1 Introduction
The directional Hilbert transform and directional maximal operator in the plane have been recurrently studied in the literature, see [Reference Bateman1–Reference Bateman and Thiele3, Reference Guo6–Reference Nagel, Stein and Wainger11] and the references therein. A prominent question concerns suitable assumptions on the direction field, under which these operators are bounded in some
$L^{p}(\mathbb{R}^{2})$
.
To be specific, define directional operators in the following form, parametrized by a measurable function
$u:\mathbb{R}^{2}\rightarrow (0,1]$
, and originally defined for functions
$f$
in the Schwartz class
${\mathcal{S}}(\mathbb{R}^{2})$
:
$$\begin{eqnarray}\displaystyle H_{u}f(x,y) & = & \displaystyle p.v.\int _{-1}^{1}f(x-t,y-u(x,y)t)\,\frac{dt}{t},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle M_{u}f(x,y) & = & \displaystyle \sup _{\unicode[STIX]{x1D716}<1}\frac{1}{2\unicode[STIX]{x1D716}}\int _{-\unicode[STIX]{x1D716}}^{\unicode[STIX]{x1D716}}|f(x-t,y-u(x,y)t)|\,dt.\end{eqnarray}$$
One natural assumption is that
$u$
is Lipschitz with sufficiently small Lipschitz constant. One of the main open questions in the area is whether
$H_{u}$
and
$M_{u}$
are bounded in
$L^{2}(\mathbb{R}^{2})$
in this case. No
$L^{p}(\mathbb{R}^{2})$
bounds other than the trivial
$L^{\infty }(\mathbb{R}^{2})$
bound for
$M_{u}$
are known. An extensive discussion of this pair of conjectures appears in the work of Lacey and Li [Reference Lacey and Li9, Reference Lacey and Li10].
Another natural assumption is that
$u$
takes values in a lacunary set. With such a mere assumption on the range of
$u$
, the truncation of the integral in
$H_{u}$
to
$[-1,1]$
and the constraint
$\unicode[STIX]{x1D716}<1$
in
$M_{u}$
are ineffective as they can be transformed by scaling. The following theorem addresses
$M_{u}$
and is a special case of a theorem proven by Nagel et al [Reference Nagel, Stein and Wainger11]. For simplicity we shall restrict attention to a particular lacunary set.
Theorem 1.1 [Reference Nagel, Stein and Wainger11].
For all
$1<p\leqslant \infty$
there is a constant
$C_{p}$
such that for all measurable functions
$u$
on
$\mathbb{R}^{2}$
with values in the set
$\{2^{-j},j\in \mathbb{N}\}$
we have
$$\begin{eqnarray}\displaystyle \biggl\|\sup _{\unicode[STIX]{x1D716}}\frac{1}{2\unicode[STIX]{x1D716}}\int _{-\unicode[STIX]{x1D716}}^{\unicode[STIX]{x1D716}}|f(x-t,y-u(x,y)t)|\,dt\biggr\|_{p}\leqslant C_{p}\Vert f\Vert _{p}. & & \displaystyle\end{eqnarray}$$
It is known that a suitable generalization of lacunarity is the precise assumption on the range of
$u$
to make
$M_{u}$
bounded, see the work of Katz [Reference Katz8] and Bateman [Reference Bateman1]. In contrast, Karagulyan [Reference Karagulyan7] proved that
$H_{u}$
is not bounded for a lacunary set of directions from
$L^{2}(\mathbb{R}^{2})$
to
$L^{2,\infty }(\mathbb{R}^{2})$
.
The purpose of this paper is to look at
$H_{u}$
for a combination of a Lipschitz assumption and a lacunarity assumption. Let
$[x]$
denote the largest integer less than
$x$
.
Theorem 1.2. Let
$u:\mathbb{R}^{2}\rightarrow (0,1]$
be Lipschitz with
$\Vert u\Vert _{Lip}\leqslant 1$
and let
$v:\mathbb{R}^{2}\rightarrow (0,1]$
be defined by
$$\begin{eqnarray}\displaystyle \log _{2}v=[\log _{2}u]. & & \displaystyle\end{eqnarray}$$
Then for every Schwartz function
$f$
and every
$p\in (1,\infty )$
we have
$$\begin{eqnarray}\displaystyle \Vert H_{v}f\Vert _{p}\leqslant C_{p}\Vert f\Vert _{p}. & & \displaystyle\end{eqnarray}$$
Here
$C_{p}$
is a constant depending only on
$p$
.
Note that the range of
$v$
is contained in the set of integer powers of two, a lacunary set. The function
$v$
itself cannot be Lipschitz, unless it is constant.
The above theorem has a dyadic model that we now present. For two points
$p_{1}$
and
$p_{2}$
in the unit line segment
$(0,1]$
(or the unit square
$(0,1]^{2}$
) we define the dyadic distance to be the length of the smallest dyadic interval
$(a,b]$
(or the side length of the smallest dyadic square) that contains both points. For a dyadic interval
$I\subset (0,1]$
, denote by
$h_{I}$
its
$L^{2}$
normalized Haar function. For a map
$v$
from
$(0,1]^{2}$
to
$(0,1]$
define the operator
$$\begin{eqnarray}\displaystyle H_{v,D}f(x,y)=\mathop{\sum }_{|J|/|I|\leqslant v(x,y)}\langle f,h_{I}\otimes h_{J}\rangle h_{I}(x)h_{J}(y), & & \displaystyle\end{eqnarray}$$
where the sum runs over all dyadic rectangles
$I\times J$
in the unit square with the stated bound on the eccentricity. We then have the following theorem.
