Proving a function $p:V→[0,\infty)$ forms a norm on V if and only if the unit ball $\{{x∈V|ρ(x)<1}\}$ is convex I have a function $p:V→[0,\infty)$ where V is a vector space and we know $p$ to be positive, absolutely homogeneous and non-degenerate (i.e.: we know $p$ satisfies all norm conditions other than the triangle inequality). I need to prove that $p$ forms a norm on $V$ (i.e.: satisfies the triangle inequality) if and only if the unit ball $\{{x∈V|p(x)<1}\}$ is convex. I've shown that if the unit ball is convex, we have $tx+(1−t)y<1$ for $x,y∈V,t∈[0,1]$. Buy 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 $V=R^2$,with the proof taking $x∈V$ to be a scalar, and I don't know if I can do that for a general vector space.
Any help is appreciated.
 A: 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 $\rho(tx + (1-t)y) < \frac{1}{\epsilon}$.
A: (i). We must have $p(Ax)=|A|p(x)$ for every scalar $A$ and every vector $x.$
(ii).If $p(Ax+By)\le 1$ whenever $p(x)=p(y)=1$ and $A,B\in [0,1]$ with $A+B=1$ then $$p(Cx+Dy)\le C+D$$ whenever $p(x)=p(y)=1$ and $C,D\ge 0.$
This is trivial by (i) if $C=0$ or $D=0.$
If $C> 0< D$ let $A=C/(C+D)$ and $B=D/(C+D).$ Then $1\ge p(Ax+By).$ So $$(C+D)\ge (C+D)\cdot p(Ax+By)=$$ $$= p((C+D)(Ax+By))=$$ $$=p(Cx+Dy).$$
(iii). The triangle inequality is equivalent to : 
$p(u+v)\le p(u)+p(v)$ for all $u,v.$
This is trivial if $u=0$ or $v=0.$ 
We must have $p(w)\ne 0$ whenever $w\ne 0.$ So if $p(u)\ne 0\ne p(v)$ let $x=u/p(u)$ and $y=v/p(v).$
Then $p(x)=p(y)=1$ by (i). And let $C=p(u)$ and $D=p(y).$ By (ii) we have $$p(u+v)=p(Cx+Dy)\le C+D=p(u)+p(v).$$
