1 Introduction and main results
Suppose \(f\in\mathcal{S}(\mathbb{R}^{n})\), the Schwartz class on \(\mathbb{R}^{n}\), denote
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
In 1979, Carleson [1] proposed a problem: if \(f\in H^{s}(\mathbb{R}^{n})\) for which s does
where \(H^{s}(\mathbb{R}^{n})\) (\(s\in\mathbb{R}\)) denotes the non-homogeneous Sobolev space, which is defined by
$$ S_{t}f(x)=(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it|\xi|^{2}}\hat{f}(\xi)\, d\xi,\quad (x,t)\in\mathbb{R}^{n}\times \mathbb{R}, $$
$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u-\Delta u=0, \quad (x,t)\in\mathbb{R}^{n}\times\mathbb {R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$
(1.1)
$$ \lim_{t\rightarrow0}u(x,t)=f(x),\quad \text{a.e. } x\in \mathbb{R}^{n}, $$
(1.2)
$$H^{s}\bigl(\mathbb{R}^{n}\bigr)= \biggl\{ f\in \mathcal{S}^{\prime}: \|f\|_{H^{s}}= \biggl(\int _{\mathbb{R}^{n}} \bigl(1+|\xi|^{2}\bigr)^{s}\bigl\vert \hat{f}(\xi)\bigr\vert ^{2}\,d\xi \biggr)^{1/2}< \infty \biggr\} . $$
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
Dim.
|
Range of
s
|
Authors
|
|---|---|---|
n = 1 |
\(s\geq\frac{1}{4}\)
| Carleson [1] in 1979 |
n ≥ 2 |
\(s>\frac{1}{2}\)
| |
n = 2 | for some \(s<\frac{1}{2}\)
| Bourgain [5] in 1992 |
n = 2 |
s>κ with \(\frac{20}{41}<\kappa<\frac{41}{84}\)
| Moyua, Vargas and Vega [6] in 1996 |
n = 2 |
\(s> \frac{15}{32}\)
| Tao and Vargas [7] in 2000 |
n = 2 |
\(s> \frac{2}{5}\)
| Tao [8] in 2003 |
n = 2 |
\(s>\frac{3}{8}\)
| Lee [9] in 2006 |
n ≥ 3 |
\(s>\frac{1}{2}-\frac{1}{4n}\)
| Bourgain [10] in 2013 |
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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
Naturally, the maximal estimates have been well studied associated with the following oscillatory integral:
which is the solution of the fractional Schrödinger equation:
Define the local maximal operator associated with the family of operators \(\{S_{t,a}\}_{0< t<1}\) by
Obviously, the following estimate (1.4) can be applied to discuss the pointwise convergence problem on the solution of Schrödinger equation (1.3):
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).
$$S^{\ast}f(x)= \sup_{0< t< 1}\bigl\vert S_{t}f(x)\bigr\vert ,\quad x\in\mathbb{R}^{n}. $$
$$S_{t,a}f(x)=\frac{1}{(2\pi)^{n}} \int_{\mathbb{R}^{n}}e^{ix\cdot\xi} e^{it|\xi|^{a}}\hat{f}(\xi)\,d\xi, \quad t\in\mathbb{R} \text{ and } a>0, $$
$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u+(-\Delta)^{a/2} u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$
(1.3)
$$S^{\ast}_{a}f(x)= \sup_{0< t< 1}\bigl\vert S_{t,a}f(x)\bigr\vert , \quad x\in \mathbb{R}^{n}. $$
$$ \bigl\Vert S^{\ast}_{a}f\bigr\Vert _{L^{2} (\mathbb{R}^{n})}\leq C\|f\|_{H^{s}(\mathbb{R}^{n})}, $$
(1.4)
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
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 \(s>\frac{a}{4}\).
$$ \bigl\Vert S^{*}f\bigr\Vert _{L^{1}(B)}\le c_{n} \|f\|_{H^{s}} $$
(1.5)
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: The operator family \(\{S_{t,\phi}\}_{t\in\mathbb{R}}\) is defined by
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
Now we state our main results in this paper as follows.
(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 \(r\geq1\).
$$ S_{t,\phi}f(x)=(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it\phi(|\xi|)}\hat{f}(\xi)\,d\xi, \quad x\in\mathbb{R}^{n}, $$
(1.6)
$$S^{\ast}_{\phi}f(x)= \sup_{0< t< 1} \bigl\vert S_{t,\phi}f(x)\bigr\vert ,\quad x\in\mathbb{R}^{n}. $$
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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
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 \(f(|\cdot|)P(\cdot)\in L^{2}(\mathbb{R}^{n})\).
$$L^{2}\bigl(\mathbb{R}^{n}\bigr)=\sum _{k=0}^{\infty}\oplus\mathfrak{D}_{k}. $$
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:
and
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).
$$ f(x)=\sum_{j=1}^{a_{k}}f_{j} \bigl(\vert x\vert \bigr)P_{j}(x) $$
(1.9)
$$\int_{\mathbb{R}^{n}}\bigl\vert f(x)\bigr\vert ^{2}\,dx= \sum_{j=1}^{a_{k}} \int_{0}^{\infty} \bigl\vert f_{j}(r)\bigr\vert ^{2}r^{n+2k-1}\,dr. $$
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
gives a formal solution of the following general dispersive equation with initial data function f:
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.
$$u(x,t)=e^{it\phi(\sqrt{-\Delta})}f(x)=(2\pi)^{-n} \int_{\mathbb{R}^{n}}e^{ix\cdot\xi+it\phi(|\xi|)} \hat{f}(\xi )\,d\xi=S_{t,\phi}f(x) $$
$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u+\phi(\sqrt{-\Delta})u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$
(1.10)
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
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
Denote
and
Therefore, by (2.1), we obtain
By (2.2) and Minkowski’s inequality, we get
Now let us recall a result which will be used in our proof of Theorem 1.1.
