Ann. Funct. Anal. 2 (2011), no. 2, 22–33
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL: www.emis.de/journals/AFA/
SOME GEOMETRIC CONSTANTS OF ABSOLUTE
NORMALIZED NORMS ON R2
HIROYASU MIZUGUCHI AND KICHI-SUKE SAITO∗
Communicated by J. I. Fujii
Abstract. We consider the Banach space X = (R2 , k · k) with a normalized,
absolute norm. Our aim in this paper is to calculate the modified Neumann0
Jordan constant CN
J (X) and the Zbăganu constant CZ (X).
1. Introduction and preliminaries
Let X be a Banach space with the unit ball BX = {x ∈ X : kxk ≤ 1} and the
unit sphere SX = {x ∈ X : kxk = 1}. Many geometric constants for a Banach
space X have been investigated. In this paper we shall consider the following
constants;
kx + yk2 + kx − yk2 CN J (X) = sup
(x, y) 6= (0, 0) ,
2(kxk2 + kyk2 )
kx + yk2 + kx − yk2 0
CN J (X) = sup
x, y ∈ SX ,
4
kx + ykkx − yk CZ (X) = sup
x, y ∈ X, (x, y) 6= (0, 0) .
kxk2 + kyk2 The constant CN J (X), called the von Neumann-Jordan constant (hereafter referred to as NJ constant) have been considered in many papers ([3, 8, 10, 12]
and so on). The constant CN0 J (X), called the modified von Neumann-Jordan
constant (shortly, modified NJ constant) was introduced by Gao in [5] and does
not necessarily coincide with CN J (X) (cf. [1, 4, 7]). The constant CZ (X) was
introduced by Zbăganu ([15]) and was conjectured that CZ (X) coincides with
Date: Received: 31 March 2011; Accepted: 14 June 2011.
∗
Corresponding author.
2010 Mathematics Subject Classification. Primary 46B20; Secondary 46B25.
Key words and phrases. Zbăganu constant, absolute norm, von Neumann-Jordan constant.
22
GEOMETRIC CONSTANTS OF R2
23
the von Neumann-Jordan constant CN J (X), but Alonso and Martin [2] gave an
example that CN J (X) 6= CZ (X) (cf.[6, 9]).
A norm k · k on R2 is said to be absolute if k(a, b)k = k(|a|, |b|)k for any
(a, b) ∈ R2 , and normalized if k(1, 0)k = k(0, 1)k = 1. Let AN2 denote the family
of all absolute normalized norm on R2 , and Ψ2 denote the family of all continuous
convex function ψ on [0, 1] such that ψ(0) = ψ(1) = 1 and max{1 − t, t} ≤ ψ(t) ≤
1 for all 0 ≤ t ≤ 1. As in [11], it is well known that AN2 and Ψ2 are in a one-toone correspondence under the equation ψ(t) = k(1 − t, t)k (0 ≤ t ≤ 1). Denote
k · kψ be an absolute normalized norm associated with a convex function ψ ∈ Ψ2 .
For ψ, ϕ ∈ Ψ2 , we denote ψ ≤ ϕ if ψ(t) ≤ ϕ(t) for any t in [0, 1]. Let
ψ(t)
ψ2 (t)
and M2 = max
,
0≤t≤1
ψ2 (t)
ψ(t)
p
where ψ2 (t) = k(1 − t, t)k2 =
(1 − t)2 + t2 corresponds to the l2 -norm. In
[11], Saito, Kato and Takahashi proved that, if ψ ≥ ψ2 (resp. ψ ≤ ψ2 ), then
CN J (C2 , k · kψ ) = M12 (resp. M22 ).
We put X = (R2 , k · kψ ) for ψ ∈ Ψ2 . Our aim in this paper is to consider the
conditions of ψ that CN J (X) = CZ (X) or CN J (X) = CN0 J (X).
In §2, we consider the modified von Neumann-Jordan constant. We prove
that if ψ ≤ ψ2 , then CN0 J (X) = CN J (X) = M22 . If ψ ≥ ψ2 , then we present the
necessarily and sufficient condition that CN0 J (R2 , k · kψ ) = CN J (R2 , k · kψ ) = M12 .
