# Circular average relative to
fractal measures

Seheon Ham  ,  Hyerim Ko  and  Sanghyuk Lee  Department of Mathematical
Sciences and RIM, Seoul National University, Seoul 08826, Republic of Korea
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We prove new $L^{p}$–$L^{q}$ estimates for averages over dilates of the circle
with respect to fractal measures, which unify different types of maximal
estimates for the circular average. Our results are consequences of
$L^{p}$–$L^{q}$ smoothing estimates for the wave operator relative to fractal
measures. We also discuss similar results concerning the spherical averages.

###### Key words and phrases:

circular average, general measures

###### 2020 Mathematics Subject Classification:

42B25

##  1\. introduction

Let $d\geq 2$. We consider the average

$\mathcal{A}f(x,t)=\int_{\mathbb{S}^{d-1}}f(x-ty)d\sigma(y),\quad
x\in\mathbb{R}^{d}.$

Here $d\sigma$ denotes the normalized measure on the unit sphere
$\mathbb{S}^{d-1}$. We study estimates for $\mathcal{A}$ relative to fractal
measures. To motivate our study, we briefly review some of the previous
results.

Stein [26] ($d\geq 3$) and Bourgain [4] ($d=2$) proved that the maximal
operator $f\mapsto\sup_{t>0}|\mathcal{A}f(\cdot,t)|$ is bounded on $L^{p}$ if
and only if $\frac{d}{d-1}<p\leq\infty$. If the supremum is taken over a
compact interval $I$ contained in $(0,\infty)$, the consequent maximal
function has the $L^{p}$ improving property ([24, 25, 14]). That is to say,
the estimate

(1.1) $\|\sup_{t\in I}|\mathcal{A}f(\cdot,t)|\|_{L^{q}(\mathbb{R}^{d})}\leq
C_{d}\|f\|_{L^{p}(\mathbb{R}^{d})}$

holds for some $p<q$. Except for a few endpoint cases, we now have a complete
characterization of $p,q$ for which (1.1) holds. Let
$\mathcal{P}_{d}\subset[0,1]\times[0,1]$ be the convex hull of the set
consisting of the points
$O:=(0,0),~{}Q_{1}:=\big{(}\frac{d(d-1)}{d^{2}+1},\frac{d-1}{d^{2}+1}\big{)},$
and
$Q_{2}:=\big{(}\frac{d-1}{d},\frac{1}{d}\big{)},~{}Q_{3}:=\big{(}\frac{d-1}{d},\frac{d-1}{d}\big{)}$.
Schlag [24] ($d=2$), and Schlag and Sogge [25] ($d\geq 3$) showed that the
estimate (1.1) holds for
$(p^{-1},q^{-1})\in\mathcal{P}_{d}\setminus((O,Q_{1})\cup[Q_{1},Q_{2})\cup[Q_{2},Q_{3}])$
and it fails if
$(p^{-1},q^{-1})\not\in({{\mathcal{P}_{d}}}\setminus\\{Q_{3}\\})$. The third
named author [14] proved the borderline case
$(p^{-1},q^{-1})\in(O,Q_{1})\cup(Q_{1},Q_{2})\cup(Q_{2},Q_{3})$, however the
case $(p^{-1},q^{-1})=Q_{1},Q_{2}$ remain open. See also [23, 1] for recent
developments regarding the circular and spherical maximal functions.

Different forms of maximal estimates have also been of interest. In
particular, the estimate

(1.2) $\|\sup_{x\in\mathbb{R}^{d}}|\mathcal{A}f(x,\cdot)|\|_{L^{q}_{t}(I)}\leq
C_{d}\|f\|_{L^{p}_{\epsilon}(\mathbb{R}^{d})}$

for some $\epsilon>0$ was used to study packing problems for the circle and
sphere [27, 13, 18, 20]. Here
$\|f\|_{L^{p}_{s}(\mathbb{R}^{d})}:=\|(1-\Delta)^{\frac{s}{2}}f\|_{L^{p}(\mathbb{R}^{d})}$.
When $d\geq 3$, Kolasa and Wolff [13] showed (1.2) with $p=q=2$ for any
$\epsilon>0$. However, a similar estimate in $\mathbb{R}^{2}$ turned out to be
more difficult. Wolff [27] obtained (1.2) with $p=q=3$ for $\epsilon>0$.
Precisely, results in [27, 13] are given in a different form in which averages
over the sphere (circle) were replaced by the average over annuli of thickness
$\delta$. However, it is easy to see that those estimates are equivalent to
the abovementioned estimates in the form of (1.2). It is also known that (1.2)
fails if $\epsilon$ is removed (see [3, 12] or [13, Proposition 2.2] for
example). Wolff’s result was extended to a variable coefficient setting by
Zahl [29].

######  Definition 1.1.

For $\alpha\in(0,d+1]$, we say a non-negative Borel measure $\nu$ on
$\mathbb{R}^{d+1}$ is $\alpha$-dimensional if there is a constant $C_{\nu}$
such that

(1.3) $\nu(\mathbb{B}^{d+1}(z,\rho))\leq
C_{\nu}\rho^{\alpha},\quad\forall(z,\rho)\in\mathbb{R}^{d+1}\times\mathbb{R}_{+},$

where
$\mathbb{B}^{d+1}(z,\rho)=\\{z^{\prime}\in\mathbb{R}^{d+1}:|z-z^{\prime}|<\rho\\}$.
By $\mathfrak{C}^{d+1}(\alpha)$ we denote the class of $\alpha$-dimensional
measure $\nu$. For $\nu\in\mathfrak{C}^{d+1}(\alpha)$, we also define

$\langle\nu\rangle_{\alpha}:=\sup_{z\in\mathbb{R}^{d+1},\,\rho>0}\rho^{-\alpha}\nu(\mathbb{B}^{d+1}(z,\rho)).$

For the rest of the paper, we fix $I=(1,2)$. We consider the estimate

(1.4) $\|\mathcal{A}f\|_{L^{q}(\mathbb{R}^{d}\times I;d\nu)}\leq
C\langle\nu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{L^{p}(\mathbb{R}^{d})}.$

Remarkably, the estimate (1.4) unifies the maximal estimates (1.1) and (1.2)
in a single framework. Indeed, it is not difficult to see that (1.4) implies
the seemingly unrelated estimates (1.1) and (1.2). One can deduce the estimate
(1.1) from (1.4) if the latter holds with a uniform $C$ for all
$d$-dimensional measures with $\langle\nu\rangle_{d}\leq 1$. This can be shown
by the Kolmogorov-Seliverstov-Plessner linearization argument and the Riesz
representation theorem (see Lemma 3.2). Similarly, (1.2) follows if (1.4) with
$L^{p}$ replaced by $L^{p}_{\epsilon}$ holds uniformly for all $1$-dimensional
measures.

Since the averaging operator $\mathcal{A}$ is translation-invariant in $x$, it
is not possible to have (1.4) unless $1\leq p\leq q\leq\infty$. Also, taking a
specific $\alpha$-dimensional measure, one can show (1.4) holds only if
$\alpha p\geq q$ (see $(i)$ in Proposition 4.1). Thus it is natural for (1.4)
to assume $\alpha>1$ and $p\leq q$.

When $d\geq 3$ the estimate (1.4) is relatively easier to obtain. There are
various estimates which are straightforward consequences of the fractal
Strichartz estimates for the wave equation ([7, 10]). We discuss the matter in
Section 3. However, when $d=2$, such estimates are not enough to prove the
estimate (1.4). In this paper we focus on the more interesting case $d=2$.

### The circular average

Let $1<\alpha\leq 3$. We define $\mathcal{P}_{2}(\alpha)\subset[0,1]^{2}$. If
$1<\alpha\leq 2$, we set

$\mathcal{P}_{2}(\alpha):=\Big{\\{}\Big{(}\frac{1}{p},\frac{1}{q}\Big{)}\in[0,1]^{2}:\frac{1}{q}\leq\frac{1}{p}<\frac{\alpha}{q},\
\ \frac{3}{p}<1+\frac{\alpha-1}{q}\Big{\\}},$

and for $2<\alpha\leq 3$,

$\mathcal{P}_{2}(\alpha):=\Big{\\{}\Big{(}\frac{1}{p},\frac{1}{q}\Big{)}\in[0,1]^{2}:\frac{1}{q}\leq\frac{1}{p}<\frac{\alpha}{q},\
\ \frac{3}{p}<1+\frac{2\alpha-3}{q},\ \
\frac{2}{p}<1+\frac{\alpha-2}{q}\Big{\\}}.$

In Section 4, we show that (1.4) fails if
$(\,{p}^{-1},\,{q}^{-1})\notin\overline{\mathcal{P}_{2}(\alpha)}$ when $d=2$.
It seems to be reasonable to conjecture that $\mathcal{P}_{2}(\alpha)$
determines, possibly except for the borderline cases, the optimal range of
$p,q$ on which (1.4) holds.111 As seen in the above, some of the borderline
cases are actually not true.

######  Conjecture 1.2.

Let $d=2$ and $1<\alpha\leq 3$. The estimate (1.4) holds for
$\nu\in\mathfrak{C}^{3}(\alpha)$ if
$(\,{p}^{-1},\,{q}^{-1})\in\mathcal{P}_{2}(\alpha)$.

The following is our main result, which verifies the conjecture for
$\alpha\in[3-\sqrt{3},3]$.

######  Theorem 1.3.

Let $d=2$, $1<\alpha\leq 3$, and $\nu\in\mathfrak{C}^{3}(\alpha)$. Then, the
estimate (1.4) holds if

(1.5) $(\,{p}^{-1},\,{q}^{-1})\in\mathcal{P}_{2}(\alpha)~{}\text{ and
}~{}p>(6-2\alpha)/\alpha.$

Note that $4-\alpha\geq(6-2\alpha)/\alpha$ if $\alpha\geq 3-\sqrt{3}$ and
$p>4-\alpha$ whenever $(1/p,1/q)\in\mathcal{P}_{2}(\alpha)$. As a corollary we
obtain $L^{p}(\mathbb{R}^{2})$–$L^{q}(d\mu)$ estimate for the circular maximal
function relative to fractal measures.

######  Corollary 1.4.

Let $1<\alpha\leq 2$ and $\mu\in\mathfrak{C}^{2}(\alpha)$. For $p,q$
satisfying (1.5), we have

(1.6) $\big{\|}\sup_{t\in I}|\mathcal{A}f(\cdot,t)|\big{\|}_{L^{q}(d\mu)}\leq
C\langle\mu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{L^{p}(\mathbb{R}^{2})}.$

A modification of the examples which give the necessary conditions of (1.4)
(Proposition 4.1) shows that (1.6) holds only if
$(\,{p}^{-1},\,{q}^{-1})\in\overline{\mathcal{P}_{2}(\alpha)}$. Consequently,
Corollary 1.4 establishes boundedness on sharp range for
$\alpha\in(3-\sqrt{3},2]$.

$L^{p}(\mathbb{R}^{d})$–$L^{q}(d\mu)$ estimate for $\mathfrak{M}f:=\sup_{t\in
I}|\mathcal{A}f(\cdot,t)|$ has been utilized to study problems in geometric
measure theory. $L^{2}(\mathbb{R}^{d})$–$L^{2}(d\mu)$ estimate was studied by
Mitsis [18] for $1<\alpha\leq d$, $d\geq 3$. D. Oberlin and R. Oberlin [21]
obtained $L^{2}_{(1-\alpha)/2+}(\mathbb{R}^{d})$–$L^{q}(d\mu)$ bound on
$\mathfrak{M}f$ when $1<q<2$ and $0<\alpha<(d-1)/2$. Also, for $d\geq 3$,
$L^{p}(d\mu_{1})$–$L^{p}(d\mu_{2})$ estimate was discussed in Iosevich et al.
[11] with $\alpha_{i}$-dimensional measure $\mu_{i}$, $1<\alpha_{i}\leq d$,
$i=1,2$. Those estimates were used to study the sphere packing problem,
Hausdorff dimension of the pinned distance set, and the divergence set of the
solution to the wave equation (see also [9]).

Compared with the previous results, it is remarkable that Corollary 1.4
establishes the sharp $L^{p}$ improving property for the circular maximal
function relative to some $\alpha$-dimensional measures. As far as the authors
are aware, no such results have appeared before. In [11] non-sharp local
smoothing estimates were combined with a duality argument. However, the
estimates in Corollary 1.4 can not be obtained by the approach in [11] even if
combined with the sharp local smoothing estimate [8]. Nevertheless, such
estimates are easier to prove (see Corollary 3.5) in higher dimensions.

Furthermore, one can use the estimate (1.6) (with $p=q$) to prove the circle
packing theorem which was shown in [28]: _Suppose $F\subset\mathbb{R}^{2}$ is
a Borel set of Hausdorff dimension exceeding $1$ and $E$ is a compact set such
that $E\subset\mathbb{R}^{2}$ contains circles centered at each point in $F$.
Then $E$ has positive Lebesgue measure._

Corollary 1.4 with $\alpha=2$ recovers the results on the circular maximal
function except for the endpoint cases ([24, 25, 14]).

#### Local smoothing estimate relative to ${\nu}\in\mathfrak{C}^{3}(\alpha)$.

In order to prove Theorem 1.3, we consider $L^{p}_{\gamma}$–$L^{q}(d\nu)$
estimate for the wave operator

$e^{it\sqrt{-\Delta}}f(x)=\int e^{i(x\cdot\xi+t|\xi|)}\widehat{f}(\xi)\,d\xi.$

For $2<\alpha\leq 3$, we obtain the estimate (1.4) by relying mainly on
$L_{\gamma}^{2}$–$L^{q}(d\nu)$ estimate for $e^{it\sqrt{-\Delta}}$, which is
to be discussed in Section 3. However, for $1<\alpha\leq 2$, such
$L_{\gamma}^{2}$–$L^{q}(d\nu)$ estimates are not enough for our purpose.
Instead, we exploit estimates for the wave operator relative to
$\alpha$-dimensional measure $\nu$:

(1.7) $\Big{(}\int_{\mathbb{B}^{2}(0,1)\times
I}\big{|}e^{it\sqrt{{-}\Delta}}f(x)\big{|}^{q}d\nu(x,t)\Big{)}^{1/q}\leq
C\langle\nu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{L^{p}_{\gamma}(\mathbb{R}^{2})}.$

Sogge conjectured that (1.7) holds with $d\nu(x,t)=dxdt$ for
$p=q\geq{2d}/(d-1)$ if $\gamma>(d-1)(\frac{1}{2}-\frac{1}{p})-\frac{1}{p}$.
The conjecture was studied by various authors [28, 16, 5]. Recently, Guth,
Wang, and Zhang [8] confirmed the conjecture in two dimensions while it
remains open in higher dimensions. See also [19, 25] for earlier results.

For $1<\alpha\leq 2$, Theorem 1.3 follows from the next theorem, which we
prove by combining the sharp $L^{p}$ local smoothing estimate [8] and the
trilinear restriction estimate for the cone (see [2, 6, 16]).

######  Theorem 1.5.

Let $1<\alpha\leq 2$ and $\nu\in\mathfrak{C}^{3}(\alpha)$. If $p,q\geq 1$
satisfy (1.5), then the estimate (1.7) holds for some $\gamma<1/2$.

For the proof of Theorem 1.3 we need only to obtain the estimate (1.7) for
$\gamma<1/2$ (see Section 3). However, it is of independent interest to
characterize $p,q$ and $\gamma$ for which (1.7) holds. In this regard, we can
obtain the sharp estimate on a restricted range of $p,q$. To state our result,
we set

(1.8) $\kappa(\alpha)=\begin{cases}\,3,~{}&~{}\hfill 2<\alpha\leq 3,\\\
\,\alpha+1,~{}&~{}\hfill 1<\alpha\leq 2,\\\ \,2\alpha,~{}&~{}\hfill
0<\alpha\leq 1.\end{cases}$

######  Theorem 1.6.

Let $0<\alpha\leq 3$ and $\nu\in\mathfrak{C}^{3}(\alpha)$. If
$\frac{1}{q}\leq\min\big{(}\frac{1}{p},~{}\frac{1}{3p}+\frac{1}{6},~{}\frac{2}{3p^{\prime}}\big{)}$
and $\frac{1}{p}+\frac{\kappa(\alpha)}{q}\leq 1$, then (1.7) holds for
$\gamma>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$.

