Journal of Inequalities in Pure and
Applied Mathematics
REFINEMENTS OF REVERSE TRIANGLE INEQUALITIES IN
INNER PRODUCT SPACES
volume 6, issue 3, article 64,
2005.
ARSALAN HOJJAT ANSARI AND MOHAMMAD SAL MOSLEHIAN
Department of Mathematics
Ferdowsi University
P.O. Box 1159, Mashhad 91775, Iran.
Received 01 February, 2005;
accepted 20 May, 2005.
Communicated by: S.S. Dragomir
EMail: msalm@math.um.ac.ir
URL: http://www.um.ac.ir/∼moslehian/
Abstract
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c
2000
Victoria University
ISSN (electronic): 1443-5756
026-05
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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.
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
5
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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1.
Introduction
It is interesting to know under which conditions the triangle inequality reverses
in a normed space X; in other words,
P we would like
P to know if there is a constant c with the property that c nk=1 kxk k ≤ k nk=1 xk k for some finite set
x1 , . . . , xn ∈ X. M. Nakai and T. Tada [7] proved that the normed spaces with
this property for any finite set x1 , . . . , xn ∈ 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 vectors xk ∈ H, k ∈ {1, . . . , n} satisfy
0≤r≤
Rehxk , ai
,
kxk k
Then
r
n
X
k=1
k ∈ {1, . . . , n}.
n
X
kxk k ≤ xk ,
k=1
where equality holds if and only if
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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n
X
k=1
xk = r
n
X
kxk ka.
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k=1
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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 referred to [9], [5] for the terminology of inner product spaces.
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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2.
Main Results
The following theorem is an improvement of Theorem 2.1 of [4] in which the
real numbers r1 , r2 are 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 vectors xk ∈ H, k ∈ {1, . . . , n} satisfy
(2.1)
0 ≤ r12 kxk k ≤ Rehxk , r1 ai,
0 ≤ r22 kxk k ≤ Imhxk , r2 ai
for some r1 , r2 ∈ [−1, 1]. Then we have the inequality
n
n
X
X
1
2
2 2
(2.2)
(r1 + r2 )
kxk k ≤ xk .
k=1
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
k=1
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The equality holds in (2.2) if and only if
n
X
(2.3)
xk = (r1 + ir2 )
k=1
Contents
n
X
JJ
J
kxk ka.
k=1
Proof. If r12 + r22 = 0, the theorem is trivial. Assume that r12 + r22 6= 0. Summing
inequalities (2.1) over k from 1 to n, we have
* n
+
* n
+
n
X
X
X
(r12 + r22 )
kxk k ≤ Re
xk , r1 a + Im
xk , r2 a
k=1
k=1
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Triangle Inequalities in Inner
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= Re
* n
X
+
xk , (r1 + ir2 )a
k=1
*
+
n
X
≤
xk , (r1 + ir2 )a k=1 n
X
≤
xk k(r1 + ir2 )ak
k=1
n
X
1 xk .
= (r12 + r22 ) 2 Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
k=1
Hence (2.2) holds.
If (2.3) holds, then
n
n
n
X
X
X
2
2 12
xk = (r1 + ir2 )
kxk k.
kxk ka = (r1 + r2 )
k=1
k=1
k=1
Conversely, if the equality holds in (2.2), we have
n
n
X
X
1 (r12 + r22 ) 2 xk = (r12 + r22 )
kxk k
k=1
+
* n k=1
X
≤ Re
xk , (r1 + ir2 )a
* k=1
+
n
X
1
≤
xk , (r1 + ir2 )a ≤ (r12 + r22 ) 2
k=1
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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n
X
xk .
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k=1
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From this we deduce
*
+ n
n
X X
xk , (r1 + ir2 )a = xk k(r1 + ir2 )ak.
k=1
k=1
Consequently there exists η ≥ 0 such that
n
X
xk = η(r1 + ir2 )a.
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
k=1
From this we have
n
n
X X
2
2
2
2 12
(r1 + r2 ) η = kη(r1 + ir2 )ak = xk = (r1 + r2 )
kxk k.
