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

  • Thread starter snoble
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In summary, the reason why l^1 is not reflexive is because its dual space, C_0, is not reflexive. This can be justified by the statement that a Banach space is reflexive if and only if its dual is reflexive. To show this, we need to prove that the natural embedding of E' into E''' is a surjection, which can be done by defining a map from E''' to E' and showing that it is a surjection. Finally, we need to show that this implies the natural embedding of E into E'' is also a surjection.
  • #1
snoble
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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.

Thanks,
Steven
 
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  • #2
Does anyone know how I can move this thread to the analysis forum. It clearly more appropriately belongs there.
 
  • #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;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 [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.

Steven
 
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  • #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.
 

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

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

The space [tex]l^1[/tex] is not reflexive because it does not satisfy the definition of reflexivity. For a space to be reflexive, every element in the space must have a unique functional that maps it to itself. However, in [tex]l^1[/tex], there are elements that do not have a unique functional that maps them to themselves. Therefore, [tex]l^1[/tex] does not meet the criteria for reflexivity.

2. Can you provide an example of an element in [tex]l^1[/tex] that does not have a unique functional mapping it to itself?

Yes, an example of such an element is the sequence [tex](1, 0, 0, 0, ...)[/tex]. This element can be mapped to itself by multiple functionals, such as [tex]f(x) = x[/tex] and [tex]g(x) = x^2[/tex]. Therefore, it does not have a unique functional that maps it to itself.

3. How does the lack of reflexivity in [tex]l^1[/tex] affect its properties?

The lack of reflexivity in [tex]l^1[/tex] affects its properties in several ways. For example, it means that [tex]l^1[/tex] is not a Hilbert space, since reflexivity is a necessary condition for a space to be a Hilbert space. Additionally, the lack of reflexivity can also affect the existence of solutions to certain problems in functional analysis.

4. Is [tex]l^1[/tex] the only space that is not reflexive?

No, [tex]l^1[/tex] is not the only space that is not reflexive. There are many other examples of non-reflexive spaces, such as [tex]l^\infty[/tex] and [tex]L^p[/tex] spaces with [tex]p \neq 2[/tex]. In fact, most Banach spaces are not reflexive.

5. Is there any way to make [tex]l^1[/tex] reflexive?

No, there is no way to make [tex]l^1[/tex] reflexive without changing its definition. The lack of reflexivity is inherent in the definition of [tex]l^1[/tex] as a Banach space. However, there are ways to construct reflexive spaces that are isomorphic to [tex]l^1[/tex], such as the space of absolutely convergent series.

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