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

Abstract

We construct a class of continuous quasi-distances in a product of metric spaces and show that, generally, when the parameter 𝜆 (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.

1. Introduction

In this paper a quasi-distance 𝑑 on a set 𝐸 is defined as a function 𝑑∶𝐸²↦[0;+∞[ with the usual properties of a metric and a weaker version of the triangle inequality: 𝑑𝑥,𝑦≤𝑘𝑑(𝑥,𝑧)+𝑑𝑦,𝑧, (1) where 𝑘≥1.

This function is not always continuous with respect to the 𝑑-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-distance 𝑑, 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-distance 𝑑, 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(𝑥₁,𝑥₂,…,𝑥_𝑛) and 𝑄(𝑦₁,𝑦₂,…,𝑦_𝑛), respectively, by 𝑑(𝑃,𝑄)=^𝑛𝑖=1𝑎_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖if^𝑛_𝑖=1𝛽_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖≥0,_𝑛𝑖=1𝑎^′_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖if^𝑛_𝑖=1𝛽_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖𝑑()=<0,𝑃,𝑄^𝑛𝑖=1𝑎_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖1+^𝑛_𝑖=1𝑎_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖if^𝑛_𝑖=1𝛽_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖≥0,_𝑛𝑖=1𝑎^′_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖1+^𝑛_𝑖=1𝑎^′_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖if^𝑛_𝑖=1𝛽_𝑖𝑑_𝑖𝑥_𝑖,𝑦_𝑖<0, (2) with 𝑎_𝑖−𝑎^′_𝑖=𝜆𝛽_𝑖 for all 𝑖∈𝐼, are distances when 𝜆>0 but only continuous quasi-distances when 𝜆<0.

In this paper we revisit this problem for another family of functions in a product space defined by 𝑑(𝑃,𝑄)=^𝑛𝑖=1𝑎²_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖if^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖≥0,_𝑛𝑖=1𝑎^′2_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖if^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖<0, (3) with 𝑎²_𝑖−𝑎^′2_𝑖=𝜆𝛽_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, 𝑑𝑃₁,𝑃_𝑛≤𝑘𝑑𝑃₁,𝑃₂+𝑑𝑃₂,𝑃₃+⋯+𝑑𝑃_𝑛−1,𝑃_𝑛, (4) which we are calling generalized improved quasi-triangle inequality, improve the general one: 𝑑𝑃₁,𝑃_𝑛≤𝑘𝑑𝑃₁,𝑃₂+𝑘²𝑑𝑃₂,𝑃₃+⋯+𝑘^𝑛−3𝑑𝑃_𝑛−3,𝑃_𝑛−2+𝑘^𝑛−2𝑑𝑃_𝑛−2,𝑃_𝑛−1+𝑘^𝑛−2𝑑𝑃_𝑛−1,𝑃_𝑛 (5) (the two last coefficients are 𝑘^𝑛−2; it is not a mistake).

In Section 2 we introduce the notion of a density condition in metric spaces for the purposes of our main result.

In Section 4 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 (Section 3) we will need metric spaces satisfying an additional property of density that we are calling (𝛼,𝑘)-density condition.

Definition 1. For an 𝛼∈]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 𝑑(𝑥,𝑧)=𝛼𝑑𝑥,𝑦,𝑑𝑦,𝑧=(1−𝛼)𝑑𝑥,𝑦 (6) and for every 𝛽∈]𝛼;1[ there exist at least three distinct points 𝑥^′,𝑦^′,𝑧^′∈𝐸^′, satisfying the relations 𝑑^′𝑥^′,𝑧^′=𝛽𝑑^′𝑥^′,𝑦^′,𝑑^′𝑦^′,𝑧^′=1−𝛽𝑑^′𝑥^′,𝑦^′,𝑑^′𝑥^′,𝑦^′=𝑘𝑑𝑥,𝑦. (7)

In terms of segments, Definition 1 may be expressed as follows.

For an 𝛼∈]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 points 𝑥,𝑦 in a metric space (𝐸,𝑑), the segment [𝑥;𝑦] is the set of points 𝑧∈𝐸 satisfying 𝑑𝑥,𝑦=𝑑(𝑥,𝑧)+𝑑𝑧,𝑦. (8)
It is obvious that the extremities 𝑥 and 𝑦 belong to the segment [𝑥;𝑦], but it may only be reduced at its two extremities.

