L¹ Boundedness of a class of rough Fourier integral operators

Abstract

In this note, we consider a class of Fourier integral operators with rough amplitudes and rough phases. When the index of symbols in some range, we prove that they are bounded on L^1 and construct an example to show that this result is sharp in some cases. This result is a generalization of the corresponding theorems of Kenig-Staubach and Dos Santos Ferreira-Staubach.

Keywords

Fourier integral operators / amplitude / phase

1 Introduction

In this note we consider a Fourier integral operator defined by

T_{\varphi ,a}f(x)={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi )\hat{f}(\xi )\mathrm{d}\xi ,

where a is called the symbol and \varphi is the phase. In particular, it is a pseudo-differential operator if \varphi (x,\xi )=x\cdot \xi.

Fourier integral operators have been used widely in the theory of partial differential equations and micro-local analysis. For example, the solution of an initial value problem for a variable coefficient hyperbolic equation can be well approximated by an FIO of the initial value. A systematic study of these operators was initiated by Hörmander. One can also see [6, 10, 11].

For the symbols and phases in Fourier integral operators, some most important definitions are as follows.

a belongs to the Hörmander class S_{\rho ,\delta }^m(m\in \mathbb{R},0⩽\rho ,\delta ⩽1) if it satisfies

for all multi-indices \alpha ,\beta.

For the phase \varphi, if it is homogeneous of degree 1 in the frequency variable \xi and satisfies

\underset{(x,\xi )\in {\mathbb{R}}^n\times {\mathbb{R}}^n\setminus \{0\}}{sup}|\xi {|}^{-1+|\alpha |}|{\mathrm{\partial }}_x^{\beta }{\mathrm{\partial }}_{\xi }^{\alpha }\varphi (x,\xi )|⩽C_{\alpha ,\beta }

for all multi-indices \alpha ,\beta with |\alpha |+|\beta |⩾k, then we say that \varphi \in {\mathrm{\Phi }}^k. In general, we can assume that k=2.

Furthermore, a real-valued function \varphi \in C^2({\mathbb{R}}^n\times ({\mathbb{R}}^n\setminus \{0\})) satisfies the strong non-degeneracy (SND) condition, if there exists a constant \lambda >0 such that

\left|\mathrm{d}\mathrm{e}\mathrm{t}\frac{{\mathrm{\partial }}^2\varphi (x,\xi )}{\mathrm{\partial }{x}_j\mathrm{\partial }{\xi }_k}\right|⩾\lambda .

For example, \varphi (x,\xi )=x\cdot \xi +|\xi | is in {\mathrm{\Phi }}^2 and satisfies the strong non-degeneracy condition. It matches the wave operator and is a typical Fourier integral operator.

The boundedness of Fourier integral operators on Lebesgue space has been extensively studied and there are numerous results, such as [1-4, 7-11, 15]. One can find more details in a survey of Dos Santos Ferreira and Staubach [5]. Especially, for the best result on L^2 so far, when a\in S_{\rho ,\delta }^m and \varphi \in {\mathrm{\Phi }}^2 satisfying the SND condition, if m⩽\mathrm{m}\mathrm{i}\mathrm{n}\{0,\frac{n}{2}(\rho -\delta )\}, , or , \delta =1, the Fourier integral operator is bounded on L^2. One can see [5, Theorem 2.7]. The bound on m is sharp. Rodino in [17] constructed an a\in S_{\rho ,1}^{n(\rho -1)/2} such that the pseudo-differential operator is unbounded on L^2.

For the endpoint estimate, Seeger et al., in [18] proved that when a\in S_{1,0}^{\frac{1-n}{2}} and \varphi \in {\mathrm{\Phi }}^2 satisfying the SND condition, the Fourier integral operator is local H^1-L^1 bounded. And they say that when , the corresponding operator is bounded on L^1. Their result is generalized to global boundedness. One can see [5]. Tao [20] subsequent proved that when a\in S_{1,0}^{\frac{1-n}{2}} and \varphi \in {\mathrm{\Phi }}^2 satisfying the SND condition, the operator is also of weak type (1,1). In the same paper, Tao pointed out the operator is not bounded on L^1.

Kenig and Staubach in [12] defined the rough Hörmander class L^{\mathrm{\infty }}{S}_{\rho }^m(0⩽\rho ⩽1). It consists all functions a that satisfy

For a\in L^{\mathrm{\infty }}{S}_{\rho }^m, Kenig−Staubach proved the pseudo-differential operator is bounded on L^1 if and on L^{\mathrm{\infty }} if .

