# Hausdorff distance via support function

Let $A$ and $B$ be convex compact sets in $\mathbb{R}^n$. Define $$h_{+}(A,B) = \inf \left\{ \varepsilon > 0 \mid A \subseteq B+\mathbb{B}_{\varepsilon} \right\}$$ where $\mathbb{B}_{\varepsilon}$ is an $\varepsilon$-ball centered at origin. Hausdorff distance between $A$ and $B$ is $$h(A,B) = \max \left\{ h_{+}(A,B), h_{+}(B,A) \right\}$$ Support function of a compact convex set $K$ is defined as $$c(y\mid K) = \max\limits_{x \in K} \langle y, x\rangle$$ How to show that $$h(A,B) = \max\limits_{ | y | \leq 1} | c(y \mid A) - c (y \mid B) |$$ I tried to use the Legendre transform but without success.

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I recently sketched a proof in my blog: calculus7.org/2012/05/06/pick-a-point-any-point – user31373 Jun 13 '12 at 22:20
Yeah, thanks, I did it. – Nimza Jun 14 '12 at 16:08

## 1 Answer

To prove the above statement we need an additional statement.

Lemma. $h_{+}(A,B) = \sup\limits_{a \in A}\; \mathop{\mathrm{dist}}{(a,B)}$.

Proof. $$h_{+}(A,B) \leq \varepsilon \Leftrightarrow A \subseteq B+\mathbb{B}(\varepsilon,0) \Leftrightarrow \forall a \in A \; (a \in B+\mathbb{B}(\varepsilon,0)) \\ \Leftrightarrow \forall a \in A \; \exists b\in B \colon|b-a|\leq\varepsilon \Leftrightarrow \forall a \in A \; \mathop{\mathrm{dist}}{(a,B)} \leq \varepsilon \\ \Leftrightarrow \sup\limits_{a \in A} \; \mathop{\mathrm{dist}}(a,B) \leq \varepsilon.$$ Since $h_{+}(A,B) \leq \varepsilon$ iff $\sup_{a \in A} \; \mathop{\mathrm{dist}}(a,B) \leq \varepsilon$ they are equal. $\blacksquare$

Hence we have an equality $$h(A,B) = \max \left\{ \sup\limits_{a \in A} \; \mathop{\mathrm{dist}}(a,B), \; \sup\limits_{b \in B} \; \mathop{\mathrm{dist}}(b,A) \right\}.$$ Recall that convex conjugate function of $x \mapsto \mathop{\mathrm{dist}}(x,B)$ is a support function of compact convex set $B$, i.e. $$d(a,B) = \sup\limits_{\|l\| \leq 1} \left( \langle l, a \rangle - c(l \mid B) \right). \tag{1}$$ Now we are ready to proove our main formula. We have $$\sup\limits_{a \in A} \;\mathop{\mathrm{dist}}(a,B) = \sup\limits_{a \in A} \sup\limits_{\|l\| \leq 1} ( \langle l, a \rangle - c(l \mid B) ) = \sup\limits_{\|l\| \leq 1} ( c(l \mid A) - c (l \mid B) ).$$ We have changed the order of supremums in the latter equality. Now since $$\sup\limits_{\|l\| \leq 1} | c(l \mid A) -c (l \mid B) | = \max \{ \sup\limits_{\|l\| \leq 1} ( c(l \mid A) - c (l \mid B) ), \sup\limits_{\|l\| \leq 1} ( c(l \mid B) - c (l \mid A) ) \}$$ we obtain the needed formula: $$h(A,B) = \sup\limits_{\|l\| \leq 1} | c(l \mid A) -c (l \mid B) |.$$

Added. As concerns the proof of (1). Put $f(x) = \mathop{\mathrm{dist}} (x,B)$. Then $$f^*(l) = \sup_x \bigl( \langle l, x\rangle - f(x) \bigr) \\ = \sup_{b \in B} \sup_x \bigl( \langle l, x\rangle - \|x-b\| \bigr) \\ = \sup_{b \in B} \sup_{\alpha > 0} \sup_{\|x-b\|=\alpha} \bigl( \bigl( \langle l,x-b\rangle - \alpha \bigr) + \langle l, b \rangle \bigr) \\ = \sup_{b \in B} \sup_{\alpha > 0} \bigl( \alpha(\|l\|-1) + \langle l, b\rangle \bigr) \\ = \sup_{b \in B} \bigl( \langle l, b\rangle + \delta_1(\|l\|) \bigr) \\ = c(l|B) + \delta_1(\|l\|),$$ where $\delta_1(t) = 0$ if $t\leq 1$ and $\delta_1(t) = +\infty$ otherwise. Hence, $$\mathop{\mathrm{dist}} d(x,B) = \sup_l \bigl( \langle l,x\rangle - c(l|B) - \delta_1(\|l\|) \bigr) \\ = \sup_{\|l\|\leq 1} \bigl( \langle l, x \rangle - c(l|B) \bigr).$$

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how did you derive the convex conjugation of $x\longmapsto dist(x, B)$ ? – Chival Mar 18 '15 at 12:15
@Chival I edited my answer. – Nimza Mar 18 '15 at 18:45
in the last step, did you use the biconjugate theorem(Fenchel-Moreau theorem) ? If so, $f(x)$ should be specified to be convex in particular. In fact, maybe we could use the additivity of support functions to conclude the proof (easier approach): $c(\ell, K + L) = c(\ell, K) + c(\ell, L)$. But your argument works nicely. – Chival Mar 19 '15 at 21:11