Definition 13.3.2 (Hausdorff Distance).label Let $A\suf\R^{n}$ be a nonempty set and $x\in\R^{n}$ then
\begin{align*}d(x,A)\define \INF{a\in A}\norm{x-a}{}\end{align*}
When $A$ is compact, the infimum is attained. For compact $A$, the closed $\e$-neighborhood of $A$ is $A_{\e}\define A+\e B^{n}_{2}=\curl{x:d(x,A)\leq \e}$ where $B_{2}^{n}=\curl{x\in\R^n:\norm{x}{2}\leq 1}$ is the Euclidean unit ball.
Let $C,D\suf\R^{n}$ be nonempty compact sets. Their Hausdorff distance is
\begin{align*}d_{H}(C,D)\define\MAX{}\curl{\SUP{x\in D}d(x,D),\SUP{y\in D}d(y,C)}=\INF{}\curl{\e>0:C\suf D+\e B^n_2,D\suf C+\e B^n_2}\end{align*}
If in addition $C,D$ are convex then
\begin{align*}d_{H}(C,D)=\MAX{u\in \mathbb{S}^{n-1}}\abs{h_C(u)-h_D(u)}\end{align*}
Proof. It suffices to show $d_{H}(C,D)\leq \e\iff C\suf D+\e B^{n}_{2},D\suf C+\e B^{n}_{2}$. But $\SUP{x\in D}d(x,D)\leq\e$ is equivalent to $C\suf D+\e B^{n}_{2}$.$\square$
Post a Comment