11footnotetext: 2000 Mathematics Subject Classification: 05D10, 46B03

# More $\ell_{r}$ saturated $\mathcal{L}^{\infty}$ spaces

I. Gasparis  Department of Mathematics, Aristotle University of Thessaloniki,
Thessaloniki 54124, Greece<EMAIL_ADDRESS>,  M.K. Papadiamantis
National Technical University of Athens, Faculty of Applied Sciences,
Department of Mathematics, Zografou Campus, 157 80, Athens, Greece
<EMAIL_ADDRESS>and  D.Z. Zisimopoulou  National Technical University
of Athens, Faculty of Applied Sciences, Department of Mathematics, Zografou
Campus, 157 80, Athens, Greece<EMAIL_ADDRESS>
###### Abstract.

Given $r\in(1,\infty)$, we construct a new $\mathcal{L}^{\infty}$ separable
Banach space which is $\ell_{r}$ saturated .

###### Key words and phrases:

Banach theory, $\ell_{p}$ saturated, $\mathcal{L}^{\infty}$ spaces

##  1\. Introduction

The Bourgain-Delbaen spaces [7] are examples of separable
$\mathcal{L}^{\infty}$ spaces containing no isomorphic copy of $c_{0}$. They
have played a key role in the solution of the scalar-plus-compact problem by
Argyros and Haydon [3], where a Hereditarily Indecomposable
$\mathcal{L}^{\infty}$ space is presented with the property that every
operator on the space is a compact perturbation of a scalar multiple of the
identity.

There has recently been an interest in the study $\mathcal{L}^{\infty}$ spaces
of the Bourgain-Delbaen type. Freeman, Odell and Schlumprecht [8] showed that
every Banach space with separable dual is isomorphic to a subspace of a
$\mathcal{L}^{\infty}$ space having a separable dual. The aim of this paper is
to present a method of constructing, for every $1<r<\infty$, a new
$\mathcal{L}^{\infty}$ space which is $\ell_{r}$ saturated. Our approach
shares common features with the Argyros-Haydon work. More precisely we
combine, as in [3], the Bourgain-Delbaen method [7] yielding exotic
$\mathcal{L}^{\infty}$ spaces, with the Tsirelson type norms that are
equivalent to some $\ell_{r}$ norm (see [2], [4], [5]). Recall that in [9],
the original Bourgain-Delbaen spaces $\mathfrak{X}_{a,b}$ with $a<1$,
$b<\frac{1}{2}$ and $a+b>1$ where shown to be $\ell_{p}$ saturated for $p$
determined by the formulas $\frac{1}{p}+\frac{1}{q}=1$ and $a^{q}+b^{q}=1$.

This paper is organized as follows. In the second section, for a given
$r\in(1,\infty)$, we construct a Banach space $\mathfrak{X}_{r}$. To do this,
we first choose $n\in\operatorname{\mathbb{N}}$, $n>1$, and a finite sequence
$\overline{b}=(b_{1},b_{2},\ldots,b_{n})$ of positive real numbers with
$b_{1}<1$, $b_{2},b_{3},\ldots,b_{n}<\frac{1}{2}$ such that
$\sum_{i=1}^{n}b_{i}^{r^{\prime}}=1$ and $\frac{1}{r}+\frac{1}{r^{\prime}}=1$.
The definition of $\mathfrak{X}_{r}$ combines the Bourgain-Delbaen method with
the Tsirelson type space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$
which will be later proved to be isomorphic to $\ell_{r}$. In particular, if
$b_{1}=b_{2}=\ldots=b_{n}=\theta$,
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$
coincides with
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\theta)$ and the
latter is known to be isomorphic to $\ell_{p}$ for some $p\in(1,\infty)$ (see
[4]). It is worth noticing that for $n=2$ the spaces $\mathfrak{X}_{r}$
essentially coincide with the original Bourgain-Delbaen spaces
$\mathfrak{X}_{a,b}$. Thus, our construction of $\mathcal{L}^{\infty}$ spaces
which are $\ell_{r}$ saturated spaces, can be considered as a generalization
of the Bourgain-Delbaen method. We must point out here that when $n=2$, our
proof of the fact that $\mathfrak{X}_{r}$ is $\ell_{r}$ saturated, differs
from Haydon’s (see [9]) corresponding one for $\mathfrak{X}_{a,b}$. To be more
specific, $\mathfrak{X}_{r}$ has a natural FDD $(M_{k})$. Given a normalized
skipped block basis $(u_{k})$ of $(M_{k})$ with the supports of the $u_{k}$’s
lying far enough apart, then it is not hard to check that $(u_{k})$ dominates
$(e_{k})$, the natural basis of
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$. The
same holds for every normalized block basis of $(u_{k})$. To obtain a
normalized block basis of $(u_{k})$ equivalent to $(e_{k})$, we select a
sequence $I_{1}<I_{2}<\dots$ of successive finite subsets of $\mathbb{N}$ such
that $\lim_{k}\|\sum_{i\in I_{k}}u_{i}\|=\infty$. Such a choice is possible by
the domination of $(e_{k})$ by $(u_{k})$. We set $v_{k}=\|\sum_{i\in
I_{k}}u_{i}\|^{-1}\sum_{i\in I_{k}}u_{i}$ and show that some subsequence of
$(v_{k})$ is dominated by $(e_{k})$. To accomplish this we adapt the method of
the analysis of the members of a finite block basis of $(e_{k})$ with respect
to a functional in the natural norming set of
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ (see
[6]), to the context of the present construction. We believe that this
approach yields a more transparent proof than Haydon’s, at least for the upper
$\ell_{r}$ estimate.

The rest of the paper is devoted to the proof of the main property, namely
that $\mathfrak{X}_{r}$ is $\ell_{r}$ saturated. In Section 3, we define the
tree analysis of the functionals $\\{e_{\gamma}^{*}:\gamma\in\Gamma\\}$ which
is a 1-norming subset of the unit ball of $\mathfrak{X}_{r}^{*}$. The tree
analysis is similar to the corresponding one used in the Tsirelson and mixed
Tsirelson spaces [4]. In the following two sections we establish the lower and
upper norm estimates for certain block sequences in the space
$\mathfrak{X}_{r}$.

In the final section we show that every block basis of $(M_{k})$ admits a
further normalized block basis $(x_{k})$ such that every normalized block
basis of $(x_{k})$ is equivalent to the natural basis of the space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$.
Zippin’s theorem [12] yields the desired result.

##  2\. Preliminaries

In this section we define the space $\mathfrak{X}_{r}$ combining the Bourgain-
Delbaen construction [7] and the Tsirelson type constructions [2], [4].

Before proceeding, we recall some notation and terminology from [3]. Let
$n\in\operatorname{\mathbb{N}}$ and $0<b_{1},b_{2},...,b_{n}<1$ with
$\sum_{i=1}^{n}b_{i}>1$ and there exists $r^{\prime}\in(1,\infty)$ such that
$\sum_{i=1}^{n}{b_{i}}^{r^{\prime}}=1$. We may also assume without loss of
generality that $b_{1}>b_{2}>\ldots>b_{n}$. We define
$W[(\operatorname{\mathcal{A}}_{n},\overline{b})]$ to be the smallest subset
$W$ of $c_{00}(\operatorname{\mathbb{N}})$ with the following properties:

  1. (1)

$\pm e^{*}_{k}\in W$ for all $k\in\mathbb{N}$,

  2. (2)

whenever $f_{i}\in W$ and
$\max\operatorname{supp}f_{i}<\min\operatorname{supp}f_{i+1}$ for all $i$, we
have $\sum_{i\leq a}b_{i}f_{i}\in W$, provided that $a\leq n$,

We say that an element $f$ of
$W[(\operatorname{\mathcal{A}}_{n},\overline{b})]$ is of Type $0$ if $f=\pm
e_{k}^{*}$ for some $k$ and of Type I otherwise; an element of Type I is said
to have weight $b_{a}$ for some $a\leq n$ if $f=\sum_{i=1}^{a}f_{i}$ for a
suitable sequence $(f_{i})$ of successive elements of
$W[\operatorname{\mathcal{A}}_{n},\overline{b}]$.
The Tsirelson space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ is
defined to be the completion of $c_{00}$ with respect to the norm

$\|x\|=\sup\\{\langle f,x\rangle:f\in
W[\operatorname{\mathcal{A}}_{n},\overline{b}]\\}.$

We may also characterize the norm of this space implicitly as being the
smallest function $x\mapsto\|x\|$ satisfying

$\|x\|=\max\bigg{\\{}\|x\|_{\infty},\sup\sum_{i=1}^{n}b_{i}\|E_{i}x\|\bigg{\\}},$

where the supremum is taken over all sequences of finite subsets
$E_{1}<E_{2}<\cdots<E_{n}$.

We shall now present the fundamental aspects related to the Bourgain-Delbaen
construction.
For the interested readers we mention that the following method can be
characterized as the ”dual” construction of the construction presented in [3].
This characterization is based on the fact that in [3] a particular kind of
basis is given to $\ell_{1}(\Gamma)$ and the Bourgain-Delbaen type space $X$
is seen as the predual of its dual, which is $\ell_{1}(\Gamma)$.

