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I want to show $c_0$ is not reflexive by showing the canonical embedding $c_0\rightarrow c_0^{**}$ is not surjective.

I saw a reference said: since the inclusion map $c_0\rightarrow\ell_\infty$ is not surjective, the canonical embedding $c_0\rightarrow {c_0}^{**}$ is not surjective.

I know $c_0$ is a closed linear subspace of $\ell_\infty$ and ${c_0}^{**}\cong{\ell_1}^*\cong\ell_\infty$. But how the canonical embedding is related to the inclusion map $c_0\rightarrow\ell_\infty$?

And my attempt is: since ${c_0}^{**}\cong{\ell_1}^*\cong\ell_\infty$, there exists a bijection $T:\ell_\infty\rightarrow {c_0}^{**}$. Then they have the same cardinality. Since $c_0\rightarrow\ell_\infty$ is not surjective, the cardinality of $c_0$ is less than $\ell_\infty$ and hence less than ${c_0}^{**}$. Thus the canonical embedding $c_0\rightarrow {c_0}^{**}$ is not surjective. But I am not sure it is true and I feel that I missed something.

Can anyone give me a precise proof by using the reference idea? Thank you!

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  • $\begingroup$ You should aim to show that $\ell_\infty$ is canonically embedded in $c_0^{**}$. $\endgroup$ – Ian Apr 10 '18 at 21:56
  • $\begingroup$ @Ian How to show is $\ell_\infty$ is canonically embedded in ${c_0}^{**}$? I only know they are isometrically isomorphic. $\endgroup$ – Answer Lee Apr 10 '18 at 22:00
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You have to use what the canonical embedding means. And, that you can identify $c_0^*$ with $\ell_1$.

If $a\in c_0$, you see it as an element of $c_0^{**}$ via $\hat a(b)=b(a)$, for each $b\in\ell_1$. That is, $$ \hat a(b)=\sum_na_nb_n. $$ This is precisely the duality $\ell_\infty=\ell_1^*$. In other words, if you are working with the characterizations $c_0^*=\ell_1$, and $\ell_1^*=\ell_\infty$, then the canonical embedding of $c_0$ in $c_0^{**}$ is precisely the embedding $c_0\subset\ell_\infty$.

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  • $\begingroup$ I am sorry I don't get it. We only have $\ell_1\cong {c_0}^*$ and ${\ell_1}^*\cong\ell_\infty$ and they are not equal. How to get the canonical embedding of $c_0$ in $c_0^{**}$ is precisely the embedding $c_0\subset\ell_\infty$? Thank you! $\endgroup$ – Answer Lee Apr 10 '18 at 23:49
  • $\begingroup$ It's not "only". The dual of $c_0$ is $\ell_1$. The dual of $\ell_1$ is $\ell_\infty$. They are not isomorphic in a mysterious way. They are isometrically isomorphic, via an isomorphism that preserves any data you may want: if you take $a\in c_0$ and $f\in c_0^*$, then there is a $b\in \ell_1$, with $\|b\|_1=\|f\|$, such that $f(a)=\sum_n a_nb_n$. And any such $b$ gives rise to an $f_b\in \ell_1$. $\endgroup$ – Martin Argerami Apr 11 '18 at 0:25
  • $\begingroup$ But why the canonical embedding of $c_0$ in $c_0^{**}$ is precisely the embedding $c_0\subset\ell_\infty$. Technically, we have isometrical isomorphism but they are not equal. How to replace $c_0^{**}$ with $\ell_\infty$? Thank you! $\endgroup$ – Answer Lee Apr 11 '18 at 1:15
  • $\begingroup$ By the way can I say there exists a bijection $T:\ell_\infty\rightarrow {c_0}^{**}$. Then they have the same cardinality. But $c_0\rightarrow\ell_\infty$ is not surjective, the cardinality of $c_0$ is less than $\ell_\infty$ and less than ${c_0}^{**}$. So, the canonical embedding $c_0\rightarrow {c_0}^{**}$ is not surjective. $\endgroup$ – Answer Lee Apr 11 '18 at 1:16
  • $\begingroup$ Cardinality has nothing to do with this. As I said in my answer, the embedding of $c_0$ in $c_0^{**}$ is explicit. And the embedding agrees precisely with the representation of $\ell_\infty$ as the dual of $\ell_1$. $\endgroup$ – Martin Argerami Apr 11 '18 at 1:35

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