Suppose I have a compact set $\mathcal A\subset \mathbb R^n$ that is not convex, and denote $\mathcal B = \{y \in \mathbb R^p : \|y\|_\infty \leq 1\}$ the unit infinity norm ball. For $p \geq n+1$, is the following set convex?
$$ \mathcal S = \left\{\sum_{i=1}^p y_i x_i : y\in \mathcal B, \; x_i \in \mathcal A\right\}. $$
If $\mathcal B$ is the unit simplex, then this is true via Cartheodory's theorem. And, if $p \to \infty$, then there is a "smearing" like action going on here, and $\mathcal S$ will be convex. But I'm not sure about this more general case.
Also what happens of $\mathcal B$ is a more general (convex, compact) set? Can we make the argument based on dimensionality alone (and including $0$ in the interior)?
Edit: An example where $\mathcal S$ is not convex when $p < n$, take $\mathcal A = \{(0,1), (0,1)\}$ and $\mathcal B = [0,1]$. Then $\mathcal S = \{(0,s) : 0\leq s \leq 1\}\cup \{(s,0) : 0\leq s \leq 1\}$, e.g. two rays along the axis, but nothing in the middle (it's spiky).
If I now allow $p = 2$, then the points are allowed to "mix", and you can get $\mathcal S = \{[s_1,s_2] : 0 \leq s_1,s_2 \leq 1\}$, which is convex.
Actually I think the set is convex; I just can't prove it for general $\mathcal A$, nor see exactly how to use Cartheodory (which feels important...)
Let $n=2$, $p > 3$, and $\mathcal A = \{(1,0), (0,1)\}$. If we take $x_i = (1,0)$ $k$ times and $(0,1)$ $p-k$ times, $\{\sum_i y_i x_i: y \in {\mathcal B}\}$ is the convex hull of $(k,0)$, $(-k,0)$, $(0,p-k)$ and $(0,-p+k)$. $\mathcal S$ is the union of these for $k$ from $0$ to $p$. But this is not convex: e.g. it contains $(p,0)$ and $(0,p)$ but not $(p/2, p/2)$.