for any $\mathbb{P}$ finite subcollection of ${{\mathbb{P}}_{\textrm{BHT}}}$
the collection $\mathcal{I}_{\bar{\mathbb{P}}}$ of spatial intervals

(374) $\mathcal{I}_{\bar{\mathbb{P}}}:=\\{I\text{ dyadic }:I=I_{P}\text{ for
some }P\in\mathbb{P}\text{ or }I=I_{\mathbf{T}}\text{ for some tree
}\mathbf{T}\subset\mathbb{P}\\}.$

######  Proposition 7.11 (Size estimates).

Let $\delta_{1},\mu>0$ be as obtained in Section 5 and as used in Definition
6.3 when introducing the indices sets $\mathfrak{l},\mathfrak{u},\mathfrak{c}$
corresponding to each tile $P\in{{\mathbb{P}}_{\textrm{BHT}}}$.
For any locally $L^{2}$-integrable functions $f,g,h$ and any subcollection
$\mathbb{P}\subset{{\mathbb{P}}_{\textrm{BHT}}}$, we have

(375)
$\displaystyle\mathrm{Size\,}^{\mathbb{P}}_{1}((f(P_{1}))_{P\in\mathbb{P}})$
$\displaystyle\lesssim\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}}}\frac{1}{|I|^{\frac{1}{2}}}\|f\cdot\tilde{\chi}_{I}\|_{L^{2}},$
$\displaystyle\mathrm{Size\,}^{\mathbb{P}}_{2}((g(P_{2}))_{P\in\mathbb{P}})$
$\displaystyle\lesssim\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}}}\frac{1}{|I|^{\frac{1}{2}}}\|g\cdot\tilde{\chi}_{I}\|_{L^{2}},$

and

(376)
$\displaystyle\mathrm{Size\,}^{\mathbb{P}}_{3}((h_{\mathfrak{l}}(P_{3}))_{P\in\mathbb{P}})$
$\displaystyle\lesssim 2^{-\frac{\delta_{1}am}{2}}$
$\displaystyle\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}}}\frac{1}{|I|^{\frac{1}{2}}}\|h\cdot\tilde{\chi}_{I}\|_{L^{2}},$
(377)
$\displaystyle\mathrm{Size\,}^{\mathbb{P}}_{3}((h_{\mathfrak{u}}(P_{3}))_{P\in\mathbb{P}})$
$\displaystyle\lesssim$
$\displaystyle\sup_{P\in\mathcal{I}_{\bar{\mathbb{P}}}}\frac{1}{|I|^{\frac{1}{2}}}\|h\cdot\tilde{\chi}_{I}\|_{L^{2}},$
(378)
$\displaystyle\mathrm{Size\,}^{\mathbb{P}}_{3}((h_{\mathfrak{c}}(P_{3}))_{P\in\mathbb{P}})$
$\displaystyle\lesssim 2^{-\frac{\mu am}{4}}$
$\displaystyle\sup_{P\in\mathcal{I}_{\bar{\mathbb{P}}}}\frac{1}{|I|^{\frac{1}{2}}}\|h\cdot\tilde{\chi}_{I}\|_{L^{2}}.$

In the above proposition, only estimates (376)-(378) require some work; the
$L^{2}$-based (375) is standard and follows from the almost orthogonality of
the wave-packets. For completeness, we treat all of them.

###### Proof of estimate (375).

We let $\mathbf{T}$ be a $1$-lacunary tree and we aim to prove that

$\displaystyle\sum_{P\in\mathbf{T}}\sum_{\hat{s}=P}\big{|}\langle
f,\phi_{s_{1}}\rangle\big{|}^{2}\lesssim\|f\cdot\tilde{\chi}_{I_{\mathbf{T}}}\|_{L^{2}}^{2}.$

When $f$ is supported close to $I_{\mathbf{T}}$, this is a simple consequence
of orthogonality; on the contrary, for $f$ supported away from
$I_{\mathbf{T}}$ we rely on the fast decay of the wave-packets.

Assume that $f$ is supported on $3I_{\mathbf{T}}$; then it suffices to prove

$\displaystyle\sum_{P\in\mathbf{T}}\sum_{\hat{s}=P}\big{|}\langle
f,\phi_{s_{1}}\rangle\big{|}^{2}\lesssim\|f\|_{L^{2}}^{2}.$

This follows from the disjointness of the collection of tiles
$\displaystyle\\{s_{1}\\}_{\hat{s}\in\mathbf{T}}$ (a weak consequence of the
fact that $\mathbf{T}$ is $1$-lacunary) and from the almost orthogonality of
the associated collection of wave-packets
$\displaystyle\\{\phi_{s_{1}}\\}_{\hat{s}\in\mathbf{T}}$.

Next we assume that $f$ is supported in
$I_{\mathbf{T}}^{\nu}:=I_{\mathbf{T}}+\nu|I_{\mathbf{T}}|$, for $|\nu|\geq 2$
an integer, and we want to prove

(379) $\displaystyle\sum_{P\in\mathbf{T}}\sum_{\hat{s}=P}\big{|}\langle
f,\phi_{s_{1}}\rangle\big{|}^{2}\lesssim|\nu|^{-10}\|f\|_{L^{2}}^{2}.$

First, we notice that the left-hand side of the estimate above writes as

$\displaystyle\sum_{\kappa\geq
0}\sum_{\begin{subarray}{c}\hat{s}\in\mathbf{T}\\\ |I_{s_{1}}|=2^{-{am\over
2}}2^{-\kappa}|I_{\mathbf{T}}|\end{subarray}}\big{|}\langle
f,\phi_{s_{1}}\rangle\big{|}^{2},$

and also

$\displaystyle\big{|}\langle f,\phi_{s_{1}}\rangle\big{|}^{2}$
$\displaystyle\lesssim\|f\|_{L^{2}}^{2}\,\int_{I_{\mathbf{T}}^{\nu}}\frac{1}{|I_{s_{1}}|}\Big{(}1+\frac{\text{dist}(I_{s_{1}},x)}{|I_{s_{1}}|}\Big{)}^{-20}dx$
$\displaystyle\lesssim\|f\|_{L^{2}}^{2}\int_{I_{\mathbf{T}}^{\nu}}\frac{1}{|I_{s_{1}}|}\Big{(}\frac{\text{dist}(I_{s_{1}},x)}{2^{-{am\over
2}}2^{-\kappa}|I_{\mathbf{T}}|}\Big{)}^{-20}dx$
$\displaystyle\lesssim\|f\|_{L^{2}}^{2}\>\frac{2^{am\over
2}2^{\kappa}}{|I_{\mathbf{T}}|}\int_{I_{\mathbf{T}}^{\nu}}2^{-10am}2^{-20\kappa}\big{(}\frac{\text{dist}(I_{\mathbf{T}},x)}{|I_{\mathbf{T}}|}\big{)}^{-20}dx$
$\displaystyle\lesssim\|f\|_{L^{2}}^{2}\,2^{-9am}2^{-10\kappa}|\nu|^{-10}.$

For a fixed scale $\kappa\geq 0$, there are at most $2^{\kappa}\cdot 2^{am}$
tiles $s$ so that $\hat{s}\in\mathbf{T}$, with $|I_{s_{1}}|=2^{-{am\over
2}}2^{-\kappa}|I_{\mathbf{T}}|$. This rough estimate is enough for concluding
(379); by standard arguments, (375) follows. ∎

Notice that the estimates for the light-mass and the heavy-mass/clustered
sizes $\mathrm{Size\,}^{\mathbb{P}}_{3}(h_{\mathfrak{l}}(P_{3}))$ and
$\mathrm{Size\,}^{\mathbb{P}}_{3}(h_{\mathfrak{c}}(P_{3}))$ feature additional
decay in the parameter $m$. First we will prove estimate (377), for which we
will not make use of the fact that we are summing over triples
$(p^{\prime},n,v)\in\mathfrak{u}$. Next, we prove (378), in which the
additional decay is a result of the fact that for any fixed $p^{\prime}\sim
2^{am}$, only a small fraction of the allowed values of $n$ is such that
$(p^{\prime},n,v)\in\mathfrak{c}$ (see Remark 6.4). Lastly, for (376), the
decay comes from the assumption that the indices
$(p^{\prime},n,v)\in\mathfrak{l}$ are associated to light averages of the
weight $w_{|P|,n,v}(\lambda)$.

###### Proof of estimate (377).

Let $\mathbf{T}$ be a $3$-lacunary tree; it will be enough to prove that

$\displaystyle\sum_{P\in\mathbf{T}}h_{\mathfrak{u}}(P_{3})^{2}\lesssim\|h\cdot\tilde{\chi}_{I_{\mathbf{T}}}\|_{L^{2}}^{2}.$

By definition (363), this amounts to proving

$\displaystyle\frac{1}{2^{am\over
2}}\sum_{P\in\mathbf{T}}\sum_{(p^{\prime},n,v)\in\mathfrak{u}}\big{|}\langle
h\,\rho_{|P|}(\lambda(\cdot))w_{|P|,n,v}^{e}(\lambda),\Psi_{P_{3}}^{p^{\prime}}\rangle\big{|}^{2}\lesssim\|h\cdot\tilde{\chi}_{I_{\mathbf{T}}}\|_{L^{2}}^{2}.$

We will see that orthogonality presents itself in disjointness across
different scales, because of the presence of the stopping time
$\rho_{|P|}(\lambda(\cdot))$. First, by using Cauchy-Schwarz and properties
0(WP1)-0(WP2), one has

(380) $\displaystyle\big{|}\langle
h\,\rho_{|P|}(\lambda(\cdot))w_{|P|,n,v}^{e}(\lambda),\Psi_{P_{3}}^{p^{\prime}}\rangle\big{|}^{2}$
$\displaystyle\lesssim\Big{(}\int|h(x)|^{2}\,\rho_{|P|}(\lambda(x))^{2}|w_{|P|,n,v}(\lambda)(x)|^{2}|\Psi_{P_{3}}^{p^{\prime}}(x)|\,dx\Big{)}\cdot\Big{(}\int|\Psi_{P_{3}}^{p^{\prime}}(x)|\,dx\Big{)}$
$\displaystyle\lesssim\Big{(}\int|h(x)|^{2}\,\rho_{|P|}(\lambda(x))^{2}|w_{|P|,n,v}(\lambda)(x)|^{2}|\underline{I}_{P}^{p^{\prime}}|^{-{1\over
2}}\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)\,dx\Big{)}|\underline{I}_{P}^{p^{\prime}}|^{1\over
2}$

and our desired estimate reduces to

(381)
$\sum_{P\in\mathbf{T}}\sum_{(p^{\prime},n,v)\in\mathfrak{u}}\frac{1}{2^{am\over
2}}\|h\,\rho_{|P|}(\lambda)w_{|P|,n,v}(\lambda)\|_{L^{2}(\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}})}^{2}\lesssim\|h\cdot\tilde{\chi}_{I_{\mathbf{T}}}\|_{L^{2}}^{2}.$

Using (140) to bound $w_{|P|,n,v}(\lambda)$ and extending the summation to all
admissible triples $(p^{\prime},n,v)$ we have

$\displaystyle\qquad\sum_{P\in\mathbf{T}}h_{\mathfrak{u}}(P_{3})^{2}$
$\displaystyle\lesssim\sum_{k\in\mathbb{Z}}\sum_{\begin{subarray}{c}P\in\mathbf{T}:\\\
|I_{P}|=2^{k}\end{subarray}}\sum_{(n,v,p^{\prime})\in\mathfrak{u}}\frac{1}{2^{am\over
2}}\|h\,\rho_{am-
ak}(\lambda(\cdot))w_{k,n,v}(\lambda)\|_{L^{2}(\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}})}^{2}$
$\displaystyle\lesssim\sum_{k\in\mathbb{Z}}\sum_{n\sim 2^{am\over
2}}\frac{1}{2^{am\over
2}}\int_{\mathbb{R}}|h(x)|^{2}\Big{(}\sum_{\begin{subarray}{c}v\sim 2^{am\over
2}\\\ p^{\prime}\sim
2^{am}\end{subarray}}\sum_{\begin{subarray}{c}P\in\mathbf{T}\\\
|I_{P}|=2^{k}\end{subarray}}\frac{\rho_{am-
ak}(\lambda(x))}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{4}}\cdot\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)\Big{)}dx.$

We claim that the following pointwise bound holds:

(382) $\sum_{k\in\mathbb{Z}}\sum_{\begin{subarray}{c}P\in\mathbf{T}:\\\
|I_{P}|=2^{k}\end{subarray}}\sum_{\begin{subarray}{c}v\sim 2^{am\over 2}\\\
p^{\prime}\sim 2^{am}\end{subarray}}\frac{\rho_{am-
ak}(\lambda(x))}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{4}}\cdot\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)\lesssim\tilde{\chi}_{I_{\mathbf{T}}}(x).$

If the above is true then this will be enough to conclude (381), since the
left/right-hand side in (382) is independent of $n$ (notice that the sum over
$n\sim 2^{am/2}$ is compensated by the factor of $2^{-am/2}$).

To see the claim, first notice that after summing in $v$ the left-hand side of
(382) is bounded above by

$\displaystyle\sum_{k\in\mathbb{Z}}\sum_{\begin{subarray}{c}P\in\mathbf{T}:\\\
|I_{P}|=2^{k}\end{subarray}}\sum_{p^{\prime}\sim 2^{am}}\rho_{am-
ak}(\lambda(x))\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)=\sum_{k\in\mathbb{Z}}\rho_{am-
ak}(\lambda(x))\sum_{\begin{subarray}{c}P\in\mathbf{T}\\\
|I_{P}|=2^{k}\end{subarray}}\tilde{\chi}_{I_{P}}(x).$

For every $x\in\mathbb{R}$, there are only $O(1)$ many $k\in\mathbb{Z}$ for
which $\rho_{am-ak}(\lambda(x))\neq 0$, and for every such $k$

$\sum_{\begin{subarray}{c}P\in\mathbf{T}:\\\
|I_{P}|=2^{k}\end{subarray}}\tilde{\chi}_{I_{P}}(x)\lesssim\tilde{\chi}_{I_{\mathbf{T}}}(x),$

which allows us to conclude (382). ∎

###### Proof of estimate (378).

The overall argument will be similar to the previous one. We start again with
$\mathbf{T}$, a $3$-lacunary tree, and observe that it will be enough to prove
that

$\displaystyle\sum_{P\in\mathbf{T}}h_{\mathfrak{c}}(P_{3})^{2}\lesssim
2^{-\mu{am\over 2}}\|h\cdot\tilde{\chi}_{I_{\mathbf{T}}}\|_{L^{2}}^{2}.$

As before, we have

$\displaystyle\qquad\sum_{P\in\mathbf{T}}h_{\mathfrak{c}}(P_{3})^{2}$
$\displaystyle\lesssim\sum_{k\in\mathbb{Z}}\sum_{\begin{subarray}{c}P\in\mathbf{T}:\\\
|I_{P}|=2^{k}\end{subarray}}\frac{1}{2^{am\over
2}}\int_{\mathbb{R}}|h(x)|^{2}\Big{(}\sum_{(p^{\prime},n,v)\in\mathfrak{c}}\frac{\rho_{am-
ak}(\lambda(x))}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{4}}\cdot\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)\Big{)}dx.$

The desired estimate will be a consequence of the pointwise bound

(383)
$\displaystyle\sum_{k\in\mathbb{Z}}\sum_{\begin{subarray}{c}P\in\mathbf{T}:\\\
|I_{P}|=2^{k}\end{subarray}}\frac{1}{2^{am\over
2}}\sum_{(p^{\prime},n,v)\in\mathfrak{c}}\frac{\rho_{am-
ak}(\lambda(x))}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{4}}$
$\displaystyle\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)$
$\displaystyle\lesssim
2^{-\mu\frac{am}{2}}\cdot\tilde{\chi}_{I_{\mathbf{T}}}(x),$

which in turn follows from

(384) $\frac{1}{2^{am\over
2}}\sum_{(p^{\prime},n,v)\in\mathfrak{c}}\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)\frac{1}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{4}}\lesssim
2^{-\mu\frac{am}{2}}\tilde{\chi}_{I_{P}}(x),$

for any fixed $k\in\mathbb{Z}$ and $P\in\mathbf{T}$.

