We prove several versions of reverse triangle inequality in HilbertC∗-modules

REVERSE TRIANGLE INEQUALITY IN HILBERT C^∗-MODULES

MARYAM KHOSRAVI, HAKIMEH MAHYAR, AND MOHAMMAD SAL MOSLEHIAN DEPARTMENT OFMATHEMATICS

TARBIATMOALLEMUNIVERSITY

TAHRAN, IRAN.

khosravi₋m@saba.tmu.ac.ir mahyar@saba.tmu.ac.ir DEPARTMENT OFPUREMATHEMATICS

CENTRE OFEXCELLENCE INANALYSIS ONALGEBRAICSTRUCTURES(CEAAS) FERDOWSIUNIVERSITY OFMASHHAD

P.O. BOX1159, MASHHAD91775, IRAN. 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

ABSTRACT. We prove several versions of reverse triangle inequality in HilbertC^∗-modules. We show that ife₁, . . . , e_mare vectors in a Hilbert moduleXover aC^∗-algebraAwith unit 1 such thathe_i, e_ji= 0 (1≤i6=j ≤m)andke_ik= 1 (1≤i≤m), and alsor_k, ρ_k ∈R(1 ≤k≤ m)andx₁, . . . , x_n∈Xsatisfy

0≤r²_kkxjk ≤Rehrkek, xji, 0≤ρ²_kkxjk ≤Imhρkek, xji, then

"_m X

k=1

r_k²+ρ²_k

#¹_{2 n} X

j=1

kxjk ≤

n

X

j=1

xj

,

and the equality holds if and only if

n

X

j=1

x_j=

n

X

j=1

kxjk

m

X

k=1

(r_k+iρ_k)e_k.

Key words and phrases: Triangle inequality, Reverse inequality, HilbertC^∗-module,C^∗-algebra.

2000 Mathematics Subject Classification. Primary 46L08; Secondary 15A39, 26D15, 46L05, 51M16.

1. INTRODUCTION ANDPRELIMINARIES

The triangle inequality is one of the most fundamental inequalities in mathematics. Several mathematicians have investigated its generalizations and its reverses.

264-09

~

In 1917, Petrovitch [17] proved that for complex numbersz₁, . . . , z_n, (1.1)

n

X

j=1

zj

≥cosθ

n

X

j=1

|zj|,

where0< θ < ^π₂ andα−θ < arg z_j < α+θ (1≤j ≤n)for a given real numberα.

The first generalization of the reverse triangle inequality in Hilbert spaces was given by Diaz and Metcalf [5]. They proved that forx₁, . . . , x_nin a Hilbert spaceH, ifeis a unit vector ofH such that0≤r ≤ ^Rehx_kx^j,ei

jk for somer∈Rand each1≤j ≤n, then

(1.2) r

n

X

j=1

kxjk ≤

n

X

j=1

xj

.

Moreover, the equality holds if and only ifPn

j=1x_j =rPn

j=1kx_jke.

Recently, a number of mathematicians have presented several refinements of the reverse tri- angle 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 Dragomir in Hilbert spaces to the framework of Hilbert C^∗-modules. For this purpose, we first recall some fundamental definitions in the theory of HilbertC^∗-modules.

Suppose thatAis aC^∗-algebra andXis a linear space, which is an algebraic rightA-module.

The spaceXis called a pre-HilbertA-module (or an inner productA-module) if there exists an A-valued inner producth·,·i:X×X→Awith the following properties:

(i) hx, xi ≥0andhx, xi= 0if and only ifx= 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 inequalityhy, xihx, yi ≤ khx, xikhy, yi[11, Page 5] (see also [14]), it follows that kxk = khx, xik¹² is a norm on X making it a right normed module. The pre-Hilbert module X is called a HilbertA-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, xi^1/2. Clearly, k |x| k= kxkfor each x ∈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 property that ifa≤ b thena^1/2 ≤b^1/2, wherea, bare positive elements of aC^∗-algebraA. We also repeatedly apply the following known relation:

(1.3) 1

2(aa^∗+a^∗a) = (Rea)²+ (Ima)²,

whereais an arbitrary element ofA. For details onC^∗-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 writeλforλ1.

