On pointwise convergence of sequential Boussinesq operator

1.   Introduction

1.1.   The pointwise convergence of the Schrödinger operator

The formal solution to the free Schrödinger equation

is defined by

where

Carleson ^[6] considered the following problem: Determine the optimal $s$ for which

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

In 1979, Carleson ^[6] first showed that the almost everywhere convergence (1.1) holds for any $f\in H^{\frac14}(\mathbb{R})$. Dahlberg-Kenig ^[10] proved (1.1) fails for $s < \frac14$ when $n\geq1$. For the situation in higher dimensions, many researchers such as Carbery ^[5] and Cowling ^[9] considered this problem, and Sjölin ^[31] and Vega ^[38] proved independently that (1.1) holds when $s > \frac12$ in any dimensions. The sufficient condition of (1.1) has been obtained in ^{[1,2,7,11,13,15,20,21,23,28,29,30,37]}. Bourgain ^[3] gave counterexamples demonstrating that (1.1) fails provided $s < \frac n{2(n+1)}$. The best sufficient condition was improved by Du-Guth-Li ^[14] and Du-Zhang ^[16] in general dimension $n\geq2$. Hence, the Carleson problem was essentially solved except for the endpoint.

1.2.   The pointwise convergence of Boussinesq operator

As a nonlinear variant of (1.1), the Boussinesq operator acting on $f\in S(\mathbb{R}^n)$ is given by

which occurs in many physical situations. The name of this operator comes from the Boussinesq equation (cf. ^[4])

modelling the propagation of long waves on the surface of water with small amplitude.

We are motivated by Section 1.1 and the similarity between the Schrödinger operator and the Boussinesq operator to study the pointwise convergence of $\mathcal{B}f(x, t)$: Evaluate the optimal $s_{c}$ such that

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

Cho-Ko ^[7] improved the convergence result on the Schrödinger operator to some generalized dispersive operators excluding the Boussinesq operator. Li-Li ^[22] proved that almost everywhere convergence (1.2) holds for any $f\in H^{\frac14}(\mathbb{R})$ and Li-Li ^[22] also proved the condition $s\geq\frac14$ is sharp. Li-Wang ^[25] obtained almost everywhere convergence (1.2) holds for the optimal $s_{c} = \frac13$ when $n = 2$.

In this paper, we are interested in a related problem: To study the pointwise convergence of $\mathcal{B}f(x, t_n)$, where $\{t_n\}_{n = 1}^\infty$ is a decreasing sequence with $\lim\limits_{n\rightarrow \infty} t_n = 0$. One may expect less regularity on $f$ is enough to obtain convergence in the discrete case. Let's review the convergence of the Schrödinger operator. When $t_n = \frac1n, n = 1, 2, \cdot\cdot\cdot$, Carleson ^[6] proved that the convergence result holds provided that $s > \frac14$ but fails for $s < \frac18$ in one dimension. Indeed, it actually fails for $s < \frac14$ by the counterexample in Dahlberg-Kenig ^[10]; see Lee-Rogers ^[21] for more details. Recently, this problem was further studied by ^{[8,12,24,26,32,33]}. In particular, under the assumption that $\{t_n\}_{n = 1}^\infty$ belongs to Lorentz space $l^{r, \infty}(\mathbb{N})$, $0 < r < \infty$, i.e.,

Dimou-Seeger ^[12] considered the fractional Schrödinger operator, which is defined by

and obtained a characterization of convergence for all functions in $H^s(\mathbb{R})$ when $0 < s < \min\left\{\frac a4, \frac14\right\}$ and $a\neq1$. Li-Wang-Yan ^[26] and Cho-Ko-Koh-Lee ^[8] extended the result of Dimou-Seeger ^[12] to higher dimensions by different methods.

In this paper, we study the almost everywhere pointwise convergence problem of the sequential Boussinesq operator. The fractional Schrödinger operator and Boussinesq operator are different operators, and the result from Dimou-Seeger ^[12] cannot cover our work. We obtain a characterization of convergence almost everywhere for any $f\in H^s(\mathbb{R})$ when $0 < s < \frac12$. Our main results are as follows:

Theorem 1.1. Suppose $0 < s < \frac12$. Let $\{t_n\}_{n = 1}^\infty$ be a decreasing sequence with $\lim\limits_{n\rightarrow \infty}t_n = 0$, and suppose that $\{t_n-t_{n+1}\}_{n = 1}^\infty$ is also decreasing. Then the following four statements are equivalent.

