Volume 10 (2009), Issue 4, Article 110, 11 pp.
REVERSE TRIANGLE INEQUALITY IN HILBERT C ∗ -MODULES
MARYAM KHOSRAVI, HAKIMEH MAHYAR, AND MOHAMMAD SAL MOSLEHIAN
D EPARTMENT OF M ATHEMATICS
TARBIAT M OALLEM U NIVERSITY
TAHRAN , I RAN .
khosravi− m@saba.tmu.ac.ir
mahyar@saba.tmu.ac.ir
D EPARTMENT OF P URE M ATHEMATICS
C ENTRE OF E XCELLENCE IN A NALYSIS ON A LGEBRAIC S TRUCTURES (CEAAS)
F ERDOWSI U NIVERSITY OF M ASHHAD
P.O. B OX 1159, M ASHHAD 91775, I RAN .
moslehian@ferdowsi.um.ac.ir
URL: http://profsite.um.ac.ir/~moslehian/
Received 16 October, 2009; accepted 13 November, 2009
Communicated by S.S. Dragomir
A BSTRACT. 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
"
m
X
# 12
2
rk2 + ρk
k=1
0 ≤ ρ2k kxj k ≤ Imhρk ek , xj i ,
X
n
kxj k ≤ xj ,
j=1 j=1
n
X
and the equality holds if and only if
n
n
m
X
X
X
xj =
kxj k
(rk + iρk )ek .
j=1
j=1
k=1
Key words and phrases: Triangle inequality, Reverse inequality, Hilbert C ∗ -module, C ∗ -algebra.
2000 Mathematics Subject Classification. Primary 46L08; Secondary 15A39, 26D15, 46L05, 51M16.
1. I NTRODUCTION AND P RELIMINARIES
The triangle inequality is one of the most fundamental inequalities in mathematics. Several
mathematicians have investigated its generalizations and its reverses.
264-09
2
M ARYAM K HOSRAVI , H AKIMEH M AHYAR ,
AND
M OHAMMAD S AL M OSLEHIAN
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 H, if e is a unit vector of H
Rehx ,ei
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:
(i)
(ii)
(iii)
(iv)
hx, xi ≥ 0 and hx, xi = 0 if and only if x = 0;
hx, λy + zi = λhx, yi + hx, zi;
hx, yai = hx, yia;
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 ≤ khx, xikhy, yi [11,
1
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:
(1.3)
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.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
http://jipam.vu.edu.au/
R EVERSE T RIANGLE I NEQUALITY
3
2. M ULTIPLICATIVE R EVERSE OF THE T RIANGLE I NEQUALITY
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
(k12 + k22 )
(2.1)
1
2
n
X
j=1
n
X
kxj k ≤ xj .
j=1
Proof. Applying the Cauchy–Schwarz inequality, we get
*
2 2
+2
n
n
n
X
X
X
xj ≤ kek2 xj ≤ xj ,
e,
j=1
and
j=1
j=1
*
2
+2 n
n
n
X
X 2
X
2
xj , e ≤ xj |e| ≤ xj ,
j=1
j=1
j=1
whence
 *
n
+2 * n
+2 
n
X 2 1 X
X


