# A degree one Carleson operator along the paraboloid

Lars Becker  Mathematical Institute, University of Bonn, Endenicher Allee 60,
53115, Bonn, Germany<EMAIL_ADDRESS>
###### Abstract.

We prove $L^{p}$ bounds, $\frac{d^{2}+4d+2}{(d+1)^{2}}<p<2(d+1)$, for maximal
linear modulations of singular integrals along paraboloids with frequencies in
certain subspaces of $\mathbb{R}^{d+1}$, for $d\geq 2$.

###### 2020 Mathematics Subject Classification:

42B20

##  1\. Introduction

This paper reports on progress in the program of Pierce and Yung [PY19] to
study maximally modulated singular Radon transforms along paraboloids. While
their work focuses on certain polynomial modulations without linear terms and
uses $TT^{*}$ methods, our result is the first instance in this program for
linear polynomials featuring symmetries that mandate the use of time-frequency
analysis. This should be compared with the result in [GPR+17], which uses
time-frequency analysis for variants with domain constrained linearizing
functions of the objects in this program. Further interesting estimates in the
program remain open, including general linear polynomials or more generally
higher degree polynomials including linear terms, as well as any estimates for
$d=1$.

We now state our main result. A $m$-Calderón-Zygmund kernel on
$\mathbb{R}^{d}$ is a function $K:\mathbb{R}^{d}\setminus\\{0\\}\to\mathbb{C}$
satisfying the estimates

$|\partial^{\alpha}K(x)|\leq|x|^{-d-|\alpha|}\,,\quad|\alpha|\leq m\,,$ (1)

and the cancellation property

$\int_{B(0,R)\setminus B(0,r)}K(x)\,\mathrm{d}x=0\,,\quad 0<r<R\,.$ (2)

Let $V\subset\mathbb{R}^{d+1}$ be a linear subspace. We consider the Carleson-
type operator defined a priori on Schwartz functions $f$ on $\mathbb{R}^{d+1}$
by

$T_{V}f(x)=\sup_{N\in
V}\sup_{r<R}\left\lvert\int_{r<|y|<R}f(x^{\prime}-y,x_{d+1}-|y|^{2})e^{iN\cdot(y,|y|^{2})}K(y)\,\mathrm{d}y\right\rvert\,,$
(3)

where $x=(x^{\prime},x_{d+1})\in\mathbb{R}^{d}\times\mathbb{R}$ and $K$ is a
Calderón-Zygmund kernel.

######  Theorem 1.

Let $d\geq 2$ and let $m>\frac{d}{2}$. Suppose that
$V=\\{0\\}^{d}\times\mathbb{R}$ or $V$ is a proper subspace of
$\mathbb{R}^{d}\times\\{0\\}$. Then for all $p$ with

$\frac{d^{2}+4d+2}{(d+1)^{2}}<p<2(d+1)$ (4)

there exists $C>0$ such that for all $m$-Calderón-Zygmund kernels $K$ and all
Schwartz functions $f$, we have with $T_{V}$ as defined in (3)

$\|T_{V}f\|_{L^{p}(\mathbb{R}^{d+1})}\leq C\|f\|_{L^{p}(\mathbb{R}^{d+1})}\,.$

The operators $T_{V}$ have the symmetry $T_{V}(e^{iN\cdot y}f(y))=T_{V}f$ for
all $N\in V$, which makes it necessary to use time-frequency analysis to prove
Theorem 1. To our knowledge, Theorem 1 is the first $L^{p}$ estimate for
maximal modulations without domain constraints of an operator with kernel
supported on a submanifold whose proof requires time-frequency analysis. We
note that Theorem 1 is stronger than the Carleson-Sjölin theorem [Car66]
[Sjö71] for maximal modulations of singular integral operators, in the sense
that one can recover the Carleson-Sjölin theorem from Theorem 1 using
projection and limiting arguments.

The restrictions on $V$ in Theorem 1 guarantee that the operator $T_{V}$ has
only linear modulation symmetries. If $\mathbb{R}^{d}\times\\{0\\}\subset V$,
then $T_{V}$ has, in addition to the linear modulation symmetries discussed
above, the symmetry $T_{V}(e^{it(|y^{\prime}|^{2}+y_{d+1})}f(y))=T_{V}f$ for
$t\in\mathbb{R}$, where $y=(y^{\prime},y_{d+1})$. The linear time-frequency
analysis used in the present paper is not suited for such symmetries. The
occurence of higher order symmetries is not the only obstacle in extending
Theorem 1 to other subspaces: An interesting intermediate case is
$V=\mathbb{R}^{d-1}\times\\{0\\}\times\mathbb{R}$, which our methods, in
particular the square function argument in Subsection 5.3, cannot handle.

The study of the operator $T_{V}$ and the larger program by Pierce and Yung
add to a line of research starting from Carleson’s theorem [Car66] on
convergence of Fourier series of functions in $L^{2}$. Carleson’s theorem is
equivalent to weak type $L^{2}$ boundedness of the maximally modulated Hilbert
transform. Other proofs of Carleson’s theorem were given by Fefferman [Fef73]
and Lacey and Thiele [LT00]. Hunt [Hun68] improved Carleson’s result to
$L^{p}$ boundedness for $1<p<\infty$, and Sjölin [Sjö71] generalized it to
maximally modulated Calderón-Zygmund operators in arbitrary dimension.

More recently, Lie [Lie09][Lie20] proved bounds in $L^{p}$ of the maximally
polynomially modulated Hilbert transform, and Zorin–Kranich [Zor21]
generalized this result to less regular kernels in higher dimensions. Lie’s
work answered a question by Stein and Wainger, who had shown earlier [SW01]
that maximal modulations with polynomials without linear term of Calderón-
Zygmund operators are bounded on $L^{p}$. Motivated by Stein and Wainger’s
result, Pierce and Yung [PY19] considered maximal polynomial modulations of
singular Radon transforms along paraboloids. They showed that for certain
subspaces $W$ of the polynomials without linear term the operators

$T_{\mathrm{PY}}f(x):=\sup_{P\in
W}\left|\mathrm{p.v.}\\!\int_{\mathbb{R}^{d}}f(x^{\prime}-y,x_{d+1}-|y|^{2})e^{iP(y)}K(y)\,\mathrm{d}y\right|$

with $K$ a Calderón-Zygmund kernel on $\mathbb{R}^{d}$, are bounded on
$L^{p}(\mathbb{R}^{d+1})$ for $1<p<\infty$. Their result was extended to other
quadratic surfaces in [AMP22]. It is now very natural to ask whether general
maximal polynomial modulations of singular Radon transforms along paraboloids
are bounded on $L^{p}$. Earlier results motivated by this question have been
obtained in [GPR+17], [Roo19], [Ram21], [Bec22], making partial progress or
showing bounds under simplifying assumptions.

Our proof of Theorem 1 follows the basic structure of Fefferman’s proof of
Carleson’s theorem [Fef73], in some technical arguments we also follow
[Zor21]. It starts with a discretization of the operator in Section 2 and a
combinatorial decomposition into so called forest operators and antichain
operators in Section 4, see Sections 2 and 3 for the definitions of these
operators. These two steps are a straightforward adaption to the anisotropic
paraboloid setting of the corresponding steps in Fefferman’s proof. We modify
the decomposition slightly by testing the operator with the indicator function
of a set $F$ and using a decomposition adapted to $F$. This allows us to
directly prove weak type $L^{2}$ bounds.

After the decomposition of the operator, the argument proceeds by proving
bounds for antichain operators in Section 5, for tree operators in Section 6
and finally for forest operators in Section 7, see Propositions 2, 3 and 4. In
these steps we introduce techniques related to the fact that the kernel is
supported on a curved submanifold.

The first such ingredient is a square function estimate for parts of the
antichain operators in Section 5. This is needed because the classical
argument relies on the smoothness of the kernel. In our argument, we decompose
the kernel at each scale into a low frequency part and a high frequency part,
use the classical argument for the low frequency part, and then use the square
function estimate for the high frequency part. We rely on oscillatory integral
estimates to get good decay of the Fourier transform of the kernel, and a
careful decomposition to ensure that the Fourier transform of the high
frequency part vanishes near a certain subspace depending on $V$.

The second such ingredient are sparse bounds for singular Radon transforms,
see Lemma 8, which were proven by Oberlin [Obe19], see also [Lac19]. The
sparse bounds are used in Subsection 6.4 to obtain good bounds for tree
operators with thin spatial support. In Fefferman’s proof, these estimates
follow from the fact that convolution with a truncated singular integral
kernel makes a function essentially constant at the scale of the lower
truncation parameter. This principle fails for singular Radon transforms, but
it can be rectified by using the Sobolev smoothening of the truncated Radon
transform, which we do via the sparse bounds.

In Section 7 we follow Fefferman’s argument to prove almost orthogonality
estimates between tree operators, and combine the estimates for tree operators
to an estimate for forest operators. We make use of square function arguments
as in [SW78], [DR86] to bound maximal averages and maximally truncated
operators associated to the singular Radon transform, and of estimates for
certain oscillatory integrals along paraboloids. This again allows us to avoid
all arguments of Fefferman that use smoothness of the kernel.

We state the tree estimates in Section 6 for very general singular Radon
transforms, similar to the setting of [DR86]. The arguments in Section 5 and
Section 7 do not generalize to that extent, it is an interesting question to
what extent they generalize.

After proving weak-type $L^{2}$ estimates for $T_{V}$ in Sections 2 to 7, we
deduce $L^{p}$ bounds using a localization argument as in [Zor21] in Section
8. The restriction (4) on $p$ comes from the sparse bounds, and is related to
the $L^{p}$-improving range for averages along paraboloids. To obtain the full
range $1<p<\infty$ with our argument, one would need the averages to be
$L^{1}\to L^{\infty}$ improving, which they are for smooth kernels, but not
for averages along paraboloids. We do not know whether the range of $p$ can be
improved by other methods.

Acknowledgement. The author thanks Christoph Thiele for various helpful
discussions about the organization of this paper. The author was supported by
the Collaborative Research Center 1060 funded by the Deutsche
Forschungsgemeinschaft (DFG, German Research Foundation) and the Hausdorff
Center for Mathematics, funded by the DFG under Germany’s Excellence Strategy
- GZ 2047/1, ProjectID 390685813.

##  2\. Reduction to a discretized operator

We may assume that $V=\\{0\\}^{d}\times\mathbb{R}$ or
$V=\mathbb{R}^{d-1}\times\\{0\\}^{2}$. Define anisotropic dilations
$\delta_{s}(x^{\prime},x_{d+1}):=(2^{s}x^{\prime},2^{2s}x_{d+1})$. Define the
dyadic cubes of scale $0$ to be

$\mathbf{D}_{0}:=\\{k+[0,1)^{d+1}\,:\,k\in\mathbb{Z}^{d+1}\\}\,,$

and the dyadic cubes of scale $s$ to be
$\mathbf{D}_{s}:=\delta_{s}(\mathbf{D}_{0})\,.$ The collection of all dyadic
cubes is denoted $\mathbf{D}=\cup_{s\in\mathbb{Z}}\mathbf{D}_{s}$. Given a
dyadic cube $I\in\mathbf{D}_{s}$, we denote by $s(I)=s$ its scale.

We define dyadic frequency cubes of scale $0$ as

$\mathbf{\Omega}_{0}:=\\{k+\\{0\\}^{d}\times[0,1)\,:\,k\in\\{0\\}^{d}\times\mathbb{Z}\\}$

if $V=\\{0\\}^{d}\times\mathbb{R}$ and as

$\mathbf{\Omega}_{0}:=\\{k+[0,1)^{d-1}\times\\{0\\}^{2}\,:\,k\in\mathbb{Z}^{d-1}\times\\{0\\}^{2}\\}\,,$

if $V=\mathbb{R}^{d-1}\times\\{0\\}^{2}$. The dyadic frequency cubes of scale
$s$ are $\mathbf{\Omega}_{s}:=\delta_{-s}(\mathbf{\Omega}_{0})$ and the
collection of all dyadic frequency cubes is defined by
$\mathbf{\Omega}:=\bigcup_{s\in\mathbb{Z}}\mathbf{\Omega}_{s}$. A tile is a
pair $p=(I,\omega)$, where $I\in\mathbf{D}_{s}$ and
$\omega\in\mathbf{\Omega}_{s}$ for some $s=s(p)$ called the scale of $p$. The
collection of all tiles is denoted by

$\mathbf{P}:=\bigcup_{s\in\mathbb{Z}}\\{(I,\omega)\,:\,I\in\mathbf{D}_{s},\,\omega\in\mathbf{\Omega}_{s}\\}\,.$

We decompose the kernel $K$ into pieces localized in dyadic annuli. Fix a
smooth function $\eta$ supported in $[1/8,1/3]$ such that

$\sum_{s\in\mathbb{Z}}\eta(2^{-s}t)=1\,,\quad t\in(0,\infty)\,,$

and define $K_{s}(x)=K(x)\eta(2^{-s}|x|)$. Then (1) and (2) imply that there
exists a constant $C=C(\eta)$ with

$|\partial^{\alpha}K_{s}(x)|\leq C2^{-s(d+|\alpha|)}\,,\quad|\alpha|\leq
m\,,\,s\in\mathbb{Z}\,,$ (5)

and that

$\int K_{s}(x)\,\mathrm{d}x=0\,,\quad s\in\mathbb{Z}\,.$ (6)

It suffices to estimate the operator

$\sup_{N\in
V}\sup_{\underline{s}<\overline{s}}\left\lvert\sum_{s=\underline{s}}^{\overline{s}}\int
f(x^{\prime}-y,x_{d+1}-|y|^{2})e^{iN\cdot(y,|y|^{2})}K_{s}(y)\,\mathrm{d}y\right\rvert\,,$
(7)

since the difference to $T_{V}$ is controlled by the maximal average along the
paraboloid, by (1). Since the sum of integrals in (7) is continuous in $N$, we
may restrict the supremum in $N$ to a countable dense subset of $V$. By
monotone convergence, we may further restrict both suprema to finite subsets
of $V$ and $\mathbb{Z}\times\mathbb{Z}$. Choosing maximizers for each $x$, it
suffices to bound the operator

$\sum_{s=\underline{s}(x)}^{\overline{s}(x)}\int
f(x^{\prime}-y,x_{d+1}-|y|^{2})e^{-iN(x)\cdot(y,|y|^{2})}K_{s}(y)\,\mathrm{d}y$
(8)

uniformly over all measurable functions $N:\mathbb{R}^{d+1}\to
V,\underline{s}:\mathbb{R}^{d+1}\to\mathbb{Z}$ and
$\overline{s}:\mathbb{R}^{d+1}\to\mathbb{Z}$ with finite range. We fix such
functions $N,\underline{s},\overline{s}$.

For a tile $p\in\mathbf{P}$, we define

$E(p)=\\{x\in I(p)\,:\,N(x)\in\omega(p),\,\underline{s}(x)\leq
s(p)\leq\overline{s}(x)\\}$

and the operator associated to the tile $p$

$T_{p}f(x)=\mathbf{1}_{E(p)}\int
f(x^{\prime}-y,x_{d+1}-|y|^{2})e^{-iN(x)\cdot(y,|y|^{2})}K_{s(p)}(y)\,\mathrm{d}y\,.$

For a subset $\mathbf{C}\subset\mathbf{P}$ we write
$T_{\mathbf{C}}=\sum_{p\in\mathbf{C}}T_{p}$. The operator (8) is then simply
$T_{\mathbf{P}}$.

Finally, we remove some tiles with unfavourable properties, using an argument
due to Fefferman [Fef73]. A tile is called admissible if
$3\omega(p)\subset\hat{\hat{\omega}}(p)$, where $\hat{\omega}$ denotes the
unique frequency cube of scale $s(\omega)-1$ containing $\omega$. Then it
suffices to show the estimate $\|T_{\mathbf{P}_{\text{ad}}}f\|_{L^{p}}\leq
C\|f\|_{L^{p}}$, where $\mathbf{P}_{\text{ad}}$ is the set of admissible
tiles, as follows from an averaging argument analogous to the one in Section 5
of [Fef73]. From now on by tile we always mean an admissible tile.

##  3\. Outline of the proof of weak type $L^{2}$-bounds for (8)

By duality and the reductions in the previous section, it suffices to show
that there exists $C>0$ such that for each compact set
$F\subset\mathbb{R}^{d+1}$, there exists a set $\tilde{F}\subset F$ with
$|\tilde{F}|\leq\frac{1}{2}|F|$ and

$\|\mathbf{1}_{F\setminus\tilde{F}}T_{\mathbf{P}_{\text{ad}}}\|_{2\to 2}\leq
C\,.$ (9)

We fix $F$. The operators $\mathbf{1}_{F}T_{p}$ do not change upon replacing
$E(p)$ by $E(p)\cap F$, so we will assume from now on that $E(p)\subset F$ for
all tiles $p$. Note that since $F$ is compact and $\underline{s},\overline{s}$
and $N$ have finite range, only finitely many operators $T_{p}$ are not zero.