2. MULTIPLICATIVEREVERSE OF THE TRIANGLE INEQUALITY

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

Theorem 2.1. LetAbe aC^∗-algebra, letXbe a HilbertA-module and let x₁, . . . , x_n ∈ X. If there exist real numbersk₁, k₂ ≥0with

0≤k₁kx_jk ≤Rehe, x_ji, 0≤k₂kx_jk ≤Imhe, x_ji, for somee∈Xwith|e| ≤1and all1≤j ≤n, then

(2.1) (k₁²+k₂²)¹²

n

X

j=1

kx_jk ≤

n

X

j=1

x_j .

Proof. Applying the Cauchy–Schwarz inequality, we get

* e,

n

X

j=1

x_j +

2

≤ kek²

n

X

j=1

x_j

2

≤

n

X

j=1

x_j

2

, and

* _n X

j=1

xj, e +

2

≤

n

X

j=1

xj

2

|e|² ≤

n

X

j=1

xj

2

, whence

n

X

j=1

xj

2

≥ 1 2





* e,

n

X

j=1

xj

+

2

+

* _n X

j=1

xj, e +

2



= 1 2

* e,

n

X

j=1

x_j +∗*

e,

n

X

j=1

x_j +

+

* _n X

j=1

x_j, e

+∗* _n X

j=1

x_j, e +!

= Re

* e,

n

X

j=1

x_j +!2

+ Im

* e,

n

X

j=1

x_j +!2

by(1.3)

= Re

n

X

j=1

he, x_ji

!2

+ Im

n

X

j=1

he, x_ji

!2

≥k₁²

n

X

j=1

kx_jk

!2

+k₂²

n

X

j=1

kx_jk

!2

= (k²₁+k₂²)

n

X

j=1

kx_jk

!2

.

Using the same argument as in the proof of Theorem 2.1, one can obtain the following result, wherek₁, k₂are hermitian elements ofA.

Theorem 2.2. If the vectorsx₁, . . . , x_n∈Xsatisfy the conditions

0≤k²₁kxjk² ≤(Rehe, xji)², 0≤k₂²kxjk² ≤(Imhe, xji)²,

for some hermitian elementsk₁, k₂ inA, some e ∈ Xwith|e| ≤ 1and all1 ≤ j ≤ n then the inequality (2.1) holds.

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

Corollary 2.3. Suppose thatXis a Hilbert A-module andf : [a, b] → Xis strongly measur- able such that the Lebesgue integralRb

akf(t)kdt exists and is finite. If there exist self-adjoint elementsa₁, a₂inAwith

a²₁kf(t)k² ≤Rehf(t), ei², a²₂kf(t)k² ≤Imhf(t), ei² (a.e. t∈[a, b]), wheree∈Xwith|e| ≤1, then

(a²₁+a²₂)¹² Z b

a

kf(t)kdt≤

Z 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. LetXbe a HilbertA-module and letx, y ∈X. If|hx, yi|=kxkkyk, then y= xhx, yi

kxk² .

Proof. Forx, y ∈Xwe have 0≤

y− xhx, yi kxk²

2

=

y− xhx, yi

kxk² , y−xhx, yi kxk²

=hy, yi − 1

kxk²hy, xihx, yi+ 1

kxk⁴hy, xihx, xihx, yi − 1

kxk²hy, xihx, yi

≤ |y|²− 1

kxk²|hx, yi|² =|y|²− 1

kxk²kxk²kyk²

=|y|²− kyk² ≤0, whence

y− ^xhx,yi_kxk2

= 0. Hencey= ^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, . . . , e_m be a family of vectors in a Hilbert moduleXover aC^∗-algebra A such that he_i, e_ji = 0 (1 ≤ i 6= j ≤ m) and ke_ik = 1 (1 ≤ i ≤ m). Suppose that rk, ρk∈R (1≤k≤m)and that the vectorsx1, . . . , xn∈Xsatisfy

0≤r²_kkx_jk ≤Rehr_ke_k, x_ji, 0≤ρ²_kkx_jk ≤Imhρ_ke_k, x_ji, Then

(2.2)

" _m X

k=1

(r²_k+ρ²_k)

#¹_{2 n} X

j=1

kx_jk ≤

n

X

j=1

x_j ,

and the equality holds if and only if (2.3)

n

X

j=1

x_j =

n

X

j=1

kx_jk

m

X

k=1

(r_k+iρ_k)e_k.

