REVERSE TRIANGLE INEQUALITY IN HILBERT C -MODULES · 2009-12-22 · Reverse Triangle Inequality Maryam Khosravi, Hakimeh Mahyar and Mohammad Sal Moslehian vol. 10, iss. 4, art. 110,

Reverse Triangle InequalityMaryam Khosravi, Hakimeh Mahyar

and Mohammad Sal Moslehian

vol. 10, iss. 4, art. 110, 2009

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REVERSE TRIANGLE INEQUALITY IN HILBERTC∗-MODULES

MARYAM KHOSRAVI, HAKIMEH MAHYARDepartment of MathematicsTarbiat Moallem UniversityTahran, Iran.

EMail: { khosravi_m, mahyar}@saba.tmu.ac.ir

MOHAMMAD SAL MOSLEHIANDepartment of Pure MathematicsCentre of Excellence in Analysis on Algebraic Structures (CEAAS)Ferdowsi University of MashhadP.O. Box 1159, Mashhad 91775, Iran.

EMail: [email protected]

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, HilbertC∗-module,C∗-algebra.

http://jipam.vu.edu.au

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Abstract: We prove several versions of reverse triangle inequality in HilbertC∗-modules. We show that ife1, . . . , em are vectors in a Hilbert moduleXover aC∗-algebraA with unit 1 such that〈ei, ej〉 = 0 (1 ≤ i 6= j ≤ m)and‖ei‖ = 1 (1 ≤ i ≤ m), and alsork, ρk ∈ R (1 ≤ k ≤ m) andx1, . . . , xn ∈ X satisfy

0 ≤ r2k‖xj‖ ≤ Re〈rkek, xj〉 , 0 ≤ ρ2

k‖xj‖ ≤ Im〈ρkek, xj〉 ,

then [m∑

k=1

(r2k + ρ2

k

)] 12 n∑

j=1

‖xj‖ ≤

∥∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥∥ ,

and the equality holds if and only if

n∑j=1

xj =n∑

j=1

‖xj‖m∑

k=1

(rk + iρk)ek .


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Contents

1 Introduction and Preliminaries 4

2 Multiplicative Reverse of the Triangle Inequality 6

3 Additive Reverse of the Triangle Inequality 16


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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 numbersz1, . . . , zn,

(1.1)

∣∣∣∣∣n∑

j=1

zj

∣∣∣∣∣ ≥ cos θn∑

j=1

|zj| ,

where0 < θ < π2

andα− θ < arg zj < α + θ (1 ≤ j ≤ n) for a given real numberα.

The first generalization of the reverse triangle inequality in Hilbert spaces wasgiven by Diaz and Metcalf [5]. They proved that forx1, . . . , xn in a Hilbert spaceH, if e is a unit vector ofH such that0 ≤ r ≤ Re〈xj ,e〉

‖xj‖ for somer ∈ R and each1 ≤ j ≤ n, then

(1.2) rn∑

j=1

‖xj‖ ≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥ .

Moreover, the equality holds if and only if∑n

j=1 xj = r∑n

j=1 ‖xj‖e.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 ofC∗-valued triangle inequalities in HilbertC∗-modules was given in [3]. Our aim is to generalize some of the results of Dragomirin Hilbert spaces to the framework of HilbertC∗-modules. For this purpose, we firstrecall some fundamental definitions in the theory of HilbertC∗-modules.

Suppose thatA is a C∗-algebra andX is a linear space, which is an algebraicright A-module. The spaceX is called a pre-HilbertA-module (or an inner productA-module) if there exists anA-valued inner product〈·, ·〉 : X × X → A with thefollowing properties:


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(i) 〈x, x〉 ≥ 0 and〈x, x〉 = 0 if and only if x = 0;

(ii) 〈x, λy + z〉 = λ〈x, y〉+ 〈x, z〉;

(iii) 〈x, ya〉 = 〈x, y〉a;