$$\sum_{k=-\infty}^{\infty} \varphi \bigl(2^{-k}\xi\bigr)=1,\quad \xi\neq0. $$
$$\begin{aligned} S_{t,\phi}f(x) =&(2\pi)^{-1} \int _{\mathbb{R}}e^{ix\cdot\xi+it\phi(|\xi|)} \varphi_{0}(\xi)\hat{f}( \xi)\,d\xi \\ &{} +(2\pi)^{-1}\sum_{k=1}^{\infty} \int_{\mathbb{R}}e^{ix\cdot\xi+it\phi(|\xi|)} \varphi\bigl(2^{-k}\xi \bigr)\hat{f}(\xi)\,d\xi \\ =:&S_{t,\phi,0}f(x)+\sum_{k=1}^{\infty}S_{t,\phi,k}f(x). \end{aligned}$$
(2.1)
$$S^{\ast}_{\phi,0}f(x)= \sup_{0< t< 1} \bigl\vert S_{t,\phi,0}f(x)\bigr\vert ,\quad x\in\mathbb{R} $$
$$S^{\ast}_{\phi,k}f(x)= \sup_{0< t< 1} \bigl\vert S_{t,\phi,k}f(x)\bigr\vert ,\quad x\in\mathbb{R}. $$
$$ S^{\ast}_{\phi}f(x)\leq S^{\ast}_{\phi,0}f(x)+ \sum_{k=1}^{\infty }S^{\ast}_{\phi,k}f(x). $$
(2.2)
$$ \bigl\Vert S^{\ast}_{\phi}f\bigr\Vert _{L^{2}(\mathbb{R})}\leq\bigl\Vert S^{\ast}_{\phi,0}f\bigr\Vert _{L^{2}(\mathbb{R})}+ \sum_{k=1}^{\infty}\bigl\Vert S^{\ast}_{\phi,k}f\bigr\Vert _{L^{2}(\mathbb{R})}. $$
(2.3)
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
Then there is a number
C
independent of
f
such that
$$\bigl\vert m(t,\xi)\bigr\vert \leq C\omega_{1}(t), \qquad \biggl\vert \frac{\partial (m(t,\xi))}{\partial t}\biggr\vert \leq C\bigl(\omega_{1}(t)+ \omega_{2}(t)|\xi |^{a}\bigr), \quad a>0. $$
$$\biggl(\int_{\mathbb{R}^{n}} \sup_{0< t< 1}\biggl\vert \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}m(t,\xi)\hat{f}(\xi)\,d\xi\biggr\vert ^{2}\,dx \biggr)^{1/2}\leq C\|f\|_{L^{2}(\mathbb{R}^{n})},\quad \operatorname{supp}\hat{f}\subseteq\bigl\{ \xi,|\xi|< 2\bigr\} . $$
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
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
where \(m(t,\xi)=(2\pi)^{-1}e^{it\phi(|\xi|)} (1+|\xi|^{2})^{-s/2}\). Define the maximal operator \(R^{\ast}_{0}\) by
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
it follows that
By the condition (K1), \(|\phi(|\xi|)|\leq C\max\{|\phi(1)|,1\}\leq C\) for \(0\leq|\xi|<1\). By (K2), we have, for \(|\xi|\geq1\),
Hence, combining with (2.5) we get, for \(\xi\in\mathbb{R}\),
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
We have
where \(\mathcal{F}^{-1}\) denotes the Fourier inverse transform. Note that
Thus, by (2.7) and (2.6), we have
which is just (2.4). Now we define the operator \(R_{N}\) by
where \(t(x)\) is a measurable function in \(\mathbb{R}\) with \(0< t(x)<1\).
$$ \bigl\Vert S^{\ast}_{\phi,0}f\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|f\|_{H^{s}(\mathbb{R})}. $$
(2.4)
$$R_{0,t}g(x)=(2\pi)^{-1} \int_{\mathbb{R}}e^{ix\cdot\xi}e^{it\phi(|\xi|)} \bigl(1+|\xi|^{2}\bigr)^{-s/2}\hat{g}(\xi)\,d\xi =:\int _{\mathbb{R}}e^{ix\cdot\xi}m(t,\xi)\hat{g}(\xi)\,d\xi, $$
$$R^{\ast}_{0}g(x)= \sup_{0< t< 1}\bigl\vert R_{0,t}g(x)\bigr\vert , \quad x\in\mathbb{R}. $$
$$\frac{\partial(m(t,\xi))}{\partial t}=\frac{i}{2\pi}e^{it\phi (|\xi|)}\phi\bigl(\vert \xi \vert \bigr) \bigl(1+|\xi|^{2}\bigr)^{-s/2}, $$
$$ \biggl\vert \frac{\partial(m(t,\xi))}{\partial t}\biggr\vert \leq \chi_{(0,1)}(t)\bigl\vert \phi\bigl(\vert \xi \vert \bigr)\bigr\vert \quad \text{for } \xi\in\mathbb{R} \text{ and } 0< t< 1. $$
(2.5)
$$ \bigl\vert \phi\bigl(\vert \xi \vert \bigr)\bigr\vert \leq \left \{ \textstyle\begin{array}{l@{\quad}l} C|\xi|^{m_{1}}, & m_{1}>0, \\ C|\xi|^{m_{1}}\leq C, & m_{1}\leq0. \end{array}\displaystyle \right . $$
$$ \biggl\vert \frac{\partial(m(t,\xi))}{\partial t} \biggr\vert \leq \left \{ \textstyle\begin{array}{l@{\quad}l} C(\chi_{(0,1)}(t)+\chi_{(0,1)}(t)|\xi|^{m_{1}}), & m_{1}>0, \\ C(\chi_{(0,1)}(t)+\chi_{(0,1)}(t)|\xi|), & m_{1}\leq0, \end{array}\displaystyle \right . $$
$$ \bigl\Vert R^{\ast}_{0}g\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|g\|_{L^{2}(\mathbb{R})}. $$
(2.6)
$$ S_{t,\phi,0}f(x)=R_{0,t} \bigl( \mathcal{F}^{-1} \bigl(\varphi_{0}(\cdot) \bigl(1+| \cdot|^{2}\bigr) ^{\frac{s}{2}}\hat{f}(\cdot) \bigr) \bigr) (x), $$
(2.7)
$$\operatorname{supp}\varphi_{0}(\cdot) \bigl(1+|\cdot|^{2} \bigr) ^{\frac{s}{2}}\hat{f}(\cdot)\subseteq\bigl\{ \xi; \vert \xi \vert < 2\bigr\} . $$
$$\begin{aligned} \bigl\Vert S^{\ast}_{\phi,0}f\bigr\Vert _{L^{2}(\mathbb{R})} &= \bigl\Vert R^{\ast}_{0} \bigl(\mathcal{F}^{-1} \bigl(\varphi_{0}(\cdot) \bigl(1+|\cdot|^{2}\bigr) ^{\frac{s}{2}}\hat{f}(\cdot) \bigr) \bigr)\bigr\Vert _{L^{2}(\mathbb{R})} \\ &\leq C \bigl\Vert \mathcal{F}^{-1} \bigl(\varphi_{0}( \cdot) \bigl(1+|\cdot |^{2}\bigr)^{\frac{s}{2}} \hat{f}(\cdot) \bigr) \bigr\Vert _{L^{2}(\mathbb{R})}\leq C\Vert f\Vert _{H^{s}(\mathbb{R})}, \end{aligned}$$
$$ R_{N}f(x)=N^{-s}\int_{\mathbb{R}}e^{ix\cdot\xi+it(x) \phi(|\xi|)} \varphi\biggl(\frac{\xi}{N}\biggr)\hat{f}(\xi)\,d\xi,\quad N\geq2, $$
(2.8)
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 \(t(x)\),
By (2.9) and (2.10), for \(k\geq1\), we have
From this we get
Summing up the estimates of (2.3), (2.4), and (2.11), we have
Therefore, to finish the proof of Theorem 1.1, it remains to show Lemma 2.2.
