# The dyadic Riesz vector II

Komla Domelevo  Stefanie Petermichl Note: The second author is supported by
the European Research Council project CHRiSHarMa no. DLV-682402 and by the
Alexander von Humboldt Foundation

###### Abstract

We derive a dyadic model operator for the Riesz vector. We show linear upper
$\displaystyle L^{p}$ bounds for $\displaystyle 1<p<\infty$ between this model
operator and the Riesz vector, when applied to functions with values in Banach
spaces. By an upper bound we mean that the boundedness of the dyadic Riesz
vector implies the boundedness of the Riesz vector. The same holds for single
dyadic Riesz transforms and their continuous counterparts. The linear
dependence is with constant one.

##  1 Introduction

Let $\displaystyle X$ be a UMD Banach space and $\displaystyle f\in
L^{p}(\mathbb{R}^{d},X)$ be a $\displaystyle L^{p}$ integrable $\displaystyle
X$–valued function, that is

$\|f\|_{L^{p}(\mathbb{R}^{d},X)}:=\left(\int_{\mathbb{R}^{d}}|f|_{X}^{p}\mathrm{d}x\right)^{1/p}<\infty.$
(1)

We are interested in comparing the $\displaystyle L^{p}$ operator norm the
Riesz transforms $\displaystyle\mathcal{R}_{i}$, $\displaystyle i\in[1,d]$
with their dyadic counterparts, the so–called dyadic Riesz transforms
$\displaystyle\mathcal{S}_{i}$, $\displaystyle i\in[1,d]$ defined below. In
the case $\displaystyle n=1$, the authors have recently proved in [arxiv] that
the Hilbert transform $\displaystyle\mathcal{H}$ on the unit disc and the
so–called dyadic Hilbert transform $\displaystyle\mathcal{S}$ have operator
norms in linear dependence. The natural question arising at that point is
whether or not the Riesz transforms and dyadic Riesz transforms have
comparable $\displaystyle L^{p}$ norms. We note further
$\displaystyle\overrightarrow{\mathcal{R}}f:=(\mathcal{R}_{1}f,\ldots,\mathcal{R}_{d}f)$
and
$\displaystyle\overrightarrow{\mathcal{S}}f:=(\mathcal{S}_{1}f,\ldots,\mathcal{S}_{d}f)$
respectively the Riesz and dyadic Riesz vectors. Their $\displaystyle L^{p}$
norms are

$\displaystyle\|\overrightarrow{\mathcal{R}}f\|_{p}:=\left(\int_{\mathbb{R}^{d}}|\overrightarrow{\mathcal{R}}f|_{\ell^{2}}^{p}\mathrm{d}x\right)^{1/p},\quad\|\overrightarrow{\mathcal{S}}f\|_{p}:=\left(\int_{\mathbb{R}^{d}}|\overrightarrow{\mathcal{S}}f|_{\ell^{2}}^{p}\mathrm{d}x\right)^{1/p}$

with
$\displaystyle|\vec{u}|_{\ell^{2}}:=\left(\sum_{i=1}^{d}|u_{i}|_{X}^{2}\right)^{1/2}$.
In the companion paper [arxiv], the authors prove linear lower bounds

$\displaystyle\forall i\in[1,d],\quad\|\mathcal{S}_{i}\|_{p\rightarrow
p}\leqslant c_{0}\|\mathcal{R}_{i}\|_{p\rightarrow p},$

as well as

$\displaystyle\|\overrightarrow{\mathcal{S}}\|_{p\rightarrow p}\leqslant
c_{0}\|\overrightarrow{\mathcal{R}}\|_{p\rightarrow p},$

for a universal constant $\displaystyle c_{0}$. The main goal of the present
paper is to prove the dimensionless upper bounds

######  Theorem 1 (Upper bound for a single Riesz transform)

We have for all $\displaystyle i\in[1,d]$,

$\displaystyle\|\mathcal{R}_{i}\|_{p\rightarrow
p}\leqslant\|\mathcal{S}_{i}\|_{p\rightarrow p}.$

######  Theorem 2 (Upper bound for the Riesz vector)

We have

$\displaystyle\|\overrightarrow{\mathcal{R}}\|_{p\rightarrow
p}\leqslant\|\overrightarrow{\mathcal{S}}\|_{p\rightarrow p}.$

As compared to the onedimensional case studied in [3], several new
difficulties arise, that we detail later. To summarize, (i) the Riesz
transforms are directly defined as stochastic integrals but as conditional
expectations upon arrival, (ii) convergence of weak formulations are needed,
(iii) the domain and its boundary are unbounded, (iv) the Riesz vector
estimate above requires a specific weak formulation, and finally (v) the
discrete random walks associated to the different dyadic Riesz transforms are
more intricate.

We introduce now the dyadic setting and define the dyadic Riesz transforms.

#### Dyadic system

Let $\displaystyle I_{0}=[0,1)$ and $\displaystyle\mathcal{D}$ the dyadic
intervals

$\displaystyle\mathcal{D}:=\left\\{[m2^{-i},(m+1)2^{-i});\quad 0\leqslant
i,0\leqslant m\leqslant 2^{i}-1\right\\}=\bigcup_{i\geqslant
0}\mathcal{D}_{i}$

where $\displaystyle\mathcal{D}_{i}:=\left\\{[m2^{-i},(m+1)2^{-i});\quad
0\leqslant m\leqslant 2^{i}-1\right\\}$ is the set of dyadic intervals of the
$\displaystyle i$–th generation having therefore length $\displaystyle
2^{-i}$. We note further $\displaystyle\hat{I}$ the parent of $\displaystyle
I$, and $\displaystyle I_{-}$ (resp. $\displaystyle I_{+}$) the left (resp.
right) children of $\displaystyle I$,
$\displaystyle\mathcal{D}^{-}\subset\mathcal{D}$ the subset of left children
and $\displaystyle\mathcal{D}^{+}\subset\mathcal{D}$ the subset of right
children. We have therefore
$\displaystyle\mathcal{D}=\\{I_{0}\\}\cup\mathcal{D}^{-}\cup\mathcal{D}^{+}$.
We also note
$\displaystyle\mathcal{D}_{i}^{-}:=\mathcal{D}_{i}\cap\mathcal{D}^{-}$ the
left children of $\displaystyle\mathcal{D}_{i}$, and
$\displaystyle\mathcal{D}_{i}^{+}:=\mathcal{D}_{i}\cap\mathcal{D}^{+}$ the
right children of $\displaystyle\mathcal{D}_{i}$.

Let the $\displaystyle X$–valued function $\displaystyle f$ locally integrable
on $\displaystyle I_{0}$. Its Haar decomposition writes.

$\displaystyle f(x)=\langle
f\rangle_{I_{0}}+\sum_{I\in\mathcal{D}}(f,h_{I})h_{I},$

where $\displaystyle
h_{I}:=(\boldsymbol{1}_{I_{+}}-\boldsymbol{1}_{I_{-}})/\sqrt{|I|}$ is the
$\displaystyle L^{2}$–normalized Haar function associated to the interval
$\displaystyle I$, and $\displaystyle\langle
f\rangle_{I}:=(f,\boldsymbol{1}_{I})/|I|$ is the average of $\displaystyle f$
on $\displaystyle I$. The dyadic Hilbert transform $\displaystyle\mathcal{S}$
studied in [2, 3] is the operator sending

$\displaystyle\langle f\rangle_{I_{0}}\mapsto 0,\quad h_{I_{0}}\mapsto 0,\quad
h_{I_{\pm}}\mapsto\pm h_{\mp}.$

Writing the Haar decomposition of $\displaystyle f$ in the equivalent forms

$\displaystyle\begin{array}[]{lll}f&=&\langle
f\rangle_{I_{0}}+\sum_{I\in\mathcal{D}}(f,h_{I})h_{I}\\\ &=&\langle
f\rangle_{I_{0}}+(f,h_{I_{0}})h_{I_{0}}+\sum_{k=1}^{\infty}\sum_{I\in{\mathcal{D}_{k}}}(f,h_{I})h_{I}\\\
&=&\langle
f\rangle_{I_{0}}+(f,h_{I_{0}})h_{I_{0}}+\sum_{k=0}^{\infty}\sum_{I\in{\mathcal{D}_{k}}}(f,h_{I_{-}})h_{I_{-}}+(f,h_{I_{+}})h_{I_{+}},\end{array}$

yields the Haar decomposition of the dyadic Hilbert transform
$\displaystyle\mathcal{S}f$ applied to $\displaystyle f$

$\displaystyle\begin{array}[]{lll}\mathcal{S}f&=&\sum_{k=1}^{\infty}\sum_{I\in{\mathcal{D}_{k}}}(f,h_{I})\mathcal{S}h_{I}\\\
&=&\sum_{k=0}^{\infty}\sum_{I\in{\mathcal{D}_{k}}}-(f,h_{I_{-}})h_{I_{+}}+(f,h_{I_{+}})h_{I_{-}}.\end{array}$

#### Dyadic Riesz transforms

We now generalize the construction of the dyadic Hilbert transfrom above to
higher dimensional settings. Let
$\displaystyle\mathcal{D}_{i}=\\{I\in\mathcal{D};|I|=2^{-i}|I_{0}|\\}$ the
$\displaystyle i$–th generation of dyadic intervalls, and
$\displaystyle\mathcal{D}^{(i)}=\cup_{k=0}^{\infty}\mathcal{D}_{kd+i}$ the
$\displaystyle i$–th slice of the dyadic system, that is all dyadic intervals
whose generation number equals $\displaystyle i$ modulo $\displaystyle d$. The
operator $\displaystyle\mathcal{S}_{i}$, $\displaystyle 1\leqslant i\leqslant
d$, is defined as

