1 Introduction

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

$$\begin{aligned} {\widehat{f}}(\xi ) = \int \limits _{{\mathbb {R}}^{n}}^{}e^{-i\xi \cdot x}f(x)dx, \quad \xi \in {\mathbb {R}}^{n}, \end{aligned}$$

and

$$\begin{aligned} S_{t}f(x) = \int \limits _{{\mathbb {R}}^{n}}^{}e^{i\xi \cdot x}e^{it|\xi |^{a}}{\widehat{f}}(\xi )d\xi ,\quad x \in {\mathbb {R}}^{n},\quad t \geqslant 0. \end{aligned}$$

For \(a=2\) and f belonging to the Schwartz class \({\mathscr {S}}({\mathbb {R}}^{n})\) we set \(u(x,t) = S_{t}f(x)/(2\pi )^{n}\). 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

$$\begin{aligned} H_{s} = \{ f\in {\mathscr {S}}' ; \Vert f\Vert _{H_{s}} < \infty \} ,\quad s\in {\mathbb {R}}, \end{aligned}$$

where

$$\begin{aligned} \Vert f\Vert _{H_{s}} = \left( \int \limits _{{\mathbb {R}}^{n}}{}{(1+ |\xi |^{2})^{s}}|{\widehat{f}}(\xi )|^{2}d\xi \right) ^{1/2}. \end{aligned}$$

In the case n = 1 it is well-known (see Sjölin [7] and Vega [9] and in the case \(a = 2\) Carleson [3] and Dahlberg and Kenig [4]) that

$$\begin{aligned} \lim _{t\rightarrow 0} (2\pi )^{-n}S_{t}f(x) = f(x) \end{aligned}$$
(1)

almost everywhere if \(f\in H_{1/4}\). Also it is known that \(H_{1/4}\) can not be replaced by \(H_{s} \) if \(s < 1/4\).

Assuming \(n\geqslant 2\) and \(a=2\) Bourgain [1] has proved that (1) holds almost everywhere if \(f\in H_{s}\) and \(s>1/2-1/4n\). On the other hand Bourgain [2] has proved that \(s\geqslant n/2(n+1)\) is necessary for convergence for \(a=2\) and all \(n \geqslant 2\). In the case \(n=2\) and \(a = 2\), Du, Guth, and Li [5] proved that the condition \(s>1/3\) is sufficient. Recently Du and Zhang [6] proved that the condition \(s > n/2(n+1)\) is sufficient for \(a = 2\) and all \(n \geqslant 3\).

In the case \(a > 1\), \(n = 2\), it is known that (1) holds almost everywhere if \(f \in H_{1/2}\) and in the case \(a> 1\), \(n\geqslant 3\), convergence has been proved for \(f \in H_{s}\) with \(s>1/2\) (see [7] and [9]).

If \(f\in L^{2}({\mathbb {R}}^{n})\) then \((2\pi )^{-n}S_{t}f \rightarrow f\) in \(L^{2}\) as \(t \rightarrow 0\). It follows that there exists a sequence \((t_{k})_{1}^{\infty }\) satisfying

$$\begin{aligned} 1>t_{1}>t_{2}>t_{3}>\cdots >0 \text{ and } \lim _{k \rightarrow \infty } t_{k} = 0 \end{aligned}$$
(2)

such that

$$\begin{aligned} \lim _{k \rightarrow \infty } (2\pi )^{-n}S_{t_{k}}f(x) = f(x) \end{aligned}$$

almost everywhere.

We shall here study the problem of deciding for which sequences \((t_{k})_{1}^{\infty }\) one has

$$\begin{aligned} \lim _{k \rightarrow \infty } (2\pi )^{-n}S_{t_{k}}f(x) = f(x) \end{aligned}$$

almost everywhere if \(f \in H_{s}\). We have the following result.

Theorem 1

Assume \(n\geqslant 1\) and \(a>1\) and \(s>0\). We assume that (2) holds and that \(\sum _{k = 1}^{\infty } t_{k}^{2s/a} < \infty \) and \(f\in H_{s}({\mathbb {R}}^{n})\). Then

$$\begin{aligned} \lim _{k \rightarrow \infty } (2\pi )^{-n}S_{t_{k}}f(x) = f(x) \end{aligned}$$

for almost every x in \({\mathbb {R}}^{n}\).