In Remark 6.4, we introduced a larger set
$\bar{\mathfrak{c}}=\bar{\mathfrak{c}}_{1}\times\\{v\sim 2^{am\over 2}\\}$
containing $\mathfrak{c}$ which allowed us to separate the $n$ and $v$
parameters; thus the left-hand side in the expression above is bounded by

(385) $\frac{1}{2^{am\over
2}}\sum_{(p^{\prime},n)\in\bar{\mathfrak{c}}_{1}}\sum_{v\sim 2^{am\over
2}}\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)\frac{1}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{4}}.$

With $p^{\prime}$ and $n$ fixed, we first sum in $v$ to get

(386) $\sum_{v\sim 2^{am\over
2}}\frac{1}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{4}}\lesssim
1.$

Now we can bound (385) by

$\frac{1}{2^{am\over
2}}\sum_{(p^{\prime},n)\in\bar{\mathfrak{c}}_{1}}\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)\leq\frac{1}{2^{am\over
2}}\sum_{p^{\prime}\sim
2^{am}}\sum_{n:(p^{\prime},n)\in\bar{\mathfrak{c}}_{1}}\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x).$

We recall the observation in Remark 6.4: once $p^{\prime}\sim 2^{am}$ is
fixed, there are at most $O(2^{(1-\mu)\frac{am}{2}})$ values of $n$ so that
$(p^{\prime},n)\in\bar{\mathfrak{c}}_{1}$. That produces as upper bound for
(385)

$\frac{1}{2^{am\over 2}}\sum_{p^{\prime}\sim
2^{am}}2^{(1-\mu)\frac{am}{2}}\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x),$

which is of course bounded by the right-hand side of (384).

Estimate (383) follows immediately from (384) and the straightforward
pointwise estimate

(387) $\sum_{k\in\mathbb{Z}}\sum_{\begin{subarray}{c}P\in\mathbf{T}:\\\
|I_{P}|=2^{k}\end{subarray}}\rho_{am-
ak}(\lambda(x))\tilde{\chi}_{I_{P}}(x)\lesssim\tilde{\chi}_{I_{\mathbf{T}}}(x).$

∎

###### Proof of estimate (376).

In this situation, taking $\mathbf{T}$ a $3$-lacunary tree, our task is to
show that

(388) $\sum_{P\in\mathbf{T}}h_{\mathfrak{l}}(P_{3})^{2}\lesssim
2^{-\delta_{1}am}\|h\cdot\tilde{\chi}_{I_{\mathbf{T}}}\|_{L^{2}}^{2}.$

We will use the fact that for a fixed tri-tile $P$, we are summing over the
indices $(p^{\prime},n,v)\in\mathfrak{l}$ in the “light-mass regime”. Indeed,
by Cauchy-Schwarz and 0(WP3) we can bound

$\displaystyle\big{|}\langle
h\,\rho_{|P|}(\lambda)w_{|P|,n,v}^{e}(\lambda),\Psi_{P_{3}}^{p^{\prime}}\rangle\big{|}^{2}$
$\displaystyle\lesssim\big{\|}h^{2}\rho_{|P|}(\lambda)^{2}w_{|P|,n,v}|\Psi_{P_{3}}^{p^{\prime}}|\big{\|}_{L^{1}}\;\big{\|}w_{k,n,v}|\Psi_{P_{3}}^{p^{\prime}}|\big{\|}_{L^{1}}$
$\displaystyle\lesssim
2^{-\delta_{1}am}\big{\|}h^{2}\rho_{|P|}(\lambda)^{2}w_{|P|,n,v}\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}\big{\|}_{L^{1}}$

(this is the natural “weighted” replacement of (380)). For any
$P\in\mathbf{T}$ we then have

$\displaystyle\qquad h_{\mathfrak{l}}(P_{3})^{2}$ $\displaystyle\lesssim
2^{-\delta_{1}am}\frac{1}{2^{\frac{a}{2}m}}\sum_{(p^{\prime},n,v)\in\mathfrak{l}}\int_{\mathbb{R}}|h(x)|^{2}\rho_{|P|}(\lambda(x))^{2}w_{|P|,n,v}(\lambda)(x)\tilde{\chi}_{\underline{I}_{P}^{p^{\prime}}}(x)dx$
$\displaystyle\lesssim
2^{-\delta_{1}am}\int_{\mathbb{R}}|h(x)|^{2}\rho_{|P|}(\lambda(x))^{2}\tilde{\chi}_{I_{P}}(x)dx.$

Estimate (388) follows now by essentially the same reasoning used for (381)
since the information $(p^{\prime},n,v)\in\mathfrak{l}$ is no longer needed. ∎

###  7.4. Estimates for energies

Now we turn towards the energies, which are naturally $L^{2}$-based quantities
as they capture the almost orthogonality of strongly disjoint trees associated
to a fixed size level.

######  Proposition 7.12 (Energy estimates).

For any $L^{2}$ functions $f,g,h$ and any subcollection
$\mathbb{P}\subset{{\mathbb{P}}_{\textrm{BHT}}}$, we have

(389)
$\displaystyle\mathrm{Energy\,}^{\mathbb{P}}_{1}((f(P_{1}))_{P\in\mathbb{P}})$
$\displaystyle\lesssim\|f\|_{L^{2}},$
$\displaystyle\mathrm{Energy\,}^{\mathbb{P}}_{2}((g(P_{2}))_{P\in\mathbb{P}})$
$\displaystyle\lesssim\|g\|_{L^{2}}$

and for $\ast\in\\{\mathfrak{l},\mathfrak{u},\mathfrak{c}\\}$

(390)
$\mathrm{Energy\,}^{\mathbb{P}}_{3}((h_{\ast}(P_{3}))_{P\in\mathbb{P}})\lesssim\|h\|_{L^{2}}.$

######  Remark 7.13.

We point out that in establishing the energy estimates (390) we will not make
use of the partitioning of the triples $(p^{\prime},n,v)$ into subsets
$\mathfrak{l},\mathfrak{u},\mathfrak{c}$. This partition was only important
for the size estimates of Proposition 7.11.

###### Proof of estimate (390).

Let $\ast$ stand for any of the $\mathfrak{l},\mathfrak{u}$ or $\mathfrak{c}$
and let $\mathscr{F}$ be a collection of mutually disjoint $3$-lacunary trees
satisfying for some integer $d$

$2^{2d}<\frac{1}{|I_{\mathbf{T}}|}\sum_{P\in\mathbf{T}}h_{*}(P_{3})^{2}\qquad\text{for
all }\mathbf{T}\in\mathscr{F}.$

It is then enough to show that

(391)
$\sum_{\mathbf{T}\in\mathscr{F}}\sum_{P\in\mathbf{T}}h_{*}(P_{3})^{2}\lesssim\|h\|_{L^{2}}^{2},$

which will in turn be a direct consequence142142142Recall that the factor
$2^{-am/2}$ in the definition of $h_{\ast}(P_{3})$ allows one to safely sum in
$n\sim 2^{am/2}$. of the estimate, uniform in the parameter $n\sim 2^{am\over
2}$,

$\displaystyle\sum_{\mathbf{T}\in\mathscr{F}}\sum_{P\in\mathbf{T}}\sum_{\begin{subarray}{c}v\sim
2^{am\over 2}\\\ p^{\prime}\sim 2^{am}\end{subarray}}\big{|}\langle
h\,\rho_{|P|}(\lambda(\cdot))w_{|P|,n,v}^{e}(\lambda),\Psi_{P_{3}}^{p^{\prime}}\rangle\big{|}^{2}\lesssim\|h\|_{L^{2}}^{2}.$

There are two key observations to make:

  1. (i)

Given $I$ a dyadic interval, we denote by
$\boldsymbol{\omega}_{\mathscr{F}}(I)$ the collection of frequency intervals
$\boldsymbol{\omega}$ for which $I\times\boldsymbol{\omega}$ corresponds to a
$P_{3}$ tile:

$\boldsymbol{\omega}_{\mathscr{F}}(I):=\\{\boldsymbol{\omega}:|I||\boldsymbol{\omega}|=2^{am},I\times\boldsymbol{\omega}\in\bigcup_{\mathbf{T}\in\mathscr{F}}\bigcup_{P\in\mathbf{T}}\\{P_{3}\\}\\}.$

Then notice that $\boldsymbol{\omega}_{\mathscr{F}}(I)$ consists of mutually
disjoint intervals of the same length.

  2. (ii)

$w_{|P|,n,v}^{e}$ and $\underline{I}_{P}^{p^{\prime}}$ do not depend on the
frequency interval of the tile $P$ but only depend on the spatial interval
$I_{P}$ and on the triple of parameters $(p^{\prime},n,v)$.

Hence, for any fixed interval $I$ with $|I|=2^{k}$ we can use property 0(WP4)
for the collection
$\big{\\{}\Psi_{I\times\boldsymbol{\omega}}^{p^{\prime}}\big{\\}}_{\boldsymbol{\omega}\in\boldsymbol{\omega}_{\mathscr{F}}(I),p^{\prime}\sim
2^{am}}$ and function $F:=h\,\rho_{I}(\lambda(\cdot))w_{I,n,v}^{e}(\lambda)$
to deduce

$\displaystyle\sum_{\boldsymbol{\omega}\in\boldsymbol{\omega}_{\mathscr{F}}(I)}\
\sum_{\begin{subarray}{c}v\sim 2^{\frac{a}{2}m},\\\ p^{\prime}\sim
2^{am}\end{subarray}}\big{|}\langle
h\,\rho_{I}(\lambda(\cdot))w_{I,n,v}^{e}(\lambda),\Psi_{I\times\boldsymbol{\omega}}^{p^{\prime}}\rangle\big{|}^{2}$
$\displaystyle\lesssim\sum_{v\sim 2^{\frac{a}{2}m}}\|h\rho_{am-
ak}(\lambda)w_{k,n,v}\tilde{\chi}_{I}\|_{L^{2}}^{2}.$

Estimate (391) follows once we notice that for every fixed $n$ (and uniformly
so)

$\sum_{k}\ \sum_{\begin{subarray}{c}I\in\mathscr{D}:\\\
|I|=2^{k}\end{subarray}}\ \sum_{v\sim 2^{\frac{a}{2}m}}\rho_{am-
ak}(\lambda(x))w_{k,n,v}(x)^{2}\tilde{\chi}_{I}(x)\lesssim 1.$

This follows from (386) and an unlocalized version of (387). ∎

It remains to prove energy estimate (389), which relies on more subtle
orthogonality properties. Although similar to the classical argument for BHT
(see [82, Lemma 6.6]), we will

  * •

rely on the extra separation condition in (369);

  * •

use the smaller tri-tiles $s=(s_{1},s_{2},s_{3})$ of area 1 (to which wave-
packets are naturally adapted), in order to capture the orthogonality between
the strongly disjoint trees through a $TT^{*}$ argument.

###### Proof of estimate (389).

By symmetry, we consider only the energy for $f$. Fix an integer
$d\in\mathbb{Z}$ and let $\mathscr{F}$ be a collection of strongly disjoint
$1$-lacunary trees having the property that for any $\mathbf{T}\in\mathscr{F}$

(392)
$2^{d}<\Big{(}\frac{1}{|I_{\mathbf{T}}|}\sum_{\hat{s}\in\mathbf{T}}|\langle
f,\phi_{s_{1}}\rangle|^{2}\Big{)}^{\frac{1}{2}}$

and

(393)
$\Big{(}\frac{1}{|I_{\mathbf{T}^{\prime}}|}\sum_{\hat{s}\in\mathbf{T}^{\prime}}|\langle
f,\phi_{s_{1}}\rangle|^{2}\Big{)}^{\frac{1}{2}}\leq 2^{d+1}\quad\text{for any
subtree }\mathbf{T}^{\prime}\subseteq\mathbf{T}.$

It will suffice to prove

(394)
$2^{2d}\sum_{\mathbf{T}\in\mathscr{F}}|I_{\mathbf{T}}|\lesssim\|f\|_{L^{2}}^{2},$

which, based on (392) and a standard $TT^{*}$ argument, follows from

(395)
$\Big{\|}\sum_{\mathbf{T}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\langle
f,\phi_{s_{1}}\rangle\phi_{s_{1}}\Big{\|}_{L^{2}}^{2}\lesssim
2^{2d}\sum_{\mathbf{T}\in\mathscr{F}}|I_{\mathbf{T}}|.$

Expanding the square, it suffices to prove

(396)
$\sum_{\mathbf{T},\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\sum_{\hat{s}^{\prime}\in\mathbf{T}^{\prime}}\big{|}\langle
f,\phi_{s_{1}}\rangle\langle
f,\phi_{s^{\prime}_{1}}\rangle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle\big{|}\lesssim
2^{2d}\sum_{\mathbf{T}\in\mathscr{F}}|I_{\mathbf{T}}|.$

Now notice that in order for
$\displaystyle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle$ to be non-zero
it must be necessarily that
$\omega_{s_{1}}\cap\omega_{s^{\prime}_{1}}\neq\emptyset$ and hence one of them
is included in the other.143143143This property is inherited from the parent
collection
$\\{\boldsymbol{\omega}_{P_{1}}\\}_{P\in{{\mathbb{P}}_{\textrm{BHT}}}}$, which
forms a lattice.

It is thus enough to prove the following two claims:

  * •

Claim $1$:

(397)
$\sum_{\mathbf{T}\in\mathscr{F}}\sum_{s:\hat{s}\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime},\\\
\omega_{s_{1}}=\omega_{s^{\prime}_{1}}\end{subarray}}\big{|}\langle
f,\phi_{s_{1}}\rangle\langle
f,\phi_{s^{\prime}_{1}}\rangle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle\big{|}\lesssim
2^{2d}\sum_{\mathbf{T}\in\mathscr{F}}|I_{\mathbf{T}}|;$

  * •

Claim $2$:

(398)
$\sum_{\mathbf{T}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime},\\\
\omega_{s_{1}}\subset\omega_{s^{\prime}_{1}}\\\
|\omega_{s_{1}}|\ll|\omega_{s^{\prime}_{1}}|\end{subarray}}\big{|}\langle
f,\phi_{s_{1}}\rangle\langle
f,\phi_{s^{\prime}_{1}}\rangle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle\big{|}\lesssim
2^{2d}\sum_{\mathbf{T}\in\mathscr{F}}|I_{\mathbf{T}}|.$

We start with the proof of Claim 1. First, by symmetry,

$\displaystyle\sum_{\mathbf{T}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime},\\\
\omega_{s_{1}}=\omega_{s^{\prime}_{1}}\end{subarray}}\big{|}\langle
f,\phi_{s_{1}}\rangle\langle
f,\phi_{s^{\prime}_{1}}\rangle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle\big{|}$
$\displaystyle\lesssim\sum_{\mathbf{T}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime},\\\
\omega_{s_{1}}=\omega_{s^{\prime}_{1}}\end{subarray}}\big{|}\langle
f,\phi_{s_{1}}\big{\rangle}|^{2}\big{|}\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle\big{|}$
$\displaystyle\lesssim\sum_{\mathbf{T}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\big{|}\langle
f,\phi_{s_{1}}\big{\rangle}|^{2}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime},\\\
\omega_{s_{1}}=\omega_{s^{\prime}_{1}}\end{subarray}}\Big{(}1+\frac{\text{dist}(I_{s_{1}},I_{s^{\prime}_{1}})}{|I_{s_{1}}|}\Big{)}^{-10}.$

Now, for $\omega_{s_{1}}$ fixed, the tiles $s^{\prime}$ such that
$\hat{s}^{\prime}\in\cup_{\mathbf{T}^{\prime}}\mathbf{T}^{\prime}$ having the
property that $\omega_{s_{1}}=\omega_{s^{\prime}_{1}}$ are mutually disjoint
and satisfy $|I_{s^{\prime}_{1}}|=|I_{s_{1}}|$; hence

$\displaystyle\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime},\\\
\omega_{s_{1}}=\omega_{s^{\prime}_{1}}\end{subarray}}\Big{(}1+\frac{\text{dist}(I_{s_{1}},I_{s^{\prime}_{1}})}{|I_{s_{1}}|}\Big{)}^{-10}$
$\displaystyle\lesssim\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime},\\\
\omega_{s_{1}}=\omega_{s^{\prime}_{1}}\end{subarray}}\frac{1}{|I_{s_{1}}|}\int_{I_{s^{\prime}_{1}}}\Big{(}1+\frac{\text{dist}(I_{s_{1}},x)}{|I_{s_{1}}|}\Big{)}^{-10}dx$
$\displaystyle\lesssim\frac{1}{|I_{s_{1}}|}\int_{\mathbb{R}}\Big{(}1+\frac{\text{dist}(I_{s_{1}},x)}{|I_{s_{1}}|}\Big{)}^{-10}dx\lesssim
1.$

In combination with (393) this implies Claim 1, since we have

$\displaystyle\sum_{\mathbf{T}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}\in\mathbf{T}^{\prime}\\\
\omega_{s_{1}}=\omega_{s^{\prime}_{1}}\end{subarray}}\big{|}\langle
f,\phi_{s_{1}}\rangle\langle
f,\phi_{s^{\prime}_{1}}\rangle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle\big{|}$
$\displaystyle\lesssim\sum_{\mathbf{T}\in\mathscr{F}}\sum_{\hat{s}\in\mathbf{T}}\big{|}\langle
f,\phi_{s_{1}}\big{\rangle}|^{2}$ $\displaystyle\lesssim
2^{2d}\sum_{\mathbf{T}\in\mathscr{F}}|I_{\mathbf{T}}|.$

Let us then focus on the second claim, which will be more technical as
interactions between tiles of different frequency scales will need to be
counted in. First, note that the left-hand side of (398) can be written as

