Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 60, 2004
REFINEMENTS OF THE SCHWARZ AND HEISENBERG INEQUALITIES IN
HILBERT 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 VIC 8001
AUSTRALIA .
sever.dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 24 August, 2004; accepted 17 September, 2004
Communicated by S. Saitoh
A BSTRACT. Some new refinements of the Schwarz inequality in inner product spaces are given.
Applications for discrete and integral inequalities including the Heisenberg inequality for vectorvalued functions in Hilbert spaces are provided.
Key words and phrases: Schwarz inequality, Triangle inequality, Heisenberg Inequality.
2000 Mathematics Subject Classification. Primary 46C05; Secondary 26D15.
1. I NTRODUCTION
Let (H; h·, ·i) be an inner product space over the real or complex number field K. One of the
most important inequalities in inner product spaces with numerous applications, is the Schwarz
inequality
(1.1)
|hx, yi|2 ≤ kxk2 kyk2 ,
x, y ∈ H
with equality iff x and y are linearly dependent.
In 1966, S. Kurepa [1] established the following refinement of the Schwarz inequality in inner
product spaces that generalises de Bruijn’s result for sequences of real and complex numbers
[2].
Theorem 1.1. Let H be a real Hilbert space and HC the complexification of H. Then for any
pair of vectors a ∈ H, z ∈ HC
1
(1.2)
|hz, ai|2 ≤ kak2 kzk2 + |hz, z̄i| ≤ kak2 kzk2 .
2
In 1985, S.S. Dragomir [3, Theorem 2] obtained a different refinement of (1.1), namely:
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
156-04
2
S.S. D RAGOMIR
Theorem 1.2. Let (H; h·, ·i) be a real or complex inner product space and x, y, e ∈ H with
kek = 1. Then we have the inequality
kxk kyk ≥ |hx, yi − hx, ei he, yi| + |hx, ei he, yi| ≥ |hx, yi| .
(1.3)
In the same paper [3, Theorem 3], a further generalisation for orthonormal families has been
given (see also [4, Theorem 3]).
Theorem 1.3. Let {ei }i∈H be an orthonormal family in the Hilbert space H. Then for any
x, y ∈ H
X
X
(1.4)
kxk kyk ≥ hx, yi −
hx, ei i hei , yi +
|hx, ei i hei , yi|
i∈I
i∈I
X
X
≥ hx, yi −
hx, ei i hei , yi + hx, ei i hei , yi
i∈I
i∈I
≥ |hx, yi| .
The inequality (1.3) has also been obtained in [4] as a particular case of the following result.
Theorem 1.4. Let x, y, a, b ∈ H be such that
kak2 ≤ 2 Re hx, ai ,
kbk2 ≤ 2 Re hy, bi .
Then we have:
kxk kyk ≥ 2 Re hx, ai − kak2
(1.5)
21
2 Re hy, bi − kbk2
12
+ |hx, yi − hx, bi − ha, yi + ha, bi| .
Another refinement of the Schwarz inequality for orthornormal vectors in inner product
spaces has been obtained by S.S. Dragomir and J. Sándor in [5, Theorem 5].
Theorem 1.5. Let {ei }i∈{1,...,n} be orthornormal vectors in the inner product space (H; h·, ·i).
Then
! 21 n
n
n
X
X
X
|hy, ei i|2
(1.6)
kxk kyk − |hx, yi| ≥
|hx, ei i|2
−
hx, ei i hei , yi ≥ 0
i=1
i=1
i=1
and
(1.7)
kxk kyk − Re hx, yi ≥
n
X
n
X
2
|hx, ei i|
|hy, ei i|2
i=1
i=1
! 12
−
n
X
Re [hx, ei i hei , yi] ≥ 0.
i=1
For some properties of superadditivity, monotonicity, strong superadditivity and strong monotonicity of Schwarz’s inequality, see [6]. Here we note only the following refinements of the
Schwarz inequality in its different variants for linear operators [6]:
a) Let H be a Hilbert space and A, B : H → H two selfadjoint linear operators with
A ≥ B ≥ 0, then we have the inequalities
1
1
1
1
hAx, xi 2 hAy, yi 2 − |hAx, yi| ≥ hBx, xi 2 hBy, yi 2 − |hBx, yi| ≥ 0
(1.8)
and
(1.9)
hAx, xi hAy, yi − |hAx, yi|2 ≥ hBx, xi hBy, yi − |hBx, yi|2 ≥ 0
for any x, y ∈ H.
