Proof Choose for every $r\in \mathbb {N}$ a norming functional $f_{r}\in W$ so that $f_{r}(x_{r})=1$ and so that $\text {supp}(f_{r})\subset \text {supp}(x_{r})$ . Then, there are two cases to consider.

1. (i) $\limsup _{r\to \infty }w(f_{r})=L<\infty $

Here, we may pass to an infinite subset $\Omega _{1}\subset \mathbb {N}$ so that $w(f_{r})=m_{j_{1}}\cdots m_{j_{l}}$ for each $r\in \Omega _{1}$ . Note that each $f_{r}$ with $r\in \Omega _{1}$ is then $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda _{r})$ -packed for some $\lambda _{r}\in \overline {\mathcal {D}}$ . We may then pass to a further infinite subset $\Omega _{2}\subset \Omega _{1}$ so that $d(\lambda _{r},\lambda )\leq \frac {1}{2}\cdot 2^{-\max \text {supp}(x_{r-1})}$ for each $r\in \Omega _{2}\setminus \{\min (\Omega _{2})\}$ and for some $\lambda \in \overline {\mathcal {D}}$ . Now, let $\Omega =\Omega _{2}$ and consider the subsequence $(x_{r})_{r\in \Omega _{2}}$ . Let $\Gamma \subset \Omega $ be a subset such that the block vectors $(x_{r})_{r\in \Gamma }$ are $S_{n_{j_{0}}}$ -admissible and let

$$ \begin{align*} f_{r}=\frac{1}{m_{j_{1}}\cdots m_{j_{l}}}\sum_{q=1}^{d_{r}}f^{r}_{q} \qquad r\in\Gamma \end{align*} $$

be the associated norming functionals. For each $r\in \Gamma $ , let $F^{r}_{1}<\ldots <F^{r}_{M_{r}}$ be the partition of $\{1,\ldots ,d_{r}\}$ and let $\lambda ^{r}_{1},\ldots ,\lambda ^{r}_{M_{r}}\in \overline {\mathcal {D}}$ be the corresponding nodes so that Definition 3.2 is satisfied because the functionals $f^{r}_{1}<\ldots <f^{r}_{d_{r}}$ are $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda _{r})$ -packed. We will now show that the concatenation

$$ \begin{align*} f^{\min(\Gamma)}_{\min(F^{\min(\Gamma)}_{2})}<\ldots<f^{\min(\Gamma)}_{d_{\min(\Gamma)}}<\ldots<f^{\max(\Gamma)}_{\min(F_{2}^{\max(\Gamma)})}<\ldots<f^{\max(\Gamma)}_{d_{\max(\Gamma)}} \end{align*} $$

is in W. First, note that there is some $\kappa \in \mathbb {N}$ so that each functional in this concatenation is a member of $W_{\kappa }$ . It therefore remains to show that these functionals are $(n_{j_{1}},\ldots ,n_{j_{l}},n_{j_{0}},\lambda )$ -packed. Indeed, it follows immediately that they are $S_{n_{j_{1}}+\cdots +n_{j_{l}}+n_{j_{0}}}$ -admissible and are very fast-growing in the sense of Definition 3.2. Then, let $F_{\min (\Gamma )}<\ldots <F_{\max (\Gamma )}$ be the partition of the number of functionals in the above concatenation such that $(f_{i})_{i\in F_{r}}=\{f^{r}_{\min (F^{r}_{2})},\ldots ,f^{r}_{d_{r}}\}$ and note that $\{f_{\min (F_{r})}\}_{r\in \Gamma }=\{f^{r}_{\min (F^{r}_{2})}\}_{r\in \Gamma }$ is $S_{n_{j_{0}}}$ -admissible. Moreover,

1. (1) $(f_{i})_{i\in F_{1}}=\{f^{1}_{\min (F^{1}_{2})},\ldots ,f^{1}_{d_{1}}\}$ is $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda _{1})$ -packed.

2. (2) $(f_{i})_{i\in F_{r}}=\{f^{r}_{\min (F^{r}_{2})},\ldots ,f^{r}_{d_{r}}\}$ is $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda _{r})$ -packed for each $r\in \Gamma \setminus \{\min (\Gamma )\}$ and, in addition, for each $f_{i}$ with $i\in F^{r}_{k}$ (where $2\leq k\leq M_{r}$ ), there is

   $$ \begin{align*} d(m(f_{i}),\lambda_{r}) &\leq d(m(f_{i}),\lambda^{r}_{k})+d(\lambda^{r}_{k},\lambda_{r}) \\ &\leq 2\cdot 2^{-\max\text{supp}(f_{\max(F^{r}_{k-1})})}\leq 2^{-\max\text{supp}(f_{\max(F_{r-1})})}\end{align*} $$
   so that, in fact, $(f_{i})_{i\in F_{r}}$ is $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda _{r}, 2^{-\max \text {supp}(f_{\max (F_{r-1})})})$ -packed.

3. (3) $d(\lambda _{r},\lambda )\leq \frac {1}{2}\cdot 2^{-\max \text {supp}(x_{r-1})}\leq 2^{-\max \text {supp}(f_{\max (F_{r-1})})}$ for each $r\in \Gamma \setminus \{\min (\Gamma )\}$ .

