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I know that the space $\ell^1$ is not reflexive as the dual space $\ell^\infty$ is not separable but I don't know the dual space of $\ell^\infty$. Someone, please provide me with the details of the dual space of $\ell^\infty$. (I have searched for it on Google but I couldn't understand it. So, if not the details, suggest some prerequisites to understand it !)

And does there exist a linear map from $\ell^1$ to its bi-dual which is isometrically isomorphic? (Because the existence of such a linear isometrically isomorphic map between a normed space and its bi-dual doesn't ensure the reflexivity of the normed space.)

Rudra
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    The dual of l infinity is known as the "ba" space - it is a space of finitely additive measures (charges). For your second question, I am fairly sure it is not true (someone can confirm) as to create spaces which are not reflexive but still isometric to their bidual takes a lot of work (see the so called James' Tree Space). – rubikscube09 Apr 06 '24 at 15:44

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The bidual of $\ell^1$ is not isometrically isomorphic to $\ell^1$. If it was, then $\ell^1\cong (\ell^1)^{**}\cong (\ell^\infty)^*$ would be separable (as $\ell^1$ is). However, a Banach space is separable if its dual is separable (see Proving that a Banach space is separable if its dual is separable), which would imply that $\ell^\infty$ is separable, which is not the case.

Severin Schraven
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  • The canonical map from X to X'' needs to be surjective for the normed space X to be reflexive, and then we can show that the canonical map is indeed a isometric isomorphism from X to X'' (if X is reflexive). But if X is not reflexive , that means the canonical map is not surjective and I think that doesn't imply that there doesn't exist a linear map which is isometric isomorphism (except the canonical map) from X to X". Correct me if I'm wrong. – Rudra Apr 06 '24 at 16:56
  • @Rudra I am not making use of any reflexivity. All I am doing is to prove that if $\ell^1$ is isometrically isomorphic to its bidual, then $\ell^\infty$ is separable, which is not the case. Note, the definition of reflexive spaces is that the canonical map is an isomorphism of Banach spaces, in particular it would mean that the canonical map is surjective. – Severin Schraven Apr 06 '24 at 17:07
  • It seems that you have difficulties processing my answer. Can you pinpoint which step is causing problems? – Severin Schraven Apr 06 '24 at 17:09
  • Okay. I have got it. Thanks a lot. – Rudra Apr 06 '24 at 17:15