On Schrödinger Oscillatory Integrals Associated with the Dunkl Transform

Abstract

In the paper we study the Schrödinger oscillatory integrals $T^t_{\lambda ,a}f(x)$ ($\lambda \ge 0$, $a>1$) associated with the one-dimensional Dunkl transform ${\mathscr {F}}_{\lambda }$. If $a=2$, the function $u(x,t):=T^t_{\lambda ,2}f(x)$ solves the free Schrödinger equation associated to the Dunkl operator, with f as the initial data. It is proved that, if f is in the Sobolev spaces $H^s_{\lambda }({\mathbb {R}})$ associated with the Dunkl transform, with the exponents s not less than 1 / 4, then $T^t_{\lambda ,a}f$ converges almost everywhere to f as $t\rightarrow 0$. A counterexample is constructed to show that 1 / 4 can not be improved for $a=2$, and when $1/4\le s\le 1/2$, the Hausdorff dimension of the divergence set of $T^t_{\lambda ,a}f$ for $f\in H_{\lambda }^s({\mathbb {R}})$ is proved to be $1-2s$ at most.

1 Introduction and Main Results

For $\lambda \ge 0$ and $1\le p<\infty$, we denote by $L_{\lambda }^p({\mathbb {R}})$ the space of measurable functions $\phi$ on ${\mathbb {R}}$ satisfying $\Vert \phi \Vert _{L_{\lambda }^p}:=\left\{ c_{\lambda }\int _{{\mathbb {R}}}|\phi (x)|^p|x|^{2\lambda }dx\right\} ^{1/p}<\infty$ with $c_{\lambda }^{-1}=2^{\lambda +1/2}\Gamma (\lambda +1/2)$. The Dunkl operator on the line ${\mathbb {R}}$ is

$$\begin{aligned} (D\phi )(x)=\phi '(x)+\frac{\lambda }{x}\left[ \phi (x)-\phi (-x)\right] , \end{aligned}$$

involving a reflection part, and the Dunkl transform of a function $\phi \in L_{\lambda }^1({\mathbb {R}})$ is defined by

$$\begin{aligned} ({\mathscr {F}}_{\lambda }\phi )(\rho ):=c_{\lambda }\int _{{\mathbb {R}}}\phi (x)E_{\lambda }(-ix\rho )|x|^{2\lambda }dx,\quad \rho \in {\mathbb {R}}, \end{aligned}$$

(1)

where $E_{\lambda }$ is the Dunkl kernel

$$\begin{aligned} E_{\lambda }(z)=j_{\lambda -1/2}(iz)+\frac{z}{2\lambda +1}j_{\lambda +1/2}(iz),\quad z\in {\mathbb {C}}, \end{aligned}$$

(2)

and $j_{\alpha }(z)$ is the normalized Bessel function

$$\begin{aligned} j_{\alpha }(z)=2^{\alpha }\Gamma (\alpha +1)\frac{J_{\alpha }(z)}{z^{\alpha }} =\Gamma (\alpha +1)\sum _{n=0}^{\infty }\frac{(-1)^n(z/2)^{2n}}{n!\Gamma (n+\alpha +1)}. \end{aligned}$$

(3)

Since $j_{-1/2}(z)=\cos z$, $j_{1/2}(z)=z^{-1}\sin z$, it follows that $E_0(iz)=e^{iz}$ and ${\mathscr {F}}_0$ agrees with the usual Fourier transform on ${\mathbb {R}}$.

For $s\ge 0$, the Sobolev space $H^s_{\lambda }({\mathbb {R}})$ associated with the Dunkl transform is the collection of all $f\in {\mathscr {S}}'({\mathbb {R}})$ (the space of tempered distributions on ${\mathbb {R}}$) satisfying

$$\begin{aligned} \Vert f\Vert _{H^s_{\lambda }}:=\left\{ c_{\lambda }\int _{{\mathbb {R}}}\left| ({\mathscr {F}}_{\lambda }f)(\rho )\right| ^2(1+|\rho |^2)^s|\rho |^{2\lambda }d\rho \right\} ^{1/2}<\infty . \end{aligned}$$

The homogeneous analog of $H^s_{\lambda }({\mathbb {R}})$ is denoted by $\dot{H}^s_{\lambda }({\mathbb {R}})$, which consists of the elements $f\in {\mathscr {S}}'({\mathbb {R}})$ satisfying

$$\begin{aligned} \Vert f\Vert _{\dot{H}^s_{\lambda }}:=\left\{ c_{\lambda }\int _{{\mathbb {R}}}\left| ({\mathscr {F}}_{\lambda }f)(\rho )\right| ^2|\rho |^{2(s+\lambda )}d\rho \right\} ^{1/2}<\infty . \end{aligned}$$

It is obvious that, for $s\ge 0$, $H^s_{\lambda }({\mathbb {R}})=L^2_{\lambda }({\mathbb {R}})\bigcap \dot{H}^s_{\lambda }({\mathbb {R}})$ and $\Vert f\Vert _{H^s_{\lambda }}\asymp \Vert f\Vert _{L^2_{\lambda }}+\Vert f\Vert _{\dot{H}^s_{\lambda }}$.

Given $a>0$, we concern with the family of operators $T^t_{\lambda ,a}$ with $t\in {\mathbb {R}}$, initially defined for $f\in {\mathscr {S}}({\mathbb {R}})$ (the space of Schwartz functions on ${\mathbb {R}}$) by

$$\begin{aligned} T^t_{\lambda ,a}f(x)=c_{\lambda }\int _{{\mathbb {R}}}({\mathscr {F}}_{\lambda }f)(\rho )e^{it|\rho |^a}E_{\lambda }(ix\rho )|\rho |^{2\lambda }d\rho ,\quad x\in {\mathbb {R}}, \end{aligned}$$

(4)

and call $T^t_{\lambda ,a}f$ the Schrödinger oscillatory integral associated with the Dunkl transform.

If $a=2$ and $f\in {\mathscr {S}}({\mathbb {R}})$, then the function $u(x,t):=T^t_{\lambda ,a}f(x)$ is the solution to the free Schrödinger equation associated to the Dunkl operator, with f as the initial data, that is,

$$\begin{aligned} \left\{ \begin{array}{l} i\frac{\partial u}{\partial t}(x,t)=\Delta _{\lambda }u(x,t),\qquad (x,t)\in {\mathbb {R}}\times {\mathbb {R}},\\ u(x,0)=f(x), \end{array}\right. \end{aligned}$$

(5)

where $\Delta _{\lambda }=D_x^2$, or explicitly, for $\phi \in C^2({\mathbb {R}})$,

$$\begin{aligned} \left( \Delta _{\lambda }\phi \right) (x)=\frac{\partial ^2}{\partial x^2}\phi (x)+\frac{2\lambda }{x}\frac{\partial }{\partial x}\phi (x)-\frac{\lambda }{x^2}[\phi (x)-\phi (-x)]. \end{aligned}$$

(6)

For given f, the solution to the problem (5) may be in symbol written as $u(x,t)=e^{-it\Delta _{\lambda }}f(x)$.

We remark that, as a family of multiplier operators with multipliers $e^{it|\rho |^a}$, uniformly bounded for $t\in {\mathbb {R}}$, $T^t_{\lambda ,a}$ can be extended to bounded operators on $L_{\lambda }^2({\mathbb {R}})$, by the Plancherel theorem for the Dunkl transform (see Proposition 2.3(v) in Sect. 2).

One of the fundamental problems is, for $f\in H_{\lambda }^s({\mathbb {R}})$ with suitable s, to identify the exponents s for which,

$$\begin{aligned} \lim _{t\rightarrow 0}T^t_{\lambda ,a}f(x)=f(x)\qquad \hbox {for a.e.}\,\,x\in {\mathbb {R}}. \end{aligned}$$

(7)

The answer to the above question is contained in the following theorem.

Theorem 1.1

Let $a>1$ and $\lambda \ge 0$. If $s\ge 1/4$, then for all $f\in H_{\lambda }^s({\mathbb {R}})$, the assertion (7) is true in the sense that, for a.e. $x\in {\mathbb {R}}$, the function $t\mapsto T^t_{\lambda ,a}f(x)$, $t\in {\mathbb {R}}\setminus \{0\}$, will, after modification on a null set, be continuous with limit f(x) as $t\rightarrow 0$.

The proof of Theorem 1.1 is essentially based on an $L^q$ estimate for the relevant maximal function, with which, an almost sharp result is given in the next theorem.

Theorem 1.2

Let $a>1$, $\lambda \ge 0$ and $q=\frac{8\lambda +4}{4\lambda +1}$. Then for all $f\in {\mathscr {S}}({\mathbb {R}})$, we have the following norm estimate

$$\begin{aligned} \left( \int _{{\mathbb {R}}}\left( \sup _{0<|t|<\infty }\left| T^t_{\lambda ,a}f(x)\right| \right) ^q|x|^{2\lambda }dx\right) ^{\frac{1}{q}}\lesssim \Vert f\Vert _{\dot{H}^{1/4}_{\lambda }}. \end{aligned}$$

(8)

When $\lambda =0$, (5) is identical with the one spatial dimension case of the initial value problem for the classical Schrödinger equation

$$\begin{aligned} \left\{ \begin{array}{lll} i\frac{\partial u}{\partial t}(x,t)=\left( \Delta u\right) (x,t),\qquad (x,t)\in {\mathbb {R}}^d\times {\mathbb {R}},\\ u(x,0)=f(x),\qquad x\in {\mathbb {R}}^d. \end{array}\right. \end{aligned}$$

(9)

For a Schwartz function f on ${\mathbb {R}}^d$, if $\hat{f}$ denotes its Fourier transform, i.e., $\hat{f}(\xi )=(2\pi )^{-d/2}\int _{{\mathbb {R}}^d}f(x)e^{-i\langle x,\xi \rangle }dx$, the solution to (9), formally written as $u(x,t)=e^{-it\Delta }f(x)$, is given by

$$\begin{aligned} u(x,t)=\frac{1}{(2\pi )^{d/2}}\int _{{\mathbb {R}}^d}\hat{f}(\xi )e^{i\left( \langle x,\xi \rangle +t|\xi |^2\right) }d\xi . \end{aligned}$$

(10)

As usual, the Sobolev space $H^s({\mathbb {R}}^d)$ for $s\ge 0$ consists of all $\phi \in L^2({\mathbb {R}}^d)$ satisfying $\Vert \phi \Vert ^2_{H^s}:=\int _{{\mathbb {R}}^d}|\hat{\phi }(\xi )|^2(1+|\xi |^2)^sd\xi <\infty$. For $f\in H^s({\mathbb {R}}^d)$, the problem to find the exponents s for which,

$$\begin{aligned} \lim _{t\rightarrow 0}e^{-it\Delta }f(x)=f(x)\qquad \hbox {for a.e.}\,\,x\in {\mathbb {R}}^d, \end{aligned}$$

(11)

has been studied by various authors. Carleson [7] yielded the first conclusion about the problem, proving convergence for $s\ge 1/4$ when $d=1$. Dahlberg and Kenig [8] showed that this result is sharp. In dimensions $d\ge 2$, Sjölin [23] and Vega [28] established independently convergence for $s>1/2$, while similar examples as considered in [8] show failure of convergence for $s<1/4$.

The problem for $d=2$ has been made more progress. The proof of convergence for some $s<1/2$ appears first in Bourgain’s papers [3, 4]. Subsequently, the required threshold was lowered by Moyua et al. [18] and Tao and Vargas [26, 27]; and in Lee [13], $s>3/8$ was proved to be sufficient. The strongest result to date appears in Du et al. [10] which asserts a.e. convergence for $f\in H^s({\mathbb {R}}^2)$, $s>1/3$.

If the problem (11) is restricted to radial initial data f on ${\mathbb {R}}^d$, $d\ge 2$, Prestini [19] proved that $s\ge 1/4$ is sufficient to guarantee a.e. convergence of $e^{it\Delta }f(x)$, and Sjölin [24] obtained a sharp maximal estimate like (8) with $q=4d/(2d-1)$.

One seemed to have reason to speculate that, the correct condition in arbitrary dimension d should be $s\ge 1/4$. After a longer time when no advances were achieved, the problem for $d>2$ was driven by progress in Bourgain’s paper [5], where, that $f\in H^s({\mathbb {R}}^d)$, $s>1/2-1/(4d)$, was proved to be sufficient for a.e. convergence. However, another result in the same paper [5] of Bourgain breaks the illusion of the exponent $s=1/4$ in higher dimension, that is, the requirement $s>1/2-1/d$ is necessary when $d>4$. In this direction, the range was improved by ${\mathrm{Luc\grave{a}}}$, Rogers [15], to $s>d/2(d+2)$ for all $d\ge 2$, and furthermore, Bourgain [6] proved that $s\ge d/2(d+1)$ is necessary for $d\ge 2$.

Combining the assertions in Bourgain [6] and Du et al. [10] shows that $s=1/3$ is optimal for $d=2$ up to the endpoint. In the higher dimensions, there is some evidence to suggest that the exponent $s=d/2(d+1)$ could be sharp too.

The paper is organized as follows. Section 2 contains some preliminaries about the Dunkl transform and the Dunkl operator which is necessary subsequently; the key lemmas about the estimates for an associated oscillatory integral involving the Dunkl kernel are given in Sect. 3; the main results, i.e., Theorems 1.1 and 1.2, are proved in Sect. 4; and in Sect. 5, we construct a function $f_0$ to show that Theorem 1.1 with $a=2$ is not true for any $s<1/4$, and when $1/4\le s\le 1/2$, we prove that the Hausdorff dimension of the divergence set of $T^t_{\lambda ,a}f$ for $f\in H_{\lambda }^s({\mathbb {R}})$ is $1-2s$ at most. There are closing comments in Sect. 6, about higher dimensional counterparts of the free Schrödinger equation associated to the Dunkl operators.

Throughout the paper, $X\lesssim Y$ or $Y\gtrsim X$ means that $X\le cY$ for some positive constant c independent of variables, functions, etc., but possibly dependent of some fixed parameters $\lambda ,s,a$ and q.

