$\partial_{\zeta}\big{\langle}\partial_{z}\widetilde{\Phi}_{z_{2d}}(z,\zeta),\,\mathrm{N}(z,z^{\prime})\big{\rangle}=0.$

Since
$\langle\partial_{z}\widetilde{\Phi}(z,z^{\prime}),\,\mathrm{N}(z,z^{\prime})\rangle=\partial_{z}^{\intercal}\Phi_{\\!\mathcal{L}}(\rho\mathbf{R}_{0}z+z_{0},\rho\mathbf{R}_{0}z^{\prime}+z^{\prime}_{0})\nu(\rho\mathbf{R}_{0}z+z_{0},\rho\mathbf{R}_{0}z^{\prime}+z^{\prime}_{0})$,
we have

$\partial_{z^{\prime}}\partial_{z^{\prime}}^{\intercal}\big{\langle}\partial_{z}\widetilde{\Phi}(0,z^{\prime}),\,\mathrm{N}(0,0)\big{\rangle}|_{z^{\prime}=0}=\rho^{2}\mathbf{R}_{0}^{\intercal}\partial_{w}\partial^{\intercal}_{w}\langle\partial_{z}\Phi_{\\!\mathcal{L}}(z_{0},w),\nu(z_{0},z^{\prime}_{0})\rangle|_{w=z^{\prime}_{0}}\mathbf{R}_{0}.$

Thus, using (3.43), we get

$\partial_{z^{\prime}}\partial_{z^{\prime}}^{\intercal}\big{\langle}\partial_{z}\widetilde{\Phi}(0,z^{\prime}),\,\mathrm{N}(0,0)\big{\rangle}|_{z^{\prime}=0}=\frac{-\rho^{2}}{\cos^{2}\mathfrak{S}_{c}(z_{0},z_{0}^{\prime})|z_{0}-z_{0}^{\prime}|^{4}}\mathbf{R}_{0}^{\intercal}\mathbf{M}(z_{0},z_{0}^{\prime})\mathbf{R}_{0}.$

From Lemma 3.10,
$\mathbf{R}_{0}^{\intercal}\mathbf{M}(z_{0},z_{0}^{\prime})\mathbf{R}_{0}$ is
a diagonal matrix with its nonzero diagonal entries $\sim\rho^{2}$. Thus,
$\mathbf{N}=\partial_{z^{\prime}}\partial_{z^{\prime}}^{\intercal}\big{\langle}\partial_{z}\widetilde{\Phi}(0,z^{\prime}),\,\mathrm{N}(0,0)\big{\rangle}|_{z^{\prime}=0}$
is a diagonal matrix such that $-\mathbf{N}_{i,i}\sim 1$ for $1\leq i\leq
2d-1$ and $\mathbf{N}_{2d,2d}=0$. Here we use
$\cos^{2}\mathfrak{S}_{c}(z_{0},z_{0}^{\prime})\gtrsim c_{0}$ and
$|z_{0}-z_{0}^{\prime}|\sim\rho$. Therefore, the matrix
$\partial_{\zeta}\partial_{\zeta}^{\intercal}\big{\langle}\partial_{z}\widetilde{\Phi}_{z_{2d}}(0,\zeta),\,\mathrm{N}(0,0)\big{\rangle}$
is a nonsingular diagonal matrix of which diagonal entries are all negative
and their absolute values are comparable to $1$. It is clear that
$|\partial_{z}^{\alpha}\partial_{z}^{\beta}\mathrm{N}(z,z^{\prime})|\leq
C_{\alpha,\beta}$. By this and (3.47) we note that
$\widetilde{\Phi}(z,z^{\prime})$ is a smooth function with bounded
derivatives. Thus, now taking $\epsilon_{0}$ small enough, we see that the
matrix

$\partial_{\zeta}\partial_{\zeta}^{\intercal}\big{\langle}\partial_{z}\widetilde{\Phi}_{z_{2d}}(z,\zeta),\,\mathrm{N}(z,z^{\prime})\big{\rangle},\quad
z^{\prime}_{2d}\in(-\epsilon_{0},\epsilon_{0})$

has $2d-1$ negative eigenvalues of which absolute values are comparable to $1$
for all $(z,\zeta)\in B(0,\epsilon_{0})\times B(0,\epsilon_{0})$. This and
(3.50) verify the conditions (C1) – (C3). Therefore, Theorem 2.11 gives the
desired estimate (3.49). ∎

##  4\. Lower bound on the summability index of
$S_{\lambda}^{\delta}(\mathcal{H})$

In this section we obtain a new lower bound on the summability index $\delta$
for uniform boundedness of $S_{\lambda}^{\delta}(\mathcal{H})$ on $L^{p}$.

