International Journal of Analysis and Applications
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 Hilbert C∗-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 space H. If e is a unit vector
of H such that 0 ≤ r ≤ Re〈xi,e〉||xi|| for some r ∈ R and each 1 ≤ i ≤ n, then

r

n∑
j=1

‖xi‖≤

∥∥∥∥∥
n∑
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 Hilbert C∗-modules.

Theorem 1.2. Let X be a Hilbert A-module and e1, ...,em ∈ X be a family of
vectors with 〈ei,ej〉 = 0 (1 ≤ i 6= j ≤ m) and ||ei|| = 1 (1 ≤ i ≤ m). If the vectors
x1, . . . ,xn in X satisfy the conditions

Re〈ek,xj〉≥ ρk ‖xj‖ , Im〈ek,xj〉≥ µk ‖xj‖
for j ∈{1, . . . ,n} , k ∈{1, . . . ,m} , where ρk,µk ∈ [0,∞) k ∈{1, . . . ,m} , then

(

m∑
k=1

(ρ2k + µ
2
k))

1
2

n∑
j=1

‖xj‖≤

∥∥∥∥∥∥
n∑
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 Hilbert C∗-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 Hilbert C∗-modules.

2010 Mathematics 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.

29



30 BOUNADER, CHAHBI AND KABBAJ

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 for C∗-algebras as [15].

Let A be a C∗-algebra (not necessarily unitary) and X be a complex linear space.

Definition 2.1. A pre-Hilbert A-module is a right A-module X equipped with a
sesquilinear map 〈., .〉 : X ×X →A satisfying

(1) 〈x,x〉≥ 0;〈x,x〉 = 0 if and only if x = 0 for all x in X ,
(2) 〈x,αy + βz〉 = α〈x,y〉 + β(〈x,z)〉 for all x,y,z in X ,α,β in C,
(3) 〈x,y〉 = 〈y,x〉∗ for all x,y in X ,
(4) 〈x,y.a〉 = 〈x,y〉a for all x,y in X , a in A.

The map 〈., .〉 is called an A-valued inner product of X and for x ∈X , we define
||x|| = ||〈x,x〉||

1
2 as a norm on X , where the latter is a norm in the C∗−algebra A.

This norm makes X into a right normed module over A. A pre-Hilbert module X
is called a Hilbert A-module if it is complete with respect to its norm. Two typical
examples of Hilbert C∗-modules are as follows:
(I) Every Hilbert space is a Hilbert C∗-module.
(II) Every C∗algebra A is a Hilbert A -module via 〈a,b〉 = 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| =
〈x,x〉

1
2 . Clearly, ‖x‖ = ‖|x|‖ for each x ∈X .

It is known that |.| does not satisfy the triangle inequality in general (see [[13],
p.4]). We also use the elementary C∗-algebra theory, we use the following property:

if a ≤ b then a
1
2 ≤ b

1
2 ,where a, b are positive elements of C∗-algebra A, and the

relation 1
2
(aa∗ + a∗a) = Re(a)2 + Im(a)2 where a is an arbitrary element of A

3. Main result
Let X be a right Hilbert A-module, which is an algebraic left A-module satisfy-

ing:

〈x,ay〉 = a〈x,y〉 for all x,y ∈X and 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

〈x,ay〉 = x∗(ay) = ax∗y (x,y ∈I,a ∈A).

For a reverse of triangle inequality, we use the following lemma.

Lemma 3.1. Let X be a right Hilbert A- module which is an algebraic left A-module
and y1, . . . ,ym be a non zero vectors in X. If x ∈X then

(3.1)

m∑
j=1

|〈yj,x〉|
2∑m

k=1 ‖〈yj,yk〉‖
≤ |x|2

and

(3.2)

m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
≤ |x|2



REVERSE OF THE TRIANGLE INEQUALITY 31

Proof. Inequality (3.1) is proved in [[3], theorem 3.1 ].
Next we prove the inequality (3.2). Let αj ∈A,j = 1, . . . ,n. We know that