In the maximal wave operator and maximal sphere average operator, the phase is not smooth in the spatial variable x. In these cases, \varphi (x,\xi )=x\cdot \xi +t(x)|\xi |,t\in L^{\mathrm{\infty }}. Dos Santos Ferreira and Staubach in [5] introduced the class L^{\mathrm{\infty }}{\mathrm{\Phi }}^2 and rough non-degeneracy condition. \varphi \in L^{\mathrm{\infty }}{\mathrm{\Phi }}^2 if \varphi is homogeneous of degree 1 in the frequency variable \xi and satisfies

for all multi-indices \alpha ,\beta with |\alpha |+|\beta |⩾2.

A real-valued function \varphi \in C^2({\mathbb{R}}^n\times ({\mathbb{R}}^n\setminus \{0\})) satisfies the rough non-degeneracy condition (RND), if there exists a constant \lambda >0 such that

|{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )-{\mathrm{\nabla }}_{\xi }\varphi (y,\xi )|⩾\lambda |x-y|

for any x,y\in {\mathbb{R}}^n,\xi \in {\mathbb{R}}^n\setminus \{0\}.

For a\in L^{\mathrm{\infty }}{S}_{\rho }^m,\varphi \in L^{\mathrm{\infty }}{\mathrm{\Phi }}^2 satisfying the RND condition, in [5], Dos Santos Ferreira and Staubach systematic studied the boundedness of Fourier integral operators on various Lebesgue spaces. For the boundedness on L^1, their main result can be stated as follows.

Theorem A　 Suppose that 0⩽\rho ⩽1,a\in L^{\mathrm{\infty }}{S}_{\rho }^m and \varphi \in L^{\mathrm{\infty }}{\mathrm{\Phi }}^2 satisfying the RND condition. If , then T_{\varphi ,a} is bounded on L^1, i.e., there exists a constant C>0 such that

\Vert T_{\varphi ,a}f{\Vert }_{L^1}⩽C\Vert f{\Vert }_{L^1}

for any f\in L^1.

In this note, we extend conditions on \varphi. We assume that there exist C>0 and \delta \in (0,1], such that

(1)\begin{array}{lll}& & |\{x:{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )\in E\}|⩽C|E|;\end{array}

(2)\begin{array}{lll}& & |{\mathrm{\nabla }}_{\xi }^k\varphi (x,\xi )|⩽C|\xi {|}^{\delta -k},k⩾2\end{array}

for any x,\xi \in {\mathbb{R}}^n and E\subset {\mathbb{R}}^n.

Remark 1　At first, we show that the RND condition yields (1). For any small ball B(x_0,r), since |{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )-{\mathrm{\nabla }}_{\xi }\varphi (y,\xi )|⩾\lambda |x-y|, the diameter of the set \{x:{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )\in B(x_0,r)\} is no more than \frac{2r}{\lambda }. Therefore, we can get that |\{x:{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )\in B(x_0,r)\}|⩽(\frac{2}{\lambda }{)}^n|B(x_0,r)|. For any measurable set E, by using the definition of outer measure, one can derive (1).

Secondly, if \varphi \in C^{\mathrm{\infty }}({\mathbb{R}}^n\times ({\mathbb{R}}^n\setminus \{0\})), by using [5, Proposition 1.11], we know that the SND condition is equivalent to the RND condition. For any fixed \xi, set F(x)={\mathrm{\nabla }}_{\xi }\varphi (x,\xi ). Then for any measurable set E, we have E\subset F^{-1}(F(E)). From (1) we can get that |E|⩽|F^{-1}(F(E))|⩽CF(E)=C{\int }_E|\mathrm{d}\mathrm{e}\mathrm{t}F|\mathrm{d}x. As E is arbitrary, by Lebesgue's differential theorem, the Jacobian determinant |\mathrm{d}\mathrm{e}\mathrm{t}F|=|\mathrm{d}\mathrm{e}\mathrm{t}\frac{{\mathrm{\partial }}^2\varphi (x,\xi )}{\mathrm{\partial }x\mathrm{\partial }\xi }| is no less than \frac{1}{C} almost everywhere. Thanks to the continuity of \varphi, we can derive the SND condition. So in this case, the SND condition, the RND condition and (1) are equivalent. At last, let n=1 and \varphi (x,\xi )=|x|\xi. It is easy to see that {\mathrm{\nabla }}_{\xi }\varphi (x,\xi )=|x| satisfies (1) but does not satisfy the RND condition. So (1) is a generalization of the RND condition.

In this note, we mainly prove the following result.

Theorem 1　 If a\in L^{\mathrm{\infty }}{S}_{\rho }^m and \varphi satisfies (1) and (2), then T_{\varphi ,a} is bounded on L^1 when . Furthermore, when \rho ⩽1-\frac{\delta }{2}, the bound on m is sharp.

Remark 2　When it is a pseudo-differential operator, i.e., \varphi (x,\xi )=x\cdot \xi. For any , we can find \delta such that . It is easy to see that x\cdot \xi satisfies (1) and (2). By using this theorem, we immediately show that the pseudo-differential operator is bounded on L^1. Therefore, this theorem is a generalization of the result of L^1-boundedness of pseudo-differential operators proved by Kenig and Staubach [12].

