\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 139, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/139\hfil 
Schr\"odinger equations with spatial variable coefficient]
{Maximal estimates for fractional Schr\"odinger equations with spatial
variable coefficient}

\author[B.-W. Zheng \hfil EJDE-2018/139\hfilneg]
{Bo-Wen Zheng}

\address{Bo-wen Zheng \newline
College of Sciences,
China Jiliang University,
Hangzhou 310018, China}
\email{bwen\_zj1516@126.com, 17a0802126@cjlu.edu.cn}

\dedicatory{Communicated by Jerome A. Goldstein}

\thanks{Submitted May 28, 2017. Published July 3, 2018.}
\subjclass[2010]{35B65, 35Q40, 35Q55}
\keywords{Schr\"odinger equation with spatial variable coefficient;
\hfill\break\indent maximal estimate; Hankel-Sobolev space}

\begin{abstract}
 Let $v(r,t)=\mathcal{T}_tv_0(r)$ be the solution to a fractional
 Schr\"odinger equation where the coefficient of Laplacian depends
 on the spatial variable. We prove some weighted $L^q$ estimates for
 the maximal operator generated by $\mathcal{T}_t$ with initial data
 in a Sobolev-type space.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}\label{intro}

In this article, we  study the maximal estimates of solutions
for the fractional Schr\"odinger equation with spatial variable coefficient,
\begin{equation} \label{e4}
\begin{gathered}
 i\partial_tv(r,t)+[-r^{p_0}(\partial_{rr}+\frac{p_1}{r}
 \partial_r-\frac{p_2}{r^2}) ]^{\alpha/2} v(r,t)=0,\\
 (r,t)\in\mathbb{R}^{+}\times\mathbb{R},\;
 \alpha\in\mathbb{R}^{+},\\
 v(r,0)=v_0(r),\quad r\in\mathbb{R}^{+},
 \end{gathered}
 \end{equation}
 where $v$ is a complex-valued function, $r=|x|$, $(x\in\mathbb{R}^n)$
is the radius, and the array $(p_0,p_1,p_2)$ satisfies the assumptions
\begin{equation}
p_0<2,\quad p_1>1,\quad p_2=(\frac{2-p_0}{2}\mu)^2-(\frac{p_1-1}{2})^2,\quad
 \mu\geq0.
\end{equation}

The difficulty in this equation comes from the spatial variable coefficient
term $r^{p_0}$ in front of the Laplacian operator.
Such a $r^{p_0}$-factor arises in the problem of the integrability of
the inhomogeneous spherically symmetric Heisenberg ferromagnetic spin system (HFSS)
\begin{equation}\label{e54}
\vec{S}_t(r,t)=\rho(r)\vec{S}\times[\vec{S}_{rr}
+\frac{n-1}{r}\vec{S}_r]+\rho_r(r)[\vec{S}\times\vec{S}_r],
\end{equation}
where the spin $\vec{S}=(S^x, S^y, S^z)$ is constrained by
$\vec{S}^2=1$, $\rho(r)$ is a scalar function, $r= |x|$, $0< r<\infty$.

By a known geometrical process \cite{DPL, PL}, the spin evolution equation
\eqref{e54} is equivalent to the following generalized nonlinear
Schr\"odinger equation
\begin{equation} \label{e55}
\begin{aligned}
&iv_t+\rho(v_{rr}+\frac{n-1}{r}v_r-\frac{n-1}{r^2}v+2|v|^2v)+2\rho_rv_r \\
&+[\rho_{rr}+\frac{n-1}{r}\rho_r+2\int_0^r\rho_{r'}|v|^2dr'+4(n-1)
\int_0^r\frac{\rho}{r'}|v|^2dr']v=0,
\end{aligned}
\end{equation}
and the integrability of \eqref{e54} holds for the conditions
 $\rho(r)=\epsilon_1r^{-2(n-1)}+\epsilon_2r^{-(n-2)}$,
where $\epsilon_1, \epsilon_2$ are arbitrary constants.
Obviously, the factor $r^{p_0}$ corresponds to the term $\rho(r)$
in the \eqref{e55}.

In the case of the non-fractional (i.e. $\alpha=2$) Schr\"odinger equation
without the spatial variable coefficient (i.e. $p_0=0$), the \eqref{e4}
 reduces to the classical Schr\"odinger equation with(out) the
inverse-square potential under the assumption of the spherical symmetry:
\begin{equation} \label{e1}
\begin{gathered}
 i\partial_tu(x,t)-\Delta u(x,t)+\frac{a}{|x|^2}u(x,t)=0,\quad
(x,t)\in\mathbb{R}^n\times\mathbb{R},\\
 u(x,0)=f(x),\quad x\in\mathbb{R}^n.
\end{gathered}
\end{equation}
As we know, when $a=0$,
there is a large body of literature studying values of $s$ for which the
estimates
\begin{equation}\label{e56}
\|S^{\ast}f\|_{L^q(wdx)}\leq C\|f\|_{H^s(\mathbb{R}^n)}, \quad (S^\ast f)(x):=\sup_{t\in\mathbb{R}}|e^{it\Delta}f(x)|
\end{equation}
holds for some $q$ and  weight $w(x)$. This has implications for the
 existence almost everywhere of $\lim_{t\to0}u(x,t)$ for its
solution $u(x,t)=e^{it\Delta}f(x)$, which can be formally expressed as
\begin{equation}
e^{it\Delta}f(x)=\int_{\mathbb{R}^n}e^{i(x\cdot\xi-t|\xi|^2)}(\mathcal{F}f)(\xi)d\xi,
\end{equation}
where $\mathcal{F}$ is the usual spatial Fourier transform defined by
$\mathcal{F}f=\int_{\mathbb{R}^n}e^{-ix\cdot\xi}f(x)dx$.

The maximal estimate \eqref{e56} and related questions were raised by
 Carleson \cite{LC} who proved convergence for $s\geq\frac{1}{4}$ when $n=1$.
Dahlberg and Kenig \cite{DK} showed that this result is sharp.
In higher dimension, the question of identifying the optimal exponent $s$
has been studied by several authors and our state of knowledge may be
summarized as follows. For $n=2$, the strongest result to date appears
in \cite{SL} for $s>3/8$. For $n\geq2$, the convergence is shown to hold
for $s>\frac{2n-1}{4n}$ (see \cite{JB, JBB}). More generally, it should also
be observed that the maximal estimates \eqref{e56} developed for \eqref{e1}
with $a=0$ can be extended to the case of fractional Schr\"odinger
equation without the spatial variable coefficient (i.e. $\alpha>0,\ p_0=0$).
Some positive partial results were obtained by  Sj\"olin \cite{SSP2},
Heinig-Wang \cite{HW}, Cho-Lee-Shim \cite{YY2, YY} and  Bourgain \cite{JB}.

In the case when $p_0\neq0$ and $\alpha>0$, equation \eqref{e4} can be viewed
as the general fractional Schr\"odinger equation with spatial variable
coefficient proposed by authors in \cite{ZZ}, which is a simplified
version of \eqref{e55}. Inspired by the results of the papers
\cite{ZZB, ZZ} and equation \eqref{e1}, we try to explore the maximal
estimate for the more general equation \eqref{e4}, which seems that there
is no previous literature on it. In this paper, we try to derive
some maximal estimates of solution to the general equation \eqref{e4}.