The regularity assumption $\gamma>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$ is
sharp in that (1.7) fails in general if
$\gamma<\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$ (see Section 2). It seems to
be natural to conjecture that the same remains valid provided that
$\frac{1}{q}\leq\frac{1}{p}$ and $\frac{1}{p}+\frac{\kappa(\alpha)}{q}\leq 1$.
This in turn implies Conjecture 1.2. When $2<\alpha\leq 3$,
$\frac{1}{q}\leq\frac{1}{p}$, and $\frac{1}{p}+\frac{3}{q}\leq 1$, the
estimate (1.7) is a consequence of the sharp local smoothing estimate [8]. In
higher dimensions ($d\geq 3$), the local smoothing estimate in [5] and Lemma
2.7 yield an analogue of (1.7) for $d\leq\alpha\leq d+1$ whenever
$\gamma>\frac{d-1}{2}+\frac{1}{p}-\frac{\alpha}{q}$,
$\frac{1}{p}+\frac{d+3}{d-1}\frac{1}{q}\leq 1$, and $p\leq q$.

#### Estimate for the spherical average

In higher dimensions one can obtain results similar to Theorem 1.3 for the
spherical average in $\mathbb{R}^{d+1}$, $d\geq 3$ (see Theorem 3.4). Such
results can be shown by making use of the Strichartz estimates for the wave
equation relative to fractal measure (see Theorem 3.3). The spherical average
estimates in Theorem 3.4 seem to be sharp for $d=3$, while we only manage to
verify the optimality for integer $\alpha$. See Section 4.

#### Organization of the paper

In Section 2, we obtain the local smoothing estimates relative to
$\alpha$-dimensional measure (Proposition 2.1 and Proposition 2.12), which we
use to prove Theorem 1.5 and Theorem 1.6. In Section 3 we prove Theorem 1.5,
Theorem 1.6, Theorem 1.3, and Corollary 1.4, and obtain the estimates in
higher dimensions. We discuss the sharpness of the estimate (1.4) in Section
4.

##  2\. Weighted local smoothing estimates

In this section, we obtain some preparatory results which we use in proving
Theorem 1.5 and Theorem 1.6. The main object is to prove Proposition 2.1 and
Proposition 2.12. To this end, we adapt the argument due to Vargas and one of
the authors [16] which is based on the trilinear restriction estimate.

We decompose $e^{it\sqrt{-\Delta}}f$ into two parts: a sum of operators with
small frequency supports and a product of operators with angularly separated
frequency supports. For the latter we apply the multilinear restriction
estimate due to Bennett, Carbery, and Tao [2] by making use of the fact that
angular separation implies transversality on the trilinear operator. For the
first part, we rescale the operator and apply an induction argument (see
(2.5)). However, the support of a rescaled measure is no longer contained in a
bounded set, so the induction can not be carried out on a bounded set. To get
around this issue, we consider a class of weight functions which are not
necessarily supported in a bounded set.

###  2.1. Weighted local smoothing estimates in $\mathbb{R}^{2+1}$

For $0<\alpha\leq 3$, we denote by $\Omega(\alpha)$ the class of nonnegative
measurable functions $\omega:\mathbb{R}^{2}\times I\mapsto\mathbb{R}$ which
satisfy

(2.1) $\int_{\mathbb{B}^{3}(z,r)}\omega(y)dy\leq
r^{\alpha},\quad\forall(z,r)\in\mathbb{R}^{3}\times\mathbb{R}_{+}.$

By Littlewood-Paley decomposition, it is enough to consider the estimate

(2.2) $\|e^{it\sqrt{-\Delta}}P_{\lambda}f\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}\leq C\lambda^{\gamma}\|f\|_{p}.$

Here $P_{\lambda}$ is the standard Littlewood-Paley projection operator given
by $\widehat{P_{\lambda}f}(\xi)=\beta(\lambda^{-1}|\xi|)\widehat{f}(\xi)$ for
$\beta\in\mathrm{C}_{c}^{\infty}((1/2,2))$ and
$\|f\|_{L^{q}(E;\omega)}^{q}:=\int_{E}|f(x)|^{q}\omega(x)dx$.

Instead of proving (2.2) directly, we consider a modified operator to
facilitate rescaling. We consider

$\psi_{{0}}(\theta):=\sqrt{1+\theta^{2}}-1.$

Note that $\psi_{{0}}$ can be regarded as a small perturbation of
$\theta^{2}/2$.

For $\lambda\geq 1$ and $J\subset[-1,1]$, let

$A_{\lambda}(J)=\\{(\eta,\rho):\lambda/4\leq\eta\leq
4\lambda,~{}\theta:=\rho/\eta\in J\\}.$

We define

(2.3)
$T_{\lambda}^{\phi}g(x,t)=\int_{A_{\lambda}([-1,1])}e^{i(x,t)\cdot(\eta,\rho,\eta\phi(\rho/\eta))}\beta(\lambda^{-1}\eta)\beta_{0}(|\rho|/\eta)\widehat{g}(\eta,\rho)d\eta
d\rho,$

where $\beta_{0}$ is a smooth function supported in $[0,2]$.

######  Proposition 2.1.

Let $1<\alpha\leq 2$ and $\omega\in\Omega(\alpha)$. If $p,q\geq 1$ satisfy
(1.5) and $\lambda\gg 1$, then for some $\gamma<1/2$ we have

(2.4) $\|T_{\lambda}^{\psi_{0}}g\|_{L^{q}(\mathbb{R}^{2}\times I;\omega)}\leq
C\lambda^{\gamma}\|g\|_{p}.$

We deduce the estimate (2.2) from (2.4) via a change of variable (see the
proof of Theorem 1.5 in Section 3).

#### Normalization of the phase function.

For our induction argument, we begin with normalizing the phase function of
$T_{\lambda}g$. Let

$\phi_{\circ}(\theta)=\frac{1}{2}\theta^{2}.$

For a given $\epsilon>0$, we define a class $\mathfrak{S}(\epsilon)$ of
$\mathrm{C}^{3}$ functions by

$\mathfrak{S}(\epsilon)=\\{\phi\in\mathrm{C}^{3}([-1,1]):\phi(0)=\phi^{\prime}(0)=0,~{}\|\phi-\phi_{\circ}\|_{\mathrm{C}^{3}([-1,1])}\leq\epsilon\\},$

where $\|\phi\|_{\mathrm{C}^{n}([-1,1])}=\sum_{j\leq
n}\sup_{\theta\in[-1,1]}\big{|}\partial_{\theta}^{j}\phi(\theta)\big{|}$. For
$\theta_{\circ}\in(-1,1)$, let $h$ be a real number such that
$[\theta_{\circ}-h,\theta_{\circ}+h]\subset[-1,1]$. For
$\phi\in\mathrm{C}^{3}([-1,1])$ satisfying
$\phi^{\prime\prime}(\theta_{\circ})\neq 0$, we define

$\phi_{\theta_{\circ},h}(\theta)=(\phi^{\prime\prime}(\theta_{\circ})h^{2})^{-1}\big{(}\phi(h\theta+\theta_{\circ})-\phi(\theta_{\circ})-\phi^{\prime}(\theta_{\circ})h\theta\big{)}.$

######  Lemma 2.2.

Suppose $\phi\in\mathfrak{S}(\epsilon_{\circ})$ for some
$0<\epsilon_{\circ}<1/8$. Then, there exists
$0<h_{\circ}=h_{\circ}(\epsilon_{\circ})$ such that
$\phi_{\theta_{\circ},h}\in\mathfrak{S}(\epsilon_{\circ})$ whenever $0<h\leq
h_{\circ}$ and $[\theta_{\circ}-h,\theta_{\circ}+h]\subset[-1,1]$.

Throughout the paper, $\epsilon_{\circ}$ denotes a positive constant such that
$0<\epsilon_{\circ}<1/8$.

###### Proof.

By Taylor series expansion, we have

$\displaystyle\phi(h\theta+\theta_{\circ})-\phi(\theta_{\circ})-\phi^{\prime}(\theta_{\circ})h\theta=\frac{\phi^{\prime\prime}(\theta_{\circ})h^{2}\theta^{2}}{2}+\mathcal{E}(\theta_{\circ},h,\theta),$

where $\|\mathcal{E}(\theta_{\circ},h,\cdot)\|_{\mathrm{C}^{3}([-1,1])}\leq
Ch^{3}$ with a constant $C$ independent of $\theta_{\circ}\in[-1,1]$. Since
$\phi\in\mathfrak{S}(\epsilon_{\circ})$,
$\phi^{\prime\prime}(\theta_{\circ})\in[3/4,5/4]$. So, we see

$\|\phi_{\theta_{\circ},h}-\phi_{\circ}\|_{\mathrm{C}^{3}([-1,1])}\leq(\phi^{\prime\prime}(\theta_{\circ})h^{2})^{-1}\|\mathcal{E}(\theta_{\circ},h,\cdot)\|_{\mathrm{C}^{3}([-1,1])}\leq
Ch.$

Thus, there exists $h_{\circ}\in(0,\epsilon_{\circ}]$ such that
$\phi_{\theta_{\circ},h}\in\mathfrak{S}(\epsilon_{\circ})$ for $0<h\leq
h_{\circ}$. ∎

#### Induction quantity.

As mentioned before, our proof is based on an induction on scale argument. We
introduce a quantity which makes it possible to carry out the induction
argument.

For $1\leq p\leq q\leq\infty$, we define
$Q(\lambda)=Q(\lambda,\epsilon_{\circ},p,q,\alpha)$ by

(2.5)
$\begin{split}Q(\lambda)=\sup\Big{\\{}\|T_{\lambda}^{\phi}g\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}:\ &{\rm supp}\,{\widehat{g}}\subset
A_{\lambda}([-1,1]),~{}\|g\|_{p}\leq 1,\\\ \
&\phi\in\mathfrak{S}(\epsilon_{\circ}),~{}\omega\in\Omega(\alpha)\Big{\\}}.\end{split}$

It is easy to see $Q(\lambda)<\infty$. Indeed, one can show a rough bound

(2.6) $Q(\lambda)\lesssim\lambda^{2},\quad\lambda\geq 1$

for $1\leq p\leq q\leq\infty$. We write
$T_{\lambda}^{\phi}g(x,t)=K_{\lambda}(\cdot,t)\ast g$, where the kernel is
given by

(2.7) $\displaystyle K_{\lambda}(x,t)=\int
e^{i(x,t)\cdot(\eta,\rho,\eta\phi(\rho/\eta))}\beta(\lambda^{-1}\eta)\beta_{0}(|\rho/\eta|)d\eta
d\rho.$

In order to show (2.6), we use an easy estimate:

(2.8)
$|K_{\lambda}(x,t)|\lesssim\mathcal{K}(x):=\lambda^{2}\big{(}1+|x|\big{)}^{-M},\quad
t\in I$

for any $M\geq 1$, which follows by routine integration by parts (for example,
see (3.6) below). Using (2.1), it is easy to see

(2.9)
$\displaystyle\Big{\|}\int\mathcal{K}(x-y)g(y)\,dy\Big{\|}_{L^{q}(\mathbb{R}^{2}\times
I,\omega)}\lesssim\lambda^{2}\|g\|_{p}$

for $1\leq p\leq q$. Indeed, since
$\int_{I}\int\mathcal{K}(x-y)\,\omega(x,t)\,dxdt\lesssim\lambda^{2}$ and
$\int\mathcal{K}(x-y)dy\lesssim\lambda^{2}$, (2.9) follows for $p=q$ by
Schur’s test. Also, by Hölder’s inequality, we get (2.9) for $q=\infty$ and
$1\leq p\leq\infty$. Interpolating those estimates, we have (2.9) and
therefore (2.6) for $1\leq p\leq q$.

In what follows, we show $Q(\lambda)\leq
C_{\epsilon}\lambda^{\frac{1}{2}-\epsilon}$ for some $\epsilon>0$.

#### Rescaling

We first observe that $L^{p}$–$L^{q}$ bound on $T_{\lambda}^{\phi}$ is
improved on a certain range of $p,q$ if $\widehat{g}$ is supported in a narrow
conic neighborhood. More precisely, we have the following.

######  Lemma 2.3.

Let $0<\alpha\leq 3$, $\omega\in\Omega(\alpha)$, and
$\phi\in\mathfrak{S}(\epsilon_{\circ})$. For $h\geq\lambda^{-\frac{1}{2}}$,
let $J_{h}=[\theta_{\circ}-h,\theta_{\circ}+h]\subset[-1,1]$. Suppose ${\rm
supp}\,\widehat{g}\subset A_{\lambda}(J_{h})$. Then, there is an
$h_{\circ}=h_{\circ}(\epsilon_{\circ})$ such that

(2.10) $\|T_{\lambda}^{\phi}g\|_{L^{q}(\mathbb{R}^{2}\times I;\omega)}\leq
Ch^{\frac{2\alpha-\kappa(\alpha)}{q}-\frac{3}{p}}Q(\phi^{\prime\prime}(\theta_{\circ})h^{2}\lambda)\|g\|_{p}$

whenever $h\leq h_{\circ}$.

We note $\kappa(\alpha)=\min\\{3,\alpha+1,2\alpha\\}$ (see (1.8)).

###### Proof.

Since $\widehat{g}(\eta,\rho)$ is supported in $A_{\lambda}(J_{h})$, we have
$\lambda/4\leq\eta\leq 4\lambda$ and $\theta=\rho/\eta\in J_{h}$ i.e.,
$|\theta-\theta_{\circ}|\leq h$. We set

$D_{\theta_{\circ},h}=\frac{1}{\phi^{\prime\prime}(\theta_{\circ})}\begin{bmatrix}h^{-2}&h^{-2}\theta_{\circ}&h^{-2}\phi(\theta_{\circ})\\\
0&h^{-1}&h^{-1}\phi^{\prime}(\theta_{\circ})\\\
0&0&\phi^{\prime\prime}(\theta_{\circ})\end{bmatrix},$

and

$L_{\theta_{\circ},h}(\eta,\rho)=\frac{1}{\phi^{\prime\prime}(\theta_{\circ})}(h^{-2}\eta,h^{-1}\rho+h^{-2}\theta_{\circ}\eta).$

Let $z=(x,t)$. Note that $z\cdot(\eta,\rho,\eta\phi(\rho/\eta))\mapsto
D_{\theta_{\circ},h}\,z\cdot(\eta,\rho,\eta\phi_{\theta_{\circ},h}(\rho/\eta))$
under the transformation $(\eta,\rho)\mapsto L_{\theta_{\circ},h}(\eta,\rho)$.
Changing variables $(\eta,\rho)\mapsto L_{\theta_{\circ},h}(\eta,\rho)$, we
obtain

$T_{\lambda}^{\phi}g(z)=T_{\lambda_{1}}^{\phi_{\theta_{\circ},h}}g_{\theta_{\circ},h}(D_{\theta_{\circ},h}z),\qquad\lambda_{1}=\phi^{\prime\prime}(\theta_{\circ})h^{2}\lambda,$

where

$\widehat{g_{\theta_{\circ},h}}(\eta,\rho)={\det(L_{\theta_{\circ},h})}\widehat{g}(L_{\theta_{\circ},h}(\eta,\rho)).$

Clearly, $\widehat{g_{\theta_{\circ},h}}$ is supported in
$A_{\lambda_{1}}([-1,1])$. By Lemma 2.2 there exists
$h_{\circ}=h_{\circ}(\epsilon_{\circ})$ such that
$\phi_{\theta_{\circ},h}\in\mathfrak{S}(\epsilon_{\circ})$ for $h\leq
h_{\circ}$.

Let us define

$\omega_{\theta_{\circ},h}(z)=(1+\widetilde{C})^{-1}h^{\kappa(\alpha)-2\alpha+3}\omega(D_{\theta_{\circ},h}^{-1}z),$

where $\widetilde{C}>0$ is a constant to be chosen later. We claim
$\omega_{\theta_{\circ},h}\in\Omega(\alpha)$ if $\widetilde{C}>0$ is large
enough. To show this, we first observe

$\int_{\mathbb{B}^{3}(y_{\circ},r)}\omega(D_{\theta_{\circ},h}^{-1}z)dz=h^{-3}\int_{D_{\theta_{\circ},h}^{-1}\mathbb{B}^{3}(y_{\circ},r)}\omega(z)dz.$

So, it is sufficient to show

(2.11)
$\displaystyle\int_{D_{\theta_{\circ},h}^{-1}\mathbb{B}^{3}(y_{\circ},r)}\omega(z)dz\leq
Cr^{\alpha}h^{2\alpha-\kappa(\alpha)}.$

Note that $D_{\theta_{\circ},h}^{-1}\mathbb{B}^{3}(y_{\circ},r)$ is contained
in a box of dimensions $ch^{2}r\times chr\times cr$ for a constant $c>0$. We
can cover $D_{\theta_{\circ},h}^{-1}\mathbb{B}^{3}(y_{\circ},r)$ with as many
as $Ch^{-3}$, $Ch^{-1}$, and $C$ balls of radii $h^{2}r$, $hr$, and $r$,
respectively. By (2.1) the left hand side of (2.11) is bounded above by
$C\min\\{r^{\alpha}h^{2\alpha-3},r^{\alpha}h^{\alpha-1},r^{\alpha}\\}=Cr^{\alpha}h^{2\alpha-\min\\{3,\alpha+1,2\alpha\\}}$
for some $C$.