J. Inequal. Pure and Appl. Math., 5(3) Art. 60, 2004
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S CHWARZ AND H EISENBERG I NEQUALITIES IN H ILBERT S PACES
3
b) Let A : H → H be a bounded linear operator on H and let kAk = sup {kAxk , kxk = 1}
the norm of A. Then one has the inequalities
kAk2 (kxk kyk − |hx, yi|) ≥ kAxk kAyk − |hAx, Ayi| ≥ 0
(1.10)
and
kAk4 kxk2 kyk2 − |hx, yi|2 ≥ kAxk2 kAyk2 − |hAx, Ayi|2 ≥ 0.
(1.11)
c) Let B : H → H be a linear operator with the property that there exists a constant
m > 0 such that kBxk ≥ m kxk for any x ∈ H. Then we have the inequalities
kBxk kByk − |hBx, Byi| ≥ m2 (kxk kyk − |hx, yi|) ≥ 0
(1.12)
and
(1.13)
kBxk2 kByk2 − |hBx, Byi|2 ≥ m4 kxk2 kyk2 − |hx, yi|2 ≥ 0.
For other results related to Schwarz’s inequality in inner product spaces, see Chapter XX of
[8] and the references therein.
Motivated by the results outlined above, it is the aim of this paper to explore other avenues
in obtaining new refinements of the celebrated Schwarz inequality. Applications for vectorvalued sequences and integrals in Hilbert spaces are mentioned. Refinements of the Heisenberg
inequality for vector-valued functions in Hilbert spaces are also given.
2. S OME N EW R EFINEMENTS
The following result holds.
Theorem 2.1. Let (H; h·, ·i) be an inner product space over the real or complex number field
K and r1 , r2 > 0. If x, y ∈ H satisfy the property
(2.1)
kx − yk ≥ r2 ≥ r1 ≥ |kxk − kyk| ,
then we have the following refinement of Schwarz’s inequality
1 2
(2.2)
kxk kyk − Re hx, yi ≥
r2 − r12 (≥ 0) .
2
1
The constant 2 is best possible in the sense that it cannot be replaced by any larger quantity.
Proof. From the first inequality in (2.1) we have
(2.3)
kxk2 + kyk2 ≥ r22 + 2 Re hx, yi .
Subtracting in (2.3) the quantity 2 kxk kyk , we get
(2.4)
(kxk − kyk)2 ≥ r22 − 2 (kxk kyk − Re hx, yi) .
Since, by the second inequality in (2.1) we have
(2.5)
r12 ≥ (kxk − kyk)2 ,
hence from (2.4) and (2.5) we deduce the desired inequality (2.2).
To prove the sharpness of the constant 21 in (2.2), let us assume that there is a constant C > 0
such that
(2.6)
kxk kyk − Re hx, yi ≥ C r22 − r12 ,
provided that x and y satisfy (2.1).
Let e ∈ H with kek = 1 and for r2 > r1 > 0, define
r2 + r1
r 1 − r2
(2.7)
x=
· e and y =
· e.
2
2
J. Inequal. Pure and Appl. Math., 5(3) Art. 60, 2004
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4
S.S. D RAGOMIR
Then
kx − yk = r2 and |kxk − kyk| = r1 ,
showing that the condition (2.1) is fulfilled with equality.
If we replace x and y as defined in (2.7) into the inequality (2.6), then we get
r22 − r12
≥ C r22 − r12 ,
2
1
which implies that C ≤ 2 , and the theorem is completely proved.
The following corollary holds.
Corollary 2.2. With the assumptions of Theorem 2.1, we have the inequality:
√
√ q
2
2
(2.8)
kxk + kyk −
kx + yk ≥
r22 − r12 .
2
2
Proof. We have, by (2.2), that
(kxk + kyk)2 − kx + yk2 = 2 (kxk kyk − Re hx, yi) ≥ r22 − r12 ≥ 0
which gives
2
2
(kxk + kyk) ≥ kx + yk +
(2.9)
By making use of the elementary inequality
2 α2 + β 2 ≥ (α + β)2 ,
q
2
r22
−
r12
.
