# Strichartz estimates for the Schrödinger equation on compact manifolds with
nonpositive sectional curvature

Xiaoqi Huang  Department of Mathematics, Louisiana State University, Baton
Rouge, LA 70808<EMAIL_ADDRESS>and  Christopher D. Sogge  Department of
Mathematics, Johns Hopkins University, Baltimore, MD 21218<EMAIL_ADDRESS>
###### Abstract.

We obtain improved Strichartz estimates for solutions of the Schrödinger
equation on compact manifolds with nonpositive sectional curvatures which are
related to the classical universal results of Burq, Gérard and Tzvetkov [11].
More explicitly, we are able refine the arguments in the recent work of Blair
and the authors [3] to obtain no-loss $L^{p}_{t}L^{q}_{x}$-estimates on
intervals of length $\log\lambda\cdot\lambda^{-1}$ for all admissible pairs
$(p,q)$ when the initial data have frequencies comparable to $\lambda$, which,
given the role of the Ehrenfest time, is the natural analog in this setting of
the universal results in [11]. We achieve this log-gain over the universal
estimates by applying the Keel-Tao theorem along with improved global kernel
estimates for microlocalized operators which exploit the geometric
assumptions.

###### Key words and phrases:

Schrödinger’s equation, curvature, Schrödinger curves

###### 2010 Mathematics Subject Classification:

58J50, 35P15

The first author was supported in part by an AMS-Simons travel grant. The
second author was supported in part by the NSF (DMS-1665373 and DMS-2348996)
and a Simons Fellowship. This research was partly carried out while the first
author was at the University of Maryland.

In memoriam: Steve Zelditch (1953-2022)

##  1\. Introduction

The purpose of this paper is to improve the Strichartz estimates for the
Schrödinger equation on negatively curved compact manifolds of Blair and the
authors [3], while at the same time simplifying the arguments there.

Let us first recall the universal estimates of Burq, Gérard and Tzvetkov [11].
If $(M^{d},g)$ is a compact Riemannian manifold of dimension $d\geq 2$, then
the main estimate in [11] is that if $\Delta_{g}$ is the associated Laplace-
Beltrami operator and

(1.1) $u(x,t)=\bigl{(}e^{-it\Delta_{g}}f\bigr{)}(x)$

is the solution of the Schrödinger equation on $M^{d}\times{\mathbb{R}}$,

(1.2) $i\partial_{t}u(x,t)=\Delta_{g}u(x,t),\quad u(x,0)=f(x),$

then if we define

$\|u\|_{L^{p}_{t}L^{q}_{x}(M^{d}\times[0,1])}=\bigr{(}\,\int_{0}^{1}\,\|u(\,\cdot\,,t)\|_{L^{q}_{x}(M^{d})}^{p}\,dt\,\bigr{)}^{1/p},$

one has the mixed-norm Strichartz estimates

(1.3)
$\|u\|_{L^{p}_{t}L^{q}_{x}(M^{d}\times[0,1])}\lesssim\|f\|_{H^{1/p}(M^{d})}$

for all admissible pairs $(p,q)$. By the latter we mean, as in Keel and Tao
[17],

(1.4) $d(\tfrac{1}{2}-\tfrac{1}{q})=\tfrac{2}{p}\,\,\,\text{and
}\,\,2<q\leq\tfrac{2d}{d-2}\,\,\text{if }\,d\geq 3,\,\,\,\text{or
}\,2<q<\infty\,\,\text{if }\,\,d=2.$

Also, in (1.3) $H^{\mu}$ denotes the standard Sobolev space

(1.5)
$\|f\|_{H^{\mu}(M^{d})}=\bigl{\|}\,(I+P)^{\mu}f\,\bigr{\|}_{L^{2}(M^{d})},\quad\text{with
}\,\,P=\sqrt{-\Delta_{g}},$

and “$\lesssim$” in (1.3) and, in what follows, denotes an inequality with an
implicit, but unstated, constant $C$ which can change at each occurrence.

Note that if $e_{\lambda}$ is an eigenfunction of $P$ with eigenvalue
$\lambda$, i.e.,

(1.6) $-\Delta_{g}e_{\lambda}=\lambda^{2}e_{\lambda},$

then

(1.7) $u(x,t)=e^{it\lambda^{2}}e_{\lambda}(x)$

solves (1.2) with initial data $f=e_{\lambda}$. Thus, unlike Euclidean space,
or hyperbolic space, one cannot replace $[0,1]$ by ${\mathbb{R}}$ in (1.3) due
to the existence of eigenfunctions. Also, for the endpoint Strichartz
estimates where $p=2$ and $q=\tfrac{2d}{d-2}$ with $d\geq 3$, the derivative
loss of $\frac{1}{2}$ in (1.3) can not be improved on the standard round
sphere $S^{d}$, by taking the initial data $f$ to be zonal eigenfunctions.
See, e.g., [11] and [3] for more details.

However, on compact manifolds with other types of geometries, (1.3) can be
improved, see e.g., [10], [12] and [13] for the torus case and [3] for general
compact manifolds with non-positive sectional curvatures. Also, for admissible
pairs $(p,q)$ other than $(2,\frac{2d}{d-2})$, one can have better estimates
than (1.3) on the sphere as well using the specific arithmetic properties of
the distinct eigenvalues of the Laplacian on $S^{d}$. See, e.g., Theorem 4 in
[11].

As in [3], to align with the numerology in related earlier results involving
eigenfunction and spectral projection estimates, in what follows, we shall
always take $d=n-1$. Our main result which improves on estimates in [3] then
is the following

######  Theorem 1.1.

Let $M^{n-1}$ be a $d=n-1\geq 2$ dimensional compact manifold all of whose
sectional curvatures are nonpositive. Then for all admissible pairs $(p,q)$,

(1.8)
$\|u\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,1])}\lesssim\bigl{\|}(I+P)^{1/p}\,(\log(2I+P))^{-\frac{1}{p}}f\bigr{\|}_{L^{2}(M^{n-1})}.$

If we let the initial data $f=e_{\lambda}$, then by (1.7) and the special case
$(p,q)=(2,\frac{2d}{d-2})$ of (1.8), we have

(1.9)
$\|e_{\lambda}\|_{L^{\frac{2d}{d-2}}(M^{d})}\lesssim\lambda^{1/2}\,(\log\lambda)^{-1/2}\|e_{\lambda}\|_{L^{2}(M^{d})},$

which gives a $(\log\lambda)^{-1/2}$ gain compared with the universal
eigenfunction estimates of the second author [19]. For manifolds with
nonpositive curvature, (1.9) was first proved by Hassell and Tacy [14], with
similar results for all $q>\tfrac{2(d+1)}{d-1}$. Thus (1.8) is a natural
generalization of (1.9) for solutions of Schrödinger equation and also
provides a novel approach to get improved eigenfunction estimates.

By the Littlewood-Paley theory, we may reduce (1.8) to proving certain dyadic
estimates. More explicitly, let us fix a Littlewood-Paley bump function
$\beta$ satisfying

(1.10) $\beta\in C^{\infty}_{0}((1/2,2))\quad\text{and
}\,\,1=\sum_{k=-\infty}^{\infty}\beta(2^{-k}s),\,\,s>0.$

Then, if we set $\beta_{0}(s)=1-\sum_{k=1}^{\infty}\beta(2^{-k}s)\in
C^{\infty}_{0}({\mathbb{R}}_{+})$ and $\beta_{k}(s)=\beta(2^{-k}s)$,
$k=1,2,\dots$, we have (see, e.g., [20])

(1.11)
$\|h\|_{L^{q}(M^{n-1})}\approx\bigl{\|}\,(\,\sum_{k=0}^{\infty}|\beta_{k}(P)h|^{2}\,)^{1/2}\,\bigr{\|}_{L^{q}(M^{n-1})},\,\,\,1<q<\infty.$

Trivially,
$\|\beta_{0}(P)e^{-it\Delta_{g}}\|_{L^{2}(M^{n-1})\to{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,1])}}=O(1)$,
and, similarly such results where $k=0$ is replaced by a small fixed
$k\in{\mathbb{N}}$ are also standard. So, as noted in [11], one can use (1.11)
and Minkowski’s inequality to see that (1.3) follows from the uniform bounds

(1.12)
$\|e^{-it\Delta_{g}}\beta(P/\lambda)f\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,1])}\leq
C\lambda^{\frac{1}{p}}\,\|f\|_{L^{2}(M^{n-1})},\quad\lambda\gg 1.$

Burq, Gérard and Tzvetkov proved this estimate in [11] by showing that one
always has the following uniform dyadic estimates over very small intervals:

(1.13)
$\|e^{-it\Delta_{g}}\beta(P/\lambda)f\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,\lambda^{-1}])}\leq
C\,\|f\|_{L^{2}(M^{n-1})},\quad\lambda\gg 1.$

It is not hard to see that (1.13) yields (1.12), since one can write $[0,1]$
as the union of $\approx\lambda$ intervals of length $\lambda^{-1}$ and thus
obtain (1.12) by adding up the uniform estimates on each of these subintervals
that (1.13) affords. Also, the bounds in (1.13) cannot be improved on any
manifold by taking $f(x)=f_{\lambda}(x)=\beta(P/\lambda)(x,x_{0})$ with
$\beta(P/\lambda)(x,y)$ being the kernel of the Littlewood-Paley operators. As
a result, to obtain improvements such as those in (1.8), larger time intervals
must be used. Specifically, we shall show that if $M^{n-1}$ is as in Theorem
1.1 then we have the uniform bounds

(1.14)
$\|e^{-it\Delta_{g}}\beta(P/\lambda)f\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,\,\log\lambda\cdot\lambda^{-1}])}\leq
C\,\|f\|_{L^{2}(M^{n-1})},\quad\lambda\gg 1,$

which is a natural extension of the uniform small-time scale estimates (1.13),
and perhaps the largest one can hope to obtain such estimates in the geometry
we are focusing on using available techniques, due to the role of the
Ehrenfest time. Also, by the above counting arguments, one obtains (1.8) from
(1.14) since $[0,1]$ can be covered by $\approx\lambda/\log\lambda$ intervals
of length $\log\lambda\cdot\lambda^{-1}$.

The bound in (1.14) improves the result in [3] in two aspects, first, it
removes the power of $\log\lambda$ loss there, and second, it includes all
admissible pairs $(p,q)$. The main ideas in the proof of (1.14) are similar to
those in [3], both of which involve a height decomposition. The larger values
can be dealt with using kernel estimates for certain global operators, while
the smaller values involves the use of a Whitney type decomposition and
bilinear oscillatory integral estimates of Lee [18]. Compared to [3], we are
able to improve the arguments used for the diagonal term in the Whitney
decomposition, which previously relied on a microlocalized version of improved
Strichartz estimates. We avoid this by applying the abstract theorem of Keel-
Tao [17] to some new Banach space with $L^{p}$ norm depending on the
$\ell^{p}$ norm of different microlocalized pieces. The analogous dispersive
estimate adapted to the new space, which is necessary to apply the Keel-Tao
theorem, is obtained by using the kernel estimates involving the
microlocalized operators as proved in [3].

If $\chi_{[\lambda,\lambda+(\log\lambda)^{-1}]}$ denotes the spectral
projection operator for $P=\sqrt{-\Delta_{g}}$ associated with the interval
$[\lambda,\lambda+(\log\lambda)^{-1}]$, then a simple consequence of (1.14) is
that for $q_{e}=\tfrac{2d}{d-2}$, $d=n-1\geq 3$,

(1.15)
$\|\chi_{[\lambda,\lambda+(\log\lambda)^{-1}]}f\|_{L^{q_{e}}(M^{d})}\leq
C\lambda(\log\lambda)^{-1/2}\|f\|_{L^{2}(M^{d})},$

assuming that $M^{d}$ has nonpositive sectional curvatures. This in turn,
implies the natural sharp spectral projection estimates for these operators
for all $q\geq q_{e}$ under these curvature assumptions, which are due to
Bérard [1] for $q=\infty$ and later generalized by Hassell and Tacy [14] to
$q>\tfrac{2(d+1)}{d-1}$ and $d\geq 2$.

To prove (1.15) fix $a\in{\mathcal{S}}({\mathbb{R}})$ satisfying $a(0)=1$ and
$\hat{a}(\tau)=0$ if $|\tau|\geq 1$. Then for $q_{e}$ and $d=n-1$ as in
(1.15), by Bernstein’s inequality and (1.14),

$\displaystyle\|$
$\displaystyle\chi_{[\lambda,\lambda+(\log\lambda)^{-1}]}f\|_{L^{q_{e}}(M^{n-1})}$
$\displaystyle\leq\|a(\lambda(\log\lambda)^{-1}t)\,e^{-it\Delta_{g}}\beta(P/\lambda)\chi_{[\lambda,\lambda+(\log\lambda)^{-1}]}f\|_{L^{\infty}_{t}L^{q_{e}}_{x}(M^{d}\times{\mathbb{R}})}$
$\displaystyle\leq
C\lambda^{\frac{1}{2}}(\log\lambda)^{-\frac{1}{2}}\|a(\lambda(\log\lambda)^{-1}t)\,e^{-it\Delta_{g}}\beta(P/\lambda)\chi_{[\lambda,\lambda+(\log\lambda)^{-1}]}f\|_{L^{2}_{t}L^{q_{e}}_{x}(M^{d}\times{\mathbb{R}})}$
$\displaystyle\leq
C\lambda^{\frac{1}{2}}(\log\lambda)^{-\frac{1}{2}}\|f\|_{L^{2}(M^{n-1})}.$

Note that this argument implies that any improvements of (1.14) to include
uniform bounds for intervals of size $\delta(\lambda)$ with
$\delta(\lambda)/[\lambda^{-1}\cdot\log\lambda]\nearrow\infty$ would imply
sharp spectral projection estimates associated with spectral windows of length
$o((\log\lambda)^{-1})$ for $q\geq q_{e}$ under the assumption of nonpositive
curvature, which, even for the case of $q=\infty$, seems difficult. There has
been no such improvement of the sup-norm estimates of Bérard [1] in the last
five decades, and progress of this nature has been elusive due to the role of
the Ehrenfest time.