Let $v(r,t)=\mathcal{T}_tv_0(r)$ be the solution to \eqref{e4},
we define the maximal operator $\mathcal{T}^{\ast}$ as
\begin{equation}
(\mathcal{T}^{\ast}v_0)(r)=\sup_{t\in\mathbb{R}}|\mathcal{T}_tv_0(r)|.
\end{equation}
Our aim is to investigate the mapping properties of $\mathcal{T}^{\ast}$,
which are from a Sobolev-type space $X$ to a weighted $L^q$ space.
 The estimates are of the form
\begin{equation}\label{equ:{15}}
\|\mathcal{T}^{\ast}v_0\|_{L^q_{\omega,\varrho}(\mathbb{R}^{+})}\leq C \|v_0\|_{X},
\quad X=\mathcal{W}^{s,2}(\mathbb{R}^{+}) \text{ or }
\dot{H}_{\rm rad}^s(\mathbb{R}^n),
\end{equation}
where $\mathcal{W}^{s,2}(\mathbb{R}^{+})$ is the inhomogeneous Hankel-Sobolev
space in Definition \ref{def1.2} and $\dot{H}_{\rm rad}^s(\mathbb{R}^n)$
is the usual homogeneous Sobolev space
\begin{equation}
\dot{H}_{\rm rad}^s(\mathbb{R}^n)=\{f \text{ is radial},\
 \|f\|^2_{\dot{H}_{\rm rad}^s(\mathbb{R}^n)}
=\int_{\mathbb{R}^n}|\xi|^{2s}|(\mathcal{F}f)(\xi)|^2d\xi<\infty\}.
\end{equation}
We also note that the norm $\|F\|_{L^q_{\omega,\varrho}(\mathbb{R}^{+})}$
is abbreviated by
\begin{equation}
\|F\|_{L^q_{\omega,\varrho}(\mathbb{R}^{+})}
:=\Big(\int_{\mathbb{R}^{+}}|F(r)|^q\varrho(r)d\omega_r\Big)^{1/q},
\end{equation}
where $d\omega_r=r^{p_1-p_0}dr$ is the Lebesgue measure.
For simplicity,
$\|F\|_{L^q_\omega(\mathbb{R}^{+})}:=\|F\|_{L^q_{\omega,1}(\mathbb{R}^{+})}$
and $\|F\|_{L^q(\mathbb{R}^{+})}:=(\int_{\mathbb{R}^{+}}|F(r)|^qdr)^{1/q}$.

For \eqref{e4}, the presence of the factor $r^{p_0}$ makes it difficult
to give the expression of the solution by using the usual Fourier transform,
which is only a well-suited tool to analyze constant coefficient
 Schr\"odinger equation such as \eqref{e1}.
Inspired by \cite{ZZB, ZZ}, we introduce a suitable Hankel transform.

\begin{definition}\label{def1.1} \rm
Suppose $f(r)$ is an integrable function in $\mathbb{R}^{+}$,
we define the Hankel transform
\begin{equation}\label{e5}
(\mathcal{H}_\mu f)(\lambda)=\int_{\mathbb{R}^{+}} (\lambda r)^{\frac{1-p_1}{2}}
\mathcal{J}_\mu(\frac{2}{2-p_0}(\lambda r)^\frac{2-p_0}{2})f(r)d\omega_r,
\end{equation}
where $\mathcal{J}_\mu(z)$ is the first Bessel function of order $\mu$
defined as
\begin{equation*}
\mathcal{J}_\mu(z)=\frac{(z/2)^\mu}{\Gamma(\mu+\frac{1}{2})\pi^{1/2}}
\int_{-1}^1e^{izy}(1-y^2)^{\mu-\frac{1}{2}}dy.
\end{equation*}
\end{definition}

We define the fractional power of the second-order operator
 $\mathcal{A}_\mu:=-r^{p_0}(\partial_{rr}+\frac{p_1}{r}\partial_r-\frac{p_2}{r^2})$
in \eqref{e4} by
\begin{equation}\label{e16}
\mathcal{A}_\mu^{\alpha/2}g(r)
=\mathcal{H}_\mu[\lambda^{\frac{2-p_0}{2}\alpha}(\mathcal{H}_\mu g)(\lambda)](r).
\end{equation}
It should be noticed that the definition of $\mathcal{A}_\mu^{\alpha/2}$
can be referred in \cite{PST, ZZB} and makes sense.

For our purpose, we also introduce the Hankel-Sobolev space via the
Hankel transform.

\begin{definition}\label{def1.2} \rm
The homogeneous \emph{Hankel-Sobolev space} $\dot{\mathcal{W}}^{s,2}(\mathbb{R}^{+})$
consists of tempered distributions $f$ for which
 $\mathcal{H}_\mu[\lambda^{\frac{2-p_0}{2}s}(\mathcal{H}_\mu f)(\lambda)](r)$
exists and is in $L_\omega^2(\mathbb{R}^{+})$ function. That is,
\[
\dot{\mathcal{W}}^{s,2}(\mathbb{R}^{+})
=\big\{f\in \mathcal{S}'(\mathbb{R}^{+}),
\|f\|^2_{\dot{\mathcal{W}}^{s,2}(\mathbb{R}^{+})}
=\int_{\mathbb{R}^{+}}|\lambda^{\frac{2-p_0}{2}s}
(\mathcal{H}_\mu f)(\lambda)|^2d\omega_\lambda<\infty\big\}.
\]
We also define the \emph{inhomogeneous Hankel-Sobolev space}
 $\mathcal{W}^{s,2}(\mathbb{R}^{+})$ as
\[
\mathcal{W}^{s,2}(\mathbb{R}^{+})
=\big\{f\in \mathcal{S}'(\mathbb{R}^{+}),\,
 \|f\|^2_{\mathcal{W}^{s,2}(\mathbb{R}^{+})}
=\int_{\mathbb{R}^{+}}(1+\lambda^{2-p_0})^{s}|
(\mathcal{H}_\mu f)(\lambda)|^2d\omega_\lambda<\infty\big\}.
\]
\end{definition}

Note that the space $\dot{H}_{\rm rad}^s(\mathbb{R}^n)$ is the special case of
$\dot{\mathcal{W}}^{s,2}(\mathbb{R}^{+})$ when $(p_0, p_1, p_2)=(0, n-1, 0)$.
Our first result is to derive an weighted $L^2$ estimate for the maximal
function $\mathcal{T}^{\ast}v_0$, which is stated as follows.

\begin{theorem}\label{thm1.3}
Suppose $p_2=0$. Let $b\in(\frac{2-p_0}{2},2-p_0)$,
$2\leq n<\frac{2|p_1-1|}{2-p_0}+2$ and $s\in(1/2,1)$. Then
\begin{equation}\label{e6}
\|\mathcal{T}^{\ast}v_0\|_{L^2_{\omega,\varrho}(\mathbb{R}^{+})}
\leq C(p_0,p_1,\alpha)\|v_0\|_{\dot{H}_{\rm rad}^s(\mathbb{R}^n)},
\end{equation}
where $\varrho(r)=(1+r)^{-b}$.
\end{theorem}

As a consequence, we obtain the almost convergence result for
$v_0\in{\dot{H}_{\rm rad}^s(\mathbb{R}^n)}$.

\begin{corollary}\label{coro1.4}
Let $v_0\in{\dot{H}_{\rm rad}^s(\mathbb{R}^n)}$ with $s\in(\frac{1}{2},1)$
and $2\leq n<\frac{2|p_1-1|}{2-p_0}+2$. Then
\[
\lim_{t\to 0}v(r,t)=v_0(r),\quad \text{a.e. } r\in\mathbb{R}^{+}.
\]
\end{corollary}

If the initial data $v_0$ lies in the space $\mathcal{W}^{s,2}(\mathbb{R}^{+})$,
we improve the integrability of the maximal function $\mathcal{T}^{\ast}v_0$
for  \eqref{e4}.

\begin{theorem}\label{thm1.5}
For $0<\alpha\neq1$. If the initial data $v_0\in \mathcal{W}^{s,2}(\mathbb{R}^{+})$
with $s\in[\frac{1}{4}, \frac{1}{2})$, Then the estimates
\begin{gather}\label{e26}
\|\mathcal{T}^{\ast}v_0\|_{L^q_{\omega}(\mathbb{R}^{+})}\leq C(p_0,p_1)
\|v_0\|_{\mathcal{W}^{s,2}(\mathbb{R}^{+})}, \\
\label{e37}
\|\mathcal{T}^{\ast}v_0\|_{L^q_{\omega,\varrho}(\mathbb{R}^{+})}
\leq C(p_0,p_1)\|v_0\|_{\mathcal{W}^{s,2}(\mathbb{R}^{+})},
\end{gather}
hold for
\[
\frac{8(p_1-p_0+1)}{4p_1-3p_0+2}\leq q<\frac{2(p_1-p_0+1)}{p_1-p_0+1-(2-p_0)s}
\quad\text{and}\quad
q=\frac{2(p_1-p_0+1)}{p_1-p_0+1-(2-p_0)s}
\]
 respectively, where $\varrho(r)=r^b(1+r)^{-b},\ b>0$.
\end{theorem}

