#  $L^{p}$ estimates for the Caffarelli-Silvestre extension operators

G. Metafune L. Negro C. Spina Dipartimento di Matematica e Fisica “Ennio De
Giorgi”, Università del Salento, C.P.193, 73100, Lecce, Italy. -mail:
<EMAIL_ADDRESS>di Matematica e Fisica “Ennio De
Giorgi”, Università del Salento, C.P.193, 73100, Lecce, Italy. email:
<EMAIL_ADDRESS>di Matematica e Fisica“Ennio De Giorgi”,
Università del Salento, C.P.193, 73100, Lecce, Italy. e-mail:
<EMAIL_ADDRESS>
###### Abstract

We study elliptic and parabolic problems governed by the singular elliptic
operators

$\mathcal{L}=\Delta_{x}+D_{yy}+\frac{c}{y}D_{y}-\frac{b}{y^{2}}$

in the half-space $\mathbb{R}^{N+1}_{+}=\\{(x,y):x\in\mathbb{R}^{N},y>0\\}$.

Mathematics subject classification (2010): 47D07, 35J70.

Keywords: elliptic operators, discontinuous coefficients, kernel estimates,
maximal regularity.

##  1 Introduction

In this paper we study solvability and regularity of elliptic and parabolic
problems associated to the degenerate operators

$\mathcal{L}=\Delta_{x}+D_{yy}+\frac{c}{y}D_{y}-\frac{b}{y^{2}}\quad{\rm
and}\quad D_{t}-\mathcal{L}$

in the half-space $\mathbb{R}^{N+1}_{+}=\\{(x,y):x\in\mathbb{R}^{N},y>0\\}$ or
$(0,\infty)\times\mathbb{R}^{N+1}_{+}$.

Here $b,\ c$ are constant real coefficients and we use
$L_{y}=D_{yy}+\frac{c}{y}D_{y}-\frac{b}{y^{2}}.$ Note that singularities in
the lower order terms appear when either $b$ or $c$ is different from 0.

The operators $\Delta_{x}$, $L_{y}$ commute and the whole operator
$\mathcal{L}$ satisfies the scaling property
$I_{s}^{-1}\mathcal{L}I_{s}=s^{2}\mathcal{L}$, if $I_{s}u(x,y)=u(sx,sy)$.

When $b=0$, then $L_{y}$ is a Bessel operator (we shall denote it by $B_{y}$)
and both $\mathcal{L}=\Delta_{x}+B_{y}$ and $D_{t}-\mathcal{L}$ play a major
role in the investigation of the fractional powers $(-\Delta_{x})^{s}$ and
$(D_{t}-\Delta_{x})^{s}$, $s=(1-c)/2$, through the “extension procedure” of
Caffarelli and Silvestre [4], after the pioneering work by Muckenhoupt and
Stein, [14]. For this reason, ${\mathcal{L}}$ and $D_{t}-\mathcal{L}$ are
named the “extension operators”. We refer the reader to [8, Section 10] for an
exposition of the theory in the language of semigroups and for the references
to the wide literature on the extension problem, both in the elliptic and
parabolic case.

Here we study unique solvability of the problems $\lambda u-\mathcal{L}u=f$
and $D_{t}v-\mathcal{L}v=g$ in $L^{p}$ spaces under appropriate boundary
conditions, and initial conditions in the parabolic case, together with the
regularity of $u,v$. In the language of semigroup theory, we prove that
$\mathcal{L}$ generates an analytic semigroup, characterize its domain and
show that it has maximal regularity, which means that both $D_{t}v$ and
$\mathcal{L}v$ have the same regularity as $g$.

Both the domains of $\Delta_{x}$ and $L_{y}$ are known, in their corresponding
$L^{p}$-spaces. Clearly $D(\Delta_{x})=W^{2,p}(\mathbb{R}^{N})$, $1<p<\infty$.
However $D(L_{y})\subset L^{p}(0,\infty)$ is more delicate, the boundary
conditions and the regularity up to $y=0$ depend on the coefficients $c,b$. We
shall devote Sections 2, 3, 4 to a careful study of the 1d operator $L_{y}$,
starting form the Bessel case where $b=0$, this last both under Dirichlet and
Neumann boundary conditions at $y=0$. We shall provide in both cases the
description of the domain, pointwise estimates for the heat kernel and its
gradient. The general case is reduced to the Bessel one, through a change of
variables. We study $L_{y}$ also in weighted spaces
$L^{p}((0,\infty),y^{m}dy)$; the cases $m=0$ and $m=c$ are the most important:
the first corresponds to the Lebesgue measure, the second to the symmetrizing
one. However we need general $m$ also for technical reasons. This makes the
exposition slightly heavier, but it is unavoidable in our approach. Not all
the results in these sections are completely new. A description of the domain
under Dirichlet boundary conditions is in [22] but here we have more precise
results; we are not aware of a similar description in the case of Neumann
boundary conditions for $B_{y}$. The heat kernel is known among probabilists
but a purely analytic derivation can be found in [9] for Neumann boundary
conditions. Here we prefer to give an analytic proof in both cases and provide
manageable and precise estimates.

The elliptic operator $\mathcal{L}$ is studied through estimates like

$\|\Delta_{x}u\|_{p}+\|L_{y}u\|_{p}\leq C\|\mathcal{L}u\|_{p}$ (1)

where the $L^{p}$ norms are taken over $\mathbb{R}_{+}^{N+1}$. This kind of
estimates are quite natural in this context but not easy to prove. Of course
they are equivalent to $\|D_{x_{i}x_{j}}u\|_{p}\leq C\|\mathcal{L}u\|_{p}$, by
the Calderón-Zygmund inequalities in the $x$-variables, and can be restated by
saying that $\mathcal{L}$ is closed on $D(\Delta_{x})\cap D(L_{y})$ or that
$\Delta_{x}\mathcal{L}^{-1}$ is bounded. Note that the weaker inequality (1)
with $\|\mathcal{L}u\|_{p}+\|u\|_{p}$ on the right hand side implies the
stronger one as stated, by the scaling properties of the operators involved.

Estimates for $D_{yy}u$ follow if and only if they hold for the one
dimensional operator $L_{y}$ but those for the mixed derivatives $D_{x_{i}y}u$
are more subtle. They are certainly true when $D_{yy}L_{y}^{-1}$ is bounded,
by Calderón-Zygmund with respect to all $x,y$ variables, but we shall prove
that they hold if (and only if) $D_{y}(I-L_{y})^{-1}$ is bounded, which was
quite unexpected for us.

Let us explain how to obtain (1) when $p=2$ and introduce our approach for
general $p$. Assuming that $\Delta_{x}u+L_{y}u=f$ and taking the Fourier
transform with respect to $x$ (with covariable $\xi$) we obtain
$-|\xi|^{2}\hat{u}(\xi,y)+L_{y}\hat{u}(\xi,y)=\hat{f}(\xi,y)$ and then
$|\xi|^{2}\hat{u}(\xi,y)=-|\xi|^{2}(|\xi|^{2}-L_{y})^{-1}\hat{f}(\xi,y)$.
Assuming that $L_{y}$ generates a bounded semigroup in $L^{2}(0,\infty)$, then
$|\xi|^{2}\|(|\xi|^{2}-L_{y})^{-1}\|\leq C$ and

$\int_{0}^{\infty}|\xi|^{4}|\hat{u}(\xi,y)|^{2}dy\leq
C^{2}\int_{0}^{\infty}|\hat{f}(\xi,y)|^{2}dy$

which gives, after integration with respect to $\xi$ and Plancherel equality,

$\|\Delta_{x}u\|_{2}=\||\xi|^{2}\hat{u}\|_{2}\leq C\|f\|_{2}.$

When $p\neq 2$ and denoting by ${\cal F}$ the Fourier transform with respect
to $x$ we get, formally,

$\Delta_{x}\mathcal{L}^{-1}=-{\cal
F}^{-1}\left(|\xi|^{2}(|\xi|^{2}-L_{y})^{-1}\right){\cal F}$

and the boundedness of $\Delta_{x}\mathcal{L}^{-1}$ is equivalent to say that
the operator valued map $\xi\in\mathbb{R}^{N}\to
M(\xi)=|\xi|^{2}(|\xi|^{2}-L_{y})^{-1}\in B(L^{p}(0,\infty))$ is a bounded
Fourier multiplier in
$L^{p}(\mathbb{R}^{N};L^{p}(0,\infty))=L^{p}(\mathbb{R}_{+}^{N+1})$. Here we
use a vector valued Mikhlin multiplier theorem which relies on the
$\mathcal{R}$-boundedness of the family $M(\xi)$ and its derivatives, which we
deduce from heat kernel estimates. We use a similar strategy for
$\nabla_{x}D_{y}\mathcal{L}^{-1}$ which this time rests on estimates for the
gradient of the heat kernel of $L_{y}$.

It is important to note that the closedness of $\Delta_{x}+L_{y}$ on the
intersection of the corresponding domains does not follow from general
results. In fact, $e^{tL_{y}}$ is not contractive and does not admit Gaussian
estimates, except for special cases; moreover it is bounded in
$L^{p}(0,\infty)$ only for certain intervals of $p$ depending on the
coefficients $c,b$.

The strategy for proving the parabolic estimates

$\|D_{t}v\|_{p}+\|\mathcal{L}v\|_{p}\leq C\|(D_{t}-\mathcal{L})v\|_{p}$

($L^{p}$ norms on $(0,\infty)\times\mathbb{R}^{N+1}_{+}$), is similar after
taking the Fourier transform with respect to $t$.

Both the elliptic and parabolic estimates rely on a vector valued Mikhlin
multiplier theorem and share the name “maximal regularity” even though this
term is often restricted to the parabolic case.

The functional analytic approach for maximal regularity is widely described in
[13] and in the new books [10], [11]. The whole theory relies on a deep
interplay between harmonic analysis and structure theory of Banach spaces but
largely simplifies when the underlying Banach spaces are $L^{p}$ spaces, by
using classical square function estimates. This last approach has been
employed extensively in [5], showing that uniformly parabolic operators have
maximal regularity, under very general boundary conditions.

We deduce the boundedness of vector valued multipliers by the
$\mathcal{R}$-boundedness of a family of integral operators, which we prove
through an extrapolation result in [2] which involves a family of Muckenhoupt
weighted estimates. Here we adopt the same strategy as T. A. Bui, see [3], in
the case of Schrödinger operators with inverse square potentials. Section 7 is
really the core of the paper, while Section 6 contains all relevant
definitions and results for the subsequent proofs.

We work in $L^{p}(\mathbb{R}^{N+1}_{+},y^{m}dxdy)$ not just for the sake of
generality but because our proof relies on weighted estimates: we are unable
to obtain the result just fixing the Lebesgue measure or the symmetrizing one
$y^{c}dxdy$ but we have to work simultaneously in different homogeneous
spaces.

As an application of our results, in Section 9 we deduce Rellich inequalities
for $\mathcal{L}=\Delta_{x}+L_{y}$ by the analogous results for the one
dimensional operator $L_{y}$, using the closedness of $\mathcal{L}$ on the
intersection of the domains of $\Delta_{x}$ and $L_{y}$.

Notation. For $N\geq 0$,
$\mathbb{R}^{N+1}_{+}=\\{(x,y):x\in\mathbb{R}^{N},y>0\\}$. For
$m\in\mathbb{R}$ we consider the measure $d\mu_{m}=y^{m}dxdy$ in
$\mathbb{R}^{N+1}_{+}$. We write $L^{p}_{m}(\mathbb{R}^{N+1})$ for
$L^{p}(\mathbb{R}_{+}^{N+1};y^{m}dxdy)$ and often only $L^{p}_{m}$ when
$\mathbb{R}^{N+1}_{+}$ is understood. Similarly
$W^{k,p}_{m}(\mathbb{R}^{N+1}_{+})=\\{u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+}):\partial^{\alpha}u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+})\quad|\alpha|\leq k\\}$. We use often
$W^{k,p}_{m}$ thus omitting $\mathbb{R}^{N+1}_{+}$ and $W^{k,p}_{0,m}$ for the
closure of $C_{c}^{\infty}(\mathbb{R}^{N+1}_{+})$ in $W^{k,p}_{m}$ and we use
$H^{k}_{m}$ for $W^{k,2}_{m}$.

Acknowledgements. The authors thank S. Fornaro, N. Garofalo, D. Pallara and V.
Vespri for several comments on a previous version of the manuscript.

##  2 Bessel operators in $1d$

In this section we state and prove the main properties of the degenerate
operator

$B=D_{yy}+\frac{c}{y}D_{y}=y^{-c}D_{y}\left(y^{c}D_{y}\right)$

on the half line $\mathbb{R}_{+}=]0,\infty[$ needed for our purposes.

###  2.1 Weighted $L^{2}$ spaces and Bessel operators

We use the Sobolev spaces defined in Appendix B, for $p=2$ and $N=0$.
According to the above notation, for $c\in\mathbb{R}$ we use
$L^{2}_{c}=\\{u:\mathbb{R}_{+}\to\mathbb{C}:\int_{0}^{\infty}|u(y)|^{2}y^{c}dy<\infty\\}$,
$H^{1}_{c}=\\{u\in L^{2}_{c},u^{\prime}\in L^{2}_{c}\\}$, where $u^{\prime}$
is understood as a distribution in the open interval $]0,\infty[$. Both
$L^{2}_{c}$ and $H^{1}_{c}$ are Hilbert spaces under their canonical inner
products; moreover $C_{c}^{\infty}(0,\infty)$ is contained in both and dense
in $L^{2}_{c}$. We denote by $H^{1}_{0,c}$ the closure of
$C_{c}^{\infty}(0,\infty)$ in $H^{1}_{c}$. We need the following properties
proved in greater generality in Appendix B.

###### Lemma 2.1

  * (i)

If $|c|\geq 1$, then $H^{1}_{0,c}=H^{1}_{c}$. When $c\leq-1$, then
$\displaystyle\lim_{y\to 0}u(y)=0$ for every $u\in H^{1}_{c}$.

  * (ii)

If $|c|<1$ and $u\in H^{1}_{c}$, then $\displaystyle\lim_{y\to
0}u(y)=\ell\in\mathbb{C}$. Moreover, $\ell=0$ if and only if $u\in
H^{1}_{0,c}$.

$B$ is associated to the symmetric form in $L^{2}_{c}$

$\displaystyle\mathfrak{a}(u,v)$
$\displaystyle:=\int_{0}^{\infty}D_{y}uD_{y}\overline{v}\,y^{c}dy=\int_{0}^{\infty}(Bu)\,\overline{v}\,y^{c}dy.$

For any $c\in\mathbb{R}$ we may consider $H^{1}_{0,c}$ as domain of the form
and, accordingly, define the Bessel operator with Dirichlet boundary
conditions $B^{d}$ by

$D(B^{d})=\\{u\in H^{1}_{0,c}:\exists f\in L^{2}_{c}\ {\rm such\ that}\
\mathfrak{a}(u,v)=\int_{0}^{\infty}f\overline{v}y^{c}\,dy\ {\rm for\ every}\
v\in H^{1}_{0,c}\\},\quad B^{d}u=-f$ (2)

Similarly, by considering $H^{1}_{c}$ we obtain the Bessel operator with
Neumann Boundary conditions $B^{n}$ defined as

$D(B^{n})=\\{u\in H^{1}_{c}:\exists f\in L^{2}_{c}\ {\rm such\ that}\
\mathfrak{a}(u,v)=\int_{0}^{\infty}f\overline{v}y^{c}\,dy\ {\rm for\ every}\
v\in H^{1}_{c}\\},\quad B^{n}u=-f$ (3)

$B^{d},B^{n}$ are non-positive self-adjoint operators and $u\in
H^{2}_{loc}(0,\infty)$ with $B^{d}u=B^{n}u=u_{yy}+\frac{c}{y}u_{y}$ for $y>0$,
if $u\in D(B^{d})$ or $u\in D(B^{n})$, by standard arguments.

###### Lemma 2.2

If $c>-1$ and $u\in D(B^{n})$, then $\displaystyle\lim_{y\to
0}y^{c}u^{\prime}(y)=0$.

Proof.  By assumption for $v\in H^{1}_{c}$

$\displaystyle\int_{0}^{\infty}u_{y}v_{y}y^{c}dy$
$\displaystyle=-\int_{0}^{\infty}(B^{n}u)vy^{c}dy=-\lim_{\varepsilon\to
0}\int_{\varepsilon}^{\infty}\frac{d}{dy}(y^{c}u_{y})vdy$
$\displaystyle=\lim_{\varepsilon\to
0}\left(\int_{\varepsilon}^{\infty}u_{y}v_{y}y^{c}dy-\varepsilon^{c}u_{y}(\varepsilon)v(\varepsilon)\right)=\int_{0}^{\infty}u_{y}v_{y}y^{c}dy-\lim_{\varepsilon\to
0}\varepsilon^{c}u_{y}(\varepsilon)v(\varepsilon).$

Choosing $v\equiv 1$ near $0$, which is possible since $c>-1$, we get the
result.

Observe that:

  * •

when $|c|\geq 1$ then $B^{d}=B^{n}$ and, when $c\leq-1$, $u(0)=0$ for every
$u\in D(B^{d})$ by Lemma 2.1 (i);

  * •

when $|c|<1$ then $B^{d}$ and $B^{n}$ are different and $u\in D(B^{d})$
fulfils $u(0)=0$, by Lemma 2.1 (ii).

Even though $B^{d}$ and $B^{n}$ are defined for every $c\in\mathbb{R}$, we
shall use $B^{d}$ when $c<1$ and $B^{n}$ when $c>-1$, according to the
literature. This allows to unify some formulas.

###  2.2 The resolvents and the heat kernels of $B^{d}$ and $B^{n}$

We start by recalling some well-known facts about the modified Bessel
functions $I_{\nu}$ and $K_{\nu}$ which constitute a basis of solutions of the
modified Bessel equation

$z^{2}\frac{d^{2}v}{dz^{2}}+z\frac{dv}{dz}-(z^{2}+\nu^{2})v=0,\quad\textrm{\emph{Re}\,}z>0.$

We recall that for $\textrm{\emph{Re}\,}z>0$ one has

$I_{\nu}(z)=\left(\frac{z}{2}\right)^{\nu}\sum_{m=0}^{\infty}\frac{1}{m!\,\Gamma(\nu+1+m)}\left(\frac{z}{2}\right)^{2m},\quad
K_{\nu}(z)=\frac{\pi}{2}\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin\pi\nu},$

where limiting values are taken for the definition of $K_{\nu}$ when $\nu$ is
an integer. The basic properties of these functions we need are collected in
the following lemma, see e.g., [1, Sections 9.6 and 9.7].

###### Lemma 2.3

For $\nu>-1$, $I_{\nu}$ is increasing and $K_{\nu}$ is decreasing (when
restricted to the positive real half line). Moreover they satisfy the
following properties if $z\in\Sigma_{\pi/2-\varepsilon}$.

  * (i)

$I_{\nu}(z)\neq 0$ for every $\textrm{\emph{Re}\,}z>0$.

  * (ii)

$I_{\nu}(z)\approx\frac{1}{\Gamma(\nu+1)}\left(\frac{z}{2}\right)^{\nu},\quad\text{as
}|z|\to 0,\qquad I_{\nu}(z)\approx\frac{e^{z}}{\sqrt{2\pi
z}}(1+O(|z|^{-1}),\quad\text{as }|z|\to\infty$.

  * (iii)

If $\nu\neq 0$,
$K_{\mu}(z)\approx\frac{\nu}{|\nu|}\frac{1}{2}\Gamma(|\nu|)\left(\frac{z}{2}\right)^{-|\nu|},\qquad
K_{0}(z)\approx-\log z,\qquad\text{as }|z|\to 0$
$K_{\mu}(z)\approx\sqrt{\frac{\pi}{2z}}e^{-z},\quad\text{as }|z|\to\infty$.

  * (iv)

$I_{\nu}^{\prime}(z)=I_{\nu+1}(z)+\frac{\nu}{z}I_{\nu}(z)$,
$K_{\nu}^{\prime}(z)=K_{\nu+1}(z)+\frac{\nu}{z}K_{\nu}(z)$, for every
$\textrm{\emph{Re}\,}z>0$.

Note that

$|I_{\nu}(z)|\simeq
C_{\nu,\epsilon}(1\wedge|z|)^{\nu+\frac{1}{2}}\frac{e^{Rez}}{\sqrt{|z|}},\qquad
z\in\Sigma_{\frac{\pi}{2}-\epsilon}$ (4)

for suitable constants $C_{\nu,\epsilon}>0$ which may be different in lower an
in the upper estimate.

Let us compute the resolvent operator of $B^{n}$. When we write $\sqrt{z}$ we
mean the square root of $z$ having positive real part.

###### Proposition 2.4

Let $c>-1$ and $\lambda\in\mathbb{C}\setminus(-\infty,0]$. Then, for every
$f\in L^{2}_{c}$,

$(\lambda-B^{n})^{-1}f=\int_{0}^{\infty}G^{n}(\lambda,y,\rho)f(\rho)\rho^{c}d\rho$

with

$G^{n}(\lambda,y,\rho)=\begin{cases}y^{\frac{1-c}{2}}\rho^{\frac{1-c}{2}}\,I_{\frac{c-1}{2}}(\sqrt{\lambda}\,y)K_{{\frac{|1-c|}{2}}}(\sqrt{\lambda}\,\rho)\quad
y\leq\rho\\\\[6.45831pt]
y^{\frac{1-c}{2}}\rho^{\frac{1-c}{2}}\,I_{\frac{c-1}{2}}(\sqrt{\lambda}\,\rho)K_{\frac{|1-c|}{2}}(\sqrt{\lambda}\,y)\quad
y\geq\rho,\end{cases}$ (5)

Proof. Let us first consider the case $\lambda=\omega^{2}$, $|\omega|=1$. By
setting $u(y)=y^{\nu}v(\omega y)$, $\nu=(1-c)/2$, the homogeneous equation

$D_{yy}u+\frac{c}{y}D_{y}u-\omega^{2}u=0$

transforms into the complex equation

$z^{2}\frac{d^{2}v}{dz^{2}}+z\frac{dv}{dz}-(z^{2}+\nu^{2})v=0,\quad\textrm{\emph{Re}\,}z>0.$

Assume first that $-1<c\leq 1$ so that $0\leq\nu<1$. Then
$u_{1}(z)=z^{\nu}I_{-\nu}(\omega z)$ and $u_{2}(z)=z^{\nu}K_{\nu}(\omega z)$
constitute a basis of solutions. Since the Wronskian of $K_{\nu}$, $I_{-\nu}$
is $1/r$, see [1, 9.6 and 9.7], that of $u_{1}$,$u_{2}$ is $r^{-c}$. It
follows that every solution of

$D_{yy}u+\frac{c}{y}D_{y}u-\omega^{2}u=f$

is given by

$u(y)=\int_{0}^{\infty}G^{n}(\omega^{2},y,\rho)f\left(\rho\right)\rho^{c}d\rho+c_{1}y^{\nu}I_{-\nu}(\omega
y)+c_{2}y^{\nu}K_{\nu}(\omega y),$ (6)

with $c_{1},\ c_{2}\in\mathbb{C}$ and

$G^{n}(\omega^{2},y,\rho)=\begin{cases}y^{\nu}\rho^{\nu}\,I_{-\nu}(\omega
y)K_{\nu}(\omega\rho)\quad y\leq\rho\\\\[4.30554pt]
y^{\nu}\rho^{\nu}\,I_{-\nu}(\omega\rho)K_{\nu}(\omega y)\quad
y\geq\rho\end{cases}$

Next we use Lemma 2.3 to show that

$\sup_{y\in(0,+\infty)}\int_{0}^{\infty}|G^{n}(\omega,y,\rho)|\rho^{c}d\rho<+\infty.$

Indeed, for $y\leq 1$, recalling that $|\omega|=1$, one has

$\displaystyle\int_{0}^{\infty}|G^{n}(\omega^{2},y,\rho)|\rho^{c}d\rho=$
$\displaystyle\int_{0}^{y}y^{\nu}\rho^{\nu}\,|I_{-\nu}(\omega\rho)||K_{\nu}(\omega
y)|\,\rho^{c}d\rho+\int_{y}^{\infty}y^{\nu}\rho^{\nu}\,|I_{-\nu}(\omega
y)||K_{\nu}(\omega\rho)|\,\rho^{c}d\rho$ $\displaystyle\leq$ $\displaystyle
C\left(\int_{0}^{y}\rho^{c}d\rho+\int_{y}^{1}\rho^{c}d\rho+\int_{1}^{\infty}\rho^{\frac{1+c}{2}}(\sqrt{\rho})^{-1}e^{-\textrm{\emph{Re}\,}{\omega}\rho}\right)\leq
C$

and similarly for $y>1$. By the symmetry of the kernel and Young’s inequality
the integral operator $T$ defined by $G^{n}(\omega^{2},\cdot,\cdot)$ is
therefore bounded in $L^{2}_{c}$.

