On Generalizations and Refinements of Triangle Inequalities: Tamotsu Izumida

On generalizations and refinements
of triangle inequalities
Tamotsu Izumida
Doctoral Program in Fundamental Sciences
Graduate School of Science and Technology
Niigata University
March 2015
Contents
Preface 2
Acknowledgements 5
1 On generalized triangle inequalities 6

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 Absolute normalized norms . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.1 Absolute normalized norms on Cn and their characterizations 8
n
1.2.2 The dual space of C with an absolute normalized norm . . . 17
1.2.3 The set Ψ
e n of all continuous concave functions on ∆n . . . . . 20
1.3 Characterizations of generalized triangle inequalities . . . . . . . . . . 24
1.3.1 Generalizations of preceding results . . . . . . . . . . . . . . . 24
1.3.2 Applications to inequalities involved with the Euler-Lagrange
type identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
1.3.3 An application to the parallelogram inequality . . . . . . . . . 33
1.4 A generalization of `p -spaces . . . . . . . . . . . . . . . . . . . . . . . 34
1.4.1 `ψ -spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.4.2 `ψ̃ -spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2 On sharp triangle inequalities 41

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.2 Sharp triangle inequalities in Banach spaces . . . . . . . . . . . . . . 42
2.2.1 Sharp triangle inequalities and their reverse inequalities . . . . 42
2.2.2 Sharper triangle inequalities . . . . . . . . . . . . . . . . . . . 46
2.3 Characterization of the intermediate values of the triangle inequality . 50
2.3.1 A simple proof of the preceding result . . . . . . . . . . . . . . 50
2.3.2 Equality attainedness in a strictly convex Banach space . . . . 57
2.4 Applications of sharp triangle inequalities . . . . . . . . . . . . . . . . 66
2.4.1 The strict convexity and the uniform convexity . . . . . . . . 66
2.4.2 The uniform non-squareness . . . . . . . . . . . . . . . . . . . 67
References 70
1
Preface
Let (X, k · k) be a normed space. The triangle inequality
kx + yk ≤ kxk + kyk (x, y ∈ X)
is one of the most significant inequalities in mathematics. This inequality and the
relevant topics have been studied by many authors (cf. [2], [3], [4], [6], [9], [10], [11],
[12], [13], [19], [20], [21], [22], [24], [27], [29], [30], [31], [33], [36], [37], [38], [40], [42]
etc. ).
In this thesis, we study the triangle inequality in two directions. We first study
a generalized triangle inequality of the following type. For a Hilbert space H, we
recall the parallelogram law
kx + yk2 + kx − yk2 = 2 kxk2 + kyk2

(x, y ∈ H).
This implies that the parallelogram inequality
kx + yk2 ≤ 2 kxk2 + kyk2

(x, y ∈ H). (1)
Saitoh in [38] noted the inequality (1) may be more suitable than the usual triangle
inequality and used the inequality (1) to the setting of a natural sum Hilbert space
for two arbitrary Hilbert spaces. This inequality actually holds in any normed space.
The following inequality which is a generalization of (1)
kx + ykq ≤ 2q−1 kxkq + kykq

(x, y ∈ X), (2)
where q ≥ 1 also holds in any normed space. Motivated by this, Belbacir et al.
[6] considered that replacing the triangle inequality in the definition of a norm by
(2). On the other hand, for a Hilbert space H, we recall the Euler-Lagrange type
identity
kxk2 kyk2 kax + byk2 kνbx − µayk2
+ − = (x, y ∈ H),
µ ν λ λµν
where a, b, λ, µ, ν ∈ R with λ = µa2 + νb2 . If λµν > 0, this implies the following
inequality
kax + byk2 kxk2 kyk2
≤ + (x, y ∈ H).
λ µ ν
2
This inequality can be considered a generalization of (1). In this direction, for a
normed space X, Takahasi et al. [40] considered the following type inequality
kax + bykp kxkp kykp
≤ + (x, y ∈ X), (3)
λ µ ν
where p ∈ R with p > 0, a, b ∈ C and λ, µ, ν ∈ R. They gave necessary and
sufficient conditions which the inequality (3) and its reverse inequality hold. In [9],
Dadipour et al. characterized all (µ1 , . . . , µn ) ∈ Rn which satisfy a general case of
this inequality:
kx1 kp kxn kp
kx1 + · · · + xn kp ≤ + ··· + (x1 , . . . , xn ∈ X) (4)
µ1 µn
and its reverse inequality. Our main aim in chapter 1 is to present a generalization
of (4) by using ψ-direct sums of Banach spaces (cf. [18]). Therefore, we give another
approach to characterizations of all (µ1 , . . . , µn ) ∈ Rn which satisfy (4) and its
reverse inequality (cf. [9, Theorems 2.4–2.7]).
Next we study the inequalities which are sharper than the usual triangle inequal-
ity for a Banach space X. We consider the usual triangle inequality for n-elements

Xn X n
xj ≤ kxj k (x1 , · · · , xn ∈ X).

j=1 j=1
In [19], Kato, Saito and Tamura proved the sharp triangle inequality and reverse
inequality as follows: for all nonzero elements x1 , · · · , xn in a Banach space X,
!
n n
X X x j

xj + n − min kxj k

kxj k 1≤j≤n

j=1 j=1
n
X
≤ kxj k
j=1
!
n n
X X xj
≤ xj + n− max kxj k

j=1
kxj k 1≤j≤n
j=1
hold. In the case of two elements, above inequalities are following: for all nonzero
elements x, y ∈ X,

x y
kx + yk + 2 −
+ min{kxk, kyk}
kxk kyk
≤ kxk + kyk

x y
≤ kx + yk + 2 −
kxk kyk max{kxk, kyk}
+
3
hold (cf. [14] and [20]). These sharp triangle inequalities are applied to the char-
acterizations of geometric properties of Banach spaces like the strict convexity, the
uniform convexity and so on (cf. [19] and [37]). After that, Mitani et al. [29] suc-
ceeded in the further extension of above inequalities as follows: for all nonzero
elements x1 , x2 , . . . , xn in a Banach space X with kx1 k ≥ kx2 k ≥ · · · ≥ kxn k, n ≥ 2,
n n
k !
X X X x
j
xj + k− (kxk k − kxk+1 k)

kxj k

j=1 k=2 j=1
n
X
≤ kxj k
j=1
 
n n n
X X X xj
≤ xj − k −  (kxn−k k − kxn−(k−1) k)

j=1

k=2
kxj k
j=n−(k−1)
hold, where x0 = xn+1 = 0. In the case of n = 3, above inequalities are following:

for all nonzero elements x, y, z ∈ X with kxk ≥ kyk ≥ kzk,

x y z
kx + y + zk + 3 − + + kzk
kxk kyk kzk

x y
+ 2− kxk + kyk (kyk − kzk)

≤ kxk + kyk + kzk

x y z
≤ kx + y + zk + 3 − + + kxk
kxk kyk kzk

y z
− 2− kyk + kzk (kxk − kyk)

hold. They also studied equality attainedness on these inequalities in a strictly

convex Banach space (cf. [19] and [27]). In this direction, Mineno, Nakamura and
Ohwada [24] studied the problem that characterize all the intermediate value C
which satisfies
n
X Xn
0≤C≤ kxj k − xj

j=1 j=1
by using x1 , x2 , . . . , xn in a Banach space X. Our main aims in Section 2 are to give

the simple proof of the result of Mineno et al. and study when this norm inequality
attains equality in a strictly convex Banach space.
4
Acknowledgements
I would like to express the deepest gratitude to my supervisor Professor Kichi-Suke
Saito at Niigata University for his great guidance and useful advices. I have studied
under him during a master’s course from 1997 to 1999 and a doctoral course. He
provided much support and instructions. I also wish to thank all my colleagues in
Professor Saito’s laboratory.
Moreover I would like to express his thanks to Professor Tomoyoshi Ohwada at
Shizuoka University for his many instructive suggestions in Chapter 2 of this thesis.
5
1 On generalized triangle inequalities
1.1 Introduction
In this chapter, for a normed space (X, k · k), we consider the following generalized
triangle inequality which is involved with the Euler-Lagrange type identity: for any
fixed n ∈ N with n ≥ 2 and fixed p ∈ R with p > 0,
ka1 x1 + · · · + an xn kp kx1 kp kxn kp
≤ + ··· + (x1 , . . . , xn ∈ X) (5)
λ µ1 µn
where (a1 , . . . , an , λ, µ1 , . . . , µn ) ∈ Cn × R × Rn . In [40], Takahasi et al. gave a
necessary and sufficient condition in order that a special case of (5):
≤ + (x, y ∈ X)
λ µ ν
and its reverse inequality hold. For simplicity, we shall write, for example,
{λ > 0, µ > 0, ν > 0}
for
{(λ, µ, ν) ∈ R3 : λ > 0, µ > 0, ν > 0}.
Theorem 1.1.1 (cf. [40, Theorem 1.1 and 4.1]). Let X be a normed space. Let
p ∈ R, p ≥ 1, a, b ∈ C, λ, µ, ν ∈ R. Put

+
Dp = (a, b, λ, µ, ν) : ≤ + (x, y ∈ X) .
λ µ ν
If p > 1, then
(i) Dp+ ∩ {λ > 0, µ > 0, ν > 0}

0 0
= {λ > 0, µ > 0, ν > 0, |λ|1/(p−1) ≥ |µ|1/(p−1) |a|p + |ν|1/(p−1) |b|p }.
(ii) Dp+ ∩ {λ < 0, µ < 0, ν > 0}

0 0
= {λ < 0, µ < 0, ν > 0, |λ|1/(p−1) ≤ |µ|1/(p−1) |a|p − |ν|1/(p−1) |b|p }.
(iii) Dp+ ∩ {λ < 0, µ > 0, ν < 0}

0 0
= {λ < 0, µ > 0, ν < 0, |λ|1/(p−1) ≤ −|µ|1/(p−1) |a|p + |ν|1/(p−1) |b|p }.
6
(iv) Dp+ ∩ {λ < 0, µ < 0, ν < 0} = ∅.
(v) Dp+ ∩ {λ > 0, µ > 0, ν < 0} = ∅.
(vi) Dp+ ∩ {λ > 0, µ < 0, ν > 0} = ∅.
(vii) Dp+ ∩ {λ > 0, µ < 0, ν < 0} = ∅.
(viii) Dp+ ∩ {λ < 0, µ > 0, ν > 0} = C × C × R3 \ {λµν = 0}.

where p0 = (1 − 1/p)−1 .
If p = 1, then
D1+ = {λ < 0, µ > 0, ν > 0}

∪ {λ > 0, µ > 0, ν > 0, λ ≥ µ|a|, λ ≥ ν|b|}
∪ {λ < 0, µ < 0, ν > 0, λ ≥ µ|a|, −ν|b| ≥ µ|a|}
∪ {λ < 0, µ > 0, ν < 0, λ ≥ ν|b|, −µ|a| ≥ ν|b|}.
In [9], Dadipour et al. characterized all (µ1 , . . . , µn ) ∈ Rn which satisfy a special

case of (5):
kx1 kp kxn kp
kx1 + · · · + xn kp ≤ + ··· + (x1 , . . . , xn ∈ X) (6)
µ1 µn
and its reverse inequality. Our aim is to present a generalization of (6) by using
ψ-direct sums of Banach spaces (cf. [18]). Therefore, we give another approach to
characterizations of all (µ1 , . . . , µn ) ∈ Rn which satisfy (6) and its reverse inequality
(cf. [9, Theorems 2.4–2.7]).
In §1.2, we summarize basic results of absolute normalized norms by [7], [26],
[28], [32], [34] and [35].
In §1.3, we present a generalization of (6) by using ψ-direct sums of Banach
spaces (cf. [18]). Therefore, we give another approach to characterizations of all
(µ1 , . . . , µn ) ∈ Rn which satisfy (6) and its reverse inequality.
In §1.4, we generalized a inequality (6) for infinite sequences {xn }∞
n=1 ⊂ X by
using generalized `p -spaces (cf. [25]).
1.2 Absolute normalized norms

In this section, we summarize basic results of absolute normalized norms on Cn by
[7], [26], [28], [32], [34] and [35].
7
1.2.1 Absolute normalized norms on Cn and their characterizations
A norm k · k on Cn is absolute if
k(z1 , z2 , . . . , zn )k = k(|z1 |, |z2 |, . . . , |zn |)k ((z1 , z2 , . . . , zn ) ∈ Cn )
and normalized if
k(1, 0, . . . , 0)k = k(0, 1, 0, . . . , 0)k = · · · = k(0, . . . , 0, 1)k = 1.
An `p -norm on Cn :
( 1
(|z1 |p + · · · + |zn |p ) p (1 ≤ p < ∞)
k(z1 , . . . , zn )kp =
max{|z1 |, . . . , |zn |} (p = ∞)
is such an example. We first show some basic facts about these norms. Let ANn be
the set of all absolute normalized norms on Cn .
Proposition 1.2.1 (cf. [35, Lemma 3.1]). Let k · k ∈ ANn . Then
k(0, z2 , z3 , . . . , zn )k ≤ k(z1 , z2 , . . . , zn )k (B1 )

k(z1 , 0, z3 , . . . , zn )k ≤ k(z1 , z2 , . . . , zn )k (B2 )
..
.
k(z1 , . . . , zn−1 , 0)k ≤ k(z1 , z2 , . . . , zn )k. (Bn )
In particular,
k · k ∞ ≤ k · k ≤ k · k1 .
Proof. For any (z1 , z2 , . . . , zn ) ∈ Cn , we have

1
k(0, z2 , z3 , . . . , zn )k = k(z1 , z2 , z3 , . . . , zn ) + (−z1 , z2 , z3 , . . . , zn )k
2
1
≤ k(z1 , z2 , z3 , . . . , zn )k + k(−z1 , z2 , z3 , . . . , zn )k
2
1
= k(z1 , z2 , z3 , . . . , zn )k + k(| − z1 |, |z2 |, |z3 |, . . . , |zn |)k
2
1
= k(z1 , z2 , z3 , . . . , zn )k + k(z1 , z2 , z3 , . . . , zn )k
2
= k(z1 , z2 , z3 , . . . , zn )k.
8
Then we have (B1 ). Similarly, we have (B2 ), . . . , (Bn ). Therefore, for any
(z1 , z2 , . . . , zn ) ∈ Cn , we have
k(z1 , z2 , . . . , zn )k∞ = max{|z1 |, |z2 |, . . . , |zn |}

= max k(z1 , 0, . . . , 0)k, k(0, z2 , 0, . . . , 0)k, . . . , k(0, . . . , 0, zn )k
≤ k(z1 , z2 , . . . , zn )k
= k(z1 , 0, . . . , 0) + (0, z2 , 0, . . . , 0) + · · · + (0, . . . , 0, zn )k
≤ k(z1 , 0, . . . , 0)k + k(0, z2 , 0, . . . , 0)k + · · · + k(0, . . . , 0, zn )k
= |z1 | + |z2 | + · · · + |zn |
= k(z1 , z2 , . . . , zn )k1 .
This completes the proof.
We remark that every absolute norm is monotone:
Proposition 1.2.2 (cf. [7, Proposition IV.1.1]). Let k · k ∈ ANn . If |zj | ≤ |wj | for
any j = 1, . . . , n, then
k(z1 , . . . , zn )k ≤ k(w1 . . . , wn )k.
Proof. We may suppose that zj , wj ≥ 0 (j = 1, . . . , n). Since z1 ≤ w1 , there exists

t, 0 ≤ t ≤ 1 such that z1 = tw1 . Then we have
k(z1 , z2 , . . . , zn )k
= k(tw1 , z2 , . . . , zn )k

1+t 1−t 1+t 1−t 1+t 1−t
= w1 − w1 , z2 + z2 , . . . , zn + zn
2 2 2 2 2 2

1+t 1 + t 1 + t 1 − t 1 − t 1 − t
= w1 , z2 , . . . , zn + − w1 , z2 , . . . , zn
2 2 2 2 2 2

1 + t 1−t
= 2 (w 1 , z 2 , . . . , z n ) + (−w1 , z2 , . . . , z n )
2
1+t 1−t
≤ k(w1 , z2 , . . . , zn )k + k(−w1 , z2 , . . . , zn )k
2 2
1+t 1−t
= k(w1 , z2 , . . . , zn )k + k(w1 , z2 , . . . , zn )k
2 2
= k(w1 , z2 , . . . , zn )k.
9
Similarly, we have
k(w1 , z2 , z3 , . . . , zn )k ≤ k(w1 , w2 , z3 , . . . , zn )k
k(w1 , w2 , z3 , . . . , zn )k ≤ k(w1 , w2 , w3 , z4 , . . . , zn )k
..
.
k(w1 , w2 , . . . , wn−1 , zn )k ≤ k(w1 , w2 , . . . , wn−1 , wn )k.
Then we have
k(z1 , z2 , . . . , zn−1 , zn )k ≤ k(w1 , w2 , . . . , wn−1 , wn )k.
Bonsal and Duncan [8] showed the following characterization of absolute nor-
malized norms on C2 . Let Ψ2 denote the set of all continuous convex functions ψ
on [0, 1] with ψ(0) = ψ(1) = 1 satisfying
max{1 − t, t} ≤ ψ(t) ≤ 1 (t ∈ [0, 1]).
Let k · k ∈ AN2 and let
ψ(t) = k(1 − t, t)k (t ∈ (0, 1)). (7)
Then ψ ∈ Ψ2 . Conversely, for given ψ ∈ Ψ2 define


 (|z | + |z |)ψ |z2 |
1 2 |z1 |+|z2 |
((z1 , z2 ) 6= (0, 0))
k(z1 , z2 )kψ =
0 ((z1 , z2 ) = (0, 0)).
Then k · kψ ∈ AN2 , and k · kψ satisfies (7). In fact, AN2 and Ψ2 are in one-to-one
correspondence under (7).
In [35], Saito et al. characterized absolute normalized norms on Cn by means of
the corresponding convex function as follows. For each n ∈ N with n ≥ 2, put
( n
)
X
∆n = (t1 , . . . , tn ) ∈ Rn : t1 , . . . , tn ≥ 0, tj = 1 ,
j=1
and let Ψn denote the set of all continuous convex functions ψ on ∆n satisfying the
following conditions:
ψ(1, 0, . . . , 0) = ψ(0, 1, 0, . . . , 0) = · · · = ψ(0, . . . , 0, 1) = 1 (A0 )
10

t2 tn
ψ(t1 , . . . , tn ) ≥ (1 − t1 )ψ 0, ,..., (A1 )
1 − t1 1 − t1

t1 t3 tn
ψ(t1 , . . . , tn ) ≥ (1 − t2 )ψ , 0, ,..., (A2 )
1 − t2 1 − t2 1 − t2
..
.

t1 tn−1
ψ(t1 , . . . , tn ) ≥ (1 − tn )ψ ,..., ,0 . (An )
1 − tn 1 − tn
Then ANn and Ψn are in one-to-one correspondence as follows.
Theorem 1.2.3 (cf. [35, Theorems 3.4]).

