I Why Does Weak-* Topology Use Finite Neighborhoods?

[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:

Thus, $U\subseteq H^\times$ is a neighborhood of $\Psi\in H^\times$ iff there are finitely many $\phi_k\in H$ such that $U$ contains all $X\in H^\times$ with $|X(\phi_k) - \Psi(\phi_k)| \le 1$ for all $k$. As a consequence, $\phi_\ell \in H^\times$ converges to $\phi\in H^\times$ iff $\phi_\ell(\psi) \to \phi(\psi)$ for all $\psi\in H$.

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.

 

[Samy_A]

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.

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.

Note that even if there are infinitely many seminorms in our family, when constructing the base we only intersect finitely many strips at a time.

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 $\psi\in H$.

(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 $\phi_\ell \to 0$.

 

[strangerep]

The "finite" is by definition
[...]
It is the weakest topology that makes all seminorms continous. [...]

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.

 

[Samy_A]

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.

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

 

[strangerep]

I wanted to like your post, but realize it would be ambiguous.

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?

 

[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).

 

[Samy_A]

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?

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.

 

[S.G. Janssens]

Each element of $H^\times$ can be represented by an element of the completion of H.

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?

 

[WWGD]

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?

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.

 

[S.G. Janssens]

Sorry if I miss something, 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.

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.

 

[Samy_A]

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?

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.

 

[S.G. Janssens]

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.

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?

 

[WWGD]

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.

Ah my bad, I had my inclusions reversed, Duh myself, sorry.

 

[Samy_A]

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?

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

 

[S.G. Janssens]

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

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.

 

[strangerep]

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.

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.]

 

[Samy_A]

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

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

 

[S.G. Janssens]

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

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 $\phi \in H$.

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?

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.

 

[strangerep]

Thank you again for your answers and patience.

You wrote in your OP that $H$ not necessarily complete, but you didn't exclude the possibility.

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

[...]
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.

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.

Such a net exists for any $\Psi$ (in fact, there is a canonical such net). First, note that if $F\subseteq H$ is a finite-dimensional subspace, then there is a unique $\phi_F \in F$ such that $\Psi(\psi) = \langle \phi_F, \psi\rangle$ for all $\psi\in F$. The collection of such $F$ form a directed set under inclusion, and the net $(\phi_F)$ will have the desired property.

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?

 

[Samy_A]

Thank you again for your answers and patience.

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?

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.

 

[Samy_A]

Quite frustrating actually, as the proof was staring me in the face.

To prove that $H$ (canonically embedded, I'll be sloppy on this from now on) is dense in $H^\times$ in the weak-* topology, one has to prove that every open set in $H^\times$ includes an element of H.
It is sufficient to prove this for each of the (now familiar) sets $U=\{ X \in H^\times|X(\phi_k) - \Psi(\phi_k)| \le 1 \}$ where $(\phi_k)_k$ is a finite set of elements in $H$ and $\Psi \in H^\times$.

Now, $E=\text{span}(\phi_1,...,\phi_n)$ is a finite-dimensional subspace of $H$, and therefore an Hilbert space.
Hence, $\Psi_E: E \to \mathbb C: x \mapsto \Psi(x)$ is an antilinear continuous function, so by the Riesz representation theorem, $\exists y \in E: \forall x \in E: \Psi_E(x)=\Psi(x)=<x,y>$. As $y \in U$, this concludes the proof.

 

[S.G. Janssens]

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.

Yes!

Incidentally, you can see in my post #18 that if you replace the sequence $(\Psi_n)_n$ by a net $(\Psi_{\alpha})_{\alpha}$ it is in general no longer true that
$$
\sup_{\alpha}{|\Psi_{\alpha}(\phi)|} < \infty \qquad \text{(false)}
$$
since, unlike sequences, convergent nets of real or complex numbers need not be bounded. It is the second time in my life that I have been fooled by this very fact.

Moral (for me): When proving statements about the weak$^{\ast}$ topology, one really needs to either work with general nets or directly with the neighborhoods, as @Samy_A did in his post #21. In fact, I prefer the latter, because nets seduce me too easily into thinking they are sequences.

My apologies to both of you for leading you astray with my Banach-Steinhaus (BS..?) argument.

 

[Samy_A]

Yes!

Incidentally, you can see in my post #18 that if you replace the sequence $(\Psi_n)_n$ by a net $(\Psi_{\alpha})_{\alpha}$ it is in general no longer true that
$$
\sup_{\alpha}{|\Psi_{\alpha}(\phi)|} < \infty \qquad \text{(false)}
$$
since, unlike sequences, convergent nets of real or complex numbers need not be bounded. It is the second time in my life that I have been fooled by this very fact.