$\displaystyle\mathcal{S}_{i}:\quad
h_{I}\mapsto\left\\{\begin{array}[]{ll}\mathcal{S}h_{I}&,I\in\mathcal{D}^{(i)}\\\
0&,I\not\in\mathcal{D}^{(i)}.\end{array}\right.$

In other words it acts on the $\displaystyle i$–th slice in the same way as
the dyadic Hilbert transform does. However, it sends to zero all elements of
the other $\displaystyle d-1$ slices. Correspondingly, the Haar decomposition
of $\displaystyle f$, written as blocks of $\displaystyle d$ generations,
reads

$\displaystyle\begin{array}[]{lll}f&=&\langle
f\rangle_{I_{0}}+(f,h_{I_{0}})h_{I_{0}}+\sum_{k=0}^{\infty}\sum_{j=1}^{d}\sum_{I\in{\mathcal{D}_{kd+j}}}(f,h_{I})h_{I}\\\
&=&\langle
f\rangle_{I_{0}}+(f,h_{I_{0}})h_{I_{0}}+\sum_{k=0}^{\infty}\sum_{j=1}^{d}\sum_{I\in{\mathcal{D}_{kd+j-1}}}(f,h_{I_{-}})h_{I_{-}}+(f,h_{I_{+}})h_{I_{+}},\end{array}$

and the $\displaystyle i$–th dyadic Riesz transform
$\displaystyle\mathcal{S}_{i}f$ defined above writes as

$\displaystyle\mathcal{S}_{i}f=\sum_{I\in\mathcal{D}^{(i)}}(f,h_{I})\mathcal{S}h_{I}=\sum_{k=0}^{\infty}\sum_{I\in\mathcal{D}_{kd+i-1}}-(f,h_{I_{-}})h_{I_{+}}+(f,h_{I_{+}})h_{I_{-}}.$

Inspired by the onedimensional case, we wish to approximate the Riesz
transforms in $\displaystyle\mathbb{R}^{d}$ with the help of
$\displaystyle(d+1)$–dimensional discrete random walks built upon the dyadic
filtration. However, a dyadic tree is naturally suited for coding a
onedimensional discrete random walk consisting of random $\displaystyle\pm 1$
steps, i.e. left/right steps. For our task, we need to use the steps provided
by the dyadic tree in order to produce steps along $\displaystyle(d+1)$
different directions. In the case $\displaystyle d=1$, this was done by
assigning left children of the dyadic tree to horizontal steps of the
twodimensional discrete random walk, and right children to vertical steps. For
the general case, we need different assignments and interpretations of the
tosses offered by the dyadic tree.

Let us first review the random objects involved in the continuous setting.

#### Stochastic representations of Riesz transforms

The stochastic representation of Riesz transforms in
$\displaystyle\mathbb{R}^{d}$ follows the work of Gundy–Varopoulos [4], where
the Riesz transforms are conditional expectations of suitable martingales
transforms. See also Bañuelos and Wang [1]. Let
$\displaystyle\Omega:=\mathbb{R}^{d+1}_{+}=\\{(x_{0},x_{1},\ldots,x_{d});x_{0}>0,x=(x_{1},\ldots,x_{d})\in\mathbb{R}^{d}\\}$
the upper–half space. We call $\displaystyle x_{0}\in\mathbb{R}_{+}$ the
“vertical” component of $\displaystyle x\in\Omega$. Given $\displaystyle y>0$,
let $\displaystyle W^{y}=(W^{y,0},W^{y,1},\ldots,W^{y,d})$ the
$\displaystyle(d+1)$–dimensional Brownian motion started at time
$\displaystyle 0$ at the point $\displaystyle(y,0,\ldots,0)\in\Omega$ above
the origin. Let $\displaystyle\tau$ the stopping time, almost surely finite,
defined as

$\displaystyle\tau:=\inf\\{t>0;W_{t}\not\in\Omega\\}.$

This is also the hitting time of $\displaystyle W^{y}$ hitting the boundary
$\displaystyle\partial\Omega=\mathbb{R}^{d}$ of the upper–half space. We
further denote by $\displaystyle(W_{t}^{y,\tau})_{t\geqslant
0}:=(W^{y}_{t\wedge\tau})_{t\geqslant 0}$ the stopped process. By construction
$\displaystyle W^{\tau,y}_{t}\in\Omega$ for all $\displaystyle t>0$ and
$\displaystyle W^{\tau,y}_{t}\in\partial\Omega$ for all $\displaystyle
t\geqslant\tau$. In particular $\displaystyle
W^{\tau,y}_{\infty}\in\partial\Omega$.

Let now $\displaystyle f\in L^{p}(\partial\Omega,X)$ a smooth function (see
the precise definition of smooth below at the end of the section). We also
denote by $\displaystyle f:\Omega\rightarrow X$ its harmonic extension in the
upper–half space. Given $\displaystyle y>0$ and $\displaystyle f$ as above,
Ito formula ensures that the stochastic process
$\displaystyle\mathcal{M}_{t}^{y,f}:=f(W^{y,\tau}_{t})$ is a martingale
satisfying

$\displaystyle\mathcal{M}_{t}^{y,f}:=f(W^{y,\tau}_{t})=f(y)+\int_{0}^{t}\nabla
f(W^{y,\tau}_{t})\cdot\mathrm{d}W^{y,\tau}_{t},$

and its $\displaystyle i$–th martingale transform
$\displaystyle\mathcal{M}_{t}^{i}$ is defined as

$\mathcal{M}_{t}^{y,i}:=\int_{0}^{t}(A_{i}\nabla
f)(W^{y,\tau}_{t})\cdot\mathrm{d}W^{y,\tau}_{t}=\int_{0}^{t}\nabla
f(W^{y,\tau}_{t})\cdot(A_{i}^{\ast}\mathrm{d}W^{y,\tau}_{t}),\quad i\in[1,d],$
(2)

where $\displaystyle A_{i}\in\mathbb{R}^{(d+1)\times(d+1)}$ is the matrix
defining the $\displaystyle i$–th martingale transform. Namely if
$\displaystyle(e_{0},e_{1},\ldots,e_{d})$ denotes the standard basis of
$\displaystyle\mathbb{R}^{d+1}$, we have set

$\displaystyle A_{i}e_{0}=-e_{i},\quad A_{i}e_{i}=e_{0},\quad
A_{i}e_{j}=0,\quad j\in[1,d],j\neq i,$

and we observe that its transpose $\displaystyle A_{i}^{\ast}$ obeys

$\displaystyle A^{\ast}_{i}e_{0}=e_{i},\quad A_{i}^{\ast}e_{i}=-e_{0},\quad
A^{\ast}_{i}e_{j}=0,\quad j\in[1,d],j\neq i.$

To summarize,

$A_{i}\nabla f=(-\partial_{i}f,0,\ldots,\partial_{0}f,0,\ldots,0),$ (3)

and for the transposed counterpart, where we let the transform act on the
random walk,

$A_{i}^{\ast}\mathrm{d}W_{t}=(\mathrm{d}W^{i}_{t},0,\ldots,0,-\mathrm{d}W^{0}_{t},0,\ldots,0).$
(4)

Following [4], see also [1], the weak formulation for the $\displaystyle i$–th
Riesz transforms writes

$(\mathcal{R}_{i}f,g)=\mathbb{E}\int_{0}^{\infty}(A_{i}\nabla
f)(W^{y,\tau}_{t})\cdot\nabla g(W^{y,\tau}_{t})\mathrm{d}t,$ (5)

for any $\displaystyle X^{\ast}$–valued harmonic test function $\displaystyle
g:\Omega\rightarrow X^{\ast}$ with smooth boundary values $\displaystyle g\in
L^{q}(\partial\Omega,X^{\ast})$. Equivalently, the $\displaystyle i$–th Riesz
transform is recovered pointwise thanks to the conditional expectation (see
[4]),

$\forall
x\in\partial\Omega,\quad\mathcal{R}_{i}f(x)=\lim_{y\rightarrow+\infty}\mathbb{E}(\mathcal{M}_{\infty}^{y,i}|W^{y,\tau}_{\infty}=(0,x)).$
(6)

#### Summary of notations and functional setting

Recall that we denote $\displaystyle\Omega=\mathbb{R}^{d+1}_{+}$ the upper
half space and $\displaystyle\partial\Omega=\mathbb{R}^{d}$ its boundary. We
assume without loss of generality that $\displaystyle f\in
L^{p}(\partial\Omega,X)$ and its harmonic extension (also noted $\displaystyle
f$) $\displaystyle f\in L^{p}(\Omega,X)$ are smooth Frechet differentiable
functions, that is $\displaystyle f\in\mathcal{C}^{k}$ for all $\displaystyle
k\geqslant 0$. We note $\displaystyle|f(x)|:=|f(x)|_{X}$ the Banach space norm
of $\displaystyle f(x)$. The $\displaystyle L^{p}$–norm of $\displaystyle f$
is defined as in (1).

Further if $\displaystyle Y$ is an $\displaystyle L^{p}$ integrable random
variable, the stochastic $\displaystyle L^{p}$ norm is defined as

$\displaystyle\|Y\|_{p}:=\mathbb{E}(|Y|^{p})^{1/p}.$

If $\displaystyle Y$ is real valued, then $\displaystyle|Y|$ denotes the
absolute value of $\displaystyle Y$ whereas if $\displaystyle Y$ is
$\displaystyle X$–valued, then $\displaystyle|Y|:=|Y|_{X}$ denotes the Banach
space norm of $\displaystyle Y$. For a real random variable, we note its
variance $\displaystyle\mathbb{V}(Y):=\mathbb{E}(Y-\mathbb{E}(Y))^{2}$.