$$\begin{aligned} S^{\ast}_{\phi,k}f(x) \leq& \frac{1}{2\pi} \biggl\vert \int_{\mathbb{R}}e^{ix\cdot\xi+it(x) \phi(|\xi|)}\varphi\biggl( \frac{\xi}{2^{k}}\biggr)\hat{f}(\xi)\,d\xi \biggr\vert \\ =&\frac{1}{2\pi}\bigl\vert R_{2^{k}} \bigl(\mathcal{F}^{-1} \bigl(\chi_{\{2^{k-1}< |\xi|< 2^{k+1}\}} 2^{ks}\hat{f}\bigr) \bigr) (x)\bigr\vert . \end{aligned}$$
(2.10)
$$\begin{aligned} \bigl\Vert S^{\ast}_{\phi,k}f\bigr\Vert _{L^{2}(\mathbb{R})} &\leq \bigl\Vert R_{2^{k}} \bigl(\mathcal{F}^{-1}\bigl( \chi_{\{2^{k-1}< |\xi |< 2^{k+1}\}} 2^{ks}\hat{f}\bigr) \bigr)\bigr\Vert _{L^{2}(\mathbb{R})} \\ &\leq C2^{-k\delta} \bigl\Vert \mathcal{F}^{-1}\bigl( \chi_{\{2^{k-1}< |\xi |< 2^{k+1}\}}2^{ks}\hat{f}\bigr) \bigr\Vert _{L^{2}(\mathbb{R})}. \end{aligned}$$
$$ \bigl\Vert S^{\ast}_{\phi,k}f\bigr\Vert _{L^{2}(\mathbb{R})} \leq C2^{-k\delta} \|f\|_{H^{s}(\mathbb{R})}. $$
(2.11)
$$\begin{aligned} \bigl\Vert S^{\ast}_{\phi}f\bigr\Vert _{L^{2}(\mathbb{R})}&\leq \bigl\Vert S^{\ast}_{\phi ,0}f\bigr\Vert _{L^{2}(\mathbb{R})}+ \sum _{k=1}^{\infty}\bigl\Vert S^{\ast}_{\phi,k}f \bigr\Vert _{L^{2}(\mathbb{R})} \\ &\leq C\|f\|_{H^{s}(\mathbb{R})}+C\sum_{k=1}^{\infty}2^{-k\delta} \| f\|_{H^{s}(\mathbb{R})} \\ &\leq C\|f\|_{H^{s}(\mathbb{R})}. \end{aligned}$$
2.2 The proof of Lemma 2.2
Write
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
and
For \(N\geq2\), \(M>1\), and \(0<\varepsilon<1\), the corresponding operators \(R_{N,M}\) and \(R_{N,M,\varepsilon}\) are defined by
and
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
and it follows that
where \(R'_{N,M}\) denotes the adjoint operator of \(R_{N,M}\). Since
and
By (2.13), (2.14), and a similar calculation as [3], p.708, we have
Therefore, invoking (2.12) and by Fatou’s lemma, we obtain
It is easy to check that the constant C is independent of N and M. Now define
and
We have the following conclusion.
$$R_{N}f(x)=\int_{\mathbb{R}}e^{ix\cdot\xi}p_{N}(x, \xi) \hat{f}(\xi)\,d\xi, \quad N\geq2, $$
$$p_{N,M}(x,\xi)=\rho\biggl(\frac{x}{M}\biggr)p_{N}(x, \xi), \quad M>1 $$
$$p_{N,M,\varepsilon}(x,\xi)=\psi\biggl(\frac{\xi}{\varepsilon}\biggr) p_{N,M}(x,\xi), \quad 0< \varepsilon< 1. $$
$$R_{N,M}f(x)=\int_{\mathbb{R}}e^{ix\cdot\xi} p_{N,M}(x,\xi)\hat{f}(\xi)\,d\xi $$
$$R_{N,M,\varepsilon}f(x)=\int_{\mathbb{R}} e^{ix\cdot\xi}p_{N,M,\varepsilon}(x, \xi) \hat{f}(\xi)\,d\xi. $$
$$R'_{N,M,\varepsilon}g(x)=\iint e^{i(x-y)\cdot\xi}\overline {p_{N,M,\varepsilon}(y,\xi)} g(y)\, dy\, d\xi $$
$$ \lim_{\varepsilon\rightarrow0}R'_{N,M,\varepsilon }g(x)=R'_{N,M}g(x), \quad g\in\mathcal{S}(\mathbb{R}), $$
(2.12)
$$ \int\bigl\vert R^{\prime}_{N,M,\varepsilon}g(x)\bigr\vert ^{2}\,dx= \lim_{L\rightarrow\infty}\int_{|x|< L} \bigl\vert R^{\prime}_{N,M,\varepsilon}g(x)\bigr\vert ^{2}\,dx $$
(2.13)
$$\begin{aligned} \int_{|x|< L} \bigl\vert R^{\prime}_{N,M,\varepsilon}g(x) \bigr\vert ^{2}\,dx =&\int_{|x|< L} R^{\prime}_{N,M,\varepsilon}g(x) \overline{R^{\prime}_{N,M,\varepsilon}g(x)} \,dx \\ =&\int_{|x|< L} \biggl(\iint e^{i(x-y)\cdot\xi}\overline {p_{N,M,\varepsilon}(y,\xi)} g(y)\, dy\, d\xi \biggr) \\ &{}\times \biggl(\overline{\iint e^{i(x-z)\cdot\eta}\overline {p_{N,M,\varepsilon}(z, \eta)} g(z)\, dz\, d\eta} \biggr)\,dx. \end{aligned}$$
(2.14)
$$\begin{aligned}& \int\bigl\vert R^{\prime}_{N,M,\varepsilon}g(x)\bigr\vert ^{2}\,dx \\& \quad = 2\pi\iint \biggl(\int e^{i(z-y)\xi}\overline{p_{N,M,\varepsilon }(y, \xi)} p_{N,M,\varepsilon}(z,\xi)\,d\xi \biggr) g(y)\overline{g(z)}\,dy\,dz \\& \quad = 2\pi\iint \biggl(\int e^{i(z-y)\xi}\rho\biggl(\frac{y}{M} \biggr)\rho\biggl(\frac {z}{M}\biggr)\psi^{2} \biggl( \frac{\xi}{\varepsilon}\biggr)\overline{p_{N}(y,\xi)} p_{N}(z, \xi)\,d\xi \biggr) \\& \qquad {} \times g(y)\overline{g(z)}\,dy\,dz. \end{aligned}$$
(2.15)
$$\begin{aligned}& \int\bigl\vert R'_{N,M}g(x)\bigr\vert ^{2}\,dx \\& \quad \leq \liminf_{\varepsilon\rightarrow0}\int\bigl\vert R'_{N,M,\varepsilon }g(x) \bigr\vert ^{2}\,dx \\& \quad = 2\pi\lim_{\varepsilon\rightarrow0} \iint \biggl(\int e^{i(z-y)\xi}\rho \biggl(\frac{y}{M}\biggr)\rho\biggl(\frac{z}{M}\biggr) \psi^{2} \biggl(\frac{\xi}{\varepsilon}\biggr)\overline{p_{N}(y, \xi)} p_{N}(z,\xi)\,d\xi \biggr) \\& \qquad {}\times g(y)\overline{g(z)}\,dy\,dz \\& \quad \leq C \iint\biggl\vert \int e^{i[(z-y)\xi+(t(z)-t(y))\phi(\vert \xi \vert )]}\varphi^{2} \biggl(\frac{\xi}{N}\biggr)\,d\xi N^{-2s}\biggr\vert \bigl\vert g(y)\bigr\vert \bigl\vert g(z)\bigr\vert \,dy\,dz. \end{aligned}$$
(2.16)
$$I_{N}(x,\omega)=N^{-2s}\int e^{i[x\xi+\omega\phi(|\xi|)]}\varphi ^{2}\biggl(\frac{\xi}{N}\biggr)\,d\xi\quad \text{for } x\in \mathbb{R}, -1< \omega< 1, N\geq2 $$
$$J_{N}(x)=\sup_{|\omega|< 1}\bigl\vert I_{N}(x,\omega)\bigr\vert , \quad x\in\mathbb{R}. $$
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
From this we get
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
It follows that (2.9) holds, and we complete the proof of Lemma 2.2 based on Lemma 2.3.