We now make some definitions to state a decomposition for the operator on the
left hand side of (9). Denote by $\dim_{h}V$ the homogeneous dimension of $V$
with respect to the dilations $\delta_{s}$, that is

$\dim_{h}(\mathbb{R}^{d-1}\times\\{0\\}^{2})=d-1\,,\qquad\qquad\dim_{h}(\\{0\\}^{d}\times\mathbb{R})=2\,,$

and define the convex cylinder associated to a collection $\mathbf{C}$ of
tiles by

$C(\mathbf{C})=\\{p\in\mathbf{P}_{\text{ad}}\,:\,\exists
p^{\prime},p^{\prime\prime}\in\mathbf{C},I(p^{\prime})\subset I(p)\subset
I(p^{\prime\prime})\\}\,.$

Then the density of a collection $\mathbf{C}$ of tiles is defined as

$\operatorname{\operatorname{dens}}(\mathbf{C})=\sup_{p\in\mathbf{C}}\sup_{\begin{subarray}{c}\lambda\geq
1\\\ \text{odd}\end{subarray}}\sup_{\begin{subarray}{c}p^{\prime}\in
C(\mathbf{C}):I(p)\subset I(p^{\prime})\\\
\lambda\omega(p^{\prime})\subset\lambda\omega(p)\end{subarray}}\lambda^{-\dim_{h}V}\frac{|E(\lambda,p^{\prime})|}{|I(p^{\prime})|}\,,$

where we use the notation
$\lambda\omega=c(\omega)+\delta_{\log_{2}\lambda}(\omega-c(\omega))$ where
$c(\omega)$ is the center of $\omega$ and

$E(\lambda,p^{\prime})=\\{x\in I(p)\,:\,N(x)\in\lambda\omega(p)\\}\,.$

We define a partial order on the set of tiles by

$(I,\omega)\leq(I^{\prime},\omega^{\prime})\iff I\subset I^{\prime}\
\text{and}\ \omega^{\prime}\subset\omega\,.$

An antichain is a set of tiles that are pairwise not comparable with respect
to this order. A set of tiles $\mathbf{C}$ is called convex, if
$p,p^{\prime}\in\mathbf{C}$ and $p\leq p^{\prime\prime}\leq p^{\prime}$
implies $p^{\prime\prime}\in\mathbf{C}$. A tree is a convex collection
$\mathbf{T}$ of tiles together with an upper bound
$\operatorname{\operatorname{top}}(\mathbf{T})$, i.e. a tile $m$ such that for
all $p\in\mathbf{T}$ we have $p\leq m$. We denote
$\omega(\mathbf{T})=\omega(\operatorname{\operatorname{top}}(\mathbf{T}))$ and
$I(\mathbf{T})=I(\operatorname{\operatorname{top}}(\mathbf{T}))$. A tree is
called normal if $3I(p)\subset I(\mathbf{T})$ for all $p\in\mathbf{T}$.

A pair of trees $\mathbf{T}_{1},\mathbf{T}_{2}$ is called $\Delta$-separated,
if for $\\{i,j\\}=\\{1,2\\}$ and each tile $p_{i}\in\mathbf{T}_{i}$ with
$I(p_{i})\subset I(\mathbf{T}_{j})$ we have
$\Delta\omega(p_{i})\cap\omega(\mathbf{T}_{j})=\emptyset$. An $n$-forest is a
collection of pairwise $2^{10dn}$-separated, normal trees of density at most
$2^{-n}$ that satisfy the overlap estimate

$\sum_{\mathbf{T}\in\mathbf{F}}\mathbf{1}_{I(\mathbf{T})}\leq
2^{n}\log(n+2)\,.$ (10)

######  Proposition 1.

There exists a constant $C=C(d)$ and an exceptional set $\tilde{F}$ with
$|\tilde{F}|\leq|F|/2$, such that the set

$\mathbf{P}_{F\setminus\tilde{F}}:=\\{p\in\mathbf{P}_{\mathrm{ad}}\,:\,I(p)\cap(F\setminus\tilde{F})\neq\emptyset\\}$

can be decomposed as a disjoint union

$\mathbf{P}_{F\setminus\tilde{F}}=\bigcup_{n\geq
0}\left(\bigcup_{l=1}^{C(n+1)^{2}}\mathbf{F}_{n,l}\cup\bigcup_{l=1}^{C(n+1)^{3}}\mathbf{A}_{n,l}\right)\,,$

where each $\mathbf{F}_{n,l}$ is an $n$-forest and each $\mathbf{A}_{n,l}$ is
an antichain of density at most $2^{-n}$.

Estimate (9), and therefore weak type $L^{2}$ boundedness of the operator
defined in (3) then follows from the following estimates for antichain and
forest operators.

######  Proposition 2.

There exists $\varepsilon=\varepsilon(d)$ and $C>0$ such that the following
holds. Let $\mathbf{A}$ be an antichain of density $\delta$. Then

$\|T_{\mathbf{A}}\|_{2\to 2}\leq C\delta^{\varepsilon}\,.$

######  Proposition 3.

For each $\varepsilon<\frac{1}{2}-\frac{1}{2(d+1)}$ there exist $C>0$ such
that the following holds. Let $\mathbf{F}$ be an $n$-forest. Then

$\|T_{\mathbf{F}}\|_{2\to 2}\leq C2^{-\varepsilon n}\,.$

We will independently prove Proposition 1 in Section 4 and Proposition 2 in
Section 5. Proposition 3 is proven using an estimate for trees and an almost
orthogonality argument. The almost orthogonality argument and the deduction of
Proposition 3 are carried out in Section 7. The estimate for single trees is
proven in Section 6, and is a mild generalization of the following.

######  Proposition 4.

For each $\varepsilon<\frac{1}{2}-\frac{1}{2(d+1)}$ there exist $C>0$ such
that the following holds. Let $\mathbf{T}$ be a tree of density $\delta$. Then

$\|T_{\mathbf{T}}\|_{2\to 2}\leq C\delta^{\varepsilon}\,.$

##  4\. Tile organization: Proof of Proposition 1

Let $k\geq 0$ and let $\mathbf{D}_{k}(F)$ be the set of maximal dyadic cubes
$Q$ with $|Q\cap F|/|Q|\geq 2^{-k-1}$. Let $\tilde{\mathbf{P}}_{\leq k}$ be
the set of tiles $p\in\mathbf{P}_{\text{ad}}$ such that $I(p)$ is contained in
some $Q\in\mathbf{D}_{k}(F)$ and let
$\tilde{\mathbf{P}}_{k}:=\tilde{\mathbf{P}}_{\leq
k}\setminus\tilde{\mathbf{P}}_{\leq k-1}$. Then we have
$\mathbf{1}_{F}T_{\mathbf{P}_{\text{ad}}}=\sum_{k\geq
0}\mathbf{1}_{F}T_{\tilde{\mathbf{P}}_{k}}$, and each
$p\in\tilde{\mathbf{P}}_{k}$ satisfies that $|I(p)\cap F|/|I(p)|<2^{-k}$. We
define

$\overline{E}(p):=\\{x\in I(p)\cap F\,:\,N(x)\in\omega(p)\\}\,,$

and define $\tilde{\mathbf{M}}_{n,k}$ to be the set of maximal tiles $p$ in
$\tilde{\mathbf{P}}_{k}$ such that $|\overline{E}(p)|/|I(p)|\geq 2^{-n-1}$.

######  Lemma 1.

The exceptional set

$\tilde{F}_{1}:=F\cap\bigcup_{k\geq
0}\bigcup_{Q\in\mathbf{D}_{k}(F)}\bigcup_{n\geq k}\\{x\in
Q\,:\,\sum_{\begin{subarray}{c}p\in\tilde{\mathbf{M}}_{n,k}\\\ I(p)\subset
Q\end{subarray}}\mathbf{1}_{I(p)}\geq 1000\cdot 2^{n}\log(n+2)\\}$

satisfies $|\tilde{F}_{1}|\leq|F|/4$.

###### Proof.

We have for each $J\in\mathbf{D}$

$\sum_{\begin{subarray}{c}p\in\tilde{\mathbf{M}}_{n,k}\\\ I(p)\subset
J\end{subarray}}|I(p)|\leq
2^{n+1}\sum_{\begin{subarray}{c}p\in\tilde{\mathbf{M}}_{n,k}\\\ I(p)\subset
J\end{subarray}}|\overline{E}(p)|\leq 2^{n+1}|J|\,,$

since tiles in $\tilde{\mathbf{M}}_{n,k}$ are pairwise not comparable and
hence the sets $\overline{E}(p)$, $p\in\tilde{\mathbf{M}}_{n,k}$ are pairwise
disjoint. We estimate for each $Q\in D_{k}(F)$ and each $n\geq k$ using the
John-Nirenberg inequality:

$|\\{x\in Q\,:\,\sum_{\begin{subarray}{c}p\in\tilde{\mathbf{M}}_{n,k}\\\
I(p)\subset Q\end{subarray}}\mathbf{1}_{I(p)}\geq 1000\cdot
2^{n}\log(n+2)\\}|\\\ \leq c_{1}\exp(-c_{2}\frac{1000\cdot
2^{n}\log(n+2)}{2^{n+1}})|Q|\leq(n+2)^{-100}|Q|\,.$ (11)

For the last inequality we have used that in this version of the John-
Nirenberg inequality, one can choose $c_{1}=e^{2}$ and $c_{2}=(2e)^{-1}$.

Note that the set on the left hand side of (11) is a disjoint union of cubes
$I(p)$ each of which satisfies $|I(p)\cap F|/|I(p)|\leq 2^{-k}$, hence

$|F\cap\\{x\in Q\,:\,\sum_{\begin{subarray}{c}p\in\tilde{\mathbf{M}}_{n,k}\\\
I(p)\subset Q\end{subarray}}\mathbf{1}_{I(p)}\geq 1000\cdot
2^{n}\log(n+2)\\}|\\\ \leq 2^{-k}(n+2)^{-100}|Q|\leq 2(n+2)^{-100}|Q\cap
F|\,.$

Summing up, we obtain

$\sum_{k\geq 0}\sum_{Q\in\mathbf{D}_{k}(F)}\sum_{n\geq k}|F\cap\\{x\in
Q\,:\,\sum_{\begin{subarray}{c}p\in\tilde{\mathbf{M}}_{n,k}\\\ I(p)\subset
Q\end{subarray}}\mathbf{1}_{I(p)}\geq 1000\cdot 2^{n}\log(n+2)\\}|\\\ \leq
2\sum_{k\geq 0}\sum_{n\geq k}(n+2)^{-100}|F|\leq\frac{1}{4}|F|\,.$

This completes the proof. ∎

After removing the exceptional set $\tilde{F}_{1}$, only the tiles in

$\mathbf{P}_{k}:=\\{p\in\tilde{\mathbf{P}}_{k}\,:\,I(p)\not\subset\tilde{F}_{1}\\}$

contribute, i.e. we have
$\mathbf{1}_{\mathbb{R}^{d+1}\setminus\tilde{F}_{1}}T_{\tilde{\mathbf{P}}_{k}}=\mathbf{1}_{\mathbb{R}^{d+1}\setminus\tilde{F}_{1}}T_{\mathbf{P}_{k}}$.
Thus it suffices to decompose the sets $\mathbf{P}_{k}$ into forests and
antichains.

We define the $k$-density of a tile $p\in\mathbf{P}_{k}$ to be

$\operatorname{\operatorname{dens}}_{k}(p):=\sup_{\begin{subarray}{c}\lambda\geq
1\\\
\text{odd}\end{subarray}}\sup_{\begin{subarray}{c}p^{\prime}\in\mathbf{P}_{k}:I(p)\subset
I(p^{\prime})\\\
\lambda\omega(p^{\prime})\subset\lambda\omega(p)\end{subarray}}\lambda^{-\dim_{h}V}\frac{|E(\lambda,p^{\prime})|}{|I(p^{\prime})|}\,.$

Then we split each $\mathbf{P}_{k}$ into sets
$\mathbf{H}_{n,k}:=\\{p\in\mathbf{P}_{k}\,:\,2^{-n-1}<\operatorname{\operatorname{dens}}_{k}(p)\leq
2^{-n}\\}$, and decompose each of them separately into forests and antichains.
Note that
$\operatorname{\operatorname{dens}}(\mathbf{C})\leq\sup_{p\in\mathbf{C}}\operatorname{\operatorname{dens}}_{k}(p)$
for each $\mathbf{C}\subset\mathbf{P}_{k}$, so all subsets forests and
antichains obtained in the decomposition of $\mathbf{H}_{n,k}$ have density at
most $2^{-n}$.

######  Lemma 2.

For each $k\geq 0$ and $n\geq k$, there exists an exceptional set
$\tilde{F}_{n,k}$ such that $|\tilde{F}_{n,k}|\leq 2^{-k-n-2}|F|$, and such
that the set of tiles $p\in\mathbf{H}_{n,k}$ with
$I(p)\not\subset\tilde{F}_{n,k}$ can be decomposed as a disjoint union of
$O(n+1)$ many $n$-forests and $O_{d}((n+1)^{2})$ many antichains.

###### Proof.

We first note that $\mathbf{H}_{n,k}$ is convex: Since
$\operatorname{\operatorname{dens}}$ is a decreasing function on tiles with
respect to $\leq$, $\mathbf{H}_{n,k}$ is the difference of two down-sets, and
as such convex.

Next, we prune the top $n+2$ layers off $\mathbf{H}_{n,k}$: Let
$\mathbf{H}_{n,k}^{+}$ be the set of tiles $p\in\mathbf{H}_{n,k}$ for which
there exists no chain $p<p_{1}<\dotsb<p_{n+2}$ with all
$p_{i}\in\mathbf{H}_{n,k}$, where we use $p<p^{\prime}$ to say that $p\leq
p^{\prime}$ and $p\neq p^{\prime}$. Clearly, $\mathbf{H}_{n,k}^{+}$ is the
union of at most $n+2$ antichains, so it suffices to decompose
$\mathbf{H}_{n,k}^{0}:=\mathbf{H}_{n,k}\setminus\mathbf{H}_{n,k}^{+}$, and
this set is still convex.

For $n\geq k\geq 0$, let $\mathbf{M}_{n,k}$ to be the set of maximal tiles $p$
in $\mathbf{P}_{k}$ that satisfy $|\overline{E}(p)|/|I(p)|\geq 2^{-n-1}$. By
Lemma 1 and the definition of $\mathbf{P}_{k}$, we then have the overlap
estimate

$\sum_{p\in\mathbf{M}_{n,k}}\mathbf{1}_{I(p)}\leq 1000\cdot 2^{n}\log(n+2)\,.$
(12)

We claim that for each $p\in\mathbf{H}_{n,k}^{0}$, there exists
$m\in\mathbf{M}_{n,k}$ with $p\leq m$. Since $p\in\mathbf{H}_{n,k}^{0}$, there
exists a chain $p<p_{1}<\dotsb<p_{n+2}$ with $p_{n+2}\in\mathbf{H}_{n,k}$.
Since all tiles are admissible, we have
$3^{\lfloor(n+2)/2\rfloor}\omega(p_{n+2})\subset\omega(p)$. Since
$p_{n+2}\in\mathbf{H}_{n,k}$, there exists an odd $\lambda\geq 1$ and a tile
$p^{\prime}\in\mathbf{P}_{k}$ with $I(p_{n+2})\subset I(p^{\prime})$ and
$\lambda\omega(p^{\prime})\subset\lambda\omega(p_{n+2})$ such that

$\frac{|E(\lambda,p^{\prime})|}{|I(p^{\prime})|}\geq\lambda^{\dim_{h}V}2^{-n-1}\,.$
(13)

Since $\lambda$ is odd, the set $\lambda\omega(p^{\prime})$ is the disjoint
union of $\lambda^{\dim_{h}V}$ cubes $\omega(p^{\prime\prime})$ of tiles
$p^{\prime\prime}$ with $I(p^{\prime\prime})=I(p^{\prime})$, thus there exists
one such $p^{\prime\prime}\in\mathbf{P}_{k}$ with
$|\overline{E}(p^{\prime\prime})|/|I(p^{\prime\prime})|\geq 2^{-n-1}$. By
definition of $\mathbf{M}_{n,k}$, there is a tile $m\in\mathbf{M}_{n,k}$ with
$p^{\prime\prime}\leq m$. Equation (13) implies that $\lambda\leq
2^{(n+1)/\dim_{h}V}$, so that

$\omega(m)\subset\omega(p^{\prime\prime})\subset
2^{(n+1)/\dim_{h}V}\omega(p_{n+2})\subset\omega(p)\,.$

Combining all of this we obtain that $p\leq m$, so the claim holds.