(iv) 〈x, y〉∗ = 〈y, x〉for all x, y, z ∈ X, a ∈ A, λ ∈ C. By (ii) and (iv), 〈·, ·〉 is conjugate lin-ear in the first variable. Using the Cauchy–Schwartz inequality〈y, x〉〈x, y〉 ≤‖〈x, x〉‖〈y, y〉 [11, Page 5] (see also [14]), it follows that ‖x‖ = ‖〈x, x〉‖ 1

2 isa norm onX making it a right normed module. The pre-Hilbert moduleX iscalled a HilbertA-module if it is complete with respect to this norm. Noticethat the inner structure of aC∗-algebra is essentially more complicated thanthat for complex numbers. For instance, properties such as orthogonality andtheorems such as Riesz’ representation in complex Hilbert space theory cannotsimply be generalized or transferred to the theory of HilbertC∗-modules.

One may define an “A-valued norm”| · | by |x| = 〈x, x〉1/2. Clearly,‖ |x| ‖ = ‖x‖for eachx ∈ X. It is known that| · | does not satisfy the triangle inequality in general.See [11, 12] for more information on HilbertC∗-modules.

We also use elementaryC∗-algebra theory, in particular we utilize the propertythat if a ≤ b thena1/2 ≤ b1/2, wherea, b are positive elements of aC∗-algebraA.We also repeatedly apply the following known relation:

(1.3)1

2(aa∗ + a∗a) = (Re a)2 + (Im a)2 ,

wherea is an arbitrary element ofA. For details onC∗-algebra theory, we referreaders to [15].

Throughout the paper, we assume thatA is a unitalC∗-algebra with unit1 andfor everyλ ∈ C, we writeλ for λ1.


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2. Multiplicative Reverse of the Triangle Inequality

Utilizing someC∗-algebraic techniques we present our first result as a generalizationof [7, Theorem 2.3].

Theorem 2.1.LetA be aC∗-algebra, letX be a HilbertA-module and letx1, . . . , xn ∈X. If there exist real numbersk1, k2 ≥ 0 with

0 ≤ k1‖xj‖ ≤ Re〈e, xj〉 , 0 ≤ k2‖xj‖ ≤ Im〈e, xj〉 ,for somee ∈ X with |e| ≤ 1 and all1 ≤ j ≤ n, then

(2.1) (k21 + k2

2)12

n∑j=1

‖xj‖ ≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥ .

Proof. Applying the Cauchy–Schwarz inequality, we get∣∣∣∣∣⟨

e,n∑

j=1

xj

⟩∣∣∣∣∣2

≤ ‖e‖2

∣∣∣∣∣n∑

j=1

xj

∣∣∣∣∣2

≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2

,

and ∣∣∣∣∣⟨

n∑j=1

xj, e

⟩∣∣∣∣∣2

≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2

|e|2 ≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2

,

whence∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2

≥ 1

2

∣∣∣∣∣⟨

e,n∑

j=1

xj

⟩∣∣∣∣∣2

+

∣∣∣∣∣⟨

n∑j=1

xj, e

⟩∣∣∣∣∣2

=1

2

(⟨e,

n∑j=1

xj

⟩∗⟨e,

n∑j=1

xj

⟩+

⟨n∑

j=1

xj, e

⟩∗⟨ n∑j=1

xj, e

⟩)


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=

(Re

⟨e,

n∑j=1

xj

⟩)2

+

(Im

⟨e,

n∑j=1

xj

⟩)2 (by (1.3)

)=

(Re

n∑j=1

〈e, xj〉

)2

+

(Im

n∑j=1

〈e, xj〉

)2

≥ k21

(n∑

j=1

‖xj‖

)2

+ k22

(n∑

j=1

‖xj‖

)2

= (k21 + k2

2)

(n∑

j=1

‖xj‖

)2

.

Using the same argument as in the proof of Theorem2.1, one can obtain thefollowing result, wherek1, k2 are hermitian elements ofA.

Theorem 2.2. If the vectorsx1, . . . , xn ∈ X satisfy the conditions

0 ≤ k21‖xj‖2 ≤ (Re〈e, xj〉)2 , 0 ≤ k2

2‖xj‖2 ≤ (Im〈e, xj〉)2 ,

for some hermitian elementsk1, k2 in A, somee ∈ X with |e| ≤ 1 and all1 ≤ j ≤ nthen the inequality(2.1) holds.

