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Let $X$ be a Banach space with a Schauder basis, and $Y$ a Banach space. Let $P_N$ denote the coordinate projections relative to the basis of $X$, and let $X_N$ denote their ranges. Specifically, $\{x_n\}_{n=1}^{\infty}$ is a basis in $X$ and $$P_N(x)=\sum_{n=1}^Na_nx_n,\quad\hbox{with}\quad x=\sum_{n=1}^{\infty}a_nx_n$$ Assume that for each $N$, the finite dimensional space $X_N$ is isometric to a subspace of $Y$. Does it follow that $X$ is isometric to a subspace of $Y$?

If $Y=L_p$ with $1\leq p\leq 2$ then the answer is affirmative, but the general case escapes me.

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It is false for $L_p$. Consider the Haar basis. The span of the first $n$ basis vectors is isometrically isomorphic $\ell_p^n$, but $L_p$ does not embed even isomorphically into $\ell_p$ if $p\not=2$.

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  • $\begingroup$ Thank you, When $Y=L_p$ with $1\leq p\leq 2$ I argued that $\exp(-|\!|x|\!|^p)$ is positive definite, and concluded from the Bretagnolle - Dacunha Castelle - Krivine theorem that $X$ does embed isometrically into $L_p$ if all the basis projections do, and I was wondering whether I was using a much too heavy tool. Apparently I wasn't . $\endgroup$
    – Yossi Lonke
    Nov 19, 2018 at 21:40
  • $\begingroup$ Another way to prove it for $Y=L_p$ is to see that $X$ must isometrically embed into an ultra power of $L_p$. An ultra power of $L_p$ is $L_p(\mu)$ for some measure $\mu$ (generalized Kakutani representation theorem) and separable subspaces of $L_p(\mu)$ embed isometrically into $L_p$. This argument works for $p>2$ as well. $\endgroup$
    – Bill Johnson
    Nov 19, 2018 at 23:55
  • $\begingroup$ I see. Is that how corollary 1 ("$L_p(0,1)$ contains an isometric copy of $\ell_2$") in Lindenstrauss-Pelczynski (1968) is proved for $X=\ell_2$? In proposition 7.5 they first embed $\ell_2^n$ isometrically into $L_p(0,1)$ by integrating $|\langle x,\cdot\rangle|^p$ on the unit sphere $S^{n-1}$ with the invariant measure, and then deduce corollary 1, with no proof. Is the argument you brought above behind this? Why didn't they look at an infinite sequence of i.d. normalized gaussians? Is it because they wanted to avoid an abstract probability space hosting these gaussians? $\endgroup$
    – Yossi Lonke
    Nov 20, 2018 at 8:46
  • $\begingroup$ I don't know, Yossi. They certainly knew some version of the argument I gave. Of course they knew the argument using gaussians. It does look like a strange choice. Maybe that is why they did not give a proof. :) $\endgroup$
    – Bill Johnson
    Nov 20, 2018 at 19:43
  • $\begingroup$ Thanks again for this discussion, Bill. (It's been a while). $\endgroup$
    – Yossi Lonke
    Nov 20, 2018 at 21:21

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