Now assume \(n = 1\), \(a > 1\), and \(0 \leqslant s < 1/4\). In Sjölin [8] we studied the problem if there is localization or localization almost everywhere for the above operators \(S_{t}\) and the functions \(f \in H_{s}\) with compact support, that is, do we have

$$\begin{aligned} \lim _{t \rightarrow 0} S_{t}f(x) = 0 \end{aligned}$$

everywhere or almost everywhere in \({\mathbb {R}}^{n}\backslash \)(suppf)?

It is proved in [8] that there is no localization or localization almost everywhere of this type for \(0\leqslant s < 1/4\). In fact the following theorem was proved in Sjölin [8].

Theorem A

There exist two disjoint compact intervals I and J in \({\mathbb {R}}\) and a function f which belongs to \(H_{s}\) for all \(s < 1/4\), with the properties that (suppf) \(\subset I\) and for every \(x \in J\) one does not have

$$\begin{aligned} \lim _{t \rightarrow 0} S_{t}f(x) = 0. \end{aligned}$$

Let \(\omega \) be a continuous and decreasing function on \([0,\infty )\). Assume that \(\omega (t) \rightarrow 0\) as \(t \rightarrow \infty \). Set

$$\begin{aligned} H_{\omega } = \{ f\in {\mathscr {S}}' ; \Vert f\Vert _{H_{\omega }} < \infty \} \end{aligned}$$

where

$$\begin{aligned} \Vert f\Vert _{H_{\omega }} = \left( \int \limits _{{\mathbb {R}}}{}|{\widehat{f}}(\xi )|^{2}{(1+ \xi ^{2})^{1/4}}\omega (|\xi |)d\xi \right) ^{1/2} \end{aligned}$$

We have the following result.

Theorem 2

The function f in theorem A can be chosen so that \(f \in H_{\omega }\).

Theorem 2 shows that the sufficient condition \(f\in H_{1/4}\) for convergence almost everywhere and localization almost everywhere of Schrödinger means is very sharp. In the case a = 2 Theorem 2 was obtained in 2009 (unpublished). After proving Theorem 2 we shall use Theorem 1 to make a remark on the Schrödinger means \(S_{t}f(x)\) for the function f which was constructed in [8] to prove Theorem A.

2 Proofs

In the proof of Theorem 1 we shall need the following lemma.

Lemma 1

Assume \(n \geqslant 1\), \(a>1\), \(0<s<1\), and \(0<\delta <1\). Set

$$\begin{aligned} m(\xi ) = \frac{e^{i\delta |\xi |^{a}}-1}{(1+|\xi |^{2})^{s/2}}, \quad \xi \in {\mathbb {R}}^{n}. \end{aligned}$$

Then one has

$$\begin{aligned} \Vert m\Vert _{\infty } \leqslant C\delta ^{s/a} \end{aligned}$$

where the constant C does not depend on \(\delta \), and \(\Vert m\Vert _{\infty }\) denotes the norm of m in \(L^{\infty }({\mathbb {R}}^{n})\).

Proof of Lemma 1

We shall write \( A\lesssim B\) if there is a constant C such that \(A\leqslant CB\).

In the case \(|\xi | \geqslant \delta ^{-1/a}\) one has

$$\begin{aligned} |\xi |^{s} \geqslant \delta ^{-s/a} \text{ and } |m(\xi )| \lesssim \frac{1}{|\xi |^{{s}}} \leqslant \delta ^{s/a}. \end{aligned}$$

Then assume \(0 \leqslant |\xi | \leqslant 1\). We obtain

$$\begin{aligned} |m(\xi )| \lesssim \delta |\xi |^{a} \leqslant \delta \leqslant \delta ^{s/a}. \end{aligned}$$

In the remaining case \(1<|\xi |<\delta ^{-1/a}\) one obtains

$$\begin{aligned} |m(\xi )| \lesssim \frac{\delta |\xi |^{a}}{|\xi |^{{s}}} = \delta |\xi |^{a-s} \lesssim \delta \delta ^{-(a-s)/a} = \delta \delta ^{-1+s/a} = \delta ^{s/a} \end{aligned}$$

and the proof of Lemma 1 is complete. \(\square \)

We shall then give the proof of Theorem 1.

