(q, r) boundedness of general rough pseudo-differential operators

Abstract

In this paper, we consider the (q,r) boundedness of the pseudo-differential operators with the amplitude a\in L^p{S}_{\rho }^m(p\ge 1,m\in \mathbb{R},0\le \rho \le 1). When , 1⩽p, q\le \mathrm{\infty }, r\le p, \frac{1}{r}\le \frac{1}{p}+\frac{1}{q}, we prove that if

　　　　　　　　　　

then for any a\in L^p{S}_{\rho }^m, the pseudo-differential operator T_a is bounded from L^q to L^r. It is a generalization and improvement of the known theorems and in general the conditions on r, m are sharp.

Keywords

Pseudo-differential operator / rough Hrmander class / (q, r) boundedness

1 Introduction

For the function f in Schwartz class \mathcal{S}\left({\mathbb{R}}^n\right), the pseudo-differential operator T_a is defined as

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

where \hat{f}\left(\xi \right)=\left(2\pi {)}^{-n}{\int }_{{\mathbb{R}}^n} f\right(x){\mathrm{e}}^{-\mathrm{i}x\cdot \xi }\mathrm{d}x represents the Fourier transformation of f, with the amplitude a satisfying certain conditions. One of the most important classes of function regarding amplitude, introduced by Hörmander in [6], is called the Hörmander class.

Definition 1.1　 m\in \mathbb{R}, 0\le \rho, \delta \le 1, if function a\in C^{\mathrm{\infty }}({\mathbb{R}}^n\times {\mathbb{R}}^n), and for any multi-index α, β, there exists the constant C_{\alpha ,\beta }>0 such that

then a falls into the Hörmander class S_{\rho ,\delta }^m.

Pseudo-differential operators have been widely applied in various fields of modern mathematics, especially in hyperbolic partial differential equation problems. Kohn-Nernberg [10] and Hörmander [5] were among the first to systematically study them. Its boundedness in Lebesgue space is a very important problem in partial differential equations and harmonic analysis.

There have been relatively systematic results on the boundedness of the amplitude pseudo-differential operator with an amplitude a\in S_{\rho ,\delta }^m in Lebesgue space.

When , for all a\in S_{\rho ,\delta }^m, the necessary and sufficient condition for the pseudo-differential operator T_ato be L² bounded is m\le \mathrm{m}\mathrm{i}\mathrm{n}\{0,\frac{n(\rho -\delta )}{2}\} (refer to [2, 7, 8]). When \delta =1, this result will no longer be established. Rodino [13] constructed an a\in S_{\rho ,1}^{\frac{n(\rho -\delta )}{2}}, such that T_a is not L² bounded (refer to [3, 15] for counterexample when \rho =1). Meanwhile, he proved when , T_ais L² bounded for all a\in S_{\rho ,\delta }^m.

When or , for any a\in S_{\rho ,\delta }^{\frac{n(\rho -1)}{2}}, the pseudo-differential operator T_ais bounded from L^∞to BMO (refer to [1]). When \delta >\rho or \delta =1, it remains unknown whether the conclusion will still hold.

Stein proved in unpublished lectures that when or , a\in S_{\rho ,\delta }^{\frac{n(\rho -1)}{2}}, corresponding pseudo-differential operator was weakly \left(1,1\right) bounded, and bounded from H^1 to L^1. Álvarez-Hounie [1] generalized the result by proving when , ,a\in S_{\rho ,\delta }^m, T_a was weakly \left(1,1\right) bounded, and bounded from H¹ to L¹ if m\le \frac{n(\rho -1)}{2}+\mathrm{m}\mathrm{i}\mathrm{n}\{0,\frac{n(\rho -\delta )}{2}\}. When \delta =1, Guo and Zhu [4] conducted systematic researches and constructed various counterexamples (for example, when a\in S_{\rho ,1}^{n(\rho -1)}, it still holds for T_a to be bounded from H¹ to L¹, but not weakly \left(1,1\right) bounded).

For general (q,r) boundedness, Álvarez-Hounie [1] obtained the following theorems according to the above endpoint results as well as analytic interpolation and fractional integral theory.

Theorem 1.1 [1, Theorem 3.5]　 Assume a\in S_{\rho ,\delta }^m, , , and , \lambda =\mathrm{m}\mathrm{i}\mathrm{n}\{0,\frac{n(\rho -\delta )}{2}\},

(a) and

m\le -n\left(\frac{1}{q}-\frac{1}{r}\right)+\lambda ;

or

(b) 2\le q\le r and

m\le -n\left(\frac{1}{q}-\frac{1}{r}+(1-\rho )\left(\frac{1}{2}-\frac{1}{q}\right)\right)+\lambda ;

or

(c) q\le r\le 2 and

m\le -n\left(\frac{1}{q}-\frac{1}{r}+(1-\rho )\left(\frac{1}{r}-\frac{1}{2}\right)\right)+\lambda ,

then T_ais bounded from L^qto L^r.

