I am misunderstanding the definition of a barrier cone:
Let $C$ be some convex set. Then the barrier cone of $C$ is the set of all vectors $x^*$ such that, for some $\beta \in \mathbb{R}$, $\langle x, x^* \rangle \leq \beta$ for every $x\in C$.
So the barrier cone of $C$ is the set of all vectors $x^*$ where $\langle x,x^*\rangle$ is bounded above over $x\in C$.
I read that a vector $x^*$ is in the barrier cone of $C$ if and only if $x^*$ is normal to some half-space, $H = \{x\,\vert \, \langle x,y \rangle \leq \beta \}$ (for some $\beta \in \mathbb{R}$ and vector $y$ ), containing $C$. But to me, these statements do not seem equivalent. To demonstrate my issue:
Consider the (convex) ray in $\mathbb{R}^2$, $C=\{ (a,0)\, |\, a\geq 0 \}$. Then $C\subset H$ where $H$ is the half-space defined by \begin{align} H=\bigg\{x\in \mathbb{R}^2 \,\bigg\vert\, \bigg\langle x,\begin{pmatrix}-1\\0\end{pmatrix}\bigg\rangle\leq0 \bigg\} = \big\{(x,y)\in \mathbb{R}^2\,\big\vert\, x\geq0\big\}. \end{align} Then by definition of $H$, \begin{pmatrix}-1\\0\end{pmatrix} is normal to $H$ so we should expect this vector to be in the barrier cone... but letting $\beta\in \mathbb{R}$ we have $(0,0)\in C$ and $(|\beta|,0) \in C$ so, (this is probably my misunderstanding) don't we have $y=(0,0)-(|\beta|,0)\in C$, a vector with the property that \begin{align} \bigg\langle y,\begin{pmatrix}-1\\0\end{pmatrix} \bigg\rangle \geq \beta \quad? \end{align} And thus this vector would not be in the barrier cone by our original definition. Any help is appreciated!
And if the vectors in $C$ are not the subtractions of points in $C$, then what are they?