Analysis of Schr\"odinger means

TL;DR

This paper investigates integral estimates of maximal functions associated with Schr"odinger means, providing insights into their behavior and bounds in harmonic analysis.

Contribution

It introduces new integral estimates for maximal functions of Schr"odinger means, advancing understanding of their boundedness properties.

Findings

Established new bounds for maximal Schr"odinger means

Improved understanding of their integral estimates

Contributed to harmonic analysis theory

Abstract

We study integral estimates of maximal functions for Schr\"odinger means.

Equations

$$\hat{f}(\xi )={\int }_{{\mathbb{R}}^n}\text{mbox}{e}^{-i\xi \cdot x}f(x)dx,\xi \in {\mathbb{R}}^n,$$

$$S_{t}f(x)=(2\pi {)}^{-n}{\int }_{{\mathbb{R}}^n}\text{mbox}{e}^{i\xi \cdot x}\text{mbox}{e}^{it|\xi {|}^a}\hat{f}(\xi )d\xi ,x\in {\mathbb{R}}^n,t\ge 0.$$

$$H_s=\{f\in {\mathcal{S}}^{\prime };\Vert f{\Vert }_{H_s}<\infty \},s\in \mathbb{R},$$

$$\Vert f{\Vert }_{H_s}={\left({\int }_{{\mathbb{R}}^n}(1+[\xi {[}^2{)}^s|\hat{f}(\xi ){|}^{2}d\xi \right)}^{1/2}.$$

$$S_E^{\ast }f(x)=\underset{t\in E}{\sup }|S_{t}f(x)|,x\in {\mathbb{R}}^n.$$

$${\int }_{{\mathbb{R}}^n}|S_E^{\ast }f(x){|}^{2}dx\lesssim \left(\sum _{m=0}^{\infty }{N}_E(2^{-m})2^{-2ms/a}\right)\Vert f{\Vert }_{H_s}^2.$$

$$\Vert S_E^{\ast }f{\Vert }_2\lesssim \Vert f{\Vert }_{H_s}$$

$$1>t_1>t_2>t_3>\cdots >0\text{mbox}and\underset{k\to \infty }{\lim }{t}_k=0.$$

$$\#A_j\lesssim 2^{bj}\text{mbox}forj=1,2,3,\dots ,$$

$$\underset{k\to \infty }{\lim }{S}_{t_k}f(x)=f(x)$$

$$S_E^{\ast }f(x)=\underset{(y,t)\in E}{\sup }|S_{t}f(x+y)|,x\in {\mathbb{R}}^n.$$

$${\int }_{{\mathbb{R}}^n}|S_E^{\ast }f(x){|}^{2}dx\lesssim \left(\sum _{m=0}^{\infty }{N}_{E,a}(2^{-m})2^{-2ms}\right)\Vert f{\Vert }_{H_s}^2.$$

$$\sum _{n=0}^{\infty }{N}_{E,a}(2^{-m})2^{-2ms}<\infty ,$$

$$\Vert S_E^{\ast }f{\Vert }_2\lesssim \Vert f{\Vert }_{H_s}.$$

$$E\subset \{(\Gamma (t),t);0\le t\le 1\}.$$

$$|\Gamma (t_1)-\Gamma (t_2|\lesssim |t_1-t_2{|}^{\beta }\text{mbox}for{t}_1,t_2\in E_0,$$

$$\Vert S_E^{\ast }f{\Vert }_{L^2(B)}\le C_B\Vert f{\Vert }_{H_s}\text{mbox}ifs>\max (1/2-\beta ,1/4),\text{mbox}and0<\beta \le 1.$$

$$E=\{(y,t)\in {\mathbb{R}}^{n+1};y_{0,j}\le y_j\le y_{0,j}+r\text{mbox}for1\le j\le n,\text{mbox}and{t}_0\le t\le t_0+r^a\}$$

$$S_{t}f(x+y)=c\int \text{mbox}{e}^{i\xi \cdot x}\text{mbox}{e}^{i{\xi }_1{y}_1}\dots \text{mbox}{e}^{i{\xi }_n{y}_n}\text{mbox}{e}^{it|\xi {|}^a}\hat{f}(\xi )d\xi$$