Kenig and Staubach [9] introduced a rough function class L^{\mathrm{\infty }}{S}_{\rho }^m. For this class of function, the property of frequency variable ξ is similar to S_{\rho ,\delta }^m, but there is no regularity in spatial variable x.

Definition 1.2　Assume m\in \mathbb{R}, 0\le \rho \le 1, if function a is ξ-smooth with respect to frequency variable and essentially bounded with respect to spatial variable x, and for any k=0,1,2,\dots , there is

Then, function a falls into L^{\mathrm{\infty }}{S}_{\rho }^m.

Obviously, S_{\rho ,1}^m is the subset of L^{\mathrm{\infty }}{S}_{\rho }^m. About the L^q boundedness of this class of pseudo-differential operator, Kenig-Staubach [9] proved the following results.

Theorem 1.2 [9, Proposition 2.3 and Theorem 2.7]　 Assume a\in L^{\mathrm{\infty }}{S}_{\rho }^m, , if , then T_a is bounded from L^q to L^r, where 1\le q\le \mathrm{\infty }.

Besides, Michalowski et al. [11] introduced the more general rough function class L^p{S}_{\rho }^m.

Definition 1.3　Assume 1\le p\le \mathrm{\infty }, m\in \mathbb{R}, 0\le \rho \le 1. If function a\left(x,\xi \right)\in C^{\mathrm{\infty }}\left({\mathbb{R}}_{\xi }^n\right)\mathrm{a}.\mathrm{e}.x\in {\mathbb{R}}^n, and for any k=0,1,2,\dots , there exists the constant C_ksuch that

function a falls into L^p{S}_{\rho }^m. The corresponding seminorm is defined as

|a{|}_{p,m,s}=\sum _{\left|\alpha \right|\le s} \underset{\xi \in {\mathbb{R}}^n}{\mathrm{s}\mathrm{u}\mathrm{p}} (1+\left|\xi \right|{)}^{\rho \left|\alpha \right|-m}\Vert {\mathrm{\partial }}_{\xi }^{\alpha }a(\cdot ,\xi ){\Vert }_{L^p\left({\mathbb{R}}^n\right)}.

For the pseudo-differential operator T_aof a\in L^p{S}_{\rho }^m, Rodríguez-López and Staubach [14] obtained the following(q,r) boundedness.

Theorem 1.3 [14, Theorem 4.10]　 Assume , 1\le p, q\le \mathrm{\infty } satisfy \frac{1}{r}=\frac{1}{q}+\frac{1}{p}, a\in L^p{S}_{\rho }^m\left(0\le \rho \le 1\right), and

then pseudo-differential operator T_a is bounded from L^q to L^r.

In this paper, we extend the results of Theorem 1.3 to general r,p,q, and prove the following theorem. This theorem can be seen as an extension of the three known theorems mentioned above, and in some cases even improves the results of Theorem 1.1.

Theorem 1.4　 Assume 1\le p, q\le \mathrm{\infty }, , satisfying \frac{1}{r}\le \frac{1}{p}+\frac{1}{q}. If a\in L^p{S}_{\rho }^m, 0\le \rho \le 1 and

then the corresponding pseudo-differential operator T_asatisfies

\Vert T_{a}f{\Vert }_r\le C\Vert f{\Vert }_q,

where constant C only depends on partial seminorms of n, m, ρ, p, q, r, and a in L^p{S}_{\rho }^m.

Note 1.1　When \frac{1}{r}=\frac{1}{p}+\frac{1}{q}, we obtain Theorem 1.2. Let p=\mathrm{\infty }, q=r\in [1,\mathrm{\infty }], then min{2, p, q}=min{2, q}, and we get Theorem 1.3.

Note 1.2　When p=\mathrm{\infty }, obviously there is S_{\rho ,\delta }^m\subset S_{\rho ,1}^m\subset L^{\mathrm{\infty }}{S}_{\rho }^m. In Theorem 1.1, assume 1\le q\le r\le 2, , then we can obtain after proper arrangement that when

m\le \frac{n(\rho -1)}{q}-n\rho \left(\frac{1}{q}-\frac{1}{r}\right)+\frac{n(1-\delta )}{2},

T_a is(q,r)bounded. Therefore, when p=\mathrm{\infty }, r\le 2, Theorem 1.4 can be seen as the generalization of Theorem 1.1 when \delta =1, and be noted that here normally we don’t take the equal sign (we will explain when presenting certain examples in Section 2).

