A.e. convergence and 2-weight inequalities for Poisson-Laguerre semigroups

Abstract

We find optimal decay estimates for the Poisson kernels associated with various Laguerre-type operators L. From these, we solve two problems about the Poisson semigroup $e^{-t\sqrt{L}}$. First, we find the largest space of initial data f so that $e^{-t\sqrt{L}}f(x)\rightarrow f(x)$ at ${\,a.e.\,}x$. Secondly, we characterize the largest class of weights w which admit 2-weight inequalities of the form $\Vert \sup _{0<t\le t_0}|e^{-t\sqrt{L}}f|\,\Vert _{L^p(v)}\lesssim \Vert f\Vert _{L^p(w)}$, for some other weight v.

1 Introduction

In this paper, we continue the research, started in [6, 7], about Poisson integrals associated with certain differential operators L, say symmetric and positive in $L^2(\Omega ,\mu )$. Namely, we are interested in the behavior of

$$\begin{aligned} u(t,x)=e^{-t\sqrt{L}}f(x) \end{aligned}$$

as a solution of the elliptic differential equation

$$\begin{aligned} \mathrm{(P)}\qquad \left\{ \begin{array}{ll}u_{tt}-Lu=0 &{} \\ u(0,x)=f(x),&{} \end{array}\right. \text{ in } \text{ the } \text{ half-plane } (0,\infty )\times \Omega .\end{aligned}$$

We shall study two questions which are closely related

• (i) find the largest class of functions f for which $\lim _{t\rightarrow 0^+}u(t,x)=f(x)$, ${\,a.e.\,}x\in \Omega$;

• (ii) establish 2-weight inequalities of the form

  $$\begin{aligned} \big \Vert \sup _{0<t\le t_0}|u(t,x)|\,\big \Vert _{L^p(v)}\lesssim \Vert f\Vert _{L^p(w)} \end{aligned}$$

  for the largest class of weights w for which a suitable v with this property exists.

In [6], we considered these questions in full detail when L is the Hermite operator. In this paper, we intend to do the same for the various Laguerre operators. We remark that to solve (i) or (ii) we shall need optimal decay estimates of the Poisson kernels $e^{-t\sqrt{L}}(x,y)$, at least in the y variable. These are new in the literature and in particular produce new results on a.e. convergence compared to those of Muckenhoupt [10] and Stempak [15], and also larger weight classes compared to those of Nowak in [11]. The kernel estimates may also have an independent interest in other problems; see, e.g., recent work by Liu and Sjögren [9].

We now state our results. For ${\alpha }>-1$ (fixed throughout the paper), we first consider the (Hermite-type) Laguerre operator in $L^2({{\mathbb R}_+}):=L^2((0,\infty ), \mathrm{d}y)$ given by

$$\begin{aligned} L=-\partial _{yy} + \Big [y^2 + \frac{{\alpha }^2-\tfrac{1}{4}}{y^2}\Big ] \,+\,2\mu , \quad \text{ where } \mu \ge -({\alpha }+1). \end{aligned}$$

(1.1)

The parameter $\mu$ will be useful later, when transferring the results to other Laguerre operators; see Table 1. We shall also consider a slightly more general family of partial differential equations, namely

$$\begin{aligned} \left\{ \begin{array}{ll}u_{tt} \,+\,\frac{1-2\nu }{t}\,u_t\,=\,L u &{} \\ u(t,0)=f,&{}\end{array}\right. \quad \text{ where } \nu >0. \end{aligned}$$

(1.2)

These pde’s appear in relation to the fractional powers of L, $f\mapsto L^\nu f$ (see, e.g., [16]). When $\nu =1/2$, we recover the original equation (P).

As discussed in [16], a candidate solution to (1.2) is given by the Poisson-like integral

$$\begin{aligned} P_tf(x):=\tfrac{t^{2\nu }}{4^\nu \Gamma (\nu )}\,\int _0^\infty e^{-\frac{t^2}{4u}}\,\big [e^{-uL}f\big ](x)\,\frac{\mathrm{d}u}{u^{1+\nu }},\quad t>0, \end{aligned}$$

(1.3)

which is subordinated to the “heat” semigroup $\{e^{-uL}\}_{u>0}$. Our first goal is to find the most general conditions on a function $f:{{\mathbb R}_+}\rightarrow {\mathbb C}$ so that $P_tf$ is a meaningful solution of (1.2). These are determined by the following key function

$$\begin{aligned} \Phi (y)=\left\{ \begin{array}{lll}\displaystyle \frac{y^{{\alpha }+\frac{1}{2}} \; e^{-\frac{y^2}{2}}}{(1+y)^{\mu +{\alpha }+1}[\ln (e+y)]^{1+\nu }} &{} &{} \,\text{ if } \mu >-({\alpha }+1)\\ &{} &{} \\ \displaystyle \frac{y^{{\alpha }+\frac{1}{2}}\; e^{-\frac{y^2}{2}}}{\,[\ln (e+y)]^{\nu }\,} &{} &{} \text{ if } \mu =-({\alpha }+1).\end{array}\right. \end{aligned}$$

(1.4)

We write $L^1(\Phi )$ for the space of measurable functions f in ${{\mathbb R}_+}$ with $\int _{{{\mathbb R}_+}}|f|\Phi <\infty$.

Theorem 1.1

Let ${\alpha }>-1$, $\nu >0$, L as in (1.1) and $\Phi$ as in (1.4). Then, for every $f\in L^1(\Phi )$ the function $u(t,x)=P_tf(x)$ in (1.3) is defined by an absolutely convergent integral such that

• (i) $u(t,x)\in C^\infty ((0,\infty )\times {{\mathbb R}_+})$ and satisfies the pde (1.2)

• (ii) $\lim _{t\rightarrow 0^+}u(t,x)=f(x)$ at every Lebesgue point x of f.

Conversely, if a function $f\ge 0$ is such that the integral in (1.3) is finite for some $(t,x)\in (0,\infty )\times {{\mathbb R}_+}$, then f must necessarily belong to $L^1(\Phi )$.

Our second question concerns more “quantitative” bounds for the solutions of (1.2), expressed in terms of the following local maximal operators

$$\begin{aligned} P_{t_0}^*f(x):=\sup _{0<t\le t_0}\big |P_tf(x)\big |\;, \quad \text{ with } t_0>0 \text{ fixed. }\end{aligned}$$

(1.5)

We shall prove estimates of the Poisson kernels, as in (1.9) below, which lead to local bounds of the form

$$\begin{aligned} P^*_{t_0}:L^1(\Phi )\rightarrow L^s_\mathrm{loc}\quad \text{ if } s<1 , {\quad \text{ and }\quad }P^*_{t_0}:L^1(\Phi )\cap L^p_\mathrm{loc}\rightarrow L^p_\mathrm{loc}\quad \text{ if } p>1.\end{aligned}$$

However, our main interest is to obtain global bounds in x, which we shall phrase through the following problem.

Problem 1

A 2-weight problem for the operator $P^*_{t_0}$. Given $1<p<\infty$, characterize the class of weights $w(x)>0$ such that $P^*_{t_0}$ maps $L^p(w)\rightarrow L^p(v)$ boundedly, for some other weight $v(x)>0$.

Our second main result gives a complete answer to Problem 1. For $p\in (1,\infty )$, we define the class of weights

$$\begin{aligned} D_p(\Phi ) \,=\,\Big \{ w(y)>0 {\,\,\,:\,\,\,}\big \Vert w^{-\frac{1}{p}}\,\Phi \,\big \Vert _{L^{p'}({{\mathbb R}_+})}<\infty \Big \}. \end{aligned}$$

(1.6)

Observe that $L^p(w)\subset L^1(\Phi )$ if and only if $w\in D_p(\Phi )$, so in view of Theorem 1.1, this is a necessary condition for Problem 1. Our second theorem shows that it is also sufficient.

Theorem 1.2

Let $1<p<\infty$ and $t_0>0$ be fixed. Then, for a weight $w(x)>0$ the condition $w\in D_p(\Phi )$ is equivalent to the existence of some other weight $v(x)>0$ such that

$$\begin{aligned} P^*_{t_0}:L^p(w)\rightarrow L^p(v) \quad \text{ boundedly }. \end{aligned}$$

(1.7)

Moreover, for every ${\varepsilon }>0$, we can choose an explicit weight $v\in D_{p+{\varepsilon }}(\Phi )$ satisfying (1.7).

We remark that Problem 1 is only a “one-side” problem, in contrast to the (more difficult) question of characterizing all pairs of weights (w, v) for which (1.7) holds. One-side problems were considered in the early 80s by Rubio de Francia [13] and Carleson and Jones [2] for various classical operators. Here, we shall follow the approach by the latter, which has the advantage of giving explicit expressions for the second weight v(x) (see Remark 6.2 below). This is also a novelty compared to [6], which was based on the non-constructive method of Rubio de Francia.

We can now briefly describe our approach to the proofs in this paper. Most of our work will be employed in deriving precise decay estimates for the kernel $P_t(x,y)$ of the operator in (1.3). First, we shall show that, for fixed t and x

$$\begin{aligned} c_1(t,x)\,\Phi (y)\,\le P_t(x,y)\,\le \,c_2(t,x)\,\Phi (y), \end{aligned}$$

(1.8)

for suitable $c_1$ and $c_2$; see Proposition 5.1. In a second step, we give more precise bounds, uniform in t (see Proposition 5.2), which lead to the following control of the operator $P^*_{t_0}$

$$\begin{aligned} P^*_{t_0}f(x)\,\lesssim \,C(x)\,\Big [\mathcal {M}^\mathrm{loc}(f\Phi )(x)\,+\,\int _{{{\mathbb R}_+}}|f|\,\Phi \;\Big ], \end{aligned}$$

(1.9)

for a reasonably well behaved C(x). Here, $\mathcal {M}^\mathrm{loc}=\mathcal {M}^\mathrm{loc}_M$ denotes a local Hardy–Littlewood maximal operator in ${{\mathbb R}_+}$, given by

$$\begin{aligned} \mathcal {M}^\mathrm{loc}_M f(x)\,:=\,\sup _{r>0}\,\frac{1}{|I(x,r)|}\,\int _{I(x,r)} |f(y)|\,\chi _{\{\frac{x}{2}\le y\le Mx\}}\,\mathrm{d}y \end{aligned}$$

(1.10)

for a suitable $M>1$. We also use the notation $I(x,r)=(x-r,x+r)\cap {{\mathbb R}_+}$.

Concerning the kernel estimates, they are necessarily more technical than in the Hermite setting treated in [5, 6] (which is essentially the case ${\alpha }=-1/2$). The term $1/y^2$ in (1.1) produces an additional singularity when $y\rightarrow 0$ which must be handled separately from $y\rightarrow \infty$. This dual behavior already appears in the Bessel function $I_{\alpha }$ which is part of the kernel expression of $e^{-uL}$; see (2.2). New difficulties also arise when ${\alpha }\in (-1,-1/2)$, related to the fact that the associated Laguerre functions blow up when $y\rightarrow 0$.

Finally, we consider the same problem for other classical families of Laguerre operators. These are listed in Table 1 below, together with their eigenfunctions. As we shall see in Sect. 7, using suitable transformations (as, e.g., in [1]) we can “transfer” the statements of Theorems 1.1 and 1.2 from L to each of these differential operators, provided we choose the function $\Phi$ as in Table 1. More precisely, we have the following

Theorem 1.3

Let ${\alpha }>-1$, $\mu \ge -({\alpha }+1)$ and $\nu >0$. Then, if L is replaced by any of the operators listed in Table 1, both Theorems 1.1 and 1.2 hold with the expression of the function $\Phi$ given in the table.

Table 1 Table of Laguerre-type operators, and their corresponding $\Phi$-functions

As an application, in the case of the standard Laguerre operator ${\mathbb L}$, with $m=0$ and $\nu =1/2$, we recall that Muckenhoupt, in his classical paper [10], proved the pointwise convergence of $P_tf$ for all data f in the space $L^1(y^{\alpha }e^{-y}\,\mathrm{d}y)$. Our results show that one can enlarge the space to

$$\begin{aligned} L^1(y^{\alpha }e^{-y}/\sqrt{{{\text {ln}}\,}(e+y)}\,\mathrm{d}y), \end{aligned}$$

and include new initial data such as $f(y)=e^y/[(1+y)^{{\alpha }+1}{{\text {ln}}\,}(e+y)]$.

The outline of the paper will be the following. In Sect. 2, we consider a version of Theorem 1.1 for heat integrals $u(t,x)=e^{-tL}f(x)$, which are solutions of the equation

$$\begin{aligned} u_t\,+\,Lu=0\quad \text{ in } (0,\mathtt {T})\times {{\mathbb R}_+},\quad \text{ with }\quad u(0,x)=f(x). \end{aligned}$$

Heat integrals are easier to handle, and the explicit expression of the heat kernel, $e^{-tL}(x,y)$, makes more transparent the behavior we shall later encounter in Poisson kernels. In Sect. 3, we study 2-weight inequalities for the local maximal operator $\mathcal {M}^\mathrm{loc}$. In Sect. 4, we apply these to prove a version of Theorem 1.2 for heat integrals. In Sect. 5, we take up the study of Poisson integrals, splitting in various subsections the detailed kernel estimates leading to (1.9). In Sect. 6, we shall give the proof of Theorems 1.1 and 1.2 for the operator L. Finally, in Sect. 7 we show how to transfer the results to each of the Laguerre operators in Table 1.

Throughout the paper ${\alpha }\!>\!-1$ is fixed, as are the parameters $\mu \ge -({\alpha }+1)$ and $\nu >0$ in the differential operators. The notation $A\lesssim B$ will mean $A\,\le \, c\,B$, for a constant $c>0$ which may depend on ${\alpha },\mu , \nu$ and other parameters like $p,M,t_0,{\varepsilon }$, but not on t, x, y. If needed, we shall stress the latter dependence by c(x), c(t, x), ... Finally, if $1<p<\infty$ we set $p'=p/(p-1)$.

2 The simpler model of heat integrals

In this section, we set $\mu =0$ in (1.1) and consider

$$\begin{aligned} L=-\partial _{yy} + \left[ y^2 + \frac{{\alpha }^2-\tfrac{1}{4}}{y^2}\right] .\end{aligned}$$

(2.1)

The corresponding eigenfunctions $\{{\varphi }^\alpha _n\}_{n=0}^\infty$ satisfy

$$\begin{aligned}L{\varphi }^\alpha _n= (4n+2{\alpha }+2){\varphi }^\alpha _n,\quad n=0,1,2,\ldots \end{aligned}$$

and form an orthonormal basis of $L^2(0,\infty )$; see [17, (5.1.2)]. The kernel of the associated heat semigroup $e^{-tL}$, in terms of the new variable $s={{\text {th}}\,}t$, has the explicit expression¹

$$\begin{aligned} e^{-tL}(x,y)= & {} \sum _{n=0}^\infty e^{-(4n+2{\alpha }+2)t}{\varphi }^\alpha _n(x){\varphi }^\alpha _n(y)\nonumber \\= & {} \sqrt{\frac{1-s^2}{2s}}\,{\bar{I}}_{\alpha }\left( \frac{(1-s^2)xy}{2s}\right) \,e^{-\frac{(x-y)^2}{4s}}\,e^{-\frac{s(x+y)^2}{4}}. \end{aligned}$$

(2.2)

Here, we have used the convenient notation ${\bar{I}}_{\alpha }(z) = \sqrt{z} e^{-z} I_{\alpha }(z)$, so that from known properties of the Bessel functions $I_{\alpha }$ (see, e.g., [8, (5.7.1), (5.11.10)]), we have ${\bar{I}}_{\alpha }(z) \approx \left\langle {z}\right\rangle ^{{\alpha }+\frac{1}{2}}$, with $\langle {z}\rangle =\min \{z,1\}$.

2.1 A.e. convergence of heat integrals

We wish to establish the pointwise convergence of $e^{-tL}f(x)$ with the weakest possible conditions on f. We claim that

$$\begin{aligned} e^{-tL}(x,y)\,\lesssim \,\Big \langle \frac{xy}{s}\Big \rangle ^{{\alpha }+\frac{1}{2}}\,\,\frac{e^{-\frac{(x-y)^2}{4s}}}{\sqrt{s}}\, e^{-\frac{sy^2}{4}}. \end{aligned}$$

(2.3)

To produce this bound from (2.2), one disregards x in the last exponential and uses $1-s^2\le 1$ when ${\alpha }\ge -\frac{1}{2}$. If ${\alpha }\in (-1,-\frac{1}{2})$, note that $\langle {{\lambda }z}\rangle \ge {\lambda }\langle {z}\rangle$ for ${\lambda }\le 1$, so one can leave outside a power $(1-s^2)^{{\alpha }+1}\le 1$.