######  Proposition 4.1.

Let $2<p\leq\infty$. The uniform bound
$\|S_{\lambda}^{\delta}(\mathcal{H})\|_{p}\leq C$ holds only if
$\delta>\delta(d,p)$ and

$\delta\geq\gamma(d,p):=-\frac{1}{3p}+\frac{d}{3}\Big{(}\frac{1}{2}-\frac{1}{p}\Big{)}.$

In particular, when $d=1$ the uniform bound (1.1) holds only if $\delta\geq 0$
for $2\leq p\leq 4$ and
$\delta\geq-\frac{1}{3p}+\frac{1}{3}(\frac{1}{2}-\frac{1}{p})$ for $p\geq 4$.
This coincides with Thangavelu’s result [29, Theorem 2.1]. In higher
dimensions, i.e., $d\geq 2$, it is necessary for (1.1) that
$\delta>\delta(d,p)$ if $\frac{2(2d-1)}{2d-3}\leq p\leq\infty$;
$\delta\geq\gamma(d,p)>\delta(d,p)$ if $\frac{2(d+1)}{d}\leq
p\leq\frac{2(2d-1)}{2d-3}$; $\delta>0$ if $2<p\leq\frac{2(d+1)}{d}$. (See
Figure 1.)

The second lower bound $\delta\geq\gamma(d,p)$ in Proposition 4.1 is a
consequence of the following two lemmas.

######  Lemma 4.2.

Let $1\leq p,q\leq\infty$ and $\lambda\gg 1$. Then, for each $\lambda\gg 1$
there exists $f_{\lambda}\in\mathcal{S}(\mathbb{R}^{d})$ such that

(4.1) $\|\Pi_{\lambda}^{\mathcal{H}}f_{\lambda}\|_{q}\geq
C\lambda^{-\frac{1}{3p}+\frac{d}{6}(1-\frac{1}{p}-\frac{1}{q})}\|f_{\lambda}\|_{p}$

with $C$ independent of $\lambda$.

######  Lemma 4.3.

Let $1\leq p\leq\infty$, $\lambda\gg 1$, and let $E,F$ be measurable subsets
of $\mathbb{R}^{d}$. Then, we have

$\|\chi_{E}\Pi_{\lambda}^{\mathcal{H}}\chi_{F}\|_{p}\leq
C\lambda^{\delta}\sup_{t>0}\|\chi_{E}S^{\delta}_{t}(\mathcal{H})\chi_{F}\|_{p}.$

###### Proof of Proposition 4.1.

Suppose that $\|S_{\lambda}^{\delta}(\mathcal{H})\|_{p}\leq C$ holds for some
$p\in(2,\infty]$ and $\delta$ and $E$ is a compact subset of $\mathbb{R}^{d}$
which contains the origin as a point of density. Since
$\|\chi_{E}S_{\lambda}^{\delta}(\mathcal{H})\chi_{E}\|_{p}\leq\|S_{\lambda}^{\delta}(\mathcal{H})\|_{p}$,
$\|\chi_{E}S_{\lambda}^{\delta}(\mathcal{H})\chi_{E}\|_{p}\leq C$ with a
constant $C$ independent of $\lambda$. Since the principal part of
$\mathcal{H}$ is $-\Delta$, the transplantation theorem of Kenig, Stanton, and
Tomas [16, Theorem 3] gives $\|(1-\Delta)_{+}^{\delta}\|_{p}\leq C$. It is
well known that $\|(1-\Delta)_{+}^{\delta}\|_{p}\leq C$ only if
$\delta>\max\\{0,-\frac{1}{2}+d(\frac{1}{2}-\frac{1}{p})\\}$ when $p\neq 2$
(see for example [10, 7]).