1
2
k=1
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
k=1
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Hence
η=
n
X
kxk k.
k=1
The next theorem
p is a refinement
pof Corollary 1 of [4] since, in the notation
of Theorem 2.1, 2 − p21 − p22 ≤ α12 + α22 .
Theorem 2.2. Let a be a unit vector in the complex inner product space (H; h·, ·i).
Suppose the vectors xk ∈ H − {0}, k ∈ {1, . . . , n}, are such that
√
(2.4)
kxk − ak ≤ p1 , kxk − iak ≤ p2 , p1 , p2 ∈ 0, α2 + 1 ,
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where α = min1≤k≤n kxk k. Let
kxk k2 − p21 + 1
:1≤k≤n ,
α1 = min
2kxk k
kxk k2 − p22 + 1
α2 = min
:1≤k≤n .
2kxk k
Then we have the inequality
(α12
+
α22 )
1
2
n
X
k=1
n
X kxk k ≤ xk ,
k=1
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
where the equality holds if and only if
n
X
k=1
xk = (α1 + iα2 )
n
X
kxk ka.
k=1
hxk − a, xk − ai ≤ p21 ,
kxk k2 + 1 − p21 ≤ 2 Rehxk , ai,
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Proof. From the first inequality in (2.4) we have
and
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
k = 1, . . . , n,
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kxk k2 − p21 + 1
kxk k ≤ Rehxk , ai.
2kxk k
Consequently,
α1 kxk k ≤ Rehxk , ai.
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Similarly from the second inequality we obtain
α2 kxk k ≤ Rehxk , iai = Imhxk , ai.
Now apply Theorem 2.1 for r1 = α1 , r2 = α2 .
Corollary 2.3. Let a be a unit vector in the complex inner product space (H; h·, ·i).
Suppose that the vectors xk ∈ H − {0}, k ∈ {1, . . . , n} such that
kxk − ak ≤ 1,
Then
kxk − iak ≤ 1.
n
n
X
α X
√
kxk k ≤ xk ,
2 k=1
k=1
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
in which α = min1≤k≤n kxk k. The equality holds if and only if
n
X
α
2
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n
(1 + i) X
xk = α
kxk ka.
2
k=1
k=1
Proof. Apply Theorem 2.2 for α1 =
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
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J
= 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 vectors xk ∈ H − {0},
k ∈ {1, . . . , n} satisfy
√
kxk − ak ≤ p, p ∈ 0, α2 + 1 , α = min kxk k.
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1≤k≤n
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Then we have the inequality
α1
n
X
k=1
where
n
X kxk k ≤ xk ,
k=1
kxk k2 − p2 + 1
α1 = min
:1≤k≤n .
2kxk k
The equality holds if and only if
n
n
X
X
kxk ka.
xk = α1
k=1
k=1
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Proof. The proof is similar to Theorem 2.2 in which we use Theorem 2.1 with
r2 = 0.
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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.
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Theorem 2.5. Let a1 , . . . , am be orthonormal vectors in the complex inner
product space (H; h·, ·i). Suppose that for 1 ≤ t ≤ m, rt , ρt ∈ R and that
the vectors xk ∈ H, k ∈ {1, . . . , n} satisfy
(2.5)
0 ≤ rt2 kxk k ≤ Rehxk , rt at i,
0 ≤ ρ2t kxk k ≤ Imhxk , ρt at i,
t ∈ {1, . . . , m}.
(2.6)
t=1
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! 21
rt2
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Then we have the inequality
m
X
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+
ρ2t
n
n
X X
kxk k ≤ xk .
k=1
k=1
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The equality holds in (2.7) if and only if
n
X
(2.7)
xk =
k=1
n
X
kxk k
m
X
(rt + iρt )at .
