# An endpoint estimate of the bilinear paraboloid restriction operator

Jungjin Lee  Department of Mathematical Sciences, School of Natural Science,
Ulsan National Institute of Science and Technology, UNIST-gil 50, Ulsan 44919,
Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

In Fourier restriction problems, a cone and a paraboloid are model surfaces.
The sharp bilinear cone restriction estimate was first shown by Wolff, and
later the endpoint was obtained by Tao. For a paraboloid, the sharp $L^{2}$
bilinear restriction estimate was obtained by Tao, but the endpoint was
remained open. In this paper we prove the endpoint $L^{2}$ bilinear
restriction estimate for a paraboloid.

###### Key words and phrases:

endpoint Fourier restriction estiamtes

###### 2010 Mathematics Subject Classification:

42B15, 42B20

This research was supported by Basic Science Research Program through the
National Research Foundation of Korea(NRF) funded by the Ministry of Education
(NRF-2020R1I1A1A01062209)

##  1\. Introduction

Fix $n\geq 2$, let $\Sigma$ be a hypersurface defined by
$\Sigma=\\{(\xi,s(\xi)):\xi\in\mathbb{R}^{n}\\}$. Then the (adjoint) Fourier
restriction operator $\mathcal{R}_{\Sigma}f$ for the hypersurface $\Sigma$ can
be defined by

$\mathcal{R}_{\Sigma}f(x,t):=\int e^{2\pi
i(x\cdot\xi+ts(\xi))}f(\xi)a(\xi)d\xi,$

where $a(\xi)$ is a smooth cut-off function.

The (adjoint) restriction estimate $\mathds{R}^{*}_{\Sigma}(p,q)$ for $1\leq
p,q\leq\infty$ is of the form

$\|\mathcal{R}_{\Sigma}f\|_{q}\leq C_{p,q,\Sigma}\|f\|_{p},$ (1.1)

and the restriction problem is to determine $(p,q)$ for which the estimate
$\mathds{R}^{*}_{\Sigma}(p,q)$ holds. There are two representative model
hypersurfaces. One is a cone
$\Sigma_{cone}=\\{(\xi,|\xi|):\xi\in\mathbb{R}^{n}\\}$, and the other is a
paraboloid
$\Sigma_{parab}=\\{(\xi,-\frac{|\xi|^{2}}{2}):\xi\in\mathbb{R}^{n}\\}$. For
these two surfaces the restriction operators $\mathcal{R}_{\Sigma_{cone}}f$
and $\mathcal{R}_{\Sigma_{parab}}f$ are related to other problems such as the
Bochner-Riesz conjecture, Kakeya conjecture and Sogge’s local smoothing
conjecture, see [20, 12, 22, 17, 28, 2]. Moreover, they are also connected to
the wave and Schrödinger equations because
$\mathcal{R}_{\Sigma_{cone}}\hat{f}$ and $\mathcal{R}_{\Sigma_{parab}}\hat{f}$
are the solutions to the free wave equation $u_{tt}-\Delta u=0$ and the free
Schrödinger equation $4\pi i\partial_{t}u-\Delta u=0$, respectively, see [22,
23, 15].

The bilinear restriction estimate
$\mathds{R}^{*}_{\Sigma_{1},\Sigma_{2}}(p\times p,q)$, $1\leq p,q\leq\infty$
is of the form

$\|\mathcal{R}_{\Sigma_{1}}f\mathcal{R}_{\Sigma_{2}}g\|_{q}\leq
C_{p,q,\Sigma_{1},\Sigma_{2}}\|f\|_{p}\|g\|_{p},$

where $\Sigma_{1},\Sigma_{2}$ are two compact subsets of $\Sigma$ such that
the set of unit normal vectors of $\Sigma_{1}$ are separated by a non-zero
distance from the set of unit normal vectors of $\Sigma_{2}$. This bilinear
restriction estimate $\mathds{R}^{*}_{\Sigma_{1},\Sigma_{2}}(p\times p,q/2)$
was used to improve the linear restriction estimate
$\mathds{R}^{*}_{\Sigma}(p,q)$. (The restriction estimate have been improved
further by Bourgain–Guth [4], Guth [5, 6], Wang [26], Hickman–Rogers [8].) In
addition, as the relation between the Stein-Tomas restriction theorem and the
Strichartz estimate, the bilinear restriction estimates
$\mathds{R}^{*}_{\Sigma_{1},\Sigma_{2}}(2\times 2,q)$ lead to the
corresponding bilinear estimates applied to null form estimates for the
relevant dispersive equations, see [16, 22, 23, 7, 9, 10].

The $L^{2}$ bilinear restriction estimate is based on the argument of Wolff
[27] for a cone. His arguments are roughly composed of two steps. One is to
use induction to avoid some critical case of the Kakeya set. The other is to
deal with the remaining relaxed Kakeya set by utilizing some geometrical
observation as follows:

The union $\Lambda(x_{0})$ of all lines passing through a point $x_{0}$ and of
direction normal to $\Sigma_{cone}$ becomes a cone. (1.2)

We can see that if a line $\ell$ passes through $\Lambda(x_{0})$, then
$\ell\cap\Lambda(x_{0})$ has at most $O(1)$ points, which is the key to obtain
the sharp bilinear cone restriction.

However, in a paraboloid the analogous property does not hold. Specifically,
the union $\Lambda(x_{0})$ of all lines passing through $x_{0}$ and of
direction normal to $\Sigma_{parab}$ does not contained in a hypersurface. The
reason is that while the cone has one vanishing principle curvature, the
paraboloid has non-vanishing Gaussian curvature. Thus,
$\ell\cap\Lambda(x_{0})$ may have infinitely many points. Because of this
difference, Wolff’s argument cannot be directly applied to the paraboloid
case.

This difficulty was resolved by Tao [24] who used a kind of orthogonality due
to the non-vanishing curvature. Such an orthogonality was first observed in
the proof of the 2-dimensional restriction theorem by Fefferman and Córdoba.
By combining Wolff’s arguments with the orthogonality Tao obtained the sharp
bilinear restriction estimate $\mathds{R}^{*}_{\Sigma_{1},\Sigma_{2}}(2\times
2,p)$, $p>\frac{n+3}{n+1}$ for a paraboloid.

It is natural to ask whether the endpoint bilinear estimate
$\mathds{R}^{*}_{\Sigma_{1},\Sigma_{2}}(2\times 2,\frac{n+3}{n+1})$ is valid
or not. Since the Kakeya example does not work in the endpoint $L^{2}$
bilinear restriction estimate, we can expect it. Tao [23] obtained the
endpoint bilinear cone restriction estimate by exploring energy concentrations
and the geometric observation (1.2). If one makes an attempt to prove the
endpoint bilinear restriction estimate for a paraboloid, it is reasonable,
first of all, to apply Tao’s arguments, but the geometric observation (1.2)
does not hold for the paraboloid restriction operator. However, it still seems
to have the $L^{2}$ bilinear paraboloid restriction estimate because the
Kakeya example does not work.

In this paper we will prove the endpoint estimate
$\mathds{R}^{*}_{\Sigma_{1},\Sigma_{2}}(2\times 2,\frac{n+3}{n+1})$ for a
paraboloid. To state more explicitly, let
$\mathit{\Sigma}=\mathit{\Sigma}_{parab}$ and

$\displaystyle\mathit{\Sigma}_{j}$
$\displaystyle=\\{(\xi,\tau)\in\mathit{\Sigma}:1<|\xi|<2,\
\angle(\xi,(-1)^{j-1}e_{1})<\pi/8\\},\qquad j=1,2,$

where $e_{1}\in\mathbb{R}^{n}$ is a standard unit vector. We define the
operator $\mathcal{U}_{j}$ by
$\mathcal{U}_{j}f=\mathcal{R}_{\Sigma_{j}}\hat{f}$ for $j=1,2$;

$\mathcal{U}_{j}f(x,t):=\int e^{2\pi
i(x\cdot\xi-\frac{1}{2}t|\xi|^{2})}\widehat{f}(\xi)a_{j}(\xi)d\xi,$ (1.3)

where $a_{j}$ is a smooth function which is equal to $1$ on

$\Xi_{j}=\\{\xi\in\mathbb{R}^{n}:(\xi,\tau)\in\mathit{\Sigma}_{j}\\}$

and supported in

$\tilde{\Xi}_{j}=\\{\xi:1/2\leq|\xi|\leq 4,\
\angle(\xi,(-1)^{j-1}e_{1})\leq\pi/4\\}.$

######  Theorem 1.1.

For $\frac{n+3}{n+1}\leq p\leq 2$, the estimate

$\|\mathcal{U}_{1}f_{1}\mathcal{U}_{2}f_{2}\|_{L^{p}(\mathbb{R}^{n}\times\mathbb{R})}\leq
C_{p}\|f_{1}\|_{L^{2}(\Xi_{1})}\|f_{2}\|_{L^{2}(\Xi_{2})}$ (1.4)

holds for all $f_{1}\in L^{2}(\Xi_{1})$ and $f_{2}\in L^{2}(\Xi_{2})$.

To obtain the endpoint we basically follow the arguments in [23]. We first
reprove the sharp bilinear restriction estimate for a paraboloid without
dyadic pigeonholing. Next we use an energy concentration argument. But, as
mentioned above, we cannot directly apply Tao’s endpoint argument to the
paraboloid problem because of the lack of (1.2). To get around this we will
devise a new energy concentration argument where the dispersiveness of
$\mathcal{U}_{j}f$ and the transversality between $\Sigma_{1}$ and
$\Sigma_{2}$ play a crucial role instead of the geometric observation (1.2).

The Fourier restriction operator can be generalized to some oscillatory
integral operator. The cone restriction operator is generalized to the
oscillatory integrals satisfying the cinematic curvature condition, and
similarly the paraboloid one generalized to the oscillatory integrals with the
Carleson–Sjölin condition, see [18, 19, 14, 13, 25]. It was shown by S. Lee
[14] that these two classes of oscillatory integral operators also satisfy the
estimate (1.4) for $p>\frac{n+3}{n+1}$, provided a suitable transversality
condition is assumed, see [14, 13, 11, 25]. For the oscillatory integrals with
the cinematic curvature condition, the endpoint bilinear estimate was obtained
by the author [11]. It is likely that the oscillatory integral operators with
the Carleson–Sjölin condition have the endpoint bilinear estimate, too.

Notation. Let $N>1$ be a large integer depending only on the dimension $n$,
which is used as a large exponent of the error terms. Let $C_{0}$ be an
integer much larger than $N$.

We use $C$ to denote various large numbers which vary each time. It may depend
on $N$ but not depend on $C_{0}$. The notation $A\lesssim B$ or $A=O(B)$
implies $A\leq CB$. If $A\lesssim B$ and $B\lesssim A$ we write $A\sim B$.

When $\phi(x,t)$ is a space-time function, let $\phi(t)$ denote the spatial
function $\phi(t)(x)=\phi(x,t)$. The hat $~{}\widehat{}~{}$ notation is used
for the Fourier transform

$\widehat{f}(\xi)=\int_{\mathbb{R}^{n}}e^{-2\pi ix\cdot\xi}f(x)dx.$

Let a spacetime cube $Q(x_{Q},t_{Q};R_{Q})$ be an $(n+1)$-dimensional cube in
$\mathbb{R}^{n+1}$ of side-length $R_{Q}$ and centered at
$(x_{Q},t_{Q})\in\mathbb{R}^{n}\times\mathbb{R}$ with sides parallel to the
axes. Let $D(x_{D},t_{D};r_{D})$ denote an $n$-dimensional disc at time
$t_{D}$ of the form

$D(x_{D},t_{D};r_{D})=\\{(x,t_{D})\in\mathbb{R}^{n}\times\mathbb{R}:|x-x_{D}|\leq
r_{D}\\}.$

For any compact subset $\pi\subset\mathbb{R}^{n}$ we define a conic set
$\Lambda_{\pi}(x_{0},t_{0})$ with vertex
$(x_{0},t_{0})\in\mathbb{R}^{n}\times\mathbb{R}$ by

$\Lambda_{\pi}(x_{0},t_{0})=\left\\{(x,t)\in\mathbb{R}^{n}\times\mathbb{R}:x=x_{0}+(t-t_{0})w,\
w\in\pi,\ t\in\mathbb{R}\right\\},$ (1.5)

and let $\Lambda_{\pi}(x_{0},t_{0};r)$ be the $r$-neighborhood of
$\Lambda_{\pi}(x_{0},t_{0})$. Briefly we write

$\Lambda_{j}(x_{0},t_{0})=\Lambda_{\tilde{\Xi}_{j}}(x_{0},t_{0})$ (1.6)

for $j=1,2$, and

$\Lambda_{\cup}(x_{0},t_{0})=\Lambda_{1}(x_{0},t_{0})\cup\Lambda_{2}(x_{0},t_{0}).$

If $Q=Q(x_{Q},t_{Q};r_{Q})$ and $c>0$ the $cQ$ is defined by
$Q(x_{Q},t_{Q};cr_{Q})$. Similarly, $cD$ is defined.

Let $\eta$ be a nonnegative Schwartz function on $\mathbb{R}^{n}$ with
$\int\eta=1$ and whose Fourier transform is supported on the unit disc. By the
Poisson summation formula we may have

$\sum_{k\in\mathbb{Z}^{n}}\eta(x-k)=1.$ (1.7)

We define $\eta_{r}$ for $r>0$ by

$\eta_{r}(x)=r^{-n}\eta(x/r).$ (1.8)

##  2\. Proof of Theorem 1.1

In this section we state some propositions and using them we prove Theorem
1.1. The proof of propositions are given in next sections.

We denote by

$\phi_{j}(x,t)=\mathcal{U}_{j}f_{j}(x,t),$

and define the energy $E(\phi_{j})$ by

$E(\phi_{j})=\|\phi_{j}(t)\|_{2}^{2}$ (2.1)

where $t\in\mathbb{R}$ is arbitrary. It makes sense by Plancherel’s theorem
and

$\widehat{\phi_{j}(t)}(\xi)=e^{-\pi it|\xi|^{2}}\hat{f}_{j}(\xi)a_{j}(\xi).$
(2.2)

Using these notations we rewrite Theorem 1.1 as follows.

######  Theorem 2.1.

For $\frac{n+3}{n+1}\leq p\leq 2$, the estimate

$\|\phi_{1}\phi_{2}\|_{p}\leq C_{p}E(\phi_{1})^{1/2}E(\phi_{2})^{1/2}$ (2.3)

holds for all $\phi_{1}$ and $\phi_{2}$ whose Fourier transforms are supported
in $\mathit{\Sigma}_{1}$ and $\mathit{\Sigma}_{2}$ respectively.

The estimate (2.3) for $p=2$ is well known. Thus, by interpolation it suffices
to prove the theorem for

$p:=\frac{n+3}{n+1}.$

######  Definition 2.2.

For any $R\geq C_{0}$ we define a constant $\mathcal{K}(R)$ to be the best
constant for which the estimate

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q_{R})}\leq\mathcal{K}(R)E(\phi_{1})^{1/2}E(\phi_{2})^{1/2}$
(2.4)

holds for all spacetime cubes $Q_{R}$ of sidelength $R$ and all $\phi_{1}$,
$\phi_{2}$ of which Fourier transforms are supported in $\mathit{\Sigma}_{1}$
and $\mathit{\Sigma}_{2}$ respectively and satisfy

$\mathrm{marg}(\phi_{1}),\mathrm{marg}(\phi_{2})\geq 1/100-R^{-1/N},$ (2.5)

where the margin $\mathrm{marg}(\phi_{j})$ is defined by

$\mathrm{marg}(\phi_{j}):=\mathrm{dist}(\operatorname{supp}(\widehat{\phi}_{j}),\mathit{\Sigma}\setminus\mathit{\Sigma}_{j}).$

Note that the margin condition can be removed by partitioning both $x$-space
and $\xi$-space and some Lorentz transforms, see [23]. By the above definition
it suffices to show

$\mathcal{K}(R)\leq 2^{CC_{0}}.$ (2.6)

We may assume that

$E(\phi_{1})=E(\phi_{2})=1.$ (2.7)

By some trivial estimates it follows

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q_{R})}\lesssim R^{C}.$ (2.8)

Thus we see that

$\mathcal{K}(R)\lesssim R^{C}.$ (2.9)

By this estimate we may assume $R\geq 2^{C_{0}}$. Let

$\overline{\mathcal{K}}(R):=\sup_{2^{C_{0}}\leq R^{\prime}\leq
R}\mathcal{K}(R^{\prime}).$

######  Proposition 2.3.

Let $R\geq 2^{C_{0}}$ and $0<c\leq 2^{-C_{0}}$. Suppose that $\phi_{1}$,
$\phi_{2}$ are Fourier supported in $\mathit{\Sigma}_{1}$ and
$\mathit{\Sigma}_{2}$ respectively and satisfy the margin condition

$\mathrm{marg}(\phi_{j})\geq 1/100-2R^{-1/N},\qquad j=1,2.$ (2.10)

Then,

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q_{R})}\leq\Big{(}(1+Cc)\overline{\mathcal{K}}(R/C_{0})+c^{-C}\Big{)}E(\phi_{1})^{1/2}E(\phi_{2})^{1/2},$
(2.11)

for all cubes $Q_{R}$ of sidelength $R$.

We will prove this proposition in Section 5. It is obtained by refining the
proof of [24]. Technically the constant $(1+Cc)$ is important to obtain the
endpoint. Note that the estimate (2.11) implies
${\mathcal{K}}(R)\leq(1+Cc)\overline{\mathcal{K}}(R/C_{0})+c^{-C}.$ By
iterating this estimate it follows $\overline{\mathcal{K}}(R)\leq
C_{\varepsilon}R^{\varepsilon}$ for all $\varepsilon>0$.

