# Another Class of Distances and Continuous Quasi-Distances in Product Spaces

Authors: Catherine Peppo
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We construct a class of continuous quasi-distances in a product of metric spaces and show that, generally, when the parameterλ (as shown in the paper) is positive, d is a distance and when λ
<
0, d is only a continuous quasi-distance, but not a distance. It is remarkable that the same result in relation to the sign of λ was found for two other classes of continuous quasi-distances (see Peppo (2010a, 2010b) and Peppo (2011)). This conclusion is due to the fact that E is a product space. For the purposes of our main result, a notion of density in metric spaces is introduced.


---

## Body

## 1. Introduction

In this paper a quasi-distanced on a set E is defined as a function d
:
E
2
↦
[
0
;
+
∞
[ with the usual properties of a metric and a weaker version of the triangle inequality:
(1)
d
(
x
,
y
)
≤
k
[
d
(
x
,
z
)
+
d
(
y
,
z
)
]
,
where k
≥
1.This function is not always continuous with respect to thed-topology generated by itself in the same manner as by a distance. This is because the “open” balls B
(
a
,
r
)
=
{
x
∈
E
:
d
(
x
,
a
)
<
r
}, which form a base for a complete system of neighbourhoods of a
∈
E, are not always open sets in the d-topology. Examples of “open” balls that are not open sets can be found in [1].It is known that for every quasi-distanced, there exists another quasi-distance d
′, whose topology and uniformity are the same as those of d and the open d
′-balls are open sets [2].But if any condition or relation is satisfied with respect to a quasi-distanced, we do not know if, in general, the same condition or relation will be satisfied with respect to the “new” quasi-distance d
′, even if d is equivalent to d
′.For this reason, the continuity of a quasi-distance cannot be omitted without “danger.”In fixed-point theory, some authors require the continuity of the used quasi-distance as complementary condition (see, among others, [3]).The most interesting fact about the quasi-metric spaces is that, in many applications, they constitute a more general setting than the metric spaces without losing the good properties of these last spaces (see, among others, [4]).In [5–7] we proved that, generally, the functions defined in a product space for two points, P
(
x
1
,
x
2
,
…
,
x
n
) and Q
(
y
1
,
y
2
,
…
,
y
n
), respectively, by
(2)
d
(
P
,
Q
)
=
{
∑
i
=
1
n
‍
a
i
d
i
(
x
i
,
y
i
)
if
∑
i
=
1
n
‍
β
i
d
i
(
x
i
,
y
i
)
≥
0
,
∑
i
=
1
n
‍
a
i
′
d
i
(
x
i
,
y
i
)
if
∑
i
=
1
n
‍
β
i
d
i
(
x
i
,
y
i
)
<
0
,
d
(
P
,
Q
)
=
{
∑
i
=
1
n
‍
a
i
d
i
(
x
i
,
y
i
)
1
+
∑
i
=
1
n
‍
a
i
d
i
(
x
i
,
y
i
)
if
∑
i
=
1
n
β
i
d
i
(
x
i
,
y
i
)
≥
0
,
∑
i
=
1
n
‍
a
i
′
d
i
(
x
i
,
y
i
)
1
+
∑
i
=
1
n
‍
a
i
′
d
i
(
x
i
,
y
i
)
if
∑
i
=
1
n
β
i
d
i
(
x
i
,
y
i
)
<
0
,
with a
i
-
a
i
′
=
λ
β
i for all i
∈
I, are distances when λ
>
0 butonly continuous quasi-distances when λ
<
0.In this paper we revisit this problem for another family of functions in a product space defined by(3)
d
(
P
,
Q
)
=
{
∑
i
=
1
n
‍
a
i
2
d
i
2
(
x
i
,
y
i
)
if
∑
i
=
1
n
β
i
d
i
2
(
x
i
,
y
i
)
≥
0
,
∑
i
=
1
n
‍
a
i
′
2
d
i
2
(
x
i
,
y
i
)
if
∑
i
=
1
n
β
i
d
i
2
(
x
i
,
y
i
)
<
0
,
with a
i
2
-
a
i
′
2
=
λ
β
i and we find the same result.This makes us think that the conclusions reached are not coincidental.We have reason to believe that a similar result holds for other classes of quasi-distances in product spaces.Additionally, we notice that the iterated quasi-triangle constants of our quasi-distances,(4)
d
(
P
1
,
P
n
)
≤
k
[
d
(
P
1
,
P
2
)
+
d
(
P
2
,
P
3
)
+
⋯
+
d
(
P
n
-
1
,
P
n
)
]
,
which we are callinggeneralized improved quasi-triangle inequality, improve the general one:
(5)
d
(
P
1
,
P
n
)
≤
k
d
(
P
1
,
P
2
)
+
k
2
d
(
P
2
,
P
3
)
+
⋯
+
k
n
-
3
d
(
P
n
-
3
,
P
n
-
2
)
+
k
n
-
2
d
(
P
n
-
2
,
P
n
-
1
)
+
k
n
-
2
d
(
P
n
-
1
,
P
n
)
(the two last coefficients are k
n
-
2; it is not a mistake).In Section2 we introduce the notion of a density condition in metric spaces for the purposes of our main result.In Section4 we give a counterexample proving the importance of the (
α
,
-
β
i
/
β
j
)-density condition in part III of our main result.


