"Minkowski Multiplication" of Convex Sets? Apologies if this question might be trivial or has been asked already (haven't found an equivalent post), but I am trying to figure out whether the following is true:
Given two convex sets $\mathcal{X} \subseteq \mathbb{R}^n$ and $\mathcal{Y} \subseteq \mathbb{R}^n$, is $\mathcal{Z} := \{x \odot y ~|~ x \in \mathcal{X}, y \in \mathcal{Y}\}$ convex? 
(Here $\odot$ represents the elementwise multiplication of both vectors.)
If I had to guess I'd say "yes", because if $\mathcal{X}$ is just a point, then $\mathcal{Z}$ is a re-scaled/flipped version of $\mathcal{Y}$ which is convex, and if $\mathcal{X}$ is a line segment (i.e. we only vary a single entry) then $\mathcal{Z}$ is the union of convex sets that can only grow/shrink in one direction, which is also convex. Beyond that I am not sure...
Thank you!
 A: No.  Consider $X=Y=\{(u,u+1): u\in R^{\ge 0}\}$.  Then $(0^2,1^2)$ and $(2^2,3^2)$ are both in $Z$.  If $Z$ is convex then their average $(2,5)$ must also be in $Z$.  But $uv=2, (u+1)(v+1)=5$ has no real solutions, so $(2,5)$ is not in $Z$ and $Z$ is not convex.
UPDATE: Even if $X$ and $Y$ are required to contain the origin, the answer is still no.  Consider the homogeneous version of the above sets, $X=Y=\{(x,y,y-x): y \ge x \ge 0\}$.  Then $X$ is clearly convex.
Now the homogeneous version of the above example works too:  $(0,1,1)^2$ and $(2,3,1)^2$ are both in $Z$.  If $Z$ is convex, then their average $(2,5,1)$ must also be in $Z$.  But $xa=2$, $yb=5$, $(y-x)(b-a)=1$ has no real solutions, so $(2,5,1)$ is not in $Z$ and $Z$ is not convex.
A: The natural "multiplication" for convex bodies (and also for their Minkowski differences, the so-called "hedgehogs") is the convolution product of support functions restricted to the unit sphere. See e.g.:
https://hal.archives-ouvertes.fr/hal-01885039/document
pages 31 and 32.