Changing variables $z\mapsto D_{\theta_{\circ},h}^{-1}z$, we have

(2.12) $\displaystyle\|T_{\lambda}^{\phi}g\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}^{q}$
$\displaystyle=(1+\widetilde{C})h^{2\alpha-\kappa(\alpha)}\int_{\mathbb{R}^{2}\times
I}\big{|}T_{\lambda_{1}}^{\phi_{\theta_{\circ},h}}g_{\theta_{\circ},h}(z)\big{|}^{q}\omega_{\theta_{\circ},h}(z)dz.$

Since $\phi_{\theta_{\circ},h}\in\mathfrak{S}(\epsilon_{\circ})$,
$\omega_{\theta_{\circ},h}\in\Omega(\alpha)$, and
$\widehat{g_{\theta_{\circ},h}}$ is supported in $A_{\lambda_{1}}([-1,1])$, by
the definition of $Q$ we have

$\int_{\mathbb{R}^{2}\times
I}\big{|}T_{\lambda_{1}}^{\phi_{\theta_{\circ},h}}g_{\theta_{\circ},h}(z)\big{|}^{q}\omega_{\theta_{\circ},h}(z)dz\leq\big{(}Q(\phi^{\prime\prime}(\theta_{\circ})h^{2}\lambda)\|g_{\theta_{\circ},h}\|_{p}\big{)}^{q}.$

Therefore, we get (2.10) since
$\|g_{\theta_{\circ},h}\|_{p}=h^{-\frac{3}{p}}\|g\|_{p}$. ∎

###  2.2. Trilinear estimate.

In this section, we obtain a weighted local $L^{p}$–$L^{q}$ estimate for the
trilinear operator $\prod_{i=1}^{3}T_{\lambda}^{\phi}g_{i}$ by interpolating
the multilinear restriction estimate due to Bennett, Carbery and Tao [2], the
local smoothing estimate in [8], and an easy $L^{1}$–$L^{\infty}$ estimate.
Subsequently, we extend the local estimate to a global one by using decay of
the kernel $K_{\lambda}$. For simplicity, we write

$T_{\lambda}=T_{\lambda}^{\phi}.$

For $\phi\in\mathfrak{S}(\epsilon_{\circ})$ and an interval $J\subset[-1,1]$,
we consider a truncated conic surface

$\Gamma_{\lambda}(J):=\\{(\eta,\rho,\tau)\in\mathbb{R}^{3}:\tau=\eta\phi(\rho/\eta),~{}(\eta,\rho)\in
A_{\lambda}(J)\\}.$

Let $\mathrm{N}(\theta)$ be the normal vector to $\Gamma_{\lambda}(J)$ at
$\eta(1,\theta,\phi(\theta))$. A computation gives

(2.13)
$\mathrm{N}(\theta)=(\phi(\theta)-\theta\phi^{\prime}(\theta),\phi^{\prime}(\theta),-1).$

The following is a consequence of the multilinear restriction estimate due to
Bennett-Carbery-Tao [2]. We denote by $d\sigma_{\lambda}$ the surface measure
on $\Gamma_{\lambda}([-1,1])$.

######  Theorem 2.4.

Let $\phi\in\mathfrak{S}(\epsilon_{\circ})$ and
$J_{1},J_{2},J_{3}\subset[-1,1]$ be intervals. Suppose

(2.14)
$\det(\mathrm{N}(\theta_{1}),\mathrm{N}(\theta_{2}),\mathrm{N}(\theta_{3}))\geq\delta$

holds for some $\delta>0$ whenever $\theta_{i}\in J_{i}$, $i=1,2,3$. If
$\epsilon_{\circ}$ is sufficiently small and $\lambda\gg\delta^{-1}$, then for
$\epsilon>0$

$\Big{\|}\prod_{i=1}^{3}\widehat{f_{i}d\sigma_{\lambda}}\Big{\|}_{L^{1}(\mathbb{B}^{3}(0,10))}\leq
C_{\epsilon}\lambda^{\epsilon}\prod_{i=1}^{3}\|f_{i}\|_{L^{2}(\Gamma_{\lambda}(J_{i}))}$

whenever ${\rm supp}\,f_{i}\subset\Gamma_{\lambda}(J_{i})$, $i=1,2,3$.

By interpolation with the local smoothing estimate and an easy
$L^{1}-L^{\infty}$ estimate, we get the following.

######  Proposition 2.5.

Let $\phi\in\mathfrak{S}(\epsilon_{\circ})$, and let
$J_{1},J_{2},J_{3}\subset[-1,1]$ be intervals. Also let
$\frac{1}{q}\leq\min(\frac{1}{p},~{}\frac{1}{3p}+\frac{1}{6},~{}\frac{2}{3p^{\prime}})$
for $1\leq p\leq\infty$. Suppose (2.14) holds whenever $\theta_{i}\in J_{i}$,
$i=1,2,3$. If $\epsilon_{\circ}$ is sufficiently small and
$\lambda\gg\delta^{-1}$, then we have

(2.15)
$\Big{\|}\prod_{i=1}^{3}T_{\lambda}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{B}^{2}(0,1)\times
I)}\leq C\lambda^{3\gamma_{0}}\prod_{i=1}^{3}\|g_{i}\|_{p},$

for $\gamma_{0}>\frac{1}{2}+\frac{1}{p}-\frac{3}{q}$ whenever ${\rm
supp}\,\widehat{g_{i}}\subset A_{\lambda}(J_{i})$, $i=1,2,3$.

###### Proof.

By interpolation it is enough to show (2.15) for
$(\frac{1}{p},\frac{1}{q})=(0,0)$, $(\frac{1}{4},\frac{1}{4})$,
$(\frac{1}{2},\frac{1}{3})$, and $(1,0)$ if
$\gamma_{0}>\frac{1}{2}+\frac{1}{p}-\frac{3}{q}$. For the first two cases,
(2.15) follows from the local smoothing estimate (1.7) with $d\nu=dxdt$ which
holds for $4\leq p=q\leq\infty$ and $\gamma_{0}>\frac{1}{2}-\frac{2}{p}$. By
Theorem 2.4 and Plancherel’s theorem, we have (2.15) with
$(\frac{1}{p},\frac{1}{q})=(\frac{1}{2},\frac{1}{3})$ for $\gamma_{0}>0$.
Also, (2.7) and van der Corput’s lemma yield
$\|K_{\lambda}\|_{\infty}\leq\lambda^{3/2}$ (e.g. (3.6)), which gives
$\|T_{\lambda}g_{i}\|_{L^{\infty}(\mathbb{B}^{2}(0,1)\times
I)}\lesssim\lambda^{3/2}\|g_{i}\|_{1}$. Thus, we have (2.15) with
$(\frac{1}{p},\frac{1}{q})=(1,0)$ for $\gamma_{0}\geq 3/2$. ∎

From (2.8), we see that $K_{\lambda}(\cdot,t)$ decays rapidly away from
$\mathbb{B}^{2}(0,1)$. Using this, we can extend the local estimate (2.15) to
a global one.

######  Lemma 2.6.

Under the same assumptions as in Proposition 2.5, we have

(2.16)
$\Big{\|}\prod_{i=1}^{3}T_{\lambda}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{2}\times
I)}\leq C\lambda^{3\gamma_{0}}\prod_{i=1}^{3}\|g_{i}\|_{p}$

for $\gamma_{0}>\frac{1}{2}+\frac{1}{p}-\frac{3}{q}$.

###### Proof.

We follow a standard localization argument (e.g., see [15, Proposition 2.10]).
For a constant $r$ satisfying $\lambda^{-1}\ll r\ll\delta$, let
$N_{r\lambda}(A_{\lambda}(J_{i}))$ denote the $r\lambda$-neighborhood of
$A_{\lambda}(J_{i})$, $i=1,2,3$. Let $\varphi_{i}$ be a smooth function
supported in $N_{r\lambda}(A_{\lambda}(J_{i}))$ such that $\varphi_{i}=1$ on
$A_{\lambda}(J_{i})$ and
$|\partial_{\xi}^{m}\varphi_{i}(\xi)|\lesssim|\xi|^{-|m|}$.

Let us denote by $\mathcal{F}_{x}$ the Fourier transform in $x$ and define
$\mathcal{F}_{x}(\widetilde{K}_{\lambda,i}(\cdot,t))=\mathcal{F}_{x}(K_{\lambda}(\cdot,t))\varphi_{i}$.
Then $T_{\lambda}g_{i}(x,t)=\widetilde{K}_{\lambda,i}(\cdot,t)\ast g_{i}(x)$.
We consider a collection $\\{B\\}$ of finitely overlapping unit balls which
cover $\mathbb{R}^{2}$. Fixing $\varepsilon>0$, let $\widetilde{B}$ be a ball
of radius $\lambda^{\varepsilon}$ with the same center as $B$. For each $i$,
we decompose $g_{i}=g_{i,B}+g_{i,B^{\mathrm{c}}}$ where
$g_{i,B}:=g_{i}\chi_{\widetilde{B}}$ and
$g_{i,B^{\mathrm{c}}}=g_{i}\chi_{\widetilde{B}^{\mathrm{c}}}$. We get

$\displaystyle\Big{\|}\prod_{i=1}^{3}T_{\lambda}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{2}\times
I)}^{\frac{q}{3}}\lesssim\sum_{B}\sum_{\begin{subarray}{c}(m_{1},m_{2},m_{3}):\\\
m_{j}\in\\{B,B^{\mathrm{c}}\\}\end{subarray}}\int_{I}\int_{B}\prod_{i=1}^{3}\big{|}\widetilde{K}_{\lambda,i}(\cdot,t)\ast(g_{i,m_{i}})(x)\big{|}^{\frac{q}{3}}dxdt.$

We consider the case $m_{1}=m_{2}=m_{3}=B$ first. Since the Fourier transform
of $\widetilde{K}_{\lambda,i}(\cdot,t)\ast(g_{i,B})$ is supported in
$N_{r\lambda}(A_{\lambda}(J_{i}))$, the transversality condition (2.14) holds
for $2^{-1}\delta$ replacing $\delta$ if we take a sufficiently small $r$.
Applying Proposition 2.5, we get

$\displaystyle\sum_{B}\int_{B\times
I}\prod_{i=1}^{3}\big{|}\widetilde{K}_{\lambda,i}(\cdot,t)\ast(g_{i,B})(x)\big{|}^{\frac{q}{3}}dxdt\lesssim\lambda^{q\gamma_{0}}\sum_{B}\prod_{i=1}^{3}\|g_{i}\chi_{\widetilde{B}}\|_{p}^{\frac{q}{3}}\lesssim\lambda^{q\gamma_{0}+\tilde{c}\varepsilon}\prod_{i=1}^{3}\|g_{i}\|_{p}^{\frac{q}{3}}$

for $\gamma_{0}>\frac{1}{2}+\frac{1}{p}-\frac{3}{q}$ and $\tilde{c}>0$. The
second inequality follows by Hölder’s inequality and the inclusion
$\ell^{p}\subset\ell^{q}$ for $p\leq q$ since the balls $\widetilde{B}$
overlap at most $C\lambda^{2\varepsilon}$.

If $m_{i}=B^{\mathrm{c}}$ for some $i$, we use decay of
$\widetilde{K}_{\lambda,i}$. As before, it is easy to show
$|\widetilde{K}_{\lambda,i}(x,t)|\lesssim\lambda^{2}(1+|x|)^{-M}$ for any
$M\geq 1$ (c.f., (2.8)). Setting $\mathscr{K}(x)=(1+|x|)^{-3}$, we see
$|\widetilde{K}_{\lambda,i}(\cdot,t)\ast
g_{i,B}(x)|\lesssim\lambda^{2}\mathscr{K}\ast|g_{i}|(x)$ and
$\big{|}\widetilde{K}_{\lambda,i}(\cdot,t)\ast
g_{i,B^{\mathrm{c}}}(x)\big{|}\lesssim\lambda^{2}\lambda^{\varepsilon(3-M)}\mathscr{K}\ast|g_{i}|(x)$
if $x\in B$. Thus we have

$\prod_{i=1}^{3}\big{|}\widetilde{K}_{\lambda,i}(\cdot,t)\ast(g_{i,m_{i}})(x)\big{|}\lesssim\lambda^{c_{1}-c_{2}\varepsilon
M}\prod_{i=1}^{3}\mathscr{K}\ast|g_{i}|(x),\quad(x,t)\in B\times I,$

for some constants $c_{1},c_{2}>0$. By Hölder’s inequality and Young’s
convolution inequality, there are positive constants $c_{1},c_{2}$ such that

$\displaystyle\sum_{B}\sum_{(m_{1},m_{2},m_{3})\neq(B,B,B)}\iint_{B\times
I}\prod_{i=1}^{3}\big{|}\widetilde{K}_{\lambda,i}(\cdot,t)\ast(g_{i,m_{i}})(x)\big{|}^{\frac{q}{3}}dxdt\lesssim\lambda^{c_{1}-c_{2}\varepsilon
M}\prod_{i=1}^{3}\|g_{i}\|_{p}^{\frac{q}{3}}$

for $p\leq q$. Combining the estimates above, we obtain

$\Big{\|}\prod_{i=1}^{3}T_{\lambda}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{2}\times
I)}^{\frac{q}{3}}\lesssim\big{(}\lambda^{q\gamma_{0}+\tilde{c}\varepsilon}+\lambda^{c_{1}-c_{2}\varepsilon
M}\big{)}\prod_{i=1}^{3}\|g_{i}\|_{p}^{\frac{q}{3}}$

for $\gamma_{0}>\frac{1}{2}+\frac{1}{p}-\frac{3}{q}$. Taking $\varepsilon>0$
small enough and then a large $M$, we obtain the desired estimate (2.16). ∎

We now obtain (2.16) with weights $\omega\in\Omega(\alpha)$ via the following
lemma.

######  Lemma 2.7.

Let $\omega\in\Omega(\alpha)$, $0<\alpha\leq 3$. Suppose $\widehat{F}$ is
supported in $\mathbb{B}^{3}(0,\lambda)$. Then, we have
$\|F\|_{L^{q}(\mathbb{R}^{3};\omega)}\lesssim\lambda^{(3-\alpha)/q}\|F\|_{L^{q}(\mathbb{R}^{3})}$
for $q\geq 1$.

###### Proof.

Let $\varphi$ be a Schwartz function such that $\widehat{\varphi}=1$ on
$\mathbb{B}^{3}(0,1)$, and set
$\varphi_{\lambda}=\lambda^{3}\varphi(\lambda\cdot)$. Since
$F=F\ast\varphi_{\lambda}$, we have
$\|F\|_{L^{q}(\mathbb{R}^{3};\omega)}\leq\|F\|_{L^{q}(\mathbb{R}^{3})}\||\varphi_{\lambda}|\ast\omega\|_{\infty}^{1/q}$.
By rapid decay of $\varphi$, we see

$\displaystyle|\varphi_{\lambda}|\ast\omega(y)\leq
C_{N}\lambda^{3}\sum_{\ell\geq
0}2^{-N\ell}\int_{\mathbb{B}^{3}(y,2^{\ell}\lambda^{-1})}\omega(z)dz$

for any $N>3$. Since $\omega\in\Omega(\alpha)$, it follows by (2.1) that
$\||\varphi_{\lambda}|\ast\omega\|_{\infty}\lesssim\lambda^{3-\alpha}$. Hence,
we get the desired bound. ∎

######  Proposition 2.8.

Let $\phi\in\mathfrak{S}(\epsilon_{\circ})$, $\omega\in\Omega(\alpha)$,
$0<\alpha\leq 3$, and $\lambda^{-1}\ll\delta\ll 1$. Suppose (2.14) holds
whenever $\theta_{i}\in J_{i}$, $i=1,2,3$. If (2.16) holds for some
$\gamma_{0}$, then

(2.17)
$\Big{\|}\prod_{i=1}^{3}T_{\lambda}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{2}\times
I;\omega)}\leq C\lambda^{3\gamma}\prod_{i=1}^{3}\|g_{i}\|_{p}$

holds for $\gamma=\gamma_{0}+\frac{3-\alpha}{q}$ whenever ${\rm
supp}\,\widehat{g_{i}}\subset A_{\lambda}(J_{i})$.