Moral (for me): When proving statements about the weak$^{\ast}$ topology, one really needs to either work with general nets or directly with the neighborhoods, as @Samy_A did in his post #21. In fact, I prefer the latter, because nets seduce me too easily into thinking they are sequences.

My apologies to both of you for leading you astray with my Banach-Steinhaus (BS..?) argument.

Yes, same here. I'm never really at ease with nets, and as I have been drilled early in topology, I prefer the neighborhood approach. That's why I rewrote the proof in these terms, more for myself than for the advancement of science.

My apologies to both of you for leading you astray with my Banach-Steinhaus (BS..?) argument.

No apologies needed, but lol at the BS.

 

[strangerep]

Thank you again @Samy_A and @Krylov. It has taken me several days to understand SamyA's version of the proof in post #21.

To make sure I do indeed understand properly, I'll now try to collect the essential content of this thread, elaborating (perhaps tediously) on points I initially failed to follow. (Please tell me if I omit something, or say something wrong.)
-----------------------------------

We assume $H \subseteq H^\times$. The weak-* topology on $H^\times$ is "generated" (explained below) by the family of seminorms:$$\|\Psi\|_\phi ~:=~ |\Psi(\phi)| ~\equiv~ |\langle \phi,\, \Psi \rangle| ~,$$ for $\Psi\in H^\times$ and any $\phi\in H$.

To "generate" the topology, we start with the sets:$$B^\phi_r(\Psi) ~:=~ \{X\in H^\times ~:~ \|\Psi-X\|_\phi < r \} ~,~~~~~ (\phi\in H),$$which are known as "open strips". (To see that this is reasonable name, consider a 2D case.) We then form a class of finite intersections of open strips as follows:$$B ~:=~ \left\{~ \bigcap_{j=1}^n B^{\phi_j}_r(\Psi) ~:~ n\in N, \phi_j\in H, r>0, \Psi\in H^\times ~\right\} ~,$$and declare that the sets in $B$ are a base for the topology, which simply means that the open sets in the topology are those which can be formed by taking arbitrary unions of sets from $B$.

That this gives a Hausdorff topology (i.e., separates elements of $H^\times$) follows easily from the properties of the seminorms. (See the Heil notes linked earlier by Samy_A).

We are interested in the case where $H$ is infinite-dimensional, and therefore concentrate on the case of an uncountable family of seminorms. This means we must use nets rather than sequences. This may be seen from the fact that a space with a countable family of seminorms is metrizable, since the function $$d(X,Y) ~:=~ \sum_{n=1}^\infty 2^{-n} \, \frac{\|X-Y\|_n}{1 + \|X-Y\|_n}$$ satisfies the usual properties required of a (translation-invariant) metric -- see Heil's notes for details -- and we are interested in the case where $H^\times$ is not metrizable in general. (Heil also gives examples of weak-* topological spaces where working with sequences instead of nets gives wrong answers.)

In applications of the above, we work with nets consisting of sets from $B$, ordered by reverse inclusion. (In practice, experts seem to take this as understood, and work with particular representative sets -- the reverse-inclusions and hence directed-set net structure being understood.)

We now wish to show that $H$ is weak-* dense in $H^\times$. (Heuristically, this means we can approximate any $\Psi\in H^\times$ arbitrarily well (though not necessarily exactly) by elements of $H$.) This is true if, given any $\Psi\in H^\times$ we can prove that every weak-* open set $U$ containing $\Psi$ also contains an element $\psi_U\in H$, that coincides with $\Psi$ on $U$ as an antilinear functional. (The idea is that we could then construct a net of successively nested open sets (all containing $\Psi$) and thus have a net converging to $\Psi$. [No, this is not the "idea". See Samy_A's next post.])

Now, every open set containing $\Psi$ is (by construction) a union of sets from $B$. But every set in $B$ is finite dimensional, being a finite intersection of $B^\phi_r(\Psi)$ sets. [This is wrong, I think. We need to select the individual $\phi_j$'s. I'll try to fix the rest later.]
So take any set in $E\in B$. Clearly, $E \subseteq H^\times$, but, being finite-dimensional, it is a Hilbert space. The Riesz representation theorem is therefore applicable on $E$. The main trick in the proof is therefore to find an antilinear functional $\Psi_E$ on $E$, which coincides with $\Psi|_E$ (i.e., when $\Psi$'s domain is restricted to $E$). But this is trivial, as we can simply define $\Psi_E(\nu) := \Psi(\nu)$, for any $\nu\in E$. The Riesz representation theorem then guarantees that there exists a $\xi\in E$ such that $$\Psi_E(\nu) ~=~ \langle \nu ,\, \xi\rangle ~,$$for all $\nu\in E$. (Also, since $\nu$ is a linear combination of a finite number of the $\phi_j \in H$, this ensures that $\nu$ is also in $H$.)

