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Given a set $\mathscr{M}$ such that $\# \mathscr{M}\geq 2$, we know that there is, at least, an uncountable number of metrics defined on $\mathscr{M}$: $$\begin{array}{cccc}d_a:&\mathscr{M}\times \mathscr{M}&\longrightarrow &\mathbb{R}\\ & (x,y)&\longmapsto & a\cdot \delta_{x,y}\end{array}$$ with $a\in \mathbb{R}_+$ and $\delta_{x,y}=0$ if $x=y$ and $\delta_{x,y}=1$ otherwise.

Now, my question is: given a metric $d$ on a set $\mathscr{M}$ is there some sufficient and necessary conditions over a function $f:A\subset \mathbb{R} \longrightarrow \mathbb{R}$ in order to $$D_f=f\circ d:\mathscr{M}\times \mathscr{M} \longrightarrow \mathbb{R}$$ also be a metric over $\mathscr{M}$?

I found that

  1. $I\supset d(\mathscr{M}\times \mathscr{M})$;
  2. $f(0)=0$ and $f(x)> 0,\forall x \in d(\mathscr{M}\times \mathscr{M})-\{0\}$.

are obviously sufficient and necessary to be possible to compose and to have $D_f(x,y)\geq 0$ with $D_f(x,y)=0\iff x=y$. We don't need to request anything from $f$ in order to gain $D_f(x,y)=D_f(y,x)$, since $d$ is symmetric and $f$ is a function. The trouble arises at the triangular inequality. Obviously, $$D_f(x,y)\leq D_f(x,z)+D_f(z,y)$$ is a sufficient and necessary condition, but I don't want to suppose anything about $D_f$, only about $f$ itself.

Well, I found that

  1. $f$ is monotonically increasing, i.e. $x\leq y \implies f(x)\leq f(y)$, when restricted to $d(\mathscr{M}\times \mathscr{M})$;
  2. $f(x+y)\leq f(x)+f(y), \forall x,y\in d(\mathscr{M}\times \mathscr{M})$

are SUFFICIENT conditions for this, but seems far from being necessary (at least, I couldn't show it).

Note, for example, that given $a>0$ a real number and $f:\mathbb{R}\longrightarrow \mathbb{R}$, $f(x)=\min \{a,x\}$, then $f$ satisfies 1, 2, 3 and 4.

Note that my question is not duplicate to

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  • $\begingroup$ Why $M\times M$? You have $f\circ d : M\to\mathbb R$. $\endgroup$ – Friedrich Philipp Mar 10 '16 at 12:45
  • $\begingroup$ @FriedrichPhilipp: $f\circ d$ is a metric on $M$. $\endgroup$ – user 170039 Mar 10 '16 at 12:54
  • $\begingroup$ Uh, damn... I'm sorry. Just got up. ;o) $\endgroup$ – Friedrich Philipp Mar 10 '16 at 12:58
  • $\begingroup$ No problem, @FriedrichPhilipp! $\endgroup$ – Ders Mar 10 '16 at 23:05
  • 1
    $\begingroup$ Relevant. $\endgroup$ – user 170039 Mar 12 '16 at 15:50
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And, curiously (found by my teacher again) there is a simple example that shows that 3 IS NOT necessary (now if we consider again, as in the beginning, $d$ a fixed metric).

Let $\mathscr{M}$ be $\{0,1,2\}$ and $d:\mathscr{M}\times \mathscr{M} \longrightarrow \mathbb{R}$ be given by $d(x,y)=|x-y|$ and $f:\{0,1,2\}\to \mathbb{R}$ given by $$f(0)=0, \hspace{5mm} f(1)=2, \hspace{5mm} f(2)=1.$$ $f$ is clearly decreasing, although $D_f=f\circ d$ defines a metric on $\mathscr{M}$ (we verify for example the triangular inequality case by case, since the set is finite).

Note that, curiously, in this case, $f$ is sub-additive! (Note that this is not necessary, at least not with the proof below where we allow $d$ to be ANY metric, not a fixed one).
In fact, $f(1+1)=f(2)=1<2+2=f(1)=f(1)$, the cases with $0$ are trivial and $1+2\notin d(\mathscr{M}\times \mathscr{M})$, in such a way that there are not $x,y\in d(\mathscr{M}\times \mathscr{M})$ that contradict $f(x+y)\leq f(x)+f(y)$.

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  • $\begingroup$ One Q I wonder about is, for metric$ e$ , if $e(x,y<1$ for all $x,y,$ what conditions (nec.\suff.) that $d=e/(1-e)$ is a metric. Of course if $d$ is a metric then $e=d/(1+d)$ is an equivalent one. But $d$ may fail to be a metric. $\endgroup$ – DanielWainfleet Mar 23 '16 at 20:29
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My teacher at university was able to figure out that, under some adaptation, 4 is also necessary, as follows.

We see that if $d$ is any metric on $\mathscr{M}$ then, if $f$ satisfies 1, 2, 3 and 4, we have that $f\circ d$ is also a metric on $\mathscr{M}$.

Suppose now that $D_f=f\circ d$ is a metric for ANY metric $d$ on ANY set $\scr{M}$. Then, consider the following example:

$$\mathscr{M}=\mathbb{R}, \hspace{5mm}d(x,y)=|x-y|$$

and suppose that there are $x_0,y_0\in d(\mathscr{M}\times \mathscr{M})=\mathbb{R}_+\cup \{0\}$ such that $$f(x_0+y_0)>f(x_0)+f(y_0).$$ And, in this case, we have $$f(|x_0-(-y_0)|)>f(|x_0-0|)+f(|0-y_0|)$$ $$\implies D_f(x_0,-y_0)>D_f(x_0,0)+D_f(0,-y_0)$$ and this contradicts the fact that $D_f$ is supposed to be a metric.

Thus, we must have $f(x+y)\leq f(x)+f(y),\forall x,y \in d(\mathscr{M}\times \mathscr{M})$ ($f$ must be sub-additive in $d(\mathscr{M}\times \mathscr{M})$), if we request $f\circ d$ to be a metric for any $d$.

This is not a complete converse to my statement, but it is a n important improvement in the direction to build a class of functions $\mathscr{F}$ such that if $f\in \mathscr{F}$ then $f\circ d$ is a metric, for any metric $d$ on any set $\mathscr{M}$.

After this little adaptation, what do you guys think about the item 3?? We weren't able to find nothing yet...

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  • $\begingroup$ You had a typo ."Thus we must have $f(x+y)\leq f(x)=f(y)$" which was my only edit. $\endgroup$ – DanielWainfleet Mar 23 '16 at 20:15

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