A note on Boussinesq maximal estimate

1.   Introduction

1.1.   The pointwise convergence of Schrödinger operator

We first introduce the free Schrödinger equation

The formal solution of (1.1) is defined by

where $\hat{f}(\xi) = \int_{\mathbb R^n} e^{-ix\cdot\xi}f(x) \, \text{d}x$.

Carleson ^[6] first posed the problem: Determine the optimal $s$ such that

holds whenever $f\in H^s(\mathbb{R}^n)$, where $H^s(\mathbb{R}^n)$ is the $L^2$ Sobolev space, which is defined by

We call the above problem as Carleson's problem.

Carleson ^[6] first showed that the almost everywhere convergence (1.2) holds for all $s\geq\frac14$ in $\mathbb{R}$. Dahlberg-Kenig ^[9] proved (1.2) fails for $s < \frac14$ when $n\geq1$. Thus, the Carleson problem was solved in one dimension. For the situation in higher dimensions, many researchers are interested in Carleson's problem. The sufficient condition of Carleson's problem has been obtained by many references ^{[1,2,5,7,8,10,11,13,15,16,18,20,21,22,23,27,28]} and references therein. Bourgain ^[3] gave counterexamples demonstrating that (1.2) fails when $s < \frac n{2(n+1)}$. The best sufficient condition was improved by Du-Guth-Li ^[12] when $n = 2$ and Du-Zhang ^[14] when $n\geq3$. Hence, the Carleson problem was essentially solved, except for the endpoint.

1.2.   The Boussinesq maximal estimate

As a nonlinear variant of (1.2), the Boussinesq operator acting on $f\in \mathcal{S}(\mathbb{R}^n)$ is given by

which occurs in many physical situations. The name of this operator comes from the Boussinesq equation

see ^[4] for more details.

We are motivated by subsection 1.1 to study the pointwise convergence of $\mathcal{B}f(x, t)$: Evaluate the optimal $s$ so that

holds for any $f\in H^s(\mathbb{R}^n)$.

Cho-Ko ^[7] improved the convergence on the Schrödinger operator to generalized dispersive operators excluding the Boussinesq operator. Li-Li ^[17] obtained the optimal $s = \frac14$ in one dimension including the endpoint. Li-Wang ^[19] obtained the almost everywhere convergence (1.3) that holds for the optimal $s = \frac13$ when $n = 2$, except for the endpoint.

In this paper, we are interested in a more general problem. Let $E$ be a bounded set in $\mathbb{R}^{n+1}$. For $f\in \mathcal{S}(\mathbb{R}^n)$, we introduce the maximal function

Let's review the fractional Schrödinger operator, which is defined by

and its maximal function, which is given by

Sjölin-Strömberg ^[24] obtained maximal function $S_E^\ast f$ is bounded from $L^2(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n)$ when $n\geq1$; see ^[25] for more studies. The Boussinesq maximal function is different from the fractional Schrödinger maximal function and they have different properties. Thus, we consider the Boussinesq maximal function in this paper. Our main result is as follows.

Theorem 1.1. Assume $n\geq1, \; \lambda\geq1$. Let the interval $J\subset[0, 1]$. Suppose $B$ is a ball in $\mathbb{R}^n$ with radius $r$ and set $E = B\times J = \{(y, t):y\in B, t\in J\}$. Let $f\in L^2(\mathbb{R}^n)$ with $\text{supp}\; \hat f\subset B(0, \lambda)$, then one has

and

In section two we give the proof of Theorem 1.1. We use the methods of frequency decomposition, linearization of the maximal operator, $TT^\ast$ and so on. In fact, in subsection 2.1 we first introduce our main lemma. In order to prove our main lemma, we shall introduce two lemmas, which are proved in section three, then we give the proof of our main lemma. In subsection 2.2 we prove Theorem 1.1.

