Some Inequalities in Inner Product Spaces Related to the Generalized Triangle Inequality
This is the Published version of the following publication
Dragomir, Sever S, Cho, Yeol Je and Kim, Seong Sik (2009) Some Inequalities in Inner Product Spaces Related to the Generalized Triangle Inequality.
Research report collection, 12 (3).
The publisher’s official version can be found at
Note that access to this version may require subscription.
Downloaded from VU Research Repository https://vuir.vu.edu.au/17758/
SOME INEQUALITIES IN INNER PRODUCT SPACES RELATED TO THE GENERALIZED TRIANGLE INEQUALITY
S.S. DRAGOMIR, Y.J. CHO?, AND S.S. KIM
Abstract. In this paper we obtain some inequalities related to the general- ized triangle and quadratic triangle inequalities for vectors in inner product spaces. Some results that employ the Ostrowski discrete inequality for vectors in normed linear spaces are also obtained.
1. Introduction
In Mathematical Analysis one of the most important inequalities is the (gener- alized) triangle inequality which states that, in a normed linear space (X;k·k) we have the inequality
(1.1)
n
X
i=1
xi
≤
n
X
i=1
kxik for any vectorsxi ∈X, i∈ {1, . . . , n}.
In 1966, J.B. Diaz and F.T. Metcalf [3] proved the following reverse of the triangle inequality in inner product spaces:
Theorem 1 (Diaz-Metcalf, 1966, [3]). Let abe a unit vector in the inner product space(H;h·,·i) over the real or complex number field K. Suppose that the vectors xi∈H\ {0}, i∈ {1, . . . , n} satisfy
(1.2) 0≤r≤Rehxi, ai
kxik , i∈ {1, . . . , n}. Then
(1.3) r
n
X
i=1
kxik ≤
n
X
i=1
xi
,
where equality holds if and only if
(1.4)
n
X
i=1
xi=r
n
X
i=1
kxik
! a.
The following result with a natural geometrical meaning also holds:
Date: July 10, 2009.
1991Mathematics Subject Classification. 46D05, 46B05.
Key words and phrases. Triangle inequality, Inner product spaces, Normed spaces, Ostrowski inequality.
The second author was supported by the Korea Research Foundation Grant funded by the Korean Government (KRF-2008-313-C00050).
?Corresponding author.
1
Theorem 2(Dragomir, 2004, [5]). Letabe a unit vector in the inner product space (H;h·,·i)andρ∈(0,1).If xi∈H, i∈ {1, . . . , n} are such that
(1.5) kxi−ak ≤ρ for each i∈ {1, . . . , n}, then we have the inequality
(1.6) p
1−ρ2
n
X
i=1
kxik ≤
n
X
i=1
xi
, with equality if and only if
(1.7)
n
X
i=1
xi=p 1−ρ2
n
X
i=1
kxik
! a.
The following result that provides a more intuitive condition for the vectors involved may be stated as well:
Theorem 3(Dragomir, 2004, [5]). Letabe a unit vector in the inner product space (H;h·,·i)andM ≥m >0. If xi∈H, i∈ {1, . . . , n} are such that either
(1.8) RehM a−xi, xi−mai ≥0 or, equivalently,
(1.9)
xi−M +m 2 ·a
≤ 1
2(M −m) holds for each i∈ {1, . . . , n}, then we have the inequality
(1.10) 2√
mM m+M
n
X
i=1
kxik ≤
n
X
i=1
xi
, or, equivalently,
(1.11) (0≤)
n
X
i=1
kxik −
n
X
i=1
xi
≤
√ M −√
m2 2√
mM
n
X
i=1
xi .
The equality holds in (1.10) (or in (1.11)) if and only if
(1.12)
n
X
i=1
xi= 2√ mM m+M
n
X
i=1
kxik
! a.
In inner product spaces (H;h·,·i) another important inequality for sequences of vectors is the following one
(1.13)
n
X
i=1
pixi
2
≤
n
X
i=1
pikxik2,
wherexi∈H,i∈ {1, . . . , n}andpi≥0 withPn
i=1pi= 1.This can be seen as the quadratic version of the generalized triangle inequality.
