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$\newcommand\R{\Bbb R}$Counterexample: $X=\R^2$, $\|x\|_1=|u|+|v|$ and $\|x\|_2=\max(|u|,|v|)$ for $x=(u,v)\in\R^2$. Then $\|\cdot\|_2\le\|\cdot\|_1$ and the map $\R^2\ni(u,v)\mapsto\frac12\,(u-v,u+v) …

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 …

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 …

(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 …

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 …

Take any $x\in S$ and $y\in E$. We have to show that \begin{equation*} \tau(x,y)\overset{\text{(?)}}=\rho(x,y):=\sup_{f\in T(x)}f(y). \tag{1}\label{1} \end{equation*} Generalize the definition of $T(\ …

$\newcommand{\om}{\omega}\newcommand{\Om}{\Omega} \newcommand{\de}{\delta}\renewcommand{\th}{\theta}$For each real $t>0$ and some $\om_t\in\Om$, \begin{equation*} \|x+ty\| =|(x+ty)(\om_t)|=|x( …

We have the following definitions: \begin{equation*} V^*:=\{f\in S^*\colon\exists x\in S\ f(x)=1\}, \end{equation*} where $S$ and $S^*$ are the unit spheres in the underlying Banach space $E$ and …

A function $f$ from a linear (or, more generally, affine) space $A$ is affine (see e.g. this post) if for all $a$ and $b$ in $A$ and all real $t$ one has $f((1-t)a+tb)=(1-t)f(a)+tf(b)$. If $f$ is affi …

This norm is not Gateaux differentiable for $n\ge2$. Indeed, let $x:=(2,4,0,\ldots,0)$ and $u:=(1,1,0,\ldots,0)$. Then $$g(t):=\|x+tu\|_0=\max(|1+\tfrac t4|,|1+\tfrac t2|) \\ =-(1+\tfrac t2)\,1(t\le-\ …

$\newcommand\la\lambda$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 …

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 $\|f_n-f\|_{\tilde L^p} \not\to0$.

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 …

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

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 $ …