x
≥
e,
x
+
x
,
e
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
* n
+!2
* n
+!2
X
X
by (1.3)
= Re e,
xj
+ Im e,
xj
j=1
=
Re
n
X
j=1
!2
he, xj i
+
Im
j=1
≥ k12
n
X
n
X
!2
he, xj i
j=1
!2
kxj k
+ k22
j=1
= (k12 + k22 )
n
X
!2
kxj k
j=1
n
X
!2
kxj k
.
j=1
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.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
http://jipam.vu.edu.au/
4
M ARYAM K HOSRAVI , H AKIMEH M AHYAR ,
AND
M OHAMMAD S AL M OSLEHIAN
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 measurRb
able 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]) ,
where e ∈ X with |e| ≤ 1, then
(a21
1
a22 ) 2
+
b
Z
a
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
y=
xhx, yi
.
kxk2
Proof. For x, y ∈ X we have
2 xhx, yi xhx, yi
xhx, yi
0 ≤ y −
= y−
,y −
kxk2 kxk2
kxk2
= hy, yi −
≤ |y|2 −
1
1
1
hy,
xihx,
yi
+
hy,
xihx,
xihx,
yi
−
hy, xihx, yi
kxk2
kxk4
kxk2
1
1
|hx, yi|2 = |y|2 −
kxk2 kyk2
2
kxk
kxk2
= |y|2 − kyk2 ≤ 0 ,
xhx,yi whence y − kxk2 = 0. Hence y =
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 ≤ rk2 kxj k ≤ Rehrk ek , xj i ,
0 ≤ ρ2k kxj k ≤ Imhρk ek , xj i ,
Then
"
(2.2)
m
X
# 12
(rk2 + ρ2k )
n
X
j=1
k=1
n
X
kxj k ≤ xj ,
j=1
and the equality holds if and only if
(2.3)
n
X
j=1
xj =
n
X
j=1
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
kxj k
m
X
(rk + iρk )ek .
k=1
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R EVERSE T RIANGLE I NEQUALITY
5
P
Pm
2
2
2
2
Proof. There is nothing to prove if m
k=1 (rk + ρk ) = 0. Assume that
k=1 (rk + ρk ) 6= 0. From
the hypothesis, by Im(a) = Re(ia∗ ), Re(a∗ ) = Re(a) (a ∈ A), we have
!2
!2
* m
+
* m
+!2
n
n
n
m
X
X
X
X
X
X
2
2
(rk + ρk )
kxj k ≤ Re
rk ek ,
xj + Im
ρk ek ,
xj
j=1
k=1
j=1
k=1
=
Re
* n
X
xj ,
j=1
m
X
k=1
j=1
+!2
(rk + iρk )ek
.
k=1
By (1.3),
Re
* n
X
j=1
xj ,
m
X
+!2
(rk + iρk )ek
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
k=1
and since |a| ≤ kak (a ∈ A),
2 2 2 *
+
n
m
n
m
m
X
X
X
X
X
(rk + iρk )ek xj (rk + iρk )ek ≤ xj (rk + iρk )ek ,
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
Hence
"
m
X
#
(rk2 + ρ2k )
n
X
!2
kxj k
j=1
k=1
n
X 2
≤
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
Re
xj ,
(rk + iρk )ek =
xj ,
(rk + iρk )ek ,
j=1
and
j=1
k=1
k=1
*
+ n
m
n
m
X
X X
X
(rk + iρk )ek ,
xj = xj (rk + iρk )ek .
k=1
j=1
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
j=1
k=1
http://jipam.vu.edu.au/
6
M ARYAM K HOSRAVI , H AKIMEH M AHYAR ,
AND
M OHAMMAD S AL M OSLEHIAN
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
(r
+
iρ
)e
k
k
k
k
k=1
j=1
j=1
k=1
* m
+
Pm
n
X
X
(r
+
iρ
)e
k
k
k
k=1
= P
Re
(rk + iρk )ek ,
xj
m
2
2
k=1 (rk + ρk )
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
=
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
hx, ayi = x∗ (ay) = ax∗ y = ahx, yi (x, y ∈ I, a ∈ A) .
The next theorem is a refinement of [7, Theorem 2.1]. To prove it we need the following lemma.
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
n
n
X
X
2
2
2
|x| ≥
|hek , xi| and |x| ≥
|hx, ek i|2 .
k=1
k=1
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
j=1
k=1
= hx, xi +
= hx, xi +
≤ |x|2 +
= |x|2 −
n X
n
X
∗
hek , xi hek , ej ihej , xi − 2
k=1 j=1
n
X
hek , xi∗ hek , xi − 2
k=1
n
X
|hek , xi|2
k=1
hek , xi∗ hek , ek ihek , xi − 2
k=1
n
X
n
X
n
X
|hek , xi|2
k=1
n
X
|hek , xi|2
k=1
|hek , xi|2 .
k=1
By considering |x −
Pn
2
k=1 hek , xiek |
, we similarly obtain the second one.
Now we will prove the next theorem without using the Cauchy–Schwarz inequality.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
http://jipam.vu.edu.au/
R EVERSE T RIANGLE I NEQUALITY
7
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
0 ≤ rk kxj k ≤ Rehek , xj i ,
(2.4)
0 ≤ ρk kxj k ≤ Imhek , xj i ,
for 1 ≤ j ≤ n, 1 ≤ k ≤ m, where rk , ρk ∈ [0, ∞) (1 ≤ k ≤ m), then
" m
# 21 n
n
X
X
X
(2.5)
(rk2 + ρ2k )
kxj k ≤ xj .
j=1
j=1
k=1
P
Proof. Applying the previous lemma for x = nj=1 xj , we obtain