## 2. Density in Metric Spaces

In the third part of our main result (Section3) we will need metric spaces satisfying an additional property of density that we are calling (
α
,
k
)-density condition.Definition 1.
For anα
∈
]
0
;
1
[ and a positive number k, the pair of metric spaces (
E
,
d
) and (
E
′
,
d
′
) satisfies the (
α
,
k
)-density condition if the space E contains at least three distinct points, x
,
y
,
z, satisfying the relations
(6)
d
(
x
,
z
)
=
α
d
(
x
,
y
)
,
d
(
y
,
z
)
=
(
1
-
α
)
d
(
x
,
y
)
and for every β
∈
]
α
;
1
[ there exist at least three distinct points x
′
,
y
′
,
z
′
∈
E
′, satisfying the relations
(7)
d
′
(
x
′
,
z
′
)
=
β
d
′
(
x
′
,
y
′
)
,
d
′
(
y
′
,
z
′
)
=
(
1
-
β
)
d
′
(
x
′
,
y
′
)
,
d
′
(
x
′
,
y
′
)
=
k
d
(
x
,
y
)
.In terms of segments, Definition1 may be expressed as follows.For anα
∈
]
0
;
1
[ and a positive number k, the pair of metric spaces (
E
,
d
) and (
E
′
,
d
′
) satisfies the (
α
,
k
)-density condition if E contains at least a segment, [
x
,
y
], containing a point z dividing it at the ratio α
:
1
-
α and, for every β
∈
]
α
;
1
[, E
′ contains at least a segment [
x
′
,
y
′
], k time “longer” in (
E
′
,
d
′
) than [
x
,
y
] in (
E
,
d
) and containing a point z
′ dividing it at the ratio β
:
1
-
β.By segment we mean the set defined by the following:Definition 2.
For two pointsx
,
y in a metric space (
E
,
d
), the segment [
x
;
y
] is the set of points z
∈
E satisfying
(8)
d
(
x
,
y
)
=
d
(
x
,
z
)
+
d
(
z
,
y
)
.
It is obvious that the extremitiesx and y belong to the segment [
x
;
y
], but it may only be reduced at its two extremities.We note also that, despite appearances, ifu
∈
[
x
;
y
], v
∈
[
x
;
y
], and d
(
x
,
u
)
=
d
(
x
,
v
), this does not mean, in general, that u
=
v, as in the example below (of course, in the particular case when d
(
x
,
u
)
=
d
(
x
,
v
)
=
0, we have u
=
v
=
x).Example 3.
Leta
,
b be two strictly positive numbers satisfying max
⁡
(
a
,
b
)
≤
min
⁡
(
2
a
,
2
b
), c
=
a
+
b, c
1 a number satisfying max
⁡
(
a
,
b
)
≤
c
1
≤
min
⁡
(
2
a
,
2
b
), and E a set containing whatever four points A
(
x
1
,
y
1
), B
(
x
2
,
y
2
), C
(
x
3
,
y
3
), and D
(
x
4
,
y
4
) of R
2. Define the function d by
(9)
d
(
A
,
B
)
=
d
(
B
,
A
)
=
c
,
d
(
A
,
C
)
=
d
(
C
,
A
)
=
b
,
d
(
B
,
C
)
=
d
(
C
,
B
)
=
a
,
d
(
A
,
D
)
=
d
(
D
,
A
)
=
b
,
d
(
B
,
D
)
=
d
(
D
,
B
)
=
a
,
d
(
C
,
D
)
=
d
(
D
,
C
)
=
c
1
,
d
(
A
,
A
)
=
d
(
B
,
B
)
=
d
(
C
,
C
)
=
d
(
D
,
D
)
=
0
.
The two first properties of a distance are obviously satisfied.So, to prove thatd is a distance, it suffices to prove the triangle inequality in all possible “triangles” formed with the points of E, that is, A
B
C, A
B
D, A
C
D, and B
C
D (Figure 1).Figure 1In the “triangles”A
B
C and A
B
D, the “sides” are a
,
b
,
c and the greatest one is c
=
a
+
b, so the “triangle” inequality is satisfied.In the “triangles”A
C
D and B
C
D, the “sides” are, respectively, b
,
b
,
c
1 and a
,
a
,
c
1 and the greatest one is, by our choice, c
1; or c
1
≤
min
⁡
(
2
a
,
2
b
)
≤
2
a, and c
1
≤
min
⁡
(
2
a
,
2
b
)
≤
2
b, so the “triangle” inequality is satisfied.The pointsC and D belong to [
A
,
B
], because d
(
C
,
A
)
+
d
(
C
,
B
)
=
b
+
a
=
c
=
d
(
A
,
B
) and d
(
D
,
A
)
+
d
(
D
,
B
)
=
b
+
a
=
c
=
d
(
A
,
B
). On the other hand d
(
C
,
A
)
=
d
(
D
,
A
)
=
b, but, generally, C
≠
D.We note that the(
α
,
k
)-density condition, for an α
∈
]
0
;
1
[ and a positive number k, as it is required in the third part of our main result (Section 3), is not a very restrictive condition. For example, if d is the usual distance in R
2, α
=
1
/
3, and k
=
2, in order that the pair ((
E
,
d
); (
E
′
,
d
′
)) satisfies the (
1
/
3
,
2
)-density condition, it suffices that (
E
,
d
) contains the set {
(
1
;
1
)
,
(
2
;
1
)
,
(
4
;
1
)
}, while (
E
′
,
d
′
) contains the set {
(
1
;
2
)
}
⋃
{
(
x
;
2
)
:
3
≤
x
≤
7
} as in Figure 2.Figure 2Clearly, every pair of normed vector spaces overR satisfies the (
α
,
k
)
-density condition for each α
∈
]
0
;
1
[ and for each positive number k.


