# Dimension-free estimates on $l^{2}(\mathbb{Z}^{d})$ for discrete dyadic
maximal function over $l^{1}$ balls: small scales

Jakub Niksiński  Institute of Mathematics
University of Wroclaw
Plac Grunwaldzki 2
50-384 Wrocław
Poland<EMAIL_ADDRESS>
###### Abstract.

We give a dimension-free bound on $l^{p}(\mathbb{Z}^{d})$ for discrete Hardy-
Littlewood operator over $l^{1}$ balls in $\mathbb{Z}^{d}$ with small dyadic
radii, where $p\in[2,\infty]$.

###### Key words and phrases:

discrete maximal function, $l^{1}$ balls, dimension-free estimates

###### 2020 Mathematics Subject Classification:

42B15, 42B25

##  1\. Introduction

Let $G$ be a bounded, closed symmetric subset of $\mathbb{R}^{d}$ with non-
empty interior, we will call such set $G$ a symmetric convex body. Natural
examples are $l^{q}$ balls:

$B^{q}=\Big{\\{}x\in\mathbb{R}^{d}:\|x\|_{q}=\Big{(}\sum_{i=1}^{d}|x_{i}|^{q}\Big{)}^{1/q}\Big{\\}}\
\text{for}\ q\in[1,\infty),$
$B^{\infty}=\Big{\\{}x\in\mathbb{R}^{d}:\|x\|_{\infty}=\max_{1\leq i\leq
d}|x_{i}|\Big{\\}}.$

With each symmetric convex body one can associate the corresponding Hardy-
Littlewood averaging operator. For any $t>0$ and $x\in\mathbb{R}^{d}$ we
define

$M_{t}^{G}f(x)=\frac{1}{|t\cdot G|}\int_{t\cdot G}f(x-y)\ dy,$

for $f$ locally integrable function, where $t\cdot G=\\{tx:x\in G\\}$ and
$|t\cdot G|$ denotes its Lebesgue measure. Now for any $p$ let $C_{p}(d,G)>0$
be the smallest number such that the following inequality

$\|\sup_{t>0}|M_{t}^{G}f|\|_{L^{p}(\mathbb{R}^{d})}\leq
C_{p}(d,G)\|f\|_{L^{p}(\mathbb{R}^{d})}$

holds for every $f\in L^{p}(\mathbb{R}^{d})$. It is well known that
$C_{p}(d,G)<\infty$ for all $p\in(1,\infty]$ and all symmetric convex bodies
$G$.

In 1980s dependency of $C_{p}(d,G)$ on dimension $d$ has begun to be studied.
Various results were obtained, but as of now the major conjecture in this
topic is still open, namely that $C_{p}(d,G)$ can be bounded from the above by
a number independent of dimension $d$ for all $p\in(1,\infty]$. We recommend
survey article [3] for a detailed exposition of the subject, which contains
results that we skipped mentioning.

Similar questions can be considered for discrete analogue of the operator
$M_{t}^{G}$. For every $t>0$ and every $x\in\mathbb{Z}^{d}$ we define the
discrete Hardy-Littlewood averaging operator

$\mathcal{M}_{t}^{G}=\frac{1}{|(t\cdot G)\cap\mathbb{Z}^{d}|}\sum_{y\in t\cdot
G\cap\mathbb{Z}^{d}}f(x-y),\ \ f\in l^{1}(\mathbb{Z}^{d}),$

where $|(t\cdot G)\cap\mathbb{Z}^{d}|$ is the number of elements of the set
$(t\cdot G)\cap\mathbb{Z}^{d}$. Similarly as before we define
$\mathcal{C}_{p}(d,G)>0$ to be the smallest number such that the following
inequality

$\|\sup_{t>0}|\mathcal{M}_{t}^{G}f|\|_{l^{p}(\mathbb{Z}^{d})}\leq\mathcal{C}_{p}(d,G)\|f\|_{l^{p}(\mathbb{Z}^{d})}$

holds for every $f\in l^{p}(\mathbb{Z}^{d})$. Using similar methods as in the
continuous case one can show that for any $p\in(1,\infty]$ and any symmetric
convex body we have $\mathcal{C}_{p}(d,G)<\infty$.

What about dependency of $\mathcal{C}_{p}(d,G)$ on the dimension $d$? We can
ask similar questions as in the continuous setup, yet it turns out that the
situation is much more delicate. Indeed in [1] it was shown that there exists
a family of ellipsoids $E(d)\subseteq\mathbb{R}^{d}$, such that for each
$p\in(1,\infty]$ there exists $C_{p}>0$ such that for every $d\in\mathbb{N}$
we have

$\mathcal{C}_{p}(d,E(d))\geq C_{p}(\log d)^{1/p}.$

This is means that if we want to establish dimension-free estimates for
$\mathcal{C}_{p}(d,G)$ we need to restrict ourselves to specific sets $G$,
which contain more symmetries, one of the simpler options are $B^{q}$ balls.

Literature in the discrete setting is not as fruitful, nevertheless there are
some papers and positive results in this regard, for example:

  * •

In [1] it was proved that for every $p\in(\frac{3}{2},\infty]$ there exists a
constant $C_{p}>0$ such that for every $d\in\mathbb{N}$ and every $f\in
l^{p}(\mathbb{Z}^{d})$ we have

$\|\sup_{t>0}|\mathcal{M}_{t}^{B^{\infty}}f|\|_{l^{p}(\mathbb{Z}^{d})}\leq
C_{p}\|f\|_{l^{p}(\mathbb{Z}^{d})}.$

This result is as strong as the ones in continuous case. Unfortunately in the
case of sets $B^{q}$ for $q\neq\infty$ authors of papers [1],[2],[6] could
only obtain way weaker conclusions.

  * •

In [2] it was proved that for every $p\in[2,\infty)$ there exists $C_{p}>0$
such that for every $d\in\mathbb{N}$ and every $f\in l^{p}(\mathbb{Z}^{d})$ we
have

$\|\sup_{t\in\mathbb{D}}|\mathcal{M}_{t}^{B^{2}}f|\|_{l^{p}(\mathbb{Z}^{d})}\leq
C_{p}\|f\|_{l^{p}(\mathbb{Z}^{d})},$

where $\mathbb{D}=\\{2^{n}:n\in\mathbb{N}\\}.$

  * •

In [6] by extending methods of [2] it was proved that for any $q\in(2,\infty)$
and any $p\in[2,\infty)$ there exists constant $C(p,q)>0$ such that for every
$d\in\mathbb{N}$ and every $f\in l^{p}(\mathbb{Z}^{d})$ we have

$\|\sup_{t\in\mathbb{D},t\geq
d^{1/q}}|\mathcal{M}_{t}^{B^{q}}f|\|_{l^{p}(\mathbb{Z}^{d})}\leq
C(p,q)\|f\|_{l^{p}(\mathbb{Z}^{d})},$

where $\mathbb{D}=\\{2^{n}:n\in\mathbb{N}\\}.$ The paper [6] did not cover the
range $t<d^{1/q}$ nor $q<2$.

In this paper we will prove the following result

######  Theorem 1.1.

There exists $C>0$ such that for any $p\in[2,\infty)$, $d\in\mathbb{N}$,
$d\geq 4$ and any $f\in l^{p}(\mathbb{Z}^{d})$ we have

$\|\sup_{t\in\mathbb{D},\
t\leq\sqrt{d}}|\mathcal{M}_{t}^{B^{1}}f|\|_{l^{p}(\mathbb{Z}^{d})}\leq
30^{4/p}\|f\|_{l^{p}(\mathbb{Z}^{d})},$

where $\mathbb{D}=\\{2^{n}:n\in\mathbb{N}_{0}\\}$ is the set of dyadic
integers.

As far as the author is aware the result above is new. It is a small step
beyond the results of [6] in the case $q=1$. Additionally we give explicit
values to various constants. In the proof of Theorem 1.1 we will rely on
methods shown in the section 3 of [2]. Right at the beginning we will replace
$t\cdot B^{1}$ by the simpler set
$S_{t}=\\{x\in\\{-1,0,1\\}^{d}:\sum_{i=1}^{d}|x_{i}|=t\\}$ and then consider
only $S_{t}$. Later on we merely use tools from section 3 of [2]. This
replacement by a simpler set is easy due to the fact that there exists a
formula for the number of lattice points in $t\cdot B^{1}$. This is not the
case for other sets of form $t\cdot B^{q}$, where $q\in(1,\infty)$. We
restrict ourselves to $t\leq\sqrt{d}$ due to the fact that then $(t\cdot
B^{1})\cap\mathbb{Z}^{d}$ consists mostly of point from $S_{t}$ (see Lemma
2.4). This is not the case when $t$ is sufficiently big in terms of $d$ and we
would need different argument in that range. Throughout the rest of this paper
we will consider $B^{q}$ only for $q=1$.

