On characterization of functions preserving metric-type conditions via triangular and polygonal structures

We begin by defining $d$. Put

From (1) we may infer that $d_{|X_{i}\times X_{i}}=d_{i}$ for $i=1,2$, thus it is an extension of both $d_{1}$ and $d_{2}$. The definition is correct thanks to $X_{1}$ and $X_{2}$ being disjoint.

It is easily seen that the pair $(X,d)$ satisfies the conditions of a semimetric space, as having at least three elements guarantees that $\max\{r_{1},r_{2}\}>0$. Moreover, the fact, that the resulting space diameter equals the greater one of the diameters of composing spaces is a direct consequence of the definition of $d$. What is left to check is whether the strong $b$-metric (or just $b$-metric) condition holds. Consider any three distinct points $x,y,z\in X$. We have the following cases:

1. (1)

   $x,y,z\in X_{1}$ or $x,y,z\in X_{2}$. In such cases $d$ coincides with either $d_{1}$ or $d_{2}$, so the desired inequality holds.

2. (2)

   $x,y\in X_{1}$ and $z\in X_{2}$. We need to check three variants of a strong $b$-metric inequality.

   	$\displaystyle d(x,y)$	$\displaystyle=$	$\displaystyle d_{1}(x,y)\leqslant r_{1}$	$\displaystyle\leqslant$	$\displaystyle\frac{\max\{r_{1},r_{2}\}}{1+K}\cdot(1+K)=Kd(x,z)+d(z,y),$
   	$\displaystyle d(x,z)$	$\displaystyle=$	$\displaystyle\frac{\max\{r_{1},r_{2}\}}{1+K}$	$\displaystyle\leqslant$	$\displaystyle K\cdot 0+\frac{\max\{r_{1},r_{2}\}}{1+K}\leqslant Kd_{1}(x,y)+d(y,z),$
   	$\displaystyle d(y,z)$	$\displaystyle=$	$\displaystyle\frac{\max\{r_{1},r_{2}\}}{1+K}$	$\displaystyle\leqslant$	$\displaystyle K\cdot 0+\frac{\max\{r_{1},r_{2}\}}{1+K}\leqslant Kd_{1}(y,x)+d(x,z).$

   One can easily notice, that the $b$-metric inequality follows from this reasoning as well.

3. (3)

   $x\in X_{1}$ and $y,z\in X_{2}$. The desired inequalities in this case are obtained analogously as above.

Therefore, $(X,d)$ is a strong $b$-metric space (or $b$-metric space). ∎

Since $\mathcal{F}$ is increasing then $d(x,y)$ is well defined for all $x,y\in X$. We can then proceed to checking axioms of semimetric

• (S1)

  $d(x,y)=0$ implies that $d_{t_{0}}(x,y)=0$ for some $t_{0}\in\mathbb{T}$. Therefore $x=y$. The reverse implication is obvious;

• (S2)

  $d(x,y)=d_{t}(x,y)=d_{t}(y,x)=d(y,x)$ for some $t\in\mathbb{T}$.

Lastly, we need to prove the (S) condition (or (B) condition) for $(X,d)$. Take any three distinct points $x,y,z\in X$. There exists $t_{0}\in\mathbb{T}$ such that $x,y,z\in X_{t_{0}}$. Since $(X_{t_{0}},d_{t_{0}})$ satisfies (S) (or (B), respectively), we have

Consequently, $(X,d)$ satisfies (S) condition. The proof of $b$-metric case goes analogously. ∎

Due to the fact, that each version of this lemma is proved by the same method, we will simply prove the version for particular subclass of $P_{B,S}$.

Let $f$ be a function satisfying the assumptions of our lemma, i.e. for every $b$-metric space $(X,d)$ with relaxation constant $K\geqslant 1$ we have that $(X,f\circ d)$ is a strong $b$-metric space.

Suppose the contrary, i.e., for every $K^{\prime}\geqslant 1$ (in particular for every $n\in\mathbb{N}$) there exists a $K$-triangle triplet $\{a_{n},b_{n},c_{n}\}$ which is mapped to $\left\{f(a_{n}),f(b_{n}),f(c_{n})\right\}$, the latter one not forming a strong $n$-triangle triplet. Without the loss of generality, we may assume that $f(a_{n})>n\cdot f(b_{n})+f(c_{n})$ (this fact becomes clear in the further part of the proof).

