E extracta mathematicae Vol. 32, Núm. 2, 173 – 208 (2017)
The Differences Between Birkhoff and Isosceles
Orthogonalities in Radon Planes
Hiroyasu Mizuguchi
Student Affairs Department-Shinnarashino Educational Affairs Section,
Chiba Institute of Technology, Narashino, Japan
hiroyasu.mizuguchi@p.chibakoudai.jp
Presented by Javier Alonso Received May 22, 2017
Abstract: The notion of orthogonality for vectors in inner product spaces is simple, interest-
ing and fruitful. When moving to normed spaces, we have many possibilities to extend this
notion. We consider Birkhoff orthogonality and isosceles orthogonality. Recently the con-
stants which measure the difference between these orthogonalities have been investigated.
The usual orthognality in inner product spaces and isosceles orthogonality in normed spaces
are symmetric. However, Birkhoff orthogonality in normed spaces is not symmetric in gen-
eral. A two-dimensional normed space in which Birkhoff orthogonality is symmetric is called
a Radon plane. In this paper, we consider the difference between Birkhoff and isosceles or-
thogonalities in Radon planes.
Key words: Birkhoff orthogonality, Isosceles orthogonality, Minkowski plane, Minkowski ge-
ometry, Radon plane.
AMS Subject Class. (2010): 46B20, 51B20, 52A21, 26D20.
1.. Introduction
We denote by X a real normed space with the norm ∥ · ∥, the unit ball
BX and the unit sphere SX. Throughout this paper, we assume that the
dimension of X is at least two. In case of that X is an inner product space,
an element x ∈ X is said to be orthogonal to y ∈ X (denoted by x ⊥ y)
if the inner product ⟨x,y⟩ is zero. In the general setting of normed spaces,
many notions of orthogonality have been introduced by means of equivalent
propositions to the usual orthogonality in inner product spaces. For example,
Roberts [20] introduced Roberts orthogonality: for any x,y ∈ X, x is said to
be Roberts orthogonal to y (denoted by x ⊥R y) if
∥x + ty∥ = ∥x − ty∥ for all t ∈ R.
Birkhoff [4] introduced Birkhoff orthogonality: x is said to be Birkhoff orthog-
onal to y (denoted by x ⊥B y) if
∥x + ty∥ ≥ ∥x∥ for all t ∈ R.
173
174 h. mizuguchi
James [4] introduced isosceles orthogonality: x is said to be isosceles orthog-
onal to y (denoted by x ⊥I y) if
∥x + y∥ = ∥x − y∥.
These generalized orthogonality types have been studied in a lot of papers
([1], [2], [8] and so on).
Recently, quantitative studies of the difference between two orthogonality
types have been performed:
D(X) = inf
{
inf
λ∈R
∥x + λy∥ : x,y ∈ SX,x ⊥I y
}
,
D′(X) = sup{∥x + y∥ − ∥x − y∥ : x,y ∈ SX,x ⊥B y},
BR(X) = sup
α>0
{
∥x + αy∥ − ∥x − αy∥
α
: x,y ∈ SX,x ⊥B y
}
= sup
{
∥x + y∥ − ∥x − y∥
∥y∥
: x,y ∈ X,x,y ̸= 0,x ⊥B y
}
,
BI(X) = sup
{
∥x + y∥ − ∥x − y∥
∥x∥
: x,y ∈ X,x,y ̸= 0,x ⊥B y
}
,
IB(X) = inf
{
infλ∈R ∥x + λy∥
∥x∥
: x,y ∈ X,x,y ̸= 0,x ⊥I y
}
.
(see [10], [14], [19]).
An orthogonality notion“⊥ ” is called symmetric if x ⊥ y implies y ⊥ x.
The usual orthogonality in inner product spaces is, of course symmetric. By
the definition, isosceles orthogonality in normed spaces is symmetric, too.
However Birkhoff orthogonality is not symmetric in general. Birkhoff [4]
proved that if Birkhoff orthogonality is symmetric in a strictly convex normed
space whose dimension is at least three, then the space is an inner product
space. Day [6] and James [9] showed that the assumption of strict convexity
in Birkhoff’s result can be released.
Theorem 1.1. ([2], [6], [9] ) A normed space X whose dimension is at
least three is an inner product space if and only if Birkhoff orthogonality is
symmetric in X.
The assumption of the dimension of the space in the above theorem cannot
be omitted. A two-dimensional normed space in which Birkhoff orthogonality
is symmetric is called a Radon plane.
difference between orthogonality in radon planes 175
In this paper, we consider the constant IB(X) in Radon planes. The
inequality 1/2 ≤ IB(X) ≤ 1 holds for any normed space X. Under the
assumption that the space X is a Radon plane, an inequality 8/9 ≤ IB(X) ≤ 1
is proved, and the Radon plane in which IB(X) = 8/9 is characterized. On
the other hand, a Radon plane is made by connecting the unit sphere of a
two-dimensional normed space and its dual ([6], [12], [13]). A collection of
normed spaces in which IB(X) < 8/9 holds and that constant of the induced
Radon plane is equal to 8/9 is obtained.
2.. The difference between two orthogonality types
in Radon planes
To consider the difference between Birkhoff and isosceles orthogonalities,
the results obtained by James in [7] are important.
Proposition 2.1. ([7])
(i) If x (̸= 0) and y are isosceles orthogonal elements in a normed space,
then ∥x + ky∥ > 1
2
∥x∥ for all k.
(ii) If x (̸= 0) and y are isosceles orthogonal elements in a normed space,
and ∥y∥ ≤ ∥x∥, then ∥x + ky∥ ≥ 2(
√
2 − 1)∥x∥ for all k.
From this, one can has 1/2 ≤ IB(X) ≤ 1 and 2(
√
2 − 1) ≤ D(X) ≤ 1 for
any normed space.
For two elements x,y in the unit sphere in a normed space X, the sine
function s(x,y) is defined by
s(x,y) = inf
t∈R
∥x + ty∥
([22]). V. Balestro, H. Martini, and R. Teixeira [3] showed the following
Proposition 2.2. ([3]) A two dimensional normed space X is a Radon
plane if and only if its associated sine function is symmetric.
Thus for elements x,y in the unit sphere in a Radon plane X with x ⊥I y
we have infλ∈R ∥x+λy∥ = infµ∈R ∥y +µx∥. Hence the inequality 2(
√
2 − 1) ≤
IB(X) ≤ 1 holds for a Radon plane X.
Using Proposition 2.2 again, we start to consider the lower bound of IB(X)
in a Radon plane.
176 h. mizuguchi
Proposition 2.3. Let X be a Radon plane, an element x ∈ SX be
isosceles orthogonality to αy for another element y ∈ SX and a real num-
ber α ∈ R. Take numbers k,l ∈ R such that ∥x + ky∥ = minλ∈R ∥x + λy∥ =
minµ∈R ∥y + µx∥ = ∥y + lx∥. Then, in the estimation of the constant IB(X),
we may only consider the situation 0 ≤ α ≤ 1, 0 ≤ k and 0 ≤ l. In this case,
k ≤ min{1/2,α} and l ≤ 1/2 hold.
Proof. Since x ⊥I αy implies x ⊥I −αy and y ⊥I x/α, we can suppose
0 ≤ α ≤ 1. From the assumption ∥x + ky∥ = minλ∈R ∥x + λy∥ = minµ∈R ∥y +
µx∥ = ∥y + lx∥, we can also suppose 0 ≤ k and 0 ≤ l. Then it follows from
x ⊥I αy and ∥x + ky∥ = minλ∈R ∥x + λy∥ that k ≤ α.
The assumption ∥x+ky∥ = minλ∈R ∥x+λy∥ implies that x+ky is Birkhoff
orthogonal to y. From the symmetry of Birkhoff orthogonality in a Radon
plane, y is Birkhoff orthogonal to x + ky. Using this fact, one has
α + k ≤ ∥x + ky − (α + k)y∥
= ∥x − αy∥ = ∥x + αy∥
= ∥x + ky + (α − k)y∥
≤ ∥x + ky∥ + α − k
and hence 2k ≤ ∥x + ky∥ = minλ∈R ∥x + λy∥ ≤ 1.
In a similar way, from the fact that x is Birkhoff orthogonal to y + lx, we
have 2l ≤ ∥y + lx∥ ≤ 1.