Let $(\Gamma_{q})_{q\in{\operatorname{\mathbb{N}}}}$ be a strictly increasing
sequence of finite sets and denote their union by $\Gamma$;
$\Gamma=\cup_{q\in{\operatorname{\mathbb{N}}}}\Gamma_{q}$.
We set $\Delta_{0}=\Gamma_{0}$ and
$\Delta_{q}=\Gamma_{q}\backslash\Gamma_{q-1}$ for $q=1,2,\ldots$
Assume furthermore that to each $\gamma\in\Delta_{q}$, $q\geq 1$, we have
assigned a linear functional
$c_{\gamma}^{*}:\ell^{\infty}(\Gamma_{q-1})\rightarrow\operatorname{\mathbb{R}}$.
Next, for $n<m$ in $\operatorname{\mathbb{N}}$, we define by induction, a
linear operator
$i_{n,m}:\ell^{\infty}(\Gamma_{n})\rightarrow\ell^{\infty}(\Gamma_{m})$ as
follows:
For $m=n+1$, we define
$i_{n,n+1}:\ell^{\infty}(\Gamma_{n})\rightarrow\ell^{\infty}(\Gamma_{n+1})$ by
the rule

$(i_{n,n+1}(x))(\gamma)=\begin{cases}x(\gamma),\ \text{if
}\gamma\in\Gamma_{n}\\\ c_{\gamma}^{*}(x),\ \text{if
}\gamma\in\Delta_{n+1}\end{cases}$

for every $x\in\ell^{\infty}(\Gamma_{n})$.
Then assuming that $i_{n,m}$ has been defined, we set
$i_{n,m+1}=i_{m,m+1}\circ i_{n,m}$. A direct consequence of the above
definition is that for $n<l<m$ it holds that $i_{n,m}=i_{l,m}\circ i_{n,l}$.
Finally we denote by
$i_{n}:\ell^{\infty}(\Gamma_{n})\rightarrow\operatorname{\mathbb{R}}^{\Gamma}$
the direct limit $i_{n}=\lim_{m\rightarrow\infty}i_{n,m}$.

We assume that there exists a $C>0$ such that for every $n<m$ we have
$\|i_{n,m}\|\leq C$. This implies that $\|i_{n}\|\leq C$ and therefore
$i_{n}:\ell^{\infty}(\Gamma_{n})\to\ell^{\infty}(\Gamma)$ is a bounded linear
map. In particular, setting $X_{n}=i_{n}[\ell^{\infty}(\Gamma_{n})]$, we have
that $X_{n}\stackrel{{\scriptstyle
C}}{{\thickapprox}}\ell^{\infty}(\Gamma_{n})$ and furthermore
$(X_{n})_{n\in\operatorname{\mathbb{N}}}$ is an increasing sequence of
subspaces of $\ell^{\infty}(\Gamma)$. We also set
$\mathfrak{X}_{BD}=\overline{\bigcup\limits_{n\in\operatorname{\mathbb{N}}}X_{n}}\hookrightarrow\ell^{\infty}(\Gamma)$
equipped with the supremum norm. Evidently, $\mathfrak{X}_{BD}$ is an
$\mathcal{L}^{\infty}$ space.

Let us denote by $r_{n}:\ell^{\infty}(\Gamma)\to\ell^{\infty}(\Gamma_{n})$ the
natural restriction map, i.e. $r_{n}(x)=x|_{\Gamma_{n}}$. We will also abuse
notation and denote by
$r_{n}:\ell^{\infty}(\Gamma_{m})\to\ell^{\infty}(\Gamma_{n})$ the restriction
function from $\ell^{\infty}(\Gamma_{m})$ to $\ell^{\infty}(\Gamma_{n})$ for
$n<m$.

######  Notation 2.1.

We denote by $e_{\gamma}^{*}$ the restriction of the unit vector
$e_{\gamma}\in\ell^{1}(\Gamma)$ on the space $\mathfrak{X}_{BD}$.

We also extend the functional
$c_{\gamma}^{*}:\ell^{\infty}(\Gamma_{n})\to\operatorname{\mathbb{R}}$ to a
functional $c_{\gamma}^{*}:\mathfrak{X}_{BD}\to\operatorname{\mathbb{R}}$ by
the rule $c_{\gamma}^{*}(x)=(c_{\gamma}^{*}\circ r_{q-1})(x)$ when
$\gamma\in\Delta_{q}$.

As it is well known from [3] and [7], instead of the Schauder basis of
$\mathfrak{X}_{BD}$, it is more convenient to work with a FDD naturally
defined as follows:

For each $q\in\operatorname{\mathbb{N}}$ we set
$M_{q}=i_{q}[\ell^{\infty}(\Delta_{q})]$.
We briefly establish this fact in the following proposition and then continue
with the details of the construction of $\mathfrak{X}_{r}$.

######  Proposition 2.2.

The sequence $(M_{q})_{q\in\operatorname{\mathbb{N}}}$ is a FDD for
$\mathfrak{X}_{BD}$.

###### Proof.

For $q\geq 0$ we define the maps $P_{\\{q\\}}:\mathfrak{X}_{BD}\to M_{q}$ with

$P_{\\{q\\}}(x)=i_{q}(r_{q}(x))-i_{q-1}(r_{q-1}(x))$

It is easy to check that each $P_{\\{q\\}}$ is a projection onto $M_{q}$ and
that for $q_{1}\neq q_{2}$ and $x\in M_{q_{2}}$ we have
$P_{\\{q_{1}\\}}(x)=0$. Also we have that $\|P_{q}\|\leq 2C$. We point out
that in a similar manner one can define projections on intervals of the form
$I=(p,q]$ so that $P_{I}(x)=\sum_{i=p+1}^{q}P_{\\{i\\}}(x)$ for which we can
readily verify the formula

$P_{I}(x)=i_{q}(r_{q}(x))-i_{p}(r_{p}(x))$

Note that $\|P_{I}\|\leq 2C$. This shows that indeed $(M_{q})_{q}$ is a FDD
generating $\mathfrak{X}_{BD}$. ∎

For $x\in\mathfrak{X}_{BD}$ we denote by $\operatorname{supp}x$ the set
$\operatorname{supp}x=\\{q:P_{\\{q\\}}(x)\neq 0\\}$ and by
$\operatorname{ran}x$ the minimal interval of $\operatorname{\mathbb{N}}$
containing $\operatorname{supp}x$.

######  Definition 2.3.

A block sequence $(x_{i})_{i=1}^{\infty}$ in $\mathfrak{X}_{BD}$ is called
skipped (with respect to $(M_{q})_{q\in\operatorname{\mathbb{N}}}$), if there
is a subsequence $(q_{i})_{i=1}^{\infty}$ of $\operatorname{\mathbb{N}}$ so
that for all $i\in\operatorname{\mathbb{N}}$,
$\operatorname{maxsupp}{x_{i}}<q_{i}<\operatorname{minsupp}{x_{i+1}}$.

In the sequel, when we refer to a skipped block sequence, we consider it to be
with respect to the FDD $(M_{q})_{q\in\operatorname{\mathbb{N}}}$.

Let $q\geq 0$. For all $\gamma\in\Delta_{q}$ we set
$d_{\gamma}^{*}=e_{\gamma}\circ P_{\\{q\\}}.$ Then the family
$(d^{*}_{\gamma})_{\gamma\in\Gamma}$ consists of the biorthogonal functionals
of the FDD $(M_{q})_{q\geq 0}$. Notice that for $\gamma\in\Delta_{q}$,

$\displaystyle d_{\gamma}^{*}(x)$ $\displaystyle=$ $\displaystyle
P_{q}(x)(\gamma)=i_{q}(r_{q}(x))(\gamma)-i_{q-1}(r_{q-1}(x))(\gamma)=$
$\displaystyle=$ $\displaystyle
r_{q}(x)(\gamma)-c_{\gamma}^{*}(r_{q-1}(x))=x(\gamma)-c_{\gamma}^{*}(x)=$
$\displaystyle=$ $\displaystyle e^{*}_{\gamma}(x)-c^{*}_{\gamma}(x).$

The sequences $(\Delta_{q})_{q\in\operatorname{\mathbb{N}}}$ and
$(c_{\gamma}^{*})_{\gamma\in\Gamma}$ are determined as in [3], section 4 and
Theorem 3.5.
We give some useful notation. For fixed $n\in\operatorname{\mathbb{N}}$ and
$\overline{b}=(b_{1},b_{2},\ldots,b_{n})$ with $0<b_{1},b_{2},\ldots,b_{n}<1$,
for each $\gamma\in\Delta_{q}$ we assign

  1. (a)

$\operatorname{rank}\gamma=q$

  2. (b)

age of $\gamma$ denoted by $a(\gamma)=a$ such that $2\leq a\leq n$

  3. (c)

weight of $\gamma$ denoted by $w(\gamma)=b_{a}$

In order to proceed to the construction, we first need to fix a positive
integer $n$ and a descending sequence of positive real numbers
$b_{1},\ldots,b_{n}$ such that $b_{1}<1$, $b_{i}<\frac{1}{2}$, for every
$i=2,\ldots,n$ and $\sum_{i=1}^{n}b_{i}>1$. Let $r\in(1,\infty)$ be such that
$\sum_{i=1}^{n}b_{i}^{r^{\prime}}=1$ and $\frac{1}{r}+\frac{1}{r^{\prime}}=1$.
Now we shall define the space $\mathfrak{X}_{r}$ by using the Bourgain-Delbaen
construction that was presented in the preceding paragraphs.

We set $\Delta_{0}=\emptyset$, $\Delta_{1}=\\{0\\}$ and recursively define for
each $q>1$ the set $\Delta_{q}$.
Assume that $\Delta_{p}$ have been defined for all $p\leq q$. We set

$\displaystyle\Delta_{q+1}$ $\displaystyle=$
$\displaystyle\big{\\{}(q+1,a,p,\eta,\varepsilon e_{\xi}^{*}):\ 2\leq a\leq
n,p\leq q,\ \varepsilon=\pm 1,\ e_{\xi}^{*}\in S_{\ell^{1}(\Gamma_{q})},\
\xi\in\Gamma_{q}\setminus\Gamma_{p},$ $\displaystyle\ \ \eta\in\Gamma_{p},\
b_{a-1}=w(\eta)\big{\\}}$

For $\gamma\in\Delta_{q+1}$ it is clear that the first coordinate is the
$\operatorname{rank}$ of $\gamma$, while the second is the age $a(\gamma)$ of
$\gamma$. The functionals $(c_{\gamma}^{*})_{\gamma\in\Delta_{q+1}}$ are
defined in a way that depends on $\gamma\in\Delta_{q+1}$. Namely, let
$x\in\ell^{\infty}(\Gamma_{q})$.