Since such a $\Psi_E$ exists for any set in $B$, and coincides with $\Psi$ on every such set, this proves that a net of elements in $H$ exists, converging to $\Psi\in H^\times$ -- since we can simply take progressively nested sets from $B$. (It is not necessary to exhibit an explicit net -- the progressive nesting is sufficient to establish existence of such a net.)

This proves that $H$ is dense in $H^\times$, as claimed.

Comments/corrections/additions are welcome.

 

[Samy_A]

A few remarks (mainly nitpicks):

We now wish to show that $H$ is weak-* dense in $H^\times$. (Heuristically, this means we can approximate any $\Psi\in H^\times$ arbitrarily well (though not necessarily exactly) by elements of $H$.) This is true if, given any $\Psi\in H^\times$ we can prove that every weak-* open set $U$ containing $\Psi$ also contains an element $\psi_U\in H$, that coincides with $\Psi$ on $U$ as an antilinear functional. (The idea is that we could then construct a net of successively nested open sets (all containing $\Psi$) and thus have a net converging to $\Psi$.)

While the last sentence about nets is correct, it's not exactly the "idea".
In a topological space $X$, as subset $E \subseteq X$ is dense if its closure $\overline E=X$. Now assume that there is some open set $U \neq \varnothing$ for which $U \cap E=\varnothing$. That implies that $E \subseteq X \setminus U$. As $X \setminus U$ is a closed set, it means that $\overline {E} \subseteq X \setminus U \neq X$, so that $E$ is not dense.

Now, every open set containing $\Psi$ is (by construction) a union of sets from $B$. But every set in $B$ is finite dimensional, being a finite intersection of $B^\phi_r(\Psi)$ sets. So take any set in $E\in B$. Clearly, $E \subseteq H^\times$, but, being finite-dimensional, it is a Hilbert space. The Riesz representation theorem is therefore applicable on $E$. The main trick in the proof is therefore to find an antilinear functional $\Psi_E$ on $E$, which coincides with $\Psi|_E$ (i.e., when $\Psi$'s domain is restricted to $E$). But this is trivial, as we can simply define $\Psi_E(\phi) := \Psi(\phi)$, for any $\phi\in E$. The Riesz representation theorem then guarantees that there exists a $\xi\in E$ such that $$\Psi_E(\phi) ~=~ \langle \phi ,\, \xi\rangle ~,$$for all $\phi\in E$.

A set in $B$ is not finite dimensional, $E=span \{\phi_1,...,\phi_n\}$ is finite dimensional.
In my version of the proof I used the Riesz representation theorem, as you do here. That is correct, but a little overkill. For a finite dimensional space, we know that the (anti)dual has the same dimension. That means that the (anti)dual only consists of the canonically embedded elements of the original space. So the existence of $\xi\in E$ is trivial.

Since such a $\Psi_E$ exists for any set in $B$, and coincides with $\Psi$ on every such set, this proves that a net of elements in $H$ exists, converging to $\Psi\in H^\times$ -- since we can simply take progressively nested sets from $B$. (It is not necessary to exhibit an explicit net -- the progressive nesting is sufficient to establish existence of such a net.)

As mentioned above, the mere fact that you found an element of $H$ in any set in $B$ is enough to prove that $H$ is dense in $H^{\times}$.

(Did you notice I don't like nets? )

 

[strangerep]

A few remarks (mainly nitpicks):

Again, it's taken me a few days to digest your remarks. (I'll think they are far more valuable than the word "nitpick" would suggest.) I'll split my followup questions into several subsequent posts.

While the last sentence about nets is correct, it's not exactly the "idea".

In that case, I had indeed got hold of the wrong end of the stick. I was thinking in terms of a constructive proof, but it seems I should be thinking in terms of a proof by contradiction?

In a topological space $X$, as subset $E \subseteq X$ is dense if its closure $\overline E=X$. Now assume that there is some open set $U \neq \varnothing$ for which $U \cap E=\varnothing$. That implies that $E \subseteq X \setminus U$. As $X \setminus U$ is a closed set, it means that $\overline {E} \subseteq X \setminus U \neq X$, so that $E$ is not dense.

So... sketching a proof by contradiction... Either $H$ is dense in $H^\times$, or it isn't. If we can find a nonempty open set $U$ s.t. $U\bigcap H = \emptyset$ then $H$ is not dense in $H^\times$. Therefore, if we can prove there is no such $U$, we have proved (by contradiction) that $H$ is dense in $H^\times$.

