$G_{2}=\cup_{k=1}^{n}\cup\\{\operatorname{range}(x_{k})\cap\operatorname{range}(f_{\alpha}):\alpha\in\mathcal{B}_{k}^{2}\\},$

so that $g_{2}=g|_{G_{2}}$. We further split $G_{2}$ as follows

$G_{2}^{1}=\cup_{k=1}^{n}\cup\\{\textnormal{supp}(x_{k})\cap\textnormal{supp}(f_{\alpha}):\alpha\in\mathcal{B}_{k}^{2}\text{
and }I^{x_{k}}_{f_{\alpha}}=\emptyset\\}\text{ and }G_{2}^{2}=G_{2}\setminus
G_{2}^{1}.$

Proposition 13.4 (ii) implies that for $k\in\\{1,\ldots,n\\}$,

$g_{2}|_{G_{2}^{1}}(x_{k})\leq\frac{18}{w(f_{k})}$

and thus

$g_{2}|_{G_{2}^{1}}(\frac{1}{\sqrt{n}}\sum_{k\in
K_{1}}x_{k})\leq\frac{18}{\sqrt{n}}.$ (14.7)

To complete the computation, we need to evaluate the action of
$g_{2}|_{G_{2}^{2}}$. To that end, for $s=m+1,m+2,\ldots$ and
$k\in\\{1,\ldots,n\\}$ put

$\mathcal{B}_{k,s}^{2}=\\{\alpha\in\mathcal{B}_{k}^{2}:|\alpha|=s\\},$

so that for each $s>m$, the sets $\mathcal{B}_{k,s}^{2}$,
$k\in\\{1,\ldots,n\\}$ are pairwise disjoint and the set
$\cup_{k=1}^{n}\mathcal{B}_{k,s}^{2}$ is pairwise incomparable. We use Lemma
14.2 and the definition of $G_{2}^{2}$ to calculate

$\displaystyle g_{2}|_{G_{2}^{2}}(\frac{1}{\sqrt{n}}\sum_{k=1}^{n}x_{k})$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\sum_{\alpha\in\mathcal{B}^{2}_{k}}\frac{\lambda_{f_{\alpha}}}{w_{f}(f_{\alpha})}f_{\alpha}|_{G^{2}_{2}}(x_{k})$
$\displaystyle\leq\frac{4\sqrt{2}+1}{\sqrt{n}}\sum_{k=1}^{n}(\sum_{\alpha\in\mathcal{B}^{2}_{k}}\frac{\lambda^{2}_{f_{\alpha}}}{w_{f}(f_{\alpha})^{2}})^{1/2}$
$\displaystyle\leq(4\sqrt{2}+1)(\sum_{k=1}^{n}\sum_{\alpha\in\mathcal{B}_{k}^{2}}\frac{\lambda^{2}_{f_{\alpha}}}{w_{f}(f_{\alpha})^{2}})^{1/2}$
$\displaystyle\leq(4\sqrt{2}+1)(\sum_{s=m+1}^{\infty}\frac{1}{4^{s}}\sum_{\alpha\in\cup_{k=1}^{n}\mathcal{B}_{k,s}^{2}}{\lambda^{2}_{f_{\alpha}}})^{1/2}$
$\displaystyle\leq(4\sqrt{2}+1)(\sum_{s=m+1}^{\infty}\frac{1}{4^{s}})^{1/2}=\frac{4\sqrt{2}+1}{2^{m}\sqrt{3}}\leq\frac{4}{2^{m}}.$
(14.8)

Then, (14.5), (14.6), (14.7), (14) yield that

$|f(x)|\leq\frac{36+48\sqrt{2\ell}}{\sqrt{n}}+\frac{4}{2^{m}}\leq\frac{36+48\sqrt{2\ell}}{\sqrt{n}}+\frac{c}{2}$

and thus for $d$ such that

$\frac{36+48\sqrt{2\ell}}{\sqrt{d}}<\frac{c}{2},$

we have the desired result. ∎

######  Proposition 14.4.

The space $\mathfrak{X}_{\text{awi}}^{(2)}$ does not contain Asymptotic
$\ell_{2}$ subspaces.

###### Proof.

It is an immediate consequence of Proposition 14.3, using similar arguments as
in Proposition 9.7. ∎

######  Remark 14.5.

Unlike the case of $\ell_{1}$, for every $1<p<\infty$, it is in fact possible
to define a reflexive Banach space with a Schauder basis, admitting a unique
$\ell_{p}$ asymptotic model with respect to the family of normalized block
sequences, whose any block subspace contains an $\ell_{1}$ block tree of
height $\omega^{\xi}$. Such a space can be defined using the attractors
method, which was first introduced in [3] and later used in [10].

###  15\. Appendix A

In this section we prove the properties of standard exact pairs in
$\mathfrak{X}_{\text{awi}}^{(1)}$, given in Proposition 8.11. This requires
three steps. First, we need to define an auxiliary space which is also a Mixed
Tsirelson space. Then, on the special convex combinations of its basis, we
give upper bounds on the evaluations of the functionals in its norming set
$W_{\text{aux}}^{(1)}$. Finally for a standard exact pair $(x,f)$, via the
basic inequality, we reduce the upper bounds of the evaluations of functionals
in $W_{(1)}$ acting on $x$, to the corresponding one of a functional $g$ in
$W_{\text{aux}}^{(1)}$ on a normalized special convex combination of the basis
of the auxiliary space.

####  15.1. The Auxiliary Space

######  Definition 15.1.

Let $W_{\text{aux}}^{(1)}$ be the minimal subset of $c_{00}(\mathbb{N})$ such
that

  * (i)

$\pm e_{i}$ is in $W_{\text{aux}}^{(1)}$ for all $i\in\mathbb{N}$ and

  * (ii)

for every $j\in\mathbb{N}$ and every $\mathcal{S}_{n_{j}+1}$-admissible
sequence of functionals $(f_{i})_{i=1}^{d}$ in $W_{\text{aux}}^{(1)}$ we have
that $f=m_{j}^{-1}\sum_{i=1}^{d}f_{i}$ is in $W_{\text{aux}}^{(1)}$.

