We considered the Boussinesq maximal estimate when n≥1. We obtained the Boussinesq maximal operator B∗Ef is bounded from L2(Rn) to L2(Rn) when f∈L2(Rn) and suppˆf⊂B(0,λ).
Citation: Dan Li, Xiang Li. A note on Boussinesq maximal estimate[J]. AIMS Mathematics, 2024, 9(1): 1819-1830. doi: 10.3934/math.2024088
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We considered the Boussinesq maximal estimate when n≥1. We obtained the Boussinesq maximal operator B∗Ef is bounded from L2(Rn) to L2(Rn) when f∈L2(Rn) and suppˆf⊂B(0,λ).
We first introduce the free Schrödinger equation
{i∂tu+Δxu=0,(x,t)∈Rn×R;u(x,0)=f(x),x∈Rn. | (1.1) |
The formal solution of (1.1) is defined by
eit△f(x)=1(2π)n∫Rnei(x⋅ξ+t|ξ|2)ˆf(ξ)dξ, |
where ˆf(ξ)=∫Rne−ix⋅ξf(x)dx.
Carleson [6] first posed the problem: Determine the optimal s such that
limt→0eit△f(x)=f(x),a. e. x∈Rn | (1.2) |
holds whenever f∈Hs(Rn), where Hs(Rn) is the L2 Sobolev space, which is defined by
Hs(Rn):={f∈S′:‖ |
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.
As a nonlinear variant of (1.2), the Boussinesq operator acting on f\in \mathcal{S}(\mathbb{R}^n) is given by
\begin{align*} \mathcal{B}f(x, t) = (2\pi)^{-n}\int_{\mathbb{R}^n} e^{i(x\cdot\xi+t|\xi|\sqrt{1+|\xi|^2})} \hat{f}(\xi) \, \text{d}\xi, \end{align*} |
which occurs in many physical situations. The name of this operator comes from the Boussinesq equation
u_{tt}-u_{xx}\pm u_{xxxx} = (u^2)_{xx}, \ \ \forall\; (x, t)\in \mathbb R\times [0, +\infty); |
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
\begin{align} \lim\limits_{t\rightarrow0}\mathcal{B}f(x, t) = f(x), \quad\text{a. e.}\ \ x\in\mathbb{R}^n \end{align} | (1.3) |
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
\mathcal{B}_E^\ast f(x): = \sup\limits_{(y, t)\in E}\left| \mathcal{B} f(x+y, t)\right|, \quad x\in \mathbb{R}^n. |
Let's review the fractional Schrödinger operator, which is defined by
\begin{align*} S f(x, t) = (2\pi)^{-n}\int_{\mathbb{R}^n} e^{i(x\cdot\xi+t|\xi|^a)} \hat{f}(\xi) \, \text{d}\xi, \quad a > 0, \end{align*} |
and its maximal function, which is given by
S_E^\ast f(x): = \sup\limits_{(y, t)\in E}\left| S f(x+y, t)\right|, \quad x\in \mathbb{R}^n. |
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
\left\| \mathcal{B}_E^{\ast} f \right\|_{L^2(\mathbb{R})} \lesssim \left( |J|^{\frac14} \lambda^{\frac 12} +r^{\frac12} \lambda^{\frac12}+1 \right) \left\| f \right\|_{L^2(\mathbb{R})} \; when\, n = 1 |
and
\left\| \mathcal{B}_E^{\ast} f \right\|_{L^2(\mathbb{R}^n)} \lesssim \left( |J|^{\frac12} \lambda +r \lambda+1 \right) \left( r \lambda+1 \right)^{ \frac{n-2}{2}} \left\| f \right\|_{L^2(\mathbb{R}^n)}\; when\, n\geq2. |
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 .