If $\displaystyle f:=f(x_{1},\ldots,x_{m})$ is a $\displaystyle X$–valued
function of $\displaystyle m$ variables defined on the open set $\displaystyle
U\subset\mathbb{R}^{m}$, we note
$\displaystyle\mathrm{D}f:=(\partial_{1}f,\ldots,\partial_{m}f)$ its
derivatives in the Frechet sense, where
$\displaystyle\mathrm{D}f:U\times\mathbb{R}^{m}\rightarrow X$ is continuous.
Further, given a $\displaystyle m$–multiindex
$\displaystyle\alpha:=(\alpha_{1},\alpha_{2},\ldots,\alpha_{m})$ with
$\displaystyle|\alpha|=k$ we note as usual
$\displaystyle\mathrm{D}^{\alpha}f:=\partial_{1}^{\alpha_{1}}\ldots\partial_{m}^{\alpha_{m}}f$
its partial derivatives of order $\displaystyle k$ in the Frechet sense, where
$\displaystyle\mathrm{D}^{\alpha}f:U\times(\mathbb{R}^{m})^{k}\rightarrow X$
is continuous. Finally,using again multiindices, monomials of the from
$\displaystyle x^{\alpha}$, where $\displaystyle x:=(x_{1},\ldots,x_{m})$, are
a shorthand for $\displaystyle x^{\alpha}:=x_{1}^{\alpha_{1}}\ldots
x_{m}^{\alpha_{m}}$.

The dual space of $\displaystyle X$ is noted $\displaystyle X^{\ast}$, and the
dual space of $\displaystyle L^{p}(U,X)$ is noted $\displaystyle
L^{q}(U,X^{\ast})$ with $\displaystyle 1/p+1/q=1$.

For the convergence results, our main parameters are
$\displaystyle\varepsilon>0$, a small number, and $\displaystyle T>0$, a large
number. We note $\displaystyle c(\varepsilon)$ a generic function tending to
$\displaystyle 0$ uniformly in $\displaystyle T$ when
$\displaystyle\varepsilon$ goes to zero, $\displaystyle c_{T}(\varepsilon)$ a
generic function tending to $\displaystyle 0$ when $\displaystyle\varepsilon$
goes to zero for any fixed $\displaystyle T$, $\displaystyle c(T)$ a generic
function tending to $\displaystyle 0$ when $\displaystyle T$ goes to infinity,
uniformely in $\displaystyle\varepsilon$. It is implicit that those functions
all depend on the fixed function $\displaystyle f$ and its derivatives.
Further dependences will be mentionned when needed.

##  2 Discrete random walks and martingales

Following the strategy used for the dyadic Hilbert transform, we wish to
approximate continuous martingales such as $\displaystyle\mathcal{M}_{t}^{f}$
that are driven by $\displaystyle W^{y,\tau}_{t}$ with discrete martingales
$\displaystyle M_{k}^{f}$ driven by a $\displaystyle(d+1)$–dimensional
discrete random walk $\displaystyle(B_{k})_{k\in\mathbb{N}}$ living on a
dyadic filtered probability space. At difference with the Hilbert case, the
construction of a single discrete random walk does not allow us to estimate
the different Riesz transforms. We build instead a set of discrete random
walks $\displaystyle(B^{(i)}_{k})_{k\in\mathbb{N}}$ for $\displaystyle
i\in[1,d]$, each of which is adapted to the study of one single Riesz
transform. For a given $\displaystyle i\in[1,d]$, we are aiming at mimicking
the action of the $\displaystyle i$–th Riesz martingale transform
$\displaystyle\mathrm{d}W_{t}\rightarrow A_{i}^{\ast}\mathrm{d}W_{t}$ as seen
in the expression (2) of $\displaystyle\mathcal{M}^{i}_{t}$ and in (4). In
other words, we want

$\displaystyle\mathcal{S}_{i}\mathrm{d}B_{k}^{(i)}=A_{i}^{\ast}\mathrm{d}B_{k}^{(i)},\quad\operatorname{all}k\geqslant
1,$

where $\displaystyle\mathrm{d}B_{k}^{(i)}:=B_{k}^{(i)}-B_{k-1}^{(i)}$ is the
$\displaystyle k$–th increment of the discrete random walk $\displaystyle
B^{(i)}$ and $\displaystyle\mathcal{S}_{i}$ the $\displaystyle i$–th dyadic
Riesz transform defined earlier. We set the starting point of $\displaystyle
B^{(i)}$ to coincide with that of $\displaystyle W^{y,\tau}$, namely

$\displaystyle B_{0}^{(i)}=W_{0}^{y,\tau}=(y,0,\ldots,0),$

and we will omit the superscript $\displaystyle y$ for the discrete random
walks. Let us split $\displaystyle B^{(i)}$ into its vertical component
$\displaystyle B^{(i),0}\in\mathbb{R}^{+}$, a one–dimensional discrete random
walk started at $\displaystyle y>0$, and its horizontal component
$\displaystyle B^{H}\in\mathbb{R}^{d}$,

$\displaystyle
B_{k}^{(i)}=(B_{k}^{(i),0},B^{H}_{k}),\quad\operatorname{all}k\geqslant 0,$

where $\displaystyle B^{H}_{k}=(B^{H,1}_{k},B^{H,2}_{k},\ldots,B^{H,d}_{k})$
is a $\displaystyle d$–dimensional discrete random walk started at
$\displaystyle 0\in\mathbb{R}^{d}$.

We associate nodes of the dyadic tree to certain increments of $\displaystyle
B^{(i)}$. We start with the horizontal part. Here, for any slice
$\displaystyle\mathcal{D}^{(j)}$, $\displaystyle j\in[1,d]$, we associate each
left interval of that slice to an increment of $\displaystyle B^{H}_{k}$ in
the $\displaystyle j$–th direction. Schematically, for $\displaystyle d=2$,
those are the nodes and directions involved, marked with $\displaystyle x_{1}$
for producing an increment of $\displaystyle B^{H}$ parallel to $\displaystyle
e_{1}$ and marked with $\displaystyle x_{2}$ for producing an increment of
$\displaystyle B^{H}$ parallel to $\displaystyle e_{2}$:

Figure 1: Nodes contributing to $\displaystyle B^{H}$

Now the definition of the vertical component $\displaystyle B^{(i),0}$ depends
on the direction $\displaystyle i$ of the chosen dyadic Riesz transform
$\displaystyle\mathcal{S}_{i}$ under study. Given $\displaystyle i\in[1,d]$,
we will collect increments of $\displaystyle B^{(i),0}$ only on intervals that
are on the one hand right children and on the other hand that belong to the
$\displaystyle i$–th slice $\displaystyle\mathcal{D}^{(i)}$. Schematically,
for $\displaystyle d=2$, those are the nodes involved respectively for
$\displaystyle B^{(1),0}$ and $\displaystyle B^{(2),0}$:

Figure 2: Nodes involved for $\displaystyle B^{(1),0}$

Figure 3: Nodes involved for $\displaystyle B^{(2),0}$

The construction of all these objects obeys a period equal to $\displaystyle
d$ generations. Let us note
$\displaystyle\mathcal{L}_{k}:=\cup_{j=1}^{d}\mathcal{D}_{(k-1)d+j}$,
$\displaystyle k\geqslant 1$, the $\displaystyle k$–th layer of the dyadic
tree, made of $\displaystyle d$ successive generations of intervals. Notice
that the first layer starts at the first generation
$\displaystyle\mathcal{D}_{1}$ of dyadic intervals, i.e.
$\displaystyle\mathcal{L}_{1}=\mathcal{D}_{1}\cup\ldots\cup\mathcal{D}_{d}$.
The root $\displaystyle\mathcal{D}_{0}=\\{I_{0}\\}$ does not belong to any
layer. This is because we only consider intervals with siblings. Introduce the
tosses $\displaystyle\varepsilon_{I}(x)=\pm 1$ defined as the $\displaystyle
L^{\infty}$ normalized Haar functions

$\displaystyle\varepsilon_{I}(x):=\sqrt{|I|}h_{I}(x)=\boldsymbol{1}_{I_{+}}(x)-\boldsymbol{1}_{I_{-}}(x)=\left\\{\begin{array}[]{ll}1,&x\in
I_{-}\\\ -1,&x\in I_{+}\\\ 0,&\operatorname{otherwise}\end{array}\right.$

We now define tosses parametrized with generation and left/right position. For
the generation $\displaystyle i=0$ we have simply

$\displaystyle\varepsilon_{0}(x)=\varepsilon_{I_{0}}(x).$

Similarly the $\displaystyle i$–th toss can be defined as

$\displaystyle\varepsilon_{i}(x)=\sum_{I\in\mathcal{D}^{i}}\varepsilon_{I}(x).$

Notice that only one term is non zero in the sum above, namely
$\displaystyle\varepsilon_{i}(x)=\varepsilon_{I^{x}_{i}}(x)$ where
$\displaystyle I^{x}_{i}$ denotes the unique dyadic interval in
$\displaystyle\mathcal{D}_{i}$ that contains $\displaystyle x$. For our
purpose, we need to split those tosses into two groups,

$\displaystyle\varepsilon_{i}^{-}(x)=\left\\{\begin{array}[]{ll}\varepsilon_{i}(x),&x\in\mathcal{D}_{i}^{-}\\\
0,&x\in\mathcal{D}_{i}^{+}\end{array}\right.,\quad\varepsilon_{i}^{+}(x)=\left\\{\begin{array}[]{ll}0,&x\in\mathcal{D}_{i}^{-}\\\
\varepsilon_{i}(x),&x\in\mathcal{D}_{i}^{+}\end{array}\right.,$

or equivalently

$\displaystyle\varepsilon^{\pm}_{i}(x)=\boldsymbol{1}(\varepsilon_{i-1}(x)=\pm
1)\varepsilon_{i}(x)=\sum_{I\in\mathcal{D}_{i}^{\pm}}\varepsilon_{I}(x).$

Here is a summary of the different tosses carried by the dyadic tree up to
generation $\displaystyle\mathcal{D}_{4}$.

