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.