ISSN 2291-8639
Volume 9, Number 1 (2015), 29-38 http://www.etamaths.com
REVERSE OF THE TRIANGLE INEQUALITY IN HILBERT C∗-MODULES
NORDINE BOUNADER, ABDELLATIF CHAHBI∗AND SAMIR KABBAJ
Abstract. In this paper we prove the reverse of triangle inequality via Sel-berg’s inequalities in the framework of HilbertC∗-modules.
1. Introduction
In 1966, Diaz and Matcalf [4] proved the following reverse triangle inequality in setting of Hilbert spaces as follows .
Theorem 1.1. Let x1, ..., xn be vectors in a Hilbert spaceH. If eis a unit vector
of Hsuch that0≤r≤ Rehxi,ei
||xi|| for somer∈Rand each 1≤i≤n, then
r
n X
j=1
kxik ≤
n X
i=1
xi
A number of mathematicians have represented several refinements of the reverse triangle inequality in Hilbert spaces and normed spaces, see[1, 2, 5, 8, 9, 12, 13]
Recently, M. Khosravi, H. Mahyar and M.S. Moslehian [12] obtained the follow-ing reverse of the triangle inequality in the framework of HilbertC∗-modules.
Theorem 1.2. Let X be a Hilbert A-module and e1, ..., em ∈ X be a family of
vectors with hei, eji= 0 (1≤i=6 j≤m) and ||ei||= 1 (1≤i≤m). If the vectors
x1, . . . , xn in X satisfy the conditions
Rehek, xji ≥ρkkxjk , Imhek, xji ≥µkkxjk
forj∈ {1, . . . , n}, k∈ {1, . . . , m}, whereρk, µk∈[0,∞)k∈ {1, . . . , m},then
( m X
k=1
(ρ2k+µ2k))12 n X
j=1
kxjk ≤
n X
j=1
xj
.
In [3] we obtained an extension of Selberg’s inequality in the framework of Hilbert
C∗-modules.
The goal of this paper is to show the reverse of triangle inequality via a extension of Selberg’s inequality in the framework of HilbertC∗-modules. Our results are ex-tensions of theorem 2.1 and Corollary 2.3 obtained by Dragomir in [5] and theorem 9 obtained by Fujii and Nakamoto see [9] in the setting of HilbertC∗-modules.
2010Mathematics Subject Classification. Primary 46L08; Secondary 26D15, 46L05.
Key words and phrases. triangle inequality; reverse inequality; Hilbert C∗−module;
C∗−algebra.
c
2015 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.
2. Preliminaries
In this section we briefly recall the definitions and examples of Hilbert C∗ -modules. For information about Hilbert C∗-module, we refer to [10, 13]. Our
reference forC∗-algebras as [15].
LetAbe aC∗-algebra (not necessarily unitary) andX be a complex linear space.
Definition 2.1. A pre-HilbertA-module is a right A-moduleX equipped with a sesquilinear maph., .i:X × X → Asatisfying
(1) hx, xi ≥0;hx, xi= 0 if and only ifx= 0 for allxin X,
(2) hx, αy+βzi=αhx, yi+β(hx, z)ifor allx, y, z inX, α, β inC,
(3) hx, yi=hy, xi∗ for allx, yinX,
(4) hx, y.ai=hx, yiafor allx, yin X,ainA.
The maph., .iis called anA-valued inner product ofX and forx∈ X, we define ||x||=||hx, xi||12 as a norm onX , where the latter is a norm in theC∗−algebraA. This norm makesX into a right normed module over A.A pre-Hilbert moduleX is called a HilbertA-module if it is complete with respect to its norm. Two typical examples of HilbertC∗-modules are as follows:
(I) Every Hilbert space is a HilbertC∗-module.
(II) EveryC∗algebra Ais a HilbertA-module viaha, bi=a∗b (a, b∈ A).
Notice that the inner product structure of a C∗-algebra is essentially more com-plicated than complex numbers. One may define an A -valued norm |.| by |x|= hx, xi12. Clearly,kxk=k|x|kfor eachx∈ X.