For a tile $p\in\mathbf{H}_{n,k}^{0}$, let $\mathbf{B}(p)$ be the set of tiles
$m\in\mathbf{M}_{n,k}$ with $p\leq m$. Decompose $\mathbf{H}_{n,k}^{0}$ into
$2n+10$ collections

$\mathbf{C}_{j}:=\\{p\in\mathbf{H}_{n,k}\,:\,2^{j-1}\leq|\mathbf{B}(p)|<2^{j}\\}\,,\qquad
j=1,\dotsc,2n+10\,.$

The collections $\mathbf{C}_{j}$ are convex, since $p\leq p^{\prime}\leq
p^{\prime\prime}$ implies $B(p^{\prime\prime})\subset B(p^{\prime})\subset
B(p)$, and by the overlap estimate (12) and the claim, their union is
$\mathbf{H}_{n,k}^{0}$.

Let $\mathbf{U}_{j}$ be the set of maximal tiles in $\mathbf{C}_{j}$, clearly
$\mathbf{U}_{j}$ also has overlap bounded by $100\cdot 2^{n}\log(n+2)$. For
each $m\in\mathbf{U}_{j}$ let

$\tilde{\mathbf{T}}(m):=\\{p\in\mathbf{C}_{j}\,:\,p\leq m\\}\,.$

Then the sets $\tilde{\mathbf{T}}(m)$ are disjoint: If
$p\in\tilde{\mathbf{T}}(m)\cap\tilde{\mathbf{T}}(m^{\prime})$ for $m\neq
m^{\prime}$, then
$\mathbf{B}(m)\cup\mathbf{B}(m^{\prime})\subset\mathbf{B}(p)$. But the sets
$\mathbf{B}(m)$ and $\mathbf{B}(m^{\prime})$ are disjoint: Else there would be
$m^{\prime\prime}$ with $p\leq m,m^{\prime}\leq m^{\prime\prime}$, which
implies that $m,m^{\prime}$ are comparable. But they are both maximal in
$\mathbf{C}_{j}$, so they cannot be comparable. Hence
$2^{j+1}>|\mathbf{B}(p)|\geq|\mathbf{B}(m)|+|\mathbf{B}(m^{\prime})|\geq
2^{j}+2^{j}$, a contradiction. In particular, tiles
$p\in\tilde{\mathbf{T}}(m)$, $p^{\prime}\in\tilde{\mathbf{T}}(m^{\prime})$ for
$m\neq m^{\prime}$ are not comparable.

The sets $\tilde{\mathbf{T}}(m)$ are of course also convex, so they are trees
with top $m$. To obtain the separation property, we prune the bottom $20dn$
layers of the trees: We define $\mathbf{T}^{-}(m)$ to be the set of tiles
$p\in\tilde{\mathbf{T}}(m)$ for which there exists no chain
$p_{20dn}<\dotsb<p_{1}<p$. Clearly, $\mathbf{T}^{-}(m)$ is the union of at
most $20dn$ antichains. As tiles in different $\tilde{\mathbf{T}}(m)$ are
never comparable, $\cup_{m\in\mathbf{U}_{j}}\mathbf{T}^{-}(m)$ is still a
union of at most $20dn$ antichains. Let
$\tilde{\mathbf{T}}^{0}(m):=\tilde{\mathbf{T}}(m)\setminus\mathbf{T}^{-}(m)$,
this is still a convex tree with top $m$. If $p\in\tilde{\mathbf{T}}^{0}(m)$,
then there exists a chain $p_{20dn}<\dotsb<p_{1}<p$ in
$\tilde{\mathbf{T}}(m)$. If $m^{\prime}\neq m$ is such that $I(p)\subset
I(m^{\prime})=I(\tilde{\mathbf{T}}(m^{\prime}))$, then by the last paragraph
we must have $\omega(p_{20dn})\cap\omega(m^{\prime})=\emptyset$. Since all
tiles are admissible, it follows that
$3^{10dn}\omega(p)\cap\omega(m^{\prime})=\emptyset$. Hence the trees
$\tilde{\mathbf{T}}^{0}(m)$ are $3^{10dn}$-separated.

Finally, we make the trees normal. For this purpose, let $r=100+d+6n$. We
first prune the top $r$ layers $\tilde{\mathbf{T}}^{0+}(m)$ off each
$\tilde{\mathbf{T}}^{0}(m)$ similarly as in the last paragraph. This produces
another $r=O_{d}(n+1)$ antichains. Then we define the exceptional set

$\tilde{F}_{n,k}=\bigcup_{j}\bigcup_{m\in\mathbf{U}_{j}}(I(m)\setminus(1-2^{-r})I(m))\,,$

where $aQ=c(Q)+\delta_{\log_{2}a}(Q-c(Q))$ is the anisotropic dilate of $Q$ by
a factor $a$ about its center $c(Q)$. We have, using that $1\leq j\leq 2n+10$,
the Bernoulli inequality and (12)

$\displaystyle|\tilde{F}_{n,k}|$
$\displaystyle\leq(2n+10)\cdot\sum_{m\in\mathbf{U}_{j}}(d+2)2^{-r}|I(m)|$
$\displaystyle\leq(2n+10)\cdot 1000\cdot 2^{n}\log(n+2)\cdot(d+2)\cdot
2^{-r}\cdot\sum_{Q\in\mathbf{D}_{k}(F)}|Q|$ $\displaystyle\leq
2^{-3n-3}\sum_{Q\in\mathbf{D}_{k}(F)}|Q|\,.$

Using the definition of $\mathbf{D}_{k}(F)$ and that $n\geq k$ we estimate
this by

$\leq 2^{-3n-2}2^{k}|F|\leq 2^{-n-k-2}|F|\,.$

We finally define

$\mathbf{T}(m):=\\{p\in\tilde{\mathbf{T}}^{0}(m)\setminus\tilde{\mathbf{T}}^{0+}(m)\,:\,I(p)\not\subset\tilde{F}_{n,k}\\}\,.$

Then $\mathbf{T}(m),m\in\mathbf{U}_{j}$ is still a collection of
$2^{10dn}$-separated trees, and they are now normal: If $p\in\mathbf{T}(m)$
then $s(p)\leq s(m)-r$ and $I(p)\subset(1-2^{-r})I(m)$, and therefore
$3I(p)\subset I(m)$. By the overlap estimate (12), the collection
$\\{\mathbf{T}(m)\,:\,m\in\mathbf{U}_{j}\\}$ is the union of at most $1000$
$n$-forests. Thus $\mathbf{C}_{j}$ can be decomposed into $1000$ $n$-forests
and $O_{d}(n+1)$ antichains, which completes the proof. ∎

Recall that all antichains in Lemma 2 have density at most $2^{-n}$, since
they are contained in $\mathbf{H}_{n,k}$. Taking into account all $k\geq 0$,
Lemma 2 then yields a total of $O_{d}((n+1)^{3})$ antichains of density at
most $2^{-n}$, a total of $O((n+1)^{2})$ many $n$-forests, and an exceptional
set $\tilde{F}_{2}=\bigcup_{n,k}\tilde{F}_{n,k}$ with
$|\tilde{F}_{2}|\leq|F|/4$. Combining this with the estimate for the measure
of $\tilde{F}_{1}$ from Lemma 1, we obtain Proposition 1 with
$\tilde{F}:=\tilde{F}_{1}\cup\tilde{F}_{2}$.

##  5\. Antichains: Proof of Proposition 2

###  5.1. Decomposition of the kernel

We fix an antichain $\mathbf{A}$ of density $\delta$. We will decompose
$T_{\mathbf{A}}$, based on a decomposition of the kernel of the singular Radon
transform. Let $\mu_{s}$ be the measure defined by

$\int
f\,\mathrm{d}\mu_{s}=\int_{\mathbb{R}^{d}}f(y,|y|^{2})K_{s}(y)\,\mathrm{d}y\,.$

Then the operator $T_{p}$ can be expressed as

$T_{p}f(x)=\mathbf{1}_{E(p)}(x)\int f(x-y)e^{iN(x)\cdot
y}\mathrm{d}\mu_{s(p)}(y)\,.$ (14)

We now decompose the measures $\mu_{s}$ into a smooth part and a high-
frequency part, choosing different decompositions depending on $V$. For this,
we let $\varepsilon_{0}>0$ be a small enough positive number such that the
exponent of $\delta$ in Lemma 5 below is positive.

In the case $V=\mathbb{R}^{d-1}\times\\{0\\}^{2}$, we let $\varphi^{1}$ be a
smooth bump function supported on $B(0,100^{-1})\subset\mathbb{R}$ with
integral $1$, and let

$\varphi_{s,\varepsilon_{0}}^{1}(x)=\delta^{-2\varepsilon_{0}}2^{-2s}\varphi^{1}(2^{-2s}\delta^{-2\varepsilon_{0}}x_{d+1})\delta(x^{\prime})\,.$
(15)

We define the low frequency part of $\mu_{s}$ by

$\mu_{s}^{l}(x):=\mu_{s}*\varphi_{s,\varepsilon_{0}}^{1}(x)=\delta^{-2\varepsilon_{0}}2^{-2s}K_{s}(x^{\prime})\varphi^{1}(2^{-2s}\delta^{-2\varepsilon_{0}}(x_{d+1}-|x^{\prime}|^{2}))$
(16)

and the high frequency part by $\mu_{s}^{h}=\mu_{s}-\mu_{s}^{l}$. Note that
$\varphi_{s,\varepsilon_{0}}^{1}$ is a measure supported on the line
$x^{\prime}=0$, so $\mu_{s}^{l}$ is essentially frequency localized near the
hyperplane $\xi_{d+1}=0$.

In the case $V=\\{0\\}^{d}\times\mathbb{R}$, we let $\varphi^{d}$ be a smooth
bump function supported on $B(0,100^{-1})\subset\mathbb{R}^{d}$ with integral
$1$, and let

$\varphi_{s,\varepsilon_{0}}^{d}(x)=\delta^{-d\varepsilon_{0}}2^{-ds}\varphi^{d}(2^{-s}\delta^{-\varepsilon_{0}}x^{\prime})\delta(x_{d+1})\,.$
(17)

We define the low frequency part of $\mu_{s}$ by

$\displaystyle\mu_{s}^{l}(x)$
$\displaystyle:=\mu_{s}*\varphi_{s,\varepsilon_{0}}^{d}(x)$
$\displaystyle=\delta^{-d\varepsilon_{0}}2^{-ds}\int\delta(|y^{\prime}|^{2}-x_{d+1})K_{s}(y^{\prime})\varphi^{d}(2^{-s}\delta^{-\varepsilon_{0}}(x^{\prime}-y^{\prime}))\,\mathrm{d}y^{\prime}$
(18)

and the high frequency part by $\mu_{s}^{h}=\mu_{s}-\mu_{s}^{l}$. In this
case, $\varphi_{s}^{d}$ is a measure supported on the hyperplane $x_{d+1}=0$,
and $\mu_{s}^{l}$ is essentially frequency localized near the line
$\xi^{\prime}=0$.

In both cases the low frequency part $\mu_{s}^{l}$ is a function, and one can
check using (16) or (18) and (5) that

$|\mu_{s}^{l}(x)|\lesssim\delta^{-d\varepsilon_{0}}|I(p)|^{-1}$ (19)

and

$|\partial_{i}\mu_{s(p)}^{l}(x)|\lesssim\begin{cases}\delta^{-(d+2)\varepsilon_{0}}2^{-s}|I(p)|^{-1}&\text{if
$i=1,\dotsc,d$,}\\\ \delta^{-(d+2)\varepsilon_{0}}2^{-2s}|I(p)|^{-1}&\text{if
$i=d+1$.}\\\ \end{cases}$ (20)

We define operators $T_{p}^{l},T_{p}^{h}$ by replacing $\mu_{s(p)}$ with
$\mu_{s(p)}^{l},\mu_{s(p)}^{h}$ in (14), and we define
$T_{\mathbf{A}}^{l},T_{\mathbf{A}}^{h}$ as the sum of $T_{p}^{l},T_{p}^{h}$
over all $p\in\mathbf{A}$. Then we have
$T_{\mathbf{A}}=T_{\mathbf{A}}^{l}+T_{\mathbf{A}}^{h}$. We prove estimates for
$T_{\mathbf{A}}^{l}$ and $T_{\mathbf{A}}^{h}$ separately in the next two
subsections.

###  5.2. The smooth part

Here we estimate the smooth part $T_{\mathbf{A}}^{l}$. The argument is the
same for $V=\mathbb{R}^{d-1}\times\\{0\\}^{2}$ and
$V=\\{0\\}^{d}\times\mathbb{R}$. While $\mu_{s}^{l}$ is defined differently in
these two cases, we will only use the properties (19) and (20) of
$\mu_{s}^{l}$, which hold in both cases.

We define the separation $\Delta(p,p^{\prime})$ of a pair of tiles
$p,p^{\prime}$ to be $1$ if $\omega(p)\cap\omega(p^{\prime})\neq\emptyset$,
and else we define it as

$\Delta(p,p^{\prime}):=\sup\\{\lambda>0\,:\,\lambda\omega(p)\cap\omega(p^{\prime})=\emptyset\
\text{and}\ \omega(p)\cap\lambda\omega(p^{\prime})=\emptyset\\}\,.$

Then we have the following almost orthogonality estimate.

######  Lemma 3.

There exists a constant $C>0$ such that for all tiles $p,p^{\prime}$ with
$s(p)\geq s(p^{\prime})$, we have

$\lvert\langle T_{p}^{l*}g,T_{p^{\prime}}^{l*}g\rangle\rvert\leq
C\delta^{-(2d+2)\varepsilon_{0}}\frac{\Delta(p,p^{\prime})^{-1}}{|I(p)|}\int_{E(p)}|g|\int_{E(p^{\prime})}|g|\,.$
(21)

###### Proof.

We abbreviate $\Delta=\Delta(p,p^{\prime})$. We have, by Fubini and the
definition of $T_{p}^{l}$

$\lvert\langle
T_{p}^{l*}g,T_{p^{\prime}}^{l*}g\rangle\rvert\leq\int\int|\mathbf{1}_{E(p)}g(x_{1})||\mathbf{1}_{E(p^{\prime})}g(x_{2})|\\\
\left\lvert\int e^{i(N(x_{1})-N(x_{2}))\cdot
y}\mu_{s(p)}^{l}(x_{1}-y)\overline{\mu_{s(p^{\prime})}^{l}(x_{2}-y)}\,\mathrm{d}y\right|\,\mathrm{d}x_{1}\,\mathrm{d}x_{2}\,.$
(22)

The inner integral in (22) is bounded by
$C\delta^{-2d\varepsilon_{0}}|I(p)|^{-1}$, by (19). This implies (21) if
$\Delta\leq 3$, so we will assume from now on that $\Delta>3$. Fix $x_{1}\in
E(p),x_{2}\in E(p^{\prime})$ and let

$\Phi(y)=\mu_{s(p)}^{l}(x_{1}-y)\overline{\mu_{s(p^{\prime})}^{l}(x_{2}-y)}\,.$

By (19) and (20), we have the bounds

$|\partial_{i}\Phi(y)|\lesssim\delta^{-(2d+2)\varepsilon_{0}}\frac{1}{|I(p)||I(p^{\prime})|}2^{-s(p^{\prime})}\,,\quad
i=1,\dotsc,d$

and

$|\partial_{i}\Phi(y)|\lesssim\delta^{-(2d+2)\varepsilon_{0}}\frac{1}{|I(p)||I(p^{\prime})|}2^{-2s(p^{\prime})}\,,\quad
i=d+1\,.$

Since $\Delta=\Delta(p,p^{\prime})>3$, we have by definition

$\max_{i=1,\dotsc,d}2^{s(p)}|N_{i}(x_{1})-N_{i}(x_{2})|\geq(\Delta-1)/2\geq\Delta/3$

or

$2^{2s(p)}|N_{d+1}(x_{1})-N_{d+1}(x_{2})|\geq(\Delta^{2}-1)/2\geq\Delta/3\,.$

Integrating by parts in the corresponding direction, we find that

$\left\lvert\int e^{i(N(x_{1})-N(x_{2}))\cdot
y}\mu_{s(p)}^{l}(x_{1}-y)\overline{\mu_{s(p^{\prime})}^{l}(x_{2}-y)}\,\mathrm{d}y\right|\\\
\lesssim\Delta^{-1}\delta^{-(2d+2)\varepsilon_{0}}|I(p)|^{-1}\,,$

which together with (22) gives the claimed estimate (21). ∎

######  Lemma 4.

Let $1\leq q\leq\infty$. There exists $C>0$ such that for every antichain
$\mathbf{A}$ of density $\delta$ and every tile $p$

$\Bigg{\|}\sum_{p^{\prime}\in\mathbf{A},s(p^{\prime})\leq
s(p)}\Delta(p,p^{\prime})^{-1}\mathbf{1}_{E(p^{\prime})}\Bigg{\|}_{q}\leq
C\delta^{\frac{\kappa}{q}}\Bigg{|}\bigcup_{p^{\prime}\in\mathbf{A},s(p^{\prime})\leq
s(p)}I(p^{\prime})\Bigg{|}^{1/q}\,,$

where $\kappa=(1+\dim_{h}V)^{-1}$.