Note 1.3　Assume p=\mathrm{\infty }, r>2, , . Obviously there is and S_{\rho ,\delta }^m\subset L^{\mathrm{\infty }}{S}_{\rho }^m. It can be obtained from Theorem 1.1 that

m\le -n\left(\frac{1}{q}-\frac{1}{r}\right)+\frac{n\left(\rho -\delta \right)}{2},q\le 2,

or

m\le -n\left(\frac{1}{q}-\frac{1}{r}+\left(1-\rho \right)\left(\frac{1}{2}-\frac{1}{q}\right)\right)+\frac{n\left(\rho -\delta \right)}{2},r\ge q>2,

T_ais (q,r) bounded. After simple calculation we can get

Hence, when p=\mathrm{\infty }, r>2, , , a\in S_{\rho ,\delta }^m, Theorem 1.4 has improved the results of Theorem 1.1.

Finally, we need to point out that if r>p or \frac{1}{r}>\frac{1}{p}+\frac{1}{q}, for any m\in \mathbb{R}, there exists a\in L^p{S}_{\rho }^m such that T_a is not (q,r) bounded, so r\le p and \frac{1}{r}\le \frac{1}{p}+\frac{1}{q} are both necessary when considering the (q,r) boundedness.

Take \eta \in C_c^{\mathrm{\infty }}\left(B\right(0,2\left)\right) satisfying \eta \ge 0 and \eta \left(\xi \right)=1, \mathrm{\forall }\xi \in B(0,1). For any \epsilon >0, take a_{\epsilon }(x,\xi ){=}_{\epsilon }^{-\frac{n}{p}}{\chi }_{B(0,\epsilon )}\left(x\right)\eta \left(\xi \right), then for any m\in \mathbb{R}, a_{\epsilon }\in L^p{S}_{\rho }^m, and all seminorms are not related to ε. Take \hat{f}\left(\xi \right)=\eta \left(\frac{\xi }{2}\right), then for any q>0, there is \Vert f{\Vert }_q\le C_q. It can be seen from the selection of η,

T_{a_{\epsilon }}f\left(x\right)={\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}x\cdot \xi }{a}_{\epsilon }(x,\xi )\hat{f}\left(\xi \right)\mathrm{d}\xi ={\epsilon }^{-\frac{n}{p}}{\chi }_{B(0,\epsilon )}\left(x\right){\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}x\cdot \xi }\eta \left(\xi \right)\mathrm{d}\xi .

When \epsilon \le \frac{\pi }{6}, there is

|\mathrm{R}\mathrm{e}{T}_{a_{\epsilon }}f(x\left)\right|={\epsilon }^{-\frac{n}{p}}{\chi }_{B(0,\epsilon )}\left(x\right){\int }_{\left|\xi \right|\le 2} \mathrm{c}\mathrm{o}\mathrm{s}(x\cdot \xi )\eta \left(\xi \right)\mathrm{d}\xi \ge \frac{v\left(n\right)}{2}{\epsilon }^{-\frac{n}{p}}{\chi }_{B(0,\epsilon )}\left(x\right),

where v(n) represents the volume of unit ball. Let ε→0. It can be obtained from direct estimation whenr>p,

\Vert T_{a_{\epsilon }}f{\Vert }_r\ge \frac{v\left(n\right)}{2}{\epsilon }^{-\frac{n}{p}}\Vert {\chi }_{B(0,\epsilon )}{\Vert }_r=c{\epsilon }^{n(\frac{1}{r}-\frac{1}{p})}\to \mathrm{\infty }.

Hence, whenr>p, for any m\in \mathbb{R}, there must exist a\in L^p{S}_{\rho }^m such that T_a is not (q,r)bounded. Or otherwise for any k\in \mathbb{N}, a_k\in L^p{S}_{\rho }^m can be selected, such that all seminorms of a_kin L^p{S}_{\rho }^m are not related to k, and

\Vert T_{a_1}{\Vert }_{L^q\to L^r}\ge 1,\Vert T_{a_k}{\Vert }_{L^q\to L^r}\ge 3\Vert T_{a_{k-1}}{\Vert }_{L^q\to L^r}.

It’s not difficult to see a=\sum _{k=1}^{\mathrm{\infty }} 2^{-k}{a}_k\in L^p{S}_{\rho }^m, but T_a is not (q,r) bounded.

