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Global maximal inequality to a class of oscillatory integrals

Journal of Inequalities and Applications20182018:355

https://doi.org/10.1186/s13660-018-1946-x

  • Received: 28 June 2018
  • Accepted: 13 December 2018
  • Published:

Abstract

In the present paper, we give the global \(L^{q}\) estimates for maximal operators generated by multiparameter oscillatory integral \(S_{t,\varPhi}\), which is defined by
$$S_{t,\varPhi}f(x)=(2\pi)^{-n} \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1} \phi_{1}(|\xi_{1}|)+t_{2}\phi_{2}(|\xi_{2}|)+ \cdots+t_{n}\phi_{n}(|\xi_{n}|))}\hat{f}(\xi)\,d\xi,\quad x\in\mathbb{R}^{n}, $$
where \(n\geq2\) and f is a Schwartz function in \(\mathcal{S}(\mathbb {R}^{n})\), \(t=(t_{1},t_{2},\ldots,t_{n})\), \(\varPhi=(\phi_{1},\phi_{2},\ldots,\phi_{n})\), \(\phi_{i}\) \((i=1,2,3,\ldots, n)\) is a function on \(\mathbb {R}^{+}\rightarrow\mathbb{R}\), which has a suitable growth condition. These estimates are apparently good extensions to the results of Sjölin and Soria (J. Math. Anal. Appl 411:129–143, 2014) for the multiparameter fractional Schrödinger equation.

Keywords

  • Maximal operator
  • Local estimate
  • Global estimate
  • Multiparameter oscillatory integrals

MSC

  • 42B15
  • 42B25

1 Introduction and main results

Let f be a Schwartz function in \(\mathcal{S}(\mathbb{R}^{n})\) and
$$S_{t}f(x)=u(x,t)=(2\pi)^{-n} \int_{\mathbb{R}^{n}} e^{ix\cdot\xi+it|\xi|^{a}}\hat{f}(\xi)\,d\xi,\quad (x,t)\in \mathbb{R}^{n}\times\mathbb{R}. $$
It is well known that \(S_{t}f(x)\) is the solution of the fractional Schrödinger equation
$$\begin{aligned} \textstyle\begin{cases} i\partial_{t}u+(-\Delta)^{a/2} u=0,\quad (x,t)\in\mathbb {R}^{n}\times\mathbb{R},\\ u(x,0)=f(x). \end{cases}\displaystyle \end{aligned}$$
(1.1)
Here denotes the Fourier transform of f defined by \(\hat{f}(\xi)=\int_{\mathbb{R}^{n}}e^{-i\xi\cdot x}f(x)\,dx\).
We recall the homogeneous Sobolev space \(\dot{H}^{s}(\mathbb{R}^{n})\ (s\in\mathbb{R})\), which is defined by
$$\dot{H}^{s}\bigl(\mathbb{R}^{n}\bigr)= \biggl\{ f\in \mathcal{S}^{\prime}: \Vert f \Vert _{H^{s}}= \biggl( \int_{\mathbb{R}^{n}} \vert \xi \vert ^{2s} \bigl\vert \hat{f}(\xi) \bigr\vert ^{2}\,d\xi \biggr)^{1/2} < \infty \biggr\} , $$
and the non-homogeneous Sobolev space \(H^{s}(\mathbb{R}^{n})\ (s\in \mathbb{R})\), which is defined by
$$H^{s}\bigl(\mathbb{R}^{n}\bigr)= \biggl\{ f\in \mathcal{S}^{\prime}: \Vert f \Vert _{H^{s}}= \biggl( \int_{\mathbb{R}^{n}} \bigl(1+ \vert \xi \vert ^{2} \bigr)^{s} \bigl\vert \hat{f}(\xi) \bigr\vert ^{2}\,d\xi \biggr)^{1/2}< \infty \biggr\} . $$
Maximal operator \(S^{\ast}f\) associated with the family of operators \(\{S_{t}\}_{0< t<1}\) is defined by
$$S^{\ast}f(x)= \sup_{0< t< 1} \bigl\vert S_{t}f(x) \bigr\vert ,\quad x\in\mathbb{R}^{n}. $$
It is well known that if \(a=2\), u is the solution of the Schrödinger equation
$$\begin{aligned} \textstyle\begin{cases} i\partial_{t}u-\Delta u=0,\quad (x,t)\in\mathbb{R}^{n}\times\mathbb {R},\\ u(x,0)=f(x). \end{cases}\displaystyle \end{aligned}$$
(1.2)
In 1979, Carleson [4] proposed a problem: if \(f\in H^{s}(\mathbb{R}^{n})\) for which s does
$$\begin{aligned} \lim_{t\rightarrow0}u(x,t)=f(x),\quad \mbox{a.e. }x\in \mathbb{R}^{n}. \end{aligned}$$
(1.3)
Carleson first considered this problem for dimension \(n=1\) in [4] and showed that the convergence (1.3) holds for \(f\in H^{s}(\mathbb{R})\) with \(s \geq\frac{1}{4}\), which is sharp was shown by Dahlberg and Kenig [8]. The higher dimensional case of convergence (1.3) has been studied by several authors, see [1, 2, 9, 1113, 24, 25, 30, 34, 35] for example. In fact, by a standard argument, for \(f\in H^{s}(\mathbb {R}^{n})\), the pointwise convergence (1.3) follows from the local estimate
$$ \bigl\Vert S^{\ast}f \bigr\Vert _{L^{q}(\mathbb {B}^{n})}\leq C \Vert f \Vert _{H^{s}(\mathbb{R}^{n})},\quad f\in H^{s}\bigl( \mathbb{R}^{n}\bigr), $$
(1.4)
for some \(q\geq1\) and \(s\in\mathbb{R}\). Here \(\mathbb{B}^{n}\) is the unit ball centered at the origin in \(\mathbb{R}^{n}\). On the other hand, the global estimates are of independent interest since they reveal global regularity properties of the corresponding oscillatory integrals. Next, we recall the global estimate
$$ \bigl\Vert S^{\ast}f \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \leq C \Vert f \Vert _{H^{s}(\mathbb{R}^{n})}. $$
(1.5)
Estimate (1.5) and related questions have been well studied in literature, see, e.g., Carbery [3], Cowling [7], Kenig and Ruiz [21], Kenig, Ponce, and Vega [20], Rogers and Villarroya [29], Rogers [28], Sjölin [3032], and so on.
For \(n\geq2\) and a multiindex \(a=(a_{1},a_{2},\ldots, a_{n})\), with \(a_{j}>1\) and f being a Schwartz function in \(\mathcal {S}(\mathbb{R}^{n})\), we set
$$S_{t}f(x)=(2\pi)^{-n} \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1} |\xi_{1}|^{a_{1}}+t_{2}|\xi_{2}|^{a_{2}}+ \cdots+t_{n}|\xi_{n}|^{a_{n}})}\hat{f}(\xi)\,d\xi,\quad x\in\mathbb{R}^{n}, $$
where \(t=(t_{1},t_{2},\ldots, t_{n})\in\mathbb{R}^{n}\). For \(n\geq2\), the local maximal operator \(M^{\ast}\) is defined by
$$M^{\ast}f(x)= \sup_{0< t_{i}< 1} \bigl\vert S_{t}f(x) \bigr\vert ,\quad x\in\mathbb{R}^{n}, $$
and the global maximal operator \(M^{\ast\ast}\) is defined by
$$M^{\ast\ast}f(x) = \sup_{t_{i}\in\mathbb{R}} \bigl\vert S_{t}f(x) \bigr\vert ,\quad x\in\mathbb{R}^{n}. $$
The global estimate
$$ \bigl\Vert M^{\ast\ast}f \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \leq C \Vert f \Vert _{\dot{H}^{s}(\mathbb{R}^{n})} $$
(1.6)
and
$$ \bigl\Vert M^{\ast}f \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \leq C \Vert f \Vert _{H^{s}(\mathbb{R}^{n})}. $$
(1.7)

In 2014, Sjolin and Soria [32] obtained the following results.

Theorem A

([32])

Assume \(n\geq2\). Then, for every a, inequality (1.6) holds if and only if \(4\leq q<\infty\) and \(s=n(\frac{1}{2}-\frac{1}{q})\).

