Let $|| \cdot ||$ be a sublinear functional on a linear vector space $X$. Then the sublevel set $$K=\{ x \in K: ||x|| \leq 1\}$$ is an absorbing convex set and $0 \in K$.
What I've got so far:
It is easy to see that a set $K$ is convex since it contains balls $||x||\leq 1$ that are convex as well.
Definition: A subset $K$ of a linear vector space $X$ is called absorbing if $$X= \bigcup_{t\geq 0} tK$$ where $tK=\{tx:x\in K\}$.
Any help would be greatly appreciated.
$K$ is absorbing if for every $x\in X$ you have a $t>0$ and an $a\in K$ with $t\cdot a= x$. So look at some $x\in X$ and try to find the relevant $t$ and $a$.
First take care of the case $\|x\|=0$. If this is the case then clearly $\|x\|≤1$ and then $x\in K$, so take $a= x$ and $t=1$.
Whats left is $\|x\|>0$, in this case note by sub-linearity that $\left\|\frac x{\|x\|}\right\|=1$ and hence $\frac{x}{\|x\|}\in K$, let this be your $"a"$, now choose $t=\|x\|$, then: $$t\cdot a= \|x\|\cdot \frac{x}{\|x\|}=x$$ and the condition is verified also in the case that $\|x\|\neq 0$, hence $K$ is absorbing.