α, β ≥ 0;
we get
q
2
2
q
1
2
2
2
2
r 2 − r1 ≥
kx + yk + r2 − r1 .
(2.10)
kx + yk +
2
Utilising (2.9) and (2.10), we deduce the desired inequality (2.8).
2
If (H; h·, ·i) is a Hilbert space and {ei }i∈I is an orthornormal family in H, i.e., we recall
that hei , ej i = δij for any i, j ∈ I, where δij is Kronecker’s delta, then we have the following
inequality which is well known in the literature as Bessel’s inequality
X
(2.11)
|hx, ei i|2 ≤ kxk2 for each x ∈ H.
i∈I
Here, the meaning of the sum is
X
|hx, ei i|2 = sup
F ⊂I
i∈I
(
X
)
|hx, ei i|2 , F is a finite part of I
.
i∈F
The following result providing a refinement of the Bessel inequality (2.11) holds.
Theorem 2.3. Let (H; h·, ·i) be a Hilbert space and {ei }i∈I an orthornormal family in H. If
x ∈ H, x 6= 0, and r2 , r1 > 0 are such that:
! 21
X
X
(2.12)
hx, ei i ei ≥ r2 ≥ r1 ≥ kxk −
|hx, ei i|2
(≥ 0) ,
x −
i∈I
i∈I
then we have the inequality
! 12
(2.13)
kxk −
X
|hx, ei i|2
i∈I
J. Inequal. Pure and Appl. Math., 5(3) Art. 60, 2004
≥
1
r2 − r12
· P 2
1 (≥ 0) .
2 2
2
i∈I |hx, ei i|
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S CHWARZ AND H EISENBERG I NEQUALITIES IN H ILBERT S PACES
The constant
5
1
2
is best possible.
P
Proof. Consider y := i∈I hx, ei i ei . Obviously, since H is a Hilbert space, y ∈ H. We also
note that
2 s
v
X
X
u
X
u
t
hx, ei i ei =
|hx, ei i|2 ,
kyk = hx, ei i ei = i∈I
i∈I
i∈I
and thus (2.12) is in fact (2.1) of Theorem 2.1.
Since
! 21
kxk kyk − Re hx, yi = kxk
X
2
*
− Re x,
|hx, ei i|
+
X
i∈I
hx, ei i ei
i∈I
! 12 
=
X
|hx, ei i|2
! 12 
kxk −
i∈I
X
|hx, ei i|2
,
i∈I
hence, by (2.2), we deduce the desired result (2.13).
We will prove the sharpness of the constant for the case of one element, i.e., I = {1} ,
e1 = e ∈ H, kek = 1. For this, assume that there exists a constant D > 0 such that
(2.14)
kxk − |hx, ei| ≥ D ·
r22 − r12
|hx, ei|
provided x ∈ H\ {0} satisfies the condition
(2.15)
kx − hx, ei ek ≥ r2 ≥ r1 ≥ kxk − |hx, ei| .
Assume that x = λe + µf with e, f ∈ H, kek = kf k = 1 and e ⊥ f. We wish to see if there
exists positive numbers λ, µ such that
(2.16)
kx − hx, ei ek = r2 > r1 = kxk − |hx, ei| .
Since (for λ, µ > 0)
kx − hx, ei ek = µ
and
p
kxk − |hx, ei| = λ2 + µ2 − λ
hence, by (2.16), we get µ = r2 and
q
λ2 + r22 − λ = r1
giving
λ2 + r22 = λ2 + 2λr1 + r12
from where we get
λ=
r22 − r12
> 0.
2r1
With these values for λ and µ, we have
kxk − |hx, ei| = r1 ,
|hx, ei| =
r22 − r12
2r1
and thus, from (2.14), we deduce
r1 ≥ D ·
giving D ≤ 21 . This proves the theorem.
J. Inequal. Pure and Appl. Math., 5(3) Art. 60, 2004
r22 − r12
r22 −r12
2r1
,
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6
S.S. D RAGOMIR
The following corollary is obvious.
Corollary 2.4. Let x, y ∈ H with hx, yi =
6 0 and r2 ≥ r1 > 0 such that
hx, yi (2.17)
kyk x − kyk · y ≥ r2 kyk ≥ r1 kyk
≥ kxk kyk − |hx, yi| (≥ 0) .