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
\left\|\mathcal{B}_E^\ast f \right\|_{L^2(\mathbb{R}^n)} \lesssim \left( |J|^{\frac n4} \lambda^{\frac n2} + r^{\frac n2} \lambda^{\frac n2} +1 \right) \left\| f \right\|_{L^2(\mathbb{R}^n)}. |
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
\left\|\mathcal{B}_E^\ast f \right\|_{L^2(\mathbb{R}^n)} \lesssim \left( |J|^{\frac n4} \lambda^{\frac n2} + r^{\frac n2} \lambda^{\frac n2} +1 \right) \left\| f \right\|_{L^2(\mathbb{R}^n)}. |
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
\left\|\sup\limits_{t\in J} \left| \mathcal{B} f (\cdot, t) \right| \right\|_{L^2(\mathbb{R}^n)} \lesssim \left( |J|^{\frac n4} \lambda^{\frac n2} +1 \right) \left\| f \right\|_{L^2(\mathbb{R}^n)}. |
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
E = \left\{(y, t)\in \mathbb{R}^{n+1}: y_{y_0, j} \leq y_j\leq y_{y_0, j} +r \quad {for} \; 1\leq j\leq n \; {and} \; t_0\leq t\leq t_0+r^2\right\}, |
then
\left\| \mathcal{B}_E^\ast f \right\|_{L^2(\mathbb{R}^n)} \lesssim (1+r^2 \lambda^2) (1+r\lambda)^n \left\| f \right\|_{L^2(\mathbb{R}^n)}. |
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
\begin{align} \left\| \mathcal{B}_E^\ast f_N \right\|_{L^2(\mathbb{R}^n)} \lesssim (1+|J|2^{-2N}\lambda^2)(1+r2^{-N}\lambda)^n \left\| f_N \right\|_{L^2(\mathbb{R}^n)} \lesssim \left\| f \right\|_{L^2(\mathbb{R}^n)}. \end{align} | (2.1) |
On the other hand, according to Lemma 2.2 we have
\left\| \mathcal{B}_E^\ast f_j \right\|_{L^2(\mathbb{R}^n)} \lesssim \left(2^{-\frac{jn}{2}} |J|^{\frac n4} \lambda^{\frac n2} + r^{\frac n2} \lambda^{\frac n2} 2^{-\frac{jn}{2}} \right) \left\| f \right\|_{L^2(\mathbb{R}^n)} |
for 0\leq j\leq N-1 , which implies that
\begin{align} \left\| \mathcal{B}_E^\ast \left( \sum\limits_{j = 0}^{N-1} f_j \right) \right\|_{L^2(\mathbb{R}^n)} \lesssim \left(|J|^{\frac n4} \lambda^{\frac n2} + r^{\frac n2} \lambda^{\frac n2} \right) \left\| f \right\|_{L^2(\mathbb{R}^n)}. \end{align} | (2.2) |
(2.1) and (2.2) yield that
\left\| \mathcal{B}_E^\ast f \right\|_{L^2(\mathbb{R}^n)} \leq \sum\limits_{j = 0}^N \left\| \mathcal{B}_E^\ast f_j \right\|_{L^2(\mathbb{R}^n)} \lesssim \left(|J|^{\frac n4} \lambda^{\frac n2} + r^{\frac n2} \lambda^{\frac n2} +1 \right) \left\| f \right\|_{L^2(\mathbb{R}^n)}. |
This completes the proof of Lemma 2.1.
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
\left\| \mathcal{B}_E^{\ast} f \right\|^2_{L^2(\mathbb{R}^n)} \lesssim \left( |J|^{\frac n2} \lambda^n +r^n \lambda^n +1 \right) \left\| f \right\|^2_{L^2(\mathbb{R}^n)}. |
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
\begin{align*} \left\| \mathcal{B}_{E_i}^{\ast} f \right\|^2_{L^2(\mathbb{R}^n)} & \lesssim \left( \left( |J_i| \lambda^2\right)^{\frac n2} + \left( r^2 \lambda^2 +1 \right)^{\frac n2} \right) \left\| f \right\|^2_{L^2(\mathbb{R}^n)}\\ & = 2 \left( r^2 \lambda^2 +1 \right)^{\frac n2} \left\| f \right\|^2_{L^2(\mathbb{R}^n)}, \end{align*} |
which implies that
\begin{align*} \left\| \mathcal{B}_E^{\ast} f \right\|^2_{L^2(\mathbb{R}^n)} &\leq \sum\limits_{i = 1}^N \left\| \mathcal{B}_{E_i}^{\ast} f \right\|^2_{L^2(\mathbb{R}^n)}\\ &\lesssim N \left( r^2 \lambda^2 +1 \right)^{\frac n2} \left\| f \right\|^2_{L^2(\mathbb{R}^n)}\\ &\leq \left(\frac{|J|}{|J_i|}+1\right) \left( r^2 \lambda^2 +1 \right)^{\frac n2} \left\| f \right\|^2_{L^2(\mathbb{R}^n)} \\ & = \left(|J| \lambda^2 \left( r^2 \lambda^2 +1 \right)^{-1} +1 \right) \left( r^2 \lambda^2 +1 \right)^{\frac n2} \left\| f \right\|^2_{L^2(\mathbb{R}^n)} \\ & = \left(|J| \lambda^2 + r^2 \lambda^2 +1\right) \left( r^2 \lambda^2 +1 \right)^{\frac {n-2}2} \left\| f \right\|^2_{L^2(\mathbb{R}^n)}\\ &\leq\left(|J|^{\frac12} \lambda+ r \lambda +1\right)^2 \left( r \lambda +1 \right)^{n-2} \left\| f \right\|^2_{L^2(\mathbb{R}^n)}, \end{align*} |
which gives the desired estimate.
