1
vote
0answers
32 views

Weak sequential compactness in a reflexive space

Let $\{X, \| \cdot \|\}$ be a normed space, $B$ is the unit ball of $X$. If $\{X, \| \cdot \|\}$ is reflexive, then is $B$ weakly sequentially compact? If it's not true, are there any counterexamples ...
3
votes
2answers
40 views

Why is the image of a compact operator separable?

Let $A$ and $B$ be normed vector spaces and let $S\in \mathscr{K}(A,B)$ be a compact operator. Question: How does it follow that the image of $S$ is separable? Thanks for the help.
2
votes
4answers
112 views

Why is $C_c^\infty(\Omega)$ not a normed space?

I am watching a Coursera video on Théorie des Distributions and I am trying to understand one of the slides. Let $\Omega \subset \mathbb{R}^N$ be an open set and $C_K^\infty(\Omega) = \{ \phi \in ...
3
votes
2answers
90 views

Completeness/Compactness of a subset in a normed linear space

Let $(X,\|\cdot\|)$ be the normed linear space consisting of the sequences $a=(a_n)_{n=1}^{\infty}$, for which the corresponding series $\sum_{n=1}^{\infty} a_n$ converges absolutely, with norm ...
1
vote
2answers
35 views

Negative exponential distance

Let $X := \left\{(a_k)_{k \in \mathbb N}, a_k \in \mathbb C\right\}$. Let $d\left( (a_k)_{k \in \mathbb N}, (b_k)_{k \in \mathbb N} \right) := e^{-u}$ with $u$ the smallest integer $k$ such that $a_k ...
0
votes
1answer
140 views

$X$ normed linear space separable $\Longleftrightarrow$ $\exists K \subset X$ compact s.t. $\overline{ \text{span}\{K\}}= X$

Let $X$ be a normed linear space. Show that $X$ is separable if and only if there is a compact subset $K$ of $X$ for which $\overline{ \text{span}\{K\}}= X$ I can't figure out how to solve this ...
0
votes
1answer
62 views

Finiteness of the dimension of a normed space and compactness

I am studying functional analysis, and in the setting of normed spaces I have seen the theorem that states that the unit ball is compact iff the space is finite dimensional. I also saw an exercise: ...
3
votes
1answer
421 views

Weakly compact implies bounded in norm [duplicate]

The weak topology on a normed vector space $X$ is the weakest topology making every bounded linear functionals $x^*\in X^*$ continuous. If a subset $C$ of $X$ is compact for the weak topology, then ...
0
votes
2answers
115 views

Are these sets compact?

I've some problems concerning this question: Are the following sets compact in $C_{[0, 1]}$ where $ d(x(t), y(t))=\sup_{[0,1]}|x(t)-y(t)|$: $${\{x(t) \mid x(t)=e^{t-a}, a>0\}},~~{ \{ x(t) ...
1
vote
1answer
492 views

Weakly sequentially compact sets

From Peter Lax Functional analysis page 104: Show that a weakly sequentially compact set is bounded. Definition. A subset $C$ of a Banach space $X$ is called weakly sequentially compact if ...
5
votes
1answer
522 views

Equivalence of reflexive and weakly compact

In a normed space $X$ is there an equivalence between these two proposition? $1)$ $X$ is reflexive; $2)$ $B$, the unit ball of $X$, is weakly compact.