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The claim is

A subspace of a normed vector space is normed closed iff it is weakly closed.

I can show one direction. Strong convergence implies weak convergence, so it is weakly closed. But I have no idea about the other direction.

Thanks in advance!

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This question seems to have been posted before: math.stackexchange.com/questions/449301/… –  echoone Sep 13 '13 at 3:34
    
@echoone, thanks. But the link only proved one direction. –  Falang Sep 13 '13 at 3:42
    
(1) "so it is weakly closed." On what assumption? Do you mean to say, "If it is norm closed, then it is weakly closed?" (2) If that is what you mean, then you must check your reasoning. If $A$-convergence implies $B$-convergence (of a net, say), then $B$-closed implies $A$-closed, not conversely. (3) The provided link proves the difficult direction. –  Jonas Meyer Sep 13 '13 at 4:40
    
@JonasMeyer, thank you! –  Falang Sep 13 '13 at 5:04
    
@JonasMeyer, could you post this as answer? –  userNaN Sep 14 '13 at 9:03

1 Answer 1

  1. "so it is weakly closed." On what assumption? Do you mean to say, "If it is norm closed, then it is weakly closed?"
  2. If that is what you mean, then you must check your reasoning. If A-convergence implies B-convergence (of a net, say), then B-closed implies A-closed, not conversely.
  3. "But the link only proved one direction." That is the difficult direction.
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