2.1 Symbols associated to subcurves

We begin with some notations. Let $N\ge 2$ , and let $\delta $ and $B'$ denote numbers such that

$$ \begin{align*} 2^{-k /N}\le \delta\le 2^{-7dN}B^{-6N}, \qquad B \le B' \le B^C \end{align*} $$

for a large constant $C\ge 3d+1$ . We note that ${\mathfrak V({N-1}, {B'})}$ holds for some $B'$ . In fact, ${\mathfrak V({N-1}, {B^2})}$ follows by (2.1) and ${\mathfrak V(N, B)}$ .

For $s\in I$ , we define a linear map $ \widetilde {\mathcal L}^\delta _s:\mathbb R^{d}\mapsto \mathbb R^{d}$ as follows:

(2.4) $$ \begin{align} \begin{aligned} (\widetilde{\mathcal L}_{s}^\delta)^\intercal \gamma^{(j)}(s) &=\delta^{N-j}\gamma^{(j)}(s), \qquad && j=1,\dots,N-1, \\ (\widetilde{\mathcal L}_{s}^\delta)^\intercal v &=v, \qquad \qquad \quad~ && v\in \big(\mathrm V_s^{\gamma,N-1}\big)^{\perp}, \end{aligned} \end{align} $$

where $\mathrm V_s^{\gamma ,\ell } = \operatorname {\mathrm {span}} \big \{ \gamma ^{(j)}(s) : j=1,\dots ,\ell \big \}$ . $\widetilde {\mathcal L}_{s}^\delta $ is well defined since ${\mathfrak V({N-1}, {B^2})}$ holds for $\gamma $ . The linear map $\widetilde {\mathcal L}_{s}^\delta $ naturally appears when we rescale a subcurve of length about $\delta $ (see the proofs of Lemma 2.7 and 2.8). We denote

(2.5) $$ \begin{align} \mathcal L_{s}^\delta(\tau, \xi)= \big(\delta^N \tau - \gamma(s)\cdot \widetilde{\mathcal L}_s^\delta \xi,\,\, \widetilde{\mathcal L}_s^\delta \xi\big), \qquad (\tau, \xi)\in \mathbb R\times \mathbb R^d. \end{align} $$

We set $G(s)=(1,\gamma (s))$ and define

$$ \begin{align*} \Lambda_k(s,\delta, B')= \bigcap_{0\le j\le N-1} \big\{(\tau,\xi) \in \mathbb R \times \mathbb A_k & : |\langle G^{(j)}(s),(\tau,\xi)\rangle| \le B'2^{k+5} \delta^{N-j} \big \}. \end{align*} $$

Definition 2.4. Let such that . Then, by we denote the set of $\mathfrak a \in \mathrm C^{d+N+2}(\mathbb R^{d+3})$ such that

(2.6)

(2.7)

We define $\operatorname {\mathrm {supp}}_{\xi } \mathfrak a=\bigcup _{s,t,\tau } \operatorname {\mathrm {supp}} \mathfrak a(s,t,\tau ,\cdot )$ and $\operatorname {\mathrm {supp}}_{s, \xi } \mathfrak a=\bigcup _{t,\tau } \operatorname {\mathrm {supp}} \mathfrak a(\cdot ,t,\tau ,\cdot )$ , and $\operatorname {\mathrm {supp}}_{s} \mathfrak a$ and $\operatorname {\mathrm {supp}}_{\tau , \xi } \mathfrak a$ are defined likewise. Note that a statement $S(s, \xi )$ , depending on $(s, \xi )$ , holds on $\operatorname {\mathrm {supp}}_{s,\xi } \mathfrak a$ if and only if $S(s,\xi )$ holds whenever $(s,t, \tau ,\xi ) \in \operatorname {\mathrm {supp}} \mathfrak a$ for some t, $\tau $ .

Denote $\mathrm V_s^{G,\ell } = \operatorname {\mathrm {span}} \{(1,0), G'(s), \dots , G^{(\ell )}(s) \}.$ We take a close look at the map ${\mathcal L}_s^\delta $ . By the equations (2.4) and (2.5) we have

(2.8) $$ \begin{align} \begin{cases} \begin{aligned} (\mathcal L_s^\delta)^\intercal G(s) &= \delta^{N}(1,0), \\[2pt] (\mathcal L_s^\delta)^\intercal G^{(j)}(s) &= \delta^{N-j}G^{(j)}(s), \qquad &&j=1,\dots, N-1, \\ ({\mathcal L}_s^\delta)^\intercal v &=v, \qquad&& v\in (\mathrm V_s^{G,N-1} )^\perp. \end{aligned} \end{cases} \end{align} $$

The first identity is clear since $(\mathcal L_s^\delta )^\intercal (\tau , \xi )=(\delta ^N \tau , (\widetilde {\mathcal L}_s^\delta )^\intercal \xi -\tau (\widetilde {\mathcal L}_s^\delta )^\intercal \gamma (s))$ . The second and the third follow from (2.4) since $G^{(j)}\in \{0\} \times \mathbb R^d$ , $1 \le j \le N-1$ , $\big ( \mathrm V_s^{G,N-1} \big )^\perp \subset \{0\} \times \mathbb R^d$ , and $(\mathcal L_s^\delta )^\intercal (0,\xi )=(0,(\widetilde {\mathcal L}_s^\delta )^\intercal \xi )$ . Furthermore, there is a constant $C=C(B)$ , independent of s and $\delta $ , such that

(2.9) $$ \begin{align} | \mathcal L_{s}^{\delta} (\tau, \xi)| \le C|(\tau,\xi)|. \end{align} $$

Note that (2.9) is equivalent to $ |( \mathcal L_{s}^{\delta })^\intercal (\tau , \xi )| \le C|(\tau ,\xi )|$ . The inequality is clear from (2.4) because ${\mathfrak V({N-1}, {B^2})}$ holds and all the eigenvalues of $(\widetilde {\mathcal L}_{s}^\delta )^\intercal $ are contained in the interval $(0,1]$ .

Lemma 2.5. Let $ \mathcal L_{s}^\delta (\tau ,\xi )\in \Lambda _k(s,\delta , B')$ and ${\mathfrak V({N-1}, {B'})}$ hold for $\gamma $ . Then, there exists a constant $C=C(B')$ such that

(2.10) $$ \begin{align} C^{-1} |(\tau,\xi)|\le 2^k \le C|\xi|. \end{align} $$

The following shows the matrices $ \mathcal L_{s}^{\delta }$ , are close to each other if so are .

Lemma 2.6. Let and $\gamma $ satisfy ${\mathfrak V({N-1}, {B'})}$ . If , then there exists a constant $C=C(B')\ge 1$ such that

(2.11)

Proof. It suffices to prove that (2.11) holds if for a constant $c>0$ , independent of s and . Applying this finitely many times, we can remove the additional assumption. Moreover, it is enough to show

(2.12)

when . Here, $\| \cdot \|$ denotes a matrix norm. Taking $c>0$ sufficiently small, we get (2.11).

By (2.8),

for $j=1,\dots ,N-1$ . Let

, $|c'|\le c$ . Expanding $G^{(j)}$ in Taylor series at s, by the condition (2.1) we have

for $j=1,\dots ,N-1$ . By (2.8) and the mean value theorem, we get

From (2.8), we also have

. A similar argument also shows

.

Let $\{ v_N,\dots , v_d\}$ denote an orthonormal basis of . By ${\mathfrak V({N-1}, {B'})}$ and (2.1), it follows that , $j=1, \dots , N-1$ . Since , there is an orthonormal basis $\{v_N(s),\dots ,v_d(s)\}$ of $( \mathrm V_s^{G,N-1})^\perp $ such that $|v_j(s)-v_j| \lesssim _{B'}\! c\delta $ , $j=N,\dots ,d$ . So, we have $|(\mathcal L_{s}^\delta )^\intercal v_j -v_j|\lesssim _{B'}\! c\delta $ by (2.9). Since , it follows that , $j=N,\dots ,d$ .

We denote by $\mathrm M$ the matrix . Then, combining all together, we have . Note that ${\mathfrak V({N-1}, {B'})}$ gives $|\mathrm M^{-1} v| \lesssim _{B'}\! |v|$ for $v\in \mathbb R^{d+1}$ . Therefore, we obtain (2.12).

For a continuous function $\mathfrak a$ supported in $I\times [1/2, 4]\times \mathbb R\times \mathbb A_k$ , we set

(2.13) $$ \begin{align} m[\mathfrak a](\tau,\xi)&=\iint e^{-it'(\tau+ \gamma(s)\cdot \xi)}\mathfrak a(s,t',\tau,\xi)dsdt', \end{align} $$

(2.14) $$ \begin{align} \mathcal T[\mathfrak a]f(x,t)&=(2\pi)^{-d-1} \iint e^{i(x\cdot\xi+t\tau)}m[\mathfrak a](\tau,\xi) \widehat f (\xi)\,d\xi d\tau. \end{align} $$

Lemma 2.7. Suppose $\mathfrak a\in \mathrm{C}^{d+3}(\mathbb R^{d+3})$ satisfies (2.6) and (2.7) for $j=l=0$ and $|\alpha | \le d+3$ . Then, there is a constant $C=C(B)$ such that

(2.15) $$ \begin{align} \| \mathcal T[\mathfrak a]f\|_{L^\infty(\mathbb R^{d+1})} &\le C\delta \|f\|_{L^\infty(\mathbb R^d)}, \end{align} $$

(2.16) $$ \begin{align} \| (1-\tilde\chi)\mathcal T[\mathfrak a]f\|_{L^p(\mathbb R^{d+1})} &\le C2^{-k}\delta^{1-N} \|f\|_{L^p(\mathbb R^d)}, \quad p> 1, \end{align} $$

where $\tilde \chi \in \mathrm C_c^\infty ((2^{-2},2^3))$ such that $\tilde \chi =1$ on $[3^{-1},6]$ .

Proof. We first note

(2.17) $$ \begin{align} \mathcal T[\mathfrak a]f (x,t) = \int K[\mathfrak a](s,t, \cdot)\ast f(x) \,ds, \end{align} $$

where

(2.18) $$ \begin{align} K[\mathfrak a](s,t,x)=(2\pi)^{-d-1} \iiint e^{i(t-t',x-t'\gamma(s))\cdot (\tau, \xi)}\mathfrak a(s,t',\tau,\xi) \, d\xi d\tau dt'. \end{align} $$

Since , to prove the estimate (2.15) we need only to show

(2.19)

for some $C=C(B)>0$ . To this end, changing variables $(\tau ,\xi ) \rightarrow 2^k \mathcal L_{s}^{\delta }(\tau ,\xi )$ in the right-hand side of (2.18) and noting $|\!\det \mathcal L_{s}^{\delta }|= \delta ^N |\!\det \widetilde {\mathcal L}_{s}^\delta | =\delta ^{{N(N+1)}/2}$ , we get

$$ \begin{align*} K[\mathfrak a](s,t,x)=C_\ast \iiint e^{i 2^k(t-t',x-t\gamma(s))\cdot (\delta^N \tau,\, \widetilde{\mathcal L}_{s}^\delta \xi)} \mathfrak a(s,t',2^k\mathcal L_{s}^{\delta}(\tau,\xi)) \, d\xi d\tau dt', \end{align*} $$

where $C_\ast =(2\pi )^{-d-1} \delta ^{{N(N+1)}/2} 2^{k(d+1)}$ . Since $\mathfrak a$ satisfies (2.6), by (2.11) and Lemma 2.5 we have $\operatorname {\mathrm {supp}}\, \mathfrak a(s,t,2^k\mathcal L_{s}^\delta \cdot )\subset \{(\tau , \xi ): |(\tau ,\xi )| \lesssim _B\! 1\}$ . Besides, by (2.7) and (2.11) it follows that $|\partial _{\tau ,\xi }^{\alpha } \big (\mathfrak a(s,t, 2^k\mathcal L_{s}^\delta (\tau ,\xi ))\big )| \lesssim _B \!1 $ for $|\alpha |\le d+3$ . Thus, repeated integration by parts in $\tau ,\xi $ yields

$$\begin{align*}|K[\mathfrak a](s,t,x)| \lesssim C_\ast \int_{1/2}^{4} \Big(1+2^k \big|\big (\delta^N(t-t'), (\widetilde{\mathcal L}_{s}^\delta)^\intercal(x-t\gamma(s))\big)\big| \Big)^{-d-3}\,dt',\end{align*}$$

by which we obtain (2.19) as desired.

It is easy to show the estimate (2.16). The above estimate for $K[\mathfrak a]$ gives

$$\begin{align*}\|(1-\tilde\chi)K[\mathfrak a](s,t,\cdot)\|_{L^1_x} \lesssim \delta^{-N} 2^{-k}|t-1|^{-1}|1-\tilde\chi(t)|.\end{align*}$$

Since , (2.16) for $p>1$ follows by (2.17) and Minkowski’s and Young’s convolution inequalities.