For other propositions we define an energy concentration. We first introduce
several constants relevant to the conic sets $\Lambda_{1}(0)$ and
$\Lambda_{2}(0)$ defined as (1.5).

  * •

Let $A_{w}$ be the maximum angle between two lines $L_{j}$ and
$L_{j}^{\prime}$ passing through the origin and contained in $\Lambda_{j}(0)$
for $j=1,2$.

  * •

Let $A_{d}$ be the minimum angle between two lines
$L_{1}\subset\Lambda_{1}(0)$ and $L_{2}\subset\Lambda_{2}(0)$ passing through
the origin.

  * •

Define the constant $A_{*}$ by

$A_{*}:=4A_{w}A_{d}^{-1}+C.$ (2.12)

######  Definition 2.4.

Let $\phi_{1},\phi_{2}$ be Fourier supported in $\mathit{\Sigma}_{1}$ and
$\mathit{\Sigma}_{2}$ respectively. For $0\leq\varepsilon<1$, $r>0$ and
$t\in\mathbb{R}$, let $\mathfrak{D}_{t}^{\varepsilon}(\phi_{1},\phi_{2})$ be
the collection of discs $D=D(x_{D},t_{D};r_{D})$ in
$\mathbb{R}^{n}\times\\{t\\}$ with $r_{D}\geq C_{0}$ such that

$\|\phi_{1}\|_{L^{2}(D(x_{D},t_{D};\frac{C_{0}}{800A_{*}}))}\|\phi_{2}\|_{L^{2}(D(x_{D},t_{D};\frac{C_{0}}{800A_{*}}))}\geq\varepsilon
E(\phi_{1})^{1/2}E(\phi_{1})^{1/2}.$ (2.13)

We define the energy concentration
$E^{\varepsilon}_{r,\mathring{r},t}(\phi_{1},\phi_{2})$ of $\phi_{1}$ and
$\phi_{1}$ at time $t$ by

$E^{\varepsilon}_{r,\mathring{r},t}(\phi_{1},\phi_{2})=\max\Big{(}\frac{1}{2}E(\phi_{1})^{1/2}E(\phi_{1})^{1/2},\sup_{\begin{subarray}{c}D\in\mathfrak{D}_{t}^{\varepsilon}(\phi_{1},\phi_{2})\\\
:r_{D}=r\end{subarray}}\sup_{\begin{subarray}{c}D_{1},D_{2}\subset\mathcal{N}(D)\\\
:r_{D_{1}}=r_{D_{2}}=\mathring{r}\end{subarray}}\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}\Big{)}$

where $r_{D}$ denotes the radius of $D$ and $\mathcal{N}(D)$ denotes the
$A_{*}$-dilated disc of $D$. (See Figure 1.)

The condition (2.13) is a technical thing to handle errors. Since
$\phi_{1},\phi_{2}$ are compactly Fourier supported, they have Schwartz tails,
and by using the Paley–Wiener theorem we can see that for any proper disc $D$,
the $\|\phi_{j}\|_{L^{2}(D)}$ is nonzero for $j=1,2$.

$\mathcal{N}(D)$$D_{1}$$D_{2}$ Figure 1. Energy concentration

######  Definition 2.5.

Let $R\geq 2^{C_{0}}$, $r>0$ and $0\leq\varepsilon<1$. We define
$\mathcal{K}_{\varepsilon}(R,r,\mathring{r})$ to be the best constant for
which the estimate

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q_{R})}\leq\mathcal{K}_{\varepsilon}(R,r,\mathring{r})(E(\phi_{1})^{1/2}E(\phi_{1})^{1/2})^{1/p}E^{\varepsilon}_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2})^{1/p^{\prime}}$
(2.14)

holds for all spacetime cube $Q_{R}$ of sidelength $R$, all
$t_{e}\in\mathbb{R}$ and all $\phi_{1},\phi_{2}$ whose Fourier transforms are
supported in $\mathit{\Sigma}_{1}$ and $\mathit{\Sigma}_{2}$ respectively and
satisfy the margin condition (2.5).

From the definitions of $\mathcal{K}(R)$ and
$\mathcal{K}_{\varepsilon}(R,r,\mathring{r})$ it immediately follows

$\displaystyle\mathcal{K}(R)$
$\displaystyle\leq\mathcal{K}_{\varepsilon}(R,r,\mathring{r}),$
$\displaystyle\mathcal{K}_{\varepsilon}(R,r,\mathring{r})$ $\displaystyle\leq
2^{1/p^{\prime}}\mathcal{K}(R).$

By the dispersive property of $\phi_{1},\phi_{2}$ the above estimates are
further improved under certain circumstances.

######  Proposition 2.6.

For any $R\geq 2^{C_{0}}$ and $0<\varepsilon\ll 1$,

$\mathcal{K}(R)\leq(1-C_{0}^{-C})\sup_{\begin{subarray}{c}r\geq
C_{0}^{-C}R,\\\
r/100<\mathring{r}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r\end{subarray}}\mathcal{K}_{\varepsilon}(R,r,\mathring{r}).$
(2.15)

This proposition will be proven in Section 7. In the above estimate, the
constant $(1-C_{0}^{-C})$ is crucial for closing an induction. The supremum
condition $r\geq C_{0}^{-C}R$ is also important, which prevents a loss caused
by iteration. Because of this condition we need only $O(1)$ iteration.

######  Proposition 2.7.

Let $R\geq 2^{C_{0}}$. If $R^{-N/4}\leq\varepsilon\ll 1$, $r\geq
C_{0}^{C}A_{*}R$ and $r/100<\mathring{r}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r$
then

$\mathcal{K}_{\varepsilon}(R,r,\mathring{r})\leq{(1+Cc)}{\overline{\mathcal{K}}(R)}+2^{CC_{0}}$

for any $0<c\leq 2^{-C_{0}}$.

We will prove this in Section 8. In the above estimate, the constant $(1+Cc)$
is crucial. Because of the condition $r\geq C_{0}^{C}R$ in the above
proposition, we can combine this with the previous one. To resolve it we use
the following recursive estimate.

######  Proposition 2.8.

Let $R\geq 2^{C_{0}}$. If $0<\varepsilon\ll 1$, $C_{0}^{-C}R\leq r\leq
C_{0}^{C}R$ and $\mathring{r}>0$,

$\mathcal{K}_{\varepsilon}(R,r,\mathring{r})\leq(1+Cc)\mathcal{K}_{\varepsilon}(R/C_{0},r_{\natural},\mathring{r}_{\natural})+c^{-C}$

for any $0<c\leq 2^{-C_{0}}$, where $r_{\natural}:=r(1-Cr^{-1/3N})$ and
$\mathring{r}_{\natural}:=(\mathring{r})_{\natural}$.

We will prove this in Section 9. The above recursive estimate is obtained by
modifying the proof of Proposition 2.3.

To prove (2.6) we combine the above three propositions. From Proposition 2.7
and Proposition 2.8 it follows that if $R^{-N/4}\leq\varepsilon\ll 1$, $r\geq
C_{0}^{-C}R$ and $\mathring{r}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r$, then

$\mathcal{K}_{\varepsilon}(R,r,\mathring{r})\leq(1+Cc)\overline{\mathcal{K}}(R)+2^{CC_{0}}+c^{-C}.$
(2.16)

Indeed, by Proposition 2.7 we may assume that $C_{0}^{-C}R\leq r\leq
C_{0}^{C}R$. Let $r_{0}:=r$, $r_{j+1}:=(r_{j})_{\natural}$ and
$\mathring{r}_{j+1}:=(\mathring{r}_{j})_{\natural}$ for $j=0,1,2,\cdots$. Then
if $\mathring{r}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r$ then
$\mathring{r}_{\natural}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r_{\natural}$. We
take $J$ as the smallest integer such that $r\geq C_{0}^{C-J}A_{*}R$. Because
of the condition $r\geq C_{0}^{-C}R$, we have $J=O(1)$. From Proposition 2.8
it follows that

$\mathcal{K}_{\varepsilon}(R/C_{0}^{j},r_{j},\mathring{r}_{j})\leq(1+Cc)\mathcal{K}_{\varepsilon}(R/C_{0}^{j+1},r_{j+1},\mathring{r}_{j+1})+c^{-C}.$

By iteration we have

$\mathcal{K}_{\varepsilon}(R,r,\mathring{r})\leq(1+Cc)^{J}\mathcal{K}_{\varepsilon}(R/C_{0}^{J},r_{J},\mathring{r}_{J})+2^{J}c^{-C}.$

Since $r\geq C_{0}^{-C}R\geq C_{0}^{-C}2^{C_{0}}$ and $C_{0}$ is very large,
we see that $r_{J}$ is comparable to $r$. By Proposition 2.7,

$\mathcal{K}_{\varepsilon}(R,r,\mathring{r})\leq(1+Cc)^{{J+1}}{\overline{\mathcal{K}}(R/C_{0}^{J})}+2^{CC_{0}}+2^{{J+1}}c^{-C}.$

Since $\overline{\mathcal{K}}(R/C_{0}^{J})\leq\overline{\mathcal{K}}(R)$, this
estimate implies (2.16).

Combine Proposition 2.6 with (2.16). Then,

$\mathcal{K}(R)\leq(1-C_{0}^{-C})(1+Cc)\overline{\mathcal{K}}(R)+2^{CC_{0}}+c^{-C}.$

If we set $c=2^{-C_{0}}$ then
$\overline{\mathcal{K}}(R)\leq(1-C_{0}^{-C})\overline{\mathcal{K}}(R)+2^{CC_{0}}.$
By rearranging we obtain $\overline{\mathcal{K}}(R)\leq
C_{0}^{C}2^{CC_{0}}\leq 2^{C^{\prime}C_{0}}$, which implies (2.6).

##  3\. Preliminaries for the bilinear restriction estimate

Let $R\geq 2^{C_{0}}$, $0<c\leq 2^{-C_{0}}$ and $\kappa>0$ an integer such
that

$r:=2^{-\kappa}R\sim R^{1/2}.$ (3.1)

###  3.1. A wave packet decomposition

Let $L=c^{-2}r\mathbb{Z}^{n}$ and $V_{j}$ be a maximal $r^{-1}$-separated
subset of $\tilde{\Xi}_{j}$ for $j=1,2$. For each $(x_{0},v_{j})\in L\times
V_{j}$, we define a tube $T_{j}=T_{j}^{(x_{0},v_{j})}$ with initial position
$x_{0}$ and direction $v_{j}$ by

$T_{j}=\big{\\{}(x,t)\in\mathbb{R}^{n}\times\mathbb{R}:|t|\leq
R,~{}|(x-x_{0})-tv_{j}|\leq r\big{\\}},$

and let $\mathbf{T}_{j}$ denote the collection of these tubes. We denote by
$x(T_{j})=x_{0}$ the position of $T_{j}$ and $v(T_{j})=v_{j}$ the velocity of
$T_{j}$.

Now we decompose $\phi_{j}$ into wave packets essentially supported on tubes
$T_{j}$. To partition $\mathbb{R}^{n}$ into cubes of sidelength ${c^{-2}r}$,
we set

$\eta^{x_{0}}(x):=\eta\Big{(}\frac{x-x_{0}}{c^{-2}r}\Big{)}.$ (3.2)

Then,

$\sum_{x_{0}\in L}\eta^{x_{0}}(x)=1.$ (3.3)

To partition $\tilde{\Xi}_{j}$, let $B_{v_{j}}$ be a neighborhood of $v_{j}\in
V_{j}$ with pairwise disjoint interiors so that

$\tilde{\Xi}_{j}=\bigcup_{v_{j}\in V_{j}}B_{v_{j}}.$

For each $\omega\in D(0;1/r)$, we define a map
$\Omega_{\omega}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ by
$\Omega_{\omega}(v)=v+w$. Let $G$ be the set of these maps, and define
$d\Omega$ by

$\int_{G}F(\Omega)d\Omega=\frac{1}{|D(0;1/r)|}\int_{D(0;1/r)}F(\Omega_{w})dw.$

For each $\Omega\in G$ and $v_{j}\in V_{j}$, we define the multiplier
$P_{\Omega,v_{j}}$ by

$\widehat{P_{\Omega,v_{j}}f}=\chi_{\Omega(B_{v_{j}})}\widehat{f},$

where $\chi$ denotes a characteristic function. For each
$T_{j}=T_{j}^{(x_{0},v_{j})}\in\mathbf{T}_{j}$, we define a function
$f_{T_{j}}$ by

$f_{T_{j}}(x):=\eta^{x_{0}}(x)\int_{G}P_{\Omega,v_{j}}f_{j}(x){d\Omega}.$
(3.4)

We define a wave packet $\phi_{T_{j}}$ as

$\phi_{T_{j}}:=\mathcal{U}_{j}f_{T_{j}}.$ (3.5)

Then by the linearity of Fourier transform,

$\phi_{j}(x,t)=\sum_{T_{j}\in\mathbf{T}_{j}}\phi_{T_{j}}(x,t).$ (3.6)

######  Lemma 3.1 (Properties of the wave packets).

Suppose that $\phi_{j}$ has a Fourier support in $\mathit{\Sigma}_{j}$ and a
margin

$\mathrm{marg}(\phi_{j})\geq Cr^{-1}$ (3.7)

for $j=1,2$. Let $T_{j}=T_{j}^{(x_{0},v_{j})}$. We define a constant
$h_{T_{j}}$ by

$h_{T_{j}}:=r^{n/2}\mathcal{M}\Big{(}\int_{G}P_{\Omega,v_{j}}f_{j}{d\Omega}\Big{)}(x_{0}),$
(3.8)

where $\mathcal{M}$ denotes the Hardy–Littlewood maximal operator. Then we
have the followings.

  * •

The margin of $\phi_{T_{j}}$ satisfies

$\mathrm{marg}(\phi_{T_{j}})\geq\mathrm{marg}(\phi_{j})-Cr^{-1}.$ (3.9)

  * •

For $(x,t)\in\mathbb{R}^{n}\times[-CR,CR]$,

$|\phi_{T_{j}}(x,t)|\leq
C_{M}c^{-C}r^{-n/2}h_{T_{j}}\left(1+\frac{\mathrm{dist}(T_{j},(x,t))}{r}\right)^{-M},\qquad\forall
M>0.$ (3.10)

  * •

$\Big{(}\sum_{T_{j}\in\mathbf{T}_{j}}h_{T_{j}}^{2}\Big{)}^{1/2}\lesssim
c^{-C}\|f_{j}\|_{2}.$ (3.11)

###### Proof.

Consider (3.9). From the definition of $f_{T_{j}}$ we can see that the Fourier
support of $f_{T_{j}}$ is contained in a $O(r^{-1})$-neighborhood of $v_{j}$.
So, the spacetime Fourier transform of $\phi_{T_{j}}$ is supported in a
$O(r^{-1})$-neighborhood of the spacetime Fourier support of $\phi_{j}$. From
this we have (3.9).

Consider (3.10). If $\rho$ is a smooth bump function supported on a
$O(1)$-neighborhood of the origin and
$\rho^{v_{j}}(\xi):=\rho\big{(}4^{-1}r(\xi-v_{j})\big{)}$ then we may replace
$\widehat{f}_{T_{j}}$ with $\rho^{v_{j}}\widehat{f}_{T_{j}}$. By interchanging
the integrals we may write

$\phi_{T_{j}}(x,t)=\int K_{v_{j}}(x-y,t)f_{T_{j}}(y)dy,$

where

$K_{v_{j}}(x,t)=\int e^{2\pi
i(x\cdot\xi-\frac{1}{2}t|\xi|^{2})}\rho^{v_{j}}(\xi)d\xi.$ (3.12)

By integration by parts, if $|t|\lesssim R$ then

$|K_{v_{j}}(x,t)|\leq
C_{M}r^{-n}\left(1+\frac{|x-tv_{j}|}{r}\right)^{-M},\qquad\forall M>0.$ (3.13)

Indeed, let $\delta=r^{-1}$ and
$\Psi(x,t,\xi)=2\pi(x\cdot(\delta\xi+v_{j})-\frac{1}{2}t|\delta\xi+v_{j}|^{2})$.
We rewrite as

$K_{v_{j}}(x,t)=\delta^{n}\int e^{i\psi(x,t,\xi)}\rho(\xi)d\xi.$ (3.14)

Suppose that $|x-tv_{j}|\geq C\delta^{-1}$. Then we have
$|\nabla_{\xi}\Psi(x,t,\xi)|\geq C\delta|x-tv_{j}|,$ since
$|\delta^{2}t\xi|\lesssim 1$. By integration by parts,

$|K_{v_{j}}(x,t)|\leq C_{M}\delta^{n}(\delta|x-tv_{j}|)^{-M},\qquad\forall
M>0.$

On the other hand, we have a trivial estimate
$|K_{v_{j}}(x,t)|\lesssim\delta^{n}.$ By combining these two estimates we have
(3.13). Thus,

$\displaystyle|\phi_{T_{j}}$ $\displaystyle(x,t)|$ $\displaystyle\leq
C_{M}r^{-n}\int\left(1+\frac{|(x-y)-tv_{j}|}{r}\right)^{-M}\eta^{x_{0}}(y)\bigg{|}\int_{G}P_{\Omega,v_{j}}f_{j}(y){d\Omega}\bigg{|}dy$
$\displaystyle\leq
C_{M}c^{-C}\left(1+\frac{|(x-x_{0})-tv_{j}|}{r}\right)^{-M}\mathcal{M}\Big{(}\int_{G}P_{\Omega,v_{j}}f_{j}{d\Omega}\Big{)}(x_{0})$
$\displaystyle\leq
C_{M}c^{-C}r^{-n/2}h_{T_{j}}\left(1+\frac{\mathrm{dist}(T_{j},(x,t))}{r}\right)^{-M}.$

