How can one show that a norm-preserving map $T: X \rightarrow X’$ where $X,X’$ are vector spaces and $T(0) = 0$ is linear? Thanks in advance.

The supremum norm is defined as $$\|f\|_\infty=\sup\limits_X|f|$$ This induces a topology: $$ \begin{align}f_n\overset\infty\to g&\Leftrightarrow \sup\limits_X|f_n-g|\overset{\mathbb{R}}\to 0\\ &\Leftrightarrow\forall\epsilon>0\exists n:\sup\limits_X|f_n-g|<\epsilon\\ &\Leftrightarrow\forall\epsilon>0\exists n\forall x \in X:g(x)-\epsilon<f_n(x)<g(x)+\epsilon \end{align}$$ I am now in the following very similar situation: $$\begin{align}\phantom{f_n\overset\infty\to g}&\phantom{\Leftrightarrow}\forall a>1\exists n\forall x \in X:g(x)\cdot a^{-1}<f_n(x)<g(x)\cdot a\end{align}$$ Question: Does a topology $\tau$ exists such that: $$\begin{align}f_n\overset\tau\to g&\Leftrightarrow\forall a>1\exists n\forall x […]

I am looking for examples on double dual spaces. As I know $\ell_p $ is the double dual of $\ell_p$ for $1<p,q<\infty$, $\mathcal L_p $ is the double dual for $\mathcal L_p$ for $1<p,q<\infty$ and $\ell_\infty$ is the double dual of $c_0$, where $c_0$ is the space of sequences of numbers that converge to 0. […]

Assume $(X,\|\cdot\|)$ is a normed space with the following property: if $x \neq y \in X$ have norm 1 then $\|\frac{x+y}{2}\|<1$. (We then say that $X$ is strictly convex) Prove that if $C$ is a convex (though not necessarily close) subset of $X$ and $x_0 \notin C$ and put $r=d(x_0,C)$ then $\{y\in X|\|y-x_0\|\leq r\}\cap C$ […]

let’s see if someone can help me with this proof. Let $\Omega\subset\mathbb{R}^n$ be a bounded domain. And let $L^2\left(\Omega\right)$ be the space of equivalence classes of square integrable functions in $\Omega$ given by the equivalence relation $u\sim v \iff u(x)=v(x)\, \text{a.e.}$ being a.e. almost everywhere, in other words, two functions belong to the same equivalence […]

I’m trying to solve the following problem but I have no clue how to do it. Let $(X,||.||)$ be a normed $\mathbb C$-vector space. Prove: For any sequence of linearly independent elements $y_j, 1 \leq j \leq N$, in $X$ and any sequence $(a_j)_{1 \leq j \leq N}$ in $\mathbb R$ there exists an element […]

Let $(X, \|.\|)$be a real normed space. Let $A$ be a closed convex suset of $X$ and $\mathbb{B}$ a unit ball in X, i.e. $$ \mathbb{B}=\{x\in X: \|x\|\leq 1\}. $$ I would like to ask whether $A+\mathbb{B}$ is still closed.

Let $X$ be a Banach Space and $Y$ be a normed linear space. Show that if $T$ is an isometry then $T(X)$ is closed in $Y$. Let me have some idea to solve this. Thank you for your help.

Let U be a normed space, $w \in U$ and $\|w\| = 1$. Show that $U$ has a closed subspace $V$ such that $d(w, V) = 1$ and $U = V \oplus\{\lambda w :\lambda \in \mathbb{F}\} $.

If we define the (Riemann) integral of an abstract function, i.e. a function $f:[a,b]\to Y$ where $Y$ is a Banach space, as$$\int_a^b F(t)dt:=\lim_{\max(t_{k+1}-t_k)\to 0}\sum_{k=0}^{n-1}(t_{k+1}-t_k)F(\xi_k)$$where $a=t_0<t_1<\ldots<t_n=b$ is one of the partitions whose intervals’ maximum length approaches $0$, I read two unproved statement in Kolmogorov-Fomin’s (p. 486 here) which I find very interesting: If $F$ is continuous, […]

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