$${\Delta }_j=\text{mbox}{e}^{i{\xi }_j{y}_j}-\text{mbox}{e}^{i{\xi }_j{y}_{0,j}}.$$

$${\Delta }_{n+1}=\text{mbox}{e}^{it|\xi {|}^a}-\text{mbox}{e}^{i{t}_0|\xi {|}^a}.$$

$$\begin{gathered} S_{t}f(x+y)=c\int \text{mbox}{e}^{i\xi \cdot x}\left({\Delta }_1+\text{mbox}{e}^{i{\xi }_1{y}_{0,1}}\right)\dots \left({\Delta }_n+\text{mbox}{e}^{i{\xi }_n{y}_{0,n}}\right) \\ \left({\Delta }_{n+1}+\text{mbox}{e}^{i{t}_0|\xi {|}^a}\right)\hat{f}(\xi )d\xi , \end{gathered}$$

$$c\int \text{mbox}{e}^{i\xi \cdot x}\left(\prod _{j\in D}{\Delta }_j\right)\left(\prod _{j\in B}\text{mbox}{e}^{i{\xi }_j{y}_{0,j}}\right){\Delta }_{n+1}\hat{f}(\xi )d\xi ,$$

$$c\int \text{mbox}{e}^{i\xi \cdot x}\left(\prod _{j\in D}{\Delta }_j\right)\left(\prod _{j\in B}\text{mbox}{e}^{i{\xi }_j{y}_{0,j}}\right)\text{mbox}{e}^{i{t}_0|\xi {|}^a}\hat{f}(\xi )d\xi ,$$

$${\Delta }_j=i{\xi }_j{\int }_{y_{0,j}}^{y_j}\text{mbox}{e}^{i{\xi }_j{s}_j}d{s}_j,$$

$${\Delta }_{n+1}=i|\xi {|}^a{\int }_{t_0}^t\text{mbox}{e}^{i|\xi {|}^a{s}_{n+1}}d{s}_{n+1}.$$

$$\begin{array}[]{ll}S^{\prime}_{t}f(x,y)=\int_{{\mathbb{R}}^{n}}\int_{y_{0,k_{1}}}^{y_{k_{1}}}\int_{y_{0,k_{2}}}^{y_{k_{2}}}\dots\int_{y_{0,k_{p}}}^{y_{k_{p}}}\int_{t_{0}}^{t}&\mbox{e}^{i\xi\cdot x}\left(\prod_{j\in D}i\xi_{j}\mbox{e}^{i\xi_{j}s_{j}}\right)\left(\prod_{j\in B}\mbox{e}^{i\xi_{j}y_{0,j}}\right)\\[5.69046pt] &\cdot i|\xi|^{a}\mbox{e}^{i|\xi|^{a}s_{n+1}}\hat{f}(\xi)\,ds_{k_{1}}ds_{k_{2}}\dots ds_{k_{p}}ds_{n+1}d\xi.\end{array}$$

$$\begin{array}[]{ll}|S^{\prime}_{t}f(x,y)|\leq c\int_{y_{0,k_{1}}}^{y_{k_{1}}}\int_{y_{0,k_{2}}}^{y_{k_{2}}}\dots\int_{y_{0,k_{p}}}^{y_{k_{p}}}\int_{t_{0}}^{t}&\left|\int_{{\mathbb{R}}^{n}}\mbox{e}^{i\xi\cdot x}\left(\prod_{j\in D}i\xi_{j}\mbox{e}^{i\xi_{j}s_{j}}\right)\left(\prod_{j\in B}\mbox{e}^{i\xi_{j}y_{0,j}}\right)\right.\\[5.69046pt] &\left.\cdot i|\xi|^{a}\mbox{e}^{i|\xi|^{a}s_{n+1}}\hat{f}(\xi)\,d\xi\right|\,ds_{k_{1}}ds_{k_{2}}\dots ds_{k_{p}}ds_{n+1},\end{array}$$

