# Tensor Product of Convex Sets?

I was wondering if such a concept was used anywhere. What I was thinking of is this. Consider two vectors spaces $V,W$ and convex sets $C_1 \subseteq V$ and $C_2 \subseteq W$ if we define $C_1 \otimes C_2 := \text{Convex Hull}(\{c_1 \otimes c_2 \in C_1 \otimes C_2 : c_1 \in C_1,c_2 \in C_2 \})$. If $C_1$ and $C_2$ are bounded convex sets then so is $C_1 \otimes C_2$. Also, I believe that it is the case that if $a$ is a vertex of $C_1$ and $b$ is a vertex of $C_2$ then $a \otimes b$ is a vertex of $C_1 \otimes C_2$ and all vertices of $C_1 \otimes C_2$ can be constructed in this way.

Does anybody know of a construction like this actually being used anywhere? In particularly interested in the case where $C_1$ is given by as the solution to a list of inequalities $\alpha_i(v) \leq 1$, $\alpha_i \in V^*$, and $C_2$ is likewise given by a list $\beta_j(w) \leq 1$, $\beta_j \in W^*$. Then I can see that $C_1 \otimes C_2$ is contained in the convex set given by the inequalities $\alpha_i \otimes \beta_j \leq 1$. I've been try and failing at showing that $C_1 \otimes C_2$ is in fact exactly equal to the solution of these inequalities. Any advice that may be helpful would be appreciated.

• Possibly related: mathoverflow.net/questions/115677 – darij grinberg Mar 30 '16 at 0:44
• In the definition of $C_1\otimes C_2$ you mean to write $c_1\otimes c_2\in C_1\times C_2$? Also, $c_1\otimes c_2$ is the same as $(c_1,c_2)$, right? – Dirk Feb 2 '18 at 9:18
• No, I meant the tensor product of the vectors c1 and c2 – Onye Feb 3 '18 at 9:47

This is well-documented in the case where $C_1$, $C_2$ are unit balls for some norm, and in that case your $C_1 \otimes C_2$ is the unit ball for the so-called projective norm. The set you compare to in the last paragraph is the unit ball for the so-called injective norm, which indeed is a different norm. Googling these keywords should give many results.