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.
 A: 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,
$$
   \mathop{\mathrm{dist}} d(x,B) = \sup_l \bigl( \langle l,x\rangle - c(l|B) - \delta_1(\|l\|) \bigr) \\
   = \sup_{\|l\|\leq 1} \bigl( \langle l, x \rangle - c(l|B) \bigr).
$$
