Prove this function defined for a convex, bounded, open, symmetric set $V$ is a norm on $\mathbb{R}^n$

Question

$V\subset \mathbb{R}^n $ be open, symmetric and contains 0 . Define $||\cdot||:\mathbb{R}^n \rightarrow \mathbb{R}$ as:

\begin{equation} ||x||=\inf\{ \lambda > 0: \lambda^{-1}x\in V \} \end{equation} for all x $\in \mathbb{R}^n$.

Prove this is a norm $\iff$ $V$ is convex and bounded with respect to the euclidean metric.

Approach:

$||x||\ge 0$ (Done)
$||\alpha x||= |\alpha|||x||$ (not done)
triangle in equality (not done)

https://en.wikipedia.org/wiki/Minkowski_functional#Convexity_of_K — Luiz Cordeiro, Sep 16 '16 at 18:41
I am not sure how the symmetric property of $V$ is necessary. — mmcrjx, Sep 16 '16 at 19:26
Also assuming that it is a norm how to show that V is convex and bounded? — mmcrjx, Sep 16 '16 at 19:44
Thanks Luiz Cordeiro. This direction of the problem is solved. — mmcrjx, Sep 16 '16 at 19:58
Possible duplicate of Norm induced by convex, open, symmetric, bounded set in $\Bbb R^n$. — Ayutac, Jan 03 '17 at 21:02

score 1 · Accepted Answer · answered Mar 03 '17 at 06:39

I guess it depends what you mean by symmetric? Inversion symmetric (i.e. contains $x$ iff it contains $-x$)? This is not a norm depending on what you mean. Consider the open set $(-10,10)\times (-1,1)$. This is inversion symmetric, contains the origin, etc. Let $z_1=(1/\sqrt{2},1/\sqrt{2})$, and $z_2=(1/\sqrt{2},-1/\sqrt{2})$. We have:

\begin{align} \|z_1\| &= \|z_2\| = 2\\ \|z_1+z_2\| &= 10/\sqrt{2} \end{align}

Property 2 appears to hold though: That is, this functional is homogeneous.

Prove this function defined for a convex, bounded, open, symmetric set $V$ is a norm on $\mathbb{R}^n$

1 Answers1