Figure 4: Tosses on the dyadic tree.

Let $\displaystyle\delta$ a small time step. For each $\displaystyle
k\geqslant 1$, we construct the step $\displaystyle\mathrm{d}B_{k}^{(i)}$ of
the discrete random walk $\displaystyle B^{(i)}_{k}$ using information from
the layer $\displaystyle\mathcal{L}_{k}$:

$\displaystyle\mathrm{d}B_{k}^{H}(x)=\sqrt{2\delta}(\varepsilon_{(k-1)d+1}^{-}(x),\varepsilon_{(k-1)d+2}^{-}(x),\ldots,\varepsilon_{(k-1)d+d}^{-}(x)),\quad\mathrm{d}B_{k}^{(i),0}(x)=\sqrt{2\delta}\varepsilon_{(k-1)d+i}^{+}(x).$

It follows

$\displaystyle\displaystyle\mathcal{S}_{i}\mathrm{d}B^{(i)}_{k}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\mathcal{S}_{i}(\mathrm{d}B^{(i),0}_{k},\mathrm{d}B_{k}^{H})$
(7) $\displaystyle\displaystyle=$
$\displaystyle\displaystyle\sqrt{2\delta}\mathcal{S}_{i}(\varepsilon_{(k-1)d+i}^{+},\varepsilon_{(k-1)d+1}^{-},\varepsilon_{(k-1)d+2}^{-},\ldots,\varepsilon_{(k-1)d+d}^{-})$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\sqrt{2\delta}(\varepsilon^{-}_{(k-1)d+i},0,\ldots,0,-\varepsilon_{(k-1)d+i}^{+},0,\ldots,0)$
$\displaystyle\displaystyle=$ $\displaystyle\displaystyle
A_{i}^{\ast}\mathrm{d}B^{(i)}_{k}.$

This achieves our first goal of expressing the action of
$\displaystyle\mathcal{S}_{i}$ as a generalized Cauchy–Riemann relation on the
dyadic system.

#### Coarse graining and stopping times

Let $\displaystyle T>0$ a fixed time, and $\displaystyle N\in\mathbb{N}$
large. Define the time step $\displaystyle\delta$ through $\displaystyle
T=N^{5}\delta$. In order to denote the corresponding discrete times, we will
use the indices $\displaystyle k,l\in[0,N^{5}]$, typically $\displaystyle
t_{k}:=k\delta$, $\displaystyle t_{l}:=l\delta$. We introduce a coarser time-
step $\displaystyle\theta$ defined as $\displaystyle\theta:=N\delta$. Notice
that $\displaystyle T=N^{4}\theta$, so that $\displaystyle\theta$ tends to
zero as $\displaystyle N$ goes to infinity for fixed $\displaystyle T$. The
discrete times corresponding to this time–step will use indices $\displaystyle
n,m\in[0,N^{4}]$, typically $\displaystyle t_{n}:=n\theta$, $\displaystyle
t_{m}:=m\theta$. For a given $\displaystyle i\in[1,d]$, we define the discrete
random walk $\displaystyle X^{(i)}$ by sampling $\displaystyle B^{(i)}$ at
times that are multiples of $\displaystyle\theta=N\delta$, therefore with
indices that are multiples of $\displaystyle N$,

$\displaystyle\forall n\geqslant 0,\quad X^{(i)}_{n}:=B^{(i)}_{nN}.$

Notice further that, for all $\displaystyle n\geqslant 1$,

$\displaystyle\mathrm{d}X^{(i)}_{n}:=X^{(i)}_{n}-X^{(i)}_{n-1}=B^{(i)}_{nN}-B^{(i)}_{(n-1)N}=\sum_{l=1}^{N}\mathrm{d}B^{(i)}_{(n-1)N+l}.$

In order to stop $\displaystyle X^{(i)}$ before it leaves the upper–half
space, we set $\displaystyle\varepsilon=1/N$ and introduce

$\displaystyle\Omega_{\varepsilon}:=\left\\{x=(x_{0},x_{1},\ldots,x_{d})\in\Omega;\quad
x_{0}>\varepsilon\right\\},$

as well as the stopping time, for each $\displaystyle i\in[1,d]$,

$\displaystyle\tau_{\varepsilon}^{(i)}:=\inf\left\\{t_{n};\quad
X^{(i)}_{n}\not\in\Omega_{\varepsilon}\right\\}.$

We will denote by $\displaystyle n_{\varepsilon}^{(i)}$ the corresponding
random index such that
$\displaystyle\tau_{\varepsilon}^{(i)}:=n_{\varepsilon}^{(i)}\theta$. One
observes that for small enough $\displaystyle\varepsilon>0$ (i.e. large enough
$\displaystyle N$), we have for all $\displaystyle n$,
$\displaystyle|\mathrm{d}X^{(i)}_{n}|\leqslant N\delta\leqslant
TN^{-4}<\varepsilon/2$. Therefore since $\displaystyle
X_{n_{\varepsilon}^{(i)}-1}\in\Omega_{\varepsilon}$ we have that
$\displaystyle X_{n_{\varepsilon}^{(i)}}$ is in the band
$\displaystyle\Omega\backslash\Omega_{\varepsilon}$. Equivalently,
$\displaystyle\tau^{(i)}_{\varepsilon}$ is the the first time $\displaystyle
t_{n}$ at which $\displaystyle X^{(i)}$ is at a distance less or equal to
$\displaystyle\varepsilon$ from the boundary $\displaystyle\partial\Omega$ of
the upper–half space. This stopping time is also a stopping time for
$\displaystyle B^{(i)}$. We note $\displaystyle
B^{(i),\tau_{\varepsilon}^{(i)}}$ the corresponding stopped process. Notice
that $\displaystyle B^{(i),\tau_{\varepsilon}^{(i)}}\in\Omega_{\varepsilon/2}$
for all times. In all what follows, all random walks, continuous or discrete,
are meant as stopped process. For convenience we might write $\displaystyle
X^{(i)}_{n}$ instead of $\displaystyle X^{(i),\tau^{(i)}_{\varepsilon}}_{n}$,
or $\displaystyle B^{(i)}_{k}$ instead of $\displaystyle
B^{(i),\tau^{(i)}_{\varepsilon}}_{k}$. By construction we have always
$\displaystyle W_{t}^{y},X_{n}^{(i)},B_{k}^{(i)}\in\Omega$.

#### Discrete martingales and their transforms

Given $\displaystyle f\in L^{p}(\mathbb{R}^{d},X)$ and its harmonic extension
also noted $\displaystyle f$, we now aim at approximating both
$\displaystyle\mathcal{M}_{t}^{f}$ and $\displaystyle\mathcal{M}^{i}_{f}$ by
discrete martingales adapted to the action of the dyadic Riez transforms. An
important difference with the onedimensional case is that the discretization
of not only $\displaystyle\mathcal{M}^{i}_{f}$ but also
$\displaystyle\mathcal{M}^{f}$ depends on the single Riesz transform under
study. For each $\displaystyle i\in[1,d]$, we define the approximate
$\displaystyle M^{(i),f}$ of $\displaystyle\mathcal{M}^{f}$ as

$\displaystyle M^{(i),f}_{k}:=\mathcal{M}_{0}^{f}+\sum_{\ell=1}^{k}\nabla
f(B_{k-1}^{(i)})\cdot\mathrm{d}B_{k}^{(i)},$

and the approximate $\displaystyle M^{(i),i}$ of
$\displaystyle\mathcal{M}^{i}_{t}$ as

$\displaystyle M_{k}^{(i),i}:=\sum_{\ell=1}^{k}A_{i}\nabla
f(B_{k-1}^{(i)})\cdot\mathrm{d}B_{k}^{(i)}.$

The superscript $\displaystyle(i)$ refers to the fact that the discrete
process is driven by the $\displaystyle i$–th discrete random walk
$\displaystyle B^{(i)}$, whereas the superscript $\displaystyle i$ refers to
the fact that we are approximating the $\displaystyle i$–th martingale
transform $\displaystyle\mathcal{M}^{i}$. Our first goal is achieved:

###### Lemma 1

For all $\displaystyle f$, all $\displaystyle i\in[1,d]$, we have
$\displaystyle M_{k}^{(i),i}=\mathcal{S}_{i}M^{(i),f}_{k}$ and therefore

$\|M_{k}^{(i),i}\|_{p}\leqslant\|\mathcal{S}_{i}\|_{p\rightarrow
p}\|M^{(i),f}_{k}\|_{p}.$ (8)

###### Proof.

By construction of the discrete random walks and property (7), we have
immediately

$\displaystyle\begin{array}[]{lll}\mathcal{S}_{i}M^{(i),f}_{k}&=&\sum_{\ell=1}^{k}\nabla
f(B_{k-1}^{(i)})\cdot\mathcal{S}_{i}\mathrm{d}B_{k}^{(i)}=\sum_{\ell=1}^{k}\nabla
f(B_{k-1}^{(i)})\cdot A_{i}^{\ast}\mathrm{d}B_{k}^{(i)}\\\
&=&\sum_{\ell=1}^{k}A_{i}\nabla
f(B_{k-1}^{(i)})\cdot\mathrm{d}B_{k}^{(i)}=M_{k}^{(i),i}\end{array}$

and the result follows. ∎

###### Remark 1

If $\displaystyle X^{(i)}_{n}:=B^{(i)}_{nN}$ is the coarse–grained random
walk, we could also define the stochastic integrals based on $\displaystyle
X^{(i)}$ instead of $\displaystyle B^{(i)}$, i.e.