  1. (i)

For $\gamma=(q+1,2,p,\eta,\varepsilon e_{\xi}^{*})$ we set

$c_{\gamma}^{*}(x)=b_{1}x(\eta)+b_{2}\varepsilon
e_{\xi}^{*}\big{(}x-i_{p,q}(r_{p}(x))\big{)}.$

  2. (ii)

For $\gamma=(q+1,a,p,\eta,\varepsilon e_{\xi}^{*})$ with $a>2$ we set

$c_{\gamma}^{*}(x)=x(\eta)+b_{a}\varepsilon
e_{\xi}^{*}\big{(}x-i_{p,q}(r_{p}(x))\big{)}.$

We may now define sequences $(i_{q})$, $(\Gamma_{q})$, $(X_{q})$ in a similar
manner as before and set
$\mathfrak{X}_{r}=\overline{\bigcup\limits_{q\in\operatorname{\mathbb{N}}}X_{q}}$.
Assuming that $(i_{q})$ is uniformly bounded by a constant C, we conclude that
the space $\mathfrak{X}_{r}$ is a subspace of $\ell_{\infty}(\Gamma)$. The
constant C is determined as in [3] Theorem 3.4, by taking
$C=\frac{1}{1-2b_{2}}$. Thus, for every $m\in\operatorname{\mathbb{N}}$,
$\|i_{m}\|\leq C$. This implies that $\|P_{I}\|\leq 2C$ for every $I$
interval.

######  Remark 2.4.

In the case of $n=2$, i.e. $\overline{b}=(b_{1},b_{2})$, the space
$\mathfrak{X_{r}}$ essentially coincides with the Bourgain-Delbaen space
$\mathfrak{X}_{b_{1},b_{2}}$, since every $\gamma\in\Gamma$ is of age 2.

######  Remark 2.5.

As it is shown in Proposition 6.2, the choice of r, based on the fixed $n$ and
$\overline{b}$, yields that
$\operatorname{\mathcal{T}}(\mathcal{A}_{n},\overline{b})\cong\ell_{r}$.
Moreover, the ingredients of the ”Tsirelson type spaces” theory that are used
throughout this paper are essentially the same with the corresponding ones in
[3]. The basic difference in our approach is that we use only one family
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ for
some appropriate $n$ and $\overline{b}$.

##  3\. The Tree Analysis of $e_{\gamma}^{*}$ for $\gamma\in\Gamma$

We begin by recalling the analysis of $e_{\gamma}^{*}$ in [3] section 4. The
only difference is that in our case all the functionals $e_{\gamma}^{*}$ have
weight depending on their age which is greater or equal to 2.

###  3.1. The evaluation Analysis of $e_{\gamma}^{*}$ for $\gamma\in\Gamma$

First we point out that for $q\in\operatorname{\mathbb{N}}$ every
$\gamma\in\Delta_{q+1}$ admits a unique analysis as follows:
Let $a(\gamma)=a\leq{n}$. Then using backwards induction we determine a
sequence of sets
$\\{p_{i},q_{i},\varepsilon_{i}e_{\xi_{i}}^{*}\\}_{i=1}^{a}\cup\\{\eta_{i}\\}_{i=2}^{a}$
with the following properties.

  1. (i)

$p_{1}<q_{1}<\cdots<p_{a}<q_{a}=q$.

  2. (ii)

$\varepsilon_{i}=\pm 1,\ \operatorname{rank}\xi_{i}\in(p_{i},q_{i}]$ for
$1\leq i\leq a$ and $\operatorname{rank}\eta_{i}=q_{i}+1$ for $2\leq i\leq a$.

  3. (iii)

$\eta_{a}=\gamma$,
$\eta_{i}=(\operatorname{rank}\eta_{i},i,p_{i},\eta_{i-1},\varepsilon_{i}e_{\xi_{i}}^{*})$
for every $i>2$
$\eta_{2}=(\operatorname{rank}\eta_{2},2,p_{2},\varepsilon_{1}\xi_{1},\varepsilon_{2}e_{\xi_{2}}^{*})$
and $(p_{1},q_{1}]=(1,\operatorname{rank}\xi_{1}]$.

######  Definition 3.1.

Let $q\in\operatorname{\mathbb{N}}$ and $\gamma\in\Gamma_{q}$. Then the
sequence
$\\{p_{i},q_{i},\varepsilon_{i}e_{\xi_{i}}^{*}\\}_{i=1}^{a}\cup\\{\eta_{i}\\}_{i=2}^{a}$
satisfying all the above properties will be called the analysis of $\gamma$.

Moreover, following similar arguments as in [3] Proposition 4.6 it holds that,

$e_{\gamma}^{*}=\sum_{i=2}^{a}d_{\eta_{i}}^{*}+\sum_{i=1}^{a}b_{i}\varepsilon_{i}e_{\xi_{i}}^{*}\circ
P_{(p_{i},q_{i}]}=\sum_{i=2}^{a}e_{\eta_{i}}^{*}\circ
P_{\\{q_{i}+1\\}}+\sum_{i=1}^{a}b_{i}\varepsilon_{i}e_{\xi_{i}}^{*}\circ
P_{(p_{i},q_{i}]}.$

We set $g_{\gamma}=\sum\limits_{i=2}^{a}d_{\eta_{i}}^{*}$ and
$f_{\gamma}=\sum\limits_{i=1}^{a}b_{i}\varepsilon_{i}e_{\xi_{i}}^{*}\circ
P_{(p_{i},q_{i}]}$.

###  3.2. The r-Analysis of the functional $e_{\gamma}^{*}$

Let $r\in{\operatorname{\mathbb{N}}}$ and $\gamma\in\Delta_{q+1}$.Let
$a(\gamma)=a\leq{n}$ and
$\\{p_{i},q_{i},\varepsilon_{i}e_{\xi_{i}}^{*}\\}_{i=1}^{a}\bigcup\\{\eta_{i}\\}_{i=2}^{a}$
the evaluation analysis of $\gamma$. We define the r-analysis of
$e_{\gamma}^{*}$ as follows:

  1. (a)

If $r\leq{p_{1}}$, then the r-analysis of $e_{\gamma}^{*}$ coincides with the
evaluation analysis of $e_{\gamma}^{*}$.

  2. (b)

If $r\geq{q_{a}}$, then we assign no r-analysis to $e_{\gamma}^{*}$ and we say
that $e_{\gamma}^{*}$ is r-indecomposable.

  3. (c)

If $p_{1}<r<q_{a}$, we define $i_{r}=\min\\{i:r<q_{i}\\}$. Note that this is
well-defined. The r-analysis of $e_{\gamma}^{*}$ is the following triplet

$\\{(p_{i},q_{i}]\\}_{i\geq i_{r}},\\{\varepsilon_{i}\xi_{i}\\}_{i\geq
i_{r}},\\{\eta_{i}\\}_{i\geq\max\\{2,i_{r}\\}}.$

where $p_{i_{r}}$ is either the same or $r$ in the case that $r>p_{i_{r}}$.

Next we introduce the tree analysis of $e_{\gamma}^{*}$ which is similar to
the tree analysis of a functional in a Mixed Tsirelson space (see [4] Chapter
II.1). Notice that the evaluation analysis and the r-analysis of
$e_{\gamma}^{*}$ form the first level of the tree analysis that we are about
to present.

We start with some notation. We denote by
$(\operatorname{\mathcal{T}},"\preceq")$ a finite partially ordered set which
is a tree. Its elements are finite sequences of natural numbers ordered by the
initial segment partial order. For every $t\in\operatorname{\mathcal{T}}$,we
denote by $S_{t}$ the immediate successors of $t$

Assume now that $(p_{t},q_{t}]_{t\in\operatorname{\mathcal{T}}}$ is a tree of
intervals of $\operatorname{\mathbb{N}}$ such that $t\preceq s$ iff
$(p_{t},q_{t}]\supset(p_{s},q_{s}]$ and $t,s$ are incomparable iff
$(p_{t},q_{t}]\cap(p_{s},q_{s}]=\emptyset$. For such a family
$(p_{t},q_{t}]_{t\in\operatorname{\mathcal{T}}}$ and $t,s$ incomparable we
shall denote by $t<s$ iff $(p_{t},q_{t}]<(p_{s},q_{s}]$ (i.e. $q_{t}<p_{s}$).

###  3.3. The Tree Analysis of the functional $e_{\gamma}^{*}$

Let $\gamma\in\Delta_{q+1}$ with $a(\gamma)=a\leq{n}$. A family of the form
$\mathcal{F}_{\gamma}=\\{\xi_{t},(p_{t},q_{t}]\\}_{t\in\operatorname{\mathcal{T}}}$
is called the tree analysis of $e_{\gamma}^{*}$ if the following are
satisfied:

  1. (1)

$\operatorname{\mathcal{T}}$ is a finite tree with a unique root denoted as
$\emptyset$.

  2. (2)

We set $\xi_{\emptyset}=\gamma$,$(p_{\emptyset},q_{\emptyset}]=(1,q]$ and let
$\\{p_{i},q_{i},\varepsilon_{i}e_{\xi_{i}}^{*}\\}_{i=1}^{a}\bigcup\\{\eta_{i}\\}_{i=2}^{a}$
the evaluation analysis of $\xi_{\emptyset}$. Set
$S_{\emptyset}=\\{(1),(2),\ldots,(a)\\}$ and for every $s=(i)\in
S_{\emptyset}$, $\\{\xi_{s},(p_{s},q_{s}]\\}=\\{\xi_{i},(p_{i},q_{i}]\\}$.

