# Equivalence of Sobolev norms in Lebesgue spaces for Hardy operators in a
half-space

The Anh Bui  The Anh Bui
School of Mathematical and Physical Sciences, Macquarie University, NSW 2109,
Australia<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Konstantin Merz
Konstantin Merz
Institut für Analysis und Algebra, Technische Universität Braunschweig,
Universitätsplatz 2, 38106 Braunschweig, Germany, and Department of
Mathematics, Graduate School of Science, Osaka University, Toyonaka, Osaka
560-0043, Japan<EMAIL_ADDRESS>
(Date: September 03, 2023)

###### Abstract.

We consider Hardy operators, i.e., homogeneous Schrödinger operators
consisting of the ordinary or fractional Laplacian in a half-space plus a
potential, which only depends on the appropriate power of the distance to the
boundary of the half-space. We compare the scales of homogeneous
$L^{p}$-Sobolev spaces generated by these Hardy operators with and without
potential with each other. To that end, we prove and use new square function
estimates for operators with slowly decaying heat kernels. Our results hold
for all admissible coupling constants in the local case and for repulsive
potentials in the fractional case, and extend those obtained recently in
$L^{2}$. They also cover attractive potentials in the fractional case, once
expected heat kernel estimates are available.

T. A. Bui is supported by the Australian Research Council via the grant ARC
DP220100285. K. Merz is supported through the PRIME programme of the German
Academic Exchange Service (DAAD) with funds from the German Federal Ministry
of Education and Research (BMBF)

###### Contents

  1. 1 Introduction
    1. 1.1 Statement of the problem and main result
    2. 1.2 Comments on Theorem 1.1
    3. 1.3 Proof of Theorem 1.1 and Corollary 1.2
  2. 2 Estimates involving the heat kernel of $L_{\lambda}$
    1. 2.1 Pointwise estimates
    2. 2.2 Spatially averaged estimates
  3. 3 Proof of the square function estimates (Theorem 1.4)
  4. 4 Bounds for differences of kernels
  5. 5 Proof of the reversed Hardy inequality (Theorem 1.6)
  6. 6 Proof of the generalized Hardy inequality (Theorem 1.8)
  7. 7 Density of $(L_{\lambda})^{s/2}C_{c}^{\infty}(\mathbb{R}^{d}_{+})$ in $L^{p}(\mathbb{R}_{+}^{d})$

##  1\. Introduction

###  1.1. Statement of the problem and main result

For $d\in\mathbb{N}=\\{1,2,...\\}$, we consider the following generalized
Schrödinger operators in $L^{2}(\mathbb{R}^{d}_{+})$,

(1) $L_{\lambda}^{(\alpha)}:=(-\Delta)^{\alpha/2}_{\mathbb{R}^{d}_{+}}+\lambda
x_{d}^{-\alpha},$

where $\mathbb{R}_{+}^{d}:=\mathbb{R}^{d-1}\times\mathbb{R}_{+}$,
$\mathbb{R}_{+}:=(0,\infty)$, $x=(x^{\prime},x_{d})\in\mathbb{R}_{+}^{d}$ with
$x^{\prime}\in\mathbb{R}^{d-1}$ and $x_{d}\in\mathbb{R}_{+}$,
$\alpha\in(0,2]$, $(-\Delta)_{\mathbb{R}_{+}^{d}}$ is the Laplacian with a
Dirichlet boundary condition on $(\mathbb{R}_{+}^{d})^{c}$, and, when
$\alpha\in(0,2)$, $(-\Delta)^{\alpha/2}_{\mathbb{R}^{d}_{+}}$ denotes the
regional fractional Laplacian, introduced by Bogdan, Budzy, and Chen [BBC03]:
It is the self-adjoint Friedrichs extension of the quadratic form

(2) $\displaystyle\langle
u,(-\Delta)^{\frac{\alpha}{2}}_{\mathbb{R}_{+}^{d}}u\rangle_{L^{2}(\mathbb{R}_{+}^{d})}\\!=\\!\frac{\alpha}{2^{2-\alpha}\pi^{\frac{d}{2}}}\
\frac{\Gamma(\frac{d+\alpha}{2})}{\Gamma(1-\frac{\alpha}{2})}\iint_{\mathbb{R}_{+}^{d}\times\mathbb{R}_{+}^{d}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{d+\alpha}}\,dx\,dy,\
u\in C_{c}^{1}(\mathbb{R}_{+}^{d})$

and its pointwise action is given by

$\displaystyle(-\Delta)^{\alpha/2}_{\mathbb{R}_{+}^{d}}u(x)=\frac{\alpha}{2^{2-\alpha}\pi^{d/2}}\
\frac{\Gamma(\frac{d+\alpha}{2})}{\Gamma(1-\frac{\alpha}{2})}\int_{\mathbb{R}_{+}^{d}}\frac{u(x)-u(y)}{|x-y|^{d+\alpha}}\,dy,\quad
x\in\mathbb{R}_{+}^{d}.$

By the decomposition
$(-\Delta)_{\mathbb{R}_{+}^{d}}=(-\Delta)_{\mathbb{R}^{d-1}}+(-\Delta)_{\mathbb{R}_{+}}$
and the classical Hardy inequality [Har19, Har20, OK90, KMP06],
$L_{\lambda}^{(2)}$ is bounded from below if and only if $\lambda\geq-1/4$.
Moreover, if $\lambda\geq-1/4$, then $L_{\lambda}^{(2)}$ is bounded from below
by zero. Bogdan and Dyda [BD11] showed the corresponding Hardy inequality for
$\alpha\in(0,2)$. That is,

(3) $\displaystyle L_{\lambda}^{(\alpha)}\geq 0$

if and only if $\lambda\geq\lambda_{*}$ with

$\lambda_{*}(\alpha):=-\frac{\Gamma(\frac{1+\alpha}{2})}{\pi}\Big{(}\Gamma(\frac{1+\alpha}{2})-\frac{2^{\alpha-1}\sqrt{\pi}}{\Gamma(1-\frac{\alpha}{2})}\Big{)}.$

Since $\lambda_{*}(2)=-1/4$ and we regard $\alpha\in(0,2]$ as fixed, we write
$\lambda_{*}$ instead of $\lambda_{*}(\alpha)$ and $L_{\lambda}$ instead of
$L_{\lambda}^{(\alpha)}$ from now on. Throughout the paper, we assume
$\lambda\geq\lambda_{*}$, and we will call $L_{\lambda}$ a Hardy operator.

The homogeneity of $L_{\lambda}$ motivates to ask in how far $L_{\lambda}$ and
$L_{0}$ are comparable with each other. Our main result in this paper, Theorem
1.1 below, says that the scales of Sobolev norms
$\|(L_{\lambda})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}$ are equivalent to the
norms $\|(L_{0})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}$ for $s>0$ belonging to
some interval depending on all parameters $d,\alpha,\lambda$. This extends
[FM23], where Frank and the second-named author proved this equivalence of
Sobolev norms for $p=2$. To give a convenient presentation of our result, we
introduce the following parameterization of the coupling constant $\lambda$ of
the Hardy potential. For $\alpha\in(0,2]$, let $M:=\alpha$ if $\alpha<2$ and
$M:=\infty$ if $\alpha=2$, and

(4) $(-1,M)\ni\sigma\mapsto
C(\sigma):=\frac{1}{\pi}\Big{(}\Gamma(\alpha)\sin\frac{\pi\alpha}{2}+\Gamma(1+\sigma)\Gamma(\alpha-\sigma)\sin\frac{\pi(2\sigma-\alpha)}{2}\Big{)}.$

As noted in [FM23] and the references therein, the function $\sigma\mapsto
C(\sigma)$ is continuous and symmetric with respect to
$\sigma=\frac{\alpha-1}{2}$, strictly increasing on $[\frac{\alpha-1}{2},M)$
and $C(\frac{\alpha-1}{2})=\lambda_{*}$. Moreover, $\lim_{\sigma\to
M}C(\sigma)=+\infty$. Therefore, for any $\lambda\in[\lambda_{*},\infty)$
there is a unique

(5) $\sigma\in\left[\frac{\alpha-1}{2},M\right)\ \text{with}\ \
C(\sigma)=\lambda.$

In particular, $C(\alpha-1)=C(0)=0$. Hence, the case $\lambda=0$ corresponds
to $\sigma=\max\\{0,\alpha-1\\}$ and the case $\lambda>0$ corresponds to
$\sigma>(\alpha-1)_{+}$. For $\alpha=2$ one finds
$\sigma=\frac{1}{2}\big{(}1+\sqrt{1+4\lambda}\big{)}$.

In the following, we always write $C$ and $c$ to denote positive constants
that are independent of the main parameters involved but whose values may
differ from line to line. We write $A\lesssim B$ for $A,B\in\mathbb{R}_{+}$
whenever there is a constant $c>0$ such that $A\leq cB$. The notation $A\simeq
B$ means $A\lesssim B\lesssim A$ and in this case we say that $A$ and $B$ are
comparable. If we want to emphasize that the constant $c$ may depend on some
parameter, say $\tau$, we write $A\lesssim_{\tau}B$. The dependence on fixed
parameters, like $d,\alpha,p$, is usually suppressed. Moreover, we use
notations $A\wedge B:=\min\\{A,B\\}$, $A\vee B:=\max\\{A,B\\}$, and
$A_{+}:=\max\\{A,0\\}$, $p^{\prime}:=p/(p-1)$ for $p\in[1,\infty]$, and the
convention $1/0\equiv\infty$. The following is our main result.

######  Theorem 1.1.

Let $p\in(1,\infty)$, $\alpha\in(0,2]$, and let $\lambda\geq 0$ when
$\alpha<2$ and $\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by
(5), and let $s\in(0,2]\cap(0,\frac{2d}{\alpha})$.

  1. (1)

If $\frac{1}{1+\sigma\wedge 0}<p<\frac{1}{(\alpha s/2-\sigma)_{+}}$, then

(6)
$\displaystyle\|(L_{0})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim_{d,\alpha,\lambda,s,p}\|(L_{\lambda})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}\quad\text{for
all}\ u\in C_{c}^{\infty}(\mathbb{R}_{+}^{d}).$

  2. (2)

If $\frac{1}{1+\sigma\wedge 0}<p<\frac{1}{\max\\{\alpha
s/2-(\alpha-1)_{+},-\sigma\\}}$, then

(7)
$\displaystyle\|(L_{\lambda})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim_{d,\alpha,\lambda,s,p}\|(L_{0})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}\quad\text{for
all}\ u\in C_{c}^{\infty}(\mathbb{R}_{+}^{d}).$

In particular, if $\frac{1}{1+\sigma\wedge
0}<p<\min\left\\{\frac{1}{\max\\{\alpha
s/2-(\alpha-1)_{+},-\sigma\\}},\frac{1}{(\alpha s/2-\sigma)_{+}}\right\\}$,
then we have the equivalence

$\|(L_{\lambda})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}\simeq_{d,\alpha,\lambda,s,p}\|(L_{0})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}\quad\text{for
all}\ u\in C_{c}^{\infty}(\mathbb{R}_{+}^{d}).$

Theorem 1.1 and a density argument (Theorem 1.10), yield the following result
for the Riesz transforms associated to $L_{0}$ and $L_{\lambda}$ with
$\lambda\neq 0$.

######  Corollary 1.2.

Let $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha<2$ and
$\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by (5), and let
$s\in(0,2]\cap(0,\frac{2d}{\alpha})$. If $\frac{1}{1+\sigma\wedge
0}<p<\frac{1}{(\alpha s/2-\sigma)_{+}}$ and $s<2(1/p+\sigma)/\alpha$, then

(8)
$\displaystyle\|(L_{0})^{s/2}(L_{\lambda})^{-s/2}f\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim_{d,\alpha,\lambda,s,p}\|f\|_{L^{p}(\mathbb{R}_{+}^{d})}.$

In the next subsection, we comment on Theorem 1.1 and put it into context with
previous works on this subject. In the following subsection, we give its
proof.

###  1.2. Comments on Theorem 1.1

(1) Frank and the second-named author proved (6)–(7) in [FM23, Theorem 1] for
$p=2$. As noted in this work, the assumptions $\lambda\geq 0$ for $\alpha=1$
and $\lambda\geq-1/4$ for $\alpha=2$ in Theorem 1.1 are best possible. For
$\alpha\in(0,2)\setminus\\{1\\}$ we do not expect $\lambda\geq 0$ to be
necessary. This assumption comes from our use of heat kernel bounds for
$L_{\lambda}$, which are currently known only for $\lambda\geq 0$ when
$\alpha<2$. Since we expect these bounds to extend to
$\lambda\in[\lambda_{*},0)$, we accompany all results whose proofs use bounds
for $\mathrm{e}^{-tL_{\lambda}}$ with a remark stating the potential
extension. In particular, we make the following remark.

######  Remark 1.3.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the upper bound in (17) below with
$\sigma$ defined by (5). Then Theorem 1.1 and Corollary 1.2 remain valid for
this value of $\lambda$. This follows by the same arguments as in the proof in
Section 1.3 below, taking into account Remarks 1.5, 1.7, and 1.9.

(2) Theorem 1.1 joins a line of recent research [KVZ16b, KMV+18, FMS21, Mer21,
BD23, BN22, FM23] on equivalence of Sobolev norms involving Hardy operators.
In the groundbreaking work [KVZ16b], Killip, Visan, and Zhang showed the
equivalence of $L^{p}$-Sobolev norms generated by the ordinary Laplacian and
the Dirichlet–Laplacian on the complement of a convex set. In [KMV+18],
Killip, Miao, Visan, Zhang, and Zheng considered
$(-\Delta)^{\alpha/2}+\lambda/|x|^{\alpha}$ and proved the equivalence of
$L^{p}$-Sobolev norms generated by this operator with $\lambda\neq 0$ and
$\lambda=0$ when $\alpha=2$. D’Ancona, Frank, Nader, Siedentop, and the
authors of the present paper extended their result in [FMS21, Mer21, BD23,
BN22] and successively treated all admissible $\alpha\in(0,2)$, $\lambda$ and
$p$. For an alternative proof and an extension of this result when $\alpha=2$,
see Miao, Su, and Zheng [MSZ23]. In the most recent work [FM23], Frank and the
second author showed the equivalence of $L^{2}(\mathbb{R}_{+}^{d})$-Sobolev
norms generated by $L_{0}$ and $L_{\lambda}$ with $\lambda\geq-1/4$ when
$\alpha=2$ and $\lambda\geq 0$ when $\alpha\in(0,2)$, i.e., Theorem 1.1 with
$p=2$. All of the above results crucially relied on heat kernel bounds for the
operators in question. The restriction $\lambda>0$ in [FM23] is because heat
kernel bounds for $L_{\lambda}$ with $\alpha\in(0,2)$ are so far only
available when $\lambda\geq 0$. However, as noted above, once it is shown that
these bounds extend to all $\lambda\in[\lambda_{*},0)$, the results in [FM23]
automatically extend to $\lambda\in[\lambda_{*},0]$.

(3) As noted in the above-mentioned works on the equivalence of Sobolev norms,
homogeneous operators appear frequently as scaling limits or model operators
of more complicated problems, e.g., in quantum mechanics, general relativity,
engineering, and finance. Thus, besides being of pure mathematical interest,
the equivalences of Sobolev norms are powerful tools, because Hardy operators
can be replaced with the easier to handle and better-understood fractional
Laplacians. For instance, Killip, Visan, and Zhang [KVZ16a] and Killip, Miao,
Murphy, Visan, Zhang, and Zheng [KMV+17, KMVZ17] used the equivalence of
Sobolev norms in [KVZ16b, KMV+18] to study global well-posedness and
scattering for nonlinear Schrödinger equations with the Dirichlet–Laplacian on
the complement of a convex obstacle, or the Hardy operator
$-\Delta+\lambda/|x|^{2}$. In the fractional case, the equivalence of Sobolev
norms [FMS21] played in important role in the proof of the strong Scott
conjecture for relativistic atoms [FMSS20] by Frank, Siedentop, Simon, and the
second author; see also [FMS23a] for a shorter proof and [FMS23b] for a
review. This problem is a genuine many-body problem, but one can reduce it to
a one-body problem involving the operator
$C_{\lambda}:=\sqrt{1-\Delta}-1+\lambda/|x|$ and its powers
$(C_{\lambda})^{s/2}$ in $L^{2}(\mathbb{R}^{3})$. Since
$\sqrt{1-\Delta}-\sqrt{-\Delta}$ is bounded, the equivalence of Sobolev norms
allowed us to deduce estimates for $(C_{\lambda})^{s/2}$ from estimates for
$(-\Delta)^{s/4}$, which are significantly easier to prove. In view of the
applications of the equivalence of Sobolev norms in [KMV+17, KMVZ17], we are
optimistic that the results in [FM23] and the present paper will be useful to
study, e.g., nonlinear Schrödinger equations involving $L_{\lambda}$.

###  1.3. Proof of Theorem 1.1 and Corollary 1.2

The proof of Theorem 1.1 follows the ideas laid out in [KMV+18, FMS21, Mer21,
BD23, FM23]. Compared with the case $p=2$, a new difficulty in the proof of
Theorem 1.1 for $p\neq 2$ arises from the question how to make sense of
$(L_{\lambda})^{s/2}f$ in $L^{p}(\mathbb{R}_{+}^{d})$. A priori, it is not
clear what the domain of $(L_{\lambda})^{s/2}$, understood as an operator in
$L^{p}$, is. A classical way to tackle this question is to try to express
$(L_{\lambda})^{s/2}$ using Littlewood–Paley square functions, i.e., to prove
square function estimates such as
$\|(L_{\lambda})^{s/2}f\|_{L^{p}(\mathbb{R}_{+}^{d})}\simeq\|\left(\sum_{N\in
2^{\mathbb{Z}}}|N^{s/2}\Psi(L_{\lambda}/N)f|^{2}\right)^{1/2}\|_{L^{p}(\mathbb{R}_{+}^{d})}$
for a bump function $\Psi\in C_{c}^{\infty}([1/2,3/2]:[0,1])$, say. As in the
earlier works [KMV+18, Mer21, BD23], we will work with square functions, where
the Littlewood–Paley projection $\Psi(L_{\lambda}/N)$ is replaced with the
well-studied heat kernel $\mathrm{e}^{-tL_{\lambda}}$. See Sections 2–2.2
below for estimates involving $\mathrm{e}^{-tL_{\lambda}}$ that are most
relevant for us.

One standard approach to prove Littlewood–Paley square function estimates
relies on a Mikhlin–Hörmander spectral multiplier theorem, i.e., the
$L^{p}$-boundedness of sufficiently smooth functions of $L_{\lambda}$. This
approach was used in the works [KMV+18, Mer21], which considered
$(-\Delta)^{\alpha/2}+\lambda/|x|^{\alpha}$ with $\alpha=2$ and
$\lambda\geq-(d-2)^{2}/4$ or $\alpha\in(0,2)$ and $\lambda\geq 0$, and relies
heavily on the $L^{p}-L^{q}$ off-diagonal estimates with exponential decay or
the Poisson upper bounds for the heat kernels. Unfortunately, neither of these
two crucial ingredients is available when $\lambda<0$ due to the emergence of
singular weights in the sharp heat kernel bound. To address this challenge,
D’Ancona and the first author [BD23] introduced the “continuous square
functions” approach, which is based on the boundedness of the square function
associated to the main operator $(-\Delta)^{\alpha/2}+\lambda/|x|^{\alpha}$.
Since the heat kernel of $(-\Delta)^{\alpha/2}+\lambda/|x|^{\alpha}$ lacks
regularity conditions, the renowned classical Calderón-Zygmund theory can not
be applied in this context. To circumvent this limitation, [BD23] employed the
theory of singular integrals beyond the Calderón-Zygmund theory as outlined in
[Aus07, Chapter 6], ultimately establishing the boundedness of the square
function and thereby demonstrating the equivalence of Sobolev norms for
$\lambda\leq 0$, and resolving the previously open case in [Mer21]. Encouraged
by [BD23], we construct similar continuous square functions, and prove
corresponding square function estimates for $L_{\lambda}$. More precisely, for
$\gamma>0$, we define the continuous square function associated to
$L_{\lambda}$ as

(9)
$\displaystyle(S_{L_{\lambda},\gamma}f)(x):=\Big{(}\int_{0}^{\infty}|(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}f|^{2}\frac{dt}{t}\Big{)}^{1/2}.$

In Section 3, we prove the following square function estimates.

######  Theorem 1.4.

Let $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha<2$ and
$\lambda\geq-1/4$ when $\alpha=2$. Let $\gamma\in(0,1)$. Then for all
$1<p<\infty$,

(10)
$\displaystyle\|S_{L_{\lambda},\gamma}f\|_{L^{p}(\mathbb{R}_{+}^{d})}\simeq\|f\|_{L^{p}(\mathbb{R}_{+}^{d})}.$

In particular, for $s\in(0,2]$ and $1<p<\infty$,

(11)
$\displaystyle\left\|\Big{(}\int_{0}^{\infty}t^{-s}|tL_{\lambda}e^{-tL_{\lambda}}f|^{2}\frac{dt}{t}\Big{)}^{1/2}\right\|_{L^{p}(\mathbb{R}_{+}^{d})}\simeq\|(L_{\lambda})^{s/2}f\|_{L^{p}(\mathbb{R}_{+}^{d})}.$

######  Remark 1.5.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then Theorem 1.4 remains valid for
$\frac{1}{1+\sigma}=:(p_{-\sigma})^{\prime}<p<p_{-\sigma}:=-\frac{1}{\sigma}$.
This follows by the same arguments as in the proof below, taking into account
Remark 2.5.

With Theorem 1.4 at hand, we prove the next ingredient in the proof of Theorem
1.1, which we call a reversed Hardy inequality. It gives a lower bound on the
Hardy potential in terms of the difference of the square functions associated
to $L_{\lambda}$ and $L_{0}$ on the left-hand side of (11).

