Proof It suffices to prove the case $p<\infty $ since $L^{\infty }\left ( U\right ) \subset L^{p}\left ( U\right ) $ . We have

(3.5) $$ \begin{align} \delta_{m,p}\left( \bigtriangleup_{I;\kappa}^{\eta,\mathcal{G}}f\right) \left( x\right) & =\delta_{m,p}\left( \left\langle \left( S_{\kappa ,\eta}^{\mathcal{G}}\right) ^{-1}f,h_{I;\kappa}\right\rangle h_{I;\kappa }^{\eta}\right) \left( x\right) \\ & =\left\langle \left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1} f,h_{I;\kappa}\right\rangle \delta_{m,p}h_{I;\kappa}^{\eta}\left( x\right) ,\nonumber \end{align} $$

and

(3.6) $$ \begin{align} \bigtriangleup_{2^{m}I;\kappa}^{\eta,2^{m}\mathcal{G}}\delta_{m,p}f\left( x\right) & =\left\langle \left( S_{\kappa,\eta}^{2^{m}\mathcal{G}}\right) ^{-1}\delta_{m,p}f,h_{2^{m}I;\kappa}\right\rangle h_{2^{m}I;\kappa}^{\eta }\left( x\right) \\ & =\left\langle \delta_{m,p}\left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1}f,h_{2^{m}I;\kappa}\right\rangle h_{2^{m}I;\kappa}^{\eta}\left( x\right) ,\nonumber \end{align} $$

since $2^{m}I\in 2^{m}\mathcal {G}$ , and since $\left ( S_{\kappa ,\eta } ^{2^{m}\mathcal {G}}\right ) ^{-1}\delta _{m,p}=\delta _{m,p}\left ( S_{\kappa ,\eta }^{\mathcal {G}}\right ) ^{-1}$ by (2.3).

We first compute the factor $\left \langle \delta _{m,p}\left ( S_{\kappa ,\eta }^{\mathcal {G}}\right ) ^{-1}f,h_{2^{m}I;\kappa }\right \rangle $ in (3.6) in terms of $I$ . We have

(3.7) $$ \begin{align} & \left\langle \delta _{m,p}\left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1}f,h_{2^{m}I;\kappa }\right\rangle =\int\frac{1}{2^{\frac{2}{p}m}}\left( S_{\kappa,\eta }^{\mathcal{G}}\right) ^{-1}f\left( \frac{y}{2^{m}}\right) h_{2^{m} I;\kappa}\left( y\right) dy\\& \quad =\int\frac{1}{2^{\frac{2}{p}m}}\left( S_{\kappa,\eta}^{\mathcal{G} }\right) ^{-1}f\left( z\right) h_{2^{m}I;\kappa}\left( 2^{m}z\right) 2^{2m}dz\nonumber\\& \quad =2^{2m}\int\left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1}f\left( z\right) \frac{1}{2^{\frac{2}{p}m}}\frac{h_{2^{m}I;\kappa}\left( 2^{m}z\right) }{h_{I;\kappa}\left( z\right) }h_{I;\kappa}\left( z\right) dz\nonumber\\& \quad =2^{2m}\!\!\int\!\!\left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1}f\left( z\right) \frac{1}{2^{\frac{2}{p}m}}\sqrt{\frac{\left\vert I\right\vert }{\left\vert 2^{m}I\right\vert }}h_{I;\kappa}\left( z\right) dz=\frac{2^{m} }{2^{\frac{2}{p}m}}\!\int\!\!\left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1}\! f\left( z\right) h_{I;\kappa}\left( z\right) dz\nonumber\\& \quad =2^{m\left( 1-\frac{2}{p}\right) }\left\langle \left( S_{\kappa,\eta }^{\mathcal{G}}\right) ^{-1}f,h_{I;\kappa}\right\rangle .\nonumber \end{align} $$

Next, we compute the factor $\delta _{m,p}h_{I;\kappa }^{\eta }\left ( x\right ) $ in (3.5) in terms of $2^{m}I$ . We have

(3.8) $$ \begin{align} \delta_{m,p}h_{I;\kappa}^{\eta}\left( x\right) & =\frac{1}{2^{\frac{2} {p}m}}h_{I;\kappa}^{\eta}\left( \frac{x}{2^{m}}\right) =\frac{1}{2^{\frac {2}{p}m}}\frac{h_{I;\kappa}\left( \frac{x}{2^{m}}\right) }{h_{2^{m}I;\kappa }\left( x\right) }h_{2^{m}I;\kappa}^{\eta}\left( x\right) \\ & =\frac{1}{2^{\frac{2}{p}m}}\sqrt{\frac{\left\vert 2^{m}I\right\vert }{\left\vert I\right\vert }}h_{2^{m}I;\kappa}^{\eta}\left( x\right) =2^{m\left( 1-\frac{2}{p}\right) }h_{2^{m}I;\kappa}^{\eta}\left( x\right) .\nonumber \end{align} $$

Using first (3.5), then (3.8), then (3.7), and finally (3.6) gives (3.4),

$$ \begin{align*} \delta_{m,p}\left( \bigtriangleup_{I;\kappa}^{\eta,\mathcal{G}}f\right) \left( x\right) & =\left\langle \left( S_{\kappa,\eta}^{\mathcal{G} }\right) ^{-1}f,h_{I;\kappa}\right\rangle \delta_{m,p}h_{I;\kappa}^{\eta }\left( x\right) \\ & =\left\langle \left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1} f,h_{I;\kappa}\right\rangle 2^{m\left( 1-\frac{2}{p}\right) }h_{2^{m} I;\kappa}^{\eta}\left( x\right) \\ & =\left\langle \delta_{m,p}\left( S_{\kappa,\eta}^{\mathcal{G}}\right) ^{-1}f,h_{2^{m}I;\kappa}\right\rangle h_{2^{m}I;\kappa}^{\eta}\left( x\right) \\ & =\bigtriangleup_{2^{m}I;\kappa}^{\eta,2^{m}\mathcal{G}}\delta_{m,p}f\left( x\right).\\[-37pt] \end{align*} $$

Proof of Theorem 3.1

We begin the argument as in [Reference Bourgain and Guth1], with notation as in [Reference Rios and Sawyer4]. We let the surface $S$ be a compact smooth piece of the paraboloid $\mathbb {P}^{2}$ given by ${z_{3}=\left \vert z^{\prime }\right \vert ^{2}=z_{1}^{2}+z_{2}^{2}}$ in $\mathbb {R}^{3}$ , and for $f\in L^{\infty }\left ( S\right ) $ with $\left \Vert f\right \Vert _{L^{\infty }\left ( S\right ) }=1$ , we consider the oscillatory integral:

$$ \begin{align*} Tf\left( \xi\right) & =\int_{U}e^{i\phi\left( \xi,y\right) }f\left( y\right) dy=\int_{U}e^{i\left\{ \xi_{1}\cdot y_{1}+\xi_{2}\cdot y_{2} +\xi_{3}\left( y_{1}^{2}+y_{2}^{2}\right) \right\} }f\left( y\right) dy\\& =\int_{U}e^{i\xi\cdot\left( y,\left\vert y\right\vert ^{2}\right) }f\left( y\right) dy=\widehat{f^{\Phi}}\left( \xi\right) =\widehat{{\Phi _{\ast}\left[ fdy\right] }}\left( \xi\right) ,\ \ \ \ \ \text{for }\xi \in\mathbb{R}^{3}, \end{align*} $$

where $f^{\Phi }\left ( z\right ) =f\left ( z_{1},z_{2}\right ) \left ( 1+4\left ( z_{1}^{2}+z_{2}^{2}\right ) \right ) ^{-\frac {1}{2}}d\sigma \left ( z\right ) $ is the push-forward of $f\left ( y\right ) \left ( 1+4\left \vert y\right \vert ^{2}\right ) ^{-\frac {1}{2}}\approx f\left ( y\right ) $ to $S$ , and

$$\begin{align*}\phi\left( \xi,y\right) =\xi\cdot\Phi\left( y\right) \text{ and } \Phi\left( y\right) \equiv\left( y_{1},y_{2},y_{1}^{2}+y_{2}^{2}\right). \end{align*}$$

For $\lambda \geq 1$ , let $f=\sum _{I\in \mathcal {G}_{\lambda }\left [ U\right ] }\boldsymbol {1}_{I}f=\sum _{I\in \mathcal {G}_{\lambda }\left [ U\right ] }f_{I}$ and write

$$ \begin{align*} Tf\left( \xi\right) & =\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] }\int_{S}e^{i\xi\cdot\left( y,\left\vert y\right\vert ^{2}\right) } f_{I}\left( y\right) dy \\ & =\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] }e^{i\phi\left( \xi,c_{I}\right) }\int e^{i\left\{ \phi\left( \xi,y\right) -\phi\left( \xi,c_{I}\right) \right\} }f_{I}\left( y\right) dy\\ & =\sum_{I\in\mathcal{G}_{\lambda}\left[ S\right] }e^{i\phi\left( \xi,c_{I}\right) }T_{I}f\left( \xi\right) , \end{align*} $$

where

(3.9) $$ \begin{align} T_{I}f\left( \xi\right) & \equiv\int e^{i\left\{ \phi\left( \xi,y\right) -\phi\left( \xi,c_{I}\right) \right\} }f_{I}\left( y\right) dy=e^{-i\phi\left( \xi,c_{I}\right) }\int e^{i\phi\left( \xi,y\right) }f_{I}\left( y\right) dy\\ & =e^{-i\phi\left( \xi,c_{I}\right) }\widehat{f_{I}^{\Phi}}\left( \xi\right) =\widehat{\tau_{-\Phi\left( c_{I}\right) }f_{I}^{\Phi}}\left( \xi\right) ,\nonumber \end{align} $$

where $\tau _{\Phi \left ( c_{I}\right ) }g\left ( z\right ) \equiv g\left ( z-\Phi \left ( c_{I}\right ) \right ) $ is translation of a function $g$ by the vector $\Phi \left ( c_{I}\right ) $ .