Theorem 2.1

Let ${\alpha }>-1$ be fixed, and f be such that

$$\begin{aligned} \int _0^\infty |f(y)| e^{-a y^2}\langle {y}\rangle ^{\alpha +\frac{1}{2}}\,\mathrm{d}y <\infty , \quad \text{ for } \text{ some } \text{(possibly } \text{ large) } a>0. \end{aligned}$$

(2.4)

Then,

$$\begin{aligned} \lim _{t\rightarrow 0^+} e^{-tL}f(x) = f(x), {\quad a.e.\,}x\in {\mathbb R}_+.\end{aligned}$$

Proof

For each fixed $N\ge 2$, it suffices to show that $\lim _{t\rightarrow 0^+} e^{-tL}f(x) = f(x)$ for ${\,a.e.\,}x\in (1/N,N)$. We split

$$\begin{aligned} f= f\chi _{\{0<y\le 2N\}}\,+\,f\chi _{\{y>2N\}}\,=\,f_1\,+f_2.\end{aligned}$$

The function $f_1$ has bounded support and belongs to $L^1(y^{{\alpha }+\frac{1}{2}}e^{-\frac{y^2}{2}}\mathrm{d}y)$, so we can apply the results of Muckenhoupt [10] (with a suitable change of variables², as indicated by Stempak [15]) to obtain

$$\begin{aligned} \lim _{t\rightarrow 0^+} e^{-tL}f_1(x) = f_1(x)= f(x), {\quad a.e.\,}x\in \big [\tfrac{1}{N},N\big ]. \end{aligned}$$

Next, we shall show that, under the hypothesis (2.4),

$$\begin{aligned} \lim _{t\rightarrow 0^+} e^{-tL}f_2(x)=0, \quad \forall x\in \big [\tfrac{1}{N},N\big ]. \end{aligned}$$

Since $t\rightarrow 0$, we may assume that $s={{\text {th}}\,}t \le s_0$ for some $s_0<\frac{1}{10}$ (which we shall make precise below). Note that $\tfrac{1}{N}\le x\le N$ and $y>2N$ imply that

$$\begin{aligned} \Big \langle \frac{xy}{s}\Big \rangle ^{{\alpha }+\frac{1}{2}} = 1,\quad \forall s<1. \end{aligned}$$

So, by (2.3), in this region we have a gaussian bound for the kernel

$$\begin{aligned} e^{-tL}(x,y)\, \lesssim \, s^{-\frac{1}{2}}{e^{-\frac{(x-y)^2}{4s}}}\,\le \,s^{-\frac{1}{2}}{e^{-\frac{y^2}{16s}}},\end{aligned}$$

using in the last step that $|x-y|\ge y/2$. Choosing $s_0<\frac{1}{32a}$ (with a as in (2.4)), we see that for all $y>2N$,

$$\begin{aligned} e^{-tL}(x,y)\, \lesssim \, \,s^{-\frac{1}{2}}{e^{-\frac{y^2}{32s}}}\,e^{-a{y^2}}\,\le \,s^{-\frac{1}{2}}{e^{-\frac{N^2}{8s}}}\,e^{-a{y^2}}\end{aligned}$$

and therefore

$$\begin{aligned} e^{-tL}f_2(x)\, \lesssim \, s^{-\frac{1}{2}}e^{-\frac{N^2}{8s}}\,\int _{y>2N}|f(y)|\,e^{-a{y^2}}\,\mathrm{d}y\longrightarrow 0,\quad \text{ as } s\rightarrow 0^+.\end{aligned}$$

$\square$

2.2 Heat kernel estimates

The elementary bound (2.3) was enough for Theorem 2.1. Below, in order to handle maximal operators, we shall need the following more precise bound.

Proposition 2.2

Let ${\alpha }>-1$. Then, for every ${\gamma }>1$ there is some $M=M_{\gamma }>1$ such that

$$\begin{aligned} e^{-tL}(x,y)\, \le \, C_{{\gamma }} \left\{ \begin{array}{lll} \displaystyle \frac{e^{-\frac{|x-y|^2}{4s}}}{\sqrt{s}}\;\Big \langle {\frac{xy}{s}}\Big \rangle ^{{\alpha }+\frac{1}{2}} &{} \mathrm{if} &{} \frac{x}{2}\le y\le Mx\\ &{} &{} \\ c(x)\, \left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}\,e^{-\frac{y^2}{2{\gamma }{{\text {th}}\,}(2t)}} &{} \mathrm{if} &{} y<\frac{x}{2} \quad \mathrm{or}\quad y>Mx\end{array}\right. ,\end{aligned}$$

(2.5)

for all $x,y\in {{\mathbb R}_+}$ and $s={{\text {th}}\,}t\in (0,1)$. Here we can set $c(x)= 1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}$.

Proof

Clearly, (2.3) implies the estimate in the local part $y\in [\frac{x}{2},Mx]$, so we shall look at the complementary range. Given x, y and s, for simplicity we write $z=\frac{xy}{s}$, so that (2.3) becomes

$$\begin{aligned} e^{-tL}(x,y)\,\lesssim \,\left\langle {z}\right\rangle ^{{\alpha }+\frac{1}{2}}\,s^{-\frac{1}{2}}\,{e^{-\frac{(x-y)^2}{4s}}}\,e^{-\frac{sy^2}{4}}. \end{aligned}$$

(2.6)

Below we shall show that

$$\begin{aligned} A_s(x,y):=\,\left\langle {z}\right\rangle ^{{\alpha }+\frac{1}{2}}\,\frac{e^{-\frac{(x-y)^2}{4s}}}{\sqrt{s}}\,\le \,C_{\gamma }\,\frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}\,e^{-\frac{y^2}{4{\gamma }s}}, \qquad \text{ if } y<\frac{x}{2} \text{ or } y>Mx , \end{aligned}$$

(2.7)

for some $M=M_{\gamma }$. This estimate inserted into (2.6) establishes (2.5), since

$$\begin{aligned} e^{-\frac{y^2}{4{\gamma }s}}\,e^{-\frac{sy^2}{4}}\le \,e^{-\frac{y^2}{4{\gamma }}\left( s+\frac{1}{s}\right) }\,=\,e^{-\frac{y^2}{2{\gamma }{{\text {th}}\,}(2t)}}, \end{aligned}$$

as $s+s^{-1}=2/{{\text {th}}\,}(2t)$ when $s={{\text {th}}\,}t$.

To prove (2.7), we need to separate the cases $z\le 1$ and $z\ge 1$. We begin with $z\ge 1$. In the region $y>Mx$, we may use $|x-y|\ge (1-\frac{1}{M}) y$ to obtain

$$\begin{aligned} A_s(x,y)\, \lesssim \, \frac{e^{-\left( \frac{M-1}{M}\right) ^2\,\frac{y^2}{4s}}}{\sqrt{s}}\,\le \, c_M\,\frac{e^{-\left( \frac{M-1}{M}\right) ^3\,\frac{y^2}{4s}}}{y}\,\le \,c_M\,{e^{-\frac{y^2}{4s{\gamma }}}}\, \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}, \end{aligned}$$

(2.8)

where in the last step we select $M=M_{\gamma }$ sufficiently large so that $(\frac{M}{M-1})^3\le {\gamma }$, and have used the trivial estimate

$$\begin{aligned} \frac{1}{y}\,\le \, \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {y}\rangle ^{{\alpha }+\frac{3}{2}}} \,\le \, \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}},\quad \text{ if } y\ge x.\end{aligned}$$

(2.9)

On the other hand, if $y<x/2$ we have $|x-y|\ge x/2$, which leads to

$$\begin{aligned} A_s(x,y)\, \lesssim \, \frac{e^{-\frac{(x/2)^2}{4s}}}{\sqrt{s}}\,\le \, c_{\gamma }\,\frac{e^{-\frac{(x/2)^2}{4s{\gamma }}}}{x}\,\le \, c_{\gamma }\,\frac{e^{-\frac{y^2}{4{\gamma }s}}}{x}\,. \end{aligned}$$

(2.10)

In the case ${\alpha }\in (-1,-\frac{1}{2})$, this can be combined with

$$\begin{aligned} \frac{1}{x}\,\le \, \frac{\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}} \,\le \, \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}, \quad \text{ since } x\ge y.\end{aligned}$$

(2.11)

If on the contrary ${\alpha }\ge -\frac{1}{2}$, we can insert $1\le z^{{\alpha }+\frac{1}{2}}$ in the first step of (2.10) to obtain

$$\begin{aligned} A_s(x,y)\, \lesssim \, \frac{e^{-\frac{(x/2)^2}{4s}}}{\sqrt{s}}\,\big (\tfrac{xy}{s}\big )^{{\alpha }+\frac{1}{2}}\,= \, \big (\tfrac{x^2}{s}\big )^{{\alpha }+1}\,{e^{-\frac{(x/2)^2}{4s}}}\,\frac{y^{{\alpha }+\frac{1}{2}}}{x^{{\alpha }+\frac{3}{2}}}\,\le \, c_{\gamma }\,{e^{-\frac{y^2}{4s{\gamma }}}}\,\frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}\,,\end{aligned}$$

(2.12)

where in the last step we have used

$$\begin{aligned} \frac{y^{{\alpha }+\frac{1}{2}}}{x^{{\alpha }+\frac{3}{2}}}\,\le \, \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}},\quad \text{ if } x\ge y \end{aligned}$$

(this is clear if $y\le 1$, and follows from ${y^{{\alpha }+\frac{1}{2}}}/{x^{{\alpha }+\frac{3}{2}}}\le 1/y\le 1$ if $y\ge 1$). This completes the proof of (2.7) when $z\ge 1$.

We turn to the case $z\le 1$, for which we have to estimate

$$\begin{aligned} A_s(x,y)\, =\, s^{-\frac{1}{2}}{e^{-\frac{(x-y)^2}{4s}}}\,z^{{\alpha }+\frac{1}{2}}. \end{aligned}$$

(2.13)

When ${\alpha }\ge -\frac{1}{2}$, the last factor helps, so some of the previous arguments also lead to (2.7); namely one can disregard z in the region $y>Mx$, and must keep it when $y<\frac{x}{2}$ and argue as in (2.12). We are left with the case ${\alpha }\in (-1,-\frac{1}{2})$, which makes $z^{{\alpha }+\frac{1}{2}}\ge 1$. In the region $y>Mx$, this can be absorbed by the exponentials as follows

$$\begin{aligned} A_s(x,y)\lesssim & {} \big (\tfrac{xy}{s}\big )^{{\alpha }+\frac{1}{2}}\,s^{-\frac{1}{2}}\, {e^{-\left( \frac{M-1}{M}\right) ^2\,\frac{y^2}{4s}}}\,=\, \big (\tfrac{y^2}{s}\big )^{{\alpha }+1}\, {e^{-\left( \frac{M-1}{M}\right) ^2\,\frac{y^2}{4s}}}\,\,\frac{x^{{\alpha }+\frac{1}{2}}}{y^{{\alpha }+\frac{3}{2}}}\\\le & {} c_M\,{e^{-\left( \frac{M-1}{M}\right) ^3\frac{y^2}{4s}}}\, \frac{x^{{\alpha }+\frac{1}{2}}}{y^{{\alpha }+\frac{3}{2}}}\,\le \, c_M\,{e^{-\frac{y^2}{4s{\gamma }}}}\, \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}\,, \end{aligned}$$

where in the last step we first argue as in (2.8), and then use the elementary inequalities

$$\begin{aligned} \frac{x^{{\alpha }+\frac{1}{2}}}{y^{{\alpha }+\frac{3}{2}}}\, \le \, \frac{\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {y}\rangle ^{2{\alpha }+2}}\,\le \,\frac{\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{2{\alpha }+2}},\qquad \text{ if } y\ge x \end{aligned}$$

(the first one should be clear if $x\le 1$, and follows from ${x^{{\alpha }+\frac{1}{2}}}/{y^{{\alpha }+\frac{3}{2}}}\le 1/x\le 1$ if $x\ge 1$). Finally, in the region $y<\frac{x}{2}$ the inequalities in (2.12) remain also valid, so we have completed the proof of (2.7), and hence of Proposition 2.2. $\square$

Using the local Hardy–Littlewood maximal function $\mathcal {M}^\mathrm{loc}_M$ defined in (1.10), we can easily deduce from Proposition 2.2 the following

Corollary 2.3

Let ${\alpha }>-1$ and ${\gamma }>1$. Then, there is some $M=M_{\gamma }>1$ such that

$$\begin{aligned} \sup _{0<t\le t_0} \big |e^{-tL}f(x)\big |\, \lesssim \, \mathcal {M}^\mathrm{loc}_M f(x)\,+\, c(x)\,\int _{{\mathbb R}_+}|f(y)|\, \left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}\,e^{-\frac{y^2}{2{\gamma }{{\text {th}}\,}{(2t_0)}}}\,\mathrm{d}y ,\end{aligned}$$

(2.14)

for every $x,t_0>0$ and $c(x)= 1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}$.

Proof

Let $f=f_1+f_2$ with $f_1=f\chi _{[\frac{x}{2}, Mx]}$. From Proposition 2.2, (2.14) is clear for $f_2$. To treat $f_1$, we first assume $s={{\text {th}}\,}t\le x^2$. Then, $z=\frac{xy}{s}\gtrsim 1$ (since $x\approx y$ when $y\in \mathop {\mathrm{supp}}f_1$), so the first bound in (2.5) and a standard slicing argument easily lead to

$$\begin{aligned} \big |e^{-tL}f_1(x)\big |\, \lesssim \,\frac{1}{\sqrt{s}}\,\int _{{{\mathbb R}_+}}e^{-\frac{(x-y)^2}{4s}}\,|f(y)|\chi _{\left\{ \frac{x}{2}\le y\le Mx\right\} }\,\mathrm{d}y\,\lesssim \, \mathcal {M}^\mathrm{loc}_M f(x)\,. \end{aligned}$$

If we assume instead $s={{\text {th}}\,}t\ge x^2$, then using again $x\approx y$ and (2.5) we have

$$\begin{aligned} \big |e^{-tL}f_1(x)\big |\, \lesssim \,\frac{x^{2{\alpha }+1}}{s^{{\alpha }+1}}\,\int _{\frac{x}{2}}^{Mx}|f(y)|\,\mathrm{d}y\,\lesssim \, \frac{x^{2{\alpha }+1}}{x^{2{\alpha }+2}}\,\int _{\frac{x}{2}}^{Mx}|f(y)|\,\mathrm{d}y\,\lesssim \,\mathcal {M}^\mathrm{loc}_Mf(x). \end{aligned}$$

$\square$

Remark 2.4

An estimate quite similar to (2.14), with a slightly worse bound for the exponential inside the integral, was obtained by Chicco-Ruiz and Harboure in [3, $\S 5$].

3 2-weight inequalities for $\mathcal {M}^\mathrm{loc}$

In this section, we prove 2-weight inequalities for the local maximal operator in ${{\mathbb R}_+}$

$$\begin{aligned} \mathcal {M}^\mathrm{loc}f(x):=\sup _{r>0}\,\frac{1}{|I(x,r)|}\,\int _{I(x,r)} |f(y)|\,\chi _{\{\frac{x}{2}<y<Mx\}}\,\mathrm{d}y, \end{aligned}$$

where $M>1$ is a fixed parameter (which for simplicity we avoid in the notation). The 1-weight theory of (a related version of) this operator was considered in [12, $\S 6$].

3.1 The Carleson–Jones theorem for $\mathcal {M}^\mathrm{loc}$

For each $p\in (1,\infty )$, we consider the following class of weights in ${{\mathbb R}_+}$

$$\begin{aligned} D_p^\mathrm{loc} \, = \,\Big \{W(x)>0{\,\,\,:\,\,\,}\int _J W^{-\frac{p'}{p}}\, <\infty ,\quad \forall \;J\Subset (0,\infty )\Big \}. \end{aligned}$$

Associated with each $W\in D_p^\mathrm{loc}$, we define a collection of weights $\{V_{\varepsilon }\}_{{\varepsilon }>0}$ by

$$\begin{aligned} V_{\varepsilon }(x) \,=\,V(x)\,\rho _{\varepsilon }\big [V(x)\big ], \quad \text{ where } V(x):=\,\Big [\mathcal {M}^\mathrm{loc}\left( W^{-\frac{p'}{p}}\right) (x)\Big ]^{-\frac{p}{p'}}, \end{aligned}$$

(3.1)

and we set $\rho _{\varepsilon }(x):=\min \big \{x^{\varepsilon },x^{-{\varepsilon }}\big \}$. Observe that $V_{{\varepsilon }_2} \le V_{{\varepsilon }_1}\le V\le W$ if ${\varepsilon }_1\le {\varepsilon }_2$. This is a slight variant of the weight family used by Carleson and Jones in [2] (for the classical Hardy–Littlewood operator).

Theorem 3.1

Let $1<p<\infty$ and $W\in D_p^\mathrm{loc}$. Then, for every ${\varepsilon }>0$

$$\begin{aligned} \mathcal {M}^\mathrm{loc}:L^p(W)\rightarrow L^p(V_{\varepsilon })\quad \text{ boundedly, } \end{aligned}$$

where $V_{\varepsilon }$ is defined as in (3.1).

Proof

We use the ideas in [2], with some modifications required for the local operator $\mathcal {M}^\mathrm{loc}$; see also [5, Prop. 4.2]. Call $E_n=\{x\in {{\mathbb R}_+}{\,\,\,:\,\,\,}\mathcal {M}^\mathrm{loc}(W^{-\frac{p'}{p}})(x)<2^n\}$, $n\in {\mathbb Z}$, and define the operators

$$\begin{aligned} T_ng(x):=\chi _{E_n}\,\mathcal {M}^\mathrm{loc}\left( W^{-\frac{p'}{p}}g\right) (x). \end{aligned}$$

(3.2)

Note that $T_n:L^1(W^{-\frac{p'}{p}})\rightarrow L^{1,\infty }$, with a uniform bound in n, since

$$\begin{aligned} \Big |\Big \{T_ng(x)>R\Big \}\Big |\le \Big |\Big \{\mathcal {M}\left( W^{-\frac{p'}{p}}g\right) (x)>R\Big \}\Big |\le \frac{c_0}{R}\int _{{{\mathbb R}_+}}W^{-\frac{p'}{p}}|g|, \end{aligned}$$

(3.3)

using in the last step the weak-1 boundedness of the classical Hardy–Littlewood maximal operator $\mathcal {M}$. Similarly, $T_n:L^\infty (W^{-\frac{p'}{p}})\rightarrow L^\infty$ with $\Vert T_n\Vert \le 2^n$, since

$$\begin{aligned} \big \Vert T_ng\big \Vert _\infty =\sup _{x\in E_n}\big |\mathcal {M}^\mathrm{loc}\left( W^{-\frac{p'}{p}}g\right) (x)\big |\,\le 2^n\,\Vert g\Vert _\infty .\end{aligned}$$

(3.4)

Thus, by the Marcinkiewicz interpolation theorem we obtain

$$\begin{aligned} \int _{E_n}|T_n(g)|^p\le \, c_p\,2^{\frac{np}{p'}}\,\int _{{{\mathbb R}_+}}|g|^p\,W^{-\frac{p'}{p}},\quad n\in {\mathbb Z}.\end{aligned}$$

(3.5)

Setting $g=fW^{\frac{p'}{p}}$ in the above inequality, this is the same as

$$\begin{aligned} \int _{E_n}|\mathcal {M}^\mathrm{loc}(f)|^p\le \, c_p\,2^{\frac{np}{p'}}\,\int _{{{\mathbb R}_+}}|f|^p\,W,\quad n\in {\mathbb Z}.\end{aligned}$$

(3.6)

Now, modulo null sets ${{\mathbb R}_+}=\cup _{n\in {\mathbb Z}}\big [E_n\setminus E_{n-1}\big ]$ (since $0<\mathcal {M}^\mathrm{loc}(W^{-\frac{p'}{p}})(x)<\infty$ at ${\,a.e.\,}x$), and we have

$$\begin{aligned} V_{\varepsilon }(x)\approx 2^{-\frac{np}{p'}} 2^{-\frac{{\varepsilon }|n|p}{p'}}, \text{ if } x\in E_n\setminus E_{n-1}. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \int _{{{\mathbb R}_+}}|\mathcal {M}^\mathrm{loc}f|^p\,V_{\varepsilon }\lesssim & {} \, \sum _{n\in {\mathbb Z}}2^{-\frac{np}{p'}}2^{-\frac{{\varepsilon }|n|p}{p'}}\int _{E_n}|\mathcal {M}^\mathrm{loc}f|^p\\ \hbox {(by (3.6))}\lesssim & {} \left( \sum _{n\in {\mathbb Z}}2^{-\frac{{\varepsilon }|n|p}{p'}}\right) \,\int _{{{\mathbb R}_+}}|f|^pW,\end{aligned}$$

as we wished to show. $\square$

3.2 Properties of the weights $V_{\varepsilon }$

The weights $V_{\varepsilon }$ inherit some of the integrability behavior of W if ${\varepsilon }$ is sufficiently small.

