# On the UMD constants for a class of iterated $L_{p}(L_{q})$ spaces

Yanqi QIU  Inst. Math. Jussieu, Équipe d’Analyse Fonctionnelle  Université
Paris VI, 75252 Paris Cedex 05, France<EMAIL_ADDRESS>
###### Abstract.

Let $1<p\neq q<\infty$ and $(D,\mu)=(\\{\pm
1\\},\frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1})$. Define by recursion:
$X_{0}=\mathbb{C}$ and $X_{n+1}=L_{p}(\mu;L_{q}(\mu;X_{n}))$. In this paper,
we show that there exist $c_{1}=c_{1}(p,q)>1$ depending only on $p,q$ and
$c_{2}=c_{2}(p,q,s)$ depending on $p,q,s$, such that the $\text{UMD}_{s}$
constants of $X_{n}$’s satisfy $c_{1}^{n}\leq C_{s}(X_{n})\leq c_{2}^{n}$ for
all $1<s<\infty$. Similar results will be showed for the analytic UMD
constants. We mention that the first super-reflexive non-UMD Banach lattices
were constructed by Bourgain. Our results yield another elementary
construction of super-reflexive non-UMD Banach lattices, i.e. the inductive
limit of $X_{n}$, which can be viewed as iterating infinitely many times
$L_{p}(L_{q})$.

###### Key words and phrases:

UMD property, analytic UMD property, iterated $L_{p}(L_{q})$ spaces, super-
reflexive non-UMD Banach lattices

The author was partially supported by ANR grant 2011 BS01 008 01

##  1\. Introduction

A Banach space $X$ is UMD if for all (or equivalently, for some) $1<s<\infty$
there is a constant $C>0$ depending only on $s$ and $X$ such that

(1)
$\displaystyle\sup_{\varepsilon_{k}\in\\{-1,1\\}}\|\sum_{k=0}^{n}\varepsilon_{k}df_{k}\|_{L_{s}(X)}\leq
C\|\sum_{k=0}^{n}df_{k}\|_{L_{s}(X)}$

for all $n\geq 0$ and all $X$-valued martingale difference sequences
$(df_{k})_{k=0}^{n}$. The best such $C$ is called the $\text{UMD}_{s}$
constant of $X$ and will be denoted by $C_{s}(X)$ in the sequel. It is well-
known that in the above definition, we can restrict to the dyadic martingale
differences and the best constant remains the same. The UMD property for
Banach spaces was introduced by Maurey and Pisier. The reader is refered to
Burkholder’s papers [5, 7] for the details of the UMD property.

Let $\mathbb{T}=\\{z\in\mathbb{C}:|z|=1\\}$ be the one dimensional torus
equipped with the normalised Haar measure $m$. Consider the canonical
filtration on the probability space
$(\mathbb{T}^{\mathbb{N}},m^{\otimes\mathbb{N}})$ defined by

$\sigma(z_{0})\subset\sigma(z_{0},z_{1})\subset\cdots\subset\sigma(z_{0},z_{1},\cdots,z_{n})\subset\cdots.$

By definition, a Hardy martingale in $L_{s}(\mathbb{T}^{\mathbb{N}};X)$ is a
martingale $f=(f_{n})_{n\geq 0}$ with respect to the canonical filtration such
that $\sup_{n}\|f_{n}\|_{L_{s}}<\infty$, and such that the martingale
difference $df_{n}=f_{n}-f_{n-1}$ (by convention, $df_{0}:=f_{0}$) is analytic
in the last variable $z_{n}$, i.e., $df_{n}$ has the form:

$df_{n}(z_{0},\cdots,z_{n-1},z_{n})=\sum_{k\geq
1}\phi_{n,k}(z_{0},\cdots,z_{n-1})z_{n}^{k}.$

In the above definition of UMD spaces, if the Banach space is over the complex
field $\mathbb{C}$, and if we restrict to the Hardy martingales, then a
different class of Banach spaces is defined, i.e. the analytic UMD class (AUMD
by abreviation). The best constant is called the $\text{AUMD}_{s}$ constant of
$X$ and will be denoted by $C_{s}^{a}(X)$. Note that UMD implies AUMD but not
conversely, for instance, $L_{1}(\mathbb{T},m)$ is an AUMD space which is not
UMD (cf. [9]).

It is well-known that UMD implies super-reflexivity but not conversely. The
first super-reflexive non-UMD Banach space was constructed by Pisier in [11].
Super-reflexive non-UMD Banach lattices were later constructed by Bourgain in
[2, 3]. We refer to Rubio de Francia’s paper [13] for some open problems
related to the super-reflexive non-UMD Banach lattices.

The main topic of this paper is the investigation of the UMD constants of a
family of iterated $L_{p}(L_{q})$-spaces. As a consequence of our results, we
give an elementary construction of super-reflexive non-UMD Banach lattices.

##  2\.  Some elementary inequalities

We will use the following lemma.

######  Lemma 2.1.

Let $(\Omega,\nu)$ be a measure space such that $\nu$ is finite. Suppose that
$\alpha\neq 1$ and $0<\alpha<\infty$. If $F,f\in L_{\alpha}(\Omega,\nu)\bigcap
L_{1}(\Omega,\nu)$ satisfy

$\int(|F|+|g|)^{\alpha}d\nu\leq\int(|f|+|g|)^{\alpha}d\nu$

for all $g\in L_{\infty}(\Omega,\nu)$. Then $|F|\leq|f|$ a.e..

###### Proof.

Consider first those $g\in L_{\infty}(\Omega,\nu)$ such that there exists
$\delta>0$ and $|g|\geq\delta$ a.e.. If $F,f$ satisfy the condition in the
statement, then for all $\varepsilon>0$, we have

(2)
$\displaystyle\int(\varepsilon|F|+|g|)^{\alpha}d\nu\leq\int(\varepsilon|f|+|g|)^{\alpha}d\nu.$

By the mean value theorem, there exists $\theta=\theta_{\varepsilon}\in(0,1)$,
such that

$\frac{(\varepsilon|f|+|g|)^{\alpha}-|g|^{\alpha}}{\varepsilon}=\alpha(\theta\varepsilon|f|+|g|)^{\alpha-1}|f|.$

If $\alpha<1$, then
$(\theta\varepsilon|f|+|g|)^{\alpha-1}|f|\leq|g|^{\alpha-1}|f|\in
L_{1}(\Omega,\nu)$ and if $\alpha>1$, then for $0<\varepsilon<1$, we have
$0<\theta\varepsilon<1$ and hence
$(\theta\varepsilon|f|+|g|)^{\alpha-1}|f|\leq
2^{\alpha-1}(|f|^{\alpha}+|g|^{\alpha-1}|f|)\in L_{1}(\Omega,\nu)$. By the
dominated convergence theorem, we have

$\lim_{\varepsilon\to
0^{+}}\frac{\int(\varepsilon|f|+|g|)^{\alpha}d\nu-\int|g|^{\alpha}d\nu}{\varepsilon}=\alpha\int|f||g|^{\alpha-1}d\nu.$

The same equality holds for $F$. Combining this with (2), we get

$\int|F||g|^{\alpha-1}d\nu\leq\int|f||g|^{\alpha-1}d\nu.$

Replacing $g$ by $|g|^{\frac{1}{\alpha-1}}$ yields

$\int|F||g|d\nu\leq\int|f||g|d\nu.$

By approximation, the above inequality holds for all $g\in
L_{\infty}(\Omega,\nu)$. Hence $|F|\leq|f|$ a.e., as announced. ∎

######  Proposition 2.2.

Let $(\Omega,\nu)$ be a measure space such that $\nu$ is finite. Suppose that
$1\leq p\neq q<\infty$. If $F,f\in L_{p}(\Omega,\nu)\bigcap L_{q}(\Omega,\nu)$
satisfy

$\int(|F|^{q}+|g|^{q})^{p/q}d\nu\leq\int(|f|^{q}+|g|^{q})^{p/q}d\nu$

for all $g\in L_{\infty}(\Omega,\nu)$. Then $|F|\leq|f|$ a.e..

###### Proof.

This is just a reformulation of Lemma 2.1. ∎

Let $D=\\{-1,1\\}$ be the Bernoulli probability space equipped with the
measure $\mu=\frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$. For any $1\leq
q\leq\infty$, the 2-dimensional $\ell_{q}$-space will be denoted by
$\ell_{q}^{(2)}$.

######  Proposition 2.3.

Suppose that $1\leq p\neq q\leq\infty$. Let $P$ be the projection on
$L_{p}(\mu;\ell_{q}^{(2)})$ defined by

$\begin{array}[]{cccc}P:&L_{p}(\mu;\ell_{q}^{(2)})&\rightarrow&L_{p}(\mu;\ell_{q}^{(2)})\\\
&(f,g)&\mapsto&(\mathbb{E}f,g)\end{array},$

where $\mathbb{E}$ is the expectation. Then $P$ is not contractive.

###### Proof.

Assume first that both $p,q$ are finite. If $P$ is contractive, then for any
two functions $f$ and $g$, we have

$\int(|\mathbb{E}f|^{q}+|g|^{q})^{p/q}d\mu\leq\int(|f|^{q}+|g|^{q})^{p/q}d\mu.$

By Proposition 2.2, it follows that $|\mathbb{E}(f)|\leq|f|$, which is a
contradiction, hence $P$ is not contractive.