######  Theorem 1.6 (Reversed Hardy inequality).

Let $p\in(1,\infty)$, $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha<2$
and $\lambda\geq-1/4$ when $\alpha=2$. Then, for
$q=\min\\{\sigma,(\alpha-1)_{+}\\}$, $-r=\min\\{0,q\\}$,
$p\in(\frac{1}{1-r},\frac{1}{r})$ and $s\in(0,2)$, we have

(12)
$\displaystyle\left\|\left(\int_{0}^{\infty}t^{-s}\left|\left(tL_{\lambda}e^{-tL_{\lambda}}-tL_{0}e^{-tL_{0}}\right)f\right|^{2}\frac{dt}{t}\right)^{1/2}\right\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim\left\|\frac{f}{x_{d}^{\alpha
s/2}}\right\|_{L^{p}(\mathbb{R}_{+}^{d})}.$

We prove Theorem 1.6 in Section 5 below.

######  Remark 1.7.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then the assertions of Theorem 1.6 remain valid. This follows
by the same arguments as in the proof below, taking into account Remark 4.5.

The final ingredient to prove Theorem 1.1 is the following generalized Hardy
inequality giving an upper bound on powers of the Hardy potential in terms of
powers of the Hardy operator $L_{\lambda}$.

######  Theorem 1.8 (Generalized Hardy inequality).

Let $1<p<\infty$, $\alpha\in(0,2]$, and let $\lambda\geq 0$ when $\alpha<2$
and $\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by (5) and
suppose $s\in(0,2\wedge\frac{2d}{\alpha}]$ and $\tfrac{\alpha
s}{2}<1+2\sigma$. Then,

(13) $\displaystyle\|x^{-\alpha
s/2}_{d}(L_{\lambda})^{-s/2}f\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim\|f\|_{L^{p}(\mathbb{R}_{+}^{d})}\quad\text{for
all}\ f\in L^{p}(\mathbb{R}_{+}^{d})$

holds if and only if $(\frac{\alpha
s}{2}-\sigma)_{+}<\frac{1}{p}<1+\sigma\wedge 0$. In that case, we have

(14) $\displaystyle\|x^{-\alpha
s/2}_{d}g\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim\|L_{\lambda}^{s/2}g\|_{L^{p}(\mathbb{R}_{+}^{d})}\quad\text{for
all}\ g\in C_{c}^{\infty}(\mathbb{R}_{+}^{d}).$

######  Remark 1.9.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then the assertions of Theorem 1.8 remain valid. This follows
by the same arguments as in the proof below, taking into account Remark 6.2.

Our proof of Theorem 1.8 (see Section 6) is similar to that in [FM23, Theorems
3 and 13], where $p=2$ was considered. With the above ingredients at hand, we
give the

###### Proof of Theorem 1.1.

We only consider the proof of (6); the proof of (7) is analogous. For
$s\in(0,2)$, the square function estimates in (11), the triangle inequality,
the reversed and generalized Hardy inequalities (Theorems 1.6–1.8) yield

(15)
$\displaystyle\begin{split}\|(L_{0})^{\frac{s}{2}}u\|_{L^{p}(\mathbb{R}_{+}^{d})}&\simeq\left\|\Big{(}\int_{0}^{\infty}t^{-s}|tL_{0}e^{-tL_{0}}f|^{2}\frac{dt}{t}\Big{)}^{1/2}\right\|_{L^{p}(\mathbb{R}_{+}^{d})}\\\
&\leq\left\|\Big{(}\int_{0}^{\infty}t^{-s}|tL_{\lambda}e^{-tL_{\lambda}}f|^{2}\frac{dt}{t}\Big{)}^{1/2}\right\|_{L^{p}(\mathbb{R}_{+}^{d})}\\\
&\quad+\left\|\Big{(}\int_{0}^{\infty}t^{-s}|(tL_{0}e^{-tL_{0}}-tL_{\lambda}\mathrm{e}^{-tL_{\lambda}})f|^{2}\frac{dt}{t}\Big{)}^{\frac{1}{2}}\right\|_{L^{p}(\mathbb{R}_{+}^{d})}\\\
&\lesssim\|(L_{\lambda})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}+\left\|\frac{f}{x_{d}^{\alpha
s/2}}\right\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim\|(L_{\lambda})^{s/2}u\|_{L^{p}(\mathbb{R}_{+}^{d})}.\end{split}$

For $s=2$, the square function estimates are not necessary; we simply use the
triangle inequality and the generalized Hardy inequality to get the desired
estimate. ∎

To prove Corollary 1.2, we use the following density result, which was proved
in [FM23, Theorem 24] for $p=2$.

######  Theorem 1.10.

Let $p\in(1,\infty)$, $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha<2$
and $\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by (5), and let
$s\in(0,2]$. Assume that $s<2(1/p+\sigma)/\alpha$. Then for any $f\in
L^{p}(\mathbb{R}^{d}_{+})$ there is a sequence $(\phi_{n})\subset
C^{\infty}_{c}(\mathbb{R}^{d}_{+})$ such that

$L_{\lambda}^{s/2}\phi_{n}\to f\ \text{in}\ L^{p}(\mathbb{R}^{d}_{+})\,.$

######  Remark 1.11.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the upper bound in (17) with
$\sigma$ defined by (5). Then Theorem 1.10 remains valid for this value of
$\lambda$. This follows by the same arguments as in the proof below, since
Lemma 7.1 remains valid for this value of $\lambda$.

We can now give the

###### Proof of Corollary 1.2.

It suffices to show (8) for a dense set of $f\in L^{p}(\mathbb{R}_{+}^{d})$.
By Theorem 1.10, we may choose $f=(L_{\lambda})^{s/2}g$ with $g\in
C_{c}^{\infty}(\mathbb{R}_{+}^{d})$. Then, by Theorem 1.1,

(16)
$\displaystyle\|(L_{0})^{s/2}g\|_{L^{p}(\mathbb{R}_{+}^{d})}\lesssim_{d,\alpha,\lambda,s,p}\|(L_{\lambda})^{s/2}g\|_{L^{p}(\mathbb{R}_{+}^{d})}=\|f\|_{L^{p}(\mathbb{R}_{+}^{d})},$

which concludes the proof of (8). ∎

### Organization

In Section 2, we recall kernel bounds for $\mathrm{e}^{-tL_{\lambda}}$ and use
them, together with a Phragmén–Lindelöf argument, to prove bounds for complex-
time heat kernels. We use these bounds to prove new bounds for the kernels of
$(tL_{\lambda})^{k}\mathrm{e}^{-tL_{\lambda}}$ and prove corresponding
$L^{p}\to L^{q}$-estimates. We use these estimates in Section 3 to prove the
square function estimates for $(L_{\lambda})^{s/2}$ in Theorem 1.4. In Section
4, we extend bounds for the kernel of the difference
$e^{-tL_{\lambda}}-e^{-tL_{0}}$ considered in [FM23] and prove bounds for the
kernel of the difference $tL_{0}e^{-tL_{0}}-tL_{\lambda}e^{-tL_{\lambda}}$.
These bounds are crucial to prove the reversed Hardy inequality in Section 5.
In Section 6, we prove the generalized Hardy inequality (Theorem 1.8).
Finally, in Section 7, we show the density of
$(L_{\lambda})^{s/2}C_{c}^{\infty}(\mathbb{R}^{d}_{+})$ in
$L^{p}(\mathbb{R}_{+}^{d})$ (Theorem 1.10).

##  2\. Estimates involving the heat kernel of $L_{\lambda}$

In this section, we collect known pointwise estimates for the heat kernel
$\mathrm{e}^{-tL_{\lambda}}$ and derive novel bounds for the complex-time heat
kernel and $(tL_{\lambda})^{k}\mathrm{e}^{-tL_{\lambda}}$, $k\in\mathbb{N}$.
In the first subsection, we discuss pointwise and in the following subsection
$L^{p}\to L^{q}$-estimates.

###  2.1. Pointwise estimates

We recall pointwise estimates for the kernel $p_{t}(x,y)$ of
$\mathrm{e}^{-tL_{\lambda}}$.

######  Theorem 2.1.

  1. (a)

Let $\alpha\in(0,2)$, $\lambda\geq 0$, and let $\sigma$ be defined by (5).
Then for all $t>0$ and $x,y\in\mathbb{R}^{d}_{+}$,

(17) $\displaystyle p_{t}(x,y)\ {\simeq}\
\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}.$

  2. (b)

Let $\alpha=2$, $\lambda\geq-\frac{1}{4}$, and let $\sigma$ be defined by (5).
Then for all $t>0$ and $x,y\in\mathbb{R}^{d}_{+}$,

(18) $\displaystyle p_{t}(x,y)\ {\simeq}\
\Big{(}1\wedge\frac{x_{d}}{\sqrt{t}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{\sqrt{t}}\Big{)}^{\sigma}t^{-d/2}\exp\Big{(}-\frac{|x-y|^{2}}{ct}\Big{)}.$

For $\lambda\\!=\\!0$ and $\alpha\\!\leq\\!1$ the bounds were proved in
[CK03], while [CKS10] proved the estimates for $\alpha\\!\in\\!(1,2)$. The
bounds for $\alpha\in(0,2)$ and $\lambda\geq 0$ were recently proved in
[CKSV20] and extended in [SWW22]. For $\alpha=2$, the bounds follow from the
factorization of the Gaussian heat kernel with respect to
$\mathbb{R}_{+}^{d}\\!=\\!\mathbb{R}^{d-1}\\!\times\\!\mathbb{R}_{+}$ and
bounds for the one-dimensional heat kernel, i.e., the Bessel heat kernel.
Bounds for Bessel heat kernels can be found, e.g., in [BS02, p. 75]. The bound
stated in [BS02] does not have the product structure as stated in (18), but
can be obtained from it; see, e.g., [FM23, Appendix B] and [BM23, GT21] for
bounds for the related $\tfrac{\alpha}{2}$-stable subordinated Bessel heat
kernels.

Next, by a Phragmén–Lindelöf argument, we extend the upper bound for
$p_{t}(x,y)$ to complex times
$z\in\mathbb{C}_{\theta}:=\\{z\in\mathbb{C}:|\arg z|<\theta\\}$,
$\theta\in[0,\pi/2]$. These bounds are the key to estimate the kernels of
$(tL_{\lambda})^{k}\mathrm{e}^{-tL_{\lambda}}$, $k\in\mathbb{N}$.

######  Proposition 2.2.

  1. (a)

Let $\alpha\in(0,2)$, $\lambda\geq 0$, and let $\sigma$ be defined by (5). Let
$p_{z}(x,y)$ be the kernels associated to the semigroups $e^{-zL_{\lambda}}$
with $z\in\mathbb{C}_{\pi/2}=\\{z\in\mathbb{C}:|\arg z|<\pi/2\\}$. Then for
any $\epsilon\in(0,1)$, there exists a constant $C=C_{\epsilon}$ such that

(19) $|p_{z}(x,y)|\leq
C\Big{(}1\wedge\frac{x_{d}}{|z|^{1/\alpha}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{|z|^{1/\alpha}}\Big{)}^{\sigma}|z|^{-d/\alpha}\Big{(}\frac{|z|^{1/\alpha}}{|z|^{1/\alpha}+|x-y|}\Big{)}^{(d+\alpha)(1-\epsilon)}$

for all $x,y\in\mathbb{R}^{d}_{+}$ and $z\in\mathbb{C}_{\epsilon\pi/4}$.

  2. (b)

Let $\alpha=2$, $\lambda\geq-1/4$, and let $\sigma$ be defined by (5). Let
$p_{z}(x,y)$ be the kernels associated to the semigroups $e^{-zL_{\lambda}}$
with $z\in\mathbb{C}_{\pi/2}$. Then there exist constants $C,c>0$ such that

(20) $|p_{z}(x,y)|\leq
C\Big{(}1\wedge\frac{x_{d}}{\sqrt{|z|}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{\sqrt{|z|}}\Big{)}^{\sigma}|z|^{-d/2}\exp\Big{(}-\frac{|x-y|^{2}}{c|z|}\Big{)}$

for all $x,y\in\mathbb{R}^{d}_{+}$ and $z\in\mathbb{C}_{\pi/4}$.

######  Remark 2.3.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then (19) remains valid. This follows by the same arguments as
in the proof below.

###### Proof.

We only to give the proof for (19) since the proof of (20) can be done
similarly. To that end, we use a Phragmén–Lindelöf argument as in [Dav90,
Section 3.4]. We set

$w_{z}(x)=\Big{(}1+\frac{{|z|^{1/\alpha}}}{x_{d}}\Big{)}^{\sigma},\quad
x\in\mathbb{R}^{d}_{+},\ z\in\mathbb{C}$

and first claim that there exists $C>0$ such that

(21) $\left|w_{z}(x)p_{z}(x,y)w_{z}(y)\right|\leq\frac{C}{|z|^{d/\alpha}}$

holds for all $x,y\in\mathbb{R}^{d}_{+}$ and $z\in\mathbb{C}_{\pi/4}$. To
prove this, we define, for $f:\mathbb{R}^{d}_{+}\rightarrow\mathbb{R}$, the
norms

$|f|_{L_{w}^{\infty}(\mathbb{R}^{d}_{+})}=\sup_{x\in\mathbb{R}^{d}_{+}}\left|f(x)w_{z}(x)\right|$

and

$|f|_{L_{w}^{1}(\mathbb{R}^{d}_{+})}=\int_{\mathbb{R}^{d}_{+}}|f(x)|w_{z}(x)dx.$

Hence, the inequality (21) is equivalent to

$\|e^{-zL_{\lambda}}\|_{L^{1}_{w_{z}^{-1}}(\mathbb{R}^{d}_{+})\rightarrow
L^{\infty}_{w_{z}}(\mathbb{R}^{d}_{+})}\leq\frac{C}{|z|^{d/\alpha}}.$

For $z\in\mathbb{C}_{\pi/4}$ we can write $z=2t+is$ for some $t\geq 0$ and
$s\in\mathbb{R}$ such that $t\simeq|z|$. Then,

$\displaystyle\|e^{-zL_{\lambda}}\|_{L^{1}_{w_{z}^{-1}}(\mathbb{R}^{d}_{+})\rightarrow
L^{\infty}_{w_{z}}(\mathbb{R}^{d}_{+})}$
$\displaystyle\quad\leq\|e^{-tL_{\lambda}}\|_{L^{2}(\mathbb{R}^{d}_{+})\rightarrow
L^{\infty}_{w_{z}}(\mathbb{R}^{d}_{+})}\|e^{-isL_{\lambda}}\|_{L^{2}(\mathbb{R}^{d}_{+})\rightarrow
L^{2}(\mathbb{R}^{d}_{+})}\|e^{-tL_{\lambda}}\|_{L^{1}_{{w_{z}}^{-1}}(\mathbb{R}^{d}_{+})\rightarrow
L^{2}(\mathbb{R}^{d}_{+})}.$

We first estimate
$\|e^{-tL_{\lambda}}\|_{L^{1}_{{w_{z}}^{-1}}(\mathbb{R}^{d}_{+})\rightarrow
L^{2}(\mathbb{R}^{d}_{+})}$. For $f\in
L^{1}_{{w_{z}}^{-1}}(\mathbb{R}^{d}_{+})$ we have, by Theorem 2.1 and
$\displaystyle\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}\simeq\Big{(}1\wedge\frac{y_{d}}{|z|^{1/\alpha}}\Big{)}^{\sigma}\simeq
w_{z}(y)^{-1}$,

$\displaystyle\|e^{-tL_{\lambda}}f\|_{L^{2}(\mathbb{R}^{d}_{+})}$
$\displaystyle\lesssim\Big{[}\int_{\mathbb{R}^{d}_{+}}\Big{|}\int_{\mathbb{R}^{d}_{+}}{t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}w_{z}(y)^{-1}|f(y)|}dy\Big{|}^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{2\sigma}\,dx\Big{]}^{\frac{1}{2}}.$

By Minkowski’s inequality,

$\displaystyle\|e^{-tL_{\lambda}}f\|_{L^{2}(\mathbb{R}^{d}_{+})}$
$\displaystyle\lesssim\int_{\mathbb{R}^{d}}\Big{[}\int_{\mathbb{R}^{d}_{+}}\Big{|}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}\Big{|}^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{2\sigma}\,dx\Big{]}^{\frac{1}{2}}w_{z}(y)^{-1}|f(y)|dy$
$\displaystyle\lesssim{\sup_{y\in\mathbb{R}^{d}}}\Big{[}\int_{\mathbb{R}^{d}}\Big{|}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}\Big{|}^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{2\sigma}\,dx\Big{]}^{\frac{1}{2}}\|f\|_{L^{1}_{w_{z}^{-1}}}.$

This, along with the fact that

(22)
${\sup_{y\in\mathbb{R}^{d}}}\Big{[}\int_{\mathbb{R}^{d}}\Big{|}t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}\Big{|}^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{2\sigma}\,dx\Big{]}^{1/2}\lesssim
t^{-\frac{d}{2\alpha}},$

which can be seen by distinguishing between $x_{d}\lessgtr t^{1/\alpha}$ and
using $\sigma>-1/2$, implies

$\|e^{-tL_{\lambda}}f\|_{L^{2}(\mathbb{R}^{d}_{+})}\lesssim
t^{-\frac{d}{2\alpha}}\|f\|_{L^{1}_{w_{z}^{-1}}(\mathbb{R}_{+}^{d})}.$

It follows that

$\|e^{-tL_{\lambda}}\|_{L^{1}_{{w_{z}}^{-1}}(\mathbb{R}^{d}_{+})\rightarrow
L^{2}(\mathbb{R}^{d}_{+})}\lesssim t^{-\frac{d}{2\alpha}}.$

Similarly,

$\|e^{-tL_{\lambda}}\|_{L^{2}(\mathbb{R}^{d}_{+})\to
L^{\infty}_{{w_{z}}}(\mathbb{R}^{d}_{+})}\lesssim t^{-\frac{d}{2\alpha}}.$

Moreover, since $L_{\lambda}$ is a (non-negative) self-adjoint operator,
$\|e^{-isL_{\lambda}}\|_{L^{2}(\mathbb{R}^{d}_{+})\rightarrow
L^{2}(\mathbb{R}^{d}_{+})}\leq 1$. Taking these three estimates into account,
we get

$\|e^{-zL_{\lambda}}\|_{L^{1}_{w_{z}^{-1}}(\mathbb{R}^{d}_{+})\rightarrow
L^{\infty}_{w_{z}}(\mathbb{R}^{d}_{+})}\leq
t^{-\frac{d}{\alpha}}\simeq\frac{1}{|z|^{d/\alpha}},$

which implies (21), i.e.,

$|p_{z}(x,y)|\lesssim\frac{1}{|z|^{d/\alpha}}\Big{(}1\wedge\frac{x_{d}}{|z|^{1/\alpha}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{|z|^{1/\alpha}}\Big{)}^{\sigma}$

for all $x,y\in\mathbb{R}^{d}_{+}$ and $z\in\mathbb{C}_{\pi/4}$. By the
Phragmén–Lindelöf argument in [DR96, Proposition 3.3], we get (19). ∎

For each $k\in\mathbb{N}$, we denote by $p_{t,k}(x,y)$ the kernel of
$(tL_{\lambda})^{k}e^{-tL_{\lambda}}$. As a direct consequence of Proposition
2.2 and Cauchy’s formula, we obtain the following result.

######  Proposition 2.4.

  1. (a)

Let $\alpha\in(0,2)$, $\lambda\geq 0$, and let $\sigma$ be defined by (5).
Then for any $\epsilon>0$ and $k\in\mathbb{N}$, there exists a constant
$C=C(k,\epsilon)$ such that

(23) $|p_{t,k}(x,y)|\leq
C\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}t^{-(k+d/\alpha)}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha+\epsilon}$

for all $x,y\in\mathbb{R}^{d}_{+}$ and $t>0$.

  2. (b)

Let $\alpha=2$ and let $\sigma$ be defined by (5). Then for each
$k\in\mathbb{N}$ there exist constants $C,c>0$ such that

(24) $|p_{t,k}(x,y)|\leq
C\Big{(}1\wedge\frac{x_{d}}{\sqrt{t}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{\sqrt{t}}\Big{)}^{\sigma}t^{-(k+d/2)}\exp\Big{(}-\frac{|x-y|^{2}}{ct}\Big{)}$

for all $x,y\in\mathbb{R}^{d}_{+}$ and $t>0$.

######  Remark 2.5.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then (23) remains valid. This follows by the same arguments as
in the proof below, taking into account Remark 2.3.

###### Proof.

Applying Cauchy’s formula, for every $t>0$ and $k\in\mathbb{N}$,

$L_{\lambda}^{k}e^{-tL_{\lambda}}=\frac{(-1)^{k}k!}{2\pi i}\int_{|\xi-t|=\eta
t}e^{-\xi L_{\lambda}}\frac{d\xi}{(\xi-t)^{k}},$

where $\eta>0$ is small enough so that $\\{\xi:|\xi-t|=\eta
t\\}\subset\mathbb{C}_{\tilde{\epsilon}\pi/4}$ with
$\tilde{\epsilon}=\frac{\epsilon}{d+\alpha}$, and the integral does not depend
on the choice of $\eta$. Thus, by Proposition 2.2 and
$|\xi|\simeq|\xi-t|\simeq t$,

$\displaystyle|p_{t,k}(x,y)|$ $\displaystyle\leq
C_{k,\tilde{\epsilon}}\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}t^{-(k+d/\alpha)}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-(d+\alpha)(1-\tilde{\epsilon})}$
$\displaystyle\leq
C_{k,\epsilon}\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}t^{-(k+d/\alpha)}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-(d+\alpha-\epsilon)}$

holds for all $x,y\in\mathbb{R}^{d}$ and $t>0$. ∎

###  2.2. Spatially averaged estimates

In this section, we use the pointwise estimates from the previous section and
tools from singular integral theory to prove spatially weighted estimates
involving the heat kernel bounds of $L_{\lambda}$. We will use them to prove
the square function estimates in Section 3 below. To do so, we introduce
further notation. We write $\|f\|_{p}:=\|f\|_{L^{p}(\mathbb{R}_{+}^{d})}$ and
denote, for $\theta<1/2$,

$p_{\theta}=\begin{cases}\frac{1}{\theta},\ \ &\text{if $\theta>0$},\\\
\infty,\ \ &\text{if $\theta\leq 0$}.\end{cases}$

We denote the average of a measurable function $f$ over a measurable set
$E\subseteq\mathbb{R}^{N}$ with $N$-dimensional Lebesgue measure
$|E|\in(0,\infty)$ by

$\fint_{E}f(x)dx=\frac{1}{|E|}\int_{E}f(x)dx.$

For any ball $B\subseteq\mathbb{R}^{d}$, we associate annuli
$S_{j}(B):=2^{j}B\backslash 2^{j-1}B$, $j\in\mathbb{N}$, and write
$S_{0}(B)=B$.