Proposition 2.4. Let X be a Radon plane, an element x ∈ SX be
isosceles orthogonality to αy for another element y ∈ SX and a number
α ∈ [0,1]. Take numbers k ∈ [0,min{1/2,α}] and l ∈ [0,1/2] such that
∥x + ky∥ = minλ∈R ∥x + λy∥ = minµ∈R ∥y + µx∥ = ∥y + lx∥. Then
∥x + ky∥ ≥ max
{
(α + k)(1 − kl)
(α + k)(1 − kl) + k(1 − l)(α − k)
,
(1 + αl)(1 − kl)
(1 + αl)(1 − kl) + l(1 − k)(1 − αl)
}
.
Proof. It follows from
x =
α(x + ky) + k(x − αy)
α + k
difference between orthogonality in radon planes 177
and x ⊥I αy that
α + k ≤ α∥x + ky∥ + k∥x − αy∥ = α∥x + ky∥ + k∥x + αy∥.
For
c =
α − k
1 + α − k − αl
and d =
1 − kl
1 + α − k − αl
,
the equality d(x + αy) = (1 − c)(x + ky) + c(y + lx) holds, and hence one has
∥x + αy∥ ≤
∥x + ky∥
d
=
1 + α − k − αl
1 − kl
∥x + ky∥.
Thus, we obtain
α + k ≤
(
α + k ·
1 + α − k − αl
1 − kl
)
∥x + ky∥
=
(α + k)(1 − kl) + k(α − k − αl + kl)
1 − kl
∥x + ky∥
=
(α + k)(1 − kl) + k(1 − l)(α − k)
1 − kl
∥x + ky∥.
Meanwhile, from the equality
y =
l(−x + αy) + y + lx
1 + αl
,
we obtain
1 + αl ≤ ∥y + lx∥ + l∥ − x + αy∥
= ∥x + ky∥ + l∥x + αy∥
≤
(
1 + l ·
1 + α − k − αl
1 − kl
)
∥x + ky∥
=
(1 + αl)(1 − kl) + l(1 − k)(1 − αl)
1 − kl
∥x + ky∥.
Let
F(α,k,l) =
k(1 − l)(α − k)
(α + k)(1 − kl)
and G(α,k,l) =
l(1 − k)(1 − αl)
(1 + αl)(1 − kl)
.
178 h. mizuguchi
From the above proposition, the inequality
∥x + ky∥−1 ≤ 1 + min
{
F(α,k,l), G(α,k,l)
}
(2.1)
holds.
It follows from
1 − l
1 − kl
=
1
k
+
1 − k
k(kl − 1)
that the function F(α,k,l) is decreasing on l in the interval [0,1]. In a similar
way, G(α,k,l) is decreasing on k in the interval [0,1].
Let us consider the upper bound of min{F(α,k,l), G(α,k,l)}.
Lemma 2.5. Let 0 ≤ α ≤ 1, 0 ≤ k ≤ min{α,1/2} and k ≤ l ≤ 1/2. Then
min
{
F(α,k,l), G(α,k,l)
}
= F(α,k,l) ≤
k(1 − k)
(1 + k)2
.
Proof. Let 0 ≤ α ≤ 1, 0 ≤ k ≤ min{α,1/2} and k ≤ l ≤ 1/2. For the
function
H(α,k,l) :=
(
G(α,k,l) − F(α,k,l)
)
(1 − kl),
we have
H(α,k,l) = l(1 − k)
1 − αl
1 + αl
− k(1 − l)
α − k
α + k
and hence
∂H
∂α
= l(1 − k)
∂
∂α
(
1 − αl
1 + αl
)
− k(1 − l)
∂
∂α
(
α − k
α + k
)
= −
2l2(1 − k)
(1 + αl)2
−
2k2(1 − l)
(α + k)2
≤ 0.
This implies that H is decreasing on α. Thus we obtain the inequality
H(α,k,l) ≥ H(1,k, l) = l(1 − k)
1 − l
1 + l
− k(1 − l)
1 − k
1 + k
=
(1 − k)(1 − l)(l − k)
(1 + k)(1 + l)
≥ 0,
and so F(α,k,l) ≤ G(α,k,l) holds.
difference between orthogonality in radon planes 179
Using the fact that F(α,k,l) is a decreasing function on l,
min
{
F(α,k,l), G(α,k,l)
}
= F(α,k,l) ≤ F(α,k,k) =
k(α − k)
(1 + k)(α + k)
.
From the fact that the function (α − k)/(α + k) is increasing on α, it follows
that
k(α − k)
(1 + k)(α + k)
≤
k(1 − k)
(1 + k)2
,
which completes the proof.
Lemma 2.6. Let 0 ≤ α ≤ 1, 0 ≤ k ≤ min{α,1/3} and 0 ≤ l < k. Then
min
{
F(α,k,l), G(α,k,l)
}
≤
k(1 − k)
(1 + k)2
.
Proof. Let 0 ≤ α ≤ 1, 0 ≤ k ≤ min{α,1/3} and 0 ≤ l < k. Then
min
{
F(α,k,l), G(α,k,l)
}
≤
(1 − k)lF(α,k,l) + (1 − l)k G(α,k,l)
(1 − k)l + (1 − l)k
=
2α(1 − k)k(1 − l)l
(α + k)(1 + αl)
(
(1 − k)l + (1 − l)k
) = (∗).
We have that (∗) ≤ k(1−k)
(1+k)2
if and only if the function
f(α,k,l) := 2α(1 + k)2(1 − l)l − (α + k)(1 + αl)
(
(1 − k)l + (1 − l)k
)
is negative. One can has
f(α,k,l) =
(
(2 + α)l − 1
)
k2
+
(
α(1 − l)(4l − 1 − αl) − l(1 + αl)(1 − α)
)
k
+ αl
(
1 − (2 + α)l
)
and hence
∂f
∂k
= 2Ak + B, where A = (2 + α)l − 1 and
B = α(1 − l)(4l − 1 − αl) − l(1 + αl)(1 − α).
From the fact l < k ≤ 1/3, we obtain A ≤ (3l − 1) ≤ 0 and
B ≤ α(1 − l)(4l − 1 − αl) − l(1 − α)
≤ α(1 − l)(l − αl) − l(1 − α)
= l(1 − α)
(
α(1 − l) − 1
)
≤ 0.
180 h. mizuguchi
Thus the function f is decreasing with respect to k and hence
f(α,k,l) ≤ f(α,l, l)
= 2α(1 + l)2(1 − l)l − 2(α + l)(1 + αl)(1 − l)l
= 2(1 − l)l
(
α(1 + l)2 − (α + l)(1 + αl)
)
= −2(1 − l)l2(1 − α)2 ≤ 0.
This completes the proof.
Under the assumption 1/3 < k and l < k, we consider the upper bound of
(∗), too.
Lemma 2.7. Let 0 ≤ α ≤ 1, 1/3 < k ≤ min{α,1/2} and 0 ≤ l < k. Then
min
{
F(α,k,l), G(α,k,l)
}
≤
2k(1 − k)
(√
2(1 − k) −
√
k
)2
(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2 .
Proof. As in the above lemma, min{F(α,k,l), G(α,k,l)} is less than
2α(1 − k)k(1 − l)l
(α + k)(1 + αl)
(
(1 − k)l + (1 − l)k
) = (∗).
The inequality
(∗) ≤
2k(1 − k)
(√
2(1 − k) −
√
k
)2
(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2
is equivalent to
g(α,k,l) : =
α(1 − l)l
(α + k)(1 + αl)
(
(1 − k)l + (1 − l)k
)
≤
(√
2(1 − k) −
√
k
)2
(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2 .
On this function g, one can see
∂g
∂α
=
(1 − l)l
l(1 − k) + k(1 − l)
×
∂
∂α
(
α
(α + k)(1 + αl)
)
=
(1 − l)l
l(1 − k) + k(1 − l)
×
k − α2l
(α + k)2(1 + αl)2
.
difference between orthogonality in radon planes 181
From the assumption k > l, the function g is increasing on α and so
g(α,k,l) ≤ g(1,k, l) =
(1 − l)l
(1 + k)(1 + l)
(
l(1 − k) + k(1 − l)
).