###### Proof.

Let $\widetilde{\chi}\in\mathrm{C}_{0}^{\infty}((1/2,4))$ such that
$\widetilde{\chi}=1$ on $I=[1,2]$. We only have to show (2.17) with
$T_{\lambda}$ replaced by $\widetilde{\chi}\,T_{\lambda}$. Since
$\widetilde{\chi}$ is compactly supported, the support of the space-time
Fourier transform of $\widetilde{\chi}(t)T_{\lambda}g_{i}(x,t)$ is unbounded.
So, in order to apply Lemma 2.7, we decompose $\widetilde{\chi}\,T_{\lambda}f$
in such a way that the Fourier supports of the consequent operators are
contained in either a bounded set or its complement. Let us define a frequency
localized operator $T_{\lambda}^{1}$ by

$\displaystyle\mathcal{F}_{x,t}\big{(}T_{\lambda}^{1}g\big{)}(\eta,\rho,\tau)=\mathcal{F}_{x,t}\big{(}\widetilde{\chi}\,T_{\lambda}g\big{)}(\eta,\rho,\tau)\beta_{0}((c\lambda)^{-1}|\tau|),$

where $c>0$ is a constant to be chosen later. We also set
$T_{\lambda}^{2}g=(\widetilde{\chi}\,T_{\lambda}-T_{\lambda}^{1})g$. Then, we
have

$\displaystyle\Big{\|}\prod_{i=1}^{3}\widetilde{\chi}\,T_{\lambda}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{3};\omega)}\leq{\rm
I}+{\rm I\\!I},$

where

$\displaystyle{\rm
I}=\Big{\|}\prod_{i=1}^{3}T_{\lambda}^{1}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{3};\omega)},\quad{\rm
I\\!I}=\sum_{\begin{subarray}{c}(m_{1},m_{2},m_{3})\neq(1,1,1):\\\
m_{j}\in\\{1,2\\}\end{subarray}}\Big{\|}\prod_{i=1}^{3}T_{\lambda}^{m_{i}}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{3};\omega)}.$

First, we show ${\rm I\\!I}\lesssim\lambda^{-M}\prod_{i=1}^{3}\|g_{i}\|_{p}$
for any $M>1$. It is enough to show

(2.18)
$\displaystyle\|T_{\lambda}^{2}g\|_{L^{q}(\mathbb{R}^{3};\omega)}\lesssim\lambda^{-M}\|g\|_{p}$

for any $M>0$ if $p\leq q$. Once we have (2.18), it follows that
$\|T_{\lambda}^{1}g\|_{L^{q}(\mathbb{R}^{3};\omega)}\lesssim\lambda^{2}\|g\|_{p}$
since
$|T_{\lambda}^{1}g|\leq|\widetilde{\chi}\,T_{\lambda}g|+|T_{\lambda}^{2}g|$
and
$\|\widetilde{\chi}\,T_{\lambda}g\|_{L^{q}(\mathbb{R}^{3};\omega)}\lesssim\lambda^{2}\|g\|_{p}$
for $p\leq q$ (see (2.8) and (2.9)). Thus, the desired bound on
$\mathrm{I\\!I}$ follows by Hölder’s inequality.

To prove (2.18), let us set

$\Phi(\eta,\rho,\tau)=\frac{1}{2\pi}\int
e^{-it(\tau-\eta\phi(\rho/\eta))}\widetilde{\chi}(t)dt\,\beta(\lambda^{-1}\eta)\beta_{0}(|\rho/\eta|).$

Recalling the definition of $T_{\lambda}$ ((2.3)), we have
$\mathcal{F}_{x,t}(\widetilde{\chi}\,T_{\lambda}g)(\eta,\rho,\tau)=\Phi(\eta,\rho,\tau)\widehat{g}(\eta,\rho)$.
Note $T_{\lambda}^{2}=(\widetilde{\chi}\,T_{\lambda}-T_{\lambda}^{1})$, so
$T_{\lambda}^{2}g(x,t)=\mathfrak{K}(\cdot,t)\ast g(x),$ where

$\mathfrak{K}(x,t)=\iiint\big{(}1-\beta_{0}((c\lambda)^{-1}|\tau|)\big{)}\Phi(\eta,\rho,\tau)e^{i(x\cdot(\eta,\rho)+t\tau)}d\eta
d\rho d\tau.$

Taking a sufficiently large $c>0$, by integration by parts, we get

(2.19) $|\mathfrak{K}(x,t)|\lesssim\lambda^{-M}(1+|x|)^{-M}(1+|t|)^{-M}$

for $M\geq 1$. We see
$\partial_{\eta,\rho,\tau}^{m}\big{(}(1-\beta_{0}((c\lambda)^{-1}|\tau|))\Phi(\eta,\rho,\tau)\big{)}$
are bounded by $C_{M}\lambda^{-M}(1+|\tau|)^{-M}$ for any $M\geq 1$ since
$\partial_{\eta,\rho}^{m}\phi$ are bounded by a constant depending only on the
multi index $m$. Repeated integration by parts in $\eta,\rho,\tau$ gives
(2.19). Using (2.19), we get (2.18) with $p=q$ by Schur’s test. Also, (2.18)
holds with $q=\infty$ by Young’s convolution inequality. By interpolation, we
get (2.18) for any $1\leq p\leq q\leq\infty$.

We now turn to the term ${\rm I}$. Since the Fourier transform of
$T_{\lambda}^{1}g_{i}$ is supported in a ball of radius $2c\lambda$, we can
apply Lemma 2.7 to $\rm I$. So, we get

$\displaystyle{\rm
I}\lesssim\lambda^{\frac{3(3-\alpha)}{q}}\Big{\|}\prod_{i=1}^{3}T_{\lambda}^{1}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{R}^{3})}.$

Since
$|T_{\lambda}^{1}g|\leq|\widetilde{\chi}T_{\lambda}g|+|T_{\lambda}^{2}g|$,
using
$\|\widetilde{\chi}\,T_{\lambda}g\|_{L^{q}(\mathbb{R}^{3})}\lesssim\lambda^{2}\|g\|_{p}$
and (2.18), we see that
$\big{\|}\prod_{i=1}^{3}T_{\lambda}^{1}g_{i}\big{\|}_{q/3}$ is bounded by
$\|\prod_{i=1}^{3}\widetilde{\chi}\,T_{\lambda}g_{i}\|_{q/3}+C\lambda^{-N}\prod_{i=1}^{3}\|g_{i}\|_{p}$
for any $N>0$. Now we apply Lemma 2.6 to
$\|\prod_{i=1}^{3}\widetilde{\chi}\,T_{\lambda}g_{i}\|_{q/3}$ expanding the
interval $I$ slightly. Choosing $N$ large enough, we obtain

$\displaystyle{\rm
I}\lesssim\lambda^{3(\gamma_{0}+\frac{3-\alpha}{q})}\prod_{i=1}^{3}\|g_{i}\|_{p}.$

Combining all the bounds on ${\rm I}$ and ${\rm I\\!I}$, we get the desired
estimate (2.17) by taking $M$ sufficiently large. ∎

The estimate (2.17) plays a crucial role in proving Theorem 1.6. However, in
order to prove Theorem 1.5 (or Proposition 2.1), it suffices to obtain the
following estimate which is not necessarily sharp.

######  Corollary 2.9.

Let $\phi\in\mathfrak{S}(\epsilon_{\circ})$, $\omega\in\Omega(\alpha)$ for
$1<\alpha\leq 2$, and $\lambda^{-1}\ll\delta\ll 1$. Suppose (2.14) holds
whenever $\theta_{i}\in J_{i}$, $i=1,2,3$, and let

(2.20) $\frac{1}{q}\leq\frac{1}{p}<\frac{\alpha}{q}~{}\text{ and
}~{}p>\begin{cases}\,\frac{2\alpha+3}{2\alpha},~{}&\quad\hfill\frac{3}{2}<\alpha\leq
2,\\\ \,\frac{6-2\alpha}{\alpha},~{}&\quad\hfill
1<\alpha\leq\frac{3}{2}.\end{cases}$

Then we have (2.17) for some $\gamma<\frac{1}{2}$ whenever ${\rm
supp}\,\widehat{g_{i}}\subset A_{\lambda}(J_{i})$.

###### Proof.

By Proposition 2.5 and Proposition 2.8, we have

(2.21)
$\Big{\|}\prod_{i=1}^{3}T_{\lambda}g_{i}\Big{\|}_{L^{\frac{q}{3}}(\mathbb{B}^{2}(0,1)\times
I;\omega)}\leq C\lambda^{3\gamma}\prod_{i=1}^{3}\|g_{i}\|_{p}$

for $\gamma^{\prime}>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$ provided that
$\frac{1}{q}\leq\min(\frac{1}{p},~{}\frac{1}{3p}+\frac{1}{6},~{}\frac{2}{3p^{\prime}})$.
By Hölder’s inequality, (2.21) continues to be true for some
$\gamma^{\prime}<\frac{1}{2}$ if $p,q$ satisfy (2.20). Indeed, there is a $q$
such that $\frac{1}{p}=\frac{\alpha}{q}$ for $p>\frac{6-2\alpha}{\alpha}$ when
$1<\alpha\leq\frac{3}{2}$, and for $p>\frac{2\alpha+3}{2\alpha}$ when
$\frac{3}{2}<\alpha\leq 2$.

Repeating the same argument as in the proof of Lemma 2.6, one can extend
(2.21) to the global estimate (2.17) as long as $q\geq p$. We omit the
details. ∎

###  2.3. Proof of Proposition 2.1

In this subsection, we prove Proposition 2.1 by combining Lemma 2.3 and
Corollary 2.9. To this end, we decompose the frequency support of
$T_{\lambda}^{\phi}g$ independent of particular choice of $\phi$.

#### Decomposition

Let $K_{i}$, $i=0,1,2$ be dyadic numbers such that

$\lambda^{-1/3}\ll K_{2}\ll K_{1}\ll K_{0}=1.$

For each $K_{i}$, we consider a collection $\mathfrak{J}^{i}$ of dyadic
intervals $J^{i}=[(j-1)K_{i},(j+1)K_{i}]$ for $j\in\mathbb{Z}$, $|j|\leq
K_{i}^{-1}$ such that union of $J^{i}$ covers $[-1,1]$.

Let $\varrho\in\mathrm{C}_{0}^{\infty}((-1,1))$ satisfying
$\sum_{j\in\mathbb{Z}}\varrho(\cdot-j)=1$. For an interval
$J=[\theta-K,\theta+K]$, we denote
$\varrho_{J}=\varrho(K^{-1}(\cdot-\theta))$. Note that
$\sum_{J^{i}\in\mathfrak{J}^{i}}\varrho_{J^{i}}=1$ on $[-1,1]$. We set
$\widehat{g_{J^{i}}}(\eta,\rho)=\widehat{g}(\eta,\rho)\,\varrho_{J^{i}}(\rho/\eta)$.
Then, we have

$T_{\lambda}^{\phi}g(x,t)=\sum_{J^{i}\in\mathfrak{J}^{i}}T_{\lambda}^{\phi}g_{J^{i}}(x,t),\quad
i=0,1,2.$

Note that $\widehat{g_{\\!J^{i}}}$ is supported in a rectangle of dimensions
$c\lambda\times c\lambda K_{i}$ for a constant $c>0$.

Following the argument in [16, Section 3], we have the next lemma.

######  Lemma 2.10.

For each $(x,t)\in\mathbb{R}^{2}\times I$, there exists a constant $C>0$,
independent of $K_{i}$ and $(x,t)$, such that

(2.22)
$\displaystyle|T_{\lambda}^{\phi}g(x,t)|\leq\sum_{i=1}^{2}CK_{i-1}^{-2}$
$\displaystyle\max_{J^{i}\in\mathfrak{J}^{i}}|T_{\lambda}^{\phi}g_{J^{i}}(x,t)|$
$\displaystyle+$ $\displaystyle
CK_{2}^{-4}\max_{(J^{2}_{1},J^{2}_{2},J^{2}_{3})\in\mathfrak{J}^{2}(K_{2})}\prod_{i=1}^{3}|T_{\lambda}^{\phi}g_{J^{2}_{i}}(x,t)|^{\frac{1}{3}},$

where
$\mathfrak{J}^{2}(K_{2}):=\\{(J^{2}_{1},J^{2}_{2},J^{2}_{3}):J^{2}_{1},J^{2}_{2},J^{2}_{3}\in\mathfrak{J}^{2},~{}\min_{k\neq
l}{\rm dist}\,(J^{2}_{k},J^{2}_{l})\geq K_{2}\\}$.

If $(J^{2}_{1},J^{2}_{2},J^{2}_{3})\in\mathfrak{J}^{2}(K_{2})$, then (2.14)
holds for $\theta_{i}\in J_{i}^{2}$, $i=1,2,3$. More precisely, we have the
following.

######  Lemma 2.11.

Let $J_{1},J_{2},J_{3}$ be subintervals of $[-1,1]$ such that $\min_{k\neq
l}{\rm dist}\,(J_{k},J_{l})\gtrsim\delta^{\frac{1}{3}}$. Suppose that
$\phi\in\mathfrak{S}(\epsilon_{\circ})$. If $\epsilon_{\circ}\in(0,1/8)$, then
$|\det\big{(}\mathrm{N}(\theta_{1}),\mathrm{N}(\theta_{2}),\mathrm{N}(\theta_{3})\big{)}|\gtrsim\delta$
for $\theta_{i}\in J_{i}$, i=1,2,3.

###### Proof.

By the generalized mean value theorem (see [22, Part V, Problem 95]), there
are $u_{j}$ for $j=1,2,3$ such that $\min_{1\leq i\leq
3}\theta_{i}<u_{j}<\max_{1\leq i\leq 3}\theta_{i}$ and

$\det(\mathrm{N}(\theta_{1}),\mathrm{N}(\theta_{2}),\mathrm{N}(\theta_{3}))=\det(\mathrm{N}(u_{1}),\mathrm{N}^{\prime}(u_{2}),\mathrm{N}^{\prime\prime}(u_{3}))\prod_{1\leq
k<\ell\leq 3}|\theta_{k}-\theta_{\ell}|.$

From (2.13) we have

$\displaystyle\det(\mathrm{N}(u_{1}),\mathrm{N}^{\prime}(u_{2}),\mathrm{N}^{\prime\prime}(u_{3}))=\det\begin{pmatrix}u_{2}\phi^{\prime\prime}(u_{2})&\phi^{\prime\prime}(u_{3})+u_{3}\phi^{(3)}(u_{3})\\\
\phi^{\prime\prime}(u_{2})&\phi^{(3)}(u_{3})\end{pmatrix}.$

Since $\phi\in\mathfrak{S}(\epsilon_{\circ})$ and
$\epsilon_{\circ}\in(0,1/8)$, $7/8\leq|\phi^{\prime\prime}(\theta)|\leq 9/8$
and $|\phi^{(3)}(\theta)|\leq 1/8$ for $\theta\in[-1,1]$. Thus, it follows
that
$|\det(\mathrm{N}(u_{1}),\mathrm{N}^{\prime}(u_{2}),\mathrm{N}^{\prime\prime}(u_{3}))|\geq
27/8^{2}$. So we obtain the desired bound since $\prod_{1\leq k<\ell\leq
3}|\theta_{k}-\theta_{\ell}|\gtrsim\delta$. ∎

We are now ready to prove Proposition 2.1.

