# Harmonic analysis for a multidimensional discrete Laplacian

Óscar Ciaurri  Departamento de Matemáticas y Computación, Universidad de La
Rioja, Complejo Científico-Tecnológico, Calle Madre de Dios 53, 26006 Logroño,
Spain<EMAIL_ADDRESS>
###### Abstract.

In this paper we analyze some classical operators in harmonic analysis
associated to the multidimensional discrete Laplacian

$\Delta_{N}f(\mathbf{n})=\sum_{i=1}^{N}(f(\mathbf{n}+\mathbf{e}_{i})-2f(\mathbf{n})+f(\mathbf{n}-\mathbf{e}_{i})),\qquad\mathbf{n}\in\mathbb{Z}^{N}.$

We deal with the heat and Poisson semigroups, the fractional integrals, the
Riesz transforms, the fractional powers of the Laplacian, and the
$g_{k}$-square functions.

###### Key words and phrases:

Discrete harmonic analysis, heat semigroup, Poisson semigroup, fractional
integral, Riesz transform, fractional Laplacian, $g_{k}$-function.

###### 2010 Mathematics Subject Classification:

Primary: 42C10.

The author was supported by grant PID2021-124332NB-C22 AEI, from Spanish
Government.

###### Contents

  1. 1 Introduction
  2. 2 Some facts about the modified Bessel functions of first kind
    1. 2.1 An AM-GM type inequality for the modified Bessel functions of first kind
    2. 2.2 Some inequalities for the difference of modified Bessel functions of first kind
    3. 2.3 A uniform bound for the modified Bessel functions of first kind
    4. 2.4 Further properties of the modified Bessel functions of first kind
  3. 3 The heat semigroup and beyond
    1. 3.1 Decay of the heat semigroup
    2. 3.2 Maximal operators for the heat and the Poisson semigroups
  4. 4 The fractional integrals and the Riesz transforms
    1. 4.1 Basic properties of the fractional integrals
    2. 4.2 Discrete Schauder estimates for the fractional integrals
    3. 4.3 The Riesz transforms
  5. 5 Fractional powers of the Laplacian
    1. 5.1 Basic properties of the fractional powers of the Laplacian
    2. 5.2 The fractional powers of the Laplacian on the Hölder classes
  6. 6 The discrete square functions
    1. 6.1 The $g_{k}$-square functions associated to the heat semigroup
    2. 6.2 The $\mathfrak{g}_{k}$-square functions associated to the Poisson semigroup
    3. 6.3 The Laplace type multipliers

##  1\. Introduction

Our target is to analyze mapping properties of some operators related to the
multidimensional discrete Laplacian

$\Delta_{N}f(\mathbf{n})=\sum_{i=1}^{N}\Delta_{N,i}f(\mathbf{n}),$

where

$\Delta_{N,i}f(\mathbf{n})=f(\mathbf{n}+\mathbf{e}_{i})-2f(\mathbf{n})+f(\mathbf{n}-\mathbf{e}_{i}),$

with $\\{\mathbf{e}_{i}\\}_{i=1,\dots,N}$ being $N$-dimensional vectors each
of whose components are all zero, except the $i$-th that equals one. This
operator is the natural extension to higher dimensions of the one-dimensional
discrete Laplacian studied in [6] and [7]. We have to observe that

$\Delta_{N,i}f(\mathbf{n})=\delta_{i}^{-}\delta_{i}^{+}f(\mathbf{n})$

with

$\delta_{i}^{+}f(\mathbf{n})=f(\mathbf{n}+\mathbf{e}_{i})-f(\mathbf{n})\qquad\text{and}\qquad\delta_{i}^{-}f(\mathbf{n})=f(\mathbf{n})-f(\mathbf{n}-\mathbf{e}_{i}).$

Note that $\delta_{i}^{-}\delta_{i}^{+}=\delta_{i}^{+}\delta_{i}^{-}$. When we
consider sequences of one variable we write $\delta^{+}f(n)=f(n+1)-f(n)$ and
similarly for $\delta^{-}f$ and $\Delta f$.

We will focus in this paper on the study of the heat and Poisson semigroups,
the fractional integrals, the Riesz transforms, the fractional powers of the
Laplacian, and the $g_{k}$-square functions related to $\Delta_{N}$.

The heat semigroup $W_{t}=e^{t\Delta_{N}}$ is the solution of the
$N$-dimensional heat equation

$\begin{cases}\Delta_{N}u(\mathbf{n},t)=\dfrac{d}{dt}u(\mathbf{n},t),\\\\[3.0pt]
u(\mathbf{n},t)=f(\mathbf{n}),\end{cases}$

and it is given by (see [11] for the one-dimensional case)

(1)
$W_{t}f(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}G_{t,N}(\mathbf{n}-\mathbf{k})f(\mathbf{k}),$

where, for $\mathbf{n}=(n_{1},\dots,n_{N})$,

$G_{t,N}(\mathbf{n})=\prod_{k=1}^{N}(e^{-2t}I_{n_{k}}(2t)),$

with $I_{a}$ being the modified Bessel function of first kind and order $a$.
This semigroup will be the main tool to analyze our discrete multidimensional
operators.

M. Riesz initiated the study of discrete operators in harmonic analysis in
[19], where he proved the boundedness of the Hilbert transform in
$L^{p}(\mathbb{R})$ and its discrete analogue in $\ell^{p}(\mathbb{Z})$.
Later, in [4] A. P. Calderon and A. Zygmund obtained the boundedness in
$\ell^{p}(\mathbb{Z}^{N})$ of the discrete analogues of the singular integrals
from his result in $L^{p}(\mathbb{R}^{N})$. Weighted inequalities for the
discrete Hilbert and discrete maximal operator in the one-dimensional case
were proved by R. Hunt, B. Muckenhoupt, and R. Wheeden in [12]. In the last
few decades, several discrete analogues of continuous operators have been
analyzed; some significative contributions are due to I. Arkhipov and K. I.
Oskolkov [2], J. Bourgain [3], and E. M. Stein and S. Wainger [24, 25]. In
[13], we can find a brief history and a nice exposition on discrete analogues
of classical operators in harmonic analysis.

In general, along the paper will work on weighted Lebesgue spaces of sequences
on $\mathbb{Z}^{N}$. More precisely, we define

$\ell^{p}(\mathbb{Z}^{n},w)=\left\\{f=\\{f(\mathbf{n})\\}_{\mathbf{n}\in\mathbb{Z}^{N}}:\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)}<\infty\right\\},\qquad
1\leq p\leq\infty,$

where the weight $w=\\{w(\mathbf{n})\\}_{\mathbf{n}\in\mathbb{Z}^{N}}$ is a
sequence of positive numbers and

$\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)}=\left(\sum_{\mathbf{n}\in\mathbb{Z}^{N}}|f(\mathbf{n})|^{p}w(\mathbf{n})\right)^{1/p},\qquad
1\leq p<\infty,$

and

$\|f\|_{\ell^{\infty}(\mathbb{Z}^{N},w)}=\sup\\{|f(\mathbf{n})|:\mathbf{n}\in\mathbb{Z}^{N}\\}.$

When $w(\mathbf{n})=1$ for all $\mathbf{n}\in\mathbb{Z}^{N}$ we will write
$\ell^{p}(\mathbb{Z}^{N})$ only.

The paper is organized as follows. In the next section we present the main
tools to deal with the kernel $G_{t,N}$. Section 3 will be focused on the
analysis of the heat and Poisson semigroups. Section 4 contains our results
about the fractional integrals and the Riesz transforms. In Section 5 we
present some results on the fractional powers of $\Delta_{N}$. Finally, in
Section 6 we analyze the mapping properties of the $g_{k}$-square functions.

##  2\. Some facts about the modified Bessel functions of first kind

The modified Bessel function of first kind and order $a$, with
$a\in\mathbb{R}$, is given by

$I_{a}(x)=\sum_{k=0}^{\infty}\frac{c_{k,a}}{k!}\left(\frac{x}{2}\right)^{2k+a},$

where

$c_{k,a}=\frac{1}{\Gamma(a+k+1)},$

understanding that for $a\leq-1$

$c_{k,a}=\frac{(a+k+1)\cdots(a+[-a])}{\Gamma(a+[-a]+1)},\qquad 0\leq
k<\lfloor-a\rfloor.$

In this section we compile some relevant inequalities for the modified Bessel
functions of first kind. They will be the main tools to obtain our results.

###  2.1. An AM-GM type inequality for the modified Bessel functions of first
kind

In [26], we find the identity

(2)
$I_{a}(x)I_{b}(x)=\sum_{k=0}^{\infty}\binom{a+b+2k}{k}c_{k,a}c_{k,b}\left(\frac{x}{2}\right)^{2k+a+b},$

which can be proved by applying the Cauchy product of two absolutely
convergent series. From this fact, we can obtain the next inequalities for the
modified Bessel functions.

######  Proposition 1.

For $a_{1},\dots,a_{n}>-1$ and $x\geq 0$ the inequalities

(3)
$\frac{\left(\Gamma\left(\frac{a_{1}+\cdots+a_{n}}{n}+1\right)\right)^{n}}{\Gamma(a_{1}+1)\cdots\Gamma(a_{n}+1)}\left(I_{\frac{a_{1}+\cdots+a_{n}}{n}}(x)\right)^{n}\leq
I_{a_{1}}(x)\cdots
I_{a_{n}}(x)\leq\left(I_{\frac{a_{1}+\cdots+a_{n}}{n}}(x)\right)^{n}$

hold.

###### Proof.

We will provide an à la Cauchy proof, obtaining the cases $n=2^{m}$ first and
extending the result for general $n$. To prove the case $n=2$ we are going to
check that

(4)
$\frac{c_{0,a_{1}}c_{0,a_{2}}}{\big{(}c_{0,\frac{a_{1}+a_{2}}{2}}\big{)}^{2}}\leq\frac{c_{k,a_{1}}c_{k,a_{2}}}{\big{(}c_{k,\frac{a_{1}+a_{2}}{2}}\big{)}^{2}}\leq
1.$

From the Weiertrass’ product formula for the Gamma function

$\Gamma(z)=\frac{e^{-\gamma
z}}{z}\prod_{n=1}^{\infty}\left(\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\right),\qquad
z\in\mathbb{C}\setminus\\{0,-1,-2,\dots\\},$

we have

$\frac{c_{k,a_{1}}c_{k,a_{2}}}{\big{(}c_{k,\frac{a_{1}+a_{2}}{2}}\big{)}^{2}}=\prod_{n=1}^{\infty}\frac{(n+a_{1}+k+1)(n+a_{2}+k+1)}{\left(n+\frac{a_{1}+a_{2}}{2}+k+1\right)^{2}}.$

Then, by using that

$\frac{(n+a_{1}+1)(n+a_{2}+1)}{\left(n+\frac{a_{1}+a_{2}}{2}+1\right)^{2}}\leq\frac{(n+a_{1}+k+1)(n+a_{2}+k+1)}{\left(n+\frac{a_{1}+a_{2}}{2}+k+1\right)^{2}}\leq
1$

the inequalities in (4) follow.

Now, from (2), (4) and the identity

$\frac{c_{0,a_{1}}c_{0,a_{2}}}{\big{(}c_{0,\frac{a_{1}+a_{2}}{2}}\big{)}^{2}}=\frac{\left(\Gamma\left(\frac{a_{1}+a_{2}}{2}+1\right)\right)^{2}}{\Gamma(a_{1}+1)\,\Gamma(a_{2}+1)},$

we have

$\frac{\left(\Gamma\left(\frac{a_{1}+a_{2}}{2}+1\right)\right)^{2}}{\Gamma(a_{1}+1)\,\Gamma(a_{2}+1)}\binom{(a_{1}+a_{2})/2+(a_{1}+a_{2})/2+2k}{k}\big{(}c_{k,\frac{a_{1}+a_{2}}{2}}\big{)}^{2}\\\
\leq\binom{a_{1}+a_{2}+2k}{k}c_{k,a_{1}}c_{k,a_{2}}\leq\binom{(a_{1}+a_{2})/2+(a_{1}+a_{2})/2+2k}{k}\big{(}c_{k,\frac{a_{1}+a_{2}}{2}}\big{)}^{2}$

and

$\frac{\left(\Gamma\left(\frac{a_{1}+a_{2}}{2}+1\right)\right)^{2}}{\Gamma(a_{1}+1)\,\Gamma(a_{2}+1)}\left(I_{\frac{a_{1}+a_{2}}{2}}(x)\right)^{2}\leq
I_{a_{1}}(x)I_{a_{2}}(x)\leq\left(I_{\frac{a_{1}+a_{2}}{2}}(x)\right)^{2}.$

With an elementary induction process, it can be deduced (3) for $n=2^{m}$.
When $n$ is not a power of two, we can find $m$ such that $n<2^{m}$. Taking
the values $b_{1}=a_{1}$, …, $b_{n}=a_{n}$, and

$b_{k}=\frac{a_{1}+\cdots+a_{n}}{n},\qquad n+1\leq k\leq 2^{m},$

from (3) with $n=2^{m}$, we have

$\frac{\left(\Gamma\left(\frac{b_{1}+\cdots+b_{2^{m}}}{2^{m}}+1\right)\right)^{2^{m}}}{\Gamma(b_{1}+1)\cdots\Gamma(b_{2^{m}}+1)}\left(I_{\frac{b_{1}+\cdots+b_{2^{m}}}{2^{m}}}(x)\right)^{2^{m}}\\\
\leq I_{b_{1}}(x)I_{b_{2}}(x)\cdots I_{b_{n}}(x)I_{b_{n+1}}(x)\cdots
I_{b_{2^{m}}}(x)\leq\left(I_{\frac{b_{1}+\cdots+b_{2^{m}}}{2^{m}}}(x)\right)^{2^{m}}.$

Finally, applying that

$\frac{b_{1}+\cdots+b_{2^{m}}}{2^{m}}=\frac{a_{1}+\cdots+a_{n}}{n}$

and

$I_{b_{n+1}}(x)\cdots
I_{b_{2^{m}}}(x)=\left(I_{\frac{a_{1}+\cdots+a_{n}}{n}}(x)\right)^{2^{m}-n}$

the general case of (3) follows. ∎

The result in the previous proposition is a particular case of [27, Theorem
2.7]. We have included this new proof because it is elementary and to do this
paper self-contained.

Taking the normalized modified Bessel functions of the first kind

$\mathcal{I}_{a}(x)=\frac{2^{a}\Gamma(a+1)}{x^{a}}I_{a}(x),$

that verify $\mathcal{I}_{a}(0)=1$, the left inequality in (3) reads

$\left(\mathcal{I}_{\frac{a_{1}+\cdots+a_{n}}{n}}(x)\right)^{n}\leq\mathcal{I}_{a_{1}}(x)\cdots\mathcal{I}_{a_{n}}(x).$

###  2.2. Some inequalities for the difference of modified Bessel functions of
first kind

It is easy to check that the modified Bessel functions decrease with the
order; i.e., $I_{a}(x)<I_{b}(x)$ when $-1<b<a$ and $x>0$. We will use this
fact sometimes along the paper without explicit mention to it.

In our next result we analyze the difference of two consecutive modified
Bessel functions of first kind.

######  Lemma 2.

For $a>-1$ and $x>0$ the inequalities

(5) $0<I_{a}(x)-I_{a+1}(x)<\frac{a+1}{x}I_{a}(x)$

hold.

###### Proof.

The lower bound is a consequence of the monotonicity respect to the order of
the functions $I_{a}(x)$. To obtain the upper bound, we use the inequalities
[22, Theorem 2]

(6) $f_{a+1}(x)<\frac{I_{a+1}(x)}{I_{a}(x)}<f_{a+1/2}(x),$

where

$f_{\alpha}(x)=\frac{x}{\alpha+\sqrt{\alpha^{2}+x^{2}}}$

and the upper bound holds for $a\geq-1/2$ and the lower one for $a\geq-1$.
Then, it is clear that

(7) $g_{a+1/2}(x)<1-\frac{I_{a+1}(x)}{I_{a}(x)}<g_{a+1}(x),$

with

$g_{\alpha}(x)=1-f_{\alpha}(x)=\frac{2\alpha}{\alpha+x+\sqrt{\alpha^{2}+x^{2}}}$

and now the upper bound holds for $a\geq-1$ and the lower one for $a\geq-1/2$.
Then, by using the upper bound in (7), for $a>-1$, we have

$I_{a}(x)-I_{a+1}(x)=\left(1-\frac{I_{a+1}(x)}{I_{a}(x)}\right)I_{a}(x)<g_{a+1}(x)I_{a}(x)\leq\frac{a+1}{x}I_{a}(x).\qed$

######  Lemma 3.

For $a\geq-1/2$ and $x>0$ the inequality

(8)
$\left|I_{a}(x)-2I_{a+1}(x)+I_{a+2}(x)\right|<\left(\frac{3}{2x}+\frac{(a+1)(a+2)}{x^{2}}\right)I_{a}(x)$

holds.

###### Proof.

We estimate

$\left|1-2\frac{I_{a+1}(x)}{I_{a}(x)}+\frac{I_{a+2}(x)}{I_{a}(x)}\right|.$

To do this we use the obvious identity

$1-2\frac{I_{a+1}(x)}{I_{a}(x)}+\frac{I_{a+2}(x)}{I_{a}(x)}=S_{1}+S_{2}$

with

$S_{1}=\frac{I_{a+2}(x)}{I_{a+1}(x)}-\frac{I_{a+1}(x)}{I_{a}(x)}$

and

$S_{2}=\left(1-\frac{I_{a+1}(x)}{I_{a}(x)}\right)\left(1-\frac{I_{a+2}(x)}{I_{a+1}(x)}\right).$

By using (6) and that $f_{\alpha}(x)<f_{\beta}(x)$ when $\beta<\alpha$, for
$a\geq-1/2$ we have

$|S_{1}|<\max\\{f_{a+1}(x)-f_{a+3/2}(x),f_{a+1/2}(x)-f_{a+2}(x)\\}=f_{a+1/2}(x)-f_{a+2}(x).$

Now, with the bound

(9)
$f_{a}(x)-f_{a+p}(x)=\frac{p+\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+(a+p)^{2}}}{x}\leq\frac{p}{x},\qquad
p>0$

we have $|S_{1}|<3/(2x)$.

For $S_{2}$ we use (7) to obtain that

$S_{2}<g_{a+1}(x)g_{a+2}(x)\leq\frac{(a+1)(a+2)}{x^{2}}$

and the proof is finished. ∎

Finally, we estimate an alternating sum of four consecutive modified Bessel
functions of first kind.

######  Lemma 4.

For $a\geq-1/2$ and $x>0$ the inequality

(10) $\left|I_{a}(x)-3I_{a+1}(x)+3I_{a+2}(x)-I_{a+3}(x)\right|<\\\
C\left(\frac{a+2}{x^{2}}+\frac{(a+1)(a+2)(a+3)}{x^{3}}\right)I_{a}(x)$

holds.

###### Proof.

As in the previous proof, we focus on an appropriate bound for

$\left|1-3\frac{I_{a+1}(x)}{I_{a}(x)}+3\frac{I_{a+2}(x)}{I_{a}(x)}-\frac{I_{a+3}(x)}{I_{a}(x)}\right|.$

In this time, we have

$1-3\frac{I_{a+1}(x)}{I_{a}(x)}+3\frac{I_{a+2}(x)}{I_{a}(x)}-\frac{I_{a+3}(x)}{I_{a}(x)}=T_{1}+T_{2}+T_{3},$

with

$T_{1}=-\frac{I_{a+1}(x)}{I_{a}(x)}+2\frac{I_{a+2}(x)}{I_{a+1}(x)}-\frac{I_{a+3}(x)}{I_{a+2}(x)},$
$T_{2}=\left(1-\frac{I_{a+1}(x)}{I_{a}(x)}\right)\left(\frac{I_{a+3}(x)}{I_{a+2}(x)}-\frac{I_{a+2}(x)}{I_{a+1}(x)}\right)\\\
+\left(1-\frac{I_{a+2}(x)}{I_{a+1}(x)}\right)\left(\frac{I_{a+3}(x)}{I_{a+2}(x)}-\frac{I_{a+1}(x)}{I_{a}(x)}\right)$

and

$T_{3}=\left(1-\frac{I_{a+1}(x)}{I_{a}(x)}\right)\left(1-\frac{I_{a+2}(x)}{I_{a+1}(x)}\right)\left(1-\frac{I_{a+3}(x)}{I_{a+2}(x)}\right).$

We are going to estimate $|T_{1}|$ by using continued fractions. Let

$\xi:=\langle
q_{0},q_{1},q_{2},\dots\rangle=q_{0}+\frac{1}{q_{1}+\frac{1}{q_{2}+\frac{1}{q_{3}+\cdots}}}$

be a continued fraction. If

$\langle q_{0},q_{1},\dots,q_{n}\rangle=\frac{P_{n}}{Q_{n}},$

it is known [14, Ch. 2, Theorem 9] that

(11) $\left|\xi-\frac{P_{n}}{Q_{n}}\right|\leq\frac{1}{Q_{n}Q_{n+1}},\qquad
n\geq 0.$

In [17, 10.33.1], we find the expansion

$\frac{I_{a+1}(x)}{I_{a}(x)}=\langle
q_{0}(a,x),q_{1}(a,x),q_{2}(a,x),\dots\rangle,\qquad x>0,\quad a>-1,$

with $q_{0}(a,x)=0$ and $q_{k}(a,x)=2(a+k)/x$ for $k\geq 1$. Then, taking
$n=2$ in (11),

$|T_{1}|\leq\frac{1}{Q_{2}(a,x)Q_{3}(a,x)}+\frac{2}{Q_{2}(a+1,x)Q_{3}(a+1,x)}+\frac{1}{Q_{2}(a+2,x)Q_{3}(a+2,x)}\\\
+\left|\frac{P_{2}(a,x)}{Q_{2}(a,x)}-2\frac{P_{2}(a+1,x)}{Q_{2}(a+1,x)}+\frac{P_{2}(a+2,x)}{Q_{2}(a+2,x)}\right|.$

By using that

$Q_{2}(a,x)=4(a+1)(a+2)+x^{2}\qquad\text{and}\qquad
Q_{3}(a,x)=4(a+2)(2(a+1)(a+3)+x^{2})$

and applying the inequalities

$(a+1)^{2}+x^{2}\leq Q_{2}(a,x)\leq Q_{2}(a+1,x)$

and

$4(a+2)((a+1)^{2}+x^{2})\leq Q_{3}(a,x)\leq Q_{3}(a+1,x),$

for $a>-1$ and $x>0$, we have

(12)
$\frac{1}{Q_{2}(a,x)Q_{3}(a,x)}+\frac{2}{Q_{2}(a+1,x)Q_{3}(a+1,x)}+\frac{1}{Q_{2}(a+2,x)Q_{3}(a+2,x)}\\\
\leq\frac{1}{(a+2)((a+1)^{2}+x^{2})^{2}}.$

Moreover, due to $P_{2}(a,x)=2(a+2)x$, for $a\geq-1/2$

$\frac{P_{2}(a,x)}{Q_{2}(a,x)}-2\frac{P_{2}(a+1,x)}{Q_{2}(a+1,x)}+\frac{P_{2}(a+2,x)}{Q_{2}(a+2,x)}\\\
=\frac{16x(4(a+2)(a+3)(a+4)-x^{2}(8+3a))}{Q_{2}(a,x)Q_{2}(a+1,x)Q_{2}(a+2,x)}$

and

(13)
$\left|\frac{P_{2}(a,x)}{Q_{2}(a,x)}-2\frac{P_{2}(a+1,x)}{Q_{2}(a+1,x)}+\frac{P_{2}(a+2,x)}{Q_{2}(a+2,x)}\right|\\\
\leq
C\frac{x(a+3)((a+2)(a+4)+x^{2})}{((a+1)^{2}+x^{2})^{3}}\leq\frac{C}{(a+1)^{2}+x^{2}}.$

In this way, from (12) and (13), we obtain that

$|T_{1}|\leq\frac{C}{x^{2}},\qquad a\geq-1/2.$

Now, from (6) and (7), it is clear that

$|T_{2}|\leq
g_{a+1}(x)(f_{a+3/2}(x)-f_{a+3}(x))+g_{a+2}(x)(f_{a+1/2}(x)-f_{a+3}(x))$

and applying (9) we have

$|T_{2}|\leq\frac{C}{x}(g_{a+1}(x)+g_{a+2}(x))\leq C\frac{a+2}{x^{2}}.$

Finally, from (7), the bound

$T_{3}<g_{a+1}(x)g_{a+2}(x)g_{a+3}(x)\leq\frac{(a+1)(a+2)(a+3)}{x^{3}}$

can be deduced and the proof of (10) is completed. ∎

###  2.3. A uniform bound for the modified Bessel functions of first kind

Now, we present a uniform bound respect to the order for the function
$I_{a}(x)$. It will be fundamental all along the paper.