We have that
g(1,k, l) ≤
(√
2(1 − k) −
√
k
)2
(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2
if and only if
Pk(l) :=
(√
2(1 − 2k) +
√
k(1 − k)
)2
(1 − l)l
−
(√
2(1 − k) −
√
k
)2
(1 + l)
(
l(1 − k) + k(1 − l)
)
≤ 0.
Letting
lk =
k
k +
√
2k(1 − k)
,
we have
lk(1 − k) + k(1 − lk) = k + (1 − 2k)lk =
1 − k +
√
2k(1 − k)
k +
√
2k(1 − k)
k,
and hence
(1 + lk)
(
k + (1 − 2k)lk
)
(1 − lk)lk
=
(
2k +
√
2k(1 − k)
)
{1 − k +
√
2k(1 − k)}√
2k(1 − k)
= 2
√
2k(1 − k) + 1 + k
=
(√
1 − k +
√
2k
)2
.
Meanwhile one can easily check(√
1 − k +
√
2k
)(√
2(1 − k) −
√
k
)
=
√
2(1 − 2k) +
√
k(1 − k).
Thus we obtain
(1 + lk)
(
lk(1 − k) + k(1 − lk)
)
(1 − lk)lk
=
(√
1 − k +
√
2k
)2
=
(√
2(1 − 2k) +
√
k(1 − k)
)2(√
2(1 − k) −
√
k
)2 ,
182 h. mizuguchi
which implies Pk(lk) = 0.
We consider the derivation
P ′k(l) =
(√
2(1 − 2k) +
√
k(1 − k)
)2
(1 − 2l)
−
(√
2(1 − k) −
√
k
)2(
(1 − k) + 2(1 − 2k)l
)
,
too. For lk, we have
1 − k + 2(1 − 2k)lk = 1 − k +
2(1 − 2k)k
k +
√
2k(1 − k)
=
3k − 5k2 + (1 − k)
√
2k(1 − k)
k +
√
2k(1 − k)
,
and hence
1 − k + 2(1 − 2k)lk
1 − 2lk
=
3k − 5k2 + (1 − k)
√
2k(1 − k)
−k +
√
2k(1 − k)
.
On the other hand, a equality(√
1 − k +
√
2k
)2( − k + √2k(1 − k))
=
(
1 + k + 2
√
2k(1 − k)
)(
− k +
√
2k(1 − k)
)
= 3k − 5k2 + (1 − k)
√
2k(1 − k)
holds. Thus we have
1 − k + 2(1 − 2k)lk
1 − 2lk
=
(√
1 − k +
√
2k
)2
=
(√
2(1 − 2k) +
√
k(1 − k)
)2(√
2(1 − k) −
√
k
)2 .
This implies P ′k(lk) = 0.
Combining the fact Pk(0) = −k
(√
2(1 − k) −
√
k
)2 ≤ 0 with Pk(lk) = 0
and P ′k(lk) = 0, one can see that Pk(l) ≤ 0 for any real number l. Therefore
the inequality
min
{
F(α,k,l), G(α,k,l)
}
≤
2k(1 − k)
(√
2(1 − k) −
√
k
)2
(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2
holds.
difference between orthogonality in radon planes 183
A fundamental derivation implies that the function k(1−k)/(1+k)2 takes
maximum 1/8 at k = 1/3. Now we let
h(k) =
k(1 − k)(
√
2(1 − k) −
√
k)2
(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2
=
k
(√
2(1 − k) −
√
k(1 − k)
)2
(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2
and consider the maximum of h(k).
Lemma 2.8. The function h(k) in the interval [0,1/2] takes maximum
1/16 at k = 1/3.
Proof. We can consider the derivation h′(k) as follows:
(1 + k)2
(√
2(1 − 2k) +
√
k(1 − k)
)4
h′(k)
=
[(√
2(1 − k) −
√
k(1 − k)
)2
+ 2k
(√
2(1 − k) −
√
k(1 − k)
)(
−
√
2 −
1 − 2k
2
√
k(1 − k)
)]
× (1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)2 − k(√2(1 − k) − √k(1 − k))2
×
[(√
2(1 − 2k) +
√
k(1 − k)
)2
+ 2(1 + k)
(√
2(1 − 2k) +
√
k(1 − k)
)(
− 2
√
2 +
1 − 2k
2
√
k(1 − k)
)]
.
Thus we obtain√
k(1 − k)(1 + k)2
(√
2(1 − k) +
√
k(1 − k)
)−1
×
(√
2(1 − 2k) +
√
k(1 − k)
)3
h′(k)
=
(
(1 − k)(2 − 5k) +
√
2k
√
k(1 − k)
)√
k(1 − k)
+ k(1 + k)
(
4
√
k(1 − k) −
√
2(2 − k)
)
= (9k2 − 3k + 2)
√
k(1 − k) − 2
√
2k
184 h. mizuguchi
and hence
√
1 − k(1 + k)2
(√
2(1 − k) +
√
k(1 − k)
)−1(√
2(1 − 2k) +
√
k(1 − k)
)3
h′(k)
= (9k2 − 3k + 2)
√
1 − k − 2
√
2k.
We note that (9k2 − 3k + 2)
√
1 − k − 2
√
2k is positive if and only if (9k2 −
3k + 2)2(1 − k) − 8k is so. Meanwhile, one has
(9k2 − 3k + 2)2(1 − k) − 8k
=
(
3k(3k − 1) + 2
)2
(1 − k) − 8k
= 9k2(3k − 1)2(1 − k) + 12k(3k − 1)(1 − k) + 4(1 − k) − 8k
= (3k − 1)
(
9k2(3k − 1)(1 − k) + 12k(3k − 1)(1 − k) − 4
)
= −(3k − 1)(1 + 3k2)(2 − 3k)2.
Therefore we obtain that the function h(k) takes maximum at k = 1/3.
One can easily have h(1/3) = 1/16, which completes the proof.
From the inequality (2.1) and the above lemmas we have
Theorem 2.9. Let X be a Radon plane. Then 8/9 ≤ IB(X) ≤ 1.
In addition, we are able to characterize a Radon plane X satisfying
IB(X) = 8/9. For simplicity, we use the notation ẑ as z/∥z∥ for any nonzero
z ∈ X.
Theorem 2.10. Let X be a Radon plane. Then IB(X) = 8/9 if and only
if its unit sphere is an affine regular hexagon.
Proof. Suppose that X is a Radon plane and the equality IB(X) = 8/9
holds. Then there exist elements x,y ∈ SX and a real number α such that
∥x + αy∥ = ∥x − αy∥ and minλ∈R ∥x + λy∥ = minµ∈R ∥y + µx∥ = 8/9. For
k and l in the above lemmas, all inequalities in the proofs have to turn into
equalities and hence k = l = 1/3. As one of them, the inequality
α + k ≤ α∥x + ky∥ + k∥x − αy∥ = α∥x + ky∥ + k∥x + αy∥
also becomes an equality for α = 1 and k = 1/3. This implies
4
3
=
∥∥∥x + 1
3
y
∥∥∥ + 1
3
∥x − y∥ =
8
9
+
1
3
∥x − y∥
difference between orthogonality in radon planes 185
and hence ∥x + y∥ = ∥x − y∥ = 4/3.
Using these facts, one has
x̂ + y =
3
4
(x + y) =
9
16
((
x +
1
3
y
)
+
(
y +
1
3
x
))
=
1
2
(
̂
x +
1
3
y +
̂
y +
1
3
x
)
.
This implies ∥∥∥∥∥12
(
̂
x +
1
3
y +
̂
y +
1
3
x
)∥∥∥∥∥ = ∥x̂ + y∥ = 1.
On the other hand, for
x =
3
4
(
x +
1
3
y
)
+
1
4
(x − y),
from ∥∥∥x + 1
3
y
∥∥∥ = 8
9
and ∥x − y∥ =
4
3
we have
x =
2
3
( ̂
x +
1
3
y
)
+
1
3
(x̂ − y)
and hence ∥∥∥∥23
( ̂
x +
1
3
y
)
+
1
3
(x̂ − y)
∥∥∥∥ = ∥x∥ = 1.
In a similar way, the equality∥∥∥∥23
( ̂
y +
1
3
x
)
+
1
3
(−̂x + y)
∥∥∥∥ = ∥y∥ = 1
holds. Thus the three segments[
x̂ − y,
̂
x +
1
3
y
]
,
[
̂
x +
1
3
y,
̂
y +
1
3
x
]
and
[
̂
y +
1
3
x, −̂x + y
]
are contained in the unit sphere SX.