The purpose of the following two lemmas is to provide upper bounds for the
norms of linear combinations of certain vectors in the auxiliary space.

######  Lemma 15.2.

Let $j\in\mathbb{N}$ with $j>1$ and $\varepsilon>0$ with $\varepsilon\leq
m_{j}^{-1}$. For every $(n_{j},\varepsilon)$-basic s.c.c. $x=\sum_{k\in
F}c_{k}e_{k}$ the following hold.

  * (i)

For every $f\in W_{\text{aux}}^{(1)}$

$\big{|}f(x)\big{|}\leq\begin{cases}\varepsilon,\quad\quad&f=\pm
e_{i}^{*}\text{ for some }i\in\mathbb{N}\\\ \frac{1}{w(f)},\quad\quad&w(f)\geq
m_{j}\\\ \frac{2}{w(f)m_{j}},\quad\quad&w(f)<m_{j}\end{cases}$

  * (ii)

If $f\in W_{\text{aux}}^{(1)}$ with a tree analysis
$(f_{\alpha})_{\alpha\in\mathcal{A}}$ such that $w(f_{\alpha})\neq m_{j}$ for
all $\alpha\in\mathcal{A}$ and $\varepsilon<m_{j}^{-2}$ then
$|f(x)|<2m_{j}^{-2}$.

###### Proof.

We may assume that $\textnormal{supp}(f)\subset F$ and $f(e_{i})\geq 0$ for
every $i\in\mathbb{N}$. If $f=\pm e_{i}^{*}$ for some $i\in F$ then
$|f(x)|=c_{i}<\varepsilon$, since $x$ is an $(n_{j},\varepsilon)$-basic s.c.c.
and $\\{i\\}\in\mathcal{S}_{0}$.

Suppose that $m_{j}\leq w(f)$. Then $\|f\|_{\infty}\leq 1/w(f)$ and hence

$\big{|}f(x)\big{|}\leq\|f\|_{\infty}\|x\|_{1}\leq\frac{1}{w(f)}.$

In the case where $w(f)=m_{i}<m_{j}$, let $f=m^{-1}_{i}\sum_{l=1}^{d}f_{l}$
with $(f_{l})_{l=1}^{d}$ an $\mathcal{S}_{n_{i}+1}$-admissible sequence in
$W_{\text{aux}}^{(1)}$. For $l=1,\ldots,d$, define $D_{l}=\\{k\in
F:f_{l}(e_{k})>m^{-1}_{j}\\}$ and $D=\cup_{l=1}^{d}D_{l}$. Then, [13, Lemma
3.16] implies that $D_{l}\in\mathcal{S}_{(\log_{2}(m_{j})-1)(n_{j-1}+1)}$ for
each $l=1,\ldots,d$ and hence, since $(f_{l})_{l=1}^{d}$ is
$S_{n_{j-1}+1}$-admissible (recall that $i<j$ since $m_{i}<m_{j}$) and
$D_{l}\subset\textnormal{supp}(f_{l})$, $l=1,\ldots,d$, we conclude that the
sequence $(D_{l})_{l=1}^{d}$ is $S_{n_{j-1}+1}$-admissible and

$D=\cup_{l=1}^{d}D_{l}\in
S_{n_{j-1}+1}*\mathcal{S}_{(\log_{2}(m_{j})-1)(n_{j-1}+1)}=\mathcal{S}_{\log_{2}(m_{j})(n_{j-1}+1)}$

Since $x$ is an $(n_{j},\varepsilon)$-basic s.c.c. and
$\log_{2}(m_{j})(n_{j-1}+1)<n_{j}$, the above implies that $\sum_{k\in
D}c_{k}<\varepsilon$ and thus

$\displaystyle f(x)$
$\displaystyle=\frac{1}{m_{i}}\sum_{l=1}^{d}f_{l}(\sum_{k\in
F}c_{k}e_{k})=\frac{1}{m_{i}}(\sum_{l=1}^{d}f_{l}|_{D}(\sum_{k\in
F}c_{k}e_{k})+\sum_{l=1}^{d}f_{l}|_{\mathbb{N}\setminus D}(\sum_{k\in
F}c_{k}e_{k}))$ $\displaystyle\leq\frac{1}{m_{i}}(\sum_{k\in
D}c_{k}+\frac{1}{m_{j}})\leq\frac{1}{m_{i}}(\varepsilon+\frac{1}{m_{j}})\leq\frac{2}{m_{i}m_{j}}.$

Finally, if there is a tree analysis $(f_{\alpha})_{\alpha\in\mathcal{A}}$ of
$f$ with $w(f_{\alpha})\neq m_{j}$ for every $\alpha\in\mathcal{A}$, [13,
Lemma 3.16] implies that $D=\\{k\in
F:f(e_{k})>m_{j}^{-2}\\}\in\mathcal{S}_{(2\log_{2}(m_{j})-1)(n_{j-1}-1)}$ and
since $(2\log_{2}(m_{j})-1)(n_{j-1}-1)<n_{j}$ we have that $\sum_{k\in
D}c_{i}<\varepsilon$. Hence, we conclude that

$f(x)=\sum_{k\in D}c_{k}f(x_{k})+\sum_{k\in F\setminus
D}c_{k}f(x_{k})\leq\varepsilon+\frac{1}{m^{2}_{j}}<\frac{2}{m_{j}^{2}}.$

∎

####  15.2. The Basic Inequality

######  Proposition 15.3 (Basic Inequality).

Let $(x_{k})_{k\in I}$ be a $(C,(j_{k})_{k\in I})$-RIS in
$\mathfrak{X}_{\text{awi}}^{(1)}$ with $4\leq\min\textnormal{supp}(x_{\min
I})$, $(a_{k})_{k\in I}$ be a sequence of nonzero scalars and $f\in W_{(1)}$
with $I_{f}\neq\emptyset$. Define $t_{k}=\max\textnormal{supp}(x_{k})$, $k\in
I$. Then there exist

  * (i)

$g\in W_{\text{aux}}^{(1)}\cup\\{0\\}$ with $w(g)=w(f)$ if $g\neq 0$ and
$\\{k:t_{k}\in\textnormal{supp}(g)\\}\subset I_{f}$,

  * (ii)