Then we have the following refinement of the Schwarz’s inequality:
kxk kyk − |hx, yi| ≥
(2.18)
The constant
1
2
kyk2
1 2
r2 − r12
(≥ 0) .
2
|hx, yi|
is best possible.
The following lemma holds.
Lemma 2.5. Let (H; h·, ·i) be an inner product space and R ≥ 1. For x, y ∈ H, the subsequent
statements are equivalent:
(i) The following refinement of the triangle inequality holds:
(2.19)
kxk + kyk ≥ R kx + yk ;
(ii) The following refinement of the Schwarz inequality holds:
1 2
(2.20)
kxk kyk − Re hx, yi ≥
R − 1 kx + yk2 .
2
Proof. Taking the square in (2.19), we have
(2.21)
2 kxk kyk ≥ R2 − 1 kxk2 + 2R2 Re hx, yi + R2 − 1 kyk2 .
Subtracting from both sides of (2.21) the quantity 2 Re hx, yi , we obtain
2 (kxk kyk − Re hx, yi) ≥ R2 − 1 kxk2 + 2 Re hx, yi + kyk2
= R2 − 1 kx + yk2 ,
which is clearly equivalent to (2.20).
By the use of the above lemma, we may now state the following theorem concerning another
refinement of the Schwarz inequality.
Theorem 2.6. Let (H; h·, ·i) be an inner product space over the real or complex number field
and R ≥ 1, r ≥ 0. If x, y ∈ H are such that
1
(2.22)
(kxk + kyk) ≥ kx + yk ≥ r,
R
then we have the following refinement of the Schwarz inequality
1 2
(2.23)
kxk kyk − Re hx, yi ≥
R − 1 r2 .
2
1
The constant 2 is best possible in the sense that it cannot be replaced by a larger quantity.
Proof. The inequality (2.23) follows easily from Lemma 2.5. We need only prove that
best possible constant in (2.23).
Assume that there exists a C > 0 such that
(2.24)
kxk kyk − Re hx, yi ≥ C R2 − 1 r2
1
2
is the
provided x, y, R and r satisfy (2.22).
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S CHWARZ AND H EISENBERG I NEQUALITIES IN H ILBERT S PACES
Consider r = 1, R > 1 and choose x =
1−R
e,
2
y=
1+R
e
2
7
with e ∈ H, kek = 1. Then
kxk + kyk
=1
R
and thus (2.22) holds with equality on both sides.
From (2.24), for the above choices, we have 21 (R2 − 1) ≥ C (R2 − 1) , which shows that
C ≤ 12 .
x + y = e,
Finally, the following result also holds.
Theorem 2.7. Let (H; h·, ·i) be an inner product space over the real or complex number field
K and r ∈ (0, 1]. For x, y ∈ H, the following statements are equivalent:
(i) We have the inequality
|kxk − kyk| ≤ r kx − yk ;
(2.25)
(ii) We have the following refinement of the Schwarz inequality
1
(2.26)
kxk kyk − Re hx, yi ≥
1 − r2 kx − yk2 .
2
1
The constant 2 in (2.26) is best possible.
Proof. Taking the square in (2.25), we have
kxk2 − 2 kxk kyk + kyk2 ≤ r2 kxk2 − 2 Re hx, yi + kyk2
which is clearly equivalent to
1 − r2 kxk2 − 2 Re hx, yi + kyk2 ≤ 2 (kxk kyk − Re hx, yi)
or with (2.26).
Now, assume that (2.26) holds with a constant E > 0, i.e.,
(2.27)
kxk kyk − Re hx, yi ≥ E 1 − r2 kx − yk2 ,
provided (2.25) holds.
Define x = r+1
e, y =
2
r−1
e
2
with e ∈ H, kek = 1. Then
|kxk − kyk| = r,
kx − yk = 1
showing that (2.25) holds with equality.
If we replace x and y in (2.27), then we get E (1 − r2 ) ≤
1
.
