Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 323609, 17 pages
doi:10.1155/2010/323609
Research Article
On Sharp Triangle Inequalities in Banach Spaces II
Ken-Ichi Mitani1 and Kichi-Suke Saito2
1
Department of Applied Chemistry and Biotechnology, Faculty of Engineering,
Niigata Institute of Technology, Kashiwazaki 945-1195, Japan
2
Department of Mathematics, Faculty of Science, Niigata University, Niigata 950-2181, Japan
Correspondence should be addressed to Kichi-Suke Saito, saito@math.sc.niigata-u.ac.jp
Received 4 November 2009; Accepted 2 March 2010
Academic Editor: Vy Khoi Le
Copyright q 2010 K.-I. Mitani and K.-S. Saito. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Sharp triangle inequality and its reverse in Banach spaces were recently showed by Mitani et
al. 2007. In this paper, we present equality attainedness for these inequalities in strictly convex
Banach spaces.
1. Introduction
In recent years, the triangle inequality and its reverse inequality have been treated in 1–5
see also 6, 7.Kato et al. 8 presented the following sharp triangle inequality and its reverse
inequality with n elements in a Banach space X.
Theorem 1.1 see 8. For all nonzero elements x1 , x2 , . . . , xn in a Banach space X,
⎞
⎛
n xj n xj ⎝n − ⎠ min xj x 1≤j≤n
j1 j j1 ≤
n xj j1
1.1
⎛
⎞
n xj n ⎠maxxj .
⎝
≤ xj n − 1≤j≤n
j1 xj j1 These inequalities are useful to treat geometrical structure of Banach spaces, such as uniform
non-1n -ness see 8. Moreover,Hsu et al. 9 presented these inequalities for strongly
integrable functions with values in a Banach space.
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Journal of Inequalities and Applications
Mitani et al. 10 showed the following inequalities which are sharper than Inequality
1.1 in Theorem 1.1.
Theorem 1.2 see 10. For all nonzero elements x1 , x2 , . . . , xn in a Banach space X with x1 ≥
x2 ≥ · · · ≥ xn , n ≥ 2,
⎞
⎛
n
n k xj xj ⎝k − ⎠
x xk − xk1 j1 k2
j1 j n xj ≤
1.2
j1
⎞
⎛
n
n n
xj ⎠ xn−k − xn−k−1 ,
⎝
−
≤
k
−
x
j
j1 k2
jn−k−1 xj 1.3
where x0 xn1 0.
In this paper we first present a simpler proof of Theorem 1.2. To do this we consider
the case x1 > x2 > · · · > xn , as follows.
Theorem 1.3. For all nonzero elements x1 , x2 , . . . , xn in a Banach space X with x1 > x2 > · · · >
xn , n ≥ 2,
⎞
⎛
n
n k xj xj ⎝k − ⎠xk − xk1 x j1 k2
j1 j n xj ≤
1.4
j1
⎞
⎛
n
n n
xj ⎠ xn−k − xn−k−1 ,
⎝
−
≤
k
−
x
j
j1 k2
jn−k−1 xj 1.5
where x0 xn1 0.
From this result we can easily obtain Theorem 1.2.
Moreover we consider equality attainedness for sharp triangle inequality and its
reverse inequality in strictly convex Banach spaces. Namely, we characterize equality
attainedness of Inequalities 1.4 and 1.5 in Theorem 1.3.
