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INNER PRODUCT SPACES

A. H. ANSARI AND M. S. MOSLEHIAN

Received 8 February 2005 and in revised form 17 May 2005

Refining some results of Dragomir, several new reverses of the generalized triangle in-equality in inner product spaces are given. Among several results, we establish some re-verses for the Schwarz inequality. In particular, it is proved that ifais a unit vector in a real or complex inner product space (H;·,·),r,s >0, p∈(0,s], D= {x∈H,rx− sa ≤p},x1,x2∈D− {0}, andαr,s=min{(r2xk2−p2+s2)/2rsxk: 1≤k≤2}, then

(x1x2 −Rex1,x2)/(x1+x2)2α

r,s.

1. Introduction

It is interesting to know under which conditions the triangle inequality went the other way in a normed spaceX; in other words, we would like to know if there is a positive constantcwith the property thatcnk=1xk ≤

n

k=1xkfor any finite setx1,...,xn∈X. Nakai and Tada [7] proved that the normed spaces with this property are precisely those of finite dimensional.

The first authors investigating reverse of the triangle inequality in inner product spaces were Diaz and Metcalf [2] by establishing the following result as an extension of an in-equality given by Petrovich [8] for complex numbers.

Theorem1.1 (Diaz-Metcalf theorem). Letabe a unit vector in an inner product space (H;·,·). Suppose the vectorsxk∈H,k∈ {1,...,n}satisfy

0≤r≤Re

xk,a

xk , k∈ {1,...,n}. (1.1)

Then

rn

k=1

xk

n

k=1 xk

, (1.2)

Copyright©2005 Hindawi Publishing Corporation

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where equality holds if and only if

n

k=1 xk=r

n

k=1

xka. (1.3)

Inequalities related to the triangle inequality are of special interest (cf. [6, Chapter XVII]). They may be applied to get interesting inequalities in complex numbers or to study vector-valued integral inequalities [4,5].

Using several ideas and following the terminology of [4,5], we modify or refine some results of Dragomir and ours [1] and get some new reverses of triangle inequality. Among several results, we show that if ais a unit vector in a real or complex inner product space (H;·,·),xk∈H− {0}, 1≤k≤n,α=min{xk: 1≤k≤n}, p∈(0,√α2+ 1),

max{xk−a: 1≤k≤n} ≤p, andβ=min{(xk2−p2+ 1)/2xk: 1≤k≤n}, then

n

k=1

xk

n

k=1 xk

1ββRe n

k=1 xk,a

. (1.4)

We also examine some reverses for the celebrated Schwarz inequality. In particular, it is proved that ifais a unit vector in a real or complex inner product space (H;·,·),r,s >0, p∈(0,s],D= {x∈H,rx−sa ≤p},x1,x2∈D− {0}, andαr,s=min{(r2xk2−p2+

s2)/2rsx

k: 1≤k≤2}, then

x1x2Re x1,x2

x1+x2 2 ≤αr,s. (1.5)

Throughout the paper, (H;·,·) denotes a real or complex inner product space. We use repeatedly the Cauchy-Schwarz inequality without mentioning it. The reader is re-ferred to [3,9] for the terminology on inner product spaces.

2. Reverse of triangle inequality

We start this section by pointing out the following theorem of [1] which is a modification of [5, Theorem 3].

Theorem 2.1. Leta1,...,am be orthonormal vectors in the complex inner product space

(H;·,·). Suppose that for1≤t≤m,rt,ρt∈R, and that the vectorsxk∈H,k∈ {1,...,n}

satisfy

0≤rt2xk≤Rexk,rtat, 0≤ρ2txk≤Imxk,ρtat, 1≤t≤m. (2.1)

Then

m

t=1

r2

t +ρ2t

1/2 n

k=1

xk

n

k=1 xk

(3)

and the equality holds in (2.2) if and only if

n

k=1 xk=

n

k=1 xkm

t=1

rt+iρt at. (2.3)

The following theorem is a strengthen of [5, Corollary 1] and a generalization of [1, Theorem 2].

Theorem2.2. Letabe a unit vector in the complex inner product space(H;·,·). Suppose that the vectorsxk∈H− {0},k∈ {1,...,n}satisfy

maxrxk−sa: 1≤k≤n≤p, maxrxk−isa: 1≤k≤n≤q, (2.4)

wherer,r,s,s>0and

p≤()2+s2 1/2, q(rα)2+s2 1/2, α=minxk: 1≤k≤n.