Take R>10 and a_R(x,\xi )=R^{-\frac{n}{p}}{\chi }_{B(0,R)}\left(x\right)\hat{\eta }\left(\xi \right). Obviously, for any m\in \mathbb{R}, a_R\in L^p{S}_{\rho }^m, and all seminorms are unrelated to R. For any q>0, take f_R=R^{-\frac{n}{q}}{\chi }_{B(0,R)}, then \Vert f{\Vert }_q\le C, and

\begin{array}{rl}\left|T_{a_R}f\right(x\left)\right|& =\left|{\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}x\cdot \xi }{a}_R(x,\xi )\hat{f}\left(\xi \right)\mathrm{d}\xi \right|\\ & =R^{-\frac{n}{p}-\frac{n}{q}}{\chi }_{B(0,R)}\left(x\right)\left|{\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}x\cdot \xi }\hat{\eta }(\xi )\widehat{{\chi }_{B(0,R)}}(\xi )d\xi \right|\\ & =c{R}^{-\frac{n}{p}-\frac{n}{q}}{\chi }_{B(0,R)}\left(x\right)\left|\right(\eta \ast {\chi }_{B(0,R)}\left)\right(x\left)\right|.\end{array}

It is easy to see from the selection of η when \left|x\right|\le R-1, there is

\left(\eta \ast {\chi }_{B(0,R)}\right)\left(x\right)\ge {\int }_{|y-x|\le 1} {\chi }_{B(0,R)}\left(y\right)\eta (x-y)\mathrm{d}y\ge c.

Therefore, for any r>0, we have

\Vert T_{a_R}{f}_R{\Vert }_r\ge c{R}^{-\frac{n}{p}-\frac{n}{q}}{\left({\int }_{\left|x\right|\le R-1} \mathrm{d}x\right)}^{\frac{1}{r}}\ge c{R}^{n(\frac{1}{r}-\frac{1}{p}-\frac{1}{q})}.

Similar process indicates that when \frac{1}{r}>\frac{1}{p}+\frac{1}{q}, for any m\in \mathbb{R}, there must exist a\in L^p{S}_{\rho }^m, such that T_ais not (q,r)bounded.

We proved the main theorems in Section 1. In Section 2, we will construct some examples under certain situations, to demonstrate that the equal sign cannot be taken for in the theorem, i.e., the theorem is optimal.

According to common practice, there are no other special notes in this paper. C, c is used to represent the positive constant of partial seminorms solely depending on n,\rho ,p,q,r,m, and a\in L^p{S}_{\rho }^m, and takes different values at different positions. In addition, if there exist multiple variables, such as x, y, we denote the L^pnorm of corresponding variable x by \Vert \cdot {\Vert }_{L^P\left({\mathbb{R}}_x^n\right)}.

2 Proof of Theorem 1.4

First, we have a brief review of the Littlewood-Paley theory. Take \eta \in C_c^{\mathrm{\infty }}\left(B\right(0,2\left)\right) satisfying \eta \left(\xi \right)=1, \mathrm{\forall }\xi \in B(0,1). Then it is easy to see that \mathrm{\forall }\xi \in {\mathbb{R}}^n, \eta \left(\xi \right)+\sum _{j=1}^{\mathrm{\infty }} \left[\eta \right(2^{-j}\xi )-\eta (2^{1-j}\xi \left)\right]=1. Hence, a can be decomposed into

a(x,\xi )=\eta \left(\xi \right)a(x,\xi )+\sum _{j=1}^{\mathrm{\infty }} \left[\eta \right(2^{-j}\xi )-\eta (2^{1-j}\xi \left)\right]a(x,\xi )=a_0(x,\xi )+\sum _{j=1}^{\mathrm{\infty }} a_j(x,\xi ).

It’s easy to see that for any j\ge 0, the support set of a_j(x,\cdot ) is contained in . Furthermore, when ξ falls into the support set of a_j(x,\cdot ), there is 1+\left|\xi \right|\approx 2^j, the fact of which will be used repeatedly later. According to the definition of L^p{S}_{\rho }^m, when ξ falls into the support set of a_j(x,\cdot ), for any multi-index α and j\ge 0, we have

(2.1)\begin{array}{rl}\Vert {\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(\cdot ,\xi ){\Vert }_p& \le C\sum _{\beta +\gamma =\alpha } 2^{-j\left|\beta \right|}\Vert {\mathrm{\partial }}_{\xi }^{\gamma }a(\cdot ,\xi ){\Vert }_p\le C\sum _{\beta +\gamma =\alpha } 2^{-j\left|\beta \right|}(1+|\xi |{)}^{m-\rho \left|\gamma \right|}\\ & \le C{2}^{j(m-\rho |\alpha \left|\right)}.\end{array}

Let

A_j(x,\xi )=\sum _{\left|\alpha \right|\le n+1} (1+|\xi |{)}^{\rho |\alpha |-m}|{\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(x,\xi )|.

With (2.1) and the property of the support set of a_j(x,\cdot ), we can immediately get

(2.2)\underset{\xi \in {\mathbb{R}}^n}{\mathrm{s}\mathrm{u}\mathrm{p}} \Vert A_j(\cdot ,\xi ){\Vert }_p\le C,

where constant C is not related to j.