Theorem 1.3. Let
$v:(0,1]^{2}\rightarrow (0,1]$
have Lipschitz constant (with respect to dyadic metric both on domain and target space) at most
$1/2$
and assume it is lacunary in the sense it takes values in
$\{2^{-k},k\in \mathbb{N}\}$
. Then, for all
$p\in (1,\infty )$
, we have
$$\begin{eqnarray}\displaystyle \Vert H_{v,D}f\Vert _{p}\leqslant C_{p}\Vert f\Vert _{p}. & & \displaystyle\end{eqnarray}$$
Here
$C_{p}$
is a constant which depends only on
$p$
.
We round up the discussion by recalling a third type of assumptions on the direction field, where a Lipschitz assumption recently surfaced naturally, namely bi-parameter assumptions. Bateman [Reference Bateman2] and Bateman along with the second author [Reference Bateman and Thiele3] proved
$L^{p}$
bounds for
$H_{u}$
under the assumption that
$u$
depends only on one variable,
$u(x,y)=u(x,0)$
for all
$y$
, and
$3/2<p<\infty$
. The first author [Reference Guo6] generalized this to direction fields constant along families of Lipschitz curves, highlighting the role of Lipschitz assumptions in this context.
2 The dyadic model: Proof of Theorem 1.3
We write
$s=I\times J$
for the dyadic rectangles below and we write
$h_{s}(x,y)=h_{I}(x)h_{J}(y)$
. We write the
$p$
th power of the
$L^{p}$
norm of (1.6) as an iterated integral:
$$\begin{eqnarray}\displaystyle \int _{0}^{1}\biggl[\int _{0}^{1}\biggl|\mathop{\sum }_{J}\mathop{\sum }_{I:|J|/|I|\leqslant v(x,y)}\langle f,h_{s}\rangle h_{s}(x,y)\biggr|^{p}\,dy\biggr]\,dx. & & \displaystyle\end{eqnarray}$$
Lemma 2.1. Assume
$v$
is as in Theorem 1.3. Let
$I\times J$
be a dyadic rectangle in
$(0,1]^{2}$
and let
$x\in I$
. If for some
$y\in J$
we have
$|J|/|I|\leqslant v(x,y)$
, then we have
$|J|/|I|\leqslant v(x,y^{\prime })$
for all
$y^{\prime }\in J$
.
Proof. Assume to get a contradiction that
$v(x,y^{\prime })<|J|/|I|$
for some
$y^{\prime }\in J$
. Since
$v(x,y)\geqslant |J|/|I|$
, the smallest dyadic interval containing both
$v(x,y^{\prime })$
and
$v(x,y)$
is
$(0,v(x,y)]$
and we have
$d(v(x,y),v(x,y^{\prime }))=v(x,y)$
. This gives a contradiction, since the Lipschitz assumption implies
$$\begin{eqnarray}\displaystyle d(v(x,y),v(x,y^{\prime }))\leqslant d(y,y^{\prime })/2\leqslant |J|/2\leqslant |J|/(2|I|). & & \displaystyle\end{eqnarray}$$
This finishes the proof of Lemma 2.1.◻
Denote by
$\mathbf{I}(x,J)$
the set of all dyadic intervals
$I\subset (0,1]$
such that there exists
$y\in J$
with
$|J|/|I|\leqslant v(x,y)$
. Only the intervals in
$\mathbf{I}(x,J)$
have non-zero contribution to the inner sum of (2.1). By Lemma 2.1, the condition
$|J|/|I|\leqslant v(x,y)$
becomes void for intervals in
$\mathbf{I}(x,J)$
and we may write for (2.1):
$$\begin{eqnarray}\displaystyle \int _{0}^{1}\biggl[\int _{0}^{1}\biggl|\mathop{\sum }_{J}\biggl(\mathop{\sum }_{I\in \mathbf{I}(x,J)}\langle f,h_{s}\rangle h_{s}\biggr)\biggr|^{p}\,dy\biggr]\,dx. & & \displaystyle\end{eqnarray}$$
We have gained independence of the inner summation constraint in
$y$
and may use Littlewood–Paley theory in this variable to estimate the last display by
$$\begin{eqnarray}\displaystyle \lesssim \int _{0}^{1}\biggl[\int _{0}^{1}\biggl(\mathop{\sum }_{J}\biggl|\mathop{\sum }_{I\in \mathbf{I}(x,J)}\langle f,h_{s}\rangle h_{s}\biggr|^{2}\biggr)^{p/2}\,dy\biggr]\,dx. & & \displaystyle\end{eqnarray}$$
The set
$\mathbf{I}(x,J)$
is convex in the sense that if it contains two intervals
$I$
and
$I^{\prime }$
, then it also contains all
$I^{\prime \prime }$
with
$I\subset I^{\prime \prime }\subset I$
. Hence, we can telescope the Haar sum in the variable
$x$
into a difference of two martingale averages and estimate these averages by the maximal operator. Hence, the last display is bounded by
$$\begin{eqnarray}\displaystyle \lesssim \int _{0}^{1}\biggl[\int _{0}^{1}\biggl(\mathop{\sum }_{J}|M_{1}(\langle f,h_{J}\rangle _{2}h_{J})|^{2}\biggr)^{p/2}\,dy\biggr]\,dx, & & \displaystyle\end{eqnarray}$$
where
$M_{1}$
denotes the one-dimensional maximal operator in the
$x$
-variable and
$\langle f,h_{J}\rangle _{2}$
denotes the inner product of
$f$
in the second variable with
$h_{J}$
. We interchange the order of integration in
$x$
and
$y$
and apply the Fefferman–Stein maximal inequality to estimate the last display by
$$\begin{eqnarray}\displaystyle \lesssim \int _{0}^{1}\biggl[\int _{0}^{1}\biggl(\mathop{\sum }_{J}|\langle f,h_{J}\rangle _{2}h_{J}|^{2}\biggr)^{p/2}\,dx\biggr]\,dy. & & \displaystyle\end{eqnarray}$$
Changing the integration order back and applying the Littlewood–Paley theory in the second variable again estimate the last display by
${\lesssim}\Vert f\Vert _{p}^{p}$
, as desired.