$\displaystyle M^{(i),f}_{n}:=\mathcal{M}_{0}^{f}+\sum_{m=1}^{n}\nabla
f(X_{m-1}^{(i)})\cdot\mathrm{d}X_{m}^{(i)},\quad
M_{n}^{(i),f_{i}}:=\sum_{m=1}^{n}A_{i}\nabla
f(X_{m-1}^{(i)})\cdot\mathrm{d}X_{m}^{(i)}.$

and we would still have by linearity of $\displaystyle\mathcal{S}_{i}$ that
$\displaystyle M_{n}^{(i),f_{i}}=\mathcal{S}_{i}M^{(i),f}_{n}$. It is not a
big simplification to the proof of convergence.

#### Strategy and plan

Our main goal is now to show that one can pass to the limit in inequality (8)
of Lemma 1. As compared to the onedimensional case $\displaystyle d=1$ studied
in [3], a few new difficulties arise.

First, the Riesz transforms are not directly defined by a stochastic integral,
but rather through a projection of a stochastic integral. This projection is
the conditional expectation upon arrival as in (6). This projection would not
be easy to handle as a limit of a discrete process. For this reason, part of
our proof relies on the weak formulation (5) and related convergence results
from discrete weak formulations to continuous weak formulations.

Second, the estimate of the Riesz vector requires a specific weak formulation
that we can relate to the dyadic Riesz transforms.

Finally, the upper half space as well as its boundary are unbounded, and the
random processes involved in the stochastic representation need to start at
“infinity”, which involves a supplementary limiting process as compared to the
onedimensional case on the circle.

##  3 Auxiliairy results

Recall that all discrete random walks $\displaystyle B^{(i)}$, or
$\displaystyle X^{(i)}$for $\displaystyle i\in[1,d]$ are started at
$\displaystyle(y,0,\ldots,0)\in\Omega$ and stopped at time
$\displaystyle\tau_{\varepsilon}^{(i)}$. We note $\displaystyle W^{t,x,\tau}$
the $\displaystyle(d+1)$–dimensional Brownian motion started at time
$\displaystyle t$ at point $\displaystyle x\in\Omega$ and stopped at time
$\displaystyle\tau$ when hitting $\displaystyle\partial\Omega$. We define,

###### Definition 1

We say that $\displaystyle X$ is weakly consistent with $\displaystyle
W^{\tau}$, iff there exists a function $\displaystyle c:=c(\varepsilon)$
tending to zero when $\displaystyle\varepsilon$ goes to zero, such that for
all $\displaystyle n$,

$\displaystyle|\mathbb{E}(X_{n+1}-X_{n}|\widetilde{\mathcal{F}}_{n})-\mathbb{E}(W^{t_{n},X_{n},\tau}_{t_{n+1}}-W_{t_{n}}^{t_{n},X_{n},\tau}|\widetilde{\mathcal{F}}_{n})|\leqslant\theta
c(\varepsilon),$

and such that for all $\displaystyle(d+1)$–multiindex
$\displaystyle\alpha:=\left(\alpha_{0},\alpha_{1}{,\ldots,\alpha_{d}}\right)$
with $\displaystyle|\alpha|=2$, there holds

$\displaystyle|\mathbb{E}((X_{n+1}-X_{n})^{\alpha}|\widetilde{\mathcal{F}}_{n})-\mathbb{E}((W^{t_{n},X_{n},\tau}_{t_{n+1}}-W_{t_{n}}^{t_{n},X_{n},\tau})^{\alpha}|\widetilde{\mathcal{F}}_{n})|\leqslant\theta
c(\varepsilon),$

and claim,

###### Lemma 2

Let $\displaystyle i\in[1,d]$. The discrete stopped process $\displaystyle
X^{(i)}$ is weakly consistent with the continuous stopped process
$\displaystyle W^{\tau}$.

###### Proof.

This is an easy consequence of the following two lemmas. ∎

######  Lemma 3 (Continuous moments)

Let $\displaystyle W^{t,x,\tau}$ as above. We have

$\displaystyle\forall t\geqslant 0,\forall
x\in\Omega,\quad\mathbb{E}(W_{t+\theta}^{\tau,t,x}-W_{t}^{\tau,t,x})=0.$

Moreover there exists a function $\displaystyle c_{T}(\varepsilon)$ tending to
zero with $\displaystyle\varepsilon$ for all fixed $\displaystyle T$, such
that for all coordinates $\displaystyle j,k\in[0,d]$

$\displaystyle\forall t\geqslant 0,\forall
x\in\Omega_{\varepsilon},\quad\mathbb{E}(W_{t+\theta}^{t,x,\tau,j}-W_{t}^{t,x,\tau,j})^{2}=\theta(1+c_{T}(\varepsilon)),$
$\displaystyle\forall t\geqslant 0,\forall
x\in\Omega_{\varepsilon},\quad\mathbb{E}(W_{t+\theta}^{t,x,\tau,j}-W_{t}^{t,x,\tau,j})(W_{t+\theta}^{t,x,\tau,k}-W_{t}^{t,x,\tau,k})=\theta
c_{T}(\varepsilon)\quad\operatorname{for}j\neq k.$

Finally for all $\displaystyle p\geqslant 2$, there holds for all coordinates
$\displaystyle j\in[0,d]$

$\displaystyle\forall t\geqslant 0,\forall x\in\Omega,\forall p\geqslant
2,\quad\mathbb{E}(|W_{t+\theta}^{\tau,t,x,j}-W_{t}^{\tau,t,x,j}|^{p})\lesssim\theta^{p/2}.$

###### Proof.

The proof is the same as in the case $\displaystyle d=1$. See [3]. ∎

######  Lemma 4 (Discrete Moments)

Let $\displaystyle i\in[1,d]$. Let $\displaystyle X^{(i)}$ the $\displaystyle
i$–th stopped discrete random walk defined above. We have

$\displaystyle\forall
n\in\mathbb{N},\quad\mathbb{E}\left(\mathrm{d}X^{(i)}_{n+1}|{\widetilde{\mathcal{F}}_{n}}\right)=0$

Moreover there exists a function $\displaystyle c(\varepsilon)$, tending to
zero with $\displaystyle\varepsilon$, such that for all coordinates
$\displaystyle j,k\in[0,d]$

$\displaystyle\forall
n\in\mathbb{N},\quad\mathbb{E}((\mathrm{d}X^{(i),j}_{n+1})^{2}|\widetilde{\mathcal{F}}_{n})=\theta(1+c(\varepsilon)),$
$\displaystyle\forall
n\in\mathbb{N},\quad\mathbb{E}((\mathrm{d}X^{(i),j}_{n+1})(\mathrm{d}X^{(i),k}_{n+1})|\widetilde{\mathcal{F}}_{n})=\theta
c(\varepsilon),\quad\operatorname{for}j\neq k.$

Finally for all $\displaystyle p\geqslant 2$, there holds for all coordinates
$\displaystyle j\in[0,d]$

$\displaystyle\mathbb{E}(|\mathrm{d}X_{n+1}^{(i),j}|^{p}|\widetilde{\mathcal{F}}_{n})\lesssim\theta^{p/2}.$

###### discrete moments.

Let $\displaystyle T>0$ and $\displaystyle N\in\mathbb{N}$. Recall that the
discretisation parameters $\displaystyle\delta$, $\displaystyle\theta$, and
$\displaystyle\varepsilon$ are defined as

$\displaystyle T=:N^{5}\delta,\quad\theta:=N\delta,\quad
T:=N^{4}\theta,\quad\varepsilon:=1/N.$

The increment

$\displaystyle\mathrm{d}X_{n+1}^{(i)}:=X^{(i)}_{n+1}-X_{n}^{(i)}=\sum_{k=1}^{N}\mathrm{d}B^{(i)}_{nN+k}$

involves $\displaystyle N$ consecutive steps of $\displaystyle B^{(i)}$,
therefore $\displaystyle N$ layers of the dyadic tree, hence $\displaystyle
Nd$ generations. The layers involved are
$\displaystyle\cup_{k=1}^{N}\mathcal{L}_{nN+k}$. Precisely,

$\displaystyle\begin{array}[]{ll}\mathrm{d}X_{n+1}^{(i)}(x)&=\sqrt{2\delta}\left(\sum_{k=1}^{N}\varepsilon_{(nN+k-1)d+i}^{+}(x),\sum_{k=1}^{N}\varepsilon_{(nN+k-1)d+1}^{-}(x),\right.\\\
&\left.\sum_{k=1}^{N}\varepsilon_{(nN+k-1)d+2}^{-}(x),\ldots,\sum_{k=1}^{N}\varepsilon_{(nN+k-1)d+d}^{-}(x)\right).\end{array}$

we have immediately
$\displaystyle\mathbb{E}(\mathrm{d}X_{n+1}^{(i)}|\mathcal{F}_{nN})=(0,\ldots,0)$.
For the coordinate $\displaystyle j=0$, we calculate the variance

$\displaystyle\begin{array}[]{lll}\mathbb{V}(\mathrm{d}X_{n+1}^{(i),0}|\widetilde{\mathcal{F}}_{n})&=&2\delta\mathbb{V}\left(\sum_{k=1}^{N}\varepsilon_{(nN+k-1)d+j}^{+}|\widetilde{\mathcal{F}}_{n}\right)\\\
&=&2\delta(\boldsymbol{1}(\varepsilon_{(nN+k-1)d}=+1))^{2}+2\delta\sum_{k=1}^{N-1}\frac{1}{2}[0^{2}+1^{2}]\\\
&=&\frac{2\delta N}{2}(1+c(\varepsilon))=\theta(1+c(\varepsilon))\end{array}$

and similarly for $\displaystyle j=1$,

$\displaystyle\mathbb{V}(\mathrm{d}X_{n+1}^{(i),1}|\widetilde{\mathcal{F}}_{n})=\theta(1+c(\varepsilon)).$

Now for $\displaystyle 2\leqslant j\leqslant d$, we have

$\displaystyle\mathbb{V}(\mathrm{d}X_{n+1}^{(i),j}|\widetilde{\mathcal{F}}_{n})=2\delta\mathbb{V}\left(\sum_{k=1}^{N}\varepsilon_{(nN+k-1)d+j}^{\pm}|\widetilde{\mathcal{F}}_{n}\right)=2\delta\sum_{k=1}^{N}\frac{1}{2}[0^{2}+1^{2}]=\theta.$

Regarding the covariation of the different coordinates of
$\displaystyle\mathrm{d}X_{n+1}^{(i)}$, we have simply

$\displaystyle\forall\alpha,\beta\in[0,d],\alpha\neq\beta,\quad\mathbb{E}(\mathrm{d}X_{n+1}^{(i),\alpha}\mathrm{d}X_{n+1}^{(i),\beta}|\widetilde{\mathcal{F}}_{n})=0.$

Indeed, increments of different coordinates involve tosses (i.e. Haar
functions) with either pairwise–disjoint or included supports.

