Global $L^{2}$ estimates for a class of maximal operators associated to general dispersive equations

Abstract

For a function ϕ satisfying some suitable growth conditions, consider the general dispersive equation defined by $\bigl\{ \scriptsize{ \begin{array}{l} i\partial_{t}u+\phi(\sqrt{-\Delta})u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R}, \\ u(x,0)=f(x), \quad f\in\mathcal{S}(\mathbb{R}^{n}). \end{array} }\bigr.$ (∗) In the present paper, we give some global $L^{2}$ estimate for the maximal operator $S_{\phi}^{*}$, which is defined by $S^{\ast}_{\phi}f(x)= \sup_{0< t<1} |S_{t,\phi}f(x)|$, $x\in\mathbb{R}^{n}$, where $S_{t,\phi}f$ is a formal solution of the equation (∗). Especially, the estimates obtained in this paper can be applied to discuss the properties of solutions of the fractional Schrödinger equation, the fourth-order Schrödinger equation and the beam equation.

1 Introduction and main results

Suppose $f\in\mathcal{S}(\mathbb{R}^{n})$, the Schwartz class on $\mathbb{R}^{n}$, denote

$$ S_{t}f(x)=(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it|\xi|^{2}}\hat{f}(\xi)\, d\xi,\quad (x,t)\in\mathbb{R}^{n}\times \mathbb{R}, $$

where $\hat{f}(\xi)=\int_{\mathbb{R}^{n}}e^{-i\xi\cdot x}f(x)\, dx$. It is well known that $u(x,t):=S_{t}f(x)$ is the solution of the Schrödinger equation

$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u-\Delta u=0, \quad (x,t)\in\mathbb{R}^{n}\times\mathbb {R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$

(1.1)

In 1979, Carleson [1] proposed a problem: if $f\in H^{s}(\mathbb{R}^{n})$ for which s does

$$ \lim_{t\rightarrow0}u(x,t)=f(x),\quad \text{a.e. } x\in \mathbb{R}^{n}, $$

(1.2)

where $H^{s}(\mathbb{R}^{n})$ ($s\in\mathbb{R}$) denotes the non-homogeneous Sobolev space, which is defined by

$$H^{s}\bigl(\mathbb{R}^{n}\bigr)= \biggl\{ f\in \mathcal{S}^{\prime}: \|f\|_{H^{s}}= \biggl(\int _{\mathbb{R}^{n}} \bigl(1+|\xi|^{2}\bigr)^{s}\bigl\vert \hat{f}(\xi)\bigr\vert ^{2}\,d\xi \biggr)^{1/2}< \infty \biggr\} . $$

Carleson first studied this problem for dimension $n=1$ in [1]. He proved that the convergence (1.2) holds for $f\in H^{s}(\mathbb{R})$ with $s \geq\frac{1}{4}$. This result is sharp, which was shown by Dahlberg and Kenig [2]. See Table 1 for the results on the convergence (1.2) when $f\in H^{s}(\mathbb{R}^{n})$.

Table 1 Convergence ( 1.2 ) holds for $\pmb{f\in H^{s}(\mathbb{R}^{n})}$

Moreover, the convergence (1.2) fails if $s<\frac {1}{4}$ (see [2] for $n=1$ and [4] for $n\ge2$). Recently, Bourgain [10] showed that the necessary condition of convergence (1.2) is $s\geq\frac{1}{2}-\frac{1}{n}$ when $n>4$.

It is well known that the pointwise convergence (1.2) is related closely to the local estimate of the local maximal operator $S^{\ast}$ defined by

$$S^{\ast}f(x)= \sup_{0< t< 1}\bigl\vert S_{t}f(x)\bigr\vert ,\quad x\in\mathbb{R}^{n}. $$

Naturally, the maximal estimates have been well studied associated with the following oscillatory integral:

$$S_{t,a}f(x)=\frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^{n}}e^{ix\cdot\xi} e^{it|\xi|^{a}}\hat{f}(\xi)\,d\xi, \quad t\in\mathbb{R} \text{ and } a>0, $$

which is the solution of the fractional Schrödinger equation:

$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u+(-\Delta)^{a/2} u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$

(1.3)

Define the local maximal operator associated with the family of operators $\{S_{t,a}\}_{0< t<1}$ by

$$S^{\ast}_{a}f(x)= \sup_{0< t< 1}\bigl\vert S_{t,a}f(x)\bigr\vert , \quad x\in \mathbb{R}^{n}. $$

Obviously, the following estimate (1.4) can be applied to discuss the pointwise convergence problem on the solution of Schrödinger equation (1.3):

$$ \bigl\Vert S^{\ast}_{a}f\bigr\Vert _{L^{2} (\mathbb{R}^{n})}\leq C\|f\|_{H^{s}(\mathbb{R}^{n})}, $$

(1.4)

which is called the global $L^{2}$ estimate of the maximal operator $S^{\ast}_{a}$ sometimes. These estimates have also independent interest since they reveal global regularity properties of the corresponding oscillatory integrals. Table 2 shows some main results in studying (1.4).

Table 2 Global $\pmb{L^{2}}$ estimate ( 1.4 ) for $\pmb{f\in H^{s}(\mathbb {R}^{n})}$

On the other hand, in 1990, Prestini [16] proved that, if $f\in H^{s}(\mathbb{R}^{n})$ ($n\geq2$) is a radial function, then the local maximal estimate

$$ \bigl\Vert S^{*}f\bigr\Vert _{L^{1}(B)}\le c_{n} \|f\|_{H^{s}} $$

(1.5)

holds if and only if $s\ge\frac{1}{4}$. In 1997, Sjölin [17] proved (1.4) holds for $a>1$ and $s>\frac{a}{4}$. In 2012, Walther [18] showed (1.4) holds for $0< a<1$ and $s>\frac{a}{4}$.

In the present paper, we will discuss some global $L^{2}$ maximal estimates like (1.4) for a local maximal operator $S^{\ast}_{\phi}$ associated with the operator family $\{S_{t,\phi}\}_{t\in\mathbb{R}}$. Let us first give some definitions as follows: Suppose the function $\phi: \mathbb{R}^{+}\rightarrow\mathbb{R}$ satisfies:

• (K1) there exists $l_{1}\geq0$ such that $|\phi(r)|\lesssim r^{l_{1}}$ for all $0< r<1$;

• (K2) there exists $m_{1}\in\mathbb{R}$ such that $|\phi (r)|\lesssim r^{m_{1}}$ for all $r\geq1$;

• (K3) there exists $m_{2}\in\mathbb{R}$ such that $|\phi^{\prime }(r)|\lesssim r^{m_{2}-1}$ for all $r\geq1$;

• (K4) there exists $m_{3}\in\mathbb{R}$ such that $|\phi^{\prime \prime}(r)|\sim r^{m_{3}-2}$ for all $r\geq1$;

• (K5) there exists $m_{4}\in\mathbb{R}$ such that $|\phi^{ (3)}(r)|\lesssim r^{m_{4}-3}$ for all $r\geq1$.

The operator family $\{S_{t,\phi}\}_{t\in\mathbb{R}}$ is defined by

$$ S_{t,\phi}f(x)=(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it\phi(|\xi|)}\hat{f}(\xi)\,d\xi, \quad x\in\mathbb{R}^{n}, $$

(1.6)

where $f\in\mathcal{S}(\mathbb{R}^{n})$ and the local maximal operator $S^{\ast}_{\phi}$ associated with $\{ S_{t,\phi}\}_{t\in\mathbb{R}}$ is defined by

$$S^{\ast}_{\phi}f(x)= \sup_{0< t< 1} \bigl\vert S_{t,\phi}f(x)\bigr\vert ,\quad x\in\mathbb{R}^{n}. $$

Now we state our main results in this paper as follows.

Theorem 1.1

For $n= 1$ and ϕ satisfies (K1)-(K5) with $l_{1}\geq0$, $m_{i}\in\mathbb{R}$ ($1\le i\le4$), and $m_{2}=m_{3}\geq m_{4}$. If $f\in H^{s}(\mathbb{R})$ with $s>\frac{m_{2}}{4}$ for $m_{2}>0$ or $s>\frac{-m_{2}}{2}$ for $m_{2}\leq0$, then

$$ \bigl\Vert S^{\ast}_{\phi}f(x)\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|f\| _{H^{s}(\mathbb{R})}. $$

(1.7)

Theorem 1.2

For $n\geq2$ and ϕ satisfying (K1)-(K5) with $l_{1}\geq 0$, $m_{i}\in\mathbb{R}$ ($1\le i\le4$), and $m_{2}=m_{3}\geq m_{4}$. If $f\in H^{s}(\mathbb{R}^{n})$ is radial with $s>\frac{m_{2}}{4}$ for $m_{2}>0$ or $s>\frac{-m_{2}}{2}$ for $m_{2}\leq0$, then

$$ \bigl\Vert S^{\ast}_{\phi}f(x)\bigr\Vert _{L^{2}(\mathbb{R}^{n})}\leq C\|f\| _{H^{s}(\mathbb{R}^{n})}. $$

(1.8)

Now let us turn to the other result obtained in the present paper, which involves the functions class formed by the radial function and the functions in ${\mathscr{A}}_{k}$, the set of all solid spherical harmonics of degree k. It is well known (see [19], p.151) that there exists a direct sum decomposition

$$L^{2}\bigl(\mathbb{R}^{n}\bigr)=\sum _{k=0}^{\infty}\oplus\mathfrak{D}_{k}. $$

The subspace $\mathfrak{D}_{k}$ is the space of all finite linear combinations of functions of the form $f(|x|)P(x)$, where f ranges over the radial functions and P over ${\mathscr{A}}_{k}$ such that $f(|\cdot|)P(\cdot)\in L^{2}(\mathbb{R}^{n})$.