3 Proof of Theorem 1.2: Initial reductions
We write the Fourier transform of
$f$
as
$$\begin{eqnarray}\widehat{f}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})=\iint f(x,y)\text{e}^{-2\unicode[STIX]{x1D70B}\text{i}(x\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}y)}\,dx\,dy.\end{eqnarray}$$
Our first step is to pass from
$H_{v}$
to a directional Fourier multiplier operator. The directional Fourier multiplier for a bounded measurable function
$m:\mathbb{R}\rightarrow \mathbb{R}$
is defined as
$$\begin{eqnarray}\displaystyle T[m]f(x,y):=\iint \widehat{f}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})m(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}v(x,y))\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(x\unicode[STIX]{x1D709}+y\unicode[STIX]{x1D702})}\,d\unicode[STIX]{x1D709}\,d\unicode[STIX]{x1D702}. & & \displaystyle\end{eqnarray}$$
Let
$\unicode[STIX]{x1D719}$
be a real even Schwartz function with
$\unicode[STIX]{x1D719}(0)=1$
, whose Fourier transform vanishes on
$[-1,1]$
and outside
$[-2,2]$
:
$$\begin{eqnarray}\displaystyle \text{supp}(\widehat{\unicode[STIX]{x1D719}})\subset [-2,2]\setminus [-1,1]. & & \displaystyle\end{eqnarray}$$
A calculation gives
$$\begin{eqnarray}\displaystyle T[\widehat{\unicode[STIX]{x1D719}}\ast (\text{i}\unicode[STIX]{x1D70B}\text{sign})]f(x,y)=p.v.\int f(x-t,y-v(x,y)t)\unicode[STIX]{x1D719}(t)\,\frac{dt}{t}. & & \displaystyle\end{eqnarray}$$
Comparing with
$H_{v}$
, since
$t^{-1}(1_{[-1,1]}-\unicode[STIX]{x1D719}(t))$
is bounded and rapidly decaying, we obtain by a standard superposition argument
$$\begin{eqnarray}|(H_{v}-T[\widehat{\unicode[STIX]{x1D719}}\ast (\text{i}\unicode[STIX]{x1D70B}\text{sign})])f(x,y)|\leqslant C\sup _{\unicode[STIX]{x1D716}}\frac{1}{\unicode[STIX]{x1D716}}\int _{-\unicode[STIX]{x1D716}}^{\unicode[STIX]{x1D716}}|f(x-t,y-v(x,y)t)|\,dt.\end{eqnarray}$$
By Theorem 1.1, the right-hand side is bounded in
$L^{p}$
. Hence, it suffices to prove Theorem 1.2 with
$T[\widehat{\unicode[STIX]{x1D719}}\ast \text{sign}]$
in place of
$H_{v}$
. In the following, we will denote
$$\begin{eqnarray}\displaystyle m=\widehat{\unicode[STIX]{x1D719}}\ast \text{sign}. & & \displaystyle\end{eqnarray}$$
Note that
$T[m]$
is well defined for functions whose Fourier transform is rapidly decaying. Given a Schwartz function
$f$
, we split it as a sum of two functions, one whose Fourier transform is supported in
$\unicode[STIX]{x1D702}\geqslant 0$
(that is, the set
$\{(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}):\unicode[STIX]{x1D702}\geqslant 0\}$
), and one whose Fourier transform is supported in
$\unicode[STIX]{x1D702}\leqslant 0$
. It suffices to prove the bound of Theorem 1.2 separately for the two functions. By symmetry it suffices to consider the first function, we may thus without loss of generality assume
$\widehat{f}$
is supported in
$\unicode[STIX]{x1D702}\geqslant 0$
.
Next we split the function as a sum of a function whose Fourier transform is supported in
$|\unicode[STIX]{x1D709}|\leqslant 100$
and one whose Fourier transform is supported in
$|\unicode[STIX]{x1D709}|\geqslant 100$
. We consider the two functions separately. First assume
$\widehat{f}$
is supported in
$|\unicode[STIX]{x1D709}|\leqslant 100$
.
Let
$\unicode[STIX]{x1D719}$
be an even real Schwartz function as in (3.2). Moreover, we assume the normalization
$$\begin{eqnarray}\displaystyle \int _{0}^{\infty }\widehat{\unicode[STIX]{x1D719}}(t)\,\frac{dt}{t}=1. & & \displaystyle\end{eqnarray}$$
Set
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}_{t}(x):=t\unicode[STIX]{x1D719}(tx) & & \displaystyle\end{eqnarray}$$
and define the Littlewood–Paley operator in the second variable
$$\begin{eqnarray}\displaystyle P_{t}f(x,y)=\int f(x,y-z)\unicode[STIX]{x1D719}_{t}(z)\,dz. & & \displaystyle\end{eqnarray}$$
We have by the classical Calderòn reproducing formula
$$\begin{eqnarray}f=\int _{0}^{\infty }P_{t}f\,\frac{dt}{t}.\end{eqnarray}$$
Hence,
$$\begin{eqnarray}\displaystyle T[m]f(x,y) & = & \displaystyle \int _{0}^{\infty }T_{m}[P_{t}f](x,y)\,\frac{dt}{t}\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{1000/v(x,y)}T_{m}[P_{t}f](x,y)\,\frac{dt}{t}\nonumber\\ \displaystyle & & \displaystyle +\,\int _{1000/v(x,y)}^{\infty }P_{t}f(x,y)\,\frac{dt}{t}.\end{eqnarray}$$
Here we have dropped the operator
$T[m]$
in the second integral since for
$t>1000/v(x,y)$
the support of
$\widehat{P_{t}f}$
is contained in
$|\unicode[STIX]{x1D709}|\leqslant 100$
and
$\unicode[STIX]{x1D702}\geqslant 1000/v(x,y)$
, where we have
$m(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}v(x,y))=1$
.