######  Lemma 5.

For $a>-1/2$, let $\alpha$ be a real number such that $-1/2\leq\alpha<a$. Then
the inequality

(14) $x^{-\alpha}e^{-x}I_{a}(x)\leq\frac{C}{(a+1)^{2\alpha+1}}$

holds with a constant $C$ independent of $a$.

###### Proof.

From the identity [17, 10.32.2]

$I_{a}(x)=\frac{x^{a}}{\sqrt{\pi}2^{a}\Gamma(a+1/2)}\int_{-1}^{1}e^{-xs}(1-s^{2})^{a-1/2}\,ds,\qquad
a>-1/2,$

by using the change of variable $x(1+s)=2w$, we can deduce that

$\displaystyle x^{-\alpha}e^{-x}I_{a}(x)$
$\displaystyle=\frac{2^{a}x^{-\alpha-1/2}}{\sqrt{\pi}\Gamma(a+1/2)}\int_{0}^{x}e^{-2w}w^{a-1/2}\left(1-\frac{w}{x}\right)^{a-1/2}\,dw$
$\displaystyle=\frac{2^{a}}{\sqrt{\pi}\Gamma(a+1/2)}\int_{0}^{x}e^{-2w}w^{a-1-\alpha}\left(\frac{w}{x}\right)^{\alpha+1/2}\left(1-\frac{w}{x}\right)^{a-1/2}\,dw.$

We start with the case $a\geq 1/2$. Using the inequality

$z^{p}(1-z)^{q}\leq\frac{p^{p}q^{q}}{(p+q)^{p+q}},\qquad p,q\geq 0,\quad
0<z<1,$

where in the cases $p=0$, $q=0$, and $p=q=0$ the bound has to be understood as
one, we have

$\left(\frac{w}{x}\right)^{\alpha+1/2}\left(1-\frac{w}{x}\right)^{a-1/2}\leq\frac{C}{a^{\alpha+1/2}},$

with $C$ depending on $\alpha$ but not on $a$, and

$\displaystyle x^{-\alpha}e^{-x}I_{a}(x)$ $\displaystyle\leq
C\frac{2^{a}}{\Gamma(a+1/2)a^{\alpha+1/2}}\int_{0}^{t}e^{-2w}w^{a-1-\alpha}\,dw$
$\displaystyle\leq\frac{C}{\Gamma(a+1/2)a^{\alpha+1/2}}\int_{0}^{\infty}e^{-r}r^{a-1-\alpha}\,dr$
$\displaystyle=C\frac{\Gamma(a-\alpha)}{\Gamma(a+1/2)a^{\alpha+1/2}}\leq\frac{C}{a^{2\alpha+1}},$

where in the last step we have applied that $\Gamma(r+p)/\Gamma(r+q)\sim
r^{p-q}$, and this is enough for our purpose.

For $-1/2<a<1/2$ we have to prove that $x^{-\alpha}e^{-x}I_{a}(x)\leq C$. We
consider the decomposition

$x^{-\alpha}e^{-x}I_{a}(x)=\frac{2^{a}}{\sqrt{\pi}\Gamma(a+1/2)}\\\
\times\left(\int_{0}^{x/2}+\int_{x/2}^{x}e^{-2w}w^{a-1-\alpha}\left(\frac{w}{x}\right)^{\alpha+1/2}\left(1-\frac{w}{x}\right)^{a-1/2}\,dw\right).$

The first integral can be controlled easily. In fact,

$\int_{0}^{x/2}e^{-2w}w^{a-1-\alpha}\left(\frac{w}{x}\right)^{\alpha+1/2}\left(1-\frac{w}{x}\right)^{a-1/2}\,dw\\\
\leq\frac{1}{2^{\alpha+a}}\int_{0}^{x/2}e^{-2w}w^{a-1-\alpha}\,dw\leq\frac{\Gamma(a-\alpha)}{2^{2a}}.$

For the second integral, with the change of variable $w=x(1-r)$, we have

$\int_{x/2}^{x}e^{-2w}w^{a-1-\alpha}\left(\frac{w}{x}\right)^{\alpha+1/2}\left(1-\frac{w}{x}\right)^{a-1/2}\,dw\\\
\begin{aligned}
&\leq\left(\frac{x}{2}\right)^{a-1-\alpha}\int_{x/2}^{x}e^{-2w}\left(1-\frac{w}{x}\right)^{a-1/2}\,dw\\\
&=\frac{x^{a-\alpha}}{2^{a-1-\alpha}}\int_{0}^{1/2}e^{-2x(1-r)}r^{a-1/2}\,dr\\\
&\leq\frac{x^{a-\alpha}}{2^{a-1-\alpha}}e^{-x}\int_{0}^{1/2}r^{a-1/2}\,dr\leq\frac{(a-\alpha)^{a-\alpha}e^{-a+\alpha}}{2^{a-1-\alpha}(a+1/2)},\end{aligned}$

where in the last step we have applied that $x^{b}e^{-x}\leq b^{b}e^{-b}$ for
$x,b>0$. Then, using that $0<a-\alpha<1$

$x^{-\alpha}e^{-x}I_{a}(x)\leq\frac{2^{a}}{\sqrt{\pi}\Gamma(a+1/2)}\left(\frac{\Gamma(a-\alpha)}{2^{2a}}+\frac{(a-\alpha)^{a-\alpha}e^{-a+\alpha}}{2^{a-1-\alpha}(a+1/2)}\right)\leq
C$

and the proof is finished. ∎

###  2.4. Further properties of the modified Bessel functions of first kind

In this subsection we collect some known properties of the modified Bessel
functions which will be used in this paper.

For each $k\in\mathbb{Z}$, it is verified that (see [17, 10.27.1])

(15) $I_{-k}(t)=I_{k}(t).$

Moreover, from its definition it is clear that $I_{k}(t)>0$ for $t>0$,

(16) $I_{k}(0)=0,\qquad k\not=0,$

and $I_{0}(0)=1$.

We have the generating function (see [17, 10.35.1])

(17) $e^{-t(u+u^{-1})/2}=\sum_{k\in\mathbb{Z}}u^{k}I_{k}(t)$

and it implies, taking $u=1$,

(18) $\sum_{k\in\mathbb{Z}}e^{-t}I_{k}(t)=1.$

Another easy consequence of the given generating function is the Neumann’s
identity (see [10, Chapter II, 7.10])

(19) $I_{n}(t_{1}+t_{2})=\sum_{k\in\mathbb{Z}}I_{k}(t_{1})I_{n-k}(t_{2}).$

It can be obtained from (17) by using the Cauchy product of two series.

The asymptotic expansion for large order and $t$ fix (see [17, 10.41.1])

(20)
$I_{\nu}(t)\sim\frac{1}{\sqrt{2\pi\nu}}\left(\frac{et}{2\nu}\right)^{\nu}$

will be used in some points.

##  3\. The heat semigroup and beyond

Basic properties of the operator (1) can be deduced easily. First, by using
(18), we have

(21) $\sum_{\mathbf{k}\in\mathbb{Z}^{N}}G_{t,N}(\mathbf{k})=1$

and, due to the positivity of $G_{t,N}$,

$|W_{t}f(\mathbf{n})|\leq\|f\|_{\ell^{\infty}(\mathbb{Z}^{N})}\sum_{\mathbf{k}\in\mathbb{Z}^{N}}G_{t,N}(\mathbf{n}-\mathbf{k})\leq\|f\|_{\ell^{\infty}(\mathbb{Z}^{N})}.$

So $W_{t}$ is well defined for any sequence in
$\ell^{\infty}(\mathbb{Z}^{N})$. Moreover, from (16) and (19), we deduce the
properties

$W_{0}f(\mathbf{n})=f(\mathbf{n})\qquad\text{and}\qquad
W_{t_{1}}W_{t_{2}}f(\mathbf{n})=W_{t_{1}+t_{2}}f(\mathbf{n})$

and the family of operators $W_{t}$ is a semigroup. By using again the
positive of the heat kernel, we obtain that $W_{t}$ is a positive operator (it
is positive for positive sequences) and from (18) it is possible to prove that
it is Markovian; i. e.,

$W_{t}\mathbf{1}(\mathbf{n})=\mathbf{1}(\mathbf{n}),$

where $\mathbf{1}(\mathbf{n})$ is the sequence having one in all its entries.

Defining the convolution on $\mathbb{Z}^{N}$ by

$f\ast
g(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{n}-\mathbf{k})g(\mathbf{k}),$

we can observe that

$W_{t}f(\mathbf{n})=G_{t,N}\ast f(\mathbf{n}).$

Then, by Young’s inequality

(22) $\|f\ast
g\|_{\ell^{p}(\mathbb{Z}^{N})}\leq\|f\|_{\ell^{q}(\mathbb{Z}^{N})}\|g\|_{\ell^{r}(\mathbb{Z}^{N})},$

with $1/q+1/r=1+1/p$ and $1\leq p,q,r\leq\infty$, we obtain that

(23)
$\|W_{t}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq\|G_{t,N}\|_{\ell^{1}(\mathbb{Z}^{N})}\|f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq\|f\|_{\ell^{p}(\mathbb{Z}^{N})},\qquad
1\leq p\leq\infty$

where in the last step we have applied (21).

For a sequence $f\in\ell^{1}(\mathbb{Z})$, its Fourier transform is given by

$\mathcal{F}f(x)=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})e^{2\pi
i\langle\mathbf{k},x\rangle},\qquad x\in[-1/2,1/2]^{N}.$

It is well known that $\mathcal{F}$ can be extended as an isometry from the
space $\ell^{2}(\mathbb{Z}^{N})$ into $L^{2}([-1/2,1/2]^{N})$ with inverse

$\mathcal{F}^{-1}f(\mathbf{n})=\int_{[-1/2,1/2]^{N}}f(x)e^{-2\pi
i\langle\mathbf{n},x\rangle}\,dx,\qquad f\in L^{2}([-1/2,1/2]^{N}).$

Moreover, the Parseval’s identity

$\|\mathcal{F}f\|_{L^{2}([-1/2,1/2]^{N})}=\|f\|_{\ell^{2}(\mathbb{Z}^{N})}$

holds.

Now, from the integral representation [17, 10.32.3]

(24)
$I_{n}(t)=\frac{1}{\pi}\int_{0}^{\pi}e^{t\cos\theta}\cos(n\theta)\,d\theta,\qquad
n\in\mathbb{Z},$

we get

$e^{-t}I_{n}(t)=\int_{-1/2}^{1/2}e^{-2t\sin^{2}(\pi\theta)}e^{-2\pi
in\theta}\,d\theta.$

and

(25)
$G_{t,N}(\mathbf{n})=\int_{[-1/2,1/2]^{N}}e^{-4t\sum_{k=1}^{N}\sin^{2}(\pi
x_{k})}e^{-2\pi i\langle\mathbf{n},x\rangle}\,dx.$

Then

$\mathcal{F}(G_{t,N})(x)=e^{-4t\sum_{k=1}^{N}\sin^{2}(\pi x_{k})}.$

Hence, using

$\mathcal{F}(f\ast g)=\mathcal{F}f\cdot\mathcal{F}g,$

we deduce that

$\lim_{t\to 0}\|W_{t}f-f\|_{\ell^{2}(\mathbb{Z}^{N})}=\lim_{t\to
0}\left\|(e^{-4t\sum_{k=1}^{N}\sin^{2}(\pi
x_{k})}-1)\mathcal{F}f\right\|_{L^{2}([-1/2,1/2]^{N})}=0$

and we have the convergence in $\ell^{2}(\mathbb{Z}^{N})$ of the heat
semigroup. Moreover, due to the structure of the spaces
$\ell^{p}(\mathbb{Z}^{N})$, we have the pointwise convergence

$W_{t}f(\mathbf{n})\longrightarrow
f(\mathbf{n}),\qquad\mathbf{n}\in\mathbb{Z}^{N}.$

Note that, by using the inequality $1-e^{-s}\leq s$ with $s\geq 0$, it is
verified that

(26) $|W_{t}f(\mathbf{n})-f(\mathbf{n})|\leq
Ct\|f\|_{\ell^{2}(\mathbb{Z}^{N})}.$

###  3.1. Decay of the heat semigroup

Now we focus on the study of the decay of the heat semigroup in time. This is
a natural question when we are deling with this operator. We start with a
$\ell^{q}-\ell^{p}$ estimate. The analogous for $\mathbb{R}^{N}$ can be seen,
for example, in [5, Ch. 3, p. 44]. Our discrete result can be found in [13]
but we include here it to do this paper self-contained and to provide a new
and simple proof.

######  Theorem 6.

For $1\leq q\leq p\leq\infty$, the inequality

(27) $\|W_{t}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
Ct^{-N/2(1/q-1/p)}\|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

holds.

###### Proof.

From the Young’s inequality for the convolution, it is enough to prove that

(28) $\|G_{t,N}\|_{\ell^{r}(\mathbb{Z}^{N})}\leq Ct^{-N/2(1-1/r)}$

and to do that we use Littlewood’s inequality

(29)
$\|f\|_{\ell^{r}(\mathbb{Z}^{N})}\leq\|f\|_{\ell^{1}(\mathbb{Z}^{N})}^{1/r}\|f\|_{\ell^{\infty}(\mathbb{Z}^{N})}^{1-1/r},\qquad
1\leq r\leq\infty,$

the identity (21) and the bound $G_{t,N}(\mathbf{n})\leq Ct^{-N/2}$. This last
estimate can be deduced from (3) and (14) with $\alpha=-1/2$. Indeed, using
the notations $\mathbf{n}_{+}=(|n_{1}|,\dots,|n_{N}|)$ and
$|\mathbf{n}|=|n_{1}|+\cdots+|n_{N}|$ and taking into account that
$G_{t,N}(\mathbf{n})=G_{t,N}(\mathbf{n}_{+})$, it is verified that

(30)
$G_{t,N}(\mathbf{n})\leq\left(e^{-2t}I_{|\mathbf{n}|/N}(2t)\right)^{N}=t^{-N/2}\left(t^{1/2}e^{-2t}I_{|\mathbf{n}|/N}(2t)\right)^{N}\leq
Ct^{-N/2}.\qed$

######  Remark 1.

(a) By using (3) and (14) with $\alpha=0$, it is easy to check that

(31) $G_{t,N}(\mathbf{n})\leq C(1+|\mathbf{n}|)^{-N}\leq C.$

Then, for $0<t<T$, with $T$ finite, we deduce the bound

(32) $\|G_{t,N}\|_{\ell^{r}(\mathbb{Z}^{N})}\leq C$

and the estimate

$\|W_{t}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq C\|f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

Moreover, combining (27) and (32), it is possible to obtain the inequality

$\|W_{t}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
C\min\\{1,t^{-N/2(1/q-1/p)}\\}\|f\|_{\ell^{q}(\mathbb{Z}^{N})},\qquad 1\leq
q\leq p\leq\infty.$

(b) From the previous theorem it is possible to prove that for $t>1$

(33) $\|G_{t,N}\|_{\ell^{r}(\mathbb{Z}^{N})}\sim t^{-N/2(1-1/r)},\qquad 1\leq
r\leq 2.$

The upper bound is given in (28). To prove the lower one, we consider the
sequence $f(\mathbf{k})=G_{t,N}(\mathbf{k})$ to obtain

$W_{t}f(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}G_{t,N}(\mathbf{k})G_{t,N}(\mathbf{n-k})=G_{2t,N}(\mathbf{n}),$

where in the las step we have applied (19). Then, from (27), we deduce that

$\|G_{2t,N}\|_{\ell^{2}(\mathbb{Z}^{N})}\leq
t^{-N/2(1/r-1/2)}\|G_{t,N}\|_{\ell^{r}(\mathbb{Z}^{N})},\qquad 1\leq r\leq 2.$

In this way, proving that

(34) $\|G_{2t,N}\|_{\ell^{2}(\mathbb{Z}^{N})}\geq Ct^{-N/4}$

the proof of the lower bound in (33) will be completed. From Parseval’s
identity

$\displaystyle\|G_{2t,N}\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}$
$\displaystyle=\int_{[-1/2,1/2]^{N}}e^{-8t\sum_{k=1}^{N}\sin^{2}(\pi
x_{k})}\,dx=\left(\int_{-1/2}^{1/2}e^{-8t\sin^{2}(\pi s)}\,ds\right)^{N}$
$\displaystyle\geq\left(\int_{-1/(2\sqrt{t})}^{1/(2\sqrt{t})}e^{-8t\sin^{2}(\pi
s)}\,ds\right)^{N}\geq Ct^{-N/2},$

where we have used that $e^{-8t\sin^{2}(\pi s)}\geq K$ for
$s\in[-1/(2\sqrt{t}),1/(2\sqrt{t})]$, and the proof of (34) is finished.

The next result contains the main novelty of this subsection. For the
classical heat equation in $\mathbb{R}^{N}$, with an initial data having
polynomial decay at infinity, in [8] the authors study how the mass of the
solution is distributed for large value of $t$. In fact, it is proved the
inequality

$\left\|H_{t}f-\left(\int_{\mathbb{R}^{N}}f(x)\,dx\right)G_{t}\right\|_{L^{p}(\mathbb{R}^{N})}\leq
Ct^{-1/2-N/2(1/q-1/p)}\||\cdot|f\|_{L^{q}(\mathbb{R}^{N})},$

where $H_{t}$ is the solution of the heat equation, $G_{t}$ its kernel (the
Gaussian function), $1\leq q\leq p\leq\infty$, $1\leq q<N/(N-1)$, $f\in
L^{1}(\mathbb{R}^{N})$, and $|\cdot|f\in L^{q}(\mathbb{R}^{N})$. Remember that
the mass of the solution $H_{t}$ is a conservative quantity; i.e.,

$\int_{\mathbb{R}^{N}}H_{t}f(x)\,dx=\int_{\mathbb{R}^{N}}f(x)\,dx$

Note that (21) implies that

$\sum_{\mathbf{k}\in\mathbb{Z}^{N}}W_{t}f(\mathbf{k})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k}),$

so in the discrete setting the mass of the solution of the heat equation is
conservative also. The discrete analogous of this result is given in [13] only
in the case $q=1$, with $1\leq p\leq\infty$, and the case $1<q\leq p$ is left
as an open problem. Our next result closes this problem completely.

######  Theorem 7.

Let $N\geq 1$, $1\leq q\leq p\leq\infty$, $1\leq q<N/(N-1)$, and
$|\cdot|f\in\ell^{q}(\mathbb{Z}^{N})$. Then the inequality

(35)
$\left\|W_{t}f-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)G_{t,N}\right\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
Ct^{-1/2-N/2(1/q-1/p)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

holds.

To prove our result we need some preliminart lemmas.

######  Lemma 8.

Let $N\geq 1$, $\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}$ such that
$|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|$, and

$H_{t,N}(z)=\frac{z}{t}\left(G_{t,1}\left(z\right)\right)^{N},\qquad z>0.$

Then the inequality

(36) $|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m})|\leq
C|\mathbf{n}-\mathbf{m}|H_{t,N}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{K}\right)$

holds with constants $C$ and $K>1$ independent of $\mathbf{n}$ and
$\mathbf{m}$.