Thus, it suffices to show that $\delta\geq\gamma(d,p)$. From Lemma 4.2, it
follows that $\lambda^{\gamma(d,p)}\leq C\|\Pi_{\lambda}^{\mathcal{H}}\|_{p}.$
On the other hand, by Lemma 4.3 we have
$\|\Pi_{\lambda}^{\mathcal{H}}\|_{p}\leq C\lambda^{\delta}$. Thus, we obtain
$\lambda^{\gamma(d,p)}\leq C\lambda^{\delta}$ for a constant $C$ independent
of $\lambda$. Letting $\lambda$ tend to infinity, it follows that
$\delta\geq\gamma(d,p)$ as desired. ∎

Let $\Gamma$ denote the Gamma function and denote $t_{+}=\max(t,0)$ and
$t_{-}=-\min(t,0)$ for $t\in\mathbb{R}$. We consider the distributions

$\chi_{\pm}^{\nu}=\frac{x_{\pm}^{\nu}}{\Gamma(\nu+1)},\quad\nu\in\mathbb{C}$

which are given by analytic continuation of the functions
$\displaystyle{\frac{x_{\pm}^{\nu}}{\Gamma(\nu+1)}}$, ${Re}(\nu)>-1$. See [12,
Section 3.2] for more about the distribution. In order to prove Lemma 4.3 we
use the identity

(4.2) $\displaystyle F(\mathcal{H})=\int
F^{(\nu)}(t)\frac{(t-\mathcal{H})_{+}^{\nu-1}}{{\Gamma(\nu)}}dt=\int
F^{(\nu)}(t)t^{\nu-1}\frac{S_{t}^{\nu-1}(\mathcal{H})}{{\Gamma(\nu)}}dt,$

which holds for any $\nu\in\mathbb{C}$ and $F\in C_{c}^{\infty}([0,\infty))$.
Here $F^{(\nu)}$ denotes the Weyl fractional derivative of $F$ of order $\nu$
which is defined by $F^{(\nu)}=F\ast\chi_{-}^{-\nu-1}$ for $F$ supported in
$[0,\infty)$. Note that $t$ in the integrand can be assumed to be positive
because $(t-\mathcal{H})_{+}^{\delta}=0$ if $t\leq 0$. The identity (4.2) can
be shown using the convolution property of the distribution $\chi_{-}^{\nu}$,
see [5, pp. 3235-3236] for the detail.

###### Proof of Lemma 4.3.

Now let $\eta$ be a smooth bump function such that $\eta(t)=1$ if
$|t|\leq\frac{1}{2}$ and $\eta(t)=0$ if $|t|\geq 1$ and $\lambda\in
2\mathbb{N}_{0}+d$. Since $\lambda\in 2\mathbb{N}_{0}+d$, taking
$\nu=\delta+1$ and $F=\eta(\cdot-\lambda)$ in (4.2), we get

$\Pi_{\lambda}^{\mathcal{H}}=\eta(H-\lambda)=\frac{1}{{\Gamma(\delta+1)}}\int(\eta(\cdot-\lambda))^{(\delta+1)}(t)t^{\delta}S_{t}^{\delta}(\mathcal{H})dt.$

A simple calculation shows $|(\eta(\cdot-\lambda))^{(\delta+1)}(t)|\leq
C(1+|t-\lambda|)^{-(\delta+1)}\chi_{\\{t\leq\lambda+10\\}}$ for a constant
$C$. Thus, combining this with the above identity, we have

$\displaystyle\|\chi_{E}\Pi_{\lambda}^{\mathcal{H}}\chi_{F}\|_{p}$
$\displaystyle\leq
C\int|(\eta(\cdot-\lambda))^{(\delta+1)}(t)|t^{\delta}\|\chi_{E}S_{t}^{\delta}(\mathcal{H})\chi_{F}\|_{p}dt$
$\displaystyle\lesssim\lambda^{\delta}\sup_{t>0}\|\chi_{E}S_{t}^{\delta}(\mathcal{H})\chi_{F}\|_{p}.\qed$

$\delta$$\frac{1}{p}$$\frac{1}{6}$$\frac{1}{2}$$\frac{3}{4}$$\frac{1}{4}$(i)
$d=1$

$\delta$$\frac{1}{p}$$\frac{d-1}{2}$$\frac{1}{2}$$\delta_{1}$
$\frac{1}{p_{1}}$$\frac{1}{p_{2}}$(ii) $d\geq 2$

Figure 1. Lower bound on the summability index $\delta$:
$p_{1}=\frac{2(2d-1)}{2d-3}$, $p_{2}=\frac{2(d+1)}{d}$, and
$\delta_{1}=\frac{1}{2(2d-1)}$.