2.2 Rescaling

Let . Suppose that

(2.20) $$ \begin{align} \sum_{j=1}^{N-1} \delta^{j} |\langle \gamma^{(j)}(s), \xi \rangle |\ge {2^k \delta^N}/{B'} \end{align} $$

holds on $\operatorname {\mathrm {supp}}_{s,\xi } \mathfrak a$ for some $B'>0$ . Then, via decomposition and rescaling, we can bound the $L^p$ norm of $\mathcal T[\mathfrak a]f$ by those of the operators given by symbols of type $(j, N-1, \tilde B)$ relative to a curve for some $\tilde B$ and j (see Lemma 2.8 below).

To do so, we define a rescaled curve by

(2.21)

As $\delta \to 0$ , the curves get close to a nondegenerate curve in an N-dimensional vector space, so the curves behave in a uniform manner. In particular, (2.1) and ${\mathfrak V(N, B)}$ hold for some B for if $\delta <\delta '$ for a constant $\delta '=\delta '(B)$ small enough.

Note

, $1\le j\le N-1$ , and

, $N+1\le j\le 3d+1$ . Thus, Taylor series expansion and (2.4) give

for $j=1, \dots , N-1$ . By (2.21), we have

. We write

, where

and

So,

. Since

and $|v'|\le B$ ,

for some $C=C(B)$ . Thus,

satisfies (2.1) with B replaced by $3B$ if $\delta <\delta '$ .

An elementary argument (elimination) shows

since

and

. Taking $\delta '$ small enough, from ${\mathfrak V(N, B)}$ for $\gamma $ we see that ${\mathfrak V({N}, {3B})}$ holds for

if $0<\delta <\delta '$ .

The next lemma (cf. [Reference Ko, Lee and Oh14, Lemma 2.9]) plays a crucial role in what follows.

Lemma 2.8. Let $2\le N\le d$ ,

, and $j_\ast =\log (2^k \delta ^N)$ . Suppose (2.20) holds on $\operatorname {\mathrm {supp}}_{s,\xi } \mathfrak a$ . Then, there exist constants $C, l_\ast $ , $\tilde B\ge 1$ and $\delta '>0$ depending on B, and symbols $a_{1}, \dots , a_{l_\ast }$ of type $(j, N-1, \tilde B)$ relative to

such that

$\|\tilde f_l\|_p= \|f\|_p$ , and $j\in [j_\ast -C, j_\ast +C]$ as long as $0<\delta <\delta '$ .

Proof. We set

Combining the identities (2.13) and (2.14), we write $\mathcal T[\mathfrak a] f$ as an integral (e.g., see (2.17) and (2.18)). Subsequently, the change of variables

and

gives

where

Let $\tilde f$ be given by

where $\mathcal F( \tilde f\,)$ denotes the Fourier transform of $ \tilde f.$ Then, $ \|\tilde f\|_p=\|f\|_p$ . Changing variables

gives

where

. This leads us to set

(2.22)

It is easy to check $\tilde a\in \mathrm{C}^{d+N+2}(\mathbb R^{d+2})$ , since so is $\mathfrak a$ and $\gamma \in \mathrm C^{3d+1}$ . By (2.21) and (2.5), we note

. Therefore,

and a change of variables gives

(2.23)

We shall obtain symbols of type $(j, N-1,\tilde B)$ from $\tilde a$ via decomposition and rescaling. To this end, we first note

(2.24) $$ \begin{align} \operatorname{\mathrm{supp}}_\xi \tilde a \subset \big\{ \xi \in \mathbb R^d: C^{-1}\delta^N 2^k \le |\xi | \le C \delta^N 2^{k} \big\} \end{align} $$

for a constant $C=C(B)\ge 1$ . This follows by Lemma 2.5 since there exists $\tau $ such that if $\xi \in \operatorname {\mathrm {supp}}_\xi \tilde a$ . We claim

(2.25) $$ \begin{align} |\partial_s^{j} \partial^{l}_t\partial_\xi^\alpha \tilde a(s,t,\xi) | \lesssim_B\! |\xi|^{-|\alpha|}, \qquad (j,l,\alpha)\in \mathcal I_{N-1}. \end{align} $$

To show (2.25), let us set

Note that $0\le j\le 1$ . Taking derivatives on both sides of the equation (2.22), we have

where

with $0 \le u_1 \le 1$ , $0 \le |\alpha _1| \le u_1$ , and constants $C_{\alpha , u}$ satisfying $|C_{\alpha , u}|=1$ . Integration by parts $u_1+|\alpha _2|$ times in $\tau $ gives $\partial _s^{j}\partial ^{l}_t\partial _\xi ^\alpha \tilde a=\mathcal I[\mathfrak b_2]$ , where

with constants $C_{\alpha , u}'$ satisfying $|C_{\alpha , u}'|=1$ . We decompose $\mathcal I[\mathfrak b_2]= \mathcal I[\chi _E \mathfrak b_2]+ \mathcal I[\chi _{E^c}\mathfrak b_2]$ , where

. Then, integrating by parts in $t'$ for $ \mathcal I[\chi _{E^c}\mathfrak b_2]$ , we obtain

Since

, $ |\partial _s^{j'} \partial _t^{l'} \partial _{\tau ,\xi }^{\alpha '} \mathfrak b| \lesssim _B\! |\xi |^{-|\alpha '|} $ for $(j', l', \alpha ')\in \mathcal I_N$ . It is also clear that

if $\delta < \delta ' $ . Thus, $|\mathfrak b_2|=O(|\xi |^{-|\alpha |})$ , and $|\partial _{t'}^2\mathfrak b_2|=O(|\xi |^{-|\alpha |})$ for $l\le 2(N-1) $ . Since $\partial _s^{j}\partial ^{l}_t\partial _\xi ^\alpha \tilde a=\mathcal I[\mathfrak b_2]$ , we obtain the inequality (2.25).

Now, we decompose $\tilde a$ . Let $\tilde \chi _1, \tilde \chi _2$ and $\tilde \chi _3\in \mathrm{C}_c^\infty (\mathbb R)$ such that $\tilde \chi _1 +\tilde \chi _2+\tilde \chi _3 =1 $ on $\operatorname {\mathrm {supp}} \tilde \chi $ and $\operatorname {\mathrm {supp}}\tilde \chi _\ell \subset [2^{\ell -3}, 2^{\ell }]$ for $\ell =1,2,3$ . Also, let $\beta \in \mathrm C_c^\infty ((2^{-1}, 2))$ such that $\sum \beta (2^{-k}\cdot ) =1$ on $\mathbb R_{+}$ . We set

$$\begin{align*}a_{\ell, j} (s,t,\xi) = \tilde \chi_\ell (t) \beta(2^{-j}|\xi|) \tilde a(s,t,\xi), \end{align*}$$

so $\sum _{\ell , j} a_{\ell , j}= \tilde a$ . By (2.24), $a_{\ell , j} =0$ if $|j-j_\ast |> C$ for some $C>0$ .

Denoting $(a)_\rho (s,t,\xi )=a(s,\rho t,\rho ^{-1}\xi )$ , via rescaling we observe

Thus, changes of variables yield

where $\tilde f_\ell = 2^{(\ell -2)d/p } \tilde f(2^{\ell -2}\cdot )$ . Since

, by (2.23) we get

To complete the proof, we only have to relabel $(a_{\ell , j})_{2^{\ell -2}}$ , $\ell =1,2,3$ , $j_\ast -C\le j\le j_\ast +C$ . Indeed, since $\tilde a\in \mathrm{C}^{d+N+2}$ , $(a_{\ell , j})_{2^{\ell -2}}\in \mathrm{C}^{d+N+2}$ , which is supported in $I \times [2^{-1},4] \times \mathbb A_{j+\ell -2}$ . Obviously, (2.25) holds for $\tilde a=(a_{\ell , j})_{2^{\ell -2}}$ because $\ell =1,2,3$ . Changing variables and in (2.20), by the identity (2.21) we see that (2.20) on $\operatorname {\mathrm {supp}}_{s,\xi } \mathfrak a$ is equivalent to for . Note that . So, the same holds on $\operatorname {\mathrm {supp}}_{s,\xi } \tilde a$ and hence on $\operatorname {\mathrm {supp}}_{s,\xi } (a_{\ell , j})_{2^{\ell -2}}$ if $B'$ replaced by $2B'$ . Therefore, $ C^{-1}(a_{\ell , j})_{2^{\ell -2}}$ is of type $(j+\ell -2, N-1, \tilde B)$ relative to for a large constant $C=C(B)$ .

2.3 Preliminary decomposition and reduction

For the proof of Proposition 2.3, we make some reductions by decomposing the symbol a. We fix a sufficiently small positive constant

(2.26) $$ \begin{align} \delta_*< \min \{ 2^{-10} B^{-3}\delta', \,\, (2^{7d}B^6)^{-N}\}, \end{align} $$

which is to be specified in what follows. Here, $\delta '$ is the number given in Lemma 2.8.

Recall $\gamma $ satisfies the condition (2.1), ${\mathfrak V(N, B)}$ and $ a$ is of type $(k, N, B)$ relative to $\gamma $ . We set

(2.27) $$ \begin{align} \eta_N(s,\xi)=\prod_{1\le j\le N-1} \beta_0 \Big( B2^{-k-1}\delta_*^{j-N} \langle\gamma^{(j)}(s),\, \xi \rangle \Big), \end{align} $$

where $\beta _0\in \mathrm C_c^\infty ((-1,1))$ such that $\beta _0= 1$ on $[-1/2,1/2]$ . It is easy to see $|\partial _s^{j} \partial ^{l}_t \partial _{\xi }^{\alpha } (a\eta _N)| \le C |\xi |^{-|\alpha |}$ for $(j,l,\alpha )\in \mathcal I_N$ , and the same holds for $a(1-\eta _N)$ .

Note $\sum _{j=1}^{N-1}|\gamma ^{(j)}(s)\cdot \xi |\ge (2B)^{-1} \delta _*^N|\xi |$ on $\operatorname {\mathrm {supp}}_{s,\xi }(a(1-\eta _N))$ . So, we see $a(1-\eta _N)$ is a symbol of type $(k, N-1, B')$ for $B'=CB^2 \delta _*^{-C}$ with a large C. Applying the assumption (Theorem 2.2 with $L=N-1$ and $B=B'$ ), we obtain

$$\begin{align*}\| \mathcal A_t[\gamma, a(1-\eta_N)]f\|_{L^p(\mathbb R^{d+1})}\le C 2^{(-\frac{2}p+\epsilon) k}\|f\|_{L^p(\mathbb R^d)}, \quad p\ge 4N-6. \end{align*}$$

Thus, it suffices to consider $\mathcal A_t[\gamma , a\eta _N]$ . Since ${\mathfrak N({N}, B)}$ holds on $\operatorname {\mathrm {supp}}_{s,\xi } a$ ,

(2.28) $$ \begin{align} | \gamma^{(N)}(s)\cdot \xi |\ge (2B)^{-1}|\xi| \end{align} $$

holds whenever $(s,t,\xi )\in \operatorname {\mathrm {supp}} a\eta _N$ for some t.

Basic assumption. Before we continue to prove the estimate for $\mathcal A_t[\gamma , a\eta _N]$ , we make several assumptions which are clearly permissible by elementary decompositions.

Decomposing a, we may assume that $\operatorname {\mathrm {supp}}_\xi a$ is contained in a narrow conic neighborhood and for some . Let us set

$$\begin{align*}\Gamma_k = \big\{\xi\in \mathbb A_k: \text{dist}\big( {|\xi|^{-1}}{\xi}, |\xi'|^{-1}{\xi'}\big)< \delta_\ast \ \text{ for some } \ \xi'\in \operatorname{\mathrm{supp}}_\xi (a\eta_N) \big\}. \end{align*}$$

We may also assume $\gamma ^{(N-1)}(s') \cdot \xi '=0$ for some . Otherwise, $|\gamma ^{(N-1)}(s) \cdot \xi |\gtrsim |\xi |$ on $\operatorname {\mathrm {supp}}_{s,\xi } a\eta _N$ and hence $a\eta _N=0$ if we take B large enough. By (2.28) and the implicit function theorem, there exists $\sigma $ such that

(2.29) $$ \begin{align} \gamma^{(N-1)}(\sigma(\xi)) \cdot \xi=0 \end{align} $$

on a narrow conic neighborhood of $\xi '$ where $\sigma \in \mathrm{C}^{2d+2}$ , since $\gamma \in \mathrm C^{3d+1}(I)$ . So, decomposing a further and taking $\delta _\ast $ small enough, we may assume that $\sigma \in \mathrm C^{2d+2}(\Gamma _k)$ and for $\xi \in \Gamma _k$ . Moreover, since $\sigma $ is homogeneous of degree zero, we have

(2.30) $$ \begin{align} |\partial_\xi^\alpha \sigma(\xi)| \le C |\xi|^{-|\alpha|}, \qquad \xi\in \Gamma_k \end{align} $$

for a constant $C=C(B)$ if $|\alpha |\le 2d+2$ . Any symbol which appears in what follows is to be given by decomposing the symbol a with appropriate cutoff functions. So, the $s,\xi $ -supports of the symbols are assumed to be contained in .