It is known that|.| does not satisfy the triangle inequality in general (see [[13], p.4]). We also use the elementaryC∗-algebra theory, we use the following property: if a≤b then a12 ≤b21,where a, b are positive elements ofC∗-algebra A, and the relation 12(aa∗+a∗a) =Re(a)2+Im(a)2 whereais an arbitrary element ofA
3. Main result
LetX be a right HilbertA-module, which is an algebraic leftA-module satisfy-ing:
hx, ayi=ahx, yi f or all x, y∈ X and a∈ A.
For example if A is a unitalC∗-algebra and I is a commutative right ideal of A,
thenI is a right Hilbert module overAand
hx, ayi=x∗(ay) =ax∗y (x, y∈ I, a∈ A).
For a reverse of triangle inequality, we use the following lemma.
Lemma 3.1. LetXbe a right HilbertA- module which is an algebraic leftA-module andy1, . . . , ymbe a non zero vectors inX. Ifx∈ X then
(3.1)
m X
j=1
|hyj, xi| 2
Pm
k=1khyj, ykik
≤ |x|2
and
(3.2)
m X
j=1
|hx, yji| 2
Pm
k=1khyj, ykik
Proof. Inequality (3.1) is proved in [[3], theorem 3.1 ].
Next we prove the inequality (3.2). Letαj∈ A, j= 1, . . . , n. We know that
0 ≤ x− m X j=1
αjyj 2 = * x− m X j=1
αjyj, x− m X
j=1
αjyj +
= hx, xi −
*
x,
m X
j=1
αjyj +
−
* m X
j=1
αjyj, x +
+
* m X
j=1
αjyj, m X
j=1
αjyj +
= |x|2−
m X
j=1
αjhx, yji − m X
j=1
hyj, xiα∗j+ m X
j,k=1
αjhyk, yjiα∗k
= |x|2−
m X
j=1
αjhx, yji − m X
j=1
hyj, xiα∗j
+ 1
2 m X
j,k=1
(αjhyk, yjiα∗k+αkhyj, ykiα∗j).
It follows from [[6], lemma 3.2] that
αjhyj, ykiα∗k+αkhyk, yjiα∗j ≤ α∗j
2
khyj, ykik+|α∗k| 2
khyj, ykik,
so x− m X j=1
αjyj 2
≤ |x|2−
m X
j=1
αjhx, yji − m X
j=1
hyj, xiα∗j
+ 1
2 m X
j,k=1
(|α∗j|2| hy
j, yki |+|α∗k| 2khy
k, yjik),
Take
αj=
hyj, xi Pm
k=1khyj, ykik
,
A simple calculation shows that
x− m X j=1
αjyj 2
≤ |x|2−
m X
j=1
|hx, yji|2 Pm
k=1khyj, ykik
−
m X
j=1
|hx, yji|2 Pm
k=1khyj, ykik
+ 1
2 m X
j,k=1
|hx, yji| 2
khyj, ykik (Pm
k=1khyj, ykik)2 +1
2 n X
j,k=1
|hx, yji| 2
khyj, ykik (Pm
k=1khyj, ykik)2
Since
|x|2−
m X
j=1
|hx, yji|2 Pm
k=1khyj, ykik
−
m X
j=1
|hx, yji|2 Pm
k=1khyj, ykik
+ 1
2 m X
j,k=1
|hx, yji| 2
khyj, ykik (Pm
k=1khyj, ykik)2 +1
2 m X
j,k=1
|hx, yji| 2
khyj, ykik (Pm
k=1khyj, ykik)2
= |x|2−2 m X
j=1
|hx, yji|2 Pm
k=1khyj, ykik +
n X
j=1
|hx, yji|2 Pm
k=1khyj, ykik
= |x|2−
m X
j=1
|hx, yji| 2
Pm
k=1khyj, ykik
.
It follows that
x−
m X
j=1
αjyj 2
≤ |x|2−
m X
j=1
|hx, yji| 2
Pm
k=1khyj, ykik
,
hence
|x|2−
m X
j=1
|hx, yji| 2
Pm
k=1khyj, ykik
≥0.
The proof is then completed.
Corollary 3.2. Let X be a right Hilbert A− module which is an algebraic left
A-module, then :
| hy, xi |2≤ kyk2
|x|2
and
| hx, yi |2≤ kyk2
|x|2.