Further, we consider the conditions that CN0 J (R2 , k · kψ ) = CN J (R2 , k · kψ ) =
M12 M22 . In §3, we study the Zbăganu constant. First, we show that, if ψ ≥ ψ2 ,
then CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) = M12 . If ψ ≤ ψ2 , then we give the
necessarily and sufficient condition for that CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) =
M22 . Further we study the conditions that CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) =
M12 M22 . In §4, we calculate the modified NJ-constant CN0 J (X) and the Zbăganu
constant CZ (X) for some normed liner spaces.
M1 = max
0≤t≤1
2. The modified NJ constant of R2
In this section, we consider the Banach space X = (R2 , k · kψ ). From the
definition of the modified NJ constant, it is clear that CN0 J (X) ≤ CN J (X). In
this section, we consider the condition that CN0 J (X) = CN J (X).
Proposition 2.1. Let ψ ∈ Ψ2 . If ψ ≤ ψ2 , then CN0 J (X) = CN J (X) = M22 .
Proof. For any x, y ∈ SX , by [11, Lemma 3],
kx + yk2ψ + kx − yk2ψ ≤ kx + yk22 + kx − yk22
= 2 kxk22 + kyk22
≤ 2M22 kxk2ψ + kyk2ψ = 4M22 .
Now let ψ2 /ψ attain the maximum at t = t0 (0 ≤ t0 ≤ 1), and put
x=
1
1
(1 − t0 , t0 ), y =
(1 − t0 , −t0 ).
ψ(t0 )
ψ(t0 )
24
H. MIZUGUCHI, K.-S. SAITO
Then x, y ∈ SX and
4(1 − t0 )2 + 4t20
ψ(t0 )2
ψ2 (t0 )2
=4
= 4M22 ,
2
ψ(t0 )
kx + yk2ψ + kx − yk2ψ =
which implies that CN0 J (X) = M22 . By [11, Theorem 1], we have this proposition.
If ψ ≥ ψ2 , by [11, Theorem 1], then CN J (X) = M12 . We now give the necessarily
and sufficient condition of CN0 J (X) = M12 .
Theorem 2.2. Let ψ ∈ Ψ2 such that ψ ≥ ψ2 . Then CN0 J (X) = M12 if and only
if there exist s, t ∈ [0, 1] (s < t) satisfying one of the following conditions:
(1) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r = ψ(s)t+ψ(t)s
, then ψψ(r)
=
ψ(s)+ψ(t)
2 (r)
ψ(1−r)
ψ2 (1−r)
= M1 .
(2) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r =
ψ(1−r)
ψ2 (1−r)
ψ(t)s+ψ(s)t
,
ψ(t)+ψ(s)(2t−1)
then
ψ(r)
ψ2 (r)
=
= M1 .
Proof. (=⇒) Suppose that CN0 J (X) = M12 . First, for any x, y ∈ SX , by [11,
Lemma 3], we have
kx + yk2ψ + kx − yk2ψ ≤ M12 (kx + yk22 + kx − yk22 )
= 2M12 kxk22 + kyk22
≤ 2M12 kxk2ψ + kyk2ψ = 4M12 .
Since X = (R2 , k · kψ ) is finite dimensional,
kx + yk2ψ + kx − yk2ψ
0
CN J (X) = max
4
x, y ∈ SX .
Therefore, CN0 J (X) = M12 if and only if there exist x, y ∈ SX (x 6= y) such that
kx + yk2ψ + kx − yk2ψ = 4M12 .
From the above inequality, the elements x, y ∈ SX (x 6= y) satisfy kxkψ = kxk2 =
1, kykψ = kyk2 = 1 and
kx + ykψ
kx − ykψ
=
= M1 .
kx + yk2
kx − yk2
Since k · kψ is absolute and x, y ∈ SX (x 6= y) satisfy kxk2 = kyk2 = 1, it is
sufficient to consider the following three cases:
(i) There exist s, t ∈ [0, 1] (s 6= t) satisfying x = ψ21(s) (1 − s, s) and y =
1
(1 − t, t).