$$\begin{aligned} \int\bigl\vert R'_{N,M}g(x)\bigr\vert ^{2} \,dx &\leq C\iint\bigl\vert I_{N}\bigl(z-y,t(z)-t(y)\bigr)\bigr\vert \bigl\vert g(y)\bigr\vert \bigl\vert g(z)\bigr\vert \,dy\,dz \\ &\leq C\iint J_{N}(z-y)\bigl\vert g(y)\bigr\vert \bigl\vert g(z)\bigr\vert \,dy\,dz \\ &=C\int\bigl(J_{N}\ast|g|\bigr) (z)\bigl\vert g(z)\bigr\vert \,dz \\ &\leq C\bigl\Vert J_{N}\ast \vert g\vert \bigr\Vert _{2}\|g\|_{2} \\ &\leq C\|J_{N}\|_{1}\|g\|_{2}^{2}\leq CN^{-2\delta}\|g\| _{2}^{2}. \end{aligned}$$
$$\bigl\Vert R'_{N,M}g\bigr\Vert _{2}\leq CN^{-\delta}\|g\|_{2}. $$
$$\|R_{N}g\|_{2}\leq CN^{-\delta}\|g\|_{2}. $$
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
where
\(h_{k}\)
is a linear combination of functions of the form
with
\(0\leq s \leq k\), \(0\leq r \leq k\), and
\(2\leq j_{q} \leq k+1\).
$$\int_{I} e^{iF(x)}g(x)\,dx=\int _{I} e^{iF(x)}h_{k}(x)\,dx, $$
$$g^{(s)}\bigl(F^{\prime}\bigr)^{-k-r}\prod _{q=1}^{r}F^{(j_{q})} $$
We now return to the proof of Lemma 2.3. Recall that
Performing a change of variable, we have
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 \(N\geq2\),
Below we give more estimates of \(|I_{N}(x,\omega)|\).
$$I_{N}(x,\omega)=N^{-2s}\int e^{i[x\cdot\xi+\omega\phi(|\xi |)]} \varphi^{2}\biggl(\frac{\xi}{N}\biggr)\,d\xi, \quad x\in\mathbb{R}, -1< \omega< 1, N\geq2. $$
$$I_{N}(x,\omega)=N^{1-2s}\int e^{i(Nx\xi+\omega\phi(N|\xi|))}G(\xi )\,d\xi, $$
$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \le CN^{1-2s}. $$
(2.18)
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
Now we give the following estimates of \(I_{N}(x,\omega)\) for \(x\in \mathbb{R}\), \(-1<\omega<1\), and \(N\geq2\):
Let \(F(\xi)=Nx\xi+\omega\phi(N|\xi|)\). We have
and
Noting \(N|\xi|>1\) by \(N\geq2\) and \(\frac{1}{2}<|\xi|<2\), by (K3) we get
When \(|\omega|<\frac{N|x|}{2C_{3}N^{m_{2}}}\) (equivalently, \(C_{3} N^{m_{2}}|\omega|< \frac{1}{2}N|x|\)), we have
Therefore,
Since ϕ satisfies (K4) and (K5) with \(m_{4}\leq m_{3}=m_{2}\), we have
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
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 \(\frac{1}{2}<|\xi|<2\),
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
This is just the second estimate in (2.19).
$$C_{2}=\max_{\frac{1}{2}\leq|\xi|\leq2}\bigl\{ |\xi|^{m_{2}-1}\bigr\} \quad \text{and} \quad C_{3}=\max\{C_{1}C_{2},1\}. $$
$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq \left \{ \textstyle\begin{array}{l@{\quad}l} C(N|x|)^{-2}N^{1-2s}, & |\omega|< \frac {N|x|}{2C_{3}N^{m_{2}}}, \\ C(N|x|)^{-\frac{1}{2}}N^{1-2s}, & |\omega|\ge\frac {N|x|}{2C_{3}N^{m_{2}}}. \end{array}\displaystyle \displaystyle \right . $$
(2.19)
$$\begin{aligned}& F^{\prime}(\xi)=Nx+N \operatorname{sgn}(\xi) \omega\phi^{\prime} \bigl(N\vert \xi \vert \bigr), \\& F^{\prime\prime}(\xi)=N^{2}\omega\phi^{\prime\prime}\bigl(N\vert \xi \vert \bigr) \end{aligned}$$
$$F^{(3)}(\xi)=N^{3}\operatorname{sgn}(\xi) \omega \phi^{(3)}\bigl(N\vert \xi \vert \bigr). $$
$$\bigl\vert N \operatorname{sgn}(\xi)\omega\phi^{\prime}\bigl(N\vert \xi \vert \bigr)\bigr\vert \leq C_{1} N\vert \omega \vert \bigl(N\vert \xi \vert \bigr)^{m_{2}-1}\leq C_{1}C_{2} N^{m_{2}}|\omega|\leq C_{3}N^{m_{2}}|\omega|. $$
$$\bigl\vert N \operatorname{sgn}(\xi) \omega\phi^{\prime}\bigl(N\vert \xi \vert \bigr)\bigr\vert < \frac{1}{2}N|x|. $$
$$ \bigl\vert F^{\prime}(\xi)\bigr\vert \geq N|x|-\bigl\vert N \operatorname{sgn}(\xi) \omega\phi^{\prime }\bigl(N\vert \xi \vert \bigr)\bigr\vert > \frac{1}{2}N|x|. $$
(2.20)
$$ \bigl\vert F^{(j)}(\xi)\bigr\vert \leq C N^{m_{2}}|\omega|\quad \text{for } j=2,3. $$
(2.21)
$$\begin{aligned}& \biggl\vert \int e^{iF(\xi)}G(\xi)\,d\xi\biggr\vert \\& \quad \leq C\int_{\frac{1}{2}< |\xi|< 2}\frac{1}{|F^{\prime}(\xi)|^{2}} \biggl(1+ \frac{|F^{\prime\prime}(\xi)|}{|F^{\prime}(\xi)|}+ + \biggl(\frac{|F^{\prime\prime}(\xi)|}{ |F^{\prime}(\xi)|} \biggr)^{2}+ \frac{|F^{(3)}(\xi)|}{ |F^{\prime}(\xi)|} \biggr)\,d\xi \\& \quad \leq C\bigl(N\vert x\vert \bigr)^{-2}\sum _{r=0}^{2} \biggl(\frac{N^{m_{2}} |\omega|}{N|x|} \biggr)^{r} \\& \quad \leq C\bigl(N\vert x\vert \bigr)^{-2}, \end{aligned}$$
$$\bigl\vert F^{\prime\prime}(\xi)\bigr\vert \geq C N^{2}|\omega| \bigl(N\vert \xi \vert \bigr)^{m_{2}-2}>CN^{m_{2}}|\omega|>0 . $$
