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 $\ell_1$.
I 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.
 A: To see this, recall that a basic sequence $(x_n)_{n=1}^\infty$ is subsymmetric if it is unconditional and equivalent to all its subsequences. If $(x_n)_{n=1}^\infty$ contains a basic sequence equivalent to the standard basis for $\ell_1$, by subsymmetry, it is itself equivalent to the basis for $\ell_1$ and we are done. 
Consider then the case where $(x_n)_{n=1}^\infty$ does not contain any $\ell_1$-sequences. Without loss of generality we may suppose that $(x_n)_{n=1}^\infty$ is (semi-)normalised, because otherwise, it would have a null subsequence thence it would have been null itself. I claim that $(x_n)_{n=1}^\infty$ is weakly null. 
Assume not. Then passing to a subsequence if necessary (which does not hurt us as the basis is subsymmetric), we infer the existence of a norm-one linear functional $x^*$ and $\delta > 0$ such that $\langle x^*, x_n\rangle \geqslant \delta$ for all $n$. For any finite sequence of scalars $(a_k)_{k=1}^N$ we obtain
$$\left\|\sum_{k=1}^N a_k x_k \right\| \geqslant \langle x^*, \sum_{k=1}^N a_k x_k \rangle \geqslant {\delta} \sum_{k=1}^N a_k.$$
Let $K\geqslant 1$ be the unconditionality constant of $(x_n)_{n=1}^\infty$. We have
$$K\cdot \left\|\sum_{k=1}^N a_k x_k \right\| \geqslant \left\|\sum_{k=1}^N |a_k| x_k \right\| \geqslant {\delta} \sum_{k=1}^N |a_k|$$
hence
$$\left\|\sum_{k=1}^N a_k x_k \right\| \geqslant \frac{1}{K}\cdot\left\|\sum_{k=1}^N a_k x_k \right\| \geqslant \frac{\delta}{K} \sum_{k=1}^N |a_k|,$$
so $(x_n)_{n=1}^\infty$ is equivalent to the standard basis for $\ell_1$; a contradiction. $\square$
On a related note, it is good to bear in mind that if $(x_n)_{n=1}^\infty$ is a basic sequence and $x$ is its cluster point with respect to the weak topology then $x=0$. Indeed, by Mazur's theorem, $x$ belongs to the closed linear span of $(x_n)_{n=1}^\infty$ so $$x = \sum_{n=1}^\infty \langle x_n^*, x\rangle x_n,$$ where $(x_n^*)_{n=1}^\infty$ denotes the sequence of associated coordinate functionals. For every $n$ the number $\langle x_n^*, x\rangle$ is a cluster point of $(\langle x_n^*, x_m\rangle)_{m=1}^\infty$, so by the $n^{{\rm th}}$ term test it must be zero. Thus $x=0$.