(i) Let k · k ∈ ANn and let
ψ(t1 , · · · , tn ) = k(t1 , . . . , tn )k ((t1 , . . . , tn ) ∈ ∆n ). (8)
Then ψ ∈ Ψn . Conversely,
(ii) For given ψ ∈ Ψn define
k(z1 , . . . , zn )kψ

 (|z | + · · · + |z |)ψ |z1 | |zn |
1 n |z1 |+···+|zn |
, . . . , |z1 |+···+|zn |
((z1 , . . . , zn ) 6= (0, . . . , 0))
=
0 ((z1 , . . . , zn ) = (0, . . . , 0)).
(9)
Then k · kψ ∈ ANn , and k · kψ satisfies (8).
To show this theorem, we prepare some lemmas. We remark that the case of an
`p -norm on Cn , by the Hölder inequality, ψ = ψp ∈ Ψn is
( 1
(t1p + · · · + tnp ) p (1 ≤ p < ∞)
ψp (t1 , . . . , tn ) =
max{t1 , . . . , tn } (p = ∞).
Lemma 1.2.4 (cf. [35, Lemma 3.2]). Let ψ ∈ Ψn . Then

1
≤ ψ∞ (t1 , · · · , tn ) ≤ ψ(t1 , · · · , tn ) ≤ 1
n
for all (t1 , . . . , tn ) ∈ ∆n .
11
Proof. Let ψ ∈ Ψn . By (A0 ) and the convexity of ψ, it is clear that
ψ(t1 , t2 , · · · , tn ) ≤ 1. Next, from (A0 ), (A2 ), . . . , (An ), we have

t1 t3 tn
ψ(t1 , . . . , tn ) ≥ (1 − t2 )ψ , 0, ,...,
1 − t2 1 − t2 1 − t2
t1 t4 tn
!
t3 1−t2 1−t2 1−t2
≥ (1 − t2 ) 1 − ψ t3 , 0, 0, t3 , . . . , t3
1 − t2 1 − 1−t 2
1 − 1−t 2
1 − 1−t
2
t1 t4 tn
= (1 − t2 − t3 )ψ , 0, 0, ,...,
1 − t2 − t3 1 − t2 − t3 1 − t2 − t3
≥ ···

t1
≥ (1 − t2 − t3 − · · · − tn )ψ , 0, . . . , 0
1 − t2 − t3 − · · · − tn−1
= t1 ψ(1, 0, . . . , 0)
= t1 .
Next, from (A0 ), (A1 ), (A3 ), . . . , (An ), we have

t2 t3 tn
ψ(t1 , . . . , tn ) ≥ (1 − t1 )ψ 0, , ,...,
1 − t1 1 − t1 1 − t1
t2 t4 tn
!
t3
≥ (1 − t1 ) 1 − ψ 0, 1−t1t3 , 0, 1−t1t3 , . . . , 1−t1t3
1 − t1 1 − 1−t1 1 − 1−t1 1 − 1−t1

t2 t4 tn
= (1 − t1 − t3 )ψ 0, , 0, ,...,
1 − t1 − t3 1 − t1 − t3 1 − t1 − t3
≥ ···

t2
≥ (1 − t1 − t3 − · · · − tn )ψ 0, , 0, . . . , 0
1 − t1 − t3 − · · · − tn
= t2 ψ(0, 1, 0, . . . , 0)
= t2 .
Similarly, we can show that for 3 ≤ i ≤ n − 1,
ψ(t1 , . . . , tn ) ≥ ti .
12
Finally, from (A0 ), (A1 ), . . . , (An−1 ), we have

t2 tn
ψ(t1 , . . . , tn ) ≥ (1 − t1 )ψ 0, ,...,
1 − t1 1 − t1
t3 tn
!
t2 1−t1 1−t1
≥ (1 − t1 ) 1 − ψ 0, 0, t2 , . . . , t2
1 − t1 1 − 1−t 1
1 − 1−t
1
t3 tn
= (1 − t1 − t2 )ψ 0, 0, ,...,
1 − t1 − t2 1 − t1 − t2
≥ ···

tn
≥ (1 − t1 − · · · − tn−1 )ψ 0, . . . , 0,
1 − t1 − · · · − tn−1
= tn ψ(0, . . . , 1)
= tn .
Then for any 1 ≤ i ≤ n, we have
ψ(t1 , . . . , tn ) ≥ ti .
Thus we have
1
ψ(t1 , . . . , tn ) ≥ max{t1 , · · · , tn } ≥ ,
n
so,
1
ψ(t1 , . . . , tn ) ≥ ψ∞ (t1 , . . . , tn ) ≥ .
n
Lemma 1.2.5 (cf. [35, Theorem 3.5]). For any (p1 , . . . , pn ), (a1 , . . . , an ) ∈ Cn such
that 0 ≤ pi ≤ ai (i = 1, . . . , n), we have that
k(p1 , . . . , pn )kψ ≤ k(a1 , . . . , an )kψ . (10)
Proof. We first show that, if 0 ≤ p1 < a1 , then
k(p1 , p2 , . . . , pn )kψ ≤ k(a1 , p2 . . . , an )kψ . (11)
This is, we show that if 0 ≤ p1 < a1

p1 pn
(p1 +p2 + · · · + pn )ψ ,...,
p1 + p2 + · · · + pn p1 + p2 + · · · + pn
≤ (a1 + p2 + · · · + pn )

a1 p2 pn
×ψ , ..., .
a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n
13
Take any (t1 , . . . , tn ) ∈ ∆n , and consider the line segment

t2 tn
(1, 0, . . . , 0), 0, ,...,
1 − t1 1 − t1
in ∆n . For any real number λ such that 1 < λ ≤ 1/(1 − t1 ), we put
(t01 ,t02 , . . . , t0n )

= (1, 0, . . . , 0) + λ {(t1 , t2 , . . . , tn ) − (1, 0, . . . , 0)}

= 1 + λ(t1 − 1), λt2 , . . . , λtn

(1 − t1 )(λ − 1) t2 tn 1 − λ(1 − t1 )
= 0, ,..., + (t1 , . . . , tn ) .
t1 1 − t1 1 − t1 t1
By the convexity of ψ,
ψ(t01 , t02 , . . . , t0n )

(1 − t1 )(λ − 1) t2 tn 1 − λ(1 − t1 )
≤ ψ 0, ,..., + ψ(t1 , . . . , tn ).
t1 1 − t1 1 − t1 t1
Then we have from (A1 )
ψ(t1 , . . . , tn ) ψ(t01 , . . . , t0n )
−
1 − t1 1 − t01
ψ(t1 , . . . , tn ) 1
≥ −
1 − t1 1 − t01

(1 − t1 )(λ − 1) t2 tn 1 − λ(1 − t1 )
× ψ 0, ,..., + ψ(t1 , . . . , tn )
t1 1 − t1 1 − t1 t1
ψ(t1 , . . . , tn ) 1
= −
1 − t1 λ(1 − t1 )

(1 − t1 )(λ − 1) t2 tn 1 − λ(1 − t1 )
× ψ 0, ,..., + ψ(t1 , . . . , tn )
t1 1 − t1 1 − t1 t1

λ−1 t2 tn
= ψ(t1 , . . . , tn ) − (1 − t1 )ψ 0, ,··· , ≥ 0.
λt1 (1 − t1 ) 1 − t1 1 − t1
Thus, we have
ψ(t1 , t2 , . . . , tn ) ψ(t01 , t02 , . . . , t0n )
≥ . (12)
1 − t1 1 − t01
Since 0 ≤ p1 < a1 , we put

a1 p2 pn
(t1 , . . . , tn ) = , ,...,
a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n

0 0 p1 p2 pn
(t1 , . . . , tn ) = , ,...,
p1 + p2 + · · · + pn p1 + p2 + · · · + pn p1 + p2 + · · · + pn
a1 + p 2 + · · · + p n
λ= >1
p1 + p2 + · · · + pn
14
respectively in (12). Thus, we have

p2 pn
ψ a1 +p2a+···+p
1
,
n a1 +p2 +···+pn
, . . . , a1 +p2 +···+pn
p1
a1 +p2 +···+pn

p1 p2
ψ ,
p1 +p2 +···+pn p1 +p2 +···+pn
, . . . , p1 +p2p+···+p
n
n
≥ p1 .
p1 +p2 +···+pn
So,
(a1 +p2 + · · · + pn )

p1 p2 pn
×ψ , ,...,
a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n
≥ (p1 + p2 + · · · + pn )

p1 p2 pn
×ψ , ,..., .
p1 + p2 + · · · + pn p1 + p2 + · · · + pn p1 + p2 + · · · + pn
This implies (11). Similarly, we can show that for 2 ≤ i ≤ n,
k(a1 , . . . , ai−1 , pi , pi+1 , . . . , pn )kψ ≤ k(a1 , . . . , ai−1 , ai , pi+1 , . . . , pn )kψ .
Therefor, we have (10). This completes the proof.
Proof of Theorem 1.2.3.

(i) A property of the continuation is clear. We check a property of the convexity.
For any (t1 , . . . , tn ), (t01 , . . . , t0n ) ∈ ∆n and any ν ∈ [0, 1], we have
ψ (1 − ν)(t1 , . . . , tn ) + ν(t01 , . . . , t0n )

= ψ (1 − ν)t1 + νt01 , . . . , (1 − ν)tn + νt0n

= (1 − ν)t1 + νt01 , . . . , (1 − ν)tn + νt0n

= k(1 − ν)(t1 , . . . , tn ) + ν(t01 , . . . , t0n )k

≤ (1 − ν)k(t1 , . . . , tn )k + νk(t01 , . . . .t0n )k
= (1 − ν)ψ(t1 , . . . , tn ) + νψ(t01 , . . . , t0n ).
Then ψ is a convex function on ∆n . Next, we check properties of (A0 ), (A1 ), . . . ,
15
(An ). (A0 ) is clear. By (B1 ), we have
ψ(t1 , . . . , tn )
= k(t1 , t2 , . . . , tn )k
≥ k(0, t2 , . . . , tn )k

t2 tn
= (t2 + · · · + tn ) 0, ,...,
t2 + · · · + tn t2 + · · · + tn

t2 tn
= (1 − t1 ) 0,
,...,
1 − t1 1 − t1

t2 tn
= (1 − t1 )ψ 0, ,..., .
1 − t1 1 − t1
Thus we have (A1 ). We similarly have (A2 ), . . . , (An ).

(ii) By Lemma 1.2.4, all properties of an absolute normalized norm are clear except
the triangle inequality
k(z1 , . . . , zn ) + (w1 , . . . , wn )kψ ≤ k(z1 , . . . , zn )kψ + k(w1 , . . . , wn )kψ (13)
for any (z1 , . . . , zn ), (w1 , . . . , wn ) ∈ Cn . From Lemma 1.2.5 and the convexity of ψ,
for any (z1 , . . . , zn ), (w1 , . . . , wn ) ∈ Cn , we have
k(z1 , . . . , zn ) + (w1 , . . . , wn )kψ

= k(z1 + w1 , . . . , zn + wn )kψ
= k(|z1 + w1 |, . . . , |zn + wn |)kψ
≤ k(|z1 | + |w1 |, . . . , |zn | + |wn |)kψ
= (|z1 | + |w1 | + · · · + |zn | + |wn |)

|z1 | + |w1 | |zn | + |wn |
×ψ ,...,
|z1 | + |w1 | + · · · + |zn | + |wn | |z1 | + |w1 | + · · · + |zn | + |wn |
= (|z1 | + · · · + |zn | + |w1 | + · · · + |wn |)

|z1 | |w1 |
×ψ + ,...
|z1 | + · · · + |wn | |z1 | + · · · + |wn |

|zn | |wn |
..., +
|z1 | + · · · + |wn | |z1 | + · · · + |wn |
16
= (|z1 | + · · · + |zn | + |w1 | + · · · + |wn |)

|z1 | |zn |
×ψ ,...,
|z1 | + · · · + |wn | |z1 | + · · · + |wn |

|w1 | |wn |
+ ,...,
|z1 | + · · · + |wn | |z1 | + · · · + |wn |
= (|z1 | + · · · + |zn | + |w1 | + · · · + |xn |)

|z1 | + · · · + |zn |
×ψ
|z1 | + · · · + |zn | + |w1 | + · · · + |wn |

|z1 | |zn |
,...,
|zn | + · · · + |zn | |z1 | + · · · + |zn |
|w1 | + · · · + |wn |
+
|z1 | + · · · + |zn | + |w1 | + · · · + |wn |

|w1 | |wn |
,...,
|w1 | + · · · + |wn | |w | + · · · + |wn |
1
|z1 | |zn |
≤ (|z1 | + · · · + |zn |)ψ ,...,
|z1 | + · · · + |zn | |z1 | + · · · + |zn |

|w1 | |wn |
+ (|w1 | + · · · + |wn |)ψ ,...,
|w1 | + · · · + |wn | |w1 | + · · · + |wn |
= k(z1 , . . . , zn )kψ + k(w1 , . . . , wn )kψ .
The inequality (13) is called the generalized Minkowski inequality.
1.2.2 The dual space of Cn with an absolute normalized norm
In this subsection, we consider the dual space of Cn with absolute normalized norm
(cf. [28]). Let ψ ∈ Ψn . For any (z1 , z2 , . . . , zn ) ∈ Cn , the dual norm k · k∗ψ of k · kψ is
defined by
( n )
X
k(z1 , z2 , . . . , zn )k∗ψ = sup zj wj : k(w1 , w2 , . . . , wn )kψ = 1 . (14)

j=1
Following properties hold.
Theorem 1.2.6 (cf. [28, Section 3]). For k · kψ ∈ ANn ,

(i) k · k∗ψ ∈ ANn .
17
(ii) A corresponding convex function of k · k∗ψ in Ψn is given by
Pn
j=1 tj sj
ψ ∗ (s1 , . . . , sn ) = sup
(t1 ,...,tn )∈∆n ψ(t1 , . . . , tn )
for (s1 , . . . , sn ) ∈ ∆n .
(iii) k · k∗ψ = k · kψ∗ .
Proof.
(i) All properties of an absolute normalized norm are clear except the normalization.
k(1, 0, . . . , 0)k∗ψ

= sup |y1 | : k(y1 , . . . , yn )kψ = 1

= sup y1 : k(y1 , . . . , yn )kψ = 1, y1 , . . . , yn ≥ 0, (y1 , . . . , yn ) 6= (0, . . . , 0)

s1
= sup : (s1 , . . . , sn ) ∈ ∆n
ψ(s1 , . . . , sn )
1 − 0 − ··· − 0
=
ψ(0, . . . , 0)
= 1,
where
y1 yn
s1 = , . . . , sn = .
y1 + · · · + yn y1 + · · · + yn
Similarly, we have the others.
(ii) For all (s1 , . . . , sn ) ∈ ∆n .
ψ ∗ (s1 , . . . , sn )
= k(s1 , . . . , sn )k∗ψ