The plan of this paper is as follows:
Section 2 is devoted to the preliminaries, including the properties
of Bessel function and the relation between
$\dot{\mathcal{W}}^{s,2}(\mathbb{R}^{+})$ and
$\dot{H}_{\rm rad}^s(\mathbb{R}^n)$.
In Section 3, through delicate computation, we give the complete argument
 about the weighted $L^q$ maximal estimates of the \eqref{e4}.
If not specified, throughout this paper, the notations $M\ll N$ and
$M\sim N$ denote $M\leq C^{-1}N$ and $CM\leq N\leq \tilde{C}M$
respectively for some large constants $C$ and $\tilde{C}$.
We also denote $\leq_{\beta}$ as $\leq C(\beta)$, where
$C(\beta)$ denotes various constant that only depends on $\beta$.
We abbreviate by writing $A+\epsilon$ as $A+$ or $A-\epsilon$ as $A-$
for $0<\epsilon\ll1$.

\section{Preliminaries}

In this section, we collect some basic facts which will be used in the later context.
We recall some asymptotic properties of the first Bessel function
$\mathcal{J}_\mu(z)$ (see \cite{W}). For fixed $\mu$, if $z\ll1$,
a simple calculation gives the rough estimate
\begin{equation}\label{e23}
|\mathcal{J}_\mu(z)|
\leq \frac{Cz^\mu}{2^\mu\Gamma(\mu+\frac{1}{2})\Gamma(1/2)}(1+\frac{1}{\mu+1/2}),
\end{equation}
where $C$ is a absolute constant. Another well known asymptotic expansion about
the Bessel function is
\begin{equation}\label{e25}
\mathcal{J}_\mu(z)=z^{-1/2}(b_{+}e^{iz}+b_{-}e^{-iz})+\Phi_\mu(z),\quad z\gg 1,
\end{equation}
where $|\Phi_\mu(z)|\leq Cz^{-1}, \ |b_{\pm}|\leq C$ and the constant $C$
is independent of $\mu$. As pointed out in \cite{MS}, if one seeks
a uniform bound for large $\mu$ and $z$, then the best one can do is
\begin{equation}\label{e24}
|\mathcal{J}_\mu(z)|\leq Cz^{-1/3},\quad z\geq 1.
\end{equation}
A simple consequence of the above properties is the following Lemma.

\begin{lemma}\label{lem2.1}
For $R\gg 1$, there exists a constant $C(p_0)$ independent of $\mu, R$ such that
\[
\int_R^{2R}|\mathcal{J}_\mu(r^{\frac{2-p_0}{2}})|^2dr
\leq C(p_0)R^{p_0/2}.
\]
\end{lemma}

Next we review some properties of the Hankel transform,
which appear in \cite{BPST, ZZ}.

\begin{lemma}\label{lem2.2}
The Hankel transform $\mathcal{H}_\mu$ satisfies:
\begin{itemize}
\item[(i)] $\mathcal{H}_{\mu}=\mathcal{H}_{\mu}^{-1}$,

\item[(ii)] $\mathcal{H}_\mu$ is an $L^2$ isometry, i.e.
 $\|\mathcal{H}_\mu\phi\|_{L^2_{\omega}(\mathbb{R}^{+})}
 =\|\phi\|_{L_{\omega}^2(\mathbb{R}^{+})}$,

\item[(iii)] $\mathcal{H}_\mu(\mathcal{A}_\mu \phi)(\lambda)
=\lambda^{2-p_0}(\mathcal{H}_\mu \phi)(\lambda)$,
where the operator $\mathcal{H}_{\mu}^{-1}$ is the inverse operator of
$\mathcal{H}_{\mu}$.
\end{itemize}
\end{lemma}

For the Hankel-Sobolev space $\dot{\mathcal{W}}^{\sigma,2}(\mathbb{R}^{+})$,
there exists the following embedding theorem with
$\dot{H}_{\rm rad}^\sigma(\mathbb{R}^n)$, which is proved in the paper \cite{ZZB}.

\begin{lemma}\label{lem2.3}
Let $n\geq2$ and $\mu>\frac{n-2}{2}$.
If $f\in \dot{H}_{\rm rad}^\sigma(\mathbb{R}^n),\; 0\leq\sigma<\frac{n}{2}$, then
\begin{equation}
\|f\|_{\dot{\mathcal{W}}^{\sigma,2}(\mathbb{R}^{+})}
\leq C(\sigma,\mu,n)\|f\|_{\dot{H}_{\rm rad}^\sigma(\mathbb{R}^n)}.
\end{equation}
\end{lemma}

\begin{proof} We  give only an outline of the proof.
From the definition of Hankel transform, \eqref{e16} and using
the integral formula of Bessel function  \cite[p. 385]{W}, we obtain
\begin{equation}\label{e2}
\begin{aligned}
&\mathcal{M}[\mathcal{A}_\mu^{\sigma/2} f](z) \\
&=(2-p_0)^\sigma\frac{\Gamma(\frac{2z -p_1+1}{2(2-p_0)}
 +\frac{\mu}{2})}{\Gamma(1-\frac{2z -p_1+1}{2(2-p_0)}
 +\frac{\mu}{2})}\frac{\Gamma(1-\frac{2z -p_1+1}{2(2-p_0)}
 +\frac{\sigma+\mu}{2})}{\Gamma(\frac{2z -p_1+1}{2(2-p_0)}
 -\frac{\sigma-\mu}{2})}\mathcal{M}[f](z-\frac{2-p_0}{2}\sigma)
\end{aligned}
\end{equation}
where  $\mathcal{M}[f(r)](z)=\int_{\mathbb{R}^{+}} r^{z-1}f(r)dr$
is the Mellin transform.

Denote $B_{\mu,w}^\sigma:=\mathcal{A}_{\mu}^{\sigma/2}
\mathcal{A}_{w}^{-\sigma/2}$. Writing $\tilde{z}=\frac{2z}{2-p_0}$ and
$\tilde{\kappa}=\frac{p_1-1}{2-p_0}$, by \eqref{e2}, we obtain
\begin{align*}
\mathcal{M}[B_{\mu,w}^\sigma f](z)
&= \frac{\Gamma((\tilde{z}-\tilde{\kappa}+\mu)/2)
 \Gamma(1-(\tilde{z}-\sigma-\tilde{\kappa}-\mu)/2)}
 {\Gamma(1-(\tilde{z}-\tilde{\kappa}-\mu)/2)
 \Gamma((\tilde{z}-\sigma-\tilde{\kappa}+\mu)/2)}\\
&\quad\times\frac{\Gamma((\tilde{z}-\sigma-\tilde{\kappa}+w)/2)
 \Gamma(1-(\tilde{z}-\tilde{\kappa}-w)/2)}
 {\Gamma(1-(\tilde{z}-\sigma-\tilde{\kappa}-w)/2)
 \Gamma((\tilde{z}-\tilde{\kappa}+w)/2)}\mathcal{M}[f](z)\\
&:= F(z)\mathcal{M}[f](z).
\end{align*}

