Why is $$l^1$$ not reflexive

1. Apr 12, 2005

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

2. Apr 13, 2005

snoble

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

3. Apr 13, 2005

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...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 $$C_0$$ 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

4. Apr 20, 2005

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.