= sup |s1 y1 + · · · + sn yn | : k(y1 , . . . , yn )kψ = 1

= sup |s1 y1 + · · · + sn yn | : k(y1 , . . . , yn )kψ = 1,

y1 , . . . , yn ≥ 0, (y1 , . . . , yn ) 6= (0, . . . , 0)
= ψ ∗ (s1 , . . . , sn )
Pn
j=1 tj sj
= sup ,
(t1 ,...,tn )∈∆n ψ(t1 , . . . , tn )
where
y1 yn
t1 = , . . . , tn = .
y1 + · · · + yn y1 + · · · + yn
18
(iii) If (x1 , . . . , xn ) 6= (0, . . . , 0), then
k(x1 , . . . , xn )kψ∗

∗ |x1 | |xn |
= (|x1 | + · · · + |xn |)ψ ,...,
|x1 | + · · · + |xn | |x1 | + · · · + |xn |
|x1 | |xn |
t1 |x1 |+···+|x n|
+ · · · + tn |x1 |+···+|x n|
= (|x1 | + · · · + |xn |) sup
(t1 ,...,tn )∈∆n ψ(t1 , . . . , tn )
t1 |x1 | + · · · + tn |xn |
= sup
(t1 ,...,tn )∈∆n ψ(t1 , . . . , tn )

= sup |x1 |y1 + · · · + |xn |yn : k(y1 , . . . , yn )kψ = 1,

y1 , . . . , yn ≥ 0, (y1 , . . . , yn ) 6= (0, . . . , 0)

= sup |x1 y1 + · · · + xn yn | : k(y1 , . . . , yn )kψ = 1
= k(x1 , . . . , xn )k∗ψ ,
where
t1 tn
y1 = , . . . , yn = .
ψ(t1 , . . . , tn ) ψ(t1 , . . . , tn )
If (x1 , . . . , xn ) = (0, . . . , 0), it is clear. This completes the proof.
The case of an `p -norm on Cn , its dual norm k · k∗p = k · k∗ψp is

( 1
(|z1 |q + · · · + |zn |q ) q (1 < p < ∞)
k(z1 , . . . , zn )k∗p = (15)
max{|z1 |, . . . , |zn |} (p = 1),
where 1/p + 1/q = 1. Moreover, we have the generalized Hölder inequality.
Proposition 1.2.7 (cf. [7, Exercise IV.1.14]). Let ψ ∈ Ψn . Then we have

n
X
|zj wj | ≤ k(z1 , . . . , zn )k∗ψ k(w1 , . . . , wn )kψ (16)
j=1
for any (z1 , . . . , zn ), (w1 , . . . , wn ) ∈ Cn .
Proof. For (14), we have

Xn
zj yj ≤ k(z1 , . . . , zn )k∗ψ

j=1
19
for any (y1 , . . . , yn ) ∈ Cn such that k(y1 , . . . , yn )kψ = 1. Put yi = wi /k(w1 , . . . , wn )kψ
for all i = 1, . . . , n, where (w1 , . . . , wn ) 6= (0, . . . , 0) ∈ Cn . Then we have
n
X
zj wj ≤ k(z1 , . . . , zn )k∗ψ k(w1 , . . . , wn )kψ .

j=1
Put zi , wi for |zi |, |wi | respectively, we have (16). This complete the proof.
1.2.3 The set Ψ

e n of all continuous concave functions on ∆n
In this subsection, we define the set Ψ

e n of all continuous concave functions on ∆n
satisfying
ψ̃(1, 0, . . . , 0) = ψ̃(0, 1, 0, . . . , 0) = · · · = ψ̃(0, . . . , 0, 1) = 1
for ψ̃ ∈ Ψ
e n (cf. [32]). Just as ψ ∈ Ψn (9), for any ψ̃ ∈ Ψ
e n , we define a mapping k · k
ψ̃
on Cn as
k(z1 , . . . , zn )kψ̃

 (|z | + · · · + |z |)ψ̃ |z1 | |zn |
1 n |z1 |+···+|zn |
, . . . , |z1 |+···+|zn |
((z1 , . . . , zn ) 6= (0, . . . , 0))
=
0 ((z1 , . . . , zn ) = (0, . . . , 0)).
(17)
This mapping is not a norm, however we have the generalized inverse Minkowski
inequality.
Proposition 1.2.8 (cf. [32, Proposition 6]). Let ψ̃ ∈ Ψ

e n . Then we have
k(|z1 |, . . . , |zn |) + (|w1 |, . . . , |wn |)kψ̃ ≥ k(|z1 |, . . . , |zn |)kψ̃ + k(|w1 |, . . . , |wn |)kψ̃ (18)
for any (z1 , . . . , zn ), (w1 , . . . , wn ) ∈ Cn .
Proof. From the concavity of ψ̃, for any (z1 , . . . , zn ), (z1 , . . . , yn ) ∈ Cn , we have
k(|z1 |, . . . , |zn |) + (|w1 |, . . . , |wn |)kψ̃

= k(|z1 | + |w1 |, . . . , |zn | + |wn |)kψ̃
= (|z1 | + |w1 | + · · · + |zn | + |wn |)

|z1 | + |w1 | |zn | + |wn |
× ψ̃ ,...,
|z1 | + |w1 | + · · · + |zn | + |wn | |z1 | + |w1 | + · · · + |zn | + |wn |
20
= (|z1 | + · · · + |zn | + |w1 | + · · · + |wn |)

|z1 | |w1 |
× ψ̃ + ,...
|z1 | + · · · + |wn | |z1 | + · · · + |wn |

|zn | |wn |
..., +
|z1 | + · · · + |wn | |z1 | + · · · + |wn |
= (|z1 | + · · · + |zn | + |w1 | + · · · + |wn |)

|z1 | |zn |
× ψ̃ ,...,
|z1 | + · · · + |wn | |z1 | + · · · + |wn |

|w1 | |wn |
+ ,...,
|z1 | + · · · + |wn | |z1 | + · · · + |wn |
= (|z1 | + · · · + |zn | + |w1 | + · · · + |xn |)

|z1 | + · · · + |zn |
× ψ̃
|z1 | + · · · + |zn | + |w1 | + · · · + |wn |

|z1 | |zn |
,...,
|zn | + · · · + |zn | |z1 | + · · · + |zn |
|w1 | + · · · + |wn |
+
|z1 | + · · · + |zn | + |w1 | + · · · + |wn |

|w1 | |wn |
,...,
|w1 | + · · · + |wn | |w | + · · · + |wn |
1
|z1 | |zn |
≥ (|z1 | + · · · + |zn |)ψ̃ ,...,
|z1 | + · · · + |zn | |z1 | + · · · + |zn |

|w1 | |wn |
+ (|w1 | + · · · + |wn |)ψ̃ ,...,
|w1 | + · · · + |wn | |w1 | + · · · + |wn |
= k(|z1 |, . . . , |zn |)kψ̃ + k(|w1 |, . . . , |wn |)kψ̃ .
Especially, for all p ∈ R with 0 < p ≤ 1,

1
ψ˜p (t1 , . . . , tn ) = (tp1 + · · · + tpn ) p
e n by the inverse Hölder inequality, and k(z1 , . . . , zn )k = (|z1 |p +
is an element of Ψ ψ̃p
1
· · · + |zn |p ) p .
This mapping k · kψ̃p is monotone since the following proposition holds.
Proposition 1.2.9. For any (p1 , . . . , pn ), (a1 , . . . , an ) ∈ Cn such that 0 ≤ pi ≤

ai (i = 1, . . . , n), we have that
k(p1 , . . . , pn )kψ̃ ≤ k(a1 , . . . , an )kψ̃ . (19)
21
Proof. We first show that, if 0 ≤ p1 < a1 , then
k(p1 , p2 , . . . , pn )kψ̃ ≤ k(a1 , p2 . . . , pn )kψ̃ . (20)
This is, we show that if 0 ≤ p1 < a1

p1 pn
(p1 + p2 + · · · + pn )ψ̃ ,...,
p1 + p2 + · · · + pn p 1 + p2 + · · · + pn

a1 pn
≤ (a1 + p2 + · · · + pn )ψ̃ ,..., .
a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n
Take any (s1 , . . . , sn ) ∈ ∆n such that s1 + · · · + sn = 1, and consider the line segment

s2 sn
(1, 0, . . . , 0), 0, ,...,
1 − s1 1 − s1
in ∆n . For any real number λ such that 1 < λ ≤ 1/(1 − s1 ), we put
(s01 , s02 , . . . , s0n ) = (1, 0, . . . , 0) + λ{(s1 , s2 , . . . , sn ) − (1, 0, . . . , 0)}.
Then we have

1 0 0 0 1
(s1 , s2 , . . . , sn ) = (s1 , s2 , . . . , sn ) + 1 − (1, 0, . . . , 0).
λ λ
By the concavity of ψ̃,

1 0 0 0 1
ψ̃(s1 , s2 , . . . , sn ) ≥ ψ̃(s1 , s2 , . . . , sn ) + 1 − ψ̃(1, 0, . . . , 0)
λ λ
1
≥ ψ̃(s01 , s02 , . . . , s0n )
λ
1 − s1
= ψ̃(s01 , s02 , . . . , s0n ).
1 − s01
Thus, we have
ψ̃(s1 , s2 , . . . , sn ) ψ̃(s01 , s02 , . . . , s0n )
≥ . (21)
1 − s1 1 − s01
Since 0 ≤ p1 < a1 , we put
(s1 , s2 , . . . , sn )

a1 p2 pn
= , ,..., ,
a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n a1 + p 2 + · · · + p n
(s01 , s02 , . . . , s0n )

p1 p2 pn
= , ,..., ,
p1 + p2 + · · · + p n p1 + p 2 + · · · + pn p1 + p2 + · · · + pn
a1 + p 2 + · · · + p n
λ= >1
p1 + p2 + · · · + pn
22
respectively in (21). Thus, we have

p2 pn
ψ̃ a1 +p2a+···+p
1
,
n a1 +p2 +···+pn
, . . . , a1 +p2 +···+pn
a1
1− a1 +p2 +···+pn

p1 p2
ψ̃ ,
p1 +p2 +···+pn p1 +p2 +···+pn
, . . . , p1 +p2p+···+p
n
n
≥ p1 .
1 − p1 +p2 +···+pn
This implies (20). Similarly, we can show that for 2 ≤ i ≤ n,
k(a1 , . . . , ai−1 , pi , pi+1 , . . . , pn )kψ̃ ≤ k(a1 , . . . , ai−1 , ai , pi+1 , . . . , pn )kψ̃ .
Therefor, we have (19). This completes the proof.
Let ψ̃ ∈ Ψ
e n . Denote
Pn
j=1 tj sj
ψ̃∗ (s1 , . . . , sn ) = inf . (22)
(t1 ,...,tn )∈∆n ψ̃(t1 , . . . , tn )
The corresponding map k · kψ̃∗ is defined by (17). Then the following generalized
inverse Hölder inequality holds.
Proposition 1.2.10 (cf. [32, Proposition 9]). Let ψ̃ ∈ Ψ

e n . Then we have
n
X
|zj wj | ≥ k(z1 , . . . , zn )kψ̃∗ k(w1 , . . . , wn )kψ̃ (23)
j=1
for any (z1 , . . . , zn ), (w1 , . . . , wn ) ∈ Cn .
Proof. By (22), we get that

n
X
sj tj ≥ ψ̃∗ (s1 , . . . , sn ) ψ̃(t1 , . . . , tn ).
j=1
Put
|zj | |wj |
sj = , tj =
|z1 | + · · · + |zn | |w1 | + · · · + |wn |
for j = 1, . . . , n, where (z1 , . . . , zn ), (w1 , . . . , wn ) ∈ Cn , then we have (23). This
completes the proof.
23
1.3 Characterizations of generalized triangle inequalities
In this section, by using ψ-dirct sum of Banach space, we show generalizations of a
following obvious fact: let X be a normed space and a1 , a2 ∈ C, then
ka1 x1 + a2 x2 k ≤ kx1 k + kx2 k (x1 , x2 ∈ X)
if and only if
max{|a1 |, |a2 |} ≤ 1.
1.3.1 Generalizations of preceding results
Let (X, k · k) be a normed space. For any n ∈ N with n ≥ 2 and ψ ∈ Ψn , we can

consider a product space X n with the norm
k(x1 , x2 , . . . , xn )kψ = k(kx1 k, kx2 k, . . . , kxn k)kψ (x1 , . . . , xn ∈ X). (24)
We call (X n , k · kψ ) ψ-direct sum of X (cf. [18]), and denote it by `nψ (X). We first
prove the following result.
Theorem 1.3.1. Let X be a normed space and ψ ∈ Ψn . Let (a1 , . . . , an ) ∈ Cn .

Then n
X
aj xj ≤ k(x1 , . . . , xn )kψ (x1 , . . . , xn ∈ X) (25)

j=1
if and only if
k(a1 , . . . , an )kψ∗ ≤ 1.
Proof. Assume that k(a1 , . . . , an )kψ∗ ≤ 1. Let x1 , . . . , xn ∈ X. From the general-

ized Hölder inequality (16), we have

Xn X n
aj x j ≤ |aj |kxj k

j=1 j=1
≤ k(|a1 |, . . . , |an |)kψ∗ k(kx1 k, . . . , kxn k)kψ

= k(a1 , . . . , an )kψ∗ k(x1 , . . . , xn )kψ
≤ k(x1 , . . . , xn )kψ .
Hence (a1 , . . . , an ) satisfies (25).
24
Conversely, assume that (a1 , . . . , an ) satisfies (25). Take any e ∈ X with kek = 1,
and any (α1 , . . . , αn ) ∈ Cn such that k(α1 , . . . , αn )kψ = 1, we have
n n
X X
aj α j = aj αj e

j=1 j=1
≤ k(α1 e, . . . , αn e)kψ
= k(kα1 ek, . . . , kαn ek)kψ
= k(|α1 |, . . . , |αn |)kψ
= k(α1 , . . . , αn )kψ
= 1.
Thus ( n )
X
sup aj αj : k(α1 , . . . , αn )kψ = 1 ≤ 1,

j=1
and so we have k(a1 , . . . , an )kψ∗ ≤ 1. This completes the proof.
In the above theorem, (a1 , . . . , an ) is an element in the unit ball of (`nψ )∗ = `nψ∗ ,
where (`nψ )∗ is a dual space of `nψ .
From Theorem 1.3.1, we have the following corollary by putting ψ = ψp and
using (15).
Corollary 1.3.2. Let X be a normed space and p ∈ R with p > 1. Let (a1 , . . . , an ) ∈
Cn . Then n p
X
aj xj ≤ kx1 kp + · · · + kxn kp (x1 , . . . , xn ∈ X)

j=1
if and only if
1
(|a1 |q + · · · + |an |q ) q ≤ 1,
where 1/p + 1/q = 1.
e n , we define k · k as in (24), and consider (X n , k · k ) which is denoted
For ψ̃ ∈ Ψ ψ̃ ψ̃
by `nψ̃ (X). Note that `nψ̃ (X) is not a normed space. In this case, we have the following
result.
e n . Let (a1 , . . . , an ) ∈ Cn .
Theorem 1.3.3. Let X be a normed space and ψ̃ ∈ Ψ
Then n
X
aj xj ≤ k(x1 , . . . , xn )kψ̃ (x1 , . . . , xn ∈ X) (26)

j=1
25
if and only if
max{|a1 |, . . . , |an |} ≤ 1.
Proof. Assume that max{|a1 |, . . . , |an |} ≤ 1. Let x1 , . . . , xn ∈ X. From the gener-
alized inverse Minkowski inequality (18), we have

Xn X n
aj x j ≤ |aj |kxj k

j=1 j=1
Xn
≤ kxj k
j=1
= k(kx1 k, 0, . . . , 0)kψ̃ + · · · + k(0, . . . , 0, kxn k)kψ̃

≤ k(kx1 k, 0, . . . , 0) + · · · + (0, . . . , 0, kxn k)kψ̃
= k(kx1 k, . . . , kxn k)kψ̃
= k(x1 , . . . , xn )kψ̃ .
Hence (26) holds.

