I have the following conjecture which, as of now, I can neither prove nor disprove:
Let $H$ be a hilbert space, and let $C \subset H$ be a closed (convex) cone with the property that $\overline{\operatorname{span}(C)} = H$. Let $C^*$ denote the dual cone to $C$.
Then, either $C \subseteq C^*$, or $C^* \subseteq C$.
If this is well known, I would like a reference. If it is not, then I'd like to prove or disprove it. I would be very interested in results that apply to the finite dimensional case $H = \Bbb R^n$ (or $H = \Bbb C^n$), if such results are easier to obtain.
It is notable that the spanning property of $S$ is important. In particular, if $S$ is a proper subspace, then $S^* = S^\perp$.
Any feedback here is appreciated.
A partial result that may be useful here:
Let $C_0$ be a self-dual (convex) cone (that is, $C_0 = C_0^*$). If $C$ is a closed convex cone with $C \subset C_0$ or $C_0 \subset C$, then $C$ satisfies the conjecture.
This is a fairly immediately consequence of the following property of dual cones:
If $C \subset D$ are subsets of $H$, then $D^* \subset C^*$.
It would be sufficient, then, to show that any cone with the spanning property must be contained in or contain some self-dual cone.
Your claim seems to be true in $\mathbb{R}^n$ for $n \in \{0,1,2\}$. In three dimensions one can construct counterexamples.
We define the vectors \begin{equation*} a = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} ,\quad b = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} ,\quad c = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} . \end{equation*} Then, these vectors span $\mathbb{R}^3$. Now, let $C$ be the conic hull of these vectors. Since $b^\top a = -1$, $a$ and $b$ do not belong to $C^*$. However, we have $d = (1, 2, -1)^\top \in C^* \setminus C$, since $a^\top d, b^\top d, c^\top d > 0$.
The key is that the cone $C$ is rather slim in one direction, but fat in a direction orthogonal to the slim direction.
An other counterexample should be the Lorentz cone \begin{equation*} \{ (t, x) \in \mathbb{R} \times \mathbb{R}^2 : \|x\|_A \le t \} \end{equation*} with $\|x\|_A^2 = x^\top A \, x$, if the symmetric, positive definit matrix $A$ is carefully chosen (two different eigenvalues).
Finally, one can produce a whole zoo of counterexamples. Assume the closed convex cones $C_1, C_2 \subset Y$ span $Y$ and $C_1 \subsetneq C_1^*$, $C_2^* \subsetneq C_2$. Then, the cone $C = C_1 \times C_2$ spans $Y^2$, but neither $C \subset C^*$ nor $C^* \subset C$ hold.