The last estimate for moments of order $\displaystyle p\geqslant 2$ is first
proved for even values of $\displaystyle p$ and then extended using Hölder to
all values of $\displaystyle p\geqslant 2$. See [3]. ∎

##  4 Auxiliary convergence results

The following two auxiliary convergence results are needed.

###### Lemma 5

(Weak convergence) Let $\displaystyle y>0$ fixed. Let $\displaystyle
i\in[1,d]$ fixed. Let $\displaystyle T>0$. Let $\displaystyle\psi$ smooth on
$\displaystyle\Omega$. Assume weak consistency. Then we have

$\displaystyle\mathbb{E}\psi(X_{T}^{(i),\tau_{\varepsilon}})=\mathbb{E}\psi(W_{T}^{y,\tau})+c_{\psi,T}(\varepsilon).$

###### Proof.

This is a consequence of the weak consistency from Lemma 2. See e.g. [6, 5]
where a slightly different definition of weak consistency is used. We sketch
the proof adapted to our situation and notations. The function defined for all
$\displaystyle t\in[0,T]$, $\displaystyle x\in\Omega$ as

$\displaystyle u(t,x):=\mathbb{E}(\psi(W^{y,\tau}_{T})|W^{y,\tau}_{t}=x),$

is a solution to the backward heat equation

$\displaystyle\partial_{t}u+\frac{1}{2}\Delta u=0,\quad u(T,x)=\psi(x),$

and Ito formula reads

$\displaystyle\psi(W^{y,\tau}_{T})=u(T,W^{y,\tau}_{T})=u(0,y)+\int_{0}^{T}\nabla
u(\psi(W^{y,\tau}_{s}))\cdot\mathrm{d}W^{y,\tau}_{s}+\int_{0}^{T}\left(\partial_{t}u+\frac{1}{2}\Delta
u\right)(s,W^{y,\tau}_{s})\mathrm{d}s,$

where the last term is null. Taking expectation allows to compare with the
discrete process

$\displaystyle\mathbb{E}\psi(X_{T}^{(i),\tau_{\varepsilon}})-\mathbb{E}\psi(W_{T}^{y,\tau})=\mathbb{E}u(T,X_{T}^{(i),\tau_{\varepsilon}})-u(0,y)=\mathbb{E}[u(T,X_{T}^{(i),\tau_{\varepsilon}})-u(0,X_{0}^{(i),\tau_{\varepsilon}})].$

The last term is written as a telescopic sum, with summands

$\displaystyle\displaystyle\mathbb{E}[u(t_{n+1},X_{t_{n+1}}^{(i),\tau_{\varepsilon}})-u(t_{n},X_{t_{n}}^{(i),\tau_{\varepsilon}})]$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\mathbb{E}u(t_{n+1},X_{t_{n+1}}^{(i),\tau_{\varepsilon}})-\mathbb{E}u\left(t_{n+1},W^{t_{n},X_{t_{n}}}_{t_{n+1}}\right)$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\mathbb{E}\\{u(t_{n+1},X_{t_{n+1}}^{(i),\tau_{\varepsilon}})-u(t_{n},X_{t_{n}})\\}$
$\displaystyle\displaystyle-\mathbb{E}\left\\{u\left(t_{n+1},W^{t_{n},X_{t_{n}}}_{t_{n+1}}\right)-u(t_{n},X_{t_{n}})\right\\},$

where we have used again Ito formula to replace $\displaystyle
u(t_{n},X_{t_{n}}^{(i),\tau_{\varepsilon}})$ by
$\displaystyle\mathbb{E}u\left(t_{n+1},W^{t_{n},X_{t_{n}}}_{t_{n+1}}\right)$.
A Taylor expansion of the last two terms around the common point
$\displaystyle(t_{n},X_{t_{n}})$ exhibits the differences of increments
$\displaystyle\\{X_{t_{n+1}}-X_{t_{n}}\\}-\left\\{W^{t_{n},X_{t_{n}}}_{t_{n+1}}-W^{t_{n},X_{t_{n}}}_{t_{n}}\right\\}$
and higher moments. Those are estimated thanks to weak consistency as in the
onedimensional case, yielding the desired result. ∎

###### Lemma 6

(Weak convergence of discrete martingale transforms) Let $\displaystyle y>0$
fixed. Let $\displaystyle i\in[1,d]$ fixed. Let $\displaystyle T>0$. Let
$\displaystyle f$ harmonic as above.

$\displaystyle\|f(X_{T}^{(i)})-M_{T}^{(i),f}\|_{p}=c_{T}(\varepsilon)$

###### Proof.

The estimates are handeled slightly differently as in the onedimensional case
in order to take advantage of the multidimensional harmonicity of
$\displaystyle f$. Recalling that

$\displaystyle\begin{array}[]{lll}M_{T}^{(i),f}-M_{0}^{(i),f}&:=&\sum_{k=1}^{N^{5}}\sum_{j=1,d}\partial_{j}f(B^{(i)}_{k-1})\mathrm{d}B_{k}^{(i),j},\end{array}$

we compare with the Taylor expansion

$\displaystyle\displaystyle f(X_{T}^{(i)})-f(X_{0}^{(i)})$
$\displaystyle\displaystyle=$ $\displaystyle\displaystyle
f(B_{T}^{(i)})-f(B^{(i)}_{0})=\sum_{k=1}^{N^{5}}[f(B_{k}^{(i)})-f(B_{k-1}^{(i)})]$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\sum_{k=1}^{N^{5}}\sum_{j=1,d}\partial_{j}f(B^{(i)}_{k-1})\mathrm{d}B_{k}^{(i),j}+\frac{1}{2}\sum_{k=1}^{N^{5}}\sum_{j_{1},j_{2}=1,d}\partial^{2}_{j_{1}j_{2}}f(B^{(i)}_{k-1})\mathrm{d}B^{(i),j_{1}}_{k}\mathrm{d}B^{(i),j_{2}}_{k}$
$\displaystyle\displaystyle+\sum_{k=1}^{N^{5}}R_{3}^{f}(B^{(i)}_{k-1},\mathrm{d}B^{(i)}_{k}).$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\sum_{k=1}^{N^{5}}\sum_{j=1,d}\partial_{j}f(B^{(i)}_{k-1})\mathrm{d}B_{k}^{(i),j}+\frac{1}{2}\sum_{k=1}^{N^{5}}\sum_{j=1,d}\partial^{2}_{j}f(B^{(i)}_{k-1})(\mathrm{d}B^{(i),j}_{k})^{2}$
$\displaystyle\displaystyle+\sum_{k=1}^{N^{5}}R_{3}^{f}(B^{(i)}_{k-1},\mathrm{d}B^{(i)}_{k}).$

where the last terms are Taylor remainders. Recalling that $\displaystyle
M_{0}^{(i),f}=f(B^{(i)}_{0})$, it remains to estimate

$\displaystyle
f(X_{T}^{(i)})-M_{T}^{(i),f}=\frac{1}{2}\sum_{k=1}^{N^{5}}\sum_{j_{1},j_{2}=1,d}\partial^{2}_{j_{1}j_{2}}f(B^{(i)}_{k-1})\mathrm{d}B^{(i),j_{1}}_{k}\mathrm{d}B^{(i),j_{2}}_{k}+\sum_{k=1}^{N^{5}}R_{3}^{f}(B^{(i)}_{k-1},\mathrm{d}B^{(i)}_{k})=:A+B$

We have
$\displaystyle|R_{3}^{f}(B_{k-1}(x),\mathrm{d}B_{k}(x))|\lesssim\|D^{3}f\|_{\infty}\delta^{3/2}$
therefore $\displaystyle\|B\|_{p}\lesssim
N^{5}\delta^{3/2}=c_{T}(\varepsilon)$.