###### Proof.

Without lost of generality we suppose that $n_{i}\not=m_{i}$ for
$i=1,\dots,N$. In the opposite case the situation can be obviously reduced to
this one easily. Indeed, let us suppose that $n_{i}=m_{i}$ for $j$ values of
$i$ with $0<j<N$, then we consider the decompositions

$\mathbf{n}=\mathbf{n}_{1}\cup\mathbf{n}_{2}\qquad\text{and}\qquad\mathbf{m}=\mathbf{n}_{1}\cup\mathbf{m}_{2},$

where $\mathbf{n}_{1}$ are the common values of $\mathbf{n}$ and $\mathbf{m}$
and $\mathbf{n}_{2}$ and $\mathbf{m}_{2}$ are the other components. Then, if
we suppose that the result has been proved for different values, we have

$\displaystyle|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m})|$
$\displaystyle=G_{t,j}(\mathbf{n}_{1})|G_{t,N-j}(\mathbf{n}_{2})-G_{t,N-j}(\mathbf{m}_{2})|$
$\displaystyle\leq
CG_{t,j}(\mathbf{n}_{1})|\mathbf{n}_{2}-\mathbf{m}_{2}|H_{t,N-j}\left(\frac{|\mathbf{n}_{2}|+|\mathbf{m}_{2}|}{K}\right)$
$\displaystyle\leq
C|\mathbf{n}-\mathbf{m}|H_{t,N}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{2KN}\right).$

To check the last inequality we use (3) to have

$G_{t,j}(\mathbf{n}_{1})\left(G_{t,1}\left(\frac{|\mathbf{n}_{2}|+|\mathbf{m}_{2}|}{K}\right)\right)^{N-j}\\\
\leq\left(G_{t,1}\left(\frac{(N-j)(|\mathbf{n}_{2}|+|\mathbf{m}_{2}|)/K+|\mathbf{n}_{1}|}{N}\right)\right)^{N},$

then the inequality follows applying the monotonicity of the functions $I_{a}$
and the estimate

$\displaystyle\frac{(N-j)(|\mathbf{n}_{2}|+|\mathbf{m}_{2}|)}{K}+|\mathbf{n}_{1}|$
$\displaystyle\geq\frac{(N-j)(|\mathbf{n}_{2}|+|\mathbf{m}_{2}|)+|\mathbf{n}_{1}|}{K}$
$\displaystyle\geq\frac{|\mathbf{n}_{2}|+|\mathbf{m}_{2}|+2|\mathbf{n}_{1}|}{2K}=\frac{|\mathbf{n}|+|\mathbf{m}|}{2K}.$

From (15), it is clear that $G_{t,N}(\mathbf{n})=G_{t,N}(\mathbf{n}_{+})$.
Moreover, we have to observe that $G_{t,N}(\mathbf{n})$ is invariant under
permutations of $\mathbf{n}$. With these previous reductions, we can consider
$\mathbf{n}=(n_{1},\dots,n_{N})$ and $\mathbf{m}=(m_{1},\cdots,m_{N})$ with

(37) $0\leq n_{1}\leq n_{2}<\cdots\leq n_{N}\qquad\text{and}\qquad 0\leq
m_{N}\leq m_{N-1}\leq\cdots\leq m_{1}.$

An important tool in the proof will be the identity

(38)
$G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m})=\sum_{i=1}^{N}\left(G_{t,1}(n_{i})-G_{t,1}(m_{i})\right)\prod_{j=1}^{i-1}G_{t,1}(m_{j})\prod_{k=i+1}^{N}G_{t,1}(n_{k}),$

where the empty products has to be understood as one. The proof of this
relation we can done induction on $N$. To do that we consider
$\tilde{\mathbf{n}}=(\mathbf{n},n_{N+1})$ and
$\tilde{\mathbf{m}}=(\mathbf{m},m_{N+1})$ and we apply the identity

$G_{t,N+1}(\tilde{\mathbf{n}})-G_{t,N+1}(\tilde{\mathbf{m}})=(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))G_{t,1}(n_{N+1})\\\
+G_{t,N}(\mathbf{m})(G_{t,1}(n_{N+1})-G_{t,1}(m_{N+1})).$

For $n_{i}<m_{i}$, applying (5) and the monotonicity of the modified Bessel
with respect to order, we have

$\displaystyle|G_{t,1}(n_{i})-G_{t,1}(m_{i})|$
$\displaystyle\leq\sum_{j=n_{i}}^{m_{i}-1}|\delta^{+}G_{t,1}(j)|$
$\displaystyle\leq\frac{1}{t}\sum_{j=n_{i}}^{m_{i}-1}(j+1)G_{t,1}(j)\leq\frac{(m_{i}-n_{i})(m_{i}+n_{i}+1)}{t}G_{t,1}(n_{i}).$

Then, in general, we obtain the bound

(39)
$|G_{t,1}(n_{i})-G_{t,1}(m_{i})|\leq\frac{|m_{i}-n_{i}|(m_{i}+n_{i}+1)}{2t}\max\\{G_{t,1}(n_{i}),G_{t,1}(m_{i})\\}$

and, from (38),

(40) $|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m})|\\\
\leq\sum_{i=1}^{N}\frac{|m_{i}-n_{i}|(m_{i}+n_{i}+1)}{2t}\max\\{G_{t,1}(n_{i}),G_{t,1}(m_{i})\\}\prod_{j=1}^{i-1}G_{t,1}(m_{i})\kern-2.5pt\prod_{k=i+1}^{N}\kern-4.5ptG_{t,1}(n_{k}).$

Now, applying (3), we deduce the estimate

$\max\\{G_{t,1}(n_{i}),G_{t,1}(m_{i})\\}\prod_{j=1}^{i-1}G_{t,1}(m_{j})\prod_{k=i+1}^{N}G_{t,1}(n_{k})\\\
\leq\left(G_{t,1}\left(\frac{m_{1}+\cdots+m_{i-1}+\min\\{m_{i},n_{i}\\}+n_{i+1}+\cdots+n_{N}}{N}\right)\right)^{N}.$

From the condition $|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|$, we can deduce that
$|\mathbf{n}|/2<|\mathbf{m}|<3|\mathbf{n}|/2$. Then if $n_{1}<m_{1}$

$\min\\{m_{1},n_{1}\\}+n_{2}+\cdots+n_{N}=|\mathbf{n}|>\frac{2}{5}(|\mathbf{n}|+|\mathbf{m}|),$

and if $m_{1}<n_{1}$, by our supposition (37), we can prove easily that

$\min\\{m_{1},n_{1}\\}+n_{2}+\cdots+n_{N}\geq\frac{|\mathbf{n}|+|\mathbf{m}|}{N+1}.$

Similarly, for $i=2,\dots,N-1$, by applying again (37), we have

$m_{1}+\cdots+m_{i-1}+\min\\{m_{i},n_{i}\\}+n_{i+1}+\cdots+n_{N}\geq\frac{|\mathbf{n}|+|\mathbf{m}|}{N+1}.$

Moreover, for $n_{N}<m_{N}$ it is verified that

$m_{1}+m_{2}+\cdots+\min\\{m_{N},n_{N}\\}\geq\frac{|\mathbf{n}|+|\mathbf{m}|}{N+1},$

and for $m_{N}<n_{N}$

$m_{1}+m_{2}+\cdots+\min\\{m_{N},n_{N}\\}=|\mathbf{m}|>\frac{|\mathbf{n}|+|\mathbf{m}|}{3}.$

As conclusion,

$\frac{m_{1}+\cdots+m_{i-1}+\min\\{m_{i},n_{i}\\}+n_{i+1}+\cdots+n_{N}}{N}\geq\frac{|\mathbf{n}|+|\mathbf{m}|}{K}$

with $K>1$. In this way

$\max\\{G_{t,1}(n_{i}),G_{t,1}(m_{i})\\}\prod_{j=1}^{i-1}G_{t,1}(m_{j})\prod_{k=i+1}^{N}G_{t,1}(n_{k})\leq\left(G_{t,1}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{K}\right)\right)^{N}$

and by (40) the proof is completed. ∎

######  Remark 2.

(a) From our proof of (36) it is clear that it holds for
$|\mathbf{n}|\sim|\mathbf{m}|$ also.

(b) From (39) and (40), using (14) with $\alpha=-1/2$, it is easy to deduce
that

(41) $|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m})|\leq
C|\mathbf{m}-\mathbf{n}|(|\mathbf{n}|+|\mathbf{m}|)t^{-N/2-1},\qquad\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}.$

(c) Moreover, we can check that for $|\mathbf{n}|\geq B|\mathbf{m}|$, for some
positive constant $B$, the inequality

(42) $|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m})|\leq
C\frac{|\mathbf{n}-\mathbf{m}|(|\mathbf{n}|+|\mathbf{m}|)}{t}\left(G_{t,1}\left(\frac{|\mathbf{m}|}{K}\right)\right)^{N},$

holds for some constant $K$.

The following estimate will be useful in the proof of Theorem 7.

######  Lemma 9.

The inequality

(43)
$\frac{|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{k})|}{|\mathbf{n}-\mathbf{k}|}\leq
Ct^{-1/2-N/2}$

holds

###### Proof.

From (25) we have

$\displaystyle\frac{|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{k})|}{|\mathbf{n}-\mathbf{k}|}$
$\displaystyle\leq\int_{[-1/2,1/2]^{N}}e^{-4t\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})}\frac{|e^{-2\pi
i\langle\mathbf{n}-\mathbf{k},x\rangle}-1|}{|\mathbf{n}-\mathbf{k}|}\,dx$
$\displaystyle\leq C\int_{[-1/2,1/2]^{N}}e^{-4t\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})}|x|\,dx,$

where we have applied that $|(e^{-iz}-1)/z|\leq C$ for $z\in\mathbb{R}$. Now,
with the estimates

$\int_{-1/2}^{1/2}e^{-4t\sin^{2}(\pi x)}\,dx\leq
Ct^{-1/2}\qquad\text{and}\qquad\int_{-1/2}^{1/2}e^{-4t\sin^{2}(\pi
x)}|x|\,dx\leq Ct^{-1}$

the result is obtained. ∎

The function $H_{t,N}$ in Lemma (8) will appear in convolution with a specific
sequence in the proof of Theorem (7), so we need the size of its norm.

######  Lemma 10.

For $N\geq 1$ and $1\leq r\leq\infty$, it is verified that

(44)
$\left\|H_{t,N}\left(\frac{|\cdot|}{K}\right)\right\|_{\ell^{r}(\mathbb{Z}^{N})}\leq
Ct^{-1/2-N/2(1-1/r)},$

with the constant $C$ being independent of $t$.

###### Proof.

From (14) with $\alpha=-1/2+1/(2N)$, we have

(45) $\left|H_{t,N}\left(\frac{|\mathbf{n}|}{K}\right)\right|\leq
Ct^{-N/2-1/2}|\mathbf{n}|\left(t^{1/2-1/(2N)}G_{t,1}\left(\frac{|\mathbf{n}|}{K}\right)\right)^{N}\leq
Ct^{-N/2-1/2}$

and

$\left\|H_{t,N}\left(\frac{|\cdot|}{K}\right)\right\|_{\ell^{\infty}(\mathbb{Z}^{N})}\leq
Ct^{-N/2-1/2},$

By using again (14) with $\alpha=1/N$ we deduce that

(46) $\left|H_{t,N}\left(\frac{|\mathbf{n}|}{K}\right)\right|\leq
C|\mathbf{n}|\left(t^{-1/N}G_{t,1}\left(\frac{|\mathbf{n}|}{K}\right)\right)^{N}\leq\frac{C}{(|\mathbf{n}|+1)^{N+1}}.$

Now from the previous estimate, for $0<t\leq T$ with $T<100$, for example, it
is obtained that

$\left\|H_{t,N}\left(\frac{|\cdot|}{K}\right)\right\|_{\ell^{1}(\mathbb{Z}^{N})}\leq
C\sum_{\mathbf{n}\in\mathbb{Z}^{N}}\frac{1}{(|\mathbf{n}|+1)^{N+1}}\leq
C\sum_{m=1}^{\infty}\frac{1}{m^{2}}\leq C.$

For $t>T$ we consider the decomposition

$\left\|H_{t,N}\left(\frac{|\cdot|}{K}\right)\right\|_{\ell^{1}(\mathbb{Z}^{N})}=S_{1}+S_{2},$

where

$S_{1}=\sum_{|\mathbf{n}|\leq
n_{t}}\left|H_{t,N}\left(\frac{|\mathbf{n}|}{K}\right)\right|\qquad\text{and}\qquad
S_{2}=\sum_{|\mathbf{n}|>n_{t}}\left|H_{t,N}\left(\frac{|\mathbf{n}|}{K}\right)\right|,$

where $n_{t}=\lfloor\sqrt{t}\rfloor$ (this notation will used in some points
along the paper without explicit mention to it). For $S_{1}$, applying (45),
we have

$S_{1}\leq Ct^{-N/2-1/2}\sum_{|\mathbf{n}|\leq n_{t}}1\leq
Ct^{-N/2-1/2}\sum_{m=0}^{n_{t}}(m+1)^{N-1}\leq Ct^{-1/2}.$

Now, by using (46), for $S_{2}$ we obtain the estimate

$S_{2}\leq C\sum_{|\mathbf{n}|>n_{t}}\frac{1}{|\mathbf{n}|^{N+1}}\leq
C\sum_{m=n_{t}}^{\infty}\frac{1}{m^{2}}\leq Ct^{-1/2}.$

In this way

$\left\|H_{t,N}\left(\frac{|\cdot|}{K}\right)\right\|_{\ell^{1}(\mathbb{Z}^{N})}\leq
Ct^{-1/2}.$

and the result follows from (29). ∎

In some points we will need the weighted estimate of the heat kernel in the
following lemma.

######  Lemma 11.

For $N\geq 1$, $1\leq r\leq\infty$, and $N(1-r)<\gamma<N$, it is verified that

(47)
$\left\|\left(G_{t,1}\left(\frac{|\cdot|}{N}\right)\right)^{N}\right\|_{\ell^{r}(\mathbb{Z}^{N},(|\mathbf{n}|+1)^{-\gamma})}\leq
Ct^{-1/2((\gamma-N)/r+N)},$

with the constant $C$ being independent of $t$.

###### Proof.

From the bound (30) the result for $r=\infty$ is clear, so we analyze the
estimate for $1\leq r<\infty$. From (14) with $\alpha=-1/2$ and $\alpha=0$, we
get

$\left\|\left(G_{t,1}\left(\frac{|\cdot|}{N}\right)\right)^{N}\right\|^{r}_{\ell^{r}(\mathbb{Z}^{N},(|\mathbf{n}|+1)^{-\gamma})}\\\
\begin{aligned} &\leq C\left(t^{-Nr/2}\sum_{|\mathbf{n}|\leq
n_{t}}(|\mathbf{n}|+1)^{-\gamma}+\sum_{|\mathbf{n}|\geq
n_{t}}(|\mathbf{n}|+1)^{-\gamma-Nr}\right)\\\ &\leq
C\left(t^{-Nr/2}\sum_{m=0}^{n_{t}}(m+1)^{-\gamma+N-1}+\sum_{m=n_{t}}^{\infty}(m+1)^{-\gamma-
Nr+N-1}\right)\\\ &\leq Ct^{-1/2(\gamma-N+Nr)}\end{aligned}$

finishing the proof. ∎

The proof of our main result will be divided in the cases $N=1$ and $N>1$ and
to deal with the first one we will apply the next estimate.

######  Lemma 12.

For $1\leq r\leq\infty$, the inequality

(48) $\|\delta^{+}G_{t,1}\|_{\ell^{r}(\mathbb{Z})}\leq Ct^{-1/2-(1-1/r)/2}$

holds.

###### Proof.

As usual, we prove the inequality for $r=1$ and $r=\infty$ and the result
follows from (29). From (5) and (14) with $\alpha=0$, we have

$|\delta^{+}G_{t,1}(n)|\leq Ct^{-1}|n|G_{t,1}(n)\leq Ct^{-1}$

and this is enough for $r=\infty$. For $r=1$ we use (5) and decompose the sum

$\|\delta^{+}G_{t,1}\|_{\ell^{1}(\mathbb{Z})}\leq Ct^{-1}(B_{1}+B_{2}),$

where

$B_{1}=\sum_{|n|\leq n_{t}}|n|G_{t,1}(n)\qquad\text{and}\qquad
B_{2}=\sum_{|n|>n_{t}}|n|G_{t,1}(n).$

To treat $B_{1}$, by using (14) with $\alpha=-1/2$, we deduce that

$t^{-1}B_{1}\leq t^{-3/2}\sum_{|n|\leq n_{t}}|n|\leq Ct^{-1/2}$

and this is enough for our purpose. The factor $B_{2}$ can be analyzed with
the estimate (14) taking $\alpha=1$. Indeed,

$t^{-1}B_{2}\leq C\sum_{|n|>n_{t}}\frac{1}{n^{2}}\leq Ct^{-1/2}$

and the proof is completed. ∎

###### Proof of the Theorem 7.

From the restriction $1\leq q<N/(N-1)$, applying Hölder inequality, we have

$\|f\|_{\ell^{1}(\mathbb{Z}^{N})}\leq
C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})},$

so, in particular $f\in\ell^{1}(\mathbb{Z}^{N})$ and the left hand side of
(35) is finite.

We start showing the inequality (35) for $N=1$. In [1, Theorem 6.1] we find
(with our notation) the identity

$W_{t}f(n)-\left(\sum_{n\in\mathbb{Z}}f(n)\right)G_{t,1}(n)=\left(\delta^{+}G_{t,1}\right)\ast
F(n),$

where

$F(n)=\sum_{j=-\infty}^{n-1}f(j),\quad\text{for $n\leq
0$},\qquad\text{and}\qquad F(n)=-\sum_{j=n}^{\infty}f(j),\quad\text{for $n\geq
1$}.$

Then, by using (22) and (48), we have

$\left\|W_{t}f-\left(\sum_{k\in\mathbb{Z}}f(k)\right)G_{t,1}\right\|_{\ell^{p}(\mathbb{Z})}\leq
Ct^{-1/2-(1/q-1/p)/2}\|F\|_{\ell^{q}(\mathbb{Z})}.$

Finally, applying the well known inequality

$\sum_{n=1}^{\infty}\left|\sum_{j=n}^{\infty}f(j)\right|^{q}\leq
C\sum_{n=1}^{\infty}|nf(n)|^{q},\qquad 1\leq q<\infty,$

which is a consequence of the classical Hardy’s inequality for $q>1$ and for
$q=1$ is obvious, the result follows in this case because

$\displaystyle\|F\|_{\ell^{q}(\mathbb{Z})}^{q}$
$\displaystyle\leq\sum_{n=1}^{\infty}\left(\left(\sum_{j=n}^{\infty}|f(-j)|\right)^{q}+\left(\sum_{j=n}^{\infty}|f(j)|\right)^{q}\right)$
$\displaystyle\leq
C\sum_{n=1}^{\infty}n^{q}(|f(-n)|^{q}+|f(n)|^{q})=C\||\cdot|f\|_{\ell^{q}(\mathbb{Z})}^{q}.$

Now, we consider $N>1$. Firstly, we are going to prove the inequality

(49)
$\left\|W_{t}f-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)G_{t,N}\right\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
Ct^{-1/2}\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}$

for $1\leq p<N/(N-1)$. To do that, for a fix $\mathbf{n}\in\mathbb{Z}^{N}$, we
consider the sets
$A_{1}=\\{\mathbf{k}\in\mathbb{Z}^{N}:|\mathbf{k}|\leq|\mathbf{n}|/2\\}$,
$A_{2}=\\{\mathbf{k}\in\mathbb{Z}^{N}:|\mathbf{n}|/2<|\mathbf{k}|<2|\mathbf{n}|\\}$,
and $A_{3}=\\{\mathbf{k}\in\mathbb{Z}^{N}:|\mathbf{k}|\geq 2|\mathbf{n}|\\}$.
With them we do the decomposition

$\left|W_{t}f(\mathbf{n})-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)G_{t,N}(\mathbf{n})\right|\leq
P_{1}f(\mathbf{n})+P_{2}f(\mathbf{n})+P_{3}f(\mathbf{n}),$

with

$P_{i}f(\mathbf{n})=\left|\sum_{\mathbf{k}\in\mathbb{Z}^{N}}\chi_{A_{i}}(\mathbf{n},\mathbf{k})f(\mathbf{n}-\mathbf{k})(G_{t,N}(\mathbf{k})-G_{t,N}(\mathbf{n}))\right|,\qquad
i=1,2,3.$

From (3), in $A_{1}$ it is verified that
$|G_{t,N}(\mathbf{k})-G_{t,N}(\mathbf{n})|\leq C(G_{t,1}(|\mathbf{k}|/N))^{N}$
and

(50) $\displaystyle P_{1}f(\mathbf{n})$ $\displaystyle\leq
C\sum_{|\mathbf{k}|\leq|\mathbf{n}|/2}|f(\mathbf{n}-\mathbf{k})|(G_{t,1}(|\mathbf{k}|/N))^{N}$
$\displaystyle\leq
C\sum_{|\mathbf{k}|\leq|\mathbf{n}|/2}|\mathbf{n}-\mathbf{k}||f(\mathbf{n}-\mathbf{k})|\frac{(G_{t,1}(|\mathbf{k}|/N))^{N}}{|\mathbf{k}|+1}.$

Then, by Minkowsky integral inequality and Lemma 11 (with $r=\gamma=1$), we
obtain that

$\|P_{1}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
C\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}\sum_{k\in\mathbb{Z}^{N}}\frac{(G_{t,1}(|\mathbf{k}|))^{N}}{|\mathbf{k}|+1}\leq
Ct^{-1/2}\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}.$

For $P_{2}f$, by using that $|\mathbf{k}|\sim|\mathbf{n}|$, we can apply
Remark 2 (a) to get

$P_{2}(\mathbf{n})\leq
C\sum_{\mathbf{k}\in\mathbb{Z}^{N}}|\mathbf{n}-\mathbf{k}||f(\mathbf{n}-\mathbf{k})|H_{t,N}(\mathbf{k})=CF\ast
H_{t,N}(\mathbf{n}),$

where $F(\mathbf{n})=|\mathbf{n}||f(\mathbf{n})|$. Then, by applying (44) and
the Young inequality (22) for the convolution, we have

$\left\|P_{2}f\right\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
Ct^{-1/2}\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}.$

Finally, we treat $P_{3}f$. First, we have to observe that for $p<N/(N-1)$, by
Hölder’s inequality, we obtain that

$\displaystyle\sum_{|\mathbf{k}|\geq 2|\mathbf{n}|}|f(\mathbf{n}-\mathbf{k})|$
$\displaystyle\leq\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}\left(\sum_{|\mathbf{k}|\geq
2|\mathbf{n}|}\frac{1}{|\mathbf{n}-\mathbf{k}|^{p^{\prime}}}\right)^{1/p^{\prime}}$
$\displaystyle\leq
C\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}(|\mathbf{n}|+1)^{N/p^{\prime}-1}.$

Then, by using that in $A_{3}$ it is satisfied the estimate
$|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{k})|\leq
C(G_{t,N}(|\mathbf{n}|/N))^{N}$,

(51) $P_{3}f(\mathbf{n})\leq
C(G_{t,1}(|\mathbf{n}|/N))^{N}(|\mathbf{n}|+1)^{N/p^{\prime}-1}\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}.$

In this way, applying Lemma 11 (with $r=p$ and $\gamma=p(1-N/p^{\prime})$) the
inequality

$\left\|P_{2}f\right\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
Ct^{-1/2}\||\cdot|f\|_{\ell^{p}(\mathbb{Z}^{N})}$

is attained and the proof of (49) is completed.

Now, we are going to check that

(52)
$\left\|W_{t}f-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)G_{t,N}\right\|_{\ell^{\infty}(\mathbb{Z}^{N})}\leq
Ct^{-1/2-N/(2q)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

for $1\leq q<N/(N-1)$. We consider the same decomposition than in the previous
case. To analyze $P_{1}$, we consider the decomposition

$\|P_{1}f\|_{\ell^{\infty}(\mathbb{Z}^{N})}=P_{1,1}+P_{1,2}$

where

$P_{1,1}=\sup_{|\mathbf{n}|/2\leq
n_{t}}|P_{1}f(\mathbf{n})|\qquad\text{and}\qquad
P_{1,2}=\sup_{|\mathbf{n}|/2>n_{t}}|P_{1}f(\mathbf{n})|.$

So, from (41) and Hölder’s inequality, it is verified that

$\displaystyle P_{1,1}$ $\displaystyle\leq Ct^{-N/2-1}\sup_{|\mathbf{n}|/2\leq
n_{t}}|\mathbf{n}|\sum_{|\mathbf{k}|\leq|\mathbf{n}|/2}|\mathbf{n}-\mathbf{k}||f(\mathbf{n}-\mathbf{k})|$
$\displaystyle\leq
Ct^{-N/2-1}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\sup_{|\mathbf{n}|/2\leq
n_{t}}|\mathbf{n}|\left(\sum_{|\mathbf{k}|\leq|\mathbf{n}|/2}1\right)^{1/q^{\prime}}$
$\displaystyle\leq
Ct^{-N/2-1}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\sup_{|\mathbf{n}|/2\leq
n_{t}}|\mathbf{n}|^{1+N/q^{\prime}}\leq
Ct^{-1/2-N/(2q)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

and this is enough. Now, to deal with $P_{1,2}$ we split the sum defining
$P_{1}f$ in two pieces: one for $|\mathbf{k}|\leq n_{t}$, denoted
$P_{1,2}^{\prime}$, and another one for
$n_{t}<|\mathbf{k}|\leq|\mathbf{n}|/2$, with the notation
$P_{1,2}^{\prime\prime}$ for this one. Then, from (43) and using Hölder’s
inequality, we get

$\displaystyle P_{1,2}^{\prime}$ $\displaystyle\leq
Ct^{-1/2-N/2}\sup_{|\mathbf{n}|/2>n_{t}}\sum_{|\mathbf{k}|\leq
n_{t}}|\mathbf{n}-\mathbf{k}||f(\mathbf{n}-\mathbf{k})|$ $\displaystyle\leq
Ct^{-1/2-N/2}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\left(\sum_{|\mathbf{k}|\leq
n_{t}}1\right)^{1/q^{\prime}}\leq
Ct^{-1/2-N/(2q)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

Now, proceeding as in the proof of (50), we deduce that

$\displaystyle P_{1,2}^{\prime\prime}$ $\displaystyle\leq
C\sup_{|\mathbf{n}|/2>n_{t}}\sum_{n_{t}<|\mathbf{k}|\leq|\mathbf{n}|/2}|\mathbf{n}-\mathbf{k}||f(\mathbf{n}-\mathbf{k})|\frac{(G_{t,1}(|\mathbf{k}|/N))^{N}}{|\mathbf{k}|+1}$
$\displaystyle\leq
C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\left(\sum_{n_{t}<|\mathbf{k}|}\frac{(G_{t,1}(|\mathbf{k}|/N))^{q^{\prime}N}}{(|\mathbf{k}|+1)^{q^{\prime}}}\right)^{1/q^{\prime}}.$

Finally, the required estimate is obtained by using that
$G_{t,1}(|\mathbf{k}|/N)\leq C/(|\mathbf{k}|+1)$. Indeed,

$\displaystyle P_{1,2}^{\prime\prime}$ $\displaystyle\leq
C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\left(\sum_{|\mathbf{k}|>n_{t}}\frac{1}{(|\mathbf{k}|+1)^{q^{\prime}(N+1)}}\right)^{1/q^{\prime}}$
$\displaystyle\leq Ct^{-1/2-N/(2q)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

To obtain the estimate

$\|P_{2}f\|_{\ell^{\infty}(\mathbb{Z}^{N})}\leq
Ct^{-1/2-N/(2q)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

we proceed as in the proof of (49), so we omit the details.