$h\in\\{\text{sign}(a_{k})e^{*}_{t_{k}}:k\in I_{f}\\}\cup\\{0\\}$ with
$k_{0}\in I_{f}$ and $k_{0}<\min\textnormal{supp}(g)$ if
$h=\text{sign}(a_{k_{0}})e^{*}_{t_{k_{0}}}$ and

  * (iii)

$j_{0}\geq\min\\{j_{k}:k\in I_{f}\\}$

such that

$|f(\sum_{k\in I_{f}}a_{k}x_{k})|\leq
C(1+\frac{1}{\sqrt{m_{j_{0}}}})[h+g(\sum_{k\in I_{f}}a_{k}e_{t_{k}})].$ (15.1)

###### Proof.

Recall that $W_{(1)}$ is the increasing union of the sequence
$(W^{n}_{(1)})_{n=0}^{\infty}$ defined in Remark 5.17. We prove the statement
by induction on $n=0,1,\ldots$ for every $f\in W^{n}_{(1)}$ and every RIS.

For $n=0$ and $f\in W^{0}_{(1)}$, the fact that $I_{f}\neq\emptyset$ implies
that $I_{f}=\\{k_{0}\\}$, i.e., $f=e^{*}_{t_{k_{0}}}$ or
$f=-e^{*}_{t_{k_{0}}}$ for some $k_{0}\in I$. In either case, it is immediate
to check that $h=\text{sign}(a_{k_{0}})e^{*}_{t_{k_{0}}}$, $g=0$ and
$j_{0}=j_{k_{0}}$ are as desired.

Fix $n\in\mathbb{N}$ and assume that the conclusion holds for every $f\in
W^{n}_{(1)}$ and every RIS. Pick an $f\in W^{n+1}_{(1)}$ with
$f={m_{i}}^{-1}\sum_{l=1}^{d}f_{l}$, where $(f_{l})_{l=1}^{d}$ is an
$\mathcal{S}_{n_{i}}$-admissible sequence in $W^{n}_{(1)}$. We will first
treat the two extreme cases, namely, the cases where $i\geq\max\\{j_{k}:k\in
I_{f}\\}$ and $i<\min\\{j_{k}:k\in I_{f}\\}$.

For the first case, set $k_{0}=\max I_{f}$ and $j_{0}=j_{k_{0}}$ and choose
$k_{1}\in I_{f}$ that maximizes the quantity $|a_{k}|$ for $k\in I_{f}$. Then,
since $(x_{k})_{k\in I}$ is a RIS, items (i) and (ii) of Definition 8.5 yield
that

$\displaystyle|f(\sum_{k\in I_{f}\setminus\\{k_{0}\\}}a_{k}x_{k})|$
$\displaystyle\leq\frac{1}{m_{i}}\max\textnormal{supp}(x_{k_{0}-1})\|\sum_{k\in
I_{f}\setminus\\{k_{0}\\}}a_{k}x_{k}\|_{\infty}$
$\displaystyle\leq\frac{\max\textnormal{supp}(x_{k_{0}-1})}{m_{j_{k_{0}}}}C|a_{k_{1}}|\leq\frac{C}{\sqrt{m_{j_{k_{0}}}}}|a_{k_{1}}|$

and thus

$\displaystyle|f(\sum_{k\in I_{f}}a_{k}x_{k})|$
$\displaystyle\leq\frac{C}{\sqrt{m_{j_{k_{0}}}}}|a_{k_{1}}|+|f(a_{k_{0}}x_{k_{0}})|$
(15.2)
$\displaystyle\leq\frac{C}{\sqrt{m_{j_{k_{0}}}}}|a_{k_{1}}|+C|a_{k_{1}}|=C(1+\frac{1}{\sqrt{m_{j_{k_{0}}}}})|a_{k_{1}}|$
$\displaystyle=C(1+\frac{1}{\sqrt{m_{j_{k_{0}}}}})\text{sign}(a_{k_{1}})e^{*}_{t_{k_{1}}}(\sum_{k\in
I_{f}}a_{k}e_{t_{k}}).$

That is, $h=\text{sign}(a_{k_{1}})e^{*}_{t_{k_{1}}}$, $g=0$ and $j_{k_{0}}$
yield the conclusion.

For the second case, the inductive hypothesis implies that, for every
$l=1,\ldots,d$ with $I_{f_{l}}\neq\emptyset$, there are $g_{l}$, $h_{l}$ and
$j_{0,l}$ as in (i) - (iii) of the statement, that satisfy the conclusion for
the functional $f_{l}$. Define $J_{f}=\\{k\in I_{f}:f(x_{k})\neq
0\\}\setminus\cup_{l=1}^{d}I_{f_{l}}$. Then, for every $k\in J_{f}$,
Definition 8.5 (iii) yields that

$\displaystyle|f(a_{k}x_{k})|\leq\frac{C}{m_{i}}|a_{k}|=\frac{C}{m_{i}}\text{sign}(a_{k})e^{*}_{t_{k}}(\sum_{k\in
I_{f}}a_{k}e_{t_{k}}).$

and hence we calculate

$\displaystyle|f(\sum_{k\in I_{f}}a_{k}x_{k})|$
$\displaystyle\leq|f(\sum_{k\in\cup_{l=1}^{d}I_{f_{l}}}a_{k}x_{k})|+|f(\sum_{k\in
J_{f}}a_{k}x_{k})|$ (15.3) $\displaystyle\leq\frac{C}{m_{i}}\sum_{k\in
J_{f}}\text{sign}(a_{k})e^{*}_{t_{k}}(\sum_{k\in
I_{f}}a_{k}e_{t_{k}})+\frac{C}{m_{i}}\sum_{l=1}^{d}[(1+\frac{1}{\sqrt{m_{j_{0,l}}}})(h_{l}+g_{l})](\sum_{k\in
I_{f_{l}}}a_{k}e_{t_{k}})$ $\displaystyle\leq C(1+\frac{1}{\sqrt{m_{j_{\min
I_{f}}}}})[\frac{1}{m_{i}}(\sum_{k\in
J_{f}}\text{sign}(a_{k})e^{*}_{t_{k}}+\sum_{l=1}^{d}h_{l}+g_{l})](\sum_{k\in
I_{f}}a_{k}e_{t_{k}}).$

