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Indeed, this can be proved more simply, and in greater generality -- assuming only that the support of $P$ is contained in $C$ (rather than in $\mathcal X$). Indeed, without loss of generality the aff …

$\newcommand\Om\Omega\newcommand\R{\Bbb R}$The answer is no. Indeed, for real $k>0$, let $$G_k:=\{(x,y)\in\Bbb R^2\colon x>1,|y|<k\sqrt x\}. $$ For $(x,y)\in G_2$, let $$f_0(x,y):=y^2/x-1.$$ For all $ …

A counterexample is given by $n=2$, $C=\{(0,0)\}$, and $C_k=\{(t,kt)\colon0\le t\le1\}$ (or one can instead take $C_k=\{(t,t/k)\colon0\le t\le1\}$). Another counterexample, in the same spirit, is give …

The tautological answer to the question "what is the centroid $x_C$ of $C$?" is "the centroid $x_C$ of $C$ is the centroid $x_C$ of $C$". It is hardly possible to give a better answer to this questio …

$\newcommand\R{\Bbb R}$If $X=S\cup T(X)$, then $$T(X)=T(S)\cup T^2(X),\quad X=S\cup T(S)\cup T^2(X), \ldots, \\ X=S\cup T(S)\cup\cdots\cup T^{n-1}(S)\cup T^n(X) \\ \subseteq Y:=S\cup T(S)\cup\cdots\ …

$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Let $Y$ denote the interior of $X$. Then indeed $g:=\na f$ is a homeomorphism of $Y$ onto $g(Y) …

A counterexample is given by the following conditions: $n=185$, $$a_k=\sum_{j=k}^n b_j,\quad b_j:=\frac{c_j}{j+1}, \quad c_j:=\frac{1000}7\,1(j=47)+\frac{302}7\,1(j=185).$$ Indeed, then $a_0\ge\cdots …

$\newcommand\R{\mathbb R}$It suffices that $f$ just be strongly concave (your conditions 1 and 2 then hold automatically). Indeed, then $$f(y)\le f(x)+h\cdot(y-x)-m|y-x|^2/2$$ for some $x\in\R^2$, som …

$\newcommand{\la}{\lambda}\newcommand{\de}{\delta}\newcommand{\ep}{\delta}$The answer is as follows: Yes, if the special point is allowed to be not in $C$, and then the moments of inertia of $C$ abou …

A counterexample is given by $n=2$, $[a_1,b_1]=[-2,1]$, $[a_2,b_2]=[-1,2]$. (Make a picture.) Even if the $n$-box $S$ is required to be a subset of $[0,\infty)^n$, the answer will still be no. E.g., …

$\newcommand\conv{\operatorname{conv}}\newcommand\ext{\operatorname{ext}}\newcommand\p{\partial}$The answer is yes. Indeed, let $K:=\conv P$ (the convex hull of $P$), let $\p K$ be the boundary of $K$ …

$\newcommand{\R}{\mathbb R}\renewcommand{\th}{\theta}\newcommand{\ep}{\varepsilon} $This indeed follows immediately from Theorem 15 of Talagrand, Regularity of Gaussian processes, Acta Math. 159, No. …

$\newcommand\de\delta\newcommand\R{\mathbb R}$As in the previous answer, assume that $C$ is a compact convex subset of $\R^2$. The condition that all the width bisectors of $C$ are concurrent means th …

We should assume that $C$ is closed; otherwise, there are easy counterexamples. If $C$ is unbounded, then there is at most one width bisector, and hence easy counterexamples again. So, assume that $C$ …

$\newcommand{\R}{\mathbb R}\newcommand{\inter}{\operatorname{int}}$The answer is yes. Here is the proof. The asymptotic cone of $C$ is \begin{equation*} K:=C^{as}=\{y\in\R^n\colon \exists(x_k) \te …