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Let $p\in [1,\infty]$ and let $X$ be a separable Banach space. Then $X$ is said to be a $\mathscr{L}_p$-space if there exists an increasing union of finite dimensional Banach spaces $F_n\subset F_{n+1}\subset \ldots$ with union dense in $X$ such that for some $\lambda>0$ the Banach-Mazur distance satisfies $d(F_n, \ell_p^{\dim F_n}) \leqslant \lambda$ for all $n$.

Many authors use without explanation the following definition:

$X$ is a $\mathscr{L}_p$-space if there exists an increasing net $(F_\alpha)$ of finite-dimensional subspaces with $X=\bigcup_\alpha F_\alpha$ such that for some $\lambda>0$ the Banach-Mazur distance satisfies $d(F_\alpha, \ell_p^{\dim F_\alpha}) \leqslant \lambda$ for all $\alpha$.

Why do these definitions turn out to be equivalent?

Secondly, is there anything special about the choice of $\ell_p$ here? Let us generalise this in the following manner.

Let $E_1 \subset E_2 \ldots $ be an increasing sequence of isometric embedding of finite-dimensional Banach spaces. Let $E$ be the direct limit of this sequence. Let us sat that $X$ is an $\mathscr{L}_E$-space if there exists an increasing net $(F_\alpha)$ of finite-dimensional subspaces with $X=\bigcup_\alpha F_\alpha$ such that for some $\lambda>0$ the Banach-Mazur distance satisfies $d(F_\alpha, E_{\dim F_\alpha}) \leqslant \lambda$ for all $\alpha$.

Suppose that $X$ is a separable Banach space. Suppose moreover that there exists an increasing union of finite dimensional Banach spaces $F_n\subset F_{n+1}\subset \ldots$ with dense union such that for some $\lambda>0$ the Banach-Mazur distance satisfies $d(F_n, E_{\dim F_n}) \leqslant \lambda$ for all $n$.

Is $X$ a $\mathscr{L}_E$-space?

If so,

Are ultrapowers of $\mathscr{L}_E$-spaces $\mathscr{L}_E$ as well?

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    $\begingroup$ It is really, really impolite to ask a question, get an answer, and vanish. $\endgroup$ Nov 21, 2015 at 20:06

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The equivalence of the two definitions of $\mathscr{L}_p$-space involves an elementary perturbation argument that is probably contained in the original article of Lindenstrauss and Pelczynski. That kind of argument is certainly contained in my paper with Rosenthal and Zippin.

Your definition of $\mathscr{L}_E$-space is not a good definition. You can write $L_p$ for $1\le p < n$ as (the closure of) an increasing union of subspaces $E_n$ with $E_n$ uniformly isomorphic to $\ell_p^n \oplus_p \ell_r^n$ with $p<r<2$. Then $L_p \oplus L_r$ is $\mathscr{L}_{L_p}$ according to your definition but it is not $\mathscr{L}_p$. That is, your definition of $\mathscr{L}_E$ depends on the paving of $E$ that you take.

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  • $\begingroup$ It is worth mentioning that the definitions are equivalent "up to an $\varepsilon$". For example, $C(0,1)$ is an $\mathcal{L}_{\infty,1}$-space in the first sense and only $\mathcal{L}_{\infty,1+\varepsilon}$-space in the second sense. $\endgroup$ Nov 21, 2015 at 23:45

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