Define

$g=\frac{1}{m_{i}}(\sum_{k\in
J_{f}}\text{sign}(a_{k})e^{*}_{t_{k}}+\sum_{l=1}^{d}h_{l}+g_{l}).$

Moreover, for each $l=1,\ldots,d$, define

$K_{l}=\big{\\{}k\in
J_{f}:\min\\{l^{\prime}=1,\ldots,d:\textnormal{supp}(x_{k})\cap\textnormal{range}(f_{l^{\prime}})\neq\emptyset\\}=l\big{\\}}$

and

$I_{l}=\\{t_{k}:k\in
K_{l}\\}\cup\\{\textnormal{supp}(h_{l})\\}\cup\\{\min\textnormal{supp}(g_{l})\\}.$

Let us make the following remarks. First, observe that $\\#K_{l}\leq 2$. In
particular, consider the case where $K_{l}=\\{k_{1},k_{2}\\}$ for some
$l=1,\ldots,d$. Then, $k_{1}<\min I_{f_{l}}\leq\max I_{f_{l}}<k_{2}$ and since
$\textnormal{supp}(h_{l})\cup\textnormal{supp}(g_{l})$ is a subset of
$\\{t_{k}:k\in I_{f_{l}}\\}$ we have
$t_{k_{1}}<\textnormal{supp}(h_{l})<\textnormal{supp}(g_{l})<t_{k_{2}}$.
Moreover, if $l<d$ and
$\textnormal{range}(f_{l+1})\cap\textnormal{supp}(x_{k_{2}})\neq\emptyset$,
then $k_{2}\notin K_{l+1}$ and clearly $k_{2}<I_{l+1}$. In the case where
$K_{l}$ is a singleton for some $l=1,\ldots,d$, then either
$\textnormal{supp}(h_{l})<\textnormal{supp}(g_{l})<k$ or
$k<\textnormal{supp}(h_{l})<\textnormal{supp}(g_{l})$ holds for
$K_{l}=\\{k\\}$. Hence we conclude that $I_{1}<\cdots<I_{d}$. Moreover, let us
finally note that $\min\textnormal{supp}(f_{l})\leq I_{l}$ and $\\#I_{l}\leq
4$ for every $l=1,\ldots,d$. For each $l=1,\ldots,d$, let
$K_{l}=\\{k_{1}^{l},k^{l}_{2}\\}$, where $k^{l}_{2}$ or $k^{l}_{2}$ can be
ommited if necessary. Then,

$g=\frac{1}{m_{i}}\sum_{l=1}^{d}\Big{(}\text{sign}(a_{k^{l}_{1}})e^{*}_{t_{k^{l}_{1}}}+h_{l}+g_{l}+\text{sign}(a_{k^{l}_{2}})e^{*}_{t_{k^{l}_{2}}}\Big{)}.$
(15.4)

We will show that the sequence $(e^{*}_{t_{k}})_{k\in
J_{f}}{}^{\frown}(h_{l})_{l=1}^{d}{}^{\frown}(g_{l})_{l=1}^{d}$ is
$\mathcal{S}_{n_{i}+1}$-admissible, when the functionals ordered as implied by
(15.4), i.e., according to the minimum of their supports. This yields that
$g\in W_{\text{aux}}^{(1)}$ and thus $h=0$, $g$ and $j_{0}=j_{\min I_{f}}$
satisfy the conclusion, as follows from (15.3). More specifically, we will
show that $\cup_{l=1}^{d}I_{l}\in\mathcal{S}_{n_{i}+1}$. To this end, note
that $(I_{l})_{l=1}^{d}$ is $\mathcal{S}_{n_{i}}$-admissible, since
$(f_{l})_{l=1}^{d}$ is $\mathcal{S}_{n_{i}}$-admissible, $I_{1}<\cdots<I_{d}$
and $\min\textnormal{supp}(f_{l})\leq I_{l}$ for every $l=1,\ldots,d$. Thus
$\cup_{l=1}^{d}I_{l}\in\mathcal{S}_{n_{i}}*\mathcal{A}_{4}$, since
$\\#I_{l}\leq 4$ for all $l=1,\ldots,d$. Using item (ii) of Lemma 2.1 and the
fact that $4\leq\min\textnormal{supp}(x_{\min I})$ we conclude that
$\cup_{l=1}^{d}I_{l}\in\mathcal{S}_{n_{i}+1}$.

Finally, in the remaining case where $\min\\{j_{k}:k\in I_{f}\\}\leq
i<\max\\{j_{k}:k\in I_{f}\\}$, define $I^{1}_{f}=\\{k\in I_{f}:j_{k}\leq i\\}$
and $I^{2}_{f}=\\{k\in I_{f}:j_{k}>i\\}$ and observe that $I_{f}=I_{f}^{1}\cup
I^{2}_{f}$, $\max\\{j_{k}:k\in I^{1}_{f}\\}\leq i$ and $i<\min\\{j_{k}:k\in
I^{2}_{f}\\}$. Applying the result of the first case for $(x_{k})_{k\in
I^{1}_{f}}$ and that of the second for $(x_{k})_{k\in I^{2}_{f}}$ we have

$\displaystyle|f(\sum_{k\in I_{f}}a_{k}x_{k})|$
$\displaystyle\leq|f(\sum_{k\in I^{1}_{f}}a_{k}x_{k})|+|f(\sum_{k\in
I^{2}_{f}}a_{k}x_{k})|$ (15.5) $\displaystyle\leq
C(1+\frac{1}{\sqrt{m_{j_{\max I^{1}_{f}}}}})h(\sum_{k\in
I_{f}}a_{k}e_{t_{k}})+|f(\sum_{k\in I^{2}_{f}}a_{k}x_{k})|$ $\displaystyle\leq
C(1+\frac{1}{\sqrt{m_{j_{\max I^{1}_{f}}}}})h(\sum_{k\in
I_{f}}a_{k}e_{t_{k}})+C(1+\frac{1}{\sqrt{m_{j_{\min I^{2}_{f}}}}})g(\sum_{k\in
I_{f}}a_{k}e_{t_{k}})$ $\displaystyle\leq C(1+\frac{1}{\sqrt{m_{j_{\max
I^{1}_{f}}}}})[h+g(\sum_{k\in I_{f}}a_{k}e_{t_{k}})]$

where $h=\text{sign}(a_{k_{1}})e^{*}_{t_{k_{1}}}$, $k_{1}\in I^{1}_{f}$
maximizes the quantity $|a_{k}|$ for $k\in I^{1}_{f}$ and $g\in
W_{\text{aux}}^{(1)}$ with $w(g)=w(f)$. ∎

######  Remark 15.4.

Let $(x_{k})_{k\in I}$ and $f$ be as in the statement of Proposition 15.3.

