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Weak and weak* convergence

  1. Jun 23, 2009 #1
    Hi All,

    I would like to appreciate functional analysis in a rigorous setting, to enrich my “engineering” understanding of the matter.

    So far I have firmly understood that:

    1) for any Banach space a dual space can be defined
    2) this brings with itself the notion of weak and weak* convergence.

    I think I understand weak convergence.
    In vulgar terms, instead of “monitoring” how the series converges in the original space E, it has to be assured that for any element of the dual, the value in R to which the mapping form the dual maps the element in the original space has to converge.


    With regards to weak* convergence, I understand that the sequence is defined in the dual space, and the convergence has to be valid when the mapping defined by the sequence is applied to any element of the original space E.

    But then, I am not sure I appreciate the consequences or the benefits of this definition.

    To start with, I would be the most obliged if somebody could explain ,even with heuristics terms or showing examples, the following qualitative statements I found in literature:

    - “<..> to check the weak convergence for a sequence of E, one needs to know what is the space E’. {fair enough}. It may happen that E’ is too big a space. This renders the verification of the weak convergence condition too difficult. Moreover in this case there are too few weakly convergent sequences”
    - “The weak topology on X is the weakest topology (the topology with the fewest open sets) such that all elements of X* remain continuous”


    Thank you very much

    Best Regards

    Muzialis
     
  2. jcsd
  3. Jun 23, 2009 #2

    quasar987

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    Like you say near the end of your post, the weak topology on E is the smallest topology for which the elements of E' remain continuous. This means more precisely that if [itex]\mathcal{A}:=\{f^{-1}(U) \ : \ U\subset\mathbb{R}, \ f\in E'\}[/itex], then the weak topology on E is the topology generated by [itex]\mathcal{A}[/itex]. (Consult wiki or a topology textbook for the meaning of this.) It can then be seen that a sequence x_n in E converge weakly to x iff f(x_n)-->f(x) for all f in E'.

    On finite dimensional spaces, the weak and strong topology coincide. But on infinite dimensional ones, the weak topology is always strictly weaker. This means it has strictly less open sets. As a consequence, it has more compact sets! And this can be seen as the main reason for considering the weak topology because on the one hand we like compact sets (for instance, because continuous maps meet their sup and inf on compact sets), and on the other hand, in infinite dimensions, compact sets are a rare thing. For instance, the closed unit ball is never compact in infinite dimension.

    Now, on E', one could put the weak topology. That is, the smallest topology for which every element of (E')'=E'' is continuous. Or, one could put yet an even weaker topology on E': the weak-* topology. This is done like so:

    First observe that there is a natural embedding (read linear injection) J of E into E'' that send x in E to Jx in E'' defined by Jx(f)=f(x) for all f in E'. In general, this is not surjective, so that J(E) is proper subset of E''. The weak-* topology on E' is defined as the smallest topology on E' for which all the elements of J(E) remain continuous. It can then be seen that f_n converges to f in the weak-* topology iff f_n(x)-->f(x) for all x in E.

    The consequence of this definition is that the closed unit ball in E' is compact in the weak-*topology, and consequently, so is any strongly closed and bounded set.

    And the benefit of this definition is that the above consequence has repercussions for the weak topology on E via the map J. For instance, it can be shown that the closed unit ball in E is weakly compact iff the map J is surjective. Spaces for which J is surjective are called reflexive. For instance, every Hilbert space is reflexive.
     
  4. Jun 27, 2009 #3

    morphism

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    Weak-* convergence is a kind of "freshman's dream" mode of convergence -- it's just pointwise convergence. Is this clear to you?
     
  5. Jun 28, 2009 #4
    LOL, I wish my freshmen had those sorts of dreams. I'm stuck convincing them you can't distribute a square root.
     
  6. Jun 29, 2009 #5

    Clear? Not at all. I can see this for separable Hilbert spaces--though often people prefer to look at them as related to spaces of functions R->C (or R->R). For other spaces, I'm having difficulty seeing what you mean at all.
     
  7. Jun 29, 2009 #6
    f_n converges to f (weak-star) if and only if f_n(x) converges to f(x) for all x. That's just the definition of pointwise convergence.
     
  8. Jun 30, 2009 #7
    So it is. Silly me.
     
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