Proposition 3.2

Let $1<p<\infty$ and $W\in D_p^\mathrm{loc}$. Then, for each ${\varepsilon }\in (0,1)$, the weight defined in (3.1) satisfies $V_{\varepsilon }\in D^\mathrm{loc}_q$ for all $\;q>p+{{\varepsilon }p}/{p'}$.

Proof

Observe that, for each $x\in {{\mathbb R}_+}$,

$$\begin{aligned} V_{\varepsilon }(x)^{-\frac{q'}{q}}=\max _{\pm }\left[ \mathcal {M}^\mathrm{loc}\left( W^{-\frac{p'}{p}}\right) (x)\right] ^{\frac{p-1}{q-1}(1\pm {\varepsilon })}. \end{aligned}$$

(3.7)

The assumption $q>p+\frac{{\varepsilon }p}{p'}$ implies that $s=\frac{(p-1)(1+{\varepsilon })}{q-1}<1$. Then, given $J=[a,b]\Subset {{\mathbb R}_+}$,

$$\begin{aligned}\int _J\left[ \mathcal {M}^\mathrm{loc}\left( W^{-\frac{p'}{p}}\right) \right] ^s\lesssim & {} |J|^{1-s}\,\left\| \mathcal {M}\left( W^{-\frac{p'}{p}}\chi _{J^*}\right) \right\| _{L^{1,\infty }}^s\\\lesssim & {} c_J\,\left( \int _{J^*}W^{-\frac{p'}{p}}\right) ^s<\infty , \end{aligned}$$

with $J^*=[a/2,Mb]\Subset (0,\infty )$. The same applies if we set $s=\frac{(p-1)(1-{\varepsilon })}{q-1}$ (which is also $<1$), so we deduce from (3.7) that $\int _JV_{\varepsilon }^{-{q'}/{q}}<\infty$. $\square$

Our later applications suggest to define the following classes of weights

$$\begin{aligned} D_p^{0}(\beta )= & {} \left\{ W\in D_p^\mathrm{loc}{\,\,\,:\,\,\,}\int _0^1 W^{-\frac{p'}{p}}(y)\,\langle {y}\rangle ^{\beta p'}\,\mathrm{d}y<\infty \right\} ,\quad \text{ for } \beta>-1, \\ D_p^\mathrm{exp_{\lambda }}(a)= & {} \left\{ W\in D_p^\mathrm{loc}{\,\,\,:\,\,\,}\int _1^\infty W^{-\frac{p'}{p}}(y)\, e^{-ay^{\lambda }p'}\,\mathrm{d}y <\infty \right\} ,\quad \text{ for } a,{\lambda }>0. \end{aligned}$$

Proposition 3.3

Let $1<p<\infty$, ${\varepsilon }\in (0,1)$ and $\;q>p+{{\varepsilon }p}/{p'}$. Then, for $\beta ,{\beta _1}>-1$ and $a,a_1,{\lambda }>0$ we have the following

• (i) $W\in D^0_p(\beta )$ implies $V_{\varepsilon }\in D^0_q({\beta _1})$, provided $q>\frac{1+\beta }{1+{\beta _1}}p+{\varepsilon }\,\frac{p}{p'}\,\frac{|1+\beta p'|}{1+{\beta _1}}$

• (ii) $W\in D^\mathrm{exp_{\lambda }}_p(a)$ implies $V_{\varepsilon }\in D^\mathrm{exp_{\lambda }}_q(a_1)$, provided $q>p(1+{\varepsilon }) M^{\lambda }a/a_1$.

Proof

With the same notation in the proof of Proposition 3.2, we set $s=\frac{(p-1)(1+{\varepsilon })}{q-1}<1$. Then, denoting $I_j=[2^{-j-1},2^{-j}]$, we have

$$\begin{aligned} \int _0^1\left[ \mathcal {M}^\mathrm{loc}\left( W^{-\frac{p'}{p}}\right) \right] ^s\,\langle {y}\rangle ^{{\beta _1}q'}\,\mathrm{d}y\lesssim & {} \sum _{j=0}^\infty 2^{-j{\beta _1}q'}\,|I_j|^{1-s}\, \left\| \mathcal {M}\left( W^{-\frac{p'}{p}}\chi _{I_j^*}\right) \right\| _{L^{1,\infty }}^s\\\lesssim & {} \sum _{j=0}^\infty 2^{-j{\beta _1}q'}2^{-j(1-s)} \,\left( \int _{2^{-j-2}}^{M2^{-j}}W^{-\frac{p'}{p}}\right) ^s\\\lesssim & {} \sum _{j=0}^\infty 2^{-j[{\beta _1}q'-\beta p's+1-s]} \,\left( \int _{0}^{M}W^{-\frac{p'}{p}}(y)\,\langle {y}\rangle ^{\beta p'}\,\mathrm{d}y\right) ^s. \end{aligned}$$

This is a finite expression when $W\in D^0_p(\beta )$, provided

$$\begin{aligned} {\beta _1}q'-\beta p's+1-s>0\;. \end{aligned}$$

Using the value of $s=\frac{(p-1)(1+{\varepsilon })}{q-1}$ and solving for q, this is equivalent to

$$\begin{aligned} q>\frac{1+\beta }{1+{\beta _1}}p\,+\,\frac{{\varepsilon }p(1+\beta p')}{p'(1+{\beta _1})}.\end{aligned}$$

In order to have $\int _0^1V_{\varepsilon }^{-q'/q}\,\langle {y}\rangle ^{{\beta _1}q'}\,\mathrm{d}y<\infty$, the previous relation must also hold with ${\varepsilon }$ replaced by $-{\varepsilon }$, so a sufficient condition is

$$\begin{aligned} q>\frac{1+\beta }{1+{\beta _1}}p\,+\,\frac{{\varepsilon }p\,|1+\beta p'|}{p'(1+{\beta _1})},\end{aligned}$$

as we wished to show.

We now prove (ii). Let ${\gamma }>1$ (to be precised later), and as before set $I_j=[{\gamma }^j,{\gamma }^{j+1}]$ and $s=\frac{(p-1)(1+{\varepsilon })}{q-1}<1$. Then

$$\begin{aligned}&\int _1^\infty \left[ \mathcal {M}^\mathrm{loc}\left( W^{-\frac{p'}{p}}\right) \right] ^s\,e^{-a_1y^{\lambda }q'}\,\mathrm{d}y\\&\quad \lesssim \sum _{j=0}^\infty \,e^{-a_1{\gamma }^{{\lambda }j}q'}\,{\gamma }^{(1-s)j}\, \left\| \mathcal {M}\left( W^{-\frac{p'}{p}}\chi _{I_j^*}\right) \right\| _{L^{1,\infty }}^s\\&\quad \lesssim \sum _{j=0}^\infty e^{-a_1{\gamma }^{{\lambda }j}q'}\,{\gamma }^{(1-s)j}\, \,\left( \int _{{\gamma }^{j}/2}^{M{\gamma }^{j+1}}W^{-\frac{p'}{p}}\,\mathrm{d}y\right) ^s\,\\&\quad \le \sum _{j=0}^\infty \,{\gamma }^{(1-s)j}\,e^{-{\gamma }^{{\lambda }j}[a_1q'-p'aM^{\lambda }{\gamma }^{{\lambda }}s]} \,\left( \int _{1/2}^{\infty }W^{-\frac{p'}{p}}\,e^{-p'ay^{\lambda }}\,\mathrm{d}y\right) ^s. \end{aligned}$$

This is now a finite expression provided

$$\begin{aligned} a_1q'> p'aM^{\lambda }{\gamma }^{{\lambda }}s, \end{aligned}$$

which using the value of s and solving for q gives

$$\begin{aligned} q>p(1+{\varepsilon })M^{\lambda }{\gamma }^{{\lambda }}a/a_1.\end{aligned}$$

Clearly, we can choose a ${\gamma }>1$ with this property under the assumption

$$\begin{aligned} q>p(1+{\varepsilon })M^{\lambda }a/a_1.\end{aligned}$$

Since this also implies the validity of the estimates with ${\varepsilon }$ replaced by $-{\varepsilon }$, we may conclude that $V_{\varepsilon }\in D^\mathrm{exp_{\lambda }}_q(a_1)$, as desired. $\square$

4 2-weight inequalities for local maximal heat operators

Let L be as in (2.1), and for each $t_0>0$, consider

$$\begin{aligned} h^*_{t_0}f(x):=\sup _{0<t\le t_0} \big |e^{-tL}f(x)\big |.\end{aligned}$$

By Corollary 2.3, given any $\mathtt {T}>t_0$, this operator is well defined for all functions $f\in L^1({\varphi }_\mathtt {T})$, where

$$\begin{aligned} {\varphi }_\mathtt {T}(y)=\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}\,e^{-\frac{y^2}{2{{\text {th}}\,}(2\mathtt {T})}}. \end{aligned}$$

We wish to study 2-weight inequalities for $h^*_{t_0}$ over subspaces $L^p(w)\subset L^1({\varphi }_\mathtt {T})$. By duality, the class of weights for which such inclusion holds is given by

$$\begin{aligned} D_p({\varphi }_\mathtt {T}):=\left\{ w>0{\,\,\,:\,\,\,}\left\| w^{-\frac{1}{p}}{\varphi }_\mathtt {T}\right\| _{p'}<\infty \right\} .\end{aligned}$$

Here, we show that for all such weights the operator $h^*_{t_0}$ satisfies a 2-weight inequality.

Theorem 4.1

Let $\mathtt {T}>t_0>0$ and $1<p<\infty$. Then, for every $w\in D_p({\varphi }_\mathtt {T})$ there exists another weight $v(x)>0$ such that

$$\begin{aligned} h^*_{t_0}:L^p(w)\rightarrow L^p(v),\quad \text{ boundedly }. \end{aligned}$$

Moreover, if $q>p$ and $t_0$ is sufficiently small (depending on q / p and $\mathtt {T}$), then we can select $v\in D_q({\varphi }_\mathtt {T})$.

Remark 4.2

The second weight v(x) will be constructed explicitly; see (4.2), (4.5) and (4.6) below. Observe that v depends on ${\alpha }, p, t_0, \mathtt {T}$ and of course w.

Proof of Theorem 4.1

The crucial estimate was already given in Corollary 2.3. We shall use it with the parameter ${\gamma }={{\text {th}}\,}(2\mathtt {T})/{{\text {th}}\,}(2t_0)>1$, which produces a suitable $M=M_{\gamma }>1$ such that

$$\begin{aligned} h^*_{t_0}f(x)\, \lesssim \, \mathcal {M}^\mathrm{loc}_M f(x)\,+\, c(x)\,\int _0^\infty |f(y)|\,{\varphi }_\mathtt {T}(y)\,\mathrm{d}y. \end{aligned}$$

(4.1)

The last integral is bounded by $\Vert f\Vert _{L^p(w)}\Vert w^{-\frac{1}{p}}{\varphi }_\mathtt {T}\Vert _{p'}$, so the second term will be fine for any weight v(x) such that $c(x)=1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}\in L^p(v)$. For instance, we may take

$$\begin{aligned} v_2(x)=\frac{\langle {x}\rangle ^{\left( {\alpha }+\frac{3}{2}\right) p-1}}{[{{\text {ln}}\,}(e/\langle {x}\rangle )]^2\,(1+x)^p} \end{aligned}$$

(4.2)

which clearly satisfies

$$\begin{aligned} \int _0^\infty |c(x)|^p\,v_2(x)\,\mathrm{d}x\,=\,\int _0^\infty \frac{\mathrm{d}x}{\langle {x}\rangle [{{\text {ln}}\,}(e/\langle {x}\rangle )]^2\,(1+x)^p}\,<\,\infty . \end{aligned}$$

Further, we claim that $v_2\in D_q({\varphi }_\mathtt {T})$ iff $q>p$. Indeed, since ${\varphi }_\mathtt {T}(x)$ decays exponentially, it suffices to check the integrability for x near 0. Writing $\beta ={\alpha }+\frac{1}{2}$ so that

$$\begin{aligned} {\varphi }_\mathtt {T}(x)\approx \langle {x}\rangle ^\beta {\quad \text{ and }\quad }v_2(x)\approx \frac{\langle {x}\rangle ^{(\beta +1)p-1}}{[{{\text {ln}}\,}(e/\langle {x}\rangle )]^2}, \end{aligned}$$

we easily see that

$$\begin{aligned} \int _0^1v_2(x)^{-\frac{q'}{q}}\,\langle {x}\rangle ^{\beta q'}\,\mathrm{d}x\,\approx \,\int _0^1\frac{[{{\text {ln}}\,}(e/x)]^{2q'/q}}{x^{1-q'(\beta +1)\left( 1-\frac{p}{q}\right) }}\,\mathrm{d}x\,<\,\infty .\end{aligned}$$

(4.3)

For the first term in (4.1), we shall use the results in Sect. 3. We first note that

$$\begin{aligned} w\in D_p({\varphi }_\mathtt {T}) \quad \Longleftrightarrow \quad w\in D^0_p({\alpha }+\tfrac{1}{2})\cap D^\mathrm{exp_2}_p(a), \quad \text{ with } a=1/(2\,{{\text {th}}\,}2\mathtt {T}),\end{aligned}$$

(4.4)

where the weight classes $D^0_p(\beta )$ and $D^\mathrm{exp_2}_p(a)$ were defined in §3.2. Then, for every ${\varepsilon }>0$ Theorem 3.1 gives

$$\begin{aligned} \big \Vert \mathcal {M}^\mathrm{loc}_M f\big \Vert _{L^p(v_{1,{\varepsilon }})}\,\lesssim \,\big \Vert f\big \Vert _{L^p(w)}, \end{aligned}$$

provided

$$\begin{aligned} v_{1,{\varepsilon }}(x)=\mathscr {V}(x)\rho _{{\varepsilon }}\Big ({\mathscr {V}(x)}\Big ),\quad \text{ where } \mathscr {V}(x)=\left[ \mathcal {M}^\mathrm{loc}_M\left( w^{-\frac{p'}{p}}\right) (x)\right] ^{-\frac{p}{p'}} \end{aligned}$$

(4.5)

(or $v_{1,{\varepsilon }}=V_{{\varepsilon }}$ in the notation of (3.1)). Hence, setting

$$\begin{aligned} v(x)=\min \{v_{1,{\varepsilon }}(x),v_2(x)\}\end{aligned}$$

(4.6)

with $v_{1,{\varepsilon }}$ and $v_2$ defined as in (4.5) and (4.2), we have proved that $h^*_{t_0}:L^p(w)\rightarrow L^p(v)$.

It remains to verify the last statement in Theorem 4.1. We already know that, for every $q>p$, we have $v_2\in D_q({\varphi }_\mathtt {T})$. Concerning $v_{1,{\varepsilon }}$, from the equivalence in (4.4) it suffices to prove that $V_{\varepsilon }\in D^0_{q}({\alpha }+\frac{1}{2})\cap D^\mathrm{exp_2}_{q}(a)$ for a sufficiently small ${\varepsilon }$ and $a=1/(2{{\text {th}}\,}2\mathtt {T})$. The first assertion is immediate from (i) in Proposition 3.3. However, (ii) in the same proposition only gives $V_{\varepsilon }\in D^\mathrm{exp_2}_{\rho }(a)$ if $\rho >p(1+{\varepsilon })M^2$, where $M=M_{\gamma }$ is the parameter obtained in Proposition 2.2 by the rule $\left( \frac{M}{M-1}\right) ^3={\gamma }={{\text {th}}\,}(2\mathtt {T})/{{\text {th}}\,}(2t_0)$. If we allow both ${\varepsilon }$ and $t_0$ to be sufficiently small (so that M becomes close enough to 1), then we can set $\rho =q$ and hence conclude that $v_{1,{\varepsilon }}\in D_q({\varphi }_\mathtt {T})$ as desired. $\square$

5 Poisson kernel estimates

In this section, we fix ${\alpha }>-1$ and $\mu \ge -({\alpha }+1)$ and recall that

$$\begin{aligned} L=-\partial _{yy} + \left[ y^2 + \frac{{\alpha }^2-\tfrac{1}{4}}{y^2}\right] \,+\,2\mu .\end{aligned}$$

(5.1)

The eigenfunctions $\{{\varphi }^\alpha _n\}_{n=0}^\infty$ form an orthonormal basis of $L^2(0,\infty )$, and satisfy

$$\begin{aligned} L{\varphi }^\alpha _n= \left( 4n+2({\alpha }+1+\mu )\right) {\varphi }^\alpha _n,\quad n=0,1,2,\ldots \end{aligned}$$

They can be expressed in terms of the (normalized) Laguerre polynomials $L^{\alpha }_n$ by

$$\begin{aligned} {\varphi }^\alpha _n(y)\,=\,\sqrt{2}\,y^{{\alpha }+\frac{1}{2}}\,e^{-\frac{y^2}{2}}\,L^{\alpha }_n(y^2), \end{aligned}$$

(5.2)

although we shall not use this formula here. The kernel of the associated heat semigroup, $e^{-tL}$, can be written explicitly in various forms

$$\begin{aligned} e^{-tL}(x,y)= & {} \sum _{n=0}^\infty e^{-[4n+2({\alpha }+1+\mu )]t}{\varphi }^\alpha _n(x){\varphi }^\alpha _n(y)\nonumber \\ (r=e^{-2t})= & {} r^\mu \,\sqrt{\tfrac{2r}{1-r^2}}\,{\bar{I}}_{\alpha }\left( \frac{2rxy}{1-r^2}\right) \, e^{-\frac{(x-ry)^2}{1-r^2}}\,e^{\frac{x^2-y^2}{2}} \end{aligned}$$

(5.3)

$$\begin{aligned} (s={{\text {th}}\,}t)= & {} \Big (\tfrac{1-s}{1+s}\Big )^\mu \,\sqrt{\tfrac{1-s^2}{2s}}\,{\bar{I}}_{\alpha }\left( \frac{(1-s^2)xy}{2s}\right) \,e^{-\frac{(x-y)^2}{4s}}\,e^{-\frac{s(x+y)^2}{4}} \end{aligned}$$

(5.4)

where as before we have set ${\bar{I}}_{\alpha }(z) = \sqrt{z} e^{-z} I_{\alpha }(z)$. Thus, using the notation $\langle {z}\rangle =\min \{z,1\}$, we shall have ${\bar{I}}_{\alpha }(z) \approx \left\langle {z}\right\rangle ^{{\alpha }+\frac{1}{2}}$. Both expressions of the heat kernel will be useful in our later computations. For instance, (5.3) is good when $r\approx 0$, as it isolates correctly the decaying factor $\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}e^{-y^2/2}$. On the other hand, (5.4) will be useful when $s\approx 0$ (hence $r\approx 1$), since it makes transparent the gaussian behavior of the singularity $s^{-\frac{1}{2}}e^{-\frac{(x-y)^2}{4s}}$.