(399)
$\displaystyle\sum_{\mathbf{T}\in\mathscr{F}}\sum_{P\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{P^{\prime}\in\mathbf{T}^{\prime}}\sum_{\begin{subarray}{c}\hat{s}=P\end{subarray}}\sum_{\begin{subarray}{c}\hat{s}^{\prime}=P^{\prime},\\\
\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}\end{subarray}}\,\big{|}\langle
f,\phi_{s_{1}}\rangle\langle
f,\phi_{s^{\prime}_{1}}\rangle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle\big{|}$

and by Cauchy-Schwarz, we have

(400)
$\displaystyle\sum_{\hat{s}=P}\,\sum_{\genfrac{}{}{0.0pt}{}{\hat{s}^{\prime}=P^{\prime}}{\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}}}\big{|}\langle
f,\phi_{s_{1}}\rangle\langle
f,\phi_{s^{\prime}_{1}}\rangle\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle|\leq$
$\displaystyle\big{(}\sum_{\hat{s}=P}|\langle
f,\phi_{s_{1}}\rangle|^{2}\big{)}^{\frac{1}{2}}\,\big{(}\sum_{\hat{s}^{\prime}=P^{\prime}}|\langle
f,\phi_{s^{\prime}_{1}}\rangle|^{2}\big{)}^{\frac{1}{2}}$
$\displaystyle\cdot\big{(}\sum_{\hat{s}=P}\,\sum_{\genfrac{}{}{0.0pt}{}{\hat{s}^{\prime}=P^{\prime}}{\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}}}|\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle|^{2}\big{)}^{\frac{1}{2}}.$

The subtree condition (393) applied to the singleton subtrees
$\\{P\\}\subset\mathbf{T}$ and $\\{P^{\prime}\\}\subset\mathbf{T}^{\prime}$
provides the following upper bound for (399):

$\displaystyle\sum_{\mathbf{T}\in\mathscr{F}}\sum_{P\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{P^{\prime}\in\mathbf{T}^{\prime}}2^{2d+2}|I_{P}|^{\frac{1}{2}}|I_{P^{\prime}}|^{\frac{1}{2}}\Big{(}\sum_{\hat{s}=P}\,\sum_{\genfrac{}{}{0.0pt}{}{\hat{s}^{\prime}=P^{\prime}}{\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}}}|\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle|^{2}\Big{)}^{\frac{1}{2}}.$

The anticipated estimate (396) follows if we prove, for any tree
$\mathbf{T}\in\mathscr{F}$, the inequality

(401)
$\sum_{P\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{P^{\prime}\in\mathbf{T}^{\prime}}|I_{P}|^{\frac{1}{2}}|I_{P^{\prime}}|^{\frac{1}{2}}\Big{(}\sum_{\hat{s}=P}\,\sum_{\genfrac{}{}{0.0pt}{}{\hat{s}^{\prime}=P^{\prime}}{\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}}}|\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle|^{2}\Big{)}^{\frac{1}{2}}\lesssim|I_{\mathbf{T}}|.$

Before moving on, we take a closer look at the geometry of the tiles in
${{\mathbb{P}}_{\textrm{BHT}}}$.

  1. (i)

if $P\in\mathbf{T}$ and $P^{\prime}\in\mathbf{T}^{\prime}$ with
$\mathbf{T},\mathbf{T}^{\prime}\in\mathscr{F}$, and $s,s^{\prime}$ are so that
$\hat{s}=P,\hat{s}^{\prime}=P^{\prime}$ with
$\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}$ then

     * •

$\boldsymbol{\omega}_{P_{1}}\subsetneq\boldsymbol{\omega}_{P^{\prime}_{1}}$
and the sparseness of the collection ${{\mathbb{P}}_{\textrm{BHT}}}$ implies
that $|\boldsymbol{\omega}_{P_{1}}|\ll|\boldsymbol{\omega}_{P^{\prime}_{1}}|$
and $|I_{P}|\gg|I_{P^{\prime}}|$;

     * •

the trees $\mathbf{T}$ and $\mathbf{T}^{\prime}$ cannot coincide: due to their
$1$-lacunarity, tiles of different scales belonging to the same tree cannot
overlap in frequency;

     * •

since $\mathscr{F}$ consists of strongly disjoint $1$-lacunary trees, we must
have

(402) $I_{P^{\prime}}\cap 3I_{\mathbf{T}}=\emptyset.$

  2. (ii)

notice that once $\mathbf{T}\in\mathscr{F}$ and $P\in\mathbb{P}$ are fixed,
the spatial supports of the tiles $P^{\prime}$ (contained in any of the trees
$\mathbf{T}^{\prime}\in\mathscr{F}$) having the property that
$\boldsymbol{\omega}_{P_{1}}\subsetneq\boldsymbol{\omega}_{P^{\prime}_{1}}$
and $|\boldsymbol{\omega}_{P_{1}}|\ll|\boldsymbol{\omega}_{P^{\prime}_{1}}|$,
are mutually disjoint – otherwise such tiles would intersect both in space and
in frequency and the strong disjointness property would be disobeyed.

We return to understanding

(403)
$\Big{(}\sum_{\hat{s}=P}\,\sum_{\genfrac{}{}{0.0pt}{}{\hat{s}^{\prime}=P^{\prime}}{\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}}}|\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle|^{2}\Big{)}^{\frac{1}{2}},$

where $P\in\mathbf{T},P^{\prime}\in\mathbf{T}^{\prime}$ are such that
$\boldsymbol{\omega}_{P_{1}}\subsetneq\boldsymbol{\omega}_{P^{\prime}_{1}},|I_{P}|\gg|I_{P^{\prime}}|$
and $I_{P^{\prime}}\cap 3I_{P}\neq\emptyset$. This quantity measures the
correlation between the tiles $P$ and $P^{\prime}$, as perceived144144144If
the wave-packets $\phi_{s_{1}}$ were supported both in space and frequency on
the tile $s_{1}$, then (403) would amount to $0$. through the wave-packets
$\displaystyle\\{\phi_{s_{1}}\\}_{\hat{s}=P},\\{\phi_{s^{\prime}_{1}}\\}_{\hat{s}^{\prime}=P^{\prime}}$.

Our claim is that

(404)
$\Big{(}\sum_{\hat{s}=P}\,\sum_{\genfrac{}{}{0.0pt}{}{\hat{s}^{\prime}=P^{\prime}}{\omega_{s_{1}}\subsetneq\omega_{s^{\prime}_{1}}}}|\langle\phi_{s_{1}},\phi_{s^{\prime}_{1}}\rangle|^{2}\Big{)}^{\frac{1}{2}}\lesssim\frac{|I_{P^{\prime}}|^{\frac{1}{2}}}{|I_{P}|^{\frac{1}{2}}}\Big{(}1+\frac{\text{dist}(I_{P},I_{P^{\prime}})}{|I_{P}|}\Big{)}^{-5}.$

Before proving this estimate, we show how it allows to conclude (401) and
consequently (398). Using (404), the left-hand side of (401) is bounded by

$\sum_{P\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}P^{\prime}\in\mathbf{T}^{\prime}\\\
\boldsymbol{\omega}_{P_{1}}\subsetneq\boldsymbol{\omega}_{P^{\prime}_{1}}\end{subarray}}|I_{P^{\prime}}|\Big{(}1+\frac{\text{dist}(I_{P},I_{P^{\prime}})}{|I_{P}|}\Big{)}^{-5}.$

For $P\in\mathbf{T}$ fixed, we invoke (ii) above, to deduce

$\displaystyle\sum_{P\in\mathbf{T}}\sum_{\mathbf{T}^{\prime}\in\mathscr{F}}\sum_{\begin{subarray}{c}P^{\prime}\in\mathbf{T}^{\prime}\\\
\boldsymbol{\omega}_{P_{1}}\subsetneq\boldsymbol{\omega}_{P^{\prime}_{1}}\end{subarray}}|I_{P^{\prime}}|\Big{(}1+\frac{\text{dist}(I_{P},I_{P^{\prime}})}{|I_{P}|}\Big{)}^{-5}$
$\displaystyle\lesssim\sum_{P\in\mathbf{T}}\int_{(3I_{\mathbf{T}})^{C}}\Big{(}1+\frac{\text{dist}(I_{P},x)}{|I_{P}|}\Big{)}^{-5}dx$
$\displaystyle\lesssim\sum_{P\in\mathbf{T}}|I_{P}|\Big{(}1+\frac{\text{dist}(I_{P},(3I_{\mathbf{T}})^{C})}{|I_{P}|}\Big{)}^{-3}$
$\displaystyle\lesssim|I_{\mathbf{T}}|.$

Now we return to proving (404) for $P$ and $P^{\prime}$ fixed, as above. This
inequality will be itself a consequence of

$\displaystyle\sum_{\hat{I}=I_{P}}\sum_{\hat{I^{\prime}}=I_{P^{\prime}}}\sum_{\hat{\omega}=\boldsymbol{\omega}_{P_{1}}}\sum_{\hat{\omega}^{\prime}=\boldsymbol{\omega}_{P^{\prime}_{1}}}|\langle\phi_{I\times\omega},\phi_{I^{\prime}\times\omega^{\prime}}\rangle|^{2}\lesssim\frac{|I_{P^{\prime}}|}{|I_{P}|}\Big{(}1+\frac{\text{dist}(I_{P},I_{P^{\prime}})}{|I_{P}|}\Big{)}^{-10}.$

Here we will mostly rely on the spatial decay: given that
$|I_{P}|\gg|I_{P^{\prime}}|$, $|I|=2^{-{am\over 2}}|I_{P}|$ and
$|I^{\prime}|=2^{-{am\over 2}}|I_{P^{\prime}}|$, we have

$|\langle\phi_{I\times\omega},\phi_{I^{\prime}\times\omega^{\prime}}\rangle|^{2}\lesssim\frac{|I^{\prime}|}{|I|}\Big{(}1+\frac{\text{dist}(I,I^{\prime})}{|I|}\Big{)}^{-50}=\frac{|I_{P^{\prime}}|}{|I_{P}|}\Big{(}1+\frac{\text{dist}(I,I^{\prime})}{|I|}\Big{)}^{-50}.$

For fixed intervals $I$ and $I^{\prime}$ with $\hat{I}=I_{P}$ and
$\hat{I^{\prime}}=I_{P^{\prime}}$, there are precisely $2^{am\over 2}$
intervals $\omega$ and $2^{am\over 2}$ intervals $\omega^{\prime}$ so that
$\hat{I}\times\hat{\omega}=P$ and
$\hat{I^{\prime}}\times\hat{\omega^{\prime}}=P^{\prime}$ respectively. This
implies

$\displaystyle\sum_{\hat{I}=I_{P}}\sum_{\hat{I^{\prime}}=I_{P^{\prime}}}\sum_{\hat{\omega}=\boldsymbol{\omega}_{P_{1}}}\sum_{\hat{\omega}^{\prime}=\boldsymbol{\omega}_{P^{\prime}_{1}}}|\langle\phi_{I\times\omega},\phi_{I^{\prime}\times\omega^{\prime}}\rangle|^{2}\lesssim
2^{am}\frac{|I_{P^{\prime}}|}{|I_{P}|}\sum_{\hat{I}=I_{P}}\sum_{\hat{I^{\prime}}=I_{P^{\prime}}}\Big{(}1+\frac{\text{dist}(I,I^{\prime})}{|I|}\Big{)}^{-50}.$

We focus on the last part. When $I$ and $I^{\prime}$ are so that
$\hat{I}=I_{P}$ and $\hat{I^{\prime}}=I_{P^{\prime}}$, with
$\boldsymbol{\omega}_{P_{1}}\subsetneq\boldsymbol{\omega}_{P^{\prime}_{1}}$,
the separation condition (402) has as a consequence

$\displaystyle\sum_{\hat{I}=I_{P}}\Big{(}1+\frac{\text{dist}(I,I^{\prime})}{|I|}\Big{)}^{-50}$
$\displaystyle\lesssim\Big{(}\frac{\text{dist}(I_{P},I^{\prime})}{2^{-{am\over
2}}|I_{P}|}\Big{)}^{-40}$ $\displaystyle\lesssim
2^{-20am}\frac{1}{|I^{\prime}|}\int_{I^{\prime}}\Big{(}\frac{\text{dist}(I_{P},x)}{|I_{P}|}\Big{)}^{-40}dx.$

Now we sum in $I^{\prime}$ to get

$\displaystyle\sum_{\hat{I}=I_{P}}\sum_{\hat{I^{\prime}}=I_{P^{\prime}}}\sum_{\hat{\omega}=\boldsymbol{\omega}_{P_{1}}}\sum_{\hat{\omega}^{\prime}=\boldsymbol{\omega}_{P^{\prime}_{1}}}|\langle\phi_{I\times\omega},\phi_{I^{\prime}\times\omega^{\prime}}\rangle|^{2}$
$\displaystyle\lesssim
2^{-10am}\frac{|I_{P^{\prime}}|}{|I_{P}|}\frac{1}{|I_{P^{\prime}}|}\int_{I_{P^{\prime}}}\Big{(}\frac{\text{dist}(I_{P},x)}{|I_{P}|}\Big{)}^{-40}dx$
$\displaystyle\lesssim
2^{-10am}\frac{|I_{P^{\prime}}|}{|I_{P}|}\Big{(}\frac{\text{dist}(I_{P},I_{P^{\prime}})}{|I_{P}|}\Big{)}^{-40}.$

This proves (404), finishing thus our proof. ∎

######  Remark 7.14.

The size estimates (376)–(378) continue to hold, with an extra
$(1+|j|)^{-{M\over 2}}$ decay, when we factor in the $j$-shifts discussed in
Remark 6.9. Indeed, although the wave-packets
$\Phi_{am-k,2\ell-1,j}^{2^{am}(r-1)+p^{\prime}}$ take into account the
information in $\underline{I}_{P}^{p^{\prime}}+j2^{am\over
2}|\underline{I}_{P}^{p^{\prime}}|$ – thus supported within $I_{P}+j|I_{P}|$ –
thanks to the decaying factor $(1+|j|)^{-{M}}$ we can regard it as adapted to
$I_{P}$ through the bump function $\tilde{\chi}_{I_{P}}$.

The same reasoning applies to the energy estimate (390), and further, allows
to deduce the same localized estimate (409) when taking the shifts into
account.

The size and energy estimates together with Proposition 7.9 allow us to prove
Proposition 7.2.

###  7.5. Establishing the boundedness range for $T$

Thanks to the estimates worked in subsections 7.2-7.4 we first establish the
$m$-decay boundedness of $T_{m}$ within the strictly local-$L^{2}$ case
($2<p,q,r^{\prime}<\infty$). This is the content of Corollary 7.3 which
follows from Proposition 7.2 via decomposition (368). Notice that, due to the
summability in $m$, the same boundedness range is implied for $T$.

Next, with Proposition 7.2 settled, we present a refinement of the above time-
frequency analysis, which, based on an intermediate interpolation step
executed at a localized level, implies ($m$-exponential decaying) estimates
for $T_{m}$ in the wider range $\\{2<p,q\leq\infty\ ;\ 1<r<\infty\\}$ thus
proving Theorem 3.3. Finally, as it was explained in Section 3.4, the full
range of Theorem 3.1 follows subsequently from standard interpolation
techniques.

####  7.5.1. Boundedness of $T$ in the local-$L^{2}$ range

As mentioned above, this is an immediate consequence of Corollary 7.3, which,
in turn, follows from Proposition 7.2. Thus our focus will be on proving the
latter.

###### Proof of Proposition 7.2.

We first establish restricted-type estimates: for any finite measure sets
$E_{1},E_{2},E_{3}\subset\mathbb{R}$ and any triple of functions $f,g,h$
satisfying

$|f(x)|\leq\mathbf{1}_{E_{1}}(x)\text{
a.e.},\quad|g(x)|\leq\mathbf{1}_{E_{2}}(x)\text{
a.e.},\quad|h(x)|\leq\mathbf{1}_{E_{3}}(x)\text{ a.e.},\quad$

we claim to have

(405)
$\displaystyle|\Lambda_{{{\mathbb{P}}_{\textrm{BHT}}}}^{\mathfrak{u}}(f,g,h)|$
$\displaystyle\lesssim$
$\displaystyle|E_{1}|^{\frac{1}{p}}\,|E_{2}|^{\frac{1}{q}}\,|E_{3}|^{\frac{1}{r^{\prime}}},$
(406)
$\displaystyle|\Lambda_{{{\mathbb{P}}_{\textrm{BHT}}}}^{\mathfrak{l}}(f,g,h)|$
$\displaystyle\lesssim 2^{-\delta_{2}^{\prime}am}$
$\displaystyle|E_{1}|^{\frac{1}{p}}\,|E_{2}|^{\frac{1}{q}}\,|E_{3}|^{\frac{1}{r^{\prime}}},$
(407)
$\displaystyle|\Lambda_{{{\mathbb{P}}_{\textrm{BHT}}}}^{\mathfrak{c}}(f,g,h)|$
$\displaystyle\lesssim 2^{-\delta_{2}^{\prime}am}$
$\displaystyle|E_{1}|^{\frac{1}{p}}\,|E_{2}|^{\frac{1}{q}}\,|E_{3}|^{\frac{1}{r^{\prime}}},$

for any $p,q,r^{\prime}\in(2,\infty)$ satisfying
$\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$ and some exponent
$\delta_{2}^{\prime}:=\delta_{2}^{\prime}(p,q)>0$.