t=1
k=1
P
Pm 2
2
2
Proof. If m
t=1 (rt + ρt ) = 0, the theorem is trivial. Assume that
t=1 (rt +
2
ρt ) 6= 0. Summing inequalities (2.6) over k from 1 to n and again over t from 1
to m, we get
* n
+
* n
+
m
n
m
m
X
X
X
X
X
X
(rt2 + ρ2t )
kxk k ≤ Re
xk ,
rt at + Im
xk ,
ρ t at
t=1
k=1
= Re
= Re
* k=1
n
X
* k=1
n
X
k=1
xk ,
xk ,
t=1
m
X
t=1
m
X
+
r t at
+ Re
k=1
+
(rt + iρt )at
t=1
*
+
n
m
X
X
≤
xk ,
(rt + iρt )at k=1 t=1
n
m
X
X
≤
xk (rt + iρt )at t=1
k=1
m
n
! 12
X X
=
xk (rt2 + ρ2t )
.
k=1
* k=1
n
X
t=1
m
X
xk , i
t=1
+
Refinements of Reverse
Triangle Inequalities in Inner
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Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
ρ t at
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Then
m
X
(2.8)
! 21
(rt2 + ρ2t )
t=1
n
X
k=1
n
X
kxk k ≤ xk .
k=1
If (2.8) holds, then
n
n
! 21
m
m
n
X X
X
X
X
xk = kxk k
(rt + iρt )at =
kxk k
(rt2 + ρ2t )
.
k=1
k=1
t=1
t=1
k=1
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Conversely, if the equality holds in (2.7), we obtain from (2.6) that
! 12 n
m
n
X X
X
2
2
2
2
(rt + ρt )
xk =
(rt + ρt )
kxk k
t=1
t=1
k=1
k=1
* n
+
m
X
X
≤ Re
xk ,
(rt + iρt )at
m
X
t=1
* k=1
+
n
m
X
X
≤
xk ,
(rt + iρt )at nk=1 t=1
m
X X
≤
xk (rt + iρt )at t=1
k=1
! 12
n
m
X
X
.
=
xk (rt2 + ρ2t )
k=1
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Mohammad Sal Moslehian
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Thus we have
* n
m
+ n
m
X
X X
X
xk ,
(rt + iρt )at = xk (rt + iρt )at .
k=1
t=1
t=1
k=1
Consequently there exists η ≥ 0 such that
n
X
xk = η
m
X
(rt + iρt )at
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
t=1
k=1
from which we have
η
m
X
t=1
! 12
(rt2 + ρ2t )
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
m
X
= η
(rt + iρt )at t=1
n
X
=
xk Title Page
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k=1
=
n
X
kxk k
t=1
k=1
Hence
η=
n
X
k=1
m
X
! 12
(rt2 + ρ2t )
.
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kxk k.
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Corollary 2.6. Let a1 , . . . , am 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 xk ∈ H, k ∈ {1, . . . , n} satisfy
0 ≤ rt2 kxk k ≤ Rehxk , rt at i.
Then we have the inequality
m
X
! 21
rt2
t=1
n
X
k=1
n
X kxk k ≤ xk .
k=1
The equality holds if and only if
n
n
m
X
X
X
xk =
kxk k
r t at .
k=1
k=1
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
t=1
Proof. Apply Theorem 2.5 for ρt = 0.
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Theorem 2.7. Let a1 , . . . , am be orthornormal vectors in the complex inner
product space (H; h·, ·i). Suppose that the vectors xk ∈ H−{0}, k ∈ {1, . . . , n}
satisfy
√
kxk − at k ≤ pt , kxk − iat k ≤ qt , pt , qt ∈ 0, α2 + 1 , 1 ≤ t ≤ m,
Contents
where α = min1≤k≤n kxk k. Let
kxk k2 − p2t + 1
αt = min
:1≤k≤n ,
2kxk k
kxk k2 − qt2 + 1
βt = min
:1≤k≤n .
2kxk k
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Then we have the inequality
m
X
! 12
αt2 + βt2
t=1
n
X
k=1
n
X kxk k ≤ xk ,
k=1
where equality holds if and only if
n
X
k=1
xk =
n
X
k=1
kxk k
m
X
(αt + iβt )at .
t=1
Proof. For 1 ≤ t ≤ m, 1 ≤ k ≤ n it follows from kxk − at k ≤ pt that
hxk − at i, xk − at i ≤ p2t ,
kxk k2 − p2t + 1
kxk k ≤ Rehxk , at i,
2kxk k
αt kxk k ≤ Rehxk , at i,
and similarly
βt kxk k ≤ Rehxk , iat i = Imhxk , at i.