One may observe an integral version of inequality (2.1) as follows:

Corollary 2.3. Suppose thatX is a HilbertA-module andf : [a, b] → X is stronglymeasurable such that the Lebesgue integral

∫ b

a‖f(t)‖dt exists and is finite. If there

exist self-adjoint elementsa1, a2 in A with

a21‖f(t)‖2 ≤ Re〈f(t), e〉2 , a2

2‖f(t)‖2 ≤ Im〈f(t), e〉2 (a.e. t ∈ [a, b]) ,


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wheree ∈ X with |e| ≤ 1, then

(a21 + a2

2)12

∫ b

a

‖f(t)‖dt ≤∥∥∥∥∫ b

a

f(t)dt

∥∥∥∥ .

Now we prove a useful lemma, which is frequently applied in the next theorems(see also [3]).

Lemma 2.4. Let X be a HilbertA-module and letx, y ∈ X. If |〈x, y〉| = ‖x‖‖y‖,then

y =x〈x, y〉‖x‖2

.

Proof. Forx, y ∈ X we have

0 ≤∣∣∣∣y − x〈x, y〉

‖x‖2

∣∣∣∣2 =

⟨y − x〈x, y〉

‖x‖2, y − x〈x, y〉

‖x‖2

⟩= 〈y, y〉 − 1

‖x‖2〈y, x〉〈x, y〉+

1

‖x‖4〈y, x〉〈x, x〉〈x, y〉 − 1

‖x‖2〈y, x〉〈x, y〉

≤ |y|2 − 1

‖x‖2|〈x, y〉|2 = |y|2 − 1

‖x‖2‖x‖2‖y‖2

= |y|2 − ‖y‖2 ≤ 0 ,

whence∣∣∣y − x〈x,y〉

‖x‖2

∣∣∣ = 0. Hencey = x〈x,y〉‖x‖2 .

Using the Cauchy–Schwarz inequality, we have the following theorem for Hilbertmodules, which is similar to [1, Theorem 2.5].


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Theorem 2.5. Let e1, . . . , em be a family of vectors in a Hilbert moduleX over aC∗-algebraA such that〈ei, ej〉 = 0 (1 ≤ i 6= j ≤ m) and‖ei‖ = 1 (1 ≤ i ≤ m).Suppose thatrk, ρk ∈ R (1 ≤ k ≤ m) and that the vectorsx1, . . . , xn ∈ X satisfy

0 ≤ r2k‖xj‖ ≤ Re〈rkek, xj〉 , 0 ≤ ρ2

k‖xj‖ ≤ Im〈ρkek, xj〉 ,

Then

(2.2)

[m∑

k=1

(r2k + ρ2

k)

] 12 n∑

j=1

‖xj‖ ≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥ ,

and the equality holds if and only if

(2.3)n∑

j=1

xj =n∑

j=1

‖xj‖m∑

k=1

(rk + iρk)ek .

Proof. There is nothing to prove if∑m

k=1(r2k + ρ2

k) = 0. Assume that∑m

k=1(r2k +

ρ2k) 6= 0. From the hypothesis, byIm(a) = Re(ia∗), Re(a∗) = Re(a) (a ∈ A), we

have (m∑

k=1

(r2k + ρ2

k)

)2( n∑j=1

‖xj‖

)2

≤

(Re

⟨m∑

k=1

rkek,n∑

j=1

xj

⟩+ Im

⟨m∑

k=1

ρkek,n∑

j=1

xj

⟩)2

=

(Re

⟨n∑

j=1

xj,m∑

k=1

(rk + iρk)ek

⟩)2

.