######  Theorem 2.6.

Let $\\{T_{t}\\}_{t>0}$ be a family of linear integral operators on
$L^{2}(\mathbb{R}^{d}_{+})$ with their associated kernels $T_{t}(x,y)$. Assume
that there exist $0<\beta<\alpha$ and $\theta<1/2$ such that for all $t>0$ and
$x,y\in\mathbb{R}^{d}_{+}$,

(25) $|T_{t}(x,y)|\leq
t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}.$

Assume that $(p_{\theta})^{\prime}<p\leq q<p_{\theta}$. Then for any ball $B$
with radius $r_{B}$, for every $t>0$ and $j\in\mathbb{N}$, we have

(26)
$\Big{(}\fint_{S_{j}(B)}|T_{t}f|^{q}\Big{)}^{\frac{1}{q}}\leq\max\Big{\\{}\Big{(}\frac{r_{B}}{t^{\frac{1}{\alpha}}}\Big{)}^{d-\frac{1}{p^{\prime}}},\Big{(}\frac{r_{B}}{t^{1/\alpha}}\Big{)}^{d}\Big{\\}}\Big{(}1+\frac{t^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{\frac{1}{q}}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{\frac{1}{p}}$

for all $f\in L^{p}(\mathbb{R}^{d}_{+})$ supported in $B$, and

(27)
$\Big{(}\fint_{B}|T_{t}f|^{q}\Big{)}^{\frac{1}{q}}\leq\max\Big{\\{}\Big{(}\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{d-\frac{1}{p^{\prime}}},\Big{(}\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{d}\Big{\\}}\Big{(}1+\frac{t^{1/\alpha}}{r_{B}}\Big{)}^{\frac{1}{q}}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{S_{j}(B)}|f|^{p}\Big{)}^{\frac{1}{p}}$

for all $f\in L^{p}(\mathbb{R}_{+}^{d})$ supported in $S_{j}(B)$.

###### Proof.

For $\theta\leq 0$, the proof is immediate. The following more elaborate
arguments cover this case nevertheless. In the following, we only prove (26);
the proof of (27) is analogous. Let $B=B(x_{B},r_{B})$ be a ball in
$\mathbb{R}^{d}_{+}$ with $x_{B}=(x_{B}^{\prime},x_{B}^{d})$ with
$x_{B}^{\prime}\in\mathbb{R}^{d-1}$ and $x_{B}^{d}\in(0,\infty)$. We write

$B^{\prime}:=\\{y^{\prime}\in\mathbb{R}^{d-1}:\|y^{\prime}-x_{B}^{\prime}\|<r_{B}\\}.$

With the abbreviation

$D_{\theta}(x_{d},t):=\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\quad\text{for
all}\ x_{d},t>0,\ \theta\in\mathbb{R},$

we have

$\displaystyle\|T_{t}f\|_{L^{q}(S_{j}(B))}$
$\displaystyle\leq\left\\{\int_{S_{j}(B)}\Big{[}\int_{B}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}}$
$\displaystyle\lesssim I_{1}+I_{2}+I_{3}+I_{4},$

where

$\displaystyle I_{1}$ $\displaystyle=\left\\{\int_{S_{j}(B)\cap\\{x_{d}\leq
t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{B\cap\\{y_{d}\leq
t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}},$
$\displaystyle I_{2}$ $\displaystyle=\left\\{\int_{S_{j}(B)\cap\\{x_{d}\leq
t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{B\cap\\{y_{d}>t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}},$
$\displaystyle I_{3}$
$\displaystyle=\left\\{\int_{S_{j}(B)\cap\\{x_{d}>t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{B\cap\\{y_{d}\leq
t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}},$
and $\displaystyle I_{4}$
$\displaystyle=\left\\{\int_{S_{j}(B)\cap\\{x_{d}>t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{B\cap\\{y_{d}>t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}}.$

We first consider $I_{1}$. Here, we have

$D_{\theta}(x_{d},t)=\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta},\qquad
D_{\theta}(y_{d},t)=\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}.$

By Hölder’s inequality,

$\displaystyle I_{1}$ $\displaystyle\leq
t^{-d/\alpha}\Big{(}1+\frac{d(B,S_{j}(B))}{t^{1/\alpha}}\Big{)}^{-d-\beta}\|f\|_{p}$
$\displaystyle\ \ \ \times\Big{[}\int_{S_{j}(B)\cap\\{x_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
q}dx\Big{]}^{1/q}\Big{[}\int_{B\cap\\{y_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
p^{\prime}}dy\Big{]}^{1/p^{\prime}}.$

Using $d(B,S_{j}(B))\simeq 2^{j}r_{B}$,

(28) $\displaystyle\Big{[}\int_{S_{j}(B)\cap\\{x_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
q}dx\Big{]}^{1/q}$
$\displaystyle\leq\Big{[}\int_{2^{j}B^{\prime}}\int_{0}^{t^{1/\alpha}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
q}dx_{d}dx^{\prime}\Big{]}^{1/q}$ $\displaystyle\lesssim t^{\frac{1}{\alpha
q}}|2^{j}B^{\prime}|^{1/q}.$

since $\theta q<1$, and

(29) $\displaystyle\Big{[}\int_{B\cap\\{y_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
p^{\prime}}dy\Big{]}^{1/p^{\prime}}$ $\displaystyle\lesssim t^{\frac{1}{\alpha
p^{\prime}}}|B^{\prime}|^{1/p^{\prime}},$

we get, for some $c>0$,

(30) $\displaystyle|S_{j}(B)|^{-1/q}\times I_{1}$ $\displaystyle\lesssim
t^{-\frac{d}{\alpha}+\frac{1}{\alpha q}+\frac{1}{\alpha
p^{\prime}}}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}|2^{j}B|^{-1/q}|2^{j}B^{\prime}|^{1/q}|B^{\prime}|^{1/p^{\prime}}\|f\|_{L^{p}(B)}$
$\displaystyle=ct^{-\frac{d}{\alpha}+\frac{1}{\alpha q}+\frac{1}{\alpha
p^{\prime}}}(2^{j}r_{B})^{-1/q}r_{B}^{d-1/p^{\prime}}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}$
$\displaystyle\lesssim{\Big{(}\frac{r_{B}}{t^{1/\alpha}}\Big{)}^{d-1/p^{\prime}}\Big{(}1+\frac{t^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{1/q}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}}.$

This concludes the estimate for $I_{1}$. We now bound $I_{2}$. Here, we have

$D_{\theta}(x_{d},t)=\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta},\quad
D_{\theta}(y_{d},t)=1.$

Hence, using Hölder’s inequality and (28), we have

$\displaystyle I_{2}$
$\displaystyle\lesssim\left\\{\int_{S_{j}(B)\cap\\{x_{d}\leq
t^{1/\alpha}\\}}\Big{[}\int_{B\cap\\{y_{d}>t^{1/\alpha}\\}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}|f(y)|dy\Big{]}^{q}dx\right\\}^{1/q}$
$\displaystyle\lesssim{t^{-\frac{d}{\alpha}}}\Big{(}1+\frac{d(B,S_{j}(B))}{t^{1/\alpha}}\Big{)}^{-d-\beta}\|f\|_{L^{1}(B)}\Big{(}\int_{S_{j}(B)\cap\\{x_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
q}dx\Big{)}^{1/q}$ $\displaystyle\lesssim t^{-\frac{d}{\alpha}+\frac{1}{\alpha
q}}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}|2^{j}B^{\prime}|^{1/q}|B|\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}.$

It follows that

(31) $\displaystyle|S_{j}(B)|^{-1/q}\times I_{2}$ $\displaystyle\lesssim
t^{-\frac{d}{\alpha}+\frac{1}{\alpha
q}}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}|2^{j}B|^{-1/q}|2^{j}B^{\prime}|^{1/q}|B|\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}$
$\displaystyle\lesssim{\Big{(}\frac{r_{B}}{t^{1/\alpha}}\Big{)}^{d}\Big{(}1+\frac{t^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{1/q}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}}.$

Considering the term $I_{3}$, we have

$D_{\theta}(x_{d},t)=1,\ \ \
D_{\theta}(y_{d},t)=\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}.$

Hence,

$\displaystyle I_{3}$
$\displaystyle\lesssim\left\\{\int_{S_{j}(B)\cap\\{x_{d}>t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{B\cap\\{y_{d}\leq
t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}}$
$\displaystyle\lesssim
t^{-\frac{d}{\alpha}}\Big{(}1+\frac{d(B,S_{j}(B))}{t^{1/\alpha}}\Big{)}^{-d-\beta}|2^{j}B|^{1/q}\int_{B\cap\\{y_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}|f(y)|dy.$

By Hölder’s inequality and (29),

$\displaystyle\int_{B\cap\\{y_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}|f(y)|dy$
$\displaystyle\lesssim\Big{(}\int_{B\cap\\{y_{d}\leq
t^{1/\alpha}\\}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
p^{\prime}}dy\Big{)}^{1/p^{\prime}}\|f\|_{L^{p}(B)}$ $\displaystyle\lesssim
t^{\frac{1}{p^{\prime}\alpha}}|B^{\prime}|^{1/p^{\prime}}\|f\|_{L^{p}(B)}.$

Consequently,

(32) $\displaystyle|S_{j}(B)|^{-1/q}\times I_{3}$ $\displaystyle\lesssim\
t^{-\frac{d}{\alpha}+\frac{1}{\alpha
p^{\prime}}}|B^{\prime}|^{1/p^{\prime}}|B|^{1/p}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}$
$\displaystyle{\simeq}\
\Big{(}\frac{r_{B}}{t^{1/\alpha}}\Big{)}^{d-1/p^{\prime}}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}{.}$

It remains to estimate the term $I_{4}$. Since in this case

$D_{\theta}(x_{d},t)=D_{\theta}(y_{d},t)=1,$

we have,

$\displaystyle I_{4}$
$\displaystyle\lesssim\left\\{\int_{S_{j}(B)}\Big{[}\int_{B}t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}|f(y)|dy\Big{]}^{q}dx\right\\}^{1/q}$
$\displaystyle\lesssim
t^{-d/\alpha}\Big{(}1+\frac{d(B,S_{j}(B))}{t^{1/\alpha}}\Big{)}^{-d-\beta}|S_{j}(B)|^{1/q}\|f\|_{L^{1}(B)}.$

Thus, by Hölder’s inequality, we obtain

(33) $\displaystyle|S_{j}(B)|^{-1/q}\times I_{4}$ $\displaystyle\lesssim
t^{-\frac{d}{\alpha}}|B|\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}$
$\displaystyle\lesssim\Big{(}\frac{r_{B}}{t^{1/\alpha}}\Big{)}^{d}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{p}\Big{)}^{1/p}.$

This completes the proof of (26). ∎

In the following, we denote the operator norm of a bounded linear operator
$T:L^{p}(\mathbb{R}_{+}^{d})\to L^{q}(\mathbb{R}_{+}^{d})$ by $\|T\|_{p\to
q}$.

######  Lemma 2.7.

Let $\theta<1/2$ and $\\{T_{t}\\}_{t>0}$ be as in Theorem 2.6. Then for
$(p_{\theta})^{\prime}<p\leq q<p_{\theta}$ we have

$\|T_{t}\|_{p\to q}\lesssim t^{-\frac{d}{\alpha}(\frac{1}{p}-\frac{1}{q})}$

for all $t>0$.

###### Proof.

For $\theta\leq 0$, we use

$\displaystyle|T_{t}(x,y)|\leq
Ct^{-\frac{d}{\alpha}}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}$

and that, for $1\leq p\leq q\leq\infty$,

(34)
$\displaystyle\begin{split}\left\|\int_{\mathbb{R}^{N}}t^{-\frac{N}{\alpha}}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-N-\beta}f(y)\,dy\right\|_{L^{q}(\mathbb{R}^{N},dx)}\lesssim
t^{-\frac{N}{\alpha}(\frac{1}{p}-\frac{1}{q})}\|f\|_{L^{p}(\mathbb{R}^{N})}\end{split}$

holds for all $N\in\mathbb{N}$ and $t>0$. This concludes the case $\theta\leq
0$.

Thus, from now on, we consider $\theta\in(0,1/2)$. We use the same notations
as in the proof of Theorem 2.6. Similarly as in that proof, we write

$\displaystyle\|T_{t}f\|_{q}$
$\displaystyle\leq\left\\{\int_{\mathbb{R}^{d}_{+}}\Big{[}\int_{\mathbb{R}^{d}_{+}}t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\alpha}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{1/q}$
$\displaystyle\lesssim I_{1}+I_{2}+I_{3}+I_{4},$

where

$\displaystyle I_{1}$
$\displaystyle=\left\\{\int_{\mathbb{R}^{d}_{+}\cap\\{x_{d}\leq
t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{\mathbb{R}^{d}_{+}\cap\\{y_{d}\leq
t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}},$
$\displaystyle I_{2}$
$\displaystyle=\left\\{\int_{\mathbb{R}^{d}_{+}\cap\\{x_{d}\leq
t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{\mathbb{R}^{d}_{+}\cap\\{y_{d}>t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}},$
$\displaystyle I_{3}$
$\displaystyle=\left\\{\int_{\mathbb{R}^{d}_{+}\cap\\{x_{d}>t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{\mathbb{R}^{d}_{+}\cap\\{y_{d}\leq
t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}},$
and $\displaystyle I_{4}$
$\displaystyle=\left\\{\int_{\mathbb{R}^{d}_{+}\cap\\{x_{d}>t^{\frac{1}{\alpha}}\\}}\Big{[}\int_{\mathbb{R}^{d}_{+}\cap\\{y_{d}>t^{\frac{1}{\alpha}}\\}}t^{-\frac{d}{\alpha}}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x-y|}{t^{1/\alpha}}\Big{)}^{-d-\beta}D_{\theta}(x_{d},t)D_{\theta}(y_{d},t)|f(y)|dy\Big{]}^{q}dx\right\\}^{\frac{1}{q}}.$

We only estimate $I_{1}$ and $I_{2}$ since $I_{3}$ and $I_{4}$ can be treated
similarly as the corresponding terms in the proof of Theorem 2.6 with minor
modifications. To estimate $I_{1}$, we note

$D_{\theta}(x_{d},t)=\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta},\quad
D_{\theta}(y_{d},t)=\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}.$

Then, applying (34) with $N=d-1$ and Minkowski’s inequality,

$\displaystyle I_{1}$ $\displaystyle\lesssim
t^{-\frac{d-1}{\alpha}(\frac{1}{p}-\frac{1}{q})}\left\\{\int_{0}^{t^{1/\alpha}}\Big{[}\int_{0}^{t^{1/\alpha}}t^{-1/\alpha}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\|f(\cdot,y_{d})\|_{L^{p}(\mathbb{R}^{d-1})}dy_{d}\Big{]}^{q}dx_{d}\right\\}^{1/q}$
$\displaystyle\lesssim
t^{-\frac{d-1}{\alpha}(\frac{1}{p}-\frac{1}{q})-\frac{1}{\alpha}}\Big{[}\int_{0}^{t^{1/\alpha}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
q}dx_{d}\Big{]}^{1/q}\int_{0}^{t^{1/\alpha}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\|f(\cdot,y_{d})\|_{L^{p}(\mathbb{R}^{d-1})}dy_{d}$
$\displaystyle\lesssim
t^{-\frac{d-1}{\alpha}(\frac{1}{p}-\frac{1}{q})-\frac{1}{\alpha}}t^{\frac{1}{\alpha
q}}\int_{0}^{t^{1/\alpha}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\|f(\cdot,y_{d})\|_{L^{p}(\mathbb{R}^{d-1})}dy_{d}.$

By Hölder’s inequality,

$\displaystyle\int_{0}^{t^{\frac{1}{\alpha}}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\|f(\cdot,y_{d})\|_{L^{p}(\mathbb{R}^{d-1})}dy_{d}$
$\displaystyle\lesssim\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}\Big{[}\int_{0}^{t^{\frac{1}{\alpha}}}\Big{(}\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
p^{\prime}}dy_{d}\Big{]}^{\frac{1}{p^{\prime}}}\lesssim t^{\frac{1}{\alpha
p^{\prime}}}\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}.$

Therefore,

$\displaystyle I_{1}\lesssim
t^{-\frac{d}{\alpha}(\frac{1}{p}-\frac{1}{q})}\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}.$

We now consider $I_{2}$. Here, we have

$D_{\theta}(x_{d},t)=\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta},\quad
D_{\theta}(y_{d},t)=1.$

By (34) with $N=d-1$, we have, for any $\beta^{\prime}\in(0,\beta)$,

$\displaystyle I_{2}$ $\displaystyle\lesssim
t^{-\frac{d-1}{\alpha}(\frac{1}{p}-\frac{1}{q})}\left\\{\int_{0}^{t^{1/\alpha}}\Big{[}\int_{0}^{t^{1/\alpha}}t^{-\frac{1}{\alpha}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta}\Big{(}\frac{t^{\frac{1}{\alpha}}+|x_{d}-y_{d}|}{t^{1/\alpha}}\Big{)}^{-1-\beta^{\prime}}\|f(\cdot,y_{d})\|_{L^{p}(\mathbb{R}^{d-1})}dy_{d}\Big{]}^{q}dx_{d}\right\\}^{\frac{1}{q}}.$

By Hölder’s inequality and (34) with $N=1$,

$\displaystyle\int_{0}^{t^{1/\alpha}}\Big{(}\frac{t^{1/\alpha}+|x_{d}-y_{d}|}{t^{1/\alpha}}\Big{)}^{-1-\beta^{\prime}}\|f(\cdot,y_{d})\|_{L^{p}(\mathbb{R}^{d-1})}dy_{d}$
$\displaystyle\quad\lesssim\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}\Big{[}\int_{0}^{\infty}\Big{(}\frac{t^{1/\alpha}+|x_{d}-y_{d}|}{t^{1/\alpha}}\Big{)}^{-(1+\beta^{\prime})p^{\prime}}dy_{d}\Big{]}^{1/p^{\prime}}\lesssim
t^{\frac{1}{\alpha p^{\prime}}}\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}.$

Therefore,

$\displaystyle I_{2}$ $\displaystyle\lesssim
t^{-\frac{d-1}{\alpha}(\frac{1}{p}-\frac{1}{q})-\frac{1}{\alpha}+\frac{1}{\alpha
p^{\prime}}}\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}\Big{[}\int_{0}^{t^{1/\alpha}}t^{-\frac{1}{\alpha}}\Big{(}\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{-\theta
q}dx_{d}\Big{]}^{1/q}$ $\displaystyle\lesssim
t^{-\frac{d-1}{\alpha}(\frac{1}{p}-\frac{1}{q})-\frac{1}{\alpha}+\frac{1}{\alpha
p^{\prime}}+\frac{1}{\alpha q}}\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}\lesssim
t^{-\frac{d}{\alpha}(\frac{1}{p}-\frac{1}{q})}\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}.$

Similarly, we also have

$I_{3}+I_{4}\lesssim
t^{-\frac{d}{\alpha}(\frac{1}{p}-\frac{1}{q})}\|f\|_{L^{p}(\mathbb{R}^{d}_{+})}.$

This completes our proof. ∎

The following two lemmas will be important to prove Theorem 1.10.

######  Lemma 2.8.

Let $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha<2$ and
$\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by (5). Let
$\gamma\in(0,1]$. Then for $1<p<\infty$ we have

(35) $\displaystyle\|L_{\lambda}^{\gamma}e^{-tL_{\lambda}}\|_{p\to p}\lesssim
t^{-\gamma},$

and

(36)
$\displaystyle\|L_{\lambda}^{-\gamma}\big{[}e^{-tL_{\lambda}}-e^{-sL_{\lambda}}\big{]}\|_{p\to
p}\lesssim(t^{\gamma}-s^{\gamma})$

for all $t\geq s>0$.

######  Remark 2.9.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then Lemma 2.8 remains valid for $p\in(1,\infty)$ obeying
$\frac{1}{1+\sigma}=(p_{-\sigma})^{\prime}<p<p_{-\sigma}=-\frac{1}{\sigma}$.
This follows by the same arguments as in the proof below, taking into account
Remark 2.5.

