# Vertical and horizontal Square Functions on a Class of Non-Doubling
Manifolds

Julian Bailey, Adam Sikora  Julian Bailey, Department of Mathematics,
Macquarie University, NSW 2109, Australia<EMAIL_ADDRESS>Adam
Sikora, Department of Mathematics, Macquarie University, NSW 2109, Australia
<EMAIL_ADDRESS>
###### Abstract.

We consider a class of non-doubling manifolds $\mathcal{M}$ that are the
connected sum of a finite number of $N$-dimensional manifolds of the form
$\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$. Following on from the work of
Hassell and the second author [19], a particular decomposition of the
resolvent operators $(\Delta+k^{2})^{-M}$, for $M\in\mathbb{N}^{*}$, will be
used to demonstrate that the vertical square function operator

$Sf(x):=\left(\int^{\infty}_{0}\left|t\nabla(I+t^{2}\Delta)^{-M}f(x)\right|^{2}\frac{dt}{t}\right)^{\frac{1}{2}}$

is bounded on $L^{p}(\mathcal{M})$ for $1<p<n_{min}=\min_{i}n_{i}$ and weak-
type $(1,1)$. In addition, it will be proved that the reverse inequality
$\left\|f\right\|_{p}\lesssim\left\|Sf\right\|_{p}$ holds for
$p\in(n_{min}^{\prime},n_{min})$ and that $S$ is unbounded for $p\geq n_{min}$
provided $2M<n_{min}$.

Similarly, for $M>1$, this method of proof will also be used to ascertain that
the horizontal square function operator

$sf(x):=\left(\int^{\infty}_{0}\left|t^{2}\Delta(I+t^{2}\Delta)^{-M}f(x)\right|^{2}\,\frac{dt}{t}\right)^{\frac{1}{2}}$

is bounded on $L^{p}(\mathcal{M})$ for all $1<p<\infty$ and weak-type $(1,1)$.
Hence, for $p\geq n_{min}$, the vertical and horizontal square function
operators are not equivalent and their corresponding Hardy spaces $H^{p}$ do
not coincide.

###### Key words and phrases:

Non-doubling spaces; square functions; resolvent estimates.

###### 1991 Mathematics Subject Classification:

42B20 (primary), 47F05, 58J05 (secondary).

##  1\. Introduction

In any historical account of the development of harmonic analysis, the
doubling condition will certainly appear as a central actor. In each step of
its genesis, from the minds of many great mathematicians in the 1960’s and
1970’s, the doubling condition was interwoven, seemingly inextricably,
throughout the entirety of the body of work that embodied harmonic analysis.
In a metric measure space $(X,d,\mu)$, the doubling condition reads that there
must exist a constant $C>0$ such that

$\mu(B(x,2r))\leq C\mu(B(x,r))$

for all $x\in X$ and $r>0$, where the notation $B(x,r)$ is used to denote the
ball of radius $r$ and centered at the point $x$. If this condition is
satisfied then $(X,d,\mu)$ is said to be a space of homogeneous type, in the
sense of Coifman and Weiss [11], whilst any space that does not satisfy this
condition is called non-homogeneous.

Although this condition aligns with our real world intuition, with the
continued progress of mathematics as a whole there are now many different
situations and applications that depart from this idealised world and demand
the consideration of non-homogeneous spaces. Moreover, it has become
increasingly apparent that the doubling condition is not quite as critically
indispensable for many harmonic analytic results as previously believed. As
such, and acting as a reversal to the assimilation of the doubling condition
in the adolescence of the field, there is now an intensive research effort
underway that aims to unthread the doubling strand from this body of work,
where possible, and push the boundaries of harmonic analysis beyond this
condition. This process must often be approached with the utmost care since
one is certain to encounter behaviours that depart very far from the
prototypical doubling example of Euclidean space $\mathbb{R}^{d}$. Some
examples of notable results in this area include: the extension of Calderón-
Zygmund theory to non-homogeneous spaces through the work of Nazarov, Treil
and Volberg [22, 23, 24] and Tolsa [33, 34, 31]; the generalisation of $BMO$
and $H^{1}$ theory by Bui and Duong [5] and Tolsa [32]; the consideration of
weighted norm inequalities through the work of Orobitg and Pérez [26]; and
non-homogeneous Tb type theorems by Hytönen and Martikainen [20].

The non-doubling spaces that are of interest to us in this article consist of
a particular class of non-doubling manifolds formed as connected sums.

######  Definition 1.1.

A manifold $\mathcal{V}$ is said to be formed as the connected sum of a finite
number of complete and connected manifolds
$\mathcal{V}_{1},\cdots,\mathcal{V}_{l}$ of the same dimension, denoted

$\mathcal{V}=\mathcal{V}_{1}\\#\cdots\\#\mathcal{V}_{l},$

if there exists some compact subset with non-empty interior
$K\subset\mathcal{V}$ for which $\mathcal{V}\setminus K$ can be expressed as
the disjoint union of open subsets $E_{i}\subset\mathcal{V}$ for
$i=1,\cdots,l$, where each $E_{i}$ is isometric to $\mathcal{V}_{i}\setminus
K_{i}$ for some compact $K_{i}\subset\mathcal{V}_{i}$.

Fix dimension $N\in\mathbb{N}^{*}:=\mathbb{N}\setminus\left\\{0\right\\}$,
$l\geq 1$ and let $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$ for $i=1,\cdots,l$
be a collection of manifolds with $\mathcal{M}_{i}$ compact and
$n_{i}+\mathrm{dim}\,\mathcal{M}_{i}=N$. We will be interested in smooth
Riemannian manifolds $\mathcal{M}$ that are of the form

$\mathcal{M}:=(\mathbb{R}^{n_{1}}\times\mathcal{M}_{1})\\#\cdots\\#(\mathbb{R}^{n_{l}}\times\mathcal{M}_{l}).$

As in the previous definition, it is possible to choose open subsets
$E_{i}\subset\mathcal{M}$ and compact $K\subset\mathcal{M}$ with non-empty
interior for which $\mathcal{M}\setminus K$ can be expressed as the disjoint
union of the $E_{i}$. The subsets $E_{i}$ are referred to as the ends of
$\mathcal{M}$, $K$ the center of $\mathcal{M}$ and the entire manifold
$\mathcal{M}$ itself is fittingly said to be a manifold with ends.

The constituent manifolds $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$ for $1\leq
i\leq l$ each have topological dimension $N$, but asymptotic dimension $n_{i}$
at infinity. That is, for any ball
$B(x,r)\subset\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$, it will be true that
the volume of the ball will satisfy

$V(B(x,r))\simeq\left\\{\begin{array}[]{c c}r^{N}&for\ r\leq 1,\\\
r^{n_{i}}&for\ r>1.\end{array}\right.$

If the values of $n_{i}$ differ, then the manifold with ends $\mathcal{M}$
will have varying asymptotic dimension on these ends. This will lead to a
violation of the doubling condition.

In the landmark article by Grigoryan and Saloff-Coste [17], the authors
effectively computed, using probabilistic methods, two-sided estimates for the
heat kernel generated by the Laplacian $\Delta$ on this prototypical
collection of non-doubling spaces. Although not the first to study this
remarkable class of manifolds in detail (see [25] for a detailed historical
account), this article acted as an inflection point for interest in this class
and had a pronounced effect on the non-homogeneous community. Indeed, it
essentially designated this class of manifolds as a battlefront for the
advancement of non-homogeneous harmonic analysis. In point of fact, the
sustained interest in these model spaces has led to investigations into the
boundedness of the heat maximal operators [14], the functional calculus of
$\Delta$ [6] and Littlewood-Paley decompositions [4]. Recently, and of
particular significance to our article, Hassell in collaboration with the
second named author considered the $L^{p}(\mathcal{M})$-boundedness of the
Riesz transforms operator $\nabla\Delta^{-\frac{1}{2}}$ on such manifolds
[19]. This paper in tern is a generalisation to the non-doubling setting of
the result obtained by Carron, Coulhon and Hassell in [8]. For other relevant
results we refer the reader to [7, 13] and references therein.

In this article, it is our aim to extend the classical theory of square
functions to this class of non-doubling manifolds. Consider a general complete
Riemannian manifold $\mathcal{M}$ and let $\Delta$ denote the Laplace-Beltrami
operator for this manifold. For $M\in\mathbb{N}^{*}$, the vertical square
function operator is defined:

(1)
$Sf(x):=\left(\int^{\infty}_{0}\left|t\nabla(I+t^{2}\Delta)^{-M}f(x)\right|^{2}\frac{dt}{t}\right)^{\frac{1}{2}}.$

The notion of square functions forms an essential part of harmonic analysis
and has numerous applications, from the definition of Hardy spaces [27] to
providing an equivalent characterisation of the bounded holomorphic functional
calculus of a sectorial operator [12]. As such, the above operators have been
extensively studied, either in this form or defined with the semigroup
replacing the higher-order resolvent (c.f. Remark 1.1), since the formation of
harmonic analysis, and a great deal is known about their behaviour when the
manifold under consideration is doubling.

  1. $\bullet$

For the classical case when $\mathcal{M}$ is simply Euclidean space
$\mathbb{R}^{d}$, $S$ is bounded on $L^{p}(\mathbb{R}^{d})$ for any
$p\in(1,\infty)$ and weak-type $(1,1)$.

  2. $\bullet$

For general complete Riemannian manifolds, the work of Coulhon, Duong and Li
[10] states that the semigroup variation of the vertical square function is
bounded on $L^{p}(\mathcal{M})$ for all $p\in(1,2]$, whether doubling or not,
and weak-type $(1,1)$ if the manifold is doubling and if Gaussian upper bounds
are satisfied by the heat kernel.

  3. $\bullet$

For doubling manifolds $\mathcal{M}$ whose heat kernel satisfies Gaussian
upper bounds bounds, one can define

$q_{+}:=\sup\left\\{p\in(1,\infty):\left\|\left|\nabla\Delta^{-\frac{1}{2}}f\right|\right\|_{p}\lesssim\left\|f\right\|_{p}\right\\}.$

It is then known that $q_{+}\geq 2$ [9] and that the semigroup variation of
$S$ is bounded on $L^{p}(\mathcal{M})$ for any $p\in(1,q_{+})$. A sparse proof
of this was shown in [3, Prop. 3.8].

Nothing is currently known about the $L^{p}(\mathcal{M})$-boundedness of the
vertical square function for $p>2$ on non-doubling manifolds. Our aim in this
article is to prove the following theorem.

######  Theorem 1.1.

Let
$\mathcal{M}=(\mathbb{R}^{n_{1}}\times\mathcal{M}_{1})\\#\cdots\\#(\mathbb{R}^{n_{l}}\times\mathcal{M}_{l})$
be a manifold with ends. For $M\in\mathbb{N}^{*}$, the vertical square
function operator $S$, as defined in (1), will satisfy the following
properties:

  1. (i)

$S$ is bounded on $L^{p}(\mathcal{M})$ for all $p\in(1,n_{min})$, where
$n_{min}:=\min_{i}\,n_{i}$, and weak-type $(1,1)$;

  2. (ii)

If $2M<n_{min}$ then $S$ is unbounded on $L^{p}(\mathcal{M})$ for $p\geq
n_{min}$; and

  3. (iii)

For $p\in(n_{min}^{\prime},n_{min})$, there exists $c>0$ for which

$\left\|f\right\|_{p}\leq c\left\|Sf\right\|_{p}$

for all $f\in L^{p}(\mathcal{M})$.

Let us briefly discuss the proof of the $L^{p}(\mathcal{M})$-boundedness
portion of this result. Notice that through a change of variables $S$ has the
representation

$\displaystyle\begin{split}Sf(x)&=\left(\int^{\infty}_{0}\left|\nabla(t^{-2}+\Delta)^{-M}f(x)\right|^{2}t^{1-4M}\,dt\right)^{\frac{1}{2}}\\\
&=\left(\int^{\infty}_{0}\left|\nabla(k^{2}+\Delta)^{-M}f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.\end{split}$

This can be controlled from above by the corresponding high and low energy
parts of this square function,

$Sf(x)\leq S_{>}f(x)+S_{<}f(x),$

where

$S_{>}f(x):=\left(\int^{\infty}_{1}\left|\nabla(k^{2}+\Delta)^{-M}f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}$

and

$S_{<}f(x):=\left(\int^{1}_{0}\left|\nabla(k^{2}+\Delta)^{-M}f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

The $L^{p}(\mathcal{M})$-boundedness of these two parts will be proved
separately. The high energy component is local in nature and does not see the
large scale non-doubling character of the manifold. It can thus be treated in
a manner analogous to the classical Calderón-Zygmund case. This will be
accomplished in Section 5.

The low energy component, on the other hand, will prove to be more challenging
since it involves extensive interaction between the different ends of the
manifold. Indeed, ultimately it is the low energy component that will prove to
be solely responsible for the unboundedness of the operator on the range
$p\geq n_{min}$. Another major difficulty that appears is that, in contrast to
many classical cases, it will not be possible to treat this term using
estimates for the spatial derivative of the heat kernel $\nabla e^{-t\Delta}$.
This is due to the pronounced abscence of these estimates in the literature.
Instead, we rely on a decomposition of the resolvent operator
$(\Delta+k^{2})^{-1}$ that was constructed in [19] using a parametrix style
argument. In Section 3 it will be proved that this decomposition can be
generalised to the higher-order resolvent operators $(\Delta+k^{2})^{-M}$.
Following this, in Section 4, this higher-order resolvent decomposition will
be applied to the low energy component and boundedness will ensue.

For the unboundedness portion of Theorem 1.1, observe that a constraint on the
order of the resolvent is required, $2M<n_{min}$. This constraint is a
consequence of our method of proof. Indeed, our proof that $S$ is unbounded on
$L^{p}(\mathcal{M})$ for $p\geq n_{min}$ relies heavily on the use of the
Riesz potential operators $\Delta^{-M}$. In an analogous manner to classical
theory on Euclidean space, these operators are not well-defined when the order
$M$ is too large in comparison to the dimension, namely $2M\geq n_{min}$. As a
result, our method of proof will not be applicable for this range. We do not
study the range $2M\geq n_{min}$ here. Therefore, we do not study the case
$n_{min}=2$ which was investigated in [18, 25].

######  Remark 1.1.

Through a careful study of the literature, one will notice that the term
vertical square function is often used to refer to the operator

$\left(\int^{\infty}_{0}\left|t\nabla
e^{-t\Delta}f(x)\right|^{2}\frac{dt}{t}\right)^{\frac{1}{2}},$

with the semigroup $e^{-t\Delta}$ taking on the role held by the higher-order
resolvent operator for $S$. Although the semigroup form is more frequently
encountered, our consideration of square functions defined using higher-order
resolvent operators is by no means rare. For instance, in the article [15] by
Frey, McIntosh and Portal, resolvent based conical square function estimates
were proved for perturbations of Dirac-type operators on
$L^{p}(\mathbb{R}^{d})$.

Indeed, the two different forms of square function are seen to be morally
equivalent. For if one can prove that $S$ is bounded on $L^{p}(\mathcal{M})$,
together with a similar estimate for $\nabla\mathrm{div}$, then one can obtain
the boundedness of the holomorphic functional calculus of the corresponding
Dirac-type operator on $L^{p}(\mathcal{M})$ (c.f. [12, Cor. 6.8]). The
boundedness of the semigroup square function would then follow immediately as
a corollary. This argument can also be reversed in order to obtain the
$L^{p}(\mathcal{M})$-boundedness of the resolvent square function from the
$L^{p}(\mathcal{M})$-boundedness of the semigroup square function.

As an alternative to the vertical square function $S$, one can also consider
the horizontal square function for $M>1$:

(2)
$sf(x):=\left(\int^{\infty}_{0}\left|t^{2}\Delta(I+t^{2}\Delta)^{-M}f(x)\right|^{2}\frac{dt}{t}\right)^{\frac{1}{2}}.$

For this operator, we will prove the following theorem.

######  Theorem 1.2.

Let
$\mathcal{M}=(\mathbb{R}^{n_{1}}\times\mathcal{M}_{1})\\#\cdots\\#(\mathbb{R}^{n_{l}}\times\mathcal{M}_{l})$
be a manifold with ends and fix $M>1$. For any $p\in(1,\infty)$ the square
function operator $s$, as defined in (2), is bounded on $L^{p}(\mathcal{M})$
and there exists $c,\,C>0$ such that for all $f\in L^{p}(\mathcal{M})$

$c\left\|f\right\|_{p}\leq\left\|sf\right\|_{p}\leq C\left\|f\right\|_{p}.$

In addition, the operator $s$ is of weak-type $(1,1)$.

This result will be proved in Section 8 and, similar to the operator $S$, it
will be achieved by decomposing $s$ into low and high energy components and
then by proving $L^{p}(\mathcal{M})$-boundedness for both of these components
separately. For the low energy component, we will once again make extensive
use of the decomposition for the higher-order resolvent operators
$(\Delta+k^{2})^{-M}$.

The boundedness of $s$ on $L^{p}(\mathcal{M})$ for $1<p<\infty$, as stated in
Theorem 1.2, is a result that already exists in the literature. Indeed, the
$L^{p}(\mathcal{M})$-boundedness of the semigroup based square function,

$\left(\int^{\infty}_{0}\left|t\Delta
e^{-t\Delta}f(x)\right|^{2}\frac{dt}{t}\right)^{\frac{1}{2}},$

was proved to hold in the general symmetric Markov semigroup setting (see [28,
pg. 111]). This implies that $\Delta$ possesses a bounded
$H^{\infty}(S^{o}_{\mu})$-functional calculus on $L^{p}(\mathcal{M})$ for any
$\mu\in[0,\pi)$, which immediately leads to the boundedness of $s$ on
$L^{p}(\mathcal{M})$ (c.f. [12, Cor. 6.8]). A new proof of this result has
been included here to illustrate the applicability of our methods to resolvent
based operators and to obtain the weak-type $(1,1)$ bounds for $s$ which, to
the best of our knowledge, is a result that is new. It is also particularly
illuminating to compare the proofs for the vertical and horizontal square
functions in order to glean some intuition as to why one is bounded on the
full reflexive range, while the other fails for $p\geq n_{min}$.

######  Remark 1.2.

Our results can also be viewed through the lens of Hardy spaces. In an
analogous manner to the classical case $\mathcal{M}=\mathbb{R}^{d}$, one can
define, for $p>0$, Hardy spaces $H_{\nabla}^{p}$ and $H_{\Delta}^{p}$
associated with $s$ and $S$ through the norms

$\left\|f\right\|_{H^{p}_{\nabla}}:=\left\|Sf\right\|_{p}\quad
and\quad\left\|f\right\|_{H^{p}_{\Delta}}:=\left\|sf\right\|_{p}.$

Theorem 1.1 tells us that for $p\in(n_{min}^{\prime},n_{min})$ it will be true
that $H^{p}_{\nabla}=L^{p}$, but $H^{p}_{\nabla}\neq L^{p}$ for
$p\in[n_{min},\infty)$ when $2M<n_{min}$. In contrast, $H^{p}_{\Delta}=L^{p}$
for all $p\in(1,\infty)$, and thus the two square functions define distinct
Hardy spaces for $p\geq n_{min}$. It remains an open problem to check that
$H^{p}_{\nabla}=H^{p}_{\Delta}$ for $p\in(1,n_{min}^{\prime}]$.

######  Remark 1.3.

Comparing Theorem 1.1 with the main result of [19] we note a posteriori that
in the considered setting boundedness of the Riesz Transform and vertical
square function are equivalent for any $L^{p}$ space, including weak type
$(1,1)$ estimates. It is an interesting question whether such equivalence
could be verified in some more general setting in the form of the abstract
statement. It seems that the implication from Riesz Transform bounds to the
square functions estimates can be approach using [21, Theorem 9.5.1 Section
9]. We do not know how to approach the opposite implication. We expect any
result which includes weak type $(1,1)$ estimates in any directions to be
especially challenging. Here we are more interested with understanding the
difference between horizontal and vertical square functions and we do not
attempt to answer this question.

