# Subordinated Bessel heat kernels

Krzysztof Bogdan  Department of Pure and Applied Mathematics, Wrocław
University of Science and Technology, Hoene-Wrońskiego 13C, 50-376 Wrocław,
Poland<EMAIL_ADDRESS>and  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: August 29, 2023)

###### Abstract.

We prove new bounds for Bessel heat kernels and Bessel heat kernels
subordinated by stable subordinators. In particular, we provide a 3G
inequality in the subordinated case.

###### Key words and phrases:

Bessel heat kernel, stable subordinator, 3G inequality

K.B. was supported through the DFG-NCN Beethoven Classic 3 programme, contract
no. 2018/31/G/ST1/02252 (National Science Center, Poland) and SCHI-419/11–1
(DFG, Germany). K.M. was 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 and main result
  2. 2 Fundamental properties, bounds, and explicit expressions
  3. 3 3G inequality for $\alpha\in(0,2)$
  4. 4 Comparability results for $p_{\zeta}^{(\alpha)}$
  5. A Proof of Theorem 2.1 for $\alpha\in(0,2)$
    1. A.1 Auxiliary bounds
      1. A.1.1 Case $rs\vee(r-s)^{2}<1$
      2. A.1.2 Case $(r-s)^{2}<1<rs$
      3. A.1.3 Case $rs<1<(r-s)^{2}$
      4. A.1.4 Case $rs\wedge(r-s)^{2}>1$
    2. A.2 Proof of (2.9)

##  1\. Introduction and main result

The Bessel differential operators
$-\tfrac{d^{2}}{dr^{2}}-\tfrac{2\zeta}{r}\,\tfrac{d}{dr}$ are of fundamental
importance in mathematics, physics, finance, and engineering. They are used to
describe heat and wave propagation under spherical symmetry. The corresponding
Bessel heat kernels are crucial for harmonic analysis and stochastic processes
related to the Brownian motion, see, e.g., Revuz and Yor [RY99]. Our goal is
to present estimates for the $\alpha/2$-subordinated Bessel heat kernels
$p_{\zeta}^{(\alpha)}$, $\alpha\in(0,2]$. We are motivated by the appearance
of $p_{\zeta}^{(\alpha)}$ in the ground state representation given in our work
[BM23a]. In fact, the estimates will be essential in our forthcoming papers
[BM23b, BM23c], but since they are of independent wide interest, we present
them in this separate note.

We now introduce our setting. For $\zeta\in(-1/2,\infty)$, we define the
Bessel heat kernel

$\displaystyle\begin{split}p_{\zeta}^{(2)}(t,r,s)&:=\frac{(rs)^{1/2-\zeta}}{2t}\exp\left(-\frac{r^{2}+s^{2}}{4t}\right)I_{\zeta-1/2}\left(\frac{rs}{2t}\right),\quad
r,s,t>0.\end{split}$ (1.1)

Here and below, for $z\in\mathbb{C}\setminus(-\infty,0]$, $I_{\nu}(z)$ denotes
the modified Bessel function of the first kind of order $\nu\in\mathbb{C}$
[DLMF, (10.25.2)].

The kernel $p_{\zeta}^{(2)}(t,r,s)$ with the reference (speed) measure
$r^{2\zeta}dr$ on $\mathbb{R}_{+}$ is the transition density of the Bessel
process of order $\zeta-1/2$ _reflected at the origin_. We remark that the
Bessel process of order $\zeta-1/2$ _killed at the origin_ has the transition
density (1.1) with $I_{\zeta-1/2}(\cdot)$ replaced with
$I_{|\zeta-1/2|}(\cdot)$, see Borodin and Salminen [BS02, Appendix 1.21, p.
133–134]. The distinction is superfluous when $\zeta\geq 1/2$ because on the
one hand $|\zeta-1/2|=\zeta-1/2$ and on the other hand the Bessel process does
not hit the origin, so no conditions (reflecting or killing) are to be imposed
at the origin. See also Malecki, Serafin, and Żórawik [MSŻ16]. Recall that
$p_{(d-1)/2}^{(2)}(t,r,s)$ is the transition density of the radial part of the
Brownian motion in $\mathbb{R}^{d}$ with the clock $2t$. For further
information on $p_{\zeta}^{(2)}$, we refer, e.g., to the textbooks [BS02, Part
I, Section IV.6 or Appendix 1.21] or [RY99, Chapter XI].

We now define the $\tfrac{\alpha}{2}$-subordinated Bessel heat kernels for
$\alpha\in(0,2)$. Recall that for $\alpha\in(0,2)$ and $t>0$, by Bernstein’s
theorem, the completely monotone function
$[0,\infty)\ni\lambda\mapsto\mathrm{e}^{-t\lambda^{\alpha/2}}$ is the Laplace
transform of a probability density function
$\mathbb{R}_{+}\ni\tau\mapsto\sigma_{t}^{(\alpha/2)}(\tau)$. Thus,

$\displaystyle\mathrm{e}^{-t\lambda^{\alpha/2}}=\int_{0}^{\infty}\mathrm{e}^{-\tau\lambda}\,\sigma_{t}^{(\alpha/2)}(\tau)\,d\tau,\quad
t>0,\,\lambda\geq 0,$ (1.2)

see, e.g., Schilling, Song, and Vondraček [SSV12, (1.4) and Chapter 5]. In
[BM23a, Appendix B] we list some useful properties of and sharp estimates for
$\sigma_{t}^{(\alpha/2)}(\tau)$ and references. We define the
$\tfrac{\alpha}{2}$-subordinated Bessel heat kernel with the reference measure
$r^{2\zeta}dr$ on $\mathbb{R}_{+}$ as

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(t,r,s)&:=\int_{0}^{\infty}p_{\zeta}^{(2)}(\tau,r,s)\,\sigma_{t}^{(\alpha/2)}(\tau)\,d\tau,\quad
r,s,t>0.\end{split}$ (1.3)

We should note that more general subordination and references are discussed in
Grzywny and Trojan [GT21] and [SSV12], but our main motivation for this study
is the fact that $p_{\zeta}^{(\alpha)}$ arises when considering the
$d$-dimensional fractional Laplacian $(-\Delta)^{\alpha/2}$ on the space of
multiples of solid harmonics, i.e., functions of the form
$[u]_{\ell,m}(x):=u(|x|)|x|^{\ell}Y_{\ell,m}(x/|x|)$, where $Y_{\ell,m}$ is a
$L^{2}(\mathbb{S}^{d-1})$-normalized spherical harmonic, $u\in
L^{2}(\mathbb{R}_{+},r^{2\zeta}dr)$, and $\zeta=(d-1+2\ell)/2$. Namely,

$\displaystyle\begin{split}\langle[u]_{\ell,m},\mathrm{e}^{-t(-\Delta)^{\alpha/2}}(t,\cdot,\cdot)[u]_{\ell,m}\rangle_{L^{2}(\mathbb{R}^{d})}&=\langle
u,p_{(d-1+2\ell)/2}^{(\alpha)}(t,\cdot,\cdot)u\rangle_{L^{2}(\mathbb{R}_{+},r^{2\zeta}dr)},\quad
t>0.\end{split}$ (1.4)

Furthermore, the following equality holds pointwise,

$\displaystyle p_{(d-1+2\ell)/2}^{(\alpha)}(t,r,s)$ (1.5)
$\displaystyle\quad=\iint\limits_{\mathbb{S}^{d-1}\times\mathbb{S}^{d-1}}\overline{[1]_{\ell,m}(r\omega_{x})}[1]_{\ell,m}(s\omega_{y})\mathrm{e}^{-t(-\Delta)^{\alpha/2}}(t,r\omega_{x},s\omega_{y})\,d\omega_{x}\,d\omega_{y},\quad
r,s,t>0,$

where $\mathrm{e}^{-t(-\Delta)^{\alpha/2}}$ is the heat kernel of
$(-\Delta)^{\alpha/2}$ on $\mathbb{R}^{d}$. See [BM23a] for details.

### Organization and notation

In Section 2, we recall and prove sharp upper and lower bounds for
$p_{\zeta}^{(\alpha)}(t,r,s)$ (Theorem 2.1), discuss
$p_{\zeta}^{(\alpha)}(t,r,s)$ as probability transition density and kernel of
a strongly continuous contraction semigroup on
$L^{2}(\mathbb{R}_{+},r^{2\zeta}dr)$ (Proposition 2.2), and recall an explicit
expression for $p_{\zeta}^{(1)}(t,r,s)$ in (2.5). In Section 3, we prove a 3G
inequality when $\alpha\in(0,2)$ (Theorem 3.1). In Section 4, we prove further
pointwise bounds, called comparability results (Theorem 4.1). The technical
part of the proof of Theorem 2.1, when $\alpha\in(0,2)$, is given in Appendix
A.

Below we denote generic constants, i.e., numbers in $(0,\infty)$ by $c$. The
values of constants may change from place to place. We may mark the dependence
of $c$ on some parameter $\tau$ by the notation $c_{\tau}$ or $c(\tau)$. For
functions $f,g\geq 0$, we write $f\lesssim g$ to indicate that there is a
constant $c$ such that $f\leq cg$. If $c$ depends on $\tau$, we may write
$f\lesssim_{\tau}g$. The notation $f\sim g$ means that $f\lesssim g\lesssim
f$; we say _$f$ is comparable to $g$_. We abbreviate $a\wedge
b:=\min\\{a,b\\}$ and $a\vee b:=\max\\{a,b\\}$. The regularized hypergeometric
function [DLMF, (15.2.1)] is denoted by
${}_{2}\tilde{F}_{1}(a,b;c;z):=\,_{2}F_{1}(a,b;c;z)/\Gamma(c)$, with
$a,b,c\in\mathbb{C}$ and $z\in\\{w\in\mathbb{C}:\,|w|<1\\}$. We introduce
further notation as we proceed.