## 3. Main Result

Theorem 4.
Let(
E
i
,
d
i
) for i
∈
{
1,2
,
…
,
n
} be metric spaces, λ and β
i real numbers, a
i and a
i
′ positive numbers satisfying a
i
2
-
a
i
′
2
=
λ
β
i, and P
(
x
1
,
x
2
,
…
,
x
n
) and Q
(
y
1
,
y
2
,
…
,
y
n
) two arbitrary points of E
=
∏
i
=
1
n
E
i. Then(i)
the functiond
:
E
2
↦
R
+ defined by
(10)
d
(
P
,
Q
)
=
{
∑
i
=
1
n
‍
a
i
2
d
i
2
(
x
i
,
y
i
)
i
f
∑
i
=
1
n
β
i
d
i
2
(
x
i
,
y
i
)
≥
0
∑
i
=
1
n
‍
a
i
′
2
d
i
2
(
x
i
,
y
i
)
i
f
∑
i
=
1
n
β
i
d
i
2
(
x
i
,
y
i
)
<
0
is a continuous quasi-distance on E satisfying thegeneralized improved triangle inequality for every m points;
(ii)
ifλ
≥
0 or β
i
s are all nonnegative or all nonpositive, d is a distance;
(iii)
ifλ
<
0 and there exist two indexes (
i
,
j
)
∈
{
1,2
,
…
,
n
} with β
i
<
0, β
j
>
0 and an α
∈
]
0
;
1
[ such that the spaces E
i and E
j satisfy the (
α
,
-
β
i
/
β
j
)-density condition, the continuous quasi-distance d is not a distance.Proof.
(i) First, we will prove thatd is a quasi-distance and then that d is a continuous function of its two variables with respect to the d-topology.
It is clear thatd
(
P
,
Q
)
≥
0, that d is symmetric, and that d
(
P
,
Q
)
=
0
⇔
P
=
Q.
LetK
1
=
min
⁡
{
a
i
,
a
i
′
}
1
≤
i
≤
n, K
2
=
max
⁡
{
a
i
,
a
i
′
}
1
≤
i
≤
n, and D
(
P
,
Q
)
=
∑
i
=
1
n
d
i
2
(
x
i
,
y
i
).
Clearly, for whatever pointsP and Q, we have
(11)
K
1
D
(
P
,
Q
)
≤
d
(
P
,
Q
)
≤
K
2
D
(
P
,
Q
)
.
AsD is a distance, we can write, for whatever points P
1
(
x
1
(
1
)
,
x
2
(
1
)
,
…
,
x
n
(
1
)
), P
2
(
x
1
(
2
)
,
x
2
(
2
)
,
…
,
x
n
(
2
)
)
,
…, P
i
(
x
1
(
i
)
,
x
2
(
i
)
,
…
,
x
n
(
i
)
),
(12)
d
(
P
1
,
P
i
)
≤
K
2
D
(
P
1
,
P
i
)
≤
K
2
[
D
(
P
1
,
P
2
)
+
D
(
P
2
,
P
3
)
+
⋯
+
D
(
P
i
-
1
,
P
i
)
]
≤
K
2
K
1
[
d
(
P
1
,
P
2
)
+
d
(
P
2
,
P
3
)
+
⋯
+
d
(
P
i
-
1
,
P
i
)
]
.
This means thatd is a quasi-distance with constant K
=
K
2
/
K
1, and the same constant holds for whatever m points. Let us show now that d is continuous with respect to the d-topology.
Inequalities (11) show that d and D are topologically equivalent. Hence to prove that d is continuous, we will show that for every two sequences, P
k
(
x
1
(
k
)
,
x
2
(
k
)
,
…
,
x
n
(
k
)
) and Q
k
(
y
1
(
k
)
,
y
2
(
k
)
,
…
,
y
n
(
k
)
), converging, respectively, to P
(
x
1
,
x
2
,
…
,
x
n
) and Q
(
y
1
,
y
2
,
…
,
y
n
) for k
→
+
∞, d
(
P
k
,
Q
k
) converges to d
(
P
,
Q
).
For the pointsP
(
x
1
,
x
2
,
…
,
x
n
) and Q
(
y
1
,
y
2
,
…
,
y
n
), in the following, we will denote g
(
P
,
Q
)
=
∑
i
=
1
n
a
i
2
d
i
2
(
x
i
,
y
i
), h
(
P
,
Q
)
=
∑
i
=
1
n
a
i
′
2
d
i
2
(
x
i
,
y
i
), and A
=
∑
i
=
1
n
β
i
d
i
2
(
x
i
,
y
i
).
IfA
>
0
(
A
<
0
), an integer p exists such that for k
∈
N, k
≥
p we have ∑
i
=
1
n
β
i
d
i
2
(
x
i
(
k
)
,
y
i
(
k
)
)
>
0 (∑
i
=
1
n
β
i
d
i
2
(
x
i
(
k
)
,
y
i
(
k
)
)
<
0) (because the sequence ∑
i
=
1
n
β
i
d
i
2
(
x
i
(
k
)
,
y
i
(
k
)
) converges to A).
Fork
∈
N, k
≥
p, we have d
(
P
k
,
Q
k
)
=
g
(
P
k
,
Q
k
) (d
(
P
k
,
Q
k
)
=
h
(
P
k
,
Q
k
)) and, consequently, d
(
P
k
,
Q
k
) converges to g
(
P
,
Q
)
=
d
(
P
,
Q
) and (h
(
P
,
Q
)
=
d
(
P
,
Q
)).
IfA
=
0, from a
i
2
-
a
i
′
2
=
λ
β
i we deduce
(13)
∑
i
=
1
n
‍
a
i
2
d
i
2
(
x
i
,
y
i
)
=
∑
i
=
1
n
(
a
i
′
2
+
λ
β
i
)
d
i
2
(
x
i
,
y
i
)
=
∑
i
=
1
n
‍
a
i
′
2
d
i
2
(
x
i
,
y
i
)
+
λ
A
=
∑
i
=
1
n
‍
a
i
′
2
d
i
2
(
x
i
,
y
i
)
.
That is, g
(
P
,
Q
)
=
h
(
P
,
Q
).
It follows thatg
(
P
k
,
Q
k
) and h
(
P
k
,
Q
k
) converge to the same limit g
(
P
,
Q
)
=
h
(
P
,
Q
)
=
d
(
P
,
Q
). We can conclude that d
(
P
k
,
Q
k
), formed by the terms of the two sequences above, also converges to d
(
P
,
Q