Consider (3.11). By the uncertainty principle, if $|x-x_{0}|\lesssim c^{-2}r$
then

$\mathcal{M}\Big{(}\int P_{\Omega,v_{j}}f_{j}{d\Omega}\Big{)}(x_{0})\lesssim
c^{-C}\mathcal{M}\Big{(}\int P_{\Omega,v_{j}}f_{j}{d\Omega}\Big{)}(x).$

Thus,

$\sum_{T_{j}\in\mathbf{T}_{j}}h_{T_{j}}^{2}\lesssim c^{-C}\sum_{v_{j}\in
V_{j}}\int\Big{|}\mathcal{M}\Big{(}\int
P_{\Omega,v_{j}}f_{j}{d\Omega}\Big{)}(x)\Big{|}^{2}dx.$

By the Hardy–Littlewood maximal theorem and Minkowski’s inequality,

$\displaystyle\big{(}\sum_{T_{j}\in\mathbf{T}_{j}}h_{T_{j}}^{2}\big{)}^{1/2}$
$\displaystyle\lesssim c^{-C}\Big{(}\sum_{v_{j}\in V_{j}}\Big{\|}\int
P_{\Omega,v_{j}}f_{j}{d\Omega}\Big{\|}_{2}^{2}\Big{)}^{1/2}$
$\displaystyle\leq c^{-C}\int\big{(}\sum_{v_{j}\in
V_{j}}\|P_{\Omega,v_{j}}f_{j}\|_{2}^{2}\big{)}^{1/2}{d\Omega}.$

By Plancherel’s theorem and orthogonality,

$\sum_{v_{j}\in
V_{j}}\|P_{\Omega,v_{j}}f_{j}\|_{2}^{2}\lesssim\|f_{j}\|_{2}^{2}.$

Inserting this into the previous integral we have (3.11). ∎

###  3.2. Estimates on a light conic set.

We define a kernel $K_{j,t}$ by

$K_{j,t}(x)=\int e^{2\pi i(x\cdot\xi-\frac{1}{2}t|\xi|^{2})}a_{j}(\xi)d\xi.$

Then $\phi_{j}$ is written as

$\phi_{j}(x,t)=\mathcal{U}_{j}f(x,t)=K_{j,t}\ast f(x).$ (3.15)

######  Lemma 3.2.

Let $\Lambda_{j,t}=\\{x\in\mathbb{R}^{n}:(x,t)\in\Lambda_{j}(0)\\}$ where
$\Lambda_{j}(0)$ is defined as in (1.6). Then,

$\big{|}K_{j,t}(x)\big{|}\leq
C_{M}(1+\mathrm{dist}(\Lambda_{j,t},x))^{-M},\qquad\forall M>0.$ (3.16)

###### Proof.

If $\mathrm{dist}(\Lambda_{j,t},x)=0$, by a trivial estimate we have

$\big{|}K_{j,t}(x)\big{|}\lesssim 1.$

Suppose that $\mathrm{dist}(\Lambda_{j,t},x)>0$. The $\xi$-derivative of the
phase $x\cdot\xi-\frac{1}{2}t|\xi|^{2}$ has

$\Big{|}\nabla_{\xi}\Big{(}x\cdot\xi-\frac{1}{2}t|\xi|^{2}\Big{)}\Big{|}\geq\mathrm{dist}(\Lambda_{j,t},x)$

for all $\xi\in\tilde{\Xi}_{j}$. So, using integration by parts we obtain

$\big{|}K_{j,t}(x)\big{|}\leq
C_{M}\mathrm{dist}(\Lambda_{j,t},x)^{-M},\qquad\forall M>0.$

Thus we have (3.16). ∎

######  Lemma 3.3.

Let $Q$ be a cube of sidelength $R^{C}$. Let $\epsilon>0$ and $j,k=1,2$ but
$j\neq k$. Let $\pi_{k}\subset\Xi_{k}$ be a hypersurface in $\mathbb{R}^{n}$.
Suppose that there is $0\leq\varepsilon_{0}<1$ such that for any $v\in\Xi_{j}$
and any $w,w^{\prime}\in\pi_{k}$ with $w\neq w^{\prime}$,

$\Big{|}\Big{\langle}\frac{v-w}{|v-w|},\frac{w^{\prime}-w}{|w^{\prime}-w|}\Big{\rangle}\Big{|}\leq\varepsilon_{0}.$
(3.17)

Then for any $r\gtrsim R^{\epsilon}$,

$\|\phi_{j}\|_{L^{2}(Q\cap\Lambda_{\pi_{k}}(z_{0};r))}\lesssim
r^{1/2}E(\phi_{j})^{1/2}.$

###### Proof.

It suffices to show

$\int\big{\|}\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(t)[\mathcal{U}_{j}(t)]f\big{\|}_{L^{2}(\mathbb{R}^{n})}^{2}dt\lesssim
r\|f\|_{L^{2}(\mathbb{R}^{n})}^{2},$

where $\Lambda_{\pi_{k}}^{Q}=\Lambda_{\pi_{k}}\cap Q$. By duality this is
equivalent to

$\left\|\int\mathcal{U}_{j}(t)^{*}\big{(}\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(t)g(t)\big{)}dt\right\|_{L^{2}(\mathbb{R}^{n})}^{2}\lesssim
r\|g\|_{L^{2}(\mathbb{R}^{n+1})}^{2}.$ (3.18)

The left side of the above estimate is written as

$\iint\left\langle[\mathcal{U}_{j}(s)\mathcal{U}_{j}(t)^{\ast}]\big{(}\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(t)g(t)\big{)},~{}\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(s)g(s)\right\rangle
dtds.$

Let

$G_{k}(x,t):=\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(x,t)g(x,t)$

and

$\mathcal{I}_{j}f(s,t):=\left\langle[\mathcal{U}_{j}(s)\mathcal{U}_{j}(t)^{\ast}]f(t),f(s)\right\rangle.$

The previous integral is divided into two parts

$\iint\mathcal{I}_{j}G_{k}(s,t)dtds=\iint_{|s-t|\leq
Cr}\mathcal{I}_{j}G_{k}(s,t)dtds+\iint_{|s-t|\geq
Cr}\mathcal{I}_{j}G_{k}(s,t)dtds.$

To show (3.18) it suffices to prove

$\displaystyle\bigg{|}\iint_{|s-t|\geq
Cr}\mathcal{I}_{j}G_{k}(s,t)dtds\bigg{|}$ $\displaystyle\lesssim
r^{-N}\|g\|_{2}^{2}$ (3.19) and $\displaystyle\bigg{|}\iint_{|s-t|\leq
Cr}\mathcal{I}_{j}G_{k}(s,t)dtds\bigg{|}$ $\displaystyle\lesssim
r\|g\|_{2}^{2}.$ (3.20)

Consider (3.19). If we set

$K_{j,t}(x)=\int e^{2\pi
i(x\cdot\xi-\frac{1}{2}t|\xi|^{2})}a_{j}^{2}(\xi)d\xi$

then

$[\mathcal{U}_{j}(s)\mathcal{U}_{j}(t)^{\ast}]f(x)=\int K_{j,s-t}(x-y)f(y)dy.$

We rewrite $\mathcal{I}_{j}G_{k}$ as

$\mathcal{I}_{j}G_{k}=\iint
K_{j,s-t}(x-y)\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(y,t)\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(x,s)g(y,t)\overline{g(x,s)}dxdy.$

We divide

$K_{j,t}=\chi_{\Lambda_{j,t}(0;r)}K_{j,t}+(1-\chi_{\Lambda_{j,t}(0;r)})K_{j,t}.$

From Lemma 3.2 it follows that

$(1-\chi_{\Lambda_{j,s-t}(0;r)}(x))|K_{j,s-t}(x)|\lesssim r^{-M},\qquad\forall
M>0.$

Using this we have

$\displaystyle\bigg{|}\iiiint((1-\chi_{\Lambda_{j,s-t}(0;r)})K_{j,s-t})(x-y)$
$\displaystyle\qquad\qquad\times\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(y,t)\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(x,s)g(y,t)\overline{g(x,s)}dxdydtds\bigg{|}$
$\displaystyle\qquad\qquad\qquad\lesssim
r^{-M}\|\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}g\|_{1}^{2}$
$\displaystyle\qquad\qquad\qquad\lesssim r^{-M}R^{C}\|g\|_{2}^{2}$
$\displaystyle\qquad\qquad\qquad\lesssim r^{-M}\|g\|_{2}^{2},\qquad\forall
M>0,$

where the last line follows from $r\gtrsim R^{\epsilon}$. Now, to show (3.19)
it suffices to show

$\begin{split}&\iiiint_{|s-t|\geq
Cr}\chi_{\Lambda_{j,s-t}(0;r)}(x-y)K_{j,s-t}(x-y)\\\
&\qquad\qquad\times\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(y,t)\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(x,s)g(y,t)\overline{g(x,s)}dxdydtds=0.\end{split}$

It is enough to show that for $s,t\in\mathbb{R}$ with $|s-t|\geq Cr$, the
equation

$\chi_{\Lambda_{j,s-t}(0;r)}(x-y)\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(y,t)\chi_{\Lambda^{Q}_{\pi_{k}}(z_{0};r)}(x,s)$
(3.21)

vanishes. Consider the contrapositive statement that if $\eqref{crosK}$ is
nonzero then one has $|s-t|\lesssim r$. Suppose that $\eqref{crosK}$ is
nonzero. Then, from the characteristic function
$\chi_{\Lambda_{j,s-t}(0;r)}(x-y)$ we can restrict ourselves to the case

$x-y=(s-t)v+{O}(r)$ (3.22)

for some $v\in\Xi_{j}$. On the other hands, from
$\chi_{\Lambda_{\pi_{k}}(z_{0};r)}(y,t)$ and
$\chi_{\Lambda_{\pi_{k}}(z_{0};r)}(x,s)$ we also have

$\begin{split}x-x_{0}&=(s-t_{0})w+{O}(r)\\\
y-x_{0}&=(t-t_{0})w^{\prime}+{O}(r)\end{split}$ (3.23)

for some $w,w^{\prime}\in\pi_{k}$. By combining (3.22) and (3.23) it follows
that

$(s-t)(v-w)+(t-t_{0})(w^{\prime}-w)=O(r).$ (3.24)

If $w=w^{\prime}$, then we have $|s-t|\lesssim r$. Otherwise, from (3.17) we
can see that there exists a unit vector $u$ such that $(v-w)\cdot u\neq 0$ but
$(w^{\prime}-w)\cdot u=0$. By taking inner product with such $u$ for (3.24),
we have $|s-t|\lesssim r$.

Consider (3.20). By the Cauchy–Schwarz inequality and Plancherel’s theorem it
follows that

$|\mathcal{I}_{j}f(s,t)|\lesssim\|f(s)\|_{2}\|f(t)\|_{2}.$

By this and the Hardy–Littlewood–Sobolev inequality,

$\displaystyle\bigg{|}\iint_{|s-t|\lesssim
r}\mathcal{I}_{j}G_{k}(s,t)dtds\bigg{|}$
$\displaystyle\lesssim\iint_{|s-t|\lesssim
r}\|g(s)\|_{L^{2}(\mathbb{R}^{n})}\|g(t)\|_{L^{2}(\mathbb{R}^{n})}dtds$
$\displaystyle\lesssim r\|g\|_{2}^{2}.$

Thus we have (3.20). ∎

###  3.3. A basic bilinear restriction estimate

Let $\psi$ be a nonnegative Schwartz function on $\mathbb{R}^{n+1}$ with
$\int\psi=1$ such that $\widehat{\psi}$ is supported in the unit ball and

$\sum_{k\in\mathbb{Z}^{n+1}}\psi^{4}(z-k)=1$ (3.25)

where $z=(x,t)\in\mathbb{R}^{n+1}$. If $q$ is a cube of sidelength $r_{q}$
with center $z_{q}$, we define

$\psi_{q}(z):=\psi(r_{q}^{-1}(z-z_{q})).$

For convenience, we use the notations $\int_{\psi}\cdot$ and
$\|\cdot\|_{L^{2}(\psi)}$ to denote

$\int_{\psi}f:=\int
f\psi\qquad\text{and}\qquad\|f\|_{L^{2}(\psi)}:=\|f\psi\|_{2}.$

######  Lemma 3.4.

Let $q$ be a cube of side-length $R^{1/2}$. Suppose that $\phi_{T_{1}}$ and
$\phi_{T_{2}}$ are wave packets defined as (3.5). Then

$\|\phi_{T_{1}}\phi_{T_{2}}\|_{L^{2}({\psi_{q}^{2}})}\lesssim
R^{-\frac{n+1}{4}}\|\phi_{T_{1}}\|_{L^{2}({\psi_{q}})}\|\phi_{T_{2}}\|_{L^{2}({\psi_{q}})}.$
(3.26)

###### Proof.

By Plancherel’s theorem the estimate (3.26) is equivalent to

$\|\widehat{\psi_{q}\phi_{T_{1}}}\ast\widehat{\psi_{q}\phi_{T_{2}}}\|_{2}\lesssim
R^{-\frac{n+1}{4}}\|\widehat{\psi_{q}\phi_{T_{1}}}\|_{2}\|\widehat{\psi_{q}\phi_{T_{2}}}\|_{2}.$

By interpolation it suffices to show the following two estimates:

$\displaystyle\|\widehat{\psi_{q}\phi_{T_{1}}}\ast\widehat{\psi_{q}\phi_{T_{2}}}\|_{1}$
$\displaystyle\leq\|\widehat{\psi_{q}\phi_{T_{1}}}\|_{1}\|\widehat{\psi_{q}\phi_{T_{2}}}\|_{1},$
$\displaystyle\|\widehat{\psi_{q}\phi_{T_{1}}}\ast\widehat{\psi_{q}\phi_{T_{2}}}\|_{\infty}$
$\displaystyle\lesssim
R^{-\frac{n+1}{2}}\|\widehat{\psi_{q}\phi_{T_{1}}}\|_{\infty}\|\widehat{\psi_{q}\phi_{T_{2}}}\|_{\infty}.$

By Young’s inequality the first one is easily obtained. Consider the second
one. Observe that the Fourier support of $\psi_{q}\phi_{T_{j}}$ is contained
in a ball $B(v(T_{j}),-{|v(T_{j})|^{2}}/{2};CR^{-1/2})$. Let
$\chi_{j}:=\chi_{B(v(T_{j}),-{|v(T_{j})|^{2}}/{2};CR^{-1/2})}$ be a
characteristic function. Then,

$\|\widehat{\psi_{q}\phi_{T_{1}}}\ast\widehat{\psi_{q}\phi_{T_{2}}}\|_{\infty}\lesssim\|\chi_{1}\ast\chi_{2}\|_{\infty}\|\widehat{\psi_{q}\phi_{T_{1}}}\|_{\infty}\|\widehat{\psi_{q}\phi_{T_{2}}}\|_{\infty}.$

Simple computation gives $\|\chi_{1}\ast\chi_{2}\|_{\infty}\lesssim
R^{-\frac{n+1}{2}}$. Thus we have the second estimate. ∎

##  4\. Refining the proof of the sharp bilinear restriction estimate

In this section we refine the proof of the sharp bilinear paraboloid
restriction estimate due to Tao [24].

###  4.1. Decomposition for bilinear estimates

Let $Q$ be the cube of sidelength $CR$ and centered at the origin. We
decompose $Q$ into subcubes $\Delta$ of sidelength $2^{-C_{0}}CR$. For any
integer $l$, we define $\mathcal{Q}_{l}(Q)$ to be the collection of subcubes
of sidelength $2^{-l}CR$ which are obtained by dividing the sides of $Q$.

Let

$\Psi_{D(x_{D};r_{D})}(x):=\left(1+\frac{|x-x_{D}|}{r_{D}}\right)^{-N^{100}}.$

and for each tube $T_{j}=T_{j}^{(x_{0},v_{j})}$,

$\Psi_{T_{j}}(x,t):=\Psi_{D(x_{0}-tv_{j};r)}(x).$ (4.1)

For each $2^{-C_{0}}CR$-cube $\Delta\in\mathcal{Q}_{C_{0}}(Q)$ and each
$T_{j}\in\mathbf{T}_{j}$, we define

$m_{\Delta,T_{j}}:=\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\kappa}(Q)\\\
:q\subset\Delta\end{subarray}}\sum_{T_{i}\in\mathbf{T}_{i}}\|\phi_{T_{i}}\Psi_{T_{j}}\|_{L^{2}(\psi_{q})}^{2}+R^{-10n}E(\phi_{i})$
(4.2)

and

$\quad
m_{T_{j}}:=\sum_{{q\in\mathcal{Q}_{\kappa}(Q)}}\sum_{T_{i}\in\mathbf{T}_{i}}\|\phi_{T_{i}}\Psi_{T_{j}}\|_{L^{2}(\psi_{q})}^{2}+R^{-10n}2^{(n+1)C_{0}}E(\phi_{i})$
(4.3)

for $i\neq j$. Then,

$\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\frac{m_{\Delta,T_{j}}}{m_{T_{j}}}=1.$
(4.4)

We now define $\Phi_{j}^{(\Delta)}$ for each $\Delta\in\mathcal{Q}_{C_{0}}(Q)$
by

$\Phi_{j}^{(\Delta)}(z):=\sum_{T_{j}\in\mathbf{T}_{j}}\frac{m_{\Delta,T_{j}}}{m_{T_{j}}}\phi_{T_{j}}(z).$
(4.5)

Then,

$\phi_{j}(z)=\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\Phi_{j}^{(\Delta)}(z).$
(4.6)

We define a function $[\Phi_{j}]$ by

$[\Phi_{j}](z):=\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\Phi_{j}^{(\Delta)}(z)\chi_{\Delta}(z).$
(4.7)

The main proposition of this section is as follows.

