Let $X$ be a Banach space and $Y$ be a separable closed subspace of $X^{*}$. Is there a separable closed subspace $Z$ of $X$ such that $Y$ is isomorphic to a subspace of $Z^{*}$? Thank you!
$\begingroup$
$\endgroup$
asked Aug 26, 2016 at 21:15
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Yes. Take a countable dense subset $A$ in $Y$. For each element $a\in A$ take a countable sequence $x_n$ in a unit ball of $X$ such that $a(x_n)$ tends to $\|a\|$. Let $Z$ be a closed span of all these sequences.
answered Aug 26, 2016 at 21:27
-
$\begingroup$ Thanks, Fedor! Indeed, we can choose $Z$ so that $Y$ is $(1+\epsilon)$-isomorphic to a subspace of $Z^{*}$. $\endgroup$ Commented Aug 26, 2016 at 22:26
-
-
$\begingroup$ I am not sure that it is isometric. Could you give a detailed proof? $\endgroup$ Commented Aug 26, 2016 at 22:34
-
$\begingroup$ The norm of each element in the unit ball of $Y$ exceeds $1-\varepsilon$ in $Z^{*}$, yes? $\endgroup$ Commented Aug 26, 2016 at 23:39
-