Let $f\in C_{c}^{\infty}((0,\infty))$ with support in $(a,b)$ with $a>0$ and
$u=(\omega^{2}-B^{n})^{-1}f\in D(B^{n})$. Then $u$ is given by (6) with
$c_{1}=0$, since $T$ is bounded in $L^{2}_{c}$, $K_{\nu}$ is exponentially
decreasing and $I_{-\nu}$ is exponentially increasing near $\infty$. Since

$\displaystyle u(y)=$
$\displaystyle\int_{0}^{y}y^{\nu}\rho^{\nu}K_{\nu}(\omega
y)I_{\nu}(\omega\rho)f\left(\rho\right)\,\rho^{c}d\rho+\int_{y}^{b}y^{\nu}\rho^{\nu}K_{\nu}(\omega\rho)I_{-\nu}(\omega
y)f\left(\rho\right)\,\rho^{c}d\rho+c_{2}y^{\nu}K_{\nu}(\omega y)$

we have for $y<a$

$\displaystyle u(y)=$
$\displaystyle\int_{a}^{b}y^{\nu}\rho^{\nu}K_{\nu}(\omega\rho)I_{-\nu}(\omega
y)f\left(\rho\right)\,\rho^{c}d\rho+c_{2}y^{\nu}K_{\nu}(\omega
y)=c_{1}y^{\nu}I_{-\nu}(\omega y)+c_{2}y^{\nu}K_{\nu}(\omega y)$

for some $c_{1},c_{2}\in\mathbb{C}$. From Lemma 2.3 it follows that
$v(y)=y^{\nu}I_{-{\nu}}(\omega y)$ satisfies the Neumann condition $\lim_{y\to
0}y^{c}v^{\prime}(y)=0$ whereas $y^{\nu}K_{\nu}(\omega y)$ does not. Since
$u\in D(B^{n})$, by Lemma 2.2 $y^{c}u^{\prime}(y)\to 0$ and hence $c_{2}=0$.
By density, $(\omega^{2}-B^{n})^{-1}=T$, since both operators are bounded and
coincide on compactly supported functions.

Finally let us compute the resolvent for a general
$\lambda\not\in(-\infty,0]$. If $M_{s}u(y)=u(sy)$, then
$M_{\sqrt{|\lambda|}}B^{n}M_{\sqrt{|\lambda|}^{-1}}=\frac{1}{|\lambda|}B^{n}$;
setting $\lambda=|\lambda|\omega$ we get using the previous step

$\displaystyle(\lambda-B^{n})^{-1}f$
$\displaystyle=|\lambda|^{-1}M_{\sqrt{|\lambda|}}(\omega-B^{n})^{-1}M_{\sqrt{|\lambda|}^{-1}}f=\frac{1}{|\lambda|}\int_{0}^{\infty}G^{n}(\omega,y\sqrt{|\lambda|},\rho)f\left(\frac{\rho}{\sqrt{|\lambda|}}\right)\rho^{c}d\rho$
$\displaystyle=|\lambda|^{\frac{c-1}{2}}\int_{0}^{\infty}G^{n}(\omega,y\sqrt{|\lambda|},\rho\sqrt{|\lambda|})f\left(\rho\right)\,\rho^{c}d\rho$

which gives (7) when $-1<c\leq 1$. When $c>1$, we use $I_{|\nu|}$, $K_{|\nu|}$
as a basis of solutions of Bessel equation and proceed as before.

A similar proof gives the resolvent of $B^{d}$. We omit the details, see also
[16, Section 4.2].

###### Proposition 2.5

Let $c<1$ and $\lambda\in\mathbb{C}\setminus(-\infty,0]$. Then, for every
$f\in L^{2}_{c}$,

$(\lambda-B^{d})^{-1}f=\int_{0}^{\infty}G^{d}(\lambda,y,\rho)f(\rho)\rho^{c}d\rho$

with

$G^{d}(\lambda,y,\rho)=\begin{cases}y^{\frac{1-c}{2}}\rho^{\frac{1-c}{2}}\,I_{\frac{1-c}{2}}(\sqrt{\lambda}\,y)K_{\frac{1-c}{2}}(\sqrt{\lambda}\,\rho)\quad
y\leq\rho\\\\[8.61108pt]
y^{\frac{1-c}{2}}\rho^{\frac{1-c}{2}}\,I_{\frac{1-c}{2}}(\sqrt{\lambda}\,\rho)K_{\frac{1-c}{2}}(\sqrt{\lambda}\,y)\quad
y\geq\rho.\end{cases}$ (7)

Note that when $|c|<1$ the resolvent of $B^{n}$ uses $I_{\frac{c-1}{2}}$
whereas $B^{d}$ is constructed with $I_{\frac{1-c}{2}}$.

Next we compute the heat kernel of $e^{zB^{n}},e^{zB^{d}}$, proceeding as in
[16, Section 4.2]. These heat kernels are known and usually computed by
probabilistic methods. Instead we provide a purely analytical proof and refer
also to [8] for a similar approach in the Neumann case.

For $z\in C_{+},y,\rho>0$ we denote now by $p(z,y,\rho)$ the heat kernel of
the operator $B$ and argue first for positive $t$. We look for a smooth
function $p(t,y,\rho)$ such that, for every $f\in L^{2}_{c}$

$e^{tB}f(y)=\int_{0}^{\infty}p(t,y,\rho)f(\rho)\,d\rho.$

Note that the kernel is written with respect to the Lebesgue measure rather
than $y^{c}dy$. The function $p$ should then satisfy

$\begin{cases}p_{t}(t,y,\rho)=D_{yy}p(t,y,\rho)+\frac{c}{y}D_{y}p(t,y,\rho)\\\
p(0,y,\rho)=\delta_{\rho}.\end{cases}$

It follows that
$\tilde{p}(t,y,\rho)=y^{\frac{c}{2}}p(t,y,\rho)\rho^{-\frac{c}{2}}$ satisfies
with $\nu^{2}=(c-1)^{2}/4$

$\begin{cases}{\tilde{p}}_{t}(t,y,\rho)=D_{yy}{\tilde{p}}(t,y,\rho)-\frac{1}{y^{2}}\left(\nu^{2}-\frac{1}{4}\right){\tilde{p}}(t,y,\rho)\\\
{\tilde{p}}(0,y,\rho)=\delta_{\rho}.\end{cases}$ (8)

Since $\lambda^{2}B=M_{\lambda}^{-1}BM_{\lambda}$ we obtain
$e^{t\lambda^{2}B}=M_{\lambda}^{-1}e^{tB}M_{\lambda}$. Rewriting this identity
using the kernel $\tilde{p}$ and setting $\lambda^{2}t=1$ we obtain

${\tilde{p}}(t,y,\rho)=\frac{1}{\sqrt{t}}{\tilde{p}}\left(1,\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right):=\frac{1}{\sqrt{t}}F\left(\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right).$

Then (8) becomes

$\displaystyle D_{yy}F\left(\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right)$
$\displaystyle-\frac{1}{y^{2}}\left(\nu^{2}-\frac{1}{4}\right)tF\left(\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right)+$
$\displaystyle+\frac{1}{2}F\left(\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right)+\frac{1}{2}\frac{y}{\sqrt{t}}D_{y}F\left(\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right)+\frac{1}{2}\frac{\rho}{\sqrt{t}}D_{\rho}F\left(\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right)=0$

that is

$\displaystyle D_{yy}F\left(y,\rho\right)$
$\displaystyle-\frac{1}{y^{2}}\left(\nu^{2}-\frac{1}{4}\right)F\left(y,\rho\right)+\frac{1}{2}F\left(y,\rho\right)+\frac{1}{2}yD_{y}F\left(y,\rho\right)+\frac{1}{2}\rho
D_{\rho}F\left(\frac{y}{\sqrt{t}},\frac{\rho}{\sqrt{t}}\right)=0.$

Since for large $y$ the operator $B$ behaves like $D^{2}$, having in mind the
gaussian kernel, we look for a solution of the form

$F(y,\rho)=\frac{1}{\sqrt{4\pi}}\exp\left\\{-\frac{(y-\rho)^{2}}{4}\right\\}H(y\rho)$

with $H$ depending only on the product of the variables. By straightforward
computations, we deduce

$\rho^{2}D_{yy}H(y\rho)+\rho^{2}D_{y}H(y\rho)-\frac{1}{y^{2}}\left(\nu^{2}-\frac{1}{4}\right)H(y\rho)=0$

or

$DH(x)+D_{x}H(x)-\frac{1}{x^{2}}\left(\nu^{2}-\frac{1}{4}\right)H(x)=0.$

Setting $H(x)=u(x)e^{-\frac{x}{2}}$, $u$ solves

$D_{xx}u-\frac{1}{4}u(x)-\frac{1}{x^{2}}\left(\nu^{2}-\frac{1}{4}\right)u(x)=0$

and $v(x)=u(2x)$ satisfies

$D_{xx}v-v(x)-\frac{1}{x^{2}}\left(\nu^{2}-\frac{1}{4}\right)v(x)=0.$

It follows that $v(x)=c_{1}\sqrt{x}I_{\nu}(x)+c_{2}\sqrt{x}K_{|\nu|}(x)$.
Since the function $H$ captures the behaviour of the heat kernel near the
origin (the behaviour at infinity is governed by the gaussian factor) and
since the resolvents of $B^{n},B^{d}$ are constructed with $\nu=(c-1)/2$,
$\nu=(1-c)/2$, respectively, we choose $c_{2}=0$, $c_{1}=1$ and $\nu$
accordingly. Therefore in the case of Neumann boundary conditions,
$u(x)=v\left(\frac{x}{2}\right)=\kappa\sqrt{\frac{x}{2}}I_{\frac{c-1}{2}}\left(\frac{x}{2}\right)$,
$H(y\rho)=u(y\rho)e^{-\frac{y\rho}{2}}=\kappa\sqrt{\frac{y\rho}{2}}I_{\frac{c-1}{2}}\left(\frac{y\rho}{2}\right)e^{-\frac{y\rho}{2}}$,

$F(y,\rho)=\frac{\kappa}{\sqrt{4\pi}}\exp\left\\{-\frac{(y-\rho)^{2}}{4}\right\\}\sqrt{\frac{y\rho}{2}}I_{\frac{c-1}{2}}\left(\frac{y\rho}{2}\right)e^{-\frac{y\rho}{2}}=\frac{\kappa}{\sqrt{4\pi}}\sqrt{\frac{y\rho}{2}}\exp\left\\{-\frac{y^{2}+\rho^{2}}{4}\right\\}I_{\frac{c-1}{2}}\left(\frac{y\rho}{2}\right)$

and

$\tilde{p}(t,y,\rho)=\frac{1}{\sqrt{4\pi
t}}H\left(\frac{y\rho}{t}\right)\exp\left\\{-\frac{(y-\rho)^{2}}{4t}\right\\}=\frac{\kappa}{t\sqrt{4\pi}}\sqrt{\frac{y\rho}{2}}\exp\left\\{-\frac{y^{2}+\rho^{2}}{4t}\right\\}I_{\frac{c-1}{2}}\left(\frac{y\rho}{2t}\right).$

It follows that

$p^{n}(t,y,\rho)=y^{-\frac{c}{2}}\tilde{p}(t,y,\rho)\rho^{\frac{c}{2}}=\frac{\kappa}{t\sqrt{4\pi}}\left(y\rho\right)^{\frac{1-c}{2}}\rho^{c}\exp\left\\{-\frac{y^{2}+\rho^{2}}{4t}\right\\}I_{\frac{c-1}{2}}\left(\frac{y\rho}{2t}\right).$
(9)

In the case of Dirichlet boundary conditions it is sufficient to change
$I_{\frac{c-1}{2}}$ with $I_{\frac{1-c}{2}}$ to obtain the corresponding
kernel.

$p^{d}(t,y,\rho)=y^{-\frac{c}{2}}\tilde{p}(t,y,\rho)\rho^{\frac{c}{2}}=\frac{\kappa}{t\sqrt{4\pi}}\left(y\rho\right)^{\frac{1-c}{2}}\rho^{c}\exp\left\\{-\frac{y^{2}+\rho^{2}}{4t}\right\\}I_{\frac{1-c}{2}}\left(\frac{y\rho}{2t}\right).$
(10)

Finally, we give a formal proof and compute the constant $\kappa$.

###### Theorem 2.6

For $z\in\mathbb{C}_{+}$ the heat kernels of the operators $B^{n},B^{d}$ are
given by

$p_{B^{n}}(z,y,\rho)=\frac{1}{2z}\rho^{c}(y\rho)^{\frac{1-c}{2}}\exp\left\\{-\frac{y^{2}+\rho^{2}}{4z}\right\\}I_{\frac{c-1}{2}}\left(\frac{y\rho}{2z}\right),\qquad
c>-1.$
$p_{B^{d}}(z,y,\rho)=\frac{1}{2z}\rho^{c}\left(y\rho\right)^{\frac{1-c}{2}}\exp\left\\{-\frac{y^{2}+\rho^{2}}{4z}\right\\}I_{\frac{1-c}{2}}\left(\frac{y\rho}{2z}\right),\qquad
c<1.$

Proof. Let us consider $B^{n}$ and $t\in\mathbb{R}^{+}$, first. The Laplace
transform of the right hand side of (9) is given by, see [7, p.200],

$\begin{cases}\frac{2\kappa}{\sqrt{4\pi}}\rho^{c}(y\rho)^{\frac{1-c}{2}}I_{-\sqrt{D}}(y\sqrt{\lambda})K_{\sqrt{D}}(\rho\sqrt{\lambda})\quad
y\leq\rho\\\\[4.30554pt]
\frac{2\kappa}{\sqrt{4\pi}}\rho^{c}(y\rho)^{\frac{1-c}{2}}I_{-\sqrt{D}}(\rho\sqrt{\lambda})K_{\sqrt{D}}(y\sqrt{\lambda})\quad
y\geq\rho.\end{cases}$

For $\kappa=\sqrt{\pi}$ it coincides with the kernel
$\rho^{c}G^{n}(\lambda,y,\rho)$ of the resolvent operator
$(\lambda-B^{n})^{-1}$, see Proposition 2.4. Let $S(t)$ be the operator
defined through the kernel $p_{B^{n}}$, that is (9) with $\kappa=\sqrt{\pi}$.
By Lemma 3.1 below, $\|S(t)\|\leq C$, $t\geq 0$, in $L^{2}_{c}$. Given $f\in
C_{c}^{\infty}((0,\infty))$, let $u(t,y)=S(t)f(y)$. By the construction of the
kernel $p$ we have $u_{t}=Bu$ pointwise. Finally, for $\lambda>0$,

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}u(t,y)\,dt$
$\displaystyle=\int_{0}^{\infty}e^{-\lambda
t}\,dt\int_{0}^{\infty}p_{B^{n}}(t,y,\rho)f(\rho)\,d\rho=\int_{0}^{\infty}f(\rho)\,d\rho\int_{0}^{\infty}e^{-\lambda
t}p_{B^{n}}(t,y,\rho)\,dt$
$\displaystyle=\int_{0}^{\infty}G^{n}(\lambda,y,\rho)\rho^{c}f(\rho)\,d\rho.$

It follows that the Laplace transform of $S(t)f$ coincides with the resolvent
of $B^{n}$, hence, by uniqueness, $S(t)$ is the generated semigroup and
$p_{B^{n}}(t,\cdot,\cdot)$ its kernel.

For complex times we argue in a similar way; we fix
$0\leq|\theta|<\frac{\pi}{2}$ and $\omega=e^{i\theta}\in\mathbb{C}_{+}$. Then
for $t>0$, $p(t\omega,\cdot,\cdot)$ is the heat kernel of the scaled semigroup
$T_{\omega}(t)=e^{t\omega B^{n}}$ whose generator is $A_{\omega}=\omega
B^{n}$. Its resolvent is then given, for $\lambda>0$, by $(\lambda-
A_{\omega})^{-1}=\omega^{-1}\left(\omega^{-1}\lambda-B^{n}\right)^{-1}$ and
its integral kernel is $\omega^{-1}G^{n}(\omega^{-1}\lambda,y\rho)$. The same
argument above applied to $T_{\omega}(t)$ proves then the assertion for
$z=t\omega$. The proof for $B^{d}$ is similar.

The following result is proved in [9].

###### Proposition 2.7

If $c>-1$, then $e^{zB^{n}}1=1$.

Proof. The proof follows using the explicit expression of $p_{B^{n}}$ of
Theorem 2.6 and the identity

$\displaystyle\int_{0}^{\infty}e^{-\alpha\rho^{2}}\rho^{1+\nu}I_{\nu}(z\rho)\,d\rho=\frac{z^{\nu}}{(2\alpha)^{\nu+1}}\,e^{\frac{z^{2}}{4\alpha}},\qquad\nu>-1$

which holds for every $z,\alpha\in\mathbb{C}_{+}$. See [1, Formula 11.4.29,
page 486] where the latter equality is expressed in terms of the Bessel
functions $J_{\nu}$ which satisfies $I_{\nu}(z)=e^{-\frac{1}{2}\nu\pi
i}J_{\nu}\left(e^{\frac{1}{2}\pi i}z\right)$.

###  2.3 Heat kernel bounds

The asymptotic behaviour of Bessel functions allows to deduce explicit bounds
for the heat kernels $p_{B^{n}}$ and $p_{B^{d}}$.

###### Proposition 2.8

Let $p_{B^{n}},p_{B^{d}}$ be the kernels defined in Theorem 2.6. Then for
every $\varepsilon>0$, there exist $C_{\varepsilon},\kappa_{\varepsilon}>0$
such that for $z\in\Sigma_{\frac{\pi}{2}-\varepsilon}$

$\displaystyle|p_{B^{d}}(z,y,\rho)|$ $\displaystyle\leq
C_{\varepsilon}|z|^{-\frac{1}{2}}\left(\frac{y}{|z|^{\frac{1}{2}}}\wedge
1\right)^{1-c}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right),\quad
c<1$

and

$\displaystyle|p_{B^{n}}(z,y,\rho)|$ $\displaystyle\leq
C_{\varepsilon}|z|^{-\frac{1}{2}}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{c}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right),\quad
c>-1$

Proof. Using (4) we get

$\displaystyle|p_{B^{d}}(z,y,\rho)|$ $\displaystyle\leq
C_{\varepsilon}\,|z|^{-1}\rho^{c}\left(y\rho\right)^{\frac{1-c}{2}}\exp\left\\{-\frac{Rez}{4|z|^{2}}(y^{2}+\rho^{2})\right\\}\left(\frac{y\rho}{2|z|}\wedge
1\right)^{\frac{1-c}{2}+\frac{1}{2}}\left(\frac{2|z|}{y\rho}\right)^{\frac{1}{2}}\exp\left\\{\frac{Rez}{2|z|^{2}}y\rho\right\\}.$
$\displaystyle\leq
C_{\varepsilon}\,|z|^{-1}\rho^{c}\left(y\rho\right)^{\frac{1-c}{2}}\left(\frac{y\rho}{|z|}\wedge
1\right)^{1-\frac{c}{2}}\left(\frac{2|z|}{y\rho}\right)^{\frac{1}{2}}\exp\left(-\frac{|y-\rho|^{2}}{\kappa^{\prime}_{\varepsilon}|z|}\right)$
$\displaystyle=C_{\varepsilon}|z|^{-\frac{1}{2}}\left(\frac{y}{\rho}\right)^{-\frac{c}{2}}\left(\frac{y\rho}{|z|}\wedge
1\right)^{1-\frac{c}{2}}\exp\left(-\frac{|y-\rho|^{2}}{\kappa^{\prime}_{\varepsilon}|z|}\right)$
$\displaystyle\leq
C^{\prime}_{\varepsilon}|z|^{-\frac{1}{2}}\left(\frac{y}{|z|^{-\frac{1}{2}}}\wedge
1\right)^{1-c}\left(\frac{\rho}{|z|^{-\frac{1}{2}}}\wedge
1\right)\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right)$

by Lemmas 10.1, 10.2. The proof for $p_{B^{n}}$ is similar.

Note that the constant $\kappa_{\varepsilon}$ above is explicit. For example,
for $z\geq 0$, that is $\varepsilon=\pi/2$, then
$\kappa^{\prime}_{\varepsilon}=4$ in the above proof and we can take
$\kappa_{\varepsilon}=4+\delta$ for every $\delta>0$.

Next we prove bounds for the gradients of the heat kernels.

###### Proposition 2.9

For every $\varepsilon>0$, there exist
$C_{\varepsilon},\kappa_{\varepsilon}>0$ such that for
$z\in\Sigma_{\frac{\pi}{2}-\varepsilon}$

$\displaystyle|D_{y}p_{B^{d}}(z,y,\rho)|\leq
C_{\varepsilon}|z|^{-1}\left(\frac{y}{|z|^{\frac{1}{2}}}\wedge
1\right)^{-c}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right),\quad
c<1;$

and

$\displaystyle|D_{y}p_{B^{n}}(z,y,\rho)|$ $\displaystyle\leq
C_{\varepsilon}|z|^{-1}\left(\frac{y}{|z|^{\frac{1}{2}}}\wedge
1\right)\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{c}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right),\quad
c>-1.$

Proof. We give a proof first for $B^{d}$. Differentiating $p_{B^{d}}$ with
respect to $y$ we obtain

$\displaystyle
D_{y}p_{B^{d}}(z,y,\rho)=\Bigg{[}\frac{1-c}{2y}-\frac{y}{2z}+\frac{\rho}{2z}\frac{I^{\prime}_{\nu}\left(\frac{y\rho}{2z}\right)}{I_{\nu}\left(\frac{y\rho}{2z}\right)}\Bigg{]}p_{B^{d}}(z,y,\rho),$

where $\nu=(1-c)/2$. We recall now that, see Lemma 2.3,

$\displaystyle I_{\nu}^{\prime}(s)=I_{\nu+1}(s)+\frac{\nu}{s}I_{\nu}(s).$

This implies

$\displaystyle D_{y}p_{B^{d}}(z,y,\rho)$
$\displaystyle=\Bigg{[}\frac{1-c}{y}-\frac{y}{2z}+\frac{\rho}{2z}\frac{I_{\nu+1}\left(\frac{y\rho}{2z}\right)}{I_{\nu}\left(\frac{y\rho}{2z}\right)}\Bigg{]}p_{B^{d}}(z,y,\rho)$
$\displaystyle=\frac{1}{\sqrt{z}}\left[(1-c)\frac{\sqrt{z}}{y}+\frac{y}{2\sqrt{z}}\left(\frac{I_{\nu+1}\left(\frac{y\rho}{2z}\right)}{I_{\nu}\left(\frac{y\rho}{2z}\right)}-1\right)-\frac{y-\rho}{2\sqrt{z}}\frac{I_{\nu+1}\left(\frac{y\rho}{2z}\right)}{I_{\nu}\left(\frac{y\rho}{2z}\right)}\right]p_{B^{d}}(z,y,\rho)$

The boundedness of $\frac{I_{\nu+1}}{I_{\nu}}$ gives

$\left|\frac{y-\rho}{2\sqrt{z}}\frac{I_{\nu+1}\left(\frac{y\rho}{2z}\right)}{I_{\nu}\left(\frac{y\rho}{2z}\right)}\right|\leq
C\left|\frac{y-\rho}{2\sqrt{z}}\right|\leq
Ce^{\varepsilon\frac{|y-\rho|^{2}}{|z|}}$

Next we use the estimate $\left|\frac{I_{\nu+1}(w)}{I_{\nu}(w)}-1\right|\leq
C(1\wedge|w|^{-1})$ for $w\in\Sigma_{\frac{\pi}{2}-\varepsilon}$ to bound

$K(\xi,\eta)=\xi\left(\frac{I_{\nu+1}\left(\frac{\xi\eta}{2}\right)}{I_{\nu}\left(\frac{\xi\eta}{2}\right)}-1\right),$

where $\xi=\frac{y}{\sqrt{z}}$ and $\eta=\frac{\rho}{\sqrt{z}}$.

