2. Proof of theorems 1.2 and 1.5

Proof of theorem 1.2 Let $s_{1}=\frac {r}{\frac {N+1}{N}r+1} + \epsilon$ for some constant $\epsilon >0$. We decompose $f$ as $f=\sum _{k=0}^{\infty }{f_{k}},$ where ${\rm supp} \hat {f_{0}} \subset B(0,\,1)$, ${\rm supp} \hat {f_{k}} \subset \{\xi : |\xi | \sim 2^{k}\},\, k \ge 1$. Then, we have

(2.1)\begin{equation} \biggl\|\sup_{n \in \mathbb{N}} |e^{it_{n}\Delta}f|\biggl\|_{L^{2}(B(0,1))} \le \sum_{k=0}^{\infty}{\biggl\|\sup_{n \in \mathbb{N}} |e^{it_{n}\Delta}f_{k}|\biggl\|_{L^{2}(B(0,1))}}. \end{equation}

For $k \lesssim 1$ and arbitrary $x \in B(0,\,1)$, $|e^{it_{n}\Delta }f_{k}(x)| \lesssim \|f_{k}\|_{L^{2}(\mathbb {R}^{N})}$, it is obvious that

(2.2)\begin{equation} \biggl\|\sup_{n \in \mathbb{N}} |e^{it_{n}\Delta}f_{k}|\biggl\|_{L^{2}(B(0,1))} \lesssim \|f\|_{H^{s_{1}}(\mathbb{R}^{N})}. \end{equation}

For each $k \gg 1$, we decompose $\{t_{n}\}_{n=1}^{\infty }$ as

\[ A^{1}_{k}:= \biggl\{t_{n}: t_{n} \ge 2^{-\frac{2k}{\frac{N+1}{N}r+1}} \biggl\} \]

and

\[ A^{2}_{k}:= \biggl\{t_{n}: t_{n} < 2^{-\frac{2k}{\frac{N+1}{N}r+1}} \biggl\}. \]

Then, we have

(2.3)\begin{align} & \biggl\|\sup_{n \in \mathbb{N}} |e^{it_{n}\Delta}f_{k}|\biggl\|_{L^{2}(B(0,1))}\nonumber\\ & \quad \le \biggl\|\sup_{n \in \mathbb{N}: t_{n} \in A^{1}_{k}} |e^{it_{n}\Delta}f_{k}|\biggl\|_{L^{2}(B(0,1))} + \biggl\|\sup_{n \in \mathbb{N}: t_{n} \in A^{2}_{k}} |e^{it_{n}\Delta}f_{k}|\biggl\|_{L^{2}(B(0,1))} \nonumber\\ & : =I + II. \end{align}

We first estimate $I$. Since $\{t_{n}\}_{n=1}^{\infty } \in {\ell }^{r,\infty }(\mathbb {N})$, we have

(2.4)\begin{equation} \sharp A^{1}_{k} \le C 2^{\frac{2rk}{\frac{N+1}{N}r+1}}, \end{equation}

which implies that

(2.5)\begin{equation} I \le \biggl(\sum_{n \in \mathbb{N}: t_{n} \in A^{1}_{k}}{ \biggl\|e^{it_{n}\Delta}f_{k}\biggl\|^{2}_{L^{2}(B(0,1))}}\biggl)^{1/2} \lesssim 2^{\frac{rk}{\frac{N+1}{N}r+1}} \|f_{k}\|_{L^{2}(\mathbb{R}^{N})} \lesssim 2^{-\epsilon k}\|f\|_{H^{s_1}(\mathbb{R}^{N})}. \end{equation}

For $II$, since

\[ A^{2}_{k} \subset \biggl(0, 2^{-\frac{2k}{\frac{N+1}{N}r+1}}\biggl). \]

By previous discussion, we have $k<\frac {2k}{\frac {N+1}{N}r+1}<2k.$ Then it follows from theorem 1.5 that,

(2.6)\begin{equation} II \lesssim 2^{(\frac{r}{\frac{N+1}{N}r+1}+\frac{\epsilon}{2})k}\|f_{k}\|_{L^{2}(\mathbb{R}^{N})} \le 2^{-\frac{\epsilon}{2} k}\|f\|_{H^{s_1}(\mathbb{R}^{N})}. \end{equation}

Inequalities (2.3), (2.5), and (2.6) yield for $k \gg 1$,

(2.7)\begin{align} \biggl\|\sup_{n \in \mathbb{N}} |e^{it_{n}\Delta}f_{k}|\biggl\|_{L^{2}(B(0,1))} & \lesssim 2^{-\frac{\epsilon k}{2}}\|f\|_{H^{s_1}(\mathbb{R}^{N})}. \end{align}

Combining inequalities (2.1), (2.2), and (2.7), inequality (1.7) holds true for $s_1$. Because $\epsilon > 0$, we have finished the proof of theorem 1.2. It remains to prove theorem 1.5.

Proof of theorem 1.5 We use a wave packets decomposition and an orthogonality argument to prove theorem 1.5.

$\bullet$ Wave packets decomposition.