2 The Dunkl Transform and the Dunkl Operator on ${\mathbb {R}}$

Lemma 2.1

Assume that $\lambda \ge 0$.

(i) The Dunkl operator D leaves ${\mathscr {D}}({\mathbb {R}})$ (the space of $C^{\infty }$-functions on ${\mathbb {R}}$ with compact supports) and ${\mathscr {S}}({\mathbb {R}})$ invariant.

(ii) If $\phi ,\psi \in C^1[a,b]$, then

$$\begin{aligned} \int _a^b(D\phi )(x)\psi (x)|x|^{2\lambda }dr=&\left. \phi (x)\psi (x)|x|^{2\lambda }\right| _{x=a}^b-\int _a^b\phi (x)(D\psi )(x)|x|^{2\lambda }dx\nonumber \\&\quad -\int _a^b\frac{\lambda }{x}\left( \phi (x)\psi (-x)+\phi (-x)\psi (x)\right) |x|^{2\lambda }dx; \end{aligned}$$

(12)

and in particular, if $\phi ,\psi \in C^1[{\mathbb {R}}]$ and $\phi (x)\psi (x)|x|^{2\lambda }$ vanishes at infinity, then

$$\begin{aligned} c_{\lambda }\int _{{\mathbb {R}}}(D\phi )(x)\psi (x)|x|^{2\lambda }dx=-c_{\lambda }\int _{{\mathbb {R}}}\phi (x)(D\psi )(x)|x|^{2\lambda }dx \end{aligned}$$

(13)

provided one of the two integral exists.

(iii) For $\phi ,\psi \in C^1({\mathbb {R}})$,

$$\begin{aligned}{}[D(\phi \psi )](x)&=(D\phi )(x)\psi (x)+\phi (x)(D\psi )(x)\nonumber \\&\quad -\frac{\lambda }{x}[\phi (x)-\phi (-x)][\psi (x)-\psi (-x)]; \end{aligned}$$

(14)

and in particular, if $\psi$ is even, then

$$\begin{aligned}{}[D(\phi \psi )](x)=(D\phi )(x)\psi (x)+\phi (x)(D\psi )(x). \end{aligned}$$

(15)

Part (i) follows from the equalities

$$\begin{aligned} (Df)(x)=f'(x)+\lambda \int _{|s|\le 1}f'(xs)ds=f'(x)-\lambda \int _{|s|\ge 1}f'(xs)ds. \end{aligned}$$

For part (ii), we begin at the left hand side of (13) and split the integral into a sum of $\int _{-\infty }^0$ and $\int _0^{\infty }$. After integrating each term by parts, the equality (13) follows from a simple change of variables. Part (iii) is verified by direct calculations.

Lemma 2.2

Assume that $\lambda \ge 0$.

(i) $E_{\lambda }(ix\rho )$ is the eigenfunction of $D_x$ with $i\rho$ as its eigenvalue, that is

$$\begin{aligned} D_x[E_{\lambda }(ix\rho )]=i\rho E_{\lambda }(ix\rho ), \ \ \ \ \hbox {and} \ \ \ \ E_{\lambda }(ix\rho )|_{x=0}=1. \end{aligned}$$

(16)

(ii) $\left| E_{\lambda }(ix)\right| \lesssim \left( 1+|x|\right) ^{-\lambda },\,\, x\in (-\infty ,\infty )$.

To show part (i), we apply the formula $\left( z^{\alpha +1}J_{\alpha +1}(z)\right) '=z^{\alpha +1}J_{\alpha }(z)$ (see [12, 7-2(50)]) to obtain

$$\begin{aligned} \left( z^{-\alpha }J_{\alpha +1}(z)\right) '&=\left( z^{-2\alpha -2}\cdot z^{\alpha +1}J_{\alpha +1}(z)\right) '\\&=z^{-\alpha }J_{\alpha }(z)-(2\alpha +1)z^{-\alpha -1}J_{\alpha +1}(z), \end{aligned}$$

which may be written as

$$\begin{aligned} \left( zj_{\alpha +1}(z)\right) '+(2\alpha +1)j_{\alpha +1}(z)=2(\alpha +1)j_{\alpha }(z). \end{aligned}$$

Furthermore, from [12, 7-2(51)] we have $\left( j_{\alpha }(z)\right) '=-zj_{\alpha +1}(z)/(2\alpha +2)$. Taking action of $D_x$ to the real and the imagine parts of $E_{\lambda }(ix\rho )$ respectively, and applying the two equalities above, we prove the first formula in part (i). The second one is obvious, by (2) and (3).

From [12, 7-3(4)], $|j_{\alpha }(x)|\lesssim 1$ for $|x|\le 1$, and from [12, 7-13(3)], $|j_{\alpha }(x)|\lesssim |x|^{-\alpha -1/2}$ for $|x|\ge 1$, so that

$$\begin{aligned} |j_{\alpha }(x)|\lesssim (1+|x|)^{-\alpha -1/2},\qquad \hbox {for}\quad x\in (-\infty ,\infty ). \end{aligned}$$

(17)

Thus part (ii) follows immediately.

The Dunkl transform shares many of the important properties with the usual Fourier transform, part of which are listed as follows.

Proposition 2.3

(cf. [14]) Assume that $\lambda \ge 0$.

(i) If $\phi \in L_{\lambda }^1({\mathbb {R}})$, then ${\mathscr {F}}_{\lambda }\phi \in C_0({\mathbb {R}})$ and $\Vert {\mathscr {F}}_{\lambda }\phi \Vert _{\infty }\le \Vert \phi \Vert _{L_{\lambda }^1}$.

(ii) (Inversion) If $\phi \in L_{\lambda }^1({\mathbb {R}})$ such that ${\mathscr {F}}_{\lambda }\phi \in L_{\lambda }^1({\mathbb {R}})$, then $\phi (x)=[{\mathscr {F}}_{\lambda }({\mathscr {F}}_{\lambda }\phi )](-x)$ for a.e. $x\in {\mathbb {R}}$.

(iii) For $\phi \in {\mathscr {S}}({\mathbb {R}})$ or $\phi \in C^1({\mathbb {R}})$ having compact support, we have

$$\begin{aligned}{}[{\mathscr {F}}_{\lambda }(D\phi )](\rho )=i\rho ({\mathscr {F}}_{\lambda }\phi )(\rho ),\quad [{\mathscr {F}}_{\lambda }(x\phi )](\rho )=i[D_{\rho }({\mathscr {F}}_{\lambda }\phi )](\rho ),\quad \rho \in {\mathbb {R}}; \end{aligned}$$

and ${\mathscr {F}}_{\lambda }$ is a topological automorphism on $\mathscr {S}({\mathbb {R}})$.

(iv) (product formula) For all $\phi ,\psi \in L_{\lambda }^1({\mathbb {R}})$, we have

$$\begin{aligned} c_{\lambda }\int _{{\mathbb {R}}}({\mathscr {F}}_{\lambda }\phi )(x)\psi (x)|x|^{2\lambda }dx=c_{\lambda }\int _{{\mathbb {R}}}\phi (x)({\mathscr {F}}_{\lambda }\psi )(x)|x|^{2\lambda }dx. \end{aligned}$$

(v) (Plancherel) There exists a unique extension of ${\mathscr {F}}_{\lambda }$ to $L_{\lambda }^2({\mathbb {R}})$ with $\Vert {\mathscr {F}}_{\lambda }\phi \Vert _{L_{\lambda }^2}=\Vert \phi \Vert _{L_{\lambda }^2}$ and $({\mathscr {F}}_{\lambda }^{-1}\phi )(x)=({\mathscr {F}}_{\lambda }\phi )(-x)$.

(vi) (Hausdorff-Young) If $1\le p\le 2$, then ${\mathscr {F}}_{\lambda }\phi$ for $\phi \in L_{\lambda }^p({\mathbb {R}})$ is well defined and $\Vert {\mathscr {F}}_{\lambda }\phi \Vert _{L_{\lambda }^{p'}}\le \Vert \phi \Vert _{L_{\lambda }^p}$, where $1/p+1/p'=1$.

It follows from Lemmas 2.1(i), 2.2(i), and Proposition 2.3(ii) that, for $f\in {\mathscr {S}}({\mathbb {R}})$, the function $u(x,t):=T^t_{\lambda ,2}f(x)$ is the solution to the initial value problem (5).

If $(x,y)\ne (0,0)$, the generalized translation $(\tau _yf)(x)$ of $f\in L_{\lambda }^p({\mathbb {R}})$ ($1\le p<\infty$) associated to the Dunkl transform ${\mathscr {F}}_{\lambda }$ is defined by (cf. [20])

$$\begin{aligned} (\tau _yf)(x)=c'_{\lambda }\int _0^\pi \left( f_e(\langle x,y\rangle _\theta )+ f_o(\langle x,y\rangle _\theta )\frac{x+y}{\langle x,y\rangle _\theta }\right) (1+\cos \theta )\sin ^{2\lambda -1}\theta d\theta ,\nonumber \\ \end{aligned}$$

(18)

where $c'_{\lambda }=\Gamma (\lambda +1/2)/(\Gamma (\lambda )\Gamma (1/2))$, $f_e(x)=(f(x)+f(-x))/2$, $f_o(x)=(f(x)-f(-x))/2$, $\langle x,y\rangle _\theta =\sqrt{x^2+y^2+2xy\cos \theta }$; and if $(x,y)=(0,0)$, $(\tau _yf)(x)=f(0)$. For two appropriate functions f and g on ${\mathbb {R}}$, their generalized convolution $f*_{\lambda }g$ is defined by

$$\begin{aligned} (f*_{\lambda } g)(x)=c_{\lambda }\int _{{\mathbb {R}}}f(y)(\tau _xg)(-y)|y|^{2\lambda }dy,\qquad x\in {\mathbb {R}}. \end{aligned}$$

(19)

The associated Poisson integral of a function f, for $\epsilon >0$, is given by $(P_{\epsilon }f)(x)=(f*_{\lambda }P_{\epsilon })(x)$, where $P_{\epsilon }(x)=m_{\lambda }\epsilon (\epsilon ^2+x^2)^{-\lambda -1}$ with $m_\lambda =2^{\lambda +1/2}\Gamma (\lambda +1)/\sqrt{\pi }$.

We shall also need the Dunkl-Stieltjies transform for $d\mu \in {\mathfrak {B}}_{\lambda }({\mathbb {R}})$ (Borel measure on ${\mathbb {R}}$ with $\Vert d\nu \Vert _{{\mathfrak {B}_{\lambda }}} :=c_{\lambda }\int _{{\mathbb {R}}}|x|^{2\lambda }|d\nu (x)|<\infty$) defined by

$$\begin{aligned}{}[{\mathscr {F}}_{\lambda }(d\mu )](\xi ):=c_{\lambda }\int _{{\mathbb {R}}}E_\lambda (-ix\xi )|x|^{2\lambda }d\mu (x),\quad \xi \in {\mathbb {R}}, \end{aligned}$$

and the generalized convolution $f*_{\lambda }(d\mu )$ of $f\in C_0({\mathbb {R}})$ and $d\mu \in {\mathfrak {B}}_{\lambda }({\mathbb {R}})$, by

$$\begin{aligned}{}[f*_{\lambda }(d\mu )](x)=c_{\lambda }\int _{{\mathbb {R}}}(\tau _xf)(-y)|y|^{2\lambda }d\mu (y),\quad \quad x\in {\mathbb {R}}. \end{aligned}$$

(20)

Proposition 2.4

(cf. [14]) Assume that $\lambda \ge 0$. The following statements are valid for $\tau$ and $*_{\lambda }$:

(i) If $f\in L_{\lambda ,{\mathrm{loc}}}({\mathbb {R}})$, then for all $x,y\in {\mathbb {R}}$, $(\tau _yf)(x)=(\tau _xf)(y)$, $(\tau _y\tilde{f})(x)=(\widetilde{\tau _{-y}f})(x)$, where $\tilde{f}(x)=f(-x)$.

(ii) For all $1\le p\le \infty$ and $f\in L_{\lambda }^p({\mathbb {R}})$, $\Vert \tau _yf\Vert _{L^p_{\lambda }}\le 4\Vert f\Vert _{L^p_{\lambda }}$ with $y\in {\mathbb {R}}$.

(iii) If $f\in L_{\lambda }^p({\mathbb {R}})$, $1\le p\le 2$, and $y\in {\mathbb {R}}$, then $[{\mathscr {F}}_{\lambda }(\tau _y f)](\xi )=E_{\lambda }(iy\xi )({\mathscr {F}}_{\lambda }f)(\xi )$ for almost every $\xi \in {\mathbb {R}}$.

(iv) (Young inequality) If $p,q,r\in [1,\infty ]$ and $1/p+1/q=1+1/r$, then $\Vert f*_{\lambda }g\Vert _{L^r_{\lambda }}\le 4\Vert f\Vert _{L^p_{\lambda }} \Vert g\Vert _{L^q_{\lambda }}$ for $f\in L^p_{\lambda }({\mathbb {R}})$, $g\in L^q_{\lambda }({\mathbb {R}})$.

(v) Assume that $p,q,r\in [1,2]$ and $1/p+1/q=1+1/r$. Then for $f\in L^p_{\lambda }({\mathbb {R}})$, $g\in L^q_{\lambda }({\mathbb {R}})$, $[{\mathscr {F}}_{\lambda }(f*_{\lambda }g)](\xi )=({\mathscr {F}}_{\lambda }f)(\xi )({\mathscr {F}}_{\lambda }g)(\xi )$. In particular $*_{\lambda }$ is associative in $L^1_{\lambda }({\mathbb {R}})$.

(vi) For $1\le p<\infty$ and $d\mu \in {\mathfrak {B}}_{\lambda }({\mathbb {R}})$, the mapping $f\mapsto f*_{\lambda }(d\mu )$ has an extension from $L^p_{\lambda }({\mathbb {R}})$ to itself and satisfies $\Vert f*_{\lambda }(d\mu )\Vert _{L^p_{\lambda }}\le 4\Vert f\Vert _{L^p_{\lambda }} \Vert d\mu \Vert _{{\mathfrak {B}_{\lambda }}}$. If f is even, then the constant 4 in (ii), (iv) and (vi) may be replaced by 1.