######  Proposition 4.1.

Let $R\geq 2^{C_{0}}$ and $0<c\leq 2^{-C_{0}}$. For any cube $Q$ we define a
set $X(Q)$ by

$X(Q):=\bigcup_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}(1-c)\Delta.$

Suppose that $\phi_{1}$, $\phi_{2}$ have Fourier supports in
$\mathit{\Sigma}_{1}$ and $\mathit{\Sigma}_{2}$ respectively which satisfy the
normalization (2.7) and the relaxed margin condition (2.10). Then,

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q_{R})}\leq(1+Cc)\|[\Phi_{1}][\Phi_{2}]\|_{L^{p}(X(Q))}+c^{-C},$
(4.8)

where $Q$ is a cube of sidelength $CR$ contained in $C^{2}Q_{R}$ and

$\mathrm{marg}(\Phi_{j})\geq\frac{1}{100}-\Big{(}\frac{2^{C_{0}}}{R}\Big{)}^{1/N}.$
(4.9)

In the remaining parts of this section we will prove the above proposition.
Consider the margin (4.9). From (3.9) and (4.5) it follows that

$\displaystyle\mathrm{marg}(\Phi_{j})$
$\displaystyle\geq\mathrm{marg}(\phi_{j})-CR^{-1/2}$
$\displaystyle\geq\frac{1}{100}-2\Big{(}\frac{1}{R}\Big{)}^{1/N}-C\Big{(}\frac{1}{R}\Big{)}^{1/2}$
$\displaystyle\geq\frac{1}{100}-\Big{(}\frac{2^{C_{0}}}{R}\Big{)}^{1/N}.$

The proof of (4.8) is accomplished through many steps. We begin with the
following averaging lemma.

######  Lemma 4.2 (Lemma 6.1 in [23]).

Let $R>0$, $0<c\leq 2^{-C_{0}}$, and let $Q_{R}$ be a cube of sidelength $R$.
If $f$ is a smooth function, then there exists a cube $Q$ of sidelength $CR$
contained in $C^{2}Q_{R}$ such that

$\|f\|_{L^{p}(Q_{R})}\leq(1+Cc)\|f\|_{L^{p}(X(Q))}.$

By this lemma,

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q_{R})}\leq(1+Cc)\|\phi_{1}\phi_{2}\|_{L^{p}(X(Q))}.$

Using the triangle inequality we divide $\|\phi_{1}\phi_{2}\|_{L^{p}(X(Q))}$
into three parts:

$\displaystyle\|\phi_{1}\phi_{2}\|_{L^{p}(X(Q))}$
$\displaystyle\leq\|[\Phi_{1}][\Phi_{2}]\|_{L^{p}(X(Q))}$
$\displaystyle\qquad+\|(\phi_{1}-[\Phi_{1}])\phi_{2}\|_{L^{p}(X(Q))}+\|[\Phi_{1}](\phi_{2}-[\Phi_{2}])\|_{L^{p}(X(Q))}.$

To prove (4.8) it suffices to show

$\displaystyle\|(\phi_{1}-[\Phi_{1}])\phi_{2}\|_{L^{p}(X(Q))}$
$\displaystyle\lesssim c^{-C},$ (4.10)
$\displaystyle\|[\Phi_{1}](\phi_{2}-[\Phi_{2}])\|_{L^{p}(X(Q))}$
$\displaystyle\lesssim c^{-C}.$

Since these two estimates are similarly obtained, we will consider only the
first one.

###  4.2. $L^{1}$-bilinear estimates

######  Lemma 4.3.

$\|(\phi_{1}-[\Phi_{1}])\phi_{2}\|_{L^{1}(X(Q))}\lesssim c^{-C}R.$

###### Proof.

By the Cauchy–Schwarz inequality it suffices to show

$\displaystyle\|\phi_{j}\|_{L^{2}(Q)}$ $\displaystyle\lesssim R^{1/2},$ (4.11)
$\displaystyle\|\phi_{j}-[\Phi_{j}]\|_{L^{2}(Q)}$ $\displaystyle\lesssim
c^{-C}R^{1/2},$ (4.12)

for $j=1,2.$ Consider (4.11). We have

$\|\phi_{j}\|_{L^{2}(Q)}^{2}\leq\int_{-CR}^{CR}\|\phi_{j}(t)\|_{2}^{2}dt\lesssim
RE(\phi_{j})\leq R.$

Thus (4.11) follows.

Consider (4.12). By the triangle inequality and (4.11) it suffices to show

$\|[\Phi_{j}]\|_{L^{2}(Q)}\lesssim c^{-C}R^{1/2}.$ (4.13)

We have that

$\|\Phi_{j}^{(\Delta)}\|_{L^{2}(\Delta)}^{2}\leq\int_{t_{\Delta}-C2^{-C_{0}}R}^{t_{\Delta}+C2^{-C_{0}}R}\|\Phi_{j}^{(\Delta)}(t)\|_{2}^{2}dt\lesssim
2^{-C_{0}}RE(\Phi_{j}^{(\Delta)}).$

By using (3.10),

$\displaystyle\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E(\Phi_{j}^{(\Delta)})$
$\displaystyle=\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}\frac{m_{\Delta,T_{j}}}{m_{T_{j}}}\phi_{T_{j}}(0)\Big{\|}_{2}^{2}$
$\displaystyle\lesssim
c^{-C}\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\sum_{T_{j}\in\mathbf{T}_{j}}\Big{(}\frac{m_{\Delta,T_{j}}}{m_{T_{j}}}\Big{)}^{2}h_{T_{j}}^{2}.$

By (4.4) and (3.11),

$\displaystyle\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E(\Phi_{j}^{(\Delta)})$
$\displaystyle\lesssim c^{-C}\sum_{T_{j}\in\mathbf{T}_{j}}h_{T_{j}}^{2}$
$\displaystyle\lesssim c^{-C}E(\phi_{j}).$

By the above estimates,

$\|[\Phi_{j}]\|^{2}_{L^{2}(Q)}=\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\|\Phi_{j}^{(\Delta)}\|_{L^{2}(\Delta)}^{2}\lesssim
2^{-C_{0}}c^{-C}RE(\phi_{j})\leq c^{-C}R.$

Thus we have (4.13). ∎

###  4.3. Orthogonality

By interpolation it now suffices to show

$\|(\phi_{1}-[\Phi_{1}])\phi_{2}\|_{L^{2}(X(Q))}\lesssim
c^{-C}R^{-\frac{n-1}{4}}.$ (4.14)

By (4.6), (4.7) and the triangle inequality,

$\|(\phi_{1}-[\Phi_{1}])\phi_{2}\|_{L^{2}(X(Q))}\leq\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\|\Phi_{1}^{(\Delta)}\phi_{2}\|_{L^{2}(X(Q)\setminus\Delta)}.$

Since the number of cubes in $\mathcal{Q}_{C_{0}}(Q)$ is
$2^{(n+1)C_{0}}\lesssim c^{-C}$, it is reduced to showing

$\|\Phi_{1}^{(\Delta)}\phi_{2}\|_{L^{2}(X(Q)\setminus\Delta)}\lesssim
c^{-C}R^{-\frac{n-1}{4}}.$ (4.15)

Observe that if $q\in\mathcal{Q}_{\kappa}(Q)$ meets $X(Q)\setminus\Delta$,
then $\mathrm{dist}(q,\Delta)\geq cR$. By using (3.25), (3.6) and (4.5),

$\displaystyle\|\Phi_{1}^{(\Delta)}\phi_{2}\|_{L^{2}(X(Q)\setminus\Delta)}^{2}$
$\displaystyle\lesssim\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\kappa}(Q):\\\
\mathrm{dist}(q,\Delta)\geq
cR\end{subarray}}\|\Phi_{1}^{(\Delta)}\phi_{2}\|_{L^{2}(\psi_{q}^{2})}^{2}$
$\displaystyle=\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\kappa}(Q):\\\
\mathrm{dist}(q,\Delta)\geq
cR\end{subarray}}\int_{\psi_{q}^{4}}\bigg{|}\sum_{T_{1}\in\mathbf{T}_{1}}\frac{m_{\Delta,T_{1}}}{m_{T_{1}}}\phi_{T_{1}}(z)\sum_{T_{2}\in\mathbf{T}_{2}}\phi_{T_{2}}(z)\bigg{|}^{2}dz$
(4.16)

where $z$ denotes $(x,t)$. We write the integral in the above equation as

$\sum_{\begin{subarray}{c}T_{1},T_{1}^{\prime}\in\mathbf{T}_{1},\\\
T_{2},T_{2}^{\prime}\in\mathbf{T}_{2}\end{subarray}}\int_{\psi_{q}^{4}}\bigg{(}\frac{m_{\Delta,T_{1}}}{m_{T_{1}}}\phi_{T_{1}}(z)\phi_{T_{2}}(z)\bigg{)}\bigg{(}\frac{m_{\Delta,T_{1}^{\prime}}}{m_{T_{1}^{\prime}}}\overline{\phi}_{T_{1}^{\prime}}(z)\overline{\phi}_{T_{2}^{\prime}}(z)\bigg{)}dz.$
(4.17)

We define $S$ to be the set of $(v_{1},v_{1}^{\prime},v_{2},v_{2}^{\prime})\in
V_{1}\times V_{1}\times V_{2}\times V_{2}$ such that

$\displaystyle v_{1}+v_{2}$
$\displaystyle=v_{1}^{\prime}+v_{2}^{\prime}+O(r^{-1}),$ (4.18)
$\displaystyle|v_{1}|^{2}+|v_{2}|^{2}$
$\displaystyle=|v_{1}^{\prime}|^{2}+|v_{2}^{\prime}|^{2}+O(r^{-1}).$

######  Lemma 4.4.

Let $S^{c}=V_{1}\times V_{1}\times V_{2}\times V_{2}\setminus S$ and let $q$
be a cube of side-length $R^{1/2}$. Suppose that
$T_{1},T_{1}^{\prime}\in\mathbf{T}_{1}$,
$T_{2},T_{2}^{\prime}\in\mathbf{T}_{2}$ satisfy
$(v(T_{1}),v(T_{1}^{\prime}),v(T_{2}),v(T_{2}^{\prime}))\in S^{c}$. Then,

$\bigg{|}\int_{\psi_{q}^{4}}\phi_{T_{1}}(z)\phi_{T_{2}}(z)\overline{\phi}_{T_{1}^{\prime}}(z)\overline{\phi}_{T_{2}^{\prime}}(z)dz\bigg{|}=0.$
(4.19)

###### Proof.

By Parseval’s formula the left side of (4.19) equals to

$\Big{|}\langle\widehat{\psi_{q}\phi_{T_{1}}}\ast\widehat{\psi_{q}\phi_{T_{2}}},~{}\widehat{\psi_{q}\phi_{T_{1}^{\prime}}}\ast\widehat{\psi_{q}\phi_{T_{2}^{\prime}}}\rangle\Big{|},$

where the hat $\widehat{\quad}\,$ denotes the spacetime Fourier transform.

By the construction of wave packets $\phi_{T_{j}}$, we see that
$\widehat{\psi_{q}\phi_{T_{j}}}$ is supported on a $O(r^{-1})$-neighborhood of
$(v(T_{j}),-|v(T_{j})|^{2}/2)$, and that
$\widehat{\psi_{q}\phi_{T_{1}}}\ast\widehat{\psi_{q}\phi_{T_{2}}}$ is
supported on a $O(r^{-1})$-neighborhood of
$\big{(}v(T_{1})+v(T_{2}),~{}-(|v(T_{1})|^{2}+|v(T_{2})|^{2})/2\big{)}$. Thus
we can see that if $(v(T_{1}),v(T_{1}^{\prime}),v(T_{2}),v(T_{2}^{\prime}))$
does not satisfy (4.18) then the supports of
$\widehat{\psi_{q}\phi_{T_{1}}}\ast\widehat{\psi_{q}\phi_{T_{2}}}$ and
$\widehat{\psi_{q}\phi_{T_{1}^{\prime}}}\ast\widehat{\psi_{q}\phi_{T_{2}^{\prime}}}$
are disjoint, so we have (4.19). ∎

By the above lemma,

$(\ref{innerForm})\leq\sum_{(T_{1},T_{1}^{\prime},T_{2},T_{2}^{\prime})\in\mathbf{S}}\bigg{|}\int_{\psi_{q}^{4}}\bigg{(}\frac{m_{\Delta,T_{1}}}{m_{T_{1}}}\phi_{T_{1}}\phi_{T_{2}}\bigg{)}\bigg{(}\frac{m_{\Delta,T_{1}^{\prime}}}{m_{T_{1}^{\prime}}}\overline{\phi}_{T_{1}^{\prime}}\overline{\phi}_{T_{2}^{\prime}}\bigg{)}dz\bigg{|}$
(4.20)

where

$\mathbf{S}=\\{(T_{1},T_{1}^{\prime},T_{2},T_{2}^{\prime})\in\mathbf{T}_{1}\times\mathbf{T}_{1}\times\mathbf{T}_{2}\times\mathbf{T}_{2}:(v(T_{1}),v(T_{1}^{\prime}),v(T_{2}),v(T_{2}^{\prime}))\in
S\\}.$

Using the arithmetic-geometric mean inequality, we divide the integral of the
above estimate as follows:

$\displaystyle\bigg{|}\int_{\psi_{q}^{4}}\bigg{(}\frac{m_{\Delta,T_{1}}}{m_{T_{1}}}\phi_{T_{1}}(z)\phi_{T_{2}}(z)\bigg{)}\bigg{(}\frac{m_{\Delta,T_{1}^{\prime}}}{m_{T_{1}^{\prime}}}\overline{\phi}_{T_{1}^{\prime}}(z)\overline{\phi}_{T_{2}^{\prime}}(z)\bigg{)}dz\bigg{|}$
$\displaystyle\qquad\qquad\lesssim\int_{\psi_{q}^{2}}\bigg{|}\frac{m_{\Delta,T_{1}^{\prime}}}{m_{T_{1}}}\Psi_{T_{1}^{\prime}}(z_{q})\frac{\phi_{T_{1}}(z)\phi_{T_{2}}(z)}{\Psi_{T_{1}}(z_{q})}\frac{}{}\bigg{|}^{2}dz$
$\displaystyle\qquad\qquad\qquad\qquad+\int_{\psi_{q}^{2}}\bigg{|}\frac{m_{\Delta,T_{1}}}{m_{T_{1}^{\prime}}}{\Psi_{T_{1}}(z_{q})}\frac{\overline{\phi}_{T_{1}^{\prime}}(z)\overline{\phi}_{T_{2}^{\prime}}(z)}{\Psi_{T_{1}^{\prime}}(z_{q})}\bigg{|}^{2}dz,$

where $z_{q}$ is the center of $q$. The two integrals of the right side in the
above equation are of the same form. By combining (4.3), (4.20) and the above
estimate, the estimate (4.15) is reduced to showing

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\kappa}(Q):\\\
\mathrm{dist}(q,\Delta)\geq
cR\end{subarray}}\sum_{(T_{1},T_{1}^{\prime},T_{2},T_{2}^{\prime})\in\mathbf{S}}\int_{\psi_{q}^{2}}\bigg{|}\frac{m_{\Delta,T_{1}^{\prime}}}{m_{T_{1}}}\Psi_{T_{1}^{\prime}}(z_{q})\frac{\phi_{T_{1}}(z)\phi_{T_{2}}(z)}{\Psi_{T_{1}}(z_{q})}\bigg{|}^{2}dz\lesssim
c^{-C}R^{-\frac{n-1}{2}}.$ (4.21)

We separate the summation
$\sum_{(T_{1},T_{1}^{\prime},T_{2},T_{2}^{\prime})\in\mathbf{S}}$ into two
parts

$\sum_{(T_{1},T_{1}^{\prime},T_{2},T_{2}^{\prime})\in\mathbf{S}}=\sum_{T_{1}\in\mathbf{T}_{1},T_{2}\in\mathbf{T}_{2}}\sum_{\begin{subarray}{c}T_{1}^{\prime}\in\mathbf{T}_{1},T_{2}^{\prime}\in\mathbf{T}_{2}:\\\
(v(T_{1}),v(T_{1}^{\prime}),v(T_{2}),v(T_{2}^{\prime}))\in S\end{subarray}}.$

By rearranging the left side of (4.21) is bounded by

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\kappa}(Q):\\\
\mathrm{dist}(q,\Delta)\geq
cR\end{subarray}}\sum_{\begin{subarray}{c}T_{1}\in\mathbf{T}_{1},\\\
T_{2}\in\mathbf{T}_{2}\end{subarray}}\Bigg{(}\frac{1}{m_{T_{1}}}\int_{\psi_{q}^{2}}\bigg{|}\frac{\phi_{T_{1}}(z)\phi_{T_{2}}(z)}{\Psi_{T_{1}}(z_{q})}\bigg{|}^{2}dz\sum_{\begin{subarray}{c}T_{1}^{\prime}\in\mathbf{T}_{1},T_{2}^{\prime}\in\mathbf{T}_{2}:\\\
(v(T_{1}),v(T_{1}^{\prime}),v(T_{2}),v(T_{2}^{\prime}))\in
S\end{subarray}}{m_{\Delta,T_{1}^{\prime}}}\Psi_{T_{1}^{\prime}}^{2}(z_{q})\Bigg{)},$

where ${(\frac{m_{\Delta,T}}{m_{T}})^{2}\leq\frac{m_{\Delta,T}}{m_{T}}}$ is
used. To show (4.21) it suffices to prove the following two estimates:

