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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 $ …

Actually, your desired conclusion does "follow just from $D$ having full measure". Indeed, without loss of generality, $U=(-1,1)^2$. Let $$X:=\{x\in(-1,1)\colon|D_x|=2\},$$ where $D_x:=\{y\in(-1,1)\co …

$\newcommand\ep\varepsilon$The answer is yes, and we only need the uniform convergence of $f_k$ to $f$. Moreover, then $co(f_k)\to co(f)$ uniformly as well. Indeed, for each real $x$, $$co(f)(x)=\sup\ …

The answer is no. Indeed, consider the case when $n=2$ and $$f(x,y)=\max(| x|,2 | y|)$$ for all $(x,y)\in\mathbb R^2$. Then for $(x,y)$ near $(\pm1,0)$ we have $f(x,y)=|x|$ and hence $\partial f(\pm1, …

$\newcommand\S{\mathbb S}$Let $|\cdot|$ denote the Euclidean distance and let $k\ge1$ denote the cardinality of $K$. Then for any $x\in\S^n$ and $y\in\S_n(K)$ $$|y-x|^2=\sum_{j\notin K}x_j^2+\sum_{j\i …

The answer is yes if $p=2$ (by the Pythagoras theorem: $\|x\|_2^2=\|x-u\|_2^2+\|u\|_2^2$, because $x-u\perp u$) and, obviously, if $p=1$ or $n\le2$. Otherwise, the answer is no, because then the value …

Yes: Just take $u(x,y):=v(x)$, which will be assumed in what follows. Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer. More generally, the result holds for a …

$\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}\newcommand\si{\sigma}\newcommand\Si{\Sigma}$ You should have defined the meaning of the integrals $\int f(x)\,dx$ and $\int f(x)C(x)\,dx$. These sho …

In part (P3) of Definition 2.1 in the paper you linked, it is also required that $g$ be $\partial$-differentiable at $x$ (meaning that both $\partial g(x)$ and $\partial(-g)(x)$ are nonempty), which i …

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

$\newcommand\la\lambda$The answer is no. E.g., suppose that $X=\ell^\infty$, $z_1=e_1$, $z_2=-e_1+e_2$, and $z_k=e_2/2+e_k/k$ for $k\ge3$, where $(e_1,e_2,\dots)$ is the standard basis of $\ell^\infty …

This conjecture is true. Indeed, for $h:=\ln(f_a-f_b)$ and real $x\ne0$ we have \begin{equation} h'(x)=R(x):=\frac{F(x)}{G(x)}, \end{equation} where \begin{equation} F(x):=\sqrt{\pi } \left …

This conjecture is true. Indeed, we have to show that for $x$ and $t$ in $(0,\pi/2)$ we have \begin{equation} h_2(p;x,t):=\partial_x\partial_t\,\ln(|\cos(x-t)|^p-|\cos(x+t)|^p) \end{equation} is $ …

$\newcommand{\ep}{\varepsilon}$The "convex" part of this conjecture is not true in general. Indeed, suppose it is true. Then (letting $x:=\phi$, $t:=\theta_1$, and $\theta_2\downarrow\theta_1=t$) we s …