Remark 3　We construct an example to show that the pseudo-differential operator is unbounded on L^1 when a\in L^{\mathrm{\infty }}{S}_{\rho }^{n(\rho -1)} in general. Therefore, the bound of the index m in the result of Kenig and Staubach is sharp. At the same time, it also shows that when \rho ⩽1-\frac{\delta }{2}, the bound on m in Theorem 1 is sharp.

Remark 4　When \delta =1 and , we prove that T_{\varphi ,a} is bounded on L^1. When \rho ⩽\frac{1}{2}, the range of m is sharp. In addition, when \rho ⩽1-\frac{1}{2n}, Theorem 1 is a generalization of Theorem A. Finally, when , Theorem 1 is weaker than Theorem A, but here we do not assume that the phase is homogeneous of degree 1 in the frequency variable.

In Section 2 we prove the low frequency part. In the proof, we get an estimate of Fourier transform (Lemma 2) which has a separate meaning. In Section 3 we prove the high frequency part. At last we construct an example a\in L^{\mathrm{\infty }}{S}_{\rho }^{n(\rho -1)} such that the pseudo-differential operator is unbounded on L^1.

Throughout this note, we use C and c to denote some positive constants, which may vary from line to line, that depends only on \varphi ,n,\delta ,\lambda and some quasi-norms of a.

2 Low frequency part

In the low frequency part, we need the following lemma.

Lemma 1　 If u is supported in B_1 and satisfies

|{\mathrm{\nabla }}^{n+1}u(x)|⩽A|x{|}^{\delta -1-n}\ln \frac{3}{|x|}

for some \delta \in (0,1] and A>0, then for any , we have

\left|{\int }_{B_1}{\mathrm{e}}^{-\mathrm{i}yx}u(x)\mathrm{d}x\right|⩽CA(1+|y|{)}^{-n-\mu },

where C depends only on n,\delta ,\mu.

The proof of this lemma is almost identical to the proof of Lemma 1.17 in [5], with only slight differences. In this note we prove the following lemma which a generalization of Lemma 1.

Lemma 2　 If u is supported in B_1 and there exists some such that

{\int }_{B_1}|{\mathrm{\nabla }}^{n+1}u||x{|}^{1-\mu }\mathrm{d}x⩽A,

then we have

\left|{\int }_{B_1}{\mathrm{e}}^{-\mathrm{i}yx}u(x)\mathrm{d}x\right|⩽C_{n,\mu }A(1+|y|{)}^{-n-\mu },

where C depends only on n,\mu.

In the proof we need the following lemma, a special case of the weighted Sobolev inequality. It may be found in some literature. For completeness, we give a simple proof here.

Lemma 3　 Suppose that u is supported in B_1. For k=1,2,\dots ,n-1 and b>k-n, we have

(3){\left({\int }_{B_1}(|u(x)||x{|}^b{)}^{\frac{n}{n-k}}\mathrm{d}x\right)}^{\frac{n-k}{n}}⩽C_{n,k,b}{\int }_{B_1}|{\mathrm{\nabla }}^{k}u(x)||x{|}^b\mathrm{d}x.

Additionally, when b>0 we have

(4)\underset{x\in B_1}{sup}|u(x)||x{|}^b⩽C_{n,b}{\int }_{B_1}|{\mathrm{\nabla }}^{n}u(x)||x{|}^b\mathrm{d}x.

Proof　When b>1-n, as u is supported in B_1, by using simple computations, we can get that

\begin{array}{rl}{\int }_{B_1}|u(x)||x{|}^{b-1}\mathrm{d}x& ={\int }_{S^{n-1}}{\int }_0^1|u(r\theta )|r^{b+n-2}\mathrm{d}r\mathrm{d}\theta \\ & ⩽{\int }_{S^{n-1}}{\int }_0^1{\int }_r^1|{\mathrm{\partial }}_{r}u(t\theta )|\mathrm{d}t{r}^{b+n-2}\mathrm{d}r\mathrm{d}\theta \\ & =\frac{1}{b+n-1}{\int }_{S^{n-1}}{\int }_0^1|{\mathrm{\partial }}_{r}u(t\theta )|t^{b+n-1}\mathrm{d}t\mathrm{d}\theta \\ & ⩽\frac{1}{b+n-1}{\int }_{B_1}|\mathrm{\nabla }u(x)||x{|}^b\mathrm{d}x.\end{array}

By iterating, for j\in \mathbb{N} and b>j-n, we have

(5){\int }_{B_1}|u(x)||x{|}^{b-j}\mathrm{d}x⩽C_{n,j,b}{\int }_{B_1}|{\mathrm{\nabla }}^{j}u(x)||x{|}^b\mathrm{d}x.