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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Now applying Theorem 2.4 with rt = αt , ρt = βt we deduce the desired inequality.
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Corollary 2.8. Let a1 , . . . , am be orthornormal vectors in the complex inner
product space (H; h·, ·i). Suppose that the vectors xk ∈ H, k ∈ {1, . . . , n}
satisfy
kxk − at k ≤ 1, kxk − iat k ≤ 1, 1 ≤ t ≤ m.
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Then
n
n
X
X
√
α
√ m
kxk k ≤ xk .
2
k=1
k=1
The equality holds if and only if
n
X
n
xk = α
k=1
m
X
(1 + i) X
kxk k
at .
2 k=1
t=1
Proof. Apply Theorem 2.7 for αt =
α
2
= βt .
Remark 1. It is interesting to note that
P
k nk=1 xk k
α √
√ m ≤ Pn
≤ 1,
2
k=1 kxk k
where
r
2
.
m
Corollary 2.9. Let a be a unit vector in the complex inner product space (H; h·, ·i).
Suppose that the vectors xk ∈ H − {0}, k ∈ {1, . . . , n} satisfy
α≤
kxk − ak ≤ p1 ,
kxk − iak ≤ p2 ,
p1 , p2 ∈ (0, 1].
Let
kxk k2 − p21 + 1
α1 = min
:1≤k≤n ,
2kxk k
kxk k2 − p22 + 1
α2 = min
:1≤k≤n .
2kxk k
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Triangle Inequalities in Inner
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Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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1
1
If α1 6= (1 − p21 ) 2 , or α2 6= (1 − p22 ) 2 , then we have the following strictly
inequality
n
n
X
X
1
(2 − p21 − p22 ) 2
kxk k < xk .
k=1
k=1
Proof. If equality holds, then by Theorem 2.2 we have
n
n
n
X
X
X
1
2
2 12
2
2 2
(α1 + α2 )
kxk k ≤ xk = (2 − p1 − p2 )
kxk k
k=1
k=1
and so
k=1
1
1
(α12 + α22 ) 2 ≤ (2 − p21 − p22 ) 2 .
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
On the other hand for 1 ≤ k ≤ n,
1
kxk k2 − p21 + 1
≥ (1 − p21 ) 2
2kxk k
and so
α1 ≥ (1 −
Close
1
2
(2 − p21 − p22 ) ≤ (α12 + α22 ) .
Thus
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1
α2 ≥ (1 − p22 ) 2 .
1
2
Contents
JJ
J
1
p21 ) 2 .
Similarly
Hence
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q
1
α12 + α22 = (2 − p21 − p22 ) 2 .
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Therefore
1
α1 = (1 − p21 ) 2
and
1
α2 = (1 − p22 ) 2 ,
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 xk ∈ H − {0}, k ∈ {1, . . . , n}
such that
RehM a − xk , xk − mai ≥ 0,
RehLia − xk , xk − `iai ≥ 0,
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
or equivalently,
kxk −
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
M −m
m+M
ak ≤
,
2
2
kxk −
L+`
L−`
iak ≤
.
2
2
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Let
αm,M = min
and
α`,L = min
2
kxk k + mM
:1≤k≤n
(m + M )kxk k
kxk k2 + `L
:1≤k≤n .
(` + L)kxk k
Then we have the inequlity
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1
2
2
(αm,M
+ α`,L
)2
n
X
k=1
n
X kxk k ≤ xk .
k=1
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The equality holds if and only if
n
X
xk = (αm,M + iα`,L )
k=1
n
X
kxk ka.
k=1
Proof. For each 1 ≤ k ≤ n, it follows from
kxk −
that
m+M
M −m
ak ≤
2
2
2
M −m
m+M
m+M
xk −
a, xk −
≤
.