Clearly $|K(\xi,\eta)|\leq C$ if $|\xi|\leq 2$ and $|K(\xi,\eta)|\leq
Ce^{\varepsilon|\xi-\eta|^{2}}$ if $|\xi|\geq 2$ and $|\eta|\leq 1$. Finally,
if $|\xi|\geq 2$, $|\eta|\geq 1$, then $|K(\xi,\eta)|\leq
C\frac{|\xi|}{|\xi\eta|}\leq\frac{C}{|\eta|}\leq C$. Then

$|D_{y}p_{B^{d}}(z,y\rho)|\leq
C\frac{1}{\sqrt{|z|}}\left(1+\frac{|(1-c)\sqrt{z}|}{y}\right)e^{\varepsilon\frac{|y-\rho|^{2}}{|z|}}|p_{B^{d}}(z,y,\rho)|$
(11)

and the thesis follows from Proposition 2.8.

Concerning $B^{n}$ we first note that $\nu=(c-1)/2$ in the above notation.
Then we get

$D_{y}p_{B^{n}}(z,y,\rho)=\Bigg{[}-\frac{y}{2z}+\frac{\rho}{2z}\frac{I_{\nu+1}\left(\frac{y\rho}{2z}\right)}{I_{\nu}\left(\frac{y\rho}{2z}\right)}\Bigg{]}p_{B^{n}}(z,y,\rho)=\frac{y}{2z}\Bigg{[}-1+\frac{\rho}{y}\frac{I_{\nu+1}\left(\frac{y\rho}{2z}\right)}{I_{\nu}\left(\frac{y\rho}{2z}\right)}\Bigg{]}p_{B^{n}}(z,y,\rho).$

Proceeding as before we get (11) without the term $(1-c)\sqrt{z}/y$ and we
only look for a better estimate in the region $\frac{y}{\sqrt{|z|}}<1$.
Setting $\xi=\frac{y}{\sqrt{z}}$ and $\eta=\frac{\rho}{\sqrt{z}}$ as before,
we prove a bound for

$K_{0}(\xi,\eta)=-1+\frac{\eta}{\xi}\frac{I_{\nu+1}\left(\frac{\eta\xi}{2}\right)}{I_{\nu}\left(\frac{\eta\xi}{2}\right)}$

in the case $|\xi|<1$. Using the estimate
$\frac{|I_{\nu+1}(w)|}{|I_{\nu}(w)|}\leq C(1\wedge|w|)$, we get
$|K_{0}(\xi,\eta)|\leq
1+C\left|\frac{\eta}{\xi}\right|\left(1\wedge|\eta\xi|\right)$. Assume first
$|\xi\eta|\leq 1$. Then $|K_{0}(\xi,\eta)|\leq C\left(1+|\eta|^{2}\right)\leq
C$ if $|\eta|\leq 1$ and $|K_{0}(\xi,\eta)|\leq
Ce^{\varepsilon|\xi-\eta|^{2}}$ if $|\eta|>1$. Let now $|\xi\eta|>1$. Then
$\left|\frac{\eta}{\xi}\right|\leq|\eta|^{2}$ and $|K_{0}(\xi,\eta)|\leq
C\left(1+|\eta|^{2}\right)\leq Ce^{\varepsilon|\xi-\eta|^{2}}$. It follows
that, when $\frac{y}{\sqrt{|z|}}<1$,

$|D_{y}p_{B^{n}}(z,y\rho)|\leq\frac{C}{\sqrt{|z|}}\frac{y}{\sqrt{|z|}}e^{\varepsilon|\xi-\eta|^{2}}|p_{B^{n}}(z,y,\rho)|$

and the thesis follows from Proposition 2.8.

##  3 The semigroups $e^{zB^{d}}$ and $e^{zB^{n}}$

In this section we show that the operators defined through the kernels
$p_{B^{d}}$ and $p_{B^{n}}$ are strongly continuous semigroups in
$L^{p}_{m}=L^{p}(\mathbb{R}_{+};y^{m}dy)$. We define
$\\{e^{zB}\\}_{z\in\mathbb{C}_{+}}$, $\\{D_{y}e^{zB}\\}_{z\in\mathbb{C}_{+}}$
for $f\in C_{c}^{\infty}(0,\infty)$ by

$[e^{zB}f](y):=\int_{0}^{\infty}p(z,y,\rho)f(\rho)\,d\rho,\quad[D_{y}e^{zB}f](y):=\int_{0}^{\infty}D_{y}p(z,y,\rho)f(\rho)\,d\rho$

where $p=p_{B^{d}}$ or $p=p_{B^{n}}$ and, accordingly, we write $e^{zB^{d}}$
and $e^{zB^{n}}$.

The following lemma is consequence of the heat kernel estimates of
Propositions 2.8, 2.9 and Proposition 12.2.

###### Lemma 3.1

Let $\theta\geq 0$, $\delta=\pi/2-\varepsilon$, $\varepsilon>0$. The following
properties hold for $z\in\Sigma_{\delta}$.

  * (i)

If $c<1$ and $c-1+\theta<\frac{m+1}{p}<2$, then $\|e^{zB^{d}}\|_{L^{p}_{m}\to
L^{p}_{m-p{\theta}}}\leq C|z|^{-\frac{\theta}{2}}.$

  * (ii)

If $c<1$ and $c+\theta<\frac{m+1}{p}<2$, then
$\|\sqrt{z}D_{y}e^{zB^{d}}\|_{L^{p}_{m}\to L^{p}_{m-p{\theta}}}\leq
C|z|^{-\frac{\theta}{2}}.$

  * (iii)

If $c>-1$ and $\theta<\frac{m+1}{p}<c+1$, then $\|e^{zB^{n}}\|_{L^{p}_{m}\to
L^{p}_{m-p{\theta}}}\leq C|z|^{-\frac{\theta}{2}}.$

  * (iv)

If $c>-1$ and $\theta-1<\frac{m+1}{p}<c+1$, then
$\|\sqrt{z}D_{y}e^{zB^{n}}\|_{L^{p}_{m}\to L^{p}_{m-p{\theta}}}\leq
C|z|^{-\frac{\theta}{2}}.$

###### Proposition 3.2

  * (i)

If $c<1$ and $c-1<\frac{m+1}{p}<2$, then $\\{e^{zB^{d}}\\}$ is a bounded
analytic semigroup of angle $\pi/2$ in $L^{p}_{m}$.

  * (ii)

If $c>-1$ and $0<\frac{m+1}{p}<c+1$, then $\\{e^{zB^{n}}\\}$ is a bounded
analytic semigroup of angle $\pi/2$ in $L^{p}_{m}$.

Proof.  The boundedness of the families
$\\{e^{zB^{d}}\\}_{z\in\Sigma_{\delta}}$,
$\\{e^{zB^{n}}\\}_{z\in\Sigma_{\delta}}$ follows from the previous lemma with
$\theta=0$; the semigroup law is inherited from the one of $L^{2}_{c}$ via a
density argument and we have only to prove the strong continuity at $0$. Let
$f,g\in C_{c}^{\infty}(0,\infty)$. Then as $z\to 0$, $z\in\Sigma_{\delta}$,

$\int_{0}^{\infty}(e^{zB}f)\,g\,y^{m}dy=\int_{0}^{\infty}(e^{zB}f)\,g\,y^{m-c}y^{c}dy\to\int_{0}^{\infty}fgy^{m-c}y^{c}dy=\int_{0}^{\infty}fgy^{m}dy,$

by the strong continuity of $e^{zB}$ in $L^{2}_{c}$. By density and uniform
boundedness of the family $(e^{zB})_{z\in\Sigma_{\delta}}$ this holds for
every $f\in L^{p}_{m}$, $g\in L^{p^{\prime}}_{m}$. The semigroup is then
weakly continuous, hence strongly continuous.

We denote by $B_{m,p}^{d},B_{m,p}^{n}$ the generators of
$e^{zB^{d}},e^{zB^{n}}$ in $L^{p}_{m}$ and characterize their domain. Observe
that, since the heat kernels of these semigroups are given by Theorem 2.6,
their resolvents are those of Propositions 2.5, 2.4.
We recall that the Sobolev spaces $W^{k,p}_{m}$ are studied in detail in
Appendix B and that traces at the boundary $y=0$ are well-defined when
$(m+1)/p<1$.

It is useful to define $D(B_{m,p,max})=\\{u\in L^{p}_{m}\cap
W^{2,p}_{loc}(\mathbb{R}_{+}):Bu\in L^{p}_{m}\\}$. We start with $B^{n}$.

###### Proposition 3.3

If $c>-1$ and $0<\frac{m+1}{p}<c+1$, then

$D(B_{m,p}^{n})=\\{u\in D(B_{m,p,max}):\;\frac{D_{y}u}{y},\;D_{yy}u\in
L^{p}_{m}\\}.$

Moreover, $D(B_{m,p}^{n})=\\{u\in W^{2,p}_{m}:D_{y}u(0)=0\\}$ when
$\frac{m+1}{p}<1$ and $D(B_{m,p}^{n})=W^{2,p}_{m}$ when $\frac{m+1}{p}>1$.

Proof.  Let $D$ be the right-hand side above, $u\in D(B^{d}_{m,p})$ and let
$f:=u-B^{n}_{m,p}u$. Then
$u=(I-B^{n}_{m,p})^{-1}f=\int_{0}^{\infty}e^{-t}e^{tB^{n}}f\,dt$ and
$D_{y}u=\int_{0}^{\infty}e^{-t}D_{y}e^{tB^{d}}f\,dt$. Using Minkowski’s
inequality and Lemma 3.1 (iv) with $\theta-1<0<\frac{m+1}{p}<c+1$, we get

$\displaystyle\|y^{-\theta}D_{y}u\|_{L^{p}_{m}}=\|D_{y}u\|_{L^{p}_{m-\theta
p}}\leq\int_{0}^{\infty}e^{-t}\frac{1}{\sqrt{t}}\|\sqrt{t}D_{y}e^{tB^{d}}f\|_{L^{p}_{m-\theta
p}}\,dt\leq
C\|f\|_{L^{p}_{m}}\int_{0}^{\infty}e^{-t}t^{-\frac{\theta+1}{2}}\,dt\,$

and then $y^{-\theta}D_{y}u\in L^{p}_{m}$ for every $0\leq\theta<1$. To reach
$\theta=1$ let $v=D_{y}u$ and $g=D_{y}v+c\frac{v}{y}=Bu$. Then

$v(y)=y^{-c}\int_{0}^{y}g(s)s^{c}\,ds+Ky^{-c}:=w(y)+Ky^{-c}$ (12)

(note that the integral converges, by Hölder inequality, since
$(c+1)>\frac{m+1}{p}$). By Hardy inequality, see Lemma 10.3,
$\|\frac{w}{y}\|_{L^{p}_{m}}\leq C\|g\|_{L^{p}_{m}}$ but $y^{-c-\theta}\not\in
L^{p}_{m}(0,1)$ for $\theta<1$, sufficiently close to $1$. It follows that
$K=0$, $v=w$ and then $D_{y}u/y\in L^{p}_{m}$ and, by difference, $D_{yy}u\in
L^{p}_{m}$, too.

This shows that $D(B^{n}_{m,p})\subset D$ and we have only to show that $I-B$
is injective on $D$. Assume that $u\in D$ and that $u-Bu=0$, then
$u(y)=c_{1}y^{\frac{1-c}{2}}I_{\frac{c-1}{2}}+c_{2}y^{\frac{1-c}{2}}K_{\frac{|1-c|}{2}}$.
However $c_{1}=0$, since $I_{\frac{c-1}{2}}$ is exponentially increasing at
$\infty$. Concerning $u_{2}(y)=y^{\frac{1-c}{2}}K_{\frac{|1-c|}{2}}$ we note
that its derivative, as $y\to 0$, behaves like $y^{-1}$ when $c\leq 1$ and
like $y^{-c}$ when $c>1$. In both cases $(D_{y}u_{2})/y\not\in L^{p}_{m}$.
Then $c_{2}=0$ and $u=0$.

The last part follows from Proposition 11.4 applied to $D_{y}u$, taking into
account that $W^{1,p}_{m}=W^{1,p}_{0,m}$ when $(m+1)/p>1$. In the case
$(m+1)/p<1$ observe that $D_{y}u$ has a finite limit as $y\to 0$, by Lemma
11.1, which must be equal to $0$, otherwise $(D_{y}u)/y\not\in L^{p}_{m}$.

###### Corollary 3.4

If $c>-1$, $0<\frac{m+1}{p}<c+1$ and $u\in D(B^{n}_{m,p})$, then $\lim_{y\to
0}y^{\frac{m+1}{p}-1}D_{y}u=0$, hence $\lim_{y\to 0}y^{c}D_{y}u=0$. Moreover

  * (i)

if $\frac{m+1}{p}<2$, then $D_{y}u\in L^{1}(0,1)$ and therefore $\lim_{y\to
0}u(y)$ exists finite;

  * (ii)

if $\frac{m+1}{p}=2$, then $\frac{u}{y^{2\theta}}\in L^{p}_{m}$ if
$2\theta<2$;

  * (iii)

if $\frac{m+1}{p}>2$ then $\frac{u}{y^{2}}\in L^{p}_{m}$.

Proof.  By (12) with $K=0$ and Hölder inequality we get

$|D_{y}u|\leq y^{-c}\int_{0}^{y}|g(s)|s^{\frac{m}{p}}s^{c-\frac{m}{p}}\,ds\leq
y^{1-\frac{m+1}{p}}\left(\int_{0}^{y}|g(s)|^{p}s^{m}\,ds\right)^{\frac{1}{p}}$

and the first statement follows and yields $D_{y}u\in L^{1}(0,1)$ when
$(m+1)/p<2$. This proves $(i)$. The proof of $(ii)$ is similar since
$D_{y}u=o(y^{-1})$, hence $u=o(\log y)$ as $y\to 0$.

Assume now that $(iii)$ holds and write, by (12) again,

$D_{y}u=y^{-c}\int_{0}^{y}g(s)s^{c}\,ds=y\int_{0}^{1}g(ty)t^{c}\,dt.$

Then

$\frac{|u(y)-u(1)|}{y^{2}}\leq\frac{1}{y^{2}}\int_{y}^{1}s\,ds\int_{0}^{1}|g(ts)|t^{c}\,dt\leq\int_{1}^{\infty}\eta\,d\eta\int_{0}^{1}|g(t\eta
y)|t^{c}\,dt$

and Minkowski’s inequality gives

$\left\|\frac{|u(y)-u(1)|}{y^{2}}\right\|_{L^{p}_{m}}\leq\int_{1}^{\infty}\eta\,d\eta\int_{0}^{1}t^{c}\|g(t\eta\cdot)\|_{L^{p}_{m}}\,dt=\|g\|_{L^{p}_{m}}\int_{1}^{\infty}\eta^{1-\frac{m+1}{p}}d\eta\int_{0}^{1}t^{c-\frac{m+1}{p}}dt.$

Since also $y^{-2}u(1)\in L^{p}_{m}(0,1)$, the proof of $(iii)$ is complete.

Next we consider $B^{d}$.

###### Proposition 3.5

Let $c<1$ and $c-1<\frac{m+1}{p}<2$. Then

$\displaystyle D(B_{m,p}^{d})=$ $\displaystyle\\{u\in
D(B_{m,p,max}):\;y^{-2\theta}u\in L^{p}_{m}{\rm\ for\ every\ }0\leq\theta\leq
1$ $\displaystyle\ {\rm such\ that\ }c-1+2\theta<\frac{m+1}{p}<2\\}.$

Moreover

  * (i)

$(1\wedge u)^{2-2\theta}D_{yy}u,(1\wedge u)^{1-2\theta}D_{y}u\in L^{p}_{m}$
for every $u\in D(B_{m,p}^{d})$ and $\theta$ as above;

  * (ii)

$D_{y}u\in L^{1}(0,1)$ and $\lim_{y\to 0}u(y)=0$ for every $u\in
D(B_{m,p}^{d})$.

Proof. Let $u\in D(B^{d}_{m,p})$ and let $f:=u-B^{d}_{m,p}u$. Then
$u=(I-B^{d}_{m,p})^{-1}f=\int_{0}^{\infty}e^{-t}e^{tB^{d}}f\,dt$. Then using
Minkowski’s inequality and Lemma 3.1 we get when $c-1+2\theta<\frac{m+1}{p}<2$
and $0\leq\theta<1$,

$\displaystyle\|y^{-2\theta}u\|_{L^{p}_{m}}=\|u\|_{L^{p}_{m-2\theta
p}}\leq\int_{0}^{\infty}e^{-t}\|e^{tB^{d}}f\|_{L^{p}_{m-2\theta p}}\,dt\leq
C\|f\|_{L^{p}_{m}}\int_{0}^{\infty}e^{-t}t^{-\theta}\,dt\,$

which yields $D(B_{m,p}^{d})\subset D$ where $D=\\{u\in
D(B_{m,p,max}):\;y^{-2\theta}u\in L^{p}_{m}\\}$ for every $0\leq\theta<1$ such
that $c-1+2\theta<\frac{m+1}{p}<2$. The equality $D(B_{m,p}^{d})=D$ follows
from the injectivity of $I-B$ on $D$, as in Proposition 3.3, since the
function $u_{2}(y)=y^{\frac{1-c}{2}}K_{\frac{1-c}{2}}\approx c\ \neq 0$ does
not belong to $D$ (choosing $2\theta$ sufficiently close to
$\frac{m+1}{p}-(c-1)$ or to $2$).

To reach the case when $\theta=1$ and to add the integrability of
$D_{y}u,D_{yy}u$ we argue as in the proposition above. If $g=Bu$ then

$D_{y}u(y)=y^{-c}\int_{1}^{y}g(s)s^{c}\,ds+Ky^{-c}:=w(y)+Ky^{-c}.$

Hölder inequality gives $|w(y)|\leq
C\|g\|_{L^{p}_{m}}(y^{-c}+y^{1-\frac{m+1}{p}})$ and then the assumption $c<1$
and $(m+1)/p<2$ show that $D_{y}u\in L^{1}(0,1)$ (with respect to the Lebesgue
measure). It follows that $\lim_{y\to 0}u(y)$ exists finite and must be 0, by
the same argument for $u_{2}$. Then $|u(y)|\leq
C\|g\|_{L^{p}_{m}}(y^{1-c}+y^{2-\frac{m+1}{p}})$. At this point the estimate
for $(1\wedge y)^{-2\theta+1}D_{y}u$ is elementary when $\theta<1$ (and that
for $D_{yy}u$ follows from the equation, multiplying by $y^{2-2\theta}$). If
$\theta=1$, that is when $c+1<(m+1)/p<2$, then $Ky^{-c-1}\in L^{p}_{m}(0,1)$
and $\frac{w}{y}\in L^{p}_{m}$ by Hardy inequality, see Lemma 10.3. The
integrability of $\frac{u}{y^{2}}$ is proved as in Corollary 3.4 (iii). Using
$u(0)=0$ we obtain

$\frac{|u(y)|}{y^{2}}\leq\int_{0}^{1}\eta\,d\eta\int_{1}^{\infty}|g(t\eta
y)|t^{c}\,dt+Ky^{-c-1}.$

Since $y^{-c-1}\in L^{p}_{m}(0,1)$ we may assume that $K=0$ and Minkowski
inequality gives

$\left\|\frac{u}{y^{2}}\right\|_{L^{p}_{m}}\leq\int_{0}^{1}\eta\,d\eta\int_{1}^{\infty}t^{c}\|g(t\eta\cdot)\|_{L^{p}_{m}}\,dt=\|g\|_{L^{p}_{m}}\int_{0}^{1}\eta^{1-\frac{m+1}{p}}d\eta\int_{1}^{\infty}t^{c-\frac{m+1}{p}}dt.$

Note that when $c<(m+1)/p<2$, $\theta=1/2$ is allowed and
$D(B^{d}_{m,p})\subset W^{1,p}_{m}$. The embeddings above do not hold outside
the indicated ranges: just take $u(y)=y^{1-c}$ near $0$.

For the next corollary we recall that $\lim_{y\to 0}u(y)$ and $\lim_{y\to
0}D_{y}u(y)$ exists finite if $u\in W^{2,p}_{m}$ when $(m+1)/p<1$, see Lemma
11.1, and that both are equal to $0$, when $m\leq-1$, see Lemma 11.2.

When $1\leq(m+1)/p<2$, Hölder inequality gives $|D_{y}u|\leq
C\|D_{yy}u\|_{L^{p}_{m}}(1+y^{1-\frac{m+1}{p}})$ if $1<(m+1)/p<2$, or
$|D_{y}u|\leq C\|D_{yy}u\|_{L^{p}_{m}}(1+|\log y|)$ if $m=p-1$. In both cases
$D_{y}u\in L^{1}(0,1)$ and $u(0)$ exists finite.

###### Corollary 3.6

Assume that $c<1$ and $c+1<\frac{m+1}{p}<2$. Then $D(B_{m,p}^{d})\subset
W^{2,p}_{m}$ and

  * (i)

If $m\leq-1$, then $D(B_{m,p}^{d})=W^{2,p}_{m};$

  * (ii)

If $0<\frac{m+1}{p}<1$ then $D(B_{m,p}^{d})=\\{u\in
W^{2,p}_{m}:u(0)=D_{y}u(0)=0\\}$

  * (iii)

If $1\leq\frac{m+1}{p}<2$, then $D(B_{m,p}^{d})=\\{u\in
W^{2,p}_{m}:u(0)=0\\}$.

Proof.  By Proposition 3.5 the inclusion $D(B_{m,p}^{d})\subset\\{u\in
W^{2,p}_{m}:u(0)=0\\}$ follows. When $(m+1)/p<1$, then also $D_{y}u(0)=0$
otherwise $D_{y}u/y\not\in L^{P}_{m}$. This shows that in all cases
$D(B_{m,p}^{d})$ is contained in the right hand sides. To show the equality it
suffices, therefore, to note that the function
$u_{2}(y)=y^{\frac{1-c}{2}}K_{\frac{1-c}{2}}\approx c\ \neq 0$ (see the proof
of Proposition 3.5) does not belong to the right hand sides (in case (iii)
observe also that $D_{y}u\in L^{1}(0,1)$ so that $u(0)$ exists finite).