Proof. There is nothing to prove ifPm

k=1(r_k²+ρ²_k) = 0. Assume thatPm

k=1(r²_k+ρ²_k)6= 0. From the hypothesis, byIm(a) = Re(ia^∗),Re(a^∗) = Re(a) (a∈A),we have

m

X

k=1

(r_k²+ρ²_k)

!2 n

X

j=1

kxjk

!2

≤ Re

* _m X

k=1

rkek,

n

X

j=1

xj

+ + Im

* _m X

k=1

ρkek,

n

X

j=1

xj

+!2

= Re

* _n X

j=1

xj,

m

X

k=1

(rk+iρk)ek

+!2

.

By (1.3), Re

* _n X

j=1

x_j,

m

X

k=1

(r_k+iρ_k)e_k +!2

≤ 1 2





* _n X

j=1

x_j,

m

X

k=1

(r_k+iρ_k)e_k +

2

+

* _m X

k=1

(r_k+iρ_k)e_k,

n

X

j=1

x_j +

2



≤ 1 2





n

X

j=1

xj

2

m

X

k=1

(rk+iρk)ek

2

+

m

X

k=1

(rk+iρk)ek

2

n

X

j=1

xj

2



≤

n

X

j=1

x_j

2

m

X

k=1

(r_k+iρ_k)e_k

2

and since|a| ≤ kak (a∈A),

n

X

j=1

x_j

2

m

X

k=1

(r_k+iρ_k)e_k

2

≤

n

X

j=1

x_j

2

* _m X

k=1

(r_k+iρ_k)e_k,

m

X

k=1

(r_k+iρ_k)e_k +

=

n

X

j=1

x_j

2 m

X

k=1

|r_k+iρ_k|²ke_kk²

=

n

X

j=1

x_j

2 m

X

k=1

(r_k²+ρ²_k). Hence

" _m X

k=1

(r_k²+ρ²_k)

# _n X

j=1

kxjk

!2

≤

n

X

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 above should be equality. Therefore

r_k²kx_jk= Rehr_ke_k, x_ji, ρ²_kkx_jk= Imhρ_ke_k, x_ji, Re

* _n X

j=1

x_j,

m

X

k=1

(r_k+iρ_k)e_k +

=

* _n X

j=1

x_j,

m

X

k=1

(r_k+iρ_k)e_k +

,

and

* _m X

k=1

(r_k+iρ_k)e_k,

n

X

j=1

x_j +

=

n

X

j=1

x_j

m

X

k=1

(r_k+iρ_k)e_k .

From Lemma 2.4 and the above equalities we have

n

X

j=1

x_j = Pm

k=1(r_k+iρ_k)e_k kPm

k=1(r_k+iρ_k)e_kk²

* _m X

k=1

(r_k+iρ_k)e_k,

n

X

j=1

x_j +

= Pm

k=1(r_k+iρ_k)e_k Pm

k=1(r²_k+ρ²_k) Re

* _m X

k=1

(r_k+iρ_k)e_k,

n

X

j=1

x_j +

= Pm

k=1(rk+iρk)ek

Pm

k=1(r²_k+ρ²_k)

m

X

k=1 n

X

j=1

(r²_kkx_jk+ρ²_kkx_jk)

=

n

X

j=1

kx_jk

m

X

k=1

(r_k+iρ_k)e_k,

which is the desired result.

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

hx, ayi=ahx, yi (x, y ∈X, a∈A). (†) For example ifAis a unitalC^∗-algebra andIis a commutative right ideal ofA, thenIis a right Hilbert module overAand

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. LetXbe a HilbertA-module ande₁, . . . , e_n ∈Xbe a family of vectors such that he_i, e_ji= 0 (i6=j)andke_ik= 1. Ifx∈X, then

|x|² ≥

n

X

k=1

|he_k, xi|² and |x|² ≥

n

X

k=1

|hx, e_ki|². Proof. The first result follows from the following inequality:

0≤

x−

n

X

k=1

e_khe_k, xi

2

=

* x−

n

X

k=1

e_khe_k, xi, x−

n

X

j=1

e_jhe_j, xi +

=hx, xi+

n

X

k=1 n

X

j=1

he_k, xi^∗he_k, e_jihe_j, xi −2

n

X

k=1

|he_k, xi|²

=hx, xi+

n

X

k=1

he_k, xi^∗he_k, e_kihe_k, xi −2

n

X

k=1

|he_k, xi|²

≤ |x|²+

n

X

k=1

he_k, xi^∗he_k, xi −2

n

X

k=1

|he_k, xi|²

=|x|²−

n

X

k=1

|he_k, xi|². By considering|x−Pn

k=1he_k, xie_k|², we similarly obtain the second one.