(vii) For $f\in L^p_{\lambda }({\mathbb {R}})$, $1\le p\le 2$, and $d\mu \in {\mathfrak {B}}_{\lambda }({\mathbb {R}})$, $[{\mathscr {F}}_{\lambda }(f*_{\lambda }(d\mu ))](\xi )=({\mathscr {F}}_{\lambda }f)(\xi )[{\mathscr {F}}_{\lambda }(d\mu )](\xi )$.

Proposition 2.5

(cf. [14]) Assume that $\lambda \ge 0$. Then for $f\in {\mathscr {S}}({\mathbb {R}})$ and given $m>0$,

$$\begin{aligned} |(\tau _y|f|)(x)|\le \frac{c_m(1+x^2+y^2)^{-\lambda }}{(1+||x|-|y||^2)^m},\qquad x,y\in {\mathbb {R}}. \end{aligned}$$

Proposition 2.6

(cf. [14]) Assume that $\lambda \ge 0$ and $\epsilon >0$. If $f\in L^p_{\lambda }({\mathbb {R}})$, $1\le p\le 2$, then $[{\mathscr {F}}_{\lambda }(P_{\epsilon }f)](\xi )=e^{-\epsilon |\xi |}({\mathscr {F}}_{\lambda }f)(\xi )$, and for $x\in {\mathbb {R}}$,

$$\begin{aligned} (P_{\epsilon }f)(x)=c_{\lambda }\int _{{\mathbb {R}}}e^{-\epsilon |\xi |}({\mathscr {F}}_{\lambda }f)(\xi )E_{\lambda }(ix\xi )|\xi |^{2\lambda }d\xi . \end{aligned}$$

3 The Key Lemmas

The key to the proofs of Theorems 1.1 and 1.2 is the following lemma. The lemma contains more information than what is need in the paper, but may be used elsewhere.

Lemma 3.1

Suppose that $a>1$, $1/2\le s<1$, and $\psi \in C^{\infty }({\mathbb {R}})$ is even and with compact support. Then for $b\in {\mathbb {R}}$, $x,y\in {\mathbb {R}}\setminus \{0\}$, $x\ne y$,

$$\begin{aligned}&\left| \int _{-\infty }^{\infty }e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{2\lambda -s}d\rho \right| \lesssim |xy|^{-\lambda }|x-y|^{s-1}. \end{aligned}$$

(21)

The proof of Lemma 3.1 is based on the following two lemmas.

Lemma 3.2

Suppose that $a>1$, $0<k_1<1<k_2$, and $\theta \in C^{\infty }({\mathbb {R}})$ is even, nonnegative, bounded, with support contained in ${\mathbb {R}}\setminus \{0\}$ and satisfies $\int _{-\infty }^{\infty }|\theta '(\rho )|d\rho <\infty$. Then for $b,x,y\in {\mathbb {R}}\setminus \{0\}$, $x\ne y$, satisfying

$$\begin{aligned} \left[ k_1c,k_2c\right] \bigcap \hbox { supp}\,\theta =\emptyset \qquad \hbox {with}\quad c=\left( \frac{|x-y|}{ab}\right) ^{\frac{1}{a-1}}, \end{aligned}$$

we have, for all $h>0$,

$$\begin{aligned}&\left| \int _{-h}^{h}\theta (\rho )e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda }d\rho \right| \lesssim \frac{|xy|^{-\lambda }}{|x-y|}. \end{aligned}$$

(22)

Lemma 3.3

Suppose that $a>1$, and $\theta \in C^{\infty }({\mathbb {R}})$ is even, nonnegative, bounded, and satisfies $\int _{-\infty }^{\infty }|\theta '(\rho )|d\rho <\infty$. Then for $b,x,y\in {\mathbb {R}}\setminus \{0\}$, $x\ne y$, and for all $h>0$,

$$\begin{aligned}&\left| \int _{-h}^{h}\theta (\rho )e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda }d\rho \right| \lesssim \frac{|xy|^{-\lambda }}{|x-y|}\left( \frac{|x-y|^a}{b}+1\right) ^{\frac{1}{2(a-1)}}. \end{aligned}$$

(23)

Proof of Lemma 3.2

Let $\Omega$ denote the integral on the left hand side in (22) and assume that $b>0$ without loss of generality. We shall need the following equality

$$\begin{aligned}&i(ab|\rho |^{a-2}\rho +x-y)e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho ) =D_{\rho }\left[ e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\right] \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad +\frac{4\lambda e^{ib|\rho |^a}}{(2\lambda +1)^2}\rho xyj_{\lambda +\frac{1}{2}}(x\rho )j_{\lambda +\frac{1}{2}}(y\rho ), \quad \hbox {for}\,\rho \ne 0. \end{aligned}$$

(24)

Indeed, by (15),

$$\begin{aligned}&D_{\rho }\left[ e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\right] \\&\quad =iab|\rho |^{a-2}\rho e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho ) +e^{ib|\rho |^a}D_{\rho }\left[ E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\right] , \end{aligned}$$

and by (14) and Lemma 2.2(i),

$$\begin{aligned}&D_{\rho }\left[ E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\right] =i(x-y)E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\\&\qquad \qquad \qquad -\frac{\lambda }{\rho }\left[ E_{\lambda }(ix\rho )-E_{\lambda }(-ix\rho )\right] \left[ E_{\lambda }(-iy\rho )-E_{\lambda }(iy\rho )\right] . \end{aligned}$$

Combining the two equalities proves (24).

Applying (24) to $\Omega$ and appealing to (12), then $\Omega$ may be written as the sum of $\Omega _1$, $\Omega _{2}$, and $\Omega _{3}$, where

$$\begin{aligned} \Omega _{1}&:=i\int _{-h}^{h}D_{\rho }\left[ \frac{\theta (\rho )}{ab|\rho |^{a-2}\rho +x-y}\right] \cdot e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda }d\rho ,\\ \Omega _{2}&:=\frac{-4i\lambda }{(2\lambda +1)^2}\int _{-h}^{h}\frac{\theta (\rho )e^{ib|\rho |^a}}{ab|\rho |^{a-2}\rho +x-y}\,\rho xy j_{\lambda +\frac{1}{2}}(x\rho )j_{\lambda +\frac{1}{2}}(y\rho )|\rho |^{2\lambda }d\rho ,\\ \Omega _{3}&:=-i\left( \left. \frac{\theta (\rho )e^{ib|\rho |^a}}{ab|\rho |^{a-2}\rho +x-y}\cdot E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda }\right) \right| _{-h}^h. \end{aligned}$$

It is noted that for $\rho \in \hbox {supp}\,\theta$, if $|\rho |\le k_1c$, then

$$\begin{aligned} \left| ab|\rho |^{a-2}\rho +x-y\right| \ge |x-y|-ab|\rho |^{a-1}\ge \left( 1-k_1^{a-1}\right) |x-y|; \end{aligned}$$

(25)

and if $|\rho |\ge k_2c$, then

$$\begin{aligned} \left| ab|\rho |^{a-2}\rho +x-y\right| \ge ab|\rho |^{a-1}-|x-y|\ge \left( k_2^{a-1}-1\right) |x-y|. \end{aligned}$$

(26)

It follows from Lemma 2.2(ii), (25) and (26) that $|\Omega _3|\lesssim |xy|^{-\lambda }|x-y|^{-1}$.

As for $\Omega _1$, by Lemma 2.2(ii) and (15) we have

$$\begin{aligned} |\Omega _{1}|&\lesssim |xy|^{-\lambda }\int _{-h}^{h}\left| D_{\rho }\left[ \frac{\theta (\rho )}{ab|\rho |^{a-2}\rho +x-y}\right] \right| d\rho \\&\lesssim |xy|^{-\lambda }\left( \int _{-h}^{h}\left| D_{\rho }\left[ \frac{1}{ab|\rho |^{a-2}\rho +x-y}\right] \right| \theta (\rho )d\rho \right. \\&\quad \left. +\int _{-h}^{h}\frac{|\theta '(\rho )|d\rho }{|ab|\rho |^{a-1}-|x-y||}\right) . \end{aligned}$$

It follows from (25) and (26) that the later integral is bounded by a multiple of

$$\begin{aligned} |x-y|^{-1}\int _{-\infty }^{\infty }|\theta '(\rho )|d\rho \lesssim |x-y|^{-1}; \end{aligned}$$

furthermore, since

$$\begin{aligned} \left| D_{\rho }\left[ \frac{1}{ab|\rho |^{a-2}\rho +x-y}\right] \right| \lesssim \frac{ab|\rho |^{a-2}}{(ab|\rho |^{a-1}-|x-y|)^2}, \end{aligned}$$

we have

$$\begin{aligned} |\Omega _{1}|&\lesssim |xy|^{-\lambda }\int _{0}^{h}\frac{ab\rho ^{a-2}\theta (\rho )}{(ab\rho ^{a-1}-|x-y|)^2}d\rho +\frac{|xy|^{-\lambda }}{|x-y|}. \end{aligned}$$

(27)

Taking partial integration to the integral above and in virtue of (25) and (26), we obtain

$$\begin{aligned} |\Omega _{1}|&\lesssim |xy|^{-\lambda }\int _{0}^{h}\frac{|\theta '(\rho )|}{|ab\rho ^{a-1}-|x-y||}d\rho +\frac{|xy|^{-\lambda }}{|x-y|}\\&\lesssim |xy|^{-\lambda }|x-y|^{-1}. \end{aligned}$$

For $\Omega _2$, taking the substitution $\rho \rightarrow -\rho$ and summing with the original form, by (17) we get

$$\begin{aligned} |\Omega _{2}|&\lesssim |xy|^{-\lambda }\int _{-h}^{h} \frac{ab|\rho |^{a-2}\theta (\rho )}{|(ab|\rho |^{a-1})^2-(x-y)^2|}d\rho \\&\lesssim |xy|^{-\lambda }\int _{0}^{h}\frac{ab\rho ^{a-2}\theta (\rho )}{\left( ab\rho ^{a-1}-|x-y|\right) ^2}d\rho , \end{aligned}$$

which is the same as the first term on the right hand side of (27). Therefore

$$\begin{aligned} |\Omega _{2}|\lesssim |xy|^{-\lambda }|x-y|^{-1}. \end{aligned}$$

Finally combining the estimates for $\Omega _1,\Omega _2$ and $\Omega _3$ proves (22). The proof of Lemma 3.2 is completed. $\square$

Proof of Lemma 3.3

Assume that $b>0$. Since, by Lemma 2.2(ii), the integrand above is dominated in absolute value by a multiple of $|xy|^{-\lambda }$, the desired estimate for $0<h\lesssim |x-y|^{-1}$ follows immediately.

To evaluate the integral in (23) for larger h, take an even function $\psi _0\in C^{\infty }({\mathbb {R}})$ satisfying

$$\begin{aligned} \hbox {supp}\,\psi _0\subset ]-1,1[,\quad 0\le \psi _0(\rho )\le 1, \quad \hbox {and}\quad \psi _0(\rho )\equiv 1\,\,\hbox {for}\,\,\rho \in ]-2,2[. \end{aligned}$$

It suffices to prove that

$$\begin{aligned}&\left| \int _{-h}^{h}\theta (\rho )\psi _0(|x-y|\rho )e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho ) |\rho |^{2\lambda }d\rho \right| \nonumber \\&\quad \lesssim \frac{|xy|^{-\lambda }}{|x-y|}\left( \frac{|x-y|^a}{b}+1\right) ^{\frac{1}{2(a-1)}}. \end{aligned}$$

(28)

Let $\Omega$ denote the integral on the left hand side above. If $|x-y|^a\le b/2$, then for $|x-y||\rho |\ge 1$,

$$\begin{aligned} |\rho |^{a-1}\ge \frac{1}{|x-y|^{a-1}}\frac{2|x-y|^a}{b}=2ac^{a-1}, \end{aligned}$$

(29)

where c is given in Lemma 3.2, so that the support of the function

$$\begin{aligned} \theta _0(\rho ):=\theta (\rho )\psi _0(|x-y|\rho ) \end{aligned}$$

is contained in $[k_2c,\infty )\cup (-\infty ,-k_2c]$ with $k_2=(2a)^{1/(a-1)}$. Thus by Lemma 3.2, $\left| \Omega \right| \lesssim |xy|^{-\lambda }|x-y|^{-1}$.