(i) Let $r(s) = \frac{s}{1-2s}$, the sequence $\{t_n\}\in l^{r(s), \infty}(\mathbb{N})$.

(ii) There exists a constant $C_1$ such that for any $f\in H^s(\mathbb{R})$ and for all sets $B$ with $diam(B)\leq1$, we obtain

(iii) There exists a constant $C_2$ such that for any $f\in H^s(\mathbb{R})$, for all sets $B$ with $diam(B)\leq1$, and for any $\alpha > 0$, we obtain

(iv) For all $f\in H^s(\mathbb{R})$, we have

It is easy to see that (ⅱ)$\Rightarrow$(ⅲ). However, the opposite result (ⅲ)$\Rightarrow$(ⅱ) seems nontrivial, so we do not have a direct proof for it. Next, we introduce the outline of proving Theorem 1.1 briefly, as follows: We prove the following five statements: (ⅰ)$\Rightarrow$(ⅱ), (ⅰ)$\Rightarrow$(ⅳ), (ⅱ)$\Rightarrow$(ⅲ), (ⅲ)$\Rightarrow$(ⅰ), and (ⅳ)$\Rightarrow$(ⅲ).

Remark 1.1. We can drop the convexity assumption in Theorem 1.1. In fact, statements $(ⅱ)$, $(ⅲ)$, and $(ⅳ)$ hold whenever $t_n$ is decreasing and $\{t_n\}\in l^{\frac{s}{1-2s}, \infty}(\mathbb{N})$, see Proposition 2.1 for more details.

We also have a global version of the maximal function inequalities, as follows:

Theorem 1.2. Suppose $0 < s < \frac12$. Let $\{t_n\}_{n = 1}^\infty$ be a decreasing sequence with $\lim\limits_{n\rightarrow \infty}t_n = 0$, and suppose that $\{t_n-t_{n+1}\}_{n = 1}^\infty$ is also decreasing. Then the following three statements are equivalent.

(i) The sequence $\{t_n\}\in l^{\frac{s}{1-2s}, \infty}(\mathbb{N})$.

(ii) There exists a constant $C_1$ such that for any $f\in H^s(\mathbb{R})$, we have

(iii) There exists a constant $C_2$ such that for any $f\in H^s(\mathbb{R})$ and for any $\alpha > 0$,

The proof of Theorem 1.2 is similar to that of Theorem 1.1. It suffices to prove the following three statements: (ⅰ)$\Rightarrow$(ⅱ), (ⅱ)$\Rightarrow$(ⅲ), (ⅲ)$\Rightarrow$(ⅰ).

Throughout this paper, we always use $C$ to denote a positive constant, independent of the main parameters involved, 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 $f\lesssim g\lesssim f$. We shall use the notation $f\gg g$, which means that there is a sufficiently large constant $C$, which does not depend on the relevant parameters arising in the context in which the quantities $f$ and $g$ appear, such that $f\geq C g$. If the function $f$ has compact support, we use ${\rm{supp}} f$ to denote the support of $f$. We write $|A|$ for the Lebesgue measure of $A\subset \mathbb{R}$. We use $S(\mathbb{R}^n)$ to denote the Schwartz functions on $\mathbb{R}^n$. The notation ${\rm{diam}}(B)$ denotes the diameter of set $B$.

2.   The estimates of maximal functions

In this part, we have proved the boundedness of the maximal function provided $\{t_n\}\in l^{\frac{s}{1-2s}, \infty}(\mathbb{N})$, which implies that (i)$\Rightarrow$(ii) and (i)$\Rightarrow$(iv) in Theorem 1.1. At the same time, we also obtain (i)$\Rightarrow$(ii) in Theorem 1.2.