We note also that, despite appearances, if 𝑢∈[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. Let 𝑎,𝑏 be two strictly positive numbers satisfying max(𝑎,𝑏)≤min(2𝑎,2𝑏), 𝑐=𝑎+𝑏, 𝑐₁ a number satisfying max(𝑎,𝑏)≤𝑐₁≤min(2𝑎,2𝑏), and 𝐸 a set containing whatever four points 𝐴(𝑥₁,𝑦₁), 𝐵(𝑥₂,𝑦₂), 𝐶(𝑥₃,𝑦₃), and 𝐷(𝑥₄,𝑦₄) of 𝑅². Define the function 𝑑 by 𝑑()()()()𝑑()()𝑑()()()()𝑑()()𝐴,𝐵=𝑑𝐵,𝐴=𝑐,𝑑𝐴,𝐶=𝑑𝐶,𝐴=𝑏,𝐵,𝐶=𝑑𝐶,𝐵=𝑎,𝐴,𝐷=𝑑𝐷,𝐴=𝑏,𝑑𝐵,𝐷=𝑑𝐷,𝐵=𝑎,𝐶,𝐷=𝑑𝐷,𝐶=𝑐₁,𝑑(𝐴,𝐴)=𝑑(𝐵,𝐵)=𝑑(𝐶,𝐶)=𝑑(𝐷,𝐷)=0. (9)
The two first properties of a distance are obviously satisfied.

So, to prove that 𝑑 is a distance, it suffices to prove the triangle inequality in all possible “triangles” formed with the points of 𝐸, that is, 𝐴𝐵𝐶, 𝐴𝐵𝐷, 𝐴𝐶𝐷, and 𝐵𝐶𝐷 (Figure 1).

Figure 1 

In the “triangles” 𝐴𝐵𝐶 and 𝐴𝐵𝐷, the “sides” are 𝑎,𝑏,𝑐 and the greatest one is 𝑐=𝑎+𝑏, so the “triangle” inequality is satisfied.

In the “triangles” 𝐴𝐶𝐷 and 𝐵𝐶𝐷, the “sides” are, respectively, 𝑏,𝑏,𝑐₁ and 𝑎,𝑎,𝑐₁ and the greatest one is, by our choice, 𝑐₁; or 𝑐₁≤min(2𝑎,2𝑏)≤2𝑎, and 𝑐₁≤min(2𝑎,2𝑏)≤2𝑏, so the “triangle” inequality is satisfied.

The points 𝐶 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 𝑅², 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 2 

