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This is indeed easy. Let $p$ be a point that you want to represent, $m$ the barycenter of all vertices and $\varepsilon>0$ so small that the point $q=(1+\varepsilon)p-\varepsilon m=p+\varepsilon(p-m)$ …

Yes this is true. Let me handle the inner ellipse, the outer one is similar. For brevity, denote $\kappa/s^3$ by $a$. It is easy to see that $$ a = \frac{\dot\gamma\wedge\ddot\gamma}{(\gamma\wedge\d …

The answer is no in dimension 2 and yes in dimension 3 and higher. The property that the nearest-point projection to a line does not increase the norm is equivalent to the symmetry of orthogonality re …

The shadow boundary can be any $C^\infty$ curve with (quadratically) strictly convex projection to the $xy$-plane. For simplicity, let me stick to the case when the projection is a circle. So conside …

Here is a non-constant counter-example to the original version (with weak monotonicity). First observe that $f(\theta) $ (up to a factor $\sqrt2$ or so) equals the maximum area of intersection of $K$ …

For the Lipschitz-only part, the answer is yes. More generally, if $\alpha$ and $\beta$ are any two convex curves and $\beta$ fits inside an $L$-homothetic copy of $\alpha$, then there exist an $L$-Li …

Yes, if the convex body is "sufficiently round". If it is not, the resulting "closeness" to the boundary of a convex set is in absolute terms rather than relative. I don't know whether it can be impro …

No, the Loewner ellipsoid is not monotone w.r.t. inclusion. Let $K$ be a square, whose Loewner ellipsoid is its circumcircle. Let $L$ be any other ellipse through the four vertices of $K$. The Loewner …

The answers are no to Question 1, and yes to Question 2 (assuming $n\ge 2$). Let $h$ be the support function of $K$. The projection of $K$ to a linear subspace $L$ is central symmetric iff the restri …

It is not compact unless you allow the origin at the boundary (or maybe impose some kind of uniform strict convexity). Consider a rectangle $K=[-1,1]\times[-\delta,1]$ in the plane, where $\delta$ is …

I believe that the statement you want is not true. In $X=\mathbb R^3$, begin with the standard cone $x^2+y^2<z^2$ and perturb it so that the resulting cone $K$ is symmetric to its Euclidean dual throu …

No, here is a counter-example (to revision 9). Let $A$ be the linear map that sends the vector $(1,1,1)$ to $V:=(100,100,100)$ and is the identity on the orthogonal complement of this vector. Then an …

No. Consider $f(x,y)=e^x+y^2$, then $\varphi(x,y)=(e^x,2y)/(e^x+y^2)$. The image of $\varphi$ has only one point $(1,0)$ on the axis $y=0$. The points $a:=\varphi(0,1)=(\frac12,1)$ and $b:=\varphi(0,- …

No, even if you assume that $f$ is convex on $\mathbb R^{d+n}$. Take $n=2$ and let $y_1,\dots,y_{d+1}\in\mathbb R^2$ be vertices of a convex polygon. Let $X_1,\dots,X_{d+1}\subset\mathbb R^n$ be your …

You are essentially asking whether a non-expanding linear map $A:(\pi,\|\cdot\|_\infty)\to(\mathbb R^3,\|\cdot\|_\infty)$ can be extended to a non-expanding linear map $B:(\mathbb R^3,\|\cdot\|_\infty …