I am trying to represent the two dimensional closed unit ball, $\mathbb{B} \in \mathbb{R^2}$ as the intersection of a collection of closed half spaces, and find the half spaces explicitly.
Definitions that I believe I have to use are the followings.
Given a nonzero vector $p\in \mathbb{R^2}$ and a real number $\alpha$, Hyperplane: $H=\{ x\in \mathbb{R^2} | p \cdot x = \alpha\}$
Its associated closed half spaces: $H^+=\{ x\in \mathbb{R^2} | p\cdot x \geq \alpha\}$ and $H^-=\{ x\in \mathbb{R^2} | p\cdot x \leq \alpha\}$
I tried to make a line that has a intersecting point on the unit circle. But I am not sure if I am right.
Please give me some advice.
For every point $y$ on the boundary sphere (i.e. $\|y\| = 1$), the halfspace $$\{x \in \Bbb{R}^2 : x \cdot y \le 1\}$$ contains the unit ball, by Cauchy-Schwarz inequality (if $\|x\| \le 1$, then $x \cdot y \le \|x\| \|y\| = \|x\| \le 1$). Thus, $$B[0; 1] \subseteq \bigcap_{y \in S[0; 1]} \{x \in \Bbb{R}^2 : x \cdot y \le 1\} \tag{$\star$}.$$ On the other hand, if $x \notin B[0;1]$, i.e. $\|x\| > 1$, and we let $y = x/\|x\|$, then $$x \cdot y = x \cdot x / \|x\| = \|x\|^2 / \|x\| = \|x\| > 1,$$ hence $x \notin \{x \in \Bbb{R}^2 : x \cdot y \le 1\}$. Hence, we get equality in $(\star)$.