Split now $\displaystyle A=\sum_{n=1}^{N^{4}}A_{n}$ in $\displaystyle N^{4}$
blocks $\displaystyle(A_{n})_{n=1,N^{4}}$ of $\displaystyle N$ terms each, and
split further $\displaystyle A_{n}=B_{n}+C_{n}$ with

$\displaystyle
B_{n}:=\sum_{l=1}^{N}\sum_{j=1}^{d}[\partial^{2}_{j}f(B^{(i)}_{(n-1)N+l-1})-\partial^{2}_{j}f(B^{(i)}_{(n-1)N})](\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2},$
$\displaystyle
C_{n}:=\sum_{j=1}^{d}\partial^{2}_{j}f(B^{(i)}_{(n-1)N})\sum_{l=1}^{N}(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}.$

We estimate first $\displaystyle|B_{n}|\lesssim
N\|D^{3}f\|_{\infty}\left(N\sqrt{\delta}\right)\delta\lesssim
N^{2}\delta^{3/2}$. Therefore

$\displaystyle\left\|\sum_{1}^{N^{4}}B_{n}\right\|_{p}\lesssim
N^{4}(N^{2}\delta^{3/2})=c_{T}(\varepsilon).$

Recalling that $\displaystyle f$ is harmonic we can write equivalently

$\displaystyle
C_{n}:=\sum_{j=1}^{d}\partial^{2}_{j}f(B^{(i)}_{(n-1)N})\left[\sum_{l=1}^{N}(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}-\theta\right].$

Notice that in the multidimensional case, we have that for a given fixed
$\displaystyle j$, successive increments
$\displaystyle\mathrm{d}B^{(i),j}_{k}$ and
$\displaystyle\mathrm{d}B^{(i),j}_{m}$, $\displaystyle m\neq k$ are
independent random variables, and the same holds for the sequence
$\displaystyle(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}$, $\displaystyle 1\leqslant
l\leqslant N$, with
$\displaystyle\mathbb{E}(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}=\delta$. Hence

$\displaystyle\sum_{l=1}^{N}(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}-\theta=\sum_{l=1}^{N}[(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}-\delta]=:\delta\sum_{l=1}^{N}\alpha_{(n-1)N+l-1},$

where $\displaystyle\alpha_{k}=\pm 1$, since
$\displaystyle(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}=0$ or
$\displaystyle(\mathrm{d}B^{(i),j}_{(n-1)N+l})^{2}=2\delta$ with equal
probability. In other words $\displaystyle\sum_{l=1}^{N}\alpha_{(n-1)N+l-1}$
is a discrete $\displaystyle N$ step random walk with $\displaystyle\pm 1$
steps. It follows, recalling moments of discrete random walks, that

$\displaystyle\|C_{n}\|_{p}\lesssim\|D^{2}f\|_{\infty}\delta\left\|\sum_{l=1}^{N}\alpha_{(n-1)N+l-1}\right\|_{p}\lesssim\|D^{2}f\|_{\infty}\delta(\theta^{p/2})^{1/p}.$

Finally

$\displaystyle\left\|\sum_{n=1}^{N^{4}}C_{n}\right\|_{p}\lesssim
N^{4}\delta\|D^{2}f\|_{\infty}(N\delta)^{1/2}=(N^{5}\delta)\|D^{2}f\|_{\infty}N^{-1/2}\delta^{1/2}=c_{T}(\varepsilon).$

This concludes the proof of Lemma 6. ∎

##  5 Proof of the main results

###  5.1 Upper bound for a single Riesz transform

######  Lemma 7 (Convergence of martingales in $\displaystyle L^{p}$)

Let $\displaystyle y>0$ fixed. Let $\displaystyle i\in[1,d]$ fixed. Let
$\displaystyle f\in L^{p}(\partial\Omega)$ and harmonic in
$\displaystyle\Omega$. We have

$\displaystyle\lim_{T\rightarrow\infty}\lim_{\varepsilon\rightarrow
0}\mathbb{E}|M_{T}^{f}|^{p}=\mathbb{E}|f(W^{y,\tau}_{\infty})|^{p},$

###### Proof.

The proof is the same as the onedimensional case [3]. It is merely a
combination of Lemmas 5 & 6, where we compare successively $\displaystyle
f(W^{y,\tau}_{T})$ to $\displaystyle f(X^{(i)}_{T})$, and $\displaystyle
f(X^{(i)}_{T})$to $\displaystyle M^{(i),f}_{T}$. ∎

######  Lemma 8 (Convergence in the weak formulations)

Let $\displaystyle y>0$ fixed. Let $\displaystyle i\in[1,d]$ fixed. Let
$\displaystyle f\in L^{p}(\partial\Omega,X)$ and $\displaystyle g\in
L^{p}(\partial\Omega,X^{\ast})$ both harmonic. We have

$\displaystyle\lim_{T\rightarrow\infty}\lim_{\varepsilon\rightarrow
0}\mathbb{E}\langle
M_{T}^{(i),f},M_{T}^{(i),g}\rangle=\mathbb{E}\langle\mathcal{M}^{y,f}_{\infty},\mathcal{M}^{y,g}_{\infty}\rangle.$
$\displaystyle\lim_{T\rightarrow\infty}\lim_{\varepsilon\rightarrow
0}\mathbb{E}\langle
M_{T}^{(i),i},M_{T}^{(i),g}\rangle=\mathbb{E}\langle\mathcal{M}^{y,i}_{\infty},\mathcal{M}^{y,g}_{\infty}\rangle.$

###### Proof.

For the first limit we can use the same strategy as for the proof of Lemma 7.
This is because the first limit involves the discretisation of the harmonic
function $\displaystyle f$. However the second limit involves the martingale
transform $\displaystyle M_{T}^{(i),i}$ of $\displaystyle M_{T}^{(i),f}$, or
in the limit the martingale transform
$\displaystyle\mathcal{M}^{y,i}_{\infty}$ of
$\displaystyle\mathcal{M}^{y,f}_{\infty}$. Those are not martingales directly
associated to $\displaystyle\mathcal{R}_{i}f$ or its discretisation, since a
projection (6) is needed to recover $\displaystyle\mathcal{R}_{i}f$.

Indeed for the first limit, using Lemma 6 and recalling that all functions are
smooth with compact support, we have successively

$\displaystyle\mathbb{E}\langle
M_{T}^{(i),f},M_{T}^{(i),g}\rangle=\mathbb{E}\langle
f(X_{T}^{(i),\tau_{\varepsilon}}),M_{T}^{(i),g}\rangle+c_{T}(\varepsilon),$

and similarly

$\displaystyle\mathbb{E}\langle
f(X_{T}^{(i),\tau_{\varepsilon}}),M_{T}^{(i),g}\rangle=\mathbb{E}\langle
f(X_{T}^{(i),\tau_{\varepsilon}}),g(X_{T}^{(i),\tau_{\varepsilon}})\rangle+c_{T}(\varepsilon).$

Applying the weak convergence of Lemma 5 to $\displaystyle\psi(x):=\langle
f(x),g(x)\rangle$ then letting $\displaystyle T$ go to infinity yields the
first limit.

For the second limit, explicit functions of $\displaystyle X^{(i)}$ are only
accessible after expanding the product martingales. Indeed, given
$\displaystyle T>0$, we have

$\displaystyle\mathbb{E}\langle\mathcal{M}^{y,i}_{T},\mathcal{M}^{y,g}_{T}\rangle=\mathbb{E}\langle\mathcal{M}^{y,i}_{0},\mathcal{M}^{y,g}_{0}\rangle+\mathbb{E}\int_{0}^{T}\langle
A_{i}\nabla f(W^{y,\tau}_{t}),\nabla g(W^{y,\tau}_{t})\rangle\mathrm{d}t,$

and similarly

$\displaystyle\mathbb{E}\langle
M_{T}^{(i),i},M_{T}^{(i),g}\rangle=\mathbb{E}\langle
M_{0}^{(i),i},M_{0}^{(i),g}\rangle+\mathbb{E}\sum_{n=1}^{N^{4}}\langle
A_{i}\nabla f(X_{t_{n}}),\nabla g(X_{t_{n}})\rangle(t_{n+1}-t_{n}).$

Notice that the first terms on the right hand sides are null. Choose now a
large integer $\displaystyle K$. We know that we can approximate the
continuous stochastic integral as

$\displaystyle\mathbb{E}\int_{0}^{T}\langle A_{i}\nabla
f(W^{y,\tau}_{t}),\nabla
g(W^{y,\tau}_{t})\rangle\mathrm{d}t=\mathbb{E}\sum_{k=1}^{K}\langle
A_{i}\nabla f(W^{y,\tau}_{T_{k-1}}),\nabla
g(W^{y,\tau}_{T_{k-1}})\rangle(T_{k}-T_{k-1})+c(K),$

where $\displaystyle(T_{k})_{k\in[0,K]}$ is a regular subdivision of
$\displaystyle[0,T]$, i.e. $\displaystyle T_{k}=kT/K$, and $\displaystyle
c(K)$ is a function that goes to zero when $\displaystyle K$ goes to infinity.
Similarly, it is now an exercise to show that we can approximate

$\displaystyle\mathbb{E}\sum_{n=1}^{N^{4}}\langle A_{i}\nabla
f(X_{t_{n}}),\nabla
g(X_{t_{n}})\rangle(t_{n+1}-t_{n})=\mathbb{E}\sum_{k=1}^{K}\langle A_{i}\nabla
f(X_{T_{k-1}}),\nabla g(X_{T_{k-1}})\rangle(T_{k}-T_{k-1})+c(K),$

where we assume without loss of generality that $\displaystyle N$ is a
multiple of $\displaystyle K$. We can now pass to the limit for each
$\displaystyle k\in[1,K]$ so that $\displaystyle\mathbb{E}\langle A_{i}\nabla
f(X_{T_{k-1}}),\nabla g(X_{T_{k-1}})\rangle\rightarrow\mathbb{E}\langle
A_{i}\nabla f(W^{y,\tau}_{T_{k-1}}),\nabla g(W^{y,\tau}_{T_{k-1}})\rangle$
when $\displaystyle\varepsilon$ goes to zero, using again weak convergence.
This allows to prove the second limit, therefore concludingg the proof of
Lemma 8. ∎

Finally, we prove our main result.