### Acknowledgments.

We thank Volker Bach, Kamil Bogus, Jacek Dziubański, Tomasz Grzywny, Jacek
Małecki, Haruya Mizutani, Adam Nowak, Marcin Preisner, and Grzegorz Serafin
for discussion and references.

##  2\. Fundamental properties, bounds, and explicit expressions

We recall the following properties of $p_{\zeta}^{(\alpha)}(t,r,s)$ for
$\zeta\in(-1/2,\infty)$, $\alpha\in(0,2]$ proved in [BM23a, Section 2]. For
all $t,t^{\prime},r,s>0$, we have
$p_{\zeta}^{(\alpha)}(t,r,s)=p_{\zeta}^{(\alpha)}(t,s,r)>0$,

$\displaystyle\int_{0}^{\infty}p_{\zeta}^{(\alpha)}(t,r,s)s^{2\zeta}\,ds=1,$
(2.1)
$\displaystyle\int_{0}^{\infty}p_{\zeta}^{(\alpha)}(t,r,z)p_{\zeta}^{(\alpha)}(t^{\prime},z,s)z^{2\zeta}\,dz=p_{\zeta}^{(\alpha)}(t+t^{\prime},r,s),\quad\text{and}$
(2.2) $\displaystyle
p_{\zeta}^{(\alpha)}(t,r,s)=t^{-\frac{2\zeta+1}{\alpha}}p_{\zeta}^{(\alpha)}\left(1,\frac{r}{t^{1/\alpha}},\frac{s}{t^{1/\alpha}}\right).$
(2.3)

Using the Lévy distribution

$\displaystyle\sigma_{t}^{(1/2)}(\tau)=\frac{1}{2\sqrt{\pi}}\cdot\frac{t}{\tau^{3/2}}\mathrm{e}^{-t^{2}/(4\tau)},\quad
t,\tau>0,$ (2.4)

see, e.g., Stein and Weiss [SW71, p. 6], it is possible to give an explicit
expression for $p_{\zeta}^{(\alpha)}(t,r,s)$ in the physically important case
$\alpha=1$. One obtains, for $\zeta\in(-1/2,\infty)$ and $r,s,t>0$,

$\displaystyle\begin{split}p_{\zeta}^{(1)}(t,r,s)&=\frac{2\Gamma(\zeta+1)}{\sqrt{\pi}}\cdot\frac{t}{\left(r^{2}+s^{2}+t^{2}\right)^{\zeta+1}}\\\
&\quad\times\,_{2}\tilde{F}_{1}\left(\frac{\zeta+1}{2},\frac{\zeta+2}{2};\zeta+\frac{1}{2};\frac{4r^{2}s^{2}}{\left(r^{2}+s^{2}+t^{2}\right)^{2}}\right),\end{split}$
(2.5)

see, e.g., Betancor, Harboure, Nowak, and Viviani [BHNV10, p. 136] for a
computation. In particular,

$\displaystyle p_{0}^{(1)}(t,r,s)$
$\displaystyle=\frac{2t}{\pi\left(r^{2}+s^{2}+t^{2}\right)\left(1-4r^{2}s^{2}\cdot\left(r^{2}+s^{2}+t^{2}\right)^{-2}\right)},$
(2.6a) $\displaystyle p_{1}^{(1)}(t,r,s)$
$\displaystyle=\frac{4}{\pi}\,\frac{t}{(r^{2}-s^{2})^{2}+t^{2}(t^{2}+2r^{2}+2s^{2})}.$
(2.6b)

Since explicit expressions for the $\alpha/2$-stable subordination density
with rational $\alpha/2$ (see, e.g., Penson and Górska [PG10]) are available,
one could also compute $p_{\zeta}^{(\alpha)}(t,r,s)$ for such $\alpha$, but
the resulting expressions are rather involved when $\alpha\neq 1$. Using
bounds for hypergeometric functions, one obtains the following upper and lower
bounds

$\displaystyle
p_{\zeta}^{(1)}(t,r,s)\sim_{\zeta}\frac{t}{(r^{2}+s^{2}+t^{2})^{\zeta}[(r-s)^{2}+t^{2}]},$
(2.7)

see, e.g., [BHNV10, Proposition 6.1]. In the following theorem, we give sharp
upper and lower bounds for $p_{\zeta}^{(\alpha)}(t,r,s)$ and all
$\alpha\in(0,2)$. We remark that the case of $\alpha=1$ in Theorem 2.1 is
resolved in [BHNV10, Proposition 6.1]; see also Dziubański and Preisner [DP18,
Proposition 6] and Betancor, Castro, and Stinga [BCS14]. The case of general
$\alpha\in(0,2)$ seems unknown, although similar estimates were obtained by
analogous techniques in various settings, see, e.g., Bogdan, Stós, and Sztonyk
[BSS03, Theorem 3.1] for ${\mathbb{R}^{d}}$ and unbounded fractals; see also
remarks in the proof below.

######  Theorem 2.1.

Let $\zeta\in(-1/2,\infty)$. Then, there are $c,c^{\prime}>0$ such that

$\displaystyle p_{\zeta}^{(2)}(t,r,s)$
$\displaystyle\asymp_{\zeta}t^{-\frac{1}{2}}\frac{\exp\left(-\frac{(r-s)^{2}}{ct}\right)}{(rs+t)^{\zeta}}$
(2.8a)
$\displaystyle\asymp_{\zeta}\left(1\wedge\frac{r}{t^{1/2}}\right)^{\zeta}\,\left(1\wedge\frac{s}{t^{1/2}}\right)^{\zeta}\,\left(\frac{1}{rs}\right)^{\zeta}\cdot
t^{-\frac{1}{2}}\cdot\exp\left(-\frac{(r-s)^{2}}{c^{\prime}t}\right),$ (2.8b)

for all $r,s,t>0$. Moreover, for all $\alpha\in(0,2)$ and all $r,s,t>0$,

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(t,r,s)&\sim_{\zeta,\alpha}\frac{t}{|r-s|^{1+\alpha}(r+s)^{2\zeta}+t^{\frac{1+\alpha}{\alpha}}(t^{\frac{1}{\alpha}}+r+s)^{2\zeta}}.\end{split}$
(2.9)

Here and below the notation $\asymp_{\zeta}$ combines an upper bound and a
lower bound similarly as $\sim_{\zeta}$, but the displayed constants in
exponential factors (i.e., the constant $c$ in (2.8a) and $c^{\prime}$ in
(2.8b)) may be different in the upper and the lower bounds. Furthermore, as
suggested by the notation $\sim_{\zeta}$, we allow the constants in the
exponential factors to depend on $\zeta$, too. Thus, for instance, (2.8a) is
equivalent to the statement that there are $c_{j,\zeta}$, $j\in\\{1,2,3,4\\}$
such that

$\displaystyle\begin{split}c_{1,\zeta}t^{-\frac{1}{2}}\frac{\exp\left(-\frac{(r-s)^{2}}{c_{2,\zeta}t}\right)}{(rs+t)^{\zeta}}\leq
p_{\zeta}^{(2)}(t,r,s)\leq
c_{3,\zeta}t^{-\frac{1}{2}}\frac{\exp\left(-\frac{(r-s)^{2}}{c_{4,\zeta}t}\right)}{(rs+t)^{\zeta}}.\end{split}$
(2.10)

###### Proof of Theorem 2.1 for $\alpha=2$.

The two-sided estimates in (2.8a) follow from the following asymptotics [DLMF,
(10.30.1), (10.30.4)],

$\displaystyle
I_{\rho}(z)\sim\frac{1}{\Gamma(\rho+1)}\cdot\left(\frac{z}{2}\right)^{\rho}\mathbf{1}_{z\leq
1}+\frac{\mathrm{e}^{z}}{\sqrt{2\pi z}}\mathbf{1}_{z\geq 1},\quad z\geq
0,\,\rho\notin\\{-1,-2,...\\}.$ (2.11)

The two-sided estimates in (2.8b) were proved, e.g., in Frank and Merz [FM23,
Theorem 10].

The upper bound in (2.9) for $\alpha<2$ can be deduced from [GT21, Corollary
3.8]. Private communication with the authors of that remarkable work indicates
that their arguments should, however, be modified to prove the lower bound in
(2.9). So, for the sake of completeness, in Appendix A below, we give a self-
contained proof of the two-sided estimates in (2.9) for $\alpha\in(0,2)$. ∎

Using the pointwise bounds in Theorem 2.1, we show that $p_{\zeta}^{(\alpha)}$
is a strongly continuous contraction semigroup on
$L^{2}(\mathbb{R}_{+},r^{2\zeta}dr)$.

######  Proposition 2.2.

Let $\zeta\in(-1/2,\infty)$ and $\alpha\in(0,2]$. Then,
$\\{p_{\zeta}^{(\alpha)}(t,\cdot,\cdot)\\}_{t>0}$ is a strongly continuous
contraction semigroup on $L^{2}(\mathbb{R}_{+},r^{2\zeta}dr)$.