###  1.1. Notation

  1. (1)

For $t>0$ we define
$B_{t}=\\{(x_{1},...,x_{d})\in\mathbb{R}^{d}:\sum_{i=1}^{d}|x_{i}|\leq t\\}$.

  2. (2)

$\mathbb{N}=\\{1,2,...\\}$ will denote the set of positive integers and
$\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$ will denote the set of non-negative
integers. $\mathbb{D}=\\{2^{n}:n\in\mathbb{N}_{0}\\}$ is the set of dyadic
integers. We define $\mathbb{N}_{N}=\\{1,2,...,N\\}$.

  3. (3)

We define $e(x)=e^{2\pi ix}$ for any $x\in\mathbb{R}^{d}$.

  4. (4)

We have standard scalar product on $\mathbb{R}^{d}$

$x\cdot y=\sum_{k=1}^{d}x_{k}y_{k},$

where $x,y\in\mathbb{R}^{d}.$

  5. (5)

For $f,g\in l^{2}(\mathbb{Z}^{d})$ we define $f\ast g\in
l^{2}(\mathbb{Z}^{d})$ by series, which is absolutely convergent for each
$x\in\mathbb{Z}^{d}$:

$f\ast
g(x)=\sum_{y\in\mathbb{Z}^{d}}f(y)g(x-y)=\sum_{y\in\mathbb{Z}^{d}}f(x-y)g(y).$

  6. (6)

If $f\in l^{1}(\mathbb{Z}^{d})$ we define discrete Fourier transform by the
formula

$\widehat{f}(\xi)=\sum_{x\in\mathbb{Z}^{d}}f(x)e(x\cdot\xi),\ \text{ for }\
\xi\in\mathbb{T}^{d}.$

One can extend discrete Fourier transform to $f\in l^{2}(\mathbb{Z}^{d})$,
then it turns out that $\widehat{f}\in L^{2}(\mathbb{T}^{d})$, moreover we
have the following Parseval identity: for any $f,g\in l^{2}(\mathbb{Z}^{d})$

$\|\widehat{f}\|_{L^{2}(\mathbb{T}^{d})}=\|f\|_{l^{2}(\mathbb{Z}^{d})}$

and

$\widehat{f\ast g}(\xi)=\widehat{f}(\xi)\widehat{f}(\xi).$

$\mathcal{F}^{-1}$ will denote inverse operation to discrete Fourier
transform, that is

$\mathcal{F}^{-1}(G)(x)=\int_{\mathbb{T}^{d}}G(\xi)e(-x\cdot\xi)\ d\xi,$

where $G\in L^{2}(\mathbb{T}^{d})$.

  7. (7)

We let $m_{t}$ be the multiplier symbol

$m_{t}(\xi)=\frac{1}{|B_{t}\cap\mathbb{Z}^{d}|}\sum_{x\in
B_{t}\cap\mathbb{Z}^{d}}e(x\cdot\xi).$

  8. (8)

$\mathbb{T}^{d}$ will denote $d$-dimensional torus, which will be identified
with the set $[-\frac{1}{2},\frac{1}{2})^{d}.$

  9. (9)

$Sym(d)$ will denote the permutation group of $\\{1,2,...,d\\}$.

######  Definition 1.2.

The discrete Hardy-Littlewood averaging operator of the $l^{1}$ ball is
defined for any function $f:\mathbb{Z}^{d}\to\mathbb{C}$ by the formula

$\mathcal{M}_{t}f(x)=\frac{1}{|B_{t}\cap\mathbb{Z}^{d}|}\sum_{y\in
B_{t}\cap\mathbb{Z}^{d}}f(x-y).$

Our goal is to prove the following theorem

######  Theorem 1.3.

For any $d\in\mathbb{N}$, $d\geq 4$ and any $f\in l^{2}(\mathbb{Z}^{d})$ we
have

$\|\sup_{t\in\mathbb{D},\
t\leq\sqrt{d}}|\mathcal{M}_{t}f|\|_{l^{2}(\mathbb{Z}^{d})}\leq
900\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

Using a complex interpolation argument one can show that Theorem 1.3 implies
Theorem 1.1. Notice that the operator $\mathcal{M}_{t}$ is given by
convolution, that is

$\mathcal{M}_{t}f(x)=f\ast K_{t}(x),$

where

$K_{t}(y)=\frac{1}{|B_{t}\cap\mathbb{Z}^{d}|}\mathds{1}_{B_{t}\cap\mathbb{Z}^{d}}(y).$

Since we’re considering $l^{2}$ norms, using Fourier theory we can reduce our
problem to understanding pointwise bounds of the following function

$m_{t}(\xi)=\widehat{K_{t}}(\xi)=\frac{1}{|B_{t}\cap\mathbb{Z}^{d}|}\sum_{x\in
B_{t}\cap\mathbb{Z}^{d}}e(x\cdot\xi),$

it will be explained in detail in the next sections.

Acknowledgements: Author was supported by National Science Centre, Poland,
grant Sonata Bis nr. 2022/46/E/ST1/00036. This paper constitutes author’s
master thesis. The author is grateful to his advisor Błażej Wróbel for
suggesting the topic and helpful remarks during the preparation of this paper.

##  2\. Estimate for $|B_{t}\cap\mathbb{Z}^{d}|$ and its consequences.

The goal of this section is to approximate function $m_{t}(\xi)$ by a simpler
function, it will be obtained by approximating $|B_{t}\cap\mathbb{Z}^{d}|$ by
the number of lattice points in a simpler set.

######  Definition 2.1.

For $t\in\mathbb{N}$ we define

$S_{t}=\\{x\in\\{-1,0,1\\}^{d}:\sum_{i=1}^{d}|x_{i}|=t\\},$

moreover we define function $s_{t}:\mathbb{T}^{d}\to\mathbb{C}$ by formula

$s_{t}(\xi)=\frac{1}{|S_{t}|}\sum_{x\in S_{t}}e(x\cdot\xi),$

next we introduce averaging operator $\mathcal{S}_{t}$ related to the above
multiplier by the formula

$\mathcal{S}_{t}f(x)=\mathcal{F}^{-1}(s_{t}\hat{f})(x),\ \text{for any}f\in
l^{2}(\mathbb{Z}^{d}).$

Notice that $S_{t}\subseteq B_{t}\cap\mathbb{Z}^{d}$.

######  Lemma 2.2.

For any $t\in\mathbb{N},t\leq d$ the following formulas hold

$|S_{t}|=2^{t}\binom{d}{t}$ (2.1)
$|B_{t}\cap\mathbb{Z}^{d}|=\sum_{l=0}^{t}2^{l}\binom{d}{l}\binom{t}{l}.$ (2.2)

These formulas are well known.We give a proof for completeness.

###### Proof.

For equality (2.1) we notice that there are $\binom{d}{t}$ options to choose
$t$ coordinates out of $d$ coordinates, on which we have value $-1$ or $1$,
next we can choose each sign in $2$ ways, thus we obtain
$|S_{t}|=2^{t}\binom{d}{t}.$
For equality (2.2) we consider the following disjoint decomposition

$B_{t}\cap\mathbb{Z}^{d}=\bigcup_{k=0}^{t}A_{k},$

where

$A_{k}=\\{(x_{1},...,x_{d})\in\mathbb{Z}^{d}:\sum_{i=1}^{d}|x_{i}|=k\\}.$

Now for each $k\in\\{0,1,...,t\\}$ we will compute $|A_{k}|$. We have
$|A_{0}|=1$, consider $k\geq 1$ then

$A_{k}=\bigcup_{l=1}^{k}\\{\textbf{x}\in\mathbb{Z}^{d}:\sum_{i=1}^{d}|x_{i}|=k,\
\textbf{x}\text{ has $l$ exactly nonzero coordinates }\\}.$

In the union above each $l$-th individual set has size

$2^{l}\binom{d}{l}\cdot|\\{(y_{1},...,y_{l})\in\mathbb{N}^{l}:y_{1}+...+y_{l}=k\\}|=2^{l}\binom{d}{l}\cdot\binom{k-1}{l-1},$

this is because we have $\binom{d}{l}$ options of choosing nonzero coordinates
and then $2^{l}$ options of choosing signs. The last equality above is
classical "stars and bars" problem from combinatorics, described on page 38 of
[4]. Hence for $k\geq 1$

$|A_{k}|=\sum_{l=1}^{k}2^{l}\binom{d}{l}\cdot\binom{k-1}{l-1},$

from which it follows

$|B_{t}\cap\mathbb{Z}^{d}|=\sum_{k=0}^{t}|A_{k}|=1+\sum_{k=1}^{t}\sum_{l=1}^{k}2^{l}\binom{d}{l}\cdot\binom{k-1}{l-1}$
$=1+\sum_{l=1}^{t}2^{l}\binom{d}{l}\sum_{k=l}^{t}\binom{k-1}{l-1}=1+\sum_{l=1}^{t}2^{l}\binom{d}{l}\binom{t}{l}=\sum_{l=0}^{t}2^{l}\binom{d}{l}\binom{t}{l}.$

