Hilbert space with weak topology is not $\alpha_4$

[Moniker1998]

Finishes currently added properties for this space.

[felixpernegger]

This looks very creative nice! Unfortunately I have never done any functional analysis so take the following with a grain of salt:

First why does $\ell^2$ even have a (countable) orthonormal basis? Standard vectors do not work since we need finite linear combinations, no? So maybe replace basis with countable orthnormal system.

Then I think you are missing squares in Bessel's inequality and actually turn it into an inequality and not equality (considering my comments above).

Change If there existed countably infinite $S\subseteq X$ such that $S_n\cap S\neq\emptyset$ for infinitely many $n$, then $S$ would be unbounded. to something like
Any countably infinite $S\subseteq X$ such that $S_n\cap S\neq\emptyset$ for infinitely many $n$ is unbounded. (Since such S actually exist)

And finally (maybe this is a standard fact in functional analysis) it would be nice to have some comment that covergence in the topology corresponds to ||y-x|| converging or something.

[felixpernegger]

By the way I really appreciate your current work on old PR's!

[Moniker1998]

This looks very creative nice! Unfortunately I have never done any functional analysis so take the following with a grain of salt:

That's okay, I wrote it sloppily so that someone could tell me how to rewrite it better.

First why does ℓ 2 even have a (countable) orthonormal basis? Standard vectors do not work since we need finite linear combinations, no? So maybe replace basis with countable orthnormal system.

An orthonormal basis for a Hilbert space does not require finite sums. We require infinite sums. So any vector $x\in X$ should be an infinite sum $x = \sum_{k = 1}^\infty x_k e_k$.

Then I think you are missing squares in Bessel's inequality and actually turn it into an inequality and not equality (considering my comments above).

That's right, I'm missing squares, and I'm actually using something called Perseval's identity, I forgot that

Change If there existed countably infinite $S\subseteq X$ such that $S_n\cap S\neq\emptyset$ for infinitely many $n$, then $S$ would be unbounded. to something like
Any countably infinite $S\subseteq X$ such that $S_n\cap S\neq\emptyset$ for infinitely many $n$ is unbounded. (Since such S actually exist)

Yes, I think I also need to mention that I mean norm bounded here which will be equivalent to weakly bounded

And finally (maybe this is a standard fact in functional analysis) it would be nice to have some comment that covergence in the topology corresponds to ||y-x|| converging or something.

In this case we don't have anything like that because it's the weak topology. So it doesn't have a norm.

By the way I really appreciate your current work on old PR's!

Thank you

[felixpernegger]

An orthonormal basis for a Hilbert space does not require finite sums. We require infinite sums. So any vector x ∈ X should be an infinite sum x = ∑ k = 1 ∞ x k e k .

Alright, I just thought "basis" is passed down from normal linear algebra where it is finite sums of course.

In this case we don't have anything like that because it's the weak topology. So it doesn't have a norm.

I meant, some sequence (of sequences that is) $x_n$ congerges to $x$ in the topology iff $||x_n - x||$ goes to zero, if that makes sense

[Moniker1998]

Alright, I just thought "basis" is passed down from normal linear algebra where it is finite sums of course.

I get how that might be confusing but it's the terminology in functional analysis.

I meant, some sequence (of sequences that is) x n congerges to x in the topology iff | | x n − x | | goes to zero, if that makes sense

If my answer doesn't answer you then I don't know what you mean

[felixpernegger]

If my answer doesn't answer you then I don't know what you mean

Like in topology, we say some sequence x_n converges to x, iff every neighborhood of x contains almost all x_n.
In analysis (in a normed space, like l^2), we say x_n converges to x iff ||x_n-x|| gets arbitrarily small for n big enough.

My question/remark is, that in this particular case (Hilbert space with weak topology) these (formally different) notions of convergence coincide.

[Moniker1998]

@felixpernegger are you asking if the norm convergence and weak convergence coincide? No they don't coincide. In this example the orthonormal basis $(e_k)$ converges weakly to $0$ but note how $||e_k|| = 1$

[felixpernegger]

I think its best just to ignore my comment regarding this then

[Moniker1998]

I haven't got around to update this PR before I will do so now

[Moniker1998]

@prabau would you like to add any references?

[felixpernegger]

Maybe just add some general standard introductory functional analysis book, I'm sure you know some...

[Moniker1998]

@felixpernegger I do, but not sure if they cover this, and I'm not sure if I need a reference.

[felixpernegger]

@felixpernegger I do, but not sure if they cover this, and I'm not sure if I need a reference.

Otherwise, I didnt know about any of the terms while reviewing this, but all the terms were very easy to find by googling, so whatever