REVERSE TRIANGLE INEQUALITY IN HILBERT
C ∗-MODULES
MARYAM KHOSRAVI, HAKIMEH MAHYAR
Department of Mathematics
Tarbiat Moallem University
Tahran, Iran.
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
EMail: {khosravi_m, mahyar}@saba.tmu.ac.ir
Title Page
MOHAMMAD SAL MOSLEHIAN
Contents
Department of Pure Mathematics
Centre of Excellence in Analysis on Algebraic Structures (CEAAS)
Ferdowsi University of Mashhad
P.O. Box 1159, Mashhad 91775, Iran.
EMail: moslehian@ferdowsi.um.ac.ir
Received:
16 October, 2009
Accepted:
13 November, 2009
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
Primary 46L08; Secondary 15A39, 26D15, 46L05, 51M16.
Key words:
Triangle inequality, Reverse inequality, Hilbert C ∗ -module, C ∗ -algebra.
JJ
II
J
I
Page 1 of 21
Go Back
Full Screen
Close
Abstract:
We prove several versions of reverse triangle inequality in Hilbert C ∗ modules. We show that if e1 , . . . , em are vectors in a Hilbert module X
over a C ∗ -algebra A with unit 1 such that hei , ej i = 0 (1 ≤ i 6= j ≤ m)
and kei k = 1 (1 ≤ i ≤ m), and also rk , ρk ∈ R (1 ≤ k ≤ m) and
x1 , . . . , xn ∈ X satisfy
0 ≤ rk2 kxj k ≤ Rehrk ek , xj i ,
then
0 ≤ ρ2k kxj k ≤ Imhρk ek , xj i ,
# 12 n
"m
X
n
X
X
kxj k ≤ x
rk2 + ρ2k
j ,
j=1 j=1
k=1
and the equality holds if and only if
n
X
j=1
xj =
n
X
j=1
kxj k
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
m
X
k=1
Contents
(rk + iρk )ek .
JJ
II
J
I
Page 2 of 21
Go Back
Full Screen
Close
Contents
1
Introduction and Preliminaries
4
2
Multiplicative Reverse of the Triangle Inequality
6
3
Additive Reverse of the Triangle Inequality
16
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
Contents
JJ
II
J
I
Page 3 of 21
Go Back
Full Screen
Close
1.
Introduction and Preliminaries
The triangle inequality is one of the most fundamental inequalities in mathematics.
Several mathematicians have investigated its generalizations and its reverses.
In 1917, Petrovitch [17] proved that for complex numbers z1 , . . . , zn ,
n n
X X
(1.1)
zj ≥ cos θ
|zj | ,
j=1
j=1
where 0 < θ < π2 and α − θ < arg zj < α + θ (1 ≤ j ≤ n) for a given real number
α.
The first generalization of the reverse triangle inequality in Hilbert spaces was
given by Diaz and Metcalf [5]. They proved that for x1 , . . . , xn in a Hilbert space
Rehx ,ei
H, if e is a unit vector of H such that 0 ≤ r ≤ kxjjk for some r ∈ R and each
1 ≤ j ≤ n, then
n
n
X
X
(1.2)
r
kxj k ≤ xj .
j=1
j=1
Pn
P
Moreover, the equality holds if and only if j=1 xj = r nj=1 kxj ke.
Recently, a number of mathematicians have presented several refinements of the
reverse triangle inequality in Hilbert spaces and normed spaces (see [1, 2, 4, 7, 8,
10, 13, 16]). Recently a discussion of C ∗ -valued triangle inequalities in Hilbert C ∗ modules was given in [3]. Our aim is to generalize some of the results of Dragomir
in Hilbert spaces to the framework of Hilbert C ∗ -modules. For this purpose, we first
recall some fundamental definitions in the theory of Hilbert C ∗ -modules.