Now we can decompose the operator T_ainto

\begin{array}{rl}{T}_{a}f\left(x\right)& ={\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}x\cdot \xi }a(x,\xi )\hat{f}\left(\xi \right)\mathrm{d}\xi =\sum _{j=0}^{\mathrm{\infty }} {\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}x\cdot \xi }{a}_j(x,\xi )\hat{f}\left(\xi \right)\mathrm{d}\xi \\ & =\sum _{j=0}^{\mathrm{\infty }} {\int }_{{\mathbb{R}}^n} {\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}(x-y)\cdot \xi }{a}_j(x,\xi )f\left(y\right)\mathrm{d}y\mathrm{d}\xi =\sum _{j=0}^{\mathrm{\infty }} {\int }_{{\mathbb{R}}^n} k_j(x,x-y)f\left(y\right)\mathrm{d}y\\ & =\sum _{j=0}^{\mathrm{\infty }} T_{a_j}f\left(x\right),\end{array}

where k_j(x,z)={\int }_{{\mathbb{R}}^n}{\mathrm{e}}^{\mathrm{i}z\cdot \xi }{a}_j(x,\xi )\mathrm{d}\xi.

Let {\sigma }_j\left(z\right)=(1+2^{j\rho }|z|{)}^{-n-1}. Take t=\mathrm{m}\mathrm{i}\mathrm{n}\{2,p,q\} and lett^{\prime } be the conjugate index of t, then 1\le t\le 2. When t>1, for anyx\in {\mathbb{R}}^n, with simple calculation, we can get

\begin{array}{rl}{\left({\int }_{{\mathbb{R}}^n} {\left|\frac{k_j(x,x-y)}{{\sigma }_j(x-y)}\right|}^{t^{\prime }}\mathrm{d}y\right)}^{\frac{1}{t^{\prime }}}& ={\left({\int }_{{\mathbb{R}}^n} \left|k_j\right(x,z\left)\right(1+2^{j\rho }\left|z\right|{)}^{n+1}{|}^{t^{\prime }}\mathrm{d}z\right)}^{\frac{1}{t^{\prime }}}\\ & \le C{\sum }_{\left|\alpha \right|\le n+1} {\left({\int }_{{\mathbb{R}}^n} \left|2^{j\rho \left|\alpha \right|}{z}^{\alpha }{k}_j\right(x,z){|}^{t^{\prime }}\mathrm{d}z\right)}^{\frac{1}{t^{\prime }}}\\ & =C{\sum }_{\left|\alpha \right|\le n+1} {\left({\int }_{{\mathbb{R}}^n} {\left|2^{j\rho \left|\alpha \right|}{z}^{\alpha }{\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}z\cdot \xi }{a}_j(x,\xi )\mathrm{d}\xi \right|}^{t^{\prime }}\mathrm{d}z\right)}^{\frac{1}{t^{\prime }}}\\ & =C{\sum }_{\left|\alpha \right|\le n+1} {\left({\int }_{{\mathbb{R}}^n} {\left|{\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}z\cdot \xi }(2^{j\rho \left|\alpha \right|}{\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(x,\xi ))\mathrm{d}\xi \right|}^{t^{\prime }}\mathrm{d}z\right)}^{\frac{1}{t^{\prime }}}.\end{array}

Since t\le 2, according to Hausdorff-Young inequation, there is

{\left({\int }_{{\mathbb{R}}^n} {\left|{\int }_{{\mathbb{R}}^n} {\mathrm{e}}^{\mathrm{i}z\cdot \xi }(2^{j\rho |\alpha |}{\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(x,\xi ))\mathrm{d}\xi \right|}^{t^{\prime }}\mathrm{d}z\right)}^{\frac{1}{t^{″}}}\le {\left({\int }_{{\mathbb{R}}^n} (2^{j\rho |\alpha |}|{\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(x,\xi )|{)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}.

So

\begin{array}{rl}{\left({\int }_{{\mathbb{R}}^n} {\left|\frac{k_j(x,x-y)}{{\sigma }_j(x-y)}\right|}^{t^{\prime }}\mathrm{d}y\right)}^{\frac{1}{t^{\prime }}}& \le C\sum _{\left|\alpha \right|\le n+1} {\left({\int }_{{\mathbb{R}}^n} (2^{j\rho \left|\alpha \right|}|{\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(x,\xi )|{)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}\\ & \le C{2}^{jm}\sum _{\left|\alpha \right|\le n+1} {\left({\int }_{{\mathbb{R}}^n} {((1+|\xi |{)}^{\rho |\alpha |-m}|{\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(x,\xi )|)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}\\ & \le C{2}^{jm}{\left({\int }_{{\mathbb{R}}^n} {\left(\sum _{|\alpha |\le n+1} (1+|\xi |{)}^{\rho |\alpha |-m}|{\mathrm{\partial }}_{\xi }^{\alpha }{a}_j(x,\xi )|\right)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}\\ & \le C{2}^{jm}{\left({\int }_{{\mathbb{R}}^n} A_j(x,\xi {)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}.\end{array}

When t=1, with the same calculation, we can get

\underset{y\in {\mathbb{R}}^n}{\mathrm{s}\mathrm{u}\mathrm{p}} \left|\frac{k_j(x,x-y)}{{\sigma }_j(x-y)}\right|\le C{2}^{jm}\underset{{\mathbb{R}}^n}{\int } A_j(x,\xi )\mathrm{d}\xi .