(2.5)

Let

αr,s=min

r2x

k2−p2+s2

2rsxk : 1≤k≤n

,

βr,s=min

r2x

k2−q2+s2

2rsxk : 1≤k≤n

.

(2.6)

Then

α2

r,s+β2r,s 1/2 n

k=1

xk

n

k=1 xk

(2.7)

and the equality holds if and only if

n

k=1

xk=αr,s+iβr,s n

k=1

xka. (2.8)

Proof. From the first inequality above, we infer that

rxk−sa,rxk−sa≤p2,

r2x

k2+s2−p22 Rerxk,sa.

(2.9)

Then

r2x

k2+s2−p2

2rsxk

xkRe

xk,a. (2.10)

Similarly,

r2x

k2−q2+s2

2rsxk

xkIm

(4)

consequently,

αr,sxk≤Rexk,a,

βr,sxkImxk,a. (2.12)

ApplyingTheorem 2.1form=1,r1=αr,s, andρ1=βr,s, we deduce the desired

inequal-ity.

The next result is an extension of [1, Corollary 3].

Corollary2.3. Letabe a unit vector in the complex inner product space(H;·,·). Sup-pose thatxk∈H,k∈ {1,...,n},max{rxk−sa: 1≤k≤n} ≤r,max{rxk−isa: 1 k≤n} ≤s, wherer >0,s >0, andα=min{xk: 1≤k≤n}. Then

s√2

n

k=1

xk

n

k=1 xk

. (2.13)

The equality holds if and only if

n

k=1

xk=rα(1 +2si) n

k=1

xka. (2.14)

Proof. ApplyTheorem 2.2withr=r,s=s,p=r,q=s. Note thatαr,s=rα/2s=βr,s.

Theorem2.4. Letabe a unit vector in(H;·,·). Suppose that the vectorsxk∈H,k∈ {1,...,n}satisfy

maxrxk−sa: 1≤k≤n≤p <()2+s2 1/2, (2.15)

wherer >0,s >0and

α= min

1≤k≤nxk. (2.16)

Let

αr,s=min

r2x

k2−p2+s2

2rsxk : 1≤k≤n

. (2.17)

Then

αr,s n

k=1

xk

n

k=1 xk

. (2.18)

Moreover, the equality holds if and only if

n

k=1

xk=αr,s

n

k=1

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Proof. Proof is similar to that ofTheorem 2.2in which we useTheorem 2.1withm=1,

ρ1=0.

Theorem2.5. Letabe a unit vector in(H;·,·). Suppose thatr,s >0, and vectorsxk∈

H− {0},k∈ {1,...,n}satisfy

n

k=1

xk=0. (2.20)

Then

r2α2+s2maxrx

k−sa: 1≤n, (2.21)

where

α= min

1≤k≤nxk. (2.22)

Proof. Letp=max{rxk−sa: 1≤k≤n}. Ifp <√r2α2+s2, then, usingTheorem 2.4, we get

αr,s n

k=1

xk

n

k=1 xk

=0. (2.23)

Henceαr,s=0. On the other hand, (p2−s2)/r2< α2, so

αr,s=min

r2x

k2−p2+s2

2rsxk : 1≤k≤n

>0 (2.24)

holds a contradiction.

Theorem 2.6. Leta1,...,am be orthonormal vectors in the complex inner product space

(H;·,·),Mt≥mt>0,Lt≥t>0,1≤t≤m, andxk∈H− {0},k∈ {1,...,n}such that

ReMtat−xk,xk−mtat≥0, ReLtiat−xk,xk−tiat≥0, (2.25)

or equivalently

xk−mt+2Mtat

≤Mt−2mt,

xk−Lt+2tiat

≤Lt−2 t, (2.26)

for all1≤k≤nand1≤t≤m. Let

αmt,Mt=min

xk2+mtMt

mt+Mt xk: 1≤k≤n

, 1≤t≤m,

αt,Lt=min

xk2+tLt

mt+Mt xk: 1≤k≤n

, 1≤t≤m,

(6)

then

m

t=1 α2

mt,Mt+α 2

t,Lt 1/2 n

k=1

xk

n

k=1 xk

. (2.28)