Theorem B

([32])

Assume \(n\geq2\). Then, for every a and for \(2< q<4\), inequality (1.7) holds if and only if \(s\geq\frac {n}{2}-\frac{|a|}{4}+\frac{|a|}{q}-\frac{n}{q}\).

Multiparameter singular integrals and related operators have been well studied and raised considerable attention in harmonic analysis, which can been seen in the work of Stein and Fefferman in [1417], and so on. In the present paper, we consider the maximal estimates associated with multiparameter oscillatory integral \(S_{t,\varPhi}\) defined by
$$S_{t,\varPhi}f(x)=(2\pi)^{-n} \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1} \phi_{1}(|\xi_{1}|)+t_{2}\phi_{2}(|\xi_{2}|)+ \cdots+t_{n}\phi_{n}(|\xi_{n}|))}\hat{f}(\xi)\,d\xi,\quad x\in\mathbb{R}^{n}. $$
Here, \(n\geq2\) and f is a Schwartz function in \(\mathcal{S}(\mathbb {R}^{n})\), \(\varPhi=(\phi_{1},\phi_{2},\ldots,\phi_{n})\), \(\phi_{i}\) \((i=1,2,3,\ldots, n)\) is a function on \(\mathbb {R}^{+}\rightarrow\mathbb{R}\). For \(n\geq2\), the local maximal operator \(M_{\varPhi}^{\ast}\) is defined by
$$M_{\varPhi}^{\ast}f(x)= \sup_{0< t_{i}< 1} \bigl\vert S_{t,\varPhi}f(x) \bigr\vert ,\quad x\in\mathbb{R}^{n}, $$
and the global maximal operator \(M_{\varPhi}^{\ast\ast}\) is defined by
$$M_{\varPhi}^{\ast\ast}f(x)= \sup_{t_{i}\in\mathbb{R}} \bigl\vert S_{t,\varPhi}f(x) \bigr\vert ,\quad x\in\mathbb{R}^{n}. $$
The global estimates of maximal operators \(M_{\varPhi}^{\ast}\) and \(M_{\varPhi}^{\ast\ast}\) are defined by
$$ \bigl\Vert M_{\varPhi}^{\ast\ast}f \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \leq C \Vert f \Vert _{\dot{H}^{s}(\mathbb{R}^{n})} $$
(1.8)
and
$$ \bigl\Vert M_{\varPhi}^{\ast}f \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \leq C \Vert f \Vert _{H^{s}(\mathbb{R}^{n})}. $$
(1.9)
Assume that \(\phi: \mathbb{R}^{+}\rightarrow\mathbb{R}\) satisfies:
  1. (H1)

    There exists \(m_{1}>1\) such that \(|\phi^{\prime}(r)|\sim r^{m_{1}-1}\) and \(|\phi^{\prime \prime}(r)|\gtrsim r^{m_{1}-2}\) for all \(0< r<1\);

     
  2. (H2)

    There exists \(m_{2}>1\) such that \(|\phi^{\prime}(r)|\sim r^{m_{2}-1}\) and \(|\phi^{\prime \prime}(r)|\gtrsim r^{m_{2}-2}\) for all \(r\geq1\);

     
  3. (H3)

    Either \(\phi^{\prime\prime}(r)>0\) or \(\phi^{\prime \prime}(r)<0\) for all \(r>0\).

     

Now we state our main results as follows.

Theorem 1.1

Assume that \(n\geq2\) and \(\phi_{i}\) \((i=1,2,3,\ldots, n)\) satisfies (H1)(H3). If \(4\leq q<\infty\) and \(s=n(\frac{1}{2}-\frac{1}{q})\), then the global estimate (1.8) holds.

Theorem 1.2

Let \(m=(m_{1,2},m_{2,2},\ldots,m_{n,2})\) and set \(|m|=m_{1,2}+m_{2,2}+\cdots+m_{n,2}\). Assume that \(n\geq2\) and \(\phi _{i}\) \((i=1,2,3,\ldots, n)\) satisfies (H1)(H3) with \(m_{i,1}>1\), \(m_{i,2}>1\). Then, for every m, inequality (1.9) holds if \(2< q<4\) and \(s\geq\frac{n}{2}-\frac{|m|}{4}+\frac{|m|}{q}-\frac{n}{q}\).

Remark 1.1

There are many elements ϕ satisfying conditions (H1)–(H3), for instance, the fractional Schrödinger equation (\(\phi(r)= r^{a}\)), or \(( \phi(r)=(1+r^{2})^{\frac{a}{2}}), (a\geq1)\), the Beam equation \((\phi(r)=\sqrt{1+r^{4}})\), the fourth-order Schrödinger equation \((\phi(r)=r^{2}+r^{4})\), iBq \((\phi(r)=r\sqrt{1+r^{2}})\), and so on (see [5, 6, 18, 19, 22, 23, 27], and the references therein). Hence, Theorem 1.1 and Theorem 1.2 imply the sufficiency part of Theorem A and Theorem B, respectively. However, due to the complexity of the symbol ϕ, we cannot obtain the necessities of the range of q in Theorem 1.1 and Theorem 1.2.

This paper is organized as follows. The proofs of Theorem 1.1 and Theorem 1.2 are given in Sect. 2 and Sect. 3, respectively. To prove Theorem 1.1 and Theorem 1.2, we next need the following important lemmas, which play a key role in proving Theorem 1.1 and Theorem 1.2, respectively. The proof of Lemma 1.4 is given in Sect. 4.

Lemma 1.3

([26])

Assume that ϕ satisfies (H1)(H3) with \(m_{1}>1\), \(m_{2}>1\). \(\frac{1}{2}\leq s<1\) and \(\mu\in C_{0}^{\infty}(\mathbb {R})\). Then
$$\biggl\vert \int_{\mathbb{R}} e^{ix\xi+it\phi( \vert \xi \vert )} \vert \xi \vert ^{-s} \mu \biggl(\frac{\xi}{N} \biggr) \,d\xi \biggr\vert \leq C \frac{1}{ \vert x \vert ^{1-s}} $$
for \(x\in\mathbb{R}\setminus\{0\}\), \(t\in\mathbb{R}\), and \(N=1,2,3,\ldots\) . Here the constant C may depend on s and \(m_{1}\), \(m_{2}\), and μ but not on x, t, or N.

Remark 1.2

The proof of Lemma 1.3 is similar to that of Lemma 2.1 in [10].

Lemma 1.4

Assume that ϕ satisfies (H1)(H3) with \(m_{1}>1\), \(m_{2}>1\). \(\frac{1}{2}\leq\alpha\leq\frac{m_{2}}{2}\), \(-1< d<1\), and \(\mu\in C_{0}^{\infty}(\mathbb{R})\). Then
$$ \biggl\vert \int_{\mathbb{R}} \frac{e^{i(d\phi( \vert \xi \vert )-x\xi)}}{(1+\xi ^{2})^{\frac{\alpha}{2}}} \mu\biggl(\frac{\xi}{N} \biggr)\,d\xi \biggr\vert \leq C\frac{1}{ \vert x \vert ^{\beta}} $$
(1.10)
for \(x\in\mathbb{R}\setminus\{0\}\) and \(N=1,2,3,\ldots\) , where \(\beta=\frac{\alpha+\frac{m_{2}}{2}-1}{m_{2}-1}\). Here the constant C may depend on α and \(m_{1}\), \(m_{2}\), and μ but not on x, d, and N.

Remark 1.3

Applying the result of Lemma 1.3, the proof of Lemma 1.4 is similar to that of Lemma 2.2 in [32]. The proof of Lemma 1.4 will be given in Sect. 4.