Proof of Theorem 1

We may assume \(0<s<1\). We set

$$\begin{aligned} h_{k}(x) = (2\pi )^{-n}S_{t_{k}}f(x) -f(x),\quad x\in {\mathbb {R}}^{n}, \text{ for } k = 1,2,3,... \end{aligned}$$

We have \(f\in H_{s}\) and we define g by taking

$$\begin{aligned} {\widehat{g}}(\xi ) = {\widehat{f}}(\xi )(1+|\xi |^{2})^{s/2}. \end{aligned}$$

It then follows that \(g \in L^{2}({\mathbb {R}}^{n})\).

We have

$$\begin{aligned} S_{t_{k}}f(x) = \int e^{ix\cdot \xi }e^{it_{k}|\xi |^{a}}(1+|\xi |^{2})^{-s/2}{\widehat{g}}(\xi )d\xi \end{aligned}$$

and

$$\begin{aligned} f(x) = (2\pi )^{-n}\int e^{ix\cdot \xi }(1+|\xi |^{2})^{-s/2}{\widehat{g}}(\xi )d\xi . \end{aligned}$$

Hence

$$\begin{aligned}&h_{k}(x) = (2\pi )^{-n}\int e^{ix\cdot \xi }(e^{it_{k}|\xi |^{a}}-1)(1+|\xi |^{2})^{-s/2}{\widehat{g}}(\xi )d\xi \\&\quad = (2\pi )^{-n}\int e^{ix\cdot \xi }m(\xi ){\widehat{g}}(\xi )d\xi , \end{aligned}$$

where

$$\begin{aligned} m(\xi ) = (e^{it_{k}|\xi |^{a}}-1)(1+|\xi |^{2})^{-s/2}. \end{aligned}$$

According to Lemma 1 we have \(\Vert m\Vert _{\infty } \lesssim t_{k}^{s/a}\) and applying the Plancherel theorem we obtain

$$\begin{aligned} \Vert h_{k}\Vert _{2}^{2} = c\int |m(\xi ){\widehat{g}}(\xi )|^{2}d\xi \lesssim \Vert m\Vert _{\infty }^{2}\int |{\widehat{g}}(\xi )|^2d\xi \lesssim t_{k}^{2s/a}\Vert f\Vert _{H_{s}}^{2}. \end{aligned}$$

It follows that

$$\begin{aligned} \sum _{1}^{\infty }\int |h_{k}|^{2}dx \lesssim (\sum _{1}^{\infty }t_{k}^{2s/a})\Vert f\Vert _{H_{s}}^{2} < \infty \end{aligned}$$

and applying the theorem on monotone convergence one also obtains

$$\begin{aligned} \int \left( \sum _{1}^{\infty }|h_{k}|^2\right) dx < \infty . \end{aligned}$$

We conclude that \(\sum _{1}^{\infty }|h_{k}|^2\) is convergent almost everywhere and hence \(\lim _{k\rightarrow \infty }h_{k}(x) = 0\) and

$$\begin{aligned} \lim _{k\rightarrow \infty }(2\pi )^{-n}S_{t_{k}}f(x) = f(x) \end{aligned}$$

almost everywhere. \(\square \)

Now assume \(n = 1\) and \(a>1\). We set

$$\begin{aligned} m(\xi ) = e^{i|\xi |^{a}}, \quad \xi \in {\mathbb {R}}, \end{aligned}$$

and let K denote the Fourier transform of m so that \(K \in {\mathscr {S}}'({\mathbb {R}})\). According to Sjölin [8] p.142, \(K\in C^{\infty }({\mathbb {R}})\) and there exists a number \(\alpha \geqslant 0\) such that

$$\begin{aligned} |K(x)| \lesssim 1 + |x|^{\alpha } \text{ for } x\in {\mathbb {R}} \end{aligned}$$