Putting these pieces together, it follows that the functional

$$ \begin{align*} \mathscr{F}=\frac{1}{m_{j_{1}}\cdots m_{j_{l}}m_{j_{0}}}\sum_{r\in\Gamma}\sum_{q=\min(F^{r}_{2})}^{d_{r}}f^{r}_{q} \end{align*} $$

is $(n_{j_{1}},\ldots ,n_{j_{l}},n_{j_{0}},\lambda )$ -packed and is therefore a member of $W_{\kappa +1}\subset W$ . Now, because $(x_{r})_{r\in \Omega }$ is itself $1$ -unconditional, there is the estimate

$$ \begin{align*} \left\Vert \sum_{r\in \Gamma}c_{r}x_{r}\right\Vert = \left\Vert \sum_{r\in\Gamma}|c_{r}|x_{r}\right\Vert &\geq\mathscr{F}\left(\sum_{r\in\Gamma}|c_{r}|x_{r}\right) \\ &=\frac{1}{m_{j_{1}}\cdots m_{j_{l}}m_{j_{0}}}\sum_{r\in\Gamma}\sum_{q=\min(F^{r}_{2})}^{d_{r}}f^{r}_{q}\left(\sum_{p\in\Gamma}|c_{p}|x_{p}\right) \\ &=\frac{1}{m_{j_{0}}}\sum_{r\in\Gamma}\frac{|c_{r}|}{m_{j_{1}}\cdots m_{j_{l}}}\sum_{q=\min(F^{r}_{2})}^{d_{r}}f^{r}_{q}(x_{r}) \\ &=\frac{1}{m_{j_{0}}}\sum_{r\in\Gamma}|c_{r}|\left(f_{r}(x_{r})-\frac{1}{m_{j_{1}}\cdots m_{j_{l}}}\sum_{q\in F^{r}_{1}}f^{r}_{q}(x_{r})\right) \\ &\geq \frac{1}{m_{j_{0}}}\sum_{r\in\Gamma}|c_{r}|\left(1 - \frac{1}{m_{j_{l}}}\cdot\frac{1}{m_{j_{1}}\cdots m_{j_{l-1}}}\sum_{q\in F^{r}_{1}}f^{r}_{q}(x_{r})\right) \\ &\geq \frac{1}{m_{j_{0}}}\sum_{r\in\Gamma}|c_{r}|\left(1-\frac{1}{m_{1}}\right) = \frac{m_{1}-1}{m_{1}m_{j_{0}}}\sum_{r\in\Gamma}|c_{r}|,\end{align*} $$

where we have used the fact that $\frac {1}{m_{j_{1}}\cdots m_{j_{l-1}}}\sum _{q\in F^{r}_{1}}f^{r}_{q}\in W$ , and this completes the proof for the first case.

1. (ii) $\limsup _{r\to \infty }w(f_{r})=\infty $

In this case, there exists an infinite subset $\Omega _{1}\subset \mathbb {N}$ such that $w(f_{r})$ for $r\in \Omega _{1}\setminus \{\min (\Omega _{1})\}$ is very fast-growing in the sense of Definition 3.1. Then, let $\lambda _{r}$ be the node in $\overline {\mathcal {D}}$ that corresponds to $m(f_{r})$ for each $r\in \Omega _{1}$ . We may then pass to a further infinite subset $\Omega _{2}\subset \Omega _{1}$ so that $d(\lambda _{r},\lambda )\leq \frac {1}{2}\cdot 2^{-\max \text {supp}(x_{r-1})}$ for some $\lambda \in \overline {\mathcal {D}}$ and for each $r\in \Omega _{2}\setminus \{\min (\Omega _{2})\}$ . Now, let $\Omega =\Omega _{2}$ and consider the subsequence $(x_{r})_{r\in \Omega }$ . Let $\Gamma \subset \Omega $ be a subset such that the block vectors $(x_{r})_{r\in \Gamma }$ are $S_{n_{j_{0}}}$ -admissible and let $f_{\min (\Gamma )}<\ldots <f_{\max (\Gamma )}$ be the corresponding norming functionals. Note that there is some $\kappa \in \mathbb {N}$ so that $f_{r}\in W_{\kappa }$ for each $r\in \Gamma $ . Note also that, by construction, these norming functionals are $S_{n_{j_{0}}}$ -admissible and are very fast-growing in the sense of Definition 3.1. Moreover, there is the estimate

$$ \begin{align*} d(\lambda_{r},\lambda)\leq 2^{-\max\text{supp}(x_{r-1})}\leq 2^{-\max\text{supp}(f_{r-1})} \end{align*} $$

for each $r\in \Gamma \setminus \{\min (\Gamma )\}$ so that these functionals are $(n_{j_{0}},\lambda )$ -packed. Then, we have that $\frac {1}{m_{j_{0}}}\sum _{r\in \Gamma }f_{r}\in W_{\kappa +1}\subset W$ so that

$$ \begin{align*} \left\Vert \sum_{r\in\Gamma}c_{r}x_{r}\right\Vert =\left\Vert \sum_{r\in\Gamma}|c_{r}|x_{r}\right\Vert &\geq \frac{1}{m_{j_{0}}}\sum_{r\in\Gamma}f_{r}\left(\sum_{p\in\Gamma}|c_{p}|x_{p}\right) \\ &=\frac{1}{m_{j_{0}}}\sum_{r\in\Gamma}|c_{r}|f_{r}(x_{r})=\frac{1}{m_{j_{0}}}\sum_{r\in\Gamma}|c_{r}|\geq\frac{m_{1}-1}{m_{1}m_{j_{0}}}\sum_{r\in\Gamma}|c_{r}|, \end{align*} $$

and this completes the proof of the theorem.

Proof We proceed by induction on the length l of the tuple $(m_{j_{1}},\ldots ,m_{j_{l}})$ . If $l=1$ , then $f=\frac {1}{m_{j_{1}}}\sum _{i=1}^{d}f_{i}\in W^{\text {aux},\delta }_{k}$ for some $k\geq 1$ , and we put

$$ \begin{align*} E_{0}&=\{1\}\setminus\{ i \mid f_{i}\in W^{\text{aux},\delta}_{0}\} \\ E_{1}&=\{2,\ldots,d\}\setminus \{i \mid f_{i}\in W^{\text{aux},\delta}_{0}\} \end{align*} $$

and note that, if $k=1$ , then $E_{0}=E_{1}=\emptyset $ so that the result holds vacuously. Assume then that $k> 1$ and note that $f_{1}<\ldots <f_{d}$ are in $W^{\text {aux},\delta }_{k-1}$ and are $(n_{j_{1}},\lambda )_{\delta }$ -packed for some $\lambda \in \overline {\mathcal {D}}$ . It therefore follows by definition that $(f_{i})_{i\in E_{1}}$ are $S_{n_{j_{1}}}\ast \mathcal {A}_{3}$ -admissible and that $\max _{i\in E_{1}}d(m(f_{i}),\lambda )\leq \delta $ .

