSS T YPE I NEQUALITIES
9
Remark 4.4. 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)
R
√
(4.7)
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.
R
Proposition 4.5. If f, g, h ∈ L2 (Ω, K) and γ, Γ ∈ K are such that Ω |h (s)|2 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)
! 12
2
Z γ
+
Γ
1
λf (s) + (1 − λ) g (s) −
h (s) dµ (s)
≤ |Γ − γ| ,
2
2
Ω
then we have the inequality
Z
h
i
(4.10)
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
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
(4.11)
|I| ≤
1
1
·
|Γ − γ|2 .
16 λ (1 − λ)
Remark 4.6. It is obvious that a sufficient condition for (4.8) to hold is
n
h
io
(4.12) Re [Γh (s) − (λf (s) + (1 − λ) g (s))] · 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.
Corollary 4.7. If Z, z ∈ K, µ (Ω) < ∞ and f, g ∈ L2 (Ω, K) are such that
h
i
(4.14)
Re (Z − (λf (s) + (1 − λ) g (s))) λf (s) + (1 − λ) g (s) − z ≥ 0
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10
S.S. D RAGOMIR
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
− Re
f (s) dµ (s) ·
g (s)dµ (s)
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
1
1
(4.16)
|J| ≤
·
|Z − z|2 .
16 λ (1 − λ)
Remark 4.8. 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.
R EFERENCES
[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].
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http://jipam.vu.edu.au/