Note that

$$\begin{align*}\left\vert \nabla_{\xi}\left\{ \xi\cdot\left( y-c_{I},\left\vert y\right\vert ^{2}-\left\vert c_{I}\right\vert ^{2}\right) \right\} \right\vert =\left\vert \left( y-c_{I},\left\vert y\right\vert ^{2} -\left\vert c_{I}\right\vert ^{2}\right) \right\vert \lesssim\frac {1}{2^{\lambda}},\ \ \ \ \ \text{for }y\in I, \end{align*}$$

implies

$$ \begin{align*} \nabla_{\xi}T_{I}f\left( \xi\right) & =\nabla_{\xi}\int e^{i\xi \cdot\left( y-c_{I},\left\vert y-c_{I}\right\vert ^{2}\right) }f_{I}\left( y\right) dy=\int\nabla_{\xi}e^{i\xi\cdot\left( y-c_{I},\left\vert y-c_{I}\right\vert ^{2}\right) }f_{I}\left( y\right) dy\\ & =\int ie^{i\xi\cdot\left( y-c_{I},\left\vert y-c_{I}\right\vert ^{2}\right) }\nabla_{\xi}\left\{ \xi\cdot\left( y-c_{I},\left\vert y\right\vert ^{2}-\left\vert c_{I}\right\vert ^{2}\right) \right\} f_{I}\left( y\right) dy, \end{align*} $$

which implies

(3.10) $$ \begin{align} \left\vert \nabla_{\xi}T_{I}f\left( \xi\right) \right\vert & \leq \int\left\vert \nabla_{\xi}\left\{ \xi\cdot\left( y-c_{I},\left\vert y\right\vert ^{2}-\left\vert c_{I}\right\vert ^{2}\right) \right\} \right\vert \left\vert f_{I}\left( y\right) \right\vert dy \\ & \lesssim\frac{1}{2^{\lambda}}\left\Vert f_{I}\right\Vert _{L^{1}\left( U\right) }\lesssim\frac{1}{2^{3\lambda}}\left\Vert f_{I}\right\Vert _{L^{\infty}\left( U\right) },\nonumber \end{align} $$

since $\ell \left ( I\right ) =\frac {1}{2^{\lambda }}$ . We will use the estimate (3.10) in (3.13) below.

Now let $\rho $ be a smooth rapidly decreasing bump function such that $\widehat {\rho }\left ( \xi \right ) =1$ for $\left \vert \xi \right \vert \leq 1$ , and set

$$ \begin{align*} \rho_{\lambda}\left( z\right) & \equiv\frac{1}{2^{3\lambda}}\rho\left( \frac{z}{2^{\lambda}}\right) ,\ \ \ \ \ \widehat{\rho_{\lambda}}\left( \xi\right) =\widehat{\rho}\left( 2^{\lambda}\xi\right) =1\text{ on }B\left( 0,2^{-\lambda}\right) \text{ and }\\ \rho_{\lambda}\left( z\right) & \approx\frac{1}{2^{3\lambda}}\text{ on }B\left( 0,2^{\lambda}\right). \end{align*} $$

Then, from (3.9), we obtain

$$\begin{align*}T_{I}f\left( \xi\right) =T_{I}f\ast\rho_{\lambda}\left( \xi\right) ,\ \ \ \ \ \text{for }I\in\mathcal{G}_{\lambda}\left[ S\right] \text{ and }\xi\in\mathbb{R}^{3}, \end{align*}$$

since $\tau _{-\Phi \left ( c_{I}\right ) }f_{I}^{\Phi }\subset B\left ( 0,2^{-\lambda }\right ) $ and $\widehat {T_{I}f}\left ( z\right ) =\tau _{-\Phi \left ( c_{I}\right ) }f_{I}^{\Phi }\left ( z\right ) $ imply

$$\begin{align*}\widehat{T_{I}f\ast\rho_{\lambda}}\left( z\right) =\widehat{T_{I}f}\left( z\right) \widehat{\rho_{\lambda}}\left( z\right) =\tau_{-\Phi\left( c_{I}\right) }f_{I}^{\Phi}\left( z\right) \widehat{\rho_{\lambda}}\left( z\right) =\tau_{-\Phi\left( c_{I}\right) }f_{I}^{\Phi}\left( z\right) =\widehat{T_{I}f}\left( z\right). \end{align*}$$

Now fix a point $a\in \Gamma _{\lambda }\left ( R\right ) $ , where $R=2^{\frac {s}{1-\delta }}$ ,

$$\begin{align*}\Gamma_{\lambda}\left( R\right) \equiv2^{\lambda}\mathbb{Z}^{3}\cap B_{R},\ \ \ \ \ \text{and }B_{R}\equiv B\left( 0,R\right) , \end{align*}$$

and restrict $\xi $ to the ball $B\left ( a,2^{\lambda }\right ) $ . We will write $B_{R}$ as apposed to $B\left ( 0,R\right ) $ above in order to emphasize the different roles played by $\lambda $ and $R$ , namely, $R\nearrow \infty $ as $s\nearrow \infty $ , while $\lambda $ remains a fixed sufficiently large integer to be chosen.

Then, for $\xi \in B\left ( a,2^{\lambda }\right ) $ and $I\in \mathcal {G} _{\lambda }\left [ U\right ] $ , we have

(3.11) $$ \begin{align} & \left\vert T_{I}f\left( \xi\right) \right\vert =\left\vert T_{I}f\ast \rho_{\lambda}\left( \xi\right) \right\vert =\left\vert \int_{\mathbb{R} ^{3}}T_{I}f\left( z\right) \rho_{\lambda}\left( \xi-z\right) dz\right\vert \\ & \leq\int_{\mathbb{R}^{3}}\left\vert T_{I}f\left( z\right) \right\vert \left\vert \rho_{\lambda}\left( \xi-z\right) \right\vert dz\leq \int_{\mathbb{R}^{3}}\left\vert T_{I}f\left( z\right) \right\vert \sup_{\omega\in B\left( a,2^{\lambda}\right) }\left\vert \rho_{\lambda }\left( z-\omega\right) \right\vert dz\nonumber\\ & =\int_{\mathbb{R}^{3}}\left\vert T_{I}f\left( z\right) \right\vert \zeta_{\lambda}\left( z-a\right) dz,\nonumber \end{align} $$

where $\zeta _{\lambda }\left ( w\right ) \equiv \sup _{\omega -a\in B\left ( 0,2^{\lambda }\right ) }\left \vert \rho _{\lambda }\left ( \omega \right ) \right \vert $ , since $\rho $ can be chosen radial and

$$ \begin{align*} \sup_{\omega\in B\left( a,2^{\lambda}\right) }\left\vert \rho_{\lambda }\left( \omega-z\right) \right\vert & =\frac{1}{2^{3\lambda}}\sup _{\omega\in B\left( a,2^{\lambda}\right) }\left\vert \rho\left( \frac{\left( \omega-a\right) -\left( z-a\right) }{2^{\lambda}}\right) \right\vert \\ & =\frac{1}{2^{3\lambda}}\sup_{\gamma\in B\left( 0,1\right) }\left\vert \rho\left( \frac{z-a}{2^{\lambda}}-\gamma\right) \right\vert =\zeta _{\lambda}\left( z-a\right) ,\\ \text{where }\zeta\left( w\right) & \equiv\sup_{\left\vert w-w^{\prime }\right\vert \leq1}\left\vert \rho\left( w^{\prime}\right) \right\vert. \end{align*} $$