2.1.   The main lemma

Without loss of generality, we can suppose that $t_n\in(0, 1)$ for all $n\in \mathbb{N}$. Next, we study the frequency truncated operator

where $\chi\in C^\infty$ is a real-value, smooth function, ${\rm{supp}}\; \chi\subset\left\{\frac12\leq|\xi|\leq1\right\}$. We can obtain the following result by ^[19], and the following conclusion will play a crucial role in our proof.

Lemma 2.1. Let $J\subset[0, 1]$ be an interval, then

In order to prove Lemma 2.1, we review the following three lemmas first: Oscillatory integrals have played a key role in harmonic analysis. So we introduce the following well-known variant of Van der Corput's lemma:

Lemma 2.2. (Van der Corput's lemma ^[36]) For $a < b$, let $F\in C^\infty([a, b])$ be real valued and $\psi\in C^\infty([a, b])$.

(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 $[a, b]$.

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

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

Schur's lemma, which is described as follows, provides sufficient conditions for linear operators to be bounded on $L^p(\mathbb R^n)$.

Lemma 2.3. (Schur's lemma ^[18]) Assume that $K(x, y)$ is a locally integral function on a product of two $\sigma$-finite measure spaces $(X, \mu)\times(Y, \nu)$, and let $T$ be a linear operator defined by

when $f$ is bounded and compactly supported. Suppose

Then the operator $T$ extends to a bounded operator from $L^p(Y)$ to $L^p(X)$ with norm $A^{1-\frac1p}B^{\frac1p}$ for $1\leq p\leq\infty$.

The following lemma is well known.

Lemma 2.4. (Lemma 2.4.2 ^[34] or ^[27]) Suppose that $F$ is $C^1(\mathbb{R})$. Then, if $q > 1$ and $\frac1q+\frac1 {q'} = 1$,

Proof of Lemma 2.1. Let us take the first $\lambda > 1$. The proof follows from the idea of Kolmogorov-Seliverstov-Plessner method. By linearizing the maximal operator, that is, let $x\rightarrow t(x)$ be a measurable function and $t(x)\in J$. It suffices to prove

where the constant $C$ does not depend on $t(\cdot)$ and $f$. Denote

where

Since $\|\hat{f}(\lambda\cdot)\|_{L^2(\mathbb{R})} = \lambda^{-\frac12} \|f\|_{L^2(\mathbb{R})}$, our goal translates into proving the following inequality:

which can further transform into demonstrating

We use the idea of $T T^\ast$ to complete the proof. After some computation, the kernel of $T_\lambda (T_\lambda)^\ast$ is

Denote

where $\xi > 0$. Thus,

On the one hand, if $|x-y|\geq 100\lambda |t(x)-t(y)|$, we have

The first inequality follows from $\frac{1+2\lambda^2 \xi^2}{\sqrt{1+\lambda^2 \xi^2}}$ is increasing on $[\frac12, 1]$, and the second inequality follows from $\lambda > 1$.

Therefore, we have

By the definition of $K_\lambda(x, y)$, we have

Since $|K_\lambda(x, y)| \lesssim (\lambda |x-y|)^{-N}$ when $|x-y|\geq 100\lambda |t(x)-t(y)|$, we have

when $|x-y|\geq 100\lambda |t(x)-t(y)|$.

On the other hand, if $|x-y|\leq 100\lambda |t(x)-t(y)|$, we have

The first inequality follows from the fact that $\frac{3\lambda^2 \xi+2\lambda^4\xi^3}{(1+\lambda^2 \xi^2)^{\frac32}}$ is increasing on $[\frac12, 1]$. And the second inequality follows from $\lambda > 1$.

Using Lemma 2.2, we obtain

The case $\xi\in [-1, -\frac12]$ is similar to case $\xi\in [\frac12, 1]$, and we neglect the details here. (2.2) and (2.3) imply that

which implies that

By symmetry, we have the same upbound for $\sup_{y\in \mathbb R } {\int}_{\mathbb R} |K_\lambda(x, y)|dx$. By Lemma 2.3, we obtain the desired conclusion (2.1).

The case $\lambda\leq1$ follows from Lemma 2.4. By Lemma 2.4, we obtain

By Hölder's inequality, we have

Plancherel's theorem implies that

where we use $J\subset[0, 1]$ in the second inequality.