##  2\. Preliminaries

Throughout this article, we fix a manifold with ends
$\mathcal{M}=(\mathbb{R}^{n_{1}}\times\mathcal{M}_{1})\\#\cdots\\#(\mathbb{R}^{n_{l}}\times\mathcal{M}_{l})$.
The notation $K$ will be used to denote the center of $\mathcal{M}$ and the
sets $K_{i}$ and $E_{i}$ will be as given in Definition 1.1. In particular,
these sets will have the property that $\mathcal{M}\setminus K$ can be
expressed as the disjoint union of the ends $E_{i}$ and
$E_{i}\simeq\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$. In this
manner, the ends $E_{i}$ will be identified with the sets
$\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ so that a point $x\in
E_{i}$ can also be viewed as belonging to the space
$\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$.

######  Notation.

For estimates concerning two quantities $a,\,b\in\mathbb{R}$, the notation
$a\lesssim b$ will be employed to signify the existence of a constant $C>0$
such that $a\leq C\cdot b$. Similarly, $a\simeq b$ will denote that $a\lesssim
b$ and $b\lesssim a$ both hold. The dependence of the constant $C$ on certain
parameters should be clear from the context of the argument under
consideration.

For $x\in\mathbb{R}^{d}$, define $\langle
x\rangle:=(1+\left|x\right|^{2})^{\frac{1}{2}}$. We employ the notation
$d(x,y)$ to denote the intrinsic distance between two points $x$ and $y$ in
some ambient Riemannian manifold. When the space under consideration is the
entire space $\mathcal{M}$, we use the shorthand notation $L^{p}$ to denote
the Lebesgue space $L^{p}(\mathcal{M})$. Finally, for a function $g(x,y)$ of
two variables, the notation $\nabla_{x}g(x,y)$ will be understood to denote
the gradient with respect to the first variable.

Before delving into the substance of our proof, it will first be beneficial to
record some useful estimates satisfied by the higher-order resolvent operators
on the constituent manifolds $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$.
Following this, we recall a vital decomposition of the first-order resolvent
on the entire space $\mathcal{M}$ that was introduced in [19]. This will form
the foundation upon which our proofs of both Theorem 1.1 and 1.2 will be
built.

###  2.1. Higher-Order Resolvents on $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$

Fix $1\leq i\leq l$ and consider the higher-order resolvents
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}$ for
$j\in\mathbb{N}^{*}$ and $k>0$, where
$\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}$ denotes the Laplacian on
$\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$. Recall the definition of the Bessel
kernel $G_{a}^{d}:\mathbb{R}\rightarrow(0,\infty)$ for $a,\,d>0$,

$G_{a}^{d}(s):=\frac{1}{(4\pi)^{\frac{a}{2}}\Gamma(a/2)}\int^{\infty}_{0}\frac{e^{-\frac{\pi
s^{2}}{t}-\frac{t}{4\pi}}}{t^{1+\frac{d-a}{2}}}\,dt.$

The Bessel kernels are well-known to satisfy the estimates,

(3) $\displaystyle\begin{split}G^{d}_{a}(s)&\simeq\left\\{\begin{array}[]{c
c}\frac{e^{-cs}}{s^{d-a}}&if\ a<d,\\\ &\\\
\max\left(1,\ln(s^{-1})\right)e^{-cs}&if\ d=a,\\\ &\\\ e^{-cs}&if\
a>d,\end{array}\right.\end{split}$

where the value of $c$ is allowed to differ in each case and for the upper and
lower estimates. So, when $a<d$ for instance,

$\frac{e^{-c_{1}s}}{s^{d-a}}\lesssim
G^{d}_{a}(s)\lesssim\frac{e^{-c_{2}s}}{s^{d-a}},$

for some $c_{1}>c_{2}>0$. Refer to [1] for a detailed proof.

######  Proposition 2.1.

For $j\in\mathbb{N}^{*}$, there exists $c_{1},\,c_{2},\,c_{3}>0$ for which

$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\lesssim
k^{n_{i}-2j}\cdot G^{n_{i}}_{2j}(c_{1}d(x,y)k)+k^{N-2j}\cdot
G^{N}_{2j}(c_{1}d(x,y)k),$
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\gtrsim
k^{n_{i}-2j}\cdot G^{n_{i}}_{2j}(c_{2}d(x,y)k)+k^{N-2j}\cdot
G^{N}_{2j}(c_{2}d(x,y)k)$

and

$\left|\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\right|\lesssim
k^{n_{i}+1-2j}\cdot
G^{n_{i}}_{2j-1}(c_{3}d(x,y)k)+k^{N+1-2j}G^{N}_{2j-1}(c_{3}d(x,y)k)$

for all $x,\,y\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$ and $k>0$.

###### Proof.

Recall that the higher-order resolvent can be expressed in terms of the heat
operator through the integral relation

(4)
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}=\frac{1}{(j-1)!}\int^{\infty}_{0}t^{j-1}e^{-tk^{2}}e^{-t\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}}\,dt.$

On applying the heat kernel estimate (11) of [19],

$\displaystyle\begin{split}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)&=\frac{1}{(j-1)!}\int_{0}^{\infty}t^{j-1}e^{-tk^{2}}e^{-t\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}}(x,y)\,dt\\\
&\lesssim\int_{0}^{\infty}t^{j-1}(t^{-n_{i}/2}+t^{-N/2})e^{-tk^{2}}e^{-c\frac{d(x,y)^{2}}{t}}\,dt\\\
&=\int_{0}^{\infty}t^{-1-\frac{n_{i}-2j}{2}}e^{-tk^{2}}e^{-c\frac{d(x,y)^{2}}{t}}\,dt+\int_{0}^{\infty}t^{-1-\frac{N-2j}{2}}e^{-tk^{2}}e^{-c\frac{d(x,y)^{2}}{t}}.\end{split}$

On applying a change of variables,

(5)
$\displaystyle\begin{split}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)&\lesssim
k^{n_{i}-2j}\int^{\infty}_{0}t^{-1-\frac{n_{i}-2j}{2}}e^{-\frac{t}{4\pi}}e^{-\pi\frac{4cd(x,y)^{2}k^{2}}{t}}\,dt\\\
&\qquad\qquad+k^{N-2j}\int^{\infty}_{0}t^{-1-\frac{N-2j}{2}}e^{-\frac{t}{4\pi}}e^{-\pi\frac{4cd(x,y)^{2}k^{2}}{t}}\,dt\\\
&\simeq k^{n_{i}-2j}\cdot G^{n_{i}}_{2j}(2\sqrt{c}d(x,y)k)+k^{N-2j}\cdot
G^{N}_{2j}(2\sqrt{c}d(x,y)k).\end{split}$

The lower estimate for
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)$ and the
upper estimate for
$\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)$
follow in an identical manner to the upper estimate through an application of
(13) and (12) respectively of [19]. ∎

Proposition 2.1, when combined with the Bessel kernel estimates (3),
immediately yield the following corollary.

######  Corollary 2.1.

For any $j\in\mathbb{N}^{*}$ with $j\neq\frac{n_{i}}{2}$ and
$j\neq\frac{N}{2}$, there exists $c>0$ such that

(6)
$\displaystyle\begin{split}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-j}(x,y)&\lesssim\left(d(x,y)^{\min(2j-N,0)}k^{-\max(2j-N,0)}\right.\\\
&\qquad\qquad\qquad\left.+d(x,y)^{\min(2j-n_{i},0)}k^{-\max(2j-n_{i},0)}\right)e^{-ckd(x,y)}.\end{split}$

If $j=\frac{n_{i}}{2}<\frac{N}{2}$,

(7)
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\lesssim\left(d(x,y)^{2j-N}+\max\left[1,\ln\left(\frac{1}{kd(x,y)}\right)\right]\right)e^{-ckd(x,y)}.$

If $j=\frac{n_{i}}{2}=\frac{N}{2}$,

(8)
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\lesssim\max\left[1,\ln\left(\frac{1}{kd(x,y)}\right)\right]e^{-ckd(x,y)}.$

Finally, if $\frac{n_{i}}{2}<j=\frac{N}{2}$,

(9)
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\lesssim\left(\max\left[1,\ln\left(\frac{1}{kd(x,y)}\right)\right]+k^{n_{i}-2j}\right)e^{-ckd(x,y)}.$

Similarly, for the operators
$\nabla(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}$, the
following estimates follow from Proposition 2.1.

######  Corollary 2.2.

For any $j\in\mathbb{N}^{*}$ with $j\neq\frac{n_{i}+1}{2}$ and
$j\neq\frac{N+1}{2}$, there exists $c>0$ such that

(10)
$\displaystyle\begin{split}\left|\nabla_{x}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-j}(x,y)\right|&\lesssim\left(d(x,y)^{\min(2j-1-N,0)}k^{-\max(2j-1-N,0)}\right.\\\
&\qquad\qquad\qquad\left.+d(x,y)^{\min(2j-1-n_{i},0)}k^{-\max(2j-1-n_{i},0)}\right)e^{-ckd(x,y)}.\end{split}$

If $j=\frac{n_{i}+1}{2}<\frac{N+1}{2}$,

(11)
$\left|\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\right|\lesssim\left(d(x,y)^{2j-1-N}+\max\left[1,\ln\left(\frac{1}{kd(x,y)}\right)\right]\right)e^{-ckd(x,y)}.$

If $j=\frac{n_{i}+1}{2}=\frac{N+1}{2}$,

(12)
$\left|\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\right|\lesssim\max\left[1,\ln\left(\frac{1}{kd(x,y)}\right)\right]e^{-ckd(x,y)}.$

Finally, if $\frac{n_{i}+1}{2}<j=\frac{N+1}{2}$,

(13)
$\left|\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\right|\lesssim\left(\max\left[1,\ln\left(\frac{1}{kd(x,y)}\right)\right]+k^{n_{i}+1-2j}\right)e^{-ckd(x,y)}.$

The below proposition follows almost immediately from the two previous
corollaries. We provide an alternative proof, that follows directly from the
first-order resolvent estimates, for the first estimate in the statement.

######  Proposition 2.2.

Let $j\in\mathbb{N}^{*}$. There exists $c>0$ such that

(14)
$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\lesssim
k^{-2(j-1)}\left(d(x,y)^{2-N}+d(x,y)^{2-n_{i}}\right)e^{-ckd(x,y)}$

and

(15)
$\left|\nabla_{x}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-j}(x,y)\right|\lesssim
k^{-2(j-1)}\left(d(x,y)^{1-N}+d(x,y)^{1-n_{i}}\right)e^{-ckd(x,y)}$

for all $x,\,y\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$ and $k>0$.

###### Proof.

For any $k>0$, the higher-order resolvent operator is given by the formula

$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}=\frac{1}{(j-1)!}\int^{\infty}_{0}t^{j-1}e^{-tk^{2}}e^{-t\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}}\,dt.$

Let $e^{-t\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}}(x,y)$ denote the
heat kernel on $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$. Since
$x^{j-1}\lesssim e^{\epsilon x}$ for any $0<\epsilon<1$,

$\displaystyle\begin{split}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)&\simeq
k^{-2(j-1)}\int^{\infty}_{0}(tk^{2})^{(j-1)}e^{-tk^{2}}e^{-t\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}}(x,y)\,dt\\\
&\lesssim
k^{-2(j-1)}\int^{\infty}_{0}e^{-tk^{2}(1-\epsilon)}e^{-t\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}}(x,y)\,dt\\\
&=k^{-2(j-1)}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}(1-\epsilon))^{-1}(x,y).\end{split}$

On applying the known estimates for the first-order resolvent, equation (17)
from [19], we obtain

$(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-j}(x,y)\lesssim
k^{-2(j-1)}\left(d(x,y)^{2-N}+d(x,y)^{2-n_{i}}\right)e^{-c^{\prime}\sqrt{1-\epsilon}kd(x,y)},$

for some $c^{\prime}>0$. This proves (14) with
$c=\sqrt{1-\epsilon}c^{\prime}$. ∎

Finally, when we come to consider the unboundedness of the square function $S$
for $p\geq n_{min}$, the following lower bounds will prove useful. This lower
bound follows directly from Proposition 2.1 and the lower bound for the Bessel
kernel given in (3).

######  Corollary 2.3.

For $j<\frac{n_{i}}{2}$, there must exist $c>0$ such that

$\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-j}(x,y)\gtrsim\left(d(x,y)^{2j-N}+d(x,y)^{2j-n_{i}}\right)e^{-ckd(x,y)},$

for all $x,\,y\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$ and $k>0$.

###  2.2. A Decomposition for the Resolvent

Recall from [19] that in order to prove the boundedness of the low energy part
of the Riesz transforms operator, the resolvent was separated into four
separate components. That is, for $0<k\leq 1$ the resolvent is given by

(16) $(\Delta+k^{2})^{-1}=\sum_{j=1}^{4}G_{j}(k).$

Let us recall the definitions of each of these components. For each
$i=1,\cdots,l$, choose a point $x_{i}^{\circ}$ in the interior of $K_{i}$. Let
$\phi_{i}\in C^{\infty}(\mathcal{M})$ be a function with support entirely
contained in $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ that is
identically equal to $1$ everywhere outside of a compact set. Define
$v_{i}:=-\Delta\phi_{i}$ and let $u_{i}$ be the function whose existence is
asserted by [19, Lem. 2.7] for $v_{i}$. The term $G_{1}(k)$ is entirely
supported on the diagonal ends and is defined through

$G_{1}(k)(x,y):=\sum_{i=1}^{l}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}(x,y)\phi_{i}(x)\phi_{i}(y).$

Let $G_{int}(k)$ be an interior parametrix for the resolvent that is supported
close to the compact subset

$K_{\Delta}:=\left\\{(x,x):x\in K\right\\}\subset\mathcal{M}^{2},$

and agreeing with the resolvent of
$\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}$ in a smaller neighbourhood
of $K_{\Delta}$ intersected with the support of
$\nabla\phi_{i}(x)\phi_{i}(y)$. $G_{2}(k)$ is an operator with kernel that is
compactly supported in $\mathcal{M}^{2}$,

$G_{2}(k)(x,y):=G_{int}(k)(x,y)\left(1-\sum_{i=1}^{l}\phi_{i}(x)\phi_{i}(y)\right).$

$G_{3}(k)$ has the nice property that its kernel is multiplicatively separable
into functions of $x$ and $y$,

$G_{3}(k)(x,y)=\sum_{i=1}^{l}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}(x_{i}^{\circ},y)u_{i}(x,k)\phi_{i}(y).$

For the final term $G_{4}(k)$, first the error term is defined by

$(\Delta+k^{2})(G_{1}(k)+G_{2}(k)+G_{3}(k))=I+E(k).$

Then the operator $G_{4}(k)$ is given by

$G_{4}(k)(x,y):=-\left(\Delta+k^{2}\right)^{-1}v_{y}(x),$

where $v_{y}(x):=E(k)(x,y)$. As computed in [19], it is useful to note that
the error term $E(k)$ has the representation

(17) $E(k)=\sum_{i=1}^{l}(E_{1}^{i}(k)+E_{2}^{i}(k))+E_{3}(k).$

Here

$E_{1}^{i}(k)(x,y):=-2\nabla\phi_{i}(x)\phi_{i}(y)\left[\nabla_{x}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}(x,y)-\nabla_{x}G_{int}(k)(x,y)\right],$
$E_{2}^{i}(k)(x,y):=\phi_{i}(y)v_{i}(x)\left(-\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}(x,y)+G_{int}(k)(x,y)+\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}(x_{i}^{\circ},y)\right)$

for $i=1,\cdots,l$, and

$E_{3}(k)(x,y):=\left((\Delta+k^{2})G_{int}(k)(x,y)-\delta_{y}(x)\right)\left(1-\sum_{i=1}^{l}\phi_{i}(x)\phi_{i}(y)\right),$

where $\delta_{y}$ is the Dirac-delta function centered at $y$.

##  3\. Higher-Order Resolvents on $\mathcal{M}$

In this section, we investigate various properties of the higher-order
resolvent operators $(\Delta+k^{2})^{-j}$ for $\mathcal{M}$. For
$a\in\mathbb{N}$ and $c>0$, define the weight functions
$\omega_{a}^{c}:\mathcal{M}\times[0,1]\rightarrow(0,\infty)$ through

$\omega_{a}^{c}(x,k):=\left\\{\begin{array}[]{c c}1,&x\in K,\\\ \langle
d(x_{i}^{\circ},x)\rangle^{-(n_{i}-a)}e^{-ckd(x_{i}^{\circ},x)},&x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i},\ 1\leq i\leq l.\end{array}\right.$

The dependence of the functions $\omega_{a}^{c}$ on the constant $c$ will
often be kept implicit through the use of the shorthand notation $\omega_{a}$,
and the value of $c$ will then be understood to change from line to line.
Higher-order analogues of the key lemma from [19] will now be proved.

######  Proposition 3.1.

Let $v\in L^{\infty}(\mathcal{M})$ be compactly supported in $K$. Let
$u:\mathcal{M}\times\mathbb{R}_{+}\rightarrow\mathbb{R}$ be a function, whose
existence is asserted by [19, Lem. 2.7], that satisfies $(\Delta+k^{2})u=v$,

$\displaystyle\begin{split}&\left|u(x,k)\right|\lesssim\left\|v\right\|_{\infty}\omega_{2}(x,k),\
and\\\ &\left|\nabla
u(x,k)\right|\lesssim\left\|v\right\|_{\infty}\omega_{1}(x,k),\end{split}$

for all $x\in\mathcal{M}$ and $0\leq k\leq 1$. For $j\in\mathbb{N}$, define
$u^{(j)}:=\partial_{k^{2}}^{(j)}u$. Then

(18) $\left|u^{(j)}(x,k)\right|\lesssim
k^{-2j}\left\|v\right\|_{\infty}\omega_{2}(x,k),$

for all $x\in\mathcal{M}$ and $0<k\leq 1$.

###### Proof.

It will first be proved that for any $j\in\mathbb{N}$, the derivative

(19)
$u^{(j)}(x,k)=\partial_{k^{2}}^{(j)}(\Delta+k^{2})^{-1}v(x)=(-1)^{j}j!(\Delta+k^{2})^{-(j+1)}v(x)$

satisfies the integral identity

(20)
$u^{(j)}(x,k)=(-1)^{j}\int^{\infty}_{0}t^{j}e^{-tk^{2}}e^{-t\Delta}v(x)\,dt$

for any $k>0$ and $x\in\mathcal{M}$. This will be proved by induction. The
base case,

$u^{(0)}(x,k)=u(x,k)=\int^{\infty}_{0}e^{-tk^{2}}e^{-t\Delta}v(x)\,dt,$

was shown to hold for any $k\geq 0$ in [19, Lem. 2.7]. For the inductive step,
assume that (20) holds for a particular $j\in\mathbb{N}$. Then,

(21)
$\displaystyle\begin{split}u^{(j+1)}(x,k)&=\partial_{k^{2}}u^{(j)}(x,k)\\\
&=\partial_{k^{2}}(-1)^{j}\int^{\infty}_{0}t^{j}e^{-tk^{2}}e^{-t\Delta}v(x)\,dt\\\
&=(-1)^{j}\lim_{h\rightarrow
0}\int^{\infty}_{0}t^{j}\left(\frac{e^{-t(k^{2}+h)}-e^{-tk^{2}}}{h}\right)e^{-t\Delta}v(x)\,dt.\end{split}$

The mean value theorem tells us that there must exist some $c(h)\in[0,h]$ for
which

$\displaystyle\begin{split}\left|\frac{e^{-t(k^{2}+h)}-e^{-tk^{2}}}{h}\right|&=te^{-t(k^{2}+c(h))}\\\
&\leq te^{-tk^{2}}.\end{split}$

As the semigroup generated by the Laplace-Beltrami operator is Markovian,
$\left\|e^{-t\Delta}v\right\|_{\infty}\leq\left\|v\right\|_{\infty}$ for all
$t>0$. In addition, from [17, Cor. 4.9], we know that
$\left\|e^{-t\Delta}v\right\|_{\infty}\lesssim\left\|v\right\|_{1}t^{-\frac{n_{min}}{2}}\lesssim\left\|v\right\|_{1}t^{-\frac{3}{2}}$
for $t\geq 1$. Thus,

$\displaystyle\begin{split}\int^{\infty}_{0}t^{j+1}e^{-tk^{2}}\left|e^{-t\Delta}v(x)\right|\,dt&=\int^{1}_{0}t^{j+1}e^{-tk^{2}}\left|e^{-t\Delta}v(x)\right|\,dt+\int^{\infty}_{1}t^{j+1}e^{-tk^{2}}\left|e^{-t\Delta}v(x)\right|\,dt\\\
&\lesssim\left\|v\right\|_{\infty}\int^{1}_{0}t^{j+1}e^{-tk^{2}}\,dt+\left\|v\right\|_{1}\int^{\infty}_{1}t^{j-\frac{1}{2}}e^{-tk^{2}}\,dt\\\
&<\infty.\end{split}$

The dominated convergence theorem applied to (21) then yields

$u^{(j+1)}(x,k)=(-1)^{j+1}\int^{\infty}_{0}t^{j+1}e^{-tk^{2}}e^{-t\Delta}v(x)\,dt,$

thereby completing the inductive proof of (20) for general $j\in\mathbb{N}$,
$x\in\mathcal{M}$ and $k>0$.