Based on the above, when 1\le t\le 2, there is

(2.3){\Vert \frac{k_j(x,x-y)}{{\sigma }_j(x-y)}\Vert }_{L^{t^{\prime }}\left({\mathbb{R}}_y^n\right)}\le C{2}^{jm}{\left({\int }_{{\mathbb{R}}^n} A_j(x,\xi {)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}},\mathrm{\forall }x\in {\mathbb{R}}^n.

Here, Hausdorff-Young inequation is used, so 1\le t\le 2 is necessary.

Now we estimate the norm \Vert T_{a}f{\Vert }_r. Through Hölder Inequation and Formula (2.3), there is

(2.4)\begin{array}{rl}\Vert T_{a_j}f{\Vert }_r& ={\Vert {\int }_{{\mathbb{R}}^n} \frac{k_j(x,x-y)}{{\sigma }_j(x-y)}{\sigma }_j(x-y)f\left(y\right)\mathrm{d}y\Vert }_{L^r\left({\mathbb{R}}_x^n\right)}\\ & \le {\Vert {\Vert \frac{k_j(x,x-y)}{{\sigma }_j(x-y)}\Vert }_{L^{t^{\prime }}\left({\mathbb{R}}_y^n\right)}{\left({\int }_{{\mathbb{R}}^n} {\sigma }_j^t(x-y)\left|f\right(y){|}^t\mathrm{d}y\right)}^{\frac{1}{t}}\Vert }_{L^r\left({\mathbb{R}}_x^n\right)}\\ & \le C{2}^{jm}{\Vert {\left({\int }_{{\mathbb{R}}^n} A_j(x,\xi {)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}({\sigma }_j^t\ast |f{|}^t{)}^{\frac{1}{t}}\Vert }_{L^r({\mathbb{R}}_r^n)}.\end{array}

Since r\le p, there exists p_1\ge r, such that

\frac{1}{r}=\frac{1}{p}+\frac{1}{p_1}.

By applying Hölder inequation, there is

(2.5)\begin{array}{rl}& {\Vert {\left({\int }_{{\mathbb{R}}^n} A_j(x,\xi {)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}{\left({\sigma }_j^t\ast |f{|}^t\right)}^{\frac{1}{t}}\Vert }_{L^r\left({\mathbb{R}}_x^n\right)}\\ & \le {\Vert {\left({\int }_{{\mathbb{R}}^n} A_j(x,\xi {)}^t\mathrm{d}\xi \right)}^{\frac{1}{t}}\Vert }_{L^p\left({\mathbb{R}}_x^n\right)}{\Vert {\left({\sigma }_j^t\ast |f{|}^t\right)}^{\frac{1}{t}}\Vert }_{p_1}\\ & ={\Vert {\int }_{{\mathbb{R}}^n} A_j(\cdot ,\xi {)}^t\mathrm{d}\xi \Vert }_{\frac{p}{t}}^{\frac{1}{t}}{\Vert {\sigma }_j^t\ast |f{|}^t\Vert }_{\frac{p_1}{t}}^{\frac{1}{t}}.\end{array}

Because \frac{1}{r}\le \frac{1}{p}+\frac{1}{q}, \frac{1}{p_1}\le \frac{1}{q}, t\le q, there exists p_2\ge t, such that

\frac{t}{p_2}=1+\frac{t}{p_1}-\frac{t}{q}\le 1.

According to relation 1+\frac{t}{p_1}=\frac{t}{q}+\frac{t}{p_2}, by applying Young inequation to {\sigma }_j^t\ast |f{|}^t, we can get

(2.6){\Vert {\sigma }_j^t\ast |f{|}^t\Vert }_{\frac{p_1}{t}}^{\frac{1}{t}}\le \Vert {\sigma }_j^t{\Vert }_{\frac{p_2}{t}}^{\frac{1}{t}}\Vert |f{|}^t{\Vert }_{\frac{q}{t}}^{\frac{1}{t}}=\Vert {\sigma }_j{\Vert }_{p_2}\Vert f{\Vert }_q.

