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$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\C}{\mathbb C}$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 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 …

$\newcommand{\E}{\operatorname{\mathsf{E}}}$ 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 …

No, in general $I-S$ will not be injective. Indeed, suppose e.g. that $E=\mathbb R^2$ and $$T=\begin{bmatrix}2&1\\ -1&0 \end{bmatrix};$$ here we will identify the linear operators with their matrices …

$\newcommand{\R}{\mathbb R}\newcommand{\C}{\mathbb C} $To attach a meaning to the intersection $C:=B\cap L^2\big((0,T)\times(0,1)\big)$ of the subset $B$ of the space $L^{\infty}\big(0,T;L^1(0,1)\big) …

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 $ …

The answer is no. E.g., let $x_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x_{2n}=x\to x$ and $x_{2n+1}=y\to y$ in a …

$\newcommand{\E}{\mathsf{E}}$ $\newcommand{\P}{\mathsf{P}}$ 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 …

A counterexample: $x=(1,1,0,0,\dots)$, $x_n=(1,0,0,0,\dots)$ for all $n$, and $f(x^1,x^2,\dots)=x^2$ for all $(x^1,x^2,\dots)\in\ell^2$. Then all your conditions on $x,x_n,f$ hold, but $f(x_n)=0\not\t …

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 …

For all natural $k$, we have $\|D^kf\|_1\le\|D^kf\|_2\le k!$, where $\|\cdot\|_p:=\|\cdot\|_{L^p(0,1)}$. So, for all $x\in(0,1)$ we have $$|(D^kf)(x)|\le\int_0^x |(D^{k+1}f)(u)|\,du\le\|D^{k+1}f\|_1\ …

Your functions $u_1$ and $u_2$ are not completely defined. For instance, $u_1(c_1)$ is undefined. If you do not need to distinguish functions differing only on a set of Lebesgue measure $0$, then yo …

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

As noted by Mikael de la Salle, $L(H,K)$ contains an isometric copy of $\ell^\infty$, which is not of martingale type $p$ for any $p\in(1,2]$, and hence $L(H,K)$ is not of martingale type $p$ for any …