For the term |u(x)||x{|}^b, by using (5) and the Sobolev embedding theorem , we can show that

\begin{array}{rl}{\left({\int }_{B_1}(|u(x)||x{|}^b{)}^{\frac{n}{n-k}}\mathrm{d}x\right)}^{\frac{n-k}{n}}& ⩽C_{n,k}{\int }_{B_1}|{\mathrm{\nabla }}^k(|u(x)||x{|}^b)|\mathrm{d}x\\ & ⩽C_{n,k}\sum _{j=0}^k{\int }_{B_1}|{\mathrm{\nabla }}^{k-j}u(x)||x{|}^{b-j}\mathrm{d}x\\ & ⩽C_{n,k,b}{\int }_{B_1}|{\mathrm{\nabla }}^{k}u(x)||x{|}^b\mathrm{d}x.\end{array}

On the other hand, by using (5) and the Sobolev embedding theorem W^{n,1}\subset C^0, one can get that

\begin{array}{rl}\underset{x\in B_1}{sup}|u(x)||x{|}^b& ⩽C_n{\int }_{B_1}|{\mathrm{\nabla }}^n(|u(x)||x{|}^b)|\mathrm{d}x\\ & ⩽C_{n,b}\sum _{j=0}^n{\int }_{B_1}|{\mathrm{\nabla }}^{n-j}u(x)||{\mathrm{\nabla }}^j|x{|}^b|\mathrm{d}x\\ & ⩽C_{n,b}\sum _{j=0}^n{\int }_{B_1}|{\mathrm{\nabla }}^{n-j}u(x)||x{|}^{b-j}\mathrm{d}x\\ & ⩽C_{n,b}{\int }_{B_1}|{\mathrm{\nabla }}^{n}u(x)||x{|}^b\mathrm{d}x.\end{array}

This finishes the proof.□

Now we turn to prove Lemma 2.

Proof　Firstly, as u is supported in B_1 and , from (4) we get that

\begin{array}{rl}|\mathrm{\nabla }u(x)|& ⩽|x{|}^{\mu -1}\underset{y\in B_1}{sup}|\mathrm{\nabla }u(y)||y{|}^{1-\mu }⩽C_{n,\mu }|x{|}^{\mu -1}{\int }_{B_1}|{\mathrm{\nabla }}^{n+1}u(y)||y{|}^{1-\mu }\mathrm{d}y\\ & ⩽C_{n,\mu }A|x{|}^{\mu -1},\end{array}

which implies that

|u(r\theta )|⩽{\int }_r^1|{\mathrm{\partial }}_{r}u(t\theta )|\mathrm{d}t⩽C_{n,\mu }A{\int }_r^1{t}^{\mu -1}\mathrm{d}t⩽C_{n,\mu }A.

When |y|⩽3, it is easy to see that

(6)\left|{\int }_{B_1}{\mathrm{e}}^{-\mathrm{i}yx}u(x)\mathrm{d}x\right|⩽C{\int }_{B_1}|u(x)|\mathrm{d}x⩽C⩽C(1+|y|{)}^{-n-\mu }.

When |y|>3, choose a cut function \eta \in C_c^{\mathrm{\infty }}(B(0,2|y{|}^{-1})) such that

Without loss of generality, we can assume that |y_1|⩾\frac{|y|}{n}. By using integration by parts and (5), we have

(7)

In the penultimate inequality we use Lemma 3. In conclusion, for all y\in {\mathbb{R}}^n, we complete the proof of Lemma 2.□

Below we give the boundedness of L^1 for the low frequency part of the Fourier integral operator.

Theorem 2　 If a\in L^{\mathrm{\infty }}{S}_{\rho }^m and \varphi satisfies (1) and (2), then for any \eta \in C_c^{\mathrm{\infty }}(B_1), the operator

T_{0,\varphi ,a}f(x)={\int }_{R^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi )\eta (\xi )\hat{f}(\xi )\mathrm{d}\xi

is bounded on L^1.

Proof　Take any point {\xi }_0\in S^{n-1}. Set w(x,\xi )=\varphi (x,\xi )-{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_0)\cdot \xi. By using some simple calculations, one have

\begin{array}{rl}& T_{0,\varphi ,a}f(x)\\ & ={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi )\eta (\xi )\hat{f}(\xi )\mathrm{d}\xi \\ & ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}(\varphi (x,\xi )-y\cdot \xi )}a(x,\xi )\eta (\xi )f(y)\mathrm{d}y\mathrm{d}\xi \\ & ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_0)-y)\cdot \xi }{\mathrm{e}}^{\mathrm{i}w(x,\xi )}a(x,\xi )\eta (\xi )f(y)\mathrm{d}\xi \mathrm{d}y\\ & ={\int }_{{\mathbb{R}}^n}{k}_0(x,y)f(y)\mathrm{d}y,\end{array}

where k_0(x,y)={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_0)-y)\cdot \xi }{\mathrm{e}}^{\mathrm{i}w(x,\xi )}a(x,\xi )\eta (\xi )\mathrm{d}\xi .