We now turn to the proof of Lemma 4.2 which is similar to that of [13,
Proposition 7.10]. Let $Q$ be the cube which is given by

$Q=\\{x\in\mathbb{R}^{d}:\lambda^{\frac{1}{2}}/200\leq|x_{1}|\leq\lambda^{\frac{1}{2}}/100,\
|x_{i}|\leq 10^{2}\lambda^{\frac{1}{6}},\ 2\leq i\leq d\,\\},$

and set

$x_{\ast}=(\lambda^{\frac{1}{2}}-10^{2}\lambda^{-\frac{1}{6}})\mathbf{e}_{1}.$

We first claim that there is a point $x_{0}\in
B(x_{\ast},10^{2}\lambda^{-\frac{1}{6}})$ such that

(4.3)
$\displaystyle\int_{Q}|\Pi_{\lambda}^{\mathcal{H}}(x_{0},y)|^{2}dy\gtrsim\lambda^{\frac{d-2}{6}}.$

Assuming this for the moment, we consider

$f_{\lambda}(x)=\chi_{Q}(x)\Pi_{\lambda}^{\mathcal{H}}(x_{0},x).$

Then, we shall show the following two estimates:

(4.4) $\displaystyle\|\Pi_{\lambda}^{\mathcal{H}}f_{\lambda}$
$\displaystyle\|_{L^{q}(B(x_{\ast},c\lambda^{-\frac{1}{6}}))}\gtrsim\lambda^{\frac{d-2}{6}-\frac{d}{6q}},$
(4.5)
$\displaystyle\|f_{\lambda}\|_{p}\lesssim\lambda^{-\frac{1}{3}+\frac{d+2}{6p}}$

with the implicit constants independent of $\lambda$ and the constant $c>0$ to
be chosen sufficiently small. Combining (4.4) and (4.5) immediately yields the
desired property (4.1).

Thus, it remains to show our claim (4.3) and the inequalities (4.4) and (4.5).
Compared with the typical construction for the Laplacian, the proofs of them
are somewhat involved.

###  4.1. Proof of (4.3)

To show (4.3) we use the following which can be found in [18, Lemma 5.1].

######  Lemma 4.4.

Let $\mu=\sqrt{2k+1}$ and define

$s^{-}_{\mu}(t)=\int_{0}^{t}\sqrt{|\tau^{2}-\mu^{2}|}d\tau\quad\text{and}\quad
s^{+}_{\mu}(t)=\int^{t}_{\mu}\sqrt{|\tau^{2}-\mu^{2}|}d\tau.$

Then the following hold:

$h_{2k}(t)=\begin{cases}a_{2k}^{-}(\mu^{2}-t^{2})^{-\frac{1}{4}}\big{(}\\!\cos
s^{-}_{\mu}(t)+\mathcal{E}\big{)},&\qquad\qquad\quad\
|t|<\mu-\mu^{-\frac{1}{3}},\\\ \qquad\qquad O(\mu^{-\frac{1}{6}}),&\
\,\mu-\mu^{-\frac{1}{3}}<|t|<\mu+\mu^{-\frac{1}{3}},\\\
a^{+}_{2k}e^{-s^{+}_{\mu}(|t|)}(t^{2}-\mu^{2})^{-\frac{1}{4}}\big{(}\\!1+\mathcal{E}\big{)},&\,\
\mu+\mu^{-\frac{1}{3}}<|t|,\end{cases}$
$h_{2k+1}(t)=\begin{cases}a_{2k+1}^{-}(\mu^{2}-t^{2})^{-\frac{1}{4}}\big{(}\\!\sin\,s^{-}_{\mu}(t)+\,\mathcal{E}\big{)},&\qquad\qquad\
\,\ |t|<\mu-\mu^{-\frac{1}{3}},\\\ \qquad\qquad
O(\mu^{-\frac{1}{6}}),&\mu-\mu^{-\frac{1}{3}}<|t|<\mu+\mu^{-\frac{1}{3}},\\\
a^{+}_{2k+1}e^{-s^{+}_{\mu}(|t|)}(t^{2}-\mu^{2})^{-\frac{1}{4}}\big{(}\\!1+\mathcal{E}\big{)},&\mu+\mu^{-\frac{1}{3}}<|t|,\end{cases}$

where $|a^{\pm}_{k}|\sim 1$ and
$\mathcal{E}=O\big{(}|t^{2}-\mu^{2}|^{-\frac{1}{2}}\left||t|-\mu\right|^{-1}\big{)}.$