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
\begin{align*} \left|\int_a^b e^{iF(x)}\psi(x)\, \text{d}x\right| \leq\frac{C}{\lambda}\left(\left|\psi(b)\right|+\int_a^b \left|\psi'(x)\right|\, \text{d}x\right), \end{align*} |
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
\begin{align*} \left|\int_a^b e^{iF(x)}\psi(x)\, \text{d}x\right| \leq \frac{C}{\lambda^{\frac{1}{2}}}\left(\left|\psi(b)\right|+\int_a^b \left|\psi'(x)\right|\, \text{d}x\right), \end{align*} |
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:
\begin{align*} \mathcal{B}_\lambda f(x, t) : = (2\pi)^{-n}\int_{\mathbb{R}^n} e^{i(x\cdot\xi+t|\xi|\sqrt{1+|\xi|^2})} \hat{f}(\xi) \chi\left( \frac{\xi}{\lambda}\right)\, \text{d}\xi. \end{align*} |
Let t:\mathbb{R}^n\rightarrow J and b: \mathbb{R}^n \rightarrow B be measurable functions. By linearizing the maximal operator, we have
\begin{align*} \mathcal{B}_\lambda f(x+b(x), t(x)) & = (2\pi)^{-n}\int_{\mathbb{R}^n} e^{i((x+b(x))\cdot\xi+t(x)|\xi|\sqrt{1+|\xi|^2})} \hat{f}(\xi) \chi\left( \frac{\xi}{\lambda}\right)\, \text{d}\xi \\ & = (2\pi)^{-n} \lambda^n \int_{\mathbb{R}^n} e^{i(\lambda(x+b(x))\cdot\eta+t(x)|\lambda\eta|\sqrt{1+|\lambda\eta|^2})} \hat{f}(\lambda\eta) \chi(\eta)\, \text{d}\eta \\ & = :\lambda^n T_\lambda(\hat f(\lambda\cdot))(x), \end{align*} |
where
T_\lambda g(x) : = \int_{\mathbb{R}^n} e^{i(\lambda(x+b(x))\cdot\xi+t(x)|\lambda\xi|\sqrt{1+|\lambda\xi|^2})}g(\xi) \chi(\xi)\, \text{d}\xi. |
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
K_\lambda(x, y) : = \int_{\mathbb{R}^n} e^{i(\lambda(x-y+b(x)-b(y))\cdot\xi+(t(x)-t(y))|\lambda\xi|\sqrt{1+|\lambda\xi|^2})} \chi^2(\xi) \, \text{d}\xi. |
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
\left|K_\lambda(x, y) \right| \lesssim 1 |
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
\begin{align*} K_\lambda(x, y) & = \int_0^\infty e^{i\lambda r(t(x)-t(y)) \sqrt{1+\lambda^2 r^2} } \chi^2(r) \left(\int_{\mathbb{S}^{n-1}} e^{i\lambda r (x-y+b(x)-b(y)) \cdot\xi'} \, \text{d}\sigma(\xi') \right) r^{n-1} \, \text{d}r \\ & = \int_0^\infty e^{i\lambda r(t(x)-t(y)) \sqrt{1+\lambda^2 r^2} } \chi^2(r) \hat \sigma(\lambda(x-y +b(x)-b(y))r) r^{n-1} \, \text{d}r . \end{align*} |
Stein [26] implies that
\hat \sigma(\xi) = (2\pi)^{-n}|\xi|^{1-\frac n2} J_{\frac{n-2}{2}}(|\xi|), |
where J_{\frac{n-2}{2}}(|\xi|) is a Bessel function, which is defined by