In the case of flat tori, if we fix $q=\frac{2(n+1)}{n-1}$, recall the near
optimal results of Bourgain and Demeter [10] for $n\geq 4$ and Bourgain [9]
for $n=2,3$

(1.16)
$\|e^{-it\Delta_{g}}\beta(P/\lambda)f\|_{L^{q}_{t,x}(M^{n-1}\times[0,\,1])}\leq
C_{\varepsilon}\lambda^{\varepsilon}\,\|f\|_{L^{2}(M^{n-1})},\,\lambda\gg
1,\,\,\forall\varepsilon>0.$

The proof of (1.16) is based on number-theoretic methods for $n=2,3$, while
for $n\geq 4$, the estimates were derived via $\ell^{2}$-decoupling methods.
More recently, for $n=3$, Herr and Kwak [15] obtained a lossless version of
(1.16) for any time intervals of length $(\log\lambda)^{-1}$ by using a new
method based on incidence geometry, and the length of the interval can not be
extended further by testing against
$f=\sum_{k\in[-\lambda,\lambda]^{2}\cap\mathbb{Z}^{2}}e^{ik\cdot x}$. See e.g,
[9] for more details.

By using the favorable properties of the universal cover ${\mathbb{R}}^{n}$
and the fact that the types of microlocal cutoffs we shall employ commute well
with Schrödinger propagators, it seems likely that we shall be able to modify
the arguments in the proof of Theorem 1.1 to obtain no loss dyadic estimates
on tori ${\mathbb{T}}^{n}$ on intervals of length $\lambda^{-1+\delta_{n}}$
for some $0<\delta_{n}<1$, which would be a natural generalization of the
results in [15] for all admissible pairs $(p,q)$ in any dimension $n\geq 3$.
We hope to explore this problem as well as possible improved Strichartz
estimates for spheres in a later work.

This paper is organized as follows. In the next section we present the main
arguments that allow us to prove Theorem 1.1. The proof requires local
bilinear arguments from harmonic analysis, which are discussed in Section 3.

##  2\. Main arguments

To start, let $\beta$ be the Littlewood-Paley bump function in (1.10), and
also fix

(2.1) $\eta\in C^{\infty}_{0}((-1,1))\quad\text{with
}\,\,\eta(t)=1,\,\,\,|t|\leq 1/2.$

We then shall consider the dyadic time-localized dilated Schrödinger operators

(2.2) $S_{\lambda}=\eta(t/T)e^{-it\lambda^{-1}\Delta_{g}}\beta(P/\lambda),$

and claim that the estimates in Theorems 1.1 is a consequence of the
following.

######  Proposition 2.1.

Let $M^{d}$, $d=n-1\geq 2$ be a fixed compact manifold all of whose sectional
curvatures are nonpositive, and $(p,q)$ be as in (1.4). Then we can fix
$c_{0}>0$ so that for large $\lambda\gg 1$ we have the uniform bounds

(2.3) $\|S_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,T])}\leq
C\lambda^{\frac{1}{p}}\,\|f\|_{L^{2}(M^{n-1})},\quad\text{if
}\,\,T=c_{0}\log\lambda.$

We claim that (2.3) implies Theorem 1.1. First note that, by changing scales,
(2.1) and (2.3) imply that for large enough $\lambda$ we have the analog of
(1.14) where the interval $[0,\log\lambda\cdot\lambda^{-1}]$ in the left is
replaced by $[0,\tfrac{1}{2}c_{0}\log\lambda\cdot\lambda^{-1}]$, and this of
course implies (1.14) at the expense of including an additional factor of
$(c_{0}/2)^{-1/q_{c}}$ in the constant in the right if $c_{0}<2$. As we
indicated before, the estimate (1.14) for large $\lambda$ and Littlewood-Paley
theory yield Theorem 1.1, which verifies our claim regarding (2.3).

Also note that if we replace $[0,T]$ by $[0,1]$, then by (1.13) and a
rescaling argument we have

(2.4) $\|S_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,1])}\leq
C\lambda^{\frac{1}{p}}\,\|f\|_{L^{2}(M^{n-1})},$

which hold on a general smooth compact manifold.

To prove (2.3), note that since the case $(p,q)=(\infty,2)$ is trivial, by
interpolation, it suffices to consider the $(p,q)$ pairs which satisfy

(2.5) $(p,q)=(2,\tfrac{2d}{d-2})\,\,\text{if }\,d\geq 3,\,\,\,\text{or
}\,d(\tfrac{1}{2}-\tfrac{1}{q})=\tfrac{2}{p},\,\,4<q<\infty\,\,\text{if
}\,\,d=2.$

The condition $q>4$ is equivalent to $q>p$ when $d=2$, this will allow us to
simplify some of the calculations to follow.

As in [3], we need to introduce a few auxiliary operators that allow us to use
bilinear techniques. First of all, we need to compose the “global operators”
$S_{\lambda}$ with related local ones. Motivated by the recent work of the
authors [16], our “local” auxiliary operators will be the following
“quasimode” operators adapted to the scaled Schrödinger operators $\lambda
D_{t}+\Delta_{g}$,

(2.6)
$\sigma_{\lambda}=\sigma\bigl{(}\lambda^{1/2}|D_{t}|^{1/2}-P\bigr{)}\,\tilde{\beta}(D_{t}/\lambda),$

where

(2.7) $\sigma\in{\mathcal{S}}({\mathbb{R}})\,\,\text{satisfies
}\,\,\sigma(0)=1\,\,\text{and }\,\,\text{supp
}\hat{\sigma}\subset\delta\cdot[1-\delta_{0},1+\delta_{0}]=[\delta-\delta_{0}\delta,\delta+\delta_{0}\delta],$

with $0<\delta,\delta_{0}<1/8$ to be specified later, and, also here

(2.8) $\tilde{\beta}\in C^{\infty}_{0}((1/8,8))\quad\text{satisfies
}\,\,\tilde{\beta}=1\,\,\text{on }\,\,[1/6,6].$

Here the properties of $\sigma$, as well as the small constants $\delta$ and
$\delta_{0}$ are the same as those in [3], which allows us to use the bilinear
oscillatory integral estimates in [3] in the next section. The smallness of
$\delta$ is also related to another initial microlocalization that is needed
for the bilinear arguments, as we shall describe below.

Let us write

(2.9) $I=\sum_{j=1}^{N}B_{j}(x,D),$

where each $B_{j}\in S^{0}_{1,0}(M^{n-1})$ is a standard pseudo-differential
operator with symbol supported in a small conic neighborhood of some
$(x_{j},\xi_{j})\in S^{*}M$. The size of the support will be described later;
however, these operators will not depend on our parameter $\lambda\gg 1$.
Next, if $\tilde{\beta}$ is as in (2.8) then the dyadic operators

(2.10) $B=B_{j,\lambda}=B_{j}\circ\tilde{\beta}(P/\lambda)$

are uniformly bounded on $L^{p}$, i.e.,

(2.11) $\|B\|_{L^{p}(M^{n-1})\to L^{p}(M^{n-1})}=O(1)\quad\text{for }\,\,1\leq
p\leq\infty.$

Also, note that since $\sigma\in{\mathcal{S}}({\mathbb{R}})$ a simple
calculation shows that if $\lambda_{k}$ is an eigenvalue of $P$

$(1-\tilde{\beta}(\lambda_{k}/\lambda))\,\sigma(\lambda^{1/2}|\tau|^{1/2}-\lambda_{k})\,\tilde{\beta}(\tau/\lambda)=O(\lambda^{-N}(1+\lambda_{k}+|\tau|)^{-N})\,\,\forall\,N.$

Consequently,

$\|\sigma_{\lambda}-\tilde{\beta}(P/\lambda)\circ\sigma_{\lambda}\|_{L^{2}(M^{n-1}\times[0,T])\to{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,T])}}=O(\lambda^{-N})\quad\forall\,N.$

Thus, if $B_{j}$ is as in (2.9) and $B_{j,\lambda}$ is the corresponding
dyadic operator in (2.10)

(2.12)
$\|B_{j}\sigma_{\lambda}-B_{j,\lambda}\sigma_{\lambda}\|_{L^{2}(M^{n-1}\times[0,T])\to{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,T])}}=O(\lambda^{-N})\quad\forall\,N,$

since operators in $S^{0}_{1,0}(M^{n-1})$ are bounded on $L^{p}$ for
$1<p<\infty$.

We need one more result for now about these local operators:

######  Lemma 2.2.

If $S_{\lambda}$ as in (2.2), $\sigma_{\lambda}$ is as in (2.6) and $(p,q)$
are as in (1.4), then

(2.13) $\|(I-\sigma_{\lambda})\circ
S_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,T])}\leq
CT^{\frac{1}{p}-\frac{1}{2}}\lambda^{\frac{1}{p}}\|f\|_{2}.$

We also have

(2.14) $\bigl{\|}\sigma_{\lambda}F\bigr{\|}_{L^{p}_{t}L^{q}_{x}}\leq
C\lambda^{\frac{1}{p}}\|F\|_{L^{2}_{t,x}}.$

Lemma 2.2 is a simple generalization of the results in Lemma 2.2 and 2.3 of
[3] to all admissible pairs $(p,q)$. Using the the local dyadic Strichartz
estimates (1.13), the proof of Lemma 2.2 follows from the same arguments as in
[3], so we skip the details here.

For a given $B=B_{j,\lambda}$ as in (2.10) let us define the microlocalized
variant of $\sigma_{\lambda}$ as follows

(2.15) $\tilde{\sigma}_{\lambda}=B\circ\sigma_{\lambda},\quad
B=B_{j,\lambda},$

and the associated “semi-global” operators

(2.16) $\tilde{S}_{\lambda}=\tilde{\sigma}_{\lambda}\circ S_{\lambda}.$

By (2.9), (2.12) and (2.13), in order to prove Proposition 2.1, it suffices to
show that if $T=c_{0}\log\lambda$ with $c_{0}>0$ sufficiently small (depending
on $M^{n-1}$), then, if all the sectional curvatures of $M^{n-1}$ are
nonpositive, for $(p,q)$ be as in (2.5), we have

(2.17) $\|\tilde{S}_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,T])}\leq
C\lambda^{\frac{1}{p}}\|f\|_{2}.$

As we shall see, in order to prove (2.17) we shall need to take $\delta$ and
$\delta_{0}$ in (2.7) and (2.8) to be sufficiently small for each $j$;
however, since, by the compactness of $M^{n-1}$ and the arguments to follow,
the sum in (2.9) can be taken to be finite, we can take these two parameters
to be the minimum over what is needed for $j=1,\dots,N$.

2.1. Height Decomposition.

Next we set up a variation of an argument of Bourgain [8] originally used to
study Fourier transform restriction problems, and, more recently, to study
eigenfunction problems in [2], [7] and [21]. This involves splitting the
estimates in Proposition 2.1 into two heights involving relatively large and
small values of $|\tilde{S}_{\lambda}f(x,t)|$.

To describe this, here, and in what follows we shall assume, as we just did,
that $f$ is $L^{2}$-normalized, that is

(2.18) $\|f\|_{2}=1.$

Then, we shall prove the estimates in Proposition 2.1, using very different
techniques by estimating $L^{p}_{t}L^{q}_{x}$ bounds over the two regions

(2.19) $A_{+}=\\{(x,t)\in
M^{n-1}\times[0,T]:\,|\tilde{S}_{\lambda}f(t,x)|\geq\lambda^{\frac{n-1}{4}+\varepsilon_{1}}\\},\\\
\text{and }\,\,A_{-}=\\{(x,t)\in
M^{n-1}\times[0,T]:\,|\tilde{S}_{\lambda}f(x,t)|<\lambda^{\frac{n-1}{4}+\varepsilon_{1}}\\}.$

Due to the numerology of the powers of $\lambda$ arising, the splitting occurs
at height $\lambda^{\frac{n-1}{4}+\varepsilon_{1}}$, where $\varepsilon_{1}>0$
is a small constant that may depend on the dimension $d=n-1$. As we shall see
later in (3.11) and (3.43), we can take $\varepsilon_{1}=\frac{1}{100}$ for
$n-1\geq 3$ while for $n-1=2$, the choice of $\varepsilon_{1}$ depends on the
exponent $q$ for admissible pairs $(p,q)$, with $\varepsilon_{1}\rightarrow 0$
as $q\rightarrow\infty$. The transition occurring at, basically,
$\lambda^{\frac{n-1}{4}}$ is natural and arises due to Knapp-type phenomena,
both in Euclidean problems, as well as the geometric ones that we are
considering here.

Thus, to prove Proposition 2.1, it suffices to prove the analog (2.17) for the
two regions in (2.19).