Assume now that the result holds for functionals of the form $f=\frac {1}{m_{j_{1}}\cdots m_{j_{l}}}\sum _{i=1}^{d}f_{i}\in W^{\text {aux},\delta }$ and let $g=\frac {1}{m_{j_{1}}\cdots m_{j_{l}}m_{j_{l+1}}}\sum _{i=1}^{d}g_{i}\in W^{\text {aux},\delta }$ . As above, the result holds vacuously if $g\in W^{\text {aux},\delta }_{1}$ , so assume that $g\in W^{\text {aux},\delta }_{k}$ for some $k>1$ . In this case, we have that $g_{1}<\ldots <g_{d}$ are in $W^{\text {aux},\delta }_{k-1}$ and are $(n_{j_{1}},\ldots ,n_{j_{l}},n_{j_{l+1}},\lambda )_{\delta }$ -packed for some $\lambda \in \overline {\mathcal {D}}$ . It follows that there is a partition $F_{1}<\ldots <F_{M}$ of $\{1,\ldots ,d\}$ as in Definition 4.2. Note that in particular that $(g_{i})_{i\in F_{1}}$ are $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda _{1})_{\delta }$ -packed for some $\lambda _{1}\in \overline {\mathcal {D}}$ so that the functional $\tilde {g}=\frac {1}{m_{j_{1}}\cdots m_{j_{l}}}\sum _{i\in F_{1}}g_{i}\in W^{\text {aux},\delta }$ . Then, by the inductive hypothesis, there are sets $E_{0}\cup E_{1}\cup \ldots \cup E_{l}\subset F_{1}$ so that the required claim holds for $\tilde {g}$ . Now, let $E_{l+1}=\left (\bigcup _{r=2}^{M}F_{r}\right )\setminus \{ i \mid g_{i}\in W^{\text {aux},\delta }_{0}\}$ , and it follows that the required claim holds for g itself so that the proof is complete.

Proof Let $\Omega \subset \{1,\ldots ,2^{N}\}$ be given. We will proceed by induction on the level k norming sets $W^{\text {aux},\delta }_{k}$ , and the inductive hypothesis will include that $\left \vert f\left (\sum _{i\in A_{f}(\Omega )}a_{i}x_{i}\right )\right \vert <\frac {4\gamma }{\sqrt {w(f)}}$ if $f\notin W_{0}^{\text {aux},\delta }$ , where $\gamma =\max _{i\in \Omega }|a_{i}|$ and where $A_{f}(\Omega ) :=\{i\in \Omega \mid m_{t_{i}}\neq m(f)\}$ . The base case is the level $k=1$ norming set, and we first let $f\in W^{\text {aux},\delta }_{0}$ so that

$$ \begin{align*} \left\vert f\left(\sum_{i\in\Omega}a_{i}x_{i}\right)\right\vert = \left\vert e^{*}_{i_{0}}\left(\sum_{i\in\Omega}a_{i}m_{t_{i}}\sum_{r\in F_{i}}c^{i}_{r}e_{r} \right)\right\vert \leq \max_{i\in\Omega}\varepsilon_{i}m_{t_{i}}|a_{i}| < \max_{i\in\Omega}|a_{i}| \end{align*} $$

for some $i_{0}\in \mathbb {N}$ . It therefore remains to check functionals of the form

$$ \begin{align*} f=\frac{1}{m_{j_{1}}\cdots m_{j_{l}}}\sum_{q=1}^{d}e^{*}_{i_{q}}\in W^{\text{aux},\delta}_{1}\setminus W^{\text{aux},\delta}_{0}, \end{align*} $$

and we write $A_{f}(\Omega )=\{ i\in \Omega \mid m_{t_{i}}\neq m(f)\}=A_{1}\cup A_{2}\cup A_{3}$ , where

$$ \begin{align*} A_{1}&=\{ i\in\Omega \mid m_{t_{i}}<m(f)\} \\ A_{2}&=\{ i\in\Omega \mid m(f)<m_{t_{i}}\text{ and } n_{j_{1}}+\cdots+n_{j_{l}}<n_{t_{i}}\} \\ A_{3}&=\{ i\in\Omega \mid m(f)<m_{t_{i}} \text{ and } n_{j_{1}}+\cdots+n_{j_{l}}\geq n_{t_{i}}\} \end{align*} $$

so that $\Omega \setminus A_{f}(\Omega )$ is at most a singleton $\{i_{0}\}$ with $m_{t_{i_{0}}}=m(f)$ . Then, we have that

$$ \begin{align*} \left\vert f\left(\sum_{i\in\Omega}a_{i}x_{i}\right)\right\vert\leq \sum_{p=1}^{3}\left\vert f\left(\sum_{i\in A_{p}}a_{i}x_{i}\right)\right\vert + |f(a_{i_{0}}x_{i_{0}})|\leq \sum_{p=1}^{3}\left\vert f\left(\sum_{i\in A_{p}}a_{i}x_{i}\right)\right\vert + \max_{i\in\Omega}|a_{i}| \end{align*} $$

because $|f(x_{i_{0}})|\leq \frac {m_{t_{i_{0}}}}{w(f)}\leq 1$ . Now, recall that $\gamma =\max _{i\in \Omega }|a_{i}|$ and note that, for $A_{1}$ , there is the estimate

$$ \begin{align*} \left\vert f\left(\sum_{i\in A_{1}}a_{i}x_{i}\right)\right\vert&\leq \sum_{i\in A_{1}}|f(a_{i}x_{i})|\leq \frac{\gamma}{w(f)}\sum_{i\in A_{1}}m_{t_{i}} \\ &\leq \frac{\gamma}{w(f)}\text{card}(A_{1})m_{t_{\max(A_{1})}} \leq \frac{\gamma}{w(f)}t_{\max(A_{1})}m_{t_{\max(A_{1})}} \\ &\leq \frac{\gamma}{w(f)}(\text{ind}(m(f))-1)m_{\text{ind}(m(f))-1} \leq \frac{\gamma m_{\text{ind}(m(f))}}{w(f)\sqrt{m_{\text{ind}(m(f))}}} \leq \frac{\gamma}{\sqrt{w(f)}},\end{align*} $$

