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This is impossible, because for any complete normed space $X$ and any nonincreasing sequence $(B_n)$ of nonempty closed balls in $X$ we have $\bigcap_n B_n\ne\emptyset$. Indeed, take any nonincreasi …

By the Riesz–Markov theorem, your $C^*$ is the space of all finite signed Borel measures on $Q:=[0,1]^2$ endowed with the total variation norm $\|\cdot\|$, which is the same as your dual norm $\|\cdot …

Apparently, there are no such nice properties. Even $K=[0,1]$ will not be such a compact Hausdorff space. Indeed, consider the countable double-indexed family $(U_{n,k}\colon n\in\Bbb N,k\in\{0,\dots, …

$\newcommand\P{\mathcal P}\newcommand\X{\mathcal X}$A counterexample is as follows: $n=m=1$, $\X=[-1,1]$, $\P=[0,2]$, $f(x)=\sqrt2-1/\sqrt2+|x|^{3/2}$, $g(x)=1/\sqrt2$, and $h(x)=1=S(x)$ (for all $x\i …

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 …

$\newcommand\de\delta\newcommand\ep\varepsilon$You wrote Could you help me to show that under Ass1 and Ass2 $$d_H(A, A_n)\rightarrow_{a.s.} 0$$ Of course, this is not true in such generality. For in …

$\newcommand{\R}{\mathbb R}$Here is an elementary (albeit somewhat longish) solution, without using the Jordan curve theorem. Let us borrow the inversion idea from Olivier Bégassat, so that the two ar …

$\newcommand\R{\mathbb R}$No. E.g., suppose that $n=k=1$ and $$F(f)=\min\Big(1,\int_0^2 |f(x)|\,dx\Big)$$ for $f\in C(\R;\R)$. Then $F$ is continuous and bounded on $C(\R;\R)$. For natural $m$, let $$ …

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[ …

$\newcommand{\R}{\mathbb R}\newcommand{\tU}{\tilde U}$Suppose that $U$ is connected. Then all its projections are connected. So, all one-dimensional projections of $U$ will be convex. So, $\tilde U$ w …

$\newcommand\M{\mathcal M}\newcommand\F{\mathcal F}\newcommand\si{\sigma}\newcommand\th{\theta}$I think the answer is no. Here is the idea of how to show this: Suppose that such a map $\phi$ exists. I …

$\newcommand\LL{\mathcal L}$Brief answer: The Scott topology is not Hausdorff, and therefore we have to deal here with the set of limits, rather than with the (unique) limit. Here, for the set $\LL_D$ …

$\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}$A counterexample is as follows. Let $X=(0,\infty)$ and $\tau:=\{\emptyset\}\cup\{(t,\infty)\colon t\in[0,\infty)\}$. Then $\F$ is the usual Borel $\s …

$\newcommand{\Z}{\mathbb Z}\newcommand{\PP}{\mathcal D}\newcommand{\R}{\mathbb R}$Your function $d$ is not a metric, for two reasons: (i) there may be many processes $(X_t)_{t\in\Z}$ with the same dis …

$\newcommand\abs[1]{\lvert#1\rvert}$It already suffices that $f$ and $g$ be even and nondecreasing on $[0,1]$ (which of course is the case if $f$ and $g$ are even and convex). Indeed, then the identit …