By combining formulas (2.2) and (2.4)‒(2.6), using Minkowski inequation and some simple calculations, we can get

By substituting \frac{1}{p_2}=\frac{1}{t}+\frac{1}{r}-\frac{1}{p}-\frac{1}{q}, t=\mathrm{m}\mathrm{i}\mathrm{n}\{2,p,q\} into the above formula, there is

\Vert T_{a}f{\Vert }_r\le \sum _{j=0}^{\mathrm{\infty }} \Vert T_{a_j}f{\Vert }_r\le C\sum _{j=0}^{\mathrm{\infty }} 2^{j\left(m+\frac{n(1-\rho )}{\mathrm{m}\mathrm{i}\mathrm{n}\{2,p,q\}}+n\rho \left(\frac{1}{p}+\frac{1}{q}-\frac{1}{r}\right)\right)}\Vert f{\Vert }_q.

, so

\Vert T_{a}f{\Vert }_r\le C\Vert f{\Vert }_q.

3 Some counterexamples

In this section, we will construct some examples to illustrate that the condition in Theorem 1.4 generally cannot take the equal sign.

We first introduce a well-known conclusion, and for completeness, here is a simple proof.

Lemma 3.1　 Assume , p, q\le \mathrm{\infty } satisfying \frac{1}{r}=\frac{1}{p}+\frac{1}{q}, if g\notin L^q, then there exists f\in L^p, such that fg\notin L^r.

Proof　If r=\mathrm{\infty }, p=q=\mathrm{\infty }. Now the Lemma is obvious.

If , with simple transformation it’s easy to see that only the result of when r=1 needs to be proved.

When q=1, g\notin L^1, directly take f=1, there is {\int }_{{\mathbb{R}}^n}\mid f\left(x\right)g\left(x\right)\mid \mathrm{d}x=\mathrm{\infty }.

When q=\mathrm{\infty }, g\notin L^{\mathrm{\infty }}, let E_k=\{x\in {\mathbb{R}}^n:|g\left(x\right)|>3^k\}. Then for any k\in \mathbb{N}, there is m\left(E_k\right)>0. Take f\left(x\right)=\sum _{k=1}^{\mathrm{\infty }} \frac{{\chi }_{E_k}\left(x\right)}{2^k\left|E_k\right|}, obviously there is f\in L^1 and

{\int }_{{\mathbb{R}}^n} \left|f\right(x\left)g\right(x\left)\right|\mathrm{d}x={\int }_{{\mathbb{R}}^n} \sum _{k=1}^{\mathrm{\infty }} \frac{{\chi }_{E_k}\left(x\right)}{2^k\left|E_k\right|}\left|g\right(x\left)\right|\mathrm{d}x=\sum _{k=1}^{\mathrm{\infty }} {\int }_{E_k} \frac{\left|g\right(x\left)\right|}{2^k\left|E_k\right|}\mathrm{d}x\ge \sum _{k=1}^{\mathrm{\infty }} {\left(\frac{3}{2}\right)}^k=\mathrm{\infty }.

When , and since g\notin L^q, \mathrm{\forall }k\in \mathbb{N}, there exists g_k such that \left|g_k\right|\le \left|g\right| and \Vert g_k{\Vert }_q=3^k. Let

f\left(x\right)=\sum _{k=1}^{\mathrm{\infty }} 2^{-k}{\left(\frac{\left|g_k\right(x\left)\right|}{\Vert g_k{\Vert }_q}\right)}^{\frac{q}{p}}.

With direct calculation we can get

\Vert f{\Vert }_p\le \sum _{k=1}^{\mathrm{\infty }} 2^{-k}{\Vert {\left(\frac{\left|g_k\right|}{\Vert g_k{\Vert }_q}\right)}^{\frac{q}{p}}\Vert }_p=\sum _{k=1}^{\mathrm{\infty }} 2^{-k}\Vert g_k{\Vert }_q^{\frac{q}{p}}\Vert g_k{\Vert }_q^{-\frac{q}{p}}=1.

Now, according to assumption \left|g_k\right|\le \left|g\right|, there is

\begin{array}{rl}{\int }_{{\mathbb{R}}^n} \left|f\right(x\left)g\right(x\left)\right|\mathrm{d}x& ={\int }_{{\mathbb{R}}^n} \sum _{k=1}^{\mathrm{\infty }} 2^{-k}{\left(\frac{\left|g_k\right(x\left)\right|}{\Vert g_k{\Vert }_q}\right)}^{\frac{q}{p}}\left|g\right(x\left)\right|\mathrm{d}x\\ & \ge \sum _{k=1}^{\mathrm{\infty }} 2^{-k}\Vert g_k{\Vert }_q^{-\frac{q}{p}}{\int }_{{\mathbb{R}}^n} \left|g_k\right(x){|}^{\frac{q}{p}+1}\mathrm{d}x\\ & =\sum _{k=1}^{\mathrm{\infty }} 2^{-k}\Vert g_k{\Vert }_q^{-\frac{q}{p}}\Vert g_k{\Vert }_q^q=\sum _{k=1}^{\mathrm{\infty }} 2^{-k}\Vert g_k{\Vert }_q=\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{3}{2}\right)}^k=\mathrm{\infty }.\end{array}

Hence, we proved the Lemma. □

Now, with some known examples, we illustrate that index is optimal under certain circumstances.