$\max_{\begin{subarray}{c}q\in\mathcal{Q}_{\kappa}(Q):\\\
\mathrm{dist}(q,\Delta)\geq
cR\end{subarray}}\max_{\begin{subarray}{c}T_{1}\in\mathbf{T}_{1},\\\
T_{2}\in\mathbf{T}_{2}\end{subarray}}\sum_{\begin{subarray}{c}T_{1}^{\prime}\in\mathbf{T}_{1},T_{2}^{\prime}\in\mathbf{T}_{2}:\\\
(v(T_{1}),v(T_{1}^{\prime}),v(T_{2}),v(T_{2}^{\prime}))\in
S\end{subarray}}{m_{\Delta,T_{1}^{\prime}}}\Psi_{T_{1}^{\prime}}^{2}(z_{q})\lesssim
c^{-C}R^{1/2}$ (4.22)

and

$\sum_{q\in\mathcal{Q}_{\kappa}(Q)}\sum_{\begin{subarray}{c}T_{1}\in\mathbf{T}_{1},\\\
T_{2}\in\mathbf{T}_{2}\end{subarray}}\frac{1}{m_{T_{1}}}\int_{\psi_{q}^{2}}\bigg{|}\frac{\phi_{T_{1}}(z)\phi_{T_{2}}(z)}{\Psi_{T_{1}}(z_{q})}\bigg{|}^{2}dz\\\
\lesssim c^{-C}R^{-n/2}.$ (4.23)

###  4.4. Proof of the estimate (4.22)

We take a close look at the condition

$(v(T_{1}),v(T_{1}^{\prime}),v(T_{2}),v(T_{2}^{\prime}))\in S.$

From the first equation of (4.18) we have

$v_{2}^{\prime}=v_{1}+v_{2}-v_{1}^{\prime}+O(r^{-1}).$ (4.24)

Inserting this into the second equation of (4.18), we have

$|v_{1}|^{2}-|v_{1}+v_{2}-v_{1}^{\prime}|^{2}=|v_{1}^{\prime}|^{2}-|v_{2}|^{2}+O(r^{-1}),$

which is equivalent to

$(v_{1}^{\prime}-v_{1})\cdot(v_{1}^{\prime}-v_{2})=O(r^{-1}).$

Let $\sigma(v_{1},v_{2})\subset\mathbb{R}^{n}$ be the sphere of radius
$\frac{|v_{1}-v_{2}|}{2}$ with center $\frac{v_{1}+v_{2}}{2}$. Then the above
equation means that $v_{1}^{\prime}$ lies in the $O(r^{-1})$-neighborhood
$\sigma(v_{1},v_{2};Cr^{-1})$ of the sphere $\sigma(v_{1},v_{2})$. Thus, if
$v_{1},~{}v_{2}$ are given then one has
$v_{1}^{\prime}\in\sigma(v_{1},v_{2};Cr^{-1})$. Also, if
$v_{1},~{}v_{2},~{},v_{1}^{\prime}$ are given then by (4.24) we see that
$v_{2}^{\prime}$ is contained in a ball
$B(v_{1}+v_{2}-v_{1}^{\prime};Cr^{-1})$.

Now we use this observation. For given $T_{1}$, $T_{2}$, we have that
$v(T_{1}^{\prime})$ is contained in

$\sigma(v(T_{1}),v(T_{2});Cr^{-1}).$

If $v(T_{1}),v(T_{2}),v(T_{1}^{\prime})$ is determined then
$v(T_{2}^{\prime})$ has $O(1)$ choices, and if $v(T_{2}^{\prime})$ is
determined then the number of $T_{2}^{\prime}$ passing through $z_{q}$ is only
one. Thus to show (4.22) it suffices to show that for given
$T_{1}\in\mathbf{T}_{1}$, $T_{2}\in\mathbf{T}_{2}$ and
$q\in\mathcal{Q}_{\kappa}(Q)$ with $\mathrm{dist}(q,\Delta)\geq cR$,

$\sum_{\begin{subarray}{c}T_{1}^{\prime}\in\mathbf{T}_{1}:\\\
v(T_{1}^{\prime})\in
V_{1}\cap\sigma(v_{1},v_{2};Cr^{-1})\end{subarray}}m_{\Delta,T_{1}^{\prime}}\Psi_{T_{1}^{\prime}}^{2}(z_{q})\lesssim
c^{-C}R^{1/2},$ (4.25)

where $v_{1}=v(T_{1})$ and $v_{2}=v(T_{2})$. By (4.2), we have

$\sum_{\begin{subarray}{c}T_{1}^{\prime}\in\mathbf{T}_{1}:\\\
v(T_{1}^{\prime})\in
V_{1}\cap\sigma(v_{1},v_{2};Cr^{-1})\end{subarray}}m_{\Delta,T_{1}^{\prime}}\Psi_{T_{1}^{\prime}}^{2}(z_{q})\lesssim\sum_{T_{2}\in\mathbf{T}_{2}}\int_{4Q}|\phi_{T_{2}}(z)|^{2}\Gamma_{q}(z)dz+R^{-C},$
(4.26)

where

$\Gamma_{q}(z):=\sum_{\begin{subarray}{c}T_{1}^{\prime}\in\mathbf{T}_{1}:\\\
v(T_{1}^{\prime})\in
V_{1}\cap\sigma(v_{1},v_{2};Cr^{-1})\end{subarray}}\psi_{\Delta}(z)\Psi_{T_{1}^{\prime}}^{2}(z)\Psi_{T_{1}^{\prime}}^{2}(z_{q}).$

Consider $\Gamma_{q}(z)$. The union of $T_{1}^{\prime}$ passing through
$z_{q}$ with $v(T_{1}^{\prime})\in V_{1}\cap\sigma(v_{1},v_{2};Cr^{-1})$ forms
a conic set $\Lambda_{\pi_{1}}(z_{q};Cr)$ where

$\pi_{1}:={\sigma(v_{1},v_{2})\cap\tilde{\Xi}_{1}}.$

From $\mathrm{dist}(\Delta,q)\geq cR$ it follows that the tubes
$T_{1}^{\prime}$ passing through $z_{q}$ can overlap at most $O(1)$ times on
$\Delta$. Thus,

$\int_{4Q}|\phi_{T_{2}}(z)|^{2}\Gamma_{q}(z)dz\lesssim
c^{-C}\int_{4Q}|\phi_{T_{2}}(z)|^{2}\bigg{(}1+\frac{\mathrm{dist}(z,\Lambda_{\pi_{1}}(z_{q}))}{r}\bigg{)}^{-N^{10}}dz.$

By combining this with (4.26),

$\sum_{\begin{subarray}{c}T_{1}^{\prime}\in\mathbf{T}_{1}:\\\
v(T_{1}^{\prime})\in
V_{1}\cap\sigma(v_{1},v_{2};Cr^{-1})\end{subarray}}m_{\Delta,T_{1}^{\prime}}\Psi_{T_{1}^{\prime}}^{2}(z_{q})\lesssim
c^{-C}\sum_{T_{2}\in\mathbf{T}_{2}}\int_{4Q}|\phi_{T_{2}}(z)|^{2}\bigg{(}1+\frac{\mathrm{dist}(z,\Lambda_{\pi_{1}}(z_{q}))}{r}\bigg{)}^{-N^{10}}dz+R^{-C}.$

By a dyadic decomposition, to prove (4.25) it suffices to show

$\sum_{T_{2}\in\mathbf{T}_{2}}\|\phi_{T_{2}}\|^{2}_{L^{2}(4Q\cap\Lambda_{\pi_{1}}(z_{q};Cr))}\lesssim
c^{-C}r.$ (4.27)

We observe that for any $w_{2}\in\Xi_{2}$ and $w_{1},w_{1}^{\prime}\in\pi_{1}$
with $w_{1}\neq w_{1}^{\prime}$, there exists $0\leq\varepsilon_{0}<1$ such
that

$\Big{\langle}\frac{w_{2}-w_{1}}{|w_{2}-w_{1}|},\frac{w_{1}^{\prime}-w_{1}}{|w_{1}^{\prime}-w_{1}|}\Big{\rangle}\leq\varepsilon_{0}.$
(4.28)

Indeed, if we take $\xi_{2}\in\sigma(v_{1},v_{2})$ such that
$w_{1}^{\prime}+\xi_{2}=v_{1}+v_{2}$, that is, $(w_{1}^{\prime}+\xi_{2})/2$ is
the center of $\sigma(v_{1},v_{2})$, then $\langle
w_{1}-\xi_{2},w_{1}^{\prime}-w_{1}\rangle=0.$ Using this we have

$\text{the left side of
\eqref{transv}}=\Big{\langle}\frac{w_{2}-\xi_{2}}{|w_{2}-w_{1}|},\frac{w_{1}^{\prime}-w_{1}}{|w_{1}^{\prime}-w_{1}|}\Big{\rangle}.$

So the above equation is bounded by ${|w_{2}-\xi_{2}|}/{|w_{2}-w_{1}|}.$ From
the definition of $\Xi_{1}$ and $\Xi_{2}$, we can see that there is
$0\leq\varepsilon_{0}<1$ such that
${|w_{2}-\xi_{2}|}/{|w_{2}-w_{1}|}\leq\varepsilon_{0}$. Thus we have (4.28).
By Lemma 3.3,

$\text{the left side of \eqref{trans_app}}\lesssim
r\sum_{T_{2}\in\mathbf{T}_{2}}E(\phi_{T_{2}}).$

By (3.10), (3.11) and (2.7),

$\sum_{T_{2}\in\mathbf{T}_{2}}E(\phi_{T_{2}})\lesssim
c^{-C}\sum_{T_{2}\in\mathbf{T}_{2}}h_{T_{2}}^{2}\lesssim c^{-C}.$ (4.29)

By combining these two estimates we obtain (4.27).

∎

###  4.5. Proof of the estimate (4.23)

Consider the integral in the left side of (4.23). By applying Lemma 3.4 to the
left side of (4.23) it suffices to show

$\sum_{q\in\mathcal{Q}_{\kappa}(Q)}\sum_{\begin{subarray}{c}T_{1}\in\mathbf{T}_{1},\\\
T_{2}\in\mathbf{T}_{2}\end{subarray}}\frac{1}{m_{T_{1}}}\frac{\|\phi_{T_{1}}\|_{L^{2}({\psi_{q}})}^{2}}{\Psi_{T_{1}}^{2}(z_{q})}\|\phi_{T_{2}}\|_{L^{2}({\psi_{q}})}^{2}\lesssim
c^{-C}R^{1/2}.$ (4.30)

The left side of the above equation is written as

$\sum_{T_{1}\in\mathbf{T}_{1}}\Bigg{(}\sup_{q\in\mathcal{Q}_{\kappa}(Q)}\frac{\|\phi_{T_{1}}\|_{L^{2}({\psi_{q}})}^{2}}{\Psi_{T_{1}}^{4}(z_{q})}\Bigg{)}\Bigg{(}\frac{1}{m_{T_{1}}}\sum_{q\in\mathcal{Q}_{\kappa}(Q)}\sum_{T_{2}\in\mathbf{T}_{2}}\Psi_{T_{1}}^{2}(z_{q})\|\phi_{T_{2}}\|_{L^{2}({\psi_{q}})}^{2}\Bigg{)}.$

Consider the inner summand. Since both the width of $T$ and the sidelength of
$q$ are $\sim r$, by some basic estimates it follows that for any tube
$T\in\mathbf{T}_{1}\cup\mathbf{T}_{2}$ and $q\in\mathcal{Q}_{\kappa}$,

$\Psi_{T}(z)\psi_{q}^{1/2}(z)\sim\Psi_{T}(z_{q}).$ (4.31)

Using this equation we have

$\sum_{q\in\mathcal{Q}_{\kappa}(Q)}\sum_{T_{2}\in\mathbf{T}_{2}}\Psi_{T_{1}}^{2}(z_{q})\|\phi_{T_{2}}\|_{L^{2}({\psi_{q}})}^{2}\lesssim\sum_{q\in\mathcal{Q}_{\kappa}(Q)}\sum_{T_{2}\in\mathbf{T}_{2}}\|\Psi_{T_{1}}\phi_{T_{2}}\|_{L^{2}(\psi_{q}^{3/2})}^{2},$

which is $\lesssim m_{T_{1}}$ by (4.3). Now to prove (4.30) it is enough to
show

$\sum_{T_{1}\in\mathbf{T}_{1}}\sup_{q\in\mathcal{Q}_{\kappa}(Q)}\frac{\|\phi_{T_{1}}\|_{L^{2}({\psi_{q}})}^{2}}{\Psi_{T_{1}}^{4}(z_{q})}\lesssim
c^{-C}R^{1/2}.$

By (3.10) and (4.31),

$\frac{\|\phi_{T_{1}}\|_{L^{2}({\psi_{q}})}^{2}}{\Psi_{T_{1}}^{4}(z_{q})}\lesssim
c^{-C}r^{-n}h_{T_{1}}^{2}\frac{\|\Psi_{T_{1}}^{2}\|_{L^{2}(\psi_{q})}^{2}}{\Psi_{T_{1}}^{4}(z_{q})}\lesssim
c^{-C}rh_{T_{1}}^{2}.$

Therefore, by (3.11)

$\sum_{T_{1}\in\mathbf{T}_{1}}\sup_{q\in\mathcal{Q}_{\kappa}(Q)}\frac{\|\phi_{T_{1}}\|_{L^{2}({\psi_{q}})}^{2}}{\Psi_{T_{1}}^{4}(z_{q})}\lesssim
c^{-C}r\sum_{T_{1}\in\mathbf{T}_{1}}h_{T_{1}}^{2}\lesssim c^{-C}R^{1/2}.$

∎

##  5\. Proof of Proposition 2.3

By Proposition 4.1 it suffices to show

$\|[\Phi_{1}][\Phi_{2}]\|_{L^{p}(Q)}\leq(1+Cc)\overline{\mathcal{K}}(R/C_{0})E(\phi_{1})^{1/2}E(\phi_{2})^{1/2}.$
(5.1)

By (4.7),

$\|[\Phi_{1}][\Phi_{2}]\|_{L^{p}(Q)}\leq\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\|\Phi_{1}^{(\Delta)}\Phi_{2}^{(\Delta)}\|_{L^{p}(\Delta)}.$

From Proposition 4.1 we see that $\mathrm{marg}(\Phi_{j}^{(\Delta)})\geq
1/100-(2^{C_{0}}/R)^{1/N}$. By Definition 2.2 the right side of the above
estimate is bounded by

$\overline{\mathcal{K}}(R/C_{0})\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E(\Phi_{1}^{(\Delta)})^{1/2}E(\Phi_{2}^{(\Delta)})^{1/2}.$

By the Cauchy–Schwarz inequality, this is bounded by

$\overline{\mathcal{K}}(R/C_{0})\bigg{(}\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E(\Phi_{1}^{(\Delta)})\bigg{)}^{1/2}\bigg{(}\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E(\Phi_{2}^{(\Delta)})\bigg{)}^{1/2}.$

Thus, to show (5.1) it suffices to show

$\bigg{(}\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E(\Phi_{j}^{(\Delta)})\bigg{)}^{1/2}\leq(1+Cc)E(\phi_{j})^{1/2}$

for $j=1,2$.

######  Lemma 5.1.

Let $\mathcal{Q}$ be a finite index set. Suppose that $m_{q,T_{j}}$ are non-
negative numbers such that

$\sum_{q\in\mathcal{Q}}m_{q,T_{j}}\leq 1$ (5.2)

for all $T_{j}\in\mathbf{T}_{j}$. Then,

$\bigg{(}\sum_{q\in\mathcal{Q}}E\Big{(}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}\Big{)}\bigg{)}^{1/2}\leq(1+Cc)E(\phi_{j})^{1/2}.$
(5.3)

###### Proof.

To get the constant $(1+Cc)$, we need to consider the Fourier support of
$f_{T_{j}}$. Let

$Y_{j}=\bigcup_{v_{j}\in V_{j}}\\{\xi\in
B_{v_{j}}:\mathrm{dist}(\xi,\tilde{\Xi}_{j}\setminus
B_{v_{j}})>Cc^{2}R^{-1/2}\\}.$

We define the operator $P_{\Omega(Y)}$ by

$\widehat{P_{\Omega(Y)}f}=\chi_{\Omega(Y)}\widehat{f}.$

By Minkowski’s inequality,

$\displaystyle\bigg{(}\sum_{q\in\mathcal{Q}}E\big{(}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}\big{)}\bigg{)}^{1/2}$
$\displaystyle=\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}(0)\Big{\|}_{2}^{2}\bigg{)}^{1/2}$
$\displaystyle=\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\int_{G}\eta^{x(T_{j})}P_{\Omega,v(T_{j})}\phi_{j}(0)d\Omega\Big{\|}_{2}^{2}\bigg{)}^{1/2}$
$\displaystyle\leq\int_{G}\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\eta^{x(T_{j})}P_{\Omega,v(T_{j})}\phi_{j}(0)\Big{\|}_{2}^{2}\bigg{)}^{1/2}{d\Omega},$

which is less than or equal to the sum of

$\int\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\eta^{x(T_{j})}P_{\Omega,v(T_{j})}P_{\Omega(Y_{j})}\phi_{j}(0)\Big{\|}_{2}^{2}\bigg{)}^{1/2}{d\Omega}$
(5.4)

and

$\int\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\eta^{x(T_{j})}P_{\Omega,v(T_{j})}(1-P_{\Omega(Y_{j})})\phi_{j}(0)\Big{\|}_{2}^{2}\bigg{)}^{1/2}{d\Omega}.$
(5.5)

To prove (5.3), it suffices to show that

$\eqref{B'v}\leq E(\phi_{j})^{1/2}\qquad\text{and}\qquad\eqref{Bv}\leq
CcE(\phi_{j})^{1/2}.$

Consider (5.4). From (3.2) we see that the Fourier transform of $\eta^{x_{0}}$
is supported in $D(0;c^{2}R^{-1/2})$. So, the Fourier support of
$\eta^{x_{0}}P_{\Omega,v_{j}}P_{\Omega(Y_{j})}f$ is contained in $B_{v_{j}}$.
By orthogonality, (5.4) is bounded by

$\int\bigg{(}\sum_{q\in\mathcal{Q}}\sum_{v_{j}\in V_{j}}\Big{\|}\sum_{x_{0}\in
L}m_{q,T_{j}^{(x_{0},v_{j})}}\eta^{x_{0}}P_{\Omega,v_{j}}P_{\Omega(Y_{j})}\phi_{j}(0)\Big{\|}_{2}^{2}\bigg{)}^{1/2}{d\Omega}.$

By rearranging, it is equal to

$\int\bigg{(}\sum_{v_{j}\in
V_{j}}\int|P_{\Omega,v_{j}}P_{\Omega(Y_{j})}\phi_{j}(x,0)|^{2}\sum_{q\in\mathcal{Q}}\bigg{|}\sum_{x_{0}\in
L}{m_{q,T_{j}^{(x_{0},v_{j})}}}\eta^{x_{0}}(x)\bigg{|}^{2}dx\bigg{)}^{1/2}{d\Omega},$

which is bounded by

$\int\bigg{(}\sum_{v_{j}\in
V_{j}}\int|P_{\Omega,v_{j}}P_{\Omega(Y_{j})}\phi_{j}(x,0)|^{2}\bigg{|}\sum_{q\in\mathcal{Q}}\sum_{x_{0}\in
L}{m_{q,T_{j}^{(x_{0},v_{j})}}}\eta^{x_{0}}(x)\bigg{|}^{2}dx\bigg{)}^{1/2}{d\Omega}.$

By (3.3) and (5.2), this is bounded by

$\int\bigg{(}\sum_{v_{j}\in
V_{j}}\int|P_{\Omega,v_{j}}P_{\Omega(Y_{j})}\phi_{j}(x,0)|^{2}dx\bigg{)}^{1/2}{d\Omega}.$

By orthogonality, the above is bounded by

$\int\|P_{\Omega(Y_{j})}\phi_{j}(0)\|_{2}{d\Omega}.$

Since $\|P_{\Omega(Y_{j})}\phi_{j}(0)\|_{2}\leq E(\phi_{j})^{1/2}$, we have
that $\eqref{B'v}\leq E(\phi_{j})^{1/2}$.