We first decompose $e^{it\Delta }f$ on $B(0,\,1) \times (0,\,2^{-j})$ in a standard way. For this goal, we decompose the annulus $\{\xi : |\xi | \sim 2^{k}\}$ into almost disjoint $2^{j-k}$-cubes $\theta$ with sides parallel to the coordinate axes in $\mathbb {R}^{N}$. Let $2^{k-j}$-cube $\nu$ be dual to $\theta$ and cover $\mathbb {R}^{N}$ by almost disjoint cubes $\nu$. Denote the centre of $\theta$ by $c(\theta )$ and the centre of $\nu$ by $c(\nu )$. We notice that if $\nu \neq \nu ^{\prime }$, then $|c(\nu ) - c(\nu ^{\prime })| \ge 2^{k-j}$.

Let $\varphi$ be a Schwartz function defined on $\mathbb {R}^{N}$ whose fourier transform is non-negative and supported in a small neighbourhood of the origin, and identically equal to $1$ in another smaller ball centred at the origin. Let $\widehat {\varphi _{\theta }}(\xi ) = 2^{-\frac {(j-k)N}{2}} \hat {\varphi }(\frac {\xi - c(\theta )}{2^{j-k}})$ and $\widehat {\varphi _{\theta, \nu }}(\xi )= e^{-ic(\nu ) \cdot \xi } \widehat {\varphi _{\theta }}(\xi )$. Then $f$ can be decomposed by

\[ f = \sum_{\nu} \sum_{\theta} f_{\theta, \nu} = \sum_{\nu} \sum_{\theta} \langle f, \varphi_{\theta, \nu}\rangle \varphi_{\theta, \nu}, \]

and

\[ \|f\|_{L^{2}}^{2} \sim \sum_{\nu} \sum_{\theta} |\langle f, \varphi_{\theta, \nu}\rangle|^{2}. \]

When $t \in (0,\,2^{-j})$, integration by parts implies

\[ |e^{it\Delta} \varphi_{\theta,\nu}(x)| \le \frac{C_{M} 2^{\frac{(j-k)N}{2}}}{(1+ 2^{j-k}|x-c(\nu) + 2tc(\theta)|)^{M}}. \]

Here, $M$ can be sufficiently large. Therefore, $e^{it\Delta } \varphi _{\theta,\nu }(x)$ is essentially supported in a tube

\[ T_{\theta, \nu}: = \{(x,t), |x-c(\nu) + 2tc(\theta)| \le 2^{(j-k)({-}1+\delta)}, 0 \le t \le 2^{{-}j} \}, \]

where $\delta =\epsilon ^3$. The direction of $T_{\theta, \nu }$ is parallel to the vector $(-2c(\theta ),\, 1)$, and the angle between $(-2c(\theta ),\, 1)$ and the $x$-plane is approximately $2^{-k}$.

$\bullet$ Orthogonality argument.

We decompose $B(0,\,1)$ by $B(0,\,1) = \cup _{\nu ^{\prime }} B(c( \nu ^{\prime }),\, 2^{k-j})$ with $|c(\nu ^{\prime })| \lesssim 1$. Then

(2.8)\begin{equation} \biggl\|\sup_{t \in (0,2^{{-}j})}|e^{it\Delta}f(x)|\biggl\|_{L^2(B(0,1))}^{2} \leq \sum_{ \nu^{\prime}} \biggl\|\sup_{t \in (0,2^{{-}j})}|e^{it\Delta}f(x)|\biggl\|_{L^2(B(c(\nu^{\prime}),2^{k-j}))}^{2}. \end{equation}

We will consider two cases: (i) $j < k+\frac {\epsilon k}{N}$ and (ii) $j \geq k+\frac {\epsilon k}{N}$, respectively. In case (i), let $j=k+\epsilon _0k$, $0<\epsilon _0<\frac {\epsilon }{N}$, by lemma 1.6,

(2.9)\begin{align} \biggl\|\sup_{t \in (0,2^{{-}j})}|e^{it\Delta}f(x)|\biggl\|_{L^2(B(0,1))} & \leq \biggl(\sum_{\nu^{\prime}}\biggl\|\sup_{t \in (0,2^{{-}j})}|e^{it\Delta}f(x)|\biggl\|^2_{L^2(B(c(\nu^{\prime}), 2^{k-j}))}\biggl)^{1/2} \nonumber\\ & \lesssim 2^{(2k-j)\frac{N}{2(N+1)}+\epsilon k}\|f\|_{L^2}. \end{align}

We use an orthogonality argument in the proof of case (ii). Fix $c(\nu ^{\prime })$, we divide $f$ into two terms

\[ f_{1}= \sum_{\theta} \sum_{\nu: |c(\nu)- c(\nu^{\prime})| \le 2^{(j-k)({-}1+10 \delta)}} f_{\theta, \nu}, \]

and

\[ f_{2}= \sum_{\theta} \sum_{\nu: |c(\nu)- c(\nu^{\prime})| > 2^{(j-k)({-}1+ 10 \delta)}} f_{\theta, \nu}. \]