###### Proof.

Since

$L_{\lambda}^{\gamma}=\frac{1}{\Gamma(1-\gamma)}\int_{0}^{\infty}u^{1-\gamma}L_{\lambda}e^{-uL_{\lambda}}\frac{du}{u},$

we get

$L_{\lambda}^{\gamma}e^{-tL_{\lambda}}=\frac{1}{\Gamma(1-\gamma)}\int_{0}^{\infty}\frac{u^{1-\gamma}}{u+t}(u+t)L_{\lambda}e^{-(u+t)L_{\lambda}}\frac{du}{u}.$

From Lemma 2.7 and Proposition 2.4,
$\|(u+v)L_{\lambda}e^{-(u+v)L_{\lambda}}\|_{p\to p}\lesssim 1$. Therefore,

$\displaystyle\|L_{\lambda}^{\gamma}e^{-tL_{\lambda}}\|_{p\to p}$
$\displaystyle\lesssim\int_{0}^{\infty}\frac{u^{1-\gamma}}{u+t}\frac{du}{u}\lesssim
t^{-\gamma}.$

This shows (35). To show (36), we use, for $\gamma\in(0,1)$,

$L_{\lambda}^{-\gamma}=\frac{1}{\Gamma(\gamma)}\int_{0}^{\infty}u^{\gamma}e^{-uL_{\lambda}}\frac{du}{u}$

and Duhamel’s formula

$e^{-tL_{\lambda}}-e^{-sL_{\lambda}}=-\int_{s}^{t}vL_{\lambda}e^{-vL_{\lambda}}\frac{dv}{v}$

to write

$L_{\lambda}^{-\gamma}\big{[}e^{-tL_{\lambda}}-e^{-sL_{\lambda}}\big{]}=-\frac{1}{\Gamma(\gamma)}\int_{s}^{t}\int_{0}^{\infty}\frac{vu^{\gamma}}{u+v}(u+v)L_{\lambda}e^{-(u+v)L_{\lambda}}\frac{du}{u}\frac{dv}{v}.$

This, together with $\|(u+v)L_{\lambda}e^{-(u+v)L_{\lambda}}\|_{p\to
p}\lesssim 1$, implies

$\displaystyle\big{\|}L_{\lambda}^{-\gamma}\big{[}e^{-tL_{\lambda}}-e^{-sL_{\lambda}}\big{]}\big{\|}_{p\to
p}$
$\displaystyle\lesssim\int_{s}^{t}\int_{0}^{\infty}\frac{vu^{\gamma}}{u+v}\frac{du}{u}\frac{dv}{v}$
$\displaystyle\lesssim\int_{s}^{t}\int_{0}^{\infty}\frac{vu^{\gamma}}{u+v}\frac{du}{u}\frac{dv}{v}$
$\displaystyle\lesssim\int_{s}^{t}v^{\gamma}\frac{dv}{v}\simeq
t^{\gamma}-s^{\gamma}.$

This concludes the proof. ∎

######  Lemma 2.10.

Let $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha<2$ and
$\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by (5). Then for
$f\in L^{p}(\mathbb{R}^{d}_{+}),1<p<\infty$, we have

  1. (a)

$\displaystyle\lim_{t\to 0}\|(I-e^{-tL_{\lambda}})f\|_{p}=0$ and

  2. (b)

$\displaystyle\lim_{t\to\infty}\|e^{-tL_{\lambda}}f\|_{p}=0$.

######  Remark 2.11.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then Lemma 2.10 remains valid for $p\in(1,\infty)$ obeying
$\frac{1}{1+\sigma}=(p_{-\sigma})^{\prime}<p<p_{-\sigma}=-\frac{1}{\sigma}$.
This follows by the same arguments as in the proof below, taking into account
Remark 2.5.

###### Proof.

(a) It suffices to prove for $p\leq 2$ since the case $p>2$ can be obtained by
duality. Since $\|e^{-tL_{\lambda}}\|_{p\to p}\lesssim 1$ for all $t$ and
$C^{\infty}_{c}(\mathbb{R}^{d}_{+})$ is dense in $L^{p}$, it suffices to prove
the lemma for $f\in C^{\infty}_{c}(\mathbb{R}^{d}_{+})$. Assume that
$\operatorname{supp}f\subset B$ for some ball $B\subset\mathbb{R}^{d}_{+}$. By

$I-e^{-tL_{\lambda}}=-\int_{0}^{t}L_{\lambda}e^{-sL_{\lambda}}ds,$

we have

$\|(I-e^{-tL_{\lambda}})f\|_{p}\leq\Big{\|}\int_{0}^{t}L_{\lambda}e^{-sL_{\lambda}}fds\Big{\|}_{L^{p}(4B)}+\Big{\|}\int_{0}^{t}L_{\lambda}e^{-sL_{\lambda}}fds\Big{\|}_{L^{p}(\mathbb{R}^{d}_{+}\backslash
4B)}.$

By Hölder’s inequality,

$\displaystyle\Big{\|}\int_{0}^{t}L_{\lambda}e^{-sL_{\lambda}}fds\Big{\|}_{L^{p}(4B)}\leq|4B|^{\frac{2-p}{2p}}\Big{\|}\int_{0}^{t}L_{\lambda}e^{-sL_{\lambda}}fds\Big{\|}_{L^{2}(4B)}.$

Applying Minkowski’s inequality,

$\displaystyle\|(I-e^{-tL_{\lambda}})f\|_{L^{p}(4B)}$
$\displaystyle\leq|4B|^{\frac{2-p}{2p}}\Big{\|}\int_{0}^{t}e^{-sL_{\lambda}}L_{\lambda}fds\Big{\|}_{L^{2}(4B)}\leq
t|4B|^{\frac{2-p}{2p}}\|L_{\lambda}f\|_{2}.$

Since $L_{\lambda}f\in L^{2}(\mathbb{R}^{d}_{+})$ by [FM23, Lemma 15], we have

$\lim_{t\to 0}\|(I-e^{-tL_{\lambda}})f\|_{L^{p}(4B)}=0.$

For the second term we write

$\displaystyle\Big{\|}\int_{0}^{t}L_{\lambda}e^{-sL_{\lambda}}fds\Big{\|}_{L^{p}(\mathbb{R}^{d}_{+}\backslash
4B)}$
$\displaystyle\leq\int_{0}^{t}\|sL_{\lambda}e^{-sL_{\lambda}}f\|_{L^{p}(\mathbb{R}^{d}_{+}\backslash
4B)}\frac{ds}{s}$ $\displaystyle\leq\int_{0}^{t}\sum_{j\geq
2}\|sL_{\lambda}e^{-sL_{\lambda}}f\|_{L^{p}(S_{j}(B))}\frac{ds}{s}.$

Then by Proposition 2.4 and Theorem 2.6, we have for $t<r_{B}^{\alpha}$,

$\displaystyle\sum_{j\geq
2}\|L_{\lambda}e^{-sL_{\lambda}}f\|_{L^{p}(S_{j}(B))}$
$\displaystyle\leq\sum_{j\geq
2}|2^{j}B|^{1/p}\Big{(}\frac{r_{B}}{s^{1/\alpha}}\Big{)}^{d}\Big{(}\frac{2^{j}r_{B}}{s^{1/\alpha}}\Big{)}^{-d-\beta}\Big{(}\fint_{B}|f|^{r}\Big{)}^{1/r}$
$\displaystyle\leq C(|B|,p,r,\|f\|_{r})s^{\beta/\alpha}.$

It follows that

$\displaystyle\Big{\|}\int_{0}^{t}L_{\lambda}e^{-sL_{\lambda}}fds\Big{\|}_{L^{p}(\mathbb{R}^{d}_{+}\backslash\int_{0}^{t}4B)}$
$\displaystyle\lesssim\int_{0}^{t}s^{\beta/\alpha}\frac{ds}{s}\simeq
t^{\beta/\alpha}\to 0\ \ \text{as $t\to 0$}.$

(b) Similarly as in (a), we assume $f\in C^{\infty}_{c}(\mathbb{R}^{d}_{+})$
without loss of generality. Fix $1<r<p$. Then,

$\|e^{-tL_{\lambda}}f\|_{p}\lesssim
t^{-\frac{d}{\alpha}(\frac{1}{r}-\frac{1}{p})}\|f\|_{r},$

which implies

$\displaystyle\lim_{t\to\infty}\|e^{-tL_{\lambda}}f\|_{p}=0.$

This completes our proof. ∎

##  3\. Proof of the square function estimates (Theorem 1.4)

For $\gamma>0$, recall the square function

$\displaystyle(S_{L_{\lambda},\gamma}f)(x):=\Big{(}\int_{0}^{\infty}|(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}f|^{2}\frac{dt}{t}\Big{)}^{1/2}.$

We recall two criteria for singular integrals to be bounded on Lebesgue
spaces. These will play an important role in the proof of the boundedness of
the square functions. The first theorem gives a criterion on the boundedness
on $L^{p}(\mathbb{R}^{d})$ spaces with $p\in(1,2)$, while the second one
covers the range $p>2$. The latter also involves, for $r>0$, the
Hardy–Littlewood maximal operator $\mathcal{M}_{r}$, which acts as

$(\mathcal{M}_{r}f)(x):=\sup_{B\ni
x}\Big{(}\frac{1}{|B|}\int_{B}|f(y)|^{r}\,dy\Big{)}^{1/r},\quad
x\in\mathbb{R}^{d}.$

Here the supremum is taken over all balls $B\subseteq\mathbb{R}^{d}$
containing $x$.

######  Theorem 3.1 ([Aus07, Theorem 1.1]).

Let $1\leq p_{0}<2$ and let $T$ be a sublinear operator which is bounded on
$L^{2}(\mathbb{R}^{d})$. Assume that there exists a family of operators
$\\{\mathcal{A}_{t}\\}_{t>0}$ satisfying that for $j\geq 2$ and every ball
$B\subseteq\mathbb{R}^{d}$

(37)
$\Big{(}\fint_{S_{j}(B)}|T(I-\mathcal{A}_{r_{B}})f|^{2}\Big{)}^{1/2}\leq\alpha(j)\Big{(}\fint_{B}|f|^{p_{0}}\Big{)}^{1/p_{0}},$

and

(38)
$\Big{(}\fint_{S_{j}(B)}|\mathcal{A}_{r_{B}}f|^{2}\Big{)}^{1/2}\leq\alpha(j)\Big{(}\fint_{B}|f|^{p_{0}}\Big{)}^{1/p_{0}},$

for all $f$ supported in $B$. If $\sum_{j\geq 2}\alpha(j)2^{jd}<\infty$, then
$T$ is bounded on $L^{p}(\mathbb{R}^{d})$ for all $p\in(p_{0},2)$.

######  Theorem 3.2 ([Aus07, Theorem 1.2]).

Let $2<q_{0}\leq\infty.$ Let $T$ be a bounded sublinear operator on
$L^{2}(\mathbb{R}^{d})$. Assume that there exists a family of operators
$\\{\mathcal{A}_{t}\\}_{t>0}$ satisfying that

(39)
$\displaystyle\Big{(}\fint_{B}\big{|}T(I-\mathcal{A}_{r_{B}})f\big{|}^{2}dx\Big{)}^{1/2}\leq
C\mathcal{M}_{2}(f)(x),$

and

(40)
$\displaystyle\Big{(}\fint_{B}\big{|}T\mathcal{A}_{r_{B}}f\big{|}^{q_{0}}dx\Big{)}^{1/q_{0}}\leq
C\mathcal{M}_{2}(Tf)(x),$

for all balls $B\subseteq\mathbb{R}^{d}$ with radius $r_{B}$, all $f\in
C^{\infty}_{c}(\mathbb{R}^{d})$ and all $x\in B$. Then $T$ is bounded on
$L^{p}(\mathbb{R}^{d})$ for all $2<p<q_{0}$.

See [Aus07, p. 2] for references, where these theorems were originally proved.
Using Theorems 3.1–3.2 with the $T=S_{L_{\lambda},\gamma}$ on the left-hand
side of (9), we give the

###### Proof of Theorem 1.4.

We begin with some reductions. First, Formula (11) immediately follows from
(10). Since the statement for $p=2$ follows from the spectral theorem, it
suffices to consider $p\in(1,\infty)\setminus\\{2\\}$. By a duality argument,
the lower bound in (10) follows from the upper bound; see the end of this
proof for an argument. Thus, it suffices to prove the upper bound. To that
end, we fix $\beta\in(0,\alpha)$ such that $\beta/\alpha>1-\gamma$. We now
distinguish between $p<2$ and $p>2$.

Step 1: The proof of the $L^{p}$-boundedness for $p<2$

Fix $1<p<2$. We use Theorem 3.1 with $T=S_{L_{\lambda},\gamma}$ and

$\mathcal{A}_{r_{B}}=I-(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m},\ \ \
m>\beta+1.$

Thus, it suffices to prove

(41)
$\Big{(}\fint_{S_{j}(B)}|S_{L_{\lambda},\gamma}(I-\mathcal{A}_{r_{B}})f(x)|^{2}dx\Big{)}^{1/2}\lesssim
2^{-(d+\beta)j}\Big{(}\fint_{B}|f(x)|^{p}dx\Big{)}^{1/p}$

and

(42)
$\Big{(}\fint_{S_{j}(B)}|\mathcal{A}_{r_{B}}f(x)|^{2}dx\Big{)}^{1/2}\lesssim
2^{-(d+\beta)j}\Big{(}\fint_{B}|f(x)|^{p}dx\Big{)}^{1/p}$

for $j\geq 2$ and for every function $f$ supported in $B$. Since

$\mathcal{A}_{r_{B}}=\sum_{k=1}^{m}C^{m}_{k}e^{-kr_{B}^{\alpha}L_{\lambda}}$

for combinatorial factors $C_{k}^{m}\in\mathbb{R}$ with $C_{0}^{m}=1$, the
estimate (42) follows directly from Theorem 2.1 and Theorem 2.6. It remains to
prove (41). To do this, we bound

(43) $\displaystyle\Big{(}\fint_{S_{j}(B)}|S_{L_{\lambda},\gamma}$
$\displaystyle(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f|^{2}dx\Big{)}^{1/2}$
$\displaystyle\lesssim\Big{(}\int_{0}^{r_{B}^{\alpha}}\left\|(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f\right\|_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\quad+\Big{(}\int_{r_{B}^{\alpha}}^{\infty}\left\|(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f\right\|_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle=:E_{1}+E_{2}.$

We first estimate $E_{1}$. Using

$L_{\lambda}^{\gamma}=\frac{1}{\Gamma(1-\gamma)}\int_{0}^{\infty}u^{1-\gamma}L_{\lambda}e^{-uL_{\lambda}}\frac{du}{u}.$

Using this and Minkowski’s inequality, we have

$\displaystyle E_{1}$
$\displaystyle\lesssim\Big{(}\int_{0}^{r_{B}^{\alpha}}\Big{[}\int_{0}^{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\left\|tL_{\lambda}e^{-(t+u)L_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f\right\|_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\ \
+\Big{(}\int_{0}^{r_{B}^{\alpha}}\Big{[}\int_{r_{B}^{\alpha}}^{\infty}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\Big{\|}tL_{\lambda}e^{-(t+u)L_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f\Big{\|}_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle=:E_{11}+E_{12}.$

Using the identity

$(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}=\sum_{k=0}^{m}(-1)^{k}C^{m}_{k}e^{-kr_{B}^{\alpha}L_{\lambda}},$

we have

$\displaystyle
E_{11}\lesssim\Big{(}\int_{0}^{r_{B}^{\alpha}}\Big{[}\int_{0}^{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{t+u}\Big{\|}(t+u)L_{\lambda}e^{-(t+u)L_{\lambda}}f\Big{\|}_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\
+\sum_{k=1}^{m}|C_{k}^{m}|\Big{(}\int_{0}^{r_{B}^{\alpha}}\Big{[}\int_{0}^{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{t+u+kr_{B}^{\alpha}}\Big{\|}(t+u+kr_{B}^{\alpha})L_{\lambda}e^{-(t+u+kr_{B}^{\alpha})L_{\lambda}}f\Big{\|}_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}.$

By Proposition 2.4 and Theorem 2.6, together with $t+u\lesssim r_{B}^{\alpha}$
and $(t+u+kr_{B}^{\alpha})^{1/\alpha}\simeq_{k}r_{B}$ for
$u,t\in(0,r_{B}^{\alpha}]$ and $k\geq 1$, we get, for some
$\tilde{C}_{k}^{m}\gtrsim|C_{k}^{m}|$,

$\displaystyle E_{11}$
$\displaystyle\lesssim\Big{(}\int_{0}^{r_{B}^{\alpha}}\Big{[}\int_{0}^{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{t+u}\Big{(}\frac{r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{d}\Big{(}\frac{2^{j}r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{-d-\beta}\|f\|_{L^{p}(B,\frac{dx}{|B|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\ \
+\sum_{k=1}^{m}\tilde{C}_{k}^{m}\Big{(}\int_{0}^{r_{B}^{\alpha}}\Big{[}\int_{0}^{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{r_{B}^{\alpha}}2^{-j(d+\beta)}\|f\|_{L^{p}(B,\frac{dx}{|B|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}.$

Using also $\beta/\alpha>1-\gamma$, we get

$\displaystyle E_{11}$ $\displaystyle\lesssim
2^{-j(d+\beta)}\Big{(}\fint_{B}|f|^{p}dx\Big{)}^{1/p},$

as desired. To estimate $E_{12}$, we use

(44)
$(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}=\int_{0}^{r_{B}^{\alpha}}\dots\int_{0}^{r_{B}^{\alpha}}L_{\lambda}^{m}e^{-(s_{1}+\dots+s_{m})L_{\lambda}}d\vec{s},$

where $d\vec{s}=ds_{1}\dots ds_{m}$, and estimate

$\displaystyle E_{12}$
$\displaystyle\lesssim\Big{(}\int_{0}^{r_{B}^{\alpha}}\Big{[}\int_{[0,r_{B}^{\alpha}]^{m}}\int^{\infty}_{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\Big{\|}tL_{\lambda}^{m+1}e^{-(t+u+s_{1}+\ldots+s_{m})L_{\lambda}}f\Big{\|}_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}\frac{du}{u}d\vec{s}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}.$

In this case, $u\simeq t+u+s_{1}+\ldots+s_{m}\geq r_{B}^{\alpha}$. Hence, by
Proposition 2.4 and Theorem 2.6, we obtain

(45) $\displaystyle E_{12}$
$\displaystyle\lesssim\Big{(}\int^{r_{B}^{\alpha}}_{0}\Big{[}\int_{[0,r_{B}^{\alpha}]^{m}}\int^{\infty}_{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{u^{m+1}}\Big{(}\frac{r_{B}}{u^{1/\alpha}}\Big{)}^{d-1/p^{\prime}}\Big{(}1+\frac{u^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{1/2}\Big{(}1+\frac{2^{j}r_{B}}{u^{1/\alpha}}\Big{)}^{-(d+\beta)}$
$\displaystyle\quad\quad\quad\quad\times\|f\|_{L^{p}(B,\frac{dx}{|B|})}\frac{du}{u}d\vec{s}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}.$

We can see that

$\displaystyle\int^{\infty}_{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}$
$\displaystyle\frac{t}{u^{m+1}}\Big{(}\frac{r_{B}}{u^{1/\alpha}}\Big{)}^{d-1/p^{\prime}}\Big{(}1+\frac{u^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{1/2}\Big{(}1+\frac{2^{j}r_{B}}{u^{1/\alpha}}\Big{)}^{-(d+\beta)}\frac{du}{u}$
$\displaystyle\lesssim\int^{\infty}_{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{u^{m+1}}\Big{(}\frac{r_{B}}{u^{1/\alpha}}\Big{)}^{d-1/2}\Big{(}\frac{u^{1/\alpha}}{r_{B}}\Big{)}^{1/2}\Big{(}\frac{2^{j}r_{B}}{u^{1/\alpha}}\Big{)}^{-(d+\beta)}\frac{du}{u}$
$\displaystyle\lesssim
2^{-j(d+\beta)}\Big{(}\frac{r_{B}^{\alpha}}{t}\Big{)}^{1-\gamma}\frac{t}{r_{B}^{\alpha(m+1)}},$

for all $m>\beta+1$. Plugging this into (45) we obtain

$\displaystyle E_{12}$ $\displaystyle\lesssim
2^{-j(d+\beta)}\|f\|_{L^{p}(B,\frac{dx}{|B|})}\Big{(}\int^{r_{B}^{\alpha}}_{0}\Big{[}\int_{[0,r_{B}^{\alpha}]^{m}}\Big{(}\frac{r_{B}^{\alpha}}{t}\Big{)}^{1-\gamma}\frac{t}{r_{B}^{\alpha(m+1)}}d\vec{s}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\lesssim 2^{-j(d+\beta)}\Big{(}\fint_{B}|f|^{p}dx\Big{)}^{1/p}.$

Collecting the estimates of $E_{11}$ and $E_{12}$, we have

$E_{1}\lesssim 2^{-j(d+\beta)}\Big{(}\fint_{B}|f|^{p}dx\Big{)}^{1/p}.$

We now estimate $E_{2}$. To do this, we bound

$\displaystyle E_{2}$
$\displaystyle\lesssim\Big{(}\int^{\infty}_{r_{B}^{\alpha}}\Big{[}\int_{0}^{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\Big{\|}tL_{\lambda}e^{-(t+u)L_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f\Big{\|}_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\ \
+\Big{(}\int^{\infty}_{r_{B}^{\alpha}}\Big{[}\int_{r_{B}^{\alpha}}^{\infty}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\Big{\|}tL_{\lambda}e^{-(t+u)L_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f\Big{\|}_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}\frac{du}{u}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle=:E_{21}+E_{22}.$

Similarly to (45),

$\displaystyle E_{21}$
$\displaystyle\lesssim\Big{(}\int_{r_{B}^{\alpha}}^{\infty}\Big{[}\int_{[0,r_{B}^{\alpha}]^{m}}\int_{0}^{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{t^{m+1}}\Big{(}\frac{r_{B}}{t^{1/\alpha}}\Big{)}^{d-1/p^{\prime}}\Big{(}1+\frac{t^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{1/2}\Big{(}1+\frac{2^{j}r_{B}}{t^{1/\alpha}}\Big{)}^{-(d+\beta)}$
$\displaystyle\quad\quad\quad\quad\times\|f\|_{L^{p}(B,\frac{dx}{|B|})}\frac{du}{u}d\vec{s}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2},$

Thus,

$E_{21}\lesssim 2^{-j(d+\beta)}\Big{(}\fint_{B}|f|^{p}dx\Big{)}^{1/p}$

for all $m>\beta+1$. By the same reason,

$\displaystyle E_{22}$
$\displaystyle\lesssim\Big{(}\int_{r_{B}^{\alpha}}^{\infty}\Big{[}\int_{[0,r_{B}^{\alpha}]^{m}}\int^{\infty}_{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{(t+u)^{m+1}}\Big{(}\frac{r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{d-1/p^{\prime}}\Big{(}1+\frac{(t+u)^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{1/2}$
$\displaystyle\quad\quad\quad\quad\times\Big{(}1+\frac{2^{j}r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{-(d+\beta)}\|f\|_{L^{p}(B,\frac{dx}{|B|})}\frac{du}{u}d\vec{s}\Big{]}^{2}\frac{dt}{t}\Big{)}^{1/2}.$

In addition,

$\displaystyle\int^{\infty}_{r_{B}^{\alpha}}$
$\displaystyle\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{(t+u)^{m+1}}\Big{(}\frac{r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{d-1/p^{\prime}}\Big{(}1+\frac{(t+u)^{1/\alpha}}{2^{j}r_{B}}\Big{)}^{1/2}\Big{(}1+\frac{2^{j}r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{-(d+\beta)}\frac{du}{u}$
$\displaystyle\lesssim\int^{\infty}_{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{(t+u)^{m+1}}\Big{(}\frac{r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{d-1/2}\Big{(}\frac{(t+u)^{1/\alpha}}{r_{B}}\Big{)}^{1/2}\Big{(}\frac{2^{j}r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{-(d+\beta)}\frac{du}{u}$
$\displaystyle\lesssim
2^{-j(d+\beta)}\int^{\infty}_{r_{B}^{\alpha}}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{t}{(t+u)^{m+1}}\Big{(}\frac{r_{B}}{(t+u)^{1/\alpha}}\Big{)}^{-\beta-1}\frac{du}{u}$
$\displaystyle\lesssim
2^{-j(d+\beta)}\int^{\infty}_{0}\Big{(}\frac{u}{t}\Big{)}^{1-\gamma}\frac{tr_{B}^{-\beta-1}}{(t+u)^{m-\beta}}\frac{du}{u}\lesssim
2^{-j(d+\beta)}\frac{tr_{B}^{-\beta-1}}{t^{m-\beta}}$

for all $m>\beta+1$. Consequently,

$E_{22}\lesssim 2^{-j(d+\beta)}\Big{(}\fint_{B}|f|^{p}dx\Big{)}^{1/p}.$

Combining the estimates for $E_{21},E_{22}$, and $E_{1}$ yields

$\Big{(}\int_{S_{j}(B)}|S_{L_{\lambda},\gamma}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f|^{2}dx\Big{)}^{1/2}\lesssim
2^{-j(d+\beta)}\Big{(}\fint_{B}|f|^{p}dx\Big{)}^{1/p}.$

This completes the proof of (41).