The hypothesized estimate for $u^{(j)}$ can now be proved using our freshly
minted integral identity. Utilising the bound $x^{j}\lesssim e^{\epsilon x}$
for any $0<\epsilon<1$ leads to

$\displaystyle\begin{split}\left|u^{(j)}(x,k)\right|&\leq
k^{-2j}\int^{\infty}_{0}(tk^{2})^{j}e^{-tk^{2}}e^{-t\Delta}\left|v(x)\right|\,dt\\\
&\lesssim
k^{-2j}\int^{\infty}_{0}e^{-tk^{2}(1-\epsilon)}e^{-t\Delta}\left|v(x)\right|\,dt\\\
&=k^{-2j}\left(\Delta+k^{2}(1-\epsilon)\right)^{-1}\left|v\right|(x).\end{split}$

Apply [19, Lem. 2.7] to obtain,

$\left|u^{(j)}(x,k)\right|\lesssim
k^{-2j}\left\|v\right\|_{\infty}\omega_{2}^{c^{\prime}}(x,k\sqrt{1-\epsilon}),$

for some $c^{\prime}>0$. Thus (18) holds with constant
$c=c^{\prime}\sqrt{1-\epsilon}$. ∎

Notice that in the previous proposition, the proof of the integral identity
(20) relied upon the key condition $k>0$. For $k=0$, if $j$ is too large then
the integral in (20) will no longer be guaranteed to converge absolutely and,
consequently, the derivative will not exist. This is analogous to the well-
known fact that on standard Euclidean space $\mathbb{R}^{d}$, the Riesz
potentials $\Delta_{\mathbb{R}^{d}}^{-\alpha}$ are not defined for
$\alpha\geq\frac{d}{2}$. Acting as a converse to this, the below proposition
states that if $j$ is small enough then the derivative at $k=0$ will exist
and, moreover, satisfy some useful estimates.

######  Proposition 3.2.

Let $v\in L^{\infty}(\mathcal{M})$ be compactly supported in $K$ and $u$ be as
given in [19, Lem. 2.7]. Then for any $0\leq j<\frac{n_{min}}{2}-1$,

(22)
$\displaystyle\begin{split}&\left|u^{(j)}(x,k)\right|\lesssim\left\|v\right\|_{\infty}\omega_{2j+2}(x,k),\
and\\\ &\left|\nabla
u^{(j)}(x,k)\right|\lesssim\left\|v\right\|_{\infty}\omega_{2j+1}(x,k),\end{split}$

for all $k\in[0,1]$ and $x\in\mathcal{M}$. Moreover, for $k>0$,

(23)
$\left\|u^{(j)}(\cdot,k)-u^{(j)}(\cdot,0)\right\|_{L^{\infty}(\mathcal{M})}\lesssim
k\left\|v\right\|_{L^{\infty}(\mathcal{M})}$

and

(24) $\left\|\nabla u^{(j)}(\cdot,k)-\nabla
u^{(j)}(\cdot,0)\right\|_{L^{\infty}(\mathcal{M})}\lesssim
k^{1-2j}\left\|v\right\|_{L^{\infty}(\mathcal{M})}.$

###### Proof.

From the previous proposition, we know that the derivative $u^{(j)}(x,k)$
exists for $k>0$ and is given by the integral identity (20). With the addition
of the assumption $j<\frac{n_{min}}{2}-1$, it is clear that $u^{(j)}(x,0)$
will also exist and that (20) will be equally applicable. Indeed, this follows
in an identical inductive manner as the proof for $k>0$ from Proposition 3.1.
However, in order to apply the dominated convergence theorem at the inductive
step of the proof, it will be required that

$\int^{\infty}_{0}t^{i+1}e^{-t\cdot 0}\left|e^{-t\Delta}v(x)\right|\,dt$

is absolutely convergent for when $1\leq i+1\leq j$. This follows from

$\displaystyle\begin{split}\int^{\infty}_{0}t^{i+1}\left|e^{-t\Delta}v(x)\right|\,dt&=\int^{1}_{0}t^{i+1}\left|e^{-t\Delta}v(x)\right|\,dt+\int^{\infty}_{1}t^{i+1}\left|e^{-t\Delta}v(x)\right|\,dt\\\
&\lesssim\left\|v\right\|_{\infty}\int^{1}_{0}t^{i+1}\,dt+\left\|v\right\|_{1}\int^{\infty}_{1}t^{i+1-\frac{n_{min}}{2}}\,dt.\end{split}$

The second integral will converge if and only if $i+1<\frac{n_{min}}{2}-1$,
which is implied by our assumption $j<\frac{n_{min}}{2}-1$.

Notice that the restriction $0\leq j<\frac{n_{min}}{2}-1$ not only implies the
existence of $u^{(j)}(x,0)$, but it also tells us that
$\frac{(-1)^{j}}{j!}u^{(j)}$ is a solution of the equation
$(\Delta+k^{2})^{j+1}\tilde{u}=v$ with $u^{(j)}\in L^{\infty}$ uniformly in
$k\in[0,1]$. Also observe that from the identity (19),

$(\Delta+k^{2})u^{(j)}=-ju^{(j-1)}.$

Recall from the proof of [19, Lem. 2.7] that
$\left\|\Delta^{m}u^{(0)}\right\|_{L^{\infty}(V)}\lesssim 1$ for any compact
$V\subset\mathcal{M}\setminus K$ and $m\in\mathbb{N}$. Then, for any $1\leq
j<\frac{n_{min}}{2}-1$,

$\displaystyle\begin{split}\left\|\Delta
u^{(j)}\right\|_{L^{\infty}(V)}&\leq\left\|(\Delta+k^{2})u^{(j)}\right\|_{L^{\infty}(V)}+k^{2}\left\|u^{(j)}\right\|_{L^{\infty}(V)}\\\
&\simeq\left\|u^{(j-1)}\right\|_{L^{\infty}(V)}+k^{2}\left\|u^{(j)}\right\|_{L^{\infty}(V)}\\\
&\lesssim 1.\end{split}$

We also have

$\displaystyle\begin{split}\left\|\Delta^{2}u^{(j)}\right\|_{L^{\infty}(V)}&\leq\left\|\Delta(\Delta+k^{2})u^{(j)}\right\|_{L^{\infty}(V)}+k^{2}\left\|\Delta
u^{(j)}\right\|_{L^{\infty}(V)}\\\ &\simeq\left\|\Delta
u^{(j-1)}\right\|_{L^{\infty}(V)}+k^{2}\left\|\Delta
u^{(j)}\right\|_{L^{\infty}(V)}\\\ &\lesssim 1,\end{split}$

for all $1\leq j<\frac{n_{min}}{2}-1$. This process can be repeated to
arbitrarily high order to obtain
$\left\|\Delta^{m}u^{(j)}\right\|_{L^{\infty}(V)}\lesssim 1$ for any
$m,\,j\in\mathbb{N}$ and compact subset $V\subset\mathcal{M}\setminus K$. This
proves that $u^{(j)}\in C^{\infty}$ uniformly in $k\in[0,1]$ on any compact
subset outside of $K$. The argument for (LABEL:eqtn:DerivativeAt0) then
follows in an identical manner to the corresponding estimates in [19, Lem.
2.7]. In particular, for $\zeta_{i}\in C^{\infty}(\mathcal{M})$ with support
contained in $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$,
therefore satisfying $\mathrm{supp}\,v\subset(\mathrm{supp}\,\zeta_{i})^{c}$,
and such that $(1-\zeta_{i})$ is compactly supported when viewed as a function
on $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}$. Define

$\tilde{u}_{i}^{(j)}(x,k):=\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}\left((\Delta+k^{2})^{(j+1)}(\zeta_{i}u^{(j)}(x,k))\right).$

It can be reasoned that $\tilde{u}_{i}^{(j)}=u^{(j)}\zeta_{i}$, at which point
Corollaries 2.1 and 2.2 can be applied to produce (LABEL:eqtn:DerivativeAt0).

Let’s now prove (23). From the relation (20) in combination with the estimates
$\left\|e^{-t\Delta}v\right\|_{\infty}\lesssim\left\|v\right\|_{\infty}$ for
$t\leq 1$ and
$\left\|e^{-t\Delta}v\right\|_{\infty}\lesssim\left\|v\right\|_{1}t^{-\frac{n_{min}}{2}}$
for $t>1$,

$\left|u^{(j)}(x,k)-u^{(j)}(x,0)\right|\leq\left\|v\right\|_{\infty}\int^{1}_{0}t^{j}(1-e^{-tk^{2}})\,dt+\left\|v\right\|_{1}\int^{\infty}_{1}t^{j}(1-e^{-tk^{2}})t^{-\frac{n_{min}}{2}}\,dt.$

For the first term, the estimate $1-e^{-x}\leq x$ for $x\in(0,1)$ implies

$\displaystyle\begin{split}\int^{1}_{0}t^{j}(1-e^{-tk^{2}})\,dt&\leq
k^{2}\int^{1}_{0}t^{j+1}\,dt\\\ &\leq k^{2}.\end{split}$

For the second term, notice that since $j<\frac{n_{min}}{2}-1$ we must have
$j\leq\frac{n_{min}}{2}-\frac{3}{2}$. A change of variables then leads to

$\displaystyle\begin{split}\int^{\infty}_{1}t^{j}(1-e^{-tk^{2}})t^{-\frac{n_{min}}{2}}\,dt&\leq\int^{\infty}_{1}t^{-\frac{3}{2}}(1-e^{-tk^{2}})\,dt\\\
&=k\int^{\infty}_{k^{2}}t^{-\frac{3}{2}}(1-e^{-t})\,dt\\\ &\leq
k\int^{\infty}_{0}t^{-\frac{3}{2}}(1-e^{-t})\,dt\\\ &\lesssim k.\end{split}$

Finally, let us consider the validity of (24). This estimate has already been
proved for the case $j=0$ in [19, Lem. 2.7]. For $1\leq
j<\frac{n_{min}}{2}-1$, observe that

$(\Delta+k^{2})u^{(j)}(x,k)=-ju^{(j-1)}(x,k)\quad and\quad\Delta
u^{(j)}(x,0)=-ju^{(j-1)}(x,0).$

Therefore,

$\left|\Delta u^{(j)}(x,k)-\Delta u^{(j)}(x,0)\right|\lesssim
k^{2}\left|u^{(j)}(x,k)\right|+\left|u^{(j-1)}(x,k)-u^{(j-1)}(x,0)\right|.$

Proposition 3.1 and (23) then imply

$\displaystyle\begin{split}\left|\Delta u^{(j)}(x,k)-\Delta
u^{(j)}(x,0)\right|&\lesssim\left(k^{2-2j}+k\right)\left\|v\right\|_{\infty}\\\
&\lesssim k^{1-2j}\left\|v\right\|_{\infty}.\end{split}$

Remark 2.8 of [19] states that

$\left\|\nabla g\right\|_{L^{\infty}(\mathcal{M})}\lesssim\left\|\Delta
g\right\|_{L^{\infty}(\mathcal{M})}+\left\|g\right\|_{L^{\infty}(\mathcal{M})}.$

Therefore,

$\displaystyle\begin{split}\left\|\nabla(u^{(j)}(\cdot,k)-u^{(j)}(\cdot,0))\right\|_{\infty}&\lesssim\left\|\Delta(u^{(j)}(\cdot,k)-u^{(j)}(\cdot,0))\right\|_{\infty}+\left\|u^{(j)}(\cdot,k)-u^{(j)}(\cdot,0)\right\|_{\infty}\\\
&\lesssim k^{1-2j}\left\|v\right\|_{\infty},\end{split}$

which completes our proof. ∎

######  Proposition 3.3.

Let $v\in L^{\infty}(\mathcal{M})$ be compactly supported in $K$. Then, for
any $j\in\mathbb{N}^{*}$,

(25) $\left|\nabla(\Delta+k^{2})^{-j}v(x)\right|\lesssim
k^{-2(j-1)}\left\|v\right\|_{\infty}\omega_{1}(x,k)$

for all $x\in\mathcal{M}$ and $0<k\leq 1$.

###### Proof.

Before beginning the proof, note that in the notation of Propositions 3.1 and
3.2, $\nabla(\Delta+k^{2})^{-j}v=\nabla u^{(j-1)}$. We use resolvent notation
in this proposition to simplify our computations.

This estimate has already been proved for $j=1$ in [19, Lem. 2.7]. For $j>1$,

$\displaystyle\begin{split}\left|\nabla(\Delta+k^{2})^{-j}v(x)\right|&=\left|\nabla(\Delta+k^{2})^{-1}(\Delta+k^{2})^{-(j-1)}v(x)\right|\\\
&\leq\sum_{i=1}^{4}\left|\nabla G_{i}(k)(\Delta+k^{2})^{-(j-1)}v(x)\right|\\\
&\leq\sum_{i=1}^{4}\int_{\mathcal{M}}\left|\nabla_{x}G_{i}(k)(x,y)\right|\left|(\Delta+k^{2})^{-(j-1)}v(y)\right|\,dy\\\
&=:\sum_{i=1}^{4}Y_{i}(k)(x).\end{split}$

Our desired estimate will be proved for each of the above terms separately.

The Term $Y_{1}(k)$.

The product rule implies that

$\nabla_{x}G_{1}(k)(x,y)=\sum_{i=1}^{l}\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-1}(x,y)\phi_{i}(x)\phi_{i}(y)+(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-1}(x,y)\nabla\phi_{i}(x)\phi_{i}(y).$

This leads to the splitting

$Y_{1}(k)(x)\leq Y_{1}^{1}(k)(x)+Y^{2}_{1}(k)(x),$

where

$Y^{1}_{1}(k)(x)=\int_{\mathcal{M}}\left|\sum_{i=1}^{l}\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-1}(x,y)\phi_{i}(x)\phi_{i}(y)\right|\left|(\Delta+k^{2})^{-(j-1)}v(y)\right|\,dy$

and

$Y^{2}_{1}(k)(x)=\int_{\mathcal{M}}\left|\sum_{i=1}^{l}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-1}(x,y)\nabla\phi_{i}(x)\phi_{i}(y)\right|\left|(\Delta+k^{2})^{-(j-1)}v(y)\right|\,dy.$

If $x\in K$, then both of these terms will clearly vanish, trivially leading
to

$Y_{1}(k)(x)\lesssim k^{-2(j-1)}\left\|v\right\|_{\infty}\omega_{1}(x,k).$

Let us consider $x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$
for some $1\leq i\leq l$. For the first term, let $D>0$ be such that
$d(x_{i}^{\circ},y)\geq 2D$ for all
$y\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$. Corollary 2.2
and Proposition 3.1 then produce,

$\displaystyle\begin{split}&Y_{1}^{1}(k)(x)\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left|\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-1}(x,y)\right|\left|(\Delta+k^{2})^{-(j-1)}v(y)\right|\,dy\\\
&\ \ \lesssim
k^{-2(j-2)}\left\|v\right\|_{\infty}\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left[d(x,y)^{1-n_{i}}+d(x,y)^{1-N}\right]\langle
d(x_{i}^{\circ},y)\rangle^{2-n_{i}}e^{-ck(d(x,y)+d(x_{i}^{\circ},y))}\,dy\\\
&=:I_{1}+I_{2},\end{split}$

where the above integral has been separated into two distinct integrals
$I_{1}$ and $I_{2}$ over the respective regions

$\mathcal{A}_{1}:=(\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i})\cap
B(x,d(x_{i}^{\circ},x)/2)$

and

$\mathcal{A}_{2}:=(\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i})\cap
B(x,d(x_{i}^{\circ},x)/2)^{c}.$

For $I_{1}$, it is evident that $d(x_{i}^{\circ},y)\gtrsim d(x_{i}^{\circ},x)$
for any $y\in\mathcal{A}_{1}$. Therefore,

$\displaystyle\begin{split}I_{1}&=k^{-2(j-2)}\left\|v\right\|_{\infty}\int_{\mathcal{A}_{1}}\left[d(x,y)^{1-n_{i}}+d(x,y)^{1-N}\right]\langle
d(x_{i}^{\circ},y)\rangle^{2-n_{i}}\exp(-ck(d(x,y)+d(x_{i}^{\circ},y)))\,dy\\\
&\lesssim k^{-2(j-2)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}e^{-ckd(x_{i}^{\circ},x)}\int_{\mathcal{A}_{1}}\left[d(x,y)^{1-n_{i}}+d(x,y)^{1-N}\right]\exp(-ckd(x,y))\,dy\\\
&\lesssim k^{-2(j-2)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}e^{-ckd(x_{i}^{\circ},x)}\left(\int_{B(x,D)}d(x,y)^{1-N}\,dy+\int_{B(x,d(x_{i}^{\circ},x)/2)\setminus
B(x,D)}d(x,y)^{1-n_{i}}\,dy\right)\\\ &\lesssim
k^{-2(j-2)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}e^{-ckd(x_{i}^{\circ},x)}\left(1+d(x_{i}^{\circ},x)\right)\\\
&\lesssim k^{-2(j-1)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}e^{-\frac{c}{2}kd(x_{i}^{\circ},x)},\end{split}$

where the last line follows on absorbing the term $(kd(x_{i}^{\circ},x))^{2}$
into the exponential. For the integral $I_{2}$,

$\displaystyle\begin{split}I_{2}&=k^{-2(j-2)}\left\|v\right\|_{\infty}\int_{\mathcal{A}_{2}}\left[d(x,y)^{1-n_{i}}+d(x,y)^{1-N}\right]\langle
d(x_{i}^{\circ},y)\rangle^{2-n_{i}}e^{-ck(d(x,y)+d(x_{i}^{\circ},y))}\,dy\\\
&\lesssim
k^{-2(j-2)}\left\|v\right\|_{\infty}d(x_{i}^{\circ},x)^{1-n_{i}}e^{-ckd(x_{i}^{\circ},x)}\int_{B(x,d(x_{i}^{\circ},x)/2)^{c}}\langle
d(x_{i}^{\circ},y)\rangle^{2-n_{i}}e^{-ckd(x_{i}^{\circ},y)}\,dy\\\ &\lesssim
k^{-2(j-2)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}e^{-ckd(x_{i}^{\circ},x)}\int_{\mathbb{R}^{n_{i}}}\frac{1}{\left|y_{1}\right|^{n_{i}-2}}e^{-ck\left|y_{1}\right|}\,dy_{1}\\\
&\simeq k^{-2(j-2)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}e^{-ckd(x_{i}^{\circ},x)}\int_{0}^{\infty}re^{-ckr}\,dr\\\
&\lesssim k^{-2(j-1)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}e^{-ckd(x_{i}^{\circ},x)}.\end{split}$

For the term $Y_{1}^{2}(k)$, Corollary 2.1 and Proposition 3.1 imply

$\displaystyle\begin{split}&Y^{2}_{1}(k)(x)=\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-1}(x,y)\nabla\phi_{i}(x)\phi_{i}(y)\right|\left|(\Delta+k^{2})^{-(j-1)}v(y)\right|\,dy\\\
&\ \
\lesssim\left|\nabla\phi_{i}(x)\right|k^{-2(j-2)}\left\|v\right\|_{\infty}\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left[d(x,y)^{2-n_{i}}+d(x,y)^{2-N}\right]\langle
d(x_{i}^{\circ},y)\rangle^{2-n_{i}}e^{-ck(d(x,y)+d(x_{i}^{\circ},y))}\,dy.\end{split}$

Using the same argument as for the term $Y^{1}_{1}(k)$, we will obtain

$\displaystyle\begin{split}Y^{2}_{1}(k)(x)&\lesssim\left|\nabla\phi_{i}(x)\right|k^{-2(j-1)}\left\|v\right\|_{\infty}\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}e^{-ckd(x_{i}^{\circ},x)}\\\ &\lesssim
k^{-2(j-1)}\left\|v\right\|_{\infty}\omega_{1}(x,k),\end{split}$

where we used the fact that $\nabla\phi_{i}$ is compactly supported to obtain
the last line.