2. Simple Proofs of Theorems 1.2 and 1.3
Proof of Theorem 1.3. According to Theorem 1.1 Inequalities 1.4 and 1.5 hold for the case
n 2 cf. 3. Therefore let n ≥ 3. We first prove 1.4 by the induction. Assume that 1.4
Journal of Inequalities and Applications
3
holds true for all n − 1 elements in X. Let x1 , x2 , . . . , xn be any n elements in X with x1 >
x2 > · · · > xn > 0. Let
xj
uj xj − xn ,
xj 2.1
for all positive numbers j with 1 ≤ j ≤ n. Then
n
xj xn j1
n
n−1
xj
uj
xj j1
j1
2.2
and u1 > u2 > · · · > un−1 > 0. By assumption,
⎞
⎛
n−1
n−1
k uj n−1 uj ≤
uj − ⎝k − ⎠uk − uk1 u j1 j j1 j1
k2
2.3
holds, where un 0. Since uk − uk1 xk − xk1 , from 2.2 and 2.3,
n
n−1 n xj
xj xn uj j1 j1 xj
j1 n−1 n xj ≤ xn uj xj j1
j1
⎞
⎛
n−1
n−1
n xj k uj uj − ⎝k − ⎠uk − uk1 ≤ xn x u j1 j j1
j1 j k2
n−1
n xj xj − xn xn x j1 j j1
⎞
⎛
n−1
k xj ⎠
− ⎝k − x xk − xk1 j
j1
k2
⎞
⎛
n n
k xj ⎠xk − xk1 ⎝
xj −
k − xj j1
j1
k2
2.4
and hence 1.4. Thus 1.4 holds true for all finite elements in X.
Next we show Inequality 1.5. Let
xn−j1
,
vj x1 − xn−j1 xn−j1 1 ≤ j ≤ n − 1.
2.5
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Journal of Inequalities and Applications
Then
n
n
n−1
xj
xj x1 − vj
j1
j1 xj
j1
2.6
and v1 > · · · > vn−1 > 0. Applying Inequality 1.4 to v1 , . . . , vn−1 ,
n−1 n xj n xj ≥ x1 − vj x j1 j j1 j1 ⎞
⎛
n−1
n−1
k vj n xj − vj ⎝k − ⎠vk − vk1 ≥ x1 v x j j j1
j1
k2
j1
n−1
n xj x1 x1 − xn−j1 x −
j1 j j1
⎞
⎛
n−1
k xn−j1 ⎠xn−k1 − xn−k ⎝k − x
n−j1
j1
k2
⎞
⎛
n n
n
xj ⎠xn−k1 − xn−k ,
⎝
xj −
k−
xj j1
k2
jn−k−1
2.7
where vn 0. Thus we obtain 1.5. This completes the proof.
Proof of Theorem 1.2. Let x1 , x2 , . . . , xn be any nonzero elements in X with x1 ≥ · · · ≥ xn .
For all positive numbers m with m > n let
xk,m k
xk ,
1−
m
k 1, 2, . . . , n.
2.8
Then x1,m > x2,m > · · · > xn,m > 0. Applying Theorem 1.3 to x1,m , . . . , xn,m ,
⎞
⎛
k xj,m n
n
⎠xk,m − xk1,m xj,m ⎝k − x j1 j,m k2
j1
≤
n xj,m j1
2.9
⎞
⎛
n
n
n
xj,m ⎠ xn−k,m − xn−k−1,m ,
− ⎝k − ≤
x
j,m
jn−k−1 xj,m k2
j1
where x0,m xn1,m 0 for all positive numbers m with m > n. As m → ∞, we have
Inequalities 1.4 and 1.5.
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5
3. Equality Attainedness in a Strictly Convex Banach Space
In this section we consider equality attainedness for sharp triangle inequality and its reverse
inequality in a strictly convex Banach space. Kato et al. in 8 showed the following.
Theorem 3.1 see 8. Let X be a strictly convex Banach space and x1 , x2 , . . . , xn nonzero elements
in X. Let xj0 min{xj : 1 ≤ j ≤ n} and xj1 max{xj : 1 ≤ j ≤ n}. Let J0 {j : xj xj0 , 1 ≤ j ≤ n}. Then
⎞
⎛
n
n xj n xj ⎝n − ⎠ min xk 3.1
xk x 1≤j≤n
j j1
j1
j1
if and only if either
a x1 x2 · · · xn or
b xj /xj xj1 /xj1 for all j ∈ J0c and
n
j1 xj /xj n
j1 xj /xj xj1 /xj1 .