Conversely, assume that (a1 , . . . , an ) satisfies (26). Take any e ∈ X with kek = 1.
(i)
From (26), |ai | = kai ek ≤ k(0, . . . , 0, e , 0, · · · 0)kψ̃ = 1 for all i = 1, . . . , n. So we
have max{|a1 |, . . . , |an |} ≤ 1. This completes the proof.
From Theorem 1.3.3, we have the following corollary by putting ψ̃ = ψ̃p , where
0 < p ≤ 1.
Corollary 1.3.4. Let X be a normed space and p ∈ R with 0 < p ≤ 1. Let
(a1 , . . . , an ) ∈ Cn . Then
n p
X
aj xj ≤ kx1 kp + · · · + kxn kp (x1 , . . . , xn ∈ X)

j=1
if and only if
max{|a1 |, . . . , |an |} ≤ 1.
1.3.2 Applications to inequalities involved with the Euler-Lagrange type

identity
In this subsection, we give another proof of characterizations of (µ1 , . . . , µn ) ∈ Rn

which satisfies
kx1 kp kxn kp
kx1 + · · · + xn kp ≤ + ··· + (x1 , . . . , xn ∈ X) , (27)
µ1 µn
26
where p ∈ R with p > 0, proved by [9, Theorems 2.4 and 2.5]. We put
X p X
( n n
)
p
kx j k
F (p) = (µ1 , . . . , µn ) ∈ Rn : xj ≤ (x1 , . . . , xn ∈ X) ,

j=1

j=1
µj
and also for each k = 0, 1, . . . , n, we place in correspondence F (p ; k) as the subset

of F (p) consisting of all n-tuples (µ1 , . . . , µn ) ∈ Rn for which inequality (27) holds
and exactly k numbers of µ1 , . . . , µn are negative. We note that
n
[
F (p) = F (p ; k).
k=0
Theorem 1.3.5 (cf. [9, Theorems 2.4 and 2.5]). Let X be a normed space and p ∈ R
with p > 0. Then the following assertions hold:
(i) F (p
 ;(0) )
Xn 1
n
µjp−1 ≤ 1

 (µ1 , . . . , µn ) ∈ R : µ1 , . . . , µn > 0, (p > 1)



j=1
=




(0, 1] × · · · × (0, 1] (0 < p ≤ 1);

(ii) F (p ; k) = ∅ for all k = 1, . . . , n;
(iii) F (p) = F (p ; 0).
Proof. (i) If (µ1 , . . . , µn ) satisfies (27), then

p 1 1
p
xn kp
1 p p
p 1
p

µ1 x1 + · · · + µn xn ≤ kµ 1 x 1 k kµ n
+ · · · +
µ1 µn
= kx1 kp + · · · + kxn kp (x1 , . . . , xn ∈ X). (28)
In the case where p > 1, from Corollary 1.3.2, we have

1 q 1 q 1
p q
(28) ⇔ µ1 + · · · + µnp ≤1

n
X 1
⇔ µjp−1 ≤ 1,
j=1
where 1/p + 1/q = 1. In the case where 0 < p ≤ 1, from Corollary 1.3.4, we have
1 p 1 p
p
(28) ⇔ max µ1 , . . . , µnp ≤ 1
⇔ (µ1 , . . . , µn ) ∈ (0, 1] × · · · × (0, 1].
27
(ii) Suppose µi < 0. In (27), take any e ∈ X with kek = 1, and put xj = δij e
for all j = 1, . . . , n, where δij is a Kronecker’s delta. Hence 1 ≤ 1/µi . This is a
contradiction, and so F (p ; k) = ∅.
(iii) is clear. This completes the proof.
Similarly, we give another proof of characterizations of (µ1 , . . . , µn ) ∈ Rn which

holds a reverse inequality of (27):
kx1 kp kxn kp
kx1 + · · · + xn kp ≥ + ··· + (x1 , . . . , xn ∈ X) , (29)
µ1 µn
where p ∈ R with p > 0, proved by [9, Theorems 2.6 and 2.7]. We put
X p X
( n n
)
p
kx j k
G(p) = (µ1 , . . . , µn ) ∈ Rn : xj ≥ (x1 , . . . , xn ∈ X) ,

j=1

j=1
µ j
and also for each k = 0, 1, . . . , n, we place in correspondence G(p ; k) as the subset

of G(p) consisting of all n-tuples (µ1 , . . . , µn ) ∈ Rn for which inequality (29) holds
and exactly k numbers of µ1 , . . . , µn are negative. We note that
n
[
G(p) = G(p ; k).
k=0
Theorem 1.3.6 (cf. [9, Theorems 2.6 and 2.7]). Let X is a normed space and p ∈ R
with p > 0. Then the following assertions hold:
(i) G(p ; k) = ∅ for all k = 0, 1, . . . , n − 2;
 ;n − 1)
(ii) G(p



 


µi > 0, µj < 0 (j 6= i) 

  1 1

n
 n

(µ1 , . . . , µn ) ∈ R p−1
P
µi ≥ 1 + j=1, j6=i |µj | p−1 (p > 1)





 
 

for some i ∈ {1, . . . , n}

  

=  


 



µi > 0, µj < 0 (j 6= i) 



  
µi ≥ max (1, |µj |)

 (µ1 , . . . , µn ) ∈ Rn

 (0 < p ≤ 1);

  j∈{1,...,n}\{i} 
  
 for some i ∈ {1, . . . , n}
  


(iii) G(p ; n) = (−∞, 0) × · · · × (−∞, 0);
(iv) G(p) = G(p ; n − 1) ∪ G(p ; n).
28
Proof. (i) In (29), put xj = 0 for all j ∈ {1, . . . , n} such that µj < 0. Next, for all
i ∈ {1, . . . , n} such that µi > 0, select xi such that xi 6= 0 and x1 + · · · + xn = 0.
Hence the left-hand side of (29) vanishes, and the right-hand side of (29) is strictly
positive. This is a contradiction, and so G(p ; k) = ∅.
(ii) If (µ1 , . . . , µn ) satisfies (29), there exists i ∈ {1, . . . , n} such that µi > 0.
Thus (29) is equivalent to
n
X kxj kp kx1 + · · · + xn kp
µ + 1 ≥ kxi kp (x1 , . . . , xn ∈ X). (30)
j=1
− µji µi
j6=i
In (30), put yj = −xj for all j ∈ {1, . . . , n} \ {i}. Next, put yi = x1 + · · · + xn .

Hence it is equivalent to
n
X kyj kp kyi kp
µ + 1 ≥ ky1 + · · · + yn kp (y1 , . . . , yn ∈ X). (31)
j=1
− µji µi
j6=i
From Theorem 1.3.5 (i), in the case where p > 1, we have

n 1 1
X µj p−1 1 p−1
(31) ⇔ − µi
+ ≤1
j=1
µ i
j6=i
1
n
X 1
p−1
⇔ µi ≥1+ |µj | p−1 ,
j=1
j6=i
and in the case 0 < p ≤ 1, we have

1 µj
(31) ⇔ ∈ (0, 1] and − ∈ (0, 1] (j ∈ {1, . . . , n} \ {i})
µi µi
⇔ µi ≥ max (1, |µj |).
j∈{1,...,n}\{i}
(iii) and (iv) are clear. This completes the proof.
At the end of this subsection, we give another proof of characterizations of

(a1 , . . . , an , λ, µ1 , . . . , µn ) ∈ Cn × R × Rn which holds

≤ + ··· + (x1 , . . . , xn ∈ X) ,
λ µ1 µn
where p ∈ R with p > 0.
29
Theorem 1.3.7 (cf. [2, Theorems 2]). Let X be a normed space and p, q ∈ R with
p > 1, 1/p + 1/q = 1. Let (a1 , . . . , an , λ, µ1 , . . . , µn ) ∈ Cn × R × Rn . Then

≤ + ··· + (x1 , . . . , xn ∈ X) (32)
λ µ1 µn
holds if:
(i) λ > 0, µj > 0 for all j ∈ {1, . . . , n} and

n 1
1 X
λ p−1 ≥ µjp−1 |aj |q ,
j=1
(ii) λ < 0, µi < 0 for some i ∈ {1, . . . , n}, µj > 0 for all j ∈ {1, . . . , n} \ {i} and
n 1
1 1 X
q
(−µi ) p−1 |ai | ≥ (−λ) p−1 + µjp−1 |aj |q ,
j=1
j6=i
(iii) λ < 0, µj > 0 for all j ∈ {1, . . . , n} and any a1 , . . . , an ∈ C.
Proof. (i) Remark that

1 1 1
λ p−1 ≥ µ1p−1 |a1 |q + · · · + µnp−1 |an |q
1 q 1 q 1q
µ 1 + · · · + an µ n p
p
⇔ a1 ≤ 1. (33)
λ λ
From Corollary 1.3.2, we have
µ p1 − 1 p X

n n
− p1 p
X
j p
(33) ⇔ aj µ j xj ≤ µ xj (x1 , . . . , xn ∈ X),

λ j
j=1 j=1
hence we have (32).

(ii) The case where ai 6= 0, remark that
n 1
1 1 X
q
(−µi ) p−1 |ai | ≥ (−λ) p−1 + µjp−1 |aj |q
j=1
j6=i
q X n q
1 1 1 1
p−1
aj
⇔ (−µi ) p−1 ≥ (−λ) p−1 +
ai µ j
.
ai (34)
j=1
j6=i
30
From (i), we have
p
xi + Pnj=1 aj
1
x
ai j n
ai j6=i kxi kp X kxj kp
(34) ⇔ ≤ + (x1 , . . . , xn ∈ X). (35)
−µi −λ j=1
µj
j6=i
1
Pn aj
In (35), put yj = −xj for any j = 2, 3, . . . , n. Next, put y1 = x
a1 1
+ j=1 ai xj .
j6=i
Hence we have (32).
The case where ai = 0, on one hand there exist no (a1 , . . . , an , λ, µ1 , . . . , µn ) such
that n
1 X 1
0 ≥ (−λ) p−1 + µjp−1 |aj |q .
j=1
j6=i
On the other hand, in (32), put xj = 0 for j ∈ {1, . . . , n} \ {i}, then we have
0 ≤ kxi kp /µi (xi ∈ X), however there exist no µi > 0 such that this inequality
holds. That is not a contradiction.
(iii) is clear. This completes the proof.
By using Corollary 1.3.4, we have the following theorem:
Theorem 1.3.8. Let X be a normed space and p ∈ R with 0 < p ≤ 1. Let

(a1 , . . . , an , λ, µ1 , . . . , µn ) ∈ Cn × R × Rn . Then

≤ + ··· + (x1 , . . . , xn ∈ X)
λ µ1 µn
holds if:
(i) λ > 0, µj > 0 for all j ∈ {1, . . . , n} and
λ ≥ max{µ1 |a1 |p , . . . , µn |a1 |p },
(ii) λ < 0, µi < 0 for some i ∈ {1, . . . , n}, µj > 0 for all j ∈ {1, . . . , n} \ {i} and
−µi |ai |p ≥ max {µj |aj |p , −λ},

j∈{1,...,n}\{i}
(iii) λ < 0, µj > 0 for all j ∈ {1, . . . , n} and any a1 , . . . , an ∈ C.
31
By using these theorems, we have characterizations of (a1 , . . . , an , λ, µ1 , . . . , µn ) ∈
C × R × Rn which holds
n

≥ + ··· + (x1 , . . . , xn ∈ X) ,
λ µ1 µn
where p ∈ R with p > 0.
Theorem 1.3.9. Let X be a normed space and p, q ∈ R with p > 1, 1/p + 1/q = 1.
Let (a1 , . . . , an , λ, µ1 , . . . , µn ) ∈ Cn × R × Rn . Then
≥ + ··· + (x1 , . . . , xn ∈ X)
λ µ1 µn
holds if:
(i) λ < 0, µj < 0 for all j ∈ {1, . . . , n} and

n
1 X 1
(−λ) p−1 ≥ (−µj ) p−1 |aj |q ,
j=1
(ii) λ > 0, µi > 0 for some i ∈ {1, . . . , n}, µj < 0 for all j ∈ {1, . . . , n} \ {i} and
1
n
1 X 1
q
µip−1
|ai | ≥ λ p−1 + (−µj ) p−1 |aj |q ,
j=1
j6=i
(iii) λ > 0, µj < 0 for all j ∈ {1, . . . , n} and any a1 , . . . , an ∈ C.
Theorem 1.3.10. Let X be a normed space and p ∈ R with 0 < p ≤ 1. Let

(a1 , . . . , an , λ, µ1 , . . . , µn ) ∈ Cn × R × Rn . Then
≥ + ··· + (x1 , . . . , xn ∈ X)
λ µ1 µn
holds if:
(i) −λ < 0, µj < 0 for all j ∈ {1, . . . , n} and
λ ≥ max{−µ1 |a1 |p , . . . , −µn |a1 |p },
(ii) λ > 0, µi > 0 for some i ∈ {1, . . . , n}, µj < 0 for all j ∈ {1, . . . , n} \ {i} and
µi |ai |p ≥ max {−µj |aj |p , λ},

j∈{1,...,n}\{i}
(iii) λ > 0, µj < 0 for all j ∈ {1, . . . , n} and any a1 , . . . , an ∈ C.
32
1.3.3 An application to the parallelogram inequality
In this subsection, we consider another aspect of classical triangle inequality kx +

yk ≤ kxk + kyk. For a Hilbert space H, we recall the parallelogram law
kx + yk2 + kx − yk2 = 2 kxk2 + kyk2

(x, y ∈ H).
This implies that the parallelogram inequality
kx + yk2 ≤ 2 kxk2 + kyk2

(x, y ∈ H).
Saitoh in [38] noted the parallelogram inequality may be more suitable than the
classical triangle inequality. Motivated by this, Belbachir et al. introduced the notion
of q-norm: let X be a vector space over K(= R or C). For 1 ≤ q < ∞, a mapping
k · k from X into R+ (= {a ∈ R : a ≥ 0}) is called a q-norm if it satisfies the following
conditions:
(i) kxk = 0 ⇔ x = 0,
(ii) kαxk = |α|kxk (x ∈ X, α ∈ K),

(iii) kx + ykq ≤ 2q−1 kxkq + kykq (x, y ∈ X).
They proved that for all q with 1 ≤ q < ∞, every norm is a q-norm, and conversely,
every q-norm is a norm in the usual sense (cf. [6]).
Saito et al. in [36] generalized the notion of q-norm, that is, the notion of ψ-norm
by considering function ψ ∈ Ψ2 : let X be a vector space over K(= R or C) and
ψ ∈ Ψ2 . A mapping k · k from X into R+ (= {a ∈ R : a ≥ 0}) is called ψ-norm if it
satisfies the following conditions:
(i) kxk = 0 ⇔ x = 0,
(ii) kαxk = |α|kxk (x ∈ X, α ∈ K),
1

(iii) kx + yk ≤ min0≤t≤1 ψ(t)
(kxk, kyk)
ψ
(x, y ∈ X).
Note that the function ψ(t) = ψq (t) = {(1−t)q +tq }1/q take the minimum at t = 1/2
and q q 1q
1 1 1 1
ψq = + = 2 q −1 ,
2 2 2
33
then the condition (iii) of a ψ-norm implies
1 (kxk, kyk) = 21− 1q kxkq + kykq .

kx + yk ≤
ψq ( 12 ) ψq
Then we have kx + ykq ≤ 2q−1 (kxkq + kykq ) and so k · k becomes a q-norm. They
proved that for all ψ ∈ Ψ2 , every norm is a ψ-norm, and conversely, every ψ-norm
is a norm in the usual sense (cf. [36]).
By using Theorem 1.3.1, we can proof that the constant 1/min0≤t≤1 ψ(t) of the
condition (iii) in a ψ-norm is the best constant of the condition.
Theorem 1.3.11. Let X be a normed space and ψ ∈ Ψ2 . If there exist a positive

constant C satisfying

kx + yk ≤ C (kxk, kyk) ψ (x, y ∈ X). (36)
Then
1
C≥ .
min0≤t≤1 ψ(t)
Proof. Remark that

1 1
(36) ⇔
C x + y ≤ (kxk, kyk)
ψ
(x, y ∈ X). (37)
C
From Theorem 1.3.1, we have

1 1
(37) ⇔
, ≤ 1.
C C ψ∗
Then
∗1 1 1
C ≥ k(1, 1)kψ∗ = 2ψ = sup = .
2 0≤t≤1 ψ(t) min0≤t≤1 ψ(t)
1.4 A generalization of `p -spaces

In this section, our aim is to present generalizations of Theorem 1.3.1 and Theorem
1.3.3 for infinite sequences {xn }∞
n=1 ⊂ X by using generalized `p -spaces.
34
1.4.1 `ψ -spaces
In this subsection, we summarize basic results of `ψ -spaces which is a generalization

of `p -spaces by [25].
Let `0 denote the set of all infinite sequences of complex numbers with only
finitely many non-zero elements. A norm k · k on `0 is called absolute if k{zn }∞
n=1 k =
k{|zn |}∞ ∞
n=1 k for all {zn }n=1 ∈ `0 , and normalized if ken k = 1 for all n = 1, 2, . . ., where
(n)
en = (0, . . . , 0, 1 , 0, . . .) ∈ `0 . We remark that every absolute normalized norm is
monotone: if |zi | ≤ |wi | for every i = 1, 2, · · · , then k{zn }∞ ∞
n=1 k ≤ k{wn }n=1 k, where
{zn }∞ ∞
n=1 , {wn }n=1 ∈ `0 .
Let AN∞ is the family of all absolute normalized norms on `0 , and put
( ∞
)
X
∆∞ = t = {tn }∞ n=1 ∈ `0 : tn ≥ 0, tn = 1 .
n=1
For every k · k ∈ AN∞ , we define the function on ∆∞ such that
ψ(t) = ktk (t = {tn }∞

n=1 ∈ ∆∞ ), (38)
then ψ is a continuous convex function on ∆∞ satisfying the following conditions:
ψ(en ) = 1 (A00 )

t1 tn−1 tn+1
ψ(t) ≥ (1 − tn )ψ ,..., , 0, ,... (A0n )
1 − tn 1 − tn 1 − tn
(n)
for all n = 1, 2, . . . and every t = {tn }∞
n=1 ∈ ∆∞ with tn 6= 1, where en = (0, . . . , 0, 1
, 0, . . .) ∈ `0 .
Conversely, we define the set Ψ∞ of all continuous convex functions on ∆∞
satisfying the conditions (A0n ) for all n = 0, 1, 2, . . .. For any ψ ∈ Ψ∞ , we define the
mapping on `0 :
k{zn }∞
n=1 kψ

 P∞ |z | ψ P∞|z1 | , . . . , P∞|zn | , . . . ({zn }∞
j=1 j j=1 |zj | j=1 |zj |
n=1 6= 0)
=
0 ({zn }∞
n=1 = 0),
then k · kψ ∈ AN∞ and it satisfies (38).