2
1
2
(1 − r2 ) , implying that E ≤
3. D ISCRETE I NEQUALITIES
Assume that (K; (·, ·))Pis a Hilbert space over the real or complex number field. Assume also
that pi ≥ 0, i ∈ H with ∞
i=1 pi = 1 and define
(
)
∞
X
2
`2p (K) := x := (xi ) xi ∈ K, i ∈ N and
pi kxi k < ∞ .
i∈N
i=1
It is well known that `2p (K) endowed with the inner product h·, ·ip defined by
hx, yip :=
∞
X
pi (xi , yi )
i=1
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8
S.S. D RAGOMIR
and generating the norm
kxkp :=
∞
X
! 21
pi kxi k
2
i=1
is a Hilbert space over K.
We may state the following discrete inequality improving the Cauchy-Bunyakovsky-Schwarz
classical result.
P
Proposition 3.1. Let (K; (·, ·)) be a Hilbert space and pi ≥ 0 (i ∈ N) with ∞
i=1 pi = 1.
Assume that x, y ∈ `2p (K) and r1 , r2 > 0 satisfy the condition
kxi − yi k ≥ r2 ≥ r1 ≥ |kxi k − kyi k|
(3.1)
for each i ∈ N. Then we have the following refinement of the Cauchy-Bunyakovsky-Schwarz
inequality
! 12
∞
∞
∞
X
X
X
1 2
(3.2)
pi kxi k2
pi kyi k2
−
pi Re (xi , yi ) ≥
r2 − r12 ≥ 0.
2
i=1
i=1
i=1
The constant
1
2
is best possible.
Proof. From the condition (3.1) we simply deduce
(3.3)
∞
X
2
pi kxi − yi k ≥
r22
≥
r12
i=1
≥
∞
X
pi (kxi k − kyi k)2
i=1

≥
∞
X
i=1
! 12
pi kxi k2
−
∞
X
! 12 2
pi kyi k2
 .
i=1
In terms of the norm k·kp , the inequality (3.3) may be written as
(3.4)
kx − ykp ≥ r2 ≥ r1 ≥ kxkp − kykp .
2
Utilising Theorem 2.1 for the Hilbert space `p (K) , h·, ·ip , we deduce the desired inequality
(3.2).
For n = 1 (p1 = 1) , the inequality (3.2) reduces to (2.2) for which we have shown that 21 is
the best possible constant.
By the use of Corollary 2.2, we may state the following result as well.
Corollary 3.2. With the assumptions of Proposition 3.1, we have the inequality
! 21
! 12 √
! 12
√ q
∞
∞
∞
X
X
X
2
2
2
2
2
(3.5)
pi kxi k
pi kxi + yi k
+
pi kyi k
−
≥
r22 − r12 .
2
2
i=1
i=1
i=1
The following proposition also holds.
Proposition 3.3. Let (K; (·, ·)) be a Hilbert space and pi ≥ 0 (i ∈ N) with
Assume that x, y ∈ `2p (K) and R ≥ 1, r ≥ 0 satisfy the condition
(3.6)
P∞
i=1
pi = 1.
1
(kxi k + kyi k) ≥ kxi + yi k ≥ r
R
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S CHWARZ AND H EISENBERG I NEQUALITIES IN H ILBERT S PACES
9
for each i ∈ N. Then we have the following refinement of the Schwarz inequality
! 12
∞
∞
∞
X
X
X
1 2
(3.7)
pi kxi k2
pi kyi k2
−
pi Re (xi , yi ) ≥
R − 1 r2 .
2
i=1
i=1
i=1
The constant
1
2
is best possible in the sense that it cannot be replaced by a larger quantity.
Proof. By (3.6) we deduce
"∞
# 12
1 X
2
≥
(3.8)
pi (kxi k + kyi k)
R i=1
∞
X
! 12
pi kxi + yi k
2
≥ r.
i=1
By the classical Minkowsky inequality for nonnegative numbers, we have
! 12
! 12 " ∞
# 12
∞
∞
X
X
X
(3.9)
pi kxi k2
+
pi kyi k2
≥
pi (kxi k + kyi k)2 ,
i=1
i=1
i=1
and thus, by utilising (3.8) and (3.9), we may state in terms of k·kp the following inequality
1
(3.10)
kxkp + kykp ≥ kx + ykp ≥ r.
R
Employing Theorem 2.6 for the Hilbert space `2p (K) and the inequality (3.10), we deduce the
desired result (3.7).
Since, for p = 1, n = 1, (3.7) is reduced to (2.23) for which we have shown that 21 is the best
constant, we conclude that 21 is the best constant in (3.7) as well.