From (2), when \xi \in B_1 and , it is easy to check that

\begin{array}{rl}|{\mathrm{\nabla }}_{\xi }w(x,\xi )|& =|{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )-{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_0)|\\ & ⩽\left|{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )-{\mathrm{\nabla }}_{\xi }\varphi \left(x,\frac{\xi }{|\xi |}\right)\right|+\left|{\mathrm{\nabla }}_{\xi }\varphi \left(x,\frac{\xi }{|\xi |}\right)-{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_0)\right|\\ & ⩽{\int }_{|\xi |}^1|{\mathrm{\nabla }}_{\xi }^2\varphi \left(x,t\frac{\xi }{|\xi |}\right)|dt+\left|{\xi }_0-\frac{\xi }{|\xi |}\right|\underset{\zeta \in S^{n-1}}{sup}|{\mathrm{\nabla }}_{\xi }^2\varphi (x,\zeta )|\\ & ⩽C\left({\int }_{|\xi |}^1{r}^{\delta -2}dr+1\right)\\ & ⩽C|\xi {|}^{\delta -1}\ln \frac{3}{|\xi |}.\end{array}

Here the term \ln \frac{3}{|\xi |} is used when \delta =1.

On the other hand, when k⩾2, we have

|{\mathrm{\nabla }}_{\xi }^{k}w(x,\xi )|=|{\mathrm{\nabla }}_{\xi }^k\varphi (x,\xi )|⩽C_k|\xi {|}^{\delta -k}.

So, for any \xi \in B_1 and k\in \mathbb{N}, we can get that

|{\mathrm{\nabla }}_{\xi }^{k}w(x,\xi )|⩽C|\xi {|}^{\delta -k}\ln \frac{3}{|\xi |}.

As , by using direct computations we get that

\begin{array}{rl}& |{\mathrm{\nabla }}_{\xi }^{n+1}[{\mathrm{e}}^{\mathrm{i}w(x,\xi )}a(x,\xi )\eta (\xi )]|\\ & ⩽C\sum _{k_0=0}^{n+1}\sum _{t=1}^{n+1-k_0}|{\mathrm{\nabla }}_{\xi }^{k_0}[a(x,\xi )\eta (\xi )]|\sum _{k_1+\cdots +k_t=n+1-k_0,k_s>0}\prod _{s=1}^t|{\mathrm{\nabla }}_{\xi }^{k_s}w(x,\xi )|\\ & ⩽C\sum _{k_0=0}^{n+1}\sum _{t=1}^{n+1-k_0}\sum _{k_1+\cdots +k_t=n+1-k_0,k_s>0}\prod _{s=1}^t\left(|\xi {|}^{\delta -k_s}\ln \frac{3}{|\xi |}\right)\\ & ⩽C\sum _{k_0=0}^{n+1}\sum _{t=1}^{n+1-k_0}|\xi {|}^{t\delta +k_0-n-1}{\ln }^t\frac{3}{|\xi |}\\ & ⩽C\sum _{k_0=0}^{n+1}\sum _{t=1}^{n+1-k_0}{\left(|\xi {|}^{\delta }\ln \frac{3}{|\xi |}\right)}^t|\xi {|}^{k_0-n-1}\\ & ⩽C|\xi {|}^{\delta }\ln \frac{3}{|\xi |}|\xi {|}^{-n-1}=C|\xi {|}^{\delta -n-1}\ln \frac{3}{|\xi |}.\end{array}

In the second inequality, we use the fact that |{\mathrm{\nabla }}_{\xi }^{k_0}[a(x,\xi )\eta (\xi )]|⩽C when , while in the last inequality, we use the fact that |\xi {|}^{\delta }\ln \frac{3}{|\xi |}⩽C_{\delta } when .

For , we have

By using Lemma 2, we get that

(8)\begin{array}{rl}|k_0(x,y)|& =\left|{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_0)-y)\cdot \xi }{\mathrm{e}}^{\mathrm{i}w(x,\xi )}a(x,\xi )\eta (\xi )\mathrm{d}\xi \right|\\ & ⩽C(1+|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_0)-y|{)}^{-n-\mu }.\end{array}

On the other hand, for y\in {\mathbb{R}}^n, we set E=B(y,r) in (1) and show that

(9)

where C is independent of r,y,{\xi }_0.

For any y\in {\mathbb{R}}^n, from (8) and (9) we can get that

The constant C is independent of y. So we can immediately show that T_{0,\varphi ,a} is bounded on L^1.□

3 The high frequency part

In this section, we consider the high frequency part of T_{\varphi ,a}.