We break a to have further localization on the Fourier side. Let

$$ \begin{align*} \mathfrak a_1 (s,t,\tau,\xi)= a\eta_N \, \beta_0 \big( 2^{-2k}\delta_*^{-2N}|\tau+ \langle \gamma(s),\xi \rangle|^2\big) \end{align*} $$

and $\mathfrak a_0=a\eta _N-\mathfrak a_1$ . Then, by Fourier inversion

$$\begin{align*}\mathcal A_t[\gamma,a\eta_N]f=\mathcal T[ \mathfrak a_1] f+ \mathcal T [ \mathfrak a_0]f. \end{align*}$$

It is easy to show $ \| \mathcal T[ \mathfrak a_0]f \|_{p} \lesssim _B\! 2^{-2k} \|f\|_{p}$ for $1 \le p \le \infty .$ Indeed, we consider $ \tilde {\mathfrak a}_0 = -(\tau +\gamma (s)\cdot \xi )^{-2} {\partial _{t}^2\mathfrak a_0}. $ By (2.13) and integration by parts in $t'$ , $m[\mathfrak a_0]=m[\tilde {\mathfrak a}_0]$ and hence $\mathcal T [ \mathfrak a_0 ]= \mathcal T [\tilde {\mathfrak a}_0]$ . Thanks to (2.17), it is sufficient to show

$$ \begin{align*} | K [ \tilde{\mathfrak a}_0 ](s,t,x) \big| &\le C\, 2^{k(d-1)} \! \int \! \big(1+2^k |t-t'| +2^k |x-t'\gamma(s)|\big)^{-d-3}\,dt' \end{align*} $$

for a constant $C=C(B, \delta _\ast )$ . Note $|\tau +\langle \gamma (s),\xi \rangle |\gtrsim 2^k$ on $\operatorname {\mathrm {supp}} \tilde {\mathfrak a}_0$ , and recall (2.18). Rescaling and integration by parts in $\tau ,\xi $ , as in the proof of Lemma 2.7, show the estimate.

The difficult part is to obtain the estimate for $\mathcal T[\mathfrak a_1]$ . Since $\delta _*$ is a fixed constant, it is obvious that for some $C=C(B,\delta _\ast )$ . So, the desired estimate for $\mathcal T[\mathfrak a_1]$ follows once we have the next proposition.

Proposition 2.9. Let with $\operatorname {\mathrm {supp}}_{\xi } \mathfrak a \subset \Gamma _k$ . Suppose Theorem 2.2 holds for $L=N-1$ . Then, if $p\ge 4N-2$ , for $\epsilon>0$ , we have

$$\begin{align*}\big\| \mathcal T[\mathfrak a]f \big\|_{L^p(\mathbb R^{d+1})} \le C_\epsilon 2^{-\frac{2} pk+ \epsilon k}\|f\|_{L^p(\mathbb R^d)}. \end{align*}$$

Therefore, the proof of Proposition 2.3 is completed if we prove Proposition 2.9. For the purpose, we use Proposition 2.10 below, which allows us to decompose $ \mathcal T[\mathfrak a]$ into the operators given by symbols with smaller s-supports while the consequent minor parts have acceptable bounds. A similar argument was used in [Reference Pramanik and Seeger24] when $L=2$ .

Let $\delta _0$ and $\delta _1$ be positive numbers such that

(2.31) $$ \begin{align} 2^{7d}B^6\delta_0^{(N+1)/N} \le \delta_1\le \delta_0 \le \delta_*, \qquad 2^{-k/N}\le \delta_1. \end{align} $$

Then, it is clear that

(2.32) $$ \begin{align} B^{6N}\delta_0^{j+1} \le 2^{-7dN} \delta_1^{j}, \qquad j=1,\dots, N. \end{align} $$

For $n\ge 0$ , we denote $ \mathfrak J_n^\mu =\{ \nu \in \mathbb Z: |2^n\delta _1 \nu - \delta _0 \mu | \le \delta _0\}. $

Proposition 2.10. For $\mu $ such that , let $\mathfrak a^\mu \in \mathfrak A_k(\delta _0\mu ,\delta _0)$ with . Suppose Theorem 2.2 holds for $L=N-1$ . Then, if $p\ge 4N-2$ , for $\epsilon>0$ there exist a constant $C_\epsilon =C_\epsilon (B) \ge 2$ and symbols $\mathfrak a_{\nu } \in \mathfrak A_k(\delta _1\nu ,\delta _1)$ with , $\nu \in \cup _\mu \mathfrak J_0^\mu $ , such that

$$ \begin{align*} \!\!\left( \sum_\mu \| \mathcal T[ \mathfrak a^\mu]f\|_p^p\right)^{\frac1p}\! \le C_\epsilon & \big({\delta_1}/{\delta_0} \big)^{\frac {2N}{p}-1-\epsilon} \left( \sum_\nu \| \mathcal T[\mathfrak a_{\nu}]f\|_p^p \right)^{\frac1p} \!+C_\epsilon \delta_0^{-\frac {2N}p +1+\epsilon } 2^{-\frac{2}pk+2\epsilon k}\|f\|_p. \end{align*} $$

Assuming Proposition 2.10, we prove Proposition 2.9.

2.4 Proof of Proposition 2.9

Let . We may assume for some $\mu \in \mathbb Z$ . To apply Proposition 2.10 iteratively, we need to choose an appropriate decreasing sequence of positive numbers since the decomposition is subject to the condition (2.31).

Let $\delta _0=\delta _*$ , so $(2^{7d}B^6)^{N}\delta _0< 1$ . Let J be the largest integer such that

$$\begin{align*}(2^{7d}B^6)^{N(\frac{N+1}N)^{J-1}-N} \delta_0^{(\frac{N+1}N)^{J-1}}>2^{-\frac kN}. \end{align*}$$

So, $ J \le C_1 \log k$ for a constant $C_1\ge 1$ . We set

(2.33) $$ \begin{align} \delta_{\!J}=2^{-\frac kN}, \qquad \delta_j=(2^{7d}B^6)^{N(\frac{N+1}N)^{j}-N} &\delta_0^{(\frac{N+1}N)^{j}} \end{align} $$

for $j=J-1,\dots ,1$ . Thus, it follows that

(2.34) $$ \begin{align} &2^{7d}B^6\delta_j^{(N+1)/N}\le \delta_{j+1} <\delta_j, \qquad \, j=0, \dots, J-1. \end{align} $$

For a given $\epsilon>0$ , let $\tilde \epsilon =\epsilon /4$ . Since $\mathfrak a\in \mathfrak A_k( \delta _0\mu ,\delta _0)$ and (2.31) holds for $\delta _0$ and $\delta _1$ , applying Proposition 2.10 to $\mathcal T[\mathfrak a]$ , we have

$$ \begin{align*} \| \mathcal T[\mathfrak a]f\|_p \le C_{\tilde \epsilon} & \big({\delta_1}/{\delta_0}\big)^{\frac {2N}{p}-1-\tilde \epsilon} \left(\sum_{\nu_1} \| \mathcal T[ \mathfrak a_{\nu_1}]f\|_p^p\,\right)^{\frac1p} +C_{\tilde \epsilon} \delta_0^{-\frac {2N}p+1 +\tilde \epsilon } 2^{-\frac{2}pk+2\tilde \epsilon k}\|f\|_p, \end{align*} $$

where $ \mathfrak a_{\nu _1} \in \mathfrak A_k(\delta _1\nu _1,\delta _1)$ , $\nu _1 \in \mathfrak J_0^\mu $ . Thanks to (2.34), we may again apply Proposition 2.10 to $\mathcal T[\mathfrak a_{\nu _1}]$ while $\delta _0$ , $\delta _1$ replaced by $\delta _1, \delta _2$ , respectively. Repeating this procedure up to J-th step yields symbols $\mathfrak a_\nu \in \mathfrak A_k(\delta _{\!J}\nu , \delta _{\!J})$ , $\delta _{\!J} \nu \in \delta _{\!J}\mathbb Z\cap I(\delta _0\mu , \delta _0)$ , such that

$$\begin{align*}\| \mathcal T[\mathfrak a]f\|_p \le C_{\tilde \epsilon}^J \delta_{\!J}^{\frac {2N}{p} -1-\tilde \epsilon} \left(\sum_\nu \|\mathcal T[\mathfrak a_\nu]f\|_p^p \,\right)^{\frac1p} +\sum_{0\le j\le J-1} C_{\tilde \epsilon}^{j+1} \delta_{0}^{-\frac{2N}p+1+\tilde \epsilon} 2^{-\frac{2} pk+ 2\tilde \epsilon k} \|f\|_p \end{align*}$$

for $p \ge 4N-2$ . Now, assuming

(2.35) $$ \begin{align} \left( \sum_\nu\| \mathcal T[ \mathfrak a_{\nu} ]f\|_p^p \,\right)^{1/p} \lesssim_B 2^{-k/N} \|f\|_p, \qquad 2\le p\le \infty \end{align} $$

for the moment, we can finish the proof of Proposition 2.9. Since $C_{\tilde \epsilon }\ge 2$ , combining the above inequalities, we get

$$\begin{align*}\| \mathcal T[\mathfrak a]f\|_p \lesssim_B \!C_{\tilde \epsilon}^{J+1} \big( 2^{-\frac 2p k+\frac{\tilde \epsilon }{N}k} +2^{-\frac{2} pk +2\tilde \epsilon k}\big) \|f\|_p. \end{align*}$$

Since $J \le C_1 \log k$ , $C_{\tilde \epsilon }^{J+1} \le C' 2^{\epsilon k/2}$ for some $C'$ if k is sufficiently large. Therefore, the right-hand side is bounded by $C 2^{- 2k/p+\epsilon k} \|f\|_p$ .

It remains to show the estimate (2.35) for $2\le p\le \infty $ . By interpolation, it is enough to obtain (2.35) for $p=\infty $ and $p=2$ . The case $p=\infty $ follows by (2.15) since $\mathfrak a_\nu \in \mathfrak A_k(\delta _{\!J}\nu ,\delta _{\!J})$ . So, we need only to prove the estimate (2.35) for $p=2$ . To do this, we first observe the following, which shows $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu }$ are finitely overlapping.

Lemma 2.11. For $b\ge 1$ , and $0<\delta \le \delta _*$ , let us set

(2.36) $$ \begin{align} {\Lambda}^{\prime}_k(s,\delta, b) = \bigcap_{1\le j\le N-1} \big\{ \xi \in \Gamma_k : | \langle \gamma^{(j)}(s), \xi \rangle| \le b 2^{k} \delta^{N-j} \big \}. \end{align} $$

If ${\Lambda }^{\prime}_k(s_1, \delta , b) \cap {\Lambda }^{\prime}_k(s_2, \delta , b)\neq \emptyset $ for some , then there is a constant $C=C(B)$ such that $|s_1-s_2|\le Cb\delta $ .

We recall (2.13). Since (2.28) holds on $\operatorname {\mathrm {supp}}_{s,\xi } \mathfrak a_\nu $ , by van der Corput’s lemma (e.g., see [Reference Stein32, p. 334]) we have

$$ \begin{align*} |m[\mathfrak a_{\nu}](\tau,\xi)| \lesssim 2^{-k/N} \big( \| \mathfrak a_{\nu} (\cdot,t,\tau,\xi) \|_\infty +\| \partial_s \mathfrak a_{\nu} (\cdot, t,\tau,\xi) \|_1\big) \lesssim_B\! 2^{-k/N}. \end{align*} $$

The second inequality is clear since $\mathfrak a_{\nu } \in \mathfrak A_k(\delta _{\!J}\nu , \delta _{\!J})$ . From (2.14), note $\mathcal F(\mathcal T[\mathfrak a_{\nu }] f)= m[\mathfrak a_{\nu }]\widehat f$ . Since $\operatorname {\mathrm {supp}} \mathfrak a_{\nu }\subset \Lambda _k(\delta _{\!J}\nu , \delta _{\!J}, B)$ , $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu }\subset S_\nu := {\Lambda }^{\prime}_k(\delta _{\!J}\nu , \delta _{\!J}, 2^5 B)$ . So, $\operatorname {\mathrm {supp}}_\xi \mathcal F(\mathcal T[\mathfrak a_{\nu }] f)\subset S_\nu $ (see (2.13)). By Lemma 2.11, it follows that the sets $S_\nu $ overlap at most $C=C(B)$ times. Therefore, Plancherel’s theorem and the estimate above give

$$\begin{align*}\| \sum_\nu \mathcal T[\mathfrak a_{\nu}] f\|_2^2 \lesssim_B\! 2^{-2k/N} \sum_\nu \int_{S_\nu}\int_{\{\tau: | \tau + \gamma(\delta_{\!J}\nu)\cdot \xi|\le 2^5 B\}} \,d\tau\, |\widehat f(\xi)|^2 \,d\xi, \end{align*}$$

by which we get (2.35) for $p=2$ .