Moreover we obtain
(x̂ − y) +
( ̂
y +
1
3
x
)
=
3
4
(x − y) +
9
8
(
y +
1
3
x
)
=
9
8
(
x +
1
3
y
)
=
̂
x +
1
3
y.
186 h. mizuguchi
Therefore, the unit sphere SX is an affine regular hexagon.
Conversely, suppose that SX is an affine regular hexagon (and therefore X
is a Radon plane). Then there exist u,v ∈ SX such that ±u, ±v and ±(u+v)
are the vertices of SX. Letting
x = u +
1
3
v and y = −
1
3
u − v,
we have
x + y =
2
3
(u − v) and x − y =
4
3
(u + v) .
Thus ∥x + y∥ = 4/3 = ∥x − y∥ and hence x ⊥I y.
Meanwhile, one has
x +
1
3
y = u +
1
3
v +
1
3
(
−
1
3
u − v
)
=
8
9
u.
Therefore, the inequality
IB(X) = inf
{
infλ∈R ∥x + λy∥
∥x∥
: x,y ∈ X, x,y ̸= 0, x ⊥I y
}
≤
8
9
holds. This implies IB(X) = 8/9.
3.. Practical Radon planes and a calculation
A Radon plane is made by connecting the unit sphere of a normed plane
and its dual ([6]). Hereafter, we make a collection of the space X in which the
unit sphere SX is a hexagon, the constant IB(X) is less than 8/9 and that of
the induced Radon plane coincides with 8/9.
A norm ∥ · ∥ on R2 is said to be absolute if ∥(a,b)∥ = ∥(|a|, |b|)∥ for any
(a,b) ∈ R2, and normalized if ∥(1,0)∥ = ∥(0,1)∥ = 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 max{1 − t, t} ≤ ψ(t) ≤ 1
for all t ∈ [0,1]. As in [5, 21], it is well known that AN2 and Ψ2 are in a
one-to-one correspondence under the equation ψ(t) = ∥(1 − t,t)∥ for t ∈ [0,1]
and
∥(a,b)∥ψ =
(|a| + |b|)ψ
(
|b|
|a| + |b|
)
if (a,b) ̸= (0,0),
0 if (a,b) = (0,0).
Let ∥ · ∥ψ denote an absolute normalized norm associated with a convex func-
tion ψ ∈ Ψ2.
difference between orthogonality in radon planes 187
For ψ ∈ Ψ2, the dual function ψ∗ on [0,1] is defined by
ψ∗(s) = sup
{
(1 − t)(1 − s) + ts
ψ(t)
: t ∈ [0,1]
}
for s ∈ [0,1]. It is known that ψ∗ ∈ Ψ2 and that ∥·∥ψ∗ ∈ AN2 is the dual norm
of ∥ · ∥ψ, that is, (R2,∥ · ∥ψ)∗ is identified with (R2,∥ · ∥ψ∗) (cf. [16, 17, 18]).
Meanwhile, for ψ ∈ Ψ2, the function ψ̃ ∈ Ψ2 is defined by ψ̃(t) = ψ(1 − t)
for any t ∈ [0,1]. One can easily check (̃ψ∗) =
(
ψ̃
)∗
. So we write it ψ̃∗.
According to [6], [12] and [13], for any ψ ∈ Ψ2, the Day-James space ℓψ-ℓψ̃∗
becomes a Radon plane.
For any c ∈ [0,1], let
ψc(t) =
{
−ct + 1 if 0 ≤ t ≤ (1 + c)−1,
t if (1 + c)−1 ≤ t ≤ 1.
Then the norm of (a,b) ∈ R2 is computed by
∥(a,b)∥ψc =
{
|a| + (1 − c)|b| if |a| ≥ c|b|,
|b| if |a| ≤ c|b|.
The dual function is calculated as follows:
Proposition 3.1. Let c ∈ [0,1]. Then
ψ∗c(s) =
1 − s if 0 ≤ s ≤
1 − c
2 − c
,
(1 − c)s + c if
1 − c
2 − c
≤ s ≤ 1.
Proof. Fix s ∈ [0,1]. We define the function fc,s(t) from [0,1] into R by
fc,s(t) =
(1 − t)(1 − s) + ts
ψc(t)
.
We note that ψ∗c(s) = max{fc,s(t) : 0 ≤ t ≤ 1} and calculate the maximum of
fc,s on [0,1]. By the definition of ψc, we have
fc,s(t) =
1 − s + (2s − 1)t
−ct + 1
if 0 ≤ t ≤ (1 + c)−1,
2s − 1 +
1 − s
t
if (1 + c)−1 ≤ t ≤ 1.
188 h. mizuguchi
The function 2s − 1 + (1 − s)/t is clearly decreasing on t.
If 0 ≤ s ≤ (1 − c)/(2 − c), then the function fc,s(t) is decreasing on
[0,(1 + c)−1]. Hence we have ψ∗c(s) = fc,s(0) = 1 − s.
Suppose that (1−c)/(2−c) ≤ s ≤ 1. Then the function fc,s(t) is increasing
on [0,(1 + c)−1]. Thus we have
ψ∗c(s) = fc,s
(
1
1 + c
)
= (1 − c)s + c.
Therefore we obtain this proposition.
From this result, one has
Proposition 3.2. Let c ∈ [0,1]. Then
∥(a,b)∥ψ∗c =
{
|a| if |b| ≤ (1 − c)|a|,
c|a| + |b| if (1 − c)|a| ≤ |b|.
Thus the Radon plane ℓψc-ℓψ̃∗c
induced by ψc is the space R2 with the norm
∥(a,b)∥
ψc,ψ̃∗c
=
|a| + (1 − c)|b| if c|b| ≤ |a| and ab ≥ 0,
|b| if − (1 − c)|b| ≤ a ≤ c|b| and b ≥ 0,
|b| if − (1 − c)|b| ≤ −a ≤ c|b| and b ≤ 0,
|a| + c|b| if (1 − c)|b| ≤ |a| and ab ≤ 0.
Therefore the unit sphere of this space is an affine regular hexagon with the
vertices ±(1,0), ±(1 − c,1), ±(−c,1) and hence the constant IB(ℓψc-ℓψ̃∗c )
coincide with 8/9 by the Theorem 2.10.
Next, we calculate the constants IB((R2,∥ · ∥ψc)) and IB((R
2,∥ · ∥
ψ̃∗c
)).
Then we obtain that the values are smaller than IB(ℓψc-ℓψ̃∗c
) = 8/9 and equal
to 8/9 only when c = 1/2. We note that ψ̃∗c = ψ1−c and it is enough to
calculate IB((R2,∥ · ∥ψc)) for c ∈ [0,1]. To do this, we need to recall the
Dunkl-Williams constant defined in [11]:
DW(X) = sup
{
∥x∥ + ∥y∥
∥x − y∥
∥∥∥∥ x∥x∥ − y∥y∥
∥∥∥∥ : x,y ∈ X, x,y ̸= 0, x ̸= y
}
= sup
{
∥u + v∥
∥(1 − t)u + tv∥
: u,v ∈ SX, 0 ≤ t ≤ 1
}
.
difference between orthogonality in radon planes 189
The unit sphere of (R2,∥ · ∥ψc) and (R
2,∥ · ∥
ψ̃∗c
).
The unit sphere of Radon plane ℓψc-ℓψ̃∗c
.
For any normed space, the equality 2 ≤ DW(X) ≤ 4 holds. In [14], it is
shown that the equality IB(X)DW(X) = 2 holds for any normed space X.
One can find a formula to calculate this constant in the paper [15]. For each
x ∈ SX and for each y ∈ X with x ⊥B y, we put
m(x,y) = sup
{∥∥∥∥x + λ + µ2 y
∥∥∥∥ : λ ≤ 0 ≤ µ, ∥x + λy∥ = ∥x + µy∥
}
.
We define the positive number M(x) by
M(x) = sup
{
m(x,y) : x ⊥B y
}
.