Set \lambda =min\{\rho ,1-\frac{\delta }{2}\}. As \delta >0, we get that . For j\in \mathbb{N}, consider the ball B=B({\xi }_B,2^{j\lambda -1}) with . It is easy to see that B is contained in the ring . For any {\eta }_B\in C_c^{\mathrm{\infty }}(B) satisfying |{\mathrm{\nabla }}^k{\eta }_B|⩽C_k{2}^{-k\lambda },k\in \mathbb{N}, we define

T_{B,\varphi ,a}f(x)={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi ){\eta }_B(\xi )\hat{f}(\xi )\mathrm{d}\xi .

The main estimate in this section is given below.

Theorem 3　 If a\in L^{\mathrm{\infty }}{S}_{\rho }^m and \varphi satisfies (1) and (2), then we have

\Vert T_{B,\varphi ,a}f{\Vert }_1⩽C{2}^{jm}\Vert f{\Vert }_1.

Proof　Set w_B(x,\xi )=\varphi (x,\xi )-{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)\cdot \xi. By using simple computations we show that

\begin{array}{rl}& T_{B,\varphi ,a}f(x)\\ & ={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi ){\eta }_B(\xi )\hat{f}(\xi )\mathrm{d}\xi \\ & ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}(\varphi (x,\xi )-y\cdot \xi )}a(x,\xi ){\eta }_B(\xi )f(y)\mathrm{d}y\mathrm{d}\xi \\ & ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y)\cdot \xi }{\mathrm{e}}^{\mathrm{i}{w}_B(x,\xi )}a(x,\xi ){\eta }_B(\xi )f(y)\mathrm{d}\xi \mathrm{d}y\\ & ={\int }_{{\mathbb{R}}^n}{k}_B(x,y)f(y)\mathrm{d}y,\end{array}

where k_B(x,y)={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y)\cdot \xi }{\mathrm{e}}^{\mathrm{i}{w}_B(x,\xi )}a(x,\xi ){\eta }_B(\xi )\mathrm{d}\xi .

Define the operator L=1-2^{2j\lambda }{\mathrm{\nabla }}_{\xi }^2. It is obvious a self-adjoint operator. By using simple computations, one can get that

L({\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y)\cdot \xi })=(1+2^{2j\lambda }|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y{|}^2){\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y)\cdot \xi }.

Since \lambda =min\{\rho ,1-\frac{\delta }{2}\}⩽\rho, for any k=0,1,\dots, we have

(10)\begin{array}{rl}|{\mathrm{\nabla }}_{\xi }^k[a(x,\xi ){\eta }_B(\xi )]|& ⩽C\sum _{k_1+k_2=k}|{\mathrm{\nabla }}_{\xi }^{k_1}a{\mathrm{\nabla }}_{\xi }^{k_2}{\eta }_B|\\ & ⩽C\sum _{k_1+k_2=k}C{2}^{j(m-k_1\rho )}{2}^{-j{k}_2\lambda }⩽C{2}^{j(m-k\lambda )}.\end{array}

Also due to \lambda =min\{\rho ,1-\frac{\delta }{2}\}⩽1-\frac{\delta }{2}, mean value theorem and (2), we can obtain that

\begin{array}{rl}|{\mathrm{\nabla }}_{\xi }{w}_B(x,\xi )|& =|{\mathrm{\nabla }}_{\xi }\varphi (x,\xi )-{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)|⩽|\xi -{\xi }_B|\underset{\zeta \in B}{sup}|{\mathrm{\nabla }}_{\zeta }^2\varphi (x,\zeta )|\\ & ⩽C{2}^{j(\lambda +\delta -2)}⩽C{2}^{-j\lambda }.\end{array}

On the other hand, for any k⩾2, from (2), one have

|{\mathrm{\nabla }}_{\xi }^k{w}_B(x,\xi )|=|{\mathrm{\nabla }}_{\xi }^k\varphi (x,\xi )|⩽C{2}^{j(\delta -k)}=C{2}^{jk(\frac{\delta }{k}-1)}⩽C{2}^{-jk\lambda }.

So, for any k=0,1,\dots, we can get that

(11)\begin{array}{l}|{\mathrm{\nabla }}_{\xi }^k{w}_B(x,\xi )|⩽C{2}^{-jk\lambda }.\end{array}