Now we will prove the next theorem without using the Cauchy–Schwarz inequality.

Theorem 2.7. Lete₁, . . . , e_m ∈ Xbe a family of vectors with he_i, e_ji = 0 (1 ≤ i 6= j ≤ m) andke_ik= 1 (1≤i≤m). If the vectorsx₁, . . . , x_n∈Xsatisfy the conditions

(2.4) 0≤r_kkx_jk ≤Rehe_k, x_ji, 0≤ρ_kkx_jk ≤Imhe_k, x_ji, for1≤j ≤n,1≤k ≤m, wherer_k, ρ_k ∈[0,∞) (1≤k ≤m), then (2.5)

" _m X

k=1

(r²_k+ρ²_k)

#¹_{2 n} X

j=1

kx_jk ≤

n

X

j=1

x_j . Proof. Applying the previous lemma forx=Pn

j=1xj, we obtain

n

X

j=1

xj

2

≥ 1 2





m

X

k=1

* ek,

n

X

j=1

xj

+

2

+

m

X

k=1

* _n X

j=1

xj, ek

+

2



=

m

X

k=1

1 2

* e_k,

n

X

j=1

x_j +∗*

e_k,

n

X

j=1

x_j +

+

* _n X

j=1

x_j, e_k

+∗* _n X

j=1

x_j, e_k +!

=

m

X

k=1

Re

* e_k,

n

X

j=1

x_j +!2

+ Im

* e_k,

n

X

j=1

x_j +!2

(by (1.3))

=

m

X

k=1

Re

n

X

j=1

he_k, x_ji

!2

+ Im

n

X

j=1

he_k, x_ji

!2

≥

m

X

k=1



r²_k

n

X

j=1

kxjk

!2

+ρ²_k

n

X

j=1

kxjk

!2

 (by (2.4))

=

m

X

k=1

(r²_k+ρ²_k)

n

X

j=1

kx_jk

!2

.

Proposition 2.8. In Theorem 2.7, ifhe_k, e_ki= 1, then the equality holds in (2.5) if and only if (2.6)

n

X

j=1

x_j =

n

X

j=1

kx_jk

! _m X

k=1

(r_k+iρ_k)e_k.

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

n

X

j=1

x_j

2

=

m

X

k=1

* e_k,

n

X

j=1

x_j +

2

and

n

X

j=1

x_j

2

=

m

X

k=1

* _n X

j=1

x_j, e_k +

2

, which is equivalent to

n

X

j=1

x_j =

m

X

k=1 n

X

j=1

e_khe_k, x_ji=

m

X

k=1 n

X

j=1

he_k, x_jie_k, and

r_kkx_jk= Rehe_k, x_ji, ρ_kkx_jk= Imhe_k, x_ji.

So

n

X

j=1

x_j =

m

X

k=1 n

X

j=1

e_khe_k, x_ji

=

m

X

k=1 n

X

j=1

e_k(r_k+iρ_k)kx_jk

=

n

X

j=1

kx_jk

! _m X

k=1

(r_k+iρ_k)e_k.

3. ADDITIVEREVERSE OF THE TRIANGLEINEQUALITY

We now present some versions of the additive reverse of the triangle inequality. In [6], Dragomir established the following theorem:

Theorem 3.1. Let {e_k}^m_k=1 be a family of orthonormal vectors in a Hilbert space H and M_jk ≥0 (1 ≤j ≤n,1≤k ≤m)such that

kx_jk −Rehe_k, x_ji ≤M_jk, for each1≤j ≤nand1≤k ≤m. Then

n

X

j=1

kx_jk ≤ 1

√m

n

X

j=1

x_j

+ 1 m

n

X

j=1 m

X

k=1

M_jk;

and the equality holds if and only if

n

X

j=1

kxjk ≥ 1 m

n

X

j=1 m

X

k=1

Mjk,

and

n

X

j=1

x_j =

n

X

j=1

kx_jk − 1 m

n

X

j=1 m

X

k=1

M_jk

! _m X

k=1

e_k.

