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:
(1)d(x,y)≤k[d(x,z)+d(y,z)],
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,…,xn) and Q(y1,y2,…,yn), respectively, by
(2)d(P,Q)={∑i=1naidi(xi,yi)if∑i=1nβidi(xi,yi)≥0,∑i=1nai′di(xi,yi)if∑i=1nβidi(xi,yi)<0,d(P,Q)={∑i=1naidi(xi,yi)1+∑i=1naidi(xi,yi)if∑i=1nβidi(xi,yi)≥0,∑i=1nai′di(xi,yi)1+∑i=1nai′di(xi,yi)if∑i=1nβidi(xi,yi)<0,
with ai-ai′=λβi 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
(3)d(P,Q)={∑i=1nai2di2(xi,yi)if∑i=1nβidi2(xi,yi)≥0,∑i=1nai′2di2(xi,yi)if∑i=1nβidi2(xi,yi)<0,
with ai2-ai′2=λβi 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(P1,Pn)≤k[d(P1,P2)+d(P2,P3)+⋯+d(Pn-1,Pn)],
which we are calling generalized improved quasi-triangle inequality, improve the general one:
(5)d(P1,Pn)≤kd(P1,P2)+k2d(P2,P3)+⋯+kn-3d(Pn-3,Pn-2)+kn-2d(Pn-2,Pn-1)+kn-2d(Pn-1,Pn)
(the two last coefficients are kn-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,-βi/βj)-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
(6)d(x,z)=αd(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′)=βd′(x′,y′),d′(y′,z′)=(1-β)d′(x′,y′),d′(x′,y′)=kd(x,y).
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
(8)d(x,y)=d(x,z)+d(z,y).
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
(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)=c1,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 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).
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.
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 ResultTheorem 4.
Let (Ei,di) for i∈{1,2,…,n} be metric spaces, λ and βi real numbers, ai and ai′ positive numbers satisfying ai2-ai′2=λβi, and P(x1,x2,…,xn) and Q(y1,y2,…,yn) two arbitrary points of E=∏i=1nEi. Then
the function d:E2↦R+ defined by
(10)d(P,Q)={∑i=1nai2di2(xi,yi)if∑i=1nβidi2(xi,yi)≥0∑i=1nai′2di2(xi,yi)if∑i=1nβidi2(xi,yi)<0
is a continuous quasi-distance on E satisfying the generalized improved triangle inequality for every m points;
if λ≥0 or βis are all nonnegative or all nonpositive, d is a distance;
if λ<0 and there exist two indexes (i,j)∈{1,2,…,n} with βi<0, βj>0 and an α∈]0;1[ such that the spaces Ei and Ej satisfy the (α,-βi/βj)-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{ai,ai′}1≤i≤n, K2=max{ai,ai′}1≤i≤n, and D(P,Q)=∑i=1ndi2(xi,yi).
Clearly, for whatever points P and Q, we have
(11)K1D(P,Q)≤d(P,Q)≤K2D(P,Q).
As D is a distance, we can write, for whatever points P1(x1(1),x2(1),…,xn(1)), P2(x1(2),x2(2),…,xn(2)),…, Pi(x1(i),x2(i),…,xn(i)),
(12)d(P1,Pi)≤K2D(P1,Pi)≤K2[D(P1,P2)+D(P2,P3)+⋯+D(Pi-1,Pi)]≤K2K1[d(P1,P2)+d(P2,P3)+⋯+d(Pi-1,Pi)].
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, Pk(x1(k),x2(k),…,xn(k)) and Qk(y1(k),y2(k),…,yn(k)), converging, respectively, to P(x1,x2,…,xn) and Q(y1,y2,…,yn) for k→+∞, d(Pk,Qk) converges to d(P,Q).
For the points P(x1,x2,…,xn) and Q(y1,y2,…,yn), in the following, we will denote g(P,Q)=∑i=1nai2di2(xi,yi), h(P,Q)=∑i=1nai′2di2(xi,yi), and A=∑i=1nβidi2(xi,yi).
If A>0(A<0), an integer p exists such that for k∈N, k≥p we have ∑i=1nβidi2(xi(k),yi(k))>0 (∑i=1nβidi2(xi(k),yi(k))<0) (because the sequence ∑i=1nβidi2(xi(k),yi(k)) converges to A).
For k∈N, k≥p, we have d(Pk,Qk)=g(Pk,Qk) (d(Pk,Qk)=h(Pk,Qk)) and, consequently, d(Pk,Qk) converges to g(P,Q)=d(P,Q) and (h(P,Q)=d(P,Q)).