In the penultimate equality above we used the following fact: for any $t\geq
l\geq 1$ we have

$\binom{t}{l}=\sum_{k=l}^{t}\binom{k-1}{l-1}.$

One can prove this equality by segregating $l$ element subsets of the set
$\\{1,...,t\\}$ based on biggest number which they contain. For each $k$ there
are $\binom{k-1}{l-1}$ subsets of $\\{1,...,t\\}$ with $l$ elements, whose
biggest number is $k$. ∎

######  Remark 2.3.

Notice that $|S_{t}|$ is equal to term corresponding to $k=t$ in the sum
$\eqref{eq:2.2}$. When $t$ is small in terms of $d$ then $\binom{d}{k+1}$ is
roughly $d$ times bigger than $\binom{d}{k}$ for each $k\leq t$; beacause of
that we expect $|B_{t}\cap\mathbb{Z}^{d}|$ to be similar in size to $|S_{t}|$
when $t$ is sufficiently small in terms of $d$. Lemma 2.4 quantifies our
expectations.

######  Lemma 2.4.

Let $t,d\in\mathbb{N}$ be numbers such that $d\geq 4$, $1\leq t\leq\sqrt{d}$.
Then the following inequalities hold

$\frac{t^{2}}{2d}\leq\frac{|B_{t}\cap\mathbb{Z}^{d}|-|S_{t}|}{|S_{t}|}\leq
e\frac{t^{2}}{d}.$

###### Proof.

Assumptions $d\geq 4$ and $t\leq\sqrt{d}$ imply that $2t\leq d$.
Using formulas from Lemma 2.2 we get that

$\frac{|B_{t}\cap\mathbb{Z}^{d}|-|S_{t}|}{|S_{t}|}=\frac{\sum_{k=0}^{t-1}2^{k}\binom{t}{k}\binom{d}{k}}{2^{t}\binom{d}{t}}=\sum_{k=0}^{t-1}2^{k-t}\binom{t}{k}\frac{t!}{k!}\frac{(d-t)!}{(d-k)!}$
$=\sum_{k=0}^{t-1}2^{k-t}\binom{t}{k}\frac{t!}{k!}\frac{1}{(d-k)(d-k-1)...(d-t+1)}<\sum_{k=0}^{t-1}2^{k-t}\binom{t}{k}\frac{t!}{k!}(d-t)^{k-t}$
$=\sum_{k=0}^{t-1}2^{k-t}\binom{t}{k}^{2}(t-k)!(d-t)^{k-t}=\sum_{u=1}^{t}(2(d-t))^{-u}\binom{t}{u}^{2}u!=\sum_{u=1}^{t}a_{u}.$

In the penultimate equality above we changed summation index by substitution
$u=t-k$ and used the fact that $\binom{t}{k}=\binom{t}{t-k}$, in the last
equality we denoted each term of the sum by $a_{u}$.
Then for each $0\leq u\leq t-1$ we have that

$a_{u+1}=(2(d-t))^{-u-1}\binom{t}{u+1}^{2}(u+1)!=(2(d-t))^{-u-1}\frac{t!^{2}}{(u+1)!(t-u-1)!^{2}}$
$=(2(d-t))^{-u-1}\frac{t!^{2}}{u!(t-u)!^{2}}\cdot\frac{(t-u)^{2}}{u+1}=\frac{(t-u)^{2}}{2(d-t)(u+1)}\cdot(2(d-t))^{-u}\binom{t}{u}^{2}u!$
$=\frac{(t-u)^{2}}{2(d-t)(u+1)}a_{u}\leq\frac{t^{2}}{2(d-t)(u+1)}a_{u}\leq\frac{t^{2}}{d}\frac{1}{u+1}a_{u}.$

Iterating the above estimate we get

$a_{u+1}\leq\frac{t^{2}}{d}\frac{1}{u+1}a_{u}\leq\Big{(}\frac{t^{2}}{d}\Big{)}^{2}\frac{1}{(u+1)\cdot
u}a_{u-1}\leq...$
$\leq\Big{(}\frac{t^{2}}{d}\Big{)}^{u}\cdot\frac{1}{(u+1)\cdot u\cdot...\cdot
2}a_{1}=\Big{(}\frac{t^{2}}{d}\Big{)}^{u}\cdot\frac{1}{(u+1)!}a_{1}.$

However, using the fact that $2t\leq d$ we obtain that

$a_{1}=\frac{t^{2}}{2(d-t)}\leq\frac{t^{2}}{d},$

which implies

$a_{u+1}\leq\Big{(}\frac{t^{2}}{d}\Big{)}^{u+1}\cdot\frac{1}{(u+1)!}$

for any $0\leq u\leq t-1$,hence by $t\leq\sqrt{d}$ we get

$\sum_{u=1}^{t}a_{u}\leq\sum_{u=1}^{t}\frac{\Big{(}\frac{t^{2}}{d}\Big{)}^{u}}{u!}<e\frac{t^{2}}{d}.$

Thus we proved the upper bound.
On the other hand the lower bound follows from estimating undermentioned sum
by the term $k=t-1$.

$\frac{|B_{t}\cap\mathbb{Z}^{d}|-|S_{t}|}{|S_{t}|}=\frac{\sum_{k=0}^{t-1}2^{k}\binom{t}{k}\binom{d}{k}}{2^{t}\binom{d}{t}}=\sum_{k=0}^{t-1}2^{k-t}\binom{t}{k}\frac{t!}{k!}\frac{(d-t)!}{(d-k)!}$
$>\frac{1}{2}\binom{t}{t-1}\frac{t!}{(t-1)!}\frac{(d-t)!}{(d-t+1)!}=\frac{t^{2}}{2(d-t+1)}\geq\frac{t^{2}}{2d}.$

∎

In the paper we will use only the upper bound of Lemma 2.4. Its main
application is given in the Corollary 2.5 below.

######  Corollary 2.5.

For $d,t\in\mathbb{N}$ such that $d\geq 4$, $1\leq t\leq\sqrt{d}$ and any
$\xi\in\mathbb{T}^{d}$ the following inequality holds

$|m_{t}(\xi)-s_{t}(\xi)|\leq 2e\frac{t^{2}}{d}.$

###### Proof.

Using the disjoint decomposition
$B_{t}\cap\mathbb{Z}^{d}=S_{t}\cup\Big{(}B_{t}\cap\mathbb{Z}^{d}\cap
S_{t}\Big{)}$ and the upper bound from Lemma 2.4 we obtain that

$|m_{t}(\xi)-s_{t}(\xi)|=\Big{|}\frac{1}{|B_{t}\cap\mathbb{Z}^{d}|}\sum_{x\in
B_{t}\cap\mathbb{Z}^{d}}e(x\cdot\xi)-\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}e(x\cdot\xi)\Big{|}$ $\leq\Big{|}\sum_{x\in
S_{t}}e(x\cdot\xi)\Big{|}\cdot\Big{(}\frac{1}{|S_{t}|}-\frac{1}{|B_{t}\cap\mathbb{Z}^{d}|}\Big{)}+\frac{1}{|B_{t}\cap\mathbb{Z}^{d}|}\Big{|}\sum_{x\in
B_{t}\cap\mathbb{Z}^{d}\setminus S_{t}}e(x\cdot\xi)\Big{|}$ $\leq
2\frac{|B_{t}\cap\mathbb{Z}^{d}|-|S_{t}|}{|B_{t}\cap\mathbb{Z}^{d}|}=2\frac{|S_{t}|}{|B_{t}\cap\mathbb{Z}^{d}|}\cdot\frac{|B_{t}\cap\mathbb{Z}^{d}|-|S_{t}|}{|S_{t}|}\leq
2e\frac{t^{2}}{d}$

∎

The simple bound above has consequences in terms of norm estimates for the
corresponding maximal dyadic operator. Proof of the next lemma is based on
Lemma 2.4 and a square function estimate.