For $f_{1}$, by lemma 1.6 and the $L^{2}$-orthogonality, we have

(2.10)\begin{align} & \sum_{ \nu^{\prime}} \biggl\|\sup_{t \in (0,2^{{-}j})}|e^{it\Delta}f_{1}(x)|\biggl\|_{L^2(B(c(\nu^{\prime}),2^{k-j}))}^{2} \nonumber\\ & \sim C_{\epsilon} 2^{(2k-j)\frac{N}{N+1} +\epsilon k} \sum_{ \nu^{\prime}} \sum_{\theta} \sum_{\nu: |c(\nu)- c(\nu^{\prime})| \le 2^{(j-k)({-}1+10 \delta)}} \|f_{\theta, \nu}\|_{L^{2}}^{2} \nonumber\\ & \lesssim C_{\epsilon} 2^{(2k-j)\frac{N}{N+1} +2\epsilon k} \|f\|_{L^{2}}^{2}. \end{align}

We will complete the proof by showing that the contribution from $|e^{it\Delta }f_{2}|$ is negligible when $(x,\,t)$ belongs to $B(c(\nu ^{\prime }),\, 2^{k-j}) \times (0,\,2^{-j})$.

Indeed, by the Cauchy–Schwarz inequality and the $L^{2}$-orthogonality, there holds

\begin{align*} |e^{it\Delta}f_{2}| & \le \|f\|_{L^{2}} \biggl( \sum_{\theta} \sum_{\nu: |c(\nu)- c(\nu^{\prime})| > 2^{(j-k)({-}1+10 \delta)}} |e^{it\Delta} \varphi_{\theta, \nu}|^{2} \biggl)^{1/2} \nonumber\\ & \le \|f\|_{L^{2}} C_{M} 2^{\frac{(j-k)N}{2}}\nonumber\\ & \quad\times \biggl( \sum_{\theta} \sum_{\nu: |c(\nu)- c(\nu^{\prime})| > 2^{(j-k)({-}1+10 \delta)}} \frac{1}{(1+ 2^{j-k}|x-c(\nu) + 2tc(\theta)|)^{2M}}\biggl)^{1/2}. \end{align*}

For each $\theta$, $|x- c(\nu ) + 2t c(\theta )| \ge |c(\nu ) -c(\nu ^{\prime })|/2$, then we have

\begin{align*} & \sum_{\nu: |c(\nu)- c(\nu^{\prime})| > 2^{(j-k)({-}1+10 \delta)}} \frac{1}{(1+ 2^{j-k}|x-c(\nu) + 2tc(\theta)|)^{2M}} \\ & \quad \le 2^{2M} \sum_{\substack{l \in \mathbb{N}^+\\ l \ge 2^{10\delta\epsilon k /N} }}\sum_{\nu: l2^{k-j} \le |c(\nu)- c(\nu^{\prime})| < (l+1)2^{k-j}} \frac{1}{(1+ 2^{j-k}|c(\nu)-c(\nu^{\prime})|)^{2M}} \\ & \quad \le 2^{2M} \sum_{\substack{l \in \mathbb{N}^+\\ l \ge 2^{10\delta\epsilon k /N} }}\frac{C_{N} l^{N}}{(1+l)^{2M}} \\ & \quad \le C_{M,N} 2^{{-}M\epsilon^4 k}. \end{align*}

Notice that the number of $\theta$'s is dominated by $2^{Nk}$. So by choosing $M$ sufficiently large, for each $(x,\,t) \in B(c(\nu ^{\prime }),\, 2^{k-j}) \times (0,\,2^{-j})$, we have

\[ |e^{it\Delta}f_{2}| \le C_{N} 2^{{-}1000k} \|f\|_{L^{2}}. \]

Then, the proof is finished since

\[ \sum_{ \nu^{\prime}} \biggl\|\sup_{t \in (0,2^{{-}j})}|e^{it\Delta}f_{2}(x)|\biggl\|_{L^2(B(c(\nu^{\prime}),2^{k-j}))}^{2} \le C_{N}^{2} 2^{{-}2000k} \|f\|_{L^{2}}^{2}.\]

3. A counterexample: theorem 1.3

We notice that the counterexample for $r=\frac {N}{N+1}$ can be also applied to the case when $r > \frac {N}{N+1}$, since ${\ell }^{N/(N+1),\infty }(\mathbb {N}) \subset {\ell }^{r,\infty }(\mathbb {N})$ and $\min \{\frac {r}{\frac {N+1}{N}r+1},\, \frac {N}{2(N+1)}\} = \frac {N}{2(N+1)}$ when $r > \frac {N}{N+1}$. Therefore, next we always assume $r \in (0,\, \frac {N}{N+1}]$.

Fix $r \in (0,\, \frac {N}{N+1}]$, we first construct a sequence which belongs to ${\ell }^{r,\infty }(\mathbb {N})$. Put $\beta = \frac {2}{\frac {N+1}{N}r+1}$. Let $R_{1}=2$ and for each positive integer $n$, $R_{n+1}^{-\beta } \le \frac {1}{2} R_{n}^{-\beta (r+1)}$. Denote the intervals $I_{n}=[R_{n}^{-\beta (r+1)},\, R_{n}^{-\beta })$, $n \in \mathbb {N}^+$. On each $I_{n}$, we get an equidistributed subsequence $t_{n_{j}},\, j =1,\, 2,\,...,\, j_{n}$ such that

\[ \{t_{n_{j}}, 1 \le j \le j_{n}\} =: R_{n}^{-\beta(r+1)}\mathbb{Z} \cap I_{n} , \]

and $t_{n_{j}} - t_{n_{j+1}} = R_{n}^{-\beta (r+1)}$. We claim that the sequence $t_{n_{j}},\, j=1,\,2,\,...,\,j_{n},\, n=1,\,2,\,...$ belongs to ${\ell }^{r,\infty }(\mathbb {N})$.