Proof of Theorem 1 (Upper bound for a single riesz transform) Recall that we
know from Lemma 1 that

$\displaystyle\|M_{k}^{(i),i}\|_{p}\leqslant\|\mathcal{S}_{i}\|_{p\rightarrow
p}\|M^{(i),f}_{k}\|_{p}.$

We have equivalently

$\displaystyle\mathbb{E}\langle
M_{k}^{(i),i},M^{(i),g}\rangle\leqslant\|\mathcal{S}_{i}\|_{p\rightarrow
p}\|M^{(i),f}_{k}\|_{p}\|M^{(i),g}_{k}\|_{p}$

for any test function $\displaystyle g\in L^{q}(\partial\Omega,X^{\ast})$.
Using Lemmas 7 & 8, we pass to the limit $\displaystyle T\rightarrow\infty$
and $\displaystyle\varepsilon\rightarrow 0$ and obtain, for any fixed
$\displaystyle y>0$,

$\displaystyle\mathbb{E}\langle\mathcal{M}^{y,i}_{\infty},\mathcal{M}^{y,g}_{\infty}\rangle\leqslant\|\mathcal{S}_{i}\|_{p\rightarrow
p}(\mathbb{E}|f(W^{y,\tau}_{\infty})|^{p})^{1/p}(\mathbb{E}|g(W^{y,\tau}_{\infty})|^{q})^{1/q}$

Notice that

$\displaystyle\mathbb{E}|f(W^{y,\tau}_{\infty})|^{p}=\int_{\partial\Omega}|f(x)|^{p}\mathrm{d}\omega^{y}(x)$

where $\displaystyle\mathrm{d}\omega^{y}$ is the harmonic measure associated
to the Brownian $\displaystyle W^{y,\tau}$ started at $\displaystyle y$ and
stopped upon hitting $\displaystyle\partial\Omega$. Due to translation
invariance of the domain, we have
$\displaystyle\mathrm{d}\omega^{y}(x)=\mathrm{d}\omega^{0}(x-y)=G(x-y)\mathrm{d}x$,
where $\displaystyle G$ is the normalized Gaussian. It follows that

$\displaystyle\int_{y}\mathbb{E}|f(W^{y,\tau}_{\infty})|^{p}=\int_{y}\int_{\partial\Omega}|f(x)|^{p}\mathrm{d}\omega^{y}(x)=\left(\int_{\partial\Omega}|f(x)|^{p}\mathrm{d}x\right)\left(\int_{y}G(y)\mathrm{d}y\right)=\|f\|_{p},$

hence

$\displaystyle\int_{y}(\mathbb{E}|f(W^{y,\tau}_{\infty})|^{p})^{1/p}(\mathbb{E}|g(W^{y,\tau}_{\infty})|^{q})^{1/q}\leqslant\|f\|_{p}\|g\|_{q}.$

Following [4], we observe that

$\displaystyle\lim_{y\rightarrow\infty}\int_{y}\mathbb{E}\langle\mathcal{M}^{y,i}_{\infty},\mathcal{M}^{y,g}_{\infty}\rangle=\int_{\Omega}\langle
A_{i}\nabla f,\nabla g\rangle
2y\mathrm{d}y\mathrm{d}x=\langle\mathcal{R}_{i}f,g\rangle,$

yielding

$\displaystyle\langle\mathcal{R}_{i}f,g\rangle\leqslant\|\mathcal{S}_{i}\|_{p\rightarrow
p}\|f\|_{p}\|g\|_{q},$

hence the desired result. $\displaystyle\Box$

###  5.2 Upper bound for the vector Riesz transform

Let
$\displaystyle\mathcal{R}f:=\left(\sum_{i=1}^{d}|\mathcal{R}_{i}f|_{X}^{2}\right)^{1/2}$
and
$\displaystyle\mathcal{S}f:=\left(\sum_{i=1}^{d}|\mathcal{S}_{i}f|_{X}^{2}\right)^{1/2}$.
Our goal is to prove Theorem 2, namely

$\displaystyle\|\mathcal{R}\|_{p}\leqslant\|\mathcal{S}\|_{p}.$

Let $\displaystyle g=(g_{i})_{i=1,\ldots,d}$ a vector consisting of
$\displaystyle d$ test functions. The usual weak formulation for the estimate
of the Riesz vector reads

$\displaystyle\begin{array}[]{rcl}\|\mathcal{R}f\|_{p}&=&\sup_{\|g\|_{q}\leqslant
1}\langle\mathcal{R}f,g\rangle=\sup_{\|g\|_{q}\leqslant
1}\sum_{i=1}^{d}\langle\mathcal{R}_{i}f,g_{i}\rangle=\sup_{\|g\|_{q}\leqslant
1}\mathbb{E}\sum_{i=1}^{d}\int_{0}^{\tau}\langle A_{i}\nabla f,\nabla
g_{i}\rangle\mathrm{d}t\end{array}$

The last term is the stochastic representation involving the
$\displaystyle(d+1)$–dimensional Brownian motion $\displaystyle W^{\tau}$,
namely

$\displaystyle\langle\mathcal{R}f,g\rangle=\mathbb{E}\sum_{i=1}^{d}\int_{0}^{\tau}\langle
A_{i}\nabla f,\nabla
g_{i}\rangle=\mathbb{E}\sum_{i=1}^{d}\int_{0}^{\tau}A_{i}\nabla
f(W_{s}^{\tau})\cdot\nabla g_{i}(W_{s}^{\tau})\mathrm{d}t.$

Notice that all summands involve the same realization of the Brownian motion.

#### Adapted stochastic representation

In order for us to compare the dyadic Riesz vector and the usual Riesz vector,
we need to adapt the stochastic representation in a simple (but effective!)
way. Observing that

$\displaystyle\langle\mathcal{R}f,g\rangle=\mathbb{E}\sum_{i=1}^{d}\int_{0}^{\tau}A_{i}\nabla
f(W_{s}^{\tau})\cdot\nabla
g_{i}(W_{s}^{\tau})\mathrm{d}t=\sum_{i=1}^{d}\mathbb{E}\int_{0}^{\tau}A_{i}\nabla
f(W_{s}^{\tau})\cdot\nabla g_{i}(W_{s}^{\tau})\mathrm{d}t,$

one realizes that the expectation operator $\displaystyle\mathbb{E}$ is
applied $\displaystyle d$ times in the last expression, and that the Brownian
motion is a dummy variable. We can therefore go further and reformulate

$\begin{array}[]{lll}\langle\mathcal{R}f,g\rangle&=&\sum_{i=1}^{d}\mathbb{E}^{i}\int_{0}^{\tau_{i}}A_{i}\nabla
f(W_{s}^{(i),\tau_{i}})\cdot\nabla g_{i}(W_{s}^{(i),\tau_{i}})\mathrm{d}t,\\\
&=&\mathbb{E}\sum_{i=1}^{d}\int_{0}^{\tau_{i}}A_{i}\nabla
f(W_{s}^{(i),\tau_{i}})\cdot\nabla
g_{i}(W_{s}^{(i),\tau_{i}})\mathrm{d}t,\end{array}$ (9)

where the $\displaystyle W^{(i),\tau_{i}}$’s are $\displaystyle d$ different
realizations of Brownian motions in $\displaystyle\mathbb{R}^{d+1}$,
$\displaystyle\mathbb{E}^{i}$ the corresponding expectations and
$\displaystyle\tau_{i}$ the corresponding stopping times. Those realizations
do not need to be independent. This is reminiscent of the $\displaystyle d$
discrete random walks $\displaystyle B^{(i)}$, each stopped at different times
but sharing the same horizontal component.

Proof of Theorem 2 (Upper bound for the Riesz vector) For all $\displaystyle
i\in[1,d]$, we approximate $\displaystyle\mathcal{M}_{t}^{f}$ as well as the
test function $\displaystyle\mathcal{M}_{t}^{g_{i}}$, by

$\displaystyle M^{(i),f}_{k}:=\mathcal{M}_{0}^{f}+\sum_{\ell=1}^{k}\nabla
f(B_{k-1}^{s(i)})\cdot\mathrm{d}B_{k}^{(i)},\quad
M^{(i),g_{i}}_{k}:=\mathcal{M}_{0}^{g_{i}}+\sum_{\ell=1}^{k}\nabla
g_{i}(B_{k-1}^{(i)})\cdot\mathrm{d}B_{k}^{(i)}.$

We have by construction, for any fixed $\displaystyle y$, any fixed numerical
parameters, that

$\displaystyle\sum_{i=1}^{d}\mathbb{E}^{i}M^{(i),\mathcal{S}_{i}f}_{T}M^{(i),g_{i}}_{T}=\sum_{i=1}^{d}\langle\mathcal{S}_{i}f,g_{i}\rangle\leqslant\|\mathcal{S}\|_{p\rightarrow
p}\|f\|_{p}\|g\|_{q}.$

An immediate consequence of Lemma 8 reads

$\displaystyle\lim_{T\rightarrow\infty}\lim_{\varepsilon\rightarrow
0}\sum_{i=1}^{d}\mathbb{E}^{i}M^{(i),f}_{T}M^{(i),g_{i}}_{T}=\sum_{i=1}^{d}\mathbb{E}^{i}\mathcal{M}^{(i),f}_{\infty}M^{(i),g_{i}}_{\infty}$
$\displaystyle\lim_{T\rightarrow\infty}\lim_{\varepsilon\rightarrow
0}\sum_{i=1}^{d}\mathbb{E}^{i}M^{(i),\mathcal{S}_{i}f}_{T}M^{(i),g_{i}}_{T}=\sum_{i=1}^{d}\mathbb{E}^{i}\mathcal{M}^{(i),\mathcal{R}_{i}f}_{\infty}M^{(i),g_{i}}_{\infty}$

Passing to the limit as for the proof of Theorem 1 yields

$\displaystyle\langle\mathcal{R}f,g\rangle\leqslant\|\mathcal{S}\|_{p\rightarrow
p}\|f\|_{p}\|g\|_{q},$

hence the desired result
$\displaystyle\|\mathcal{R}\|_{p}\leqslant\|\mathcal{S}\|_{p}$.
$\displaystyle\Box$

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