Is that what you meant? (If so, I need to rewrite quite a lot.)

 

[Samy_A]

Again, it's taken me a few days to digest your remarks. (I'll think they are far more valuable than the word "nitpick" would suggest.) I'll split my followup questions into several subsequent posts.

In that case, I had indeed got hold of the wrong end of the stick. I was thinking in terms of a constructive proof, but it seems I should be thinking in terms of a proof by contradiction?

The two proofs are very similar (which is not a surprise since I wrote mine after reading the one you posted).

The construction of the net in the proof you posted starts by stating that for every finite-dimensional subspace$F \subset H$ there is a $\phi_F \in F$ that is "equal" to $\Psi$ when restricted to $F$.
The proof then goes on to prove that $(\phi_F)_F$ is a net converging to $\Psi$.
In one sentence: "For each $\Psi \in H^{\times}$ there is a net in $H$ converging to $\Psi$."

"My" proof starts with an element $U\in B$, associates with it a finite-dimensional subspace$F \subset H$, and shows the existence of $\phi_F \in F\cap U$ as above. Then I invoque the definition of densesness in topology to immediately conclude that $H$ is weak-* dense in $H^{\times}$.
In one sentence: "Every non-empty open set in $H^{\times}$ contains an element of $H$."

The key element in both proofs is the existence of $\phi_F$ for every finite-dimensional subspace$F \subset H$. How to go from this to denseness is more a matter of taste than of deep Mathematics.

So... sketching a proof by contradiction... Either $H$ is dense in $H^\times$, or it isn't. If we can find a nonempty open set $U$ s.t. $U\bigcap H = \emptyset$ then $H$ is not dense in $H^\times$. Therefore, if we can prove there is no such $U$, we have proved (by contradiction) that $H$ is dense in $H^\times$.
Is that what you meant? (If so, I need to rewrite quite a lot.)

My only quibble would be with calling this a proof by contradiction. That a dense subset has a non-empty intersection with every non-empty open set is an immediate corollary to the definition of denseness.
I'd rather write:
Let $U$ be a non-empty open set in $H^{\times}$, then we show that $U\cap H \neq \varnothing$, proving that $H$ is dense.

 

[strangerep]

(I'm still working on this stuff, but my revised writeup of the above is not yet finished.)

In the meantime, I'd like to check something else...

The paper I mentioned in my opening post also contains the following proposition (slightly paraphrased):

Proposition: For every weak-* continuous antilinear functional $\Psi$ on $H^\times$ there is a vector $\psi\in H$ such that $$\Psi(\phi) ~=~ \langle\phi, \psi\rangle ~,~~~~~~ \mbox{for all}~ \phi\in H^\times ~.$$
[Edit: changed final $H$ to $H^\times$]

ISTM that this is an empty proposition -- just circular reasoning -- since the definition of weak-* topology is specified by requiring that all elements of $H$ be continuous as functionals when canonically embedded in $H^{\times\times}$.

I.e., the proposition is "proving" something which is essentially just the original definition of weak-* topology.

Or am I missing something?

 

[Samy_A]

(I'm still working on this stuff, but my revised writeup of the above is not yet finished.)

In the meantime, I'd like to check something else...

The paper I mentioned in my opening post also contains the following proposition (slightly paraphrased):

Proposition: For every weak-* continuous antilinear functional $\Psi$ on $H^\times$ there is a vector $\psi\in H$ such that $$\Psi(\phi) ~=~ \langle\phi, \psi\rangle ~,~~~~~~ \mbox{for all}~ \phi\in H ~.$$
ISTM that this is an empty proposition -- just circular reasoning -- since the definition of weak-* topology is specified by requiring that all elements of $H$ be continuous as functionals when canonically embedded in $H^{\times\times}$.

I.e., the proposition is "proving" something which is essentially just the original definition of weak-* topology.

Or am I missing something?

The original definition of the weak-* topology requires that the elements of H are continuous as functionals. But there could (in principle) be more weak-* continuous functionals than these corresponding to elements of H.

Compare with $H^{*}$, the usual (I mean the space of norm-continuous functionals) dual of a non complete pre-Hilbert space H. Elements in the completion of H also define weak-* functionals on $H^{*}$.
EDIT: the last sentence is wrong. Elements in the completion of H that are not in H do not define weak-* functionals on $H^{*}$.

(all in sloppy mode).

 

[strangerep]

Thanks.

Compare with $H^{*}$, the usual (I mean the space of norm-continuous functionals) dual of a non complete pre-Hilbert space H. Elements in the completion of H also define weak-* functionals on $H^{*}$.