ψ2 (t)
(ii) There exist s, t ∈ [0, 1] (s < t) satisfying x = ψ21(s) (1 − s, s) and y =
1
(−1 + t, t).
ψ2 (t)
GEOMETRIC CONSTANTS OF R2
25
(iii) There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ21(s) (1 − s, s) and y =
1
(−1 + t, t).
ψ2 (t)
Case (i). We may suppose that s < t. Then there exist α, β ∈ [0, π2 ] (α < β) such
that
1
1
x=
(1 − s, s) = (cos α, sin α), y =
(1 − t, t) = (cos β, sin β).
ψ2 (s)
ψ2 (t)
Since kxk2 = kyk2 = 1, we have
x+y =(
s
t
α+β
α+β
1−s 1−t
+
,
+
) = ||x + y||2 (cos
, sin
).
ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t)
2
2
By [13, Propositions 2a and 2b], we remark that
1−s
1−t
s
t
≥
,
≤
.
ψ2 (s)
ψ2 (t) ψ2 (s)
ψ2 (t)
Since x − y is orthogonal to x + y in the Euclidean space (R2 , k · k2 ), we have
1−s
1−t
s
t
x−y =(
−
,
−
)
ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t)
α+β−π
α+β−π
= ||x − y||2 (cos
, sin
)
2
2
α+β
α+β
= ||x − y||2 (sin
, − cos
).
2
2
Thus we have
α+β
α+β
, sin
)||ψ
||x + y||ψ = ||x + y||2 ||(cos
2
2
sin α+β
α+β
α+β
2
= ||x + y||2 (cos
+ sin
)ψ(
).
2
2
cos α+β
+
sin α+β
2
2
Since ||x + y||ψ = M1 ||x + y||2 , we have
sin α+β
α+β
α+β
2
M1 = (cos
+ sin
)ψ(
).
α+β
2
2
cos 2 + sin α+β
2
Putting r =
We also have
cos
sin α+β
2
α+β
+sin α+β
2
2
, then it is clear that r =
||x − y||ψ = ||x − y||2 (sin
ψ(s)t+ψ(t)s
ψ(s)+ψ(t)
and M1 =
ψ(r)
.
ψ2 (r)
cos α+β
α+β
α+β
2
+ cos
)ψ(
).
α+β
2
2
cos 2 + sin α+β
2
Since ||x − y||ψ = M1 ||x − y||2 , we similarly have
M1 = (sin
cos α+β
α+β
α+β
ψ(1 − r)
2
+ cos
)ψ(
)=
.
α+β
α+β
2
2
ψ2 (1 − r)
sin 2 + cos 2
Case (ii). Then there exist α ∈ [0, π2 ] and β ∈ [ π2 , π] such that
x=
1
1
(1 − s, s) = (cos α, sin α), y =
(−1 + t, t) = (cos β, sin β).
ψ2 (s)
ψ2 (t)
26
H. MIZUGUCHI, K.-S. SAITO
Since kxk2 = kyk2 = 1, we have
x+y =(
1−s
1−t
s
t
α+β
α+β
−
,
+
) = ||x + y||2 (cos
, sin
).
ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t)
2
2
By [13, Propositions 2a and 2b], we remark that
1−t
s
t
1−s
≥
,
≤
.
ψ2 (s)
ψ2 (t) ψ2 (s)
ψ2 (t)
Since x − y is orthogonal to x + y in the Euclidean space (R2 , k · k2 ), we have
1−s 1−t
s
t
x−y =(
+
,
−
)
ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t)
α+β−π
α+β−π
= ||x − y||2 (cos
, sin
)
2
2
α+β
α+β
= ||x − y||2 (sin
, − cos
).
2
2
Since cos α+β
≥ 0 and sin α+β
≥ 0, we have
2
2
α+β
α+β
, sin
)||ψ
2
2
sin α+β
α+β
α+β
2
= ||x + y||2 (cos
+ sin
)ψ(
).
