# Embedding of a Banach space in $\ell_\infty(\Gamma)$ with a subspace embedded in a copy of $\ell_\infty$

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.

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.

• 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! Commented Jun 27 at 18:02
• Now that I've look at it more carefully I was wondering the same thing about $\hat T(\ell_\infty(\Gamma_1))$ Commented Jun 28 at 7:05
• 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. Commented Jun 28 at 7:27
• 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 $\ell_\infty(\Gamma)$. Commented Jun 28 at 7:53
• 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 $\ell_\infty(\Gamma)/c_0(\Gamma)$. Commented Jun 28 at 12:07