0 ≤

∣∣∣∣∣∣x−
m∑
j=1

αjyj

∣∣∣∣∣∣
2

=

〈
x−

m∑
j=1

αjyj,x−
m∑
j=1

αjyj

〉

= 〈x,x〉−

〈
x,

m∑
j=1

αjyj

〉
−

〈
m∑
j=1

αjyj,x

〉
+

〈
m∑
j=1

αjyj,

m∑
j=1

αjyj

〉

= |x|2 −
m∑
j=1

αj 〈x,yj〉−
m∑
j=1

〈yj,x〉α∗j +
m∑

j,k=1

αj 〈yk,yj〉α∗k

= |x|2 −
m∑
j=1

αj 〈x,yj〉−
m∑
j=1

〈yj,x〉α∗j

+
1

2

m∑
j,k=1

(αj 〈yk,yj〉α∗k + αk 〈yj,yk〉α
∗
j ).

It follows from [[6], lemma 3.2] that

αj 〈yj,yk〉α∗k + αk 〈yk,yj〉α
∗
j ≤

∣∣α∗j∣∣2 ‖〈yj,yk〉‖ + |α∗k|2 ‖〈yj,yk〉‖ ,
so

∣∣∣∣∣∣x−
m∑
j=1

αjyj

∣∣∣∣∣∣
2

≤ |x|2 −
m∑
j=1

αj 〈x,yj〉−
m∑
j=1

〈yj,x〉α∗j

+
1

2

m∑
j,k=1

(|α∗j|
2| 〈yj,yk〉 | + |α∗k|

2 ‖〈yk,yj〉‖),

Take

αj =
〈yj,x〉∑m

k=1 ‖〈yj,yk〉‖
,

A simple calculation shows that

∣∣∣∣∣∣x−
m∑
j=1

αjyj

∣∣∣∣∣∣
2

≤ |x|2 −
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
−

m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖

+
1

2

m∑
j,k=1

|〈x,yj〉|
2 ‖〈yj,yk〉‖

(
∑m
k=1 ‖〈yj,yk〉‖)2

+
1

2

n∑
j,k=1

|〈x,yj〉|
2 ‖〈yj,yk〉‖

(
∑m
k=1 ‖〈yj,yk〉‖)2

.



32 BOUNADER, CHAHBI AND KABBAJ

Since

|x|2 −
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
−

m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖

+
1

2

m∑
j,k=1

|〈x,yj〉|
2 ‖〈yj,yk〉‖

(
∑m
k=1 ‖〈yj,yk〉‖)2

+
1

2

m∑
j,k=1

|〈x,yj〉|
2 ‖〈yj,yk〉‖

(
∑m
k=1 ‖〈yj,yk〉‖)2

= |x|2 − 2
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
+

n∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖

= |x|2 −
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
.

It follows that ∣∣∣∣∣∣x−
m∑
j=1

αjyj

∣∣∣∣∣∣
2

≤ |x|2 −
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
,

hence

|x|2 −
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
≥ 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 :

| 〈y,x〉 |2 ≤‖y‖2 |x|2

and

| 〈x,y〉 |2 ≤‖y‖2 |x|2 .

Corollary 3.3. Let X be a right Hilbert A− module which is an algebraic left A-
module. If y1, . . . ,yn is a sequence of unit vectors in X such that 〈yj,yk〉 = 0 for
1 ≤ j 6= k ≤ n. Then

m∑
j=1

|〈yj,x〉|
2 ≤ |x|2 .

and
m∑
j=1

|〈x,yj〉|
2 ≤ |x|2 .

Theorem 3.4. Let X be a right Hilbert A- module which is an algebraic left A-
module, x1, . . . ,xn and y1, . . . ,ym be a non zero vectors in X such that there exist
the non-negative real numbers ρj,µj,j ∈{1, . . . ,m} with

(3.3) Re〈yj,xi〉≥ ρj ‖xi‖‖yj‖ , Im〈yj,xi〉≥ µj ‖xi‖‖yj‖

for each i ∈{1, . . . ,n} , j ∈{1, . . . ,m} . Then

(

m∑
j=1

(ρ2j + µ
2
j)‖yj‖

2∑m
k=1 ‖〈yj,yk〉‖

)
1
2

n∑
i=1

|xi| ≤

∣∣∣∣∣
n∑
i=1

xi

∣∣∣∣∣ .



