Show that $\min\{\lambda≥0:x\in\lambda K\}=h(K^{*},x)$

Question

Let $K\in\mathscr{C}^n$ with $\overrightarrow{0}\in$ int$K$ and let $h(K,\cdot)=\sup\{<x,y>:y∈ K\}$ be the support function. Also, for $\overrightarrow{x}\in\mathbb{R}^n$ we define $$|\overrightarrow{x}|_K=\min\{\lambda≥0:\overrightarrow{x}\in\lambda K\}$$ as the counter of K. $\textbf{Show that for every}$ $ \overrightarrow{x}\in\mathbb{R}^n$ $\textbf{the following holds}:$ $$|\overrightarrow{x}|_K=h(K^{*},\overrightarrow{x})$$

1. $\mathscr{C}^n$ means convex in $\mathbb{R}^n$,
2. <•,•> is the inner product in $\mathbb{R}^n$,
3. $K^{*}=\{y∈\mathbb{R}^n: <x,y>\leq 1,\quad ∀ x\in K\}$.

Answer 1

So you want to show that $$\inf\{\lambda \geq 0 : x\in\lambda K\}=\sup\{\langle x,y\rangle : y\in K^*\}.$$ We can suppose $\lambda \neq 0$ otherwise $x=0$ and it's $0=0$.

• $\geq $ : If $x\in \lambda K$, then $x/\lambda \in K$ so that $\langle x/\lambda ,y\rangle\leq 1$, that is $\langle x,y\rangle\leq \lambda$. So $$\langle x,y\rangle \leq \inf\{\lambda \geq 0 : x\in\lambda K\}$$ for all $y\in K^*$, hence the inequality.
• $\leq$ : Now if $y\in K^*$, there exists $0\leq \lambda\leq \langle x,y\rangle\leq 1$ such that $x\in \lambda K$ because $0\in \text{int} K$ and $K$ is convex.