  * (i)

Define $E=\textnormal{range}(f)$ and note that the sequence
$(x^{\prime}_{k})_{k\in I^{\prime}_{f}}$ where $x^{\prime}_{k}=x_{k}|_{E}$,
$k\in I^{\prime}_{f}$, is also a $(C,(j_{k})_{k\in I^{\prime}_{f}})$-RIS.
Then,

$f(\sum_{k\in I}a_{k}x_{k})=f(\sum_{k\in I^{\prime}_{f}}a_{k}x^{\prime}_{k})$

and $\\{k\in
I^{\prime}_{f}:\textnormal{supp}(x^{\prime}_{k})\subset\textnormal{range}(f)\\}=I^{\prime}_{f}$.
Hence, the basic inequality yields $h,g$ and $j_{0}$ as in items (i) - (iii)
such that

$|f(\sum_{k\in I}a_{k}x_{k})|\leq
C(1+\frac{1}{\sqrt{m_{j_{0}}}})[h+g(\sum_{k\in
I^{\prime}_{f}}a_{k}e_{z_{k}})].$

where $z_{k}=\max\textnormal{supp}(x^{\prime}_{k})$, $k\in I^{\prime}_{f}$.

  * (ii)

Let $j\in\mathbb{N}$. It follows from the proof of Proposition 15.3 that if
$f$ has a tree analysis $(f_{\alpha})_{\alpha\in\mathcal{A}}$ such that
$I_{f_{\alpha}}=\emptyset$ for every $\alpha\in\mathcal{A}$ with
$w(f_{\alpha})=m_{j}$, then the functional $g\in
W_{\text{aux}}^{(1)}\cup\\{0\\}$ that the basic inequality yields for
$(x_{k})_{k\in I}$ and $f$ has a tree analysis
$(g_{\beta})_{\beta\in\mathcal{B}}$ with $w(g_{\beta})\neq m_{j}$ for every
$\beta\in\mathcal{B}$, whenever $g\neq 0$.

####  15.3. Evaluations on Standard Exact Pairs

We prove the following lemma, which yields Proposition 8.11 as an immediate
corollary.

######  Lemma 15.5.

For every $(C,m_{j_{0}})$-SEP $(x,f)$ the following hold.

  * (i)

For every $f^{\prime}\in W_{(1)}$

$\big{|}f^{\prime}(x)\big{|}\leq\begin{cases}\frac{C}{m_{j_{0}}}(1+\frac{1}{\sqrt{m_{j_{0}}}}),\quad\quad&f^{\prime}=\pm
e_{i}^{*}\text{ for some }i\in\mathbb{N}\\\
C(1+\frac{1}{\sqrt{m_{j_{0}}}})[\frac{1}{m_{j_{0}}}+\frac{m_{j_{0}}}{w(f^{\prime})}],\quad\quad&w(f^{\prime})\geq
m_{j_{0}}\\\
C(1+\frac{1}{\sqrt{m_{j_{0}}}})[\frac{1}{m_{j_{0}}}+\frac{2}{w(f^{\prime})}],\quad\quad&w(f^{\prime})<m_{j_{0}}\end{cases}$

  * (ii)

If $f^{\prime}\in W_{(1)}$ with a tree analysis
$(f^{\prime}_{\alpha})_{\alpha\in\mathcal{A}}$ such that
$I_{f^{\prime}_{\alpha}}=\emptyset$ for every $\alpha\in\mathcal{A}$ with
$w(f^{\prime}_{\alpha})=m_{j_{0}}$, then

$|f^{\prime}(x)|\leq\frac{3C}{m_{j_{0}}}(1+\frac{1}{\sqrt{m_{j_{0}}}}).$

###### Proof.

Let $(x_{k})_{k=1}^{n}$ be a $(C,(j_{k})_{k=1}^{n})$-RIS witnessing that
$(x,f)$ is a $(C,m_{j_{0}})$-SEP. Applying Proposition 15.3 , we obtain $h$
and $g$ as in items (i) and (ii) respectively, that satisfy (15.1) for $x$ and
$f^{\prime}$, namely,

$|f^{\prime}(x)|\leq
Cm_{j_{0}}(1+\frac{1}{\sqrt{m_{j_{0}}}})[h(\tilde{x})+g(\tilde{x})],$

where $\tilde{x}=\sum_{k\in I}a_{k}e_{z_{k}}$,
$z_{k}=\max\textnormal{supp}(x_{k}|_{\textnormal{range}(f^{\prime})})$ and
$I=\\{k=1,\ldots,n:\textnormal{supp}(x_{k})\cap\textnormal{range}(f^{\prime})\neq\emptyset\\}$.
Note that $\tilde{x}$ is a $(n_{j_{0}},m^{-2}_{j_{0}})$-b.s.c.c. and hence,
since $\textnormal{supp}(h)\in\mathcal{S}_{0}$, we have
$h(\tilde{x})<m_{j_{0}}^{-2}$.

To prove (i), first observe that if $g=0$, which is the case e.g. when
$f^{\prime}=\pm e^{*}_{i}$ for some $i\in\mathbb{N}$, then we already have
established a valid upper bound for $|f^{\prime}(x)|$. Hence, suppose that
$g\neq 0$. Then, using Lemma 15.2 and the fact that $w(g)=w(f^{\prime})$, we
obtain the following upper bounds for $g(\tilde{x})$

$g(\tilde{x})\leq\begin{cases}\frac{1}{w(f^{\prime})},\quad\quad&w(f^{\prime})\geq
m_{j_{0}}\\\
\frac{2}{w(f^{\prime})m_{j_{0}}},\quad\quad&w(f^{\prime})<m_{j_{0}}\end{cases}$

which yield the desired upper bounds for $|f^{\prime}(x)|$.