###### Proof.

The estimate holds for $q=\infty$ with $C=1$, since $\Delta(p,p^{\prime})\geq
1$ and the sets $E(p^{\prime})$, $p^{\prime}\in\mathbf{A}$ are disjoint. By
Hölder’s inequality it therefore only remains to show the estimate for $q=1$:

$\sum_{p^{\prime}\in\mathbf{A},s(p^{\prime})\leq
s(p)}\Delta(p,p^{\prime})^{-1}|E(p^{\prime})|\lesssim\delta^{\kappa}\Bigg{|}\bigcup_{p^{\prime}\in\mathbf{A},s(p^{\prime})\leq
s(p)}I(p^{\prime})\Bigg{|}\,.$

The contribution of tiles with $\Delta(p,p^{\prime})\geq\delta^{-\kappa}$ is
clearly bounded by the right hand side, since $E(p^{\prime})\subset
I(p^{\prime})$ and the sets $E(p^{\prime})$ are disjoint. So it remains only
to estimate the contribution of the tiles

$\mathbf{A}^{\prime}:=\\{p^{\prime}\in\mathbf{A}\,:\,s(p^{\prime})\leq
s(p)\,,\ \Delta(p,p^{\prime})<\delta^{-\kappa}\\}\,.$

Let $\mathbf{L}$ be the set of maximal dyadic cubes $L$ for which there exists
$p^{\prime}\in\mathbf{A}^{\prime}$ with $L\subsetneq I(p^{\prime})$, and there
exists no $p^{\prime}\in\mathbf{A}^{\prime}$ with $I(p^{\prime})\subset L$.
The set $\mathbf{L}$ is a partition of
$\cup_{p^{\prime}\in\mathbf{A}^{\prime}}I(p^{\prime})$, so it will be enough
to show for each $L\in\mathbf{L}$ the estimate

$\bigg{|}L\cap\bigcup_{p^{\prime}\in\mathbf{A}^{\prime}}E(p^{\prime})\bigg{|}\leq\delta^{1-\kappa\dim_{h}V}|L|\,.$
(23)

Fix $L\in\mathbf{L}$. There exists a tile $p^{\prime}\in\mathbf{A}^{\prime}$
with $I(p^{\prime})\subset\hat{L}$. If $I(p^{\prime})=\hat{L}$, define
$p_{L}=p^{\prime}$, and else let $p_{L}$ be the unique tile with
$I(p_{L})=\hat{L}$ and $\omega(p)\subset\omega(p_{L})$. If $\lambda$ is the
smallest odd number such that $\lambda\geq 5\delta^{-\kappa}$, then we have in
both cases that $\lambda\omega(p^{\prime})\supset\lambda\omega(p_{L})$. For
each $p^{\prime\prime}\in\mathbf{A}^{\prime}$ with $L\cap
I(p^{\prime\prime})\neq\emptyset$, it holds that $L\subsetneq
I(p^{\prime\prime})$ and
$\omega(p^{\prime\prime})\subset\lambda\omega(p_{L})$. It follows that for all
such $p^{\prime\prime}$, we have that $L\cap E(p^{\prime\prime})\subset
E(\lambda,p_{L})$. Thus

$|L\cap\bigcup_{p^{\prime\prime}\in\mathbf{A}^{\prime}}E(p^{\prime\prime})|\leq|E(5\delta^{-\kappa}p_{L})|\\\
\leq\lambda^{-\dim_{h}V}\operatorname{\operatorname{dens}}(\mathbf{A})|L|\lesssim\delta^{1-\kappa\dim_{h}V}|L|\,,$

giving (23) and hence the lemma. ∎

Using the last two lemmas, we can prove our main estimate for
$T_{\mathbf{A}}^{l}$.

######  Lemma 5.

For all $1\leq q<2$, there exists a constant $C>0$ such that

$\|T_{\mathbf{A}}^{l}\|_{2\to 2}\leq
C\delta^{\frac{\kappa}{2q^{\prime}}-(d+1)\varepsilon_{0}}\,,$

where $\kappa=1/(1+\dim_{h}V)$.

###### Proof.

Define for each $p\in\mathbf{A}$ the set

$\mathbf{A}(p)=\\{p^{\prime}\in\mathbf{A}\,:\,s(p^{\prime})\leq
s(p),\,3I(p)\cap 3I(p^{\prime})\neq\emptyset\\}\,.$

Since $T_{p}^{l*}g$ is always supported in $3I(p)$, we have that

$\lvert\langle T_{\mathbf{A}}^{l*}g,T_{\mathbf{A}}^{l*}g\rangle\rvert\leq
2\sum_{p\in\mathbf{A}}\sum_{p^{\prime}\in\mathbf{A}(p)}\lvert\langle
T_{p}^{l*}g,T_{p^{\prime}}^{l*}g\rangle\rvert\,.$

By Lemma 3, this is bounded up to constant factor by

$\delta^{-(2d+2)\varepsilon_{0}}\sum_{p\in\mathbf{A}}\int\mathbf{1}_{E(p)}|g|\frac{1}{|I(p)|}\int|g|\sum_{p^{\prime}\in\mathbf{A}(p)}\Delta(p,p^{\prime})^{-1}\mathbf{1}_{E(p^{\prime})}\,.$

Cubes $I(p^{\prime})$ associated to $p^{\prime}\in\mathbf{A}(p)$ are contained
in $5I(p)$, so we may apply Hölder’s inequality in the inner integral with
$q<2$ to bound this by a constant times

$\delta^{-(2d+2)\varepsilon_{0}}\sum_{p\in\mathbf{A}}\int\mathbf{1}_{E(p)}|g|M^{q}|g|\frac{\|\sum_{p^{\prime}\in\mathbf{A}(p)}\Delta(p,p^{\prime})^{-1}\mathbf{1}_{E(p^{\prime})}\|_{L^{q^{\prime}}}}{|I(p)|^{1/q^{\prime}}}\,,$

where $M^{q}g=M(g^{q})^{1/q}$ is the $q$-maximal function. Using Lemma 4 to
estimate the $L^{q^{\prime}}$-norm, we estimate this by a constant times

$\delta^{\frac{\kappa}{q^{\prime}}-(2d+2)\varepsilon_{0}}\sum_{p\in\mathbf{A}}\int\mathbf{1}_{E(p)}|g|M^{q}|g|\\\
\leq\delta^{\frac{\kappa}{q^{\prime}}-(2d+2)\varepsilon_{0}}\int|g|M^{q}|g|\lesssim\delta^{\frac{\kappa}{q^{\prime}}-(2d+2)\varepsilon_{0}}\|g\|_{2}^{2}\,.$

Here we used disjointness of the sets $E(p)$ and $L^{2}$-boundedness of
$M^{q}$ for $q<2$. This completes the proof. ∎

###  5.3. The high frequency part

Here we estimate the high frequency part $T_{\mathbf{A}}^{h}$.

We start by discretizing modulation frequencies. For this, we let
$\varepsilon_{1}>0$ be a small positive number, much smaller than
$\varepsilon_{0}$, it will be chosen at the end of this section. We fix finite
subsets $M(\omega)\subset\omega$ of each dyadic frequency cube $\omega$, such
that the following holds with $\rho=\delta^{\varepsilon_{1}}$:

  1. (i)

$|M(\omega)|\lesssim\rho^{-\dim_{h}V}$,

  2. (ii)

For each $N\in\omega$, there exists $N^{\prime}\in M(\omega)$ with

$d_{\text{par}}(N,N^{\prime})\leq\rho 2^{-s(\omega)}\,,$ (24)

where $d_{\text{par}}$ denotes the parabolic distance

$d_{\text{par}}(N,N^{\prime})=\max\\{\max_{1\leq i\leq
d}|N_{i}-N_{i}^{\prime}|,|N_{d+1}-N_{d+1}^{\prime}|^{1/2}\\}\,.$

  3. (iii)

If $\omega=a+b\omega^{\prime}$ then $M(\omega)=a+bM(\omega^{\prime})$.

For each tile $p$, we partition the set $E(p)$ into $E(p^{c})$, $c\in
M(\omega(p))$, such that for all $x\in E(p^{c})$

$d_{\text{par}}(N(x),c)\leq\delta^{\varepsilon_{1}}2^{-s(p)}\,.$ (25)

We define for a tile $p$ and $c\in M(\omega(p))$ the operator

$T_{p}^{h,c}f(x):=\mathbf{1}_{E(p^{c})}\int f(x-y)e^{ic\cdot
y}\,\mathrm{d}\mu_{s(p)}^{h}(y)\,,$

and we define also

$T_{\mathbf{A}}^{h,d}=\sum_{p\in\mathbf{A}}\sum_{c\in
M(\omega(p))}T_{p}^{h,c}\,.$

By (25), we have for $x\in E(p^{c})$ and all $y$ in the support of
$\mu_{s(p)}^{h}$ that

$|c\cdot y-N(x)\cdot y|\leq\delta^{\varepsilon_{1}}\,,$

hence

$|T_{p}f(x)-\sum_{c\in
M(\omega(p))}T_{p}^{h,c}f(x)|\leq\delta^{\varepsilon_{1}}\mathbf{1}_{E(p)}(x)\int|f(x-y)|\,\mathrm{d}\lvert\mu_{s(p)}^{h}\rvert(y)\,.$

Summing this over all $p\in\mathbf{A}$ and using that the corresponding sets
$E(p)$ are disjoint, it follows that
$|T_{\mathbf{A}}^{h}-T_{\mathbf{A}}^{h,d}|$ is dominated by
$\delta^{\varepsilon_{1}}$ times a mollified maximal average along the
parabola. Thus

$\|T_{\mathbf{A}}^{h}-T_{\mathbf{A}}^{h,d}\|_{2\to
2}\lesssim\delta^{\varepsilon_{1}}\,.$ (26)

It now remains to estimate the discretized operator $T_{\mathbf{A}}^{h,d}$. By
disjointness of the sets $E(p^{c})$ and the third property of the sets
$M(\omega)$, we have

$T_{\mathbf{A}}^{h,d}f(x)\leq Sf(x)\,,$ (27)

where the square function $Sf$ is defined depending on $V$ as follows: In the
case $V=\mathbb{R}^{d-1}\times\\{0\\}^{2}$ it is defined by

$(Sf(x))^{2}=\sum_{s\in\mathbb{Z}}\sum_{N\in\mathbb{Z}^{d-1}\times\\{0\\}^{2}}\sum_{c\in
M(\omega_{0})}|f*(e^{i\delta_{s}(c+N)\cdot y}\mu_{s}^{h}(y))(x)|^{2}$ (28)

with $\omega_{0}=[0,1)^{d-1}\times\\{0\\}^{2}$, and in the case
$V=\\{0\\}^{d}\times\mathbb{R}$ it is defined by

$(Sf(x))^{2}=\sum_{s\in\mathbb{Z}}\sum_{N\in\\{0\\}^{d}\times\mathbb{Z}}\sum_{c\in
M(\omega_{0})}|f*(e^{i\delta_{s}(c+N)\cdot y}\mu_{s}^{h}(y))(x)|^{2}$ (29)

with $\omega_{0}=\\{0\\}^{d}\times[0,1)$.

######  Lemma 6.

There exist $C>0$ such that the following holds. Let $S$ be defined by (28) or
by (29). Then we have

$\|S\|_{2\to 2}\leq
C\delta^{\frac{2}{d+2}\varepsilon_{0}-\frac{\dim_{h}V}{2}\varepsilon_{1}}\,.$

###### Proof.

We first treat the case (28). We have

$\hat{\mu}_{0}(\xi)=\int_{1/8<|y|<1/3}e^{i(\xi^{\prime}\cdot
y+\xi_{d+1}|y|^{2})}K_{0}(y)\,\mathrm{d}y\,.$

Thus we obtain from the method of stationary phase (see e.g. [Hör03], Lemma
7.7.3) and (5):

$|\hat{\mu}_{0}(\xi)|^{2}\lesssim\min\\{1,|\xi_{d+1}|^{-d}\\}\,.$ (30)

Combining this with (15) yields

$|\hat{\mu}_{0}^{h}(\xi)|^{2}=|(1-\hat{\varphi}^{1}(\delta^{2\varepsilon_{0}}\xi_{d+1}))\hat{\mu}_{0}(\xi)|^{2}\lesssim\min\\{|\delta^{2\varepsilon_{0}}\xi_{d+1}|^{2},|\xi_{d+1}|^{-d}\\}\,.$
(31)

Summing up (31) gives

$\sum_{\begin{subarray}{c}N\in\mathbb{Z}^{d-1}\times\\{0\\}^{2}\\\
|\xi^{\prime}+c^{\prime}+N^{\prime}|\leq|\xi_{d+1}|\end{subarray}}|\hat{\mu}_{0}^{h}(\xi+c+N)|^{2}\lesssim|\xi_{d+1}|^{d-1}\min\\{|\delta^{2\varepsilon_{0}}\xi_{d+1}|^{2},|\xi_{d+1}|^{-d}\\}\,.$
(32)

If $|\xi^{\prime}|>|\xi_{d+1}|$ then we have

$|\nabla(\xi^{\prime}\cdot
y+\xi_{d+1}|y|^{2})|=|\xi^{\prime}+\xi_{d+1}y|>\frac{1}{2}|\xi^{\prime}|\,.$

Integrating by parts (see e.g. [Hör03], Theorem 7.7.1), we obtain

$|\hat{\mu}_{0}(\xi)|^{2}\lesssim\min\\{1,|\xi^{\prime}|^{-d-1}\\}\,,$ (33)

and hence

$|\hat{\mu}_{0}^{h}(\xi)|^{2}\lesssim\min\\{|\delta^{2\varepsilon_{0}}\xi_{d+1}|\min\\{1,|\xi^{\prime}|^{-d-1}\\},|\xi^{\prime}|^{-d-1}\\}\,.$
(34)

Summing up estimate (34) gives

$\sum_{\begin{subarray}{c}N\in\mathbb{Z}^{d-1}\times\\{0\\}^{2}\\\
|\xi^{\prime}+c^{\prime}+N^{\prime}|>|\xi_{d+1}|\end{subarray}}|\hat{\mu}_{0}^{h}(\xi+c+N)|^{2}\lesssim\min\\{|\delta^{2\varepsilon_{0}}\xi_{d+1}|^{2},|\xi_{d+1}|^{-1}\\}\,.$
(35)

Combining (32) and (35) and using
$\min\\{a,b\\}\leq\min\\{(a^{3}b^{d-1})^{1/(d+2)},b\\}$ yields

$\sum_{N\in\mathbb{Z}^{d-1}\times\\{0\\}^{2}}|\hat{\mu}_{0}^{h}(\xi+c+N)|^{2}\lesssim\min\\{\delta^{2\varepsilon_{0}/d}|\xi_{d+1}|^{2},\,|\xi_{d+1}|^{-1}\\}\,.$

Using dilation invariance of all assumptions on $K$ this implies

$\sum_{N\in\mathbb{Z}^{d-1}\times\\{0\\}^{2}}|\hat{\mu}_{s}^{h}(\xi+\delta_{-s}(c+N))|^{2}\lesssim\min\\{\delta^{\frac{12}{d+2}\varepsilon_{0}}|2^{2s}\xi_{d+1}|^{2},\,|2^{2s}\xi_{d+1}|^{-1}\\}\,.$

Summing in $s$, and also summing in the
$\lesssim\delta^{-\varepsilon_{1}\dim_{h}V}$ choices of $c$, we obtain

$\sum_{s\in\mathbb{Z}}\sum_{N\in\mathbb{Z}^{d-1}\times\\{0\\}^{2}}\sum_{c\in
M(\omega_{0})}|\hat{\mu}_{s}^{h}(\xi+\delta_{-s}(c+N))|^{2}\lesssim\delta^{\frac{4}{d+2}\varepsilon_{0}-\varepsilon_{1}\dim_{h}V}\,.$

This completes the proof in the case (28) by Plancherel.

Now we deal with the vertical modulation case (29). The stationary phase
estimate (30) combined with (17) yields

$|\hat{\mu}_{0}^{h}(\xi)|^{2}\lesssim\min\\{1,|\xi_{d+1}|^{-d}\\}|\delta^{\varepsilon_{0}}\xi^{\prime}|^{2}\,,$

and hence

$\sum_{N\in\\{0\\}^{d}\times\mathbb{Z}}|\hat{\mu}_{0}^{h}(\xi+c+N)|^{2}\lesssim|\delta^{\varepsilon_{0}}\xi^{\prime}|^{2}\,.$

Combining the estimates (33) in the range $|\xi^{\prime}|>|(\xi+c+N)_{d+1}|$
and (30) in the range $|\xi^{\prime}|\leq|(\xi+c+N)_{d+1}|$ gives

$\sum_{N\in\\{0\\}^{d}\times\mathbb{Z}}|\hat{\mu}_{0}^{h}(\xi+c+N)|^{2}\lesssim|\xi^{\prime}|^{1-d}\,.$

Therefore we have

$\sum_{N\in\\{0\\}^{d}\times\mathbb{Z}}|\hat{\mu}_{0}^{h}(\xi+c+N)|^{2}\lesssim\min\\{\delta^{2\varepsilon_{0}}|\xi^{\prime}|^{2},|\xi^{\prime}|^{1-d}\\}\,.$

As before, this implies the square function estimate (with better dependence
on $\delta$) using dilation invariance of all assumptions to sum in $s$, the
bound on the number of summands to sum in $c$, and then Plancherel. ∎

Combining the estimate for $T_{\mathbf{A}}^{l}$ in Lemma 5, for
$T_{\mathbf{A}}^{h}-T_{\mathbf{A}}^{h,d}$ in (26) and for
$T_{\mathbf{A}}^{h,d}$ in (27) and Lemma 6, and choosing $\varepsilon_{0}$ and
then $\varepsilon_{1}$ sufficiently small, we obtain Proposition 2.