  3. (3)

Assume that for a $t\in\operatorname{\mathcal{T}}$
$\\{\xi_{t},(p_{t},q_{t}]\\}$ has been defined. There are two cases:

    1. (a)

If $e_{\xi_{t}}^{*}$ is $p_{t}$-decomposable, let

$\\{(p_{i},q_{i}]\\}_{i\geq i_{p_{t}}},\\{\varepsilon_{i}\xi_{i}\\}_{i\geq
i_{p_{t}}},\\{\eta_{i}\\}_{i\geq\max\\{2,i_{p_{t}}\\}}$

the $p_{t}$ analysis of $e_{\xi_{t}}^{*}$. We set
$S_{t}=\\{(t^{\smallfrown}i):i\geq i_{p_{t}}\\}$ and

$S_{t}^{p_{t}}=\begin{cases}S_{t},\ \text{if }\eta_{i_{p_{t}}}\
\text{exists}\\\ S_{t}\diagdown\\{(t^{\smallfrown}i_{p_{t}})\\},\
\text{otherwise}\end{cases}$

Then, for every $s=(t^{\smallfrown}i)\in S_{t}$, we set
$\\{\xi_{s},(p_{s},q_{s}]\\}=\\{\xi_{i},(p_{i},q_{i}]\\}$ where
$\\{\varepsilon_{i}\xi_{i},(p_{i},q_{i}]\\}$ is a member of the $p_{t}$
analysis of $e_{\xi_{t}}^{*}$.

    2. (b)

$e_{\xi_{t}}^{*}$ is $p_{t}$-indecomposable, then $\xi_{t}$ consists a maximal
node of $\operatorname{\mathcal{F}}_{\gamma}$.

######  Notation 3.2.

For later use we need the following:
For every $t\in\operatorname{\mathcal{T}}$ $e_{\xi_{t}}^{*}=f_{t}+g_{t}$,
where $f_{t}=\sum_{s\in S_{t}}b_{s}\varepsilon_{s}e_{\xi_{s}}^{*}\circ
P_{(p_{s},q_{s}]}$ and $g_{t}=\sum_{s\in S_{t}^{p_{t}}}d_{\eta_{s}}^{*}$ and
for $s=(t^{\smallfrown}i)\in S_{t}^{p_{t}}$,
$\eta_{(t^{\smallfrown}i)}=(\operatorname{rank}\eta_{(t^{\smallfrown}i)},i,p_{(t^{\smallfrown}i)},\eta_{(t^{\smallfrown}i-1)},\varepsilon_{(t^{\smallfrown}i)}e_{\xi_{(t^{\smallfrown}i)}}^{*})$.
In the rest of the paper, we set $f_{t}=f_{\xi_{t}}$ and $g_{t}=g_{t}$.

######  Lemma 3.3.

Let $x\in\mathfrak{X}_{r}$ and $\gamma\in\Gamma$. Then,

$e_{\gamma}^{*}(x)=\prod_{\emptyset\preceq s\preceq
t_{x}}(\varepsilon_{s}b_{s})(f_{t_{x}}+g_{t_{x}})(x),$

where $t_{x}=\max\\{t:\operatorname{ran}x\subseteq(p_{t},q_{t}]\\}$.

###### Proof.

Let
$\operatorname{\mathcal{F}}_{\gamma}=\\{\xi_{t},(p_{t},q_{t}]\\}_{t\in\operatorname{\mathcal{T}}}$
a tree analysis of $\gamma$.
If $\\{t:\operatorname{ran}x\subseteq(p_{t},q_{t}]\\}=\emptyset$, then
$e_{\gamma}^{*}(x)=f_{\emptyset}(x)+g_{\emptyset}(x)$ and the equality holds.
If $\\{t:\operatorname{ran}x\subseteq(p_{t},q_{t}]\\}\neq\emptyset$, we can
find $\\{t_{1}\prec t_{2}\prec\ldots\prec
t_{m}\\}\in\operatorname{\mathcal{T}}$ such that $t_{1}\in S_{\emptyset}$ and
$t_{m}=t_{x}$.
For every $t\in\operatorname{\mathcal{T}}$ with $t\prec t_{x}$, $g_{t}(x)=0$.
Indeed, for every $s\in S_{t}^{p_{t}}$,
$d_{\eta_{s}}^{*}(x)=e_{\eta_{s}}^{*}\circ P_{\\{q_{s}+1\\}}(x)=0$ because
$\operatorname{ran}x\subseteq(p_{t_{x}},q_{t_{x}}]\subseteq(p_{s},q_{s}]$.
So, we have that

$\displaystyle e_{\gamma}^{*}(x)$ $\displaystyle=$ $\displaystyle
f_{\emptyset}(x)=\sum_{s\in
S_{\emptyset}}b_{s}\varepsilon_{s}e_{\xi_{s}}^{*}\circ
P_{(p_{s},q_{s}]}(x)=b_{t_{1}}\varepsilon_{t_{1}}e_{\xi_{t_{1}}}^{*}(x)$
$\displaystyle=$ $\displaystyle
b_{t_{1}}\varepsilon_{t_{1}}f_{t_{1}}(x)=b_{t_{1}}\varepsilon_{t_{1}}b_{t_{2}}\varepsilon_{t_{2}}e_{\xi_{t_{2}}}^{*}\circ
P_{(p_{t_{2}},q_{t_{2}}]}(x)=b_{t_{1}}b_{t_{2}}\varepsilon_{t_{1}}\varepsilon_{t_{2}}e_{\xi_{t_{2}}}^{*}(x)$
$\displaystyle=$ $\displaystyle
b_{t_{1}}b_{t_{2}}\varepsilon_{t_{1}}\varepsilon_{t_{2}}f_{t_{2}}(x)=\ldots=\prod_{\emptyset\preceq
s\preceq t_{x}}(\varepsilon_{s}b_{s})(f_{t_{x}}+g_{t_{x}})(x)$

setting $\varepsilon_{\emptyset}=b_{\emptyset}=1$. ∎

######  Corollary 3.4.

If $(f_{t_{x}},(p_{t_{x}},q_{t_{x}}])$ is a maximal node, then
$e_{\gamma}^{*}(x)=0$.

###### Proof.

Let $(f_{t_{x}},(p_{t_{x}},q_{t_{x}}])$ be a maximal node. Then
$f_{t_{x}}(x)=0$ and $g_{t_{x}}(x)=0$ and from Lemma 3.3 we deduce that
$e_{\gamma}^{*}(x)=0$. ∎

##  4\. The lower estimate

######  Definition 4.1.

An $\phi\in W(\operatorname{\mathcal{A}}_{n},\overline{b})$ is said to be a
proper functional if it admits a tree analysis
$(\phi_{t})_{t\in\operatorname{\mathcal{T}}}$ such that for every non-maximal
node $t\in\operatorname{\mathcal{T}}$ the set $\\{\phi_{s}:s\in S_{t}\\}$ has
at least two non-zero elements.

We denote by $W_{pr}(\operatorname{\mathcal{A}}_{n},\overline{b})$ to be the
subset of $W(\operatorname{\mathcal{A}}_{n},\overline{b})$ consisting of all
proper functionals. For every $t\in\operatorname{\mathcal{T}}$ it holds that
$\phi_{t}=\sum_{s\in S_{t}}b_{s}\phi_{s}$ with $\\{b_{s}\\}_{s\in
S_{t}}\subseteq\\{b_{1},b_{2},\ldots,b_{n}\\}$ and $b_{\emptyset}=1$.

######  Lemma 4.2.

The set $W_{pr}(\operatorname{\mathcal{A}}_{n},\overline{b})$ 1-norms the
space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$.

###### Proof.

We shall show that for every $\phi\in
W(\operatorname{\mathcal{A}}_{n},\overline{b})$ there exists $g\in
W_{pr}(\operatorname{\mathcal{A}}_{n},\overline{b})$ such that $|\phi(m)|\leq
g(m)\ \forall m\in{\operatorname{\mathbb{N}}}$. Since the basis is
1-unconditional the previous statement yields the result.

To this end, let $\phi\in W(\operatorname{\mathcal{A}}_{n},\overline{b})$.
Then using a tree analysis $\\{\phi_{t}\\}_{t\in\operatorname{\mathcal{T}}}$
of $\phi$ we easily see that for every $m\in\operatorname{supp}f$, there
exists a maximal node $t_{m}\in\operatorname{\mathcal{T}}$ with
$\phi_{t_{m}}=\varepsilon_{m}e_{m}^{*}$ and
$\phi(m)=\varepsilon_{m}\prod\limits_{t<t_{m}}b_{t}$.

For every $m\in\operatorname{supp}\phi$ we set
$K_{m}=\\{t\in\operatorname{\mathcal{T}}:t<t_{m}\ \text{and}\ \\#S_{t}>1\\}$.
Then it is easy to see that the functional
$g=\sum\limits_{m\in\operatorname{supp}\phi}(\prod\limits_{t\in
K_{m}}b_{t})e_{m}^{*}$ is a functional belonging to
$W_{pr}(\operatorname{\mathcal{A}}_{n},\overline{b})$. Moreover, since
$b_{t}<1$ for every $t\in\operatorname{\mathcal{T}}$ we get that
$|\phi(m)|\leq g(m)\ \forall m\in{\operatorname{\mathbb{N}}}$. ∎

######  Lemma 4.3.

Let $\phi\in W_{pr}(\operatorname{\mathcal{A}}_{n},\overline{b})$ and
$l\in\operatorname{\mathbb{N}}$. If $\operatorname{maxsupp}\phi=l$, then
$h(\operatorname{\mathcal{T}}_{\phi})\leq l$.

###### Proof.

Let $\theta_{n}$ be the amount of nodes at the $n$ level of
$\operatorname{\mathcal{T}}_{\phi}$. Since $\phi$ is proper, it holds that
$\theta_{n+1}>\theta_{n}$ for every $n\in\operatorname{\mathbb{N}}$. Assume to
the contrary that $h(\operatorname{\mathcal{T}}_{\phi})>l$, i.e.
$h(\operatorname{\mathcal{T}}_{\phi})=l+k$ for some
$k\in\operatorname{\mathbb{N}}$. Then,

$\theta_{1}=1,\ \theta_{2}\geq 2,\ \ldots\ ,\ \theta_{l+k}\geq l+k$

Since, the $l+k$ level of $\operatorname{\mathcal{T}}_{\phi}$ consists of
functionals of the form $e_{i}^{*}$, we deduce that
$\operatorname{maxsupp}\phi\geq l+k>l$, which leads to a contradiction. ∎

######  Proposition 4.4.

Let $(x_{k})_{k\in\operatorname{\mathbb{N}}}$ be a normalized skipped block
sequence in $\mathfrak{X}_{r}$ and $(q_{k})_{k\in\operatorname{\mathbb{N}}}$ a
strictly increasing sequence of integers such that
$\operatorname{supp}x_{k}\subset(q_{k}+k,q_{k+1})$. Then, for every sequence
of positive scalars $(a_{k})_{k\in\operatorname{\mathbb{N}}}$ and for every
$l\in\operatorname{\mathbb{N}}$, it holds that

(1)
$\|\sum_{k=1}^{l}a_{k}e_{k}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}\leq
C\|\sum_{k=1}^{l}a_{k}x_{k}\|_{\infty}$

where
$(e_{k})_{k\in\operatorname{\mathbb{N}}}\subseteq\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$
and C is an upper bound for the norms of the operators $i_{m}$ in
$\mathfrak{X}_{r}$.