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By (1.3),(Re

⟨n∑

j=1

xj,m∑

k=1

(rk + iρk)ek

⟩)2

≤ 1

2

∣∣∣∣∣⟨

n∑j=1

xj,m∑

k=1

(rk + iρk)ek

⟩∣∣∣∣∣2

+

∣∣∣∣∣⟨

m∑k=1

(rk + iρk)ek,n∑

j=1

xj

⟩∣∣∣∣∣2

≤ 1

2

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2 ∣∣∣∣∣

m∑k=1

(rk + iρk)ek

∣∣∣∣∣2

+

∥∥∥∥∥m∑

k=1

(rk + iρk)ek

∥∥∥∥∥2 ∣∣∣∣∣

n∑j=1

xj

∣∣∣∣∣2

≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2 ∥∥∥∥∥

m∑k=1

(rk + iρk)ek

∥∥∥∥∥2

and since|a| ≤ ‖a‖ (a ∈ A),∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2 ∥∥∥∥∥

m∑k=1

(rk + iρk)ek

∥∥∥∥∥2

≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2 ∥∥∥∥∥⟨

m∑k=1

(rk + iρk)ek,m∑

k=1

(rk + iρk)ek

⟩∥∥∥∥∥=

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2 m∑

k=1

|rk + iρk|2‖ek‖2

=

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2 m∑

k=1

(r2k + ρ2

k) .


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Hence [m∑

k=1

(r2k + ρ2

k)

](n∑

j=1

‖xj‖

)2

≤

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2

.

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 aboveshould be equality. Therefore

r2k‖xj‖ = Re〈rkek, xj〉 , ρ2

k‖xj‖ = Im〈ρkek, xj〉 ,

Re

⟨n∑

j=1

xj,m∑

k=1

(rk + iρk)ek

⟩=

⟨n∑

j=1

xj,m∑

k=1

(rk + iρk)ek

⟩,

and ∣∣∣∣∣⟨

m∑k=1

(rk + iρk)ek,n∑

j=1

xj

⟩∣∣∣∣∣ =

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥∥∥∥∥∥

m∑k=1

(rk + iρk)ek

∥∥∥∥∥ .

From Lemma2.4and the above equalities we have

n∑j=1

xj =

∑mk=1(rk + iρk)ek

‖∑m

k=1(rk + iρk)ek‖2

⟨m∑

k=1

(rk + iρk)ek,n∑

j=1

xj

⟩

=

∑mk=1(rk + iρk)ek∑m

k=1(r2k + ρ2

k)Re

⟨m∑

k=1

(rk + iρk)ek,n∑

j=1

xj

⟩

=

∑mk=1(rk + iρk)ek∑m

k=1(r2k + ρ2

k)

m∑k=1

n∑j=1

(r2k‖xj‖+ ρ2

k‖xj‖)


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=n∑

j=1

‖xj‖m∑

k=1

(rk + iρk)ek ,

which is the desired result.

In the next results of this section, we assume thatX is a right HilbertA-module,which is an algebraic leftA-module subject to

〈x, ay〉 = a〈x, y〉 (x, y ∈ X, a ∈ A) . (†)

For example ifA is a unitalC∗-algebra andI is a commutative right ideal ofA, thenI is a right Hilbert module overA and

〈x, ay〉 = x∗(ay) = ax∗y = a〈x, y〉 (x, y ∈ I, a ∈ A) .

The next theorem is a refinement of [7, Theorem 2.1]. To prove it we need thefollowing lemma.

Lemma 2.6. LetX be a HilbertA-module ande1, . . . , en ∈ X be a family of vectorssuch that〈ei, ej〉 = 0 (i 6= j) and‖ei‖ = 1. If x ∈ X, then

|x|2 ≥n∑

k=1

|〈ek, x〉|2 and |x|2 ≥n∑

k=1

|〈x, ek〉|2 .