In fact, AN∞ and Ψ∞ are in a one-to-one correspondence under the equation
(38).
35
Using this, we introduce the `ψ -spaces. Let `∞ be the Banach space of all
bounded infinite sequences of complex numbers. For {zn }∞
n=1 ∈ `∞ and ψ ∈ Ψ∞ , by
Proposition 1.2.2, {k(z1 , . . . , zn , 0, 0, . . .)kψ }∞
n=1 is an increasing sequence. Thus we
define the space `ψ by
n o
`ψ = {zn }∞
n=1 ∈ `∞ : lim k(z1 , . . . , z n , 0, 0, . . .)kψ < ∞ . (39)
n→∞
Proposition 1.4.1 (cf. [25, Proposition 2.4]). `ψ is a Banach space with the norm
k{zn }∞
n=1 kψ = lim k(z1 , . . . , zn , 0, 0, . . .)kψ .
n→∞
(k)
Proof. Let {yk }∞ ∞
k=1 ⊂ `ψ be any Cauchy sequence of `ψ . We put yk = {zn }n=1 for
every k ∈ N. By Proposition 1.2.2, for each n ∈ N, for any k, l ∈ N, we have
|zn(k) − zn(l) | = k(0, . . . , 0, zn(k) − zn(l) , 0, . . .)kψ

(k) (l)
≤ k(z1 − z1 , . . . , zn(k) − zn(l) , 0, . . .)kψ ,
(k)
then {zn }∞
k=1 ⊂ C is a Cauchy sequence in C. Thus there exists zn ∈ C such that
(k)
zn = limk→∞ zn for each n ∈ N. By Proposition 1.2.2, we have
(k)
k(z1 − z1 , . . . , zn(k) − zn , 0, 0, . . .)kψ
(k) (m)
= lim k(z1 − z1 , . . . , zn(k) − zn(m) , 0, 0, . . .)kψ
m→∞
(k) (m)
= lim inf k(z1 − z1 , . . . , zn(k) − zn(m) , 0, 0, . . .)kψ
m→∞
≤ lim inf kyk − ym kψ .

m→∞
As n → ∞, we have y = {zn }∞
n=1 ∈ `ψ and kyk − ykψ → 0. This completes the
proof.
Next, we consider the dual space of `ψ . Let ψ ∈ Ψ∞ . For any {zn }∞

n=1 ∈ `0 , the
dual norm of k · kψ is defined by following:
( ∞ )
X
k{zn }k∗ψ = sup zn wn : w = {wn }∞
n=1 ∈ `0 , kwkψ = 1 .

n=1
Then k · k∗ψ ∈ AN∞ and the corresponding convex function in Ψ∞ is given by

P∞
∗ n=1 sn tn
ψ (s) = sup (s = {sn }∞
n=1 ∈ ∆∞ ),
t∈∆∞ ψ(t)
36
and k · k∗ψ = k · kψ∗ . Then
n o
∞
`ψ = {wn }n=1 ∈ `∞ : lim k(w1 , . . . , wn , 0, 0, . . .)kψ < ∞
∗ ∗
n→∞
is also a Banach space with the norm
k{wn }∞
n=1 kψ ∗ = lim k(w1 , . . . , wn , 0, 0, . . .)kψ ∗ .
n→∞
Moreover we have the Generalized Hölder inequality:

∞
X
|zn wn | ≤ k{zn }∞ ∞
n=1 kψ k{wn }n=1 kψ ∗ (40)
n=1
for any {zn }∞ ∞

n=1 ∈ `ψ and any {wn }n=1 ∈ `ψ ∗ .
Now we note the `p -norm which is a good example of absolute normalized norms.
For any {xn }∞
n=1 ∈ `0 , it is
( P 1
( ∞ p p
n=1 |zn | ) (1 ≤ p < ∞)
k{zn }∞
n=1 kp =
max1≤n<∞ |zn | (p = ∞),
and also for every k · k ∈ AN∞ , we have k · k∞ ≤ k · k ≤ k · k1 . In this case,

ψ = ψp ∈ Ψ∞ is ( P 1
( ∞ p p
n=1 tn ) (1 ≤ p < ∞)
ψp (t) =
max1≤n<∞ tn (p = ∞)
for any t = {tn }∞ ∞
n=1 ∈ ∆∞ . For any {zn }n=1 ∈ `ψ , a norm k · kp = k · kψp is
( P 1
∞ ( ∞ p p
n=1 |zn | ) (1 ≤ p < ∞)
k{zn }n=1 kp =
sup1≤n<∞ |zn | (p = ∞),
and a dual norm k · k∗p = k · k∗ψp is

( P 1
∞ ∗ ( ∞ q q
n=1 |zn | ) (1 < p < ∞)
k{zn }n=1 kp = (41)
sup1≤n<∞ |zn | (p = 1),
where 1/p + 1/q = 1. Thus `ψ is a generalization of `p .

Let (X, k · k) be a Banach space. For any ψ ∈ Ψ∞ , we define the ψ-direct sums
of X to be the space
`ψ (X) = {xn }∞ ∞

n=1 ⊂ X : {kxn k}n=1 ∈ `ψ ,
where `ψ is (39). Then it is a Banach space with the norm k{xn }∞ ∞

n=1 kψ = k{kxn k}n=1 kψ
(cf. [41]). We first prove the following result.
37
Theorem 1.4.2. Let X be a Banach space, ψ ∈ Ψ∞ and {an }∞
n=1 ∈ `∞ . Then
following conditions are equivalent:
P∞
(i) for all {xn }∞
n=1 ∈ `ψ (X), n=1 an xn converges in X and satisfies

X ∞
an xn ≤ k{xn }∞
n=1 kψ ;

n=1
(ii) {an }∞
n=1 ∈ `ψ ∗ and satisfies
k{an }∞
n=1 kψ ∗ ≤ 1.
Proof. If {an }∞
n=1 satisfies (ii), we remark that k(a1 , . . . , an )kψ ∗ ≤ 1 for all n ∈ N.
From the Generalized Hölder inequality (40), we have
Xn
kaj xj k ≤ k(x1 , . . . , xn )kψ
j=1
for all x1 , . . . , xn ∈ X. Then we have (i).

Conversely, assume that {an }∞
n=1 ∈ `∞ satisfies (i). For all fixed n ∈ N, put
xn+1 = xn+2 = · · · = 0, then we have

Xn
aj xj ≤ k(x1 , . . . , xn )kψ .

j=1
From Theorem 1.3.1, we have k(a1 , . . . , an )kψ∗ ≤ 1 for all n ∈ N. Hence {an }∞
n=1
holds (ii).
In this theorem, k{an }∞ ∗

n=1 kψ ∗ ≤ 1 is an element in the unit ball of (`ψ (X)) =
`ψ∗ (X), where (`ψ (X))∗ is a dual space of `ψ (X).
From this theorem, we have a following corollary by putting ψ = ψp and using
(41).
Corollary 1.4.3. Let X be a Banach space, n ∈ N with n ≥ 2, p ∈ R with p > 1,
and {an }∞n=1 ∈ `∞ . Then following conditions are equivalent:
P∞
n=1 ∈ `p (X), n=1 an xn converges in X and satisfies
p
X ∞
an xn ≤ k{xn }∞ p
n=1 k ;

n=1
(ii) {an }∞
n=1 ∈ `q and satisfies
k{an }∞
n=1 kq ≤ 1,
where 1/p + 1/q = 1.
38
1.4.2 `ψ̃ -spaces
We define the set Ψ e ∞ of all continuous concave functions on ∆∞ satisfying the

following conditions: if ψ̃ ∈ Ψ
e ∞ , then ψ̃(en ) = 1 for all n = 1, 2, . . ., where en =
(n)
(0, . . . , 0, 1 , 0, . . .) ∈ `0 . For any ψ̃ ∈ Ψ
e ∞ , we define a mapping k · k on `0 such
ψ̃
that
k{zn }∞
n=1 kψ̃

 (P∞ |z |)ψ̃ P∞|z1 | , . . . , P∞|zn | , . . . ({zn }∞
j=1 j j=1 |zj | j=1 |zj |
n=1 6= 0)
=
0 ({zn }∞
n=1 = 0).
Using this, we introduce the `ψ̃ -spaces. Let `∞ is the Banach space of all bounded
infinite sequences of complex numbers. For any ψ̃ ∈ Ψ e ∞ , we define the space ` by
ψ̃
n o
∞
`ψ̃ = {zn }n=1 ∈ `∞ : lim k(z1 , . . . , zn , 0, 0, . . .)kψ̃ < ∞ .
n→∞
For any {zn }∞

n=1 ∈ `ψ̃ , we define the mapping
k{zn }∞
n=1 kψ̃ = lim k(z1 , . . . , zn , 0, 0, . . .)kψ̃ .
n→∞
This mapping is not a norm, however, we have the generalized inverse Minkowski
inequality:
k{|zn | + |wn |}∞ ∞ ∞
n=1 kψ̃ ≥ k{|zn |}n=1 kψ̃ + k{|wn |}n=1 kψ̃ (42)
for any {zn }∞ ∞

n=1 , {wn }n=1 ∈ `ψ̃ . For all p ∈ R with 0 < p ≤ 1,
∞
! p1
X
ψ˜p (t) = tnp
n=1
1
is a element of Ψe ∞ and k{zn }∞ k e = k{zn }∞ kp = (P∞ |zn |p ) p .
n=1 ψp n=1 n=1
Let (X, k · k) be a Banach space. For any ψ̃ ∈ Ψ∞ , we define the ψ̃-direct sums
e
of X to be the space
`ψ̃ (X) = {xn }∞ ∞

n=1 ⊂ X : {kxn k}n=1 ∈ `ψ̃ ,
with the mapping k{xn }∞ ∞

n=1 kψ̃ = k{kxn k}n=1 kψ̃ . We have a following result.
e ∞ and {an }∞ ∈ `∞ . Then

Theorem 1.4.4. Let X be a Banach space, ψ̃ ∈ Ψ n=1
following conditions are equivalent:
39
P∞
n=1 ∈ `ψ̃ (X), an xn converges in X and satisfies
n=1
∞
X
∞
a n n ≤ k{xn }n=1 kψ̃ ;
x

n=1
(ii) sup1≤n<∞ |an | ≤ 1.
Proof. If {an }∞
n=1 satisfies (ii), we remark that max{|a1 |, . . . , |an |} ≤ 1 for all n ∈ N.
As in the proof of Theorem 1.3.3, from the generalized inverse Minkowski inequality
(42), we have
n
X
kaj xj k ≤ k(x1 , . . . , xn )kψ̃
j=1
for all x1 , . . . , xn ∈ X. Then we have (i).

Conversely, assume that {an }∞
n=1 ∈ `∞ satisfies (i). For all fixed n ∈ N, put
xn+1 = xn+2 = · · · = 0, then we have

Xn
aj xj ≤ k(x1 , . . . , xn )kψ̃ .

j=1
From Theorem 1.3.3, we have max{|a1 |, . . . , |an |} ≤ 1 for all n ∈ N. Hence {an }∞
n=1
holds (ii).
From this theorem, we have a following corollary by putting ψ̃ = ψ̃p , where

0 < p ≤ 1.
Corollary 1.4.5. Let X be a Banach space, n ∈ N with n ≥ 2, p ∈ R with 0 < p ≤ 1

and {an }∞
n=1 ∈ `∞ . Then following conditions are equivalent:
P∞
n=1 ∈ `p (X), n=1 an xn converges in X and satisfies
p
X ∞
an xn ≤ k{xn }∞ p
n=1 k ;

n=1
(ii) sup1≤n<∞ |an | ≤ 1.
40
2 On sharp triangle inequalities
2.1 Introduction
In this chapter, we consider the inequalities which are sharper than the usual triangle
inequality for a Banach space X. In [14], Hudizik and Landes remarked a sharp
triangle inequality for two elements. In [19], Kato, Saito and Tamura proved the
sharp triangle inequality and reverse inequality as follows: for all nonzero elements
x1 , · · · , xn ∈ X,
!
n n
X X xj
xj + n− min kxj k

kxj k 1≤j≤n

j=1 j=1
n
X
≤ kxj k (43)
j=1
!
n n
X X x j

≤ xj + n− max kxj k

j=1
kxj k 1≤j≤n
j=1
hold. In [29], Mitani et al. succeeded in the further extensions of above inequalities
as follows: for all nonzero elements x1 , x2 , . . . , xn ∈ X, n ≥ 2,
!
n n k
X X X x∗j
xj + k− (kx∗k k − kx∗k+1 k)

∗
kxj k

j=1 k=2 j=1
n
X
≤ kxj k (44)
j=1
 
n n n
x∗j
X X
X
 (kx∗n−k k − kx∗n−(k−1) k)

≤ xj − k −

j=1

k=2
kx∗j k
j=n−(k−1)
hold, where x∗1 , x∗2 , . . . , x∗n are the rearrangement of x1 , x2 , . . . , xn satisfying kx∗1 k ≥
kx∗2 k ≥ · · · ≥ kx∗n k, and x∗0 = x∗n+1 = 0. Moreover they studied equality attained-
ness on these inequalities in a strictly convex Banach space (cf. [19] and [27]). In
[24], Mineno, Nakamura and Ohwada studied the problem that characterize all the
intermediate value C which satisfies

n
X Xn
0≤C≤ kxj k − xj

j=1 j=1
41
by using x1 , x2 , . . . , xn in a Banach space X. Inequalities (43) and (44) give partial
solutions of this problem: for all x1 , x2 , . . . , xn ∈ X,
n ! n
n
X x X
j
X
0≤ n− min kxj k ≤ kxj k − xj

j=1
kx j k 1≤j≤n
j=1

j=1

and !
n k ∗ n n
X X x j ∗ ∗
X X
0≤ k− (kxk k − kxk+1 k) ≤ kxj k − xj

∗
k=2
kxj k
j=1

j=1

j=1
hold. Our main aims in this chapter are to give the simple proof of the result of
Mineno et al. and study when this norm inequality attains equality in a strictly
convex Banach space.
In §2.2, we summarize basic results of sharp triangle inequalities and their reverse
inequalities in Banach spaces.
In §2.3, we present the simple proof of the result of Mineno et al. and study when
this norm inequality attains equality in strictly convex Banach spaces.
In §2.4, by using sharp triangle inequalities, we characterize some geometrical
properties of Banach spaces.
2.2 Sharp triangle inequalities in Banach spaces

2.2.1 Sharp triangle inequalities and their reverse inequalities
In this section, we summarize basic results of sharp triangle inequalities and their
reverse inequalities in Banach spaces. In [19], Kato et al. proved the sharp triangle
inequality and reverse inequalities for an arbitrary number of finitely many nonzero
elements x1 , x2 , . . . , xn ∈ X as follows:
Theorem 2.2.1 (cf. [19, Theorem 1]). For all nonzero elements x1 , x2 , . . . , xn in a
Banach space X,
!
n n
X X xj
xj + n− min kxj k

kxj k 1≤j≤n

j=1 j=1
n
X
≤ kxj k (45)
j=1
!
n n
X X x j

≤ xj + n− max kxj k (46)

j=1
kxj k 1≤j≤n
j=1
42
hold.
In the case n = 2, we have the following inequalities which will be used to prove
the general n elements case.
Theorem 2.2.2 (cf. [14] and [20]). For all nonzero elements x, y in a Banach space
X,

x y
kx + yk + 2 −
kxk kyk min{kxk, kyk}
+
≤ kxk + kyk (47)

x y
kxk kyk max{kxk, kyk}
≤ kx + yk + 2 − + (48)
hold.
Proof. If kxk = kyk, both inequalities (47) and (48) hold with equality. Therefore
we may assume this is not the case. Let us see the first inequality. Let kxk > kyk.
Then we have

x y x y x x
kxk + kyk = kyk + kyk − kyk + kxk

x y 1 1
= + − − x
kyk kyk kyk kxk

x y 1 1
≥ kyk + kyk − kyk − kxk kxk

kx + yk 1 1 1 1
= − − kxk − − kyk
kyk kyk kxk kyk kyk

kx + yk kxk + kyk
= − −2 ,
kyk kyk
for which it follows that

kxk + kyk kx + yk x y
≥ + 2− + .
kyk kyk kxk kyk
Hence we obtain

x y
kxk + kyk ≥ kx + yk + 2 −
+ kyk,
kxk kyk
43
or the inequality (47). For the second inequality as in the above proof, let kxk > kyk.
Then we have

x y x y y y
+ = + −
kxk kyk kxk kxk kxk kyk +

x y 1 1
= + + − y
kxk kxk kyk kxk

x y 1 1
≤ kxk + kyk + kyk − kxk kyk

kx + yk 1 1 1 1
= + − kyk + − kxk
kxk kyk kxk kxk kxk

kx + yk kxk + kyk
= + 2− ,
kxk kxk

kxk + kyk kx + yk x y
≤ + 2−
kxk kyk .
+
kxk kxk
Hence we obtain

x y
kxk + kyk ≤ kx + yk + 2 −
+ kxk,
kxk kyk
or the inequality (48). Thus we have the conclusion.
Proof of Theorem 2.2.1. If kx1 k = kx2 k = · · · = kxn k, both inequalities (45)

and (46) hold with equality. Therefore we may assume this is not the case. Let us
see the first inequality. Let kxj0 k = min{kxj k : 1 ≤ j ≤ n} and J0 = {j : kxj k =
kxj0 k, 1 ≤ j ≤ n}. Then for any nonzero x1 , . . . , xn ∈ X we have

n
X xj X xj X xj
= +

kx k kx k kx k

j j j
j∈J0c

j=1 j∈J0

n
X xj X xj X xj
= − +
j=1 kxj0 k j∈J0c kxj0 k j∈J0c kxj k

n
X xj X 1 1
= − − xj
j=1 kxj0 k j∈J0c kxj0 k kxj k

n
X xj X 1 1
≥ − − kxj k

j=1
kx j0 k
j∈J c
kx j 0 k kx j k
0
44

n n
X xj X 1 1
= − − kxj k

j=1
kxj0 k j=1 kxj0 k kxj k
!
n n
X x j
X kx j k
= − −n ,

kxj0 k
j=1
kxj0 k
j=1

n !
n
k nj=1 xj k
P
X kxj k X x
j
≥ + n− .

j=1
kx j 0 k kx j 0 k kxj k
j=1
Hence we obtain
!
n n n
X X X x j

kxj k ≥ xj + n− kxj0 k,

j=1

j=1
kxj k
j=1
or the inequality (45).