Fix $k\geq0$ and let $P_{1},P_{2},\ldots,P_{a_{k}}$ denote an orthonormal basis in ${\mathscr{A}}_{k}$. Every element in $\mathfrak {D}_{k}$ can be written in the following form:

$$ f(x)=\sum_{j=1}^{a_{k}}f_{j} \bigl(\vert x\vert \bigr)P_{j}(x) $$

(1.9)

and

$$\int_{\mathbb{R}^{n}}\bigl\vert f(x)\bigr\vert ^{2}\,dx= \sum_{j=1}^{a_{k}} \int_{0}^{\infty} \bigl\vert f_{j}(r)\bigr\vert ^{2}r^{n+2k-1}\,dr. $$

Denote by $\mathcal{H}_{0}(\mathbb{R}^{n})$ the class of all radial functions in $\mathcal{S}(\mathbb{R}^{n})$, and $\mathcal{H}_{k}$ ($k\in\mathbb{N}$) the set of functions defined by (1.9) with $f_{j}\in \mathcal{H}_{0}(\mathbb{R}^{n})$ and $P_{j}\in{\mathscr{A}}_{k}$ for $j=1,2,\ldots,a_{k}$. Sjölin obtained the following result (see [20], p.397).

Theorem A

Suppose that $n\geq2$, $a>1$, and $f\in\mathcal{H}_{k}$ ($k\ge 0$). If $s>\frac{a}{4}$ then (1.4) holds.

We give the global $L^{2}$ estimate of the maximal operator $S^{\ast }_{\phi}$ for $f\in\mathcal{H}_{k}$.

Theorem 1.3

For $n\geq2$ and ϕ satisfies (K1)-(K5) with $l_{1}\geq0$, $m_{i}\in\mathbb{R}$ ($1\le i\le4$), and $m_{2}=m_{3}\geq m_{4}$. If $f\in \mathcal{H}_{k}$ ($k\ge0$) with $s>\frac{m_{2}}{4}$ for $m_{2}>0$ or $s>\frac{-m_{2}}{2}$ for $m_{2}\leq0$, then (1.8) holds.

Note that

$$u(x,t)=e^{it\phi(\sqrt{-\Delta})}f(x)=(2\pi)^{-n} \int_{\mathbb{R}^{n}}e^{ix\cdot\xi+it\phi(|\xi|)} \hat{f}(\xi )\,d\xi=S_{t,\phi}f(x) $$

gives a formal solution of the following general dispersive equation with initial data function f:

$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u+\phi(\sqrt{-\Delta})u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$

(1.10)

Hence, the inequalities (1.7) and (1.8) imply the convergence almost everywhere of the solution of (1.10) in one dimension and higher dimension, respectively.

The proofs of Theorems 1.1-1.3 are given in Sections 2-4, respectively. In the last section, we will give some examples of (1.10).

2 Proof of Theorem 1.1

2.1 Proof of Theorem 1.1 based on Lemma 2.2

In this subsection, we give the proof of Theorem 1.1 by using Lemma 2.2, which will be proved in the next subsection.

Choose a nonnegative function $\varphi\in C_{0}^{\infty}(\mathbb {R})$ such that $\operatorname{supp}\varphi\subset\{\xi: \frac{1}{2}<|\xi|<2\}$ and

$$\sum_{k=-\infty}^{\infty} \varphi \bigl(2^{-k}\xi\bigr)=1,\quad \xi\neq0. $$

Set $\varphi_{0}(\xi)=1-\sum_{k=1}^{\infty} \varphi(2^{-k}\xi)$ and $\psi(\xi)=\sum_{k=1}^{\infty} \varphi(2^{-k}\xi)$. It follows that $\varphi_{0}\in C_{0}^{\infty}(\mathbb{R})$. Rewrite

$$\begin{aligned} S_{t,\phi}f(x) =&(2\pi)^{-1} \int _{\mathbb{R}}e^{ix\cdot\xi+it\phi(|\xi|)} \varphi_{0}(\xi)\hat{f}( \xi)\,d\xi \\ &{} +(2\pi)^{-1}\sum_{k=1}^{\infty} \int_{\mathbb{R}}e^{ix\cdot\xi+it\phi(|\xi|)} \varphi\bigl(2^{-k}\xi \bigr)\hat{f}(\xi)\,d\xi \\ =:&S_{t,\phi,0}f(x)+\sum_{k=1}^{\infty}S_{t,\phi,k}f(x). \end{aligned}$$

(2.1)

Denote

$$S^{\ast}_{\phi,0}f(x)= \sup_{0< t< 1} \bigl\vert S_{t,\phi,0}f(x)\bigr\vert ,\quad x\in\mathbb{R} $$

and

$$S^{\ast}_{\phi,k}f(x)= \sup_{0< t< 1} \bigl\vert S_{t,\phi,k}f(x)\bigr\vert ,\quad x\in\mathbb{R}. $$

Therefore, by (2.1), we obtain

$$ S^{\ast}_{\phi}f(x)\leq S^{\ast}_{\phi,0}f(x)+ \sum_{k=1}^{\infty }S^{\ast}_{\phi,k}f(x). $$

(2.2)

By (2.2) and Minkowski’s inequality, we get

$$ \bigl\Vert S^{\ast}_{\phi}f\bigr\Vert _{L^{2}(\mathbb{R})}\leq\bigl\Vert S^{\ast}_{\phi,0}f\bigr\Vert _{L^{2}(\mathbb{R})}+ \sum_{k=1}^{\infty}\bigl\Vert S^{\ast}_{\phi,k}f\bigr\Vert _{L^{2}(\mathbb{R})}. $$

(2.3)

Now let us recall a result which will be used in our proof of Theorem 1.1.

Lemma 2.1

(see [18])

Assume that the functions $\omega_{1}$ and $\omega_{2}$ belong to $L^{2}(\mathbb{R})$ and that the function m satisfies the following assumption: there is a number C independent of $(t,\xi)$ such that

$$\bigl\vert m(t,\xi)\bigr\vert \leq C\omega_{1}(t), \qquad \biggl\vert \frac{\partial (m(t,\xi))}{\partial t}\biggr\vert \leq C\bigl(\omega_{1}(t)+ \omega_{2}(t)|\xi |^{a}\bigr), \quad a>0. $$

Then there is a number C independent of f such that

$$\biggl(\int_{\mathbb{R}^{n}} \sup_{0< t< 1}\biggl\vert \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}m(t,\xi)\hat{f}(\xi)\,d\xi\biggr\vert ^{2}\,dx \biggr)^{1/2}\leq C\|f\|_{L^{2}(\mathbb{R}^{n})},\quad \operatorname{supp}\hat{f}\subseteq\bigl\{ \xi,|\xi|< 2\bigr\} . $$

We first prove that if $s>\frac{m_{2}}{4}$ for $m_{2}>0$ or $s>\frac {-m_{2}}{2}$ for $m_{2}\leq0$, then

$$ \bigl\Vert S^{\ast}_{\phi,0}f\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|f\|_{H^{s}(\mathbb{R})}. $$

(2.4)

For $g\in\mathcal{S}(\mathbb{R})$ and $\operatorname{supp}\hat{g}\subseteq\{\xi,|\xi|<2\}$, $s>\frac {m_{2}}{4}$ for $m_{2}>0$ or $s>\frac{-m_{2}}{2}$ for $m_{2}\leq0$ and $0< t<1$, let

$$R_{0,t}g(x)=(2\pi)^{-1} \int_{\mathbb{R}}e^{ix\cdot\xi}e^{it\phi(|\xi|)} \bigl(1+|\xi|^{2}\bigr)^{-s/2}\hat{g}(\xi)\,d\xi =:\int _{\mathbb{R}}e^{ix\cdot\xi}m(t,\xi)\hat{g}(\xi)\,d\xi, $$

where $m(t,\xi)=(2\pi)^{-1}e^{it\phi(|\xi|)} (1+|\xi|^{2})^{-s/2}$. Define the maximal operator $R^{\ast}_{0}$ by

$$R^{\ast}_{0}g(x)= \sup_{0< t< 1}\bigl\vert R_{0,t}g(x)\bigr\vert , \quad x\in\mathbb{R}. $$

On the one hand, it is obvious that $|m(t,\xi)|\leq\chi_{(0,1)}(t)$ for $\xi\in\mathbb{R}$ and $0< t<1$. On the other hand, by

$$\frac{\partial(m(t,\xi))}{\partial t}=\frac{i}{2\pi}e^{it\phi (|\xi|)}\phi\bigl(\vert \xi \vert \bigr) \bigl(1+|\xi|^{2}\bigr)^{-s/2}, $$

it follows that

$$ \biggl\vert \frac{\partial(m(t,\xi))}{\partial t}\biggr\vert \leq \chi_{(0,1)}(t)\bigl\vert \phi\bigl(\vert \xi \vert \bigr)\bigr\vert \quad \text{for } \xi\in\mathbb{R} \text{ and } 0< t< 1. $$

