Proving a function $\rho:V\to[0,\infty)$ forms a norm on $V$ if and only if the unit ball $\{x \in V\mid ρ(x)<1\}$ is convex

Question

I have a function $\rho:V\to[0,\infty)$ where $V$ is a vector space and we know $\rho$ to be positive, absolutely homogeneous and non-degenerate (i.e.: we know $\rho$ satisfies all norm conditions other than the triangle inequality).

I need to prove that $\rho$ forms a norm on $V$ (i.e.: satisfies the triangle inequality) if and only if the unit ball $\{x\in V\mid \rho(x)<1\}$ is convex.

I've shown that if the unit ball is convex, we have $tx+(1−t)y<1$ for $x,y\in V,t\in[0,1],$ but I am unsure as to where to go from there, or if that's even useful.

I've seen a similar question like this on here, but it is for $V=\mathbb{R}^2$, with the proof taking $x\in V$ to be a scalar, and I don't know if I can do that for a general vector space.

Any help is appreciated.

Answer 1

Some hints: If you knew that the closed unit ball $\{ x \in V \mid \rho(x) \le 1 \}$ was convex, then for nonzero vectors $x$ and $y$ you could use the convexity to conclude: $$ \rho \left( \frac{\rho(x)}{\rho(x) + \rho(y)} \cdot \frac{x}{\rho(x)} + \frac{\rho(y)}{\rho(x) + \rho(y)} \cdot \frac{y}{\rho(y)} \right) \le 1. $$

Now, as for how you show the closed unit ball is convex if the open unit ball is: suppose $\rho(x), \rho(y) \le 1$. Then for any $\epsilon \in (0, 1)$, we have $\rho(\epsilon x) < 1$ and $\rho(\epsilon y) < 1$, so $\epsilon \rho(tx+(1-t)y) = \rho(t \cdot \epsilon x + (1-t) \cdot \epsilon y) < 1$, and so $\rho(tx + (1-t)y) < \frac{1}{\epsilon}$.

Answer 2

(i). We must have $\rho(Ax)=|A|\rho(x)$ for every scalar $A$ and every vector $x.$

(ii).If $\rho(Ax+By)\le 1$ whenever $\rho(x)=\rho(y)=1$ and $A,B\in [0,1]$ with $A+B=1$ then $$\rho(Cx+Dy)\le C+D$$ whenever $\rho(x)=\rho(y)=1$ and $C,D\ge 0.$

This is trivial by (i) if $C=0$ or $D=0.$

If $C,D>0,$ let $A=C/(C+D)$ and $B=D/(C+D).$ Then $1\ge \rho(Ax+By).$ So $$(C+D)\ge (C+D)\cdot \rho(Ax+By)=$$ $$= \rho((C+D)(Ax+By))=$$ $$=\rho(Cx+Dy).$$

(iii). The triangle inequality is equivalent to :

$\rho(u+v)\le \rho(u)+\rho(v)$ for all $u,v.$

This is trivial if $u=0$ or $v=0.$

We must have $\rho(w)\ne 0$ whenever $w\ne 0.$ So if $\rho(u)\ne 0\ne \rho(v)$ let $x=u/\rho(u)$ and $y=v/\rho(v).$

Then $\rho(x)=\rho(y)=1$ by (i). And let $C=\rho(u)$ and $D=\rho(y).$ By (ii) we have $$\rho(u+v)=\rho(Cx+Dy)\le C+D=\rho(u)+\rho(v).$$