Consider (5.5). Apply the previous arguments but using almost orthogonality
instead of orthogonality. Then we have

$\eqref{Bv}\lesssim\int\|(1-P_{\Omega(Y_{j})})\phi_{j}(0)\|_{2}{d\Omega}.$

By the Cauchy–Schwarz inequality this is bounded by

$\Big{(}\int\|(1-P_{\Omega(Y_{j})})\phi_{j}(0)\|_{2}^{2}{d\Omega}\Big{)}^{1/2}.$

By Plancherel’s theorem and rearranging the integrals, this is equal to

$\Big{(}\int\|(1-\chi_{\Omega(Y_{j})})\widehat{\phi_{j}(0)}\|_{2}^{2}{d\Omega}\Big{)}^{1/2}=\Big{(}\int\Big{(}\int(1-\chi_{\Omega(Y_{j})})(\xi){d\Omega}\Big{)}|\widehat{\phi_{j}(0)}(\xi)|^{2}d\xi\Big{)}^{1/2}.$

By a direct calculation we have that for any $\xi\in\tilde{\Xi}_{j}$,

$\displaystyle\int 1-\chi_{\Omega(Y_{j})}(\xi){d\Omega}$
$\displaystyle=\frac{1}{|D(0;CR^{-1/2})|}\int_{D(0;CR^{-1/2})}1-\chi_{Y_{j}}(\xi+w)dw$
$\displaystyle\lesssim c^{2}.$

Inserting this into the previous, we obtain that $\eqref{Bv}\lesssim
c^{2}E(\phi_{j})$.

∎

##  6\. A localization operator

In this section we introduce a localization operator and state some relevant
basic estimates. When exploiting energy concentrations, the localization
operator is used as a tool.

By (2.2) we have $\widehat{\phi_{j}(t)}(\xi)=e^{-\pi
i(t-t_{0})|\xi|^{2}}\widehat{\phi_{j}(t_{0})}(\xi),$ which is written as

$\phi_{j}(t)=\mathcal{U}_{j}[\phi_{j}(t_{0})](t-t_{0}).$ (6.1)

######  Definition 6.1.

Let $D=D(x_{D},t_{D};r)$ be a disc. We define an operator $P_{D}\phi_{j}$ by

$P_{D}\phi_{j}(t)=\mathcal{U}_{j}[(\chi_{D}\ast\eta_{r^{1-1/N}})\phi_{j}(t_{D})](t-t_{D})$
(6.2)

where $\eta_{r}$ is defined as (1.8).

######  Lemma 6.2.

Let $r\geq C_{0}$, $D=D(x_{D},t_{D};r)$ and

$D^{\pm}:=D(x_{D},t_{D};r(1\pm r^{-1/2N})).$

Suppose that $\phi_{j}$ satisfies that $\mathrm{marg}(\phi_{j})\geq
C_{0}r^{-1+1/N}$ for $j=1,2$. Then,

$\mathrm{marg}(P_{D}\phi_{j})\geq\mathrm{marg}(\phi_{j})-C_{0}r^{-1+1/N}$
(6.3)

and

$\displaystyle\|P_{D}\phi_{j}(t_{D})\|_{L^{2}(\mathbb{R}^{n}\setminus D^{+})}$
$\displaystyle\lesssim r^{-N}E(\phi_{j})^{1/2},$ (6.4)
$\displaystyle\|(1-P_{D})\phi_{j}\|_{L^{2}(D^{-})}$ $\displaystyle\lesssim
r^{-N}E(\phi_{j})^{1/2},$ (6.5) $\displaystyle E(P_{D}\phi_{j})$
$\displaystyle\leq\|\phi_{j}\|_{L^{2}(D^{+})}^{2}+Cr^{-N}E(\phi_{j}),$ (6.6)
$\displaystyle E((1-P_{D})\phi_{j})$
$\displaystyle\leq\|\phi_{j}(t_{D})\|^{2}_{L^{2}(\mathbb{R}^{n}\setminus
D^{-})}+Cr^{-N}E(\phi_{j}).$ (6.7)

###### Proof.

Consider (6.3). Observe that the size of
$\operatorname{supp}\widehat{\phi}_{j}$ is comparable to that of
$\operatorname{supp}\widehat{\phi_{j}(0)}$. Similarly, the size of
$\operatorname{supp}\widehat{P_{D}\phi_{j}}$ and
$\operatorname{supp}\widehat{P_{D}\phi_{j}(t_{D})}$ is comparable. From (6.2)
we have

$\widehat{P_{D}\phi_{j}(t_{D})}=(\widehat{\chi_{D}}\widehat{\eta_{r^{1-1/N}}})\ast\widehat{\phi_{j}(t_{D})}.$
(6.8)

Since $\widehat{\eta_{r^{1-1/N}}}(\xi)=\widehat{\eta}(r^{1-1/N}\xi)$ is
supported on $D(0;r^{-1+1/N})$, the Fourier support of $P_{D}\phi_{j}(t_{D})$
is expanded $O(r^{-1+1/N})$ more than that of $\phi_{j}(t_{D})$. Thus we have
(6.3).

We have that $\frac{1}{2}D\subset D^{-}\subset D\subset D^{+}\subset 2D.$ From
this relation it follows that

$\displaystyle 0\leq\chi_{D}\ast\eta_{r^{1-1/N}}\leq 1,$
$\displaystyle\chi_{D}\ast\eta_{r^{1-1/N}}(x)\lesssim r^{-N}\quad\text{ for
$x\in\mathbb{R}^{n}\setminus D^{+}$},$ (6.9) $\displaystyle
1-\chi_{D}\ast\eta_{r^{1-1/N}}(x)\lesssim r^{-N}\quad\text{ for $x\in D^{-}$
}.$ (6.10)

Indeed, the first one is trivial. Consider (6.9). We have that

$\chi_{D}\ast\eta_{r^{1-1/N}}(x)\lesssim\bigg{(}1+\frac{\mathrm{dist}(x,D)}{r^{1-1/N}}\bigg{)}^{-M},\qquad\forall
M>0.$ (6.11)

If $x\in\mathbb{R}^{n}\setminus D^{+}$ then

$\mathrm{dist}(x,D)\geq r(1+r^{-1/2N})-r=r^{1-1/2N}.$

By inserting this into the previous inequality we can obtain (6.9).

Consider (6.10). We have that

$1-\chi_{D}\ast\eta_{r^{1-1/N}}(x)\lesssim\bigg{(}1+\frac{\mathrm{dist}(x,\mathbb{R}^{n}\setminus
D)}{r^{1-1/N}}\bigg{)}^{-M},\qquad\forall M>0.$ (6.12)

If $x\in D^{-}$, then

$\mathrm{dist}(x,\mathbb{R}^{n}\setminus D)\geq r-r(1-r^{-1/2N})=r^{1-1/2N}.$

Thus we have (6.10).

Now consider from (6.4) to (6.7). By (6.2) and (6.9) it follows that

$\|P_{D}\phi_{j}(t_{D})\|_{L^{2}(\mathbb{R}^{n}\setminus
D^{+})}=\|(\chi_{D}\ast\eta_{r^{1-1/N}})\phi_{j}(t_{D})\|_{L^{2}(\mathbb{R}^{n}\setminus
D^{+})}\lesssim r^{-N}E(\phi_{j})^{1/2}.$

So we have (6.4). Similar arguments also give (6.5).

From (6.4) it follows that

$\displaystyle E(P_{D}\phi_{j})$
$\displaystyle=\|P_{D}\phi_{j}(t_{D})\|_{2}^{2}$
$\displaystyle\leq\|P_{D}\phi_{j}\|^{2}_{L^{2}(D^{+})}+Cr^{-N}E(\phi_{j})^{1/2}$
$\displaystyle\leq\|\phi_{j}\|^{2}_{L^{2}(D^{+})}+Cr^{-N}E(\phi_{j})^{1/2}.$

So we have (6.6). Analogously we have (6.7). ∎

Now we consider some properties of $P_{D}\phi_{j}$ in
$\mathbb{R}^{n}\times\mathbb{R}$.

######  Lemma 6.3.

Let $r\geq C_{0}$ and $D=D(x_{D},t_{D};r)$. Suppose that $\phi_{j}$ satisfies
that $\mathrm{marg}(\phi_{j})\geq C_{0}r^{-1+1/N}$. Then,

$\displaystyle|P_{D}\phi_{j}(x,t)$ $\displaystyle|\leq
C_{M}r^{n/2}\bigg{(}1+\frac{\mathrm{dist}\big{(}(x,t),\Lambda_{j}(x_{D},t_{D};r)\big{)}}{r^{1-1/N}}\bigg{)}^{-M}E(\phi_{j})^{1/2},\qquad\forall
M>0$ (6.13) and
$\displaystyle\|(1-P_{D})\phi_{j}\|_{L^{\infty}(Q(x_{D},t_{D};r/4))}\lesssim
r^{-N}E(\phi_{j})^{1/2}.$ (6.14)

###### Proof.

Consider (6.13). By (6.2) and (3.15),

$P_{D}\phi_{j}(x,t)=\int
K_{j,t-t_{D}}(x-y)(\chi_{D}\ast\eta_{r^{1-1/N}})(y)\phi_{j}(y,t_{D})dy.$

By Lemma 3.2 and (6.11),

$|K_{j,t-t_{D}}(x-y)(\chi_{D}\ast\eta_{r^{1-1/N}})^{1/2}(y)|\leq
C_{M}\bigg{(}1+\frac{\mathrm{dist}\big{(}(x,t),\Lambda_{j}(x_{D},t_{D};r)\big{)}}{r^{1-1/N}}\bigg{)}^{-M}$

for any $M>0$. Thus,

$\displaystyle|P_{D}\phi_{j}(x,t)|$ $\displaystyle\leq
C_{M}\bigg{(}1+\frac{\mathrm{dist}\big{(}(x,t),\Lambda_{j}(x_{D},t_{D};r)\big{)}}{r^{1-1/N}}\bigg{)}^{-M}\int(\chi_{D}\ast\eta_{r^{1-1/N}})^{1/2}(y)|\phi_{j}(y,t_{D})|dy$
$\displaystyle\leq
C_{M}r^{n/2}\bigg{(}1+\frac{\mathrm{dist}\big{(}(x,t),\Lambda_{j}(x_{D},t_{D};r)\big{)}}{r^{1-1/N}}\bigg{)}^{-M}E(\phi_{j})^{1/2},$

where the last line follows from the Cauchy–Schwarz inequality.

Consider (6.14). Similarly, by (6.2) and (3.15), it is written as

$(1-P_{D})\phi_{j}(x,t)=\int
K_{j,t-t_{D}}(x-y)(1-\chi_{D}\ast\eta_{r^{1-1/N}})(y)\phi_{j}(y,t_{D})dy.$

By the Cauchy–Schwarz inequality,

$|(1-P_{D})\phi_{j}(x,t)|\leq\bigg{(}\int\big{|}K_{j,t-t_{D}}(x-y)(1-\chi_{D}\ast\eta_{r^{1-1/N}})(y)\big{|}^{2}dy\bigg{)}^{1/2}E(\phi_{j})^{1/2}.$
(6.15)

By Lemma 3.2 and (6.12) we have that for $(x,t)\in Q(x_{D},t_{D};r/4)$,

$\displaystyle\int\big{|}K_{j,t-t_{D}}(x-y)(1-\chi_{D}\ast\eta_{r^{1-1/N}})(y)\big{|}^{2}dy$
$\displaystyle\lesssim r^{-M}\int|K_{j,t-t_{D}}(x-y)|dy$
$\displaystyle\lesssim r^{-M}$

for any $M>0$. Substituting this in (6.15) we can obtain (6.14).

∎

######  Lemma 6.4.

Let $r\geq C_{0}$ and $D=D(x_{D},t_{D};r)$. Then, for each
$t_{0}\in\mathbb{R}$ there is a disc $\mathcal{G}_{j,t_{0}}(D)$ of radius
$A_{w}|t_{D}-t_{0}|/2+4r$ in $\mathbb{R}^{n}\times\\{t_{0}\\}$ such that
$\mathcal{G}_{j,t_{0}}(D)$ contains
$\Lambda_{j}(x_{D},t_{D};r)\cap(\mathbb{R}^{n}\times\\{t_{0}\\})$ and

$E(P_{D}\phi_{j})\leq\|\phi_{j}\|_{L^{2}(\mathcal{G}_{j,t_{0}}(D))}+Cr^{-N}E(\phi_{j}).$

###### Proof.

By (6.1) and (6.2),

$\displaystyle P_{D}\phi_{j}(x,t_{D})$
$\displaystyle=(\chi_{D}\ast\eta_{r^{1-1/N}})(x)\mathcal{U}_{j}[\phi_{j}(t_{0})](x,t_{D}-t_{0})$
$\displaystyle=\int(\chi_{D}\ast\eta_{r^{1-1/N}})(x)K_{j,t_{D}-t_{0}}(x-y)\phi_{j}(y,t_{0})dy.$
(6.16)

If we ignore Schwarz tails, the equation $K_{j,t_{D}-t_{0}}(x-y)$ implies that
$x-y$ is contained in $\Lambda_{j,t_{D}-t_{0}}$ by Lemma 3.2. The
$\Lambda_{j,t_{D}-t_{0}}$ is contained in a disc of radius
$A_{w}|t_{D}-t_{0}|/2+C$, so $x-y$ is contained in a disc of radius
$A_{w}|t_{D}-t_{0}|/2+C$. Since $x$ is contained in $D$, we see that that $y$
is contained in a disc $D_{j}^{\star}:=D(x_{j},t_{0};A_{w}|t_{D}-t_{0}|/2+2r)$
for some $x_{j}$. Using the symmetric property of $\Lambda_{j}$ about the
origin, we also have that $y-x$ is contained in $\Lambda_{j,t_{0}-t_{D}}$,
which implies $(y,t_{0})\in\Lambda_{j}(x,t_{D})$. So, we can see that
$D_{j}^{\star}$ contains
$\Lambda_{j}(x_{D},t_{D};r)\cap\mathbb{(}R^{n}\times\\{t_{0}\\})$. Using this
observation we have

$|(\chi_{D}\ast\eta_{r^{1-1/N}})^{1/2}(x)K_{j,t_{D}-t_{0}}(x-y)|\leq
C_{M}\bigg{(}1+\frac{\mathrm{dist}(y,D_{j}^{\star})}{r^{1-1/N}}\bigg{)}^{-M},\quad\forall
M>0.$

Let
$\tilde{D}_{j}^{\star}:=D(x_{j},t_{0};A_{w}|t_{D}-t_{0}|/2+2r(1+r^{-1/2N}))$.
By the above estimate,

$|(\chi_{D}\ast\eta_{r^{1-1/N}})^{1/2}(x)K_{j,t_{D}-t_{0}}(x-y)(1-\chi_{\tilde{D}_{j}^{\star}}(y))|\leq
C_{M}r^{-M},\quad\forall M>0.$

We divide
$\phi_{j}(t_{0})=\chi_{\tilde{D}_{j}^{\star}}\phi_{j}(t_{0})+(1-\chi_{\tilde{D}_{j}^{\star}})\phi_{j}(t_{0})$
and insert it into (6.16). Then,

$\displaystyle\|P_{D}\phi_{j}(t_{D})\|_{2}$
$\displaystyle\leq\|\mathcal{U}_{j}[\chi_{\tilde{D}_{j}^{\star}}\phi_{j}(t_{0})](t_{D}-t_{0})\|_{2}+Cr^{-N}E(\phi_{j})^{1/2}$
$\displaystyle\leq\|\chi_{\tilde{D}_{j}^{\star}}\phi_{j}(t_{0})\|_{2}+Cr^{-N}E(\phi_{j})^{1/2}$
$\displaystyle\leq\|\phi_{j}\|_{L^{2}(D(x_{j},t_{0};A_{w}|t_{D}-t_{0}|/2+4r))}+Cr^{-N}E(\phi_{j})^{1/2}.$

If we take $\mathcal{G}_{j,t_{0}}(D):=D(x_{j},t_{0};A_{w}|t_{D}-t_{0}|/2+4r)$
then we have the desired estimate. Since $D_{j}^{\star}$ contains
$\Lambda(x_{D},t_{D};r)\cap\mathbb{(}R^{n}\times\\{t_{0}\\})$, the
$\mathcal{G}_{j,t_{0}}(D)$ also contains
$\Lambda_{j}(x_{D},t_{D};r)\cap(\mathbb{R}^{n}\times\\{t_{0}\\})$. ∎

##  7\. Proof of Proposition 2.6

Let $\phi_{1}$, $\phi_{2}$ satisfy the margin (2.5) and the normalization
(2.7). Let $Q=Q(x_{Q},t_{Q};R)$ be a cube of sidelength $R$ and centered at
$(x_{Q},t_{Q})$ and let $I_{Q}=[t_{Q}-R/2,t_{Q}+R/2]$ be the time interval of
$Q$.