Step 2: The proof of the $L^{p}$-boundedness for $p>2$

We use Theorem 3.2 with $T=S_{L_{\lambda},\gamma}$ and
$\mathcal{A}_{r_{B}}=I-(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}$,
$m>d+\beta+\alpha(1-\gamma)$. Thus, for any $q\in(2,\infty)$ it suffices to
prove

(46)
$\displaystyle\Big{(}\fint_{B}\left|S_{L_{\lambda},\gamma}(I-\mathcal{A}_{r_{B}})f\right|^{2}dx\Big{)}^{1/2}\leq
C\mathcal{M}_{2}(f)(x)$

and

(47)
$\displaystyle\Big{(}\fint_{B}\big{|}S_{L_{\lambda},\gamma}\mathcal{A}_{r_{B}}f\big{|}^{q}dx\Big{)}^{1/q}\leq
C\mathcal{M}_{2}(S_{L_{\lambda},\gamma}f)(x)$

for all balls $B$ with radius $r_{B}$, all $f\in
C^{\infty}_{c}(\mathbb{R}^{d})$ and all $x\in B$. To prove (46), we write

$\displaystyle\begin{aligned}
\Big{(}\fint_{B}|S_{L_{\lambda},\gamma}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f|^{2}dx\Big{)}^{\frac{1}{2}}&\leq\sum_{j\geq
0}\Big{(}\fint_{B}|S_{L_{\lambda},\gamma}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f_{j}|^{2}dx\Big{)}^{\frac{1}{2}}=:\sum_{j=0}^{\infty}F_{j},\end{aligned}$

where $f_{j}=f\mathbf{1}_{S_{j}(B)}$. For $j=0,1$, we use the
$L^{2}$-boundedness of $S_{L_{\lambda},\gamma}$ and $\mathcal{A}_{r_{B}}$ and
obtain

$F_{j}\lesssim\mathcal{M}_{2}(f)(x).$

Hence, it suffices to prove

(48) $F_{j}\lesssim 2^{-j\beta}\Big{(}\fint_{S_{j}(B)}|f|^{2}dx\Big{)}^{1/2}$

for all $j\geq 2$. To do this, we write

$\displaystyle\Big{(}\fint_{B}|S_{L_{\lambda},\gamma}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f_{j}|^{2}dx\Big{)}^{1/2}$
$\displaystyle\quad\leq\Big{(}\int_{0}^{r_{B}^{\alpha}}\left\|(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f_{j}\right\|_{L^{2}(B,\frac{dx}{|B|})}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\qquad+\Big{(}\int_{r_{B}^{\alpha}}^{\infty}\left\|(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}(I-e^{-r_{B}^{\alpha}L_{\lambda}})^{m}f_{j}\right\|_{L^{2}(B,\frac{dx}{|B|})}^{2}\frac{dt}{t}\Big{)}^{1/2}.$

At this stage, we can argue as in the proof of (41) in Step 1. However, in
this case, we use (27) instead of (26). By doing so, we arrive at the
expression (48). As the proof follows a similar structure, we omit the
details.

It remains to prove (47). We first write

$\displaystyle\Big{(}\int_{B}|S_{L_{\lambda},\gamma}\mathcal{A}_{r_{B}}f(x)|^{q}dx\Big{)}^{\frac{1}{q}}=\Big{[}\int_{B}\Big{(}\int_{0}^{\infty}\Big{|}\sum_{k=1}^{m}C^{m}_{k}e^{-kr_{B}^{\alpha}L_{\lambda}}(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}f(x)\Big{|}^{2}\frac{dt}{t}\Big{)}^{\frac{q}{2}}dx\Big{]}^{\frac{1}{q}}$
$\displaystyle\quad\lesssim\sum_{j\geq
0}\Big{[}\int_{B}\Big{(}\int_{0}^{\infty}\Big{|}\sum_{k=1}^{m}C^{m}_{k}e^{-kr_{B}^{\alpha}L_{\lambda}}[(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}f\mathbf{1}_{S_{j}(B)}](x)\Big{|}^{2}\frac{dt}{t}\Big{)}^{q/2}dx\Big{]}^{\frac{1}{q}}$

which, along with Minkowski’s inequality, Proposition 2.4, and (27) in Theorem
2.6, gives

$\displaystyle\Big{(}\fint_{B}$
$\displaystyle|S_{L_{\lambda},\gamma}\mathcal{A}_{r_{B}}f(x)|^{q}dx\Big{)}^{1/q}$
$\displaystyle\lesssim\sum_{j\geq
0}\Big{(}\int_{0}^{\infty}\Big{\|}\sum_{k=1}^{m}e^{-kr_{B}^{\alpha}L_{\lambda}}[(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}f\mathbf{1}_{S_{j}(B)}]\Big{\|}_{L^{q}(B,\frac{dx}{|B|})}^{2}\frac{dt}{t}\Big{)}^{1/2}$
$\displaystyle\lesssim\sum_{j\geq
0}2^{-j\beta}\left(\int_{0}^{\infty}\left\|(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}f\right\|_{L^{2}(S_{j}(B),\frac{dx}{|S_{j}(B)|})}^{2}\frac{dt}{t}\right)^{1/2}$
$\displaystyle\lesssim\sum_{j\geq
0}2^{-j\beta}\Big{(}\fint_{2^{j}B}|S_{L_{\lambda},\gamma}f(x)|^{2}dx\Big{)}^{1/2}.$

This implies (47). Hence the proof of Step 2 is completed.

Thus we have proved that the square function $S_{L_{\lambda},\gamma}$ is
bounded on $L^{p}(\mathbb{R}^{d}_{+})$ for all $1<p<\infty$, i.e.,

$\|S_{L_{\lambda},\gamma}f\|_{p}\lesssim\|f\|_{p}.$

The reverse inequality follows from this and duality. We provide the details.
By functional calculus, for any $g\in L^{p^{\prime}}(\mathbb{R}^{d}_{+})$, we
have

$\displaystyle\int_{\mathbb{R}^{d}_{+}}f(x)g(x)dx$
$\displaystyle=c(\gamma)\int_{\mathbb{R}^{d}_{+}}\int_{0}^{\infty}(tL_{\lambda})^{2\gamma}e^{-2tL_{\lambda}}f(x)g(x)\frac{dt}{t}dx,$

where $c(\gamma)=\int_{0}^{\infty}t^{2\gamma}e^{-2t}\frac{dt}{t}$. By Hölder’s
inequality, we obtain

$\displaystyle\left|\int_{\mathbb{R}^{d}_{+}}f(x)g(x)dx\right|$
$\displaystyle=c(\gamma)\left|\int_{\mathbb{R}^{d}_{+}}\int_{0}^{\infty}(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}f(x)(tL_{\lambda})^{\gamma}e^{-tL_{\lambda}}g(x)\frac{dt}{t}dx\right|$
$\displaystyle\lesssim\int_{\mathbb{R}^{d}_{+}}S_{L_{\lambda},\gamma}f(x)S_{L_{\lambda},\gamma}g(x)dx$
$\displaystyle\lesssim\|S_{L_{\lambda},\gamma}f\|_{p}\|S_{L_{\lambda},\gamma}g\|_{p^{\prime}}.$

By the upper bound
$\|S_{L_{\lambda},\gamma}g\|_{p^{\prime}}\lesssim\|g\|_{p^{\prime}}$,

$\left|\int_{\mathbb{R}^{d}_{+}}f(x)g(x)dx\right|\lesssim\|S_{L_{\lambda},\gamma}f\|_{p}\|g\|_{p^{\prime}},$

and consequently,

$\|f\|_{p}\lesssim\|S_{L_{\lambda},\gamma}f\|_{p}.$

This completes the proof of Theorem 1.4. ∎

##  4\. Bounds for differences of kernels

In this section, we prepare the proof of the reversed Hardy inequality
(Theorem 1.6), which relies on Schur tests and bounds for differences of heat
kernels and derivatives of heat kernels.

Let $\alpha\in(0,2]$ and $\sigma$ as in (5). Set
$q:=\min\\{\sigma,(\alpha-1)_{+}\\}$. For each $\beta\in(0,\alpha)$, we define

$\displaystyle L^{\alpha,\beta}_{t}(x,y):=$
$\displaystyle\big{(}\mathbf{1}_{x_{d}\vee y_{d}\leq
t^{1/\alpha}}+\mathbf{1}_{x_{d}\vee y_{d}\geq
t^{1/\alpha}}\mathbf{1}_{|x-y|\geq(x_{d}\wedge
y_{d})/2}\big{)}\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{q}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{q}$
$\displaystyle\times
t^{-d/\alpha}\Big{[}\Big{(}\frac{t^{1/\alpha}}{t^{1/\alpha}+|x-y|}\Big{)}^{d+\beta}\mathbf{1}_{\alpha<2}+\exp\Big{(}-\frac{|x-y|^{2}}{t}\Big{)}\mathbf{1}_{\alpha=2}\Big{]}$

and

$\displaystyle M^{\alpha,\beta}_{t}(x,y):=$
$\displaystyle\mathbf{1}_{x_{d}\vee y_{d}\geq
t^{1/\alpha}}\mathbf{1}_{|x-y|\leq(x_{d}\wedge y_{d})/2}$
$\displaystyle\times\frac{t^{1-d/\alpha}}{(x_{d}\vee
y_{d})^{\alpha}}\Big{[}\Big{(}\frac{t^{1/\alpha}}{t^{1/\alpha}+|x-y|}\Big{)}^{d+\beta}\mathbf{1}_{\alpha<2}+\exp\Big{(}-\frac{|x-y|^{2}}{t}\Big{)}\mathbf{1}_{\alpha=2}\Big{]}.$

These two functions were used in [FM23, Theorem 17] in the proof of the upper
bound for the kernel of the difference $e^{-tL_{\lambda}}-e^{-tL_{0}}$. In
this section, we will prove new bounds for the difference
$tL_{0}e^{-tL_{0}}-tL_{\lambda}e^{-tL_{\lambda}}$, see Proposition 4.4 below.
We will often use the following auxiliary bound.

######  Lemma 4.1.

Let $N\in\mathbb{N}$. Then for all $\beta\in(0,2]$ and all $s,t>0$ we have

$\int_{\mathbb{R}^{N}}\frac{(st)^{\beta}}{(s+|x-z|)^{N+\beta}(s+|z-y|)^{N+\beta}}dz\simeq\frac{(s+t)^{\beta}}{[(s+t)+|x-y|]^{N+\beta}}$

for all $x,y\in\mathbb{R}^{N}$.

###### Proof.

The proof of the inequality “$\lesssim$” was given in [FM23, Lemma 22]. Here,
we prove “$\simeq$” directly. It is well-known that

$e^{-t(-\Delta_{\mathbb{R}^{N}})^{\beta/2}}(x,y)\simeq\frac{t}{(t^{1/\beta}+|x-y|)^{N+\beta}}.$

Hence,

$\displaystyle\int_{\mathbb{R}^{N}}\frac{(st)^{\beta}}{(s+|x-z|)^{N+\beta}(s+|z-y|)^{N+\beta}}dz$
$\displaystyle\simeq\int_{\mathbb{R}^{N}}e^{-s^{\beta}(-\Delta_{\mathbb{R}^{N}})^{\beta/2}}(x,z)e^{-t^{\beta}(-\Delta_{\mathbb{R}^{N}})^{\beta/2}}(z,y)dz$
$\displaystyle=e^{-(s^{\beta}+t^{\beta})(-\Delta_{\mathbb{R}^{N}})^{\beta/2}}(x,z)$
$\displaystyle\simeq\frac{(s+t)^{\beta}}{[(s+t)+|x-y|]^{N+\beta}}$

for all $s,t>0$ and $x,y\in\mathbb{R}^{N}$. ∎

The following technical lemma is inspired from [FM23].

######  Lemma 4.2.

Let $\alpha\in(0,2)$, $\min\\{0,\alpha-1\\}<\sigma<\alpha$, and
$\beta\in((\alpha-1)_{+},\alpha)$. For $t>0$ and $x,y\in\mathbb{R}_{+}^{d}$,
let

$T^{\alpha,\beta}_{t}(x,y):=\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{\sigma}t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}}{t^{1/\alpha}+|x-y|}\Big{)}^{d+\beta},$

and

$H^{\alpha,\beta}_{t}(x,y):=\Big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\Big{)}^{(\alpha-1)_{+}}\Big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\Big{)}^{(\alpha-1)_{+}}t^{-d/\alpha}\Big{(}\frac{t^{1/\alpha}}{t^{1/\alpha}+|x-y|}\Big{)}^{d+\beta}.$

Then there exists $C>0$ such that

(49)
$\displaystyle\begin{split}U^{\alpha,\beta}_{t}(x,y)&:=\mathbf{1}_{x_{d}\vee
y_{d}\geq t^{1/\alpha}}\mathbf{1}_{|x-y|\leq(x_{d}\wedge
y_{d})/2}\int_{\mathbb{R}^{d}_{+}}\int_{0}^{t}H^{\alpha,\beta}_{t-s}(x,z)z_{d}^{-\alpha}T^{\alpha,\beta}_{s}(z,y)dsdz\\\
&\leq CM^{\alpha,\beta}_{t}(x,y)\end{split}$

for all $x,y\in\mathbb{R}^{d}_{+}$ and $t>0$.

Note that $\sigma\geq(\alpha-1)/2$ implies $\sigma>\min\\{0,\alpha-1\\}$ when
$\alpha\neq 1$.

###### Proof.

Without loss of generality, we assume $y_{d}\geq x_{d}$ and $t=1$. Set

$S:=\\{(x,y)\in\mathbb{R}_{+}^{d}\times\mathbb{R}_{+}^{d}:x_{d}\vee y_{d}\geq
1,|x-y|\leq(x_{d}\wedge y_{d})/2\\}.$

Since $|x_{d}-y_{d}|\leq|x-y|\leq(x_{d}\wedge y_{d})/2$, we have $x_{d}\simeq
y_{d}$. Hence, $x_{d}\simeq y_{d}\gtrsim 1$ for $(x,y)\in S$. We now write

$\displaystyle U^{\alpha,\beta}_{t}(x,y)=$
$\displaystyle\mathbf{1}_{S}(x,y)\int_{\mathbb{R}^{d-1}}\int_{0}^{x_{d}/2}\int_{0}^{1}H^{\alpha,\beta}_{1-s}(x,z)z_{d}^{-\alpha}T^{\alpha,\beta}_{s}(z,y)dsdz$
$\displaystyle+\mathbf{1}_{S}(x,y)\int_{\mathbb{R}^{d-1}}\int_{x_{d}/2}^{\infty}\int_{0}^{1}H^{\alpha,\beta}_{1-s}(x,z)z_{d}^{-\alpha}T^{\alpha,\beta}_{s}(z,y)dsdz$
$\displaystyle=:$ $\displaystyle E_{1}+E_{2}.$

For the term $E_{2}$, since $z_{d}\geq x_{d}/2$ and $x_{d}\simeq y_{d}\gtrsim
1$, by Lemma 4.1, we have

$\displaystyle E_{2}$
$\displaystyle\lesssim\frac{1}{x_{d}^{\alpha}}\int_{0}^{1}\int_{\mathbb{R}^{d}_{+}}(1-s)^{-d/\alpha}\Big{(}\frac{(1-s)^{1/\alpha}}{(1-s)^{1/\alpha}+|x-z|}\Big{)}^{d+\beta}s^{-d/\alpha}\Big{(}\frac{s^{1/\alpha}}{s^{1/\alpha}+|z-y|}\Big{)}^{d+\beta}dzds$
$\displaystyle\lesssim\frac{1}{x_{d}^{\alpha}}\int_{0}^{1}\Big{(}\frac{1}{1+|x-y|}\Big{)}^{d+\beta}ds\lesssim\frac{1}{x_{d}^{\alpha}}\Big{(}\frac{1}{1+|x-y|}\Big{)}^{d+\beta}\lesssim
M_{1}^{\alpha,\beta}(x,y).$

For the term $E_{1}$, we note that since $y_{d}\geq x_{d}\geq 2z_{d}$, we have
$|y_{d}-z_{d}|\simeq y_{d}\simeq x_{d}\simeq|x_{d}-z_{d}|$. Therefore, in this
case for $s\in(0,1)$,

$(1-s)^{1/\alpha}+|x-z|\simeq x_{d}+|x^{\prime}-z^{\prime}|\ \ \text{and}\ \
s^{1/\alpha}+|y-z|\simeq x_{d}+|y^{\prime}-z^{\prime}|.$

This and Lemma 4.1 imply

$\displaystyle\int_{\mathbb{R}^{d-1}}\frac{(1-s)^{\beta/\alpha}}{((1-s)^{1/\alpha}+|x-z|)^{d+\beta}}\frac{s^{\beta/\alpha}}{(s^{1/\alpha}+|x-z|)^{d+\beta}}dz^{\prime}$
$\displaystyle\quad\simeq\int_{\mathbb{R}^{d-1}}\frac{(1-s)^{\beta/\alpha}}{(x_{d}+|x^{\prime}-z^{\prime}|)^{d+\beta}}\frac{s^{\beta/\alpha}}{(x_{d}+|y^{\prime}-z^{\prime}|)^{d+\beta}}dz^{\prime}$
$\displaystyle\quad\lesssim\frac{x_{d}^{-\beta-1}s^{\beta/\alpha}(1-s)^{\beta/\alpha}}{(x_{d}+|x^{\prime}-y^{\prime}|)^{d+\beta}}\simeq\frac{x_{d}^{-\beta-1}(1-s)^{\beta/\alpha}s^{\beta/\alpha}}{(1+|x-y|)^{d+\beta}}.$

It follows that

$\displaystyle E_{1}$
$\displaystyle\lesssim\frac{x_{d}^{-\beta-1}}{(1+|x-y|)^{d+\beta}}\int_{0}^{x_{d}/2}\int_{0}^{1}z_{d}^{-\alpha}(1-s)^{\beta/\alpha}s^{\frac{\beta}{\alpha}}\Big{(}1\wedge\frac{z_{d}}{(1-s)^{\frac{1}{\alpha}}}\Big{)}^{(\alpha-1)_{+}}\Big{(}1\wedge\frac{z_{d}}{s^{1/\alpha}}\Big{)}^{\sigma}dsdz_{d}.$

By a simple calculation,

$\displaystyle\int_{0}^{1}(1-s)^{\beta/\alpha}s^{\beta/\alpha}\Big{(}1\wedge\frac{z_{d}}{(1-s)^{1/\alpha}}\Big{)}^{(\alpha-1)_{+}}\Big{(}1\wedge\frac{z_{d}}{s^{1/\alpha}}\Big{)}^{\sigma}ds=\int_{0}^{1/2}\ldots+\int_{1/2}^{1}\ldots$
$\displaystyle\quad\lesssim\Big{(}1\wedge
z_{d}\Big{)}^{\sigma+(\alpha-1)_{+}}.$

Consequently,

$\displaystyle E_{1}$
$\displaystyle\lesssim\frac{x_{d}^{-\beta-1}}{(1+|x-y|)^{d+\beta}}\int_{0}^{x_{d}/2}z_{d}^{-\alpha}\Big{(}1\wedge
z_{d}\Big{)}^{\sigma+(\alpha-1)_{+}}dz_{d}$
$\displaystyle\lesssim\frac{1}{x_{d}^{\beta+1}}\frac{1}{(1+|x-y|)^{d+\beta}}\int_{0}^{x_{d}/2}z_{d}^{-\alpha}\Big{(}1\wedge
z_{d}\Big{)}^{\sigma+(\alpha-1)_{+}}dz_{d}$
$\displaystyle=\frac{1}{x_{d}^{\beta+1}}\frac{1}{(1+|x-y|)^{d+\beta}}\Big{[}x_{d}^{-\alpha+1}\cdot\mathbf{1}_{\alpha<1}+\ln(x_{d})\cdot\mathbf{1}_{\alpha=1}+\mathbf{1}_{\alpha>1}\Big{]}.$

Since $x_{d}\gtrsim 1$, we use the fact $\ln(x_{d})\leq
c_{\epsilon}x_{d}^{\epsilon}$ for every $\epsilon>0$ to further obtain

$\displaystyle E_{1}$
$\displaystyle\lesssim\frac{1}{(1+|x-y|)^{d+\beta}}\Big{[}\Big{(}\frac{1}{x_{d}}\Big{)}^{\beta+\alpha}\cdot\mathbf{1}_{\alpha<1}+\Big{(}\frac{1}{x_{d}}\Big{)}^{\beta+\alpha-\epsilon}\cdot\mathbf{1}_{\alpha=1}+\Big{(}\frac{1}{x_{d}}\Big{)}^{\beta+1}\cdot\mathbf{1}_{\alpha>1}\Big{]}.$

This completes our proof. ∎

######  Lemma 4.3.

Let $\alpha=2$ and $\sigma$ as in (5). For $x,y\in\mathbb{R}_{+}^{d}$ and
$t>0$, let

$T_{t}(x,y):=\Big{(}1\wedge\frac{x_{d}}{\sqrt{t}}\Big{)}^{\sigma}\Big{(}1\wedge\frac{y_{d}}{\sqrt{t}}\Big{)}^{\sigma}t^{-d/2}\exp\Big{(}-\frac{|x-y|^{2}}{t}\Big{)}$

and

$H_{t}(x,y):=\Big{(}1\wedge\frac{x_{d}}{\sqrt{t}}\Big{)}\Big{(}1\wedge\frac{y_{d}}{\sqrt{t}}\Big{)}t^{-d/2}\exp\Big{(}-\frac{|x-y|^{2}}{t}\Big{)}.$

Then there exist $C,c>0$ such that

$\displaystyle U_{t}(x,y)$ $\displaystyle:=\mathbf{1}_{x_{d}\vee
y_{d}\geq\sqrt{t}}\mathbf{1}_{|x-y|\leq(x_{d}\wedge
y_{d})/2}\int_{\mathbb{R}^{d}_{+}}\int_{0}^{t}H_{t-s}(x,z)z_{d}^{-\alpha}T_{s}(z,y)dsdz$
$\displaystyle\leq CM^{\alpha,\beta}_{ct}(x,y)$

for all $x,y\in\mathbb{R}^{d}_{+}$ and $t>0$.