Proposition 3.1　 For 0\le \rho \le 1, if q=2, p>2, and \frac{1}{r}=\frac{1}{p}+\frac{1}{2}, then there exists a\in L^p{S}_{\rho }^{\frac{n(\rho -1)}{2}}, and T_a is not bounded from L² to L^r.

Proof　According to [13], there exists b\in S_{\rho ,1}^{\frac{n(\rho -1)}{2}} such that T_b is not L² bounded, i.e., there exists f_b\in L^2 such that T_b{f}_b\notin L^2; with Lemma 3.1, there exists a₁∈L^p, a_1{T}_b{f}_b\notin L^r. Let a(x,\xi )=a_1\left(x\right)b(x,\xi ). Then there is a\in L^p{S}_{\rho }^{\frac{n(\rho -1)}{2}} but T_a{f}_b\notin L^r, i.e., T_a is not (2, r) bounded. □

Proposition 3.2　 For 0\le \rho \le 1, if q=1 and \frac{1}{r}=\frac{1}{p}+1, then there exists a\in L^p{S}_{\rho }^{n(\rho -1)}, and T_a is not bounded from L¹ to L^r.

Proof　According to [4], there exists b\in S_{\rho ,1}^{n(\rho -1)}, and T_b is not L¹ bounded. Similar to Proposition 3.1, it can be seen that there exists a_1\in L^p, a(x,\xi )=a_1\left(x\right)b(x,\xi )\in L^p{S}_{\rho }^{n(\rho -1)}, such that T_a is not (1, r) bounded. □

In the same way, for \rho =1, by combining the example in [4], we can reach the following conclusions.

Proposition 3.3　 If \rho =1, , and \frac{1}{r}=\frac{1}{p}+\frac{1}{q}, then there exists a\in L^p{S}_1^0, and T_a is not bounded from L^qto L^r. Besides, we can also get the following proposition.

Proposition 3.4　 For , when q=\mathrm{\infty }, p=r\ge 2, there exists a\in L^p{S}_{\rho }^{\frac{n(\rho -1)}{2}}, and T_a is not bounded from L^∞ to L^r.

Proof　Take the Fourier multiplier in [12], b\left(\xi \right)=(1-\eta (\xi \left)\right)|\xi {|}^{\frac{n(\rho -1)}{2}}{\mathrm{e}}^{\mathrm{i}|\xi {|}^{1-\rho }}. When , b\in S_{\rho ,0}^{\frac{n(\rho -1)}{2}}, and T_b is not L^∞ bounded. Similarly, we know from Lemma 3.1 that there exists a_1\in L^p, for a(x,\xi )=a_1\left(x\right)b\left(\xi \right), there is a\in L^p{S}_{\rho }^{\frac{n(\rho -1)}{2}}, and T_a is not bounded from L^∞ to L^r. □

In the above propositions, we have all required \frac{1}{r}=\frac{1}{p}+\frac{1}{q}. Finally, for another special case where \rho =0, we demonstrate that Theorem 1.4 is optimal in more general cases. For the sake of convenience, we only calculate the case where n=1, and the same result holds for general n=1 of the calculation results in [4] without proof.

Proposition 3.5　 Assume n=1, \rho =0, then as long as q>1, there must exist a\in L^p{S}_0^{-\frac{1}{q}}, and T_a is not (q, r) bounded.

Proof　Take a non-negative function \mathrm{\Phi }\in C_c^{\mathrm{\infty }}(-\frac{3}{4},\frac{3}{4}), and when \xi \in (-\frac{2}{3},\frac{2}{3}), \mathrm{\Phi }\left(\xi \right)=1. Assume \hat{\varphi }=\mathrm{\Phi }, then \varphi \in \mathcal{S} and \Vert \varphi {\Vert }_1\ge \mathrm{\Phi }\left(0\right)=1. Let

For any positive integer N>8, take f_N satisfying

We get \left|T_b{f}_N\right(x\left)\right|\ge CN\left|\varphi \right(x\left)\right| from the calculation in [4, p12]. Let a(x,\xi )=\varphi \left(x\right)b(x,\xi ), it’s easy to prove that for any p\ge 1, there is a\in L^p{S}_0^{-\frac{1}{q}}, and

\Vert T_a{f}_N{\Vert }_r\ge CN\Vert {\varphi }^2{\Vert }_r\ge cN.

With [4, (4.2)], \Vert f_N{\Vert }_q\le C{N}^{\frac{1}{q}}. Hence, we readily know T_a is not (q,r) bounded. □

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