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Let $A$ and $B$ are convex, compact sets in $\mathbb{R}^n$. We have known that

$$\max_{a\in A}\min_{b \in B} \|a-b\|=\sup_{\|g\|\le1}(\sigma_A(g)-\sigma_B(g)),$$ where $\sigma_M(x)=\sup_{u\in M}\langle u,x\rangle$ is the support function of $M$.

(For more detail of the equality, please see Hausdorff distance via support function.)

Assume now that the quantity $\max_{a\in A}\min_{b \in B} \|a-b\|$ attains at $a^0\in A$ and $b^0\in B$, i.e., $$\max_{a\in A}\min_{b \in B} \|a-b\|=\|a^0-b^0\|.$$ Put $g^0=(a^0-b^0)/\|a^0-b^0\|$. My question is whether we have the equality $$\sup_{\|g\|\le1}(\sigma_A(g)-\sigma_B(g))=\sigma_A(g^0)-\sigma_B(g^0),$$

or not?

Thanks in advance.

Community
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abc1234
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1 Answers1

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We have $\|a^0-b\|\ge\|a^0-b^0\|\, \forall b\in B$. Hence $$\langle a^0-b^0,\, (a^0-b)-(a^0-b^0)\rangle=\langle a^0-b^0,\, b^0-b\rangle\ge 0\, \forall b\in B.$$ Thus,

$$\langle g^0,b^0\rangle\ge\langle g^0,b\rangle\, \forall b\in B.$$

This implies that $\langle g^0,b^0\rangle=\sigma_B(g^0)$. Similarly, we also have $\langle g^0,a^0\rangle=\sigma_A(g^0)$. Consequently, $$\sigma_A(g^0)-\sigma_B(g^0)=\|a^0-b^0\|.$$

The equality in the question is verified.

hntuan
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