We first note that

$\sup_{x\in
B(x_{\ast},10^{2}\lambda^{-1/6})}\int_{Q}\Pi_{\lambda}(\mathcal{H})(x,y)^{2}dy=\|\chi_{B(x_{\ast},10^{2}\lambda^{-1/6})}\Pi_{\lambda}(\mathcal{H})\chi_{Q}\|_{2\to\infty}^{2}.$

Thus (4.3) follows once we find a function $g$ such that

(4.6)
$\|\chi_{B(x_{\ast},10^{2}\lambda^{-1/6})}\Pi_{\lambda}(\mathcal{H})\chi_{Q}g\|_{\infty}\gtrsim\lambda^{\frac{d-2}{12}}\|g\|_{2}.$

Let $k:=\frac{\lambda-d}{2}\in\mathbb{N}_{0}$. Then, we define the set of
indices $J$ by

$J=\Big{\\{}\alpha\in\mathbb{N}_{0}^{d}:|\alpha|=k,\,{\lambda^{\frac{1}{3}}}/{d}\leq\alpha_{i}\leq{2\lambda^{\frac{1}{3}}}/{d},\,2\leq
i\leq d\Big{\\}}.$

We note that $\lambda-2\lambda^{\frac{1}{3}}\leq\alpha_{1}\leq\lambda$ for
$\alpha\in J$. Thus, using Lemma 4.4, we see

$\displaystyle\int_{\lambda^{1/2}-20\lambda^{-1/6}}^{\lambda^{1/2}-10\lambda^{-1/6}}h_{\alpha_{1}}(t)^{2}dt\sim\lambda^{-\frac{1}{3}},\quad\int_{-\frac{\lambda^{-1/6}}{\sqrt{d}}}^{\frac{\lambda^{-1/6}}{\sqrt{d}}}h_{\alpha_{i}}(t)^{2}dt\sim\lambda^{-\frac{1}{3}},\quad
2\leq i\leq d$

whenever $\alpha\in J$. Denote
$D=[-\frac{\lambda^{-\frac{1}{6}}}{\sqrt{d}},\frac{\lambda^{-\frac{1}{6}}}{\sqrt{d}}]^{d-1}$.
Since $|J|\sim\lambda^{\frac{d-1}{3}}$ and
$\Phi_{\alpha}(x)=\prod_{i=1}^{d}h_{\alpha_{i}}(x_{i})$, by using the
estimates in the above we get

$\displaystyle\sum_{\alpha\in
J}\int_{\lambda^{1/2}-20\lambda^{-1/6}}^{\lambda^{1/2}-10\lambda^{-1/6}}\int_{D}\Phi_{\alpha}(x_{1},x^{\prime})^{2}dx^{\prime}dx_{1}\sim\lambda^{-\frac{1}{3}}.$

Thus there exists a point
$\widetilde{x}_{*}\in[\lambda^{1/2}-20\lambda^{-1/6},\lambda^{1/2}-10\lambda^{-1/6}]\times
D\subset B(x_{\ast},10^{2}\lambda^{-\frac{1}{6}})$ such that $\sum_{\alpha\in
J}\Phi_{\alpha}^{2}(\widetilde{x}_{*})\sim\lambda^{\frac{d-2}{6}}$. Now we set

$g(x)=\sum_{\alpha\in J}\Phi_{\alpha}(\widetilde{x}_{*})\Phi_{\alpha}(x).$

We proceed to prove that $g$ satisfies (4.6). By orthogonality, we have
$\|g\|_{2}=\big{(}\sum_{\alpha\in
J}\Phi_{\alpha}(\widetilde{x}_{*})^{2}\big{)}^{\frac{1}{2}}\lesssim\lambda^{\frac{d-2}{12}}$,
so (4.6) follows if we show