######  Lemma 2.6.

For $d\in\mathbb{N}$, $d\geq 4$ and any $f\in l^{2}(\mathbb{Z}^{d})$ the
following inequality holds

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f|\|_{l^{2}(\mathbb{Z}^{d})}\leq
6\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

###### Proof.

Take any $f\in l^{2}(\mathbb{Z}^{d})$, then we have

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f|\|_{l^{2}(\mathbb{Z}^{d})}\leq\|\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f|^{2}\Big{)}^{1/2}\|_{l^{2}(\mathbb{Z}^{d})}$
$=\Big{(}\sum_{x\in\mathbb{Z}^{d}}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f(x)|^{2}\Big{)}^{1/2}=\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\sum_{\
x\in\mathbb{Z}^{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f(x)|^{2}\Big{)}^{1/2}$
$=\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\||(\mathcal{M}_{t}-\mathcal{S}_{t})f|\|^{2}_{l^{2}(\mathbb{Z}^{d})}\Big{)}^{1/2}=\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\int_{\mathbb{T}^{d}}|m_{t}(\xi)-s_{t}(\xi)|^{2}|\widehat{f}|^{2}d\xi\Big{)}^{1/2}$
$\leq
2e\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\frac{t^{4}}{d^{2}}\int_{\mathbb{T}^{d}}|\widehat{f}|^{2}d\xi\Big{)}^{1/2}=\|\widehat{f}\|_{L^{2}(\mathbb{T}^{d})}\cdot\frac{2e}{d}\Big{(}\sum_{n=1}^{\lfloor\log_{2}(\sqrt{d})\rfloor}2^{4n}\Big{)}^{1/2}$
$=\|\widehat{f}\|_{L^{2}(\mathbb{T}^{d})}\cdot\frac{2e}{d}\Big{(}2^{4}\cdot\frac{2^{4\lfloor\log_{2}(\sqrt{d})\rfloor}-1}{2^{4}-1}\Big{)}^{1/2}\leq\|\widehat{f}\|_{L^{2}(\mathbb{T}^{d})}\cdot\frac{2e}{d}\Big{(}\frac{16}{15}\cdot
d^{2}\Big{)}^{1/2}$ $\leq
6\|\widehat{f}\|_{L^{2}(\mathbb{T}^{d})}=6\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

The second inequality above comes from Corollary 2.5. In the above reasoning
multiple times we made use of Parseval identity and the fact that
$\mathcal{M}_{t}(f)=\mathcal{F}^{-1}(m_{t}\cdot\widehat{f})$,
$\mathcal{S}_{t}(f)=\mathcal{F}^{-1}(s_{t}\cdot\widehat{f})$. ∎

##  3\. Estimates for the multiplier $s_{t}$.

In the proof of Lemma 2.6 we’ve seen how pointwise bound of
$|m_{t}-s_{t}(\xi)|$ impacted
$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f|\|_{l^{2}(Z^{d})}$.
The bound in corollary 2.5 is fairly strong and uniform in
$\xi\in\mathbb{T}^{d}$. Unfortunately we won’t be that lucky in the future.
Our goal for now is to obtain bounds on $|s_{t}(\xi)|$ for big $\xi$ (then we
expect some decay in $\xi$) and $|s_{t}(\xi)-1|$ for small $\xi$.
Proof of the next lemma exploits in a simple way invariance of the set $S_{t}$
under both permutation of coordinates and sign chances of each coordinate.

######  Lemma 3.1.

For every $d,t\in\mathbb{N}$, $t\leq d$ and every $\xi\in\mathbb{T}^{d}$ we
have

$|s_{t}(\xi)-1|\leq 2\frac{t}{d}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j}).$

###### Proof.

Notice that for any $\epsilon\in\\{-1,1\\}^{d}$ we have that

$x\in S_{t}\iff\epsilon\cdot x\in S_{t},$

this implies that

$s_{t}(\xi)=\frac{1}{|S_{t}|}\sum_{\epsilon\cdot x\in S_{t}}e(\epsilon\cdot
x\cdot\xi)=\frac{1}{|S_{t}|}\sum_{x\in S_{t}}e(\epsilon\cdot x\cdot\xi).$

Hence

$s_{t}(\xi)=\frac{1}{2^{d}}\sum_{\epsilon\in\\{-1,1\\}^{d}}s_{t}(\xi)=\frac{1}{2^{d}}\sum_{\epsilon\in\\{-1,1\\}^{d}}\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}e(\epsilon\cdot x\cdot\xi)$ $=\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}\frac{1}{2^{d}}\sum_{\epsilon\in\\{-1,1\\}^{d}}e(\epsilon\cdot
x\cdot\xi)=\frac{1}{|S_{t}|}\sum_{x\in S_{t}}\prod_{j=1}^{d}\cos(2\pi
x_{j}\xi_{j}).$

Note that for any sequences of complex numbers
$\\{a_{j}\\}_{j=1}^{d},\\{b_{j}\\}_{j=1}^{d}$ such that $\max_{1\leq j\leq
d}|a_{j}|\leq 1$ and $\max_{1\leq j\leq d}|b_{j}|\leq 1$ we have

$\Big{|}\prod_{j=1}^{d}a_{j}-\prod_{j=1}^{d}b_{j}\Big{|}\leq\sum_{j=1}^{d}|a_{j}-b_{j}|,$
(3.1)

This follows from an easy induction argument. Using (3.1) and the formula
$\cos(2x)=1-2\sin^{2}(x)$ we obtain

$|s_{t}(\xi)-1|\leq\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}\Big{|}\prod_{j=1}^{d}\cos(2\pi
x_{j}\xi_{j})-1\Big{|}\leq\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}\sum_{j=1}^{d}|\cos(2\pi x_{j}\xi_{j})-1|$ $=\frac{2}{|S_{t}|}\sum_{x\in
S_{t}}\sum_{j=1}^{d}\sin^{2}(\pi
x_{j}\xi_{j})\leq\frac{2}{|S_{t}|}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j})\cdot\sum_{x\in
S_{t}}x_{j}^{2}.$

In the last inequality we used the fact that $|\sin(xy)|\leq|x||\sin(y)|$ for
any $x\in\mathbb{Z},y\in\mathbb{R}$ (this can be proved by only considering
$x>0$ and induction with respect to $x$ using formula for $\sin(a+b)$).
Now to understand the sum $\sum_{x\in S_{t}}x_{j}^{2}$ we make use of the fact
that $S_{t}$ is closed under permutations of coordinates, that is for any
$\sigma\in Sym(d)$ we have

$x\in S_{t}\iff\sigma\cdot x\in S_{t},$

where $\sigma\cdot x=(x_{\sigma(1)},x_{\sigma(2)},...,x_{\sigma(d)})$. Using
similar argument as in the beginning of the proof one can show that for any
$j\in\\{1,...,d\\}$ we have

$\sum_{x\in S_{t}}x_{j}^{2}=\frac{1}{d}\sum_{x\in S_{t}}\|x\|_{l^{2}}^{2}.$

Thus

$|s_{t}(\xi)-1|\leq\frac{2}{|S_{t}|}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j})\cdot\sum_{x\in
S_{t}}x_{j}^{2}=\frac{2}{|S_{t}|}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j})\cdot\frac{1}{d}\sum_{x\in
S_{t}}\|x\|_{l^{2}}^{2}$
$=\frac{2}{d}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j})\cdot\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}\|x\|_{l^{1}}=2\frac{t}{d}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j}).$

Above we used the fact that any for any $x\in S_{t}$ we have
$\|x\|_{l^{2}}^{2}=\|x\|_{l^{1}}$ (since $x$ has coordinates in the set
$\\{-1,0,1\\}$) and that $\|x\|_{l^{1}}=t$. ∎

###  3.1. Krawtchouk polynomials

######  Definition 3.2.

(Krawtchouk polynomial). For every $n\in\mathbb{N}_{0}$, $k\in\\{0,1,...,n\\}$
and $x\in\mathbb{R}$ we define $k$-th Krawtchouk polynomial by the formula

$\mathds{kr}_{k}^{(n)}(x)=\frac{1}{\binom{n}{k}}\sum_{j=0}^{k}(-1)^{j}\binom{x}{j}\binom{n-x}{k-j},$

if $x$ is not integer, then $\binom{x}{j}=\frac{x(x-1)...(x-j+1)}{j!}$.

Next theorem describes important facts regarding Krawtchouk polynomials.

######  Theorem 3.3.

For every $n\in\mathbb{N}_{0}$ and integers $x,k\in[0,n]$ we have

  1. (1)

Symmetry: $\mathds{kr}_{k}^{(n)}(x)=\mathds{kr}_{x}^{(n)}(k)$.

  2. (2)

Reflection symmetry:
$\mathds{kr}^{(n)}_{k}(n-x)=(-1)^{k}\mathds{kr}_{k}^{(n)}(x)$.