2.2.   The boundedness of a maximal function

Proposition 2.1. Let $\{t_n\}\in l^{r, \infty}(\mathbb{N})$ be decreasing. Then

Moreover, $\mathcal{B} f(x, t_n) \rightarrow f(x), \quad$ a. e. $x\in \mathbb R$ whenever $f\in H^\kappa(\mathbb R)$ for $\kappa\geq \min\left \{\frac14, \frac{r}{1+2r} \right\}$.

Proof of Proposition 2.1. We use the idea of ^[12] to complete the proof of Proposition 2.1. We use a standard inhomogeneous frequency decomposition, that is, $\sum\limits_{k\geq0}P_kf = f$, where

Obviously, $P_k P_k = P_k$.

For each integer $l\geq0$, we define

Since $\{t_n\}\in l^{r, \infty}(\mathbb{N})$, there exists $C > 0$ such that

where $s = \frac{r}{1+2r}$.

According to the frequency, we divide our proof into three parts, that is,

where

By the definition of $\# \mathfrak{n}_l$, we have

where $n\in \mathfrak{n}_l$.

Firstly, we consider the term $\|A_1\|_{L^2(\mathbb{R})}$. By Minkowski's inequality, we obtain

For $n\in \cup_{l > k(1+2r)} \mathfrak{n}_l$, $t_n$ lies in an interval of length $O(2^{-2k})$, that is, $J_{l(k)}$ with $l(k) = \lfloor k(1+2r) \rfloor$. We use Lemma 2.1 and take $J = J_{l(k)}$ and $\lambda = 2^k$. Thus

The last inequality follows from the fact that $2^{\frac k2}|J_{l(k)}|^{\frac14} \lesssim1$. Therefore, we obtain

Secondly, we study the term $\|A_2\|_{L^2(\mathbb{R})}$. For simplicity of notation, we take the change of variables $k = l-j$. By Minkowski's inequality, we have

Since $l\geq j \frac {1+2r}{2r}$, we have

By Minkowski's inequality and using Lemma 2.1 again with $J = J_l$ and $\lambda = 2^{l-j}$, we then obtain

where $s = \frac12\left(1-\frac{1}{1+2r}\right) = \frac r{1+2r}$.

Finally, we consider the estimate $\|A_3\|_{L^2(\mathbb{R})}$. We also make the change of variable $k = l+m$. By Minkowski's inequality, we have

By Minkowski's inequality and Plancherel's theorem, we obtain

From (2.5), it follows that

We are ready to combine all our ingredients and finish the proof.

which means that the maximal inequality (2.4) is established. For any $f\in S(\mathbb R)$, we have $\lim\limits_{t\rightarrow0} \mathcal{B} f(x, t) = f(x)$ for all $x\in\mathbb{R}$. By ^[22], it holds

Since Schwartz functions are dense in $H^\kappa(\mathbb{R})$, by (2.4) and (2.6), we obtain

whenever $f\in H^\kappa(\mathbb R), \kappa\geq \min\left \{\frac14, \frac{r}{1+2r} \right\}$.

3.   Necessary conditions

In this part, we use ideas from Nikishin-Stein theory to prove necessity in Theorems 1.1 and 1.2; that is, we obtain the following statements: (ⅲ)$\Rightarrow$(ⅰ), (ⅳ)$\Rightarrow$(ⅲ) in Theorem 1.1, and (ⅲ)$\Rightarrow$(ⅰ) in Theorem 1.2.

3.1.   The proofs of Theorems 1.1 and 1.2

Firstly, we introduce the following proposition:

Proposition 3.1. Suppose that for all $f\in H^s(\mathbb R)$, the limit $\lim\limits_{n\rightarrow \infty} \mathcal{B} f(x, t_n)$ exists for almost every $x\in\mathbb{R}$. Then for all compact sets $K\subset \mathbb{R}$, there exists a constant $C_k$, such that for any $\alpha > 0$,

We postpone the proof of Proposition 3.1 here, and the details will be shown in Section 3.3.