Suppose that A is a C ∗ -algebra and X is a linear space, which is an algebraic
right A-module. The space X is called a pre-Hilbert A-module (or an inner product
A-module) if there exists an A-valued inner product h·, ·i : X × X → A with the
following properties:
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
Contents
JJ
II
J
I
Page 4 of 21
Go Back
Full Screen
Close
(i) hx, xi ≥ 0 and hx, xi = 0 if and only if x = 0;
(ii) hx, λy + zi = λhx, yi + hx, zi;
(iii) hx, yai = hx, yia;
(iv) hx, yi∗ = hy, xi
for all x, y, z ∈ X, a ∈ A, λ ∈ C. By (ii) and (iv), h·, ·i is conjugate linear in the first variable. Using the Cauchy–Schwartz inequality hy, xihx, yi ≤
1
khx, xikhy, yi [11, Page 5] (see also [14]), it follows that kxk = khx, xik 2 is
a norm on X making it a right normed module. The pre-Hilbert module X is
called a Hilbert A-module if it is complete with respect to this norm. Notice
that the inner structure of a C ∗ -algebra is essentially more complicated than
that for complex numbers. For instance, properties such as orthogonality and
theorems such as Riesz’ representation in complex Hilbert space theory cannot
simply be generalized or transferred to the theory of Hilbert C ∗ -modules.
One may define an “A-valued norm” |·| by |x| = hx, xi1/2 . Clearly, k |x| k = kxk
for each x ∈ X. It is known that |·| does not satisfy the triangle inequality in general.
See [11, 12] for more information on Hilbert C ∗ -modules.
We also use elementary C ∗ -algebra theory, in particular we utilize the property
that if a ≤ b then a1/2 ≤ b1/2 , where a, b are positive elements of a C ∗ -algebra A.
We also repeatedly apply the following known relation:
1
(aa∗ + a∗ a) = (Re a)2 + (Im a)2 ,
2
where a is an arbitrary element of A. For details on C ∗ -algebra theory, we refer
readers to [15].
Throughout the paper, we assume that A is a unital C ∗ -algebra with unit 1 and
for every λ ∈ C, we write λ for λ1.
(1.3)
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
Contents
JJ
II
J
I
Page 5 of 21
Go Back
Full Screen
Close
2.
Multiplicative Reverse of the Triangle Inequality
Utilizing some C ∗ -algebraic techniques we present our first result as a generalization
of [7, Theorem 2.3].
Theorem 2.1. Let A be a C ∗ -algebra, let X be a Hilbert A-module and let x1 , . . . , xn ∈
X. If there exist real numbers k1 , k2 ≥ 0 with
0 ≤ k1 kxj k ≤ Rehe, xj i ,
0 ≤ k2 kxj k ≤ Imhe, xj i ,
for some e ∈ X with |e| ≤ 1 and all 1 ≤ j ≤ n, then
n
n
X
X
1
(2.1)
(k12 + k22 ) 2
kxj k ≤ xj .
j=1
j=1
Proof. Applying the Cauchy–Schwarz inequality, we get
*
2 2
+2
n
n
n
X
X
X
2
e,
x
≤
kek
x
≤
x
j j
j ,
j=1
and
j=1
j=1
*
2
+2 n
n
n
X
X 2
X
xj , e ≤ xj |e|2 ≤ xj ,
j=1
j=1
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
Contents
JJ
II
J
I
Page 6 of 21
Go Back
j=1
whence
 *
n
+2 * n
+2 
n
X 2 1 X
X


e,
x
+
x
,
e
x
≥
j j
j
2
j=1
j=1
j=1
* n
+∗ * n
+ * n
+∗ * n
+!
X
X
X
X
1
e,
xj
e,
xj +
xj , e
=
xj , e
2
j=1
j=1
j=1
j=1
Full Screen
Close
*
=
Re e,
n
X
+!2
xj
*
+
n
X
Im e,
j=1
=
Re
n
X
≥ k12
by (1.3)
xj
j=1
!2
he, xj i
+
Im
j=1
n
X
+!2
n
X
!2
he, xj i
j=1
!2
kxj k
+ k22
j=1
= (k12 + k22 )
n
X
!2
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
kxj k
j=1
n
X
vol. 10, iss. 4, art. 110, 2009
!2
kxj k
.
Title Page
j=1
Contents
Using the same argument as in the proof of Theorem 2.1, one can obtain the
following result, where k1 , k2 are hermitian elements of A.
Theorem 2.2. If the vectors x1 , . . . , xn ∈ X satisfy the conditions
0 ≤ k12 kxj k2 ≤ (Rehe, xj i)2 ,
0 ≤ k22 kxj k2 ≤ (Imhe, xj i)2 ,
for some hermitian elements k1 , k2 in A, some e ∈ X with |e| ≤ 1 and all 1 ≤ j ≤ n
then the inequality (2.1) holds.