We start with (405). Estimates (375) and (377) imply for
$\mathbb{P}={{\mathbb{P}}_{\textrm{BHT}}}$

$\mathrm{Size\,}^{\mathbb{P}}_{1}(f(P_{1}))\lesssim
1,\qquad\mathrm{Size\,}^{\mathbb{P}}_{2}(g(P_{2}))\lesssim
1,\qquad\mathrm{Size\,}^{\mathbb{P}}_{3}(h_{\mathfrak{u}}(P_{3}))\lesssim 1;$

similarly, due to (389) and (390),

$\mathrm{Energy\,}^{\mathbb{P}}_{1}(f(P_{1}))\lesssim|E_{1}|^{\frac{1}{2}},\quad\mathrm{Energy\,}^{\mathbb{P}}_{2}(g(P_{2}))\lesssim|E_{2}|^{\frac{1}{2}},\quad\mathrm{Energy\,}^{\mathbb{P}}_{3}(h_{\mathfrak{u}}(P_{3}))\lesssim|E_{3}|^{\frac{1}{2}}.$

Applying Proposition 7.9 with the size/energy estimates and the choice
$\theta_{1}=1-\frac{2}{p}\in(0,1),\theta_{2}=1-\frac{2}{q}\in(0,1)$ and
$\theta_{3}=-1+\frac{2}{r}\in(0,1)$, one has

$\displaystyle\Lambda_{{{\mathbb{P}}_{\textrm{BHT}}}}^{\mathfrak{u}}(f,g,h)$
$\displaystyle\lesssim|E_{1}|^{{1-\theta_{1}}\over
2}\cdot|E_{2}|^{{1-\theta_{2}}\over 2}\cdot|E_{3}|^{{1-\theta_{3}}\over 2}$
$\displaystyle\lesssim|E_{1}|^{1\over p}\cdot|E_{2}|^{1\over
q}\cdot|E_{3}|^{1\over{r^{\prime}}},$

which is (405).

For (406), the corresponding size and energy estimates are

$\mathrm{Size\,}^{\mathbb{P}}_{1}(f(P_{1}))\lesssim
1,\qquad\mathrm{Size\,}^{\mathbb{P}}_{2}(g(P_{2}))\lesssim
1,\qquad\mathrm{Size\,}^{\mathbb{P}}_{3}(h_{\mathfrak{l}}(P_{3}))\lesssim
2^{-\frac{\delta_{1}am}{2}};$

and

$\mathrm{Energy\,}^{\mathbb{P}}_{1}(f(P_{1}))\lesssim|E_{1}|^{\frac{1}{2}},\quad\mathrm{Energy\,}^{\mathbb{P}}_{2}(g(P_{2}))\lesssim|E_{2}|^{\frac{1}{2}},\quad\mathrm{Energy\,}^{\mathbb{P}}_{3}(h_{\mathfrak{l}}(P_{3}))\lesssim|E_{3}|^{\frac{1}{2}}.$

We apply again Proposition 7.9 with the same choice of exponents
$\theta_{1},\theta_{2},\theta_{3}$, obtaining

$\displaystyle\Lambda_{{{\mathbb{P}}_{\textrm{BHT}}}}^{\mathfrak{l}}(f,g,h)$
$\displaystyle\lesssim|E_{1}|^{{1-\theta_{1}}\over
2}\cdot|E_{2}|^{{1-\theta_{2}}\over 2}\cdot
2^{-\frac{\delta_{1}}{2}am\theta_{3}}|E_{3}|^{{1-\theta_{3}}\over 2}$
$\displaystyle\lesssim
2^{-\delta_{1}am(-\frac{1}{2}+\frac{1}{r})}|E_{1}|^{1\over
p}\cdot|E_{2}|^{1\over q}\cdot|E_{3}|^{1\over{r^{\prime}}};$

this is estimate (406), where the decaying factor is
$2^{-\delta_{1}am(-\frac{1}{2}+\frac{1}{r})}$. The proof of (407) follows
analogously (with the only difference consisting in the estimate
$\mathrm{Size\,}^{\mathbb{P}}_{3}(h_{\mathfrak{c}}(P_{3}))\lesssim
2^{-\frac{\mu am}{4}}$), which concludes the proof of the three restricted-
type estimates with
$\delta_{2}^{\prime}:=\min\\{\delta_{1},\frac{\mu}{2}\\}(-\frac{1}{2}+\frac{1}{r})>0$.

Finally, by invoking restricted-type interpolation (see for example [81] or
[44] for a reference), we obtain the strong-type estimates of Proposition 7.2
with, say,
$\delta_{2}(p,q):=\frac{1}{2}\min\\{\delta_{1},\frac{\mu}{2}\\}(-\frac{1}{2}+\frac{1}{r})>0$.
∎

####  7.5.2. Boundedness of $T$ in the wider range
$\\{(p,q,r):2<p,q\leq\infty\ ;\ 1<r<\infty\\}$

To deal with the extended range and, in particular, to be able to prove some
$L^{p}\times L^{\infty}\to L^{p}$ endpoint estimates, we need to move outside
of the local-$L^{2}$ range. As it turns out, it will be sufficient to remove
the local-$L^{2}$ constraints on the third function $h$. To this aim, we will
use the helicoidal method of [4], as follows:

  1. (a)

first we transform the time-frequency information obtained in Sections 7.2-7.4
into exclusively spatial information (in the form of maximal averages over
arbitrary intervals);

  2. (b)

next, we interpolate between this localized information and the localized,
exclusively spatial information for an operator resembling a linearized
version of the bilinear maximal function from (111);

  3. (c)

lastly, we re-organise this information in the form of a Fefferman-Stein
inequality from which the extended range of boundedness will follow.

###### Proof of Theorem 3.3.

To implement the strategy outlined above, we rely on the initial
discretization performed in Section 4.2.6 which allows us to represent
$\underline{\mathcal{L}}_{m}(f,g,h)$ as
$\sum_{P\in{{\mathbb{P}}_{\textrm{BHT}}}}\underline{\mathcal{L}}_{m,P}(f,g,h)$
– see (134).

For any $\mathbb{P}\subset{{\mathbb{P}}_{\textrm{BHT}}}$ and $I_{0}$ a dyadic
interval

$\mathbb{P}(I_{0}):=\\{P\in\mathbb{P}:I_{P}\subseteq I_{0}\\}$

denotes the subcollection of $\mathbb{P}$ with spatial intervals contained in
$I_{0}$. Recalling the definition (374), we also define

(408) $\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}:=\\{I\subseteq I_{0}\text{ dyadic
}:I\in\mathcal{I}_{\bar{\mathbb{P}}}\text{ or }I=I_{0}\\}.$

In this way, we only keep track of the spatial intervals included in $I_{0}$
corresponding to tiles or trees in $\mathbb{P}$, together with the spatial
interval $I_{0}$ itself.

Item (a) above indicates that
$|\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)|$ will be
estimated using the strategy highlighted in Sections 7.2-7.4, which also
produces an exponential decay in $m$; an upper bound for the same expression
$|\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)|$ can be
obtained directly from (134) and its equivalent (180), although this time the
exponential decay in $m$ is not available. Interpolating between the two – as
item (b) suggests, will produce estimates for
$|\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)|$ that fall
outside the local-$L^{2}$ range and have exponential decay in $m$. It will
remain to reconstruct the global information for
$|\sum_{P\in\mathbb{P}}\underline{\mathcal{L}}_{m,P}(f,g,h)|$ from the local
one, and this is achieved via Fefferman-Stein inequalities – this is the last
step (c).

Propositions 7.9, 7.11, and 7.12, together with the observation that the
energies localize well,145145145Using the decay of the wave-packets it is not
difficult to see that
$\mathrm{Energy\,}^{\mathbb{P}(I_{0})}_{1}((f(P_{1}))_{P\in\mathbb{P}(I_{0})})\lesssim\|f\cdot\tilde{\chi}_{I_{0}}\|_{L^{2}},$
and similarly for $g,h$. This resembles somehow the proof of the size estimate
(375); see also [3, Lemma 39].  imply that for some $\tilde{\delta}>0$

(409)
$\displaystyle\Big{|}\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)\Big{|}$
$\displaystyle\lesssim
2^{-\tilde{\delta}am}\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|f\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}\,\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|g\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}$
$\displaystyle\hskip
120.00018pt\cdot\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I_{P}|^{\frac{1}{2}}}\|h\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}\,\cdot|I_{0}|.$

Notice that the above estimate represents a localized multi-scale version of
the $L^{2}\times L^{2}\times L^{2}\to\mathbb{C}$ estimate (184) presented in
Section 5. In order to pass from the latter to the former, all the time-
frequency discussion performed in Subsections 7.2 and 7.3, as well as the
refinement in Section 6 are necessary.

If we were now to invoke Theorem 12 of [4], we could deduce from (409) that

(410) $|\underline{\mathcal{L}}_{m}(f,g,h)|\lesssim
2^{-\tilde{\delta}am}\int_{\mathbb{R}}M_{2}f(x)\cdot M_{2}g(x)\cdot
M_{2}h(x)\,dx$

(where we recall that $M_{2}f=(M|f|^{2})^{1/2}$). This inequality is clearly
sufficient to deduce the summability of the operator norms of
$\underline{\mathcal{L}}_{m}$ (and hence of $T_{m}$) in the strict
local-$L^{2}$ range – in this way we would recover the result in Corollary
7.3. However, $L^{p}\times L^{\infty}\to L^{p}$ bounds cannot be concluded
from the estimate above.

Identity (409) illustrates step (a) of the strategy described at the beginning
of the present subsection. We continue now with step (b), which consists of
obtaining a different type of local estimate for
$\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)$.

For this, we rely on the initial definition (181) of
$\underline{\mathcal{L}}_{m,P}(f,g,h)$, but will momentarily return to the
parametric formulation based on indices $k,\ell,r,n,p,u,v$. With this,
$|\underline{\mathcal{L}}_{m,P}(f,g,h)|$ becomes

(411) $\displaystyle\frac{1}{2^{\frac{am+2k}{4}}}\Big{|}\sum_{n,p\sim
2^{\frac{am}{2}}}$ $\displaystyle\sum_{u,v\sim 2^{\frac{am}{2}}}\langle
f_{\ell+1},\check{\varphi}_{\frac{am}{2}-k,\ell}^{u-v,p_{r}-n}\rangle\langle
g_{\ell-1},\check{\varphi}_{\frac{am}{2}-k,\ell}^{u+v,p_{r}+n}\rangle$
$\displaystyle\cdot\int_{\mathbb{R}}\rho_{am-
ak}(\lambda(x))\,\,\check{\varphi}_{\frac{am}{2}-k,2\ell-1}^{2u,p_{r}}(x)w_{k,n,v}^{e}(\lambda)(x)h(x)dx\Big{|}.$

Recalling the definition of $w_{k,n,v}^{e}(\lambda)$, the decaying estimate
(140), and the pointwise inequality

$\big{|}\check{\varphi}_{\frac{am}{2}-k,2\ell-1}^{2u,p_{r}}(x)\big{|}\lesssim
2^{\frac{am-2k}{4}}\tilde{\chi}_{I_{k}^{p_{r}}}(x),$

we bound (411) by

(412) $\displaystyle\frac{1}{2^{k}}\Big{|}\sum_{n,p\sim 2^{\frac{am}{2}}}$
$\displaystyle\sum_{u,v\sim 2^{\frac{am}{2}}}\langle
f_{\ell+1},\check{\varphi}_{\frac{am}{2}-k,\ell}^{u-v,p_{r}-n}\rangle\langle
g_{\ell-1},\check{\varphi}_{\frac{am}{2}-k,\ell}^{u+v,p_{r}+n}\rangle\Big{|}$
$\displaystyle\cdot\int_{\mathbb{R}}\rho_{am-
ak}(\lambda(x))\,\,\tilde{\chi}_{I_{k}^{p_{r}}}(x)\frac{1}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{2}}|h(x)|dx.$

Notice that the second line does not depend on the $u$ parameter anymore since
we disregarded the frequency information contained in the wave-packet
$\check{\varphi}_{\frac{am}{2}-k,2\ell-1}^{2u,p_{r}}$. We can use Cauchy-
Schwarz in $u$ to obtain, for fixed parameters $v,n,p$,

$\displaystyle\sum_{u\sim 2^{am\over 2}}\big{|}\langle
f_{\ell+1},\check{\varphi}_{\frac{am}{2}-k,\ell}^{u-v,p_{r}-n}\rangle\langle
g_{\ell-1},\check{\varphi}_{\frac{am}{2}-k,\ell}^{u+v,p_{r}+n}\rangle\big{|}$
$\displaystyle\lesssim\big{(}\sum_{u_{1}\sim 2^{\frac{am}{2}}}\big{|}\langle
f,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{1},p_{r}-n}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}\big{(}\sum_{u_{2}\sim
2^{\frac{am}{2}}}\big{|}\langle
g,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{2},p_{r}+n}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}.$

So that $|\underline{\mathcal{L}}_{m,P}(f,g,h)|$ is further bounded by

(413) $\displaystyle\frac{1}{2^{k}}\sum_{n,p\sim 2^{\frac{am}{2}}}$
$\displaystyle\sum_{v\sim 2^{\frac{am}{2}}}\big{(}\sum_{u_{1}\sim
2^{\frac{am}{2}}}\big{|}\langle
f,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{1},p_{r}-n}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}\big{(}\sum_{u_{2}\sim
2^{\frac{am}{2}}}\big{|}\langle
g,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{2},p_{r}+n}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}$
$\displaystyle\cdot\int_{\mathbb{R}}\rho_{am-
ak}(\lambda(x))\,\,\tilde{\chi}_{I_{k}^{p_{r}}}(x)\frac{1}{1+\Big{|}\bar{c}_{a}\frac{n^{a-1}\lambda(x)}{2^{a(\frac{am}{2}-k)}}+2v\Big{|}^{2}}|h(x)|dx.$

Now the expression in the first line does not depend on $v$ anymore, and thus,
summing in $v$ (see (386)), we get

(414)
$\displaystyle|\underline{\mathcal{L}}_{m,P}(f,g,h)|\lesssim\frac{1}{2^{k}}\sum_{n,p\sim
2^{\frac{am}{2}}}$ $\displaystyle\big{(}\sum_{u_{1}\sim
2^{\frac{am}{2}}}\big{|}\langle
f,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{1},p_{r}-n}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}\big{(}\sum_{u_{2}\sim
2^{\frac{am}{2}}}\big{|}\langle
g,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{2},p_{r}+n}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}$
$\displaystyle\cdot\int_{\mathbb{R}}\rho_{am-
ak}(\lambda(x))\,\,\tilde{\chi}_{I_{k}^{p_{r}}}(x)|h(x)|dx.$

We apply Cauchy-Schwarz with respect to $n$ – which only affects the first
line – and thus, modulo considering $O(1)$ similar terms, we have

(415)
$\displaystyle|\underline{\mathcal{L}}_{m,P}(f,g,h)|\lesssim\frac{1}{2^{k}}$
$\displaystyle\big{(}\sum_{n_{1},u_{1}\sim 2^{\frac{am}{2}}}\big{|}\langle
f,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{1},n_{1}}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}\big{(}\sum_{u_{2},n_{2}\sim
2^{\frac{am}{2}}}\big{|}\langle
g,\check{\varphi}_{\frac{am}{2}-k,\ell}^{u_{2},n_{2}}\rangle\big{|}^{2}\big{)}^{\frac{1}{2}}$
$\displaystyle\cdot\int_{\mathbb{R}}\rho_{am-
ak}(\lambda(x))\,\,\big{(}\sum_{p\sim
2^{\frac{am}{2}}}\tilde{\chi}_{I_{k}^{p_{r}}}(x)\big{)}|h(x)|dx.$

At this point, we can return to the notation involving only the tile
formulation: for any $P\in{{\mathbb{P}}_{\textrm{BHT}}}$, we have proved that

$\displaystyle|\underline{\mathcal{L}}_{m,P}(f,g,h)|\lesssim\frac{1}{|I_{P}|}f(P_{1})g(P_{2})\int_{\mathbb{R}}\rho_{|P|}(\lambda(x))\,\tilde{\chi}_{I_{P}}(x)|h(x)|dx,$

where we recall the notations (361) and (362).