2
*
*
+2
+2 
n
m n
m X
n
X
X
X
X
1
xj ≥ 
xj +
xj , ek 
ek ,
2
j=1
j=1
j=1
k=1
k=1
*
+∗ *
+ * n
+∗ * n
+!
m
n
n
X
X
X
X
X
1
=
ek ,
xj
ek ,
xj +
xj , ek
xj , ek
2
j=1
j=1
j=1
j=1
k=1
*
+!
*
+!
2
2
n
m
n
X
X
X
=
Re ek ,
xj
+ Im ek ,
xj
(by (1.3))
j=1
k=1
=
m
X
Re
m
X

rk2
+
Im
m
X
n
X
(rk2 + ρ2k )
n
X
!2
hek , xj i
j=1
!2
n
X
+ ρ2k
kxj k
j=1
k=1
=
hek , xj i
j=1
k=1
≥
n
X
j=1
!2
!2 
kxj k
(by (2.4))

j=1
n
X
!2
kxj k
.
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
(2.6)
xj =
kxj k
(rk + iρk )ek .
j=1
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 of Theorem
2.7 should be equalities. Hence
2
*
2
*
+2
+2
n
n
m n
n
m X
X
X
X
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
j=1
xj =
m X
n
X
ek hek , xj i =
k=1 j=1
m X
n
X
hek , xj iek ,
k=1 j=1
and
rk kxj k = Rehek , xj i ,
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
ρk kxj k = Imhek , xj i .
http://jipam.vu.edu.au/
8
M ARYAM K HOSRAVI , H AKIMEH M AHYAR ,
AND
M OHAMMAD S AL M OSLEHIAN
So
n
X
xj =
j=1
n
m X
X
ek hek , xj i
k=1 j=1
=
m X
n
X
ek (rk + iρk )kxj k
k=1 j=1
n
X
=
!
kxj k
j=1
m
X
(rk + iρk )ek .
k=1
3. A DDITIVE R EVERSE OF THE T RIANGLE I NEQUALITY
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
n X
m
1 X
X
X
1 kxj k ≤ √ xj +
Mjk ;
m j=1 m j=1 k=1
j=1
and the equality holds if and only if
n
X
n
kxj k ≥
j=1
m
1 XX
Mjk ,
m j=1 k=1
and
n
X
xj =
j=1
n
X
n
m
1 XX
kxj k −
Mjk
m j=1 k=1
j=1
!
m
X
ek .
k=1
We can prove this theorem for Hilbert C ∗ -modules using some different techniques.
∗
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),
(3.1)
kxj k − Rehek , xj i ≤ Mjk
(1 ≤ j ≤ n, 1 ≤ k ≤ m) ,
then
(3.2)
n
n X
m
1 X
X
1 kxj k ≤ √ xj +
Mjk .
m
m
j=1
j=1
j=1 k=1
n
X
Moreover, if |ek | = 1 (1 ≤ k ≤ m), then the equality in (3.2) holds if and only if
(3.3)
n
X
n
m
1 XX
kxj k ≥
Mjk ,
m j=1 k=1
j=1
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
http://jipam.vu.edu.au/
R EVERSE T RIANGLE I NEQUALITY
9
and
(3.4)
n
X
n
n
X
m
1 XX
Mjk
kxj k −
m
j=1 k=1
j=1
xj =
j=1
!
m
X
ek .
k=1
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
j=1
for each k ∈ {1, . . . , m}. Summing these inequalities over k from 1 to m, we deduce
* m
+
n
m
n
n
X X
X
1 XX
1
Mjk .
kxj k ≤
Re
ek ,
xj +
(3.5)
m
m
j=1
j=1
j=1
k=1
k=1
Using the Cauchy–Schwarz we obtain
 *
* m
+!2
+2 * m
+∗ 2 
n
m
n
n
X X
X
X
X
X
1 (3.6)
Re
ek ,
xj
≤ 
ek ,
xj + ek ,
xj 
2 k=1
j=1
j=1
j=1
k=1
k=1

2 2 2 2 
m
n
m
n
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
j=1
since
2 *
+
m
m
m
X
X
X
ek = ek ,
ek k=1
k=1
k=1
m X
m
m
X
X
=
hek , el i = |ek |2 ≤ m .
k=1 l=1
k=1
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 k=1
j=1
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 ,
k=1
j=1
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 110, 11 pp.
k=1
j=1
http://jipam.vu.edu.au/
10
M ARYAM K HOSRAVI , H AKIMEH M AHYAR ,
AND
M OHAMMAD S AL M OSLEHIAN
and
*
+ m n
m
n
X X X
X
ek ,
ek xj = xj .
k=1
j=1
j=1
k=1
It follows from Lemma 2.4 and the previous relations that
* m
+
Pm
n
n
X X
X
e
k
ek ,
xj
xj = Pmk=1 2
k k=1 ek k
j=1
j=1
k=1
* m
+
Pm
n
X X
e
k
= k=1
Re
ek ,
xj
m
j=1
k=1
Pm
=
=
k=1 ek
m
m X
n
X
(kxj k − Mjk )
k=1 j=1
n
X
n
m
1 XX
kxj k −
Mjk
m
j=1
j=1 k=1
!
m
X
ek .
k=1
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