$$S_t^{\prime }f(x,y)={\int }_{y_{0,k_1}}^{y_{k_1}}\cdots {\int }_{y_{0,k_p}}^{y_{k_p}}{\int }_{t_0}^t{F}_D(x;s_{k_1},,\dots ,s_{k_p},s_{n+1})d{s}_{k_1}d{s}_{k_2}\dots d{s}_{k_p}d{s}_{n+1},$$

$$F_D(x;s_{k_1},s_{k_2},\dots ,s_{k_p},s_{n+1})=c{\int }_{{\mathbb{R}}^n}\text{mbox}{e}^{i\xi \cdot x}\left(\prod _{j\in D}i{\xi }_j\text{mbox}{e}^{i{\xi }_j{s}_j}\right)\left(\prod _{j\in B}\text{mbox}{e}^{i{\xi }_j{y}_{0,j}}\right)i|\xi {|}^a\text{mbox}{e}^{i|\xi {|}^a{s}_{n+1}}\hat{f}(\xi )d\xi .$$

Full text

††Mathematics Subject Classification (2010):42B99.Key Words and phrases: Schrödinger equation, maximal functions, integral estimates, Sobolev spaces

1. Introduction

For $f\in L^{2}({\mathbb{R}}^{n}),n\geq 1$ and $a>0$ we set

[TABLE]

and

[TABLE]

For $a=2$ and $f$ belonging to the Schwartz class $\mathscr{S}({\mathbb{R}}^{n})$ we set $u(x,t)=S_{t}f(x)$. It then follows that $u(x,0)=f(x)$ and $u$ satisfies the Schrödinger equation $i\partial u/\partial t=\Delta u$.

We introduce Sobolev spaces $H_{s}=H_{s}({\mathbb{R}}^{n})$ by setting

[TABLE]

where

[TABLE]

Let $E$ denote a bounded set in ${\mathbb{R}}$. For $r>0$ we let $N_{E}(r)$ denote the minimal number $N$ of intervals $I_{l},l=1,2,\dots N$, of length $r$ such that $E\subset\bigcup_{1}^{N}I_{l}$. For $f\in\mathscr{S}$ we introduce the maximal function

[TABLE]

In Sjölin and Strömberg [2] we proved the following theorem.

Theorem A Assume that $n\geq 1$ and $a>0$ and $s>0$. if $f\in\mathscr{S}$ then one has

[TABLE]

Here we write $A\lesssim B$ if there is a constant $C$ such that $A\leq CB$.

In the case $E=[0,1]$ it is easy to see that Theorem A implies the estimate

[TABLE]

if $s>a/2$.

Let $(t_{k})_{1}^{\infty}$ be a sequence satisfying

[TABLE]

Set $A_{j}=\{t_{k};2^{-j-1}<t_{k}\leq 2^{-j}\}\mbox{ for }j=1,2,3,\dots.$ Let $\#A$ denote the number of elements in a set $A$.

In [2] we used Theorem A to obtain the following results.

Theorem B *Assume that $n\geq 1$ and $a\geq 2s$ and $0<s\leq 1/2$, and $1/b>(a-2s)/2s$.

Assume also that*

[TABLE]

and that $f\in H_{s}$. Then

[TABLE]

*almost everywhere.

Theorem C *Assume that $n\geq 1$ and $a\geq 2s$ and $0<s\leq 1/2$, and that $\sum_{1}^{\infty}t_{k}^{\gamma}>\infty$, where $1/\gamma>(a-2s)/2s$.

If also $f\in H_{s}$ then (3) holds almost everywhere.

Now let $E$ denote a bounded set in ${\mathbb{R}}^{n+1}$ and let $E_{0}$ denote the projection of $E$ onto the $t$-axis, i.e. $E_{0}=\{t;\mbox{ there extists }x\in{\mathbb{R}}^{n}\mbox{ such that }(x,t)\in E\}$.

