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).(adsbygoogle = window.adsbygoogle || []).push({});

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

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