Note that there is a Schwartz function
$\unicode[STIX]{x1D711}(\cdot )$
which coincides with
$$\begin{eqnarray}\displaystyle \int _{0}^{1}\unicode[STIX]{x1D719}_{t}(\cdot )\,\frac{dt}{t} & & \displaystyle\end{eqnarray}$$
outside the origin. Moreover, we denote
$\unicode[STIX]{x1D711}_{t}(\cdot )=t\unicode[STIX]{x1D711}(t\cdot )$
. We estimate the second term in (3.8) by
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl|\int _{1000/v(x,y)}^{\infty }P_{t}f(x,y)\,\frac{dt}{t}-f(x,y)\biggr|=\biggl|\int _{0}^{1000/v(x,y)}P_{t}f(x,y)\,\frac{dt}{t}\biggr|\nonumber\\ \displaystyle & & \displaystyle \quad =\biggl|1000/v(x,y)\int \unicode[STIX]{x1D711}_{1000/v(x,y)}(z)f(x,y-z)\,dz\biggr|\lesssim M_{2}f(x,y),\nonumber\end{eqnarray}$$
where
$M_{2}$
is the maximal operator in the second variable,
$$\begin{eqnarray}M_{2}f(x,y)=\sup _{\unicode[STIX]{x1D716}>0}\frac{1}{2\unicode[STIX]{x1D716}}\int _{-\unicode[STIX]{x1D716}}^{\unicode[STIX]{x1D716}}|f(x,y-z)|\,dz.\end{eqnarray}$$
The classical bound on
$M_{2}$
in
$L^{p}$
takes care of the second term in (3.8).
To estimate the first term in (3.8), we note
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{1000/v(x,y)}T[m]P_{t}f(x,y)\,\frac{dt}{t}\nonumber\\ \displaystyle & & \displaystyle \quad =\iint \widehat{f}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\widehat{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D702}v(x,y)/1000)m(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}v(x,y))\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(x\unicode[STIX]{x1D709}+y\unicode[STIX]{x1D702})}\,d\unicode[STIX]{x1D709}\,d\unicode[STIX]{x1D702}.\nonumber\end{eqnarray}$$
Note that
$\widehat{f}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\widehat{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D702}v(x,y)/1000)$
is non-zero only if
$|\unicode[STIX]{x1D709}|\leqslant 100$
and
$\unicode[STIX]{x1D702}v(x,y)\leqslant 2000$
. Moreover, we have for
$n\geqslant 0$
the symbol estimates
$$\begin{eqnarray}\displaystyle \begin{array}{@{}c@{}}\displaystyle |\unicode[STIX]{x2202}_{\unicode[STIX]{x1D709}}^{n}(\widehat{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D702}v(x,y)/1000)m(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}v(x,y)))|\leqslant C_{n},\\ \displaystyle |\unicode[STIX]{x2202}_{\unicode[STIX]{x1D702}}^{n}(\widehat{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D702}v(x,y)/1000)m(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}v(x,y)))|\leqslant C_{n}v(x,y)^{n}.\end{array} & & \displaystyle\end{eqnarray}$$
This means that if we dilate the multiplier
$\widehat{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D702}v(x,y)/100)m(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}v(x,y))$
by a factor
$v(x,y)$
in direction
$\unicode[STIX]{x1D702}$
we obtain a (
$(x,y)$
-dependent) multiplier with uniform symbol estimates, which therefore can be controlled pointwise by the Hardy–Littlewood maximal function. The undilated multiplier can then be estimated by the strong maximal function. Hence, we obtain for the second term in (3.8) the estimate
$$\begin{eqnarray}\displaystyle \int _{0}^{1000/v(x,y)}T[m]P_{t}f(x,y)\,\frac{dt}{t}\lesssim M_{1}M_{2}f(x,y), & & \displaystyle\end{eqnarray}$$
where
$M_{1}$
and
$M_{2}$
are the maximal operators in the first and second variable.
This concludes the case that
$\widehat{f}$
is supported in
$|\unicode[STIX]{x1D709}|\leqslant 100$
. Henceforth, we assume that
$\widehat{f}$
is supported in
$|\unicode[STIX]{x1D709}|\geqslant 100$
. We note that for
$\unicode[STIX]{x1D709}>100$
and
$\unicode[STIX]{x1D702}>0$
we have
$m(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}v(x,y))=1$
and, hence,
$T[m]f=f$
if
$f$
is supported in
$\unicode[STIX]{x1D709}>100$
. We may thus assume
$f$
is supported in
$\unicode[STIX]{x1D709}<-100$
.
In what follows, the truncation of the directional Hilbert transform turns out unnecessary and we shall for simplicity remove it as follows. By the fundamental theorem of calculus we write
$$\begin{eqnarray}\displaystyle m(\unicode[STIX]{x1D709})=-1+\int _{-10}^{10}m^{\prime }(\unicode[STIX]{x1D70F})1_{(\unicode[STIX]{x1D70F},\infty )}(\unicode[STIX]{x1D709})\,d\unicode[STIX]{x1D70F}, & & \displaystyle\end{eqnarray}$$
where we have used that
$m^{\prime }$
vanishes outside
$[-10,10]$
. Since
$T[-1]$
is clearly bounded, it suffices to prove bounds on
$T[1_{(\unicode[STIX]{x1D70F},\infty )}]$
for any
$|\unicode[STIX]{x1D70F}|\leqslant 10$
.