Corollary 3.3. Let X be a right Hilbert A−module which is an algebraic leftA -module. If y1, . . . , yn is a sequence of unit vectors in X such that hyj, yki= 0for 1≤j 6=k≤n. Then
m X
j=1
|hyj, xi| 2
≤ |x|2.
and
m X
j=1
|hx, yji| 2
≤ |x|2.
Theorem 3.4. Let X be a right Hilbert A- module which is an algebraic left A -module,x1, . . . , xn andy1, . . . , ym be a non zero vectors inX such that there exist
the non-negative real numbersρj, µj, j∈ {1, . . . , m} with
(3.3) Rehyj, xii ≥ρjkxik kyjk , Imhyj, xii ≥µjkxik kyjk
for eachi∈ {1, . . . , n}, j∈ {1, . . . , m}.Then
( m X
j=1
(ρ2j+µ2j)kyjk 2
Pm
k=1khyj, ykik )12
n X
i=1
|xi| ≤
n X
i=1
xi
Proof. By (3.3), we have
( n X
i=1
Rehyj, xii)2+ ( n X
i=1
Imhyj, xii)2 ≥ ρ2j|yj|2( n X
i=1
|xi|)2+µ2kjkyjk2( n X
i=1
kxik)2
= (ρ2j+µ2j)|yj|2( n X
i=1
|xi|)2.
By combining the above inequality and this equality
1 2(|h
Pn
i=1xi, yji| 2
+|hyj,Pni=1xii| 2
) = (Pn
i=1Rehyj, xii)
2+ (Pn
i=1Imhyj, xii) 2, we deduce 1 2 Pm j=1
|hyj,Pni=1xii| 2
Pm
k=1khyj,ykik+
1 2
Pm j=1
|hPn i=1xi,yji|
2
Pm
k=1khyj,ykik ≥(
Pm j=1
(ρ2
j+µ2j)kyjk2 Pm
k=1khyj,ykik)(
Pn i=1kxik)
2.
Apply lemma 3.1, we get
1 2
m X
j=1
|hyj,P n i=1xii|
2
Pm
k=1khyj, ykik +1
2 m X
j=1
|hPn
i=1xi, yji| 2
Pm
k=1khyj, ykik
≤ |
n X
i=1
xi|2.
Then
( m X
j=1
(ρ2j+µ2j)kyjk 2
Pm
k=1khyj, ykik
)( n X
i=1
||xi| |)2≤ n X i=1 xi 2 .
And since|x| ≤ kxk and|x|2≤ kxk2for allx∈ X,then
( m X
j=1
(ρ2
j+µ2j)kyjk 2
Pm
k=1khyj, ykik
)( n X
i=1
|xi|)2≤ n X i=1 xi 2 .
The desired result follows by taking the square roots.
Remark 3.5. If the first condition of (3.3) is the only one available, then
( m X
j=1
ρ2jkyjk 2
Pm
k=1khyj, ykik )12
n X
i=1
|xi| ≤ n X i=1 xi .
Corollary 3.6. Let X be a right Hilbert A−module which is an algebraic left A -module,x1, . . . , xn andy1, . . . , ym be a non zero vectors inX such that there exist
the non-negative real numbersρj, µj, j∈ {1, . . . , m} with
Rehyj, xii ≥ρjkxik kyjk , Imhyj, xii ≥µjkxik kyjk.
Then
( m X
j=1
(ρ2
j+µ2j)|yj|2
max1≤j≤m|yj|2+ (m−1) maxk6=j| hyj, yki | )12
n X
i=1
|xi| ≤ n X i=1 xi .
Proof. It is easy to show that m
X
k=1
khyj, ykik ≤ max 1≤j≤mkyjk
2
+ (m−1) max
j6=k khyj, ykik. We thus have that
1
max1≤j≤mkyjk2+ (m−1) maxj6=kkhyj, ykik
≤ Pm 1
k=1khyj, ykik
and
(ρ2
j+µ2j)kyjk 2
max1≤j≤mkyjk 2
+ (m−1) maxj6=kkhyj, ykik
≤ (ρ
2
j+µ2j)kyjk 2
Pm
k=1khyj, ykik
.