Using the subordination formula in (1.3), the Poisson-like kernel associated with L becomes

$$\begin{aligned} P_t(x,y):=\tfrac{t^{2\nu }}{4^\nu \Gamma (\nu )}\,\int _0^\infty e^{-\frac{t^2}{4u}}\,\left[ e^{-uL}(x,y)\right] \,\frac{\mathrm{d}u}{u^{1+\nu }},\quad t>0. \end{aligned}$$

(5.5)

Changing variables $r=e^{-2u}$ (i.e., $u=\frac{1}{2}\ln \frac{1}{r}$), one sees that

$$\begin{aligned} P_t(x,y)\,\approx \,t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _0^1 e^{-\frac{t^2}{2\ln \frac{1}{r}}}\, r^{\mu +\frac{1}{2}}\, \frac{e^{-\frac{(x-ry)^2}{1-r^2}}}{\sqrt{1-r}}\, \frac{\left\langle {\frac{rxy}{1-r^2}}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\left( \ln \frac{1}{r}\right) ^{1+\nu }}\, \frac{dr}{r}\,. \end{aligned}$$

We shall consider two regions of integration according to the behavior of $z:=\displaystyle \frac{rxy}{1-r^2}$. The regions will be separated by the number

$$\begin{aligned} r_0(xy)\,=\,\left\{ \begin{array}{ll}\frac{1}{2xy} &{} , \text{ if } xy\ge 1\\ 1-\frac{xy}{2} &{}, \text{ if } xy\le 1.\end{array}\right. \end{aligned}$$

Indeed, it is elementary to check that

• (1) If $\;0<r\le r_0(xy)\;$ then $\;z\le 1$.

• (2) If $\;r_0(xy)\le r<1\;$ then $\;z\ge 1/2$.

Thus we can write

$$\begin{aligned} P_t(x,y)\approx & {} t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\left[ \int _0^{r_0(xy)} \,\cdots \,\left( \frac{rxy}{1-r^2}\right) ^{{\alpha }+\frac{1}{2}}\, \frac{dr}{r}\,+\,\int _{r_0(xy)}^1 \,\cdots \,\frac{dr}{r}\;\right] \\= & {} B_t(x,y)\,+\,A_t(x,y),\end{aligned}$$

where in $A_t(x,y)$ we have used that $\left\langle {z}\right\rangle ^{{\alpha }+\frac{1}{2}}\approx 1$. The next two propositions summarize the estimates we shall need to handle these kernels. We shall make extensive use of the function $\Phi$ in (1.4), which we can equivalently write as

$$\begin{aligned} \Phi (y)\,\approx \,\frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}\,e^{-y^2/2}}{(1+y)^{\mu +\frac{1}{2}}\,[{{\text {ln}}\,}(y+e)]^{1+\nu }}, \end{aligned}$$

(5.6)

with the agreement that in the extreme case $\mu =-({\alpha }+1)$ the logarithm in the denominator is just $[{{\text {ln}}\,}(y+e)]^{\nu }$. The first result gives, for fixed t and x, the optimal decay of $y\mapsto P_t(x,y)$ in terms of the function $\Phi (y)$.

Proposition 5.1

Given $t,x>0$, there exist $c_1(t,x)>0$ and $c_2(t,x)>0$ such that

$$\begin{aligned} c_1(t,x)\, \Phi (y) \,\le \,P_t(x,y)\,\le \, c_2(t,x)\, \Phi (y) \;,\quad y\in {{\mathbb R}_+}. \end{aligned}$$

(5.7)

The second result is a refinement of the upper bound in (5.7) with a few advantages: it is uniform in the variable t, isolates in the “local part” the singularities of the kernel $P_t(x,y)$, and finally provides “reasonable” bounds for the constant’s dependence on x.

Proposition 5.2

There exists $M>1$ such that the following holds for all $t,x,y>0$

$$\begin{aligned} P_t(x,y) \, \lesssim \; C_1(x)\,\frac{t^{2\nu }\,e^{-{y^2}/2}}{\big (t+|x-y|\big )^{1+2\nu }} \,\chi _{\big \{\frac{x}{2}<y< Mx\big \}}\;+\; C_2(x)\,(t\vee 1)^{2\nu }\,\Phi (y), \end{aligned}$$

(5.8)

where $C_1(x) \,=\,(1+x)^{2\nu }e^{\frac{x^2}{2}}$ and $C_2(x)= [{{\text {ln}}\,}(e+x)]^{1+\nu }\,{(1+x)^{|\mu +\frac{1}{2}|}\,e^{\frac{x^2}{2}}}/{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}$.

If we consider, for fixed $M>1$, the local maximal operator $\mathcal {M}^\mathrm{loc}_M$ in (1.10), then we may express (5.8) as follows.

Corollary 5.3

Let $t_0>0$ be fixed. Then, there is some $M>1$ such that

$$\begin{aligned} P^*_{t_0}f(x) \,\lesssim \, C_1(x)\,\mathcal {M}^\mathrm{loc}_M \big (fe^{-\frac{y^2}{2}}\big )(x)\,+\,C_2(x)\,\big \Vert f\big \Vert _{L^1(\Phi )} ,\quad x\in {{\mathbb R}_+}, \end{aligned}$$

(5.9)

with $C_1(x)$ and $C_2(x)$ as in Proposition 5.2.

This is the key estimate from which we shall deduce the theorems claimed in Sect. 1 for the operator L. The reader willing to skip the technical proofs of the propositions in the next subsections may pass directly to Sect. 6 for the proof of the theorems.

5.1 Estimates from below for $B_t(x,y)$

Recall that

$$\begin{aligned} B_t(x,y)\,\approx \,t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\,e^{\frac{x^2-y^2}{2}}\;\int _0^{r_0(xy)} \frac{r^{{\alpha }+\mu +1}}{(1-r)^{{\alpha }+1}\,(\ln \frac{1}{r})^{1+\nu }}\, e^{-\frac{t^2}{2\ln \frac{1}{r}}}\,e^{-\frac{(x-ry)^2}{1-r^2}}\, \frac{dr}{r}\,. \end{aligned}$$

(5.10)

The lower bound in Proposition 5.1 will be a consequence of $P_t(x,y)\gtrsim B_t(x,y)$ and the estimate in the next lemma.

Lemma 5.4

For fixed $t,x>0$, it holds

$$\begin{aligned} B_t(x,y)\,\ge \,c_1(t,x)\,\Phi (y),\quad y\in {{\mathbb R}_+}, \end{aligned}$$

for a suitable function $c_1(t,x)>0$.

Proof

We first look at $y<x\wedge \frac{1}{x}$. Then, $xy<1$, and hence $r_0(xy)>1/2$. So, we can estimate $B_t(x,y)$ by an integral over $0<r<\frac{1}{2}$, which disregarding irrelevant terms becomes

$$\begin{aligned} B_t(x,y)\gtrsim \,t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\,e^{\frac{x^2-y^2}{2}}\,\int _0^{\frac{1}{2}}\frac{r^{{\alpha }+\mu +1}}{(\ln \frac{1}{r})^{1+\nu }}\, e^{-\frac{t^2}{2\ln \frac{1}{r}}}\, e^{-\frac{(x-ry)^2}{1-r^2}}\,\frac{dr}{r}. \end{aligned}$$

We can get rid of the first two exponentials using

$$\begin{aligned} e^{\frac{x^2-y^2}{2}}\ge 1\quad (\hbox {since } y\le x){\quad \text{ and }\quad }e^{-\frac{t^2}{2\ln \frac{1}{r}}}\ge e^{-\frac{t^2}{2\ln 2}} \quad \left( \text{ since } r\le \frac{1}{2}\right) . \end{aligned}$$

For the last exponential, notice that $0<x-ry\le x$, and hence $e^{-\frac{(x-ry)^2}{1-r^2}}\ge e^{-\frac{4}{3}x^2}$. This leaves a convergent integral in r, so we conclude that

$$\begin{aligned} B_t(x,y)\gtrsim \,c_1(t,x)\,\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}, \end{aligned}$$

with $c_1(t,x)=t^{2\nu }x^{{\alpha }+\frac{1}{2}}e^{-\frac{t^2}{2\ln 2}}e^{-\frac{4}{3}x^2}$. Notice that $y\le 1$ in this range, so we find the required expression for $\Phi (y)$.

Suppose now that $y\ge x\vee \frac{1}{x}$. Then, $xy\ge 1$, and hence $r_0(xy)=\frac{1}{2xy}\le 1/2$. Arguing as before, we can estimate $B_t(x,y)$ by

$$\begin{aligned} B_t(x,y)\gtrsim \,t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\,e^{-\frac{y^2}{2}}\,e^{-\frac{t^2}{2\ln 2}}\, \int _0^{\frac{1}{2xy}}\frac{r^{{\alpha }+\mu +1}}{\left( \ln \frac{1}{r}\right) ^{1+\nu }}\,\, e^{-\frac{(x-ry)^2}{1-r^2}}\,\frac{dr}{r}. \end{aligned}$$

This time, we get rid of the exponential inside the integral using

$$\begin{aligned} |x-ry|\le x+ry\le x+\frac{1}{2x} \quad \left( \text{ since } r\le \frac{1}{2xy}\right) , \end{aligned}$$

which implies $e^{-\frac{(x-ry)^2}{1-r^2}}\ge e^{-\frac{4}{3}(x+\frac{1}{x})^2}$. We can easily compute the integral

$$\begin{aligned} \int _0^{\frac{1}{2xy}}\frac{r^{{\alpha }+\mu +1}}{ \left( \ln \frac{1}{r}\right) ^{1+\nu }}\,\frac{dr}{r}\;\approx \;\frac{1}{(xy)^{{\alpha }+\mu +1}\,[{{\text {ln}}\,}2xy]^{1+\nu }},\quad \text{ if } {{\alpha }+\mu +1>0}, \end{aligned}$$

with the right hand side becoming $1/[{{\text {ln}}\,}2xy]^{\nu }$ in the extreme case ${\alpha }+1+\mu =0$. Since $y>\max \{x,1\}$, note that

$$\begin{aligned} {{\text {ln}}\,}(2xy)\,\le \,{{\text {ln}}\,}(2y^2)\,\approx {{\text {ln}}\,}(y+e).\end{aligned}$$

Thus, combining all the previous estimates we conclude that

$$\begin{aligned} B_t(x,y)\gtrsim \,c_1(t,x)\,\frac{e^{-\frac{y^2}{2}}}{y^{\mu +\frac{1}{2}}\,[{{\text {ln}}\,}(y+e)]^{1+\nu }},\end{aligned}$$

which, since $y\ge 1$, is the required expression for $\Phi (y)$ (with the usual agreement when $\mu +{\alpha }+1=0$). In this part we have set $c_1(t,x)=t^{2\nu }e^{-\frac{t^2}{2\ln 2}}x^{-(\mu +\frac{1}{2})}\,e^{-\frac{4}{3}(x+\frac{1}{x})^2}$.

Finally, since the function $y\mapsto B_t(x,y)/\Phi (y)$ is continuous and positive, it is also bounded from below by some $c_1(t,x)$ when y belongs to the compact set $[x\wedge \frac{1}{x},x\vee \frac{1}{x}]$. $\square$

5.2 Estimates from above for $B_t(x,y)$

The next lemma, combined with the previous one, shows that for fixed t and x, the function $B_t(x,y)$ essentially behaves like $\Phi (y)$.

Lemma 5.5

For every $t,x, y>0$, it holds

$$\begin{aligned} B_t(x,y)\,\lesssim \,c(x)\,\max \{t^{2\nu },1\}\,\Phi (y), \end{aligned}$$

(5.11)

with $c(x)=1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}$.

Proof

We first notice that the two exponential factors in (5.3) can be written as

$$\begin{aligned} e^{-\frac{(x-ry)^2}{1-r^2}}\,e^{\frac{x^2-y^2}{2}}\,=\,e^{-\frac{1+r^2}{1-r^2}\frac{x^2+y^2}{2}}\,e^{\frac{2rxy}{1-r^2}} \,\lesssim \,e^{-\frac{x^2+y^2}{2}}, \end{aligned}$$

(5.12)

since $\frac{1+r^2}{1-r^2}\ge 1$ and in the region of integration of $B_t(x,y)$ the exponent $z=\frac{2rxy}{1-r^2}\lesssim 1$. We now separate cases.

(i) Case $xy\ge 1$: then $r_0(xy)=\frac{1}{2xy}\le \frac{1}{2}$ and

$$\begin{aligned} B_t(x,y)\,\lesssim \,t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\, e^{-\frac{x^2+y^2}{2}}\,\int _0^{\frac{1}{2xy}} \frac{r^{{\alpha }+\mu +1}}{\left( \ln \frac{1}{r}\right) ^{\nu +1}}\,\frac{dr}{r}. \end{aligned}$$

(5.13)

The last integral is approximately given by

$$\begin{aligned} \int _0^{\frac{1}{2xy}} \frac{r^{{\alpha }+\mu +1}}{\left( \ln \frac{1}{r}\right) ^{\nu +1}}\,\frac{dr}{r}\,\approx \,\left( \frac{1}{xy}\right) ^{{\alpha }+\mu +1}\,\frac{1}{[{{\text {ln}}\,}(2xy)]^{1+\nu }} \end{aligned}$$

(with the usual convention when ${\alpha }+\mu +1=0$ of reducing one power of the logarithm). This is a good estimate if we assume that $x\ge 1/2$, since we may use

$$\begin{aligned} {{\text {ln}}\,}(2xy)\,\gtrsim \, {{\text {ln}}\,}(y\vee 2)\,\approx \,{{\text {ln}}\,}(y+e), \end{aligned}$$

and overall obtain

$$\begin{aligned} B_t(x,y)\,\lesssim \,t^{2\nu }\,x^{-(\mu +\frac{1}{2})}\,e^{-\frac{x^2}{2}}\, \frac{y^{{\alpha }+\frac{1}{2}}e^{-y^2/2}}{y^{{\alpha }+\mu +1}[{{\text {ln}}\,}(y+e)]^{1+\nu }}\,\lesssim \,t^{2\nu }\,\Phi (y). \end{aligned}$$

When $x\le 1/2$, we need a refinement to obtain the c(x) in the statement of the lemma. We split the integral bounding $B_t(x,y)$ in (5.13) as

$$\begin{aligned} B_t(x,y)\,\lesssim \,t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\, e^{-\frac{x^2+y^2}{2}}\,\left[ \int _0^{\frac{2x}{y}}\cdots \;+\;\int _{\frac{2x}{y}}^{r_0(xy)}\cdots \right] \;=I\,+\,II, \end{aligned}$$

(5.14)

noticing that the partition point $\frac{2x}{y}\le r_0(xy)=\frac{1}{2xy}$. Since $x\le \frac{1}{2}$ and $xy\ge 1$, we also have $y\ge 2$. Now, the first summand can be bound as above by

$$\begin{aligned} I\lesssim & {} t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\, e^{-\frac{x^2+y^2}{2}}\,\left( \frac{x}{y}\right) ^{{\alpha }+\mu +1}\,\frac{1}{\left[ {{\text {ln}}\,}\left( \frac{y}{2x}\right) \right] ^{1+\nu }}\\\lesssim & {} t^{2\nu }\,\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\,x^{{\alpha }+\mu +1}\,e^{-\frac{x^2}{2}}\, \frac{e^{-y^2/2}}{y^{\mu +\frac{1}{2}}[{{\text {ln}}\,}y]^{1+\nu }}\,\lesssim \,t^{2\nu }\,\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\,\Phi (y). \end{aligned}$$

since in this range $y\ge 2$. This implies the stated estimate because $\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\le 1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}=c(x)$. To handle II, we need a different bound for the exponentials in (5.12), noticing that

$$\begin{aligned} r>\tfrac{2x}{y}\quad \Rightarrow \quad |x-ry|= ry-x\,\ge \,\tfrac{ry}{2}\quad \Rightarrow \quad e^{-\frac{(x-ry)^2}{1-r^2}}\,\le \,e^{-\frac{r^2y^2}{4}}. \end{aligned}$$

(5.15)

Thus

$$\begin{aligned} II \; \lesssim \; t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\, e^{\frac{x^2-y^2}{2}}\, \int _{\frac{2x}{y}}^{\frac{1}{2}}\frac{r^{{\alpha }+\mu +1}\,e^{-\frac{(ry)^2}{4}}}{\left( \ln \frac{1}{r}\right) ^{\nu +1}}\,\frac{dr}{r}. \end{aligned}$$

(5.16)

Changing variables $ry=u$, the latter integral can be estimated by

$$\begin{aligned} y^{-({\alpha }+\mu +1)}\, \int _{0}^{\frac{y}{2}}\frac{u^{{\alpha }+\mu +1}\,e^{-\frac{u^2}{4}}}{\left( \ln \frac{y}{u}\right) ^{\nu +1}}\,\frac{\mathrm{d}u}{u}\; \approx \;\frac{1}{y^{{\alpha }+\mu +1}\,[{{\text {ln}}\,}y]^{1+\nu }} \end{aligned}$$

since the major contribution happens when $u\approx 1$ (with the usual convention of reducing one power of the logarithm if ${\alpha }+\mu +1=0$). Inserting this into (5.16) (and using $y\ge 2$ and $x\le 1/2$), we obtain once again

$$\begin{aligned} II\;\lesssim \;t^{2\nu }\,\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\,\Phi (y). \end{aligned}$$

This concludes the proof of the case $xy\ge 1$.

