# I Weak-* topology & Neighborhoods

1. Jan 22, 2016

### strangerep

I'm trying to understand a paper which uses weak-* topology. (Unfortunately, the paper was given to me confidentially, so I can't provide a link.) My specific question concerns a use of weak-* topology, and interpretation/use of neighborhoods in that topology.

First, I'll summarize the context:

Let $H$ be a complex vector space with Hermitian form (inner product), denoted $\langle \psi,\phi\rangle = \overline{\langle \phi,\psi\rangle} ~,$ (where $\psi,\phi\in H$). $H$ has a norm induced by the inner product, but is not necessarily complete in norm topology.

Denote by $H^\times$ the vector space of all antilinear functionals on $H$, i.e., the algebraic antidual of $H$, and identify $\psi\in H$ with the antilinear functional on $H$ defined by $$\psi(\phi) := \langle \phi,\psi\rangle ~,~~~ (\phi\in H) ~.$$ With this identification, $H \subseteq H^\times$.

Now give $H^\times$ the structure of a locally convex space with the weak-* topology induced by the family of seminorms $\|\Psi\|_\phi := |\Psi(\phi)|$, with $\Psi\in H^\times,\; \phi\in H$.

This much I understand. I also understand (I think) the standard meaning of weak-* topology on $H^\times$, and the associated pointwise convergence of sequences of elements $\{\Psi_k\} \to \Psi$ in $H^\times$.

But now the paper says:

I understand why this $U$ can be considered to be a neighborhood in the sense of weak-* topology, but the restriction to a "finite" number of $\phi_k\in H$mystifies me.

I also don't understand why this definition of neighborhoods means that, "as a consequence", $\phi_\ell \to \phi$ iff $\phi_\ell(\psi) \to \phi(\psi)$ for all $\psi\in H$.

I would have thought that the last bit, i.e., $\phi_\ell \to \phi$ iff $\phi_\ell(\psi) \to \phi(\psi)$ for all $\psi\in H$ is simply a statement of pointwise convergence in ordinary weak-* topology. But I don't get the relevance of the neighborhoods $U$ defined via a restriction to finite numbers of $\phi_k\in H$.

I sure hope someone can help me out. My knowledge of general topology is proving insufficient for me to figure it out for myself.

2. Jan 23, 2016

### Samy_A

The "finite" is by definition: that is how the topology induced by a family of seminorms is defined.
Stated differently, one defines the topology induced by a family of seminorms as the the topology generated by the seminorm "balls" or "strips". The two definitions are equivalent. (pages 3 to 5 in this pdf: http://people.math.gatech.edu/~heil/6338/summer08/section9d.pdf)
It is the weakest topology that makes all seminorms continous.
(out of laziness I take $\phi=0$)
For $\Rightarrow$: the seminorms are by definition continous.
For $\Leftarrow$:
Assume $\phi_\ell(\psi) \to 0 \ \forall \psi \in H$.
Take a neighborhood $U$ of $0 \in H^\times$.
Then $\exists \psi_1, ... ,\psi_k \ \in H: V= \{X \in H^\times, |X(\psi_i)| \leq 1, i=1,...,k \} \subseteq U$
As $\phi_\ell(\psi_j) \to 0$ for all $j=1,...,k$, all $\phi_l \in V \subseteq U$ for $l$ sufficiently large, proving $\phi_\ell \to 0$.

Last edited: Jan 23, 2016
3. Jan 23, 2016

### strangerep

Thank you.

I think there's something I was/am missing about weak(-*) topologies, but I'll
study the Heil reference before I ask any more silly questions.

4. Jan 24, 2016

### Samy_A

You are most welcome.
I wanted to like your post, but realize it would be ambiguous.

5. Jan 24, 2016

### strangerep

Huh?

I think there was very little to "like" about my post, especially now that I've read some of Heil's course material. I see now there was rather a lot that I didn't have a clue about. It's unfortunate that his notes are still incomplete -- I find I enjoy his way of explaining things and would like to read a more comprehensive version.

One more question, if I may...

The paper then goes on to "prove" that $H$ is dense in $H^\times$ wrt the weak* topology induced by that family of seminorms, but the proof is so brief that I don't understand it. I looked for a similar result in Heil's notes, but didn't find one. Do you know whether such a result is/isn't generally valid?

6. Jan 24, 2016

### Samy_A

Yes. I'm on the phone now. If no one chimes in, I'll try to find a good reference this afternoon (Belgian time).

7. Jan 24, 2016

### Samy_A

Hmm, this proved more tricky than I first thought.

The way I see it is that $H$ being dense in $H^\times$ wrt the weak* topology is a consequence of the Riesz representation theorem (http://www.math.umn.edu/~garrett/m/fun/Notes/02_hsp.pdf, page 9).
Each element of $H^\times$ can be represented by an element of the completion of H. H is by definition dense in the completion wrt to the norm topology, and thus certainly dense in the weaker weak* topology.
There must be an easier way to do this, but I don't see it yet.

8. Jan 24, 2016

### Krylov

Sorry for interrupting, but I was reading along a little bit and I wondered whether "Riesz" also applies to the algebraic dual of a Hilbert space? (I read that $H^{\times}$ was defined by the OP as an algebraic antidual space.) Every infinite-dimensional Banach space admits a discontinuous linear functional. Doesn't this cause problems when applying Riesz' theorem?

9. Jan 24, 2016

### WWGD

Sorry if I miss something obvious, but isn't the algebraic dual a subset of the continuous dual? If Riesz applies to the continuous dual, why would it not apply to the algebraic dual? And I did not see any restrictions on the dimension in Riesz Rep.

10. Jan 24, 2016

### Krylov

Indeed, there are no restrictions on the dimension in Riesz' theorem.