In what follows, we assume that $|x-y|^a>b/2$. $\square$

Put

$$\begin{aligned} \gamma =\frac{1}{|x-y|}\left( \frac{|x-y|^a}{b}\right) ^{\frac{1}{2(a-1)}}. \end{aligned}$$

We split the integral $\Omega$ on the left hand side of (28) into two parts: $\Omega =\Omega _4+\Omega _5$, where

$$\begin{aligned} \Omega _4&=\int _{-h}^{h}\left( 1-\psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \right) \theta _0(\rho )e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho ) |\rho |^{2\lambda }d\rho ;\\ \Omega _5&=\int _{-h}^{h}\psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \theta _0(\rho )e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho ) |\rho |^{2\lambda }d\rho . \end{aligned}$$

It is easy to see that the support of the integrand in $\Omega _4$ is contained in $[-c-2\gamma ,-c+2\gamma ]\cup [c-2\gamma ,c+2\gamma ]$; and then by Lemma 2.2(ii), we have

$$\begin{aligned} |\Omega _4|\lesssim \gamma |xy|^{-\lambda }. \end{aligned}$$

As in the proof of Lemma 3.2, applying (24) to $\Omega _5$ and appealing to (12), then $\Omega _5$ may be written as the sum of $\Omega _6$, $\Omega _{7}$, and $\Omega _{8}$, where

$$\begin{aligned} \Omega _{6}&:=i\int _{-h}^{h}D_{\rho }\left[ \psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \right. \\&\quad \left. \times \frac{\theta _0(\rho )}{ab|\rho |^{a-2}\rho +x-y} \right] \cdot e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda }d\rho ,\\ \Omega _{7}&:=\frac{-4i\lambda }{(2\lambda +1)^2}\int _{-h}^{h}\psi _0 \left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \\&\quad \times \frac{\theta _0(\rho )e^{ib|\rho |^a}}{ab|\rho |^{a-2}\rho +x-y}\,\rho xy\, j_{\lambda +\frac{1}{2}}(x\rho )j_{\lambda +\frac{1}{2}}(y\rho )|\rho |^{2\lambda }d\rho ,\\ \Omega _{8}&:=-i\left( \left. \psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \right. \right. \\&\quad \left. \left. \times \frac{\theta _0(\rho )e^{ib|\rho |^a}}{ab|\rho |^{a-2}\rho +x-y}\cdot E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda }\right) \right| _{-h}^h. \end{aligned}$$

By Lemma 2.2(ii),

$$\begin{aligned} \left| \Omega _{6}\right| \lesssim |xy|^{-\lambda } \int _{-h}^{h}\left| D_{\rho }\left[ \psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \frac{\theta _0(\rho )}{ab|\rho |^{a-2}\rho +x-y}\right] \right| d\rho , \end{aligned}$$

and from (15), $\Omega _{6}$ is dominated by a multiple of the sum of $\Omega _{6}'$ and $\Omega _{6}''$, where

$$\begin{aligned} \Omega _{6}'&=|xy|^{-\lambda }\int _{-h}^{h}\left| D_{\rho }\left[ \frac{1}{ab|\rho |^{a-2}\rho +x-y}\right] \right| \psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \theta _0(\rho )d\rho ,\\ \Omega _{6}''&=|xy|^{-\lambda }\int _{-h}^{h}\left| D_{\rho }\left[ \psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \theta _0(\rho )\right] \right| \frac{d\rho }{|ab|\rho |^{a-1}-|x-y||}. \end{aligned}$$

Since $\left| D_{\rho }\left[ (ab|\rho |^{a-2}\rho +x-y)^{-1}\right] \right| \lesssim ab|\rho |^{a-2}/\left( ab|\rho |^{a-1}-|x-y|\right) ^2$, it follows that

$$\begin{aligned} \Omega _{6}'&\lesssim |xy|^{-\lambda }\left( \int _{|x-y|^{-1}}^{c-\gamma }+\int _{c+\gamma }^{h}\right) \frac{ab\rho ^{a-2}}{\left( ab\rho ^{a-1}-|x-y|\right) ^2}d\rho \end{aligned}$$

(30)

$$\begin{aligned}&\lesssim \frac{|xy|^{-\lambda }}{|x-y|-ab(c-\gamma )^{a-1}}+\frac{|xy|^{-\lambda }}{ab(c+\gamma )^{a-1}-|x-y|}. \end{aligned}$$

(31)

We note that, if $\gamma \ge c$, the first term above disappears and

$$\begin{aligned} ab(c+\gamma )^{a-1}-|x-y|\ge \left( 2^{a-1}-1\right) |x-y|, \end{aligned}$$

so that $\Omega _{6}'\lesssim |xy|^{-\lambda }|x-y|^{-1}\lesssim |xy|^{-\lambda }\gamma$; if $0<\gamma <c$, then

$$\begin{aligned} |x-y|-ab(c-\gamma )^{a-1}&=|x-y|\left( 1-\left( 1-\gamma /c\right) ^{a-1}\right) \gtrsim |x-y|\gamma /c,\\ ab(c+\gamma )^{a-1}-|x-y|&=|x-y|\left( \left( 1+\gamma /c\right) ^{a-1}-1\right) \gtrsim |x-y|\gamma /c, \end{aligned}$$

so that $\Omega _{6}'\lesssim |xy|^{-\lambda }|x-y|^{-1}c/\gamma \lesssim |xy|^{-\lambda }\gamma$.

For $\Omega _{6}''$, we first note that the value of $\rho$ in the support of the integrand satisfies $|\rho +c|\ge \gamma ,\,|\rho -c|\ge \gamma$, so that $|ab|\rho |^{a-1}-|x-y||$ is bounded below by $|x-y|-ab(c-\gamma )^{a-1}\gtrsim |x-y|\gamma /c$ or $ab(c+\gamma )^{a-1}-|x-y|\gtrsim |x-y|\gamma /c$. Thus we have

$$\begin{aligned} \Omega _{6}''&\lesssim \frac{|xy|^{-\lambda }}{|x-y|}\frac{c}{\gamma } \left( \int _{-2|x-y|^{-1}}^{2|x-y|^{-1}}|x-y|d\rho +\int _{-c-\gamma }^{-c+\gamma }\gamma ^{-1}d\rho +\int _{c-\gamma }^{c+\gamma }\gamma ^{-1}d\rho \right) \\&\lesssim \frac{|xy|^{-\lambda }}{|x-y|}\frac{c}{\gamma }\lesssim |xy|^{-\lambda }\gamma , \end{aligned}$$

and so

$$\begin{aligned} \Omega _{6}\lesssim |xy|^{-\lambda }\gamma . \end{aligned}$$

For $\Omega _7$, taking the substitution $\rho \rightarrow -\rho$ and summing with the original form, we get

$$\begin{aligned} |\Omega _{7}|&\lesssim |xy|^{-\lambda }\int _{-h}^{h}\psi _0\left( \frac{\rho +c}{\gamma }\right) \psi _0\left( \frac{\rho -c}{\gamma }\right) \frac{\psi _0(|x-y|\rho )ab|\rho |^{a-2}}{|(ab|\rho |^{a-1})^2-(x-y)^2|}d\rho \\&\lesssim |xy|^{-\lambda }\left( \int _{|x-y|^{-1}}^{c-\gamma }+\int _{c+\gamma }^h\right) \frac{ab\rho ^{a-2}}{\left( ab\rho ^{a-1}-|x-y|\right) ^2}d\rho , \end{aligned}$$

which is the same as in (30) for $\Omega _6'$. Therefore

$$\begin{aligned} |\Omega _{7}|\lesssim |xy|^{-\lambda }\gamma . \end{aligned}$$

Finally for $\Omega _8$, by Lemma 2.2(ii) it is easy to see that it has a bound similar to (31) for $\Omega _6'$, and so

$$\begin{aligned} |\Omega _8|\lesssim |xy|^{-\lambda }\gamma . \end{aligned}$$

Combining the estimates for $\Omega _6,\Omega _7$ and $\Omega _8$ shows that $|\Omega _5|\lesssim |xy|^{-\lambda }\gamma$, and in conjunction with that for $\Omega _4$, one has $|\Omega |\lesssim |xy|^{-\lambda }\gamma$ which proves (28) when $|x-y|^a>b/2$.

Proof of Lemma 3.1

Assume that $b>0$. Denote by $\Lambda$ the integral on the left hand side of (21) and define the function $\psi _0\in C^{\infty }({\mathbb {R}})$ as in the proof of Lemma 3.3

We then write $\Lambda$ into the sum of $\Lambda _0$ and $\Lambda _1$, where

$$\begin{aligned} \Lambda _0&=\int _{-\infty }^{\infty }\left( 1-\psi _0(|x-y|\rho )^2\right) e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{2\lambda -s}d\rho ,\nonumber \\ \end{aligned}$$

(32)

$$\begin{aligned} \Lambda _1&=\int _{-\infty }^{\infty }\psi _0(|x-y|\rho )^2 e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{2\lambda -s}d\rho . \end{aligned}$$

(33)

The estimate for $\Lambda _0$ is trivial, since by Lemma 2.2(ii),

$$\begin{aligned} |\Lambda _0|\lesssim |xy|^{-\lambda }\int _{0}^{2|x-y|^{-1}}\rho ^{-s}d\rho \lesssim |xy|^{-\lambda }|x-y|^{s-1}. \end{aligned}$$

To evaluate $\Lambda _1$, Two cases are to be considered separately. $\square$

Case I:   $|x-y|^a\le b/2$.

As in the former part of the proof of Lemma 3.3, in this case, the deduction in (29) for $|x-y||\rho |\ge 1$ implies that the support of the function

$$\begin{aligned} \theta _1(\rho ):=\psi _0(|x-y|\rho ) \end{aligned}$$

is contained in $[k_2c,\infty )\cup (-\infty ,-k_2c]$ with $k_2=(2a)^{1/(a-1)}$. Thus by Lemma 3.2, the function

$$\begin{aligned} \Omega (\rho ):=\int _{-\rho }^{\rho }\theta _1(\eta )e^{ib|\eta |^a}E_{\lambda }(ix\eta )E_{\lambda }(-iy\eta )|\eta |^{2\lambda }d\eta \end{aligned}$$

satisfies, for all $\rho >0$, $\left| \Omega (\rho )\right| \lesssim |xy|^{-\lambda }|x-y|^{-1}$. Now taking partial integration

$$\begin{aligned} \Lambda _1&=\frac{1}{2}\int _{-\infty }^{\infty } \theta _1(\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{-s}d\Omega (\rho )\\&=-\frac{1}{2}\int _{-\infty }^{\infty }\Omega (\rho ) \left( \theta _1(\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{-s}\right) '_{\rho }d\rho , \end{aligned}$$

and hence

$$\begin{aligned} \left| \Lambda _1\right|&\lesssim \frac{|xy|^{-\lambda }}{|x-y|}\left( |x-y|^s\int _{|x-y|^{-1}}^{\infty } \left| \left( \theta _1(\rho )\psi \left( \frac{\rho }{N}\right) \right) '_{\rho }\right| d\rho \right. \\&\quad \left. +\int _{|x-y|^{-1}}^{\infty }\rho ^{-s-1}d\rho \right) \\&\lesssim |xy|^{-\lambda }|x-y|^{s-1}. \end{aligned}$$

Case II:   $|x-y|^a>b/2$.

Since

$$\begin{aligned} 1\equiv \psi _0\left( \frac{\rho }{2c}\right) \left( 2-\psi _0\left( \frac{\rho }{2c}\right) \right) +\left( 1-\psi _0\left( \frac{4\rho }{c}\right) ^2\right) + \left( 1-\psi _0\left( \frac{\rho }{2c}\right) \right) ^2\psi _0\left( \frac{4\rho }{c}\right) ^2, \end{aligned}$$

we split the integral for $\Lambda _1$ into two parts: $\Lambda _1=\Lambda _{11}+\Lambda _{12}$, where

$$\begin{aligned} \Lambda _{11}&=\int _{-\infty }^{\infty }\theta _1(\rho )\theta _2(\rho )e^{ib|\rho |^a} E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{2\lambda -s}d\rho ,\\ \Lambda _{12}&=\int _{-\infty }^{\infty }\theta _3(\rho )^2 e^{ib|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{2\lambda -s}d\rho , \end{aligned}$$

with

$$\begin{aligned} \theta _2(\rho )&=\left[ \psi _0\left( \frac{\rho }{2c}\right) \left( 2-\psi _0\left( \frac{\rho }{2c}\right) \right) +\left( 1-\psi _0\left( \frac{4\rho }{c}\right) ^2\right) \right] \psi _0(|x-y|\rho ),\\ \theta _3(\rho )&=\left( 1-\psi _0\left( \frac{\rho }{2c}\right) \right) \psi _0\left( \frac{4\rho }{c}\right) \psi _0(|x-y|\rho ). \end{aligned}$$

For $\Lambda _{11}$, we take partial integration to get

$$\begin{aligned} \Lambda _{11}&=\frac{1}{2}\int _{-\infty }^{\infty } \theta _1(\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{-s}d\Omega (\rho )\\&=-\frac{1}{2}\int _{-\infty }^{\infty }\Omega (\rho ) \left( \theta _1(\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{-s}\right) '_{\rho }d\rho , \end{aligned}$$

here

$$\begin{aligned} \Omega (\rho ):=\int _{-\rho }^{\rho }\theta _2(\eta )e^{ib|\eta |^a}E_{\lambda }(ix\eta )E_{\lambda }(-iy\eta )|\eta |^{2\lambda }d\eta . \end{aligned}$$

It is easy to see that

$$\begin{aligned} \hbox {supp}\,\theta _2\subset \{\rho :\,|\rho |\le \frac{c}{2}\}\cup \{\rho :\,|\rho |\ge 2c\}, \end{aligned}$$

and $\theta _2$ satisfies the conditions in Lemma 3.2. Hence $\left| \Omega (\rho )\right| \lesssim |xy|^{-\lambda }|x-y|^{-1}$ for all $\rho >0$ and

$$\begin{aligned} \left| \Lambda _{11}\right| \lesssim |xy|^{-\lambda }|x-y|^{s-1}. \end{aligned}$$

As for $\Lambda _{12}$, we again take partial integration to get

$$\begin{aligned} \Lambda _{12}&=\frac{1}{2}\int _{-\infty }^{\infty } \theta _3(\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{-s}d\Omega (\rho )\nonumber \\&=-\frac{1}{2}\int _{-\infty }^{\infty }\Omega (\rho ) \left( \theta _3(\rho )\psi \left( \frac{\rho }{N}\right) |\rho |^{-s}\right) '_{\rho }d\rho , \end{aligned}$$

(34)

where

$$\begin{aligned} \Omega (\rho ):=\int _{-\rho }^{\rho }\theta _3(\eta )e^{ib|\eta |^a}E_{\lambda }(ix\eta )E_{\lambda }(-iy\eta )|\eta |^{2\lambda }d\eta . \end{aligned}$$

It is noted that

$$\begin{aligned} \hbox {supp}\,\theta _3\subset \left\{ \rho :\,\frac{c}{4}\le |\rho |\le 4c\right\} . \end{aligned}$$

Now we apply Lemma 3.3 to obtain

$$\begin{aligned} \left| \Lambda _{12}\right|&\lesssim \frac{|xy|^{-\lambda }}{|x-y|}\left( \frac{|x-y|^a}{b}\right) ^{\frac{1}{2(a-1)}} \left( c^{-s}\int _{\frac{c}{4}}^{4c} \left| \left( \theta _3(\rho )\psi \left( \frac{\rho }{N}\right) \right) '_{\rho }\right| d\rho +\int _{\frac{c}{4}}^{4c}\rho ^{-s-1}d\rho \right) \\&\lesssim \frac{|xy|^{-\lambda }}{|x-y|}\left( \frac{|x-y|^a}{b}\right) ^{\frac{1}{2(a-1)}}c^{-s}, \end{aligned}$$

so that, for $1/2\le s<1$, $|x-y|^a>b/2$,

$$\begin{aligned} \left| \Lambda _{12}\right| \lesssim \frac{|xy|^{-\lambda }}{|x-y|^{1-s}}\left( \frac{b}{|x-y|^a}\right) ^{\left( s-\frac{1}{2}\right) \frac{1}{a-1}} \lesssim \frac{|xy|^{-\lambda }}{|x-y|^{1-s}}. \end{aligned}$$

(35)

The proof of Lemma 3.1 is finished.