For the second inequality let kxj1 k = max{kxj k : 1 ≤ j ≤ n} and J1 = {j :
kxj k = kxj1 k, 1 ≤ j ≤ n}. Then we have

n
X xj X xj X xj
= +

kx k kx k kx k

j j j
j∈J1c

j=1 j∈J1

n
X xj X xj X xj
= − +
j=1 kxj1 k j∈J1c kxj1 k j∈J1c kxj k

n
X xj X 1 1
= + − xj
j=1 kxj1 k j∈J1c kxj k kxj1 k

n
X xj X 1 1
≤ + − kxj k

j=1
kx j 1 k
j∈J1c
kx j k kx j 1 k

n n
X xj X 1 1
= + − kxj k

j=1
kxj1 k j=1 kxj k kxj1 k

n n
X x j
X kxj k
= +n− ,

j=1
kx j 1 k
j=1
kx j 1 k
and hence !
n n n
X X X xj
kxj k ≤ xj + n− kxj1 k.

j=1

j=1
kxj k
j=1
Thus we have the conclusion.
45
2.2.2 Sharper triangle inequalities
In [29], Mitani et al. succeeded in the further extension of above inequalities.
Theorem 2.2.3 (cf. [29, Theorem 1]). For all nonzero elements x1 , x2 , . . . , xn in a
Banach space X, n ≥ 2,
!
n n k ∗
X X X x j
xj + k− (kx∗k k − kx∗k+1 k)

kx∗j k

j=1 k=2 j=1
n
X
≤ kxj k (49)
j=1
 
n n n
x∗j
X X
X
 (kx∗n−k k − kx∗n−(k−1) k)

≤ xj − k − (50)

j=1

k=2
kx∗j k
j=n−(k−1)
kx∗2 k ≥ · · · ≥ kx∗n k, and x∗0 = x∗n+1 = 0.
As the case n = 2 the above theorem includes Theorem 2.2.2. To see explicitly
that the Theorem 2.2.3 refines Theorem 2.2.1, we reformulate Theorem 2.2.3 as
follows: for all nonzero elements x1 , x2 , . . . , xn in a Banach space X, n ≥ 3,
n n !
X X x∗
j
xj + n − kx∗n k

∗
kx k

j=1

j=1 j
!
n−1 k
X X x∗j
+ k− (kx∗k k − kx∗k+1 k)

∗

k=2
kxj k
j=1
n
X
≤ kxj k
j=1
!
n n ∗
X X x j
≤ xj + n− kx∗1 k

∗

j=1

j=1
kx j k
 
n−1 n ∗
X X xj
− k −  (kx∗n−k k − kx∗n−(k−1) k)
∗
k=2
kx
j=n−(k−1) j k
kx∗2 k ≥ · · · ≥ kx∗n k.
46
In the case of n = 3 we have the following inequalities: for all nonzero elements
x, y, z in a Banach space X with kxk ≥ kyk ≥ kzk,

x y z
kx + y + zk + 3 −
kxk + kyk + kzk kzk

x y
+ 2− kxk kyk (kyk − kzk)
+
≤ kxk + kyk + kzk

x y z
≤ kx + y + zk + 3 − kxk kyk kzk kxk
+ +

y z
− 2− kyk kzk (kxk − kyk)
+
hold.
In [27], Mitani and Saito gave a simple proof of these inequalities.
Proof of Theorem 2.2.3. To prove this theorem we shall prove that for all nonzero
elements x1 , x2 , . . . , xn in a Banach space X with kx1 k ≥ kx2 k ≥ · · · ≥ kxn k, n ≥ 2,
!
n n k
X X X xj
xj + k− (kxk k − kxk+1 k)

kxj k

j=1 k=2 j=1
n
X
≤ kxj k (51)
j=1
 
n n n
X X X xj
≤ xj − k −  (kxn−k k − kxn−(k−1) k) (52)

j=1

k=2
kxj k
j=n−(k−1)
hold, where x0 = xn+1 = 0. According to Theorem 2.2.1, inequalities (51) and (52)
hold for the case n = 2. Therefore let n ≥ 3. We first prove the case kx1 k > kx2 k >
· · · > kxn k. We prove inequality (51) by the induction. Assume that the inequality
holds true for all n − 1 elements in X. Let x1 , x2 , . . . , xn be any n elements in X
with kx1 k > kx2 k > · · · > kxn k > 0. Let
xj
uj = (kxj k − kxn k) ,
kxj k
for all positive j with 1 ≤ j ≤ n. Then
n n n−1
X X xj X
xj = kxn k + uj (53)
j=1 j=1
kxj k j=1
47
and ku1 k > ku2 k > · · · > kun−1 k > 0. By assumption,
!
n−1 n−1 n−1 k
X X X X uj
uj ≤ kuj k − k− (kuk k − kuk+1 k) (54)

kuj k

j=1 j=1 k=2 j=1
holds, where un = 0. Since kuk k − kuk+1 k = kxk k − kxk+1 k, from (53) and (54),

n n n−1
X X xj X
xj = kxn k + uj

kxj k j=1

j=1 j=1

n n−1
X x j
X
≤ kxn k + u

j
kxj k

j=1 j=1

n n−1
X xj X
≤ kxn k + kuj k

j=1
kxj k j=1
n−1
k !
X X u
j
− k− (kuk k − kuk+1 k)

k=2

j=1
kuj k
n n−1
X x X
j
= kxn k + (kxj k − kxn k)

j=1
kxj k j=1
!
n−1 k
X X xj
− k− (kxk k − kxk+1 k)

k=2

j=1
kxj k
!
n n k
X X X xj
= kxj k − k− (kxk k − kxk+1 k)

j=1 k=2

j=1
kxj k
and hence (51). Thus (51) holds true for all finite elements in X.
Next we show inequality (52). Let
xn−j+1
vj = (kx1 k − kxn−j+1 k)
kxn−j+1 k
for all positive j with 1 ≤ j ≤ n − 1. Then

n n n−1
X X xj X
xj = kx1 k − vj
j=1 j=1
kx j k j=1
48
and kv1 k > kv2 k > · · · > kvn−1 k > 0. Applying Inequality (51) to v1 , v2 , . . . , vn−1 ,
n n n−1
X X x X
j
xj ≥ kx1 k − vj

kxj k j=1

j=1 j=1
n n−1
X x X
j
≥ kx1 k − kvj k

j=1
kx j k
j=1
!
n−1 k
X X vj
+ k− (kvk k − kvk+1 k)

k=2

j=1
kvj k

n n−1
X x j
X
= kx1 k − (kx1 k − kxn−j+1 k)

j=1
kx j k
j=1
!
n−1 k
X X xn−j+1
+ k− (kxn−k+1 k − kxn−k k)

k=2

j=1
kxn−j+1 k
 
n n n
X X X x j
= kxj k − k −  (kxn−k+1 k − kxn−k k)
j=1 k=2
kx
j=n−(k−1) j k
hold, where vn = 0. Thus we obtain (52).

Now we show the case of kx1 k ≥ kx2 k ≥ · · · ≥ kxn k > 0. For all positive
numbers m with m > n, let

k
xk,m = 1− xk , (k = 1, 2, . . . , n).
m
Then kx1,m k > kx2,m k > · · · > kxn,m k > 0. Applying inequalities (51) and (52) to
x1,m , x2,m , . . . , xn,m ,
!
n n k
X X X xj,m
xj,m + k− (kxk,m k − kxk+1,m k)

kxj,m k

j=1 k=2 j=1
n
X
≤ kxj,m k
j=1
 
n n n
X X X xj,m
≤ xj,m − k−  (kxn−k,m k − kxn−(k−1),m k)


j=1

k=2
kxj,m k
j=n−(k−1)
hold, where x0,m = xn+1,m = 0 for all positive numbers m with m > n. As m → +∞,
we have Inequalities (51) and (52) for the case of kx1 k ≥ kx2 k ≥ · · · ≥ kxn k > 0.
49
2.3 Characterization of the intermediate values of the tri-
angle inequality
2.3.1 A simple proof of the preceding result
In [24], Mineno et al. characterised the intermediate values of the triangle inequality
in normed spaces as follows. For positive integer n ≥ 2, let Mn ([0, 1]) be the set of
all n × n matrices whose all elements belong to the interval [0, 1], and Ln denote the
set of all lower triangular matrices of Mn ([0, 1]), i.e.,
n o
Ln = a = (aij ) ∈ Mn ([0, 1]) : aij = 0, i < j .
Let 1 ≤ m ≤ n. For each a = (aij ) in Ln , we set àmj (m) = amj , 1 ≤ j ≤ m, and if

2 ≤ n, then, for each m with 2 ≤ m ≤ n, we put
m
Y
àij (m) = aij (1 − akj ), 1 ≤ i ≤ m − 1, 1 ≤ j ≤ m.
k=i+1
Lemma 2.3.1 (cf. [24, Lemma 3.1]). Keep the notation as above. Let a = (aij ), b =
(bij ) in Ln . Then the following statements hold.
(i) Let 1 ≤ m ≤ n. For each i with 1 ≤ i ≤ m,
0 ≤ àij (m) ≤ aij ≤ 1 (1 ≤ j ≤ n),

and n × n matrix àij (n) belongs to Ln .
(ii) Let n ≥ 3. For each m with 3 ≤ m ≤ n,

(
àij (m − 1)(1 − amj ) (2 ≤ i ≤ m − 1, 1 ≤ j ≤ m − 1)
àij (m) =
amj (i = m, 1 ≤ j ≤ m).
(iii) For each i with 1 ≤ i ≤ n,

m
a X
ìj (m) − `bij (m) ≤ |aik − bik | (1 ≤ j ≤ m).
k=j
Proof. We only prove (iii). It is clear in case of n = 1 and 2. Let n ≥ 3. It is also

clear in case of i = 1 and m. By using induction on m, we see that

m
Y m
Y m
X
(1 − akj ) − (1 − bkj ) ≤ |akj − bkj |.

k=i+1 k=i+1 k=i+1
50
Thus we have
a
ìj (m) − `bij (m)

m
Y Ym
= aij (1 − akj ) − bij (1 − bkj )

k=i+1 k=i+1
m m

Y Y
≤ aij (1 − akj ) − aij (1 − bkj )

k=i+1 k=i+1

Ym Ym
+ aij (1 − bkj ) − bij (1 − bkj )

k=i+1 k=i+1

Ym Ym
≤ (1 − akj ) − (1 − bkj ) + |aij − bij |

k=i+1 k=i+1
m
X
≤ |akj − bkj | + |aij − bij |
k=i+1
Xm
≤ |aij − bij |.
k=i
Theorem 2.3.2 (cf. [24, Lemma 3.2]). Let n ≥ 2. With the above notation, take
any a = (aij ) in Ln . For any elements x1 , x2 , · · · , xn in a normed space X, the
following inequalities hold:
n i
i ! n
n
X X X X X
0≤ kàij (n)xj k − àij (n)xj ≤ kxj k − xj . (55)

i=1 j=1 j=1 j=1 j=1
In [24], the above theorem was established for à1j (m) = a1j , 1 ≤ j ≤ m, which
values do not influence (55). We made a modification in order to consider the
equality attainability in (55) in a particular space.

Take any a = (aij ) ∈ Ln and fix it. Considering àij (n) as the matrix acting on
| ⊕X ⊕
a Banach space X {z· · · ⊕ X} , we have
n times
    
à11 (n) x à (n)x1
  1  P2 11
0
 
 à (n) à (n)   x2  a
 j=1 `2j (n)xj 
 
 21 22
 .  =  ,
 
 .
 .. .
.. . ..   ..   ..
    P . 

a a a n a
`n1 (n) ... `nn−1 (n) `nn (n) x n `
j=1 nj (n)x j
51
where x1 , x2 , . . . , xn ∈ X. For each entries, we have the triangle inequalities

X i X i
a
ìj (n)xj ≤ kàij (n)xj k (1 ≤ i ≤ n). (56)

j=1 j=1
Theorem 2.3.2 means that the sum of all differences of (56) is less than the difference
of the triangle inequality
Xn X n
xj ≤ kxj k.

j=1 j=1
To present a simple proof of Theorem 2.3.2, we would need the following lemma.
Lemma 2.3.3. Let n ≥ 2 and a = (aij ) ∈ Ln . Take m with 1 ≤ m ≤ n. For each j

with m ≤ j ≤ n, the following identity holds:
n
X n
Y
àim (n) + (1 − aim ) = 1.
i=j i=j
Proof. We shall prove this lemma by using induction on j. If j = n, then
(1 − anm ) + ànm (n) = (1 − anm ) + anm = 1.
Next let j = n − 1. Then

n
Y n
X
(1 − ai m ) + àim (n)
i=n−1 i=n−1
= (1 − an−1 m )(1 − an m ) + àn−1 m (n) + àn m (n)

= (1 − an−1 m )(1 − an m ) + an−1 m (1 − an m ) + an m
= (1 − an−1 m )(1 − an m ) + {1 − (1 − an−1 m )}(1 − an m ) + an m
= (1 − an m ) + an m
= 1.
Now if we assume that, for each j with m + 1 ≤ j ≤ n,

n
Y n
X
(1 − ai m ) + àim (n) = 1,
i=j i=j
52
then we have
Yn n
X
(1 − ai m ) + àim (n)
i=j−1 i=j−1
n
Y n
X
= (1 − ai m ) + àj−1 m (n) + àj m (n)
i=j−1 i=j
Yn n
Y n
X
= (1 − ai m ) + aj−1 m (1 − ai m ) + àj m (n)
i=j−1 i=j i=j
Yn n
Y n
X
= (1 − ai m ) + {1 − (1 − aj−1 m )} (1 − ai m ) + àj m (n)
i=j−1 i=j i=j
Yn n
X
= (1 − ai m ) + àj m (n)
i=j i=j
= 1.
Thus, by induction on j, we obtain the lemma.
Proof of Theorem 2.3.2. For each x1 , . . . , xn ∈ X and a = (aij ) ∈ Ln , the fol-

lowing equations hold:
n n n
! n X
n
X X X X
xj = 1− àij (n) xj + àij (n)xj
j=1 j=1 i=j j=1 i=j
n n
! n X
i
X X X
= 1− àij (n) xj + àij (n)xj .
j=1 i=j i=1 j=1
By Lemma 2.3.3, we see that

n
Y n
X
0≤ (1 − aij ) = 1 − àij (n).
i=j i=j
Moreover, 0 ≤ àij (n) ≤ 1 for all i, j ∈ {1, . . . , n}. Thus, applying the triangle
inequality, we have
n n n
! n X i

X X X X
a a
x = 1 − ` (n) x + ` (n)x

j ij j ij j

j=1 j=1 i=j i=1 j=1
!
Xn X n X n X i
≤ 1− àij (n) xj + àij (n)xj (57)

j=1 i=j i=1 j=1
!
Xn Xn X n X i
= 1− àij (n) kxj k + à
(n)x

ij j

j=1 i=j i=1 j=1
53
n n
! i
n X

X X a X
= kxj k − ìj (n)xj + àij (n)xj

j=1 i=j i=1 j=1

Xn Xn X n n X i
a X a
= kxj k − ìj (n)xj + ìj (n)xj

j=1 j=1 i=j i=1 j=1

Xn Xn X i n X i
a X
= kxj k − ìj (n)xj + àij (n)xj

j=1 i=1 j=1 i=1 j=1
n n i
i
!
X X X a X a

= kxj k − ìj (n)xj − ìj (n)xj .

j=1 i=1 j=1 j=1
Let x1 , x2 , · · · , xn ∈ X. For each a ∈ Ln , if we put

!
n i i
X X a X
a

f (a) = ìj (n)xj −
` ij (n)x j ,

i=1 j=1 j=1
then we see that

n
X Xn
f (a0 ) = 0, f (a1 ) = kxj k − xj ,

j=1 j=1
where a0 , a1 ∈ Ln with
 
a11 0 ··· ··· 0
   .. .. .. 
0 ··· 0  a21 a22 . . .
. .
.. . . ...  ,
 
  .. .. .. .. 
a0 = 
  a1 =  . . . 0 . .
 