One may observe an integral version of inequality (2.1) as follows:
Corollary 2.3. Suppose that X is a Hilbert A-module and f : [a, b] → X is strongly
Rb
measurable such that the Lebesgue integral a kf (t)kdt exists and is finite. If there
exist self-adjoint elements a1 , a2 in A with
a21 kf (t)k2 ≤ Rehf (t), ei2 ,
a22 kf (t)k2 ≤ Imhf (t), ei2
(a.e. t ∈ [a, b]) ,
JJ
II
J
I
Page 7 of 21
Go Back
Full Screen
Close
where e ∈ X with |e| ≤ 1, then
(a21 + a22 )
1
2
Z
a
b
Z b
.
kf (t)kdt ≤ f
(t)dt
a
Now we prove a useful lemma, which is frequently applied in the next theorems
(see also [3]).
Lemma 2.4. Let X be a Hilbert A-module and let x, y ∈ X. If |hx, yi| = kxkkyk,
then
xhx, yi
.
y=
kxk2
Proof. For x, y ∈ X we have
2 xhx,
yi
xhx,
yi
xhx,
yi
= y−
0 ≤ y −
,y −
kxk2 kxk2
kxk2
= hy, yi −
≤ |y|2 −
1
1
1
hy, xihx, yi +
hy, xihx, xihx, yi −
hy, xihx, yi
2
4
kxk
kxk
kxk2
1
1
|hx, yi|2 = |y|2 −
kxk2 kyk2
2
kxk
kxk2
= |y|2 − kyk2 ≤ 0 ,
whence y − xhx,yi
= 0. Hence y =
2
kxk Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
Contents
JJ
II
J
I
Page 8 of 21
Go Back
Full Screen
Close
xhx,yi
.
kxk2
Using the Cauchy–Schwarz inequality, we have the following theorem for Hilbert
modules, which is similar to [1, Theorem 2.5].
Theorem 2.5. Let e1 , . . . , em be a family of vectors in a Hilbert module X over a
C ∗ -algebra A such that hei , ej i = 0 (1 ≤ i 6= j ≤ m) and kei k = 1 (1 ≤ i ≤ m).
Suppose that rk , ρk ∈ R (1 ≤ k ≤ m) and that the vectors x1 , . . . , xn ∈ X satisfy
0 ≤ ρ2k kxj k ≤ Imhρk ek , xj i ,
0 ≤ rk2 kxj k ≤ Rehrk ek , xj i ,
Then
"
(2.2)
m
X
# 21
(rk2 + ρ2k )
n
X
j=1
k=1
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
n
X kxj k ≤ xj ,
vol. 10, iss. 4, art. 110, 2009
j=1
and the equality holds if and only if
n
X
(2.3)
xj =
j=1
n
X
kxj k
j=1
m
X
Title Page
(rk + iρk )ek .
Contents
k=1
Pm
P
2
Proof. There is nothing to prove if k=1 (rk2 + ρ2k ) = 0. Assume that m
k=1 (rk +
ρ2k ) 6= 0. From the hypothesis, by Im(a) = Re(ia∗ ), Re(a∗ ) = Re(a) (a ∈ A), we
have
!2
!2
m
n
X
X
(rk2 + ρ2k )
kxj k
j=1
k=1
≤
Re
* m
X
rk ek ,
=
Re
j=1
xj
+ Im
j=1
k=1
* n
X
n
X
+
xj ,
m
X
k=1
* m
X
k=1
+!2
(rk + iρk )ek
.
ρk ek ,
n
X
j=1
+!2
xj
JJ
II
J
I
Page 9 of 21
Go Back
Full Screen
Close
By (1.3),
+!2
* n
m
X
X
(rk + iρk )ek
Re
xj ,
j=1
k=1
 *
+2 * m
+2 
n
m
n
X
X
X
X
1 ≤ 
xj ,
(rk + iρk )ek + (rk + iρk )ek ,
xj 
2 j=1
j=1
k=1
k=1

2 2 2 2 
n
m
m
n
X
X
X
X
1 
≤
xj (rk + iρk )ek + (rk + iρk )ek xj 
2 j=1 k=1
j=1
k=1
2 2
n
m
X
X
≤
xj (rk + iρk )ek j=1
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
Contents
k=1
and since |a| ≤ kak (a ∈ A),
n
m
2 n
* m
+
m
X 2 X
X 2 X
X
x
(r
+
iρ
)e
≤
x
(r
+
iρ
)e
,
(r
+
iρ
)e
j k
k k
j k
k k
k
k k j=1
j=1
k=1
k=1
k=1
2
n
m
X
X
=
xj |rk + iρk |2 kek k2
j=1
k=1
2
n
m
X X
=
xj (rk2 + ρ2k ) .