##  6\. Trees: Proof of Proposition 4

Here we prove a mild generalization of Proposition 4. We will estimate what we
call generalized tree operators, with general kernels satisfying conditions
described below. We need the estimates in Section 7 both for the singular
Radon transform and for a version of it with a smoothened kernel.

###  6.1. General setup

Let in the following $\mu_{s},\lambda_{s}$, $s\in\mathbb{Z}$ be any finite
measures satisfying the following conditions:

  1. i)

$\mu_{s}$, $\lambda_{s}$ are supported in $\delta_{s}(B(0,1/2))$,

  2. ii)

$|\mu_{s}|\leq\lambda_{s}$,

  3. iii)

$\mu_{s}(\mathbb{R}^{d+1})=0$,

  4. iv)

we have

$\max\\{|\widehat{D_{-s}\lambda_{s}}(\xi))|,|\widehat{D_{-s}\mu_{s}}(\xi)|\\}\leq|\xi|^{-1}\,,$
(36)

where $D_{-s}\mu_{s}:=2^{-(d+2)s}\mu_{s}\circ\delta_{-s}$,

  5. v)

and there exist $\gamma>0$ and $p,q<2$ such that convolution with
$D_{-s}\lambda_{s}$ and $D_{-s}\mu_{s}$ is bounded from $L^{p}$ to
$L^{q^{\prime}+\gamma}$ with norm at most $1$ for each $s$.

Relevant for the proof of Theorem 1 are the measures defined by

$\int
f\,\mathrm{d}\mu_{s}=\int_{\mathbb{R}^{d}}f(y,|y|^{2})K_{s}(y)\,\mathrm{d}y\,,$
(37) $\int
f\,\mathrm{d}\lambda_{s}=\int_{\mathbb{R}^{d}}f(y,|y|^{2})2^{-ds}\eta(2^{-s}y)\,\mathrm{d}y\,,$
(38)

and smoothened versions thereof. They clearly satisfy conditions i) to iii),
condition iv) follows from standard stationary phase estimates (e.g. [Hör03],
Lemma 7.7.3) and for condition v) one can take any $(p,q)$ such that
$(\frac{1}{p},\frac{1}{q})$ is in the interior of the convex hull of
$(1,0),(0,1),(\frac{d+1}{d+2},\frac{d+1}{d+2})$ (see [Lit73]).

###  6.2. Generalized tree operators

With the choice of $\mu_{s}$ from (37), the tile operators defined in Section
2 can be written as

$T_{p}f(x)=\mathbf{1}_{E(p)}(x)\int f(x-y)e^{iN(x)\cdot
y}\mathrm{d}\mu_{s}(y)\,,$

and tree operators can be written as

$T_{\mathbf{T}}f(x)=\sum_{p\in\mathbf{T}}T_{p}f(x)=\sum_{s\in\sigma(x)}\int
f(x-y)e^{iN(x)\cdot y}\mathrm{d}\mu_{s}(y)\,,$

where $\sigma(x):=\\{s(p)\,:\,p\in\mathbf{T},x\in E(p)\\}$. Since $\mathbf{T}$
is a subset of the set of all admissible tiles, the set $\sigma(x)$ is always
contained in

$J=J(\mathbf{T}):=\\{s\in\mathbb{Z}\,:\,\exists\omega\
\text{admissible,}\,s(\omega)=s,\,\omega(\mathbf{T})\subset\omega\\}\,,$

and convexity of the tree $\mathbf{T}$ implies that there are
$\underline{\sigma}(x)\leq\overline{\sigma}(x)$ with

$\sigma(x)=J\cap[\underline{\sigma}(x),\overline{\sigma}(x)]\,.$

Motivated by this, we define a generalized tree to be a pair
$(\mathbf{T},\sigma^{\prime})$, where $\mathbf{T}$ is a tree and
$\sigma^{\prime}$ is a function associating to each $x$ a set

$\sigma^{\prime}(x)=J\cap[\underline{\sigma}^{\prime}(x),\overline{\sigma}^{\prime}(x)]\subset\sigma(x)\,.$

The generalized tree operator associated to $(\mathbf{T},\sigma^{\prime})$ is
defined by

$T_{\mathbf{T},\sigma^{\prime}}f(x)=\sum_{s\in\sigma^{\prime}(x)}\int
f(x-y)e^{iN(x)\cdot y}\mathrm{d}\mu_{s}(y)\,.$

The above discussion shows that this includes in particular the tree operators
defined in Section 3, by choosing $\mu_{s}$ as in (37) and
$\sigma^{\prime}(x)=\sigma(x)$ for each $x$.

###  6.3. Single tree estimate

We now dominate the generalized tree operators
$T_{\mathbf{T},\sigma^{\prime}}$ by a sum of two simpler operators
$T_{*\mathbf{T},\sigma^{\prime}}$ and $M_{\mathbf{T}}$, resembling a maximally
truncated singular integral and a maximal average along the paraboloid.

Define the unmodulated operator associated to $(\mathbf{T},\sigma^{\prime})$
by

$\displaystyle
T_{*\mathbf{T},\sigma^{\prime}}f(x):=\sum_{s\in\sigma^{\prime}(x)}f*\mu_{s}(x)=\sum_{s=\underline{\sigma}^{\prime}(x)}^{\overline{\sigma}^{\prime}(x)}f*(\mathbf{1}_{J}(s)\mu_{s})(x)\,.$

This is a maximally truncated singular integral along the paraboloid, and it
follows from standard square function arguments as in [SW78] that
$T_{*\mathbf{T}}$ is bounded on $L^{2}$.

Define also the maximal average $M_{\mathbf{T}}$ associated to $\mathbf{T}$:

$M_{\mathbf{T}}f(x):=\sup_{s\in\sigma(x)}|f|*\lambda_{s}(x)\,.$

The operator $M_{\mathbf{T}}$ is bounded above pointwise by the maximal
average associated to $(\lambda_{s})_{s\in\mathbb{Z}}$, which is bounded on
$L^{2}$ under our assumptions.

We have the following pointwise estimate for $T_{\mathbf{T},\sigma^{\prime}}$
in terms of $T_{*\mathbf{T},\sigma^{\prime}}$ and $M_{\mathbf{T}}$.

######  Lemma 7.

There exists $C>0$ such that for each generalized tree
$(\mathbf{T},\sigma^{\prime})$

$|T_{\mathbf{T},\sigma^{\prime}}f(x)|\leq|T_{*\mathbf{T},\sigma^{\prime}}(e^{-iN(\mathbf{T})(y)}f(y))(x)|+CM_{\mathbf{T}}f(x)\,,$

where $N(\mathbf{T})$ is the smallest element, in lexicographic order, of
$\omega(\mathbf{T})$.

###### Proof.

Suppose that $p\in\mathbf{T}$ with $s(p)=\max\sigma(x)-k$. Then there exists
some $p^{\prime}\in\mathbf{T}$ with $s(p^{\prime})=s(p)+k$ and $x\in
E(p^{\prime})$. Since $N(x)\in\omega(p^{\prime})$ and
$N(\mathbf{T})\in\omega(p^{\prime})$, it follows that

$|(N(\mathbf{T})-N(x))\cdot y|\leq d2^{-k}$

for all $y$ in the support of $\mu_{s(p)}$. Hence we have

$\left|T_{p}f(x)-e^{iN(\mathbf{T})\cdot x}\mathbf{1}_{E(p)}(x)\int
f(x-y)e^{-iN(\mathbf{T})\cdot(x-y)}\mathrm{d}\mu_{s(p)}(y)\right|\\\
=\left|\mathbf{1}_{E(p)}(x)\int f(x-y)(e^{iN(x)\cdot y}-e^{iN(\mathbf{T})\cdot
y})\mathrm{d}\mu_{s(p)}(y)\right|\\\ \lesssim
2^{-k}\mathbf{1}_{E(p)}(x)|f|*\lambda_{s}(x)\,,$

using that $|\mu_{s}|\leq\lambda_{s}$ and $|e^{iN(x)\cdot
y}-e^{iN(\mathbf{T})\cdot y}|\lesssim 2^{-k}$ on the support of $\mu_{s(p)}$.
Summing over all tiles $p\in\mathbf{T}$, we obtain:

$|T_{\mathbf{T},\sigma^{\prime}}f(x)-e^{N(\mathbf{T})\cdot
x}T_{*\mathbf{T},\sigma^{\prime}}(e^{-N(\mathbf{T})\cdot y}f(y))(x)|\\\
\leq\sum_{p\in\mathbf{T}}\left|T_{p}f(x)-e^{iN(\mathbf{T})\cdot
x}\mathbf{1}_{E(p)}(x)\int
f(x-y)e^{-iN(\mathbf{T})\cdot(x-y)}\mathrm{d}\mu_{s(p)}(y)\right|\\\
\lesssim\sum_{k\geq 0}2^{-k}\sum_{\begin{subarray}{c}p\in\mathbf{T}\\\
s(p)=\max\sigma(x)-k\end{subarray}}\mathbf{1}_{E(p)}(x)|f|*\lambda_{s}(x)\,.$

By disjointness of the sets $E(p)$ for tiles $p$ of a fixed scale, the inner
sum is bounded by $M_{\mathbf{T}}$. This completes the proof. ∎

###  6.4. Low density trees and sparse bounds

Lemma 7 implies that the operators $T_{\mathbf{T},\sigma^{\prime}}$ are
bounded on $L^{2}$ uniformly over all generalized trees
$(\mathbf{T},\sigma^{\prime})$. We will now improve this estimate for trees of
small density $\delta$, using sparse bounds for the operators
$T_{*\mathbf{T},\sigma^{\prime}}$ and $M_{\mathbf{T}}$. These sparse bounds
are variants of bounds for prototypical singular Radon transforms that were
shown by Oberlin [Obe19]. Our setting is slightly more general than in
[Obe19], and we need a more precise estimate, but it still follows from
Oberlin’s proof with only minor modifications.

A collection $\mathcal{S}$ of dyadic cubes is called sparse if for each
$Q\in\mathcal{S}$ there exists a subset $U(Q)\subset Q$ with
$|U(Q)|\geq|Q|/2$, such that the sets $U(Q)$ are pairwise disjoint. For every
cube $Q$ and $p\geq 1$, we denote by

$\langle
f\rangle_{Q,p}:=\left(\frac{1}{|Q|}\int_{Q}|f|^{p}\,\mathrm{d}x\right)^{1/p}$

the $p$-average of a function $f$ over $Q$. Finally, given a dyadic cube $Q$
and a tree $\mathbf{T}$, we define

$E(Q)=E(\mathbf{T},Q):=\bigcup_{\begin{subarray}{c}p\in\mathbf{T}\\\
I(p)\subset 3Q\end{subarray}}E(p)\,.$

######  Lemma 8 ([Obe19]).

Suppose that $(p,q)$ are as in condition v) in Subsection 6.1. Then there
exists a constant $C>0$ such that for every generalized tree
$(\mathbf{T},\sigma^{\prime})$ and all $f,g$ there exists a sparse collection
of cubes $\mathcal{S}\subset\mathbf{D}$ such that

$\left\lvert\int
T_{*\mathbf{T},\sigma^{\prime}}f(x)g(x)\,\mathrm{d}x\right\rvert\leq
C\sum_{Q\in\mathcal{S}}|Q|\langle
f\rangle_{Q,p}\langle\mathbf{1}_{E(Q)}g\rangle_{3Q,q}\,.$ (39)

The same statement holds with $T_{*\mathbf{T},\sigma^{\prime}}$ replaced by
$M_{\mathbf{T}}$.

###### Proof.

This follows from the proof of Theorems 1.3 and 1.4 in [Obe19] (note that our
$q$ is his $q^{\prime}$), with the following modifications. Firstly, Oberlin
considers operators

$\sum_{s\in\mathbb{Z}}\varepsilon_{s}(x)f*\mu_{s}\,,$

with $\varepsilon_{s}(x)\in[-1,1]$, where $\mu_{s}$ is an isotropic dilate of
a fixed measure $\mu_{0}$. In our setting the $\mu_{s}$ are _anisotropic_
dilates of $D_{-s}\mu_{s}$, but Oberlin’s proof goes through in the
anisotropic setting as well. Furthermore, in our setting the measures
$D_{-s}\mu_{s}$ are not identical, they depend on $s$. However, they satisfy
all assumptions of Oberlin’s theorems uniformly in $s$, by i) - v) above, and
the proof still goes through with this assumption.

It remains to explain why we can insert $\mathbf{1}_{E(Q)}$ in the $q$-average
over $3Q$ in (39). We explain it for $T_{*\mathbf{T},\sigma^{\prime}}$, the
argument for $M_{\mathbf{T}}$ is very similar. Oberlin constructs the sparse
collection $\mathcal{S}$ by an iterative argument starting from a large cube
$Q_{0}$, such that $f$, $g$ are supported in $Q_{0}$, $3Q_{0}$ respectively.
The expression on the left hand side of (39) does not change if $f$ is
restricted to $I(\mathbf{T})$ and $g$ to $3I(\mathbf{T})$, so we can choose
$Q_{0}=I(\mathbf{T})$. Oberlin then defines operators $T_{Q}$, which in our
notation are

$T_{Q}f(x)=\sum_{\begin{subarray}{c}p\in\mathbf{T}\\\ s(p)\leq
s(Q)\end{subarray}}\mathbf{1}_{E(p)}(x)\mathbf{1}_{s(p)\in\sigma^{\prime}(x)}\cdot\mu_{s(p)}*(\mathbf{1}_{Q}f)(x)\,,$
(40)

so that $T_{*\mathbf{T},\sigma^{\prime}}=T_{Q_{0}}$. Note that in the sum in
(40) only tiles $p$ with $Q\cap 3I(p)\neq\emptyset$ contribute. But if
$s(p)\leq s(Q)$ and $Q\cap 3I(p)\neq\emptyset$, then $I(p)\subset 3Q$. Thus,
$T_{Q}=\mathbf{1}_{E(Q)}T_{Q}$, and hence

$\langle T_{Q}f,g\rangle=\langle T_{Q}f,\mathbf{1}_{E(Q)}g\rangle\,.$ (41)

Our goal is to estimate

$|\langle T_{Q_{0}}f,g\rangle|=|\langle
T_{Q_{0}}f,\mathbf{1}_{E(Q_{0})}g\rangle|\,.$

Oberlin shows that for every dyadic cube $Q$ and every $f$ there exists a
collection $Q_{1}(Q)$ of dyadic cubes such that

$|\langle T_{Q}f,g\rangle|\leq|Q|\langle f\rangle_{Q,p}\langle
g\rangle_{3Q,q}+\sum_{\begin{subarray}{c}Q^{\prime}\in Q_{1}(Q)\\\
Q^{\prime}\subset Q\end{subarray}}\langle T_{Q^{\prime}}f,g\rangle\,.$ (42)

This follows from his equations (3.2) and (3.3) and the claims below them. The
sparse collection $\mathcal{S}$ is then constructed by starting with $Q_{0}$
and iteratively adding for all $Q\in\mathcal{S}$ the cubes $Q_{1}(Q)$ to
$\mathcal{S}$. Combining (41) and (42) one obtains (39) for this
$\mathcal{S}$, and Oberlin shows that $\mathcal{S}$ is sparse. This completes
the proof. Note that the only change in our argument compared to Oberlin’s is
that we use (41) to insert $\mathbf{1}_{E(Q)}$ into the $q$-averages. ∎

######  Corollary 1.

Let $q$ be as in v) in Subsection 6.1. Then for each
$\varepsilon<\frac{1}{q}-\frac{1}{2}$ there exists $C>0$ such that for each
generalized tree $(\mathbf{T},\sigma^{\prime})$ with $\mathbf{T}$ of density
$\delta$, we have

$\|T_{*\mathbf{T},\sigma^{\prime}}\|_{2\to 2}\leq C\delta^{\varepsilon}\,.$

The same statement holds with $T_{*\mathbf{T},\sigma^{\prime}}$ replaced by
$M_{\mathbf{T}}$.

