Converting a bounded metric into an unbounded metric

Question

Suppose $d$ is a bounded metric on $X$, i.e. $d(x,y)< K<\infty$ for all $x,y\in X$. Is there a standard way to convert $d$ into another metric $\widetilde{d}$ on $X$ with the property that $\widetilde{d}(x,y)\to\infty$ if and only if $d(x,y)\to K$? One way would be to find some function $f$ such that $\widetilde{d}(x,y)=f(d(x,y))$ satisfies the given conditions, but it is not obvious that this is always possible.

The following properties are additionally useful, but not necessary:

• $\widetilde{d}$ preserves the topology of $(X,d)$
• $d(x,y)>d(x',y')\implies \widetilde{d}(x,y)>\widetilde{d}(x',y')$

Also, if this is possible when $(X,d)$ satisfies certain extra assumptions but not in general, answers in this direction are welcome.

Answer 1

I do not know of any procedure for constructing such metric space in the general setting. On the contrary, if you restrict your class of metric spaces to the smaller class of complete locally compact path-metric spaces then the Hopf-Rinow theorem, see the First Chapter of [Gr], will prevent the construction of your $(X,\tilde{d})$.

If $(X,d)$ is in the intermediate class of (non-complete and non-compact) locally compact path-metric spaces, then I recall having seen the construction of a $(X,\tilde{d})$ satisfying

1. $(X,\tilde{d})$ is a complete locally compact path-metric space
2. The topology of $X$ is preserved.
3. $d(x,y) \leq \tilde{d}(x,y)$.

The procedure will be something like completing $(X,d)$, embedding $X$ inside its metric completion $\bar{X}$ and then enlarging the metric of $X$ around the points in $\bar{X} \setminus X$ (think of this as sending the extra points "to infinity"). I ignore whether your second condition can be preserved under this construction

[Gr] Gromov, Misha, Metric structures for Riemannian and non-Riemannian spaces. Progress in Mathematics (Boston, Mass.). 152. Boston, MA: Birkhäuser. xix, 585 p. (1999). ZBL0953.53002.

Answer 2

I don't think there's any standard procedure, but here is one way to do this if there is a "base point" $e\in X$ satisfying $d(e,p) < K/2$ for all $p\in X$. I am not sure whether such a point can always be added. Wlog take $K=2$ and fix $e$. For any $p,q \in X$ find $p', q'$ in the open unit disc equipped with the euclidean metric such that the distances between $p'$, $q'$, and $0$ equal the distances between $p$, $q$, and $e$. There should be only one way to do this up to rotations and reflections. Then define $\tilde{d}(p,q)$ to be the distance between $p'$ and $q'$ for the hyperbolic metric on the disc.

Answer 3

Not unless you're willing to lose the triangle inequality.

Nik Weaver's counterexample in the comments doesn't require $\tilde{d}$ to be of the form $f \circ d$: If $(X,d) = [0,1]$ with the usual metric, then you'd be requiring $\tilde{d}(0,1) = \infty$ while $\tilde{d}(0,0.5)$ and $\tilde{d}(0.5,1)$ are finite.

For the specific case of $f \circ d$ on $[0,1]$, the triangle inequality appears to be related to concavity of $f$.

The closest thing I can think of that respects the triangle inequality is dividing a Riemannian metric by the distance* to some set that you're about to delete.

*If you want the result to be Riemannian too, you still have to worry about smoothness, which probably requires constructing a different function to divide by; e.g. assume the deleted set is a submanifold and modify the metric only on a tubular neighborhood.