Let $A\subset \mathbb{R}^n$ be any bounded, open, convex, and the centre symmetry set having centre at $0$, that is if $x\in A$, then $-x\in A$.
Show that
$$\|x\| = \inf \{k>0 : x/k \in A \}$$
is a norm on $\mathbb{R^n}$ and the induced open ball $B$ centered at $0$ and radius $1$ covers $A$.
I don't know how to show triangle inequality and $\|tx\|=|t| \cdot \|x\|$ and also how to show that the induced unit ball covers $A$.
Best Answer
Let $A\subset \Bbb R^n$ be a bounded, symmetric, convex, open set.
Let $\|x\|=\inf\{k>0 : x/k\in A\}$.
$\bullet$ $\|0\|=0$
$0\in A$ because $A$ is symmetric and convex (because $0=x/2-x/2\in A$). Thus $0/k \in A$ for every $k>0$ implying that $\|0\|=0$.
$\bullet$ $\|t x\|=|t|\|x\|\quad \forall t\in \Bbb R$
Let $t>0$, then it holds $$ \|tx\| = \inf\Big\{k>0 : \frac{tx}{k}\in A\Big\}= t\inf\Big\{\frac{l}{t}>0 : \frac{x}{\frac{l}{t}}\in A\Big\}=t\|x\|.$$ Now, as $A$ is symmetric, we have $x/k\in A$ if and only if $-x/k\in A$ which implies that $\|x\|=\|-x\|$. Hence, if $t<0$, then $$\|tx\|=\||t|(-x)\|=|t|\|-x\|=|t|\|x\|.$$
$\bullet$ $\|x+y\|\leq \|x\| + \|y\|$
Let $x,y\in \Bbb R^n$, then for every $k,l>0$ such that $x/k\in A$ and $y/l\in A$ it holds $$ \frac{k}{k+l}\frac{x}{k}+\frac{l}{k+l}\frac{y}{l}=\frac{x+y}{k+l}\in A$$ by convexity of $A$. It follows that \begin{align*} \|x+y\| &= \inf\Big\{m>0 : \frac{x+y}{m}\in A\Big\}\\ &\leq \inf\Big\{k>0 : \frac{x}{k}\in A\Big\}+\inf\Big\{l>0 : \frac{y}{l}\in A\Big\}\\ &=\|x\|+\|y\|.\end{align*}
$\bullet$ $\|x\|=0 \ \implies \ x=0$
If $\|x\|=0$, then $x/k\in A$ for every $k>0$. But $A$ is bounded, and so, if $x\neq 0$, we get a contradiction by letting $k\to 0$. Hence, we must have $x=0$.
Combining the first and last point we get $\|x\|=0$ if and only if $x=0$ which shows that $\|\cdot \|$ is a norm on $\Bbb R^n$.
Remark: The reverse direction of the exercise can be shown. Indeed, if $\|\cdot\|'$ is a norm on $\Bbb R^n$, then $A'=\{x\mid \|x\|'< 1\}$ is a bounded, symmetric, convex, open set.
For the last part, let $B=\{x\in \Bbb R^n\mid \|x\|<1\}$ and $z\in A$. Then, we have $z=\frac{z}{1}\in A$ implying that $\|z\| <1$, i.e. $z\in B$. It follows that $A\subset B$.