Proof. The first result follows from the following inequality:

0 ≤

∣∣∣∣∣x−n∑

k=1

ek〈ek, x〉

∣∣∣∣∣2

=

⟨x−

n∑k=1

ek〈ek, x〉, x−n∑

j=1

ej〈ej, x〉

⟩

= 〈x, x〉+n∑

k=1

n∑j=1

〈ek, x〉∗〈ek, ej〉〈ej, x〉 − 2n∑

k=1

|〈ek, x〉|2


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= 〈x, x〉+n∑

k=1

〈ek, x〉∗〈ek, ek〉〈ek, x〉 − 2n∑

k=1

|〈ek, x〉|2

≤ |x|2 +n∑

k=1

〈ek, x〉∗〈ek, x〉 − 2n∑

k=1

|〈ek, x〉|2

= |x|2 −n∑

k=1

|〈ek, x〉|2 .

By considering|x−∑n

k=1〈ek, x〉ek|2, we similarly obtain the second one.

Now we will prove the next theorem without using the Cauchy–Schwarz inequal-ity.

Theorem 2.7. Let 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 vectorsx1, . . . , xn ∈ X satisfy theconditions

(2.4) 0 ≤ rk‖xj‖ ≤ Re〈ek, xj〉 , 0 ≤ ρk‖xj‖ ≤ Im〈ek, xj〉 ,

for 1 ≤ j ≤ n, 1 ≤ k ≤ m, whererk, ρk ∈ [0,∞) (1 ≤ k ≤ m), then

(2.5)

[m∑

k=1

(r2k + ρ2

k)

] 12 n∑

j=1

‖xj‖ ≤

∣∣∣∣∣n∑

j=1

xj

∣∣∣∣∣ .

Proof. Applying the previous lemma forx =∑n

j=1 xj, we obtain∣∣∣∣∣n∑

j=1

xj

∣∣∣∣∣2

≥ 1

2

m∑k=1

∣∣∣∣∣⟨

ek,

n∑j=1

xj

⟩∣∣∣∣∣2

+m∑

k=1

∣∣∣∣∣⟨

n∑j=1

xj, ek

⟩∣∣∣∣∣2


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=m∑

k=1

1

2

(⟨ek,

n∑j=1

xj

⟩∗⟨ek,

n∑j=1

xj

⟩+

⟨n∑

j=1

xj, ek

⟩∗⟨ n∑j=1

xj, ek

⟩)

=m∑

k=1

(Re

⟨ek,

n∑j=1

xj

⟩)2

+

(Im

⟨ek,

n∑j=1

xj

⟩)2

(by (1.3))

=m∑

k=1

(Re

n∑j=1

〈ek, xj〉

)2

+

(Im

n∑j=1

〈ek, xj〉

)2

≥m∑

k=1

r2k

(n∑

j=1

‖xj‖

)2

+ ρ2k

(n∑

j=1

‖xj‖

)2 (by (2.4))

=m∑

k=1

(r2k + ρ2

k)

(n∑

j=1

‖xj‖

)2

.

Proposition 2.8. In Theorem2.7, if 〈ek, ek〉 = 1, then the equality holds in (2.5) ifand only if

(2.6)n∑

j=1

xj =

(n∑

j=1

‖xj‖

)m∑

k=1

(rk + iρk)ek .

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


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of Theorem2.7should be equalities. Hence∣∣∣∣∣n∑

j=1

xj

∣∣∣∣∣2

=m∑

k=1

∣∣∣∣∣⟨

ek,n∑

j=1

xj

⟩∣∣∣∣∣2

and

∣∣∣∣∣n∑

j=1

xj

∣∣∣∣∣2

=m∑

k=1

∣∣∣∣∣⟨

n∑j=1

xj, ek

⟩∣∣∣∣∣2

,

which is equivalent to

n∑j=1

xj =m∑

k=1

n∑j=1

ek〈ek, xj〉 =m∑

k=1

n∑j=1

〈ek, xj〉ek ,

andrk‖xj‖ = Re〈ek, xj〉 , ρk‖xj‖ = Im〈ek, xj〉 .

Son∑

j=1

xj =m∑

k=1

n∑j=1

ek〈ek, xj〉

=m∑

k=1

n∑j=1

ek(rk + iρk)‖xj‖

=

(n∑

j=1

‖xj‖

)m∑

k=1

(rk + iρk)ek .