##  4 Degenerate operators in weighted spaces

In this section we add a potential term to $B$ and study the whole operator

$L=D_{yy}+\frac{c}{y}D_{y}-\frac{b}{y^{2}}$

on the (open) half line $\mathbb{R}_{+}=]0,\infty[$. However, we shall
consider $L$ only with Dirichlet boundary conditions at $0$, hence $L=B^{d}$,
when $b=0$, with the understanding that $B^{n}=B^{d}$ when $c\geq 1$.

If $1<p<\infty$, we define the maximal operator $L_{p,max}$ through the domain

$D(L_{m,p,max})=\\{u\in L_{m}^{p}\cap W^{2,p}_{loc}(\mathbb{R}_{+}):Lu\in
L_{m}^{p}\\}.$ (13)

The equation $Lu=0$ has solutions $y^{-s_{1}}$, $y^{-s_{2}}$ where
$s_{1},s_{2}$ are the roots of the indicial equation $f(s)=-s^{2}+(c-1)s+b=0$
given by

$s_{1}:=\frac{c-1}{2}-\sqrt{D},\quad s_{2}:=\frac{c-1}{2}+\sqrt{D}$ (14)

where

$D:=b+\left(\frac{c-1}{2}\right)^{2}.$ (15)

The above numbers are real if and only if $D\geq 0$. When $D<0$ the equation
$u-Lu=f$ cannot have positive distributional solutions for certain positive
$f$, see [22].When $b=0$, then $\sqrt{D}=|c-1|/2$ and $s_{1}=0,s_{2}=c-1$ for
$c\geq 1$ and $s_{1}=c-1,s_{2}=0$ for $c<1$.

A multiplication operator transforms $L$ into a Bessel operator and allows to
transfer the results of the previous section to this more general situation.

###### Lemma 4.1

For $k\in\mathbb{R}$, let $(T_{k}u)(y):=y^{k}u(y),y>0.$ Then $T_{k}$ maps
isometrically $L^{p}_{m+kp}$ onto $L_{m}^{p}$ and for every $u\in
W^{2,1}_{loc}\left(\mathbb{R}_{+}\right)$ one has

$T_{-k}LT_{k}u=\tilde{L}u:=D_{yy}u+\frac{\tilde{c}}{y}D_{y}u-\frac{\tilde{b}}{y^{2}}u$

$\tilde{b}=b-k\left(c+k-1\right),\qquad\tilde{c}=c+2k.$ (16)

Moreover the discriminant $\tilde{D}$ and the parameter $\tilde{\gamma}$,
$\tilde{s}_{1,2}$ of $\tilde{L}$ defined in (15), (14) are given by

$\displaystyle\tilde{D}=D,\quad\tilde{s}_{1,2}=s_{1,2}+k,\quad{\tilde{s}}^{\ast}_{1,2}=s^{\ast}_{1,2}-k.$
(17)

Observe now that, choosing $k=-s_{i}$, $i=1,2$, we get $\tilde{b}=0$,
$\tilde{c}_{i}=c-2s_{i}$. The operators

$T_{s_{i}}LT_{-s_{i}}=B_{i}=D_{yy}+\frac{c-2s_{i}}{y}D_{y}$

are therefore Bessel operators. The following two results can be found also in
[16] and [22] when $m=0$ or $m=c$.

###### Proposition 4.2

Assume that $D>0$. If $s_{1}<\frac{m+1}{p}<s_{2}+2$ then
$L_{m,p}=(L,D(L_{m,p}))$ where

$\displaystyle D(L_{m,p})$ $\displaystyle=\bigg{\\{}u\in
D(L_{m,p,max})\;:\;y^{-2\theta}u\in L_{m}^{p}\quad$ $\displaystyle\textrm{for
every }\ \theta\in(0,1]\ \textrm{such that}\
s_{1}+2\theta<\frac{m+1}{p}<s_{2}+2.\bigg{\\}}$ (18)

generates a bounded positive analytic semigroup of angle $\pi/2$ on
$L_{m}^{p}$. Moreover

  * (i)

$(1\wedge y)^{2-2\theta}D_{yy}u,(1\wedge y)^{1-2\theta}D_{y}u\in L^{p}_{m}$
for every $u\in D(L_{m,p})$ and $\theta$ as above;

  * (ii)

$\lim_{y\to 0}y^{s_{2}}u(y)=0$ for every $u\in D(L_{m,p}^{d})$.

Proof. We use the identity
$T_{s_{2}}LT_{-s_{2}}=D_{yy}+\frac{c-2s_{2}}{y}D_{y}:=B^{d}$ and apply
Proposition 3.5 in $L^{p}_{m-s_{2}p}$. Note that $c-2s_{2}=1-2\sqrt{D}<1$ and
that $s_{1}+2\theta<\frac{m+1}{p}<s_{2}+2$ is equivalent to
$c-2s_{2}-1+2\theta<\frac{m-s_{2}p+1}{p}<2$. Since, by definition,
$D(L_{m,p})=T_{-s_{2}}D(B^{d}_{m-s_{2}p,p})$, (4.2) is immediate. The
verification of $(i)$ and $(ii)$ is similar. If $u=y^{-s_{2}}v\in D(L_{m,p})$,
then $\lim_{y\to 0}y^{s_{2}}u(y)=\lim_{y\to 0}v(y)=0$ and
$y^{1-2\theta}D_{y}u=y^{-s_{2}}(y^{1-2\theta}D_{y}v-s_{2}y^{-2\theta}v)\in
L^{p}_{m}$, by Proposition 3.5, again.

Let us now turn to the case $D=0$, where $s_{1}=s_{2}$.

###### Proposition 4.3

Assume that $D=0$. If $s_{1}<\frac{m+1}{p}<s_{1}+2$ then
$L_{m,p}=(L,D(L_{m,p}))$ where

$\displaystyle D(L_{m,p})$ $\displaystyle=\bigg{\\{}u\in
D(L_{m,p,max})\;:\;\exists\lim_{y\to 0}y^{s_{1}}u(y)\in\mathbb{C}\bigg{\\}}$
$\displaystyle=\left\\{u\in D(L_{m,p,max})\;:\;y^{-2\theta_{0}}|\log
y|^{-\frac{2}{p}}u\in L^{p}_{m}\Bigl{(}0,\frac{1}{2}\Bigr{)}\right\\}$

with $\theta_{0}=\frac{1}{2}(\frac{m+1}{p}-s_{1})\in(0,1)$ generates a bounded
positive analytic semigroup of angle $\pi/2$ on $L_{m}^{p}$. Moreover,

$(1\wedge y)^{2-2\theta}D_{yy}u,(1\wedge y)^{1-2\theta}D_{y}u\in L^{p}_{m}$

for every $u\in D(L_{m,p})$ and $s_{1}+2\theta<\frac{m+1}{p}<s_{1}+2$.

Proof. Let us write

$\displaystyle D_{1}\\!$ $\displaystyle=\\!\\{u\in D(L_{m,p,max}):\lim_{y\to
0}y^{s_{1}}u(y)\in\mathbb{C}\\},$ $\displaystyle D_{2}\\!$
$\displaystyle=\\!\\{u\in D(L_{m,p,max}):y^{-2\theta_{0}}|\log
y|^{-\frac{2}{p}}u\in L^{p}_{m}(0,\frac{1}{2})\\}$

and note that $D_{1}\subset D_{2}$, by the choice of $\theta_{0}$.

We use the identity
$T_{s_{1}}LT_{-s_{1}}=D_{yy}+\frac{c-2s_{1}}{y}D_{y}=D_{yy}+\frac{1}{y}D_{y}=B^{n}$
and apply Proposition 3.3 in $L^{p}_{m-s_{1}p}$ since
$c-2s_{1}=1-2\sqrt{D}=1$. Note that $s_{1}+2\theta<\frac{m+1}{p}<s_{1}+2$ is
equivalent to $2\theta<\frac{m-s_{1}p+1}{p}<2$. If $u=y^{-s_{1}}v\in
D(L_{m,p})=T_{-s_{1}}D(B^{n}_{m-s_{1}p,p})$, then $\lim_{y\to
0}y^{s_{1}}u(y)=\lim_{y\to 0}v(y)\in\mathbb{C}$, by Corollary 3.4 (i) (and
similarly
$y^{1-2\theta}D_{y}u=y^{-s_{1}}(y^{1-2\theta}D_{y}v-s_{1}y^{-2\theta}v)\in
L^{p}_{m}$). This shows that $D(L_{m,p})\subset D_{1}\subset D_{2}$ and the
equality follows since $I-L$ is injective on $D_{2}$. In fact, if $u-Lu=0$,
then $v=y^{s_{1}}u$ solves $v-Bv=0$, hence $v(y)=c_{1}I_{0}+c_{2}K_{0}$.
However, $c_{1}=0$, since $I_{0}$ grows exponentially at $\infty$ and
$c_{2}=0$ since $K_{0}\approx-\log y$, as $y\to 0$, hence does not satisfy the
integrability condition required by $D_{2}$ near $y=0$.

An alternative description of the domain is contained in the following
proposition, where we do not need to distinguish between $D>0$ and $D=0$.

###### Proposition 4.4

If $D\geq 0$ and $s_{1}<\frac{m+1}{p}<s_{2}+2$, then

$\displaystyle D(L_{m,p})$ $\displaystyle=\bigg{\\{}u\in
D(L_{m,p,max})\;:\;s_{1}\frac{u}{y^{2}}+\frac{D_{y}u}{y}\in
L^{p}_{m}\bigg{\\}}.$ (19)

Proof. As in the case $D=0$ we use the identity
$T_{s_{1}}LT_{-s_{1}}=D_{yy}+\frac{c-2s_{1}}{y}D_{y}=\tilde{B}^{n}$ (this last
is in $L^{p}_{m-s_{1}p}$), after observing that $c-2s_{1}=1+2\sqrt{D}\geq 1$.
Note that the conditions $s_{1}+2\theta<\frac{m+1}{p}<s_{2}+2$ and
$2\theta<\frac{m-s_{1}p+1}{p}<c-2s_{1}+1$ are equivalent.

This definition yields the same operator as in Proposition 4.2 if (and only
if) $T_{-s_{2}}D(B^{d}_{m-s_{2}p,p})=T_{-s_{1}}D(\tilde{B}^{n}_{m-s_{1}p,p})$
but this holds since $L$ endowed with both domains generates a semigroup and
the first contains the second.

Indeed, let $u\in T_{s_{2}-s_{1}}D(\tilde{B}^{n}_{m-s_{1}p,p})$, that is
$y^{s_{1}-s_{2}}u\in D(\tilde{B}^{n}_{m-s_{1}p,p})$. Then by construction
$\tilde{B}^{n}_{m-s_{1}p,p}(y^{s_{1}-s_{2}}u)\in L^{p}_{m-s_{1}p}$ is
equivalent to $B^{d}_{m-s_{2}p,p}u\in L^{p}_{m-s_{2}p}$. Analogously,
Corollary 3.4 applied to $y^{s_{1}-s_{2}}u$ yields $y^{-2\theta}u\in
L^{p}_{m-s_{2}p}$ for every $0\leq\theta\leq 1$ such that
$c-s_{2}-1+2\theta<\frac{m-s_{2}p+1}{p}<2$. By Proposition 4.2 this proves
that $u\in D(\tilde{B}^{n}_{m-s_{1}p,p})$ i.e.
$T_{s_{2}-s_{1}}D(\tilde{B}^{n}_{m-s_{1}p,p})\subseteq
D(\tilde{B}^{n}_{m-s_{1}p,p})$.

Applying now Proposition 3.3 and Corollary 3.4 to $v=y^{s_{1}}u$ we get, in
addition, that $u\in D(L_{m,p})$ if and only if $u\in D(L_{m,p,max})$ and

$(a)\quad s_{1}\frac{u}{y^{2}}+\frac{D_{y}u}{y}\in L^{p}_{m},\quad(b)\quad
s_{1}(s_{1}-1)\frac{u}{y^{2}}+2s_{1}\frac{D_{y}u}{y}+D_{yy}u\in L^{p}_{m}.$

However, $(b)$ follows from $(a)$ and $u\in D(L_{m,p,max})$ since

$\displaystyle
Lu-\left(s_{1}(s_{1}-1)\frac{u}{y^{2}}+2s_{1}\frac{D_{y}u}{y}+D_{yy}u\right)$
$\displaystyle=\left((2-c)s_{1}-2b\right)\frac{u}{y^{2}}+(1+2\sqrt{D})\frac{D_{y}u}{y}$
$\displaystyle=(1+2\sqrt{D})\left[s_{1}\frac{u}{y^{2}}+\frac{D_{y}u}{y}\right].$

We now deduce the estimates for the heat kernel and its derivative for the
operator $L$ from those for $B$. We shall consider only the case $s_{1}\neq
0$; in fact, if $s_{1}=0$, then $b=0$ and $c\geq 1$, hence $L=B^{n}=B^{d}$ and
the estimates are those of Propositions 2.8, 2.9.

###### Proposition 4.5

Let $1<p<\infty$ such that $0\neq s_{1}<\frac{m+1}{p}<s_{2}+2$. Then for
$z\in\mathbb{C}_{+}$

$\displaystyle
e^{zL}f(y)=\int_{\mathbb{R}^{+}}p_{L}(z,y,\rho)f(\rho)d\rho,\quad f\in
L_{m}^{p}$ (20)

where

$p_{L}(z,y,\rho)=\frac{1}{2z}y^{\frac{1-c}{2}}\rho^{\frac{1+c}{2}}\exp\left\\{-\frac{y^{2}+\rho^{2}}{4z}\right\\}I_{\sqrt{D}}\left(\frac{y\rho}{2z}\right).$

For every $\varepsilon>0$, there exist $C_{\varepsilon}>0$ and
$\kappa_{\varepsilon}>0$ such that for
$z\in\Sigma_{\frac{\pi}{2}-\varepsilon}$

$|p_{L}(z,y,\rho)|\leq
C_{\varepsilon}|z|^{-\frac{1}{2}}\left(\frac{y}{|z|^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}+c}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right)$

and

$y^{-1}|p_{L}(z,y,\rho)|+|D_{y}p_{L}(z,y,\rho)|\leq
C_{\varepsilon}|z|^{-1}\left(\frac{y}{|z|^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}-1}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}+c}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right).$

Proof. Let us consider the isometry $T_{-s_{1}}:L^{p}_{m}\to
L^{p}_{m-s_{1}p}$. Then for every $u\in L^{p}_{m}$

$e^{zL}u=T_{-s_{1}}\,e^{zB}\left(T_{s_{1}}u\right)$

where $B$ is the pure Bessel operator $D_{yy}+\frac{c-2s_{1}}{y}$ on
$L^{p}_{m-s_{1}p}$. Let $p_{L}(z,y,\rho)$ be the heat kernel of
$\left(e^{zL}\right)_{z\in\mathbb{C}_{+}}$ on $L_{m}^{p}$. The last relation
between the semigroups translate into the analogous equality for the heat
kernels:

$\displaystyle p_{L}(z,y,\rho)=y^{-s_{1}}p_{B^{n}}(z,y,\rho)\rho^{s_{1}}.$

From Proposition 2.8 and from Lemma 10.2 it follows that

$\displaystyle|p_{L}(z,y,\rho)|$
$\displaystyle=y^{-s_{1}}|p_{B}^{n}(z,y,\rho)|\rho^{s_{1}}\leq\frac{C_{\varepsilon}}{|z|^{\frac{1}{2}}}y^{-s_{1}}\rho^{s_{1}}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{c-2s_{1}}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right)$
$\displaystyle=\frac{C_{\varepsilon}}{|z|^{\frac{1}{2}}}\left(\frac{y}{|z|^{\frac{1}{2}}}\right)^{-s_{1}}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\right)^{s_{1}}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{c-2s_{1}}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right)$
$\displaystyle\leq\frac{C_{\varepsilon}}{|z|^{\frac{1}{2}}}\left(\frac{y}{|z|^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{c-s_{1}}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right).$

Concerning the derivative with respect to $y$, we get

$\displaystyle
D_{y}p_{L}(z,y,\rho)=-s_{1}y^{-s_{1}-1}p_{B^{n}}(z,y,\rho)\rho^{s_{1}}+y^{-s_{1}}D_{y}p_{B^{n}}(z,y,\rho)\rho^{s_{1}}.$

From Proposition 2.9, Proposition 2.8 and Lemma 10.2 it follows that

$\displaystyle D_{y}p_{L}(z,y,\rho)$
$\displaystyle=-s_{1}y^{-s_{1}-1}p_{B^{n}}(z,y,\rho)\rho^{s_{1}}+y^{-s_{1}}D_{y}p_{B^{n}}(z,y,\rho)\rho^{s_{1}}$
$\displaystyle\leq\frac{C_{\varepsilon}}{|z|}\left(\frac{y}{|z|^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}-1}\left(\frac{\rho}{|z|^{\frac{1}{2}}}\wedge
1\right)^{c-s_{1}}\exp\left(-\frac{|y-\rho|^{2}}{\kappa_{\varepsilon}|z|}\right).$

The estimate for $y^{-1}p_{L}$ follows easily from that of $p_{L}$.

##  5 Remarks on domain characterization and uniqueness

The domain characterizations for $B^{n},B^{d},L$ can be stated by adding
explicit estimates. For example, in Proposition 3.3, one can add

$\|D_{yy}u\|_{L^{p}_{m}}+\|y^{-1}D_{y}u\|_{L^{p}_{m}}\leq
C\|Bu\|_{L^{p}_{m}},\quad u\in D(B^{n}_{m,p})$

(the additional term $\|u\|_{L^{p}_{m}}$ does not appear, by scaling). This
follows from the proof (actually, this is the proof) but can be also deduced
from the statement by the closed graph theorem. This remark applies to all the
domain characterizations including those of the next sections; we decided not
to write down them explicitly for exposition reasons but we shall use in
Section 8.

As already pointed out, the assumption $D\geq 0$ is crucial for positivity and
always satisfied in the case of Bessel operators. When $D<0$ the equation
$u-Lu=f$ cannot have positive distributional solutions for certain positive
$f$, see [22]. However, $L$ can be the generator of a semigroup even when
$D<0$, see [18] for Schrödinger operators with inverse square potential
$\Delta-b|y|^{-2}$ with $b<-1/4$.

We note that the results for $B^{d}$ can be deduced from those for
$\tilde{B}^{n}$ (here we denote by $B,\tilde{B}$ two different but related
Bessel operators). This simple (but surprising) fact is actually the core of
the proof of Proposition 4.4 where the operators $L_{m,p}$ are transformed via
similarity to pure Bessel operators with $c\geq 1$. This approach, however,
needs a change in the reference measures and the knowledge of Bessel operators
in $L^{p}_{m}$ for every admissible $m$. We prefer to start with a form in
$L^{2}$ of the symmetrizing measure, as it is usually done. Moreover, using
both the direct approach and the transformation above, one gets different and
complementing descriptions of $D(L_{m,p})$, see Propositions 4.4, 4.2 and
subsequent corollaries, which are likely difficult to discover simultaneously
using only one method.

A natural question arises if different boundary conditions can be imposed to
produce different semigroups. This is the case, for example, for the Bessel
operators of Section 3, in the range $-1<c<1$ where $B^{n}\neq B^{d}$. To
state the uniqueness question more precisely we define $L_{m,p,min}$ as the
closure, in $L^{p}_{m}$ of $(L,C_{c}^{\infty}(\mathbb{R}_{+}))$ (the closure
exists since this operator is contained in the closed operator $L_{m,p,max}$)
and it is clear that $L_{m,p,min}\subset L_{m,p,max}$. We look at realizations
$L_{D}=(L,D)$, such that $L_{m,p,min}\subset L_{D}\subset L_{m,p,max}$. The
following results can be found in [17, Propositions 2.4, 2.5] and [22,
Propositions 3.12, 3.28, 3.30] in the $N$-dimensional case and for $m=0$. The
generalization to any $m$ is straightforward, through the transformation
$T_{k}$.

###### Proposition 5.1

  * (i)

If $\frac{m+1}{p}\not\in(s_{1},s_{2}+2)$, then no realization
$L_{m,p,min}\subset L_{D}\subset L_{m,p,max}$ generates a semigroup in
$L^{p}_{m}$;

  * (ii)

$L_{m,p,max}$ generates a semigroup if and only if $s_{1}<\frac{m+1}{p}\leq
s_{2}$;

  * (iii)

$L_{m,p,min}$ generates a semigroup if and only if
$s_{1}+2\leq\frac{m+1}{p}<s_{2}+2$.

In particular $L$ generates a unique semigroup in cases $(ii)$ and $(iii)$ and
$L_{m,p}=L_{m,p,max}$ or $L_{m,p}=L_{m,p,min}$, respectively.

Therefore if the intervals $(s_{1},s_{2}]$ and $[s_{1}+2,s_{2}+2)$ overlap,
that is if $s_{1}+2\leq s_{2}$ or equivalently $D\geq 1$, we have uniqueness
in all $L^{p}_{m}$ for which there is generation and, moreover,
$L_{m,p,max}=L_{m,p,min}$ if $(m+1)/p\in[s_{1}+2,s_{2}]$.

Uniqueness fails if $s_{2}<s_{1}+2$, i.e. $0\leq D<1$, and
$(m+1)/p\in(s_{2},s_{1}+2)$, as we show below but only for $D>0$.

###### Proposition 5.2

If $0<D<1$ and $s_{2}<\frac{m+1}{p}<s_{1}+2$, then $(L,D(L))$ where

$\displaystyle D(L)$ $\displaystyle=\bigg{\\{}u\in
D(L_{m,p,max})\;:\;s_{2}\frac{u}{y^{2}}+\frac{D_{y}u}{y}\in
L^{p}_{m}\bigg{\\}}$ (21)

generates a bounded positive analytic semigroup of angle $\pi/2$ on
$L_{m}^{p}$.

Proof. We proceed as in the proof of Proposition 4.4 but in place of the
isometry $T_{-s_{1}}$ we use the identity
$T_{s_{2}}LT_{-s_{2}}=D_{yy}+\frac{c-2s_{2}}{y}D_{y}=\tilde{B}^{n}$ (this last
is in $L^{p}_{m-s_{2}p}$), after observing that, under the given hypotheses,
$c-2s_{2}=1-2\sqrt{D}>-1$. Note that the conditions
$s_{2}<\frac{m+1}{p}<s_{1}+2$ and $0<\frac{m-s_{2}p+1}{p}<c-2s_{2}+1$ are
equivalent. The generation result follows then by similarity from Proposition
3.2. The description of $D(L)$ follows by applying Proposition 3.3 to
$v=y^{s_{2}}u\in D(\tilde{B}^{n}_{m-s_{2}p,p})$, $u\in D(L)$, as in the proof
of Proposition 4.4.

We point out that in the range $s_{2}<\frac{m+1}{p}<s_{1}+2$ the operators
$L_{m,p}$ of Section 4 and $(L,D(L))$ just constructed are different. In fact,
a compactly supported function $u$ which is equal to $y^{-s_{1}}$ in a
neighborhood of $0$, belongs to $D(L_{m,p})$ but not to $D(L)$ (otherwise
$y^{-2}u$ would be in $L^{p}_{m}$ and this is not the case since
$(m+1)/p<s_{1}+2$).

When $c=0$, then $L$ is a 1d-Schrödinger operator with inverse square
potential and the condition $s_{2}<s_{1}+2$ becomes $-\frac{1}{4}\leq
b<\frac{3}{4}$. That uniqueness really does not occur is proved for example in
[21] where different positive and analytic semigroups are exhibited in
$L^{2}$. In the case of Bessel operators, the condition $s_{2}<s_{1}+2$
becomes $-1<c<3$, and not $-1<c<1$ as one could guess.