REVERSE OF THE TRIANGLE INEQUALITY 33

Proof. By (3.3), we have

(

n∑
i=1

Re〈yj,xi〉)2 + (
n∑
i=1

Im〈yj,xi〉)2 ≥ ρ2j|yj|
2(

n∑
i=1

|xi|)2 + µ2kj ‖yj‖
2

(

n∑
i=1

‖xi‖)2

= (ρ2j + µ
2
j)|yj|

2(

n∑
i=1

|xi|)2.

By combining the above inequality and this equality

1
2
(|〈
∑n
i=1 xi,yj〉|

2
+ |〈yj,

∑n
i=1 xi〉|

2
) = (

∑n
i=1 Re〈yj,xi〉)

2 + (
∑n
i=1 Im〈yj,xi〉)

2,

we deduce

1
2

∑m
j=1
|〈yj,

∑n
i=1 xi〉|

2∑
m
k=1
‖〈yj,yk〉‖

+ 1
2

∑m
j=1
|〈∑ni=1 xi,yj〉|2∑

m
k=1
‖〈yj,yk〉‖

≥ (
∑m
j=1

(ρ2j +µ
2
j )‖yj‖

2∑
m
k=1
‖〈yj,yk〉‖

)(
∑n
i=1 ‖xi‖)

2.

Apply lemma 3.1, we get

1

2

m∑
j=1

|〈yj,
∑n
i=1 xi〉|

2∑m
k=1 ‖〈yj,yk〉‖

+
1

2

m∑
j=1

|〈
∑n
i=1 xi,yj〉|

2∑m
k=1 ‖〈yj,yk〉‖

≤ |
n∑
i=1

xi|2.

Then

(

m∑
j=1

(ρ2j + µ
2
j)‖yj‖

2∑m
k=1 ‖〈yj,yk〉‖

)(

n∑
i=1

||xi| |)2 ≤

∥∥∥∥∥
n∑
i=1

xi

∥∥∥∥∥
2

.

And since |x| ≤ ‖x‖ and |x|2 ≤‖x‖2 for all x ∈X , then

(

m∑
j=1

(ρ2j + µ
2
j)‖yj‖

2∑m
k=1 ‖〈yj,yk〉‖

)(

n∑
i=1

|xi|)2 ≤

∣∣∣∣∣
n∑
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∑
j=1

ρ2j ‖yj‖
2∑m

k=1 ‖〈yj,yk〉‖
)

1
2

n∑
i=1

|xi| ≤

∣∣∣∣∣
n∑
i=1

xi

∣∣∣∣∣ .
Corollary 3.6. Let X be a right Hilbert A−module which is an algebraic left A-
module, x1, . . . ,xn and y1, . . . ,ym be a non zero vectors in X such that there exist
the non-negative real numbers ρj,µj,j ∈{1, . . . ,m} with

Re〈yj,xi〉≥ ρj ‖xi‖‖yj‖ , Im〈yj,xi〉≥ µj ‖xi‖‖yj‖ .
Then

(

m∑
j=1

(ρ2j + µ
2
j)|yj|

2

max1≤j≤m |yj|2 + (m− 1) maxk 6=j | 〈yj,yk〉 |
)

1
2

n∑
i=1

|xi| ≤

∣∣∣∣∣
n∑
i=1

xi

∣∣∣∣∣ .
Proof. It is easy to show that

m∑
k=1

‖〈yj,yk〉‖≤ max
1≤j≤m

‖yj‖
2

+ (m− 1) max
j 6=k
‖〈yj,yk〉‖ .

We thus have that
1

max1≤j≤m‖yj‖
2

+ (m− 1) maxj 6=k ‖〈yj,yk〉‖
≤

1∑m
k=1 ‖〈yj,yk〉‖

,



34 BOUNADER, CHAHBI AND KABBAJ

and
(ρ2j + µ

2
j)‖yj‖

2

max1≤j≤m‖yj‖
2

+ (m− 1) maxj 6=k ‖〈yj,yk〉‖
≤

(ρ2j + µ
2
j)‖yj‖

2∑m
k=1 ‖〈yj,yk〉‖

.