The Term $Y_{2}(k)$.

The kernel $\nabla_{x}G_{2}(k)(x,y)$ is compactly supported in some
neighbourhood of

$K_{\Delta}:=\left\\{(x,x)\in\mathcal{M}^{2}:x\in K\right\\}.$

There must then exist some compact set $V$ containing $K$ such that
$\nabla_{x}G_{2}(k)(x,y)$ is supported in $V\times V$. This implies that
$Y_{2}(k)$ must also be compactly supported in $V$. For $x\in V^{c}$, the
estimate

(26) $Y_{2}(k)(x)\lesssim k^{-2(j-1)}\left\|v\right\|_{\infty}\omega_{1}(x,k)$

will then be trivially satisfied. It remains to ascertain the validity of this
estimate for $x\in V$. The operators $\left\\{\nabla
G_{2}(k)\right\\}_{k\in(0,1)}$ constitute a family of pseudodifferential
operators of order $-1$. At the scale of an individual chart on $\mathcal{M}$,
it is evident that in local coordinates the symbol of $\nabla G_{2}(k)$,
denoted $a_{2}(k)(x,\xi)$, will satisfy

$\left|\partial^{\alpha}_{\xi}a_{2}(k)(x,\xi)\right|\lesssim(\left|\xi\right|^{2}+k^{2})^{\frac{-1-\left|\alpha\right|}{2}}$

for all multi-indices $\alpha\geq 0$ and $k\in(0,1)$. From this, standard
pseudodifferential operator theory (c.f. [29, Sec. 0.2] for instance) implies

$\left|\nabla_{x}G_{2}(k)(x,y)\right|\lesssim d(x,y)^{1-N},$

uniformly in $k\in(0,1)$. This, in combination with Proposition 3.1, implies

$\displaystyle\begin{split}Y_{2}(k)(x)&\lesssim
k^{-2(j-2)}\left\|v\right\|_{\infty}\int_{V}d(x,y)^{1-N}\langle
d(x_{i}^{\circ},y)\rangle^{2-n_{i}}\exp(-ckd(x_{i}^{\circ},y))\,dy\\\
&\lesssim k^{-2(j-1)}\left\|v\right\|_{\infty},\end{split}$

which completes the proof of (26) for $x\in V$.

The Term $Y_{3}(k)$.

Recall from [19] that the kernel $\nabla_{x}G_{3}(k)(x,y)$ satisfies the
estimate

$\left|\nabla_{x}G_{3}(k)(x,y)\right|\lesssim\omega_{1}(x,k)\omega_{2}(y,k).$

Proposition 3.1 then implies,

$\displaystyle\begin{split}Y_{3}(k)(x)&=\int_{\mathcal{M}}\left|\nabla_{x}G_{3}(k)(x,y)\right|\left|(\Delta+k^{2})^{-(j-1)}v(y)\right|\,dy\\\
&\lesssim
k^{-2(j-2)}\left\|v\right\|_{\infty}\int_{\mathcal{M}}\omega_{1}(x,k)\omega_{2}(y,k)\omega_{2}(y,k)\,dy\\\
&=k^{-2(j-2)}\left\|v\right\|_{\infty}\omega_{1}(x,k)\int_{\mathcal{M}}\omega_{2}(y,k)^{2}\,dy.\end{split}$

Therefore, in order to prove the desired estimate for $Y_{3}(k)$, it is
sufficient to show that

$\int_{\mathcal{M}}\omega_{2}(y,k)^{2}\,dy\lesssim k^{-2}.$

Considering the integral over the center $K$ first,

$\int_{K}\omega_{2}(y,k)^{2}\,dy\lesssim\int_{K}1\,dy\lesssim 1\leq k^{-2}.$

On the ends $\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ for
$1\leq i\leq l$ we have,

$\displaystyle\begin{split}\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\omega_{2}(y,k)^{2}\,dy&\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\langle
d(x_{i}^{\circ},y)\rangle^{4-2n_{i}}\exp(-2ckd(x_{i}^{\circ},y))\,dy\\\
&\lesssim\int^{\infty}_{0}\langle
r\rangle^{4-2n_{i}}\exp(-2ckr)r^{n_{i}-1}\,dr\\\ &\lesssim k^{-1}\leq
k^{-2}.\end{split}$

The Term $Y_{4}(k)$.

This can be handled in an identical manner to the term $Y_{3}(k)$ since
$\nabla_{x}G_{4}(k)(x,y)$ satisfies even stronger estimates than
$\nabla_{x}G_{3}(k)(x,y)$ (c.f. [19, Sec. 3]).

∎

In [19], the decomposition (16) allowed for the Riesz transform to be
separated into four corresponding components. The $L^{p}$-boundedness of each
part was then proved independently. In this article, we will also make use of
this decomposition. Notice that

$\left(\Delta+k^{2}\right)^{-M}=\frac{(-1)^{M-1}}{(M-1)!}\partial_{k^{2}}^{(M-1)}(\Delta+k^{2})^{-1}.$

On combining this with the splitting (16), we have the relation

(27) $\left(\Delta+k^{2}\right)^{-M}=\sum_{i=1}^{4}H_{i}^{(M)}(k),$

where
$H_{i}^{(M)}(k):=\frac{(-1)^{M-1}}{(M-1)!}\partial_{k^{2}}^{(M-1)}G_{i}(k)$.
To simplify notation, when the integer $M>0$ is understood, the shorthand
notation $H_{i}(k)$ will be employed. For the remainder of this section we
will investigate various properties of these operators and, in particular,
obtain asymptotic estimates for the kernels of $H_{3}(k)$ and $H_{4}(k)$.

######  Proposition 3.4.

The kernel of the operator $H_{3}(k)$ satisfies

$\left|H_{3}(k)(x,y)\right|\lesssim k^{-2(M-1)}\omega_{2}(x,k)\omega_{2}(y,k)$

and

$\left|\nabla_{x}H_{3}(k)(x,y)\right|\lesssim
k^{-2(M-1)}\omega_{1}(x,k)\omega_{2}(y,k),$

for all $x,\,y\in\mathcal{M}$ and $0<k\leq 1$.

###### Proof.

We have

$\displaystyle\begin{split}&H_{3}(k)(x,y)=\frac{(-1)^{M-1}}{(M-1)!}\partial^{(M-1)}_{k^{2}}G_{3}(k)(x,y)\\\
&\quad=\frac{(-1)^{M-1}}{(M-1)!}\partial^{(M-1)}_{k^{2}}\sum_{i=1}^{l}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}(x_{i}^{\circ},y)u_{i}(x,k)\phi_{i}(y)\\\
&\quad=\frac{(-1)^{M-1}}{(M-1)!}\sum_{i=1}^{l}\sum_{j=0}^{M-1}\left(\begin{array}[]{c}M-1\\\
j\end{array}\right)\partial^{(j)}_{k^{2}}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-1}(x_{i}^{\circ},y)\cdot
u_{i}^{(M-1-j)}(x,k)\phi_{i}(y)\\\
&\quad=\sum_{i=1}^{l}\sum_{j=0}^{M-1}\frac{(-1)^{M+j-1}}{(M-j-1)!}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-(j+1)}(x_{i}^{\circ},y)u_{i}^{(M-1-j)}(x,k)\phi_{i}(y).\end{split}$

Proposition 3.1 then implies that

$\displaystyle\begin{split}\left|H_{3}(k)(x,y)\right|&\lesssim\sum_{i=1}^{l}\sum_{j=0}^{M-1}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x_{i}^{\circ},y)\left|u_{i}^{(M-1-j)}(x,k)\right|\phi_{i}(y)\\\
&\lesssim\omega_{2}(x,k)\sum_{j=0}^{M-1}k^{-2(M-1-j)}\sum_{i=1}^{l}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x_{i}^{\circ},y)\phi_{i}(y).\end{split}$

Proposition 2.2, when combined with the fact that $\mathrm{supp}\
\phi_{i}\subset\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ and
$x_{i}^{\circ}$ is in the interior of $K_{i}$, leads to

$\left|H_{3}(k)(x,y)\right|\lesssim\omega_{2}(x,k)\omega_{2}(y,k)\sum_{j=0}^{M-1}k^{-2(M-1-j)}k^{-2j}\simeq
k^{-2(M-1)}\omega_{2}(x,k)\omega_{2}(y,k),$

which proves our claim. The estimate for $\nabla_{x}H_{3}(k)(x,y)$ follows via
an identical argument, except the use of Proposition 3.1 must be replaced with
Proposition 3.3. ∎

######  Lemma 3.1.

For any $j\in\mathbb{N}$, all $0<k\leq 1$ and $y\in\mathcal{M}$,

$\left\|\partial_{k^{2}}^{(j)}E(k)(\cdot,y)\right\|_{\infty}\lesssim
k^{-2j}\omega_{1}(y,k).$

###### Proof.

This lemma has already been proved for the case $j=0$ in [19], and so it can
be assumed that $j>0$. From the splitting (17), it is sufficient to prove this
bound for the kernels of each of the components $E_{1}^{i}(k)$, $E_{2}^{i}(k)$
for $i=1,\cdots,l$ and $E_{3}(k)$.

The Error Term $E^{i}_{1}(k)$.

On differentiating the expression for $E^{i}_{1}$ with respect to $k^{2}$,

$\partial^{(j)}_{k^{2}}E^{i}_{1}(k)(x,y)=-2\nabla\phi_{i}(x)\phi_{i}(y)\left[(-1)^{j}j!\nabla_{x}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x,y)-\nabla_{x}\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right].$

$G_{int}(k)$ agrees with the resolvent
$\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}$
near the diagonal and on the support of $\nabla\phi_{i}(x)\cdot\phi_{i}(y)$.
Therefore $E_{1}^{i}$ will vanish on this set.

For $(x,y)$ away from the diagonal and on the support of
$\nabla\phi_{i}(x)\phi_{i}(y)$, we have

(28)
$\displaystyle\begin{split}\left|\partial^{(j)}_{k^{2}}E^{i}_{1}(k)(x,y)\right|&\lesssim\left|\nabla_{x}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x,y)\right|+\left|\nabla_{x}\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|\\\
&\lesssim
k^{-2j}\omega_{1}(y,k)+\left|\nabla_{x}\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|,\end{split}$

where the last line follows from Proposition 2.2. The operators
$\left\\{\nabla\partial^{(j)}_{k^{2}}G_{int}(k)\right\\}_{k\in(0,1)}$
constitute a family of pseudodifferential operators of order $-1-2j$. At the
scale of an individual chart on $\mathcal{M}$, it is evident that in local
coordinates the symbol of $\nabla\partial^{(j)}_{k^{2}}G_{int}(k)$, denoted
$a_{int}(k)(x,\xi)$, will satisfy

$\left|\partial^{\alpha}_{\xi}a_{int}(k)(x,\xi)\right|\lesssim(\left|\xi\right|^{2}+k^{2})^{\frac{-1-2j-\left|\alpha\right|}{2}}$

for all multi-indices $\alpha\geq 0$ and $k\in(0,1)$. From this, standard
pseudodifferential operator theory (c.f. [29, Sec. 0.2]) implies that for any
$b>-1-2j+N$,

$\left|\nabla_{x}\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|\lesssim
k^{N-1-2j-b}d(x,y)^{-b}$

uniformly in $k\in(0,1)$, for all $x,\,y\in\mathcal{M}$. Setting $b=N-1$ gives

$\left|\nabla_{x}\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|\lesssim
k^{-2j}d(x,y)^{1-N}$

uniformly in $k\in(0,1)$, for all $x,\,y\in\mathcal{M}$. As we are considering
$(x,y)$ away from the diagonal and $G_{int}(k)(x,y)$ is compactly supported in
$\mathcal{M}^{2}$, this estimate implies

$\displaystyle\begin{split}\left|\nabla_{x}\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|&\lesssim
k^{-2j}\langle d(x,y)\rangle^{1-N}\\\ &\lesssim
k^{-2j}\omega_{1}(y,k),\end{split}$

thereby implying our desired estimate.

The Error Term $E^{i}_{2}(k)$.

For the second term,

$\displaystyle\begin{split}\partial^{(j)}_{k^{2}}E^{i}_{2}(k)(x,y)&=\phi_{i}(y)v_{i}(x)\left[(-1)^{j}j!\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x,y)+\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right.\\\
&\qquad\left.+(-1)^{j}j!\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x_{i}^{\circ},y)\right].\end{split}$

The first two terms will vanish on the diagonal and on the support of
$\phi_{i}(y)v_{i}(x)$, since this is where $G_{int}(k)$ coincides with the
resolvent. The desired estimate would then follow from Proposition 2.2.

For $(x,y)$ away from the diagonal and on the support of
$\phi_{i}(y)v_{i}(x)$,

$\displaystyle\begin{split}\left|\partial_{k^{2}}^{(j)}E^{i}_{2}(k)(x,y)\right|&\lesssim\left|\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x_{i}^{\circ},y)-\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x,y)\right|\\\
&\qquad\qquad+\left|\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|\\\
&\lesssim\left|x_{i}^{\circ}-x\right|\left|\nabla_{x}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(\tilde{x},y)\right|+\left|\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|\\\
&\lesssim
k^{-2j}\omega_{1}(y,k)+\left|\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|,\end{split}$

for some $\tilde{x}$ close to the set $K_{i}$, where the last line follows
from Proposition 2.2. Similar pseudodifferential reasoning as for the term
$E_{1}^{i}(k)$ then implies
$\left|\partial^{(j)}_{k^{2}}G_{int}(k)(x,y)\right|\lesssim
k^{-2j}\omega_{1}(y,k)$, thereby proving the desired estimate.

The Error Term $E_{3}(k)$.

Finally, from the form of $\partial_{k^{2}}^{(j)}E_{3}(k)$, it is clear that
in an analogous manner to the previous two terms, pseudodifferential reasoning
can be applied to obtain the estimate
$\left\|\partial^{(j)}_{k^{2}}E_{3}(k)(\cdot,y)\right\|_{\infty}\lesssim
k^{-2j}\omega_{1}(y,k)$. ∎

######  Proposition 3.5.

The kernel of the operator $H_{4}(k)$ satisfies

$\left|H_{4}(k)(x,y)\right|\lesssim k^{-2(M-1)}\omega_{2}(x,k)\omega_{1}(y,k)$

and

$\left|\nabla_{x}H_{4}(k)(x,y)\right|\lesssim
k^{-2(M-1)}\omega_{1}(x,k)\omega_{1}(y,k)$

for all $x,\,y\in\mathcal{M}$ and $0<k\leq 1$.

###### Proof.

Recall that

$\displaystyle\begin{split}G_{4}(k)(x,y)&=-(\Delta+k^{2})^{-1}v_{y}(x)\\\
&=-\int^{\infty}_{0}e^{-tk^{2}}e^{-t\Delta}v_{y}(x)\,dt,\end{split}$

where $v_{y}(x):=E(k)(x,y)$. If we define
$v^{(j)}_{y}:=\partial_{k^{2}}^{(j)}v_{y}$ for $j\in\mathbb{N}$, then Lemma
3.1 implies

$\left\|v^{(j)}_{y}\right\|_{\infty}=\left\|\partial_{k^{2}}^{(j)}E(k)(\cdot,y)\right\|_{\infty}\lesssim
k^{-2j}\omega_{1}(y,k).$

Therefore $e^{-t\Delta}v_{y}^{(j)}$ is well-defined and we have by the
dominated convergence theorem

$\partial_{k^{2}}^{(j)}e^{-t\Delta}v_{y}(x)=e^{-t\Delta}v_{y}^{(j)}(x).$

Thus

$\displaystyle\begin{split}\partial_{k^{2}}^{(M-1)}G_{4}(k)(x,y)&=\partial_{k^{2}}^{(M-1)}\int^{\infty}_{0}e^{-tk^{2}}e^{-t\Delta}v_{y}(x)\,dt\\\
&=\int^{\infty}_{0}\partial^{(M-1)}_{k^{2}}\left(e^{-tk^{2}}e^{-t\Delta}v_{y}(x)\right)\,dt\\\
&=\sum_{j=0}^{M-1}\left(\begin{array}[]{c}M-1\\\
j\end{array}\right)(-1)^{j}\int^{\infty}_{0}t^{j}e^{-tk^{2}}e^{-t\Delta}v_{y}^{(M-1-j)}(x)\,dt.\end{split}$

The integral identity (20) then implies

(29)
$\partial_{k^{2}}^{(M-1)}G_{4}(k)(x,y)=\sum_{j=0}^{M-1}\left(\begin{array}[]{c}M-1\\\
j\end{array}\right)(\Delta+k^{2})^{-(j+1)}v_{y}^{(M-1-j)}(x).$

An application of Proposition 3.1 leads to,

$\displaystyle\begin{split}\left|\partial_{k^{2}}^{(M-1)}G_{4}(k)(x,y)\right|&\lesssim\sum_{j=0}^{M-1}k^{-2j}\omega_{2}(x,k)\left\|v_{y}^{(M-1-j)}\right\|_{\infty}\\\
&\lesssim\omega_{2}(x,k)\omega_{1}(y,k)\sum_{j=0}^{M-1}k^{-2j}k^{-2(M-1-j)}\\\
&\simeq k^{-2(M-1)}\omega_{2}(x,k)\omega_{1}(y,k).\end{split}$

The definition
$H_{4}(k):=\frac{(-1)^{M-1}}{(M-1)!}\partial_{k^{2}}^{(M-1)}G_{4}(k)$ then
allows us to immediately obtain

$\left|H_{4}(k)(x,y)\right|\lesssim
k^{-2(M-1)}\omega_{2}(x,k)\omega_{1}(y,k).$

Estimates for $\left|\nabla_{x}H_{4}(k)(x,y)\right|$ follow in an identical
manner from (29), except the use of Proposition 3.1 must be replaced with an
application of Proposition 3.3. ∎

##  4\. The Low Energy Square Function

Recall from Section 3 that the higher-order resolvent has the splitting

$(\Delta+k^{2})^{-M}=\sum_{i=1}^{4}H_{i}(k),$

where $H_{i}(k):=\frac{(-1)^{M-1}}{(M-1)!}\partial_{k^{2}}^{(M-1)}G_{i}(k)$.
The low energy square function can then be controlled from above by

$S_{<}f(x)\lesssim\sum_{i=1}^{4}S_{<}^{i}f(x),$

where $S_{<}^{i}$ is the part of the low energy square function corresponding
to $H_{i}(k)$,

$S_{<}^{i}f(x):=\left(\int^{1}_{0}\left|\nabla
H_{i}(k)f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

The $L^{p}$-boundedness for $p\in(1,n_{min})$ and weak-type $(1,1)$ property
of $S_{<}$ will be proved by demonstrating that each component $S_{<}^{i}$ is
itself bounded on $L^{p}$ and weak-type $(1,1)$ for $i=1,\cdots,4$.