Indeed, according to Lemma 3.2 from [Reference Dimou and Seeger7], it suffices to show that

(3.1)\begin{equation} \sup_{b>0}b^{r}\sharp \biggl\{(n,j): b< t_{n_{j}} \le 2b\biggl\} \lesssim 1. \end{equation}

Notice that we only need to consider $0 < b <1$ because $t_{n_{j}} \in (0,\,1)$ for each $n$ and $j$. Assume that $(b,\, 2b] \cap I_{n} \neq \emptyset$ for some $n$, then we have $b < R_{n}^{-\beta }$, $2b \ge R_{n}^{-\beta (r+1)}.$ Therefore,

\[ 2b < 2R_{n}^{-\beta} \le R_{n-1}^{-\beta(r+1)}, \quad b \ge \frac{1}{2} R_{n}^{-\beta(r+1)} \ge R_{n+1}^{-\beta}. \]

This yields $(b,\, 2b] \cap I_{n^{\prime }} = \emptyset$ for any $n^{\prime } \neq n$, hence

\[ b^{r}\sharp \biggl\{(n,j): b< t_{n_{j}} \le 2b\biggl\} \le b^{r+1}R_{n}^{\beta(r+1)} < 1. \]

Then (3.1) follows by the arbitrariness of $b$.

Our counterexample comes from the following lemma.

Lemma 3.1 Let $R \gg 1$ and $I=[R^{-\beta (r+1)},\, R^{-\beta })$. Assume that the sequence $\{ t_{j} : 1 \le j \le j_{0}\} = R^{-\beta (r+1)}\mathbb {Z} \cap I$ and $t_{j} - t_{j+1} = R^{-\beta (r+1)}$ for each $1 \le j \le j_{0}-1$. Then there exists a function $f$ with supp $\hat {f} \subset B(0,\,2R)$ such that

(3.2)\begin{equation} \biggl\|\sup_{1 \le j \le j_{0}}|e^{i\frac{t_{j}}{2 \pi}\Delta}f| \biggl \|_{L^{2}(B(0,1))} \gtrsim R^{\frac{1-\beta}{2}} R^{\frac{\beta}{2}} R^{(N-1)(1-\frac{(r+1)\beta}{2})- \epsilon}, \end{equation}

and

(3.3)\begin{equation} \|f \|_{H^{s}(\mathbb{R}^{N})} \lesssim R^{s} R^{\frac{\beta}{4}} R^{\frac{N-1}{2}(1-\frac{(r+1)\beta}{2})}. \end{equation}

Here, $\epsilon$ is any positive number.

We use lemma 3.1 to show the counterexample here and prove the lemma a moment later. Assume that the maximal estimate

(3.4)\begin{equation} \biggl\|\sup_{n}\sup_{j} |e^{i\frac{t_{n_{j}}}{2\pi}\Delta}f|\biggl\|_{L^{2}(B(0,1))} \leq C\|f\|_{H^s(\mathbb{R}^N)} \end{equation}

holds for some $s >0$ and each $f \in H^{s}(\mathbb {R}^{N})$, then for each $n \in \mathbb {N}^+$, we have

(3.5)\begin{equation} \biggl\|\sup_{j} |e^{i\frac{t_{n_{j}}}{2\pi}\Delta}f|\biggl\|_{L^{2}(B(0,1))} \leq C\|f\|_{H^s(\mathbb{R}^N)} \end{equation}

whenever $f \in H^{s}(\mathbb {R}^{N})$. Lemma 3.1 and inequality (3.5) yield

(3.6)\begin{equation} R_{n}^{\frac{2-\beta}{4}} R_{n}^{\frac{N-1}{2}(1-\frac{(r+1)\beta}{2})- \epsilon} \le C R_{n}^{s}. \end{equation}

Then, we have $s \ge \frac {r}{\frac {N+1}{N}r +1}$, since $R_{n}$ can be sufficiently large and $\epsilon$ is arbitrarily small. Finally, we obtain a sequence $\frac {t_{n_{j}}}{2\pi },\, j=1,\,2,\,...,\,j_{n},\, n=1,\,2,\,... \in {\ell }^{r,\infty }(\mathbb {N})$ such that the maximal estimate (3.4) holds only if $s \ge \frac {r}{\frac {N+1}{N}r +1}$.

