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Weak convergence

  1. Sep 23, 2007 #1

    HMY

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    Take the sequence x_n = (1-1/n)e_n in l^2
    Consider the map l^2 to R given by x |--> ||x||^2

    The set A = {x_n} in l ^2 is closed & its image is not closed in R
    under the norm topology (it doesn't contain its accumulation point 1).
    So ultimately the above map is not closed.

    What I'm not sure about is whether the map is closed if I consider the
    weak topology on R instead of the norm topology? I think my misunderstandings are arising from the weak convergence.
     
  2. jcsd
  3. Dec 21, 2007 #2

    morphism

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    Because R is finite-dimensional its weak topology coincides with the norm topology.

    (I realize this post is really old, but I was bored and decided to browse the forum.)
     
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