where we have used the first condition on the sequence $(m_{j})$ and the fact that $t_{i}<\text {ind}(m(f)):=\max _{1\leq z\leq l}j_{z}$ for each $i\in A_{1}$ . Next, for $A_{2}$ , note that

$$ \begin{align*} \left\vert f\left(\sum_{i\in A_{2}}a_{i}x_{i}\right)\right\vert &\leq \frac{\gamma}{w(f)}\sum_{i\in A_{2}}m_{t_{i}}\left\vert\sum_{q=1}^{d}e^{*}_{i_{q}}\left(\sum_{r\in F_{i}}c^{i}_{r}e_{r}\right) \right\vert \\ &\leq \frac{\gamma}{w(f)}\sum_{i\in A_{2}} 3m_{t_{i}}\varepsilon_{i} \leq\frac{\gamma}{w(f)}\sum_{i\in A_{2}}\frac{1}{2^{N}}\leq\frac{\gamma}{w(f)}\leq\frac{\gamma}{\sqrt{w(f)}}, \end{align*} $$

where we have used the observation that immediately precedes Proposition 2.1 and the fact that $e^{*}_{i_{1}}<\ldots <e^{*}_{i_{d}}$ are $S_{n_{j_{1}}+\cdots +n_{j_{l}}}\ast \mathcal {A}_{3}$ -admissible with $n_{j_{1}}+\cdots +n_{j_{l}}<n_{t_{i}}$ for each $i\in A_{2}$ . Lastly, for $A_{3}$ , note that

$$ \begin{align*} \left\vert f\left(\sum_{i\in A_{3}}a_{i}x_{i}\right)\right\vert &\leq \frac{\gamma}{w(f)}\sum_{i\in A_{3}}m_{t_{i}} \leq \frac{\gamma}{w(f)}|A_{3}|m_{t_{\max(A_{3})}} \\ &\leq\frac{\gamma m^{2}_{t_{\max(A_{3})}}}{w(f)}\leq \frac{\gamma\sqrt{w(f)}}{w(f)}=\frac{\gamma}{\sqrt{w(f)}}\end{align*} $$

because $m_{t_{i}}^{2}\leq \sqrt {w(f)}$ for each $i\in A_{3}$ , by Remark 4.2. It now follows by putting these pieces together that

$$ \begin{align*} \left\vert f\left(\sum_{i\in\Omega}a_{i}x_{i}\right)\right\vert \leq \gamma+\frac{3\gamma}{\sqrt{w(f)}}\end{align*} $$

for each $f\in W^{\text {aux},\delta }_{1}$ , and, in particular, we have that $\left \vert f\left (\sum _{i\in A_{f}(\Omega )}a_{i}x_{i}\right )\right \vert < \frac {4\gamma }{\sqrt {w(f)}}$ , as is required as part of the inductive hypothesis.

Assume now that the statement of Proposition 4.3 holds for every functional $f\in W^{\text {aux},\delta }_{k}$ , let $\Omega \subset \{1,\ldots ,2^{N}\}$ be given, and let

$$ \begin{align*} f=\frac{1}{m_{j_{1}}\cdots m_{j_{l}}}\sum_{q=1}^{d}f_{q}\in W^{\text{aux},\delta}_{k+1}\setminus W^{\text{aux},\delta}_{k}, \end{align*} $$

where $f_{q}\in W^{\text {aux},\delta }_{k}$ for each $1\leq q\leq d$ . Define the sets $B=\{q \mid f_{q}\in W^{\text {aux},\delta }_{0}\}$ and $C=\{3,\ldots ,d\}\setminus B$ . We may then write $f=\sum _{\alpha =1}^{4}g_{\alpha }$ , where

$$ \begin{align*} g_{1}=\frac{1}{w(f)}f_{1},\text{ }g_{2}=\frac{1}{w(f)}f_{2},\text{ }g_{3}=\frac{1}{w(f)}\sum_{q\in B}f_{q}, \text{ and } g_{4}=\frac{1}{w(f)}\sum_{q\in C}f_{q}, \end{align*} $$

and it follows that

$$ \begin{align*} \left\vert f\left(\sum_{i\in\Omega}a_{i}x_{i}\right)\right\vert \leq \left\vert f\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert + |f(a_{i_{0}}x_{i_{0}})| \leq \max_{i\in\Omega}|a_{i}|+\left\vert f\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert, \end{align*} $$

where, as before, $A_{f}(\Omega )=\{ i\in \Omega \mid m_{t_{i}}\neq m(f)\}$ so that $\Omega \setminus A_{f}(\Omega )$ is at most a singleton $\{i_{0}\}$ with $m_{t_{i_{0}}}=m(f)$ , in which case, $|f(x_{i_{0}})|\leq \frac {m_{t_{i_{0}}}}{w(f)}\leq 1$ . Let $\gamma =\max _{i\in \Omega }|a_{i}|$ as before, and it now remains to estimate

$$ \begin{align*} \left\vert f\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert&\leq \sum_{\alpha=1}^{4}\left\vert g_{\alpha}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert \\ &\leq\frac{1}{w(f)}\left\vert f_{1}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert + \frac{1}{w(f)}\left\vert f_{2}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert \\ &\qquad + \frac{1}{w(f)}\left\vert \sum_{q\in B}f_{q}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert + \frac{1}{w(f)}\left\vert\sum_{q\in C}f_{q}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert, \end{align*} $$

where, after applying the inductive hypothesis to the $g_{1}$ and $g_{2}$ terms and using the same argument as is used in the base case for the $g_{3}$ term, this quantity is bounded above by

$$ \begin{align*} &\frac{\gamma}{w(f)}\left(1+\frac{3}{\sqrt{w(f_{1})}}\right)+\frac{\gamma}{w(f)}\left(1+\frac{3}{\sqrt{w(f_{2})}}\right) +\frac{3\gamma}{\sqrt{w(f)}}+\frac{1}{w(f)}\left\vert \sum_{q\in C}f_{q}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert \\ &\qquad\leq \frac{\frac{13}{5}\gamma}{w(f)}+\frac{3\gamma}{\sqrt{w(f)}}+\frac{1}{w(f)}\left\vert\sum_{q\in C}f_{q}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert \\ &\qquad\leq \frac{(3+\frac{13}{50})\gamma}{\sqrt{w(f)}}+ \frac{1}{w(f)}\left\vert\sum_{q\in C}f_{q}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert \end{align*} $$