(4.7)
$\|\chi_{B(x_{\ast},10^{2}\lambda^{-1/6})}\Pi_{\lambda}(\mathcal{H})\chi_{Q}g\|_{\infty}\gtrsim\lambda^{\frac{d-2}{6}}.$

We write

$\displaystyle\Pi_{\lambda}(\mathcal{H})\chi_{Q}g(x)$
$\displaystyle=\sum_{\alpha:|\alpha|=k}\sum_{\beta\in
J}\Phi_{\alpha}(x)\Phi_{\beta}(\widetilde{x}_{*})\int_{Q}\Phi_{\alpha}(y)\Phi_{\beta}(y)dy$
$\displaystyle=:\mathbf{I}(x)+\mathbf{II}(x),$

where

$\displaystyle\mathbf{I}(x)=\sum_{\alpha\in
J}\Phi_{\alpha}(x)\Phi_{\alpha}(\widetilde{x}_{*})\int_{Q}\Phi_{\alpha}(y)\Phi_{\alpha}(y)dy,$
$\displaystyle\mathbf{II}(x)=\sum_{\alpha\neq\beta}\Phi_{\alpha}(x)\Phi_{\beta}(\widetilde{x}_{*})\int_{Q}\Phi_{\alpha}(y)\Phi_{\beta}(y)dy.$

We first handle $\mathbf{I}$. By the choice of $J$,
$\lambda-2\lambda^{\frac{1}{3}}\leq\alpha_{1}\leq\lambda$ and
${\lambda^{\frac{1}{3}}}/{d}\leq\alpha_{i}\leq{2\lambda^{\frac{1}{3}}}/{d}$
for $i=2,\dots,d$. Thus, using Lemma 4.4 we see that

$\int_{\lambda^{\frac{1}{2}}/200\leq|t|\leq\lambda^{\frac{1}{2}}/100}h_{\alpha_{1}}(t)^{2}dt\sim
1,\quad\int_{|t|\leq 10^{2}\lambda^{\frac{1}{6}}}h_{\alpha_{i}}(t)^{2}dt\sim
1,\quad 2\leq i\leq d$

for any $\alpha\in J$. Thus, we have $\int_{Q}\Phi_{\alpha}(x)^{2}dx\gtrsim 1$
for $\alpha\in J$, so $\mathbf{I}(x)$ satisfies the estimate

$|\mathbf{I}(x_{*})|\gtrsim\lambda^{\frac{d-2}{6}}$

because $\sum_{\alpha\in
J}\Phi_{\alpha}^{2}(x_{*})\sim\lambda^{\frac{d-2}{6}}$. For $\mathbf{II}(x)$,
we make use of the following formula

$\int_{-l}^{l}h_{u}(t)h_{v}(t)dt=\frac{1+(-1)^{u+v}}{\sqrt{2}(u-v)}\big{(}\sqrt{u+1}\,h_{u+1}(l)h_{v}(l)-\sqrt{v+1}\,h_{u}(l)h_{v+1}(l)\big{)},\quad
u\neq v.$

See [13, pp.24–25] for its proof. By Lemma 4.4 it follows that
$|h_{v}(10^{2}\lambda^{\frac{1}{6}})|\lesssim e^{-c\lambda^{\frac{1}{3}}}$ for
some $c>0$ when $v\leq 2\lambda^{\frac{1}{3}}d^{-1}$. Also, if
$\alpha\neq\beta$, there exists at least one $i$’s such that $2\leq i\leq d$
and $\alpha_{i}\neq\beta_{i}$ because $|\alpha|=|\beta|=k$. Therefore, we have

$\bigg{|}\int_{Q}\Phi_{\alpha}(y)\Phi_{\beta}(y)dy\bigg{|}\leq\prod_{i=2}^{d}\int_{-10^{2}\lambda^{\frac{1}{6}}}^{10^{2}\lambda^{\frac{1}{6}}}h_{\alpha_{i}}(t)h_{\beta_{i}}(t)dt\lesssim\lambda^{N}e^{-c\lambda^{\frac{1}{3}}}$

for some constants $c>0$ and $N$. From Lemma 4.4 we see
$\mathbf{II}(x_{*})=O(\lambda^{N}e^{-c\lambda^{\frac{1}{3}}})$. Combining this
and the lower bound for $\mathbf{I}(x_{*})$ together with the estimate for
$\|g\|_{2}$, we obtain (4.7) taking $\lambda$ large enough. ∎

###  4.2. Proof of (4.4)

To show (4.4), we make use of the following Lemma, which we prove later.