Now, for $I\in \mathcal {G}_{\lambda }\left [ S\right ] $ , define the right-hand side of (3.11) to be

$$ \begin{align*} w_{I}^{a}\left( f\right) &\equiv\int_{\mathbb{R}^{3}}\left\vert T_{I}f\left( z\right) \right\vert \zeta_{\lambda}\left( z-a\right) dz=\int_{\mathbb{R}^{3}}\left\vert T_{I}f\left( z\right) \right\vert \zeta\left( \frac{z-a}{2^{\lambda}}\right) \frac{dz}{2^{3\lambda}}\\& =\int_{\mathbb{R}^{3}}\left\vert \widehat{f_{I}^{\Phi}}\left( z\right) \right\vert \zeta\left( \frac{z-a}{2^{\lambda}}\right) \frac{dz} {2^{3\lambda}}\approx\frac{1}{\left\vert B\left( a,2^{\lambda}\right) \right\vert }\int_{B\left( a,2^{\lambda}\right) }\left\vert \widehat{f_{I} ^{\Phi}}\left( z\right) \right\vert , \end{align*} $$

and refer to $w_{I}^{a}\left ( f\right ) $ as the “weight” of $\widehat {f_{I} ^{\Phi }}$ relative to the ball $B\left ( a,2^{\lambda }\right ) $ , which represents the average contribution of $\left \vert \widehat {f_{I}^{\Phi } }\left ( z\right ) \right \vert $ over the ball $B\left ( a,2^{\lambda }\right ) $ . Note that $w_{I}^{a}\left ( f\right ) \lesssim \left \Vert \widehat {f_{I} ^{\Phi }}\right \Vert _{L^{\infty }}\lesssim \left \Vert \textbf {1}_{I}f\right \Vert _{L^{1}}\leq \left \vert I\right \vert \left \Vert f\right \Vert _{L^{\infty } }=2^{-2\lambda }$ .

Summarizing, we have

(3.12) $$ \begin{align} \left\vert T_{I}f\left( \xi\right) \right\vert \leq\int_{\mathbb{R}^{3} }\left\vert T_{I}f\left( z\right) \right\vert \zeta_{\lambda}\left( z-a\right) dz=w_{I}^{a}\left( f\right) ,\ \ \ \ \ \text{for }\xi\in B\left( a,2^{\lambda}\right) , \end{align} $$

and

(3.13) $$ \begin{align} \int_{\mathbb{R}^{3}}\left\vert T_{I}f\left( z\right) \right\vert \zeta_{\lambda}\left( z-\xi\right) dz=w_{I}^{\xi}\left( f\right) \approx w_{I}^{a}\left( f\right) \text{ },\ \ \ \ \ \text{for }\xi\in B\left( a,2^{\lambda}\right) , \end{align} $$

by (3.10) and since for $\xi \in B\left ( a,2^{\lambda }\right ) $ , we have $\zeta _{\lambda }\left ( \xi -z\right ) \approx \zeta _{\lambda }\left ( a-z\right ) =\zeta _{\lambda }\left ( z-a\right ) $ .

Now set

$$\begin{align*}w_{\ast}^{a}\left( f\right) \equiv\max_{I\in\mathcal{G}_{\lambda}\left[ S\right] }w_{I}^{a}\left( f\right) =\max_{I\in\mathcal{G}_{\lambda}\left[ S\right] }\int_{\mathbb{R}^{3}}\left\vert T_{I}f\left( z\right) \right\vert \zeta_{\lambda}\left( z-a\right) dz, \end{align*}$$

and fix $I_{\ast }^{a}$ such that

$$\begin{align*}w_{I_{\ast}^{a}}^{a}=w_{\ast}^{a}. \end{align*}$$

Now we replace the function $f$ above with a function in $\mathcal {F} _{\kappa ,\infty }\left ( s\right ) $ (see (3.2)) for the remainder of the proof, which with a small abuse of notation we write as $\mathsf {Q} _{\kappa ,s,U}f$ with $\left \Vert f\right \Vert _{L^{\infty }\left ( U\right ) }\leq 1$ . For $\mathsf {Q}_{\kappa ,s,U}f\in \mathcal {F}_{\kappa ,\infty }\left ( s\right ) $ , $1<\lambda \in \mathbb {N}$ chosen appropriately, we will estimate the contributions to the norm $\left \Vert T\mathsf {Q}_{\kappa ,s,U}f\right \Vert _{L^{q}\left ( \mathbb {R}^{3}\right ) }$ in two exhaustive cases in turn. The first case will yield the growth factor $2^{\varepsilon s}$ , while the next case will be absorbed. In fact, we show at the end of the proof that we may take $\lambda>\frac {3q}{q-3}$ .

Proof continued

In Case 1, we assume the following property. There exists a triple of squares $I_{0}^{a},J_{0}^{a},K_{0}^{a}\in \mathcal {G} _{\lambda }\left [ U\right ] $ such that

$$ \begin{align*} w_{I_{0}^{a}},w_{J_{0}^{a}},w_{K_{0}^{a}} &>2^{-2\lambda}w_{\ast} ^{a},\ \ \ \ \ \text{and }\left\vert \textbf{c}_{I_{0}^{a}}-\textbf{c} _{J_{0}^{a}}\right\vert ,\left\vert \textbf{c}_{J_{0}^{a}}-\textbf{c} _{K_{0}^{a}}\right\vert ,\left\vert \textbf{c}_{K_{0}^{a}}-\textbf{c} _{I_{0}^{a}}\right\vert >2^{10}2^{-\lambda},\\ & \begin{array} [c]{|ccccc|} \textbf{c}_{I_{0}^{a}} & & & & \\ & \cdot & & & \\ & & \textbf{c}_{K_{0}^{a}} & \leftrightarrows & \textbf{c}_{K_{0}^{a}}\\ & & & \cdot & \\ & & & & \textbf{c}_{J_{0}^{a}}\\ \end{array} , \end{align*} $$

where $\textbf {c}_{Q}$ denotes the center of the cube $Q$ ; that is, there exists a “ $2^{10}2^{-\lambda }$ -separated” triple $I_{0}^{a},J_{0}^{a},K_{0}^{a}$ of squares of side-length $2^{-\lambda }$ , such that each of $I_{0}^{a}$ , $J_{0}^{a} $ , and $K_{0}^{a}$ has near maximal weight. In Case 1, we will use the $\nu $ -disjoint trilinear estimate $\mathcal {A} _{\mathrm {disj}\nu }^{s,\kappa ,\delta }\left ( \otimes _{3}L^{\infty }\rightarrow L^{\frac {q}{3}};\varepsilon \right ) $ with $\nu =2^{10} 2^{-\lambda }$ . For $\xi \in B\left ( a,2^{\lambda }\right ) $ , we throw away the unimodular function $e^{-i\Phi \left ( c_{I}\right ) \cdot \xi }$ , and using (3.12), we estimate that for $\xi \in B\left ( a,2^{\lambda }\right ) $ ,

(3.14) $$ \begin{align} \left\vert T\mathsf{Q}_{\kappa,s,U}f\left( \xi\right) \right\vert & =\left\vert \sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] }e^{i\phi\left( \xi,c_{I}\right) }T_{I}\mathsf{Q}_{\kappa,s,U}f\left( \xi\right) \right\vert \leq\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] }\left\vert T_{I}\mathsf{Q}_{\kappa,s,U}f\left( \xi\right) \right\vert \\ & \lesssim\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] }w_{I} ^{a}<2^{2\lambda}w_{\ast}^{a}<2^{4\lambda}\left( w_{I_{1}^{a}}^{a} w_{I_{2}^{a}}^{a}w_{I_{3}^{a}}^{a}\right) ^{\frac{1}{3}},\nonumber \end{align} $$

since the fixed triple $\left ( I_{0}^{a},J_{0}^{a},K_{0}^{a}\right ) $ satisfies the near maximal weight condition in Case 1:

$$\begin{align*}w_{\ast}^{a}<\min\left\{ 2^{2\lambda}w_{I_{0}^{a}}^{a},2^{2\lambda} w_{J_{0}^{a}}^{a},2^{2\lambda}w_{K_{0}^{a}}^{a}\right\} \leq2^{2\lambda }\left( w_{I_{0}^{a}}^{a}\right) ^{\frac{1}{3}}\left( w_{J_{0}^{a}} ^{a}\right) ^{\frac{1}{3}}\left( w_{K_{0}^{a}}^{a}\right) ^{\frac{1}{3}}. \end{align*}$$