Throughout this paper, we always use $C$ to denote a positive constant independent of the main parameters involved, but whose value may change at each occurrence. The positive constants with subscripts, such as $C_1$ and $C_2$, do not change in different occurrences. For two real functions $f$ and $g$, we always use $f\lesssim g$ or $g\gtrsim f$ to denote that $f$ is smaller than a positive constant $C$ times $g$, and we always use $f\sim g$ as shorthand for $f\lesssim g\lesssim f$. If the function $f$ has compact support, we use $\text{supp} f$ to denote the support of $f$. We write $|A|$ for the Lebesgue measure of $A\subset \mathbb{R}$. We use $\mathcal{S}(\mathbb{R}^n)$ to denote the Schwartz function on $\mathbb{R}^n$. We use $B(c, r)$ to represent the ball centered at $c$ with radius $r$ in $\mathbb{R}^n$.

2.   The Boussinesq maximal estimate

2.1.   The main lemma

In order to prove Theorem 1.1, we give our main lemma as follows.

Lemma 2.1. Assume $n\geq1, \; \lambda\geq1$. Let the interval $J\subset[0, 1]$. Suppose $B$ is a ball in $\mathbb{R}^n$ with radius $r$ and $E = B\times J = \{(y, t): y\in B, t\in J\}$. If $f\in L^2(\mathbb{R}^n)$ with $\text{supp}\; \hat f \subset B(0, \lambda)$, then

Remark 2.1. In fact, Lemma 2.1 contains the Theorem 1.1 when $n = 1$ and $n = 2$, so it suffices to prove Theorem 1.1 when $n\geq3$.

Lemma 2.1 plays a key role in the proof of Theorem 1.1. In order to prove Lemma 2.1, we shall use the following Lemmas 2.2 and 2.3. We postpone the proofs of Lemmas 2.2 and 2.3 here and the details will be shown in section three.

Lemma 2.2. Assume $n\geq1, \; \lambda\geq1$. Let the interval $J\subset[0, 1]$. Suppose $B$ is a ball in $\mathbb{R}^n$ with radius $r$ and $E = B\times J = \{(y, t): y\in B, t\in J\}$. If $f\in L^2(\mathbb{R}^n)$ with $\text{supp}\; \hat f\subset \{\xi\in \mathbb{R}^n: \frac{\lambda}{2} \leq |\xi|\leq \lambda\}$, then

The only difference between Lemma 2.1 and Lemma 2.2 is the support of $\hat f$ and that the condition of Lemma 2.1 is weaker than that of Lemma 2.2.

Remark 2.2. If we take $B = B(0, \epsilon)$ with $\epsilon > 0$ small enough in Lemma 2.2, then we have

Lemma 2.3. Let $y_0\in \mathbb{R}^n, \; t_0\in \mathbb{R}, \; 0 < r\leq1$, $f\in L^2(\mathbb{R}^n)$ with $\text{supp}\; \hat f \subset B(0, \lambda)$ and $\lambda\geq1$. Set

then

Proof of Lemma 2.1. Let $N$ be the smallest integer so that $|J|2^{-2N}\lambda^2+r 2^{-N}\lambda < 2$. We write $f = \sum\limits_{j = 0}^N f_j$ where $\text{supp}\; \widehat{f_j}\subset\{\xi\in \mathbb{R}^n: 2^{-j-1}\lambda\leq|\xi|\leq2^{-j}\lambda\}$ for $0\leq j\leq N-1$ and $\text{supp}\; \widehat{f_N}\subset B(0, 2^{-N}\lambda)$. We make the following two-fold analysis:

On the one hand, we take $E = B\times J = \{(y, t): y\in B, t\in J\}$ in Lemma 2.3, where $B$ is the same as in Lemma 2.1, which implies that

On the other hand, according to Lemma 2.2 we have

for $0\leq j\leq N-1$, which implies that

(2.1) and (2.2) yield that

This completes the proof of Lemma 2.1.

2.2.   The proof of Theorem 1.1

We now are ready to combine our main Lemma 2.1 and finish our proof.