because $m_{1}\geq 100$ . Now, note that there are sets $E_{0}\cup E_{1}\cup \ldots \cup E_{l}$ from Lemma 4.1 such that $\{q \mid f_{q}\in \mathcal {C}\}=E_{0}\cup E_{1}\cup \ldots \cup E_{l}$ and such that, for each $1\leq z\leq l$ , we have that $(f_{q})_{q\in E_{z}}$ is $S_{n_{j_{1}}+\cdots +n_{j_{z}}}\ast \mathcal {A}_{3}$ -admissible and $\max _{q\in E_{z}}d(m(f_{q}),\lambda _{z})\leq \delta $ for some $\lambda _{z}\in \overline {\mathcal {D}}$ . Next, let $\Gamma _{z}=\{ i\in A_{f}(\Omega ) \mid m_{t_{i}}=m(f_{q}) \text { for some } q\in E_{z}\}$ for each $1\leq z\leq l$ and note that if $i_{1},i_{2}\in \Gamma _{z}$ , then we have that

$$ \begin{align*} 2^{-N} &\leq d(m_{t_{i_{1}}},m_{t_{i_{2}}}) \\ &= d(m(f_{q_{1}}),m(f_{q_{2}}))\leq d(m(f_{q_{1}}),\lambda_{z})+d(\lambda_{z},m(f_{q_{2}})) \leq 2\delta \end{align*} $$

for the corresponding $q_{1},q_{2}\in E_{z}$ , and this cannot happen. It follows that $\Gamma _{z}$ is at most a singleton, say $i_{z}$ , for each $1\leq z\leq l$ so that

$$ \begin{align*}\left\vert \sum_{q\in\mathcal{C}}f_{q}\left(\sum_{i\in A_{f}(\Omega)}a_{i}x_{i}\right)\right\vert &\leq \sum_{z=0}^{l} \sum_{q\in E_{z}} \left\vert f_{q}\left(\sum_{i \in A_{f}(\Omega)\setminus\{i_{z}\}}a_{i}x_{i}\right)\right\vert + \sum_{z=0}^{l}\sum_{q\in E_{z}}|f_{q}(a_{i_{z}}x_{i_{z}})| \\ &\leq \sum_{z=0}^{l}\sum_{q\in E_{z}}\frac{4\gamma}{\sqrt{w(f_{q})}} + \gamma\sum_{z=0}^{l}\sum_{q\in E_{z}}\frac{m_{t_{i_{z}}}}{w(f_{q})} \\ &\leq (l+1)\gamma + \sum_{z=0}^{l}\sum_{q\in E_{z}}\frac{4\gamma}{\sqrt{w(f_{q})}} \leq (l+5)\gamma, \end{align*} $$

where again, we have used the inductive hypothesis, and, by putting all of these pieces together, there is the estimate

$$ \begin{align*} \left\vert f\left(\sum_{i\in\Omega}a_{i}x_{i}\right)\right\vert &\leq \gamma+\frac{(3+\frac{13}{50})\gamma}{\sqrt{w(f)}}+\frac{(l+5)\gamma}{w(f)} \\ &\leq \gamma+\frac{(3+\frac{13}{50})\gamma}{\sqrt{w(f)}}+\frac{(l+5)\gamma}{10^{l}\sqrt{w(f)}} \leq \gamma + \frac{(3+\frac{13}{50}+\frac{3}{5})\gamma}{\sqrt{w(f)}} \leq \gamma+ \frac{4\gamma}{\sqrt{w(f)}},\end{align*} $$

where, in particular, $\left \vert f\left (\sum _{i\in A_{f}(\Omega )}a_{i}x_{i}\right )\right \vert \leq \frac {4\gamma }{\sqrt {w(f)}} $ so that the proof is complete.

Proof We proceed by induction on the length l of the tuple $(m_{j_{1}},\ldots ,m_{j_{l}})$ . Let $l=1$ and assume that $f_{1}<\ldots <f_{d}\in W$ with $2^{-\max \text {supp}(f_{1})}\leq \delta $ are $(n_{j_{1}},\lambda )$ -packed for some $\lambda \in \overline {\mathcal {D}}$ . Then, $(g^{q}_{r})_{r=1,q=1}^{3,d}\subset W^{\text {aux},\delta }_{k}$ for some $k\in \mathbb {N}$ are $S_{n_{j_{1}}}\ast \mathcal {A}_{3}$ -admissible, and we have that

$$ \begin{align*} w(g^{q}_{\tilde{r}})=w(f_{q})\geq 2^{\max\text{supp}(f_{q-1})}\geq 2^{\max\text{supp}(\varphi)} \end{align*} $$

for each $g^{q}_{\tilde {r}}\in \mathcal {F}_{q}$ with at least two preceding nonzero functionals, where $\varphi $ is the nonzero functional that is two before $g^{q}_{\tilde {r}}$ . It follows that $(g^{q}_{r})_{r=1,q=1}^{3,d}$ are skipped very fast-growing in the sense of Definition 4.1. Moreover, we have that

$$ \begin{align*} d(m(g^{q}_{\tilde{r}}),\lambda)=d(m(f_{q}),\lambda) \leq 2^{-\max\text{supp}(f_{q-1})} \leq 2^{-\max\text{supp}(f_{1})}\leq\frac{\delta}{2}< \delta \end{align*} $$

for $2\leq q\leq d$ so that $(g^{q}_{r})_{q=1,r=1}^{d,3}$ are $(n_{j_{1}},\lambda )_{\delta }$ -packed.