###### Proof.

Let $\phi\in W(\operatorname{\mathcal{A}}_{n},\overline{b})$. From Lemma 4.2
we may assume that $\phi$ is proper. We will use induction on the height of
the tree $\operatorname{\mathcal{T}}_{\phi}$.

If $h(\operatorname{\mathcal{T}}_{\phi})=0$ (i.e. $f$ is maximal), then $\phi$
is of the form $\phi=\varepsilon_{k}e_{k}^{*}$ with $\varepsilon_{k}=\pm 1$.
We observe that, $|\phi(\sum_{k=1}^{l}a_{k}e_{k})|=|a_{k}|=a_{k}$. From [3]
Proposition 4.8, we can choose
$\gamma\in\Gamma_{q_{k+1}-1}\backslash\Gamma_{q_{k}+k}$ such that
$|x_{k}(\gamma)|\geq\frac{1}{C}\|x_{k}\|=\frac{1}{C}$. Then,
$|\phi(\sum_{k=1}^{l}a_{k}e_{k})|=a_{k}\leq
C|a_{k}||x_{k}(\gamma)|=C|e_{\gamma}^{*}(a_{k}x_{k})|\leq
C|e_{\gamma}^{*}(\sum_{k=1}^{l}a_{k}x_{k})|$.

We assume that for every $\phi\in
W(\operatorname{\mathcal{A}}_{n},\overline{b})$ with
$h(\operatorname{\mathcal{T}}_{\phi})=h>0$ and
$\operatorname{maxsupp}\phi=l_{0}$, there exists $\gamma\in\Gamma$, such that:

  1. (1)

$\gamma\in\Gamma_{q_{l_{0}+1}+h}\backslash\Gamma_{q_{l_{0}+1}}$

  2. (2)

$h(\operatorname{\mathcal{T}}_{\phi})=h(\operatorname{\mathcal{F}}_{\gamma})\leq
l_{0}$

  3. (3)

$|\phi(\sum_{k=1}^{l}a_{k}e_{k})|\leq C|\sum_{k=1}^{l}a_{k}x_{k}(\gamma)|$ for
every $l\geq l_{0}$

Observe that assumption (1) yields
$x_{l_{0}}<\operatorname{rank}\gamma<x_{l_{0}+1}$, while assumption (2) gives
us that
$\operatorname{minsupp}x_{l_{0}+1}-\operatorname{maxsupp}x_{l_{0}}>h(\operatorname{\mathcal{T}}_{\phi})$.
Indeed,

$x_{l_{0}}<q_{l_{0}+1}<\operatorname{rank}\gamma\leq q_{l_{0}+1}+h\leq
q_{l_{0}+1}+l_{0}<q_{l_{0}+1}+(l_{0}+1)<x_{l_{0}+1}$ $\text{and
}\operatorname{minsupp}x_{l_{0}+1}-\operatorname{maxsupp}x_{l_{0}}>l_{0}+1>l_{0}\geq
h(\operatorname{\mathcal{F}}_{\gamma}).$

Let $\phi\in W(\operatorname{\mathcal{A}}_{n},\overline{b})$ with
$h(\operatorname{\mathcal{T}}_{\phi})=h+1$, $l_{0}=\operatorname{maxsupp}\phi$
and let $(\phi_{t})_{t\in\operatorname{\mathcal{T}}}$ the tree analysis of
$\phi$. Then, $\phi$ is of the form $\phi=\sum_{s\in
S_{\emptyset}}b_{s}\phi_{s}$, $\\#S_{\emptyset}\leq n$. We observe that for
every $s\in S_{\emptyset}$, $h(\operatorname{\mathcal{T}}_{\phi_{s}})=h$. We
set $p_{1}=1$, for every $s\in S_{\emptyset}\diagdown\\{1\\}$
$p_{s}=\min\\{q_{k}+k:k\in\operatorname{supp}\phi_{s}\\}$ and for every $s\in
S_{\emptyset}$, $r_{s}=q_{l_{s}+1}+h$ where
$l_{s}=\operatorname{maxsupp}\phi_{s}$.
We next apply the inductive hypothesis to obtain
$\xi_{s}\in\Gamma_{r_{s}}\backslash\Gamma_{q_{l_{s}}+1}$ with
$h(\operatorname{\mathcal{T}}_{\phi_{s}})=h(\operatorname{\mathcal{F}}_{\xi_{s}})$
such that

$\displaystyle|\phi_{s}(\sum_{k=1}^{l}a_{k}e_{k})|$ $\displaystyle=$
$\displaystyle|\phi_{s}(\sum_{k\in\operatorname{supp}\phi_{s}}a_{k}e_{k})|\leq
C\varepsilon_{s}\sum_{k\in\operatorname{supp}\phi_{s}}a_{k}x_{k}(\xi_{s})$
$\displaystyle=$ $\displaystyle
C\varepsilon_{s}e_{\xi_{s}}^{*}(\sum_{k\in\operatorname{supp}\phi_{s}}a_{k}x_{k})=C\varepsilon_{s}e_{\xi_{s}}^{*}\circ
P_{(p_{s},r_{s}]}(\sum_{k=1}^{l}a_{k}x_{k}),$

with $\varepsilon_{s}$ such that
$\varepsilon_{s}e_{\xi_{s}}^{*}(\sum_{k\in\operatorname{supp}\phi_{s}}a_{k}x_{k})=|\sum_{k\in\operatorname{supp}\phi_{s}}a_{k}x_{k}(\xi_{s})|$.

Let $\gamma\in\Gamma$ have analysis
$\\{p_{s},r_{s},\varepsilon_{s}e_{\xi_{s}}^{*}\\}_{s\in
S_{\emptyset}}\bigcup\\{\eta_{s}\\}_{s\in S_{\emptyset}\diagdown\\{1\\}}$
where $\eta_{s}\in\Delta_{r_{s}+1}$. Observe that
$\operatorname{rank}\xi_{s}\in{(q_{l_{s}+1},r_{s}]}\subset{(p_{s},r_{s}]}$. It
is clear that for every $s\in S_{\emptyset}\diagdown\\{1\\}$,
$d_{\eta_{s}}^{*}(\sum_{k=1}^{l}a_{k}x_{k})=0$. Indeed,

$\operatorname{supp}x_{l_{s}}<q_{l_{s}+1}<q_{l_{s}+1}+{(h+1)}=r_{s}+1\leq
q_{l_{s}+1}+(l_{s}+1)<\operatorname{supp}x_{l_{s}+1}.$

Therefore,

$\displaystyle|\phi(\sum_{k=1}^{l}a_{k}e_{k})|$ $\displaystyle\leq$
$\displaystyle\sum_{s\in
S_{\emptyset}}|b_{s}\phi_{s}(\sum_{k\in\operatorname{supp}\phi_{s}}a_{k}e_{k})|\leq
C\sum_{s\in S_{\emptyset}}b_{s}\varepsilon_{s}e_{\xi_{s}}^{*}\circ
P_{(p_{s},r_{s}]}(\sum_{k=1}^{l}a_{k}x_{k})$ $\displaystyle\leq$
$\displaystyle C|\sum_{k=1}^{l}a_{k}x_{k}(\gamma)|$

It is clear that
$h(\operatorname{\mathcal{T}}_{\phi})=h(\operatorname{\mathcal{F}}_{\gamma})\leq
l_{0}$ and $x_{l_{0}}<\operatorname{rank}\gamma<x_{l_{0}+1}$. ∎

######  Corollary 4.5.

For every block sequence in $\mathfrak{X}_{r}$ there exists a further block
sequence satisfying inequality (1).

##  5\. The upper estimate

Let $(y_{l})_{l\in\operatorname{\mathbb{N}}}$ be a normalized skipped block
sequence in $\mathfrak{X}_{r}$. From Corollary 4.5, we can find a further
block sequence of $(y_{l})_{l}$, still denoted by $(y_{l})_{l}$, satisfying
inequality (1).
Therefore, we have that

$\|\sum_{l=1}^{m}y_{l}\|_{\infty}\geq\frac{1}{C}\|\sum_{l=1}^{m}e_{l}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}$

For every $j\in\operatorname{\mathbb{N}}$, set $M_{j}=\\{1,2,\ldots,n\\}^{j}$.
It is easily checked, after identifying $M_{j}$ with $\\{1,\dots,n^{j}\\}$ for
every $j$, that the functional $f_{j}=\sum_{s\in
M_{j}}(\prod_{i=1}^{j}b_{s_{i}})e_{s}^{*}$ belongs to
$W(\operatorname{\mathcal{A}}_{n},\overline{b})$ where $s_{i}$ is the $i$-th
coordinate of $s$, for each $i=1,2,\ldots,n$ and $\sum_{s\in
M_{j}}\prod_{i=1}^{j}b_{s_{i}}=(\sum_{i=1}^{n}b_{i})^{j}$. Using the fact that
$\\#M_{j}=n^{j}$, we obtain that
$\|\sum_{l=1}^{n^{j}}e_{l}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}=\|\sum_{s\in
M_{j}}e_{s}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}\geq
f_{j}(\sum_{l=1}^{n^{j}}e_{l})=(\sum_{i=1}^{n}b_{i})^{j}$.