For $z=\frac{p_1-p_0+1}{2}+iy$ and $\tilde{z}=\tilde{\kappa}+1+\frac{2}{2-p_0}iy$,
 using the following properties of Gamma function $\Gamma(z)$:
\begin{gather*}
\overline{\Gamma(z)}=\Gamma(\bar{z}),\quad \forall z\in\mathbb{C},\\
|\Gamma(x+iy)|=\Gamma(x)\prod_{k=0}^\infty(1+\frac{y^2}{(x+k)^2})^{-1/2}, \quad
 \forall x>0,\; \forall y\in\mathbb{R},
\end{gather*}
we obtain
\begin{align*}
|F(\frac{p_1-p_0+1}{2}+iy)|
&= |\frac{\Gamma((\mu+\sigma+1-\frac{2}{2-p_0}iy)/2)}
 {\Gamma((\mu-\sigma+1+\frac{2}{2-p_0}iy)/2)}
 \frac{\Gamma((w-\sigma+1+\frac{2}{2-p_0}iy)/2)}
 {\Gamma((w+\sigma+1-\frac{2}{2-p_0}iy)/2)}|\\
&= |\frac{\Gamma((\mu+\sigma+1)/2)}{\Gamma((\mu-\sigma+1)/2)}
 \frac{\Gamma((w-\sigma+1)/2)}
 {\Gamma((w+\sigma+1)/2)}|\prod_0^\infty[R_k(\tilde{y})]^{1/2},
\end{align*}
where
\begin{align*}
R_k(\tilde{y})
&=\frac{(1+\tilde{y}^2/(\mu-\sigma+1+2k)^2)(1+\tilde{y}^2
 /(w+\sigma+1+2k)^2)}{(1+\tilde{y}^2/(\mu+\sigma+1+2k)^2)
 (1+\tilde{y}^2/(w-\sigma+1+2k)^2)}\\
&=\frac{(1+\tilde{y}^2/(M_k-\sigma)^2)(1+\tilde{y}^2/(N_k+\sigma)^2)}
 {(1+\tilde{y}^2/(M_k+\sigma)^2)(1+\tilde{y}^2/(N_k-\sigma)^2)}\leq 1.
\end{align*}
and $\tilde{y}=\frac{2y}{2-p_0}$, $M_k=\mu+1+2k, N_k=w+1+2k$.
Therefore, for $n>2\sigma$, we have
\[
\sup_y |F(\frac{p_1-p_0+1}{2}+iy)|<\infty.
\]
Hence, using \cite[Lemma 2.5]{ZZB}, we obtain
\[
\|B_{\mu,\kappa}^\sigma f\|_{L_{\omega}^2(\mathbb{R}^{+})}
\leq C\|f\|_{L_{\omega}^2(\mathbb{R}^{+})},
\]
which is the desired result.
\end{proof}

At the end of this section, we show the oscillatory integral estimate
 \cite{YY, SSP}.

\begin{lemma}\label{lem2.4}
Suppose $\varphi\in C^2(\mathbb{R}^n\backslash\{0\})$ is a radial function such that $|\varphi^{(k)}(\xi)|\sim|\xi|^{a-k},\ k=0,1,2$ for $0<a\neq1$.  Let $A, B, \sigma$ be the real numbers such that $A, B\neq0,\ \sigma\in[1/2,1)$, then there exists a constant $C(a,\sigma)$, independent of $A, B$, such that
\begin{equation}
\big|\int_{\mathbb{R}} e^{i(A\varphi(\xi)+B\xi)}|\xi|^{-\sigma}d\xi\big|
\leq C|B|^{-(1-\sigma)}.
\end{equation}
\end{lemma}

\vspace{0.6cm}

\section{Proof of main results}

Applying the Hankel transform \eqref{e5} to the \eqref{e4}, by 
Definition \ref{def1.1}, \eqref{e16} and Lemma \ref{lem2.2} (i), we have
\begin{gather*}
 i\partial_t\tilde{v}+\lambda^{\frac{2-p_0}{2}\alpha} \tilde{v}=0\\
 \tilde{v}(\lambda,0)=\tilde{v}_0(\lambda),
\end{gather*}
where
\[
\tilde{v}(\lambda,t)=(\mathcal{H}_\mu v)(\lambda,t),\quad 
\tilde{v}_0(\lambda)=(\mathcal{H}_\mu v_0)(\lambda).
\]
Solving the ODE and inverting the Hankel transform, we obtain the formal solution
\begin{equation}\label{e20}
\mathcal{T}_tv_0(r)=\int_{\mathbb{R}^{+}} (\lambda r)^{\frac{1-p_1}{2}}
\mathcal{J}_\mu(\frac{2}{2-p_0}(\lambda r)^\frac{2-p_0}{2})
e^{it\lambda^{\frac{2-p_0}{2}\alpha}}\tilde{v}_0(\lambda)d\omega_\lambda.
\end{equation}

\subsection*{Proof of Theorem \ref{thm1.3}}

The key ingredients are the asymptotic behavior of the Bessel 
function, and the properties of Hankel transform.

By the continuity of the embedding 
$\dot{H}^{\frac{1}{2}-}(\mathbb{R})\cap\dot{H}^{\frac{1}{2}+}
(\mathbb{R})\hookrightarrow L^\infty(\mathbb{R})$, it suffices to prove

\begin{proposition}\label{prop3.1}
Let $p_2=0$. For $b\in(\frac{2-p_0}{2}, 2-p_0)$, and 
$a\in[\frac{1}{2}-\frac{b}{(2-p_0)\alpha},\frac{1}{2}+\frac{n}{2\alpha}
-\frac{b}{(2-p_0)\alpha})$, there exists a constant $C$ independent of 
$v_0$ such that
\[
\int_{\mathbb{R}}\int_{\mathbb{R}^{+}}|\partial_t^a (\mathcal{T}_tv_0(r))|^2
\frac{d\omega_rdt}{(1+r)^b}\leq C(p_0,p_1,a,\alpha)
\|v_0\|_{\dot{\mathcal{W}}^{\sigma,2}(\mathbb{R}^{+})}^2,
\]
where $\sigma:=\frac{(2a-1)\alpha}{2}+\frac{b}{2-p_0}.$
\end{proposition}

This proposition, NS Lemma \ref{lem2.3}, yield Theorem \ref{thm1.3}.
 Indeed, under the assumption of $a, b$ above, we have 
$\sigma\in[0,\frac{n}{2})$ and $\frac{n-2}{2}<\frac{|p_1-1|}{2-p_0}$, then
\begin{equation}\label{e9}
\int_{\mathbb{R}}\int_{\mathbb{R}^{+}}|\partial_t^a 
(\mathcal{T}_tv_0(r))|^2\frac{d\omega_rdt}{(1+r)^b}
\leq C(p_0,p_1,a,\alpha)\|v_0\|_{\dot{H}_{\rm rad}^\sigma(\mathbb{R}^n)}^2.
\end{equation}
Choosing $a=\frac{1}{2}+$ and $a=\frac{1}{2}-$ in \eqref{e9}, we obtain
\[
\int_{\mathbb{R}^{+}}|\mathcal{T}^{\ast}v_0(r)|^2
\frac{d\omega_r}{(1+r)^b}\leq C(p_0,p_1,\alpha)
\|v_0\|_{\dot{H}_{\rm rad}^{\frac{b}{2-p_0}}(\mathbb{R}^n)}^2.
\]
Hence, Theorem \ref{thm1.3} is proved.