Finally, we may state and prove the following result incorporated in
Proposition 3.4. Let (K; (·, ·)) be a Hilbert space and pi ≥ 0 (i ∈ N) with
Assume that x, y ∈ `2p (K) and r ∈ (0, 1] such that
P∞
i=1
pi = 1.
|kxi k − kyi k| ≤ r kxi − yi k for each i ∈ N,
(3.11)
holds true. Then we have the following refinement of the Schwarz inequality
! 12
∞
∞
∞
∞
X
X
X
X
1
2
2
2
(3.12)
pi kxi k
pi kyi k
−
pi Re (xi , yi ) ≥
1−r
pi kxi − yi k2 .
2
i=1
i=1
i=1
i=1
The constant
1
2
is best possible in (3.12).
Proof. From (3.11) we have
# 12
# 21
"∞
"∞
X
X
2
2
≤r
pi kxi − yi k
.
pi (kxi k − kyi k)
i=1
i=1
Utilising the following elementary result
! 12 ! 12
∞
∞
X
X
2
2
≤
−
p
ky
k
p
kx
k
i
i
i
i
i=1
i=1
∞
X
! 12
pi (kxi k − kyi k)2
,
i=1
we may state that
kxkp − kykp ≤ r kx − ykp .
Now, by making use of Theorem 2.7, we deduce the desired inequality (3.12) and the fact that
1
is the best possible constant. We omit the details.
2
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10
S.S. D RAGOMIR
4. I NTEGRAL I NEQUALITIES
Assume that (K; (·, ·)) is a Hilbert space over the real or complex number field K. If ρ :
Rb
[a, b] ⊂ R → [0, ∞) is a Lebesgue integrable function with a ρ (t) dt = 1, then we may
consider the space L2ρ ([a, b] ; K) of all functions f : [a, b] → K, that are Bochner measurable
Rb
and a ρ (t) kf (t)k2 dt < ∞. It is known that L2ρ ([a, b] ; K) endowed with the inner product
h·, ·iρ defined by
Z b
hf, giρ :=
ρ (t) (f (t) , g (t)) dt
a
and generating the norm
Z
kf kρ :=
b
21
ρ (t) kf (t)k dt
2
a
is a Hilbert space over K.
Now we may state and prove the first refinement of the Cauchy-Bunyakovsky-Schwarz integral inequality.
Proposition 4.1. Assume that f, g ∈ L2ρ ([a, b] ; K) and r2 , r1 > 0 satisfy the condition
kf (t) − g (t)k ≥ r2 ≥ r1 ≥ |kf (t)k − kg (t)k|
(4.1)
for a.e. t ∈ [a, b] . Then we have the inequality
b
Z
2
b
Z
ρ (t) kf (t)k dt
(4.2)
a
21
ρ (t) kg (t)k dt
2
a
Z
−
b
ρ (t) Re (f (t) , g (t)) dt ≥
a
The constant
1
2
1 2
r2 − r12 (≥ 0) .
2
is best possible in (4.2).
Proof. Integrating (4.1), we get
b
Z
12
ρ (t) (kf (t) − g (t)k) dt
2
(4.3)
a
Z
≥ r2 ≥ r1 ≥
b
12
ρ (t) (kf (t)k − kg (t)k) dt .
2
a
Utilising the obvious fact
Z
(4.4)
a
b
12
ρ (t) (kf (t)k − kg (t)k)2 dt
Z
21 21 Z b
b
−
ρ (t) kg (t)k2 dt ,
≥
ρ (t) kf (t)k2 dt
a
a
we can state the following inequality in terms of the k·kρ norm:
(4.5)
kf − gkρ ≥ r2 ≥ r1 ≥ kf kρ − kgkρ .
Employing Theorem 2.1 for the Hilbert space L2ρ ([a, b] ; K) , we deduce the desired inequality
(4.2).
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S CHWARZ AND H EISENBERG I NEQUALITIES IN H ILBERT S PACES
11
To prove the sharpness of 21 in (4.2), we choose a = 0, b = 1, f (t) = 1, t ∈ [0, 1] and
f (t) = x, g (t) = y, t ∈ [a, b] , x, y ∈ K. Then (4.2) becomes
kxk kyk − Re hx, yi ≥
1 2
r2 − r12
2
provided
kx − yk ≥ r2 ≥ r1 ≥ |kxk − kyk| ,
1
2
which, by Theorem 2.1 has the quantity
as the best possible constant.