2.5 Decoupling inequalities

We denote $\mathbf r_\circ ^N(s)=(s,s^2/2!,\,\dots, \, s^N/N!)$ and consider a collection of curves from I to $\mathbb R^N$ which are small perturbations of $\mathbf r_\circ ^N$ :

$$\begin{align*}\mathfrak C(\epsilon_\circ;N) =\{ \mathbf r \in \mathrm C^{2N+1}(I): \|\mathbf r-\mathbf r_\circ^N \|_{\mathrm C^{2N+1}(I)} < \epsilon_\circ\}. \end{align*}$$

For $\mathbf r\in \mathfrak C(\epsilon _\circ ;N) $ and $s\in I$ , we define an anisotropic neighborhood by

$$\begin{align*}{\mathcal N}_{\mathbf r}(s,\delta)= \Big\{ \mathbf r(s)+ \sum_{1\le j\le N} u_j {{\mathbf r}^{(j)}(s)}: |u_j|\le \delta^j, \quad~j=1,\dots,N \Big\}. \end{align*}$$

Let $s_1, \dots , s_l\in I$ be $\delta $ -separated points, that is, $|s_n-s_j|\ge \delta $ if $n\neq j$ such that $\bigcup _{j=1}^l(s_j-\delta , s_j+\delta )\supset I$ . Then, we set

$$\begin{align*}\theta_j=\mathcal N_{\mathbf r}(s_j,\delta), \qquad 1\le j\le l. \end{align*}$$

The following is due to Bourgain, Demeter and Guth [Reference Bourgain, Demeter and Guth5] (also see [Reference Guo, Li, Yung and Zorin-Kranich10]).

Theorem 2.12. Let $0<\delta \ll 1$ . Suppose $\mathbf r\in \mathfrak C(\epsilon _\circ ;N) $ for a small enough $\epsilon _\circ>0$ . Then, if $2 \le p\le N(N+1)$ , for $\epsilon>0$ we have

(2.37) $$ \begin{align} \big\| \sum_{1\le j\le l} f_j \big\|_{L^p(\mathbb R^{N})}\le C_\epsilon \delta^{-\epsilon} \left( \sum_{1\le j\le l}\|f_j \|_{L^p(\mathbb R^{N})}^2 \right)^{1/2} \end{align} $$

whenever $\operatorname {\mathrm {supp}} \widehat f_j \subset \theta _j$ for $1\le j\le l$ .

The constant $C_\epsilon $ can be taken to be independent of particular choices of the $\delta $ -separated points $s_1, \dots , s_l$ . One can obtain a conical extension of the inequality (2.37) by modifying the argument in [Reference Bourgain and Demeter4] which deduces the decoupling inequality for the cone from that for the paraboloid (see [Reference Beltran, Guo, Hickman and Seeger2, Proposition 7.7]). Let us consider the conical sets

$$\begin{align*}{\bar{\theta}}_j=\{ (\eta, \rho)\in \mathbb R^N\times [1,2]: \eta/\rho\in \theta_j\}, \quad 1\le j\le l.\end{align*}$$

Corollary 2.13. Let $0<\delta \le 1$ , and let $\mathbf r\in \mathfrak C(\epsilon _\circ ;N) $ for a small enough $\epsilon _\circ>0$ . Then, if $2 \le p\le N(N+1)$ , for $\epsilon>0$ we have

(2.38) $$ \begin{align} \big\|\sum_{1\le j\le l} F_j \big\|_{L^p(\mathbb R^{N+1})}\le C_\epsilon \delta^{-\epsilon} \left(\sum_{1\le j\le l} \| F_j\|_{L^p(\mathbb R^{N+1})}^2 \,\right)^{1/2} \end{align} $$

whenever $\operatorname {\mathrm {supp}} \widehat F_j \subset {\bar {\theta }}_j$ for $1\le j\le l$ .

The inequality (2.38) does not fit with the symbols to appear when we decompose $\mathfrak a$ (see Section 3.1 and Section 4.2). As to be seen, those symbols are associated with the slabs of the following form.

Definition 2.14. Let $N\ge 2$ and $\tilde {\mathbf r}\in \mathfrak C(\epsilon _\circ ;N+1)$ . For $s\in I$ , we denote by $\mathbf S(s,\delta , \rho ;\tilde {\mathbf r})$ the set of $(\tau ,\eta ) \in \mathbb R\times \mathbb R^{N}$ which satisfies

$$ \begin{align*} & \rho^{-1}\le|\langle \tilde{\mathbf r}^{(N+1)}(s), (\tau,\eta) \rangle|\le 2\rho, \\[3pt] &\quad |\langle \tilde{\mathbf r}^{(j)}(s), (\tau,\eta) \rangle| \le \delta^{N+1-j}, \qquad \ \ j=N,\dots,1. \end{align*} $$

The same form of decoupling inequality remains valid for the slabs $\mathbf S(s_1,\delta ,1; \tilde {\mathbf r}), \dots , \mathbf S(s_l,\delta ,1; \tilde {\mathbf r})$ . Beltran et al. [Reference Beltran, Guo, Hickman and Seeger2, Theroem 4.4] showed, using the Frenet–Serret formulas, that those slabs can be generated by conical extensions of the anisotropic neighborhoods given by a nondegenerate curve in $\mathbb R^N$ . Therefore, the following is a consequence of Corollary 2.13 and a simple manipulation using decomposition and rescaling.

Corollary 2.15. Let $0<\delta \le 1$ , $\rho \ge 1$ and $\tilde {\mathbf r} \in \mathfrak C(\epsilon _\circ ;N+1)$ for a small enough $\epsilon _\circ>0$ . Denote $\mathbf S_j=\mathbf S(s_j,\delta , \rho ;\tilde {\mathbf r})$ for $1\le j\le l$ . Then, if $2 \le p\le N(N+1)$ , for $\epsilon>0$ there is a constant $C_\epsilon =C_\epsilon (\rho )$ such that

(2.39) $$ \begin{align} \big\| \sum_{1\le j\le l} F_j \big\|_{L^p(\mathbb R^{N+1})}\le C_\epsilon \delta^{-\epsilon} \left( \sum_{1\le j\le l} \|F_j \|_{L^p(\mathbb R^{N+1})}^2 \right)^{1/2} \end{align} $$

whenever $\operatorname {\mathrm {supp}} \widehat {F_j} \subset \mathbf S_j$ for $1\le j\le l$ .

For our purpose of proving Proposition 2.10, we use a modified form. If $p_\ast \in [2 , N(N+1)] $ , then we have

$$\begin{align*}\big\| \sum_{1\le j\le l} F_j \big\|_{L^p(\mathbb R^{N+1})}\le C_\epsilon \delta^{-1+\frac{2+p_\ast}{2p}-\epsilon} \left( \sum_{1\le j\le l} \|F_j \|_{L^p(\mathbb R^{N+1})}^p \right)^{1/p} \end{align*}$$

for $p\ge p_\ast $ . The case $p=p_\ast $ follows by the inequality (2.39) and Hölder’s inequality. Interpolation with the trivial $\ell ^\infty L^\infty $ – $L^\infty $ estimate gives the estimate for $p> p_\ast $ . One may choose different $p_\ast $ for the particular purposes. In fact, for the local smoothing estimate we take $p_\ast = 4N-2$ to get

(2.40) $$ \begin{align} \big\| \sum_{1\le j\le l} F_j \big\|_{L^p(\mathbb R^{N+1})}\le C_\epsilon \delta^{-1+\frac{2N}{p}-\epsilon} \left( \sum_{1\le j\le l} \|F_j \|_{L^p(\mathbb R^{N+1})}^p \right)^{1/p} \end{align} $$

for $p\ge 4N-2$ (see Section 3.2). For the $L^p$ Sobolev regularity estimate, we observe that

(2.41) $$ \begin{align} \big\| \sum_{1\le j\le l} F_j \big\|_{L^p(\mathbb R^{N+1})}\le C_{\epsilon_0} \delta^{-1+\frac{N+1}{p}+\epsilon_0} \left( \sum_{1\le j\le l} \|F_j \|_{L^p(\mathbb R^{N+1})}^p \right)^{1/p} \end{align} $$

holds for some $\epsilon _0= \epsilon _0(p)>0$ if $2N<p<\infty $ . Indeed, we need only to take $p_\ast> 2N$ close enough to $2N$ . The presence of $\epsilon _0$ in (2.41) is crucial for proving the optimal Sobolev regularity estimate (see Proposition 4.5).

The inequalities (2.40) and (2.41) obviously extend to cylindrical forms via the Minkowski inequality. For example, set $ \tilde {\mathbf S}_j=\big \{ (\xi ,\eta )\in \mathbb R^{N+1}\times \mathbb R^M: \xi \in \mathbf S_j \big \}$ for $1\le j\le l$ . Using (2.40), we have

(2.42) $$ \begin{align} \big\| \sum_{1\le j\le l} G_j \big\|_{L^p(\mathbb R^{N+M+1})}\le C_\epsilon \delta^{-1+\frac{2N}{p}-\epsilon} \left( \sum_{1\le j\le l} \| G_j \|_{L^p(\mathbb R^{N+M+1})}^2 \right)^{1/2} \end{align} $$

whenever $\widehat {G}_j$ is supported in $\tilde {\mathbf S}_j$ . Clearly, we also have a similar extension of the inequality (2.41).

3.1 Decomposition of the symbol $\mathfrak a^\mu $

We begin by introducing some notations.

Fixing $\mu \in \mathbb Z$ such that , we consider the linear maps

$$\begin{align*}y_\mu^j(\tau,\xi)= \langle G^{(j)}(\delta_0\mu ),(\tau,\xi)\rangle, \qquad j=0, 1,\dots,N. \end{align*}$$

In particular, $y_\mu ^j(\tau ,\xi )=\langle \gamma ^{(j)}(\delta _0\mu ),\xi \rangle $ if $1\le j\le N.$ By (2.28), it follows that

(3.1) $$ \begin{align} |y_\mu^N(\tau,\xi)| \ge (2B)^{-1}|\xi|. \end{align} $$

We denote

$$\begin{align*}\omega_\mu(\xi)=\frac{y_\mu^{N-1}(\tau,\xi)}{y_\mu^{N}(\tau,\xi)}, \end{align*}$$

which is close to $\delta _0\mu -\sigma (\xi )$ (see (3.5) below). Then, we define $\mathfrak g_\mu ^N, \mathfrak g_\mu ^{N-1},\dots , \mathfrak g_\mu ^{0}$ recursively, by setting $ \mathfrak g_\mu ^N=y_\mu ^N,$ and

(3.2) $$ \begin{align} \mathfrak g_\mu^{j}(\tau,\xi)=y^{j}_\mu(\tau,\xi)-\sum_{\ell=j+1}^{N}\frac{\mathfrak g^{\ell}_\mu(\tau,\xi)}{(\ell-j)!} (\omega_\mu(\xi))^{\ell-j}, \quad j=N-1, \dots, 0. \end{align} $$

Note that $\mathfrak g_\mu ^{N-1}=0$ and (3.2) can be rewritten as follows:

(3.3) $$ \begin{align} y^{m}_\mu(\tau,\xi) =\sum_{\ell=m}^{N}\frac{\mathfrak g^{\ell}_\mu(\tau,\xi)}{(\ell-m)!} (\omega_\mu(\xi))^{\ell-m}, \quad m=0,\dots,N. \end{align} $$

The identity continues to hold for $m=N$ since $\mathfrak g_\mu ^N=y_\mu ^N$ . Apparently, $\mathfrak g_\mu ^1,\dots , \mathfrak g_\mu ^N$ are independent of $\tau $ since so are $y_\mu ^1, \dots , y_\mu ^N$ .

For $j=1,\dots ,N$ , set

(3.4) $$ \begin{align} \mathcal E_j(\xi):=(y_\mu^N(\tau,\xi))^{-1}\int_{\sigma(\xi)}^{\delta_0\mu}\frac{\langle\gamma^{(N+1)}(r),\xi\rangle}{j!}(\sigma(\xi) -r)^{j}dr. \end{align} $$

By (3.4) with $j=1$ and integration by parts, we have

(3.5) $$ \begin{align} \mathcal E_1(\xi) = \sigma(\xi) -\delta_0\mu+\omega_\mu(\xi). \end{align} $$

Lemma 3.1. Let $0\le j\le N-1$ . Then, we have

(3.6) $$ \begin{align} \langle G^{(j)}(\sigma(\xi)),(\tau,&\xi) \rangle = \sum_{\ell=j}^{N}\frac{\,\, \mathfrak g^{\ell}_\mu\,\,(\mathcal E_1)^{\ell-j}}{(\ell-j)!} -y^N_\mu\mathcal E_{N-j}. \end{align} $$

Proof. When $j=N-1$ , the equation (3.6) is clear. To show (3.6) for $j=0,1,\dots ,N-2$ , by Taylor’s theorem with integral remainder we have

$$\begin{align*}\langle G^{(j)}(\sigma(\xi)),(\tau,\xi)\rangle =\sum_{m=j}^{N}y^{m}_\mu (\tau,\xi) \frac{(\sigma(\xi)-\delta_0\mu)^{m-j}}{(m-j)!}-y^N_\mu(\tau,\xi)\mathcal E_{N-j}(\xi).\end{align*}$$

Using (3.3) and then changing the order of the sums, we see

$$\begin{align*}\langle G^{(j)}(\sigma(\xi)),(\tau,\xi)\rangle= \sum_{\ell=j}^{N} \mathfrak g^{\ell}_\mu \left(\sum_{m=j}^{\ell} \frac{(\sigma(\xi)-\delta_0\mu)^{m-j}}{(\ell-m)!(m-j)!} (\omega_\mu)^{\ell-m} \right) -y^N_\mu\,\mathcal E_{N-j}. \end{align*}$$

The sum over m equals $(\sigma (\xi )-\delta _0\mu + \omega _\mu )^{\ell -j} /(\ell -j)!$ . So, (3.6) follows by (3.5).

