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volume 6, issue 3, article 64, 2005.

Received 01 February, 2005;

accepted 20 May, 2005.

Communicated by:S.S. Dragomir

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Journal of Inequalities in Pure and Applied Mathematics

REFINEMENTS OF REVERSE TRIANGLE INEQUALITIES IN INNER PRODUCT SPACES

ARSALAN HOJJAT ANSARI AND MOHAMMAD SAL MOSLEHIAN

Department of Mathematics Ferdowsi University

P.O. Box 1159, Mashhad 91775, Iran.

EMail:msalm@math.um.ac.ir URL:http://www.um.ac.ir/∼moslehian/

c

2000Victoria University ISSN (electronic): 1443-5756 026-05

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Refinements of Reverse Triangle Inequalities in Inner

Product Spaces

Arsalan Hojjat Ansari and Mohammad Sal Moslehian

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J. Ineq. Pure and Appl. Math. 6(3) Art. 64, 2005

http://jipam.vu.edu.au

Abstract

Refining some results of S.S. Dragomir, several new reverses of the triangle inequality in inner product spaces are obtained.

2000 Mathematics Subject Classification:Primary 46C05; Secondary 26D15.

Key words: Triangle inequality, Reverse inequality, Diaz-Metkalf inequality, Inner product space.

1. Introduction

It is interesting to know under which conditions the triangle inequality reverses in a normed spaceX; in other words, we would like to know if there is a con- stant c with the property that cPn

k=1kx_kk ≤ kPn

k=1x_kk for some finite set x₁, . . . , x_n ∈X. M. Nakai and T. Tada [7] proved that the normed spaces with this property for any finite setx₁, . . . , x_n ∈ X are only those of finite dimen- sion.

The first authors to investigate the reverse of the triangle inequality in inner product spaces were J. B. Diaz and F. T. Metcalf [2]. They did so by establishing the following result as an extension of an inequality given by M. Petrovich [8]

for complex numbers:

Theorem 1.1 (Diaz-Metcalf Theorem). Let a be a unit vector in the inner product space(H;h·,·i). Suppose the vectorsx_k ∈H, k ∈ {1, . . . , n}satisfy

0≤r ≤ Rehx_k, ai

kxkk , k ∈ {1, . . . , n}.

Then

r

n

X

k=1

kx_kk ≤

n

X

k=1

x_k , where equality holds if and only if

n

X

k=1

xk=r

n

X

k=1

kxkka.

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Inequalities related to the triangle inequality are of special interest; cf. Chap- ter XVII of [6] and may be applied to obtain inequalities in complex numbers or to study vector-valued integral inequalities [3], [4].

Using several ideas and the notation of [3], [4] we modify or refine some results of S.S. Dragomir to procure some new reverses of the triangle inequality (see also [1]).

We use repeatedly the Cauchy-Schwarz inequality without mentioning it.

The reader is referred to [9], [5] for the terminology of inner product spaces.

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2. Main Results

The following theorem is an improvement of Theorem 2.1of [4] in which the real numbers r1, r2 are not neccesarily nonnegative. The proof seems to be different as well.

Theorem 2.1. Letabe a unit vector in the complex inner product space(H;h·,·i).

Suppose that the vectorsxk∈H, k ∈ {1, . . . , n}satisfy

(2.1) 0≤r₁²kx_kk ≤Rehx_k, r₁ai, 0≤r₂²kx_kk ≤Imhx_k, r₂ai for somer₁, r₂ ∈[−1,1].Then we have the inequality

(2.2) (r²₁+r²₂)¹²

n

X

k=1

kx_kk ≤

n

X

k=1

x_k . The equality holds in (2.2) if and only if

(2.3)

n

X

k=1

x_k = (r₁+ir₂)

n

X

k=1

kx_kka.

Proof. Ifr²₁+r₂² = 0, the theorem is trivial. Assume thatr²₁+r²₂ 6= 0. Summing inequalities (2.1) overkfrom1ton, we have

(r²₁+r²₂)

n

X

k=1

kx_kk ≤Re

* _n X

k=1

x_k, r₁a +

+ Im

* _n X

k=1

x_k, r₂a +

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= Re

* _n X

k=1

x_k,(r₁+ir₂)a +

≤

* _n X

k=1

xk,(r1+ir2)a +

≤

n

X

k=1

x_k

k(r₁+ir₂)ak

= (r²₁ +r₂²)¹²

n

X

k=1

x_k . Hence (2.2) holds.

If (2.3) holds, then

n

X

k=1

x_k

=

(r₁+ir₂)

n

X

k=1

kx_kka

= (r₁²+r₂²)¹²

n

X

k=1

kx_kk.