). Thus, d is continuous and the “open” balls B
(
P
0
,
r
)
=
{
P
∈
E
:
d
(
P
,
P
0
)
<
r
} really are, in fact, open sets for the d-topology.
(ii) Ifβ
is are all nonnegative, d
(
P
,
Q
)
=
g
(
P
,
Q
) for every P and Q; if β
is are all nonpositive, d
(
P
,
Q
)
=
h
(
P
,
Q
) for every P and Q (recall that if ∑
i
=
1
n
β
i
d
i
2
(
x
i
,
y
i
)
=
0, d
(
P
,
Q
)
=
g
(
P
,
Q
)
=
h
(
P
,
Q
)). As g and h are distances, d is a distance too.
Ifλ
=
0, a
i
=
a
i
′ for every i
∈
{
1,2
,
…
,
n
}, so d
(
P
,
Q
)
=
g
(
P
,
Q
)
=
h
(
P
,
Q
) for every P and Q; that is, d is a distance.
Suppose now thatλ
>
0; for two arbitrary points P
(
x
1
,
x
2
,
…
,
x
n
) and Q
(
y
1
,
y
2
,
…
,
y
n
), denoting again A
=
∑
i
=
1
n
β
i
d
i
2
(
x
i
,
y
i
) we have
(14)
g
(
P
,
Q
)
=
∑
i
=
1
n
‍
a
i
2
d
i
2
(
x
i
,
y
i
)
=
∑
i
=
1
n
(
a
i
′
2
+
λ
β
i
)
d
i
2
(
x
i
,
y
i
)
=
∑
i
=
1
n
‍
a
i
′
2
d
i
2
(
x
i
,
y
i
)
+
λ
A
.
So, forλ
>
0 it follows that
(15)
A
≥
0
⟹
g
(
P
,
Q
)
≥
h
(
P
,
Q
)
,
A
<
0
⟹
g
(
P
,
Q
)
<
h
(
P
,
Q
)
.This means that ifλ
>
0, d
=
max
⁡
(
g
,
h
), and since the maximum of two distances is a distance, d is a distance.Notice that in this part of the theorem we did not need any supplementary density condition on the spaces(
E
i
,
d
i
).(iii) Suppose now thatλ
<
0 and that there exist an α
∈
]
0
;
1
[ and two indexes i
,
j
∈
{
1,2
,
…
,
n
} such that β
i
<
0, β
j
>
0, and the pair (
E
i
,
E
j
) satisfies the (
α
,
-
β
i
/
β
j
)-density condition.We choose aβ
∈
]
α
;
1
[ satisfying the relation
(16)
-
β
i
β
j
a
j
2
a
i
2
(
β
-
α
)
<
λ
2
β
i
2
β
j
2
(
β
-
α
)
(
1
-
α
+
β
2
)
2
+
2
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
λ
β
i
β
j
(
1
-
α
+
β
2
)
.
Such a β
>
α exists because, passing into limit in (16) as β
→
α
+, we obtain
(17)
0
<
2
λ
β
i
β
j
(
1
-
α
)
α
(
β
j
a
i
2
-
β
i
a
j
2
)
that is clearly true.As(
E
i
,
E
j
) satisfies the (
α
,
-
β
i
/
β
j
)-density condition, there exist three points x
i
,
y
i
,
z
i
∈
E
i, satisfying the relations d
i
(
x
i
,
z
i
)
=
α
d
i
(
x
i
,
y
i
) and d
i
(
y
i
,
z
i
)
=
(
1
-
α
)
d
i
(
x
i
,
y
i
), and three points x
j
,
y
j
,
z
j
∈
E
j, satisfying the relations
(18)
d
j
(
x
j
,
z
j
)
=
β
d
j
(
x
j
,
y
j
)
,
d
j
(
y
j
,
z
j
)
=
(
1
-
β
)
d
j
(
x
j
,
y
j
)
,
d
j
(
x
j
,
y
j
)
=
-
β
i
β
j
d
i
(
x
i
,
y
i
)
.With the pointsx
i
,
y
i
,
z
i
∈
E
i and x
j
,
y
j
,
z
j
∈
E
j we construct the points P
(
…
,
x
i
,
…
,
x
j
,
…
), Q
(
…
,
y
i
,
…
,
y
j
,
…
), and R
(
…
,
z
i
,
…
,
z
j
,
…
) of E (except the ith and the jth coordinates, all the other coordinates are equal in P
,
Q, and R) and will prove that
(19)
d
(
P
,
Q
)
>
d
(
P
,
R
)
+
d
(
Q
,
R
)
.For the pointsP and Q we have from (18) that
(20)
∑
k
=
1
n
β
k
d
k
2
(
x
k
,
y
k
)
=
β
i
d
i
2
(
x
i
,
y
i
)
+
β
j
d
j
2
(
x
j
,
y
j
)
=
0
,
so d
(
P
,
Q
)
=
a
i
2
d
i
2
(
x
i
,
y
i
)
+
a
j
2
d
j
2
(
x
j
,
y
j
)
=
(
a
i
2
-
(
β
i
/
β
j
)
a
j
2
)
d
i
(
x
i
,
y
i
).For the pointsP and R,
(21)
∑
k
=
1
n
β
k
d
k
2
(
x
k
,
z
k
)
=
β
i
d
i
2
(
x
i
,
z
i
)
+
β
j
d
j
2
(
x
j
,
z
j
)
=
β
i
α
2
d
i
2
(
x
i
,
y
i
)
+
β
j
β
2
(
-
β
i
β
j
)
d
i
2
(
x
i
,
y
i
)
=
β
i
d
i
2
(
x
i
,
y
i
)
(
α
2
-
β
2
)
>
0
,
because β
i
<
0 and 0
<
α
<
β, so d
(
P
,
R
)
=
a
i
2
d
i
2
(
x
i
,
z
i
)
+
a
j
2
d
j
2
(
x
j
,
z
j
)
=
(
a
i
2
α
2
-
a
j
2
β
2
(
β
i
/
β
j
)
)
d
i
(
x
i
,
y
i
).For the pointsQ and R,
(22)
∑
k
=
1
n
β
k
d
k
2
(
x
k
,
y
k
)
=
β
i
d
i
2
(
y
i
,
z
i
)
+
β
j
d
j
2
(
y
j
,
z
j
)
=
β
i
(
1
-
α
)
2
d
i
2
(
x
i
,
y
i
)
-
β
j
(
1
-
β
)
2
β
i
β
j
d
i
2
(
x
i
,
y
i
)
=
β
i
d
i
2
(
x
i
,
y
i
)
[
(
1
-
α
)
2
-
(
1
-
β
)
2
]
<
0
,
because β
i
<
0 and (
1
-
α
)
2
>
(
1
-
β
)
2, so d
(
Q
,
R
)
=
a
i
′
2
d
i
2
(
y
i
,
z
i
)
+
a
