# Extending a metric in a bi-Lipschitz way

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.

Yes. Up to multiply $$d$$ with a scalar, we can suppose that for some $$0 we have $$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.

• @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 $y=x$, $y'=x'$.
– YCor
Sep 30 at 9:48
• (My previous comment justifies that $d'$ extends $d$ on $Y\times Y$.)
– YCor
Sep 30 at 10:36