\begin{proof}[Proof of Proposition \ref{prop3.1}]
Using the Plancherel theorem of the usual Fourier transform with respect to 
time $t$, we have
\begin{equation}\label{e3}
\int_{\mathbb{R}}\int_{\mathbb{R}^{+}}|\partial_t^a (\mathcal{T}_tv_0)|^2
\frac{d\omega_rdt}{(1+r)^b}=\int_{\mathbb{R}^{+}}
\int_{\mathbb{R}}|\tau^a \int_{\mathbb{R}}e^{-it\tau} (\mathcal{T}_tv_0)(r,t)dt|^2
\frac{d\tau d\omega_r}{(1+r)^b}.
\end{equation}
Using \eqref{e20}, we obtain
\begin{align*}
&\text{left-hand side  of \eqref{e3}} \\
&=\int_{\mathbb{R}^{+}}\int_{\mathbb{R}}\Big|\tau^a \int_{\mathbb{R}}
 \int_{\mathbb{R}^{+}}(\lambda r)^{\frac{1-p_1}{2}}\mathcal{J}_\mu(\frac{2}{2-p_0}
(\lambda r)^\frac{2-p_0}{2})e^{it(\lambda^{\frac{2-p_0}{2}\alpha}-\tau)}
 \tilde{v}_0(\lambda)d\omega_\lambda dt\Big|^2 \\
&\quad\times \frac{d\tau d\omega_r}{(1+r)^b}\\
&\leq\int_{\mathbb{R}^{+}}\int_{\mathbb{R}}\Big|\tau^a\int_{\mathbb{R}^{+}}
 (\lambda r)^{\frac{1-p_1}{2}}\mathcal{J}_\mu(\frac{2}{2-p_0}
 (\lambda r)^\frac{2-p_0}{2})\delta(\lambda^{\frac{2-p_0}{2}\alpha}-\tau)
 \tilde{v}_0(\lambda)d\omega_\lambda\Big|^2 \\
 &\quad \frac{d\tau d\omega_r}{(1+r)^b}\\ 
&\leq_{p_0,\alpha}\int_{\mathbb{R}^{+}}
 \int_{\mathbb{R}}\Big|\tau^ar^{\frac{1-p_1}{2}}
 \int_{\mathbb{R}^{+}}\lambda^{\frac{p_1-2p_0+3}{(2-p_0)\alpha}-1} 
 \mathcal{J}_\mu(\frac{2 r^\frac{2-p_0}{2}}{2-p_0}\lambda^{\frac{1}{\alpha}} )
 \tilde{v}_0(\lambda^{\frac{2}{(2-p_0)\alpha}})\delta(\lambda-\tau)d\lambda\Big|^2\\
&\quad  \frac{d\tau d\omega_r}{(1+r)^b}\\
&\leq_{p_0,\alpha}\int_{\mathbb{R}^{+}}
\int_{\mathbb{R}}\Big|r^{\frac{1-p_1}{2}}\lambda^{\frac{p_1-2p_0+3}{(2-p_0)
 \alpha}-1+a} \mathcal{J}_\mu(\frac{2 r^\frac{2-p_0}{2}}{2-p_0}\lambda^{\frac{1}{\alpha}} )
 \tilde{v}_0(\lambda^{\frac{2}{(2-p_0)\alpha}})\Big|^2\frac{d\lambda d\omega_r}{(1+r)^b},
\end{align*}
where the above inequality can be rewritten as
\[
C(p_0,\alpha)\iint_{\mathbb{R}^{+}}\big|r^{\frac{1-p_1}{2}}
 \lambda^{\frac{p_1-2p_0+2}{2}+(2a-1)\frac{(2-p_0)\alpha}{4}}
 \mathcal{J}_\mu(\frac{2 }{2-p_0}(\lambda r)^\frac{2-p_0}{2})
\tilde{v}_0(\lambda)\big|^2\frac{d\lambda d\omega_r}{(1+r)^b}.
\]

We make a dyadic decomposition by choosing $\chi$, which is a smoothing 
function supported in $[\frac{1}{2},2]$, and change the variable 
$\frac{\lambda}{M}\mapsto \lambda,\ Mr\mapsto r$,
\begin{equation} \label{e12}
\begin{aligned}
&\text{left-hand side of \eqref{e3}}\\
&\leq_{p_0,\alpha}\sum_{M\in2^{\mathbb{Z}}}\iint_{\mathbb{R}^{+}}
 \big|\lambda^{\tilde{d}+\frac{2p_1-2p_0+1}{2}}(\lambda r)^{\frac{1-p_1}{2}}
 \mathcal{J}_\mu(\frac{2}{2-p_0}(\lambda r)^\frac{2-p_0}{2})\tilde{v}_0
 (\lambda)\chi(\frac{\lambda}{M})\big|^2\frac{ d\omega_r d\lambda}{(1+r)^b}\\
&\leq_{a,\alpha,p_0}\sum_{M\in2^{\mathbb{Z}}}M^{2\tilde{d}+p_1-p_0+1}
 \iint_{\mathbb{R}^{+}}\big|(\lambda r)^{\frac{1-p_1}{2}}
 \mathcal{J}_\mu(\frac{2}{2-p_0}(\lambda r)^\frac{2-p_0}{2})\tilde{v}_0
 (M\lambda)\chi(\lambda)\big|^2 \\
&\quad\times \frac{d\omega_rd\omega_\lambda}{(1+\frac{r}{M})^b}\\
&\leq_{a,\alpha,p_0}\sum_{M\in2^{\mathbb{Z}}}
 \sum_{R\in2^{\mathbb{Z}}}M^{2\tilde{d}+p_1-p_0+1}R^{p_1-p_0}\mathcal{K}_R^M,
\end{aligned}
\end{equation}
where $\tilde{d}:=(2a-1)\frac{(2-p_0)\alpha}{4}$. By integrating,
\[
\mathcal{K}_R^M:=\int_{\mathbb{R}^{+}}\int_R^{2R}|(\lambda r)^{\frac{1-p_1}{2}}
\mathcal{J}_\mu(\frac{2}{2-p_0}(\lambda r)^\frac{2-p_0}{2})
\tilde{v}_0(M\lambda)\chi(\lambda)|^2\frac{dr d\omega_\lambda}{(1+\frac{r}{M})^b}.
\]
Now we need to prove the bound of $\mathcal{K}_R^M$, which is divided
in two cases:
\smallskip

\noindent\textbf{Case 1:} $R\ll1$. 
Since $\lambda\sim1$, we have $(r\lambda)^{\frac{2-p_0}{2}}\ll1$.
 Using the property of Bessel function \eqref{e23}, we have
\begin{equation}  \label{e17}
\begin{aligned}
\mathcal{K}_R^M
&\leq_{p_0,\mu} \int_{\mathbb{R}^{+}}\int_R^{2R}|(\lambda r
 )^{\frac{1-p_1+(2-p_0)\mu}{2}}\tilde{v}_0(M\lambda)\chi(\lambda)|^2
 \frac{dr d\omega_\lambda}{(1+\frac{r}{M})^b}\\
&\leq_{p_0,\mu} R^{2-p_1+(2-p_0)\mu}\min\{1,(\frac{M}{R})^b\}
\int_{\mathbb{R}^{+}}|\tilde{v}_0(M\lambda)\chi(\lambda)|^2 d\omega_\lambda.
\end{aligned}
\end{equation}

\noindent\textbf{Case 2:} $R\gg1$. 
Since $\lambda\sim1$, we obtain $(r\lambda)^{\frac{2-p_0}{2}}\gg1$.
Then $\mathcal{K}_R^M$ can be bounded by
\[
C(p_1)R^{1-p_1}\int_{\mathbb{R}^{+}}|\tilde{v}_0(M\lambda)
\chi(\lambda)|^2(\int_R^{2R}|\mathcal{J}_\mu(\frac{2}{2-p_0}
(\lambda r)^\frac{2-p_0}{2})|^2\frac{dr }{(1+\frac{r}{M})^b})d\omega_\lambda.
\]
Noticing that $\lambda\sim1$, so
\begin{align*}
&\int_R^{2R}|\mathcal{J}_\mu(\frac{2}{2-p_0}(\lambda r)^\frac{2-p_0}{2})|^2
\frac{dr }{(1+\frac{r}{M})^b} \\
&\leq \min\{1,(\frac{M}{R})^b\}\int_R^{2R}|\mathcal{J}_\mu(\frac{2}{2-p_0}
 (\lambda r)^\frac{2-p_0}{2})|^2dr\\
&\leq_{p_0} \min\{1,(\frac{M}{R})^b\}R^{p_0/2} ,
\end{align*}
where the last inequality is obtained by Lemma \ref{lem2.1}. Then
\begin{equation}\label{e18}
\mathcal{K}_R^M\leq_{p_0,p_1} R^{\frac{p_0-2p_1+2}{2}}
\min\{1,(\frac{M}{R})^b\}\int_{\mathbb{R}^{+}}|\tilde{v}_0(M\lambda)
\chi(\lambda)|^2d\omega_\lambda.
\end{equation}