2.2. Estimates for relatively large values: Proof of (2.17) on the set $A_{+}$
.

We first note that, by Lemma 2.2 and (2.11) we have

$\|\tilde{S}_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(A_{+})}\leq\|BS_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(A_{+})}+CT^{\frac{1}{p}-\frac{1}{2}}\lambda^{\frac{1}{p}},$

and, since $p\geq 2$ for $(p,q)$ as in (2.5), (2.17) would follow from

(2.20) $\|BS_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(A_{+})}\leq
C\lambda^{\frac{1}{p}}+\tfrac{1}{2}\|\tilde{S}_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(A_{+})}.$

To prove this we shall adapt an argument of Bourgain [8] and more recent
variants in [2] and [21] . Specifically, choose $g(x,t)$ such that

$\|g\|_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}(A_{+})}=1\quad\text{and
}\,\,\|BS_{\lambda}f\|_{L^{p}_{t}L^{q}_{x}(A_{+})}=\iint
BS_{\lambda}f\cdot\overline{\bigl{(}\mathbb{1}_{A_{+}}\cdot g\bigr{)}}\,dxdt.$

Then, since we are assuming that $\|f\|_{2}=1$, by the Schwarz inequality

(2.21) $\displaystyle\|BS_{\lambda}f\|^{2}_{L^{p}_{t}L^{q}_{x}(A_{+})}$
$\displaystyle=\Bigl{(}\,\int
f(x)\,\cdot\,\overline{\bigl{(}S^{*}B^{*}\bigr{)}(\mathbb{1}_{A_{+}}\cdot
g\bigr{)}(x)}\,dx\,\Bigr{)}^{2}$
$\displaystyle\leq\int|S^{*}_{\lambda}B^{*}(\mathbb{1}_{A_{+}}\cdot
g)(x)|^{2}\,dx$
$\displaystyle=\iint\bigl{(}BS_{\lambda}S^{*}_{\lambda}B^{*}\bigr{)}(\mathbb{1}_{A_{+}}\cdot
g)(x,t)\,\overline{(\mathbb{1}_{A_{+}}\cdot g)(x,t)}\,dxdt$
$\displaystyle=\iint\bigl{(}B\circ L_{\lambda}\circ
B^{*}\bigr{)}(\mathbb{1}_{A_{+}}\cdot
g)(x,t)\,\overline{(\mathbb{1}_{A_{+}}\cdot g)(x,t)}\,dxdt$
$\displaystyle\qquad+\iint\bigl{(}B\circ G_{\lambda}\circ
B^{*}\bigr{)}(\mathbb{1}_{A_{+}}\cdot
g)(x,t)\,\overline{(\mathbb{1}_{A_{+}}\cdot g)(x,t)}\,dxdt$
$\displaystyle=I+II,$

where $L_{\lambda}$ is the integral operator with kernel equaling that of
$S_{\lambda}S^{*}_{\lambda}$ if $|t-s|\leq 1$ and $0$ otherwise, i.e,

(2.22) $L_{\lambda}(x,t;y,s)=\\\
\begin{cases}\bigl{(}S_{\lambda}S_{\lambda}^{*}\bigr{)}(x,t;y,s)=\eta(t/T)\eta(s/T)\bigl{(}\,\beta^{2}(P/\lambda)e^{-i(t-s)\lambda^{-1}\Delta_{g}}\bigr{)}(x,y),\,\,\text{if
}\,\,|t-s|\leq 1,\\\ 0\,\,\,\text{otherwise}.\end{cases}$

Since $p\geq 2$, it is straightforward to see that (2.4) yields

(2.23) $\|L_{\lambda}\|_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}\to
L^{p}_{t}L^{q}_{x}}=O(\lambda^{\frac{2}{p}}).$

If we use this, along with Hölder’s inequality and (2.11), we obtain for the
term $I$ in (2.21)

(2.24) $\displaystyle|I|$
$\displaystyle\leq\|BL_{\lambda}B^{*}(\mathbb{1}_{A_{+}}\cdot
g)\|_{L^{p}_{t}L^{q}_{x}}\cdot\|\mathbb{1}_{A_{+}}\cdot
g\|_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}}$
$\displaystyle\lesssim\|L_{\lambda}B^{*}(\mathbb{1}_{A_{+}}\cdot
g)\|_{L^{p}_{t}L^{q}_{x}}\cdot\|\mathbb{1}_{A_{+}}\cdot
g\|_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}}$
$\displaystyle\lesssim\lambda^{\frac{2}{p}}\|B^{*}(\mathbb{1}_{A_{+}}\cdot
g)\|_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}}\cdot\|\mathbb{1}_{A_{+}}\cdot
g\|_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}}$
$\displaystyle\lesssim\lambda^{\frac{2}{p}}\|g\|^{2}_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}(A_{+})}=\lambda^{\frac{2}{p}}.$

To estimate the other term in (2.21), $II$, we need the following kernel bound
which is Proposition 4.1 in [3],

(2.25) $|(S_{\lambda}S_{\lambda}^{*})(x,t;y,s)|\leq
C\lambda^{\frac{n-1}{2}}\,|t-s|^{-\frac{n-1}{2}}\exp(C_{M}|t-s|),\quad\text{if
}\,\,|t-s|\leq 2T.$

The proof of (2.25) follows from arguments in Bérard [1], and also in [6],
[22] and other related works, which use the Hadamard parametrix and the
Cartan-Hadamard theorem to lift the calculations that will be needed up to the
universal cover $({\mathbb{R}}^{n-1},\tilde{g})$ of $(M^{n-1},g)$.

If we choose $c_{0}$ small enough so that if $C_{M}$ is the constant in (2.25)

$\exp(2C_{M}T)\leq\lambda^{\varepsilon_{1}},\quad\text{if
}\,T=c_{0}\log\lambda\,\,\text{and }\,\,\lambda\gg 1.$

Then, since $\eta(t)=0$ for $|t|\geq 1$, it follows from (2.22) and (2.25)
that

$\|G_{\lambda}\|_{L^{1}(M^{n-1}\times{\mathbb{R}})\to
L^{\infty}(M^{n-1}\times{\mathbb{R}})}\leq
C\lambda^{\frac{n-1}{2}+{\varepsilon_{1}}}.$

As a result, since, by (2.11), the dyadic operators $B$ are bounded on $L^{1}$
and $L^{\infty}$, we can repeat the arguments to estimate $I$ and use Hölder’s
inequality to see that

$|II|\leq
C\lambda^{\frac{n-1}{2}}\lambda^{\varepsilon_{1}}\|\mathbb{1}_{A_{+}}\cdot
g\|^{2}_{1}\leq
C\lambda^{\frac{n-1}{2}}\lambda^{\varepsilon_{1}}\|g\|^{2}_{L^{p^{\prime}}_{t}L^{q^{\prime}}_{x}}\cdot\|\mathbb{1}_{A_{+}}\|_{L^{p}_{t}L^{q}_{x}}^{2}=C\lambda^{\frac{n-1}{2}}\lambda^{\varepsilon_{1}}\|\mathbb{1}_{A_{+}}\|_{L^{p}_{t}L^{q}_{x}}^{2}.$

If we recall the definition of $A_{+}$ in (2.19), we can estimate the last
factor:

$\|\mathbb{1}_{A_{+}}\|_{L^{p}_{t}L^{q}_{x}}^{2}\leq\bigl{(}\lambda^{\frac{n-1}{4}+\varepsilon_{1}}\bigr{)}^{-2}\|\tilde{S}_{\lambda}f\|^{2}_{{L^{p}_{t}L^{q}_{x}}(A_{+})}.$

Therefore,

$|II|\lesssim\lambda^{-\varepsilon_{1}}\|\tilde{S}_{\lambda}f\|^{2}_{{L^{p}_{t}L^{q}_{x}}(A_{+})}\leq\bigl{(}\tfrac{1}{2}\|\tilde{S}_{\lambda}f\|_{{L^{p}_{t}L^{q}_{x}}(A_{+})}\bigr{)}^{2},$

assuming, as we may, that $\lambda$ is large enough.

If we combine this bound with the earlier one, (2.24) for $I$, we conclude
that (2.20) is valid, which completes the proof of (2.17) on the set $A_{+}$.
∎

2.3. Estimates for relatively small values: Proof of (2.17) on the set
$A_{-}$.

We now turn to the proving the ${L^{p}_{t}L^{q}_{x}}(A_{-})$ estimates in
(2.17). To do this we need to borrow the bilinear estimates from [3], which
replies on the results from bilinear harmonic analysis in [18] and [23].

We need to utilize a microlocal decomposition as in [3]. Recall that the
symbol $B(x,\xi)$ of $B$ in (2.10) is supported in a small conic neighborhood
of some $(x_{0},\xi_{0})\in S^{*}M^{n-1}$. We may assume that its symbol has
small enough support so that we may work in a coordinate chart $\Omega$ and
that $x_{0}=0$, $\xi_{0}=(0,\dots,0,1)$ and $g_{jk}(0)=\delta^{j}_{k}$ in the
local coordinates. So, we shall assume that $B(x,\xi)=0$ when $x$ is outside a
small relatively compact neighborhood of the origin or $\xi$ is outside of a
small conic neighborhood of $(0,\dots,0,1)$.

Next, let us define the microlocal cutoffs that we shall use. We fix a
function $a\in C^{\infty}_{0}({\mathbb{R}}^{2(n-2)})$ supported in
$\\{z:\,|z_{j}|\leq 1,\,\,1\leq j\leq 2(n-2)\\}$ which satisfies

(2.26) $\sum_{j\in{\mathbb{Z}}^{2(n-2)}}a(z-j)\equiv 1.$

We shall use this function to build our microlocal cutoffs. By the above, we
shall focus on defining them for $(y,\eta)\in S^{*}\Omega$ with $y$ near the
origin and $\eta$ in a small conic neighborhood of $(0,\dots,0,1)$. We shall
let

$\varPi=\\{y:\,y_{n-1}=0\\}$

be the points in $\Omega$ whose last coordinate vanishes. Let
$y^{\prime}=(y_{1},\dots,y_{n-2})$ and
$\eta^{\prime}=(\eta_{1},\dots,\eta_{n-2})$ denote the first $n-2$ coordinates
of $y$ and $\eta$, respectively. For $y\in\varPi$ near $0$ and $\eta$ near
$(0,\dots,0,1)$ we can just use the functions
$a(\theta^{-1}(y^{\prime},\eta^{\prime})-j)$, $j\in{\mathbb{Z}}^{2(n-2)}$ to
obtain cutoffs of scale $\theta$.

We can then extend the definition to a neighborhood of $(0,(0,\dots,0,1))$ by
setting for $(x,\xi)\in S^{*}\Omega$ in this neighborhood

(2.27)
$a^{\theta}_{j}(x,\xi)=a(\theta^{-1}(y^{\prime},\eta^{\prime})-j)\quad\text{if
}\,\,\chi_{s}(x,\xi)=(y^{\prime},0,\eta^{\prime},\eta_{n-1})\,\,\,\text{with
}\,\,\,s=d_{g}(x,\varPi).$

Here $\chi_{s}$ denotes geodesic flow in $S^{*}\Omega$. Thus,
$a^{\theta}_{j}(x,\xi)$ is constant on all geodesics $(x(s),\xi(s))\in
S^{*}\Omega$ with $x(0)\in\varPi$ near $0$ and $\xi(0)$ near $(0,\dots,0,1)$.
As a result,

(2.28) $a^{\theta}_{j}(\chi_{s}(x,\xi))=a^{\theta}_{j}(x,\xi)$

for $s$ near $0$ and $(x,\xi)\in S^{*}\Omega$ near $(0,(0,\dots,0,1))$.

We then extend the definition of the cutoffs to a conic neighborhood of
$(0,(0,\dots,0,1))$ in $T^{*}\Omega\,\backslash\,0$ by setting

(2.29) $a^{\theta}_{j}(x,\xi)=a^{\theta}_{j}(x,\xi/p(x,\xi)).$

Notice that if $(y^{\prime}_{j},\eta^{\prime}_{j})=\theta j$ and $\gamma_{j}$
is the geodesic in $S^{*}\Omega$ passing through
$(y^{\prime}_{j},0,\eta_{j})\in S^{*}\Omega$ with $\eta_{j}\in
S^{*}_{(y^{\prime}_{j},0)}\Omega$ having $\eta^{\prime}_{j}$ as its first
$(n-2)$ coordinates then

(2.30) $a^{\theta}_{j}(x,\xi)=0\quad\text{if }\,\,\,\text{dist
}\bigl{(}(x,\xi),\gamma_{j}\bigr{)}\geq C_{0}\theta,$

for some fixed constant $C_{0}>0$. Also, $a^{\theta}_{j}$ satisfies the
estimates

(2.31)
$\bigl{|}\partial_{x}^{\sigma}\partial_{\xi}^{\gamma}a^{\theta}_{j}(x,\xi)\bigr{|}\lesssim\theta^{-|\sigma|-|\gamma|},\,\,\,(x,\xi)\in
S^{*}\Omega$

related to this support property.

The $a^{\theta}_{j}$ provide “directional” microlocalization. We also need a
“height” localization since the characteristics of the symbols of our scaled
Schrödinger operators lie on paraboloids. The variable coefficient operators
that we shall use of course are adapted to our operators and are analogs of
ones that are used in the study of Fourier restriction problems involving
paraboloids.