Secondly, we also need the following key lemma, which is proved in ^[12].

Lemma 3.1. ^[12] Let $\{t_n\}$ be a sequence of positive numbers in $[0, 1]$, let $0 < r < \infty$, and suppose that

Then $\{t_n\}\in l^{r, \infty}(\mathbb{N})$.

Thirdly, we are now ready to prove the necessity of the $l^{r, \infty}(\mathbb{N})$ condition in Theorems 1.1 and 1.2. In fact, we summarize the necessity of the $l^{r, \infty}(\mathbb{N})$ condition into the following Proposition 3.2 which plays a key role in this paper.

Proposition 3.2. Suppose that $\{t_n\}_{n = 1}^\infty$ is a decreasing sequence such that $\{t_n-t_{n+1}\}_{n = 1}^\infty$ is decreasing and $\lim\limits_{n\rightarrow \infty}t_n = 0$. For $0 < s < \frac12$, let $r(s) = \frac{s}{1-2s}$.

(i) If $s < \frac14$ and

holds for any $f\in H^s(\mathbb R)$, then $\{t_n\}\in l^{r(s), \infty}(\mathbb{N})$.

(ii) If $s < \frac12$ and the global weak type inequality

holds for any $f\in H^s(\mathbb R)$, then $\{t_n\}\in l^{r(s), \infty}(\mathbb{N})$.

We will prove Proposition 3.2 in Section 3.2. We are now ready to combine all our ingredients and finish the proofs of Theorems 1.1 and 1.2.

Proof of Theorem 1.1. By Proposition 2.1, we can prove the implications of (ⅰ)$\Rightarrow$(ⅱ) and (ⅰ)$\Rightarrow$(ⅳ). By Tshebyshev's inequality, we have the result (ⅱ)$\Rightarrow$(ⅲ). By the first part of Proposition 3.2, we obtain the implication (ⅲ)$\Rightarrow$(ⅰ). Finally, using Proposition 3.1, we obtain the conclusion (ⅳ)$\Rightarrow$(ⅲ). Thus, the four statements (ⅰ), (ⅱ), (ⅲ), and (ⅳ) are equivalent.

Proof of Theorem 1.2. From Proposition 2.1, we have the implication (ⅰ)$\Rightarrow$(ⅱ). It is easy to obtain the implication (ⅱ)$\Rightarrow$(ⅲ) by Tshebyshev's inequality. From the second part of Proposition 3.2, we see the implication (ⅲ)$\Rightarrow$(ⅰ). Thus, the three statements (ⅰ), (ⅱ), and (ⅲ) are equivalent.

3.2.   The proof of Proposition 3.2

We can divide the proof of part (ⅱ) of Proposition 3.2 into two cases: $s < \frac14$ and $\frac14\leq s < \frac12$. Furthermore, since (3.2) yields (3.1), we have (ⅰ) implies (ⅱ) when $s < \frac14$. So we only need to consider $\frac14\leq s < \frac12$ when we prove part (ⅱ).

We use a contradiction argument. Assume that $\{t_n\}\notin l^{r(s), \infty}(\mathbb{N})$, while (3.1) holds for $s < \frac14$ or (3.2) holds in the case $\frac14\leq s < \frac12$. By Lemma 3.1, we obtain

Thus, there exists an increasing sequence $\{K_j\}_{j = 1}^\infty$ with $\lim\limits_{j\rightarrow \infty} K_j = \infty$ and a sequence of positive numbers with $\lim\limits_{j\rightarrow \infty} b_j = 0$, such that

We choose another sequence $L_j \leq K_j$ with $\lim\limits_{j\rightarrow \infty} L_j = \infty$ so that in the case where $s < \frac14$

In the case $\frac14\leq s < \frac12$, we let $L_j = K_j$.