(2.5)

By the condition (K1), $|\phi(|\xi|)|\leq C\max\{|\phi(1)|,1\}\leq C$ for $0\leq|\xi|<1$. By (K2), we have, for $|\xi|\geq1$,

$$ \bigl\vert \phi\bigl(\vert \xi \vert \bigr)\bigr\vert \leq \left \{ \textstyle\begin{array}{l@{\quad}l} C|\xi|^{m_{1}}, & m_{1}>0, \\ C|\xi|^{m_{1}}\leq C, & m_{1}\leq0. \end{array}\displaystyle \right . $$

Hence, combining with (2.5) we get, for $\xi\in\mathbb{R}$,

$$ \biggl\vert \frac{\partial(m(t,\xi))}{\partial t} \biggr\vert \leq \left \{ \textstyle\begin{array}{l@{\quad}l} C(\chi_{(0,1)}(t)+\chi_{(0,1)}(t)|\xi|^{m_{1}}), & m_{1}>0, \\ C(\chi_{(0,1)}(t)+\chi_{(0,1)}(t)|\xi|), & m_{1}\leq0, \end{array}\displaystyle \right . $$

where C is independent of $(t,\xi)$. It follows that $m(t,\xi)$ satisfies the assumptions of Lemma 2.1. Therefore, when $s>\frac{m_{2}}{4}$ for $m_{2}>0$ or $s>\frac {-m_{2}}{2}$ for $m_{2}\leq0$, we obtain

$$ \bigl\Vert R^{\ast}_{0}g\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|g\|_{L^{2}(\mathbb{R})}. $$

(2.6)

We have

$$ S_{t,\phi,0}f(x)=R_{0,t} \bigl( \mathcal{F}^{-1} \bigl(\varphi_{0}(\cdot) \bigl(1+| \cdot|^{2}\bigr) ^{\frac{s}{2}}\hat{f}(\cdot) \bigr) \bigr) (x), $$

(2.7)

where $\mathcal{F}^{-1}$ denotes the Fourier inverse transform. Note that

$$\operatorname{supp}\varphi_{0}(\cdot) \bigl(1+|\cdot|^{2} \bigr) ^{\frac{s}{2}}\hat{f}(\cdot)\subseteq\bigl\{ \xi; \vert \xi \vert < 2\bigr\} . $$

Thus, by (2.7) and (2.6), we have

$$\begin{aligned} \bigl\Vert S^{\ast}_{\phi,0}f\bigr\Vert _{L^{2}(\mathbb{R})} &= \bigl\Vert R^{\ast}_{0} \bigl(\mathcal{F}^{-1} \bigl(\varphi_{0}(\cdot) \bigl(1+|\cdot|^{2}\bigr) ^{\frac{s}{2}}\hat{f}(\cdot) \bigr) \bigr)\bigr\Vert _{L^{2}(\mathbb{R})} \\ &\leq C \bigl\Vert \mathcal{F}^{-1} \bigl(\varphi_{0}( \cdot) \bigl(1+|\cdot |^{2}\bigr)^{\frac{s}{2}} \hat{f}(\cdot) \bigr) \bigr\Vert _{L^{2}(\mathbb{R})}\leq C\Vert f\Vert _{H^{s}(\mathbb{R})}, \end{aligned}$$

which is just (2.4). Now we define the operator $R_{N}$ by

$$ R_{N}f(x)=N^{-s}\int_{\mathbb{R}}e^{ix\cdot\xi+it(x) \phi(|\xi|)} \varphi\biggl(\frac{\xi}{N}\biggr)\hat{f}(\xi)\,d\xi,\quad N\geq2, $$

(2.8)

where $t(x)$ is a measurable function in $\mathbb{R}$ with $0< t(x)<1$.

Lemma 2.2

Suppose that ϕ satisfies the conditions in Theorem  1.1. If $s>\frac{m_{2}}{4}$ for $m_{2}>0$ or $s>\frac{-m_{2}}{2}$ for $m_{2}\leq0$, then there exist $\delta>0$ and $C>0$, such that, for all $N\ge2$,

$$ \|R_{N}f\|_{L^{2}(\mathbb{R})}\leq C N^{-\delta}\|f \|_{L^{2}(\mathbb{R})}. $$

(2.9)

The proof of Lemma 2.2 will be given in the next subsection. Now we finish the proof of Theorem 1.1 by using Lemma 2.2. By linearizing the maximal operator, we have, for some real-valued function $t(x)$,

$$\begin{aligned} S^{\ast}_{\phi,k}f(x) \leq& \frac{1}{2\pi} \biggl\vert \int_{\mathbb{R}}e^{ix\cdot\xi+it(x) \phi(|\xi|)}\varphi\biggl( \frac{\xi}{2^{k}}\biggr)\hat{f}(\xi)\,d\xi \biggr\vert \\ =&\frac{1}{2\pi}\bigl\vert R_{2^{k}} \bigl(\mathcal{F}^{-1} \bigl(\chi_{\{2^{k-1}< |\xi|< 2^{k+1}\}} 2^{ks}\hat{f}\bigr) \bigr) (x)\bigr\vert . \end{aligned}$$

(2.10)

By (2.9) and (2.10), for $k\geq1$, we have

$$\begin{aligned} \bigl\Vert S^{\ast}_{\phi,k}f\bigr\Vert _{L^{2}(\mathbb{R})} &\leq \bigl\Vert R_{2^{k}} \bigl(\mathcal{F}^{-1}\bigl( \chi_{\{2^{k-1}< |\xi |< 2^{k+1}\}} 2^{ks}\hat{f}\bigr) \bigr)\bigr\Vert _{L^{2}(\mathbb{R})} \\ &\leq C2^{-k\delta} \bigl\Vert \mathcal{F}^{-1}\bigl( \chi_{\{2^{k-1}< |\xi |< 2^{k+1}\}}2^{ks}\hat{f}\bigr) \bigr\Vert _{L^{2}(\mathbb{R})}. \end{aligned}$$

From this we get

$$ \bigl\Vert S^{\ast}_{\phi,k}f\bigr\Vert _{L^{2}(\mathbb{R})} \leq C2^{-k\delta} \|f\|_{H^{s}(\mathbb{R})}. $$

(2.11)

Summing up the estimates of (2.3), (2.4), and (2.11), we have

$$\begin{aligned} \bigl\Vert S^{\ast}_{\phi}f\bigr\Vert _{L^{2}(\mathbb{R})}&\leq \bigl\Vert S^{\ast}_{\phi ,0}f\bigr\Vert _{L^{2}(\mathbb{R})}+ \sum _{k=1}^{\infty}\bigl\Vert S^{\ast}_{\phi,k}f \bigr\Vert _{L^{2}(\mathbb{R})} \\ &\leq C\|f\|_{H^{s}(\mathbb{R})}+C\sum_{k=1}^{\infty}2^{-k\delta} \| f\|_{H^{s}(\mathbb{R})} \\ &\leq C\|f\|_{H^{s}(\mathbb{R})}. \end{aligned}$$

Therefore, to finish the proof of Theorem 1.1, it remains to show Lemma 2.2.

2.2 The proof of Lemma 2.2

Write

$$R_{N}f(x)=\int_{\mathbb{R}}e^{ix\cdot\xi}p_{N}(x, \xi) \hat{f}(\xi)\,d\xi, \quad N\geq2, $$

where $f\in\mathcal{S}(\mathbb{R})$ and $p_{N}(x,\xi)=e^{it(x)\phi (|\xi|)}\varphi(\frac{\xi}{N})N^{-s}$. Take the function $\rho\in C_{0}^{\infty}(\mathbb{R})$ such that $\rho(x)=1$ if $|x|<1$, and $\rho(x)=0$ if $|x|\geq2$, and set $\psi=1-\rho$. Denote

$$p_{N,M}(x,\xi)=\rho\biggl(\frac{x}{M}\biggr)p_{N}(x, \xi), \quad M>1 $$

and

$$p_{N,M,\varepsilon}(x,\xi)=\psi\biggl(\frac{\xi}{\varepsilon}\biggr) p_{N,M}(x,\xi), \quad 0< \varepsilon< 1. $$

For $N\geq2$, $M>1$, and $0<\varepsilon<1$, the corresponding operators $R_{N,M}$ and $R_{N,M,\varepsilon}$ are defined by

$$R_{N,M}f(x)=\int_{\mathbb{R}}e^{ix\cdot\xi} p_{N,M}(x,\xi)\hat{f}(\xi)\,d\xi $$

and

$$R_{N,M,\varepsilon}f(x)=\int_{\mathbb{R}} e^{ix\cdot\xi}p_{N,M,\varepsilon}(x, \xi) \hat{f}(\xi)\,d\xi. $$