2 The proof of Theorem 1.1

Assume that \(n\geq2\), \(\phi_{i}\) \((i=1,2,3,\ldots ,n)\) satisfies (H1)–(H3). For \(i=1,2,3,\ldots ,n\), let \(t_{i}(x)\) be a measurable function on \(\mathbb{R}^{n}\) with \(t_{i}(x)\in\mathbb{R}\). Denote \(t(x)=(t_{1}(x),t_{2}(x),\ldots,t_{n}(x))\), we set
$$\begin{aligned} S_{t(x),\varPhi}f(x)={}&(2\pi)^{-n} \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1}(x) \phi_{1}(|\xi_{1}|)+t_{2}(x)\phi_{2}(|\xi_{2}|)+ \cdots+t_{n}(x)\phi_{n}(|\xi_{n}|))}\hat{f}(\xi)\,d\xi,\\ & x\in\mathbb{R}^{n}, f\in\mathcal{S}\bigl(\mathbb{R}^{n}\bigr). \end{aligned}$$
For \(4\leq q<\infty\) and \(s=n(\frac{1}{2}-\frac{1}{q})\), that is, \(\frac{n}{4}\leq s<\frac{n}{2}\) and \(q=\frac{2n}{n-2s}\). By linearizing the maximal operator (see [30]) to prove the global estimate (1.8) holds, it suffices to show that
$$ \Vert S_{t(x),\varPhi}f \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \Vert f \Vert _{\dot {H}^{s}}=C \biggl( \int_{\mathbb{R}^{n}} \vert \xi \vert ^{2s} \bigl\vert \hat{f}(\xi) \bigr\vert ^{2}\,d\xi \biggr)^{1/2}. $$
(2.1)
To prove (2.1) it suffices to prove that
$$ \Vert S_{t(x),\varPhi}f \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \biggl( \int_{\mathbb{R}^{n}} \vert \xi_{1} \vert ^{\frac{2s}{n}} \vert \xi_{2} \vert ^{\frac{2s}{n}} |\cdots \vert \xi_{n} \vert ^{\frac{2s}{n}} \bigl\vert \hat{f}(\xi) \bigr\vert ^{2} \,d\xi \biggr)^{1/2}. $$
(2.2)
Let \(g(\xi)=|\xi_{1}|^{\frac{s}{n}}|\xi_{2}|^{\frac{s}{n}} \cdots|\xi_{n}|^{\frac{s}{n}}\hat{f}(\xi)\), then we have
$$\begin{aligned} S_{t(x),\varPhi}f(x)={}& \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1}(x) \phi_{1}( \vert \xi_{1} \vert )+t_{2}(x)\phi_{2}( \vert \xi_{2} \vert )+ \cdots+t_{n}(x)\phi_{n}( \vert \xi_{n} \vert ))} \vert \xi_{1} \vert ^{-\frac{s}{n}} \vert \xi_{2} \vert ^{-\frac{s}{n}} \cdots \vert \xi_{n} \vert ^{-\frac{s}{n}}g(\xi)\,d \xi \\ ={}&R_{\varPhi}g(x), \end{aligned}$$
(2.3)
where
$$R_{\varPhi}g(x)= \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1}(x) \phi_{1}( \vert \xi_{1} \vert )+t_{2}(x)\phi_{2}( \vert \xi_{2} \vert )+ \cdots+t_{n}(x)\phi_{n}( \vert \xi_{n} \vert ))} \vert \xi_{1} \vert ^{-\frac{s}{n}} \vert \xi_{2} \vert ^{-\frac{s}{n}} \cdots \vert \xi_{n} \vert ^{-\frac{s}{n}}g(\xi)\,d \xi. $$
To prove (2.2) it suffices to prove that
$$ \Vert R_{\varPhi}g \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \Vert g \Vert _{L^{2}(\mathbb{R}^{n})} $$
(2.4)
for g continuous and rapidly decreasing at infinity. We take a real-valued function \(\rho\in C_{0}^{\infty}(\mathbb{R}^{n})\) such that \(\rho(x)=1\) if \(|x|\leq1\) and \(\rho(x)=0\) if \(|x|\geq2\). And we choose a real-valued function \(\psi\in C_{0}^{\infty}(\mathbb{R})\) such that \(\psi(x)=1\) if \(|x|\leq1\) and \(\psi(x)=0\) if \(|x|\geq2\), and set \(\sigma(\xi )=\psi(\xi_{1})\psi(\xi_{2}) \cdots\psi(\xi_{n})\). For \(\xi\in\mathbb{R}^{n}\) and \(N=1,2,3,\ldots\) , we set \(\rho_{N}(x)=\rho(\frac{x}{N})\) and \(\sigma_{N}(\xi)=\sigma(\frac{\xi}{N})\). For \(x\in\mathbb{R}^{n}\), \(g\in L^{2}(\mathbb{R}^{n})\), and for \(N=1,2,3,\ldots\) , we define
$$\begin{aligned} R_{N,\varPhi}g(x)={}&\rho_{N}(x) \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1}(x) \phi_{1}( \vert \xi_{1} \vert )+t_{2}(x)\phi_{2}( \vert \xi_{2} \vert )+ \cdots+t_{n}(x)\phi_{n}( \vert \xi_{n} \vert ))} \vert \xi_{1} \vert ^{-\frac{s}{n}} \vert \xi_{2} \vert ^{-\frac{s}{n}} \cdots\\ &{}\times \vert \xi_{n} \vert ^{-\frac{s}{n}} \sigma_{N}(\xi)g(\xi)\,d\xi. \end{aligned}$$
The adjoint of \(R_{N,\varPhi}\) is given by
$$\begin{aligned} R^{\prime}_{N,\varPhi}h(\xi)={}&\sigma_{N}(\xi) \vert \xi_{1} \vert ^{-\frac{s}{n}} \vert \xi_{2} \vert ^{-\frac{s}{n}} \cdots\\ &{}\times \vert \xi_{n} \vert ^{-\frac{s}{n}} \int_{\mathbb{R}^{2}} e^{-ix\cdot\xi}e^{-i(t_{1}(x) \phi_{1}( \vert \xi_{1} \vert )+t_{2}(x)\phi_{2}( \vert \xi_{2} \vert )+ \cdots+t_{n}(x)\phi_{n}( \vert \xi_{n} \vert ))} \rho_{N}(x)h(x)\,dx, \end{aligned}$$
where \(\xi\in\mathbb{R}^{n}\) and \(h\in L^{2}(\mathbb{R}^{n})\). To prove (2.4) it is sufficient to prove that
$$ \Vert R_{N,\varPhi}g \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \Vert g \Vert _{L^{2}(\mathbb{R}^{n})}. $$
(2.5)
By duality, to prove (2.5) it suffices to show that
$$ \bigl\Vert R_{N,\varPhi}^{\prime}h \bigr\Vert _{L^{2}(\mathbb{R}^{n})}\leq C \Vert h \Vert _{L^{q^{\prime}}(\mathbb{R}^{n})}, $$
(2.6)
where \(\frac{1}{q}+\frac{1}{q^{\prime}}=1\). Thus, we have
$$ \bigl\Vert R_{N,\varPhi}^{\prime}h \bigr\Vert _{L^{2}(\mathbb{R}^{n})}^{2} = \int \bigl\vert R^{\prime}_{N,\varPhi}h(\xi) \bigr\vert ^{2}\,d\xi = \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} K_{N}(x,y)\rho_{N}(x) \rho_{N}(y)h(x)\overline{h(y)}\,dx\,dy, $$
(2.7)
where
$$ K_{N}(x,y)=K^{1}_{N}(x,y) K^{2}_{N}(x,y)\cdots K^{n}_{N}(x,y) $$
(2.8)
and
$$ K^{i}_{N}(x,y)= \int_{\mathbb{R}} \vert \xi_{i} \vert ^{-\frac{2s}{n}} e^{i(y_{i}-x_{i})\xi_{i}} e^{i(t_{i}(y)-t_{i}(x))\phi( \vert \xi_{i} \vert )} \psi_{N}(\xi_{i})^{2} \,d\xi_{i}, $$
(2.9)
where \(i=1,2,\ldots,n\) and \(N=1,2,\ldots\) . Since \(\frac{n}{4}\leq s <\frac{n}{2}\), we have \(\frac{1}{2}\leq \frac{2s}{n}<1\). Therefore, by Lemma 1.3, (2.9), and (2.8), we obtain
$$ \bigl\vert K_{N}(x,y) \bigr\vert \leq C \frac{1}{ \vert x_{1}-y_{1} \vert ^{1-\frac{2s}{n}}} \frac{1}{ \vert x_{2}-y_{2} \vert ^{1-\frac{2s}{n}}} \cdots\frac{1}{ \vert x_{n}-y_{n} \vert ^{1-\frac{2s}{n}}}. $$
(2.10)
We define
$$P_{i}f(x_{1},x_{2},\ldots, x_{n})= \int_{\mathbb{R}} \frac{1}{ \vert x_{i}-y_{i} \vert ^{1-\frac{2s}{n}}} f(x_{1}, \ldots,x_{i-1},y_{i}, x_{i+1},\ldots,x_{n}) \,dy_{i}, $$
\(i=1,2,\ldots,n\). Thus, by (2.7) and (2.10), we obtain
$$\begin{aligned} &\int \bigl\vert R_{N,\varPhi}^{\prime}h(\xi) \bigr\vert ^{2}\,d\xi \\ &\quad \leq C \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} \frac{1}{ \vert x_{1}-y_{1} \vert ^{1-\frac{2s}{n}}} \frac{1}{ \vert x_{2}-y_{2} \vert ^{1-\frac{2s}{n}}} \cdots \frac{1}{ \vert x_{n}-y_{n} \vert ^{1-\frac{2s}{n}}} \bigl\vert h(x) \bigr\vert \bigl\vert h(y) \bigr\vert \,dx\,dy \\ &\quad = C \int_{\mathbb{R}^{n}} \biggl( \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}}\frac{1}{ \vert x_{n}-y_{n} \vert ^{1-\frac{2s}{n}}} \frac{1}{ \vert x_{n-1}-y_{n-1} \vert ^{1-\frac{2s}{n}}}\cdots \frac {1}{ \vert x_{3}-y_{3} \vert ^{1-\frac{2s}{n}}} \\ &\qquad{}\times\frac {1}{ \vert x_{2}-y_{2} \vert ^{1-\frac{2s}{n}}} \biggl( \int\frac {1}{ \vert x_{1}-y_{1} \vert ^{1-\frac{2s}{n}}} \bigl\vert h(y_{1},y_{2}, \ldots, y_{n}) \bigr\vert \,dy_{1} \biggr) \,dy_{2}\,dy_{3}\cdots \,dy_{n} \biggr) \bigl\vert h(x) \bigr\vert \,dx \\ &\quad =C \int_{\mathbb{R}^{n}}P_{n}P_{n-1}\cdots P_{2}P_{1} \vert h \vert (x) \bigl\vert h(x) \bigr\vert \,dx. \end{aligned}$$
(2.11)
Invoking Hölder’s inequality, we get
$$ \int \bigl\vert R_{N,\varPhi}^{\prime}h(\xi) \bigr\vert ^{2}\,d\xi\leq C \bigl\Vert P_{n}P_{n-1}\cdots P_{2}P_{1} \vert h \vert \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \Vert h \Vert _{L^{q^{\prime}}(\mathbb{R}^{n})}. $$
(2.12)
Since \(q=\frac{2n}{n-2s}\), it follows that \(q^{\prime}=\frac {2n}{n+2s}\) and the fact \(\frac{1}{q}=\frac{1}{q^{\prime}}-\frac{2s}{n}\). Denote by \(I_{\sigma}\) the Riesz potential of order σ, which is defined by
$$I_{\sigma}(f) (u)= \int_{\mathbb{R}}\frac{f(v)}{ \vert u-v \vert ^{1-\sigma}}\,dv. $$
Applying the fact \(I_{s}\) is bounded from \(L^{q^{\prime}}(\mathbb{R})\) to \(L^{q}(\mathbb{R})\), we have
$$ \biggl( \int_{\mathbb{R}} \bigl\vert P_{j}h(x) \bigr\vert ^{q}\,dx_{j} \biggr) ^{1/q}\leq C \biggl( \int_{\mathbb{R}} \bigl\vert h(x) \bigr\vert ^{q^{\prime}} \,dx_{j} \biggr) ^{1/q^{\prime}}, $$
(2.13)
where \(j=1,2,\ldots,n\). By (2.13) and Minkowski’s inequality, we have
$$ \bigl\Vert P_{n}P_{n-1}\cdots P_{2}P_{1} \vert h \vert \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \leq C \Vert h \Vert _{L^{q^{\prime}}(\mathbb{R}^{n})}. $$
(2.14)
Therefore, (2.6) follows from (2.12) and (2.14). Now we complete the proof of Theorem 1.1.