Finally, item (ii) of Remark 15.4 implies that $g$ admits a tree analysis
$(g_{\beta})_{\beta\in\mathcal{B}}$ such that $w(g_{\beta})\neq m_{j_{0}}$ for
every $\beta\in\mathcal{B}$. We derive the desired upper bound using item (ii)
of Lemma 15.2, which yields that $|g(\tilde{x})|\leq 2m^{-2}_{j_{0}}$. ∎

###  16\. Appendix B

We prove another version of the basic inequality that reduces evaluations on
standard exact pairs of $\mathfrak{X}^{(2)}_{\text{awi}}$ to evaluations on
the basis of an auxiliary space. The results are almost identical those of
Appendix A and we include them for completeness.

####  16.1. The Auxiliary Space

######  Definition 16.1.

Let $W_{\text{aux}}^{(2)}$ be the minimal subset of $c_{00}(\mathbb{N})$ such
that

  * (i)

$\pm e^{*}_{i}$ is in $W_{\text{aux}}^{(2)}$ for all $i\in\mathbb{N}$ and

  * (ii)

whenever $j\in\mathbb{N}$, $(f_{i})_{i=1}^{d}$ is an
$\mathcal{S}_{n_{j}+1}$-admissible sequence in $W_{\text{aux}}^{(2)}$ and
$\lambda_{1},\ldots,\lambda_{d}\in\mathbb{Q}$ with
$\sum_{i=1}^{d}\lambda_{i}^{2}\leq 1$, then
$f=2m_{j}^{-1}\sum_{i=1}^{d}\lambda_{i}f_{i}$ is in $W_{\text{aux}}^{(2)}$.

######  Remark 16.2.

For each $f\in W_{\text{aux}}^{(2)}$, the weight of $f$ is defined as $w(f)=0$
if $f=\pm e^{*}_{i}$ for some $i\in\mathbb{N}$ and $w(f)=m_{j}/2$ in the case
where $f=2m_{j}^{-1}\sum_{i=1}^{d}\lambda_{i}f_{i}$.

The following lemma is a slightly modified version of [13, Lemma 3.16]. We use
it to prove Lemma 16.4.

######  Lemma 16.3.

Let $f\in W_{\text{aux}}^{(2)}$ with a tree analysis
$(f_{\alpha})_{\alpha\in\mathcal{A}}$.

  * (i)

For all $j\in\mathbb{N}$ we have

$\\{k\in\textnormal{supp}(f):w_{f}(e^{*}_{k})<m_{j}\\}\in\mathcal{S}_{(\log_{2}(m_{j})-1)(n_{j-1}+1)}.$

  * (ii)

If $j\in\mathbb{N}$ is such that $w(f_{\alpha})\neq m_{j}$ for each
$\alpha\in\mathcal{A}$ then

$\\{k\in\textnormal{supp}(f):w_{f}(e^{*}_{k})<m_{j}^{2}\\}\in\mathcal{S}_{(2\log_{2}(m_{j})-1)(n_{j-1}+1)}.$

###### Proof.

The proof is similiar to [13, Lemma 3.16]. ∎

Next, we prove a lemma similar to 15.2, for the evaluations of functionals in
$W^{(2)}_{\text{aux}}$ on the $\ell_{2}$ version of basic special convex
combinations.

######  Lemma 16.4.

Let $j\in\mathbb{N}$ and $\varepsilon>0$ with $\varepsilon\leq m_{j}^{-2}$.
For every $(2,n_{j},\varepsilon)$-basic s.c.c. $x=\sum_{k\in F}c_{k}e_{k}$ the
following hold.

  * (i)

For every $f\in W_{\text{aux}}^{(2)}$

$\big{|}f(x)\big{|}\leq\begin{cases}\frac{1}{w(f)},\quad\quad&w(f)\geq
m_{j}/2\\\ \frac{2}{w(f)m_{j}},\quad\quad&w(f)<m_{j}/2\end{cases}$

  * (ii)

If $f\in W^{(2)}_{\text{aux}}$ with a tree analysis
$(f_{\alpha})_{\alpha\in\mathcal{A}}$ such that $w(f_{\alpha})\neq m_{j}$ for
all $\alpha\in\mathcal{A}$ and $\varepsilon<m_{j}^{-4}$ then
$|f(x)|<2m_{j}^{-2}$.

###### Proof.

Without loss of generality, we may assume that $\textnormal{supp}(f)\subset F$
and $f(e_{k})\geq 0$ for every $k\in F$. If $m_{j}/2\leq w(f)$, then
$\|f\|_{2}\leq 1/w(f)$ and hence

$\big{|}f(x)\big{|}\leq\|f\|_{2}\|x\|_{2}\leq\frac{1}{w(f)}.$

Suppose now that $m_{i}<m_{j}$ and let
$f=2m^{-1}_{i}\sum_{l=1}^{d}\lambda_{l}f_{l}$, where $(f_{l})_{l=1}^{d}$ is an
$\mathcal{S}_{n_{i}+1}$-admissible sequence in $W^{(2)}_{\text{aux}}$. For
$l=1,\ldots,d$, define

$D_{l}=\\{k\in\textnormal{supp}(f_{l}):w_{f_{l}}(e^{*}_{k})<m_{j}\\},\quad
F_{l}=\textnormal{supp}(f_{l})\setminus D_{l}.$

Then, Lemma 16.3 (i) implies that
$D_{l}\in\mathcal{S}_{(2\log_{2}(m_{j})-1)(n_{j-1}+1)}$ for each
$l=1,\ldots,d$ and hence, since $(f_{l})_{l=1}^{d}$ is
$S_{n_{j-1}+1}$-admissible (recall that $i<j$ since $m_{i}<m_{j}$) and
$D_{l}\subset\textnormal{supp}(f_{l})$, $l=1,\ldots,d$, we have

$D=\cup_{l=1}^{d}D_{l}\in
S_{n_{j-1}+1}*\mathcal{S}_{(2\log_{2}(m_{j})-1)(n_{j-1}+1)}=\mathcal{S}_{2\log_{2}(m_{j})(n_{j-1}+1)}.$