###  4.1. The Operator $S_{<}^{1}$

From the definition of $H_{1}(k)$, we have

$\displaystyle\begin{split}H_{1}(k)(x,y)&=\frac{(-1)^{M-1}}{(M-1)!}\partial^{(M-1)}_{k^{2}}\sum_{i=1}^{l}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-1}(x,y)\phi_{i}(x)\phi_{i}(y)\\\
&=\sum_{i=1}^{l}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(x)\phi_{i}(y).\end{split}$

As $H_{1}(k)$ consists of finitely many terms, in order to prove the
boundedness of the operator $S^{1}_{<}$, we need only prove boundedness of the
operators

$S^{1,i}_{<}f(x):=\left(\int^{1}_{0}\left|\int_{\mathcal{M}}\nabla_{x}\left[(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(x)\phi_{i}(y)\right]f(y)\,dy\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}$

for all $1\leq i\leq l$. From the product rule, this operator is in turn
controlled by

$S^{1,i}_{<}f(x)\leq\Lambda^{i}f(x)+\Pi^{i}f(x),$

where

$\Lambda^{i}f(x):=\left(\int^{1}_{0}\left|\int_{\mathcal{M}}\nabla_{x}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(x)\phi_{i}(y)f(y)\,dy\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}$

and

$\Pi^{i}f(x):=\left(\int^{1}_{0}\left|\int_{\mathcal{M}}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\nabla\phi_{i}(x)\phi_{i}(y)f(y)\,dy\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

First consider the operator $\Lambda^{i}$. It is well-known from classical
theory that the square function

$\left(\int^{\infty}_{0}\left|\nabla(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}g(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}=\left(\int^{\infty}_{0}\left|t\nabla(t^{2}\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+I)^{-M}g(x)\right|^{2}\,\frac{dt}{t}\right)^{\frac{1}{2}}$

is bounded on $L^{p}(\mathbb{R}^{n_{i}}\times\mathcal{M}_{i})$ for all
$p\in(1,\infty)$ and weak-type $(1,1)$. Therefore,

$\displaystyle\begin{split}\left\|\Lambda_{i}f\right\|_{p}&=\left\|\phi_{i}\cdot\left(\int^{1}_{0}\left|\nabla(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(\phi_{i}\cdot
f)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\right\|_{p}\\\
&\leq\left\|\left(\int^{\infty}_{0}\left|\nabla(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(\phi_{i}\cdot
f)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\right\|_{p}\\\
&\lesssim\left\|f\right\|_{p}\end{split}$

for any $p\in(1,\infty)$. Weak-type $(1,1)$ bounds for $\Lambda_{i}$ follow in
the same manner.

Next, consider the operator $\Pi^{i}$. For $r>0$, define the set
$D_{r}:=\left\\{(x,y)\in\mathcal{M}^{2}:d(x,y)\leq r\right\\}$. Minkowski’s
integral inequality implies that

$\displaystyle\begin{split}\Pi^{i}f(x)&\leq\int_{\mathcal{M}}\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\nabla\phi_{i}(x)\phi_{i}(y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\left|f(y)\right|\,dy\\\
&=\int_{\mathcal{M}}\left|\nabla\phi_{i}(x)\right|\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\left|f(y)\right|\,dy\\\
&\leq\int_{\mathcal{M}}\pi_{1}^{i}(x,y)\left|f(y)\right|\,dy+\int_{\mathcal{M}}\pi_{2}^{i}(x,y)\left|f(y)\right|\,dy\\\
&=:\Pi_{1}^{i}f(x)+\Pi_{2}^{i}f(x),\end{split}$

where $\pi_{1}^{i}(x,y)$ and $\pi_{2}^{i}(x,y)$ are the kernels defined by

$\pi_{1}^{i}(x,y):=\left|\nabla\phi_{i}(x)\right|\chi_{D_{r}}(x,y)\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y)$

and

$\pi_{2}^{i}(x,y):=\left|\nabla\phi_{i}(x)\right|\left|1-\chi_{D_{r}}(x,y)\right|\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y).$

Let’s prove that $\Pi_{1}^{i}$ is $L^{p}$-bounded for all $p\in[1,\infty]$.
From Proposition 2.2,

$\displaystyle\begin{split}&\left|\nabla\phi_{i}(x)\right|\chi_{D_{r}}(x,y)\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y))^{2}\right|k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\hskip
144.54pt\lesssim\chi_{D_{r}}(x,y)\left(\int^{1}_{0}\left|d(x,y)^{2-N}\exp\left(-ckd(x,y)\right)\right|^{2}k\,dk\right)^{\frac{1}{2}}\\\
&\hskip
144.54pt=\chi_{D_{r}}(x,y)d(x,y)^{2-N}\left(\int^{1}_{0}k\exp\left(-2ckd(x,y)\right)\,dk\right)^{\frac{1}{2}}\\\
&\hskip 144.54pt\lesssim\chi_{D_{r}}(x,y)d(x,y)^{1-N}.\end{split}$

It is obvious that an operator with this kernel will be bounded on $L^{p}$ for
all $p\in[1,\infty]$ since the local decay $d(x,y)^{1-N}$ is stronger than
$d(x,y)^{-N}$, and globally it is cut off by the function $\chi_{D_{r}}$.
Therefore $\Pi_{1}^{i}$ is bounded on $L^{p}$ for all $p\in[1,\infty]$.

Next, consider the operator $\Pi_{2}^{i}$. From an application of Hölder’s
inequality for $p\geq 1$,

$\displaystyle\begin{split}\left\|\Pi^{i}_{2}f\right\|_{p}^{p}&=\int_{\mathcal{M}}\left(\int_{\mathcal{M}}\pi_{2}^{i}(x,y)\left|f(y)\right|\,dy\right)^{p}\,dx\\\
&\leq\int_{\mathcal{M}}\left\|\pi^{i}_{2}(x,\cdot)\right\|^{p}_{L^{p^{\prime}}(\mathcal{M})}\,dx\cdot\left\|f\right\|^{p}_{L^{p}(\mathcal{M})}.\end{split}$

On applying Proposition 2.2 for
$x,\,y\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ with
$d(x,y)>r$,

$\displaystyle\begin{split}\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}&\lesssim
d(x,y)^{2-n_{i}}\left(\int^{1}_{0}k\exp(-2ckd(x,y))\,dk\right)^{\frac{1}{2}}\\\
&\lesssim d(x,y)^{1-n_{i}}.\end{split}$

It then follows from the fact that $\nabla\phi_{i}$ is compactly supported,

$\displaystyle\begin{split}\int_{\mathcal{M}}\left\|\pi^{i}_{2}(x,\cdot)\right\|_{L^{p^{\prime}}(\mathcal{M})}^{p}\,dx&\lesssim\int_{\mathrm{supp}\,\nabla\phi_{i}}\left\|d(x,\cdot)^{1-n_{i}}\right\|_{L^{p^{\prime}}(B(x,r)^{c})}^{p}\,dx\\\
&\lesssim\sup_{x\in\mathrm{supp}\,\nabla\phi_{i}}\left\|d(x,\cdot)^{1-n_{i}}\right\|_{L^{p^{\prime}}(B(x,r)^{c})}^{p}.\end{split}$

For $p=1$, this quantity is obviously finite and thus $\Pi_{2}^{i}$ is bounded
on $L^{1}$. For $p>1$,

$\left\|d(x,\cdot)^{1-n_{i}}\right\|_{L^{p^{\prime}}(B(x,r)^{c})}=\left(\int_{d(x,y)>r}d(x,y)^{(1-n_{i})p^{\prime}}\,dy\right)^{\frac{1}{p^{\prime}}}.$

This quantity will be bounded from above by a constant, uniformly in $x$,
provided that

$p^{\prime}(1-n_{i})<-n_{i}\quad\Leftrightarrow\quad
p^{\prime}>\frac{n_{i}}{n_{i}-1}=n_{i}^{\prime}\quad\Leftrightarrow\quad
p<n_{i}.$

This proves that $\Pi^{i}_{2}$ is bounded on $L^{p}$ for $1\leq p<n_{i}$. This
completes the proof of the $L^{p}$-boundedness of $\Pi^{i}$ for all
$p\in[1,n_{i})$, thereby demonstrating that the operator $S_{<}^{1}$ is
$L^{p}$-bounded for $p\in(1,n_{min})$ and weak-type $(1,1)$.

###  4.2. The Operator $S_{<}^{2}$

The operator $S^{2}_{<}$ is given by

$S^{2}_{<}f(x)=\left(\int^{1}_{0}\left|\int_{\mathcal{M}}\nabla_{x}H_{2}(k)(x,y)f(y)\,dy\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

Minkowski’s integral inequality implies that

$\displaystyle\begin{split}S^{2}_{<}f(x)&\leq\int_{\mathcal{M}}\left(\int^{1}_{0}\left|\nabla_{x}H_{2}(k)(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\left|f(y)\right|\,dy\\\
&=\int_{\mathcal{M}}h_{2}(x,y)\left|f(y)\right|\,dy,\end{split}$

where $h_{2}(x,y)$ is the kernel

$h_{2}(x,y):=\left(\int^{1}_{0}\left|\nabla_{x}H_{2}(k)(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

First observe that the kernel $h_{2}$ is compactly supported in
$\mathcal{M}^{2}$. Next, notice that the operators $\left\\{\nabla
H_{2}(k)\right\\}_{k\in(0,1)}$ constitute a family of pseudodifferential
operators of order $1-2M$. At the scale of an individual chart on
$\mathcal{M}$, it is evident that in local coordinates the symbol of $\nabla
H_{2}(k)$, denoted $a_{2}(k)(x,\xi)$, will satisfy

$\left|\partial^{\alpha}_{\xi}a_{2}(k)(x,\xi)\right|\lesssim(\left|\xi\right|^{2}+k^{2})^{\frac{1-2M-\left|\alpha\right|}{2}}$

for all multi-indices $\alpha\geq 0$ and $k\in(0,1)$. From this, standard
pseudodifferential operator theory (c.f. [29, Sec. 0.2] for instance) implies

$\left|\nabla_{x}H_{2}(k)(x,y)\right|\lesssim k^{N+1-2M-b}d(x,y)^{-b},$

for any $b>N+1-2M$. Setting $b=N-\frac{1}{2}$ then yields

$\left|\nabla_{x}H_{2}(k)(x,y)\right|\lesssim
k^{\frac{3}{2}-2M}d(x,y)^{\frac{1}{2}-N}.$

Therefore,

$\displaystyle\begin{split}h_{2}(x,y)&=\left(\int^{1}_{0}\left|\nabla_{x}H_{2}(k)(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\left(\int^{1}_{0}dk\right)^{\frac{1}{2}}d(x,y)^{\frac{1}{2}-N}\\\
&=d(x,y)^{\frac{1}{2}-N}.\end{split}$

Thus $S_{<}^{2}$ is pointwise bounded from above by an operator with kernel
that is compactly supported in $\mathcal{M}^{2}$ and controlled by
$d(x,y)^{\frac{1}{2}-N}$. It follows immediately that $S_{<}^{2}$ must be
bounded on $L^{p}$ for all $p\in[1,\infty]$.

###  4.3. The Operator $S_{<}^{3}$

In an identical manner to the operator $S_{<}^{2}$, Minkowski’s integral
inequality allows us to control $S_{<}^{3}$ from above by

$S^{3}_{<}f(x)\leq\int_{\mathcal{M}}h_{3}(x,y)\left|f(y)\right|\,dy,$

where $h_{3}(x,y)$ is the kernel defined through

$h_{3}(x,y):=\left(\int^{1}_{0}\left|\nabla_{x}H_{3}(k)(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

Hölder’s inequality implies that

$\displaystyle\begin{split}\left\|S^{3}_{<}f\right\|^{p}_{p}&\leq\int_{\mathcal{M}}\left(\int_{\mathcal{M}}h_{3}(x,y)\left|f(y)\right|\,dy\right)^{p}\,dx\\\
&\leq\int_{\mathcal{M}}\left(\int_{\mathcal{M}}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\cdot\left\|f\right\|^{p}_{p}.\end{split}$

Thus if it can be proved that

(30)
$\int_{\mathcal{M}}\left(\int_{\mathcal{M}}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx<\infty$

then the operator $S^{3}_{<}$ will be bounded on $L^{p}$. In order to prove
(30) it is sufficient to prove the below four separate conditions,

(31)
$\int_{K}\left(\int_{K}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx<\infty,$
(32) $\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left(\int_{K}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx<\infty\quad
for\ any\ 1\leq i\leq l,$ (33)
$\int_{K}\left(\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx<\infty\quad
for\ any\ 1\leq j\leq l,$

and

(34) $\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left(\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx<\infty\quad
for\ any\ 1\leq i,\,j\leq l.$

For the first estimate (31), observe that Proposition 3.4 implies
$\nabla_{x}H_{3}(k)(x,y)\lesssim k^{2-2M}$ for all $x,\,y\in\mathcal{M}$. This
estimate, when applied to the definition of $h_{3}(x,y)$, gives

(35) $h_{3}(x,y)\lesssim 1\quad\forall\ x,\,y\in\mathcal{M},$

which immediately implies the validity of (31).

Let us consider (32). According to Proposition 3.4, for
$x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ and $y\in K$,

$\left|\nabla_{x}H_{3}(k)(x,y)\right|\lesssim k^{2-2M}\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}\exp\left(-ckd(x_{i}^{\circ},x)\right).$

This implies that

(36)
$\displaystyle\begin{split}h_{3}(x,y)&=\left(\int^{1}_{0}\left|\nabla_{x}H_{3}(k)(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}\left(\int^{1}_{0}k\exp(-2ckd(x_{i}^{\circ},x))\,dk\right)^{\frac{1}{2}}\\\
&\simeq\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}\left(\frac{1}{d(x_{i}^{\circ},x)^{2}}\right)^{\frac{1}{2}}\\\
&\simeq\langle d(x_{i}^{\circ},x)\rangle^{-n_{i}}.\end{split}$

On applying this estimate to (32),

$\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left(\int_{K}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\langle d(x_{i}^{\circ},x)\rangle^{-n_{i}p}\,dx.$

This will clearly be finite provided that $p>1$.

Let us now consider (33). For $x\in K$ and
$y\in\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus K_{j}$, Proposition 3.4
tells us that

$\left|\nabla_{x}H_{3}(k)(x,y)\right|\lesssim k^{2-2M}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\exp(-ckd(x_{j}^{\circ},y)).$

This leads to the pointwise estimate

$\displaystyle\begin{split}h_{3}(x,y)&\lesssim\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\left(\int^{1}_{0}k\exp(-2ckd(x_{j}^{\circ},y))\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle d(x_{j}^{\circ},y)\rangle^{1-n_{j}}.\end{split}$

Applying this to (33),

$\int_{K}\left(\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\lesssim\int_{K}\left(\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}\langle
d(x_{j}^{\circ},y)\rangle^{(1-n_{j})p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx.$

This will be finite provided that

$(n_{j}-1)p^{\prime}>n_{j}\quad\Leftrightarrow\quad
p^{\prime}>\frac{n_{j}}{n_{j-1}}=n_{j}^{\prime}\quad\Leftrightarrow\quad
p<n_{j}.$

Finally, it remains to prove estimate (34). For
$x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ and
$y\in\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus K_{j}$, Proposition 3.4
implies that

$\left|\nabla_{x}H_{3}(k)(x,y)\right|\lesssim k^{2-2M}\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\exp(-ck(d(x_{i}^{\circ},x)+d(x_{j}^{\circ},y))).$

This leads to the pointwise bound

(37) $\displaystyle\begin{split}\left|h_{3}(x,y)\right|&\lesssim\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\left(\int^{1}_{0}k\exp(-2ck(d(x_{i}^{\circ},x)+d(x_{j}^{\circ},y)))\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle d(x_{i}^{\circ},x)\rangle^{1-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\left(\frac{1}{(d(x_{i}^{\circ},x)+d(x_{j}^{\circ},y))^{2}}\right)^{\frac{1}{2}}\\\
&\leq\min\left(\langle d(x_{i}^{\circ},x)\rangle^{-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}},\langle
d(x_{i}^{\circ},x)\rangle^{1-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{1-n_{j}}\right).\end{split}$

Applying this to (34),

$\displaystyle\begin{split}\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}&\left(\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}h_{3}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\\\
&\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left(\int_{D_{j}^{1}(x)}\langle
d(x_{i}^{\circ},x)\rangle^{(1-n_{i})p^{\prime}}\langle
d(x_{j}^{\circ},y)\rangle^{(1-n_{j})p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\\\
&\qquad+\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left(\int_{D_{j}^{2}(x)}\langle
d(x_{i}^{\circ},x)\rangle^{-n_{i}p^{\prime}}\langle
d(x_{j}^{\circ},y)\rangle^{(2-n_{j})p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\\\
&=:I_{1}+I_{2},\end{split}$

where

$\displaystyle\begin{split}&D_{j}^{1}(x):=\left\\{y\in\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}:d(x_{j}^{\circ},y)\geq d(x_{i}^{\circ},x)\right\\},\\\
&D_{j}^{2}(x):=\left\\{y\in\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}:d(x_{j}^{\circ},y)<d(x_{i}^{\circ},x)\right\\}.\end{split}$

For the first term,

$I_{1}=\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle
d(x_{i}^{\circ},x)\rangle^{(n_{i}-1)p}}\left(\int_{D_{j}^{1}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{(n_{j}-1)p^{\prime}}}\right)^{\frac{p}{p^{\prime}}}\,dx.$

The interior integral is given by

(38) $\displaystyle\begin{split}\int_{D^{1}_{j}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{(n_{j}-1)p^{\prime}}}&=\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}\mathbbm{1}_{d(x_{j}^{\circ},y)\geq d(x_{i}^{\circ},x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{(n_{j}-1)p^{\prime}}}\\\
&\lesssim\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}}\frac{dy}{(d(x_{j}^{\circ},y)+d(x_{i}^{\circ},x))^{(n_{j}-1)p^{\prime}}}\\\
&\lesssim\int_{\mathbb{R}^{n_{j}}}\frac{dy_{1}}{(|y_{1}-x_{j,1}^{\circ}|+d(x_{i}^{\circ},x))^{(n_{j}-1)p^{\prime}}},\end{split}$

where the notation $x_{j,1}^{\circ}$ denotes the Euclidean component of
$x_{j}^{\circ}$ in $\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}$. This will be
integrable when

$(n_{j}-1)p^{\prime}>n_{j}\quad\Leftrightarrow\quad
p^{\prime}>\frac{n_{j}}{n_{j}-1}=n_{j}^{\prime}\quad\Leftrightarrow\quad
p<n_{j}.$

In which case,

$\displaystyle\begin{split}\int_{D^{1}_{j}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{(n_{j}-1)p^{\prime}}}&\lesssim\int^{\infty}_{d(x_{i}^{\circ},x)}\frac{r^{n_{j}-1}}{r^{(n_{j}-1)p^{\prime}}}\,dr\\\
&\simeq\left[-r^{-(n_{j}-1)p^{\prime}+n_{j}}\right]^{\infty}_{d(x_{i}^{\circ},x)}\\\
&\lesssim\langle
d(x_{i}^{\circ},x)\rangle^{-(n_{j}-1)p^{\prime}+n_{j}},\end{split}$

where we used the fact that $p<n_{j}$ to deduce that
$-(n_{j}-1)p^{\prime}+n_{j}<0$ when performing the integration. Applying this
estimate to $I_{1}$ gives

$\displaystyle\begin{split}I_{1}&\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle d(x_{i}^{\circ},x)\rangle^{(n_{i}-1)p}}\cdot\langle
d(x_{i}^{\circ},x)\rangle^{-(n_{j}-1)p+n_{j}\frac{p}{p^{\prime}}}\,dx\\\
&=\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{dx}{\langle
d(x_{i}^{\circ},x)\rangle^{(n_{i}-1)p+(n_{j}-1)p-n_{j}(p-1)}}.\end{split}$

This will be finite provided that

$(n_{i}-1)p+(n_{j}-1)p-n_{j}(p-1)>n_{i}\quad\Leftrightarrow\quad
p>\frac{n_{i}-n_{j}}{n_{i}-2}.$

Since $\frac{n_{i}-n_{j}}{n_{i}-2}<\frac{n_{i}-2}{n_{i}-2}=1$, this will be
satisfied when $p>1$. It has therefore been proved that $I_{1}$ is finite when
$1<p<n_{j}$.

It remains to consider the term $I_{2}$,

(39) $I_{2}=\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle
d(x_{i}^{\circ},x)\rangle^{n_{i}p}}\left(\int_{D^{2}_{j}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)^{(n_{j}-2)p^{\prime}}}\right)^{\frac{p}{p^{\prime}}}\,dx.$

It will be proved that this term is finite for all $1<p<\infty$. The interior
integral is given by

$\displaystyle\begin{split}\int_{D^{2}_{j}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{(n_{j}-2)p^{\prime}}}&=\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}\mathbbm{1}_{d(x_{j}^{\circ},y)<d(x_{i}^{\circ},x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{(n_{j}-2)p^{\prime}}}\\\
&\lesssim\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}}\mathbbm{1}_{d(x_{j}^{\circ},y)<d(x_{i}^{\circ},x)}\frac{dy}{(d(x_{j}^{\circ},y)+D_{K})^{(n_{j}-2)p^{\prime}}}\\\
&\lesssim\int_{\mathbb{R}^{n_{j}}}\mathbbm{1}_{|y_{1}-x_{j,1}^{\circ}|<d(x_{i}^{\circ},x)}\frac{dy_{1}}{(|y_{1}-x_{j,1}^{\circ}|+D_{K})^{(n_{j}-2)p^{\prime}}}\\\
&\lesssim\int^{2d(x_{i}^{\circ},x)}_{D_{K}}r^{n_{j}-1-(n_{j}-2)p^{\prime}}\,dr,\end{split}$

where $D_{K}>0$ is some constant that satisfies $d(x_{k}^{\circ},z)>D_{K}$ for
all $z\in\mathcal{M}\setminus K$ and $k=1,\cdots,l$. Let’s estimate this
integral based on three distinct cases.