0 ··· 0 an−11 an−12 · · ·

an−1n−1 0

1 1 ··· 1 1
Moreover, by Lemma 2.3.1, we see that, for each a = (aij ), b = (bij ) ∈ Ln ,

n X
X i
a
|f (a) − f (b)| ≤ 2 ìj (n) − `bij (n) · kxj k
i=2 j=1
n X
X i X
n
≤ 2 max kxj k · |aki − bki |.
1≤j≤n
i=2 j=1 k=i
Qn(n+1)/2 Qn(n+1)/2
Thus, considering f to a function on i=1 [0, 1], f is continuous on i=1 [0, 1].
Therefor, we obtain the following.
54
Corollary 2.3.4 (cf. [24, Corollary 3.3]). Let n ≥ 2 and x
1 , x2 , · · · , xn in a Banach
Pn Pn
space X. For each ω with 0 ≤ ω ≤ j=1 kxj k − j=1 xj , there exists a ∈ Ln such

that !
n i i
X X a X
a

ω= ìj (n)xj − ìj (n)xj .

i=1 j=1 j=1
Of course inequality (55) contains (45) and (49).
Theorem 2.3.5 (cf. [24, Corollary 3.4]). Let n ≥ 2. For all elements x1 , x2 , · · · , xn
in a Banach space X, the following inequalities hold
!
n n n
X x
j
X X
0≤ n− min kxj k ≤ kxj k − xj .

kxj k 1≤j≤n
j=1

j=1

j=1
Proof. We may assume that kxn k = min1≤j≤n kxj k. If we take a ∈ Ln as

 
0
0
 .
...

 .
 .

a= ,

 0 ... 0 
 
kxn k kxn k kxn k
kx1 k
. . . kxn−1 k kxn k
then it is clear that
kxn k
àij (n) = 0 (1 ≤ i ≤ n − 1), ànj (n) = (1 ≤ j ≤ n).
kxj k
In this case, we see that
n i
i !
X X X
f (a) = kàij (n)xj k − àij (n)xj

i=1 j=1 j=1
n
n
X X
a a
= k`nj (n)xj k − `nj (n)xj

j=1 j=1
n n

X kxn k X kxn k

= x
kxj k j
− x j
kx k

j=1 j=1 j
n !
X x
j
= n− min kxj k.

j=1
kx j k 1≤j≤n
Thus, by Theorem 2.3.2, we have

!
n n n
X x
j
X X
0≤ n− min kx k ≤ kx k − x j .

j j
kxj k 1≤j≤n

j=1 j=1 j=1
55
Theorem 2.3.6 (cf. [24, Corollary 3.5]). Let n ≥ 2. For all elements x1 , x2 , · · · , xn
in a Banach space X, the following inequalities hold
n
i ! n
n
X X x∗ X X
j
0≤ i− (kx∗i k − kx∗i+1 k) ≤ kxj k − xj ,

∗

i=2
kxj k j=1

j=1 j=1
where x∗1 , x∗2 , . . . , x∗n are the rearrangement of x1 , x2 , . . . , xn which satisfies kx∗1 k ≥
kx∗2 k ≥ · · · ≥ kx∗n k and x∗n+1 = 0.
Proof. Let us first show in the case kx∗1 k > kx∗2 k > · · · > kx∗n k. If we take a =
(aij ) ∈ Ln as
kx∗i k − kx∗i+1 k
aij = (1 ≤ j ≤ i ≤ n),
kx∗j k − kx∗i+1 k
then, for all j ≤ i with 2 ≤ i ≤ n, we have àij (n) = (kx∗i k − kx∗i+1 k)/kx∗j k. Indeed,
when 2 ≤ j ≤ n − 1, for all i with 1 ≤ j ≤ i, we have
n
Y
àij (n) = aij (1 − akj )
k=i+1
n
kx∗i k − kx∗i+1 k Y kx∗k k − kx∗k+1 k

= ∗ · 1− ∗
kxj k − kx∗i+1 k k=i+1 kxj k − kx∗k+1 k
n
kx∗i k − kx∗i+1 k Y kx∗j k − kx∗k k
= ∗ ·
kxj k − kx∗i+1 k k=i+1 kx∗j k − kx∗k+1 k
kx∗i k − kx∗i+1 k kx∗j k − kx∗i+1 k kx∗j k − kx∗i+2 k kx∗j k − kx∗n k
= · · · · ·
kx∗j k − kx∗i+1 k kx∗j k − kx∗i+2 k kx∗j k − kx∗i+3 k kx∗j k − kx∗n+1 k
kx∗ k − kx∗i+1 k
= i .
kx∗j k
Moreover, when j = n, we see that àin (n) = kx∗n k/kx∗j k = (kx∗n k − kx∗n+1 k)/kx∗j k. In
this case, we see that
!
n
X i
X X i
f (a) = kàij (n)x∗j k − àij (n)x∗j

i=1 j=1 j=1
n i i !
X X kx∗i k − kx∗i+1 k ∗ X kx∗i k − kx∗i+1 k
= xj − x∗j

∗ ∗
j=2 j=1
kxj k
j=1
kxj k
n
( i
i
)
x∗j ∗

X X
∗ ∗
X x j
= kx∗ k (kxi k − kxi+1 k) − (kx∗i k − kx∗i+1 k)

∗
i=2 j
j=1
kx j k
j=1

56
n
( i i )
X X X x∗
j
= (kx∗i k − kx∗i+1 k) − (kx∗i k − kx∗i+1 k)

∗
i=2 j=1

j=1
kx j k
!
n i
X X x∗j
= i− (kx∗i k − kx∗i+1 k).

∗
i=1

j=1
kxj k
Applying Theorem 2.3.2, we have

n
i ! n
n
X X x∗ X X
j
0≤ i− (kx∗i k − kx∗i+1 k) ≤ kx∗j k − x∗j .

∗
i=2

j=1
kx j k
j=1

j=1

We next prove a general case by using a technique in the proof of Theorem 2.2.3.
Let kx∗1 k ≥ kx∗2 k ≥ · · · ≥ kx∗n k. For each fixed integer m with m > n, we set

∗ i
xi,m = 1 − x∗i (1 ≤ i ≤ n).
m
Then we see that kx∗1,m k > kx∗2,m k > · · · > kx∗n,m k > 0, and so we have
!
n i ∗ n n
X X x j,m

∗ ∗
X
∗
X
∗

0≤ i− (kxi,m k − kxi+1,m k) ≤ kxj,m k − xj,m ,

i=2
kx∗j,m k
j=1 j=1

j=1
where x∗n+1,m = 0. Since x∗i,m → x∗i (m → ∞), we have the conclusion. This
completes the proof.
2.3.2 Equality attainedness in a strictly convex Banach space
In this subsection, we consider equality attainedness for (55) in a strictly convex

Banach space. For a Banach space X, let SX be a unit sphere in X: SX = {x ∈
X : kxk = 1}. A Banach space X is strictly convex if for all x, y ∈ SX with x 6= y,
kx+yk < 2 holds. For the case of the usual triangle inequality, equality attainedness
is following.
Lemma 2.3.7 (cf. [1, Problem 11.1]). Let (X, k · k) be a strictly convex Banach
space. For x1 , . . . , xn ∈ X, the following assertions are equivalent:

Xn X n
(i) xj = kxj k;

j=1 j=1
(ii) kxi kxj = kxj kxi f or all i, j ∈ {1, · · · , n}.
57
In [19] and [27], Kato et al. consider equality attainedness for the inequality
(45) and (49). 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}.
+ − + −
|Im (α)| and |Im (α)| are the cardinal numbers of Im (α) and Im (α) respectively.
Theorem 2.3.8 (cf. [27, Theorem 3.7]). Let X be a strictly convex Banach space
and x1 , x2 , . . . , xn nonzero elements in X with kx1 k > kx2 k > · · · > kxn k. Then the
equality
!
n n k n
X X X x j
X
x + k − (kx k − kx k) = kxj k

j k k+1
kxj k

j=1 k=2 j=1 j=1
holds if and only if there exists α = (α1 , α2 , . . . , αn ) ∈ Rn with 1 = α1 > |α2 | >
|α3 | > · · · > |αn | such that
xm = αm x1 ,
+ −
|Im (α)| ≥ |Im (α)|
for every m with 1 ≥ m ≥ n.
For two or three elements cases are following.
and x, y nonzero elements in X with kxk > kyk. Then the equality

x y
kx + yk + 2 − kxk kyk kyk = kxk + kyk
+
holds if and only if there exists a real number α with 0 < α < 1 satisfying y = ±αx.
and x, y, z nonzero elements in X with kxk > kyk > kzk. Then the equality

x y z
kx + y + zk + 3 − kxk kyk kzk kzk
+ +

x y
+ 2− + (kyk − kzk) = kxk + kyk + kzk
kxk kyk
holds if and only if there exist α, β with 0 < β < α < 1 satisfying one of the following
conditions:
58
(a) y = αx, z = ±βx,
(b) y = −αx, z = βx.
Now we consider equality attainedness for (55). For a = (aij ) ∈ Ln and

x1 , . . . , xn ∈ X, put
( n
)
X
J= j ∈ {1, . . . , n} : àij (n) 6= 1 ,
i=j
( i
)
X
I= i ∈ {1, . . . , n} : àij (n)xj 6= 0 .
j=1
Lemma 2.3.11. Take a = (aij ) ∈ Ln . Then
(i) J = {1, . . . , n} if and only if aij ∈ [0, 1) for all i, j ∈ {1, . . . , n}; and
(ii) J = ∅ if and only if, for each j ∈ {1, . . . , n}, there exists i with j ≤ i ≤ n
such that aij = 1
Proposition 2.3.12. Let n ≥ 2 and a = (aij ) in Ln . For x1 , . . . , xn in a strictly

convex Banach space X, the equality
!
Xn Xi Xi n
X Xn
kàij (n)xj k − àij (n)xj = kxj k − xj (58)

i=1 j=1 j=1 j=1 j=1
holds if and only if either I = ∅ and J = ∅ or, for each g, h, k, m ∈ {1, . . . , n}, the
following assertions hold:
(i) if J 6= ∅, then, for each g, h, k ∈ J and m ∈ {1, . . . , n},
kxh kxg = kxg kxh , (59)

Xm m
X
àmj (n)xj xk = kxk k àmj (n)xj ; (60)

j=1 j=1
(ii) if J = ∅, then, for each k, m ∈ {1, . . . , n},

X m Xk X k Xm
a a a
`mj (n)xj `kj (n)xj = `kj (n)xj àmj (n)xj . (61)

j=1 j=1 j=1 j=1
59
Proof. Let I = ∅ and J = ∅. In this case, for each i, j ∈ {1, . . . , n},
i
X n
X
àij (n)xj = 0 and àij (n) = 1.
j=1 i=j
Hence
n X
X i n X
X n n
X n
X n
X
0= àij (n)xj = àij (n)xj = xj àij (n) = xj .
i=1 j=1 j=1 i=j j=1 i=j j=1
Therefore we have
!
Xn Xi Xi n X
X i
kàij (n)xj k − àij (n)xj = àij (n)kxj k

i=1 j=1 j=1 i=1 j=1
Xn X n
= àij (n)kxj k
j=1 i=j
Xn n
X
= kxj k àij (n)
j=1 i=j
Xn
= kxj k
j=1

n
X Xn
= kxj k − xj .

j=1 j=1
Next let J 6= ∅. If (58) holds, then the equality (57) also holds. Thus, by Lemma
2.3.7, the following equalities hold for any g, h, k, m ∈ {1, . . . , n}:
! !
X n X n
1− àih (n) xh 1 − àig (n) xg

i=h i=g
! !
Xn Xn
a a
= 1− ìg (n) xg 1 − ìh (n) xh ,

i=g i=h
!
X m X n
àmj (n)xj 1 − àik (n) xk

j=1 i=k
! m
Xn X
= 1− àik (n) xk àmj (n)xj .

i=k j=1
Pn a
By Lemma 2.3.3, we see that 0 < 1 − i=j ìj (n) for all j ∈ J. Thus the above
equalities equivalent to kxh kxg = kxg kxh , g, h ∈ J, and

X m m
X
a
`mj (n)xj xk = kxk k àmj (n)xj , k ∈ J, m ∈ {1, . . . , n},

j=1 j=1
60
respectively. The converse is clear. Finally let J = ∅ and I 6= ∅. In this case if
(58) holds, then, for each k, m ∈ I, we have
m k k m
X X X X
a a a
`mj (n)xj `kj (n)xj = `kj (n)xj àmj (n)xj .

j=1 j=1 j=1 j=1
The converse is clear, and the proof is completed.
Next, we consider equality attainedness of (58) in more detail.
Theorem 2.3.13. Let n ≥ 2 and a = (aij ) in Ln . Assume that J 6= ∅ and put

j0 = minj∈J j. For nonzero elements x1 , . . . , xn in a strictly convex Banach space
X, the equality (58) holds if and only if there exist positive real numbers αj , j ∈ J,
such that xj = αj xj0 , and, for each i ∈ {1, . . . , n}, there are real numbers βi such
that
X X
0≤ αj àij (n) + βi and àij (n)xj = βi xj0 ,
j∈Ji j∈Jic
where Ji = {j ∈ J : j ≤ i} and Jic is a complement of Ji in {1, . . . , n}.
Proof. (⇒) Assume that (58) holds. By Proposition 2.3.12 (i), if we put αj =
kxj k/kxj0 k, j ∈ J, then αj > 0 and xj = αj xj0 , j ∈ J. Moreover, for each i ∈
{1, . . . , n},
i
X Xi x
j0
àij (n)xj = àij (n)xj .

kxj0 k
j=1 j=1
Since
i
X X X
àij (n)xj = àij (n)xj + àij (n)xj
j=1 j∈Ji j∈Jic
X X
= αj àij (n)xj0 + àij (n)xj , i ∈ {1, . . . , n},
j∈Ji j∈Jic
we have i !
X 1 X a
X
àij (n)xj = ` (n)xj − αj àij (n) xj0 .

kxj0 k j=1 ij

j∈Jic

j∈J i
Put
i
1 X a
X
βi = ìj (n)xj − αj àij (n).

kxj0 k j=1

j∈J i
61
αj àij (n) + βi and
P
Then we see that 0 ≤ j∈Ji
X
àij (n)xj = βi xj0 , i ∈ {1, . . . , n}. (62)
j∈Jic
(⇐) Assume that there exist positive real numbers αj and real numbers βi such that
xj = αj xj0 , j ∈ J, 0 ≤ j∈Ji αj àij (n) + βi and (62) holds. Since kxj k = αj kxj0 k for
P
all j ∈ J, it is clear that (59) is valid. So, by Proposition 2.3.12 (i), we need only to
prove (60). Since
i
!
X X
àij (n)xj = αj àij (n) + βi xj 0 , i ∈ {1, . . . , n},
j=1 j∈Ji
we have
!
Xi X
a
ìj (n)xj = αj àij (n) + βi kxj0 k, i ∈ {1, . . . , n}.

j=1 j∈Ji
Thus
i i
X
a 1 X
a

ìj (n)xj = ìj (n)xj xj0 , i ∈ {1, . . . , n},

kxj0 k

j=1 j=1
and the proof is completed.
By Lemma 2.3.11, for a = (aij ) ∈ Ln , recall that aij ∈ [0, 1), i, j ∈ {1, . . . , n}, is
equivalent to J = {1, . . . , n}. Thus we immediately have the following corollary.
Corollary 2.3.14. Let n ≥ 2 and a = (aij ) in Ln . For nonzero elements x1 , . . . , xn

in a strictly convex Banach space X, if aij ∈ [0, 1) for all i, j ∈ {1, . . . , n}, then the
equality (58) holds if and only if there exist positive real numbers αj , j ∈ {1, . . . , n},
such that xj = αj x1 , j ∈ {1, . . . , n}.
Next, we consider the case J = ∅ for a = (aij ) ∈ Ln . In this case we note that
ann = 1. Take a = (aij ) ∈ Ln with anj = 1 for all j ∈ {1, . . . , n}. Then we see that
 
0 ... ... 0
. .. .. 
. ...
. . .