Clearly, every pair of normed vector spaces over 𝑅 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 𝑎²_𝑖−𝑎^′2_𝑖=𝜆𝛽_𝑖, and 𝑃(𝑥₁,𝑥₂,…,𝑥_𝑛) and 𝑄(𝑦₁,𝑦₂,…,𝑦_𝑛) two arbitrary points of 𝐸=∏^𝑛_𝑖=1𝐸_𝑖. Then(i)the function 𝑑∶𝐸²↦𝑅⁺ defined by 𝑑(𝑃,𝑄)=^𝑛𝑖=1𝑎²_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖𝑖𝑓^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖≥0_𝑛𝑖=1𝑎^′2_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖𝑖𝑓^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖<0 (10)  is a continuous quasi-distance on 𝐸 satisfying the generalized improved triangle inequality for every 𝑚 points;(ii)if 𝜆≥0 or 𝛽_𝑖𝑠 are all nonnegative or all nonpositive, 𝑑 is a distance;(iii)if 𝜆<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 that 𝑑 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 that 𝑑(𝑃,𝑄)≥0, that 𝑑 is symmetric, and that 𝑑(𝑃,𝑄)=0⇔𝑃=𝑄.
Let 𝐾₁=min{𝑎_𝑖,𝑎^′_𝑖}_{1≤𝑖≤𝑛}, 𝐾₂=max{𝑎_𝑖,𝑎^′_𝑖}_{1≤𝑖≤𝑛}, and 𝐷(𝑃,𝑄)=∑^𝑛_𝑖=1𝑑²_𝑖(𝑥_𝑖,𝑦_𝑖).
Clearly, for whatever points 𝑃 and 𝑄, we have 𝐾₁𝐷(𝑃,𝑄)≤𝑑(𝑃,𝑄)≤𝐾₂𝐷(𝑃,𝑄). (11)
As 𝐷 is a distance, we can write, for whatever points 𝑃₁(𝑥₁⁽¹⁾,𝑥₂⁽¹⁾,…,𝑥_𝑛⁽¹⁾), 𝑃₂(𝑥₁⁽²⁾,𝑥₂⁽²⁾,…,𝑥_𝑛⁽²⁾),…, 𝑃_𝑖(𝑥₁^(𝑖),𝑥₂^(𝑖),…,𝑥_𝑛^(𝑖)), 𝑑𝑃₁,𝑃_𝑖≤𝐾₂𝐷𝑃₁,𝑃_𝑖≤𝐾₂𝐷𝑃₁,𝑃₂+𝐷𝑃₂,𝑃₃+⋯+𝐷𝑃_𝑖−1,𝑃_𝑖≤𝐾₂𝐾₁𝑑𝑃₁,𝑃₂+𝑑𝑃₂,𝑃₃+⋯+𝑑𝑃_𝑖−1,𝑃_𝑖. (12)
This means that 𝑑 is a quasi-distance with constant 𝐾=𝐾₂/𝐾₁, 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(𝑥₁^(𝑘),𝑥₂^(𝑘),…,𝑥_𝑛^(𝑘)) and 𝑄_𝑘(𝑦₁^(𝑘),𝑦₂^(𝑘),…,𝑦_𝑛^(𝑘)), converging, respectively, to 𝑃(𝑥₁,𝑥₂,…,𝑥_𝑛) and 𝑄(𝑦₁,𝑦₂,…,𝑦_𝑛) for 𝑘→+∞, 𝑑(𝑃_𝑘,𝑄_𝑘) converges to 𝑑(𝑃,𝑄).
For the points 𝑃(𝑥₁,𝑥₂,…,𝑥_𝑛) and 𝑄(𝑦₁,𝑦₂,…,𝑦_𝑛), in the following, we will denote 𝑔(𝑃,𝑄)=∑^𝑛_𝑖=1𝑎²_𝑖𝑑²_𝑖(𝑥_𝑖,𝑦_𝑖), ℎ(𝑃,𝑄)=∑^𝑛_𝑖=1𝑎^′2_𝑖𝑑²_𝑖(𝑥_𝑖,𝑦_𝑖), and 𝐴=∑^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖(𝑥_𝑖,𝑦_𝑖).
If 𝐴>0(𝐴<0), an integer 𝑝 exists such that for 𝑘∈𝑁, 𝑘≥𝑝 we have ∑^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖(𝑥_𝑖^(𝑘),𝑦_𝑖^(𝑘))>0 (∑^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖(𝑥_𝑖^(𝑘),𝑦_𝑖^(𝑘))<0) (because the sequence ∑^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖(𝑥_𝑖^(𝑘),𝑦_𝑖^(𝑘)) converges to 𝐴).
For 𝑘∈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)).
If 𝐴=0, from 𝑎²_𝑖−𝑎^′2_𝑖=𝜆𝛽_𝑖 we deduce ^𝑛_𝑖=1𝑎²_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖=^𝑛_𝑖=1𝑎^′2_𝑖+𝜆𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖=^𝑛_𝑖=1𝑎^′2_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖=+𝜆𝐴_𝑛𝑖=1𝑎^′2_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖. (13) That is, 𝑔(𝑃,𝑄)=ℎ(𝑃,𝑄).
It follows that 𝑔(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(𝑃₀,𝑟)={𝑃∈𝐸∶𝑑(𝑃,𝑃₀)<𝑟} really are, in fact, open sets for the 𝑑-topology.
(ii) If 𝛽_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 ∑^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖(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.
If 𝜆=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 that 𝜆>0; for two arbitrary points 𝑃(𝑥₁,𝑥₂,…,𝑥_𝑛) and 𝑄(𝑦₁,𝑦₂,…,𝑦_𝑛), denoting again 𝐴=∑^𝑛_𝑖=1𝛽_𝑖𝑑²_𝑖(𝑥_𝑖,𝑦_𝑖) we have 𝑔(𝑃,𝑄)=^𝑛_𝑖=1𝑎²_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖=^𝑛_𝑖=1𝑎^′2_𝑖+𝜆𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖=^𝑛_𝑖=1𝑎^′2_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖+𝜆𝐴. (14)
So, for 𝜆>0 it follows that ()(),()().𝐴≥0⟹𝑔𝑃,𝑄≥ℎ𝑃,𝑄𝐴<0⟹𝑔𝑃,𝑄<ℎ𝑃,𝑄 (15)