By Definition 2.2 it suffices to show that for each $0<\varepsilon\ll 1$

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq(1-C_{0}^{-C})\sup_{\begin{subarray}{c}r\geq
C_{0}^{-C}R,\\\
r/100<\mathring{r}\leq(A_{w}A_{d}^{-1}+C_{0}^{-C})r\end{subarray}}\mathcal{K}_{\varepsilon}(R,r,\mathring{r}).$

By Definition 2.5 we have

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq\mathcal{K}_{\varepsilon}(R,r,\mathring{r})E_{r,\mathring{r},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})^{1/p^{\prime}}.$

It suffices to show the following proposition.

######  Proposition 7.1.

Let $\phi_{1}$, $\phi_{2}$ be the same as described above and $R\geq
2^{C_{0}}$. If $0<\delta\leq C_{0}^{-C}$, then for $0<\varepsilon\ll 1$, there
exist $t_{e}\in I_{Q}$, $r\geq C_{0}^{-C}R$ and
$r/100<\mathring{r}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r$ such that

$E^{\varepsilon}_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2})\leq 1-\delta.$

###### Proof.

Let $t\in I_{Q}$ and let $\mathfrak{D}_{t}^{\varepsilon}(\phi_{1},\phi_{2})$
and $\mathcal{N}(D)$ be defined as in Definition 2.4. We define
$\mathbf{D}^{\varepsilon}_{t}(\delta)$ to be the collection of discs
$D\in\mathfrak{D}_{t}^{\varepsilon}(\phi_{1},\phi_{2})$ such that there exist
$D_{1},D_{2}\subset\mathcal{N}(D)$ of radius
$(2A_{w}A_{d}^{-1}+C_{0}^{-C})r_{D}$ satisfying

$\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}\geq 1-\delta.$

For a disc $D$ we define $\mathring{r}_{D}$ to be the infimum of the radii of
the discs $D_{1},D_{2}\subset\mathcal{N}(D)$ satisfying the above inequality,
that is,

$\mathring{r}_{D}:=\inf\\{r:\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}\geq
1-\delta\text{ for }D_{1},D_{2}\subset\mathcal{N}(D)\text{ with
}r_{D_{1}}=r_{D_{2}}=r\\}.$

Then for $D\in\mathbf{D}_{t}^{\varepsilon}(\delta)$,

$\mathring{r}_{D}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r_{D}.$ (7.1)

Since $\mathring{r}_{D}$ is the infimum, we have

$\sup_{\begin{subarray}{c}D_{1},D_{2}\subset\mathcal{N}(D)\\\
:r_{D_{1}}=r_{D_{2}}\leq\mathring{r}_{D}\end{subarray}}\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}\leq
1-\delta.$

In fact, since $\phi_{1}$ and $\phi_{2}$ are smooth, we have the equality

$\sup_{\begin{subarray}{c}D_{1},D_{2}\subset\mathcal{N}(D)\\\
:r_{D_{1}}=r_{D_{2}}\leq\mathring{r}_{D}\end{subarray}}\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}=1-\delta.$
(7.2)

Let

$r^{\delta}(t):=\inf_{D\in\mathbf{D}_{t}^{\varepsilon}(\delta)}r_{D}.$ (7.3)

Since $\phi_{1}$ and $\phi_{2}$ are smooth, there is a disc
$D\in\mathbf{D}_{t}^{\varepsilon}(\delta)$ of radius $r^{\delta}(t)$. Let
$\mathring{r}^{\delta}(t)$ is the infimum of radii $\mathring{r}_{D}$ for
$D\in\mathbf{D}_{t}^{\varepsilon}(\delta)$ with radius $r^{\delta}(t)$, i.e.,

$\mathring{r}^{\delta}(t):=\inf_{D\in\mathbf{D}_{t}^{\varepsilon}(\delta):r_{D}=r^{\delta}(t)}\mathring{r}_{D}.$
(7.4)

Then from (7.1) it follows that for each $t\in I_{Q}$,

$\mathring{r}^{\delta}(t)\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})r^{\delta}(t).$
(7.5)

To show that $\mathring{r}^{\delta}(t)\geq r^{\delta}(t)/100$, let
$D\in\mathbf{D}_{t}^{\varepsilon}(\delta)$ be a disc of radius $r^{\delta}(t)$
with $\mathring{r}_{D}=\mathring{r}^{\delta}(t)$, and let
$D_{1},\,D_{2}\subset\mathcal{N}(D)$ be the discs of radius
$\mathring{r}^{\delta}(t)$. If we suppose
$\mathring{r}^{\delta}(t)<r^{\delta}(t)/100$, then by (7.3) the distance
between $D_{1}$ and $D_{2}$ is larger than $C^{-1}r^{\delta}(t)$. Since the
Fourier transform of $\phi_{j}(t)$ is compactly supported, for any proper disc
$D\subset\mathbb{R}^{n}\times\\{t\\}$ we have $\|\phi_{j}\|_{L^{2}(D)}\gneq
0$. Thus, there is a disc $D^{\prime}$ of radius $\lneq r^{\delta}(t)$ such
that

$\|\phi_{1}\|_{L^{2}(\mathcal{N}(D^{\prime})}\|\phi_{2}\|_{L^{2}(\mathcal{N}(D^{\prime}))}=1-\delta.$

This implies $D^{\prime}\in\mathbf{D}_{t}^{\varepsilon}(\delta)$ but it
contradicts $\eqref{radiusECen}$. Thus we have $\mathring{r}^{\delta}(t)\geq
r^{\delta}(t)/100$.

We also have that for each $t\in I_{Q}$,

$E^{\varepsilon}_{r^{\delta}(t),\mathring{r}^{\delta}(t),t}(\phi_{1},\phi_{2})=1-\delta.$
(7.6)

Indeed, let $D$ be a disc of radius $r^{\delta}(t)$ in
$\mathbb{R}^{n}\times\\{t\\}$. If $D\in\mathbf{D}_{t}^{\varepsilon}(\delta)$
then by (7.2) and (7.4),

$\sup_{\begin{subarray}{c}D_{1},D_{2}\subset\mathcal{N}(D)\\\
:r_{D_{1}}=r_{D_{2}}=\mathring{r}^{\delta}(t)\end{subarray}}\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}=1-\delta.$

If $D\notin\mathbf{D}_{t}^{\varepsilon}(\delta)$ then from the definition of
$\mathbf{D}_{t}^{\varepsilon}(\delta)$ it follows that for any discs
$D_{1},D_{2}\subset\mathcal{N}(D)$ of radius
$(2A_{w}A_{d}^{-1}+C_{0}^{-C})r^{\delta}(t)$,

$\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}<1-\delta.$

Thus we have (7.6).

We choose a time $t_{e}\in I_{Q}$ such that

$\frac{1}{2}\sup_{t\in I_{Q}}r^{\delta}(t)\leq r^{\delta}(t_{e})\leq\sup_{t\in
I_{Q}}r^{\delta}(t).$ (7.7)

By (7.5) and (7.6), to prove the proposition it suffices to show that if
$0<\delta\leq C_{0}^{-C}$ then

$r^{\delta}(t_{e})\geq C_{0}^{-C}R.$ (7.8)

Since $\phi_{1}$ and $\phi_{2}$ are smooth, from (7.6) it follows that for
each $t\in I_{Q}$, there exist discs
$D_{t}^{\delta}\subset\mathbb{R}^{n}\times\\{t\\}$ of radius $r^{\delta}(t)$
and $D_{1},D_{2}\subset\mathcal{N}(D_{t}^{\delta})$ of radius
$\mathring{r}^{\delta}(t)$ such that

$\|\phi_{1}\|_{L^{2}(D_{1})}\|\phi_{2}\|_{L^{2}(D_{2})}=1-\delta.$ (7.9)

Let $(x_{e},t_{e})$ be the center of $D_{t_{e}}^{\delta}$,

$r_{e}=r^{\delta}(t_{e})+C_{0}$

and

$\Lambda_{j,e}:=\Lambda_{j}(x_{e},t_{e};C_{0}^{2}A_{*}r_{e}).$

To have (7.8) it is enough to show

$\bigcup_{t\in I_{Q}}D_{t}^{\delta}\subset\bigcap_{j=1,2}C\Lambda_{j,e}.$
(7.10)

Indeed, since $\Lambda_{1,e}$ and $\Lambda_{2,e}$ meet transversely, the union
$\bigcup_{t\in I_{Q}}D_{t}^{\delta}$ is contained in a ball
$B(x_{e},t_{e};C_{0}^{C}r_{e})$. By comparing the length of $I_{Q}$ with the
radius $C_{0}^{C}r_{e}$, we have $r_{e}\geq C_{0}^{-C}R$, which implies (7.8)
because $R\geq 2^{C_{0}}$.

To show (7.10), by (7.7) it suffices to prove that for each $t\in I_{Q}$, the
disc $\mathcal{N}(D_{t}^{\delta})$ intersects both $\Lambda_{1,e}$ and
$\Lambda_{2,e}$. Suppose for contradiction that there exists
$\mathcal{N}(D_{t}^{\delta})$ contained in
$\mathbb{R}^{n+1}\setminus\Lambda_{j,e}$ for some $j=1,2$. Let

$D_{e}:=D(x_{e},t_{e};C_{0}A_{*}r_{e}).$

We decompose

$\|\phi_{j}\|_{L^{2}(\mathcal{N}(D_{t}^{\delta}))}^{2}\leq
4\|P_{D_{e}}\phi_{j}\|_{L^{2}(\mathcal{N}(D^{\delta}_{t}))}^{2}+4\|(1-P_{D_{e}})\phi_{j}\|_{L^{2}(\mathcal{N}(D^{\delta}_{t}))}^{2}.$
(7.11)

From
$\mathcal{N}(D_{t}^{\delta})\subset\mathbb{R}^{n+1}\setminus\Lambda_{j,e}$,
one has
$\mathrm{dist}(\mathcal{N}(D_{t}^{\delta}),\Lambda_{j}(x_{e},t_{e};C_{0}A_{*}r_{e}))\geq
C_{0}A_{*}r_{e}$. Since $r_{e}\geq C_{0}$ and the radius of
$\mathcal{N}(D_{t}^{\delta})$ is $\leq A_{*}r_{e}$, by (6.13)

$\displaystyle\|P_{D_{e}}\phi_{j}\|_{L^{2}(\mathcal{N}(D_{t}^{\delta}))}$
$\displaystyle\lesssim
r_{e}^{n/2}(C_{0}A_{*}r_{e}^{1/N})^{-N^{10}}E(\phi_{j})^{1/2}|\mathcal{N}(D_{t}^{\delta})|^{1/2}$
$\displaystyle\lesssim C_{0}^{-C}.$ (7.12)

From (6.7) and $r_{e}\geq C_{0}$ it follows that

$\displaystyle\|(1-P_{D_{e}})\phi_{j}\|^{2}_{L^{2}(\mathcal{N}(D_{t}^{\delta}))}$
$\displaystyle\leq E((1-P_{D_{e}})\phi_{j})$
$\displaystyle\leq\|\phi_{j}(t_{e})\|_{L^{2}(\mathbb{R}^{n}\setminus
D_{e}^{-})}^{2}+C_{0}^{-C}.$ (7.13)

Since $\mathcal{N}(D_{t_{e}}^{\delta})\subset\frac{1}{2}D_{e}\subset
D_{e}^{-}$, we have

$\|\phi_{j}(t_{e})\|_{L^{2}(\mathbb{R}^{n}\setminus D_{e}^{-})}^{2}\leq
1-\|\phi_{j}\|_{L^{2}(D_{e}^{-})}^{2}\leq
1-\|\phi_{j}\|_{L^{2}(\mathcal{N}(D_{t_{e}}^{\delta}))}^{2}.$

By (2.7) and (7.9),

$\|\phi_{j}\|_{L^{2}(\mathcal{N}(D_{t}^{\delta}))}\geq 1-\delta.$ (7.14)

Combining the above two estimates we have

$\|\phi_{j}\|_{L^{2}(\mathbb{R}^{n}\setminus D_{e}^{-})}^{2}\leq 2\delta.$

Substituting this in (7.13) we have

$\|(1-P_{D_{e}})\phi_{j}\|_{L^{2}(\mathcal{N}(D_{t}^{\delta}))}^{2}\leq
2\delta+C_{0}^{-C}.$

By inserting this and (7.12) into (7.11), it follows that

$\|\phi_{j}\|_{L^{2}(\mathcal{N}(D_{t}^{\delta}))}^{2}\leq
8\delta+8C_{0}^{-C}.$

Comparing this with (7.14) we have $1-8C_{0}^{-C}\leq 10\delta$. However,
since $0<\delta\leq C_{0}^{-C}$ is small, this is a contradiction. ∎

##  8\. Proof of Proposition 2.7

Let $\phi_{1},\phi_{2}$ satisfy the margin requirement (2.5) and the energy
normalization (2.7). Let $Q$ be a cube of sidelength $R$. By Definition 2.5 we
may assume that $\phi_{1}$, $\phi_{2}$ satisfy

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}=\mathcal{K}_{\varepsilon}(R,r,\mathring{r})E_{r,\mathring{r},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})^{1/p^{\prime}}.$
(8.1)

It suffices to show

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq(1+Cc)E_{r,\mathring{r},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})^{1/p^{\prime}}\overline{\mathcal{K}}(R)+2^{CC_{0}}.$
(8.2)

We may assume that

$E_{r,\mathring{r},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})=\sup_{\begin{subarray}{c}D\in\mathfrak{D}_{t}^{\varepsilon}(\phi_{1},\phi_{2})\\\
:r_{D}=r\end{subarray}}\sup_{\begin{subarray}{c}\mathring{D}_{1},\mathring{D}_{2}\subset\mathcal{N}(D)\\\
:r_{\mathring{D}_{1}}=r_{\mathring{D}_{2}}=\mathring{r}\end{subarray}}\big{(}\|\phi_{1}\|_{L^{2}(\mathring{D}_{1})}\|\phi_{2}\|_{L^{2}(\mathring{D}_{2})}\big{)}.$

Indeed, if $E_{r,\mathring{r},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})=1/2$, we
can take $\tilde{r}\geq r$ and
$\mathring{r}\leq\mathring{\tilde{r}}\leq(2A_{w}A_{d}^{-1}+C_{0}^{-C})\tilde{r}$
such that

$1/2=\sup_{\begin{subarray}{c}D\in\mathfrak{D}_{t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})\\\
:r_{D}=\tilde{r}\end{subarray}}\sup_{\begin{subarray}{c}\mathring{D}_{1},\mathring{D}_{2}\subset\mathcal{N}(D)\\\
:r_{\mathring{D}_{1}}=r_{\mathring{D}_{2}}=\mathring{\tilde{r}}\end{subarray}}\big{(}\|\phi_{1}\|_{L^{2}(\mathring{D}_{1})}\|\phi_{2}\|_{L^{2}(\mathring{D}_{2})}\big{)}.$

Thus, in (8.2) we can replace
$E_{r,\mathring{r},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})$ with
$E_{\tilde{r},\mathring{\tilde{r}},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})$.

By the smoothness of $\phi_{1}$, $\phi_{2}$, there exists a disc
$D_{e}\in\mathfrak{D}_{t_{e}}^{\varepsilon}(\phi_{1},\phi_{2})$ of radius $r$
such that

$\sup_{\begin{subarray}{c}\mathring{D}_{1},\mathring{D}_{2}\subset\mathcal{N}(D_{e})\\\
:r_{\mathring{D}_{1}}=r_{\mathring{D}_{2}}=\mathring{r}\end{subarray}}\big{(}\|\phi_{1}\|_{L^{2}(\mathring{D}_{1})}\|\phi_{2}\|_{L^{2}(\mathring{D}_{2})}\big{)}=E_{r,\mathring{r},t_{e}}^{\varepsilon}(\phi_{1},\phi_{2}).$
(8.3)

Set

$\Lambda_{j}^{[e]}:=\Lambda_{j}(x_{e},t_{e};r^{\prime}/2)$

for $j=1,2$ where $(x_{e},t_{e})$ is the center of $D_{e}$ and
$r^{\prime}:=\frac{\mathring{r}}{A_{*}}(1+(\frac{\mathring{r}}{A_{*}})^{-1/2N})$.

