Given a function $\rho:X\to\mathbb{R}$ on a vector space $X$ which satisfies the following properties:
- $\rho(x)=0$ if and only if $x=0$
- $\rho(x+y)\leq\rho(x)+\rho(y)$
- $\rho(-x)=\rho(x)$
for any $x,y\in X$. Is the space belonging to the metric induced by $\rho$, i.e. $d_\rho(x,y)=\rho(x-y)$, normable?
Background: This proof shows that the metric induced by a norm is in fact a metric, using only the special case $\rho(-x)=\rho(x)$ of the absolute homogeneity requirement of a norm. So I was wondering what happens if we leave it out.
Own attempt: Since a topological vector space is normable if and only if it is Hausdorff and has a convex bounded neighborhood of $0$, and since all metric spaces are Haussdorf, it suffices to show that it has a convex bounded neighborhood of $0$.
Could somebody help me prove or disprove the latter? If $\rho(\alpha x)\leq|\alpha|\rho(x)$, for all $x\in X$ and all $0\leq\alpha\leq 1$, the unit ball is convex by the triangle inequality, but what about the general case above?
The topology induced by the metric $d_\rho$ may be not even a topology of a topological vector space. For instance, let $X=\mathbb R^2$, $\rho(x_1,x_2)=1$ if $x_2\ne 0$, and $\min\{1,|x_1|\} $ otherwise. Then the scalar multiplication $\Bbb R\times X\to X$ is not continuous at the point $(0, (0,1))$.