4 Proofs of the Main Results

We shall prove a more general theorem than Theorem 1.2.

Theorem 4.1

Let $a>1$, $\lambda \ge 0$, $1/4\le s<1/2$, and $q=\frac{4\lambda +2}{2\lambda +1-2s}$. Then for all $f\in {\mathscr {S}}({\mathbb {R}})$, we have the following norm estimate

$$\begin{aligned} \left( \int _{{\mathbb {R}}}\left( \sup _{0<|t|<\infty }\left| T^t_{\lambda ,a}f(x)\right| \right) ^q|x|^{2\lambda }dx\right) ^{\frac{1}{q}}\lesssim \Vert f\Vert _{\dot{H}^{s}_{\lambda }}. \end{aligned}$$

(36)

Obviously Theorem 1.2 is the particular case of Theorem 4.1 with $s=1/4$.

Proof of Theorem 4.1

We first note that, to show (36), it is sufficient to prove the inequality taking integration over $[-M,M]$ for each $M>0$, instead of over ${\mathbb {R}}$. Further, it suffices to prove that, for all $A>0$,

$$\begin{aligned} \left( \int _{-M}^M\left( \sup _{0<t\le A}\left| T^t_{\lambda ,a}f(x)\right| \right) ^q|x|^{2\lambda }dx\right) ^{\frac{1}{q}}\lesssim \Vert f\Vert _{\dot{H}^{s}_{\lambda }}, \end{aligned}$$

since $\sup _{0<t\le A}|T^t_{\lambda ,a}f(x)|$ increases and tends to $\sup _{0<t<\infty }|T^t_{\lambda ,a}f(x)|$ as $A\rightarrow \infty$. Moreover, for $f\in {\mathscr {S}}({\mathbb {R}})$, one may manipulate the linearization of the maximal function $\sup _{0<t\le A}|T^t_{\lambda ,a}f(x)|$ instead of itself, in terms of a measurable function $t=t(x)$ with $0<t(x)\le A$, that is, we shall show that

$$\begin{aligned} \left( \int _{-M}^M\left| T^{t(x)}_{\lambda ,a}f(x)\right| ^q|x|^{2\lambda }dx\right) ^{\frac{1}{q}}\lesssim \Vert f\Vert _{\dot{H}^{s}_{\lambda }},\qquad q=\frac{4\lambda +2}{2\lambda +1-2s}. \end{aligned}$$

(37)

If we set $g(\rho )=|\rho |^{s}({\mathscr {F}}_{\lambda }f)(\rho )$, then (37) is equivalent to

$$\begin{aligned} \left( \int _{-M}^M\left| U_{\lambda ,a}g(x)\right| ^q|x|^{2\lambda }dx\right) ^{\frac{1}{q}}\lesssim \Vert g\Vert _{L^2_{\lambda }}, \end{aligned}$$

(38)

where

$$\begin{aligned} U_{\lambda ,a}g(x)=c_{\lambda }\int _{{\mathbb {R}}}g(\rho )e^{it(x)|\rho |^a}E_{\lambda }(ix\rho )|\rho |^{-s}|\rho |^{2\lambda }d\rho ,\quad x\in {\mathbb {R}}. \end{aligned}$$

By duality, (38) is further equivalent to

$$\begin{aligned} \left( c_{\lambda }\int _{{\mathbb {R}}}\left| U^*_{\lambda ,a}h(\rho )\right| ^2|\rho |^{2\lambda }d\rho \right) ^{\frac{1}{2}}\lesssim \left( \int _{-M}^M\left| h(x)\right| ^{q'}|x|^{2\lambda }dx\right) ^{\frac{1}{q'}},\quad q'=\frac{4\lambda +2}{2\lambda +1+2s}, \end{aligned}$$

(39)

where

$$\begin{aligned} U^*_{\lambda ,a}h(\rho )=|\rho |^{-s}\int _{-1}^1h(x)e^{-it(x)|\rho |^a}E_{\lambda }(-ix\rho )|x|^{2\lambda }dx,\quad \rho \in {\mathbb {R}}. \end{aligned}$$

(40)

We take an even function $\psi \in C^{\infty }({\mathbb {R}})$ satisfying

$$\begin{aligned} \hbox {supp}\,\psi \subset [-2,2],\quad 0\le \psi (\rho )\le 1, \quad \hbox {and}\quad \psi (\rho )\equiv 1\,\,\hbox {for}\,\,\rho \in [-1,1]. \end{aligned}$$

(41)

It follows that, for large $N>0$,

$$\begin{aligned}&\int _{{\mathbb {R}}}\psi \left( \frac{\rho }{N}\right) \left| U^*_{\lambda ,a}h(\rho )\right| ^2|\rho |^{2\lambda }d\rho =\int _{{\mathbb {R}}}\psi \left( \frac{\rho }{N}\right) U^*_{\lambda ,a}h(\rho )\overline{U^*_{\lambda ,a}h(\rho )}|\rho |^{2\lambda }d\rho \\&\quad =\int _{-M}^M\int _{-M}^M\left[ \int _{{\mathbb {R}}}e^{i(t(x)-t(y))|\rho |^a}E_{\lambda } (ix\rho )E_{\lambda }(-iy\rho )\right. \\&\quad \quad \left. \times \psi \left( \frac{\rho }{N}\right) |\rho |^{2\lambda -2s}d\rho \right] \overline{h(x)}h(y)|x|^{2\lambda }|y|^{2\lambda }\,dx\,dy, \end{aligned}$$

and then by Lemma 3.1, we have

$$\begin{aligned} \int _{{\mathbb {R}}}\psi \left( \frac{\rho }{N}\right) \left| U^*_{\lambda ,a}h(\rho )\right| ^2|\rho |^{2\lambda }d\rho \lesssim \int _{-M}^M\int _{-M}^M\frac{|h(x)h(y)|}{|x-y|^{1-2s}}|x|^{\lambda }|y|^{\lambda }\,dx\,dy. \end{aligned}$$

Now letting $N\rightarrow \infty$, Fatou’s lemma asserts that

$$\begin{aligned} \int _{{\mathbb {R}}}\left| U^*_{\lambda ,a}h(\rho )\right| ^2|\rho |^{2\lambda }d\rho \lesssim \int _{-M}^M\int _{-M}^M\frac{|h(x)h(y)|}{|x-y|^{1-2s}}|x|^{\lambda }|y|^{\lambda }\,dx\,dy. \end{aligned}$$

(42)

Finally we need to apply a lemma from [25] to the right hand side above.

Lemma 4.2

([25, Lemma 1.7]) Assume that $0<\beta <1$, $2/(2-\beta )\le p\le 2$, $\gamma =1-\beta /2-1/p$. Then for measurable functions h,

$$\begin{aligned} \int _{{\mathbb {R}}}\int _{{\mathbb {R}}}\frac{|h(x)||h(y)|}{|x-y|^{\beta }|x|^{\gamma }|y|^{\gamma }}dx\,dy\lesssim \left( \int _{{\mathbb {R}}}|h(x)|^pdx\right) ^{2/p}. \end{aligned}$$

We rewrite the right hand side of (42) as

$$\begin{aligned} \int _{-M}^M\int _{-M}^M\frac{|h(x)||x|^{2\lambda /q'}\cdot |h(y)||y|^{2\lambda /q'}}{|x-y|^{1-2s}|x|^{2\lambda /q'-\lambda }|y|^{2\lambda /q'-\lambda }}\,dx\,dy, \end{aligned}$$

and then, putting $\beta =1-2s$, $p=q'$ (given in (39)), and $\gamma =1-\beta /2-1/p=2\lambda /q'-\lambda$, this double integral is bounded by a multiple of

$$\begin{aligned} \left[ \int _{-M}^M\left( |h(x)||x|^{2\lambda /q'}\right) ^{q'}\,dx\right] ^{2/q'}=\left( \int _{-M}^M|h(x)|^{q'}|x|^{2\lambda }\,dx\right) ^{2/q'}. \end{aligned}$$

Thus the inequality (39) is proved, and then, the proof of Theorem 4.1 is completed. $\square$

Now we turn to the proof of Theorem 1.1.

We restrict the discussion to $t\in [0,\infty )$.

An immediate consequence of Theorem 1.2 is that, for each $M>0$ and $A>0$,

$$\begin{aligned} \int _{-M}^M\sup _{0<t\le A}\left| T^t_{\lambda ,a}f(x)\right| |x|^{2\lambda }dx\le c_M\Vert f\Vert _{H^{1/4}_{\lambda }},\qquad f\in {\mathscr {S}}({\mathbb {R}}), \end{aligned}$$

(43)

where the constant $c_M>0$ is independent of f. Since $H^s_{\lambda }({\mathbb {R}})\subseteq H^{1/4}_{\lambda }({\mathbb {R}})$ when $s\ge 1/4$, it suffices to assume $f\in H^{1/4}_{\lambda }({\mathbb {R}})$ in proving Theorem 1.1. In what follows, we work with a procedure as in [22]. Given such an f, we take a sequence $\{f_k:\,k\ge 1\}\subset {\mathscr {S}}({\mathbb {R}})$ satisfying

$$\begin{aligned} \Vert f_k-f\Vert _{H^{1/4}_{\lambda }}\le 2^{-k}. \end{aligned}$$

It then follows from (43) that

$$\begin{aligned} \int _{-M}^M\sum _{k=1}^{\infty }\sup _{0<t\le A}\left| T^t_{\lambda ,a}f_k(x)-T^t_{\lambda ,a}f_{k+1}(x)\right| |x|^{2\lambda }dx\le c_M\sum _{k=1}^{\infty }\Vert f_k-f_{k-1}\Vert _{H^{1/4}_{\lambda }}\le 2c_M. \end{aligned}$$

It is noted that each function $(x,t)\mapsto T^t_{\lambda ,a}f_k(x)$ is continuous in ${\mathbb {R}}\times {\mathbb {R}}$ with values $f_k(x)$ when $t=0$, so that the supremum can be taken over $t\in [0,A]$. We have that, for almost all $x\in [-M,M]$,

$$\begin{aligned} \sum _{k=1}^{\infty }\sup _{0\le t\le A}\left| T^t_{\lambda ,a}f_k(x)-T^t_{\lambda ,a}f_{k+1}(x)\right| <\infty , \end{aligned}$$

which implies, for x with the property, that the sequence of functions $t\mapsto T^t_{\lambda ,a}f_k(x)$ converges uniformly in $t\in [0,A]$, to a continuous function, $v_x(t)$, says. Obviously $v_x(t)$ is well defined for a.e. $x\in {\mathbb {R}}$, and continuous in $t\in [0,A]$. However, since $f_k$ tends to f as $k\rightarrow \infty$ in $L^2_{\lambda }({\mathbb {R}})$ norm, it follows that

$$\begin{aligned} \lim _{k\rightarrow \infty }\int _{{\mathbb {R}}}|x|^{2\lambda }\int _0^A\left| T^t_{\lambda ,a}f_k(x)-T^t_{\lambda ,a}f(x)\right| ^2dt\,dx=0, \end{aligned}$$

and hence, there is a subsequence $k_j\rightarrow \infty$ such that, for a.e. $x\in {\mathbb {R}}$, the sequence of functions $t\mapsto T^t_{\lambda ,a}f_{k_j}(x)$ converges in the $L^2([0,A])$ norm to the function $t\mapsto T^t_{\lambda ,a}f(x)$. We conclude that, except x in a null set in ${\mathbb {R}}$, $T^t_{\lambda ,a}f(x)=v_x(t)$ for a.e. $t\in [0,A]$. By choosing $A_{\ell }\rightarrow \infty$, the functions $t\mapsto v_x(t)$ are continuously extended to $[0,\infty )$, for a.e. $x\in {\mathbb {R}}$.

It remains to show that $v_x(0)=f(x)$ for a.e. $x\in {\mathbb {R}}$. Indeed, since $\Vert T^t_{\lambda ,a}f-f\Vert _{L^2_{\lambda }}\rightarrow 0$ as $t\rightarrow 0$, then, for a sequence $t_j\rightarrow 0$, $T^{t_j}_{\lambda ,a}f(x)\rightarrow f(x)$ almost everywhere, and so does $v_x(t_j)\rightarrow f(x)=v_x(0)$.

5 The Divergence Set of $T^t_{\lambda ,a}f$

5.1 Another Representation of $T^t_{\lambda ,2}f(x)$

We shall need the following equality (cf. [29, §§13.31])

$$\begin{aligned}&2c_{\lambda }\int _0^{\infty }e^{-z\xi ^2}j_{{\lambda -1/2}}(x\xi )j_{{\lambda -1/2}}(y\xi )\xi ^{2\lambda }d\xi \nonumber \\&\quad =\frac{1}{(2z)^{\lambda +1/2}}\exp \left( -\frac{x^2+y^2}{4z}\right) j_{{\lambda -1/2}}\left( i\frac{xy}{2z}\right) \end{aligned}$$

(44)

for $\lambda >-1/2$ and $\arg z\in (-\pi /2,\pi /2)$.

