Determine the convex hull of
$$V=\{(x,y,0): x^2 + y^2 = 1\}\cup \{(1,0,z): |z| \leq 1\}$$
By drawing a picture, I conjectured that
$$Conv(V) =\{(x,y,z): (x-|z|)^2 + y^2 = (1-|z|)^2, |z| \le 1\}=: K$$
Is there an easy way to verify that equality actually holds? In particular, I have trouble showing that the set I conjectured is actually convex itself. Clearly, $V \subseteq K$ so if I can show $K$ is convex I already get $Conv (V) \subseteq K$ and then half the work is done.
So, is there an easy way to see that $K$ is convex (if it even is convex, my guess might be wrong!)? I tried to do the explicit computations but they got nasty (because of the absolute value signs) and I couldn't continue. Maybe I should consider cases where $z \leq 0$ and $z \geq 0$ and show my set is closed under convex combinations but even then the computations become quite long.
Best Answer
It is easy to verify that $V \subset K$ but the set $K$ is not convex. Note that $0 \in \operatorname{co} V$ but $0 \notin K$.
I think $K= \partial \operatorname{co} V$, you need to add the interior as well.
Here is one way to visualise the convex hull. Pick some $z' \in [-1,1]$ then $\operatorname{co} V \cap \{(x,y,z') | x,y \in \mathbb{R}\} = \bar{B}((|z'|,0),1-|z'|) \times \{z'\}$, where the ball is the two dimensional ball.
Hence $\operatorname{co} V = \{ (x,y,z) | \sqrt{(x-|z|)^2+y^2} \le 1-|z|\}$. The $\sqrt{}$ avoids the restriction $|z| \le 1$.