I was given this exercise
Let $\lbrace e_n \rbrace $ be the canonical basis of $\mathit {l}^2 $ (the space of square sommable series with the usual norm) and set $$ A:= \lbrace \sum_{n \ in \mathbb{N}} a_ne_n : a_n\geq 0 \forall n, 1 \leq \sum_{n \in \mathbb{N}}\mid a_n \mid ^2 \leq 2 \rbrace $$
(i) Is A strong closed? I think yes, because it is the intersection of 2 strong closed subsets, namely the closed ball with Radius 2 and the complementary of the open ball with radius 1. I didn't found a way to manage the condition about positivity of the coefficients :/
(ii) Is A strong compact? Here i wanted to found an absurd using a proof of Riesz theorem that started with a finite cover of the unit ball and ended creating a set of finite element whose span is the entire space. But i didn't found the proof to check his suitability.
(iii) What is the strong-interior of A? I suppose that (i) is true and again according to me is the set with $<,> $ replacing $\leq, \geq $
(iv) Is A sequentially weakly compact? Here my attempt is to make heavy use of Riesz representation theorem and reflexivity of the space. So if I prove the weak closure, i can conclude by Banach-Alaoglu. If (i) is true, but without Mazur (A is not convex) i don't have any idea to prove weak-closureness.
(v) What is the weak interior of A? According to me is empty. Argued by absurd, suppose that A contained a weak-neighborhood of a point. Then it would contain at least an entire affine subspace passing throught that point, but A is bounded so this is impossible.
Just an observation, what is the canonical base in $\mathit{l}^2 $ in this case? It can't be the Hamel one because it seems to be numerable and my space is banach therefore complete. I think my teacher means the hilbert base (even though hilbert spaces are introduced later than this exercise)
I need help for all my doubts please. Maybe is just the fact that i need to make mine the correct reasoning and observations that on must do when solving this kind of exercises. Thanks in advance
Yes, the basis is the Hilbert basis.
Let $P=\{\sum_na_ne_n\in\ell^2:\ a_n\geq0\}$. Then $P$ is closed; indeed, $\ell^2$-convergence implies pointwise convergence, so the entries of the limit will again be non-negative.
(i) We have $A=P\cap C$, where $C$ is a closed set as you say. So $A$ is closed.
(ii) The set $A$ cannot be compact, because it contains the set $\{e_n\}$, where the distance between any two elements is $\sqrt2$.
(iii) No. If $a\in A$ with $1<\|a\|<\sqrt2$ then any $b$ sufficiently close to $a$ will satisfy $1<\|b\|<\sqrt2$. But you can choose $b$ carefully to have some negative coefficient, thus putting outside $A$. This shows that the interior of $A$ is empty.
(iv) Note that $A$ is not even weakly closed. Indeed, the sequence $\{e_n\}$ is in $A$ and it converges weakly to zero.
(v) Correct.