We can prove this theorem for HilbertC^∗-modules using some different techniques.

Theorem 3.2. Let{ek}^m_k=1 be a family of vectors in a Hilbert module Xover aC^∗-algebraA with unit1,|e_k| ≤ 1 (1 ≤k ≤ m),he_i, e_ji= 0 (1 ≤i 6=j ≤m)andx_j ∈X (1≤j ≤n). If for some scalarsM_jk ≥0 (1 ≤j ≤n,1≤k ≤m),

(3.1) kx_jk −Rehe_k, x_ji ≤M_jk (1≤j ≤n,1≤k ≤m), then

(3.2)

n

X

j=1

kx_jk ≤ 1

√m

n

X

j=1

x_j

+ 1 m

n

X

j=1 m

X

k=1

M_jk.

Moreover, if|e_k|= 1 (1≤k ≤m), then the equality in (3.2) holds if and only if (3.3)

n

X

j=1

kx_jk ≥ 1 m

n

X

j=1 m

X

k=1

M_jk,

and (3.4)

n

X

j=1

x_j =

n

X

j=1

kx_jk − 1 m

n

X

j=1 m

X

k=1

M_jk

! _m X

k=1

e_k.

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

n

X

j=1

kx_jk ≤Re

* e_k,

n

X

j=1

x_j +

+

n

X

j=1

M_jk,

for eachk∈ {1, . . . , m}. Summing these inequalities overk from 1 tom, we deduce (3.5)

n

X

j=1

kx_jk ≤ 1 m Re

* _m X

k=1

e_k,

n

X

j=1

x_j +

+ 1 m

m

X

k=1 n

X

j=1

M_jk.

Using the Cauchy–Schwarz we obtain Re

* _m X

k=1

e_k,

n

X

j=1

x_j +!2

≤ 1 2





* _m X

k=1

e_k,

n

X

j=1

x_j +

2

+

* _m X

k=1

e_k,

n

X

j=1

x_j +∗

2 (3.6) 

≤ 1 2





m

X

k=1

ek

2

n

X

j=1

xj

2

+

m

X

k=1

ek

2

n

X

j=1

xj

2



≤

m

X

k=1

e_k

2

n

X

j=1

x_j

2

≤m

n

X

j=1

x_j

2

, since

m

X

k=1

e_k

2

=

* _m X

k=1

e_k,

m

X

k=1

e_k +

=

m

X

k=1 m

X

l=1

he_k, e_li

=

m

X

k=1

|e_k|²

≤m . Using (3.6) in (3.5), we deduce the desired inequality.

If (3.3) and (3.4) hold, then

√1 m

n

X

j=1

x_j

= 1

√m

n

X

j=1

kx_jk − 1 m

n

X

j=1 m

X

k=1

M_jk

!

m

X

k=1

e_k

=

n

X

j=1

kx_jk − 1 m

n

X

j=1 m

X

k=1

M_jk, 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

kx_jk −Rehe_k, x_ji=M_jk, Re

* _m X

k=1

e_k,

n

X

j=1

x_j +

=

* _m X

k=1

e_k,

n

X

j=1

x_j +

,

and

* _m X

k=1

e_k,

n

X

j=1

x_j +

=

m

X

k=1

e_k

n

X

j=1

x_j . It follows from Lemma 2.4 and the previous relations that

n

X

j=1

x_j = Pm

k=1ek

kPm k=1ekk²

* _m X

k=1

e_k,

n

X

j=1

x_j +

= Pm

k=1e_k

m Re

* _m X

k=1

e_k,

n

X

j=1

x_j +

= Pm

k=1ek

m

m

X

k=1 n

X

j=1

(kx_jk −M_jk)

=

n

X

j=1

kx_jk − 1 m

n

X

j=1 m

X

k=1

M_jk

! _m X

k=1

e_k.

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