J_\nu(t) = \frac{\left(\frac t 2\right)^\nu}{\Gamma\left(\nu+\frac12\right)\Gamma\left(\frac12\right)} \int_{-1}^1 e^{its} (1-s^2)^\nu \frac{\, \text{d}s}{\sqrt{1-s^2}}. |
We take C_0 large enough such that
\begin{align*} J_{\frac{n-2}{2}}(r) = a_0 \frac{e^{ir}}{r^{\frac12}} +a_1 \frac{e^{ir}}{r^{\frac32}} +\cdot\cdot\cdot+a_N \frac{e^{ir}}{r^{N+\frac 12}} +b_0 \frac{e^{-ir}}{r^{\frac12}} +b_1 \frac{e^{-ir}}{r^{\frac32}} +\cdot\cdot\cdot+b_N \frac{e^{-ir}}{r^{N+\frac 12}} +R(r), \end{align*} |
for r\geq C_0 , where |R(r)| \lesssim \frac{1}{r^{N+\frac 32}} (see [26]). This yields that
\begin{align} K_\lambda(x, y) & = \int_0^\infty e^{i\lambda r(t(x)-t(y)) \sqrt{1+\lambda^2 r^2} } \chi^2(r)r^{n-1} \Bigg ( a_0 \frac{e^{i\lambda|x-y+b(x)-b(y)|r}}{(\lambda|x-y+b(x)-b(y)|r)^{\frac n2-\frac 12}} \\ &\quad+\cdot\cdot\cdot +b_N \frac{e^{-i\lambda|x-y+b(x)-b(y)|r}}{(\lambda|x-y+b(x)-b(y)|r)^{N+\frac n2-\frac 12}} +R_1(\lambda|x-y+ b(x)-b(y)|r) \Bigg ) \, \text{d}r, \end{align} | (3.1) |
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
R_1(\lambda|x-y+b(x)-b(y)|r) \lesssim \frac{1}{(\lambda|x-y+b(x)-b(y)|)^{N+\frac n2+\frac12}} . |
Observing that b(x), b(y)\in B , we have |b(x)-b(y)|\leq d , which yields
|x-y|\left( 1-\frac{d}{|x-y|} \right) = |x-y| -d < |x-y + b(x)-b(y)| < |x-y| + d = |x-y|\left( 1+\frac{d}{|x-y|} \right). |
Note that \lambda|x-y| > C_0+2\lambda d . It follows that
\frac12|x-y| < |x-y+b(x)-b(y)| < \frac32|x-y|. |
Furthermore, we conclude
R_1(\lambda|x-y+b(x)-b(y)|r) \lesssim \frac{1}{(\lambda|x-y |)^{N+\frac n2+\frac12}} . |
Henceforth, we establish the estimate of the remainder term
\left|K_{\lambda, \text{rem}}(x, y)\right| \lesssim (\lambda|x-y |)^{-N-\frac n2-\frac12}. |
In order to obtain the upbound of \left|K_\lambda(x, y)\right| , it suffices to estimate the main term, which is defined by
K_{\lambda, \text{main}}(x, y) : = \frac{a_0}{(\lambda|x-y+b(x)-b(y)|)^{\frac n2-\frac12}} \int_0^\infty e^{i\Phi_\lambda(r)} \chi^2(r) r^{\frac n2-\frac 12} \, \text{d}r, |
where
\Phi_\lambda(r) : = \lambda r (t(x)-t(y)) \sqrt{1+\lambda^2 r^2} +\lambda|x-y+b(x)-b(y)|r. |
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
\left| \Phi'_\lambda(r)\right| \geq \lambda |x-y+b(x)-b(y)| -\lambda |t(x)-t(y)| \frac {1+2\lambda^2 r^2}{\sqrt{1+\lambda^2 r^2} } \gtrsim \lambda |x-y |. |