Putting these together we obtain the local estimate:

$\displaystyle\big{|}\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)\big{|}$
$\displaystyle\lesssim\sum_{\kappa:2^{\kappa}\leq|I_{0}|}\sum_{\begin{subarray}{c}I\subseteq
I_{0}\\\
|I|=2^{\kappa}\end{subarray}}\frac{1}{|I|}\sum_{\begin{subarray}{c}P\in\mathbb{P}\\\
I_{P}=I\end{subarray}}\|f_{P}\|_{L^{2}}\|g_{P}\|_{L^{2}}\int_{\mathbb{R}}\rho_{am-a\kappa}(\lambda(x))\,\tilde{\chi}_{I}(x)|h(x)|dx$

The tiles in $\mathbb{P}(I_{0})$ sharing the same spatial interval correspond
to mutually disjoint frequency intervals and due to orthogonality and Cauchy-
Schwartz we have

$\sum_{\begin{subarray}{c}P\in\mathbb{P}(I_{0})\\\
I_{P}=I\end{subarray}}\|f_{P}\|_{L^{2}}\|g_{P}\|_{L^{2}}\lesssim\|f\cdot\tilde{\chi}_{I}\|_{L^{2}}\|g\cdot\tilde{\chi}_{I}\|_{L^{2}}.$

This observation produces

$\displaystyle\big{|}\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)\big{|}$
$\displaystyle\lesssim\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|f\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}\,\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|g\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}$
$\displaystyle\sum_{\kappa:2^{\kappa}\leq|I_{0}|}\sum_{\begin{subarray}{c}J\subseteq
I_{0}\\\
|J|=2^{\kappa}\end{subarray}}\frac{1}{|J|}\int_{\mathbb{R}}\rho_{am-a\kappa}(\lambda(x))\,\tilde{\chi}_{J}(x)|h(x)|dx.$

Since the intervals have the same scale,

$\sum_{\begin{subarray}{c}J\subseteq I_{0}\\\
|J|=2^{\kappa}\end{subarray}}\tilde{\chi}_{J}(x)\lesssim\tilde{\chi}_{I_{0}}(x);$

in addition, thanks to the presence of the stopping-time $\lambda(x)$, for
every $x$ there are only $O(1)$ scales $\kappa$ for which
$\rho_{am-a\kappa}(\lambda(x))\neq 0$; and therefore

$\sum_{\kappa:2^{\kappa}\leq|I_{0}|}\sum_{\begin{subarray}{c}J\subseteq
I_{0}\\\
|J|=2^{\kappa}\end{subarray}}\int_{\mathbb{R}}\rho_{am-a\kappa}(\lambda(x))\,\tilde{\chi}_{J}(x)|h(x)|dx\lesssim\int_{\mathbb{R}}\tilde{\chi}_{I_{0}}(x)|h(x)|dx.$

We have therefore proven the following bound:

(416)
$\displaystyle\big{|}\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)\big{|}$
$\displaystyle\lesssim\big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|f\cdot\tilde{\chi}_{I}\|_{L^{2}}\big{)}\cdot\big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|g\cdot\tilde{\chi}_{I}\|_{L^{2}}\big{)}$
$\displaystyle\hskip
10.00002pt\cdot\big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|}\int_{\mathbb{R}}|h(x)|\tilde{\chi}_{I}(x)dx\big{)}\,|I_{0}|.$

Although the expression involving $h$ is a genuine average, for symmetry
reasons we express it as a maximal average. Similarly to (410), [4, Theorem
12] implies the global Fefferman-Stein-type inequality

(417)
$|\underline{\mathcal{L}}_{m}(f,g,h)|\lesssim\int_{\mathbb{R}}M_{2}f(x)\cdot
M_{2}g(x)\cdot Mh(x)\,dx.$

At this point, we would like to apply a suitable interpolation procedure
between (410) and (417), in a way that avoids canonical interpolation for the
operator $T_{m}$.146146146Estimates of the type $L^{p}\times L^{\infty}\to
L^{p}$ cannot be obtained through classical interpolation since they
correspond to boundary points.

This is achieved by interpolating between the two local estimates (409) and
(416), in the particular situation when $h$ is a restricted-type function:
i.e. $h$ is such that $|h(x)|\leq\mathbf{1}_{H}(x)$ for some measurable set
$H\subset\mathbb{R}$ of finite measure. The maximal averages of $h$ in (409)
are then bounded by

$\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|}\int_{\mathbb{R}}\mathbf{1}_{H}(x)\tilde{\chi}_{I}(x)dx\Big{)}^{{1\over
2}}$

and in (416) by

$\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|}\int_{\mathbb{R}}\mathbf{1}_{H}(x)\tilde{\chi}_{I}(x)dx\Big{)}.$

Using trivial interpolation (geometric average), we deduce that

(418)
$\displaystyle\Big{|}\sum_{P\in\mathbb{P}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)\Big{|}$
$\displaystyle\lesssim$ $\displaystyle
2^{-\tilde{\delta}am\theta}\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I_{P}|^{\frac{1}{2}}}\|f\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|g\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}$
$\displaystyle\hskip
10.00002pt\cdot\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|}\int_{\mathbb{R}}\tilde{\chi}_{I}(x)\mathbf{1}_{H}(x)dx\Big{)}^{{\theta\over
2}+(1-\theta)}\,|I_{0}|,$

for any $0<\theta<1$. Then a “mock interpolation” result – see Proposition 14
of [4] – implies that

(419)
$\displaystyle\Big{|}\sum_{P\in{{\mathbb{P}}_{\textrm{BHT}}}(I_{0})}\underline{\mathcal{L}}_{m,P}(f,g,h)\Big{|}$
$\displaystyle\lesssim$ $\displaystyle
2^{-\tilde{\delta}am\theta}\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|f\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|^{\frac{1}{2}}}\|g\cdot\tilde{\chi}_{I}\|_{L^{2}}\Big{)}$
$\displaystyle\hskip
10.00002pt\cdot\Big{(}\sup_{I\in\mathcal{I}_{\bar{\mathbb{P}}(I_{0})}}\frac{1}{|I|}\int_{\mathbb{R}}\tilde{\chi}_{I}(x)|h(x)|^{\tau}dx\Big{)}^{1\over\tau}\,|I_{0}|,$

for any $\tau>0$ such that $\frac{1}{\tau}<1-\frac{\theta}{2}$. We note that
the implicit constant blows up as $\tau$ nears $\frac{2}{2-\theta}$.

For the last part (c) of the strategy, we appeal again to [4, Theorem 12],
which allows to put back together the information obtained at the local level
and deduce – via a sparse domination argument – that

(420) $|\underline{\mathcal{L}}_{m}(f,g,h)|\lesssim
2^{-\tilde{\delta}am\theta}\int_{\mathbb{R}}M_{2}f(x)\cdot M_{2}g(x)\cdot
M_{\tau}h(x)\,dx.$

This is indeed an interpolation of (410) and (417), but it is achieved through
local control of $\underline{\mathcal{L}}_{m}$.

Finally, this implies by duality that with $\bar{\delta}=\tilde{\delta}\theta$

$\|\underline{\mathcal{L}}_{m}\|_{L^{p}\times L^{q}\times
L^{r^{\prime}}\to{\mathbb{C}}}=\|T_{m}\|_{L^{p}\times L^{q}\to L^{r}}\lesssim
2^{-\bar{\delta}am}$

for any $p,q,r$ satisfying

$\frac{1}{p}<\frac{1}{2},\quad\frac{1}{q}<\frac{1}{2},\quad\frac{1}{r^{\prime}}<1-{\theta\over
2},$

additionally to the usual Hölder scaling condition $1/p+1/q=1/r$. Since
$\theta\in(0,1)$ was arbitrary, these conditions reduce to $2<p,q\leq\infty$
and $r^{\prime}>1$. ∎

######  Remark 7.15.

  1. (1)

The inequality

$\displaystyle|\underline{\mathcal{L}}_{m,P}(f,g,h)|\lesssim\frac{1}{|I_{P}|}f(P_{1})g(P_{2})\int_{\mathbb{R}}\rho_{|P|}(\lambda(x))\,\tilde{\chi}_{I_{P}}(x)|h(x)|dx,$

is a discretized, dualized version of the estimate

$\displaystyle\Big{|}\rho_{a}(2^{-a(m-k)}\lambda(x))\int
f(x-t)g(x+t)e^{i\lambda(x)t^{a}}\rho(2^{-k}t)\frac{dt}{t}\Big{|}$
$\displaystyle\lesssim\rho_{a}(2^{-a(m-k)}\lambda(x))M(f,g)(x)$

used in Section 3 to bound pointwise $T_{m}$ by the bilinear maximal function
$M$ of (111). The Gabor frame decomposition used for the higher resolution
model naturally induces $L^{2}$ norms for the functions $f$ and $g$.

  2. (2)

If in (376)-(378) the size of $h$ was an $L^{\tau}$ quantity for some
$1<\tau<2$, some $L^{p}\times L^{\infty}\to L^{p}$ estimates could have been
obtained by applying (a) directly. The difference between (409) and (416)
consists in the fact that (409) presents a $2^{-\tilde{\delta}am}$ decay, but
we have to use $L^{2}$ maximal averages for $h$, while (416) presents no decay
in $m$, but the (maximal) averages of $h$ are measured relative to the better
(smaller) $L^{1}$ norm. This dichotomy is explained as follows: The
compatibility step in Section 6, which allows to extract spatial and frequency
information for the third function $h$ as well, requires a careful
distribution of the information on the level set of the oscillatory factor
carried by $\rho_{|P|}(\lambda(x))w_{|P|,n,v}^{e}(\lambda)(x)$. Using
$\ell^{2}$ estimates in the parameters $n,v$ and $p^{\prime}$, this
information is split between the bilinear term corresponding to $f$ and $g$,
and the term corresponding to the dualizing function $h$, respectively.
Because of this, the sizes of $f,g$ and $h$ used above in the previous
Subsection 7.3 are naturally $L^{2}$-adapted. In obtaining (416) however, one
may disregard $w_{|P|,n,v}^{e}(\lambda)(x)$ altogether exchanging the lack of
$m$-decay with the use of $L^{1}$ averages for $h$.

##  Appendix A Lack of $m$-decaying absolute summability for the discrete
phase-linearized wave-packet model

As stated in the Introduction, in this first appendix we prove the lack of the
absolute single scale $m$-decay for the discrete phase-linearized wave-packet
model defined by (143).

We start by recalling the formulations in (180) and (181) and focus our
attention on the expression of this model localized within a fixed tri-tile
$P=P(k,\ell,r)\in{{\mathbb{P}}_{\textrm{BHT}}}$ given by some prescribed
parameters $k,\ell,r\in\mathbb{N}$ and a fixed $m\in\mathbb{N}$:

$\displaystyle\underline{\mathcal{L}}_{m,P}(f,g,h)$
$\displaystyle=\underline{\mathcal{L}}_{m,k,\ell,r}(f,g,h)$
$\displaystyle=\sum_{n,p\sim 2^{am\over 2}}\sum_{u,v\sim 2^{am\over
2}}\frac{1}{2^{am+2k\over 4}}\langle f_{\ell+1},\check{\varphi}_{{am\over
2}-k,\ell}^{u-v,p_{r}-n}\rangle\langle g_{\ell-1},\check{\varphi}_{{am\over
2}-k,\ell}^{u+v,p_{r}+n}\rangle$
$\displaystyle\int_{\mathbb{R}}\check{\varphi}_{{am\over
2}-k,2\ell-1}^{2u,p_{r}}(x)h(x)\rho_{am-
ak}(\lambda(x))w_{k,n,v}^{e}(\lambda)(x)dx.$

For convenience we take $P=P_{0}:=P(0,1,1)$ so that $k=0$, $\ell=1$, $r=1$,
$I_{P^{0}}=[1,2]$, $\boldsymbol{\omega}_{P^{0}_{1}}=2^{am}\,[1,2]$ and
$\boldsymbol{\omega}_{P^{0}_{2}}=2^{am}\,[-1,0]$. In what follows we show that
if we take absolute values inside the summation signs appearing in the
definition of $\underline{\mathcal{L}}_{m,P}\big{|}_{P=P_{0}}$ above then no
analogue of Proposition 5.1 can hold. Moreover, one can not obtain a single
scale $m$-decay estimate even within the Hölder scaling $L^{2}\times
L^{2}\times L^{\infty}$:

######  Proposition A.1.

Fix147147147The case $a\in\\{1,2\\}$ is superfluous. $a>0$. There are no
absolute constants $C,\delta>0$ for which the following holds148148148Here we
recall that since $r=1$ we must have that $p_{r}=p$. uniformly in the choices
of $m\in\mathbb{N}$ (large enough), phase function $\lambda$ and input
functions $f,g\in L^{2}$, $h\in L^{\infty}$:

(421) $\displaystyle\underline{\mathcal{L}}_{m,P^{0}}^{abs}(f,g,h)$
$\displaystyle:=\sum_{n,p\sim 2^{am\over 2}}\sum_{u,v\sim 2^{am\over
2}}\frac{1}{2^{am\over 4}}|\langle f_{2},\check{\varphi}_{{am\over
2},1}^{u-v,p-n}\rangle|\cdot|\langle g_{0},\check{\varphi}_{{am\over
2},1}^{u+v,p+n}\rangle|$ $\displaystyle\hskip
50.00008pt\cdot\Big{|}\int_{\mathbb{R}}\check{\varphi}_{{am\over
2},1}^{2u,p}(x)h(x)\rho_{am}(\lambda(x))w_{0,n,v}^{e}(\lambda)(x)dx\Big{|}$
$\displaystyle\leq C2^{-\delta
am}\|f\|_{L^{2}}\|g\|_{L^{2}}\|h\|_{L^{\infty}}.$

######  Remark A.2.

(a) The above proposition remains true if one modifies the norm in which $h$
is measured: one can substitute the $L^{\infty}$ norm with $L^{2}$ or even
with Sobolev type norms. The existence of a counterexample at (421) is not
specific to the control/norm that we put on $h$ but rather depends on the
interaction between, on the one hand, the information carried by $f$ and $g$
and, on the other hand, the stricture of the linearizing function $\lambda$;
(b) In fact, our proof reveals the existence of an absolute constant $K>0$ so
that for any $m\in\mathbb{N}$ (large enough), there exist nontrivial $f,g\in
L^{2}$, $h\in L^{\infty}$ and a real measurable phase function $\lambda$ for
which

$\underline{\mathcal{L}}_{m,P^{0}}^{abs}(f,g,h)>K\|f\|_{L^{2}}\|g\|_{L^{2}}\|h\|_{L^{\infty}}.$

###### Proof.

Our proof will proceed via reductio ad absurdum. Thus we assume that for any
$f,g\in L^{2}$ and $h\in L^{\infty}$ there exist $C,\delta>0$ absolute
constants such that (421) holds uniformly in $m\in{\mathbb{N}}$ large enough.

Take now

$f:=\sum_{\begin{subarray}{c}2\cdot 2^{am\over 2}\leq|n_{1}|\leq 3\cdot
2^{am\over 2}\\\ |p|\leq 3\cdot 2^{am\over
2}\end{subarray}}\check{\varphi}_{{am\over
2},1}^{n_{1},p}\qquad\textrm{and}\qquad
g:=\sum_{\begin{subarray}{c}1\leq|n_{2}|\leq 2^{am\over 2}\\\ |p|\leq 3\cdot
2^{am\over 2}\end{subarray}}\check{\varphi}_{{am\over 2},1}^{n_{2},p}\,.$

Thus $f=f_{2}$, $g=g_{0}$ and for $n,p,u,v\sim 2^{am\over 2}$ (suitably
understood) we have

$\langle f,\check{\varphi}_{{am\over 2},1}^{u-v,p-n}\rangle\sim 1\sim\langle
g,\check{\varphi}_{{am\over 2},1}^{u+v,p+n}\rangle$

with

$\|f\|_{L^{2}}\simeq 2^{am\over 2}\simeq\|g\|_{L^{2}}.$

For this specific choice of $f$ and $g$ we have that (421) is equivalent with

(422) $\sum_{n,p\sim 2^{am\over 2}}\sum_{u,v\sim 2^{am\over
2}}\Big{|}\int_{I_{P_{0}}}\check{\varphi}_{{am\over
2},1}^{2u,p}(x)h(x)\rho_{am}(\lambda(x))w_{0,n,v}^{e}(\lambda)(x)dx\Big{|}\leq
C2^{-\delta am}2^{\frac{5}{4}am}\|h\|_{L^{\infty}}.$

Next, we are going to construct the essential item in what will become our
counterexample to (421) – the phase function $\lambda$. For this we decompose

(423) $\lambda(x)=\lambda_{u}(x)\,+\lambda_{s}(x)\,,$

where here

  * •

the uniform component is defined as $\lambda_{u}(x):=2^{am}$ for every $x\in
I_{P^{0}}$;

  * •

the structured component has both the input and the output correlated with
some suitable (generalized) arithmetic progressions.149149149The fact that
such a generalized arithmetic progressions structure plays the key rôle in the
construction of our present counterexample suggests a natural connection
between the (limiting) behavior of $BC^{a}$ and that of the Carleson operator
$C$. Indeed, same type of structures are essential in providing
counterexamples for the behavior of the pointwise convergence of Fourier
Series near $L^{1}$ – for the latter see [74] and the bibliography therein.
Indeed, for some $L\in\mathbb{N}$ large enough, we take

(424) $\textrm{supp}\,\lambda_{s}=A=\bigcup_{{j\simeq 2^{am\over 2}}\atop{j\in
L\mathbb{N}}}A_{j}\,,$

with

(425) $A_{j}:=\bigsqcup_{{q\simeq 2^{am\over 2}}\atop{q\in
L\mathbb{N}}}(2^{-am}[j,j+1]+q2^{-\frac{am}{2}})=:\bigsqcup_{{q\simeq
2^{am\over 2}}\atop{q\in L\mathbb{N}}}I_{j,q}\,,$

and let

(426) $\lambda_{s}(x):=\sum_{{j\sim 2^{am\over 2}}\atop{j\in
L\mathbb{N}}}(2^{am\over 2}\,(j-2^{am\over 2}))\mathbf{1}_{A_{j}}(x)\,.$

With these, let us record the following facts:

  * •

by properly adjusting the support of $\rho$ we have from the above that
$\rho_{am}(\lambda(x)))=1$ for every $x\in I_{P^{0}}$;

  * •

the phase function $\lambda$ is constant on each of the subsets $A_{j}$ as
well as on $B:=I_{P_{0}}\setminus A$;

  * •

since $w_{0,n,v}^{e}(\lambda_{u})$ is constant on the whole interval
$I_{P^{0}}$ (it does not depend on $x$) it is not hard to see that (421) holds
for $\lambda=\lambda_{u}$.150150150This should become transparent once the
reader mimics the reasoning presented in (427)–(430) for the case in which
$x\in B$.