Case 1. Suppose $n_{j}-1-(n_{j}-2)p^{\prime}>-1$, which itself is equivalent
to the condition $p>\frac{n_{j}}{2}$. Then

$\displaystyle\begin{split}\int^{2d(x_{i}^{\circ},x)}_{D_{K}}r^{n_{j}-1-(n_{j}-2)p^{\prime}}\,dr&\simeq\left[r^{n_{j}-(n_{j}-2)p^{\prime}}\right]^{2d(x_{i}^{\circ},x)}_{D_{K}}\\\
&\leq\langle
d(x_{i}^{\circ},x)\rangle^{n_{j}-(n_{j}-2)p^{\prime}}.\end{split}$

Substituting this back into $I_{2}$,

$\displaystyle\begin{split}I_{2}&\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle d(x_{i}^{\circ},x)\rangle^{n_{i}p}}\cdot\langle
d(x_{i}^{\circ},x)\rangle^{n_{j}\frac{p}{p^{\prime}}-(n_{j}-2)p}\,dx\\\
&=\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{dx}{\langle
d(x_{i}^{\circ},x)\rangle^{n_{i}p+(n_{j}-2)p-n_{j}\frac{p}{p^{\prime}}}}.\end{split}$

This will be integrable when

$n_{i}p+(n_{j}-2)p-n_{j}(p-1)>n_{i}\quad\Leftrightarrow\quad
p>\frac{n_{i}-n_{j}}{n_{i}-2},$

which will be satisfied when $p>1$ since
$\frac{n_{i}-n_{j}}{n_{i}-2}<\frac{n_{i}-2}{n_{i}-2}=1$.

Case 2. Suppose that $n_{j}-1-(n_{j}-2)p^{\prime}<-1$, which itself is
equivalent to the condition $p<\frac{n_{j}}{2}$. We would then have

$\displaystyle\begin{split}\int^{2d(x_{i}^{\circ},x)}_{D_{K}}r^{n_{j}-1-(n_{j}-2)p^{\prime}}\,dr&\simeq\left[-r^{n_{j}-(n_{j}-2)p^{\prime}}\right]^{2d(x_{i}^{\circ},x)}_{D_{K}}\\\
&\lesssim 1.\end{split}$

Thus

$I_{2}\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle d(x_{i}^{\circ},x)\rangle^{n_{i}p}}\,dx,$

which will be finite when $p>1$.

Case 3. Suppose $n_{j}-1-(n_{j}-2)p^{\prime}=-1$, which itself is equivalent
to the condition $p=\frac{n_{i}}{2}$. We would then have

$\displaystyle\begin{split}\int^{2d(x_{i}^{\circ},x)}_{D_{K}}r^{n_{j}-1-(n_{j}-2)p^{\prime}}\,dr&=\int^{2d(x_{i}^{\circ},x)}_{D_{K}}r^{-1}\,dr\\\
&=\frac{1}{D_{K}^{\epsilon}}\int^{2d(x_{i}^{\circ},x)}_{D_{K}}\frac{D_{K}^{\epsilon}}{r}\,dr\\\
&\lesssim\int^{2d(x_{i}^{\circ},x)}_{D_{K}}r^{\epsilon-1}\,dr\\\
&\lesssim\langle d(x_{i}^{\circ},x)\rangle^{\epsilon}.\end{split}$

Therefore,

$I_{2}\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle
d(x_{i}^{\circ},x)\rangle^{n_{i}p-\epsilon\frac{p}{p^{\prime}}}}\,dx.$

This will be finite when

$n_{i}p-\epsilon(p-1)>n_{i}\quad\Leftrightarrow\quad p>1.$

This completes our proof of estimate (34) and thus shows that the operator
$S^{3}_{<}$ is bounded on $L^{p}$ for all $p\in(1,\min_{i}n_{i})$.

It remains to prove that $S^{3}_{<}$ is weak-type $(1,1)$. On combining (35),
(36) and (37), it is evident that

$\sup_{y\in\mathcal{M}}\left|h_{3}(x,y)\right|\lesssim\left\\{\begin{array}[]{c
c}\langle d(x_{i}^{\circ},x)\rangle^{-n_{i}}&for\
x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i},\ 1\leq i\leq l,\\\
1&for\ x\in K.\end{array}\right.$

This implies that for $f\in C^{\infty}_{c}$,

$\displaystyle\begin{split}\left|S_{<}^{3}f(x)\right|&\leq\int_{\mathcal{M}}h_{3}(x,y)\left|f(y)\right|\,dy\\\
&\leq\sup_{y\in\mathcal{M}}\left|h_{3}(x,y)\right|\int_{\mathcal{M}}\left|f(y)\right|\,dy\\\
&\lesssim\left\\{\begin{array}[]{c c}\langle
d(x_{i}^{\circ},x)\rangle^{-n_{i}}\left\|f\right\|_{L^{1}}&for\
x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i},\ 1\leq i\leq l,\\\
\left\|f\right\|_{L^{1}}&for\ x\in K.\end{array}\right.\end{split}$

This function is clearly in $L^{1,\infty}$, and thus $S_{<}^{3}$ is weak-type
$(1,1)$.

###  4.4. The Operator $S_{<}^{4}$

The previous proof of the $L^{p}$-boundedness and weak-type $(1,1)$ property
of the operator $S^{3}_{<}$ was derived entirely from the pointwise estimates
for the kernel of $\nabla H_{3}(k)$ provided by Proposition 3.4. Since the
kernels of $\nabla H_{4}(k)$ satisfy even stronger pointwise estimates by
Proposition 3.5, an identical argument can be applied to obtain the
$L^{p}$-boundedness and weak-type $(1,1)$ property of the operator
$S^{4}_{<}$. Moreover, due to the increased strength of the pointwise
estimates for $\nabla H_{4}(k)$, one will obtain $L^{p}$-boundedness for all
$p$ in the full range $(1,\infty)$ instead of the restricted range
$(1,n_{min})$. This improved range of boundedness will prove to be an
essential part of our proof of the unboundedness of $S$ for $p\geq n_{min}$ in
Section 7. As such, the reader is advised keep it in mind.

##  5\. The High Energy Square Function

Recall that the high energy part of the square function is defined by

$S_{>}f(x):=\left(\int^{\infty}_{1}\left|\nabla(k^{2}+\Delta)^{-M}f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

The main aim of this section is to verify that the operator $S_{>}$ is bounded
on all $L^{p}$ spaces for $1<p<\infty$ and is weak type $(1,1)$. The argument
we describe below has the same structure as in [19, Section 5] which we adapt
to the square function setting.

######  Proposition 5.1.

The high energy square function operator $S_{>}$ is bounded on
$L^{p}(\mathcal{M})$ for any $p\in(1,\infty)$. That is, for every $1<p<\infty$
there exists $c>0$ such that

$\left\|S_{>}f\right\|_{p}\leq c\left\|f\right\|_{p}$

for all $f\in L^{p}(\mathcal{M})$. In addition the operator $S_{>}$ satisfies
weak type $(1,1)$ estimates.

###### Proof.

Denote by $F_{M}$ the Fourier transform of the function
$\lambda\to(\lambda+1)^{-M}$. Note that

$\int_{-\infty}^{\infty}e^{-i\xi\lambda}(\lambda^{2}+k^{2})^{-M}d\lambda=k^{1-2M}F_{M}(k|\xi|).$

In fact $F_{1}(|\xi|)=ce^{-|\xi|}$ and
$k^{1-2(M+1)}F_{M+1}(k|\xi|)=\partial_{k^{2}}k^{1-2M}F_{M}(k|\xi|)$. Note also
that there exist positive constants $C,c>0$ (depending on $M$) such that

$F_{M}(|\xi|)\leq Ce^{-c|\xi|}.$

Now, let $\eta\in C_{c}^{\infty}(\mathbb{R})$ be an even compactly supported
function such that $0\leq\eta(\xi)\leq 1$, $\eta(\xi)=1$ for all $|\xi|\leq
1/2$ and $\eta(\xi)=0$ for $|\xi|\geq 1$. For any $r>0$ we define the function
${G}^{M}_{r,k}$ as the inverse Fourier transform of

$k^{1-2M}F_{M}(k|\xi|)\eta(\xi/r).$

Then we set

(40) $H^{M}_{r,k}(\lambda)=(\lambda^{2}+k^{2})^{-M}-{G}^{M}_{r,k}(\lambda)$

so that the Fourier transform of $H^{M}_{r,k}$ is equal to

$k^{1-2M}F_{M}(k|\xi|)\Big{(}1-\eta(\xi/r)\Big{)}.$

Hence for some positive constants $C,c>0$

(41) $\sup_{\lambda}|H^{M}_{r,k}(\lambda)|\leq Ce^{-ckr}$

Proceeding as in [19, (42) p. 1089], we have (using the evenness of
$\widehat{G}^{M}_{r,k}(\xi)$)

(42) $\displaystyle
G^{M}_{r,k}(\sqrt{\Delta})=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\sqrt{\Delta}}\widehat{G^{M}_{r,k}}(t)\,dt=\frac{1}{\pi}\int_{0}^{r}\cos(t\sqrt{\Delta})\widehat{G^{M}_{r,k}}(t)\,dt.$

By (42) and by the finite speed of propagation of $\cos(t\sqrt{\Delta})$ it
follows that $G^{M}_{r,k}(\sqrt{\Delta})$ has Schwartz kernel supported in the
set $\\{(z,z^{\prime})\mid d(z,z^{\prime})\leq r\\}$. Next, set
$r=\tilde{r}>0$ to be half the injectivity radius of $\mathcal{M}$. Note that
the injectivity radius of $\mathcal{M}$ is strictly positive and, without loss
of generality, it can be assumed that $\tilde{r}=1$.

The operator $\nabla G^{M}_{\tilde{r},k}(\sqrt{\Delta})=\nabla
G^{M}_{1,k}(\sqrt{\Delta})$ is a pseudodifferential operator of order $1-2M$
since the function $G^{M}_{1,k}(\lambda)$ is a symbol of order $-2M$ see [29,
Chapter XII, Section 1]. This implies that for any $b\geq N+1-2M$, see [30,
Poposition 2.2 p. 6]

$\displaystyle|\nabla G^{M}_{1,k}(\sqrt{\Delta})(x,y)|\leq
Ck^{N-2M-b+1}d(x,y)^{-b}.$

Hence, taking $b_{1}=N-\frac{1}{2}$ and $b_{2}=N+\frac{1}{2}$,

$\displaystyle\int^{\infty}_{1}|\nabla
G^{M}_{1,k}(\sqrt{\Delta})(x,y)|^{2}k^{4M-3}\leq\int_{1}^{d(x,y)^{-1}}Ck^{2N-4M-2b_{1}+2}d(x,y)^{-2b_{1}}k^{4M-3}dk$
$\displaystyle+\int_{d(x,y)^{-1}}^{\infty}Ck^{2N-4M-2b_{2}+2}d(x,y)^{-2b_{2}}k^{4M-3}\leq
Cd(x,y)^{-2N}.$

Note also that if $X$ is $C^{\infty}$ vector field on $\mathcal{M}$, uniformly
bounded in every $C^{m}$ norm then

$\int^{\infty}_{1}|X\nabla
G^{M}_{1,k}(\sqrt{\Delta})(x,y)|^{2}k^{4M-3}\,dk\leq Cd(x,y)^{-2(N+1)}.$

Hence we can use vector-valued standard Calderon-Zygmund approach (with
respect to $L^{2}([1,\infty),k^{4M-3}dk)$ to conclude the argument for $\nabla
G^{M}_{1,k}(\sqrt{\Delta})$ and $k\geq 1$, see for example [27, §6.4, p. 28 ].

Next we consider the family of operators $\nabla H^{M}_{r,k}(\sqrt{\Delta})$.
By Minkowski’s integral inequality

$\displaystyle\bigg{(}\int_{1}^{\infty}\bigg{|}\int_{\mathcal{M}}K_{\nabla
H^{M}_{1,k}(\sqrt{\Delta})}(x,y)f(y)dy\bigg{|}^{2}k^{4M-3}dk\bigg{)}^{\frac{1}{2}}$
$\displaystyle\hskip
144.54pt\leq\int_{\mathcal{M}}\int_{1}^{\infty}\bigg{(}\bigg{|}\ K_{\nabla
H^{M}_{1,k}(\sqrt{\Delta})}(x,y)\bigg{|}^{2}k^{4M-3}dk\bigg{)}^{\frac{1}{2}}|f(y)|dy$
$\displaystyle\hskip 144.54pt=\int_{\mathcal{M}}h_{*}(x,y)|f(y)|dy,$

where

$h_{*}(x,y)=\bigg{(}\int_{1}^{\infty}\bigg{|}\ K_{\nabla
H^{M}_{1,k}(\sqrt{\Delta})}(x,y)\bigg{|}^{2}k^{4M-3}dk\bigg{)}^{\frac{1}{2}}$

By Schur’s integral test, to conclude the proof it is now enough to show that
there exists a constant $C$ such that

(43) $\displaystyle\sup_{y}\int_{\mathcal{M}}h_{*}(x,y)dx\leq C$

and

(44) $\displaystyle\sup_{x}\int_{\mathcal{M}}h_{*}(x,y)dy\leq C.$

Here we again use the approach described in [19]. First we note that finite
propagation speed implies that

$K_{\nabla H^{M}_{{r_{1}},k}(\sqrt{\Delta})}(x,y)=K_{\nabla
H^{M}_{{r_{2}},k}(\sqrt{\Delta})}(x,y)\text{ if }x\notin B(y,r_{2}),\quad
r_{2}\geq{r_{1}}>0.$

It follows that for every $r>1$

(45) $\displaystyle\int_{x\notin B(y,r)}|K_{\nabla
H^{M}_{{1},k}(\sqrt{\Delta})}(x,y)|^{2}dx$ $\displaystyle=\int_{x\notin
B(y,r)}|K_{\nabla H^{M}_{{r},k}(\sqrt{\Delta})}(x,y)|^{2}dx$
$\displaystyle\leq\int_{x\in M}|K_{\nabla
H^{M}_{{r},k}(\sqrt{\Delta})}(x,y)|^{2}dx$
$\displaystyle=\Big{\langle}\nabla_{x}K_{H^{M}_{{r},k}(\sqrt{\Delta})}(\cdot,y),\nabla_{x}K_{H^{M}_{{r},k}(\sqrt{\Delta})}(\cdot,y)\Big{\rangle}$
$\displaystyle=\Big{\langle}\Delta
K_{H^{M}_{{r},k}(\sqrt{\Delta})}(\cdot,y),K_{H^{M}_{{r},k}(\sqrt{\Delta})}(\cdot,y)\Big{\rangle}$
$\displaystyle=\int_{x\in
M}|K_{\sqrt{\Delta}H^{M}_{r,k}(\sqrt{\Delta})}(x,y)|^{2}dx.$

A straightforward argument, see [19, Proposition 2.4], shows that
$\|(I+\Delta)^{-n}\|^{2}_{2\to\infty}<\infty$ for any $n\geq[N/4]+1$. Hence by
(41) for some $c>0$

$\displaystyle\sup_{y}\int_{x\in
M}|K_{\sqrt{\Delta}H^{M}_{r,k}(\sqrt{\Delta})}(x,y)|^{2}dx$ $\displaystyle=$
$\displaystyle\big{\|}\sqrt{\Delta}H^{M}_{{r},k}(\sqrt{\Delta})\big{\|}^{2}_{1\to
2}=\big{\|}\sqrt{\Delta}H^{M}_{{r},k}(\sqrt{\Delta})\big{\|}^{2}_{2\to\infty}$
$\displaystyle\leq$
$\displaystyle\big{\|}(I+\Delta)^{n}\sqrt{\Delta}H^{M}_{{r},k}(\sqrt{\Delta})\big{\|}^{2}_{2\to
2}\big{\|}(I+\Delta)^{-n}\big{\|}^{2}_{2\to\infty}$ $\displaystyle\leq$
$\displaystyle C\sup_{\lambda\geq
0}\big{|}(1+\lambda^{2})^{n}{\lambda}H^{M}_{{r},k}(\lambda)\big{|}^{2}$
$\displaystyle\leq$ $\displaystyle Ce^{-ckr}.$

Now we can conclude that

(46) $\sup_{y}\int_{x\notin B(y,r)}\int_{1}^{\infty}\bigg{|}\ K_{\nabla
H^{M}_{1,k}(\sqrt{\Delta})}(x,y)\bigg{|}^{2}k^{4M-3}dkdx\lesssim e^{-cr}$

This proves (43) since it implies in particular that

$\sup_{y}\int_{r\leq d(x,y)\leq 2r}h_{*}(x,y)^{2}dx\lesssim e^{-cr}.$

The measure of the set $\\{x\in M\mid r\leq d(x,y)\leq 2r\\}$ is bounded by
$Cr^{N}$, $r\geq\tilde{r}=1$, where $N$ is the dimension of $\mathcal{M}$,
uniformly in $y\in\mathcal{M}$. So we can apply Hölder’s inequality to find
that

$\sup_{y}\int_{r\leq d(x,y)\leq 2r}h_{*}(x,y)dx\lesssim e^{-cr}.$

Then the above estimates can be summed over a sequence of dyadic annuli to
obtain (43).

To prove (44) we use the notion of the Hodge Laplacian. Recall that for any
bounded Borel function $F$ if we write $\Delta_{q}$ for the Hodge Laplacian
acting on $q$-forms then we can write
$dF(\sqrt{\Delta_{0}})=F(\sqrt{\Delta_{1}})d$. This means that to verify (44)
it is enough to prove (43) for $H^{M}_{{r},k}(\sqrt{\Delta_{1}})$. This can be
achieved by essentially repeating the above argument for $\Delta_{0}=\Delta$.
See [19] for more detailed calculations. Now the continuity of the off-
diagonal part of high energy square function follows from (43), (44) and
Minkowski integral inequality. ∎

##  6\. The Reverse Inequality

In this section, the reverse inequality portion of Theorem 1.1 will be proved.
That is, it will be shown that
$\left\|f\right\|_{p}\lesssim\left\|Sf\right\|_{p}$ for any
$p\in(n_{min}^{\prime},n_{min})$. This will be achieved using, by now,
standard arguments that can be found in [2, Thm. 7.1]. By the functional
calculus of $\Delta$ on $L^{2}$, for $f\in L^{p}\cap L^{2}$ we have the
resolution of the identity,

$f=c_{M}\int^{\infty}_{0}t^{2}\Delta(1+t^{2}\Delta)^{-2M}f\,\frac{dt}{t}$

for some constant $c_{M}>0$ that is dependent on $M$. Then for any $g\in
L^{p^{\prime}}$, Fubini-Tonelli’s theorem, whose application will be justified
retrospectively, implies

$\displaystyle\begin{split}\int_{\mathcal{M}}f\cdot\overline{g}&\simeq\int_{\mathcal{M}}\left(\int^{\infty}_{0}t^{2}\Delta(1+t^{2}\Delta)^{-2M}f(x)\,\frac{dt}{t}\right)\overline{g(x)}\,dx\\\
&=\int^{\infty}_{0}\int_{\mathcal{M}}t^{2}\Delta(1+t^{2}\Delta)^{-2M}f(x)\cdot\overline{g(x)}\,dx\,\frac{dt}{t}\\\
&=\int^{\infty}_{0}\int_{\mathcal{M}}t\nabla(1+t^{2}\Delta)^{-M}f(x)\overline{t\nabla(1+t^{2}\Delta)^{-M}g(x)}\,dx\,\frac{dt}{t}.\end{split}$

On applying Fubini-Tonelli once more, followed by Cauchy-Schwarz and Hölder’s
inequality,

$\displaystyle\begin{split}\int_{\mathcal{M}}f\cdot\overline{g}&=\int_{\mathcal{M}}\int^{\infty}_{0}t\nabla(1+t^{2}\Delta)^{-M}f(x)\overline{t\nabla(1+t^{2}\Delta)^{-M}g(x)}\,\frac{dt}{t}\,dx\\\
&\leq\int_{\mathcal{M}}Sf(x)\cdot Sg(x)\,dx\\\
&\leq\left\|Sf\right\|_{p}\left\|Sg\right\|_{p^{\prime}}.\end{split}$

Since $S$ is bounded on both $L^{p}$ and $L^{p^{\prime}}$ for
$p\in(n_{min}^{\prime},n_{min})$ it follows that this quantity is finite,
thereby retrospectively justifying our two previous applications of the
Fubini-Tonelli theorem. Moreover, the $L^{p^{\prime}}$-boundedness of $S$ in
particular implies

$\int_{\mathcal{M}}f\cdot\overline{g}\lesssim\left\|Sf\right\|_{p}\left\|g\right\|_{p^{\prime}},$

and thus

$\left\|f\right\|_{p}\lesssim\left\|Sf\right\|_{p}$

for all $f\in L^{p}\cap L^{2}$. As $S$ is bounded on $L^{p}$, this estimate
must also hold true for all $f\in L^{p}$ by density.