$$\bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N^{m_{2}}|\omega|\bigr)^{-\frac{1}{2}} \bigl(\|G\|_{\infty}+ \bigl\Vert G^{\prime}\bigr\Vert _{1}\bigr)N^{1-2s} \leq C \bigl(N\vert x\vert \bigr)^{-\frac {1}{2}}N^{1-2s}. $$
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
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
As for \(E_{2}\), we first prove that if \(\frac{1}{N}<|x|\leq 2C_{3}N^{m_{2}-1}\), then
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
In fact, if \(|\omega|<\frac{N|x|}{2C_{3}N^{m_{2}}}\), by the first estimate in (2.19) and \(N|x|>1\), then
If \(|\omega|\geq\frac{N|x|}{2C_{3}N^{m_{2}}}\), by the second estimate in (2.19), we obtain
Thus (2.24) holds and so (2.23). Hence, by (2.23), we get
Finally, we consider \(E_{3}\). We first show that if \(|x|>2C_{3}N^{m_{2}-1}\), then
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
and (2.26) follows from this. By (2.26), we obtain
Since \(m_{2}>0\), by (2.22), (2.25), and (2.27), we have
where \(2\delta=2s-\frac{m_{2}}{2}>0\) since \(s>\frac{m_{2}}{4}\) and \(m_{2}>0\).
$$\begin{aligned} \int\bigl\vert J_{N}(x)\bigr\vert \,dx =&\int _{0< |x|\leq\frac{1}{N}} \bigl\vert J_{N}(x)\bigr\vert \,dx + \int_{\frac{1}{N}< |x|\leq2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ &{} +\int_{|x|>2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ =:&E_{1}+E_{2}+E_{3}. \end{aligned}$$
$$ E_{1}\leq C\int_{0< |x|\leq\frac {1}{N}} N^{1-2s}\,dx\leq C N^{-2s}. $$
(2.22)
$$ J_{N}(x)\leq C \bigl(N\vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}. $$
(2.23)
$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N\vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}. $$
(2.24)
$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N \vert x\vert \bigr)^{-2} N^{1-2s}\leq C \bigl(N\vert x \vert \bigr)^{-\frac {1}{2}} N^{1-2s}. $$
$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C \bigl(N \vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}. $$
$$ E_{2}\leq C\int_{|x|\leq2C_{3}N^{m_{2}-1}} \bigl(N\vert x\vert \bigr)^{-\frac{1}{2}} N^{1-2s}\,dx\leq C N^{\frac{m_{2}}{2}-2s}. $$
(2.25)
$$ \bigl\vert J_{N}(x)\bigr\vert \leq C\bigl(N\vert x\vert \bigr)^{-2}N^{1-2s}. $$
(2.26)
$$ \bigl\vert I_{N}(x,\omega)\bigr\vert \leq C\bigl(N \vert x\vert \bigr)^{-2}N^{1-2s}, $$
$$ E_{3}\leq C\int_{|x|>2C_{3}N^{m_{2}-1}} \bigl(N \vert x\vert \bigr)^{-2}N^{1-2s}\,dx \leq C N^{-m_{2}-2s}. $$
(2.27)
$$\bigl\vert J_{N}(x)\bigr\vert \leq C N^{\frac{m_{2}}{2}-2s}=:CN^{-2\delta}, $$
First we consider the case where \(2C_{3}N^{m_{2}-1}>\frac{1}{N}\). Write
Since \(m_{2}\leq0\), by (2.22), (2.25), and (2.27), we have
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
Since \(m_{2}\leq0\), by (2.22) and (2.27), we have
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.
$$\begin{aligned} \int\bigl\vert J_{N}(x)\bigr\vert \,dx =&\int _{0< |x|\leq\frac{1}{N}} \bigl\vert J_{N}(x)\bigr\vert \,dx + \int_{\frac{1}{N}< |x|\leq2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ &{} +\int_{|x|>2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ =:&E_{1}+E_{2}+E_{3}. \end{aligned}$$
$$\bigl\vert J_{N}(x)\bigr\vert \leq C N^{-m_{2}-2s}=:CN^{-2\delta}, $$
$$\begin{aligned} \int\bigl\vert J_{N}(x)\bigr\vert \,dx &\leq\int _{0< |x|\leq\frac{1}{N}} \bigl\vert J_{N}(x)\bigr\vert \,dx + \int_{|x|>2C_{3}N^{m_{2}-1}} \bigl\vert J_{N}(x)\bigr\vert \,dx \\ &=:E_{1}+E_{3}. \end{aligned}$$
$$\bigl\vert J_{N}(x)\bigr\vert \leq C N^{-m_{2}-2s}=:CN^{-2\delta}, $$
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
holds for \(s>\frac{m_{2}}{4}\) if \(m_{2}>0\) or \(s>\frac{-m_{2}}{2}\) if \(m_{2}<0\).
$$ \bigl\Vert S^{\ast}_{\phi}f\bigr\Vert _{L^{2}(\mathbb{R}^{n})}\leq C\|f\|_{H^{s}(\mathbb{R}^{n})} $$
(3.1)
Let \(t(x)\) is a measurable radial function with \(0< t(x)<1\). Denote
Recall the Bessel function \(J_{m}(r)\) is defined by
Since f is radial,
Therefore,
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
Denote
By (3.2) and (3.4), it follows that
Let
Thus, we have
By (3.5), to prove (3.3) it suffices to prove that
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 \(J_{m}\).
