I Why Does Weak-* Topology Use Finite Neighborhoods?

[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?

 

[Samy_A]

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 $H$.

There must be a typo here, since $\Psi(\phi)$ is antilinear in $\phi$ and $\phi(\psi)$ is linear in $\phi$.

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

I guess that means your proof must change significantly?

Well, not by much.
I already established (assuming the proof is correct) that there is a vector $\psi\in H$ such that $\Psi(\phi) ~=~ \psi(\phi) ~~~~ \mbox{for all}~ \phi\in H~.$
But $H$ is dense in $H^\times$ with the weak-* topology, so since the weak-* continuous functionals $\Psi, \psi$ are equal on a dense subset of $H^\times$, they must be equal on all of $H^\times$

 

[strangerep]

There must be a typo here, [...]

No, this time it was an actual mistake.
The original paper used $\langle\phi, \psi\rangle$, but I was trying to keep parenthesis notation.

I need more time to study your proof properly. The existing proof in the paper (which is apparently an instance of a general result: Thm 3.10 in Rudin FA) is hard to relate to yours. Later...

(Oh, and thanks yet again. )

 

[Samy_A]

No, this time it was an actual mistake.
The original paper used $\langle\phi, \psi\rangle$, but I was trying to keep parenthesis notation

I need more time to study your proof properly. The existing proof in the paper (which is apparently an instance of a general result: Thm 3.10 in Rudin FA) is hard to relate to yours. Later...

(Oh, and thanks yet again. )

Ah, Rudin's Functional Analysis (sweet memories ... )

Yes, that is definitely a faster and more elegant way to prove it. Maybe one can "map" the steps in my proof to the more general steps in Rudin's proof of theorem 3.10, but why bother.

EDIT: so I bothered anyway. After reading Rudin's proof, one can indeed see the similarities.
Where my proof diverges is that after finding the set $B \subset U$, I prove that $\Psi$ is a linear combination of $\phi_1,...,\phi_n$ by using a Pre-Hilbert space argument. Rudin uses his more general lemma 3.9 to prove that.

 

[strangerep]

I figured I should append a closing post to this thread.

I've realized that the author of the paper I originally referred to is not an FA expert. Some of the proofs therein are taken verbatim from experts without additional elaboration. Others are beyond my ability to assess without spending a vast amount of time, and even then doubts would remain.

So I think I must put the paper aside for now, and get on with other things.

Thank you again to Samy_A and Krylov. I can say that, with your help, my understanding of weak topology is now better than it was, though I still have a long way to go.