Using these notions, the Dunkl-Williams constant can be calculated as
DW(X) = 2 sup
{
M(x) : x ∈ SX
}
= 2 sup
{
M(x) : x ∈ fr(BX)
}
,
190 h. mizuguchi
where fr(BX) is the frame of unit ball. An element x ∈ SX is called an extreme
point of BX if y,z ∈ SX and x = (y + z)/2 implies x = y = z. The set of all
extreme points of BX is denoted by ext(BX). Suppose that the space X has
two-dimension. Then the above calculation method is turned into
DW(X) = 2 sup
{
M(x) : x ∈ ext(BX)
}
.
Here, we reduce the amount of calculation a little more. As in Section 2, we
use the notation ẑ.
Proposition 3.3. Let X be a two-dimensional normed space. Then
DW(X) = sup
{
∥u + v∥
∥(1 − t)u + tv∥
: u ∈ ext(BX), v ∈ SX, 0 ≤ t ≤ 1
}
.
Proof. Take arbitrary elements u,v ∈ SX \ ext(BX). If the segment [u,v]
belongs to the unit sphere SX, then
∥u + v∥
∥(1 − t)u + tv∥
= 2
for any t ∈ [0,1]. So we may assume [u,v] ̸⊂ SX. Then we have t0 ∈ [0,1]
such that
min
0≤t≤1
∥(1 − t)u + tv∥ = ∥(1 − t0)u + t0v∥.
Letting x = ̂(1 − t0)u + t0v and y = û − v, we have four elements u1, u2,
v1, v2 ∈ SX such that at least two elements among them belong to ext(BX)
and satisfying u ∈ [u1,u2] ⊂ SX, v ∈ [v1,v2] ⊂ SX and û1 − v1 = y =
û2 − v2. For these elements, from the fact that three vectors u − v, u1 − v1
and u2 −v2 are parallel each other, we can take numbers s0 ∈ (0,1) satisfying
u = (1−s0)u1+s0u2, v = (1−s0)v1+s0v2. Meanwhile there exist t1, t2 ∈ (0,1)
such that
min
0≤t≤1
∥(1 − t)u1 + tv1∥ = ∥(1 − t1)u1 + t1v1∥,
min
0≤t≤1
∥(1 − t)u2 + tv2∥ = ∥(1 − t2)u2 + t2v2∥.
It follow from x ⊥B y and û1 − v1 = y = û2 − v2 that ̂(1 − t2)u2 + t2v2 = x
and ̂(1 − t1)u1 + t1v1 = ±x. In case of ̂(1 − t1)u1 + t1v1 = −x, the ele-
ment −u1 belongs to the arc between v1 and x. Letting v3 = −u1, we
difference between orthogonality in radon planes 191
can take element u3 satisfying û3 − v3 = y, again. Hence we may consider
̂(1 − t1)u1 + t1v1 = x.
Then the equalities
(1 − t0)u + t0v = (1 − s0)
(
(1 − t1)u1 + t1v1
)
+ s0
(
(1 − t2)u2 + t2v2
)
and
∥(1 − t0)u + t0v∥ = (1 − s0)∥(1 − t1)u1 + t1v1∥ + s0∥(1 − t2)u2 + t2v2∥
holds. Thus, using triangle inequality and the fact that an inequality
(1 − α)a + αb
(1 − α)c + αd
≤ max
{
a
c
,
b
d
}
holds for α ∈ [0,1] and positive numbers a,b,c,d, we obtain
∥u + v∥
∥(1 − t0)u + t0v∥
=
∥(1 − s0)u1 + s0u2 + (1 − s0)v1 + s0v2∥
(1 − s0)∥(1 − t1)u1 + t1v1∥ + s0∥(1 − t2)u2 + t2v2∥
≤
(1 − s0)∥u1 + v1∥ + s0∥u2 + v2∥
(1 − s0)∥(1 − t1)u1 + t1v1∥ + s0∥(1 − t2)u2 + t2v2∥
≤ max
{
∥u1 + v1∥
∥(1 − t1)u1 + t1v1∥
,
∥u2 + v2∥
∥(1 − t2)u2 + t2v2∥
}
≤ sup
{
∥u + v∥
∥(1 − t)u + tv∥
: u ∈ ext(BX), v ∈ SX, 0 ≤ t ≤ 1
}
.
This completes the proof.
Thus, to obtain the value of the Dunkl-Williams constant, in the above
calculation method, for x ∈ ext(BX) and y ∈ X with x ⊥B y, the value
m(x,y) can be computed as
m(x,y) = sup
{∥∥∥∥x + λ + µ2 y
∥∥∥∥ : λ ≤ 0 ≤ µ, ∥x + λy∥ = ∥x + µy∥,
x̂ + λy ∈ ext(BX)
}
.
192 h. mizuguchi
4.. The constant IB(X) in hexagonal planes
Now, we start to compute DW((R2,∥ · ∥ψc)) and IB((R
2,∥ · ∥ψc)) for
c ∈ [0,1]. For simplicity we write Xc and ∥ · ∥ for (R2,∥ · ∥ψc) and ∥ · ∥ψc,
respectively. First we suppose 1/2 ≤ c. Let e1 = (1,0), u = (c,1). Then, by
[15, Proposition 2.5], DW(Xc) = 2 max{M(e1),M(u)}. Putting vt = (−t,1)
and
wt = (1 − t)(−e1) + t(−c,1) = (−1 + t − ct,t)
for t ∈ [0,1], we have e1 ⊥B vt for t ∈ [0,1 − c], u ⊥B vt for t ∈ [1 − c,c] and
u ⊥B wt for t ∈ [0,1]. By [15, Theorem 2.9 and Corollary 2.10], one has
M(e1) = sup
{
m(e1,vt) : t ∈ (0,1 − c)
}
and
M(u) = max
{
sup{m(u,vt) : t ∈ (1 − c,c)},
sup{m(u,wt) : t ∈ (0,1) \ {1/2}}
}
.
Lemma 4.1. Let c ∈ [1/2,1]. Then, in Xc,
M(e1) = 1 +
1 − c
(1 +
√
2c)2
.
Proof. Let t ∈ (0,1 − c). Then the norm of e1 + λvt is computed as
∥e1 + λvt∥ =
−λ if λ ≤ −(c − t)−1,
1 − (1 − c + t)λ if − (c − t)−1 ≤ λ ≤ 0,
1 + (1 − c − t)λ if 0 ≤ λ ≤ (c + t)−1,
λ if (c + t)−1 ≤ λ.
From the inequality∥∥∥∥e1 + 1c + tvt
∥∥∥∥ = 1 + 1 − c − tc + t < 1 + 1 − c + tc − t =
∥∥∥∥e1 − 1c − tvt
∥∥∥∥ ,
we can find real numbers pt ∈ (−(c − t)−1,0) and qt more than (c + t)−1 such
that
∥e1 + ptvt∥ =
∥∥∥∥e1 + 1c + tvt
∥∥∥∥ and ∥e1 + qtvt∥ =
∥∥∥∥e1 − 1c − tvt
∥∥∥∥ ,
difference between orthogonality in radon planes 193
respectively. To obtain m(e1,vt), it is enough to consider∥∥∥∥e1 + 12
(
pt +
1
c + t
)
vt
∥∥∥∥ and
∥∥∥∥e1 + 12
(
−
1
c + t
+ qt
)
vt
∥∥∥∥ .
Since the equality
qt = ∥e1 + qtvt∥ =
∥∥∥∥e1 − 1c − tvt
∥∥∥∥ = 1 + 1 − c + tc − t = 1c − t
holds, one has
(
−(c − t)−1 + qt
)
/2 = 0. On the other hand, from the equality
1 − (1 − c + t)pt = ∥e1 + ptvt∥ =
∥∥∥∥e1 + 1c + tvt
∥∥∥∥ = 1 + 1 − c − tc + t ,
we have
pt = −
1 − c − t
(1 − c + t)(c + t)
and hence
1
2
(
pt +
1
c + t
)
=
t
(1 − c + t)(c + t)
.
It follows from
0 <
t
(1 − c + t)(c + t)
=
1
2
(
pt +
1
c + t
)
<
1
c + t
that ∥∥∥∥e1 + 12
(
pt +
1
c + t
)
vt
∥∥∥∥ = 1 + (1 − c − t)t(1 − c + t)(c + t).
This implies that
m(e1,vt) = 1 +
(1 − c − t)t
(1 − c + t)(c + t)
.