For convenience, we suppress dependence on $\kappa $ and $U$ and write

$$\begin{align*}f^{s}\equiv\mathsf{Q}_{\kappa,s,U}f. \end{align*}$$

Let

$$\begin{align*}d\mu_{\lambda}\left( z_{1},z_{2},z_{3}\right) \equiv\zeta_{\lambda}\left( z_{1}\right) \zeta_{\lambda}\left( z_{2}\right) \zeta_{\lambda}\left( z_{3}\right) dz_{1}dz_{2}dz_{3}, \end{align*}$$

so $d\mu _{\lambda }$ is a bounded multiple of a probability measure, that is, $\mu _{\lambda }\left ( \mathbb {R}^{3}\right ) \approx 1$ . Let ${\nu =2^{10}2^{-\lambda }}$ . Then for $q>3$ and $\xi \in B\left ( a,2^{\lambda }\right ) $ , we have from (3.13) and (3.14), followed by Hölder’s inequality, that

$$ \begin{align*} & \left\vert Tf^{s}\left( \xi\right) \right\vert ^{q}\lesssim2^{4q\lambda }\left( w_{I_{0}^{a}}^{a}w_{J_{0}^{a}}^{a}w_{K_{0}^{a}}^{a}\right) ^{\frac{q}{3}}\\& \quad \approx2^{4q\lambda}\left( \int_{\mathbb{R}^{3}}\left\vert T_{I_{0}^{a} }f^{s}\left( z_{1}\right) \right\vert \zeta_{\lambda}\left( z_{1}-a\right) dz_{1}\right) ^{\frac{q}{3}}\left( \int_{\mathbb{R}^{3}}\left\vert T_{J_{0}^{a}}f^{s}\left( z_{2}\right) \right\vert \zeta_{\lambda}\left( z_{2}-a\right) dz_{2}\right) ^{\frac{q}{3}}\\& \qquad \times\left( \int_{\mathbb{R}^{3}}\left\vert T_{K_{0}^{a}}f^{s}\left( z_{3}\right) \right\vert \zeta_{\lambda}\left( z_{3}-a\right) dz_{3}\right) ^{\frac{q}{3}}\\& \quad \lesssim2^{4q\lambda}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}}\left\vert T_{I_{0}^{a}}f^{s}\left( \xi-z_{1}\right) T_{J_{0}^{a}}f^{s}\left( \xi-z_{2}\right) T_{K_{0}^{a}}f^{s}\left( \xi-z_{3}\right) \right\vert ^{\frac{q}{3}}\\& \qquad\qquad\qquad\qquad\qquad\times\zeta_{\lambda}\left( z_{1}\right) \zeta_{\lambda}\left( z_{2}\right) \zeta_{\lambda}\left( z_{3}\right) ~dz_{1}dz_{2}dz_{3}\\& \quad \lesssim2^{4q\lambda}\sum_{\left( I,J,K\right) \in\mathcal{G}_{\lambda }^{\nu}\left[ U\right] }\int_{\mathbb{R}^{9}}\left\vert T_{I}f^{s}\left( \xi-z_{1}\right) T_{J}f^{s}\left( \xi-z_{2}\right) T_{K}f^{s}\left( \xi-z_{3}\right) \right\vert ^{\frac{q}{3}}\\& \qquad d\mu_{\lambda}\left( z_{1},z_{2},z_{3}\right) , \end{align*} $$

where

$$\begin{align*}\mathcal{G}_{\lambda}^{\nu}\left[ U\right] \equiv\left\{ \left( I,J,K\right) \in\mathcal{G}_{\lambda}:\left( I,J,K\right) \text{ is } \nu\text{-separated as in (1.2)}\right\}. \end{align*}$$

Then we have

$$ \begin{align*} & \int_{B\left( a,2^{\lambda}\right) }\left\vert Tf^{s}\left( \xi\right) \right\vert ^{q}d\xi\\& \quad \lesssim2^{4q\lambda}\int_{B\left( a,2^{\lambda}\right) }\sum_{\left( I,J,K\right) \in\mathcal{G}_{\lambda}^{\nu}\left[ U\right] }\int _{\mathbb{R}^{9}}\left\vert T_{I}f^{s}\left( \xi-z_{1}\right) T_{J} f^{s}\left( \xi-z_{2}\right) T_{K}f^{s}\left( \xi-z_{3}\right) \right\vert ^{\frac{q}{3}}\\& \qquad d\mu_{\lambda }\left( z_{1},z_{2},z_{3}\right) d\xi\\& \quad =2^{4q\lambda}\sum_{\left( I,J,K\right) \in\mathcal{G}_{\lambda}^{\nu }\left[ U\right] }\int_{\mathbb{R}^{9}}\left\{ \int_{B\left( a,2^{\lambda }\right) }\left\vert T_{I}f^{s}\left( \xi-z_{1}\right) T_{J}f^{s}\left( \xi-z_{2}\right) T_{K}f^{s}\left( \xi-z_{3}\right) \right\vert ^{\frac {q}{3}}d\xi\right\} \\& \qquad d\mu_{\lambda }\left( z_{1},z_{2},z_{3}\right). \end{align*} $$

Now consider those $a\in \Gamma _{\lambda }\left ( R\right ) $ for which Case 1 is in effect for the ball $B\left ( a,2^{\lambda }\right ) $ and denote by $\Gamma _{\lambda }\left ( {\mathbf {Case 1}}\right ) $ the union of all the balls $B\left ( a,2^{\lambda }\right ) $ for which a is in Case 1. Summing over points $a\in \Gamma _{\lambda }\left ( R\right ) $ such that Case 1 is in effect for the ball $B\left ( a,2^{\lambda }\right ) $ , we obtain

$$ \begin{align*} & \sum_{a\in\Gamma_{\lambda}\left( {\mathbf{Case 1}}\right) } \int_{B\left( a,2^{\lambda}\right) }\left\vert Tf^{s}\left( \xi\right) \right\vert ^{q}d\xi\\& \quad \lesssim2^{4q\lambda}\sum_{\left( I,J,K\right) \in\mathcal{G}_{\lambda }^{\nu}\left[ U\right] }\int_{\mathbb{R}^{9}}\left\{ \sum_{a\in \Gamma_{\lambda}\left( {\mathbf{Case 1}}\right) }\right. \\& \left. \qquad\int_{B\left( a,2^{\lambda}\right) }\left\vert T_{I} f^{s}\left( \xi-z_{1}\right) T_{J}f^{s}\left( \xi-z_{2}\right) T_{K} f^{s}\left( \xi-z_{3}\right) \right\vert ^{\frac{q}{3}}d\xi\right\} d\mu_{\lambda}\left( z_{1},z_{2},z_{3}\right) \\& \quad \lesssim2^{4q\lambda}\sum_{\left( I,J,K\right) \in\mathcal{G}_{\lambda }^{\nu}\left[ U\right] }\\& \qquad \int_{\mathbb{R}^{9}}\left\{ \int_{B_{R}}\left\vert T_{I}f^{s}\left( \xi-z_{1}\right) T_{J}f^{s}\left( \xi-z_{2}\right) T_{K}f^{s}\left( \xi-z_{3}\right) \right\vert ^{\frac{q}{3}}d\xi\right\} d\mu_{\lambda }\left( z_{1},z_{2},z_{3}\right). \end{align*} $$

This is now dominated by

$$ \begin{align*} & 2^{4q\lambda}\left( \#\mathcal{G}_{\lambda}^{\nu}\left[ U\right] \right) ^{3}\int_{\mathbb{R}^{9}}\left\{ C_{\delta,\kappa,q,\varepsilon ,\lambda}^{q}2^{q\varepsilon s}\left\Vert f_{1}\right\Vert _{L^{\infty} }\left\Vert f_{2}\right\Vert _{L^{\infty}}\left\Vert f_{3}\right\Vert _{L^{\infty}}\right\} d\mu_{\lambda}\left( z_{1},z_{2},z_{3}\right) \\& \quad \leq2^{4q\lambda}\left( \#\mathcal{G}_{\lambda}^{\nu}\left[ U\right] \right) ^{3}\int_{\mathbb{R}^{9}}\left\{ C_{\varepsilon}^{q}2^{q\varepsilon s}\left( 2^{\frac{2}{r}s}\left\Vert f\right\Vert _{L^{\infty}}\right) ^{3}\right\} d\mu_{\lambda}\left( z_{1},z_{2},z_{3}\right) \\& \quad \lesssim C_{\varepsilon}^{q}2^{4q\lambda}2^{6\lambda}2^{q\varepsilon s}2^{\frac{6}{r}s}, \end{align*} $$

by (2.2). This gives $\left \Vert f_{k}\right \Vert _{L^{\infty }}\leq C_{r}2^{\frac {2}{r}s}\left \Vert f\right \Vert _{L^{\infty }}=C_{r}2^{\frac {2} {r}s}$ for $k=1,2,3$ here, upon appealing to the local single scale $\nu $ -disjoint trilinear assumption $\mathcal {A}_{\mathrm {disj}\nu }^{s,\kappa ,\delta }( \otimes _{3}L^{\infty }\rightarrow L^{\frac {q}{3} };\varepsilon ) $ with $\nu =2^{10}2^{-\lambda }$ , and with

$$\begin{align*}f_{1}=\widetilde{\mathsf{M}_{z_{1}}}f_{I}^{s},\ \ \ f_{2} =\widetilde{\mathsf{M}_{z_{1}}}f_{J}^{s},\ \ \ \ f_{3}=\widetilde{\mathsf{M} _{z_{1}}}f_{K}^{s}, \end{align*}$$