(ii) Case $xy\le 1$: this time $r_0(xy)=1-\frac{xy}{2}\ge \frac{1}{2}$, so we can write

$$\begin{aligned} B_t(x,y)\,\lesssim \;t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\, e^{-\frac{x^2+y^2}{2}}\,\left[ \int _0^{\frac{1}{2}}\cdots \;+\;\int _{\frac{1}{2}}^{r_0(xy)}\cdots \right] \;=\;B_1\;+\,B_2\end{aligned}$$

The first term can be handled essentially as in the previous case. Namely, if $y\le 2$ we use a similar bound to (5.13)

If $y\ge 2$, then $x\le \frac{1}{2}$ and $\frac{2x}{y}\le \frac{1}{2}$, so we may split

$$\begin{aligned} B_1\;\le \;\int _0^{\frac{2x}{y}}\cdots \;+\;\int _{\frac{2x}{y}}^{\frac{1}{2}}\cdots \; \end{aligned}$$

and exactly the same computations we used in (5.14), give us the bound

$$\begin{aligned} B_1\lesssim \;t^{2\nu }\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\Phi (y).\end{aligned}$$

Thus we are left with the integral corresponding to $B_2$, that is the range $\frac{1}{2}<r<1-\frac{xy}{2}$. First of all, observe that

$$\begin{aligned} \ln \frac{1}{r} \,\approx \,1-r,\quad r\in [1/2,1]\quad \Rightarrow \quad e^{-\frac{t^2}{2\ln \frac{1}{r}}}\le \,e^{-\frac{ct^2}{1-r}}, \end{aligned}$$

for a suitable $c>0$. Next, we need once again more precise bounds for the exponentials in (5.12). We claim that if $r\in [1/2,1]$, then

$$\begin{aligned} e^{-\frac{1+r^2}{1-r^2}\frac{x^2+y^2}{2}}\,\le \,e^{-{\gamma }\frac{x^2+y^2}{1-r}}\,e^{-(1+{\gamma })\frac{x^2+y^2}{2}}\, \end{aligned}$$

for a small constant ${\gamma }>0$. This is easily obtained using the fact that $\frac{1+r^2}{1-r^2}\ge \frac{5}{3}$ in this interval. With these exponential bounds, we can control the integral $B_2$ as follows

$$\begin{aligned} B_2\lesssim & {} t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\, e^{-(1+{\gamma })\frac{x^2+y^2}{2}}\,\int _{1/2}^{1-\frac{xy}{2}} \frac{e^{-\frac{ct^2+{\gamma }(x^2+y^2)}{1-r}}}{(1-r)^{{\alpha }+\nu +1}} \,\frac{dr}{1-r}\nonumber \\\lesssim & {} \frac{t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}}{[t^2+x^2+y^2]^{{\alpha }+\nu +1}}\,\, e^{-(1+{\gamma })\frac{x^2+y^2}{2}}\int _0^\infty e^{-u}\,u^{{\alpha }+\nu +1}\,\frac{\mathrm{d}u}{u}, \end{aligned}$$

(5.17)

after changing variables $u=[ct^2+{\gamma }(x^2+y^2)]/(1-r)$. The last integral is a finite constant (because ${\alpha }+\nu +1>0$), so we observe two possible cases:

• (1) if $\max \{y,t,x\}\ge 1$, we can disregard the denominator and obtain

  $$\begin{aligned}B_2\lesssim t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}\,e^{-(1+{\gamma })\frac{x^2+y^2}{2}}\,\lesssim \,t^{2\nu }\,\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}\,\Phi (y),\end{aligned}$$

  since the exponential decay in y is actually better than $\Phi (y)$.

• (2) if $\max \{y,t,x\}\le 1$, we bound the denominator in the two obvious ways to obtain

  $$\begin{aligned} B_2\,\lesssim \,\frac{t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}}{t^{2\nu }\,x^{2({\alpha }+1)}}\,=\,\frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}},\end{aligned}$$

  (5.18)

  which is precisely the upper bound stated in (5.11). Observe that when $x\rightarrow 0$ this piece gives the largest contribution to $B_t(x,y)$.

$\square$

Remark 5.6

It is also possible to obtain a bound

$$\begin{aligned} B_t(x,y)\,\lesssim \,c'(x)\,t^{2\nu }\,\Phi (y), \end{aligned}$$

(5.19)

with perhaps a worse function $c'(x)$, but without the loss produced by $\max \{1,t^{2\nu }\}$. This loss appeared when $t,x,y\le 1$ in (5.18) above. Looking at (5.17), we may replace that bound by

$$\begin{aligned} B_2\,\lesssim \,\frac{t^{2\nu }\,(xy)^{{\alpha }+\frac{1}{2}}}{x^{2({\alpha }+\nu +1)}},\end{aligned}$$

which implies (5.19) with $c'(x)=1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}+2\nu }$. This estimate will also be useful later.

5.3 Upper estimates for $A_t(x,y)$: integrals over $r\le 1/2$

Recall that

$$\begin{aligned} A_t(x,y)\,\approx \,t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _{r_0(xy)}^1 \frac{r^{\mu +\frac{1}{2}}}{\sqrt{1-r}\,\left( \ln \frac{1}{r}\right) ^{1+\nu }}\, e^{-\frac{t^2}{2\ln \frac{1}{r}}}\,e^{-\frac{(x-ry)^2}{1-r^2}}\, \frac{dr}{r}\,. \end{aligned}$$

(5.20)

It will be convenient to split

$$\begin{aligned} A_t(x,y)\,=\,\int _{r_0(xy) \wedge \frac{1}{2}}^{\frac{1}{2}}\cdots \;+\;\int _{r_0(xy) \vee \frac{1}{2}}^1\cdots \;=\,A1\,+\,A2. \end{aligned}$$

When $xy<1$, since $r_0(xy)=1-\frac{xy}{2}>\frac{1}{2}$, we have $A1=0$ (and $A_t(x,y)=A2$). When $xy\ge 1$, since $r_0(xy)=\frac{1}{2xy}\le \frac{1}{2}$, both terms A1 and A2 play a role. In this section, we shall only estimate A1.

Lemma 5.7

If $xy\ge 1$, then

$$\begin{aligned} A1\,\lesssim \,t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _{\frac{1}{2xy}}^\frac{1}{2} \frac{r^{\mu +\frac{1}{2}}}{\left( \ln \frac{1}{r}\right) ^{1+\nu }}\,e^{-\frac{(x-ry)^2}{1-r^2}}\, \frac{dr}{r}\, \,\lesssim \,c(x)\,t^{2\nu }\,\Phi (y),\end{aligned}$$

(5.21)

where $c(x)=\,\left[ {{\text {ln}}\,}(e+x)\right] ^{\nu +1}\,(1+x)^{|\mu +\frac{1}{2}|}\,\exp (x^2/2)$.

Proof

We shall distinguish cases

(i) Case $y\le 4x$. In this region, we essentially disregard the exponential factor $e^{-\frac{(x-ry)^2}{1-r^2}}$ inside the integral, and directly estimate

$$\begin{aligned} A1\,\lesssim \,t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _{\frac{1}{2xy}}^\frac{1}{2} \frac{r^{\mu +\frac{1}{2}}}{(\ln \frac{1}{r})^{1+\nu }}\, \frac{dr}{r}\,. \end{aligned}$$

(5.22)

Notice however that when $y\le x$ such exponential produces an additional gain, due to

$$\begin{aligned} ry\le \tfrac{x}{2}\quad \Rightarrow \quad |x-ry|\,\ge \,\tfrac{x}{2}\quad \Rightarrow \quad e^{-\frac{(x-ry)^2}{1-r^2}}\,\le \,e^{-\frac{x^2}{4}}. \end{aligned}$$

(5.23)

In particular, we can use it when $y\le 1$ (since then $x\ge xy\ge 1$) to control some of the powers of x appearing below.

We now evaluate the integral in (5.22), depending on the sign of $\mu +\frac{1}{2}$.

• (1) If $\mu +\frac{1}{2}\ge 0$ the integral is bounded by a constant, and hence

  $$\begin{aligned} A1\,\lesssim \,t^{2\nu }\,e^{\frac{x^2-y^2}{2}}. \end{aligned}$$

  We shall enlarge this value to match (5.21) as follows. Since $xy\ge 1$, in this range we have $x\ge 1/2$. So if $1\le y\le 4x$ we may use

  $$\begin{aligned} 1\,\lesssim \,\frac{(1+x)^{\mu +\frac{1}{2}}[{{\text {ln}}\,}(e+x)]^{\nu +1}}{(1+y)^{\mu +\frac{1}{2}}[{{\text {ln}}\,}(e+y)]^{\nu +1}}. \end{aligned}$$

  If $\frac{1}{x}\le y\le 1$, we use instead

  $$\begin{aligned} 1\,\lesssim \,\max \left\{ x^{{\alpha }+\frac{1}{2}},1\right\} \,\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}, \end{aligned}$$

  which in this region can be combined with the extra exponential in factor in (5.23). In both cases, we obtain $A1\lesssim t^{2\nu }c(x)\Phi (y)$, as wished.

• (2) If $\mu +\frac{1}{2}<0$, the integral diverges near 0, but we still obtain

  $$\begin{aligned} \int _{\frac{1}{2xy}}^\frac{1}{2} \frac{r^{\mu +\frac{1}{2}}}{\left( \ln \frac{1}{r}\right) ^{1+\nu }}\, \frac{dr}{r}\,\lesssim \,\frac{1}{(xy)^{\mu +\frac{1}{2}}({{\text {ln}}\,}2xy)^{\nu +1}}. \end{aligned}$$

  Thus, using the inequality ${{\text {ln}}\,}(2xy)\gtrsim \max \left\{ {{\text {ln}}\,}y,{{\text {ln}}\,}2\right\}$ we arrive at

  $$\begin{aligned} A1\lesssim & {} t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\,\frac{(1+x)^{|\mu +\frac{1}{2}|}}{(1+y)^{\mu +\frac{1}{2}}\,[{{\text {ln}}\,}(e+y)]^{\nu +1}}\;\le \;t^{2\nu }c(x)\Phi (y),\quad \text{ if } \;1\le y\le 4x\\ A1\lesssim & {} t^{2\nu }\,\frac{e^{\frac{x^2}{4}}\,y^{{\alpha }+\frac{1}{2}}}{x^{\mu +\frac{1}{2}}\,y^{\mu +{\alpha }+1}}\,\le \,t^{2\nu }\,e^{\frac{x^2}{4}}x^{{\alpha }+\frac{1}{2}}\,\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}\;\le \;t^{2\nu }c(x)\Phi (y),\quad \text{ if } \,\frac{1}{x}\le y\le 1, \end{aligned}$$

  after inserting in the first step of this last case the additional exponential gain in (5.23).

(ii) Case $y\ge 4x$. This is the same as $\frac{2 x}{y}\le \frac{1}{2}$, and remember from (5.15) that when $r\in \left[ \frac{2x}{y},\frac{1}{2}\right]$ a better bound for the exponential is available, namely

$$\begin{aligned} e^{-\frac{(x-ry)^2}{1-r^2}}\,\le \,e^{-\frac{r^2y^2}{4}}.\end{aligned}$$

(5.24)

Thus we may consider two subcases, depending on whether $\frac{2x}{y}$ is above or below $r_0(xy)$.

$\bullet$ Subcase $\frac{2x}{y}\le r_0(xy)=\frac{1}{2xy}\le \frac{1}{2}$. Using (5.24) we obtain

$$\begin{aligned} A1\lesssim & {} t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _{\frac{1}{2xy}}^\frac{1}{2} \frac{r^{\mu +\frac{1}{2}}\,e^{-\frac{(ry)^2}{4}}}{(\ln \frac{1}{r})^{1+\nu }}\, \frac{dr}{r}\,\\ (ry=u)= & {} \frac{t^{2\nu }\,e^{\frac{x^2-y^2}{2}}}{y^{\mu +\frac{1}{2}}} \;\int _{\frac{1}{2x}}^{y/2} \frac{u^{\mu +\frac{1}{2}}\,e^{-\frac{u^2}{4}}}{\left( \ln \frac{y}{u}\right) ^{1+\nu }}\, \frac{\mathrm{d}u}{u}\,\end{aligned}$$

Observe that $x\le \frac{1}{2}$ (and $y\ge 2$), so the latter integral reaches its major contribution at $u=\frac{1}{2x}$

$$\begin{aligned} \int _{\frac{1}{2x}}^{y/2} \frac{u^{\mu +\frac{1}{2}}\,e^{-\frac{u^2}{4}}}{\left( \ln \frac{y}{u}\right) ^{1+\nu }}\, \frac{\mathrm{d}u}{u}\,\lesssim \frac{(1/x)^{\mu -\frac{1}{2}}e^{-c/x^2}}{({{\text {ln}}\,}2xy)^{1+\nu }}\;\lesssim \;\frac{1}{({{\text {ln}}\,}y)^{1+\nu }}, \end{aligned}$$

using in the last step the elementary bound of logarithms

$$\begin{aligned} {{\text {ln}}\,}(2xy)\,\gtrsim \,\frac{{{\text {ln}}\,}(y+e)}{{{\text {ln}}\,}\left( \frac{1}{x}+e\right) }, \quad \text{ if } y\ge \max \{1,1/x\}, \end{aligned}$$

which is easily verified considering the cases $x>1/\sqrt{y}$ and $x\le 1/\sqrt{y}$. Thus we conclude that

$$\begin{aligned} A1\,\lesssim \,t^{2\nu }\,e^{\frac{x^2}{2}}\,\Phi (y). \end{aligned}$$

$\bullet$ Subcase $r_0(xy)<\frac{2x}{y}\le \frac{1}{2}$. Here we split

$$\begin{aligned} A1\,= \,\int ^\frac{1}{2}_{\frac{2x}{y}}\cdots \;+\;\int ^{\frac{2x}{y}}_{r_0(xy)}\cdots \;=I\,+\,II. \end{aligned}$$

The first term is similar to the previous subcase, except that now $x>\frac{1}{2}$ (and $y\ge 4x\ge 2$)

$$\begin{aligned} I\lesssim & {} \frac{t^{2\nu }\,e^{\frac{x^2-y^2}{2}}}{y^{\mu +\frac{1}{2}}}\;\int _{2x}^{y/2} \frac{u^{\mu +\frac{1}{2}}\,e^{-\frac{u^2}{4}}}{\left( \ln \frac{y}{u}\right) ^{1+\nu }}\, \frac{\mathrm{d}u}{u}\,\end{aligned}$$

and the last integral is bounded by constant times

$$\begin{aligned} \frac{x^{\mu -\frac{1}{2}}\,e^{-cx^2}}{\left( \ln \frac{y}{2x}\right) ^{1+\nu }}\, \lesssim \,x^{\mu -\frac{1}{2}}\,e^{-cx^2}\,\frac{[{{\text {ln}}\,}(e+x)]^{1+\nu }}{[{{\text {ln}}\,}(e+y)]^{1+\nu }}\, \lesssim \,\frac{1}{[{{\text {ln}}\,}(e+y)]^{1+\nu }}. \end{aligned}$$

Finally, we consider II. Here, there is no exponential gain, and similarly to (5.22) we have

$$\begin{aligned} II\,\lesssim \,t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _{\frac{1}{2xy}}^\frac{2x}{y} \frac{r^{\mu +\frac{1}{2}}}{\left( \ln \frac{1}{r}\right) ^{1+\nu }}\, \frac{dr}{r}\,=\,\frac{t^{2\nu }\,e^{\frac{x^2-y^2}{2}}}{y^{\mu +\frac{1}{2}}}\;\int _{\frac{1}{2x}}^{2x} \frac{u^{\mu +\frac{1}{2}}}{\left( \ln \frac{y}{u}\right) ^{1+\nu }}\, \frac{\mathrm{d}u}{u}\,.\end{aligned}$$

Now, the last integral can easily be analyzed (depending on the sign of $\mu +\frac{1}{2}$) to obtain

$$\begin{aligned} \int _{\frac{1}{2x}}^{2x} \frac{u^{\mu +\frac{1}{2}}}{\left( \ln \frac{y}{u}\right) ^{1+\nu }}\, \frac{\mathrm{d}u}{u}\,\lesssim \,\frac{x^{|\mu +\frac{1}{2}|}[{{\text {ln}}\,}(x+e)]^{1+\nu }}{[{{\text {ln}}\,}(y+e)]^{1+\nu }}. \end{aligned}$$

Thus, overall we conclude that in this subcase

$$\begin{aligned} A1\;\lesssim \;I\,+\,II\;\lesssim \,t^{2\nu }\,(1+x)^{|\mu +\frac{1}{2}|}[{{\text {ln}}\,}(x+e)]^{1+\nu }\,e^{\frac{x^2}{2}}\, \Phi (y). \end{aligned}$$

$\square$

5.4 Upper estimates for A2 when $y\le x/2$ or $y\ge Mx$

In view of the previous subsection, it only remains to estimate

$$\begin{aligned} A2\approx & {} t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _{\max \left\{ r_0(xy),\frac{1}{2}\right\} }^1 \frac{e^{-\frac{t^2}{2\ln \frac{1}{r}}}}{\sqrt{1-r}\,\left( \ln \frac{1}{r}\right) ^{1+\nu }}\, \,e^{-\frac{(x-ry)^2}{1-r^2}}\,{dr}\,\nonumber \\\lesssim & {} t^{2\nu }\,e^{\frac{x^2-y^2}{2}}\;\int _{\max \left\{ 1-\frac{xy}{2},\frac{1}{2}\right\} }^1 \frac{e^{-\frac{ct^2}{1-r}}}{(1-r)^{\nu +\frac{3}{2}}}\, \,e^{-\frac{(x-ry)^2}{1-r^2}}\,{dr}\, \end{aligned}$$

(5.25)

noticing that $\ln \frac{1}{r}\approx 1-r$ when $r\in [\frac{1}{2},1]$. In this region, however, it is more convenient to use the write-up for the heat kernel in (5.4), in terms of the parameter s. This gives a more reasonable expression for the exponentials, namely

$$\begin{aligned} e^{\frac{x^2-y^2}{2}}\,e^{-\frac{(x-ry)^2}{1-r^2}}\,= \,e^{-\frac{1}{4}\left[ \frac{(x-y)^2}{s}+s(x+y)^2\right] }. \end{aligned}$$

Since the parameters r and s are related by $s=\frac{1-r}{1+r}$ (or $r=\frac{1-s}{1+s}$), either from (5.4) or directly from (5.25), we obtain that

$$\begin{aligned} A2\;\lesssim \;t^{2\nu } \int _0^{\min \left\{ \frac{1}{3},\frac{xy}{3}\right\} }\frac{e^{-\frac{ct^2}{s}}\,e^{-\frac{1}{4} \left[ \frac{(x-y)^2}{s}+s(x+y)^2\right] }}{s^{\nu +\frac{3}{2}}}\,ds,\end{aligned}$$

(5.26)

after perhaps slightly enlarging the range of integration. Our first result shows that when y is far from x this can also be controlled by the function $\Phi (y)$.