2
2
cos α+β
+
sin α+β
2
2
||x + y||ψ = ||x + y||2 ||(cos
Since ||x + y||ψ = M1 ||x + y||2 , we have
sin α+β
α+β
α+β
2
+ sin
)ψ(
M1 = (cos
).
α+β
2
2
cos 2 + sin α+β
2
Putting r =
cos
We also have
sin α+β
2
α+β
α+β
+sin
2
2
, then it is clear that r =
||x − y||ψ = ||x − y||2 (sin
ψ(t)s+ψ(s)t
ψ(t)+ψ(s)(2t−1)
and M1 =
ψ(r)
.
ψ2 (r)
cos α+β
α+β
α+β
2
).
+ cos
)ψ(
α+β
2
2
cos 2 + sin α+β
2
Since ||x − y||ψ = M1 ||x − y||2 , we similarly have
M1 = (sin
cos α+β
α+β
α+β
ψ(1 − r)
2
+ cos
)ψ(
)=
.
α+β
α+β
2
2
ψ2 (1 − r)
sin 2 + cos 2
Case (iii). There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ21(s) (1 − s, s) and
y = ψ21(t) (−1 + t, t). Then, we put s0 = t and t0 = s. We define x0 , y0 in SX by
x0 =
1
1
(1 − s0 , s0 ), y0 =
(−1 + t0 , t0 ).
ψ(s0 )
ψ(t0 )
Then we can reduce Case (ii).
(⇐=). If we suppose (1) (resp. (2)), then we put x = ψ21(s) (1 − s, s) (resp.
x = ψ21(s) (1 − s, s)) and y = ψ21(t) (1 − t, t) (resp. y = ψ21(t) (−1 + t, t)). Then
we have ||x||ψ = ||x||2 = 1, ||y||ψ = ||y||2 = 1, ||x + y||ψ = M1 ||x + y||2 and
GEOMETRIC CONSTANTS OF R2
27
||x − y||ψ = M1 ||x − y||2 . Hence it is clear to prove that CN0 J (X) = M12 . This
completes the proof.
We next study the modified NJ constant in the general case. If ψ ∈ Ψ, then
by [11, Therem 3], we have
max{M12 , M22 } ≤ CN J (X) ≤ M12 M22 .
However, by Theorem 2.2, there exist many ψ ∈ Ψ satisfying ψ ≥ ψ2 such that
CN0 J (X) < max{M12 , M22 } = CN J (X).
From [11, Theorem 3], CN J (X) = M12 M22 if either ψ/ψ2 or ψ2 /ψ attains a maximum at t = 1/2. Then, we have the following
Proposition 2.3. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1 − t) for all t ∈ [0, 1]. If ψ/ψ2
attains a maximum at t = 1/2, then CN0 J (X) = CN J (X) = M12 M22 .
Proof. Suppose first M1 = ψ(1/2)/ψ2 (1/2). Take an arbitrary t ∈ [0, 1] and put
x=
1
1
(t, 1 − t) , y =
(1 − t, t).
ψ(t)
ψ(t)
Then x, y ∈ SX and
kx + ykψ =
2
1
2|2t − 1| 1
ψ( ) , kx − ykψ =
ψ( ).
ψ(t) 2
ψ(t)
2
Therefore we have
ψ(1/2)2
kx + yk2ψ + kx − yk2ψ = (2t − 1)2 + 1
4
ψ(t)2
2
2 ψ(1/2)
= 2ψ2 (t)
ψ(t)2
2
ψ2 (t)2 ψ(1/2)2
2 ψ2 (t)
=
=
M
.
1
ψ(t)2 ψ2 (1/2)2
ψ(t)2
Since t is arbitrary, we have CN0 J (X) ≥ M12 M22 which prove that CN0 J (X) =
M12 M22 .