Conversely, if the equality holds in (2.2), we have

(r₁²+r₂²)¹²

n

X

k=1

x_k

= (r²₁+r²₂)

n

X

k=1

kx_kk

≤Re

* _n X

k=1

x_k,(r₁+ir₂)a +

≤

* _n X

k=1

xk,(r1+ir2)a +

≤(r²₁+r²₂)¹²

n

X

k=1

xk

.

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From this we deduce

* _n X

k=1

x_k,(r₁+ir₂)a +

=

n

X

k=1

x_k

k(r₁+ir₂)ak.

Consequently there existsη≥0such that

n

X

k=1

x_k =η(r₁+ir₂)a.

From this we have

(r₁²+r₂²)¹²η=kη(r₁ +ir₂)ak=

n

X

k=1

x_k

= (r²₁+r²₂)¹²

n

X

k=1

kx_kk.

Hence

η=

n

X

k=1

kx_kk.

The next theorem is a refinement of Corollary 1 of [4] since, in the notation of Theorem2.1,p

2−p²₁−p²₂ ≤p

α²₁+α²₂.

Theorem 2.2. Letabe a unit vector in the complex inner product space(H;h·,·i).

Suppose the vectorsx_k ∈H− {0}, k ∈ {1, . . . , n}, are such that (2.4) kx_k−ak ≤p₁, kx_k−iak ≤p₂, p₁, p₂ ∈

0,√

α²+ 1 ,

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whereα= min1≤k≤nkx_kk. Let α₁ = min

kx_kk²−p²₁+ 1

2kxkk : 1≤k ≤n

, α₂ = min

kxkk²−p²₂+ 1

2kx_kk : 1≤k ≤n

. Then we have the inequality

(α²₁+α²₂)¹²

n

X

k=1

kxkk ≤

n

X

k=1

xk

, where the equality holds if and only if

n

X

k=1

xk = (α1+iα2)

n

X

k=1

kxkka.

Proof. From the first inequality in (2.4) we have hxk−a, xk−ai ≤p²₁,

kx_kk²+ 1−p²₁ ≤2 Rehx_k, ai, k = 1, . . . , n, and kx_kk²−p²₁+ 1

2kx_kk kxkk ≤Rehxk, ai.

Consequently,

α1kxkk ≤Rehxk, ai.

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Similarly from the second inequality we obtain

α₂kx_kk ≤Rehx_k, iai= Imhx_k, ai.

Now apply Theorem2.1forr₁ =α₁, r₂ =α₂.

Corollary 2.3. Letabe a unit vector in the complex inner product space(H;h·,·i).

Suppose that the vectorsx_k∈H− {0}, k ∈ {1, . . . , n}such that kx_k−ak ≤1, kx_k−iak ≤1.

Then

√α 2

n

X

k=1

kx_kk ≤

n

X

k=1

x_k ,

in whichα= min1≤k≤nkx_kk.The equality holds if and only if

n

X

k=1

x_k =α(1 +i) 2

n

X

k=1

kx_kka.

Proof. Apply Theorem2.2forα₁ = ^α₂ =α₂.

Theorem 2.4. Letabe a unit vector in the inner product space(H;h·,·i)over the real or complex number field. Suppose that the vectors xk ∈ H − {0}, k ∈ {1, . . . , n}satisfy

kx_k−ak ≤p, p∈ 0,√

α²+ 1

, α= min

1≤k≤nkx_kk.

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Then we have the inequality α₁

n

X

k=1

kx_kk ≤

n

X

k=1

x_k , where

α₁ = min

kx_kk²−p²+ 1

2kx_kk : 1≤k ≤n

. The equality holds if and only if

n

X

k=1

x_k=α₁

n

X

k=1

kx_kka.

Proof. The proof is similar to Theorem2.2in which we use Theorem2.1 with r₂ = 0.

The next theorem is a generalization of Theorem2.1. It is a modification of Theorem 3 of [4], however our proof is apparently different.

Theorem 2.5. Let a₁, . . . , a_m be orthonormal vectors in the complex inner product space (H;h·,·i). Suppose that for 1 ≤ t ≤ m, r_t, ρ_t ∈ R and that the vectorsxk ∈H, k ∈ {1, . . . , n}satisfy

0≤r_t²kx_kk ≤Rehx_k, r_ta_ti, (2.5)

0≤ρ²_tkx_kk ≤Imhx_k, ρ_ta_ti, t ∈ {1, . . . , m}.

Then we have the inequality

(2.6)

m

X

t=1

r²_t +ρ²_t

!¹₂ _n X

k=1

kx_kk ≤

n

X

k=1

x_k .