Using the idea originally proposed by Dahlberg-Kenig ^[10], we complete the construction of a counterexample. We introduce a family of Schwartz functions that are used to test (3.1). Take a $C^\infty$ function $g$ with ${\rm{supp}}\; g \subset \left[-\frac12, \frac12\right]$ so that ${\int}_{\mathbb R} g(\xi)d\xi = 1$ and $g(\xi)\geq0$ and study a family of functions $f_{\lambda, \rho}$, where $\lambda$ is a large number $\rho\ll\lambda$, and $\lambda, \rho$ will be given later. $f_{\lambda, \rho}$ is defined via the Fourier transform by

Thus, ${\rm{supp}}\; \widehat{f_{\lambda, \rho}}$ belongs to an interval of length $\rho\ll\lambda$ contained in $\left[-2\lambda, -\frac\lambda2\right]$. By the definition of $f_{\lambda, \rho}$, we obtain

We now study the property of $\mathcal{B}$ on $f_{\lambda, \rho}$. We have

where

For $x$ in a suitable interval $I_j\subset I$, and for suitable choices of $\lambda_j, \rho_j$ and $n(x, j)$, we obtain

In order to prove $\left| \mathcal{B}f_{\lambda_j, \rho_j}(x, t_{n(x, j)})\right|\geq \frac12$, we only need to demonstrate that

for our choices of $x, n(x, j)$ and $(\lambda_j, \rho_j)$.

From the definition of ${\Phi}_{\lambda, \rho}(\xi; x, t_n)$, it follows that

By Taylor expansion, we obtain

Noting that terms $-\lambda x+t_n \lambda\sqrt{1+\lambda^2}$ do not depend on $\xi$, we have terms $-\lambda x+t_n \lambda\sqrt{1+\lambda^2}$ do not affect our integral. We may neglect the terms $-\lambda x+t_n \lambda\sqrt{1+\lambda^2}$ and only need to consider the last three terms $I$–$III$. We consider $t_n$ with $t_n\leq \frac{b_j}2$ and let $\epsilon$ be such that $\epsilon < \frac1{100}$. We choose

Firstly, we study the upper bound of $|I|$. We consider $x$ in the interval

Observe that in the case $s < \frac14$, by (3.4), we obtain

which implies that $I_j \subset\left[0, \frac12\right]$ in this case. Each $x\in I_j$ implies $x\in \left(\frac{1+2\lambda_j^2}{\sqrt{1+\lambda_j^2}} t_{n+1}, \frac{1+2\lambda_j^2}{\sqrt{1+\lambda_j^2}} t_{n}\right]$ for a unique $n$, which we label $n(x, j)$.

We now claim that

where $t_n\leq b_j$.

We can see this as follows: Since $\{t_n-t_{n+1}\}_{n = 1}^\infty$ is decreasing, for $t_n\leq b_j$, by (3.3), we obtain

By (3.8), we have that

By (3.9) and the definitions of $\lambda_j$ and $\rho_j$, we obtain

Secondly, we consider the upper bound of $|II|$. From the definitions of $\lambda_j$ and $\rho_j$, it follows that

Finally, we provide the estimate of the last term $III$. We obtain

Using the change of variables $t = \xi s$ and the choices of $\lambda_j, \rho_j$, we obtain

Since $\epsilon < \frac1{100}$, from (3.7) and (3.10)–(3.12), we obtain

which implies that

for $x\in I_j = \left[0, \frac{b_j}{2} \frac{1+2\lambda_j^2}{\sqrt{1+\lambda_j^2}} \right]\subset [0, 1]$. By (3.1) or (3.2), we have

which implies that

Since $\lambda_j = L_j b_j^{-\frac1{2-4s}}, \ \rho_j = \epsilon b_j^{-\frac12}$, we have $b_j \lesssim \left(L_j b_j^{-\frac1{2-4s}} \right)^{2s-1} \left(\epsilon b_j^{-\frac12}\right)^{-1},$ which implies that

Since $\lim\limits_{j\rightarrow \infty} L_j = \infty$, we have $\lim\limits_{j\rightarrow \infty} L_j^{2s-1} = 0$ with $0 < s < \frac12$, which leads to a contradiction. This means that if $\{t_n\}\notin l^{r(s), \infty}(\mathbb{N})$, then (3.1) (and therefore (3.2)) fails if $s < \frac14$ and (3.2) fails if $\frac14\leq s < \frac12$. Thus we complete the proof of the proposition.