If A=0, from ai2-ai′2=λβi we deduce
(13)∑i=1nai2di2(xi,yi)=∑i=1n(ai′2+λβi)di2(xi,yi)=∑i=1nai′2di2(xi,yi)+λA=∑i=1nai′2di2(xi,yi).
That is, g(P,Q)=h(P,Q).
It follows that g(Pk,Qk) and h(Pk,Qk) converge to the same limit g(P,Q)=h(P,Q)=d(P,Q). We can conclude that d(Pk,Qk), 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 βis are all nonnegative, d(P,Q)=g(P,Q) for every P and Q; if βis are all nonpositive, d(P,Q)=h(P,Q) for every P and Q (recall that if ∑i=1nβidi2(xi,yi)=0, d(P,Q)=g(P,Q)=h(P,Q)). As g and h are distances, d is a distance too.
If λ=0, ai=ai′ 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,…,xn) and Q(y1,y2,…,yn), denoting again A=∑i=1nβidi2(xi,yi) we have
(14)g(P,Q)=∑i=1nai2di2(xi,yi)=∑i=1n(ai′2+λβi)di2(xi,yi)=∑i=1nai′2di2(xi,yi)+λA.
So, for λ>0 it follows that
(15)A≥0⟹g(P,Q)≥h(P,Q),A<0⟹g(P,Q)<h(P,Q).
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 (Ei,di).
(iii) Suppose now that λ<0 and that there exist an α∈]0;1[ and two indexes i,j∈{1,2,…,n} such that βi<0, βj>0, and the pair (Ei,Ej) satisfies the (α,-βi/βj)-density condition.
We choose a β∈]α;1[ satisfying the relation
(16)-βiβjaj2ai2(β-α)<λ2βi2βj2(β-α)(1-α+β2)2+2(-βiaj2β+βjai2α)λβiβj(1-α+β2).
Such a β>α exists because, passing into limit in (16) as β→α+, we obtain
(17)0<2λβiβj(1-α)α(βjai2-βiaj2)
that is clearly true.
As (Ei,Ej) satisfies the (α,-βi/βj)-density condition, there exist three points xi,yi,zi∈Ei, satisfying the relations di(xi,zi)=αdi(xi,yi) and di(yi,zi)=(1-α)di(xi,yi), and three points xj,yj,zj∈Ej, satisfying the relations
(18)dj(xj,zj)=βdj(xj,yj),dj(yj,zj)=(1-β)dj(xj,yj),dj(xj,yj)=-βiβjdi(xi,yi).
With the points xi,yi,zi∈Ei and xj,yj,zj∈Ej we construct the points P(…,xi,…,xj,…), Q(…,yi,…,yj,…), and R(…,zi,…,zj,…) 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 points P and Q we have from (18) that
(20)∑k=1nβkdk2(xk,yk)=βidi2(xi,yi)+βjdj2(xj,yj)=0,
so d(P,Q)=ai2di2(xi,yi)+aj2dj2(xj,yj)=(ai2-(βi/βj)aj2)di(xi,yi).
For the points P and R,
(21)∑k=1nβkdk2(xk,zk)=βidi2(xi,zi)+βjdj2(xj,zj)=βiα2di2(xi,yi)+βjβ2(-βiβj)di2(xi,yi)=βidi2(xi,yi)(α2-β2)>0,
because βi<0 and 0<α<β, so d(P,R)=ai2di2(xi,zi)+aj2dj2(xj,zj)=(ai2α2-aj2β2(βi/βj))di(xi,yi).
For the points Q and R,
(22)∑k=1nβkdk2(xk,yk)=βidi2(yi,zi)+βjdj2(yj,zj)=βi(1-α)2di2(xi,yi)-βj(1-β)2βiβjdi2(xi,yi)=βidi2(xi,yi)[(1-α)2-(1-β)2]<0,
because βi<0 and (1-α)2>(1-β)2, so d(Q,R)=ai′2di2(yi,zi)+aj′2dj2(yj,zj)=ai′2(1-α)2-aj′2(1-β)2(βi/βj)di(xi,yi).
We are ready to prove now the inequality d(P,Q)>d(P,R)+d(Q,R), transforming it equivalently to
(23)ai2-βiβjaj2di(xi,yi)>ai2α2-aj2β2βiβjdi(xi,yi)+ai′2(1-α)2-aj′2(1-β)2βiβj×di(xi,yi)⟺ai2-βiβjaj2>ai2α2-aj2β2βiβj+ai′2(1-α)2-aj′2(1-β)2βiβj⟺βjai2-βiaj2>βjai2α2-βiaj2β2+βjai′2(1-α)2-βiaj′2(1-β)2.