For $f\in {\mathscr {D}}({\mathbb {R}})$ and $x\in {\mathbb {R}}$, from (4) we have

$$\begin{aligned} T^t_{\lambda ,2}f(x)&=\lim _{\epsilon \rightarrow 0+}c_{\lambda } \int _{{\mathbb {R}}}({\mathscr {F}}_{\lambda }f)(\rho )e^{(it-\epsilon )\rho ^2}E_{\lambda }(ix\rho )|\rho |^{2\lambda }d\rho \nonumber \\&=\lim _{\epsilon \rightarrow 0+}c_{\lambda } \int _{{\mathbb {R}}}f(y)K_{\epsilon }^t(x,y)|y|^{2\lambda }dy, \end{aligned}$$

(45)

where, for $\epsilon >0$,

$$\begin{aligned} K_{\epsilon }^t(x,y)=c_{\lambda } \int _{{\mathbb {R}}} e^{(it-\epsilon )\rho ^2}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda }d\rho . \end{aligned}$$

(46)

It follows from (2) that

$$\begin{aligned} K_{\epsilon }^t(x,y)&=2c_{\lambda } \int _0^{\infty } e^{(it-\epsilon )\rho ^2}\left[ j_{{\lambda -1/2}}(x\rho )j_{{\lambda -1/2}}(y\rho )\right. \\&\quad \left. +\frac{xy\rho ^2}{(2\lambda +1)^2}j_{{\lambda +1/2}}(x\rho )j_{{\lambda +1/2}}(y\rho )\right] \rho ^{2\lambda }d\rho , \end{aligned}$$

and then, by (44),

$$\begin{aligned} K_{\epsilon }^t(x,y)=\frac{1}{(2z)^{\lambda +1/2}}\exp \left( -\frac{x^2+y^2}{4z}\right) E_{\lambda }\left( \frac{xy}{2z}\right) ,\qquad \hbox {with}\quad z=\epsilon -it. \end{aligned}$$

(47)

Substituting this expression into (45) and applying the dominated convergence theorem, we obtain, for $f\in {\mathscr {D}}({\mathbb {R}})$ and $t\ne 0$,

$$\begin{aligned} T^t_{\lambda ,2}f(x)=c_{\lambda } \int _{{\mathbb {R}}}f(y)K^t(x,y)|y|^{2\lambda }dy, \end{aligned}$$

(48)

where

$$\begin{aligned} K^t(x,y)=\left( \frac{i}{2t}\right) ^{\lambda +1/2}\exp \left( -i\frac{x^2+y^2}{4t}\right) E_{\lambda }\left( i\frac{xy}{2t}\right) . \end{aligned}$$

(49)

5.2 A Counterexample in $H_{\lambda }^s({\mathbb {R}})$ for $s<1/4$

Assume that $s<1/4$. We shall construct a function $f_0\in H_{\lambda }^s({\mathbb {R}})$ for which, in a set of positive measure, $T^t_{\lambda ,2}f$ diverges as $t\rightarrow 0+$, even after a modification on a null set of t.

Let $\phi$ be a nonzero element in ${\mathscr {D}}({\mathbb {R}})$, supported in $(-\infty ,-1)$. We consider the functions, for $0<t<1$,

$$\begin{aligned} \phi _t(x)=\frac{1}{t^{2\lambda }}\phi \left( \frac{x}{t}\right) E_{\lambda }\left( -i\frac{x}{t^2}\right) ,\qquad x\in {\mathbb {R}}. \end{aligned}$$

Since, from (1),

$$\begin{aligned} ({\mathscr {F}}_{\lambda }\phi _t)(\rho ):=c_{\lambda }t\int _{{\mathbb {R}}}\phi (-x)E_{\lambda }\left( ixt^{-1}\right) E_{\lambda }(ixt\rho )|x|^{2\lambda }dx,\quad \rho \in {\mathbb {R}}, \end{aligned}$$

then by Propositions 2.3(ii) and 2.4(iii),

$$\begin{aligned} ({\mathscr {F}}_{\lambda }\phi _t)(\rho )=t\left[ \tau _{t^{-1}}({\mathscr {F}}_{\lambda }\phi )\right] (t\rho ). \end{aligned}$$

Applying Proposition 2.5 to ${\mathscr {F}}_{\lambda }\phi$ we have

$$\begin{aligned} \Vert \phi _t\Vert _{H^s_{\lambda }}^2&=c_{\lambda }t^{1-2\lambda -2s}\int _{{\mathbb {R}}}\left| \left[ \tau _{t^{-1}}({\mathscr {F}}_{\lambda }\phi )\right] (\rho )\right| ^2(t^2+\rho ^2)^s|\rho |^{2\lambda }d\rho \\&\lesssim t^{1-2\lambda -2s}\int _{{\mathbb {R}}}\frac{(1+\rho ^2+t^{-2})^{-2\lambda }}{(1+||\rho |-t^{-1}|^2)^{2}} (t^2+\rho ^2)^s |\rho |^{2\lambda }d\rho \\&\lesssim t^{1-2s}\int _{{\mathbb {R}}}\frac{(t^2+\rho ^2)^s}{(1+||\rho |-t^{-1}|^2)^{2}} d\rho , \end{aligned}$$

so that

$$\begin{aligned} \Vert \phi _t\Vert _{H^s_{\lambda }}\lesssim t^{(1-4s)/2}\qquad \hbox {for}\quad 0<t<1. \end{aligned}$$

(50)

For $x\in (0,1)$, we choose $t(x)=t^2x/2$. From (48) and (49), we have

$$\begin{aligned} T^{t(x)}_{\lambda ,2}\phi _t(x)&=\frac{c_{\lambda }e^{i\pi (\lambda +1/2)/2}}{t^{2\lambda }x^{\lambda +1/2}} e^{-ix/(2t^2)}\nonumber \\&\quad \times \int _{{\mathbb {R}}}\phi (y) \exp \left( -i\frac{y^2}{2x}\right) E_{\lambda }\left( -i\frac{y}{t}\right) E_{\lambda }\left( i\frac{y}{t}\right) |y|^{2\lambda }dy. \end{aligned}$$

(51)

We shall need the following equality

$$\begin{aligned} E_{\lambda }(iy)E_{\lambda }(-iy)=\frac{|y|^{-2\lambda }}{2\pi c_{\lambda }^2}\left( 1+O\left( \frac{1}{y}\right) \right) , \qquad \hbox {for}\quad |x|\ge c>0. \end{aligned}$$

(52)

Indeed, from (2) we have

$$\begin{aligned} E_{\lambda }(iy)E_{\lambda }(-iy)=j_{\lambda -1/2}(y)^2+\frac{y^2}{(2\lambda +1)^2}j_{\lambda +1/2}(y)^2, \end{aligned}$$

and then, (52) follows by making use of (cf. [12, (7-13(3))])

$$\begin{aligned} j_{\lambda -1/2}(y)=\frac{|y|^{-\lambda }}{\sqrt{2\pi } c_{\lambda }} \left[ \cos \left( |y|-\frac{\pi }{2}\lambda \right) +O\left( \frac{1}{y}\right) \right] , \qquad \hbox {for}\quad |x|\ge c>0. \end{aligned}$$

Applying (52) to (51) we obtain, for $x\in (0,1)$ and $0<t<1$,

$$\begin{aligned} T^{t(x)}_{\lambda ,2}\phi _t(x)&=\frac{e^{i\pi (\lambda +1/2)/2}}{2\pi c_{\lambda }x^{\lambda +1/2}} e^{-ix/(2t^2)} \left[ \psi (x)+O\left( t\right) \right] , \end{aligned}$$

(53)

where

$$\begin{aligned} \psi (z)=\int _{{\mathbb {R}}}\phi (y) \exp \left( -i\frac{y^2}{2z}\right) dy,\qquad z\ne 0. \end{aligned}$$

Since $\psi (z)$ is a nonzero holomorphic function in ${\mathbb {C}}\setminus \{0\}$, there exist an interval $I\subset (1/2,1)$, $t_0\in (0,1)$, and a constant $c_0>0$, such that for $x\in I$ and $t\in (0,t_0)$,

$$\begin{aligned} \left| T^{t(x)}_{\lambda ,2}\phi _t(x)\right| >c_0. \end{aligned}$$

(54)

We now adopt the procedure in [22] to finish the construction of $f_0$. We shall define recursively a sequence $\left\{ t_j\right\} \subset (0,t_0)$, converging to zero so fast that it satisfies $\sum _{j=1}^{\infty }jt_j^{(1-4s)/2}<\infty$, and for all $x\in I$,

$$\begin{aligned} \left| T^{t_k(x)}_{\lambda ,2}\phi _{t_j}(x)\right| <2^{-j}, \qquad \hbox {for}\quad j\ne k, \end{aligned}$$

(55)

where $t_k(x)=t_k^2x/2$.

We start with a number $t_1\in (0,t_0)$ and assume that $t_1,\ldots ,t_{m-1}$ have been chosen so that $0<t_j<\min \{t_0,2^{-j+1}\}$ ($1\le j<m$) and (55) is satisfied for $j,k<m$. Since for each j ($1\le j<m$), $\phi _{t_j}\in {\mathscr {D}}({\mathbb {R}})$ and $\hbox {supp}\,\phi _{t_j}\subset (-\infty ,0)$, $T^{t(x)}_{\lambda ,2}\phi _{t_j}(x)$ converges to zero uniformly for $x\in I$ as $t\rightarrow 0+$, and from (48), (49), and Lemma 2.2(ii), we have, for each j ($1\le j<m$),

$$\begin{aligned} \left| T^{t_j(x)}_{\lambda ,2}\phi _{t}(x)\right| \lesssim \frac{t}{(t_j\sqrt{|x|})^{2\lambda +1}}\int _{{\mathbb {R}}}|\phi (y)||y|^{2\lambda }dy, \end{aligned}$$

which, again, converges to zero uniformly for $x\in I$ as $t\rightarrow 0+$. Hence one may choose $t_m$, $0<t_m<\min \{t_0,2^{-m+1}\}$, so that, for $j=1,\ldots ,m-1$, and for $x\in I$,

$$\begin{aligned} \left| T^{t_m(x)}_{\lambda ,2}\phi _{t_j}(x)\right|<2^{-j} \qquad \hbox {and}\qquad \left| T^{t_j(x)}_{\lambda ,2}\phi _{t_m}(x)\right| <2^{-m}. \end{aligned}$$

We define $f_0=\sum _{j=1}^{\infty }j\phi _{t_j}$. It follows from (50) that $f_0\in H_{\lambda }^s({\mathbb {R}})$ and

$$\begin{aligned} T^t_{\lambda ,2}f_0(x)=\sum _{j=1}^{\infty }jT^t_{\lambda ,2}\phi _{t_j}(x),\qquad x\in {\mathbb {R}}. \end{aligned}$$

For $x\in I$ and each k, by (54) and (55) we have

$$\begin{aligned} \left| T^{t_k(x)}_{\lambda ,2}f_0(x)\right| \ge c_0k-\sum _{j\ne k}j2^{-j}, \end{aligned}$$

(56)

which shows divergence of $T^{t_k(x)}_{\lambda ,2}f_0(x)$ as $k\rightarrow \infty$.

Finally we remark that, by continuity, (54) is also true if t(x) is replaced by any number in a small neighborhood of t(x). Similarly, (55) still holds when $t_k(x)$ is replaced by numbers close to $t_k(x)$, and so does (56).

5.3 The Divergence Set of $T^t_{\lambda ,a}f$

Assume that $a>1$ and $\lambda \ge 0$. Theorem 1.1 shows that, if $f\in H_{\lambda }^s({\mathbb {R}})$ with $s\ge 1/4$, then the set $\{x\in {\mathbb {R}}:\,T^{t}_{\lambda ,a}f(x)\nrightarrow f(x)\,\,\hbox {as}\,\,t\rightarrow 0\}$ is of zero measure in Lebesgue sense. A refinement of this question in the classical case (i.e. for $\lambda =0$) was studied in [1, 2, 21], by considering the Hausdorff dimension of the divergence sets. This subsection will extend the methods of [1, 2] to determine the Hausdorff dimension of the divergence set of $T^t_{\lambda ,a}f$.

As in [1], discrete times will be taken to avoid measurability issues. Let

$$\begin{aligned} \alpha _{\lambda }(s)=\sup _{\{t_k\},f}\hbox {dim}_H\left\{ x\in {\mathbb {R}}:\,T^{t_k}_{\lambda ,a}f(x)\nrightarrow f(x)\,\,\hbox {as}\,\,k\rightarrow \infty \right\} , \end{aligned}$$

where $\hbox {dim}_H$ denotes the Hausdorff dimension, and the supremum is taken over all sequence $t_k\rightarrow 0+$ and all $f\in H_{\lambda }^s({\mathbb {R}})$. $\alpha _{\lambda }(s)$ may have no sense when considering the space $H_{\lambda }^s({\mathbb {R}})$ itself, since the functions are only defined up to a set of zero measure of Lebesgue. However each equivalence class of the Sobolev space $H_{\lambda }^s({\mathbb {R}})$ has a representative as $f=g*_{\lambda }G_s$ with some $g\in L^2_{\lambda }({\mathbb {R}})$, where $G_s$, with the Dunkl transform $(1+|\rho |^2)^{-\frac{s}{2}}$, is the Bessel potential kernel associated to the Dunkl transform. The definition of $\alpha _{\lambda }(s)$, and also Lemma 5.1, Lemma 5.2, and Theorem 5.3 below, will be restricted to these representatives.

To make sense of $T^t_{\lambda ,a}f$ for all $f\in H_{\lambda }^s({\mathbb {R}})$, it may be defined as the pointwise limit

$$\begin{aligned} T^t_{\lambda ,a}f=\lim _{N\rightarrow \infty }T^{t,N}_{\lambda ,a}f \end{aligned}$$

whenever the limit exists, where

$$\begin{aligned} T^{t,N}_{\lambda ,a}f(x)=c_{\lambda }\int _{{\mathbb {R}}}\psi \left( \frac{|\rho |}{N}\right) ({\mathscr {F}}_{\lambda }f)(\rho )e^{it|\rho |^a}E_{\lambda }(ix\rho )|\rho |^{2\lambda }d\rho ,\quad x\in {\mathbb {R}}, \end{aligned}$$

(57)

$\psi \in C^{\infty }({\mathbb {R}})$ satisfying the conditions in (41).