Obviously, both of the operators $R_{N,M}$ and $R_{N,M,\varepsilon}$ are bounded on $L^{2}(\mathbb{R})$. On the other hand, it is easy to see that the adjoint operator $R'_{N,M,\varepsilon}$ of $R_{N,M,\varepsilon}$ is given by

$$R'_{N,M,\varepsilon}g(x)=\iint e^{i(x-y)\cdot\xi}\overline {p_{N,M,\varepsilon}(y,\xi)} g(y)\, dy\, d\xi $$

and it follows that

$$ \lim_{\varepsilon\rightarrow0}R'_{N,M,\varepsilon }g(x)=R'_{N,M}g(x), \quad g\in\mathcal{S}(\mathbb{R}), $$

(2.12)

where $R'_{N,M}$ denotes the adjoint operator of $R_{N,M}$. Since

$$ \int\bigl\vert R^{\prime}_{N,M,\varepsilon}g(x)\bigr\vert ^{2}\,dx= \lim_{L\rightarrow\infty}\int_{|x|< L} \bigl\vert R^{\prime}_{N,M,\varepsilon}g(x)\bigr\vert ^{2}\,dx $$

(2.13)

and

$$\begin{aligned} \int_{|x|< L} \bigl\vert R^{\prime}_{N,M,\varepsilon}g(x) \bigr\vert ^{2}\,dx =&\int_{|x|< L} R^{\prime}_{N,M,\varepsilon}g(x) \overline{R^{\prime}_{N,M,\varepsilon}g(x)} \,dx \\ =&\int_{|x|< L} \biggl(\iint e^{i(x-y)\cdot\xi}\overline {p_{N,M,\varepsilon}(y,\xi)} g(y)\, dy\, d\xi \biggr) \\ &{}\times \biggl(\overline{\iint e^{i(x-z)\cdot\eta}\overline {p_{N,M,\varepsilon}(z, \eta)} g(z)\, dz\, d\eta} \biggr)\,dx. \end{aligned}$$

(2.14)

By (2.13), (2.14), and a similar calculation as [3], p.708, we have

$$\begin{aligned}& \int\bigl\vert R^{\prime}_{N,M,\varepsilon}g(x)\bigr\vert ^{2}\,dx \\& \quad = 2\pi\iint \biggl(\int e^{i(z-y)\xi}\overline{p_{N,M,\varepsilon }(y, \xi)} p_{N,M,\varepsilon}(z,\xi)\,d\xi \biggr) g(y)\overline{g(z)}\,dy\,dz \\& \quad = 2\pi\iint \biggl(\int e^{i(z-y)\xi}\rho\biggl(\frac{y}{M} \biggr)\rho\biggl(\frac {z}{M}\biggr)\psi^{2} \biggl( \frac{\xi}{\varepsilon}\biggr)\overline{p_{N}(y,\xi)} p_{N}(z, \xi)\,d\xi \biggr) \\& \qquad {} \times g(y)\overline{g(z)}\,dy\,dz. \end{aligned}$$

(2.15)

Therefore, invoking (2.12) and by Fatou’s lemma, we obtain

$$\begin{aligned}& \int\bigl\vert R'_{N,M}g(x)\bigr\vert ^{2}\,dx \\& \quad \leq \liminf_{\varepsilon\rightarrow0}\int\bigl\vert R'_{N,M,\varepsilon }g(x) \bigr\vert ^{2}\,dx \\& \quad = 2\pi\lim_{\varepsilon\rightarrow0} \iint \biggl(\int e^{i(z-y)\xi}\rho \biggl(\frac{y}{M}\biggr)\rho\biggl(\frac{z}{M}\biggr) \psi^{2} \biggl(\frac{\xi}{\varepsilon}\biggr)\overline{p_{N}(y, \xi)} p_{N}(z,\xi)\,d\xi \biggr) \\& \qquad {}\times g(y)\overline{g(z)}\,dy\,dz \\& \quad \leq C \iint\biggl\vert \int e^{i[(z-y)\xi+(t(z)-t(y))\phi(\vert \xi \vert )]}\varphi^{2} \biggl(\frac{\xi}{N}\biggr)\,d\xi N^{-2s}\biggr\vert \bigl\vert g(y)\bigr\vert \bigl\vert g(z)\bigr\vert \,dy\,dz. \end{aligned}$$

(2.16)

It is easy to check that the constant C is independent of N and M. Now define

$$I_{N}(x,\omega)=N^{-2s}\int e^{i[x\xi+\omega\phi(|\xi|)]}\varphi ^{2}\biggl(\frac{\xi}{N}\biggr)\,d\xi\quad \text{for } x\in \mathbb{R}, -1< \omega< 1, N\geq2 $$

and

$$J_{N}(x)=\sup_{|\omega|< 1}\bigl\vert I_{N}(x,\omega)\bigr\vert , \quad x\in\mathbb{R}. $$

We have the following conclusion.

Lemma 2.3

Let $J_{N}$ be defined as above, ϕ satisfies the conditions in Theorem  1.1. If $s>\frac {m_{2}}{4}$ for $m_{2}>0$ or $s>\frac{-m_{2}}{2}$ for $m_{2}\leq0$, then there exist $\delta, C>0$, such that, for all $N\ge2$,

$$ \|J_{N}\|_{L^{1}(\mathbb{R})}\leq C N^{-2\delta}. $$

(2.17)

Below we first finish the proof of Lemma 2.2 by applying Lemma 2.3, whose proof will be given in the next subsection. By (2.16) and (2.17), invoking Hölder’s inequality and Young’s inequality, we have

$$\begin{aligned} \int\bigl\vert R'_{N,M}g(x)\bigr\vert ^{2} \,dx &\leq C\iint\bigl\vert I_{N}\bigl(z-y,t(z)-t(y)\bigr)\bigr\vert \bigl\vert g(y)\bigr\vert \bigl\vert g(z)\bigr\vert \,dy\,dz \\ &\leq C\iint J_{N}(z-y)\bigl\vert g(y)\bigr\vert \bigl\vert g(z)\bigr\vert \,dy\,dz \\ &=C\int\bigl(J_{N}\ast|g|\bigr) (z)\bigl\vert g(z)\bigr\vert \,dz \\ &\leq C\bigl\Vert J_{N}\ast \vert g\vert \bigr\Vert _{2}\|g\|_{2} \\ &\leq C\|J_{N}\|_{1}\|g\|_{2}^{2}\leq CN^{-2\delta}\|g\| _{2}^{2}. \end{aligned}$$

From this we get

$$\bigl\Vert R'_{N,M}g\bigr\Vert _{2}\leq CN^{-\delta}\|g\|_{2}. $$

Thus, $\|R_{N,M}g\|_{2}\leq CN^{-\delta}\|g\|_{2}$ by duality, where C is independent of N and M. Letting $M\rightarrow\infty$, we obtain

$$\|R_{N}g\|_{2}\leq CN^{-\delta}\|g\|_{2}. $$

It follows that (2.9) holds, and we complete the proof of Lemma 2.2 based on Lemma 2.3.

2.3 The proof of Lemma 2.3

Now we verify the estimate (2.17). We need the following results.

Lemma 2.4

(Van der Corput’s lemma; see [21], p.309)

Let $\psi\in C_{0}^{\infty}(\mathbb{R})$ and $\phi\in C^{\infty}(\mathbb{R})$ satisfy $|\phi^{\prime\prime}(\xi)|>\lambda>0$ on the support of ψ. Then

$$\biggl\vert \int e^{i\phi(\xi)}\psi(\xi)\,d\xi\biggr\vert \leq10\lambda ^{-\frac{1}{2}} \bigl\{ \|\psi\|_{\infty}+\bigl\Vert \psi^{\prime} \bigr\Vert _{1}\bigr\} . $$

Lemma 2.5

([22])

Let I denote an open integral in $\mathbb {R}$. For $g\in C_{0}^{\infty}(I)$ and the real-valued function $F\in C^{\infty}(I)$ with $F^{\prime}\neq0$, if $k\in\mathbb{N}$, then

$$\int_{I} e^{iF(x)}g(x)\,dx=\int _{I} e^{iF(x)}h_{k}(x)\,dx, $$

where $h_{k}$ is a linear combination of functions of the form

$$g^{(s)}\bigl(F^{\prime}\bigr)^{-k-r}\prod _{q=1}^{r}F^{(j_{q})} $$

with $0\leq s \leq k$, $0\leq r \leq k$, and $2\leq j_{q} \leq k+1$.

We now return to the proof of Lemma 2.3. Recall that

$$I_{N}(x,\omega)=N^{-2s}\int e^{i[x\cdot\xi+\omega\phi(|\xi |)]} \varphi^{2}\biggl(\frac{\xi}{N}\biggr)\,d\xi, \quad x\in\mathbb{R}, -1< \omega< 1, N\geq2. $$

Performing a change of variable, we have

$$I_{N}(x,\omega)=N^{1-2s}\int e^{i(Nx\xi+\omega\phi(N|\xi|))}G(\xi )\,d\xi, $$

where $x\in\mathbb{R}$, $-1<\omega<1$, $N\geq2$, and $G(\xi)=\varphi ^{2}(\xi)$. It is obvious that, for all $x\in\mathbb{R}$, $-1<\omega<1$, and $N\geq2$,

$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \le CN^{1-2s}. $$

(2.18)

Below we give more estimates of $|I_{N}(x,\omega)|$.

Step 1: The other estimates of $I_{N}(x,\omega)$.