We now decompose the symbol $\mathfrak a^\mu \in \mathfrak A_k(\delta _0\mu ,\delta _0)$ by making use of $\mathfrak g_\mu ^j$ , $j=0,\dots ,N-2$ . We define

$$\begin{align*}\mathfrak G_N^\mu(s,\tau,\xi)=\sum_{j=0}^{N-2}\big(2^{-k} \mathfrak g^{j}_\mu(\tau,\xi)\big)^{\frac{2N!}{N-j}}+(s-\sigma(\xi))^{2N!}. \end{align*}$$

Let $ \beta _N=\beta _0-\beta _0(2^{2N!}\cdot )$ , so we have $\sum _{\ell \in \mathbb Z}\beta _N(2^{2N!\ell }\cdot )=1$ on $\mathbb R_+$ . We also take $\zeta \in \mathrm C_c^\infty ((-1,1))$ such that $\sum _{\nu \in \mathbb Z} \zeta (\cdot -\nu )=1$ . For $n\ge 0$ and $\nu \in \mathfrak J_n^\mu $ , we set

$$\begin{align*}\mathfrak a_{\nu}^{\mu,n} = \mathfrak a^\mu \times \begin{cases} \beta_0 \big( \delta_1^{-2N!} \, \mathfrak G_N^\mu \big) \, \zeta(\delta_1^{-1}s-\nu), \quad & n=0, \\[4pt] \beta_N \big( (2^{n}\delta_1)^{-2N!} \, \mathfrak G_N^\mu \big) \, \zeta(2^{-n}\delta_1^{-1}s-\nu), \quad & n\ge 1. \end{cases} \end{align*}$$

Then, it follows that

(3.7) $$ \begin{align} \mathfrak a^\mu = \sum_{n\ge 0} \sum_{\nu \in \mathfrak J_n^\mu} \mathfrak a^{\mu,n}_{\nu}. \end{align} $$

The proof of Lemma 3.2 is elementary though it is somewhat involved. We postpone the proof until Section 3.3.

We collect some elementary facts regarding $\mathfrak a_{\nu }^{\mu ,n}$ . First, we may assume

(3.8) $$ \begin{align} 2^n\delta_1\le 2^{10} B^3\delta_0 \end{align} $$

since, otherwise, $\mathfrak a_{\nu }^{\mu ,n}=0$ . To show this, we note $|\langle \gamma ^{(N-1)}(\delta _0\mu ),\xi \rangle |\le B2^{k+5}\delta _0$ if $\xi \in \operatorname {\mathrm {supp}}_{\xi } \mathfrak a^\mu $ . Thus, (2.28), (2.29), and the mean value theorem show that

(3.9) $$ \begin{align} |\sigma(\xi)-\delta_0\mu|\le B^22^{7}\delta_0 \end{align} $$

for $\xi \in \operatorname {\mathrm {supp}}_{\xi } \mathfrak a^\mu .$ If $(\tau , \xi )\in \operatorname {\mathrm {supp}}_{\tau , \xi } \mathfrak a^\mu \subset \Lambda _k(\delta _0\mu , \delta _0, B)$ , $|y_\mu ^j(\tau ,\xi )|\le B2^{k+5}\delta _0^{N-j} $ for $0\le j\le N-1$ . Note $|\omega _\mu | \le B^22^7\delta _0$ and $|\mathfrak g_\mu ^N| \le B2^{k+1}$ . A routine computation using (3.2) gives $|\mathfrak g_\mu ^{j}| \le B2^{k-1}(B^22^8\delta _0)^{N-j} $ for $j=N-2, \dots , 0$ . Since $|s-\sigma (\xi )|\le (B^22^{7}+1)\delta _0$ , we have $ \mathfrak G_N^\mu \le 2(B^32^8)^{2N!} \delta _0^{2N!}$ on $\operatorname {\mathrm {supp}} \mathfrak a_{\nu }^{\mu ,n}$ , and (3.8) follows.

Since $\mathfrak G_N^\mu \le (2^n\delta _1)^{2N!}$ on $\operatorname {\mathrm {supp}} \mathfrak a_{\nu }^{\mu ,n}$ , the following hold:

(3.10) $$ \begin{align} |s-\sigma(&\xi)|\le 2^n\delta_1, \end{align} $$

(3.11) $$ \begin{align} 2^{-k}|\mathfrak g^{j}_\mu(\tau,\xi)&|\le (2^n\delta_1)^{N-j}, \qquad 0\le j \le N-1. \end{align} $$

Obviously, (3.11) holds true for $j=N-1$ since $\mathfrak g^{N-1}_\mu =0$ . We also have

(3.12) $$ \begin{align} &|\mathcal E_j(\xi)|\le B^2 (B^22^7\delta_0)^{j+1}, \end{align} $$

(3.13) $$ \begin{align} &|\sigma(\xi)-2^n\delta_1\nu| \le 2^{n+1}\delta_1 \end{align} $$

on $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu }^{\mu ,n}$ . By using (3.4), (3.9) and (3.1), it is easy to show (3.12). Since $|s-2^n\delta _1\nu | \le 2^n\delta _1$ on $\operatorname {\mathrm {supp}}_s \mathfrak a_{\nu }^{\mu ,n}$ , (3.13) follows by (3.10).

3.2 Proof of Proposition 2.10

By (3.7) and the Minkowski inequality, we have

(3.14) $$ \begin{align} \begin{aligned} \left( \sum_\mu \big\| \mathcal T [\mathfrak a^\mu]f \big\|_p^p \,\right)^{1/p} \le \sum_{n\ge0} \left(\sum_{\mu} \big\| \sum_{\nu \in \mathfrak J_n^\mu} \mathcal T [ \mathfrak a^{\mu,n}_{\nu}]f \big\|_p^p\,\right)^{1/p}. \end{aligned} \end{align} $$

We use the inequality (2.40) for after a suitable linear change of variables. The symbols $\mathfrak a_{\nu }^{\mu ,0}$ are to constitute the set $\{\mathfrak a_\nu \}$ while the operators associated to $\mathfrak a_{\nu }^{\mu ,n}$ , $n\ge 1$ are to be handled similarly as in Section 2.

Applying the inequality (2.40). To prove Proposition 2.10, we first show

(3.15) $$ \begin{align} \big\| \sum_{\nu\in \mathfrak J_{n}^\mu} \mathcal T [\mathfrak a_{\nu}^{\mu,n}]f \big\|_p \le C_\epsilon \big({2^n\delta_1}/{\delta_0} \big)^{\frac {2N} {p}-1-\epsilon} \left(\sum_{\nu\in \mathfrak J_{n}^\mu} \big\| \mathcal T [\mathfrak a_{\nu}^{\mu,n}]f \big\|_p^p \,\right)^{1/p} \end{align} $$

for $p \ge 4N-2$ . To use (2.40), we consider $\operatorname {\mathrm {supp}}_{\tau ,\xi } \mathfrak a_{\nu }^{\mu ,n}$ , which contains the Fourier support of $ \mathcal T [\mathfrak a_{\nu }^{\mu ,n}]f$ as is clear from (2.13) and (2.14).

We set

$$\begin{align*}\mathbf y_\mu(\tau,\xi)=\big(y_\mu^0(\tau,\xi),\dots,y^{N}_\mu(\tau,\xi)\big). \end{align*}$$

Lemma 3.3. Let $\mathbf r =\mathbf r_{\circ }^{N+1}$ and $\mathcal D_\delta $ denote the matrix $ (\delta ^{-N} e_1, \delta ^{1-N} e_2, \dots , \delta ^0 e_{N+1})$ , where $e_j$ denotes the j-th standard unit vector in $\mathbb R^{N+1}$ . On $\operatorname {\mathrm {supp}}_{\tau ,\xi } \mathfrak a^{\mu ,n}_{\nu }$ , we have

(3.16) $$ \begin{align} \Big|\Big\langle \mathcal D_{\delta_0} \mathbf y_\mu(\tau,\xi), \mathbf r^{(j)}\Big(\frac{2^n\delta_1}{\delta_0}\nu-\mu\Big)\Big\rangle\Big| \lesssim &\, 2^k \Big(\frac{2^n\delta_1}{\delta_0}\Big)^{N+1-j}, \qquad 1 \le j \le N, \end{align} $$

(3.17) $$ \begin{align} (2B)^{-1} 2^{k-1} \le \big|\big\langle \mathbf y_\mu(\tau,\xi), \mathbf r^{(N+1)}& \big\rangle\big|\le B 2^{k+1}. \end{align} $$

Proof. We write $\mathbf r=(\mathbf r_1,\dots ,\mathbf r_{N+1})$ . Note $\mathbf r^{(j)}_m(s)=s^{m-j}/(m-j)!$ for $m\ge j$ . By the equation (3.3), we have

$$ \begin{align*} y^{m-1}_\mu \mathbf r^{(j)}_m(2^n\delta_1\nu-\delta_0\mu) &=\sum_{\ell=m-1}^{N}\mathfrak g^{\ell}_\mu\frac{(2^n\delta_1\nu-\delta_0\mu)^{m-j}}{(\ell+1-m)!(m-j)!}\,\omega_\mu^{\ell+1-m} \end{align*} $$

for $m\ge j$ . Since $\mathbf r^{(j)}_m(s)=0$ for $j>m$ , taking sum over m gives

$$ \begin{align*} \big\langle \mathbf y_\mu, \mathbf r^{(j)}(2^n\delta_1\nu-\delta_0\mu)\big\rangle =\sum_{\ell=j-1}^{N}\mathfrak g^{\ell}_\mu\frac{( 2^n\delta_1\nu-\delta_0\mu+ \omega_\mu)^{\ell+1-j}}{(\ell+1-j)!}. \end{align*} $$

From the equation (3.5), we note $2^n\delta _1\nu -\delta _0\mu +\omega _\mu =2^n\delta _1\nu -\sigma (\xi )+\mathcal E_1$ . Thus, (3.13), (3.12) with $j=1$ and (2.32) with $j=1$ show $|2^n\delta _1\nu -\delta _0\mu +\omega _\mu |\lesssim 2^n\delta _1$ . By (3.11), we obtain

$$\begin{align*}\big|\big\langle \mathbf y_\mu(\tau,\xi), \mathbf r^{(j)}(2^n\delta_1\nu-\delta_0\mu)\big\rangle\big| \lesssim 2^k (2^{n}\delta_1)^{N+1-j}, \qquad 1 \le j \le N. \end{align*}$$

By homogeneity, it follows that $\langle \eta , \mathbf r^{(j)}(\delta _0 s)\rangle =\delta _0^{N+1-j}\langle \mathcal D_{\delta _0} \eta , \mathbf r^{(j)}( s)\rangle $ for $\eta \in \mathbb R^{N+1}$ . Therefore, we get (3.16). For the inequality (3.17), note that $\mathbf r^{(N+1)}=(0,\dots , 0, 1)$ . Thus, $\langle \mathbf y_\mu , \mathbf r^{(N+1)}\rangle =y^{N}_\mu $ and (3.17) follows by (3.1).

Let $\mathrm V= \operatorname {\mathrm {span}}\{\gamma '(\delta _0\mu ),\dots ,\gamma ^{(N)}(\delta _0\mu )\}$ and $\{v_{N+1},\dots , v_d\}$ be an orthonormal basis of $\mathrm V^\perp $ . Since $\gamma $ satisfies ${\mathfrak V({N}, B)}$ , for each $\xi \in \mathbb R^d$ we can write

(3.18) $$ \begin{align} \xi= \overline \xi+\sum_{N+1\le j\le d}y_j(\xi) v_j, \end{align} $$

where $\overline \xi \in \mathrm V$ and $y_j(\xi ) \in \mathbb R$ , $N+1\le j\le d$ . We define a linear map by

$$\begin{align*}\mathrm Y_\mu^{\delta_0}(\tau,\xi)= \big( 2^{-k} \mathcal D_{\delta_0} \mathbf y_\mu(\tau, \xi),\, y_{N+1}(\xi),\dots,\,y_d(\xi)\big). \end{align*}$$

From (3.16) and (3.17), we see

(3.19) $$ \begin{align} \mathrm Y_\mu^{\delta_0} ( \operatorname{\mathrm{supp}}_{\tau,\xi} \mathfrak a_{\nu}^{\mu,n} ) \subset \mathbf S\Big(\frac{2^n\delta_1}{\delta_0}\nu-\mu, C\frac{2^n\delta_1}{\delta_0}, 2^2B; \mathbf r_\circ^{N+1}\Big) \times \mathbb R^{d-N} \end{align} $$

for some $C>1$ . We now have the inequality (2.40) for $\delta =C{2^n\delta _1}/{\delta _0}$ and the slabs $\mathbf S(2^n\delta _1\nu /\delta _0-\mu , C2^{n}\delta _1/\delta _0, 2^2B; \mathbf r_\circ ^{N+1}), \nu \in \mathfrak J_{n}^\mu $ . Therefore, by cylindrical extension in $y_{N+1},\dots ,y_d$ (see (2.42)) and the change of variables $(\tau ,\xi )\to \mathrm Y_\mu ^{\delta _0}(\tau ,\xi )$ , we obtain (3.15) since the decoupling inequality is not affected by an affine change of variables in the Fourier side.