We close this section by describing cores for these degenerate operators.
Observe that by (iii) of the above proposition, $C_{c}^{\infty}(0,\infty)$ is
a core for $L_{m,p}$ if and only if $s_{1}+2\leq(m+1)/p<s_{2}+2$.

###### Proposition 5.3

  * (i)

If $c>-1$ and $0<\frac{m+1}{p}<c+1$, then $\mathcal{D}=\left\\{u\in
C_{c}^{\infty}([0,\infty)):D_{y}u(0)=0\right\\}$ is a core for $B^{n}_{m,p}$.

  * (ii)

If $s_{1}<\frac{m+1}{p}<s_{2}+2$, then $\mathcal{D}=\left\\{u=y^{-s_{1}}v:v\in
C_{c}^{\infty}([0,\infty),\ D_{y}v(0)=0\right\\}$ is a core for $D(L_{m,p})$.

Proof.  (i) By Proposition 3.3, $\mathcal{D}\subset D(B^{n}_{m,p})$. Let $u\in
D(B^{n}_{m,p})$, $f=(I-B^{n})u$, $f_{k}=f\chi_{[k^{-1},\infty)}$ and
$u_{k}=(I-B^{n}_{m,p})^{-1}f_{k}$ so that $u_{k}\to u$ with respect to the
graph norm. By Proposition 2.4,
$u_{k}(y)=c_{k}y^{\frac{1-c}{2}}I_{\frac{c-1}{2}}(y)$ for $y\leq\frac{1}{k}$,
hence $u_{k}\in C^{\infty}([0,\frac{1}{k}])$ and $D_{y}u_{k}(0)=0$ by Lemma
2.3. Now the proof is straightforward. We fix $k$ and a cut-off function
$\phi$ which is equal to $1$ in $[0\frac{1}{2k}]$ and to $0$ for
$y\geq\frac{1}{k}$; then write $u_{k}=\phi u_{k}+(1-\phi)u_{k}$ and smooth
$(1-\phi)u_{k}$ by using convolutions (plus cut-off at $\infty$ to make
everything with compact support).

The proof of (ii) is similar using the Green function of $L_{m,p}$ or using
the transormation $T_{-s_{1}}$ as in Proposition 4.4.

##  6 Sums of closed operators

The operator $\mathcal{L}=\Delta_{x}+L_{y}$ (we write $\Delta_{x}$, $L_{y}$ to
indicate the variables on which the operators act) is the sum of the Laplacian
$\Delta_{x}$ and of the degenerate one dimensional operator $L_{y}$ which
clearly commute on smooth functions. Regularity properties for $\mathcal{L}$
follow once we prove the estimate

$\|\Delta_{x}u\|_{p}+\|L_{y}u\|_{p}\leq C\|\mathcal{L}u\|_{p}$ (22)

where the $L^{p}$ norms are taken over $\mathbb{R}^{N+1}_{+}$ on a
sufficiently large set of functions $u$. This is equivalent to saying that the
domain of $\mathcal{L}$ is the intersection of the domain of $\Delta_{x}$ and
$L_{y}$ (after aprropriate tensorization) or that the operator
$\Delta_{x}{\mathcal{L}}^{-1}$ is bounded. This strategy arose first in the
study of maximal regularity of parabolic problems, that is for the equation
$u_{t}=Au+f,u(0)=0$ where $A$ is the generator of an analytic semigroup on a
Banach space $X$. Estimates like

$\|u_{t}\|_{p}+\|Au\|_{p}\leq\|f\|_{p}$

where now the $L^{p}$ norm is that of $L^{p}([0,T[;X)$ can be interpreted as
closedness of $D_{t}-A$ on the intersection of the respective domains or,
equivalently, boundedness of the operator $A(D_{t}-A)^{-1}$ in
$L^{p}([0,T[;X)$.

Nowadays this strategy is well established and relies on Mikhlin vector-valued
multiplier theorems. Let us state the relevant definitions and main results we
need, referring the reader to [5], [24] or [13].

Let ${\cal S}$ be a subset of $B(X)$, the space of all bounded linear
operators on a Banach space $X$. ${\cal S}$ is $\mathcal{R}$-bounded if there
is a constant $C$ such that

$\|\sum_{i}\varepsilon_{i}S_{i}x_{i}\|_{L^{p}(\Omega;X)}\leq
C\|\sum_{i}\varepsilon_{i}x_{i}\|_{L^{p}(\Omega;X)}$

for every finite sum as above, where $(x_{i})\subset X,(S_{i})\subset{\cal S}$
and $\varepsilon_{i}:\Omega\to\\{-1,1\\}$ are independent and symmetric random
variables on a probability space $\Omega$. It is well-known that this
definition is independent of $1\leq p<\infty$ and that
$\mathcal{R}$-boundedness is equivalent to boundedness when $X$ is an Hilbert
space. When $X$ is an $L^{p}$ space (with respect to any $\sigma$-finite
measure), testing $\mathcal{R}$-boundedness is equivalent to proving square
function estimates, see [13, Remark 2.9 ].

###### Proposition 6.1

Let ${\cal S}\subset B(L^{p}(\Sigma))$, $1<p<\infty$. Then ${\cal S}$ is
$\mathcal{R}$-bounded if and only if there is a constant $C>0$ such that for
every finite family $(f_{i})\in L^{p}(\Sigma),(S_{i})\in{\cal S}$

$\left\|\left(\sum_{i}|S_{i}f_{i}|^{2}\right)^{\frac{1}{2}}\right\|_{L^{p}(\Sigma)}\leq
C\left\|\left(\sum_{i}|f_{i}|^{2}\right)^{\frac{1}{2}}\right\|_{L^{p}(\Sigma)}.$

By the proposition above $\mathcal{R}$-boundedness follows from domination. We
formualate this simple but important fact as a corollary.

###### Corollary 6.2

Let ${\cal S},{\cal T}\subset B(L^{p}(\Sigma))$, $1<p<\infty$ and asuume that
$\cal T$ is $\mathcal{R}$ bounded and that for every $S\in\cal S$ there exists
$T\in\cal T$ such that $|Sf|\leq|Tf|$ pointwise, for every $f\in
L^{p}(\Sigma)$. Then ${\cal S}$ is $\mathcal{R}$-bounded.

Proof.  This follows since

$\left(\sum_{i}|S_{i}f_{i}|^{2}\right)^{\frac{1}{2}}\leq\left(\sum_{i}|T_{i}f_{i}|^{2}\right)^{\frac{1}{2}}$

pointwise.

Let $(A,D(A))$ be a sectorial operator in a Banach space $X$; this means that
$\rho(-A)\supset\Sigma_{\pi-\phi}$ for some $\phi<\pi$ and that
$\lambda(\lambda+A)^{-1}$ is bounded in $\Sigma_{\pi-\phi}$. The infimum of
all such $\phi$ is called the spectral angle of $A$ and denoted by $\phi_{A}$.
Note that $-A$ generates an analytic semigroup if and only if
$\phi_{A}<\pi/2$. The definition of $\mathcal{R}$-sectorial operator is
similar, substituting boundedness of $\lambda(\lambda+A)^{-1}$ with
$\mathcal{R}$-boundedness in $\Sigma_{\pi-\phi}$. As above one denotes by
$\phi^{R}_{A}$ the infimum of all $\phi$ for which this happens; since
$\mathcal{R}$-boundedness implies boundedness, we have
$\phi_{A}\leq\phi^{R}_{A}$.

The $\mathcal{R}$-boundedness of the resolvent characterizes the regularity of
the associated inhomogeneous parabolic problem, as we explain now.

An analytic semigroup $(e^{-tA})_{t\geq 0}$ on a Banach space $X$ with
generator $-A$ has maximal regularity of type $L^{q}$ ($1<q<\infty$) if for
each $f\in L^{q}([0,T];X)$ the function $t\mapsto
u(t)=\int_{0}^{t}e^{-(t-s)A})f(s)\,ds$ belongs to $W^{1,q}([0,T];X)\cap
L^{q}([0,T];D(B))$. This means that the mild solution of the evolution
equation

$u^{\prime}(t)+Au(t)=f(t),\quad t>0,\qquad u(0)=0,$

is in fact a strong solution and has the best regularity one can expect. It is
known that this property does not depend on $1<q<\infty$ and $T>0$. A
characterization of maximal regularity is available in UMD Banach spaces,
through the $\mathcal{R}$-boundedness of the resolvent in sector larger than
the right half plane or, equivalently, of the semigroup in a sector around the
positive axis. In the case of $L^{p}$ spaces it can be restated in the
following form, see [13, Theorem 1.11]

###### Theorem 6.3

Let $(e^{-tA})_{t\geq 0}$ be a bounded analytic semigroup in $L^{p}(\Sigma)$
with generator $-A$. Then $T(\cdot)$ has maximal regularity of type $L^{q}$ if
and only if there are constants $0<\phi<\pi/2$, $C>0$ such that for every
finite sequence $(\lambda_{i})\subset\Sigma_{\pi/2+\phi}$, $(f_{i})\subset
L^{p}$

$\left\|\left(\sum_{i}|\lambda_{i}(\lambda_{i}+A)^{-1}f_{i}|^{2}\right)^{\frac{1}{2}}\right\|_{L^{p}(\Sigma)}\leq
C\left\|\left(\sum_{i}|f_{i}|^{2}\right)^{\frac{1}{2}}\right\|_{L^{p}(\Sigma)}.$

or, equivalently for every finite sequence $(z_{i})\subset\Sigma_{\phi}$,
$(f_{i})\subset L^{p}$

$\left\|\left(\sum_{i}|e^{-z_{i}A}f_{i}|^{2}\right)^{\frac{1}{2}}\right\|_{L^{p}(\Sigma)}\leq
C\left\|\left(\sum_{i}|f_{i}|^{2}\right)^{\frac{1}{2}}\right\|_{L^{p}(\Sigma)}.$

Finally we state the operator-valued Mikhlin Fourier multiplier theorem in the
N-dimensional case, see [5, Theorem 3.25] or [13, Theorem 4.6].

###### Theorem 6.4

Let $1<p<\infty$, $M\in C^{N}(\mathbb{R}^{N}\setminus\\{0\\};B(L^{p}(\Sigma))$
be such that the set

$\left\\{|\xi|^{|\alpha|}D^{\alpha}_{\xi}M(\xi):\xi\in\mathbb{R}^{N}\setminus\\{0\\},\
|\alpha|\leq N\right\\}$

is $\mathcal{R}$-bounded. Then the operator $T_{M}={\cal F}^{-1}M{\cal F}$ is
bounded in $L^{p}(\mathbb{R}^{N},L^{p}(\Sigma))$, where $\cal{F}$ denotes the
Fourier transform.

###  6.1 Muckenhoupt weighted estimates

Let $\left(S,d,\nu\right)$ be a space of homogeneous type, that is a metric
space endowed with a Borel measure $\nu$ which is doubling on balls. When
$X=L^{p}\left(S,d,\nu\right)$ the square function estimate in Theorem 6.1 can
be reduced to a family of Muckenhoupt weighted estimates of the type

$\|e^{-zA}f\|_{L^{p}(w)}\leq C\|f\|_{L^{p}(w)},\quad z\in\Sigma_{\phi},$

see Theorem 6.7 below. With this in mind, we recall preliminarily, the
definition and the essential properties about Muckenhoupt weights. For the
proof of the following results as well as for further details, we refer the
reader to [2, Chapter 2 and 5] and [25, Chapter 1]. Let $w$ be a weight i.e. a
non-negative locally integrable function defined on $S$; we use the notation

$-\hskip-10.81218pt\int_{E}w=\frac{1}{\nu(E)}\int_{E}w(x)\,d\nu(x),\qquad
w(E)=\int_{E}w(x)\,d\nu(x).$

Let $\mathcal{M}_{\nu}$ denote the uncentered maximal operator over balls in
$S$ defined by

$\displaystyle\mathcal{M}_{\nu}f(x):=\sup_{B\ni
x}\,-\hskip-10.81218pt\int_{B}|f|,\quad x\in S,$ (23)

where the supremum is taken over all balls of $S$ containing $x$. We recall
that $\mathcal{M}_{\nu}$ is bounded on $L^{p}(w)$ if and only if $w\in A_{p}$,
see for example [6, Theorem 7.3].

We say that $w\in A_{p}$, $1<p<\infty$, if there exists a constant $C$ such
that for every ball $B\subseteq S$ one has

$\displaystyle\Big{(}-\hskip-10.81218pt\int_{B}w\Big{)}\,\Big{(}-\hskip-10.81218pt\int_{B}w^{1-p^{\prime}}\Big{)}^{p-1}\leq
C.$ (24)

For $p=1$, we say that $w\in A_{1}$ if there is a constant $C$ such hat
$\mathcal{M}_{\nu}w\leq C\,w$ a.e..

The weight $w$ is in the reverse Hölder class of order $q$, $w\in RH_{q}$,
$1<q\leq\infty$, if there is a constant $C$ such that for every ball
$B\subseteq S$

$\Big{(}-\hskip-10.81218pt\int_{B}w^{q}\Big{)}^{\frac{1}{q}}\leq
C\,-\hskip-10.81218pt\int_{B}w,$

with the usual modification for $q=\infty$. For $p=1$, $RH_{1}$ is the set of
all weights. The best constants appearing in the previous inequalities are
referred respectively as the $A_{p}$ and the $RH_{q}$ constants of $w$.

We sum up in the following proposition the properties we need about these
classes of weights.

###### Proposition 6.5

The following properties hold:

  * (i)

$A_{1}\subset A_{p}\subset A_{q}$ for every $1\leq p\leq q\leq\infty$;

  * (ii)

$w\in A_{p}$, $1<p<\infty$, if and only if $w^{1-p^{\prime}}\in
A_{p^{\prime}}$;

  * (iii)

If $w\in A_{p}$, $1<p<\infty$, then there exists $1<q<p$ such that $w\in
A_{q}$;

  * (iv)

$RH_{\infty}\subset RH_{q}\subset RH_{p}$ for $1<p\leq q\leq\infty$;

  * (v)

If $w\in RH_{q}$, $1<q<\infty$, then there exists $q<p<\infty$ such that $w\in
RH_{p}$;

  * (vi)

$A_{\infty}:=\bigcup_{1\leq p<\infty}A_{p}=\bigcup_{1<q\leq\infty}RH_{q}.$

  * (vii)

Let $1<p_{0}<p<q_{0}<\infty$. Then we have

$w\in A_{\frac{p}{p_{0}}}\cap RH_{\left(\frac{q_{0}}{p}\right)^{\prime}}\iff
w^{-\frac{p^{\prime}}{p}}=w^{1-p^{\prime}}\in
A_{\frac{p^{\prime}}{q_{0}^{\prime}}}\cap
RH_{\left(\frac{p^{\prime}_{0}}{p^{\prime}}\right)^{\prime}}.$

  * (viii)

If $1\leq p\leq\infty$ and $1\leq r<\infty$ then

$w\in A_{p}\cap RH_{r}\quad\Longleftrightarrow\quad
w^{r},w^{-\frac{1}{p-1}}\in A_{\infty}\quad\Longleftrightarrow\quad w^{r}\in
A_{r\,(p-1)+1}.$

Proof. Properties $(i)$-$(vi)$ can be found in [6, Chapter 7], [25, Chapter
1]. Point (vii) follows as in [2, Lemma 4.4]. The first equivalence in (viii)
is proved in [25, Lemma 11, Chapter 1]; the second follows as in [12].

A proof of the following result is in [25, Corollary 14] or [6, Chapter 7].

###### Lemma 6.6

Let $w\in A_{p}\cap RH_{r}$, $1<r,p<\infty$. Then there exists a constant
$C>1$ such that for any ball $B$ and any measurable subset $E\subset B$,

$C^{-1}\left(\frac{\nu(E)}{\nu(B)}\right)^{p}\leq\frac{w(E)}{w(B)}\leq
C\left(\frac{\nu(E)}{\nu(B)}\right)^{\frac{r-1}{r}}.$

We now state an extrapolation result originally due to Rubio de Francia,
adapted as in [2, Theorem 4.9], which allows to reduce the square function
estimate in Theorem 6.1 to a family of Muckenhoupt weighted estimates. Only
weights and pairs of functions appear and no operator is involved. In what
follows we consider families $\mathcal{F}=\\{(f,g):f,g\in L_{+}^{0}(S)\\}$,
where $L_{+}^{0}(S)$ is the set of all non-negative, measurable functions
defined on $S$.

###### Theorem 6.7

Let $\left(S,d,\nu\right)$ be a space of homogeneous type and let
$\mathcal{F}\subseteq L_{+}^{0}(S)\times L_{+}^{0}(S)$. Suppose that there
exists $p$ with $p_{0}\leq p\leq q_{0}$ (and $p<\infty$ if $q_{0}=\infty$),
such that for $(f,g)\in\mathcal{F}$,

$\|f\|_{L^{p}(w)}\leq C\|g\|_{L^{p}(w)},\qquad\mbox{for all }w\in
A_{\frac{p}{p_{0}}}\cap RH_{\left(\frac{q_{0}}{p}\right)^{\prime}},$

Then, for all $p_{0}<q<q_{0}$ and $(f,g)\in\mathcal{F}$ we have

$\|f\|_{L^{q}(w)}\leq C\,\|g\|_{L^{q}(w)},\qquad\mbox{for all }w\in
A_{\frac{q}{p_{0}}}\cap RH_{\left(\frac{q_{0}}{q}\right)^{\prime}},$

Moreover, for all $p_{0}<q,r<q_{0}$ and
$\\{(f_{j},g_{j})\\}\subset\mathcal{F}$ we have

$\Big{\|}\Big{(}\sum_{j}(f_{j})^{r}\Big{)}^{1/r}\Big{\|}_{L^{q}(w)}\leq
C\,\Big{\|}\Big{(}\sum_{j}(g_{j})^{r}\Big{)}^{1/r}\Big{\|}_{L^{q}(w)},\quad\mbox{for
all }w\in A_{\frac{q}{p_{0}}}\cap RH_{\left(\frac{q_{0}}{q}\right)^{\prime}}.$

All the constants $C$ above may vary from line to line but depend only on the
$A_{s}$ and $RH_{s}$ constants of $w$.

Combining Theorem 6.1 and Theorem 6.7 we derive the following characterization
of maximal regularity in terms of boundedness over $L^{p}(w)$ spaces.

###### Theorem 6.8

Let $\left(S,d,\nu\right)$ be a space of homogeneous type, $p_{0}\leq p\leq
q_{0}$ with $p<\infty$ if $q_{0}=\infty$ and $p_{0}<2<q_{0}$. Let
$(e^{-zA})_{z\in\Sigma_{\delta}}$ be a bounded analytic semigroup in
$L^{p}\left(S,\nu\right)$ defined in a sector $\Sigma_{\delta}$, $\delta>0$.
Suppose that such that for $f\in L^{p}\left(S,\nu\right)$,

$\|e^{-zA}f\|_{L^{p}(w)}\leq C\|f\|_{L^{p}(w)},\qquad\mbox{for all
}z\in\Sigma_{\delta},\qquad\mbox{for all }w\in A_{\frac{p}{p_{0}}}\cap
RH_{\left(\frac{q_{0}}{p}\right)^{\prime}},$

where $C$ depends only on the $A_{s}$ and $RH_{s}$ constants of $w$.

Then, for all $p_{0}<q<q_{0}$, $(e^{-tA})_{t\geq 0}$ has maximal regularity on
$L^{q}(w)$ for all $w\in A_{\frac{q}{p_{0}}}\cap
RH_{\left(\frac{q_{0}}{q}\right)^{\prime}}$.

The following three lemmas will be crucial in the proof of maximal regularity.

###### Lemma 6.9

Let $w\in A_{p}$, $p\geq 1$, and let $\nu_{w}$ be the measure $wd\nu$. Denote
by $\mathcal{M}_{\nu_{w}}$ and $\mathcal{M}_{\nu}$ the maximal function
defined by $\nu_{w}$ and $\nu$. Then $\left(S,d,\nu_{w}\right)$ is a space of
homogeneous type and

$\displaystyle\mathcal{M}_{\nu}f\leq
A_{p}(w)^{\frac{1}{p}}\left(\mathcal{M}_{\nu_{w}}|f|^{p}\right)^{\frac{1}{p}},\quad
f\in L^{1}_{loc}\left(S,\nu\right),$

where $A_{p}(w)$ is the $A_{p}$ constant of $w$.

Proof. The doubling condition for the measure $\nu_{w}$ follows from that of
$\nu$ and Lemma 6.6. To prove the second claim, let $f\in
L^{1}_{loc}\left(S,\nu\right)$. Then for every ball $B$ of $S$ one has,
applying Hölder’s inequality,

$\displaystyle\frac{1}{\nu(B)}\int_{B}|f|d\nu$
$\displaystyle=\frac{1}{\nu(B)}\int_{B}|f|w^{\frac{1}{p}}w^{-\frac{1}{p}}d\nu\leq\frac{1}{\nu(B)}\left(\int_{B}|f|^{p}wd\nu\right)^{\frac{1}{p}}\left(\int_{B}w^{1-p^{\prime}}d\nu\right)^{\frac{1}{p^{\prime}}}.$

Using (24) we get

$\displaystyle\frac{1}{\nu(B)}\int_{B}|f|d\nu\leq
A_{p}(w)^{\frac{1}{p}}\left(\frac{1}{\nu_{w}(B)}\int_{B}|f|^{p}wd\nu\right)^{\frac{1}{p}}$

which, taking the supremum over $B$, yields the required claim. The case $p=1$
follows similarly.

###### Lemma 6.10

Let $p$ be a non-negative, locally integrable function on $\mathbb{R}^{M}$ and
consider the measure $\nu=p\,dy$. Let $\mathcal{M_{\nu}}$ be the uncentered
maximal operator relative to $\nu$, defined as in (23). If $0\leq\phi\in
L^{1}\left(\mathbb{R}^{M},\nu\right)$ is radial and decreasing then

$\displaystyle|(\phi\ast
pf)(y)|\leq\|\phi\|_{L^{1}\left(\mathbb{R}^{M},\nu\right)}\mathcal{M}_{\nu}f(y),\quad
y\in\mathbb{R}^{M},\quad f\in L^{1}_{loc}\left(\mathbb{R}^{M},\nu\right).$

If $p$ is homogeneous of degree $k$ i.e. $p(ty)=t^{k}p(y)$ for all
$x\in\mathbb{R}^{N}$ and $t>0$, then setting $\phi_{t}:=t^{-M-k}\phi(t^{-1}y)$
one has

$\displaystyle\sup_{t>0}|(\phi_{t}\ast
pf)(y)|\leq\|\phi\|_{L^{1}\left(\mathbb{R}^{M},\nu\right)}\mathcal{M}_{\nu}f(y).$

Proof. Let us suppose preliminarily that $\phi$ is a simple function and let
us write, for some $a_{1},\dots,a_{k}>0$ and balls $B_{1},\dots,B_{k}$
centered at $0$,

$\displaystyle\phi(y)=\sum_{j=1}^{k}a_{j}\chi_{B_{j}}(y).$

Then, since
$\|\phi\|_{L^{1}\left(\mathbb{R}^{M},\nu\right)}=\sum_{j=1}^{k}a_{j}\,\nu(B_{j})$
and $(\chi_{B_{j}}\ast pf)(y)=\int_{y-B_{j}}f(z)d\nu$ , we get

$\displaystyle(\phi\ast
pf)(y)=\sum_{j=1}^{k}a_{j}\,\nu(B_{j})\frac{1}{\nu(B_{j})}(\chi_{B_{j}}\ast
pf)(y)\leq\|\phi\|_{L^{1}\left(\mathbb{R}^{M},\nu\right)}\mathcal{M}_{\nu}f(y).$

In the general case the first required claim follows since $\phi$ can be
approximated by a sequence of simple functions which increase to it
monotonically. To prove the second claim it is enough to observe that, under
the homogeneity assumptions on $p$, one has
$\|\phi_{t}\|_{L^{1}\left(\mathbb{R}^{M},\nu\right)}=\|\phi\|_{L^{1}\left(\mathbb{R}^{M},\nu\right)}$

###### Lemma 6.11

Let $m\in\mathbb{R}$ be such that $M+m>0$ and let $d\mu_{m}=|y|^{m}dy$. For
every $k\in\mathbb{R}$ let us consider the radial weight $w(y)=|y|^{k}$. The
following properties hold.