If $p=\infty$ and $1<q<\infty$, then $p^{\prime}=1$ and $1<q^{\prime}<\infty$.
Since the adjoint map $P^{*}$ on $L_{1}(\mu;\ell_{q^{\prime}}^{(2)})$ has the
same form as $P$, the preceding argument shows that $P^{*}$ and hence $P$ is
not contractive.

If $p=\infty$ and $q=1$. Assume $P$ is contractive, then we have

(3)
$\displaystyle\big{\|}|\mathbb{E}f|+|g|\big{\|}_{\infty}\leq\big{\|}|f|+|g|\big{\|}_{\infty}.$

Consider $f=1+\varepsilon,g=1-\varepsilon$, where $\varepsilon:D\rightarrow D$
is the identity function. Then the left hand side of (3) equals to 3 while the
right hand side equals to 2. This contradiction shows that $P$ is not
contractive.

If $1\leq p<\infty$ and $q=\infty$, then $1<p^{\prime}\leq\infty$ and
$q^{\prime}=1$, hence $P^{*}$ is not contractive. It follows that $P$ is not
contractive. ∎

The norm of $P$ on $L_{p}(\mu;\ell_{q}^{(2)})$ will be denoted by $c(p,q)$ in
the sequel. If $p=q$, then $c(p,p)=1$. If $1\leq p\neq q\leq\infty$, then

(4) $\displaystyle c(p,q)>1.$

######  Remark 2.4.

It is not difficult to check that $c(\infty,1)=c(1,\infty)=\frac{3}{2}$. But
we do not know the exact value of $c(p,q)$ for general $p\neq q$.

As usual, we set

$H_{p}(\mathbb{T})=\\{f\in L_{p}(\mathbb{T},m):\hat{f}(k)=0,\forall
k\in\mathbb{Z}_{<0}\\}.$

We will say that a measurable function $f:\mathbb{T}\rightarrow\mathbb{C}$ is
bounded from below, if there exists $\delta>0$, such that $|f|\geq\delta$ a.e.
on $\mathbb{T}$. If $f\in L_{p}(\mathbb{T})$ is bounded from below, then the
geometric mean $M(|f|)$ of $|f|$ is defined by

$\log M(|f|)=\int_{\mathbb{T}}\log|f(z)|dm(z).$

In particular, if $f:\mathbb{D}\rightarrow\mathbb{C}$ is an outer function,
then

(5) $\displaystyle M(|f|)=|f(0)|=|\mathbb{E}f|.$

The following elementary proposition will be used in §4 when we treat the
analytic UMD property.

######  Proposition 2.5.

Suppose that $1\leq p\neq q<\infty$. Define $\kappa(p,q)$ to be the best
constant $C$ satisfying the property: For any measurable partition
$\mathbb{T}=A\dot{\cup}B$ with $m(A)=m(B)=\frac{1}{2}$, for any function
$f=f_{1}\chi_{A}+f_{2}\chi_{B}$ with $f_{1}>0,f_{2}>0$ and any function
$g=g_{1}\chi_{A}+g_{2}\chi_{B}$, we have

$\int_{\mathbb{T}}(M(|f|)^{q}+|g|^{q})^{p/q}dm\leq
C^{p}\int_{\mathbb{T}}(|f|^{q}+|g|^{q})^{p/q}dm.$

Then $\kappa(p,q)>1$.

###### Proof.

Assume $k(p,q)\leq 1$. Fix any measurable partition $\mathbb{T}=A\dot{\cup}B$
such that $m(A)=m(B)=\frac{1}{2}$. Consider the 2-valued functions
$f=f_{1}\chi_{A}+f_{2}\chi_{B}$ and $g=g_{1}\chi_{A}+g_{2}\chi_{B}$ with
$f_{1},f_{2}$ positive scalars. By Proposition 2.2, $M(f)\leq f$. However, one
can easily check that $M(f)=f_{1}^{1/2}f_{2}^{1/2}$. If $f_{1}>f_{2}$, then
$M(f)>f_{2}^{1/2}f_{2}^{1/2}=f_{2}$, which contradicts to $M(f)\leq f$. Whence
the announced statement. ∎

##  3\. UMD constants of iterated $L_{p}(L_{q})$ spaces

The following definition is essential in the sequel.

######  Definition 3.1.

Consider a Banach space $X$ with a fixed family of vectors $\\{x_{i}\\}_{i\in
I}$. We define $S(X;\\{x_{i}\\})$ to be the best constant $C$ such that

(6)
$\displaystyle\Big{\|}\sum_{k=0}^{N}\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})x_{i_{k}}\Big{\|}_{L_{1}(\Omega,\mathbb{P};X)}\leq
C\Big{\|}\sum_{k=0}^{N}\theta_{k}x_{i_{k}}\Big{\|}_{L_{\infty}(\Omega,\mathbb{P};X)}$

holds for any $N\in\mathbb{N}$, any probability space
$(\Omega,\mathcal{F},\mathbb{P})$ equipped with a filtration
$\mathcal{A}_{0}\subset\mathcal{A}_{1}\subset\cdots\subset\mathcal{A}_{n}\subset\cdots\subset\mathcal{F}$,
any $N+1$ distinct indices $\\{i_{0},i_{1},\cdots,i_{N}\\}\subset I$ and any
$N+1$ functions $\theta_{0},\theta_{1},\cdots,\theta_{N}$ in
$L_{\infty}(\Omega,\mathcal{F},\mathbb{P})$.

If there does not exist such constant, we set $S(X;\\{x_{i}\\})=\infty$.

In what follows, we are mostly interested in the special case when
$\\{x_{i}\\}$ is a 1-unconditional basic sequence, since in this case we can
relate $S(X;\\{x_{i}\\})$ to the UMD constants of $X$. If $\\{x_{i}\\}$ is
clear from the context and there is no confusion, we will use the simplified
notation $S(X)$ for $S(X;\\{x_{i}\\})$. In particular, if $X$ has a natural
basis, then $S(X)$ will always mean to be calculated with this basis.

We will need the following well-known Stein inequality in UMD spaces, which
was originally proved by Bourgain [4]. For the sake of completeness, we
include the proof.

######  Theorem 3.2.

Let $X$ be a UMD space. Then for any $1<s<\infty$, any finite sequences of
functions $(F_{k})_{k\geq 0}$ in $L_{s}(\Omega,\mathbb{P};X)$ and any
filtration
$\mathcal{A}_{0}\subset\mathcal{A}_{1}\subset\cdots\subset\mathcal{A}_{n}\subset\cdots$
on $(\Omega,\mathbb{P})$, we have

(7)
$\displaystyle\Big{\|}\sum_{k}\varepsilon_{k}\mathbb{E}_{k}(F_{k})\Big{\|}_{L_{s}(\mu_{\infty}\times\mathbb{P};X)}\leq
C_{s}(X)\Big{\|}\sum_{k}\varepsilon_{k}F_{k}\Big{\|}_{L_{s}(\mu_{\infty}\times\mathbb{P};X)},$

where $\mathbb{E}_{k}=\mathbb{E}^{\mathcal{A}_{k}}$ and
$(\varepsilon_{k})_{k\geq 0}$ is the usual Rademacher sequence on
$(D^{\mathbb{N}},\mu_{\infty})$, $\mu_{\infty}=\mu^{\otimes\mathbb{N}}$.

###### Proof.

Let $f=\sum_{k}\varepsilon_{k}F_{k}$ and
$f^{\prime}=\sum_{k}\varepsilon_{k}\mathbb{E}_{k}(F_{k})$. Then if
$\mathcal{C}_{2j}=\mathcal{A}_{j}\otimes\sigma(\varepsilon_{0},\cdots,\varepsilon_{j})$
and
$\mathcal{C}_{2j-1}=\mathcal{A}_{j}\otimes\sigma(\varepsilon_{0},\cdots,\varepsilon_{j-1})$,
we have

$f^{\prime}=\sum_{j}(\mathbb{E}^{\mathcal{C}_{2j}}-\mathbb{E}^{\mathcal{C}_{2j-1}})(f).$

Indeed,
$\mathbb{E}^{\mathcal{C}_{2j}}(f)=\sum_{0}^{j}\varepsilon_{k}\mathbb{E}_{j}(F_{k})$
and
$\mathbb{E}^{\mathcal{C}_{2j-1}}(f)=\sum_{0}^{j-1}\varepsilon_{k}\mathbb{E}_{j}(F_{k})$.
Hence
$(\mathbb{E}^{\mathcal{C}_{2j}}-\mathbb{E}^{\mathcal{C}_{2j-1}})(f)=\varepsilon_{j}\mathbb{E}_{j}(F_{j})$.
It follows (see the next remark) that

$\|f^{\prime}\|_{L_{s}(\mu_{\infty}\times\mathbb{P};X)}\leq
C_{s}(X)\|f\|_{L_{s}(\mu_{\infty}\times\mathbb{P};X)},$

whence (7). ∎

######  Remark 3.3.

By an extreme point argument, we have

$\sup_{-1\leq\alpha_{k}\leq
1}\|\sum_{k=0}^{n}\alpha_{k}df_{k}\|_{L_{s}(X)}=\sup_{\varepsilon_{k}\in\\{-1,1\\}}\|\sum_{k=0}^{n}\varepsilon_{k}df_{k}\|_{L_{s}(X)}.$