$$Tf(x)=(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it(x)\phi(|\xi|)} \hat{f}(\xi)\,d\xi. $$
$$J_{m}(r)=\frac{(\frac{r}{2})^{m}}{\Gamma(m+\frac{1}{2}) \pi^{\frac{1}{2}}}\int_{-1}^{1}e^{irt} \bigl(1-t^{2}\bigr)^{m-\frac{1}{2}}\,dt, \quad m>-\frac{1}{2}. $$
$$\hat{f}(\xi)=(2\pi)^{\frac{n}{2}}|\xi|^{1-\frac{n}{2}}\int_{0}^{\infty} f(s)J_{\frac{n}{2}-1}\bigl(s\vert \xi \vert \bigr)s^{\frac{n}{2}}\, ds. $$
$$ Tf(u)=(2\pi)^{\frac{n}{2}-n}u^{1-\frac{n}{2}}\int _{0} ^{\infty}J_{\frac{n}{2}-1} (ru)e^{it(u)\phi(r)} \hat{f}(r)r^{\frac{n}{2}}\,dr, \quad u>0. $$
(3.2)
$$ \biggl(\int_{0}^{\infty}\bigl\vert Tf(u)\bigr\vert ^{2} u^{n-1}\, du \biggr)^{1/2}\leq \biggl(\int_{0}^{\infty}\bigl\vert \hat{f}(r)\bigr\vert ^{2} \bigl(1+r^{2}\bigr)^{s}r^{n-1} \,dr \biggr)^{1/2}. $$
(3.3)
$$ g(r)=\hat{f}(r) \bigl(1+r^{2}\bigr)^{\frac{s}{2}}r^{\frac{n-1}{2}}, \quad r>0. $$
(3.4)
$$\begin{aligned} Tf(u)u^{\frac{n-1}{2}} &=(2\pi)^{-\frac{n}{2}} u^{\frac{1}{2}}\int _{0}^{\infty}J_{\frac{n}{2}-1} (ru)e^{it(u)\phi(r)} \hat{f}(r)r^{\frac{n}{2}}\,dr \\ &=(2\pi)^{-\frac{n}{2}} u^{\frac{1}{2}}\int_{0}^{\infty}J_{\frac{n}{2}-1} (ru)e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}r^{\frac {1}{2}} \,dr. \end{aligned}$$
$$Pg(u)=u^{\frac{1}{2}}\int_{0}^{\infty} J_{\frac{n}{2}-1}(ru)e^{it(u)\phi(r)}g(r) \bigl(1+r^{2} \bigr)^{-\frac{s}{2}} r^{\frac{1}{2}}\,dr. $$
$$ Tf(u)u^{\frac{n-1}{2}}=(2\pi)^{-\frac{n}{2}} Pg(u). $$
(3.5)
$$ \biggl(\int_{0}^{\infty}\bigl\vert Pg(u)\bigr\vert ^{2}\, du \biggr)^{1/2}\leq C \biggl(\int _{0}^{\infty}\bigl\vert g(r)\bigr\vert ^{2}\,dr \biggr)^{1/2} $$
(3.6)
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
where \(b_{1}\) and \(b_{2}\) are the constants depending on n. In fact, by Lemma 3.1, as \(t\rightarrow\infty \), we have
It follows that, as \(t\rightarrow\infty\), we have
where
and
It follows that, when \(t>1\), we have
On the other hand, by the definition of the Bessel function
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
It follows from (3.8) and (3.9) that (3.7) holds. Invoking (3.7), we have
where
and
From [17], pp.59-61, we have
and
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
and
Therefore, we have
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
Since \(u>0\) and by (3.13) and (3.14), for \(i=1,2\), we have
$$ t^{\frac{1}{2}}J_{\frac{n}{2}-1}(t)= b_{1}e^{it}+b_{2}e^{-it}+ {O}\biggl(\min\biggl(1,\frac{1}{t}\biggr)\biggr), \quad t>0, $$
(3.7)
$$ J_{\frac{n}{2}-1}(t) =\sqrt{\frac{2}{\pi t}}\cos\biggl(t-\frac{\pi(n-1)}{4} \biggr)+ {O}\bigl(t^{-\frac{3}{2}}\bigr). $$
$$\begin{aligned} t^{\frac{1}{2}}J_{\frac{n}{2}-1}(t) &=\sqrt{\frac{2}{\pi}}\cos\biggl( \frac{\pi(n-1)}{4}\biggr)\cos t +\sqrt{\frac{2}{\pi}}\sin\biggl( \frac{\pi(n-1)}{4}\biggr)\sin t+ {O}\bigl(t^{-1}\bigr) \\ &=(b_{1}+b_{2})\cos t+i(b_{1}-b_{2}) \sin t+{O}\bigl(t^{-1}\bigr) \\ &=b_{1}e^{it}+b_{2}e^{-it}+{O} \bigl(t^{-1}\bigr), \end{aligned}$$
$$b_{1}=\frac{1}{2}\sqrt{\frac{2}{\pi}} \biggl(\cos\biggl( \frac{\pi (n-1)}{4}\biggr)+i\sin\biggl(\frac{\pi(n-1)}{4}\biggr) \biggr) $$
$$b_{2}=\frac{1}{2}\sqrt{\frac{2}{\pi}} \biggl(\cos\biggl( \frac{\pi (n-1)}{4}\biggr)-i\sin\biggl(\frac{\pi(n-1)}{4}\biggr) \biggr). $$
$$ \bigl\vert t^{\frac{1}{2}}J_{\frac {n}{2}-1}(t)- \bigl(b_{1}e^{it}+b_{2}e^{-it}\bigr)\bigr\vert \leq C t^{-1}. $$
(3.8)
$$J_{m}(t)=\frac{(\frac{t}{2})^{m}}{\Gamma(m+\frac{1}{2}) \pi^{\frac{1}{2}}}\int_{-1}^{1}e^{its} \bigl(1-s^{2}\bigr)^{m-\frac{1}{2}}\, ds, \quad m>-\frac{1}{2}, $$
$$\begin{aligned} \bigl\vert t^{\frac{1}{2}}J_{\frac {n}{2}-1}(t)- \bigl(b_{1}e^{it}+b_{2}e^{-it}\bigr)\bigr\vert \leq& \bigl\vert t^{\frac{1}{2}}J_{\frac{n}{2}-1}(t)\bigr\vert + \bigl\vert b_{1}e^{it}\bigr\vert +\bigl\vert b_{2}e^{-it}\bigr\vert \\ \leq& Ct^{\frac{1}{2}}t^{-\frac{1}{2}}+|b_{1}|+|b_{2}| \leq C. \end{aligned}$$
(3.9)
$$\begin{aligned} \begin{aligned}[b] Pg(u)={}& b_{1}\int_{0}^{\infty}e^{iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr \\ &{}+ b_{2}\int_{0}^{\infty}e^{-iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr+E(u)+F(u) \\ =:{}&b_{1}D_{1}(u)+b_{2}D_{2}(u)+E(u)+F(u), \end{aligned} \end{aligned}$$
(3.10)
$$\bigl\vert E(u)\bigr\vert \leq C\int_{0}^{\frac{1}{u}} \bigl\vert g(r)\bigr\vert \,dr $$
$$\bigl\vert F(u)\bigr\vert \leq C\frac{1}{u}\int_{\frac{1}{u}}^{\infty} \frac{1}{r}\bigl\vert g(r)\bigr\vert \,dr. $$
$$ \biggl(\int_{0}^{\infty}\bigl\vert E(u)\bigr\vert ^{2}\, du \biggr)^{1/2}\leq C\|g \|_{L^{2}(0,\infty)} $$
(3.11)
$$ \biggl(\int_{0}^{\infty}\bigl\vert F(u)\bigr\vert ^{2}\, du \biggr)^{1/2}\leq C\|g \|_{L^{2}(0,\infty)}. $$
(3.12)
$$D_{1}(u)=\int_{0}^{\infty}e^{iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr=\int_{\mathbb{R}} e^{iru}e^{it(u)\phi(r)}\hat{h}(r)\,dr $$
$$D_{2}(u)=\int_{0}^{\infty}e^{-iru}e^{it(u)\phi(r)}g(r) \bigl(1+r^{2}\bigr)^{-\frac{s}{2}}\,dr=\int_{\mathbb{R}} e^{-iru}e^{it(u)\phi(r)}\hat{h}(r)\,dr. $$
$$ \bigl\vert D_{i}(u)\bigr\vert \leq S_{\phi}^{\ast} h(u)\quad \text{for } i=1,2. $$
(3.13)
$$ \bigl\Vert S_{\phi}^{\ast} h\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|h\|_{H^{s}(\mathbb{R})}. $$
(3.14)
$$\begin{aligned} \|D_{i}\|_{L^{2}(0,\infty)} \leq& \|D_{i} \|_{L^{2}(\mathbb{R})} \leq C\bigl\Vert S_{\phi}^{\ast} h\bigr\Vert _{L^{2}(\mathbb{R})}\leq C\|h\| _{H^{s}(\mathbb{R})} \\ =& C \biggl(\int_{0}^{\infty}\bigl\vert g(r)\bigr\vert ^{2}\bigl(1+r^{2}\bigr)^{-s} \bigl(1+r^{2}\bigr)^{s}\,dr \biggr)^{1/2} \\ =& C\|g\|_{L^{2}(0,\infty)}. \end{aligned}$$
(3.15)
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
where
\(J_{m}\)
denotes the Bessel function.