Therefore, since $x$ is an $(2,n_{j},\varepsilon)$-basic s.c.c. and
$2\log_{2}(m_{j})(n_{j-1}+1)<n_{j}$, we have $\sum_{k\in
D}c^{2}_{k}<\varepsilon$. Moreover, observe that for $l=1,\ldots,d$ and $k\in
F_{l}$

$f_{l}(e_{k})=\frac{\lambda_{f_{l},\alpha_{k}}}{w_{f_{l}}(e^{*}_{k})}\leq\frac{\lambda_{f_{l},\alpha_{k}}}{m_{j}}$

where $a_{k}$ is the node in the induced tree analysis of $f_{l}$ with
$f_{l,\alpha_{k}}=e^{*}_{k}$, and

$\sum_{l=1}^{d}\lambda_{l}^{2}\sum_{k\in
F_{l}}\lambda_{f_{l},\alpha_{k}}^{2}\leq 1.$

We then calculate, using the Cauchy-Schwarz inequality

$\displaystyle f(x)$
$\displaystyle=\frac{{2}}{m_{i}}(\sum_{l=1}^{d}\lambda_{l}f_{l}|_{D}(\sum_{k\in
F}c_{k}e_{k})+\sum_{l=1}^{d}\lambda_{l}f_{l}|_{\mathbb{N}\setminus
D}(\sum_{k\in F}c_{k}e_{k}))$
$\displaystyle=\frac{{2}}{m_{i}}(\sum_{l=1}^{d}\lambda_{l}\sum_{k\in
D_{l}}\frac{c_{k}\lambda_{f_{l},\alpha_{k}}}{w_{f_{l}}(e^{*}_{k})}+\sum_{l=1}^{d}\lambda_{l}\sum_{k\in
F_{l}}\frac{c_{k}\lambda_{f_{l},\alpha_{k}}}{w_{f_{l}}(e^{*}_{k})})$
$\displaystyle\leq\frac{{2}}{m_{i}}(\sum_{l=1}^{d}\lambda_{l}\sum_{k\in
D_{l}}c_{k}\lambda_{f_{l},\alpha_{k}}+\frac{1}{m_{j}}\sum_{l=1}^{d}\lambda_{l}\sum_{k\in
F_{l}}c_{k}\lambda_{f_{l},\alpha_{k}})$
$\displaystyle\leq\frac{2}{m_{i}}(\sum_{l=1}^{d}\lambda_{l}(\sum_{k\in
D_{l}}c_{k}^{2})^{\frac{1}{2}}(\sum_{k\in
D_{l}}\lambda_{f_{l},\alpha_{k}}^{2})^{\frac{1}{2}}+\frac{1}{m_{j}}\sum_{l=1}^{d}\lambda_{l}(\sum_{k\in
F_{l}}c_{k}^{2})^{\frac{1}{2}}(\sum_{k\in
F_{l}}\lambda_{f_{l},\alpha_{k}}^{2})^{\frac{1}{2}})$
$\displaystyle\leq\frac{{2}}{m_{i}}((\sum_{l=1}^{d}\lambda^{2}_{l})^{\frac{1}{2}}(\sum_{k\in
D}c_{k}^{2})^{\frac{1}{2}}+\frac{1}{m_{j}}(\sum_{l=1}^{d}\sum_{k\in
F_{l}}c_{k}^{2})^{\frac{1}{2}}(\sum_{l=1}^{d}\lambda^{2}_{l}\sum_{k\in
F_{l}}\lambda_{f_{l},\alpha_{k}}^{2})^{\frac{1}{2}})$
$\displaystyle\leq\frac{{2}}{m_{i}}(\sqrt{\varepsilon}+\frac{1}{m_{j}})\leq\frac{4}{m_{i}m_{j}}.$

Finally, if there is a tree analysis $(f_{\alpha})_{\alpha\in\mathcal{A}}$ of
$f$ such that $w(f_{\alpha})\neq m_{j}$ for every $\alpha\in\mathcal{A}$,
Lemma 16.3 (ii) implies that

$D=\\{k\in\textnormal{supp}(f):w_{f}(e^{*}_{k})<m_{j}^{2}\\}\in\mathcal{S}_{(2\log_{2}(m_{j})-1)(n_{j-1}-1)}$

and since $(2\log_{2}(m_{j})-1)(n_{j-1}-1)<n_{j}$ we have that $\sum_{k\in
D}c^{2}_{k}<\varepsilon$. Hence, using similar arguments as above, we conclude
that

$f(x)\leq\sqrt{\varepsilon}+\frac{1}{m^{2}_{j}}<\frac{2}{m_{j}^{2}}.$

∎

####  16.2. The Basic Inequality

######  Proposition 16.5 (Basic Inequality).

Let $(x_{k})_{k\in I}$ be a $(C,(j_{k})_{k\in I})$-RIS in
$\mathfrak{X}^{(2)}_{\text{awi}}$ with $4\leq\min\textnormal{supp}(x_{\min
I})$, $(a_{k})_{k\in I}$ be a sequence of nonzero scalars and $f\in W_{(2)}$
with $I_{f}\neq\emptyset$. Define $t_{k}=\max\textnormal{supp}(x_{k})$, $k\in
I$. Then there exist

  * (i)

$g\in W_{\text{aux}}^{(2)}\cup\\{0\\}$ with $w(g)=w(f)/2$ if $g\neq 0$ and
$\\{k:t_{k}\in\textnormal{supp}(g)\\}\subset I_{f}$,

  * (ii)

$h\in\\{\text{sign}(a_{k})e^{*}_{t_{k}}:k\in I_{f}\\}\cup\\{0\\}$ with
$k_{0}\in I_{f}$ and $k_{0}<\min\textnormal{supp}(g)$ if
$h=\text{sign}(a_{k_{0}})e^{*}_{t_{k_{0}}}$ and

  * (iii)

$j_{0}\geq\min\\{j_{k}:k\in I_{f}\\}$

such that

$|f(\sum_{k\in I_{f}}a_{k}x_{k})|\leq
C(1+\frac{1}{\sqrt{m_{j_{0}}}})[h+g(\sum_{k\in I_{f}}a_{k}e_{t_{k}})].$

###### Proof.

As in Proposition 15.3, we prove the statement by induction on $n=0,1,\ldots$
for every $f\in W^{n}_{(2)}$ and every RIS. The case of $n=0$ follows easily.