Define
$$\begin{eqnarray}\displaystyle H_{\unicode[STIX]{x1D70F},j}f(x,y):=\iint \widehat{f}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})1_{(\unicode[STIX]{x1D70F},\infty )}(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}2^{-j})\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(x\unicode[STIX]{x1D709}+y\unicode[STIX]{x1D702})}\,d\unicode[STIX]{x1D709}\,d\unicode[STIX]{x1D702}. & & \displaystyle\end{eqnarray}$$
We need the Cordoba and Fefferman [Reference Cordoba and Fefferman4] inequality in the form of [Reference Demeter and Silva5, Theorem 6.1]. It states that for any
$1<p<\infty$
and any collection of functions
$f_{j}$
in
$L^{p}(\mathbb{R}^{2})$
we have
$$\begin{eqnarray}\displaystyle \biggl\|\biggl(\mathop{\sum }_{j}|H_{0,j}f_{j}|^{2}\biggr)^{1/2}\biggr\|_{p}\lesssim \biggl\|\biggl(\mathop{\sum }_{j}|f_{j}|^{2}\biggr)^{1/2}\biggr\|_{p}. & & \displaystyle\end{eqnarray}$$
This is proved in [Reference Demeter and Silva5] using the maximal function bound of Theorem 1.1. Applying the above estimate with the modulated functions
$$\begin{eqnarray}f_{j}(x,y)\text{e}^{2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D70F}x},\end{eqnarray}$$
we obtain more generally
$$\begin{eqnarray}\displaystyle \biggl\|\biggl(\mathop{\sum }_{j}|H_{\unicode[STIX]{x1D70F},j}f_{j}|^{2}\biggr)^{1/2}\biggr\|_{p}\lesssim \biggl\|\biggl(\mathop{\sum }_{j}|f_{j}|^{2}\biggr)^{1/2}\biggr\|_{p}. & & \displaystyle\end{eqnarray}$$
Let
$\unicode[STIX]{x1D719}_{s}$
be as in (3.2), (3.5), (3.6) and define the Littlewood–Paley operator in the first variable
$$\begin{eqnarray}\displaystyle P_{s,1}g(x,y):=\int g(x-z,y)\unicode[STIX]{x1D719}_{s}(z)\,dz. & & \displaystyle\end{eqnarray}$$
Since
$\widehat{f}$
is supported in
$\unicode[STIX]{x1D709}<-100$
, we have the Calderòn reproducing formula in the form
$$\begin{eqnarray}\displaystyle f=\int _{10}^{\infty }P_{s,1}f\,\frac{ds}{s}. & & \displaystyle\end{eqnarray}$$
Now let
$\unicode[STIX]{x1D713}$
be an even real Schwartz function supported in
$[-1,1]$
with
$\widehat{\unicode[STIX]{x1D713}}(0)=0$
and
$$\begin{eqnarray}\int _{0}^{\infty }(\widehat{\unicode[STIX]{x1D713}}(t))^{2}\,\frac{dt}{t}=1,\end{eqnarray}$$
and define
$$\begin{eqnarray}\displaystyle P_{t,2}g(x,y):=\int g(x,y-z)\unicode[STIX]{x1D713}_{t}(z)\,dz. & & \displaystyle\end{eqnarray}$$
Then, by the reproducing formula again,
$$\begin{eqnarray}\displaystyle f=\int _{10}^{\infty }\!\!\int _{0}^{\infty }P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}. & & \displaystyle\end{eqnarray}$$
For
$j\in \mathbb{Z}$
let
$E_{j}$
be the set of
$(x,y)$
such that
$v(x,y)=2^{-j}$
. Then
$$\begin{eqnarray}\displaystyle T[1_{(\unicode[STIX]{x1D70F},\infty )}]f & = & \displaystyle \mathop{\sum }_{j\geqslant 0}1_{E_{j}}H_{\unicode[STIX]{x1D70F},j}f\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{d\in \mathbb{Z}}\mathop{\sum }_{j\geqslant 0}1_{E_{j}}\int _{10}^{\infty }\!\!\int _{2^{j+d}s}^{2^{j+d+1}s}H_{\unicode[STIX]{x1D70F},j}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}.\end{eqnarray}$$
We consider a summand in (3.20) for fixed
$d$
. Setting
$$\begin{eqnarray}\displaystyle f_{j}=\int _{10}^{\infty }\!\!\int _{2^{d+j}s}^{2^{d+j+1}s}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}, & & \displaystyle\end{eqnarray}$$
we recognize with the Cordoba–Fefferman inequality (3.15)
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl\|\mathop{\sum }_{j}1_{E_{j}}\int _{10}^{\infty }\!\!\int _{2^{j+d}s}^{2^{j+d+1}s}H_{\unicode[STIX]{x1D70F},j}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}\biggr\|_{p}\nonumber\\ \displaystyle & & \displaystyle \quad =\biggl\|\mathop{\sum }_{j}1_{E_{j}}H_{\unicode[STIX]{x1D70F},j}f_{j}\biggr\|_{p}\leqslant \biggl\|\biggl(\mathop{\sum }_{j}|H_{\unicode[STIX]{x1D70F},j}f_{j}|^{2}\biggr)^{1/2}\biggr\|_{p}\lesssim \biggl\|\biggl(\mathop{\sum }_{j}|f_{j}|^{2}\biggr)^{1/2}\biggr\|_{p}.\nonumber\end{eqnarray}$$
To estimate the last term, by Khintchine’s inequality, we need to estimate
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{j}\unicode[STIX]{x1D716}_{j}f_{j}=\mathop{\sum }_{j}\unicode[STIX]{x1D716}_{j}\int _{10}^{\infty }\!\!\int _{2^{d+j}s}^{2^{d+j+1}s}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s} & & \displaystyle\end{eqnarray}$$
in
$L^{p}$
, uniformly in all choices of
$|\unicode[STIX]{x1D716}_{j}|\leqslant 1$
. However, the last expression is identified as a Marcinkiewicz multiplier applied to
$f$
, see [Reference Stein12, Ch. IV].
This estimates each term in (3.20) uniformly in
$d$
. To obtain summability in
$d$
, we proceed to refine this estimate, both for
$d\leqslant -5$
and for
$d\geqslant 5$
.