Consequently

(
∑n
j=1

(ρ2j +µ
2
j )‖yj‖

2

max1≤j≤n‖yj‖2+(m−1) maxj 6=k‖〈yj,yk〉‖
)

1
2

∑n
i=1 |xi| ≤ (

∑n
j=1

(ρ2j +µ
2
j )‖yj‖

2∑
n
k=1‖〈yj,yk〉‖

)
1
2

∑n
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 and ej be a unitary orthogonal vectors for j ∈ {1, . . . ,m} in X such that there
exist the real numbers ρj,µj,j ∈{1, . . . ,m} with

Re〈ρjxi,ej〉≥ ρ2j ‖xi‖ , Im〈µjxi,ej〉≥ µ
2
j ‖xi‖

for each i ∈{1, . . . ,n} , j ∈{1, . . . ,m} . Then

(

m∑
j=1

(ρ2j + µ
2
j))

1
2

n∑
j=1

‖xi‖≤

∥∥∥∥∥
n∑
i=1

xi

∥∥∥∥∥ .
Corollary 3.8. Let X be a right Hilbert A−module which is an algebraic left A-
module, x1, . . . ,xn and y1 . . . ,ym be 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 − 2Re〈yj,xi〉 + ||yj||2 ≤ p2j ≤ ||yj||
2

and

(3.5) ||xi||2 − 2 Im〈yj,xi〉 + ||yj||2 ≤ p2j ≤ ||yj||
2

for each i ∈{1, . . . ,n} , j ∈{1, . . . ,m} . Then

(

m∑
j=1

(2‖yj‖
2 −p2j −qj

2)∑m
k=1 ‖〈yj,yk〉‖

)
1
2

n∑
i=1

|xi| ≤

∣∣∣∣∣
n∑
i=1

xi

∣∣∣∣∣ .
Proof. By the inequality (3.4), we get

||xi||2 + ||yj||2 −p2j ≤ 2Re〈yj,xi〉 .

Since ||yj||2 −p2j ≥ 0, then

2||xi||
√
||yj||2 −p2j ≤ ||xi||

2 + ||yj||2 −p2j ≤ 2Re〈yj,xi〉 .

This implies that

Re〈yj,xi〉≥

√
||yj||2 −p2j
||yj||

||yj||||xi||.

Even from (3.5), we get

Im〈yj,xi〉≥

√
||yj||2 −q2j
||yj||

||yj||||xi||,

and if we let ρj =

√
||yj||2−p2j
||yj

and µj =

√
||yj||2−p2j
||yj

in theorem 3.4,

then by simple computation, we get the desired result. �



REVERSE OF THE TRIANGLE INEQUALITY 35

The following lemma gives a refinement of Selberg’s inequality in a right Hilbert
A− module which is an algebraic left A-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 in X . If x ∈X then

(3.6) | 〈y,x〉 |2 +
m∑
j=1

|〈yj,x〉|
2∑m

k=1 ‖〈yj,yk〉‖
||y||2 ≤ |x|2 ||y||2,

(3.7) | 〈x,y〉 |2 +
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
||y||2 ≤ |x|2 ||y||2,

(3.8) | 〈x,y〉 |2 +
m∑
j=1

|〈yj,x〉|
2∑m

k=1 ‖〈yj,yk〉‖
||y||2 ≤ |x|2 ||y||2,

and

(3.9) | 〈y,x〉 |2 +
m∑
j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖
||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∑
j=1

αjyj

where

αj =
〈yj,x〉∑n

k=1 ‖〈yj,yk〉‖
.

According to the proof of lemma 3.1, we have

|u|2 = |x−
m∑
j=1

αjyj|2

≤ |x|2 −
m∑
j=1

| 〈yj,x〉 |2∑m
k=1 | 〈yj,yk〉 |

.

Hence it follows that

‖y‖2

|x|2 − m∑

j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖


 ≥‖y‖2 |u|2 .

Applying Cauchy Schwartz inequality, we get

‖y‖2 |u|2 ≥ |〈u,y〉|2 .

and since 〈y,yj〉 = 0, so

|〈u,y〉|2 =

∣∣∣∣∣∣
〈
x−

m∑
j=1

αjyj,y

〉∣∣∣∣∣∣
2

= |〈x,y〉|2 .



