Is the convex hull of closed cones closed?

Question

We say $K$ is a cone if $K + K \subseteq K$ and $[0,\infty) K \subseteq K$.

Let $K_i$, $i \leq m$ be closed cones.

Is conv$(\cup_{i=1}^{m}K_i)$ closed?

Answer 1

Consider two circles $C_\pm$ in $\mathbb{R}^3$ $$ x^2+(y\pm 1)^2+(z-1)^2=\epsilon^2$$

Define a cone $K_\pm =[0,\infty)\cdot {\rm conv}\ C_\pm$. Then ${\rm conv}\ K_+\cup K_-$ is not closed.

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(1) If $x_n\in K_1,\ y_n\in K_2$, and if $t_nx_n +(1-t_n)y_n\rightarrow z,\ t_n\in [0,1]$, then $|x_n|\rightarrow \infty$ is nontrivial case.

Let $|x_n|=l_nv_n,\ |v_n|=1$. By compactness, we let $v_n \rightarrow v$. Note that since $K_1$ is a closed cone, so $[0,\infty) v $ is in $K_1$. Then $X:={\rm conv}\ K_1\cup K_2$ is closed.

(2) $X$ is cone : If $z\in X$, then $z= tx +(1-t)y,\ x\in K_1,\ y\in K_2$

So $sz =t(sx)+(1-t)(sy) \in X$ so that $[0,\infty) X\subset X$.

If $z'=t'x'+(1-t')y'$, then $$ z+z' = 2\bigg( \frac{1}{2} (tx+t'x') + \frac{1}{2} \{ (1-t)y+(1-t')y'\} \bigg) \in X$$

(3) ${\rm conv}\ K_1\cup K_2\cup K_3= {\rm conv}\ \bigg({\rm conv}\ K_1\cup K_2\bigg)\cup K_3$ so that we complete the proof.