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There does not exist any function $p:F\mapsto p_F$ as in the question, to prove it by contradiction suppose such a function $p$ exists. In the following let $k+A=\{k+a;a\in A\}$ for any $k\in\mathbb{R …

The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$. Below I show it for $p>2$, but this is enough because if we have convex sets $A\subseteq B$, …

Here are triangulations of a side $1$ square with vertices at a arbitrarily high distance of all the four vertices of the square. The sides of the side $1$ square are not edges but it is easy to see t …

Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not. Let $(v_i)_{i\i …

$K_t$ need not be a translate of $K$. Let $A=[-4,4]\times[-1,1]\subseteq\mathbb{R}^2$ and consider the convex cone $K=\{t\cdot v;t\in[0,\infty),v\in A\times\{1\}\}\subseteq\mathbb{R}^3$. Note that $\p …

This is false for any fixed $\theta\in(\frac{1}{2},1)$, we will use the balls of radius $1$ in $\mathbb{R}^d$ as a counterexample. Given $\theta$, if we want $ \theta K \subseteq \operatorname{conv}\{ …

Seems like in fact convex sets are the only ones for which $p_C$ is continuous. To prove this, we can begin by noticing that for a set $C$ with the property, $p_C$ can only take one point sets as valu …

This is not an answer to the question, but here are some upper/lower bounds. Firstly, if we let $A_i\subseteq\{0,1,\dots,k+1\}$ be the non zero coordinates of $m_i$, then we can't have $A_i\subseteq A …