###### Proof of Proposition 2.1.

We begin by making a primary decomposition by which we reduce the matter to
obtaining estimate for $T_{\lambda}^{\phi}$,
$\phi\in\mathfrak{S}(\epsilon_{\circ})$ relative to $\omega\in\Omega(\alpha).$

Let $\epsilon_{\circ}\in(0,1/8)$. Following the argument in the proof of Lemma
2.2, we can fix a small positive dyadic number
$h_{\ast}<h_{\circ}(\epsilon_{\circ})$ such that
$(\psi_{{0}})_{\theta,h_{\ast}}\in\mathfrak{S}(\epsilon_{\circ})$ for
$\theta\in[-1,1]$. Let $J_{j}=[(j-1)h_{\ast},(j+1)h_{\ast}]$, $j\in\mathbb{Z}$
such that $|j|\leq h_{*}^{-1}+2$. Then $\sum_{j}\varrho_{J_{j}}=1$ on
$[-1,1]$. Denoting
$\widehat{g_{j}}(\eta,\rho)=\widehat{g}(\eta,\rho)\varrho_{J_{j}}(\rho/\eta)$,
we get

$\|T_{\lambda}^{\psi_{{0}}}g\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}\leq\sum_{j}\|T_{\lambda}^{\psi_{{0}}}g_{j}\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}.$

Let $\theta_{j}=jh_{*}$. As before (cf. (2.12)), the change of variable
$(\eta,\rho)\rightarrow L_{\theta_{j},{h_{\ast}}}(\eta,\rho)$ gives

(2.23)
$\displaystyle\|T_{\lambda}^{\psi_{{0}}}g_{j}\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}^{q}$ $\displaystyle\leq
Ch_{\ast}^{2\alpha-\kappa(\alpha)}\big{\|}T_{\psi_{0}^{\prime\prime}(\theta_{j})h^{2}_{\ast}\lambda}^{(\psi_{{0}})_{\theta_{j},{h_{\ast}}}}(g_{j})_{\theta_{j},{h_{\ast}}}\big{\|}_{L^{q}(\mathbb{R}^{2}\times
I;\omega_{\theta_{j},{h_{\ast}}})}^{q}.$

Note $\|(g_{j})_{\theta_{j},{h_{\ast}}}\|_{p}=h_{*}^{-3/p}\|g\|_{p}$ and
$\omega_{\theta_{j},{h_{\ast}}}\in\Omega(\alpha)$. Since
$|\psi_{0}^{\prime\prime}(\theta_{\circ})-1|\leq\epsilon_{\circ}$ and
$h_{\ast}$ is a fixed constant depending only on $\epsilon_{\circ}$, in order
to prove Proposition 2.1 it suffices to show

(2.24) $\|T_{\lambda}^{\phi}g\|_{L^{q}(\mathbb{R}^{2}\times I;\omega)}\leq
C\lambda^{\gamma}\|g\|_{p}$

for some $\gamma<1/2$ provided that $\phi\in\mathfrak{S}(\epsilon_{\circ})$
and $\omega\in\Omega(\alpha).$

Since $1<\alpha\leq 2$, we have $\frac{\alpha-1}{q}-\frac{3}{p}+1>0$ from the
hypothesis. So we can choose $\gamma<\frac{1}{2}$ such that

(2.25) $\frac{\alpha-1}{q}-\frac{3}{p}+2\gamma>0.$

For such $\gamma$ and for $\lambda^{\prime}\geq 1$ we set

$\mathfrak{Q}(\lambda^{\prime})=\mathfrak{Q}(\lambda^{\prime},p,q,\alpha):=\sup_{1\leq\lambda\leq\lambda^{\prime}}{\lambda}^{-\gamma}Q({\lambda}),$

where $Q({\lambda})$ is defined by (2.5). The estimate (2.24) follows if we
show $\mathfrak{Q}(\lambda^{\prime})\leq C$.

Let $h_{\circ}$ be a small positive number so that Lemma 2.2 and Lemma 2.3
hold. Let $K_{1},$ $K_{2}$ be positive dyadic numbers such that $0<K_{2}\ll
K_{1}\leq h_{\circ}$. We also set $K_{0}=1$ as before. Using the decomposition
(2.22), we have

(2.26) $\displaystyle\|T_{\lambda}^{\phi}g\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}\leq\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3},$

where

$\displaystyle\mathrm{S}_{i}$
$\displaystyle=CK_{i-1}^{-2}\big{\|}\max_{J^{i}\in\mathfrak{J}^{i}}|T_{\lambda}^{\phi}g_{J^{i}}|\big{\|}_{L^{q}(\mathbb{R}^{2}\times
I;\omega)},\quad i=1,2$ $\displaystyle\mathrm{S}_{3}$
$\displaystyle=CK_{2}^{-4}\Big{\|}\max_{(J^{2}_{1},J^{2}_{2},J^{2}_{3})\in\mathfrak{J}^{2}(K_{2})}\prod_{i=1}^{3}|T_{\lambda}^{\phi}g_{J^{2}_{i}}|^{\frac{1}{3}}\Big{\|}_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}.$

Note that $\mathrm{S}_{i}^{q}\leq
C^{q}K_{i-1}^{-2q}\sum_{J^{i}\in\mathfrak{J}^{i}}\|T_{\lambda}^{\phi}g_{J^{i}}\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}^{q}$. By Lemma 2.3 with $\kappa(\alpha)=\alpha+1$, for $i=1,2$ we
have

$\displaystyle\mathrm{S}_{i}^{q}$ $\displaystyle\leq
C^{\prime}K_{i-1}^{-2q}\big{(}K_{i}^{\frac{\alpha-1}{q}-\frac{3}{p}}\sup_{\theta\in[-1,1]}Q(\phi^{\prime\prime}(\theta)K_{i}^{2}\lambda)\big{)}^{q}\sum_{J^{i}\in\mathfrak{J}^{i}}\|g_{J^{i}}\|_{p}^{q}.$

By the embedding $\ell^{p}\subset\ell^{q}$ ($p\leq q$), it follows that
$\sum_{J^{i}\in\mathfrak{J}^{i}}\|g_{J^{i}}\|_{p}^{q}\leq\|g\|_{p}^{q}$. Here,
we use the estimate
$\big{(}\sum_{J^{i}\in\mathfrak{J}^{i}}\|g_{J^{i}}\|_{p}^{p}\big{)}^{1/p}\lesssim\|g\|_{p}$
for $2\leq p\leq\infty$, which can be shown by interpolation between trivial
$L^{\infty}$–$\ell^{\infty}L^{\infty}$ estimate and $L^{2}$–$\ell^{2}L^{2}$
estimate via Plancherel’s theorem.

By Lemma 2.11, the transversality condition (2.14) with $\delta=(K_{2})^{3}$
holds if $(J^{2}_{1},J^{2}_{2},J^{2}_{3})\in\mathfrak{J}^{2}(K_{2})$. By
Corollary 2.9, we have $\mathrm{S}_{3}\leq
CK_{2}^{-C}\lambda^{\gamma}\|g\|_{p}$ for $\lambda\gg K_{2}^{-3}$ and $p,q$
satisfying (2.20). For the case of $\lambda\lesssim K_{2}^{-3}$, we have
$\mathrm{S}_{3}\leq CK_{2}^{-C}\lambda^{\gamma}\|g\|_{p}$ for some constant
$C>0$ from the easy estimate (e.g. (2.9)) and Hölder’s inequality. So we get

$\mathrm{S}_{3}\leq CK_{2}^{-C}\lambda^{\gamma}\|g\|_{p}.$

Combining all the estimates for $\mathrm{S}_{1}$, $\mathrm{S}_{2}$, and
$\mathrm{S}_{3}$, which hold uniformly for
$\phi\in\mathfrak{S}(\epsilon_{\circ})$ and $\omega\in\Omega(\alpha)$, we
obtain

(2.27) $\displaystyle Q(\lambda)$
$\displaystyle\leq\sum_{i=1}^{2}CK_{i-1}^{-2}K_{i}^{\frac{\alpha-1}{q}-\frac{3}{p}}\sup_{\theta\in[-1,1]}Q(\phi^{\prime\prime}(\theta)K_{i}^{2}\lambda)+CK_{2}^{-C}\lambda^{\gamma}.$

Since $|\phi^{\prime\prime}(\theta)-1|\leq\epsilon_{\circ}$, note that
$(\phi^{\prime\prime}(\theta)K_{i}^{2}\lambda)^{-\gamma}Q(\phi^{\prime\prime}(\theta)K_{i}^{2}\lambda)\leq\mathfrak{Q}(\lambda^{\prime})$
if $\lambda^{\prime}\geq 2^{-1}K_{i}^{2}\lambda\geq 1$. Otherwise, i.e., if
$1\leq\lambda\leq 2K_{i}^{-2}$, we have
$Q(\phi^{\prime\prime}(\theta)K_{i}^{2}\lambda)\leq CK_{i}^{-C}$ for some
$C>0$. Thus, we have
$(\phi^{\prime\prime}(\theta)K_{i}^{2}\lambda)^{-\gamma}Q(\phi^{\prime\prime}(\theta)K_{i}^{2}\lambda)\leq\mathfrak{Q}(\lambda^{\prime})+CK_{i}^{-C}$.

We now multiply $\lambda^{-\gamma}$ to both sides of (2.27). Then, using the
above observation we obtain

$\displaystyle\lambda^{-\gamma}Q(\lambda)\leq\sum_{i=1}^{2}CK_{i-1}^{-2}K_{i}^{\frac{\alpha-1}{q}-\frac{3}{p}+2\gamma}\mathfrak{Q}(\lambda^{\prime})+CK_{2}^{-C}$

for $\lambda\leq\lambda^{\prime}$. Taking supremum over
$\lambda\leq\lambda^{\prime}$, we get

$\mathfrak{Q}(\lambda^{\prime})\leq\sum_{i=1}^{2}CK_{i-1}^{-2}K_{i}^{\frac{\alpha-1}{q}-\frac{3}{p}+2\gamma}\mathfrak{Q}(\lambda^{\prime})+CK_{2}^{-C}.$

Since (2.25) holds, we can successively choose dyadic numbers $K_{1},K_{2}$
such that

$CK_{i-1}^{-2}K_{i}^{\frac{\alpha-1}{q}-\frac{3}{p}+2\gamma}\leq\,3^{-1},\quad
i=1,2.$

Therefore, we obtain
$\mathfrak{Q}(\lambda^{\prime})\leq\frac{2}{3}\mathfrak{Q}(\lambda^{\prime})+CK_{2}^{-C}$,
which clearly gives $\mathfrak{Q}(\lambda^{\prime})\leq CK_{2}^{-C}$ for
$\lambda^{\prime}\geq 1$ and $p,q$ satisfying (2.20) and
$\frac{\alpha-1}{q}-\frac{3}{p}+1>0$, i.e., (1.5). This completes the proof. ∎

###  2.4. Weighted local smoothing estimate with sharp regularity

In this section, modifying the proof of Proposition 2.1, we show (2.4) with
$\gamma>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$, which implies Theorem 1.6.
We also discuss sharpness of the exponent $\gamma$.

######  Proposition 2.12.

Let $0<\alpha\leq 3$ and $\omega\in\Omega(\alpha)$. If
$\frac{1}{q}\leq\min\big{(}\frac{1}{p},~{}\frac{1}{3p}+\frac{1}{6},~{}\frac{2}{3p^{\prime}}\big{)}$
and $\frac{1}{p}+\frac{\kappa(\alpha)}{q}\leq 1$, then (2.4) holds for
$\gamma>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$.

###### Proof.

The proof is essentially identical to that of Proposition 2.1, so we shall be
brief. The main difference is that we use Proposition 2.5 and Proposition 2.8
instead of Corollary 2.9. For a fixed
$\gamma>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$, we need to show
$\mathfrak{Q}(\lambda^{\prime})\leq C$.

As in the proof of Proposition 2.1, we use the decomposition (2.22). Since
$\sum_{J^{i}}\|g_{J^{i}}\|_{p}^{q}\lesssim\|g\|_{p}^{q}$, by Lemma 2.3 we have

$\displaystyle\big{\|}\max_{J^{i}\in\mathfrak{J}^{i}}|T_{\lambda}g_{J^{i}}|\big{\|}_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}\leq
CK_{i}^{\frac{2\alpha-\kappa(\alpha)}{q}-\frac{3}{p}}Q(K_{i}^{2}\lambda)\|g\|_{p}.$

By Proposition 2.5 and Proposition 2.8 we also have

$\displaystyle\Big{\|}\max_{(J^{2}_{1},J^{2}_{2},J^{2}_{3})\in\mathfrak{J}^{2}(K_{2})}\prod_{i=1}^{3}\big{|}T_{\lambda}g_{J^{2}_{i}}\big{|}^{\frac{1}{3}}\Big{\|}_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}\leq CK_{2}^{-C}\lambda^{\gamma}\|g\|_{p}$

for $\gamma>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$ if
$\frac{1}{q}\leq\min(\frac{1}{p},\frac{1}{3p}+\frac{1}{6},\frac{2}{3p^{\prime}})$
and $1\leq p\leq\infty$.

By following the same lines of argument as in the proof of Proposition 2.1 we
obtain

$\mathfrak{Q}(\lambda^{\prime})\leq
C\sum_{i=1}^{2}K_{i-1}^{-2}K_{i}^{\frac{2\alpha-\kappa(\alpha)}{q}-\frac{3}{p}+2\gamma}\mathfrak{Q}(\lambda^{\prime})+CK_{2}^{-C}$

for $\lambda^{\prime}\geq 1$. We take $\gamma$ sufficiently close to
$\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$ such that
$\frac{2\alpha-\kappa(\alpha)}{q}-\frac{3}{p}+2\gamma>1-\frac{1}{p}-\frac{\kappa(\alpha)}{q}\geq
0$. Then, we can choose $K_{1},K_{2}$ such that
$CK_{i-1}^{-2}K_{i}^{\frac{2\alpha-\kappa(\alpha)}{q}-\frac{3}{p}+2\gamma}\leq\frac{1}{3}$
for $i=1,2$. Therefore, $\mathfrak{Q}(\lambda^{\prime})\leq C$ holds provided
that
$\frac{1}{q}\leq\min(\frac{1}{p},\frac{1}{3p}+\frac{1}{6},\frac{2}{3p^{\prime}})$
and $1-\frac{1}{p}-\frac{\kappa(\alpha)}{q}\geq 0$, as desired. ∎

We discuss sharpness of the regularity exponent in Theorem 1.6. We show (1.7)
holds only if

(2.28) $\displaystyle\gamma$
$\displaystyle\geq\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q},$ (2.29)
$\displaystyle\gamma$
$\displaystyle\geq\frac{3}{2p}+\frac{\kappa(\alpha)-2\alpha}{2q}.$

Note the two lower bounds on $\gamma$ coincide if
$\frac{1}{p}+\frac{\kappa(\alpha)}{q}=1$. So, (1.7) holds for
$\gamma>\frac{1}{2}+\frac{1}{p}-\frac{\alpha}{q}$ only if
$\frac{1}{p}+\frac{\kappa(\alpha)}{q}\leq 1$.

###### Proof of (2.28).

For $\lambda\gg 1$, we consider $f$ given by
$\widehat{f}(\xi)=e^{-i|\xi|}\zeta(\lambda^{-1}\xi)$ for a Schwartz function
$\zeta$ supported on $1\leq|\xi|\leq 2$. Using the polar coordinates (see
(3.6)), we have

$|f(x)|\lesssim\lambda^{\frac{3}{2}}(1+\lambda||x|-1|)^{-N}$

for any $N\geq 1$. Since ${\rm supp}\,\widehat{f}$ is supported in
$\\{|\xi|\sim\lambda\\}$, we see
$\|f\|_{L_{\gamma}^{p}}\lesssim\lambda^{\gamma}\lambda^{\frac{3}{2}-\frac{1}{p}}.$
Note that $|e^{it\sqrt{-\Delta}}f(x)|\gtrsim\lambda^{2}$ if $|x|,|t-1|\leq
c\lambda^{-1}$ for a sufficiently small $c>0$. If we take

$d\nu(z)=\chi_{\mathbb{B}^{3}(0,3)}(z)|z|^{\alpha-3}dz,\quad
z=(x,t)\in\mathbb{R}^{3},$

then it is easy to see $\langle\nu\rangle_{\alpha}\leq C$ for some $C$. So,
(1.7) gives
$\lambda^{2-\frac{\alpha}{q}}\lesssim\lambda^{\gamma}\lambda^{\frac{3}{2}-\frac{1}{p}}$.
Taking $\lambda\to\infty$, we get (2.28). ∎

###### Proof of (2.29).

We use a Knapp type example. Let us take $f$ such that
$\widehat{f}(\xi)=\lambda^{-\frac{3}{2}}\zeta(\lambda^{-1}\xi_{1},\lambda^{-\frac{1}{2}}\xi_{2})$.
Then $|e^{it\sqrt{-\Delta}}f(x)|\gtrsim 1$ for $(x,t)\in R_{\lambda}$, where

$R_{\lambda}:=\\{(x,t):|x_{1}-t|\leq c\lambda^{-1},~{}|x_{2}|\leq
c\lambda^{-\frac{1}{2}},~{}t\in I\\}$

for a sufficiently small $c>0$. We consider

$d\nu(x,t)=\lambda^{\frac{3-2\alpha+\kappa(\alpha)}{2}}\chi_{R_{\lambda}}(x,t)\,dxdt.$

It is easy to show $\langle\nu\rangle_{\alpha}\lesssim 1$. Indeed, if
$0<r\leq\lambda^{-1}$, we have
$\nu(\mathbb{B}^{3}(z,r))\lesssim\lambda^{\frac{3-2\alpha+\kappa(\alpha)}{2}}r^{3}\leq
r^{\alpha}$ since $r^{\frac{3}{2}}\leq r^{\frac{\kappa(\alpha)}{2}}$. One can
show $\nu(\mathbb{B}^{3}(z,r))\lesssim r^{\alpha}$ for the cases
$\lambda^{-1}<r\leq\lambda^{-\frac{1}{2}}$ and $\lambda^{-\frac{1}{2}}<r\leq
1$ similarly. Thus, the inequality (1.7) gives
$\lambda^{\frac{-2\alpha+\kappa(\alpha)}{2q}}\lesssim\lambda^{\gamma}\lambda^{-\frac{3}{2p}}$,
which implies (2.29). ∎

##  3\. Proofs of Theorem 1.3, Theorem 1.5, and Theorem 1.6

In this section, we first prove Theorem 1.5 and Theorem 1.6, which we deduce
from Proposition 2.1 and Proposition 2.12, respectively. Combining Littlewood-
Paley decomposition and Theorem 1.5, we prove Theorem 1.3, and then Corollary
1.4. We also discuss results for the spherical average in higher dimensions.

