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following question jumped into my mind.

Let $X$ be a Banach space. Let $C$ be a nonempty subset of $X$. Set

$$A : = \{ x \in X \;|\; \exists \text{ a norm-bounded net } x_{i} \to x \text{ (weakly) } \, \forall i: x_i \in C \}$$

Question 1 : Is $A$ actually the entire weak-closure of C?

Question 2 : Is $A$ weakly closed set?

I'm positive that the answer of 1 is NO, but dont have explicit example. This is because in any infinite dimension space one can construct a net $x_i$ such that $\| x_i \| \to \infty$ but $x_i \to 0~(weakly)$.

Question 2 is harder, I tried to prove it, I had to go through some diagonal process regarding for index sets, which was kinda impossible !! Now I changed my mind, and start to find a counter example instead of proving it.

Any help would be greatly appreciated.

Henno Brandsma
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Red shoes
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    @copper.hat no, nets are weird see the answer here https://math.stackexchange.com/q/3262680/219176 – Red shoes Jun 20 '19 at 20:30
  • I realised that after I posted my comments. I avoid nets except for fishing and accounting. – copper.hat Jun 20 '19 at 20:53
  • Note that $C\subseteq A\subseteq\overline{C}$ and so 1 and 2 are equivalent, right? – freakish Jun 20 '19 at 21:00
  • @freakish Thanks right. good observation – Red shoes Jun 20 '19 at 21:01
  • @copper.hat I start to understand them, so I start keep asking questions here from experts. Now I feel like I asked too many dumb questions about them, that's embarrassing – Red shoes Jun 20 '19 at 21:09
  • @Redshoes see this: https://math.stackexchange.com/questions/825790/every-weak-convergent-sequence-is-bounded I knew there was something odd with your question. The link you've posted is about weak-* convergence, not weak convergence. Is that a typo in your question? – freakish Jun 20 '19 at 21:27
  • @freakish No . if that works in any infinite dimension then it also works for weak convergence as well. Just take reflexive spaces, then weak-star is actually weak topology – Red shoes Jun 20 '19 at 21:30
  • @Redshoes Yeah, you're right. Anyway the proof in the link I've posted is for sequences only. But I'm not sure why it doesn't generalize to nets... – freakish Jun 20 '19 at 21:35
  • Also the problem is that even if $(b_i)$ is a net without a bounded subnet then we still can create a bounded net from elements ${b_i}$ convergent to some point outside of ${b_i}$. Being a subnet is a stronger condition (cofinality). Note that this cannot be done with sequences. Yet another anomaly with nets. And a major problem in finding a counterexample. – freakish Jun 20 '19 at 21:49

1 Answers1

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The answer to both questions is negative. Here are some counterexamples in $\ell^2$.

For $$C := \{ e_n + n \, e_m \mathrel\mid n,m \in \mathbb N,\; n < m \}$$ one hase $$A = C \cup \{ e_n \mathrel\mid n \in \mathbb N \}$$ but the weak closure of $C$ contains $0$. The set $A$ is also not weakly closed.

For another example, you can take $$ C:= \{ \sqrt{n} \, e_n \mathrel\mid n \in \mathbb N\}.$$ Then, $A = C$, but again, the weak closure of $C$ contains $0$.

Conclusion: nets are weird.

gerw
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