Consider first
$d\leqslant -5$
. Let
$\unicode[STIX]{x1D711}$
be a Schwartz function with
$\widehat{\unicode[STIX]{x1D711}}$
supported in
$[-2,2]$
and constantly equal to one on
$[-1,1]$
. Define
$$\begin{eqnarray}\displaystyle A_{t,2}g(x,y)=\int g(x,y-z)\unicode[STIX]{x1D711}_{t}(z)\,dz. & & \displaystyle\end{eqnarray}$$
We claim that for
$d\leqslant -5$
the summand in (3.20) is equal to
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{j}1_{E_{j}}\int _{10}^{\infty }\!\!\int _{2^{j+d}s}^{2^{j+d+1}s}H_{\unicode[STIX]{x1D70F},j}(1-A_{2^{-d/2}t,2})P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}. & & \displaystyle\end{eqnarray}$$
Namely, for fixed
$j$
and
$(x,y)\in E_{j}$
and fixed
$s>10$
and
$2^{j+d}s\leqslant t\leqslant 2^{j+d+1}s$
we have
$$\begin{eqnarray}\displaystyle & & \displaystyle H_{\unicode[STIX]{x1D70F},j}A_{2^{-d/2}t,2}P_{t,2}^{2}P_{s,1}f\nonumber\\ \displaystyle & & \displaystyle \quad =\iint \widehat{P_{t,2}^{2}f}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})1_{(\unicode[STIX]{x1D70F},\infty )}(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}2^{-j})\widehat{\unicode[STIX]{x1D719}}(s^{-1}\unicode[STIX]{x1D709})\widehat{\unicode[STIX]{x1D711}}(2^{d/2}t^{-1}\unicode[STIX]{x1D702})\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(x\unicode[STIX]{x1D709}+y\unicode[STIX]{x1D702})}\,d\unicode[STIX]{x1D709}\,d\unicode[STIX]{x1D702}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
We claim the last expression vanishes, since the integrand vanishes. For
$\widehat{P_{t,2}^{2}f}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$
not to vanish we need
$\unicode[STIX]{x1D709}<-100$
. For
$\widehat{\unicode[STIX]{x1D719}}(s^{-1}\unicode[STIX]{x1D709})$
not to vanish we need
$s<|\unicode[STIX]{x1D709}|<2s$
, and for
$\widehat{\unicode[STIX]{x1D711}}(2^{d/2}t^{-1}\unicode[STIX]{x1D702})$
not to vanish we need
$\unicode[STIX]{x1D702}\leqslant 2^{1-d/2}t$
. Hence,
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}2^{-j}\leqslant -s+2^{1-d/2-j}t\leqslant -s+2^{1+d/2}s\leqslant -50. & & \displaystyle\end{eqnarray}$$
Hence,
$1_{(\unicode[STIX]{x1D70F},\infty )}(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}2^{-}j)=0$
since
$|\unicode[STIX]{x1D70F}|\leqslant 10$
and the integrand vanishes.
Applying the Cordoba–Fefferman inequality as above and observing that, by the Marcinkiewicz multiplier theorem, for any choices of
$|\unicode[STIX]{x1D716}_{j}|\leqslant 1$
we obtain
$$\begin{eqnarray}\displaystyle \biggl\|\mathop{\sum }_{j}\unicode[STIX]{x1D716}_{j}\int _{10}^{\infty }\!\!\int _{2^{d+j}s}^{2^{d+j+1}s}P_{t,2}P_{s,1}(1-A_{2^{d/2}t,2})P_{t,2}f\,\frac{dt}{t}\,\frac{ds}{s}\biggr\|\lesssim 2^{-d}, & & \displaystyle\end{eqnarray}$$
that is, we obtain the desired decay in
$d$
. Here we have used that for
$d\leqslant -5$
the portion
$(1-A_{2^{d/2}t,2})P_{t,2}$
produces very small symbol estimates for the Marcinkiewicz multiplier.
To obtain good bounds for
$d\geqslant 5$
we compare with the operator
$$\begin{eqnarray}\displaystyle Sf(x,y):=\mathop{\sum }_{d\geqslant 5}\mathop{\sum }_{j}1_{E_{j}}\int _{1}^{\infty }\!\!\int _{2^{d+j}s}^{2^{d+j+1}s}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}, & & \displaystyle\end{eqnarray}$$
which we will estimate in the next section.
We claim that
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j}1_{E_{j}}\biggl[\int _{10}^{\infty }\!\!\int _{2^{j+d}s}^{2^{j+d+1}s}H_{\unicode[STIX]{x1D70F},j}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}-\int _{10}^{\infty }\!\!\int _{2^{d+j}s}^{2^{d+j+1}s}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}\biggr]\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j}1_{E_{j}}\int _{1}^{\infty }\!\!\int _{2^{j+d}s}^{2^{j+d+1}s}H_{\unicode[STIX]{x1D70F},j}A_{2^{-d/2}t,2}P_{t,2}^{2}P_{s,1}f\,\frac{dt}{t}\,\frac{ds}{s}.\nonumber\end{eqnarray}$$
The argument is similar as before. The integrand on the right-hand side takes again the form (3.25). We need that for those
$\unicode[STIX]{x1D702}>0$
with
$\widehat{\unicode[STIX]{x1D711}}(2^{d/2}t^{-1}\unicode[STIX]{x1D702})\neq 1$
, that is
$\unicode[STIX]{x1D702}>t2^{-d/2}$
, we have that either the integrand vanishes or
$1_{(\unicode[STIX]{x1D70F},\infty )}(\unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}2^{-j})=1$
. But if the integrand does not vanish, we have
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}+\unicode[STIX]{x1D702}2^{-j}\geqslant -2s+2^{-d/2-j}t\geqslant -2s+2^{d/2}s\geqslant 50. & & \displaystyle\end{eqnarray}$$
This proves the claim.