Hence we have

$\sup_{-1\leq\alpha_{k}\leq
1}\|\sum_{k=0}^{n}\alpha_{k}df_{k}\|_{L_{s}(X)}\leq
C_{s}(X)\|\sum_{k=0}^{n}df_{k}\|_{L_{s}(X)}.$

######  Proposition 3.4.

Let $X$ be a UMD space. Assume that $\\{x_{i}\\}_{i\in I}$ is a
1-unconditional basic sequence in $X$. Then for any $1<s<\infty$, any finite
sequence of functions $(\theta_{k})_{k\geq 0}$ in $L_{s}(\Omega,\mathbb{P})$
and any filtration
$\mathcal{A}_{0}\subset\mathcal{A}_{1}\subset\cdots\subset\mathcal{A}_{n}\subset\cdots$
on $(\Omega,\mathbb{P})$, we have

(8)
$\displaystyle\Big{\|}\sum_{k}\mathbb{E}_{k}(\theta_{k})x_{i_{k}}\Big{\|}_{L_{s}(\Omega,\mathbb{P};X)}\leq
C_{s}(X)\Big{\|}\sum_{k}\theta_{k}x_{i_{k}}\Big{\|}_{L_{s}(\Omega,\mathbb{P};X)}.$

###### Proof.

For any $i_{k}$’s, consider the sequence $(F_{k})_{k\geq 0}$ in
$L_{s}(\Omega,\mathbb{P};X)$ defined by $F_{k}(w)=\theta_{k}(w)x_{i_{k}}$.
Then $\mathbb{E}_{k}(F_{k})=\mathbb{E}_{k}(\theta_{k})x_{i_{k}}$. By the
1-unconditionality of $\\{x_{i}\\}_{i\in I}$, for any fixed choice of signs
$\varepsilon_{k}\in\\{-1,1\\}$ and $w\in\Omega$, we have

$\Big{\|}\sum_{k}\varepsilon_{k}F_{k}(w)\Big{\|}_{X}=\Big{\|}\sum_{k}\varepsilon_{k}\theta_{k}(w)x_{i_{k}}\Big{\|}_{X}=\Big{\|}\sum_{k}\theta_{k}(w)x_{i_{k}}\Big{\|}_{X}.$

It follows that

$\Big{\|}\sum_{k}\varepsilon_{k}F_{k}\Big{\|}_{L_{s}(\mu_{\infty}\times\mathbb{P};X)}=\Big{\|}\sum_{k}\theta_{k}x_{i_{k}}\Big{\|}_{L_{s}(\Omega,\mathbb{P};X)}.$

Similarly, we have

$\Big{\|}\sum_{k}\varepsilon_{k}\mathbb{E}_{k}(F_{k})\Big{\|}_{L_{s}(\mu_{\infty}\times\mathbb{P};X)}=\Big{\|}\sum_{k}\mathbb{E}_{k}(\theta_{k})x_{i_{k}}\Big{\|}_{L_{s}(\Omega,\mathbb{P};X)}.$

By these equalities, (8) follows from (7). ∎

Let $X$ be as in Proposition 3.4, $\\{x_{i}\\}_{i\in I}$ is a 1-unconditional
basic sequence in $X$. Assume that $\theta_{k}\in
L_{\infty}(\Omega,\mathbb{P})$. By an application of the contractive
inclusions $L_{\infty}(\Omega,\mathbb{P};X)\subset
L_{s}(\Omega,\mathbb{P};X)\subset L_{1}(\Omega,\mathbb{P};X)$, we have

(9)
$\displaystyle\Big{\|}\sum_{k}\mathbb{E}_{k}(\theta_{k})x_{i_{k}}\Big{\|}_{L_{1}(\Omega,\mathbb{P};X)}\leq
C_{s}(X)\Big{\|}\sum_{k}\theta_{k}x_{i_{k}}\Big{\|}_{L_{\infty}(\Omega,\mathbb{P};X)}.$

Hence

(10) $\displaystyle S(X;\\{x_{i}\\})\leq C_{s}(X)$

for all $1<s<\infty$.

######  Theorem 3.5.

Let $E$ be a Banach space with a 1-unconditional basis $\\{e_{i}:i\in I\\}$,
let $F$ be another Banach space. By definition, $E(F)$ is the completion of
the algebraic tensor product $E\otimes F$ under the norm defined as follows:
if $x=\sum_{i}e_{i}\otimes x_{i}\in E\otimes F$, where $(x_{i})$ is a finite
supported sequence in $F$, then

$\|x\|_{E(F)}:=\Big{\|}\sum_{i}e_{i}\big{\|}x_{i}\big{\|}_{F}\Big{\|}_{E}.$

For any fixed family of vectors $\\{f_{j}:j\in J\\}$ in $F$, consider the
family of vectors $\\{e_{i}\otimes f_{j}:i\in I,j\in J\\}$. Then we have

$S(E(F))\geq S(E)S(F),$

where $S(E(F))$, $S(E)$ and $S(F)$ are defined with respect to the mentioned
families of vectors respectively.

###### Proof.

From the definition, for any $\varepsilon>0$, there exist finite number of
distinct indices $\\{i_{k}:1\leq k\leq N_{1}\\}\subset I$ and $\\{j_{n}:1\leq
n\leq N_{2}\\}\subset J$, and there exist functions $\theta_{k}\in
L_{\infty}(\Omega^{\prime},\mathbb{P}^{\prime}),1\leq k\leq N_{1}$ and
functions $\xi_{n}\in L_{\infty}(\Omega_{0},\mathbb{P}_{0}),1\leq n\leq N_{2}$
satisfying

$\|\sum_{k}\theta_{k}e_{i_{k}}\|_{L_{\infty}(\Omega^{\prime},\mathbb{P}^{\prime};E)}\leq
1$

and

$\|\sum_{n}\xi_{n}f_{j_{n}}\|_{L_{\infty}(\Omega_{0},\mathbb{P}_{0};F)}\leq 1$

such that

$\Big{\|}\sum_{k}\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})e_{i_{k}}\Big{\|}_{L_{1}(\Omega^{\prime},\mathbb{P}^{\prime};E)}\geq
S(E)-\varepsilon$

and

$\Big{\|}\sum_{n}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})f_{j_{n}}\Big{\|}_{L_{1}(\Omega_{0},\mathbb{P}_{0};F)}\geq
S(F)-\varepsilon.$

Let
$(\Omega,\mathbb{P})=(\Omega^{\prime}\times\Omega_{0}^{\mathbb{N}},\mathbb{P}^{\prime}\otimes\mathbb{P}_{0}^{\otimes\mathbb{N}})$,
the general element in $\Omega$ will be denoted by
$w=(w^{\prime},(w_{l})_{l\geq 0})$. Consider the $\sigma$-algebras
$\mathcal{F}_{k,n}$ defined on $(\Omega,\mathbb{P})$ by

$\mathcal{F}_{k,n}:=\mathcal{A}_{k}\otimes\underbrace{\mathcal{B}_{\infty}\otimes\cdots\otimes\mathcal{B}_{\infty}}_{k-1\text{
times }}\otimes\mathcal{B}_{n}\otimes\mathcal{C}_{\geq k+1},$

where $\mathcal{B}_{\infty}=\sigma(\mathcal{B}_{n}:n\geq 0)$ is a
$\sigma$-algebra on $(\Omega_{0},\mathbb{P}_{0})$, $\mathcal{B}_{0}$ is
assumed to be trivial and $\mathcal{C}_{\geq k+1}$ is the trivial
$\sigma$-algebra on
$(\Omega_{0}^{\mathbb{N}_{\geq{k+1}}},\mathbb{P}_{0}^{\mathbb{N}_{\geq
k+1}})$. It is easy to check that $\mathcal{F}_{k,n}$ is a filtration with
respect to the lexigraphic order, i.e. if $(k,n)<(k^{\prime},n^{\prime})$
(that is $k<k^{\prime}$ or $k=k^{\prime}$ but $n<n^{\prime}$), then
$\mathcal{F}_{k,n}\subset\mathcal{F}_{k^{\prime},n^{\prime}}$.