Finally, we have to bound $P_{3}f$ and to do this we take the decomposition

$P_{3}f(\mathbf{n})\leq P_{3,1}+P_{3,2},$

with

$P_{3,1}=\sup_{2|\mathbf{n}|\leq
n_{t}}|P_{3}f(\mathbf{n})|\qquad\text{and}\qquad
P_{3,2}=\sup_{2|\mathbf{n}|>n_{t}}|P_{3}f(\mathbf{n})|.$

For $P_{3,2}$, proceeding as in (51) and using the bound (30), we deduce that

$\displaystyle P_{3,2}$ $\displaystyle\leq
C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\sup_{2|\mathbf{n}|>n_{t}}(G_{t,1}(|\mathbf{n}|/N))^{N}(|\mathbf{n}|+1)^{N/q^{\prime}-1}$
$\displaystyle\leq Ct^{-1/2-N/(2q)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

Now, to treat $P_{3,1}$ we decompose the sum defining $P_{3}f$ in two pieces,
$P_{3,2}^{\prime}$ and $P_{3,2}^{\prime\prime}$, where
$2|\mathbf{n}|\leq|\mathbf{k}|\leq n_{t}$ and $n_{t}<|\mathbf{k}|$,
respectively. For $P_{3,2}^{\prime}$ the required bound can be obtained as for
$P_{1,2}^{\prime}$ and we omit the details. To conclude the result for
$P_{3,1}^{\prime\prime}$ we observe that, due to the restriction $q<N/(N-1)$,

$\displaystyle\sum_{|\mathbf{k}|>n_{t}}|f(\mathbf{n}-\mathbf{k})|$
$\displaystyle\leq\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\left(\sum_{|\mathbf{k}|>n_{t}}\frac{1}{|\mathbf{n}-\mathbf{k}|^{q^{\prime}}}\right)^{1/q^{\prime}}$
$\displaystyle\leq
Ct^{-1/2+N/(2q^{\prime})}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

Then, applying (30),

$\displaystyle P_{3,1}^{\prime\prime}$ $\displaystyle\leq
Ct^{-1/2+N/(2q^{\prime})}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\sup_{2|\mathbf{n}|\leq
n_{t}}(G_{t,1}(|\mathbf{n}|/N))^{N}$ $\displaystyle\leq
Ct^{-1/2-N/(2q)}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

and the proof of (52) is completed.

The proof of the theorem will be finished interpolating between (49) and (52).
∎

######  Remark 3.

By using (32) and Minkowski integral inequality, for $0<t<T$ (with $T>100$ for
example) we obtain

$\left\|W_{t}f-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)G_{t,N}\right\|_{\ell^{p}(\mathbb{Z}^{N})}\\\
\leq\|f\|_{\ell^{1}(\mathbb{Z}^{N})}\left(\sup_{\mathbf{k}\in\mathbb{Z}^{N}}\left\|G_{t,N}(\cdot-\mathbf{k})\right\|_{\ell^{p}(\mathbb{Z}^{N})}+\left\|G_{t,N}\right\|_{\ell^{p}(\mathbb{Z}^{N})}\right)\leq
C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

Then, combining this fact with (35) and with the hypotheses of Theorem 7, we
get

(53)
$\left\|W_{t}f-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)G_{t,N}\right\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
C\min\\{1,t^{-1/2-N/2(1/q-1/p)}\\}\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

###  3.2. Maximal operators for the heat and the Poisson semigroups

From the well known identity

$e^{-|\beta|t}=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-u}}{\sqrt{u}}e^{-t^{2}\beta^{2}/(4u)}\,du=\frac{t}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-t^{2}/(4v)}}{\sqrt{v}}e^{-v\beta^{2}}\frac{dv}{v},$

we can define by subordination the Poisson semigroup

(54)
$P_{t}f(\mathbf{n})=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-u}}{\sqrt{u}}W_{t^{2}/(4u)}f(\mathbf{n})\,du=\frac{t}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-t^{2}/(4v)}}{\sqrt{v}}W_{v}f(\mathbf{n})\frac{dv}{v}.$

It is an easy exercise to show that $P_{t}$ satisfies the Laplace type
equation

$\frac{d^{2}}{dt^{2}}P_{t}f(\mathbf{n})+\Delta_{N}P_{t}f(\mathbf{n})=0.$

In this part of the paper we are interested in the maximal operators

$W^{\star}f(\mathbf{n})=\sup_{t>0}|W_{t}f(\mathbf{n})|$

and

$P^{\star}f(\mathbf{n})=\sup_{t>0}|P_{t}f(\mathbf{n})|.$

In fact, we want analyze weighted inequalities for these operators with
weights in the Muckenhoupt class. We say that a positive sequence $w$ is a
weight in $A_{p}(\mathbb{Z}^{N})$ for $1<p<\infty$ when

$\left(\frac{1}{\mu(Q)}\sum_{\mathbf{n}\in
Q}w(\mathbf{n})\right)\left(\frac{1}{\mu(Q)}\sum_{\mathbf{n}\in
Q}(w(\mathbf{n}))^{-1/(p-1)}\right)^{p-1}\leq C,$

where $Q$ is any cube in $\mathbb{Z}^{N}$ and $\mu(A)=\sum_{\mathbf{n}\in A}1$
for any $A\subset\mathbb{Z}^{N}$. The weight $w$ belongs to
$A_{1}(\mathbb{Z}^{N})$ when

$\left(\frac{1}{\mu(Q)}\sum_{\mathbf{n}\in
Q}w(\mathbf{n})\right)\sup\left\\{(w(\mathbf{n})^{-1}:\mathbf{n}\in\mathbb{Z}^{N}\right\\}\leq
C.$

To obtain our result, we will use the Calderón-Zygmund theory adapts to our
setting. We say that $K$, defined on $\mathbb{Z}^{N}\setminus\\{\mathbf{0}\\}$
and taking values in a Banach space $A$, is a standard kernel when

(55) $\|K(\mathbf{n})\|_{A}\leq\frac{C}{|\mathbf{n}|^{N}}$

and

(56) $\|K(\mathbf{n})-K(\mathbf{n}+\mathbf{m})\|_{A}\leq
C\frac{|\mathbf{m}|}{|\mathbf{n}|^{N+1}},\qquad|\mathbf{n}|>2|\mathbf{m}|.$

Let $T$ be an operator bounded from $\ell^{q}(\mathbb{Z}^{N})$ to
$\ell_{A}^{q}(\mathbb{Z}^{N})$ (where we understand that
$f\in\ell_{A}^{q}(\mathbb{Z}^{N})$ when
$\|f\|_{A}\in\ell^{q}(\mathbb{Z}^{N})$) for some $1<q<\infty$ and for
$f\in\ell^{q}(\mathbb{Z}^{N})$ with compact support

$Tf(\mathbf{n})=K\ast
f(\mathbf{n}),\qquad\mathbf{n}\notin\operatorname{supp}f,$

where $K$ is a standard kernel taking values in $A$, we say that $T$ is
Calderón-Zygmund operator.

A Calderón-Zygmung operator $Tf$ is bounded from $\ell^{p}(\mathbb{Z}^{N},w)$
into $\ell^{p}_{A}(\mathbb{Z}^{N},w)$ for $1<p<\infty$ and $w\in
A_{p}(\mathbb{Z}^{N})$, and from $\ell^{1}(\mathbb{Z}^{N},w)$ into
$\ell^{1,\infty}_{A}(\mathbb{Z}^{N},w)$ for $w\in A_{1}(\mathbb{Z}^{N})$ (see
[20, 21]). As usual the space $\ell^{1,\infty}(\mathbb{Z}^{N},w)$ is defined
as the space of sequences such that
$\|f\|_{\ell^{1,\infty}(\mathbb{Z}^{N},w)}<\infty$, where

$\|f\|_{\ell^{1,\infty}(\mathbb{Z}^{N},w)}=\sup_{\lambda>0}\left(\lambda\sum_{\mathbf{n}\in
A_{\lambda}}w(\mathbf{n})\right)$

and

$A_{\lambda}=\\{\mathbf{n}\in\mathbb{Z}^{N}:|f(\mathbf{n})|>\lambda\\}.$

######  Theorem 13.

For $T$ equal $W^{\star}$ or $P^{\star}$ and $w\in A_{p}(\mathbb{Z}^{N})$, the
inequalities

$\|Tf\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
1<p<\infty,\qquad\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w),$

and

$\|Tf\|_{\ell^{1,\infty}(\mathbb{Z}^{N},w)}\leq\|f\|_{\ell^{1}(\mathbb{Z}^{N},w)},\qquad\ell^{2}(\mathbb{Z}^{N})\cap\ell^{1}(\mathbb{Z}^{N},w),$

hold.

###### Proof.

We focus on $W^{\star}$ because the result for $P^{\star}$ follows from its
definition by subordination (54). By using that

$W^{\ast}f(\mathbf{n})=\|W_{t}f(\mathbf{n})\|_{L^{\infty}((0,\infty))},$

we prove that $W^{\star}$ is a Calderón-Zygmund operator taking values in
$L^{\infty}((0,\infty))$. From (23), applying Stein’s Maximal Theorem of
diffusion semigroups (see [23, Chapter III, Section 2]), we have

$\|W^{\ast}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N})},\qquad 1<p<\infty.$

In particular we have the case $p=2$. To conclude it is enough with the
estimates

(57)
$\|G_{t,N}(\mathbf{n})\|_{L^{\infty}((0,\infty))}\leq\frac{C}{(|\mathbf{n}|+1)^{N}}$

and

(58)
$\|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{n}+\mathbf{m})\|_{L^{\infty}((0,\infty))}\leq
C\frac{|\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{n}+\mathbf{m}|)^{N+1}},\qquad|\mathbf{n}|>2|\mathbf{m}|.$

The bound (57) has been proved yet in (31).

Now, from (36) and applying (14) with $\alpha=1/N$, we deduce that for
$|\mathbf{n}|>2|\mathbf{m}|$ it is verified

$\displaystyle|G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{n}+\mathbf{m})|$
$\displaystyle\leq
C|\mathbf{m}|(|\mathbf{n}|+|\mathbf{n}+\mathbf{m}|)\left(t^{-1/N}G_{t,1}\left(\frac{|\mathbf{n}|+|\mathbf{n}+\mathbf{m}|}{K}\right)\right)^{N}$
$\displaystyle\leq
C\frac{|\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{n}+\mathbf{m}|)^{N+1}}$

and the proof of (58) is completed. ∎

##  4\. The fractional integrals and the Riesz transforms

From the identity

$\frac{1}{\lambda^{\sigma}}=\frac{1}{\Gamma(\sigma)}\int_{0}^{\infty}e^{-\lambda
t}t^{\sigma-1}\,dt,$

we define the fractional integrals (or negative powers of the Laplacian
$\Delta_{N}$) by the relation

$(-\Delta_{N})^{-\sigma}f(\mathbf{n})=\frac{1}{\Gamma(\sigma)}\int_{0}^{\infty}W_{t}f(\mathbf{n})t^{\sigma-1}\,dt.$

By using that (see [15] or take $\alpha=-1/2$ in (20))

$e^{-t}I_{k}(t)\sim t^{-1/2},\qquad t\to\infty,$

it is clear that the powers $(-\Delta_{N})^{-\sigma}$ are well defined for
$0<2\sigma<N$.

Having these operators, for $N\geq 2$ we can define the Riesz transforms as

$R_{i}f(\mathbf{n})=\delta_{i}^{+}(-\Delta_{N})^{-1/2}f(\mathbf{n}),\qquad
i=1,\dots,N.$

In [6] it was analyzed the Riesz transform for $N=1$ defining it as a limit.
In fact, it was defined as

$Rf(n)=\lim_{\sigma\to(1/2)^{-}}\delta^{+}(-\Delta_{1})^{-\sigma}f(n).\qquad
n\in\mathbb{Z}.$

From this, the identity

$Rf(n)=\sum_{k\in\mathbb{Z}}\frac{f(n)}{n-k+1/2}$

was obtained; i.e., it matches with the discrete Hilbert transform.

In this section, firstly, we show some basic properties of the fractional
integrals and we provide a Hardy-Littlewood-Sobolev inequality in our setting
and an extension of it. We continue with the study of multidimensional
discrete Schauder inequalities for $(-\Delta_{N})^{-\sigma}$. To conclude we
analyze the mapping properties of the Riesz transforms on the Hölder classes
and on $\ell^{p}(\mathbb{Z}^{N},w)$

###  4.1. Basic properties of the fractional integrals

In this part, we provide some properties of the fractional integrals, we prove
a Hardy-Littlewood-Sobolev inequality for them and an extension of it.

In some points a long the paper we will consider the spaces

(59) $\ell_{\sigma}(\mathbb{Z}^{N})=\ell^{1}(\mathbb{Z}^{N},w_{\sigma})$

with $w_{\sigma}(\mathbf{n})=(1+|\mathbf{n}|)^{2\sigma-N}$. Moreover, we will
denote
$\|f\|_{\ell_{\sigma}(\mathbb{Z}^{N})}=\|f\|_{\ell^{1}(\mathbb{Z}^{N},w_{\sigma})}$.

######  Theorem 14.

Let $0<\sigma<N/2$ and $f\in\ell_{\sigma}(\mathbb{Z}^{N})$. We have the
pointwise formula

(60)
$(-\Delta_{N})^{-\sigma}f(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{n}-\mathbf{k})K_{\sigma}(\mathbf{k})$

where

$K_{\sigma}(\mathbf{n})=\frac{1}{\Gamma(\sigma)}\int_{0}^{\infty}G_{t,N}(\mathbf{n})t^{\sigma-1}\,dt$

and it verifies that

(61) $0<K_{\sigma}(\mathbf{n})\leq\frac{C}{(|\mathbf{n}|+1)^{N-2\sigma}}.$

###### Proof.

First we are going to check that the heat operator is well defined for
$f\in\ell_{\sigma}(\mathbb{Z}^{N})$. Indeed, for each $L>0$ with $t$ fixed, by
using (3) and the asymptotic (20),

$\displaystyle\sum_{{|\mathbf{k}|>L}}G_{t,N}(\mathbf{k})|f(\mathbf{n}-\mathbf{k})|$
$\displaystyle\leq
Ce^{-2Nt}N^{N/2}\sum_{{|\mathbf{k}|>L}}\left(\frac{(Net)^{|\mathbf{k}|/N}}{|\mathbf{k}|^{|\mathbf{k}|/N+1/2}}\right)^{N}|f(\mathbf{n}-\mathbf{k})|$
$\displaystyle\leq
C\|f\|_{\ell_{\sigma}(\mathbb{Z}^{N})}e^{-2Nt}N^{N/2}\sup_{{|\mathbf{k}|>L}}\frac{(Net)^{|\mathbf{k}|}(1+|\mathbf{n}-\mathbf{k}|)^{N-2\sigma}}{|\mathbf{k}|^{|\mathbf{k}|+N/2}}$
$\displaystyle\leq C\|f\|_{\ell_{\sigma}(\mathbb{Z}^{N})},$

where the last constant depends on $t$, $N$, $L$, $\mathbf{n}$, and $\sigma$
but not on $f$.

Let us prove (60). To do it we can suppose that $f$ is a non negative sequence
because in other case we can consider its decomposition in positive and
negative parts. Then, applying Tonelli’s theorem, we have

$\displaystyle(-\Delta_{N})^{-\sigma}f(\mathbf{n})$
$\displaystyle=\frac{1}{\Gamma(\sigma)}\int_{0}^{\infty}\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{n}-\mathbf{k})G_{t}(\mathbf{k})t^{\sigma-1}\,dt$
$\displaystyle=\frac{1}{\Gamma(\sigma)}\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{n}-\mathbf{k})\int_{0}^{\infty}G_{t,N}(\mathbf{k})t^{\sigma-1}\,dt=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{n}-\mathbf{k})K_{\sigma}(\mathbf{k}).$

Now, let us go with the proof of (61). The positivity of $K_{\sigma}$ is a
simple consequence of the positivity of $G_{t,N}$. Now, to obtain the upper
bound for the kernel, we consider the decomposition

$K_{\sigma}(\mathbf{n})=I_{1}+I_{2}$

with

$I_{1}=\frac{1}{\Gamma(\sigma)}\int_{0}^{|\mathbf{n}|^{2}}G_{t,N}(\mathbf{n})t^{\sigma-1}\,dt\qquad\text{
and }\qquad
I_{2}=\frac{1}{\Gamma(\sigma)}\int_{|\mathbf{n}|^{2}}^{\infty}G_{t,N}(\mathbf{n})t^{\sigma-1}\,dt.$

We consider $\mathbf{n}\not=\mathbf{0}$, because the bound
$K_{\sigma}(\mathbf{0})\leq C$ is clear. For $I_{1}$ we use (31) to have

$I_{1}\leq\frac{C}{|\mathbf{n}|^{N}}\int_{0}^{|\mathbf{n}|^{2}}t^{\sigma-1}\,dt=\frac{C}{|\mathbf{n}|^{N-2\sigma}},$

and for $I_{2}$ we apply the bound in (30) to obtain that

$I_{2}\leq
C\int_{|\mathbf{n}|^{2}}^{\infty}\frac{dt}{t^{N/2+1-\sigma}}=\frac{C}{|\mathbf{n}|^{N-2\sigma}}.\qed$

To prove our next theorem we need the following result that we can see in [9,
Theorem 1.6].

######  Theorem 15.

Let be

$\mathcal{I}_{\lambda}f(\mathbf{n})=\sum_{\begin{smallmatrix}\mathbf{k}\in\mathbb{Z}^{N}\\\
\mathbf{k}\not=\mathbf{n}\end{smallmatrix}}\frac{f(\mathbf{k})}{|\mathbf{n}-\mathbf{k}|^{n-2\sigma}},\qquad
0<2\sigma<N,$

and $\langle\mathbf{n}\rangle=(1+|\mathbf{n}|^{2})^{1/2}$. Taking $1\leq
p,q<\infty$, $s,t\in\mathbb{R}$ and the sets of conditions

$(\mathcal{C}_{1})\qquad\begin{cases}s\leq t,\\\\[3.0pt]
\dfrac{2\sigma}{N}+\max\left\\{\dfrac{1}{p}+\dfrac{s}{N},0\right\\}\leq\max\left\\{\dfrac{1}{q}+\dfrac{t}{N},0\right\\},\\\\[3.0pt]
\dfrac{2\sigma}{N}+\dfrac{1}{p}+\dfrac{s}{N}\leq 1,\\\\[3.0pt] \end{cases}$
$(\mathcal{C}_{2})\qquad\begin{cases}s\leq t=2\sigma-N,q=1,\\\\[3.0pt]
\dfrac{1}{p}+\dfrac{s}{N}<0\end{cases}$

and

$(\mathcal{C}_{3})\qquad\begin{cases}s\leq t,\\\\[3.0pt]
\dfrac{2\sigma}{N}+\dfrac{1}{p}+\dfrac{s}{N}=\dfrac{1}{q}+\dfrac{t}{N},\\\\[3.0pt]
\dfrac{1}{p}\leq\dfrac{1}{q},\quad
0<\dfrac{1}{p}+\dfrac{s}{N},\quad\dfrac{1}{q}+\dfrac{t}{N}<1,\end{cases}$

it is verified that

$\|\langle\mathbf{\cdot}\rangle^{s}\mathcal{I}_{2\sigma}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq\|\langle\mathbf{\cdot}\rangle^{t}f\|_{\ell^{q}(\mathbb{Z}^{N})}$

if and only if $(p,q,s,t)$ satisfies one of the conditions $\mathcal{C}_{1}$,
$\mathcal{C}_{2}$, and $\mathcal{C}_{3}$.

######  Remark 4.

The previous theorem in its full version includes the case $p,q=\infty$ but
the definition of the Lebesgue for such value in [9] is different from ours
one, so we omit this particular value from the theorem.

Now, we can present the weighted Hardy-Littlewood-Sobolev for the fractional
powers $(-\Delta_{N})^{-\sigma}$.

######  Theorem 16.

Let $0<2\sigma<N$, $1\leq p,q<\infty$ and $s,t\in\mathbb{R}$. If $(q,p,s,t)$
verify one of the conditions $\mathcal{C}_{1}$, $\mathcal{C}_{2}$, and
$\mathcal{C}_{3}$, then

$\|\langle\mathbf{\cdot}\rangle^{s}(-\Delta_{N})^{-\sigma}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq\|\langle\mathbf{\cdot}\rangle^{t}f\|_{\ell^{q}(\mathbb{Z}^{N})}.$

###### Proof.

The result is immediate from the previous theorem because, by (61), it is
verified that

$|(-\Delta_{N})^{-\sigma}f(\mathbf{n})|\leq
C\left(|\mathcal{I}_{2\sigma}f(\mathbf{n})|+|f(\mathbf{n})|\right).\qed$

Taking $t=s=0$ in the previous theorem, we recover the classical Hardy-
Littlewood-Sobolev inequality in our setting. However, we can go further.

######  Corollary 17.

Let $0<2\sigma<N$, $1<q<p<\infty$ with $1/p\leq 1/q-2\sigma/N$, then the
inequality

$\|(-\Delta_{N})^{-\sigma}f\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
C\|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

holds.

Moreover, for $1<q<p\leq 2$ the Hardy-Littlewood-Sobolev implies $1/p\leq
1/q-2\sigma/N$

###### Proof.

We have to proof the second part of the result. To do this, we consider the
sequence $f(\mathbf{n})=G_{t,N}(\mathbf{n})$ with $t>1$. In this way,

$\displaystyle(-\Delta_{N})^{-\sigma}f(\mathbf{n})$
$\displaystyle=\int_{0}^{\infty}\sum_{\mathbf{k}\in\mathbb{Z}^{N}}G_{t,N}(\mathbf{k})G_{s,N}(\mathbf{n}-\mathbf{k})s^{\sigma-1}\,ds$
$\displaystyle=\int_{0}^{\infty}G_{t+s,N}(\mathbf{n})s^{\sigma-1}\,ds\geq\int_{0}^{t}G_{t+s,N}(\mathbf{n})s^{\sigma-1}\,ds\geq
G_{2t}(\mathbf{n})t^{\sigma}.$

Then, by the Hardy-Littlewood-Sobolev, we have

$t^{\sigma}\|G_{2t,N}\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
C\|G_{t,N}\|_{\ell^{q}(\mathbb{Z}^{N})}$

and, applying (33), it becomes $t^{\sigma-N/2(1/q-1/p)}\leq C$ and this
implies $1/p\leq 1/q-2\sigma/N$. ∎

To finish this subsection we show a new inequality extending the classical
Hardy-Littlewood-Sobolev one.