In the case that M2 = ψ2 (1/2)/ψ(1/2), CN0 J (X) does not necessarily coincide
with M12 M22 . However, we have the following
Theorem 2.4. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1 − t) for all t ∈ [0, 1]. Assume that
M2 = ψ2 (1/2)/ψ(1/2) and M1 > 1. Then CN0 J (X) = M12 M22 if and only if there
exist s, t ∈ [0, 1] (s < t) satisfying one of the following conditions:
, then
(1) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(s)t+ψ(t)s
ψ(s)+ψ(t)
ψ(r) = M1 ψ2 (r).
ψ(t)s+ψ(s)t
, then
(2) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(t)+ψ(s)(2t−1)
ψ(r) = M1 ψ2 (r).
28
H. MIZUGUCHI, K.-S. SAITO
Proof. (=⇒). For all x, y ∈ SX , we have
kx + yk2ψ + kx − yk2ψ ≤ M12 kx + yk22 + kx − yk22
= 2M12 kxk22 + kyk22
≤ 2M12 M22 kxk2ψ + kyk2ψ = 4M12 M22 .
From this inequality, CN0 J (X) = M12 M22 if and only if there exist x, y ∈ SX (x 6=
y) such that
kx + yk2ψ + kx − yk2ψ = 4M12 M22 .
Suppose that CN0 J (X) = M12 M22 . Then, the elements x, y ∈ SX (x 6= y) satisfy
kxk2 = kyk2 = M2 , kx + ykψ = M1 kx + yk2 , kx − ykψ = M1 kx − yk2 .
Since k · kψ is absolute, it is sufficient to consider the following three cases:
1
(1 − s, s) and y =
(i) There exist s, t ∈ [0, 1] (s 6= t) satisfying x = ψ(s)
1
(1 − t, t).
ψ(t)
1
(ii) There exist s, t ∈ [0, 1] (s < t) satisfying x = ψ(s)
(1 − s, s) and y =
1
(−1 + t, t).
ψ(t)
1
(iii) There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ(s)
(1 − s, s) and y =
1
(−1 + t, t).
ψ(t)
As in the proof of Theorem 2.2, we can prove this theorem. This completes
the proof.
3. The Zbăganu constant of R2
The Zbăganu constant CZ (X) in [15] is defined by
kx + ykkx − yk CZ (X) = sup
x, y ∈ X, (x, y) 6= (0, 0) .
kxk2 + kyk2 Then it is clear that CZ (X) ≤ CN J (X) for any Banach space X. In this section,
we consider the condition that CZ (X) = CN J (X) for X = (R2 , k · kψ ). Then, we
have the following
Proposition 3.1. Let ψ ∈ Ψ2 . If ψ ≥ ψ2 , then CZ (X) = CN J (X) = M12 .
Proof. For any x, y ∈ X,
2kx + ykψ kx − ykψ ≤ kx + yk2ψ + kx − yk2ψ
≤ M12 kx + yk22 + kx − yk22
= 2M12 kxk22 + kyk22
≤ 2M12 kxk2ψ + kyk2ψ .
Since ψ/ψ2 attains the maximum at t = t0 (0 ≤ t0 ≤ 1), we put x = (1 − t0 , 0)
and y = (0, t0 ), respectively. Then we have
kx + yk2ψ + kx − yk2ψ = 2ψ(t0 )2
= 2M12 ψ2 (t0 )2
= 2M12 kxk2ψ + kyk2ψ .
GEOMETRIC CONSTANTS OF R2
29
Since kx + ykψ = ψ(t0 ) = kx − ykψ , we have
2kx + ykψ kx − ykψ = kx + yk2ψ + kx − yk2ψ
= 2M12 kxk2ψ + kyk2ψ .
Therefore we have
kx + ykψ kx − ykψ
= M12 ,
kxk2ψ + kyk2ψ
which implies that CZ (X) = M12 .
We next consider the case that ψ ≤ ψ2 . We remark that the Zbăganu constant
CZ (X) is in the following form;
4kxkkyk
CZ (X) = sup
x, y ∈ X, (x, y) 6= (0, 0) .
kx + yk2 + kx − yk2 Then we have the following
Theorem 3.2. Let ψ ∈ Ψ2 . Assume that ψ ≤ ψ2 . Then CZ (X) = M22 if and
only if there exist s, t ∈ [0, 1] (s < t) satisfying one of the following conditions:
2 (r)
(1) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r = ψ(s)t+ψ(t)s
, then ψψ(r)
=
ψ(s)+ψ(t)
ψ(1−r)
ψ2 (1−r)
= M2 .