###  3.1. Proof of Theorem 1.5 and Theorem 1.6

By Littlewood-Paley decomposition, it suffices to consider the operator
$e^{it\sqrt{-\Delta}}P_{\lambda}f$ with $\lambda\geq 1$. We first observe that
(2.2) is a simple consequence of the estimate (2.4) via a change of variable.
Let $U(x,t)=(x_{1}-t,x_{2},t)$ where $x=(x_{1},x_{2})$. Setting
$\xi=(\eta,\rho)$, we write
$|\xi|-\eta=\eta\sqrt{1+\rho^{2}/\eta^{2}}-\eta=\eta\psi_{{0}}(\rho/\eta)$.
Then,

$U(x,t)\cdot(\xi,|\xi|)=(x,t)\cdot(\eta,\rho,\eta\psi_{{0}}(\rho/\eta)).$

By decomposition into finite angular sectors and rotation, we may assume
$\widehat{f}$ is supported in a small conic neighborhood of $e_{1}$. Changing
variables $(x,t)\to U(x,t)$, we have

(3.1) $\|e^{it\sqrt{-\Delta}}P_{\lambda}f\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega)}\leq C\|T_{\lambda}^{\psi_{{0}}}f\|_{L^{q}(\mathbb{R}^{2}\times
I;\omega^{\prime})},$

where $\omega^{\prime}(x,t):=\omega(U(x,t))$ and $\widehat{f}$ is supported on
$4^{-1}\lambda\leq\eta\leq 4\lambda$ and $|\rho/\eta|\leq 1$. Since $U$ is an
invertible linear map, it is clear that
$C^{-1}\omega^{\prime}\in\Omega(\alpha)$ for some $C>0$. Therefore, (2.2)
holds for the same $\gamma$ and $p,q$ as in Proposition 2.1.

The following lemma shows that estimates relative to $\alpha$-dimensional
measures can be deduced from weighted estimates. See [17, Section 3] for
example.

######  Lemma 3.1.

Let $0<\alpha\leq 3$ and $\nu\in\mathfrak{C}^{3}(\alpha)$. Set
$\varphi_{\lambda}=\lambda^{3}\varphi(\lambda\cdot)$ for a Schwartz function
$\varphi$ such that $\widehat{\varphi}=1$ on $\mathbb{B}^{3}(0,4)$ and ${\rm
supp}\,\widehat{\varphi}\subset\mathbb{B}^{3}(0,8)$. There is a constant
$C_{1}=C_{1}(\nu)>0$ such that
$C_{1}^{-1}|\varphi_{\lambda}|\ast\nu\in\Omega(\alpha)$.

###### Proof of Lemma 3.1.

Let us define $\nu_{\lambda}:=|\varphi_{\lambda}|\ast\nu$. By rapid decay of
$\varphi$, we have

$\nu_{\lambda}(z)\leq C_{N}\lambda^{3}\sum_{\ell\geq
0}2^{-N\ell}\nu\big{(}\mathbb{B}^{3}(z,2^{\ell}\lambda^{-1})\big{)}$

for any $N>0$. It suffices to show
$\int_{\mathbb{B}^{3}(z_{\circ},r)}\nu_{\lambda}(y)dy\leq C_{1}r^{\alpha}$ for
a constant $C_{1}$.

Let us consider two cases $r\leq\lambda^{-1}$ and $r>\lambda^{-1}$,
separately. If $r\leq\lambda^{-1}$, using $\|\nu_{\lambda}\|_{\infty}\leq
C\langle\nu\rangle_{\alpha}\lambda^{3-\alpha}$, we see
$\int_{\mathbb{B}^{3}(z_{\circ},r)}\nu_{\lambda}(z)dz\lesssim\langle\nu\rangle_{\alpha}\lambda^{3-\alpha}r^{3}\leq
C_{N}^{\prime}\langle\nu\rangle_{\alpha}r^{\alpha}$. To handle the case
$r>\lambda^{-1}$, we use

$\int_{\mathbb{B}^{3}(z_{\circ},r)}\nu\big{(}\mathbb{B}^{3}(z,2^{\ell}\lambda^{-1})\big{)}dz\leq\int_{\mathbb{B}^{3}(z_{\circ},(2^{\ell}+1)r)}\int\chi_{\mathbb{B}^{3}(z^{\prime},2^{\ell}\lambda^{-1})}(z)dzd\nu(z^{\prime}).$

Taking $N$ sufficiently large, we get

$\displaystyle\int_{\mathbb{B}^{3}(z_{\circ},r)}\nu_{\lambda}(z)dz\leq
C_{N}\lambda^{3}\sum_{\ell\geq
0}2^{-N\ell}(2^{\ell}\lambda^{-1})^{3}(2^{\ell}+1)^{\alpha}r^{\alpha}\leq
C_{N}^{\prime\prime}r^{\alpha}.$

Therefore, taking $C_{1}=\max\\{C_{N}^{\prime},C_{N}^{\prime\prime}\\}$, we
have $C_{1}^{-1}\nu_{\lambda}\in\Omega(\alpha)$. ∎

We are ready to prove Theorem 1.5 and Theorem 1.6.

###### Proof of Theorem 1.5 and Theorem 1.6.

By Littlewood-Paley decomposition, it is enough to show

(3.2) $\|e^{it\sqrt{-\Delta}}P_{\lambda}f\|_{L^{q}(\mathbb{R}^{2}\times
I;\nu)}\leq C\lambda^{\gamma}\|f\|_{p},\quad\lambda\geq 1.$

Let $\varphi$ be the Schwartz function given in Lemma 3.1. Since
$\mathcal{F}_{x,t}(e^{it\sqrt{-\Delta}}P_{\lambda}f)=\mathcal{F}_{x,t}(e^{it\sqrt{-\Delta}}P_{\lambda}f)\widehat{\varphi_{\lambda}}$,
we have
$|e^{it\sqrt{-\Delta}}P_{\lambda}f|^{q}\lesssim|e^{it\sqrt{-\Delta}}P_{\lambda}f|^{q}\ast|\varphi_{\lambda}|$
by Hölder’s inequality. Thus it follows that

(3.3) $\|e^{it\sqrt{-\Delta}}P_{\lambda}f\|_{L^{q}(\mathbb{R}^{2}\times
I;\nu)}\leq C\|e^{it\sqrt{-\Delta}}P_{\lambda}f\|_{L^{q}(\mathbb{R}^{2}\times
I;|\varphi_{\lambda}|\ast\nu)}.$

By Lemma 3.1, $C^{-1}|\varphi_{\lambda}|\ast\nu\in\Omega(\alpha)$ for some
$C>0$. Since we are assuming $\widehat{f}$ is supported in a small conic
neighborhood of $e_{1}$, using (3.3) and (3.1), we see that the estimate (2.4)
implies (3.2) for the same $p,q$ and $\gamma$ as in Proposition 2.1 and
Proposition 2.12. Therefore, Proposition 2.1 and Proposition 2.12 give Theorem
1.5 and Theorem 1.6, respectively. ∎

Next, we show Theorem 1.3 and Corollary 1.4.

###  3.2. Estimates for the circular average

Recall the average $\mathcal{A}$ in $\mathbb{R}^{d}$, $d\geq 2$. Let us set
$P_{0}=\sum_{j\leq 0}P_{2^{j}}$. The kernel $\mathcal{K}(\cdot,t)$ of
$f\to\mathcal{A}P_{0}f(\cdot,t)$ is rapidly decaying, i.e.,
$|\mathcal{K}(\cdot,t)|\leq C(1+|x|)^{-N}$ for any $N$. By Schur’s test, one
can easily see $\|\mathcal{A}P_{0}f\|_{L^{q}(\mathbb{R}^{2}\times I;d\nu)}\leq
C\|f\|_{p}$ for $1\leq p\leq q$. Therefore, it is sufficient to consider the
contribution from $\sum_{j\geq 1}\mathcal{A}P_{2^{j}}f$.

Let $d\sigma_{t}$ denote the normalized spherical measure. Then it is well
known that

(3.4)
$\widehat{d\sigma_{t}}(\xi)=e^{it|\xi|}a_{+}(t\xi)+e^{-it|\xi|}a_{-}(t\xi),$

where $a_{\pm}$ are smooth functions satisfying
$|\partial_{\xi}^{m}a_{\pm}(\xi)|\leq C_{m}(1+|\xi|)^{-\frac{d-1}{2}-|m|}$ for
any $m$. So, it is sufficient to consider $e^{it|\xi|}a_{+}(t\xi)$ since the
contribution from $e^{-it|\xi|}a_{-}(t\xi)$ can be handled similarly by
reflection $t\to-t$. We set

$\mathfrak{A}f(x,t)=\int
e^{i(x\cdot\xi+t|\xi|)}(1+|\xi|^{2})^{-\frac{d-1}{4}}\widehat{f}(\xi)d\xi$

and

$\mathfrak{A}_{j}f=\mathfrak{A}P_{2^{j}}f.$

Since $t\in I$, to prove the estimate (1.4) it suffices to show $\|\sum_{j\geq
1}\mathfrak{A}_{j}f\|_{L^{q}(\mathbb{R}^{d}\times
I;d\nu)}\lesssim\langle\nu\rangle_{\alpha}^{1/q}\|f\|_{p}.$ Thus, the matter
is reduced to showing

(3.5) $\big{\|}\mathfrak{A}_{j}f\big{\|}_{L^{q}(\mathbb{R}^{d}\times
I;d\nu)}\lesssim\langle\nu\rangle_{\alpha}^{\frac{1}{q}}2^{-\epsilon
j}\|f\|_{p},\quad j\geq 1,$

for some $\epsilon>0$.

Let $\mathbf{K}_{j}(\cdot,t)$ denote the kernel of
$f\to\mathfrak{A}_{j}f(\cdot,t)$. It is easy to see

(3.6) $\displaystyle|\mathbf{K}_{j}(x,t)|\lesssim
2^{j}\big{(}1+2^{j}\big{|}|x|-|t|\big{|}\big{)}^{-N}$

for any $N\geq 1$ (see, for example, [14]). Using the estimate we obtain

(3.7) $\displaystyle\|\mathfrak{A}_{j}f\|_{L^{\infty}(\mathbb{R}^{d}\times
I;d\nu)}$ $\displaystyle\lesssim 2^{j}\|f\|_{1},$ (3.8)
$\displaystyle\|\mathfrak{A}_{j}f\|_{L^{1}(\mathbb{R}^{d}\times I;d\nu)}$
$\displaystyle\lesssim\langle\nu\rangle_{\alpha}\min\big{\\{}2^{j},2^{(d+1-\alpha)j}\big{\\}}\|f\|_{1}.$

In fact, (3.7) is clear from (3.6). To show (3.8), we use the fact that
$\mathbf{K}_{j}$ is essentially supported in $O(2^{-j})$ neighborhood of the
truncated cone with height $\sim 1$. Since the neighborhood can be covered by
$O(2^{dj})$ balls of radius $2^{-j}$, using (2.1) we see
$\|\mathbf{K}_{j}\|_{L^{1}(d\nu)}\lesssim\langle\nu\rangle_{\alpha}2^{(d+1-\alpha)j}$.
Besides, we have $|\mathbf{K}_{j}|\lesssim 2^{j}(1+|x|)^{-N}$ since $t\in I$.
So,
$\|\mathbf{K}_{j}\|_{L^{1}(d\nu)}\lesssim\langle\nu\rangle_{\alpha}\min\\{2^{j},2^{(d+1-\alpha)j}\\}$.
We get (3.8) by Fubini’s theorem.

###### Proof of Theorem 1.3.

As discussed above, we need only to show (3.5) for some $\epsilon>0$ if $p,q$
satisfy (1.5). When $1<\alpha\leq 2$, the desired estimate (1.4) follows from
Theorem 1.5 since

(3.9)
$\mathfrak{A}_{j}f=2^{-\frac{j}{2}}e^{it\sqrt{-\Delta}}P_{2^{j}}\widetilde{f}$

for some $\widetilde{f}$ satisfying $\|\widetilde{f}\|_{p}\lesssim\|f\|_{p}$.

For $2<\alpha\leq 3$, we use (3.7), (3.8), and

(3.10) $\displaystyle\|\mathfrak{A}_{j}f\|_{L^{r}(\mathbb{R}^{2}\times
I;d\nu)}$
$\displaystyle\lesssim\langle\nu\rangle_{\alpha}^{\frac{1}{r}}2^{(\frac{3-\alpha}{r}-\frac{1}{2}+\epsilon)j}\|f\|_{r},\quad
2\leq r\leq 4,~{}\epsilon>0.$

To show (3.10), we use the estimate
$\|\mathfrak{A}_{j}f\|_{L^{r}(\mathbb{R}^{2}\times I)}\lesssim
2^{(-\frac{1}{2}+\epsilon)j}\|f\|_{L^{r}}$ for $\epsilon>0$ and $2\leq r\leq
4$, which follows from interpolation between the estimates for $r=2$ and
$r=4$. Indeed, the first is clear from Plancherel’s theorem and (3.9). The
second follows by the sharp local smoothing estimate (1.7) and (3.9) combined
with the globalization lemma (see Lemma 2.6). Then, modifying the argument in
the proof of Proposition 2.8, we get the estimate (3.10).

When $2<\alpha\leq 3$, $\mathcal{P}_{2}(\alpha)$ is the closed quadrangle with
vertices $(0,0)$,

$\quad\mathbf{P}_{1}:=\Big{(}\frac{\alpha}{\alpha+3},\frac{1}{\alpha+3}\Big{)},\quad\mathbf{P}_{2}:=\Big{(}\frac{\alpha-1}{\alpha},\frac{1}{\alpha}\Big{)},\quad\mathbf{P}_{3}:=\Big{(}\frac{1}{4-\alpha},\frac{1}{4-\alpha}\Big{)}.$

Interpolating the estimates (3.8) and (3.10) with $r=2$, we have

$\|\mathfrak{A}_{j}f\|_{L^{q}(\mathbb{R}^{2}\times
I;d\nu)}\lesssim\langle\nu\rangle_{\alpha}^{\frac{1}{q}}2^{(\frac{4-\alpha}{q}-1)j}\|f\|_{q}$

for $1\leq q\leq 2$. Since
$\|\mathfrak{A}_{j}f\|_{L^{\infty}(\mathbb{R}^{2}\times I;d\nu)}\leq
C\|f\|_{\infty}$, this gives (3.5) for $4-\alpha<{p=q}<\infty$. Therefore, to
complete the proof, by interpolation it is enough to show (3.5) for
$(1/p,1/q)$ arbitrarily close to $\mathbf{P}_{1}$ and $\mathbf{P}_{2}$.
Interpolating (3.7) and (3.10) yields

$\|\mathfrak{A}_{j}f\|_{L^{q}(\mathbb{R}^{2}\times
I;d\nu)}\lesssim\langle\nu\rangle_{\alpha}^{\frac{1}{q}}2^{(\frac{3}{2p}+\frac{3-2\alpha}{2q}-\frac{1}{2}+\epsilon)j}\|f\|_{p}$

for $\epsilon>0$ if $1-\frac{3}{q}\leq\frac{1}{p}\leq 1-\frac{1}{q}$ and
$p\leq q$. A simple computation shows that this gives (3.5) for $(1/p,1/q)$
contained in the closed quadrangle with vertices
$(1/4,1/4),\mathbf{P}_{1},\mathbf{P}_{2},$ and $(1/2,1/2)$ excluding the
closed line segment $[\mathbf{P}_{1},\mathbf{P}_{2}]$. ∎

Corollary 1.4 is a direct consequence of Theorem 1.3 and the following lemma
combined with the globalization argument (e.g., Lemma 2.6). For $0<k\leq d+1$,
let $V$ be a $k$-dimensional subspace in $\mathbb{R}^{d+1}$. We write
$(x,t)=(z_{1},z_{2})$ where $z_{1}\in V$, $z_{2}\in V^{\perp}$.