3 The proof of Theorem 1.2

Assume that \(n\geq2\), \(\phi_{i}\) \((i=1,2,3,\ldots, n)\) satisfies (H1)–(H3) with \(m_{i,1}>1\), \(m_{i,2}>1\). For every \(m=(m_{1,2},m_{2,2},\ldots,m_{n,2})\) and \(2< q<4\), we will prove that inequality (1.9) holds if \(s=\frac{n}{2}-\frac{|m|}{4}+\frac{|m|}{q}-\frac{n}{q}\), where \(|m|=m_{1,2}+m_{2,2}+\cdots+m_{n,2}\). For \(i=1,2,3,\ldots, n\), let \(t_{i}(x)\) be a measurable function on \(\mathbb{R}^{n}\) with \(0< t_{i}(x)<1\). Denote \(t(x)=(t_{1}(x),t_{2}(x),\ldots,t_{n}(x))\), we set
$$S_{t(x),\varPhi}f(x)= \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1}(x) \phi_{1}(|\xi_{1}|)+t_{2}(x)\phi_{2}(|\xi_{2}|)+ \cdots+t_{n}(x)\phi_{n}(|\xi_{n}|))}\hat{f}(\xi)\,d\xi,\quad x\in\mathbb{R}^{n}, f\in\mathcal{S}\bigl(\mathbb{R}^{n}\bigr). $$
By linearizing the maximal operator, to prove the global estimate (1.9) it suffices to show that
$$ \Vert S_{t(x),\varPhi}f \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \Vert f \Vert _{H^{s}}=C \biggl( \int_{\mathbb{R}^{n}} \bigl(1+ \vert \xi \vert ^{2} \bigr)^{s} \bigl\vert \hat{f}(\xi) \bigr\vert ^{2}\,d\xi \biggr)^{1/2}. $$
(3.1)
Since \(s=\frac{n}{2}-\frac{|m|}{4}+\frac{|m|}{q}-\frac{n}{q} =n\frac{1}{2}-\frac{m_{1,2}+m_{2,2}+\cdots+m_{n,2}}{4} +\frac{m_{1,2}+m_{2,2}+\cdots+m_{n,2}}{q}-n\frac{1}{q} =:s_{1}+s_{2}+\cdots+s_{n}\), where \(s_{i}=\frac{1}{2}-\frac{m_{i,2}}{4}+ \frac{m_{i,2}}{q}-\frac{1}{q}\), \(i=1,2,\ldots,n\). Therefore, to prove (3.1) it suffices to prove that
$$ \Vert S_{t(x),\varPhi}f \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \biggl( \int_{\mathbb{R}^{2}} \bigl(1+ \vert \xi_{1} \vert ^{2}\bigr)^{s_{1}}\bigl(1+ \vert \xi_{2} \vert ^{2}\bigr)^{s_{2}} \vert \cdots \vert \bigl(1+ \vert \xi_{n} \vert ^{2}\bigr)^{s_{n}} \bigl\vert \hat{f}(\xi) \bigr\vert ^{2} \,d\xi \biggr)^{1/2}. $$
(3.2)
Let \(g(\xi)=(1+|\xi_{1}|^{2})^{\frac{s_{1}}{2}}(1+|\xi_{2}|^{2}) ^{\frac{s_{2}}{2}} \cdots(1+|\xi_{n}|^{2})^{\frac{s_{n}}{2}}\hat{f}(\xi)\), then we have
$$ S_{t(x),\varPhi}f(x)=R_{\varPhi}g(x), $$
(3.3)
where
$$\begin{aligned} R_{\varPhi}g(x)={}& \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1}(x) \phi_{1}( \vert \xi_{1} \vert )+t_{2}(x)\phi_{2}( \vert \xi_{2} \vert )+ \cdots+t_{n}(x)\phi_{n}( \vert \xi_{n} \vert ))} \bigl(1+ \vert \xi_{1} \vert ^{2}\bigr)^{-\frac{s_{1}}{2}}\bigl(1+ \vert \xi_{2} \vert ^{2}\bigr) ^{-\frac{s_{2}}{2}} \cdots\\ &{}\times\bigl(1+ \vert \xi_{n} \vert ^{2}\bigr)^{-\frac{s_{n}}{2}} g(\xi)\,d \xi. \end{aligned}$$
By (3.3), to prove (3.2) it is sufficient to show that
$$ \Vert R_{\varPhi}g \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \Vert g \Vert _{L^{2}(\mathbb{R}^{n})} $$
(3.4)
for g continuous and rapidly decreasing at infinity. We take a real-valued function \(\rho\in C_{0}^{\infty}(\mathbb{R}^{n})\) such that \(\rho(x)=1\) if \(|x|\leq1\) and \(\rho(x)=0\) if \(|x|\geq2\). And we choose a real-valued function \(\psi\in C_{0}^{\infty}(\mathbb{R})\) such that \(\psi(x)=1\) if \(|x|\leq1\) and \(\psi(x)=0\) if \(|x|\geq2\), and set \(\sigma(\xi )=\psi(\xi_{1})\psi(\xi_{2})\cdots\psi(\xi_{n})\) for \(\xi\in \mathbb{R}^{n}\). For \(N=1,2,3,\ldots\) , we set \(\rho_{N}(x)=\rho (\frac{x}{N})\) and \(\sigma_{N}(\xi)=\sigma(\frac{\xi}{N})\). For \(x\in\mathbb{R}^{n}\), \(g\in L^{2}(\mathbb{R}^{n})\), and \(N=1,2,3,\ldots\) , we define
$$\begin{aligned} R_{N,\varPhi}g(x)={}& \rho_{N}(x) \int_{\mathbb{R}^{n}} e^{ix\cdot\xi}e^{i(t_{1}(x) \phi_{1}( \vert \xi_{1} \vert )+t_{2}(x)\phi_{2}( \vert \xi_{2} \vert )+ \cdots+t_{n}(x)\phi_{n}( \vert \xi_{n} \vert ))} \bigl(1+ \vert \xi_{1} \vert ^{2}\bigr)^{-\frac{s_{1}}{2}}\bigl(1+ \vert \xi_{2} \vert ^{2}\bigr) ^{-\frac{s_{2}}{2}} \\ &{}\times \cdots\times\bigl(1+ \vert \xi_{n} \vert ^{2} \bigr)^{-\frac{s_{n}}{2}} \sigma_{N}(\xi)g(\xi)\,d\xi. \end{aligned}$$
The adjoint of \(R_{N,\varPhi}\) is given by