  3. (3)

Orthogonality:

$\sum_{j=0}^{n}\binom{n}{j}\mathds{kr}^{(n)}_{k}(j)\mathds{kr}^{(n)}_{x}(j)=2^{n}\binom{n}{k}^{-1}\delta_{k,x}.$

  4. (4)

Roots: the roots of $\mathds{kr}^{(n)}_{k}(x)$ are real, simple and lie in the
interval
$[\frac{n}{2}-\sqrt{k(n-k)},\frac{n}{2}+\sqrt{k(n-k)}].$

  5. (5)

Uniform bound: For any $x,k\leq n/2$ we have

$|\mathds{kr}^{(n)}_{k}(x)|\leq e^{-ckx/n},$

where $c=-2\log(0.93)=0.14514...$

Proof of the first three points is contained in [8], proof of point 4 is in
theorem 8 of [7], point 5 is in lemma 2.2 of [5]. The value of the constant
$c$ is not explicitly given in [5], however one can deduce it from its proof.
From now on we define

$c=-2\log(0.93)=0.14514...$

######  Lemma 3.4.

For any $t,d\in\mathbb{N}_{0}$ such that $t\leq d/2$ we have

$|s_{t}(\xi)|\leq\exp(-\frac{ct}{2d}\sum_{i=1}^{d}\sin(\pi\xi_{i})^{2})+\exp(-\frac{ct}{2d}\sum_{i=1}^{d}\cos(\pi\xi_{i})^{2}).$
(3.2)

###### Proof.

We will proceed as in the later part of the proof of Proposition 3.3 of [2].
Using the same symmetry invariance arguments as in the proof of 3.1 we have
the following equalities

$s_{t}(\xi)=\frac{1}{|S_{t}|}\sum_{y\in S_{t}}\prod_{i=1}^{d}cos(2\pi
y_{i}\xi_{i})=\frac{1}{|S_{t}|}\sum_{y\in S_{t}}\frac{1}{d!}\sum_{\sigma\in
Sym(d)}\prod_{i=1}^{d}cos(2\pi y_{\sigma(i)}\xi_{i}).$

It is sufficient to prove that for any $y\in S_{t}$ we have that

$\Big{|}\frac{1}{d!}\sum_{\sigma\in Sym(d)}\prod_{i=1}^{d}cos(2\pi
y_{\sigma(i)}\xi_{i})\Big{|}\leq\exp(-\frac{ct}{2d}\sum_{i=1}^{d}\sin(\pi\xi_{i})^{2})+\exp(-\frac{ct}{2d}\sum_{i=1}^{d}\cos(\pi\xi_{i})^{2}).$

Fix any $y\in S_{t}$, notice that we have following disjoint decomposition.

$Sym(d)=\bigcup_{\begin{subarray}{c}J\subseteq\mathbb{N}_{d}\\\
|J|=t\end{subarray}}\\{\sigma\in Sym(d):|y_{\sigma(j)}|=1\iff j\in J\\}.$
(3.3)

Notice that each set under the union has size $(d-t)!\cdot t!$. Since cosine
is even from (3.3) we obtain

$\frac{1}{d!}\sum_{\sigma\in Sym(d)}\prod_{i=1}^{d}cos(2\pi
y_{\sigma(i)}\xi_{i})=\frac{1}{\binom{d}{t}}\sum_{\begin{subarray}{c}J\subseteq\mathbb{N}_{d}\\\
|J|=t\end{subarray}}\prod_{j\in J}\cos(2\pi\xi_{j}).$

For any $S\subseteq\mathbb{N}_{d}$ we define

$a_{S}(\xi)=\prod_{j\in\mathbb{N}_{d}\setminus
S}\cos^{2}(\pi\xi_{j})\cdot\prod_{i\in S}\sin^{2}(\pi\xi_{i}).$

For any set $J\subseteq\mathbb{N}_{d}$ such that $|J|=t$ we define
$\varepsilon(J)\in\\{-1,1\\}^{d}$ to be vector, such that
$\varepsilon(J)_{j}=-1$ exactly when $j\in J$. Then we have following
equalities

$\prod_{j\in
J}\cos(2\pi\xi_{j})=\prod_{j\in\mathbb{N}_{d}}\Big{(}\frac{1+\cos(2\pi\xi_{j})}{2}+\varepsilon(J)_{j}\frac{1-\cos(2\pi\xi_{J})}{2}\Big{)}$
$=\prod_{j\in\mathbb{N}_{d}}\big{(}\cos^{2}(\pi\xi_{j})+\varepsilon(J)_{j}\sin^{2}(\pi\xi_{j})\big{)}=\sum_{S\subset\mathbb{N}_{d}}a_{S}(\xi)w_{S}(\varepsilon(J)),$

where $w_{S}:\\{-1,1\\}^{d}\to\\{-1,1\\}$ is defined by the formula
$w_{S}(\varepsilon)=\prod_{j\in S}\varepsilon_{j}$. Now by changing the order
of summation we obtain

$\frac{1}{d!}\sum_{\sigma\in Sym(d)}\prod_{i=1}^{d}cos(2\pi
y_{\sigma(i)}\xi_{i})=\frac{1}{\binom{d}{t}}\sum_{\begin{subarray}{c}J\subseteq\mathbb{N}_{d}\\\
|J|=t\end{subarray}}\prod_{j\in J}\cos(2\pi\xi_{j})$
$=\sum_{S\subset\mathbb{N}_{d}}a_{S}(\xi)\frac{1}{\binom{d}{t}}\sum_{\begin{subarray}{c}J\subseteq\mathbb{N}_{d}\\\
|J|=t\end{subarray}}w_{S}(\varepsilon(J)).$

Notice that for fixed $S\subseteq\mathbb{N}_{d}$ the following equalities hold

$\frac{1}{\binom{d}{t}}\sum_{\begin{subarray}{c}J\subseteq\mathbb{N}_{d}\\\
|J|=t\end{subarray}}w_{S}(\varepsilon(J))=\frac{1}{\binom{d}{t}}\sum_{\begin{subarray}{c}J\subseteq\mathbb{N}_{d}\\\
|J|=t\end{subarray}}(-1)^{|S\cap
J|}=\frac{1}{\binom{d}{t}}\sum_{j=0}^{t}\sum_{\begin{subarray}{c}J\subseteq\mathbb{N}_{d}\\\
|J|=t\\\ |S\cap J|=j\end{subarray}}(-1)^{j}$
$=\frac{1}{\binom{d}{t}}\sum_{j=0}^{t}(-1)^{j}\binom{|S|}{j}\binom{d-|S|}{t-j}=\mathds{kr}_{t}^{(d)}(|S|).$

Using the above we get

$\frac{1}{d!}\sum_{\sigma\in Sym(d)}\prod_{i=1}^{d}cos(2\pi
y_{\sigma(i)}\xi_{i})=\sum_{S\subset\mathbb{N}_{d}}a_{S}(\xi)\mathds{kr}_{t}^{(d)}(|S|).$

If $|S|\leq d/2$ then by our assumption that $t\leq d/2$ and point 5 of
theorem 3.3 we get

$|\mathds{kr}_{t}^{d}(|S|)|\leq e^{-\frac{c|S|t}{d}},$

otherwise if $|S|>d/2$ then using reflection symmetry and point 5 of theorem
3.3 we obtain

$|\mathds{kr}_{t}^{d}(|S|)|=|\mathds{kr}_{t}^{d}(d-|S|)|\leq
e^{-\frac{c(d-|S|)t}{d}}.$

This gives us

$\frac{1}{d!}\Big{|}\sum_{\sigma\in Sym(d)}\prod_{i=1}^{d}cos(2\pi
y_{\sigma(i)}\xi_{i})\Big{|}\leq\sum_{S\subseteq\mathbb{N}_{d}}a_{S}(\xi)e^{-\frac{c|S|t}{d}}+\sum_{S\subseteq\mathbb{N}_{d}}a_{S}(\xi)e^{-\frac{c(d-|S|)t}{d}}$
$=\sum_{S\subseteq\mathbb{N}_{d}}\prod_{j\in\mathbb{N}_{d}\setminus
S}\cos^{2}(\pi\xi_{j})\cdot\prod_{j\in
S}e^{-\frac{ct}{d}}\sin^{2}(\pi\xi_{j})+\sum_{S\subseteq\mathbb{N}_{d}}\prod_{j\in\mathbb{N}_{d}\setminus
S}e^{-\frac{ct}{d}}\cos^{2}(\pi\xi_{j})\cdot\prod_{j\in
S}\sin^{2}(\pi\xi_{j})$
$=\prod_{j\in\mathbb{N}_{d}}\big{(}\cos^{2}(\pi\xi_{j})+e^{-\frac{ct}{d}}\sin^{2}(\pi\xi_{j})\big{)}+\prod_{j\in\mathbb{N}_{d}}\big{(}e^{-\frac{ct}{d}}\cos^{2}(\pi\xi_{j})+\sin^{2}(\pi\xi_{j})\big{)}$
$=\prod_{j\in\mathbb{N}_{d}}\big{(}1-(1-e^{-\frac{ct}{d}})\sin^{2}(\pi\xi_{j})\big{)}+\prod_{j\in\mathbb{N}_{d}}\big{(}1-(1-e^{-\frac{ct}{d}})\cos^{2}(\pi\xi_{j})\big{)}$
$\leq\exp\Big{(}-(1-e^{-\frac{ct}{d}})\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j})\Big{)}+\exp\Big{(}-(1-e^{-\frac{ct}{d}})\sum_{j=1}^{d}\cos^{2}(\pi\xi_{j})\Big{)}$
$\leq\exp\Big{(}-\frac{ct}{2d}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j})\Big{)}+\exp\Big{(}-\frac{ct}{2d}\sum_{j=1}^{d}\cos^{2}(\pi\xi_{j})\Big{)}.$

In the penultimate inequality we used inequality $1-x\leq e^{-x}$ for
$x=(1-e^{-\frac{ct}{d}})\sin(\pi\xi_{j})$ and for each $j$. In the ultimate
inequality we used $x/2\leq 1-e^{-x}$ for $x=\frac{ct}{d}$ (note that $x\leq
1/2$). This concludes the proof of Lemma 3.4 . ∎

##  4\. Introduction of new multipliers

Now we will define a pair of Fourier multipliers, whose corresponding maximal
operators are bounded independently of the dimension by a general theory. Then
we will try to approximate $s_{t}$ pointwise by these multipliers.