The equality holds if and only if

n

k=1 xk=

n

k=1

xkm

t=1

αmt,Mt+iαt,Lt at. (2.29)

Proof. Given 1≤t≤mand all 1≤k≤n, it follows fromxk−((mt+Mt)/2)at ≤(Mt−

mt)/2 that

xk2

+mtMt≤mt+Mt Rexk,at. (2.30)

Then

xk2 +mtMt

mt+Mt xk

xkRe

xk,at, (2.31)

and so

αmt,Mtxk≤Re

xk,at. (2.32)

Similarly, from the second inequality, we deduce that

αt,Ltxk≤Im

xk,at. (2.33)

ApplyingTheorem 2.5forrt=αmt,Mt andρt=αt,Lt, we obtain the required inequality.

We will need [4, Theorem 7]. We mention it for the sake of completeness.

Theorem 2.7. Let abe a unit vector in(H;·,·), and xk∈H− {0},k∈ {1,...,n}. If rk≥0,k∈ {1,...,n}such that

xkRe

xk,a≤rk, (2.34)

then

n

k=1

xk

n

k=1 xk

n

k=1

rk. (2.35)

The equality holds if and only if

n

k=1

xkn

k=1 rk,

n

k=1 xk=

n

k=1

xkn k=1

rk

a.

(7)

Theorem2.8. Letabe a unit vector in(H;·,·), andxk∈H− {0},k∈ {1,...,n}. Let α=minxk: 1≤k≤n, p∈0,α2+ 1 , maxxka: 1knp,

β=minxk 2

−p2+ 1

2xk : 1≤k≤n

.

(2.37)

Then

n

k=1

xk

n

k=1 xk

1ββRe n

k=1 xk,a

. (2.38)

The equality holds if and only if

n

k=1

xk1−β

β Re

n

k=1 xk,a

,

n

k=1 xk=

n

k=1

xk1−β

β Re

n

k=1 xk,a

a.

(2.39)

Proof. Since max{xk−a: 1≤k≤n} ≤p, we have

xk−a,xk−a≤p2, xk2+ 1−p22 Rexk,a,

xk2

−p2+ 1 2xk

xkRe

xk,a, βxk≤Rexk,a,

xk1 βRe

xk,a,

(2.40)

for allk∈ {1,...,n}. Then xkRe

xk,a≤1−β

β Re

xk,a, k∈ {1,...,n}. (2.41)

ApplyingTheorem 2.7forrk=((1−β)) Rexk,a,k∈ {1,...,n}, we deduce the desired

inequality.

As a corollary, we obtain a result similar to [4, Theorem 9].

Corollary2.9. Letabe a unit vector in(H;·,·), andxk∈H− {0},k∈ {1,...,n}. Let

maxxk−a: 1≤k≤n≤1,

α=minxk: 1≤k≤n. (2.42)

Then

n

k=1

xk

n

k=1 xk

2ααRe n

k=1 xk,a

(8)

The equality holds if and only if

n

k=1

xk2−α

α Re

n

k=1 xk,a

,

n

k=1 xk=

n

k=1

xk2−α

α Re

n

k=1 xk,a

a.

(2.44)

Proof. ApplyTheorem 2.8withβ=α/2.

Theorem2.10. Letabe a unit vector in(H;·,·),M≥m >0, andxk∈H− {0},k∈ {1,...,n}such that

ReMa−xk,xk−ma≥0 (2.45)

or equivalently

xk−m+2Ma≤M−2 m. (2.46)

Let

αm,M=min xk

2 +mM

(m+M)xk: 1≤k≤n

. (2.47)

Then

n

k=1

xk

n

k=1 xk

1αmα,Mm,MRe n

k=1 xk,a

. (2.48)

The equality holds if and only if

n

k=1

xk1−αm,M

αm,M Re

n

k=1 xk,a

,

n

k=1 xk=

n

k=1

xk1−αm,M αm,M Re

n

k=1 xk,a

a.

(2.49)

Proof. For each 1≤k≤n, it follows from the inequality

xk−m+2Ma≤M−2m (2.50)

that

xk−m+M

2 a,xk− m+M

2 a

≤M−m 2

2

. (2.51)

Hence

xk2

(9)

So that

αm,Mxk≤Re

xk,a, (2.53)

consequently

xkRe

xk,a≤1ααm,M m,M Re

xk,a. (2.54)

Now applyTheorem 2.7forrk=((1−αm,M)/αm,M) Rexk,a,k∈ {1,...,n}.