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The equality holds in (2.7) if and only if

(2.7)

n

X

k=1

x_k=

n

X

k=1

kx_kk

m

X

t=1

(r_t+iρ_t)a_t. Proof. If Pm

t=1(r²_t +ρ²_t) = 0, the theorem is trivial. Assume thatPm

t=1(r_t² + ρ²_t)6= 0.Summing inequalities (2.6) overkfrom1tonand again overtfrom1 tom,we get

m

X

t=1

(r²_t +ρ²_t)

n

X

k=1

kx_kk ≤Re

* _n X

k=1

x_k,

m

X

t=1

r_ta_t +

+ Im

* _n X

k=1

x_k,

m

X

t=1

ρ_ta_t +

= Re

* _n X

k=1

x_k,

m

X

t=1

r_ta_t +

+ Re

* _n X

k=1

x_k, i

m

X

t=1

ρ_ta_t +

= Re

* _n X

k=1

xk,

m

X

t=1

(rt+iρt)at

+

≤

* _n X

k=1

x_k,

m

X

t=1

(r_t+iρ_t)a_t +

≤

n

X

k=1

x_k

m

X

t=1

(r_t+iρ_t)a_t

=

n

X

k=1

x_k

m

X

t=1

(r²_t +ρ²_t)

!¹₂ .

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Then

(2.8)

m

X

t=1

(r_t²+ρ²_t)

!¹₂ _n X

k=1

kx_kk ≤

n

X

k=1

x_k . If(2.8)holds, then

n

X

k=1

x_k

=

n

X

k=1

kx_kk

m

X

t=1

(r_t+iρ_t)a_t

=

n

X

k=1

kx_kk

m

X

t=1

(r_t²+ρ²_t)

!¹₂ . Conversely, if the equality holds in(2.7),we obtain from (2.6) that

m

X

t=1

(r_t²+ρ²_t)

!¹₂

n

X

k=1

x_k

=

m

X

t=1

(r_t²+ρ²_t)

n

X

k=1

kx_kk

≤Re

* _n X

k=1

x_k,

m

X

t=1

(r_t+iρ_t)a_t +

≤

* _n X

k=1

x_k,

m

X

t=1

(r_t+iρ_t)a_t +

≤

n

X

k=1

x_k

m

X

t=1

(r_t+iρ_t)a_t

=

n

X

k=1

xk

m

X

t=1

(r²_t +ρ²_t)

!¹₂ .

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Thus we have

* _n X

k=1

x_k,

m

X

t=1

(r_t+iρ_t)a_t +

=

n

X

k=1

x_k

m

X

t=1

(r_t+iρ_t)a_t . Consequently there existsη≥0such that

n

X

k=1

x_k =η

m

X

t=1

(r_t+iρ_t)a_t from which we have

η

m

X

t=1

(r_t²+ρ²_t)

!¹₂

= η

m

X

t=1

(rt+iρt)at

=

n

X

k=1

x_k

=

n

X

k=1

kxkk

m

X

t=1

(r²_t +ρ²_t)

!¹₂ . Hence

η=

n

X

k=1

kxkk.

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Corollary 2.6. Let a₁, . . . , a_m be orthornormal vectors in the inner product space(H;h·,·i)over the real or complex number field. Suppose for1≤t≤m that the vectorsx_k ∈H, k ∈ {1, . . . , n}satisfy

0≤r_t²kx_kk ≤Rehx_k, r_ta_ti.

m

X

t=1

r²_t

!¹₂ _n X

k=1

kxkk ≤

n

X

k=1

xk

. The equality holds if and only if

n

X

k=1

x_k =

n

X

k=1

kx_kk

m

X

t=1

r_ta_t. Proof. Apply Theorem2.5forρ_t= 0.

Theorem 2.7. Let a₁, . . . , a_m be orthornormal vectors in the complex inner product space(H;h·,·i). Suppose that the vectorsx_k∈H−{0}, k ∈ {1, . . . , n}

satisfy

kxk−atk ≤pt, kxk−iatk ≤qt, pt, qt∈ 0,√

α²+ 1

, 1≤t ≤m, whereα= min_1≤k≤nkx_kk.Let

α_t = min

kx_kk² −p²_t + 1

2kx_kk : 1≤k ≤n

, βt = min

kx_kk² −q_t²+ 1

2kx_kk : 1≤k≤n

.

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m

X

t=1

α²_t +β_t²

!¹₂ _n X

k=1

kxkk ≤

n

X

k=1

xk

, where equality holds if and only if

n

X

k=1

x_k=

n

X

k=1

kx_kk

m

X

t=1

(α_t+iβ_t)a_t.