Letting
Fc(t) =
(1 − c − t)t
(1 − c + t)(c + t)
,
one can figure out
(1 − c + t)2(c + t)2F ′c(t)
= (−2t + 1 − c)(1 − c + t)(c + t) − (2t + 1)(1 − c − t)t
= −(2 − c)t2 − 2c(1 − c)t + c(1 − c)2.
194 h. mizuguchi
Let t0 be the larger solution of the equation −(2−c)t2−2c(1−c)t+c(1−c)2 = 0.
Then
t0 =
c(1 − c)
√
2c + c
∈ (0,1 − c)
and Fc takes maximum at t0. This t0 satisfies the equality
(−2t0 + 1 − c)(1 − c + t0)(c + t0) = (1 − c − t0)t0(2t0 + 1),
too. Thus we obtain
M(e1) = 1 +
(1 − c − t0)t0
(1 − c + t0)(c + t0)
= 1 +
−2t0 + 1 − c
2t0 + 1
= 1 +
−2c(1 − c) + (
√
2c + c)(1 − c)
2c(1 − c) +
√
2c + c
= 1 +
1 − c(
1 +
√
2c
)2 .
Lemma 4.2. Let c ∈ [1/2,1]. Then, in Xc,
sup
{
m(u,vt) : t ∈ (1 − c,c)
}
= 2c.
Proof. Let t ∈ (1 − c,c). Then the norm of u + λvt is calculated by
∥u + λvt∥ =
−(1 + λ) if λ ≤ −2c/(c − t),
2c − 1 − {t + (1 − c)}λ if − 2c/(c − t) ≤ λ ≤ −1,
1 − {t − (1 − c)}λ if − 1 ≤ λ ≤ 0,
1 + λ if 0 ≤ λ.
There exist two real numbers αt, βt satisfying 0 < αt < βt, ∥u + αtvt∥ =
∥u − vt∥ and
∥u + βtvt∥ =
∥∥∥∥u − 2cc − tvt
∥∥∥∥ .
It is enough to consider ∥u + 1
2
(−1 + αt)vt∥ and∥∥∥∥u + 12
(
−
2c
c − t
+ βt
)
vt
∥∥∥∥ .
difference between orthogonality in radon planes 195
From the equality
1 + αt = ∥u + αtvt∥ = ∥u − vt∥ = 1 +
(
t − (1 − c)
)
,
we have αt = t−(1−c) and hence (−1+αt)/2 = −
(
2−(t+c)
)
/2. Meanwhile,
it follows from
1 + βt = ∥u + βtvt∥ =
∥∥∥∥u − 2cc − tvt
∥∥∥∥ = −
(
1 −
2c
c − t
)
that
1
2
(
−
2c
c − t
+ βt
)
= −1.
By the inequality
1
2
(
−
2c
c − t
+ βt
)
= −1 < −
(
2 − (t + c)
)
/2 = (−1 + αt)/2 < 0,
we obtain m(u,vt) = ∥u−vt∥ = t+c and hence sup{m(u,vt) : t ∈ (1−c,c)} =
2c.
Next, for t ∈ (0,1), the norm of u + λwt is calculated by
∥u + λwt∥ =
2c − 1 − λ if λ ≤ −1/t,
1 − {1 − 2(1 − c)t}λ if − 1/t ≤ λ ≤ 0,
1 + tλ if 0 ≤ λ ≤ 2c/(1 − t),
−(2c − 1) + λ if 2c/(1 − t) ≤ λ.
In particular we have ∥∥∥∥u + 2c1 − twt
∥∥∥∥ = 1 + 2c1 − tt,∥∥∥∥u − 1twt
∥∥∥∥ = 1 + 1 − 2(1 − c)tt ,
and hence∥∥∥∥u − 1twt
∥∥∥∥ −
∥∥∥∥u + 2c1 − twt
∥∥∥∥ = (1 − t)
(
1 − 2(1 − c)t
)
− 2ct2
t(1 − t)
=
(1 − 2t)
(
1 + (2c − 1)t
)
t(1 − t)
.
196 h. mizuguchi
From this equality, we obtain that if t ∈ (0,1/2) then∥∥∥∥u + 2c1 − twt
∥∥∥∥ <
∥∥∥∥u − 1twt
∥∥∥∥
and that if t ∈ (1/2,1) then∥∥∥∥u − 1twt
∥∥∥∥ <
∥∥∥∥u + 2c1 − twt
∥∥∥∥ .
Lemma 4.3. Let c ∈ [1/2,(1 +
√
5)/4]. Then, in Xc,
sup
{
m(u,wt) : t ∈ (0,1/2)
}
= max
{
1
2
+ c, 1 +
c
{1 +
√
2(1 − c)}2
}
.
Proof. Let t ∈ (0,1/2). Then there exist two numbers γt ∈ (−1/t,0) and
δt greater than 2c/(1 − t) satisfying
∥u + γtwt∥ =
∥∥∥∥u + 2c1 − twt
∥∥∥∥ and ∥u + δtwt∥ =
∥∥∥∥u − 1twt
∥∥∥∥ ,
respectively. To obtain m(u,wt) it is enough to consider∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥ and
∥∥∥∥u + 12
(
−
1
t
+ δt
)
wt
∥∥∥∥ .
From the equality
−(2c − 1) + δt = ∥u + δtwt∥ =
∥∥∥∥u − 1twt
∥∥∥∥ = 2c − 1 −
(
−
1
t
)
,
one has
1
2
(
−
1
t
+ δt
)
= 2c − 1.
It is easy to check 2c − 1 < 2c/(1 − t) and hence we obtain∥∥∥∥u + 12
(
−
1
t
+ δt
)
wt
∥∥∥∥ = 1 + (2c − 1)t.
Under the assumption t ∈ (0,1/2), this function takes the supremum 1/2 + c.
Meanwhile, it follows from
1 −
(
1 − 2(1 − c)t
)
γt = ∥u + γtwt∥ =
∥∥∥∥u + 2c1 − twt
∥∥∥∥ = 1 + 2ct1 − t
difference between orthogonality in radon planes 197
that
γt = −
2ct(
1 − 2(1 − c)t
)
(1 − t)
.
Hence we have
1
2
(
γt +
2c
1 − t
)
=
c
(
1 − (3 − 2c)t
)
(1 − t)
(
1 − 2(1 − c)t
)
and ∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥ = 1 + ct
(
1 − (3 − 2c)t
)
(1 − t)
(
1 − 2(1 − c)t
).
We note that 1 − (3 − 2c)t > 1 − (3 − 2c)/2 = c − 1/2 > 0.
Letting
Gc(t) =
t
(
1 − (3 − 2c)t
)
(1 − t)
(
1 − 2(1 − c)t
)
in the interval [0,1/2], we have(
1 − 2(1 − c)t
)2
(1 − t)2G′c(t)
=
(
− 2(3 − 2c)t + 1
)
(1 − t)
(
1 − 2(1 − c)t
)
−
(
4(1 − c)t − (3 − 2c)
)
t
(
1 − (3 − 2c)t
)
=
(
(3 − 2c)2 − 2(1 − c)
)
t2 − 2(3 − 2c)t + 1
We note that (3−2c)2 −2(1−c) = 4c2 −10c+7 = 4(c−5/4)2 +3/4 > 0. Let t1
be the smaller solution of equality
(
(3−2c)2 −2(1−c)
)
t2 −2(3−2c)t+1 = 0,
i.e.,
t1 =
1
3 − 2c +
√
2(1 − c)
.
If c < (1 +
√
5)/4, then this t1 belongs to the interval (0,1/2). Thus Gc(t)
takes the maximum
Gc(t1) =
1(
1 +
√
2(1 − c)
)2 .
This implies that ∥∥∥∥u + 12
(
γ +
2c
1 − t
)
wt
∥∥∥∥
takes maximum
1 +
c(
1 +
√
2(1 − c)
)2 .
198 h. mizuguchi
In case of (1 +
√
5)/4 ≤ c, the solution t1 is more than 1/2 and hence the
function Gc(t) takes the maximum at t = 1/2. One can follow the above proof
except for this part, and obtain the following:
Lemma 4.4. Let c ∈ [(1 +
√
5)/4,1]. Then, in Xc,
sup
{
m(u,wt) : t ∈ (0,1/2)
}
= 1/2 + c.
Next we consider sup{m(u,wt) : t ∈ (1/2,1)}.
