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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?

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If it's not complicated, can you please add the definition of an absorbing set -- or at least link to one? –  Asaf Karagila Aug 31 '12 at 22:20
    
A convex set $K$ in vector space $X$ is absorbing if every $x \in X$ is in $tK$ for some $t > 0$. –  Robert Israel Aug 31 '12 at 22:29
    
@AsafKaragila: Of course, done. –  Sam Aug 31 '12 at 22:32
    
@AsafKaragila: then $\lambda(-1,0)$ is not in $K$ for any $\lambda > 0$, so $K$ is not absorbing. –  Robert Israel Aug 31 '12 at 22:38
    
@RobertIsrael: Yes, thanks. :-) –  Asaf Karagila Aug 31 '12 at 22:39

1 Answer 1

up vote 2 down vote accepted

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 $X$.

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The fact that you can construct a discontinuous linear operator into a normed space already proves that some choice is used (since in some models there is automatic continuity for linear operators from a Banach space to a normed space), but I'm not sure the appeal to the Uniform Boundness Principle is valid here, in the sense that it might not hold in a particular model (where automatic continuity fails, of course). Regardless to the above, very nice answer. –  Asaf Karagila Aug 31 '12 at 23:20
    
This is very nice! I think it already answers my question mostly (in the sense that I should not expect to see an explicit example of such a convex $K$ in, say, $\ell^p$, $c_0$, $L^p(\mathbb R)$ or any other well-known Banach-space). Thank you! –  Sam Aug 31 '12 at 23:39
    
Thank you again for your answer. =) –  Sam Sep 1 '12 at 11:04

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