Proof of Theorem 1.1. By Remark 2.1, it suffices to consider the case $n\geq3$. By Lemma 2.1 we have

Cover $J$ with intervals $J_i, \; i = 1, 2, \cdot\cdot\cdot, N$, of intervals of equal length $|J_i|$ such that $|J_i| \lambda^2 = r^2 \lambda^2 +1$ with $N\leq \frac{|J|}{|J_i|}+1$. Set $E_i: = B\times J_i$, then we have

which implies that

which gives the desired estimate.

3.   The proofs of Lemmas 2.2 and 2.3

3.1.   The proof of Lemma 2.2

In order to finish the proof of Lemma 2.2, we will need the following lemma, known as Van der Corput's lemma.

Lemma 3.1. (Van der Corput's lemma ^[26]) For $a < b$, let $F\in C^\infty([a, b])$ be real valued and $\psi\in C^\infty([a, b])$.

$({\rm{i}})$ If $|F'(x)|\geq \lambda > 0, \ \forall\ x\in [a, b]$ and $F'(x)$ is monotonic on $[a, b]$, then

where $C$ does not depend on $F$, $\psi$ or $[a, b]$.

$({\rm{ii}})$ If $|F''(x)|\geq \lambda > 0, \ \forall\ x\in [a, b]$, then

where $C$ does not depend on $F$, $\psi$ or $[a, b]$.

Proof of Lemma 2.2. Assume that $\chi$ is a smooth nonnegative function on $\mathbb{R}$, $\text{supp}\; \chi\subset[\frac13, \frac43]$ and $\chi\equiv1$ on $[\frac12, 1]$. We also use the same notation for the radial function on $\mathbb{R}^n$ with $\chi(\xi) = \chi(|\xi|)$, then we get the following truncated Boussinesq operator:

Let $t:\mathbb{R}^n\rightarrow J$ and $b: \mathbb{R}^n \rightarrow B$ be measurable functions. By linearizing the maximal operator, we have

where

We use the method of $TT^\ast$ to finish the proof of Lemma 2.2. After some computation, we get that the kernel of $T_\lambda T_\lambda^\ast$ is

We need to control $K_\lambda(x, y)$. However, it is difficult to estimate $K_\lambda(x, y)$, which leads us to majorize the kernel $K_\lambda$ by a convolution kernel $G_\lambda$; that is $|K_\lambda(x, y)|\lesssim G_\lambda(x-y)$. Next, we divide the proof into two parts in order to obtain the expression of function $G_\lambda$.

On the one hand, we have that the trivial estimate

holds for any $x$ and $y$. We shall use this estimate when $\lambda|x-y| \leq C_0+2\lambda d$, where $d = 2r$.

On the other hand, we discuss the case $\lambda|x-y| > C_0+2\lambda d$. Let $\sigma$ be the surface measure on the unit sphere in $\mathbb{R}^n$. Clearly, polar coordinates yield that

Stein ^[26] implies that

where $J_{\frac{n-2}{2}}(|\xi|)$ is a Bessel function, which is defined by

We take $C_0$ large enough such that

for $r\geq C_0$, where $|R(r)| \lesssim \frac{1}{r^{N+\frac 32}}$ (see ^[26]). This yields that

where $R_1(r) = r^{1-\frac n2} R(r)$.

We first consider the remainder term. Since $|R(r)| \lesssim \frac{1}{r^{N+\frac 32}}$, we obtain

Observing that $b(x), b(y)\in B$, we have $|b(x)-b(y)|\leq d$, which yields

Note that $\lambda|x-y| > C_0+2\lambda d$. It follows that

Furthermore, we conclude

Henceforth, we establish the estimate of the remainder term

In order to obtain the upbound of $\left|K_\lambda(x, y)\right|$, it suffices to estimate the main term, which is defined by

where

Next, we make the following two-fold analysis:

Case 1. $|x-y| \gg \lambda |t(x)-t(y)|$. The definition of $\Phi_\lambda(r)$ implies that $\Phi'_\lambda(r) = \lambda (t(x)-t(y)) \frac {1+2\lambda^2 r^2}{\sqrt{1+\lambda^2 r^2} } +\lambda |x-y+b(x)-b(y)|,$ which yields