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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}mk=1 be a family of orthonormal vectors in a Hilbert spaceH

andMjk ≥ 0 (1 ≤ j ≤ n, 1 ≤ k ≤ m) such that

‖xj‖ − Re〈ek, xj〉 ≤ Mjk ,

for each1 ≤ j ≤ n and1 ≤ k ≤ m. Then

n∑j=1

‖xj‖ ≤1√m

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥+1

m

n∑j=1

m∑k=1

Mjk ;

and the equality holds if and only ifn∑

j=1

‖xj‖ ≥1

m

n∑j=1

m∑k=1

Mjk ,

andn∑

j=1

xj =

(n∑

j=1

‖xj‖ −1

m

n∑j=1

m∑k=1

Mjk

)m∑

k=1

ek .

We can prove this theorem for HilbertC∗-modules using some different tech-niques.

Theorem 3.2. Let {ek}mk=1 be a family of vectors in a Hilbert moduleX over aC∗-

algebraA with unit 1, |ek| ≤ 1 (1 ≤ k ≤ m), 〈ei, ej〉 = 0 (1 ≤ i 6= j ≤ m) andxj ∈ X (1 ≤ j ≤ n). If for some scalarsMjk ≥ 0 (1 ≤ j ≤ n, 1 ≤ k ≤ m),

(3.1) ‖xj‖ − Re〈ek, xj〉 ≤ Mjk (1 ≤ j ≤ n, 1 ≤ k ≤ m) ,


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then

(3.2)n∑

j=1

‖xj‖ ≤1√m

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥+1

m

n∑j=1

m∑k=1

Mjk .

Moreover, if|ek| = 1 (1 ≤ k ≤ m), then the equality in (3.2) holds if and only if

(3.3)n∑

j=1

‖xj‖ ≥1

m

n∑j=1

m∑k=1

Mjk ,

and

(3.4)n∑

j=1

xj =

(n∑

j=1

‖xj‖ −1

m

n∑j=1

m∑k=1

Mjk

)m∑

k=1

ek .

Proof. Taking the summation in (3.1) overj from 1 ton, we obtain

n∑j=1

‖xj‖ ≤ Re

⟨ek,

n∑j=1

xj

⟩+

n∑j=1

Mjk ,

for eachk ∈ {1, . . . ,m}. Summing these inequalities overk from 1 tom, we deduce

(3.5)n∑

j=1

‖xj‖ ≤1

mRe

⟨m∑

k=1

ek,n∑

j=1

xj

⟩+

1

m

m∑k=1

n∑j=1

Mjk .


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Using the Cauchy–Schwarz we obtain(Re

⟨m∑

k=1

ek,n∑

j=1

xj

⟩)2

(3.6)

≤ 1

2

∣∣∣∣∣⟨

m∑k=1

ek,n∑

j=1

xj

⟩∣∣∣∣∣2

+

∣∣∣∣∣⟨

m∑k=1

ek,n∑

j=1

xj

⟩∗∣∣∣∣∣2

≤ 1

2

∥∥∥∥∥m∑

k=1

ek

∥∥∥∥∥2 ∣∣∣∣∣

n∑j=1

xj

∣∣∣∣∣2

+

∣∣∣∣∣m∑

k=1

ek

∣∣∣∣∣2 ∥∥∥∥∥

n∑j=1

xj

∥∥∥∥∥2

≤

∥∥∥∥∥m∑

k=1

ek

∥∥∥∥∥2 ∥∥∥∥∥

n∑j=1

xj

∥∥∥∥∥2

≤ m

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥2

,

since∥∥∥∥∥m∑

k=1

ek

∥∥∥∥∥2

=

∥∥∥∥∥⟨

m∑k=1

ek,m∑

k=1

ek

⟩∥∥∥∥∥ =

∥∥∥∥∥m∑

k=1

m∑l=1

〈ek, el〉

∥∥∥∥∥ =

∥∥∥∥∥m∑

k=1

|ek|2∥∥∥∥∥ ≤ m .