######  Theorem 18.

Let $0<2\sigma<N$, $1\leq q\leq p\leq\infty$ with $1\leq q<N/(N-1)$ and
$1/p<1/q-(2\sigma-1)/N$, and $|\cdot|f\in\ell^{q}(\mathbb{Z}^{N})$, then the
inequality

$\left\|(-\Delta_{N})^{-\sigma}f-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)K_{\sigma}\right\|_{\ell^{p}(\mathbb{Z}^{N})}\leq
C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}$

holds.

###### Proof.

The result follows from Minkowsky integral inequality and (53). Indeed,

$\left\|(-\Delta_{N})^{-\sigma}f-\left(\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\right)K_{\sigma}\right\|_{\ell^{p}(\mathbb{Z}^{N})}\\\
\begin{aligned}
&\leq\int_{0}^{\infty}\left\|W_{t}f-\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})G_{t,N}\right\|_{\ell^{p}(\mathbb{Z}^{N})}t^{\sigma-1}\,dt\\\
&\leq
C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}\int_{0}^{\infty}\min\\{1,t^{-1/2-N/2(1/q-1/p)}\\}t^{\sigma-1}\,dt\\\
&\leq C\||\cdot|f\|_{\ell^{q}(\mathbb{Z}^{N})}.\end{aligned}$

∎

###  4.2. Discrete Schauder estimates for the fractional integrals

Now, we prove multidimensional discrete Schauder estimates for the fractional
integrals and to do that we need to define the Hölder spaces. Given a sequence
$f=\\{f(\mathbf{k})\\}_{\mathbf{k}\in\mathbb{Z}^{N}}$ and $0<\alpha\leq 1$, we
say that $f$ belongs to the Hölder space $C^{0,\alpha}(\mathbb{Z}^{N})$ when

$[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}:=\sup_{\mathbf{n},\mathbf{m}\in\mathbb{Z}}\frac{|f(\mathbf{n})-f(\mathbf{m})|}{|\mathbf{n}-\mathbf{m}|^{\alpha}}<\infty.$

For a natural number $k$ such that $k\geq 1$ we say that $f\in
C^{k,\alpha}(\mathbb{Z}^{N})$ when it is verified that

$[f]_{C^{k,\alpha}(\mathbb{Z}^{N})}:=\sum_{m_{1}+\cdots+m_{N}=k}[(\delta_{1}^{+})^{m_{1}}(\delta_{2}^{+})^{m_{2}}\cdots(\delta_{N}^{+})^{m_{N}}f]_{C^{0,\alpha}(\mathbb{Z}^{N})}<\infty.$

Sometimes it is common write the definition of Hölder classes combining
$\delta_{i}^{+}$ and $\delta_{i}^{-}$ but such definitions are equivalents to
the our one due to the relation
$\delta_{i}^{-}f(\mathbf{n})=\delta_{i}^{+}f(\mathbf{n}-\mathbf{e}_{i})$.

The next result is the multidimensional version of [7, Theorem 1.6].

######  Theorem 19.

Let $k\geq 0$, $0<\alpha\leq 1$, $0<\sigma<1/2$ and
$f\in\ell_{\sigma}(\mathbb{Z}^{N})$.

  1. a)

If $f\in C^{k,\alpha}(\mathbb{Z}^{N})$ and $2\sigma+\alpha<1$ then
$(-\Delta)^{-\sigma}f\in C^{k,2\sigma+\alpha}(\mathbb{Z}^{N})$ and

$[(-\Delta)^{-\sigma}f]_{C^{k,2\sigma+\alpha}(\mathbb{Z}^{N})}\leq
C[f]_{C^{k,\alpha}(\mathbb{Z}^{N})}.$

  2. b)

If $f\in C^{k,\alpha}(\mathbb{Z}^{N})$ and $2\sigma+\alpha>1$ then
$(-\Delta)^{-\sigma}f\in C^{k+1,2\sigma+\alpha-1}(\mathbb{Z}^{N})$ and

$[(-\Delta)^{-\sigma}f]_{C^{k+1,2\sigma+\alpha-1}(\mathbb{Z}^{N})}\leq
C[f]_{C^{k,\alpha}(\mathbb{Z}^{N})}.$

  3. c)

If $f\in\ell^{\infty}(\mathbb{Z}^{N})$ then $(-\Delta)^{-\sigma}f\in
C^{0,2\sigma}(\mathbb{Z}^{N})$ and

$[(-\Delta)^{-\sigma}f]_{C^{0,2\sigma}(\mathbb{Z}^{N})}\leq
C\|f\|_{\ell^{\infty}(\mathbb{Z}^{N})}.$

Before starting with the proof of the Schauder estimates, we need the four
following lemmas.

######  Lemma 20.

For $N\geq 1$ and $0<\sigma<1/2$, we have

$\sum_{k\in\mathbb{Z}^{N}}((-\Delta_{N})^{-\sigma}(\mathbf{n}-\mathbf{k})-(-\Delta_{N})^{-\sigma}(\mathbf{k}))=0.$

###### Proof.

From the obvious identity

$\sum_{k\in\mathbb{Z}^{N}}(G_{t,N}(\mathbf{n}-\mathbf{k})-G_{t,N}(\mathbf{k}))=0,$

we deduce that

$\int_{0}^{\infty}\sum_{k\in\mathbb{Z}^{N}}(G_{t,N}(\mathbf{n}-\mathbf{k})-G_{t,N}(\mathbf{k}))t^{\sigma-1}\,dt=0,$

Let us see now that, for $0<\sigma<1/2$, we can apply Fubini theorem. From
(21), it is clear that

$\int_{0}^{|\mathbf{n}|^{2}}\sum_{k\in\mathbb{Z}^{N}}|G_{t,N}(\mathbf{n}-\mathbf{k})-G_{t,N}(\mathbf{k})|t^{\sigma-1}\,dt\leq
2\int_{0}^{|\mathbf{n}|^{2}}t^{\sigma-1}\,dt\leq
C|\mathbf{n}|^{2\sigma}<\infty,$

so we have to focus on the integral in the interval
$(|\mathbf{n}|^{2},\infty)$. To do that we consider

$I_{1}=\int_{|\mathbf{n}|^{2}}^{\infty}\sum_{|\mathbf{k}|\leq
2|\mathbf{n}|}|G_{t,N}(\mathbf{n}-\mathbf{k})-G_{t,N}(\mathbf{k})|t^{\sigma-1}\,dt$

and

$I_{2}=\int_{|\mathbf{n}|^{2}}^{\infty}\sum_{|\mathbf{k}|>2|\mathbf{n}|}|G_{t,N}(\mathbf{n}-\mathbf{k})-G_{t,N}(\mathbf{k})|t^{\sigma-1}\,dt.$

For $I_{1}$ we use the bound (41) to obtain that

$\displaystyle I_{1}$ $\displaystyle\leq
C|\mathbf{n}|^{2}\int_{|\mathbf{n}|^{2}}^{\infty}\left(\sum_{|\mathbf{k}|\leq
2|\mathbf{n}|}1\right)t^{\sigma-N/2-2}\,dt$ $\displaystyle\leq
C|\mathbf{n}|^{2+N}\int_{|\mathbf{n}|^{2}}^{\infty}t^{\sigma-N/2-2}\,dt\leq
C|\mathbf{n}|^{2\sigma}<\infty.$

To analyze $I_{2}$ we apply the identity
$G_{t,N}(\mathbf{n}-\mathbf{k})=G_{t,N}(\mathbf{k}-\mathbf{n})$, (36) and (44)
to deduce the estimate

$\displaystyle I_{2}$ $\displaystyle\leq
C|\mathbf{n}|\int_{|\mathbf{n}|^{2}}^{\infty}\left(\sum_{|\mathbf{k}|>2|\mathbf{n}|}H_{t,N}\left(\frac{|\mathbf{k}|}{K}\right)\right)t^{\sigma-1}\,dt$
$\displaystyle\leq
C|\mathbf{n}|\int_{|\mathbf{n}|^{2}}^{\infty}t^{\sigma-3/2}\,dt\leq
C|n|^{2\sigma}<\infty.$

In this way,

$\sum_{k\in\mathbb{Z}^{N}}((-\Delta_{N})^{-\sigma}(\mathbf{n}-\mathbf{k})-(-\Delta_{N})^{-\sigma}(\mathbf{k}))\\\
=\int_{0}^{\infty}\sum_{k\in\mathbb{Z}^{N}}(G_{t,N}(\mathbf{n}-\mathbf{k})-G_{t,N}(\mathbf{k}))t^{\sigma-1}\,dt=0$

and the proof is completed. ∎

######  Lemma 21.

Let $N\geq 1$, $\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}$ such that
$|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|$, and $0<2\sigma<N$. Then the
inequality

(62)
$|K_{\sigma}(\mathbf{n})-K_{\sigma}(\mathbf{m})|\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1-2\sigma}}$

holds with a constant $C$ independent of $\mathbf{n}$ and $\mathbf{m}$.

###### Proof.

It is clear that

$|K_{\sigma}(\mathbf{n})-K_{\sigma}(\mathbf{m})|\leq\int_{0}^{\infty}|G_{t,N}(\mathbf{n})-G_{t,n}(\mathbf{m})|t^{\sigma-1}\,dt=J_{1}+J_{2},$

where

$J_{1}=\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}|G_{t,N}(\mathbf{n})-G_{t,n}(\mathbf{m})|t^{\sigma-1}\,dt$

and

$J_{2}=\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}|G_{t,N}(\mathbf{n})-G_{t,n}(\mathbf{m})|t^{\sigma-1}\,dt.$

To estimate $J_{1}$ we consider Lemma 8 and apply (14) with $\alpha=1/N$.
Indeed,

$\displaystyle J_{1}$ $\displaystyle\leq
C|\mathbf{n}-\mathbf{m}|(|\mathbf{n}|+|\mathbf{m}|)\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}\frac{1}{t}\left(G_{t,1}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{K}\right)\right)^{N}t^{\sigma-1}\,dt$
$\displaystyle\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1}}\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}t^{\sigma-1}\,dt\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1-2\sigma}}.$

The bound for $J_{2}$ is obtained from (41) in the following way

$J_{2}\leq|\mathbf{n}-\mathbf{m}|(|\mathbf{n}|+|\mathbf{m}|)\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}t^{\sigma-N/2-2}\,dt\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1-2\sigma}}.\qed$

######  Lemma 22.

Let $N\geq 1$, $\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}$ such that
$|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|$, and

$\mathcal{H}_{t,N}(|\mathbf{n}|)=\left(\frac{1}{t}+\frac{z^{2}}{t^{2}}\right)\left(G_{t,1}\left(z\right)\right)^{N},\qquad
z>0.$

Then the inequality

(63) $|\delta_{i}^{+}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|\leq
C|\mathbf{n}-\mathbf{m}|\mathcal{H}_{t,N}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{K}\right),\qquad
i=1,\dots,N,$

holds with constants $C$ and $K>1$ independent of $\mathbf{n}$ and
$\mathbf{m}$.

###### Proof.

We consider $i=1$ because the other cases can be done in the same way. We can
prove the result for $n_{i}\not=m_{i}$ for $i=1,\dots,N$ because this
situation implies the general case. In fact, to deduce the general case from
this one, as in Lemma 8, let us suppose that $n_{i}=m_{i}$ for $j$ values of
$i$, with $1<j<N$ and consider the decompositions

$\mathbf{n}=\mathbf{n}_{1}\cup\mathbf{n}_{2}\qquad\text{and}\qquad\mathbf{m}=\mathbf{n}_{1}\cup\mathbf{m}_{2},$

where $\mathbf{n}_{1}$ are the common values of $\mathbf{n}$ and $\mathbf{m}$
and $\mathbf{n}_{2}$ and $\mathbf{m}_{2}$ are the other components. Now, we
have to distinguish two possibilities $n_{1}\in\mathbf{n}_{1}$ and
$n_{1}\notin\mathbf{n}_{1}$. In the first situation, by using (5) and Lemma 8,
we have

$|\delta_{1}^{+}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|\leq
C\frac{|n_{1}|+1}{t}G_{t,j}(\mathbf{n}_{1})|G_{t,N-j}(\mathbf{n}_{2})-G_{t,N-j}(\mathbf{m}_{2})|\\\
\begin{aligned} &\leq
C|\mathbf{n}_{2}-\mathbf{m_{2}}|\frac{(|\mathbf{n}_{1}|+1)(|\mathbf{n_{2}}|+|\mathbf{m_{2}}|)}{t^{2}}G_{t,j}(\mathbf{n}_{1})\left(G_{t,1}\left(\frac{|\mathbf{n}_{2}|+|\mathbf{m}_{2}|}{K}\right)\right)^{N-j}\\\
&\leq
C|\mathbf{n}-\mathbf{m}|\left(\frac{1}{t}+\frac{(|\mathbf{n}|+|\mathbf{m}|)^{2}}{t^{2}}\right)G_{t,j}(\mathbf{n}_{1})\left(G_{t,1}\left(\frac{|\mathbf{n}_{2}|+|\mathbf{m}_{2}|}{K}\right)\right)^{N-j}\end{aligned}$

and we conclude as in Lemma 8. When $n_{1}\notin\mathbf{n}_{1}$, if we
consider proved the case with different components, it is verified that

$\displaystyle|\delta_{1}^{+}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|$
$\displaystyle=G_{t,j}(\mathbf{n}_{1})|\delta_{1}^{+}(G_{t,N-j}(\mathbf{n}_{2})-G_{t,N-j}(\mathbf{m}_{2}))|$
$\displaystyle\leq
CG_{t,j}(\mathbf{n}_{1})|\mathbf{n}_{2}-\mathbf{m}_{2}|\mathcal{H}_{t,N-j}\left(\frac{|\mathbf{n}_{2}|+|\mathbf{m}_{2}|}{K}\right)$

and the proof of this case is finished again as in Lemma 8.

Now, we assume the restrictions (37) on $\mathbf{n}$ and $\mathbf{m}$ as in
the proof of Lemma 8. Then, using (38), we get

(64)
$\delta_{1}^{+}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))=\delta^{+}(G_{t,1}(n_{1})-G_{t,1}(m_{1}))\prod_{k=2}^{N}G_{t,1}(n_{k})\\\
+\delta^{+}G_{t,1}(m_{1})\sum_{i=2}^{N}(G_{t,1}(n_{i})-G_{t,1}(m_{i}))\prod_{j=2}^{i-1}G_{t,1}(m_{j})\prod_{k=i+1}^{N}G_{t,1}(n_{k}).$

Proceeding as in Lemma 8 and using (5), we have

(65) $|\delta^{+}G_{t,1}(m_{1})(G_{t,1}(n_{i})-G_{t,1}(m_{i}))|\\\ \leq
C\frac{(m_{1}+1)|n_{i}-m_{i}|(m_{i}+n_{i}+1)}{t^{2}}G_{t,1}(m_{1})\max\\{G_{t,1}(n_{i}),G_{t,1}(m_{i})\\}.$

Now, taking $n_{1}<m_{1}$, it is obtained that

$\displaystyle\delta^{+}(G_{t,1}(n_{1})-G_{t,1}(m_{1}))$
$\displaystyle=-\sum_{k=n_{1}}^{m_{1}-1}(\delta^{+}G_{t,1}(k+1)-\delta^{+}G_{t,1}(k))$
$\displaystyle=-\sum_{k=n_{1}}^{m_{1}-1}(G_{t,1}(k+2)-2G_{t,1}(k+1)+G_{t,1}(k))$

and, applying (8),

$\displaystyle|\delta^{+}(G_{t,1}(n_{1})-G_{t,1}(m_{1}))|$ $\displaystyle\leq
C\sum_{k=n_{1}}^{m_{1}-1}\left(\frac{1}{t}+\frac{(k+1)(k+2)}{t^{2}}\right)G_{t,1}(k)$
$\displaystyle\leq
C(m_{1}-n_{1})\left(\frac{1}{t}+\frac{(m_{1}+n_{1})^{2}}{t^{2}}\right)G_{t,1}(n_{1}).$

In general,

(66) $|\delta^{+}(G_{t,1}(n_{1})-G_{t,1}(m_{1}))|\\\ \leq
C|m_{1}-n_{1}|\left(\frac{1}{t}+\frac{(m_{1}+n_{1})^{2}}{t^{2}}\right)\max\\{G_{t,1}(n_{1}),G_{t,1}(m_{1})\\}.$

With (64), (65), and (66) the result can be concluded as in Lemma 8. ∎

######  Remark 5.

Again, following the proof of the previous lemma, it is easy to check that

(67) $|\delta_{i}^{+}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|\leq
C|\mathbf{n}-\mathbf{m}|\left(\frac{1}{t}+\frac{(|\mathbf{n}|+|\mathbf{m}|)^{2}}{t^{2}}\right)t^{-N/2},\qquad\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}.$

######  Lemma 23.

Let $N\geq 1$, $\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}$ such that
$|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|$, and $0<2\sigma<N$. Then the
inequality

(68)
$|\delta_{i}^{+}(K_{\sigma}(\mathbf{n})-K_{\sigma}(\mathbf{m}))|\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+2-2\sigma}},\qquad
i=1,\dots,N,$

holds with a constant $C$ independent of $\mathbf{n}$ and $\mathbf{m}$.

###### Proof.

It is clear that

$|\delta_{i}^{+}(K_{\sigma}(\mathbf{n})-K_{\sigma}(\mathbf{m}))|\leq\int_{0}^{\infty}|\delta_{i}^{+}(G_{t,N}(\mathbf{n})-G_{t,n}(\mathbf{m}))|t^{\sigma-1}\,dt=Q_{1}+Q_{2},$

with

$Q_{1}=\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}|\delta_{i}^{+}(G_{t,N}(\mathbf{n})-G_{t,n}(\mathbf{m}))|t^{\sigma-1}\,dt$

and

$Q_{2}=\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}|\delta_{i}^{+}(G_{t,N}(\mathbf{n})-G_{t,n}(\mathbf{m}))|t^{\sigma-1}\,dt.$

To obtain the required bound for $Q_{1}$ we consider Lemma 22 and apply (14)
with $\alpha=1/N$ and $\alpha=2/N$. Indeed,

$\displaystyle Q_{1}$ $\displaystyle\leq
C|\mathbf{n}-\mathbf{m}|\left(\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}\frac{1}{t}\left(G_{t,1}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{K}\right)\right)^{N}t^{\sigma-1}\,dt\right.$
$\displaystyle\kern
10.0pt\left.+(|\mathbf{n}|+|\mathbf{m}|)^{2}\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}\frac{1}{t^{2}}\left(G_{t,1}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{K}\right)\right)^{N}t^{\sigma-1}\,dt\right)$
$\displaystyle\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+2}}\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}t^{\sigma-1}\,dt\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+2-2\sigma}}.$

The estimate for $Q_{2}$ can be deduced from (67) in the following way

$\displaystyle Q_{2}$
$\displaystyle\leq|\mathbf{n}-\mathbf{m}|\left(\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}t^{\sigma-N/2-2}\,dt+(|\mathbf{n}|+|\mathbf{m}|)^{2}\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}t^{\sigma-N/2-3}\,dt\right)$
$\displaystyle\leq\frac{C|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+2-2\sigma}}.\qed$

###### Proof of Theorem 19.

a) By using that

$(\delta_{i}^{+})^{k}(-\Delta_{N})^{-\sigma}f(\mathbf{n})=(-\Delta_{N})(\delta_{i}^{+})^{k}f(\mathbf{n}),$

it is enough to prove the result for $k=0$. Now, from Lemma 20, it is verified
that

$(-\Delta_{N})^{-\sigma}f(\mathbf{n})-(-\Delta_{N})^{-\sigma}f(\mathbf{m})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}(K_{\sigma}(\mathbf{n}-\mathbf{k})-K_{\sigma}(\mathbf{m}-\mathbf{k}))(f(\mathbf{k})-f(\mathbf{n})).$

Now, we split the sum on $0<|\mathbf{n}-\mathbf{k}|\leq
2|\mathbf{n}-\mathbf{m}|$ and its complementary. By using (61) we get

$\sum_{0<|\mathbf{n}-\mathbf{k}|\leq
2|\mathbf{n}-\mathbf{m}|}K_{\sigma}(\mathbf{n}-\mathbf{k})|f(\mathbf{k})-f(\mathbf{n})|\\\
\begin{aligned} &\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\sum_{0<|\mathbf{n}-\mathbf{k}|\leq
2|\mathbf{n}-\mathbf{m}|}\frac{|\mathbf{n}-\mathbf{k}|^{\alpha}}{|\mathbf{n}-\mathbf{k}|^{N-2\sigma}}\\\
&\leq C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\sum_{0<|\mathbf{j}|\leq
2|\mathbf{n}-\mathbf{m}|}|\mathbf{j}|^{\alpha+2\sigma-N}\\\ &\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha+2\sigma}.\end{aligned}$

For the other term in the first part of the decomposition, by using that
$|\mathbf{n}-\mathbf{k}|\leq 2|\mathbf{n}-\mathbf{m}|$ implies
$|\mathbf{m}-\mathbf{k}|\leq 3|\mathbf{n}-\mathbf{m}|$ and (61), we have

$\sum_{0<|\mathbf{n}-\mathbf{k}|\leq
2|\mathbf{n}-\mathbf{m}|}K_{\sigma}(\mathbf{m}-\mathbf{k})|f(\mathbf{k})-f(\mathbf{n})|\\\
\begin{aligned} &\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\sum_{|\mathbf{m}-\mathbf{k}|\leq
3|\mathbf{n}-\mathbf{m}|}\frac{|\mathbf{n}-\mathbf{k}|^{\alpha}}{(|\mathbf{m}-\mathbf{k}|+1)^{N-2\sigma}}\\\
&\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\left(\sum_{|\mathbf{m}-\mathbf{k}|\leq
3|\mathbf{n}-\mathbf{m}|}\frac{|\mathbf{n}-\mathbf{m}|^{\alpha}}{(|\mathbf{m}-\mathbf{k}|+1)^{N-2\sigma}}\right.\\\
&\kern 20.0pt\left.+\sum_{|\mathbf{m}-\mathbf{k}|\leq
3|\mathbf{n}-\mathbf{m}|}(|\mathbf{m}-\mathbf{k}|+1)^{\alpha+2\sigma-N}\right)\\\
&\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha+2\sigma}.\end{aligned}$

Finally, applying (62), we have

$\sum_{|\mathbf{n}-\mathbf{k}|\geq
2|\mathbf{n}-\mathbf{m}|}|K_{\sigma}(\mathbf{n}-\mathbf{k})-K_{\sigma}(\mathbf{m}-\mathbf{k})||f(\mathbf{k})-f(\mathbf{n})|\\\
\begin{aligned} &\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|\sum_{|\mathbf{n}-\mathbf{k}|\geq
2|\mathbf{n}-\mathbf{m}|}\frac{|\mathbf{n}-\mathbf{k}|^{\alpha}}{|\mathbf{n}-\mathbf{k}|^{N+1-2\sigma}}\\\
&\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha+2\sigma}\end{aligned}$

and we have concluded this part.

b) Again, we can reduce the proof to the case $k=0$. Now, from Lemma 20, for
any $i=1,\dots,n$ we have

$\delta_{i}^{+}((-\Delta_{N})^{-\sigma}f(\mathbf{n})-(-\Delta_{N})^{-\sigma}f(\mathbf{m}))\\\
=\sum_{k\in\mathbb{Z}^{N}}\delta_{i}^{+}(K_{\sigma}(\mathbf{n}-\mathbf{k})-K_{\sigma}(\mathbf{m}-\mathbf{k}))(f(\mathbf{k})-f(\mathbf{n}))).$

Splitting the sum as in a), the proof follows the same steps but using in the
region $|\mathbf{n}-\mathbf{k}|\leq 2|\mathbf{n}-\mathbf{m}|$ the bounds
(deduced from (62))

$|K_{\sigma}(\mathbf{n}+\mathbf{e_{i}}-\mathbf{k})-K_{\sigma}(\mathbf{n}-\mathbf{k})|\leq
C|\mathbf{n}-\mathbf{k}|^{2\sigma-N-1},\qquad\mathbf{n},\mathbf{k}\in\mathbb{Z},$

and

$|K_{\sigma}(\mathbf{m}+\mathbf{e_{i}}-\mathbf{k})-K_{\sigma}(\mathbf{m}-\mathbf{k})|\leq
C|\mathbf{m}-\mathbf{k}|^{2\sigma-N-1},\qquad\mathbf{m},\mathbf{k}\in\mathbb{Z},$

and in the region $|\mathbf{n}-\mathbf{k}|>2|\mathbf{n}-\mathbf{m}|$ the
estimate (68).

c) The proof of this part can be done as in a) but changing the bound
$|f(\mathbf{n})-f(\mathbf{k})|\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{k}|^{\alpha}$ by
$|f(\mathbf{n})-f(\mathbf{k})|\leq 2\|f\|_{\ell^{\infty}(\mathbb{Z}^{N})}$. ∎

###  4.3. The Riesz transforms

As we have said at the beginning of this section the Riesz transform are well
defined, because the kernel

$\int_{0}^{\infty}\delta_{i}^{+}G_{t,N}(\mathbf{n})t^{-1/2}\,dt$

is absolutely convergent.

