# Subspaces of $l_p$ and Banach-Mazur distance

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 $||T||\cdot||T^{-1}||>K$.

OR

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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On MathOverflow this question would probably be tagged ask-johnson. Why not ask there? As far as I know, Bill Johnson does not read Math.SE. –  user53153 Jan 12 '13 at 21:17
Now asked and answered on mathoverflow. Theo, could you write a short summary and post that as an answer here to indicate that the matter is settled, at least after your bounty has expired? @PavelM: it didn't even take an hour after asking :) –  Martin Jan 13 '13 at 18:50
@Martin Yes, I will do that. –  Theo Jan 13 '13 at 22:22

This question was answered on mathoverflow, and the answer is the option 1) above. The construction is very non-trivial and it involves finding, for every $n$, $n$-dimensional subspaces of $l_p$ such that the $L_p$-factorization constant diverges to infinity. One way to construct such subspaces is a by-product of Szankowski's construction of subspaces of $l_p$ without the approximation property (see, for example, Theorem 1.g.4 in Lindenstrauss-Tzafriri, vol.2).