Theorem 3.2 see 8. Let X be a strictly convex Banach space and x1 , x2 , . . . , xn nonzero elements
in X. Let xj0 min{xj : 1 ≤ j ≤ n} and xj1 max{xj : 1 ≤ j ≤ n}. Let J1 {j : xj xj1 , 1 ≤ j ≤ n}. Then
⎞
⎛
n n xj n xj xj ⎝n − ⎠maxxj 3.2
x 1≤j≤n
j j1
j1
j1
if and only if either
a x1 x2 · · · xn or
b xj /xj xj0 /xj0 for all j ∈ J1c and
n
j1
xj n
j1
xj xj0 /xj0 .
We present equality attainedness for 1.4 and 1.5 in Theorem 1.2. The following
lemma given in 8 is quite powerful.
Lemma 3.3 see 8. Let X be a strictly convex Banach space. Let x1 , x2 , . . . , xn be nonzero elements
in X. Then the following are equivalent.
i nj1 αj xj nj1 αj xj holds for any positive numbers α1 , α2 , . . . , αn .
ii nj1 αj xj nj1 αj xj holds for some positive numbers α1 , α2 , . . . , αn .
iii x1 /x1 x2 /x2 · · · xn /xn .
Theorem 3.4. Let X be a strictly convex Banach space and x, y nonzero elements in X with x >
y. Then
x
y x y 2 − 3.3
y x y
x y if and only if there exists a real number α with 0 < α < 1 satisfying y ±αx.
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Journal of Inequalities and Applications
Proof. Assume that 3.3 is true. By Theorem 3.1, the equality 3.3 is equivalent to Equality
x
y
y x
x
.
3.4
x y x y x
Put
x
y
y − 1
.
α x y x
3.5
Then y αx. Since x > y, we obtain 0 < |α| < 1. Conversely, if y αx where 0 < |α| < 1,
then
y
α
x
x
1
.
3.6
|α| x
x y
By 1 α/|α| ≥ 0, we have 3.4. Thus we get 3.3.
Next we consider the case n 3.
Theorem 3.5. Let X be a strictly convex Banach space and x, y, z nonzero elements in X with x >
y > z. Then
x
x
y
y z x y z 3 − y − z
z 2 − x y x y z 3.7
x y z
if and only if there exist α, β with 0 < β < α < 1 satisfying one of the following conditions:
a y αx, z ±βx,
b y −αx, z βx.
Proof. Assume that 3.7 is true. Put
u x − z
Then u > v > 0 and
x
,
x
x y z z
y
v y − z .
y
y
x
z
x
z
y
3.8
u v.
3.9
Note that u v /
0. As in the proof of Theorem 1.2 given in 10, we have 3.7 if and only if
we have the equalities
x
y
y
z
z x
u v z
u v,
z
x y z x y z
u
v v.
u v u v − 2 − u v 3.10
3.11
Journal of Inequalities and Applications
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By Theorem 3.4, Equality 3.11 implies that
v ±γu
3.12
for some γ with 0 < γ < 1. By 3.8 we have
y ±αx
3.13
for some 0 < α < 1. On the other hand, by Lemma 3.3, Equality 3.10 implies
x
y
y
x
z
z uv
.
x
z x
z u v
y
y
3.14
Hence, by using 3.8, 3.12, and 3.13 we have z βx for some real number β. Since x >
y > z > 0, we have 0 < |β| < α < 1. We consider the following two cases
Case 1. y αx.
Equality 3.14 implies
x
x αx − 2βx
βx
βx x
x
x
.
x x βx x x βx x αx − 2βx
3.15
Hence we have
β
2 2 β β 1 α − 2β
.
β 1 α − 2β
3.16
By 2 β/|β| ≥ 0 and 1 α − 2|β| 1 − |β| α − |β| ≥ 0, Equality 3.16 is valid for all real
numbers β with β /
0.
Case 2. y −αx.
Equality 3.14 implies
x
βx
βx x
x
x
x − αx
−
−
.
x x
x x
βx
βx x − αx
3.17
β 1−α
β β
.
β β |1 − α| β 3.18
So we have
Hence β > 0. Thus ⇒ holds.
