A subsymmetric basic sequence that is not weakly null is seminormalised

Question

I'm trying to understand the proof that a subsymmetric basic sequence in a Banach space is either weakly null or equivalent to the unit vector basis of $l_1$. This question was already asked and answered in this post, and I think I understand the structure of the argument (though I wonder if it can be done without contradiction).

The part I'm stuck at is when we assume the basic sequence $(x_n)$ is not equivalent to the unit vector basis of $l_1$. Towards a contradiction, suppose that $(x_n)$ is not weakly null. Then no subsequence of $(x_n)$ is, so in particular, no subsequence of $(x_n)$ converges to $0$. So there is $m>0$ such that $m \le |\!|x_n|\!|$ for all $n$. The answerer to the other post then says here that we can assume $(x_n)$ to be seminormalised, i.e., there exists $C\ge 1$ such that $$1/C \le |\!|x_n|\!| \le C$$ for all $n$>. I believe the lower bound but not the upper bound. I think I can follow the rest of the proof, so it's really just this snag that I'm hitting.

Is this where we are supposed to use the hypothesis that $(x_n)$ is not equivalent to the unit vector basis of $l_1$? I don't think it's used anywhere else in the proof, and if it isn't, then I can't see why one can't just structure the argument as a ``Weakly null? Done. Not weakly null? Then equivalent to canonical basis of $l_1$.'' sort of thing.

I realise there are a lot of definitions here, so I'll list them below. Perhaps I have a definition wrong, and that is what is causing confusion.

Definitions. A (Schauder) basis is a sequence $(x_n)$ of vectors such that every $x$ in the Banach space $X$ has a unique expression as $$x = \sum_{n=1}^\infty a_n x_n.$$ A sequence that is a basis for its closed linear span is a basic sequence. Two bases $(x_n)$ of $X$ and $(y_n)$ of $Y$ are equivalent if the map taking $x_n\mapsto y_n$ for all $n$ extends to a linear homeomorphism from $X$ to $Y$. This is the same as saying that there exist constants $0<m\le M$ such that $$ m\Big|\!\Big| \sum_{n=1}^\infty a_n x_n\Big|\!\Big| \le \Big|\!\Big| \sum_{n=1}^\infty a_n y_n\Big|\!\Big| \le M\Big|\!\Big| \sum_{n=1}^\infty a_n x_n\Big|\!\Big|. $$ for all sequences $(a_n)$ of scalars.

A sequence $(x_n)$ is seminormalised if for some $C\ge 1$, we have $1/C\le |\!|x_n|\!| \le C$ for all $n$, and it is normalised if $C=1$. A sequence is weakly null if it converges to $0$ in the weak topology. And the unit vector basis of $l_1$ is given by $(e_n)$, where $e_n$ has a $1$ in the $n$th coordinate and zeroes elsewhere.

Answer 1

Probably a bit late to answer this, but I am relatively new to the site and I just saw your question. For the record let me give a couple of proofs that every non-trivial spreading sequence is bounded (by spreading I mean that the sequence is equivalent to all its subsequences, which is one of the conditions for it being subsymmetric). First one observation: if $(x_n)_n$ is a spreading sequence and we have $x_{n_0} = 0$ for some $n_0$, then automatically $x_n = 0$ for every $n$. This follows from the fact that, given $n$ arbitrary, the sequences $(x_k)_{k\geq n_0}$ and $(x_k)_{k\geq n}$ are equivalent, so in particular $\|x_n\| \leq M \|x_{n_0}\| = 0$ (where $M$ is the equivalence constant of these two sequences). Thus I call an spreading sequence trivial if it is identically zero, and non-trivial if some (equivalently all) $x_n$ is non zero. With this distinction we have the following.

Proposition: Every spreading sequence is bounded. If the sequence is non-trivial, then it is seminormalized.

Proof. Assume for a contradiction that it is not bounded. Find $n_0\in\mathbb{N}$ such that $\|x_{n_0}\| \geq 1$ and, once $n_k$ has been chosen, find $n_{k+1} > n_k$ such that $\|x_{n_{k+1}}\| \geq 2\|x_n\|$. Note that with these choices we get $\|x_{n_k}\| \geq 2^{k-j} \|x_{n_j}\|$ for every $j < k$. Now the subsequences $(x_{n_k})_k$ and $(x_{n_{2k}})_k$ must be equivalent, so there exists $M > 0$ such that in particular $$ \frac{1}{M} \|x_{n_k}\| \leq \|x_{n_{2k}}\| \leq M \|x_{n_k}\| $$ for every $k$. But the second inequality and the choice of $x_{n_k}$ imply that $$ 2^{2k - k} \|x_{n_k}\| \leq \|x_{n_{2k}}\| \leq M \|x_{n_k}\|, $$ so $M \geq 2^k$ for every $k$, which is absurd. hence $(x_n)_n$ is bounded.

