Is "$K$ convex + absorbing $\not\Rightarrow$ $0\in \mathrm{Int }\, K$" dependent on AC? 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?

 A: I'm assuming real scalars.  Suppose $K$ is such a set.  Then $K \cap (-K)$ is another such set which is also balanced, so we may assume $K$ is balanced.  Similarly, since the intersection of $K$ with a ball around $0$ is again absorbing, we may assume $K$ is bounded.  Let $p$ be the Minkowski functional of $K$, i.e. $p(x) = \inf \{t > 0: x/t \in K\}$.  Then $p$ is a norm on $X$, and
$\{x: p(x) < 1\} \subseteq K \subseteq \{x: p(x) \le 1\}$.  The fact that there is no ball around $0$ contained in $K$ says that $p$ is not continuous.  The identity map from $X$ (with its original norm) to $X$ with the norm $p$ (which I'll denote as $X_p$) is then a discontinuous linear operator. Let $B^*_p$ be the closed unit ball of $X_p^*$, i.e. the set of all linear functionals $\phi$ on $X$ such that $|\phi(x)| \le p(x)$ for all $x$).
If every such $\phi$ is continuous (with respect to the original norm), then by the Uniform Boundedness Principle there would be some $R$ such that $|\phi(x)| \le R \|x\|$ for all $\phi \in B_p^*$ and $x \in X$.  But then $p(x) = \sup_{\phi \in B_p^*} |\phi(x)| \le R \|x\|$ and $p$ is continuous, contradiction.  So we must conclude that some $\phi \in B^*_p$ is discontinuous, i.e. that there is a discontinuous linear functional on $X$. 
