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As written in my paper [1], the inequality $$P(f^*>r) \le 2\exp\big(-r^2/2(p-1)\big) $$ in Theorem 3 in [1] for martingales in $\mathcal{X}=L^p$ can be compared with the inequality $$ P(f^*>r) \le …

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 …

$\newcommand{\E}{\operatorname{\mathsf{E}}}$ You do not need the separability of $B$ to define $\E F$ for a random vector $F\colon\Omega\to B$; however, you need to assume that $F$ is strongly measura …

A counterexample is given by $f_n=1_{(n,\infty)}$, $g=1$, and $f=0$. Then all the conditions on $f_n,g,f$ hold, but $\|f_n-f\|_{\tilde L^p} \not\to0$.

$\newcommand\la\lambda$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 …

$\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$This is to detail and correct the answer by Onur Oktay, which is based on a …

The expression \begin{equation} \tilde W_m:=\bigcup_{n=1}^\infty T_n(V_m) \end{equation} is undefined in general for $m\ge2$, because $T_n$ is defined (and is invertible) only on $V_n$ and hence $ …

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\C}{\mathbb C}$No, in general the map $F_r$ \begin{equation*} \sum_1^n w_j b_j\mapsto \sum_1^n \Re(w_j) b_j \tag{10}\label{10} \ …

First of all, instead of $L^{\infty}(0,\infty;L^{\infty}(0,1))$, you should write $L^{\infty}([0,\infty);L^{\infty}(0,1))$ (or maybe $L^{\infty}((0,\infty);L^{\infty}(0,1))$, depending of what you wan …

(There should be only one question in one post.) Anyhow, concerning your first question: There is nothing here really to study, as the Gateaux derivative at a given point in a given direction is just …

There is no explicit expression for $\int_X \|x\|^2 \mu(dx)$ in general, even if $X=\mathbb R^d$ with $d\in\{2,3,\dots\}$ and $\mu$ is the standard normal distribution on $\mathbb R^d$. Indeed, any no …

$\newcommand{\E}{\mathsf{E}}$ $\newcommand{\P}{\mathsf{P}}$ The affirmative answer to this question is provided by Scalora, Theorem 2.1, page 354, which can be stated as follows, using the setting and …

This is a problem of real algebraic geometry. So, in principle, the dual norm can be computed purely algorithmically; however, this calculation can take too much time. In Mathematica, this algorithm i …

The answer is no in general. E.g., let $X=\mathbb R$ with the standard norm and $Y=\mathbb R^2$ with the norm given by the formula $\|y\|=|y_1|\vee|y_2-y_1|$. Let $S$ and $T$ be defined by the formula …

$\newcommand{\R}{\mathbb{R}} \newcommand{\ep}{\epsilon}$ Without loss of generality (wlog), the space is the two dimensional space spanned by $x,y$. So, wlog the space is $\R^2$, with $x=[1\ 0]^T$ an …