36 BOUNADER, CHAHBI AND KABBAJ

It follows that

‖y‖2

|x|2 − m∑

j=1

|〈x,yj〉|
2∑m

k=1 ‖〈yj,yk〉‖


 ≥ |〈x,y〉|2 ,

which completes the proof of the inequality (3.7).
Similarly, we can get inequalities (3.8) and (3.9) . �

Lemma 3.10. Let X be a Hilbert A - module, y1, · · · , ym,y be non zero vectors in
X and x1, · · · , xn ∈X such that such that there exist the real numbers ρj,µj,j ∈
{1, · · · ,m} with

(3.10) 0 ≤ ρj ‖xi‖||yj|| ≤ Re〈yj,xi〉 , 0 ≤ µj ‖xi‖||yj|| ≤ Im〈yj,xi〉

and 〈y,yj〉 = 0, for each i ∈{1, . . . ,n} , j ∈{1, . . . ,m} . Then

(3.11) |(y,
n∑
i=1

xi)|2 + (
m∑
k=1

ρ2j + µ
2
j

cj
||y2j )(

n∑
i=1

|xi|)2||y||2 ≤ |
n∑
i=1

xi|2||y||2,

and

(3.12) |(
n∑
i=1

xi,y)|2 + (
m∑
j=1

ρ2j + µ
2
j

cj
||yj||2)(

n∑
i=1

|xi|)2||y||2 ≤ |
n∑
i=1

xi|2||y||2,

where cj =
∑m
k=1 ‖〈yj,yk〉‖ .

Proof. Let x =
∑n
i=1 xi, from (3.10), we get

||y||2{|x|2 −
∑m
j=1

ρ2j +µ
2
j

cj
(
∑n
i=1 ||xi||)

2}

≥ ||y||2{|x|2 −
∑m
j=1

Re〈x,yj〉2+Im〈x,yj〉2
cj

},

Since

||y||2{|x|2−
∑m
j=1

Re〈x,yj〉2+Im〈x,yj〉2
cj

} = ‖y‖2 {|x|2−1
2

∑m
j=1

|〈x,yj〉|2
cj
−1

2

∑m
j=1

〈yj,x〉2
cj
}.

Then, from (3.6) and (3.9), we get

‖y‖2 {|x|2 −
1

2

m∑
j=1

| 〈x,yj〉 |2

cj
−

1

2

m∑
j=1

〈yj,x〉
2

cj
}≥ |〈y,x〉 |2,

it follows that

||y||2{|x|2 −
m∑
j=1

ρ2j + µ
2
j

cj
(‖x1‖ + · · · + ‖xn‖)2}≥ |〈y,x〉|

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 ≤ ρj ‖xi‖||yj|| ≤ Re〈yj,xi〉 , 0 ≤ µj ‖xi‖||yj|| ≤ Im〈yj,xi〉 ,

(3.14) 0 ≤ a‖xi‖||y|| ≤ Re〈y,xi〉 , 0 ≤ b‖xi‖||yj|| ≤ Im〈y,xi〉



REVERSE OF THE TRIANGLE INEQUALITY 37

and 〈y,yj〉 = 0, for each i ∈{1, . . . ,n} , j ∈{1, . . . ,m} . Then

(a2 + b2 +

m∑
j=1

ρ2j + µ
2
j

cj
)||y2j ||

1
2 (|x1| + · · · + |xn|) ≤ |

n∑
i=1

xi|.

where cj =
∑m
k=1 ‖〈yj,yk〉‖ .

Proof. From (3.11) and (3.12), we get

1
2
(| 〈
∑n
i=1 xi,y〉 |

2 + | 〈y,
∑n
i=1 xi〉 |

2)

+(
∑m
j=1

ρ2j +µ
2
j

cj
||y||2)(

∑n
i=1 |xi|)

2||y||2

≤ |
∑n
i=1 xi|

2||y||2,

since

1
2
(| 〈
∑n
i=1 xi,y〉 |

2 + | 〈y,
∑n
i=1 xi〉 |

2) ≥ (Re〈y,x〉)2 + (Im〈y,x〉)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∑
j=1

(ρ2j + µ
2
j))

1
2 (

n∑
i=1

|xi|) ≤

∣∣∣∣∣
n∑
i=1

xi

∣∣∣∣∣ .
This inequality is an extension of Diaz-Metcalf [4] inequality in C∗-module.

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38 BOUNADER, CHAHBI AND KABBAJ

Department of Mathematics, Faculty of Sciences, University of Ibn Tofail, Kenitra,

Morocco

∗Corresponding author