$$\begin{aligned} R^{\prime}_{N,\varPhi}h(\xi)={}& \sigma_{N}(\xi) \bigl(1+ \vert \xi_{1} \vert ^{2}\bigr)^{-\frac{s_{1}}{2}} \bigl(1+ \vert \xi_{2} \vert ^{2}\bigr) ^{-\frac{s_{2}}{2}} \cdots\bigl(1+ \vert \xi_{n} \vert ^{2}\bigr)^{-\frac{s_{n}}{2}} \int_{\mathbb{R}^{2}} e^{-ix\cdot\xi}e^{-it_{1}(x) \phi_{1}( \vert \xi_{1} \vert )} \\ &{}\times e^{-i(t_{2}(x)\phi_{2}( \vert \xi_{2} \vert )+ \cdots+t_{n}(x)\phi_{n}( \vert \xi_{n} \vert ))}\rho_{N}(x)h(x)\,dx, \end{aligned}$$
where \(\xi\in\mathbb{R}^{n}\) and \(h\in L^{2}(\mathbb{R}^{n})\). To prove (3.4) it suffices to prove that
$$ \Vert R_{N,\varPhi}g \Vert _{L^{q}(\mathbb{R}^{n})}\leq C \Vert g \Vert _{L^{2}(\mathbb{R}^{n})}. $$
(3.5)
By duality, to prove (3.5) it is sufficient to show that
$$ \bigl\Vert R_{N,\varOmega}^{\prime}h \bigr\Vert _{L^{2}(\mathbb{R}^{n})}\leq C \Vert h \Vert _{L^{q^{\prime}}(\mathbb{R}^{n})}, $$
(3.6)
where \(\frac{1}{q}+\frac{1}{q^{\prime}}=1\). Thus, we have
$$ \bigl\Vert R_{N,\varPhi}^{\prime}h \bigr\Vert _{L^{2}(\mathbb{R}^{n})}^{2}= \int \bigl\vert R^{\prime}_{N,\varPhi}h(\xi) \bigr\vert ^{2}\,d\xi = \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} K_{N}(x,y)\rho_{N}(x) \rho_{N}(y)h(x)\overline{h(y)}\,dx\,dy, $$
(3.7)
where
$$ K_{N}(x,y)=K^{1}_{N}(x,y) K^{2}_{N}(x,y)\cdots K^{n}_{N}(x,y), $$
(3.8)
and
$$ K^{i}_{N}(x,y)= \int_{\mathbb{R}}\bigl(1+\xi_{i}^{2} \bigr)^{-s_{i}} e^{i(y_{i}-x_{i})\xi_{i}} e^{i(t_{i}(y)-t_{i}(x))\phi(|\xi_{i}|)} \psi_{N}( \xi_{i})^{2}\,d\xi_{i}, $$
(3.9)
where \(i=1,2,\ldots,n\) and \(N=1,2,\ldots\) . Denote \(\alpha_{i}=2s_{i}\), since \(s_{i}=\frac{1}{2}-\frac{m_{i,2}}{4}+ \frac{m_{i,2}}{q}-\frac{1}{q}\), \(i=1,2,\ldots,n\), and \(2< q<4\), it follows that \(\frac{1}{2}<\alpha_{i}<\frac{m_{i,2}}{2}\), \(i=1,2,\ldots,n\). Therefore, by (3.9) and Lemma 1.4, we obtain
$$ \bigl\vert K^{i}_{N}(x,y) \bigr\vert \leq C\frac{1}{ \vert x_{i}-y_{i} \vert ^{\beta_{i}}}, $$
(3.10)
where \(\beta_{i}=\frac{\alpha_{i}+\frac{m_{i,2}}{2}-1}{ m_{i,2}-1}\). Denote \(\sigma_{i}=1-\beta_{i}\), we define
$$P_{i}f(x_{1},x_{2},\ldots, x_{n})= \int_{\mathbb{R}} \frac{1}{ \vert x_{i}-y_{i} \vert ^{1-\sigma_{i}}} f(x_{1}, \ldots,x_{i-1},y_{i}, x_{i+1},\ldots,x_{n}) \,dy_{i}, $$
\(i=1,2,\ldots,n\). Thus, by (3.7), (3.8), and (3.10), we obtain
$$\begin{aligned} &\int \bigl\vert R_{N,\varPhi}^{\prime}h(\xi) \bigr\vert ^{2}\,d\xi \\ &\quad \leq C \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} \frac{1}{ \vert x_{1}-y_{1} \vert ^{1-\sigma_{1}}} \frac{1}{ \vert x_{2}-y_{2} \vert ^{1-\sigma_{2}}} \cdots \frac{1}{ \vert x_{n}-y_{n} \vert ^{1-\sigma_{n}}} \bigl\vert h(x) \bigr\vert \bigl\vert h(y) \bigr\vert \,dx\,dy \\ &\quad = C \int_{\mathbb{R}^{n}} \biggl( \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}}\frac{1}{ \vert x_{n}-y_{n} \vert ^{1-\sigma_{n}}} \frac{1}{ \vert x_{n-1}-y_{n-1} \vert ^{1-\sigma_{n-1}}}\cdots \frac {1}{ \vert x_{3}-y_{3} \vert ^{1-\sigma_{3}}} \\ &\qquad{}\times\frac {1}{ \vert x_{2}-y_{2} \vert ^{1-\sigma_{2}}} \biggl( \int\frac {1}{ \vert x_{1}-y_{1} \vert ^{1-\sigma_{1}}} \bigl\vert h(y_{1},y_{2}, \ldots, y_{n}) \bigr\vert \,dy_{1} \biggr) \,dy_{2}\,dy_{3}\cdots \,dy_{n} \biggr) \bigl\vert h(x) \bigr\vert \,dx \\ &\quad = C \int_{\mathbb{R}^{n}}P_{n}P_{n-1}\cdots P_{2}P_{1} \vert h \vert (x) \bigl\vert h(x) \bigr\vert \,dx. \end{aligned}$$
(3.11)
Invoking Hölder’s inequality, we get
$$ \int \bigl\vert R_{N,\varPhi}^{\prime}h(\xi) \bigr\vert ^{2}\,d\xi\leq C \bigl\Vert P_{n}P_{n-1}\cdots P_{2}P_{1} \vert h \vert \bigr\Vert _{L^{q}(\mathbb{R}^{n})} \Vert h \Vert _{L^{q^{\prime}}(\mathbb{R}^{n})}. $$
(3.12)
Since \(\beta_{i}=\frac{\alpha_{i}+\frac{m_{i,2}}{2}-1}{m_{i,2}-1}\) and \(\alpha_{i}=2s_{i}\), \(s_{i}=\frac{1}{2}-\frac{m_{i,2}}{4}+ \frac{m_{i,2}}{q}-\frac{1}{q}\), \(i=1,2,\ldots,n\). It follows that \(\beta_{i}=\frac{2}{q}\) and \(\sigma_{i}=1-\beta_{i}=1-\frac {2}{q}\), \(\frac{1}{q}=\frac{1}{q^{\prime}}-\sigma_{i}\). Thus, estimate (3.6) follows from (3.12) and estimate (2.14) in the proof of Theorem 1.1. Now we complete the proof of Theorem 1.2.