For \(t>0\) it is then clear that

$$\begin{aligned} e^{it|\xi |^{a}} = m(t^{1/a}\xi ) \end{aligned}$$

has the Fourier transform

$$\begin{aligned} K_{t}(y) = t^{-1/a}K(t^{-1/a}y). \end{aligned}$$

It follows that \(S_{t}f(x)=K_{t}\star f(x)\) for \(f\in L^{2}({\mathbb {R}}^{m})\) with compact support. We let denote the inverse Fourier transform of g and choose \(g\in {\mathscr {S}}({\mathbb {R}})\) such that supp and . We set

According to [7], p.143, one has \({\widehat{f}}_{v}(\xi ) = vg(v\xi +1/v)\) and

$$\begin{aligned}\Vert f_{v}\Vert _{H_s} \lesssim v^{1/2-2s} \text{ for } 0<s<1/4.\end{aligned}$$

We shall state three lemmas from [8].

Lemma 2

There exist positive numbers \(c_0\), \(\delta \) and \(v_0\) such that

$$\begin{aligned} |S_{xv^{2a-2}/a}f_v(x)| \geqslant c_0 \end{aligned}$$

for \(0<v<v_0\) and \(0< x <\delta \).

In the remaining part of this paper \(\delta \) and \(v_0\) are given by Lemma 2. We may also assume that \(\delta < 1\).

Lemma 3

For \(0<v<\min (v_0,\delta /4)\), \(0<t<1\), and \(\delta /2< x < \delta \) one has

$$\begin{aligned} |S_{t}f_v(x)| \lesssim \frac{v}{t^{\gamma }} \end{aligned}$$

where \(\gamma = (1+\alpha )/a > 0\).

Lemma 4

For \(0<v<\min (v_0,\delta /4)\), \(0<t<1\), and \(\delta /2<x<\delta \) one has

$$\begin{aligned} |S_{t}f_v(x)|\lesssim \frac{t}{v^{\beta }} \end{aligned}$$

where \(\beta = 2a\).

We shall use these lemmas to prove Theorem 2.

Proof of Theorem 2

Now let \(v_1\) satisfy \(0<v_1<\min (v_0,\delta /4)\) and set \(\epsilon _k = 2^{-k}\), \(k=1,2,3,...\)

We also set \(\mu = \max ((2a-2)\gamma ,\beta /(2a-2))\) and choose \(v_k\), \(k = 2,3,4,...\), such that \(0<v_k\leqslant \epsilon _{k}v_{k-1}^{\mu }\) and

$$\begin{aligned} \sum _{k=1}^{\infty }\root \of {\omega (1/v_{k}^{1/2})}<\infty . \end{aligned}$$

We then set \(f = \sum _{k=1}^{\infty }f_{v_k}\) and shall prove that \(f\in H_\omega \).

Arguing as in [8, pp. 145–147], it follows from Lemmas 2, 3, and 4 that with \(t_k(x) = xv_{k}^{2a-2}/a\) one has

$$\begin{aligned} |S_{t_k(x)}f(x)|\geqslant c_0>0 \end{aligned}$$

for \(\delta /2<x<\delta \) and \(k\geqslant k_0\). Hence we do not have \(\lim _{t\rightarrow 0}S_tf(x)=0\) in the interval \((\delta /2,\delta )\). Taking \(I = [-v_1,v_1]\) and \( J\subset (\delta /2,\delta )\) we have supp\(f\subset I\) and for every \(x\in J\) one does not have \(\lim _{t\rightarrow 0}S_tf(x)=0\). Thus Theorem 2 follows. It remains to prove that \(f\in H_{\omega }\).

We have

$$\begin{aligned} \Vert f_v\Vert _{H_\omega }^2 = \int |{\widehat{f}}_v(\xi )|^{2}(1+\xi ^2)^{1/4}\omega (|\xi |)d\xi \lesssim I_1 + I_2, \end{aligned}$$

where

$$\begin{aligned} I_1 = \int _{-1}^{1}|{\widehat{f}}_v(\xi )|^{2}d\xi \leqslant Cv^2 \end{aligned}$$

and

$$\begin{aligned} I_2 = \int |{\widehat{f}}_v(\xi )|^2|\xi |^{1/2}\omega (|\xi |)d\xi . \end{aligned}$$