This means that if 𝜆>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 that 𝜆<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 a 𝛽∈]𝛼;1[ satisfying the relation −𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖𝛽−𝛼<𝜆²𝛽²_𝑖𝛽²_𝑗𝛽−𝛼1−𝛼+𝛽2²+2−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼𝜆𝛽_𝑖𝛽_𝑗1−𝛼+𝛽2. (16) Such a 𝛽>𝛼 exists because, passing into limit in (16) as 𝛽→𝛼⁺, we obtain 0<2𝜆𝛽_𝑖𝛽_𝑗(1−𝛼)𝛼𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗 (17) 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)=𝛼𝑑_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 𝑑_𝑗𝑥_𝑗,𝑧_𝑗=𝛽𝑑_𝑗𝑥_𝑗,𝑦_𝑗,𝑑_𝑗𝑦_𝑗,𝑧_𝑗=1−𝛽𝑑_𝑗𝑥_𝑗,𝑦_𝑗,𝑑_𝑗𝑥_𝑗,𝑦_𝑗=−𝛽_𝑖𝛽_𝑗𝑑_𝑖𝑥_𝑖,𝑦_𝑖. (18)

With the points 𝑥_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 𝑑(𝑃,𝑄)>𝑑(𝑃,𝑅)+𝑑(𝑄,𝑅). (19)

For the points 𝑃 and 𝑄 we have from (18) that ^𝑛_𝑘=1𝛽_𝑘𝑑²_𝑘𝑥_𝑘,𝑦_𝑘=𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖+𝛽_𝑗𝑑²_𝑗𝑥_𝑗,𝑦_𝑗=0, (20) so 𝑑(𝑃,𝑄)=𝑎²_𝑖𝑑²_𝑖(𝑥_𝑖,𝑦_𝑖)+𝑎²_𝑗𝑑²_𝑗(𝑥_𝑗,𝑦_𝑗)=(𝑎²_𝑖−(𝛽_𝑖/𝛽_𝑗)𝑎²_𝑗)𝑑_𝑖(𝑥_𝑖,𝑦_𝑖).

For the points 𝑃 and 𝑅, ^𝑛_𝑘=1𝛽_𝑘𝑑²_𝑘𝑥_𝑘,𝑧_𝑘=𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑧_𝑖+𝛽_𝑗𝑑²_𝑗𝑥_𝑗,𝑧_𝑗=𝛽_𝑖𝛼²𝑑²_𝑖𝑥_𝑖,𝑦_𝑖+𝛽_𝑗𝛽²−𝛽_𝑖𝛽_𝑗𝑑²_𝑖𝑥_𝑖,𝑦_𝑖=𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖𝛼²−𝛽²>0, (21) because 𝛽_𝑖<0 and 0<𝛼<𝛽, so 𝑑(𝑃,𝑅)=𝑎²_𝑖𝑑²_𝑖(𝑥_𝑖,𝑧_𝑖)+𝑎²_𝑗𝑑²_𝑗(𝑥_𝑗,𝑧_𝑗)=(𝑎²_𝑖𝛼²−𝑎²_𝑗𝛽²(𝛽_𝑖/𝛽_𝑗))𝑑_𝑖(𝑥_𝑖,𝑦_𝑖).

For the points 𝑄 and 𝑅, ^𝑛_𝑘=1𝛽_𝑘𝑑²_𝑘𝑥_𝑘,𝑦_𝑘=𝛽_𝑖𝑑²_𝑖𝑦_𝑖,𝑧_𝑖+𝛽_𝑗𝑑²_𝑗𝑦_𝑗,𝑧_𝑗=𝛽_𝑖(1−𝛼)²𝑑²_𝑖𝑥_𝑖,𝑦_𝑖−𝛽_𝑗1−𝛽²𝛽_𝑖𝛽_𝑗𝑑²_𝑖𝑥_𝑖,𝑦_𝑖=𝛽_𝑖𝑑²_𝑖𝑥_𝑖,𝑦_𝑖(1−𝛼)²−1−𝛽²<0, (22) because 𝛽_𝑖<0 and (1−𝛼)²>(1−𝛽)², so 𝑑(𝑄,𝑅)=𝑎^′2_𝑖𝑑²_𝑖(𝑦_𝑖,𝑧_𝑖)+𝑎^′2_𝑗𝑑²_𝑗(𝑦_𝑗,𝑧_𝑗)=𝑎^′2_𝑖(1−𝛼)²−𝑎^′2_𝑗(1−𝛽)²(𝛽_𝑖/𝛽_𝑗)𝑑_𝑖(𝑥_𝑖,𝑦_𝑖).