Lemma 5.8

There exists $M>1$ such that, if $\;y\le \frac{x}{2}\;$ or $\;y\ge Mx\;$, then

$$\begin{aligned} A2\;\lesssim \;t^{2\nu } \int _0^{\min \left\{ \frac{1}{3},\frac{xy}{3}\right\} }\frac{e^{-\frac{ct^2}{s}}\,e^{-\frac{1}{4} \left[ \frac{(x-y)^2}{s}+s(x+y)^2\right] }}{s^{\nu +\frac{3}{2}}}\,ds\;\lesssim \; c(x)\,\max \{t^{2\nu },1\}\,\Phi (y), \end{aligned}$$

with $c(x)=1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}$.

Proof

We claim that, in the assumed range of x and y, there is some ${\gamma }>0$ such that

$$\begin{aligned} A2\;\lesssim \;t^{2\nu }\,e^{-\left( \frac{1}{2}+{\gamma }\right) y^2}\; \int _0^{\min \left\{ \frac{1}{3},\frac{xy}{3}\right\} }{e^{-{\gamma }\frac{t^2+(x-y)^2}{s}}} {s^{-(\nu +\frac{3}{2})}}\,ds. \end{aligned}$$

(5.27)

This is just a bound of the exponentials. Indeed, if we distinguish the two cases

• (1) case $y\ge Mx$: this implies $|y-x|\ge (1-\frac{1}{M})y$, so for any $\eta <1$ we have

  $$\begin{aligned} e^{-\frac{ct^2}{s}}\,e^{-\frac{1}{4}\left[ \frac{(x-y)^2}{s}+s(x+y)^2\right] }\,\le \, e^{-\frac{ct^2+\frac{\eta }{4}(x-y)^2}{s}}\,e^{-\frac{1-\eta }{4}\left( \frac{M-1}{M}\right) ^2\left( \frac{1}{s}+s\right) y^2}\, \end{aligned}$$

  which implies the required assertion using that $\frac{1}{s}+s\ge \frac{10}{3}$, when $s\in (0,\frac{1}{3})$, and choosing M sufficiently large and $\eta$ sufficiently small.

• (2) case $y\le x/2$: this time $|x-y|\ge \frac{x}{2}\ge y$, so we have

  $$\begin{aligned} e^{-\frac{ct^2}{s}}\,e^{-\frac{1}{4}\left[ \frac{(x-y)^2}{s}+s(x+y)^2\right] }\,\le \, e^{-\frac{ct^2+\frac{\eta }{4}(x-y)^2}{s}}\,e^{-\frac{1-\eta }{4}(\frac{1}{s}+s)y^2}\, \end{aligned}$$

  which again implies the assertion using $\frac{1}{s}+s\ge \frac{10}{3}$ and choosing $\eta$ sufficiently small.

Thus (5.27) is proven, and we may change variables ${\gamma }[t^2+(x-y)^2]/s=u$ to obtain

$$\begin{aligned} A2\lesssim & {} \frac{t^{2\nu }\,e^{-\left( \frac{1}{2}+{\gamma }\right) y^2}}{[t^{2}+(x-y)^2]^{\nu +\frac{1}{2}}} \; \int ^\infty _{3{\gamma }[t^2+(x-y)^2]\max \left\{ 1,\frac{1}{xy}\right\} } {u^{\nu +\frac{1}{2}}}\,{e^{- u}}\,\frac{\mathrm{d}u}{u} \end{aligned}$$

(5.28)

$$\begin{aligned}\lesssim & {} \frac{t^{2\nu }\,e^{-\left( \frac{1}{2}+{\gamma }\right) y^2}}{(t+x+y)^{2\nu +1}} \; \int ^\infty _{{\gamma }'[x^2+y^2]\max \left\{ 1,\frac{1}{xy}\right\} } {u^{\nu +\frac{1}{2}}}\,{e^{- u}}\,\frac{\mathrm{d}u}{u}, \end{aligned}$$

(5.29)

since in the selected range of x, y we have $|x-y|\gtrsim x+y$. To finish the proof, we must distinguish some cases.

Case $y\ge 1$: then bounding the denominator and the integral in (5.29) by a constant we immediately see that

$$\begin{aligned} A2\;\lesssim \;t^{2\nu }\,e^{-\left( \frac{1}{2}+{\gamma }\right) y^2}\,\lesssim \,t^{2\nu }\,\Phi (y)\end{aligned}$$

since the exponential has a better decay.

Case $y\le 1$ and $y\ge Mx$: we again bound the integral by a constant and estimate the fraction in (5.29) as follows

$$\begin{aligned} A2\;\lesssim \;\frac{t^{2\nu }}{t^{2\nu }y}=\frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{y^{{\alpha }+\frac{3}{2}}}\,\le \,c_M\, \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}. \end{aligned}$$

(5.30)

Case $y\le 1$ and $y\le \frac{x}{2}$: this is a relevant case, since the integral in (5.29) plays actually a role. To compute the integral, we must distinguish the two subcases

• (1) If $xy\le 1$, then since also $\frac{x}{y}\ge 2$,

  $$\begin{aligned} A2\lesssim & {} \frac{t^{2\nu }}{t^{2\nu }x}\;\int ^\infty _{{\gamma }'\frac{x}{y}} {u^{\nu -\frac{1}{2}}}\,{e^{- u}}\,{\mathrm{d}u}\,\approx \,\frac{1}{x}\,\left( \frac{x}{y}\right) ^{\nu -\frac{1}{2}}e^{-c\frac{x}{y}} \\\lesssim & {} \frac{1}{x}\,\left( \frac{y}{x}\right) ^{{\alpha }+\frac{1}{2}}\;\le \; \frac{\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}}{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}.\nonumber \end{aligned}$$

  (5.31)

• (2) If $xy\ge 1$, then we have $x\ge \frac{1}{y}\ge 1$ and

  $$\begin{aligned} A2 \; \lesssim \;\frac{t^{2\nu }}{x^{1+2\nu }}\;\int ^\infty _{{\gamma }'\,x^2} {u^{\nu -\frac{1}{2}}}\,{e^{- u}}\,{\mathrm{d}u}\;\lesssim \;t^{2\nu }\,x^{-2}\,e^{-c\,x^2}. \end{aligned}$$

  Now, since $\frac{1}{x}\le y \le 1$ we can insert the estimate

  $$\begin{aligned} 1\,\lesssim \, \left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}\,\max \left\{ x^{{\alpha }+\frac{1}{2}},1\right\} , \end{aligned}$$

  to obtain $A2\lesssim t^{2\nu }\left\langle {y}\right\rangle ^{{\alpha }+\frac{1}{2}}$.$\square$

Remark 5.9

As mentioned earlier in Remark 5.6, here it is also possible to obtain a bound

$$\begin{aligned} A2\,\lesssim \,c'(x)\,t^{2\nu }\,\Phi (y), \end{aligned}$$

(5.32)

with $c'(x)=1/\langle {x}\rangle ^{{\alpha }+\frac{3}{2}+2\nu }$. The loss produced by $\max \{1,t^{2\nu }\}$ can be corrected in (5.30) and (5.31) by replacing the factor $t^{2\nu }$ in the denominator by $x^{2\nu }$, as one readily notices from (5.29). As mentioned before, this estimate will play a role later.

5.5 Upper estimates for A2 in the local part $\frac{x}{2}<y<Mx$

As in the previous subsection, our starting point is the formula (5.26), which we must estimate in the local region $\frac{x}{2}<y<Mx$. A sufficient bound for us is stated in the next lemma.

Lemma 5.10

If $\;\frac{x}{2}<y<Mx\;$, then

$$\begin{aligned} A2\,\lesssim \,t^{2\nu } \int _0^{\min \left\{ \frac{1}{3},\frac{xy}{3}\right\} }\frac{e^{-\frac{ct^2}{s}- \frac{1}{4}\left[ \frac{(x-y)^2}{s}+s(x+y)^2\right] }}{s^{\nu +\frac{3}{2}}}\,ds\;\lesssim \;C(x)\;\frac{ t^{2\nu }\,e^{-\frac{y^2}{2}}}{\big (t+|x-y|\big )^{1+2\nu }},\qquad \end{aligned}$$

(5.33)

where $C(x) \,=\,(1+x)^{2\nu }e^{\frac{x^2}{2}}$.

Proof

We shall crudely enlarge the integral in (5.33) to the range $\int _0^{1/3}$. This last integral was already estimated in [6] and [5], by a similar procedure to the one used in the last subsection. More precisely, from the estimates in [5, Lemma 3.2], formula (3.16), it follows that

$$\begin{aligned} t^{2\nu } \int _0^{\frac{1}{2}}\frac{e^{-\frac{ct^2}{s}- \frac{1}{4}\left[ \frac{(x-y)^2}{s}+s(x+y)^2\right] }}{s^{\nu +\frac{3}{2}}} \,ds\;\lesssim \;\frac{t^{2\nu }\,(1+x)^{2\nu }\,e^{\frac{x^2-y^2}{2}}}{\big (t+|x-y|\big )^{1+2\nu }}, \end{aligned}$$

which agrees with (5.33). $\square$

5.6 Proof of Propositions 5.1 and 5.2

Proposition 5.2 follows by putting together Lemmas 5.5, 5.7, 5.8 and 5.10. Concerning Proposition 5.1, the lower bound was shown in Lemma 5.4, while the upper bound also follows from Lemmas 5.5, 5.7 and 5.8, at least when $y<\frac{x}{2}$ or $y>Mx$. This actually implies the asserted result for all x and y, since when y belongs to the compact set $[\frac{x}{2},Mx]$, the continuous function $y\rightarrow P_t(x,y)/\Phi (y)$ is bounded above by a constant $c_2(t,x)$.$\square$

5.7 Proof of Corollary 5.3

By Proposition 5.2, observe that

$$\begin{aligned} P_tf(x)\,\lesssim \,\frac{C_1(x)}{t} \,\int _{{{\mathbb R}_+}}\frac{g(y)\,\mathrm{d}y}{\left( 1+\frac{|x-y|}{t}\right) ^{1+2\nu }}\;+\; C_2(x)(1\vee t_0)^{2\nu }\,\big \Vert f\big \Vert _{L^1(\Phi )},\end{aligned}$$

(5.34)

where $g(y)=f(y)e^{-\frac{y^2}{2}}\chi _{\{\frac{x}{2}<y<Mx\}}$. The first term is then controlled by the Hardy–Littlewood maximal function by a standard slicing argument.$\square$

Remark 5.11

We wrote in (1.9) a different version of (5.9) with $\mathcal {M}^\mathrm{loc}(f\Phi )$ in place of $\mathcal {M}^\mathrm{loc}\left( fe^{-\frac{y^2}{2}}\right)$. Since $x\approx y$,

$$\begin{aligned} \mathcal {M}^\mathrm{loc}(f\Phi )(x)\,\approx \, \tfrac{\left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}}{[{{\text {ln}}\,}(e+x)]^{1+\nu }\,(1+x)^{\mu +\frac{1}{2}}} \,\mathcal {M}^\mathrm{loc}\left( fe^{-\frac{y^2}{2}}\right) (x), \end{aligned}$$

so they are actually equivalent modulo x-constants. The write-up in (1.9) has the advantage of remaining valid for other Laguerre systems; see Sect. 7 below.

6 Proofs

As indicated in Sect. 1, we present the proof of Theorems 1.1 and 1.2 for the differential operator L in (1.1) and the function $\Phi$ in (1.4). We postpone to Sect. 7 the proof of the results for the other systems mentioned in Table 1.

6.1 Proof of Theorem 1.1

First of all, it is an immediate consequence of Proposition 5.1 that $P_t|f|(x)<\infty$ for some (or all) $t,x>0$ if and only if $f\in L^1(\Phi )$. This justifies that $f\in L^1(\Phi )$ is the right setting for this problem. Notice also that taking derivatives of the kernel $P_t(x,y)$ in (5.5) with respect to t does not worsen its decay in y, so $P_tf(x)$ automatically becomes infinitely differentiable in the t-variable when $f\in L^1(\Phi )$. Notice, moreover, that the kernel satisfies³

$$\begin{aligned} \left[ \partial _{xx} - x^2 - \tfrac{{\alpha }^2-\tfrac{1}{4}}{x^2}\,-\,2\mu \,+\,\partial _{tt} \,+\,\tfrac{1-2\nu }{t}\,\partial _t\right] P_t(x,y)=0. \end{aligned}$$

Thus, the (distributional) derivatives $\partial _x^{2m}[P_tf]$ are transformed into t-derivatives, and hence are continuous functions. Since this is valid for all m, it implies that $(t,x)\mapsto P_tf(x)$ is a $C^\infty$ function. We have thus completed the proof of paragraph (i) and the last statement in Theorem 1.1. We shall now prove the statement in (ii), namely that for $f\in L^1(\Phi )$

$$\begin{aligned} \lim _{t\rightarrow 0^+}P_tf(x)=f(x), \quad \forall \;x\in \mathcal L_f \end{aligned}$$

(6.1)

where $\mathcal L_f$ denotes the set of Lebesgue points of f. When $f(x)=0$, this is easily obtained from the kernel estimates in Proposition 5.2. Indeed,

$$\begin{aligned} P_tf(x) \,\lesssim \,C_1(x)\int _{\frac{x}{2}}^{Mx}\frac{t^{2\nu }|f(y)|\,\mathrm{d}y}{(t+|x-y|)^{1+2\nu }}\,+\, C_2'(x)\,t^{2\nu }\,\int _{{\mathbb R}_+}|f|\Phi , \end{aligned}$$

where in the second term we have replaced $(t\vee 1)^{2\nu }$ by $t^{2\nu }$ in view of Remarks 5.6 and 5.9. Thus, this second term vanishes as $t\rightarrow 0$ (actually for all $x\in {\mathbb R}_+$). Concerning the first term, it is given by convolution of $|f(y)|\chi _{\{\frac{x}{2}<y<Mx\}}\in L^1_c({\mathbb R}_+)$ with a radially decreasing approximate identity, so from well-known results (see, e.g., [14, p. 112]), it must vanish as $t\rightarrow 0$ at every Lebesgue point x of f with $f(x)=0$.

It remains to prove (6.1) when f(x) is not necessarily 0. To show this, we first notice that the first eigenfunction ${\varphi }={\varphi }^\alpha _0$ (with eigenvalue ${\lambda }=2(\mu +{\alpha }+1)$) satisfies

$$\begin{aligned}P_t{\varphi }=F_t({\lambda }){\varphi },\quad \text{ with }\quad \lim _{t\rightarrow 0}F_t({\lambda })=1.\end{aligned}$$

Indeed, setting $u={t^2}/{(4v)}$ in (1.3), gives

$$\begin{aligned} F_t({\lambda })=\tfrac{(t/2)^{2\nu }}{{\Gamma }(\nu )}\,\int _0^\infty e^{-\frac{t^2}{4u}-{\lambda }u}\,\frac{\mathrm{d}u}{u^{1+\nu }}=\tfrac{1}{{\Gamma }(\nu )}\int _0^\infty e^{-v-\frac{t^2{\lambda }}{4v}}\,v^{\nu -1}\,dv\longrightarrow 1,\quad \text{ as } t\rightarrow 0. \end{aligned}$$

Therefore, we can write

$$\begin{aligned} P_tf(x)-f(x)=\,P_tf(x)-F_t({\lambda })f(x) \;+\;f(x)[F_t({\lambda })-1], \end{aligned}$$

(6.2)

with the last term vanishing as $t\rightarrow 0$. Since ${\varphi }>0$, the first term can be rewritten as

$$\begin{aligned} P_tf(x)-\tfrac{f(x)}{{\varphi }(x)}\,P_t{\varphi }(x)\,=\, P_t\left( f-\tfrac{f(x)}{{\varphi }(x)}{\varphi }\right) (x). \end{aligned}$$

Setting $g=f-\tfrac{f(x)}{{\varphi }(x)}{\varphi }$, it is easily seen that $g\in L^1(\Phi )$, $g(x)=0$ and x is a Lebesgue point of g. This last assertion follows from

which vanishes as $r\rightarrow 0$. Thus we can apply our earlier case to g and conclude that $\lim _{t\rightarrow 0}P_tg(x)=0$. So the left-hand side of (6.2) goes to 0 as $t\rightarrow 0$, establishing (6.1) and completing the proof of Theorem 1.1.$\square$

Remark 6.1

A close look at the last part of the proof shows that when $f\in C([a,b])$ with $[a,b]\Subset {\mathbb R}_+$, the convergence of $P_tf(x)\rightarrow f(x)$ is uniform in $x\in [a,b]$.