j=1
k=1
JJ
II
J
I
Page 10 of 21
Go Back
Full Screen
Close
Hence
"
m
X
#
(rk2 + ρ2k )
n
X
!2
kxj k
j=1
k=1
2
n
X
≤
xj .
j=1
By taking square roots the desired result follows.
Clearly we have equality in (2.2) if condition (2.3) holds. To see the converse,
first note that if equality holds in (2.2), then all inequalities in the relations above
should be equality. Therefore
rk2 kxj k = Rehrk ek , xj i , ρ2k kxj k = Imhρk ek , xj i ,
+
+ * n
* n
m
m
X
X
X
X
(rk + iρk )ek =
xj ,
(rk + iρk )ek ,
xj ,
Re
j=1
and
j=1
k=1
j=1
j=1
vol. 10, iss. 4, art. 110, 2009
Title Page
k=1
*
+ n
m
n
m
X X
X
X
(rk + iρk )ek ,
xj = xj (rk + iρk )ek .
k=1
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
k=1
From Lemma 2.4 and the above equalities we have
* m
+
Pm
n
n
X
X
X
(r
+
iρ
)e
k
k
k
xj = Pmk=1
(rk + iρk )ek ,
xj
2
k k=1 (rk + iρk )ek k
j=1
j=1
k=1
* m
+
Pm
n
X
X
(r
+
iρ
)e
k k
k=1 k
= P
Re
(rk + iρk )ek ,
xj
m
2
2
(r
+
ρ
k)
k=1 k
j=1
k=1
Pm
m X
n
X
k=1 (rk + iρk )ek
= P
(rk2 kxj k + ρ2k kxj k)
m
2
2
(r
+
ρ
)
k
k=1 k
k=1 j=1
Contents
JJ
II
J
I
Page 11 of 21
Go Back
Full Screen
Close
=
n
X
kxj k
j=1
m
X
(rk + iρk )ek ,
k=1
which is the desired result.
In the next results of this section, we assume that X is a right Hilbert A-module,
which is an algebraic left A-module subject to
hx, ayi = ahx, yi (x, y ∈ X, a ∈ A) .
(†)
For example if A is a unital C ∗ -algebra and I is a commutative right ideal of A, then
I is a right Hilbert module over A and
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
hx, ayi = x∗ (ay) = ax∗ y = ahx, yi (x, y ∈ I, a ∈ A) .
Title Page
The next theorem is a refinement of [7, Theorem 2.1]. To prove it we need the
following lemma.
Contents
Lemma 2.6. Let X be a Hilbert A-module and e1 , . . . , en ∈ X be a family of vectors
such that hei , ej i = 0 (i 6= j) and kei k = 1. If x ∈ X, then
2
|x| ≥
n
X
2
|hek , xi|
and
k=1
2
|x| ≥
n
X
II
J
I
Page 12 of 21
2
|hx, ek i| .
Go Back
k=1
Full Screen
Proof. The first result follows from the following inequality:
2 *
+
n
n
n
X
X
X
0 ≤ x −
ek hek , xi = x −
ek hek , xi, x −
ej hej , xi
k=1
JJ
k=1
= hx, xi +
n X
n
X
k=1 j=1
Close
j=1
hek , xi∗ hek , ej ihej , xi − 2
n
X
k=1
|hek , xi|2
= hx, xi +
n
X
∗
hek , xi hek , ek ihek , xi − 2
k=1
≤ |x|2 +
= |x|2 −
n
X
k=1
n
X
n
X
|hek , xi|2
k=1
hek , xi∗ hek , xi − 2
n
X
|hek , xi|2
k=1
|hek , xi|2 .