Combining the inequalites (3.14) and (3.15), we obtain

$$ \begin{align*} \left(\sum_\mu \| \mathcal T[\mathfrak a^\mu]f\|_p^p\,\right)^{1/p} \le \sum_{n\ge 0}\, \mathbf E_n \end{align*} $$

for $p \ge 4N-2$ , where

$$ \begin{align*} \mathbf E_n= C_\epsilon \big({2^n\delta_1}/\delta_0\big)^{\frac {2N} {p}-1-\epsilon} \left(\sum_{\mu} \sum_{\nu\in \mathfrak J_n^\mu} \| \mathcal T[\mathfrak a^{\mu, n}_\nu]f\|_p^p\,\right)^{1/p}. \end{align*} $$

Since the intervals $I(\delta _0\mu ,\delta _0)$ overlap, there are at most three nonzero $\mathfrak a^{\mu , 0}_{\nu }$ for each $\nu $ . We take $\mathfrak a_\nu =\mathfrak a^{\mu , 0}_\nu $ which maximizes $\| \mathcal T[\mathfrak a^{\mu , 0}_\nu ]f\|_p$ . Then, it is clear that $\mathbf E_0\le 3^{1/p} C_\epsilon (\delta _1/\delta _0)^{\frac {2N} {p}-1-\epsilon } \left(\sum _{\nu } \| \mathcal T[\mathfrak a_\nu ]f\|_p^p\right)^{1/p}.$ By Lemma 3.2, $C^{-1}\mathfrak a_\nu \in \mathfrak A_k(\delta _1\nu , \delta _1)$ for a constant C. Thus, the proof of Proposition 2.10 is now reduced to showing

(3.20) $$ \begin{align} \sum_{n\ge1}\, \mathbf E_n \lesssim_B\! \delta_0^{-\frac{2N}p+1+\epsilon} 2^{-\frac 2pk +2\epsilon k} \|f\|_p, \quad p\ge 4N-2. \end{align} $$

Estimates for $\mathbf E_n$ when $n\ge 1$ . To prove the estimate (3.20), we decompose $\mathfrak a^{\mu , n}_\nu $ so that the inequalites (3.25) or (3.26) (see Lemma 3.5 below) holds on the $s,\xi $ -supports of the resulting symbols. If (3.25) holds, we use the assumption after rescaling, whereas we handle the other case using estimates for the kernels of the operators.

Let

(3.21) $$ \begin{align} \bar{\mathfrak G}_N^{\mu}(s,\xi) =\sum_{1\le j\le N-2}\big(2^{-k}\mathfrak g^{j}_\mu\big)^{\frac{2N!}{N-j}}+\big(s-\sigma(\xi)\big)^{2N!}. \end{align} $$

Note that the right-hand side is independent of $\tau $ since so are $\mathfrak g^{j}_\mu $ , $1\le j\le N-2$ .

Let $C_0=2^{2d}B$ . We set

(3.22) $$ \begin{align} \mathfrak a_{\nu,1}^{\mu,n} =\mathfrak a_{\nu}^{\mu,n} \, \beta_0 \Big({ (2^{-k}\mathfrak g_\mu^0)^{2(N-1)!}}/ ({C_0^{2N!} \bar{\mathfrak G}_N^{\mu}})\Big), \quad n\ge 1, \end{align} $$

and $\mathfrak a_{\nu ,2}^{\mu ,n}=\mathfrak a_{\nu }^{\mu ,n} -\mathfrak a_{\nu ,1}^{\mu ,n},$ so we have $ \mathfrak a_{\nu }^{\mu ,n} =\mathfrak a_{\nu ,1}^{\mu ,n}+\mathfrak a_{\nu ,2}^{\mu ,n}. $ Similarly as before, we have the following, which we prove in Section 3.4.

The estimate (3.20) follows if we show

(3.23) $$ \begin{align} \left( \sum_{\mu} \sum_{\nu \in \mathfrak J_n^\mu} \|\mathcal T[\mathfrak a_{\nu,1}^{\mu,n}]f\|_p^p \right)^{1/p} \le C_\epsilon 2^{-\frac 2pk+\epsilon k} (2^n\delta_1)^{-\frac{2N}p+1+\epsilon} \|f\|_p, \, \quad p\ge 4N-6 \hspace{-4pt} \end{align} $$

for any $\epsilon>0$ , and

(3.24) $$ \begin{align} \left( \sum_\mu\sum_{\nu\in \mathfrak J_{n}^{\mu}}\| \mathcal T[\mathfrak a_{\nu,2}^{\mu,n}]f\|_p^p \right)^{1/p} \lesssim_B\! 2^{-\frac {(N+2)k} {2N}}(2^n\delta_1)^{-\frac N2}\|f\|_p, \quad 2\le p\le \infty \end{align} $$

when $n\ge 1$ . Thanks to the inequality (3.8), those estimates give

$$\begin{align*}\sum_{n\ge1} \mathbf E_n\,\le C_\epsilon \delta_0^{-\frac{2N}p+1+\epsilon} \! \sum_{1\le n\le \log_2 (C\delta_0/\delta_1)} \!\!\big(2^{-\frac {2}pk + \epsilon k} + 2^{-\frac {(N+2)k} {2N}}(2^n\delta_1)^{\frac{2N}p-\frac {N+2}2-\epsilon}\big) \|f\|_p \end{align*}$$

for $p\ge 4N-2$ . Note $ \log _2 (\delta _0/\delta _1)\le Ck $ from (2.31). So, the estimate (3.20) follows since $4N-2> 4N/(N+2)$ and $\delta _1\ge 2^{-k/N}$ .

In order to prove the estimates (3.23) and (3.24), we start with the next lemma.

Lemma 3.5. Let $n\ge 1$ . For a constant $C=C(B)>0$ , we have the following $:$

(3.25) $$ \begin{align} \sum_{1\le j\le N-1} (2^n\delta_1)^{-(N-j)} |\langle\gamma^{(j)}(s),\xi\rangle| \ge C 2^k, \qquad (s,\xi)\in \operatorname{\mathrm{supp}}_{s,\xi} \mathfrak a_{\nu,1}^{\mu,n}, \end{align} $$

(3.26) $$ \begin{align} (2^n\delta_1)^{-N} |\tau+\langle\gamma(s),\xi\rangle|\ge C 2^{k}, \qquad (s,\xi)\in \operatorname{\mathrm{supp}}_{s,\xi} \mathfrak a_{\nu,2}^{\mu,n}. \end{align} $$

Proof. We first prove (3.25). Since $\mathfrak G_N^\mu \ge 2^{-2N!-1}(2^n\delta _1)^{2N!}$ on $\operatorname {\mathrm {supp}}_{s,\xi } \mathfrak a_{\nu }^{\mu ,n}$ , one of the following holds on $\operatorname {\mathrm {supp}} \mathfrak a_{\nu ,1}^{\mu ,n}$ :

(3.27) $$ \begin{align} |s-\sigma(\xi)|&\ge (2^{3}C_0B)^{-1}2^n\delta_1, \end{align} $$

(3.28) $$ \begin{align} 2^{-k}|\mathfrak g^{j}_\mu(\tau,\xi)| &\ge (2^2C_0)^{-(N-j)} (2^n\delta_1)^{N-j} \end{align} $$

for some $1\le j \le N-2$ , where $C_0=2^{2d}B$ (see (3.22)). If (3.27) holds, by (2.28) and (2.29) it follows that $ (2^n\delta _1)^{-1}|\langle \gamma ^{(N-1)}(s),\xi \rangle | \gtrsim 2^{k}.$ Thus, to show (3.25) we may assume the inequality (3.27) fails, that is, (3.28) holds for some $1\le j \le N-2$ . So, there is an integer $\ell \in [0, N-2]$ such that (3.28) fails for $\ell +1\le j \le N-2$ , whereas (3.28) holds for $j=\ell $ . By (3.6) and (3.12), we have

(3.29) $$ \begin{align} |\langle G^{(\ell)}(\sigma(\xi)),(\tau,\xi) \rangle| \ge |\mathfrak g_\mu^\ell| -\!\!\sum_{j=\ell+1}^N|\mathfrak g_\mu^j| \frac{(B^6 2^{14}\delta_0^2)^{j-\ell}}{(j-\ell)!} -2B^3 (B^2 2^7\delta_0)^{N+1-\ell}|\xi|. \end{align} $$

Thus, (2.32) gives $|\langle G^{(\ell )}(\sigma (\xi )),(\tau ,\xi ) \rangle | \ge (2^3C_0)^{-(N-\ell )} 2^k (2^n\delta _1)^{N-\ell }$ . Also, the equation (3.6) and our choice of $\ell $ yield $|\langle G^{(j)}(\sigma (\xi )),(\tau ,\xi ) \rangle | \le (2C_0)^{-(N-j)} 2^k (2^n\delta _1)^{N-j}$ for $\ell +1 \le j \le N-2$ . Combining this with $|s-\sigma (\xi )|< (2^3 C_0B)^{-1} 2^n\delta _1$ and expanding $G^{(\ell )}$ in Taylor series at $\sigma (\xi )$ , we see that $ |\langle G^{(\ell )}(s), (\tau ,\xi ) \rangle |\ge C 2^k (2^n\delta _1)^{N-\ell }$ for some $C=C(B)>0$ . This proves (3.25).

We now show (3.26), which is easier. On $\operatorname {\mathrm {supp}} \mathfrak a_{\nu ,2}^{\mu ,n}$ , $2^{-k}|\mathfrak g_\mu ^0|\ge 2^{-N-1}(2^n\delta _1)^N$ and $2^{-k}|\mathfrak g_\mu ^j| \le 2C_0^{-(N-j)}(2^n\delta _1)^{N-j}$ for $j=1, \dots , N-2$ . Using (3.29) with $\ell =0$ , by (2.32) and (2.31) we get $ (2^n\delta _1)^{-N} |\tau +\langle \gamma (\sigma (\xi )),\xi \rangle | \ge 2^{-N-2}2^k$ . We also note that $|s-\sigma (\xi )| \le 2C_0^{-1} 2^n\delta _1$ and $|\langle G^{(j)}(\sigma (\xi )),(\tau ,\xi ) \rangle | \le C_0^{-1}2^k (2^n\delta _1)^{N-j}$ for $1 \le j \le N-2$ on $\operatorname {\mathrm {supp}} \mathfrak a_{\nu ,2}^{\mu ,n}$ . Since $|\langle G^{(N)}(s),(\tau ,\xi ) \rangle | \le B2^{k+1}$ , using Taylor series expansion at $\sigma (\xi )$ as above, we see (3.26) holds true for some $C=C(B)>0$ .

Additionally, we make use of disjointness of $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu }^{\mu ,n}$ by combining Lemma 2.11 and the next.

Lemma 3.6. There is a positive constant $C=C(B)$ such that

(3.30) $$ \begin{align} |( \widetilde{\mathcal L}_s^\delta)^{-1}\xi|\le Cb2^{k} \end{align} $$

whenever $\xi \in {\Lambda }^{\prime}_k(s,\delta ,b) ($ see (2.36) $)$ . If $\xi \in \Gamma _k$ and (3.30) holds with $C=1$ , then $\xi \in {\Lambda }^{\prime}_k(s,\delta , C_1b)$ for some $C_1=C_1(B)>0$ .

Proof. Let $\eta \in \mathbb R^d$ and $\{ v_{N}, \dots , v_d\}$ be an orthonormal basis of $(\operatorname {\mathrm {span}}\{ \gamma ^{(j)}(s): 1\le j\le N-1\})^\perp $ . We write $ \eta =\sum _{j=1}^{N-1} \mathbf c_j \gamma ^{(j)}(s) + \sum _{j=N}^d \mathbf c_j v_j.$ Since ${\mathfrak V(N, B)}$ holds for $\gamma $ , $|\eta |\sim |(\mathbf c_1, \cdots ,\mathbf c_d )|$ . Let $\xi \in {\Lambda }^{\prime}_k(s,\delta ,b) $ . Then, (2.4) gives

$$\begin{align*}\langle \eta, ( \widetilde{\mathcal L}_s^\delta)^{-1}\xi \rangle= \langle ( \widetilde{\mathcal L}_s^\delta)^{-\intercal} \eta, \xi \rangle= \sum_{j=1}^{N-1} \delta^{j-N} \mathbf c_j \langle\gamma^{(j)}(s), \xi\rangle + \sum_{j=N}^d \mathbf c_j \langle v_j,\xi\rangle. \end{align*}$$

Thus, by (2.36) we get $|\langle \eta , ( \widetilde {\mathcal L}_s^\delta )^{-1}\xi \rangle |\le Cb|\eta | 2^k $ , which shows (3.30).

By (2.4), $\langle \gamma ^{(j)}(s),\xi \rangle = \delta ^{N-j} \langle \gamma ^{(j)}(s), (\widetilde {\mathcal L}_s^\delta )^{-1}\xi \rangle $ for $1\le j\le N-1$ . Therefore, (3.30) with $C=1$ gives $|\langle \gamma ^{(j)}(s),\xi \rangle | \le C_1 b \delta ^{N-j} 2^k$ for a constant $C_1>0$ when $1\le j\le N-1$ . This proves the second statement.

