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)2≤αr,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≤Rexk,a
xk , k∈ {1,...,n}. (1.1)
Then
rn
k=1
xk≤
n k=1
xk
, (1.2)
Copyright©2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:18 (2005) 2883–2893 DOI:10.1155/IJMMS.2005.2883
where equality holds if and only if n k=1
xk=rn
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
≤1−β
β Re 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)/2rsxk: 1≤k≤2}, then
x1x2−Rex1,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
rt2+ρ2t
1/2n k=1
xk≤
n k=1
xk
(2.2)
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≤
(rα)2+s2 1/2, q≤
(rα)2+s2 1/2,
α=minxk: 1≤k≤n. (2.5)
Let
αr,s=min
r2xk2−p2+s2
2rsxk : 1≤k≤n
, βr,s=min
r2xk2−q2+s2
2rsxk : 1≤k≤n
.
(2.6)
Then
α2r,s+β2r,s 1/2n
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,
r2xk2+s2−p2≤2 Rerxk,sa. (2.9) Then
r2xk2+s2−p2
2rsxk xk≤Rexk,a. (2.10) Similarly,
r2xk2−q2+s2
2rsxk xk≤Imxk,a, (2.11)
consequently,
αr,sxk≤Rexk,a,
βr,sxk≤Imxk,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
rα 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 +i) 2s
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 <(rα)2+s2 1/2, (2.15) wherer >0,s >0and
α= min
1≤k≤nxk. (2.16)
Let
αr,s=min
r2xk2−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
xka. (2.19)
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+s2≤maxrxk−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
r2xk2−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+Mt
2 at
≤Mt−mt
2 ,
xk−Lt+t
2 iat
≤Lt−t
2 , (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,
(2.27)
then
m
t=1
α2mt,Mt+α2t,Lt
1/2n 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 xkxk≤Rexk,at
, (2.31)
and so
αmt,Mtxk≤Rexk,at
. (2.32)
Similarly, from the second inequality, we deduce that αt,Ltxk≤Imxk,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
xk−Rexk,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
xk≥ n k=1
rk, n
k=1
xk= n
k=1
xk−n
k=1
rk
a.
(2.36)
Theorem2.8. Letabe a unit vector in(H;·,·), andxk∈H− {0},k∈ {1,...,n}. Let α=minxk: 1≤k≤n, p∈
0,α2+ 1 , maxxk−a: 1≤k≤n≤p, β=minxk2−p2+ 1
2xk : 1≤k≤n
.
(2.37) Then
n k=1
xk−
n k=1
xk
≤1−β
β Re n
k=1
xk,a
. (2.38)
The equality holds if and only if n k=1
xk≥1−β β Re
n
k=1
xk,a
, n
k=1
xk= n
k=1
xk−1−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−p2≤2 Rexk,a, xk2−p2+ 1
2xk xk≤Rexk,a, βxk≤Rexk,a, xk≤1
βRexk,a,
(2.40)
for allk∈ {1,...,n}. Then
xk−Rexk,a≤1−β
β Rexk,a, k∈ {1,...,n}. (2.41) ApplyingTheorem 2.7forrk=((1−β)/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
≤2−α
α Re n
k=1
xk,a
. (2.43)
The equality holds if and only if n k=1
xk≥2−α α Re
n
k=1
xk,a
, n
k=1
xk= n
k=1
xk−2−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+M
2 a≤M−m
2 . (2.46)
Let
αm,M=min xk2+mM
(m+M)xk: 1≤k≤n
. (2.47)
Then
n k=1
xk−
n k=1
xk
≤1−αm,M
αm,M Re n
k=1
xk,a
. (2.48)
The equality holds if and only if n k=1
xk≥1−αm,M
αm,M Re n
k=1
xk,a
, n
k=1
xk= n
k=1
xk−1−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+M
2 a≤M−m
2 (2.50)
that
xk−m+M
2 a,xk−m+M
2 a≤M−m 2
2
. (2.51)
Hence
xk2+mM≤(m+M) Rexk,a. (2.52)
So that
αm,Mxk≤Rexk,a, (2.53)
consequently
xk−Rexk,a≤1−αm,M
αm,M Rexk,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
x1x2−Rex1,x2 x1+x2 2 ≤1
2
1−
r2x12−p2+s2 2rsx1 2
(3.2) or
x1x2−Rex1,x2
x1+x2
2 ≤1
2
1−r2x22−p2+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
α2r,sx12+ 2x1x2+x22
≤x12+ 2 Rex1,x2
+x22. (3.5) Setα2r,s=1−t2. Then
x1x2−Rex1,x2
x1+x2
2 ≤1
2t2, (3.6)
namely,
x1x2−Rex1,x2
x1+x2 2 ≤1
2
1−α2r,s . (3.7)
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+M
2 a≤M−m
2 . (3.11)
Then
x1x2−Rex1,x2 x1+x2 2 ≤1
2
1− x12+mM (m+M)x12
(3.12) or
x1x2−Rex1,x2
x1+x2
2 ≤1
2
1− x22+mM (m+M)x22
. (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
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[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.
[5] , Some reverses of the generalised triangle inequality in complex inner product spaces, Linear Algebra Appl.402(2005), 245–254.
[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).
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A. H. Ansari: Department of Mathematics, Ferdowsi University, P.O. Box 1159, Mashhad 91775, Iran
E-mail address:arsalan h [email protected]
M. S. Moslehian: Department of Mathematics, Ferdowsi University, P.O. Box 1159, Mashhad 91775, Iran
E-mail address:[email protected]
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