4
$\begingroup$

Let $(X,d_X)$ be a compact metric space and let $Comp(X)$ be the set of closed subsets of $X$ with the Hausdorff metric: $$ D(A,B)\overset{\text{def}}{=} \, \max\left\{\sup_{b\in B}\,d_{A}(b),\sup_{a\in A}\,d_{B}(a)\right\}. $$ Recently, in my research, the following quantity popped up:

Let $\alpha>0$ and define $D_{\alpha}$ on $\text{Comp}(X)$ as follows: $ D_{\alpha}(A,B)\overset{\text{def}}{=} \, \max\left\{\sup_{b\in B}\,d_{A}^{\alpha}(b),\sup_{a\in A}\,d_{B}^{\alpha}(a)\right\}? $

The metric $D_{\alpha}$ seems to just be the Hausdorff distance on the snowflake $(X,d_{X}^{\alpha})$. But I wonder, has anyone else seen this object in the literature before?

Improve this question
$\endgroup$
3
  • $\begingroup$ Do you assume that $\alpha\leq 1$? Otherwise $d_X^\alpha$ need not be a metric. I haven't seen $D_\alpha$, but I have seen a lot of literature about $(X,d_X^\alpha)$ spaces. I guess, you are familiar with that too, since you are using a correct term "snowflake". $\endgroup$
    – Piotr Hajlasz
    May 18, 2021 at 22:28
  • $\begingroup$ Exactly; though I include the $\alpha>1$ "metric" case since it could be interesting (even if the triangle inequality degenerates). $\endgroup$
    – TomCat
    May 18, 2021 at 22:30
  • 3
    $\begingroup$ So is $D_\alpha(A,B)=D(A,B)^\alpha$ ? $\endgroup$
    – Gerald Edgar
    May 19, 2021 at 0:09

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.