2
2
2
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
Hence
kxk k2 + mM ≤ (m + M ) Rehxk , ai.
Then
kxk k2 + mM
kxk k ≤ Rehxk , ai,
(m + M )kxk k
and consequently
αm,M kxk k ≤ Rehxk , ai.
Similarly from the second inequality we deduce
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α`,L kxk k ≤ Imhxk , ai.
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Applying Theorem 2.1 for r1 = αm,M , r2 = α`,L , we infer the desired inequality.
Page 19 of 24
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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 − xk , xk − mai ≥ 0,
RehLia − xk , xk − `iai ≥ 0,
or equivalently
M −m
m
+
M
≤
xk −
a
,
2
2
L−`
L
+
`
xk −
≤
ia
.
2
2
Let
αm,M = min
and
α`,L = min
kxk k2 + mM
:1≤k≤n
(m + M )kxk k
kxk k2 + `L
:1≤k≤n .
(` + L)kxk k
√
√
mM
`L
If αm,M 6= 2 m+M
, or α`,L 6= 2 `+L
, then we have
2
`L
mM
+
2
(m + M )
(` + L)2
12 X
n
k=1
n
X
kxk k < xk .
k=1
Proof. If
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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2
mM
`L
+
2
(m + M )
(` + L)2
12 X
n
k=1
n
X
kxk k = xk k=1
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then by Theorem 2.10 we have
2
(αm,M
n
X
2
+ α`,L
)
kxk k ≤ xk k=1
k=1
12 X
n
mM
`L
=2
+
kxk k.
(m + M )2 (` + L)2
k=1
1
2
n
X
Consequently
2
(αm,M
+
1
2
α`,L
)2
≤2
On the other hand for 1 ≤ k ≤ n,
√
kxk k2 + mM
mM
≥2
,
(m + M )kxk k
m+M
mM
`L
+
2
(m + M )
(` + L)2
and
1
1
2
2
(αm,M
+ α`,L
)2 ≥ 2
Hence
`L
mM
+
2
(m + M )
(` + L)2
mM
`L
+
(m + M )2 (` + L)2
12
√
αm,M
.
12
Then
2
2
(αm,M
+ α`,L
)2 = 2
12
√
`L
kxk k2 + `L
,
≥2
`+L
(` + L)kxk k
so
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
mM
=2
m+M
.
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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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. Let a ∈ C with |a| = 1. Suppose that zk ∈ C, k ∈ {1, . . . , n}
such that
√
2
|zk − a| ≤ p1 , |zk − ia| ≤ p2 , p1 , p2 ∈ 0, α + 1 ,
where
α = min{|zk | : 1 ≤ k ≤ n}.
Let
|zk |2 − p21 + 1
α1 = min
:1≤k≤n ,
2|zk |
|zk |2 − p22 + 1
:1≤k≤n .
α2 = min
2|zk |
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
Then we have the inequality
n
n
q
X X
α12 + α22
|zk | ≤ zk .
k=1
k=1
The equality holds if and only if
n
X
k=1
zk = (α1 + iα2 )
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n
X
!
|zk | a.
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Proof. Apply Theorem 2.2 for H = C.
Corollary 2.13. Let a ∈ C with |a| = 1. Suppose that zk ∈ C, k ∈ {1, . . . , n}
such that
|zk − a| ≤ 1, |zk − ia| ≤ 1.
If α = min{|zk | : 1 ≤ k ≤ n}. Then we have the inequality
n
n
X
α X
√
|zk | ≤ zk 2 k=1
k=1
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
the equality holds if and only if
n
X
k=1
zk = α
(1 + i)
2
Proof. Apply Corollary 2.3 for H = C.
n
X
k=1
!
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
|zk | a.
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References
[1] A.H. ANSARI AND M.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-BunyakovskySchwarz Type, Nova Science Publishers, Inc., Hauppauge, NY, 2004.
[6] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and
New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/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.
Refinements of Reverse
Triangle Inequalities in Inner
Product Spaces
Arsalan Hojjat Ansari and
Mohammad Sal Moslehian
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