4 The proof of Lemma 1.4

To prove Lemma 1.4, we need to present the following lemma.

Lemma 4.1

(see [33], pp. 309–312)

Assume that \(a< b\) and set \(I=[a,b]\). Let \(F\in C^{\infty}(I)\) be real-valued and assume that \(\psi\in C^{\infty}(I)\).
  1. (i)
    Assume that \(|F^{\prime}(x)|\geq\lambda>0\) for \(x\in I\) and that \(F^{\prime}\) is monotonic on I. Then
    $$\biggl\vert \int_{a}^{b}e^{iF(x)}\psi(x)\,dx \biggr\vert \leq C\frac{1}{\lambda }\biggl\{ \bigl\vert \psi(b) \bigr\vert + \int_{a}^{b} \bigl\vert \psi^{\prime}(x) \bigr\vert \,dx\biggr\} , $$
    where C does not depend on F, ψ, or I.
     
  2. (ii)
    Assume that \(|F^{\prime\prime}(x)|\geq\lambda>0\) for \(x\in I\). Then
    $$\biggl\vert \int_{a}^{b}e^{iF(x)}\psi(x)\,dx \biggr\vert \leq C\frac{1}{\lambda ^{1/2}}\biggl\{ \bigl\vert \psi(b) \bigr\vert + \int_{a}^{b} \bigl\vert \psi^{\prime}(x) \bigr\vert \,dx\biggr\} , $$
    where C does not depend on F, ψ, or I.
     

Proof of Lemma 1.4

By conditions (H1) and (H2), there exist positive constants \(C_{i}\) (\(i=1,2,\ldots,6\)) so that for \(r\geq1\) and \(m_{2}>1\) such that
$$ C_{1}r^{m_{2}-1}\leq \bigl\vert \phi^{\prime}(r) \bigr\vert \leq C_{2}r^{m_{2}-1} \quad\mbox{and}\quad \bigl\vert \phi^{\prime\prime}(r) \bigr\vert \geq C_{3}r^{m_{2}-2}, $$
(4.1)
and for \(0< r<1\) and \(m_{1}>1\) such that
$$ C_{4}r^{m_{1}-1}\leq \bigl\vert \phi^{\prime}(r) \bigr\vert \leq C_{5}r^{m_{1}-1} \quad\mbox{and}\quad \bigl\vert \phi^{\prime\prime}(r) \bigr\vert \geq C_{6}r^{m_{1}-2}. $$
(4.2)
Set
$$J= \int_{\mathbb{R}} \frac{e^{i(d\phi(|\xi|)-x\xi)}}{(1+\xi ^{2})^{\frac{\alpha}{2}}} \mu\biggl(\frac{\xi}{N} \biggr)\,d\xi. $$
To prove Lemma 1.4, it suffices to show that there exists a constant C such that for \(x\in\mathbb{R}\setminus\{ 0\}\), \(\beta=\frac{\alpha+\frac{m_{2}}{2}-1}{m_{2}-1}\) and \(N\in \Bbb {N}\),
$$ \vert J \vert \leq C\frac{1}{ \vert x \vert ^{\beta}}, $$
(4.3)
where C depends only on α, \(m_{1}\), \(m_{2}\), \(C_{i}\) \((i=1,2,\ldots,6)\), and μ.
Without loss of generality, we may assume \(\xi, d>0\). Denote \(\psi(\xi)={(1+\xi^{2})^{-\frac{\alpha}{2}}} \mu(\frac{\xi}{N})\), then we have
$$ \max_{\xi\geq0} \bigl\vert \psi(\xi) \bigr\vert + \int_{ 0}^{\infty} \bigl\vert \psi ^{\prime}( \xi) \bigr\vert \,d\xi\leq C. $$
(4.4)
In fact, since \(\mu\in C_{0}^{\infty}(\mathbb{R})\) and \(\frac {1}{2}\leq\alpha\leq\frac{m_{2}}{2}\), we get
$$ \max_{\xi\geq0} \bigl\vert \psi(\xi) \bigr\vert \leq C. $$
(4.5)
Noting that
$$ \psi^{\prime}(\xi) =-\alpha\xi{\bigl(1+\xi^{2} \bigr)^{-\frac{\alpha}{2}-1}} \mu \biggl(\frac{\xi}{N} \biggr) +{\bigl(1+ \xi^{2}\bigr)^{-\frac{\alpha}{2}}}\frac{1}{N} \mu^{\prime} \biggl(\frac{\xi}{N} \biggr), $$
(4.6)
we have
$$\begin{aligned} \int_{0}^{\infty} \bigl\vert \psi^{\prime}(\xi) \bigr\vert \,d\xi\leq{}&\alpha \int_{0}^{\infty}\xi{\bigl(1+\xi^{2} \bigr)^{-\frac{\alpha}{2}-1}} \biggl\vert \mu \biggl(\frac{\xi}{N} \biggr) \biggr\vert \,d\xi \\ &{}+ \int_{0}^{\infty}{\bigl(1+\xi^{2} \bigr)^{-\frac{\alpha}{2}}}\frac{1}{N} \biggl\vert \mu^{\prime} \biggl( \frac{\xi}{N} \biggr) \biggr\vert \,d\xi \\ =:{}&G_{1}+G_{2}. \end{aligned}$$
(4.7)
Since \(\mu\in C_{0}^{\infty}(\mathbb{R})\) and \(\frac{1}{2}\leq \alpha\leq\frac{m_{2}}{2}\), we obtain
$$ G_{1}\leq C \int_{0}^{\infty}\xi{\bigl(1+\xi^{2} \bigr)^{-\frac{\alpha }{2}-1}}\,d\xi =C \int_{0}^{\infty}{\bigl(1+\xi^{2} \bigr)^{-\frac{\alpha}{2}-1}}\,d\bigl(1+\xi ^{2}\bigr) =C $$
(4.8)
and
$$ G_{2}\leq C \int_{0}^{\infty}\frac{1}{N} \biggl\vert \mu^{\prime} \biggl(\frac{\xi}{N} \biggr) \biggr\vert \,d\xi\leq C. $$
(4.9)
By (4.7), (4.8), and (4.9), we get
$$ \int_{ 0}^{\infty} \bigl\vert \psi^{\prime}(\xi) \bigr\vert \,d\xi\leq C. $$
(4.10)
Therefore, (4.4) follows from (4.5) and (4.10).

To estimate (4.3), we choose a positive constant M such that \(M=\max\{(\frac{1}{\delta})^{m_{2}-1},2C_{5},2\}\), where δ is a small positive constant such that \(\delta ^{m_{2}-1}C_{2}\leq\frac{1}{2}\). Below, we show (4.3) by dividing two cases \(|x|\geq M\) and \(|x|< M\).

