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Yes, $\partial C + \partial C$ is convex since it equals $2C$. Equivalently, every point in $z \in C$ is a midpoint of two boundary points. This is obvious if $z \in \partial C$. Otherwise, let $f :S^ …

The answer is yes. The fact that any polytope is affinely equivalent to a section of a simplex is well-known (see the answer by Tobias Fritz). Now in any simplex with vertices $(v_i)$ we may conside …

Here is an argument that is certainly overkill and introduces a logarithmic factor which is probably unnecessary. Let $K$ be the unit ball in $\ell_1^n \otimes_\epsilon \ell_1^n$ and $K^\circ$ the po …

There are centrally symmetric self-dual polytopes in every dimension. This follows from Proposition 3.9 in Reisner, S., Certain Banach spaces associated with graphs and CL-spaces with 1- unconditional …

This is well-documented in the case where $C_1$, $C_2$ are unit balls for some norm, and in that case your $C_1 \otimes C_2$ is the unit ball for the so-called projective norm. The set you compare to …

This question has been considered in detail, and essentially solved, in the recent preprint Half-space depth of log-concave probability measures, where the authors show the bound $c_\mathbb{P} \leq c^ …

This question has been investigated in Życzkowski, Karol; Sommers, Hans-Jürgen, Hilbert-Schmidt volume of the set of mixed quantum states, J. Phys. A, Math. Gen. 36, No. 39, 10115-10130 (2003). ZBL105 …

Let me prove the weaker bound $c_n \leq c^{\sqrt{n}}$. Since the problem is affine-invariant, we may assume that $\mathbb{P}$ is isotropic, i.e. has mean zero and covariance matrix equal to identity. …

This fails already in dimension $2$ for very unbalanced convex bodies. Given $\alpha > 0$, consider the ellipse $K_\alpha$ obtained as the image of the unit disk by the area-preserving transformation …

The answer is yes, for a $3$-dimensional convex body this is a 1915 theorem by Funk. See for example the introduction to this paper : https://sites.tufts.edu/tquinto/files/2021/01/funk2.pdf The follow …

The cube has a bounded outer volume ratio (which you denote $v_r^\circ$) while the cross-polytope (=$\ell_1$ ball, or octahedron) has an outer volume ratio of order $\sqrt{n}$. Actually the cross-poly …

Edit : the following argument does not answer the question and actually appears already in the OP's linked note. Let $H = 1_{[0,\infty)}$. Let $\Pi$ be the projector defined by functional calculus as …