$$F_{0}(r)= (2\pi)^{\frac{n}{2}}i^{-k}r^{-\frac{n}{2}-k+1} \int_{0}^{\infty}f_{0}(s)J_{\frac{n}{2}+k-1} (rs)s^{\frac{n}{2}+k}\,ds, $$
Let us return to the proof of Theorem 1.3. First we show that, for \(f\in\mathcal{H}_{k}\) (\(k\geq1\)),
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
where
By (4.2) and noting that \(\{P_{1},P_{1},\ldots ,P_{a_{k}}\}\) is an orthonormal basis in \({\mathscr{A}}_{k}\), we have
which is just (4.1). On the other hand, by (4.2), we have
Applying Lemma 4.1, we get
where \(s=|x|>0\). Therefore, we have
$$ \|f\|_{H^{s}(\mathbb{R}^{n})}= \Biggl(\sum_{j=1}^{a_{k}} \int_{0}^{\infty}\bigl\vert F_{j}(r) \bigr\vert ^{2}\bigl(1+r^{2}\bigr)^{s}r^{n+2k-1} \,dr \Biggr)^{1/2}. $$
(4.1)
$$ \hat{f}(x)=\sum_{j=1}^{a_{k}}F_{j} \bigl(\vert x\vert \bigr)P_{j}(x), $$
(4.2)
$$F_{j}(r)=(2\pi)^{\frac{n}{2}}i^{-k}r^{1-\frac{n}{2}-k}\int _{0}^{\infty}f_{j}(s) J_{\frac{n}{2}+k-1}(rs)s^{\frac{n}{2}+k} \,ds, \quad r>0. $$
$$\begin{aligned}& \int_{\mathbb{R}^{n}}\bigl(1+\vert \xi \vert ^{2} \bigr)^{s}\bigl\vert \hat{f}(\xi)\bigr\vert ^{2}\,d\xi \\& \quad = \int_{0}^{\infty} \biggl(\int _{S^{n-1}}\bigl\vert \hat{f}\bigl(r\xi^{\prime}\bigr) \bigr\vert ^{2}\,d\sigma\bigl(\xi^{\prime}\bigr) \biggr) \bigl(1+r^{2}\bigr)^{s}r^{n-1}\,dr \\& \quad = \int_{0}^{\infty} \Biggl(\int _{S^{n-1}} \Biggl(\sum_{j=1}^{a_{k}}F_{j}(r)P_{j} \bigl(r\xi^{\prime}\bigr) \Biggr) \Biggl(\sum_{i=1}^{a_{k}} \overline{F_{i}(r)} \overline {P_{i}\bigl(r \xi^{\prime}\bigr)} \Biggr)\,d\sigma\bigl(\xi^{\prime}\bigr) \Biggr) \bigl(1+r^{2}\bigr)^{s}r^{n-1}\,dr \\& \quad =\int_{0}^{\infty} \Biggl(\sum _{j=1}^{a_{k}} \bigl\vert F_{j}(r)\bigr\vert ^{2} \Biggr)r^{2k}\bigl(1+r^{2} \bigr)^{s}r^{n-1}\,dr \\& \quad =\sum_{j=1}^{a_{k}}\int _{0}^{\infty} \bigl\vert F_{j}(r)\bigr\vert ^{2}\bigl(1+r^{2}\bigr)^{s}r^{n+2k-1} \,dr, \end{aligned}$$
$$ S_{t,\phi}f(x) =(2\pi)^{-n}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi} e^{it\phi(|\xi|)}\hat{f}(\xi)\,d\xi =\sum_{j=1}^{a_{k}}(2\pi)^{-n}\int _{\mathbb{R}^{n}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j} \bigl(\vert \xi \vert \bigr)P_{j}(\xi) \bigr)\,d\xi. $$
$$\begin{aligned}& \int_{\mathbb{R}^{n}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j} \bigl(\vert \xi \vert \bigr)P_{j}(\xi) \bigr)\,d\xi \\& \quad = \bigl(e^{it\phi(|\cdot|)}F_{j}\bigl(\vert \cdot \vert \bigr)P_{j}(-\cdot) \bigr)^{\wedge}(x) \\& \quad =(2\pi)^{\frac{n}{2}}i^{-k}s^{1-\frac{n}{2}-k} \biggl(\int _{0}^{\infty}J_{\frac{n}{2}+k-1}(rs)e^{it\phi(r)} F_{j}(r)r^{\frac{n}{2}+k}\,dr \biggr)P_{j}(-x), \end{aligned}$$
$$\begin{aligned} S_{t,\phi}f(x) =&\sum_{j=1}^{a_{k}}(2 \pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j}\bigl(\vert \xi \vert \bigr)P_{j}(\xi) \bigr)\,d\xi \\ =&\sum_{j=1}^{a_{k}}(2\pi)^{-\frac{n}{2}}i^{-k}|x|^{1-\frac {n}{2}-k} \\ &{} \times \biggl(\int_{0}^{\infty}J_{\frac {n}{2}+k-1} \bigl(r\vert x\vert \bigr)e^{it\phi(r)} F_{j}(r)r^{\frac{n}{2}+k} \,dr \biggr)P_{j}(-x). \end{aligned}$$
(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
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
Obviously, \(S_{t,\phi}^{n+2k}h\) is still a radial function. Then
By (4.3) and (4.4), we have
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
Then by (4.5) and (4.6), we obtain
Using the notation \(v=|x|\) and \(r=|\xi|\), by (4.7), we have
Using the representation of polar coordinates and noting (4.6), we obtain
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
Note that \(\mathscr{F}_{n+2k}f_{j}=i^{k}F_{j}\), and we get
$$\mathscr{F}_{n+2k}h(x)=(2\pi)^{\frac{n}{2}}|x|^{1-\frac {n}{2}-k}\int _{0}^{\infty}h(r) J_{\frac{n}{2}+k-1}\bigl(r\vert x \vert \bigr)r^{\frac{n}{2}+k}\,dr. $$
$$S_{t,\phi}^{n+2k}h(x) :=(2\pi)^{-n-2k}\int _{\mathbb{R}^{n+2k}} e^{ix\cdot\xi}e^{it\phi(|\xi|)} \mathscr{F}_{n+2k}h \bigl(\vert \xi \vert \bigr)\,d\xi. $$