  * (i)

If $1\leq p\leq\infty$ then $w\in A_{p}\left(\mu_{m}\right)$ if and only if
$-(M+m)<k<(M+m)(p-1)$.

  * (ii)

If $1\leq p\leq\infty$ and $1\leq r<\infty$ then $w\in A_{p}(\mu_{m})\cap
RH_{r}(\mu_{m})$ if and only if $-\frac{M+m}{r}<k<(M+m)(p-1)$.

Proof. To prove (i), we start by considering balls of center $y_{0}$ and
radius $1$. Fix $R>1$ and assume first that $|y_{0}|\leq R$. Then both
$|y|^{k}$ and $|y|^{-\frac{k}{p-1}}$ are integrable in $B(y_{0},1)$ with
respect to the measure $\mu_{m}$ and

$\left(\frac{1}{\mu_{m}(B(y_{0},1))}\int_{B(y_{0},1)}|y|^{k}\
d\mu_{m}\right)\left(\frac{1}{\mu_{m}(B(y_{0},1))}\int_{B(y_{0},1)}|y|^{-\frac{k}{p-1}}\
d\mu_{m}\right)^{p-1}\leq C$ (25)

for some positive constant $C$ depending on $R$. On the other hand, when
$|y_{0}|>R$, then

$\left(\frac{1}{\mu_{m}(B(y_{0},1))}\int_{B(y_{0},1)}|y|^{k}\
d\mu_{m}\right)\approx|y_{0}|^{k},\quad\left(\frac{1}{\mu_{m}(B(y_{0},1))}\int_{B(y_{0},1)}|y|^{-\frac{k}{p-1}}\
d\mu_{m}\right)^{p-1}\approx|y_{0}|^{-k}$

and the left hand side in (25) is bounded from above and below by a constant.
For a general ball of radius $r$ the claim follows by scaling. Property (ii)
follows using (i) and property (viii) of Proposition 6.5.

##  7 $\mathcal{R}$-boundedness for a family of integral operators

In this section we study the ${\mathcal{R}}$-boundedness of the family of
integral operators

$\displaystyle
S^{\alpha,\beta}(t)f(y)=t^{-\frac{M}{2}}\,\left(\frac{|y|}{\sqrt{t}}\wedge
1\right)^{-\alpha}\int_{\mathbb{R}^{M}}\left(\frac{|z|}{\sqrt{t}}\wedge
1\right)^{-\beta}\exp\left(-\frac{|y-z|^{2}}{\kappa t}\right)f(z)\,dz,$ (26)

where $\kappa$ is a positive constant. We omit the dependence on $\kappa$ even
though in some proofs we need to vary it.

For $m\in\mathbb{R}$ we consider the measure $d\mu_{m}=|y|^{m}dy$ on
$\mathbb{R}^{M}$ and study the action of $S^{\alpha,\beta}(t)$ over the space
$L^{p}_{m}=L^{p}(\mathbb{R}^{M},d\mu_{m})$ for $1<p<\infty$. We prove that
when $S^{\alpha,\beta}(1)$ is bounded in $L^{p}_{m}$, then the family
$\left(S^{\alpha,\beta}(t)\right)_{t>0}$ is also $\mathcal{R}$-bounded on
$L^{p}_{m}$. Note that we write the kernel of these operators always with
respect to the Lebesgue measure, even when they act in weighted spaces.

We start by observing that the scale homogeneity of $S^{\alpha,\beta}$ is $2$
since a change of variables yields

$\displaystyle
S^{\alpha,\beta}(t)\left(I_{s}f\right)=I_{s}\left(S^{\alpha,\beta}(s^{2}t)f\right),\qquad
I_{s}f(y)=f(sy),\qquad t,s>0,$

which in particular gives

$\displaystyle
S^{\alpha,\beta}(t)f=I_{1/\sqrt{t}}\left(S^{\alpha,\beta}(1)I_{\sqrt{t}}f\right),\qquad
t>0.$

The boundedness of $S^{\alpha,\beta}(t)$ in $L^{p}_{m}$ is then equivalent to
that for $t=1$ and $\|S^{\alpha,\beta}(t)\|_{p}=\|S^{\alpha,\beta}(1)\|_{p}$
and this is equivalent to $\alpha<\frac{M+m}{p}<M-\beta$ (in particular
$\alpha+\beta<M$), see Proposition 12.2, with $\theta=0$.

For future purpose we also observe that the adjoint of $S^{\alpha,\beta}(t)$
taken with respect to the measure $\mu_{m}$ is given by the operator

$\displaystyle\left(S^{\alpha,\beta}(t)\right)^{\ast
m}f(y)=t^{-\frac{M}{2}}\,\left(\frac{|y|}{\sqrt{t}}\wedge
1\right)^{-\beta}\int_{\mathbb{R}^{N}}\left(\frac{|y|}{|z|}\right)^{-m}\left(\frac{|z|}{\sqrt{t}}\wedge
1\right)^{-\alpha}\exp\left(-\frac{|y-z|^{2}}{\kappa t}\right)f(z)\,dz.$ (27)

###  7.1 $\mathcal{R}$-boundedness when $M+m>$0

If $d$ denotes the euclidean distance, $\left(\mathbb{R}^{M},d,\mu_{m}\right)$
is of homogeneous type. In what follows we write $A_{p}(\mu_{m})$,
$RH_{p}(\mu_{m})$, $\mathcal{M}_{\mu_{m}}$ to denote respectively the class of
Muckenhoupt weights, the reverse Hölder class and the maximal function over
balls taken with respect to the measure $\mu_{m}$. When $m=0$ we write
$A_{p}$, $RH_{p}$, $\mathcal{M}$.

We observe preliminarily that (26) yields when $m\geq 0$

$\displaystyle|S^{\alpha,\beta}(t)f(y)|$ $\displaystyle\leq
Ct^{-\frac{M+m}{2}}\,\left(\frac{|y|}{\sqrt{t}}\wedge
1\right)^{-\alpha}\int_{\mathbb{R}^{M}}\left(\frac{|z|}{\sqrt{t}}\wedge
1\right)^{-\beta-m}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|\,d\mu_{m}(z).$ (28)

This follows after observing that, when $m\geq 0$, one has

$\displaystyle|z|^{-m}\left(|y|\wedge 1\right)^{-\beta}\leq\left(|z|\wedge
1\right)^{-\beta-m},\quad z\in\mathbb{R}^{M}\setminus\\{0\\}.$

When $m<0$, up to a small perturbation of the constant in the exponential
argument, estimate (28) continues to hold in the range
$\frac{|y|}{\sqrt{t}}\leq 1$, $\frac{|z|}{\sqrt{t}}\geq 1$. Indeed in this
case one has, for $\epsilon>0$ and for some $K>0$,

$\displaystyle|z|^{-m}\exp\left(-\varepsilon|y-z|^{2}\right)\leq
K,\quad|y|\leq 1,\;|z|\geq 1.$

This implies, for $\frac{|y|}{\sqrt{t}}\leq 1$, $\frac{|z|}{\sqrt{t}}\geq 1$
and $\kappa^{\prime}>\kappa$

$\displaystyle|S^{\alpha,\beta}(t)f(y)|$ $\displaystyle\leq
Ct^{-\frac{N}{2}}\,\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha}\int_{\mathbb{R}^{M}}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|\,dz$ $\displaystyle\leq
CKt^{-\frac{M+m}{2}}\,\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha}\int_{\mathbb{R}^{M}}\exp\left(-\frac{|y-z|^{2}}{\kappa^{\prime}t}\right)|f(z)|\,d\mu_{m}(z).$
(29)

We prove the $\mathcal{R}$-boundedness of the family
$\left(S^{\alpha,\beta}(t)\right)_{t\geq 0}$ using the extrapolation result of
Theorem 6.7. We follow the proof in [3, Theorem 2.9] but new complications
arise because the operator is non-symmetric and the measure $\mu_{m}$ is not
the Lebesgue one. In particular we have to distinguish between the cases
$m\geq 0$ and $-M<m<0$ and both the maximal functions with respect to the
Lebesgue measure and the weighted one appear. Note that, since $M+m>0$,
condition (iii) in Proposition 12.2 implies $\beta<M$ and $\alpha<M+m$.

For the reader’s convenience, in what follows we write for $t>0$,
$B=B(0,\sqrt{t})$ and

$\displaystyle S^{\alpha,\beta}(t)f$
$\displaystyle=\chi_{B^{c}}\left(S^{\alpha,\beta}(t)f\chi_{B^{c}}\right)+\chi_{B}\left(S^{\alpha,\beta}(t)(f\chi_{B})\right)+\chi_{B^{c}}\left(S^{\alpha,\beta}(t)f\chi_{B}\right)+\chi_{B}\left(S^{\alpha,\beta}(t)(f\chi_{B^{c}})\right)$
$\displaystyle:=S^{\alpha,\beta}_{1}(t)f+S^{\alpha,\beta}_{2}(t)f+S^{\alpha,\beta}_{3}(t)f+S^{\alpha,\beta}_{4}(t)f.$
(30)

###### Proposition 7.1

Let $M+m>0$, $1<p<\infty$ and assume that $\alpha<\frac{M+m}{p}<M-\beta$. Let

$\displaystyle q_{0}$ $\displaystyle=$ $\displaystyle\frac{M+m}{\alpha}\ {\rm
when}\ \alpha>0,\quad q_{0}=\infty\ {\rm when\ }\alpha\leq 0$ (31)
$\displaystyle p_{0}$ $\displaystyle=$
$\displaystyle\left(\frac{M+m}{\beta+m}\right)^{\prime}\ {\rm when}\
\beta+m>0,\quad p_{0}=1\ {\rm when\ }\beta+m\leq 0$

so that $p_{0}<p<q_{0}$. Then for every weight

$w\in A_{\frac{p}{p_{0}}}(\mu_{m})\cap
RH_{\left(\frac{q_{0}}{p}\right)^{\prime}}(\mu_{m})$

there exist $C>0$ depending on the $A_{\frac{p}{p_{0}}}(\mu_{m})$ and
$RH_{\left(\frac{q_{0}}{p}\right)^{\prime}}(\mu_{m})$ constants of $w$ such
that for every $t\geq 0$ one has

$\|S^{\alpha,\beta}(t)f\|_{L^{p}(w)}\leq C\|f\|_{L^{p}(w)},\quad f\in
L^{p}(\mathbb{R}^{M},wd\mu_{m})=:L^{p}(w).$

Finally, if in addition $p_{0}<2<q_{0}$ (i.e. $S^{\alpha,\beta}(1)$ is bounded
on $L^{2}(\mathbb{R}^{M},wd\mu_{m})$) then the family
$\left(S^{\alpha,\beta}(t)\right)_{t\geq 0}$ is $\mathcal{R}$-bounded on
$L^{p}(\mathbb{R}^{M},wd\mu_{m})$.

We split the proof in four lemmas according to (7.1).

###### Lemma 7.2

The estimate of Proposition 7.1 holds for $(S^{\alpha,\beta}_{1}(t))_{t\geq
0}$.

Proof. Assume first that $m\geq 0$. Then using (28) and Lemma 6.10 with
$p(y)=|y|^{m}$ we get

$\displaystyle|S^{\alpha,\beta}_{1}(t)f(y)|$ $\displaystyle\leq
Ct^{-\frac{M+m}{2}}\int_{\mathbb{R}^{M}}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|\,d\mu_{m}(z)\leq C\mathcal{M}_{\mu_{m}}f(y),$

The claim then follows since $\mathcal{M}_{\mu_{m}}$ is bounded on $L^{p}(w)$.

When $-M<m<0$ we use (26) (and Lemma 6.10 with respect to the Lebesgue
measure) to get

$\displaystyle|S^{\alpha,\beta}_{1}(t)f(y)|$ $\displaystyle\leq
Ct^{-\frac{M}{2}}\int_{\mathbb{R}^{M}}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|\chi_{B^{c}}(z)\,dy\leq C\mathcal{M}f(y),$

Since $w\in A_{\frac{p}{p_{0}}}(\mu_{m})$, by Proposition 6.5 there exists $r$
sufficiently close to $p_{0}$ such that $p_{0}<r<p<q_{0}$ and $w\in
A_{\frac{p}{r}}(\mu_{m})$. Since $-M<m<0$, Lemma 6.11 (i) gives $|y|^{m}\in
A_{r}(dy)$ and then Lemma 6.9 yields

$\displaystyle|S^{\alpha,\beta}_{1}(t)f(y)|$ $\displaystyle\leq
C\left(\mathcal{M}_{\mu_{m}}|f|^{r}(y)\right)^{\frac{1}{r}}.$

Since $w\in A_{\frac{p}{r}}(\mu_{m})$, $\mathcal{M}_{\mu_{m}}$ is bounded on
$L^{\frac{p}{r}}(w)$ and we get $\|S^{\alpha,\beta}_{1}(t)f\|_{L^{p}(w)}\leq
C\|f\|_{L^{p}(w)}$.

###### Lemma 7.3

The estimate of Proposition 7.1 holds for $(S^{\alpha,\beta}_{2}(t))_{t\geq
0}$.

Proof. Using (26) and Hölder’s inequality we get

$\displaystyle|S^{\alpha,\beta}_{2}(t)f(y)|$ $\displaystyle\leq
Ct^{-\frac{M+m}{2}}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha}\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-\beta-m}|f(z)|\,d\mu_{m}(z)$
$\displaystyle\leq
Ct^{-\frac{M+m}{2}}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha}\|f\|_{L^{p}(w)}\left(\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-(\beta+m)p^{\prime}}w(z)^{1-p^{\prime}}\
d\mu_{m}(z)\right)^{\frac{1}{p^{\prime}}}.$

Setting $v=w^{1-p^{\prime}}$ this implies

$\displaystyle\|S^{\alpha,\beta}_{2}(t)f\|_{L^{p}(w)}^{p}\leq
Ct^{-\frac{M+m}{2}p}\|f\|_{L^{p}(w)}^{p}\left(\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-(\beta+m)p^{\prime}}v(z)\
d\mu_{m}(z)\right)^{\frac{p}{p^{\prime}}}\int_{B}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha
p}w(y)\ d\mu_{m}(y).$

Let us treat the first integral. If $\beta+m>0$, then one has

$\displaystyle\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-(\beta+m)p^{\prime}}v(z)\
d\mu_{m}(z)$ $\displaystyle=\sum_{j\geq
0}\int_{2^{-j-1}\leq\frac{|z|}{\sqrt{t}}<2^{-j}}\left(\frac{|z|}{\sqrt{t}}\right)^{-(\beta+m)p^{\prime}}v(z)\
d\mu_{m}(z)$ $\displaystyle\leq C\sum_{j\geq
0}2^{j(\beta+m)p^{\prime}}v(2^{-j}B).$

By property (vii) of Proposition 6.5, $v\in
A_{\frac{p^{\prime}}{q^{\prime}_{0}}}\cap
RH_{\left(\frac{p^{\prime}_{0}}{p^{\prime}}\right)^{\prime}}$; by property (v)
of Proposition 6.5 there exists $r>p^{\prime}$ such that $v\in
RH_{\left(\frac{p^{\prime}_{0}}{r}\right)^{\prime}}$. Lemma 6.6 then implies

$v(2^{-j}B)\leq
Cv(B)\left(\frac{\mu_{m}\left(2^{-j}B\right)}{\mu_{m}\left(B\right)}\right)^{\frac{(\beta+m)r}{M+m}}=Cv(B)2^{-jr(\beta+m)}.$

Therefore since $\beta+m>0$

$\displaystyle\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-(\beta+m)p^{\prime}}v(z)\
d\mu_{m}(z)$ $\displaystyle\leq Cv(B)\sum_{j\geq
0}2^{-j(\beta+m)(r-p^{\prime})}=Cv(B).$

The last inequality holds also when $\beta+m\leq 0$, since in this case

$\displaystyle\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-(\beta+m)p^{\prime}}v(z)\
d\mu_{m}(z)\leq\int_{B}v(z)\ d\mu_{m}(y)=v(B).$

Similarly if $\alpha>0$ then

$\displaystyle\int_{B}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha p}w(y)\
d\mu_{m}(y)$ $\displaystyle=\sum_{j\geq
0}\int_{2^{-j-1}\leq\frac{|y|}{\sqrt{t}}<2^{-j}}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha
p}w(y)\ d\mu_{m}(y)$ $\displaystyle\leq C\sum_{j\geq 0}2^{j\alpha
p}w(2^{-j}B).$

Since $w\in A_{\frac{p}{p_{0}}}\cap
RH_{\left(\frac{q_{0}}{p}\right)^{\prime}}$ by property (v) of Proposition 6.5
there exists $r>p$ such that $w\in
RH_{\left(\frac{q_{0}}{r}\right)^{\prime}}$. By Lemma 6.6 then

$w(2^{-j}B)\leq Cw(B)2^{-jr\alpha}.$

Therefore

$\displaystyle\int_{B}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha p}w(y)\
d\mu_{m}(y)\leq Cw(B)\sum_{j\geq 0}2^{-j\alpha(r-p)}=Cw(B).$

The last inequality holds also when $\alpha\leq 0$, since in this case

$\displaystyle\int_{B}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha}w(y)\
d\mu_{m}(y)\leq\int_{B}w(y)\ d\mu_{m}(y)=w(B).$

Putting together the last inequalities we have in any case

$\|S^{\alpha,\beta}_{2}(t)f\|_{L^{p}(w)}^{p}\leq
C\|f\|_{L^{p}(w)}^{p}t^{-p\frac{M+m}{2}}\left(v(B)\right)^{\frac{p}{p^{\prime}}}w(B).$

Since $\beta<M$ from property (i) of Proposition 6.5 we get $w\in
A_{\frac{p}{p_{0}}}\subseteq A_{p}$ which implies, by the definition (24) of
$A_{p}$ weights,
$\sup_{t>0}t^{-p\frac{M+m}{2}}\left(v(B)\right)^{\frac{p}{p^{\prime}}}w(B)<\infty$.

###### Lemma 7.4

The estimate of Proposition 7.1 holds for $(S^{\alpha,\beta}_{3}(t))_{t\geq
0}$.

Proof. Using (26) we get

$\displaystyle|S^{\alpha,\beta}_{3}(t)f(y)|$ $\displaystyle\leq
Ct^{-\frac{M+m}{2}}\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-\beta-m}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|\,d\mu_{m}(z).$

Let us fix $r$ such that $p_{0}^{\prime}<r<p<q_{0}$. Applying Hölder’s
inequality we obtain

$\displaystyle|S^{\alpha,\beta}_{3}(t)f(y)|$ $\displaystyle\leq
C\left(t^{-\frac{M+m}{2}}\int_{\mathbb{R}^{M}}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|^{r}\ d\mu_{m}(z)\right)^{\frac{1}{r}}$ $\displaystyle\hskip
86.11084pt\times\left(t^{-\frac{M+m}{2}}\int_{B}\left(\frac{|z|}{t^{\frac{1}{2}}}\right)^{-(\beta+m)r^{\prime}}\
d\mu_{m}(z)\right)^{\frac{1}{r^{\prime}}}.$

The substitution $\xi=z/\sqrt{t}$ and Lemma 6.10 yield

$\displaystyle|S^{\alpha,\beta}_{3}(t)f(y)|$ $\displaystyle\leq
C\left(\mathcal{M}_{\mu_{m}}|f|^{r}(y)\right)^{\frac{1}{r}}\left(\int_{B(0,1)}|\xi|^{-(\beta+m)r^{\prime}+m}\
d\xi\right)^{\frac{1}{r^{\prime}}}=C\left(\mathcal{M}_{\mu_{m}}|f|^{r}(y)\right)^{\frac{1}{r}}$

Since $w\in A_{\frac{p}{p_{0}}}$, by Proposition 6.5 there exists $r$
sufficiently close to $p_{0}$ such that $p_{0}<r<p<q_{0}$ and $w\in
A_{\frac{p}{r}}$ . This implies that $\mathcal{M}_{\mu_{m}}$ is bounded on
$L^{\frac{p}{r}}(w)$ which, using the latter inequality, proves the required
claim.

Finally, in order to prove the boundedness of $S^{\alpha,\beta}_{4}(t)$, we
employ estimates (28), (7.1) which allow to equivalently prove, up to a small
modification of the constant in the exponential argument, the boundedness of
the operator

$\displaystyle F_{4}(t)f(y)$
$\displaystyle=\chi_{B}(y)\,{t}^{-\frac{M+m}{2}}\left(\frac{|y|}{\sqrt{t}}\right)^{-\alpha}\,\int_{B^{c}}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|\,d\mu_{m}(z).$

We apply a duality argument and we observe that the adjoint of $F_{4}(t)$
taken with respect to the measure $\mu_{m}$ is given by the operator

$\displaystyle\left(F_{4}(t)^{\ast m}\right)f(y)$
$\displaystyle=t^{-\frac{M+m}{2}}\int_{B}\left(\frac{|z|}{\sqrt{t}}\right)^{-\alpha}\exp\left(-\frac{|y-z|^{2}}{\kappa
t}\right)|f(z)|\,d\mu_{m}(z)$ $\displaystyle=S^{\beta,\alpha}_{3}(t)f(y)$

With this aim let us observe that $q_{0}^{\prime}<p^{\prime}<p_{0}^{\prime}$.

###### Lemma 7.5

The estimate of Proposition 7.1 holds for $(S^{\alpha,\beta}_{4}(t))_{t\geq
0}$.

Proof. We apply a duality argument. Let $g\in
L^{p^{\prime}}(\mathbb{R}^{M},w\mu_{m})$; since $(F_{4}(t))^{\ast
m}=S^{\beta,\alpha}_{3}(t)$ we obtain

$\displaystyle\int_{\mathbb{R}^{M}}F_{4}(t)fg\,w\mu_{m}=\int_{\mathbb{R}^{M}}f\,S^{\beta,\alpha}_{3}(t)(gw)\mu_{m}=\int_{\mathbb{R}^{M}}f\,\frac{S^{\beta,\alpha}_{3}(t)(gw)}{w}w\mu_{m}.$

Using Hölder’s inequality we then yield

$\displaystyle\left|\int_{\mathbb{R}^{M}}F_{4}(t)fg\,w\mu_{m}\right|\leq\|f\|_{L^{p}(\omega)}\left\|\frac{S^{\beta,\alpha}_{3}(t)(gw)}{w}\right\|_{L^{p^{\prime}}(\omega)}=\|f\|_{L^{p}(\omega)}\left\|S^{\beta,\alpha}_{3}(t)(gw)\right\|_{L^{p^{\prime}}(\omega^{1-p^{\prime}})}.$

By property (vii) of Proposition 6.5, $\omega^{1-p^{\prime}}\in
A_{\frac{p^{\prime}}{q_{0}^{\prime}}}\cap
RH_{\left(\frac{p_{0}^{\prime}}{p^{\prime}}\right)^{\prime}}$. Then using the
estimate for $S^{\beta,\alpha}_{3}$ and with $p$ replaced by $p^{\prime}$ we
get

$\displaystyle\left|\int_{\mathbb{R}^{M}}F_{4}(t)fg\,w\mu_{m}\right|\leq
C\|f\|_{L^{p}(\omega)}\left\|gw\right\|_{L^{p^{\prime}}(\omega^{1-p^{\prime}})}=C\|f\|_{L^{p}(\omega)}\left\|g\right\|_{L^{p^{\prime}}(\omega)}$

which concludes the proof.