where $\widetilde {\mathsf {M}_{z}}\left ( x\right ) =e^{i\left \langle z,\Phi \left ( x\right ) \right \rangle }$ . Indeed, recall from [Reference Rios and Sawyer4] that if $\mathsf {M}_{z}\left ( w\right ) =e^{i\left \langle z,w\right \rangle }$ , then $\mathsf {M}_{z}\Phi _{\ast }=\Phi _{\ast }\widetilde {\mathsf {M}_{z}}$ and thus,

$$ \begin{align*} \left\vert T_{I}f^{s}\left( \xi-z_{1}\right) \right\vert & =\widehat{\Phi _{\ast}f_{I}^{s}}\left( \xi-z_{1}\right) =\widehat{\mathsf{M}_{z_{1}} \Phi_{\ast}f_{I}^{s}}\left( \xi\right) =\widehat{\Phi_{\ast}\widetilde{\mathsf{M}_{z_{1}}}f_{I}^{s}}\left( \xi\right) =\mathcal{E} f_{1}\left( \xi\right) ,\\\text{and }\left\vert T_{J}f^{s}\left( \xi-z_{2}\right) \right\vert & =\mathcal{E}f_{1}\left( \xi\right) \text{ and }\left\vert T_{K}f^{s}\left( \xi-z_{1}\right) \right\vert =\mathcal{E}f_{3}\left( \xi\right) \text{,} \end{align*} $$

and

$$ \begin{align*} & \int_{B_{\delta}\left( 0,2^{s}\right) }\left\vert T_{I}f^{s}\left( \xi-z_{1}\right) T_{J}f^{s}\left( \xi-z_{2}\right) T_{K}f^{s}\left( \xi-z_{3}\right) \right\vert ^{\frac{q}{3}}d\xi \\& \quad = \int_{B_{\delta}\left( 0,2^{s}\right) }\left\vert \mathcal{E}f_{1}\left( \xi\right) \mathcal{E} f_{2}\left( \xi\right) \mathcal{E}f_{3}\left( \xi\right) \right\vert ^{\frac{q}{3}}d\xi\\& \quad =\left\Vert \prod_{j=1}^{3}\mathcal{E}_{j}f_{j}\right\Vert _{L^{^{\frac {q}{3}}}\left( B_{\delta}\left( 0,2^{s}\right) \right) }^{\frac{q}{3}} \leq\left( C_{\delta,\kappa,q,\varepsilon,\lambda}R^{\varepsilon}\right) ^{q}\prod_{j=1}^{3}\left\Vert f_{j}\right\Vert _{L^{\infty}}^{\frac{q}{3} }\lesssim\left( C_{\delta,\kappa,q,\varepsilon,\lambda,r}2^{\varepsilon s}\right) ^{q}2^{\frac{2q}{r}s}, \end{align*} $$

since the triple $\left ( I,J,K\right ) $ is $\nu $ -separated as in (1.2), and since $\left \vert f_{j}\right \vert \leq 1$ . As a consequence, from the local single scale $\nu $ -disjoint trilinear assumption and since $\nu =\nu \left ( \lambda \right ) $ , we have

$$ \begin{align*} & \sum_{\left( I,J,K\right) \in\mathcal{G}_{\lambda}^{\nu}\left[ U\right] }\int_{B_{\delta}\left( 0,2^{s}\right) }\left\vert T_{I}f^{s}\left( \xi\right) T_{J}f^{s}\left( \xi\right) T_{K}f^{s}\left( \xi\right) \right\vert ^{\frac{q}{3}}d\xi\\& \quad \leq\left( C_{\delta,\varepsilon,\kappa,\lambda,q,r}2^{\varepsilon s}\right) ^{\frac{q}{3}}\sum_{\left( I,J,K\right) \in\mathcal{G}_{\lambda }^{\nu}\left[ U\right] }\left( \left\Vert f^{s}\right\Vert _{L^{\infty }\left( I\right) }\left\Vert f^{s}\right\Vert _{L^{\infty}\left( J\right) }\left\Vert f^{s}\right\Vert _{L^{\infty}\left( K\right) }\right) ^{\frac{q}{3}}d\xi\\& \quad \leq\left( C_{\delta,\kappa,q,\varepsilon,\lambda}2^{\varepsilon s}\right) ^{\frac{q}{3}}C_{r}^{q}~2^{\frac{2q}{r}s}\left( \#\mathcal{G} _{\lambda}^{\nu}\left[ U\right] \right) ^{3}\\& \quad \lesssim C_{\delta ,\kappa,q,\varepsilon,\lambda,r}^{q}2^{6\lambda}2^{\frac{\varepsilon sq}{3} }2^{\frac{2q}{r}s}=C_{\delta,\kappa,q,\varepsilon,\lambda,r}^{q}2^{6\lambda }2^{\frac{q\varepsilon s}{3}}2^{\frac{2q}{r}s}.\\ \end{align*} $$

Altogether then, we have proved that

$$ \begin{align*} \left\Vert \mathbf{1}_{\Gamma_{\lambda}\left( {\mathbf{Case 1}}\right) }T\mathsf{Q}_{\kappa,s,U}f\right\Vert _{L^{q}\left( B_{\delta}\left( 0,2^{s}\right) \right) } & = \left\Vert \textbf{1}_{\Gamma_{\lambda}\left( {\mathbf{Case 1}}\right) }Tf^{s}\right\Vert _{L^{q}\left( B_{\delta }\left( 0,2^{s}\right) \right) } \\& \lesssim C_{\delta,\kappa,q,\varepsilon ,\lambda,r}2^{\left( \frac{6}{q}+4\right) \lambda}2^{\left( \frac {\varepsilon}{3}+\frac{2}{r}\right) s}, \end{align*} $$

where $\textbf {1}_{\Gamma _{\lambda }\left ( {\mathbf {Case 1}}\right ) }$ indicates the union of those balls $B\left ( a,2^{\lambda }\right ) $ for which Case 1 holds. We can absorb the small constant $\frac {2}{r}$ into $\varepsilon $ simply by taking $r$ large ( $r=3/\varepsilon $ ) in (2.2), and the bound we will carry forward to the end of the proof will be $C_{\delta ,\kappa ,q,\varepsilon ,\lambda }2^{\left ( \frac {6}{q}+4\right ) \lambda }2^{\varepsilon s}$ .

Proof continued

In Case 2, we assume the negation of Case 1. Then either there exists a maximal pair of cubes $\left \{ I_{1}^{a},I_{2} ^{a}\right \} $ such that $w_{I_{1}^{a}},w_{I_{2}^{a}}>2^{-2\lambda }w_{\ast }^{a} $ and $\left \vert \textbf {c}_{I_{1}^{a}}-\textbf {c}_{I_{2}^{a} }\right \vert>2^{10}2^{-\lambda }$ , in which case, we have

(3.15) $$ \begin{align} w_{I}\leq2^{-2\lambda}w_{\ast}^{a},\ \ \ \ \ \text{ if }\mathrm{dist} \left( I,I_{1}^{a}\cup I_{2}^{a}\right)>2^{10}2^{-\lambda}, \end{align} $$

or there exist at most one maximal cube $I_{1}^{a}$ such that $w_{I_{1}^{a} }>2^{-2\lambda }w_{\ast }^{a}$ and

$$\begin{align*}w_{I}\leq2^{-2\lambda}w_{\ast}^{a},\ \ \ \ \ \text{ if }\mathrm{dist}\ \mathrm{dist}\left( I,I_{1}^{a}\right)>2^{10}2^{-\lambda}. \end{align*}$$

For simplicity, we consider the former case only, as the proof for the latter is similar. Thus, we assume that (3.15) holds. We will use rescaling and recursion as in [Reference Tao, Vargas and Vega6].