Using integration by parts, we obtain
\left| K_{\lambda, \text{main}}(x, y)\right| \lesssim \frac{1}{(\lambda|x-y+b(x)-b(y)|)^{\frac n2-\frac12}} (\lambda |x-y |)^{-N} \lesssim (\lambda |x-y |)^{-N} \; \text{for}\; \forall \; N. |
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
\left| \Phi''_\lambda(r)\right| \gtrsim \lambda^2 |t(x)-t(y)|. |
Using Lemma 3.1, we have
\left| K_{\lambda, \text{main}}(x, y)\right| \lesssim \frac{1}{(\lambda|x-y|)^{\frac n2-\frac12}} \left( \lambda^2 |t(x)-t(y)| \right)^{-\frac12} \lesssim (\lambda|x-y|)^{-\frac n2}. |
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
G_\lambda(x) : = \chi_{\{|x| < C_0 \lambda^{-1}+2d \}}(x) + \lambda^{-N} \chi_{\{|x|\geq \lambda^{-1} \}}(x) |x|^{-N} + \lambda^{-\frac n2} \chi_{\{|x|\leq C \lambda |J|\}}(x) |x|^{-\frac n2}, |
which yields
\begin{align*} \left\| G_\lambda \right\|_{L^1(\mathbb{R}^n)} &\lesssim (\lambda^{-1}+d)^n +\lambda^{-n} +\int_{|x| \leq C \lambda |J| } \lambda^{-\frac n2} |x|^{-\frac n2} \, \text{d}x \\ &\lesssim (\lambda^{-1}+d)^n +\lambda^{-n} +\lambda^{-\frac n2} \int_0^{ C \lambda |J|} r^{\frac n2-1 }\, \text{d}r \\ &\lesssim (\lambda^{-1}+d)^n +\lambda^{-n} +|J|^{\frac n2}\\ &\lesssim \lambda^{-n} +d^n +|J|^{\frac n2}, \end{align*} |
where in the second inequality we used polar coordinates. This implies that
\left\| T_\lambda T_\lambda^\ast \right\|_{L^2(\mathbb{R}^n)\rightarrow L^2(\mathbb{R}^n)} \lesssim \left\| G_\lambda \right\|_{L^1(\mathbb{R}^n)} \lesssim \lambda^{-n} +d^n +|J|^{\frac n2}. |
It follows that
\left\| T_\lambda \right\|_{L^2(\mathbb{R}^n)\rightarrow L^2(\mathbb{R}^n)} \lesssim \lambda^{-\frac n2} +d^{\frac n2} +|J|^{\frac n4}. |
We combine the above estimates and get
\begin{align*} \left\| \mathcal{B}_\lambda f(x+b(x), t(x)) \right\|_{L^2(\mathbb{R}^n)} &\leq \lambda^n \left\| T_\lambda (\hat f(\lambda\cdot)) \right\|_{L^2(\mathbb{R}^n)}\\ &\leq \lambda^n \left\| T_\lambda \right\|_{L^2(\mathbb{R}^n)\rightarrow L^2(\mathbb{R}^n)} \left\| \hat f(\lambda\cdot) \right\|_{L^2(\mathbb{R}^n)} \\ &\lesssim \lambda^n \left( \lambda^{-\frac n2} +d^{\frac n2} +|J|^{\frac n4} \right) \lambda^{-\frac n2} \left\| f \right\|_{L^2(\mathbb{R}^n)}\\ & = \left( 1+ d^{\frac n2} \lambda^{\frac n2} +\lambda^{\frac n2} |J|^{\frac n4} \right) \left\| f \right\|_{L^2(\mathbb{R}^n)}. \end{align*} |
This completes the proof of Lemma 2.2.