A positive Borel measure $\mu$ on ${\mathbb {R}}$ is said to be $\alpha$-dimensional, $0\le \alpha \le 1$, if

$$\begin{aligned} c_{\alpha }(\mu ):=\sup _{x\in {\mathbb {R}},r>0}\frac{\mu \left( (x-r,x+r)\right) }{r^{\alpha }}<\infty . \end{aligned}$$

${\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$ denotes the set of $\alpha$-dimensional probability measures supported in the “annulus” ${\mathbb {A}}_m=\{x\in {\mathbb {R}}:\,1/m\le |x|\le m\}$.

Lemma 5.1

Assume that $0<s\le 1/2$ and $1-2s<\alpha \le 1$. If $f\in H_{\lambda }^s({\mathbb {R}})$ and $\mu \in {\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$, then

$$\begin{aligned} \Vert f\Vert _{L^1(d\mu )}\lesssim m^{3\lambda +(\alpha -1+2s)/2}\sqrt{c_{\alpha }(\mu )}\Vert f\Vert _{H^{s}_{\lambda }}. \end{aligned}$$

(58)

Proof

For $s>0$, put

$$\begin{aligned} G_s(x)=\frac{2^{-\lambda -1/2}}{\Gamma (s/2)} \int _0^{\infty }e^{-\delta -x^2/(4\delta )}\delta ^{\frac{s}{2}-\lambda -\frac{3}{2}}d\delta ,\qquad x\in {\mathbb {R}}. \end{aligned}$$

(59)

Applying Fubini’s theorem shows that $c_{\lambda }\int _{{\mathbb {R}}}G_s(x)|x|^{2\lambda }dx=1$; and furthermore we have

$$\begin{aligned} ({\mathscr {F}}_{\lambda }G_s)(\rho )=(1+|\rho |^2)^{-\frac{s}{2}},\qquad \rho \in {\mathbb {R}}. \end{aligned}$$

(60)

Indeed, since $({\mathscr {F}}_{\lambda }\phi )(\rho )=e^{-\delta \rho ^2}$ for $\phi (x)=(2\delta )^{-\lambda -1/2}e^{-x^2/(4\delta )}$ (taking $\epsilon =1/(4\delta )$, $t=x=0$ in (46) and (47), Proposition 2.3(iv) implies that

$$\begin{aligned} c_{\lambda }\int _{{\mathbb {R}}}e^{-\delta x^2}\psi (x)|x|^{2\lambda }dx =c_{\lambda }\int _{{\mathbb {R}}}(2\delta )^{-\lambda -1/2}e^{-x^2/(4\delta )}({\mathscr {F}}_{\lambda }\psi )(x)|x|^{2\lambda }dx, \end{aligned}$$

whenever $\psi \in {\mathscr {S}}({\mathbb {R}})$. We integrate both sides with respect to $e^{-\delta }\delta ^{\frac{s}{2}-1}d\delta$ and change the order of integration, to get

$$\begin{aligned} c_{\lambda }\int _{{\mathbb {R}}}(1+|x|^2)^{-\frac{s}{2}}\psi (x)|x|^{2\lambda }dx =c_{\lambda }\int _{{\mathbb {R}}}G_s(x)({\mathscr {F}}_{\lambda }\psi )(x)|x|^{2\lambda }dx \end{aligned}$$

for all $\psi \in {\mathscr {S}}({\mathbb {R}})$. This proves (60).

Appealing to $G_s$ and by Proposition 2.4(v), proving (58) is equivalent to showing, for $g\in L^2_{\lambda }({\mathbb {R}})$,

$$\begin{aligned} \Vert g*_{\lambda }G_s\Vert _{L^1(d\mu )}\lesssim m^{3\lambda +(\alpha -1+2s)/2}\sqrt{c_{\alpha }(\mu )}\Vert g\Vert _{L^2_{\lambda }}. \end{aligned}$$

(61)

We observe that, from (19) and (20),

$$\begin{aligned} \Vert g*_{\lambda }G_s\Vert _{L^1(d\mu )}&\le c_{\lambda }\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}|g(y)|\left| (\tau _xG)(-y)\right| |y|^{2\lambda }dy\,d\mu (x)\\&\le m^{2\lambda }c_{\lambda }^{-1}\Vert G_s*_{\lambda }(d\mu )\Vert _{L^2_{\lambda }}\Vert g\Vert _{L^2_{\lambda }}, \end{aligned}$$

where we have used the fact $(\tau _xG)(-y)=(\tau _yG)(-x)$ by Proposition 2.4(i), and so, (61) is a consequence of the inequality

$$\begin{aligned} \Vert G_s*_{\lambda }(d\mu )\Vert ^2_{L^2_{\lambda }}\lesssim m^{2\lambda +\alpha -1+2s}c_{\alpha }(\mu ). \end{aligned}$$

(62)

To prove (62), we first note that, by Propositions 2.3(v) and 2.4(vii),

$$\begin{aligned} \Vert G_s*_{\lambda }(d\mu )\Vert ^2_{L^2_{\lambda }}=\Vert {\mathscr {F}}_{\lambda }G_s\cdot {\mathscr {F}}_{\lambda }(d\mu )\Vert ^2_{L^2_{\lambda }} =\Vert {\mathscr {F}}_{\lambda }\left( G_{2s}*_{\lambda }(d\mu )\right) \cdot \overline{{\mathscr {F}}_{\lambda }(d\mu )}\Vert _{L^1_{\lambda }}. \end{aligned}$$

Applying Lebesgue’s monotone convergence theorem, Fubini’s theorem, and Proposition 2.6 successively, we obtain

$$\begin{aligned} \Vert G_s*_{\lambda }(d\mu )\Vert ^2_{L^2_{\lambda }}&=\lim _{\epsilon \rightarrow 0+}c_{\lambda }\int _{{\mathbb {R}}}e^{-\epsilon |\rho |}\left[ {\mathscr {F}}_{\lambda }\left( G_{2s}*_{\lambda }(d\mu )\right) \right] (\rho ) \cdot \overline{{\mathscr {F}}_{\lambda }(d\mu )}(\rho )|\rho |^{2\lambda }d\rho \\&=\lim _{\epsilon \rightarrow 0+}c_{\lambda }^2\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}e^{-\epsilon |\rho |} \left[ {\mathscr {F}}_{\lambda }\left( G_{2s}*_{\lambda }(d\mu )\right) \right] (\rho )\\&\quad E_\lambda (ix\rho )|\rho |^{2\lambda }d\rho \, |x|^{2\lambda }d\mu (x)\\&=\lim _{\epsilon \rightarrow 0+}c_{\lambda }\int _{{\mathbb {R}}}\left[ P_{\epsilon }\left( G_{2s}*_{\lambda }(d\mu )\right) \right] (x) |x|^{2\lambda }d\mu (x). \end{aligned}$$

In what follows, we shall show that, for $\mu \in {\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$ with $1\ge \alpha >1-2s\ge 0$,

$$\begin{aligned} \left( G_{2s}*_{\lambda }(d\mu )\right) (x) \lesssim m^{\alpha -1+2s}c_{\alpha }(\mu ),\qquad x\in {\mathbb {R}}, \end{aligned}$$

(63)

which implies, by Proposition 2.4(iv), $\left[ P_{\epsilon }\left( G_{2s}*_{\lambda }(d\mu )\right) \right] (x)\lesssim m^{\alpha -1+2s}c_{\alpha }(\mu )$ for all $x\in {\mathbb {R}}$, so that (62) is proved.

Since, by (59),

$$\begin{aligned} G_s(x)\lesssim \int _0^{\infty }e^{-x^2/(4\delta )}\delta ^{\frac{s}{2}-\lambda -\frac{3}{2}}d\delta \lesssim |x|^{s-2\lambda -1},\qquad x\ne 0, \end{aligned}$$

it follows from (18) that

$$\begin{aligned} (\tau _xG_{2s})(-y)\lesssim \int _0^\pi (x^2+y^2-2xy\cos \theta )^{s-\lambda -1/2}(1+\cos \theta )\sin ^{2\lambda -1}\theta d\theta . \end{aligned}$$

(64)

If $xy>0$, then

$$\begin{aligned} (\tau _xG_{2s})(-y)&\lesssim \int _0^{\pi /2} \left[ (x-y)^2+4xy\sin ^2\frac{\theta }{2}\right] ^{s-\lambda -\frac{1}{2}}\sin ^{2\lambda -1}\theta d\theta +(x^2+y^2)^{s-\lambda -\frac{1}{2}}\\&\lesssim \int _0^{\pi /2} \left( |x-y|+\sqrt{xy}\theta \right) ^{2s-2\lambda -1}\theta ^{2\lambda -1}d\theta +(x^2+y^2)^{s-\lambda -\frac{1}{2}}\\&\lesssim \frac{|xy|^{-\lambda }}{|x-y|^{1-2s}}\int _0^{A}(1+r)^{2s-2\lambda -1}r^{2\lambda -1}dr +(x^2+y^2)^{s-\lambda -\frac{1}{2}}, \end{aligned}$$

where $A=(\pi /2)\sqrt{|xy|}/|x-y|$. Since, for $0<s\le 1/2$,

$$\begin{aligned} \int _0^{A}(1+r)^{2s-2\lambda -1}r^{2\lambda -1}dr\lesssim \left( \frac{A}{A+1}\right) ^{2\lambda }\log \left( A+2\right) , \end{aligned}$$

it follows that, for $xy>0$,

$$\begin{aligned} (\tau _xG_{2s})(-y)&\lesssim \frac{(|x|+|y|)^{-2\lambda }}{|x-y|^{1-2s}}\log \left( \frac{|x|+|y|}{|x-y|}+1\right) . \end{aligned}$$

(65)

It is remarked that only for $s=1/2$, the factor $\log$ appears in (65). If $xy<0$, then

$$\begin{aligned} (x^2+y^2)\cos ^2\frac{\theta }{2}\le x^2+y^2-2xy\cos \theta \le 2(x^2+y^2), \end{aligned}$$

so that from (64), $(\tau _xG_{2s})(-y)\lesssim (x^2+y^2)^{s-\lambda -\frac{1}{2}}$, which shows that (65) hold also for $xy<0$.

Now for $0<s\le 1/2$, $1-2s<\alpha \le 1$, and for $\mu \in {\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$, applying (65) gives

$$\begin{aligned} \left( G_{2s}*_{\lambda }(d\mu )\right) (x)\lesssim \int _{{\mathbb {R}}} \frac{(|x|+|y|)^{-2\lambda }}{|x-y|^{1-2s}}\log \left( \frac{|x|+|y|}{|x-y|}+1\right) |y|^{2\lambda }d\mu (y). \end{aligned}$$

Choosing $\alpha '\in (1-2s,\alpha )$ and noting that $r^{\alpha '-1+2s}\log (r^{-1}+1)\lesssim 1$ for $0<r\le 1$, it follows that

$$\begin{aligned} \left( G_{2s}*_{\lambda }(d\mu )\right) (x)\lesssim \int _{{\mathbb {R}}} \frac{(|x|+|y|)^{\alpha '-1+2s}}{|x-y|^{\alpha '}}d\mu (y). \end{aligned}$$

(66)

If $|x|\le 2m$, decomposing dyadically around x shows that

$$\begin{aligned} \left( G_{2s}*_{\lambda }(d\mu )\right) (x)&\lesssim m^{\alpha '-1+2s}\sum _{j=-2}^{\infty }(m2^{-j})^{\alpha -\alpha '}\frac{\mu (x-m2^{-j},x+m2^{-j})}{(m2^{-j})^{\alpha }}\nonumber \\&\lesssim m^{\alpha -1+2s}c_{\alpha }(\mu ); \end{aligned}$$

(67)

and if $|x|>2m$, then $\left( G_{2s}*_{\lambda }(d\mu )\right) (x)\lesssim |x|^{2s-1}\mu (-m,m)\lesssim m^{\alpha -1+2s}c_{\alpha }(\mu )$. Thus (63) is proved. $\square$

Lemma 5.2

Let $a>1$, $\lambda \ge 0$, and $1/4\le s<1/2$. Then for $1-2s<\alpha \le 1$ and a real sequence $\{t_k\}$, we have

$$\begin{aligned} \left\| \sup _{k\ge 1}\sup _{N\ge 1}\left| T^{t_k,N}_{\lambda ,a}f\right| \right\| _{L^1(d\mu )}\lesssim m^{\lambda +(\alpha -1+2s)/2}\sqrt{c_{\alpha }(\mu )}\Vert f\Vert _{H^{s}_{\lambda }} \end{aligned}$$

(68)

whenever $\mu \in {\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$ and $f\in H_{\lambda }^s({\mathbb {R}})$.

Proof

By linearization of the maximal function, it suffices to prove

$$\begin{aligned} \left| \int _{{\mathbb {A}}_m}\left( T^{t(x),N(x)}_{\lambda ,a}f\right) (x)\,g(x)d\mu (x)\right| \lesssim m^{\lambda +(\alpha -1+2s)/2}\sqrt{c_{\alpha }(\mu )}\Vert f\Vert _{H^{s}_{\lambda }}\Vert g\Vert _{L^{\infty }(d\mu )}\nonumber \\ \end{aligned}$$

(69)

uniformly in the measurable functions $t=t(x)$ and $N=N(x)$, whenever $\mu \in {\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$, $f\in H_{\lambda }^s({\mathbb {R}})$, and $g\in L^{\infty }(d\mu )$.

