I am struggling with an exercice, maybe I am missing something or I don't understand something. Here's the exercice:
Let $V \subset \mathbb{R}^n$ open, convex, bounded and such that if $x \in V$ then $-x \in V$. Now consider the function defined by
$$ \mid \mid x \mid \mid_V := \inf_{r>0} \{ \frac{x}{r} \in V \}. $$
I have to show that:
a) $\mid \mid . \mid \mid_V $ defines a norm on $\mathbb{R}^n$.
b) $V=B(0,1)$, where $B(0,1)$ is the open ball with respect to the norm $\mid \mid . \mid \mid_V $.
My attempt:
I am not even sure how to show that $\mid \mid x \mid \mid_V = 0 \Leftrightarrow x =0.$ Since $V$ is convex and since $x$ and $-x$ are in $V$, $0$ has to be in $V$. Then, if $x=0$, I should have $$ \inf_{r>0} \{ \frac{0}{r} \in V \} = \inf_{r>0} \{ 0 \in V \} =0. $$ Is this correct ?
For the other direction I am unsure how to interpret
$$ \inf_{r>0} \{ \frac{x}{r} \in V \} = 0 $$
This mean that the smallest value of $r>0$ I can take such that $\frac{x}{r}$ is in $V$ is $0$. But how does this show that $x=0$ ?
For the second property of the norm, $\mid \mid \lambda x \mid \mid_V = \mid \lambda \mid \cdot \mid \mid x \mid \mid_V$, is this just the property of the infimum ? I mean something like $ \inf \{ \lambda X \} = \lambda \inf \{ X \} $? And similarly for the triangle inequality ?
To be honest I have no clue on how to prouve the part b), any help and hints will be appreciated.
Thank you.
Proof of b): if $\|x\|<1$ then there exists $r<1$ such that $\frac x r \in V$. Now $x= r(\frac x r)+(1-r)0$. This implies that $x \in V$ because $V$ is convex. [Note that $0 \in V$ because $0=\frac {y+(-y)} 2$ (where $y$ is any element of $V$) and $V$ is convex]. Conversely, suppose $x \in V$. Since $V$ is open there exists $t>0$ such that $(1+t)x \in V$. This implies $\|x\|_V \leq r$ where $r= \frac 1 {1+t}$. Hence $\|x\|_V <1$.
Triangle inequality: suppose $\frac x {r_1} \in V$ and $\frac y {r_2} \in V$. Then $\frac {x+y} {r_1+r_2}=\frac {r_1} {r_1+r_2} (\frac x {r_1})+\frac {r_2} {r_1+r_2} (\frac y {r_2}) \in V$. This gives $\|x+y\|_V \leq r_1+r_2$. Now take infimum over $r_1$ and $r_2$.