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Let $E$ be a separable Banach space with symmetric basis $\{e_i\}$ (it is also called a symmetric sequence space). Let $\{x_i\}$ be a normalized disjoint sequence in $E$, i.e., $\lVert x_i\rVert_E=1$ and $x_i =\sum_{n=n_i}^{n_{i+1}-1}a_n e_n $ for some strictly increasing sequence $\{n_i\}$.

Assume that the uniform norm $\lVert x_i\rVert_\infty \to 0$ as $i\to \infty$ ($\lVert x\rVert_\infty=\sup \lvert a_n\rvert$, $x =\sum a_n e_n $). Assume, in addition, that $\{x_i\}$ is a symmetric basic sequence in $E$. Then, I guess, if $\{x_i\}$ is equivalent to $\{e_i\}$, then $E$ is equivalent to $\ell_p$ for some $p\in [1,\infty)$ or it is equivalent to $c_0$.

Such a property holds true for $\ell_p$ but I don't know any other symmetric sequence space having such a property.

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The question is not well formulated but i will answer the way I understood it. I think you are asking if there is any space with a symmetric basis other than $c_0, \ell_p$ which contains symmetric basic sequences with sup norm tends to zero and equivalent to the basis. The answer is yes. For instance, a minimal Orlicz sequence spaces $\ell_M$ not isomorphic to $\ell_p$ have such symmetric sequences, in fact, they are some constant coefficient blocks (with increasing support).

Minimal Orlicz sequence space means something else than a minimal Banach space. If you are not familiar with Orlicz space terminology you will need to skim through Chapter 4 of Lindenstaruss-Tzafriri's book. Here is a very quick pointers: For each $\ell_M$ there is associated set $E_{M,1}$ of Orlicz functions. If $N\in E_{M,1}$ then $\ell_N\subseteq \ell_M$ and moreover the basis of $\ell_N$ is some constant coefficient block basis. $\ell_M$ minimal means for every $N\in E_{M,1}$ we have $E_{N_,1}=E_{M,1}$. So in particular, if $E_{M,1}\neq \{t^p\}$, in such a space every constant coefficient blocks basis have a further blocks sequence which is equivalent to the basis. It is clear that the sup norm of such a sequence tends to zero.

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  • $\begingroup$ Many thx!that is exactly What I wanted $\endgroup$
    – user92646
    Commented Jan 8, 2022 at 21:30

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