We then obtain similarly the decay for Marcinkiewicz multiplier estimates, using that for
$d>5$
the portion
$A_{2^{-d/2}t,2}P_{t,2}^{2}$
produces small symbol estimates since
$\widehat{\unicode[STIX]{x1D713}}(0)=0$
. This finishes all steps of initial reductions.
4 Proof of Theorem 1.2: Main argument
We rewrite the operator
$S$
as
$$\begin{eqnarray}\displaystyle \int _{10}^{\infty }\!\!\int _{32s/v(x,y)}^{\infty }P_{t,2}^{2}P_{s,1}f(x,y)\,\frac{dt}{t}\,\frac{ds}{s}. & & \displaystyle\end{eqnarray}$$
It suffices to prove for every pair of functions
$f$
,
$g$
with
$\Vert f\Vert _{p}=1$
and
$\Vert g\Vert _{p^{\prime }}=1$
that
$$\begin{eqnarray}\iiint _{10}^{\infty }\!\!\int _{32s/v(x,y)}^{\infty }P_{t,2}^{2}P_{s,1}f(x,y)\overline{g(x,y)}\,\frac{dt}{t}\frac{ds}{s}\,dx\,dy\lesssim 1.\end{eqnarray}$$
We fix
$s$
and
$x$
and consider the integrand in these variables:
$$\begin{eqnarray}\displaystyle \iint _{32s/v(x,y)}^{\infty }P_{t,2}^{2}P_{s,1}f(x,y)\overline{g(x,y)}\,\frac{dt}{t}\,dy. & & \displaystyle\end{eqnarray}$$
We write out one of the
$P_{t,2}$
convolution operators to obtain for the last display:
$$\begin{eqnarray}\displaystyle \iint _{32s/v(x,y)}^{\infty }\int P_{t,2}P_{s,1}f(x,y-z)\unicode[STIX]{x1D713}_{t}(z)\overline{g(x,y)}\,dz\,\frac{dt}{t}\,dy. & & \displaystyle\end{eqnarray}$$
We pass the
$z$
integration outside the
$t$
integration. Then we compare the value
$v(x,y)$
occurring in the specification of the integration domain with the value
$v(x,y-z)$
. The difference we consider as an error term
$$\begin{eqnarray}\displaystyle E(x,s):=\iiint _{32s/v(x,y)}^{32s/v(x,y-z)}P_{t,2}P_{s,1}f(x,y-z)\unicode[STIX]{x1D713}_{t}(z)\overline{g(x,y)}\,\frac{dt}{t}\,dz\,dy, & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
which we estimate later. In the remaining integral, with
$t$
-integration in the domain
$s/v(x,y-z)<t<\infty$
, we use the variable
${\tilde{y}}=y-z$
in place of
$y$
and obtain
$$\begin{eqnarray}\iiint _{32s/v(x,{\tilde{y}})}^{\infty }P_{t,2}P_{s,1}f(x,{\tilde{y}})\unicode[STIX]{x1D713}_{t}(z)\overline{g(x,{\tilde{y}}+z)}\,\frac{dt}{t}\,d{\tilde{y}}\,dz.\end{eqnarray}$$
The
$z$
integral is recognized as a Littlewood–Paley operator acting on
$g$
. Writing again the
$x,s$
integrations and calling
${\tilde{y}}$
again
$y$
we need to estimate
$$\begin{eqnarray}\displaystyle \iiint _{10}^{\infty }\!\!\int _{32s/v(x,y)}^{\infty }P_{t,2}P_{s,1}f(x,y)\overline{P_{t,2}g(x,y)}\,\frac{dt}{t}\,\frac{ds}{s}\,dx\,dy. & & \displaystyle\end{eqnarray}$$
We now interchange the
$s$
and
$t$
integrations and apply Cauchy–Schwarz and Hölder to estimate the last display by
$$\begin{eqnarray}\displaystyle & & \displaystyle \iiint _{320/v(x,y)}^{\infty }\int _{10}^{tv(x,y)/32}P_{t,2}P_{s,1}f(x,y)\overline{P_{t,2}g(x,y)}\,\frac{ds}{s}\,\frac{dt}{t}\,dx\,dy\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \,\iint \biggl(\int _{0}^{\infty }\biggl|\int _{10}^{tv(x,y)/32}P_{t,2}P_{s,1}f(x,y)\,\frac{ds}{s}\biggr|^{2}\,\frac{dt}{t}\biggr)^{1/2}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\biggl(\int _{0}^{\infty }|\overline{P_{t,2}g(x,y)}|^{2}\,\frac{dt}{t}\biggr)^{1/2}\,dx\,dy\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \,\biggl\|\biggl\|\biggl(\int _{0}^{\infty }\biggl|\int _{10}^{tv(x,y)/32}P_{t,2}P_{s,1}f(x,y)\,\frac{ds}{s}\biggr|^{2}\,\frac{dt}{t}\biggr)^{1/2}\biggr\|_{L^{p}(x)}\biggr\|_{L^{p}(y)}.\nonumber\end{eqnarray}$$
Here we have used the Littlewood–Paley square function estimate for the function
$g$
in the second variable. Telescoping the
$s$
integral similarly to (3.9) and using the maximal function
$M_{1}$
in the first direction, we estimate the last display by
$$\begin{eqnarray}\displaystyle \lesssim \biggl\|\biggl\|\biggl(\biggl(\int _{0}^{\infty }M_{1}(P_{t,2}f)(x,y)\biggr)^{2}\,\frac{dt}{t}\biggr)^{1/2}\biggr\|_{L^{p}(x)}\biggr\|_{L^{p}(y)}. & & \displaystyle\end{eqnarray}$$
Applying the Fefferman–Stein vector-valued inequality for the maximal function in the first variable gives the bound
$$\begin{eqnarray}\displaystyle \lesssim \biggl\|\biggl\|\biggl(\int _{0}^{\infty }P_{t,2}f(x,y)\,\frac{ds}{s}\bigg|^{2}\,\frac{dt}{t}\biggr)^{1/2}\biggr\|_{L^{p}(x)}\biggr\|_{L^{p}(y)}. & & \displaystyle\end{eqnarray}$$
Commuting the
$L^{p}$
norms and using the Littlewood–Paley square function estimate in the second variable controls the last display by
${\lesssim}\Vert f\Vert _{p}=1$
.