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Journal of Inequalities and Applications
Conversely, assume that there exist α, β with 0 < β < α < 1 satisfying one of the
conditions a and b. Then it is clear that 3.7 holds. Thus we obtain ⇐.
Moreover we consider general cases. For each m with 1 ≤ m ≤ n, we put Im {1, 2, . . . , m}. For α α1 , α2 , . . . , αn ∈ Rn and 1 ≤ m ≤ n, we define
Im
α {k ∈ Im : αk > 0},
−
Im
α {k ∈ Im : αk < 0}.
3.19
For a finite set A, the cardinal number of A is denoted by |A|.
−
Lemma 3.6. Let |α1 | > |α2 | > · · · > |αn | > 0. If |Im
α| ≥ |Im
α| for all m with 1 ≤ m ≤ n, then
n
αj > 0.
3.20
j1
Proof. Let
In α {m1 , m2 , . . . , m } Im
α ,
In− α {n1 , n2 , . . . , nk } Ink α ,
3.21
where m1 < m2 < · · · < m and n1 < n2 < · · · < nk . By the assumption, we have > k. We first
show mj < nj for all j with 1 ≤ j ≤ k. It is clear that m1 < n1 . Assume that mi < ni for all i with
1 ≤ i ≤ j. We will show mj1 < nj1 . Suppose that mj1 > nj1 . By mj < nj < nj1 < mj1 , we
have
Inj1 α Im
α m1 , m2 , . . . , mj ,
j
In−j1 α n1 , n2 , . . . , nj1 .
3.22
Hence we have |Inj1 α| < |In−j1 α|, which is a contradiction. Therefore we have mj1 < nj1 .
Namely, mj < nj for all j with 1 ≤ j ≤ k. From this result, we obtain αmj αnj |αmj | − |αnj | > 0.
Hence
n
αj j1
k
αmj αnj
j1
k j1
j1
αmj αnj 3.23
αmj > 0.
jk1
Theorem 3.7. Let X be a strictly convex Banach space and x1 , x2 , . . . , xn nonzero elements in X with
x1 > x2 > · · · > xn . Then
⎞
⎛
n
n k xj n ⎠xk − xk1 xj ⎝k − xj x j1 j j1 k2
j1
3.24
Journal of Inequalities and Applications
9
if and only if there exists α α1 , α2 , . . . , αn ∈ Rn with 1 α1 > |α2 | > |α3 | > · · · > |αn | such that
xm αm x1 ,
−
α| ≥ Im
α
|Im
3.25
3.26
for every m with 1 ≤ m ≤ n.
Proof. ⇒: According to Theorems 3.4 and 3.5, Theorem 3.7 is valid for the cases n 2, 3.
Therefore let n ≥ 4. We will prove Theorem 3.7 by the induction. Assume that this theorem
holds true for all nonzero elements in X less than n. Let x1 > x2 > · · · > xn and assume
that Equality 3.24 holds. Let
xj
uj xj − xn xj 3.27
for positive integer j with 1 ≤ j ≤ n − 1. As in the proof of Theorem 1.3, Equality 3.24 holds
if and only if
n−1 n
n−1 n xj xj
xn uj xn uj ,
j1 xj j1 j1 xj
j1 ⎞
⎛
n−1
n−1
n−1 k uj ui uj − ⎝k − ⎠uk − uk1 u j1 j1
j1 j k2
3.28
3.29
hold, where un 0. Hence, by assumption, there exists β β1 , . . . , βn−1 ∈ Rn−1 with 1 β1 >
|β2 | > · · · > |βn−1 | > 0 such that
1≤j ≤n−1 ,
uj βj u1 ,
− Im β ≥ Im β , 1 ≤ m ≤ n − 1.