From the above we deduce that it is enough to disprove (422) for
$\rho_{am}(\lambda(x)))=1$ and $I_{P_{0}}$ replaced by $A$ where here
$\lambda$ and $A$ are as constructed above.

We now fix $x\in A$ and compute the contribution of the weight defined in
(137)

$\displaystyle\mathbf{1}_{A}(x)\,w_{0,n,v}^{e}(\lambda)(x)$
$\displaystyle=\int_{\mathbb{R}}\psi(\xi+2v)e^{i\,n\,\xi}e^{ic_{a}2^{({am\over
2})a^{\prime}}\xi^{a^{\prime}}\lambda(x)^{-{a^{\prime}\over a}}}\,d\xi$
$\displaystyle=\sum_{{j\sim 2^{am\over 2}}\atop{j\in
L\mathbb{N}}}\mathbf{1}_{A_{j}}(x)\int_{\mathbb{R}}\psi(\xi+2v)e^{i\,n\,\xi}e^{ic_{a}2^{({am\over
2})a^{\prime}}\xi^{a^{\prime}}(2^{am\over 2}j)^{-{a^{\prime}\over
a}}}\,d\xi\,.$

Using now the principle of (non)stationary phase (recall that we are in the
regime in which the phase function $\nu$ in the integral above satisfies
$|\nu^{\prime\prime}(\xi)|\sim 1$ on the domain of integration), we have the
following approximation:

(427) $\displaystyle w_{0,n,v}^{e}(\lambda)(x)$
$\displaystyle\approx\sum_{{j\sim 2^{am\over 2}}\atop{j\in
L\mathbb{N}}}\mathbf{1}_{A_{j}}(x)e^{i\bar{c}_{a}n^{a}2^{am\over
2}j2^{-a\left(\frac{am}{2}\right)}}\psi\big{(}\bar{c}_{a}\frac{n^{a\over{a^{\prime}}}2^{am\over
2}j}{2^{a\left(\frac{am}{2}\right)}}+2v\big{)}$ $\displaystyle=\sum_{{j\sim
2^{am\over 2}}\atop{j\in
L\mathbb{N}}}\mathbf{1}_{A_{j}}(x)e^{i\bar{c}_{a}n^{a}j2^{-\frac{a^{2}m}{2a^{\prime}}}}\psi\big{(}\bar{c}_{a}\frac{n^{a\over{a^{\prime}}}j}{2^{\frac{a^{2}m}{2a^{\prime}}}}+2v\big{)}.$

Indeed, the above holds up to an error term which involves a decaying factor
of the form $2^{-\delta am}$ for some $\delta>0$ thus not affecting the
construction of our counterexample below.

Consequently, in order to disprove (422) it is enough to show that there are
no $C,\,\delta>0$ such that the following holds uniformly in $m\in\mathbb{N}$
and $h\in L^{\infty}$:

$\displaystyle\sum_{n,p\sim 2^{am\over 2}}\sum_{u,v\sim 2^{am\over
2}}\Big{|}\sum_{{j\sim 2^{am\over 2}}\atop{j\in
L\mathbb{N}}}\big{(}\int_{\mathbb{R}}\check{\varphi}_{{am\over
2},1}^{2u,p}(x)h(x)\mathbf{1}_{A_{j}}(x)dx\big{)}e^{i\bar{c}_{a}n^{a}j2^{-\frac{a^{2}m}{2a^{\prime}}}}\psi\big{(}\bar{c}_{a}\frac{n^{a\over{a^{\prime}}}j}{2^{\frac{a^{2}m}{2a^{\prime}}}}+2v\big{)}\Big{|}$
(428) $\displaystyle\leq C2^{-\delta
am}\,2^{\frac{5}{4}am}\,\|h\|_{L^{\infty}}.$

Recalling that $\psi\in C_{0}^{\infty}(\mathbb{R})$ and choosing appropriately
its support, we deduce that in order for the inner summation term in (428) to
be nonzero we must have

(429)
$\Big{|}\bar{c}_{a}\frac{n^{a\over{a^{\prime}}}j}{2^{\frac{a^{2}m}{2a^{\prime}}}}+2v\Big{|}\leq
1$

which, since $n\sim 2^{am\over 2}$, is equivalent to

$\Big{|}j+\frac{2^{\frac{a^{2}m}{2a^{\prime}}+1}}{\bar{c}_{a}n^{a\over{a^{\prime}}}}v\Big{|}\leq\frac{2^{\frac{a^{2}m}{2a^{\prime}}}}{\bar{c}_{a}n^{a\over{a^{\prime}}}}\sim
1\>.$

Hence for $n$ fixed, $j$ is determined by $v$ since $j\in L\mathbb{N}$ and
$L\in\mathbb{N}$ may be chosen as large as we want (but still independently of
$m$). Thus, without loss of generality we may assume that (429) describes a
$1$-to-$1$ map $v\leftrightarrow j$.

Fixing $n\sim 2^{am\over 2}$, we have from the above that

(430) $\displaystyle\sum_{v\sim 2^{am\over 2}}\Big{|}\sum_{{j\sim 2^{am\over
2}}\atop{j\in L\mathbb{N}}}\Big{(}\int_{\mathbb{R}}\check{\varphi}_{{am\over
2},1}^{2u,p}(x)h(x)\mathbf{1}_{A_{j}}(x)dx\Big{)}e^{i\bar{c}_{a}n^{a}j2^{-\frac{a^{2}m}{2a^{\prime}}}}\psi\big{(}\bar{c}_{a}\frac{n^{a\over{a^{\prime}}}j}{2^{\frac{a^{2}m}{2a^{\prime}}}}+2v\big{)}\Big{|}$
$\displaystyle\qquad=\sum_{v\sim 2^{am\over 2}}\sum_{{j\sim 2^{am\over
2}}\atop{j\in
L\mathbb{N}}}\Big{|}\Big{(}\int_{\mathbb{R}}\check{\varphi}_{{am\over
2},1}^{2u,p}(x)h(x)\mathbf{1}_{A_{j}}(x)dx\Big{)}e^{i\bar{c}_{a}n^{a}j2^{-\frac{a^{2}m}{2a^{\prime}}}}\psi\big{(}\bar{c}_{a}\frac{n^{a\over{a^{\prime}}}j}{2^{\frac{a^{2}m}{2a^{\prime}}}}+2v\big{)}\Big{|}$
$\displaystyle\qquad=\sum_{v\sim 2^{am\over 2}}\sum_{{j\sim 2^{am\over
2}}\atop{j\in
L\mathbb{N}}}\Big{|}\Big{(}\int_{\mathbb{R}}\check{\varphi}_{{am\over
2},1}^{2u,p}(x)h(x)\mathbf{1}_{A_{j}}(x)dx\Big{)}\Big{|}\Big{|}\psi\big{(}\bar{c}_{a}\frac{n^{a\over{a^{\prime}}}j}{2^{\frac{a^{2}m}{2a^{\prime}}}}+2v\big{)}\Big{|}$
$\displaystyle\qquad\gtrsim\sum_{{j\sim 2^{am\over 2}}\atop{j\in
L\mathbb{N}}}\big{|}\langle h\mathbf{1}_{A_{j}},\check{\varphi}_{{am\over
2},1}^{2u,p}\rangle\big{|}.$

Notice that the last lower bound does not depend on the parameter $n$.

So in order for (428) to be true, we must have

(431) $\sum_{u,p\sim 2^{am\over 2}}\sum_{{j\sim 2^{am\over 2}}\atop{j\in
L\mathbb{N}}}\big{|}\langle h\mathbf{1}_{A_{j}},\check{\varphi}_{{am\over
2},1}^{2u,p}\rangle\big{|}\leq C2^{-\delta
am}2^{\frac{3}{4}am}\|h\|_{L^{\infty}}.$

######  Remark A.3.

Observe that the trivial case $\delta=0$ in (431) is an immediate consequence
of the Cauchy-Schwarz inequality via the pairwise disjointness of
$\\{A_{j}\\}_{j}$ and the almost orthogonality of
$\big{\\{}\check{\varphi}_{{am\over 2},1}^{2u,p}\big{\\}}_{u,p}$. Thus, in
order to maximize the left hand-side of (431) it is natural to pursue the
equality case in the above Cauchy-Schwarz argument thus constructing
$\\{A_{j}\\}_{j}$ and $h\in L^{\infty}$ such that the local Fourier
coefficients $\big{|}\langle h\mathbf{1}_{A_{j}},\check{\varphi}_{{am\over
2},1}^{2u,p}\rangle\big{|}$ are (essentially) pairwise equal and hence
independent of the specific choice of $u,p,j$. In particular, this explains
why we design the structure of $(A_{j})_{j}$ as essentially a
$2^{-am}$-neighborhood of a generalized arithmetic progression of rank 2
(hence its “$2^{-\frac{am}{2}}$–translation invariance behavior” in (425)).

Take now151151151Or some smooth variant of it which is still equal to $1$ on
$[1,2]$. $h=\mathbf{1}_{[1,2]}$ and recall the definition of $A_{j}$ in (425).
By properly massaging the properties of $\check{\varphi}$ we can assume
without loss of generality that for every $j\sim 2^{am\over 2}$ with $j\in
L\mathbb{N}$ we have that

(432) $\check{\varphi}(2^{\frac{am}{2}}x-q)\sim 1\qquad\textrm{for any}\qquad
x\in I_{j,q}\,,$

and

(433)
$|\check{\varphi}(2^{\frac{am}{2}}x-q)|\leq|q-q^{\prime}|^{-10}\qquad\textrm{for
any}\>\>x\in I_{j,q^{\prime}}\>\>\textrm{with}\>\>\>q,q^{\prime}\in
L\mathbb{Z},\>q\not=q^{\prime}\,.$

In what follows we fix $u,p\sim 2^{am\over 2}$. From the above we then deduce
that

$\displaystyle\big{|}\langle h\mathbf{1}_{A_{j}},\check{\varphi}_{{am\over
2},1}^{2u,p}\rangle\big{|}$
$\displaystyle=2^{\frac{am}{4}}\Big{|}\int_{A_{j}}\check{\varphi}(2^{\frac{am}{2}}x-p)\,e^{i\,2^{\frac{am}{2}}x\left(2u+2^{\frac{am}{2}}\right)}dx\Big{|}$
$\displaystyle=2^{\frac{am}{4}}\Big{|}\sum_{{{q\simeq 2^{am\over 2}}\atop{q\in
L\mathbb{N}}}}\int_{I_{j,q}}\check{\varphi}(2^{\frac{am}{2}}x-p)\,e^{i\,2^{\frac{am}{2}}x\left(2u+2^{\frac{am}{2}}\right)}dx\Big{|}$
$\displaystyle\geq
2^{\frac{am}{4}}\Big{|}\int_{I_{j,p}}\check{\varphi}(2^{\frac{am}{2}}x-p)e^{i\,2^{\frac{am}{2}}x\left(2u+2^{\frac{am}{2}}\right)}dx\Big{|}$
$\displaystyle\qquad\qquad-2^{\frac{am}{4}}\,\sum_{{q\neq p}\atop{q\in
L\mathbb{Z}}}\Big{|}\int_{I_{j,q}}\check{\varphi}(2^{\frac{am}{2}}x-p)e^{i\,2^{\frac{am}{2}}x\left(2u+2^{\frac{am}{2}}\right)}dx\Big{|}$
(434) $\displaystyle=:(I)-(II).$

For the first term, applying (432) and a change of variable, we claim
that152152152Recall our earlier convention $e^{iy}=e^{2\pi iy}$.

(435) $\displaystyle(I)$
$\displaystyle=2^{-\frac{3am}{4}}\Big{|}\int_{[0,1]}\check{\varphi}(2^{-\frac{am}{2}}(j+y))e^{iy\left(2^{-\frac{am}{2}+1}u+1\right)}dy\Big{|}$
$\displaystyle\simeq
2^{-\frac{3am}{4}}\Big{|}\check{\varphi}(2^{-\frac{am}{2}}j)\int_{[0,1]}e^{iy\left(2^{-\frac{am}{2}+1}u+1\right)}dy\Big{|}\simeq
2^{-\frac{3am}{4}}$

holds for a large fraction of the possible values of $u\sim 2^{am\over 2}$.

Indeed, for this we first notice that while $u$ runs through the set
${\mathbb{Z}}\cap[2^{am\over 2},2\cdot 2^{am\over 2}]$, the expression
$2^{-\frac{am}{2}+1}u+1$ covers the set $2^{-{am\over
2}+1}{\mathbb{Z}}+1\cap[3,5]$. Thus, for $m\in\mathbb{N}$ large enough, (say
$am\geq 10$), there exists a value of $u=u_{0}$ in the range $u_{0}\sim
2^{am\over 2}$ such that

$\text{dist}(2^{-\frac{am}{2}+1}u_{0}+1,{\mathbb{Z}})\geq\frac{1}{10}.$

Then an easy argument shows that for $u\in{\mathbb{Z}}\cap[2^{am\over
2},2\cdot 2^{am\over 2}]\cap[u_{0}-\frac{1}{100}2^{am\over
2},u_{0}+\frac{1}{100}2^{am\over 2}]$ one still has

$\text{dist}(2^{-\frac{am}{2}+1}u+1,{\mathbb{Z}})\geq\frac{1}{4}$

which implies the lower bound in (435)

$|\int_{[0,1]}e^{iy\left(2^{-\frac{am}{2}+1}u+1\right)}dy|\geq\frac{1}{100}.$

Hence, for $u$ belonging to ${\mathbb{Z}}\cap[2^{am\over 2},2\cdot 2^{am\over
2}]\cap[u_{0}-\frac{1}{100}2^{am\over 2},u_{0}+\frac{1}{100}2^{am\over 2}]$
one can now verify the validity of the statement

(436) $(I)\geq 10^{-10}\,2^{-\frac{3am}{4}}.$

For the second part $(II)$ we are using (433) for, say, a choice of $L\geq
100$. With this we have

(437) $\displaystyle(II)$ $\displaystyle\lesssim
2^{-\frac{3am}{4}}\sum_{{q\neq p}\atop{q\in
L\mathbb{Z}}}\int_{[0,1]}|\check{\varphi}(2^{-\frac{am}{2}}(j+y)+(p-q))|dy$
$\displaystyle\leq
2^{-\frac{3am}{4}}\sum_{d\in\mathbb{Z}\setminus\\{0\\}}(1+|d|\,L)^{-10}\leq
2^{-\frac{3am}{4}}\,100\,L^{-10}.$

Putting now together (A)–(437) we obtain a refutation of (431) thus
contradicting the assumption that (421) holds. ∎

##  Appendix B Analysis of the multiplier

In this appendix we perform the last part of the initial reductions, the
result of which was anticipated in Subsection 3.5: namely that the initial
symbol (116) is equal to a main term of the form (117) plus an error term
which is suitably bounded. This is a rather standard non-stationary phase
argument153153153A classical reference for this type of stationary phase
analysis is provided by [95]., applied to our specific setting.

We start by recalling our symbol

$\mathfrak{M}_{m,k}(\zeta;\lambda)=\int e^{i\lambda
t^{a}}\frac{\rho(2^{-k}t)}{t}e^{-i\zeta t}\,dt.$

The analysis of the above expression proceeds as follows: first one reduces
the matters to the subset of parameters $(\zeta,\lambda)$ for which the phase
contains a stationary point (this produces controlled errors of non-stationary
nature); secondly, one specializes the amplitude of the symbol to a
sufficiently small neighbourhood of the stationary point (this produces
further controlled errors of non-stationary nature); thirdly, one applies the
stationary phase method to the remaining part, obtaining the main contribution
to the oscillatory integral $\mathfrak{M}_{m,k}$; finally, the resulting
symbol is put into a tensorized form by a standard Fourier series
decomposition argument. The errors that have arisen along the way are all
controlled by the bilinear maximal function – more precisely, the error
contribution to $T_{m}(f,g)(x)$ will be controlled pointwise by $\lesssim
2^{-\delta am}M(f,g)(x)$ for some absolute $\delta>0$ – any such $\delta$ is
good enough for our purposes.