The following corollary holds.
Corollary 4.2. With the assumptions of Proposition 4.1, we have the inequality
Z
b
12
21 Z b
2
ρ (t) kf (t)k dt
+
ρ (t) kg (t)k dt
2
(4.6)
a
a
√ Z b
√ q
12
2
2
2
−
ρ (t) kf (t) + g (t)k dt
≥
r22 − r12 .
2
2
a
The following two refinements of the Cauchy-Bunyakovsky-Schwarz (CBS) integral inequality also hold.
Proposition 4.3. If f, g ∈ L2ρ ([a, b] ; K) and R ≥ 1, r ≥ 0 satisfy the condition
1
(kf (t)k + kg (t)k) ≥ kf (t) + g (t)k ≥ r
R
(4.7)
for a.e. t ∈ [a, b] , then we have the inequality
Z
b
Z
2
ρ (t) kf (t)k dt
(4.8)
a
b
21
ρ (t) kg (t)k dt
2
a
Z
b
−
ρ (t) Re (f (t) , g (t)) dt ≥
a
1
2
The constant
1 2
R − 1 r2 .
2
is best possible in (4.8).
The proof follows by Theorem 2.6 and we omit the details.
Proposition 4.4. If f, g ∈ L2ρ ([a, b] ; K) and ζ ∈ (0, 1] satisfy the condition
|kf (t)k − kg (t)k| ≤ ζ kf (t) − g (t)k
(4.9)
for a.e. t ∈ [a, b] , then we have the inequality
Z
b
2
Z
ρ (t) kf (t)k dt
(4.10)
a
21
ρ (t) kg (t)k dt
2
a
Z
−
a
The constant
b
1
2
b
1
ρ (t) Re (f (t) , g (t)) dt ≥
1 − ζ2
2
Z
b
ρ (t) kf (t) − g (t)k2 dt.
a
is best possible in (4.10).
The proof follows by Theorem 2.7 and we omit the details.
J. Inequal. Pure and Appl. Math., 5(3) Art. 60, 2004
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12
S.S. D RAGOMIR
5. R EFINEMENTS OF H EISENBERG I NEQUALITY
It is well known that if (H; h·, ·i) is a real or complex Hilbert space and f : [a, b] ⊂ R →H
is an absolutely continuous vector-valued function, then f is differentiable almost everywhere
on [a, b] , the derivative f 0 : [a, b] → H is Bochner integrable on [a, b] and
Z t
(5.1)
f (t) =
f 0 (s) ds
for any t ∈ [a, b] .
a
The following theorem provides a version of the Heisenberg inequalities in the general setting
of Hilbert spaces.
Theorem 5.1. Let ϕ : [a, b] → H be an absolutely continuous function with the property that
b kϕ (b)k2 = a kϕ (a)k2 . Then we have the inequality:
Z b
2
Z b
Z b
2
2
2
2
(5.2)
kϕ (t)k dt ≤ 4
t kϕ (t)k dt ·
kϕ0 (t)k dt.
a
a
a
The constant 4 is best possible in the sense that it cannot be replaced by any smaller constant.
Proof. Integrating by parts, we have successively
b Z b
Z b
d
2
2
kϕ (t)k dt = t kϕ (t)k −
t
(5.3)
kϕ (t)k2 dt
dt
a
a
a
Z b
d
2
2
= b kϕ (b)k − a kϕ (a)k −
t hϕ (t) , ϕ (t)i dt
dt
a
Z b
=−
t [hϕ0 (t) , ϕ (t)i + hϕ (t) , ϕ0 (t)i] dt
a
Z b
= −2
t Re hϕ0 (t) , ϕ (t)i dt
a
Z b
=2
Re hϕ0 (t) , (−t) ϕ (t)i dt.
a
If we apply the Cauchy-Bunyakovsky-Schwarz integral inequality
Z b
12
Z b
Z b
2
2
Re hg (t) , h (t)i dt ≤
kg (t)k dt
kh (t)k dt
a
a
a
0
for g (t) = ϕ (t) , h (t) = −tϕ (t) , t ∈ [a, b] , then we deduce the desired inequality (4.5).