Remark 3.1. We explain the choices of the parameters $\lambda_j$ and $\rho_j$. We divide the estimate of $\Phi_{\lambda, \rho}(\xi; x, t_n)$ into three terms $I, II$ and $III$. From the estimate of $|I|$, we have

From the estimate of $|II|$, we have

From the estimate of $|III|$, we have

We obtain (3.14), which is the same as (3.15). By (3.13) and (3.14), we obtain

3.3.   The proof of Proposition 3.1

We use Nikishin's theorem here, whose proof can be found in ^[17,35]. Nikishin's theorem asserts that if $M: L^2(Y, \mu)\rightarrow L^0(\mathbb{R}^n, |\cdot|)$ is a continuous sublinear operator (with $(Y, \mu)$ an arbitrary measure space), then there exists a measurable function $\omega(x)$ with $\omega(x) > 0$ such that

Let $M^{\mathcal{B}} f(x) = \sup\limits_{n\in \mathbb{N}} | \mathcal{B} f(x, t_n)|$ and $T_n^{\mathcal{B}} g(x) = (2\pi)^{-1} {\int}_{\mathbb{R}} e^{i(x\cdot\xi +t_n |\xi| \sqrt{1+|\xi|^2})} g(\xi) d\xi$. We obtain

Then $T_n^{\mathcal{B}}$ acts on functions in the weighted $L^2$ space $L^2(\mu_s)$, where $d\mu_s = (1+|\xi|^2)^s d\xi$. Define the maximal operator $\widetilde{M}^{\mathcal{B}}g = \sup\limits_{n\in \mathbb{N}} |T_n^{\mathcal{B}} g|$. Since $\lim\limits_{n\rightarrow \infty} \mathcal{B}f(x, t_n)$ exists almost everywhere for every $f\in H^s(\mathbb{R})$, we have $\widetilde{M}^{\mathcal{B}}g < \infty$ almost everywhere for every $g\in L^2(\mu_s)$. Then ^[17] implies that the sublinear operator $\widetilde{M}^{\mathcal{B}}: L^2(\mu_s)\rightarrow L^0(|\cdot|)$ is continuous. By Nikishin's theorem, we obtain

for some weight $\omega(x)$, with $\omega(x) > 0$ almost everywhere. Without loss of generality, we may further assume that $\omega$ is bounded.

Next, for $f\in H^s(\mathbb{R})$, we obtain

and

(3.16)–(3.18) imply that

Using the change of variables $x\rightarrow x-y$, we have

We multiply both sides of (3.19) by $h(y)$, where $h$ is a strictly positive continuous function with ${\int}_{ \mathbb{R}} h(y) dy = 1$, we obtain

Then we integrate in $y$ to obtain that

which yields that

Since $h \ast\omega$ is continuous, it attains a minimum over any compact set. For every compact set $K$, by (3.20), we obtain

Therefore, Proposition 3.1 is established.

4.   Conclusions

In this paper, we study the almost everywhere pointwise convergence problem of the sequential Boussinesq operator. The fractional Schrödinger operator and Boussinesq operator are different operators, and the result from Dimou-Seeger ^[12] cannot cover our work. The Boussinesq operator along sequences $\{t_n\}_{n = 1}^\infty$ with $\lim\limits_{n\rightarrow \infty} t_n = 0$ in one dimension is studied. We obtain a characterization of convergence almost everywhere when $\{t_n\}\in l^{r, \infty}(\mathbb{N})$ for all $f\in H^s(\mathbb{R})$ provided $0 < s < \frac12$.

Author contributions

Dan Li: Conceptualization, Methodology, Investigation, Writing – Original Draft; Fangyuan Chen: Writing – Review and Editing. All authors have read and approved the final version of the manuscript for publication.

Use of AI tools declaration

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

Acknowledgments

The authors thank Professor Junfeng Li for helpful suggestions and discussions. 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). Fangyuan Chen^* (the Corresponding author) is supported by Young Elite Scientists Sponsorship Program by BAST (No. BYESS2023036) and Young Teachers Program of Beijing Institute of Fashion Technology (No. NHFZ20230100).

Conflict of interest

The authors declare they have no conflict of interest.