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Why is [tex]l^1[/tex] not reflexive

  1. Apr 12, 2005 #1
    I am trying to find a good explanation of why [tex]l^1[/tex] is not reflexive. It is easy to find the statement that it isn't (even mathworld has it http://mathworld.wolfram.com/ReflexiveBanachSpace.html). The reason why I wonder is that the reason given to me doesn't satisfy me. The reason is that since the dual of [tex]C_0[/tex] is [tex]l^1[/tex] and since [tex]C_0[/tex] can't be a dual of any space [tex]l^1[/tex] isn't reflexive. ([tex]C_0[/tex] not being the dual of any space is just a consequence of the Krein-Milman theorem and Alaoglu's theorem). So certainly [tex](l^1)^* \ne C_0[/tex] but I already know that since it is easy to show [tex](l^1)^* = l^\infty[/tex] (at least isomorphically).

    Can someone give me a linear continuous map on [tex]l^\infty[/tex] that isn't the image of an element in [tex]l^1[/tex] by the canonical map [tex]\theta[/tex]

    So that we are all using the same definitions here's what I use
    [tex]l^1[/tex] is the space of complex sequences who's series absolutely converge
    [tex]l^\infty[/tex] is the space of bounded complex sequences
    [tex]C_0[/tex] is space of complex sequences that converges to 0
    The canonical map [tex]\theta: X \rightarrow X^{**}[/tex] is defined as given [tex]x\in X[/tex] and [tex]x^*\in X^*[/tex] then [tex]\theta(x)(x^*) = x^*(x) [/tex]

    So that there isn't any moral dilemmas I will just mention that this is not a homework assignment for a class. The only class that this could be related has already ended.

  2. jcsd
  3. Apr 13, 2005 #2
    Does anyone know how I can move this thread to the analysis forum. It clearly more appropriately belongs there.
  4. Apr 13, 2005 #3
    God I feel like a putz replying to my own question but I found another hint that might help somebody help me.

    I found this statement on a wikipedia (http://www.answers.com/main/ntquery...eflexive+space&gwp=8&curtab=2222_1&sbid=lc02a)
    without proof
    "A Banach space is reflexive if and only if its dual is reflexive."

    Since [tex]C_0[/tex] is Banach and is not reflexive (remember not the dual of any space) then its dual [tex]l^1[/tex] is not reflexive. So I can accecpt [tex]l^1[/tex] not being reflexive if someone can justify the quoted statement to me.

    Surely someone here must know enough about functional analysis to at least make a comment on this stuff. I hope people don't think what I'm writing is just gibberish made to sound like math.

  5. Apr 20, 2005 #4
    Dunno if you're still interested in an answer, but here goes:

    If we let E be a vector space where the natural embedding [itex]J_{E}:E \rightarrow E''[/itex] is surjective. we then need to show the natural embedding [itex]J_{E'}:E' \rightarrow E'''[/itex] is a surjection. This can be done by defining a [itex] w' \in E' [/itex] given a [itex] w \in E'''[/itex] by

    [itex] <x, w'>_{E} = <J_{E}(x), w> [/itex] for all [itex]x \in E[/itex]. Then we have

    [itex]<J_{E}(x), J_{E'}(w')>_{E''}[/itex]
    [itex] = <w', J_{E}(x)>_{E'}[/itex]
    [itex] = <x, w'>_{E} [/itex]
    [itex] = <J_{E}(x),w>_{E''}[/itex]

    Now we need to show that the map [itex]J_{E'}:E' \rightarrow E'''[/itex] being a surjection implies that [itex]J_{E}:E \rightarrow E''[/itex] is a surjection.

    Given a [itex] w \in E'''[/itex] define a [itex]w' \in E'[/itex] by

    [itex]w'(x) = w(J_{E}(x))[/itex]

    for all [itex]x \in E[/itex]

    Now all we have to do is show that J_{E'}(w') = w, and then J_{E} is surjective.

    [itex]J_{E'}(w')(J_{E}(x)) = J_{E}(x)(w') = w'(x) = w(J_{E}(x))[/itex]

    Which is what we wanted.
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