But wouldn't that mean the proposition is incorrect -- since its "$H$'' is also an incomplete pre-Hilbert space... (?)

 

[Samy_A]

Thanks.
But wouldn't that mean the proposition is incorrect -- since its "$H$'' is also an incomplete pre-Hilbert space... (?)

If the proposition is correct (and I assume it is), this would mean that an element of the completion that is not in H doesn't define an antilinear weak-* continuous functional on $H^{\times}$.
It is another example where this algebraic (anti)dual behaves differently from the more familiar (anti)dual with bounded functionals.

(I forgot the "antilinear" in my previous post, but it was implied (insert whistling smiley)).

 

[strangerep]

[...]
It is another example where this algebraic (anti)dual behaves differently from
the more familiar (anti)dual with bounded functionals.

IIuc, $H^\times$ is the "more familiar (anti)dual with bounded functionals". We start with the algebraic (antidual), then turn it into a topological space by equipping it with weak-* topology.

Or am I missing something (again)?

 

[Samy_A]

IIuc, $H^\times$ is the "more familiar (anti)dual with bounded functionals". We start with the algebraic (antidual), then turn it into a topological space by equipping it with weak-* topology.

Or am I missing something (again)?

I was referring to $H^*$, the (anti)dual of H consisting of the bounded (anti)linear functionals on $H$. The "more familiar" was probably better formulated as "more familiar to me". As @Krylov has pointed out, for an infinite-dimensional $H$, $H^\times$ will be larger than $H^*$.

 

[strangerep]

I think we're using incompatible terminology. I use $H^*$ and $H^\times$ (initially) for the algebraic dual and antidual of $H$. Then I equip them with weak-* topologies, but continue to use the same symbols in each case. OTOH, I get the feeling @Krylov was using $H^\times$ for the algebraic dual, and $H^*$ for the topological dual. (?)

Anyway,... I think I now understand the point of the proposition in post #28. Although the definition of weak-* topology is in terms of continuity of functionals on $H^\times$, there is initially no transparent connection between this requirement and the specific base $B$ constructed in post #24. The proposition therefore closes the loop: showing that the only antilinear functionals continuous wrt to a topology based on $B$ are indeed elements of $H$.

 

[Samy_A]

I think we're using incompatible terminology. I use $H^*$ and $H^\times$ (initially) for the algebraic dual and antidual of $H$. Then I equip them with weak-* topologies, but continue to use the same symbols in each case. OTOH, I get the feeling @Krylov was using $H^\times$ for the algebraic dual, and $H^*$ for the topological dual. (?)

Yes, I too meant the topological (anti)dual with $H^*$. That was implied by the "bounded" in "the (anti)dual of H consisting of the bounded (anti)linear functionals on H".

Anyway,... I think I now understand the point of the proposition in post #28. Although the definition of weak-* topology is in terms of continuity of functionals on $H^\times$, there is initially no transparent connection between this requirement and the specific base $B$ constructed in post #24.

I somewhat disagree with the "no transparent connection". To make the functionals on $H^\times$ continuous, the open strips defined in post #24 have to be open sets in the to be defined weak-* topology, as $$B^\phi_r(\Psi)=\phi^{-1}(\{z \in \mathbb C\ | \ |\langle \phi,\Psi \rangle -z| \lt r\})$$
So finite intersections of these strips also have to be open sets.

The proposition therefore closes the loop: showing that the only antilinear functionals continuous wrt to a topology based on $B$ are indeed elements of $H$.

This is indeed what the proposition does.
Take a set $X$, and a set $\mathcal F$ of functions from $X \to \mathbb C$. One can define a topology on $X$ as the weakest topology that makes all the functions in $\mathcal F$ continuous. That is basically how the weak-* topology is defined.
Nothing in this definition precludes other functions from $X \to \mathbb C$ that are not in $\mathcal F$ to be continuous with that topology on $X$.

The proposition states that in you specific case, $H^\times$ with the weak-* topology, there are no additional continuous antilinear functionals.

 

[S.G. Janssens]

IIuc, $H^\times$ is the "more familiar (anti)dual with bounded functionals". We start with the algebraic (antidual), then turn it into a topological space by equipping it with weak-* topology.

Or am I missing something (again)?

I was referring to $H^*$, the (anti)dual of H consisting of the bounded (anti)linear functionals on $H$. The "more familiar" was probably better formulated as "more familiar to me". As @Krylov has pointed out, for an infinite-dimensional $H$, $H^\times$ will be larger than $H^*$.