By the condition (K3), there exist $m_{2}\in\mathbb{R}$ and $C_{1}>0$ such that $|\phi^{\prime}(r)|\leq C_{1} r^{m_{2}-1}$ for $r\geq1$. Denote

$$C_{2}=\max_{\frac{1}{2}\leq|\xi|\leq2}\bigl\{ |\xi|^{m_{2}-1}\bigr\} \quad \text{and} \quad C_{3}=\max\{C_{1}C_{2},1\}. $$

Now we give the following estimates of $I_{N}(x,\omega)$ for $x\in \mathbb{R}$, $-1<\omega<1$, and $N\geq2$:

$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq \left \{ \textstyle\begin{array}{l@{\quad}l} C(N|x|)^{-2}N^{1-2s}, & |\omega|< \frac {N|x|}{2C_{3}N^{m_{2}}}, \\ C(N|x|)^{-\frac{1}{2}}N^{1-2s}, & |\omega|\ge\frac {N|x|}{2C_{3}N^{m_{2}}}. \end{array}\displaystyle \displaystyle \right . $$

(2.19)

Let $F(\xi)=Nx\xi+\omega\phi(N|\xi|)$. We have

$$\begin{aligned}& F^{\prime}(\xi)=Nx+N \operatorname{sgn}(\xi) \omega\phi^{\prime} \bigl(N\vert \xi \vert \bigr), \\& F^{\prime\prime}(\xi)=N^{2}\omega\phi^{\prime\prime}\bigl(N\vert \xi \vert \bigr) \end{aligned}$$

and

$$F^{(3)}(\xi)=N^{3}\operatorname{sgn}(\xi) \omega \phi^{(3)}\bigl(N\vert \xi \vert \bigr). $$

Noting $N|\xi|>1$ by $N\geq2$ and $\frac{1}{2}<|\xi|<2$, by (K3) we get

$$\bigl\vert N \operatorname{sgn}(\xi)\omega\phi^{\prime}\bigl(N\vert \xi \vert \bigr)\bigr\vert \leq C_{1} N\vert \omega \vert \bigl(N\vert \xi \vert \bigr)^{m_{2}-1}\leq C_{1}C_{2} N^{m_{2}}|\omega|\leq C_{3}N^{m_{2}}|\omega|. $$

When $|\omega|<\frac{N|x|}{2C_{3}N^{m_{2}}}$ (equivalently, $C_{3} N^{m_{2}}|\omega|< \frac{1}{2}N|x|$), we have

$$\bigl\vert N \operatorname{sgn}(\xi) \omega\phi^{\prime}\bigl(N\vert \xi \vert \bigr)\bigr\vert < \frac{1}{2}N|x|. $$

Therefore,

$$ \bigl\vert F^{\prime}(\xi)\bigr\vert \geq N|x|-\bigl\vert N \operatorname{sgn}(\xi) \omega\phi^{\prime }\bigl(N\vert \xi \vert \bigr)\bigr\vert > \frac{1}{2}N|x|. $$

(2.20)

Since ϕ satisfies (K4) and (K5) with $m_{4}\leq m_{3}=m_{2}$, we have

$$ \bigl\vert F^{(j)}(\xi)\bigr\vert \leq C N^{m_{2}}|\omega|\quad \text{for } j=2,3. $$

(2.21)

By the fact $\frac{N^{m_{2}}|\omega|}{ N|x|} \leq\frac{1}{2C_{3}}$ and Lemma 2.5 for $k=2$ and (2.20), (2.21), we get

$$\begin{aligned}& \biggl\vert \int e^{iF(\xi)}G(\xi)\,d\xi\biggr\vert \\& \quad \leq C\int_{\frac{1}{2}< |\xi|< 2}\frac{1}{|F^{\prime}(\xi)|^{2}} \biggl(1+ \frac{|F^{\prime\prime}(\xi)|}{|F^{\prime}(\xi)|}+ + \biggl(\frac{|F^{\prime\prime}(\xi)|}{ |F^{\prime}(\xi)|} \biggr)^{2}+ \frac{|F^{(3)}(\xi)|}{ |F^{\prime}(\xi)|} \biggr)\,d\xi \\& \quad \leq C\bigl(N\vert x\vert \bigr)^{-2}\sum _{r=0}^{2} \biggl(\frac{N^{m_{2}} |\omega|}{N|x|} \biggr)^{r} \\& \quad \leq C\bigl(N\vert x\vert \bigr)^{-2}, \end{aligned}$$

from which follows the first estimate in (2.19). On the other hand, since ϕ satisfies (K4) with $m_{3}=m_{2}$, we get, for $\frac{1}{2}<|\xi|<2$,

$$\bigl\vert F^{\prime\prime}(\xi)\bigr\vert \geq C N^{2}|\omega| \bigl(N\vert \xi \vert \bigr)^{m_{2}-2}>CN^{m_{2}}|\omega|>0 . $$

Note that $\|G\|_{\infty}\leq C$ and $\|G^{\prime}\|_{1}\leq C$ on the support of φ. By Lemma 2.4 and noting that $|\omega|\ge\frac{N|x|}{2C_{3}N^{m_{2}}}$ (equivalently, $C_{3} N^{m_{2}}|\omega|\ge\frac{1}{2}N|x|$), we have

$$\bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N^{m_{2}}|\omega|\bigr)^{-\frac{1}{2}} \bigl(\|G\|_{\infty}+ \bigl\Vert G^{\prime}\bigr\Vert _{1}\bigr)N^{1-2s} \leq C \bigl(N\vert x\vert \bigr)^{-\frac {1}{2}}N^{1-2s}. $$

This is just the second estimate in (2.19).

Step 2: Proof of Lemma 2.3 for $s>\frac{m_{2}}{4}$ ($m_{2}>0$).

We now prove (2.17) for the case $s>\frac{m_{2}}{4}$ ($m_{2}>0$). Since $m_{2}>0$, $N\geq2$, and $2C_{3}>1$, we write

$$\begin{aligned} \int\bigl\vert J_{N}(x)\bigr\vert \,dx =&\int _{0< |x|\leq\frac{1}{N}} \bigl\vert J_{N}(x)\bigr\vert \,dx + \int_{\frac{1}{N}< |x|\leq2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ &{} +\int_{|x|>2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ =:&E_{1}+E_{2}+E_{3}. \end{aligned}$$

The estimate of $E_{1}$ is simple. Since $|I_{N}(x,\omega)|\leq C N^{1-2s}$ by (2.18), by the definition of $J_{N}$, we see that

$$ E_{1}\leq C\int_{0< |x|\leq\frac {1}{N}} N^{1-2s}\,dx\leq C N^{-2s}. $$

(2.22)

As for $E_{2}$, we first prove that if $\frac{1}{N}<|x|\leq 2C_{3}N^{m_{2}-1}$, then

$$ J_{N}(x)\leq C \bigl(N\vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}. $$

(2.23)

By the definition of $J_{N}$, to prove (2.23) it suffices to show that, if $\frac{1}{N}<|x|\leq2C_{3}N^{m_{2}-1}$ and $|\omega|<1$, then

$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N\vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}. $$

(2.24)

In fact, if $|\omega|<\frac{N|x|}{2C_{3}N^{m_{2}}}$, by the first estimate in (2.19) and $N|x|>1$, then

$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N \vert x\vert \bigr)^{-2} N^{1-2s}\leq C \bigl(N\vert x \vert \bigr)^{-\frac {1}{2}} N^{1-2s}. $$

If $|\omega|\geq\frac{N|x|}{2C_{3}N^{m_{2}}}$, by the second estimate in (2.19), we obtain

$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N \vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}. $$

Thus (2.24) holds and so (2.23). Hence, by (2.23), we get

$$ E_{2}\leq C\int_{|x|\leq2C_{3}N^{m_{2}-1}} \bigl(N\vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}\,dx\leq C N^{\frac{m_{2}}{2}-2s}. $$

(2.25)

Finally, we consider $E_{3}$. We first show that if $|x|>2C_{3}N^{m_{2}-1}$, then

$$ \bigl\vert J_{N}(x)\bigr\vert \leq C\bigl(N\vert x\vert \bigr)^{-2}N^{1-2s}. $$

(2.26)

In fact, if $|x|>2C_{3}N^{m_{2}-1}$ and $|\omega|<1$, then $|x|>2C_{3}N^{m_{2}-1}|\omega|$. Equivalently, $|\omega|<\frac{N|x|}{2C_{3}N^{m_{2}}}$. Thus, by the first inequality in (2.19), we obtain

$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C\bigl(N \vert x\vert \bigr)^{-2}N^{1-2s}, $$

and (2.26) follows from this. By (2.26), we obtain

$$ E_{3}\leq C\int_{|x|>2C_{3}N^{m_{2}-1}} \bigl(N \vert x\vert \bigr)^{-2}N^{1-2s}\,dx \leq C N^{-m_{2}-2s}. $$

(2.27)

Since $m_{2}>0$, by (2.22), (2.25), and (2.27), we have

$$\bigl\vert J_{N}(x)\bigr\vert \leq C N^{\frac{m_{2}}{2}-2s}=:CN^{-2\delta}, $$

where $2\delta=2s-\frac{m_{2}}{2}>0$ since $s>\frac{m_{2}}{4}$ and $m_{2}>0$.

Step 3: Proof of Lemma 2.3 for $s>\frac{-m_{2}}{2}$ ($m_{2}\leq0$).