Since
n−1
j1
3.30
βj > 0 by Lemma 3.6, we have
n−1
uj j1
n−1
βj u1 /
0.
j1
3.31
Since
xn 1− xj xn xj βj 1 −
x1
x1 3.32
by the definition of uj , we have
xj αj x1 ,
1≤j ≤n−1 ,
3.33
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Journal of Inequalities and Applications
where
x1 − xn xj αj βj ·
.
xj − xn x1 3.34
Since
Im
β ,
α Im
−
−
Im
β ,
α Im
3.35
we have
−
α| ≥ Im
α
|Im
3.36
for all m with 1 ≤ m ≤ n − 1. By Lemma 3.3, Equality 3.28 implies
n−1
n
n−1 n xj xj
uj uj .
j1 j1 xj j1 xj j1
3.37
Hence there exists αn ∈ R such that xn αn x1 . Also,
n
n
xj
αj x1
xj αj x1 j1
j1
−
αn
x1
In−1
,
α − In−1
α |αn | x1 ⎞
⎛
n−1
n−1
uj ⎝ βj ⎠u1 .
j1
Since
n−1
j1
3.38
3.39
j1
βj > 0, we have from 3.37 and 3.38,
I α − I − α αn I α − I − α αn ,
n−1
n−1
n−1
|αn | n−1
|αn | 3.40
I α − I − α αn ≥ 0.
n−1
n−1
|αn |
3.41
which implies
−
If |In−1
α| > |In−1
α|, then it is clear that |In α| ≥ |In− α|.
−
If |In−1 α| |In−1 α|, then, by 3.41, we have αn /|αn | ≥ 0. Hence αn > 0. Thus we
obtain |In α| > |In− α|.
Journal of Inequalities and Applications
11
⇐: Let α α1 , α2 , . . . , αn ∈ Rn with 1 α1 > |α2 | > · · · > |αn | > 0 satisfying 3.25
and 3.26, and let αn1 0. From 3.25 we have
⎞
⎛
n
k xj n xj ⎝k − ⎠xk − xk1 x j1 j j1 k2
⎞
⎛
n αj x1 n
n
⎠
⎝
k
−
α
x
j 1
α x αk x1 − αk1 x1 j1 j 1 k2
j1
⎞
⎛
n
k αj n αj ⎝k − ⎠|αk | − |αk1 | × x1 .
αj j1
3.42
j1
k2
By Lemma 3.6,
n
n αj αj −
αj .
αj j1 j1
j∈In α
j∈In− α
3.43
Let In− α {k1 , k2 , . . . , km }, where 1 < k1 < k2 < · · · < km < n. From 3.26 and |Ik α| |Ik− α| k we have
⎞
⎛
n
k αj ⎝k − ⎠|αk | − |αk1 |
α j1 j k2
n
k − Ik α Ik− α |αk | − |αk1 |
k2
n 2 Ik− α|αk | − |αk1 |
k2
k −1
k
n 1
2 −1 − −
−
I α|αk | − |αk1 | I α|αk | − |αk1 | · · · I α|αk | − |αk1 |
2
k
k
k
k2
kk1
kkm
k −1
k
n
3 −1
2
2
2|αk | − |αk1 | · · · m|αk | − |αk1 |
|αk | − |αk1 | kk1
kk2
kkm
|αk1 | − |αk2 | 2|αk2 | − |αk3 |
· · · m − 1|αkm−1 | − |αkm | m|αkm | − |αkn1 |
αj .
2
j∈In− α
3.44
Thus we obtain 3.24. This completes the proof.
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Journal of Inequalities and Applications
In what follows, we characterize the equality condition of Inequality 1.3 in
Theorem 1.2. For α α1 , α2 , . . . , αn ∈ Rn and positive integer m with 2 ≤ m ≤ n − 1 we
−
α {j ∈ Jm : αj > 0}, and Jm
α {j ∈ Jm : αj < 0}.
define Jm {n − m − 1, . . . , n − 1, n}, Jm
Lemma 3.8. Let α α1 , α2 , . . . , αn ∈ Rn with |α1 | > |α2 | > · · · > |αn−1 | > αn 1. If
−
α| ≥ Jm
α
|Jm
3.45
for all positive integers m with 2 ≤ m ≤ n − 1, then one has
n
−
α − Jn−1
α − αj ≥ 0.
