Can you tell me whether my (partial) proof that $\|v\|_B := \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \}$ where $\varnothing \neq B \subset \mathbb R^d$ is open, bounded, $B = -B$ and convex defines a norm is correct? It seems too long (especially (ii)) and is therefore either too complicated or wrong. Also, could someone complete it for me by showing me how to do (iv)? Thanks.
Claim: Let $V$ be a real vector space. Then $\|v\|_B := \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \}$ where $B \subset \mathbb R^d$ is open, bounded, $B = -B$ and convex defines a norm.
Proof:
(i) $\|v\|_B \geq 0$ by definition.
In the following let $B$ be bounded by $K$, i.e. for all $b$ in $B$, $\|b\|_2 \leq K$.
(ii) $\|v\|_B = 0 \iff v = 0$: First $\implies$: Let $\|v\|_B = 0$. Then $\inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \} = 0$, in particular, for every $\lambda = \frac1n \in \mathbb N$, $\frac{1}{\lambda}\|v\|_2 = n \|v\|_2 \leq K$. Hence $\|v\|_2 = 0$ and hence $v = 0$ since $\|\cdot\|_2$ is a norm.
$\Longleftarrow$: Let $v=0$. Then for every $n \in \mathbb N$, $n \|v\|_2 = 0$ and hence for every $\lambda = \frac{1}{n}$, $\frac{1}{\lambda} \|v\|_2 = 0 \leq K$. Hence $ \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \} = \lim_{n \to \infty} \frac{1}{n} = 0$.
(iii) $\| \alpha v \|_B = |\alpha| \|v\|_B$ for all $\alpha \in \mathbb R$: $ \| \alpha v \|_B = \inf \{\lambda > 0 \mid \frac{1}{\lambda} \alpha v \in B \} \stackrel{B = -B}{=} \inf \{\lambda > 0 \mid \frac{1}{\lambda} |\alpha| v \in B \} = \inf \{\lambda |\alpha| > 0 \mid \frac{|\alpha|}{|\alpha|\lambda} v \in B \} = |\alpha| \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \} = |\alpha| \|v\|_B$
(iv) $ \|v + w \|_B \leq \|v\|_B + \|w\|_B$: