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In this question (which I think may be interesting in its own) I was asking if we can find a copy of $\ell_\infty$ between a separable subspace $Y$ contained in $\ell_\infty(\Gamma)$ and the whole space. This was in order to complete a proof, but I've realised that even if that turns out to be false, the following lemma would also complete my proof:

Lemma: Let $X$ be a Banach space and $Y$ a separable subspace. Then there exists a non-empty set $\Gamma$, an isometry $S: X \rightarrow \ell_\infty(\Gamma)$ and a subset $W \subset \ell_\infty(\Gamma)$ such that $S(Y) \subset W$ and $W$ is linearly isometric to $\ell_\infty$.

If we suppose that $Y$ is locally complemented in $X$ (which we can because of Sims-Yost theorem), then we have that $$X^* = Y^* \oplus U$$ with $U \subset Y^\perp$. Then, it seems "reasonable" to think about finding a norming set of $X$ of the form $\{y_n^*\}_{n=1}^\infty \cup \{x_i^*\}_{i \in I}$ such that the $y_n^*$'s are also norming for $Y$ and $x_i^* \in Y^\perp$ for every $i \in I$. From there it is easy to construct the isometry $S$ in the usual way.

Any ideas of how to find the norming set? If not, any ideas about the proof (if it is true) of the lemma?

EDIT: The question linked at the beggining has been solved and it is easy to deduce the lemma from that question.

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In this reference (Definition 2.47) a Banach space $X$ is called separably automorphic if given a separable space $Y$ and isomorphic copies $A$ and $B$ of $Y$ in $X$, every bijective operator $T:A\to B$ can be extended to an automorphism (bijective operator) $\hat T$ of $X$.

The spaces $\ell_\infty(\Gamma)$ are separably automorphic by Proposition 2.52 in the reference.

Let $Y$ be a separable subspace of $\ell_\infty(\Gamma)$. Taking a numerable subset $\Gamma_1$ of $\Gamma$, we can identify $\ell_\infty(\Gamma_1)$ with a copy of $\ell_\infty$ in $\ell_\infty(\Gamma)$. Moreover $\ell_\infty(\Gamma_1)$ contains an isometric copy $Y_1$ of $Y$. Given a bijective isometry $T:Y_1\to Y$ and the automorphic extension $\hat T$ we mentioned, $\hat T(\ell_\infty(\Gamma_1))\supset Y$.

This result gives a positive answer to the initial question.

The Lemma can be proved with the same arguments.

ADDED on June 28, 2024: The above arguments prove the isomorphic version of the problem, but not the isometric version: in general $T$ isometry does not imply $\hat T$ isometry.

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  • $\begingroup$ Thanks a lot! In fact, we are working with your notion of universally separable injective spaces :) But I'm new and I hadn't notice that section of your book yet. Thanks again! $\endgroup$ Commented Jun 27 at 18:02
  • $\begingroup$ Now that I've look at it more carefully I was wondering the same thing about $\hat T(\ell_\infty(\Gamma_1))$ $\endgroup$ Commented Jun 28 at 7:05
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    $\begingroup$ I saw it yesterday. The argument does not prove the isometric case. In fact, if every isometry between separable subspaces of $X$ extends to an isometric isomorphism of $X$ then $X$ is a space of universal disposition, but$\ell_\infty(\Gamma)$ is not (Theorem 3.34). If you need the isometric case, you have to find a different argument. $\endgroup$ Commented Jun 28 at 7:27
  • $\begingroup$ In Theorem 2.40 it is proved that every separable subspace of $\ell_\infty/c_0$ is contained in an isometric copy of $\ell_\infty$. Maybe the argument can be adapted to $\ell_\infty(\Gamma)$. $\endgroup$ Commented Jun 28 at 7:53
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    $\begingroup$ My suggestion is to use the ideas in the proof for $\ell_\infty/c_0$ to prove the case $\ell_\infty(\Gamma)$, without using the quotient $\ell_\infty(\Gamma)/c_0(\Gamma)$. $\endgroup$ Commented Jun 28 at 12:07

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