###### Proof.

The proof is similar to that of Lemma 4.2. The only difference is that instead
of Lemma 4.1, we use that for $N\geq 1$ and $c_{1},c_{2}>0$, there exist
$C,c>0$ such that for all $s,t>0$ and $x,y\in\mathbb{R}^{N}$, we have

$\int_{\mathbb{R}^{N}}\frac{1}{(st)^{N/2}}\exp\Big{(}-\frac{|x-z|^{2}}{c_{1}s}\Big{)}\exp\Big{(}-\frac{|z-y|^{2}}{c_{2}t}\Big{)}dz\leq
C\frac{1}{(s+t)^{N/2}}\exp\Big{(}-\frac{|x-z|^{2}}{c(s+t)}\Big{)}.$

Hence, we omit the details. ∎

We now prove new bounds for the integral kernel $Q_{t}(x,y)$ of the difference
$tL_{0}e^{-tL_{0}}-tL_{\lambda}e^{-tL_{\lambda}}$. These bounds will be
instrumental to prove the reversed Hardy inequality expressed in terms of the
square functions in (11).

######  Proposition 4.4.

Let $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha<2$ and
$\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by (5). Then for
any $\beta\in(0,\alpha)$, there exists $C>0$ such that for all
$x,y\in\mathbb{R}^{d}$ and $t>0$,

(50) $|Q_{t}(x,y)|\leq
C\big{[}L^{\alpha,\beta}_{t}(x,y)+M^{\alpha,\beta}_{t}(x,y)\big{]}.$

######  Remark 4.5.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then (50) remains valid. This follows by the same arguments as
in the proof below.

###### Proof.

We only give the proof for $\alpha\in(0,2)$ since the case $\alpha=2$ can be
treated similarly. Let $T^{\alpha,\beta}_{t}(x,y)$ and
${H}^{\alpha,\beta}_{t}(x,y)$ be the two functions defined in Lemma 4.2. Let
$\widetilde{p}_{t}(x,y)$ and $\widetilde{p}_{t,1}(x,y)$ denote the kernels of
$e^{-tL_{0}}$ and $-L_{0}e^{-tL_{0}}$, respectively. By Proposition 2.4,

$|\widetilde{p}_{t}(x,y)|+t|\widetilde{p}_{t,1}(x,y)|\lesssim
H^{\alpha,\beta}_{t}(x,y)$

and

$|p_{t}(x,y)|+t|p_{t,1}(x,y)|\lesssim T^{\alpha,\beta}_{t}(x,y)$

for all $t>0$ and $x,y\in\mathbb{R}^{d}_{+}$. We now consider two cases.

Case 1: $x_{d}\vee y_{d}\leq t^{1/\alpha}$ OR $x_{d}\vee y_{d}\geq
t^{1/\alpha}$ and $|x-y|\geq(x_{d}\wedge y_{d})/2$.

Since in this case

$H^{\alpha,\beta}_{t}(x,y)+T^{\alpha,\beta}_{t}(x,y)\lesssim
L^{\alpha,\beta}_{t}(x,y),$

we get

$|Q_{t}(x,y)|\lesssim L^{\alpha}_{t}(x,y).$

Case 2: $x_{d}\vee y_{d}\geq t^{1/\alpha}$ and $|x-y|<(x_{d}\wedge y_{d})/2$.

By Duhamel’s formula,

$\displaystyle\tilde{p}_{t}(x,y)$ $\displaystyle-{p}_{t}(x,y)$
$\displaystyle=\lambda\int_{0}^{t}\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{t-s}(x,z){z_{d}^{-\alpha}}p_{s}(z,y)dzds$
$\displaystyle=\lambda\int_{0}^{t/2}\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{t-s}(x,z){z_{d}^{-\alpha}}p_{s}(z,y)dzds+\lambda\int_{0}^{t/2}\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{s}(x,z){z_{d}^{-\alpha}}p_{t-s}(z,y)dzds.$

Differentiating both sides with respect to $t$ and multiplying by $t$, we get

(51) $\displaystyle{Q}_{t}(x,y)$ $\displaystyle=\lambda
t\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{t/2}(x,z){z_{d}^{-\alpha}}p_{t/2}(z,y)dz+\lambda
t\int_{0}^{t/2}\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{t-s,1}(x,z){z_{d}^{-\alpha}}p_{s}(z,y)dzds$
$\displaystyle\quad+\lambda
t\int_{t/2}^{t}\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{t-s}(x,z){z_{d}^{-\alpha}}p_{s,1}(z,y)dzds$
$\displaystyle=I_{1}+I_{2}+I_{3}.$

The term $I_{1}$ can be written as

$\displaystyle\lambda
t\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{t/2}(x,z){z_{d}^{-\alpha}}p_{t/2}(z,y)dz$
$\displaystyle=6\lambda\int_{t/3}^{t/2}\int_{\mathbb{R}^{d}_{+}}\tilde{p}_{t/2}(x,z){z_{d}^{-\alpha}}p_{t/2}(z,y)dzds.$

It is easy to see that $H^{\alpha,\beta}_{t-s}(\cdot,\cdot)\simeq
H^{\alpha,\beta}_{t/2}(\cdot,\cdot)\gtrsim\tilde{p}_{t/2}(\cdot,\cdot)$ and
$T^{\alpha,\beta}_{s}(\cdot,\cdot)\simeq
T^{\alpha,\beta}_{t/2}(\cdot,\cdot)\gtrsim p_{t/2}(\cdot,\cdot)$ for
$s\in[t/3,t/2]$. This, along with Lemma 4.2, implies

$\displaystyle I_{1}$
$\displaystyle\lesssim\int_{t/3}^{t/2}\int_{\mathbb{R}^{d}}H^{\alpha,\beta}_{t-s}(x,z){z_{d}^{-\alpha}}T^{\alpha,\beta}_{s}(z,y)dzds\lesssim\int_{0}^{t}\int_{\mathbb{R}^{d}_{+}}H^{\alpha,\beta}_{t-s}(x,z){z_{d}^{-\alpha}}T^{\alpha,\beta}_{s}(z,y)dzds$
$\displaystyle\lesssim L^{\alpha,\beta}_{t}(x,y)+M^{\alpha,\beta}_{t}(x,y).$

For the second term, since $t-s\simeq t$ for ${s}\in(0,t/2)$, we have

$t|\tilde{p}_{t-s,1}(x,z)|\simeq(t-s)|\tilde{p}_{t-s,1}(x,z)|\lesssim
H^{\alpha,\beta}_{t-s}(x,z).$

Therefore, by Lemma 4.2,

$\displaystyle I_{2}$
$\displaystyle\lesssim\int_{0}^{t/2}\int_{\mathbb{R}^{d}_{+}}H^{\alpha,\beta}_{t-s}(x,z){z_{d}^{-\alpha}}T^{\alpha,\beta}_{s}(z,y)dzds\lesssim
L^{\alpha,\beta}_{t}(x,y)+M^{\alpha,\beta}_{t}(x,y).$

Similarly,

$\displaystyle I_{3}$
$\displaystyle\lesssim\int_{t/2}^{t}\int_{\mathbb{R}^{d}_{+}}H^{\alpha,\beta}_{t-s}(x,z)z_{d}^{-\alpha}T^{\alpha,\beta}_{s}(z,y)dzds\lesssim
L^{\alpha,\beta}_{t}(x,y)+M^{\alpha,\beta}_{t}(x,y).$

This completes our proof. ∎

######  Remark 4.6.

In [FM23], similar upper bounds were obtained for the kernel of the difference
$e^{-tL_{0}}-e^{-tL_{\lambda}}$. However, these are not suitable for our
purpose since the square functions used here are different.

##  5\. Proof of the reversed Hardy inequality (Theorem 1.6)

We now use the previous bounds for the difference of kernels to prove Theorem
1.6, i.e., the reversed Hardy inequality, expressed in our new square
functions.

_Step 1._ By Proposition 4.4,

(52)
$\displaystyle\begin{split}\Big{(}\int_{0}^{\infty}&t^{-s}\left|\left(tL_{\lambda}e^{-tL_{\lambda}}-tL_{0}e^{-tL_{0}}\right)f(x)\right|^{2}\frac{dt}{t}\Big{)}^{1/2}\\\
&=\Big{[}\sum_{j\in\mathbb{Z}}\int_{2^{\alpha
j}}^{2^{\alpha(j+1)}}t^{-s}\left|\left(tL_{\lambda}e^{-tL_{\lambda}}-tL_{0}e^{-tL_{0}}\right)f(x)\right|^{2}\frac{dt}{t}\Big{]}^{1/2}\\\
&\leq\left[\sum_{j\in\mathbb{Z}}\int_{2^{\alpha
j}}^{2^{\alpha(j+1)}}t^{-s}\left(\int_{\mathbb{R}^{d}_{+}}[L^{\alpha,\beta}_{t}(x,y)+M^{\alpha,\beta}_{t}(x,y)]|f(y)|dy\right)^{2}\frac{dt}{t}\right]^{1/2}\\\
&\leq\sum_{j\in\mathbb{Z}}2^{-js\alpha/2}\int_{\mathbb{R}^{d}_{+}}\left[L^{\alpha,\beta}_{2^{\alpha(j+1)}}(x,y)+M^{\alpha,\beta}_{2^{j\alpha}}(x,y)\right]y_{d}^{\alpha
s/2}\,\frac{|f(y)|}{y_{d}^{\alpha s/2}}dy\end{split}$

where in the last inequality we used the embedding
$\ell_{1}\hookrightarrow\ell_{2}$. Thus, it suffices to show the
$L^{p}(\mathbb{R}_{+}^{d})$-boundedness of the operator with kernel

(53) $\displaystyle\sum_{j\in\mathbb{Z}}2^{-j\alpha
s/2}\left[L_{2^{\alpha(j+1)}}^{\alpha,\beta}(x,y)+M_{2^{j\alpha}}^{\alpha,\beta}(x,y)\right]y_{d}^{\alpha
s/2}.$

To that end, we use Schur tests. In the following, we will always replace the
Gaussian decay for $\alpha=2$ by a polynomial decay of order $d+2$.

_Step 2._ We begin with the kernel coming from the
$M_{2^{j\alpha}}^{\alpha,\beta}(x,y)$-part in (54). As discussed in the proof
of Lemma 4.2, we have $x_{d}\simeq y_{d}$ on the support of
$M_{2^{j\alpha}}^{\alpha,\beta}(x,y)$. Thus, the kernel

(54) $\displaystyle\begin{split}\sum_{j\in\mathbb{Z}}2^{-j\alpha
s/2}M_{2^{j\alpha}}^{\alpha,\beta}(x,y)\,y_{d}^{\alpha s/2}\simeq\sum_{N\in
2^{\alpha\mathbb{Z}}}N^{-s/2}M_{N}^{\alpha,\beta}(x,y)\,(x_{d}y_{d})^{\alpha
s/4}\end{split}$

in question is replaced by a symmetric one and we only have to perform a
single Schur test instead of two. The Schur test is analogous to that in the
proof of Step 2 in [FM23, Theorem 5], where the summation over dyadic numbers
$2^{\alpha\mathbb{Z}}$ is replaced with an integral over $t$ with respect to
the Haar measure $dt/t$. Indeed, the bound follows from

(55) $\displaystyle\begin{split}&\int\limits_{y_{d}\simeq x_{d}}dy\sum_{N\in
2^{\alpha\mathbb{Z}}}N^{-\frac{s}{2}-\frac{d}{\alpha}}\,\frac{N}{x_{d}^{\alpha}}\left(\frac{N^{1/\alpha}}{N^{1/\alpha}+|x-y|}\right)^{d+\beta}\mathbf{1}_{N<x_{d}^{\alpha}}x_{d}^{\alpha
s/2}\\\ &\quad\lesssim x_{d}^{\alpha s/2-\alpha}\sum_{2^{\alpha\mathbb{Z}}\ni
N<x_{d}^{\alpha}}\int_{\mathbb{R}^{d}}dy\,\left(1\wedge\frac{N^{(d+\beta)/\alpha}}{(N^{1/\alpha}+|x-y|)^{d+\beta}}\right)\lesssim
1.\end{split}$

_Step 3._ We now consider the kernel coming from the
$L_{2^{\alpha(j+1)}}^{\alpha,\beta}(x,y)$-part in (54). As in the proof of
[FM23, Theorem 5], we will prove the stronger statement that the operator with
kernel

(56)
$\displaystyle\begin{split}\tilde{L}_{t}^{\alpha,\beta}(x,y)&:=\left(\mathbf{1}_{x_{d}\vee
y_{d}<t^{1/\alpha}}+\mathbf{1}_{x_{d}\vee
y_{d}>t^{1/\alpha}}\mathbf{1}_{|x-y|>(x_{d}\wedge
y_{d})/2}\right)\,\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{-r}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{-r}\\\
&\qquad\times
t^{-d/\alpha}\left(\frac{t^{1/\alpha}}{t^{1/\alpha}+|x-y|}\right)^{d+\beta}\end{split}$

is $L^{p}(\mathbb{R}_{+}^{d})$-bounded, where $-r:=(q\wedge 0)\leq 0$ and
$q=\min\\{\sigma,(\alpha-1)_{+}\\}$. As in the proof of [FM23, Theorem 5], we
will show that

(57) $\displaystyle\sum_{N\in
2^{\alpha\mathbb{Z}}}N^{-s/2}\tilde{L}_{N}^{\alpha,\beta}(x,y)y_{d}^{\alpha
s/2}\lesssim\left(\frac{|x-y|\vee x_{d}\vee
y_{d}}{\sqrt{x_{d}y_{d}}}\right)^{2r}\frac{(|x-y|\vee x_{d}\vee
y_{d})^{\alpha}}{(|x-y|\vee(x_{d}\wedge y_{d}))^{d+\alpha}}.$

As with the $M_{t}^{\alpha,\beta}$-part, the proof is similar to that in
[FM23]. Let us give the details. We distinguish between $N\lessgtr(x_{d}\vee
y_{d})^{\alpha}$. On the one hand,

(58) $\displaystyle\begin{split}&\sum_{2^{\alpha\mathbb{Z}}\ni N\geq(x_{d}\vee
y_{d})^{\alpha}}N^{-\frac{s}{2}}\tilde{L}_{N}^{\alpha,\beta}(x,y)y_{d}^{\alpha
s/2}\\\ &\quad\simeq y_{d}^{\frac{\alpha
s}{2}}(x_{d}y_{d})^{-r}\sum_{2^{\alpha\mathbb{Z}}\ni N\geq(x_{d}\vee
y_{d})^{\alpha}}N^{-\frac{s}{2}+\frac{2r}{\alpha}-\frac{d}{\alpha}}\left(1\wedge\frac{N^{(\beta+d)/\alpha}}{|x-y|^{d+\beta}}\right)\\\
&\quad\lesssim y_{d}^{\frac{\alpha
s}{2}}(x_{d}y_{d})^{-r}\left[(|x-y|\\!\vee\\!x_{d}\\!\vee\\!y_{d})^{-\frac{\alpha
s}{2}+2r-d}+\mathbf{1}_{x_{d}\vee
y_{d}\leq|x-y|}|x-y|^{-d-\alpha}(x_{d}\\!\vee\\!y_{d})^{2r+\alpha-\frac{\alpha
s}{2}}\right].\end{split}$

The first term here comes from the $N$-summation from $(|x-y|\vee x_{d}\vee
y_{d})^{\alpha}$ to $\infty$. This sum converges since
$-\frac{s}{2}+\frac{2r}{\alpha}-\frac{d}{\alpha}<0$. (Note that $s>0$ and
$2r\leq(1-\alpha)_{+}<1$.) The second term comes from an upper bound on the
sum between $(x_{d}\vee y_{d})^{\alpha}$ and $|x-y|^{\alpha}$, in fact, from
an upper bound on the integral between $0$ and $|x-y|^{\alpha}$. This sum
converges since $-\frac{s}{2}+\frac{2r}{\alpha}+1>0$.

We now turn to the contribution to $\tilde{L}_{N}^{\alpha,\beta}$ from
$\\{x_{d}\vee y_{d}\geq t^{1/\alpha}\\}$. We have

(59) $\displaystyle\begin{split}&\sum_{2^{\alpha\mathbb{Z}}\ni N\leq(x_{d}\vee
y_{d})^{\alpha}}N^{-\frac{s}{2}}\tilde{L}_{N}^{\alpha,\beta}(x,y)\,y_{d}^{\frac{\alpha
s}{2}}\\\ &\simeq y_{d}^{\frac{\alpha s}{2}}\sum_{2^{\alpha\mathbb{Z}}\ni
N\leq(x_{d}\vee
y_{d})^{\alpha}}N^{-\frac{s}{2}-\frac{d}{\alpha}}\left(1\wedge\frac{x_{d}\wedge
y_{d}}{N^{1/\alpha}}\right)^{-r}\left(1\wedge\frac{N^{(\beta+d)/\alpha}}{|x-y|^{d+\beta}}\right)\mathbf{1}_{|x-y|\geq(x_{d}\wedge
y_{d})/2}\\\ &\leq y_{d}^{\frac{\alpha
s}{2}}|x-y|^{-d-\alpha}\sum_{2^{\alpha\mathbb{Z}}\ni N\leq(x_{d}\vee
y_{d})^{\alpha}}N^{1-\frac{s}{2}}\left(1\wedge\frac{x_{d}\wedge
y_{d}}{N^{1/\alpha}}\right)^{-r}\mathbf{1}_{|x-y|\geq(x_{d}\wedge y_{d})/2}\\\
&\lesssim y_{d}^{\frac{\alpha s}{2}}|x-y|^{-d-\alpha}\left[(x_{d}\wedge
y_{d})^{\alpha-\frac{\alpha s}{2}}+(x_{d}\wedge y_{d})^{-r}(x_{d}\vee
y_{d})^{r+\alpha-\frac{\alpha s}{2}}\right]\mathbf{1}_{|x-y|\geq(x_{d}\wedge
y_{d})/2}\,.\end{split}$

The first term here comes from the sum from $0$ to $(x_{d}\wedge
y_{d})^{\alpha}$. This converges since $s<2$. The second term comes from an
upper bound on the sum from $(x_{d}\wedge y_{d})^{\alpha}$ to $(x_{d}\vee
y_{d})^{\alpha}$, in fact, from an upper bound on the sum between $0$ and
$(x_{d}\vee y_{d})^{\alpha}$. This sum converges since
$-\frac{s}{2}+\frac{r}{\alpha}>-1$. Combining (58)–(59) as in the proof of
[FM23, Theorem 5] eventually yields (57). Thus, the
$L^{p}(\mathbb{R}_{+}^{d})$-boundedness of
$L_{2^{\alpha(j+1)}}^{\alpha,\beta}(x,y)$-part in (54) follows from
Proposition 5.1 below. This concludes the proof of Theorem 1.6. ∎

######  Proposition 5.1.

Let $\alpha>0$ and $0\leq r<\frac{1}{2}$. Then the integral operator with
integral kernel

$\displaystyle\left(\frac{|x-y|\vee x_{d}\vee
y_{d}}{\sqrt{x_{d}y_{d}}}\right)^{2r}\frac{(|x-y|\vee x_{d}\vee
y_{d})^{\alpha}}{(|x-y|\vee(x_{d}\wedge y_{d}))^{d+\alpha}}$

is bounded on $L^{p}(\mathbb{R}_{+}^{d})$ for all
$p\in(\frac{1}{1-r},\frac{1}{r})$.

###### Proof.

The proof is similar to that of [FM23, Proposition 19]. We denote the kernel
in the proposition by $k(x,y)$. By symmetry of $k(x,y)$ and a duality
argument, it suffices to consider $p\leq 2$.

_Step 1._ As a preliminary step to the main argument, we integrate over the
$\mathbb{R}^{d-1}$-variables. By [FM23, (20)],

(60) $\int_{\mathbb{R}^{d-1}}dy^{\prime}\,k(x,y)\lesssim\left(\frac{x_{d}\vee
y_{d}}{\sqrt{x_{d}y_{d}}}\right)^{2r}\frac{(x_{d}\vee
y_{d})^{\alpha}}{(|x_{d}-y_{d}|\vee(x_{d}\wedge y_{d}))^{1+\alpha}}.$

_Step 2._ We perform weighted Schur tests for the operator with kernel given
by the right side of (60). We use the weight

$w(x,y)=\left(\frac{x_{d}}{y_{d}}\right)^{\beta}\qquad\text{with}\
\beta\in(pr,p(1-r))\cap(p^{\prime}r,p^{\prime}(1-r))\,.$

Since $r<1/2$ and $p\in(1/(1-r),1/r)$, it is possible to find such a $\beta$.