Using integration by parts, we obtain

Case 2. $|x-y| \lesssim \lambda |t(x)-t(y)|$. Since $t(x), t(y)\in J$, we get $|x-y|\lesssim \lambda|J|$. It follows from the definition of $\Phi_\lambda(r)$ that $\Phi''_\lambda(r) = \lambda (t(x)-t(y)) \frac {\lambda^2 r(3+2\lambda^2 r^2)}{(1+\lambda^2 r^2)^{\frac32} }$, which implies

Using Lemma 3.1, we have

We have established the upbound of $| K_{\lambda}(x, y)|$. In summary, by $| K_{\lambda}(x, y)|\lesssim G_\lambda (x-y)$, we may take

which yields

where in the second inequality we used polar coordinates. This implies that

It follows that

We combine the above estimates and get

This completes the proof of Lemma 2.2.

3.2.   The proof of Lemma 2.3

Proof of Lemma 2.3. We write $y = (y_1, \cdot\cdot\cdot, y_n)$ and $y_0 = (y_{0, 1}, \cdot\cdot\cdot, y_{0, n})$. For $1\leq j\leq n$, we write $\Lambda_j : = e^{i\xi_j y_j} -e^{i\xi_j y_{0, j}}$ and $\Lambda_{n+1}: = e^{it|\xi| \sqrt{1+|\xi|^2}} -e^{it_0 |\xi| \sqrt{1+|\xi|^2}}$. It follows that

Henceforth, $\mathcal{B} f(x+y, t)$ is the sum of integrals of the form

or

Here, $\Omega_1$ and $\Omega_2$ are disjoint subsets of $\{1, 2, 3, \cdot\cdot\cdot, n\}$ and $\Omega_1\cup \Omega_2 = \{1, 2, 3, \cdot\cdot\cdot, n\}$.

First, we give the discussion of $\mathcal{B}_1 f(x, y, t)$. For $j\in \Omega_1$, we have

and we also get

Assuming $\Omega_1 = \{k_1, k_2, \cdot\cdot\cdot, k_m\}$, we conclude

By changing the order of integration we get

where

It follows that

Taking $L^2$ norms of both sides of (3.4) and from Minkowski's inequality and Plancherel's theorem, we deduce

where the last inequality follows by applying the fact that $f\in L^2(\mathbb{R}^n)$ and $\text{supp}\; \hat f\subset B(0, \lambda)$.

Next, we study $\mathcal{B}_2 f(x, y, t)$ in (3.3). The estimate of $\mathcal{B}_2 f(x, y, t)$ is similar to that of $\mathcal{B}_1 f(x, y, t)$. Since $\Omega_1 = \{k_1, k_2, \cdot\cdot\cdot, k_m\}$, it follows that

Changing the order of integration again, one then obtains

where

Furthermore, we get

Using Minkowski's inequality and Plancherel's theorem, we then obtain

By summation of the above integrals, we conclude that

Thus, Lemma 2.3 is established.

4.   Conclusions

In this paper, we studied the boundedness of the Boussinesq maximal operator when $n\geq1$. We obtained the Boussinesq maximal operator is bounded from $L^2(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n)$ when $f\in L^2(\mathbb{R}^n)$ and $\text{supp}\; \hat f\subset B(0, \lambda)$ by using the methods of frequency decomposition, linearization of the maximal operator, $TT^\ast$ and so on.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

Dan Li is supported by Mathematics Research Branch Institute of Beijing Association of Higher Education and Beijing Interdisciplinary Science Society (No. SXJC-2022-032) and the Disciplinary funding of Beijing Technology and Business University (No. STKY202308). Xiang Li$^\ast$ is supported by the Scientific Research Foundation Funded Project of Chuzhou University (No. 2022qd058).

Conflict of interest

The authors declare that they have no conflicts of interest.