Assume now that the result holds for tuples of length l and let $f_{1}<\ldots <f_{d}\in W$ be $(n_{j_{1}},\ldots ,n_{j_{l}},n_{j_{l+1}},\lambda )$ -packed for some $\lambda \in \overline {\mathcal {D}}$ . Let $\mathcal {F}_{1},\ldots ,\mathcal {F}_{d}$ be sets of at most three auxiliary weighted functionals as in the statement of the lemma and let $F_{1}<\ldots <F_{M}$ be a partition of $\{1,\ldots ,d\}$ as in Definition 3.2. Note that $(f_{q})_{q\in F_{1}}$ are $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda _{1})$ -packed for some $\lambda _{1}\in \overline {\mathcal {D}}$ . It therefore follows from the inductive hypothesis that $(g^{q}_{r})_{r=1,q\in F_{1}}^{3}$ are $(n_{j_{1}},\ldots ,n_{j_{;}},\lambda _{1})_{\delta }$ -packed. Moreover, for each $g^{q}_{\tilde {r}}$ with $q\in F_{p}$ for $p>1$ , we have that

$$ \begin{align*} d(m(g^{q}_{\tilde{r}}),\lambda)&=d(m(f_{q}),\lambda) \leq d(m(f_{q}),\lambda_{p})+d(\lambda_{p},\lambda) \\ &\leq 2\cdot 2^{-\max\text{supp}(f_{\max(F_{p-1})})}\leq 2\cdot 2^{-\max\text{supp}(f_{1})}\leq \delta, \end{align*} $$

and because $(g^{q}_{r})_{q=1,r=1}^{d,3}$ are skipped very fast-growing in the sense of Definition 4.2 by the same reasoning that is used in the base case, it follows that these functionals are $(n_{j_{1}},\ldots ,n_{j_{l}},n_{j_{l+1}},\lambda )_{\delta }$ -packed so that the proof is complete.

Proof We proceed by induction on the level k norming sets $W_{k}$ , and the base case is the level $k=0$ set $W_{0}=\{\pm e_{i}^{*} \mid i\in \mathbb {N}\}$ . Indeed, if $f=e^{*}_{i_{0}}\in W_{0}$ , then

$$ \begin{align*} \left\vert e^{*}_{i_{0}}\left(\sum_{i\in I}a_{i}x_{i}\right)\right\vert \leq |a_{i_{1}}e^{*}_{i_{0}}(x_{i_{1}})| \leq C|a_{i_{1}}|= C|e^{*}_{t_{i_{1}}}(a_{t_{i_{1}}}e_{t_{i_{1}}})|, \end{align*} $$

where $i_{0}\in \text {supp}(x_{i_{1}})$ so that the desired estimate holds with $h=\{0\}$ and with $g=e^{*}_{t_{i_{1}}}$ . Note that, as part of the inductive hypothesis, we have that $\text {supp}(h),\text {supp}(g)\subset \{t_{i}\}_{i\in I}$ and that either $h=0$ or $h\in W^{\text {aux},\delta }_{0}$ , $\text {supp}(h)<\min \text {supp}(g)$ .

Assume now that the desired estimate holds for functionals in the level k norming set $W_{k}$ and consider the functional

$$ \begin{align*} f=\frac{1}{m_{j_{1}}\cdots m_{j_{l}}}\sum_{q=1}^{d}f_{q}\in W_{k+1}\setminus W_{k}, \end{align*} $$

which is to say that $f_{1}<\ldots <f_{d}$ are in $W_{k}$ and are $(n_{j_{1}},\ldots ,n_{j_{l}},\lambda )$ -packed for some $\lambda \in \overline {\mathcal {D}}$ . We omit if necessary the functionals $f_{q}$ that are below the support of $x_{\min (I)}$ . It then follows that

$$ \begin{align*} \frac{1}{\delta}\leq 2^{\min\text{supp}(x_{\min(I)})}\leq 2^{\max\text{supp}(f_{1})} \implies 2^{-\max\text{supp}(f_{1})}\leq\delta, \end{align*} $$

and we will use this observation later. First, define the quantity $i_{0}$ by

$$ \begin{align*} i_{0}=\max\{i\in I \mid m_{j_{1}}\cdots m_{j_{l}}\geq m_{j_{i}}\} \end{align*} $$

if it exists. If there does exist such an $i_{0}$ , then choose $i_{1}\in [\min (I),i_{0}]$ so that $|a_{i_{1}}|=\max \{|a_{i}|\}_{i=\min (I)}^{i_{1}}$ and define the functional $h=\text {sign}(a_{i_{1}})e^{*}_{t_{i_{1}}}\in W^{\text {aux},\delta }_{0}$ . Note that, if $\min (I)=i_{0}=i_{1}$ , then

$$ \begin{align*} \left\vert f\left(\sum_{i\leq i_{0}}a_{i}x_{i}\right)\right\vert=|f(a_{i_{1}}x_{i_{1}})|\leq C|a_{i_{1}}|=Ch(a_{i_{1}}e_{t_{i_{1}}}), \end{align*} $$

and, if $\min (I)<i_{0}$ , then

$$ \begin{align*} \left\vert f\left(\sum_{i\leq i_{0}}a_{i}x_{i}\right)\right\vert &\leq |a_{i_{0}}f(x_{i_{0}})|+\sum_{i<i_{0}}|a_{i}f(x_{i})| \leq C|a_{i_{0}}|+\sum_{i<i_{0}}\left\vert\frac{a_{i}}{w(f)}\sum_{q=1}^{d}f_{q}(x_{i})\right\vert \\ &\leq C|a_{i_{1}}|+\frac{|a_{i_{1}}|}{m_{j_{i_{0}}}}\sum_{i<i_{0}}\sum_{q=1}^{d}\left\vert f_{q}\left(\sum_{r\in\text{supp}(x_{i})}b_{r}e_{r}\right)\right\vert \\ &\leq C|a_{i_{1}}|+\frac{|a_{i_{1}}|}{m_{j_{i_{0}}}}\sum_{i<i_{0}}\sum_{r\in\text{supp}(x_{i})}|b_{r}| \leq C|a_{i_{1}}|+\frac{C|a_{i_{1}}|}{m_{j_{i_{0}}}}\max\text{supp}(x_{i_{0}-1}) \\ &\leq C\left(1+\frac{1}{\sqrt{m_{j_{i_{0}-1}}}}\right)|a_{i_{1}}| \leq C\left(1+\frac{1}{\sqrt{m_{j_{\min(I)}}}}\right)h\left(\sum_{i\leq i_{0}}a_{i}e_{t_{i}}\right).\end{align*} $$

If, however, there is no such number $i_{0}$ , then the sum $\sum _{i\leq i_{0}}a_{i}x_{i}$ is empty, so we take $h=0$ .