For the first part of the Schur test, we use (60) to bound

$\displaystyle\int_{\mathbb{R}^{d}_{+}}dy\,w(x,y)^{1/p}k(x,y)$
$\displaystyle\simeq\int_{0}^{\infty}dy_{d}\,\left(\frac{x_{d}}{y_{d}}\right)^{\beta/p}\left(\frac{x_{d}\vee
y_{d}}{\sqrt{x_{d}y_{d}}}\right)^{2r}\frac{(x_{d}\vee
y_{d})^{\alpha}}{(|x_{d}-y_{d}|\vee(x_{d}\wedge y_{d}))^{1+\alpha}}$
$\displaystyle=\int_{0}^{\infty}dt\,t^{-\beta/p-r}\frac{(1\vee
t)^{\alpha+2r}}{(|1-t|\vee(1\wedge t))^{1+\alpha}}$
$\displaystyle\simeq\int_{0}^{\infty}dt\,t^{-\beta/p-r}(1\wedge
t^{-1+2r})<\infty\,.$

The finiteness of the last integral uses the assumptions $pr<\beta<p(1-r)$.

For the second part of the Schur test, we note that, by symmetry, (60) remains
valid with $dy^{\prime}$ replaced by $dx^{\prime}$. Thus, and by
$p^{\prime}r<\beta<p^{\prime}(1-r)$,

$\displaystyle\int_{\mathbb{R}^{d}_{+}}dx\,w(x,y)^{-1/p^{\prime}}k(x,y)$
$\displaystyle\simeq\int_{0}^{\infty}dx_{d}\,\left(\frac{y_{d}}{x_{d}}\right)^{\beta/p^{\prime}}\left(\frac{x_{d}\vee
y_{d}}{\sqrt{x_{d}y_{d}}}\right)^{2r}\frac{(x_{d}\vee
y_{d})^{\alpha}}{(|x_{d}-y_{d}|\vee(x_{d}\wedge y_{d}))^{1+\alpha}}$
$\displaystyle=\int_{0}^{\infty}dt\,t^{-\beta/p^{\prime}-r}\frac{(1\vee
t)^{\alpha+2r}}{(|1-t|\vee(1\wedge t))^{1+\alpha}}<\infty\,,$

as before. The $L^{p}(\mathbb{R}^{d}_{+})$-boundedness therefore follows from
the Schur test. ∎

##  6\. Proof of the generalized Hardy inequality (Theorem 1.8)

To prove Theorem 1.8, we use the following Riesz kernel estimates.

######  Lemma 6.1 ([FM23, Theorem 11]).

Let $\alpha\in(0,2]$ and let $\lambda\geq 0$ when $\alpha\in(0,2)$ and
$\lambda\geq-1/2$ when $\alpha=2$. Let $\sigma$ be defined by (5) and let
$s\in(0,\frac{2d}{\alpha}\wedge\frac{2(d+2\sigma)}{\alpha})$. Then,

  1. (a)

For all $x,y\in\mathbb{R}^{d}_{+}$ with $|x-y|\leq x_{d}\vee y_{d}$,

(61) $\displaystyle L_{\lambda}^{-s/2}(x,y)\simeq|x-y|^{\frac{\alpha
s}{2}-d}\Big{(}1\wedge\frac{x_{d}}{|x-y|}\wedge\frac{y_{d}}{|x-y|}\Big{)}^{\sigma}.$

  2. (b)

For all $x,y\in\mathbb{R}^{d}_{+}$ with $x_{d}\vee y_{d}\leq|x-y|$,

(62)
$\displaystyle\begin{split}L_{\lambda}^{-s/2}(x,y)&\simeq|x-y|^{\frac{\alpha
s}{2}-d}\Big{(}\frac{x_{d}y_{d}}{|x-y|^{2}}\Big{)}^{\sigma}\\\
&\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\Big{(}\ln\frac{|x-y|}{x_{d}\vee
y_{d}}\Big{)}{\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+\Big{(}\frac{|x-y|}{x_{d}\vee
y_{d}}\Big{)}^{2\sigma-\alpha(1+\frac{s}{2})}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}.\end{split}$

######  Remark 6.2.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then (61) and (62) remain valid. Similarly, the upper (resp.
lower) bound in (17) implies the upper (resp. lower) bound in (61) and (62).
This was observed in [FM23, Theorem 11].

Besides these Riesz kernel bounds, we use the following extension of [FM23,
Lemma 15], where
$(L_{\lambda})^{s/2}C_{c}^{\infty}(\mathbb{R}_{+}^{d})\subseteq
L^{2}(\mathbb{R}_{+}^{d})$, i.e.,
$C_{c}^{\infty}(\mathbb{R}_{+}^{d})\subset{\operatorname{dom}}L_{\lambda}^{s/2}$
was shown for all $s,\alpha\in(0,2]$.

######  Lemma 6.3.

Let $1<p<\infty$, $\alpha\in(0,2]$, and let $\lambda\geq 0$ when $\alpha<2$
and $\lambda\geq-1/4$ when $\alpha=2$. Let $\sigma$ be defined by (5) and
suppose $s,\alpha\in(0,2]$. Then,
$(L_{\lambda})^{s/2}C_{c}^{\infty}(\mathbb{R}_{+}^{d})\subseteq
L^{p}(\mathbb{R}_{+}^{d})$.

###### Proof.

We first consider $s=2$. Let
$\hat{f}(\xi)=(2\pi)^{-d/2}\int_{\mathbb{R}^{d}}\mathrm{e}^{2\pi
ix\cdot\xi}f(x)\,dx$ denote the Fourier transform of $f$. Then, for
$\lambda_{0}=C(\alpha/2)$, we have (see, e.g., [BBC03])

$L_{\lambda}f=((-\Delta)^{\alpha/2}\widetilde{f})|_{\mathbb{R}^{d}_{+}}+(\lambda-\lambda_{0})x_{d}^{-\alpha}f,$

where $\widetilde{f}$ is the zero extension of $f$ to $\mathbb{R}^{d}$. Since
$((-\Delta)^{\alpha/2}\widetilde{f})^{\wedge}(\xi)=|\xi|^{\alpha}(\widetilde{f})^{\wedge}\
(\xi)$, $((-\Delta)^{\alpha/2}\widetilde{f})$ is a Schwartz function and hence
$((-\Delta)^{\alpha/2}\widetilde{f})\in L^{p}(\mathbb{R}^{d})$. On the other
hand, since $x_{d}^{-\alpha}$ is bounded on the support of $f$,
$x_{d}^{-\alpha}f\in L^{p}(\mathbb{R}^{d}_{+})$. Consequently,
$L_{\lambda}f\in L^{p}(\mathbb{R}^{d}_{+})$ for all $1<p<\infty$.

Now let $s\in(0,2)$. We use

$\displaystyle L_{\lambda}^{s/2}f$
$\displaystyle=\frac{1}{\Gamma(1-s/2)}\int_{0}^{\infty}u^{1-s/2}L_{\lambda}e^{-uL_{\lambda}}f\frac{du}{u}$
$\displaystyle=\frac{1}{\Gamma(1-s/2)}\int_{0}^{1}u^{1-s/2}e^{-uL_{\lambda}}L_{\lambda}f\frac{du}{u}+\frac{1}{\Gamma(1-s/2)}\int_{1}^{\infty}uL_{\lambda}e^{-uL_{\lambda}}f\frac{du}{u^{1+s/2}},$

which implies

$\displaystyle\|L_{\lambda}^{s/2}f\|_{p}$
$\displaystyle=\frac{1}{\Gamma(1-s/2)}\int_{0}^{1}u^{1-s/2}\|e^{-uL_{\lambda}}L_{\lambda}f\|_{p}\frac{du}{u}+\frac{1}{\Gamma(1-s/2)}\int_{1}^{\infty}\|uL_{\lambda}e^{-uL_{\lambda}}f\|_{p}\frac{du}{u^{1+s/2}}.$

From Lemma 2.7 and Proposition 2.4, we have

$\|e^{-uL_{\lambda}}\|_{p\to p}+\|uL_{\lambda}e^{-uL_{\lambda}}\|_{p\to
p}\lesssim 1$

uniformly in $u>0$. Therefore,

$\displaystyle\|L_{\lambda}^{s/2}f\|_{p}$
$\displaystyle=\frac{1}{\Gamma(1-s/2)}\int_{0}^{1}u^{1-s/2}\|L_{\lambda}f\|_{p}\frac{du}{u}+\frac{1}{\Gamma(1-s/2)}\int_{1}^{\infty}\|f\|_{p}\frac{du}{u^{1+s/2}}$
$\displaystyle\lesssim\|L_{\lambda}f\|_{p}+\|f\|_{p}.$

Since $f,L_{\lambda}f\in L^{p}(\mathbb{R}^{d}_{+})$, $L_{\lambda}^{s/2}f\in
L^{p}(\mathbb{R}^{d}_{+})$. ∎

######  Remark 6.4.

Let $\alpha\in(0,2)$, $\lambda\in[\lambda_{*},0)$ and assume that
$\mathrm{e}^{-tL_{\lambda}}(x,y)$ satisfies the bound in (17) with $\sigma$
defined by (5). Then Lemma 6.3 remains valid for for
$\frac{1}{1+\sigma}=:(p_{-\sigma})^{\prime}<p<p_{-\sigma}:=-\frac{1}{\sigma}$.
This follows by the same arguments as in the proof above, taking into account
Remark 2.5.

We can now give the

###### Proof of Theorem 1.8.

The necessity of $(\frac{\alpha s}{2}-\sigma)_{+}<\frac{1}{p}<1+\sigma\wedge
0$ for (13) to hold is proved exactly as in [FM23, Theorem 13], so we omit the
details.

Thus, let $(\frac{\alpha s}{2}-\sigma)_{+}<\frac{1}{p}<1+\sigma\wedge 0$ in
the following. Formula (14) follows from (13) and Lemma 6.3 by setting
$f=(L_{\lambda})^{s/2}g$ with $g\in C_{c}^{\infty}(\mathbb{R}_{+}^{d})$. Thus,
it suffices to show (13) for all $f\in L^{p}(\mathbb{R}_{+}^{d})$. We set

$K(x,y):=x_{d}^{-\alpha s/2}L_{\lambda}^{-s/2}(x,y)$

and show that the integral operator with kernel $K(x,y)$ is bounded on
$L^{p}(\mathbb{R}_{+}^{d})$. We will use weighted Schur tests for the kernel
$K(x,y)$ in four regions corresponding to

$|x-y|\leq 4(x_{d}\wedge y_{d}),4x_{d}\leq|x-y|\leq 4y_{d},4y_{d}\leq|x-y|\leq
4x_{d},$

and

$4(x_{d}\vee y_{d})\leq|x-y|.$

Case 1: $|x-y|\leq 4(x_{d}\wedge y_{d})$. The Schur test in this region was
already carried out in [FM23]. We give the proof for completeness. From Lemma
6.1 in this region,

$K(x,y)\simeq\frac{x_{d}^{-\alpha s/2}}{|x-y|^{d-\frac{\alpha s}{2}}}.$

Since $d-\alpha s/2<d$, we have

$\displaystyle\int_{|x-y|\leq 4(x_{d}\wedge y_{d})}K(x,y)dy$
$\displaystyle\lesssim\int_{|x-y|\leq 4y_{d}}\frac{x_{d}^{-\alpha
s/2}}{|x-y|^{d-\frac{\alpha s}{2}}}dy$ $\displaystyle\lesssim 1.$

For the integral with respect to $dx$, we observe $y_{d}\geq
x_{d}+|x_{d}-y_{d}|\leq x_{d}+|x-y|\leq 5x_{d}$. Therefore,

$\displaystyle\int_{|x-y|\leq 4(x_{d}\wedge y_{d})}K(x,y)dx$
$\displaystyle\lesssim\int_{|x-y|\leq 4y_{d}}\frac{y_{d}^{-\alpha
s/2}}{|x-y|^{d-\frac{\alpha s}{2}}}dx\lesssim 1.$

Case 2: $4x_{d}\leq|x-y|\leq 4y_{d}$. From Lemma 6.1,

$K(x,y)\simeq\frac{x_{d}^{\sigma-\frac{\alpha
s}{2}}}{|x-y|^{d+\sigma-\frac{\alpha s}{2}}}.$

Since $\frac{s\alpha}{2}-\sigma<\frac{1}{p}$ and $1+2\sigma-\alpha s/2>0$, we
can choose $\beta$ such that

(63) $p(\alpha s/2-\sigma)<\beta<\min\\{p^{\prime}(1+\sigma-\alpha
s/2),p(1+\sigma)\\}.$

It follows that

(64) $\sigma-\frac{\alpha s}{2}+\beta/p>0,$ (65) $\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}>-1,$

and

(66) $\sigma-\beta/p>-1.$

Moreover, since $1+\sigma-\alpha s/2>0$ for
$\sigma>\frac{\alpha}{2}\big{(}1+\frac{s}{2}\big{)}$, we can choose $\beta$
such that

(67) $\max\\{0,p(\alpha s/2-\sigma)\\}<\beta<\min\\{p^{\prime}(1+\sigma-\alpha
s/2),p(1+\sigma)\\}.$

as long as $\alpha<2$ and
$\sigma>\frac{\alpha}{2}\big{(}1+\frac{s}{2}\big{)}$. (Note that this
threshold for $\sigma$ also occurs in [FM23, Theorem 11].) We verify the
$L^{p}$-boundedness using a Schur test with weight

$w(x,y)=\Big{(}\frac{x_{d}}{|x-y|}\Big{)}^{\beta},$

i.e., we show that

$\sup_{x\in\mathbb{R}_{+}^{d}}\int_{4x_{d}\leq|x-y|\leq
4y_{d}}w(x,y)^{1/p}K(x,y)dy\lesssim 1$

and

$\sup_{y\in\mathbb{R}_{+}^{d}}\int_{4x_{d}\leq|x-y|\leq
4y_{d}}w(x,y)^{-1/p^{\prime}}K(x,y)dx\lesssim 1.$

From (64), we have

$\displaystyle\int_{4x_{d}\leq|x-y|\leq 4y_{d}}w(x,y)^{1/p}K(x,y)dy$
$\displaystyle\lesssim\int_{4x_{d}\leq|x-y|}\frac{x_{d}^{\sigma-\frac{\alpha
s}{2}+\beta/p}}{|x-y|^{d+\sigma-\frac{\alpha s}{2}+\beta/p}}dy\lesssim 1.$

For the second inequality, we have

$\displaystyle\int_{4x_{d}\leq|x-y|\leq
4y_{d}}w(x,y)^{-\beta/p^{\prime}}K(x,y)dx$
$\displaystyle\lesssim\int_{4x_{d}\leq|x-y|\leq
4y_{d}}\frac{x_{d}^{\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}{|x-y|^{d+\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}dx.$

In this region, we have $|x-y|\geq y_{d}-x_{d}\geq y_{d}-|x-y|/4$. Thus,
$|x-y|\geq 4y_{d}/5$ and, by (65),

$\displaystyle\int_{4x_{d}\leq|x-y|\leq
4y_{d}}w(x,y)^{-\beta/p^{\prime}}K(x,y)dx$
$\displaystyle\lesssim\int_{4x_{d}\leq|x-y|\leq
4y_{d}}\frac{x_{d}^{\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}{y_{d}^{d+\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}dx$
$\displaystyle\lesssim\int_{0}^{y_{d}}\int_{|x^{\prime}-y^{\prime}|\leq
4y_{d}}\frac{x_{d}^{\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}{y_{d}^{d+\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}dx^{\prime}dx_{d}$
$\displaystyle\lesssim\int_{0}^{y_{d}}\frac{x_{d}^{\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}{y_{d}^{1+\sigma-\frac{\alpha
s}{2}-\beta/p^{\prime}}}dx_{d}\lesssim 1.$

Case 3: $4y_{d}\leq|x-y|\leq 4x_{d}$. In this case,

$K(x,y)\simeq\frac{x_{d}^{-\frac{\alpha
s}{2}}y_{d}^{\sigma}}{|x-y|^{d+\sigma-\frac{\alpha s}{2}}}.$

Since $\frac{1}{p}<1+\sigma$, we can choose $\gamma$ such that

$\sigma p^{\prime}<\gamma<p(1+\sigma),$

which implies

(68) $\sigma-\gamma/p>-1\ \ \text{and}\ \ \sigma+\gamma/p^{\prime}>0.$

We will verify a Schur test with weight

$w(x,y)=\Big{(}\frac{|x-y|}{y_{d}}\Big{)}^{\gamma}.$

We first show that

$\int_{4y_{d}\leq|x-y|\leq 4x_{d}}w(x,y)^{1/p}K(x,y)dy\lesssim 1.$

To do this, we write

$\displaystyle\int_{4y_{d}\leq|x-y|\leq 4x_{d}}w(x,y)^{1/p}K(x,y)dy$
$\displaystyle\simeq\int_{4y_{d}\leq|x-y|\leq
4x_{d}}\Big{(}\frac{|x-y|}{y_{d}}\Big{)}^{\gamma/p}\frac{x_{d}^{-\frac{\alpha
s}{2}}y_{d}^{\sigma}}{|x-y|^{d+\sigma-\frac{\alpha s}{2}}}dy$
$\displaystyle\simeq\int_{4y_{d}\leq|x-y|\leq
4x_{d}}\frac{x_{d}^{-\frac{\alpha
s}{2}}y_{d}^{\sigma-\gamma/p}}{|x-y|^{d+\sigma-\gamma/p-\frac{\alpha
s}{2}}}dy$

Similarly to Case 2, in this region, $|x-y|\geq x_{d}-y_{d}\geq
x_{d}-|x-y|/4$, which implies that $|x-y|\geq 4x_{d}/5$. Hence, $|x-y|\simeq
x_{d}$ in this region. Consequently,

$\displaystyle\int_{4y_{d}\leq|x-y|\leq 4x_{d}}w(x,y)^{1/p}K(x,y)dy$
$\displaystyle\lesssim\int_{0}^{x_{d}}\int_{|x^{\prime}-y^{\prime}|\leq
4x_{d}}\frac{x_{d}^{-\frac{\alpha
s}{2}}y_{d}^{\sigma-\gamma/p}}{x_{d}^{d+\sigma-\gamma/p-\frac{\alpha
s}{2}}}dy^{\prime}dy_{d}$
$\displaystyle\lesssim\int_{0}^{x_{d}}\frac{x_{d}^{d-\frac{\alpha
s}{2}-1}y_{d}^{\sigma-\gamma/p}}{x_{d}^{d+\sigma-\gamma/p-\frac{\alpha
s}{2}}}dy_{d}$
$\displaystyle\lesssim\int_{0}^{x_{d}}x_{d}^{\gamma/p-\sigma-1}y_{d}^{\sigma-\gamma/p}dy_{d}\lesssim
1,$

where in the last inequality we used (68).

It remains to show that

$\int_{4y_{d}\leq|x-y|\leq 4x_{d}}w(x,y)^{-1/p^{\prime}}K(x,y)dx\lesssim 1.$

Since in this case $|x-y|\simeq x_{d}$, by (68) we have

$\displaystyle\int_{4y_{d}\leq|x-y|\leq 4x_{d}}w(x,y)^{-1/p^{\prime}}K(x,y)dx$
$\displaystyle\simeq\int_{|x-y|\geq
4y_{d}}\Big{(}\frac{|x-y|}{y_{d}}\Big{)}^{-\gamma/p^{\prime}}\frac{y_{d}^{\sigma}}{|x-y|^{d+\sigma}}dx$
$\displaystyle\lesssim\int_{|x-y|\geq
4y_{d}}\frac{y_{d}^{\sigma+\gamma/p^{\prime}}}{|x-y|^{d+\sigma+\gamma/p^{\prime}}}dx\lesssim
1.$

Case 4: $4(x_{d}\vee y_{d})\leq|x-y|$. In this case,

$\displaystyle K$ $\displaystyle(x,y)\simeq x_{d}^{-\alpha
s/2}|x-y|^{\frac{\alpha
s}{2}-d}\Big{(}\frac{x_{d}y_{d}}{|x-y|^{2}}\Big{)}^{\sigma}$
$\displaystyle\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\Big{(}\ln\frac{|x-y|}{x_{d}\vee
y_{d}}\Big{)}{\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+\Big{(}\frac{|x-y|}{x_{d}\vee
y_{d}}\Big{)}^{2\sigma-\alpha(1+\frac{s}{2})}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}.$

We will perform a Schur test with weight

$w(x,y)=\Big{(}\frac{x_{d}}{y_{d}}\Big{)}^{\beta},$

where $\beta$ is defined by (63).