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
\begin{align*} \mathcal{B} f(x+y, t) & = (2\pi)^{-n} \int_{\mathbb{R}^n} e^{i\xi\cdot x} e^{i\xi_1 y_1} \cdot\cdot\cdot e^{i\xi_n y_n} e^{it|\xi| \sqrt{1+|\xi|^2}} \hat {f}(\xi) \, \text{d}\xi \\ & = (2\pi)^{-n} \int_{\mathbb{R}^n} e^{i\xi\cdot x} \left(\Lambda_1+ e^{i\xi_1 y_{0, 1}}\right) \cdot\cdot\cdot \left(\Lambda_n+ e^{i\xi_n y_{0, n}}\right) \left(\Lambda_{n+1} + e^{it_0 |\xi| \sqrt{1+|\xi|^2} }\right) \hat {f}(\xi) \, \text{d}\xi. \end{align*} |
Henceforth, \mathcal{B} f(x+y, t) is the sum of integrals of the form
\begin{align} (2\pi)^{-n} \int_{\mathbb{R}^n} e^{i\xi\cdot x} \left( \prod\limits_{j\in \Omega_1} \Lambda_j \right) \left( \prod\limits_{j\in \Omega_2} e^{i\xi_j y_{0, j}} \right) \Lambda_{n+1} \hat f(\xi) \, \text{d}\xi = :\mathcal{B}_1 f(x, y, t) \end{align} | (3.2) |
or
\begin{align} (2\pi)^{-n} \int_{\mathbb{R}^n} e^{i\xi\cdot x} \left( \prod\limits_{j\in \Omega_1} \Lambda_j \right) \left( \prod\limits_{j\in \Omega_2} e^{i\xi_j y_{0, j}} \right) e^{it_0 |\xi| \sqrt{1+|\xi|^2} } \hat f(\xi) \, \text{d}\xi = :\mathcal{B}_2 f(x, y, t). \end{align} | (3.3) |
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
\Lambda_j = i\xi_j \int_{y_{0, j}}^{y_j} e^{i\xi_j s_j} \, \text{d}s_j, |
and we also get
\Lambda_{n+1} = i |\xi| \sqrt{1+|\xi|^2} \int_{t_0}^t e^{i |\xi| \sqrt{1+|\xi|^2} s_{n+1}} \, \text{d} s_{n+1}. |
Assuming \Omega_1 = \{k_1, k_2, \cdot\cdot\cdot, k_m\} , we conclude
\begin{align*} \mathcal{B}_1 f(x, y, t) & = \int_{\mathbb{R}^n} \int_{y_{0, k_1}}^{y_{k_1}} \int_{y_{0, k_2}}^{y_{k_2}}\cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{k_m}} \int_{t_0}^t e^{i\xi\cdot x} \left( \prod\limits_{j\in \Omega_1} i \xi_j e^{i\xi_j s_j} \right) \left( \prod\limits_{j\in \Omega_2} e^{i\xi_j y_{0, j}} \right) \\ &\quad \times i |\xi| \sqrt{1+|\xi|^2} e^{i |\xi| \sqrt{1+|\xi|^2} s_{n+1}} \hat f(\xi) \, \text{d} s_{k_1} \, \text{d} s_{k_2} \cdot\cdot\cdot\, \text{d} s_{k_m} \, \text{d} s_{n+1} \, \text{d} \xi. \end{align*} |
By changing the order of integration we get
\begin{align*} \left|\mathcal{B}_1 f(x, y, t) \right| &\leq \int_{y_{0, k_1}}^{y_{k_1}} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{k_m}} \int_{t_0}^t \left| F_{\Omega_1}(x;s_{k_1}, \cdot\cdot\cdot, s_{k_m}, s_{n+1})\right| \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} \, \text{d} s_{n+1} , \end{align*} |
where
\begin{align*} &F_{\Omega_1}(x;s_{k_1}, \cdot\cdot\cdot, s_{k_m}, s_{n+1})\\ : = & (2\pi)^{-n} \int_{\mathbb{R}^n} e^{i\xi\cdot x} \left( \prod\limits_{j\in \Omega_1} i \xi_j e^{i\xi_j s_j} \right) \left( \prod\limits_{j\in \Omega_2} e^{i\xi_j y_{0, j}} \right) i |\xi| \sqrt{1+|\xi|^2} e^{i |\xi| \sqrt{1+|\xi|^2} s_{n+1}} \hat f(\xi)\, \text{d} \xi . \end{align*} |
It follows that
\begin{align} \sup\limits_{(y, t)\in E} \left| \mathcal{B}_1 f(x, y, t)\right| \leq \int_{y_{0, k_1}}^{y_{0, k_1}+r} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{0, k_m}+r} \int_{t_0}^{t_0+r^2 } \left| F_{\Omega_1}(x;s_{k_1}, \cdot\cdot\cdot, s_{k_m}, s_{n+1}) \right| \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} \, \text{d} s_{n+1} . \end{align} | (3.4) |
Taking L^2 norms of both sides of (3.4) and from Minkowski's inequality and Plancherel's theorem, we deduce