A reverse of this inequality is incorporated in the following result:
Theorem 4(Dragomir, 2000, [4]). Consider an inner product space(H;h·,·i)over the real or complex number field K. Suppose that the vectors xi, x, X ∈ H, i ∈ {1, . . . , n} satisfy the condition
(1.14) RehX−xi, xi−xi ≥0 for alli∈ {1, . . . , n},
INEQUALITIES IN INNER PRODUCT SPACES 3
or, equivalently, (1.15)
xi−x+X 2
≤ 1
2kX−xk for alli∈ {1, . . . , n}, then we have the inequality
(1.16)
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
≤ 1
4kX−xk2 for any pi≥0, i∈ {1, . . . , n}, withPn
i=1pi= 1.
The constant 14 is best possible in (1.16).
For other results related to the generalised triangle inequality, see [1], [2], [7], [8]
and [9], where further references are provided.
In this paper we obtain other inequalities related to the generalized triangle and quadratic triangle inequalities for vectors in inner product spaces. Some results that employ the Ostrowski discrete inequality for vectors in normed linear spaces are also obtained.
2. The Results
The following result that encompasses a condition depending on the probability distributionpi≥0, i∈ {1, . . . , n} can be stated:
Theorem 5. Let (H;h·,·i) be an inner product space,xi ∈H,i∈ {1, . . . , n} and pi≥0 with Pn
i=1pi= 1.If there exist constants ri>0, i∈ {1, . . . , n},so that (2.1)
xi−
n
X
j=1
pjxj
≤ri for each i∈ {1, . . . , n},
then
(2.2)
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
≤
n
X
i=1
pir2i.
The inequality (2.2) is sharp.
Proof. Taking the square in (2.1) we get (2.3) kxik2−2 Re
* xi,
n
X
j=1
pjxj
+ +
n
X
j=1
pjxj
2
≤ri2
for eachi∈ {1, . . . , n}.
Now, if we multiply (2.3) bypi≥0 and sum over ifrom 1 ton,we obtain (2.4)
n
X
i=1
kxik2−2 Re
* n X
i=1
pixi,
n
X
j=1
pjxj +
+
n
X
j=1
pjxj
2
≤
n
X
i=1
pir2i
and since
Re
* n X
i=1
pixi,
n
X
j=1
pjxj
+
=
n
X
i=1
pixi
2
we deduce from (2.4) the desired inequality (2.2).
Now, assume thatp1, p2∈(0,1) withp1+p2= 1 and x1, x2∈H withx16=x2. With this assumption, (2.1) becomes
kx1−p1x1−p2x2k ≤r1, kx2−p1x1−p2x2k ≤r2 which obviously holds forr1=p2kx1−x2k andr2=p1kx1−x2k.
Now, assume that there exists a constantc >0 such that (2.2) holds withc, i.e.,
(2.5)
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
≤c
n
X
i=1
piri2. Since
p1kx1k2+p2kx2k2− kp1x1+p2x2k2
=p1(1−p1)kx1k2+p2(1−p2)kx2k2−2p1p2Rehx1, x2i
=p1p2
kx1k2+kx2k2−2 Rehx1, x2i
=p1p2kx1−x2k2 and
p1r21+p2r22= p1p22+p2p21
kx1−x2k2=p1p2kx1−x2k2, hence (2.5) becomes
p1p2kx1−x2k2≤cp1p2kx1−x2k2,
implying thatc≥1. This proves the sharpness of the inequality (2.2).
The following reverse of the generalized triangle inequality also holds.
Theorem 6. Let xi, pi andri, i∈ {1, . . . , n}be as in the statement of Theorem 5.
Then
(2.6) (0≤)
n
X
i=1
pikxik −
n
X
i=1
pixi
≤1 2 ·
Pn i=1pir2i kPn
i=1pixik, provided that Pn
i=1pixi6= 0.
Proof. From (2.3) we obviously have (2.7) kxik2+
n
X
j=1
pjxj
2
≤2 Re
* xi,
n
X
j=1
pjxj +
+ri2 for anyi∈ {1, . . . , n}.