###### Proof.

Let $\mathbf{L}=\mathbf{L}(\mathbf{T})$ be the collection of maximal dyadic
cubes $L$ such that there exists some $p\in\mathbf{T}$ with $L\subset I(p)$
but there exists no $p\in\mathbf{T}$ with $I(p)\subset L$. This is a partition
of $I(\mathbf{T})$. Define

$E(L)=L\cap\bigcup_{p^{\prime}\in\mathbf{T}}E(p^{\prime})\,.$

We claim that for each $L\in\mathbf{L}$ we have

$|E(L)|\leq\delta|L|\,.$ (43)

To prove this, fix $L\in\mathbf{L}$. By definition of $\mathbf{L}$ there
exists $p\in\mathbf{T}$ with $I(p)\subset\hat{L}$. Define $p_{L}$ to be $p$ if
$I(p)=\hat{L}$, and else let $p_{L}$ be the unique tile with
$I(p_{L})=\hat{L}$ and $\omega(\mathbf{T})\subset\omega(p_{L})$. If
$p^{\prime}\in\mathbf{T}$ with $I(p^{\prime})\cap L\neq\emptyset$ then we have
$p_{L}\leq p^{\prime}$. Thus

$|E(L)|\leq|E(p_{L})|\leq\operatorname{\operatorname{dens}}(\mathbf{T})|L|\leq\delta|L|\,,$

giving the claim (43). Since $\mathbf{L}$ forms a partition of
$I(\mathbf{T})$, we have for all $Q\in\mathbf{D}$

$|E(Q)|\leq\sum_{L\subset 3Q}|E(L)|\leq\delta\sum_{L\subset 3Q}|L|\leq\delta
3^{d+2}|Q|\,.$

Thus, for every $\tilde{q}>q$, we have by Hölder’s inequality

$\langle\mathbf{1}_{E(Q)}g\rangle_{3Q,q}\leq(3^{d+2}\delta)^{1/q-1/\tilde{q}}\langle
g\rangle_{3Q,\tilde{q}}\,.$ (44)

Now pick $(p,q)$ as in v), and pick $\tilde{q}$ with $q<\tilde{q}<2$. We
obtain from Lemma 8 and (44)

$\displaystyle\left\lvert\int
T_{*\mathbf{T},\sigma^{\prime}}f(x)g(x)\,\mathrm{d}x\right\rvert$
$\displaystyle\lesssim\delta^{1/q-1/\tilde{q}}\sum_{Q\in\mathcal{S}}|Q|\langle
f\rangle_{Q,p}\langle g\rangle_{3Q,\tilde{q}}$ $\displaystyle\leq
2\delta^{1/q-1/\tilde{q}}\sum_{Q\in\mathcal{S}}|U(Q)|\langle
f\rangle_{Q,p}\langle g\rangle_{3Q,\tilde{q}}$ $\displaystyle\leq
2\delta^{1/q-1/\tilde{q}}\sum_{Q\in\mathcal{S}}\int_{U(Q)}M_{p}fM_{\tilde{q}}g$
$\displaystyle\lesssim\delta^{1/q-1/\tilde{q}}\|f\|_{2}\|g\|_{2}\,.$

Here we used boundedness of the $p$\- and $\tilde{q}$\- maximal function on
$L^{2}$ for $p,\tilde{q}<2$. This yields the desired result for
$T_{*\mathbf{T},\sigma^{\prime}}$.

For $M_{\mathbf{T}}$ the proof is exactly the same. ∎

Combining Corollary 1 with Lemma 7 we obtain an estimate for a single tree
with decay in the density parameter.

######  Corollary 2.

For each $\varepsilon<\frac{1}{q}-\frac{1}{2}$ there exist $C>0$ such for each
generalized tree $(\mathbf{T},\sigma^{\prime})$ with $\mathbf{T}$ of density
$\delta$, we have

$\|T_{\mathbf{T},\sigma^{\prime}}\|_{2\to 2}\leq C\delta^{\varepsilon}\,.$

Recall from Subsection 6.1 that for the measures $\mu_{s}$ defined by (37),
condition v) holds for all $p,q<2$ such that $(1/p,1/q)$ is in the interior of
the convex hull of $(1,0),(0,1),(\frac{d+1}{d+2},\frac{d+1}{d+2})$. In this
case the condition in Corollary 2 becomes
$\varepsilon<\frac{1}{2}-\frac{1}{2(d+1)}$.

##  7\. Forests: Proof of Proposition 3

To prove Proposition 3, it remains to combine the bounds for the operators
$T_{\mathbf{T}}$ from Proposition 4 for all trees in a forest $\mathbf{F}$.
For this, we will show almost orthogonality estimates for tree operators
associated to separated trees.

###  7.1. Basic orthogonality estimates

As a first step, we show that the adjoint generalized tree operator
$T_{\mathbf{T},\sigma^{\prime}}^{*}$ is frequency localized near
$N_{\mathbf{T}}$. We use the definitions from Subsection 6.2.

We let $\Psi$ denote the set of Schwartz functions $\psi$ such that

  1. (i)

$\psi$ is supported in $B(0,1)$,

  2. (ii)

$\psi$ has integral $1$,

  3. (iii)

$|\partial^{\alpha}\psi(x)|\leq L$ for all $|\alpha|\leq 10d$,

where $L$ is chosen sufficiently large such that for each $\xi$ with
$|\xi|\geq 1$, there exists $\psi\in\Psi$ with $\hat{\psi}(\xi)=0$.

For $\psi\in\Psi$, we define the frequency projection
$\Pi_{R,N}=\Pi_{R,N,\psi}$ by

$\widehat{\Pi_{R,N}f}(\xi)=(1-\hat{\psi}(R^{-1}(\xi-N)))\hat{f}(\xi)\,.$

######  Lemma 9.

For each $\kappa<\frac{1}{2(d+1)}$ there exist $C$ such that the following
holds. For each generalized tree $(\mathbf{T},\sigma^{\prime})$ such that each
tile in $\mathbf{T}$ has scale at least $0$, all $R\geq 1$ and all
$\psi\in\Psi$

$\|T_{\mathbf{T},\sigma^{\prime}}\Pi_{R,N(\mathbf{T})}\|_{2\to 2}\leq
CR^{-\kappa}\,.$

###### Proof.

We start by separating the $\log_{2}(R)$ largest scales:

$\displaystyle T_{\mathbf{T},\sigma^{\prime}}f(x)$
$\displaystyle=\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s\geq\overline{\sigma}^{\prime}(x)-\log_{2}(R)\end{subarray}}\int
f(x-y)e^{iN(x)\cdot y}\mathrm{d}\mu_{s}(y)$ (45)
$\displaystyle\quad+\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s<\overline{\sigma}^{\prime}(x)-\log_{2}(R)\end{subarray}}\int
f(x-y)e^{iN(x)\cdot y}\mathrm{d}\mu_{s}(y)\,.$ (46)

The small scale contribution (46) is close to the $N(\mathbf{T})$-modulated
operator:

$\left\lvert\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s<\overline{\sigma}^{\prime}(x)-\log_{2}(R)\end{subarray}}\int
f(x-y)(e^{-iN(x)\cdot y}-e^{-iN(\mathbf{T})\cdot
y})\,\mathrm{d}\mu_{s}(y)\right\rvert\\\
\lesssim\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s<\overline{\sigma}^{\prime}(x)-\log_{2}(R)\end{subarray}}2^{s-\overline{\sigma}^{\prime}(x)}\int|f(x-y)|\,\mathrm{d}|\mu_{s}|(y)\lesssim
R^{-1}M^{\lambda}f\,,$ (47)

where $M^{\lambda}$ denotes the maximal average associated to $\lambda$. Here
we have used that if $s\in\sigma^{\prime}(x)$ with
$s=\overline{\sigma}^{\prime}(x)-k$, then there exists a frequency cube of
scale $s+k$ containing $N(\mathbf{T})$ and $N(x)$, which implies that for all
$y$ in the support of $\mu_{s}$ we have $|(N(x)-N(\mathbf{T}))\cdot y|\lesssim
2^{-k}$. So to bound (46), it suffices to show the following estimate for the
$N(\mathbf{T})$-modulated operator:

$\left\|\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s<\overline{\sigma}^{\prime}(x)-\log_{2}(R)\end{subarray}}\int
e^{iN(\mathbf{T})\cdot(x-y)}(\Pi_{R,N(\mathbf{T})}f)(x-y)\,\mathrm{d}\mu_{s}(y)\right\|_{2}\lesssim
R^{-1/2}\|f\|_{2}\,.$

Replacing $f$ by $e^{iN(\mathbf{T})\cdot x}f(x)$ and taking differences it
suffices to show

$\left\lVert\sup_{0\leq\underline{\sigma}}\left\lvert\sum_{\underline{\sigma}\leq
s}\mathbf{1}_{J}(s)\int(\Pi_{R,0}f)(x-y)\,\mathrm{d}\mu_{s}(y)\right\rvert\right\rVert_{2}\lesssim
R^{-1/2}\|f\|_{2}\,.$

This follows from a standard square function argument, using that by (36) we
have for all $\underline{\sigma}\geq 0$

$|1-\hat{\psi}(R^{-1}\xi)|\sum_{\underline{\sigma}\leq
s}|\hat{\mu}_{s}(\xi)|\lesssim R^{-1/2}2^{-\underline{\sigma}/2}\,.$

Now we treat the large scales (45), using that only logarithmically many
scales contribute at each point. We discretize modulation frequencies. Let
$\gamma>0$, to be chosen later. For each dyadic frequency cube $\omega$, we
fix finite subsets $M(\omega)$ satisfying conditions (i), (ii), (iii) in
Section 5.3, with $\rho=R^{-\gamma}$. For each $s\in\mathbb{Z}$, let
$\omega_{s}$ be the unique frequency cube of scale $s$ containing
$N(\mathbf{T})$. For each $x$ and each $s\in\sigma^{\prime}(x)$, we pick a
frequency $c(x,s)\in M(\omega_{s})$ such that $d_{\text{par}}(N(x),c(x,s))\leq
R^{-\gamma}2^{-s}$. By a similar computation as in (47)

$\left\lvert\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s\geq\overline{\sigma}^{\prime}(x)-\log_{2}(R)\end{subarray}}\int
f(x-y)(e^{-iN(x)\cdot y}-e^{ic(x,s)\cdot
y})\,\mathrm{d}\mu_{s}(y)\right\rvert\\\
\leq\log_{2}(R)R^{-\gamma}M^{\lambda}f\,.$

Thus we may replace $N(x)$ by $c(x,s)$ in (45). We bound the resulting sum by
the number of nonzero summands, which is at most $\log_{2}(R)$ at each point
$x$, times the maximal summand. The maximal summand is controlled by the
estimate

$\left\lVert\sup_{s\geq 0}\sup_{c\in
M(s)}\left\lvert\int(\Pi_{R,N}f)(x-y)e^{ic\cdot
y}\,\mathrm{d}\mu_{s}(y)\right\rvert\right\rVert_{2}\lesssim
R^{\gamma\dim_{h}V}R^{-1/2}\|f\|_{2}\,.$

This follows from
$|(1-\hat{\psi}(R^{-1}(\xi-N+c)))|\lvert\hat{\mu}_{s}(\xi)\rvert\lesssim
R^{-1/2}2^{-s/2}$ and a standard square function argument, using that there
are at most $R^{\gamma\dim_{h}V}$ choices of $c$ for each $s$. Optimizing
$\gamma$, we obtain the lemma. ∎

An immediate corollary are almost orthogonality estimate for separated trees
with the same top cube.

######  Corollary 3.

For each $\kappa<\frac{1}{2d+3}$ there exists $C>0$ such that the following
holds. Let $\mathbf{T}_{1},\mathbf{T}_{2}$ be a pair of $\Delta$-separated,
normal trees with $I(\mathbf{T}_{1})=I(\mathbf{T}_{2})=:I$. Let
$(\mathbf{T}_{1},\sigma_{1}),(\mathbf{T}_{2},\sigma_{2})$ be generalized tree
operators, with possibly different $\mu_{s}$. Then

$\left\lvert\int
T_{\mathbf{T}_{1},\sigma_{1}}^{*}g_{1}\overline{T_{\mathbf{T}_{2},\sigma_{2}}^{*}g_{2}}\right\rvert\leq
C\Delta^{-\kappa}\|g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\,.$ (48)

###### Proof.

We drop the $\sigma_{i}$ from the notation and write
$T_{\mathbf{T}_{i},\sigma_{i}}=T_{\mathbf{T}_{i}}$. By scaling, we may assume
that the minimal scale of a tile in $\mathbf{T}_{1}\cup\mathbf{T}_{2}$ is $0$.
Furthermore we may assume that $\Delta>3$, otherwise (48) follows from $L^{2}$
boundedness of generalized tree operators. Then it holds by
$\Delta$-separation that $|N(\mathbf{T}_{1})-N(\mathbf{T}_{2})|\geq\Delta/3$.

We let $\kappa^{\prime}=\kappa/(1-\kappa)<1/(2(d+1))$ and
$\gamma=1/(1+\kappa^{\prime})$, define
$\Pi_{i}=\Pi_{\Delta^{\gamma},N(\mathbf{T}_{i})}$ for some function
$\psi\in\Psi$, and split up

$\langle
T_{\mathbf{T}_{1}}^{*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=\langle\Pi_{1}T_{\mathbf{T}_{1}}^{*}g_{1},\Pi_{2}T_{\mathbf{T}_{2}}^{*}g_{2}\rangle+\langle(1-\Pi_{1})T_{\mathbf{T}_{1}}^{*}g_{1},\Pi_{2}T_{\mathbf{T}_{2}}^{*}g_{2}\rangle\\\
+\langle\Pi_{1}T_{\mathbf{T}_{1}}^{*}g_{1},(1-\Pi_{2})T_{\mathbf{T}_{2}}^{*}g_{2}\rangle+\langle(1-\Pi_{1})T_{\mathbf{T}_{1}}^{*}g_{1},(1-\Pi_{2})T_{\mathbf{T}_{2}}^{*}g_{2}\rangle\,.$

The first three terms are bounded by
$\Delta^{-\gamma\kappa^{\prime}}\|g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}$ by
Lemma 9, and the last term is bounded by

$\|(1-\Pi_{1})(1-\Pi_{2})\|_{2\to
2}\|g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\lesssim\Delta^{\gamma-1}\|g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\,,$

because $\psi\in\Psi$ and $|N(\mathbf{T}_{1})-N(\mathbf{T}_{2})|\geq\Delta/3$.
This gives the desired estimate since $\gamma\kappa^{\prime}=\gamma-1=\kappa$.
∎

###  7.2. Auxiliary estimates for oscillatory integrals

In this subsection we prepare the proof of a version of Corollary 3 for a
general pair of separated trees by showing some estimates for oscillatory
integrals along paraboloids.

From now on, $\mu_{s}$ and $\lambda_{s}$ are fixed to be the measures defined
in (37) and (38). We define

$\chi_{s,\varepsilon}=\varepsilon^{-d-1}\mathbf{1}_{B(0,\varepsilon)}*(\lambda_{s-1}+\lambda_{s}+\lambda_{s+1})$
(49)

and
$\chi_{s,\varepsilon,\delta}=\delta^{-1}\chi_{s,\varepsilon}\mathbf{1}_{|x_{d}|<\delta
2^{s}}$. We also define the associated maximal convolution operators

$M^{\chi,1}f(x):=\sup_{s\in\mathbb{Z}}\sup_{\varepsilon>0}|f|*\tilde{\chi}_{s,\varepsilon}\,,$
(50)
$M^{\chi,2}f(x):=\sup_{s\in\mathbb{Z}}\sup_{\varepsilon>0}\sup_{\delta>0}|f|*\tilde{\chi}_{s,\varepsilon,\delta}\,,$
(51)

where $\tilde{\chi}(x)=\chi(-x)$. Both $M^{\chi,1}$ and $M^{\chi,2}$ are
bounded on $L^{2}$, because they are dominated by the composition of the
Hardy-Littlewood maximal function and maximal averages along the parabolas in
the direction of the coordinate axes
$\\{te_{i}+t^{2}e_{d+1}\,:\,t\in\mathbb{R}\\}$ for $i=1,\dotsc,d$.

######  Lemma 10.

There exists $C>0$ such that the following holds. For all $a,b>0$ with $b\geq
4a$, all $N\in V$, all $\psi\in\Psi$ and all $s\geq-1$ we have

$|(e^{iN\cdot y}\mu_{s}(y))*\psi_{a}(x)|\lesssim
b\chi_{s,a,b}(x)+\frac{1}{ab|N|}\chi_{s,a}(x)\,.$ (52)

Furthermore, if $\varphi\geq 0$ has integral $1$ and
$\mu_{s}^{l}=\mu_{s}*\varphi$, then

$|(e^{iN\cdot y}\mu_{s}^{l}(y))*\psi_{a}(x)|\lesssim
b\chi_{s,a,b}*\varphi(x)+\frac{1}{ab|N|}\chi_{s,a}*\varphi(x)\,.$ (53)

Here we write $\psi_{t}(x)=t^{-d-1}\psi(t^{-1}x)$.

###### Proof.

The estimate (53) follows from (52) and the triangle inequality. For
$|x_{d}|<b2^{s}$ estimate (52) also follows directly from the triangle
inequality. For $|x_{d}|\geq b2^{s}$ we distinguish cases depending on $V$.