##  7\. Unboundedness for $p\geq n_{min}$

Throughout this section we impose the restriction $2M<n_{min}$ and provide a
proof of the unboundedness portion of Theorem 1.1. The upper restriction on
the order of the resolvent will allow us to utilize Proposition 3.2, which
will form a key component of the proof. Our proof will exploit some of the
ideas utilized in the unboundedness argument of [19, Sec. 6]. However, a
number of difficulties will arise due to the quadratic nature of the square
function and the higher-order degree of the resolvent.

In order to prove unboundedness of the square function operator on $L^{p}$ for
$p\geq n_{min}$ it is clearly sufficient to prove unboundedness of the low
energy square function. Recall from our proof of the boundedness portion of
Theorem 1.1 that the components $S^{2}_{<}$ and $S^{4}_{<}$ are bounded on
$L^{p}(\mathcal{M})$ for all $p\in(1,\infty)$. The unboundedness of the
operator $S_{<}$ will thus follow directly from the unboundedness of the
operator

$\displaystyle\begin{split}S_{<}^{1+3}f(x)&:=\left(\int^{1}_{0}\left|\nabla(H_{1}(k)+H_{3}(k))f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&=\left(\int^{1}_{0}\left[\int_{\mathcal{M}}(\nabla H_{1}(k)(x,y)+\nabla
H_{3}(k)(x,y))f(y)\,dy\right]^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.\end{split}$

Recall from the proof of the boundedness portion of Theorem 1.1 that the
operator $S^{1}_{<}$ divides into two parts, $\Lambda^{i}$ and $\Pi^{i}$,
depending on whether the gradient hits the resolvent factor or the function
$\phi_{i}$. The term where the gradient hits the resolvent $\Lambda_{i}$ was
proved to be bounded on $L^{p}$ for all $p\in(1,\infty)$. It is therefore
sufficient to prove unboundedness of the operator

(47)
$\displaystyle\begin{split}&\Bigg{(}\int^{1}_{0}\Bigg{[}\sum_{i=1}^{l}\int_{\mathcal{M}}\nabla\phi_{i}(x)(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(y)f(y)+\xi_{i}(k)(x,y)f(y)\\\
&\qquad\qquad\qquad+\nabla
u_{i}(x,k)(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\phi_{i}(y)f(y)\,dy\Bigg{]}^{2}k^{4M-3}\,dk\Bigg{)}^{\frac{1}{2}},\end{split}$

where $\xi_{i}(k)(x,y)$ is defined through

$\xi_{i}(k)(x,y):=\sum_{j=0}^{M-2}\frac{(-1)^{M+j-1}}{(M-j-1)!}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-(j+1)}(x_{i}^{\circ},y)\nabla
u_{i}^{(M-1-j)}(x,k)\phi_{i}(y).$

Let $\tau$ be a nonnegative function, not identically zero, that is compactly
supported on one of the ends. It is clear that the unboundedness of the
operator (LABEL:eqtn:Xi) will be implied by the unboundedness of the operator

(48)
$\displaystyle\begin{split}&\tau(x)\Bigg{(}\int^{1}_{0}\Bigg{[}\sum_{i=1}^{l}\int_{\mathcal{M}}\nabla\phi_{i}(x)(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(y)f(y)+\xi_{i}(k)(x,y)f(y)\\\
&\qquad\qquad\qquad+\nabla
u_{i}(x,k)(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\phi_{i}(y)f(y)\,dy\Bigg{]}^{2}k^{4M-3}\,dk\Bigg{)}^{\frac{1}{2}}.\end{split}$

Let $\xi_{i}(k)$ be the operator corresponding to the kernel $\xi_{i}(k)(x,y)$
and define

$\Xi
f(x):=\tau(x)\left(\int^{1}_{0}\left[\sum_{i=1}^{l}\xi_{i}(k)f(x)\right]^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

If it can be proved that $\Xi$ is bounded on $L^{p}$ for any $p\in(2,\infty)$
then the task of proving the unboundedness of the operator (LABEL:eqtn:Xi1) on
$L^{p}$ for $p\geq n_{min}$ will be reduced to proving the unboundedness of
the operator

(49)
$\displaystyle\begin{split}&\tau(x)\Bigg{(}\int^{1}_{0}\Bigg{[}\sum_{i=1}^{l}\int_{\mathcal{M}}\nabla\phi_{i}(x)(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(y)f(y)\\\
&\qquad\qquad\qquad+\nabla
u_{i}(x,k)(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\phi_{i}(y)f(y)\,dy\Bigg{]}^{2}k^{4M-3}\,dk\Bigg{)}^{\frac{1}{2}}.\end{split}$

To this end, the triangle inequality tells us that the $L^{p}$-boundedness of
$\Xi$ will follow from the $L^{p}$-boundedness of the operators

$\Xi^{j}_{i}f(x):=\left(\int^{1}_{0}\left|\xi^{j}_{i}(k)f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}},$

for each $1\leq i\leq l$ and $0\leq j\leq M-2$, where

$\xi^{j}_{i}(k)(x,y):=\tau(x)\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x_{i}^{\circ},y)\nabla
u_{i}^{(M-1-j)}(x,k)\phi_{i}(y).$

This operator can then itself be controlled from above by

$\displaystyle\begin{split}\Xi^{j}_{i}f(x)&\leq\left(\int^{1}_{0}\left|\xi^{j,1}_{i}(k)f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}+\left(\int^{1}_{0}\left|\xi^{j,2}_{i}(k)f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&=:\Xi^{j,1}_{i}f(x)+\Xi^{j,2}_{i}f(x),\end{split}$

where

$\xi^{j,1}_{i}(k)(x,y):=\tau(x)(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-(j+1)}(x_{i}^{\circ},y)\left|\nabla
u_{i}^{(M-1-j)}(x,k)-\nabla u_{i}^{(M-1-j)}(x,0)\right|\phi_{i}(y)$

and

$\xi^{j,2}_{i}(k)(x,y):=\tau(x)\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x_{i}^{\circ},y)\left|\nabla
u_{i}^{(M-1-j)}(x,0)\right|\phi_{i}(y).$

######  Lemma 7.1.

For any $0\leq j\leq M-1$, $1\leq i\leq l$ and $p\in(2,\infty)$, the operator
$\Xi_{i}^{j,1}$ is bounded on $L^{p}$.

###### Proof.

Minkowski’s integral inequality, that can be applied since $\frac{p}{2}>1$,
implies that

(50)
$\displaystyle\begin{split}\left\|\Xi^{j,1}_{i}f\right\|_{p}&=\left\|\int^{1}_{0}\left|\xi^{j,1}_{i}(k)f\right|^{2}k^{4M-3}\,dk\right\|^{\frac{1}{2}}_{\frac{p}{2}}\\\
&\leq\left(\int^{1}_{0}\left\|\left|\xi_{i}^{j,1}(k)f\right|^{2}\right\|_{\frac{p}{2}}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&=\left(\int^{1}_{0}\left\|\xi_{i}^{j,1}(k)\right\|^{2}_{p}k^{4M-3}\,dk\right)^{\frac{1}{2}}\left\|f\right\|_{p}.\end{split}$

For fixed $k$, $\xi^{j,1}_{i}(k)$ is a rank one operator given by
$\xi^{j,1}_{i}(k)=a_{j}\cdot\langle b_{j},\cdot\rangle$ with

$a_{j}(x)=\tau(x)\left|\nabla u_{i}^{(M-1-j)}(x,k)-\nabla
u_{i}^{(M-1-j)}(x,0)\right|$

and

$b_{j}(y):=\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-(j+1)}(x_{i}^{\circ},y)\phi_{i}(y).$

From Proposition 3.2, it is clear that

$\left\|a_{j}\right\|_{p}\lesssim k^{3+2j-2M}.$

Let $D>0$ be such that $d(x_{i}^{\circ},y)\geq D$ for all $y\in\mathrm{supp}\
\phi_{i}$. Then, in a similar manner to (52) of [19],

(51)
$\displaystyle\begin{split}\left\|b_{j}\right\|_{p^{\prime}}&=\left(\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-(j+1)}(x_{i}^{\circ},y)\phi_{i}(y)\right|^{p^{\prime}}\,dy\right)^{\frac{1}{p^{\prime}}}\\\
&\leq k^{-2j}\left(\int_{d(x_{i}^{\circ},y)\geq
D}\left|d(x_{i}^{\circ},y)^{2-n_{i}}\exp(-ckd(x_{i}^{\circ},y))\right|^{p^{\prime}}\,dy\right)^{\frac{1}{p^{\prime}}}\\\
&\lesssim\left\\{\begin{array}[]{c
c}k^{\frac{n_{i}}{p}-2-2j}&p>\frac{n_{i}}{2}\\\
k^{-2j}\log\,k&p=\frac{n_{i}}{2}\\\
k^{-2j}&p<\frac{n_{i}}{2}.\end{array}\right.\end{split}$

We therefore have

$\displaystyle\begin{split}\left\|\xi^{j,1}_{i}(k)\right\|_{p}&=\left\|a_{j}\right\|_{p}\cdot\left\|b_{j}\right\|_{p^{\prime}}\\\
&\lesssim\left\\{\begin{array}[]{c
c}k^{\frac{n_{i}}{p}+1-2M}&p>\frac{n_{i}}{2}\\\
k^{3-2M}\log\,k&p=\frac{n_{i}}{2}\\\
k^{3-2M}&p<\frac{n_{i}}{2}.\end{array}\right.\end{split}$

On applying this estimate to (50) we find that
$\left\|\Xi_{i}^{j,1}\right\|_{p}<\infty$ for any $p\in(2,\infty)$. ∎

######  Lemma 7.2.

For any $0\leq j\leq M-2$, $1\leq i\leq l$ and $p\in(2,\infty)$, the operator
$\Xi_{i}^{j,2}$ is bounded on $L^{p}$.

###### Proof.

As was the case for $\xi^{j,1}_{i}(k)$, for fixed $k$ the operator
$\xi^{j,2}_{i}(k)$ is of rank one. In particular, we have
$\xi^{j,2}_{i}(k)=a^{\prime}_{j}\cdot\langle b_{j},\cdot\rangle$ where

$a_{j}^{\prime}(x):=\tau(x)\left|\nabla u_{i}^{(M-1-j)}(x,0)\right|$

and $b_{j}$ is as defined previously for the operator $\xi^{j,1}_{i}(k)$. It
is obvious from Proposition 3.2 that
$\left\|a_{j}^{\prime}\right\|_{p}\lesssim 1$ uniformly in $k$. Estimate (51)
then implies

$\displaystyle\begin{split}\left\|\xi^{j,2}_{i}(k)\right\|_{p}&\simeq\left\|a_{j}^{\prime}\right\|_{p}\cdot\left\|b_{j}\right\|_{p^{\prime}}\\\
&\lesssim\left\|b_{j}\right\|_{p^{\prime}}\\\
&\lesssim\left\\{\begin{array}[]{c
c}k^{\frac{n_{i}}{p}-2-2j}&p>\frac{n_{i}}{2}\\\
k^{-2j}\log\,k&p=\frac{n_{i}}{2}\\\
k^{-2j}&p<\frac{n_{i}}{2}.\end{array}\right.\end{split}$

For $p>\frac{n_{i}}{2}$, once again by Minkowski’s inequality,

$\displaystyle\begin{split}\left\|\Xi^{j,2}_{i}\right\|_{p}&\leq\left(\int^{1}_{0}\left\|\xi^{j,2}_{i}(k)\right\|_{p}^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&\leq\left(\int^{1}_{0}k^{\frac{2n_{i}}{p}+4(M-j)-7}\,dk\right)^{\frac{1}{2}}.\end{split}$

This will be finite provided that

$\frac{2n_{i}}{p}+4(M-j)-7>-1,$

which is implied by the restriction $j\leq M-2$. The cases $p=\frac{n_{i}}{2}$
and $p<\frac{n_{i}}{2}$ proceed similarly thereby producing
$\left\|\Xi^{j,2}_{i}\right\|_{p}<\infty$. ∎

Now that the $L^{p}$-boundedness of $\Xi$ has been proved for
$p\in(2,\infty)$, it remains to consider the unboundedness of the operator
given in (LABEL:eqtn:XiRemoved). Notice that

$\displaystyle\begin{split}&\left(\int^{1}_{0}\left[\int_{\mathcal{M}}\nabla\phi_{i}(x)((\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)-(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y))\phi_{i}(y)f(y)\,dy\right]^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&\leq\int_{\mathcal{M}}\nabla\phi_{i}(x)\left[\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)-(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\right|^{2}k^{4M-3}\,dk\right]^{\frac{1}{2}}\phi_{i}(y)f(y)\,dy\\\
&=:\int_{\mathcal{M}}p(x,y)f(y)\,dy=:Pf(x).\end{split}$

######  Lemma 7.3.

The operator $P$ is bounded on $L^{p}$ for all $p\in(1,\infty)$.

###### Proof.

Let $V\subset\mathrm{supp}\,\phi_{i}$ be an open subset, compactly supported,
that contains $\mathrm{supp}\,\nabla\phi_{i}$. For $y\in V$, Proposition 2.2
implies

$\displaystyle\begin{split}p(x,y)&\lesssim\nabla\phi_{i}(x)\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\hskip
72.26999pt+\nabla\phi_{i}(x)\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\lesssim\nabla\phi_{i}(x)d(x,y)^{2-N}\left(\int^{1}_{0}k\exp(-2ckd(x,y))\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\hskip
72.26999pt+\nabla\phi_{i}(x)d(x_{i}^{\circ},y)^{2-n_{i}}\left(\int^{1}_{0}k\exp(-2ckd(x_{i}^{\circ},y))\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\lesssim\nabla\phi_{i}(x)\left(d(x,y)^{1-N}+d(x_{i}^{\circ},y)^{1-n_{i}}\right)\phi_{i}(y).\end{split}$

Schur’s test then tells us that the operator with kernel
$p(x,y)\mathbbm{1}_{V}(y)$ is bounded on $L^{p}$ for all $p\in(1,\infty)$.

On the other hand, for $y\in\mathrm{supp}\,\phi_{i}\setminus V$, Proposition
2.2 and the mean value theorem give

$\displaystyle\begin{split}&\nabla\phi_{i}(x)\left(\int^{1}_{0}\left|(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)-(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\qquad\qquad\lesssim\nabla\phi_{i}(x)\left(\int^{1}_{0}d(x_{i}^{\circ},x)^{2}\left[k^{4-4M}\langle
d(x_{i}^{\circ},y)\rangle^{-2(n_{i}-1)}\exp(-2ckd(x_{i}^{\circ},y))\right]k^{4M-3}\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\qquad\qquad\lesssim\nabla\phi_{i}(x)\langle
d(x_{i}^{\circ},y)\rangle^{-(n_{i}-1)}\left(\int^{1}_{0}k\exp(-2ckd(x_{i}^{\circ},y))\,dk\right)^{\frac{1}{2}}\phi_{i}(y)\\\
&\qquad\qquad\lesssim\nabla\phi_{i}(x)\langle
d(x_{i}^{\circ},y)\rangle^{-n_{i}}\phi_{i}(y).\end{split}$

The kernel $\mathbbm{1}_{\mathrm{supp}\,\phi_{i}\setminus V}(y)p(x,y)$ is
therefore compactly supported in $x$ and it decays to order $n_{i}$ in $y$.
From the argument used to prove the boundedness of $S_{<}^{3}$, it is clear
that the corresponding operator will be bounded on $L^{p}$ for all
$p\in(1,\infty)$. This demonstrates that $P$ is bounded on $L^{p}$ for all
$p\in(1,\infty)$. ∎

From the above lemma, it is clear that the unboundedness of the operator in
(LABEL:eqtn:XiRemoved) will follow from the unboundedness of the operator

$\tau(x)\left(\int^{1}_{0}\left[\sum_{i=1}^{l}\int_{\mathcal{M}}(\nabla\phi_{i}(x)+\nabla
u_{i}(x,k))(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\phi_{i}(y)f(y)\,dy\right]^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

Choose $i$ such that $n_{i}$ is minimal. Then, in an identical manner to [19],
the unboundedness of the above operator can be further reduced to proving the
unboundedness of the operator

$\tau(x)\cdot\left(\int^{1}_{0}\left[\int_{\mathcal{M}}(\nabla\phi_{i}(x)+\nabla
u_{i}(x,k))(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\phi_{i}(y)f(y)\,dy\right]^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}.$

From an application of Lemma 7.1 for the case $j=M-1$, it is evident that in
order to deduce that the above operator is unbounded on $L^{p}$ for $p\geq
n_{min}$ it is sufficient to show that

$Rf(x):=\left(\int^{1}_{0}\left|r(k)f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}$

is unbounded on $L^{p}$, where $r(k)$ is the operator with kernel defined
through

$r(k)(x,y):=\tau(x)(\nabla\phi_{i}(x)+\nabla
u_{i}(x,0))(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\phi_{i}(y).$

This final claim will be proved in the below lemma.

######  Lemma 7.4.

The operator $R$ is unbounded on $L^{p}$ for any $p\geq\min_{i}n_{i}$.