Let $\xi \in B\left ( a,2^{\lambda }\right ) $ for some $a\in \Gamma _{\lambda }\left ( R\right ) $ . Using (3.15), and with $\left \vert \cdot \right \vert _{\mathrm {sq}}$ denoting the “square” norm in $\mathbb {R}^{3}$ ,

(3.16) $$ \begin{align} & \left\vert Tf^{s}\left( \xi\right) \right\vert =\left\vert \sum _{I\in\mathcal{G}_{\lambda}\left[ U\right] }\int_{I}e^{i\phi\left( \xi,y\right) }f^{s}\left( y\right) dy\right\vert \nonumber\\& \quad \leq\sum_{m=1}^{2}\left\vert \sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] :\ \left\vert c_{I}-c_{I_{m}^{a}}\right\vert _{\mathrm{sq} }\leq2^{10}2^{-\lambda}}\int_{I}e^{i\phi\left( \xi,y\right) }f^{s}\left( y\right) dy\right\vert \nonumber\\& \qquad\qquad\qquad\qquad+\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] :\ \mathrm{dist}\left( I,\bigcup_{m=1}^{2}I_{m}^{a}\right)>2^{10}2^{-\lambda}}\left\vert T_{I}f^{s}\left( \xi\right) \right\vert \nonumber\\& \quad \leq C\max_{K\in\mathcal{G}_{\lambda-10}\left[ U\right] }\left\vert \int_{K}e^{i\phi\left( \xi,y\right) }f^{s}\left( y\right) dy\right\vert +\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] :\ \mathrm{dist} \left( I,\bigcup_{m=1}^{2}I_{m}^{a}\right) >2^{10}2^{-\lambda}}\left\vert T_{I}f^{s}\left( \xi\right) \right\vert \nonumber\\& \quad \leq C\max_{K\in\mathcal{G}_{\lambda-10}\left[ U\right] }\left\vert T_{K}f^{s}\left( \xi\right) \right\vert +\left( \#\mathcal{G}_{\lambda }\left[ U\right] \right) 2^{-2\lambda}w_{\ast}^{a}\nonumber\\& \quad \leq C\max_{K\in\mathcal{G}_{\lambda-10}\left[ U\right] }\left\vert T_{K}f^{s}\left( \xi\right) \right\vert +w_{\ast}^{a} ,\nonumber\\ \end{align} $$

since if $K_{m}^{a}$ is the unique cube in $\mathcal {G}_{\lambda -10}\left [ U\right ] $ containing $I_{m}^{a}$ , then

$$ \begin{align*} & \sum_{m=1}^{2}\left\vert \sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] :\ {\left\vert c_{I}-c_{I_{m}^{a}}\right\vert} _{\mathrm{sq}}\leq 2^{10}2^{-\lambda}}\int_{I}e^{i\phi\left( \xi,y\right) }f^{s}\left( y\right) dy\right\vert =\sum_{m=1}^{2}\left\vert \int_{K_{m}^{a}} e^{i\phi\left( \xi,y\right) }f^{s}\left( y\right) dy\right\vert \\& \quad \leq2\max_{K\in{\mathcal{G}}_{\lambda-10}\left[ U\right] }\left\vert \int_{K}e^{i\phi\left( \xi,y\right) }f^{s}\left( y\right) dy\right\vert. \end{align*} $$

We decompose $f^{s}=f_{K_{m}^{a}}^{s}+\sum _{I\in \mathcal {G}_{\lambda }\left [ S\right ] :\ I\cap K_{m}^{a}=\emptyset }f_{I}^{s}$ , where without loss of generality, we may also assume $\left \vert c_{I}-c_{I_{m}^{a}}\right \vert \geq 2^{10-\lambda }$ . Now, since $\zeta _{\lambda }^{a}\left ( z\right ) \equiv \zeta _{\lambda }\left ( z-a\right ) \approx 2^{-3\lambda }$ in $B\left ( a,2^{\lambda }\right ) $ , we have

$$ \begin{align*} \int\left\vert T_{I}f^{s}\left( \xi-z\right) \right\vert \zeta_{\lambda} ^{a}\left( z\right) dz & \leq\left( \int\left\vert T_{I}f^{s}\left( \xi-z\right) \right\vert ^{q}\zeta_{\lambda}\left( z-a\right) dz\right) ^{\frac{1}{q}}\left( \int\zeta_{\lambda}\left( z-a\right) dz\right) ^{\frac{1}{q^{\prime}}}\\& \lesssim\left( \int\left\vert T_{I}f^{s}\left( z\right) \right\vert ^{q}\zeta_{\lambda}\left( z-a\right) dz\right) ^{\frac{1}{q}}, \end{align*} $$

and so for $\xi \in B\left ( a,2^{\lambda }\right ) $ , from (3.16), we obtain

$$ \begin{align*} \left\vert Tf^{s}\left( \xi\right) \right\vert ^{q} & \leq C\sum _{K\in\mathcal{G}_{\lambda-10}\left[ U\right] }\left\vert T_{K}f^{s}\left( \xi\right) \right\vert ^{q}+C\int\left\vert T_{I_{\ast}^{a}}f^{s}\left( z\right) \right\vert ^{q}\zeta_{\lambda}\left( z-a\right) dz\\& \leq C\sum_{K\in\mathcal{G}_{\lambda-10}\left[ U\right] }\left\vert T_{K}f^{s}\left( \xi\right) \right\vert ^{q}+C\sum_{I\in\mathcal{G} _{\lambda}\left[ U\right] }\int\left\vert T_{I}f^{s}\left( z\right) \right\vert ^{q}\zeta_{\lambda}\left( z-a\right) dz, \end{align*} $$

where $C$ is independent of all the relevant parameters, and where we have added back in all $K\in \mathcal {G}_{\lambda -10}\left [ U\right ] $ rather than just $K_{\ast }^{a}$ , and all $I\in \mathcal {G}_{\lambda }\left [ U\right ] $ rather than just $I_{\ast }^{a}$ , in order that the dominating expression on the right-hand side is independent of $a\in \Gamma _{\lambda }\left ( R\right ) $ .

Summing over $a\in \Gamma _{\lambda }\left ( R\right ) $ , we see that the corresponding contribution over $B_{R}$ is at most

(3.17) $$ \begin{align} & \left\Vert \textbf{1}_{\Gamma_{\lambda}\left( {\mathbf{Case 2} }\right) }Tf^{s}\right\Vert _{L^{q}\left( B_{\delta}\left( 0,2^{s}\right) \right) }^{q}\\& \quad \lesssim\sum_{a\in\Gamma_{\lambda}}\left\{ \sum_{K\in\mathcal{G} _{\lambda-10}\left[ U\right] }\int_{B\left( a,2^{\lambda}\right) }\left\vert T_{K}f^{s}\left( \xi\right) \right\vert ^{q}d\xi+~2^{3\lambda }\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] }\int\left\vert T_{I} f^{s}\left( z\right) \right\vert ^{q}\zeta_{\lambda}^{a}\left( z\right) dz\right\} \nonumber\\& \quad \lesssim\sum_{K\in\mathcal{G}_{\lambda-10}\left[ U\right] } \int_{\mathbb{R}^{3}}\left\vert T_{K}f^{s}\left( \xi\right) \right\vert ^{q}d\xi+\sum_{I\in\mathcal{G}_{\lambda}\left[ U\right] }\int_{\mathbb{R} ^{3}}\left\vert T_{I}f^{s}\left( \xi\right) \right\vert ^{q}d\xi,\nonumber \end{align} $$

since $\sum _{a\in \Gamma _{\lambda }}\zeta _{\lambda }\left ( z-a\right ) \lesssim 2^{-3\lambda }\textbf {1}_{\mathbb {R}^{3}}\left ( z\right ) +\mathrm {rapid}\mathrm {decay}$ .

At this point, we follow [Reference Bourgain and Guth1] in using parabolic rescaling, as introduced in Tao et al. [Reference Tao, Vargas and Vega6], on the integral

$$\begin{align*}\mathrm{Int}_{\varrho}\left( \xi,\overline{y}\right) \equiv \int_{\left\vert y-\overline{y}\right\vert _{\mathrm{sq}}<\varrho }e^{i\phi\left( \xi,y\right) }f^{s}\left( y\right) dy=\int_{\left\vert y-\overline{y}\right\vert _{\mathrm{sq}}<\varrho}e^{i\left[ \xi _{1}y_{1}+\xi_{2}y_{2}+\xi_{3}\left( y_{1}^{2}+y_{2}^{2}\right) \right] }f^{s}\left( y\right) dy, \end{align*}$$

for $0<\varrho <1$ ; we substitute $y=\overline {y}+y^{\prime }$ to obtain

$$ \begin{align*} \left\vert {\mathrm{Int}}_{\varrho}\left( \xi,\overline{y}\right) \right\vert & =\left\vert \int_{{\left\vert y^{\prime}\right\vert} _{\mathrm{sq}}<\varrho}e^{i\left[ \xi_{1}\left( \overline{y} _{1}+y_{1}^{\prime}\right) +\xi_{2}\left( \overline{y}_{2}+y_{2}^{\prime }\right) +\xi_{3}\left( \left( \overline{y}_{1}+y_{1}^{\prime}\right) ^{2}+\left( \overline{y}_{2}+y_{2}^{\prime}\right) ^{2}\right) \right] }f^{s}\left( \overline{y}+y^{\prime}\right) dy^{\prime}\right\vert \\ & =\left\vert \int_{{\left\vert y^{\prime}\right\vert} _{\mathrm{sq} }<\varrho}e^{i\left[ \left( \xi_{1}+2\overline{y_{1}}\xi_{3}\right) y_{1}^{\prime}+\left( \xi_{2}+2\overline{y_{2}}\xi_{3}\right) y_{2}^{\prime }+\xi_{3}{\left\vert y^{\prime}\right\vert} ^{2}\right] }f^{s}\left( \overline{y}+y^{\prime}\right) dy^{\prime}\right\vert. \end{align*} $$

