Let us say that two sequences $(x_n)_n$ and $(y_n)_n$ in (possibly different) Banach spaces $X$ and $Y$ are $\lambda$-equivalent, with $\lambda\geq 1$, if for every sequence of coefficients $(a_n)_n\in c_{00}$ we have $$ \frac{1}{\lambda}\left\lVert\sum_{n\in\mathbb N} a_n x_n\right\rVert_X \leq \left\lVert\sum_{n\in\mathbb N} a_n y_n\right\rVert_Y \leq \lambda \left\lVert\sum_{n\in\mathbb N} a_n x_n\right\rVert_X. $$ They are said to be equivalent if they are $\lambda$-equivalent for some $\lambda\geq 1$, or equivalently, if there exists an isomorphism $T: [x_n]\to [y_n]$ (where $[x_n]$ denotes the closed linear span of the sequence) with $Tx_n = y_n$ for every $n$. In this case the equivalence constant is easily seen to be $\lambda = \max\{\|T\|, \|T^{-1}\|\}$. Note that we do not assume any of the sequences to be basic (i.e. a Schauder basis of their closed linear span).
Now a spreading sequence is a sequence $(x_n)_n$ in some Banach space which is equivalent to each one of its subsequences. My question is whether we always have uniformity in the equivalence constant, or to be more precise:
Question: If $(x_n)_n$ is a spreading sequence, is there some $\lambda\geq 1$ such that $(x_n)_n$ is $\lambda$-equivalent to each of its subsequences? And if we assume $(x_n)_n$ to be at least basic?
Not every spreading sequence is basic, for example $(e_0 + e_{n+1})_n$ in $c_0$ (where $(e_n)_n$ is the canonical basis), is easily seen to be $1$-equivalent to each of its subsequences, but it's not basic since it converges weakly to $e_0\neq 0$. However, in most books these sequences are treated under additional hypothesis, the one I've seen the most being that the sequence is also unconditional basic (in which case it is called subsymmetric). But not every spreading basic sequence is subsymmetric, as the summing basis of $c_0$ shows.
Intuition: If we assume that the sequence is subsymmetric then the answer is yes: a proof can be found on Theorem 21.2 of Bases in Banach Spaces I by I. Singer (it follows easily from the implication 1º$\implies$2º). Another proof for symmetric sequences (those which are unconditional and equivalent to all their permutations, a particular case of subsymmetric ones) can be found on Lemma 9.2.2 of Topics in Banach Space Theory by Albiac and Kalton. However, the proofs rely heavily on unconditionality for the handling of the series involved. Also, all the examples of spreading sequences I've been able to find, which are not many, satisfy this uniformity in the equivalence constant.
My small attempt: Looking at the proof in Singer's book my idea would be as follows. Fix an spreading sequence $(x_n)_n$ in $X$. For each infinite subset $M = \{m_0 < m_1 < \ldots\}\subseteq\mathbb N$ define the isomorphism $T_M: [x_n]\to [x_m : m\in M]$ by $Tx_j := x_{m_j}$. My goal is now to prove that $$ \sup\{\|T_M\| : \text{$M\subseteq\mathbb N$ infinite}\} < \infty. \tag{1}\label{eq1} $$ Assume otherwise and find, through the Uniform Boundedness Principle, some $x\in [x_n]$ such that $$ \sup\{\|T_Mx\| : \text{$M\subseteq\mathbb N$ infinite}\} = \infty. $$ I would like to find some $M$, maybe through some sort of diagonal argument, for which the series defining $T_Mx$ does not converge, which would be a contradiction. My two problems with this attempt are that:
- I am unable to construct this $M$ I'm looking for. Start finding $M^0\subseteq\mathbb N$ such that $\|T_{M^0} x\| > 1$, and approximate $x$ with finite linear combinations of the $x_n$'s to reach $$ \left\lVert T_{M^0}\!\left(\sum_{j=0}^{n_0} a_j x_j\right)\right\rVert \geq 1 $$ for some $n_0\in\mathbb N$ and some $a_j\in\mathbb K$. I would perhaps like to prove now that $$ \sup\{\|T_Mx\| : \text{$M\subseteq\mathbb N$ infinite with $\min M > n_0$}\} = \infty $$ to keep going, but I'm not even sure how to do that, even assuming that $(x_n)_n$ is basic.
- Even if I were able to prove \eqref{eq1}, that does not conclude the question. The equivalence constant between $(x_n)_n$ and $(x_{m_j})_j$ (the $\lambda$ in the definition at the beginning of the question) is $\max\{\|T_M\|, \|T_M^{-1}\|\}$, and I have no idea on how to bound $\|T_M^{-1}\|$.
Any help with completing this proof, or with any other proof, counterexample or reference to the question above, would be greatly appreciated.