Now let us define $h:\Omega\rightarrow E(F)$ by

$h(w)=h(w^{\prime},(w_{l}))=\sum_{k,n}\theta_{k}(w^{\prime})\xi_{n}(w_{k})e_{i_{k}}\otimes
f_{j_{n}}.$

Let $h_{k,n}(w)=\theta_{k}(w^{\prime})\xi_{n}(w_{k})$, then
$h=\sum_{k,n}h_{k,n}e_{i_{k}}\otimes f_{j_{n}}$. Clearly, we have

(11)
$\displaystyle\mathbb{E}^{\mathcal{F}_{k,n}}(h_{k,n})(w)=\Big{[}\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})\Big{]}(w^{\prime})\Big{[}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})\Big{]}(w_{k})\quad
a.e..$

By the 1-unconditionality of $\\{e_{i}:i\in I\\}$, for a.e. $w\in\Omega$, we
have

$\displaystyle\|h(w)\|_{E(F)}$ $\displaystyle=$
$\displaystyle\Big{\|}\sum_{k,n}\theta_{k}(w^{\prime})\xi_{n}(w_{k})e_{i_{k}}\otimes
f_{j_{n}}\Big{\|}_{E(F)}$ $\displaystyle=$
$\displaystyle\Big{\|}\sum_{k}e_{i_{k}}\big{\|}\sum_{n}\theta_{k}(w^{\prime})\xi_{n}(w_{k})f_{j_{n}}\big{\|}_{F}\Big{\|}_{E}$
$\displaystyle=$
$\displaystyle\Big{\|}\sum_{k}e_{i_{k}}|\theta_{k}(w^{\prime})|\big{\|}\sum_{n}\xi_{n}(w_{k})f_{j_{n}}\big{\|}_{F}\Big{\|}_{E}$
$\displaystyle\leq$
$\displaystyle\Big{\|}\sum_{k}e_{i_{k}}|\theta_{k}(w^{\prime})|\Big{\|}_{E}=\Big{\|}\sum_{k}e_{i_{k}}\theta_{k}(w^{\prime})\Big{\|}_{E}\leq
1.$

Hence $\|h\|_{L_{\infty}(\Omega,\mathbb{P};E(F))}\leq 1$. If we denote

$\widetilde{h}=\sum_{k,n}\mathbb{E}^{\mathcal{F}_{k,n}}(h_{k,n})e_{i_{k}}\otimes
f_{j_{n}},$

then by (11),

$\displaystyle\|\widetilde{h}(w)\|_{E(F)}$ $\displaystyle=$
$\displaystyle\Big{\|}\sum_{k}e_{i_{k}}|\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})(w^{\prime})|\big{\|}\sum_{n}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})(w_{k})f_{j_{n}}\big{\|}_{F}\Big{\|}_{E}.$

By Jensen’s inequality, we have

$\displaystyle\int\Big{\|}\sum_{k}e_{i_{k}}|\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})(w^{\prime})|\big{\|}\sum_{n}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})(w_{k})f_{j_{n}}\big{\|}_{F}\Big{\|}_{E}d\mathbb{P}_{0}^{\otimes\mathbb{N}}((w_{l}))$
$\displaystyle\geq$
$\displaystyle\Big{\|}\int\sum_{k}e_{i_{k}}|\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})(w^{\prime})|\big{\|}\sum_{n}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})(w_{k})f_{j_{n}}\big{\|}_{F}d\mathbb{P}_{0}^{\otimes\mathbb{N}}((w_{l}))\Big{\|}_{E}$
$\displaystyle=$
$\displaystyle\Big{\|}\sum_{k}e_{i_{k}}|\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})(w^{\prime})|\Big{\|}_{E}\cdot\Big{\|}\sum_{n}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})f_{j_{n}}\Big{\|}_{L_{1}(\Omega_{0},\mathbb{P}_{0};F)}$
$\displaystyle=$
$\displaystyle\Big{\|}\sum_{k}e_{i_{k}}\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})(w^{\prime})\Big{\|}_{E}\cdot\Big{\|}\sum_{n}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})f_{j_{n}}\Big{\|}_{L_{1}(\Omega_{0},\mathbb{P}_{0};F)}.$

Note that in the last equality, we used the 1-unconditionality assumption on
$\\{e_{i}:i\in I\\}$. By integrating both sides with respect to $\int
d\mathbb{P}^{\prime}(w^{\prime})$, we get

$\displaystyle\Big{\|}\sum_{k,n}\mathbb{E}^{\mathcal{F}_{k,n}}(h_{k,n})e_{i_{k}}\otimes
f_{j_{n}}\Big{\|}_{L_{1}(\Omega,\mathbb{P};E(F))}$ $\displaystyle\geq$
$\displaystyle\Big{\|}\sum_{k}\mathbb{E}^{\mathcal{A}_{k}}(\theta_{k})e_{i_{k}}\Big{\|}_{L_{1}(\Omega^{\prime},\mathbb{P}^{\prime};E)}\cdot\Big{\|}\sum_{n}\mathbb{E}^{\mathcal{B}_{n}}(\xi_{n})f_{j_{n}}\Big{\|}_{L_{1}(\Omega_{0},\mathbb{P}_{0};F)}$
$\displaystyle\geq$ $\displaystyle(S(E)-\varepsilon)(S(F)-\varepsilon).$

Therefore $S(E(F))\geq(S(E)-\varepsilon)(S(F)-\varepsilon)$. Since
$\varepsilon>0$ is arbitrary, it follows that $S(E(F))\geq S(E)S(F)$ as
desired. ∎

######  Remark 3.6.

If $E$ is a Banach lattice which is $p$-convex and $q$-concave $($see [10] for
the details$)$ with $1\leq p\leq q\leq\infty$ and $F$ is a Banach space. Then
the preceding proof is valid with $S_{q,p}(E)$ and $S_{q,p}(F)$ defined using
(6) with $L_{p}$-norm on the left hand side and $L_{q}$-norm on the right hand
side.

######  Remark 3.7.

Let $1\leq p<q\leq\infty$. If we define $C_{q,p}(X)$ as the best constant $C$
in (1) with $L_{p}$-norm on the left hand side and $L_{q}$-norm on the right
hand side, it is well-known that $X$ is in the UMD class if and only if
$C_{q,p}(X)<\infty$. The preceding argument shows that under the same
assumption of Theorem 3.5, we have $C_{\infty,1}(E(F))\geq
S(E)C_{\infty,1}(F)$. Moreover, if $E$ is $p$-convex and $q$-concave we have
$C_{q,p}(E(F))\geq S_{q,p}(E)C_{q,p}(F)$.

######  Lemma 3.8.

Suppose that $1\leq p\neq q\leq\infty$. If
$E_{1}=\ell_{p}^{(2)}(\ell_{q}^{(2)})$, then

$S(E_{1})\geq c(p,q)>1.$

###### Proof.

Denote by $\\{e_{1}^{p},e_{2}^{p}\\}$, $\\{e_{1}^{q},e_{2}^{q}\\}$ the
canonical basis of $\ell_{p}^{(2)}$ and $\ell_{q}^{(2)}$ respectively .Then
$\\{e_{1}^{p}\otimes e_{1}^{q},e_{1}^{p}\otimes e_{2}^{q},e_{2}^{p}\otimes
e_{1}^{q},e_{2}^{p}\otimes e_{2}^{q}\\}$ is the canonical 1-unconditional
basis of $\ell_{p}^{(2)}(\ell_{q}^{(2)})$. Consider the probability space
$(D,\mu)$ equipped with the filtration
$\\{\phi,D\\}\subset\sigma(\varepsilon)$, where $\varepsilon$ is the identity
function on $D$. Define a linear map $T:L_{\infty}(D;E_{1})\rightarrow
L_{1}(D;E_{1})$ by setting

$T\Big{[}a_{ij}(\varepsilon)e_{i}^{p}\otimes
e_{j}^{q}\Big{]}=\left\\{\begin{array}[]{lc}\mathbb{E}(a_{ij})e_{i}^{p}\otimes
e_{j}^{q},\text{ if }j=1\\\ a_{ij}(\varepsilon)e_{i}^{p}\otimes
e_{j}^{q},\text{ if }j=2\end{array}.\right.$

By definition of $S(E_{1})$ we have
$S(E_{1})\geq\|T\|_{L_{\infty}(D;E_{1})\rightarrow L_{1}(D;E_{1})}$. Now for
any $a,b$ two scalar functions on $D$ , consider

$f(\varepsilon)=e_{1}^{p}\otimes\Big{[}a(\varepsilon)e_{1}^{q}+b(\varepsilon)e_{2}^{q}\Big{]}+e_{2}^{p}\otimes\Big{[}a(-\varepsilon)e_{1}^{q}+b(-\varepsilon)e_{2}^{q}\Big{]}.$

Then

$(Tf)(\varepsilon)=e_{1}^{p}\otimes\Big{[}\mathbb{E}(a)e_{1}^{q}+b(\varepsilon)e_{2}^{q}\Big{]}+e_{2}^{p}\otimes\Big{[}\mathbb{E}(a)e_{1}^{q}+b(-\varepsilon)e_{2}^{q}\Big{]}.$

If $p,q$ are both finite, then for any fixed $\varepsilon\in D$, we have

$\displaystyle\|f(\varepsilon)\|_{E_{1}}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\Big{\\{}(|a(\varepsilon)|^{q}+|b(\varepsilon)|^{q})^{p/q}+(|a(-\varepsilon)|^{q}+|b(-\varepsilon)|^{q})^{p/q}\Big{\\}}^{1/p}$
$\displaystyle=$
$\displaystyle\\!\\!\Big{\\{}(|a(1)|^{q}+|b(1)|^{q})^{p/q}+(|a(-1)|^{q}+|b(-1)|^{q})^{p/q}\Big{\\}}^{1/p}$
$\displaystyle=$
$\displaystyle\\!\\!2^{1/p}\Big{\\{}\frac{1}{2}(|a(1)|^{q}+|b(1)|^{q})^{p/q}+\frac{1}{2}(|a(-1)|^{q}+|b(-1)|^{q})^{p/q}\Big{\\}}^{1/p}$
$\displaystyle=$
$\displaystyle\\!\\!2^{1/p}\Big{\\{}\int(|a(\varepsilon)|^{q}+|b(\varepsilon)|^{q})^{p/q}d\mu(\varepsilon)\Big{\\}}^{1/p}$
$\displaystyle=$
$\displaystyle\\!\\!2^{1/p}\big{\|}(a,b)\big{\|}_{L_{p}(\mu;\ell_{q}^{(2)})}.$