(2) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r =
ψ(1−r)
ψ2 (1−r)
ψ(t)s+ψ(s)t
,
ψ(t)+ψ(s)(2t−1)
then
= M2 .
Proof. For any x, y ∈ X,
4kxkψ kykψ ≤ 2 kxk2ψ + kyk2ψ
≤ 2 kxk22 + kyk22
= kx + yk22 + kx − yk22
≤ M22 kx + yk2ψ + kx − yk2ψ .
Since X = (R2 , k · kψ ) is finite dimensional,
(
4kxkψ kykψ
CZ (X) = max
kx + yk2ψ + kx − yk2ψ
)
x, y ∈ X, (x, y) 6= (0, 0) .
Then CZ (X) = M22 if and only if there exist x, y ∈ SX (x 6= y) such that
4kxkψ kykψ
= M22 .
2
2
kx + ykψ + kx − ykψ
From the above inequality, kxk2 = kxkψ = kykψ = kyk2 and
kx + yk2
kx − yk2
=
= M22 .
kx + ykψ
kx − ykψ
Hence we may assume that
kxk2 = kxkψ = kykψ = kyk2 = 1.
ψ2 (r)
ψ(r)
=
30
H. MIZUGUCHI, K.-S. SAITO
As in the proof of Theorem 2.2, it is sufficient to consider the following three
cases:
(i) There exist s, t ∈ [0, 1] (s 6= t) satisfying x = ψ21(s) (1 − s, s) and y =
1
(1 − t, t).
ψ2 (t)
(ii) There exist s, t ∈ [0, 1] (s < t) satisfying x = ψ21(s) (1 − s, s) and y =
1
(−1 + t, t).
ψ2 (t)
(iii) There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ21(s) (1 − s, s) and y =
1
(−1 + t, t).
ψ2 (t)
As in the proof of Theorem 2.2, we can similarly prove this theorem.
We next study the Zbăganu constant CZ (X) in general case. If ψ ∈ Ψ, by [11,
Theorem 3], then we have
max{M12 , M22 } ≤ CZ (X) ≤ CN J (X) ≤ M12 M22 .
However, by Theorem 3.2, there exist many ψ ∈ Ψ satisfying ψ ≥ ψ2 such that
CZ (X) < CN J (X) ≤ max{M12 , M22 }.
From [11, Theorem 3], CN J (X) = M12 M22 if either ψ/ψ2 or ψ2 /ψ attains a maximum at t = 1/2. Then, we have the following
Proposition 3.3. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1 − t) for all t ∈ [0, 1]. If
2 (1/2)
M2 = ψψ(1/2)
, then CZ (X) = CN J (X) = M12 M22 .
Proof. From the definition, we have CZ (X) ≤ CN J (X) = M12 M22 . Take an arbitrary t ∈ [0, 1] and put x = (t, 1 − t) and y = (1 − t, t). Then kxkψ = kykψ = ψ(t)
and kx + ykψ = k(1, 1)kψ = 2ψ(1/2), kx − ykψ = k(2t − 1, 1 − 2t)kψ = 2|2t −
1|ψ(1/2). Hence we have
2 kxk2ψ + kyk2ψ
4kxkψ kykψ
=
kx + yk2ψ + kx − yk2ψ
kx + yk2ψ + kx − yk2ψ
ψ(t)2
(1 + (2t − 1)2 ) ψ(1/2)2
ψ(t)2
=
2ψ2 (t)2 ψ(1/2)2
2
ψ(t)2 ψ2 (1/2)2
2 ψ(t)
=
=
M
2
ψ2 (t)2 ψ(1/2)2
ψ2 (t)2
=
Since t is arbitrary, we have CZ (X) ≥ M12 M22 . Therefore we have CZ (X) =
M12 M22 . This completes the proof.