First we consider the case where $2C_{3}N^{m_{2}-1}>\frac{1}{N}$. Write

$$\begin{aligned} \int\bigl\vert J_{N}(x)\bigr\vert \,dx =&\int _{0< |x|\leq\frac{1}{N}} \bigl\vert J_{N}(x)\bigr\vert \,dx + \int_{\frac{1}{N}< |x|\leq2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ &{} +\int_{|x|>2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ =:&E_{1}+E_{2}+E_{3}. \end{aligned}$$

Since $m_{2}\leq0$, by (2.22), (2.25), and (2.27), we have

$$\bigl\vert J_{N}(x)\bigr\vert \leq C N^{-m_{2}-2s}=:CN^{-2\delta}, $$

where $2\delta=2s+m_{2}>0$ since $s>\frac{-m_{2}}{2}$ and $m_{2}\leq 0$. On the other hand, if $2C_{3}N^{m_{2}-1}\leq\frac{1}{N}$, we have

$$\begin{aligned} \int\bigl\vert J_{N}(x)\bigr\vert \,dx &\leq\int _{0< |x|\leq\frac{1}{N}} \bigl\vert J_{N}(x)\bigr\vert \,dx + \int_{|x|>2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ &=:E_{1}+E_{3}. \end{aligned}$$

Since $m_{2}\leq0$, by (2.22) and (2.27), we have

$$\bigl\vert J_{N}(x)\bigr\vert \leq C N^{-m_{2}-2s}=:CN^{-2\delta}, $$

where $2\delta=2s+m_{2}>0$ by $s>\frac{-m_{2}}{2}$ and $m_{2}\leq0$. Thus, we complete the proof of Lemma 2.3.

3 The proof of Theorem 1.2

Assume $n\geq2$. Let f be radial and belong to $\mathcal{S}(\mathbb {R}^{n})$; we need to show that

$$ \bigl\Vert S^{\ast}_{\phi}f\bigr\Vert _{L^{2}(\mathbb{R}^{n})}\leq C\|f\|_{H^{s}(\mathbb{R}^{n})} $$

(3.1)

holds for $s>\frac{m_{2}}{4}$ if $m_{2}>0$ or $s>\frac{-m_{2}}{2}$ if $m_{2}<0$.

Let $t(x)$ is a measurable radial function with $0< t(x)<1$. Denote

$$Tf(x)=(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it(x)\phi(|\xi|)} \hat{f}(\xi)\,d\xi. $$

Recall the Bessel function $J_{m}(r)$ is defined by

$$J_{m}(r)=\frac{(\frac{r}{2})^{m}}{\Gamma(m+\frac{1}{2}) \pi^{\frac{1}{2}}}\int_{-1}^{1}e^{irt} \bigl(1-t^{2}\bigr)^{m-\frac{1}{2}}\,dt, \quad m>-\frac{1}{2}. $$

Since f is radial,

$$\hat{f}(\xi)=(2\pi)^{\frac{n}{2}}|\xi|^{1-\frac{n}{2}}\int_{0}^{\infty} f(s)J_{\frac{n}{2}-1}\bigl(s\vert \xi \vert \bigr)s^{\frac{n}{2}}\, ds. $$

Therefore,

$$ Tf(u)=(2\pi)^{\frac{n}{2}-n}u^{1-\frac{n}{2}}\int _{0} ^{\infty}J_{\frac{n}{2}-1} (ru)e^{it(u)\phi(r)} \hat{f}(r)r^{\frac{n}{2}}\,dr, \quad u>0. $$

(3.2)

Here $Tf(u)=Tf(x)$ if $u=|x|$ and $\hat{f}(r)=\hat{f}(\xi)$ if $r=|\xi|$. By linearizing the maximal operator and using polar coordinates, to prove (3.1) it suffices to prove that

$$ \biggl(\int_{0}^{\infty}\bigl\vert Tf(u)\bigr\vert ^{2} u^{n-1}\, du \biggr)^{1/2}\leq \biggl(\int_{0}^{\infty}\bigl\vert \hat{f}(r)\bigr\vert ^{2} \bigl(1+r^{2}\bigr)^{s}r^{n-1} \,dr \biggr)^{1/2}. $$

(3.3)

Denote

$$ g(r)=\hat{f}(r) \bigl(1+r^{2}\bigr)^{\frac{s}{2}}r^{\frac{n-1}{2}}, \quad r>0. $$

(3.4)

By (3.2) and (3.4), it follows that

$$\begin{aligned} Tf(u)u^{\frac{n-1}{2}} &=(2\pi)^{-\frac{n}{2}} u^{\frac{1}{2}}\int _{0}^{\infty}J_{\frac{n}{2}-1} (ru)e^{it(u)\phi(r)} \hat{f}(r)r^{\frac{n}{2}}\,dr \\ &=(2\pi)^{-\frac{n}{2}} u^{\frac{1}{2}}\int_{0}^{\infty}J_{\frac{n}{2}-1} (ru)e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}r^{\frac {1}{2}} \,dr. \end{aligned}$$

Let

$$Pg(u)=u^{\frac{1}{2}}\int_{0}^{\infty} J_{\frac{n}{2}-1}(ru)e^{it(u)\phi(r)}g(r) \bigl(1+r^{2} \bigr)^{-\frac{s}{2}} r^{\frac{1}{2}}\,dr. $$

Thus, we have

$$ Tf(u)u^{\frac{n-1}{2}}=(2\pi)^{-\frac{n}{2}} Pg(u). $$

(3.5)

By (3.5), to prove (3.3) it suffices to prove that

$$ \biggl(\int_{0}^{\infty}\bigl\vert Pg(u)\bigr\vert ^{2}\, du \biggr)^{1/2}\leq C \biggl(\int _{0}^{\infty}\bigl\vert g(r)\bigr\vert ^{2}\,dr \biggr)^{1/2} $$

(3.6)

holds for $s>\frac{m_{2}}{4}$ ($m_{2}>0$) or $s>\frac{-m_{2}}{2}$ ($m_{2}\leq0$). Let us recall a well-known estimate of $J_{m}$.

Lemma 3.1

([19], p.158)

$J_{m}(r)=\sqrt{\frac{2}{\pi r}}\cos(r-\frac{\pi m}{2}-\frac{\pi}{4})+O(r^{-\frac{3}{2}})$ as $r\rightarrow\infty$. In particular, $J_{m}(r)=O(r^{-\frac{1}{2}})$ as $r\rightarrow\infty$.

By Lemma 3.1, we may get

$$ t^{\frac{1}{2}}J_{\frac{n}{2}-1}(t)= b_{1}e^{it}+b_{2}e^{-it}+ {O}\biggl(\min\biggl(1,\frac{1}{t}\biggr)\biggr), \quad t>0, $$

(3.7)

where $b_{1}$ and $b_{2}$ are the constants depending on n. In fact, by Lemma 3.1, as $t\rightarrow\infty$, we have

$$ J_{\frac{n}{2}-1}(t) =\sqrt{\frac{2}{\pi t}}\cos\biggl(t-\frac{\pi(n-1)}{4} \biggr)+ {O}\bigl(t^{-\frac{3}{2}}\bigr). $$

It follows that, as $t\rightarrow\infty$, we have

$$\begin{aligned} t^{\frac{1}{2}}J_{\frac{n}{2}-1}(t) &=\sqrt{\frac{2}{\pi}}\cos\biggl( \frac{\pi(n-1)}{4}\biggr)\cos t +\sqrt{\frac{2}{\pi}}\sin\biggl( \frac{\pi(n-1)}{4}\biggr)\sin t+ {O}\bigl(t^{-1}\bigr) \\ &=(b_{1}+b_{2})\cos t+i(b_{1}-b_{2}) \sin t+{O}\bigl(t^{-1}\bigr) \\ &=b_{1}e^{it}+b_{2}e^{-it}+{O} \bigl(t^{-1}\bigr), \end{aligned}$$

where

$$b_{1}=\frac{1}{2}\sqrt{\frac{2}{\pi}} \biggl(\cos\biggl( \frac{\pi (n-1)}{4}\biggr)+i\sin\biggl(\frac{\pi(n-1)}{4}\biggr) \biggr) $$

and

$$b_{2}=\frac{1}{2}\sqrt{\frac{2}{\pi}} \biggl(\cos\biggl( \frac{\pi (n-1)}{4}\biggr)-i\sin\biggl(\frac{\pi(n-1)}{4}\biggr) \biggr). $$

It follows that, when $t>1$, we have

$$ \bigl\vert t^{\frac{1}{2}}J_{\frac {n}{2}-1}(t)- \bigl(b_{1}e^{it}+b_{2}e^{-it}\bigr)\bigr\vert \leq C t^{-1}. $$

(3.8)