From (10) and (11), for any M\in \mathbb{N}, we have

\begin{array}{rl}& |k_B(x,y)|\\ & =\left|{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y)\cdot \xi }{\mathrm{e}}^{\mathrm{i}{w}_B(x,\xi )}a(x,\xi ){\eta }_B(\xi )\mathrm{d}\xi \right|\\ & =(1+2^{2j\lambda }|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y{|}^2{)}^{-M}\left|{\int }_{{\mathbb{R}}^n}{L}^M({\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y)\cdot \xi }){\mathrm{e}}^{\mathrm{i}{w}_B(x,\xi )}a(x,\xi ){\eta }_B(\xi )\mathrm{d}\xi \right|\\ & =(1+2^{2j\lambda }|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y{|}^2{)}^{-M}\left|{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}({\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y)\cdot \xi }{L}^M[{\mathrm{e}}^{\mathrm{i}{w}_B(x,\xi )}a(x,\xi ){\eta }_B(\xi )]\mathrm{d}\xi \right|\\ & ⩽C(1+2^{2j\lambda }|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y{|}^2{)}^{-M}{\int }_{{\mathbb{R}}^n}\sum _{k_0=0}^{2M}\sum _{t=0}^{2M-k_0}{2}^{j{k}_0\lambda }|{\mathrm{\nabla }}_{\xi }^{k_0}[a(x,\xi ){\eta }_B(\xi )]|\\ & \cdot \sum _{k_1+\cdots +k_t⩽2M-k_0,k_s>0}\prod _{s=1}^t(2^{j{k}_s\lambda }|{\mathrm{\nabla }}_{\xi }^{k_s}{w}_B(x,\xi )|)\mathrm{d}\xi \\ & ⩽C(1+2^{2j\lambda }|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y{|}^2{)}^{-M}{\int }_B\sum _{k_0=0}^{2M}\sum _{t=0}^{2M-k_0}{2}^{j{k}_0\lambda }{2}^{j(m-k_0\lambda )}\\ & \cdot \sum _{k_1,\dots ,k_t}\prod _{s=1}^t(2^{j{k}_s\lambda }{2}^{-j{k}_s\lambda })\mathrm{d}\xi \\ & ⩽C(1+2^{2j\lambda }|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y{|}^2{)}^{-M}|B|2^{jm}\\ & ⩽C{2}^{j(m+n\lambda )}(1+2^{2j\lambda }|{\mathrm{\nabla }}_{\xi }\varphi (x,{\xi }_B)-y{|}^2{)}^{-M}.\end{array}

By using (9) and some similar calculations in the first part, for any y\in {\mathbb{R}}^n and M>\frac{n}{2}, we get that

This finishes the proof of this theorem.□

Finally, for j⩾1, one can easily check that can be covered by no more than C{2}^{jn(1-\lambda )} balls B_j^v=B({\xi }_j^v,2^{j\lambda -1}) with . On the other hand, it is easy to find functions {\eta }_0\in C_c^{\mathrm{\infty }}(B_2),{\eta }_j^v\in C_c^{\mathrm{\infty }}(B_j^v) such that

{\eta }_0+\sum _{j=1}^{\mathrm{\infty }}\sum _{v=1}{\eta }_j^v=1,|{\mathrm{\nabla }}^k{\eta }_0|⩽C_k,|{\mathrm{\nabla }}^k{\eta }_j^v|⩽C_k{2}^{-jk\lambda },k>0.

So the operator T_{\varphi ,a} can be decomposed into

\begin{array}{rl}{T}_{\varphi ,a}f(x)& ={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi )\hat{f}(\xi )\mathrm{d}\xi \\ & ={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi )\left({\eta }_0+\sum _{j=1}^{\mathrm{\infty }}\sum _{v=1}{\eta }_j^v\right)\hat{f}(\xi )\mathrm{d}\xi \\ & ={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi ){\eta }_0\hat{f}(\xi )\mathrm{d}\xi +\sum _{j=1}^{\mathrm{\infty }}\sum _{v=1}{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}\varphi (x,\xi )}a(x,\xi ){\eta }_j^v\hat{f}(\xi )\mathrm{d}\xi \\ & =T_{0,\varphi ,a}f(x)+\sum _{j=1}^{\mathrm{\infty }}\sum _{v=1}{T}_{B_j^v,\varphi ,a}f(x).\end{array}

From Theorem 2 and Theorem 3, when , we can show that

\begin{array}{rl}\Vert T_{\varphi ,a}f{\Vert }_1& ⩽\Vert T_{0,\varphi ,a}f{\Vert }_1+\sum _{j=1}^{\mathrm{\infty }}\sum _{v=1}\Vert T_{B_j^v,\varphi ,a}f{\Vert }_1\\ & ⩽C\left(1+\sum _{j=1}^{\mathrm{\infty }}\sum _{v=1}{2}^{jm}\right)\Vert f{\Vert }_1\\ & ⩽C\left(1+\sum _{j=1}^{\mathrm{\infty }}{2}^{j(m+n(1-\lambda ))}\right)\Vert f{\Vert }_1\\ & =C\Vert f{\Vert }_1.\end{array}

So, the first half of Theorem 1 is done.

4 A counterexample

In this section, we construct an example a\in L^{\mathrm{\infty }}{S}_{\rho }^{n(\rho -1)} such that T_a is unbounded on L^1.

At first, for any j\in \mathbb{N}, we use A_j to denote the set of all lattice points (each coordinate component is an integer) in the ring . It is obvious that the number of points in A_j is about O(4^{jn(1-\rho )}).