If $V=\mathbb{R}^{d-1}\times\\{0\\}^{2}$ then $N_{d}=N_{d+1}=0$, and without
loss of generality we have that $|N_{d-1}|\geq|N|/d$. We let
$\tilde{y}=(y_{1},\dotsc,y_{d-2})$ and put

$y=(\tilde{y},\sqrt{h^{2}-|\tilde{y}|^{2}}\cos(\theta),\sqrt{h^{2}-|\tilde{y}|^{2}}\sin(\theta))\,.$

We abbreviate $r=\sqrt{h^{2}-|\tilde{y}|^{2}}$ and change variables to obtain

$\int\psi_{a}(x-z)e^{iN\cdot z}\,\mathrm{d}\mu_{s}(z)\\\
=\int\psi_{a}(x^{\prime}-y,x_{d+1}-|y|^{2})e^{iN(x)\cdot(y,|y|^{2})}K_{s}(y)\,\mathrm{d}y\\\
=\int\\!\\!\int\\!\\!\int\Gamma_{h,\tilde{y}}(\theta)e^{i\phi_{h,\tilde{y}}(\theta)}\,\mathrm{d}\theta\,h\,\mathrm{d}h\,\mathrm{d}\tilde{y}\,,$
(54)

where we put $\phi_{h,\tilde{y}}(\theta)=N_{d-1}r\cos(\theta)$ and

$\Gamma_{h,\tilde{y}}(\theta)=\psi_{a}(\tilde{x}-\tilde{y},x_{d-1}-r\cos(\theta),x_{d}-r\sin(\theta),x_{d+1}-h)\\\
K_{s}(\tilde{y},r\cos(\theta),r\sin(\theta))e^{i\tilde{N}\cdot\tilde{y}}\,.$

Using (5) and the assumption $s\geq-1$, we find that

$\left|\Gamma^{\prime}_{h,\tilde{y}}(\theta)\right|\lesssim
a^{-d-2}2^{-ds}r\,.$ (55)

Since $|x_{d}|>b2^{s}$ and since $|x_{d}-r\sin(\theta)|\leq a$ on the support
of $\Gamma_{h,\tilde{y}}$, we have additionally that $|\sin(\theta)|\geq
b-2^{-s}a\geq b/2$. Thus

$\left|\phi^{\prime}_{h,\tilde{y}}(\theta)\right|\geq\frac{|N|br}{2d}\qquad\text{and}\qquad\left|\phi^{\prime\prime}_{h,\tilde{y}}(\theta)\right|\leq\frac{|N|r}{d}\,.$
(56)

Integrating by parts in (54), we obtain with (55) and (56)

$\displaystyle\eqref{eq ibp prep}$
$\displaystyle\leq\int\\!\\!\int\\!\\!\int\frac{|\Gamma_{h,\tilde{y}}^{\prime}(\theta)|}{|\phi_{h,\tilde{y}}^{\prime}(\theta)|}+\frac{|\Gamma_{h,\tilde{y}}(\theta)|}{|\phi_{h,\tilde{y}}^{\prime}(\theta)|^{2}}|\phi_{h,\tilde{y}}^{\prime\prime}(\theta)|\,\mathrm{d}\theta
h\,\mathrm{d}h\,\mathrm{d}\tilde{y}$
$\displaystyle\lesssim\int\\!\\!\int\\!\\!\int(\mathbf{1}_{B(0,1)})_{a}(\tilde{x}-\tilde{y},x_{d-1}-r\cos(\theta),x_{d}-r\sin(\theta),x_{d+1}-h)$
$\displaystyle\qquad\qquad\qquad
2^{-ds}(\frac{1}{ab|N|}+\frac{1}{b^{2}|N|})\mathbf{1}_{2^{s-3}<|y|<2^{s}}\,\mathrm{d}\theta\,h\,\mathrm{d}h\,\mathrm{d}\tilde{y}$
$\displaystyle\lesssim\int_{2^{s-3}<|y|<2^{s}}2^{-ds}\frac{1}{ab|N|}(\mathbf{1}_{B(0,1)})_{a}(x^{\prime}-y,x_{d+1}-|y|^{2})\,\mathrm{d}y$
$\displaystyle\lesssim\frac{1}{ab|N|}\chi_{s,a}\,.$

Now we assume that $V=\\{0\\}^{d}\times\mathbb{R}$. Then we have
$N^{\prime}=0$ and hence

$\int\psi_{a}(x-z)e^{iN\cdot
z}\,\mathrm{d}\mu_{s}(z)=\int\psi_{a}(x^{\prime}-y,x_{d+1}-|y|^{2})e^{iN_{d+1}|y|^{2}}K_{s}(y)\,\mathrm{d}y\\\
=\int\Gamma(t)e^{iN_{d+1}t}\,\mathrm{d}t\,,$ (57)

where this time we have set

$\Gamma(t)=\int\psi_{a}(x^{\prime}-y,x_{d+1}-t)K_{s}(y)\delta(t-|y|^{2})\,\mathrm{d}y\,.$

Using (5) and that $s\geq-1$, one finds that
$\left|\Gamma^{\prime}(t)\right|\lesssim a^{-d-2}2^{-ds}$. Integrating by
parts in (57) gives the desired estimate (52) after a similar, but simpler
computation as in the case $V=\mathbb{R}^{d-1}\times\\{0\\}^{2}$. ∎

######  Lemma 11.

There exists $C>0$ such that the following holds. Let $s\in\mathbb{Z}$, $a>0$
and $b>2^{-s}$. Let $\psi,\varphi\in\Psi$ with $\hat{\psi}(-aN)=0$. Define

$\varphi_{b,s}(x)=(2^{s}b)^{-d-2}\varphi(2^{-s}b^{-1}x^{\prime},2^{-2s}b^{-2}x_{d+1})$

and $\mu_{s}^{l}=\mu_{s}*\varphi_{b,s}$. Then we have that

$|\psi_{a}*(e^{iN\cdot y}\mu_{s}^{l}(y))|\leq
Ca2^{-s}b^{-(d+3)}2^{-(d+2)s}\mathbf{1}_{[-2^{s},2^{s}]^{d}\times[-2^{2s},2^{2s}]}(y)\,.$
(58)

###### Proof.

We have that

$\displaystyle|\psi_{a}*e^{iN\cdot y}\mu_{s}^{l}(y)|$
$\displaystyle\lesssim\int|\hat{\psi}(a\xi-
aN)||\hat{\varphi}_{b,s}(\xi)|\,\mathrm{d}\xi$ $\displaystyle\lesssim
a\int|\xi||\hat{\varphi}_{b,s}(\xi)|\,\mathrm{d}\xi$
$\displaystyle=a2^{-(d+3)s}b^{-(d+3)}\int|(\xi^{\prime},2^{-s}b^{-1}\xi_{d+1})||\hat{\varphi}(\xi)|\,\mathrm{d}\xi\,.$

The lemma now follows since $2^{-s}b^{-1}<1$ and
$\||\xi|\hat{\varphi}(\xi)\|_{1}\lesssim 1$ for $\psi\in\Psi$, and since the
left hand side of (58) is clearly supported in
$[-2^{s},2^{s}]^{d}\times[-2^{2s},2^{2s}]$. ∎

###  7.3. Main almost orthogonality estimate for separated trees

######  Lemma 12.

There exists $C>0$ such that the following holds. Let
$\mathbf{T}_{1},\mathbf{T}_{2}$ be a pair of $\Delta$-separated, normal trees.
Then

$\left\lvert\int
T_{\mathbf{T}_{1}}^{*}g_{1}\overline{T_{\mathbf{T}_{2}}^{*}g_{2}}\right\rvert\leq
C\Delta^{-\frac{1}{10d}}\|W_{\mathbf{T}_{1}}g_{1}\|_{L^{2}(I(\mathbf{T}_{1})\cap
I(\mathbf{T}_{2}))}\|W_{\mathbf{T}_{2}}g_{2}\|_{L^{2}(I(\mathbf{T}_{1})\cap
I(\mathbf{T}_{2}))}$

for certain operators $W_{\mathbf{T}_{i}}$ depending only on $\mathbf{T}_{i}$
with

$\|W_{\mathbf{T}_{i}}\|_{2\to 2}\leq 1\,.$

###### Proof.

Since the trees are normal, the left hand side vanishes if
$I(\mathbf{T}_{1})\cap I(\mathbf{T}_{2})=\emptyset$. Thus we may assume
without loss of generality that $I:=I(\mathbf{T}_{2})\subset
I(\mathbf{T}_{1})$. Note also that we can always assume that $\Delta$ is
sufficiently large, by adding $|T_{\mathbf{T}_{i}}^{*}|$ to
$W_{\mathbf{T}_{i}}$. Finally, we assume that the minimal scale of a tile in
$\mathbf{T}_{2}$ is $0$, by scaling.

We fix $\gamma:=\frac{2(d+1)}{(2d+1)(2d+7)}>\frac{2}{10d}>0$ and
$\varphi\in\Psi$ and define
$\mu_{s}^{l}:=\mu_{s}*\varphi_{\Delta^{-\gamma},s}$, where

$\varphi_{\Delta^{-\gamma},s}(x)=2^{-(d+2)s}\Delta^{\gamma(d+2)}\varphi(2^{-s}\Delta^{\gamma}x^{\prime},2^{-2s}\Delta^{2\gamma}x_{d+1})\,.$

Then we decompose $T_{\mathbf{T}_{1}}$ into a large scales part, a smoothened
small scales part and an error term:

$\displaystyle T_{\mathbf{T}_{1}}f$
$\displaystyle=\sum_{\begin{subarray}{c}s\in\sigma(x)\\\
s>\overline{\sigma}(x)-\log_{2}(\Delta)\end{subarray}}\int f(x-y)e^{iN(x)\cdot
y}\,\mathrm{d}\mu_{s}(y)$
$\displaystyle\quad+\sum_{\begin{subarray}{c}s\in\sigma(x)\\\
s\leq\overline{\sigma}(x)-\log_{2}(\Delta)\end{subarray}}\int
f(x-y)e^{iN(\mathbf{T}_{1})\cdot y}\,\mathrm{d}\mu_{s}^{l}(y)$
$\displaystyle\quad+\mathcal{E}_{\mathbf{T}_{1}}(f)$
$\displaystyle=:T_{\mathbf{T}_{1}}^{1}f+T_{\mathbf{T}_{1}}^{2}f+\mathcal{E}_{\mathbf{T}_{1}}(f)\,.$

Notice that $\mu_{s},\lambda_{s}$ are supported in $\delta_{s}(B(0,1/3))$, so
for $\Delta$ sufficiently large the measures $\mu_{s}^{l}$ and
$\lambda_{s}^{l}$ also satisfy conditions i) to v). Thus both
$T_{\mathbf{T}_{1}}^{1}$ and $T_{\mathbf{T}_{1}}^{2}$ are generalized tree
operators.

We first show that the error term $\mathcal{E}_{\mathbf{T}_{1}}$ is small. We
have

$\displaystyle|\mathcal{E}_{\mathbf{T}_{1}}(f)|$
$\displaystyle\leq\left\lvert\sum_{\begin{subarray}{c}s\in\sigma(x)\\\
s\leq\overline{\sigma}(x)-\log_{2}(\Delta)\end{subarray}}\int
f(x-y)(e^{iN(x)\cdot y}-e^{iN(\mathbf{T}_{1})\cdot
y})\,\mathrm{d}\mu_{s}(y)\right\rvert$
$\displaystyle\quad+\left\lvert\sum_{\begin{subarray}{c}s\in\sigma(x)\\\
s\leq\overline{\sigma}(x)-\log_{2}(\Delta)\end{subarray}}\int
f(x-y)e^{iN(\mathbf{T}_{1})\cdot
y}\,\mathrm{d}(\mu_{s}-\mu_{s}^{l})(y)\right\rvert$
$\displaystyle\lesssim\sum_{\begin{subarray}{c}s\in\sigma(x)\\\
s\leq\overline{\sigma}(x)-\log_{2}(\Delta)\end{subarray}}2^{s-\overline{\sigma}(x)}\int|f(x-y)|\,\mathrm{d}|\mu_{s}|(y)$
$\displaystyle\quad+\sup_{\underline{s}<\overline{s}}\left\lvert\sum_{\begin{subarray}{c}\underline{s}<s\leq\overline{s}\\\
s\in J\end{subarray}}(\mu_{s}-\mu_{s}^{l})*(f(y)e^{iN(\mathbf{T}_{1})\cdot
y})\right\rvert\,.$

Here $J=J(\mathbf{T}_{1})$ is as in Subsection 6.2. The first term is bounded
by $\Delta^{-1}M^{\lambda}f(x)$. The second term is the maximally truncated
singular integral associated to the single scale operators
$\mathbf{1}_{J}(s)(\mu_{s}-\mu_{s}^{l})$, applied to
$e^{iN(\mathbf{T}_{1})\cdot y}f(y)$. Since

$|\hat{\mu}_{s}(\xi)-\hat{\mu}_{s}^{l}(\xi)|\leq\min\\{|\xi|^{-1},|\Delta^{-\gamma}\xi|\\}\,,$

this operator has norm $\lesssim\Delta^{-\gamma/2}$ on $L^{2}$. Thus we have

$\|\mathcal{E}_{\mathbf{T}_{1}}(f)\|_{2}\lesssim\Delta^{-\gamma/2}\|f\|_{2}\leq\Delta^{-\frac{1}{10d}}\|f\|_{2}\,.$

It now remains to bound $\langle
T_{\mathbf{T}_{1}}^{1*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle+\langle
T_{\mathbf{T}_{1}}^{2*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle$. We decompose
$\mathbf{T}_{1}=\mathbf{T}_{1}^{\prime}\cup\mathbf{T}_{1}^{\prime\prime}\cup\mathbf{T}_{1}^{\prime\prime\prime}$,
where

$\mathbf{T}_{1}^{\prime}=\\{p\in\mathbf{T}_{1}\,:\,s(p)<-1\,,3I(p)\subset
I\\}\,,$
$\mathbf{T}_{1}^{\prime\prime}=\\{p\in\mathbf{T}_{1}\,:\,s(p)\geq-1,\,3I(p)\cap
I\neq\emptyset\\}$

and
$\mathbf{T}_{1}^{\prime\prime\prime}=\mathbf{T}_{1}\setminus(\mathbf{T}_{1}^{\prime}\cup\mathbf{T}_{1}^{\prime\prime})$.
Then $\mathbf{T}_{1}^{\prime}$ is a normal tree with top $I$. For all
$p\in\mathbf{T}_{1}^{\prime\prime\prime},p^{\prime}\in\mathbf{T}_{2}$ we have
$2I(p)\cap(2+\frac{1}{4})I(p^{\prime})=\emptyset$, this follows from normality
of $\mathbf{T}_{2}$ and the fact that all tiles in
$\mathbf{T}_{1}^{\prime\prime\prime}$ with $3I(p)\cap I\neq\emptyset$ have
scale at most $s(p^{\prime})-2$. Thus $\langle
T_{\mathbf{T}_{1}^{\prime\prime\prime}}^{*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=0$.