Given a Banach space $W$, the algebraic dual $W^{\times}$ consists of all linear functionals on $W$ and the continuous dual $W^{\ast}$ consists of all continuous linear functionals on $W$. So it holds that $W^{\ast} \subseteq W^{\times}$ and this inclusion is strict when $\text{dim}\,W = \infty$. Riezs says that if $W$ is Hilbert, then $W =W^{\ast}$ but not $W = W^{\times}$, as far as I know.

So, if we denote by $H$ the space from the OP and by $\overline{H}$ its completion, then we have
$$H \subseteq \overline{H} = (\overline{H})^{\ast} \subset (\overline{H})^{\times}$$
where the equality is due to Riesz. (I suppress the isos and embeddings for simplicity, which I admit is a bit sloppy.)

Actually, come to think of it, I have never seen the weak$^\ast$ topology being used on the algebraic dual space, but always on the continuous (topological) dual space. Perhaps that merely indicates a lack of knowledge on my part, however.

Last edited: Jan 24, 2016
11. Jan 24, 2016

### Samy_A

Yes, it does as far as I know. I completely lost sight of the fact that we were dealing here with not necessarily continuous functionals. My apologies.

12. Jan 24, 2016

### Krylov

No need to apologize, it is always my pleasure to "read you".

It leaves the question open, whether $H$ is weak$^{\ast}$ dense in $H^{\times}$. Maybe the OP can reproduce the (small) proof of that fact from the article he is reading, to provide a clue?

Last edited: Jan 24, 2016
13. Jan 24, 2016

### WWGD

14. Jan 24, 2016

### Samy_A

I second that. I tried all kind of sinister manipulations with the neighborhoods, but to no avail.

15. Jan 24, 2016

### Krylov

It seems to me that Banach-Steinhaus implies that if $\Psi \in H^{\times}$ is the weak$^{\ast}$ limit of a sequence $(\psi_n)_n$ in (the embedding into $H^{\times}$ of) $H$, then $\Psi$ is bounded.

16. Jan 24, 2016

### strangerep

Bounded in what sense? Do you mean $|\Psi(\phi)| < \infty$, for all $\phi\in H$ ?

Doesn't that theorem require the domain to be a Banach space (in this case, that would be $\overline H$)?
IIUC, $H$ is not Banach, since not complete. But does that make enough of a difference?

[Edit: about reproducing the short proof, I'll have to seek the author's permission first, since the draft paper was provided to me in confidence.]

Last edited: Jan 25, 2016
17. Jan 25, 2016

### Samy_A

He meant $|\Psi(\phi)| <C ||\phi||$, for all $\phi\in H$ and for some constant C.

18. Jan 25, 2016

### Krylov

No, that is always true. I don't mean $\sup_{\phi \in H}{|\Psi(\phi)|} < \infty$ either. (That only happens for the zero functional.) What is usually meant by this is that there exists $C > 0$ such that $|\Psi(\phi)| \le C \|\phi\|$ for all $\phi \in H$.
You wrote in your OP that $H$ not necessarily complete, but you didn't exclude the possibility.

What I meant in my post #15 is the following. Suppose $H$ is complete. Let $\Psi \in H^{\times}$ be arbitrary and suppose there exists a sequence $(\Psi_n)_n$ in $H$ such that $\Psi_n \xrightarrow{w*} \Psi$. Then
$$\sup_n{|\Psi_n(\phi)|} < \infty \qquad \forall\,\phi \in H$$
so Banach-Steinhaus yields
$$C:= \sup_n{\|\Psi_n\|} < \infty$$
("Weak$^{\ast}$ convergent sequences are bounded in norm.") Then, for any $\phi \in H$,
$$|\Psi(\phi)| = \lim_{n \to \infty}{|\Psi_n(\phi)|} \le C\|\phi\|$$
so $\Psi$ is bounded, or in oher words: $\Psi \in H^{\ast}$. This, together with the fact that every Banach space admits a discontinuous linear functional, shows that $H$ is not densely embedded into $H^{\times}$, at least not when $H$ is complete.

19. Jan 25, 2016

### strangerep

I see. Let us then restrict to the case where $H$ is not complete.

Hmm, I'll have study the details.

Anyway, it turns out the original version of the short proof I mentioned was already public on MathOverflow, here. (See the answer.) I tried to ask for extra clarification there, but MathOverflow wouldn't let me. So here's the same proof, with some symbols changed to match those used earlier in this thread.

I guess that the 2nd sentence is true because one can perform orthogonal decomposition(?), or perhaps by projecting from $H^\times$ to $F$?

I don't follow the final sentence at all. I guess it has something to do with how the weak-* topology is constructed in terms of finite intersections?

20. Jan 26, 2016

### Samy_A

I guess the same as you.
Take any neighborhood $U$ of $\psi$.
Then we have a finite number of $\phi_k \in H$ such that $V=\{ X \in H^\times|X(\phi_k) - \Psi(\phi_k)| \le 1 \} \subseteq U$.
Now, for any finite-dimensional subspace G containing the finite-dimensional subspace $E=\text{span}\{\phi_i\}$, $\phi_G \in V$ as $\phi_G(\phi_k)=\phi_E(\phi_k)=\Psi(\phi_k)$. This proves that the net $(\phi_F)$ converges to $\psi$ in the weak* topology.

That's my interpretation of the proof. Still very much doubting the part I put in red.

EDIT: doubts removed, looks legit. Would like a second opinion though.

EDIT 2: if the proof is indeed correct, and in view of @Krylov 's Banach-Steinhaus argument, if follows that $H^\times$ with the weak* topology is not a sequential space when H is a infinite-dimensional Hilbert space.

Last edited: Jan 26, 2016