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I Weak-* topology & Neighborhoods

  1. Jan 22, 2016 #1

    strangerep

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    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. :oldconfused:
     
  2. jcsd
  3. Jan 23, 2016 #2

    Samy_A

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    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
  4. Jan 23, 2016 #3

    strangerep

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    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. :blushing:
     
  5. Jan 24, 2016 #4

    Samy_A

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    You are most welcome.
    I wanted to like your post, but realize it would be ambiguous. :oldsmile:
     
  6. Jan 24, 2016 #5

    strangerep

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    Huh? :confused:

    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?
     
  7. Jan 24, 2016 #6

    Samy_A

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    Yes. I'm on the phone now. If no one chimes in, I'll try to find a good reference this afternoon (Belgian time).
     
  8. Jan 24, 2016 #7

    Samy_A

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    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.
     
  9. Jan 24, 2016 #8

    Krylov

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    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?
     
  10. Jan 24, 2016 #9

    WWGD

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    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.
     
  11. Jan 24, 2016 #10

    Krylov

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    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
  12. Jan 24, 2016 #11

    Samy_A

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    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.
     
  13. Jan 24, 2016 #12

    Krylov

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    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
  14. Jan 24, 2016 #13

    WWGD

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    Ah my bad, I had my inclusions reversed, Duh myself, sorry.
     
  15. Jan 24, 2016 #14

    Samy_A

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    I second that. I tried all kind of sinister manipulations with the neighborhoods, but to no avail.
     
  16. Jan 24, 2016 #15

    Krylov

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    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.
     
  17. Jan 24, 2016 #16

    strangerep

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    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
  18. Jan 25, 2016 #17

    Samy_A

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    He meant ##|\Psi(\phi)| <C ||\phi||##, for all ##\phi\in H## and for some constant C.
     
  19. Jan 25, 2016 #18

    Krylov

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    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.
     
  20. Jan 25, 2016 #19

    strangerep

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    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?
     
  21. Jan 26, 2016 #20

    Samy_A

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    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
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