Proof. For1≤t≤m,1≤k≤nit follows fromkx_k−a_tk ≤p_tthat hx_k−a_ti, x_k−a_ti ≤p²_t,

kx_kk² −p²_t + 1

2kx_kk kx_kk ≤Rehx_k, a_ti, αtkxkk ≤Rehxk, ati, and similarly

β_tkx_kk ≤Rehx_k, ia_ti= Imhx_k, a_ti.

Now applying Theorem 2.4 with r_t = α_t, ρ_t = β_t we deduce the desired inequality.

Corollary 2.8. Let a1, . . . , am be orthornormal vectors in the complex inner product space (H;h·,·i). Suppose that the vectors x_k ∈ H, k ∈ {1, . . . , n}

satisfy

kxk−atk ≤1, kxk−iatk ≤1, 1≤t≤m.

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Then

√α 2

√m

n

X

k=1

kx_kk ≤

n

X

k=1

x_k . The equality holds if and only if

n

X

k=1

xk=α(1 +i) 2

n

X

k=1

kxkk

m

X

t=1

at. Proof. Apply Theorem2.7forα_t= ^α₂ =β_t.

Remark 1. It is interesting to note that

√α 2

√m≤ kPn k=1x_kk Pn

k=1kx_kk ≤1, where

α ≤ r2

m.

Corollary 2.9. Letabe a unit vector in the complex inner product space(H;h·,·i).

Suppose that the vectorsx_k∈H− {0}, k ∈ {1, . . . , n}satisfy kx_k−ak ≤p₁, kx_k−iak ≤p₂, p₁, p₂ ∈(0,1].

Let

α₁ = min

kx_kk²−p²₁+ 1

2kx_kk : 1≤k ≤n

, α2 = min

kx_kk²−p²₂+ 1

2kx_kk : 1≤k ≤n

.

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If α₁ 6= (1 − p²₁)¹², or α₂ 6= (1− p²₂)¹², then we have the following strictly inequality

(2−p²₁−p²₂)¹²

n

X

k=1

kx_kk<

n

X

k=1

x_k . Proof. If equality holds, then by Theorem2.2we have

(α²₁+α²₂)¹²

n

X

k=1

kx_kk ≤

n

X

k=1

x_k

= (2−p²₁−p²₂)¹²

n

X

k=1

kx_kk

and so

(α²₁+α²₂)¹² ≤(2−p²₁−p²₂)¹². On the other hand for1≤k ≤n,

kx_kk²−p²₁+ 1

2kxkk ≥(1−p²₁)¹² and so

α₁ ≥(1−p²₁)¹². Similarly

α₂ ≥(1−p²₂)¹². Hence

(2−p²₁−p²₂)¹² ≤(α₁²+α₂²)¹².

Thus q

α²₁+α²₂ = (2−p²₁−p²₂)¹².

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Therefore

α1 = (1−p²₁)¹² and α2 = (1−p²₂)¹², a contradiction.

The following result looks like Corollary 2 of [4].

Theorem 2.10. Let a be a unit vector in the complex inner product space (H;h·,·i), M ≥ m > 0, L ≥ ` > 0 and x_k ∈ H − {0}, k ∈ {1, . . . , n}

such that

RehM a−xk, xk−mai ≥0, RehLia−xk, xk−`iai ≥0, or equivalently,

kx_k− m+M

2 ak ≤ M−m

2 , kx_k− L+`

2 iak ≤ L−` 2 . Let

α_m,M = min

kx_kk²+mM

(m+M)kxkk : 1≤k ≤n

and

α_`,L = min

kxkk² +`L

(`+L)kx_kk : 1≤k ≤n

. Then we have the inequlity

(α²_m,M +α²_`,L)¹²

n

X

k=1

kxkk ≤

n

X

k=1

xk

.

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The equality holds if and only if

n

X

k=1

x_k= (α_m,M +iα_`,L)

n

X

k=1

kx_kka.

Proof. For each1≤k ≤n, it follows from kx_k−m+M

2 ak ≤ M −m 2 that

x_k− m+M

2 a, x_k−m+M 2

≤

M −m 2

2

. Hence

kx_kk²+mM ≤(m+M) Rehx_k, ai.

Then kx_kk²+mM

(m+M)kx_kkkx_kk ≤Rehx_k, ai, and consequently

α_m,Mkx_kk ≤Rehx_k, ai.

Similarly from the second inequality we deduce α`,Lkxkk ≤Imhxk, ai.

Applying Theorem2.1forr₁ =α_m,M, r₂ =α_`,L, we infer the desired inequality.