For $a>0$ define an $a$-cube in ${\mathbb{R}}^{n+1}$ with side $r>0$ as an axis-parallell rectangular box with sidelength $r^{a}$ in the $t$-direction and with sidelength $r$ in the remaining directions $x_{1},\dots,x_{n}$. Thus an $a$-cube in ${\mathbb{R}}^{n+1}$ has volume $r^{n+a}$.

For $r>0$ let $N_{E,a}(r)$ denote the minimal number $N$ of $a$-cubes $Q_{l},l=1,2,\dots,N$ of side $r$, such that $E\subset\bigcup_{1}^{N}Q_{l}$.

For $f\in\mathscr{S}$ we introduce the maximal function

[TABLE]

We shall prove the following inequality

Theorem 1**.**

Assume that $n\geq 1$ and $a>0$ and $s>0$. if $f\in\mathscr{S}$ then one has

[TABLE]

The following estimate follows directly.

Corollary 1**.**

Assume $n\geq 1,a>0,s>0$ and $f\in\mathscr{S}$ . If

[TABLE]

then

[TABLE]

Now let $\Gamma:[0,1]\to({\mathbb{R}}^{n})$ and let $E$ be a subset of the graph of $\Gamma$ that is

[TABLE]

Assume that

[TABLE]

where $0<\beta$. Also set $a_{1}=1/\beta$ and $a_{2}=\max(a,a_{1})$.

We shall first study the case when $E$ is the graph of $\Gamma$. Then (5) holds for any $\beta>1$ only if $\Gamma$ is constant.

It follows from a result of Cho, Lee, and Vargas [1] that if $n=1,a=2,B$ is an interval of ${\mathbb{R}}$ and $E$ is the graph of $\Gamma$, then

[TABLE]

We have the following result.

Theorem 2**.**

Assume $n\geq 1,a>0,f\in\mathscr{S}$ and that $E$ is the graph of $\Gamma$. Then (4) holds for $2s>a_{2}$.

We shall then study the case when $E=\{(\Gamma(t_{k}),t_{k});k=1,2,3\dots\}$ where the sequence $(t_{k})_{1}^{\infty}$ satisfies (2). We have the following results.

Theorem 3**.**

*Assume $n\geq 1,a>0$, and $f\in\mathscr{S}$.

In the case $2s>a_{2}$ then (4) holds.

In the case $2s=a_{2}$ assume that $\sum_{1}^{\infty}t_{k}^{\gamma}<\infty$ for some $\gamma>0$. Then (4) holds.

In the case $2s<a_{2}$ assume that $\sum_{1}^{\infty}t_{k}^{\gamma}<\infty$ for some $\gamma$ satisfying

$\gamma<2s/(a_{2}-2s).$ Then (4) holds.*

2. Proof of Theorem 1

If $y$ and $y_{0}$ belong to to ${\mathbb{R}}^{n}$ we write $y=(y_{1},\dots,y_{n})$ and $y_{0}=(y_{0,1},\dots,y_{0,n})$. We shall give the proof of Theorem 1.

Proof.

Assume $y_{0}\in{\mathbb{R}}^{n},t_{0}\in{\mathbb{R}},0<r\leq 1$ and let

[TABLE]

We have

[TABLE]

where $c=(2\pi)^{-n}$, and for $1\leq j\leq n$ we write $\mbox{e}^{i\xi_{j}y_{j}}=\Delta_{j}+\mbox{e}^{i\xi_{j}y_{0,j}}$ where

[TABLE]

We also write $\mbox{e}^{it|\xi|^{a}}=\Delta_{n+1}+\mbox{e}^{it_{0}|\xi|^{a}}$ where

[TABLE]

Hence

[TABLE]

and it follows that $S_{t}f(x+y)$ is the sum of integrals of the form

[TABLE]

or

[TABLE]

Here $D$ and $B$ ar disjoint subsets of $\{1,2,3,\dots,n\}$ and $D\cup B=\{1,2,3,\dots,n\}$. We denote the integrals in (6) by $S^{\prime}_{t}f(x,y)$ and shall describe how they can be estimated. The same argument works also for integrals in (7).