Fix $n\in\mathbb{N}$ and assume that the conclusion holds for every $f\in
W^{n}_{(2)}$ and every RIS. Pick an $f\in W^{n+1}_{(2)}$ with
$f={m_{i}}^{-1}\sum_{l=1}^{d}\lambda_{l}f_{l}$, where $(f_{l})_{l=1}^{d}$ is
an $\mathcal{S}_{n_{i}}$-admissible AWI sequence in $W^{n}_{(2)}$ and
$\lambda_{1},\ldots,\lambda_{d}\in\mathbb{Q}$ with
$\sum_{l=1}^{d}\lambda_{l}^{2}\leq 1$. The proof of the case where
$i\geq\max\\{j_{k}:k\in I_{f}\\}$ is identical to that of Proposition 15.3.

Suppose then that $i<\min\\{j_{k}:k\in I_{f}\\}$. The inductive hypothesis
implies that, for every $l=1,\ldots,d$ with $I_{f_{l}}\neq\emptyset$, there
are $g_{l}$, $h_{l}$ and $j_{0,l}$ as in (i) - (iii) of the statement, that
satisfy the conclusion for the functional $f_{l}$. Define $J_{f}=\\{k\in
I_{f}:f(x_{k})\neq 0\\}\setminus\cup_{l=1}^{d}I_{f_{l}}$. For every $k\in
J_{f}$, since $i<j_{k}$, Definition 8.5 (iii) yields that

$\displaystyle|f(a_{k}x_{k})|\leq(\sum_{l\in
L_{k}}\lambda_{l}^{2})^{\frac{1}{2}}\frac{C}{m_{i}}|a_{k}|=(\sum_{l\in
L_{k}}\lambda_{l}^{2})^{\frac{1}{2}}\frac{C}{m_{i}}\text{sign}(a_{k})e^{*}_{t_{k}}(\sum_{k\in
I_{f}}a_{k}e_{t_{k}}).$

where

$L_{k}=\\{l\in\\{1,\ldots,d\\}:\textnormal{supp}(x_{k})\cap\textnormal{supp}(f_{l})\neq\emptyset\\}.$

Hence we calculate

$\displaystyle|f(\sum_{k\in I_{f}}a_{k}x_{k})|\leq|f(\sum_{k\in
J_{f}}a_{k}x_{k})|+|f(\sum_{k\in\cup_{l=1}^{d}I_{f_{l}}}a_{k}x_{k})|$
$\displaystyle\leq\frac{C}{m_{i}}\sum_{k\in J_{f}}(\sum_{l\in
L_{k}}\lambda_{l}^{2})^{\frac{1}{2}}\text{sign}(a_{k})e^{*}_{t_{k}}(\sum_{k\in
I_{f}}a_{k}e_{t_{k}})+\frac{C}{m_{i}}\sum_{l=1}^{d}[(1+\frac{1}{\sqrt{m_{j_{0,l}}}})\lambda_{l}(h_{l}+g_{l})](\sum_{k\in
I_{f_{l}}}a_{k}e_{t_{k}})$ $\displaystyle\leq C(1+\frac{1}{\sqrt{m_{j_{\min
I_{f}}}}})[\frac{1}{m_{i}}(\sum_{k\in J_{f}}(\sum_{l\in
L_{k}}\lambda_{l}^{2})^{\frac{1}{2}}\text{sign}(a_{k})e^{*}_{t_{k}}+\sum_{l=1}^{d}\lambda_{l}h_{l}+\lambda_{l}g_{l})](\sum_{k\in
I_{f}}a_{k}e_{t_{k}}).$

Define

$g=\frac{2}{m_{i}}(\sum_{k\in J_{f}}\frac{1}{2}(\sum_{l\in
L_{k}}\lambda_{l}^{2})^{\frac{1}{2}}\text{sign}(a_{k})e^{*}_{t_{k}}+\sum_{l=1}^{d}\frac{\lambda_{l}}{2}h_{l}+\frac{\lambda_{l}}{2}g_{l}).$

Then, observe that each $l=1,\ldots,d$, belongs to $L_{k}$ for at most two
$k\in J_{f}$ and thus, using the same arguments as in Proposition 15.3, we
have that $g\in W_{\text{aux}}^{(2)}$ and this completes the proof for case
where $i<j_{k}$ for all $k\in I_{f}$.

Finally, the proof of the remaining case is the same as in Proposition 15.3. ∎

####  16.3. Evaluations on Standard Exact Pairs

Finally, we prove the following lemma which shows that standard exact pairs
are in fact strong exact pairs.

######  Lemma 16.6.

For every $(2,C,m_{j_{0}})$-SEP $(x,f)$ the following hold.

  * (i)

For every $g\in W$

$\big{|}g(x)\big{|}\leq\begin{cases}2C(1+\frac{1}{\sqrt{m_{j_{0}}}})[\frac{1}{m_{j_{0}}}+\frac{m_{j_{0}}}{w(g)}],\quad\quad&w(g)\geq
m_{j_{0}}\\\
2C(1+\frac{1}{\sqrt{m_{j_{0}}}})[\frac{1}{m_{j_{0}}}+\frac{2}{w(g)}],\quad\quad&w(g)<m_{j_{0}}\end{cases}$

  * (ii)

If $g\in W$ with a tree analysis $(g_{\alpha})_{\alpha\in\mathcal{A}}$ such
that $I_{g_{\alpha}}=\emptyset$ for every $\alpha\in\mathcal{A}$ with
$w(g_{\alpha})=m_{j_{0}}$, then

$|g(x)|\leq\frac{3C}{m_{j_{0}}}(1+\frac{1}{\sqrt{m_{j_{0}}}}).$

###### Proof.

Apply the basic inequality and the evaluations of functionals in
$W_{\text{aux}}^{(2)}$ on $2$-b.s.c.c. from Lemma 16.4. The proof is identical
to that of Lemma 15.5. ∎

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