###### Proof.

By symmetry, the normalization (2.1), and a Schur test,
$p_{\zeta}^{(\alpha)}(t,\cdot,\cdot)$ defines a contraction on
$L^{2}(\mathbb{R}_{+},r^{2\zeta}dr)$ for every $t>0$. To prove the strong
continuity of $p_{\zeta}^{(\alpha)}(t,\cdot,\cdot)$, it suffices, by the
density of $C_{c}^{\infty}(\mathbb{R}_{+})$ in
$L^{2}(\mathbb{R}_{+},r^{2\zeta}dr)$, to show

$\displaystyle\lim_{t\searrow
0}\int_{0}^{\infty}dr\,r^{2\zeta}\left|\int_{0}^{\infty}ds\,s^{2\zeta}p_{\zeta}^{(\alpha)}(t,r,s)\varphi(s)-\varphi(r)\right|^{2}=0$
(2.12)

for every non-negative function $\varphi\in C_{c}^{\infty}(\mathbb{R}_{+})$.
Indeed, this follows from the semigroup property,
$|\varphi(r)|\lesssim_{\varphi}\mathbf{1}_{r<c}$ for some $c=c(\varphi)>0$,
the bound

$\displaystyle\int_{0}^{\infty}dr\,r^{2\zeta}\left|\int_{0}^{\infty}ds\,s^{2\zeta}p_{\zeta}^{(\alpha)}(t,r,s)\varphi(s)\right|^{2}\lesssim\int_{0}^{c}dr\int_{0}^{c}ds\,(rs)^{2\zeta}\,p_{\zeta}^{(\alpha)}(2t,r,s),$
(2.13)

the bounds for $p_{\zeta}^{(\alpha)}(t,r,s)$ in (2.8), and the dominated
convergence theorem. ∎

##  3\. 3G inequality for $\alpha\in(0,2)$

In our forthcoming work [BM23b], we use the following 3G inequality for
$p_{\zeta}^{(\alpha)}(t,r,s)$. The result is motivated and similar to that of
Bogdan and Jakubowski [BJ07, (7)–(9)].

######  Theorem 3.1.

Let $\zeta\in[0,\infty)$, $\alpha\in(0,2)$. Then, for all $r,s,z,t,\tau>0$, we
have

$\displaystyle\begin{split}&\min\left\\{\\!\left(\\!\frac{r+z}{r+s+z}\\!\right)^{2\zeta}p_{\zeta}^{(\alpha)}(t,r,z),\left(\\!\frac{s+z}{r+s+z}\\!\right)^{2\zeta}p_{\zeta}^{(\alpha)}(\tau,z,s)\\!\right\\}\\!\lesssim_{\zeta,\alpha}\\!p_{\zeta}^{(\alpha)}(t+\tau,r,s),\end{split}$
(3.1)

and

$\displaystyle p_{\zeta}^{(\alpha)}(t,r,z)\cdot
p_{\zeta}^{(\alpha)}(\tau,z,s)$ (3.2) $\displaystyle\
\lesssim_{\zeta,\alpha}p_{\zeta}^{(\alpha)}(t+\tau,r,s)\left[\\!\left(\\!\frac{r+s+z}{s+z}\\!\right)^{2\zeta}p_{\zeta}^{(\alpha)}(t,r,z)+\left(\\!\frac{r+s+z}{r+z}\\!\right)^{2\zeta}p_{\zeta}^{(\alpha)}(\tau,z,s)\\!\right]\\!.$

######  Remarks 3.2.

(1) Weighted 3G inequalities for Green’s functions (resolvent kernels) and
their application to Schrödinger perturbations were studied, e.g., by Hansen
[Han06], with quasi-metric interpretations.
(2) The weights in (3.1)–(3.2) are independent of $t$ and $\tau$, but they
involve all three spatial variables $r,s,z$, which is slightly incompatible
with the setting of [Han06].
(3) As observed in [BJ07, p. 182], there is no 3G inequality for the Brownian
motion in $\mathbb{R}^{d}$. However, there is a substitute, called 4G
inequality, in Bogdan and Szczypkowski [BS14, Theorem 1.3].
(4) We do not know of “suitable” substitutes of the statements in Theorem 3.1
for $\zeta\in(-1/2,0)$.

###### Proof of Theorem 3.1.

Define and observe

$\displaystyle\begin{split}f(r,s,z)&:=\left(\frac{s+z}{r+s}\right)^{2\zeta}\mathbf{1}_{\\{r>s\vee
z\\}}+\mathbf{1}_{\\{r<s\vee
z\\}}\sim\left(\frac{s+z}{r+s+z}\right)^{2\zeta}.\end{split}$ (3.3)

To prove (3.1), without loss of generality, we assume $r>s$. Then,
$f(r,s,z)=\left(\tfrac{s+z}{r+s}\right)^{2\zeta}\mathbf{1}_{\\{r>z\\}}+\mathbf{1}_{\\{r<z\\}}$
and $f(s,r,z)=1$. By (2.9),

$\displaystyle
p_{\zeta}^{(\alpha)}(t,r,z)\sim\min\left\\{\frac{t}{|r-z|^{1+\alpha}(r+z)^{2\zeta}},\frac{1}{t^{1/\alpha}(t^{1/\alpha}+r+z)^{2\zeta}}\right\\}.$
(3.4)

Thus, the left-hand side of (3.1) is

$\displaystyle\scriptsize\begin{split}&\left(f(s,r,z)\cdot
p_{\zeta}^{(\alpha)}(t,r,z)\right)\wedge\left(f(r,s,z)\cdot
p_{\zeta}^{(\alpha)}(\tau,z,s)\right)\\\ &\
\sim\min\left\\{\\!\frac{t}{|r-z|^{1+\alpha}(r+z)^{2\zeta}},\frac{\tau\cdot
f(r,s,z)}{|s-z|^{1+\alpha}(s+z)^{2\zeta}},\frac{1}{t^{\frac{1}{\alpha}}(t^{\frac{1}{\alpha}}+r+z)^{2\zeta}},\frac{f(r,s,z)}{\tau^{\frac{1}{\alpha}}(\tau^{\frac{1}{\alpha}}+s+z)^{2\zeta}}\right\\}.\end{split}$
(3.5)

We begin with estimating the minimum of the first two terms on the right-hand
side of (3.5). Since $(s+z)^{2\zeta}\mathbf{1}_{z>r>s}\geq
r^{2\zeta}\mathbf{1}_{z>r>s}$, we get

$\displaystyle\begin{split}&\min\left\\{\frac{t}{|r-z|^{1+\alpha}(r+z)^{2\zeta}},\frac{\tau
f(r,s,z)}{|s-z|^{1+\alpha}(s+z)^{2\zeta}}\right\\}\\\
&\quad\lesssim\frac{t+\tau}{r^{2\zeta}}\min\left\\{\frac{1}{|r-z|^{1+\alpha}},\frac{1}{|s-z|^{1+\alpha}}\right\\}\\\
&\quad\lesssim\frac{t+\tau}{r^{2\zeta}\cdot|r-s|^{1+\alpha}}\sim\frac{t+\tau}{|r-s|^{1+\alpha}(r+s)^{2\zeta}},\end{split}$
(3.6)

which is half of our desired estimate. We now consider the minimum of the last
two terms in (3.5) and claim

$\displaystyle\begin{split}&\min\left\\{\frac{1}{t^{1/\alpha}(t^{1/\alpha}+r)^{2\zeta}},\frac{f(r,s,z)}{\tau^{1/\alpha}(\tau^{1/\alpha}+s+z)^{2\zeta}}\right\\}\\\
&\quad\lesssim\frac{1}{t^{1/\alpha}(t^{1/\alpha}+r)^{2\zeta}+\tau^{1/\alpha}(\tau^{1/\alpha}+r)^{2\zeta}}.\end{split}$
(3.7)

Suppose (3.7) is true. Then, by

$\displaystyle\begin{split}(t+\tau)^{1/\alpha}((t+\tau)^{1/\alpha}+r+s)^{2\zeta}&\sim(t+\tau)^{1/\alpha}((t+\tau)^{1/\alpha}+r)^{2\zeta}\\\
&\lesssim(t+\tau)^{(1+2\zeta)/\alpha}+(t+\tau)^{1/\alpha}r^{2\zeta}\\\
&\lesssim
t^{1/\alpha}(t^{2\zeta/\alpha}+r^{2\zeta})+\tau^{1/\alpha}(\tau^{2\zeta/\alpha}+r^{2\zeta})\\\
&\sim
t^{1/\alpha}(t^{1/\alpha}+r)^{2\zeta}+\tau^{1/\alpha}(\tau^{1/\alpha}+r)^{2\zeta},\end{split}$
(3.8)

we obtain

$\displaystyle\begin{split}&\min\left\\{\frac{1}{t^{\frac{1}{\alpha}}(t^{\frac{1}{\alpha}}+r+z)^{2\zeta}},\frac{f(r,s,z)}{\tau^{\frac{1}{\alpha}}(\tau^{\frac{1}{\alpha}}+s+z)^{2\zeta}}\right\\}\lesssim\frac{1}{(t+\tau)^{\frac{1}{\alpha}}((t+\tau)^{\frac{1}{\alpha}}+r+s)^{2\zeta}},\end{split}$
(3.9)

which, together with (3.6), completes the proof of (3.1). It remains to prove
(3.7). We distinguish between $z<r$ and $z>r$. In the latter case, we have