Utilising the elementary inequalityα2+β2≥2αβ,withα, β >0, we also have
(2.8) 2kxik
n
X
j=1
pjxj
≤ kxik2+
n
X
j=1
pjxj
2
for anyi∈ {1, . . . , n},which together with (2.7) produces
(2.9) kxik
n
X
j=1
pjxj
≤Re
* xi,
n
X
j=1
pjxj
+ +1
2ri2 for anyi∈ {1, . . . , n}.
INEQUALITIES IN INNER PRODUCT SPACES 5
Now, if we multiply (2.9) bypi>0 and sum over ifrom 1 ton, we deduce (2.10)
n
X
i=1
pikxik
n
X
j=1
pjxj
≤
n
X
i=1
pixi
2
+1 2
n
X
i=1
piri2,
which, forPn
i=1pixi6= 0,is equivalent to the desired result (2.6).
Another result is as follows.
Theorem 7. Let xi, pi andri be as in the statement of Theorem 5. Then
(2.11) 0≤
n
X
i=1
pikxik2
!12
−
n
X
i=1
pixi
≤1 2 ·
Pn i=1pir2i kPn
i=1pixik.
Proof. If we multiply the inequality (2.7) bypi≥0 and sum overifrom 1 ton,we deduce
(2.12)
n
X
i=1
pikxik2+
n
X
j=1
pjxj
2
≤2
n
X
j=1
pjxj
2
+
n
X
i=1
piri2
for anyi∈ {1, . . . , n}. However,
(2.13) 2
n
X
i=1
pikxik2
!12
n
X
j=1
pjxj
≤
n
X
i=1
pikxik2+
n
X
j=1
pjxj
2
and by (2.12) and (2.13), we deduce the desired result (2.11).
Remark 1. It is an open problem of whether the constant 12 is best possible in (2.6) or in (2.11). Also, since
n
X
i=1
pikxik ≤
n
X
i=1
pikxik2
!12 ,
we observe that the inequality (2.11) is better than (2.6).
The following result also holds.
Theorem 8. With the assumptions of Theorem 5 and if, in addition, we have
(0<) max
i=1,n
{ri} ≤
n
X
j=1
pjxj
and xi6= 0 for all i∈ {1, ..., n},
then we obtain
(2.14)
n
X
j=1
pjxj
2
≤
n
X
i=1
piri2+ Re2
* n X
i=1
pi xi kxik,
n
X
j=1
pjxj +
.
This implies the inequality
(2.15) (0≤) 1−
n
X
i=1
pi xi kxik
2
≤ Pn
i=1pir2i kPn
i=1pixik2.
Proof. From (2.3), we have
(2.16) kxik2+
n
X
j=1
pjxj
2
−ri2
1 2
2
≤2 Re
* xi,
n
X
j=1
pjxj
+
for eachi∈ {1, . . . , n}and since
(2.17) 2kxik v u u u t
n
X
j=1
pjxj
2
−ri2≤ kxik2+
n
X
j=1
pjxj
2
−r2i
1 2
2
,
then by (2.16) and (2.17) we get
(2.18) kxik
n
X
j=1
pjxj
2
−ri2
1 2
≤Re
* xi,
n
X
j=1
pjxj +
for eachi∈ {1, . . . , n}.If we multiply (2.14) by kxpi
ik ≥0 and sum overifrom 1 to n,we obtain
(2.19)
n
X
i=1
n
X
j=1
pjxj
2
−ri2
1 2
pi≤Re
* n X
i=1
pi
xi kxik,
n
X
j=1
pjxj
+ .
However,
(2.20)
n
X
i=1
pi
n
X
j=1
pjxj
2
−r2i
1 2
≥
n
X
j=1
pjxj
2
−
n
X
i=1
pir2i
1 2
and by (2.19) and (2.20), we then have
n
X
j=1
pjxj
2
≤
n
X
i=1
piri2+ Re2
* n X
i=1
pi
xi
kxik,
n
X
j=1
pjxj
+ ,
which proves the desired inequality (2.14).