From βjai2>βjai2α2 and -βiaj2>-βiaj2β2, we have βjai2-βiaj2>βjai2α2-βiaj2β2, so βjai2-βiaj2-βjai2α2-βiaj2β2 is positive and we can equivalently square
(24)⟺(βjai2-βiaj2-βjai2α2-βiaj2β2)2>(βjai′2(1-α)2-βiaj′2(1-β)2)2(25)⟺βjai2-βiaj2+βjai2α2-βiaj2β2-2βjai2-βiaj2βjai2α2-βiaj2β2>βjai′2(1-α)2-βiaj′2(1-β)2
(because ai′2=ai2-λβi and aj′2=aj2-λβj),
(26)⟺-2βjai2-βiaj2βjai2α2-βiaj2β2>βjai2(1-α)2+λβiβj(1-β)2-βjai2-βjai2α2-λβiβj(1-α)2-βiaj2(1-β)2+βiaj2+βiaj2β2⟺-2βjai2-βiaj2βjai2α2-βiaj2β2>βjai2[(1-α)2-1-α2]+λβiβj[(1-β)2-(1-α)2]-βiaj2[(1-β)2-1-β2]⟺-2βjai2-βiaj2βjai2α2-βiaj2β2>βiaj22β-βjai22α+λβiβj(α-β)(2-α-β)⟺βjai2-βiaj2βjai2α2-βiaj2β2<-βiaj2β+βjai2α+λβiβj(β-α)(1-α+β2)
(we square)
(27)⟺(βjai2-βiaj2)(βjai2α2-βiaj2β2)<(-βiaj2β+βjai2α)2+[λβiβj(β-α)(1-α+β2)]2+2(-βiaj2β+βjai2α)λβiβj(β-α)(1-α+β2)⟺βj2ai4α2-βiβjaj2ai2α2-βiβjaj2ai2β2+βi2aj4β2<βi2aj4β2+βj2ai4α2-2βiβjai2aj2αβ+λ2βi2βj2(β-α)2(1-α+β2)2+2(-βiaj2β+βjai2α)λβiβj(β-α)(1-α+β2)⟺-βiβjaj2ai2α2-βiβjaj2ai2β2<-2βiβjai2aj2αβ+λ2βi2βj2(β-α)2(1-α+β2)2+2(-βiaj2β+βjai2α)λβiβj(β-α)(1-α+β2)⟺βiβjaj2ai2(2αβ-α2-β2)<λ2βi2βj2(β-α)2(1-α+β2)2+2(-βiaj2β+βjai2α)λβiβj×(β-α)(1-α+β2)⟺-βiβjaj2ai2(β-α)2<λ2βi2βj2(β-α)2(1-α+β2)2+2(-βiaj2β+βjai2α)λβiβj×(β-α)(1-α+β2).
Dividing by (β-α) we have
(28)⟺-βiβjaj2ai2(β-α)<λ2βi2βj2(β-α)(1-α+β2)2+2(-βiaj2β+βjai2α)λβiβj(1-α+β2).
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
(29)d(P,Q)={2(x1-y1)2+(x2-y2)2if2(x2-y2)2≥(x1-y1)2,(x1-y1)2+3(x2-y2)2if2(x2-y2)2<(x1-y1)2,
is a continuous quasi-distance, but not a distance. Here, λ=-1, β1=-1,β2=2, a12=2, a1′2=1, a22=1, and a2′2=3, while the function defined by
(30)d(P,Q)={(x1-y1)2+3(x2-y2)2if2(x2-y2)2≥(x1-y1)2,2(x1-y1)2+(x2-y2)2if2(x2-y2)2<(x1-y1)2,
is a distance.
Here, λ=1, β1=-1, β2=2, a12=1, a1′2=2, a22=3, and a2′2=1.
4. A Counterexample
Showing the importance of the (α,-βi/βj)-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)}, xi≠yi, βi, real numbers, and ai,ai′ positive numbers satisfying, for i=1,2, ai2-ai′2=λβi, λ<0, β1<0, β2>0, and (a12-a1′2)(a22-a2′2)+4a1′2a22≥0⇔λ2β1β2+4a1′2a22≥0.