I think we're using incompatible terminology. I use $H^*$ and $H^\times$ (initially) for the algebraic dual and antidual of $H$. Then I equip them with weak-* topologies, but continue to use the same symbols in each case. OTOH, I get the feeling @Krylov was using $H^\times$ for the algebraic dual, and $H^*$ for the topological dual. (?)

I was indeed using $H^{\times}$ for the algebraic (anti)dual and $H^{\ast}$ for the topological (anti)dual, because that is how I saw you defined $H^{\times}$ in your OP:

Denote by $H^\times$ the vector space of all antilinear functionals on $H$, i.e., the algebraic antidual of $H$

I think it is perfectly fine to first regard $H^{\times}$ as a set, then topologize it but keep the same symbol, which is what I believe you did. However, by topologizing $H^{\times}$ you do not make its discontinuous elements continuous, because their continuity depends only on the topology of $H$ (and the underlying field), which are both fixed.

Could this have caused some confusion? I hope this post helps.

P.S. If we want to be very precise, we have to use four different notations: Two for the algebraic dual and antidual and two for the topological dual and antidual. However, I think that in the arguments the difference between duals and antiduals is probably not really essential. Also, in the article they are probably only using antiduals?

 

[strangerep]

I think it is perfectly fine to first regard $H^{\times}$ as a set, then topologize it but keep the same symbol, which is what I believe you did.

Yes.

However, by topologizing $H^{\times}$ you do not make its discontinuous elements continuous, because their continuity depends only on the topology of $H$ (and the underlying field), which are both fixed.

Yes.

Could this have caused some confusion? I hope this post helps.

Yes and yes. Thank you.

P.S. If we want to be very precise, we have to use four different notations: Two for the algebraic dual and antidual and two for the topological dual and antidual. However, I think that in the arguments the difference between duals and antiduals is probably not really essential. Also, in the article they are probably only using antiduals?

The paper uses both duals and antiduals (and sets up an involution mapping between them, analogous to complex conjugation). But it only mentions the algebraic (anti)dual just once, and then immediately equips it with weak-* topology. So probably it would be better not to give "algebraic antidual" a symbol at all, and reserve the symbols $H^*, H^\times$ for the topological dual and antidual respectively.

Cheers.

 

[strangerep]

I somewhat disagree with the "no transparent connection".

I should probably have said "... not transparent to me...".

To make the functionals on $H^\times$ continuous, the open strips defined in post #24 have to be open sets in the to be defined weak-* topology, as $$B^\phi_r(\Psi)=\phi^{-1}(\{z \in \mathbb C\ | \ |\langle \phi,\Psi \rangle -z| \lt r\})$$

(Sigh.) This is what happens when someone with only basic (formal) training in analysis tries to understand something rather more advanced (wherein books, etc, tend to skip steps they consider elementary).

I see now how one can go from the continuity requirement, to the (original) specification of the $B^\phi_r(\Psi)$ sets -- one need merely re-express the set (as you've written it above) in terms of $X := \phi^{-1}(z)$, with a little help from linearity. It's obvious now that I see it.

Edit: And,... I also see now how the base sets can be set up without using a canonical embedding of $H$ in $H^{\times\times}$. In that case, the proposition is indeed much more significant than I previously thought, since it means we can ignore $H^{\times\times} \setminus H$.

 

[S.G. Janssens]

Yes.
The paper uses both duals and antiduals (and sets up an involution mapping between them, analogous to complex conjugation). But it only mentions the algebraic (anti)dual just once, and then immediately equips it with weak-* topology. So probably it would be better not to give "algebraic antidual" a symbol at all, and reserve the symbols $H^*, H^\times$ for the topological dual and antidual respectively.

Sorry for rambling on about this, but I think there may possibly still be a misunderstanding. When you start with the algebraic (anti)dual and you topologize it with the weak$^{\ast}$ topology, it surely becomes a topological space but it does not become equal to what is known in the textbooks as the topological (anti)dual. (The standard definition of "topological (anti)dual" is the vector space of all continuous (equivalently: bounded) linear functionals on $H$.)

So, when you topologize the algebraic (anti)dual, it is quite confusing to call the result "the topological (anti)dual".

 

[strangerep]

Sorry for rambling on about this, but I think there may possibly still be a misunderstanding.

Probably there's an unbounded sequence of misunderstandings, so feel free to "ramble"...

When you start with the algebraic (anti)dual and you topologize it with the weak$^{\ast}$ topology, it surely becomes a topological space but it does not become equal to what is known in the textbooks as the topological (anti)dual. (The standard definition of "topological (anti)dual" is the vector space of all continuous (equivalently: bounded) linear functionals on $H$.)

So, when you topologize the algebraic (anti)dual, it is quite confusing to call the result "the topological (anti)dual".