In case that M1 = ψ(1/2)/ψ2 (1/2), we have the following theorem as in the
proof of Theorem 2.2 and so omit the proof.
Theorem 3.4. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1−t) for all t ∈ [0, 1]. If M1 = ψψ(1/2)
2 (1/2)
and M2 > 1, then CZ (X) = M12 M22 if and only if there exist s, t ∈ [0, 1] (s < t)
satisfying one of the following conditions:
GEOMETRIC CONSTANTS OF R2
31
(1) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(s)t+ψ(t)s
, then
ψ(s)+ψ(t)
ψ(r) = M1 ψ2 (r).
ψ(t)s+ψ(s)t
(2) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(t)+ψ(s)(2t−1)
, then
ψ(r) = M1 ψ2 (r).
4. Examples
In this section, we calculate CN0 J (X) and CZ (X) of some Banach spaces X =
(R2 , k · kψ ), where ψ ∈ Ψ. First, we consider the case that ψ = ψp .
Example 4.1. Let 1 ≤ p ≤ ∞ and 1/p + 1/q = 1. We put t = min(p, q). Then
2
CN0 J (R2 , k · kp ) = CZ (R2 , || · ||p ) = CN J (R2 , k · kp ) = 2 t −1 .
2
Suppose that 1 ≤ p ≤ 2. Since ψp ≥ ψ2 , we have CZ (R2 , || · ||p ) = 2 p −1 by
Proposition 3.1. On the other hand, as in Theorem 2.2, we take s = 0 and t = 1.
1
− 12
1
p
Since r = ψ(0)·1+ψ(1)·0
=
and
M
=
ψ
(1/2)/ψ
(1/2)
=
2
, by Theorem 2.2,
1
p
2
ψ(0)+ψ(1)
2
2
we have CN0 J (R2 , k · kp ) = M12 = 2 p −1 .
If 2 ≤ p ≤ ∞, then we similarly have, by Proposition 2.1 and Theorem 3.2,
2
0
CN J (R2 , k · kp ) = CZ (R2 , || · ||p ) = CN J (R2 , k · kp ) = 2 p −1 .
In [14, Example], C. Yang and H. Li calculated the modified NJ constant of
the following normed linear space. From our theorems, we have
Example 4.2. Let λ > 0 and Xλ = R2 endowed with norm
||(x, y)||λ = (||(x, y)||2p + λ||(x, y)||2q )1/2 .
(i) If 2 ≤ p ≤ q ≤ ∞, then CN J (Xλ ) = CN0 J (Xλ ) = CZ (Xλ ) =
2(λ+1)
.
22/p +λ22/q
2/p
2/q
2 +λ2
.
2(λ+1)
(ii) If 1 ≤ p ≤ q ≤ 2, then CN J (Xλ ) = CN0 J (Xλ ) = CZ (Xλ ) =
To see this, first, we remark that (p, q) is not necessarily a Hölder pair. We
define the normalized norm || · ||0λ by
||(x, y)||λ
||(x, y)||0λ = √
.
1+λ
Then ||·||0λ is absolute and so put the corresponding function ψλ (t) = ||(1−t, t)||0λ .
(i) Suppose that 2 ≤ p ≤ q ≤ ∞. Since ψλ ≤ ψ2 , by Proposition 2.1, we have
2(λ+1)
CN J (Xλ ) = CN0 J (Xλ ) = M22 = 22/p
. On the other hand, in Theorem 3.2, we
+λ22/q
take s = 0 and t = 1. Then we have r = 1/2 and
M22
ψ2 (1/2)
ψλ (1/2)
= M2 . Thus we have
22/p +λ22/q
.
2(λ+1)
CZ (Xλ ) =
=
(ii) Suppose that 1 ≤ p ≤ q ≤ 2. Since ψλ ≥ ψ2 , by Theorem 2.2 and Proposition
3.1, we similarly have (ii).
Example 4.3. Put


ψ2 (t)
(0 ≤ t ≤ 1/2),
ψ(t) =
 2 − √2 t + √2 − 1 (1/2 ≤ t ≤ 1).