Since $|x-y|\geq 4(x_{d}\vee y_{d})$, $|x-y|\simeq|x^{\prime}-y^{\prime}|$.
Then we have

$\displaystyle\int_{\mathbb{R}^{d}_{+}}$ $\displaystyle w(x,y)^{1/p}K(x,y)dy$
$\displaystyle\lesssim\sum_{j\geq
1}\int_{0}^{2^{j}x_{d}}\int_{2^{j}x_{d}<|x^{\prime}-y^{\prime}|\leq
2^{j+1}x_{d}}w(x,y)^{1/p}K(x,y)dy^{\prime}dy_{d}$
$\displaystyle\lesssim\sum_{j\geq 1}\int_{0}^{2^{j}x_{d}}x_{d}^{-\alpha
s/2}(2^{j}x_{d})^{\frac{\alpha
s}{2}-1}\Big{(}\frac{x_{d}y_{d}}{(2^{j}x_{d})^{2}}\Big{)}^{\sigma}\Big{(}\frac{x_{d}}{y_{d}}\Big{)}^{\beta/p}$
$\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\Big{(}\ln\frac{2^{j}x_{d}}{x_{d}\vee
y_{d}}\Big{)}{\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+\Big{(}\frac{2^{j}x_{d}}{x_{d}\vee
y_{d}}\Big{)}^{2\sigma-\alpha(1+\frac{s}{2})}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}dy_{d}.$

By a straightforward calculation, we get

$\displaystyle\int_{\mathbb{R}^{d}_{+}}$ $\displaystyle w(x,y)^{1/p}K(x,y)dy$
$\displaystyle\lesssim\sum_{j\geq 1}2^{j(\alpha
s/2-1-2\sigma)}\int_{0}^{2^{j}x_{d}}x_{d}^{-1-\sigma+\beta/p}y_{d}^{\sigma-\beta/p}dy_{d}$
$\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\ln(2^{j}){\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+2^{j(2\sigma-\alpha(1+\frac{s}{2}))}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}.$

From (66),

$\displaystyle\int_{\mathbb{R}^{d}_{+}}$ $\displaystyle w(x,y)^{1/p}K(x,y)dy$
$\displaystyle\lesssim\sum_{j\geq 1}2^{j(\alpha
s/2-1-2\sigma)}2^{j(1+\sigma-\beta/p)}$ $\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\ln(2^{j}){\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+2^{j(2\sigma-\alpha(1+\frac{s}{2}))}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}$
$\displaystyle\lesssim\sum_{j\geq 1}2^{j(\alpha s/2-\sigma-\beta/p)}$
$\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\ln(2^{j}){\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+2^{j(2\sigma-\alpha(1+\frac{s}{2}))}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}.$

This, along with (64), implies

$\displaystyle\int_{\mathbb{R}^{d}_{+}}$ $\displaystyle w(x,y)^{1/p}K(x,y)dy$
$\displaystyle\lesssim 1+\sum_{j\geq 1}2^{j(\alpha
s/2-\sigma-\beta/p)}2^{j(2\sigma-\alpha(1+\frac{s}{2}))}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\mathbf{1}_{\alpha<2}$
$\displaystyle\lesssim 1+\sum_{j\geq
1}2^{j(\sigma-\alpha-\beta/p)}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\mathbf{1}_{\alpha<2}\lesssim
1,$

where in the last inequality we used (67) and the fact $\sigma<\alpha$.

For the second inequality, similarly to the first inequality

$\displaystyle\int_{\mathbb{R}^{d}_{+}}$ $\displaystyle
w(x,y)^{-1/p^{\prime}}K(x,y)dx$ $\displaystyle\lesssim\sum_{j\geq
1}\int_{0}^{2^{j}y_{d}}\int_{2^{j}y_{d}<|x^{\prime}-y^{\prime}|\leq
2^{j+1}y_{d}}w(x,y)^{-1/p^{\prime}}K(x,y)dx^{\prime}dx_{d}$
$\displaystyle\lesssim\sum_{j\geq 1}\int_{0}^{2^{j}y_{d}}x_{d}^{-\alpha
s/2}(2^{j}y_{d})^{\frac{\alpha
s}{2}-1}\Big{(}\frac{x_{d}y_{d}}{(2^{j}y_{d})^{2}}\Big{)}^{\sigma}\Big{(}\frac{x_{d}}{y_{d}}\Big{)}^{-\beta/p^{\prime}}$
$\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\Big{(}\ln\frac{2^{j}y_{d}}{x_{d}\vee
y_{d}}\Big{)}{\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+\Big{(}\frac{2^{j}y_{d}}{x_{d}\vee
y_{d}}\Big{)}^{2\sigma-\alpha(1+\frac{s}{2})}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}dx_{d}.$

By a simple calculation, we get

$\displaystyle\int_{\mathbb{R}^{d}_{+}}$ $\displaystyle w(x,y)^{1/p}K(x,y)dx$
$\displaystyle\lesssim\sum_{j\geq 1}2^{j(\alpha
s/2-1-2\sigma)}\int_{0}^{2^{j}y_{d}}x_{d}^{-\alpha
s/2+\sigma-\beta/p^{\prime}}y_{d}^{\alpha
s/2-1-\sigma+\beta/p^{\prime}}dx_{d}$ $\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\ln(2^{j}){\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+2^{j(2\sigma-\alpha(1+\frac{s}{2}))}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}.$

This, along with (65), yields

$\displaystyle\int_{\mathbb{R}^{d}_{+}}$ $\displaystyle w(x,y)^{1/p}K(x,y)dx$
$\displaystyle\lesssim\sum_{j\geq 1}2^{j(\alpha
s/2-1-2\sigma+(\gamma-\beta)/p)}2^{j(1+\sigma-\gamma/p)}$ $\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\ln(2^{j}){\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+2^{j(2\sigma-\alpha(1+\frac{s}{2}))}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}$
$\displaystyle\lesssim\sum_{j\geq 1}2^{j(\alpha s/2-\sigma-\beta/p)}$
$\displaystyle\ \
\times\Big{[}\mathbf{1}_{\alpha=2}+\Big{(}\mathbf{1}_{\sigma\leq\frac{\alpha}{2}(1+\frac{s}{2})}+\ln(2^{j}){\mathbf{1}_{\sigma=\frac{\alpha}{2}(1+\frac{s}{2})}}+2^{j(2\sigma-\alpha(1+\frac{s}{2}))}\mathbf{1}_{\sigma>\frac{\alpha}{2}(1+\frac{s}{2})}\Big{)}\mathbf{1}_{\alpha<2}\Big{]}.$

At this stage, arguing similarly to the first inequality we also come up with

$\int_{\mathbb{R}^{d}_{+}}w(x,y)^{1/p}K(x,y)dx\lesssim 1.$

This completes the proof of (13) under the assumption $(\frac{\alpha
s}{2}-\sigma)_{+}<\frac{1}{p}<1+\sigma\wedge 0$. ∎

##  7\. Density of $(L_{\lambda})^{s/2}C_{c}^{\infty}(\mathbb{R}^{d}_{+})$ in
$L^{p}(\mathbb{R}_{+}^{d})$

The strategy to prove Theorem 1.10 is the same as in [FM23, Theorem 25]. We
first prove Theorem 1.10 for $f$ of a special form, namely, $f\in
L_{\lambda}^{s/2}\mathrm{e}^{-tL_{\lambda}}C_{c}^{\infty}(\mathbb{R}^{d}_{+})$
for some $0<t<\infty$. To do so, we use the following pointwise bounds on
functions in $\mathrm{e}^{-tL_{\lambda}}C_{c}^{\infty}(\mathbb{R}^{d}_{+})$.
For a function $u$ on a set $\Omega$ and $0<\beta\leq 2$, we define its
$\beta$-th Hölder seminorm as

(69)
$[u]_{C^{\beta}(\Omega)}:=\begin{cases}\sup_{x,y\in\Omega}\frac{|u(x)-u(y)|}{|x-y|^{\beta}}&\text{if}\
0<\beta\leq 1\,,\\\ \sup_{x,y\in\Omega}\frac{|\nabla u(x)-\nabla
u(y)|}{|x-y|^{\beta-1}}&\text{if}\ 1<\beta\leq 2\,.\end{cases}$

######  Lemma 7.1 ([FM23, Lemma 27]).

Let $\alpha$, $\lambda$ and $\sigma$ be as in Theorem 1.10. Let $0<t<\infty$
and $\psi\in\mathrm{e}^{-tL_{\lambda}}C_{c}^{\infty}(\mathbb{R}^{d}_{+})$.
Then, for all $x\in\mathbb{R}^{d}_{+}$,

(70) $\displaystyle|\psi(x)|$ $\displaystyle\lesssim(1\wedge
x_{d})^{\sigma}(1\wedge|x|^{-d-\alpha})\,,$ (71)
$\displaystyle|L_{\lambda}\psi(x)|$ $\displaystyle\lesssim(1\wedge
x_{d})^{\sigma}(1\wedge|x|^{-d-\alpha})\,,$ (72)
$\displaystyle|(-\Delta)^{\alpha/2}\psi(x)|$ $\displaystyle\lesssim(1\wedge
x_{d})^{\sigma-\alpha}(1\wedge|x|^{-d-\alpha})\,,$ (73)
$\displaystyle[\psi]_{C^{\beta}(B_{\ell_{x}}(x))}$
$\displaystyle\lesssim(1\wedge
x_{d})^{\sigma-\beta}(1\wedge|x|^{-d-\alpha})\quad\text{with}\
\ell_{x}:=1\wedge\tfrac{x_{d}}{2}\,,\ 0<\beta<\alpha.$

As noted in [FM23] the decay in the bounds for $\alpha=2$ can be vastly
improved, but it is convenient for us to have a unified statement. To give the
proof of Theorem 1.10, we use, for two parameters $0<r\leq 1\leq R<\infty$,
the commutator bounds in Corollary 7.2 below. These bounds involve functions
satisfying

(74) $0\leq\chi\leq 1\,,\qquad\chi(x)=1\ \text{if}\ |x|\leq
R\,,\qquad\chi(x)=0\ \text{if}\ |x|\geq 2R\,,\qquad|\nabla\chi|\lesssim
R^{-1}\,,$

and

(75) $0\leq\theta\leq 1\,,\qquad\theta(x)=0\ \text{if}\ x_{d}\leq
r\,,\qquad\theta(x)=1\ \text{if}\ x_{d}\geq 2r\,,\qquad|\nabla\theta|\lesssim
r^{-1}\,,$

and, if $\alpha\geq 1$, also the following bounds on their Hessians, namely

(76) $|D^{2}\chi|\lesssim R^{-2},$

and

(77) $|D^{2}\theta|\lesssim r^{-2}.$

######  Corollary 7.2.

Let $p\in(1,\infty)$ and $0<\alpha<2$. Let $0<r\leq 1\leq R<\infty$, assume
that $\chi$ and $\theta$ satisfy (74) and (75) and, if $\alpha\geq 1$, also
(76) and (77). Let $\frac{\alpha-1}{2}\leq\sigma<\alpha$, assume that $v$
satisfies

(78) $|v(x)|\leq(1\wedge|x|^{-d-\alpha})(1\wedge
x_{d})^{\sigma}\qquad\text{for all}\ x\in\mathbb{R}^{d}_{+}\,$

and, if $\alpha\geq 1$, also

(79) $[v]_{C^{\beta}(B_{\ell_{x}}(x))}\leq(1\wedge|x|^{-d-\alpha})\,(1\wedge
x_{d})^{\sigma-\beta}\qquad\text{for all}\ x\in\mathbb{R}^{d}_{+}\
\text{with}\ \ell_{x}:=1\wedge\tfrac{x_{d}}{2}$

with some $\beta>\alpha-1$. Then

$\|[(-\Delta)^{\alpha/2},\chi\theta]v\|_{L^{p}(\mathbb{R}^{d}_{+})}\lesssim
r^{\sigma-\alpha+1/p}+R^{-\alpha-d/p^{\prime}}\,.$

###### Proof.

The proof is exactly as in [FM23, Corollary 23]. For an explicit constant
$c_{d,\alpha}>0$, we estimate the $L^{p}$-norm of the right-hand side of

(80) $\displaystyle
c_{d,\alpha}[(-\Delta)^{\alpha/2},\chi\theta]v(x)=\theta(x)\int_{\mathbb{R}^{d}}\frac{\chi(x)-\chi(y)}{|x-y|^{d+\alpha}}v(y)\,dy+\int_{\mathbb{R}^{d}}\frac{\theta(x)-\theta(y)}{|x-y|^{d+\alpha}}\chi(y)v(y)\,dy.$

For the first summand on the right-hand side, we estimate $|\theta(x)|\leq 1$
and use the pointwise bounds in [FM23, Lemmas 20], i.e.,

(81)
$\displaystyle\begin{split}&\left|\int_{\mathbb{R}^{d}}\frac{\chi(x)-\chi(y)}{|x-y|^{d+\alpha}}v(y)\,dy\right|\lesssim\mathbf{1}_{|x|\leq
R}R^{-d-2\alpha}+\mathbf{1}_{|x|>R}|x|^{-d-\alpha}\\\
&\quad+\mathbf{1}_{\alpha\geq 1}\mathbf{1}_{|x|\simeq
R}R^{-d-\alpha-1}\left((1\wedge
x_{d})^{-(\sigma-\alpha+1)_{-}}+\mathbf{1}_{\sigma=\alpha-1}\ln\tfrac{1}{1\wedge
x_{d}}+\mathbf{1}_{\alpha=1}\ln R\right)\\\ &\quad\text{for all}\
x\in\mathbb{R}^{d}_{+}\end{split}$

and all $v$ satisfying the assumptions of the present corollary. The
$L^{p}$-norm of this right-hand side is
$\mathcal{O}(R^{-\alpha-d/p^{\prime}})$ for $R>1$. For the second summand on
the right-hand side of (80), we use [FM23, Lemmas 21], i.e.,

(82)
$\displaystyle\begin{split}\left|\int_{\mathbb{R}^{d}}\frac{\theta(x)-\theta(y)}{|x-y|^{d+\alpha}}(\chi
v)(y)\,dy\right|\lesssim\left(r^{\sigma-\alpha}\wedge\frac{r^{\sigma+1}}{x_{d}^{1+\alpha}}\right)(1+x_{d})^{1+\alpha}(1\wedge|x|^{-d-\alpha})\quad\text{for
all}\ x\in\mathbb{R}_{+}^{d}\end{split}$

whenever $\chi v$ satisfies the assumptions of the present corollary. By
[FM23, Corollary 23] the function $\chi v$ indeed satisfies these assumptions
when $v$ does. Since the $L^{p}$-norm of the right-hand side of (82) is
$\mathcal{O}(r^{\sigma-\alpha+1/p})$ for $r<1$, the proof is concluded. ∎

###### Proof of Theorem 1.10.

_Step 1._ We first prove this theorem for
$f\in(L_{\lambda})^{s/2}\mathrm{e}^{-tL_{\lambda}}C_{c}^{\infty}(\mathbb{R}^{d}_{+})$
for some $0<t<\infty$. Let $0<t<\infty$ and let
$\psi\in\mathrm{e}^{-tL_{\lambda}}C_{c}^{\infty}(\mathbb{R}^{d})$. For
parameters $0<r\leq 1\leq R<\infty$ to be determined, let $\chi$ and $\theta$
be as in Corollary 7.2 and

$\phi:=\chi\theta\psi.$

Then, by (70),

$\|\phi-\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}\leq\|\mathbf{1}_{x_{d}\leq
2r}\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}+\|\mathbf{1}_{|x|>R}\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}\lesssim
r^{\sigma+1/p}+R^{-\alpha-d/p^{\prime}}\,.$

Moreover,

$\|L_{\lambda}(\phi-\psi)\|_{L^{p}(\mathbb{R}^{d}_{+})}\leq\|(1-\chi\theta)L_{\lambda}\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}+\|[(-\Delta)^{\alpha/2},\chi\theta]\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}$

and, by (71),

$\|(1-\chi\theta)L_{\lambda}\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}\lesssim
r^{\sigma+1/p}+R^{-\alpha-d/p^{\prime}}\,.$

For $\alpha<2$ we apply Corollary 7.2 and find

$\|[(-\Delta)^{\alpha/2},\chi\theta]\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}\lesssim
r^{\sigma-\alpha+1/p}+R^{-\alpha-d/p^{\prime}}\,.$

The same bound holds for $\alpha=2$ as well, as follows by writing

$[-\Delta,\chi\theta]\psi=-2\nabla(\chi\theta)\cdot\nabla\psi-\Delta(\chi\theta)\psi$

and using the pointwise bounds (70) and (73). Thus, for all $\alpha\leq 2$,

$\|L_{\lambda}(\phi-\psi)\|_{L^{p}(\mathbb{R}^{d}_{+})}\lesssim
r^{\sigma-\alpha+1/p}+R^{-\alpha-d/p^{\prime}}\,.$

Since $0<s\leq 2$ we have, by interpolation,

$\|L_{\lambda}^{s/2}(\phi-\psi)\|_{L^{p}(\mathbb{R}^{d}_{+})}\leq\|\phi-\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}^{1-s/2}\|L_{\lambda}(\phi-\psi)\|_{L^{p}(\mathbb{R}^{d}_{+})}^{s/2}\,.$

Inserting the above bounds, we get

$\|L_{\lambda}^{s/2}(\phi-\psi)\|_{L^{p}(\mathbb{R}^{d}_{+})}\lesssim
r^{\sigma+1/p-\alpha s/2}+R^{-\alpha-d/p^{\prime}}.$

Since, by assumption $s<2(1/p+\sigma)/\alpha$, this tends to zero as $r\to 0$
and $R\to\infty$.

_Step 2._ We now prove Theorem 1.10 in the general case. Let $f\in
L^{p}(\mathbb{R}^{d}_{+})$ and $\epsilon>0$. By Lemma 2.10, there exist
$t:=t_{1}/2$ and $T:=t_{2}/2$ such that

(83)
$\|(\mathrm{e}^{-2tL_{\lambda}}-\mathrm{e}^{-2TL_{\lambda}})f-f\|_{L^{p}(\mathbb{R}^{d}_{+})}\leq\epsilon\,.$

On the other hand,

$\mathrm{e}^{-2tL_{\lambda}}-\mathrm{e}^{-2TL_{\lambda}}f=L_{\lambda}^{s/2}(\mathrm{e}^{-tL_{\lambda}}+\mathrm{e}^{-TL_{\lambda}})L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})f.$

By Lemma 2.8,

$\|L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})f\|\lesssim(T^{s/2}-t^{s/2})\|f\|_{p}.$

Since $C_{c}^{\infty}(\mathbb{R}^{d}_{+})$ is dense in
$L^{p}(\mathbb{R}^{d}_{+})$, there is a $k\in
C^{\infty}_{c}(\mathbb{R}^{d}_{+})$ such that

(84)
$\|k-L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})f\|_{L^{p}(\mathbb{R}^{d}_{+})}\leq\frac{\epsilon}{\|L_{\lambda}^{s/2}(\mathrm{e}^{-tL_{\lambda}}+\mathrm{e}^{-TL_{\lambda}})\|_{p\to
p}}.$

We define
$\psi:=L_{\lambda}^{s/2}(\mathrm{e}^{-tL_{\lambda}}+\mathrm{e}^{-TL_{\lambda}})k$.
Then, according to Step 1 (applied both to
$L_{\lambda}^{s/2}\mathrm{e}^{-tL_{\lambda}}k$ and to
$L_{\lambda}^{s/2}\mathrm{e}^{-TL_{\lambda}}k$) there is a $\phi\in
C_{c}^{\infty}$ such that

(85)
$\|L_{\lambda}^{s/2}\phi-\psi\|_{L^{p}(\mathbb{R}^{d}_{+})}\leq\epsilon\,.$

Therefore,

$\displaystyle\|L_{\lambda}^{s/2}\phi-f\|_{p}$
$\displaystyle\leq\|L_{\lambda}^{s/2}\phi-\psi\|_{p}+\|\psi-
L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}+\mathrm{e}^{-TL_{\lambda}})L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})\mathrm{e}^{-TL_{\lambda}})f\|_{p}$
$\displaystyle\ \
+\|L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})f-f\|_{p}.$

By (85),

$\|L_{\lambda}^{s/2}\phi-\psi\|_{p}\leq\epsilon.$

Using (84) and Lemma 2.8,

$\displaystyle\|\psi-
L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}+\mathrm{e}^{-TL_{\lambda}})\mathrm{e}^{-TL_{\lambda}})\mathrm{e}^{-TL_{\lambda}})f\|_{p}$
$\displaystyle\quad=\|L_{\lambda}^{s/2}(\mathrm{e}^{-tL_{\lambda}}+\mathrm{e}^{-TL_{\lambda}})k-L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})\mathrm{e}^{-TL_{\lambda}})f\|_{p}$
$\displaystyle\quad\leq\|L_{\lambda}^{s/2}(\mathrm{e}^{-tL_{\lambda}}+\mathrm{e}^{-TL_{\lambda}})\|_{p\to
p}\|k-L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})f\|_{L^{p}(\mathbb{R}^{d}_{+})}\leq\epsilon.$

Finally, by (83),

$\displaystyle\|L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})L_{\lambda}^{-s/2}(\mathrm{e}^{-tL_{\lambda}}-\mathrm{e}^{-TL_{\lambda}})f-f\|_{p}$
$\displaystyle=\|(\mathrm{e}^{-2tL_{\lambda}}-\mathrm{e}^{-2TL_{\lambda}})f-f\|_{p}\leq\epsilon.$

Consequently, for any $\epsilon>0$ there is a $\phi\in C_{c}^{\infty}$ such
that

$\|L_{\lambda}^{-s/2}\phi-f\|_{p}\leq 3\epsilon.$

This concludes the proof. ∎

## References

  * [Aus07] Pascal Auscher.  On necessary and sufficient conditions for $L^{p}$-estimates of Riesz transforms associated to elliptic operators on $\mathbb{R}^{n}$ and related estimates.  Mem. Amer. Math. Soc., 186(871):xviii+75, 2007.
  * [BBC03] Krzysztof Bogdan, Krzysztof Burdzy, and Zhen-Qing Chen.  Censored stable processes.  Probab. Theory Related Fields, 127(1):89–152, 2003.
  * [BD11] Krzysztof Bogdan and Bartłomiej Dyda.  The best constant in a fractional Hardy inequality.  Math. Nachr., 284(5-6):629–638, 2011.
  * [BD23] The Anh Bui and Piero D’Ancona.  Generalized Hardy operators.  Nonlinearity, 36(1):171–198, 2023.
  * [BM23] Krzysztof Bogdan and Konstantin Merz.  Subordinated Bessel heat kernels.  arXiv e-prints, page arXiv:2308.15026, August 2023.
  * [BN22] The Anh Bui and Georges Nader.  Hardy spaces associated to generalized Hardy operators and applications.  NoDEA Nonlinear Differential Equations Appl., 29(4):Paper No. 40, 40, 2022.
  * [BS02] Andrei N. Borodin and Paavo Salminen.  Handbook of Brownian Motion — Facts and Formulae.  Probability and its Applications. Birkhäuser Verlag, Basel, second edition, 2002.
  * [CK03] Zhen-Qing Chen and Takashi Kumagai.  Heat kernel estimates for stable-like processes on $d$-sets.  Stochastic Process. Appl., 108(1):27–62, 2003.
  * [CKS10] Zhen-Qing Chen, Panki Kim, and Renming Song.  Two-sided heat kernel estimates for censored stable-like processes.  Probab. Theory Related Fields, 146(3-4):361–399, 2010.