### 8.1.

$r^{\prime}$$A_{d}$$\Lambda_{2}^{[e]}$$\Lambda_{1}^{[e]}$ Figure 2.
Intersection of $\Lambda_{1}$ and $\Lambda_{2}$

Consider the case that $Q$ intersects both $\Lambda_{1}^{[e]}$ and
$\Lambda_{2}^{[e]}$. Let $D_{Q}=D(x_{Q},t_{Q};4R)$ be the disc of radius $4R$
with the same center as $Q$. We decompose

$\begin{split}\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq&\|P_{D_{Q}}\phi_{1}P_{D_{Q}}\phi_{2}\|_{L^{p}(Q)}\\\
&\qquad+\|P_{D_{Q}}\phi_{1}(1-P_{D_{Q}})\phi_{2}\|_{L^{p}(Q)}+\|(1-P_{D_{Q}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}.\end{split}$

From (6.14) and Hölder’s inequality it follows that

$\displaystyle\|(1-P_{D_{Q}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}$
$\displaystyle\lesssim R^{-N+C},$
$\displaystyle\|P_{D_{Q}}\phi_{1}(1-P_{D_{Q}})\phi_{2}\|_{L^{p}(Q)}$
$\displaystyle\lesssim R^{-N+C}.$

Both $P_{D_{Q}}\phi_{1}$ and $P_{D_{Q}}\phi_{2}$ satisfy the relaxed margin
condition (2.10), so by Proposition 2.3,

$\displaystyle\|P_{D_{Q}}\phi_{1}P_{D_{Q}}\phi_{2}\|_{L^{p}(Q)}$
$\displaystyle\leq(1+Cc)E(P_{D_{Q}}\phi_{1})^{1/2}E(P_{D_{Q}}\phi_{2})^{1/2}\overline{\mathcal{K}}(R)+2^{CC_{0}}$
$\displaystyle\leq(1+Cc)(E(P_{D_{Q}}\phi_{1})^{1/2}E(P_{D_{Q}}\phi_{2})^{1/2})^{1/p^{\prime}}\overline{\mathcal{K}}(R)+2^{CC_{0}}.$

By (2.9), to prove (8.2) it suffices to show

$E(P_{D_{Q}}\phi_{1})^{1/2}E(P_{D_{Q}}\phi_{2})^{1/2}\leq
E^{\varepsilon}_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2})+CR^{-N}.$

By Lemma 6.4 there are discs $\mathcal{G}_{1,t_{e}}(D_{Q})$,
$\mathcal{G}_{2,t_{e}}(D_{Q})$ of radius

$R_{e}:=A_{w}|t_{Q}-t_{e}|/2+16R$

at time $t_{e}$ such that the $\mathcal{G}_{j,t_{e}}(D_{Q})$ contains
$\Lambda_{j}(x_{Q},t_{Q};4R)\cap(\mathbb{R}^{n}\times\\{t_{e}\\})$ and

$E(P_{D_{Q}}\phi_{j})^{1/2}\leq\|\phi_{j}\|_{L^{2}(\mathcal{G}_{j,t_{e}}(D_{Q}))}+CR^{-N},\qquad
j=1,2.$

To show

$\|\phi_{j}\|_{L^{2}(\mathcal{G}_{1,t_{e}}(D_{Q}))}\|\phi_{j}\|_{L^{2}(\mathcal{G}_{2,t_{e}}(D_{Q}))}\leq
E^{\varepsilon}_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2}),$ (8.4)

we consider a geometric property of $\Lambda_{1}^{[e]}\cap\Lambda_{2}^{[e]}$.
Since $\Lambda_{1}^{[e]}\cap\Lambda_{2}^{[e]}$ is a conic set (see Figure 2),
we can see that $A_{d}|t_{Q}-t_{e}|\leq 2r^{\prime}\leq 4\mathring{r}/A_{*}$
and so

$R_{e}\leq
2A_{w}A_{d}^{-1}\mathring{r}/A_{*}+16R\leq(2A_{w}A_{d}^{-1}+2000C_{0}^{-C})\mathring{r}/A_{*}\leq\mathring{r}.$

Since $Q$ intersects $\Lambda_{j}^{[e]}$ and the
$\mathcal{G}_{j,t_{e}}(D_{Q})$ contains
$\Lambda_{j}(x_{Q},t_{Q};4R)\cap(\mathbb{R}^{n}\times\\{t_{e}\\})$, the
$\mathcal{G}_{j,t_{e}}(D_{Q})$ intersects
$\Lambda_{j}^{[e]}\cap(\mathbb{R}^{n}\times\\{t_{e}\\})$. To show that
$\mathcal{G}_{j,t_{e}}(D_{Q})$ is contained in $\mathcal{N}(D_{e})$ it
suffices to show $r^{\prime}+2R_{e}\leq A_{*}r$. Using $\mathring{r}/A_{*}\leq
r$ we have

$r^{\prime}+2R_{e}\leq
2\mathring{r}/A_{*}+2(2A_{w}A_{d}^{-1}+2000C_{0}^{-C})\mathring{r}/A_{*}\leq
A_{*}r.$

Therefore we have (8.4).

### 8.2.

Consider the case that $Q$ is contained in
$\mathbb{R}^{n+1}\setminus\Lambda_{j}^{[e]}$ for some $j=1,2$. We only
consider the case that $Q$ is contained in
$\mathbb{R}^{n+1}\setminus\Lambda_{1}^{[e]}$, because the other case is
similar. Let $\check{D}_{e}:=D(x_{e},t_{e};\frac{r}{400A_{*}})$. By the
triangle inequality,

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq\|P_{\check{D}_{e}}\phi_{1}\phi_{2}\|_{L^{p}(Q)}+\|(1-P_{\check{D}_{e}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}.$
(8.5)

Since $Q$ is contained in $\mathbb{R}^{n+1}\setminus\Lambda_{j}^{[e]}$ and
$\frac{r^{\prime}}{2}\geq\frac{r}{200A_{*}}$, from Lemma 6.3 it follows that

$\|P_{\check{D}_{e}}\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq Cr^{-N^{4}}.$ (8.6)

By Definition 2.5,

$\displaystyle\|(1-P_{\check{D}_{e}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}$
$\displaystyle\leq\mathcal{K}(R,r)E((1-P_{\check{D}_{e}})\phi_{1})^{1/2p}E_{r,\mathring{r},t_{e}}((1-P_{\check{D}_{e}})\phi_{1},\phi_{2})^{1/p^{\prime}}$
$\displaystyle\leq\mathcal{K}(R,r)E((1-P_{\check{D}_{e}})\phi_{1})^{1/2p}E_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2})^{1/p^{\prime}}.$

By (8.1),

$\|(1-P_{\check{D}_{e}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq
E((1-P_{\check{D}_{e}})\phi_{1})^{1/2p}\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}.$

By (6.7),

$E((1-P_{\check{D}_{e}})\phi_{1})\leq
1-\|\phi(t_{e})\|^{2}_{L^{2}(\check{D}_{e}/2)}+Cr^{-N}.$

By (2.13) and (2.7),

$\|\phi(t_{e})\|_{L^{2}(\check{D}_{e}/2)}\geq\varepsilon$

By combining the above three inequalities,

$\|(1-P_{\check{D}_{e}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq\kappa\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}$

where $\kappa:=(1-\varepsilon^{2}+Cr^{-N})^{1/2p}$.
By applying the triangle inequality to the right side of the above inequality,

$\displaystyle\|(1-P_{\check{D}_{e}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}$
$\displaystyle\leq\kappa\|P_{\check{D}_{e}}\phi_{1}\phi_{2}\|_{L^{p}(Q)}+\kappa\|(1-P_{\check{D}_{e}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}.$

By rearranging,

$\|(1-P_{\check{D}_{e}})\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq\frac{\kappa}{1-\kappa}\|P_{\check{D}_{e}}\phi_{1}\phi_{2}\|_{L^{p}(Q)}.$

By inserting this estimate into (8.5),

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}\leq\frac{1}{1-\kappa}\|P_{\check{D}_{e}}\phi_{1}\phi_{2}\|_{L^{p}(Q)}.$

From $\varepsilon\geq R^{-N/4}$ and $r\geq C_{0}^{C}R$, we have
$\frac{1}{1-\kappa}\lesssim R^{CN},$ and

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q)}\lesssim
R^{CN}\|P_{\check{D}_{e}}\phi_{1}\phi_{2}\|_{L^{p}(Q)}.$

By (8.6) we thus have (8.2).

##  9\. Proof of Proposition 2.8.

Suppose that $\phi_{1}$, $\phi_{2}$ obey the margin condition (2.5). We may
assume the normalization (2.7). It suffices to show that

$\|\phi_{1}\phi_{2}\|_{L^{p}(Q_{R})}\leq(1+Cc)\mathcal{K}_{\varepsilon}(R/C_{0},r_{\natural},\mathring{r}_{\natural})E_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2})^{1/p^{\prime}}+c^{-C}.$

We apply Proposition 4.1. Then it suffices to show

$\|[\Phi_{1}][\Phi_{2}]\|_{L^{p}(Q)}\leq(1+Cc)\mathcal{K}_{\varepsilon}(R/C_{0},r_{\natural},\mathring{r}_{\natural})E_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2})^{1/p^{\prime}}+c^{-C},$
(9.1)

where the cube $Q$ is of side-length $CR$. By (4.7),

$\|[\Phi_{1}][\Phi_{2}]\|_{L^{p}(Q)}=\Big{(}\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}\|\Phi_{1}^{(\Delta)}\Phi_{2}^{(\Delta)}\|_{L^{p}(\Delta)}^{p}\Big{)}^{1/p}.$

Since the sidelength of $\Delta$ is $2^{-C_{0}}CR$, there is a cube of
sidelength $R/C_{0}$ containing $\Delta$. So, by the margin of $\Phi_{j}$
shown in Proposition 4.1 and Definition 2.5,

$\|\Phi_{1}^{(\Delta)}\Phi_{2}^{(\Delta)}\|_{L^{p}(\Delta)}\leq\mathcal{K}_{\varepsilon}(R/C_{0},r_{\natural},\mathring{r}_{\natural})E_{r_{\natural},\mathring{r}_{\natural},t_{e}}(\Phi_{1}^{(\Delta)},\Phi_{2}^{(\Delta)})^{1/p^{\prime}}\big{(}E(\Phi_{1}^{(\Delta)})^{1/2}E(\Phi_{2}^{(\Delta)})^{1/2}\big{)}^{1/p}.$

By combining the above two equations,

$\displaystyle\|[\Phi_{1}][\Phi_{2}]\|_{L^{p}(Q)}$
$\displaystyle\leq\mathcal{K}_{\varepsilon}(R/C_{0},r_{\natural},\mathring{r}_{\natural})\bigg{(}\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E_{r_{\natural},\mathring{r}_{\natural},t_{e}}(\Phi_{1}^{(\Delta)},\Phi_{2}^{(\Delta)})^{p/p^{\prime}}E(\Phi_{1}^{(\Delta)})^{1/2}E(\Phi_{2}^{(\Delta)})^{1/2}\bigg{)}^{1/p}$
$\displaystyle\leq\mathcal{K}_{\varepsilon}(R/C_{0},r_{\natural},\mathring{r}_{\natural})\sup_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E_{r_{\natural},\mathring{r}_{\natural},t_{e}}(\Phi_{1}^{(\Delta)},\Phi_{2}^{(\Delta)})^{1/p^{\prime}}\bigg{(}\sum_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E(\Phi_{1}^{(\Delta)})^{1/2}E(\Phi_{2}^{(\Delta)})^{1/2}\bigg{)}^{1/p}.$

By the Cauchy–Schwarz inequality and Lemma 5.1, it is bounded by

$(1+Cc)\mathcal{K}_{\varepsilon}(R/C_{0},r_{\natural},\mathring{r}_{\natural})\sup_{\Delta\in\mathcal{Q}_{C_{0}(Q)}}E_{r_{\natural},\mathring{r}_{\natural},t_{e}}(\Phi_{1}^{(\Delta)},\Phi_{2}^{(\Delta)})^{1/p^{\prime}}.$

Now it suffices to show

$\sup_{\Delta\in\mathcal{Q}_{C_{0}}(Q)}E_{r_{\natural},\mathring{r}_{\natural},t_{e}}(\Phi_{1}^{(\Delta)},\Phi_{2}^{(\Delta)})\leq(1+Cc)E_{r,\mathring{r},t_{e}}(\phi_{1},\phi_{2})+Cr^{-N},$

because $\mathcal{K}(R,r,\mathring{r})\lesssim\mathcal{K}(R)\lesssim R^{C}$.
By Lemma 5.1 we have $E(\Phi_{j}^{\Delta})^{1/2}\leq(1+Cc)E(\phi_{j})^{1/2}$.
Thus it is enough to show

$\|\Phi_{j}^{(\Delta)}\|_{L^{2}(D(z_{0};r_{\natural}))}\leq(1+Cc)\|\phi_{j}\|_{L^{2}(D(z_{0};r))}+Cr^{-N}$

for all $\Delta\in\mathcal{Q}_{C_{0}}(Q)$ and all $z_{0}\in\mathbb{R}^{n+1}$.
We have this estimate from the following lemma:

######  Lemma 9.1.

Let $\mathcal{Q}$ be a finite index set. Suppose that $m_{q,T_{j}}$ are non-
negative numbers with (5.2). Then, for any $r\geq 2C_{0}$ and
$z_{0}\in\mathbb{R}^{n+1}$,

$\displaystyle\bigg{(}\sum_{q\in\mathcal{Q}}\big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}\big{\|}^{2}_{L^{2}(D(z_{0};r(1-Cr^{-1/3N})))}\bigg{)}^{1/2}$
$\displaystyle\qquad\qquad\qquad\qquad\leq(1+Cc)\|\phi_{j}\|_{L^{2}(D(z_{0};r))}+r^{-N}E(\phi_{j})^{1/2}.$
(9.2)

###### Proof.

Let $\tilde{D}=D(x_{0},t_{0};r(1-2r^{-1/3N}))$,
$\tilde{D}^{\prime}=D(x_{0},t_{0};r(1-r^{-1/2N}))$ and
$\tilde{D}^{\prime\prime}=D(x_{0},t_{0};r)$. Then we have
$\tilde{D}\subsetneq\tilde{D}^{\prime}\subsetneq\tilde{D}^{\prime\prime}$.
We divide

$\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}=\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}P_{\tilde{D}^{\prime}}\phi_{T_{j}}+\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}(1-P_{\tilde{D}^{\prime}})\phi_{T_{j}}.$

Consider the first summation in the right side. We have

$\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}P_{\tilde{D}^{\prime}}\phi_{T_{j}}\Big{\|}_{L^{2}(\tilde{D})}^{2}\leq\sum_{q\in\mathcal{Q}}E\Big{(}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}P_{\tilde{D}^{\prime}}\phi_{T_{j}}\Big{)}.$

Applying Lemma 5.1 we have

$\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}P_{\tilde{D}^{\prime}}\phi_{T_{j}}\Big{\|}_{L^{2}(\tilde{D})}^{2}\leq(1+Cc)E(P_{\tilde{D}^{\prime}}\phi_{j}).$

From (6.6) we can see that

$\displaystyle E(P_{\tilde{D}^{\prime}}\phi_{j})$
$\displaystyle\leq\|\phi_{j}\|^{2}_{L^{2}(\tilde{D}^{\prime+})}+Cr^{-N}E(\phi_{j})$
$\displaystyle\leq\|\phi_{j}\|^{2}_{L^{2}(\tilde{D}^{\prime\prime})}+Cr^{-N}E(\phi_{j}),$

where
$\tilde{D}^{\prime}\subset\tilde{D}^{\prime+}\subset\tilde{D}^{\prime\prime}$
is used. Thus we obtain

$\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}P_{\tilde{D}^{\prime}}\phi_{T_{j}}\Big{\|}_{L^{2}(\tilde{D})}^{2}\bigg{)}^{1/2}\leq(1+Cc)\|\phi_{j}\|_{L^{2}(\tilde{D}^{\prime\prime})}+Cr^{-N}E(\phi_{j})^{1/2}.$

To prove (9.1) it now suffices to show

$\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}(1-P_{\tilde{D}^{\prime}})\phi_{T_{j}}\Big{\|}_{L^{2}(\tilde{D})}^{2}\bigg{)}^{1/2}\lesssim
r^{-N}E(\phi_{j})^{1/2}.$ (9.3)

Since the operator $P_{D}$ is linear, we have

$\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}(1-P_{\tilde{D}^{\prime}})\phi_{T_{j}}=(1-P_{\tilde{D}^{\prime}})\Big{(}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}\Big{)}.$

By $\tilde{D}\subset\tilde{D}^{\prime-}\subset\tilde{D}^{\prime}$ and (6.5),

$\displaystyle\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}(1-P_{\tilde{D}^{\prime}})\phi_{T_{j}}\Big{\|}_{L^{2}(\tilde{D})}$
$\displaystyle\leq\Big{\|}(1-P_{\tilde{D}^{\prime}})\Big{(}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}\Big{)}\Big{\|}_{L^{2}(\tilde{D}^{\prime-})}$
$\displaystyle\lesssim
r^{-N}E\Big{(}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}\Big{)}^{1/2}.$

Using this we have

$\bigg{(}\sum_{q\in\mathcal{Q}}\Big{\|}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}(1-P_{\tilde{D}^{\prime}})\phi_{T_{j}}\Big{\|}_{L^{2}(\tilde{D})}^{2}\bigg{)}^{1/2}\lesssim
r^{-N}\bigg{(}\sum_{q\in\mathcal{Q}}E\Big{(}\sum_{T_{j}\in\mathbf{T}_{j}}m_{q,T_{j}}\phi_{T_{j}}\Big{)}\bigg{)}^{1/2}.$

Thus, by Lemma 5.1 we obtain (9.3). ∎

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