To construct these, choose $b\in C^{\infty}_{0}({\mathbb{R}})$ supported in
$|s|\leq 1$ satisfying $\sum_{-\infty}^{\infty}b(s-\ell)\equiv 1$. We then
simply define the “height operator” as follows

(2.32)
$A_{\ell}^{\theta}(P)=b(\theta^{-1}\lambda^{-1}(P-\lambda\kappa^{\theta}_{\ell}))\,\Upsilon(P/\lambda),\quad\kappa^{\theta}_{\ell}=1+\theta\ell,\quad|\ell|\lesssim\theta^{-1},$

where if $\tilde{\beta}$ is as in (2.8)

(2.33) $\Upsilon\in C^{\infty}_{0}((1/10,10))\,\,\text{satisfies
}\,\,\,\Upsilon(r)=1\,\,\,\text{in a neighborhood of}\,\,\text{supp
}\,\tilde{\beta}.$

Thus, these operators microlocalize $P$ to intervals of size
$\approx\theta\lambda$ about “heights”
$\lambda\kappa^{\theta}_{\ell}\approx\lambda$. As we shall see below,
different “heights” will give rise to different “Schrödinger tubes” about
which the kernels of our microlocalization of the $\tilde{\sigma}_{\lambda}$
operators are highly concentrated. Also, standard arguments as in [20] show
that if $A^{\theta}_{\ell}(x,y)$ is the kernel of this operator then

(2.34) $A^{\theta}_{\ell}(x,y)=O(\lambda^{-N})\,\forall\,N,\quad\text{if
}\,d_{g}(x,y)\geq C_{0}\theta,$

for a fixed constant if $\theta\in[\lambda^{-\delta_{0}},1]$ with, as we are
assuming $\delta_{0}<1/2$.

If $\psi(x)\in C^{\infty}_{0}(\Omega)$ equals $1$ in a neighborhood of the
$x$-support of the $B(x,\xi)$ and $A^{\theta}_{j}(x,D_{x})$ is the operator
with symbol

(2.35) $A^{\theta}_{j}(x,\xi)=\psi(x)a^{\theta}_{j}(x,\xi),$

then for $\nu=(\theta j,\theta\ell)\in\theta{\mathbb{Z}}^{2(n-2)+1}$ we can
finally define the cutoffs that we shall use:

(2.36) $A^{\theta}_{\nu}=A^{\theta}_{j}(x,D_{x})\circ A^{\theta}_{\ell}(P).$

Let us collect several basic facts about the $A^{\theta}_{\nu}$ operators for
later use. First, if $A^{\theta}_{\nu}(x,\xi)$ and
$A^{\theta}_{\tilde{\nu}}(x,\xi)$ are the symbols of $A^{\theta}_{\nu}$ and
$A^{\theta}_{\tilde{\nu}}$, respectively, then

(2.37) $A^{\theta}_{\nu}(x,\xi)A^{\theta}_{\tilde{\nu}}(x,\xi)\equiv
0,\quad\text{if }\,\,\,|\nu-\tilde{\nu}|\geq C_{0}\theta,$

for some uniform constant $C_{0}$. Also, the principal symbol
$a_{\nu}^{\theta}(x,\xi)$ of $A^{\theta}_{\nu}$ are all supported in a
neighborhood of the support of $B(x,\xi)$, and satisfies

(2.38) $a^{\theta}_{\nu}(\chi_{r}(x,\xi))=a^{\theta}_{\nu}(x,\xi),\,\,\text{on
supp }B(x,\xi)\,\,\text{if }\,\,|r|\leq 2\delta,$

assuming that $\delta>0$ is small.

Besides, as operators between any $L^{p}\to L^{q}$, $1\leq p,q\leq\infty$,
spaces we have

(2.39)
$\tilde{\sigma}_{\lambda}=\sum_{\nu}\tilde{\sigma}_{\lambda}A^{\theta}_{\nu}+O(\lambda^{-N})\,\,\,\forall
N,$

and the $A^{\theta}_{\nu}$ are almost orthogonal in the sense that we have

(2.40)
$\sum_{\nu}\|A^{\theta}_{\nu}h\|_{L^{2}_{x}}^{2}\lesssim\|h\|_{L^{2}_{x}}^{2},$

with constants independent of $\theta\in[\lambda^{-\varepsilon_{0}},1]$.

Also, since for each $x$ the symbols vanish outside of cubes of sidelength
$\theta\lambda$ and
$|\partial^{\gamma}_{\xi}A^{\theta}_{\nu}(x,\xi)|=O((\lambda\theta)^{-|\gamma|})$,
we also have that their kernels are $O((\theta\lambda)^{n-1}(1+\theta\lambda
d_{g}(x,y))^{-N})$ for all $N$ and so

(2.41) $\|A_{\nu}^{\theta}\|_{L^{p}(M)\to L^{p}(M)}=O(1)\quad\forall\,1\leq
p\leq\infty.$

By interpolation, (2.40) and (2.41) imply

(2.42)
$\|A_{\nu}^{\theta}h\|_{\ell_{\nu}^{p}L^{p}(M)}\lesssim\|h\|_{L^{p}(M)}\quad\forall\,2\leq
p\leq\infty.$

In view of (2.39) we have for $\theta_{0}=\lambda^{-\varepsilon_{0}}$

(2.43)
$\bigl{(}\tilde{\sigma}_{\lambda}H\bigr{)}^{2}=\sum_{\nu,\tilde{\nu}}\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{)}\cdot\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H\bigr{)}+O(\lambda^{-N}\|H\|_{2}^{2}).$

Recall that in $A_{\nu}^{\theta_{0}}$,
$\nu\in\theta_{0}{\mathbb{Z}}^{2(n-2)+1}$ indexes a
$\lambda^{-\varepsilon_{0}}$-separated set in ${\mathbb{R}}^{2n-3}$. Here
$\varepsilon_{0}<\frac{1}{2}$ is small constant that we shall specify later,
the choice of $\varepsilon_{0}$ depends on the dimension $d=n-1$.

We need to organize the pairs of indices $\nu,\tilde{\nu}$ in (2.43) as in
many earlier works (see [18] and [23]). To this end, consider dyadic cubes,
$\tau^{\theta}_{\mu}$ in ${\mathbb{R}}^{2n-3}$ of sidelength
$\theta=2^{k}\theta_{0}$, with $\tau^{\theta}_{\mu}$ denoting translations of
the cube $[0,\theta)^{2n-3}$ by $\mu\in\theta{\mathbb{Z}}^{2n-3}$. Two such
dyadic cubes of sidelength $\theta$ are said to be close if they are not
adjacent but have adjacent parents of length $2\theta$, and, in this case, we
write $\tau^{\theta}_{\mu}\sim\tau^{\theta}_{\tilde{\mu}}$. We note that close
cubes satisfy
$\text{dist}(\tau^{\theta}_{\mu},\tau^{\theta}_{\tilde{\mu}})\approx\theta$,
and so each fixed cube has $O(1)$ cubes which are “close” to it. Moreover, as
noted in [23, p. 971], any distinct points
$\nu,\tilde{\nu}\in{\mathbb{R}}^{2n-3}$ must lie in a unique pair of close
cubes in this Whitney decomposition. So, there must be a unique triple
$(\theta=\theta_{0}2^{k},\mu,\tilde{\mu})$ such that
$(\nu,\tilde{\nu})\in\tau^{\theta}_{\mu}\times\tau^{\theta}_{\tilde{\mu}}$ and
$\tau^{\theta}_{\mu}\sim\tau^{\theta}_{\tilde{\mu}}$. We remark that by
choosing $B$ to have small support we need only consider
$\theta=2^{k}\theta_{0}\ll 1$.

Taking these observations into account, as in earlier works, we conclude that
the bilinear sum (2.43) can be organized as follows:

(2.44) $\sum_{\\{k\in{\mathbb{N}}:\,k\geq 10\,\,\text{and
}\,\theta=2^{k}\theta_{0}\ll
1\\}}\sum_{\\{(\mu,\tilde{\mu}):\,\tau^{\theta}_{\mu}\sim\tau^{\theta}_{\tilde{\mu}}\\}}\sum_{\\{(\nu,\tilde{\nu})\in\tau^{\theta}_{\mu}\times\tau^{\theta}_{\tilde{\mu}}\\}}\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{)}\cdot\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H\bigr{)}\\\
+\sum_{(\tau,\tilde{\tau})\in\Xi_{\theta_{0}}}\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{)}\cdot\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H\bigr{)},$

where $\Xi_{\theta_{0}}$ indexes the remaining pairs such that
$|\nu-\tilde{\nu}|\lesssim\theta_{0}=\lambda^{-\varepsilon_{0}}$, including
the diagonal ones where $\nu=\tilde{\nu}$.

The key estimate that we require, which follows from bilinear harmonic
analysis arguments, then is the following.

######  Proposition 2.3.

If $H=S_{\lambda}f$ is as in (2.2) and $(p,q)$ is as in (2.5), we have

(2.45)
$\|\tilde{\sigma}_{\lambda}H\|_{L^{p}_{t}L^{q}_{x}(A_{-})}\lesssim\bigl{\|}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}(M^{n-1}\times[0,T])}+\lambda^{\frac{1}{p}-}\|H\|_{L^{2}_{t,x}(M^{n-1}\times{\mathbb{R}})}.$

The $\lambda^{\frac{1}{p}-}$ notation that we are using for the last term in
(2.45) denotes $\lambda^{\frac{1}{p}-\varepsilon}$ for some unspecified
$\varepsilon>0$. Note that since $\|H\|_{L^{2}_{t,x}}\approx T^{1/2}$ for
$H=S_{\lambda}f$ and $T\approx\log\lambda$ the log-loss afforded by having the
last term involve this norm is more than overset by the power gain $1/p-$ of
$\lambda$. (2.45) is the place where we require the mixed–norm to be taken
over the set $A_{-}$. As we shall see later in the proof, the upper bound of
$\tilde{\sigma}_{\lambda}H$ on the set $A_{-}$ will allow us to fully exploit
the gain from bilinear estimates.

We shall postpone the proof of Proposition 2.3 until the next section. Let us
now see how we can use it to prove (2.17) on the set $A_{-}$. Given (2.45), it
suffices to show that when $M^{n-1}$ has nonpositive curvature

(2.46)
$\bigl{\|}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}(M^{n-1}\times[0,T])}\leq
C\lambda^{\frac{1}{p}},$

with $T=c_{0}\log\lambda$ for $c_{0}>0$ sufficiently small.

We shall also need the following simple lemma whose proof we postpone until
the end of this subsection.

######  Lemma 2.4.

If $\delta>0$ in (2.7) is small enough and
$\theta_{0}=\lambda^{-\varepsilon_{0}}$ we have for $B$ as in (2.10)

(2.47)
$\bigl{\|}\,B\sigma_{\lambda}A^{\theta_{0}}_{\nu}-BA^{\theta_{0}}_{\nu}\sigma_{\lambda}\,\bigr{\|}_{L^{2}_{t,x}\to{L^{p}_{t}L^{q}_{x}}}=O(\lambda^{\frac{1}{p}-\frac{1}{2}+2\varepsilon_{0}}).$

By (2.47) along with the fact that $\ell^{q}\subset\ell^{2}$ if $q\geq 2$ , we
have

(2.48) $\displaystyle\bigl{\|}$
$\displaystyle\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}}$
$\displaystyle\lesssim\bigl{\|}BA^{\theta_{0}}_{\nu}\sigma_{\lambda}H\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}}+\|\,(B\sigma_{\lambda}A^{\theta_{0}}_{\nu}-BA^{\theta_{0}}_{\nu}\sigma_{\lambda})H\,\|_{\ell_{\nu}^{2}L^{p}_{t}L^{q}_{x}}$
$\displaystyle\lesssim\bigl{\|}BA^{\theta_{0}}_{\nu}\sigma_{\lambda}H\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}}+\lambda^{\frac{1}{p}-\frac{1}{2}+2\varepsilon_{0}}\|\,H\,\|_{\ell_{\nu}^{2}L^{2}_{t,x}}.$

Since the number of choices of $\nu$ is $O(\lambda^{(2n-3)\varepsilon_{0}})$
and $H$ is independent of $\nu$, we have
$\|\,H\,\|_{\ell_{\nu}^{2}L^{2}_{t,x}}\lesssim\lambda^{(n-\frac{3}{2})\varepsilon_{0}}\|\,H\,\|_{L^{2}_{t,x}}$.
Thus if we choose $\varepsilon_{0}<\tfrac{1}{2n+1}$, the second term on the
right side of (2.48) is bounded by
$\lambda^{\frac{1}{p}-}\|H\|_{L^{2}_{t,x}}$.