Lemma 4.5. Let c ∈ [1/2,1]. Then, in Xc,
sup
{
m(u,wt) : t ∈ (1/2,1)
}
=
9
8
if
1
2
≤ c ≤
9
16
,
2c if
9
16
≤ c ≤ 1.
Proof. Let t ∈ (1/2,1). Then from∥∥∥∥u − 1twt
∥∥∥∥ <
∥∥∥∥u + 2c1 − twt
∥∥∥∥
one can take γt less than −1/t and δt ∈ (0,2c/(1 − t)) satisfying
∥u + γtwt∥ =
∥∥∥∥u + 2c1 − twt
∥∥∥∥ and ∥u + δtwt∥ =
∥∥∥∥u − 1twt
∥∥∥∥ ,
respectively. From the equality
2c − 1 − γt = ∥u + γtwt∥ =
∥∥∥∥u + 2c1 − twt
∥∥∥∥ = −(2c − 1) + 2c1 − t,
we have
1
2
(
γt +
2c
1 − t
)
= 2c − 1.
The fact
0 ≤ 2c − 1 =
1
2
(
γt +
2c
1 − t
)
<
2c
1 − t
implies that∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥ = ∥u + (2c − 1)wt∥ = 1 + (2c − 1)t.
difference between orthogonality in radon planes 199
It is clear that the function 1 + (2c − 1)t takes the supremum 2c.
On the other hand, it follows from
1 + tδt = ∥u + δtwt∥ =
∥∥∥∥u − 1twt
∥∥∥∥ = 1 + 1 − 2(1 − c)tt
that
δt =
1 − 2(1 − c)t
t2
and hence
1
2
(
−
1
t
+ δt
)
=
1 − (3 − 2c)t
2t2
.
Under the assumption c ∈ [1/2,1], one can easily check (3 − 2c)−1 ∈ (1/2,1).
In case of t ∈ (1/2,(3 − 2c)−1], from the inequality
0 ≤
1 − (3 − 2c)t
2t2
=
1
2
(
−
1
t
+ δt
)
< δt <
2c
1 − t
,
we have ∥∥∥∥u + 12
(
−
1
t
+ δt
)
wt
∥∥∥∥ = 1 + 1 − (3 − 2c)t2t .
It is easy to check that this function takes the supremum 1/2 + c.
Suppose that t ∈ ((3 − 2c)−1,1). Then, from the inequality
−
1
t
<
1
2
(
−
1
t
+ δt
)
= −
(3 − 2c)t − 1
2t2
< 0,
one has ∥∥∥∥u + 12
(
−
1
t
+ δt
)
wt
∥∥∥∥ = 1 +
(
1 − 2(1 − c)t
)(
(3 − 2c)t − 1
)
2t2
.
Considering the function Hc(t) in the interval [1/2,1] defined by
Hc(t) =
(
1 − 2(1 − c)t
)(
(3 − 2c)t − 1
)
t2
,
we figure out
t4H′c(t) =
(
− 4(1 − c)(3 − 2c)t + 5 − 4c
)
t2
− 2
(
1 − 2(1 − c)t
)(
(3 − 2c)t − 1
)
t
200 h. mizuguchi
and hence
t3H′c(t) =
(
− 4(1 − c)(3 − 2c)t + 5 − 4c
)
t
− 2
(
1 − 2(1 − c)t
)(
(3 − 2c)t − 1
)
= −(5 − 4c)t + 2.
Since the function −(5 − 4c)t + 2 is decreasing, we have the following: If
c < 3/4, then one has 2/(5 − 4c) ∈ ((3 − 2c)−1,1) and hence
max
{
Hc(t) : t ∈ ((3 − 2c)−1,1)
}
= Hc
(
2
5 − 4c
)
=
(
(5 − 4c) − 4(1 − c)
)(
2(3 − 2c) − (5 − 4c)
)
4
=
1
4
.
This implies that ∥∥∥∥u + 12
(
−
1
t
+ δt
)
wt
∥∥∥∥
takes the maximum 9/8 at t = 2/(5 − 4c). Meanwhile, 9/8 is greater than 2c
only when c < 9/16.
In case of 3/4 ≤ c, from 1 ≤ 2/(5 − 4c) one has that Hc(t) is increasing.
Hence we have
max
{
Hc(t) : t ∈ ((3 − 2c)−1,1)
}
= H(1) = 2(1 − c)(2c − 1)
This implies that ∥∥∥∥u + 12
(
−
1
t
+ δt
)
wt
∥∥∥∥
takes the supremum 1 + (1 − c)(2c − 1). We note that 1 + (1 − c)(2c − 1) <
1 + (2c − 1) = 2c.
Therefore we obtain the following proposition.
difference between orthogonality in radon planes 201
Proposition 4.6. Let c ∈ [1/2,1]. Then IB(Xc)−1 = DW(Xc)/2 coin-
cide with
max
{
1 +
1 − c
{1 +
√
2c}2
, 1 +
c
{1 +
√
2(1 − c)}2
,
9
8
}
if
1
2
≤ c ≤
9
16
,
max
{
1 +
1 − c
{1 +
√
2c}2
, 1 +
c
{1 +
√
2(1 − c)}2
, 2c
}
if
9
16
< c <
1 +
√
5
4
,
max
{
1 +
1 − c
{1 +
√
2c}2
, 2c
}
if
1 +
√
5
4
≤ c ≤ 1.
Hereafter we suppose c < 1/2. Similarly to the above paragraph,
DW(Xc) = 2 max{M(e1),M(u)} holds. On the other hand, for vt and wt,
Birkhoff orthogonality relations differ from the above paragraph. We have
e1 ⊥B vt for t ∈ [0,c], e1 ⊥B wt for t ∈ [1/2(1 − c),1] and u ⊥B wt for
t ∈ [0,1/2(1 − c)].
By [15, Theorem 2.9 and Corollary 2.10], one figure out
M(e1) = max
{
sup{m(e1,vt) : t ∈ (0,c)},
sup{m(e1,wt) : t ∈ (1/2(1 − c),1)}
}
and
M(u) = sup
{
m(u,wt) : t ∈ (0,1/2(1 − c)) \ {1/2}
}
.
Lemma 4.7. Let c ∈ [0,(3 −
√
5)/4]. Then in Xc,
sup
{
m(e1,vt) : t ∈ (0,c)
}
=
3
2
− c.
Proof. Let t ∈ (0,c). Then in a similar way to the proof of Lemma 4.1, we
have
m(e1,vt) = 1 +
(1 − c − t)t
(1 − c + t)(c + t)
.
Moreover letting
Fc(t) =
(1 − c − t)t
(1 − c + t)(c + t)
,
we also have
(1 − c + t)2(c + t)2F ′c(t) = −(2 − c)t
2 − 2c(1 − c)t + c(1 − c)2
202 h. mizuguchi
again. From c ∈ [0,(3 −
√
5)/4], it is more than
−(2 − c)c2 − 2c(1 − c)c + c(1 − c)2 = c(4c2 − 6c + 1) ≥ 0.
From this fact, Fc(t) increases and hence
sup
{
m(e1,vt) : t ∈ (0,c)
}
= 1 + Fc(c) =
3
2
− c.
Suppose that c ∈ ((3 −
√
5)/4,1/2). Then for t0 defined in the same
formula to the proof of Lemma 4.1, we have t0 ∈ (0,c) and F ′c(t0) = 0. Hence
we obtain
Lemma 4.8. Let c ∈ ((3 −
√
5)/4,1/2). Then in Xc,
sup
{
m(e1,vt) : t ∈ (0,c)
}
= 1 +
1 − c(
1 +
√
2c
)2 .
Lemma 4.9. Let c ∈ [0,1/2). Then in Xc,
sup
{
m(e1,wt) : t ∈ (1/2(1 − c),1)
}
= 2(1 − c).
Proof. Let t ∈ (1/2(1 − c),1). Then the norm of e1 + λwt is calculated as
∥e1 + λwt∥ =
1 − λ if λ ≤ 0,
1 +
(
2(1 − c)t − 1
)
λ if 0 ≤ λ ≤
(
1 − (1 − 2c)t
)−1
,
tλ if
(
1 − (1 − 2c)t
)−1 ≤ λ ≤ (1 − t)−1,
−1 + λ if (1 − t)−1 ≤ λ.