If the sequence is non-trivial then the lower bound is proven similarly. Assume it is not bounded below, take $n_0\in\mathbb{N}$ such that $\|x_{n_0}\| \leq 1$ and, once $n_k$ has been chosen, find $n_{k+1} > n_k$ such that $\|x_{n_{k+1}}\| \leq \|x_{n_k}\|/2$. This time we have $\|x_{n_k}\| \leq 2^{j-k} \|x_{n_j}\|$ for every $j < k$. Since the subsequence $(x_{n_k})_k$ and $(x_{n_{2k}})_k$ are equivalent, find again $M > 0$ such that, in particular, $$ \frac{1}{M} \|x_{n_k}\| \leq \|x_{n_{2k}}\| \leq M \|x_{n_k}\| $$ for every $k$. Now the first inequality and the choice of $x_{n_k}$ give $$ \frac{1}{M} \|x_{n_k}\| \leq \|x_{n_{2k}}\| \leq 2^{k-2k} \|x_{n_k}\|. $$ Since all $x_{n_k}$ are non-zero by assumption we can cancel out $\|x_{n_k}\|$ and get $M \geq 2^k$ for all $k$, absurd again. $\square$

Note how we didn't need $(x_n)_n$ to be subsymmetric, or even basic! For basic sequences (again without unconditionality) we can use a different proof, which is an adaptation of the one in the book by I. Singer Bases in Banach Spaces I (Chapter II, Proposition 2.21.4, page 569). This one is based on the fact that two basic sequences $(x_n)_n$ and $(y_n)_n$ are equivalent if and only if whenever $\sum_n a_n x_n$ converges also $\sum_n a_n y_n$ converges and vice-versa. This is equivalent to your definition of equivalence only for basic sequences!

Alternative proof for basic sequences. Assume that $(x_n)_n$ is spreading and basic, and let us show that it is seminormalized. Clearly in this case all $x_n$ are non-zero. Again assume first that it is not bounded, so there exists a subsequence $(x_{n_k})_k$ with $\|x_{n_k}\| \geq k$. Choose an increasing sequence of integers $(k_j)_j$ such that $k_j \geq 2^j \|x_j\|$. Then the series $\sum_{n=1}^{\infty} \frac{2^{-n}}{\|x_n\|} x_n$ converges (absolutely, in fact), but the series $\sum_{j=1}^{\infty} \frac{2^{-j}}{\|x_j\|} x_{n_{k_j}}$ does not because $$ \frac{2^{-j}}{\|x_j\|} \|x_{n_{k_j}}\| \geq \frac{2^{-j}}{\|x_j\|} k_j \geq 1, $$ so its general term does not converge to $0$. This contradicts the fact that $(x_n)_n$ and $(x_{n_{k_j}})_j$ are equivalent. The lower bound is proven analogously, by contradiction, comparing $\sum_{n=0}^{\infty} \|x_n\|^{-1} x_n$ with $\sum_{j=0}^{\infty} \|x_j\|^{-1} x_{n_{k_j}}$ for appropriately chosen indices $k_j$. $\square$

Regarding your question about this post the answer is yes, you can perfectly re-write the argument into a $``$Weakly null? Done. Not weakly null? Then equivalent to $\ell_1$.$"$ In order to do so assume that $(x_n)_n$ is subsymmetric but not weakly null. This means that there exists $x^*\in X^*$, which we can assume has norm $1$, for which $(x^*(x_n))_n$ does not converge to $0$. Again this means that there exists a subsequence $(x_{n_k})_k$ and some $\delta > 0$ such that $|x^*(x_{n_k})|\geq \delta$ for every $k$. The argument of the above mentioned post shows that $(x_{n_k})_k$ is equivalent to the canonical $\ell_1$ basis, but since $(x_{n_k})_k$ is equivalent to $(x_n)_n$, this one is also equivalent to $\ell_1$.