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
k=1
By considering |x −
2
k=1 hek , xiek | ,
Pn
we similarly obtain the second one.
vol. 10, iss. 4, art. 110, 2009
Now we will prove the next theorem without using the Cauchy–Schwarz inequality.
Title Page
Theorem 2.7. Let e1 , . . . , em ∈ X be a family of vectors with hei , ej i = 0 (1 ≤
i 6= j ≤ m) and kei k = 1 (1 ≤ i ≤ m). If the vectors x1 , . . . , xn ∈ X satisfy the
conditions
Contents
(2.4)
0 ≤ rk kxj k ≤ Rehek , xj i ,
0 ≤ ρk kxj k ≤ Imhek , xj i ,
for 1 ≤ j ≤ n, 1 ≤ k ≤ m, where rk , ρk ∈ [0, ∞) (1 ≤ k ≤ m), then
n
# 12 n
" m
X X
X
kxj k ≤ xj .
(rk2 + ρ2k )
(2.5)
j=1
k=1
j=1
Pn
Proof. Applying the previous lemma for x = j=1 xj , we obtain

2
*
*
+2
+2 
n
m n
m X
n
X
X
X
X
1
xj , ek 
xj ≥ 
xj +
ek ,
2
j=1
k=1
j=1
k=1
j=1
JJ
II
J
I
Page 13 of 21
Go Back
Full Screen
Close
=
ek ,
2
k=1
=
*
m
X
1
n
X
Re ek ,
m
X
Re
m
X

rk2
=
xj
*
+
n
X
n
X
Im ek ,
(rk2
hek , xj i
+
ρ2k )
xj , ek
+∗ * n
X
+!
xj , ek
j=1
+!2
xj
(by (1.3))
j=1
!2
n
X
* n
X
j=1
+!2
+
Im
n
X
!2
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
hek , xj i
j=1
!2
kxj k
n
X
+ ρ2k
j=1
k=1
m
X
n
X
+
xj
j=1
j=1
k=1
≥
ek ,
+
j=1
k=1
=
xj
n
X
j=1
*
m
X
+∗ *
vol. 10, iss. 4, art. 110, 2009
!2 
kxj k

(by (2.4))
j=1
n
X
kxj k
Contents
.
j=1
k=1
Proposition 2.8. In Theorem 2.7, if hek , ek i = 1, then the equality holds in (2.5) if
and only if
! m
n
n
X
X
X
(rk + iρk )ek .
(2.6)
xj =
kxj k
j=1
Title Page
!2
j=1
k=1
Proof. If (2.6) holds, then the inequality in (2.5) turns trivially into equality.
Next, assume that equality holds in (2.5). Then the two inequalities in the proof
JJ
II
J
I
Page 14 of 21
Go Back
Full Screen
Close
of Theorem 2.7 should be equalities. Hence
n
*
n
* n
+2
+2
m n
m X
X 2 X
X 2 X
X
xj =
xj and xj =
xj , ek ,
ek ,
j=1
j=1
k=1
j=1
k=1
j=1
which is equivalent to
n
X
xj =
j=1
m X
n
X
ek hek , xj i =
m X
n
X
k=1 j=1
hek , xj iek ,
k=1 j=1
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
and
rk kxj k = Rehek , xj i ,
ρk kxj k = Imhek , xj i .
Title Page
So
Contents
n
X
xj =
j=1
=
m X
n
X
k=1 j=1
m X
n
X
ek hek , xj i
ek (rk + iρk )kxj k
k=1 j=1
=
n
X
j=1
II
J
I
Page 15 of 21
Go Back
!
kxj k
JJ
m
X
k=1
(rk + iρk )ek .
Full Screen
Close
3.