Thus, we conclude that

(3.18) $$ \begin{align} & {\left\Vert \mathrm{Int}_{\varrho}\right\Vert} _{L^{q}\left( B_{R}\right) }=\left( \int_{B_{R}}\left\vert \mathrm{Int}_{\varrho}\left( \xi,\overline{y}\right) \right\vert ^{q}d\xi\right) ^{\frac{1}{q} }\\& \quad =\left( \int_{B_{R}}\left\vert \int_{\left\vert y^{\prime}\right\vert _{\mathrm{sq}}<\varrho}e^{i\left[ \left( \xi_{1}+2\overline{y_{1}} \xi_{3}\right) y_{1}^{\prime}+\left( \xi_{2}+2\overline{y_{2}}\xi _{3}\right) y_{2}^{\prime}+\xi_{3}\left\vert y^{\prime}\right\vert ^{2}\right] }f^{s}\left( \overline{y}+y^{\prime}\right) dy^{\prime }\right\vert ^{q}d\xi\right) ^{\frac{1}{q}}\nonumber\\& \quad =\left(\! \int_{B_{R}}\!\left\vert \int_{\left\vert y^{\prime}\right\vert _{\mathrm{sq}}<\varrho}\! e^{i\left[ \left( \varrho\xi_{1} +2\overline{\frac{y_{1}}{\varrho}}\varrho^{2}\xi_{3}\right) \frac {y_{1}^{\prime}}{\varrho}+\left( \varrho\xi_{2}+2\overline{\frac{y_{2} }{\varrho}}\varrho^{2}\xi_{3}\right) \frac{y_{2}^{\prime}}{\varrho} +\varrho^{2}\xi_{3}\left\vert \frac{y^{\prime}}{\varrho}\right\vert ^{2}\right] }f^{s}\left( \varrho\frac{\overline{y}+y^{\prime}}{\varrho }\right) \varrho^{2}d\left( \frac{y^{\prime}}{\varrho}\right) \right\vert ^{q}\right. \nonumber\\& \qquad\quad \left. \frac{d\left( \varrho\xi^{\prime}\right) d\left( \varrho^{2}\xi_{3}\right) }{\varrho^{4}}\right) ^{\frac{1}{q} }\nonumber\\& \quad =\varrho^{2}\varrho^{-\frac{4}{q}}\left( \int_{B_{\varrho R}}\!\left\vert \int_{\left\vert y^{\prime}\right\vert _{\mathrm{sq}}<1}e^{i\left[ \left( \xi_{1}+2\overline{y_{1}}\xi_{3}\right) y_{1}^{\prime}+\left( \xi_{2}+2\overline{y_{2}}\xi_{3}\right) y_{2}^{\prime}+\xi_{3}\left\vert y^{\prime}\right\vert ^{2}\right] }f^{s}\left( \varrho\left( \overline {y}+y^{\prime}\right) \right) dy^{\prime}\right\vert ^{q}d\xi^{\prime}\! d\xi_{3}\right) ^{\frac{1}{q}}\nonumber\\& \quad \leq C\rho^{2}\rho^{-\frac{4}{q}}P_{\kappa,\delta^{\prime}}^{\left( q\right) }\left( s-m\right) ,\nonumber \end{align} $$

by taking $\varrho =2^{-m}$ , $m\geq 0$ , $s\geq m$ , and $\delta ^{\prime }$ is given by the identity $2^{\frac {s-m}{1-\delta ^{\prime }}}=2^{-m}2^{\frac {s}{1-\delta }}$ , that is,

$$\begin{align*}\text{ }\delta^{\prime}=\frac{\delta}{1-\frac{m}{s}\left( 1-\delta\right) }\text{ for }s>m, \end{align*}$$

so $\delta <\delta ^{\prime }<1$ . In (3.18), we applied Corollary 3.3 to

$$\begin{align*}f^{s}\left( \varrho\left( \overline{y}+y^{\prime}\right) \right) =\mathsf{Q}_{\kappa,s,U}f\left( 2^{-\tilde{m}}\left( \overline{y}+y^{\prime }\right) \right), \end{align*}$$

which says that $f^{s}\left ( 2^{-m}\left ( \overline {y}+y^{\prime }\right ) \right ) $ belongs to $\mathcal {F}_{\kappa ,\infty }\left ( s-m\right ) $ for $\varrho =2^{-m}$ , $m\geq 0$ . We also used that

$$\begin{align*}\delta_{m,\infty}f^{s}\left( x\right) \equiv f^{s}\left( 2^{-m}x\right) =f^{s}\left( \varrho x\right) \end{align*}$$

has the same $L^{\infty }$ norm as $f^{s}$ , and finally that the paraboloid is invariant under parabolic rescaling. Note that the factor $\varrho ^{2}$ arises from $\left \vert y^{\prime }\right \vert <\varrho $ , and that the factor $\varrho ^{-\frac {4}{q}}$ arises from parabolic rescaling. These features remain in play for an arbitrary quadratic surface of positive Gaussian curvature.

Thus, using (3.18), and the identity

$$\begin{align*}\left\vert T_{I}f^{s}\left( \xi\right) \right\vert =\left\vert \mathrm{Int}_{\frac{1}{2}\ell\left( I\right) }\left( \xi ,c_{I}\right) \right\vert , \end{align*}$$

first with $\varrho =\frac {1}{2}2^{10-\lambda }$ and then with $\varrho =\frac {1}{2}2^{-\lambda }$ , together with (3.17), we obtain that the contribution $\left \Vert \textbf {1}_{\Gamma _{\lambda }\left ( {\mathbf {Case 2}}\right ) }Tf^{s}\right \Vert _{L^{q}\left ( B_{\delta }\left ( 0,2^{s}\right ) \right ) }^{q}$ to the norm $\left \Vert Tf^{s}\right \Vert _{L^{q}\left ( B_{\delta }\left ( 0,2^{s}\right ) \right ) }$ satisfies:

(3.19) $$ \begin{align} & \left\Vert \textbf{1}_{\Gamma_{\lambda}\left( {\mathbf{Case 2} }\right) }T\mathsf{Q}_{\kappa,s,U}f\right\Vert _{L^{q}\left( B_{\delta }\left( 0,2^{s}\right) \right) }=\left\Vert \textbf{1}_{\Gamma_{\lambda }\left( {\mathbf{Case 2}}\right) }Tf^{s}\right\Vert _{L^{q}\left( B_{\delta}\left( 0,2^{s}\right) \right) }\\& \quad \leq C\left( \#\mathcal{G}_{\lambda-10}\left[ U\right] \right) ^{\frac{1}{q}}\left( 2^{10-\lambda-1}\right) ^{2-\frac{4}{q}}P_{\kappa ,\delta}^{\left( q\right) }\left( s+10-\lambda\right) \nonumber\\& \qquad +C\left( \#\mathcal{G}_{\lambda}\left[ U\right] \right) ^{\frac{1}{q} }\left( 2^{-\lambda-1}\right) ^{2-\frac{4}{q}}P_{\kappa,\delta}^{\left( q\right) }\left( s-\lambda\right) \nonumber\\& \quad =C~2^{2\left( 10-\lambda\right) \frac{q-3}{q}}P_{\kappa,\delta}^{\left( q\right) }\left( s+10-\lambda\right) +C~2^{-2\lambda\frac{q-3}{q}} P_{\kappa,\delta}^{\left( q\right) }\left( s-\lambda\right) ,\nonumber \end{align} $$

for $s>\lambda $ , since, for $q>3$ ,

$$\begin{align*}\left( \#\mathcal{G}_{\lambda-10}\left[ U\right] \right) ^{\frac{1}{q} }\left( 2^{10-\lambda-1}\right) ^{2-\frac{4}{q}}=2^{\frac{2}{q}\left( 10-\lambda\right) }2^{\left( 10-\lambda-1\right) \left( 2-\frac{4} {q}\right) }\leq2^{2\left( 10-\lambda\right) \frac{q-3}{q}}, \end{align*}$$

and

$$\begin{align*}\left( \#\mathcal{G}_{\lambda}\left[ U\right] \right) ^{\frac{1}{q} }\left( 2^{-\lambda-1}\right) ^{2-\frac{4}{q}}=2^{\frac{2}{q}\lambda }2^{-\left( \lambda+1\right) \left( 2-\frac{4}{q}\right) }\leq 2^{-2\lambda\frac{q-3}{q}}.\\[-33pt] \end{align*}$$

Proof continued

Recall from (3.1) that

$$\begin{align*}P_{\kappa,\delta}^{\left( q\right) }\left( s\right) \equiv\sup _{\mathcal{G}\in\mathbb{G}}\sup_{\left\Vert f\right\Vert _{L^{\infty}\left( U\right) }\leq1}\left( \int_{B_{\delta}\left( 0,2^{s}\right) }\left\vert \mathcal{E}\left( \Phi_{\ast}\mathsf{Q}_{\kappa,s,U}^{\eta,\mathcal{G} }f\right) \left( \xi\right) \right\vert ^{q}d\xi\right) ^{\frac{1}{q}}. \end{align*}$$