The fact that 4 is the best constant in (4.5) follows from the fact that in the (CBS) inequality,
the case of equality holds iff g (t) = λh (t) for a.e. t ∈ [a, b] and λ a given scalar in K. We omit
the details.
For details on the classical Heisenberg inequality, see, for instance, [7].
Utilising Proposition 4.1, we can state the following refinement of the Heisenberg inequality
obtained above in (5.2):
Proposition 5.2. Assume that ϕ : [a, b] → H is as in the hypothesis of Theorem 5.1. In addition,
if there exist r2 , r1 > 0 so that
kϕ0 (t) + tϕ (t)k ≥ r2 ≥ r1 ≥ |kϕ0 (t)k − |t| kϕ (t)k|
for a.e. t ∈ [a, b] , then we have the inequality
Z b
21
Z b
Z
1 b
1
2
2
2
0
−
kϕ (t)k2 dt ≥ (b − a) r22 − r12 (≥ 0) .
t kϕ (t)k dt ·
kϕ (t)k dt
2 a
2
a
a
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S CHWARZ AND H EISENBERG I NEQUALITIES IN H ILBERT S PACES
13
The proof follows by Proposition 4.1 on choosing f (t) = ϕ0 (t) , g (t) = −tϕ (t) and ρ (t) =
∈ [a, b] .
On utilising Proposition 4.3 for the same choices of f, g and ρ, we may state the following
results as well:
1
,t
b−a
Proposition 5.3. Assume that ϕ : [a, b] → H is as in the hypothesis of Theorem 5.1. In addition,
if there exist R ≥ 1 and r > 0 so that
1
(kϕ0 (t)k + |t| kϕ (t)k) ≥ kϕ0 (t) − tϕ (t)k ≥ r
R
for a.e. t ∈ [a, b] , then we have the inequality
Z
b
2
t2 kϕ (t)k dt ·
a
Z
b
a
21
Z
1 b
0
kϕ (t)k dt
−
kϕ (t)k2 dt
2 a
1
≥ (b − a) R2 − 1 r2 (≥ 0) .
2
2
Finally, we can state
Proposition 5.4. Let ϕ : [a, b] → H be as in the hypothesis of Theorem 5.1. In addition, if there
exists ζ ∈ (0, 1] so that
|kϕ0 (t)k − |t| kϕ (t)k| ≤ ζ kϕ0 (t) + tϕ (t)k
for a.e. t ∈ [a, b] , then we have the inequality
Z
a
b
2
t2 kϕ (t)k dt ·
Z
a
b
12
Z
1 b
0
kϕ (t)k dt
−
kϕ (t)k2 dt
2 a
Z
b 0
1
2
2
≥
1−ζ
kϕ (t) + tϕ (t)k dt (≥ 0) .
2
a
2
This follows by Proposition 4.4 and we omit the details.
R EFERENCES
[1] S. KUREPA, On the Buniakowsky-Cauchy-Schwarz inequality, Glasnik Mat. Ser III, 1(21) (1966),
147–158.
[2] N.G. DE BRUIJN, Problem 12, Wisk. Opgaven, 21 (1960), 12–13.
[3] S.S. DRAGOMIR, Some refinements of Schwarz inequality, Suppozionul de Matematică şi Aplicaţii,
Polytechnical Institute Timişoara, Romania, 1-2 November 1985, 13–16.
[4] S.S. DRAGOMIR AND J. SÁNDOR, Some inequalities in prehilbertian spaces, Studia Univ., BabeşBolyai, Mathematica, 32(1) (1987), 71–78 MR 89h: 46034.
[5] S.S. DRAGOMIR AND J. SÁNDOR, On Bessels’ and Gram’s inequalities in prehilbertian spaces,
Periodica Math. Hungarica, 29(3) (1994), 197–205.
[6] S.S. DRAGOMIR AND B. MOND, On the superadditivity and monotonicity of Schwarz’s inequality
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[7] G.H. HARDY, J.E. LITTLEWOOD
Cambridge, United Kingdom, 1952.
AND
G. POLYA, Inequalities, Cambridge University Press,
[8] D.S. MITRINOVĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis,
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J. Inequal. Pure and Appl. Math., 5(3) Art. 60, 2004
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