In the rest of this section, we prove lemma 3.1. Setting

\begin{align*} \Omega_{1} & = \biggl(-\frac{1}{100}R^{\frac{\beta}{2}}, \frac{1}{100}R^{\frac{\beta}{2}} \biggl),\\ \Omega_{2}& = \biggl\{ \bar{\xi}\in \mathbb{R}^{N-1}: \bar{\xi} \in 2 \pi R^{\frac{(r+1)\beta}{2}} \mathbb{Z}^{N-1} \cap B(0,R^{1-\epsilon}) \biggl\} + B\left(0, \frac{1}{1000}\right), \end{align*}

then we define $\hat {f_{1}}(\xi _{1}) = \hat {h}(\xi _{1} + \pi R)$, $\hat {f_{2}}(\bar {\xi }) = \hat {g} (\bar {\xi } + \pi R\theta )$, where $\hat {h} = \chi _{\Omega _{1} }$, $\hat {g} = \chi _{\Omega _{2}}$, and some $\theta \in \mathbb {S}^{N-2}$ (when $N=2$, $\theta \in (0,\,1)$) which will be determined later. Define $f$ by $\hat {f} = \hat {f_{1}}\hat {f_{2}}$, it is easy to check that $f$ satisfies (3.3). We are left to prove that inequality (3.2) holds for such $f$. Notice that

(3.7)\begin{equation} |e^{i \frac{t_{j}}{2\pi} \Delta}f(x_{1},\bar{x})| = |e^{i\frac{t_{j}}{2\pi} \Delta}f_{1}(x_{1})||e^{i\frac{t_{j}}{2\pi}\Delta}f_{2}(\bar{x})|. \end{equation}

We first consider $|e^{i\frac {t_{j}}{2\pi }\Delta }f_{1}(x_{1})|$. A change of variables implies

\[ |e^{i\frac{t_{j}}{2\pi} \Delta}f_{1}(x_{1})| = |e^{i\frac{t_{j}}{2\pi}\Delta}h(x_{1}-Rt_{j})| . \]

It is easy to check that $|e^{i\frac {t_{j}}{2\pi }\Delta }h(x_{1})| \gtrsim |\Omega _{1}|$ for each $j$ whenever $|x_{1}| \le R^{-\frac {\beta }{2}}$. Note that for each $x_{1} \in (0,\, R^{1-\beta })$, there exists at least one $t_{j}$ such that $|x_{1}- Rt_{j}| \le R^{1-\beta (r+1)} \le R^{-\frac {\beta }{2}}$ since $\{t_j\}_{j=1}^{j_0}\subset [R^{-\beta (r+1)},\,R^{-\beta })$ and $t_j-t_{j+1}=R^{-\beta (r+1)}$. Hence, we have

(3.8)\begin{equation} |e^{i\frac{t_{j}}{2\pi}\Delta}f_{1}(x_{1})| \gtrsim |\Omega_{1}|, \end{equation}

whenever $x_{1} \in (0,\, \frac {1}{2}R^{1-\beta })$ and $Rt_{j} \in (x_{1},\, x_{1} + R^{-\frac {\beta }{2}})$.

For $|e^{i\frac {t_{j}}{2\pi }\Delta }f_{2}(\bar {x})|$, we have

\[ |e^{i\frac{t_{j}}{2\pi} \Delta}f_{2}(\bar{x})| = |e^{i\frac{t_{j}}{2\pi} \Delta}g(\bar{x}- Rt_{j} \theta)| . \]

According to Barceló–Bennett–Carbery–Ruiz–Vilela [Reference Barceló, Bennett, Carbery, Ruiz and Vilela1], for each $j$ and $\bar {x} \in U_{0}$,

(3.9)\begin{equation} |e^{i\frac{t_{j}}{2\pi} \Delta}g(\bar{x})| \gtrsim |\Omega_{2}|, \end{equation}

here

\[ U_{0}= \biggl\{ \bar{x}\in \mathbb{R}^{N-1}: \bar{x} \in R^{-\frac{(r+1)\beta}{2}} \mathbb{Z}^{N-1} \cap B(0,2) \biggl\} + B\left(0, \frac{1}{1000}R^{{-}1+\epsilon}\right). \]

We sketch the main idea of the proof of inequality (3.9) for the reader's convenience. Indeed, for each $\bar {\xi } \in \Omega _{2}$, we write $\bar {\xi } = 2 \pi R^{\frac {(r+1)\beta }{2}}l + \bar {\eta }$, $l \in \mathbb {Z}^{N-1}$, $2\pi |l| \le R^{1-\frac {(r+1)\beta }{2}-\epsilon }$, $\bar {\eta } \in B(0,\, \frac {1}{1000})$. Then for any $\bar {x}_{m}= R^{-\frac {(r+1)\beta }{2}}m$, $m \in \mathbb {Z}^{N-1}$, $|m| \le 2 R^{\frac {(r+1)\beta }{2}}$, $t_{j} = R^{-(r+1)\beta }(j_0+1-j)$, $1 \le j \le j_{0}$, we have

\begin{align*} & e^{i \frac{t_{j}}{2\pi} \Delta }g(\bar{x}_{m}) \\ & \quad =\frac{1}{(2\pi)^{N}} \sum_{l \in \mathbb{Z}^{N-1} : 2\pi |l| \le R^{1- (r+1)\beta /2-\epsilon}} e^{2\pi i m \cdot l + 2\pi i (j_0+1-j)|l|^{2}}\\ & \qquad\times \int_{B(0,\frac{1}{1000})} e^{ i \bar{x}_{m} \cdot \bar{\eta} + 2i \frac{t_{j}}{2\pi} 2\pi R^{\frac{(r+1)\beta}{2}}l \cdot \bar{\eta} + i \frac{t_{j}}{2\pi} |\bar{\eta}|^{2} }\,{\rm d}\bar{\eta} \\ & \quad =\frac{1}{(2\pi)^{N}} \sum_{ l \in \mathbb{Z}^{N-1}:2\pi |l| \le R^{1-(r+1)\beta /2-\epsilon}}\int_{B(0,\frac{1}{1000})} e^{ i \bar{x}_{m} \cdot \bar{\eta} + 2i \frac{t_{j}}{2\pi} 2 \pi R^{\frac{(r+1)\beta}{2}}l \cdot \bar{\eta} + i \frac{t_{j}}{2\pi} |\bar{\eta}|^{2} }\,{\rm d}\bar{\eta} . \end{align*}