6.2 Proof of Theorem 1.2

We have to show that $P^*_{t_0}$ maps $L^p(w)\rightarrow L^p(v)$, for some weight $v(x)>0$, under the assumption that

$$\begin{aligned} \Vert w\Vert _{D_p(\Phi )}:=\,\left[ \int _{{{\mathbb R}_+}}w^{-\frac{p'}{p}}(x)\Phi (x)^{p'}\,\mathrm{d}x\right] ^{1/p'}\,<\,\infty , \end{aligned}$$

with $\Phi$ as in (5.6). We shall use the bound for $P^*_{t_0}$ in (5.9), namely

$$\begin{aligned} P^*_{t_0}f(x)\lesssim & {} C_1(x)\,\mathcal {M}^\mathrm{loc}_M \left( fe^{-\frac{y^2}{2}}\right) (x) \,+\, C_2(x)\,\int _{{{\mathbb R}_+}}|f(y)|\,\Phi (y)\,\mathrm{d}y\nonumber \\= & {} \, I(x)\,+\,II(x), \end{aligned}$$

(6.3)

for a suitable $M>1$, and $C_1(x), C_2(x)$ given explicitly in Proposition 5.2. We first treat the last term, which by Hölder’s inequality is bounded by

$$\begin{aligned} II(x)\,\le \, C_2(x)\,\Vert f\Vert _{L^p(w)}\,\Vert w\Vert _{D_p(\Phi )}\;. \end{aligned}$$

Thus, it suffices to choose a weight v such that $C_2(x)= [{{\text {ln}}\,}(e+x)]^{1+\nu }\,{(1+x)^{|\mu +\frac{1}{2}|}\,e^{\frac{x^2}{2}}}/{\langle {x}\rangle ^{{\alpha }+\frac{3}{2}}}$ belongs to $L^p(v)$ to conclude that

$$\begin{aligned} \Vert II\Vert _{L^p(v)} \le \,\Vert C_2\Vert _{L^p(v)}\,\Vert w\Vert _{D_p(\Phi )}\, \Vert f\Vert _{L^p(w)}\,. \end{aligned}$$

(6.4)

For instance, we may take any $v(x)\le v_2(x)$ with

$$\begin{aligned} v_2(x):=\,\frac{\langle {x}\rangle ^{({\alpha }+\frac{3}{2})p-1}}{[{{\text {ln}}\,}(e/\langle {x}\rangle )]^2}\,\frac{e^{-\frac{p}{2}x^2}}{(1+x)^{N}}\, \end{aligned}$$

(6.5)

for any $N>1+p\,|\mu +\frac{1}{2}|$. We remark that $v_2\in D_q(\Phi )$ for all $q>p$. Indeed, the local condition was already established in (4.3). For the global condition, notice that

$$\begin{aligned} \int _1^\infty v_2(x)^{-\frac{q'}{q}}\,\Phi (x)^{q'}\,\mathrm{d}x \,\lesssim \, \int _1^\infty e^{-\left( 1-\frac{p}{q}\right) \frac{q'}{2}x^2}\,(1+x)^{\frac{Nq'}{q}}\,\mathrm{d}x\,<\,\infty . \end{aligned}$$

We now consider the term I(x) in (6.3). We define a new weight $W(x)=w(x)e^{\frac{p}{2}x^2}$ and observe that

$$\begin{aligned} w\in D_p(\Phi )\quad \Longrightarrow \quad W\in D^0_p\left( {\alpha }+\tfrac{1}{2}\right) \cap D^\mathrm{exp_2}_p(a),\quad \forall \;a>0, \end{aligned}$$

(6.6)

for the weight classes defined in Sect. 3.2. Indeed, the local estimate follows from $\Phi (x)\approx \left\langle {x}\right\rangle ^{{\alpha }+\frac{1}{2}}$ when $x\in (0,1)$, and the global estimate is a consequence of

$$\begin{aligned} \big \Vert W\big \Vert _{D^\mathrm{exp_2}_p(a)} \, = \, \int _1^\infty w^{-\frac{p'}{p}}(x)\,e^{-\frac{p'}{2}x^2}\, e^{-ap'x^2}\,\mathrm{d}x\,\le \, C^{p'}_{a}\,\int _1^\infty w^{-\frac{p'}{p}}(x)\Phi (x)^{p'}\,\mathrm{d}x\,, \end{aligned}$$

(6.7)

with $C_a=\max _{x\ge 1}|{{\text {ln}}\,}(e+x)|^{1+\nu }|1+x|^{\mu +\frac{1}{2}}e^{-ax^2}$.

We shall now set

$$\begin{aligned} v_{1,{\varepsilon }}(x)=\frac{e^{-\frac{px^2}{2}}}{(1+x)^{2p\nu }}\, \mathscr {V}(x)\rho _{{\varepsilon }}\Big ({\mathscr {V}(x)}\Big ),\quad \text{ where } \mathscr {V}(x)=\left[ \mathcal {M}^\mathrm{loc}_M\left( w^{-\frac{p'}{p}}e^{-\frac{p'y^2}{2}}\right) (x)\right] ^{-\frac{p}{p'}} \end{aligned}$$

(6.8)

(or $v_{1,{\varepsilon }}(x)=(1+x)^{-2p\nu }e^{-\frac{px^2}{2}}V_{{\varepsilon }}(x)$ in the notation of (3.1)). Given $f\in L^p(w)$, we denote ${\tilde{f}}(y)=f(y)e^{-\frac{y^2}{2}}$ which is a function in $L^p(W)$. Then, using the two-weight inequality for $\mathcal {M}^\mathrm{loc}_M$ in Theorem 3.1, and the expression for $C_1(x)=(1+x)^{2\nu }e^{\frac{x^2}{2}}$, we see that, for any $v\le v_{1,{\varepsilon }}$, the term I(x) in (6.3) is controlled by

$$\begin{aligned} \Vert I(x)\Vert ^p_{L^p(v)}\le & {} \int _{{{\mathbb R}_+}}\frac{C_1(x)^p\,e^{-\frac{p}{2}x^2}}{(1+|x|)^{2p\nu }}\, \big |\mathcal {M}^\mathrm{loc}_M{\tilde{f}}(x)\big |^p\,V_{{\varepsilon }}(x)\,\mathrm{d}x\nonumber \\\lesssim & {} \Vert {\tilde{f}}\Vert _{L^p(W)}^p\,=\,\Vert f\Vert _{L^p(w)}^p. \end{aligned}$$

(6.9)

So, combining (6.3), (6.4) and (6.9), we have shown that $\Vert P^*_{t_0} f\Vert _{L^p(v)}\lesssim \Vert f\Vert _{L^p(w)}$, provided

$$\begin{aligned} v(x)= \min \{v_{1,{\varepsilon }}(x),v_2(x)\},\end{aligned}$$

(6.10)

with $v_{1,{\varepsilon }}(x)$ and $v_2(x)$ defined in (6.8) and (6.5).

We only have to verify that if $q>p$, then we can choose ${\varepsilon }$ sufficiently small so that $v_{1,{\varepsilon }}\in D_q(\Phi )$ (which implies $v\in D_q(\Phi )$). This actually follows from (6.6) and Proposition 3.3. Indeed, on the one hand, since $W\in D^0_p({\alpha }+\frac{1}{2})$

$$\begin{aligned} \int _0^1 v_{1,{\varepsilon }}(x)^{-\frac{q'}{q}}\,\Phi (x)^{q'}\,\mathrm{d}x\,\lesssim \, \int _0^1 V_{{\varepsilon }}(x)^{-\frac{q'}{q}}\,\langle {x}\rangle ^{({\alpha }+\frac{1}{2})q'}\,\mathrm{d}x, \end{aligned}$$

(6.11)

which is finite by (i) in the proposition (choosing ${\varepsilon }$ sufficiently small). On the other hand,

$$\begin{aligned} \int _1^\infty v_{1,{\varepsilon }}(x)^{-\frac{q'}{q}}\,\Phi (x)^{q'}\,\mathrm{d}x\,\lesssim \, \int _1^\infty V_{{\varepsilon }}(x)^{-\frac{q'}{q}}\,e^{-q'\left( 1-\frac{p}{q}\right) \frac{x^2}{2}}\, (1+x)^{2\nu pq'/q}\,\mathrm{d}x\,,\end{aligned}$$

(6.12)

and since $W\in D^\mathrm{exp_2}_p(a)$ for all $a>0$, we can apply part (ii) of Proposition 3.3 (for a sufficiently small ${\varepsilon }$) to conclude that this is also a finite quantity.$\square$

Remark 6.2

Alternative expression for the second weight. A slight modification of the above construction suggests to define a new weight by

$$\begin{aligned} v^{\Phi ,w}_{\varepsilon }(x)\,:=\,\min \left\{ \Phi (x)^p\,\left[ \mathcal {M}^\mathrm{loc}\left( w^{-\frac{p'}{p}}\Phi ^{p'}\right) (x)\right] ^{-\frac{p}{p'}}\,\Upsilon _{\varepsilon }(x),\;\frac{\langle {x}\rangle ^{p-1}}{[{{\text {ln}}\,}(e/\langle {x}\rangle )]^2}\,\frac{\Phi (x)^p}{(1+x)^{N_0}}\right\} \end{aligned}$$

(6.13)

with

$$\begin{aligned} \Upsilon _{\varepsilon }(x)=\, \frac{\langle {x}\rangle ^{{\varepsilon }N_1}}{(1+x)^{N_2}}\, \rho _{{\varepsilon }}\left( \left[ \mathcal {M}^\mathrm{loc}\left( w^{-\frac{p'}{p}}\Phi ^{p'}\right) (x)\right] ^{-\frac{p}{p'}}\right) ,\quad {\varepsilon }\in (0,1). \end{aligned}$$

If $N_0,N_1,N_2$ are sufficiently large, then similar arguments as above lead to the boundedness of $P^*_{t_0}:L^p(w)\rightarrow L^p(v^{\Phi ,w}_{\varepsilon })$ when $w\in D_p(\Phi )$ and ${\varepsilon }\in (0,1)$, and give also the property that $v^{\Phi ,w}_{\varepsilon }\in D_q(\Phi )$ if ${\varepsilon }$ is sufficiently small.

We only sketch the proof for the first weight inside (6.13), namely

$$\begin{aligned} {\tilde{v}}_{1,{\varepsilon }}(x)\,:=\,\Phi (x)^p\,\left[ \mathcal {M}^\mathrm{loc}\left( w^{-\frac{p'}{p}}\Phi ^{p'}\right) (x)\right] ^{-\frac{p}{p'}}\,\Upsilon _{\varepsilon }(x). \end{aligned}$$

(6.14)

We claim that if $N_1,N_2$ are sufficiently large, then

$$\begin{aligned} \,v_{1,{\varepsilon }}(x)\,\frac{\langle {x}\rangle ^{{\varepsilon }M_1}}{(1+x)^{M_2}}\,\lesssim \,{\tilde{v}}_{1,{\varepsilon }}(x)\,\lesssim \,v_{1,{\varepsilon }}(x),\quad \forall \,{\varepsilon }\in (0,1), \end{aligned}$$

(6.15)

for suitable $M_1,M_2>0$. Assuming this claim, it is not difficult to deduce that

• (i) $\Vert I(x)\Vert _{L^p({\tilde{v}}_{1,{\varepsilon }})}\lesssim \,\Vert w\Vert _{D_p(\Phi )}\Vert f\Vert _{L^p(w)}$.

• (ii) Given $q>p$, then ${\tilde{v}}_{1,{\varepsilon }}\in D_q(\Phi )$, provided ${\varepsilon }$ is sufficiently small.

Indeed, (i) is immediate from (6.9), while (ii) is not hard to obtain from (6.12) and (6.11) (here using Proposition 3.3 (i) with some $\beta _1<\beta$). Finally, to justify (6.15), first notice that, by the local restriction in the maximal function,

$$\begin{aligned} \Phi (x)^p\left[ \mathcal {M}^\mathrm{loc}\left( w^{-\frac{p'}{p}}\Phi ^{p'}\right) (x)\right] ^{-\frac{p}{p'}}\, \approx \,e^{-\frac{px^2}{2}}\mathscr {V}(x), \end{aligned}$$

with the notation in (6.8). Moreover, using the trivial inequality

$$\begin{aligned} \left( {\lambda }\wedge \tfrac{1}{{\lambda }}\right) ^{\varepsilon }\,\rho _{\varepsilon }(x)\,\le \,\rho _{\varepsilon }({\lambda }x)\,\le \,\left( {\lambda }\vee \tfrac{1}{{\lambda }}\right) ^{\varepsilon }\,\rho _{\varepsilon }(x), \end{aligned}$$

we obtain that

$$\begin{aligned} \Upsilon _{\varepsilon }(x)\le \frac{\langle {x}\rangle ^{{\varepsilon }N_1}}{(1+x)^{N_2}}\, \left[ \frac{(1+x)^{|\mu +\frac{1}{2}|}[{{\text {ln}}\,}(e+x)]^{1+\nu }}{\langle {x}\rangle ^{|{\alpha }+\frac{1}{2}|}}\right] ^{p{\varepsilon }}\, \rho _{{\varepsilon }}\Big ({\mathscr {V}(x)}\Big )\,\lesssim \,\frac{\rho _{{\varepsilon }}({\mathscr {V}(x)})}{(1+x)^{2\nu p}}, \end{aligned}$$

provided we choose $N_1\ge p|{\alpha }+\frac{1}{2}|$ and $N_2>2\nu p+{\varepsilon }p|\mu +\frac{1}{2}|$. Thus ${\tilde{v}}_{1,{\varepsilon }}(x)\lesssim v_{1,{\varepsilon }}(x)$. On the other hand, the reverse estimate

$$\begin{aligned} \Upsilon _{\varepsilon }(x)\ge \frac{\langle {x}\rangle ^{{\varepsilon }N_1}}{(1+x)^{N_2}}\, \left[ \frac{\langle {x}\rangle ^{|{\alpha }+\frac{1}{2}|}}{(1+x)^{|\mu +\frac{1}{2}|}[{{\text {ln}}\,}(e+x)]^{1+\nu }}\right] ^{p{\varepsilon }}\, \rho _{{\varepsilon }}\Big ({\mathscr {V}(x)}\Big )\,, \end{aligned}$$

implies the lower bound in (6.15) with $M_1=N_1+p|{\alpha }+\frac{1}{2}|$ and any $M_2>N_2+{\varepsilon }p|\mu +\frac{1}{2}|$.

The new weight expression in (6.13) will have the advantage of being jointly valid for all the Laguerre systems in Table 1 (with the corresponding $\Phi$ functions).

7 Transference to other Laguerre-type systems

In this section, we show how to transfer the results already proved for the system $\{{\varphi }^\alpha _n\}$ and the operator L to the other Laguerre systems and operators in Table 1. This procedure is completely general and has been used before in other instances (see, e.g., [1]).

7.1 Results for the system $\psi ^{\alpha }_n$

The starting point is the identity defining $\psi ^{\alpha }_n$, namely

$$\begin{aligned} \psi ^{\alpha }_n(y)\,=\,a(y)\,{\varphi }^\alpha _n(y),\quad \text{ with } a(y)=y^{-{\alpha }-\frac{1}{2}}.\end{aligned}$$

(7.1)

Clearly, ${\varphi }^\alpha _n$ is an eigenvector of L if and only if $\psi ^\alpha _n$ is an eigenvector of the operator

$$\begin{aligned} f\longmapsto {\Lambda }f(x)= a(x)L[a^{-1}f\;](x) \end{aligned}$$

(with the same eigenvalue ${\lambda }_n=4n+2({\alpha }+1+\mu )$). An elementary computation shows that the differential operator ${\Lambda }$ obtained in this fashion is exactly the one listed in Table 1. Remark also that $\{\psi ^\alpha _n\}$ becomes an orthonormal basis in $L^2$ with the measure $a^{-2}(y)\mathrm{d}y=y^{2{\alpha }+1}\mathrm{d}y$.

The identity in (7.1) leads to a pointwise relation of the corresponding heat kernels

$$\begin{aligned} e^{-t{\Lambda }}(x,y)\,=\,\sum _{n=0}^\infty e^{-{\lambda }_nt}\psi ^\alpha _n(x)\psi ^\alpha _n(y)\,=\, a(x)\,a(y)\,e^{-tL}(x,y), \end{aligned}$$

and by the subordination formula, also of the corresponding Poisson kernels

$$\begin{aligned} P_t^{{\Lambda }}(x,y)\,=\,a(x)\,a(y)\,P_t^{L}(x,y). \end{aligned}$$

In particular,

$$\begin{aligned} P_t^{\Lambda }f(x)\,=\,\int _{{{\mathbb R}_+}}P_t^{\Lambda }(x,y)\,f(y)\,a^{-2}(y)\mathrm{d}y\,=\,a(x)\,P^L_t[a^{-1}f](x).\end{aligned}$$

(7.2)

From this relation, it is clear that Theorem 1.1 becomes true for the operator ${\Lambda }$ with

$$\begin{aligned} \Phi ^{\Lambda }(y)=a(y)^{-1}\Phi ^L(y), \end{aligned}$$

as listed in Table 1. From (7.2), it also follows that $P^{*,{\Lambda }}_{t_0}$ maps $L^p(w)\rightarrow L^p(v)$ if and only if $P^{*,L}_{t_0}$ maps $L^p(a^pw)\rightarrow L^p(a^pv)$, and hence the necessary and sufficient condition becomes

$$\begin{aligned} a^pw\in D_p(\Phi ^L)\Longleftrightarrow \Vert a^{-1}w^{-\frac{1}{p}}\Phi ^L\Vert _{p'}<\infty \Longleftrightarrow w\in D_p(\Phi ^{\Lambda }), \end{aligned}$$

as was claimed in Theorem 1.2. For the assertions about the weight v, one may argue directly as follows. Observe from (7.2) and Corollary 5.3 that we can write

$$\begin{aligned} P^{*,{\Lambda }}_{t_0}f(x)\lesssim & {} C_1(x)\,a(x)\mathcal {M}^\mathrm{loc}_M \left( fa^{-1}e^{-\frac{y^2}{2}}\right) (x)\,+\,C_2(x)a(x)\,\int _{{{\mathbb R}_+}} |f|\Phi ^{\Lambda },\end{aligned}$$

with a cancelation in the first term due to $a(x)a(y)^{-1}\approx 1$ when $\frac{x}{2}<y<Mx$. At this point, we can apply the same arguments as in Sect. 6.2. Namely, we construct $v=\min \{v_{1,{\varepsilon }},v_2\}$ with the same choice of $v_{1,{\varepsilon }}$, and with $v_2$ in (6.5) now replaced by

$$\begin{aligned} v_2(x):=\,\frac{\langle {x}\rangle ^{(2{\alpha }+2)p-1}}{[{{\text {ln}}\,}(e/\langle {x}\rangle )]^2}\,\frac{e^{-\frac{p}{2}x^2}}{(1+x)^{N}}\,. \end{aligned}$$

The same proof will give that, for any $q>p$, there is a sufficiently small ${\varepsilon }$ so that $v\in D_q(\Phi ^{\Lambda })$ (the only difference being that, locally, this condition now becomes $v\in D^0_q(2{\alpha }+1)$). We remark that this part will work as well with the choice

$$\begin{aligned} v(x)\,=\,a^{-p}(x)\,v^{\Phi ^L, a^pw}(x)\,=\,v^{\Phi ^{\Lambda },w}(x), \end{aligned}$$

as defined in (6.13).