j
′
2
d
j
2
(
y
j
,
z
j
)
=
a
i
′
2
(
1
-
α
)
2
-
a
j
′
2
(
1
-
β
)
2
(
β
i
/
β
j
)
d
i
(
x
i
,
y
i
).We are ready to prove now the inequalityd
(
P
,
Q
)
>
d
(
P
,
R
)
+
d
(
Q
,
R
), transforming it equivalently to
(23)
a
i
2
-
β
i
β
j
a
j
2
d
i
(
x
i
,
y
i
)
>
a
i
2
α
2
-
a
j
2
β
2
β
i
β
j
d
i
(
x
i
,
y
i
)
+
a
i
′
2
(
1
-
α
)
2
-
a
j
′
2
(
1
-
β
)
2
β
i
β
j
×
d
i
(
x
i
,
y
i
)
⟺
a
i
2
-
β
i
β
j
a
j
2
>
a
i
2
α
2
-
a
j
2
β
2
β
i
β
j
+
a
i
′
2
(
1
-
α
)
2
-
a
j
′
2
(
1
-
β
)
2
β
i
β
j
⟺
β
j
a
i
2
-
β
i
a
j
2
>
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
+
β
j
a
i
′
2
(
1
-
α
)
2
-
β
i
a
j
′
2
(
1
-
β
)
2
.
From β
j
a
i
2
>
β
j
a
i
2
α
2 and -
β
i
a
j
2
>
-
β
i
a
j
2
β
2, we have β
j
a
i
2
-
β
i
a
j
2
>
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2, so β
j
a
i
2
-
β
i
a
j
2
-
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2 is positive and we can equivalently square
(24)
⟺
(
β
j
a
i
2
-
β
i
a
j
2
-
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
)
2
>
(
β
j
a
i
′
2
(
1
-
α
)
2
-
β
i
a
j
′
2
(
1
-
β
)
2
)
2
(25)
⟺
β
j
a
i
2
-
β
i
a
j
2
+
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
-
2
β
j
a
i
2
-
β
i
a
j
2
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
>
β
j
a
i
′
2
(
1
-
α
)
2
-
β
i
a
j
′
2
(
1
-
β
)
2
(because a
i
′
2
=
a
i
2
-
λ
β
i and a
j
′
2
=
a
j
2
-
λ
β
j),
(26)
⟺
-
2
β
j
a
i
2
-
β
i
a
j
2
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
>
β
j
a
i
2
(
1
-
α
)
2
+
λ
β
i
β
j
(
1
-
β
)
2
-
β
j
a
i
2
-
β
j
a
i
2
α
2
-
λ
β
i
β
j
(
1
-
α
)
2
-
β
i
a
j
2
(
1
-
β
)
2
+
β
i
a
j
2
+
β
i
a
j
2
β
2
⟺
-
2
β
j
a
i
2
-
β
i
a
j
2
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
>
β
j
a
i
2
[
(
1
-
α
)
2
-
1
-
α
2
]
+
λ
β
i
β
j
[
(
1
-
β
)
2
-
(
1
-
α
)
2
]
-
β
i
a
j
2
[
(
1
-
β
)
2
-
1
-
β
2
]
⟺
-
2
β
j
a
i
2
-
β
i
a
j
2
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
>
β
i
a
j
2
2
β
-
β
j
a
i
2
2
α
+
λ
β
i
β
j
(
α
-
β
)
(
2
-
α
-
β
)
⟺
β
j
a
i
2
-
β
i
a
j
2
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
<
-
β
i
a
j
2
β
+
β
j
a
i
2
α
+
λ
β
i
β
j
(
β
-
α
)
(
1
-
α
+
β
2
)
(we square)
(27)
⟺
(
β
j
a
i
2
-
β
i
a
j
2
)
(
β
j
a
i
2
α
2
-
β
i
a
j
2
β
2
)
<
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
2
+
[
λ
β
i
β
j
(
β
-
α
)
(
1
-
α
+
β
2
)
]
2
+
2
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
λ
β
i
β
j
(
β
-
α
)
(
1
-
α
+
β
2
)
⟺
β
j
2
a
i
4
α
2
-
β
i
β
j
a
j
2
a
i
2
α
2
-
β
i
β
j
a
j
2
a
i
2
β
2
+
β
i
2
a
j
4
β
2
<
β
i
2
a
j
4
β
2
+
β
j
2
a
i
4
α
2
-
2
β
i
β
j
a
i
2
a
j
2
α
β
+
λ
2
β
i
2
β
j
2
(
β
-
α
)
2
(
1
-
α
+
β
2
)
2
+
2
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
λ
β
i
β
j
(
β
-
α
)
(
1
-
α
+
β
2
)
⟺
-
β
i
β
j
a
j
2
a
i
2
α
2
-
β
i
β
j
a
j
2
a
i
2
β
2
<
-
2
β
i
β
j
a
i
2
a
j
2
α
β
+
λ
2
β
i
2
β
j
2
(
β
-
α
)
2
(
1
-
α
+
β
2
)
2
+
2
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
λ
β
i
β
j
(
β
-
α
)
(
1
-
α
+
β
2
)
⟺
β
i
β
j
a
j
2
a
i
2
(
2
α
β
-
α
2
-
β
2
)
<
λ
2
β
i
2
β
j
2
(
β
-
α
)
2
(
1
-
α
+
β
2
)
2
+
2
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
λ
β
i
β
j
×
(
β
-
α
)
(
1
-
α
+
β
2
)
⟺
-
β
i
β
j
a
j
2
a
i
2
(
β
-
α
)
2
<
λ
2
β
i
2
β
j
2
(
β
-
α
)
2
(
1
-
α
+
β
2
)
2
+
2
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
λ
β
i
β
j
×
(
β
-
α
)
(
1
-
α
+
β
2
)
.
Dividing by (
β
-
α
) we have
(28)
⟺
-
β
i
β
j
a
j
2
a
i
2
(
β
-
α
)
<
λ
2
β
i
2
β
j
2
(
β
-
α
)
(
1
-
α
+
β
2
)
2
+
2
(
-
β
i
a
j
2
β
+
β
j
a
i
2
α
)
λ
β
i
β
j
(
1
-
α
+
β
2
)
.
The last inequality is true from (16). The proof is over. So, d is not a distance.Example 5.
ForP
(
x
1
,
x
2
) and Q
(
y
1
,
y
2
)
∈
R
2, the function defined by
(29)
d
(
P
,
Q
)
=
{
2
(
x
1
-
y
1
)
2

