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Let $X=c_0$ and $X_2=c$ with norm $||x||_{\infty}$. Both $X_1^*$ and $X_2^*$ are isometrically isomorphic to $l_1$.

I know that $e_n\to 0$ in weak* topology in $l_1$ if I consider it as dual of $c_0$, but can I make a similar conclusion when I consider the other case.

I mean does $e_n\to 0$ in weak* topology in $l_1$, considered as dual of $c$. I am not able to prove it.

How can we prove it?

Edit: The space of convergent sequences c is a sequence space.

The subspace of null sequences c0 consists of all sequences whose limit is zero

$e_n=(0,\dots, 1\dots)$ , 1 at n_th position

Thanks.

Shweta Aggrawal
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If $x_n^{*} \to x^{*}$ in the weak topology of $X^{*}$ then $x_n^{*} (x)\to x^{*}(x)$ for every $x \in X$.

Consider $(1,1,...)$ as an element of $c$. This acts on $e_n $ to give you $1$ for every $n$ so $(e_n)$ does not converge weak*-ly in $l^{1}=c^{*}$ to $0$.

Kavi Rama Murthy
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  • I don't think that this is correct. What is the duality pairing between $\ell^1$ and $c$? It can't be the same as between $\ell^1$ and $c_0$… – Dirk Nov 27 '19 at 13:02
  • @dirk For $x\in\ell_1$ and $a\in c$, $xa= x_1\lim a + \sum_{i>1} x_ia_i$. (So, the above is correct.) – David Mitra Nov 27 '19 at 13:04
  • @Dirk For a sequence $(c_n)$ in $c_0$ the limit is $0$. There is a difference between the two cases. – Kavi Rama Murthy Nov 27 '19 at 13:05
  • @DavidMitra Ah, ok, sure! I thought that the notion of weak* convergence would be the same for different but isomorphic pre-dual space. Obviously that's wrong! – Dirk Nov 27 '19 at 13:09