Noting that $|\bar {x}_{m}| \le 2$, $|t_{j}| \le R^{-\beta }$ and $|\bar {\eta }| \le \frac {1}{1000}$ imply

\[ \biggl | \bar{x}_{m} \cdot \bar{\eta} + 2 \frac{t_{j}}{2\pi} 2 \pi R^{\frac{(r+1)\beta}{2}}l \cdot \bar{\eta} + \frac{t_{j}}{2\pi} |\bar{\eta}|^{2} \biggl| \le \frac{1}{100}, \]

then, we have

\[ |e^{i\frac{t_{j}}{2\pi} \Delta}g(\bar{x}_{m})| \ge \frac{1}{2(2\pi)^{N}} |\Omega_{2}|. \]

Moreover, for each $\bar {x} \in U_{0}$, there exits an $\bar {x}_{m}$ such that $|\bar {x} - \bar {x}_{m}| \le \frac {1}{1000} R^{-1 + \epsilon }$, by the mean value theorem and the fact that $|\bar {\xi }| \le 2R^{1-\epsilon }$,

\[ |e^{i\frac{t_{j}}{2\pi} \Delta}g(\bar{x})- e^{i\frac{t_{j}}{2\pi} \Delta}g(\bar{x}_{m})| \le \int_{\mathbb{R}^{N-1}} |\bar{x} -\bar{x}_{m}| |\bar{\xi}| \hat{g}(\bar{\xi}) \,{\rm d}\bar{\xi} \le \frac{1}{500} | \Omega_{2} |. \]

Finally, we arrive at inequality (3.9) by the triangle inequality.

Therefore, we have

(3.10)\begin{equation} |e^{i \frac{t_{j}}{2\pi} \Delta}f_{2}(\bar{x})| \gtrsim |\Omega_{2}|, \quad \quad \bar{x} \in U_{j}=U_{0} + Rt_{j}\theta. \end{equation}

Set $U_{x_{1}} = \bigcup _{j: Rt_{j} \in R^{1-(r+1)\beta }\mathbb {Z} \cap (x_{1},\, x_{1} + R^{-\beta /2}) } U_{0} + Rt_{j} \theta$. Then inequalities (3.8) and (3.10) imply that for each $x_{1} \in (0,\, \frac {1}{2}R^{1-\beta })$ and $\bar {x} \in U_{x_{1}}$, there holds

(3.11)\begin{equation} \sup_{j} |e^{i \frac{t_{j}}{2\pi} \Delta}f(x_{1}, \bar{x})| \gtrsim |\Omega_{1}||\Omega_{2}|. \end{equation}

Next, we need to select a $\theta \in \mathbb {S}^{N-2}$, such that $|U_{x_{1}}| \gtrsim 1$ for each $x_{1} \in (0,\, \frac {1}{2} R^{1-\beta })$, which follows if we can prove that there exists a $\theta \in \mathbb {S}^{N-2}$ such that $B(0,\,1/2)\subset U_{x_1}$ for all $x_{1} \in (0,\, \frac {1}{2} R^{1-\beta })$. So it remains to prove the claim that there exists a $\theta \in \mathbb {S}^{N-2}$ such that

\[ \bigcup_{j: Rt_{j} \in R^{1-\beta(r+1)}\mathbb{Z} \cap (x_{1}, x_{1} + R^{-\beta/2}) } \biggl\{ \bar{x}\in\mathbb{R}^{N-1}: \bar{x} \in R^{-\frac{(r+1)\beta}{2}} \mathbb{Z}^{N-1} \cap B(0,2) \biggl\} + Rt_{j} \theta \]

is $\frac {1}{1000}R^{-1 + \epsilon }$ dense in the ball $B(0,\, 1/2)$. In order to apply Lemma 2.1 from Lucà–Rogers [Reference Lucà and Rogers12] to get this claim, we first rescale by $R^{\frac {\beta (r+1)}{2}}$, and replace $R^{1+\frac {\beta (r+1)}{2}}t_{j}$ by $s_{j}$, replace $R^{\frac {\beta r}{2}}$ by $R^{\prime }$, recall that $\beta = \frac {2}{\frac {N+1}{N}r+1}$, then we are reduced to show

\[ \bigcup_{j: s_{j} \in (R^{\prime})^{1/N} \mathbb{Z} \cap ((R^{\prime})^{(r+1)/r}x_{1}, (R^{\prime})^{(r+1)/r}x_{1} + R^{\prime}) } \biggl\{ \bar{x}: \bar{x} \in \mathbb{Z}^{N-1} \cap B(0,2(R^{\prime})^{(r+1)/r}) \biggl\} + s_{j} \theta \]