Let $\hat{X}_{1}=X_{1}:=\{(1,1),(1,2),(1,3)\}$ and

The rest of the values, again, stems from extending $d_{1}$ to the whole $X_{1}\times X_{1}$ as a semimetric (i.e., (S1) and (S2) conditions imply the remaining function values), yielding $\hat{d}_{1}$. Thus $(\hat{X}_{1},\hat{d}_{1})$ is not a strong $b$-metric space with relaxation constant $1$. Nonetheless, it is still a strong $b$-metric space if we replace $1$ by some larger constant.

Now, we will proceed inductively. Having defined $\left(\hat{X}_{n-1},\hat{d}_{n-1}\right)$, which is a finite (thus bounded) $b$-metric space with relaxation constant $K$, we try to define $\hat{X}_{n}$. Put $X_{n}:=\{(n,1),(n,2),(n,3)\}$ and $d_{n}:X_{n}\times X_{n}\to[0,+\infty),$ as a unique semimetric satisfying

Again, $(X_{n},d_{n})$ is a bounded $b$-metric space with relaxation constant $K$. We can now use Lemma 3.1 to obtain larger $b$-metric space $\hat{X}_{n}:=X_{n}\cup\hat{X}_{n-1}$, equipped with the extended $b$-metric $\hat{d}_{n}$, which equals $d_{n}$ on $X_{n}\times X_{n}$ and $\hat{d}_{n-1}$ on $\hat{X}_{n-1}\times\hat{X}_{n-1}$. Successively, we obtain the increasing family of $b$-metric spaces $\left\{\left(\hat{X}_{n},\hat{d}_{n}\right)\ :\ n\in\mathbb{N}\right\}$. Due to Lemma 3.2 we obtain a $b$-metric space $(X,D)$ defined as

What is left to prove is that $(X,f\circ D)$ fails to be a strong $b$-metric space.

Suppose that (S) inequality holds in $(X,D)$ for some $K^{\prime}\geqslant 1$ and let $n_{0}\geqslant K^{\prime}$ be a fixed natural number. This way, the fact that $(X,D)$ satisfies the strong $b$-metric inequality with constant $K^{\prime}$ implies satisfying it for $n_{0}$. Considering $(n_{0},1),(n_{0},2),(n_{0},3)\in X$ yields

Of course $f(a_{n})$, $f(b_{n})$ and $f(c_{n})$ do not have to be aligned in the same order as $a_{n}$, $b_{n}$ and $c_{n}$, but the result stays the same as we can interchange the appropriate pairs of points in the reasoning above to obtain the proper counterexample.

This shows that $(X,f\circ D)$ fails to satisfy strong $b$-metric condition for $n_{0}$, thus it fails for $K^{\prime}$ as well. Since $K^{\prime}$ was arbitrary, $(X,f\circ D)$ is not a strong $b$-metric space despite $(X,D)$ being a $b$-metric one. This concludes the proof of both part a) and b), since it yields a contradiction with the assumption that $(X,f\circ d)$ is a strong $b$-metric space whenever $(X,d)$ is a $b$-metric space with relaxation constant $K$.

The reasoning for the remaining variants of this lemma is exactly the same – all that needs to be changed is the type of triangle triplets used in constructing the proper counterexample. ∎

Once again, we would like to remind that $f$ does not have to be monotone, thus the relation $f(a)\geqslant f(b)\geqslant f(c)$ does not have to hold (see Remark 4.1).

The necessity in each case follows as a simple conclusion from Lemma 4.2. In the last, seventh equivalence another short proof of necessity can be given, as considering a three-point space $X:=\{1,2,3\}$ is enough.

Supposing that for some $K$ a strong $K$-triangle triplet $\{a,b,c\}$ which is not mapped to a triangle one exists yields an instant contradiction if we define

and allow the assumptions (S1), (S2) to fill in the rest of values. Applying $f$ to such defined strong $b$-metric space maps it into a semimetric structure which is not a metric space. The sufficient part of this last implication follows from the same method as described in point f. above.