Similarly,

$\|(Tf)(\varepsilon)\|_{E_{1}}=2^{1/p}\big{\|}(\mathbb{E}a,b)\big{\|}_{L_{p}(\mu;\ell_{q}^{(2)})}.$

It follows that

$\|f\|_{L_{\infty}(D;E_{1})}=2^{1/p}\big{\|}(a,b)\big{\|}_{L_{p}(\mu;\ell_{q}^{(2)})}$

and

$\|Tf\|_{L_{1}(D;E_{1})}=2^{1/p}\big{\|}(\mathbb{E}a,b)\big{\|}_{L_{p}(\mu;\ell_{q}^{(2)})}.$

Hence

(12) $\displaystyle\|T\|_{L_{\infty}(D;E_{1})\rightarrow
L_{1}(D;E_{1})}\geq\frac{\|Tf\|_{L_{1}(D;E_{1})}}{\|f\|_{L_{\infty}(D;E_{1})}}=\frac{\big{\|}(\mathbb{E}a,b)\big{\|}_{L_{p}(\mu;\ell_{q}^{(2)})}}{\big{\|}(a,b)\big{\|}_{L_{p}(\mu;\ell_{q}^{(2)})}}.$

Similarly, if $q=\infty$ and $p$ is finite, then

$\|f\|_{L_{\infty}(D;E_{1})}=2^{1/p}\|(a,b)\|_{L_{p}(\mu;\ell_{\infty}^{(2)})}$

and

$\|Tf\|_{L_{1}(D;E_{1})}=2^{1/p}\big{\|}(\mathbb{E}a,b)\big{\|}_{L_{p}(\mu;\ell_{\infty}^{(2)})}.$

If $p=\infty$ and $q$ is finite, then
$\|f\|_{L_{\infty}(D;E_{1})}=\|(a,b)\|_{L_{\infty}(\mu;\ell_{q}^{(2)})}$ and
$\|Tf\|_{L_{1}(D;E_{1})}=\big{\|}(\mathbb{E}a,b)\big{\|}_{L_{\infty}(\mu;\ell_{q}^{(2)})}$.
Therefore, (12) holds in full generality. By Proposition 2.3, we have

$\|T\|_{L_{\infty}(D;E_{1})\rightarrow L_{1}(D;E_{1})}\geq\|P\|=c(p,q).$

Hence $S(E_{1})\geq c(p,q)>1$, as announced. ∎

######  Remark 3.9.

Let $(e_{k})_{k\geq 0}$ be the canonical basis of
$\ell_{p}=\ell_{p}(\mathbb{N})$, then $S(\ell_{p})=1$. Indeed, if
$(\theta_{k})_{k\geq 0}$ is a finite sequence of functions, then

$\displaystyle\Big{\|}\sum_{k}\mathbb{E}_{k}(\theta_{k})e_{k}\Big{\|}_{L_{1}(\ell_{p})}$
$\displaystyle\leq$
$\displaystyle\Big{\|}\sum_{k}\mathbb{E}_{k}(\theta_{k})e_{k}\Big{\|}_{L_{p}(\ell_{p})}=\Big{\|}(\sum_{k}|\mathbb{E}_{k}(\theta_{k})|^{p})^{1/p}\Big{\|}_{L_{p}}$
$\displaystyle=$
$\displaystyle\Big{\|}\sum_{k}|\mathbb{E}_{k}(\theta_{k})|^{p}\Big{\|}_{L_{1}}^{1/p}=(\sum_{k}\big{\|}\mathbb{E}_{k}(\theta_{k})\big{\|}_{p}^{p})^{1/p}$
$\displaystyle\leq$
$\displaystyle(\sum_{k}\big{\|}\theta_{k}\big{\|}_{p}^{p})^{1/p}=\Big{\|}\sum_{k}\theta_{k}e_{k}\Big{\|}_{L_{p}(\ell_{p})}$
$\displaystyle\leq$
$\displaystyle\Big{\|}\sum_{k}\theta_{k}e_{k}\Big{\|}_{L_{\infty}(\ell_{p})}.$

######  Theorem 3.10.

Suppose that $1\leq p,q\leq\infty$. Let $E_{1}=\ell_{p}^{(2)}(\ell_{q}^{(2)})$
and define by recursion: $E_{n+1}=\ell_{p}^{(2)}(\ell_{q}^{(2)}(E_{n}))$. Then
for any $1<s<\infty$, we have

$C_{s}(E_{n})\geq S(E_{n})\geq c(p,q)^{n},$

where $S(E_{n})$ is computed with respect to the canonical basis of $E_{n}$.
In particular, if $p\neq q$, then $C_{s}(E_{n})$ has at least an exponential
growth with respect to $n$.

###### Proof.

By Theorem 3.5,

$S(E_{n+1})\geq S(\ell_{p}^{(2)}(\ell_{q}^{(2)}))S(E_{n}).$

By Lemma 3.8, we have $S(E_{n+1})\geq c(p,q)S(E_{n})$. It follows that
$S(E_{n})\geq c(p,q)^{n}$. Since the canonical basis of $E_{n}$ is
1-unconditional, by (10), for any $1<s<\infty$, we have $C_{s}(E_{n})\geq
S(E_{n})$. ∎

The following simple observation shows that the exponential growth of
$C_{s}(E_{n})$ is optimal.

######  Proposition 3.11.

Suppose $1<p\neq q<\infty$. Let $X$ be a Banach space. Define by recursion:
$Y_{0}=X$ and $Y_{n+1}=L_{p}(\mathbb{T};L_{q}(\mathbb{T};Y_{n}))$. Then for
all $1<s<\infty$, there exists $\chi=\chi(p,q,s)$, such that

$C_{s}(Y_{n})\leq\chi^{n}C_{s}(X).$

###### Proof.

We will use the following well-known fact (see e.g. [5, 6]) about UMD
constants: for any $1<r,s<\infty$, there exist $\alpha(r,s)$ and $\beta(r,s)$
such that for all Banach space $X$,

(13) $\displaystyle\alpha(r,s)C_{s}(X)\leq C_{r}(X)\leq\beta(r,s)C_{s}(X).$

We will also use the elementary identity $C_{s}(L_{s}(X))=C_{s}(X)$. Combining
these, we have

$\displaystyle C_{s}(Y_{n+1})$ $\displaystyle=$ $\displaystyle
C_{s}(L_{p}(L_{q}(Y_{n})))\leq\beta(s,p)C_{p}(L_{p}(L_{q}(Y_{n})))$
$\displaystyle=$
$\displaystyle\beta(s,p)C_{p}(L_{q}(Y_{n}))\leq\beta(s,p)\beta(p,q)C_{q}(L_{q}(Y_{n}))$
$\displaystyle=$
$\displaystyle\beta(s,p)\beta(p,q)C_{q}(Y_{n})\leq\beta(s,p)\beta(p,q)\beta(q,s)C_{s}(Y_{n}).$

Let $\chi=\beta(s,p)\beta(p,q)\beta(q,s)$, then
$C_{s}(E_{n})\leq\chi^{n}C_{s}(X).$ ∎

######  Remark 3.12.

Even if one of $p,q$ is infinite or equals to 1, then since
$\dim(E_{n})=4^{n}$, we have $C_{s}(E_{n})\lesssim\sqrt{\dim E_{n}}=2^{n}$.
Indeed, the Banach-Mazur distance between $E_{n}$ and $\ell_{2}^{\dim E_{n}}$
is $\leq\sqrt{\dim E_{n}}$ $($cf. e.g. [14]$)$.

##  4\. Analytic UMD constants

The main idea in §3 can be easily adapted for treating the analytic UMD
property. In this section, all spaces are over $\mathbb{C}$.

Denote the general element in $\mathbb{T}^{\mathbb{N}}$ be $z=(z_{n})_{n\geq
0}$ and let $m_{\infty}=m^{\otimes\mathbb{N}}$ be the Haar measure on
$\mathbb{T}^{\mathbb{N}}$. Recall the canonical filtration on
$(\mathbb{T}^{\mathbb{N}},m_{\infty})$ defined by

$\sigma(z_{0})\subset\sigma(z_{0},z_{1})\subset\cdots\subset\sigma(z_{0},z_{1},\cdots,z_{n})\subset\cdots.$

From now on, we will denote
$\mathcal{G}_{n}=\sigma(z_{0},z_{1},\cdots,z_{n})$. Recall that
$H_{s}(\mathbb{T}^{\mathbb{N}})$ is the subspace of
$L_{s}(\mathbb{T}^{\mathbb{N}},m_{\infty})$ consisting of limit values of
Hardy martingales, i.e. $f\in H_{s}(\mathbb{T}^{\mathbb{N}})$ if and only if
$f\in L_{s}(\mathbb{T}^{\mathbb{N}},m_{\infty})$ and the associated martingale
$(\mathbb{E}^{\mathcal{G}_{n}}f)_{n\geq 0}$ is a Hardy martingale. For
convenience, we always assume $z_{0}\equiv 1$ such that $\mathcal{G}_{0}$ is a
trivial $\sigma$-algebra.