######  Lemma 3.2.

Let $V$ be a $k$-dimensional subspace in $\mathbb{R}^{d+1}$ for
$k=1,\dots,d+1$. Let $\mu$ be an $\alpha$-dimensional measure on $V$ for some
$0<\alpha\leq k$. If (1.4) holds with a uniform constant $C$, independent of
$\nu\in\mathfrak{C}^{d+1}(\alpha)$, then

$\big{\|}\sup_{z_{2}\in V^{\perp}}|\chi_{\mathbb{B}^{d}(0,1)\times
I}\mathcal{A}f(\cdot,z_{2})|\big{\|}_{L^{q}(V,d\mu)}\leq
C\langle\mu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{p}.$

By Lemma 3.2 and the globalization argument, $L^{p}$–$L^{q}(d\mu)$ estimate
for $f\to\sup_{t\in I}|\mathcal{A}f(\cdot,t)|$ can be deduced from (1.4) by
taking $V=\mathbb{R}^{d}\times\\{0\\}$ and $\nu\in\mathfrak{C}^{d+1}(d)$.

###### Proof.

By the Kolmogorov-Seliverstov-Plessner linearization it suffices to show

(3.11) $\|\chi_{\mathbb{B}^{d}(0,1)\times
I}\mathcal{A}f(\cdot\,,\mathbf{z}_{2}(\cdot))\|_{L^{q}(d\mu)}\leq
C\langle\mu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{p}$

for any measurable function $\mathbf{z}_{2}:V\to V^{\perp}$ with $C$
independent of $\mathbf{z}_{2}$. For any compactly supported continuous
function $F$, we define a linear functional $\ell$ on
$\mathrm{C}_{c}(\mathbb{R}^{d+1})$ by $\ell(F)=\int
F(z_{1},\mathbf{z}_{2}(z_{1}))d\mu(z_{1})$. Then, by the Riesz representation
theorem, there is a unique Radon measure $\nu$ on $\mathbb{R}^{d+1}$ such that

$\ell(F)=\int F(x,t)d\nu(x,t)=\int F(z_{1},\mathbf{z}_{2}(z_{1}))d\mu(z_{1}).$

Since
$\nu\big{(}\mathbb{B}^{d+1}((y_{1},y_{2}),r)\big{)}\leq\int\chi_{\mathbb{B}^{k}(y_{1},r)}(z_{1})\,d\mu(z_{1})\leq\langle\mu\rangle_{\alpha}r^{\alpha}$,
the measure $\nu$ belongs to $\mathfrak{C}^{d+1}(\alpha)$ and
$\langle\nu\rangle_{\alpha}\leq\langle\mu\rangle_{\alpha}$. Therefore,
$\|\chi_{\mathbb{B}^{d}(0,1)\times
I}\mathcal{A}f(\cdot\,,\mathbf{z}_{2}(\cdot))\|_{L^{q}(d\mu)}=\|\chi_{\mathbb{B}^{d}(0,1)\times
I}\mathcal{A}f\|_{L^{q}(d\nu)}$, and (3.11) follows by (1.4). ∎

###  3.3. Estimate for the spherical average

We close this section with results in higher dimensions. As mentioned in
Section 1, we can obtain some estimates for the spherical average from the
fractal Strichartz estimates for the wave equation in $\mathbb{R}^{d+1}$.

######  Theorem 3.3 ([7, Theorem 3.1]).

Let $d\geq 3$, $2\leq q\leq\infty$, and $1<\alpha\leq d+1$. Suppose
$\nu\in\mathfrak{C}^{d+1}(\alpha)$. Then, we have

(3.12) $\|e^{it\sqrt{-\Delta}}f\|_{L^{q}(\mathbb{R}^{d}\times
I;d\nu)}\lesssim\langle\nu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{H^{s}}$

holds with $s>s(\alpha,q,d)$, where

$s(\alpha,q,d)=\begin{cases}\,\max\big{\\{}\frac{d}{2}-\frac{\alpha}{q},~{}\frac{d+1}{4}+\frac{1-\alpha}{2q},~{}\frac{3d+1-2\alpha}{8}\big{\\}},~{}\hfill\quad
1<\alpha\leq d,\phantom{111}\\\\[4.30554pt]
\,\max\big{\\{}\frac{d}{2}-\frac{\alpha}{q},~{}\frac{d+1}{4}+\frac{d+1-2\alpha}{2q},~{}\frac{d+1-\alpha}{2}\big{\\}},~{}\hfill\quad
d<\alpha\leq d+1.\end{cases}$

In fact, the estimate in [7, Theorem 3.1] is a local one. However, it can be
easily extended to a global estimate (3.12) by the argument in Lemma 2.6. The
regularity exponents in Theorem 3.3 are sharp when $d=3$ or
$\alpha\in[1,\frac{d-1}{2}]\cup[d,d+1]$ in higher dimensions.

Using (3.7) and (3.8), we obtain an analogue of Theorem 1.3 for the spherical
averages. To state the result, let us set

${p_{d}}(\alpha)=\begin{cases}\,\frac{\alpha+1}{\alpha},~{}\hfill\quad
1<\alpha\leq\frac{d-1}{2},\\\\[3.44444pt]
\,\frac{d+3+2\alpha}{d-1+2\alpha},~{}\hfill\quad\frac{d-1}{2}<\alpha\leq
d.\end{cases}$

######  Theorem 3.4.

Let $d\geq 3$ and $1<\alpha\leq d+1$. Suppose
$\nu\in\mathfrak{C}^{d+1}(\alpha)$.

  1. ($i$)

If $1<\alpha\leq d$, then (1.4) holds for $p,q$ satisfying $p>p_{d}(\alpha)$,

(3.13) $\displaystyle q^{-1}$ $\displaystyle\leq p^{-1}<\alpha q^{-1},$
$\displaystyle(d+1)\,{p}^{-1}$ $\displaystyle<d-1+(\alpha-1)\,{q}^{-1}.$

  2. ($ii$)

If $d<\alpha\leq d+1$, then (1.4) holds for $p,q$ satisfying

$\displaystyle q^{-1}$ $\displaystyle\leq p^{-1}<\alpha q^{-1},$
$\displaystyle(d+1)p^{-1}$ $\displaystyle<d-1+(2\alpha-d-1)q^{-1},$
$\displaystyle dp^{-1}$ $\displaystyle<d-1+(\alpha-d)q^{-1}.$

When $d<\alpha\leq d+1$, the results are sharp (see Proposition 4.1). We also
expect the result to be sharp when $d=3$, but the necessity of the condition
$p>p_{3}(\alpha)$ for $1<\alpha\leq 3$ is verified only for $\alpha=2,3$ (see
4 and 5 in Proposition 4.1). Using the recent result due to Harris [10], which
improves the exponent $s$ in (3.12) for $q=2$ and $\frac{d+1}{2}<\alpha<d$,
$d\geq 4$, one can obtain a slightly better result.

###### Proof of Theorem 3.4.

As in the proof of Theorem 1.3, it suffices to show (3.5) for some
$\epsilon>0$. Note
$\mathfrak{A}_{j}f=2^{-\frac{d-1}{2}j}e^{it\sqrt{-\Delta}}P_{2^{j}}\widetilde{f}$
for some $\widetilde{f}$ satisfying $\|\widetilde{f}\|_{p}\lesssim\|f\|_{p}$
(cf. (3.9)). By (3.12), we have

(3.14) $\|\mathfrak{A}_{j}f\|_{L^{q}(\mathbb{R}^{d}\times
I;d\nu)}\lesssim\langle\nu\rangle_{\alpha}^{\frac{1}{q}}2^{(s-\frac{d-1}{2})j}\|f\|_{2},\quad
q\geq 2$

for $s>s(\alpha,q,d)$.

We first consider the case $1<\alpha\leq d$. We interpolate (3.14) and (3.7),
and then (3.14) and the trivial estimate
$\|\mathfrak{A}_{j}f\|_{L^{\infty}(\mathbb{R}^{d}\times
I;d\nu)}\lesssim\|f\|_{\infty}$. Consequently, for $p,q$ satisfying
$\,{q}^{-1}\leq\,{p}^{-1}\leq 1-\,{q}^{-1}$ we have

(3.15) $\|\mathfrak{A}_{j}f\|_{L^{q}(\mathbb{R}^{d}\times
I;d\nu)}\lesssim\langle\nu\rangle_{\alpha}^{\frac{1}{q}}2^{\tilde{s}j}\|f\|_{p}$

if

$\tilde{s}>\max\Big{\\{}\frac{1}{p}-\frac{\alpha}{q},~{}\,\frac{d+1}{2p}+\frac{1-\alpha}{2q}-\frac{d-1}{2},~{}\,\frac{d+3+2\alpha}{4p}-\frac{d-1+2\alpha}{4}\Big{\\}}.$

Therefore, (3.5) holds for some $\epsilon>0$ if $\tilde{s}<0$ i.e., $p,q$
satisfy (3.13), $p>(d+3+2\alpha)/(d-1+2\alpha)$, and $\,{p}^{-1}\leq
1-\,{q}^{-1}$.

On the other hand, if $p,q$ satisfy
$1-\,{q}^{-1}\leq\,{p}^{-1}\leq\,{q}^{-1}$, then we have (3.15) for

$\tilde{s}>\max\Big{\\{}\frac{\alpha+1}{p}-\alpha,~{}\,\frac{d+3+2\alpha}{4p}-\frac{d-1+2\alpha}{4}\Big{\\}},$

which is a consequence of interpolation between (3.7), (3.8), and (3.14) with
$q=2$ and
$s>s(\alpha,2,d)=\max\big{\\{}\frac{d-\alpha}{2},~{}\frac{3d+1-2\alpha}{8}\big{\\}}$.
So, (3.5) holds for $p,q$ satisfying
$1-\,{q}^{-1}\leq\,{p}^{-1}\leq\,{q}^{-1}$ and $p>p_{d}(\alpha)$. Combining
the results for the cases $\,{q}^{-1}\leq\,{p}^{-1}\leq 1-\,{q}^{-1}$ and
$1-\,{q}^{-1}\leq\,{p}^{-1}\leq\,{q}^{-1}$, we have (3.5) for $p,q$ satisfying
(3.13) and $p>p_{d}(\alpha)$. This proves the first part of Theorem 3.4 (when
$1<\alpha\leq d$).

We next consider the case $d<\alpha\leq d+1$. We need to show (3.5) holds for
$(1/p,1/q)$ contained in the union of the open quadrangle with vertices
$(0,0)$,

$\mathbf{Q}_{1}:=\Big{(}\frac{(d-1)\alpha}{(d-1)\alpha+(d+1)},\frac{d-1}{(d-1)\alpha+(d+1)}\Big{)},~{}\
\mathbf{Q}_{2}:=\Big{(}\frac{\alpha-1}{\alpha},\frac{1}{\alpha}\Big{)},~{}\
\mathbf{Q}_{3}:=\Big{(}\frac{d-1}{2d-\alpha},\frac{d-1}{2d-\alpha}\Big{)},$

and the line segment $[(0,0),\mathbf{Q}_{3})$. Interpolating (3.14), (3.7),
and the trivial $L^{\infty}$–$L^{\infty}(\mathbb{R}^{d}\times I;d\nu)$
estimate, we obtain (3.15) for $q^{-1}\leq p^{-1}\leq 1-q^{-1}$ if

$\tilde{s}>\max\Big{\\{}\frac{1}{p}-\frac{\alpha}{q},~{}\,\frac{d+1}{2p}+\frac{d+1-2\alpha}{2q}-\frac{d-1}{2},~{}\,\frac{\alpha}{p}-(\alpha-1)\Big{\\}}.$

Thus, (3.5) holds for some $\epsilon>0$ if $\tilde{s}<0$ i.e.,
$(\,{p}^{-1},\,{q}^{-1})$ is contained in the interior of the quadrangle with
vertices $(0,0)$, $\mathbf{Q}_{1}$, $\mathbf{Q}_{2}$, and $(1/2,1/2)$.

Finally, by interpolation we need only to obtain (3.5) for
$(1/p,1/q)\in[(1/2,1/2),\mathbf{Q}_{3})$. To see this, we interpolate two
estimates (3.8) and (3.14) with $q=2$ and $s>(d+1-\alpha)/2$ to obtain (3.15)
for $1\leq p=q\leq 2$ and $\tilde{s}>\frac{2d-\alpha}{p}-(d-1)$. Thus, (3.5)
holds for $(\frac{1}{p},\frac{1}{q})\in[(1/2,1/2),\mathbf{Q}_{3})$, as
desired. ∎

Similarly as in Corollary 1.4, $L^{p}$-$L^{q}(d\mu)$ estimate for the
spherical maximal function holds for the range of $p,q$ as in Theorem 3.4.

######  Corollary 3.5.

Let $d\geq 3$, $1<\alpha\leq d$ and $\mu\in\mathfrak{C}^{d}(\alpha)$. For
$p,q$ satisfying $p>p_{d}(\alpha)$ and (3.13), we have

(3.16) $\displaystyle\big{\|}\sup_{t\in
I}|\mathcal{A}f(\cdot,t)|\big{\|}_{L^{q}(d\mu)}\leq
C\langle\mu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{L^{p}(\mathbb{R}^{d})}.$

##  4\. Necessary conditions on $p,q$ for (1.4)

In this section, we obtain necessary conditions on $p,q,\alpha$ for (1.4) to
hold. Obviously, it is enough to consider the local estimate

(4.1) $\|\mathcal{A}f\|_{L^{q}(\mathbb{B}^{d}(0,1)\times I;d\nu)}\leq
C\langle\nu\rangle_{\alpha}^{\frac{1}{q}}\|f\|_{L^{p}(\mathbb{R}^{d})}.$

This is done by taking specific $\alpha$ dimensional measures and particular
functions. A modification of these examples also gives necessary conditions
for the maximal estimate (3.16) (also see (1.6)) when $1<\alpha\leq d$.

######  Proposition 4.1.

Let $d\geq 2$ and $1<\alpha\leq d+1$. Suppose (4.1) holds for
$\nu\in\mathfrak{C}^{d+1}(\alpha)$. Then,

  1. ($i$)

$\qquad\quad\ \,{p}^{-1}\leq\alpha\,{q}^{-1}$,

  2. ($ii$)

$(d+1)\,{p}^{-1}\leq\begin{cases}\,d-1+(\alpha-1)\,{q}^{-1},~{}\hfill\quad
1<\alpha\leq d,\phantom{111}\\\ \,d-1+(2\alpha-d-1)\,{q}^{-1},~{}\hfill\quad
d<\alpha\leq d+1,\end{cases}$

  3. ($iii$)

$\qquad d\,{p}^{-1}\leq\begin{cases}\,d-1,~{}\hfill\quad 1<\alpha\leq
d,\phantom{111}\\\ \,d-1+(\alpha-d)\,{q}^{-1},~{}\hfill\quad d<\alpha\leq
d+1.\end{cases}$

Furthermore, if $d\geq 3$ and $k<\alpha\leq k+1$ for an integer $k\in[1,d-1]$,
then the following are necessary for (4.1) to hold:

  1. (4)

$\begin{aligned} \hphantom{g}\quad(d+k)p^{-1}&\leq d+k-2+(\alpha-k)q^{-1},\\\
p^{-1}&\leq\frac{d+k-1}{d+k+1},\end{aligned}$

  2. (5)

$\begin{aligned} \qquad(k+1)p^{-1}&\leq k+(\alpha-k)q^{-1},\\\
p^{-1}&\leq\frac{k+1}{k+2}.\end{aligned}$

When $d=2$ and $1<\alpha\leq 3$, (4.1) holds only if

  1. (6)

$p\geq 4-\alpha$.

It seems to be natural to expect that (4.1) holds for
$\nu\in\mathfrak{C}^{d+1}(\alpha)$ only if, instead of 4 and 5,

(4.2) $\displaystyle
p^{-1}\leq\begin{cases}\frac{d+\alpha-2}{d+\alpha},~{}&~{}\hfill
d-2<\alpha\leq d,\\\\[3.44444pt] \frac{\alpha}{\alpha+1},~{}&~{}\hfill
1<\alpha\leq d-2.\end{cases}$

When $\alpha=k+1$ ($k=1,\dots,d-1$), (4.2) matches with the second conditions
in 4 and 5. In view of the global estimate (1.4) with $d=2$, the condition 6
is redundant since (1.4) holds only if $p\leq q$. In fact, the conditions
($i$), ($ii$), ($iii$) combined with $p\leq q$ show that (1.4) with $d=2$
fails unless 6 holds.

### Proof of Proposition 4.1

Let us assume that (4.1) holds for all $\nu\in\mathfrak{C}^{d+1}(\alpha)$ with
$1<\alpha\leq d+1$. We show the necessity of the conditions ($i$)–6 by
considering specific measures contained in $\mathfrak{C}^{d+1}(\alpha)$ and
functions which show the failure of (4.1) unless the conditions hold.