Next, let $\tilde {I}=\{i\in I \mid i>i_{0}\}$ if such an $i_{0}$ exists and put $\tilde {I}=I$ otherwise. We then define the subsets

$$ \begin{align*} A&=\{ i\in\tilde{I} \mid \text{supp}(x_{i})\cap\text{supp}(f_{q})\neq\emptyset \text{ for at most one }q\} \\ I_{q}&=\{ i\in A \mid \text{supp}(x_{i})\cap\text{supp}(f_{q})\neq\emptyset\} \\ D&=\{q\mid I_{q}\neq\emptyset\} \\ B&=\tilde{I}\setminus A \end{align*} $$

so that the intervals $I_{q}$ are pairwise disjoint and $\text {supp}(x_{i})$ intersects the support of at least two $f_{q}$ if $i\in B$ . It follows by the inductive hypothesis that, for each $q\in D$ , there exist functionals $h_{q}\in \{0\}\cup W^{\text {aux},\delta }_{0}$ and $g_{q}\in W^{\text {aux},\delta }$ with $\vec w(g_{q})=\vec w(f_{q})$ so that

$$ \begin{align*} \left\vert f_{q}\left(\sum_{i\in I_{q}}a_{i}x_{i}\right)\right\vert\leq C\left(1+\frac{1}{\sqrt{m_{j_{\min(I)}}}}\right)\left\vert (h_{q}+g_{q})\left(\sum_{i\in I_{q}}a_{i}e_{t_{i}}\right)\right\vert, \end{align*} $$

where $\text {supp}(h_{q}),\text {supp}(g_{q})\subset \text {supp}(f_{q})$ and we have that either $h_{q}=0$ or $h_{q}\in \{e_{t_{i}}\}_{i\in I_{q}}$ and $t_{i}<\min \text {supp}(g_{q})$ . Now, replace $h_{q}+g_{q}$ by $\text {sign}\left [(h_{q}+g_{q})\left (\sum _{i\in I_{q}}a_{i}e_{t_{i}}\right )\right ] (h_{q}+g_{q})$ if necessary and note that

$$ \begin{align*} \left\vert f\left(\sum_{i\in\tilde{I}}a_{i}x_{i}\right)\right\vert &\leq \sum_{i\in B}|f(x_{i})|+\frac{1}{w(f)}\sum_{q\in D}\left\vert f_{q}\left(\sum_{i\in I_{q}}a_{i}x_{i}\right)\right\vert \\ &\leq\frac{C}{w(f)}\sum_{i\in B}|a_{i}|+C\left(1+\frac{1}{\sqrt{m_{j_{\min(I)}}}}\right)\frac{1}{w(f)}\sum_{q\in D}(h_{q}+g_{q})\left(\sum_{i\in I_{q}}a_{i}e_{t_{i}}\right) \\ &\leq C\left(1+\frac{1}{\sqrt{m_{j_{\min(I)}}}}\right)\frac{1}{w(f)}\left[\sum_{i\in B}\tilde{h}_{i}+\sum_{q\in D}(h_{q}+g_{q}) \right]\left(\sum_{i\in\tilde{I}}a_{i}e_{t_{i}}\right),\end{align*} $$

where we have used the third condition from Definition 5.1 and where $\tilde {h}_{i}=\text {sign}(a_{i})e_{t_{i}}^{*}$ for each $i\in B$ . It therefore follows that

$$ \begin{align*} \left\vert f\left(\sum_{i\in I}a_{i}x_{i}\right)\right\vert \leq C\left(1+\frac{1}{\sqrt{m_{j_{\min(I)}}}}\right)\left[(h+g)\left(\sum_{i\in I}a_{i}e_{t_{i}}\right)\right], \end{align*} $$

where either $h=0$ or $h\in \{e_{t_{i}}\}_{i=\min (I)}^{i_{0}}$ and $\text {supp}(h)=t_{i_{1}}<\min \text {supp}(g)$ . Indeed, we have that

$$ \begin{align*} g=\frac{1}{w(f)}\left(\sum_{i\in B}\tilde{h}_{i}+\sum_{q\in D}(h_{q}+g_{q})\right) \end{align*} $$

so that $\vec w(f)=\vec w(g)$ . It remains to prove that $g\in W^{\text {aux},\delta }$ . To that end, define for each $i\in B$ the quantity $q_{i}=\max \{1\leq q\leq d \mid \min \text {supp}(f_{q})\leq \max \text {supp}(x_{i})=t_{i}\}$ and note that the correspondence $i\mapsto q_{i}$ is strictly increasing. If $q=q_{i}$ for some $i\in B$ , then write $\mathcal {F}_{q}=\{h_{q},g_{q},\tilde {h}_{i}\}$ . If $q\neq q_{i}$ for any $i\in B$ , then write $\mathcal {F}_{q}=\{h_{q},g_{q}\}$ . Note that $\mathcal {F}_{q}$ may consist only of $\tilde {h}_{i}$ or may be empty if $q\notin D$ . It therefore follows from Lemma 5.1 (applied to $(f_{q})_{q\in D}$ ) that $g\in W^{\text {aux},\delta }$ , and this completes the proof.

Proof Let $Y\subset X_{\mathcal {D}}$ and let $Z\subset Y$ be a block subspace. Then, fix $C> 1$ and let $(z_{i})$ be an infinite $(C,(j_{i})_{i\in \mathbb {N}})$ -RIS in Z with $1\leq \|z_{i}\|$ for each $i\in \mathbb {N}$ , which exists by Corollary 5.4. By passing to a subsequence, we may assume that the conclusion of Theorem 3.2 is satisfied. Recall that the nodes of $\mathcal {D}$ are in lexicographical order with respect to $\mathcal {M}$ and choose for each weight $m_{j}$ a number $\varepsilon _{j}\in \left (0,\frac {1}{3m_{j}2^{N_{j}}}\right )$ , where $N_j = |\phi (m_j)|$ .