On the other hand, by the definition of the Bessel function

$$J_{m}(t)=\frac{(\frac{t}{2})^{m}}{\Gamma(m+\frac{1}{2}) \pi^{\frac{1}{2}}}\int_{-1}^{1}e^{its} \bigl(1-s^{2}\bigr)^{m-\frac{1}{2}}\, ds, \quad m>-\frac{1}{2}, $$

we have $|J_{m}(t)|\leq C t^{m}$ for $m>-\frac{1}{2}$ and $t>0$. Thus, $|J_{m}(t)|\leq C t^{-\frac{1}{2}}$ when $m>-\frac{1}{2}$ and $0< t<1$. Since $n\geq2$, so $|J_{\frac{n}{2}-1}(t)|\leq C t^{-\frac{1}{2}}$ for $0< t<1$. Therefore, when $0< t<1$, we have

$$\begin{aligned} \bigl\vert t^{\frac{1}{2}}J_{\frac {n}{2}-1}(t)- \bigl(b_{1}e^{it}+b_{2}e^{-it}\bigr)\bigr\vert \leq& \bigl\vert t^{\frac{1}{2}}J_{\frac{n}{2}-1}(t)\bigr\vert + \bigl\vert b_{1}e^{it}\bigr\vert +\bigl\vert b_{2}e^{-it}\bigr\vert \\ \leq& Ct^{\frac{1}{2}}t^{-\frac{1}{2}}+|b_{1}|+|b_{2}| \leq C. \end{aligned}$$

(3.9)

It follows from (3.8) and (3.9) that (3.7) holds. Invoking (3.7), we have

$$\begin{aligned} \begin{aligned}[b] Pg(u)={}& b_{1}\int_{0}^{\infty}e^{iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr \\ &{}+ b_{2}\int_{0}^{\infty}e^{-iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr+E(u)+F(u) \\ =:{}&b_{1}D_{1}(u)+b_{2}D_{2}(u)+E(u)+F(u), \end{aligned} \end{aligned}$$

(3.10)

where

$$\bigl\vert E(u)\bigr\vert \leq C\int_{0}^{\frac{1}{u}} \bigl\vert g(r)\bigr\vert \,dr $$

and

$$\bigl\vert F(u)\bigr\vert \leq C\frac{1}{u}\int_{\frac{1}{u}}^{\infty} \frac{1}{r}\bigl\vert g(r)\bigr\vert \,dr. $$

From [17], pp.59-61, we have

$$ \biggl(\int_{0}^{\infty}\bigl\vert E(u)\bigr\vert ^{2}\, du \biggr)^{1/2}\leq C\|g \|_{L^{2}(0,\infty)} $$

(3.11)

and

$$ \biggl(\int_{0}^{\infty}\bigl\vert F(u)\bigr\vert ^{2}\, du \biggr)^{1/2}\leq C\|g \|_{L^{2}(0,\infty)}. $$

(3.12)

Thus, to prove (3.6), it remains to estimate $D_{1}$ and $D_{2}$. Denote $\hat{h}(r)=g(r) (1+r^{2})^{-\frac{s}{2}}\chi_{(0,\infty)}$, and we get

$$D_{1}(u)=\int_{0}^{\infty}e^{iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr=\int_{\mathbb{R}} e^{iru}e^{it(u)\phi(r)}\hat{h}(r)\,dr $$

and

$$D_{2}(u)=\int_{0}^{\infty}e^{-iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr=\int_{\mathbb{R}} e^{-iru}e^{it(u)\phi(r)}\hat{h}(r)\,dr. $$

Therefore, we have

$$ \bigl\vert D_{i}(u)\bigr\vert \leq S_{\phi}^{\ast} h(u)\quad \text{for } i=1,2. $$

(3.13)

Since ϕ satisfies the conditions in Theorem 1.1, by the results of Theorem 1.1, when $s>\frac{m_{2}}{4}$ ($m_{2}>0$) or $s>\frac{-m_{2}}{2}$ ($m_{2}\leq0$), we have

$$ \bigl\Vert S_{\phi}^{\ast} h\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|h\|_{H^{s}(\mathbb{R})}. $$

(3.14)

Since $u>0$ and by (3.13) and (3.14), for $i=1,2$, we have

$$\begin{aligned} \|D_{i}\|_{L^{2}(0,\infty)} \leq& \|D_{i} \|_{L^{2}(\mathbb{R})} \leq C\bigl\Vert S_{\phi}^{\ast} h\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|h\| _{H^{s}(\mathbb{R})} \\ =& C \biggl(\int_{0}^{\infty}\bigl\vert g(r)\bigr\vert ^{2}\bigl(1+r^{2}\bigr)^{-s} \bigl(1+r^{2}\bigr)^{s}\,dr \biggr)^{1/2} \\ =& C\|g\|_{L^{2}(0,\infty)}. \end{aligned}$$

(3.15)

Thus, (3.6) follows from (3.10), (3.11), (3.12), and (3.15). We hence complete the proof of Theorem 1.2.

4 The proof of Theorem 1.3

In this case $k=0$, Theorem 1.3 follows from Theorem 1.2. Hence we only give the proof of Theorem 1.3 for $k\geq1$. We first recall a well-known result.

Lemma 4.1

([19], p.158)

Suppose $n\geq2$ and $f\in L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb {R}^{n})$ has the form $f(x)=f_{0}(|x|)P(x)$, where $P(x)$ is a solid spherical harmonic of degree k, then $\hat{f}$ has the form $\hat{f}(x)=F_{0}(|x|)P(x)$, where

$$F_{0}(r)= (2\pi)^{\frac{n}{2}}i^{-k}r^{-\frac{n}{2}-k+1} \int_{0}^{\infty}f_{0}(s)J_{\frac{n}{2}+k-1} (rs)s^{\frac{n}{2}+k}\,ds, $$

where $J_{m}$ denotes the Bessel function.

Let us return to the proof of Theorem 1.3. First we show that, for $f\in\mathcal{H}_{k}$ ($k\geq1$),

$$ \|f\|_{H^{s}(\mathbb{R}^{n})}= \Biggl(\sum_{j=1}^{a_{k}} \int_{0}^{\infty}\bigl\vert F_{j}(r) \bigr\vert ^{2}\bigl(1+r^{2}\bigr)^{s}r^{n+2k-1} \,dr \Biggr)^{1/2}. $$

(4.1)

In fact, $f(x)=\sum_{j=1}^{a_{k}}f_{j}(|x|)P_{j}(x)$ where $f_{j}$ are radial functions in $\mathcal{S}(\mathbb{R}^{n})$ and $\{P_{j}\} _{1}^{a_{k}}$ is an orthonormal basis in ${\mathscr{A}}_{k}$. By Lemma 4.1 we get

$$ \hat{f}(x)=\sum_{j=1}^{a_{k}}F_{j} \bigl(\vert x\vert \bigr)P_{j}(x), $$

(4.2)

where

$$F_{j}(r)=(2\pi)^{\frac{n}{2}}i^{-k}r^{1-\frac{n}{2}-k}\int _{0}^{\infty}f_{j}(s) J_{\frac{n}{2}+k-1}(rs)s^{\frac{n}{2}+k} \,ds, \quad r>0. $$

By (4.2) and noting that $\{P_{1},P_{1},\ldots ,P_{a_{k}}\}$ is an orthonormal basis in ${\mathscr{A}}_{k}$, we have

$$\begin{aligned}& \int_{\mathbb{R}^{n}}\bigl(1+\vert \xi \vert ^{2} \bigr)^{s}\bigl\vert \hat{f}(\xi)\bigr\vert ^{2}\,d\xi \\& \quad = \int_{0}^{\infty} \biggl(\int _{S^{n-1}}\bigl\vert \hat{f}\bigl(r\xi^{\prime}\bigr) \bigr\vert ^{2}\,d\sigma\bigl(\xi^{\prime}\bigr) \biggr) \bigl(1+r^{2}\bigr)^{s}r^{n-1}\,dr \\& \quad = \int_{0}^{\infty} \Biggl(\int _{S^{n-1}} \Biggl(\sum_{j=1}^{a_{k}}F_{j}(r)P_{j} \bigl(r\xi^{\prime}\bigr) \Biggr) \Biggl(\sum_{i=1}^{a_{k}} \overline{F_{i}(r)} \overline {P_{i}\bigl(r \xi^{\prime}\bigr)} \Biggr)\,d\sigma\bigl(\xi^{\prime}\bigr) \Biggr) \bigl(1+r^{2}\bigr)^{s}r^{n-1}\,dr \\& \quad =\int_{0}^{\infty} \Biggl(\sum _{j=1}^{a_{k}} \bigl\vert F_{j}(r)\bigr\vert ^{2} \Biggr)r^{2k}\bigl(1+r^{2} \bigr)^{s}r^{n-1}\,dr \\& \quad =\sum_{j=1}^{a_{k}}\int _{0}^{\infty} \bigl\vert F_{j}(r)\bigr\vert ^{2}\bigl(1+r^{2}\bigr)^{s}r^{n+2k-1} \,dr, \end{aligned}$$

which is just (4.1). On the other hand, by (4.2), we have

$$ S_{t,\phi}f(x) =(2\pi)^{-n}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi} e^{it\phi(|\xi|)}\hat{f}(\xi)\,d\xi =\sum_{j=1}^{a_{k}}(2\pi)^{-n}\int _{\mathbb{R}^{n}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j} \bigl(\vert \xi \vert \bigr)P_{j}(\xi) \bigr)\,d\xi. $$