Using (3.6) in (3.5), we deduce the desired inequality.If (3.3) and (3.4) hold, then

1√m

∥∥∥∥∥n∑

j=1

xj

∥∥∥∥∥ =1√m

(n∑

j=1

‖xj‖ −1

m

n∑j=1

m∑k=1

Mjk

)∥∥∥∥∥m∑

k=1

ek

∥∥∥∥∥=

n∑j=1

‖xj‖ −1

m

n∑j=1

m∑k=1

Mjk ,


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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

‖xj‖ − Re〈ek, xj〉 = Mjk ,

Re

⟨m∑

k=1

ek,n∑

j=1

xj

⟩=

⟨m∑

k=1

ek,n∑

j=1

xj

⟩,

and ∣∣∣∣∣⟨

m∑k=1

ek,n∑

j=1

xj

⟩∣∣∣∣∣ =

∥∥∥∥∥m∑

k=1

ek

∥∥∥∥∥∥∥∥∥∥

n∑j=1

xj

∥∥∥∥∥ .

It follows from Lemma2.4and the previous relations that

n∑j=1

xj =

∑mk=1 ek

‖∑m

k=1 ek‖2

⟨m∑

k=1

ek,n∑

j=1

xj

⟩

=

∑mk=1 ek

mRe

⟨m∑

k=1

ek,n∑

j=1

xj

⟩

=

∑mk=1 ek

m

m∑k=1

n∑j=1

(‖xj‖ −Mjk)

=

(n∑

j=1

‖xj‖ −1

m

n∑j=1

m∑k=1

Mjk

)m∑

k=1

ek .


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References

[1] A.H. ANSARI AND M.S. MOSLEHIAN, Refinements of reverse triangle in-equality 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 inequalityin inner product spaces,Inter. J. Math. Math. Sci.,18 (2005), 2883–2893.

[3] Lj. ARAMBA CIC AND R. RAJIC, On theC∗-valued triangle equality and in-equality in HilbertC∗-modules,Acta Math. Hungar.,119(4) (2008), 373–380.

[4] I. BRNETIC, S.S. DRAGOMIR, R. HOXHAAND J. PECARIC, A reverse ofthe triangle inequality in inner product spaces and applications for polynomials,Aust. J. Math. Anal. Appl.,3(2) (2006), Art. 9.

[5] J.B. DIAZ AND F.T. METCALF, A complementary triangle inequality inHilbert 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 com-plex inner product spaces,RGMIA Res. Rep. Coll.,7(E) (2004), Art. 7.

[8] S.S. DRAGOMIR, Reverse of the continuous triangle inequality for Bochnerintegral in complex Hilbert spaces,J. Math. Anal. Appl.,329(2007), 65–76.

[9] J. KARAMATA, Teorijia i Praksa Stieltjesova Integrala (Sebro-Coratian)(Stieltjes Integral, Theory and Practice), SANU, Posebna izdanja,154, Beograd, 1949.


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[10] M. KATO, K.S. SAITO AND T. TAMURA, Sharp triangle inequality and itsreverse in Banach spaces,Math. Inequal. Appl.,10 (2007), 451–460.

[11] E.C. LANCE, Hilbert C∗-modules, London Mathematical Society LectureNote Series, 210, Cambridge University Press, Cambridge, 1995.

[12] V.M. MANUILOV AND E.V. TROITSKY,Hilbert C∗-Modules, Translations ofMathematical Monographs, 226. American Mathematical Society, Providence,RI, 2005.

[13] D.S. MITRINOVIC, J.E. PECARIC AND A.M. FINK, Classical and New In-equalities in Analysis, Kluwer Academic, Dordrecht, 1993.

[14] M.S. MOSLEHIAN AND L.-E. PERSSON, Reverse Cauchy–Schwarz inequal-ities for positive C*-valued sesquilinear forms,Math. Inequal. Appl.(to ap-pear).

[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.


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REVERSE TRIANGLE INEQUALITY IN HILBERT C -MODULES · 2009-12-22 · Reverse Triangle Inequality Maryam Khosravi, Hakimeh Mahyar and Mohammad Sal Moslehian vol. 10, iss. 4, art. 110,