We are ready to prove now the inequality 𝑑(𝑃,𝑄)>𝑑(𝑃,𝑅)+𝑑(𝑄,𝑅), transforming it equivalently to 𝑎²_𝑖−𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑑_𝑖𝑥_𝑖,𝑦_𝑖>𝑎²_𝑖𝛼²−𝑎²_𝑗𝛽²𝛽_𝑖𝛽_𝑗𝑑_𝑖𝑥_𝑖,𝑦_𝑖+𝑎^′2_𝑖(1−𝛼)²−𝑎^′2_𝑗1−𝛽²𝛽_𝑖𝛽_𝑗×𝑑_𝑖𝑥_𝑖,𝑦_𝑖⟺𝑎²_𝑖−𝛽_𝑖𝛽_𝑗𝑎²_𝑗>𝑎²_𝑖𝛼²−𝑎²_𝑗𝛽²𝛽_𝑖𝛽_𝑗+𝑎^′2_𝑖(1−𝛼)²−𝑎^′2_𝑗1−𝛽²𝛽_𝑖𝛽_𝑗⟺𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗>𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²+𝛽_𝑗𝑎^′2_𝑖(1−𝛼)²−𝛽_𝑖𝑎^′2_𝑗1−𝛽². (23) From 𝛽_𝑗𝑎²_𝑖>𝛽_𝑗𝑎²_𝑖𝛼² and −𝛽_𝑖𝑎²_𝑗>−𝛽_𝑖𝑎²_𝑗𝛽², we have 𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗>𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽², so 𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗−𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽² is positive and we can equivalently square ⟺𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗−𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²²>𝛽_𝑗𝑎^′2_𝑖(1−𝛼)²−𝛽_𝑖𝑎^′2_𝑗1−𝛽²²⟺𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗+𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²−2𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²>𝛽_𝑗𝑎^′2_𝑖(1−𝛼)²−𝛽_𝑖𝑎^′2_𝑗1−𝛽² (24)(25) (because 𝑎^′2_𝑖=𝑎²_𝑖−𝜆𝛽_𝑖 and 𝑎^′2_𝑗=𝑎²_𝑗−𝜆𝛽_𝑗), ⟺−2𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²>𝛽_𝑗𝑎²_𝑖(1−𝛼)²+𝜆𝛽_𝑖𝛽_𝑗1−𝛽²−𝛽_𝑗𝑎²_𝑖−𝛽_𝑗𝑎²_𝑖𝛼²−𝜆𝛽_𝑖𝛽_𝑗(1−𝛼)²−𝛽_𝑖𝑎²_𝑗1−𝛽²+𝛽_𝑖𝑎²_𝑗+𝛽_𝑖𝑎²_𝑗𝛽²⟺−2𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²>𝛽_𝑗𝑎²_𝑖(1−𝛼)²−1−𝛼²+𝜆𝛽_𝑖𝛽_𝑗1−𝛽²−(1−𝛼)²−𝛽_𝑖𝑎²_𝑗1−𝛽²−1−𝛽²⟺−2𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²>𝛽_𝑖𝑎²_𝑗2𝛽−𝛽_𝑗𝑎²_𝑖2𝛼+𝜆𝛽_𝑖𝛽_𝑗⟺𝛼−𝛽2−𝛼−𝛽𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²<−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼+𝜆𝛽_𝑖𝛽_𝑗𝛽−𝛼1−𝛼+𝛽2 (26) (we square) ⟺𝛽_𝑗𝑎²_𝑖−𝛽_𝑖𝑎²_𝑗𝛽_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝑎²_𝑗𝛽²<−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼²+𝜆𝛽_𝑖𝛽_𝑗𝛽−𝛼1−𝛼+𝛽2²+2−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼𝜆𝛽_𝑖𝛽_𝑗𝛽−𝛼1−𝛼+𝛽2⟺𝛽²_𝑗𝑎⁴_𝑖𝛼²−𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖𝛽²+𝛽²_𝑖𝑎⁴_𝑗𝛽²<𝛽²_𝑖𝑎⁴_𝑗𝛽²+𝛽²_𝑗𝑎⁴_𝑖𝛼²−2𝛽_𝑖𝛽_𝑗𝑎²_𝑖𝑎²_𝑗𝛼𝛽+𝜆²𝛽²_𝑖𝛽²_𝑗𝛽−𝛼²1−𝛼+𝛽2²+2−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼𝜆𝛽_𝑖𝛽_𝑗𝛽−𝛼1−𝛼+𝛽2⟺−𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖𝛼²−𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖𝛽²<−2𝛽_𝑖𝛽_𝑗𝑎²_𝑖𝑎²_𝑗𝛼𝛽+𝜆²𝛽²_𝑖𝛽²_𝑗𝛽−𝛼²1−𝛼+𝛽2²+2−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼𝜆𝛽_𝑖𝛽_𝑗𝛽−𝛼1−𝛼+𝛽2⟺𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖2𝛼𝛽−𝛼²−𝛽²<𝜆²𝛽²_𝑖𝛽²_𝑗𝛽−𝛼²1−𝛼+𝛽2²+2−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼𝜆𝛽_𝑖𝛽_𝑗×𝛽−𝛼1−𝛼+𝛽2⟺−𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖𝛽−𝛼²<𝜆²𝛽²_𝑖𝛽²_𝑗𝛽−𝛼²1−𝛼+𝛽2²+2−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼𝜆𝛽_𝑖𝛽_𝑗×𝛽−𝛼1−𝛼+𝛽2. (27) Dividing by (𝛽−𝛼) we have ⟺−𝛽_𝑖𝛽_𝑗𝑎²_𝑗𝑎²_𝑖𝛽−𝛼<𝜆²𝛽²_𝑖𝛽²_𝑗𝛽−𝛼1−𝛼+𝛽2²+2−𝛽_𝑖𝑎²_𝑗𝛽+𝛽_𝑗𝑎²_𝑖𝛼𝜆𝛽_𝑖𝛽_𝑗1−𝛼+𝛽2. (28) The last inequality is true from (16). The proof is over. So, 𝑑 is not a distance.