3. Reverses of Schwarz inequality

In this section, we provide some reverses of the Schwarz inequality. The first theorem is an extension of [4, Proposition 5.1].

Theorem3.1. Letabe a unit vector in(H;·,·). Suppose thatr,s >0,p∈(0,s], and

D=x∈H,rx−sa ≤p. (3.1)

If0 =x1∈D,0 =x2∈D, then x1x2Re

x1,x2 x1+x2 2

1 2

1

r2x12p2+s2

2rsx1

2

(3.2)

or

x1x2Re x1,x2 x1+x2 2

1 2

1

r2x22p2+s2

2rsx2

2

. (3.3)

Proof. Putαr,s=min{(r2xk2−p2+s2)/2rsxk: 1≤k≤2}. ByTheorem 2.4, we

ob-tain

αr,sx1+x2 ≤x1+x2. (3.4)

Then

α2

r,s

x12

+ 2x1x2+x22≤x12

+ 2 Rex1,x2+x22. (3.5)

Setα2

r,s=1−t2. Then

x1x2Re x1,x2 x1+x2 2

1 2t

2, (3.6)

namely,

x1x2Re x1,x2 x1+x2 2

1 2

1−α2

r,s . (3.7)

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Corollary3.2. Letabe a unit vector in(H;·,·). Suppose thatr,s >0and

D=x∈H,rx−sa ≤s. (3.8)

Ifx,y∈Dand0<x<y, then

xy −Rex,y

x+y 2 1 2

1

rx

2s 2

. (3.9)

Proof. In the notation of the proof ofTheorem 3.1, we get fromp=s,x1=x,x2=ythat

αr,s=rx/2s. Now applyTheorem 3.1.

Corollary3.3. Letabe a unit vector in(H;·,·),M≥m >0, andxk∈H− {0},k=1, 2

such that

ReMa−xk,xk−ma≥0 (3.10)

or equivalently

xk−m+2Ma≤M−2 m. (3.11)

Then

x1x2Re x1,x2 x1+x2 2

1 2

1 x1

2 +mM (m+M)x1

2

(3.12)

or

x1x2Re x1,x2 x1+x2 2

1 2

1 x2

2 +mM (m+M)x2

2

. (3.13)

Proof. Putr=1,s=(m+M)/2,p=(M−m)/2,x=x1, andy=x2inTheorem 3.1.

Acknowledgment

The authors would like to thank the referees for their valuable suggestions.

References

[1] A. H. Ansari and M. S. Moslehian,Refinements of reverse triangle inequalities in inner product spaces, JIPAM. J. Inequal. Pure Appl. Math.6(2005), no. 3, article 64.

[2] J. B. Diaz and F. T. Metcalf,A complementary triangle inequality in Hilbert and Banach spaces, Proc. Amer. Math. Soc.17(1966), no. 1, 88–97.

[3] S. S. Dragomir,Discrete Inequalities of the Cauchy-Bunyakovsky-Schwarz Type, Nova Science, New York, 2004.

[4] ,Reverses of the triangle inequality in inner product spaces, Aust. J. Math. Anal. Appl.1

(2004), no. 2, 1–14, article 7.

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[6] D. S. Mitrinovi´c, J. E. Peˇcari´c, and A. M. Fink,Classical and New Inequalities in Analysis, Math-ematics and Its Applications (East European Series), vol. 61, Kluwer Academic, Dordrecht, 1993.

[7] M. Nakai and T. Tada,The reverse triangle inequality in normed spaces, New Zealand J. Math.

25(1996), no. 2, 181–193.

[8] M. Petrovich,Module d’une somme, L’Ensignement Math´ematique19(1917), no. 1/2, 53–56 (French).

[9] Th. M. Rassias (ed.),Inner Product Spaces and Applications, Pitman Research Notes in Mathe-matics Series, vol. 376, Longman, Harlow, 1997.

A. H. Ansari: Department of Mathematics, Ferdowsi University, P.O. Box 1159, Mashhad 91775, Iran

E-mail address:arsalan h a@yahoo.com

M. S. Moslehian: Department of Mathematics, Ferdowsi University, P.O. Box 1159, Mashhad 91775, Iran

References

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