Of course the Riesz transforms can be written for $N\geq 2$ as

$R_{i}f(\mathbf{n})=\mathcal{R}_{i}\ast f(\mathbf{n}),$

with

$\mathcal{R}_{i}(\mathbf{n})=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\delta_{i}^{+}G_{t,N}(\mathbf{n})t^{-1/2}\,dt.$

To present our results about the Riesz transforms we need two preliminary
lemmas.

######  Lemma 24.

Let $N\geq 2$, then the inequalities

(69) $|\mathcal{R}_{i}(\mathbf{n})|\leq\frac{C}{(|\mathbf{n}|+1)^{N}}$

and

(70) $|\mathcal{R}_{i}(\mathbf{n})-\mathcal{R}_{i}(\mathbf{m})|\leq
C\frac{|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1}},\qquad|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|,$

hold.

###### Proof.

To prove (69) we consider $\mathbf{n}\not=\mathbf{0}$ because the bound
$|\mathcal{R}_{i}(\mathbf{0})|\leq C$ is obvious. Now, from (5) and (3), it is
deduced that

$|\delta_{i}G_{t,N}(\mathbf{n})|\leq
C\frac{|n_{i}|+1}{t}G_{t,N}(\mathbf{n})\leq
C\frac{|\mathbf{n}|}{t}\left(G_{t,1}\left(\frac{|\mathbf{n}|}{N}\right)\right)^{N}.$

Then, using (14) with $\alpha=1/N$ and $\alpha=-1/2$, we deduce that

$|\mathcal{R}_{i}(\mathbf{n})|\leq
C|\mathbf{n}|\left(\frac{1}{|\mathbf{n}|^{N+2}}\int_{0}^{|\mathbf{n}|^{2}}t^{-1/2}\,dt+\int_{|\mathbf{n}|^{2}}^{\infty}t^{-N/2-3/2}\,dt\right)\leq\frac{C}{|\mathbf{n}|^{N}}.$

Now, we decompose the difference in (70) in the following way

$|\mathcal{R}_{i}(\mathbf{n})-\mathcal{R}_{i}(\mathbf{m})|\leq\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}|\delta_{i}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|t^{-1/2}\,dt\\\
+\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}|\delta_{i}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|t^{-1/2}\,dt:=L_{1}+L_{2}.$

To analyze $L_{1}$ we apply (63) and (14) with $\alpha=1/N$ and $\alpha=2/N$
to obtain that

$|\delta_{i}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|\leq
C\frac{|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+2}}$

and

$L_{1}\leq
C\frac{|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+2}}\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}t^{-1/2}\,dt\leq
C\frac{|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1}}.$

From (67), we have

$\displaystyle L_{2}$ $\displaystyle\leq
C|\mathbf{n}-\mathbf{m}|\left(\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}t^{-N/2-3/2}\,dt+(|\mathbf{n}|+|\mathbf{m}|)^{2}\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}t^{-N/2-5/2}\,dt\right)$
$\displaystyle\leq
C\frac{|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1}}$

and the proof is finished. ∎

######  Lemma 25.

For $N\geq 2$ it is verified that

(71) $\sum_{\mathbf{k}\in\mathbb{Z}^{N}}\mathcal{R}_{i}(\mathbf{k})=0,\qquad
i=1,\dots,N,$

and, for a fix value $\ell\in\mathbb{N}$,

(72) $\left|\sum_{|\mathbf{k}|\geq\ell}\mathcal{R}_{i}(\mathbf{k})\right|\leq
C,\qquad i=1,\dots,N.$

###### Proof.

To prove the result we focus on $i=1$ because the other cases are similar. We
consider $M\in\mathbb{N}$ big enough and the partial sums

$S_{M}=\sum_{|\mathbf{k}|\leq M}\mathcal{R}_{1}(\mathbf{k}).$

Now, we observe that

(73)
$\mathcal{R}_{1}(-\mathbf{k})=-\mathcal{R}_{1}(\mathbf{k}-\mathbf{e}_{1}).$

Indeed, it is clear that

$\displaystyle\delta_{1}^{+}G_{t,N}(-\mathbf{k})$
$\displaystyle=(G_{t,1}(-k_{1}+1)-G_{t,1}(-k_{1}))\prod_{j=2}^{N}G_{t,1}(-k_{j})$
$\displaystyle=-(G_{t,1}(k_{1})-G_{t,1}(k_{1}-1))\prod_{j=2}^{N}G_{t,1}(k_{j})=-\delta_{1}^{+}G_{t,N}(\mathbf{k}-\mathbf{e}_{1})$

and (73) follows immediately. With (73) we have

$\displaystyle 2S_{M}$ $\displaystyle=\sum_{|\mathbf{k}|\leq
M}\mathcal{R}_{1}(\mathbf{k})+\sum_{|\mathbf{k}|\leq
M}\mathcal{R}_{1}(-\mathbf{k})$ $\displaystyle=\sum_{|\mathbf{k}|\leq
M}\mathcal{R}_{1}(\mathbf{k})-\sum_{|\mathbf{k}|\leq
M}\mathcal{R}_{1}(\mathbf{k}-\mathbf{e}_{1})=\sum_{|\mathbf{k}|\leq
M}\mathcal{R}_{1}(\mathbf{k})-\sum_{|\mathbf{k}+\mathbf{e}_{1}|\leq
M}\mathcal{R}_{1}(\mathbf{k}).$

Then, applying (69) we get

$|S_{M}|\leq C\sum_{M-1\leq|\mathbf{k}|\leq
M+1}|\mathcal{R}_{1}(\mathbf{k})|\leq\frac{C}{M}$

and

$\sum_{\mathbf{k}\in\mathbb{Z}^{N}}\mathcal{R}_{1}(\mathbf{k})=\lim_{M\to\infty}S_{M}=0$

proving (71).

To analyze (72), we have to note that

$\left|\sum_{|\mathbf{k}|\geq\ell}\mathcal{R}_{i}(\mathbf{k})\right|=\left|\sum_{\mathbf{k}\in\mathbb{Z}^{N}}\mathcal{R}_{i}(\mathbf{k})-S_{\ell-1}\right|=|S_{\ell-1}|\leq\frac{C}{\ell-1}$

and the result follows for $\ell>1$. The case $\ell=1$ is obvious. ∎

First, we present the behaviour of the Riesz transforms on the Hölder classes.

######  Theorem 26.

Let $N\geq 2$, $0<\alpha<1/2$, and $f\in C^{0,\alpha}(\mathbb{Z}^{N})$. Then
for $1\leq i\leq N$ the inequality

$[R_{i}f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}$

holds.

###### Proof.

By using Lemma 25, for $\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}$ it is easy to
check that

$\displaystyle R_{i}f(\mathbf{n})$
$\displaystyle=\sum_{k\in\mathbb{Z}^{N}}(f(\mathbf{k})-f(\mathbf{n}))\mathcal{R}_{i}(\mathbf{n}-\mathbf{k})$
$\displaystyle=\sum_{0<|\mathbf{n}-\mathbf{k}|\leq
2|\mathbf{n}-\mathbf{m}|}(f(\mathbf{k})-f(\mathbf{n}))\mathcal{R}_{i}(\mathbf{n}-\mathbf{k})$
$\displaystyle\kern
20.0pt+\sum_{|\mathbf{n}-\mathbf{k}|>2|\mathbf{n}-\mathbf{m}|}(f(\mathbf{k})-f(\mathbf{n}))\mathcal{R}_{i}(\mathbf{n}-\mathbf{k})$
$\displaystyle:=S_{1}(\mathbf{n})+S_{2}(\mathbf{n}).$

For the difference $S_{1}(\mathbf{n})-S_{1}(\mathbf{m})$, by using that
$|\mathbf{n}-\mathbf{k}|\leq 2|\mathbf{n}-\mathbf{m}|$ implies
$|\mathbf{m}-\mathbf{k}|\leq 3|\mathbf{n}-\mathbf{m}|$ and the estimate (69),
it is verified that

$|S_{1}(\mathbf{n})-S_{1}(\mathbf{m})|\leq|S_{1}(\mathbf{n})|+|S_{1}(\mathbf{m})|\\\
\begin{aligned} &\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\left(\sum_{0<|\mathbf{n}-\mathbf{k}|\leq
2|\mathbf{n}-\mathbf{m}|}|\mathbf{k}-\mathbf{n}|^{\alpha-N}+\sum_{0<|\mathbf{m}-\mathbf{k}|\leq
3|\mathbf{n}-\mathbf{m}|}|\mathbf{k}-\mathbf{m}|^{\alpha-N}\right)\\\ &\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha}.\end{aligned}$

To treat $S_{2}(\mathbf{n})-S_{2}(\mathbf{m})$ we add the term $\pm
f(\mathbf{m})\mathcal{R}_{i}(\mathbf{n}-\mathbf{k})$ and apply (70) and (72)
to have

$|S_{2}(\mathbf{n})-S_{2}(\mathbf{m})|\\\ \begin{aligned}
&\leq\sum_{|\mathbf{n}-\mathbf{k}|>2|\mathbf{n}-\mathbf{m}|}|f(\mathbf{k})-f(\mathbf{m})||\mathcal{R}_{i}(\mathbf{n}-\mathbf{k})-\mathcal{R}_{i}(\mathbf{m}-\mathbf{k})|\\\
&\kern
15.0pt+|f(\mathbf{n})-f(\mathbf{m})|\left|\sum_{|\mathbf{n}-\mathbf{k}|>2|\mathbf{n}-\mathbf{m}|}\mathcal{R}_{i}(\mathbf{n}-\mathbf{k})\right|\\\
&\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\left(|\mathbf{n}-\mathbf{m}|\sum_{|\mathbf{n}-\mathbf{k}|>2|\mathbf{n}-\mathbf{m}|}|\mathbf{n}-\mathbf{k}|^{\alpha-N-1}+|\mathbf{n}-\mathbf{m}|^{\alpha}\right)\\\
&\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha}\end{aligned}$

and this concludes the proof. ∎

It would be possible to define another Riesz transform by mean

$\overline{R}_{i}f(\mathbf{n})=\delta_{i}^{-}(-\Delta_{N})^{-1/2}f(\mathbf{n}).$

The identity
$\delta_{i}^{-}f(\mathbf{n})=\delta_{i}^{+}f(\mathbf{n}-\mathbf{e}_{i})$
proves the following result.

######  Corollary 27.

Let $N\geq 2$, $0<\alpha<1/2$, and $f\in C^{0,\alpha}(\mathbb{Z}^{N})$. Then
for $1\leq i\leq N$ the inequality

$[\overline{R}_{i}f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}$

holds.

To finish this section we give a result about the boundedness with weights of
the Riesz transforms.

######  Theorem 28.

Let $N\geq 2$, $w\in A_{p}(\mathbb{Z}^{N})$, and $1\leq i\leq N$. Then the
inequalities

$\|R_{i}f\|_{\ell^{p}(\mathbb{Z}^{n},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{n},w)},\qquad 1<p<\infty,\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w),$

and

$\|R_{i}f\|_{\ell^{1,\infty}(\mathbb{Z}^{n},w)}\leq
C\|f\|_{\ell^{1}(\mathbb{Z}^{n},w)},\qquad\ell^{2}(\mathbb{Z}^{N})\cap\ell^{1}(\mathbb{Z}^{N},w),$

hold.

###### Proof.

We will prove that each $R_{i}$ is Calderón-Zygmund operator.

First, we check that $R_{i}$ is bounded from $\ell^{2}(\mathbb{Z}^{N})$ into
itself. For $f\in\ell^{2}(\mathbb{Z}^{N})$, we have

$\mathcal{F}(\delta_{i}f)(x)=\sum_{\mathbf{n}\in\mathbb{Z}^{N}}f(\mathbf{n}+\mathbf{e}_{i})e^{2\pi
i\langle\mathbf{n},x\rangle}-\sum_{\mathbf{n}\in\mathbb{Z}^{N}}f(\mathbf{n})e^{2\pi
i\langle\mathbf{n},x\rangle}=(e^{-2\pi ix_{i}}-1)\mathcal{F}f(x)$

and

$\displaystyle\mathcal{F}((-\Delta_{N})^{-1/2}f)(x)$
$\displaystyle=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-4t(\sum_{k=1}^{N}\sin^{2}(\pi
x_{k}))}\mathcal{F}f(x)t^{-1/2}\,dt$
$\displaystyle=\left(4\sum_{k=1}^{N}\sin^{2}(\pi
x_{k})\right)^{-1/2}\mathcal{F}f(x).$

With this we deduce that $\delta_{i}^{+}$ and $(-\Delta_{N})^{-1/2}$ can be
seen as multipliers of the functions $e^{-2\pi ix_{i}}-1$ and
$\left(4\sum_{k=1}^{N}\sin^{2}(\pi x_{k})\right)^{-1/2}$, respectively. Then,
the operator $R_{i}$ is associated with the multiplier

$\frac{e^{-2\pi ix_{i}}-1}{\left(4\sum_{k=1}^{N}\sin^{2}(\pi
x_{k})\right)^{1/2}};$

i.e.,

$\displaystyle\mathcal{F}(R_{i}f)(x)$ $\displaystyle=\frac{e^{-2\pi
ix_{i}}-1}{\left(4\sum_{k=1}^{N}\sin^{2}(\pi x_{k})\right)^{1/2}}\mathcal{F}f$
$\displaystyle=\frac{-ie^{\pi ix_{i}}\sin(\pi
x_{i})}{\left(\sum_{k=1}^{N}\sin^{2}(\pi
x_{k})\right)^{1/2}}\mathcal{F}f,\qquad i=1,\dots,N.$

In this way, using the bound

$\frac{|\sin(\pi x_{i})|}{\left(\sum_{k=1}^{N}\sin^{2}(\pi
x_{k})\right)^{1/2}}\leq 1,$

by Plancherel identity we have

$\|R_{i}f\|_{\ell^{2}(\mathbb{Z}^{N})}=\|\mathcal{F}(R_{i}f)\|_{L^{2}\left([-1/2,1/2]^{N}\right)}\leq
C\|\mathcal{F}f\|_{L^{2}\left([-1/2,1/2]^{N}\right)}=\|f\|_{\ell^{2}(\mathbb{Z}^{N})}.$

Finally, the estimates (55) and (56) can be deduced from (69) and (70) and we
have concluded. ∎

For the Riesz transforms $\overline{R}_{i}$ we have the next result.

######  Corollary 29.

Let $N\geq 2$, $w\in A_{p}(\mathbb{Z}^{N})$, and $1\leq i\leq N$. Then the
inequalities

$\|\overline{R}_{i}f\|_{\ell^{p}(\mathbb{Z}^{n},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{n},w)},\qquad 1<p<\infty,\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w),$

and

$\|\overline{R}_{i}f\|_{\ell^{1,\infty}(\mathbb{Z}^{n},w)}\leq
C\|f\|_{\ell^{1}(\mathbb{Z}^{n},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{1}(\mathbb{Z}^{N},w),$

hold.

##  5\. Fractional powers of the Laplacian

From the identity

$\lambda^{s}=\frac{1}{\Gamma(-s)}\int_{0}^{\infty}(e^{-\lambda
t}-1)\frac{dt}{t^{s+1}},\qquad 0<s<1,$

we define the fractional powers (in fact, positive powers) of $-\Delta_{N}$ by

(74)
$(-\Delta_{N})^{s}f(\mathbf{n})=\frac{1}{\Gamma(-s)}\int_{0}^{\infty}(W_{t}f(\mathbf{n})-f(\mathbf{n}))\frac{dt}{t^{s+1}},\qquad
f\in\ell^{2}(\mathbb{Z}^{N}).$

By using (21), it is clear that

(75)
$W_{t}f(\mathbf{n})-f(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}G_{t,N}(\mathbf{n}-\mathbf{k})(f(\mathbf{k})-f(\mathbf{n})).$

Moreover, we can prove that the operators $(-\Delta_{N})^{s}$ are well defined
in $\ell^{2}(\mathbb{Z}^{N})$ by (74). Indeed, we consider the decomposition

$\int_{0}^{\infty}|W_{t}f(\mathbf{n})-f(\mathbf{n})|\frac{dt}{t^{s+1}}\leq
F_{1}+F_{2}$

where

$F_{1}=\int_{0}^{1}|W_{t}f(\mathbf{n})-f(\mathbf{n})|\frac{dt}{t^{s+1}}$

and $F_{2}$ is the integral on $[1,\infty)$ of the same function. For $F_{1}$,
applying (26), we have

$F_{1}\leq C\|f\|_{\ell^{2}(\mathbb{Z}^{N})}\int_{0}^{1}t^{-s}\,dt\leq
C\|f\|_{\ell^{2}(\mathbb{Z}^{N})}.$

Now, from Theorem 6 with $p=\infty$ and $q=2$, we deduce that

$|W_{t}f(\mathbf{n})-f(\mathbf{n})|\leq
C(t^{-N/4}+1)\|f\|_{\ell^{2}(\mathbb{Z}^{N})}$

and

$F_{2}\leq
C\|f\|_{\ell^{2}(\mathbb{Z}^{N})}\int_{1}^{\infty}(t^{-N/4}+1)t^{-s-1}\,dt\leq
C\|f\|_{\ell^{2}(\mathbb{Z}^{N})}.$

In this section we will focus on the study of some aspects of the operators
$(-\Delta_{N})^{s}$. Firstly, we will show some basic properties of the
fractional powers of the Laplacian, including a maximum principle. In the last
part of the section we will deal with the mapping properties of
$(-\Delta_{N})^{s}$ on the Hölder classes.

###  5.1. Basic properties of the fractional powers of the Laplacian

In this section we present some elementary properties of the fractional powers
$(-\Delta_{N})^{s}$. They are collected in the next theorem.

######  Theorem 30.

Let $N\geq 1$ and $0<s<1$.

  1. i)

For $f\in\ell_{-s}(\mathbb{Z}^{N})$, it is verified that

$(-\Delta_{N})^{s}f(\mathbf{n})=\sum_{\begin{smallmatrix}\mathbf{k}\in\mathbb{Z}^{N}\\\
\mathbf{k}\not=\mathbf{n}\end{smallmatrix}}\mathcal{K}_{s}(\mathbf{n}-\mathbf{k})(f(\mathbf{n})-f(\mathbf{k})),$

where

(76)
$\mathcal{K}_{s}(\mathbf{n})=\frac{1}{|\Gamma(-s)|}\int_{0}^{\infty}G_{t,N}(\mathbf{n})\frac{dt}{t^{s+1}},\qquad\mathbf{n}\not=\mathbf{0},$

and $\mathcal{K}_{s}(\mathbf{0})=0$.

  2. ii)

For $\mathbf{n}\not=\mathbf{0}$, the inequalities

(77)
$0<\mathcal{K}_{s}(\mathbf{n})\leq\frac{C}{|\mathbf{n}|^{N+2s}}\frac{1}{|\Gamma(-s)|}\left(\frac{1}{1-s}+\frac{2}{N+2s}\right)$

hold with a constant $C$ independent of $s$.

  3. iii)

For $f\in\ell_{0}(\mathbb{Z}^{N})$, it is verified that

$\lim_{s\to 0^{+}}(-\Delta_{N})^{s}f(\mathbf{n})=f(\mathbf{n}).$

  4. iv)

For $f\in\ell^{\infty}(\mathbb{Z}^{N})$, it is verified that

$\lim_{s\to 1^{-}}(-\Delta_{N})^{s}f(\mathbf{n})=-\Delta_{N}f(\mathbf{n}).$

###### Proof.

First, we observe that $W_{t}f$ is well defined on the spaces
$\ell_{-s}(\mathbb{Z}^{N})$ for $0\leq s\leq 1$. Effectively, for $M>0$ big
enough, with $t$ fix, we have

$\sum_{|\mathbf{k}|>M}G_{t,N}(\mathbf{k})f(\mathbf{n}-\mathbf{k})\leq
C\|f\|_{\ell_{-s}(\mathbb{Z}^{N})}\sup_{|\mathbf{k}|>M}G_{t,N}(\mathbf{k})(1+|\mathbf{n}-\mathbf{k}|)^{N+2s}.$

From (3) and the asymptotic expansion (20), we deduce that

$\sup_{|\mathbf{k}|>M}G_{t,N}(\mathbf{k})(1+|\mathbf{n}-\mathbf{k}|)^{N+2s}\\\
\leq
Ce^{-2Nt}N^{N/2}\sup_{{|\mathbf{k}|>M}}\frac{(Net)^{|\mathbf{k}|}(1+|\mathbf{n}|+|\mathbf{k}|)^{N+2s}}{|\mathbf{k}|^{|\mathbf{k}|+N/2}}\leq
C_{t,M,|\mathbf{n}|,s,N}$

and we conclude.

To prove i) we use (75) to get

$\displaystyle(-\Delta_{N})^{s}f(\mathbf{n})$
$\displaystyle=\frac{1}{\Gamma(-s)}\int_{0}^{\infty}\sum_{\mathbf{k}\not=\mathbf{n}}G_{t,N}(\mathbf{n}-\mathbf{k})(f(\mathbf{k})-f(\mathbf{n}))\frac{dt}{t^{s+1}}$
$\displaystyle=\sum_{\mathbf{k}\not=\mathbf{n}}\mathcal{K}_{s}(\mathbf{n}-\mathbf{k})(f(\mathbf{n})-f(\mathbf{k}))$

where $\mathcal{K}_{s}$ is given by (76). The proof will be completed
justifying the application of Fubini theorem in the second identity. To do
this we need the upper bound in ii), so we proceed with its proof.