We now define the quantity

$$\begin{align*}P_{\kappa,\delta,\ast}^{\left( q\right) }\left( s\right) \equiv \sup_{\mathcal{G}\in\mathbb{G}}\sup_{\left\Vert f\right\Vert _{L^{\infty }\left( U\right) }\leq1}\left( \int_{\mathbb{R}^{3}\setminus B_{\delta }\left( 0,2^{s}\right) }\left\vert \mathcal{E}\left( \Phi_{\ast} \mathsf{Q}_{\kappa,s,U}^{\eta,\mathcal{G}}f\right) \left( \xi\right) \right\vert ^{q}d\xi\right) ^{\frac{1}{q}}, \end{align*}$$

where we are integrating away from the ball $B\left ( 0,2^{s}\right ) $ . As a consequence, we can use the estimates in [Reference Rios and Sawyer4, see Case 2 of the proof of Theorem 10] to show that

$$\begin{align*}P_{\kappa,\delta,\ast}^{\left( q\right) }\left( s\right) \lesssim C_{\kappa,q,\delta,N}2^{-N\delta s}\leq C_{\kappa,q,p,N}, \end{align*}$$

for some large integer $N$ (arising from integration by parts in [Reference Rios and Sawyer4]). So far, remembering that we have absorbed the penalty factor $2^{\frac {2}{r} s}$ into $2^{\varepsilon s}$ in the Case 1 bound, we have shown that for $s>\lambda $ , we have

$$ \begin{align*} & \left\Vert T\mathsf{Q}_{\kappa,s,U}f\right\Vert _{L^{q}\left( B_{\delta }\left( 0,2^{s}\right) \right) } \\& \quad \leq\left\Vert \textbf{1}_{\Gamma_{\lambda }\left( {\mathbf{Case 1}}\right) }T\mathsf{Q}_{\kappa,s,U}f\right\Vert _{L^{q}\left( B_{\delta}\left( 0,2^{s}\right) \right) }+\left\Vert \textbf{1}_{\Gamma_{\lambda}\left( {\mathbf{Case 2}}\right) }T\mathsf{Q}_{\kappa,s,U}f\right\Vert _{L^{q}\left( B_{\delta}\left( 0,2^{s}\right) \right) }\\& \quad \lesssim C_{\varepsilon}2^{\left( 2+\alpha\right) \lambda}2^{\frac{6} {q}\lambda}2^{\varepsilon s}+C2^{\left( \frac{6}{q}-2\right) \lambda }P_{\kappa,2\delta}^{\left( q\right) }\left( s+10-\lambda\right) +C2^{\left( \frac{3}{q}+\alpha-4\right) \lambda}P_{\kappa,\delta^{\prime} }^{\left( q\right) }\left( s-\lambda\right) \\& \quad \lesssim C_{\varepsilon,N}2^{\left( 2+\alpha\right) \lambda}2^{\frac {6}{q}\lambda}2^{\varepsilon s}+C\left[ 2^{\left( \frac{6}{q}-2\right) \lambda}+2^{\left( \frac{3}{q}+\alpha-4\right) \lambda}\right] \max_{1\leq s^{\prime}\leq s+10-\lambda}P_{\kappa,\delta^{\prime}}^{\left( q\right) }\left( s^{\prime}\right) \\& \quad \lesssim C_{\varepsilon}2^{\left( 2+\alpha\right) \lambda}2^{\frac{6} {q}\lambda}2^{\varepsilon s}+\frac{1}{2}\max_{1\leq s^{\prime}\leq\lambda }P_{\kappa,\delta}^{\left( q\right) }\left( s^{\prime}\right) +\frac{1} {2}\max_{\lambda<s^{\prime}\leq s+10-\lambda}P_{\kappa,\delta^{\prime} }^{\left( q\right) }\left( s^{\prime}\right) \\& \quad \lesssim C_{\varepsilon}2^{\left( 2+\alpha\right) \lambda}2^{\frac{6} {q}\lambda}2^{\varepsilon s}+C_{\lambda,\kappa,\delta,q}+\frac{1}{2} \max_{\lambda\leq s^{\prime}\leq s+10-\lambda}P_{\kappa,\delta}^{\left( q\right) }\left( s^{\prime}\right) +\frac{1}{2}\max_{\lambda<s^{\prime}\leq s+10-\lambda}P_{\kappa,\delta,\ast}^{\left( q\right) }\left( s^{\prime }\right) , \end{align*} $$

provided $q>3$ and $\lambda $ is sufficiently large, namely,

(3.20) $$ \begin{align} 2^{\left( \frac{6}{q}-2\right) \lambda}+2^{\left( \frac{3}{q} +\alpha-4\right) \lambda}<\frac{1}{2C}, \end{align} $$

where $C$ is a constant arising from the early stages of the proof in Case 2. Thus, for $t>\lambda ,$ we have

$$ \begin{align*} & \max_{\lambda\leq s\leq t}P_{\kappa,\delta}^{\left( q\right) }\left( s\right) =\max_{\lambda\leq s\leq t}\sup_{\left\Vert f\right\Vert _{L^{\infty}}\leq1}\left\Vert T\mathsf{Q}_{\kappa,s,U}f\right\Vert _{L^{q}\left( B_{\delta}\left( 0,2^{s}\right) \right) }\\& \quad \lesssim C_{\varepsilon,N}2^{\left( 2+\alpha\right) \lambda}2^{\frac {6}{q}\lambda}\max_{\lambda<s\leq t}2^{\varepsilon s}+C_{\lambda,\kappa ,\delta,q}+\frac{1}{2}\max_{\lambda\leq s^{\prime}\leq s+10-\lambda} P_{\kappa,\delta}^{\left( q\right) }\left( s^{\prime}\right) \\&\qquad +\frac{1} {2}\max_{\lambda<s^{\prime}\leq s+10-\lambda}P_{\kappa,\delta,\ast}^{\left( q\right) }\left( s\right) \\& \quad \lesssim C_{\varepsilon,N}2^{\left( 2+\alpha\right) \lambda}2^{\frac {6}{q}\lambda}\max_{\lambda<s\leq t}2^{\varepsilon s}+C_{\lambda,\kappa ,\delta,q}+\frac{1}{2}\max_{\lambda\leq s^{\prime}\leq s+10-\lambda} P_{\kappa,\delta}^{\left( q\right) }\left( s\right) , \end{align*} $$

and absorption now yields the inequality,

$$\begin{align*}P_{\kappa,\delta}^{\left( q\right) }\left( t\right) \leq\max _{\lambda<s\leq t}P_{\kappa,\delta}^{\left( q\right) }\left( s\right) \lesssim C_{\lambda,\kappa,\delta,q}+C_{\varepsilon}2^{\left( 2+\alpha \right) \lambda}2^{\frac{6}{q}\lambda}2^{\varepsilon t},\ \ \ \ \ \text{for all }t>\lambda. \end{align*}$$

If we take $\alpha =2$ and write $C=2^{c-1}$ , then (3.20) becomes,

$$\begin{align*}2^{-\frac{2}{q}\left( q-3\right) \lambda}+2^{-\frac{2}{q}\left( q-\frac {3}{2}\right) \lambda}<\frac{1}{2C}=2^{-c}, \end{align*}$$

which is satisfied if both

$$ \begin{align*} \left( \frac{2}{q}\left( q-3\right) \right) \lambda &>c+1\text{ and }\left( \frac{2}{q}\left( q-\frac{3}{2}\right) \right) \lambda >c+1\text{,}\\ \text{and in particular if }\lambda & >\frac{c+1}{2}\frac{q}{q-3}\text{ and }q>3, \end{align*} $$

This shows that the “disjoint” parameter $\nu $ may be taken as

(3.21) $$ \begin{align} \nu<2^{10-\lambda}=2^{10}2^{-\frac{c+1}{2}\frac{q}{q-3}}, \end{align} $$

and so depends only on how much larger $q$ is than $3$ . In conclusion, we have

$$ \begin{align*} P_{\kappa,\delta}^{\left( q\right) }\left( s\right) & \leq C_{\lambda,\kappa,\delta,q}+C_{\varepsilon,\nu}2^{4\lambda}2^{2\left( \lambda-10\right) }2^{\varepsilon s}\leq C_{\lambda,\kappa,\delta ,q}+C_{\varepsilon,\nu}2^{6\lambda}2^{\varepsilon s}\\ & =C_{\lambda,\kappa,\delta,q}+C_{\varepsilon,\nu}2^{60}\nu^{-6} 2^{\varepsilon s}=C_{\lambda,\kappa,\delta,q}+C_{\varepsilon,\nu }2^{\varepsilon s}, \end{align*} $$

for all $s>\lambda $ and $0<\nu <2^{10}2^{-\frac {c+1}{2}\frac {q}{q-3}}$ , which completes the proof of Theorem 3.1 since

$$\begin{align*}s>\lambda>\frac{c+1}{2}\frac{q}{q-3}\equiv C_{0}\frac{q}{q-3}.\\[-33pt] \end{align*}$$