For $j\in D$ we write

[TABLE]

and we also write

[TABLE]

Assuming $D=\{k_{1},k_{2},\dots,k_{p}\}$ we then have

[TABLE]

Changing the order of integration one then obtains

[TABLE]

or

[TABLE]

where

[TABLE]

and

[TABLE]

Invoking Minkowski’s inequality and Plancherel’s formula we then obtain

[TABLE]

if $f\in L^{2}({\mathbb{R}}^{n}))$ and $\mbox{supp }\hat{f}\subset B(0,A)$, where $A\geq 1$.

With similar arguments we get the estimate $r^{p}A^{p}\|f\|_{2}$ for the integrals in (7), and by summation of the all integrals of forms (6) or (7) we get

[TABLE]

Now let $E$ be a set in ${\mathbb{R}}^{n+1}$ with the property that $E\subset\bigcup_{1}^{N}Q_{l}$ where the sets $Q_{l}$ are $a$-cubes with side $r$ with $rA\leq 1$ of the type we have just considered. One has

[TABLE]

and it follows from (8) that

[TABLE]

if $rA\leq 1$ and $\mbox{supp }\hat{f}\subset B(0,A)$.

Now let $f\in\mathscr{S}$. We write $f=\sum_{k=0}^{\infty}f_{k}$, where the functions $f_{k}$ are defined in the following way. We set $\hat{f}_{0}(\xi)=\hat{f}(\xi)$ for $|\xi|\leq 1$ and $\hat{f}_{0}(\xi)=0$ for $|\xi|>1$. For $k\geq 1$ we let $\hat{f}_{k}(\xi)=\hat{f}(\xi)$ for $2^{k-1}<|\xi|\leq 2^{k}$ and $\hat{f}_{k}(\xi)=0$ otherwise.

Choosing real numbers $g_{k}>0,k=0,1,2,\dots,$ we have

[TABLE]

and invoking inquality (9) we also have

[TABLE]

Chosing $g_{k}=N_{E,a}(2^{-k})2^{-2ks}$ we conclude that

[TABLE]

and the proof of Theorem 1 is complete. ∎

3. Proof of Theorems 2 and 3

The following two lemmas follow easily from the definition of $N_{E,a}(r)$.

Lemma 1**.**

Assume tha $0<r\leq 1$ and $0<b<b_{1}$. Then

[TABLE]

and

[TABLE]

Lemma 2**.**

*Assume that $E$ is a subset of the graph of $\Gamma$ satisfying (5) with $\beta>0$ and let $b\geq 1/\beta$

Then*

[TABLE]

($E_{0}$, $\Gamma,\beta$ are defined and equation (5) is in Section 1).

We shall then give the proofs of Theorems 2 and 3.

Proof of Theorem 2.

We have $a\leq a_{2}$. Invoking Lemma 1 and Lemma 2 we obtain

[TABLE]

and

[TABLE]

if $a_{2}-2s<0$, i.e.$s>a_{2}/2$.

Using Corollary 1 we condclude that

[TABLE]

if $s>a_{2}/2$. ∎

We shall then prove Theorem 3.

Proof of Theorem 3.

In the case $2s>a_{2}$ we can use the same argument as in the proof of Theorem 2 to prove that (4) holds.

Then assume $2s\leq a_{2}$. We have $E_{0}=\{t_{k};k=1,2,3,\dots\}$ and assuming $\sum_{1}^{\infty}t_{k}^{\gamma}<\infty$ one obtains $\#A_{j}\lesssim 2^{\gamma j}$. It then follows from Lemma 6 in [2] that $N_{E_{0}}(2^{-m})\lesssim 2^{\gamma m/(\gamma+1)}$.

We have $a\leq a_{2}$. Applying Lemma 1 and Lemma 2 one obtains

[TABLE]

It follows that

[TABLE]

if $\gamma a_{2}/(\gamma+1)<2s$, that is $a_{2}\gamma<2s\gamma+2s$ .

In the case $2s=a_{2}$ this holds for every $\gamma>0$.

In the case $2s<a_{2}$ one has to assume $\gamma<2s/(a_{2}-2s)$.

This completes the proof of Theorem 3. ∎