Classification of subsymmetric basic sequences I just encountered the following statement: any subsymmetric basic sequence is either weakly null or equivalent to the unit vector basis of $\ell_1$.
It's the first time I realize that. I do see the case in which it is equivalent to the unit vector basis of $\ell_1$. However, I cannot prove that if that's not the case, it must be weakly null. Is this "trivial"? Also, while thinking about that, I came up with a couple of questions: is every basic sequence bounded away from zero? What about a subsymmetric basic sequence? So far I cannot come up with a counterexample.
I'm not looking for a complete answer. A hint or reference would suffice.
 A: In the theorem you quote, part of the definition of subsymmetric includes the hypothesis that the sequence is unconditionally basic.  If you do not include this in the definition of subsymmetric, then the summing basis for $c_0$ is an example of a weakly Cauchy subsymmetric basis that is not equivalent to the unit vector basis of $\ell_1$.
A: I would say it is "routine" but not quite "trivial."  The key ingredient is the following.
Proposition.  If $(x_n)_{n=1}^\infty$ is a subsymmetric basic sequence and $(y_k)_{k=1}^\infty$ is a seminormalized block basic sequence of $(x_n)_{n=1}^\infty$ such that $\sup_{k\in\mathbb{N}}\#\text{supp }y_k<\infty$ then $(y_k)_{k=1}^\infty$ is equivalent to $(x_n)_{n=1}^\infty$.
Here, "$\text{supp }y_k$" (the "support" of $y_k$) means the indices of $(x_n)_{n=1}^\infty$ which have nonzero coefficients when forming $y_k$.  The proof appears in the Altshuler/Casazza/Lin paper "On symmetric basic sequences in Lorentz sequence spaces" (1973) (as "Proposition 3").  It is given there for symmetric basic sequences, but valid for subsymmetric ones too.
Now just apply Rosenthal's $\ell_1$ Theorem.  If $(x_n)_{n=1}^\infty$ is subsymmetric and not equivalent to $\ell_1$ then it has a weak Cauchy subsequence, which means we can form a "difference sequence" $(x_{n_{2k+1}}-x_{n_{2k}})_{k=1}^\infty$ which is seminormalized and weakly null.  It is equivalent to $(x_n)_{n=1}^\infty$ by the above result.
A: A more detailed version of the above proof:
Assume that $(x_n)$ is not weakly null. Then, passing to a subsequence, we get a functional $x^*$ such that $\inf_n |x^*(x_n)|>0$. By unconditionality, there are  $c_1$, $c_2$ $c_3>0$ such that 
$$
\Vert \sum_n a_n x_n\Vert \ge
c_1 \Vert \sum_n \epsilon_n a_n x_n\Vert \ge
c_2 \vert \sum_n  \epsilon_n a_n  x^*(x_n) \vert=
c_2  \sum_n  \vert a_n  x^*(x_n) \vert
 \ge c_3 \sum_n \vert a_n\vert,
$$
where $(\epsilon_n)$ is a suitable choice of signs.