7.2 Results for the system $\mathfrak {L}^\alpha _n$

Consider the following isometry of $L^2({{\mathbb R}_+},\mathrm{d}y)$

$$\begin{aligned} f\longmapsto Af(x)\,=\,\sqrt{2x}\,f(x^2). \end{aligned}$$

The systems $\mathfrak {L}^\alpha _n$ and ${\varphi }^\alpha _n$ are related by ${\varphi }^\alpha _n=A\mathfrak {L}^\alpha _n$, or equivalently

$$\begin{aligned} \mathfrak {L}^\alpha _n(y)\,=\,[A^{-1}{\varphi }^\alpha _n](y)\,=\,(4y)^{-\frac{1}{4}}\,{\varphi }^\alpha _n\left( \sqrt{y}\right) .\end{aligned}$$

(7.3)

In particular, $\mathfrak {L}^\alpha _n$ is an eigenvector of the operator

$$\begin{aligned}\mathfrak {L}\,= \frac{1}{4}\, A^{-1}\circ L\circ A,\end{aligned}$$

this time with eigenvalue ${\lambda }_n/4=n+({\alpha }+1+\mu )/2$. The factor $\frac{1}{4}$ has been inserted so that $\mathfrak {L}$ coincides with the operator listed in Table 1.

The heat kernels are now related by

$$\begin{aligned} e^{-t\mathfrak {L}}(x,y)\,=\,\sum _{n=0}^\infty e^{-t{\lambda }_n/4}\mathfrak {L}^\alpha _n(x)\mathfrak {L}^\alpha _n(y)\,=\, \frac{1}{2(xy)^{\frac{1}{4}}}\,e^{-\frac{t}{4} L}\left( \sqrt{x},\sqrt{y}\right) , \end{aligned}$$

and therefore, a substitution in the subordinated integral in (5.5) gives

$$\begin{aligned} P_t^{\mathfrak {L}}(x,y)\,=\,(16xy)^{-\frac{1}{4}}\,P_{t/2}^{L}\left( \sqrt{x},\sqrt{y}\right) . \end{aligned}$$

Thus, we obtain the formula

$$\begin{aligned} P_t^\mathfrak {L}f(x)\,=\,\int _{{{\mathbb R}_+}}P_t^\mathfrak {L}(x,y)\,f(y)\,\mathrm{d}y\,=\,(4x)^{-\frac{1}{4}}\,P^L_{t/2}[Af]\left( \sqrt{x}\right) .\end{aligned}$$

(7.4)

From this relation, one easily deduces Theorem 1.1 for the operator $\mathfrak {L}$, provided that

$$\begin{aligned} \Phi ^\mathfrak {L}(y)=[A^{-1}\Phi ^L](y), \end{aligned}$$

(7.5)

which is comparable to the function in Table 1.

To establish the second theorem, first observe from (7.4) and Corollary 5.3 that

$$\begin{aligned} P^{*,\mathfrak {L}}_{t_0}f(x)\lesssim & {} \frac{C_1(\sqrt{x})}{x^{1/4}}\,\mathcal {M}^\mathrm{loc}_M \left( \sqrt{y}f(y^2)e^{-\frac{y^2}{2}}\right) (\sqrt{x})\,+\,\frac{C_2\left( \sqrt{x}\right) }{x^{1/4}}\,\int _{{{\mathbb R}_+}} \sqrt{2y}|f(y^2)|\Phi ^L(y).\end{aligned}$$

We claim that this inequality can be rewritten as

$$\begin{aligned} P^{*,\mathfrak {L}}_{t_0}f(x)\lesssim & {} C_1\left( \sqrt{x}\right) \,\mathcal {M}^\mathrm{loc}_{\frac{1}{4},M^2} \left( fe^{-\frac{y}{2}}\right) (x)\,+\,\frac{C_2\left( \sqrt{x}\right) }{x^{1/4}}\,\int _{{{\mathbb R}_+}} |f(u)|\Phi ^\mathfrak {L}(u),\quad \end{aligned}$$

(7.6)

where $\mathcal {M}^\mathrm{loc}_{\frac{1}{4},M^2}$ stands for the local maximal function in (1.10) with the cutoff replaced by $\chi _{\{\frac{x}{4}\le y\le M^2 x\}}$. The expression for the second term is clear from (7.5) (after a change of variables $y^2=u$). To handle the first term, notice that the local region now becomes $\frac{\sqrt{x}}{2}<y<M\sqrt{x}$, which in particular gives $x^{-\frac{1}{4}}\sqrt{y}\approx 1$. We also need the following lemma for the maximal function.

Lemma 7.1

For every function $g:{{\mathbb R}_+}\rightarrow {\mathbb C}$ and every $x\in {{\mathbb R}_+}$, we have

$$\begin{aligned} \mathcal M\left( g(y^2)\chi _{\{\frac{\sqrt{x}}{2}<y<M\sqrt{x}\}}\right) \left( \sqrt{x}\right) \,\lesssim \, \mathcal M\left( g(u)\chi _{\{\frac{x}{4}<u<M^2x\}}\right) (x).\end{aligned}$$

Moreover, x is a Lebesgue point of g if and only if $\sqrt{x}$ is a Lebesgue point of $g(y^2)$.

Proof

The first assertion follows essentially from the change of variables $y^2=u$,

$$\begin{aligned} LHS\le & {} \sup _{r>0}\frac{1}{r}\int _{|y-\sqrt{x}|<r}|g(y^2)|\,\chi _{\left\{ \frac{x}{4}<y^2<M^2x\right\} }\,\mathrm{d}y\\= & {} \sup _{r>0}\frac{1}{r}\int _{|\sqrt{u}-\sqrt{x}|<r}|g(u)|\,\chi _{\left\{ \frac{x}{4}<u<M^2x\right\} }\,\frac{\hbox {d}u}{2\sqrt{u}}.\end{aligned}$$

In this local range, we have $\sqrt{u}\approx \sqrt{x}$, so may take the denominator outside the integral. The local behavior also implies that

$$\begin{aligned} |\sqrt{u}-\sqrt{x}|=\left| \int _x^u\frac{\hbox {d}s}{2\sqrt{s}}\right| \,\approx \,\frac{|u-x|}{\sqrt{x}}. \end{aligned}$$

(7.7)

Therefore, we conclude that

$$\begin{aligned} LHS\,\lesssim \, \sup _{r>0}\frac{1}{r\sqrt{x}}\int _{|u-x|<r\sqrt{x}}|g(u)|\,\chi _{\{\frac{x}{4}<u<M^2x\}}\,\mathrm{d}u\,\lesssim \,RHS. \end{aligned}$$

For the last assertion, we only show that $x\in \mathcal L_g$ implies $\sqrt{x}\in \mathcal L_{{\tilde{g}}}$ with ${\tilde{g}}(y)=g(y^2)$ (the converse is similar). As before, the change in variables $y=\sqrt{u}$ gives

$$\begin{aligned} \mathcal I_r(x):=\frac{1}{r}\int _{|y-\sqrt{x}|<r}|{\tilde{g}}(y)-{\tilde{g}}(\sqrt{x})|\,\mathrm{d}y\, = \, \frac{1}{r}\int _{|\sqrt{u}-\sqrt{x}|<r}|g(u)-g(x)|\,\frac{\hbox {d}u}{2\sqrt{u}}. \end{aligned}$$

If r is sufficiently small (e.g., $r<\sqrt{x}/2$), we have $\sqrt{u}\approx \sqrt{x}$ and also $|\sqrt{u}-\sqrt{x}|\approx |u-x|/\sqrt{x}$, by (7.7). Thus,

$$\begin{aligned} \mathcal I_r(x)\lesssim \frac{1}{r\sqrt{x}}\int _{|u-x|\le c\,r\sqrt{x}}|g(u)-g(x)|\,\mathrm{d}u \ \end{aligned}$$

which vanishes as $r\rightarrow 0$ since $x\in \mathcal L_g$. $\square$

We have thus shown (7.6). From here, one proves Theorem 1.2 (for the operator $\mathfrak {L}$) arguing once again as in Sect. 6.2. Remark that, in view of the new constants $C_1$ and $C_2$ in (7.6), the weight $v=\min \{v_{1,{\varepsilon }},v_2\}$ must be defined with

$$\begin{aligned} v_{1,{\varepsilon }}(x)=\frac{e^{-\frac{px}{2}}}{(1+x)^{p\nu }}\, \mathscr {V}(x)\rho _{{\varepsilon }}\Big ({\mathscr {V}(x)}\Big ){\quad \text{ and }\quad }v_2(x)=\,\frac{\langle {x}\rangle ^{\left( \frac{{\alpha }}{2}+1\right) p-1}}{[{{\text {ln}}\,}(e/\langle {x}\rangle )]^2}\,\frac{e^{-\frac{p}{2}x}}{(1+x)^{N}}\,, \end{aligned}$$

where $\mathscr {V}(x)=\left[ \mathcal {M}^\mathrm{loc}_{\frac{1}{4},M^2}\left( w^{-\frac{p'}{p}}e^{-\frac{p'x}{2}}\right) (x)\right] ^{1-p}$ and $N>1+\frac{p}{2}\left( |\mu +\frac{1}{2}|-\frac{1}{2}\right)$. Then, the same proof as before gives that $P^{*,\mathfrak {L}}_{t_0}:L^p(w)\rightarrow L^p(v)$ if $w\in D_p(\Phi ^\mathfrak {L})$. One can also establish (with a few obvious modifications, such as using the class $D^{\mathrm{exp}_1}_p$ in Proposition 3.3 ii) that for every given $q>p$, there is a sufficiently small ${\varepsilon }$ so that $v\in D_q(\Phi ^\mathfrak {L})$. Once again, we may also replace this weight by $v^{\Phi ^\mathfrak {L},w}(x)$, as defined in (6.13).

7.3 Results for the system $\ell ^{\alpha }_n$

Remember that these functions satisfy

$$\begin{aligned} \ell ^{\alpha }_n(y)\,=\,a(y)\,\mathfrak {L}^\alpha _n(y),\quad \text{ with } a(y)=y^{-\frac{{\alpha }}{2}}.\end{aligned}$$

(7.8)

Thus, they are eigenvectors of the differential operator

$$\begin{aligned} f\mapsto \mathscr {L}f(x)=a(x)\mathfrak {L}[a^{-1}f](x) \end{aligned}$$

(with the same eigenvalues as $\mathfrak {L}^\alpha _n$) and constitute an orthonormal system in $L^2(a(y)^{-2}\mathrm{d}y)$. One then derives Theorems 1.1 and 1.2 for the operator $\mathscr {L}$, from the known results about $\mathfrak {L}$, by repeating exactly the same arguments that we gave in Sect. 7.1. We leave the details to the reader.

7.4 Results for the Laguerre polynomials $L^{\alpha }_n$

Since ${\mathbb L}$ is the most classical Laguerre operator, we shall give a few more details here. First of all, recall that $L^{\alpha }_n$ and $\mathfrak {L}^\alpha _n$ are linked by

$$\begin{aligned} L^{\alpha }_n(y)\,=\,a(y)\,\mathfrak {L}^\alpha _n(y),\quad \text{ with } a(y)=y^{-\frac{{\alpha }}{2}}e^{y/2}.\end{aligned}$$

(7.9)

Thus, the functions $L^{\alpha }_n$ are orthonormal in $L^2(a(y)^{-2}\mathrm{d}y)=L^2(y^{\alpha }e^{-y}\mathrm{d}y)$ and are also eigenvectors of the differential operator

$$\begin{aligned} f\mapsto {\mathbb L}f(x)=a(x)\mathfrak {L}[a^{-1}f](x), \end{aligned}$$

with the same eigenvalues as $\mathfrak {L}^\alpha _n$, namely $n+({\alpha }+1+\mu )/2$. We remark that ${\mathbb L}$ coincides with the operator in Table 1. Thus, the heat and Poisson kernels of these two operators are related by

$$\begin{aligned} e^{-t{\mathbb L}}(x,y)\,=\,\sum _{n=0}^\infty e^{-\left( n+\frac{{\alpha }+\mu +1}{2}\right) t}L^{\alpha }_n(x)L^{\alpha }_n(y)\,=\, a(x)\,a(y)\,e^{-t\mathfrak {L}}(x,y), \end{aligned}$$

and

$$\begin{aligned} P_t^{{\mathbb L}}(x,y)\,=\,a(x)\,a(y)\,P_t^{\mathfrak {L}}(x,y). \end{aligned}$$

This implies the identity

$$\begin{aligned} P_t^{\mathbb L}f(x)\,=\,\int _{{{\mathbb R}_+}}P_t^{\mathbb L}(x,y)\,f(y)\,y^{\alpha }e^{-y}\,\mathrm{d}y\,= \,a(x)\,P^\mathfrak {L}_t\left[ fy^{\frac{{\alpha }}{2}}e^{-\frac{y}{2}}\right] (x),\end{aligned}$$

(7.10)

from which one deduces the validity of Theorem 1.1 for the operator ${\mathbb L}$, provided

$$\begin{aligned} \Phi ^{\mathbb L}(x)\,=\,y^{\frac{{\alpha }}{2}}e^{-\frac{y}{2}}\Phi ^\mathfrak {L}(y)\,=\, \frac{y^{\alpha }e^{-y}}{(1+y)^{\frac{1+{\alpha }+\mu }{2}}[{{\text {ln}}\,}(e+x)]^{1+\nu }}. \end{aligned}$$

Note that this coincides with the function in Table 1. Moreover, (7.10) combined with (7.6) implies the estimate

$$\begin{aligned} P^{*,{\mathbb L}}_{t_0}f(x) \, \lesssim \, C^{\mathbb L}_1(x)\,\mathcal {M}^\mathrm{loc}_{\frac{1}{4},M^2} \big (fe^{-{y}}\big )(x)\,+ \,C^{\mathbb L}_2(x)\,\int _{{{\mathbb R}_+}} |f(u)|\Phi ^{\mathbb L}(u)=\,I(x)+II(x), \end{aligned}$$

(7.11)

with the new constants

$$\begin{aligned} C^{\mathbb L}_1(x)=(1+x)^\nu e^x{\quad \text{ and }\quad }C^{\mathbb L}_2(x)=[\ln (1+x)]^{1+\nu }\,(1+x)^{\left( |\mu +\frac{1}{2}|-{\alpha }-\frac{1}{2}\right) /2} e^x/\langle {x}\rangle ^{{\alpha }+1}. \end{aligned}$$

We now apply the same arguments as in Sect. 6.2 to show that, for a suitable weight v, we have $\Vert P^{*,{\mathbb L}}_{t_0}f\Vert _{L^p(v)}\lesssim \Vert f\Vert _{L^p(w)}$, under the assumption $w\in D_p(\Phi ^{{\mathbb L}})$. Indeed, to control the second term II(x) we choose a weight $v_2$ such that $C_2^{\mathbb L}\in L^p(v_2)$, namely

$$\begin{aligned} v_2(x)=\,\frac{\langle {x}\rangle ^{({\alpha }+1)p-1}}{[{{\text {ln}}\,}(e/\langle {x}\rangle )]^2}\,\frac{e^{-px}}{(1+x)^{N}}\,, \end{aligned}$$

with say $N>1+p(\frac{1+{\alpha }+\mu }{2}+|{\alpha }+\frac{1}{2}|)$. It is not difficult to verify that $v_2\in D_q(\Phi ^{\mathbb L})$ for all $q>p$. To control the first term, we set

$$\begin{aligned} v_{1,{\varepsilon }}(x)=\frac{e^{-px}}{(1+x)^{p\nu }}\, \mathscr {V}(x)\rho _{{\varepsilon }}\Big ({\mathscr {V}(x)}\Big ) \text{ with } \mathscr {V}(x)=\left[ \mathcal {M}^\mathrm{loc}_{\frac{1}{4},M^2}\left( w^{-\frac{p'}{p}}e^{-p'x}\right) (x)\right] ^{1-p}.\end{aligned}$$

That is, if $W(x)=w(x)e^{px}$, then $v_{1,{\varepsilon }}(x)=(1+x)^{-p\nu }e^{-px}\,V_{\varepsilon }(x)$ with the notation in (3.1). So we may use Theorem 3.1 to obtain

$$\begin{aligned} \big \Vert I\big \Vert _{L^p(v_{1,{\varepsilon }})}\,\lesssim \,\left[ \int _{{\mathbb R}_+}\left| \mathcal {M}^\mathrm{loc}_{\frac{1}{4},M^2} \left( fe^{-{y}}\right) (x)\right| ^p\,V_{{\varepsilon }}(x)\mathrm{d}x\right] ^{\frac{1}{p}}\, \lesssim \,\Vert fe^{-{y}}\Vert _{L^p(W)}=\Vert f\Vert _{L^p(w)}. \end{aligned}$$

Again, it is not difficult to verify that for a sufficiently small ${\varepsilon }$ one has $v_{1,{\varepsilon }}\in D_q(\Phi ^{\mathbb L})$, arguing as in the last part⁴ of Sect. 6.2. Thus, Theorem 1.2 holds with $v=\min \{v_{1,{\varepsilon }},v_2\}$. Alternatively, with the notation in (6.13), one may as well choose the weight $v^{\Phi ^{\mathbb L},w}(x)$.