Hence, combining \eqref{e17} and \eqref{e18}, for $p_2=0$, we obtain 
the  estimate
\[
\mathcal{K}_R^M\leq_{p_0,p_1}
\begin{cases}
 R^{2-p_1+|p_1-1|}M^{p_0-p_1-1}\min\{1,(\frac{M}{R})^b\}
 \|\tilde{v}_0(\lambda)\chi(\frac{\lambda}{M})\|^2_{L_{\omega}^2(\mathbb{R}^{+})},
& R\ll1,\\
 R^{\frac{p_0-2p_1+2}{2}}M^{p_0-p_1-1}\min\{1,(\frac{M}{R})^b\}\|\tilde{v}_0
 (\lambda)\chi(\frac{\lambda}{M})\|^2_{L^2_{\omega}(\mathbb{R}^{+})}, 
& R\gg1.
\end{cases}
\]
 Now, returning  to \eqref{e12}, we have
\begin{align*}
&\text{left-hand side of \eqref{e3}}\\
&\leq_{a,\alpha,p_0,p_1}\sum_{M\in2^{\mathbb{Z}}}
 \sum_{R\in2^{\mathbb{Z}}:R\ll1}M^{2\bar{d}}R^{2-p_0+|p_1-1|}
 \min\{1,(\frac{M}{R})^b\}\|\tilde{v}_0(\lambda)
 \chi(\frac{\lambda}{M})\|^2_{L_{\omega}^2(\mathbb{R}^{+})}\\
&\quad +\sum_{M\in2^{\mathbb{Z}}}\sum_{R\in2^{\mathbb{Z}}:R\gg1}
 M^{2\bar{d}}R^{\frac{2-p_0}{2}}\min\{1,(\frac{M}{R})^b\}
 \|\tilde{v}_0(\lambda)\chi(\frac{\lambda}{M})\|^2_{L_{\omega}^2(\mathbb{R}^{+})}\\
&\leq_{a,\alpha,p_0,p_1}\sum_{M\in2^{\mathbb{Z}}}M^{2\bar{d}+b}
 [\sum_{R\ll1}R^{|p_1-1|+2-p_0-b}+\sum_{R\gg1}R^{\frac{2(1-b)-p_0}{2}}]
 \|\tilde{v}_0(\lambda)\chi(\frac{\lambda}{M})\|^2_{L_{\omega}^2(\mathbb{R}^{+})}.
\end{align*}

From the assumption $b\in(\frac{2-p_0}{2},2-p_0)$, the above inequality can 
be further controlled by
\[
\sum_{M\in2^{\mathbb{Z}}}M^{(2a-1)\frac{(2-p_0)\alpha}{2}+b}\|
\tilde{v}_0(\lambda)\chi(\frac{\lambda}{M})\|^2_{L_{\omega}^2(\mathbb{R}^{+})}.
\]

Finally, by Lemma \ref{lem2.2} (ii) and letting $M=2^j$, we have
\begin{align*}
\text{left-hand side of \eqref{e3}}
&\leq_{a,\alpha,p_0,p_1} \sum_{j\in\mathbb{Z}}2^{j(2-p_0)\sigma}
 \|\mathcal{H}_\mu[\chi(\frac{\lambda}{2^j})\mathcal{H}_\mu v_0]
 \|^2_{L_{\omega}^2(\mathbb{R}^{+})}\\
&\leq_{a,\alpha,p_0,p_1} \|\Big(\sum_{j\in\mathbb{Z}}|2^{\frac{(2-p_0)\sigma}{2}j}
 \mathcal{H}_\mu[\chi(\frac{\lambda}{2^j})\mathcal{H}_\mu v_0]|^2
 \Big)^{1/2}\|^2_{L_{\omega}^2(\mathbb{R}^{+})},
\end{align*}
where $\sigma:=\frac{(2a-1)\alpha}{2}+\frac{b}{2-p_0}$ and Proposition 
\ref{prop3.1} follows from Lemma \ref{lem3.2} below.
\end{proof}


\begin{lemma}[\cite{ZZB}]\label{lem3.2}
 Let $\varsigma\in\mathbb{R}$. Then there exists a constant $C$ that 
depends only on $\varsigma$ and $\beta$ such that for all 
$f\in \dot{\mathcal{W}}^{\frac{2\varsigma}{2-p_0},2}(\mathbb{R}^{+})$, we have
\begin{equation} 
\|\Big(\sum_{j\in\mathbb{Z}}|2^{j\varsigma}\mathcal{H}_\mu[\beta_j(\lambda)
\mathcal{H}_\mu f]|^2\Big)^{1/2}\|_{L_{\omega}^2(\mathbb{R}^{+})}
\leq C\|f\|_{\dot{\mathcal{W}}^{\frac{2\varsigma}{2-p_0},2}},
\end{equation}
where the function $\beta\in C_0^\infty(\mathbb{R}^{+})$ is supported 
in the interval $[1/2,2]$, $\beta_j(\lambda)=\beta(\frac{\lambda}{2^j})$, 
$\sum_{j\in\mathbb{Z}}\beta_j=1$.
\end{lemma}


\subsection*{Proof of Theorem \ref{thm1.5}}

We consider the solution \eqref{e20}:
\[
\mathcal{T}_tv_0(r)=\int_{\mathbb{R}^{+}} (\lambda r)^{\frac{1-p_1}{2}}
\mathcal{J}_\mu(\frac{2}{2-p_0}(\lambda r)^\frac{2-p_0}{2})
e^{it\lambda^{\frac{2-p_0}{2}\alpha}}\tilde{v}_0(\lambda)d\omega_\lambda.
\]
By changing the variable $\frac{2}{2-p_0}\lambda^\frac{2-p_0}{2}\mapsto\lambda,
 r^\frac{2-p_0}{2}\mapsto r$, the solution becomes
\begin{align*}
&\big(\frac{2-p_0}{2}\big)^\frac{p_1-p_0+1}{2-p_0}\int_{\mathbb{R}^{+}} 
\lambda^\frac{p_1-p_0+1}{2-p_0}r^\frac{1-p_1}{2-p_0}\mathcal{J}_\mu(\lambda r)
 e^{it(\frac{2-p_0}{2}\lambda)^{\alpha}}\tilde{v}_0
\big((\frac{2-p_0}{2}\lambda)^\frac{2}{2-p_0}\big)d\lambda \\
&:=\mathfrak{T}[\tilde{v}_0](r),
\end{align*}
To prove \eqref{e26} in Theorem \ref{thm1.5}, it suffices to prove that for 
$q=\frac{2(p_1-p_0+1)}{p_1-p_0+1-(2-p_0)s'}$ and $s'\in[\frac{1}{4},s)$, 
the norm $\|\mathfrak{T}[\tilde{v}_0](r) r^{\frac{2p_1-p_0}{(2-p_0)q}}
\|_{L^qL_t^\infty(\mathbb{R}^{+}\times\mathbb{R})}$ is bounded by
\[
\Big(\int_{\mathbb{R}^{+}}|\tilde{v}_0((\frac{2-p_0}{2}
 \lambda)^\frac{2}{2-p_0})[1+(\frac{2-p_0}{2}\lambda)^2]^{s/2}
 \lambda^{\frac{2p_1-p_0}{2(2-p_0)}}|^2d\lambda\Big)^{1/2},
\]
where the norm $\|h(r,t)\|_{L^qL_t^\infty(\mathbb{R}^{+}\times\mathbb{R})}
:=\sup_{t\in\mathbb{R}}\|h(\cdot,t)\|_{L^q(\mathbb{R}^{+})}$.