Now, we are ready to prove the estimates (3.23) and (3.24).

Proof of (3.23).

By Lemma 3.4, $C^{-1}\mathfrak a_{\nu ,1}^{\mu ,n}\in \mathfrak A_k(2^n\delta _1\nu , 2^n\delta _1)$ for some $C>0$ . Besides, (3.25) holds on $\operatorname {\mathrm {supp}}_{s,\xi } \mathfrak a_{\nu ,1}^{\mu ,n}$ , and we note $2^n\delta _1<\delta '$ from (3.8), (2.26), and (2.31). Thus, taking $\delta =2^n\delta _1$ and

, we may use Lemma 2.8 for $\tilde \chi \mathcal T [\mathfrak a_{\nu ,1}^{\mu ,n}] f$ to get

where $ \|\tilde f_l\|_p= \|f\|_p$ , $a_{l}$ are of type $(j, N-1, B')$ relative to

for some $B'>0$ , and $2^j\sim (2^n\delta _1)^N 2^k$ . As seen before,

satisfies ${\mathfrak V({N}, {3B})}$ and (2.1) with B replaced by $3B$ for $\delta \le \delta _*$ . So, ${\mathfrak V({N-1}, {B'})}$ with a large $B'$ holds for

.

Therefore, we may apply the assumption (Theorem 2.2 with $L=N-1$ ) to , which gives for a constant $C_\epsilon =C_\epsilon (B')$ . Consequently, we obtain

$$\begin{align*}\| \tilde\chi \mathcal T[\mathfrak a_{\nu,1}^{\mu,n}]f\|_p \le C_\epsilon 2^{-\frac 2pk+\epsilon k} (2^n\delta_1)^{1-\frac{2N}p+\epsilon}\|f\|_p\end{align*}$$

for $p\ge 4(N-1)-2$ . Besides, since $C^{-1}\mathfrak a_{\nu ,1}^{\mu ,n}\in \mathfrak A_k(2^n\delta _1\nu , 2^n\delta _1)$ , by (2.16) we have $\| (1-\tilde \chi )\mathcal T[\mathfrak a_{\nu ,1}^{\mu ,n}]f\|_{L^p(\mathbb R^{d+1})} \lesssim _ B\!2^{-k}(2^n\delta _1)^{1-N} \|f\|_{L^p(\mathbb R^d)}$ for $p>1$ . Note $2^n\delta _1\gtrsim 2^{-k/N}$ . Combining those two estimates yields

(3.31) $$ \begin{align} \| \mathcal T[\mathfrak a_{\nu,1}^{\mu,n}]f\|_p \le C_\epsilon 2^{-\frac 2pk+\epsilon k} (2^n\delta_1)^{1-\frac{2N}p+\epsilon} \|f\|_p. \end{align} $$

To exploit disjointness of $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu ,1}^{\mu ,n}$ , we define a multiplier operator by

$$\begin{align*}\mathcal F(P_s^\delta f)(\xi)= \beta_0\big(| (\widetilde {\mathcal L}_s^\delta)^{-1}\xi|/(C_12^k) \big)\widehat f(\xi)\end{align*}$$

for a constant $C_1>0$ . Since $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu ,1}^{\mu ,n}\subset {\Lambda }^{\prime}_k (2^n\delta _1\nu , 2^n\delta _1, 2^5B)$ , by Lemma 3.6 we may choose $C_1$ large enough so that $\beta _0\big (| (\widetilde {\mathcal L}_{2^n\delta _1\nu }^{2^n\delta _1})^{-1}\cdot |/(C_12^k) \big )=1 $ on $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu ,1}^{\mu ,n}$ . Thus, $\mathcal T[\mathfrak a_{\nu ,1}^{\mu ,n}]f= \mathcal T[\mathfrak a_{\nu ,1}^{\mu ,n}] P_{2^n\delta _1\nu }^{2^n\delta _1} f $ . Combining this and (3.31), we obtain

$$\begin{align*}\left(\sum_{\mu}\sum_{ \nu \in \mathfrak J_n^\mu} \| \mathcal T[\mathfrak a_{\nu,1}^{\mu,n}]f\|_p^p \,\right)^{1/p} \le C_\epsilon 2^{-\frac 2pk+\epsilon k} (2^n\delta_1)^{1-\frac{2N}p+\epsilon} \left( \sum_{\mu}\sum_{ \nu \in \mathfrak J_n^\mu} \|P_{2^n\delta_1\nu}^{2^n\delta_1} f\|_p^p\, \right)^{1/p} \end{align*}$$

for a constant $C_\epsilon =C_\epsilon (B)$ if $p\ge 4N-6$ . Therefore, the estimate (3.23) follows if we show

(3.32) $$ \begin{align} \left( \sum_{\mu}\sum_{ \nu \in \mathfrak J_n^\mu} \|P_{2^n\delta_1\nu}^{2^n\delta_1} f\|_p^p\,\right)^{1/p}\lesssim_B\! \|f\|_p, \qquad 2\le p\le\infty. \end{align} $$

By interpolation, it suffices to obtain (3.32) for $p=2, \infty $ . The case $p=\infty $ is trivial since $\|P_{2^n\delta _1\nu }^{2^n\delta _1} f\|_\infty \lesssim \|f\|_\infty $ . For $p=2$ , (3.32) follows by Plancherel’s theorem since $\operatorname {\mathrm {supp}} \beta _0\big (| (\widetilde {\mathcal L}_{2^n\delta _1\nu }^{2^n\delta _1})^{-1}\cdot |/(C_12^k) \big )\widehat f$ , $\nu \in \mathfrak J_n^\mu $ are finitely overlapping. Indeed, by Lemma 3.6 we have $\operatorname {\mathrm {supp}} \beta _0\big (| (\widetilde {\mathcal L}_{2^n\delta _1\nu }^{2^n\delta _1})^{-1}\cdot |/(C_12^k) \big )\widehat f\subset {\Lambda }^{\prime}_k (2^n\delta _1\nu , 2^n\delta _1, CB)$ for a constant C. It is clear from Lemma 2.11 that ${\Lambda }^{\prime}_k (2^n\delta _1\nu , 2^n\delta _1, C B)$ , $\nu \in \mathfrak J_n^\mu $ overlap at most $C(B)$ times.

The proof of the estimate (3.24) is much easier since we have a favorable estimate for the kernel of $\mathcal T[\mathfrak a_{\nu ,2}^{\mu ,n}]$ thanks to the inequality (3.26).

Proof of (3.24).

Let

$$\begin{align*}\mathfrak b(s,t,\tau,\xi)= i^{-1}(\tau+\langle\gamma(s),\xi\rangle)^{-1}\partial_t \mathfrak a_{\nu,2}^{\mu,n} (s,t,\tau,\xi). \end{align*}$$

Then, integration by parts in t shows $m[\mathfrak a_{\nu ,2}^{\mu ,n}]=m[\mathfrak b]$ . Note that (3.26) holds and $C^{-1}\mathfrak a_{\nu ,2}^{\mu ,n} \in \mathfrak A_k(2^n\delta _1\nu , 2^n\delta _1)$ for a constant $C\ge 1$ . Thus, $\mathfrak a:=C^{-1}2^k (2^n\delta _1)^N \mathfrak b$ satisfies, with $\delta =2^n\delta _1$ and , (2.6) and (2.7) for $0\le j\le 1$ , $0\le l\le 2N-1$ , and $|\alpha |\le d+N+2$ . Applying the estimate (2.15), we obtain $ \|\mathcal T[\mathfrak a_{\nu ,2}^{\mu ,n}]f\|_\infty \lesssim _B\! 2^{-k}(2^n\delta _1)^{1-N}\|f\|_\infty. $ Since $\delta _1\ge 2^{-k/ N}$ , this gives

$$\begin{align*}\|\mathcal T[\mathfrak a_{\nu,2}^{\mu,n}]f\|_\infty \lesssim_B \! 2^{-\frac {(N+2)k} {2N}}(2^n\delta_1)^{-\frac N2}\|f\|_\infty. \end{align*}$$

By interpolation it is sufficient to show (3.24) for $p=2$ . Note that $\|b(\cdot ,t,\tau ,\xi )\|_\infty + \|\partial _s b(\cdot ,t,\tau ,\xi )\|_1\lesssim 2^{-k} (2^n\delta _1)^{-N}.$ Thus, (2.28) and using van der Corput’s lemma in s give $ |m[\mathfrak a_{\nu ,2}^{\mu ,n}](\tau ,\xi )| \lesssim 2^{-k(1+N)/N} (2^n\delta _1)^{-N}. $ Since $\operatorname {\mathrm {supp}}_\xi \mathfrak a_{\nu ,2}^{\mu ,n} \subset {\Lambda }^{\prime}_k ( 2^n\delta _1\nu , 2^n\delta _1, 2^5B)$ , as before, we have $\mathcal T[\mathfrak a^{\mu ,n}_{2,\nu }]f= \mathcal T[\mathfrak a^{\mu ,n}_{2,\nu }] P_{2^n\delta _1\nu }^{2^n\delta _1} f $ with a positive constant $C_1$ large enough. Thus, by Plancherel’s theorem

$$ \begin{align*} \| \mathcal T[\mathfrak a_{\nu,2}^{\mu,n}] f \|_{L^2}^2 \lesssim_B\! 2^{-\frac{2(1+N)}N k} (2^n\delta_1)^{-2N}\!\!\!\iint_{\{ \tau: |\mathfrak g_\mu^0(\tau,\xi)| \le 2^{k+1}(2^n\delta_1)^N \}}\!\!\!\!\! d\tau\, |\mathcal F(P_{2^n\delta_1\nu}^{2^n\delta_1} f)(\xi)|^2\, d\xi. \end{align*} $$

Combining this and (3.32) yields (3.24) for $p=2$ .

3.3 Proof of Lemma 3.2

To simplify notations, we denote

for the rest of this section. To prove Lemma 3.2, we verify the conditions (2.6) and (2.7) with $\mathfrak a=\mathfrak a_{\nu }^{\mu ,n}$ ,

and

. The first is easy. In fact, since $\mathfrak a^\mu \in \mathfrak A_k(\delta _0\mu ,\delta _0)$ and

, we only need to show

(3.33)

on $\operatorname {\mathrm {supp}}_{\tau , \xi } \mathfrak a_{\nu }^{\mu ,n}$ . Using (3.6) and (3.11) together with (2.32) and (3.12), one can easily obtain

(3.34)

on $\operatorname {\mathrm {supp}}_{\tau , \xi } \mathfrak a_{\nu }^{\mu ,n}$ . Expanding $\langle G^{(j)}(s),(\tau ,\xi )\rangle $ in Taylor’s series at $\sigma (\xi )$ gives (3.33) since (3.13) holds.

We now proceed to show (2.7) with $\mathfrak a=\mathfrak a_{\nu }^{\mu ,n}$ ,

and

. Since $\mathfrak a_{\nu }^{\mu ,n}$ consists of three factors $\mathfrak a^\mu $ ,

, and

, by Leibniz’s rule it is sufficient to consider the derivatives of each of them. The bounds on the derivatives of

are clear. So, it is enough to show (2.7) for

with

and

whenever

.

We handle $\mathfrak a^\mu $ first. That is to say, we show

(3.35)

for

. Since $\mathfrak a^\mu \in \mathfrak A_k(\delta _0\mu , \delta _0)$ and

, we have

(3.36)

One can show this using (2.11). We consider

By (2.8), we have $|\,\mathcal U^\intercal z|\lesssim _B\! |z|$ because $|\delta _0^{-1} 2^n \delta _1| \lesssim _B 1$ . Thus, (3.36) gives

for

.

Let

. Then,

, so

by Lemma 2.5. This and (2.9) give

for

. Therefore, we obtain (3.35) since

.

We continue to show (2.7) for

. Note that $\delta _\ast ^{-2N!} \mathfrak G_N^\mu $ is a sum of

and

, $0\le j\le N-2$ . Since the exponents ${2N!}/{(N-j)}$ are even integers, for the desired bounds on

it suffices to show the same bounds on the derivatives of

The bound on

is a consequence of (2.10) and the following lemma. For simplicity, we denote

Lemma 3.7. If $\Xi \in \operatorname {\mathrm {supp}}_{\tau ,\xi } \mathfrak a_{\nu }^{\mu ,n}$ , then we have

(3.37)

Proof. By (2.29), $\gamma ^{(N-1)} ( \sigma (\tilde \Xi ) ) \cdot \tilde \Xi =0$ . Differentiation gives

(3.38)

Denote $s=\sigma (\tilde \Xi )$ . By (2.4),

. Since

, that is, (3.10), by Lemma 2.6 we have

. Besides, $|\gamma ^{(N)}(\sigma (\tilde \Xi )) \cdot \tilde \Xi | \gtrsim |\tilde \Xi | \sim 2^k$ (see (2.28)). Thus, (3.38) and (2.10) give

which proves (3.37) with $|\alpha |=1$ .

We show the bounds on the derivatives of higher orders by induction. Assume that (3.37) holds true for $|\alpha |\le L$ . Let $\alpha '$ be a multi-index such that $|\alpha '|=L+1$ . Then, differentiating the equation (3.38) and using the induction assumption, one can easily see , by which we get (3.37) for $|\alpha |=L+1$ . Since $\sigma \in \mathrm{C}^{2d+2}$ , one can continue this as far as $L\le 2d+1$ .