$\displaystyle\begin{split}&\min\left\\{\frac{1}{t^{\frac{1}{\alpha}}(t^{\frac{1}{\alpha}}+r+z)^{2\zeta}},\frac{1}{\tau^{\frac{1}{\alpha}}(\tau^{\frac{1}{\alpha}}+s+z)^{2\zeta}}\right\\}\\\
&\quad\leq\min\left\\{\frac{1}{t^{\frac{1}{\alpha}}(t^{\frac{1}{\alpha}}+r)^{2\zeta}},\frac{1}{\tau^{\frac{1}{\alpha}}(\tau^{\frac{1}{\alpha}}+r)^{2\zeta}}\right\\},\end{split}$
(3.10)

as desired. On the other hand, if $z<r$, then we have, using $s\vee z<r$,

$\displaystyle\frac{(s+z)(\tau^{1/\alpha}+r)}{(r+s)(\tau^{1/\alpha}+s+z)}\sim\frac{(s+z)\tau^{1/\alpha}+r(s+z)}{r\tau^{1/\alpha}+r(s+z)}\leq
3,$

and so

$\displaystyle\begin{split}&\min\left\\{\frac{1}{t^{1/\alpha}(t^{1/\alpha}+r+z)^{2\zeta}},\frac{[(s+z)/(r+s)]^{2\zeta}}{\tau^{1/\alpha}(\tau^{1/\alpha}+s+z)^{2\zeta}}\right\\}\\\
&\quad\lesssim\min\left\\{\frac{1}{t^{1/\alpha}(t^{1/\alpha}+r)^{2\zeta}},\frac{1}{\tau^{1/\alpha}(\tau^{1/\alpha}+r)^{2\zeta}}\right\\}\\\
&\quad\sim\frac{1}{t^{1/\alpha}(t^{1/\alpha}+r)^{2\zeta}+\tau^{1/\alpha}(\tau^{1/\alpha}+r)^{2\zeta}},\end{split}$
(3.11)

which shows (3.7) for $z<r$. Plugging (3.6) and (3.9) into (3.5) yields

$\displaystyle\begin{split}&\left(f(s,r,z)\cdot
p_{\zeta}^{(\alpha)}(t,r,z)\right)\wedge\left(f(r,s,z)\cdot
p_{\zeta}^{(\alpha)}(\tau,z,s)\right)\\\
&\quad\lesssim\min\left\\{\frac{t+\tau}{|r-s|^{1+\alpha}(r+s)^{2\zeta}},\frac{1}{(t+\tau)^{1/\alpha}((t+\tau)^{1/\alpha}+r+s)^{2\zeta}}\right\\}\\\
&\quad\sim p_{\zeta}^{(\alpha)}(t+\tau,r,s),\end{split}$ (3.12)

as claimed. Estimate (3.2) follows from (3.1). ∎

##  4\. Comparability results for $p_{\zeta}^{(\alpha)}$

The following comparability results are crucial for our forthcoming work
[BM23b]. As we will see, they follow from the bounds in Theorem 2.1.

######  Theorem 4.1.

Let $\zeta\in(-1/2,\infty)$ and $\alpha\in(0,2]$.

  1. (1)

Let $z,s>0$, $0<C\leq 1$, and $\tau\in[C,C^{-1}]$. Then, there is
$c_{j}=c_{j}(\zeta,C)$, $j\in\\{1,2\\}$ with

$\displaystyle\begin{split}p_{\zeta}^{(2)}(1,c_{1}z,c_{1}s)&\lesssim_{C,\zeta}p_{\zeta}^{(2)}(\tau,z,s)\lesssim_{C,\zeta}p_{\zeta}^{(2)}(1,c_{2}z,c_{2}s),\\\
p_{\zeta}^{(\alpha)}(\tau,z,s)&\sim_{C,\zeta,\alpha}p_{\zeta}^{(\alpha)}(1,z,s),\quad\alpha<2.\end{split}$
(4.1)

In particular, for $\alpha\in(0,2)$ and $\tau,z,s,c>0$, one has

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(\tau,cz,cs)\sim_{\zeta,\alpha,c}p_{\zeta}^{(\alpha)}(\tau,z,s),\\\
p_{\zeta}^{(\alpha)}(c\tau,z,s)\sim_{\zeta,\alpha,c}p_{\zeta}^{(\alpha)}(\tau,z,s),\quad\alpha<2.\end{split}$
(4.2)

  2. (2)

Let $C,\tau>0$ and $0<z\leq s/2<\infty$. Then, there is $c=c(\zeta,C)$ with

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(\tau,z,s)&\lesssim_{C,\zeta,\alpha}p_{\zeta}^{(\alpha)}(\tau,c,cs)\mathbf{1}_{\\{\tau>C\\}}\\\
&\quad+\left(\tau^{-\frac{1}{2}}\frac{\mathrm{e}^{-cs^{2}/\tau}}{(\tau+s^{2})^{\zeta}}\mathbf{1}_{\alpha=2}+\frac{\tau}{s^{2\zeta+1+\alpha}+\tau^{\frac{2\zeta+1+\alpha}{\alpha}}}\mathbf{1}_{\alpha\in(0,2)}\right)\mathbf{1}_{\\{\tau<C\\}},\end{split}$
(4.3a) $\displaystyle p_{\zeta}^{(\alpha)}(\tau,z,s)$
$\displaystyle\lesssim_{\zeta,\alpha}\tau^{-\frac{1}{2}}\frac{\mathrm{e}^{-cs^{2}/\tau}}{(\tau+s^{2})^{\zeta}}\mathbf{1}_{\alpha=2}+\frac{\tau}{s^{2\zeta+1+\alpha}+\tau^{(2\zeta+1+\alpha)/\alpha}}\mathbf{1}_{\alpha\in(0,2)},$
(4.3b) $\displaystyle p_{\zeta}^{(\alpha)}(\tau,z,s)$
$\displaystyle\lesssim_{\zeta,\alpha}s^{-(2\zeta+1)}.$ (4.3c)

  3. (3)

Let $0<\tau\leq 1$, $0<z\leq s/2$, and $s\geq C>0$. Then, there is
$c=c(\zeta,C)$ with

$\displaystyle
p_{\zeta}^{(\alpha)}(\tau,z,s)\lesssim_{C,\zeta,\alpha}p_{\zeta}^{(\alpha)}(1,c,cs).$
(4.4)

  4. (4)

Let $0<r\leq s$. Then, there is $c=c(\zeta,C)$ with

$\displaystyle
p_{\zeta}^{(\alpha)}(1,1,s)\lesssim_{\zeta,\alpha}p_{\zeta}^{(\alpha)}(1,cr,cs).$
(4.5)

In particular, for all $r,s>0$,

$\displaystyle\min\\{p_{\zeta}^{(\alpha)}(1,1,r),p_{\zeta}^{(\alpha)}(1,1,s)\\}\lesssim_{\zeta,\alpha}p_{\zeta}^{(\alpha)}(1,cr,cs).$
(4.6)

  5. (5)

Let $r,s,z,t>0$ with $|z-s|>|r-s|/2$. Then, there is $c=c(\zeta,C)$ with

$\displaystyle
p_{\zeta}^{(\alpha)}(t,z,s)\lesssim_{\zeta,\alpha}p_{\zeta}^{(\alpha)}(t,cr,cs).$
(4.7)

The constants $c$ in the arguments of functions in Theorem 4.1 may change from
place to place. Note that in the above estimates the point $z=1$ is a natural
reference point for the heat kernel $p_{\zeta}^{(\alpha)}(\tau,z,s)$ when
$s\gg z$ and $\tau\sim 1$ by the spatial homogeneity of
$p_{\zeta}^{(\alpha)}$. This is in contrast to the analysis of the heat kernel
$\mathrm{e}^{-\tau(-\Delta)^{\alpha/2}}(z,y)$ in $\mathbb{R}^{d}$, where $z=0$
is a natural reference point when $\tau\sim 1$ and $|y|\gg|z|$ because of the
additional translation invariance in this setting. We give some
interpretations of the bounds in Theorem 4.1 after the proof.

###### Proof.

  1. (1)

The estimates follow from (2.8) for $\alpha=2$ and from (2.9) if $\alpha<2$.

  2. (2)

We start with $\alpha=2$. We first consider (4.3a). By (2.8a),

$\displaystyle
p_{\zeta}^{(2)}(\tau,z,s)\asymp\tau^{-\frac{1}{2}}\frac{\exp(-c(s-z)^{2}/\tau)}{(\tau+sz)^{\zeta}}\asymp\tau^{-\frac{1}{2}}\frac{\exp(-cs^{2}/\tau)}{(\tau+sz)^{\zeta}}.$
(4.8)

If $\tau>C>0$, then, by $2s/\tau\leq(1+s^{2})/\tau\lesssim_{C}1+s^{2}/\tau$,

$\displaystyle\begin{split}\frac{(\tau+s)^{\zeta}}{(\tau+sz)^{\zeta}}\mathrm{e}^{-cs^{2}/\tau}&\leq\left(1+\frac{s}{\tau}\right)^{\zeta}\mathrm{e}^{-cs^{2}/\tau}\lesssim_{C}\left(1+\frac{s^{2}}{\tau}\right)^{\zeta}\mathrm{e}^{-cs^{2}/\tau}\lesssim_{\zeta}\mathrm{e}^{-cs^{2}/\tau}\\\
&\sim_{C}\mathrm{e}^{-c(s-1)^{2}/\tau},\end{split}$ (4.9)

which proves (4.3a) for $\tau>C>0$. Similarly, we have for all $\tau>0$,

$\displaystyle\frac{(\tau+s^{2})^{\zeta}}{(\tau+sz)^{\zeta}}\mathrm{e}^{-cs^{2}/\tau}\leq(1+s^{2}/\tau)^{\zeta}\mathrm{e}^{-cs^{2}/\tau}\lesssim\mathrm{e}^{-cs^{2}/\tau},$
(4.10)

which yields (4.3b) and the second part of (4.3a), where $\tau<C$. Estimate
(4.3c) follows from

$\displaystyle
p_{\zeta}^{(2)}(\tau,z,s)\lesssim\left(\frac{s^{2}}{\tau}\right)^{\zeta+\frac{1}{2}}\exp(-cs^{2}/\tau)\cdot
s^{-(2\zeta+1)}\lesssim s^{-(2\zeta+1)}.$ (4.11)

This concludes the proof of (4.3a)–(4.3c) for $\alpha=2$.