###  B.1. Non-stationary contributions to $\mathfrak{M}_{m,k}$ when there is
no critical point

First of all we tackle the non-stationary parts of the symbol. In what follows
we will always assume that $\lambda\equiv\lambda(x)\sim 2^{a(m-k)}$ because of
the presence of the factor $\tilde{\rho}(2^{-a(m-k)}\lambda(x))$ in the
expression for $T_{m}(f,g)$.

If $\varphi(t)=\varphi_{\zeta,\lambda}(t):=\lambda t^{a}-\zeta t$ denotes the
phase of $\mathfrak{M}_{m,k}(\zeta;\lambda)$, a simple dimensional analysis
shows that in the domain of integration we have $|\lambda t^{a-1}|\sim
2^{am-k}$, and therefore $\varphi^{\prime}$ will certainly be non-zero if
$|\zeta|\not\sim 2^{am-k}$; the non-stationary phase principle then suggests
that the symbol will be suitably small for such $\zeta$. To make this
intuition rigorous, we begin by observing that by integration by parts

(438) $\displaystyle\int e^{i\lambda t^{a}}$
$\displaystyle\frac{{\rho}(2^{-k}t)}{t}e^{-i\zeta t}\,dt=-\int
e^{i\varphi(t)}\frac{d}{dt}\Big{(}\frac{{\rho}(2^{-k}t)}{it\varphi^{\prime}(t)}\Big{)}\,dt$
$\displaystyle=-\int
e^{i\varphi(t)}\Big{[}\frac{2^{-k}{\rho}^{\prime}(2^{-k}t)}{it\varphi^{\prime}(t)}-\frac{{\rho}(2^{-k}t)}{it^{2}\varphi^{\prime}(t)}-\frac{{\rho}(2^{-k}t)\varphi^{\prime\prime}(t)}{it(\varphi^{\prime}(t))^{2}}\Big{]}\,dt.$

Recall that $\rho$ as given in (107) is smooth, compactly supported on
$\Big{[}\frac{1}{2},2\Big{]}$ and such that
$\sum_{j\in\mathbb{Z}}\rho(2^{-j}\zeta)=1$ for $\zeta\neq 0$. Let us then
decompose smoothly $1=\chi_{0}+\chi_{1}+\sum_{j\geq 2}\chi_{j}$, where

  1. i)

$\chi_{1}(\zeta):=\sum_{\ell=-2\lceil a\rceil}^{2\lceil
a\rceil}\rho(a^{-1}2^{-\ell}\zeta)$;

  2. ii)

$\chi_{j}(\zeta):=\rho(a^{-1}2^{-2\lceil a\rceil+1}2^{-j}\zeta)$ for any
$j\geq 2$;

  3. iii)

$\chi_{0}=1-\chi_{1}-\sum_{j\geq 2}\chi_{j}$.

Observe that $\chi_{0}$ is a bump function supported in a neighbourhood of the
origin, and more precisely it is supported in $\\{\zeta:|\zeta|<a2^{-2\lceil
a\rceil}\\}$.

We can then split

$\mathfrak{M}_{m,k}(\zeta;\lambda)=\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\mathfrak{M}_{m,k}(\zeta;\lambda)+\Big{(}1-\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\Big{)}\mathfrak{M}_{m,k}(\zeta;\lambda)$

and further split the last term (which will be shown to contribute an error
term) according to (438) into

$\Big{(}1-\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\Big{)}\mathfrak{M}_{m,k}(\zeta;\lambda)=-\mathfrak{E}^{1}_{m,k}(\zeta;\lambda)+\mathfrak{E}^{2}_{m,k}(\zeta;\lambda)+\mathfrak{E}^{3}_{m,k}(\zeta;\lambda),$

where for $i\in\\{1,2,3\\}$ the term $\mathfrak{E}^{i}_{m,k}$ is given by the
sum

$\mathfrak{E}^{i}_{m,k}(\zeta;\lambda):=\mathfrak{E}^{i,0}_{m,k}(\zeta;\lambda)+\sum_{j\geq
2}\mathfrak{E}^{i,j}_{m,k}(\zeta;\lambda)$

and for $j=0$ or $j\geq 2$ we have set

$\displaystyle\mathfrak{E}^{1,j}_{m,k}(\zeta;\lambda)$
$\displaystyle:=\chi_{j}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi(t)}\frac{{\rho}^{\prime}(2^{-k}t)}{it}\frac{2^{-k}}{\varphi^{\prime}(t)}\,dt,$
$\displaystyle\mathfrak{E}^{2,j}_{m,k}(\zeta;\lambda)$
$\displaystyle:=\chi_{j}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi(t)}\frac{{\rho}(2^{-k}t)}{it}\frac{1}{t\varphi^{\prime}(t)}\,dt,$
$\displaystyle\mathfrak{E}^{3,j}_{m,k}(\zeta;\lambda)$
$\displaystyle:=\chi_{j}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi(t)}\frac{{\rho}(2^{-k}t)}{it}\frac{\varphi^{\prime\prime}(t)}{(\varphi^{\prime}(t))^{2}}\,dt.$

Let us begin with showing that

(439)
$\displaystyle\Big{|}\sum_{k\in\mathbb{Z}}\tilde{\rho}\Big{(}\frac{\lambda(x)}{2^{a(m-k)}}\Big{)}\iint\widehat{f}(\xi)\widehat{g}(\eta)$
$\displaystyle
e^{i(\xi+\eta)x}\mathfrak{E}^{1,0}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta\Big{|},$
$\displaystyle\lesssim 2^{-am}M(f,g)(x),$

so that this contribution to $T_{m}(f,g)$ can be safely ignored (thanks to the
fact that the bilinear maximal function is bounded in the range given in
Theorem 3.1). Considering $\mathfrak{E}^{1,0}_{m,k}$, we see that for the
factor $2^{-k}/\varphi^{\prime}(t)$ we can write

$\frac{2^{-k}}{\varphi^{\prime}(t)}=2^{-am}\frac{1}{a\lambda
t^{a-1}2^{-(am-k)}-\zeta 2^{-(am-k)}},$

and if we let

$\Phi_{\lambda,t}(\zeta^{\prime})=\Phi_{m,k,\lambda,t}(\zeta^{\prime}):=\frac{\chi_{0}(\zeta^{\prime})}{a\lambda
t^{a-1}2^{-(am-k)}-\zeta^{\prime}}$

we can rewrite

$\mathfrak{E}^{1,0}_{m,k}(\zeta;\lambda)=2^{-am}\int\Phi_{\lambda,t}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}e^{-i\zeta
t}\frac{{\rho}^{\prime}(2^{-k}t)}{it}e^{i\lambda t^{a}}\,dt.$

Notice that $|a\lambda t^{a-1}2^{-(am-k)}|\sim_{a}1$ and more precisely that

$\displaystyle a2^{-2a+1}$ $\displaystyle\leq|a\lambda t^{a-1}2^{-(am-k)}|\leq
a2^{2a-1}\qquad$ when $a>1$, $\displaystyle\frac{a}{2}$
$\displaystyle\leq|a\lambda t^{a-1}2^{-(am-k)}|\leq 2a$ when $a<1$.

For $\zeta^{\prime}$ in the support of $\chi_{0}$, we then have in both cases
$a>1$ and $a<1$ that

$|a\lambda t^{a-1}2^{-(am-k)}-\zeta^{\prime}|\gtrsim_{a}1.$

From this, one can see that $\Phi_{\lambda,t}$ is a smooth bump function
adapted to an interval of length $\sim 1$ uniformly in all the parameters
$m,k,\lambda\sim 2^{a(m-k)},|t|\sim 2^{k}$, and in particular we have
uniformly

(440) $|\widehat{\Phi_{\lambda,t}}(s)|\lesssim_{a}(1+|s|)^{-100}$

for all $|t|\sim 2^{k}$. All implicit constants here are allowed to depend on
$a$, so from this point on we do not track the dependence systematically.
Fixing $m,k,\lambda(x)\sim 2^{a(m-k)}$ and using Fubini twice, we see by going
back to the spatial representation of the operator that we have

$\displaystyle\Big{|}$
$\displaystyle\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}\mathfrak{E}^{1,0}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta\Big{|}$
$\displaystyle=2^{-am}\Big{|}\int f(x-s)g(x+s)\Big{[}\int
2^{am-k}\widehat{\Phi_{\lambda,t}}(2^{am-k}(s-t))\frac{{\rho}^{\prime}(2^{-k}t)}{it}e^{i\lambda
t^{a}}\,dt\Big{]}\,ds\Big{|}$ $\displaystyle\leq
2^{-am}\int|f(x-s)||g(x+s)|\int
2^{am-k}\big{|}\widehat{\Phi_{\lambda,t}}(2^{am-k}(s-t))\big{|}\frac{|{\rho}^{\prime}(2^{-k}t)|}{|t|}\,dt\,ds.$

Using (440) we obtain for the resulting kernel

$\displaystyle\int$ $\displaystyle
2^{am-k}\big{|}\widehat{\Phi_{\lambda,t}}(2^{am-k}(s-t))\big{|}\frac{|{\rho}^{\prime}(2^{-k}t)|}{|t|}\,dt$
$\displaystyle\lesssim\int
2^{am-k}\Big{(}1+2^{am-k}|s-t|\Big{)}^{-100}\frac{|{\rho}^{\prime}(2^{-k}t)|}{|t|}\,dt\lesssim
2^{-k}(1+2^{-k}|s|)^{-50},$

where we have used the fact that
${\rho}^{\prime}(2^{-k}t)/t=2^{-k}{\rho}^{\prime}(2^{-k}t)/(2^{-k}t)$ is an
$L^{1}$-normalised bump function adapted to $|t|\sim 2^{k}$. Thus we have
shown

$\displaystyle\Big{|}$
$\displaystyle\tilde{\rho}\Big{(}\frac{\lambda(x)}{2^{a(m-k)}}\Big{)}\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}\mathfrak{E}^{1,0}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta\Big{|}$
$\displaystyle\lesssim
2^{-am}\tilde{\rho}\Big{(}\frac{\lambda(x)}{2^{a(m-k)}}\Big{)}\int|f(x-s)||g(x+s)|2^{-k}(1+2^{-k}|s|)^{-50}\,ds$
$\displaystyle\lesssim
2^{-am}\tilde{\rho}\Big{(}\frac{\lambda(x)}{2^{a(m-k)}}\Big{)}M(f,g)(x);$

summing over $k\in\mathbb{Z}$ shows the desired pointwise inequality (439).

For $\mathfrak{E}^{1,j}_{m,k}$ and $j\geq 2$ the situation is entirely similar
but we have summability in $j$ as well: one can write

$\chi_{j}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\frac{2^{-k}}{\varphi^{\prime}(t)}=2^{-am}2^{-j}\Phi_{\lambda,j,t}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}$

for $\Phi_{\lambda,j,t}$ a smooth bump function adapted to an interval of
length $\sim 2^{j}$ uniformly in $j$ and $|t|\sim 2^{k}$ (and the rest of the
parameters); in particular we have

$|\widehat{\Phi_{\lambda,j,t}}(s)|\lesssim 2^{j}(1+2^{j}|s|)^{-100}$

uniformly in $j$ and in $|t|\sim 2^{k}$, which replaces (440). The same
argument given above then shows that for any $j\geq 2$

$\displaystyle\Big{|}\sum_{k\in\mathbb{Z}}\tilde{\rho}\Big{(}\frac{\lambda(x)}{2^{a(m-k)}}\Big{)}\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}$
$\displaystyle\mathfrak{E}^{1,j}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta\Big{|}$
$\displaystyle\lesssim 2^{-am}2^{-j}M(f,g)(x),$

which is summable in $j$ and therefore we have shown that

$\displaystyle\Big{|}\sum_{k\in\mathbb{Z}}\tilde{\rho}\Big{(}\frac{\lambda(x)}{2^{a(m-k)}}\Big{)}\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}$
$\displaystyle\mathfrak{E}^{1}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta\Big{|}$
$\displaystyle\lesssim 2^{-am}M(f,g)(x),$

and this whole contribution to $T_{m}(f,g)$ can be safely ignored.

The error terms with symbols $\mathfrak{E}^{2}_{m,k},\mathfrak{E}^{3}_{m,k}$
can be treated analogously: one just repeats the arguments given above with
$2^{-k}/\varphi^{\prime}(t)$ replaced by $1/(t\varphi^{\prime}(t))$ and
$\varphi^{\prime\prime}(t)/(\varphi^{\prime}(t))^{2}$ – which can be seen to
behave in exactly the same way for the purposes of the above argument (we
stress that the presence of the factor of ${\rho}(2^{-k}t)$ forces $|t|\sim
2^{k}$). Thus we can dispense with these error terms as well and have
effectively reduced ourselves to the study of the bilinear operator with
multiplier symbol $\chi_{1}(\zeta
2^{-(am-k)})\mathfrak{M}_{m,k}(\zeta;\lambda)$.

###  B.2. Non-stationary contributions to $\mathfrak{M}_{m,k}$ close to the
critical point

At this point the phase $\varphi(t)=\lambda t^{a}-\zeta t$ can have one or two
critical points154154154In the latter case they agree in absolute value.: we
denote either one of these critical points by $t_{0}$ and observe that

$|t_{0}|=\Big{(}\frac{|\zeta|}{a|\lambda|}\Big{)}^{1/(a-1)}\sim 2^{-k},$

since in what follows we require $|\zeta|\sim 2^{am-k}$. We will further
resolve the symbol by decomposing the domain of integration in
$\mathfrak{M}_{m,k}$ according to a Whitney decomposition relative to $t_{0}$:
when we are not too close, $|\varphi^{\prime}|$ is still large enough to
produce decay while when we get too close, we split the phase into its
quadratic part and the remainder Taylor tail with the former giving rise to
the main term while the latter representing an error term.

We proceed now by splitting the function ${\rho}$ into two functions, one
supported in the positive reals and one supported in the negative reals (with
a slight abuse of notation we will continue to denote either one of them
${\rho}$). Next, we assume without loss of generality that there is a critical
point $\varphi^{\prime}(t_{0})=0$ and that $t,t_{0},\zeta>0$. In this case we
write

$t_{0}=t_{0}(\zeta;\lambda)=\Big{(}\frac{\zeta}{a\lambda}\Big{)}^{1/(a-1)}$

and notice that $t_{0}$ is smooth as a function of $\zeta$. We have
$|t_{0}|\sim 2^{k}$ as remarked before and therefore $|t-t_{0}|\lesssim 2^{k}$
for all $t$ in the support of ${\rho}(2^{-k}t)$, so if we use ${\rho}_{0}$ to
denote the same function as in (107) (a smooth bump function supported in an
annulus) we can decompose in $j$ as

$\int e^{i\varphi(t)}\frac{{\rho}(2^{-k}t)}{t}\,dt=\sum_{j:2^{j}\lesssim
2^{k}}\int
e^{i\varphi(t)}{\rho}_{0}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\frac{{\rho}(2^{-k}t)}{t}\,dt.$

We first restrict ourselves to those $j$ for which $2^{-am/3+k}\lesssim
2^{j}\lesssim 2^{k}$ (the reason for this particular choice will be clear a
posteriori) and claim that all these terms actually give rise to error terms -
by the same arguments used in B.1. Indeed, using integration by parts on

(441) $\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi(t)}{\rho}_{0}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\frac{{\rho}(2^{-k}t)}{t}\,dt$

we obtain a number of terms of similar nature; we consider here for shortness
only the term

$\mathcal{E}^{j}_{m,k}(\zeta;\lambda):=\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi(t)}\frac{{\rho}(2^{-k}t)}{it}{{\rho}_{0}}^{\prime}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\frac{2^{-j}}{\varphi^{\prime}(t)}\,dt$

which provides the largest contribution to the error term - the other terms
are all treated similarly. Applying the argument used in B.1 together with the
fact that

(442)
$\displaystyle|\varphi^{\prime}(t)|=a\lambda\Big{|}t^{a-1}-t_{0}(\zeta)^{a-1}\Big{|}$
$\displaystyle=a\lambda\Big{|}\int_{t_{0}(\zeta)}^{t}(a-1)s^{a-2}\,ds\Big{|}$
$\displaystyle\sim_{a}2^{a(m-k)}|t-t_{0}(\zeta)|2^{(a-2)k}\sim_{a}2^{am-2k+j},$

we obtain that

$\Big{|}\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}\mathcal{E}^{j}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta\Big{|}\lesssim
2^{-am}2^{2k-2j}M(f,g)(x).$

Summing over the $2^{-am/3+k}\lesssim 2^{j}\lesssim 2^{k}$ range we obtain
then an error contribution bounded by $\lesssim 2^{-am/3}M(f,g)(x)$, which is
acceptable. The other contributions arisen from the integration by parts can
be shown to be controlled by $\lesssim 2^{-2am/3}M(f,g)(x)$, and thus we have
completed this intermediate reduction.