On the other hand, since we are assuming $f$ is $L^{2}$ normalized in (2.18),
by (2.11), (2.42), (2.13) and the fact that $H=S_{\lambda}f$, we have

(2.49)
$\bigl{\|}BA^{\theta_{0}}_{\nu}\big{(}I-\sigma_{\lambda}\big{)}H\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}(M^{n-1}\times[0,T])}\leq\bigl{\|}\big{(}I-\sigma_{\lambda}\big{)}H\bigr{\|}_{L^{p}_{t}L^{q}_{x}(M^{n-1}\times[0,T])}\leq\lambda^{\frac{1}{p}}.$

Thus, by (2.48) we would have (2.46) if

(2.50)
$\bigl{\|}BA^{\theta_{0}}_{\nu}H\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}(M^{n-1}\times[0,T])}\lesssim\lambda^{\frac{1}{p}},$

which, by (2.11) and the fact that $H=S_{\lambda}f$, is a consequence of

(2.51)
$\bigl{\|}A^{\theta_{0}}_{\nu}S_{\lambda}f\bigr{\|}_{L^{p}_{t}\ell_{\nu}^{q}L^{q}_{x}(M^{n-1}\times[0,T])}\lesssim\lambda^{\frac{1}{p}}.$

To prove (2.51), let us define

$f\to\bigl{(}Wf\bigr{)}(x,t,\nu)=\eta(t/T)\bigl{(}A_{\nu}^{\theta_{0}}\circ
e^{-it\lambda^{-1}\Delta_{g}}f\bigr{)}(x).$

By applying the abstract theorem of Keel-Tao [17] and a simple rescaling
argument, we would have (2.51) if

(2.52) $\|Wf(t,\cdot)\|_{\ell_{\nu}^{2}L^{2}_{x}}\leq C\|f\|_{L^{2}_{x}},$

and

(2.53) $\|W(t)W^{*}(s)G\|_{\ell_{\nu}^{\infty}L^{\infty}_{x}}\leq
C\lambda^{\frac{n-1}{2}}\,|t-s|^{-\frac{n-1}{2}}\|G\|_{\ell_{\nu}^{1}L^{1}_{x}},$

with

(2.54) $\displaystyle WW^{*}G(x,t,\nu)=$
$\displaystyle=\eta(t/T)\sum_{\nu^{\prime}}\int_{-\infty}^{\infty}\eta(s/T)\Bigl{[}\bigr{(}A_{\nu}^{\theta_{0}}e^{-i(t-s)\lambda^{-1}\Delta_{g}}(A^{\theta_{0}}_{\nu^{\prime}})^{*}\bigr{)}G(\,\cdot\,,s,\nu^{\prime})\Bigr{]}(x)\,ds$
$\displaystyle=\sum_{\nu^{\prime}}\iint
K(x,t,\nu;y,s,\nu^{\prime})\,G(y,s,\nu^{\prime})\,dyds,$

where

(2.55)
$K(x,t,\nu;y,s,\nu^{\prime})=\eta(t/T)\bigr{(}A_{\nu}^{\theta_{0}}e^{-i(t-s)\lambda^{-1}\Delta_{g}}(A^{\theta_{0}}_{\nu^{\prime}})^{*}\bigr{)}(x,y)\,\eta(s/T).$

It is not hard to check that (2.52) follows from (2.41) and the fact that
$e^{-it\lambda^{-1}\Delta_{g}}$ is unitary, and (2.53) is a consequence of the
kernel estimates

(2.56) $|K(x,t,\nu;y,s,\nu^{\prime})|\leq
C\lambda^{\frac{n-1}{2}}\,|t-s|^{-\frac{n-1}{2}},$

which follows from Proposition 4.2 in [3].

Here compared with (2.25), there is no $e^{CT}$ loss in the above kernel
estimates due to the presence of the $A_{\nu}^{\theta_{0}}$ operators. The
proof of (2.56) follows from exploiting the directional localization in the
$A_{\nu}^{\theta_{0}}$ operators through the use of Toponogov’s triangle
comparison theorem. See section 4 in [3], and also recent works [6] and [7]
for more details. Also note that here we take
$\theta_{0}=\lambda^{-\varepsilon_{0}}$, which may be much larger than
$\theta_{0}=\lambda^{-\frac{1}{8}}$ as in [3]; however, the proof of the
microlocalized kernel estimates extends to the larger $\theta_{0}$ similarly
as long as we fix $T=c_{0}\log\lambda$ for some $c_{0}\ll\varepsilon_{0}$.

This completes the proof of (2.17) on the set $A_{-}$ up to proving the
crucial local estimates in Proposition 2.3, which we shall postpone to the
next section.

The other task remaining to complete the proofs Theorems 1.1 is to prove the
commutator estimate that we employed:

###### Proof of Lemma 2.4.

The proof follows from similar arguments as in Lemma 2.7 in [3], we include
the details here for completeness. Recall that by (2.10) the symbol
$B(x,\xi)=B_{\lambda}(x,\xi)\in S^{0}_{1,0}$ vanishes when $|\xi|$ is not
comparable to $\lambda$. In particular, it vanishes if $|\xi|$ is larger than
a fixed multiple of $\lambda$, and it belongs to a bounded subset of
$S^{0}_{1,0}$. Furthermore, if $a^{\theta_{0}}_{\nu}(x,\xi)$ is the principal
symbol of our zero-order dyadic microlocal operators, we recall that by (2.38)
we have that for $\delta>0$ small enough

(2.57)
$a^{\theta_{0}}_{\nu}(x,\xi)=a^{\theta_{0}}_{\nu}(\chi_{r}(x,\xi))\quad\text{on
supp }\,B_{\lambda}\,\,\,\text{if }\,\,|r|\leq 2\delta,$

where $\chi_{r}:\,T^{*}M^{n-1}\,\backslash 0\to T^{*}M^{n-1}\,\backslash 0$
denotes geodesic flow in the cotangent bundle.

By Sobolev estimates for $M^{n-1}\times{\mathbb{R}}$, in order to prove
(2.47), it suffices to show that

(2.58)
$\Bigl{\|}\,\Bigl{(}\sqrt{I+P^{2}+D_{t}^{2}}\,\,\Bigr{)}^{(n-1)(\frac{1}{2}-\frac{1}{q})+\frac{1}{2}-\frac{1}{p}}\,\bigl{[}B_{\lambda}\sigma_{\lambda}A^{\theta_{0}}_{\nu}-B_{\lambda}A^{\theta_{0}}_{\nu}\sigma_{\lambda}\bigr{]}\,\Bigr{\|}_{L^{2}_{t,x}\to
L^{2}_{t,x}}\\\ =O(\lambda^{\frac{1}{p}-\frac{1}{2}+2\varepsilon_{0}}).$

Note that for $(p,q)$ as in (2.5),
$(n-1)(\frac{1}{2}-\frac{1}{q})+\frac{1}{2}-\frac{1}{p}=\frac{1}{2}+\frac{1}{p}$,
and since the symbol of $B_{\lambda}$ and $\sigma_{\lambda}$ are supported in
$|\xi|,|\tau|\approx\lambda$, in order to prove (2.58), it suffices to show
that

(2.59)
$\Bigl{\|}\,B_{\lambda}\sigma_{\lambda}A^{\theta_{0}}_{\nu}-B_{\lambda}A^{\theta_{0}}_{\nu}\sigma_{\lambda}\,\Bigr{\|}_{L^{2}_{t,x}\to
L^{2}_{t,x}}=O(\lambda^{-1+2\varepsilon_{0}}).$

To prove this we recall that

$\sigma_{\lambda}=(2\pi)^{-1}\tilde{\beta}(D_{t}/\lambda)\int\hat{\sigma}(r)e^{ir\lambda^{1/2}|D_{t}|^{1/2}}\,e^{-irP}\,dr,$

and, therefore, since $e^{ir\lambda^{1/2}|D_{t}|^{1/2}}$ has $L^{2}\to L^{2}$
norm one and commutes with $B_{\lambda}$ and $A^{\theta_{0}}_{\nu}$, and since
$\hat{\sigma}(r)=0$, $|r|\geq 2\delta$, by Minkowski’s integral inequality, we
would have (2.59) if

(2.60) $\sup_{|r|\leq
2\delta}\,\Bigl{\|}\,\,\tilde{\beta}(D_{t}/\lambda)\,\bigl{[}B_{\lambda}e^{-irP}A^{\theta_{0}}_{\nu}-B_{\lambda}A^{\theta_{0}}_{\nu}e^{-irP}\bigr{]}\,\Bigr{\|}_{L^{2}_{t,x}\to
L^{2}_{t,x}}=O(\lambda^{-1+2\varepsilon_{0}}).$

Next, to be able to use Egorov’s theorem, we write

$\bigl{[}B_{\lambda}e^{-irP}A^{\theta_{0}}_{\nu}-B_{\lambda}A^{\theta_{0}}_{\nu}e^{-irP}\bigr{]}=B_{\lambda}\,\bigl{[}(e^{-irP}A^{\theta_{0}}_{\nu}e^{irP})-B_{\lambda}A^{\theta_{0}}_{\nu}]\circ
e^{-irP}.$

Since $e^{-irP}$ also has $L^{2}$-operator norm one, we would obtain (2.60)
from

(2.61)
$\Bigl{\|}\,\,\tilde{\beta}(D_{t}/\lambda)\,B_{\lambda}\,\bigl{[}(e^{-irP}A^{\theta_{0}}_{\nu}e^{irP})-A^{\theta_{0}}_{\nu}\bigr{]}\,\Bigr{\|}_{L^{2}_{t,x}\to
L^{2}_{t,x}}=O(\lambda^{-1+2\varepsilon_{0}}),$

which is a simple consequence of the Egorov’s theorem, see e.g., Taylor [24,
§VIII.1] and Lemma 2.7 in [3] for more details. ∎

##  3\. Proof of Proposition 2.3

Let us fix $p=2,q=q_{e}=\frac{2(n-1)}{n-3}$, we shall first give the arguments
for $d=n-1\geq 4$, and later modify it for $n-1=3$. The case $d=n-1=2$ follows
from similar arguments, which we shall briefly discuss at the end of this
section.

To prove (2.45), the strategy is similar to [3], which is related to the ideas
in Blair and Sogge [7] and earlier works, especially Tao, Vargas and Vega [23]
and Lee [18].

We first note that if $\delta$ as in (2.7) is small enough we have

(3.1)
$\tilde{\sigma}_{\lambda}-\sum_{\nu}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}=R_{\lambda},\,\,\text{where
}\,\,\|R_{\lambda}H\|_{L^{\infty}_{t,x}}\lesssim\lambda^{-N}\|H\|_{L^{2}_{t,x}}\,\,\forall
N.$

Thus, we have

(3.2)
$\bigl{(}\,\tilde{\sigma}_{\lambda}H\,\bigr{)}^{2}=\sum_{\nu,\tilde{\nu}}\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{)}\cdot\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H\bigr{)}+O(\lambda^{-N}\|H\|_{L^{2}_{t,x}}^{2})\,\,\forall\,N.$

As in earlier works, let

(3.3)
$\Upsilon^{\text{diag}}(H)=\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{)}\cdot\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H\bigr{)},$

and

(3.4)
$\Upsilon^{\text{far}}(H)=\sum_{(\nu,\tilde{\nu})\notin\Xi_{\theta_{0}}}\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{)}\cdot\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H\bigr{)}+O(\lambda^{-N}\|H\|_{L^{2}_{t,x}}^{2}),$

with the last term denoting the error term in (3.2). Thus,

(3.5)
$\bigl{(}\tilde{\sigma}_{\lambda}H\bigr{)}^{2}=\Upsilon^{\text{diag}}(H)+\Upsilon^{\text{far}}(H).$

Here, the summation in $\Upsilon^{\text{diag}}(H)$ is over near diagonal pairs
$(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}$ as in (2.44). In particular we have
$|\nu-\tilde{\nu}|\leq C\theta_{0}$ for some uniform constant as
$\nu,\tilde{\nu}$ range over $\theta_{0}{\mathbb{Z}}^{(2n-3)}$. The other term
$\Upsilon^{\text{far}}(H)$ is the remaining pairs, which include many which
are far from the diagonal. This sum will provide the contribution to the last
term in (2.45).

The two types of terms here are treated differently, as in analyzing parabolic
restriction problems or spectral projection estimates.

We shall treat the first term in the right of (3.5) as in [2] and [7] by using
a variable coefficient variant of Lemma 6.1 in [23] (see also Lemma 4.2 in
[7]):

######  Lemma 3.1.

If $\Upsilon^{\text{diag}}(H)$ is as in (3.5) and $d=n-1\geq 4$, then we have
the uniform bounds

(3.6)
$\|\Upsilon^{\text{diag}}(H)\|_{L^{1}_{t}L^{q_{e}/2}_{x}}\lesssim\|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\|^{2}_{L^{2}_{t}\ell_{\nu}^{q_{e}}L^{q_{e}}_{x}}\,+O(\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

We also need the following estimate for $\Upsilon^{\text{far}}(H)$ which is a
consequence of the bilinear oscillatory integral estimates of Lee [18] and
arguments of Blair and Sogge in [4], [5] and [7].

######  Lemma 3.2.

If $\Upsilon^{\text{far}}(H)$ is as in (3.4), and, as above
$\theta_{0}=\lambda^{-\varepsilon_{0}}$, then for all $\varepsilon>0$ we have
for $H=S_{\lambda}f$

(3.7)
$\iint|\Upsilon^{\text{far}}(H)|^{q/2}\,dxdt\lesssim_{\varepsilon}\lambda^{1+\varepsilon}\,\bigl{(}\lambda^{1-\varepsilon_{0}}\bigr{)}^{\frac{n-1}{2}(q-\frac{2(n+1)}{n-1}))}\,\|H\|_{L^{2}_{t,x}}^{q},\quad\text{if
}\,q=\tfrac{2(n+2)}{n}.$

Let us postpone the proofs of these two lemmas for a bit and show how they can
be used to obtain Proposition 2.3 if $d=n-1\geq 4$.