In addition, we take \phi such that

\hat{\phi }\in C_c^{\mathrm{\infty }}\left(B\left(0,\frac{1}{3}\right)\right);0⩽\hat{\phi }⩽1;\hat{\phi }(z)=1,z\in B\left(0,\frac{1}{4}\right);|{\mathrm{\nabla }}^k\hat{\phi }|⩽{20}^k.

Now we set

a(x,\xi )=\sum _{j=1}^{\mathrm{\infty }}\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}{\mathrm{e}}^{-\mathrm{i}{4}^{j\rho }x\cdot \alpha }\hat{\phi }(4^{-j\rho }(\xi -4^{j\rho }\alpha )).

As \hat{\phi }\in C_c^{\mathrm{\infty }}(B(0,\frac{1}{3})), for any \xi, there is at most one j\in \mathbb{N} and one point \alpha \in A_j such that \hat{\phi }(4^{-j\rho }(\xi -4^{j\rho }\alpha ))\ne 0 (Note that the distance of two different lattice points is no less than 1). In this case we have . It is easy to see that

|\xi {|}^{k\rho +n(1-\rho )}|{\mathrm{\partial }}_{\xi }^{\alpha }a(x,\xi )|⩽|\xi {|}^{k\rho +n(1-\rho )}{4}^{jn(\rho -1)}{4}^{-jk\rho }{20}^k⩽{\left(\frac{19}{12}\right)}^{k\rho +n(1-\rho )}{20}^k⩽2^n{40}^k

for any k,x,\xi. Thus a\in L^{\mathrm{\infty }}{S}_{\rho }^{n(\rho -1)}. In the above analysis, we can also see that a(x,\cdot ) is supported in the set .

Similar to \phi, we choose \psi such that

\hat{\psi }\in C_c^{\mathrm{\infty }}\left(B\left(0,\frac{5}{3}\right)\right);\hat{\psi }(z)=1,z\in B\left(0,\frac{19}{12}\right).

For any N\in \mathbb{N}, set

a_N(x,\xi )=a(x,\xi )\hat{\psi }(4^{-N}\xi ).

It is obvious that \Vert T_{a_N}{\Vert }_{L^1-L^1}⩽\Vert T_a{\Vert }_{L^1-L^1}.

By using some simple calculations, the kernel of T_{a_N} is

\begin{array}{rl}{k}_N(x,y)& ={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}(x-y)\xi }{a}_N(x,\xi )\mathrm{d}\xi \\ & =\sum _{j=1}^N\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}(x-y)\xi }{\mathrm{e}}^{-\mathrm{i}{4}^{j\rho }x\cdot \alpha }\hat{\phi }(4^{-j\rho }(\xi -4^{j\rho }\alpha ))\mathrm{d}\xi \\ & =\sum _{j=1}^N\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}{\mathrm{e}}^{-\mathrm{i}{4}^{j\rho }y\cdot \alpha }{\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}(x-y)(\xi -4^{j\rho }\alpha )}\hat{\phi }(4^{-j\rho }(\xi -4^{j\rho }\alpha ))\mathrm{d}\xi \\ & =\sum _{j=1}^N\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}{\mathrm{e}}^{-\mathrm{i}{4}^{j\rho }y\cdot \alpha }{4}^{jn\rho }\phi (4^{j\rho }(x-y)).\end{array}

As \hat{\phi }(0)=1, by using some simple calculations, for any y we have

\begin{array}{rl}{\int }_{{\mathbb{R}}^n}{k}_N(x,y)\mathrm{d}x& =\sum _{j=1}^N\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}{\mathrm{e}}^{-\mathrm{i}{4}^{j\rho }y\cdot \alpha }{\int }_{{\mathbb{R}}^n}{4}^{jn\rho }\phi (4^{j\rho }(x-y))\mathrm{d}x\\ & =\sum _{j=1}^N\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}{\mathrm{e}}^{-\mathrm{i}{4}^{j\rho }y\cdot \alpha }.\end{array}

When , since \alpha \in A_j,j⩽N, we get that |4^{j\rho }y\cdot \alpha |⩽\frac{5}{16}. Additionally, as the number of points in A_j is about O(4^{jn(1-\rho )}), when we have

\left|{\int }_{{\mathbb{R}}^n}{k}_N(x,y)\mathrm{d}x\right|=\left|\sum _{j=1}^N\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}{\mathrm{e}}^{-\mathrm{i}{4}^{j\rho }y\cdot \alpha }\right|⩾\left|\sum _{j=1}^N\sum _{\alpha \in A_j}{4}^{jn(\rho -1)}\cos \frac{5}{16}\right|⩾c_{n}N,

which implies that \Vert T_{a_N}{\Vert }_{L^1-L^1}>c_{n}N. Since N is arbitrary, T_a is not bounded by L^1. By synthesizing all the proofs, we get the main theorem of this note.□

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