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Theorem 2.11. Let a be a unit vector in the complex inner product space (H;h·,·i), M ≥ m > 0, L ≥ ` > 0 and xk ∈ H − {0}, k ∈ {1, . . . , n}

such that

RehM a−x_k, x_k−mai ≥0, RehLia−x_k, x_k−`iai ≥0, or equivalently

xk− m+M 2 a

≤ M−m 2 ,

xk− L+` 2 ia

≤ L−` 2 . Let

αm,M = min

kx_kk²+mM

(m+M)kx_kk : 1≤k ≤n

and

α_`,L = min

kx_kk² +`L

(`+L)kx_kk : 1≤k ≤n

. Ifαm,M 6= 2

√ mM

m+M, orα`,L 6= 2

√

`L

`+L, then we have

2

mM

(m+M)² + `L (`+L)²

¹₂ ⁿ X

k=1

kx_kk<

n

X

k=1

x_k . Proof. If

2

mM

(m+M)² + `L (`+L)²

¹₂ n

X

k=1

kx_kk=

n

X

k=1

x_k

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then by Theorem2.10we have

(α²_m,M +α²_`,L)¹²

n

X

k=1

kx_kk ≤

n

X

k=1

x_k

= 2

mM

(m+M)² + `L (`+L)²

¹₂ ⁿ X

k=1

kx_kk.

Consequently

(α²_m,M +α²_`,L)¹² ≤2

mM

(m+M)² + `L (`+L)²

¹₂ . On the other hand for1≤k ≤n,

kx_kk²+mM (m+M)kx_kk ≥2

√mM

m+M, and kx_kk²+`L (`+L)kx_kk ≥2

√`L

`+L, so

(α²_m,M +α²_`,L)¹² ≥2

mM

(m+M)² + `L (`+L)²

¹₂ . Then

(α²_m,M +α²_`,L)¹² = 2

mM

(m+M)² + `L (`+L)²

¹₂ . Hence

α_m,M = 2

√mM m+M

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and

α_`,L = 2

√`L

`+L a contradiction.

Finally we mention two applications of our results to complex numbers.

Corollary 2.12. Leta∈ Cwith|a| = 1. Suppose thatz_k ∈ C, k∈ {1, . . . , n}

such that

|z_k−a| ≤p₁, |z_k−ia| ≤p₂, p₁, p₂ ∈ 0,√

α²+ 1 , where

α= min{|z_k|: 1≤k ≤n}.

Let

α₁ = min

|zk|² −p²₁+ 1

2|z_k| : 1≤k ≤n

, α₂ = min

|z_k|² −p²₂+ 1

2|z_k| : 1≤k ≤n

. Then we have the inequality

q

α²₁+α²₂

n

X

k=1

|z_k| ≤

n

X

k=1

z_k . The equality holds if and only if

n

X

k=1

zk = (α1+iα2)

n

X

k=1

|zk|

! a.

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Proof. Apply Theorem2.2forH =C.

Corollary 2.13. Leta∈ Cwith|a| = 1. Suppose thatzk ∈ C, k∈ {1, . . . , n}

such that

|z_k−a| ≤1, |z_k−ia| ≤1.

Ifα= min{|z_k|: 1≤k ≤n}. Then we have the inequality

√α 2

n

X

k=1

|zk| ≤

n

X

k=1

zk

the equality holds if and only if

n

X

k=1

zk=α(1 +i) 2

n

X

k=1

|zk|

! a.

Proof. Apply Corollary2.3forH =C.

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References

[1] A.H. ANSARIANDM.S. MOSLEHIAN, More on reverse triangle inequality in inner product spaces, arXiv:math.FA/0506198.

[2] J.B. DIAZ AND F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces, Proc. Amer. Math. Soc., 17(1) (1966), 88–97.

[3] S.S. DRAGOMIR, Reverses of the triangle inequality in inner product spaces, arXiv:math.FA/0405495.

[4] S.S. DRAGOMIR, Some reverses of the generalized triangle inequality in complex inner product spaces, arXiv:math.FA/0405497.

[5] S.S. DRAGOMIR, Discrete Inequalities of the Cauchy-Bunyakovsky- Schwarz Type, Nova Science Publishers, Inc., Hauppauge, NY, 2004.

[6] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dor- drecht/Boston/London, 1993.

[7] M. NAKAI AND T. TADA, The reverse triangle inequality in normed spaces, New Zealand J. Math., 25(2) (1996), 181–193.

[8] M. PETROVICH, Module d’une somme, L’ Ensignement Mathématique, 19 (1917), 53–56.

[9] Th.M. RASSIAS, Inner Product Spaces and Applications, Chapman-Hall, 1997.