######  Definition 4.1.

Let $X$ be a Banach space and let $\\{x_{i}\\}_{i\in I}$ be a family of
vectors in $X$. The number $S^{a}(X;\\{x_{i}\\})$ is defined to be the best
constant $C$ such that for any $N\in\mathbb{N}$ and any finite sequence of
functions $(\theta_{k})_{k=0}^{N}$ in $H_{\infty}(\mathbb{T}^{\mathbb{N}})$,
we have

$\Big{\|}\sum_{k}\mathbb{E}^{\mathcal{G}_{k}}(\theta_{k})x_{i_{k}}\Big{\|}_{L_{1}(m_{\infty};X)}\leq
C\Big{\|}\sum_{k}\theta_{k}x_{i_{k}}\Big{\|}_{L_{\infty}(m_{\infty};X)}$

If there does not exist such constant, we set $S^{a}(X;\\{x_{i}\\})=\infty$.

If $\\{x_{i}\\}$ is clear from the context, then $S^{a}(X;\\{x_{i}\\})$ will
be simplified as $S^{a}(X)$.

The Stein type inequality still holds in this setting, more precisely, we have

######  Proposition 4.2.

Let $X$ be an AUMD space. For any $1\leq s<\infty$, let $(F_{k})_{k\geq 0}$ be
an arbitrary finite sequence in $H_{s}(\mathbb{T}^{\mathbb{N}};X)$. Then we
have

(14)
$\displaystyle\Big{\|}\sum_{k}\zeta_{k}\mathbb{E}^{\mathcal{G}_{k}}(F_{k})(z)\Big{\|}_{L_{s}(X)}\leq
C_{s}^{a}(X)\Big{\|}\sum_{k}\zeta_{k}F_{k}(z)\Big{\|}_{L_{s}(X)},$

where $\zeta=(\zeta_{k})_{k\geq 0}$ is an independent copy of
$z=(z_{k})_{k\geq 0}$ and
$L_{s}(X)=L_{s}(\mathbb{T}_{z}^{\mathbb{N}}\times\mathbb{T}_{\zeta}^{\mathbb{N}},m_{\infty}\times
m_{\infty};X)$.

###### Proof.

Consider the filtration on
$\mathbb{T}_{z}^{\mathbb{N}}\times\mathbb{T}_{\zeta}^{\mathbb{N}}$ defined by
$\mathcal{B}_{2j}=\sigma(z_{0},\cdots,z_{j})\otimes\sigma(\zeta_{0},\cdots,\zeta_{j})$
and
$\mathcal{B}_{2j-1}=\sigma(z_{0},\cdots,z_{j})\otimes\sigma(\zeta_{0},\cdots,\zeta_{j-1})$.
Then $f=\sum_{k}\zeta_{k}F_{k}(z)$ is a Hardy martingale with respect to the
above filtration. Let
$f^{\prime}=\sum_{k}\zeta_{k}\mathbb{E}^{\mathcal{G}_{k}}(F_{k})$. Then we
have
$f^{\prime}=\sum_{j}(\mathbb{E}^{\mathcal{B}_{2j}}-\mathbb{E}^{\mathcal{B}_{2j-1}})(f)$.
It follows (see Remark 3.3) that $\|f^{\prime}\|_{L_{s}(X)}\leq
C_{s}^{a}(X)\|f\|_{L_{s}(X)}$, whence (14). ∎

######  Proposition 4.3.

Let $X$ be an AUMD space. Assume that $\\{x_{i}\\}_{i\in I}$ is a
1-unconditional basic sequence in $X$. Then for any $1\leq s<\infty$ and any
finite sequence of functions $(\theta_{k})_{k\geq 0}$ in
$H_{s}(\mathbb{T}^{\mathbb{N}})$,

$\displaystyle\Big{\|}\sum_{k}\mathbb{E}^{\mathcal{G}_{k}}(\theta_{k})x_{i_{k}}\Big{\|}_{L_{s}(m_{\infty};X)}\leq
C_{s}^{a}(X)\Big{\|}\sum_{k}\theta_{k}x_{i_{k}}\Big{\|}_{L_{s}(m_{\infty};X)}.$

###### Proof.

It follows verbatim the proof of Proposition 3.4. ∎

Let $X$ be as in Proposition 4.3, $\\{x_{i}\\}$ is a 1-unconditional basic
sequence in $X$. Then for all $1\leq s<\infty$, we have

$\displaystyle\Big{\|}\sum_{k}\mathbb{E}^{\mathcal{G}_{k}}(\theta_{k})x_{i_{k}}\Big{\|}_{L_{1}(m_{\infty};X)}\leq
C_{s}^{a}(X)\Big{\|}\sum_{k}\theta_{k}x_{i_{k}}\Big{\|}_{L_{\infty}(m_{\infty};X)}.$

Hence

$S^{a}(X;\\{x_{i}\\})\leq C_{s}^{a}(X)$

for all $1\leq s<\infty$.

######  Theorem 4.4.

Let $E$ be a Banach space with a 1-unconditional basis $\\{e_{i}:i\in I\\}$,
let $F$ be another Banach space. Let $E(F)$ be defined as in Theorem 3.5. For
any fixed family of vectors $\\{f_{j}:j\in J\\}$ in $F$, consider the family
of vectors $\\{e_{i}\otimes f_{j}:i\in I,j\in J\\}$ in $E(F)$, then we have

$S^{a}(E(F))\geq S^{a}(E)S^{a}(F),$

where $S^{a}(E(F))$, $S^{a}(E)$ and $S^{a}(F)$ are defined with respect to the
mentioned families of vectors respectively.

###### Proof.

The proof is similar to the proof of Theorem 3.5. We mention the slight
difference concerning the filtration. Consider the infinite tensor product
$L_{\infty}(\mathbb{T}^{\mathbb{N}})\otimes
L_{\infty}(\mathbb{T}^{\mathbb{N}})\otimes\cdots$, define

$z_{k,n}=\underbrace{1\otimes\cdots\otimes 1}_{k\quad\text{times}}\otimes
z_{n}\otimes 1\otimes\cdots,\text{if }n\geq 1$

and

$z_{k,0}=z_{k}\otimes 1\otimes 1\otimes\cdots.$

Then the filtration defined by
$\mathcal{F}^{a}_{k,n}:=\sigma\left(z_{j}:j\leq(k,n)\right)$ is an analytic
filtration, where the order on $\mathbb{N}\times\mathbb{N}$ is the lexigraphic
order as defined in the proof of Theorem 3.5. This filtration plays the role
similar to that of $(\mathcal{F}_{k,n})_{k,n}$ in the proof of Theorem 3.5.
Note that we may restrict to the functions $\theta_{k},\xi_{n}$ depending only
on finitely many variables. Thus only a finite subset of
$\mathbb{N}\times\mathbb{N}$ is used. ∎

The following lemma requires slightly more efforts than Lemma 3.8.

######  Lemma 4.5.

Suppose that $1\leq p\neq q<\infty$. If
$E_{1}=\ell_{p}^{(2)}(\ell_{q}^{(2)})$, then

$S^{a}(E_{1})\geq\kappa(p,q)>1.$

###### Proof.

We will use the notations in the proof of Lemma 3.8. Define a linear map
$U:H_{\infty}(\mathbb{T},m;E_{1})\rightarrow H_{1}(\mathbb{T},m;E_{1})$ by

$U\Big{[}a_{ij}(z)e_{i}^{p}\otimes
e_{j}^{q}\Big{]}=\left\\{\begin{array}[]{lc}\mathbb{E}(a_{ij})e_{i}^{p}\otimes
e_{j}^{q},\text{ if }j=1\\\ a_{ij}(z)e_{i}^{p}\otimes e_{j}^{q},\text{ if
}j=2\end{array}.\right.$

If $C=\|U\|_{H_{\infty}(E_{1})\rightarrow H_{1}(E_{1})}$, then
$S^{a}(E_{1})\geq C$. By definition, for any $a,b,c,d$ functions in
$H_{\infty}(\mathbb{T})$, we have

$\displaystyle\int_{\mathbb{T}}\Big{\\{}(|\mathbb{E}a|^{q}+|b(z)|^{q})^{p/q}+(|\mathbb{E}c|^{q}+|d(z)|^{q})^{p/q}\Big{\\}}^{1/p}dm(z)$
$\displaystyle\leq$ $\displaystyle
C\sup_{z\in\mathbb{T}}\Big{\\{}(|a(z)|^{q}+|b(z)|^{q})^{p/q}+(|c(z)|^{q}+|d(z)|^{q})^{p/q}\Big{\\}}^{1/p}.$

Note that if $a,c$ are outer functions, then by (5), we have
$|\mathbb{E}a|=M(|a|)$ and $|\mathbb{E}c|=M(|c|)$. So for any functions
$a,b,c,d\in H_{\infty}(\mathbb{T})$ such that $a,c$ are outer, we have

$\displaystyle\int_{\mathbb{T}}\Big{\\{}(M(|a|)^{q}+|b(z)|^{q})^{p/q}+(M(|c|)^{q}+|d(z)|^{q})^{p/q}\Big{\\}}^{1/p}dm(z)$
$\displaystyle\leq$ $\displaystyle
C\sup_{z\in\mathbb{T}}\Big{\\{}(|a(z)|^{q}+|b(z)|^{q})^{p/q}+(|c(z)|^{q}+|d(z)|^{q})^{p/q}\Big{\\}}^{1/p}.$

