# Interior of a dual cone

Let $K$ be a closed convex cone in $\mathbb{R}^n$. Its dual cone (which is also closed and convex) is defined by $K' = \{ \phi\ |\ \phi(x) \geq 0,\ \ \forall x \in K\}$.

I know that the interior of $K'$ is exactly the set $\tilde{K} = \{ \phi\ |\ \phi(x) > 0,\ \ \forall x \in K\backslash0\}$. (see for example this question).

What happen to this statement when we are not in $\mathbb{R}^n$ any more, but in an infinite dimensional (non separable) vector space? In fact I am only interested in knowing wether $\tilde{K}$ is included into the interior of $K'$.

I tried to read about weak*-topology and Banach-Alaoglu theorem, but this is really not my cup of tea. Any answer or pointer to suitable literature would be most welcome.

• I'm not sure why it is important that your ambient vector space is non-separable. What I think can happen is that $K'$ may have empty interior (for instance, if it has compact cross section in the weak-* topology), which would answer your question in the negative. However, that statement depends strongly on the topology you put on the dual space. See here: en.wikipedia.org/wiki/Banach–Alaoglu_theorem#Generalization:_Bourbaki.E2.80.93Alaoglu_theorem – Igor Khavkine Jan 13 '14 at 23:46
• Is there an instance of this question that you are interested in and that is more concrete? – Igor Khavkine Jan 13 '14 at 23:47

The bad news is that many closed convex cones in infinite dimensions have empty interior, e.g., the natural cone of positive functions in $L^p(\Omega)$ for all $\Omega \subset \mathbb{R}^n$; or in $W^{1,p}(\Omega)$ for $\Omega \subset \mathbb{R}^n$ and $p < n$.
The positive cone in $L^2(\Omega)$ is self-dual (if you identify $L^2(\Omega)$ with its dual space) and has empty interior. However, the positive constant functions belong to $\tilde K$.