Now we verify (4.3) for $\alpha<2$. We start with (4.3a). By (2.9),

$\displaystyle
p_{\zeta}^{(\alpha)}(\tau,z,s)\sim\frac{\tau}{s^{2\zeta+1+\alpha}+\tau^{1+1/\alpha}(\tau^{1/\alpha}+s)^{2\zeta}}.$
(4.12)

Thus, (4.3a) for $\tau>C>0$ follows from

$\displaystyle\begin{split}&\frac{|s-1|^{1+\alpha}(s+1)^{2\zeta}}{s^{2\zeta+1+\alpha}+\tau^{1+1/\alpha}(\tau^{1/\alpha}+s)^{2\zeta}}+\frac{\tau^{1+1/\alpha}(\tau^{1/\alpha}+s+1)^{2\zeta}}{s^{2\zeta+1+\alpha}+\tau^{1+1/\alpha}(\tau^{1/\alpha}+s)^{2\zeta}}\\\
&\quad\lesssim\frac{s^{1+\alpha}(1+s)^{2\zeta}+(s+1)^{2\zeta}}{s^{2\zeta+1+\alpha}+1}+1+\frac{1}{(\tau^{1/\alpha}+s)^{2\zeta}}\\\
&\quad\lesssim_{C,\zeta}1.\end{split}$ (4.13)

Similarly, for all $\tau>0$,

$\displaystyle
p_{\zeta}^{(\alpha)}(\tau,z,s)\lesssim\frac{\tau}{s^{2\zeta+1+\alpha}+\tau^{\frac{2\zeta+1+\alpha}{\alpha}}},$
(4.14)

which yields (4.3b) and the second part of (4.3a). Estimate (4.3c) follows
from

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(\tau,z,s)&\lesssim\frac{\tau}{s^{2\zeta+1+\alpha}+\tau^{(1+\alpha)/\alpha}\cdot
s^{2\zeta}}\\\
&\lesssim\frac{\tau}{s^{2\zeta+1+\alpha}}\mathbf{1}_{\\{\tau<s^{\alpha}\\}}+\frac{\tau}{\tau^{(1+\alpha)/\alpha}\cdot
s^{2\zeta}}\mathbf{1}_{\\{\tau>s^{\alpha}\\}}\lesssim
s^{-(2\zeta+1)}.\end{split}$ (4.15)

This concludes the proof of (4.3a)–(4.3c) for $\alpha<2$.

  3. (3)

To prove (4.4), we begin with $\alpha=2$. By (2.8a), we have, for
$\tau\in(0,1]$, $0<z\leq s/2$, and $s\geq C>0$,

$\displaystyle\begin{split}p_{\zeta}^{(2)}(\tau,z,s)&\asymp\tau^{-\frac{1}{2}}\frac{\exp\left(-\frac{c(z-s)^{2}}{\tau}\right)}{(zs+\tau)^{\zeta}}\lesssim\frac{\exp(-cs^{2}/\tau)}{\tau^{\zeta+1/2}}\cdot\frac{s^{2\zeta+1}}{s^{2\zeta+1}}\\\
&\lesssim_{C}\frac{(s^{2}/\tau)^{\zeta+1/2}\exp(-cs^{2}/\tau)}{s^{\zeta}}\lesssim_{C}\frac{\mathrm{e}^{-cs^{2}/\tau}}{(s+1)^{\zeta}}\lesssim\frac{\mathrm{e}^{-c(s-1)^{2}}}{(s+1)^{\zeta}}.\end{split}$
(4.16)

This concludes the proof of (4.4) for $\alpha=2$.

If $\alpha<2$, we have, for $\tau\in(0,1]$, $0<z\leq s/2$, and $s\geq C>0$,

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(\tau,z,s)&\sim_{\zeta}\frac{\tau}{s^{2\zeta+1+\alpha}+\tau^{1+1/\alpha}(\tau^{1/\alpha}+s)^{2\zeta}}.\end{split}$
(4.17)

Thus, (4.4) follows from

$\displaystyle\begin{split}&\frac{\tau\left(|s-1|^{1+\alpha}(s+1)^{2\zeta}+(1+s)^{2\zeta}\right)}{s^{2\zeta+1+\alpha}+\tau^{1+1/\alpha}(\tau^{1/\alpha}+s)^{2\zeta}}\lesssim\frac{(1+s)^{2\zeta}(1+s^{1+\alpha})}{s^{2\zeta+1+\alpha}}\lesssim_{C}1.\end{split}$
(4.18)

This concludes the proof of (4.4) for $\alpha<2$.

  4. (4)

We come to (4.5), with $0<r\leq s$. We first treat $\alpha=2$. By (2.8a),

$\displaystyle\begin{split}p_{\zeta}^{(2)}(1,1,s)&\asymp_{\zeta}\frac{\mathrm{e}^{-c(s-1)^{2}}}{(s+1)^{\zeta}}\lesssim\frac{(1+rs)^{\zeta}\mathrm{e}^{-cs^{2}}}{(s+1)^{\zeta}(1+rs)^{\zeta}}\leq\frac{(1+s^{2})^{\zeta}\mathrm{e}^{-cs^{2}}}{(1+rs)^{\zeta}}\lesssim_{\zeta}\frac{\mathrm{e}^{-cs^{2}}}{(1+rs)^{\zeta}}\\\
&\lesssim\frac{\mathrm{e}^{-c(r-s)^{2}}}{(rs+1)^{\zeta}}\asymp_{\zeta}p_{\zeta}^{(2)}(1,cr,cs),\end{split}$
(4.19)

where we used $2s^{2}\geq(r-s)^{2}$. This concludes the proof of (4.5) for
$\alpha=2$.

If $\alpha<2$, then, by (2.9),

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(1,1,s)&\sim_{\zeta}\frac{1}{|s-1|^{1+\alpha}(1+s)^{2\zeta}+(1+s)^{2\zeta}}.\end{split}$
(4.20)

Since $2s^{2}\geq(r-s)^{2}$, we have

$\displaystyle\begin{split}&\frac{|r-s|^{1+\alpha}(r+s)^{2\zeta}+(1+r+s)^{2\zeta}}{|s-1|^{1+\alpha}(1+s)^{2\zeta}+(1+s)^{2\zeta}}\\\
&\quad\lesssim\frac{s^{2\zeta+1+\alpha}+(1+s)^{2\zeta}}{|s-1|^{1+\alpha}(1+s)^{2\zeta}+(1+s)^{2\zeta}}\lesssim
1,\end{split}$ (4.21)

which concludes the proof of (4.5) for $\alpha<2$. To prove (4.6), it
suffices, by symmetry, to assume $s\geq r>0$. Then, the claim follows from
(4.5) since
$\min\\{p_{\zeta}^{(\alpha)}(1,1,r),p_{\zeta}^{(\alpha)}(1,1,s)\\}\leq
p_{\zeta}^{(\alpha)}(1,1,s)$.