If we let $q=\tfrac{2(n+2)}{n}$ as in Lemma 3.2, we note that $q<q_{e}$ and
also

(3.8) $|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{q_{e}}\leq
2^{q/2}\,|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\cdot\bigl{(}\,|\Upsilon^{\text{diag}}(H)|^{q/2}+|\Upsilon^{\text{far}}(H)|^{q/2}\,\bigr{)}.$

Thus,

(3.9) $\displaystyle\|$
$\displaystyle\tilde{\sigma}_{\lambda}H\|_{L^{2}_{t}L^{q_{e}}_{x}(A_{-})}^{2}=\int\Big{(}\int\bigl{|}\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H\bigr{|}^{q_{e}/2}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\lesssim\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|\Upsilon^{\text{diag}}(H)|^{q/2}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\quad+\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|\Upsilon^{\text{far}}(H)|^{q/2}\,dx\Big{)}^{\frac{2}{q_{e}}}dt=I+II.$

To estimate $II$, first note that
$\tilde{\sigma}_{\lambda}H=\tilde{S}_{\lambda}f$ if $H=S_{\lambda}f$, we have

(3.10) $\displaystyle II$
$\displaystyle\lesssim\|\tilde{S}_{\lambda}f\|_{L^{\infty}(A_{-})}^{\frac{2(q_{e}-q)}{q_{e}}}\cdot\|\Upsilon^{\text{far}}(H)\|_{L^{\frac{q}{q_{e}}}_{t}L^{\frac{q}{2}}_{x}}^{\frac{q}{q_{e}}}$
$\displaystyle\lesssim
T^{(\frac{q_{e}}{q}-\frac{2}{q})\cdot\frac{q}{q_{e}}}\|\tilde{S}_{\lambda}f\|_{L^{\infty}(A_{-})}^{\frac{2(q_{e}-q)}{q_{e}}}\cdot\|\Upsilon^{\text{far}}(H)\|_{L^{\frac{q}{2}}_{t}L^{\frac{q}{2}}_{x}}^{\frac{q}{q_{e}}},$

where in the second line we used Hölder’s inequality on the time interval
$[0,T]$. By (3.7) and the ceiling for $A_{-}$, we have

(3.11) $\displaystyle II\leq
T^{(\frac{q_{e}}{q}-\frac{2}{q})\cdot\frac{q}{q_{e}}}\lambda^{(\frac{n-1}{4}+\varepsilon_{1})(\frac{2(q_{e}-q)}{q_{e}})}\cdot\Big{(}\lambda^{1+\varepsilon}\,\bigl{(}\lambda^{1-\varepsilon_{0}}\bigr{)}^{\frac{n-1}{2}(q-\frac{2(n+1)}{n-1})}\Big{)}^{\frac{2}{q_{e}}}\|H\|^{\frac{2q}{q_{e}}}_{L^{2}_{x}}.$

If we take $\varepsilon_{0},\varepsilon_{1}$ and $\varepsilon$ to be small
enough, e.g., $\varepsilon=\varepsilon_{1}=\frac{1}{100}$ and
$\varepsilon_{0}=\frac{1}{2n+2}$, it is straightforward to check that

(3.12) $\displaystyle
II=O(\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{\frac{2q}{q_{e}}})=O(\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

Here we also used the fact that $\|H\|_{L^{2}_{t,x}}^{2}$ dominates
$\|H\|_{L^{2}_{t,x}}^{\frac{2q}{q_{e}}}$ since $q_{e}>q$ and
$\|H\|_{L^{2}_{t,x}}\approx T$ since $H=S_{\lambda}f$, $\|f\|_{2}=1$ and
$e^{-it\lambda^{-1}\Delta_{g}}$ is a unitary operator on $L^{2}_{x}$.

Consequently, we just need to see that $I^{1/2}$ is dominated by the other
term in the right side of this inequality To estimate this term we use
Hölder’s inequality followed by Young’s inequality and Lemma 3.1 to see that

$\displaystyle I=$
$\displaystyle\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|\Upsilon^{\text{diag}}(H)|^{\frac{q}{2}}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\leq$
$\displaystyle\int\Big{(}\|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H\|_{L_{x}^{\frac{q_{e}}{2}}}^{\frac{q_{e}-q}{2}}\,\|\Upsilon^{\text{diag}}(H)\|_{L_{x}^{\frac{q_{e}}{2}}}^{\frac{q}{2}}\,\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\leq$
$\displaystyle\|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H\|_{L^{1}_{t}L_{x}^{\frac{q_{e}}{2}}}^{\frac{q_{e}-q}{q_{e}}}\,\|\Upsilon^{\text{diag}}(H)\|_{L^{1}_{t}L_{x}^{\frac{q_{e}}{2}}}^{\frac{q}{q_{e}}}$
$\displaystyle\leq$
$\displaystyle\tfrac{q_{e}-q}{q_{e}}\|\tilde{\sigma}_{\lambda}H\|^{2}_{L^{2}_{t}L_{x}^{q_{e}}}+\tfrac{q}{q_{e}}\|\Upsilon^{\text{diag}}(H)\|_{L^{1}_{t}L_{x}^{\frac{q_{e}}{2}}}$
$\displaystyle\leq$
$\displaystyle\tfrac{q_{e}-q}{q_{e}}\|\tilde{\sigma}_{\lambda}H\|^{2}_{L^{2}_{t}L_{x}^{q_{e}}}+C(\|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\|^{2}_{L^{2}_{t}\ell_{\nu}^{q_{e}}L^{q_{e}}_{x}}\,+\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

Since $\tfrac{q_{e}-q}{q_{e}}<1$, the first term in the right can be absorbed
in the left side of (3.9), and this, along with the estimate for $II$ above
yields (2.45).

Thus, if we can prove Lemma 3.1 and Lemma 3.2, the proof of Proposition 2.3
for $d=n-1\geq 4$ will be complete.

###### Proof of Lemma 3.1.

To prove (3.6), let us first define slightly wider microlocal cutoffs by
setting

(3.13) $\tilde{A}^{\theta_{0}}_{\nu}=\sum_{|\mu-\nu|\leq
C_{0}\theta_{0}}A^{\theta_{0}}_{\mu}.$

We can fix $C_{0}$ large enough so that

(3.14)
$\|A^{\theta_{0}}_{\nu}-A^{\theta_{0}}_{\nu}\tilde{A}^{\theta_{0}}_{\nu}\|_{L^{p}_{x}\to
L^{p}_{x}}=O(\lambda^{-N})\,\,\forall\,N\,\,\text{if }\,\,1\leq p\leq\infty.$

Also, like the original $A^{\theta_{0}}_{\nu}$ operators the
$\tilde{A}^{\theta_{0}}_{\nu}$ operators are almost orthogonal

(3.15)
$\sum_{\nu}\|\tilde{A}^{\theta_{0}}_{\nu}h\|^{2}_{L^{2}_{x}}\lesssim\|h\|_{L^{2}_{x}}^{2}.$

Since by (2.11) and (2.14), we have

(3.16)
$\|\tilde{\sigma}_{\lambda}F\|_{L^{2}_{t}L^{q_{e}}_{x}(M^{n-1}\times[0,T])}\leq
C\lambda^{\frac{1}{2}}\|F\|_{L^{2}_{t,x}},$

we conclude that, in order to prove (3.6), we may replace
$\Upsilon^{\text{diag}}(H)$ by $\tilde{\Upsilon}^{\text{diag}}(H)$ where the
latter is defined by the analog of (3.3) with $A^{\theta_{0}}_{\nu}$ and
$A^{\theta_{0}}_{\tilde{\nu}}$ replaced by
$A^{\theta_{0}}_{\nu}\tilde{A}^{\theta_{0}}_{\nu}$ and
$A^{\theta_{0}}_{\tilde{\nu}}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}$,
respectively.

So, it suffices to prove

(3.17)
$\displaystyle\bigl{\|}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}(\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}\tilde{A}^{\theta_{0}}_{\nu}H)\cdot(\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}$
$\displaystyle H)\bigr{\|}_{L^{1}_{t}L^{q_{e}/2}_{x}}$
$\displaystyle\lesssim\|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\|^{2}_{L^{2}_{t}\ell_{\nu}^{q_{e}}L^{q_{e}}_{x}}\,+O(\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

We shall need the following variant of (2.47),

(3.18)
$\|[\,\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}-A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\,]F\|_{L^{2}_{t}L^{q_{e}}_{x}}\lesssim\lambda^{2\varepsilon_{0}}\|F\|_{L^{2}_{t,x}}.$

This follows from the proof of Lemma 2.4, (2.47) and the fact that the
commutator $[B,A^{\theta_{0}}_{\nu}]$ is bounded on $L^{q_{e}}_{x}(M^{n-1})$
with norm $O(\lambda^{-1+\varepsilon_{0}})$. By (3.15) and (3.18) we would
have (3.17) if we could show that

(3.19)
$\displaystyle\bigl{\|}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}(A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\nu}H)\cdot(A^{\theta_{0}}_{\tilde{\nu}}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}$
$\displaystyle H)\bigr{\|}_{L^{1}_{t}L^{q_{e}/2}_{x}}$
$\displaystyle\lesssim\|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\|^{2}_{L^{2}_{t}\ell_{\nu}^{q_{e}}L^{q_{e}}_{x}}\,+O(\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

Since the $A^{\theta_{0}}_{\nu}$ operators are time independent, we claim
that, it suffices to show that for arbitrary $h_{\nu},h_{\tilde{\nu}}$, which
may depend on $\nu$ and $\tilde{\nu}$,

(3.20)
$\displaystyle\bigl{\|}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}A^{\theta_{0}}_{\nu}h_{\nu}\cdot
A^{\theta_{0}}_{\tilde{\nu}}h_{\tilde{\nu}}\bigr{\|}_{L^{q_{e}/2}_{x}}\lesssim$
$\displaystyle\Bigl{(}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|A^{\theta_{0}}_{\nu}h_{\nu}\cdot
A^{\theta_{0}}_{\tilde{\nu}}h_{\tilde{\nu}}\|_{L^{q_{e}/2}_{x}}^{q_{e}/2}\,\Bigr{)}^{2/q_{e}}$
$\displaystyle\qquad+O\big{(}\lambda^{-N}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|h_{\nu}\|_{L^{1}_{x}}\|h_{\tilde{\nu}}\|_{L^{1}_{x}}\big{)},\,\,\,\forall
N.$

To verify the claim, note that if we take
$h_{\nu}=\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\nu}H$ and
$h_{\tilde{\nu}}=\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}H$,
(3.20) implies

(3.21) $\displaystyle\bigl{\|}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}$
$\displaystyle(A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\nu}H)\cdot(A^{\theta_{0}}_{\tilde{\nu}}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}H)\bigr{\|}_{L^{1}_{t}L^{q_{e}/2}_{x}}$
$\displaystyle\lesssim\int\Bigl{(}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|(A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\nu}H)\cdot(A^{\theta_{0}}_{\tilde{\nu}}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}H)\|_{L^{q_{e}/2}_{x}}^{q_{e}/2}\,\Bigr{)}^{2/q_{e}}dt\,$
$\displaystyle\qquad+\lambda^{-N}\int\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|(A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\nu}H)\|_{L^{1}_{x}}\|(A^{\theta_{0}}_{\tilde{\nu}}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}H)\|_{L^{1}_{x}}\,dt.$
$\displaystyle\lesssim\|A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}H\|^{2}_{L^{2}_{t}\ell_{\nu}^{q_{e}}L^{q_{e}}_{x}}\,+O(\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

Here we used the fact that for fixed $\nu$, the number of choices of
$\tilde{\nu}$ is finite, and for each pair
$(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}$

(3.22)
$\int\|(A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\nu}H)\|_{L^{1}_{x}}\|(A^{\theta_{0}}_{\tilde{\nu}}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}H)\|_{L^{1}_{x}}\,dt\\\
\leq\int\|(A^{\theta_{0}}_{\nu}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\nu}H)\|_{L^{2}_{x}}\|(A^{\theta_{0}}_{\tilde{\nu}}\tilde{\sigma}_{\lambda}\tilde{A}^{\theta_{0}}_{\tilde{\nu}}H)\|_{L^{2}_{x}}\,dt=O(\|H\|_{L^{2}_{t,x}}^{2}).$

If we repeat earlier arguments and use (3.14) again, we conclude that the
first term in the right side of (3.21) is dominated by the right side of
(3.19).