+

(
x
2
-
y
2
)
2
if
2
(
x
2
-
y
2
)
2

≥

(
x
1
-
y
1
)
2
,
(
x
1
-
y
1
)
2

+

3
(
x
2
-
y
2
)
2
if
2
(
x
2
-
y
2
)
2

<

(
x
1
-
y
1
)
2
,
is a continuous quasi-distance, but not a distance. Here, λ
=
-
1, β
1
=
-
1
,
β
2
=
2, a
1
2
=
2, a
1
′
2
=
1, a
2
2
=
1, and a
2
′
2
=
3, while the function defined by
(30)
d
(
P
,
Q
)
=
{
(
x
1
-
y
1
)
2

+

3
(
x
2
-
y
2
)
2
if
2
(
x
2
-
y
2
)
2

≥

(
x
1
-
y
1
)
2
,
2
(
x
1
-
y
1
)
2

+

(
x
2
-
y
2
)
2
if
2
(
x
2
-
y
2
)
2

<

(
x
1
-
y
1
)
2
,
is a distance.Here,λ
=
1, β
1
=
-
1, β
2
=
2, a
1
2
=
1, a
1
′
2
=
2, a
2
2
=
3, and a
2
′
2
=
1.


## 4. A Counterexample

Showing the importance of the(
α
,
-
β
i
/
β
j
)-density condition in part III of the theorem.LetE
1
=
{
x
1
,
y
1
}, E
2
=
{
x
2
,
y
2
} be two metric spaces with distances, respectively, d
1
,
d
2, E
=
E
1
×
E
2
=
{
P
(
x
1
,
x
2
)
,
Q
(
x
1
,
y
2
)
,
R
(
y
1
,
x
2
)
,
T
(
y
1
,
y
2
)
}, x
i
≠
y
i, β
i, real numbers, and a
i
,
a
i
′ positive numbers satisfying, for i
=
1,2, a
i
2
-
a
i
′
2
=
λ
β
i, λ
<
0, β
1
<
0, β
2
>
0, and (
a
1
2
-
a
1
′
2
)
(
a
2
2
-
a
2
′
2
)
+
4
a
1
′
2
a
2
2
≥
0
⇔
λ
2
β
1
β
2
+
4
a
1
′
2
a
2
2
≥
0.For whateverα
∈
]
0
;
1
[, it is obvious that the pair (
(
E
1
,
d
1
)
,
(
E
2
,
d
2
)
) does not satisfy the (
α
,
-
β
i
/
β
j
)-density condition.We will show that the functiond
:
E
×
E
↦
R
+ defined for every two points M
(
z
1
,
z
2
)
,
N
(
t
1
,
t
2
)
∈
E by
(31)
d
(
M
,
N
)
=
{
a
1
2
d
1
2
(
z
1
,
z
2
)