We can finally prove Proposition 7.1.

(Proof of Proposition 7.1) The first claim follows by using (7.1) and Lemmas
7.2, 7.3, 7.4, 7.5. If $p=2$ satisfies $p_{0}<2<q_{0}$, then the
$\mathcal{R}$-boundedness of $(S^{\alpha,\beta}(t))_{t\geq 0}$ on
$L^{p}(\mathbb{R}^{M},wd\mu_{m})$ follows by Theorems 6.1, 6.7 since, in this
case, the boundedness of $S^{\alpha,\beta}(t)$ in all the Muckenhoupt weighted
spaces just proved implies the square function estimate required by Theorem
6.1.

###  7.2 $\mathcal{R}$-boundedness in the general case

Let us remove the assumptions $M+m>0$ and $p_{0}<2<q_{0}$ first in the case of
the Lebesgue measure, that is when $m=0$. Since we use different measures here
we do not shorten $L^{p}(\mathbb{R}^{M},|y|^{m}dy)$ to $L^{p}_{m}$.

###### Theorem 7.6

Let $1<p<\infty$ and let us suppose that $\alpha<\frac{M+m}{p}<M-\beta$.
Then$\left(S^{\alpha,\beta}(t)\right)_{t\geq 0}$ is $\mathcal{R}$-bounded on
$L^{p}(\mathbb{R}^{M})$.

Proof. If $\alpha<\frac{M}{2}<M-\beta$ i.e. $\alpha,\beta<\frac{M}{2}$, the
thesis is part of Proposition 7.1 with $m=0$. Let us suppose now
$\alpha\geq\frac{M}{2}$ or $\beta\geq\frac{M}{2}$; in this case
$S^{\alpha,\beta}(t)$ is not bounded on $L^{2}\left(\mathbb{R}^{M}\right)$ and
therefore $p\neq 2$.

Given $m\in\mathbb{R}$, let us consider the isometry

$\displaystyle T_{\frac{m}{p}}:L^{p}(\mathbb{R}^{M},|y|^{m}dy)\to
L^{p}(\mathbb{R}^{M},dy),\quad f\mapsto|y|^{\frac{m}{p}}f.$

A straightforward computation shows that

$\displaystyle
T_{-\frac{m}{p}}S^{\alpha,\beta}(t)T_{\frac{m}{p}}f=\tilde{S}^{\alpha,\beta,\frac{m}{p}}(t)f$

where $\tilde{S}^{\alpha,\beta,\frac{m}{p}}$ is the operator defined in (39)
with $r=m/p$. Lemma 12.1 gives

$\displaystyle|T_{-\frac{m}{p}}S^{\alpha,\beta}(t)T_{\frac{m}{p}}f|\leq
CS^{\alpha+\frac{m}{p},\beta-\frac{m}{p}}(t)|f|$ (32)

Since the $\mathcal{R}$-boundedness of a family of operators is preserved by
isometries and pointwise domination, from the equality (32) one can easily
deduce that $\left(S^{\alpha,\beta}(t)\right)_{t\geq 0}$ is
$\mathcal{R}$-bounded on $L^{p}(\mathbb{R}^{M})$ if there exists
$m\in\mathbb{R}$ such that
$\left(S^{\alpha+\frac{m}{p},\beta-\frac{m}{p}}(t)\right)_{t\geq 0}$ is
$\mathcal{R}$-bounded on $L^{p}(\mathbb{R}^{M},|y|^{m}dy)$. From Proposition
7.1 it is then sufficient to require

$\displaystyle 0$
$\displaystyle<M+m,\qquad\alpha+\frac{m}{p}<\frac{M+m}{2}<M-\beta+\frac{m}{p}.$

By elementary calculation the latter inequalities read as

$\displaystyle\begin{cases}M+m>0;\\\\[5.59721pt]
m\left(\frac{1}{p}-\frac{1}{2}\right)<\frac{M}{2}-\alpha;\\\\[5.59721pt]
m\left(\frac{1}{p}-\frac{1}{2}\right)>\beta-\frac{M}{2},\end{cases}$

If $p<2$ the system has a solution $m$ when

$\beta-\frac{M}{2}<-M\left({\frac{1}{p}-\frac{1}{2}}\right)<\frac{M}{2}-\alpha$

that is when $\alpha<\frac{M}{p}<M-\beta$. If $p>2$ the claim follows in the
same way.

The results for $S^{\alpha,\beta}(t)$ in
$L^{p}\left(\mathbb{R}^{M},d\mu_{m}\right)$ are immediate consequence of those
of $S^{\alpha-\frac{m}{p},\beta+\frac{m}{p}}(t)$ in
$L^{p}(\mathbb{R}^{M},dy)$. Note that the condition $M+m>0$ is no longer
required.

###### Theorem 7.7

If , $1<p<\infty$ and $\alpha<\frac{M+m}{p}<M-\beta$, then the family
$\left(S^{\alpha,\beta}(t)\right)_{t\geq 0}$ is $\mathcal{R}$-bounded on
$L^{p}_{m}=L^{p}(\mathbb{R}^{M},d\mu_{m})$.

Proof. Let us consider the isometry

$\displaystyle T_{-\frac{m}{p}}:L^{p}(\mathbb{R}^{M},dy)\to
L^{p}(\mathbb{R}^{M},|y|^{m}dy),\quad f\mapsto|y|^{-\frac{m}{p}}f.$

Then, as done in the previous proof, one has, using Lemma 12.1,

$\displaystyle|T_{\frac{m}{p}}S^{\alpha,\beta}(t)T_{-\frac{m}{p}}f|=|T^{\alpha,\beta,-\frac{m}{p}}f|\leq
CS^{\alpha-\frac{m}{p},\beta+\frac{m}{p}}|f|.$

By construction, the boundedness conditions for
$S^{\alpha-\frac{m}{p},\beta+\frac{m}{p}}(1)$ in $L^{p}(\mathbb{R}^{M},dy)$
are satisfied under the given hypotheses on $S^{\alpha,\beta}$. Then the
family $S^{\alpha-\frac{m}{p},\beta+\frac{m}{p}}$ is $\mathcal{R}$-bounded in
$L^{p}(\mathbb{R}^{M},dy)$ by Theorem 7.6; the same result then also holds for
$T_{\frac{m}{p}}S^{\alpha,\beta}T_{-\frac{m}{p}}$ by domination. By similarity
this yields the $\mathcal{R}$-boundedness of $S^{\alpha,\beta}$ in
$L^{p}(\mathbb{R}^{M},|y|^{m}dy)$.

##  8 Domain and maximal regularity for $\mathcal{L}=\Delta_{x}+L_{y}$

###  8.1 Basic facts

Here we deduce generation results for the whole operator
$\mathcal{L}=\Delta_{x}+L_{y}$ by standard tensor product arguments. If $X,Y$
are function spaces over $G_{1},G_{2}$ we denote by $X\otimes Y$ the algebraic
tensor product of $X,Y$, that is the set of all functions
$u(x,y)=\sum_{i=1}^{n}f_{i}(x)g_{i}(y)$ where $f_{i}\in X,g_{i}\in Y$ and
$x\in G_{1},y\in G_{2}$. If $T,S$ are linear operators on $X,Y$ we denote by
$T\otimes S$ the operator on $X\otimes Y$ defined by

$\left(\left(T\otimes
S\right)u\right)(x,y)=\sum_{i=1}^{n}(Tf_{i})(x)(Sg_{i})(y)$

and we keep the same notation to denote its bounded extension to the
completion of $X\otimes Y$, if no ambiguity can arise. The generation result
for $\mathcal{L}$ follows from well-known general facts. We start by two
preliminary lemmas where the tensor product of a semigroup $(e^{tA})_{t\geq
0}$ with the identity operator is considered. The first follows from [23, AI,
Section 3.7].

###### Lemma 8.1

Let $(e^{tA})_{t\geq 0}$ be the semigroup generated by $(A,D(A))$ in
$L^{p}(\Omega,d\mu)$. The family $(e^{tA})_{t\geq 0}\otimes I)_{t\geq 0}$ on
$L^{p}(\Omega\times\Lambda,d\mu\otimes
d\nu)=\overline{L^{p}(\Omega,d\mu)\otimes L^{p}(\Lambda,d\nu)}$ is a semigroup
generated by the closure $\overline{A\otimes I}$ of the operator $A\otimes I$
initially defined on $D(A)\otimes L^{p}(\Lambda)$.

Let us introduce the operator $(A^{\otimes},D(A^{\otimes}))$

$\displaystyle D(A^{\otimes})$ $\displaystyle:=\\{u\in
L^{p}(\Omega\times\Lambda):\ u(\,\cdot\,,y)\in D(A)\ \textrm{for almost
every}\ y\in\Lambda,\ Au(\cdot,y)\in L^{p}(\Omega\times\Lambda)\\}$
$\displaystyle A^{\otimes}u(\,\cdot\,,y)$
$\displaystyle:=Au(\,\cdot\,,y),\quad\textrm{for almost every}\ y\in\Lambda.$

We can identify the domain of the generator as follows.

###### Lemma 8.2

Keeping the notation of the previous Lemma, we have $\overline{A\otimes
I}=A^{\otimes}$.

Proof. We start by proving that $A^{\otimes}$ is a closed operator. Let
$(u_{n})\in D(A^{\otimes})$ such that $u_{n}\to u$, $A^{\otimes}u_{n}\to v$ in
$L^{p}(\Omega\times\Lambda)$. Up to considering a subsequence,
$u_{n}(\cdot,y)\to u(\cdot,y)$, $Au_{n}(\cdot,y)\to v(\cdot,y)$ for almost
every $y\in\Lambda$. Then, since $A$ is closed, $u(\cdot,y)\in D(A)$ and
$v(\cdot,y)=Au(\cdot,y)$ for almost every $y\in\Lambda$. It follows $u\in
D(A^{\otimes})$ and $A^{\otimes}u=v$. Next we prove that $\overline{A\otimes
I}\subset A^{\otimes}$. By the definition, it easily follows that $A\otimes
I\subset A^{\otimes}$. Since $A^{\otimes}$ is closed by the previous step, the
inclusion $\overline{A\otimes I}\subset A^{\otimes}$ is proved. Finally we
prove that, for $\lambda$ large enough, the operator $\lambda-A^{\otimes}$ is
injective. Indeed, if $u\in D(A^{\otimes})$ and $\lambda u-A^{\otimes}u=0$,
then $\lambda u(\cdot,y)-Au(\cdot,y)=0$ for almost every $y\in\Lambda$ and, by
the injectivity of $A$, $u(\cdot,y)=0$ for almost every $y\in\Lambda$.

We therefore deduce the following generation result, under the assumption that
$A_{m,p}$ generates in $L^{p}_{m}(0,\infty)$. Here $A_{m,p}$ denotes the
degenerate operator $L_{m,p}$ of Section 4, with Dirichlet boundary
conditions, or the Bessel operator $B^{n}_{m,p}$ of Section 3, under Neumann
boundary conditions. We still write $A_{y}$ to indicate that $A$ acts only in
the $y$ variable.

###### Lemma 8.3

Let $(e^{z\Delta_{x}})_{z\in\Sigma_{\phi}}$,
$(e^{zA_{y}})_{z\in\Sigma_{\phi}}$ be the semigroups generated by
$\left(\Delta_{x},W^{2,p}(\mathbb{R}^{N})\right)$ in $L^{p}(\mathbb{R}^{N})$,
$\left(A_{m,p},D(A_{m,p})\right)$ in $L_{m}^{p}(0,\infty)$, respectively. The
operators $\Delta_{x}^{\otimes}$ and $A_{m,p}^{\otimes}$ defined by

$\displaystyle D(\Delta_{x}^{\otimes}):=\Big{\\{}$ $\displaystyle u\in
L^{p}_{m}(\mathbb{R}_{+}^{N+1}):\ u(\,\cdot\,,y)\in W^{2,p}(\mathbb{R}^{N})\
\textrm{for a.e.}\ y\in(0,\infty),\ \nabla_{x}u(\cdot,y),\Big{.}$
$\displaystyle\Big{.}D^{2}_{x}u(\cdot,y)\in
L^{p}_{m}(\mathbb{R}_{+}^{N+1}),\quad\Delta_{x}^{\otimes}u(\,\cdot\,,y):=\Delta_{x}u(\,\cdot\,,y),\quad\textrm{for
almost every}\ y\in(0,\infty)\Big{\\}};$ $\displaystyle
D(A_{m,p}^{\otimes}):=\Big{\\{}$ $\displaystyle u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+}):\ u(x,\cdot)\in D(A_{m,p})\ \textrm{for a.e.}\
x\in\mathbb{R}^{N},\ A_{y}u(x,\cdot)\Big{.}$ . $\displaystyle\in
L^{p}_{m}(\mathbb{R}_{+}^{N+1}),\quad
A_{m,p}^{\otimes}u(x,\cdot):=A_{y}u(x,\cdot),\quad\textrm{for almost every}\
x\in\mathbb{R}^{N}\Big{\\}}$

generate the semigroups $(e^{z\Delta_{x}}\otimes I)_{z\in\Sigma_{\phi}}$,
$(I\otimes e^{zA_{y}})_{z\in\Sigma_{\phi}}$ in
$L^{p}_{m}(\mathbb{R}_{+}^{N+1})$.

###  8.2 Maximal regularity and domain characterization

We can finally prove maximal regularity and domain characterization for
$\mathcal{L}=\Delta_{x}+L_{y}$ and $\mathcal{L}=\Delta_{x}+B^{n}_{y}$. Both
cases have similar proofs but some details are different since the domain of
$B^{n}$ is more regular. In the gradient estimates for $B^{n}$, in fact, the
factor $y^{-s_{1}-1}$ does not appear, see Proposition 2.9, in contrast with
Proposition 4.5 where it is assumed that $s_{1}\neq 0$. However, if $s_{1}=0$,
then $b=0$ and $c\geq 1$ so that $L=B^{d}=B^{n}$. Therefore, we distinguish
between the cases of $L$ with $s_{1}\neq 0$ and $B^{n}$, the case of $L$ with
$s_{1}=0$ being included in the last.

###  8.3 $\mathcal{L}=\Delta_{x}+L_{y}$ with $s_{1}\neq 0$

First we state a $\mathcal{R}$-boundedness result for the heat kernel of $L$
and its gradient. We remark that the $\mathcal{R}$-boundedness of the heat
kernel has been also proved in [17].

###### Theorem 8.4

If $s_{1}<\frac{m+1}{p}<s_{2}+2$, then the family
$(e^{zL_{m,p}})_{z\in\Sigma_{\phi}}$, $\phi<\pi/2$ is $\mathcal{R}$-bounded in
$L^{p}_{m}(0,\infty)$, hence $L_{m,p}$ has maximal regularity.

If $s_{1}+1<\frac{m+1}{p}<s_{2}+2$, then families
$(\frac{\sqrt{z}}{y}e^{zL_{m,p}})_{z\in\Sigma_{\phi}}$,
$(\sqrt{z}D_{y}e^{zL_{m,p}})_{z\in\Sigma_{\phi}}$, $\phi<\pi/2$ are
$\mathcal{R}$-bounded in $L^{p}_{m}(0,\infty)$.

Proof. By Proposition 4.5 we have

$|e^{zL_{m,p}}f|\leq CS^{\alpha,\beta}(c|z|)|f|$

pointwise, for $\alpha=s_{1}$, $\beta=s_{1}-c$ and suitable positive constants
$C,c$ (note also that since $s_{1}+s_{2}=c-1$, then $1-\beta=s_{2}+2$). The
assertion for $(e^{zL_{m,p}})_{z\in\Sigma_{\phi}}$ then follow from Theorem
7.7 together with Corollary 6.2.

Those for $(\frac{\sqrt{z}}{y}e^{zL_{m,p}})_{z\in\Sigma_{\phi}}$,
$(\sqrt{z}D_{y}e^{zL_{m,p}})_{z\in\Sigma_{\phi}}$ are proved in a similar way,
using Proposition 4.5 and setting setting $\alpha=s_{1}+1$ and
$\beta=s_{1}-c$.

###### Proposition 8.5

Let $s_{1}<\frac{m+1}{p}<s_{2}+2$. Then the closure of the operator
$\mathcal{L}$, initially defined on $W^{2,p}(\mathbb{R}^{N})\otimes
D(L_{m,p})$, generates a bounded analytic semigroup of angle $\pi/2$,
$(e^{z\mathcal{L}})_{z\in C_{+}}$, in $L^{p}_{m}(\mathbb{R}_{+}^{N+1})$ which
has maximal regularity.

Proof. Observe first that

$\mathcal{L}=\Delta_{x}\otimes I+I\otimes L_{y}$

on $W^{2,p}(\mathbb{R}^{N})\otimes D(L_{m,p})$. The family
$(e^{z\Delta_{x}}\otimes e^{zL_{y}})_{z\in\mathbb{C}_{+}}$ is a semigroup and
leaves $W^{2,p}(\mathbb{R}^{N})\otimes D(L_{m,p})$ invariant. This last, being
dense, is then a core for the generator. The $\mathcal{R}$-boundedness of the
family $(e^{z\mathcal{L}})_{z\in\Sigma_{\phi}}=(e^{z\Delta_{x}}\otimes
e^{zL_{y}})_{z\in\Sigma_{\phi}}$, $\phi<\pi/2$ follows by composition writing
$(e^{z\Delta_{x}}\otimes e^{zL_{y}})=(e^{z\Delta_{x}}\otimes I)\circ(I\otimes
e^{zL_{y}})$, using the above theorem and the $\mathcal{R}$-boundedness of
$(e^{z\Delta_{x}})_{z\in\Sigma_{\phi}}$.

We note that, by construction, $e^{t\mathcal{L}}$ consists of integral
operators. For $t>0$,
$z_{1}=(x_{1},y_{1}),z_{2}=(x_{2},y_{2})\in\mathbb{R}^{N+1}_{+}$

$\displaystyle e^{t\mathcal{L}}f(z_{1})$
$\displaystyle=\int_{R^{N+1}_{+}}p(t,z_{1},z_{2})f(z_{2})dm(z_{2}),\quad f\in
L^{p}_{m}\left(\mathbb{R}^{N+1}\right)$ $\displaystyle p(t,z_{1},z_{2})$
$\displaystyle=(4\pi
t)^{-\frac{N}{2}}e^{-\frac{|x_{1}-x_{2}|^{2}}{4t}}p_{L_{y}}(t,y_{1},y_{2})$
$\displaystyle\simeq
t^{-\frac{N+1}{2}}\left(\frac{|y_{1}|}{t^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}}\left(\frac{|y_{2}|}{t^{\frac{1}{2}}}\wedge
1\right)^{-s_{1}+c}\exp\left(-\frac{|z_{1}-z_{2}|^{2}}{\kappa t}\right).$

Before describing the domain of $\mathcal{L}_{m,p}$, let us show how the
results for the $1d$ operator $L_{y}$ give easily a core.

###### Proposition 8.6

If $s_{1}<\frac{m+1}{p}<s_{2}+2$, then

$\mathcal{D}=\left\\{u=y^{-s_{1}}v:v\in
C_{c}^{\infty}(\mathbb{R}^{N}\times[0,\infty)),\ D_{y}v(x,0)=0\right\\}$

is a core for $D(\mathcal{L}_{m,p})$.

Proof. $C_{c}^{\infty}(\mathbb{R}^{N})$ is a core for $\Delta_{x}$ and
$\mathcal{D}_{1}=\left\\{u=y^{-s_{1}}v:v\in C_{c}^{\infty}[0,\infty),\
D_{y}v(0)=0\right\\}$ is a core for $L_{m,p}$, by Proposition 5.3. Then
$C_{c}^{\infty}(\mathbb{R}^{N})\otimes\mathcal{D}_{1}\subset\mathcal{D}$ is
dense in $W^{2,p}(\mathbb{R}^{N})\otimes D(L_{m,p})$ for the norm
$\|u\|=\|u\|_{L^{p}_{m}}+\|\Delta_{x}u\|_{L^{p}_{m}}+\|L_{y}u\|_{L^{p}_{m}}$,
hence for the graph norm induced by $\mathcal{L}$. Since
$W^{2,p}(\mathbb{R}^{N})\otimes D(L_{m,p})$ is a core for $\mathcal{L}_{m,p}$,
the proof is complete.

###### Theorem 8.7

Let $D\geq 0$, $s_{1}<\frac{m+1}{p}<s_{2}+2$. Then

$D(\mathcal{L}_{m,p})=\Big{\\{}u\in
W^{2,p}_{loc}(\mathbb{R}^{N+1}_{+}):u,\nabla_{x}u,D^{2}_{x}u,L_{y}u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+})\Big{\\}}.$

Proof.  Observe that, by construction,

$\mathcal{S}(\mathbb{R}^{N})\otimes D(L_{m,p})\subset
W^{2,p}(\mathbb{R}^{N})\otimes D(L_{m,p})\subset D(\Delta_{x}^{\otimes})\cap
D(L_{m,p}^{\otimes})\subset D(\mathcal{L}_{m,p})$ (33)

where $\mathcal{S}(\mathbb{R}^{N})$ denotes the Schwartz class. Note that
$\mathcal{S}(\mathbb{R}^{N})\otimes D(L_{m,p})$ is a core for
$\mathcal{L}_{m,p}$ by the above proposition (or also since it is invariant
for $(e^{z\Delta_{x}}\otimes e^{zL_{y}})_{z\in\mathbb{C}_{+}}$).

We endow $D(\mathcal{L}_{m,p})$ with the graph norm and
$Z:=D(\Delta_{x}^{\otimes})\cap D(L_{m,p}^{\otimes})$ with the norm

$\|u\|_{Z}=\|u\|_{L^{p}_{m}}+\|\Delta_{x}u\|_{L^{p}_{m}}+\|L_{y}u\|_{L^{p}_{m}},\
\ u\in Z,$

so that the embedding $Z\subset D(\mathcal{L}_{m,p})$ is continuous. Let us
show that the graph norm and the norm of $Z$ are equivalent on
$\mathcal{S}(\mathbb{R}^{N})\otimes D(L_{m,p})$. Let
$u\in\mathcal{S}(\mathbb{R}^{N})\otimes D(L_{m,p})$ and $f=\mathcal{L}u$. By
taking the Fourier transform with respect to $x$ (with co-variable $\xi$) we
obtain

$(-|\xi|^{2}+L_{y})\hat{u}(\xi,\cdot)=\hat{f}(\xi,\cdot),\qquad|\xi|^{2}\hat{u}(\xi,\cdot)=-|\xi|^{2}(|\xi|^{2}-L_{y})^{-1}\hat{f}(\xi,\cdot).$

This means $\Delta_{x}u=-{\cal F}^{-1}M(\xi){\cal F}f$, where ${\cal F}$
denotes the Fourier transform and $M(\xi)=|\xi|^{2}(|\xi|^{2}-L_{y})^{-1}$.