For whatever α∈]0;1[, it is obvious that the pair ((E1,d1),(E2,d2)) does not satisfy the (α,-βi/βj)-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
(31)d(M,N)={a12d12(z1,z2)+a22d22(t1,t2)ifβ1d12(z1,z2)+β2d22(t1,t2)≥0a1′2d12(z1,z2)+a2′2d22(t1,t2)ifβ1d12(z1,z2)+β2d22(t1,t2)<0
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 (β1d12(x1,y1)+β2d22(x2,y2)≥0)
The six “distances” are
(32)d(P,Q)=d(R,T)=a2d2(x2,y2),d(P,R)=d(Q,T)=a1′d1(x1,y1),d(Q,R)=d(P,T)=a12d12(x1,y1)+a22d22(x2,y2).
As a12-a1′2=λβ1>0, we have a1>a1′, so
(33)a12d12(x1,y1)+a22d22(x2,y2)≥a1d1(x1,y1)≥a1′d1(x1,y1),a12d12(x1,y1)+a22d22(x2,y2)≥a2d2(x2,y2).
The greatest of the six “distances” is d(Q,R)=d(P,T)=a12d12(x1,y1)+a22d22(x2,y2).
In each of the four possible “triangles,” PQR, PQT, PRT, and QRT, there is a side equal to a12d12(x1,y1)+a22d22(x2,y2), another is equal to a1′d1(x1,y1), and the third is a2d2(x2,y2).
Therefore we have to prove only one “triangle inequality,” valid for the four triangles:
(34)a12d12(x1,y1)+a22d22(x2,y2)≤a1′d1(x1,y1)+a2d2(x2,y2),
or equivalently
(35)a12d12(x1,y1)+a22d22(x2,y2)≤a1′2d12(x1,y1)+a22d22(x2,y2)+2a1′a2d1(x1,y1)d2(x2,y2)⟺(a12-a1′2)d1(x1,y1)≤2a1′a2d2(x2,y2)⟺d12(x1,y1)d22(x2,y2)≤4a1′2a22λ2β12.
But from β1d12(x1,y1)+β2d22(x2,y2)≥0 we have d12(x1,y1)/d22(x2,y2)≤β2/-β1 and from λ2β1β2+4a1′2a22≥0 we have β2/-β1≤4a1′2a22/λ2β12, so d12(x1,y1)/d22(x2,y2)≤4a1′2a22/λ2β12 and the triangle inequality is satisfied.
4.2. Second Case (β1d12(x1,y1)+β2d22(x2,y2)<0)
The six “distances” are
(36)d(P,Q)=d(R,T)=a2d2(x2,y2),d(P,R)=d(Q,T)=a1′d1(x1,y1),d(Q,R)=d(P,T)=a1′2d12(x1,y1)+a2′2d22(x2,y2).
As a22-a2′2=λβ2<0, we have a2<a2′, so a1′2d12(x1,y1)+a2′2d22(x2,y2)≥a2′d2(x2,y2)>a2d2(x2,y2), and a1′2d12(x1,y1)+a2′2d22(x2,y2)≥a1′d1(x1,y1).
The greatest of the six “distances” is d(Q,R)=d(P,T)=a1′2d12(x1,y1)+a2′2d22(x2,y2). In each of the four possible “triangles,” PQR, PQT, PRT, and QRT, there is a side equal to a1′2d12(x1,y1)+a2′2d22(x2,y2), another is equal to a1′d1(x1,y1), and the third is a2d2(x2,y2). Therefore we have to prove only one “triangle inequality,” valid for the four triangles:
(37)a1′2d12(x1,y1)+a2′2d22(x2,y2)≤a1′d1(x1,y1)+a2d2(x2,y2),
that we square equivalently
(38)a1′2d12(x1,y1)+a2′2d22(x2,y2)≤a1′2d12(x1,y1)+a22d22(x2,y2)+2a1′a2d1(x1,y1)d2(x2,y2)⟺(a2′2-a22)d2(x2,y2)≤2a1′a2d1(x1,y1)⟺d12(x1,y1)d22(x2,y2)≥λ2β224a1′2a22.
But from β1d12(x1,y1)+β2d22(x2,y2)<0 we have d12(x1,y1)/d22(x2,y2)>β2/-β1 and from λ2β1β2+4a1′2a22≥0 we have β2/-β1≥4a1′2a22/λ2β12, so d12(x1,y1)/d22(x2,y2)≥λ2β22/4a1′2a22 and the triangle inequality is satisfied. So, d is a distance.
We note that if (Ei,di) are normed vector spaces, (Ei,∥∥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.
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