Is a closed convex cone with empty interior contained in the kernel of a linear functional? (prove or disprove)

Question

Let $K$ be a closed convex cone contained in a locally convex (or maybe even Hilbert) topological vector space. If $\operatorname{int}(K) = \emptyset$, is it contained in the kernel of a continuous linear functional?

Answer 1

Using a separation theorem, it's easy to see that there exists a nontrivial continuous $f$ such that $f(x)\leq 0$, for every $x\in K$. (Just separate $K$ from a point $x_0\in X\setminus K$ and use the fact that $K$ is a cone.) But this is the most you can hope for:

If your space is (the Hilbert space) $\ell_2$ and $K=\{(x_n)_{n\in\mathbb{N}}\in \ell_2: x_n\geq 0, \forall n\in \mathbb{N}\}$ its standard cone, it's easy to see that $K$ is closed, convex and that it contains no interior points. Notice also that the usual orthonormal basis $(e_n)_{n\in \mathbb{N}}$ of $\ell_2$, is contained in $K$. Now suppose that there exists an $f\in \ell_2^*$ such that $K\subseteq \ker f$. Then $f$ should be identically zero, since $f(e_n)=0$, for every $n\in \mathbb{N}$.

If, however, your space $X$ is finite dimensional, then you can find nonzero functionals $f$ with $K\subseteq\ker f $. Just take $Y=K-K$ the space generated by the cone $K$. Then $Y$ has also an empty interior so, in particular, it is a proper closed subspace of $X$. Now you can easily find a nonzero $f$ with $Y\subseteq\ker f $.

Notice that in both cases our main concern was whether $K$ could be generating or not. Or to be more precise, whether $K-K$ could be dense in $X$.