From (3) and (14) with $\alpha=1/N$, we obtain that

$G_{t,N}(\mathbf{n})t^{-1}\leq\frac{C}{|\mathbf{n}|^{N+2}}.$

Then, with (30), we deduce that

$\displaystyle\mathcal{K}_{s}(\mathbf{n})$
$\displaystyle\leq\frac{C}{|\Gamma(-s)|}\left(\frac{1}{|\mathbf{n}|^{N+2}}\int_{0}^{|\mathbf{n}|^{2}}t^{-s}\,dt+\int_{|\mathbf{n}|^{2}}^{\infty}t^{-N/2-s-1}\,dt\right)$
$\displaystyle=\frac{C}{|\mathbf{n}|^{N+2s}}\frac{1}{|\Gamma(-s)|}\left(\frac{1}{1-s}+\frac{2}{N+2s}\right).$

The positivity of the kernel $\mathcal{K}_{s}$ is clear.

Let us justify the interchange of the sum and the integral. First, we apply
(77) and Tonelli’s theorem in the term

$\int_{0}^{\infty}\sum_{\mathbf{k}\not=\mathbf{n}}G_{t,N}(\mathbf{n}-\mathbf{k})|f(\mathbf{k})|\frac{dt}{t^{s+1}}$

to have that it is bounded by

$\sum_{\mathbf{k}\not=\mathbf{n}}\frac{|f(\mathbf{k})|}{|\mathbf{n}-\mathbf{k}|^{N+2s}}$

and this term is finite for each $\mathbf{n}$ because
$f\in\ell_{-s}(\mathbb{Z}^{N})$. Similarly for the term

$|f(\mathbf{n})|\int_{0}^{\infty}\sum_{\mathbf{k}\not=\mathbf{n}}G_{t,N}(\mathbf{n}-\mathbf{k})\frac{dt}{t^{s+1}},$

we obtain the bound

$|f(\mathbf{n})|\sum_{\mathbf{k}\not=\mathbf{n}}\frac{1}{|\mathbf{n}-\mathbf{k}|^{N+2s}}$

and this last sum is finite.

To prove iii) we check that

$\lim_{s\to 0^{+}}N_{1}=1\qquad\text{ and }\qquad\lim_{s\to 0^{+}}N_{2}=0$

where

$N_{1}=\sum_{\mathbf{k}\not=\mathbf{n}}\mathcal{K}_{s}(\mathbf{n}-\mathbf{k})\qquad\text{
and }\qquad
N_{2}=\sum_{\mathbf{k}\not=\mathbf{n}}\mathcal{K}_{s}(\mathbf{n}-\mathbf{k})|f(\mathbf{k})|.$

We start with $N_{2}$. From (77) and (30), we have

$N_{2}\leq\frac{C}{|\Gamma(-s)|}\left(\frac{1}{1-s}+\frac{2}{N+2s}\right)\sum_{\mathbf{k}\not=\mathbf{n}}\frac{|f(\mathbf{k})|}{|\mathbf{n}-\mathbf{k}|^{N+2s}}.$

Now, using that $f\in\ell_{0}(\mathbb{Z}^{N})$, applying dominated
convergence, we deduce that for each $\mathbf{n}$ it is verified that

$\sum_{\mathbf{k}\not=\mathbf{n}}\frac{|f(\mathbf{k})|}{|\mathbf{n}-\mathbf{k}|^{N+2s}}\leq
C\|f\|_{\ell_{0}(\mathbb{Z}^{N})}$

and then $\lim_{s\to 0^{+}}N_{2}=0$. To treat $N_{1}$, we consider the
decomposition $N_{1}=N_{1,1}+N_{1,2}$ where

$N_{1,1}=\frac{1}{|\Gamma(-s)|}\sum_{\mathbf{k}\not=\mathbf{n}}\int_{0}^{1}G_{t,N}(\mathbf{n}-\mathbf{k})\frac{dt}{t^{s+1}}$

and $N_{1,2}$ is the same sum but with the integral on the interval
$[1,\infty)$. To analyze $N_{1,1}$ we apply (3) and (14) with $\alpha=1/(2N)$
and we consider $0<s<1/2$. In this way,

$N_{1,1}\leq\frac{C}{|\Gamma(-s)|}\sum_{\mathbf{k}\not=\mathbf{n}}\frac{1}{|\mathbf{n}-\mathbf{k}|^{N+1}}\int_{0}^{1}t^{-1/2-s}\,dt\leq\frac{C}{|\Gamma(-s)|(1-2s)}$

and $\lim_{s\to 0^{+}}N_{1,1}=0$. Now, using that

$\sum_{\mathbf{k}\not=\mathbf{n}}G_{t,N}(\mathbf{n}-\mathbf{k})=1-G_{t,N}(\mathbf{0}),$

we obtain that

$N_{1,2}=\frac{1}{|\Gamma(-s)|}\left(\frac{1}{s}-\int_{1}^{\infty}G_{t,N}(\mathbf{0})\frac{dt}{t^{s+1}}\right).$

With the bound (30) we get

$\int_{1}^{\infty}G_{t,N}(\mathbf{0})\frac{dt}{t^{s+1}}\leq
C\int_{1}^{\infty}t^{-N/2-s-1}\,dt=\frac{C}{N+2s}$

and, with the identity $|\Gamma(-s)|s=\Gamma(1-s)$ we conclude that
$\lim_{s\to 0^{+}}N_{1,2}=1$.

To prove iv) we start observing

$(-\Delta_{N})^{s}f(\mathbf{n})=\sum_{k=1}^{N}(\mathcal{K}_{s}(\mathbf{e}_{k}))(f(\mathbf{n})-f(\mathbf{n}-\mathbf{e}_{k}))+\mathcal{K}_{s}(-\mathbf{e}_{k}))(f(\mathbf{n})-f(\mathbf{n}+\mathbf{e}_{k})))\\\
+\sum_{\begin{smallmatrix}\mathbf{k}\in\mathbb{Z}^{N}\\\
\mathbf{k}\notin\\{\mathbf{0},\pm\mathbf{e}_{1},\dots,\pm\mathbf{e}_{N}\\}\end{smallmatrix}}\mathcal{K}_{s}(\mathbf{k})(f(\mathbf{n})-f(\mathbf{n}-\mathbf{k})):=M_{1}+M_{2}.$

Taking into account that
$\mathcal{K}_{s}(-\mathbf{n})=\mathcal{K}_{s}(\mathbf{n})$, we have

$M_{1}=-\sum_{k=1}^{N}\mathcal{K}_{s}(\mathbf{e}_{k})\Delta_{N,i}f(\mathbf{n}).$

Now, using that for any continuous and bounded function on $[0,\infty)$,

$\lim_{s\to
1^{-}}\frac{1}{\Gamma(1-s)}\int_{0}^{\infty}e^{-t}f(t)\frac{dt}{t^{s}}=f(0)$

(the identity is clear for polynomials and with the hypotheses on $f$ the
general result is obtained), taking the function

$f(z)=\frac{2I_{1}(z)}{z}(I_{0}(z))^{N-1},$

we deduce that

$\displaystyle\lim_{s\to 1^{-}}\mathcal{K}_{s}(\mathbf{e}_{i})$
$\displaystyle=\lim_{s\to
1^{-}}\frac{1}{\Gamma(1-s)}\int_{0}^{\infty}e^{-2Nt}\frac{I_{1}(2t)}{t}(I_{0}(2t))^{N-1}\frac{dt}{t^{s}}$
$\displaystyle=\lim_{s\to
1^{-}}\frac{(2N)^{s-1}}{\Gamma(1-s)}\int_{0}^{\infty}e^{-w}f\left(\frac{w}{N}\right)\frac{dw}{w^{s}}=f(0)=1.$

Then,

$\lim_{s\to 1^{-}}M_{1}=-\Delta_{N}f(\mathbf{n}).$

Finally, let us check that $\lim_{s\to 1^{-}}M_{2}=0$. By using that
$G_{t,N}(\mathbf{n})t^{-3/2}\leq C|\mathbf{n}|^{-N-3}$ (which can be deduced
from (3) and (14) with $\alpha=3/(2N)$) and proceeding as in the proof of
(77), we can obtain the bound

$\mathcal{K}_{s}(\mathbf{n})\leq\frac{C}{|\mathbf{n}|^{N+2s}}\frac{1}{|\Gamma(-s)|}\left(\frac{1}{3-2s}+\frac{1}{N+2s}\right).$

Then

$M_{2}\leq\frac{C}{|\Gamma(-s)|}\|f\|_{\ell^{\infty}(\mathbb{Z}^{N})}\sum_{\mathbf{k}\not=\mathbf{0}}\frac{1}{|\mathbf{k}|^{N+2s}}\leq\frac{C}{|\Gamma(-s)|}\|f\|_{\ell^{\infty}(\mathbb{Z}^{N})}$

and $\lim_{s\to 1^{-}}M_{2}=0$. ∎

To conclude this section, we present maximum and comparison principles for the
fractional powers which are obvious consequences of the positivity of the
kernel $\mathcal{K}_{s}$.

######  Theorem 31.

Let $N\geq 1$ and $0<s<1$.

  1. i)

(Maximum principle) Let $f\in\ell^{2}(\mathbb{Z}^{N})$ be such that $f\geq 0$
and $f(\mathbf{n}_{0})=0$ for some $\mathbf{n}_{0}\in\mathbb{Z}^{N}$. Then

$(-\Delta_{N})^{s}f(\mathbf{n}_{0})\leq 0.$

Moreover, $(-\Delta_{N})^{s}f(\mathbf{n}_{0})=0$ only if $f(\mathbf{n})=0$ for
all $\mathbf{n}\in\mathbb{Z}^{N}$.

  2. ii)

(Comparison principle) Let $f,g\in\ell^{2}(\mathbb{Z}^{N})$ be such that
$f\geq g$ and $f(\mathbf{n}_{0})=g(\mathbf{n}_{0})$ for some
$\mathbf{n}_{0}\in\mathbb{Z}^{N}$. Then

$(-\Delta_{N})^{s}f(\mathbf{n}_{0})\leq(-\Delta_{N})^{s}g(\mathbf{n}_{0}).$

Moreover,
$(-\Delta_{N})^{s}f(\mathbf{n}_{0})=(-\Delta_{N})^{s}g(\mathbf{n}_{0})$ only
if $f(\mathbf{n})=g(\mathbf{n})$ for all $\mathbf{n}\in\mathbb{Z}^{N}$.

###  5.2. The fractional powers of the Laplacian on the Hölder classes

The target of this subsection is the analysis of the behaviour of
$(-\Delta_{N})^{s}$ in the Hölder classes and it is contained in the next
result.

######  Theorem 32.

Let $N\geq 1$, $k\geq 0$, $0<\alpha\leq 1$, $0<s<1$, and
$f\in\ell_{-s}(\mathbb{Z}^{N})$.

  1. i)

If $f\in C^{k,\alpha}(\mathbb{Z}^{N})$ and $2s<\alpha$ then
$(-\Delta_{N})^{s}f\in C^{k,\alpha-2s}(\mathbb{Z}^{N})$ and

$[(-\Delta_{N})^{s}f]_{C^{k,\alpha-2s}(\mathbb{Z}^{N})}\leq
C[f]_{C^{k,\alpha}(\mathbb{Z}^{N})}.$

  2. ii)

If $f\in C^{k+1,\alpha}(\mathbb{Z}^{N})$ and $\alpha<2s<\alpha+1$ then
$(-\Delta_{N})^{s}f\in C^{k,\alpha-2s+1}(\mathbb{Z}^{N})$ and

$[(-\Delta_{N})^{s}f]_{C^{k,\alpha-2s+1}(\mathbb{Z}^{N})}\leq
C[f]_{C^{k+1,\alpha}(\mathbb{Z}^{N})}.$

###### Proof.

We will prove both results for $k=0$ because the other cases can be deduced by
using that $\delta_{i}^{+}(-\Delta_{N})^{s}=(-\Delta_{N})^{s}\delta_{i}^{+}$.

To prove i), we consider the identity

$|(-\Delta_{N})^{s}f(\mathbf{n})-(-\Delta_{N})^{s}f(\mathbf{m})|=|B_{1}+B_{2}|,$

where

$B_{1}=\sum_{\begin{smallmatrix}\mathbf{k}\in\mathbb{Z}^{N}\\\
1\leq|\mathbf{k}|\leq|\mathbf{n}-\mathbf{m}|\end{smallmatrix}}(f(\mathbf{n})-f(\mathbf{n}+\mathbf{k})-f(\mathbf{m})+f(\mathbf{m}+\mathbf{k}))\mathcal{K}_{s}(\mathbf{k})$

and $B_{2}$ is the sum on $|\mathbf{k}|>|\mathbf{n}-\mathbf{m}|$. First, from
the estimate

$\displaystyle|f(\mathbf{n})-f(\mathbf{n}+\mathbf{k})-f(\mathbf{m})+f(\mathbf{m}+\mathbf{k})|$
$\displaystyle\leq|f(\mathbf{n})-f(\mathbf{n}+\mathbf{k})|+|f(\mathbf{m})-f(\mathbf{m}+\mathbf{k})|$
$\displaystyle\leq C\|f\|_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{k}|^{\alpha}$

and (77), we get

$B_{1}\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\sum_{\begin{smallmatrix}\mathbf{k}\in\mathbb{Z}^{N}\\\
1\leq|\mathbf{k}|\leq|\mathbf{n}-\mathbf{m}|\end{smallmatrix}}|\mathbf{k}|^{\alpha-N-2s}\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha-2s}$

and this is appropriate for our purpose. To treat $B_{2}$, we use that

$\displaystyle|f(\mathbf{n})-f(\mathbf{n}+\mathbf{k})-f(\mathbf{m})+f(\mathbf{m}+\mathbf{k})|$
$\displaystyle\leq|f(\mathbf{n})-f(\mathbf{m})|+|f(\mathbf{n}+\mathbf{k})-f(\mathbf{m}+\mathbf{k})|$
$\displaystyle\leq
C\|f\|_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha}$

and the bound (77), to get

$B_{2}\leq[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha}\sum_{\begin{smallmatrix}\mathbf{k}\in\mathbb{Z}^{N}\\\
|\mathbf{k}|>|\mathbf{n}-\mathbf{m}|\end{smallmatrix}}|\mathbf{k}|^{-N-2s}\leq
C[f]_{C^{0,\alpha}(\mathbb{Z}^{N})}|\mathbf{n}-\mathbf{m}|^{\alpha-2s}$

and the proof of i) is completed.

We consider $N\geq 2$, the case $N=1$ can be found in [7, Theorem 1.5]. We
consider the identity

$\displaystyle(-\Delta_{N})^{s}$
$\displaystyle=(-\Delta_{N})^{s-1/2}(-\Delta_{N})^{-1/2}(-\Delta_{N})=(-\Delta_{N})^{s-1/2}(-\Delta_{N})^{-1/2}\sum_{i=1}^{N}\delta_{i}^{-}\delta_{i}^{+}$
$\displaystyle=(-\Delta_{N})^{s-1/2}\sum_{i=1}^{N}\overline{R}_{i}\delta_{i}^{+}$

and we distinguish two cases. For $s-1/2>0$ it is verified that
$0<\alpha-2(s-1/2)<1$ and, applying i) and Corollary 27, we deduce that

$\displaystyle[(-\Delta_{N})^{s}f]_{C^{0,\alpha-2s+1}(\mathbb{Z}^{N})}$
$\displaystyle\leq
C\sum_{i=1}^{N}[\overline{R}_{i}\delta_{i}^{+}f]_{C^{0,\alpha}(\mathbb{Z}^{N})}$
$\displaystyle\leq
C\sum_{i=1}^{N}[\delta_{i}^{+}f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\leq
C[f]_{C^{1,\alpha}(\mathbb{Z}^{N})}.$

In the case $s-1/2<0$ we have $0<\alpha+2(-s+1/2)$ and we can apply a) in
Theorem 19 and Corollary 27 to conclude

$\displaystyle[(-\Delta_{N})^{s}f]_{C^{0,\alpha-2s+1}(\mathbb{Z}^{N})}$
$\displaystyle=[(-\Delta_{N})^{-(-s+1/2)}\sum_{i=1}^{N}\overline{R}_{i}\delta_{i}^{+}f]_{C^{0,\alpha-2s+1}(\mathbb{Z}^{N})}$
$\displaystyle\leq
C\sum_{i=1}^{N}[\overline{R}_{i}\delta_{i}^{+}f]_{C^{0,\alpha}(\mathbb{Z}^{N})}$
$\displaystyle\leq
C\sum_{i=1}^{N}[\delta_{i}^{+}f]_{C^{0,\alpha}(\mathbb{Z}^{N})}\leq
C[f]_{C^{1,\alpha}(\mathbb{Z}^{N})}.$

For $s=1/2$ the result is a direct consequence of Corollary 27. ∎

##  6\. The discrete square functions

In this section we will prove weighted inequalities for the discrete
Littlewood-Paley-Stein $g_{k}$-square functions associated to the heat
semigroup and for the $\mathfrak{g}_{k}$-square functions associated to the
Poisson semigroup. Moreover, we conclude with a result about Laplace type
multipliers.

###  6.1. The $g_{k}$-square functions associated to the heat semigroup

The discrete $g_{k}$-square functions associated to the heat semigroup are
defined by

$g_{k}(f)(\mathbf{n})=\left(\int_{0}^{\infty}t^{2k-1}\left|\frac{\partial^{k}}{\partial
t^{k}}W_{t}f(\mathbf{n})\right|^{2}\,dt\right)^{1/2},\qquad k=1,2,\dots.$

Note that taking the Banach space
$\mathbb{B}_{k}=L^{2}((0,\infty),t^{2k-1}\,dt)$, we have

$g_{k}f(\mathbf{n})=\|P_{t,k}f(\mathbf{n})\|_{\mathbb{B}_{k}}$

with

$P_{t,k}f(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})\mathcal{P}_{t,k}(\mathbf{n}-\mathbf{k})$

and

$\displaystyle\mathcal{P}_{t,k}(\mathbf{n})$
$\displaystyle=\frac{d^{k}}{dt^{k}}G_{t,N}(\mathbf{n})=\frac{d^{k}}{dt^{k}}\int_{[-1/2,1/2]^{N}}e^{-4t\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})}e^{-2\pi i\langle x,\mathbf{n}\rangle}\,dt$
$\displaystyle=\int_{[-1/2,1/2]^{N}}F_{t,k}(x)e^{-2\pi i\langle
x,\mathbf{n}\rangle}\,dt,$

being

$F_{t,k}(x)=(-4)^{k}\left(\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})\right)^{k}e^{-4t\sum_{i=1}^{N}\sin^{2}(\pi x_{i})}.$

Our main result about the $g_{k}$-square functions is the following one.

######  Theorem 33.

Let $N\geq 1$, $1<p<\infty$, $w\in A_{p}(\mathbb{Z}^{N})$, and
$k\in\mathbb{N}$. Then the inequalities

(78)
$C_{1}\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq\|g_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C_{2}\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w),$

hold.

To obtain the proof of the equivalence in the previous theorem, we start
proving the boundedness of the $g_{k}$-square functions from
$\ell^{2}(\mathbb{Z}^{N})$ into itself.

######  Lemma 34.

For $N\geq 1$ and $k\in\mathbb{N}$, it is verified that

(79)
$\|g_{k}(f)\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}=\frac{\Gamma(2k)}{2^{2k}}\|f\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}.$

###### Proof.

For each sequence in $\ell^{2}(\mathbb{Z}^{N})$, we have

$P_{t,k}f=\mathcal{F}^{-1}(F_{t,k}\mathcal{F}f).$

Then

$\|g_{k}(f)\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}\|P_{t,k}f(\mathbf{k})\|_{\mathbb{B}_{k}}^{2}=\left\|\sum_{\mathbf{k}\in\mathbb{Z}^{N}}(P_{t,k}f(\mathbf{k}))^{2}\right\|_{\mathbb{B}_{k}}^{2}.$

By applying Parseval’s identity,

$\sum_{\mathbf{k}\in\mathbb{Z}^{N}}(P_{t,k}f(\mathbf{k}))^{2}=\int_{[-1/2,1/2]^{N}}(F_{t,k}(x))^{2}(\mathcal{F}f(x))^{2}\,dx$

and

$\displaystyle\|g_{k}(f)\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}$
$\displaystyle=\int_{0}^{\infty}t^{2k-1}\int_{[-1/2,1/2]^{N}}(F_{t,k}(x))^{2}(\mathcal{F}f(x))^{2}\,dx\,dt$
$\displaystyle=\int_{[-1/2,1/2]^{N}}(\mathcal{F}f(x))^{2}\int_{0}^{\infty}t^{2k-1}(F_{t,k}(x))^{2}\,dt\,dx$
$\displaystyle=\frac{\Gamma(2k)}{2^{2k}}\int_{[-1/2,1/2]^{N}}(\mathcal{F}f(x))^{2}\,dx=\frac{\Gamma(2k)}{2^{2k}}\|f\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}.\qed$

Now, we proceed with two reductions to simplify the proof of Theorem 33.

######  Lemma 35.

Let $N\geq 1$, $1<p<\infty$, $w\in A_{p}(\mathbb{Z}^{N})$, and
$k\in\mathbb{N}$. Then the inequality

$\|g_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w)$

implies

$\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|g_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N}).$

###### Proof.

Polarising the identity (79), we have

$\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})h(\mathbf{k})=\frac{2^{2k}}{\Gamma(2k)}\sum_{\mathbf{k}\in\mathbb{Z}^{N}}\int_{0}^{\infty}t^{2k-1}\left(\frac{d^{k}}{dt^{k}}W_{t}f(\mathbf{k})\right)\left(\frac{d^{k}}{dt^{k}}W_{t}h(\mathbf{k})\right)\,dt$

and

$\left|\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})h(\mathbf{k})\right|\leq
C\sum_{\mathbf{k}\in\mathbb{Z}^{N}}g_{k}(f)(\mathbf{k})g_{k}(h)(\mathbf{k}).$

In this way, taking $h(\mathbf{k})=w^{1/p}(\mathbf{k})f_{1}(\mathbf{k})$ and
applying a duality argument the result follows. ∎

######  Lemma 36.

Let $N\geq 1$, $1<p<\infty$, and $w\in A_{p}(\mathbb{Z}^{N})$. Then the
inequality

$\|g_{1}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N}),$

implies

$\|g_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad k>1,\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N}).$

###### Proof.

We use an induction argument to prove the result. Let us suppose that the
operator $P_{t,k}$ is bounded from $\ell^{p}(\mathbb{Z}^{N},w)$ into
$\ell^{p}_{\mathbb{B}_{k}}(\mathbb{Z}^{N},w)$. Taking $k=1$ and applying
Krivine’s theorem (see [16, Theorem 1.f.14]), we deduce that the operator
$\overline{P}_{t,1}:\ell^{p}_{\mathbb{B}_{k}}(\mathbb{Z}^{N},w)\longrightarrow\ell^{p}_{\mathbb{B}_{k}\times\mathbb{B}_{1}}(\mathbb{Z}^{N},w)$,
given by

$\\{f_{s}(n)\\}_{s\geq 0}\longmapsto\\{P_{t,1}f_{s}\\}_{t,s\geq 0},$

is bounded. Moreover, $\overline{P}_{t,1}\circ P_{s,k}$ is a bounded operator
from $\ell^{p}(\mathbb{Z}^{N},w)$ into
$\ell^{p}_{\mathbb{B}_{k}\times\mathbb{B}_{1}}(\mathbb{Z}^{N},w)$. With the
Chapman-Kolmogorov type identity

$\sum_{\mathbf{k}\in\mathbb{Z}^{N}}G_{t,N}(\mathbf{k})G_{s,N}(\mathbf{n}-\mathbf{k})=G_{t+s,N}(\mathbf{n}),$

which can be deduced from Neumann’s identity (19), we obtain that

$\frac{\partial}{\partial t}\left(W_{t}\left(\frac{\partial^{k}}{\partial
s^{k}}W_{s}f\right)\right)=\left.\frac{\partial^{k+1}}{\partial
u^{k+1}}W_{u}f\right|_{u=s+t},$

and using this we have

$\displaystyle\left\|\overline{P}_{t,1}\circ
P_{s,k}f\right\|_{\mathbb{B}_{k}\times\mathbb{B}_{1}}^{2}$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}ts^{2k-1}\left|\left.\frac{\partial^{k+1}}{\partial
u^{k+1}}W_{u}f\right|_{u=s+t}\right|^{2}\,ds\,dt$
$\displaystyle=\int_{0}^{\infty}\int_{t}^{\infty}t(r-t)^{2k-1}\left|\left.\frac{\partial^{k+1}}{\partial
u^{k+1}}W_{u}f\right|_{u=r}\right|^{2}\,dr\,dt$
$\displaystyle=\int_{0}^{\infty}\left|\frac{\partial^{k+1}}{\partial
r^{k+1}}W_{r}f\right|^{2}\int_{0}^{r}t(r-t)^{2k-1}\,dt\,dr$
$\displaystyle=\frac{1}{(2k+1)(2k)}\int_{0}^{\infty}r^{2k+1}\left|\frac{\partial^{k+1}}{\partial
r^{k+1}}W_{r}f\right|^{2}\,dr$ $\displaystyle=\frac{g_{k+1}(f)}{(2k+1)(2k)}$

and the proof is finished. ∎

After the reductions in the previous lemmas, Theorem 33 will be a consequence
of the following result.

