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What I'm currently reading discusses the notion of spaces being well-isomorphic, specifically a Banach space containing well-isomorphic copies of $\ell_1^n$ for every $n\in\mathbb{N}$. The author doesn't define this, so I gather it's an elementary definition, but I gather that it means for every $n$ we have a linear operator $T_n$ such that $\| x\| \leq \|T_nx\|\leq D\|x\|$ for some $D\in(0,\infty)$.

However, the only place I've actually found a definition stated is in Isomorphisms Between H1 Spaces, by Paul Muller, who just gives the requirement that $\|T_n^{-1}\|\|T_n\| <\infty$. The definition that the author of the paper I'm reading seems to be using is stronger, I think, since we don't have any constant in front of the first $\|x\|$ in the inequality. It seems that we also need to know that $\|T_n^{-1}\|=1$.

What is the proper definition?

Edit: I'm reading "An Introduction to the Ribe Program", by Assaf Naor. It's publicly available. The passage I'm considering is at the bottom of page 17.

Michael Hardy
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user348871
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  • I have taken the liberty of changing your \mathcal{l} to \ell. I thought the \mathcal{l} didn't look particularly different from the ordinary lower-case form. $\qquad$ – Michael Hardy Jun 19 '16 at 18:24
  • No problem. Thanks! – user348871 Jun 19 '16 at 18:24
  • $\ldots,$Actually, it looked identical to the ordinary lower-case form $\qquad$ – Michael Hardy Jun 19 '16 at 18:25
  • If I were to guess, I would say that your partial definition has the quantifiers backwards. I.e., it ought to be “there is some $D$ so that for every $n$ …”. The lack of a constant in front of $|x|$ is likely not a problem, as you can always rescale $T_n$ so that constant becomes $1$. (The constant on the other side – $D$ – will have to pay the price.) – Harald Hanche-Olsen Jun 19 '16 at 18:39
  • @HaraldHanche-Olsen I suspect it you're correct and it must be something like that. I see how we could rescale any particular $T_n$ so that it works, but could we rescale all of them so that fixed $D$ still works? – user348871 Jun 19 '16 at 19:12
  • Well, what I am saying is that if the inequalities were given as $C|x|\le|T_nx|\le D|x|$ for all $n$, then you can rewrite that as $|x|\le|C^{-1}T_nx|\le C^{-1}D|x|$, so replacing $T_n$ by $C^{-1}T_n$ and $D$ by $C^{-1}D$, you have the inequalities stated in the question. – Harald Hanche-Olsen Jun 20 '16 at 20:40

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