Also, for every $m\in\operatorname{\mathbb{N}}$ large enough we may find
$j\in\operatorname{\mathbb{N}}$ such that $n^{j+1}>m\geq n^{j}$. From the
above and the unconditionality of the basis of the space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$, it
follows that

$\|\sum_{l=1}^{m}y_{l}\|_{\infty}\geq\frac{1}{C}\|\sum_{l=1}^{m}e_{l}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}\geq\frac{1}{C}\|\sum_{l=1}^{n^{j}}e_{l}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}=(\sum_{i=1}^{n}b_{i})^{j}$

We conclude that $\|\sum_{l=1}^{m}y_{l}\|_{\infty}\stackrel{{\scriptstyle
m\to\infty}}{{\longrightarrow}}\infty$ as $\sum_{i=1}^{n}b_{i}>1$.

We next choose a further block sequence
$(x_{k})_{k\in\operatorname{\mathbb{N}}}$ of
$(y_{l})_{l\in\operatorname{\mathbb{N}}}$ with some additional properties. Let
$\varepsilon>0$ and choose a descending sequence $(\varepsilon_{k})_{k}$ of
positive reals such that $(\sum_{k=1}^{\infty}\varepsilon_{k})<\varepsilon$.
We can also find an increasing sequence $(n_{k})_{k}$ of positive integers and
a sequence $(F_{k})_{k}$ of succesive subsets of $\operatorname{\mathbb{N}}$
such that the following are satisfied:

  1. (1)

For every $k\in\operatorname{\mathbb{N}}$, $\frac{1}{n_{k}}<\varepsilon_{k}$.

  2. (2)

For every $k\in\operatorname{\mathbb{N}}$, $\|\sum_{l\in F_{k}}y_{l}\|>n_{k}$.
This is possible, due to the above notation.

We have thus constructed a normalized skipped block sequence
$(x_{k})_{k\in\operatorname{\mathbb{N}}}$ of the form $x_{k}=\sum_{l\in
F_{k}}\lambda_{l}y_{l}$, where $\lambda_{l}=\frac{1}{\|\sum_{l\in
F_{k}}y_{l}\|}$. Notice that $|\lambda_{l}|<\varepsilon_{k}$ for every $l\in
F_{k}$.

Let $\gamma\in\Gamma$ with tree analysis
$\operatorname{\mathcal{F}}_{\gamma}=\\{\xi_{t},(p_{t},q_{t}]\\}_{t\in\operatorname{\mathcal{T}}}$.
For every $k\in\operatorname{\mathbb{N}}$, we set
$t_{k}=\max\\{t:\operatorname{ran}x_{k}\subset(p_{t},q_{t}]\\}$. Notice that
if for a given $x_{k}$, $t_{k}$ is non-maximal, then there exist at least two
immediate successors of $t_{k}$, say $s_{1}$, $s_{2}$ such that the
corresponding intervals $(p_{s_{1}},q_{s_{1}}]$, $(p_{s_{2}},q_{s_{2}}]$
intersect $\operatorname{ran}x_{k}$. For later use we shall denote by
$(p_{s_{0}},q_{s_{0}}]$ the first interval in the natural order of disjoint
segments of the natural numbers that intersects $x_{k}$. Notice that $s_{0}$
is not necessarily the first element of $S_{t}$.
For the pair $\gamma$, $(x_{k})_{k\in\operatorname{\mathbb{N}}}$ and for every
$t\in\operatorname{\mathcal{T}}$ we define the following sets:
$D_{t}=\bigcup_{s\succeq t}\\{k:s=t_{k}\\}$, $K_{t}=D_{t}\backslash\cup_{s\in
S_{t}}D_{s}\ =\\{k:t=t_{k}\\}$ and $E_{t}=\\{s\in
S_{t}:D_{s}\neq\emptyset\\}$.

We now set
$x_{k}=x^{\prime}_{k}+x^{\prime\prime}_{k}+x^{\prime\prime\prime}_{k}$ where,

$x^{\prime}_{k}=x_{k}\mid_{(p_{s_{0}},q_{s_{0}}]},\ \
x^{\prime\prime}_{k}=x_{k}\mid_{\bigcup_{s\in S_{t_{k}},s\neq
s_{0}}(p_{s},q_{s}]}\ \text{and
}x^{\prime\prime\prime}_{k}=x_{k}-x^{\prime}_{k}-x^{\prime\prime}_{k}.$

######  Remark 5.1.

  1. (1)

The sets $D_{t}$,$K_{t}$,$E_{t}$ are determined by the chosen pair
$\gamma,(x_{k})_{k}$. For a different pair, these sets may differ as well. For
example, let $k\in K_{t}$, for the pair $\gamma,(x_{k})_{k}$. Then $t=t_{k}$
for $x_{k}$. By the construction of $x^{\prime}_{k}$, there exists $s_{k}\in
S_{t}$ such that $x^{\prime}_{k}=x_{k}\mid_{(p_{s_{k}},q_{s_{k}}]}$. Thus,
taking the pair $\gamma,(x^{\prime}_{k})_{k}$ the same $k$ belongs to
$K_{s_{k}}$.

  2. (2)

For every $k\in\operatorname{\mathbb{N}}$, $|g_{t_{k}}(x_{k})|\leq
2Cn\varepsilon_{k}$.
Indeed, from the definition of $(x_{k})_{k\in\operatorname{\mathbb{N}}}$ we
have that

$\displaystyle|g_{t_{k}}(x_{k})|$ $\displaystyle\leq$ $\displaystyle\sum_{s\in
S_{t_{k}}^{p_{t_{k}}}}|d_{\eta_{s}}^{*}(x_{k})|\leq\sum_{s\in
S_{t_{k}}^{p_{t_{k}}}}|e_{\eta_{s}}^{*}\circ P_{\\{q_{s}+1\\}}(\sum_{l\in
F_{k}}\lambda_{l}y_{l})|\leq$ $\displaystyle\leq$ $\displaystyle\sum_{s\in
S_{t_{k}}^{p_{t_{k}}}}\|e_{\eta_{s}}^{*}\|\|P_{\\{q_{s}+1\\}}\||\lambda_{l}^{s}|\|y_{l}^{s}\|\leq\sum_{s\in
S_{t_{k}}^{p_{t_{k}}}}2C\varepsilon_{k}\leq$ $\displaystyle\leq$
$\displaystyle 2C\varepsilon_{k}(\sharp S_{t_{k}})\leq 2Cn\varepsilon_{k}.$

  3. (3)

It is obvious that $g_{t_{k}}(x_{k})=g_{t_{k}}(x^{\prime\prime\prime}_{k})$,
$f_{t_{k}}(x^{\prime\prime\prime}_{k})=0$ and for every $t\prec t_{k}$,
$g_{t}(x^{\prime\prime\prime}_{k})=0$.

######  Lemma 5.2.

For the pairs $\gamma,(x^{\prime}_{k})_{k\in\operatorname{\mathbb{N}}}$ and
$\gamma,(x^{\prime\prime}_{k})_{k\in\operatorname{\mathbb{N}}}$ it holds that
$\\#(K_{t}\cup E_{t})\leq n$.

###### Proof.

Let $t\in\operatorname{\mathcal{T}}$ and let $k\in K_{t}$.
We set $s_{k}=\max\\{s\in
S_{t}:(p_{s},q_{s}]\cap\operatorname{ran}x^{\prime}_{k}\neq\emptyset\\}$. From
the definition of $t_{k}$, notice that $\\#S_{t}\geq 2$. It holds that
$s_{k}\not\in E_{t}$.
Indeed, from the definition of $t_{k}$, $s_{k}$ we have that
$(p_{t_{k}},q_{t_{k}}]\cap\operatorname{ran}x^{\prime}_{k}=\operatorname{ran}x^{\prime}_{k}$
and
$(p_{s_{k}},q_{s_{k}}]\cap\operatorname{ran}x^{\prime}_{k}=(p_{s_{k}},q_{s_{k}}]$.
Since $s_{k}\in S_{t_{k}}$,
$(p_{s_{k}},q_{s_{k}}]\subseteq(p_{t_{k}},q_{t_{k}}]$. It follows that
$(p_{s_{k}},q_{s_{k}}]\subseteq\operatorname{ran}x^{\prime}_{k}$.
Therefore, we can define a one-to-one map $G:K_{t}\rightarrow S_{t}\backslash
E_{t}$, hence $\\#K_{t}+\\#E_{t}\leq\\#S_{t}\leq n$.
The proof for the pair
$\gamma,(x^{\prime\prime}_{k})_{k\in\operatorname{\mathbb{N}}}$ is similar. ∎

######  Proposition 5.3.

Let $(x_{k})_{k\in\operatorname{\mathbb{N}}}$ be as above.Then for every
$\gamma\in\Gamma$ there exist $\phi_{1},\phi_{2}\in
W(\operatorname{\mathcal{A}}_{n},\overline{b})$ such that for every sequence
$(a_{k})_{k\in\operatorname{\mathbb{N}}}$ of positive scalars, for every
$l\in\operatorname{\mathbb{N}}$ it holds that,

(2)
$|\sum_{k=1}^{l}a_{k}x_{k}(\gamma)|\leq\frac{1}{b_{n}}(\phi_{1}+\phi_{2})(\sum_{k=1}^{l}a_{k}e_{k})+2Cn\varepsilon(\sum_{k=1}^{l}a_{k}^{r})^{\frac{1}{r}}$

###### Proof.

Let $\gamma\in\Delta_{q+1}witha(\gamma)=a\leq n$. Let
$\operatorname{\mathcal{F}}_{\gamma}=\\{\xi_{t},(p_{t},q_{t}]\\}_{t\in\operatorname{\mathcal{T}}}$,
where $\xi_{\emptyset}=\gamma$, be the tree analysis of $\gamma$. We may
assume that
$\bigcup_{k=1}^{l}\operatorname{ran}x_{k}\subset(p_{\emptyset},q_{\emptyset}]$.