######  Theorem 37.

Let $N\geq 1$ and $w\in A_{p}$. Then the inequalities

$\|g_{1}(f)\|_{\ell^{p}(\mathbb{Z}^{n},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{n},w)},\qquad
1<p<\infty,\qquad\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N})$

and

$\|g_{1}(f)\|_{\ell^{1,\infty}(\mathbb{Z}^{n},w)}\leq
C\|f\|_{\ell^{1}(\mathbb{Z}^{n},w)},\qquad\ell^{2}(\mathbb{Z}^{N})\cap\ell^{1}(\mathbb{Z}^{N}),$

hold.

To prove this theorem we check that $g_{1}$ is a Calderón-Zygmund operator.
With Lemma 34 and the estimates in the following lemma this fact will be
clear.

######  Lemma 38.

Let $N\geq 1$, then inequalities

(80)
$\|\mathcal{P}_{t,1}(\mathbf{n})\|_{\mathbb{B}_{1}}\leq\frac{C}{(|\mathbf{n}|+1)^{N}}$

and

(81)
$\|\mathcal{P}_{t,1}(\mathbf{n})-\mathcal{P}_{t,1}(\mathbf{m})\|_{\mathbb{B}_{1}}\leq
C\frac{|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1}},\qquad|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|$

hold.

Before proving the previous lemma, we need another one.

######  Lemma 39.

Let $N\geq 1$, $\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}$ such that
$|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|$, and

$\mathfrak{H}_{t,N}(z)=\left(\frac{z}{t^{2}}+\frac{z^{3}}{t^{3}}\right)\left(G_{t,1}(z)\right)^{N},\qquad
z>0.$

Then the inequality

(82) $|\Delta_{N,i}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|\leq
C|\mathbf{n}-\mathbf{m}|\mathfrak{H}_{t,N}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{k}\right),\qquad
i=1,\dots,N,$

holds with constants $C$ and $K>1$ independent of $\mathbf{n}$ and
$\mathbf{m}$.

###### Proof.

Again, it is enough to analyze the case $i=1$ because the other ones work
similarly.

By using the ideas in Lemma 8 and Lemma 22, we can reduce the proof to the
case where $n_{i}\not=m_{i}$ for $i=1,\dots,N$. Moreover, the assumptions (37)
on $\mathbf{n}$ and $\mathbf{m}$ in the proof of Lemma 8 can be considered
also.

We start with the case $m_{1}\geq 2$. From (38) we deduce that

$\Delta_{N,1}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))=\Delta(G_{t,1}(n_{1})-G_{t,1}(m_{1}))\prod_{k=2}^{N}G_{t,1}(n_{k})\\\
+\sum_{j=2}^{N}\Delta
G_{t,1}(m_{1})(G_{t,1}(n_{i})-G_{t,1}(m_{i}))\prod_{j=2}^{i-1}G_{t,1}(m_{j})\prod_{k=i+1}^{N}G_{t,1}(n_{k}).$

We can control $G_{t,1}(n_{i})-G_{t,1}(m_{i})$ as in the proof of Lemma 8 and,
applying (8), we have

(83) $|\Delta G_{t,1}(m_{1})(G_{t,1}(n_{i})-G_{t,1}(m_{i}))|\\\
\begin{aligned} &\leq
C\frac{|n_{i}-m_{i}|(m_{i}+n_{i}+1)}{t}\left(\frac{1}{t}+\frac{m_{1}^{2}}{t^{2}}\right)G_{t,1}(m_{1}-1)\max\\{G_{t,1}(n_{i}),G_{t,1}(m_{i})\\}\\\
&\leq
C\frac{|n_{i}-m_{i}|(m_{i}+n_{i}+1)}{t}\left(\frac{1}{t}+\frac{m_{1}^{2}}{t^{2}}\right)G_{t,1}(m_{1}/2)\max\\{G_{t,1}(n_{i}),G_{t,1}(m_{i})\\},\end{aligned}$

where in the last step we have used that $m_{1}-1\geq m_{1}/2$ because
$m_{1}\geq 2$.

For $n_{1}<m_{1}$, we can obtain that

$\Delta(G_{t,1}(n_{1})-G_{t,1}(m_{1}))=-\sum_{k=n_{1}}^{m_{1}-1}\Delta(G_{t,1}(k+1)-G_{t,1}(k)))\\\
=-\sum_{k=n_{1}}^{m_{1}-1}(G_{t,1}(k+3)-3G_{t,1}(k+2)+3G_{t,1}(k+1)-G_{t,1}(k))$

and, applying (10),

$\displaystyle|\Delta(G_{t,1}(n_{1})-G_{t,1}(m_{1}))|$ $\displaystyle\leq
C\sum_{k=n_{1}}^{m_{1}-1}\left(\frac{k+2}{t^{2}}+\frac{(k+1)(k+2)(k+3)}{t^{3}}\right)G_{t,1}(k)$
$\displaystyle\leq
C(m_{1}-n_{1})\left(\frac{m_{1}+n_{1}}{t^{2}}+\frac{m_{1}^{3}+n_{1}^{3}}{t^{3}}\right)G_{t,1}(n_{1}).$

More generally,

$|\Delta(G_{t,1}(n_{1})-G_{t,1}(m_{1}))|\\\ \leq
C|m_{1}-n_{1}|\left(\frac{m_{1}+n_{1}}{t^{2}}+\frac{m_{1}^{3}+n_{1}^{3}}{t^{3}}\right)\max\\{G_{t,1}(n_{1}),G_{t,1}(m_{1})\\}.$

Now, the proof of this case can be concluded as in Lemma 8.

To finish we have to analyze the cases $m_{1}=0,1$. To obtain the result in
these cases we need an appropriate estimate for $\Delta G_{t,1}(m_{1})$ to
insert it in (83). For $m_{1}=0$, we have

$\Delta
G_{t,1}(0)=G_{t,1}(1)-2G_{t,1}(0)+G_{t,1}(-1)=2(G_{t,1}(1)-G_{t,1}(0))$

and, applying (5), it is verified that

$|\Delta_{t,1}(0)|\leq\frac{C}{t}G_{t,1}(0)$

and this is enough for our purpose. In the case $m_{1}=1$, for $0<t\leq 1$ it
is clear that

$|\Delta_{t,1}(1)|\leq C\leq\frac{C}{t}G_{t,1}(1)$

and for $t>1$, applying (10) and (6), we have

$\displaystyle|\Delta_{t,1}(1)|$ $\displaystyle\leq
C\left(\frac{1}{t}+\frac{1}{t^{3}}\right)G_{t,1}(0)$ $\displaystyle\leq
C\left(\frac{1}{t}+\frac{1}{t^{3}}\right)\frac{1}{f_{1}(t)}G_{t,1}(1)\leq
C\left(\frac{1}{t}+\frac{1}{t^{3}}\right)G_{t,1}(1),$

where in the last step we have used that $1/f_{1}(t)\leq C$ for $t>1$. ∎

######  Remark 6.

From the proof of the previous result, it is cleat that

(84) $|\Delta_{N,i}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))|\leq
C|\mathbf{n}-\mathbf{m}|\left(\frac{|\mathbf{n}|}{t^{2}}+\frac{|\mathbf{n}|^{3}}{t^{3}}\right)t^{-N/2},\qquad\mathbf{n},\mathbf{m}\in\mathbb{Z}^{N}.$

###### Proof of Lemma 38.

First, it is easy to check that

$\|\mathcal{P}_{t,1}(\mathbf{n})\|_{\mathbb{B}_{1}}^{2}=\left\|\frac{\partial}{\partial
t}G_{t,N}(\mathbf{n})\right\|_{\mathbb{B}_{1}}^{2}=\left\|\Delta_{N}G_{t,N}(\mathbf{n})\right\|_{\mathbb{B}_{1}}^{2}\leq\sum_{i=1}^{N}\left\|\Delta_{N,i}G_{t,N}(\mathbf{n})\right\|_{\mathbb{B}_{1}}^{2}.$

Now, for $|\mathbf{n}|\geq 1$, by using (8), we have

$|\Delta_{N,i}G_{t,N}(\mathbf{n})|\leq
C\left(\frac{1}{t}+\frac{|\mathbf{n}|^{2}}{t^{2}}\right)\left(G_{t,1}\left(\frac{|\mathbf{n}|-1}{N}\right)\right)^{N}.$

We consider the decomposition

$\left\|\Delta_{N,i}G_{t,N}(\mathbf{n})\right\|_{\mathbb{B}_{1}}^{2}\leq
C(D_{1}+D_{2}),$

where

$D_{1}=\int_{0}^{|\mathbf{n}|^{2}}t\left(\frac{1}{t}+\frac{|\mathbf{n}|^{2}}{t^{2}}\right)^{2}\left(G_{t,1}\left(\frac{|\mathbf{n}|-1}{N}\right)\right)^{2N}\,dt$

and $D_{2}$ the integral of the same function on $[|\mathbf{n}|^{2},\infty)$.
Applying (14) with $\alpha=3/2N$ we deduce that

$\displaystyle D_{1}$ $\displaystyle\leq
C\int_{0}^{|\mathbf{n}|^{2}}(t+|\mathbf{n}|^{2})^{2}\left(t^{-3/(2N)}G_{t,1}\left(\frac{|\mathbf{n}|-1}{N}\right)\right)^{2N}\,dt$
$\displaystyle\leq\frac{C}{|\mathbf{n}|^{2N+6}}\int_{0}^{|\mathbf{n}|^{2}}(t+|\mathbf{n}|^{2})^{2}\,dt=\frac{C}{|\mathbf{n}|^{2N}}.$

For $D_{2}$, using the bound $(G_{t,1}((|\mathbf{n}|-1)/N))^{N}\leq
Ct^{-N/2}$, we obtain the estimate

$D_{2}\leq
C\left(\int_{|\mathbf{n}|^{2}}^{\infty}t^{-N-1}\,dt+|\mathbf{n}|^{4}\int_{|\mathbf{n}|^{2}}^{\infty}t^{-N-3}\,dt\right)=\frac{C}{|\mathbf{n}|^{2N}}.$

The case $|\mathbf{n}|=0$ is clear and the proof of (80) is completed.

To obtain (81), it is enough to prove that

(85)
$\|\Delta_{N,i}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))\|_{\mathbb{B}_{1}}\leq
C\frac{|\mathbf{n}-\mathbf{m}|}{(|\mathbf{n}|+|\mathbf{m}|)^{N+1}},\qquad|\mathbf{n}|>2|\mathbf{n}-\mathbf{m}|.$

To do this we consider (82) and we have

$\|\Delta_{N,i}(G_{t,N}(\mathbf{n})-G_{t,N}(\mathbf{m}))\|_{\mathbb{B}_{1}}^{2}\leq
C|\mathbf{n}-\mathbf{m}|^{2}(E_{1}+E_{2}),$

with

$E_{1}=\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}t\left(\mathfrak{H}_{t,N}\left(\frac{|\mathbf{n}|+|\mathbf{m}|}{k}\right)\right)^{2}\,dt$

and $E_{2}$ equals to the integral of the same function on
$[(|\mathbf{n}|+|\mathbf{m}|)^{2},\infty)$. For $E_{1}$, we use (14) with
$\alpha=5/(2N)$ to obtain that

$\displaystyle E_{1}$
$\displaystyle\leq\frac{C}{(|\mathbf{n}|+|\mathbf{m}|)^{2N+10}}\int_{0}^{(|\mathbf{n}|+|\mathbf{m}|)^{2}}(t(|\mathbf{n}|+|\mathbf{m}|)+(|\mathbf{n}|+|\mathbf{m}|)^{3})^{2}\,dt$
$\displaystyle\leq\frac{C}{(|\mathbf{n}|+|\mathbf{m}|)^{2N+2}}.$

Finally, to estimate $E_{2}$ we use (84) to obtain the required bound. Indeed,

$\displaystyle E_{2}$ $\displaystyle\leq
C\left((|\mathbf{n}|+|\mathbf{m}|)^{2}\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}t^{-N-3}\,dt+(|\mathbf{n}|+|\mathbf{m}|)^{6}\int_{(|\mathbf{n}|+|\mathbf{m}|)^{2}}^{\infty}t^{-N-5}\,dt\right)$
$\displaystyle\leq\frac{C}{(|\mathbf{n}|+|\mathbf{m}|)^{2N+2}}$

and the proof of (85) is finished. ∎

###  6.2. The $\mathfrak{g}_{k}$-square functions associated to the Poisson
semigroup

It is very common to define some square functions in terms of the Poisson
semigroup instead of the heat semigroup. In our case such
$\mathfrak{g}_{k}$-square functions are given by

$\mathfrak{g}_{k}(f)(\mathbf{n})=\left(\int_{0}^{\infty}t^{2k-1}\left|\frac{\partial^{k}}{\partial
t^{k}}P_{t}f(\mathbf{n})\right|^{2}\,dt\right)^{1/2},\qquad k=1,2,\dots.$

By using that

$P_{t}f(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})Q_{t}(\mathbf{n}-\mathbf{k})$

with

$Q_{t}(\mathbf{n})=\int_{[-1/2,1/2]^{N}}e^{-2t\sqrt{\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})}}e^{-2\pi i\langle x,\mathbf{n}}\rangle\,dx,$

it is clear that

(86)
$\mathfrak{g}_{k}(f)(\mathbf{n})=\|R_{t,k}f(\mathbf{n})\|_{\mathbb{B}_{k}}$

where

$R_{t,k}f(\mathbf{n})=\sum_{\mathbf{k}\in\mathbb{Z}^{N}}f(\mathbf{k})Q_{t,k}(\mathbf{n}-\mathbf{k})$

being

$Q_{t,k}(\mathbf{n})=\frac{d^{k}}{dt^{k}}Q_{t}(\mathbf{n})=\int_{[-1/2,1/2]^{N}}S_{t,k}(x)e^{-2\pi
i\langle x,\mathbf{n}}\rangle\,dx$

and

$S_{t,k}(x)=(-2)^{k}\left(\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})\right)^{k/2}e^{-2t\sqrt{\sum_{i=1}^{N}\sin^{2}(\pi x_{i})}}e^{-2\pi
i\langle x,\mathbf{n}}\rangle\,dx.$

For this family of square functions we have the following result.

######  Theorem 40.

Let $N\geq 1$, $1<p<\infty$, $w\in A_{p}(\mathbb{Z}^{N})$, and
$k\in\mathbb{N}$. Then the inequalities

$C_{1}\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq\|\mathfrak{g}_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C_{2}\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w),$

hold.

To prove the previous result we need the two next lemmas.

######  Lemma 41.

For $N\geq 1$ and $k\in\mathbb{N}$, it is verified that

(87)
$\|\mathfrak{g}_{k}(f)\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}=\frac{\Gamma(2k)}{2^{2k}}\|f\|_{\ell^{2}(\mathbb{Z}^{N})}^{2}.$

The proof of this lemma can be done exactly as the proof of Lemma 34 but using
(86) so we omit the details.

######  Lemma 42.

Let $N\geq 1$, $1<p<\infty$, $w\in A_{p}(\mathbb{Z}^{N})$, and
$k\in\mathbb{N}$. Then the inequality

$\|\mathfrak{g}_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N}),$

implies

$\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|\mathfrak{g}_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N}).$

In this case, the proof of this lemma can be obtained polarising the identity
(87) and using duality. Again we omit the details.

Finally, to conclude the proof of Theorem 40 we will use the estimate

$\|g_{k}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w)$

and the next result.

######  Lemma 43.

Let $k\in\mathbb{N}$, then

$\mathfrak{g}_{k}(f)(\mathbf{n})\leq\sum_{j=0}^{\lfloor
k/2\rfloor}A_{k,j}g_{k-j}(f)(\mathbf{n}),$

where $A_{k,j}$ are some positive constants and $\lfloor\cdot\rfloor$ denotes
the floor function.

###### Proof.

First, we observe that

$\frac{\partial^{k}}{\partial
t^{k}}h\left(\frac{t^{2}}{4u}\right)=\sum_{j=0}^{[k/2]}B_{k,j}\left.\frac{\partial^{k-j}}{\partial
s^{k-j}}h(s)\right|_{s=\frac{t^{2}}{4u}}\frac{t^{k-2j}}{(4u)^{k-j}},$

for some constants $B_{k,j}$. Then, from (54), we have

$\frac{\partial^{k}}{\partial
t^{k}}P_{t}f(\mathbf{n})=\frac{1}{\sqrt{\pi}}\sum_{j=0}^{[k/2]}B_{j}\int_{0}^{\infty}\frac{e^{-u}}{\sqrt{u}}\left(\left.\frac{\partial^{k-j}}{\partial
s^{k-j}}W_{s}f(\mathbf{n})\right|_{s=\frac{t^{2}}{4u}}\right)\frac{t^{k-2j}}{(4u)^{k-j}}\,du$

and, by Minkowsky’s integral inequality,

$\mathfrak{g}_{k}(f)(\mathbf{n})\leq\sum_{j=0}^{[k/2]}B_{k,j}S_{j}(\mathbf{n})$

where

$S_{j}(\mathbf{n})\\\
=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-u}}{\sqrt{u}(4u)^{k-j}}\left(\int_{0}^{\infty}t^{4k-4j-1}\left(\left.\frac{\partial^{k-j}}{\partial
s^{k-j}}W_{s}f(\mathbf{n})\right|_{s=\frac{t^{2}}{4u}}\right)^{2}\,dt\right)^{1/2}\,du.$

Now, by using an appropriate change of variables, we have

$\displaystyle S_{j}(\mathbf{n})$
$\displaystyle=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{e^{-u}}{\sqrt{u}}\left(\int_{0}^{\infty}s^{2k-2j-1}\left(\frac{\partial^{k-j}}{\partial
s^{k-j}}W_{s}f(\mathbf{n})\right)^{2}\,ds\right)^{1/2}\,du$
$\displaystyle=\frac{1}{\sqrt{2}}g_{k-j}(f)(\mathbf{n})$

and the result follows. ∎

###  6.3. The Laplace type multipliers

Given a bounded function $M$ defined on $[0,4N]$, the multiplier associated
with $M$ is the operator, initially defined on $\ell^{2}(\mathbb{N})$, by the
identity

$T_{M}f(\mathbf{n})=\int_{[-1/2,1/2]^{N}}M\left(4\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})\right)\mathcal{F}f(x)e^{-2\pi i\langle x,\mathbf{n}\rangle}\,dx.$

We say that $T_{M}$ is a Laplace type multiplier when

$M(x)=x\int_{0}^{\infty}e^{-xt}a(t)\,dt,$

with $a$ being a bounded function.

The Laplace type multipliers were introduced by Stein in [23, Ch. 2]. There,
it is observed that they verify $|x^{k}M^{(k)}(x)|\leq C_{k}$ for
$k=0,1,\dots$, and then form a subclass of Marcinkiewicz multipliers. For the
operators $T_{M}$ we have the following result.

######  Theorem 44.

Let $1<p<\infty$ and $w\in A_{p}(\mathbb{Z}^{N})$. Then,

$\|T_{M}f\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w),$

where $C$ is a constant independent of $f$.

From the identity

$x^{i\gamma}=\frac{x}{\Gamma(1-i\gamma)}\int_{0}^{\infty}e^{-xt}t^{-i\gamma}\,dt,\qquad\gamma\in\mathbb{R},$

we deduce the following corollary.

######  Corollary 45.

Let $1<p<\infty$, $\gamma\in\mathbb{R}$, and $w\in A_{p}(\mathbb{Z}^{N})$.
Then,

$\|(-\Delta_{N})^{i\gamma}f\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)},\qquad
f\in\ell^{2}(\mathbb{Z}^{N})\cap\ell^{p}(\mathbb{Z}^{N},w),$

where $C$ is a constant independent of $f$.

###### Proof of Theorem 44.

We only need prove that

(88) $g_{1}(T_{M}f)(\mathbf{n})\leq Cg_{2}(f)(\mathbf{n}),$

since by Theorem 33 we get that

$\|T_{M}f\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|g_{1}(T_{M}f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|g_{2}(f)\|_{\ell^{p}(\mathbb{Z}^{N},w)}\leq
C\|f\|_{\ell^{p}(\mathbb{Z}^{N},w)}.$

Moreover, it is enough to prove (88) for sequences in $c_{00}$, the space of
sequences having a finite number of non-null terms. First, we have

$T_{M}f(\mathbf{n})=-\int_{0}^{\infty}a(s)\frac{\partial}{\partial
s}W_{s}f(\mathbf{n})\,ds,$

which is an elementary consequence of the relation

$\int_{[-1/2,1/2]^{N}}M\left(4\sum_{i=1}^{N}\sin^{2}(\pi x_{i})\right)e^{-2\pi
i\langle x,\mathbf{n}\rangle}\,dx\\\ \begin{aligned}
&=\int_{0}^{\infty}a(s)\int_{[-1/2,1/2]^{N}}4\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})e^{-4s\sum_{i=1}^{N}\sin^{2}(\pi x_{i})}e^{-2\pi i\langle
x,\mathbf{n}\rangle}\,dx\,ds\\\
&=-\int_{0}^{\infty}a(s)\frac{\partial}{\partial
s}\int_{[-1/2,1/2]^{N}}4\sum_{i=1}^{N}\sin^{2}(\pi
x_{i})e^{-4s\sum_{i=1}^{N}\sin^{2}(\pi x_{i})}e^{-2\pi i\langle
x,\mathbf{n}\rangle}\,dx\,ds\\\
&=-\int_{0}^{\infty}a(s)\frac{\partial}{\partial
s}G_{s,N}(\mathbf{n})\,ds.\end{aligned}$

Then, applying the semigroup property of $W_{t}$ we obtain

$W_{t}(T_{M}f)(\mathbf{n})=-\int_{0}^{\infty}a(s)\frac{\partial}{\partial
s}W_{s+t}f(\mathbf{n})\,ds$

and hence,