Dragomir, several new reverses of the triangle in- equality in inner product spaces are obtained

http://jipam.vu.edu.au/

Volume 6, Issue 3, Article 64, 2005

REFINEMENTS OF REVERSE TRIANGLE INEQUALITIES IN INNER PRODUCT SPACES

ARSALAN HOJJAT ANSARI AND MOHAMMAD SAL MOSLEHIAN DEPARTMENT OFMATHEMATICS

FERDOWSIUNIVERSITY

P.O. BOX1159, MASHHAD91775, IRAN.

msalm@math.um.ac.ir

URL:http://www.um.ac.ir/∼moslehian/

Received 01 February, 2005; accepted 20 May, 2005 Communicated by S.S. Dragomir

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

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

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

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 constantcwith 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 set x₁, . . . , x_n ∈ X are only those of finite dimension.

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

kx_kk , k∈ {1, . . . , n}.

Then

r

n

X

k=1

kx_kk ≤

n

X

k=1

x_k ,

ISSN (electronic): 1443-5756

c 2005 Victoria University. All rights reserved.

026-05

where equality holds if and only if

n

X

k=1

x_k =r

n

X

k=1

kx_kka.

Inequalities related to the triangle inequality are of special interest; cf. Chapter 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 re- ferred to [9], [5] for the terminology of inner product spaces.

2. MAINRESULTS

The following theorem is an improvement of Theorem 2.1 of [4] in which the real numbers r₁, r₂are not neccesarily nonnegative. The proof seems to be different as well.

Theorem 2.1. Let a be a unit vector in the complex inner product space(H;h·,·i). Suppose that the vectorsxk ∈H, k ∈ {1, . . . , n}satisfy

(2.1) 0≤r²₁kx_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 +

= Re

* _n X

k=1

x_k,(r₁+ir₂)a +

≤

* _n X

k=1

x_k,(r₁+ir₂)a +

≤

n

X

k=1

x_k

k(r₁+ir₂)ak

= (r₁²+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

x_k,(r₁+ir₂)a +

≤(r²₁+r²₂)¹²

n

X

k=1

x_k . 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

kxkk.

The next theorem is a refinement of Corollary 1 of [4] since, in the notation of Theorem 2.1, p2−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 , whereα= min1≤k≤nkx_kk. Let

α1 = min

kxkk²−p²₁ + 1

2kx_kk : 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

kx_kk ≤

n

X

k=1

x_k , where the equality holds if and only if

n

X

k=1

x_k = (α₁+iα₂)

n

X

k=1

kx_kka.

Proof. From the first inequality in (2.4) we have

hx_k−a, x_k−ai ≤p²₁,

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

kx_kk²−p²₁+ 1

2kx_kk kx_kk ≤Rehx_k, ai.

Consequently,

α₁kx_kk ≤Rehx_k, ai.

Similarly from the second inequality we obtain

α₂kx_kk ≤Rehx_k, iai= Imhx_k, ai.

Now apply Theorem 2.1 forr₁ =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 Theorem 2.2 forα1 = ^α₂ =α2.

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

kx_k−ak ≤p, p∈ 0,√

α²+ 1

, α = min

1≤k≤nkx_kk.

Then we have the inequality

α₁

n

X

k=1

kx_kk ≤

n

X

k=1

x_k , where

α₁ = min

kxkk²−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 Theorem 2.2 in which we use Theorem 2.1 withr₂ = 0.

The next theorem is a generalization of Theorem 2.1. It is a modification of Theorem 3 of [4], however our proof is apparently different.

Theorem 2.5. Leta1, . . . , ambe orthonormal vectors in the complex inner product space(H;h·,·i).

Suppose that for1≤t≤m, rt, ρt∈Rand that the vectorsxk∈H, k ∈ {1, . . . , n}satisfy (2.5) 0≤r_t²kx_kk ≤Rehx_k, r_ta_ti, 0≤ρ²_tkx_kk ≤Imhx_k, ρ_ta_ti, t∈ {1, . . . , m}.

Then we have the inequality (2.6)

m

X

t=1

r_t²+ρ²_t

!¹_{2 n} X

k=1

kx_kk ≤

n

X

k=1

x_k . 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. IfPm

t=1(r_t²+ρ²_t) = 0, the theorem is trivial. Assume thatPm

t=1(r_t²+ρ²_t)6= 0.Summing inequalities (2.6) overkfrom1tonand again overtfrom1tom,we get

m

X

t=1

(r_t²+ρ²_t)

n

X

k=1

kxkk ≤Re

* _n X

k=1

xk,

m

X

t=1

rtat

+ + Im

* _n X

k=1

xk,

m

X

t=1

ρtat

+

= Re

* _n X

k=1

xk,

m

X

t=1

rtat

+ + Re

* _n X

k=1

xk, i

m

X

t=1

ρtat

+

= 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 . Then

(2.8)

m

X

t=1

(r²_t +ρ²_t)

!¹_{2 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

(r²_t +ρ²_t)

!¹₂ .