According to the decomposition of $\mathbf{T}_{1}$ we split
$T_{\mathbf{T}_{1}}^{1}=T_{\mathbf{T}_{1}^{\prime}}^{1}+T_{\mathbf{T}_{1}^{\prime\prime}}^{1}+T_{\mathbf{T}_{1}^{\prime\prime\prime}}^{1}$,
and similarly for $T_{\mathbf{T}_{1}}^{2}$. We still have $\langle
T_{\mathbf{T}_{1}^{\prime\prime\prime}}^{1*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=0$,
thus

$\langle
T_{\mathbf{T}_{1}}^{1*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=\langle
T_{\mathbf{T}_{1}^{\prime}}^{1*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle+\langle
T_{\mathbf{T}_{1}^{\prime\prime}}^{1*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle\,.$
(59)

For the first summand in (59) we obtain with Corollary 3

$|\langle
T_{\mathbf{T}_{1}^{\prime}}^{1*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle|\lesssim\Delta^{-\frac{1}{10d}}\|g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\,.$
(60)

For the second summand in (59) we put $\nu=\frac{d+1}{2d+1}$ and choose
$\Pi_{2}=\Pi_{\Delta^{\nu},N(\mathbf{T}_{2})}$ for some function
$\psi\in\Psi$. If $\Delta$ is sufficiently large then
$\Pi_{2}T_{p^{\prime}}^{*}g_{2}$ is supported in
$(2+\frac{1}{4})I(p^{\prime})$ for each tile $p^{\prime}\in\mathbf{T}_{2}$,
thus we also have $\langle
T_{\mathbf{T}_{1}^{\prime\prime\prime}}^{1*}g_{1},\Pi_{2}T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=0$.
Using this, we expand

$\langle
T_{\mathbf{T}_{1}^{\prime\prime}}^{1*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=\langle
g_{1},T_{\mathbf{T}_{1}^{\prime\prime}}^{1}(1-\Pi_{2})T_{\mathbf{T}_{2}}^{*}g_{2}\rangle+\langle
T_{\mathbf{T}_{1}}^{1*}g_{1}-T_{\mathbf{T}_{1}^{\prime}}^{1*}g_{1},\Pi_{2}T_{\mathbf{T}_{2}}^{*}g_{2}\rangle\,.$
(61)

The second term in (61) is by Cauchy-Schwarz, Lemma 9 and $L^{2}$-boundedness
of
$T_{\mathbf{T}_{1}^{\prime}}^{1*}=T_{\mathbf{T}_{1}^{\prime}}^{1*}\mathbf{1}_{I(\mathbf{T}_{2})}$
bounded by

$\Delta^{-\frac{1}{10d}}\||T_{\mathbf{T}_{1}}^{1*}g_{1}|+|g_{1}|\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\,.$
(62)

For the first term in (61) we may assume that $N(\mathbf{T}_{2})=0$. Then we
have

$T_{\mathbf{T}_{1}^{\prime\prime}}^{1}(1-\Pi_{2})f(x)=\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s>\overline{\sigma}(x)-\log_{2}(\Delta)\end{subarray}}(e^{iN(x)\cdot
y}\mu_{s}(y))*\psi_{\Delta^{-\nu}}*f(x)\,,$

where $\sigma^{\prime}(x)=\\{s(p)\,:\,x\in
E(p),p\in\mathbf{T}_{1}^{\prime\prime}\\}$. Using the estimate for the
convolution kernels proven in Lemma 10 with $a=\Delta^{-\nu}$,
$b=\Delta^{-1/2}$, we obtain with the notation defined at (49)

$|(e^{iN(x)\cdot y}\mu_{s}(y))*\psi_{\Delta^{-\nu}}*f(x)|\\\
\lesssim(\Delta^{-1/2}\chi_{s,\Delta^{-\nu},\Delta^{-1/2}}+\Delta^{\nu-1/2}\chi_{s,\Delta^{-\nu}})*|f|(x)\,.$

Passing the convolutions to the other side in the inner product, we find that
the first term in (61) is bounded by

$\log(\Delta)\Delta^{\nu-1/2}\langle
M^{1,\chi}g_{1}+M^{2,\chi}g_{2},|T_{\mathbf{T}_{2}}^{*}g_{2}|\rangle\\\
\lesssim\Delta^{-\frac{1}{10d}}\|M^{1,\chi}g_{1}+M^{2,\chi}g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}$

where $M^{1,\chi}$, $M^{2,\chi}$ are defined in (50) and (51). Since
$M^{1,\chi}$ and $M^{2,\chi}$ are bounded on $L^{2}$, this gives the desired
estimate for $T_{\mathbf{T}_{1}}^{1}$.

Now we turn to $\langle
T_{\mathbf{T}_{1}}^{2*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle$. As above, we
have

$\langle
T_{\mathbf{T}_{1}}^{2*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=\langle
T_{\mathbf{T}_{1}^{\prime}}^{2*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle+\langle
T_{\mathbf{T}_{1}^{\prime\prime}}^{2*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle\,.$
(63)

For the first term in (63) we have again from Corollary 3 that

$|\langle
T_{\mathbf{T}_{1}^{\prime}}^{2*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle|\lesssim\Delta^{-\frac{1}{10d}}\|g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\,.$
(64)

For the second term in (63) we split as before

$\langle
T_{\mathbf{T}_{1}^{\prime\prime}}^{2*}g_{1},T_{\mathbf{T}_{2}}^{*}g_{2}\rangle=\langle
g_{1},T_{\mathbf{T}_{1}^{\prime\prime}}^{2}(1-\Pi_{2})T_{\mathbf{T}_{2}}^{*}g_{2}\rangle+\langle
T_{\mathbf{T}_{1}}^{2*}g_{1}-T_{\mathbf{T}_{1}^{\prime}}^{2*}g_{1},\Pi_{2}T_{\mathbf{T}_{2}}^{*}g_{2}\rangle\,.$
(65)

The second term in (65) is by Cauchy-Schwarz and Lemma 9 bounded by

$\Delta^{-\frac{1}{10d}}\||T_{\mathbf{T}_{1}}^{2*}g_{1}|+|g_{1}|\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\,.$
(66)

For the first term in (65), we assume again that $N(\mathbf{T}_{2})=0$. Then

$\displaystyle
T_{\mathbf{T}_{1}^{\prime\prime}}^{2}(1-\Pi_{2})f(x)=\sum_{\begin{subarray}{c}s\in\sigma^{\prime}(x)\\\
s\leq\overline{\sigma}(x)-\log_{2}(\Delta)\end{subarray}}\psi_{\Delta^{-\nu}}*(e^{iN(\mathbf{T}_{1})\cdot
y}\mu_{s}^{l}(y))*f(x)\,.$

By our assumptions on $\Psi$, we may then further assume that we have
$\hat{\psi}(\Delta^{-\nu}N(\mathbf{T}_{1}))=0$, since for $\Delta$
sufficiently large $|\Delta^{-\nu}N(\mathbf{T}_{1})|\gtrsim\Delta^{1-\nu}\geq
1$. By Lemma 11 with $a=\Delta^{-\nu}$, $b=\Delta^{-\gamma}$ we have

$\sum_{s>\gamma\log_{2}(\Delta)}|\psi_{\Delta^{-\nu}}*(e^{iN(\mathbf{T}_{1})\cdot
y}\mu_{s}^{l}(y))*f(x)|\lesssim\Delta^{(d+3)\gamma-\nu}\Phi*|f|(x)\,,$

where $\Phi(y)=\sum_{s\geq
0}2^{-s(d+3)}\mathbf{1}_{[-2^{s},2^{s}]^{d}\times[-2^{2s},2^{2s}]}(y)$. By
Lemma 10 with $a=\Delta^{-\nu}$, $b=\Delta^{-1/2}$ we have for $-1\leq
s\leq\gamma\log(\Delta)$ that

$|\psi_{\Delta^{-\nu}}*(e^{iN(\mathbf{T}_{1}\cdot y}\mu_{s}^{l}(y))*f(x)|\\\
\lesssim(\Delta^{-1/2}\chi_{s,\Delta^{-\gamma},\Delta^{-1/2}}+\Delta^{\nu-1/2}\chi_{s,\Delta^{-\gamma}})*M|f|(x)\,.$

Similarly as for $T_{\mathbf{T}_{1}}^{1}$, we obtain from these estimates that
the first term in (65) is bounded by

$\Delta^{-\frac{1}{10d}}\|Mg_{1}+M^{1,\chi}g_{1}+M^{2,\chi}g_{1}\|_{L^{2}(I)}\|g_{2}\|_{L^{2}(I)}\,.$

This completes the proof. ∎

###  7.4. Completing the argument for forests

Here we use Lemma 12 to complete the proof of Proposition 3. We follow the
presentation in [Zor21].

A row is a union of normal trees with pairwise disjoint top cubes.

######  Lemma 13.

There exists $C>0$ such that the following holds. Let
$\mathbf{R}_{1},\mathbf{R}_{2}$ be two rows such that the trees in
$\mathbf{R}_{1}$ are $\Delta$-separated from the trees in $\mathbf{R}_{2}$.
Then

$\left\lvert\int
T_{\mathbf{R}_{1}}^{*}g_{1}\overline{T_{\mathbf{R}_{2}}^{*}g_{2}}\right\rvert\leq
C\Delta^{-\frac{1}{10d}}\|g_{1}\|_{2}\|g_{2}\|_{2}\,.$

###### Proof.

We have by Lemma 12, with $W_{\mathbf{T}_{i}}$ as defined there:

$\left\lvert\int
T_{\mathbf{R}_{1}}^{*}g_{1}\overline{T_{\mathbf{R}_{2}}^{*}g_{2}}\right\rvert\leq\sum_{\mathbf{T}_{1}\in\mathbf{R}_{1}}\sum_{\mathbf{T}_{2}\in\mathbf{R}_{2}}\left\lvert\int
T_{\mathbf{T}_{1}}^{*}\mathbf{1}_{I(\mathbf{T}_{1})}g_{1}\overline{T_{\mathbf{T}_{2}}^{*}\mathbf{1}_{I(\mathbf{T}_{2})}g_{2}}\right\rvert\\\
\leq
C\Delta^{-\frac{1}{10d}}\sum_{\mathbf{T}_{1}\in\mathbf{R}_{1}}\sum_{\mathbf{T}_{2}\in\mathbf{R}_{2}}\prod_{i=1,2}\|W_{\mathbf{T}_{i}}\mathbf{1}_{I(\mathbf{T}_{i})}g_{i}\|_{L^{2}(I(\mathbf{T}_{1})\cap
I(\mathbf{T}_{2}))}\,.$

Using Cauchy-Schwarz, disjointness of the cubes $I(\mathbf{T}_{i})$ for
$\mathbf{T}_{i}\in\mathbf{R}_{i}$ and $\|W_{\mathbf{T}_{i}}\|_{2\to 2}\lesssim
1$, we estimate:

$\displaystyle\leq
C\Delta^{-\frac{1}{10d}}\prod_{i=1,2}\left(\sum_{\mathbf{T}_{1}\in\mathbf{R}_{2}}\sum_{\mathbf{T}_{2}\in\mathbf{R}_{2}}\|W_{\mathbf{T_{i}}}\mathbf{1}_{I(\mathbf{T}_{i})}g_{i}\|^{2}_{L^{2}(I(\mathbf{T}_{1})\cap
I(\mathbf{T}_{2}))}\right)^{1/2}$ $\displaystyle\leq
C\Delta^{-\frac{1}{10d}}\prod_{i=1,2}\left(\sum_{\mathbf{T}_{i}\in\mathbf{R}_{i}}\|W_{\mathbf{T_{i}}}\mathbf{1}_{I(\mathbf{T}_{i})}g_{i}\|^{2}_{L^{2}}\right)^{1/2}$
$\displaystyle\leq C\Delta^{-\frac{1}{10d}}\|g_{1}\|_{2}\|g_{2}\|_{2}\,.\qed$

###### Proof of Proposition 3.

Let $\mathbf{F}$ be an $n$-forest. By the overlap estimate (10), we can
decompose $\mathbf{F}$ into $2^{n}\log(n+2)$ rows $\mathbf{R}_{i}$. Since
$\mathbf{F}$ is an $n$-forest, the trees in different rows are $2^{10dn}$
separated and have density at most $2^{-n}$. Note also that separation of the
rows implies that the sets $E(\mathbf{R}_{i}):=\cup_{p\in\mathbf{R}_{i}}E(p)$
are pairwise disjoint.

By orthogonality and Corollary 2, we have that

$\|T_{\mathbf{R}_{i}}\|_{2\to 2}\lesssim_{\varepsilon}2^{-\epsilon n}$

for each $i$ and each $\varepsilon<\frac{1}{2}-\frac{1}{2(d+1)}$. Using this
and Lemma 13 yields

$\|T_{\mathbf{F}}^{*}g\|_{2}^{2}\leq\sum_{i,j}\left\lvert\int
T_{\mathbf{R}_{i}}^{*}\mathbf{1}_{E(\mathbf{R}_{i})}gT_{\mathbf{R}_{j}}^{*}\mathbf{1}_{E(\mathbf{R}_{j})}g\right\rvert\\\
\lesssim(2^{-2\varepsilon
n}+2^{n}\log(n+2)2^{-\frac{1}{10d}10dn})\sum_{i}\|\mathbf{1}_{E(\mathbf{R}_{i})}g\|_{2}^{2}\\\
\lesssim 2^{-2\varepsilon n}\|g\|_{2}^{2}\,.$ (67)

This completes the proof. ∎

##  8\. $L^{p}$-bounds: Proof of Theorem 1

Theorem 1 follows from interpolation, Proposition 1 and the following
$L^{p}$-estimates for antichain and forest operators replacing Proposition 2
and Proposition 3.

######  Proposition 5.

For each $1<p<\infty$ there exists $\varepsilon=\varepsilon(d,p)>0$ and $C>0$
such that for every antichain $\mathbf{A}$ of density $\delta$

$\|T_{\mathbf{A}}\|_{p\to p}\leq C\delta^{-\varepsilon}\,.$

###### Proof.

This follows from interpolation between Proposition 2 and the trivial bound
$\|T_{\mathbf{A}}\|_{p\to p}\leq C(p)$, which holds since $T_{\mathbf{A}}$ is
dominated pointwise by the maximal average along the paraboloid. ∎

######  Proposition 6.

Let $p$ be such that

$\frac{d^{2}+4d+2}{(d+1)^{2}}<p<2(d+1)\,.$ (68)

There exists $C>0$ such that the following holds. Let for each $n\geq 0$ and
$1\leq l\leq(n+1)^{2}$ an $n$-forest $\mathbf{F}_{n,l}$ of normal trees be
given. Then

$\bigg{\|}\sum_{n\geq
0}\sum_{l=1}^{(n+1)^{2}}T_{\mathbf{F}_{n,l}}\bigg{\|}_{p\to p}\leq C\,.$ (69)

###### Proof.

Suppose first that $p<2$. We will show that for all
$0\leq\alpha<d^{2}/(2(d^{2}+4d+2))$ there exists $\varepsilon>0$ such that for
each $G\subset\mathbb{R}^{d+1}$ with $0<|G|<\infty$ it holds with
$\tilde{G}=\\{x\,:\,M\mathbf{1}_{G}>t\\}$ that

$\|\mathbf{1}_{\mathbb{R}^{d+1}\setminus\tilde{G}}T_{\mathbf{F}_{n,l}}\mathbf{1}_{G}\|_{2\to
2}\lesssim t^{\alpha}2^{-\varepsilon n}\,.$ (70)

Using Bateman’s extrapolation argument from [BT13] (see also [Zor21], Appendix
B), (70) implies that for $p<2$ satisfying (68)

$\|T_{\mathbf{F}_{n,l}}\|_{L^{p,1}\to L^{p,\infty}}\lesssim 2^{-\varepsilon
n}\,.$

Then (69) follows by interpolation and summation in $n$.

To prove (70), recall that by Lemma 8 for each tree $\mathbf{T}$, all
$p>1+\frac{1}{d+1}$ and all $f,g$, there exists a sparse collection
$\mathcal{S}$ with

$|\langle\mathbf{1}_{\mathbb{R}^{d+1}\setminus\tilde{G}}T_{*\mathbf{T}}\mathbf{1}_{G}f,g\rangle|\lesssim\sum_{Q\in\mathcal{S}}|Q|\langle\mathbf{1}_{G}f\rangle_{p,Q}\langle\mathbf{1}_{(\mathbb{R}^{d+1}\setminus\tilde{G})\cap
E(Q)}g\rangle_{p,3Q}\,.$ (71)

If $3Q\cap(\mathbb{R}^{d+1}\setminus\tilde{G})\neq\emptyset$, then $|Q\cap
G|\lesssim t|Q|$. Using this and Hölder’s inequality, we obtain with a similar
argument as in the proof of Corollary 1 that for each
$\varepsilon<\frac{1}{2}-\frac{1}{d+2}$ and each tree $\mathbf{T}$ of density
at most $2^{-n}$

$\|\mathbf{1}_{\mathbb{R}^{d+1}\setminus\tilde{G}}T_{*\mathbf{T}}\mathbf{1}_{G}\|_{2\to
2}\lesssim_{\varepsilon}t^{\varepsilon}2^{-\varepsilon n}\,.$

By the same argument, the same holds for $M_{\mathbf{T}}$ and hence for
$T_{\mathbf{T}}$. By orthogonality, the same estimate holds for row operators
$T_{\mathbf{R}}$. Finally, each forest $\mathbf{F}_{n,l}$ can be decomposed
into at most $2^{n}\log(n+2)$ rows, so

$\|\mathbf{1}_{\mathbb{R}^{d+1}\setminus\tilde{G}}T_{\mathbf{F}_{n,l}}\mathbf{1}_{G}\|_{2\to
2}\leq 2^{n/2-\varepsilon n}\log(n+2)^{1/2}t^{\varepsilon}\,.$

We obtain (70) by taking a geometric average of this estimate and (67).

For the case $p>2$ we use that for all
$0\leq\alpha<\frac{1}{2}-\frac{1}{2(d+1)}$, there exists $C>0$ such that for
each $G\subset\mathbb{R}^{d+1}$ with $0<|G|<\infty$

$\|\mathbf{1}_{G}\sum_{n\geq
0}\sum_{l=1}^{(n+1)^{2}}T_{\mathbf{F}_{n,l}}\mathbf{1}_{\mathbb{R}^{d+1}\setminus\tilde{G}}\|_{2\to
2}\leq Ct^{\alpha}\,.$ (72)

Indeed, if we replace the sets $E(p)$ by $E(p)\setminus\tilde{G}$, then each
tile has density $\lesssim t$, so (72) follows immediatly from Proposition 3.
Using Bateman’s extrapolation argument for the adjoint operator and
interpolation, we obtain (69) for all $p>2$ satisfying (68). ∎

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