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This is a question I posted on SE, and I have been advised to post it here.

It is well-known that every subspace of $l_2$ is isometric to $l_2$. When $p\neq 2$, $l_p$ has subspaces that are not even isomorphic, let alone isometric, to $l_p$. Suppose $X$ is a subspace of $l_p$ with $p\neq 2$ such that $X$ is isomorphic to $l_p$. What can one say about the Banach-Mazur distance between $X$ and $l_p$? More precisely, which one of the following mutually exclusive options holds true:

1) Given $K$, there exists a subspace $X$ of $l_p$, isomorphic to $l_p$, such that for any isomorphism $T:X\to l_p$ one has $||T||\cdot||T^{-1}||>K$.


2) There exist a constant $K$ (possibly depending on $p$), such that for any subspace $X$ of $l_p$, isomorphic to $l_p$, there exist an isomorphism $T:X\to l_p$ such that $||T||\cdot||T^{-1}||\leq K$.

Intuitively, I very strongly suspect it is 1) but I do not have an argument to exclude 2) and, if it is indeed 1), I would like to see a concrete example of a subspace having that property.

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up vote 8 down vote accepted

(1) is correct. It follows from the fact that there is a sequence $(E_n)$ of finite dimensional subspaces of $\ell_p$ s.t. $\gamma_p(E_n) \to \infty$. Here $\gamma_p(X)$ is the factorization constant of the identity on $X$ through an $L_p$ space; that is, $\gamma_p(X)= \inf \|T\|\cdot \|S\|$, where the infimum is over all $T:X\to Y$, $S:Y:\to X$, and $Y$ an $L_p$ space.

For fixed $n$, set $X_n=\ell_p(E_n)$. Since $E_n$ is finite dimensional, the space $X_n$ is isomorphic to $\ell_p$ (and embeds isometrically into $\ell_p$) but the Banach-Mazur distance of $X_n$ to $\ell_p$ is at least $\gamma_p(E_n)$.

There remains the sticky point of producing a sequence $(E_n)$ as above. One way is to take a subspace $X$ of $\ell_p$ that fails the approximation property and let $E_n$ be a sequence of finite dimensional subspaces of $X$ with $E_1\subset E_2 \subset \dots$ and $\cup E_n$ dense in $X$. Or, for $p<2$, $E_n$ can be (Banach-Mazur arbitrarily close to) $\ell_r^n$ with $p<r<2$, because $\ell_r$ embeds isometrically into $L_p(0,1)$ and there are various ways of checking that $\gamma_p(\ell_r^n)\to \infty$. Since I do not know your background (you have, unfortunately, chosen to remain anonymous), I do not know which approach of producing $(E_n)$ is best for you.

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Thank you, I am modestly familiar with Szankowski's construction, so I think I understand what the choice of $(E_n)$ can be. But it is not clear to me why $\gamma_p(E_n)\to\infty$, when $X$ fails AP. I can see that by looking at Szankowski's concrete definition (LT(II), Theorem 1.g.4 ) and taking $E_n=\span\{z_1, z_2,\dots z_n\}$ as in the proof, but I suppose there exists a simpler, abstract argument why any $(E_n)$ as above have the factorization constant going to infinity. – Theo Jan 13 '13 at 22:17
Yes, it is not obvious. What is clear is that $(\sum_{n=1}^\infty E_p)_p$ cannot have the uniform approximation property, while $\ell_p$ does. This does not require the specifics of Szankowski's construction but uses another concept, the uniform approximation property. AFAIK, there is no really simple proof using only elementary facts about $\ell_p$. $$ $$ Another approach is to show that there are finite dimensional subspaces $F_n$ of $\ell_p$ s.t. the GL-constants of $F_n$ tend to infinity, but that also uses a non elementary concept and requires more arguments. – Bill Johnson Jan 13 '13 at 23:38

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