######  Definition 4.1.

For any $\xi\in\mathbb{T}^{d}$ and any $t\in\mathbb{N}_{0},t\leq d$ we define

$\lambda_{t}^{1}(\xi)=\exp\Big{(}-\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi)\Big{)},$
(4.1) $\lambda_{t}^{2}(\xi)=\frac{1}{|S_{t}|}\Big{(}\sum_{x\in
S_{t}}(-1)^{\sum_{i=1}^{d}x_{i}}\Big{)}\exp\Big{(}-\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)\Big{)}.$
(4.2)

Proof of the following theorem is a consequence of the theory of symmetric
diffusion semigroups, it is explained on page 73 of [9] (see also the proof of
theorem 2.2 in [2] for more details).

######  Theorem 4.2.

For any $f\in l^{2}(\mathbb{Z}^{d})$ the following inequality holds

$\|\sup_{t>0}|\mathcal{F}^{-1}(\lambda_{t}^{1}\widehat{f})|\|_{l^{2}(\mathbb{Z}^{d})}\leq
2\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

From the theorem above we can conclude similar estimate for $\lambda_{t}^{2}$.

######  Corollary 4.3.

For any $f\in l^{2}(\mathbb{Z}^{d})$ the following inequality holds

$\|\sup_{t\in\mathbb{N}_{0},1\leq t\leq
d}|\mathcal{F}^{-1}(\lambda_{t}^{2}\widehat{f})|\|_{l^{2}(\mathbb{Z}^{d})}\leq
2\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

###### Proof.

This argument is exactly the same as in proof of theorem 3.1 in [2], we give
details for completeness. Define $F(x)=(-1)^{\sum_{j=1}^{d}x_{j}}f(x)$, then
we have

$\widehat{F}(\xi)=\sum_{x\in\mathbb{Z}^{d}}e(x\cdot\xi)(-1)^{\sum_{j=1}^{d}x_{j}}f(x)=\sum_{x\in\mathbb{Z}^{d}}e(x\cdot(\xi+\frac{1}{2}))f(x)=\widehat{f}(\xi+\frac{1}{2}).$

Thus, we estimate

$\|\sup_{t\in\mathbb{N}_{0},1\leq t\leq
d}|\mathcal{F}^{-1}(\lambda_{t}^{2}\widehat{f})(x)|\|_{l^{2}(\mathbb{Z}^{d})}=$
$\|\sup_{t\in\mathbb{N}_{0},1\leq t\leq d}\Big{|}\frac{1}{|S_{t}|}\sum_{y\in
S_{t}}(-1)^{\sum_{i=1}^{d}y_{i}}\int_{\mathbb{T}^{d}}\exp\Big{(}-\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)\Big{)}\widehat{f}(\xi)e(-x\cdot\xi)d\xi\Big{|}\|_{l^{2}(\mathbb{Z}^{d})}$
$\leq\|\sup_{t\in\mathbb{N}_{0},1\leq t\leq
d}\Big{|}\int_{\mathbb{T}^{d}}\exp(-\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi))\widehat{f}(\xi)e(-x\cdot\xi)d\xi\Big{|}\|_{l^{2}(\mathbb{Z}^{d})}$
$=\|\sup_{t\in\mathbb{N}_{0},1\leq t\leq
d}\Big{|}\int_{\mathbb{T}^{d}}\exp(-\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi))\widehat{F}(\xi)e(-x\cdot\xi)d\xi\Big{|}\|_{l^{2}(\mathbb{Z}^{d})}$
$\leq\|\sup_{t>0}|\mathcal{F}^{-1}(\lambda_{t}^{1}\widehat{F})(x)|\|_{l^{2}(\mathbb{Z}^{d})}\leq
2\|F\|_{l^{2}(\mathbb{Z}^{d})}=2\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

In the penultimate equality we used the substitution
$\xi\mapsto\xi+\frac{1}{2}$, then
$e(-x\cdot\xi)\mapsto(-1)^{\sum_{i=1}^{d}x_{i}}\cdot e(-x\cdot\xi)$, but the
factor $(-1)^{\sum_{i=1}^{d}x_{i}}$ was removed, because its absolute value is
one. In the last inequality we used Theorem 4.2. ∎

We will approximate $s_{t}(\xi)$ by $\lambda_{t}^{1}(\xi)$ or
$\lambda_{t}^{2}(\xi)$ depending on which term in the right hand side of
inequality (3.2) we expect to be bigger. For any $\xi\in\mathbb{T}^{d}$ we
define

$V_{\xi}=\\{i\in\mathbb{N}_{d}:\cos(2\pi\xi_{i})<0\\}=\\{i\in\mathbb{N}_{d}:\frac{1}{4}<|\xi_{i}|\leq\frac{1}{2}\\}.$

Then we have the following crucial inequalities.

######  Lemma 4.4.

For every $t,d\in\mathbb{N}_{0}$ such that $t\leq d/2$ and every
$\xi\in\mathbb{T}^{d}$ we have

  1. (1)

if $|V_{\xi}|\leq d/2$, then

$|s_{t}(\xi)-\lambda_{t}^{1}(\xi)|\leq
7\min\Big{\\{}\exp\Big{(}-\frac{ct}{8d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)},\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{\\}},$
(4.3)

  2. (2)

if $|V_{\xi}|\geq d/2$, then

$|s_{t}(\xi)-\lambda_{t}^{2}(\xi)|\leq
7\min\Big{\\{}\exp\Big{(}-\frac{ct}{8d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i})\Big{)},\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i})\Big{\\}}.$
(4.4)

The above bounds will essentially allow us to more or less the copy proof of
lemma 2.6. This will be shown at the end of section 5.
We start with a proof of $\eqref{eq:4.3}$.

###### Proof.

Take any $\xi$ such that $|V_{\xi}|\leq d/2$, then $|\mathbb{N}_{d}\setminus
V_{\xi}|\geq d/2$. Moreover for any $i\in\mathbb{N}_{d}\setminus V_{\xi}$ we
have $\cos^{2}(\pi\xi_{i})\geq 1/2$, hence

$\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i})\geq\frac{|\mathbb{N}_{d}\setminus
V_{\xi}|}{2}\geq d/4\geq\frac{1}{4}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i}).$

Using lemma 3.4 and the above, we obtain

$|s_{t}(\xi)|\leq\exp\Big{(}-\frac{ct}{2d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}+\exp\Big{(}-\frac{ct}{2d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i})\Big{)}\leq
2\exp\Big{(}-\frac{ct}{8d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)},$

so that

$|s_{t}(\xi)-\lambda_{t}^{1}(\xi)|\leq
3\exp\Big{(}-\frac{ct}{8d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}.$

On the other hand using lemma 3.1 and inequality $1-e^{-x}\leq x$ we obtain

$|s_{t}(\xi)-\lambda_{t}^{1}(\xi)|\leq|s_{t}(\xi)-1|+|1-\lambda_{t}^{1}(\xi)|\leq
3\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i}).$

∎

Our goal now is to adapt the above proof, mainly adapt Lemma 3.1 to obtain
inequality (4.4). Reasoning in the next two lemmas will explain motivation
behind defining $\lambda_{t}^{2}$ the way we did.

