I was trying to find if the Hausdorff distance under a K-Lipschitz transformation would itself respect the K-Lipschitz constraint. Specifically,
If $f: \mathcal{X} \mapsto \mathcal{Y}$ is K-Lipschitz function, then for any $X,X' \in \mathcal{X}$ and metric spaces $(\mathcal{X}, \ell_p)$ and $(\mathcal{Y}, \ell_p)$,
$$
\Vert f(x) – f(x') \Vert \leq K\Vert x – x'\Vert \;\; \forall \;x \in X,\; x' \in X' \implies d_H(f(X), f(X')) \leq Kd_H(X, X')
$$
reading another answer here, it seems to imply this is true without proof. I have searched and I cannot find any proof. I would like to find…
- The proof of this
- Also to know if it holds only for Hausdorff distance or for any proper distance measure on the sets $X, X'$?
Best Answer
Let $A,B\subset X$ be two subsets and $a\in A$. We have that $$d(f(a), f(B))=\inf_{b\in B}d(f(a),f(b))$$ Since $d(f(a),f(b))\le K\cdot d(a,b)$ for all $b\in B$, by taking infima over $b\in B$ in this inequality, we can conclude that $d(f(a),f(B))\le K d(a,B)$. Since $a\in A$ was arbitrary, by taking suprema over $a\in A$ in this inequality, we conclude that $\sup_{a\in A}d(f(a),f(B))\le K\cdot\sup_{a\in A}d(a,B)$.
Likewise, one sees that $\sup_{b\in B}d(f(b),f(A))\le K\cdot\sup_{b\in B}d(b,A)$. Therefore, $$d(f(A),f(B)):=\max\bigg\{\sup_{a\in A}d(f(a),f(B))\le K\cdot\sup_{a\in A}d(a,B),\;\;\sup_{b\in B}d(f(b),f(A))\bigg\}\le$$ $$\le\max\bigg\{K\cdot\sup_{a\in A}d(a,B),\;\; K\cdot\sup_{b\in B}d(b,A)\bigg\}=$$ $$=K\cdot\max\bigg\{\sup_{a\in A}d(a,B),\;\;\sup_{b\in B}d(b,A)\bigg\}=K\cdot d(A,B)$$