Case (I): \(|x|\geq M\). Let \(F(\xi)=d\phi(\xi)-x\xi\), we have
$$F^{\prime}(\xi)=d\phi^{\prime}(\xi)-x,\qquad F^{\prime\prime}(\xi )=d \phi^{\prime\prime}(\xi). $$
Denote \(\rho=(\frac{|x|}{d})^{\frac{1}{m_{2}-1}}\), then we have \(\delta\rho\geq1\). In fact, noting that \(|x|\geq(\frac{1}{\delta})^{m_{2}-1}\), \(0< d<1\), \(m_{2}>1\), and \(\frac{|x|}{d}>|x|\), it follows that \(\delta\rho >\delta|x|^{\frac{1}{m_{2}-1}}\geq1\). We choose a large positive constant λ such that \(\lambda\geq\max\{(\frac{2}{C_{1}}) ^{\frac{1}{m_{2}-1}},\delta\}\). Denote
$$I_{1}=[0,\delta\rho],\qquad I_{2}=[\delta\rho,\lambda\rho],\qquad I_{3}=[\lambda\rho,\infty). $$
Thus, we obtain
$$\begin{aligned} \vert J \vert &= \biggl\vert \int_{0}^{\infty}e^{iF(\xi)}\psi(\xi)\,d\xi \biggr\vert \leq \sum_{j=1}^{3} \biggl\vert \int_{I_{j}}e^{iF(\xi)}\psi(\xi)\,d\xi \biggr\vert =:\sum _{j=1}^{3} J_{j}. \end{aligned}$$
(4.11)
Firstly, we estimate \(J_{1}\). We will show that the following estimate holds:
$$ \bigl\vert F^{\prime}(\xi) \bigr\vert \geq \frac{ \vert x \vert }{2},\quad \xi\in[0,\delta \rho]. $$
(4.12)
Now we divide the verification of (4.12) into two cases according to the value of ξ.
Case (I-a): \(\xi\in[0,1)\). Since \(m_{1}>1\) and \(0< d<1\), we have
$$ d \bigl\vert \phi^{\prime}(\xi) \bigr\vert \leq C_{5} \,d \xi^{m_{1}-1}\leq C_{5}\leq \frac {M}{2}\leq\frac{ \vert x \vert }{2}. $$
(4.13)
By (4.13), if \(\xi\in[0,1)\), we get
$$ \bigl\vert F^{\prime}(\xi) \bigr\vert \geq \vert x \vert -d \bigl\vert \phi^{\prime}(\xi) \bigr\vert \geq \frac{ \vert x \vert }{2}. $$
(4.14)
Case (I-b): \(\xi\in[1,\delta\rho]\). Since \(m_{2}>1\), we have
$$ d \bigl\vert \phi^{\prime}(\xi) \bigr\vert \leq C_{2} \,d \xi^{m_{2}-1}\leq C_{2}\,d \delta ^{m_{2}-1}\frac{ \vert x \vert }{d}\leq C_{2} \delta^{m_{2}-1} \vert x \vert \leq\frac{ \vert x \vert }{2}. $$
(4.15)
By (4.15), we get
$$ \bigl\vert F^{\prime}(\xi) \bigr\vert \geq \vert x \vert -d \bigl\vert \phi^{\prime}(\xi) \bigr\vert \geq \frac{ \vert x \vert }{2}. $$
(4.16)
Therefore (4.12) follows from (4.14) and (4.16). Since \(\phi^{\prime}\) is monotonic on \(\mathbb{R}^{+}\) by condition (H3) and \(d>0\), it follows that \(F^{\prime}\) is monotonic on \(\xi\in I_{1}\). Thus, by (i) of Lemma 4.1 and estimate (4.12), (4.4), we have
$$ \vert J_{1} \vert \leq C\frac{1}{ \vert x \vert } \leq C \frac{1}{ \vert x \vert ^{\beta}}, $$
(4.17)
where we use \(|x|\geq2\) and the fact \(\frac{1}{2}\leq\beta\leq1\). Next we prove estimate \(J_{3}\). Since \(\xi\geq\lambda(\frac{|x|}{d})^{\frac{1}{m_{2}-1}}>1\) and \(\lambda\geq(\frac{2}{C_{1}})^{\frac{1}{m_{2}-1}}\),
$$d \bigl\vert \phi^{\prime}(\xi) \bigr\vert \geq C_{1} \,d \xi^{m_{2}-1}\geq C_{1} \,d\lambda ^{m_{2}-1} \frac{ \vert x \vert }{d}\geq2 \vert x \vert , $$
it follows that
$$ \bigl\vert F^{\prime}(\xi) \bigr\vert \geq2 \vert x \vert - \vert x \vert = \vert x \vert ,\quad \xi\in[\lambda\rho ,\infty). $$
(4.18)
Thus, by (i) of Lemma 4.1 and estimate (4.18), (4.4), we have
$$ \vert J_{3} \vert \leq C\frac{1}{ \vert x \vert } \leq C \frac{1}{ \vert x \vert ^{\beta}}, $$
(4.19)
where we use \(|x|\geq2\) and the fact \(\frac{1}{2}\leq\beta\leq1\). Now, we give estimate \(J_{2}\). Since \(\xi\in I_{2}\), we have \(|\xi|\geq1\). By (4.1), we obtain
$$ \bigl\vert F^{\prime\prime}(\xi) \bigr\vert \geq d \bigl\vert \phi^{\prime\prime}(\xi) \bigr\vert \geq C_{3}\,d \xi^{m_{2}-2} \geq C_{3}\,d \biggl(\frac{ \vert x \vert }{d} \biggr)^{\frac {m_{2}-2}{m_{2}-1}}. $$
(4.20)
We first prove that the following estimate holds:
$$ \max_{I_{2}} \vert \psi \vert + \int_{I_{2}} \bigl\vert \psi^{\prime} \bigr\vert \,d\xi \leq C \biggl(\frac{ \vert x \vert }{d} \biggr)^{-\frac{\alpha}{m_{2}-1}}. $$
(4.21)
In fact, since \(\mu\in C_{0}^{\infty}(\mathbb{R})\) and \(\frac {1}{2}\leq\alpha\leq\frac{m_{2}}{2}\), we get
$$ \max_{\xi\in A_{2}} \bigl\vert \psi(\xi) \bigr\vert \leq C(\delta\rho)^{-\alpha }=C\delta^{-\alpha}(\rho)^{-\alpha}= C \delta^{-\alpha} \biggl(\frac{ \vert x \vert }{d} \biggr)^{-\frac{\alpha}{m_{2}-1}}. $$
(4.22)
By (4.6), we have
$$\begin{aligned} \int_{A_{2}} \bigl\vert \psi^{\prime}(\xi) \bigr\vert \,d\xi &\leq\alpha \int_{A_{2}}\xi{\bigl(1+\xi^{2}\bigr)^{-\frac{\alpha}{2}-1}} \biggl\vert \mu \biggl(\frac{\xi}{N} \biggr) \biggr\vert \,d\xi+ \int_{A_{2}}{\bigl(1+\xi^{2}\bigr)^{-\frac{\alpha}{2}}} \frac{1}{N} \biggl\vert \mu^{\prime} \biggl(\frac{\xi}{N} \biggr) \biggr\vert \,d\xi \\ &=:L_{1}+L_{2}. \end{aligned}$$
(4.23)
Since \(\mu\in C_{0}^{\infty}(\mathbb{R})\) and \(\frac{1}{2}\leq \alpha\leq\frac{m_{2}}{2}\), we obtain
$$ L_{1}\leq C \int_{A_{2}}\xi{\bigl(1+\xi^{2}\bigr)^{-\frac{\alpha}{2}-1}} \,d\xi \leq C \int_{\delta\rho}^{\lambda\rho}\xi^{-\alpha-1}\,d\xi =C \biggl( \frac{ \vert x \vert }{d} \biggr)^{-\frac{\alpha}{m_{2}-1}} $$
(4.24)
and
$$ L_{2}\leq C (\delta\rho)^{-\alpha} \int_{A_{2}}\frac{1}{N} \biggl\vert \mu^{\prime} \biggl(\frac{\xi}{N} \biggr) \biggr\vert \,d\xi\leq C \biggl( \frac{ \vert x \vert }{d} \biggr)^{-\frac{\alpha}{m_{2}-1}}. $$
(4.25)
By (4.23), (4.24), and (4.25), we get
$$ \int_{ 0}^{\infty} \bigl\vert \psi^{\prime}(\xi) \bigr\vert \,d\xi\leq C \biggl(\frac { \vert x \vert }{d} \biggr)^{-\frac{\alpha}{m_{2}-1}}. $$
(4.26)
Therefore, (4.21) follows from (4.22) and (4.26). Thus, by (ii) of Lemma 4.1 and estimate (4.20), (4.21), we have
$$ \vert J_{2} \vert \leq d^{-\frac{1}{2}} \biggl( \frac{ \vert x \vert }{d} \biggr)^{-\frac {m_{2}-2}{2(m_{2}-1)}} \biggl(\frac{ \vert x \vert }{d} \biggr)^{-\frac{\alpha}{m_{2}-1}} =C\frac{d^{\frac{\alpha-\frac{1}{2}}{m_{2}-1}}}{ \vert x \vert ^{\frac{\alpha+\frac{m_{2}}{2}-1}{m_{2}-1}}}\leq C\frac{1}{ \vert x \vert ^{\beta}}. $$
(4.27)
Here in the last inequality we use the fact \(\frac{\alpha-\frac{1}{2}}{m_{2}-1}\geq0\) and \(0< d<1\). Therefore, for \(|x|\geq M\), by estimates (4.11), (4.17), (4.19), and (4.27), it follows (4.3).

