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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.

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  • $\begingroup$ My questions about boundedness came to mind while remembering Rosenthal's $\ell_1$ Theorem. If every subsymmetric basic sequence is bounded, then that theorem would allow you to conclude that a subsequence is either weakly Cauchy or equivalent to the canonical basis of $\ell_1$. But a weakly Cauchy sequence need not be weakly null, right? Anyway, the conclusion would not be the same as the conclusion of the statement that originated my question. $\endgroup$
    – ragrigg
    Commented Mar 16, 2016 at 3:53
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    $\begingroup$ Not every basic sequence is bounded or bounded away from zero. In fact, for any basic sequence $(x_n)_{n=1}^\infty$ and any sequence of nonzero scalars $(a_n)_{n=1}^\infty$, the sequence $(a_nx_n)_{n=1}^\infty$ is basic. However, if $(x_n)_{n=1}^\infty$ is subsymmetric then it must be seminormalized, i.e. admit $M\in(0,\infty)$ such that $M^{-1}\leq\|x_n\|\leq M$ for every $n\in\mathbb{N}$. $\endgroup$
    – Ben W
    Commented Mar 16, 2016 at 11:55
  • $\begingroup$ Cross-posted at MSe: math.stackexchange.com/questions/1699572/… $\endgroup$
    – Tomasz Kania
    Commented Mar 19, 2016 at 20:22

3 Answers 3

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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 $\ell_1$.

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  • $\begingroup$ The definition that I am using for subsymmetric sequence requires the sequence to be unconditional. Under this definition, is a subsymmetric sequence always bounded away from zero? Do you know/have a reference/hint related to the dichotomy for subsymmetric sequences I stumbled upon? $\endgroup$
    – ragrigg
    Commented Mar 16, 2016 at 4:52
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    $\begingroup$ Yes, it is trivial that a subsymmetric basic sequence is bounded and bounded away from zero. If a basis is subsymmetric and not weakly null, then there is a bounded linear functional that is bounded away from zero on a subsequence of the basis, hence, by subsymmetry, there is another bounded linear functional that is bounded away from zero on the basis itself. Unconditionality then gives that the basis is equivalent to the unit vector basis of $\ell_1$. $\endgroup$
    – Bill Johnson
    Commented Mar 16, 2016 at 12:53
  • $\begingroup$ Oh cool, that's a much simpler proof than I had in mind below. $\endgroup$
    – Ben W
    Commented Mar 16, 2016 at 13:49
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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.

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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.

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