Now, on making use of the Schwarz inequality, we have Re2
* n X
i=1
pi
xi
kxik,
n
X
j=1
pjxj
+
≤
n
X
i=1
pi
xi
kxik
2
n
X
j=1
pjxj
2
,
which together with (2.14) produces the desired result (2.15).
Theorem 9. With the assumptions of Theorem 5 and if, in addition, we have
(0<) max
i=1,n
{ri} ≤
n
X
j=1
pjxj ,
INEQUALITIES IN INNER PRODUCT SPACES 7
then
(0≤)
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
≤ Pn
i=1pir2ikxik2
Pn j=1pjxj
(2.21) 2
≤max
i=1,n
ri2 Pn
i=1pikxik2
Pn
j=1pjxj
2
.
Proof. From (2.18) we have that
(2.22)
n
X
i=1
pikxik
n
X
j=1
pjxj
2
−r2i
1 2
≤
n
X
j=1
pjxj
2
.
However,
(2.23)
n
X
i=1
pikxik2
n
X
j=1
pjxj
2
−ri2
1 2
≤
n
X
i=1
pikxik
n
X
j=1
pjxj
2
−r2i
1 2
and since
n
X
i=1
pikxik2
n
X
j=1
pjxj
2
−ri2
=
n
X
i=1
pikxik2
n
X
j=1
pjxj
2
−
n
X
i=1
pikxik2r2i,
hence by (2.22) and (2.23) we obtain (2.21).
3. Some Results Via Ostrowski’s Inequality
In order to provide more convenient inequalities, the condition (2.1), namely
(3.1)
xi−
n
X
j=1
pjxj
≤ri for all i∈ {1, . . . , n},
where xi ∈H, pi ≥0 with i∈ {1, . . . , n} and Pn
i=1pi = 1 can be replaced with a number of discrete Ostrowski inequalities such as those obtained by Dragomir in [6].
For the sake of completeness, we recall them here in a single theorem.
Theorem 10(Dragomir, 2002, [6]). Let(X,k·k)be a normed linear space,xi∈X (i∈ {1, . . . , n}) and pi ≥ 0, i ∈ {1, . . . , n} with Pn
i=1pi = 1. Then we have the
inequalities:
xi−
n
X
j=1
pjxj
(3.2)
≤ max
k∈{1,...,n−1}k∆xkk
n
X
j=1
pj|j−i|
≤ max
k∈{1,...,n−1}k∆xkk ×
n−1 2 +
i−n+12 ; Pn
j=1|j−i|p1p Pn
j=1pqj1q , p >1, 1p+1q = 1;
hn2−1
4 + i−n+12 2i max
j∈{1,...,n}{pj} and, forα >1, α1 +β1 = 1,the inequalities:
xi−
n
X
j=1
pjxj
≤
n−1
X
k=1
k∆xkkα
!1α n
X
j=1
|i−j|β1pj
(3.3)
≤
n−1
X
k=1
k∆xkkα
!1α
n−1
2 +
i−n+12
β1
;
n
P
j=1
|i−j|δβ
!1δ
n
P
j=1
pγj
!γ1 , γ >1, γ1+1δ = 1;
n
P
j=1
|i−j|1β max
j∈{1,...,n}{pj} and
xi−
n
X
j=1
pjxj
≤
max{Pi−1,1−Pi}n−1P
k=1
k∆xkk
(1−pi) max i−1
P
k=1
k∆xkk,
n−1
P
k=i
k∆xkk (3.4)
≤(1−pi)
n−1
X
k=1
k∆xkk
for each i ∈ {1, . . . , n}, where ∆xk := xk+1 −xk and Pm := Pm
i=1pi, k ∈ {1, . . . , n−1}, m∈ {1, . . . , n} andP0:= 0.
Now, if we use Theorems 5 and 10 we can state the following result that provides upper bounds for the quantity
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
.