Well, I'm certainly not understanding something. What is the topology for what you call the "topological (anti)dual"?

And is this related to the distinction between pointwise convergence and uniform convergence? (I'm reading about "polar topology" on Wikipedia. :-)

 

[Samy_A]

Well, I'm certainly not understanding something. What is the topology for what you call the "topological (anti)dual"?

For me (and I think for Krylov too), $H^*$ is the set of (anti)linear functionals on $H$ that are continuous if we take the norm-topology on $H$.

Once defined as a set, $H^*$ can be given a weak-* topology in a similar way you gave $H^\times$ the weak-* topology.

But the set $H^\times$ is larger than the set $H^*$ if $H$ has infinite dimension.

EDIT:
It seems that what we call $H^*$ is not relevant to the paper you are treating here. I just brought it up in post #29 to "illustrate" why the proposition is not necessarily empty.

And is this related to the distinction between pointwise convergence and uniform convergence? (I'm reading about "polar topology" on Wikipedia. :-)

No. We could confuse matters further () by introducing the weak topological (anti)dual of $H$. For an infinite dimensional $H$ that is still another beast than what we call $H^*$. But let's not do that.

 

[strangerep]

No. We could confuse matters further () by introducing the weak topological (anti)dual of $H$. For an infinite dimensional $H$ that is still another beast than what we call $H^*$. But let's not do that.

Oh, I think I do need to do that.

I've been reading a little more on Wikipedia...

Wiki (Continuous Dual Space) uses $V^*$ for the algebraic dual, and $V'$ for the topological dual (aka "continuous dual space"). It then distinguishes 3 topologies on $V'$ :

1) Strong topology on $V'$ -- based on uniform convergence on bounded subsets in $V$.
2) Stereotype topology on $V'$ -- based on uniform convergence on totally bounded subsets in $V$.
3) Weak topology on $V'$ -- based on uniform convergence on finite subsets in $V$.

However, that notation is not fully consistent with this (more comprehensive) Wiki page:
Wiki(Topology of Uniform Convergence)
(See the section "G-topologies on the continuous dual induced by X").
It says the "continuous dual space [...] is denoted by $X^*$ and sometimes by $X'$". So it seems to conflate $X^*$ and $X'$.

A bit further down, under Examples, it gives "the weak topology $\sigma(X^*, X)$ or the weak-* topology", i.e., "the weak topology on $X^*$", more commonly known as "the weak-* topology, or the topology of pointwise convergence, which is denoted by $\sigma(X^*, X)$, and $X^*$ with this topology is denoted by $X^*_\sigma$, or by $X^*_{\sigma(X^*,X)}$ if there may be ambiguity.

So what was called "weak topology on $V'$" in (3) above, is the same as "weak-* topology on $V'$".

It seems there's lots of topologies based on the notion of uniform convergence -- they differ according to the type of set on which the "uniform convergence" occurs. Pointwise convergence is then just a special case of uniform convergence.

I think I need to study the Mackey-Arens thm, which appears further down that Wiki page. Apparently the usual space of tempered distributions admits the strong dual topology $b(X^*,X)$, which is finer than weak-* topology on the same space.

 

[Samy_A]

Reading all that it is understandable that you got confused by my use of $H^*$.

Out of curiosity I tried to prove the proposition in post #28. I think I succeeded in that.
But my proof only makes use of the familiar base of neighborhoods. Nowhere does the fact that $H^\times$ is the algebraic antidual, and not the topological antidual, play any role.
Conclusion: my "illustration" in post #29 is wrong.

 

[strangerep]

Reading all that it is understandable that you got confused by my use of $H^*$.

At least I now seem to have a comprehensive notation when one is dealing with continuous duals.

Out of curiosity I tried to prove the proposition in post #28. I think I succeeded in that.
But my proof only makes use of the familiar base of neighborhoods. Nowhere does the fact that $H^\times$ is the algebraic antidual, and not the topological antidual, play any role.
Conclusion: my "illustration" in post #29 is wrong.

Heh, I'm glad we're both (apparently) having fun.

I was gradually approaching a similar conclusion, since the base $B$ constructed earlier seems the same if one asks for continuity of $H^{\times\times}$.

Actually, I noticed some further interesting material in Wiki (Continuous Dual Space). E.g.,

[...] the algebraic dual space is always of larger dimension (as a cardinal number) than the original vector space. This is in contrast to the case of the continuous dual space, [...], which may be isomorphic to the original vector space even if the latter is infinite-dimensional.

And also that inf-dim Hilbert spaces are only isomorphic to their continuous duals, not their algebraic duals.