######  Lemma 4.5.

For any $t,d\in\mathbb{N}_{0}$ such that $t\leq d$ and any
$\xi\in\mathbb{T}^{d}$ we have

$\Big{|}s_{t}(\xi)-\frac{1}{|S_{t}|}\sum_{x\in S_{t}}(-1)^{\sum_{i\in
V_{\xi}}x_{i}}\Big{|}\leq 2\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi).$

###### Proof.

We will make a substitution and basically copy the proof of Lemma 3.1. For any
$\xi\in\mathbb{T}^{d}$ we define $\xi^{\prime}\in\mathbb{T}^{d}$ by the
following formula

$\xi^{\prime}_{i}=\begin{cases}\xi_{i}\ \ \ \ \ \ \ \text{if}\ i\not\in
V_{\xi}.\\\ \xi_{i}-\frac{1}{2}\ \ \text{if}\
\frac{1}{4}<\xi_{i}\leq\frac{1}{2}.\\\ \xi_{i}+\frac{1}{2}\ \ \text{if}\
-\frac{1}{2}\leq\xi_{i}<-\frac{1}{4}.\end{cases}$

Notice that for any $i\in V_{\xi}$ and any $x_{i}\in\mathbb{Z}$ we have
$\cos(2\pi x_{i}\xi_{i})=\cos(2\pi x_{i}\xi_{i}^{\prime})\cos(\pi
x_{i})=(-1)^{x_{i}}\cos(2\pi x_{i}\xi_{i}^{\prime})$. This implies

$s_{t}(\xi)=\frac{1}{|S_{t}|}\sum_{x\in S_{t}}\prod_{j=1}^{d}\cos(2\pi
x_{j}\xi_{j})$ $=\frac{1}{|S_{t}|}\sum_{x\in S_{t}}\Big{(}(-1)^{\sum_{i\in
V_{\xi}}x_{i}}\prod_{j=1}^{d}\cos(2\pi x_{j}\xi_{j}^{\prime})\Big{)}.$

Using similar arguments as in proof of Lemma 3.1 we obtain

$\Big{|}s_{t}(\xi)-\frac{1}{|S_{t}|}\sum_{x\in S_{t}}(-1)^{\sum_{i\in
V_{\xi}}x_{i}}\Big{|}\leq\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}\Big{(}1-\prod_{j=1}^{d}\cos(2\pi x_{j}\xi_{j}^{\prime})\Big{)}$
$\leq\frac{2}{|S_{t}|}\sum_{x\in S_{t}}\sum_{j=1}^{d}\sin^{2}(\pi
x_{j}\xi_{j}^{\prime})\leq\frac{2}{|S_{t}|}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j}^{\prime})\sum_{x\in
S_{t}}x_{j}^{2}$ $=2\frac{t}{d}\sum_{j=1}^{d}\sin^{2}(\pi\xi_{j}^{\prime})\leq
2\frac{t}{d}\sum_{j=1}^{d}\cos^{2}(\pi\xi_{j}).$

In the last inequality above we used the fact that
$\sin^{2}(\pi\xi_{j}^{\prime})=\cos^{2}(\pi\xi_{j})$ for $j\in V_{\xi}$ and
$\sin^{2}(\pi\xi_{j}^{\prime})\leq\cos^{2}(\pi\xi_{j})$ for $j\not\in
V_{\xi}.$ ∎

######  Lemma 4.6.

For any $t,d\in\mathbb{N}_{0}$ such that $t\leq d$ and any
$\xi\in\mathbb{T}^{d}$ we have

$\Big{|}\frac{1}{|S_{t}|}\sum_{x\in S_{t}}(-1)^{\sum_{i\in
V_{\xi}}x_{i}}-\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}(-1)^{\sum_{i=1}^{d}x_{i}}\Big{|}\leq
4\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i}).$

###### Proof.

$\Big{|}\frac{1}{|S_{t}|}\sum_{x\in S_{t}}(-1)^{\sum_{i\in
V_{\xi}}x_{i}}-\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}(-1)^{\sum_{i=1}^{d}x_{i}}\Big{|}\leq\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}|(-1)^{\sum_{i\in\mathbb{N}_{d}\setminus V_{\xi}}x_{i}}-1|$
$\leq\frac{2}{|S_{t}|}\sum_{x\in S_{t}}\sum_{i\in\mathbb{N}_{d}\setminus
V_{\xi}}x_{i}^{2}=\sum_{i\in\mathbb{N}_{d}\setminus
V_{\xi}}\frac{2}{|S_{t}|}\sum_{x\in S_{t}}x_{i}^{2}$
$=2\frac{t}{d}(d-|V_{\xi}|)\leq
4\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i}).$

In the last equality we used the fact, that for any $i\in\mathbb{N}_{d}$ we
have $\frac{1}{|S_{t}|}\sum_{x\in S_{t}}x_{i}^{2}=\frac{t}{d}$ (this was
explained in the proof of Lemma 3.1). In the last inequality we used the fact,
that $1/2\leq\cos^{2}(\pi\xi_{i})$ for any $i\in\mathbb{N}_{d}\setminus
V_{\xi}$, which implies

$d-|V_{\xi}|\leq 2\sum_{i\in\mathbb{N}_{d}\setminus
V_{\xi}}\cos^{2}(\pi\xi_{i})\leq 2\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i}).$

∎

Now we are ready to prove inequality $\eqref{eq:4.4}$ of Lemma 4.4

###### Proof.

Assume that $|V_{\xi}|\geq d/2$. Then for any $i\in V_{\xi}$ we have
$\sin^{2}(\pi\xi_{i})\geq\frac{1}{2}$, hence

$\sum_{i=1}^{d}\sin^{2}(\pi\xi)\geq\frac{|V_{\xi}|}{2}\geq\frac{d}{4}\geq\frac{1}{4}\sum_{i=1}^{d}\cos^{2}(\pi\xi).$

Using Lemma 3.4 and the above we obtain

$|s_{t}(\xi)|\leq\exp\Big{(}-\frac{ct}{2d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}+\exp\Big{(}-\frac{ct}{2d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i})\Big{)}\leq
2\exp\Big{(}-\frac{ct}{8d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i})\Big{)}.$

Thus

$|s_{t}(\xi)-\lambda_{t}^{2}(\xi)|\leq
3\exp\Big{(}-\frac{ct}{8d}\sum_{i=1}^{d}\cos^{2}(\pi\xi_{i})\Big{)}.$

On the other hand using Lemmas 4.5 and 4.6 and inequality $1-e^{-x}\leq x$ we
get

$\Big{|}s_{t}(\xi)-\frac{1}{|S_{t}|}\Big{(}\sum_{x\in
S_{t}}(-1)^{\sum_{i=1}^{d}x_{i}}\Big{)}\exp\Big{(}-\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)\Big{)}\Big{|}$
$\leq\Big{|}s_{t}(\xi)-\frac{1}{|S_{t}|}\sum_{x\in S_{t}}(-1)^{\sum_{i\in
V_{\xi}}x_{i}}\Big{|}+\Big{|}\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}(-1)^{\sum_{i\in
V_{\xi}}x_{i}}\Big{(}1-\exp\Big{(}-\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)\Big{)}\Big{)}\Big{|}$
$+\Big{|}\exp\Big{(}-\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)\Big{)}\Big{(}\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}(-1)^{\sum_{i\in V_{\xi}}x_{i}}-\frac{1}{|S_{t}|}\sum_{x\in
S_{t}}(-1)^{\sum_{i=1}^{d}x_{i}}\Big{)}\Big{|}$ $\leq
2\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)+\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)+4\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi)=7\frac{t}{d}\sum_{i=1}^{d}\cos^{2}(\pi\xi).$

This concludes proof of inequality (4.4), hence the entire Lemma 4.4, since we
have already proven inequality (4.3). ∎

##  5\. Conclusions and proof of the main theorem

We are almost ready to prove the main theorem. The next lemma is simple,
however we give a proof to calculate the explicit constant $\frac{8}{3}$.

######  Lemma 5.1.

For any $x\in\mathbb{R}_{+}$ we have

$\sum_{k=0}^{\infty}\min\\{4^{k}x,(4^{k}x)^{-1}\\}\leq\frac{8}{3}$

###### Proof.

If $x>1$ then

$\sum_{k=0}^{\infty}\min\\{4^{k}x,(4^{k}x)^{-1}\\}=\sum_{k=0}^{\infty}4^{-k}x^{-1}=\frac{4}{3}x^{-1}<\frac{4}{3}.$

If $x\leq 1$, then for any $k\in\mathbb{N}_{0}$ we have

$\min\\{4^{k}x,(4^{k}x)^{-1}\\}=4^{k}x\iff 4^{k}x\leq 1\iff
k\leq\lfloor\log_{4}(1/x)\rfloor.$

Hence we have

$\sum_{k=0}^{\infty}\min\\{4^{k}x,(4^{k}x)^{-1}\\}=\sum_{k=0}^{\lfloor\log_{4}(1/x)\rfloor}4^{k}x+\sum_{k=\lfloor\log_{4}(1/x)\rfloor+1}^{\infty}4^{-k}x^{-1}\leq$
$\frac{1}{3}\cdot
4^{\lfloor\log_{4}(1/x)\rfloor+1}x+4^{-\lfloor\log_{4}(1/x)\rfloor-1}x^{-1}\cdot\frac{4}{3}\leq\frac{8}{3}$

∎

We are finally ready to prove Theorem 1.3.

