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

[snoble]

I am trying to find a good explanation of why l^1 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 C_0 is l^1 and since C_0 can't be a dual of any space l^1 isn't reflexive. (C_0 not being the dual of any space is just a consequence of the Krein-Milman theorem and Alaoglu's theorem). So certainly (l^1)^* \ne C_0 but I already know that since it is easy to show (l^1)^* = l^\infty (at least isomorphically).

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

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

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.

Thanks,
Steven

 

[snoble]

Does anyone know how I can move this thread to the analysis forum. It clearly more appropriately belongs there.

 

[snoble]

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;jsessionid=pw9nxr2vghkk?method=4&dsid=2222&dekey=Reflexive+space&gwp=8&curtab=2222_1&sbid=lc02a )
without proof
"A Banach space is reflexive if and only if its dual is reflexive."

Since C_0 is Banach and is not reflexive (remember not the dual of any space) then its dual l^1 is not reflexive. So I can accecpt l^1 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.

Steven

 

[Lonewolf]

Dunno if you're still interested in an answer, but here goes:

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

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

<J_{E}(x), J_{E'}(w')>_{E''}
= <w', J_{E}(x)>_{E'}
= <x, w'>_{E}
= <J_{E}(x),w>_{E''}

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

Given a w \in E''' define a w' \in E' by

w'(x) = w(J_{E}(x))

for all x \in E

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

J_{E'}(w')(J_{E}(x)) = J_{E}(x)(w') = w'(x) = w(J_{E}(x))

Which is what we wanted.