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One of the characterizing properties of the canonical bases of the $\ell_p/c_0$ spaces is their perfect homogeneity (Zippin), so that every normalized block sequence in these spaces behaves like the canonical basis. This question can be thought of as a partial converse. Roughly, if we have long enough (with specific quantitative dependence on the dimensions) sequence in a space isomorphic to $\ell_p$, must it be true that in the span of this sequence, we can find a shorter sequence which is close to being a block of the basis?

Suppose that for some $1<p<\infty$ and $C\geq 1$, $X$ is a Banach space such that $d_{BM}(X,\ell_p)< C$. That is, there exists an isomorphism $T:\ell_p\to X$ with $\|T\|\|T^{-1}\|<C$. Suppose also that for such an isomorphism $T$, $x_n=Te_n$, where $(e_n)_{n=1}^\infty$ is the canonical basis of $\ell_p$. Let $b\geq 1$ be fixed. For each natural number $d$ and each $\epsilon>0$, does there exist a natural number $D=D(p,C,b,n,\epsilon)$ such that if $(u_i)_{i=1}^D\subset X$ is a sequence satisfying $$b^{-p}\sum_{i=1}^D |a_i|^p \leq \Bigl\|\sum_{i=1}^D a_iu_i\Bigr\|_X^p \leq b^p\sum_{i=1}^D |a_i|^p$$ for all scalars $(a_i)_{i=1}^D$, then there exist $0=r_0<\ldots < r_n\leq D$ and scalars $(b_j)_{j=1}^D$ such that, with $y_i=\sum_{j=r_{i-1}+1}^{r_i}b_ju_j$,

  1. $\sum_{j=r_{i-1}+1}^{r_i}|b_j|^p=1$ for all $1\leq i \leq n$,
  2. $(y_i)_{i=1}^n$ is an $\epsilon$-perturbation of a block sequence with respect to $(x_i)_{i=1}^\infty$ (that is, $\sum_{i=1}^n \|y_i-z_i\|\leq \epsilon$ for some sequence $(z_i)_{i=1}^n$ of successively supported vectors with respect to the basis $(x_i)_{i=1}^\infty$)?
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    $\begingroup$ Except for being obviously false for $p=2$, this is a good question. But I do not answer questions from new users who do not use their real names. $\endgroup$
    – Bill Johnson
    Jun 26, 2022 at 18:18
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    $\begingroup$ Take a sequence $(u_i)_1^D$ of disjointly supported but with interlacing supports. For instance, take $nD$ basis vectors for $n>D$ is arbitrary, and each $u_i$ is sum of $i'$th basis vectors from each block of $D$ vectors and normalized (each coefficient is $1/n^p$). Then any $y_i$'s as you defined are also interlacing and it won't be close to a block sequence of the basis. The point is that quantitatively you cannot distinguish disjointly supported vectors from successive blocks. $\endgroup$
    – Bunyamin Sari
    Jun 26, 2022 at 18:43

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