$$\begin{aligned} S_{t,\phi}^{n+2k}f_{j}\bigl(|x|\bigr) =&i^{k}(2\pi)^{-n-2k}\int_{\mathbb{R}^{n+2k}} e^{ix\cdot\xi} \bigl(e^{it\phi(|\xi|)}F_{j}(\xi) \bigr)\,d\xi \\ =&i^{k}(2\pi)^{-\frac{n}{2}-2k}|x|^{1-\frac{n}{2}-k} \\ &{}\times\int_{0}^{\infty}J_{\frac{n}{2}+k-1}\bigl(r \vert x\vert \bigr)e^{it\phi(r)} F_{j}(r)r^{\frac{n}{2}+k} \,dr. \end{aligned}$$
(4.4)
$$ S_{t,\phi}f(x)= i^{-2k}(2\pi)^{2k}\sum _{j}S_{t,\phi }^{n+2k}f_{j} \bigl(\vert x\vert \bigr)\cdot P_{j}(-x), \quad x\in \mathbb{R}^{n}, $$
(4.5)
$$ S_{\phi}^{n+2k,\ast}f_{j}\bigl(\vert y \vert \bigr)= \sup_{0< t< 1}\bigl\vert S_{t,\phi}^{n+2k}f_{j} \bigl(\vert y\vert \bigr)\bigr\vert ,\quad y\in\mathbb{R}^{n+2k} \text{ or } y\in\mathbb{R}^{n}. $$
(4.6)
$$ S_{\phi}^{\ast}f(x)\leq C_{n,k}\sum _{j}\bigl(S_{\phi}^{n+2k,\ast }f_{j} \bigl(\vert x\vert \bigr)\bigr)|x|^{k}. $$
(4.7)
$$ \bigl\Vert S_{\phi}^{\ast}f\bigr\Vert _{L^{2}(\mathbb{R}^{n})}^{2}\leq C\sum_{j=1}^{a_{k}} \int_{\mathbb{R}^{n}} \bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2}v^{2k}\,dx. $$
(4.8)
$$\begin{aligned}& \int_{\mathbb{R}^{n}}\bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2}v^{2k}\,dx \\& \quad = \omega_{n-1} \int _{0}^{\infty}\bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2} v^{n+2k-1}\, dv \\& \quad = \frac{\omega_{n-1}}{\omega_{n+2k-1}}\int_{\mathbb{R}^{n+2k}} \bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v)\bigr\vert ^{2} \,dx, \end{aligned}$$
(4.9)
$$ \int_{\mathbb{R}^{n+2k}}\bigl\vert S_{\phi}^{n+2k,\ast}f_{j}(v) \bigr\vert ^{2}\,dx\leq C\|f_{j}\|_{H^{s}(\mathbb{R}^{n+2k})}^{2}. $$
(4.10)
$$\begin{aligned} \|f_{j}\|_{H^{s}(\mathbb{R}^{n+2k})}^{2} =& \int _{\mathbb{R}^{n+2k}}\bigl\vert F_{j}\bigl(\vert \xi \vert \bigr)\bigr\vert ^{2}\bigl(1+|\xi|^{2}\bigr)^{s} \,d\xi \\ =&\omega_{n+2k-1}\int_{0}^{\infty} \bigl\vert F_{j}(r)\bigr\vert ^{2}\bigl(1+r^{2} \bigr)^{s}r^{n+2k-1}\,dr. \end{aligned}$$
(4.11)
Therefore, by (4.8), (4.9), (4.10), (4.11), and (4.1), we obtain
Thus, we complete the proof of Theorem 1.3.
$$ \bigl\Vert S_{\phi}^{\ast}f\bigr\Vert _{L^{2}(\mathbb{R}^{n})}^{2}\leq C\sum_{j=1}^{a_{k}} \int_{0}^{\infty}\bigl\vert F_{j}(r) \bigr\vert ^{2} \bigl(1+r^{2}\bigr)^{s}r^{n+2k-1} \,dr=C\|f\|_{H^{s}(\mathbb{R}^{n})}^{2}. $$
(4.12)
5 Some applications
We now give some examples to show that (1.10) includes some well-known equations.
Example 2
Example 3
Let \(\phi(r)=r^{2}+r^{4}\), then (1.10) is the fourth-order Schrödinger equation:
In this case, \(\phi(r)\) satisfies (K1) with \(l_{1}=2\geq0\), (K2)-(K5) with \(m_{1}=m_{2}=m_{3}=m_{4}=4>0\).
$$ \left \{ \textstyle\begin{array}{l} i\partial_{t}u+\Delta^{2}u-\Delta u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R}, \\ u(x,0)=f(x). \end{array}\displaystyle \right . $$
(5.1)
Example 4
Recall the definition of the beam equation:
Note that the solution of (5.2) can be formally written as the real part of
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
$$ \left \{ \textstyle\begin{array}{l} \partial_{tt}u+\Delta^{2}u+u=0, \quad (x,t)\in\mathbb{R}^{n}\times \mathbb{R}, \\ u(x,0)=f(x), \\ \partial_{t}u(x,0)=0. \end{array}\displaystyle \right . $$
(5.2)
$$u(x,t)=e^{it\sqrt{I+\Delta^{2}}}f(x) =(2\pi)^{-n}\int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it\sqrt{1+|\xi|^{4}}}\hat{f}(\xi)\,d\xi. $$
$$S_{t,\phi}f(x)=(2\pi)^{-n} \int_{\mathbb{R}^{n}}e^{ix\cdot\xi+it\phi(|\xi|)} \hat{f}(\xi )\,d\xi. $$
Acknowledgements
The authors would like to express their deep gratitude to the anonymous referees for their careful reading of the manuscript and their fruitful comments and suggestions. The work is supported by NSF of China (Nos. 11371057, 11471033), SRFDP of China (No. 20130003110003), and the Fundamental Research Funds for the Central Universities of China (No. 2014KJJCA10).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
YD participated in the design of the study and in the discussions of all results. YN participated in the discussions of all results and drafted the manuscript. All authors read and approved the final manuscript.