Additive Reverse of the Triangle Inequality
We now present some versions of the additive reverse of the triangle inequality. In
[6], Dragomir established the following theorem:
Theorem 3.1. Let {ek }m
k=1 be a family of orthonormal vectors in a Hilbert space H
and Mjk ≥ 0 (1 ≤ j ≤ n, 1 ≤ k ≤ m) such that
kxj k − Rehek , xj i ≤ Mjk ,
for each 1 ≤ j ≤ n and 1 ≤ k ≤ m. Then
n
n X
m
n
1 X
X
X
1 Mjk ;
kxj k ≤ √ xj +
m
m
j=1 k=1
j=1
j=1
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
vol. 10, iss. 4, art. 110, 2009
Title Page
and the equality holds if and only if
n
X
Contents
n
m
1 XX
Mjk ,
kxj k ≥
m
j=1
j=1 k=1
and
n
X
j=1
xj =
n
X
n
m
1 XX
kxj k −
Mjk
m j=1 k=1
j=1
!
ek .
k=1
∗
Theorem 3.2. Let {ek }m
k=1 be a family of vectors in a Hilbert module X over a C algebra A with unit 1, |ek | ≤ 1 (1 ≤ k ≤ m), hei , ej i = 0 (1 ≤ i 6= j ≤ m) and
xj ∈ X (1 ≤ j ≤ n). If for some scalars Mjk ≥ 0 (1 ≤ j ≤ n, 1 ≤ k ≤ m),
kxj k − Rehek , xj i ≤ Mjk
II
J
I
Page 16 of 21
m
X
We can prove this theorem for Hilbert C ∗ -modules using some different techniques.
(3.1)
JJ
(1 ≤ j ≤ n, 1 ≤ k ≤ m) ,
Go Back
Full Screen
Close
then
n
n X
m
1 X
X
1 kxj k ≤ √ xj +
Mjk .
m j=1 m j=1 k=1
j=1
n
X
(3.2)
Moreover, if |ek | = 1 (1 ≤ k ≤ m), then the equality in (3.2) holds if and only if
n
X
n
m
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
1 XX
kxj k ≥
Mjk ,
m
j=1 k=1
j=1
(3.3)
vol. 10, iss. 4, art. 110, 2009
and
(3.4)
n
X
xj =
j=1
n
X
n
m
1 XX
kxj k −
Mjk
m
j=1
j=1 k=1
!
m
X
Title Page
ek .
Proof. Taking the summation in (3.1) over j from 1 to n, we obtain
+
*
n
n
n
X
X
X
kxj k ≤ Re ek ,
xj +
Mjk ,
j=1
j=1
Contents
k=1
JJ
II
J
I
Page 17 of 21
j=1
Go Back
for each k ∈ {1, . . . , m}. Summing these inequalities over k from 1 to m, we deduce
(3.5)
* m
+
n
m
n
X X
1
1 XX
kxj k ≤
Re
ek ,
xj +
Mjk .
m
m
j=1
j=1
k=1
k=1 j=1
n
X
Full Screen
Close
Using the Cauchy–Schwarz we obtain
* m
+!2
n
X X
Re
ek ,
(3.6)
xj
j=1
k=1
 *
+2 * m
+∗ 2 
n
m
n
X
X
X
X
1 
ek ,
xj 
≤
ek ,
xj + 2 k=1
j=1
j=1
k=1

2 2 2 2 
n
m
n
m
X
X
X
X
1 
≤
ek xj + ek xj 
2 k=1 j=1 j=1
k=1
2 2
2
m
n
n
X
X
X
≤
ek xj ≤ m xj ,
j=1
k=1
k=1
k=1
vol. 10, iss. 4, art. 110, 2009
Title Page
Contents
j=1
since
m 2 * m
m
+ m m
m
X X X
X X
X
ek = ek ,
ek = hek , el i = |ek |2 ≤ m .
k=1
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
k=1 l=1
JJ
II
J
I
Page 18 of 21
k=1
Go Back
Using (3.6) in (3.5), we deduce the desired inequality.
If (3.3) and (3.4) hold, then
! m n
n
n X
m
X X
X
X
1 1
1
√ xj = √
kxj k −
Mjk ek m j=1 k=1
m j=1 m j=1
k=1
n
X
n
m
1 XX
=
kxj k −
Mjk ,
m
j=1
j=1 k=1
Full Screen
Close
and the equality in (3.2) holds true.
Conversely, if the equality holds in (3.2), then obviously (3.3) is valid and we
have equalities throughout the proof above. This means that
kxj k − Rehek , xj i = Mjk ,
* m
+ * m
+
n
n
X X
X X
Re
ek ,
xj =
ek ,
xj ,
j=1
k=1
and
k=1
j=1
* m
+ m n
n
X X
X X ek ,
xj = ek xj .
k=1
j=1
Title Page
Contents
=
n
m
1 XX
kxj k −
Mjk
m
j=1
j=1 k=1
II
J
I
Go Back
m X
n
X
k=1 ek
(kxj k − Mjk )
m
k=1 j=1
!