Case (II): \(|x|< M\). Now we divide the verification of (4.3) into three cases according to the value of α for \(|x|< M\).

Case (II-a): \(\alpha>1\). Since \(\mu\in C_{0}^{\infty}(\mathbb{R})\) and \(\alpha>1\), we get
$$ \vert J \vert = \biggl\vert \int_{0}^{\infty} \frac{e^{i(d\phi( \vert \xi \vert )-x\xi )}}{(1+\xi^{2})^{\frac{\alpha}{2}}} \mu\biggl( \frac{\xi}{N}\biggr)\,d\xi \biggr\vert \leq C \int_{0}^{\infty} \frac {1}{{(1+\xi^{2})^{\frac{\alpha}{2}}}}\,d\xi\leq C. $$
(4.28)
Noting that \(|x|< M\) and \(\frac{1}{2}\leq\beta\leq1\), by (4.28), we have
$$\vert J \vert \leq C=C \vert x \vert ^{\beta}\frac{1}{ \vert x \vert ^{\beta}} \leq CM^{\beta}\frac {1}{ \vert x \vert ^{\beta}}=C\frac{1}{ \vert x \vert ^{\beta}}, $$
which follows (4.3).
Case (II-b): \(\frac{1}{2}\leq\alpha<1\). By the mean value theorem, when \(\frac{1}{2}\leq\alpha<1\), we have
$$ 0< \bigl(1+\xi^{2}\bigr)^{\frac{\alpha}{2}}- \xi^{\alpha} =\bigl(1+\xi^{2}\bigr)^{\frac{\alpha}{2}}-\bigl( \xi^{2}\bigr)^{\frac{\alpha}{2}} \leq\frac{\alpha}{2}\bigl( \xi^{2}\bigr)^{\frac{\alpha}{2}-1}\leq\xi ^{\alpha-2}. $$
(4.29)
By (4.29), we obtain
$$ \frac{1}{\xi^{\alpha}}-\frac{1}{(1+\xi^{2})^{\frac{\alpha}{2}}} =O \biggl(\frac{1}{\xi^{\alpha+2}} \biggr),\quad \xi\rightarrow\infty. $$
(4.30)
Noting that \(\frac{1}{2}\leq\alpha<1\), by (4.30), we have
$$ \int_{0}^{\infty} \biggl\vert \frac{1}{\xi^{\alpha}}- \frac{1}{(1+\xi^{2})^{\frac{\alpha}{2}}} \biggr\vert \,d\xi \leq C $$
(4.31)
and
$$\begin{aligned} \vert J \vert = \biggl\vert \int_{0}^{\infty} \frac{e^{i(d\phi( \vert \xi \vert )-x\xi )}}{(1+\xi^{2})^{\frac{\alpha}{2}}} \mu\biggl( \frac{\xi}{N}\biggr)\,d\xi \biggr\vert \leq{}& \biggl\vert \int_{0}^{\infty}e^{i(d\phi( \vert \xi \vert )-x\xi)} \biggl( \frac{1}{(1+\xi^{2})^{\frac{\alpha}{2}}} -\frac{1}{\xi^{\alpha}} \biggr)\mu\biggl(\frac{\xi}{N}\biggr) \,d\xi \biggr\vert \\ &{} + \biggl\vert \int_{0}^{\infty}e^{i(d\phi( \vert \xi \vert )-x\xi)} \frac{1}{\xi^{\alpha}}\mu \biggl(\frac{\xi}{N}\biggr)\,d\xi \biggr\vert \\ ={}& :K_{1}+K_{2}. \end{aligned}$$
By (4.31), we have
$$ \vert K_{1} \vert \leq C=C \vert x \vert ^{\beta}\frac{1}{ \vert x \vert ^{\beta}}\leq CM^{\beta}\frac {1}{ \vert x \vert ^{\beta}}=C \frac{1}{ \vert x \vert ^{\beta}}. $$
(4.32)
By Lemma 1.3, we obtain
$$ \vert K_{2} \vert \leq C\frac{1}{ \vert x \vert ^{1-\alpha}}. $$
(4.33)
Noting that \(\frac{1}{|x|}>\frac{1}{M}\), and the fact \(\beta \geq1-\alpha\), it follows from \(\frac{1}{2}\leq\alpha<1\), \(\frac {1}{2}\leq\beta\leq1 \) that
$$ \vert x \vert ^{1-\alpha}= \vert x \vert ^{\beta} \vert x \vert ^{1-\alpha-\beta} = \vert x \vert ^{\beta} \biggl(\frac{1}{ \vert x \vert } \biggr)^{\beta-(1-\alpha)}\geq C \vert x \vert ^{\beta}. $$
(4.34)
Therefore, by (4.33) and (4.34), we have
$$ \vert K_{2} \vert \leq C\frac{1}{ \vert x \vert ^{\beta}}. $$
(4.35)
Hence, (4.3) holds from (4.32) and (4.35).
Case (II-c): \(\alpha=1\). From the proof of Lemma 1.3, noting that \(M\geq2\), we may get
$$ \vert J \vert \leq C \log\biggl(\frac{1}{ \vert x \vert }\biggr) \quad\text{if } 0< \vert x \vert \leq \frac{1}{2} $$
(4.36)
and
$$ \vert J \vert \leq C \quad\text{if } \frac{1}{2}< \vert x \vert < M. $$
(4.37)
By (4.36) and \(\frac{1}{2}\leq\beta\leq1\), for \(0<|x|\leq\frac{1}{2}\), we have
$$ \vert J \vert \leq C \log\biggl(\frac{1}{ \vert x \vert }\biggr)\leq C\frac{1}{ \vert x \vert ^{\beta}}. $$
(4.38)
By (4.37), for \(\frac{1}{2}<|x|<M\), we have
$$ \vert J \vert \leq C=C \vert x \vert ^{\beta} \frac{1}{ \vert x \vert ^{\beta}}\leq CM^{\beta}\frac {1}{ \vert x \vert ^{\beta}}=C\frac{1}{ \vert x \vert ^{\beta}}. $$
(4.39)
Thus, for \(\alpha=1\), \(|x|< M\), by (4.38) and (4.39), we get
$$\vert J \vert \leq C\frac{1}{ \vert x \vert ^{\beta}}, $$
which is just estimate (4.3).

Summing up all the above estimates, we complete the proof of estimate (4.3) and finish the proof of Lemma 1.4. □

5 Conclusion

In this paper, by linearizing the maximal operator and duality methods, and applying the results of Lemma 1.3 and Lemma 1.4, we obtain the maximal global \(L^{q}\) inequalities (1.8) and (1.9) for multiparameter oscillatory integral \(S_{t,\varPhi}\). These estimates are apparently good extensions to maximal global \(L^{q}\) inequalities (1.6) and (1.7) for the multiparameter fractional Schrödinger equation in [32].

Declarations

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.

Funding

The work is supported by NSFC (No. 11661061, No. 11561062, No. 11761054), Inner Mongolia University scientific research projects (No. NJZZ16234, NJZY17289).

Authors’ contributions

YN participated in the design of the study and in the discussions of all results. YX participated in the discussions of all results. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

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.

Authors’ Affiliations

(1)
Faculty of Mathematics, Baotou Teachers’ College of Inner Mongolia University of Science and Technology, Baotou, P.R. China

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