INEQUALITIES IN INNER PRODUCT SPACES 9
Proposition 1. Let (H;h·,·i) be an inner product space, xi ∈ H, i ∈ {1, . . . , n}
andpi≥0 with Pn
i=1pi= 1.Then we have the inequalities:
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
(3.5)
≤ max
k∈{1,...,n−1}k∆xkk2
n
X
i=1
pi
n
X
j=1
pj|j−i|
2
≤ max
k∈{1,...,n−1}k∆xkk2
Pn
i=1pi n−12 +
i−n+12
2
; Pn
j=1pqj2q Pn
i=1
Pn
j=1|j−i|p2p , p >1, 1p+1q = 1;
max
j∈{1,...,n}
p2j Pn i=1pih
n2−1
4 + i−n+12 2i2 and, forα >1, α1 +β1 = 1,the inequalities:
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
(3.6)
≤
n−1
X
k=1
k∆xkkα
!2α n X
i=1
pi
n
X
j=1
|i−j|β1 pj
2
≤
n−1
X
k=1
k∆xkkα
!2α
Pn
i=1pin−1
2 +
i−n+12
2β
; Pn
j=1pγj2γ Pn
i=1pi
Pn
j=1|i−j|βδ2δ , γ >1, γ1+1δ = 1;
j∈{1,...,n}max
p2j Pn i=1pi
Pn
j=1|i−j|β12 and
n
X
i=1
pikxik2−
n
X
i=1
pixi
2
(3.7)
≤
Pn−1
k=1k∆xkk2 Pn
i=1pimaxn
Pi−12 ,(1−Pi)2o Pn
i=1pi(1−pi)2 maxn
Pi−1
k=1k∆xkk,Pn−1
k=i k∆xkko2
≤
n−1
X
k=1
k∆xkk
!2 n X
i=1
pi(1−pi)2.
Remark 2. If one uses the other inequalities obtained in Theorems 6 – 9, then other similar results can be stated. However the details are not presented here.
References
[1] A. H. ANSARI and M. S. MOSLEHIAN, More on reverse triangle inequality in inner product spaces,Int. J. Math. Math. Sci.,18(2005), 2883–2893.
[2] A. H. ANSARI and M. S. MOSLEHIAN, Refinements of reverse triangle inequalities in inner product spaces,J. Inequal. Pure Appl. Math.,6(2005), no. 3, Article 64, 12 pp.
[3] J.B. DIAZ and F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces,Proceedings Amer. Math. Soc.,17(1) (1966), 88-97.
[4] S.S. DRAGOMIR, A Gr¨uss’ type discrete inequality in inner product spaces and applications, J. Math. Anal. Appl.,250(2000), 494-511.
[5] S.S. DRAGOMIR, Reverses of the triangle inequality in inner product spaces,Austral. J. Math.
Anal Appl.,1(2004), No. 2, Article 7, pp. 1-14. PreprintRGMIA Res. Rep. Coll., 7(2004), Supplement, Article 7,[ONLINE: http://rgmia.vu.edu.au/v7(E).html].
[6] S.S. DRAGOMIR, The discrete version of Ostrowski’s inequality in normed lin- ear spaces, J. Inequal. Pure Appl. Math., 3(2002), No. 1, Article 2, [Online http://jipam.vu.edu.au/images/042 01 JIPAM/042 01.pdf].
[7] S.S. DRAGOMIR, A survey of recent reverses for the generalized triangle inequality in inner product spaces,Kragujevac J. Math.,27(2005), 101–143.
[8] S. SAITOH, Generalizations of the triangle inequality,J. Inequal. Pure Appl. Math.,4(2003), no. 3, Article 62, 5 pp.
[9] C. WU and Y. LI, On the triangle inequality in quasi-Banach spaces,J. Inequal. Pure Appl.
Math.,9(2008), no. 2, Article 41, 4 pp.
Mathematics, School of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia.
E-mail address:[email protected]
URL:http://www.staff.vu.edu.au/rgmia/dragomir/
Department of Mathematical Education and the RINS, Gyeongsang National Uni- versity, Chinju 660-701, Korea
E-mail address:[email protected]
Department of Mathematics, Dongeu University, Busan 614-714, Korea E-mail address:[email protected]
