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](https://mathoverflow.net/questions/101562/proving-the-interior-of-a-dual-cone-is-the-set-of-vectors-whose-inner-product-is)). 



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> 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 $K'$.
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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.



Interior of a dual cone
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 $\tilde K$.


And things get worse if you work with weaker topologies on the dual space.