Example 5. For 𝑃(𝑥₁,𝑥₂) and 𝑄(𝑦₁,𝑦₂)∈𝑅², the function defined by 𝑑()=𝑃,𝑄2𝑥₁−𝑦₁²+𝑥₂−𝑦₂²if2𝑥₂−𝑦₂²≥𝑥₁−𝑦₁²,𝑥₁−𝑦₁²+3𝑥₂−𝑦₂²if2𝑥₂−𝑦₂²<𝑥₁−𝑦₁², (29) is a continuous quasi-distance, but not a distance. Here, 𝜆=−1, 𝛽₁=−1,𝛽₂=2, 𝑎²₁=2, 𝑎^′2₁=1, 𝑎²₂=1, and 𝑎^′2₂=3, while the function defined by 𝑑()=𝑃,𝑄𝑥₁−𝑦₁²+3𝑥₂−𝑦₂²if2𝑥₂−𝑦₂²≥𝑥₁−𝑦₁²,2𝑥₁−𝑦₁²+𝑥₂−𝑦₂²if2𝑥₂−𝑦₂²<𝑥₁−𝑦₁², (30) is a distance.

Here, 𝜆=1, 𝛽₁=−1, 𝛽₂=2, 𝑎²₁=1, 𝑎^′2₁=2, 𝑎²₂=3, and 𝑎^′2₂=1.

4. A Counterexample

Showing the importance of the (𝛼,−𝛽_𝑖/𝛽_𝑗)-density condition in part III of the theorem.

Let 𝐸₁={𝑥₁,𝑦₁}, 𝐸₂={𝑥₂,𝑦₂} be two metric spaces with distances, respectively, 𝑑₁,𝑑₂, 𝐸=𝐸₁×𝐸₂={𝑃(𝑥₁,𝑥₂),𝑄(𝑥₁,𝑦₂),𝑅(𝑦₁,𝑥₂),𝑇(𝑦₁,𝑦₂)}, 𝑥_𝑖≠𝑦_𝑖, 𝛽_𝑖, real numbers, and 𝑎_𝑖,𝑎^′_𝑖 positive numbers satisfying, for 𝑖=1,2, 𝑎²_𝑖−𝑎^′2_𝑖=𝜆𝛽_𝑖, 𝜆<0, 𝛽₁<0, 𝛽₂>0, and (𝑎²₁−𝑎^′2₁)(𝑎²₂−𝑎^′2₂)+4𝑎^′2₁𝑎²₂≥0⇔𝜆²𝛽₁𝛽₂+4𝑎^′2₁𝑎²₂≥0.

For whatever 𝛼∈]0;1[, it is obvious that the pair ((𝐸₁,𝑑₁),(𝐸₂,𝑑₂)) does not satisfy the (𝛼,−𝛽_𝑖/𝛽_𝑗)-density condition.