It follows that

$$\begin{aligned} I_2= & {} \int v^2|g(v\xi +1/v)|^2|\xi |^{1/2}\omega (|\xi |)d\xi \\&\quad = \int v^{1/2}|g(\eta +1/v)|^2|\eta |^{1/2}\omega (\frac{|\eta |}{v})d\eta =\\&\quad = v^{1/2}\int |g(\xi )|^2|\xi -1/v|^{1/2}\omega (\frac{|\xi -1/v|}{v})\xi \leqslant Cv^{1/2}\\&\quad \times \int \limits _{|\xi -1/v|\leqslant v^{1/2}}{}|g(\xi )|^2v^{1/4}d\xi \\&\quad + \, Cv^{1/2}\int \limits _{|\xi -1/v|\geqslant v^{1/2}}{}|g(\xi )|^2(|\xi |^{1/2}+v^{-1/2})\omega (v^{-1/2})d\xi \\&\quad \leqslant Cv^{3/4}+C\omega (v^{-1/2}). \end{aligned}$$

Hence

$$\begin{aligned} \Vert f_v\Vert _{H_\omega }^2\lesssim v^{3/4} + \omega (v^{-1/2}),\quad 0<v<1, \end{aligned}$$

and

$$\begin{aligned}\Vert f_v\Vert _{H_\omega } \lesssim v^{3/8} + \root \of {\omega (v^{-1/2})}. \end{aligned}$$

We have \(f = \sum _{1}^{\infty }f_{v_k}\) and it follows that

$$\begin{aligned} \Vert f\Vert _{H_\omega } \leqslant \sum _{1}^{\infty }\Vert f_{v_k}\Vert _{H_\omega }\lesssim \sum _{1}^{\infty }v_k^{3/8} + \sum _{1}^{\infty }\root \of {\omega (v_k^{-1/2})}<\infty \end{aligned}$$

since \(v_k \leqslant \epsilon _k\).

We conclude that \(f\in H_\omega \) and the proof of Theorem 2 is complete. \(\square \)

Remark 1

In Sjölin [8] the function f in Theorem A is given by the formula

$$\begin{aligned} f = \sum _{1}^{\infty }f_{v_k}, \end{aligned}$$

where \(v_k\) is defined by taking \(0<v_1<\min (v_0,\delta /4)\) and \(v_k = \epsilon _kv_{k-1}^{\mu }\) for \(k = 2,3,4,...\) Here \(\epsilon _k = 2^{-k}\) and \(\mu >0\) is given in the proof of Theorem 2. Also let the intervals I and J be defined as in the proof of Theorem 2. We then set \(t_k(x) = xv_k^{2a-2}/a\) for \(x\in J\) and \(k=1,2,3,...\)

It is proved in [8] that for every \(x_0 \in J\)

$$\begin{aligned} \text{ one } \text{ does } \text{ not } \text{ have } \lim _{k \rightarrow \infty } S_{t_k(x_0)}f(x_0) = 0. \end{aligned}$$
(3)

We now fix \(x_0 \in J\) and shall use Theorem 1 to prove that although (3) holds one also has

$$\begin{aligned} \lim _{k\rightarrow \infty }S_{t_k(x_0)}f(x) = 0 \text{ for } \text{ almost } \text{ every } x\in J. \end{aligned}$$
(4)

We have \(v_k<\epsilon _k\) and it follows that

$$\begin{aligned} 0<t_k(x_0) \leqslant \epsilon _k^{2a-2} \end{aligned}$$

and

$$\begin{aligned} \sum _{1}^{\infty }(t_k(x_0))^{2s/a} \leqslant \sum _{1}^{\infty }2^{-k(2a-2)2s/a}< \infty \end{aligned}$$

for \(0<s<1/4\). Also \(f \in H_s\) for \(0<s<1/4\) and (4) follows from an application of Theorem 1.

Remark 2

In the case \(a=2\) one has \(\mu = 2\) and \(v_k = \epsilon _kv_{k-1}^{2}\), and we also have \(0<v_1<1/4\). It can be proved that it follows that

$$\begin{aligned} v_k = 4\cdot 2^{k-d2^{k}} \end{aligned}$$

where d is a constant and \(d>2\).