Now, let $F_{1}<F_{2}<\ldots $ be subsets of $\mathbb {N}$ so that $\{\min \text {supp}(z_{i}) \mid i\in F_{j}\}$ is a maximal set in $S_{n_{j}}$ with respect to inclusion for each $j\in \mathbb {N}$ . It follows from Proposition 2.1 that there exist coefficients $(c^{j}_{i})_{i\in F_{j}}$ so that the vector $\sum _{i\in F_{j}}c^{j}_{i}e_{i}$ is an $(n_{j},\frac {\varepsilon _{j}}{2})$ -basic scc. In particular, we have that $x_{m_j}=\sum _{i\in F_{j}}c^{j}_{i}z_{i}$ is an $(n_{j},\frac {\varepsilon _{j}}{2})$ -scc, and we note also that

$$ \begin{align*} \|x_{m_j}\|=\left\Vert\sum_{i\in F_{j}}c^{j}_{i}z_{i}\right\Vert \geq \frac{m_{1}-1}{m_{1}m_{j}}\sum_{i\in F_{j}}c^{j}_{i}=\frac{m_{1}-1}{m_{1}m_{j}}\end{align*} $$

from Theorem 3.2. Let $\tilde {x}_{m_j}=m_{j}x_{j}$ and define, for $\lambda \in \mathcal {D}$ , $y_{\lambda }= \tilde x_{\phi ^{-1}(m_j)}/\|\tilde x_{\phi ^{-1}(m_j)}\|$ so that $(y_{\lambda })$ is a normalized sequence in Z that is block in the lexicographical order of $\mathcal {D}$ .

Now, fix $N\in \mathbb {N}$ and, for $\lambda \in \{0,1\}^N$ , let $\mu _\lambda \in \mathcal {D}$ such that $\mu _\lambda \geq \lambda $ . Because $(y_\lambda )_{\lambda \in \mathcal {D}}$ is lexicographically block, there are $2^N\leq j_1<\cdots <j_{2^N}$ such that $\phi (\{m_{j_1},\ldots ,m_{j_{2^N}}\}) = \{\mu _\lambda :\lambda \in \{0,1\}^N\}$ . In particular, $\min \text {supp}(y_{\mu _\lambda })\geq N$ , $\lambda \in \{0,1\}^N$ . Furthermore, for $\lambda \in \{0,1\}^N$ , $\mu _\lambda \geq \lambda $ , and thus, for $1\leq k\neq k'\leq 2^N$ , $d(m_{j_k},m_{j_{k'}}) \geq 2^{-N}$ . Then, it follows that

$$ \begin{align*} \left\Vert \sum_{\lambda\in\{0,1\}^N}y_{\mu_\lambda}\right\Vert =\mathscr{F}\left(\sum_{\lambda\in\{0,1\}^N}y_{\mu_\lambda}\right) =\mathscr{F}\left(\sum_{k=0}^{2^{N}-1}\frac{1}{\|\tilde{x}_{m_{j_{k}}}\|}m_{j_{k}}\sum_{i\in F_{j_{k}}}c^{j_{k}}_{i}z_{i}\right) \qquad (\#) \end{align*} $$

for some norming functional $\mathscr {F}\in W$ , and, by Proposition 5.5, this quantity is bounded above by

$$ \begin{align*} C\left\vert (h+g)\left(\sum_{k=0}^{2^{N}-1}\frac{1}{\|\tilde{x}_{m_{j_{k}}}\|}m_{j_{k}}\sum_{i\in F_{j_{k}}}c^{j_{k}}_{i}e_{\rho(i)}\right)\right\vert \qquad (\#\#), \end{align*} $$

where $h\in W^{\text {aux},\delta }_{0}\cup \{0\}$ , $g\in W^{\text {aux},\delta }$ with $\vec w(g)=\vec w(\mathscr {F})$ , and where $\rho (i)=\max \text {supp}(z_{i})$ for each $i\in \mathbb {N}$ . Note that $\sum _{i\in F_{j_{k}}}c^{j_{k}}_{i}e_{\rho (i)}$ is for each $0\leq k\leq 2^{N}-1$ an $(n_{j_{k}},\varepsilon _{j_{k}})$ -basic scc by the observation that immediately follows Proposition 2.1, and note also that

$$\begin{align*}\frac{m_{j_{k}}}{\|\tilde{x}_{m_{j_{k}}}\|}=\frac{1}{\|x_{m_{j_{k}}}\|}\leq \frac{m_{1}m_{j_{k}}}{m_{1}-1}.\end{align*}$$

It therefore follows that $(\#\#)$ is bounded above by

$$ \begin{align*} C\max_{0\leq k\leq 2^{N}-1}\frac{m_{j_{k}}\varepsilon_{j_{k}}}{\|\tilde{x}_{j_{k}}\|} +C\left\vert g\left(\sum_{k=0}^{2^{N}-1}\frac{1}{\|\tilde{x}_{j_{k}}\|}m_{j_{k}}\sum_{i\in F_{j_{k}}}c^{j_{k}}_{i}e_{\rho(i)}\right)\right\vert \qquad (\#\#\#), \end{align*} $$

and lastly, by Proposition 4.3, we note that $(\#\#\#)$ is bounded above by

$$ \begin{align*} C\frac{m_{1}m_{j_{k}}}{m_{1}-1}\frac{1}{3m_{j_{k}} 2^{N_{j_{k}}}}+C\left(1+\frac{4}{\sqrt{w(g)}}\right)\max_{0\leq k\leq 2^{N}-1}\frac{1}{\|\tilde{x}_{j_{k}}\|} \leq C\frac{100}{99}\left(1+\left(1+\frac{2}{5}\right)\right)\leq 3, \end{align*} $$

for $C>1$ sufficiently close to one. The estimate for arbitrary scalars coefficients follows from the 1-unconditionality of the basis of $X_{\mathcal {D}}$ .