Thus, it remains to prove (3.20). By duality, if we take
$r=(q_{e}/2)^{\prime}$ so that $r$ is the conjugate exponent for $q_{e}/2$,
(3.20) is equivalent to

(3.23) $\displaystyle\Bigl{|}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}$
$\displaystyle\iint A^{\theta_{0}}_{\nu}h_{\nu}\cdot
A^{\theta_{0}}_{\tilde{\nu}}h_{\tilde{\nu}}\,\cdot G\,dx\,\Bigr{|}$
$\displaystyle\lesssim\Bigl{(}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|A^{\theta_{0}}_{\nu}h_{\nu}\cdot
A^{\theta_{0}}_{\tilde{\nu}}h_{\tilde{\nu}}\|_{L^{q_{e}/2}_{x}}^{q_{e}/2}\,\Bigr{)}^{2/q_{e}}$
$\displaystyle\qquad\qquad\qquad+O\big{(}\lambda^{-N}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|h_{\nu}\|_{L^{1}_{x}}\|h_{\tilde{\nu}}\|_{L^{1}_{x}}\big{)}\,\,\,\text{if
}\,\|G\|_{L^{r}_{x}}=1.$

To prove (3.23), note that if $x$ and $\nu$ are fixed and $\xi\to
A^{\theta_{0}}_{\nu}(x,\xi)$ does not vanish identically, then this function
of $\xi$ is supported in a cube
$Q^{\theta_{0}}_{\nu}(x)\subset{\mathbb{R}}^{n-1}_{\xi}$ of sidelength
$\approx\lambda^{1-\varepsilon_{0}}$. The cubes can be chosen so that, if
$\eta_{\nu}(x)$ is its center, then
$\partial^{\gamma}_{x}\eta_{\nu}(x)=O(\lambda)$ for all multi-indices
$\gamma$. Keeping this in mind it is straightforward to construct for every
pair $(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}$ symbols
$b_{\nu,\tilde{\nu}}(x,\xi)$ belonging to a bounded subset of
$S^{0}_{1-\varepsilon_{0},\varepsilon_{0}}$ satisfying

(3.24) $b_{\nu,\tilde{\nu}}(x,\eta)=1\,\,\text{if
}\,\,\text{dist}\bigl{(}\eta,\,\text{supp}_{\xi}A^{\theta_{0}}_{\nu}(x,\xi)\,+\,\text{supp}_{\xi}A^{\theta_{0}}_{\tilde{\nu}}(x,\xi)\bigr{)}\leq\lambda^{1-\varepsilon_{0}},$

with “$+$” denoting the algebraic sum. Using this and a simple integration by
parts argument shows that for every pair
$(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}$

(3.25)
$\bigl{\|}(I-b_{\nu,\tilde{\nu}}(x,D))\bigl{[}A^{\theta_{0}}_{\nu}h_{\nu}\cdot
A^{\theta_{0}}_{\tilde{\nu}}h_{\tilde{\nu}}]\bigr{]}\bigr{\|}_{L^{\infty}_{x}}\leq
C_{N}\lambda^{-N}\|h_{\nu}\|_{L^{1}_{x}}\|h_{\tilde{\nu}}\|_{L^{1}_{x}},\quad\forall\,N.$

The symbols can also be chosen so that $b_{\nu_{1},\tilde{\nu}_{1}}(x,\xi)$
and $b_{\nu_{2},\tilde{\nu}_{2}}(x,\xi)$ have disjoint supports if
$(\nu_{j},\tilde{\nu}_{j})\in\Xi_{\theta_{0}}$, $j=1,2$ and
$\min(|(\nu_{1}-\nu_{2},\tilde{\nu}_{1}-\tilde{\nu}_{2})|,\,|(\nu_{1}-\tilde{\nu}_{2},\tilde{\nu}_{1}-\nu_{2})|)\geq
C_{2}\theta_{0}$ with $C_{2}$ being a fixed constant independent of $\lambda$
since all pairs in $\Xi_{\theta_{0}}$ are nearly diagonal. Due to this, the
adjoints, $b^{*}_{\nu,\tilde{\nu}}(x,D)$ are almost orthogonal in the sense
that we have the uniform bounds

(3.26)
$\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|b^{*}_{\nu,\tilde{\nu}}(x,D)h\|^{2}_{L^{2}_{x}}\lesssim\|h\|^{2}_{L^{2}_{x}}.$

Since
$\text{supp}_{\xi}A^{\theta_{0}}_{\nu}(x,\xi)\,+\,\text{supp}_{\xi}A^{\theta_{0}}_{\tilde{\nu}}(x,\xi)$
is contained in a cube of sidelength $\approx\lambda^{1-\varepsilon_{0}}$ and
can be chosen to have center $\eta_{\nu,\tilde{\nu}}(x)$ satisfying
$\partial^{\gamma}_{x}\eta_{\nu,\tilde{\nu}}(x)=O(\lambda)$, we can
furthermore assume that we have the uniform bounds

(3.27)
$\sup_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|b^{*}_{\nu,\tilde{\nu}}(x,D)h\|_{L^{\infty}_{x}}\lesssim\|h\|_{L^{\infty}_{x}}.$

We have now set up our variable coefficient version of the simple argument in
[23] that will allow us to obtain (3.23). First, by (3.25), it suffices to
estimate the left side of (3.23) with $G$ replaced by
$b^{*}_{\nu,\tilde{\nu}}(x,D)G$. By Hölder’s inequality,

(3.28)
$\Bigl{|}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\iint(A^{\theta_{0}}_{\nu}h_{\nu}\cdot
A^{\theta_{0}}_{\tilde{\nu}}h_{\tilde{\nu}}\,\cdot\bigl{(}b^{*}_{\nu,\tilde{\nu}}(x,D)G\bigr{)}\,dx\,\Bigr{|}\\\
\leq\Bigl{(}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|A^{\theta_{0}}_{\nu}h_{\nu}\cdot
A^{\theta_{0}}_{\tilde{\nu}}h_{\tilde{\nu}}\|_{L^{q_{e}/2}_{x}}^{q_{e}/2}\,\Bigr{)}^{2/q_{e}}\cdot\Bigl{(}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|b^{*}_{\nu,\tilde{\nu}}(x,D)G\|_{L^{r}_{x}}^{r}\Bigr{)}^{1/r}.$

Note that when $n-1\geq 4$, $q_{e}\in(2,4]$, thus $r\in[2,\infty)$. So, if we
use (3.26), (3.27) and an interpolation argument we conclude that

(3.29)
$\Bigl{(}\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\|b^{*}_{\nu,\tilde{\nu}}(x,D)G\|_{L^{r}_{x}}^{r}\Bigr{)}^{1/r}=O(1),$

for $G$ as in (3.23). ∎

###### Proof of Lemma 3.2.

To prove this let $\alpha\in C^{\infty}_{0}((-1,1))$ with $\alpha\equiv 1$ in
$(-\frac{1}{2},\frac{1}{2})$, then if $\alpha_{m}(t)=\alpha(t-m)$, up to a
$\log\lambda$ loss, it suffices to show that for $m=1,2,\dots,T$

(3.30)
$\iint|\alpha_{m}(t)\Upsilon^{\text{far}}(H)|^{q/2}\,dxdt\lesssim_{\varepsilon}\lambda^{1+\varepsilon}\,\bigl{(}\lambda^{1-\varepsilon_{0}}\bigr{)}^{\frac{n-1}{2}(q-\frac{2(n+1)}{n-1})}\,\|H\|_{L^{2}_{t,x}}^{q}.$

This follow from Lemma 3.2 in [3], where we replace $\lambda^{7/8}$ there by
$\lambda^{1-\varepsilon_{0}}$. The proof uses bilinear oscillatory integral
estimates of Lee [18] and arguments of previous work of Blair and Sogge in
[4], [5] and [7]. And this is also where we used the condition on
$\delta,\delta_{0}$ in (2.7), in order to apply the bilinear oscillatory
integral theorems of Lee [18], see section 3 in [3] for more details.

The constants in the right side of (3.30) is better than applying classical
linear estimate. Actually one can also rewrite the right side as

(3.31)
$\lambda^{\frac{n-1}{2}(\frac{q}{2}-1)}\lambda^{\varepsilon}\,\bigl{(}\lambda^{\frac{1}{2}-\varepsilon_{0}}\bigr{)}^{\frac{n-1}{2}(q-\frac{2(n+1)}{n-1})}\,\|H\|_{L^{2}_{t,x}}^{q},$

thus if $\varepsilon_{0}\leq 1/2$, we have a gain compared with the bound
$\lambda^{\frac{n-1}{2}(\frac{q}{2}-1)}$, which follows from applying linear
estimates. ∎

Modified arguments for $n-1=3$.

When $n-1=3$, note that $q_{e}=\tfrac{2(n-1)}{n-3}=6$, thus
$r=(q_{e}/2)^{\prime}=\frac{3}{2}$, which means that we do not have (3.29) and
thus Lemma 3.1 in this case, so we need to modify the arguments above. If we
repeat the previous arguments, it suffices to estimate the first term

(3.32)
$I=\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|\Upsilon^{\text{diag}}(H)|^{\frac{q}{2}}\,dx\Big{)}^{\frac{2}{q_{e}}}dt,$

where as in (3.3),

$\Upsilon^{\text{diag}}(H)=\sum_{(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\bigr{)}\cdot\bigl{(}\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H\bigr{)}.$

For later use, note that when $n-1=3$, it is not hard to check that
$\,(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}$ implies $|\nu-\tilde{\nu}|\leq
2^{13}\theta_{0}$.

Let us define

$T_{\nu}H=\sum_{\tilde{\nu}:\,(\nu,\tilde{\nu})\in\Xi_{\theta_{0}}}(\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H)(\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\tilde{\nu}}H),$

and write

(3.33) $\displaystyle(\Upsilon^{\text{diag}}(H))^{2}$
$\displaystyle=\big{(}\sum_{\nu}T_{\nu}H\big{)}^{2}$
$\displaystyle=\sum_{\nu_{1},\nu_{2}}T_{\nu_{1}}HT_{\nu_{2}}H.$

Now we can employ another Whitney-type decomposition, more explicitly, the sum
in (3.33) can be organized as

(3.34) $\displaystyle\big{(}\sum_{\\{k\in{\mathbb{N}}:\,k\geq 20\,\,\text{and
}\,\theta=2^{k}\theta_{0}\ll
1\\}}\,\,\sum_{\\{(\mu_{1},\mu_{2}):\,\tau^{\theta}_{\mu_{1}}\sim\tau^{\theta}_{\mu_{2}}\\}}\sum_{\\{(\nu_{1},\nu_{2})\in\tau^{\theta}_{\mu_{1}}\times\tau^{\theta}_{\mu_{2}}\\}}+\sum_{(\nu_{1},\nu_{2})\in\overline{\Xi}_{\theta_{0}}}\big{)}T_{\nu_{1}}hT_{\nu_{2}}h$
$\displaystyle={\overline{\Upsilon}^{\text{far}}}(H)+{\overline{\Upsilon}^{\text{diag}}}(H),$

where $\tau^{\theta}_{\mu_{1}}\sim\tau^{\theta}_{\mu_{2}}$ means they are
adjacent cubes of distance $\approx\theta$, $\overline{\Xi}_{\theta_{0}}$
denotes the remaining pairs not included in the first sum. Here the diagonal
set $\overline{\Xi}_{\theta_{0}}$ is much larger than ${\Xi}_{\theta_{0}}$,
and it is not hard to check that
$\,(\nu_{1},\nu_{2})\in\overline{\Xi}_{\theta_{0}}$ implies
$|\nu_{1}-\nu_{2}|\leq 2^{23}\theta_{0}$.

By (3.34), we have

(3.35) $\displaystyle I=\int$
$\displaystyle\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|\Upsilon^{\text{diag}}(H)|^{\frac{q}{2}}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\lesssim\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|{\overline{\Upsilon}^{\text{diag}}}(H)|^{q/4}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\quad+\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|{\overline{\Upsilon}^{\text{far}}}(H)|^{q/4}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle=A+B.$

For the diagonal term $A$, note that when $n-1=3$, $\frac{q_{e}}{4}\in[1,2]$,
if we repeat the proof of Lemma 3.1, it is not hard to show the following
analog of (3.6)

(3.36)
$\|{\overline{\Upsilon}^{\text{diag}}}(H)\|^{\frac{1}{2}}_{L^{\frac{1}{2}}_{t}L_{x}^{\frac{q_{e}}{4}}}\lesssim\|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\|^{2}_{L^{2}_{t}\ell_{\nu}^{q_{e}}L^{q_{e}}_{x}}\,+O(\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

To prove this, we just need to define the auxiliary operator
$b_{\nu_{1},\nu_{2}}(x,D)$ in (3.24) such that its frequency support is
essentially the algebraic sum of the frequency support of 4 nearby
$A^{\theta_{0}}_{\nu}(x,D)$ operators, and the remaining arguments can be
carried over in the same way.

By (3.36), if we use Hölder’s inequality followed by Young’s inequality

$\displaystyle A=$
$\displaystyle\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|{\overline{\Upsilon}^{\text{diag}}}(H)|^{\frac{q}{4}}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\leq$
$\displaystyle\int\Big{(}\|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H\|_{L_{x}^{\frac{q_{e}}{2}}}^{\frac{q_{e}-q}{2}}\,\|{\overline{\Upsilon}^{\text{diag}}}(H)\|_{L_{x}^{\frac{q_{e}}{4}}}^{\frac{q}{4}}\,\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\leq$
$\displaystyle\|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H\|_{L^{1}_{t}L_{x}^{\frac{q_{e}}{2}}}^{\frac{q_{e}-q}{q_{e}}}\,\|{\overline{\Upsilon}^{\text{diag}}}(H)\|_{L^{\frac{1}{2}}_{t}L_{x}^{\frac{q_{e}}{4}}}^{\frac{q}{2q_{e}}}$
$\displaystyle\leq$
$\displaystyle\tfrac{q_{e}-q}{q_{e}}\|\tilde{\sigma}_{\lambda}H\|^{2}_{L^{2}_{t}L_{x}^{q_{e}}}+\tfrac{q}{q_{e}}\|{\overline{\Upsilon}^{\text{diag}}}(H)\|^{\frac{1}{2}}_{L^{\frac{1}{2}}_{t}L_{x}^{\frac{q_{e}}{4}}}$
$\displaystyle\leq$
$\displaystyle\tfrac{q_{e}-q}{q_{e}}\|\tilde{\sigma}_{\lambda}H\|^{2}_{L^{2}_{t}L_{x}^{q_{e}}}+C(\|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H\|^{2}_{L^{2}_{t}\ell_{\nu}^{q_{e}}L^{q_{e}}_{x}}\,+\lambda^{1-}\|H\|_{L^{2}_{t,x}}^{2}).$

which is desired since the first term in the right can be absorbed to the left
side of (3.9).