  5. (5)

We come to (4.7). If $z>r$, the estimate follows immediately from (2.8a) if
$\alpha=2$ and from (2.9) if $\alpha<2$. So suppose $z<r$. We distinguish now
between $z>s$ and $z<s$ and start with the former case. Then the assumption
$|z-s|>|r-s|/2$ implies

$\displaystyle z=z-s+s\geq s+\frac{|r-s|}{2}=\frac{r+s}{2}\gtrsim r.$ (4.22)

Thus, $z\gtrsim r$ and we can again use (2.8a) for $\alpha=2$ and (2.9) for
$\alpha<2$ to conclude the estimate in this case. Thus, we are left to treat
$z<s$. We first let $\alpha=2$ and distinguish between the three subcases
$s>2r$, $s<r/2$, and $s\in(r/2,2r)$. To treat $s>2r$ and $s<r/2$, we estimate

$\displaystyle\begin{split}\frac{\exp\left(-\frac{c(z-s)^{2}}{t}\right)}{(zs+t)^{\zeta}}&\leq\frac{(rs/t+1)^{\zeta}}{(zs/t+1)^{\zeta}}\cdot\frac{\exp\left(-\frac{c(r-s)^{2}}{4t}\right)}{(rs+t)^{\zeta}}\\\
&\leq\left(1+\frac{rs}{t}\right)^{\zeta}\cdot\exp\left(-\frac{c(r-s)^{2}}{8t}\right)\cdot\frac{\exp\left(-\frac{c(r-s)^{2}}{8t}\right)}{(rs+t)^{\zeta}}\\\
&\lesssim_{\zeta}\frac{\exp\left(-\frac{c(r-s)^{2}}{8t}\right)}{(rs+t)^{\zeta}},\end{split}$
(4.23)

as desired. It remains to treat the case $s\in(r/2,2r)$ and $z<r\vee s$. We
distinguish between $s>2z$ and $s\in(z,2z)$. If $s>2z$, then we can argue
similarly as before and obtain

$\displaystyle\begin{split}\frac{\exp\left(-\frac{c(z-s)^{2}}{t}\right)}{(zs+t)^{\zeta}}&\leq\left(1+\frac{rs}{t}\right)^{\zeta}\cdot\exp\left(-\frac{c(z-s)^{2}}{8t}\right)\cdot\frac{\exp\left(-\frac{c(r-s)^{2}}{8t}\right)}{(rs+t)^{\zeta}}\\\
&\leq\left(1+\frac{2s^{2}}{t}\right)^{\zeta}\cdot\exp\left(-\frac{cs^{2}}{8t}\right)\cdot\frac{\exp\left(-\frac{c(r-s)^{2}}{8t}\right)}{(rs+t)^{\zeta}}\\\
&\lesssim_{\zeta}\frac{\exp\left(-\frac{c(r-s)^{2}}{8t}\right)}{(rs+t)^{\zeta}}.\end{split}$
(4.24)

Finally, for $s\in(r/2,2r)$, $z<r$, and $s\in(z,2z)$, we use $zs>s^{2}/2>rs/4$
to get the desired estimate in the denominator in (2.8a). This completes the
proof of (4.7) for $\alpha=2$.

Now we deal with $\alpha<2$ and $z<r\wedge s$. Observe that in (2.9) it
suffices to estimate $z+s\gtrsim s+r$ and $z^{2}+s^{2}\gtrsim r^{2}+s^{2}$.
These estimates follow immediately if $s>r$. So suppose $z<s<r$. Then, by the
assumption $|z-s|=s-z>|r-s|/2=(r-s)/2$, we get $3s/2>z+r/2>r/2$. This
completes the proof of (4.7).

The proof of Theorem 4.1 is concluded. ∎

We close with some interpretations of the bounds in Theorem 4.1. The numbering
of the following remarks refers to the numbering of the bounds in Theorem 4.1.

######  Remarks 4.2.

(1) For fixed time scales $\tau\sim 1$, the heat kernels
$p_{\zeta}^{(\alpha)}(\tau,z,s)$ and $p_{\zeta}^{(\alpha)}(1,r,s)$ are
comparable for all $z,s>0$.
(2) Spatial or temporal dilations on the unit order are negligible when
$\alpha<2$.
(3) Suppose $s\geq 2z$. For large times $\tau>C$, the heat kernel
$p_{\zeta}^{(\alpha)}(\tau,z,s)$ can be replaced with
$p_{\zeta}^{(\alpha)}(\tau,1,s)$, even if $z$ is small, thanks to the lower-
boundedness of $\tau$ and the large distance between $z$ and $s$. For small
times $\tau<C$, the distance $|s-z|$ can still be replaced with $s$. However,
for small $z$ and $\tau$, we cannot compare $p_{\zeta}^{(\alpha)}(\tau,z,s)$
with $p_{\zeta}^{(\alpha)}(\tau,1,s)$. At least when $\alpha<2$, we can
$p_{\zeta}^{(\alpha)}(\tau,z,s)$ compare with
$p_{\zeta}^{(\alpha)}(\tau,0,s)$.
(4) For small times $\tau\in(0,1]$ and locations $s\geq 2z$, which are
strictly away from zero, the heat kernel $p_{\zeta}^{(\alpha)}(\tau,z,s)$ can
be estimated from above by $p_{\zeta}^{(\alpha)}(1,1,s)$. This is because the
lower-boundedness of $s$ prevents the heat kernel from blowing up for small
times and small $z$, and the large separation between $s$ and $z$ allows to
replace $|z-s|$ and $z+s$ by $s$.
(5) For fixed time $\tau=1$ and location $s>0$, it is more likely to go from
$s$ to $r\leq s$ than from $s$ to $1$. This is plausible for $r>1$, while, if
$r\leq 1$, it does not really matter if we go to $1$ or to $r$ within a unit
time step.
(6) The probability for going from $s$ to $z$ is bounded by that for going
from $s$ to $r$ when $|z-s|$ is greater than $|r-s|$. Moreover, the (killing
or mass creating) effect from the boundary, i.e., the origin, is negligible.

##  Appendix A Proof of Theorem 2.1 for $\alpha\in(0,2)$

Recall our claim (2.9). Namely, if $\alpha\in(0,2)$ and $\zeta>-1/2$, then

$\displaystyle p_{\zeta}^{(\alpha)}(t,r,s)$
$\displaystyle\sim_{\zeta,\alpha}\frac{t}{|r-s|^{1+\alpha}(r+s)^{2\zeta}+t^{\frac{1+\alpha}{\alpha}}(t^{\frac{1}{\alpha}}+r+s)^{2\zeta}}\quad\mbox{
for $r,s,t>0$.}$

By the scaling (2.3), $t=1$ suffices. We split the proof into several steps.

###  A.1. Auxiliary bounds

We first prove the following, rather bulky, bounds, namely

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(1,r,s)&\sim_{\zeta,\alpha}\mathbf{1}_{rs<1}\mathbf{1}_{(r-s)^{2}<1}\\\
&\qquad+(rs)^{-\zeta}\mathbf{1}_{(r-s)^{2}<1<rs}\\\
&\qquad+|r-s|^{-(2\zeta+1+\alpha)}\mathbf{1}_{rs<1<(r-s)^{2}}\\\
&\qquad+\mathbf{1}_{rs>1}\mathbf{1}_{(r-s)^{2}>1}\left(\\!\frac{\mathbf{1}_{rs<(r-s)^{2}}}{|r-s|^{2\zeta+1+\alpha}}+\frac{\mathbf{1}_{rs>(r-s)^{2}}}{(rs)^{\zeta}|r-s|^{1+\alpha}}\\!\right).\end{split}$
(A.1)

To prove (A.1), we use the subordinator bounds

$\displaystyle\sigma_{1}^{(\alpha/2)}(\tau)\sim_{\alpha}\frac{\exp\left(-C(\alpha)\tau^{-c_{1}}\right)}{\tau^{c_{2}}}\mathbf{1}_{\tau<1}+\tau^{-1-\alpha/2}\mathbf{1}_{\tau>1},$
(A.2)

with $C(\alpha)>0$, $c_{1}=c_{1}(\alpha)=\alpha/(2-\alpha)\in(0,\infty)$, and
$c_{2}=c_{2}(\alpha)=(2-\alpha/2)/(2-\alpha)\in(1,\infty)$; see, e.g., [BM23a,
Proposition B.1]. Since $\mathbb{R}_{+}\ni x\mapsto\mathrm{e}^{-1/x^{s}}$
vanishes at zero faster than any polynomial whenever $s>0$, there are
$C_{1}(\alpha),C_{2}(\alpha)>0$ with $C_{1}(\alpha)>C_{2}(\alpha)$ such that

$\displaystyle\frac{\exp\left(-C_{1}(\alpha)\tau^{-c_{1}}\right)}{\tau^{1+\alpha/2}}\lesssim_{\alpha}\sigma_{1}^{(\alpha/2)}(\tau)\lesssim_{\alpha}\frac{\exp\left(-C_{2}(\alpha)\tau^{-c_{1}}\right)}{\tau^{1+\alpha/2}}.$
(A.3)

From the following computations, it will transpire that the precise rate of
the exponential decay of $\sigma_{1}^{(\alpha/2)}(\tau)$ at $\tau=0$ is
irrelevant. This motivates us to not distinguish between $C_{1}(\alpha)$ and
$C_{2}(\alpha)$ in (A.3) in the following and simply write

$\displaystyle\sigma_{1}^{(\alpha/2)}(\tau)\asymp_{\alpha}\frac{\exp\left(-C(\alpha)\tau^{-c_{1}}\right)}{\tau^{1+\alpha/2}}.$
(A.4)

We recall that the notation $\asymp$ means the same as $\sim$, but constants
in the argument of the exponential function (like $C(\alpha)$ in (A.4)) may be
different in the upper and lower bounds. For simplicity, in the following, we
will not explicitly indicate any parameter dependence of constants appearing
as prefactors or in exponentials anymore, so we will write, e.g., $c$ for
$c_{\zeta}$. Then, Formulae (1.2), (2.8), and (A.4) yield, for some
$c,c_{1}>0$,

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(1,r,s)&\asymp\int_{0}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\left[\frac{1}{\tau^{\zeta}}\mathbf{1}_{rs\leq\tau}+\frac{\mathbf{1}_{rs\geq\tau}}{(rs)^{\zeta}}\right]\exp\left(\\!-c\left(\frac{(r-s)^{2}}{\tau}+\frac{1}{\tau^{c_{1}}}\right)\\!\right).\end{split}$
(A.5)

We now estimate the integral on the right of (A.5). To this end, we
distinguish between four cases based on $rs\lessgtr 1$ and $|r-s|\lessgtr 1$.
Although the notation will not emphasize it, we repeat once more that the
following estimates are not uniform in $\zeta$ or $\alpha$.