 

[strangerep]

Out of curiosity I tried to prove the proposition in post #28. I think I succeeded in that.
But my proof only makes use of the familiar base of neighborhoods. Nowhere does the fact that $H^\times$ is the algebraic antidual, and not the topological antidual, play any role.

How did you proceed? I've been trying to replicate this, but I can't even get started.

Indeed, my earlier statement that:

[...] the base $B$ constructed earlier seems the same if one asks for continuity of $H^{\times\times}$

is wrong, since the $\phi_j$'s are now in $H^{\times\times}$ rather than $H$. As a consequence, the subsequent steps which reply on $\mbox{span}(\phi_j)$ being a Hilbert space are no longer valid, since that span is not even an inner product space in general.

 

[Samy_A]

How did you proceed? I've been trying to replicate this, but I can't even get started.

My proof has not been reviewed, so hopefully I won't add to the confusion by posting a wrong proof.

The proposition is:

Proposition: For every weak-* continuous antilinear functional $\Psi$ on $H^\times$ there is a vector $\psi\in H$ such that $$\Psi(\phi) ~=~ \langle\phi, \psi\rangle ~,~~~~~~ \mbox{for all}~ \phi\in H ~.$$

Proof (hopefully):

As $\Psi$ is continuous in the weak-* topology on $H^\times$, $U=\Psi^{-1}\{z \in \mathbb C \mid \ |z|<1 \}$ is an open neighborhood of $0$ in $H^\times$.
That means that we can find a set $B=\{ X \in H^\times \mid |X(\phi_k) | \le 1 \}$, where $\phi_1,...,\phi_n$ are elements of $H$, such that $B \subset U$.
Let $E=span\{\phi_1,...,\phi_n\}$.
$\Psi$ restricted to the finite dimensional Hilbert space $E$ is an antilinear functional, so $\exists \psi \in E$ such that $\forall h\in E: \Psi(h)=\langle h,\psi \rangle$.

Now take any $\phi \in H$.
$\phi$ can be decomposed as $\phi=\phi_E +\phi_{E^{\perp}}$, with $\phi_E \in E$ and $\phi_{E^{\perp}} \in E^{\perp}$.
For all $\phi_1,...,\phi_n$ (elements of $E$), we have $\langle \phi_i,\phi_{E^{\perp}} \rangle =0$. That implies that $\phi_{E^{\perp}} \in B \subset U$. So we have $|\Psi(\phi_{E^{\perp}})| \lt 1$. Since the same holds for any scalar multiple of $\phi_{E^{\perp}}$, we can conclude that $\Psi(\phi_{E^{\perp}})=0$.
That means that $$\Psi(\phi)=\Psi(\phi_E +\phi_{E^{\perp}})=\Psi(\phi_E) +\Psi(\phi_{E^{\perp}})=\langle \phi_E,\psi \rangle +0=\langle \phi_E +\phi_{E^{\perp}},\psi \rangle=\langle \phi,\psi \rangle$$.

 

[strangerep]

That means that we can find a set $B=\{ X \in H^\times|X(\phi_k) | \le 1 \}$, where $\phi_1,...,\phi_n$ are elements of $H$, [...]

Ah, that makes it easier: you're using the standard definition of weak-* topology (continuity of functionals in $H$), whereas I was trying to be more general (continuity of functionals in $H^{\times\times}$).

No wonder I was having difficulty. Thanks.

 

[Samy_A]

Ah, that makes it easier: you're using the standard definition of weak-* topology (continuity of functionals in $H$), whereas I was trying to be more general (continuity of functionals in $H^{\times\times}$).

It's not really a matter of choice. A "weak-* continuous antilinear functional $\Psi$ on $H^{\times}$" is an element of $H^{\times\times}$. But its continuity as a function $\Psi: H^\times \to \mathbb C$ depends on the topology used on $H^\times$, in this case the weak-* topology.

 

[strangerep]

Yes, but what I meant was: I was trying to use a (probably weaker) topology, starting from the basic sets: $$S^\phi_r(\Psi) ~:=~ \{X\in H^\times ~:~ |\phi(\Psi-X)| < r \} ~,~~~~~ (\phi\in H^{\times\times}) ~.$$And,... I see now why no one does this -- it's pretty much unusable.

 

[strangerep]

Aargh, sorry! I just realized I made a typo in my post #28. The proposition should have said:

Proposition:
For every weak-* continuous antilinear functional $\Psi$ on $H^\times$, there is a vector $\psi\in H$ such that $$\Psi(\phi) ~=~ \phi(\psi) ~,~~~~~ \mbox{for all}~ \phi\in H^\times ~.$$I.e., the difference is that $\phi$ can be any element of $H^\times$, not merely $H$.

I guess that means your proof must change significantly?