To control the off-diagonal term $B$, first we note that by the Cauchy-Schwarz
inequality,

(3.37) $|T_{\nu_{1}}hT_{\nu_{2}}H|\leq
C\big{(}\sum_{\tilde{\nu}_{1}:\,|\tilde{\nu}_{1}-\nu_{1}|\leq
2^{13}\theta_{0}}|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu_{1}}H|^{2}\big{)}\big{(}\sum_{\tilde{\nu}_{2}:\,|\tilde{\nu}_{2}-\nu_{2}|\leq
2^{13}\theta_{0}}|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu_{2}}H|^{2}\big{)}.$

Since for fixed $\nu_{1},\nu_{2}$, the number of choices of
$\tilde{\nu}_{1},\tilde{\nu}_{2}$ is finite, (3.37) implies that

(3.38)
$\displaystyle\sum_{\\{(\mu_{1},\mu_{2}):\,\tau^{\theta}_{\mu_{1}}\sim\tau^{\theta}_{\mu_{2}}\\}}$
$\displaystyle\sum_{\\{(\nu_{1},\nu_{2})\in\tau^{\theta}_{\mu_{1}}\times\tau^{\theta}_{\mu_{2}}\\}}|T_{\nu_{1}}HT_{\nu_{2}}H|$
$\displaystyle\lesssim\sum_{\\{(\mu_{1},\mu_{2}):\,\tau^{\theta}_{\mu_{1}}\sim\tau^{\theta}_{\mu_{2}}\\}}\sum_{\\{(\nu_{1},\nu_{2})\in\tilde{\tau}^{\theta}_{\mu_{1}}\times\tilde{\tau}^{\theta}_{\mu_{2}}\\}}|A^{\theta_{0}}_{\nu_{1}}\tilde{\sigma}_{\lambda}H|^{2}\cdot|A^{\theta_{0}}_{\nu_{2}}\tilde{\sigma}_{\lambda}H|^{2}.$

Here $\tilde{\tau}^{\theta}_{\mu_{1}}$ and $\tilde{\tau}^{\theta}_{\mu_{2}}$
are the cubes with the same centers but $11/10$ times the side length of
$\tau^{\theta}_{\mu_{1}}$ and $\tau^{\theta}_{\mu_{2}}$, respectively, we used
the fact that the side length of $\tau^{\theta}_{\mu_{1}}$ is $\geq
2^{20}\theta_{0}$, so $0.1*\text{side length}\gg 2^{13}\theta_{0}$.

Furthermore, we have for a given fixed $c_{0}=2^{-m_{0}}$,
$m_{0}\in{\mathbb{N}}$, and pair of dyadic cubes $\tau^{\theta}_{\mu_{1}}$,
$\tau^{\theta}_{\mu_{2}}$ with
$\tau^{\theta}_{\mu_{1}}\sim\tau^{\theta}_{\mu_{2}}$ and
$\theta=2^{k}\theta_{0}$

(3.39)
$\sum_{(\nu_{1},\nu_{2})\in\tilde{\tau}^{\theta}_{\mu_{1}}\times\tilde{\tau}^{\theta}_{\mu_{2}}}|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu_{1}}H|^{2}\cdot|\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu_{2}}H|^{2}\\\
=\sum_{(\nu_{1},\nu_{2})\in\tilde{\tau}^{\theta}_{\mu_{1}}\times\tilde{\tau}^{\theta}_{\mu_{2}}}\,\sum_{\begin{subarray}{c}\tau^{c_{0}\theta}_{\mu_{1}^{\prime}}\cap\overline{\tau}^{\theta}_{\mu_{1}}\neq\emptyset\\\
\tau^{c_{0}\theta}_{\mu_{2}^{\prime}}\cap\overline{\tau}^{\theta}_{\mu_{2}}\neq\emptyset\end{subarray}}|\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{1}^{\prime}}A^{\theta_{0}}_{\nu_{1}}H|^{2}\cdot|\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{2}^{\prime}}A^{\theta_{0}}_{\nu_{2}}H|^{2}+O(\lambda^{-N}\|H\|_{2}^{4}),$

if $\overline{\tau}^{\theta}_{\mu_{1}}$ and
$\overline{\tau}^{\theta}_{\mu_{2}}$ the cubes with the same centers but
$12/10$ times the side length of $\tau^{\theta}_{\mu_{1}}$ and
$\tau^{\theta}_{\mu_{2}}$, respectively, so that we have
$\text{dist}(\overline{\tau}^{\theta}_{\mu_{1}},\overline{\tau}^{\theta}_{\mu_{2}})\geq\theta/2$
when $\tau^{\theta}_{\mu_{1}}\sim\tau^{\theta}_{\mu_{2}}$. This follows from
the fact that for $c_{0}$ small enough the product of the symbol of
$A_{\mu_{1}^{\prime}}^{c_{0}\theta}$ and $A_{\nu_{1}}^{\theta_{0}}$ vanishes
identically if
$\tau_{\mu_{1}^{\prime}}^{c_{0}\theta}\cap\overline{\tau}^{\theta}_{\mu_{1}}=\emptyset$
and $\nu_{1}\in\tilde{\tau}^{\theta}_{\mu_{1}}$, since
$\theta=2^{k}\theta_{0}$ with $k\geq 20$. Also notice that we then have for
fixed $c_{0}=2^{-m_{0}}$ small enough

(3.40)
$\text{dist}(\tau^{c_{0}\theta}_{\mu^{\prime}_{1}},\tau^{c_{0}\theta}_{\mu^{\prime}_{2}})\in[4^{-1}\theta,4^{2}\theta],\quad\text{if
}\,\,\tau^{c_{0}\theta}_{\mu_{1}^{\prime}}\cap\overline{\tau}^{\theta}_{\mu_{1}}\neq\emptyset,\,\,\,\text{and
}\,\,\tau^{c_{0}\theta}_{\mu_{2}^{\prime}}\cap\overline{\tau}^{\theta}_{\mu_{2}}\neq\emptyset.$

Also, for each $\mu_{1}$ there are $O(1)$ terms $\mu_{1}^{\prime}$ with
$\tau^{c_{0}\theta}_{\mu_{1}^{\prime}}\cap\overline{\tau}^{\theta}_{\mu_{1}}\neq\emptyset$,
if $c_{0}$ is fixed.

If the cubes $\tau^{c_{0}\theta}_{\mu^{\prime}_{1}}$,
$\tau^{c_{0}\theta}_{\mu^{\prime}_{2}}$ satisfy the conditions above, we have
the following bilinear type estimate which is a consequence of Proposition 3.3
in [3].

(3.41)
$\iint\bigl{|}\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{1}}H_{1}\cdot\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{2}}H_{2}\bigr{|}^{\frac{q}{2}}\,dxdt\lesssim\,\lambda^{1+\varepsilon}\,\bigl{(}2^{k}\theta_{0}\lambda\bigr{)}^{\frac{n-1}{2}(q-\frac{2(n+1)}{n-1}))}\|H_{1}\|^{q/2}_{L^{2}_{t,x}}\,\|H_{2}\|^{q/2}_{L^{2}_{t,x}}.$

The proof of (3.41) follows from using the Hadamard parametrix to rewrite the
$\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\nu}$, $\nu=\mu_{1},\mu_{2}$
operators as oscillatory integral operators with explicit kernels and then
applying Lee’s oscillatory integral estimate [18]. The assumption that $c_{0}$
is small enough is crucial to ensure the separation conditions in Theorem 1.1
of [18], see the proof of Proposition 3.3 in [3] for more details.

Note that by (3.34) and (3.39), we have

(3.42)
$\displaystyle\big{|}{\overline{\Upsilon}^{\text{far}}}(H)\big{|}\leq\sum_{\\{k\in{\mathbb{N}}:\,k\geq
20\,\,\text{and }\,\theta=2^{k}\theta_{0}\ll
1\\}}\,\,\sum_{\\{(\mu_{1},\mu_{2}):\,\tau^{\theta}_{\mu_{1}}\sim\tau^{\theta}_{\mu_{2}}\\}}\sum_{(\nu_{1},\nu_{2})\in\tilde{\tau}^{\theta}_{\mu_{1}}\times\tilde{\tau}^{\theta}_{\mu_{2}}}\,\sum_{\begin{subarray}{c}\tau^{c_{0}\theta}_{\mu_{1}^{\prime}}\cap\overline{\tau}^{\theta}_{\mu_{1}}\neq\emptyset\\\
\tau^{c_{0}\theta}_{\mu_{2}^{\prime}}\cap\overline{\tau}^{\theta}_{\mu_{2}}\neq\emptyset\end{subarray}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\Big{(}|\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{1}^{\prime}}A^{\theta_{0}}_{\nu_{1}}H|^{2}\cdot|\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{2}^{\prime}}A^{\theta_{0}}_{\nu_{2}}H|^{2}\Big{)}+O(\lambda^{-N}\|H\|_{2}^{4}),$

It is not hard to see that the number of terms in the sum is
$O(\lambda^{(2n-3)\varepsilon_{0}})=O(\lambda^{5\varepsilon_{0}})$, and by
(3.41), we have for each term in the right side of (3.42),

(3.43) $\displaystyle\int\Big{(}$
$\displaystyle\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,\Big{|}|\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{1}^{\prime}}A^{\theta_{0}}_{\nu_{1}}H|^{2}\cdot|\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{2}^{\prime}}A^{\theta_{0}}_{\nu_{2}}H|^{2}\Big{|}^{q/4}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\lesssim\|\tilde{S}_{\lambda}f\|_{L^{\infty}(A_{-})}^{\frac{2(q_{e}-q)}{q_{e}}}\cdot\|\big{(}\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{1}^{\prime}}A^{\theta_{0}}_{\nu_{1}}H\big{)}\cdot\big{(}\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{2}^{\prime}}A^{\theta_{0}}_{\nu_{2}}H\big{)}\|_{L^{\frac{q}{q_{e}}}_{t}L^{\frac{q}{2}}_{x}}^{\frac{q}{q_{e}}}$
$\displaystyle\lesssim
T^{(\frac{q_{e}}{q}-\frac{2}{q})\cdot\frac{q}{q_{e}}}\|\tilde{S}_{\lambda}f\|_{L^{\infty}(A_{-})}^{\frac{2(q_{e}-q)}{q_{e}}}\cdot\|\big{(}\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{1}^{\prime}}A^{\theta_{0}}_{\nu_{1}}H\big{)}\cdot\big{(}\tilde{\sigma}_{\lambda}A^{c_{0}\theta}_{\mu_{2}^{\prime}}A^{\theta_{0}}_{\nu_{2}}H\big{)}\|_{L^{\frac{q}{2}}_{t}L^{\frac{q}{2}}_{x}}^{\frac{q}{q_{e}}},$
$\displaystyle\lesssim
T^{(\frac{q_{e}}{q}-\frac{2}{q})\cdot\frac{q}{q_{e}}}\lambda^{(\frac{n-1}{4}+\varepsilon_{1})(\frac{2(q_{e}-q)}{q_{e}})}\cdot\Big{(}\lambda^{1+\varepsilon}\,\bigl{(}2^{k}\lambda^{1-\varepsilon_{0}}\bigr{)}^{\frac{n-1}{2}(q-\frac{2(n+1)}{n-1})}\Big{)}^{\frac{2}{q_{e}}}\|H\|^{\frac{2q}{q_{e}}}_{L^{2}_{x}}.$

As in (3.11), if we take $\varepsilon,\varepsilon_{0}$ and $\varepsilon_{1}$
to be small enough, e.g.,
$\varepsilon=\varepsilon_{0}=\varepsilon_{1}=\frac{1}{100}$ it is
straightforward to check that the right side of (3.43) is
$O(\lambda^{1-6\varepsilon_{0}}\|H\|_{L^{2}_{t,x}}^{2}),$ using the fact that
$\|H\|_{L^{2}_{t,x}}^{2}$ dominates $\|H\|_{L^{2}_{t,x}}^{\frac{2q}{q_{e}}}$
since we are assuming $f$ is $L^{2}$ normalized. This implies that

(3.44) $\displaystyle B=$
$\displaystyle\int\Big{(}\int|\tilde{\sigma}_{\lambda}H\cdot\tilde{\sigma}_{\lambda}H|^{\frac{q_{e}-q}{2}}\,|{\overline{\Upsilon}^{\text{far}}}(H)|^{q/4}\,dx\Big{)}^{\frac{2}{q_{e}}}dt$
$\displaystyle\lesssim$
$\displaystyle\lambda^{1-6\varepsilon_{0}+5\varepsilon_{0}}\|H\|_{L^{2}_{t,x}}^{2},$

which finishes the proof of Proposition 2.3 for $n-1=3$.

Remarks about $n-1=2$. The arguments for $n-1=2$ is similar to the case
$n-1=3$. Recall that when $n-1=3$, we are essentially doing another round of
bilinear estimates within the diagonal terms in the Whitney decomposition due
to the fact that $q_{e}=\frac{2(n-1)}{n-3}=6>4$. When $n-1=2$, $q$ can be
arbitrary large, if $q\in[2^{k+1},2^{k+2}]$ for some $k\in\mathbb{N}^{+}$,
then one can repeat the above arguments $k$ times, the resulting diagonal term
will involve a product of $2^{k+1}$ terms of type
$\tilde{\sigma}_{\lambda}A^{\theta_{0}}_{\nu}H$, and it satisfies the natural
analog of Lemma 3.1 since $\frac{q}{2^{k+1}}\in[1,2]$. And there are off-
diagonal terms each time we run the Whitney decomposition, those terms can be
treated as in (3.43) by using bilinear estimates. The main difference is, as
$q\rightarrow\infty$, unlike (3.11) and (3.43), we have to take
$\varepsilon_{0}$ and $\varepsilon_{1}$ to be small enough depending on $q$,
instead of some fixed small constant.

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