Support function is the Minkowski functional of polar body?

Question

Let $K \subset R^d $ be an compact convex set with $0 \in int (K)$ let: $$h_K(x) = \max_{y \in K } \langle x,y\rangle$$ be the support function of $K$

the Minkowski functional of $K$ is define as

$$\|x\|_K = \min \{\lambda \geq 1: x \in \lambda K\}$$

and

$$K^* = \{y \in R^d : \forall x \in K, \langle x,y\rangle \leq 1 \}$$ be the polar set of $K$ then $h_K(\cdot) = \|\cdot\|_{K^*}$

how to see that?

I’m just guessing but it probably means bounded so the max (or maybe better, supremum) in the support function is well-defined (and presumably the $0$ is there to make sure it is positive). — J.G, Jul 09 '19 at 22:42
As an aside, the Minkowski functional is usually defined with $\lambda >0$. — copper.hat, Jul 09 '19 at 22:43
@copper.hat no idea ,just convey from forth part of this paper ... https://projecteuclid.org/download/pdf_1/euclid.ecp/1465316761 — ShaoyuPei, Jul 09 '19 at 22:45
There is something wrong then. It indicates that if $K$ is symmetric that $|\cdot |_K$ is a norm, but that is not possible since $|x|_K \ge 1 $ for all $x$. — copper.hat, Jul 09 '19 at 22:50

score 1 · Answer 1 · answered Jul 10 '19 at 00:03

As noted in the comments, the Minkowski functional is written with the infimum over $\lambda >0$, or else this is false. For any $\alpha>0$ and $x\in \mathbb{R}^d$, \begin{gather*} \alpha\geq h_K(x)\iff\\ \alpha\geq \langle x,y\rangle\quad \forall y\in K\iff\\1\geq \langle \frac{1}{\alpha}x,y\rangle \quad \forall y\in K\iff\\ \frac{1}{\alpha}x\in K^*\iff\\ x\in \alpha K^*\iff\\ \alpha\geq \|x\|_{K^*}. \end{gather*} It is not hard to see that $h_K(x)$ and $\|x\|_K^*$ are finite and nonnegative by the compactness and $0$ in the interior assumptions, and this shows they have the same set of upper bounds, and therefore are equal.

Support function is the Minkowski functional of polar body?

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