####  A.1.1. Case $rs\vee(r-s)^{2}<1$

We show that $p_{\zeta}^{(\alpha)}(1,r,s)\sim 1$. We consider the first
summand in (A.5) and estimate

$\displaystyle\begin{split}&\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\zeta-\frac{1+\alpha}{2}}\mathrm{e}^{-c((r-s)^{2}/\tau+1/\tau^{c_{1}})}\\\
&\quad\asymp\left[\int_{rs}^{1}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c((r-s)^{2}/\tau+\tau^{-c_{1}})}+\int_{1}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\right].\end{split}$
(A.6)

The second summand in the last line of (A.6) is $\sim 1$ as desired. Since the
first summand on the right-hand side of (A.6) is non-negative, we get that the
left-hand side of (A.6) is bounded from below by the second summand on the
right-hand side of (A.6). On the other hand, by reflecting
$\tau\mapsto\tau^{-1}$, the first summand on the right-hand side of (A.6) can
be estimated by

$\displaystyle\int_{1}^{\infty}d\tau\,\tau^{\frac{2\zeta-1+\alpha}{2}}\mathrm{e}^{-\tau^{c_{1}}}\lesssim
1$ (A.7)

from above. This gives the desired estimate. Now, we consider the second
summand in (A.5). For a lower bound, we drop it (like in the discussion of the
lower bound for (A.6)). For an upper bound, we estimate

$\displaystyle\begin{split}0&\leq\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}(rs)^{-\zeta}\,\mathrm{e}^{-c((r-s)^{2}/\tau+1/\tau^{c_{1}})}\lesssim(rs)^{-\zeta}\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c\tau^{-c_{1}}}\\\
&=(rs)^{-\zeta}\int_{(rs)^{-1}}^{\infty}\frac{d\tau}{\tau}\,\tau^{\frac{1+\alpha}{2}}\mathrm{e}^{-c\tau^{c_{1}}}\asymp(rs)^{-\zeta}\cdot(rs)^{-(\frac{\alpha+1}{2}-c_{1})}\mathrm{e}^{-c/(rs)^{c_{1}}}\lesssim
1,\end{split}$ (A.8)

as desired. This concludes the case of $rs\vee(r-s)^{2}<1$.

####  A.1.2. Case $(r-s)^{2}<1<rs$

We show that $p_{\zeta}^{(\alpha)}(1,r,s)\sim(rs)^{-\zeta}$. We consider the
first summand in (A.5). For a lower bound we drop it, while for an upper bound
we estimate

$\displaystyle\begin{split}0&\leq\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c(\frac{(r-s)^{2}}{\tau}+\tau^{-c_{1}})}\lesssim\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\\\
&\sim(rs)^{-\frac{2\zeta+1+\alpha}{2}}\lesssim(rs)^{-\zeta}.\end{split}$ (A.9)

We now consider the second summand in (A.5) and obtain

$\displaystyle\begin{split}&\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}(rs)^{-\zeta}\,\mathrm{e}^{-c((r-s)^{2}/\tau+1/\tau^{c_{1}})}\\\
&\quad\asymp(rs)^{-\zeta}\left[\int_{0}^{1}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c((r-s)^{2}/\tau+\tau^{-c_{1}})}+\int_{1}^{rs}\frac{d\tau}{\tau^{(3+\alpha)/2}}\right]\sim(rs)^{-\zeta}.\end{split}$
(A.10)

This concludes the case of $(r-s)^{2}<1<rs$.

####  A.1.3. Case $rs<1<(r-s)^{2}$

We show that $p_{\zeta}^{(\alpha)}(1,r,s)\sim|r-s|^{-(2\zeta+1+\alpha)}$. We
first consider the second summand in (A.5). For a lower bound we drop it,
while for an upper bound we estimate

$\displaystyle\begin{split}0&\leq\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}(rs)^{-\zeta}\,\mathrm{e}^{-c(\frac{(r-s)^{2}}{\tau}+\frac{1}{\tau^{c_{1}}})}\lesssim(rs)^{-\zeta}\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c\frac{(r-s)^{2}}{\tau}}\\\
&=(rs)^{-\zeta}|r-s|^{-(1+\alpha)}\int_{(r-s)^{2}/(rs)}^{\infty}d\tau\,\tau^{\frac{\alpha-1}{2}}\mathrm{e}^{-c\tau}\\\
&\asymp(rs)^{-\zeta}|r-s|^{-(1+\alpha)}\cdot\frac{|r-s|^{\alpha-1}}{(rs)^{(\alpha-1)/2}}\mathrm{e}^{-c(r-s)^{2}/(rs)}\\\
&=|r-s|^{-(2\zeta+1+\alpha)}\cdot\frac{|r-s|^{2\zeta-1+\alpha}}{(rs)^{\frac{2\zeta-1+\alpha}{2}}}\mathrm{e}^{-c\frac{(r-s)^{2}}{rs}}\lesssim|r-s|^{-(2\zeta+1+\alpha)}.\end{split}$
(A.11)

To bound the first summand in (A.5), we distinguish between $c_{1}\leq 1$ and
$c_{1}>1$. Consider first $c_{1}\leq 1$. Then we have
$\exp\left(-c((r-s)^{2}/\tau+1/\tau^{c_{1}})\right)\asymp\exp(-c(r-s)^{2}/\tau)$.
In this case, we have the two-sided estimate

$\displaystyle\begin{split}&\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\zeta-\frac{1+\alpha}{2}}\mathrm{e}^{-c(\frac{(r-s)^{2}}{\tau}+\frac{1}{\tau^{c_{1}}})}\asymp\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c\frac{(r-s)^{2}}{\tau}}\\\
&\quad\asymp\int_{rs}^{(r-s)^{2}}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c(r-s)^{2}/\tau}+\int_{(r-s)^{2}}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\\\
&\quad\asymp\frac{1}{|r-s|^{2\zeta+1+\alpha}}\left[\int_{rs/(r-s)^{2}}^{1}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c/\tau}+1\right]\sim\frac{1}{|r-s|^{2\zeta+1+\alpha}}.\end{split}$
(A.12)

In particular, this estimate suffices for an upper bound for all $c_{1}>0$
since $\mathrm{e}^{-c/\tau^{c_{1}}}\leq 1$. Thus, it remains to prove the
lower bound when $c_{1}>1$, i.e., to find lower bounds for

$\displaystyle\int_{0}^{\infty}\frac{d\tau}{\tau}\,\tau^{-\frac{2\zeta+1+\alpha}{2}}\left[\\!\exp\left(-c\frac{(r-s)^{2}}{\tau}\right)\mathbf{1}_{\tau>|r-s|^{2/(c_{1}-1)}}+\mathrm{e}^{-c/\tau^{c_{1}}}\mathbf{1}_{\tau\in[rs,|r-s|^{2/(c_{1}-1)}]}\\!\right]\\!.$
(A.13)

To this end, we distinguish $2/(c_{1}-1)\lessgtr 2$. When $2/(c_{1}-1)\leq 2$,
we drop the second summand in (A.13) for a lower bound, while the first
summand we treat as in the case $c_{1}\leq 1$ in (A.12), by splitting the
integral at $\tau=(r-s)^{2}$. The integral for $\tau<(r-s)^{2}$ can be dropped
for a lower bound, while the integral for $\tau>(r-s)^{2}$ gives the desired
contribution. Now consider $2/(c_{1}-1)>2$. In this case, we drop the first
summand in (A.13) for a lower bound. On the other hand, we split the
$\tau$-integration in the second summand at $\tau=(r-s)^{2}$, drop the
contribution for $\tau<(r-s)^{2}$ for a lower bound, and estimate

$\displaystyle\left[\int_{rs}^{(r-s)^{2}}\frac{d\tau}{\tau}\,\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c/\tau^{c_{1}}}+\int_{(r-s)^{2}}^{(r-s)^{2/(c_{1}-1)}}\frac{d\tau}{\tau}\,\tau^{-\frac{2\zeta+1+\alpha}{2}}\right]\gtrsim\frac{1}{|r-s|^{2\zeta+1+\alpha}}.$
(A.14)

This concludes the analysis of the case $rs<1<(r-s)^{2}$.