One can take two real numbers st,rt satisfying st < rt < 0, ∥e1 + rtwt∥ =∥∥e1 +(1 − (1 − 2c)t)−1wt∥∥ and ∥e1 + stwt∥ = ∥e1 + (1 − t)−1wt∥. It is enough
to consider
∥∥e1 + 12(rt +(1 − (1 − 2c)t)−1)wt∥∥ and ∥∥e1 + 12(st + (1 −t)−1)wt∥∥.
From the equality
1 − rt = ∥e1 + rtwt∥ =
∥∥e1 + (1 − (1 − 2c)t)−1wt∥∥ = 1 + 2(1 − c)t − 1
1 − (1 − 2c)t
,
one has
rt = −
2(1 − c)t − 1
1 − (1 − 2c)t
difference between orthogonality in radon planes 203
and hence
1
2
(
rt +
1
1 − (1 − 2c)t
)
=
1 − (1 − c)t
1 − (1 − 2c)t
.
It follows from
1 − st = ∥e1 + stwt∥ = ∥e1 + (1 − t)−1wt∥ = −1 +
1
1 − t
that 1
2
(st + (1 − t)−1) = 1. Since the inequality
0 <
1 − (1 − c)t
1 − (1 − 2c)t
< 1 <
1
1 − (1 − 2c)t
holds, we obtain
m(e1,wt) =
∥∥∥∥e1 +
(
st +
1
1 − t
)
wt
∥∥∥∥ = 1 + (2(1 − c)t − 1) × 1 = 2(1 − c)t.
This implies
sup
{
m(e1,wt) : t ∈ (1/2(1 − c),1)
}
= 2(1 − c).
For t ∈ (0,1/2(1−c)) the norm of u+λwt is calculated in a similar way to
the case of c ∈ [1/2,1]. Now we suppose c ∈ [0,1/2) and so 1/2 ≤ 1/2(1−c) <
1 holds. Thus we have to consider the following two cases again:
If t ∈ (0,1/2) then ∥∥∥∥u + 2c1 − twt
∥∥∥∥ <
∥∥∥∥u − 1twt
∥∥∥∥ .
If t ∈ (1/2,1) then ∥∥∥∥u − 1twt
∥∥∥∥ <
∥∥∥∥u + 2c1 − twt
∥∥∥∥ .
Lemma 4.10. Let c ∈ [0,1/2). Then, in Xc
sup
{
m(u,wt) : t ∈ (0,1/2)
}
= max
{
2(1 − c),1 +
c(
1 +
√
2(1 − c)
)2
}
.
204 h. mizuguchi
Proof. Let t ∈ (0,1/2). In a similar way to Lemma 4.3, one can take δt
and figure out that this constant satisfy ∥u + 1
2
(−1/t + δt)wt∥ = 1 + (1 −
2c)
(
1−2(1−c)t
)
and that this function of t takes the supremum 2(1−c). We
also have γt and that
1
2
(
γt +
2c
1 − t
)
=
c
(
1 − (3 − 2c)t
)
(1 − t)
(
1 − 2(1 − c)t
).
Now we are considering the case of c ∈ [0,1/2) and so 1/(3 − 2c) ∈ (0,1/2).
If t ∈ (0,1/(3 − 2c)), then we have
0 <
1
2
(
γt +
2c
1 − t
)
<
2c
1 − t
and hence ∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥ = 1 + ct
(
1 − (3 − 2c)t
)
(1 − t)
(
1 − 2(1 − c)t
).
For t1 defined by same formula to Lemma 4.3, we have t1 ∈ (0,1/(3 − 2c))
and that the function ∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥
takes maximum 1 + c/
(
1 +
√
2(1 − c)
)2
at t1.
Assume that t ∈ (1/(3 − 2c),1/2). Then from the inequality
−
1
t
< γt <
1
2
(
γt +
2c
1 − t
)
= −
c
(
(3 − 2c)t − 1
)
(1 − t)
(
1 − 2(1 − c)t
) < 0,
we obtain ∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥ = 1 + c
(
(3 − 2c)t − 1
)
1 − t
.
This function of t is increasing and hence less than
1 +
c
(
(3 − 2c)/2 − 1
)
1 − 1/2
= (1 − c)(1 + 2c) < 2(1 − c).
Thus we obtain ∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥ < 2(1 − c),
which completes the proof.
difference between orthogonality in radon planes 205
Lemma 4.11. Let c ∈ [0,1/2). Then, in Xc,
sup
{
m(u,wt) : t ∈ (1/2,1/(2−c))
}
=
(1 − c)(1 + 2c) if 0 < c ≤
1
4
,
max
{
(1 − c)(1 + 2c),
9
8
}
if
1
4
< c <
1
2
.
Proof. Let t ∈ (1/2,1/(2 − c)). In a similar way to Lemma 4.5, we take γt
less than −1/t and δt ∈ (0,2c/(1 − t)). Then we have
1
2
(
γt +
2c
1 − t
)
= −(1 − 2c)
It follows from −1/t < −2(1 − c) < −(1 − 2c) < 0 that∥∥∥∥u + 12
(
γt +
2c
1 − t
)
wt
∥∥∥∥ = 1 + (1 − 2c)(1 − 2(1 − c)t)
In the situation t ∈ (1/2,1/(2 − c)), it takes supremum 1 + c(1 − 2c) =
(1 − c)(1 + 2c). In addition, we have
−
1
t
<
1
2
(
−
1
t
+ δt
)
= −
(3 − 2c)t − 1
2t2
< 0.
Hence the equality∥∥∥∥u + 12
(
−
1
t
+ δt
)
wt
∥∥∥∥ = 1 +
(
1 − 2(1 − c)t
)(
(3 − 2c)t − 1
)
2t2
holds. As in Lemma 4.5, one can consider the following two cases:
If 0 < c < 1/4, then the above function is decreasing and hence takes the
supremum
1 +
(
1 − 2(1 − c)/2
)(
(3 − 2c)/2 − 1
)
2(1/2)2
= 1 + c(1 − 2c) = (1 − c)(1 + 2c)
When 1/4 ≤ c < 1/2, we have that the above function takes maximum
9/8 at t = 2/(5 − 4c).
Indeed, (1−c)(1+2c) is less than 2(1−c) for any c ∈ [0,1/2). Meanwhile,
it is easy to see that 2(1 − c) < 9/8 only if c > 7/16. Therefore we have
206 h. mizuguchi
Proposition 4.12. Let c ∈ [0,1/2]. Then IB(Xc)−1 = DW(Xc)/2 coin-
cide with
max
{
2(1 − c), 1 +
c(
1 +
√
2(1 − c)
)2
}
if 0 ≤ c ≤
3 −
√
5
4
,
max
{
2(1 − c), 1 +
c(
1 +
√
2(1 − c)
)2 ,
1 +
1 − c(
1 +
√
2c
)2
}
if
3 −
√
5
4
< c <
7
16
,
max
{
1 +
c(
1 +
√
2(1 − c)
)2 , 1 + 1 − c(
1 +
√
2c
)2 , 98
}
if
7
16
≤ c ≤
1
2
.
Considering the symmetry of the functions
c(
1 +
√
2(1 − c)
)2 and 1 − c(
1 +
√
2c
)2
and that these function takes value 1/8 at t = 1/2, we finally obtain
Theorem 4.13. Let c ∈ [0,1] and put d = max{c,1 − c}. Then both
DW(Xc) and DW(X
∗
c ) coincide with
2 max
{
2d, 1 +
d(
1 +
√
2(1 − d)
)2
}
.
This is more than DW(ℓψc-ℓψ̃∗c
) = 9/4 and the equality holds only when
c = 1/2.
Theorem 4.14. Let c ∈ [0,1] and put d = max{c,1 − c}. Then both
IB(Xc) and IB(X
∗
c ) coincide with
min
{
1
2d
,
(
1 +
√
2(1 − d)
)2
d +
(
1 +
√
2(1 − d)
)2
}
.
This is less than IB(ℓψc-ℓψ̃∗c
) = 8/9 and the equality holds only when c = 1/2.
difference between orthogonality in radon planes 207
The graphs of the functions y = 2x, y = 2(1 − x), y = 1 +
x(
1 +
√
2(1 − x)
)2 ,
y = 1 +
1 − x(
1 +
√
2x
)2 and y = 98 on the interval [0,1].
Acknowledgements
The author would like to thank the anonymous referee for the careful
reading, valuable comments and suggestions to improving this paper.
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