From (57), Fubini’s theorem and the Cauchy-Schwarz inequality give us

$$\begin{aligned} \left| \int _{{\mathbb {A}}_m}\left( T^{t(x),N(x)}_{\lambda ,a}f\right) (x)\,g(x)d\mu (x)\right| \le \Theta \Vert f\Vert _{H^{s}_{\lambda }}, \end{aligned}$$

(70)

where

$$\begin{aligned} \Theta ^2=c_{\lambda }\int _{{\mathbb {R}}}\left| \int _{{\mathbb {A}}_M}\psi \left( \frac{|\rho |}{N(x)}\right) e^{it(x)|\rho |^a}E_{\lambda }(ix\rho )g(x)d\mu (x)\right| ^2 |\rho |^{2\lambda -2s}d\rho . \end{aligned}$$

Again by Fubini’s theorem,

$$\begin{aligned} \Theta ^2\le c_{\lambda }\Vert g\Vert ^2_{L^{\infty }(d\mu )}\int _{{\mathbb {A}}_m}\int _{{\mathbb {A}}_m}\Omega (x,y)d\mu (x)d\mu (y), \end{aligned}$$

where

$$\begin{aligned} \Omega (x,y)=\left| \int _{{\mathbb {R}}}\psi \left( \frac{|\rho |}{N(x)}\right) \psi \left( \frac{|\rho |}{N(y)}\right) e^{i(t(x)-t(y))|\rho |^a}E_{\lambda }(ix\rho )E_{\lambda }(-iy\rho )|\rho |^{2\lambda -2s}d\rho \right| . \end{aligned}$$

But by Lemma 3.1, $\Omega (x,y)\lesssim |xy|^{-\lambda }|x-y|^{2s-1}$, and hence, in a way similar to (66) and (67),

$$\begin{aligned} \Theta ^2&\lesssim \Vert g\Vert ^2_{L^{\infty }(d\mu )}\int _{{\mathbb {A}}_m}\int _{{\mathbb {A}}_m}\frac{|xy|^{-\lambda }}{|x-y|^{1-2s}}d\mu (x)d\mu (y)\\&\lesssim \Vert g\Vert ^2_{L^{\infty }(d\mu )}m^{2\lambda +\alpha -1+2s}c_{\alpha }(\mu ). \end{aligned}$$

Substituting this into (70) proves (69) $\square$

Theorem 5.3

Let $a>1$ and $\lambda \ge 0$. Then for $1/4\le s\le 1/2$, we have

$$\begin{aligned} \alpha _{\lambda }(s)=1-2s. \end{aligned}$$

Proof

We first assume that $1/4\le s<1/2$. To show $\alpha _{\lambda }(s)\le 1-2s$, it suffices to prove that, for $1-2s<\alpha \le 1$ and each $m\ge 1$,

$$\begin{aligned} {\mathcal {H}}^{\alpha }\left( \left\{ x\in {\mathbb {A}}_m:\,T^{t_k}_{\lambda ,a}f(x)\nrightarrow f(x)\,\,\hbox {as}\,\,k\rightarrow \infty \right\} \right) =0, \end{aligned}$$

(71)

whenever $f\in H_{\lambda }^s({\mathbb {R}})$ and $t_k\rightarrow 0+$, where ${\mathcal {H}}^{\alpha }$ denotes the $\alpha$-Hausdorff measure. Furthermore, by Frostman’s lemma (cf. [16]), (71) is asserted once we prove

$$\begin{aligned} \mu \left( \left\{ x:\,T^{t_k}_{\lambda ,a}f(x)\nrightarrow f(x)\,\,\hbox {as}\,\,k\rightarrow \infty \right\} \right) =0, \end{aligned}$$

(72)

for $f\in H_{\lambda }^s({\mathbb {R}})$, $\mu \in {\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$ with $1-2s<\alpha \le 1$, and $t_k\rightarrow 0+$.

It is first noted that, if $f\in {\mathscr {S}}({\mathbb {R}})$, $T^{t_k}_{\lambda ,a}f(x)$ tends to f(x) for every $x\in {\mathbb {R}}$. For general $f\in H_{\lambda }^s({\mathbb {R}})$ and a given $\epsilon >0$, let $f_0\in {\mathscr {S}}({\mathbb {R}})$ satisfy $\Vert f-f_0\Vert _{H_{\lambda }^s}<\epsilon$. Since

$$\begin{aligned} |T^{t,N}_{\lambda ,a}f-f|\le |T^{t,N}_{\lambda ,a}(f-f_0)|+|T^{t,N}_{\lambda ,a}f_0-f_0|+|f_0-f|, \end{aligned}$$

it follows that, for $\delta >0$,

$$\begin{aligned}&\mu \left( \left\{ x:\,\limsup _{k\rightarrow \infty }\limsup _{N\rightarrow \infty }|T^{t_k,N}_{\lambda ,a}f-f|>\delta \right\} \right) \\&\quad \le \mu \left( \left\{ x:\,\sup _{k\ge 1}\sup _{N\ge 1}|T^{t_k,N}_{\lambda ,a}(f-f_0)|>\frac{\delta }{3}\right\} \right) +\mu \left( \left\{ x:\,|f_0-f|>\frac{\delta }{3}\right\} \right) . \end{aligned}$$

Thus by Lemmas 5.1 and 5.2,

$$\begin{aligned} \mu \left( \left\{ x:\,\limsup _{k\rightarrow \infty }\limsup _{N\rightarrow \infty }|T^{t_k,N}_{\lambda ,a}f-f|>\delta \right\} \right) \lesssim&\frac{1}{\delta }\Vert f-f_0\Vert _{H^{s}_{\lambda }}<\frac{\epsilon }{\delta }, \end{aligned}$$

and then, (72) is concluded by letting $\epsilon \rightarrow 0+$ and $\delta \rightarrow 0+$ successively.

The reverse inequality $\alpha _{\lambda }(s)\ge 1-2s$ is a consequence of the existence of functions like $f=g*_{\lambda }G_s$ with some $g\in L^2_{\lambda }({\mathbb {R}})$, which are singular on sets of dimension $\alpha$ when $0<\alpha <1-2s$. Indeed. as in [2], we may take $E=E_0\bigcap [1/2,3/2]$, where $E_0$ is the generalized Cantor set with $\hbox {dim}_H E_0=\alpha$, and define, for $s<\gamma <(1-\alpha )/2$,

$$\begin{aligned} g_0(x)=\chi _{(0,2)}(x)d(x,E)^{-\gamma }, \end{aligned}$$

where d(x, E) denotes the distance from x to E. We then have $\int _{{\mathbb {R}}}|g_0(x)|^2|x|^{2\lambda }dx\le 2^{2\lambda }\int _0^2d(x,E)^{-2\gamma }dx<\infty$ by [17, Lemma 3.6] (cf. [30, Lemma 1] also). It remains to show that $g_0*_{\lambda }G_s$ is singular on E.

If $|x|\le 4$, then, from (59), $G_s(x)\gtrsim \int _0^1e^{-x^2/(4\delta )}\delta ^{\frac{s}{2}-\lambda -\frac{3}{2}}d\delta \gtrsim |x|^{s-1-2\lambda }$, and so for $x,y\in (0,2)$, we have, by (18),

$$\begin{aligned} (\tau _xG_{s})(-y)&\gtrsim \int _0^\pi (x^2+y^2-2xy\cos \theta )^{\frac{s}{2}-\lambda -\frac{1}{2}}(1+\cos \theta )\sin ^{2\lambda -1}\theta d\theta \\&\gtrsim \int _0^{\pi /2} \left( |x-y|+\sqrt{xy}\theta \right) ^{s-2\lambda -1}\theta ^{2\lambda -1}d\theta \\&\gtrsim (x+y)^{-2\lambda }|x-y|^{s-1}. \end{aligned}$$

Now for $x\in (0,2)$, let $x_0\in E$ be such that $|x-x_0|=d(x,E)$. We have

$$\begin{aligned} \left( g_0*_{\lambda }G_s\right) (x)&\gtrsim \int _0^2(x+y)^{-2\lambda }|x-y|^{s-1}d(y,E)^{-\gamma }|y|^{2\lambda }dy\\&\gtrsim \int _{|y-x_0|\le |x-x_0|/8}|x-y|^{s-1}|y-x_0|^{-\gamma }dy\\&\gtrsim |x-x_0|^{s-\gamma }, \end{aligned}$$

that means, $\left( g_0*_{\lambda }G_s\right) (x)\gtrsim 1/d(x,E)^{\gamma -s}$ and $g_0*_{\lambda }G_s$ is singular on E.

If $s=1/2$ and $f\in H_{\lambda }^{1/2}({\mathbb {R}})$, we choose, for $0<\alpha \le 1$, $s'\in [1/4,1/2)$ such that $1-2s'<\alpha \le 1$. Then, since $f\in H_{\lambda }^{1/2}({\mathbb {R}})\subseteq H_{\lambda }^{s'}({\mathbb {R}})$, (72) holds for all $\mu \in {\mathcal {M}}^{\alpha }({\mathbb {A}}_{m})$ and $t_k\rightarrow 0+$. Therefore, for $0<\alpha \le 1$, (71) is true for all $f\in H_{\lambda }^{1/2}({\mathbb {R}})$ and $t_k\rightarrow 0+$, which shows that $\alpha _{\lambda }(1/2)=0$. $\square$

6 Closing Comments

6.1. Here we give a short description on the free Schrödinger equation associated to the Dunkl operators on ${\mathbb {R}}^d\times {\mathbb {R}}$ for $d\ge 2$, which is left for further work.

For given $\lambda _k\ge 0$, $k=1,\ldots ,d$, we put

$$\begin{aligned} \lambda =(\lambda _1,\ldots ,\lambda _d) \end{aligned}$$

as a multiplicity vector. For a differentiable function f on ${\mathbb {R}}^d$, the Dunkl operators are defined by

$$\begin{aligned} {\mathcal {D}}_kf(x):=\frac{\partial }{\partial x_k}f(x)+\frac{\lambda _k}{x_k}(f(x)-f(x\sigma _k)),\quad k=1,\ldots ,d, \end{aligned}$$

where $x\sigma _k=(x_1,\ldots ,-x_k\ldots ,x_d)$. The associated Laplacian is $\Delta _{\lambda }=\sum _{k=1}^d{\mathcal {D}}_k^2$, or explicitly, for a twice differentiable function f,

$$\begin{aligned} \left( \Delta _{\lambda }f\right) (x)=\Delta f(x)+\sum _{k=1}^d\frac{2\lambda _k}{x_k}\frac{\partial }{\partial x_k}f(x)-\sum _{k=1}^d\frac{\lambda _k}{x_k^2}(f(x)-f(x\sigma _k)). \end{aligned}$$

For each k, the adjoint of the Dunkl operator ${\mathcal {D}}_k$ is $-{\mathcal {D}}_k$, as a densely defined operator from the $L^2$ space on ${\mathbb {R}}^d$ to itself, which is associated with the measure $w_{\lambda }(x)dx$, where

$$\begin{aligned} w_{\lambda }(x)=\prod _{k=1}^d|x_k|^{2\lambda _k}. \end{aligned}$$

In general, for $1\le p<\infty$, we denote by $L_{\lambda }^p({\mathbb {R}}^d)$ the space of measurable functions f on ${\mathbb {R}}^d$ satisfying $\Vert f\Vert _{L_{\lambda }^p({\mathbb {R}}^d)}:=\left\{ c_{\lambda }\int _{{\mathbb {R}}^d}|f(x)|^p w_{\lambda }(x)dx\right\} ^{1/p}<\infty$, where $c_{\lambda }=\prod _{k=1}^d c_{\lambda _k}$ with $c_{\gamma }^{-1}=2^{\gamma +1/2}\Gamma (\gamma +1/2)$.

The Schrödinger equation associated to the Dunkl operators is $i\partial _t u(x,t)=\Delta _{\lambda }u(x,t)$, $(x,t)\in {\mathbb {R}}^d\times {\mathbb {R}}$, and we consider its initial value problem with an initial data f,

$$\begin{aligned} \left\{ \begin{array}{l} i\partial _t u(x,t)=\Delta _{\lambda }u(x,t),\qquad (x,t)\in {\mathbb {R}}^d\times {\mathbb {R}},\\ \quad u(x,0)=f(x),\qquad \qquad \quad \,\,\,\,\, x\in {\mathbb {R}}^d. \end{array}\right. \end{aligned}$$

(73)

If $f\in {\mathscr {S}}({\mathbb {R}}^d)$, the solution u to the problem (73) would have an explicit representation in terms of the multiple Dunkl transform. From the details, we define the Dunkl transform of a function $f\in L_{\lambda }^1({\mathbb {R}}^d)$ by

$$\begin{aligned} ({\mathscr {F}}_{\lambda }f)(\xi ):=c_{\lambda }\int _{{\mathbb {R}}^d}f(x)E_\lambda (-ix\circ \xi )w_{\lambda }(x)dx,\qquad \xi \in {\mathbb {R}}^d, \end{aligned}$$

where $x\circ \xi =(x_1\xi _1,\ldots , x_d\xi _d)$, $E_{\lambda }$ is the Dunkl kernel given by

$$\begin{aligned} E_\lambda (ix\circ \xi )=\prod _{k=1}^d E_{\lambda _k}(ix_k\xi _k), \end{aligned}$$

with $E_{\lambda _k}(z)$ is the one-dimensional Dunkl kernel given in (2). It is noted that $E_{0}(-ix\circ \xi )=e^{-i\langle x,\xi \rangle }$.

For $f\in {\mathscr {S}}({\mathbb {R}}^d)$, the solution u to (73) is given by

$$\begin{aligned} u(x,t)=c_{\lambda }\int _{{\mathbb {R}}^d}({\mathscr {F}}_{\lambda }f)(\xi )e^{it|\xi |^2}E_{\lambda }(ix\circ \xi )w_{\lambda }(\xi )d\xi ,\qquad (x,t)\in {\mathbb {R}}^d\times {\mathbb {R}}. \end{aligned}$$

6.2. The general setting of the Dunkl theory is on the study of multivariable analytic structures associated with finite reflection groups, of which the basic tools are the Dunkl transform and the Dunkl operators invariant under a given group. During the last decades, it has gained considerable interest in various fields of mathematics and also in physical applications (cf. [11]); for example, the Dunkl operators for the symmetric group $S_d$ on ${\mathbb {R}}^d$ are naturally connected with the analysis of quantum many body systems of Calogero–Moser–Sutherland type, which describe algebraically integrable systems in one dimension.

The free Schrödinger equation studied in the present paper and its higher dimensional counterparts described above are associated to the abelian groups ${\mathbb {Z}}_2$ on ${\mathbb {R}}$ and ${\mathbb {Z}}_2^d$ on ${\mathbb {R}}^d$ for $d\ge 2$ respectively.