√ √
Then CN0 J (R2 , k · kψ ) < CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) = 2 2( 2 − 1).
32
H. MIZUGUCHI, K.-S. SAITO
In fact, ψ ∈ Ψ2 and the norm of k · kψ is

p

|a|2 + |b|2
(|a| ≥ |b|)
k(a, b)kψ =
 √
2 − 1 |a| + |b| (|a| ≤ |b|) .
√ √
Since ψ ≥ ψ2 , by Proposition 3.1, we have CZ (R2 , k · kψ ) = M12 = 2 2 2 − 1 .
We assume that CN0 J (R2 , k · kψ ) = M12 . By Theorem 2.2, we can choose r ∈
[0, 1] such that ψψ(r)
= ψψ(1−r)
= M1 . This is impossible by the definition of ψ.
2 (r)
2 (1−r)
0
2
Therefore we have CN J (R , k · kψ ) < M12 .
Example 4.4. Let 1/2 ≤ β ≤ 1. We define a convex function ψβ ∈ Ψ2 by
ψβ (t) = max{1 − t, t, β}.
By [11, Example 4], we have
 2
2
 β + (1 − β)
(β ∈ [ 12 , √12 ])
2
2
β
CN J (R , k · kψβ ) =
 2
2(β + (1 − β)2 ) (β ∈ ( √12 , 1]).
Indeed,
M1 =
(
1
ψβ (1/2)
ψ2 (1/2)
=
β√
1/ 2
=
√
(β ∈ [ 12 , √12 ])
2β
(β ∈ ( √12 , 1])
and
ψ2 (β)
1
= {(1 − β)2 + β 2 }1/2 .
ψβ (β)
β
√
If 1/2 ≤ β ≤ 1/ 2, then ψβ ≤ ψ2 and so, by Proposition 2.1, we have
M2 =
β 2 + (1 − β)2
.
β2
By Theorem 3.2, we have CZ (R2 , k · kψβ ) < M22 .
√
ψ (1/2)
Assume that 1/ 2 < β ≤ 1. Since M1 = ψβ2 (1/2) , we have, by Proposition 2.3,
CN0 J (R2 , k · kψβ ) = M22 =
CN0 J (R2 , k · kψβ ) = M12 M22 = 2(β 2 + (1 − β)2 ).
On the other hand, we take s = β and t = 1 − β in Theorem 3.4. Then we have
= 1/2. By Theorem 3.4, we have
r = ψ(β)(1−β)+ψ(1−β)β
ψ(β)+ψ(1−β)
CZ (R2 , k · kψβ ) = M12 M22 = 2(β 2 + (1 − β)2 ).
√
Example 4.5. We consider ψβ in Example 4.4 in case of β = 1/ 2. Then we
have
√ √
CN0 J (R2 , k · kψβ ) = CN J (R2 , k · kψβ ) = M22 = 2 2( 2 − 1).
√ √
On the other hand, we have CZ (R2 , k · kψβ ) = M22 = 2 2( 2 − 1).
For this ψβ , define a convex function ϕ ∈ Ψ2 by
(
ψβ (t) (0 ≤ t ≤ 1/2),
ϕ(t) =
ψ2 (t) (1/2 ≤ t ≤ 1).
GEOMETRIC CONSTANTS OF R2
33
As in Example 4.2, we similarly have
√ √
CZ (R2 , k · kϕ ) < CN0 J (R2 , k · kϕ ) = CN J (R2 , k · kϕ ) = M22 = 2 2
2−1 .
Acknowledgement. The second author is supported in part by Grants-inAid for Scientific Research, Japan Society for the Promotion of Science (No.
23540189). The authors would also like to thank the referees for some helpful
comments.
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Department of Mathematical Sciences, Graduate School of Science and Technology, Niigata University, Niigata, 950-2181 Japan.
E-mail address: mizuguchi@m.sc.niigata-u.ac.jp
E-mail address: saito@math.sc.niigata-u.ac.jp