|α1 | Jn−1
3.46
j2
−
Proof. Let Jn−1
α {m1 , m2 , . . . , m , m1 , . . . , mk } and Jn−1
α {n1 , n2 , . . . , n }, where n m1 > · · · > mk ≥ 2 and n > n1 > · · · > n ≥ 2. As in the proof of Lemma 3.6, we have mj > nj
for all j. So |αmj | < |αnj | for all j. Hence
n
−
α − Jn−1
α − αj
|α1 | Jn−1
j2
⎛
−
|α1 | Jn−1 α − Jn−1 α − ⎝
αj j∈Jn−1
α
−
j∈Jn−1
α
⎛
−
αj −
|α1 | Jn−1 α − Jn−1 α − ⎝
j∈Jn−1
α
k j1
|α1 | − αj −
j∈Jn−1
α
−
j∈Jn−1
α
|α1 | − αj ⎞
αj ⎠
⎞
αj ⎠
−
j∈Jn−1
α
|α1 | − αmj −
|α1 | − αnj j1
|α1 | − αmj αnj − αmj k j1
j1
≥ 0.
3.47
This completes the proof.
Theorem 3.9. Let X be a strictly convex Banach space and x1 , x2 , . . . , xn nonzero elements in X with
x1 > x2 > · · · > xn . Then
⎞
⎛
n
n n n
xj ⎠ xn−k − xn−k−1 xj xj − ⎝k − j1 k2
jn−k−1 xj j1
3.48
Journal of Inequalities and Applications
13
holds if and only if there exists α α1 , α2 , . . . , αn ∈ Rn with |α1 | > |α2 | > · · · > |αn−1 | > αn 1,
xj αj xn for all positive integers j with 1 ≤ j ≤ n satisfying
−
α| ≥ Jm
α
|Jm
for all positive integers m with 2 ≤ m ≤ n − 1 and
n
j1
3.49
αj ≥ 0.
Proof. ⇒: Let x1 > x2 > · · · > xn and
⎞
⎛
n
n n n
xj ⎠ xn−k − xn−k−1 .
xj xj − ⎝k − j1 k2
jn−k−1 xj j1
3.50
For positive integers j with 2 ≤ j ≤ n we put
vj x1 − 1 xj .
xj 3.51
Note that 0 < v2 < · · · < vn and
n
xj x1 j1
n
n
xj
− vj .
xj j1
j2
3.52
Then Equality 3.48 holds if and only if
n n
n
n xj vj xj vj ,
j1 j2 j1
j2 3.53
⎞
⎛
n vj n vj ⎝n − 1 − ⎠v2 v j2 j j2 ⎞
⎛
n−1
n n
vj ⎠ vn−k−1 − vn−k vj .
⎝k − jn−k−1 vj j2
k2
3.54
Thus, by the equality condition of sharp triangle inequality with n − 1 elements, there exists
β β1 , . . . , βn ∈ Rn such that
vj βj vn
3.55
0 < β2 < β3 < · · · < βn−1 < βn 1,
− Jm β ≥ Jm β 3.56
for all positive integers j with 2 ≤ j ≤ n,
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Journal of Inequalities and Applications
for all positive integers m with 2 ≤ m ≤ n − 1. Since
x1 x1 − 1 xj βj
− 1 xn
xj xn 3.57
for each j with 2 ≤ j ≤ n, we have xj αj xn , where
x1 − xn xj αj βj
·
.
x1 − xj xn 3.58
−
Note that |Jm
α| |Jm
β| and |Jm
α| |Jm
β|. By 3.53,
n
n
n n xj vj vj xj .
j2 j1
j1 j2
3.59
Hence there exists α1 ∈ R such that x1 α1 xn . We also have |α1 | > |α2 | > · · · > |αn−1 | > αn 1
and
n
n
vj j2
j2
x 1 − 1 xj
xj ⎞
n α
n
j
⎝|α1 | − αj ⎠xn
αj ⎛
j2
3.60
j2
⎛
⎞
n
−
⎝|α1 | Jn−1 α − Jn−1 α − αj ⎠xn .
j2
Hence we have
⎛
⎞
n
n −
αj ⎝|α1 | J α − J α − αj ⎠
n−1
n−1
j1 j2
n
n
−
|α1 | Jn−1 α − Jn−1 α − αj αj .
j2 j1
3.61
From 3.3, we obtain nj1 αj ≥ 0. Thus we have ⇒.
