Suppose we are in the following situation: $(X,d)$ is a metric space and $Y$ is a subspace of $X$. Furthermore we have a different metric $\delta$ defined on $Y$ such that $\delta$ is bi Lipschitz equivalent to $d|_Y$. Is it possible to extend $\delta$ to e metric $\bar{\delta}$ on the whole $X$ such that $\bar{\delta}$ is bi Lipschitz equivalent to $d$?

I suspect the answer to be no in such generality, but I would also be interested in particular cases of metric spaces for which the answer is positive.


1 Answer 1


Yes. Up to multiply $d$ with a scalar, we can suppose that for some $0<c\le 1$ we have $cd \le\delta\le d$ on $Y\times Y$.

Define $$d'(x,x')=\min(d(x,x'),D(x,x'));\quad \text{where}$$ $$D(x,x')=\inf_{y,y'\in Y} d(x,y)+\delta(y,y')+d(y',x').$$

Clearly $cd\le d'\le d$ on $X\times X$. It remains to prove the triangle inequality for $d'$: $d'(x,x'')\le d'(x,x')+d'(x',x'')$.

There are four cases to consider.

  1. If $d'(x,x')=d(x,x')$ and $d'(x',x'')=d(x',x'')$, then $d'(x,x'')\le d(x,x'')\le d(x,x')+d(x',x'')=d'(x,x')+d'(x',x'')$.

  2. Suppose $d'(x,x')=D(x,x')$ and $d'(x',x'')=d(x',x'')$. Fix $\varepsilon>0$. Choose $y,y'\in Y$ such that $d'(x,x')\ge d(x,y)+\delta(y,y')+d(y',x')-\varepsilon$. Then $$d'(x,x')+d(x',x'')\ge d(x,y)+\delta(y,y')+d(y',x')+d(x',x'')-\varepsilon$$ $$\ge d(x,y)+\delta(y,y')+d(y',x'')-\varepsilon\ge d'(x,x'')-\varepsilon.$$ Since $\varepsilon$ is arbitrary, the inequality follows.

  3. Case $d'(x,x')=d'(x,x')$ and $d'(x',x'')=D(x',x'')$: reduces to the previous case by switching $x$ and $x''$.

  4. Suppose $d'(x,x')=D(x,x')$ and $d'(x',x'')=D(x',x'')$. Fix $\varepsilon>0$. Fix $y,y'_1,y'_2,y''\in Y$ such that $d'(x,x')\ge d(x,y)+\delta(y,y'_1)+d(y'_1,x')-\varepsilon$ and $d'(x',x'')\ge d(x',y'_2)+\delta(y'_2,y'')+d(y'',x'')-\varepsilon$.

So $$d'(x,x')+d'(x',x'')\ge $$ $$d(x,y)+\delta(y,y'_1)+d(y'_1,x')+d(x',y'_2)+\delta(y'_2,y'')+d(y'',x'')-2\varepsilon$$ $$\ge d(x,y)+\delta(y,y'_1)+d(y'_1,y'_2)+\delta(y'_2,y'')+d(y'',x'')-2\varepsilon$$ $$\ge d(x,y)+\delta(y,y'_1)+\delta(y'_1,y'_2)+\delta(y'_2,y'')+d(y'',x'')-2\varepsilon$$ $$\ge d(x,y)+\delta(y,y'')+d(y'',x'')-2\varepsilon$$ $$\ge d'(x,x'')-2\varepsilon.$$ Since $\varepsilon$ is arbitrary, we deduce the triangle inequality.

  • $\begingroup$ @MattF. Yes it does. If $x,x'\in Y$, then $D(x,x')\ge \delta(x,x')$ (first using $d\ge \delta$, then using the triangle inequality). Since $d\ge\delta$ it follows that $d'\ge\delta$ on $Y\times Y$. The other inequality is clear by choosing $y=x$, $y'=x'$. $\endgroup$
    – YCor
    Sep 30 at 9:48
  • $\begingroup$ (My previous comment justifies that $d'$ extends $d$ on $Y\times Y$.) $\endgroup$
    – YCor
    Sep 30 at 10:36

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