######  Claim.

For the pairs $\gamma,(x^{\prime}_{k})_{k\in\operatorname{\mathbb{N}}}$ and
$\gamma,(x^{\prime\prime}_{k})_{k\in\operatorname{\mathbb{N}}}$ there exist
$\phi_{1},\phi_{2}\in W(\operatorname{\mathcal{A}}_{n},\overline{b})$ such
that for every sequence of positive scalars
$(a_{k})_{k\in\operatorname{\mathbb{N}}}$ and for every
$l\in\operatorname{\mathbb{N}}$, it holds that

(3)
$|f_{\emptyset}(\sum_{k=1}^{l}a_{k}x^{\prime}_{k})|\leq\frac{2C}{b_{n}}\phi_{1}(\sum_{k=1}^{l}a_{k}e_{k})$
(4)
$|f_{\emptyset}(\sum_{k=1}^{l}a_{k}x^{\prime\prime}_{k})|\leq\frac{2C}{b_{n}}\phi_{2}(\sum_{k=1}^{l}a_{k}e_{k})$

###### Proof of the Claim.

We only prove inequality 3. The proof of inequality 4 requires the same
arguments. We recall that $f_{t}=\sum_{s\in
S_{t}}b_{s}\varepsilon_{s}(f_{s}+g_{s})\circ P_{(p_{s},q_{s}]}$ for every
$t\in\operatorname{\mathcal{T}}$ non maximal. From the definition of
$(x^{\prime}_{k})_{k\in\operatorname{\mathbb{N}}}$, we have that $g_{s}\circ
P_{(p_{s},q_{s}]}(x^{\prime}_{k})=0$ for every $s\in S_{t}$.Therefore,
$f_{t}(\sum_{k\in D_{t}}a_{k}x^{\prime}_{k})=(\sum_{s\in
S_{t}}b_{s}\varepsilon_{s}f_{s}\circ P_{(p_{s},q_{s}]})(\sum_{k\in
D_{t}}a_{k}x^{\prime}_{k})$. We will use backwards induction on the levels of
the tree $\operatorname{\mathcal{T}}$, i.e we shall show that for every
$t\in\operatorname{\mathcal{T}}$ there exists $\phi_{1}^{t}\in
W(\operatorname{\mathcal{A}}_{n},\overline{b})$ with
$\operatorname{supp}\phi_{1}^{t}\subseteq D_{t}$ such that

$|f_{t}(\sum_{k\in
D_{t}}a_{k}x^{\prime}_{k})|\leq\frac{2C}{b_{n}}\phi_{1}^{t}(\sum_{k\in
D_{t}}a_{k}e_{k})$

.
Let $0<h\leq\max\\{|t|:t\in\operatorname{\mathcal{T}}\\}$
We assume that the proposition has been proved for all $t$ with $|t|=h$.
Let $t\in\operatorname{\mathcal{T}}$ with $|t|=h-1$.Then we have the following
cases:

  1. (1)

If $f_{t}$ is a maximal node, $f_{t}(\sum_{k\in D_{t}}a_{k}x^{\prime}_{k})=0$,
so there is nothing to prove. Indeed, $K=D_{t}$, therefore for every $k\in
D_{t}$, from Corollary 3.4 $f_{t}(x^{\prime}_{k})=0$ since $t=t_{k}$.

  2. (2)

If $f_{t}$ is a non-maximal node, then
$f_{t}(\sum_{k\in D_{t}}a_{k}x^{\prime}_{k})=(\sum_{s\in
S_{t}}b_{s}\varepsilon_{s}f_{s}\circ P_{(p_{s},q_{s}]})(\sum_{k\in
D_{t}}a_{k}x^{\prime}_{k})=$
$=\sum_{s\in S_{t}}b_{s}\varepsilon_{s}f_{s}(\sum_{k\in
D_{s}}a_{k}x^{\prime}_{k})+\sum_{k\in K}(\sum_{s\in
S_{t}}b_{s}\varepsilon_{s}f_{s})(a_{k}x^{\prime}_{k})$.
From the fact that, for every $k\in K_{t}$, $g_{t}(x^{\prime}_{k})=0$ we get
that

$|f_{t}(x^{\prime}_{k})|=|x^{\prime}_{k}(\xi_{t})|\leq\|x^{\prime}_{k}\|\leq
2C=2Ce_{k}^{*}(e_{k}).$

Moreover, for $s\in E_{t}$ it holds that $|s|=h-1$. For every $k\in D_{s}$,
from the inductive hypothesis we obtain

$|\sum_{s\in
S_{t}}b_{s}f_{s}(x^{\prime}_{k})|=|b_{s}f_{s}(x^{\prime}_{k})|\leq
b_{s}\frac{2C}{b_{n}}\phi_{1}^{s}(e_{k}).$

with $\phi_{1}^{s}\in W(\operatorname{\mathcal{A}}_{n},\overline{b})$ and
$\operatorname{supp}\phi_{1}^{s}\subseteq D_{s}$.
We set $\phi_{1}^{t}=(\sum_{s\in E_{t}}b_{s}\phi_{1}^{s}+\sum_{k\in
K_{t}}b_{k}e_{k}^{*})$.
From Lemma 5.2, it is easily checked that $\phi_{1}^{t}\in
W(\operatorname{\mathcal{A}}_{n},\overline{b})$ and it holds that,
$|f_{t}(\sum_{k\in
D_{t}}a_{k}x^{\prime}_{k})|\leq\frac{2C}{b_{n}}\phi_{1}^{t}(\sum_{k\in
D_{t}}a_{k}e_{k})$.

∎

Recall that
$e_{\gamma}^{*}(\sum_{k=1}^{l}a_{k}x_{k})=g_{\emptyset}(\sum_{k=1}^{l}a_{k}x_{k})+f_{\emptyset}(\sum_{k=1}^{l}a_{k}x_{k})$.
The fact that
$g_{\emptyset}(\sum_{k=1}^{l}a_{k}x^{\prime}_{k})=g_{\emptyset}(\sum_{k=1}^{l}a_{k}x^{\prime\prime}_{k})=g_{\emptyset}(\sum_{k\in{\\{m:t_{m}\neq\emptyset\\}}}a_{k}x^{\prime\prime\prime}_{k})=f_{\emptyset}(\sum_{k\in{\\{m:t_{m}=\emptyset\\}}}a_{k}x^{\prime\prime\prime}_{k})=0$
implies the following:

$\displaystyle|e_{\gamma}^{*}(\sum_{k=1}^{l}a_{k}x_{k})|$ $\displaystyle\leq$
$\displaystyle|g_{\emptyset}(\sum_{k\in{\\{m:t_{m}=\emptyset\\}}}a_{k}x^{\prime\prime\prime}_{k})|+|f_{\emptyset}(\sum_{k=1}^{l}a_{k}x^{\prime}_{k})|$
$\displaystyle+$
$\displaystyle|f_{\emptyset}(\sum_{k=1}^{l}a_{k}x^{\prime\prime}_{k})|+|f_{\emptyset}(\sum_{k\in{\\{m:t_{m}\neq\emptyset\\}}}a_{k}x^{\prime\prime\prime}_{k})|$

¿From Remark 5.1 we get that,

$|g_{\emptyset}(\sum_{k\in{\\{m:t_{m}=\emptyset\\}}}a_{k}x^{\prime\prime\prime}_{k})|\leq\sum_{k\in{\\{m:t_{m}=\emptyset\\}}}a_{k}|g_{\emptyset}(x^{\prime\prime\prime}_{k})|\leq
2Cn\sum_{k\in{\\{m:t_{m}=\emptyset\\}}}a_{k}\varepsilon_{k}.$

From Lemma 3.3 and Remark 5.1 we have that,

$\displaystyle|f_{\emptyset}(\sum_{k\in{\\{m:t_{m}\neq\emptyset\\}}}a_{k}x^{\prime\prime\prime}_{k})|$
$\displaystyle\leq$
$\displaystyle\sum_{k\in{\\{m:t_{m}\neq\emptyset\\}}}a_{k}(\prod_{t<t_{k}}b_{t})|g_{t_{k}}(x^{\prime\prime\prime}_{k})|\leq$
$\displaystyle\leq$ $\displaystyle
2C\frac{1}{2}n\sum_{k\in{\\{m:t_{m}\neq\emptyset\\}}}a_{k}\varepsilon_{k}\leq
2Cn\sum_{k\in{\\{m:t_{m}\neq\emptyset\\}}}a_{k}\varepsilon_{k}.$

Finally, we conclude that

$\displaystyle|\sum_{k=1}^{l}a_{k}x_{k}(\gamma)|$ $\displaystyle\leq$
$\displaystyle
2Cn\sum_{k\in{\\{m:t_{m}=\emptyset\\}}}a_{k}\varepsilon_{k}+\frac{2C}{b_{n}}\phi_{1}(\sum_{k=1}^{l}a_{k}e_{k})$
$\displaystyle+$
$\displaystyle\frac{2C}{b_{n}}\phi_{2}(\sum_{k=1}^{l}a_{k}e_{k})+2Cn\sum_{k\in{\\{m:t_{m}\neq\emptyset\\}}}a_{k}\varepsilon_{k}$
$\displaystyle\leq$
$\displaystyle\frac{2C}{b_{n}}(\phi_{1}+\phi_{2})(\sum_{k=1}^{l}a_{k}e_{k})+2Cn\sum_{k=1}^{l}a_{k}\varepsilon_{k}$
$\displaystyle\leq$
$\displaystyle\frac{2C}{b_{n}}(\phi_{1}+\phi_{2})(\sum_{k=1}^{l}a_{k}e_{k})+2Cn\max\\{a_{k}:k\in\operatorname{\mathbb{N}}\\}(\sum_{k=1}^{l}\varepsilon_{k})$
$\displaystyle\leq$
$\displaystyle\frac{2C}{b_{n}}(\phi_{1}+\phi_{2})(\sum_{k=1}^{l}a_{k}e_{k})+2Cn\varepsilon(\sum_{k=1}^{l}a_{k}^{r})^{\frac{1}{r}}.$

where in the last inequality we used the fact that the $\ell_{r}$ norm
dominates the $c_{0}$ norm. ∎

######  Remark 5.4.

From [4] Theorem I.4, we know that $\|\sum
a_{k}e_{k}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}\geq
M(\sum a_{k}^{r})^{\frac{1}{r}}$. This result and the previous Proposition,
yield that

$|\sum_{k=1}^{l}a_{k}x_{k}(\gamma)|\leq\frac{2C}{b_{n}}(\phi_{1}+\phi_{2})(\sum_{k=1}^{l}a_{k}e_{k})+\frac{2Cn\varepsilon}{M}\|\sum_{k=1}^{l}a_{k}e_{k}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}.$

For $\varepsilon=\frac{M}{nb_{n}}$,

$|\sum_{k=1}^{l}a_{k}x_{k}(\gamma)|\leq\frac{6C}{b_{n}}\|\sum_{k=1}^{l}a_{k}e_{k}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}.$

Therefore,

(5)
$\|\sum_{k=1}^{l}a_{k}x_{k}\|_{\infty}\leq\frac{6C}{b_{n}}\|\sum_{k=1}^{l}a_{k}e_{k}\|_{\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})}.$

######  Corollary 5.5.

For every block sequence in $\mathfrak{X}_{r}$ there exists a further block
sequence satisfying inequality (5).