⇐: Assume that there exists α α1 , α2 , . . . , αn ∈ Rn with |α1 | > |α2 | > · · · > |αn−1 | >
αn 1, xj αj xn for all positive integers j with 1 ≤ j ≤ n satisfying 3.49 for all positive
Journal of Inequalities and Applications
integers m with 2 ≤ m ≤ n − 1 and
assumption,
n
n
xj αj xn ,
j1
j1
n
j1
15
αj ≥ 0. By Theorem 3.7, we have 3.54. From the
⎞
⎛
n
n
−
vj ⎝|α1 | Jn−1 α − Jn−1 α − αj ⎠xn .
3.62
j2
j1
By nj1 αj ≥ 0 and Lemma 3.8, we obtain 3.53. Thus we have ⇐. This completes the
proof.
If n 2, then we have the following corollary.
Corollary 3.10. Let X be a strictly convex Banach space and x, y nonzero elements in X with x >
y. Then
x y x y x
y 2−
x
x y 3.63
if and only if there exists a real number α with α > 1 such that x αy.
If n 3, then we have the following corollary.
Corollary 3.11. Let X be a strictly convex Banach space and x, y, z nonzero elements in X with
x > y > z. Then
x
y
z 3−
x
x y z y
z − 2−
x − y
y z x y z x y z 3.64
if and only if there exist α, β with |α| > |β| > 1 such that x αz, y βz and α β 1 ≥ 0.
4. Remark
In this section we consider equality attainedness for sharp triangle inequality in a more
general case, that is, the case without the assumption that x1 > x2 > · · · > xn . Let
us consider the case n 3.
Proposition 4.1. Let X be a strictly convex Banach space, and x, y, z nonzero elements in X.
i If x y z, then the equality
x
x
y
y
z
x y z 3 − y − z
z 2 − x y z x y x y z
always holds.
4.1
16
Journal of Inequalities and Applications
ii If x > y z, then the equality 4.1 holds if and only if there exists a real number α
satisfying α ≥ −y/x and y z αx.
iii If x y > z, then the equality 4.1 holds if and only if there exists a real number α
satisfying α ≥ −x/z and x y αz.
Proof. i Is clear.
ii Assume that 4.1 holds. By y z, 4.1 implies
x y z x
y
z 3−
z x y z.
x y z 4.2
From Theorem 3.1, we have
x
y
y
x
z
z x
.
x y z x y z x
4.3
Hence y z αx for some α ∈ R. The following
y
x
z
x y z
1α
x x
,
y x
4.4
implies
1 α x 1 α x .
y y
4.5
Hence 1 αx/y ≥ 0 and so α ≥ −y/x. The converse is clear.
iii Assume that 4.1 holds. Put u x − zx/x and v y − zy/y.
As in the proof of Theorem 3.5, we have
x
y
y
z
z x
uv
x y z x y z u v
4.6
Since x y, we have x y αz for some α ∈ R. The following
y
z
x
x
z
y
1
z
z
α
x
z
4.7
implies
1 z α 1 z α.
x x
Hence α ≥ −x/z. The converse is clear.
4.8
Journal of Inequalities and Applications
17
Conjecture 1. What is the necessary and sufficient condition when Equality 3.24 (resp. Equality
3.48) holds for n elements x1 , x2 , . . . , xn with x1 ≥ x2 ≥ · · · ≥ xn in Theorem 3.7 (resp.
Theorem 3.9)?
Acknowledgment
The second author was supported in part by Grants-in-Aid for Scientific Research no.
20540158, Japan Society for the Promotion of Science.
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