+

a
2
2
d
2
2
(
t
1
,
t
2
)
if
β
1
d
1
2
(
z
1
,
z
2
)
+
β
2
d
2
2
(
t
1
,
t
2
)
≥
0
a
1
′
2
d
1
2
(
z
1
,
z
2
)

+

a
2
′
2
d
2
2
(
t
1
,
t
2
)
if
β
1
d
1
2
(
z
1
,
z
2
)
+
β
2
d
2
2
(
t
1
,
t
2
)
<
0
is a distance.For every three points ofE, we will show that the greatest “distance” (we put quotes because we do not yet know if d is really a distance) between these points is inferior or equal to the sum of the two others.


### 4.1. First Case(
β
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
≥
0
)

The six “distances” are(32)
d
(
P
,
Q
)
=
d
(
R
,
T
)
=
a
2
d
2
(
x
2
,
y
2
)
,
d
(
P
,
R
)
=
d
(
Q
,
T
)
=
a
1
′
d
1
(
x
1
,
y
1
)
,
d
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
.Asa
1
2
-
a
1
′
2
=
λ
β
1
>
0, we have a
1
>
a
1
′, so
(33)
a
1
2
d
1
2
(
x
1
,
y
1
)

+

a
2
2
d
2
2
(
x
2
,
y
2
)

≥

a
1
d
1
(
x
1
,
y
1
)

≥

a
1
′
d
1
(
x
1
,
y
1
)
,
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
≥
a
2
d
2
(
x
2
,
y
2
)
.The greatest of the six “distances” isd
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
).In each of the four possible “triangles,”P
Q
R, P
Q
T, P
R
T, and Q
R
T, there is a side equal to a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
), another is equal to a
1
′
d
1
(
x
1
,
y
1
), and the third is a
2
d
2
(
x
2
,
y
2
).Therefore we have to prove only one “triangle inequality,” valid for the four triangles:(34)
a
1
2
d
1
2
(
x
1
,
y
1
)

+

a
2
2
d
2
2
(
x
2
,
y
2
)

≤

a
1
′
d
1
(
x
1
,
y
1
)

+

a
2
d
2
(
x
2
,
y
2
)
,
or equivalently
(35)
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
≤
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
+
2
a
1
′
a
2
d
1
(
x
1
,
y
1
)
d
2
(
x
2
,
y
2
)
⟺
(
a
1
2
-
a
1
′
2
)
d
1
(
x
1
,
y
1
)
≤
2
a
1
′
a
2
d
2
(
x
2
,
y
2
)
⟺
d
1
2
(
x
1
,
y
1
)
d
2
2
(
x
2
,
y
2
)
≤
4
a
1
′
2
a
2
2
λ
2
β
1
2
.But fromβ
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
≥
0 we have d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
≤
β
2
/
-
β
1 and from λ
2
β
1
β
2
+
4
a
1
′
2
a
2
2
≥
0 we have β
2
/
-
β
1
≤
4
a
1
′
2
a
2
2
/
λ
2
β
1
2, so d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
≤
4
a
1
′
2
a
2
2
/
λ
2
β
1
2 and the triangle inequality is satisfied.


### 4.2. Second Case(
β
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
<
0
)

The six “distances” are(36)
d
(
P
,
Q
)
=
d
(
R
,
T
)
=
a
2
d
2
(
x
2
,
y
2
)
,
d
(
P
,
R
)
=
d
(
Q
,
T
)
=
a
1
′
d
1
(
x
1
,
y
1
)
,
d
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
.Asa
2
2
-
a
2
′
2
=
λ
β
2
<
0, we have a
2
<
a
2
′, so a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≥
a
2
′
d
2
(
x
2
,
y
2
)
>
a
2
d
2
(
x
2
,
y
2
), and a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≥
a
1
′
d
1
(
x
1
,
y
1
).The greatest of the six “distances” isd
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
). In each of the four possible “triangles,” P
Q
R, P
Q
T, P
R
T, and Q
R
T, there is a side equal to a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
), another is equal to a
1
′
d
1
(
x
1
,
y
1
), and the third is a
2
d
2
(
x
2
,
y
2
). Therefore we have to prove only one “triangle inequality,” valid for the four triangles:
(37)
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≤
a
1
′
d
1
(
x
1
,
y
1
)
+
a
2
d
2
(
x
2
,
y
2
)
,
that we square equivalently
(38)
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≤
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
+
2
a
1
′
a
2
d
1
(
x
1
,
y
1
)
d
2
(
x
2
,
y
2
)
⟺
(
a
2
′
2
-
a
2
2
)
d
2
(
x
2
,
y
2
)
≤
2
a
1
′
a
2
d
1
(
x
1
,
y
1
)
⟺
d
1
2
(
x
1
,
y
1
)
d
2
2
(
x
2
,
y
2
)
≥
λ
2
β
2
2
4
a
1
′
2
a
2
2
.But fromβ
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
<
0 we have d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
>
β
2
/
-
β
1 and from λ
2
β
1
β
2
+
4
a
1
′
2
a
2
2
≥
0 we have β
2
/
-
β
1
≥
4
a
1
′
2
a
2
2
/
λ
2
β
1
2, so d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
≥
λ
2
β
2
2
/
4
a
1
′
2
a
2
2 and the triangle inequality is satisfied. So, d is a distance.We note that if(
E
i
,
d
i
) are normed vector spaces, (
E
i
,
∥
∥
i
), very slight modifications transform our main result in its normed version and that, in this case, no density condition is needed.


