Why is the barrier cone of a convex set a cone?

Question

Barier cone $L$ of a convex set C is defined as $\{x^*| \langle x, x^* \rangle \le \beta, x\in C\}$ for some $\beta \in \mathbb{R}$. However, consider a scenario when $x_1\in L$, $\beta>0$ and $\langle x, x_1 \rangle > 0$ for all $x\in C$. The we can make $\alpha x_1$ arbitrary big, so $\alpha \langle x, x_1 \rangle$ is bigger that $\beta$. Thus, $x_1 \notin L$. Where is the problem with such reasoning?

Answer 1

You've misunderstood the definition of a barrier cone. In particular, we have $$ \newcommand{ip}[1]{\left\langle #1 \right \rangle} L = \{x^* \mid \exists \beta_{x^*} \text{ such that } \ip{x,x^*} \leq \beta_{x^*} \text{ for all }x \in C\} $$

Or, in other words, $$ L = \{x^* \mid \sup_{x \in C} \ip{x,x^*} \text{ is finite}\} $$ If we take $\alpha \,x^*$ instead of $x^*$, then our old supremum is multiplied by $|\alpha|$ to get the new supremum, but this new supremum is still finite. Or, in the notation of my first definition, we can define $$ \beta_{\alpha\,x^*} = |\alpha|\, \beta_{x^*} $$ to find that $\alpha\, x^*$ is a member of $L$.

So, $L$ is indeed a cone.