Consequently
(Pn j=1
(ρ2
j+µ2j)kyjk2
max1≤j≤nkyjk2+(m−1) maxj6=kkhyj,ykik) 1 2Pn
i=1|xi| ≤(P n j=1
(ρ2
j+µ2j)kyjk2 Pn
k=1khyj,ykik) 1 2Pn
i=1|xi|.
We apply the theorem 3.4 to get the result
The next corollary is the theorem 2.5 in [12].
Corollary 3.7. Let X be a Hilbert A-module, x1, . . . , xn be a family of vectors in
X andej be a unitary orthogonal vectors for j ∈ {1, . . . , m} in X such that there
exist the real numbersρj, µj, j∈ {1, . . . , m}with
Rehρjxi, eji ≥ρ2jkxik , Imhµjxi, eji ≥µ2jkxik
for eachi∈ {1, . . . , n}, j∈ {1, . . . , m}.Then
( m X
j=1
(ρ2j+µ2j))12 n X
j=1
kxik ≤
n X
i=1
xi
.
Corollary 3.8. Let X be a right Hilbert A−module which is an algebraic left A -module,x1, . . . , xn andy1. . . , ymbe a non zero vectors in X,such that there exist
the non-negative real number in[0; 1] pj, qj, j∈ {1, . . . , m} with (3.4) ||xi||2−2Rehyj, xii+||yj||2≤p2j ≤ ||yj||2
and
(3.5) ||xi||2−2Imhyj, xii+||yj||2≤p2j ≤ ||yj||2
for eachi∈ {1, . . . , n}, j∈ {1, . . . , m}.Then
( m X
j=1
(2kyjk 2
−p2 j−qj2) Pm
k=1khyj, ykik )12
n X
i=1
|xi| ≤
n X
i=1
xi
.
Proof. By the inequality (3.4), we get
||xi||2+||yj||2−p2j ≤2Rehyj, xii. Since||yj||2−p2j ≥0,then
2||xi|| q
||yj||2−pj2≤ ||xi||2+||yj||2−p2j ≤2Rehyj, xii.
This implies that
Rehyj, xii ≥ q
||yj||2−p2j
||yj||
||yj||||xi||.
Even from (3.5), we get
Imhyj, xii ≥ q
||yj||2−q2j
||yj||
||yj||||xi||,
and if we letρj =
√
||yj||2−p2j
||yj andµj = √
||yj||2−p2j
||yj in theorem 3.4,
The following lemma gives a refinement of Selberg’s inequality in a right Hilbert A−module which is an algebraic leftA-module.
Lemma 3.9. Let X be a right Hilbert A− module which is an algebraic left A -module and y1, . . . , ym be a non zero vectors inX. Ifx∈ X then
(3.6) | hy, xi |2+
m X
j=1
|hyj, xi| 2
Pm
k=1khyj, ykik
||y||2≤ |x|2
||y||2,
(3.7) | hx, yi |2+
m X
j=1
|hx, yji| 2
Pm
k=1khyj, ykik
||y||2≤ |x|2
||y||2,
(3.8) | hx, yi |2+
m X
j=1
|hyj, xi| 2
Pm
k=1khyj, ykik
||y||2≤ |x|2
||y||2,
and
(3.9) | hy, xi |2+
m X
j=1
|hx, yji| 2
Pm
k=1khyj, ykik
||y||2≤ |x|2
||y||2.
Proof. Inequality ( 3.6) is proved in [ [3],theorem 3.3]. Now we prove the inequality (3.7), let
u=x−
m X
j=1
αjyj
where
αj=
hyj, xi Pn
k=1khyj, ykik
.
According to the proof of lemma 3.1, we have
|u|2 = |x−
m X
j=1
αjyj|2
≤ |x|2− m X
j=1
| hyj, xi |2 Pm
k=1| hyj, yki |
.
Hence it follows that
kyk2
|x| 2
−
m X
j=1
|hx, yji| 2
Pm
k=1khyj, ykik
≥ kyk 2
|u|2.
