I have encountered the following problem in Dirk Werner's "Funktionalanalysis" (English translation by me):
Definition: A convex set $K\subset X$ is called absorbing, if given $x\in X$ there exists $\lambda>0$ such that $\lambda x\in K$.
Let $X$ be a normed vector space. If $K$ is convex, is it necessarily true that "$K$ absorbing $\implies$ $0 \in \mathrm{Int }K$"?
If $X$ is only assumed to be a normed space (not Banach), then I think $X = L^1[0,1]\cap L^\infty[0,1]$ equipped with the $L^1$-norm, and $K = \{f\in X\mid \Vert f\Vert_\infty \le 1\}$ gives a counterexample. $K$ is clearly absorbing and convex, but we can find a sequence $f_n \in X$ with $\Vert f_n\Vert_1 = 1/n \to 0$ and $\Vert f_n\Vert_{\infty} = 2$ for all $n$. So no ball around $0$ can be contained in $K$.
I also tried to find a counter-example, where $X$ is Banach. This seemed much more difficult. I could so far only find a counter-example under the assumption of the existence of a non-continuous linear functional: If $f$ is such a functional on $X$, set $K = \{x\in X\mid |f(x)|\le 1\}$. Then $K$ is convex and absorbing, but $\mathrm{Int }K = \emptyset$. This leads to the question, whether the axiom of choice is necessary to construct a counter-example on Banach spaces or not.
Question: Is there an explicit example (i.e. one whose construction does not involve the axiom of choice) of a convex set $K$ which is absorbing, but does not contain $0$ in its interior?