By the classical Szegö’s condition, if
$a^{\prime},b^{\prime},c^{\prime},d^{\prime}$ are functions in
$L_{\infty}(\mathbb{T})$ which are bounded from below, then there are outer
functions $a,b,c,d\in H_{\infty}(\mathbb{T})$, such that
$|a^{\prime}|=|a|,|b^{\prime}|=|b|,|c^{\prime}|=|c|,|d^{\prime}|=|d|$ a.e..
Hence (4) still holds for any 2-valued non-vanishing functions $a,b,c,d\in
L_{\infty}(\mathbb{T})$ (note that for a function taking only two values, non-
vanishing is the same as bounded from below). By approximation, we can further
relax the non-vanishing condition on $b,d$. Now consider any measurable
partition $\mathbb{T}=A\dot{\cup}B$, such that $m(A)=m(B)=\frac{1}{2}$. If
$a=u\chi_{A}+v\chi_{B}$, $c=v\chi_{A}+u\chi_{B}$, $b=w\chi_{A}+t\chi_{B}$ and
$d=t\chi_{A}+w\chi_{B}$, then it is easy to check that for all
$z\in\mathbb{T}$, we have

$\displaystyle\Big{\\{}(|a(z)|^{q}+|b(z)|^{q})^{p/q}+(|c(z)|^{q}+|d(z)|^{q})^{p/q}\Big{\\}}^{1/p}$
$\displaystyle=$
$\displaystyle\Big{\\{}(|u|^{q}+|w|^{q})^{p/q}+(|v|^{q}+|t|^{q})^{p/q}\Big{\\}}^{1/p}$
$\displaystyle=$ $\displaystyle
2^{1/p}\Big{\\{}\int_{\mathbb{T}}(|a|^{q}+|b|^{q})^{p/q}dm\Big{\\}}^{1/p}.$

Similarly for all $z\in\mathbb{T}$, we have

$\displaystyle\Big{\\{}(M(|a|)^{q}+|b(z)|^{q})^{p/q}+(M(|c|)^{q}+|d(z)|^{q})^{p/q}\Big{\\}}^{1/p}$
$\displaystyle=$ $\displaystyle
2^{1/p}\Big{\\{}\int_{\mathbb{T}}(M(|a|)^{q}+|b|^{q})^{p/q}dm\Big{\\}}^{1/p}.$

Substituting these equalities to (4), we get

$\Big{\\{}\int_{\mathbb{T}}(M(|a|)^{q}+|b|^{q})^{p/q}dm\Big{\\}}^{1/p}\leq
C\Big{\\{}\int_{\mathbb{T}}(|a|^{q}+|b|^{q})^{p/q}dm\Big{\\}}^{1/p}.$

By Proposition 2.5, we have $C\geq\kappa(p,q)$. This completes the proof. ∎

######  Theorem 4.6.

Suppose that $1\leq p\neq q<\infty$. If $E_{n}$’s are defined as in Theorem
3.10, then for any $1\leq s<\infty$, we have

$C_{s}^{a}(E_{n})\geq S^{a}(E_{n})\geq\kappa(p,q)^{n}.$

Moreover, if $1<p,q<\infty$, then there exists $\kappa_{2}=\kappa_{2}(p,q,s)$,
such that

$C_{s}^{a}(E_{n})\leq\kappa_{2}^{n}.$

###### Proof.

The first part of proof is identical to the proof of Theorem 3.10. The second
part follows from the fact that $C_{s}^{a}(E_{n})\leq C_{s}(E_{n})$ and
Proposition 3.11. ∎

##  5\. Construction and further discussions

For the sake of clearness, we introduce the family $X_{n}(p,q)$, which is
defined as follows: Let $X_{0}(p,q)=\mathbb{R}$, and define by recursion that

$X_{n+1}(p,q)=L_{p}(D,\mu;L_{q}(D,\mu;X_{n}(p,q))).$

In the complex case, $X^{\mathbb{C}}_{n}(p,q)$ is defined similarly.

Obviously, $X_{n}(p,q)$ is isometric to $E_{n}$ defined in the previous
sections using $p,q$. Our main purpose for introducing $X_{n}$’s is the
existence of canonical isometric inclusion $X_{n}(p,q)\subset X_{n+1}(p,q)$.
By these inclusions, the union $\cup_{n}X_{n}(p,q)$ is a normed space and its
completion will be denoted by $X(p,q)$. We have

$X(p,q):=\overline{\cup_{n}X_{n}(p,q)}\simeq\lim_{\longrightarrow}X_{n}(p,q),$

where the last term is the inductive limit of $X_{n}(p,q)$’s associated to the
canonical inclusions. In the complex case, $X^{\mathbb{C}}(p,q)$ is defined
similarly.

######  Remark 5.1.

If $1\leq p=q<\infty$, then $X(p,p)$ is the real space
$L^{p}_{\mathbb{R}}(D^{\mathbb{N}},\mu^{\otimes\mathbb{N}})$ and
$X^{\mathbb{C}}(p,p)$ is the complex space
$L_{\mathbb{C}}^{p}(D^{\mathbb{N}},\mu^{\otimes\mathbb{N}})$.

We have the following complex interpolation result.

######  Proposition 5.2.

Let $1<p_{0},p_{1},q_{0},q_{1}<\infty$ and $0<\theta<1$. Then we have the
following isometric isomorphism:

$X^{\mathbb{C}}(p_{\theta},q_{\theta})=[X^{\mathbb{C}}(p_{0},q_{0}),X^{\mathbb{C}}(p_{1},q_{1})]_{\theta},$

with $\frac{1}{p}=\frac{\theta}{p_{1}}+\frac{1-\theta}{p_{0}}$ and
$\frac{1}{q}=\frac{\theta}{q_{1}}+\frac{1-\theta}{q_{0}}$.

###### Proof.

Note that $X(p,q)$ is a Banach lattice of functions on
$(D^{\mathbb{N}},\mu^{\otimes\mathbb{N}})$. Clearly, $X(p,q)$ is
$\min(p,q)$-convex and $\max(p,q)$-concave in the sense of §1.d in [10], and
hence by Theorem 1.f.1 (p. 80) and Proposition 1.e.3 (p. 61) in [10] it is
reflexive. Then the above result is a particular case of a classical formula
going back to Calderón ([8], p. 125). ∎

Recall that a Banach space $X$ over the complex field is $\theta$-Hilbertian
($0\leq\theta\leq 1$) if there exists an interpolation pair $(X_{0},X_{1})$ of
Banach spaces such that $X$ is isometric with $[X_{0},X_{1}]_{\theta}$ and
$X_{1}$ is a Hilbert space.

######  Corollary 5.3.

Let $1<p\neq q<\infty$. Then $X(p,q)$ is non-UMD and $X^{\mathbb{C}}(p,q)$ is
non-AUMD. Moreover, there exists $0<\theta<1$ such that $X^{\mathbb{C}}(p,q)$
is $\theta$-Hilbertian. In particular, $X^{\mathbb{C}}(p,q)$ and a fortiori
$X(p,q)$ is super-reflexive.

###### Proof.

It follows easily from Theorem 3.10 and Theorem 4.6 that $X(p,q)$ is non-UMD
and $X^{\mathbb{C}}(p,q)$ is non-AUMD.

For $0<\theta<1$ small enough, such that
$\max(\frac{1/p-\theta/2}{1-\theta},\frac{1/q-\theta/2}{1-\theta})<1$, we can
find $1<\tilde{p},\tilde{q}<\infty$ satisfying the equalities:

$\frac{1}{p}=\frac{\theta}{2}+\frac{1-\theta}{\tilde{p}},\quad\frac{1}{q}=\frac{\theta}{2}+\frac{1-\theta}{\tilde{q}}.$

By Proposition 5.2, we have

$X^{\mathbb{C}}(p,q)=[X^{\mathbb{C}}(\tilde{p},\tilde{q}),X^{\mathbb{C}}(2,2)]_{\theta}.$

Since
$X^{\mathbb{C}}(2,2)=L^{2}_{\mathbb{C}}(D^{\mathbb{N}},\mu^{\otimes\mathbb{N}})$
is Hilbertian, $X^{\mathbb{C}}(p,q)$ is $\theta$-Hilbertian. The super-
reflexivity of $X^{\mathbb{C}}(p,q)$ follows from the well-known fact that any
$\theta$-Hilbertian space is super-reflexive for $\theta>0$ (cf.[12]). ∎

######  Remark 5.4.

Let $1<p\neq q<\infty$. For any $0<\eta<1$, let
$\frac{1}{p_{\eta}}=\frac{1-\eta}{p}+\frac{\eta}{q}$ and
$\frac{1}{q_{\eta}}=\frac{1-\eta}{q}+\frac{\eta}{p}$. By Proposition 5.2, we
have

$X^{\mathbb{C}}(p_{\eta},q_{\eta})=[X^{\mathbb{C}}(p,q),X^{\mathbb{C}}(q,p)]_{\eta}.$

Note that in this interpolation scale, there is only one UMD space
corresponding to $\eta=\frac{1}{2}$.