The proof of Lemma 3.2 is now completed if we show

(3.39)

for $0\le \ell \le N-2$ whenever $\Xi \in \operatorname {\mathrm {supp}} \mathfrak a_{\nu }^{\mu ,n} (s,t,\cdot )$ . To this end, we use the following.

Lemma 3.8. For $ j=0,\dots ,N$ , set

If

, then for $ j=0,\dots ,N$ we have

(3.40) $$ \begin{align} |\partial_{\tau,\xi}^\alpha A_j| \lesssim_B\! |(\tau,\xi)|^{-|\alpha|}, \qquad 1\le |\alpha|\le 2d+2. \end{align} $$

Proof. When $j=N$ , the estimate (3.40) follows by Lemma 3.7 and (2.10). So, we may assume $j\le N-1$ . Differentiating $A_j$ , we have

$$\begin{align*}\nabla_{\tau,\xi}A_j= B_j+ D_j,\end{align*}$$

where

Note that

for $0\le j\le N-1$ . Since

, similarly as before, Lemma 2.6 and (2.8) give

(3.41)

By Lemma 3.7 and (3.34), $| B_j|\lesssim |\xi |^{-1}$ . Thus, for

, we have

$$\begin{align*}|\nabla_{\tau,\xi}A_j| \lesssim_B\! |\xi|^{-1}+2^{-k} \lesssim_B\! |(\tau,\xi)|^{-1}, \quad j=0,\dots,N-1. \end{align*}$$

For the second inequality we use (2.10). This gives the inequality (3.40) when $|\alpha |=1$ .

To show (3.40) for $2\le |\alpha |\le 2d+2$ , we use backward induction. By (2.29), we note $A_{N-1}=0$ , so (3.40) trivially holds when $j=N-1$ . We now assume that (3.40) holds true if $j_0+1\le j\le N-1$ for some $j_0\le N-2$ . Lemma 3.7, (2.10) and the induction assumption show $\partial ^{\alpha '}_{\tau , \xi } B_{j_0}=O(|(\tau ,\xi )|^{-1-|\alpha '|})$ for $1\le |\alpha '|\le 2d+1$ . Concerning $ D_{j_0}$ , observe that $\partial ^{\alpha '}_{\xi } ( G^{(j_0)}(\sigma (\tilde \Xi )))$ is given by a sum of the terms

$$\begin{align*}G^{(j)}(\sigma(\tilde \Xi ))\prod_{n=1}^{j-j_0}\partial^{\alpha^{\prime}_{n}}_{\xi}(\sigma(\tilde \Xi )), \end{align*}$$

where $j\ge j_0$ and $\alpha ^{\prime}_1+\cdots +\alpha ^{\prime}_{j-j_0}=\alpha '$ . Hence, Lemma 3.7, (3.41) and (2.10) give $\partial ^{\alpha '}_{ \xi } D_{j_0} =O(|(\tau ,\xi )|^{-1-|\alpha '|})$ for $1\le |\alpha '|\le 2d+1$ . Therefore, combining the estimates for $B_{j_0}$ and $D_{j_0}$ , we get $\partial ^{\alpha '}_{\tau , \xi } \nabla _{\tau ,\xi }A_{j_0}= O(|(\tau ,\xi )|^{-1-|\alpha '|})$ . This proves (3.40) for $j=j_0$ .

Before proving (3.39), we first note that

(3.42)

for $j=1, \dots , N$ . This can be shown by a routine computation. Indeed, differentiating (3.4) and using Lemma 3.7 and (2.32), one can easily see (3.42) holds since $|\sigma (\tilde \Xi )-\delta _0\mu |\lesssim \delta _0$ .

To show (3.39) for $0\le \ell \le N-2$ , we again use backward induction. Observe that (3.39) holds for $\ell =N,N-1$ . Then, we assume that (3.39) holds for $j+1\le \ell \le N$ for some $j\le N-2$ . By (3.6), we have

Thus, by Lemma 3.8 and (3.42), we get (3.39) with $\ell =j$ . This completes the proof of Lemma 3.2.

3.4 Proof of Lemma 3.4

Lemma 3.4 can be shown in the same manner as Lemma 3.2. So, we shall be brief.

By Lemma 3.2, we have

for a constant $C\ge 1$ , so it suffices to show

for some $C\ge 1$ . The support condition (2.6) is obvious, so we need only to show (2.7) with $\mathfrak a=\mathfrak a_{\nu ,1}^{\mu ,n}$ ,

, and

. Moreover, by recalling (3.22), it is enough to consider the additional factor only, that is, to show

for

. Since

on $\operatorname {\mathrm {supp}}_{s,\xi }\mathfrak a_{\nu ,1}^{\mu ,n}$ , one can obtain the estimate in the same way as in the proof of Lemma 3.2.

3.5 Sharpness of Theorem 1.3

Before closing this section, we show optimality of the regularity exponent $\alpha $ in Theorem 1.3.

Proof. We write $\gamma =(\gamma _1, \dots , \gamma _d)$ . Via an affine change of variables, we may assume $\gamma _1(0)=0$ and $\gamma _1'(s) \neq 0$ on an interval $J=[-\delta _0,\delta _0] $ for $0<\delta _0\ll 1$ . Since $\psi (0)\neq 0$ , we may also assume $\psi \ge 1$ on J.

We choose $\zeta _0 \in \mathcal S(\mathbb R)$ such that $\operatorname {\mathrm {supp}} \widehat \zeta _0 \subset [-1,1]$ and $ \zeta _0 \ge 1$ on $[-r_1,r_1]$ , where $r_1=1+2\max \{ |\gamma (s)| : s \in J\}$ . Denoting $\bar x = (x_1,\dots ,x_{d-1})$ and $\bar \gamma (t)=(\gamma _1(t),\dots ,\gamma _{d-1}(t))$ , we define

$$ \begin{align*} \bar{\mathcal A}_t h(x) =\int e^{it\lambda \gamma_d(s)} \zeta_0(x_d-t\gamma_d(s)) h(\bar x - t \bar \gamma(s)) \psi(s)\,ds. \end{align*} $$

Let $\zeta \in \mathrm C_c^\infty ((-2,2))$ be a positive function such that $\zeta =1$ on $[-1,1]$ . For a positive constant $c\ll \delta _0$ , let $g_1(\bar x)=\sum _{\nu \in \lambda ^{-1}\mathbb Z \cap [-c,c]} \zeta (\lambda |\bar x+\bar \gamma (\nu )|)$ . We consider

$$\begin{align*}g(\bar x)=e^{-i\lambda \varphi(x_1)}g_1(\bar x), \end{align*}$$

where $\varphi (s)= \gamma _d \circ (- \gamma _1)^{-1}(s)$ . We claim that, if c is small enough,

(3.43) $$ \begin{align} |\bar{\mathcal A}_t g( x)| \gtrsim1, \qquad (x,t)\in S_c, \end{align} $$

where $S_c= \{ (x,t): |\bar x| \le c \lambda ^{-1}, \, |x_d| \le c, \, |t-1| \le c \lambda ^{-1}\}$ . To show this, note

$$ \begin{align*} \bar{\mathcal A}_t g(x) =\int e^{i\lambda ( t\gamma_d(s)- \varphi (x_1-t\gamma_1(s)))} \zeta_0(x_d-t\gamma_d(s)) g_1(\bar x - t \bar \gamma(s))\,\psi(s)\,ds. \end{align*} $$

Let $(x,t)\in S_c $ . Then, $\operatorname {\mathrm {supp}} g_1(\bar x - t \bar \gamma (\cdot )) \subset [-C_1c, C_1c]$ for some $C_1>0$ . Since $\varphi (s)= \gamma _d \circ (- \gamma _1)^{-1}(s)$ , by the mean value theorem we see $ |\varphi (x_1-t\gamma _1(s)) -\gamma _d(s)|\le 2r_0c\lambda ^{-1}, $ where $r_0=10 r_1\max \{ |\partial _s \varphi (s)| : \, s \in (-\gamma _1)(J_\ast )\} $ and $J_\ast =[-(C_1+1)c, (C_1+1)c]$ . Thus, we have

(3.44) $$ \begin{align} |t \gamma_d(s)- \varphi (x_1-t \gamma_1(s))| \le 3r_0c\lambda^{-1}. \end{align} $$

Besides, if $\lambda $ is sufficiently large, $ g_1(\bar x-t\bar \gamma (s))=\sum _{\nu \in \lambda ^{-1}\mathbb Z \cap [-c,c]} \zeta (\lambda |\bar x-(t-1) \bar \gamma (s)+\bar \gamma (\nu )-\bar \gamma (s)|) \gtrsim 1 $ if $s \in [-c/2, c/2]$ . Since $ \operatorname {\mathrm {supp}} g_1(\bar x - t \bar \gamma (\cdot ))\subset J$ with c small enough and $\zeta _0 (x_d-t\gamma _d(s))\ge 1$ , we get $\int \zeta _0 (x_d-t\gamma _d(s)) g_1(\bar x-t\bar \gamma (s)) \psi (s)\,ds \gtrsim 1. $ Therefore, (3.43) follows by (3.44) if c is small enough, that is, $c\ll 1/(3r_0)$ .

We set $f(x)=e^{-i\lambda x_d} \zeta _0(x_d) g(\bar x).$ Then, $ \chi (t) \mathcal A_t f(x)= e^{-i\lambda x_d} \chi (t) \bar {\mathcal A}_t g(x). $ By our choice of $\zeta _0$ , $\operatorname {\mathrm {supp}} \widehat f\subset \{\xi : |\xi _d+\lambda |\le 1\}$ , so $\operatorname {\mathrm {supp}} \mathcal F(\chi (t)\mathcal A_t f)\subset \{(\tau , \xi ): |\xi _d+\lambda |\le 1\}$ . This gives

(3.45) $$ \begin{align} \lambda^{\alpha}\| \chi(t) \mathcal A_t f\|_{L^p(\mathbb R^{d+1})} \lesssim \| \chi(t) \mathcal A_t f \|_{L_\alpha^p(\mathbb R^{d+1})}. \end{align} $$

Indeed, $ \lambda ^\alpha \| \chi (t) \mathcal A_t f \|_{L^p(\mathbb R^{d+1})} \lesssim \| \chi (t) \mathcal A_t f \|_{L^p( \mathbb R_{t,\bar x}; L_\alpha ^p(\mathbb R_{x_d}))}$ by Mihlin’s multiplier theorem in $x_d$ . Similarly, one also sees $ \|F \|_{L^p( \mathbb R_{t,\bar x}; L_\alpha ^p(\mathbb R_{x_d}))} \le C \|F \|_{L_\alpha ^p(\mathbb R^{d+1})} $ for $\alpha \ge 0$ and any F. Combining those inequalities gives (3.45).

From (3.43), we have $\|\chi (t) \mathcal A_t f\|_p= \| \chi (t) \bar {\mathcal A}_t g \|_{p} \ge C\lambda ^{-d/p}.$ Note that $\operatorname {\mathrm {supp}} g$ is contained in a $O(\lambda ^{-1})$ -neighborhood of $-\bar {\gamma }$ , so it follows that $\|f\|_p \lesssim \lambda ^{-(d-2)/p}$ . Therefore, by (3.45) the inequality (1.4) implies $\lambda ^\alpha \lambda ^{-d/p} \lesssim \lambda ^{-(d-2)/p}$ . Taking $\lambda \to \infty $ gives $\alpha \le 2/p$ .

4.1 Proof of Theorem 4.1

The case $L=2$ is easy. Since $ \bar a $ is a symbol of type $(k,2,B)$ relative to $\gamma $ , van der Corput’s lemma and Plancherel’s theorem give (4.3) for $p=2$ . Interpolation with the $L^\infty $ estimate shows (4.3) for $p\ge 2$ . When $L\ge 3$ , we have the following, which immediately yields Theorem 4.1.

To prove the proposition, we fix $N\in [3, d]$ and $\gamma $ satisfying ${\mathfrak V(N, B)}$ , and $\bar a$ of type $(k, N, B)$ relative to $\gamma $ . For

and $\delta>0$ such that

, let

By

we denote the collection of $ \bar {\mathfrak a} \in \mathrm C^{d+1}(\mathbb R^{d+1}) $ satisfying

and

for $0 \le j \le 1$ , $|\alpha | \le d+1$ .

The next lemma which plays the same role as Lemma 2.8 can be shown by routinely following the proof of Lemma 2.8.

Lemma 4.3. Let

and $j_\ast =\log (2^k \delta ^N)$ . Suppose (2.20) holds on $\operatorname {\mathrm {supp}} \bar {\mathfrak a}$ . Then, there exist constants $C, l_\ast $ , $\tilde B\ge 1$ , and $\delta '>0$ depending on B, and symbols $\bar {\mathfrak a}_{1}, \dots , \bar {\mathfrak a}_{l_\ast }$ of type $(j, N-1, \tilde B)$ relative to

such that

$ \|\tilde f_l\|_p= \|f\|_p$ , and $j\in [j_\ast -C, j_\ast +C]$ as long as $0<\delta <\delta '$ .