We write $g(\lambda)=\tilde{v}_0((\frac{2-p_0}{2}\lambda
)^\frac{2}{2-p_0})[1+(\frac{2-p_0}{2}\lambda)^2]^{\frac{s}{2}}
\lambda^{\frac{2p_1-p_0}{2(2-p_0)}}$. Then the above estimate is equivalent to
\[
\|\mathcal{P}(g)|\|_{L^qL_t^\infty(\mathbb{R}^{+}\times\mathbb{R})}
\leq_{q,p_0,p_1}\|g\|_{L^2(\mathbb{R}^{+})},
\]
where
\begin{align*}
\mathcal{P}(g)(r,t)
&=C(p_0,p_1) r^{\frac{2p_1-p_0}{(2-p_0)q}+
\frac{1-p_1}{2-p_0}}\int_{\mathbb{R}^{+}} \mathcal{J}_\mu(\lambda r)
e^{it(\frac{2-p_0}{2}\lambda)^{\alpha}}g(\lambda) \\
&\quad\times [1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}\lambda^{1/2}d\lambda.
\end{align*}
Therefore, utilizing the dual argument, our main task is to prove that
\begin{equation}\label{e33}
\|\mathcal{P}^\ast(f)\|_{L^2(\mathbb{R}^{+})}\leq_{q,p_0,p_1}
\|f\|_{L^{p}L_t^1(\mathbb{R}^{+}\times\mathbb{R})},
\end{equation}
where $p=\frac{2(p_1-p_0+1)}{p_1-p_0+1+(2-p_0)s'}$ and
\begin{align*}
\mathcal{P}^\ast(f)(\lambda) 
&=C(p_0,p_1)[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}\lambda^{1/2}\\
&\quad\times \int_{\mathbb{R}}\int_{\mathbb{R}^{+}}
 \mathcal{J}_\mu(\lambda r)e^{-it(\frac{2-p_0}{2}\lambda)^{\alpha}}f(r,t) 
r^{\frac{2p_1-p_0}{(2-p_0)q}+\frac{1-p_1}{2-p_0}}\,dr\,dt.
\end{align*}
Now we decompose $\mathcal{P}^\ast (f)(\lambda)=\sum_{j=0,1,2}
(\mathcal{P}_j^\ast f)(\lambda)$, where
\begin{align*}
(\mathcal{P}_j^\ast f)(\lambda)
&= C(p_0,p_1)[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}\lambda^{1/2}\\
&\quad\times \int_{\mathbb{R}}\int_{\mathbb{R}^{+}} \mathcal{J}_\mu(\lambda r)
e^{-it(\frac{2-p_0}{2}\lambda)^{\alpha}}\phi_j(\frac{\lambda r}{\mu})
f( r,t) r^{\frac{1}{2}-\gamma}\,dr\,dt,
\end{align*}
and $\phi_j$ are smooth cut-off functions such that $\phi_0=1$ on 
$\{|\eta|<\frac{1}{4}\}, \ \phi_0=0$ on $\{|\eta|<1/4\}$, 
$\phi_1=1$ on $\{|\eta|\sim1\}, \ \phi_1=0$ otherwise, $\phi_2=0$ on 
$\{|\eta|<2\},\ \phi_2=1$ on $\{|\eta|>3\}$, and $\phi_0+\phi_1+\phi_2=1$. 
The symbol $\gamma=\frac{2p_1-p_0}{2-p_0}(\frac{1}{2}-\frac{1}{q})$.
\smallskip

\noindent\textbf{Non-endpoint case: $s'\in[\frac{1}{4}, s)$.}
(i) $j=0$. Using the property of Bessel function \eqref{e23}, and 
$s'\in[\frac{2-p_0}{2},s)$, we obtain
\begin{align*}
|(\mathcal{P}_0^\ast f)(\lambda)|
&\leq_{p_0,p_1} [1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}
 \int_0^{\mu/(2\lambda)}(\lambda r)^{\mu+\frac{1}{2}} 
  r^{-\gamma}\|f( r,\cdot)\|_{L_t^1}d r\\
&\leq_{p_0,p_1,\mu} \lambda^{-s'}\int_0^{\mu/(2\lambda)}r^{-\gamma}
 \|f( r,\cdot)\|_{L_t^1}d r.
\end{align*}
From the basic relation that 
$\|(\mathcal{P}_0^\ast f)(\lambda)\|_{L^2(\mathbb{R}^{+})}
=\|(\mathcal{P}_0^\ast f)(\frac{1}{\lambda})\frac{1}{\lambda}\|_{L^2(\mathbb{R}^{+})}$, we have
\begin{equation}\label{e34}
\|(\mathcal{P}_0^\ast f)(\lambda)\|_{L^2(\mathbb{R}^{+})}
\leq_{p_0,p_1,\mu}\|\mu^{1-s'}\int_0^{(\mu\lambda)/2} 
\frac{r^{-\gamma}}{(\frac{\mu\lambda}{2})^{1-s'}}
\|f(r,\cdot)\|_{L_t^1}d r\|_{L^2(\mathbb{R}^{+})}.
\end{equation}

By extending the function $ r^{-\gamma}\|f( r,\cdot)\|_{L_t^1}$ to $0$ for 
$ r\leq0$, the right hand side of \eqref{e34} can be controlled by the Riesz 
potential operator $I_\beta,\ 0<\beta<1$, where
\[
I_\beta(g)(\lambda)=C_\beta\int_{\mathbb{R}}|\lambda-r|^{\beta-1}g(r)dr,\quad 
\lambda\in\mathbb{R},
\]
and $C_\beta$ is chosen so that 
$(I_\beta)^{\wedge}(\xi)=|\xi|^{-\beta}\widehat{g}(\xi)$. So we further obtain
\begin{align}
\|\mathcal{P}_0^\ast f\|_{L^2(\mathbb{R}^{+})}
&\leq_{p_0,p_1,\mu,s'} \|I_{s'}( r^{-\gamma}\|f( r,\cdot)\|_{L_t^1})
(\frac{\mu\lambda}{2})\|_{L^2(\mathbb{R}^{+})}\\
&\leq_{p_0,p_1,\mu,s'} \Big(\int_{\mathbb{R}}|\xi|^{-2s'}|
( r^{-\gamma}\|f( r,\cdot)\|_{L_t^1})^{\wedge}(\xi)|^2d\xi\Big)^{1/2} \label{e27}\\
&\leq_{p_0,p_1,\mu,s'} \|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})} \label{e28}
\end{align}
where the last inequality is proved by using  Pitt's inequality for the
 usual Fourier transform:

\begin{lemma}[\cite{MB}]
If $l\geq p$, $0\leq a<1-1/p$, $0\leq d<1/l$ and $d=a+1/p+1/l-1$, 
then 
\[
\Big(\int_{\mathbb{R}}|\widehat{f}(\xi)|^l|\xi|^{-dl}d\xi\Big)^{1/l}
\leq C(\int_{\mathbb{R}}|f(x)|^p|x|^{ap}dx)^{1/p},
\]
 where $\widehat{f}(\xi)$ is the usual Fourier transform.
\end{lemma}



(ii) $j=1$. The estimate of $\mathcal{P}_1^\ast f$ is similar to 
$\mathcal{P}_0^\ast f$. When $|\lambda r|\sim\mu$, the property \eqref{e24} 
implies $|\mathcal{J}_\mu(\lambda r)\phi_1(\frac{\lambda r}{\mu})|
\leq C(\lambda r)^{-1/3}$, we obtain
\begin{equation} \label{e29}
\begin{aligned}
\|(\mathcal{P}_1^\ast f)(\frac{1}{\lambda})\frac{1}{\lambda}\|_{L^2(\mathbb{R}^{+})}
&\leq_{p_0,p_1,\mu,s'} \|\int_{(\mu\lambda)/2}^{2\mu\lambda}
 \frac{ r^{\frac{1}{6}-\gamma}}{(2\mu\lambda)^{\frac{7}{6}-s'}}
 \|f( r,\cdot)\|_{L_t^1}d r\|_{L^2(\mathbb{R}^{+})}\\
&\leq_{p_0,p_1,\mu,s'} \|I_{s'}( r^{-\gamma}\|f( r,\cdot)\|_{L_t^1})
 (2\mu\lambda)\|_{L^2(\mathbb{R}^{+})}\\
&\leq_{p_0,p_1,\mu,s'} \|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})}.
\end{aligned}
\end{equation}


(iii) $j=2$. Applying the asymptotic expansion of Bessel function \eqref{e25},
 we write
\begin{equation}
\begin{aligned}
(\mathcal{P}_2^\ast f)(\lambda)
&=C(p_0,p_1)b_{\pm}[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}\\
&\quad\times \int_{\mathbb{R}}\int_{\mathbb{R}^{+}}
 e^{i[\pm \lambda r-t(\frac{2-p_0}{2}\lambda)^{\alpha}]}\phi_2
 (\frac{\lambda r}{\mu})f(r,t) r^{-\gamma}\,dr\,dt\\
&\quad +C(p_0,p_1)[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}
 \int_{\mathbb{R}}\int_{\mathbb{R}^{+}}e^{-it(\frac{2-p_0}{2}\lambda)^{\alpha}}
( r\lambda)^{1/2}\Phi_\mu( r\lambda) \\
&\quad\times \phi_2(\frac{\lambda r}{\mu}) f( r,t) r^{-\gamma}\,dr\,dt\\
&:=C(p_0,p_1)[(P_{\pm}f)(\lambda)+(Qf)(\lambda)],
\end{aligned}
\end{equation}
where $|\Phi_\mu(r\lambda)|\leq C(r\lambda)^{-1}$, $|b_{\pm}|\leq C$ and the
constant $C$ is independent of $\mu$.