(àij (n)) =  .
0 . . . 0 0
 
1 ... 1 1
Thus it is clear that, for each x1 , . . . , xn in a normed space X, (58) always holds.
Thus we are interested in the case anj 6= 1 for some j ∈ {1, . . . , n − 1}. However,
generally it is complicated. We give the following special case.
62
Theorem 2.3.15. Let n ≥ 2 and a = (aij ) in Ln . For nonzero elements x1 , . . . , xn
in a strictly convex Banach space X, if aij ∈ [0, 1) for all i > j and aii = 1 for all
i ∈ {1, . . . , n}, then the equality (58) holds if and only if there exist real numbers
αj , j ∈ {1, . . . , n}, such that xj = αj x1 and ij=1 αj àij (n) ≥ 0, i ∈ {1, . . . , n}.
P
Pn a
Proof. (⇒) Since aii = 1 for all i ∈ {1, . . . , n}, by Lemma 2.3.3, i=k ìk (n) =
1, k ∈ {1, . . . , n}, and we note that à11 (n) > 0. If (58) holds, then, by Proposition
2.3.12, for all i ∈ {1, . . . , n},

Xi i
X
a a a
ìj (n)xj `11 (n)x1 = k`11 (n)x1 k àij (n)xj . (63)

j=1 j=1
P2 a
Let i = 2 and assume that = 0. Since à22 (n) > 0, if we put
j=1 `2j (n)xj
α2 = −à21 (n)/à22 (n), then x2 = α2 x1 and 2j=1 αj à2j (n) = 0. Next assume that
P
P2 a
j=1 `2j (n)xj 6= 0. By (63), we have
!
2
1 1 X a

def
x2 = a `2j (n)xj − α1 à21 (n) x1 = α2 x1 .

`22 (n) kx1 k j=1

We see that
!
X2 X2 x 2 2
a a 1 X X
αj `2j (n) x1 = `2j (n)xj = à2j (n)xj = αj à2j (n) x1 .

kx1 k

j=1 j=1 j=1 j=1
P2
Thus j=1 αj à2j (n) > 0, and hence in any case we have
2
X
αj à2j (n) ≥ 0.
j=1
Now we assume that, for each i = m with 2 ≤ m ≤ n − 1, (58) holds. Then there
exist real numbers αj , j ∈ {1, . . . , m}, such that xj = αj x1 , j ∈ {1, . . . , m}, and
Pi a
j=1 αj ìj (n) ≥ 0, i ∈ {1, . . . , m}.
If j=1 `m+1 j (n)xj = 0, then we see that xm+1 = αm+1 x1 and m+1
Pm+1 a P a
j=1 αj `m+1 j (n) =
0 for αm+1 = − j=1 αj àm+1 j (n)/àm+1 m+1 (n). In the case that m+1
Pm P a
j=1 `m+1 j (n)xj 6=
0, if we put
m+1 !
1 1 X
αm+1 = à (n)xj − α1 àm+1 1 (n) ,

àm+1 m+1 (n) kx1 k j=1 m+1 j

63
then we see that xm+1 = αm+1 x1 and, by (63),
m+1 m+1
X X x
a a 1
α ` (n) x = ` (n)x

j m+1 j 1 m+1 j j
kx1 k

j=1 j=1
m+1
X
= àm+1 j (n)xj
j=1
m+1
!
X
= αj àm+1 j (n) x1 .
j=1
Pm+1
Thus j=1 αj àm+1 j (n) > 0, and hence we have
m+1
X
αj àm+1 j (n) ≥ 0.
j=1
By using induction on i, we have desired result.

(⇐) Suppose that there exist real numbers αj , j ∈ {1, . . . , n}, such that xj =
Pi a
αj x1 , j ∈ {1, . . . , n}, and j=1 αj ìj (n) ≥ 0, i ∈ {1, . . . , n}, where α1 = 1. If
m 6∈ I, then, for each k ∈ {1, . . . , n},

Xm m
X
a
` (n)x x = 0 = kx k àmj (n)xj .

mj j k k

j=1 j=1
On the other hand, since

Xi i
X
a
ìj (n)xj = kx1 k αj àij (n)

j=1 j=1
for each i ∈ I, we have

i
X Xi x
1
àij (n)xj = àij (n)xj .

kx1 k
j=1 j=1
Therefore, by Proposition 2.3.12 (ii), the proof is completed.
As a corollary of Theorem 2.3.15, we can get [27, Theorem 3.7].
Corollary 2.3.16 (cf. [27, Theorem 3.7]). For all nonzero elements x1 , . . . , xn in a
strictly convex Banach space X with kx1 k > kx2 k > . . . > kxn k, the equality
i ! n
Xn X x n X
j
X
i− (kxi k − kxi+1 k) = kxj k − xj (64)

i=1

j=1
kxj k j=1

j=1

64
holds if and only if there exist real numbers αj , j ∈ {1, . . . , n}, with 1 = α1 > |α2 | >
+ −
. . . > |αn | such that xj = αj x1 , j ∈ {1, . . . , n}, and |Im (α)| ≥ |Im (α)| for every m
+ − +
with 1 ≤ m ≤ n, where |Im (α)| and |Im (α)| are the cardinal numbers of Im (α) =
−
{j ∈ {1, . . . , m} : αj > 0} and Im (α) = {j ∈ {1, . . . , m} : αj < 0}, respectively.
Proof. As in the proof of [24, Corollary 3.5] or the proof of Theorem 2.3.6, if we
put
kxi k − kxi+1 k
aij = , i, j ∈ {1, . . . , n},
kxj k − kxi+1 k
then a = (aij ) ∈ Ln and
kxi k − kxi+1 k
àij (n) = , i, j ∈ {1, . . . , n}.
kxj k
In this case, (63) is equivalent to (64). Thus, by (61), if (64) holds, then there exists
αj ∈ R such that
m
X
xj = αj x1 and αj àmj (n) ≥ 0, j, m ∈ {1, . . . , n}.
j=1
Since
kxm k − kxm+1 k |αm | − |αm+1 |
àmj (n) = = ,
kxj k |αj |
we see that m m
X X αj
0≤ αj àmj (n) = (|αm | − |αm+1 |) .
j=1 j=1
|αj |
Thus we have
m
X αj
0≤
j=1
|αj |
m m
X αj X αj
= +
+
|αj | −
|αj |
j∈Im (α) j∈Im (α)
Xm m
X
= 1+ (−1)
+ −
j∈Im (α) j∈Im (α)
+ −
= |Im (α)| − |Im (α)|,
m ∈ {1, . . . , n}. The converse is clear, and the proof is completed.
65
2.4 Applications of sharp triangle inequalities
By using sharp triangle inequalities, we can characterize some geometrical properties
of Banach spaces. In this section, for a Banach space X, SX is a unit sphere in X:
SX = {x ∈ X : kxk = 1}, and BX is a unit ball in X: BX = {x ∈ X : kxk ≤ 1}.
2.4.1 The strict convexity and the uniform convexity
We first consider the strict convexity. A Banach space X is strictly convex if for all
x, y ∈ SX with x 6= y, kx + yk < 2 holds (cf. [23]).
Theorem 2.4.1 (cf. [37, Proposition 8]). Let X be a Banach space. The following
conditions are equivalent:
(i) X is strictly convex.
(ii) Let x, y ∈ X such that kx + yk = kxk + kyk and x, y 6= 0. Then there exists
α > 0 such that x = αy.
Proof. Assume (i). Suppose that kx + yk = kxk + kyk and x, y 6= 0. By the shape
triangle inequality (47):

x y
kx + yk + 2 −
kxk + kyk min{kxk, kyk} ≤ kxk + kyk,

we have
x y
kxk + kyk = 2.

Since X is strictly convex, we have x/kxk = y/kyk. Thus x = αy and α > 0, we

have (ii). The opposite implication is clear.
Next we consider the uniform convexity. A Banach space X is uniformly convex

if for any ε > 0 there exists δ > 0 such that x, y ∈ SX and kx − yk ≥ ε imply
kx + yk < 2(1 − δ) (cf. [23]).
(i) X is uniformly convex.
(ii) For all {xn }, {yn } in SX such that kxn + yn k → 2, we have kxn − yn k → 0.
66
(iii) For all {xn }, {yn } in BX such that kxn + yn k → 2, we have kxn − yn k → 0.
Proof. We only show the implication (ii) ⇒ (iii). Take {xn }, {yn } in BX such that
kxn +yn k → 2. Since kxn +yn k ≤ kxn k+kyn k ≤ 2, we have kxn k → 1 and kyn k → 1.
Hence we may suppose that xn 6= 0 and yn 6= 0 for all n. Then, since

xn yn
kxn + yn k + 2 −
kxn k + kyn k min{kxn k, kyn k} ≤ kxn k + kyn k ≤ 2,

we have
xn y n
kxn k + kyn k → 2.

By the assumption,
xn yn
kxn k − kyn k → 0.

Since
x n xn yn yn
kxn − yn k ≤ x n − + − + − yn
,
kxn k kxn k kyn k kyn k
we have kxn − yn k → 0 as n → ∞.
2.4.2 The uniform non-squareness
Next we consider the uniform non-squareness. A Banach space X is uniformly non-

square if there exist ε > 0 such that for all x, y ∈ SX , we have min{kx+yk, kx−yk} ≤
2(1 − ε) (cf. [17] and [5]).
(i) X is uniformly non-square.
(ii) There exists ε > 0 such that for all x, y ∈ BX , we have min{kx+yk, kx−yk} ≤
2(1 − ε).
Proof. Assume (i). Then there exists ε > 0 such that for all x, y ∈ SX , we have
min{kx + yk, kx − yk} ≤ 2(1 − ε). Take any x, y ∈ BX . If min{kxk, kyk} ≤ 1/2,
then
1 3 1
kx ± yk ≤ kxk + kyk ≤ 1 + = = 2 1 − .
2 2 4
67
If min{kxk, kyk} ≥ 1/2, then

x y x y
min + , − ≤ 2(1 − ε).
kxk kyk kxk kyk
Let
x y x y
kxk + kyk ≤ kxk − kyk .

Then by the shape triangle inequality,

x y
kx + yk ≤ kxk + kyk − 2 − kxk kyk min{kxk, kyk}
+
2ε ε
≤2− =2 1− .
2 2
Similarly, if
x y x y
kxk + kyk ≥ kxk − kyk ,

then

x y
kx − yk ≤ kxk + kyk − 2 −
− min{kxk, kyk}
kxk kyk
2ε ε
≤2− =2 1− .
2 2
Hence min{kx + yk, kx − yk} ≤ 2(1 − ε/2). Put ε0 = min{ε/2, 1/4}, then we have
(ii). The opposite implication is clear.
In [19], Kato et al. mentioned a direct application of (45) of Theorem 2.2.1 to

a uniformly non-`n1 Banach space. A Banach space X is called uniformly non-`n1
provided there exists ε (0 < ε < 1) such that for any x1 , . . . , xn ∈ SX , there exists
θ = (θj ) of n signs ±1 for which

Xn
θj j ≤ n(1 − ε).
x (65)

j=1
When n = 2, X is uniformly non-square (cf. [17] and [5]).
Theorem 2.4.4 (cf. [19, Corollary 4]). For a Banach space X the following condi-
tions are equivalent:
(i) X is uniformly non-`n1 .
68
(ii) There exists ε (0 < ε < 1) such that for any x1 , . . . , xn ∈ BX , there exists
θ = (θj ) of n signs ±1 for which (65) holds true.
Proof. Assume (i). Then there exists ε (0 < ε < 1) such that for any x1 , . . . , xn ∈
SX , there exists θ = (θj ) of n signs ±1 for which (65) is valid. Take x1 , . . . , xn ∈ BX .
If kxj0 k := min{kx1 k, . . . , kxn k} ≤ 1/2, we have
n
X X 1 1
θj xj ≤ kxj k + kxj0 k ≤ (n − 1) + ≤ n 1 − .

2 2n

j=1 j6=j
0
Let kxj0 k ≥ 1/2. According to our assumption there exists n signs (θj ) for which
(65) is valid for x1 /kx1 k, . . . , xn /kxn k. Therefore by the first inequality of Theorem
2.2.1,
!
n n n
X X X x j

θ x ≤ kxj k − n− θj kxj0 k

j j
kxj k

j=1 j=1 j=1
nε
≤2−
2 ε
=n 1− .
2
Consequently by letting ε0 = min{ε/2, 1/(2n)} we have the conclusion.
69
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73

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On generalizations and refinements

Doctoral Program in Fundamental Sciences

Graduate School of Science and Technology

1 On generalized triangle inequalities 6

2 On sharp triangle inequalities 41

kx + yk ≤ kxk + kyk (x, y ∈ X)

kx + yk2 + kx − yk2 = 2 kxk2 + kyk2

This implies that the parallelogram inequality

kx + yk2 ≤ 2 kxk2 + kyk2

kx + ykq ≤ 2q−1 kxkq + kykq

hold, where x0 = xn+1 = 0. In the case of n = 3, above inequalities are following:

≤ kxk + kyk + kzk

hold. They also studied equality attainedness on these inequalities in a strictly

by using x1 , x2 , . . . , xn in a Banach space X. Our main aims in Section 2 are to give

{λ > 0, µ > 0, ν > 0}

(i) Dp+ ∩ {λ > 0, µ > 0, ν > 0}

(ii) Dp+ ∩ {λ < 0, µ < 0, ν > 0}

(iii) Dp+ ∩ {λ < 0, µ > 0, ν < 0}

(v) Dp+ ∩ {λ > 0, µ > 0, ν < 0} = ∅.

(vi) Dp+ ∩ {λ > 0, µ < 0, ν > 0} = ∅.

(vii) Dp+ ∩ {λ > 0, µ < 0, ν < 0} = ∅.

(viii) Dp+ ∩ {λ < 0, µ > 0, ν > 0} = C × C × R3 \ {λµν = 0}.

D1+ = {λ < 0, µ > 0, ν > 0}

In [9], Dadipour et al. characterized all (µ1 , . . . , µn ) ∈ Rn which satisfy a special

1.2 Absolute normalized norms

k(z1 , z2 , . . . , zn )k = k(|z1 |, |z2 |, . . . , |zn |)k ((z1 , z2 , . . . , zn ) ∈ Cn )

k(1, 0, . . . , 0)k = k(0, 1, 0, . . . , 0)k = · · · = k(0, . . . , 0, 1)k = 1.

Proposition 1.2.1 (cf. [35, Lemma 3.1]). Let k · k ∈ ANn . Then

k(0, z2 , z3 , . . . , zn )k ≤ k(z1 , z2 , . . . , zn )k (B1 )

Proof. For any (z1 , z2 , . . . , zn ) ∈ Cn , we have

k(z1 , z2 , . . . , zn )k∞ = max{|z1 |, |z2 |, . . . , |zn |}

This completes the proof.

We remark that every absolute norm is monotone:

k(z1 , . . . , zn )k ≤ k(w1 . . . , wn )k.

Proof. We may suppose that zj , wj ≥ 0 (j = 1, . . . , n). Since z1 ≤ w1 , there exists

k(z1 , z2 , . . . , zn−1 , zn )k ≤ k(w1 , w2 , . . . , wn−1 , wn )k.

This completes the proof.

max{1 − t, t} ≤ ψ(t) ≤ 1 (t ∈ [0, 1]).

Let k · k ∈ AN2 and let

ψ(t) = k(1 − t, t)k (t ∈ (0, 1)). (7)

Then ψ ∈ Ψ2 . Conversely, for given ψ ∈ Ψ2 define

ψ(1, 0, . . . , 0) = ψ(0, 1, 0, . . . , 0) = · · · = ψ(0, . . . , 0, 1) = 1 (A0 )

Then ANn and Ψn are in one-to-one correspondence as follows.

Theorem 1.2.3 (cf. [35, Theorems 3.4]).

ψ(t1 , · · · , tn ) = k(t1 , . . . , tn )k ((t1 , . . . , tn ) ∈ ∆n ). (8)

Then k · kψ ∈ ANn , and k · kψ satisfies (8).

Lemma 1.2.4 (cf. [35, Lemma 3.2]). Let ψ ∈ Ψn . Then

Next, from (A0 ), (A1 ), (A3 ), . . . , (An ), we have

Similarly, we can show that for 3 ≤ i ≤ n − 1,

Then for any 1 ≤ i ≤ n, we have

k(p1 , . . . , pn )kψ ≤ k(a1 , . . . , an )kψ . (10)

Proof. We first show that, if 0 ≤ p1 < a1 , then

k(p1 , p2 , . . . , pn )kψ ≤ k(a1 , p2 . . . , an )kψ . (11)

This is, we show that if 0 ≤ p1 < a1

(t01 ,t02 , . . . , t0n )

ψ(t01 , t02 , . . . , t0n )

This implies (11). Similarly, we can show that for 2 ≤ i ≤ n,

k(a1 , . . . , ai−1 , pi , pi+1 , . . . , pn )kψ ≤ k(a1 , . . . , ai−1 , ai , pi+1 , . . . , pn )kψ .