###### Proof.

Take any $f\in l^{2}(\mathbb{Z}^{d})$ we have that

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{M}_{t}f|\|_{l^{2}(\mathbb{Z}^{d})}\leq\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f|\|_{l^{2}(\mathbb{Z}^{d})}+\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{S}_{t}f|\|_{l^{2}(\mathbb{Z}^{d})}.$

By Lemma 2.6 the first term above is bounded by $6\|f\|_{l^{2}}$. For the
second term let $f=f_{1}+f_{2}$ with
$\widehat{f_{1}}(\xi)=\widehat{f}(\xi)\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|\leq
d/2\\}}(\xi)$ and
$\widehat{f_{2}}(\xi)=\widehat{f}(\xi)\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|>d/2\\}}(\xi)$.
Then we have

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{S}_{t}f|\|_{l^{2}(\mathbb{Z}^{d})}=\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}(s_{t}\widehat{f})|\|_{l^{2}(\mathbb{Z}^{d})}$
$\leq\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}(s_{t}\widehat{f_{1}})|\|_{l^{2}(\mathbb{Z}^{d})}+\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}(s_{t}\widehat{f_{2}})|\|_{l^{2}(\mathbb{Z}^{d})}$
$\leq\sum_{i=1}^{2}\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}(\lambda_{t}^{i}\widehat{f_{i}})|\|_{l^{2}(\mathbb{Z}^{d})}+\sum_{i=1}^{2}\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{i})\widehat{f_{i}}\big{)}|\|_{l^{2}(\mathbb{Z}^{d})}.$

By theorem 4.2 and corollary 4.3 the first sum is bounded by

$2\|f_{1}\|_{l^{2}(\mathbb{Z}^{d})}+2\|f_{2}\|_{l^{2}(\mathbb{Z}^{d})}=2\|\widehat{f_{1}}\|_{L^{2}(\mathbb{T}^{d})}+2\|\widehat{f_{2}}\|_{L^{2}(\mathbb{T}^{d})}$
$=2\Big{(}\int_{\mathbb{T}^{d}}|\widehat{f}|^{2}\cdot\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|\leq
d/2\\}}(\xi)d\xi\Big{)}^{1/2}+2\Big{(}\int_{\mathbb{T}^{d}}|\widehat{f}|^{2}\cdot\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|>d/2\\}}(\xi)d\xi\Big{)}^{1/2}$
$\leq
2\sqrt{2}\Big{(}\int_{\mathbb{T}^{d}}|\widehat{f}|^{2}\cdot(\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|\leq
d/2\\}}(\xi)+\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|>d/2\\}}(\xi))d\xi\Big{)}^{1/2}$
$=2\sqrt{2}\|\widehat{f}\|_{L^{2}(\mathbb{T}^{d})}=2\sqrt{2}\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

To finish the proof it remains to estimate for $i=1,2$ the following term

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{i})\widehat{f_{i}}\big{)}|\|_{l^{2}(\mathbb{Z}^{d})}.$

We will only bound the above for $i=1$, the proof for $i=2$ is basically the
same. From Lemma 4.4 we have that if $|V_{\xi}|\leq d/2$ then

$|s_{t}-\lambda_{t}^{1}(\xi)|\leq
7\min\Big{\\{}\exp\Big{(}-\frac{ct}{8d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)},\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{\\}}$
$\leq
7\min\Big{\\{}\Big{(}\frac{ct}{8d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}^{-1},\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{\\}}$
$\leq\frac{56}{c}\min\Big{\\{}\Big{(}\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}^{-1},\frac{t}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{\\}}.$

Using lemma 5.1 for any $\xi\in\mathbb{T}^{d}$ such that $|V_{\xi}|\leq d/2$
we obtain

$\sum_{t\in\mathbb{D},t\leq\sqrt{d}}|s_{t}-\lambda_{t}^{1}(\xi)|^{2}\leq\Big{(}\frac{56}{c}\Big{)}^{2}\cdot\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\min\Big{\\{}t^{-2}\Big{(}\frac{1}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}^{-2},t^{2}\Big{(}\frac{1}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}^{2}\Big{\\}}$
$\leq\Big{(}\frac{56}{c}\Big{)}^{2}\cdot\sum_{k=0}^{\infty}\min\Big{\\{}4^{-k}\Big{(}\frac{1}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}^{-2},4^{k}\Big{(}\frac{1}{d}\sum_{i=1}^{d}\sin^{2}(\pi\xi_{i})\Big{)}^{2}\Big{\\}}\leq\Big{(}\frac{56}{c}\Big{)}^{2}\cdot\frac{8}{3}.$

Finally the following is true

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{1})\widehat{f_{1}}\big{)}|\|_{l^{2}(\mathbb{Z}^{d})}\leq\|\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{1})\widehat{f_{1}}\big{)}|^{2}\Big{)}^{1/2}\|_{l^{2}(\mathbb{Z}^{d})}$
$=\Big{(}\sum_{x\in\mathbb{Z}^{d}}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{1})\widehat{f_{1}}\big{)}(x)|^{2}\Big{)}^{1/2}=\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\sum_{x\in\mathbb{Z}^{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{1})\widehat{f_{1}}\big{)}(x)|^{2}\Big{)}^{1/2}$
$=\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\||\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{1})\widehat{f_{1}}\big{)}|\|_{l^{2}(\mathbb{Z}^{d})}^{2}\Big{)}^{1/2}$
$=\Big{(}\sum_{t\in\mathbb{D},t\leq\sqrt{d}}\int_{\mathbb{T}^{d}}|s_{t}(\xi)-\lambda_{t}^{1}(\xi)|^{2}|\widehat{f}(\xi)|^{2}\cdot\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|\leq
d/2\\}}(\xi)d\xi\Big{)}^{1/2}$
$\leq\frac{56}{c}\cdot\sqrt{\frac{8}{3}}\Big{(}\int_{\mathbb{T}^{d}}|\widehat{f}(\xi)|^{2}\cdot\mathds{1}_{\\{\eta\in\mathbb{T}^{d}:|V_{\eta}|\leq
d/2\\}}(\xi)d\xi\Big{)}^{1/2}$
$=\frac{56}{c}\cdot\sqrt{\frac{8}{3}}\|\widehat{f_{1}}\|_{L^{2}(\mathbb{T}^{d})}=\frac{56}{c}\cdot\sqrt{\frac{8}{3}}\|f_{1}\|_{l^{2}(\mathbb{Z}^{d})}.$

Similarly we obtain

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{2})\widehat{f_{2}}\big{)}|\|_{l^{2}(\mathbb{Z}^{d})}\leq\frac{56}{c}\cdot\sqrt{\frac{8}{3}}\|f_{2}\|_{l^{2}(\mathbb{Z}^{d})}.$

Which gives us

$\sum_{i=1}^{2}\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{i})\widehat{f_{i}}\big{)}|\|_{l^{2}(\mathbb{Z}^{d})}\leq$
$\frac{56}{c}\cdot\sqrt{\frac{8}{3}}\cdot\Big{(}\|f_{1}\|_{l^{2}(\mathbb{Z}^{d})}+\|f_{2}\|_{l^{2}(\mathbb{Z}^{d})}\Big{)}\leq\frac{56}{c}\cdot\frac{4}{\sqrt{3}}\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

Combining all of the estimates we finally get

$\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{M}_{t}f|\|_{l^{2}(\mathbb{Z}^{d})}\leq\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f|\|_{l^{2}(\mathbb{Z}^{d})}+\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{S}_{t}f|\|_{l^{2}(\mathbb{Z}^{d})}$
$\leq\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|(\mathcal{M}_{t}-\mathcal{S}_{t})f|\|_{l^{2}(\mathbb{Z}^{d})}+\sum_{i=1}^{2}\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}(\lambda_{t}^{i}\widehat{f_{i}})|\|_{l^{2}(\mathbb{Z}^{d})}$
$+\sum_{i=1}^{2}\|\sup_{t\in\mathbb{D},t\leq\sqrt{d}}|\mathcal{F}^{-1}\big{(}(s_{t}-\lambda_{t}^{i})\widehat{f_{i}}\big{)}|\|_{l^{2}(\mathbb{Z}^{d})}$
$\leq\|f\|_{l^{2}(\mathbb{Z}^{d})}\Big{(}6+2\sqrt{2}+\frac{56}{c}\cdot\frac{4}{\sqrt{3}}\Big{)}\leq
900\|f\|_{l^{2}(\mathbb{Z}^{d})}.$

This completes the proof of Theorem 1.3. ∎

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