A sufficiency part can be shown as follows.

Assume that $f$ satisfies the right-hand side of equivalence (3) and let $(X,d)$ be an arbitrary $b$-metric space. Let us denote the relaxation constant of this space by $K$. Consider three distinct points $x_{1},x_{2},x_{3}\in X$. WLOG we can assume that

Since $(X,d)$ is a $b$-metric space, then values $d\left(x_{1},x_{3}\right),d\left(x_{1},x_{2}\right),d\left(x_{2},x_{3}\right)$ can be arranged into a $K$-triangle triplet. From our assumption there exists $K^{\prime}$ such that $\{f\circ d\left(x_{1},x_{3}\right),f\circ d\left(x_{1},x_{2}\right),f\circ d\left(x_{2},x_{3}\right)\}$ is a strong $K$-triangle triplet. From the definition of such triplet we obtain

Due to $K^{\prime}$ being independent of the choice of $x_{1},x_{2},x_{3}$ (since it depends solely on $K$) the proof of equivalence (3) is finished.

The sufficiency proofs for points (4-7) are almost identical. ∎

We will construct such space in two steps. Let $X:=\{1,\dots,n\}$ and put

• •

  $d(1,n):=a_{1}$;

• •

  for each $1\leqslant i\leqslant n$, $d({i-1},{i}):=a_{i}$;

Thus, all of the requirements (I1)-(I2) are satisfied. This definition leaves us with $n$-gon with no diagonal distances defined. Consider now $i,j\in X$ such that they are not adjacent i.e. $|i-j|\notin\{1,n\}$.

For such pair (for the sake of simplicity let us assume $i<j$) one can define the distance as follows

where the sums in the latter part of the minimum might be empty (for example when $i=1$ and $j<n$ or $j=n$ and $i>1$).

Notice, that for any two non-adjacent points $i,j$ thanks to the formulation of $d(i,j)$ the distance can be bound from above by the distance over the perimeter of the discussed polygon. Indeed, for any sequence of point $i,x_{1},x_{2},\dots,x_{k},j\in X$.

Perhaps this part of the proof can be reworked to be nicer. Now, consider any two adjacent points of the space, $i$ and $i+1$ (the case of $n$ and $1$ is analogous). Consider any point $x\in X$. WLOG we can assume that $x>i$. Then we have three posibilities:

1. (1)

   $d(i,x)=\sum_{k=i}^{x-1}d(k,k+1)$, then obviously $d(i,x)\geqslant d(i,i+1)$;

2. (2)

   $d(i,x)=d(1,n)+\sum_{k=x}^{n-1}d(k,k+1)+\sum_{k=1}^{i-1}d(k,k+1)$.
   If $d(x,i+1)=d(1,n)+\sum_{k=x}^{n-1}d(k,k+1)+\sum_{k=1}^{i}d(k,k+1)$ then $d(x,i+1)\geqslant d(i,i+1)$ and the case is once again trivial.

3. (3)

   $d(i,x)=d(1,n)+\sum_{k=x}^{n-1}d(k,k+1)+\sum_{k=1}^{i-1}d(k,k+1)$ and $d(x,i+1)=\sum_{k=i+1}^{x-1}d(k,k+1)$.

   Then, using the fact that $A$ is a $K$-relaxed polygon

   $$d(i,i+1)\leqslant K\cdot\left(d(1,n)-d(i,i+1)+\sum_{i=1}^{n}d(k,k+1)\right),$$

   which translates to

   $$d(i,i+1)\cdot(1+K)\leqslant a_{1}\leqslant K\cdot\sum_{k=1}^{n}d(k,k+1)=K\cdot\sum_{k=1}^{n}a_{k}.$$

When more points are considered, the reasoning remains the same (although it might need some more formal touch) – we either use the fact that length of one of the edges exceeds the distance $d(i,i+1)$ or the sum of the distances is greater or equal to the remaining part of the perimeter of the initial $K$-relaxed polygon. In any case, the property (P) from Definition 2. holds. ∎