Applying Cauchy Schwartz inequality, we get
kyk2|u|2≥ |hu, yi|2.
and sincehy, yji= 0,so
|hu, yi|2=
*
x−
m X
j=1
αjyj, y +
2
It follows that
kyk2
|x| 2
−
m X
j=1
|hx, yji|2 Pm
k=1khyj, ykik
≥ |hx, yi| 2
,
which completes the proof of the inequality (3.7).
Similarly, we can get inequalities (3.8) and (3.9) .
Lemma 3.10. LetX be a HilbertA- module,y1,· · · ,ym, ybe non zero vectors in
X andx1, · · · ,xn∈ X such that such that there exist the real numbersρj, µj, j∈
{1,· · ·, m} with
(3.10) 0≤ρjkxik ||yj|| ≤Rehyj, xii, 0≤µjkxik ||yj|| ≤Imhyj, xii
andhy, yji= 0, for eachi∈ {1, . . . , n}, j∈ {1, . . . , m}.Then
(3.11) |(y,
n X
i=1
xi)|2+ ( m X
k=1
ρ2 j+µ2j
cj
||yj2)( n X
i=1
|xi|)2||y||2≤ | n X
i=1
xi|2||y||2,
and
(3.12) |(
n X
i=1
xi, y)|2+ ( m X
j=1
ρ2 j+µ2j
cj
||yj||2)( n X
i=1
|xi|)2||y||2≤ | n X
i=1
xi|2||y||2,
wherecj=P m
k=1khyj, ykik.
Proof. Letx=Pn
i=1xi,from (3.10), we get
||y||2{|x|2−Pm j=1
ρ2 j+µ2j
cj (
Pn
i=1||xi||)2}
≥ ||y||2{|x|2−Pm j=1
Rehx,yji2+Imhx,yji2
cj },
Since
||y||2{|x|2−Pm j=1
Rehx,yji2+Imhx,yji2
cj }=kyk
2
{|x|2−1 2
Pm j=1
|hx,yji|2
cj −
1 2
Pm j=1
hyj,xi2
cj }. Then, from (3.6) and (3.9), we get
kyk2{|x|2−1 2
m X
j=1
| hx, yji |2
cj
−1 2
m X
j=1
hyj, xi 2
cj
} ≥ | hy, xi |2,
it follows that
||y||2{|x|2
−
m X
j=1
ρ2j+µ2j cj
(kx1k+· · ·+kxnk)2} ≥ |hy, xi| 2
.
By using (3.7) and (3.8) and by similar argument, we get (3.12).
Theorem 3.11. Let X be a Hilbert A - module, y1, · · · , ym be non zero
vec-tors in X and x1, · · · , xn ∈ X such that such that there exist the real numbers
a, b, ρj, µj, j∈ {1,· · ·, m} with
(3.13) 0≤ρjkxik ||yj|| ≤Rehyj, xii, 0≤µjkxik ||yj|| ≤Imhyj, xii,
andhy, yji= 0, for eachi∈ {1, . . . , n}, j∈ {1, . . . , m}. Then
(a2+b2+ m X
j=1
ρ2 j+µ2j
cj
)||yj2||12(|x1|+· · ·+|xn|)≤ | n X
i=1
xi|.
wherecj=P m
k=1khyj, ykik .
Proof. From (3.11) and (3.12), we get
1 2(| h
Pn
i=1xi, yi |2+| hy,P n
i=1xii |2) +(Pm
j=1 ρ2
j+µ2j
cj ||y||
2)(Pn i=1|xi|)
2||y||2
≤ |Pn
i=1xi|2||y||2, since
1 2(| h
Pn
i=1xi, yi |2+| hy,P n
i=1xii |2)≥(Rehy, xi)2+ (Imhy, xi)2
applying (3.13) and (3.14) and taking the square root, the desired result follows.
Remark 3.12. If in Theorem 3.11 y1, . . . , ym is a sequence orthonormal vectors, then
(a2+b2+ m X
j=1
(ρ2j+µ2j))12( n X
i=1
|xi|)≤
n X
i=1
xi
.
This inequality is an extension of Diaz-Metcalf [4] inequality inC∗-module.
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Department of Mathematics, Faculty of Sciences, University of Ibn Tofail, Kenitra, Morocco
∗