## 4.1. First Case(
β
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
≥
0
)

The six “distances” are(32)
d
(
P
,
Q
)
=
d
(
R
,
T
)
=
a
2
d
2
(
x
2
,
y
2
)
,
d
(
P
,
R
)
=
d
(
Q
,
T
)
=
a
1
′
d
1
(
x
1
,
y
1
)
,
d
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
.Asa
1
2
-
a
1
′
2
=
λ
β
1
>
0, we have a
1
>
a
1
′, so
(33)
a
1
2
d
1
2
(
x
1
,
y
1
)

+

a
2
2
d
2
2
(
x
2
,
y
2
)

≥

a
1
d
1
(
x
1
,
y
1
)

≥

a
1
′
d
1
(
x
1
,
y
1
)
,
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
≥
a
2
d
2
(
x
2
,
y
2
)
.The greatest of the six “distances” isd
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
).In each of the four possible “triangles,”P
Q
R, P
Q
T, P
R
T, and Q
R
T, there is a side equal to a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
), another is equal to a
1
′
d
1
(
x
1
,
y
1
), and the third is a
2
d
2
(
x
2
,
y
2
).Therefore we have to prove only one “triangle inequality,” valid for the four triangles:(34)
a
1
2
d
1
2
(
x
1
,
y
1
)

+

a
2
2
d
2
2
(
x
2
,
y
2
)

≤

a
1
′
d
1
(
x
1
,
y
1
)

+

a
2
d
2
(
x
2
,
y
2
)
,
or equivalently
(35)
a
1
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
≤
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
+
2
a
1
′
a
2
d
1
(
x
1
,
y
1
)
d
2
(
x
2
,
y
2
)
⟺
(
a
1
2
-
a
1
′
2
)
d
1
(
x
1
,
y
1
)
≤
2
a
1
′
a
2
d
2
(
x
2
,
y
2
)
⟺
d
1
2
(
x
1
,
y
1
)
d
2
2
(
x
2
,
y
2
)
≤
4
a
1
′
2
a
2
2
λ
2
β
1
2
.But fromβ
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
≥
0 we have d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
≤
β
2
/
-
β
1 and from λ
2
β
1
β
2
+
4
a
1
′
2
a
2
2
≥
0 we have β
2
/
-
β
1
≤
4
a
1
′
2
a
2
2
/
λ
2
β
1
2, so d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
≤
4
a
1
′
2
a
2
2
/
λ
2
β
1
2 and the triangle inequality is satisfied.


## 4.2. Second Case(
β
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
<
0
)

The six “distances” are(36)
d
(
P
,
Q
)
=
d
(
R
,
T
)
=
a
2
d
2
(
x
2
,
y
2
)
,
d
(
P
,
R
)
=
d
(
Q
,
T
)
=
a
1
′
d
1
(
x
1
,
y
1
)
,
d
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
.Asa
2
2
-
a
2
′
2
=
λ
β
2
<
0, we have a
2
<
a
2
′, so a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≥
a
2
′
d
2
(
x
2
,
y
2
)
>
a
2
d
2
(
x
2
,
y
2
), and a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≥
a
1
′
d
1
(
x
1
,
y
1
).The greatest of the six “distances” isd
(
Q
,
R
)
=
d
(
P
,
T
)
=
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
). In each of the four possible “triangles,” P
Q
R, P
Q
T, P
R
T, and Q
R
T, there is a side equal to a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
), another is equal to a
1
′
d
1
(
x
1
,
y
1
), and the third is a
2
d
2
(
x
2
,
y
2
). Therefore we have to prove only one “triangle inequality,” valid for the four triangles:
(37)
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≤
a
1
′
d
1
(
x
1
,
y
1
)
+
a
2
d
2
(
x
2
,
y
2
)
,
that we square equivalently
(38)
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
′
2
d
2
2
(
x
2
,
y
2
)
≤
a
1
′
2
d
1
2
(
x
1
,
y
1
)
+
a
2
2
d
2
2
(
x
2
,
y
2
)
+
2
a
1
′
a
2
d
1
(
x
1
,
y
1
)
d
2
(
x
2
,
y
2
)
⟺
(
a
2
′
2
-
a
2
2
)
d
2
(
x
2
,
y
2
)
≤
2
a
1
′
a
2
d
1
(
x
1
,
y
1
)
⟺
d
1
2
(
x
1
,
y
1
)
d
2
2
(
x
2
,
y
2
)
≥
λ
2
β
2
2
4
a
1
′
2
a
2
2
.But fromβ
1
d
1
2
(
x
1
,
y
1
)
+
β
2
d
2
2
(
x
2
,
y
2
)
<
0 we have d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
>
β
2
/
-
β
1 and from λ
2
β
1
β
2
+
4
a
1
′
2
a
2
2
≥
0 we have β
2
/
-
β
1
≥
4
a
1
′
2
a
2
2
/
λ
2
β
1
2, so d
1
2
(
x
1
,
y
1
)
/
d
2
2
(
x
2
,
y
2
)
≥
λ
2
β
2
2
/
4
a
1
′
2
a
2
2 and the triangle inequality is satisfied. So, d is a distance.We note that if(
E
i
,
d
i
) are normed vector spaces, (
E
i
,
∥
∥
i
), very slight modifications transform our main result in its normed version and that, in this case, no density condition is needed.



---

*Source: 861362-2014-03-18.xml*