Applying Lemma 4.1, we get

$$\begin{aligned}& \int_{\mathbb{R}^{n}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j} \bigl(\vert \xi \vert \bigr)P_{j}(\xi) \bigr)\,d\xi \\& \quad = \bigl(e^{it\phi(|\cdot|)}F_{j}\bigl(\vert \cdot \vert \bigr)P_{j}(-\cdot) \bigr)^{\wedge}(x) \\& \quad =(2\pi)^{\frac{n}{2}}i^{-k}s^{1-\frac{n}{2}-k} \biggl(\int _{0}^{\infty}J_{\frac{n}{2}+k-1}(rs)e^{it\phi(r)} F_{j}(r)r^{\frac{n}{2}+k}\,dr \biggr)P_{j}(-x), \end{aligned}$$

where $s=|x|>0$. Therefore, we have

$$\begin{aligned} S_{t,\phi}f(x) =&\sum_{j=1}^{a_{k}}(2 \pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j}\bigl(\vert \xi \vert \bigr)P_{j}(\xi) \bigr)\,d\xi \\ =&\sum_{j=1}^{a_{k}}(2\pi)^{-\frac{n}{2}}i^{-k}|x|^{1-\frac {n}{2}-k} \\ &{} \times \biggl(\int_{0}^{\infty}J_{\frac {n}{2}+k-1} \bigl(r\vert x\vert \bigr)e^{it\phi(r)} F_{j}(r)r^{\frac{n}{2}+k} \,dr \biggr)P_{j}(-x). \end{aligned}$$

(4.3)

Denote by $\mathscr{F}_{n}$ the Fourier transform in $\mathbb{R}^{n}$. Then ${F}_{j}=i^{-k}\mathscr{F}_{n+2k}f_{j}$. Note that for a radial function $h\in\mathcal{S}(\mathbb{R}^{n+2k})$, its Fourier transform is

$$\mathscr{F}_{n+2k}h(x)=(2\pi)^{\frac{n}{2}}|x|^{1-\frac {n}{2}-k}\int _{0}^{\infty}h(r) J_{\frac{n}{2}+k-1}\bigl(r\vert x \vert \bigr)r^{\frac{n}{2}+k}\,dr. $$

Now we define the operator $S_{t,\phi}^{n+2k}$ on the set of all radial function in $\mathcal{S}(\mathbb{R}^{n+2k})$ by

$$S_{t,\phi}^{n+2k}h(x) :=(2\pi)^{-n-2k}\int _{\mathbb{R}^{n+2k}} e^{ix\cdot\xi}e^{it\phi(|\xi|)} \mathscr{F}_{n+2k}h \bigl(\vert \xi \vert \bigr)\,d\xi. $$

Obviously, $S_{t,\phi}^{n+2k}h$ is still a radial function. Then

$$\begin{aligned} S_{t,\phi}^{n+2k}f_{j}\bigl(|x|\bigr) =&i^{k}(2\pi)^{-n-2k}\int_{\mathbb{R}^{n+2k}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j}(\xi) \bigr)\,d\xi \\ =&i^{k}(2\pi)^{-\frac{n}{2}-2k}|x|^{1-\frac{n}{2}-k} \\ &{}\times\int_{0}^{\infty}J_{\frac{n}{2}+k-1}\bigl(r \vert x\vert \bigr)e^{it\phi(r)} F_{j}(r)r^{\frac{n}{2}+k} \,dr. \end{aligned}$$

(4.4)

By (4.3) and (4.4), we have

$$ S_{t,\phi}f(x)= i^{-2k}(2\pi)^{2k}\sum _{j}S_{t,\phi }^{n+2k}f_{j} \bigl(\vert x\vert \bigr)\cdot P_{j}(-x), \quad x\in \mathbb{R}^{n}, $$

(4.5)

where we may see $S_{t,\phi}^{n+2k}f_{j}(|x|)$ as a function on $\mathbb{R}^{n}$, since $S_{t,\phi}^{n+2k}f_{j}$ is a radial function. Denote

$$ S_{\phi}^{n+2k,\ast}f_{j}\bigl(\vert y \vert \bigr)= \sup_{0< t< 1}\bigl\vert S_{t,\phi}^{n+2k}f_{j} \bigl(\vert y\vert \bigr)\bigr\vert ,\quad y\in\mathbb{R}^{n+2k} \text{ or } y\in\mathbb{R}^{n}. $$

(4.6)

Then by (4.5) and (4.6), we obtain

$$ S_{\phi}^{\ast}f(x)\leq C_{n,k}\sum _{j}\bigl(S_{\phi}^{n+2k,\ast }f_{j} \bigl(\vert x\vert \bigr)\bigr)|x|^{k}. $$

(4.7)

Using the notation $v=|x|$ and $r=|\xi|$, by (4.7), we have

$$ \bigl\Vert S_{\phi}^{\ast}f\bigr\Vert _{L^{2}(\mathbb{R}^{n})}^{2}\leq C\sum_{j=1}^{a_{k}} \int_{\mathbb{R}^{n}} \bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2}v^{2k}\,dx. $$

(4.8)

Using the representation of polar coordinates and noting (4.6), we obtain

$$\begin{aligned}& \int_{\mathbb{R}^{n}}\bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2}v^{2k}\,dx \\& \quad = \omega_{n-1} \int _{0}^{\infty}\bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2} v^{n+2k-1}\, dv \\& \quad = \frac{\omega_{n-1}}{\omega_{n+2k-1}}\int_{\mathbb{R}^{n+2k}} \bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v)\bigr\vert ^{2} \,dx, \end{aligned}$$

(4.9)

where $\omega_{n-1}$ and $\omega_{n+2k-1}$ denote the area of the unit sphere in $\mathbb{R}^{n}$ and $\mathbb{R}^{n+2k}$, respectively. Applying Theorem 1.2, when $s>\frac{m_{2}}{4}$ ($m_{2}>0$) or $s>\frac{-m_{2}}{2}$ ($m_{2}\leq 0$), we have

$$ \int_{\mathbb{R}^{n+2k}}\bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2}\,dx\leq C\|f_{j}\|_{H^{s}(\mathbb{R}^{n+2k})}^{2}. $$

(4.10)

Note that $\mathscr{F}_{n+2k}f_{j}=i^{k}F_{j}$, and we get

$$\begin{aligned} \|f_{j}\|_{H^{s}(\mathbb{R}^{n+2k})}^{2} =& \int _{\mathbb{R}^{n+2k}}\bigl\vert F_{j}\bigl(\vert \xi \vert \bigr)\bigr\vert ^{2}\bigl(1+|\xi|^{2}\bigr)^{s} \,d\xi \\ =&\omega_{n+2k-1}\int_{0}^{\infty} \bigl\vert F_{j}(r)\bigr\vert ^{2}\bigl(1+r^{2} \bigr)^{s}r^{n+2k-1}\,dr. \end{aligned}$$

(4.11)

Therefore, by (4.8), (4.9), (4.10), (4.11), and (4.1), we obtain

$$ \bigl\Vert S_{\phi}^{\ast}f\bigr\Vert _{L^{2}(\mathbb{R}^{n})}^{2}\leq C\sum_{j=1}^{a_{k}} \int_{0}^{\infty}\bigl\vert F_{j}(r) \bigr\vert ^{2} \bigl(1+r^{2}\bigr)^{s}r^{n+2k-1} \,dr=C\|f\|_{H^{s}(\mathbb{R}^{n})}^{2}. $$

(4.12)

Thus, we complete the proof of Theorem 1.3.

5 Some applications

We now give some examples to show that (1.10) includes some well-known equations.

Example 1

Let $\phi(r)=r^{2}$, then (1.10) is the classical Schrödinger equation (1.1).

Example 2

Let $\phi(r)=r^{a}$ ($a>0$, $a\neq1$), then (1.10) is the fractional Schrödinger equation (1.3). In this case, $\phi(r)$ satisfies (K1)-(K5) with $l_{1}=m_{1}=m_{2}=m_{3}=m_{4}=a$.

Example 3

Let $\phi(r)=r^{2}+r^{4}$, then (1.10) is the fourth-order Schrödinger equation:

$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u+\Delta^{2}u-\Delta u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$

(5.1)

In this case, $\phi(r)$ satisfies (K1) with $l_{1}=2\geq0$, (K2)-(K5) with $m_{1}=m_{2}=m_{3}=m_{4}=4>0$.

Example 4

Recall the definition of the beam equation:

$$ \left \{ \textstyle\begin{array}{l} \partial_{tt}u+\Delta^{2}u+u=0, \quad (x,t)\in\mathbb{R}^{n}\times \mathbb{R}, \\ u(x,0)=f(x), \\ \partial_{t}u(x,0)=0. \end{array}\displaystyle \right . $$

(5.2)

Note that the solution of (5.2) can be formally written as the real part of

$$u(x,t)=e^{it\sqrt{I+\Delta^{2}}}f(x) =(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it\sqrt{1+|\xi|^{4}}}\hat{f}(\xi)\,d\xi. $$

Thus, taking $\phi(r)=\sqrt{1+r^{4}}$, we see that $\phi(r)$ satisfies (K1) with $l_{1}=0\geq0$, (K2)-(K5), with $m_{1}=m_{2}=m_{3}=m_{4}=2>0$, and the solution of (5.2) is the real part of

$$S_{t,\phi}f(x)=(2\pi)^{-n} \int_{\mathbb{R}^{n}}e^{ix\cdot\xi+it\phi(|\xi|)} \hat{f}(\xi )\,d\xi. $$