We will show that the function 𝑑∶𝐸×𝐸↦𝑅⁺ defined for every two points 𝑀(𝑧₁,𝑧₂),𝑁(𝑡₁,𝑡₂)∈𝐸 by 𝑑(𝑀,𝑁)=𝑎²₁𝑑²₁𝑧₁,𝑧₂+𝑎²₂𝑑²₂𝑡₁,𝑡₂if𝛽₁𝑑²₁𝑧₁,𝑧₂+𝛽₂𝑑²₂𝑡₁,𝑡₂≥0𝑎^′2₁𝑑²₁𝑧₁,𝑧₂+𝑎^′2₂𝑑²₂𝑡₁,𝑡₂if𝛽₁𝑑²₁𝑧₁,𝑧₂+𝛽₂𝑑²₂𝑡₁,𝑡₂<0 (31) is a distance.

For every three points of 𝐸, 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 (𝛽₁𝑑²₁(𝑥₁,𝑦₁)+𝛽₂𝑑²₂(𝑥₂,𝑦₂)≥0)

The six “distances” are 𝑑(𝑃,𝑄)=𝑑(𝑅,𝑇)=𝑎₂𝑑₂𝑥₂,𝑦₂,𝑑(𝑃,𝑅)=𝑑(𝑄,𝑇)=𝑎^′₁𝑑₁𝑥₁,𝑦₁,𝑑(𝑄,𝑅)=𝑑(𝑃,𝑇)=𝑎²₁𝑑²₁𝑥₁,𝑦₁+𝑎²₂𝑑²₂𝑥₂,𝑦₂. (32)

As 𝑎²₁−𝑎^′2₁=𝜆𝛽₁>0, we have 𝑎₁>𝑎^′₁, so 𝑎²₁𝑑²₁𝑥₁,𝑦₁+𝑎²₂𝑑²₂𝑥₂,𝑦₂≥𝑎₁𝑑₁𝑥₁,𝑦₁≥𝑎^′₁𝑑₁𝑥₁,𝑦₁,𝑎²₁𝑑²₁𝑥₁,𝑦₁+𝑎²₂𝑑²₂𝑥₂,𝑦₂≥𝑎₂𝑑₂𝑥₂,𝑦₂. (33)

The greatest of the six “distances” is 𝑑(𝑄,𝑅)=𝑑(𝑃,𝑇)=𝑎²₁𝑑²₁(𝑥₁,𝑦₁)+𝑎²₂𝑑²₂(𝑥₂,𝑦₂).

In each of the four possible “triangles,” 𝑃𝑄𝑅, 𝑃𝑄𝑇, 𝑃𝑅𝑇, and 𝑄𝑅𝑇, there is a side equal to 𝑎²₁𝑑²₁(𝑥₁,𝑦₁)+𝑎²₂𝑑²₂(𝑥₂,𝑦₂), another is equal to 𝑎^′₁𝑑₁(𝑥₁,𝑦₁), and the third is 𝑎₂𝑑₂(𝑥₂,𝑦₂).

Therefore we have to prove only one “triangle inequality,” valid for the four triangles: 𝑎²₁𝑑²₁𝑥₁,𝑦₁+𝑎²₂𝑑²₂𝑥₂,𝑦₂≤𝑎^′₁𝑑₁𝑥₁,𝑦₁+𝑎₂𝑑₂𝑥₂,𝑦₂, (34) or equivalently 𝑎²₁𝑑²₁𝑥₁,𝑦₁+𝑎²₂𝑑²₂𝑥₂,𝑦₂≤𝑎^′2₁𝑑²₁𝑥₁,𝑦₁+𝑎²₂𝑑²₂𝑥₂,𝑦₂+2𝑎^′₁𝑎₂𝑑₁𝑥₁,𝑦₁𝑑₂𝑥₂,𝑦₂⟺𝑎²₁−𝑎^′2₁𝑑₁𝑥₁,𝑦₁≤2𝑎^′₁𝑎₂𝑑₂𝑥₂,𝑦₂⟺𝑑²₁𝑥₁,𝑦₁𝑑²₂𝑥₂,𝑦₂≤4𝑎^′2₁𝑎²₂𝜆²𝛽²₁. (35)

But from 𝛽₁𝑑²₁(𝑥₁,𝑦₁)+𝛽₂𝑑²₂(𝑥₂,𝑦₂)≥0 we have 𝑑²₁(𝑥₁,𝑦₁)/𝑑²₂(𝑥₂,𝑦₂)≤𝛽₂/−𝛽₁ and from 𝜆²𝛽₁𝛽₂+4𝑎^′2₁𝑎²₂≥0 we have 𝛽₂/−𝛽₁≤4𝑎^′2₁𝑎²₂/𝜆²𝛽²₁, so 𝑑²₁(𝑥₁,𝑦₁)/𝑑²₂(𝑥₂,𝑦₂)≤4𝑎^′2₁𝑎²₂/𝜆²𝛽²₁ and the triangle inequality is satisfied.