is $\frac {1}{1000}(R^{\prime })^{-\frac {1}{N} + \frac {(\frac {N+1}{N}r+1)\epsilon }{ r}}$ dense in the ball $B(0,\, \frac {1}{2} (R^{\prime })^{(r+1)/r})$, which is equivalent to prove that for any $y\in B(0,\,\frac {1}{2}(R')^{(r+1)/r})$, there exist

\begin{align*} & \bar{x}_y\in \mathbb{Z}^{N-1}\cap B(0,2(R')^{(r+1)/r}) \quad {\text and}\\ & s_y\in (R^{\prime})^{1/N} \mathbb{Z} \cap ((R^{\prime})^{(r+1)/r}x_{1}, (R^{\prime})^{(r+1)/r}x_{1} + R^{\prime}), \end{align*}

such that

(3.12)\begin{equation} |y-\bar{x}_y-s_y\theta|<\frac{1}{1000}(R^{\prime})^{-\frac{1}{N} + \frac{\left(\frac{N+1}{N}r+1\right)\epsilon}{ r}}, \end{equation}

for a fixed $\theta \in \mathbb {S}^{N-2}$, which is independent of $y$ and $x_1$. This is implied by the following Lemma 3.2 from Lucà–Rogers [Reference Lucà and Rogers12], but we prefer to omit the proof of (3.12), because a similar but more detailed proof can be found in Corollary 2.2 of [Reference Lucà and Rogers12].

Lemma 3.2 [Reference Lucà and Rogers12, Lemma 2.1]

Let $d\geq 2$, $0<\epsilon,\,\delta <1$ and $\kappa >\frac {1}{d+1}$. Then, if $\delta <\kappa$ and $R>1$ is sufficiently large, there is $\theta \in \mathbb {S}^{d-1}$ for which, given any $[y]\in \mathbb {T}^d$ and $a\in \mathbb {R}$, there is a $t_y\in R^{\delta }\mathbb {Z} \cap (a,\,a+R)$ such that

\begin{equation*} |[y]-[t_y\theta]|\leq \epsilon R^{(\kappa-1)/d}, \end{equation*}

where ‘$[\cdot ]$’ means taking the quotient $\mathbb {R}^d/\mathbb {Z}^d=\mathbb {T}^d$. Moreover, this remains true with $d=1$, for some $\theta \in (0,\,1)$.

Finally, it follows from (3.7) and (3.11) that

\begin{align*} & \int_{B(0,1)} \sup_{j}|e^{i \frac{t_{j}}{2\pi} \Delta}f(x_{1},\bar{x})|^{2}\,{\rm d}\bar{x}\,{\rm d}x_{1}\\ & \quad \ge \int_{0}^{\frac{R^{1-\beta}}{2}} \int_{U_{x_{1}}} \sup_{j}|e^{i \frac{t_{j}}{2\pi} \Delta}f(x_{1},\bar{x})|^{2}\,{\rm d}\bar{x}\,{\rm d}x_{1} \gtrsim R^{1-\beta}|\Omega_{1}|^{2} |\Omega_{2}|^{2}, \end{align*}

which implies inequality (3.2).

4. A counterexample for theorem 1.9

For convenience, we first set $N=2$. By changing of variables, the nonelliptic Schrödinger operator can be written as

(4.1)\begin{equation} e^{it\square}f(x):=\frac{1}{(2\pi)^{2}} \int_{\mathbb{R}^{2}}{e^{ix \cdot \xi +it\xi_{1}\xi_{2}} \hat{f} (\xi)\,{\rm d}\xi }. \end{equation}

For each $r \in (0,\,1]$, we will show that there exists $\{t_{n}\}_{n=1}^{\infty } \in \ell ^{r, \infty }(\mathbb {N})$, such that the maximal estimate

(4.2)\begin{equation} \biggl\| \sup_{n \in \mathbb{N}} |e^{it_{n}\square}f| \biggl\|_{L^{2}(B(0,1))} \le C\|f\|_{H^{s}} \end{equation}

holds for all $f \in H^{s}(\mathbb {R}^{2})$ only if $s \ge \frac {r}{r+1}$.

Indeed, we may choose $t_{n}=1/n^{\frac {1}{r} + \epsilon }$, it is clear that $\{t_{n}\}_{n=1}^{\infty } \in \ell ^{r, \infty }(\mathbb {N})$ but never belongs to $\ell ^{r-\epsilon, \infty }(\mathbb {N})$ for any small $\epsilon >0$. Moreover, $t_{n} -t_{n+1}$ is decreasing. According to Lemma 3.2 in [Reference Dimou and Seeger7], we can select $\{b_{j}\}_{j=1}^{\infty }$ and $\{M_{j}\}_{j=1}^{\infty }$ satisfying $\lim _{j \rightarrow \infty }b_{j} = 0,\, \:\ \lim _{j \rightarrow \infty }M_{j} = \infty,\,$ and

(4.3)\begin{equation} M_{j}b_{j}^{1-r + \epsilon} \le 1, \end{equation}

such that

(4.4)\begin{equation} \sharp\biggl\{n: b_{j} < t_{n} \le 2b_{j}\biggl\} \ge M_{j}b_{j}^{{-}r + \epsilon}. \end{equation}