####  A.1.4. Case $rs\wedge(r-s)^{2}>1$

We show that

$\displaystyle
p_{\zeta}^{(\alpha)}(1,r,s)\sim|r-s|^{-(2\zeta+1+\alpha)}\mathbf{1}_{rs<(r-s)^{2}}+(rs)^{-\zeta}|r-s|^{-(1+\alpha)}\mathbf{1}_{rs>(r-s)^{2}}.$

Consider first $c_{1}\leq 1$. Then, as before, we have
$\exp\left(-c((r-s)^{2}/\tau+1/\tau^{c_{1}})\right)\asymp\exp(-c(r-s)^{2}/\tau)$.
We consider the first summand in (A.5) and estimate

$\displaystyle\begin{split}&\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c((r-s)^{2}/\tau+1/\tau^{c_{1}})}\asymp\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c(r-s)^{2}/\tau}\\\
&\quad\asymp\mathbf{1}_{rs>(r-s)^{2}}\int_{rs}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\\\
&\qquad+\mathbf{1}_{rs<(r-s)^{2}}\left[\int_{rs}^{(r-s)^{2}}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\mathrm{e}^{-c(r-s)^{2}/\tau}+\int_{(r-s)^{2}}^{\infty}\frac{d\tau}{\tau}\tau^{-\frac{2\zeta+1+\alpha}{2}}\right]\\\
&\quad\sim(rs)^{-\frac{2\zeta+1+\alpha}{2}}\mathbf{1}_{rs>(r-s)^{2}}+\frac{1}{|r-s|^{2\zeta+1+\alpha}}\mathbf{1}_{rs<(r-s)^{2}}.\end{split}$
(A.15)

The second summand in this estimate is already in the desired form, while the
first summand can be dropped for a lower bound and estimated from above by
$(rs)^{-\zeta}|r-s|^{-1-\alpha}$ for the desired upper bound. We now consider
the second summand in (A.5) and obtain

$\displaystyle\begin{split}&(rs)^{-\zeta}\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c((r-s)^{2}/\tau+1/\tau^{c_{1}})}\sim(rs)^{-\zeta}\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-(r-s)^{2}/\tau}\\\
&\quad\asymp\mathbf{1}_{rs>(r-s)^{2}}(rs)^{-\zeta}\left[\int_{0}^{(r-s)^{2}}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c(r-s)^{2}/\tau}+\int_{(r-s)^{2}}^{rs}\frac{d\tau}{\tau^{(3+\alpha)/2}}\right]\\\
&\qquad+\mathbf{1}_{rs<(r-s)^{2}}(rs)^{-\zeta}\int_{0}^{rs}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c(r-s)^{2}/\tau}\\\
&\quad\asymp\mathbf{1}_{rs>(r-s)^{2}}(rs)^{-\zeta}\left[\int_{(r-s)^{-2}}^{\infty}d\tau\,\tau^{\frac{\alpha-1}{2}}\mathrm{e}^{-c\tau(r-s)^{2}}+|r-s|^{-(1+\alpha)}\right]\\\
&\qquad+\mathbf{1}_{rs<(r-s)^{2}}(rs)^{-\zeta}|r-s|^{-(1+\alpha)}\int_{0}^{rs/(r-s)^{2}}\frac{d\tau}{\tau}\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c/\tau}\\\
&\quad\asymp(rs)^{-\zeta}\cdot|r-s|^{-(1+\alpha)}\left[\mathbf{1}_{rs>(r-s)^{2}}+\frac{|r-s|^{\alpha-1}}{(rs)^{(\alpha-1)/2}}\exp\left(-c\frac{(r-s)^{2}}{rs}\right)\mathbf{1}_{rs<(r-s)^{2}}\right].\end{split}$
(A.16)

Comparing (A.15) and (A.16) for $rs>(r-s)^{2}$ shows that the term
$(rs)^{-\zeta}\cdot|r-s|^{-(1+\alpha)}$ on the right-hand side of (A.16)
dominates the term $(rs)^{-(2\zeta+1+\alpha)/2}$ on the right-hand side of
(A.15) since
$(rs)^{-\zeta}\cdot|r-s|^{-(1+\alpha)}\gtrsim(rs)^{-\frac{2\zeta+1+\alpha}{2}}$.
On the other hand, if $rs<(r-s)^{2}$, then the $|r-s|^{-(2\zeta+1+\alpha)}$
term on the right-hand side of (A.15) dominates the prefactor on the right-
hand side of (A.16) since

$\displaystyle\begin{split}&\frac{|r-s|^{-2}}{(rs)^{\frac{2\zeta-1+\alpha}{2}}}\cdot\exp\left(-c\frac{(r-s)^{2}}{rs}\right)\mathbf{1}_{rs<(r-s)^{2}}\\\
&\quad=|r-s|^{-(2\zeta+1+\alpha)}\cdot\frac{|r-s|^{2\zeta-1+\alpha}}{(rs)^{\frac{2\zeta-1+\alpha}{2}}}\exp\left(-c\frac{(r-s)^{2}}{rs}\right)\mathbf{1}_{rs<(r-s)^{2}}\\\
&\quad\lesssim|r-s|^{-(2\zeta+1+\alpha)}.\end{split}$ (A.17)

Thus, for $c_{1}\leq 1$,

$\displaystyle\begin{split}\eqref{eq:heatkernelalphasubordinatedbounds}&\sim(rs)^{-\zeta}|r-s|^{-1-\alpha}\mathbf{1}_{rs>(r-s)^{2}}+|r-s|^{-(2\zeta+1+\alpha)}\mathbf{1}_{rs<(r-s)^{2}},\end{split}$
(A.18)

as needed. Now suppose $c_{1}>1$. Then the previous analysis suffices for the
upper bound since $\mathrm{e}^{-c/\tau^{c_{1}}}\leq 1$. Thus, it suffices to
prove the lower bound when $c_{1}>1$, i.e., it remains to find lower bounds
for

$\displaystyle\begin{split}&\int_{0}^{\infty}\frac{d\tau}{\tau}\,\left[\tau^{-\frac{2\zeta+1+\alpha}{2}}\left[\mathrm{e}^{-c\frac{(r-s)^{2}}{\tau}}\mathbf{1}_{\tau>|r-s|^{2/(c_{1}-1)}\vee
rs}+\mathrm{e}^{-c/\tau^{c_{1}}}\mathbf{1}_{\tau\in[rs,|r-s|^{2/(c_{1}-1)}]}\right]\right.\\\
&\quad\left.+(rs)^{-\zeta}\tau^{-\frac{1+\alpha}{2}}\left[\mathrm{e}^{-c\frac{(r-s)^{2}}{\tau}}\mathbf{1}_{|r-s|^{2/(c_{1}-1)}<\tau<rs}+\mathrm{e}^{-c/\tau^{c_{1}}}\mathbf{1}_{\tau<rs\wedge|r-s|^{2/(c_{1}-1)}}\right]\right].\end{split}$
(A.19)

When $rs<(r-s)^{2}$, we can argue exactly as in the previously analyzed case
$rs<1<(r-s)^{2}$. Thus, we assume $(r-s)^{2}<rs$. In this case, we drop the
first line in (A.19) for a lower bound, i.e., it remains to control the second
line. If $2/(c_{1}-1)\leq 2$ then we drop the second summand in the second
line of (A.19) and estimate the first summand there precisely as in (A.16) by
splitting the $\tau$-integral at $\tau=(r-s)^{2}$ and dropping the integral
over $\\{\tau<(r-s)^{2}\\}$. On the other hand, if $2/(c_{1}-1)>2$, then we
drop the first summand in the second line of (A.19). The second summand is
bounded from below by

$\displaystyle(rs)^{-\zeta}\int_{0}^{\infty}\frac{d\tau}{\tau}\,\tau^{-\frac{1+\alpha}{2}}\mathrm{e}^{-c/\tau^{c_{1}}}\mathbf{1}_{(r-s)^{2}<\tau<rs\wedge|r-s|^{2/(c_{1}-1)}}\gtrsim(rs)^{-\zeta}|r-s|^{-(1+\alpha)}.$
(A.20)

This concludes the analysis of the case $rs\wedge(r-s)^{2}>1$ and thereby the
proof of the bounds in (A.1).

###  A.2. Proof of (2.9)

The bounds (A.1) imply

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(1,r,s)&\sim_{\zeta,\alpha}\left(1+rs\right)^{\zeta}\mathbf{1}_{(r-s)^{2}<1}+|r-s|^{-(2\zeta+1+\alpha)}\mathbf{1}_{rs<1<(r-s)^{2}}\\\
&\quad+\left(|r-s|^{-2\zeta}\mathbf{1}_{1<rs<(r-s)^{2}}+(rs)^{-\zeta}\mathbf{1}_{rs>(r-s)^{2}>1}\right)|r-s|^{-(1+\alpha)}.\end{split}$
(A.21)

Using (A.21), we now show the desired bound (2.9), i.e.,

$\displaystyle\begin{split}p_{\zeta}^{(\alpha)}(1,r,s)&\sim_{\zeta,\alpha}\frac{1}{|r-s|^{1+\alpha}\cdot(r+s)^{2\zeta}+(1+r+s)^{2\zeta}}\end{split}$

for all $r,s>0$. To that end, we distinguish the cases $(r-s)^{2}\lessgtr 1$.
(1) If $(r-s)^{2}<1$, then, by $2rs=r^{2}+s^{2}-(r-s)^{2}$, we have
$p_{\zeta}^{(\alpha)}(1,r,s)\sim(1+rs)^{-\zeta}\sim(1+r+s)^{-2\zeta}$. This
gives the claimed estimate for $p_{\zeta}^{(\alpha)}(1,r,s)$. Note that we use
$|r-s|^{1+\alpha}(r+s)^{2\zeta}\leq(r+s)^{2\zeta+1+\alpha}\lesssim 1$ for
$r+s\lesssim 1$ and $\zeta>-1/2$.
(2) If $(r-s)^{2}>1$, then
$|r-s|^{1+\alpha}(r+s)^{2\zeta}\gtrsim(1+r+s)^{2\zeta}$. We now distinguish
the cases $rs\lessgtr(r-s)^{2}$ and use $(r+s)^{2}=(r-s)^{2}+4rs$. If
$(r-s)^{2}>rs$, then the claimed estimate for $p_{\zeta}^{(\alpha)}(1,r,s)$
follows immediately. If $(r-s)^{2}<rs$, then
$|r-s|^{1+\alpha}(rs)^{\zeta}\sim|r-s|^{1+\alpha}(r+s)^{2\zeta}$ as desired.

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