JJ
Page 19 of 21
Pm
n
X
vol. 10, iss. 4, art. 110, 2009
j=1
k=1
It follows from Lemma 2.4 and the previous relations that
* m
+
Pm
n
n
X
X X
e
k
xj = Pmk=1 2
ek ,
xj
k
e
k
k
k=1
j=1
j=1
k=1
*
+
Pm
m
n
X
X
e
k
= k=1
Re
ek ,
xj
m
j=1
k=1
=
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
Full Screen
Close
m
X
k=1
ek .
References
[1] A.H. ANSARI AND M.S. MOSLEHIAN, Refinements of reverse triangle inequality in inner product spaces, J. Inequal. Pure Appl. Math., 6(3) (2005),
Art. 64. [ONLINE: http://jipam.vu.edu.au/article.php?sid=
537].
[2] A.H. ANSARI AND M.S. MOSLEHIAN, More on reverse triangle inequality
in inner product spaces, Inter. J. Math. Math. Sci., 18 (2005), 2883–2893.
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
[3] Lj. ARAMBAČIĆ AND R. RAJIĆ, On the C ∗ -valued triangle equality and inequality in Hilbert C ∗ -modules, Acta Math. Hungar., 119(4) (2008), 373–380.
vol. 10, iss. 4, art. 110, 2009
[4] I. BRNETIC, S.S. DRAGOMIR, R. HOXHA AND J. PEČARIĆ, A reverse of
the triangle inequality in inner product spaces and applications for polynomials,
Aust. J. Math. Anal. Appl., 3(2) (2006), Art. 9.
Title Page
[5] J.B. DIAZ AND F.T. METCALF, A complementary triangle inequality in
Hilbert and Banach spaces, Proc. Amer. Math. Soc., 17(1) (1966), 88–97.
[6] S.S. DRAGOMIR, Reverses of the triangle inequality in inner product spaces,
Linear Algebra Appl., 402 (2005), 245–254.
[7] S.S. DRAGOMIR, Some reverses of the generalized triangle inequality in complex inner product spaces, RGMIA Res. Rep. Coll., 7(E) (2004), Art. 7.
[8] S.S. DRAGOMIR, Reverse of the continuous triangle inequality for Bochner
integral in complex Hilbert spaces, J. Math. Anal. Appl., 329 (2007), 65–76.
[9] J. KARAMATA, Teorijia i Praksa Stieltjesova Integrala (SebroCoratian)(Stieltjes Integral, Theory and Practice), SANU, Posebna izdanja,
154, Beograd, 1949.
Contents
JJ
II
J
I
Page 20 of 21
Go Back
Full Screen
Close
[10] M. KATO, K.S. SAITO AND T. TAMURA, Sharp triangle inequality and its
reverse in Banach spaces, Math. Inequal. Appl., 10 (2007), 451–460.
[11] E.C. LANCE, Hilbert C ∗ -modules, London Mathematical Society Lecture
Note Series, 210, Cambridge University Press, Cambridge, 1995.
[12] V.M. MANUILOV AND E.V. TROITSKY, Hilbert C ∗ -Modules, Translations of
Mathematical Monographs, 226. American Mathematical Society, Providence,
RI, 2005.
Reverse Triangle Inequality
Maryam Khosravi, Hakimeh Mahyar
and Mohammad Sal Moslehian
[13] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic, Dordrecht, 1993.
vol. 10, iss. 4, art. 110, 2009
[14] M.S. MOSLEHIAN AND L.-E. PERSSON, Reverse Cauchy–Schwarz inequalities for positive C*-valued sesquilinear forms, Math. Inequal. Appl. (to appear).
Title Page
[15] J.G. MURPHY, C ∗ -Algebras and Operator Theory, Academic Press, Boston,
1990.
[16] M. NAKAI AND T. TADA, The reverse triangle inequality in normed spaces,
New Zealand J. Math., 25(2) (1996), 181–193.
[17] M. PETROVICH, Module d’une somme, L’ Ensignement Math., 19 (1917),
53–56.
Contents
JJ
II
J
I
Page 21 of 21
Go Back
Full Screen
Close