We turn to the estimate for the error term
$E$
. We distinguish the case
$v(x,y-z)<v(x,y)$
and
$v(x,y-z)>v(x,y)$
, more precisely we write
$E=E_{{<}}+E_{{>}}$
with
$E_{{<}}(x,s)$
equal to
$$\begin{eqnarray}\displaystyle & & \displaystyle \iint 1_{s/v(x,y-z)>s/v(x,y)}\int _{32s/v(x,y)}^{32s/v(x,y-z)}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,P_{t,2}P_{s,1}f(x,y-z)\unicode[STIX]{x1D713}_{t}(z)\overline{g(x,y)}\,\frac{dt}{t}\,dz\,dy.\nonumber\end{eqnarray}$$
Define
$w:\mathbb{R}^{2}\rightarrow (0,1]$
by
$$\begin{eqnarray}\displaystyle \log _{2}w(x,y)=[1/2+\log _{2}u], & & \displaystyle\end{eqnarray}$$
that is,
$w$
at any point is either equal to
$v$
or twice as large. We claim that
$E_{{<}}(x,s)$
is equal to
$$\begin{eqnarray}\displaystyle & & \displaystyle \iiint _{32s/w(x,y-z)}^{64s/w(x,y-z)}P_{t,2}P_{s,1}f(x,y-z)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\unicode[STIX]{x1D713}_{t}(z)\overline{g(x,y)}1_{2v(x,y-z)=w(x,y-z)}1_{v(x,y)=w(x,y)}\,\frac{dt}{t}\,dz\,dy.\end{eqnarray}$$
To see the claim, note that the only change between the expressions concerns the coding of the domain of integration. Thus, we need to show that the domains of integration are equal. Consider a point in the domain of integration of the defining expression for
$E_{{>}}$
. We thus have
$s/v(x,y-z)>s/v(x,y)$
. We may also assume that
$\unicode[STIX]{x1D713}_{t}(z)\neq 0$
, or else the integrand vanishes. Then
$|z|\leqslant t^{-1}$
from the support of
$\unicode[STIX]{x1D713}_{t}$
. By the Lipschitz assumption on
$u$
,
$$\begin{eqnarray}\displaystyle |u(x,y)-u(x,y-z)|\leqslant |z|\leqslant t^{-1}\leqslant u(x,y-z)/32, & & \displaystyle\end{eqnarray}$$
where in the last inequality we have used
$t\in [32s/v(x,y),32s/v(x,y-z)]$
and
$s>1$
. Thus,
$\log _{2}u(x,y)$
and
$\log _{2}u(x,y-z)$
differ by less than
$1/4$
. Since
$v(x,y)>v(x,y-z)$
with a strict inequality, then
$u(x,y)$
has to be slightly above an integer power of
$2$
and
$u(x,y-z)$
has to be slightly below the integer power. Hence, we conclude
$$\begin{eqnarray}\displaystyle 2v(x,y-z)=v(x,y)=w(x,y-z)=w(x,y). & & \displaystyle\end{eqnarray}$$
This shows that the essential domain of integration in the defining expression of
$E_{{<}}$
is contained in the essential domain of integration of (4.9). The converse implication is rather straight forward.
Setting
$\tilde{g}(x,y)=g(x,y)1_{v(x,y)=2w(x,y)}$
we thus obtain with the change of variables
${\tilde{y}}=y-z$
for
$E_{{<}}(x,s)$
$$\begin{eqnarray}\displaystyle \iiint _{32s/w(x,{\tilde{y}})}^{64s/w(x,{\tilde{y}})}P_{t,2}P_{s,1}f(x,{\tilde{y}})\unicode[STIX]{x1D713}_{t}(z)\overline{P_{t,2}\tilde{g}(x,{\tilde{y}})}1_{v(x,{\tilde{y}})=w(x,{\tilde{y}})}\,\frac{dt}{t}\,dz\,d{\tilde{y}}. & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
Integrating in
$x$
and
$s$
with the
$s$
integration inside the
$t$
integration and applying Cauchy–Schwarz in the
$t$
integration and calling
${\tilde{y}}$
again
$y$
we estimate the last display similarly to before by
$$\begin{eqnarray}\displaystyle & & \displaystyle \leqslant \iint \biggl(\int _{0}^{\infty }\biggl|\int _{32tw(x,y)}^{64tw(x,y)}P_{t,2}P_{s,1}f(x,y)\,\frac{ds}{s}\biggr|^{2}\,\frac{dt}{t}\biggr)^{1/2}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\biggl(\int _{0}^{\infty }|\overline{P_{t,2}\tilde{g}(x,y)}|^{2}\,\frac{dt}{t}\biggr)^{1/2}\,dx\,dy\nonumber\\ \displaystyle & & \displaystyle \lesssim \,\biggl\|\biggl\|\biggl(\int _{0}^{\infty }\biggl|M_{1}(P_{t,2}f)(x,y)\,\frac{ds}{s}\biggr|^{2}\,\frac{dt}{t}\biggr)^{1/2}\biggr\|_{L^{p}(x)}\biggr\|_{L^{p}(y)}\lesssim 1.\nonumber\end{eqnarray}$$
The finishes the estimate for
$E_{{<}}$
. The estimate for
$E_{{>}}$
is very similar, hence we leave it out.
Acknowledgements
The second author acknowledges support by the Hausdorff Center for Mathematics. We thank Pavel Zorin-Kranich and Olli Saari for catching a typo.