\begin{align*} &\left\| \sup\limits_{(y, t)\in E} \left| \mathcal{B}_1 f(\cdot, y, t)\right| \right\|_{L^2(\mathbb{R}^n)} \\ \leq& \int_{y_{0, k_1}}^{y_{0, k_1}+r} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{0, k_m}+r} \int_{t_0}^{t_0+r^2 } \left\| F_{\Omega_1}(\cdot;s_{k_1}, \cdot\cdot\cdot, s_{k_m}, s_{n+1}) \right\|_{L^2(\mathbb{R}^n)} \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} \, \text{d} s_{n+1} \\ = & (2\pi)^{-\frac n2} \int_{y_{0, k_1}}^{y_{0, k_1}+r} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{0, k_m}+r} \int_{t_0}^{t_0+r^2 } \left( \int_{\mathbb{R}^n} \left( \prod\limits_{j\in \Omega_1 } |\xi_j|^2 \right) \left( |\xi| \sqrt{1+|\xi|^2} \right)^2 \left| \hat f(\xi)\right|^2 \, \text{d}\xi \right)^{\frac12} \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} \, \text{d} s_{n+1} \\ \leq& r^m r^2 \lambda^m \lambda^2 \left\| f \right\|_{L^2(\mathbb{R}^n)}, \end{align*} |
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
\begin{align*} \mathcal{B}_2 f(x, y, t) & = \int_{\mathbb{R}^n} \int_{y_{0, k_1}}^{y_{k_1}} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{k_m}} e^{i\xi\cdot x} \left( \prod\limits_{j\in \Omega_1} i \xi_j e^{i\xi_j s_j} \right) \left( \prod\limits_{j\in \Omega_2} e^{i\xi_j y_{0, j}} \right) e^{it_0 |\xi| \sqrt{1+|\xi|^2} } \hat f(\xi) \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} \, \text{d} \xi. \end{align*} |
Changing the order of integration again, one then obtains
\begin{align*} \left| \mathcal{B}_2 f(x, y, t)\right| &\leq \int_{y_{0, k_1}}^{y_{k_1}} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{k_m}} \left| H_{\Omega_1}(x;s_{k_1}, \cdot\cdot\cdot, s_{k_m})\right| \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} , \end{align*} |
where
\begin{align*} H_{\Omega_1}(x;s_{k_1}, \cdot\cdot\cdot, s_{k_m}) : = (2\pi)^{-n} \int_{\mathbb{R}^n} e^{i\xi\cdot x} \left( \prod\limits_{j\in \Omega_1} i \xi_j e^{i\xi_j s_j} \right) \left( \prod\limits_{j\in \Omega_2} e^{i\xi_j y_{0, j}} \right) e^{it_0 |\xi| \sqrt{1+|\xi|^2} } \hat f(\xi)\, \text{d} \xi . \end{align*} |
Furthermore, we get
\begin{align*} \sup\limits_{(y, t)\in E} \left| \mathcal{B}_2 f(x, y, t)\right| \leq \int_{y_{0, k_1}}^{y_{0, k_1}+r} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{0, k_m}+r} \left| H_{\Omega_1}(x;s_{k_1}, \cdot\cdot\cdot, s_{k_m}) \right| \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m}. \end{align*} |
Using Minkowski's inequality and Plancherel's theorem, we then obtain
\begin{align*} \left\| \sup\limits_{(y, t)\in E} \left| \mathcal{B}_2 f(\cdot, y, t)\right| \right\|_{L^2(\mathbb{R}^n)} &\leq \int_{y_{0, k_1}}^{y_{0, k_1}+r} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{0, k_m}+r} \left\| H_{\Omega_1}(\cdot;s_{k_1}, \cdot\cdot\cdot, s_{k_m}) \right\|_{L^2(\mathbb{R}^n)} \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} \\ & = (2\pi)^{-\frac n2} \int_{y_{0, k_1}}^{y_{0, k_1}+r} \cdot\cdot\cdot \int_{y_{0, k_m}}^{y_{0, k_m}+r} \left( \int_{\mathbb{R}^n} \left( \prod\limits_{j\in \Omega_1 } |\xi_j|^2 \right) \left| \hat f(\xi)\right|^2 \, \text{d}\xi \right)^{\frac12} \, \text{d} s_{k_1} \cdot\cdot\cdot\, \text{d} s_{k_m} \\ &\leq r^m \lambda^m \left\| f \right\|_{L^2(\mathbb{R}^n)}. \end{align*} |
By summation of the above integrals, we conclude that
\left\| \mathcal{B}_E^\ast f \right\|_{L^2(\mathbb{R}^n)} \lesssim (1+r^2 \lambda^2) (1+r\lambda)^n \left\| f \right\|_{L^2(\mathbb{R}^n)}. |
Thus, Lemma 2.3 is established.
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.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
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).
The authors declare that they have no conflicts of interest.
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