In what follows, $0<\delta\ll 1$ and $c$ is a small positive constant.

###### Proof of ($i$).

Let $N_{\delta}=\\{x:||x|-1|<\delta\\}$ and set $f=\chi_{N_{\delta}}$.
Clearly, $\mathcal{A}f(x,t)\gtrsim 1$ if
$(x,t)\in\mathbb{B}^{d}(0,c\delta)\times[1,1+c\delta]$. Let
$d\nu(x,t)=\psi(x,t)|x|^{\alpha-d-1}dxdt$ for
$\psi\in\mathrm{C}_{0}^{\infty}(\mathbb{B}^{d+1}(0,3))$. It is easy to see
$\langle\nu\rangle_{\alpha}\lesssim 1$. Then (4.1) implies
$\delta^{\frac{\alpha}{q}}\lesssim\delta^{\frac{1}{p}}$. Thus, letting
$\delta\to 0$, we get $1/p\leq\alpha/q$. ∎

###### Proof of ($ii$).

Let us set

$V_{\delta}=\\{(x_{1},x^{\prime},t)\in\mathbb{R}\times\mathbb{R}^{d-1}\times[1,2]:|x_{1}-t|\leq
c\delta^{2},|x^{\prime}|\leq c\delta\\}.$

We also define a measure $\nu$ by

$d\nu(x,t)=\begin{cases}\,\delta^{\alpha-d-2}\chi_{V_{\delta}}\,dxdt,~{}\hfill\quad
1<\alpha\leq d,\phantom{111}\\\\[3.01385pt]
\,\delta^{2(\alpha-d-1)}\chi_{V_{\delta}}\,dxdt,~{}\hfill\quad d<\alpha\leq
d+1.\end{cases}$

One can easily show $\langle\nu\rangle_{\alpha}\lesssim 1$ for $1<\alpha\leq
d+1$. Indeed, when $1<\alpha\leq d$,
$\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{\alpha-d-2}r^{d+1}$ for
$0<r<\delta^{2}$, $\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{\alpha-d}r^{d}$
for $\delta^{2}<r<\delta$, and
$\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{\alpha-1}r$ for $\delta<r$. So we
have $\nu(\mathbb{B}^{d+1}(z,r))\lesssim r^{\alpha}$ in each case. The case
$d<\alpha\leq d+1$ can be handled similarly.

We consider $f=\chi_{U_{\delta}}$ for
$U_{\delta}=[0,\delta^{2}]\times[0,\delta]^{d-1}$. Then, we have
$\mathcal{A}f(x,t)\gtrsim\delta^{d-1}$ if $(x,t)\in V_{\delta}$. Note
$\nu(V_{\delta})\sim\delta^{\alpha-1}$ for $1<\alpha\leq d$, and
$\nu(V_{\delta})\sim\delta^{2\alpha-d-1}$ for $d<\alpha\leq d+1$. Since
$\|f\|_{p}=\delta^{\frac{d+1}{p}}$, (4.1) implies
$\delta^{d-1}\delta^{\frac{\alpha-1}{q}}\lesssim\delta^{\frac{d+1}{p}}$ for
$1<\alpha\leq d$, which gives ($ii$) for $1<\alpha\leq d$. When $d<\alpha\leq
d+1$, we get ($ii$) in the same manner. ∎

###### Proof of ($iii$).

Let us take $f=\chi_{\mathbb{B}^{d}(0,\delta)}$ so that
$\|f\|_{p}\lesssim\delta^{d/p}$. Then we have
$\mathcal{A}f\gtrsim\delta^{d-1}\chi_{S_{\delta}}$ where
$S_{\delta}=\\{(x,t):||x|-t|\leq c\delta,~{}t\in[1,2]\\}$ for a small $c>0$.
Let

$d\nu(x,t)=\begin{cases}\delta^{-1}\chi_{S_{\delta}}dxdt,~{}\hfill\quad
1<\alpha\leq d,\phantom{111}\\\\[3.01385pt]
\,\delta^{\alpha-d-1}\chi_{S_{\delta}}dxdt,~{}\hfill\quad d<\alpha\leq
d+1.\end{cases}$

It is not difficult to see $\langle\nu\rangle_{\alpha}\lesssim 1$. When
$d<\alpha\leq d+1$,
$\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{\alpha-d-1}r^{d+1}$ for
$0<r\leq\delta$, and
$\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{\alpha-d}r^{d}$ for $\delta<r$. The
case $1<\alpha\leq d$ can be treated similarly and we omit the detail. Since
$|S_{\delta}|\sim\delta$, we have $\nu(S_{\delta})\sim 1$ for $1<\alpha\leq
d$, and $\nu(S_{\delta})\sim\delta^{\alpha-d}$ for $d<\alpha\leq d+1$. So,
(4.1) implies $\delta^{d-1}\lesssim\delta^{d/p}$ for $1<\alpha\leq d$, and
$\delta^{d-1+(\alpha-d)/q}\lesssim\delta^{d/p}$ for $d<\alpha\leq d+1$.
Letting $\delta\to 0$ yields ($iii$). ∎

###### Proof of 4.

We modify the example in the proof of ($ii$). For a given $\alpha\in(1,d]$,
let $k\in[1,d-1]$ be an integer such that $k<\alpha\leq k+1$. Let
$U_{\delta}^{k}=[0,\delta^{2}]^{k}\times[0,\delta]^{d-k}$. We have

(4.3)
$\mathcal{A}\chi_{U_{\delta}^{k}}\gtrsim\delta^{2(k-1)+d-k}\chi_{V_{\delta}^{k}},$

where

$V_{\delta}^{k}:=\big{\\{}(x^{\prime},x^{\prime\prime},t)\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}\times[1,2]:\quad~{}||x^{\prime}|-t|\leq
c\delta^{2},\quad~{}|x^{\prime\prime}|\leq c\delta\big{\\}}.$

Let us set $d\nu=\delta^{\alpha-d-2}\chi_{V_{\delta}^{k}}dxdt$. In the same
manner as before it is easy to show $\langle\nu\rangle_{\alpha}\lesssim 1$.
Since $\nu(V_{\delta}^{k})\sim\delta^{\alpha-k}$ and
$\|\chi_{U_{\delta}^{k}}\|_{p}\lesssim\delta^{(d+k)/{p}}$, (4.1) and (4.3)
yield $\delta^{d+k-2}\delta^{(\alpha-k)/{q}}\lesssim\delta^{(d+k)/{p}}$. This
gives the first part of 4.

For the second part of 4, taking $f=\chi_{U_{\delta}^{k+1}}$ instead of
$\chi_{U_{\delta}^{k}}$, we have
$\mathcal{A}f\gtrsim\delta^{d+k-1}\chi_{V_{\delta}^{k+1}}$. We consider
$d\nu=\delta^{k-d-1}\chi_{V_{\delta}^{k+1}}dxdt$. Then, similarly as before,
one can easily see $\langle\nu\rangle_{\alpha}\lesssim 1$ and
$\nu(V_{\delta}^{k+1})\gtrsim 1$. Since
$\|\chi_{U_{\delta}^{k+1}}\|_{p}\lesssim\delta^{(d+k+1)/{p}}$, (4.1) implies
$\delta^{d+k-1}\lesssim\delta^{(d+k+1)/{p}}$. This shows the second part of 4.
∎

###### Proof of 5.

We show 5 by combining the examples considered while proving ($i$) and
($iii$). Let $U_{k}$ be the $\delta$-neighborhood of the set
$\\{(0,\tilde{z})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}:|\tilde{z}|=1\\}$. We
set $f=\chi_{U_{k}}$ so that $\|f\|_{p}\lesssim\delta^{\frac{k+1}{p}}$.
Parametrizing $\mathbb{S}^{d-1}$, we have

$\mathcal{A}f(x,t)\geq\int_{|y|\leq
1/2}\chi_{U_{k}}\big{(}x-t(y,\sqrt{1-|y|^{2}})\big{)}\,dy.$

Denoting
$x=(x^{\prime},x^{\prime\prime},x_{d})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k-1}\times\mathbb{R}$,
we set

$V_{k}=\big{\\{}(x,t):\big{|}|x^{\prime}|-\sqrt{t^{2}-1}\big{|}\leq
c\delta,~{}|x^{\prime\prime}|,|x_{d}|\leq c\delta,~{}t\in[3/2,2]\big{\\}}.$

We write
$y=(y^{\prime},y^{\prime\prime})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k-1}$.
Then, we claim

(4.4) $x-t(y^{\prime},y^{\prime\prime},\sqrt{1-|y|^{2}})\in U_{k}$

for $t\in[3/2,2]$ and $|y^{\prime\prime}|\leq 1/2$ if $(x,t)\in V_{k}$ and
$|x^{\prime}-ty^{\prime}|\leq c\delta$. Since
$|t^{-1}x^{\prime}-y^{\prime}|\leq c\delta$ and
$||x^{\prime}|-\sqrt{t^{2}-1}|\leq c\delta$, it is clear that
$||y^{\prime}|-t^{-1}{\sqrt{t^{2}-1}}|<3c\delta$. This gives
$t|(y^{\prime\prime},\sqrt{1-|y|^{2}})|=t\sqrt{1-|y^{\prime}|^{2}}=1+O(c\delta)$.
Thus, for $|x^{\prime\prime}|,|x_{d}|\leq c\delta$ we have

$|(x^{\prime\prime},x_{d})-t(y^{\prime\prime},\sqrt{1-|y^{\prime}|^{2}-|y^{\prime\prime}|^{2}})|\in(1-2^{-1}\delta,1+2^{-1}\delta)$

if $c$ is small enough. This proves (4.4). Since
$y^{\prime}=x^{\prime}/t+O(c\delta)$ and $|y^{\prime\prime}|\lesssim 1$, (4.4)
implies

$\mathcal{A}f\gtrsim\delta^{k}\chi_{V_{k}}.$

Let $d\nu=\delta^{\alpha-d-1}\chi_{V_{k}}dxdt$. Then we have
$\nu(V_{k})\gtrsim\delta^{\alpha-k}$, and $\langle\nu\rangle_{\alpha}\lesssim
1$ since $\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{\alpha-d-1}r^{d+1}$ if
$0<r\leq\delta$ and $\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{\alpha-k}r^{k}$
if $\delta<r$. So (4.1) gives
$\delta^{k}\delta^{\frac{\alpha-k}{q}}\lesssim\delta^{\frac{k+1}{p}}$. Hence
the first part of 5 follows if we take $\delta\to 0$.

To prove the second part of 5, taking $f=\chi_{U_{k+1}}$, we have
$\|f\|_{p}\lesssim\delta^{\frac{k+2}{p}}$, and
$\mathcal{A}\chi_{U_{k+1}}\gtrsim\delta^{k+1}$ whenever $(x,t)\in V_{k+1}$.
Considering $d\nu=\delta^{k-d}\chi_{V_{k+1}}dxdt$, we have
$\langle\nu\rangle_{\alpha}\lesssim 1$ and $\nu(V_{k+1})\gtrsim 1$ since
$\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta^{k-d}r^{d+1}$ for $0<r\leq\delta$,
and $\nu(\mathbb{B}^{d+1}(z,r))\lesssim\delta r^{k}$ for $\delta<r$.
Therefore, (4.1) implies $\delta^{k+1}\lesssim\delta^{\frac{k+2}{p}}$ for
$0<\delta\ll 1$. This gives the second part of 5. ∎

###### Proof of 6.

Let us consider the case $1<\alpha\leq 2$ first. Let $\\{\theta_{\ell}\\}$ be
a collection of $\sim C\delta^{1-\alpha}$ points on $\mathbb{S}^{1}$ which are
separated by $\delta^{\alpha-1}$. For each $\theta_{\ell}$, we set

$\mathrm{T}_{\ell}(\delta)=\big{\\{}(x,t)\in\mathbb{R}^{2}\times[1,2]:||x|-t|\leq
c\delta^{2},~{}\big{|}x/|x|-\theta_{\ell}\big{|}\leq c\delta\big{\\}}.$

Let $\mathrm{T}_{\delta}=\cup_{\ell}\mathrm{T}_{\ell}(\delta)$. We consider
$d\nu(x,t)=\delta^{\alpha-4}\chi_{\mathrm{T}_{\delta}}(x,t)\,dxdt$, then
$\langle\nu\rangle_{\alpha}\lesssim 1$. In fact, one can easily see

$\displaystyle\nu(\mathbb{B}^{3}(z,r))\leq C\begin{cases}\ \ \ \
\delta^{\alpha-4}r^{3},\quad\quad&\quad\ \ 0<r\leq\delta^{2},\\\ \ \ \ \
\delta^{\alpha-2}r^{2},\quad\quad&\ \ \ \,\delta^{2}<r\leq\delta,\\\ \ \ \ \
\delta^{\alpha-1}r,\quad\quad&\ \ \ \ \ \delta<r\leq\delta^{1-\alpha},\\\
\delta^{\alpha-1}r\times(r/\delta^{\alpha-1}),\quad\quad&\delta^{\alpha-1}<r<1,\\\
\delta^{\alpha-1}\times(r/\delta^{\alpha-1}),\quad\quad&\quad\ \
1<r.\end{cases}$

Therefore, we have $\nu(\mathbb{B}^{3}(z,r))\lesssim r^{\alpha}$.

Let $\mathrm{R}(\theta,\delta)$ be a rectangle of side length
$\delta^{2}\times\delta$ centered at the origin such that the short side is
parallel to $\theta\in\mathbb{S}^{1}$. Setting
$\mathrm{R}_{\delta}:=\cup_{\ell}\mathrm{R}(\theta_{\ell},\delta)$, we have
$\mathcal{A}\chi_{\mathrm{R}_{\delta}}\gtrsim\delta\chi_{\mathrm{T}_{\delta}}$.
Since $\|\chi_{\mathrm{R}_{\delta}}\|_{p}\lesssim\delta^{(4-\alpha)/p}$ and
$\nu(\mathrm{T}_{\delta})\gtrsim 1$, we get
$\delta\lesssim\delta^{(4-\alpha)/p}$ which implies $p\geq 4-\alpha$.

Now, we consider the case $2<\alpha\leq 3$. Let
$\\{\omega_{j}\\}\subset\mathbb{R}^{2}$ be a collection of $\sim
C\delta^{2-\alpha}$ points on the $x_{1}$-axis which are separated by
$\delta^{\alpha-2}$. For each $\omega_{j}$, we denote

$\mathbf{C}_{j}(\delta)=\\{(x,t)\in\mathbb{R}^{2}\times[1,2]:||x-\omega_{j}|-t|\leq
c\delta,\quad|x_{2}|\leq 2^{-3}\\}.$

We set $\mathcal{T}_{\delta}=\cup_{j}\mathbf{C}_{j}(\delta)$ and
$d\nu(x,t)=\delta^{\alpha-3}\chi_{\mathcal{T}_{\delta}}(x,t)\,dxdt.$ Then, it
is easy to see $\langle\nu\rangle_{\alpha}\lesssim 1$ and
$\nu(\mathcal{T}_{\delta})\sim 1$. Indeed,

$\displaystyle\nu(\mathbb{B}^{3}(z,r))\leq C\begin{cases}\ \ \ \
\delta^{\alpha-3}r^{3},\quad&\quad\ \ 0<r\leq\delta,\\\ \ \ \ \
\delta^{\alpha-2}r^{2}\quad&\quad\ \ \delta<r<\delta^{\alpha-2},\\\
\delta^{\alpha-2}r^{2}\times(r/\delta^{\alpha-2}),\quad&\delta^{\alpha-2}<r<1,\\\
\delta^{\alpha-2}\times(r/\delta^{\alpha-2}),\quad&\quad\ \ 1<r.\end{cases}$

So, it follows that $\nu(\mathbb{B}^{3}(z,r))\lesssim r^{\alpha}$. Also we
have

$\mathcal{A}\chi_{\cup_{j}\mathbb{B}^{2}(\omega_{j},c\delta)}(x,t)\gtrsim\delta,\quad(x,t)\in\mathcal{T}_{\delta}.$

Since $|\cup_{j}\mathbb{B}^{2}(\omega_{j},c\delta)|\lesssim\delta^{4-\alpha}$,
(4.1) implies $\delta\lesssim\delta^{\frac{4-\alpha}{p}}$. Thus, we see $p\geq
4-\alpha$. ∎

### Acknowledgement

This work was supported by the NRF (Republic of Korea) grants No.
2017R1C1B2002959 (Ham), No. 2019R1A6A3A01092525 (Ko), and No.
2021R1A2B5B02001786 (Lee).

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