By the similar argument as Proposition 3.3 in [Reference Dimou and Seeger7], when $t_{n} \le b_{j}$, we have

(4.5)\begin{equation} t_{n}-t_{n+1} \le 2M_{j}^{{-}1}b_{j}^{r-\epsilon +1}. \end{equation}

For fixed $j$, choose $\lambda _{j}=\frac {1}{1000} M_{j}^{\frac {1}{2}}b_{j}^{-\frac {r-\epsilon +1}{2}}$ and $\widehat {f_{j}}(\xi _{1},\,\xi _{2}) =\frac {1}{\lambda _{j}} \chi _{[0,\lambda _{j}]\times [{-\lambda _{j}-1,\,-\lambda _{j}}]} (\xi _{1},\,\xi _{2}).$ Therefore,

(4.6)\begin{equation} \|f_{j}\|_{H^{\frac{r-\epsilon}{r-\epsilon +1}}}\leq\lambda_{j}^{\frac{r-\epsilon}{r-\epsilon +1}-\frac{1}{2}}. \end{equation}

Let $U_{j}=(0,\,\frac {\lambda _{j}b_{j}}{2}) \times (-\frac {1}{1000},\, \frac {1}{1000}).$ Notice that $U_{j} \subset B(0,\,1)$ due to inequality (4.3). Next, we will show that for each $x \in U_{j}$,

(4.7)\begin{equation} \sup_{n \in \mathbb{N}} |e^{it_{n}\square}f_{j}| > \frac{1}{2(2\pi)^{2}}. \end{equation}

After changing variables, we have for each $n \in \mathbb {N}$,

(4.8)\begin{equation} |e^{it_{n}\square}f_{j}(x)| =\frac{1}{(2\pi)^{2}} \biggl|\int_{{-}1}^{0}\int_{0}^{1}{e^{i\lambda_{j}(x_{1}-\lambda_{j}t_{n})\eta_{1}+ix_{2}\eta_{2}+it_{n}\lambda_{j}\eta_{1}\eta_{2}}\,{\rm d}\eta_{1}\,{\rm d}\eta_{2}}\biggl|. \end{equation}

For each $x \in U_{j}$, there exists a unique $n(x,\,j)$ such that

\[ x_{1} \in (\lambda_{j}t_{n(x,j)+1}, \lambda_{j}t_{n(x,j)}]. \]

It is obvious that $t_{n(x,j)+1} \le \frac {b_{j}}{2}$, then $t_{n(x,j)} \le b_{j}$ due to inequality (4.4) and the assumption that $t_{n} -t_{n+1}$ is decreasing. Then it follows from inequality (4.5) that

\[ |\lambda_{j}(x_{1}-\lambda_{j}t_{n(x,j)})\eta_{1}| \le 2\lambda_{j}^{2} M^{{-}1}_{j}b_{j}^{r-\epsilon+1} \le \frac{1}{1000}. \]

Also, $| x_{2}\eta _{2}| \le \frac {1}{1000},$ and by inequality (4.3), we have $|\lambda _{j}t_{n(x,j)}\eta _{1}\eta _{2}| \le \lambda _{j}b_{j} \le \frac {1}{1000}.$ Therefore, if we take $n=n(x,\,j)$ in (4.8), then the phase function will be sufficiently small such that $|e^{it_{n(x,j)}\square }f_{j}(x)| > \frac {1}{2(2\pi )^{2}}$ for each $x \in U_{j}$, which implies inequality (4.7). Then, it follows from inequalities (4.6) and (4.7) that

\[ \frac{\| \sup_{n \in \mathbb{N}} |e^{it_{n}\square}f_j| \|_{L^{2}(B(0,1))}}{\|f_{j}\|_{H^{\frac{r-\epsilon}{r-\epsilon +1}}}} \ge C M_{j}^{\frac{1}{2(r-\epsilon +1)}}. \]

This implies that the maximal estimate (4.2) can not hold when $s \le \frac {r-\epsilon }{r-\epsilon +1}$, hence when $s < \frac {r}{r+1}$ by the arbitrariness of $\epsilon$.

Remark 4.1 The original idea we adopted to construct the above counterexample comes from [Reference Rogers, Vargas and Vega13]. The same idea remains valid in general dimensions. For example, in $\mathbb {R}^{3}$, by changing variables, we can write

\[ e^{itL}f(x):= \frac{1}{(2\pi)^{3}} \int_{\mathbb{R}^{3}}{e^{ix \cdot \xi +it(\xi_{1}\xi_{2}\pm \xi_{3}^{2})} \hat{f} (\xi)\,{\rm d}\xi }. \]

In order to prove the necessary condition, we only need to take

\[ U_{j}=(0,\frac{\lambda_{j}b_{j}}{2}) \times \left(-\frac{1}{1000}, \frac{1}{1000}\right) \times \left(-\frac{1}{1000}, \frac{1}{1000}\right) \]

and

\[ \widehat{f_{j}}(\xi_{1},\xi_{2}, \xi_{3}) =\frac{1}{\lambda_{j}} \chi_{[0,\lambda_{j}]\times[{-\lambda_{j}-1,-\lambda_{j}}] \times (0,1)}(\xi_{1},\xi_{2}, \xi_{3}). \]