Finally, we deal with the terms of the form (441) for those $j$ such that
$2^{j}\lesssim 2^{-am/3+k}$. These terms contain the main contribution to
$\mathfrak{M}_{m,k}$, but to properly isolate this we need to remove some
further error terms. Expanding the phase in Taylor series around the critical
point, we let

$\varphi_{0}(t):=\varphi(t_{0})+\frac{\varphi^{\prime\prime}(t_{0})}{2}(t-t_{0})^{2}$

denote the quadratic part of the phase ($\varphi^{\prime}(t_{0})=0$ by
definition) and $\widetilde{\varphi}(t):=\varphi(t)-\varphi_{0}(t)$ denote the
tail. We decompose for any $j$

$\displaystyle\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}$ $\displaystyle\int
e^{i\varphi(t)}{\rho}_{0}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\frac{{\rho}(2^{-k}t)}{t}\,dt$
(443) $\displaystyle=\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi_{0}(t)}{\rho}_{0}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\frac{{\rho}(2^{-k}t)}{t}\,dt$
(444) $\displaystyle\hskip
10.00002pt+\underbrace{\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi_{0}(t)}(e^{i\widetilde{\varphi}(t)}-1)\tilde{\rho}_{0}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\frac{\tilde{\rho}(2^{-k}t)}{t}\,dt}_{\textstyle=:\mathscr{E}^{j}_{m,k}(\zeta;\lambda)}$

and claim that (444) represents an error term. Indeed, for $t$ in the support
of ${\rho}_{0}(2^{-j}(t-t_{0}))$, we observe that155155155Here we are using
the real analyticity of $\widetilde{\varphi}$ on any given compact interval
that does not contain the origin.

$\displaystyle|\widetilde{\varphi}(t)|$ $\displaystyle\leq\sum_{\ell\geq
3}\frac{|\widetilde{\varphi}^{(\ell)}(t_{0})|}{\ell!}|t-t_{0}|^{\ell}\lesssim\sum_{\ell\geq
3}\frac{2^{a(m-k)}2^{(a-\ell)k}}{\ell!}2^{j\ell}$ $\displaystyle\lesssim
2^{am}\sum_{\ell\geq 3}2^{-(k-j)\ell}\lesssim 2^{am-3k+3j},$

and this is $\ll 1$ when $2^{j}\lesssim 2^{-am/3+k}$ (this is the reason for
our choice of cutoff in the previous argument). A similar estimate shows that
we have

$|{\widetilde{\varphi}}^{\prime}(t)|\lesssim 2^{am-4k+3j}$

as well; if we let $\nu(t):=e^{i\widetilde{\varphi}(t)}-1$, the trivial
inequality $|e^{is}-1|\leq|s|$ shows that we have

(445) $|\nu(t)|\lesssim 2^{am-3k+3j}\hskip 40.00006pt\text{ and }\hskip
40.00006pt|\nu^{\prime}(t)|\lesssim 2^{am-3k+3j}\cdot 2^{-k}$

and analogous bounds for higher derivatives. We also observe that a simple
computation shows

(446) $|{\varphi_{0}}^{\prime}(t)|\sim 2^{am-2k+j}$

(the same bound one has for $\varphi^{\prime}$; see (442)). After integrating
(444) by parts, one can feed these estimates to the – by now usual – argument
of B.1 to show that

$\displaystyle\Big{|}\sum_{k\in\mathbb{Z}}\tilde{\rho}\Big{(}\frac{\lambda(x)}{2^{a(m-k)}}\Big{)}\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}\Big{(}\sum_{j:2^{j}\lesssim
2^{-am/3+k}}\mathscr{E}^{j}_{m,k}(\xi-\eta;\lambda(x))\Big{)}\,d\xi\,d\eta\Big{|}$
$\displaystyle\lesssim 2^{-am/3}M(f,g)(x),$

which is another acceptable error term.

We see that we have effectively reduced our task to that of treating the
bilinear operator with multiplier symbol given by terms (443), or more
precisely

$\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi_{0}(t)}\Big{[}\sum_{j:2^{j}\lesssim
2^{-am/3+k}}{\rho}_{0}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\Big{]}\frac{{\rho}(2^{-k}t)}{t}\,dt.$

It is a little inconvenient to have the cut-off factor in square brackets in
the expression, and thus we remove it by writing the above as

(447) $\displaystyle\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi_{0}(t)}\frac{{\rho}(2^{-k}t)}{t}\,dt$ (448)
$\displaystyle-\sum_{j:2^{-am/3+k}\lesssim 2^{j}\lesssim
2^{k}}\chi_{1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}\int
e^{i\varphi_{0}(t)}{\rho}_{0}\Big{(}\frac{t-t_{0}(\zeta)}{2^{j}}\Big{)}\frac{{\rho}(2^{-k}t)}{t}\,dt;$

a repetition of the arguments given above156156156Notice that phases $\varphi$
and $\varphi_{0}$ satisfy the same estimates. shows that the contribution to
$T_{m}(f,g)(x)$ from (448) amounts to another error term dominated by
$\lesssim 2^{-am/3}M(f,g)(x)$. What is left is (447), which is the main
contribution that will be analysed below.

###  B.3. Stationary contributions to $\mathfrak{M}_{m,k}$

We have managed to reduce the multiplier symbol from $\mathfrak{M}_{m,k}$ to
(447), where the difference between the two amounts to the fact that the phase
is now purely quadratic and we have localised the multiplier symbol to
$|\zeta|\sim 2^{am-k}$. The advantage is that we can now appeal to the
identity157157157The formula is well-known; it is an easy consequence of the
formula for the Fourier transform of a Gaussian and of an analytic
continuation argument. Recall also our convention in footnote 4.

$\int e^{i\mu t^{2}}\phi(t)\,dt=\Big{(}\frac{i}{2\mu}\Big{)}^{1/2}\int
e^{-i\alpha^{2}/(4\mu)}\,\widehat{\phi}(\alpha)\,d\alpha$

for (say) Schwartz functions $\phi$ in order to produce explicit formulas for
the multiplier symbol. Indeed, if we let $\theta(s):={\rho}(s)/s$ and apply
the above identity to the integral expression in (447) we obtain (with a
change of variables) the expression

(449)
$e^{i\varphi(t_{0})}\Big{(}\frac{i}{2^{2k}\varphi^{\prime\prime}(t_{0})}\Big{)}^{1/2}\int
e^{-i\alpha^{2}/(2^{2k+1}\varphi^{\prime\prime}(t_{0}))}e^{i(t_{0}/2^{k})\alpha}\,\widehat{\theta}(\alpha)\,d\alpha.$

Easy computations–with $c_{a}:=a^{-a/(a-1)}-a^{-1/(a-1)}$–yield that

$\displaystyle\varphi(t_{0})$
$\displaystyle=c_{a}\zeta^{\frac{a}{a-1}}\lambda^{-\frac{1}{a-1}},$
$\displaystyle\frac{t_{0}}{2^{k}}$
$\displaystyle=\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}^{\frac{1}{a-1}}\Big{(}\frac{a\lambda}{2^{a(m-k)}}\Big{)}^{-\frac{1}{a-1}},$
$\displaystyle\frac{1}{2^{2k}\varphi^{\prime\prime}(t_{0})}$
$\displaystyle=\frac{2^{-am}}{a-1}\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}^{-\frac{a-2}{a-1}}\Big{(}\frac{a\lambda}{2^{a(m-k)}}\Big{)}^{-\frac{1}{a-1}}.$

The latter in particular shows that, in the integration region
$|\alpha|\lesssim 1$ where most of the mass of $\widehat{\theta}(\alpha)$ is
concentrated, the quadratic phase
$\alpha^{2}/(2^{2k+1}\varphi^{\prime\prime}(t_{0}))$ is extremely small (in
particular it is $\lesssim 2^{-am}$). To take advantage of this we split (449)
as

(450) $\displaystyle
e^{i\varphi(t_{0})}\Big{(}\frac{i}{2^{2k}\varphi^{\prime\prime}(t_{0})}\Big{)}^{1/2}\int
e^{i(t_{0}/2^{k})\alpha}\,\widehat{\theta}(\alpha)\,d\alpha$ (451)
$\displaystyle+e^{i\varphi(t_{0})}\Big{(}\frac{i}{2^{2k}\varphi^{\prime\prime}(t_{0})}\Big{)}^{1/2}\int\big{(}e^{-i\alpha^{2}/(2^{2k+1}\varphi^{\prime\prime}(t_{0}))}-1\big{)}e^{i(t_{0}/2^{k})\alpha}\,\widehat{\theta}(\alpha)\,d\alpha.$

Expression (450) is clearly equal to

$e^{i\varphi(t_{0})}\Big{(}\frac{i}{2^{2k}\varphi^{\prime\prime}(t_{0})}\Big{)}^{1/2}\theta\Big{(}\frac{t_{0}}{2^{k}}\Big{)},$

which will give our main term, while (451) (we claim) contributes another
error term. If we define the smooth function

$\Theta(\zeta,\lambda):=\frac{\tilde{\rho}(a^{-1}\lambda)\chi_{1}(\zeta)}{\big{(}\zeta^{\frac{a-2}{a-1}}\lambda^{\frac{1}{a-1}}\big{)}^{\frac{1}{2}}}\theta\Big{(}\Big{(}\frac{\zeta}{\lambda}\Big{)}^{\frac{1}{a-1}}\Big{)}$

(compactly supported in $|\zeta|\sim|\lambda|\sim 1$), then the main term in
$\tilde{\rho}(\lambda 2^{-a(m-k)})\mathfrak{M}_{m,k}(\zeta;\lambda)$ is (a
constant multiple of)

(452)
$\mathfrak{m}_{m,k}(\zeta;\lambda):=2^{-\frac{am}{2}}\;e^{ic_{a}\zeta^{\frac{a}{a-1}}\lambda^{-\frac{1}{a-1}}}\;\Theta\Big{(}\frac{\zeta}{2^{am-k}},\frac{a\lambda}{2^{a(m-k)}}\Big{)}.$

It remains to prove our claim about the error term nature of (451). Observe
that with a little more algebra this remaining contribution to the symbol can
be written as

$\mathfrak{e}_{m,k}(\zeta;\lambda):=2^{-\frac{am}{2}}\;e^{ic_{a}\zeta^{\frac{a}{a-1}}\lambda^{-\frac{1}{a-1}}}\;\tilde{\Theta}\Big{(}\frac{\zeta}{2^{am-k}},\frac{a\lambda}{2^{a(m-k)}}\Big{)}\Psi_{m}\Big{(}\frac{\zeta}{2^{am-k}},\frac{a\lambda}{2^{a(m-k)}}\Big{)},$

where we have defined

$\displaystyle\tilde{\Theta}(\zeta,\lambda)$
$\displaystyle:=\frac{\tilde{\rho}(a^{-1}\lambda)\chi_{1}(\zeta)}{\big{(}\zeta^{\frac{a-2}{a-1}}\lambda^{\frac{1}{a-1}}\big{)}^{\frac{1}{2}}},$
$\displaystyle\Psi_{m}(\zeta,\lambda)$
$\displaystyle:=\int\big{(}e^{ic^{\prime}_{a}2^{-am}\zeta^{-\frac{a-2}{a-1}}\lambda^{-\frac{1}{a-1}}\alpha^{2}}-1\big{)}e^{i\zeta^{\frac{1}{a-1}}\lambda^{-\frac{1}{a-1}}\alpha}\;\widehat{\theta}(\alpha)\,d\alpha$

(and $c^{\prime}_{a}$ is another constant that depends on $a$). We have that

$\displaystyle\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}\mathfrak{e}_{m,k}(\xi-\eta;\lambda)\,d\xi\,d\eta$
$\displaystyle=\int
f(x-s)g(x+s)\Big{[}\int\mathfrak{e}_{m,k}(\zeta;\lambda)e^{-i\zeta
s}\,d\zeta\Big{]}\,ds;$

the oscillatory integral expression in square brackets has phase
$c_{a}\zeta^{\frac{a}{a-1}}\lambda^{-\frac{1}{a-1}}-\zeta s$ and, thanks to
the support properties of $\tilde{\Theta}$, there will be a critical point
only if $|s|\sim 2^{k}$. Splitting the domain of integration into $|s|\sim
2^{k}$ and its complement, it is easy to show (using the trivial inequality
$|e^{i\alpha}-1|\leq|\alpha|$ and the rapid decay of
$|\widehat{\theta}(\alpha)|$) that, say,

$|\tilde{\Theta}(\zeta,\lambda)\Psi_{m}(\zeta,\lambda)|\lesssim 2^{-(3/4)am},$

and therefore one can bound in the former case

$\displaystyle\Big{|}\int_{|s|\sim
2^{k}}f(x-s)g(x+s)\Big{[}\int\mathfrak{e}_{m,k}(\zeta;\lambda)e^{-i\zeta
s}\,d\zeta\Big{]}\,ds\Big{|}$ $\displaystyle\lesssim\int_{|s|\sim
2^{k}}|f(x-s)||g(x+s)|2^{-\frac{am}{2}}2^{am-k}2^{-(3/4)am}\,ds$
$\displaystyle=2^{-am/4}\frac{1}{2^{k}}\int_{|s|\sim
2^{k}}|f(x-s)||g(x+s)|\,ds\lesssim 2^{-\frac{am}{4}}M(f,g)(x),$

an acceptable contribution. In the complementary case ($|s|\not\sim 2^{k}$) we
find ourselves with a non-stationary phase situation, which can be treated by
a dyadic decomposition in $|s|$ and by means of integration by parts. The
resulting elementary argument is quite analogous to that in B.1, albeit longer
and possibly even more tedious - hence we omit the unnecessary details.
Overall we have shown that

$\Big{|}\sum_{k\in\mathbb{Z}}\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}{\mathfrak{e}}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta\Big{|}\lesssim
2^{-\delta am}M(f,g)(x),$

for some absolute $\delta>0$ thus proving our claim that
$\mathfrak{m}_{m,k}(\xi-\eta;\lambda)$ is the main contribution to the
multiplier symbol.

###  B.4. Tensorization of the main contribution

The function $\Theta$ in (452) is very well-behaved: it is smooth and
compactly supported in $\\{(\zeta,\lambda)\,:\,|\zeta|\sim
1\>\>\textrm{and}\>\>|\lambda|\sim 1\\}$. However it is not in the most
convenient form, since currently it is defined as a function that depends on
$\zeta$ and $\lambda$ simultaneously. It will be more convenient to separate
the two variables by means of a tensorization argument. Indeed, by a standard
double Fourier series decomposition one can rewrite the function $\Theta$ as

$\Theta(\zeta,\lambda)=\sum_{j,\ell\in\mathbb{Z}}c_{j,\ell}\rho_{j}(\lambda)\psi_{\ell}(\zeta),$

where $\rho_{j}(\lambda),\psi_{\ell}(\zeta)$ are smooth bump functions
supported in $|\lambda|\sim 1$ and $|\zeta|\sim 1$ respectively which satisfy

$\|\rho_{j}^{(n)}\|_{\infty}\lesssim_{n}|j|^{n},\quad\|\psi_{\ell}^{(n)}\|_{\infty}\lesssim_{n}|\ell|^{n}$

for sufficiently many derivatives and the coefficients $c_{j,\ell}$ are fastly
decaying. By triangle/Hölder inequality it is thus sufficient158158158Provided
that the $L^{p}\times L^{q}\to L^{r}$ bounds thus obtained depend at most
polynomially on the $C^{1}$ norms of $\rho,\psi$. to prove Theorem 3.2 for the
operators

$\sum_{k\in\mathbb{Z}}{\rho}\Big{(}\frac{\lambda(x)}{2^{-a(m-k)}}\Big{)}\iint\widehat{f}(\xi)\widehat{g}(\eta)e^{i(\xi+\eta)x}\mathfrak{m}_{m,k}(\xi-\eta;\lambda(x))\,d\xi\,d\eta,$

where we have redefined $\mathfrak{m}_{m,k}$ to be

$\mathfrak{m}_{m,k}(\zeta;\lambda):=2^{-am/2}\;e^{ic_{a}\zeta^{\frac{a}{a-1}}\lambda^{-\frac{1}{a-1}}}\;\psi\Big{(}\frac{\zeta}{2^{am-k}}\Big{)}$

and $\rho,\psi$ are now _generic_ smooth bump functions supported in
$|\lambda|\sim 1,|\zeta|\sim 1$ respectively. This is precisely the form of
the multiplier symbol given in (117), as desired.

######  Remark B.1.

This last tensorization step is where the smoothness of the function
$\tilde{\rho}$ has finally come in handy. All the previous steps in the
reduction could have equally been achieved with a characteristic function in
place of $\tilde{\rho}$, but the resulting function $\Theta$ would not have
been smooth in $\lambda$.

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