##  6\. The main result

######  Proposition 6.1.

Let $(x_{k})_{k\in\operatorname{\mathbb{N}}}$ be a skipped block sequence in
$\mathfrak{X}_{r}$ satisfying
$\operatorname{minsupp}x_{k+1}>\operatorname{maxsupp}x_{k}+k$ and the
conditions of Proposition 5.3. Then $(x_{k})_{k\in\operatorname{\mathbb{N}}}$
is equivalent to the basis of the Tsirelson space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ for
$n$ and $\overline{b}$ determined as before.

###### Proof.

It is an immediate consequence of Propositions 4.4, 5.3 and Remark 5.4. ∎

######  Proposition 6.2.

The space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ is
isomorphic to $\ell_{p}$ for some $p\in(1,\infty)$.

###### Proof.

In a similar manner as in [4] Theorem I.4, one can see that for every
normalized block sequence $(x_{k})_{k}$ of the basis $(e_{j})_{j}$ and for
every scalar sequence $(a_{k})$ it holds that, $\|\sum
a_{k}x_{k}\|\leq\frac{2}{b_{n}}\|\sum a_{k}e_{k}\|$. Zippin’s Theorem [12]
yields that
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ is
isomorphic to some $\ell_{p}$ for some $p\in(1,\infty)$.∎

######  Remark 6.3.

An alternative proof could also be derived using the Results in Sections 4 and
5. Indeed, let $(y_{l})_{l\in\operatorname{\mathbb{N}}}$ be a skipped block
sequence in $\mathfrak{X}_{r}$. Then, there exists a further block sequence
$(x_{k})_{k\in\operatorname{\mathbb{N}}}$ satisfying simultaneously the
assumptions of Corollaries 4.5 and 5.5. Therefore,
$(x_{k})_{k\in\operatorname{\mathbb{N}}}$ satisfies the assumptions of
Proposition 6.1.

Let’s observe that every further block sequence $(z_{k})_{k}$ of $(x_{k})_{k}$
is also skipped block and satisfies Proposition 6.1, thus it is equivalent to
the basis of the space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$.
Hence, every block sequence $(z_{n})_{n}$ of $(x_{k})_{k}$ is equivalent to
$(x_{k})_{k}$. Zippin’s theorem [12] yields that the space
$\overline{<(x_{k})_{k}>}$ is isomorphic to some $\ell_{p}$. Therefore,
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})\cong\ell_{p}$
for some $p\in(1,\infty)$.

In order to determine the exact value of $p$, we need the following
Proposition.

######  Proposition 6.4.

The space
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ is
isomorphic to $\ell_{r}$ with $\frac{1}{r}+\frac{1}{r^{\prime}}=1$ and
$\sum_{i=1}^{n}b_{i}^{r^{\prime}}=1$.

###### Proof.

First, let observe that for every $x\in c_{00}$, $\|x\|\leq\|x\|_{r}$. We
shall use induction on the cardinality of $\operatorname{supp}x$. If
$|\operatorname{supp}x|=1$, it is trivial. Assume that it holds for every
$y\in c_{00}$ with $|\operatorname{supp}y|\leq n$ and let $x\in c_{00}$ with
$|\operatorname{supp}x|=n+1$. Then either $\|x\|=\|x\|_{\infty}$ or
$\|x\|=\sum_{i=1}^{n}b_{i}\|E_{i}x\|$ for some appropriate subsets
$E_{1}<E_{2}<\ldots<E_{n}$. In the first case, there is nothing to prove as
for every $p\in[r,\infty)$ $\|x\|_{\infty}\leq\|x\|_{p}$. Therefore we only
need to deal with the second case.

It suffices to observe that for every $i=1,2,\ldots,n$, the cardinality of
$\operatorname{supp}E_{i}x$ is less than $\operatorname{supp}x$ and thus,
using the inductive hypothesis along with $H\ddot{o}lder^{\prime}s$
inequality, we get that

$\|x\|\leq\sum_{i=1}^{n}b_{i}\|E_{i}x\|_{r}\leq(\sum_{i=1}^{n}b_{i}^{r^{\prime}})^{\frac{1}{r^{\prime}}}(\sum_{i=1}^{n}\|E_{i}x\|_{r}^{r})^{\frac{1}{r}}=\|x\|_{r}.$

By combining the preceding argument with Proposition 6.2, we conclude that
$\operatorname{\mathcal{T}}(\operatorname{\mathcal{A}}_{n},\overline{b})$ is
isomorphic to $\ell_{p}$ for some $p\in[r,\infty)$.
For every $l\in\operatorname{\mathbb{N}}$ set $M_{l}=\\{1,2,\ldots,n\\}^{l}$.
We have already mentioned that for every $l\in\operatorname{\mathbb{N}}$ the
functional $f_{l}=\sum_{s\in M_{l}}(\prod_{i=1}^{l}b_{s_{i}})e_{s}^{*}$
belongs to $W(\operatorname{\mathcal{A}}_{n},\overline{b})$ where $s_{i}$ is
the $i$-th coordinate of $s$, for each $i=1,2,\ldots,n$ and $\sum_{s\in
M_{l}}\prod_{i=1}^{l}b_{s_{i}}=(\sum_{i=1}^{n}b_{i})^{l}$. We set
$a_{s}=\prod_{i=1}^{l}b_{s_{i}}$ and $x_{l}=\sum_{s\in
M_{l}}a_{s}^{\frac{r^{\prime}}{r}}e_{s}$. It is easily seen that for every
$l\in\operatorname{\mathbb{N}}$, $\|x_{l}\|=1$. Indeed,

$\|x_{l}\|\leq\|x_{l}\|_{r}=(\sum_{s\in
M_{l}}a_{s}^{r^{\prime}})^{\frac{1}{r}}=(\sum_{i=1}^{n}b_{i}^{r^{\prime}})^{\frac{l}{r}}=1=f_{l}(x_{l})\leq\|x_{l}\|.$

We claim that for $p^{\prime}>r$ and every $\varepsilon>0$ there exists
$l\in\operatorname{\mathbb{N}}$ such that
$\|x_{l}\|_{p^{\prime}}<\varepsilon$. If the claim holds we are done as $p$
coincides with $r$.
Proof of the Claim: Notice that for $p^{\prime}>r$,
$\sum_{i=1}^{n}b_{i}^{\frac{r^{\prime}}{r}p^{\prime}}=\sum_{i=1}^{n}b_{i}^{r^{\prime}(1+\delta)}$
for some $0<\delta<1$. But for every $i=1,2,\ldots,n$ $b_{i}<1$, and therefore

$\sum_{i=1}^{n}b_{i}^{r^{\prime}(1+\delta)}<\sum_{i=1}^{n}b_{i}^{r^{\prime}}=1.$

Thus, there exists $l\in\operatorname{\mathbb{N}}$ such that
$(\sum_{i=1}^{n}b_{i}^{r^{\prime}(1+\delta)})^{l}<\varepsilon^{p^{\prime}}$.
Then for this $l$,

$\|x_{l}\|_{p^{\prime}}=(\sum_{s\in
M_{l}}a_{s}^{\frac{r^{\prime}}{r}p^{\prime}})^{\frac{1}{p^{\prime}}}=(\sum_{s\in
M_{l}}a_{s}^{r^{\prime}(1+\delta)})^{\frac{1}{p^{\prime}}}=(\sum_{i=1}^{n}b_{i}^{r^{\prime}(1+\delta)})^{\frac{l}{p^{\prime}}}<\varepsilon.$

∎

######  Theorem 6.5.

For every $r\in(1,\infty)$ the space $\mathfrak{X}_{r}$ is $\ell_{r}$
saturated.

###### Proof.

As it was mentioned in the above Remark, for every skipped block sequence in
$\mathfrak{X}_{r}$ we can find a further block sequence $(x_{k})_{k}$ such
that the space $\overline{<(x_{k})_{k}>}$ is isomorphic to $\ell_{r}$. ∎

######  Remark 6.6.

From the previous Theorem, we deduce that the space $\mathfrak{X}_{r}$ is a
separable $\mathcal{L}^{\infty}$ space which does not contain $\ell_{1}$.
Therefore, the results of D.Lewis-C.Stegall [10] and A. Pelczyński [11] yields
that $\mathfrak{X}_{r}^{*}$ is isomorphic to $\ell_{1}$. Alternatively, one
can use the corresponding argument of D. Alspach [1] and show directly that
$(M_{q})$ is a shrinking FDD for $\mathfrak{X}_{r}$. It then follows that
$(e_{\gamma}^{*})_{\gamma\in\Gamma}$ is a basis for $\mathfrak{X}_{r}^{*}$,
equivalent to the usual $\ell_{1}$-basis.

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