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

m

X

t=1

(r²_t +ρ²_t)

!¹₂

n

X

k=1

x_k

=

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_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

x_k

m

X

t=1

(r_t²+ρ²_t)

!¹₂ . 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

(r²_t +ρ²_t)

!¹₂

= η

m

X

t=1

(r_t+iρ_t)a_t

=

n

X

k=1

xk

=

n

X

k=1

kxkk

m

X

t=1

(r²_t +ρ²_t)

!¹₂ . Hence

η=

n

X

k=1

kx_kk.

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 for 1 ≤ t ≤ m that the vectors x_k ∈ H, k ∈ {1, . . . , n}satisfy

0≤r_t²kx_kk ≤Rehx_k, r_ta_ti.

Then we have the inequality

m

X

t=1

r²_t

!¹_{2 n} X

k=1

kx_kk ≤

n

X

k=1

x_k .

The equality holds if and only if

n

X

k=1

xk =

n

X

k=1

kx_kk

m

X

t=1

rtat.

Proof. Apply Theorem 2.5 forρ_t= 0.

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

kx_k−a_tk ≤p_t, kx_k−ia_tk ≤q_t, p_t, q_t∈ 0,√

α²+ 1

, 1≤t ≤m, whereα= min1≤k≤nkx_kk.Let

α_t= min

kx_kk²−p²_t + 1

2kx_kk : 1≤k ≤n

, βt= min

kx_kk²−q²_t + 1

2kx_kk : 1≤k ≤n

. Then we have the inequality

m

X

t=1

α²_t +β_t²

!¹_{2 n} X

k=1

kx_kk ≤

n

X

k=1

x_k , 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, α_tkx_kk ≤Rehx_k, a_ti, and similarly

βtkxkk ≤Rehxk, iati= Imhxk, ati.

Now applying Theorem 2.4 withrt=αt, ρt =βtwe deduce the desired inequality.

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

kx_k−a_tk ≤1, kx_k−ia_tk ≤1, 1≤t≤m.

Then

√α 2

√m

n

X

k=1

kxkk ≤

n

X

k=1

xk

. The equality holds if and only if

n

X

k=1

x_k =α(1 +i) 2

n

X

k=1

kx_kk

m

X

t=1

a_t.

Proof. Apply Theorem 2.7 forαt= ^α₂ =βt.

Remark 2.9. It is interesting to note that

√α 2

√m ≤ kPn k=1x_kk Pn

k=1kxkk ≤1, where

α≤ r2

m.

Corollary 2.10. Leta be 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

2kxkk : 1≤k ≤n

.

Ifα1 6= (1−p²₁)¹², orα2 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 Theorem 2.2 we have

(α²₁+α²₂)¹²

n

X

k=1

kxkk ≤

n

X

k=1

xk

= (2−p²₁ −p²₂)¹²

n

X

k=1

kxkk and so

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

kx_kk²−p²₁+ 1

2kx_kk ≥(1−p²₁)¹² and so

α1 ≥(1−p²₁)¹². Similarly

α₂ ≥(1−p²₂)¹². Hence

(2−p²₁−p²₂)¹² ≤(α²₁+α²₂)¹². Thus

q

α²₁+α²₂ = (2−p²₁−p²₂)¹². Therefore

α₁ = (1−p²₁)¹² and α₂ = (1−p²₂)¹²,

a contradiction.

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

Theorem 2.11. Letabe a unit vector in the complex inner product space(H;h·,·i), M ≥m >

0, L≥` >0andx_k∈H− {0}, k ∈ {1, . . . , n}such that

RehM a−x_k, x_k−mai ≥0, RehLia−x_k, x_k−`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)kx_kk : 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

kx_kk ≤

n

X

k=1

x_k . 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)kxkkkx_kk ≤Rehx_k, ai, and consequently

α_m,Mkx_kk ≤Rehx_k, ai.

Similarly from the second inequality we deduce

α_`,Lkx_kk ≤Imhx_k, ai.

Applying Theorem 2.1 forr₁ =α_m,M, r₂ =α_`,L, we infer the desired inequality.

Theorem 2.12. Letabe a unit vector in the complex inner product space(H;h·,·i), M ≥m >

0, L≥` >0andx_k∈H− {0}, k ∈ {1, . . . , n}such that

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

x_k−m+M 2 a

≤ M −m 2 ,

x_k− L+` 2 ia

≤ L−` 2 .

Let

α_m,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

. Ifα_m,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 then by Theorem 2.11 we have

(α²_m,M +α_`,L² )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

kxkk.

Consequently

(α_m,M² +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

(α_m,M² +α²_`,L)¹² ≥2

mM

(m+M)² + `L (`+L)²

¹₂ . Then

(α²_m,M +α_`,L² )¹² = 2

mM

(m+M)² + `L (`+L)²

¹₂ . Hence

α_m,M = 2

√ mM m+M and

α_`,L = 2

√`L

`+L

a contradiction.

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

Corollary 2.13. 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

z_k= (α₁+iα₂)

n

X

k=1

|z_k|

! a.

Proof. Apply Theorem 2.2 forH=C.

Corollary 2.14. 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

|z_k| ≤

n

X

k=1

z_k the equality holds if and only if

n

X

k=1

zk =α(1 +i) 2

n

X

k=1

|zk|

! a.

Proof. Apply Corollary 2.3 forH=C.

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