For the estimate of $(P_{\pm}f)(\lambda)$, it is sufficient to consider 
$(P_{+}f)(\lambda)$. We decompose
\[
(P_{+}f)(\lambda)=S_1(\lambda)+S_2(\lambda),
\]
where
\begin{gather*}
S_1(\lambda)=b_{+}[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}
 \int_{\mathbb{R}}\int_{\mathbb{R}^{+}} e^{i[ \lambda r
 -t(\frac{2-p_0}{2}\lambda)^{\alpha}]}f( r,t) r^{-\gamma}\,dr\,dt,\\
S_2(\lambda)=b_{+}[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}
 \int_{\mathbb{R}}\int_{\mathbb{R}^{+}} e^{i[ \lambda r
 -t(\frac{2-p_0}{2}\lambda)^{\alpha}]}(\phi_2(\frac{\lambda r}{\mu})-1)f( r,t)
 r^{-\gamma}\,dr\,dt.
\end{gather*}

For $S_2(\lambda)$, arguing as $\mathcal{P}_0^\ast f$, we obtain
\begin{equation}\label{equ:{30}}
\|S_2(\lambda)\|_{L^2(\mathbb{R}^{+})}
\leq \big\|\lambda^{-s'} \int_0^{\frac{3\mu}{\lambda}}\|f( r,\cdot)
 \|_{L_t^1}r^{-\gamma}\,dr\big\|_{L^2(\mathbb{R}^{+})}
\leq \|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})}.
\end{equation}
For $S_1(\lambda)$, we extend $S_1$ to $\mathbb{R}$ by setting
\[
S_1(\xi)=b_{+}[1+(\frac{2-p_0}{2}y)^2]^{-s/2}
\iint_{\mathbb{R}^2}e^{i[ry-t(\frac{2-p_0}{2}|y|)^{\alpha}]}
f(r,t) r^{-\gamma}drdt,\quad y<0,
\]
so
\[
\|S_1\|^2_{L^2(\mathbb{R}^{+})}\leq_{p_0}\iiiint K( r, r',t,t') 
r^{-\gamma}f( r,t)( r')^{-\gamma}f( r',t')\,dr\,dt\,d r'\,dt',
\]
where
\[
K( r, r',t,t')=\int e^{-i[\lambda(r'-r)-(t'-t)(\frac{2-p_0}{2}\lambda)^{\alpha}]}
\lambda^{-2s'}d\lambda.
\]

Since $s'\in[\frac{1}{4},\frac{1}{2})$, by Lemma \ref{lem2.4}, we obtain 
$|K( r, r',t,t')|\leq_{s',\alpha}| r- r'|^{2s'-1}$. Using the theory of 
Riesz potential and the Plancherel theorem of the usual Fourier transform, we have
\begin{equation} \label{e31}
\begin{aligned}
\|S_1\|_{L^2(\mathbb{R}^{+})}
&\leq_{p_0,s',\alpha} \Big(\int_{\mathbb{R}}|\xi|^{-2s'}|
(r^{-\gamma}\|f( r,\cdot)\|_{L_t^1})^{\wedge}(\xi)|^2d\xi\Big)^{1/2}\\
&\leq_{p_0,s',\alpha} \|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})}.
\end{aligned}
\end{equation}

It remains to estimate $(Qf)(\lambda)$. The uniform decay of the function 
$\Phi_\mu$ on $\mu$ shows that
\begin{equation}\label{e11}
|(Qf)(\lambda)|\leq_{p_0,p_1}[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}
\lambda^{-1/2}\int_{2\mu/\lambda}^\infty r^{-\frac{1}{2}-\gamma}\|f\|_{L_t^1}dr.
\end{equation}
Applying  H\"older's inequality to \eqref{e11}, we obtain
\begin{equation} \label{e32}
\begin{aligned}
\|Qf\|_{L^2(\mathbb{R}^{+})} 
&\leq_{p_0,p_1} \mu^{-s'}\|(\lambda^{-\frac{1}{2}+s'}\chi_{(0,\mu)}(\lambda)\\
&\quad +\lambda^{s'-s-\frac{1}{2}}\chi_{(\mu,\infty)}(\lambda))
 \|_{L^2(\mathbb{R}^{+})}\|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})}\\
&\leq_{p_0,p_1,\mu,s'} \|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})}
\end{aligned}
\end{equation}
Combining \eqref{equ:{30}}-\eqref{e32}, we conclude that
\begin{equation}\label{e35}
\|\mathcal{P}_2^\ast f\|_{L^2(\mathbb{R}^{+})}\leq C(p_0,p_1,\mu,s',\alpha)
\|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})},
\end{equation}
Therefore, the claim \eqref{e33} follows from the estimates
\eqref{e28}, \eqref{e29} and \eqref{e35}.
\smallskip


\noindent\textbf{Endpoint case: $s'=s$.}
For the endpoint case $q=\frac{2(p_1-p_0+1)}{p_1-p_0+1-(2-p_0)s}, 
s\in[\frac{1}{4},\frac{1}{2})$, we follow the almost same line as in the 
argument of \eqref{e33} by replacing $s'$ with $s$ and $ r^{-\gamma}$ 
with $ r^{-\gamma}\varrho(r^{\frac{2}{2-p_0}})^{1/q}$ except for the 
estimate \eqref{e32}, where 
$\varrho(r^{\frac{2}{2-p_0}})=r^{\frac{2b}{2-p_0}}(1+r^{\frac{2}{2-p_0}})^{-b}$. 
We only need to check the part of $\mathcal{P}_2^\ast f$.

Replacing $r^{-\gamma}$ with $r^{-\gamma}\varrho(r^{\frac{2}{2-p_0}})^{1/q}$ 
in \eqref{e11} and using H\"older inequality, we obtain
\begin{align*}
&\int_{2\mu/\lambda}^\infty r^{-\frac{1}{2}-\gamma}
 \varrho(r^{\frac{2}{2-p_0}})^{1/q}\|f\|_{L_t^1}dr \\
&\leq\Big(\int_{2\mu/\lambda}^\infty\min\{1, r^{\frac{2bq}{2-p_0}}\} 
 r^{-(\frac{1}{2}+\gamma)q}dr\Big)^{1/q}\|f\|_{L^pL_t^1(\mathbb{R}^{+}
 \times\mathbb{R})}.
\end{align*}
Then for $\gamma=\frac{2p_1-p_0}{2-p_0}(\frac{1}{2}-\frac{1}{q})$ and $b>0$,
\begin{align*}
|(Qf)(\lambda)|
&\leq_{p_0,p_1}[1+(\frac{2-p_0}{2}\lambda)^2]^{-s/2}\lambda^{-1/2}\\
&\quad \times \{((\frac{2\mu}{\lambda})^{-s}\chi_{(0,\mu)}(\lambda)
+(\frac{2\mu}{\lambda})^{\frac{2b}{(2-p_0)}-s}\chi_{(\mu,\infty)}(\lambda)\}
\|f\|_{L^pL_t^1(\mathbb{R}^{+}\times\mathbb{R})},
\end{align*}
which yields the estimate \eqref{e32}. Hence \eqref{e37} follows.
This completes the proof of Theorem \ref{thm1.5}.

\subsection*{Acknowledgments}
This work was supported by grant NSFC11671354.



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\end{document}