The estimate $\|\Delta_{x}u\|_{p}\leq C\|f\|_{p}$ (norms on
$L^{p}_{m}(\mathbb{R}^{N+1})$) follows from the boundedness of the multiplier
$M$ in $L^{p}(\mathbb{R}^{N};L^{p}_{m}(0,\infty))$ which we prove using
Theorem 6.4. In fact, since $(e^{tL_{y}})_{t\geq 0}$ is $\mathcal{R}$-bounded
by Theorem 8.4, then the family

$\Gamma(\lambda)=\lambda(\lambda-L_{y})^{-1}=\int_{0}^{\infty}\lambda
e^{-\lambda t}e^{tL_{y}}\,dt,\quad\lambda>0$

is $\mathcal{R}$-bounded by [13, Corollary 2.14] and satisfies Mikhlin
condition in Theorem 6.4 for every $N$, by the resolvent equation (or arguing
as in Lemma 8.9 below). The same then holds for $M(\xi)=\Gamma(|\xi|^{2})$, as
readily verified.

The estimate $\|L_{y}u\|_{p}\leq C\|f\|_{p}$ follows by difference and shows
the equivalence of the graph norm and of the norm of $Z$ on
$\mathcal{S}(\mathbb{R}^{N})\otimes D(L_{m,p})$. If $u\in
D(\mathcal{L}_{m,p})$, let $(u_{n})\subset\mathcal{S}(\mathbb{R}^{N})\otimes
D(L_{m,p})$ converge to $u$ with respect to the graph norm. Then $(u_{n})$ is
a Cauchy sequence with respect to the norm of $Z$, hence $u\in Z$ and the
equivalence of the corresponding norms extends to $Z=D(\mathcal{L}_{m,p})$.

A more detailed description of the domain of $\mathcal{L}$ follows immediately
from the above theorem and the domain description of $D(L_{m,p})$ of Section
4. We do not list all the results that can be obtained in this way, since this
is straightforward. See however Corollary 8.10 below for an important case.

When $(m+1)/p>s_{1}+1$ and $u\in D(\mathcal{L}_{m,p})$, the mixed derivatives
$D_{y}\nabla_{x}u$ belong to $L^{p}_{m}$ even though $D_{yy}u$ could be not in
$L^{p}_{m}$.

###### Theorem 8.8

Let $D\geq 0$, $s_{1}+1<\frac{m+1}{p}<s_{2}+2$. Then

$D(\mathcal{L}_{m,p})=\\{u\in
W^{2,p}_{loc}(\mathbb{R}^{N+1}_{+}):u,\nabla_{x}u,D^{2}_{x}u,\frac{\nabla_{x}u}{y},D_{y}\nabla_{x}u,L_{y}u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+})\\}.$

Proof. We proceed as in Theorem 8.7 to estimate $D_{y}\nabla_{x}u$ for
$u\in\mathcal{S}(\mathbb{R}^{N})\otimes D(L_{m,p})$. We have

$(-|\xi|^{2}+L_{y})\hat{u}(\xi,\cdot)=\hat{f}(\xi,\cdot),\qquad\xi
D_{y}\hat{u}(\xi,\cdot)=-\xi D_{y}(|\xi|^{2}-L_{y})^{-1}\hat{f}(\xi,\cdot).$

and this time the multiplier is

$M(\xi)=-\xi
D_{y}(|\xi|^{2}-L_{y})^{-1}=-\xi\int_{0}^{\infty}e^{-|\xi|^{2}t}D_{y}e^{tL_{y}}\,dt=-\xi\int_{0}^{\infty}\frac{e^{-|\xi|^{2}t}}{\sqrt{t}}S_{t}\,dt$

where $(S_{t})_{t\geq 0}=(\sqrt{t}D_{y}e^{tL_{y}})_{t\geq 0}$ is $R$-bounded,
by Theorem 8.4. The Mikhlin condition of Theorem 6.4 follows from [13,
Corollary 2.14] and the lemma below.

The proof for $y^{-1}\nabla_{x}u$ is similar.

###### Lemma 8.9

Let $h(\xi,t)=\frac{\xi}{\sqrt{t}}e^{-|\xi|^{2}t}$. Then if $|\alpha|=k$

$|\xi|^{k}\int_{0}^{\infty}|D^{\alpha}_{\xi}h(\xi,t)|dt\leq C_{k}.$

Proof.  Let $g(\eta)=\eta e^{-|\eta|^{2}}$ so that
$h(\xi,t)=\frac{1}{t}g(\xi\sqrt{t})$ . Let us observe that for $|\alpha|=k>0$
is $|D^{\alpha}g(\eta)|=P^{k+1}(\eta)e^{-|\eta|^{2}}$ where $P^{k+1}$ is a
vector polynomial of degree $k+1$. This implies in particular that

$|\eta|^{k-1}|D^{\alpha}_{\eta}g(\eta)|\leq
C_{k}e^{-|\eta|^{2}},\qquad\text{for}\quad k\geq 0$

which yields

$|D^{\alpha}_{\xi}h(\xi,t)|=t^{\frac{k}{2}-1}|D^{\alpha}_{\xi}g(\xi\sqrt{t})|\leq
C_{k}t^{-\frac{1}{2}}|\xi|^{1-k}e^{-|\xi|^{2}t}.$

Then one has

$\displaystyle|\xi|^{k}\int_{0}^{\infty}|D^{\alpha}_{\xi}h(\xi,t)|dt$
$\displaystyle\leq
C_{k}|\xi|\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-|\xi|^{2}t}dt=C_{k}\int_{0}^{\infty}s^{-\frac{1}{2}}e^{-s}ds=C_{k}\sqrt{\pi}.$

Finally, if $(m+1)/p>s_{1}+2$, then $D(\mathcal{L}_{m,p})$ has the maximal
regularity one can expect.

###### Corollary 8.10

Let $D>0$, $s_{1}+2<\frac{m+1}{p}<s_{2}+2$. Then

$D(\mathcal{L}_{m,p})=\\{u\in
W_{m}^{2,p}(\mathbb{R}^{N+1}_{+}):\frac{u}{y^{2}},\frac{\nabla u}{y}\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+})\\}$

.

Proof.  We have only to show that $y^{-2}u,y^{-1}D_{y}u,D_{yy}\in L^{p}_{m}$,
since the rest follows from Theorem 8.8. Using Proposition 4.2 with $\theta=1$
we get

$\int_{0}^{\infty}\left(|D_{yy}u(x,y)|^{p}+\frac{|D_{y}u(x,y)|^{p}}{y^{p}}+\frac{|u(x,y)|^{p}}{y^{2p}}\right)y^{m}dy\leq
C\int_{0}^{\infty}|L_{y}u(x,y)|^{p}y^{m}dy$

(the additional term containing $u$ on the right hand side does not appear, by
homogeneity reasons). Integrating with respect to $x\in\mathbb{R}^{N}$ and
using the estimate $\|L_{y}u\|_{L^{p}_{m}(\mathbb{R}^{N+1}_{+})}\leq
C\|\mathcal{L}u\|_{L^{p}_{m}(\mathbb{R}^{N+1}_{+})}$ the proof is complete.

###  8.4  $\mathcal{L}=\Delta_{x}+L_{y}$ with $s_{1}=0$ and
$\mathcal{L}^{n}=\Delta_{x}+B^{n}_{y}$

If $s_{1}=0$ then $b=0$, $c\geq 1$ and $L=B=D_{yy}+\frac{c}{y}D_{y}$ is a
Bessel operator. Since $c\geq 1$, then $B^{d}=B^{n}$, see Section 2, and
therefore it is sufficient to deal with $B^{n}$. Note, however, that
$s_{1}=c-1$ for $B^{n}$ when $c<1$.

###### Theorem 8.11

If $c>-1$ and $\frac{m+1}{p}\in(0,c+1)$, then the families
$(e^{zB^{n}_{m,p}})_{z\in\Sigma_{\phi}}$,
$(\sqrt{z}D_{y}e^{zB^{n}_{m,p}})_{z\in\Sigma_{\phi}}$, $\phi<\pi/2$ are
$R$-bounded in $L^{p}_{m}(0,\infty)$. In particular, $B^{n}_{m,p}$ has maximal
regularity.

Proof. All the assertions follow from Theorem 7.7 and the heat kernel
estimates of Propositions 2.8, 2.9, setting $\alpha=0$ or $\alpha=-1$ and
$\beta=-c$.

###### Proposition 8.12

If $c>-1$ and $0<\frac{m+1}{p}<c+1$, then

$\mathcal{D}=\left\\{v\in C_{c}^{\infty}(\mathbb{R}^{N}\times[0,\infty)),\
D_{y}v(x,0)=0\right\\}$

is a core for $D(\mathcal{L}^{n}_{m,p})$.

Proof. Identical to that of Proposition 8.6

###### Theorem 8.13

Let $c>-1$, $0<\frac{m+1}{p}<c+1$. Then
$\mathcal{L}^{n}_{m,p}=\Delta_{x}+B^{n}_{y}$ with domain

$D(\mathcal{L}^{n}_{m,p})=\\{u\in
W_{m}^{2,p}(\mathbb{R}^{N+1}_{+}):\frac{D_{y}u}{y}\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+})\\}$

has maximal regularity in $L^{p}_{m}(\mathbb{R}^{N+1}_{+})$. In particular, if
$\frac{m+1}{p}<1$ then

$D(\mathcal{L}^{n}_{m,p})=\\{u\in
W_{m}^{2,p}(\mathbb{R}^{N+1}_{+}):D_{y}u(x,0)=0,\quad x\in\mathbb{R}^{N}\\}$

and if $\frac{m+1}{p}>1$ then

$D(\mathcal{L}^{n}_{m,p})=W_{m}^{2,p}(\mathbb{R}^{N+1}_{+}).$

Proof.  We apply Theorem 8.11 as in Proposition 8.5 and Theorem 8.7 to prove
that $\mathcal{L}^{n}_{m,p}$ has maximal regularity and is closed on the
intersection of the domains of $\Delta_{x}$ and $B^{n}_{y}$. Then the same
argument as in Theorem 8.8 yield the $L^{p}_{m}$ boundedness of the mixed
derivatives. By the domain characterization of $B^{n}$ and the closedness of
$\Delta_{x}+B^{n}_{y}$ again we finally have $D_{yy}u,y^{-1}D_{y}u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+})$ for $u\in D(B^{n}_{m,p})$.

When $(m+1)/p>1$, by Hardy inequality of Proposition 11.6 and the equality
$W^{1,p}_{m}=W^{1,p}_{0,m}$ of Proposition 11.4, we get
$D(\mathcal{L}_{m,p})=W_{m}^{2,p}(\mathbb{R}^{N+1}_{+})$. Instead, if
$(m+1)/p<1$, then $(D_{y}u)/y\in L^{p}_{m}$ and $D_{y}u(x,0)=0$ are equivalent
for $u\in W^{2,p}_{m}$. In fact, if $D_{y}u(x,0)=0$, then $D_{y}u\in
W^{1,p}_{0,m}$ and $(D_{y}u)/y\in L^{p}_{m}$ by Proposition 11.6. Conversely,
if $(D_{y}u)/y\in L^{p}_{m}$, since
$\int_{0}^{\infty}\frac{|D_{y}u(x,y)|^{p}}{y^{p}}\,dy=\infty$ whenever
$D_{y}u(x,0)\neq 0$, we get $D_{y}u(x,0)=0$ a.e.

Note that the closedness of $\Delta_{x}+B_{y}$ and the domain characterization
of $B^{n}$ allow to conclude that $\Delta_{x}u,D_{yy}u\in
L^{p}_{m}(\mathbb{R}^{N+1})$, for $u\in D(\mathcal{L}^{n}_{m,p})$, hence
$D_{x_{i}x_{j}}u\in L^{p}_{m}(\mathbb{R}^{N+1})$ by the Calderón-Zygmund
inequality in $\mathbb{R}^{N}$. However, to deduce that the mixed derivatives
$D_{x_{i}y}u$ belongs to $L^{p}_{m}(\mathbb{R}^{N+1})$ one needs the
boundedness of singular integrals in $L^{p}_{m}(\mathbb{R}^{N+1})$ which
usually requires that the one dimensional weight $|y|^{m}$ belongs to
$A_{p}(\mathbb{R}^{N+1})$, that is $0<(m+1)/p<1$, a restriction that does not
appear in the above theorem.

##  9 Rellich inequalities

Rellich inequalities have been intensively studied in any dimension, also for
degenerate operators, but here we recall only the 1d result. All the $L^{p}$
norms in this section are taken with respect to the Lebesgue measure and
accordingly we write $L_{p},\mathcal{L}_{p}$ for $L_{m,p},\mathcal{L}_{m,p}$,
when $m=0$. We set for $1\leq p\leq\infty$

$\displaystyle\gamma_{p}:$
$\displaystyle=\left(\frac{1}{p}-2\right)\left(\frac{1}{p^{\prime}}+c\right)$

and

$\displaystyle{\cal P}_{p}:$
$\displaystyle=\left\\{\lambda=-\xi^{2}+i\xi\left(3-\frac{2}{p}+c\right)-\gamma_{p}\;;\;\xi\in\mathbb{R}\right\\}.$
(34)

Observe that ${\cal P}_{p}$ is a parabola with vertex at $-\gamma_{p}$ when
$3-\frac{2}{p}+c\neq 0$, otherwise coincides with the semiaxis
$]-\infty,-\gamma_{p}]$.

###### Theorem 9.1

([19, Section 3],[15, Section 4]) There exists a positive constant $C$ such
that

$\left\||y|^{-2}u\right\|_{p}\leq C\|Lu\|_{p}$ (35)

holds for every $u\in D(L_{p,max})$ such that $u/|y|^{2}\in L^{p}(0,\infty)$,
if and only if, $b\not\in{\cal P}_{p}$.

Observe that

$b+\gamma_{p}=\left(\frac{1}{p}-s_{1}-2\right)\left(s_{2}+2-\frac{1}{p}\right)$

so that $b\not\in{\cal P}_{p}$ means $\frac{1}{p}\neq s_{1}+2,s_{2}+2$ when
$3-\frac{2}{p}+c\neq 0$ and $\frac{1}{p}\in(s_{1}+2,s_{2}+2)$ when
$3-\frac{2}{p}+c=0$.

When $s_{1}+2<\frac{1}{p}<s_{2}+2$, independently of the value of
$3-\frac{2}{p}+c$, Rellich inequalities can be proved by integrating by parts,
see the Remark below. In the other cases the proof relies on spectral theory
and best constants are known only in special cases. We refer the reader to
[19, Section 3], again.

In the next result we show that Rellich inequalities hold for
$\mathcal{L}=\Delta_{x}+L_{y}$ ($L_{y}=L$) in the generation interval
$s_{1}<\frac{1}{p}<s_{2}+2$ if and only if they hold for $L_{y}$. In the
proof, the closedness of the sum $\Delta_{x}+L_{y}$ on the intersection of the
corresponding domains plays a major role.

Note that the theorem below (as that above) does not concern with Rellich
inequalities in the whole domain of $\mathcal{L}$, as described in the
previous section, but on the (possibly) smaller subspace of all $u$ in the
maximal domain, satisfying $u/|y|^{2}\in L^{p}$.

###### Theorem 9.2

Assume $s_{1}<\frac{1}{p}<s_{2}+2$. There exists $C>0$ such that

$\left\||y|^{-2}|u|\right\|_{p}\leq C\|\mathcal{L}u\|_{p}$ (36)

holds for every $u\in D(\mathcal{L}_{p,max})$ such that $u/|y|^{2}\in
L^{p}(\mathbb{R}_{+}^{N+1})$, if and only if $b\not\in{\cal P}_{p}$.

Proof. Assume that Rellich inequalities hold for the complete operator
$\mathcal{L}$ and let $u(x,y)=z(x)\psi(y)$ with
$\|z\|_{L^{p}(\mathbb{R}^{N})}=1$. Then

$\mathcal{L}u=\Delta_{x}u+L_{y}u=\psi\Delta_{x}z+zL_{y}\psi$

and (36) is equivalent to

$\int_{0}^{\infty}\frac{|\psi(y)|^{p}}{|y|^{2p}}\,dy=\int_{0}^{\infty}|z(x)|^{p}\,dx\int_{0}^{\infty}\frac{|\psi(y)|^{p}}{|y|^{2p}}\,dy\leq
C\int_{\mathbb{R}^{N+1}}\left(|\psi(y)\Delta_{x}z(x)+z(x)L_{y}\psi(y)\right|)^{p}\,dx\,dy.$

Let $z_{R}(x)=R^{-\frac{N}{p}}z\left(\frac{x}{R}\right)$. Then
$\|z_{R}\|_{L^{p}(\mathbb{R}^{N})}=1$, $\|\Delta_{x}z_{R}\|_{p}\to 0$ as
$R\to\infty$ and letting $R\to\infty$

$\int_{0}^{\infty}\frac{|\psi(y)|^{p}}{|y|^{2p}}\,dy\leq
C\int_{0}^{\infty}|L_{y}\psi(y)|^{p}\,dy$

so that Rellich inequalities hold for $L_{y}$.

Next, assume that Rellich inequalities hold for $L_{y}$ and let $u\in
D(\mathcal{L}_{p,max})$ be such that $u/|y|^{2}\in L^{p}(\mathbb{R}^{N+1})$.
Then $u\in D(\mathcal{L}_{p})=D(\Delta_{x})\cap D(L_{y,p})$ and for almost all
$x\in\mathbb{R}^{N}$, $u(x,\cdot)\in D(L_{p,max})$ and $u(x,\cdot)/|y|^{2}\in
L^{p}((0,\infty))$. Then

$\int_{0}^{\infty}\frac{|u(x,y)|^{p}}{|y|^{2p}}\,dy\leq
C\int_{0}^{\infty}|L_{y}u(x,y)|^{p}\,dy$

and, integrating with respect to $x\in\mathbb{R}^{N}$ and using the closedness
of $\Delta_{x}+L_{y}$ we get

$\int_{\mathbb{R}^{N+1}}\frac{|u(x,y)|^{p}}{|y|^{2p}}\,dx\,dy\leq
C\int_{\mathbb{R}^{N+1}}|L_{y}u(x,y)|^{p}\,dx\,dy\leq
C\int_{\mathbb{R}^{N+1}}|\mathcal{L}u(x,y)|^{p}\,dx\,dy.$

###### Remark 9.3

When $s_{1}+2<\frac{1}{p}<s_{2}+2$ the best constant $C$ above is
$(b+\gamma_{p})^{-1}>0$. This can be seen multiplying $\mathcal{L}u$ by
$u|u|^{p-2}y^{2-2p}$ and integrating by parts, assuming $u$ smooth and with
support faraway from $\\{y=0\\}$).

##  10 Appendix A: auxiliary inequalities

###### Lemma 10.1

For every $\varepsilon>0$ there exists $C>0$ such that for $r,s>0$

$(1\wedge r)(1\wedge s)\leq 1\wedge rs\leq C(1\wedge r)(1\wedge
s)\,e^{\epsilon|r-s|^{2}}.$

Proof. $(1\wedge rs)=(1\wedge r)(1\wedge s)$ when $r,s\leq 1$ or $r,s\geq 1$.
Assume that $s\leq 1\leq r$. Then $(1\wedge r)(1\wedge s)=s\leq 1\wedge rs$.
Conversely, if $rs\leq 1$ then $1\wedge rs=rs\leq Cse^{\varepsilon(r-s)^{2}}$
for a suitable $C>0$, since $s\leq 1\leq r$. If, instead, $rs\geq 1$, then
$1\wedge rs=1\leq Cr^{-1}e^{\varepsilon(r-s)^{2}}\leq
Cse^{\varepsilon(r-s)^{2}}$.

###### Lemma 10.2

If $\gamma_{1}\leq\gamma_{2}$ then for every $\varepsilon>0$ there exists
$C>0$ such that

$\frac{|y|^{\gamma_{1}}}{|z|^{\gamma_{2}}}\leq C\frac{(|y|\wedge
1)^{\gamma_{1}}}{(|z|\wedge
1)^{\gamma_{2}}}\exp\left(\epsilon|y-z|^{2}\right).$ (37)

Proof.  If $|y|\leq 1$ and $|z|\leq 1$ this is clearly true. Assume that
$|z|\leq 1\leq|y|$. Then $|y-z|^{2}\geq(|y|-1)^{2}$ and

$|y|^{\gamma_{1}}\leq C\exp\left\\{\epsilon(|y|-1)^{2}\right\\}\leq
C\exp\left\\{\epsilon(|y-z|)^{2}\right\\}$

and (37) holds. If $|y|\leq 1\leq|z|$ we argue in similar way. Finally, when
$|y|\geq 1$, $|z|\geq 1$ we write $y=r\omega,z=\rho\eta$ with
$|\omega|=|\eta|=1$. The left hand side of (37) is then
$(r/\rho)^{\gamma_{1}}\rho^{\gamma_{2}-\gamma_{1}}\leq(r/\rho)^{\gamma_{1}}$
which is now symmetric in $r,\rho$. Assuming that $r\geq\rho\geq 1$ we write
$r=s\rho$ with $s\geq 1$, the inequality $s^{\gamma_{1}}\leq
Ce^{\epsilon(s-1)^{2}}\leq Ce^{\epsilon(s-1)^{2}\rho^{2}}$ ($s,\rho\geq 1$)
implies that

$\frac{|y|^{\gamma_{1}}}{|z|^{\gamma_{2}}}\leq\left(\frac{r}{\rho}\right)^{\gamma_{1}}\leq
Ce^{\epsilon|r-\rho|^{2}}\leq Ce^{\epsilon|y-z|^{2}}.$

The following Hardy inequalities have been used several times throughout the
paper.

###### Lemma 10.3

  * (i)

When $c+1>\frac{m+1}{p}$ the map
$H_{1}f(y)=\frac{1}{y^{c+1}}\int_{0}^{y}f(s)s^{c}\,ds$ is bounded from
$L^{p}_{m}$ to itself.

  * (ii)

When $c+1<\frac{m+1}{p}$ the map
$H_{2}f(y)=\frac{1}{y^{c+1}}\int_{y}^{\infty}f(s)s^{c}\,ds$ is bounded from
$L^{p}_{m}$ to itself.

Proof. Take $f\geq 0$ and let $w=H_{1}f$. Then
$w(y)=\int_{0}^{1}f(ty)t^{c}\,dt$ and by Minkowski’s inequality

$\|w\|_{L^{p}_{m}}\leq\int_{0}^{1}t^{c}\left(\int_{0}^{\infty}f(ty)^{p}y^{m}dy\right)^{\frac{1}{p}}=\int_{0}^{1}t^{c-\frac{m+1}{p}}\left(\int_{0}^{\infty}f(x)^{p}y^{m}dx\right)^{\frac{1}{p}}=C\|f\|_{L^{p}_{m}}$

with $C=\left((c+1)-\frac{m+1}{p}\right)^{-1}$. The proof for $H_{2}$ is
similar.

##  11 Appendix B: Sobolev spaces with weights

For $1<p<\infty$ let $W^{k,p}_{m}(\mathbb{R}^{N+1}_{+})=\\{u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+}):\partial^{\alpha}u\in
L^{p}_{m}(\mathbb{R}^{N+1}_{+})\quad|\alpha|\leq k\\}$. We use often
$W^{k,p}_{m}$ thus omitting $\mathbb{R}^{N+1}_{+}$ and $W^{k,p}_{0,m}$ for the
closure of $C_{c}^{\infty}(\mathbb{R}^{N+1}_{+})$ in $W^{k,p}_{m}$.

###### Lemma 11.1

If $\frac{m+1}{p}<1$ then $L^{p}_{m}$ embeds into $L^{1}(Q\times(0,1))$, where
$Q$ is any cube of $\mathbb{R}^{N}$. It follows that $W^{1,p}_{m}$ embeds into