For futher discussions, let us now turn to the non-atomic case and modify
slightly the definitions. For any $1<p,q<\infty$, consider the family of
spaces $Z_{n}=Z_{n}(p,q)$ defined by recursion: $Z_{0}=\mathbb{C}$ and
$Z_{n+1}=Z_{n}(L_{p}(\mathbb{T},m;L_{q}(\mathbb{T},m))$. From the definition,
we have

$Z_{n}(p,q)\subset Z_{n+1}(p,q).$

Thus we can define

$Z(p,q)=\lim_{\longrightarrow}Z_{n}(p,q).$

To avoid ambiguity, let us emphasize the inclusions $Z_{n}(p,q)\subset
Z_{n+1}(p,q)$ used to define the inductive limit. For simplicity of notations,
we will write $L_{p_{1}}L_{p_{2}}=L_{p_{1}}(L_{p_{2}})$,
$L_{p_{1}}L_{p_{2}}L_{p_{3}}=L_{p_{1}}(L_{p_{2}}(L_{p_{3}}))$, etc. With these
notations, one can easily see the difference between $X_{n}$ and $Z_{n}$ as
follows:

$X_{n+1}=L_{p}(L_{q}(X_{n}))=L_{p}L_{q}\underbrace{L_{p}L_{q}\cdots
L_{p}L_{q}}_{X_{n}},$

where $L_{p}=L_{p}(D,\mu)$ and $L_{q}=L_{q}(D,\mu)$ are two dimensional. And

$Z_{n+1}=Z_{n}(L_{p}(L_{q}))=\underbrace{L_{p}L_{q}\cdots
L_{p}L_{q}}_{Z_{n}}L_{p}L_{q},$

where $L_{p}=L_{p}(\mathbb{T},m)$ and $L_{q}=L_{q}(\mathbb{T},m)$.

######  Remark 5.5.

The main purpose of introducing the spaces $Z_{n}(p,q)$ is that we have
lattice isometric isomorphisms $L_{p}(Z_{n}(p,q))\simeq Z_{n}(p,q)$ for all
$n$ and moreover, these isomorphisms are compatible with the inclusion of
$Z_{n}(p,q)\subset Z_{n+1}(p,q)$ (the word “compatible” will be explained by a
commutative diagram in the sequel) and this will be used to show some
additional properties for $Z(p,q)$. The family of $X_{n}(p,q)$’s shares the
property of having lattice isometric isomorphisms $L_{p}(X_{n}(p,q))\simeq
X_{n}(p,q)$ for all $n$, but the isomorphisms are not compatible with the
inclusions $X_{n}(p,q)\subset X_{n+1}(p,q)$.

The $Z(p,q)$’s are Banach lattices of functions on the infinite torus
$\mathbb{T}^{\mathbb{N}}$, they have the following properties.

######  Proposition 5.6.

Let $1<p,q<\infty$. We have isomorphisms

$Z(p,q)\simeq Z(q,p)$

and

$L_{p}(Z(p,q))\simeq L_{q}(Z(p,q)).$

If $p\neq q$, then $Z(p,q)$ does not have unconditional basis.

###### Proof.

Since $L_{p}(\mathbb{T})$ and $L_{p}(\mathbb{T}\times\mathbb{T})$ are
isometric as Banach lattices, we have isometric isomorphisms which are
compatible with the inclusions $Z_{n}\subset Z_{n+1}$, that is we have the
commutative diagram

$\begin{CD}Z_{n}(p,q)@>{\text{ inclusion }}>{}>Z_{n+1}(p,q)\\\ @V{\text{
isometric }}V{\simeq}V@V{\simeq}V{\text{ isometric}}V\\\ L_{p}(Z_{n}(p,q))@
>\text{ inclusion }>>L_{p}(Z_{n+1}(p,q)).\end{CD}$

By taking Banach space inductive limit, we have

$Z(p,q)\xrightarrow[\text{isometric}]{\simeq}L_{p}(Z(p,q)).$

If $p\neq q$, then $Z(p,q)$ and hence $L_{p}(Z(p,q))$ is non-UMD. By a result
of D.J. Aldous (see [1], Proposition 4), $Z(p,q)$ has no unconditional basis.

It is easy to see that $Z(p,q)$ and $Z(q,p)$ complementably embed into each
other. Since $\ell_{p}^{(2)}(L_{p})=L_{p}$ as Banach lattices, we have

$\ell_{p}^{(2)}(L_{p}(Z(p,q)))=L_{p}(Z(p,q)).$

Moreover, since $L_{p}(Z(p,q))=Z(p,q)$, the above isometry implies that as
Banach space $Z(p,q)=Z(p,q)\oplus Z(p,q)$. Similarly, $Z(q,p)=Z(q,p)\oplus
Z(q,p)$. By the classical Pełcyński decomposition method, we have
$Z(p,q)\simeq Z(q,p)$. Hence

$\qquad\quad L_{p}(Z(p,q))=Z(p,q)\simeq Z(q,p)=L_{q}(Z(q,p))\simeq
L_{q}(Z(p,q)).$

∎

Let $(p_{i})_{i\geq 1}$ be a sequence of real numbers such that
$1<p_{i}<\infty$. Define

$X[(p_{i})]=\lim_{\longrightarrow}L_{p_{n}}\cdots L_{p_{2}}L_{p_{1}}$

and

$Z[(p_{i})]=\lim_{\longrightarrow}L_{p_{1}}L_{p_{2}}\cdots L_{p_{n}}.$

######  Problem.

Under which condition is $X[(p_{i})]$ or $Z[(p_{i})]$ in the UMD class ?

We have the following observations on the necessary condition:

  * (i)

A trivial necessary condition is that there exist $1<p_{0},p_{\infty}<\infty$,
such that $p_{0}\leq p_{i}\leq p_{\infty}$ for all $i\geq 1$.

  * (ii)

If the above condition is satisfied, then the sequence $(p_{i})$ has at least
one cluster point $1<p<\infty$. Then a necessary condition is that the
sequence has only one cluster point, i.e. $\lim_{i\to\infty}p_{i}=p$. Indeed,
assume that the sequence $(p_{i})$ has two cluster points $1<p\neq q<\infty$,
so that there exist two subsequences of $(p_{i})$ which tend to $p,q$
respectively. Then one can easily show that by Theorem 3.10, both $X[(p_{i})]$
and $Z[(p_{i})]$ are non-UMD (they are in fact non-AUMD).

  * (iii)

Now the speed of convergence of $(p_{i})$ will play a role. Since
$\ell_{p_{1}}^{(2)}(\ell_{p_{2}}^{(2)}(\cdots(\ell_{p_{n}}^{(2)})\cdots))$
embeds isometrically into $L_{p_{1}}L_{p_{2}}\cdots L_{p_{n}}$. A necessary
condition for $Z[(p_{i})]$ to be UMD is $\prod_{i}c(p_{2i},p_{2i+1})<\infty$.
Similarly, it is necessary that $\prod_{i}c(p_{2i+1},p_{2i+2})<\infty$.
Combining these, a necessary condition for $Z[(p_{i})]$ to be in the UMD class
is

$\prod_{i}c(p_{i},p_{i+1})<\infty.$

The same statement remains true for $X[(p_{i})]$. Note that by (4),
$c(p_{i},p_{i+1})>1$ if $p_{i}\neq p_{i+1}$.

Intuitively, if $p_{i}$ tends to $p$ sufficiently fast, then both $X[(p_{i})]$
and $Z[(p_{i})]$ are in the UMD class. The author obtained some partial
results in this direction, which will be treated elsewhere.

######  Remark 5.7.

Let $1<p<q<\infty$. We have the following Banach lattices isometries

$L_{p}L_{q}=L_{p}L_{p}L_{q},\quad L_{p}L_{q}=L_{p}L_{q}L_{q}.$

Since $L_{p}L_{r}L_{q}$ is an interpolation space between $L_{p}L_{p}L_{q}$
and $L_{p}L_{q}L_{q}$ for any $p\leq r\leq q$, the $\text{UMD}_{s}$ constant
of $L_{p}L_{r}L_{q}$ is actually the same as that of $L_{p}(L_{q})$. The same
argument shows that $L_{p}L_{u}L_{r}L_{v}L_{q}$ has the same $\text{UMD}_{s}$
constant with $L_{p}L_{q}$, provided $p\leq u\leq r\leq v\leq q$. More
generally, if $(p_{i})_{i=1}^{n}$ is a finite sequence, assume that
$(p_{i})_{i=k}^{l}$ is consecutive monotone (non-increasing or non-decreasing)
subsequence, then $L_{p_{1}}\cdots L_{p_{k}}\cdots L_{p_{l}}\cdots L_{p_{n}}$
and $L_{p_{1}}\cdots L_{p_{k}}L_{p_{l}}\cdots L_{p_{n}}$ have the same
$\text{UMD}_{s}$ constant for all $1<s<\infty$.

Our results have some applications in the non-commutative setting, i.e. on the
operator space UMD property, which will appear in a future publication.

## Acknowledgements

This work was carried out while the author was visiting at Texas A&M
University. The author would like to acknowledge the hospitality provided by
Department of Mathematics of Texas A&M. He would like to thank his advisor G.
Pisier for suggesting this problem and for the constant and stimulating
discussions. The author also appreciates the careful review of the paper by
the referee who suggested many changes to enhance the readability of the
paper.

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