###### Proof.

From the definition of the kernel $r(k)$, for $f\in L^{p}$ non-negative and
$x\in\mathcal{M}$ we have

$\displaystyle\begin{split}Rf(x)&=\left(\int^{1}_{0}\left|r(k)f(x)\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&=a(x)\left(\int^{1}_{0}\left|\int_{\mathcal{M}}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x_{i}^{\circ},y)\phi_{i}(y)f(y)\,dy\right|^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}},\end{split}$

where $a(x):=\tau(x)\left|\nabla\phi_{i}(x)+\nabla u_{i}(x,0)\right|$. From
Corollary 2.3,

$\displaystyle\begin{split}Rf(x)&\gtrsim
a(x)\left(\int^{1}_{0}\left(\int_{\mathcal{M}}d(x_{i}^{\circ},y)^{2M-n_{i}}\exp(-ckd(x_{i}^{\circ},y))\phi_{i}(y)f(y)\,dy\right)^{2}k^{4M-3}\,dk\right)^{\frac{1}{2}}\\\
&=a(x)\left(\int^{1}_{0}\left(\int_{\mathcal{M}}d(x_{i}^{\circ},y)^{2M-n_{i}}\exp(-ckd(x_{i}^{\circ},y))\phi_{i}(y)f(y)\,dy\right)\right.\\\
&\qquad\qquad\qquad\left.\cdot\left(\int_{\mathcal{M}}d(x_{i}^{\circ},z)^{2M-n_{i}}\exp(-ckd(x_{i}^{\circ},z))\phi_{i}(z)f(z)\,dz\right)k^{4M-3}\,dk\right)^{\frac{1}{2}}.\end{split}$

On applying Tonelli’s theorem,

(52) $Rf(x)\gtrsim
a(x)\left(\int_{\mathcal{M}}\int_{\mathcal{M}}d(x_{i}^{\circ},y)^{2M-n_{i}}d(x_{i}^{\circ},z)^{2M-n_{i}}\eta(y,z)\phi_{i}(y)\phi_{i}(z)f(y)f(z)\,dy\,dz\right)^{\frac{1}{2}},$

with

$\eta(y,z):=\int^{1}_{0}k^{4M-3}\exp(-ck(d(x_{i}^{\circ},y)+d(x_{i}^{\circ},z)))\,dk.$

For $y,\,z\in\mathrm{supp}\,\phi_{i}$ the quantity
$d:=(d(x_{i}^{\circ},y)+d(x_{i}^{\circ},z))$ is bounded from below. Therefore,
after a change of variables it can be said that

$\displaystyle\begin{split}\eta(y,z)&=\frac{1}{d^{4M-2}}\int^{d}_{0}k^{4M-3}\exp\left(-ck\right)\,dk\\\
&\geq\frac{1}{d^{4M-2}}\int^{D}_{0}k^{4M-3}\exp\left(-ck\right)\,dk\\\
&\gtrsim\frac{1}{d^{4M-2}},\end{split}$

where $D>0$ is some constant independent of $y$ and $z$. Applying this to
(52),

$Rf(x)\gtrsim
a(x)\left(\int_{\mathcal{M}}\int_{\mathcal{M}}\frac{d(x_{i}^{\circ},y)^{2M-n_{i}}d(x_{i}^{\circ},z)^{2M-n_{i}}}{(d(x_{i}^{\circ},y)+d(x_{i}^{\circ},z))^{4M-2}}\phi_{i}(y)\phi_{i}(z)f(y)f(z)\,dy\,dz\right)^{\frac{1}{2}}.$

Due to symmetry, we can then say that

$\displaystyle\begin{split}Rf(x)&\gtrsim
a(x)\left(\int_{\mathcal{M}}\int_{\mathcal{M}}\frac{d(x_{i}^{\circ},y)^{2M-n_{i}}d(x_{i}^{\circ},z)^{2M-n_{i}}}{(d(x_{i}^{\circ},y)+d(x_{i}^{\circ},z))^{4M-2}}\mathbbm{1}_{d(x_{i}^{\circ},y)\leq
d(x_{i}^{\circ},z)}\phi_{i}(y)\phi_{i}(z)f(y)f(z)\,dy\,dz\right)^{\frac{1}{2}}\\\
&\geq
a(x)\left(\int_{\mathcal{M}}d(x_{i}^{\circ},y)^{2M-n_{i}}\left[\int_{\mathcal{M}}d(x_{i}^{\circ},z)^{2-2M-n_{i}}\mathbbm{1}_{d(x_{i}^{\circ},y)\leq
d(x_{i}^{\circ},z)}\phi_{i}(z)f(z)\,dz\right]\phi_{i}(y)f(y)\,dy\right)^{\frac{1}{2}}.\end{split}$

Set
$f_{\varepsilon}(y):=d(x_{i}^{\circ},y)^{-\frac{n_{i}}{p}(1+\varepsilon)}\phi_{i}(y)$
for $\varepsilon>0$. It is obvious that $f_{\varepsilon}\in L^{p}$. In the
argument to follow, as is frequently the case with asymptotic arguments, we
will need to keep careful track of the dependence of the constants on
$\varepsilon$. On applying the previous estimate for $Rf$ to the function
$f_{\varepsilon}$,

(53) $Rf_{\varepsilon}(x)\gtrsim
a(x)\left(\int_{\mathcal{M}}d(x_{i}^{\circ},y)^{2M-n_{i}(1+\frac{(1+\varepsilon)}{p})}\phi_{i}(y)\int_{\mathcal{M}}d(x_{i}^{\circ},z)^{2-2M-n_{i}(1+\frac{(1+\varepsilon)}{p})}\mathbbm{1}_{d(x_{i}^{\circ},y)\leq
d(x_{i}^{\circ},z)}\phi_{i}(z)\,dz\,dy\right)^{\frac{1}{2}}.$

Evidently

$\displaystyle\begin{split}\int_{\mathcal{M}}d(x_{i}^{\circ},z)^{2-2M-n_{i}(1+\frac{(1+\varepsilon)}{p})}\mathbbm{1}_{d(x_{i}^{\circ},y)\leq
d(x_{i}^{\circ},z)}\phi_{i}(z)\,dz&\gtrsim\int^{\infty}_{d(x_{i}^{\circ},y)}r^{1-2M-\frac{n_{i}(1+\varepsilon)}{p}}\,dr\\\
&=\frac{1}{\frac{n_{i}(1+\varepsilon)}{p}+2M-2}d(x_{i}^{\circ},y)^{2-2M-\frac{n_{i}(1+\varepsilon)}{p}}\\\
&\gtrsim d(x_{i}^{\circ},y)^{2-2M-\frac{n_{i}(1+\varepsilon)}{p}},\end{split}$

where the implicit constant in the final line is independent of $\varepsilon$
since for $\varepsilon$ small $(1+\varepsilon)^{-1}\gtrsim 1$. On applying
this to (53),

$\displaystyle\begin{split}Rf_{\varepsilon}(x)&\gtrsim
a(x)\left(\int_{\mathcal{M}}d(x_{i}^{\circ},y)^{2-n_{i}(1+\frac{2(1+\varepsilon)}{p})}\phi_{i}(y)\,dy\right)^{\frac{1}{2}}\\\
&\simeq
a(x)\left(\int_{D}^{\infty}r^{1-2n_{i}\frac{(1+\varepsilon)}{p}}\,dr\right)^{\frac{1}{2}},\end{split}$

for some constant $D>0$ determined by the distance of
$\mathrm{supp}\,\phi_{i}$ to $x_{i}^{\circ}$. This will blow up everywhere on
the non-zero support of $a$ when $p>(1+\varepsilon)n_{i}$, thereby forcing the
norm $\left\|Rf_{\varepsilon}\right\|_{p}$ to become infinite. Since
$\varepsilon$ was arbitrary, this implies that $R$ is an unbounded operator
for any $p>n_{i}$.

For the case $p=n_{i}$, we must carefully consider the asymptotics of
$Rf_{\varepsilon}$ as $\varepsilon\rightarrow 0$. From the previous estimate,

(54) $\displaystyle\begin{split}Rf_{\varepsilon}(x)&\gtrsim
a(x)\left(\int^{\infty}_{D}r^{1-2(1+\varepsilon)}\,dr\right)\\\ &\gtrsim
a(x)\frac{1}{\varepsilon}D^{-2\varepsilon}.\end{split}$

Also,

(55)
$\displaystyle\begin{split}\left\|f_{\varepsilon}\right\|_{p}&=\left(\int_{\mathcal{M}}d(x_{i}^{\circ},y)^{-n_{i}(1+\varepsilon)}\phi_{i}(y)\,dy\right)^{\frac{1}{p}}\\\
&\lesssim\left(\int^{\infty}_{D}r^{-1-n_{i}\varepsilon}\,dr\right)^{\frac{1}{p}}\\\
&\simeq\frac{1}{\varepsilon^{\frac{1}{p}}}D^{-\varepsilon}.\end{split}$

Suppose that $\left\|Rf\right\|_{n_{i}}\lesssim\left\|f\right\|_{n_{i}}$.
Then, from (54) and (55), we would have

$\frac{1}{\varepsilon}D^{-2\varepsilon}\lesssim\frac{1}{\varepsilon^{\frac{1}{p}}}D^{-\varepsilon}\quad\Leftrightarrow\quad\varepsilon\lesssim\varepsilon^{p}D^{\varepsilon
p},$

which is clearly not true asymptotically as $\varepsilon\rightarrow 0$. A
contradiction has been reached and we can safely conclude that $R$ is also
unbounded for $p=n_{i}$. ∎

##  8\. The Horizontal Square Function

In this section, Theorem 1.2 will be proved. In an analogous manner to the
argument from Section 4 for the operator $S$, $s$ can be expressed as

$\displaystyle\begin{split}sf(x)&=\left(\int^{\infty}_{0}\left|t^{2}\Delta(I+t^{2}\Delta)^{-M}f(x)\right|^{2}\,\frac{dt}{t}\right)^{\frac{1}{2}}\\\
&\left(\int^{\infty}_{0}\left|\Delta\left(\frac{1}{t^{2}}+\Delta\right)^{-M}f(x)\right|^{2}t^{3-4M}\,dt\right)^{\frac{1}{2}}\\\
&=\left(\int^{\infty}_{0}\left|\Delta(k^{2}+\Delta)^{-M}f(x)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}.\end{split}$

This can be controlled from above by the sum of the high and low energy
components,

$s_{<}f(x):=\left(\int^{1}_{0}\left|\Delta(k^{2}+\Delta)^{-M}f(x)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}$

and

$s_{>}f(x):=\left(\int^{\infty}_{1}\left|\Delta(k^{2}+\Delta)^{-M}f(x)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}.$

The proof of the high energy part $s_{>}$ is quite similar to our discussion
of the vertical square function. In fact it is simpler argument because the
kernel of the operator $\Delta(k^{2}+\Delta)^{-M}$ is symmetric. Hence it
suffices to prove only (43) and (44) automatically follows. Thus one does not
have to use the notion of Hodge Laplacian this time. Otherwise the proof is
unchanged and we skip it. Here we discuss only details of the low energy
component.

###  8.1. Low Energy

As a first step to proving the boundedness of the low energy square function,
observe that

$s_{<}f(x)=\left(\int^{1}_{0}\left|\left[(k^{2}+\Delta)^{-(M-1)}-k^{2}(k^{2}+\Delta)^{-M}\right]f(x)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}.$

From the formula

$(k^{2}+\Delta)^{-j}=\sum_{i=1}^{4}H_{i}^{(j)}(k)$

we then obtain

$s_{<}f(x)\lesssim\sum_{i=1}^{4}s_{<}^{i}f(x),$

where

$s^{i}_{<}f(x):=\left(\int^{\infty}_{0}\left|\left[H^{(M-1)}_{i}(k)-k^{2}H^{(M)}_{i}(k)\right]f(x)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}.$

The $L^{p}$-boundedness and weak-type $(1,1)$ property of each of these
operators will be proved separately.

####  8.1.1. The Operator $s_{<}^{1}$

Notice that

$\displaystyle\begin{split}H^{(M-1)}_{1}(k)(x,y)-k^{2}H_{1}^{(M)}(k)(x,y)&=\sum_{i=1}^{l}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-(M-1)}(x,y)\phi_{i}(x)\phi_{i}(y)\\\
&\qquad\qquad-k^{2}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(x)\phi_{i}(y)\\\
&=\sum_{i=1}^{l}\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}\left(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2}\right)^{-M}(x,y)\phi_{i}(x)\phi_{i}(y).\end{split}$

As this kernel only consists of finitely many terms, in order to prove that
$s_{<}^{1}$ is bounded on $L^{p}$ and weak-type $(1,1)$ it is sufficient to
prove for any $1\leq i\leq l$ that the operator

$s_{<}^{1,i}:=\left(\int^{1}_{0}\left|\int_{\mathcal{M}}\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(x,y)\phi_{i}(x)\phi_{i}(y)f(y)\,dy\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}$

is bounded on $L^{p}$ and weak-type $(1,1)$. Recall from classical theory that
the operator

$\displaystyle\begin{split}&\left(\int^{\infty}_{0}\left|\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}g(x)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}\\\
&\hskip
144.54pt=\left(\int^{\infty}_{0}\left|t^{2}\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}(t^{2}\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+1)^{-M}g(x)\right|^{2}\,\frac{dt}{t}\right)^{\frac{1}{2}}\end{split}$

is bounded on $L^{p}(\mathbb{R}^{n_{i}}\times\mathcal{M}_{i})$ for all
$p\in(1,\infty)$ and weak-type $(1,1)$. Therefore,

$\displaystyle\begin{split}\left\|s_{<}^{1,i}f\right\|_{p}&=\left\|\phi_{i}\cdot\left(\int^{1}_{0}\left|\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(\phi_{i}\cdot
f)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}\right\|_{p}\\\
&\lesssim\left\|\left(\int^{\infty}_{0}\left|\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}(\Delta_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}}+k^{2})^{-M}(\phi_{i}\cdot
f)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}\right\|_{p}\\\
&\lesssim\left\|f\right\|_{p}\end{split}$

for any $p\in(1,\infty)$. Weak-type $(1,1)$ bounds follow in the same manner.

####  8.1.2. The Operator $s_{<}^{2}$

The operator family
$\left\\{H_{2}^{(M-1)}(k)-k^{2}H_{2}^{(M)}(k)\right\\}_{k\in(0,1)}$
constitutes a family of pseudodifferential operators of order $2-2M$. In a
similar manner to Section 4.2, standard pseudodifferential operator theory can
be applied to yield the $L^{p}$-boundedness of $s_{<}^{2}$ for any
$p\in[1,\infty]$.

####  8.1.3. The Operator $s_{<}^{3}$

From an application of the triangle inequality followed by Minkowski’s
integral inequality, the operator $s^{3}_{<}$ can be estimated from above by

$s^{3}_{<}f(x)\lesssim
W_{1}f(x)+W_{2}f(x)=\sum_{i=1,2}\int_{\mathcal{M}}w_{i}(x,y)\left|f(y)\right|\,dy,$

where $W_{1}$ and $W_{2}$ are operators with the respective kernels

$w_{1}(x,y):=\left(\int^{1}_{0}\left|H_{3}^{(M-1)}(k)(x,y)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}$

and

$w_{2}(x,y):=\left(\int^{1}_{0}\left|H^{(M)}_{3}(k)(x,y)\right|^{2}k^{4M-1}\,dk\right)^{\frac{1}{2}}.$

The $L^{p}$-boundedness and weak-type $(1,1)$ property of the operators
$W_{1}$ and $W_{2}$ will be proved separately.

######  Lemma 8.1.

The operator $W_{1}$ is bounded on $L^{p}$ for all $p\in(1,\infty)$ and weak-
type $(1,1)$.

###### Proof.

As in Section 4.3, in order to obtain the boundedness of $W_{1}$ on $L^{p}$
for $1<p<\infty$ it suffices to demonstrate that the kernel satisfies the
estimates (31), (32), (33) and (34).

For the condition (31), observe that Proposition 3.4 implies
$\left|H_{3}^{(M-1)}(k)(x,y)\right|\lesssim k^{4-2M}$ for all
$x,\,y\in\mathcal{M}$. This estimate, when applied to the definition of
$w_{1}(x,y)$, gives

(56) $w_{1}(x,y)\lesssim 1\quad\forall\ x,\,y\in\mathcal{M},$

which immediately implies the validity of (31)

For (32), Proposition 3.4 implies that for
$x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$ and $y\in K$,

$\left|H_{3}^{(M-1)}(k)(x,y)\right|\lesssim k^{4-2M}\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}\exp(-ckd(x_{i}^{\circ},x)).$

Therefore,

(57)
$\displaystyle\begin{split}w_{1}(x,y)&=\left(\int^{1}_{0}\left|H_{3}^{(M-1)}(k)(x,y)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}\left(\int^{1}_{0}k^{3}\exp(-2ckd(x_{i}^{\circ},x))\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle d(x_{i}^{\circ},x)\rangle^{-n_{i}}.\end{split}$

This will clearly imply (32) when $p>1$.

Let us next consider (33). For $x\in K$ and
$y\in\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus K_{j}$, Proposition 3.4
implies that

$\left|H_{3}(k)(x,y)\right|\lesssim k^{4-2M}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\exp(-ckd(x_{j}^{\circ},y)).$

Therefore,

$\displaystyle\begin{split}w_{1}(x,y)&=\left(\int^{1}_{0}\left|H_{3}^{(M-1)}(k)(x,y)\right|^{2}k^{4M-5}\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\left(\int^{1}_{0}k^{3}\exp(-2ckd(x_{j}^{\circ},y))\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle d(x_{j}^{\circ},y)\rangle^{-n_{j}}.\end{split}$

It is easy to see that this implies (33) provided $p<\infty$

Finally, it remains to validate the estimate (34). Proposition 3.4 once again
implies that for $x\in\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus K_{i}$
and $y\in\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus K_{j}$,

(58) $\displaystyle\begin{split}w_{1}(x,y)&\lesssim\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\left(\int^{1}_{0}k^{3}\exp(-2ck(d(x_{i}^{\circ},x)+d(x_{j}^{\circ},y)))\,dk\right)^{\frac{1}{2}}\\\
&\lesssim\langle d(x_{i}^{\circ},x)\rangle^{2-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}}\left(\frac{1}{(d(x_{i}^{\circ},x)+d(x_{j}^{\circ},y))^{4}}\right)^{\frac{1}{2}}\\\
&\leq\min\left(\langle d(x_{i}^{\circ},x)\rangle^{-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{2-n_{j}},\langle
d(x_{i}^{\circ},x)\rangle^{2-n_{i}}\langle
d(x_{j}^{\circ},y)\rangle^{-n_{j}}\right).\end{split}$

Applying this to (34),

$\displaystyle\begin{split}\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}&\left(\int_{\mathbb{R}^{n_{j}}\times\mathcal{M}_{j}\setminus
K_{j}}w_{1}(x,y)^{p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\\\
&\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left(\int_{D^{1}_{j}(x)}\langle
d(x_{i}^{\circ},x)\rangle^{(2-n_{i})p^{\prime}}\langle
d(x_{j}^{\circ},y)\rangle^{-n_{j}p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\\\
&\qquad+\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\left(\int_{D^{2}_{j}(x)}\langle
d(x_{i}^{\circ},x)\rangle^{-n_{i}p^{\prime}}\langle
d(x_{j}^{\circ},y)\rangle^{(2-n_{j})p^{\prime}}\,dy\right)^{\frac{p}{p^{\prime}}}\,dx\\\
&=:J_{1}+J_{2},\end{split}$

where $D_{j}^{1}(x)$ and $D_{j}^{2}(x)$ are as defined in Section 4.3. For the
first term,

$J_{1}=\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle
d(x_{i}^{\circ},x)\rangle^{(n_{i}-2)p}}\left(\int_{D^{1}_{j}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{n_{j}p^{\prime}}}\right)^{\frac{p}{p^{\prime}}}\,dx.$

For the interior integral, in an analogous manner to (38),

$\int_{D^{1}_{j}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{n_{j}p^{\prime}}}\lesssim\int_{\mathbb{R}^{n_{j}}}\frac{dy_{1}}{(|y_{1}-x_{j,1}^{\circ}|+d(x_{i}^{\circ},x))^{n_{j}p^{\prime}}}.$

This will be integrable since $p^{\prime}>1$, in which case

$\displaystyle\begin{split}\int_{D^{1}_{j}(x)}\frac{dy}{\langle
d(x_{j}^{\circ},y)\rangle^{n_{j}p^{\prime}}}&\lesssim\int^{\infty}_{d(x_{i}^{\circ},x)}\frac{r^{n_{j}-1}}{r^{n_{j}p^{\prime}}}\,dr\\\
&\simeq\left[-r^{n_{j}(1-p^{\prime})}\right]^{\infty}_{d(x_{i}^{\circ},x)}\\\
&=\langle d(x_{i}^{\circ},x)\rangle^{n_{j}(1-p^{\prime})}.\end{split}$

Applying this estimate to $J_{1}$ gives

$\displaystyle\begin{split}J_{1}&\lesssim\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{1}{\langle d(x_{i}^{\circ},x)\rangle^{(n_{i}-2)p}}\cdot\langle
d(x_{i}^{\circ},x)\rangle^{-n_{j}}\,dx\\\
&=\int_{\mathbb{R}^{n_{i}}\times\mathcal{M}_{i}\setminus
K_{i}}\frac{dx}{\langle
d(x_{i}^{\circ},x)\rangle^{(n_{i}-2)p+n_{j}}}.\end{split}$

This will be finite provided that

$(n_{i}-2)p+n_{j}>n_{i}\quad\Leftrightarrow\quad
p>\frac{n_{i}-n_{j}}{n_{i}-2}.$

Since $\frac{n_{i}-n_{j}}{n_{i}-2}<\frac{n_{i}-2}{n_{i}-2}=1$, this will be
satisfied when $p>1$. It has therefore been proved that $J_{1}$ is finite for
$1<p<\infty$.

It remains to consider the term $J_{2}$. Looking back to our proof of the