4.2. Second Case (𝛽₁𝑑²₁(𝑥₁,𝑦₁)+𝛽₂𝑑²₂(𝑥₂,𝑦₂)<0)

The six “distances” are 𝑑(𝑃,𝑄)=𝑑(𝑅,𝑇)=𝑎₂𝑑₂𝑥₂,𝑦₂,𝑑(𝑃,𝑅)=𝑑(𝑄,𝑇)=𝑎^′₁𝑑₁𝑥₁,𝑦₁,𝑑(𝑄,𝑅)=𝑑(𝑃,𝑇)=𝑎^′2₁𝑑²₁𝑥₁,𝑦₁+𝑎^′2₂𝑑²₂𝑥₂,𝑦₂. (36)

As 𝑎²₂−𝑎^′2₂=𝜆𝛽₂<0, we have 𝑎₂<𝑎^′₂, so 𝑎^′2₁𝑑²₁(𝑥₁,𝑦₁)+𝑎^′2₂𝑑²₂(𝑥₂,𝑦₂)≥𝑎^′₂𝑑₂(𝑥₂,𝑦₂)>𝑎₂𝑑₂(𝑥₂,𝑦₂), and 𝑎^′2₁𝑑²₁(𝑥₁,𝑦₁)+𝑎^′2₂𝑑²₂(𝑥₂,𝑦₂)≥𝑎^′₁𝑑₁(𝑥₁,𝑦₁).

The greatest of the six “distances” is 𝑑(𝑄,𝑅)=𝑑(𝑃,𝑇)=𝑎^′2₁𝑑²₁(𝑥₁,𝑦₁)+𝑎^′2₂𝑑²₂(𝑥₂,𝑦₂). In each of the four possible “triangles,” 𝑃𝑄𝑅, 𝑃𝑄𝑇, 𝑃𝑅𝑇, and 𝑄𝑅𝑇, there is a side equal to 𝑎^′2₁𝑑²₁(𝑥₁,𝑦₁)+𝑎^′2₂𝑑²₂(𝑥₂,𝑦₂), another is equal to 𝑎^′₁𝑑₁(𝑥₁,𝑦₁), and the third is 𝑎₂𝑑₂(𝑥₂,𝑦₂). Therefore we have to prove only one “triangle inequality,” valid for the four triangles: 𝑎^′2₁𝑑²₁𝑥₁,𝑦₁+𝑎^′2₂𝑑²₂𝑥₂,𝑦₂≤𝑎^′₁𝑑₁𝑥₁,𝑦₁+𝑎₂𝑑₂𝑥₂,𝑦₂, (37) that we square equivalently 𝑎^′2₁𝑑²₁𝑥₁,𝑦₁+𝑎^′2₂𝑑²₂𝑥₂,𝑦₂≤𝑎^′2₁𝑑²₁𝑥₁,𝑦₁+𝑎²₂𝑑²₂𝑥₂,𝑦₂+2𝑎^′₁𝑎₂𝑑₁𝑥₁,𝑦₁𝑑₂𝑥₂,𝑦₂⟺𝑎^′2₂−𝑎²₂𝑑₂𝑥₂,𝑦₂≤2𝑎^′₁𝑎₂𝑑₁𝑥₁,𝑦₁⟺𝑑²₁𝑥₁,𝑦₁𝑑²₂𝑥₂,𝑦₂≥𝜆²𝛽²₂4𝑎^′2₁𝑎²₂. (38)

But from 𝛽₁𝑑²₁(𝑥₁,𝑦₁)+𝛽₂𝑑²₂(𝑥₂,𝑦₂)<0 we have 𝑑²₁(𝑥₁,𝑦₁)/𝑑²₂(𝑥₂,𝑦₂)>𝛽₂/−𝛽₁ and from 𝜆²𝛽₁𝛽₂+4𝑎^′2₁𝑎²₂≥0 we have 𝛽₂/−𝛽₁≥4𝑎^′2₁𝑎²₂/𝜆²𝛽²₁, so 𝑑²₁(𝑥₁,𝑦₁)/𝑑²₂(𝑥₂,𝑦₂)≥𝜆²𝛽²₂/4𝑎^′2₁𝑎²₂ 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.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author would like to thank Redon A. Cabej for his valuable suggestions which improved the final version of this paper.