######  Lemma 4.5.

Let $\lambda\in 2\mathbb{N}_{0}+d$ and
$\mu\in[\lambda^{-\frac{2}{3}},\frac{1}{4}]$. Suppose that
$h\in\mathcal{S}(\mathbb{R}^{d})$ is a eigenfunction of $\mathcal{H}$ with the
eigenvalue $\lambda$, i.e., $\mathcal{H}h(x)=\lambda h(x)$. If
$\lambda^{\frac{1}{2}}(1-2\mu)\leq|y_{0}|\leq\lambda^{\frac{1}{2}}(1-\mu),$
then for any $\alpha\in\mathbb{N}_{0}^{d}$ we have

(4.8) $|\partial_{y}^{\alpha}h(y_{0})|\leq
C(\lambda\mu)^{\frac{|\alpha|}{2}}\|h\|_{L^{\infty}(B(y_{0},2(\lambda\mu)^{-\frac{1}{2}}))}$

with $C$ independent of $\lambda$, $\mu$ and $h$.

We start by recalling the estimate

(4.9)
$\displaystyle\|\chi_{A_{\lambda,0}}\Pi_{\lambda}^{\mathcal{H}}\chi_{A_{\lambda,0}}\|_{1\to\infty}\lesssim\lambda^{\frac{d-2}{6}},$

which can be found in ([13, p.79], [18, p.375]). Here
$A_{\lambda,0}:=\\{x\in\mathbb{R}^{d}:||x|-\lambda^{\frac{1}{2}}|\leq
10^{3}\lambda^{-\frac{1}{6}}\\}$. This is equivalent to the estimate
$|\Pi_{\lambda}^{\mathcal{H}}(x,y)|\lesssim\lambda^{\frac{d-2}{6}}$ for any
$x,y\in A_{\lambda,0}$. Now we note that

(4.10)
$\Pi_{\lambda}^{\mathcal{H}}f_{\lambda}(x)=\int_{Q}\Pi_{\lambda}^{\mathcal{H}}(x_{0},y)\Pi_{\lambda}^{\mathcal{H}}(x,y)dy.$

Using the Cauchy-Schwartz inequality and orthogonality between the Hermite
functions $\Phi_{\alpha}$, we get

$|\Pi_{\lambda}^{\mathcal{H}}f_{\lambda}(x)|\leq\bigg{(}\int\Pi_{\lambda}^{\mathcal{H}}(x_{0},y)^{2}dy\int\Pi_{\lambda}^{\mathcal{H}}(x,y)^{2}dy\bigg{)}^{\frac{1}{2}}\leq\big{(}\Pi_{\lambda}^{\mathcal{H}}(x_{0},x_{0})\Pi_{\lambda}^{\mathcal{H}}(x,x)\big{)}^{\frac{1}{2}}.$

By (4.9), we see that
$|\Pi_{\lambda}^{\mathcal{H}}f_{\lambda}(x)|\lesssim\lambda^{\frac{d-2}{6}}$
for every $x\in A_{\lambda,0}$. Since
$\Pi_{\lambda}^{\mathcal{H}}f_{\lambda}\in\mathcal{S}(\mathbb{R}^{d})$ and
trivially is an eigenfunction of $\mathcal{H}$ with the eigenvalue $\lambda$,
by Lemma 4.5 we see

$|\nabla(\Pi_{\lambda}^{\mathcal{H}}f_{\lambda})(x)|\lesssim
C\lambda^{\frac{d-1}{6}}$

if $||x|-\lambda^{\frac{1}{2}}|\leq 10^{2}\lambda^{-\frac{1}{6}}$. On the
other hand, from (4.3) and (4.10) we have
$\Pi_{\lambda}^{\mathcal{H}}f_{\lambda}(x_{0})\gtrsim\lambda^{\frac{d-2}{6}}.$
Thus, it follows that

$\Pi_{\lambda}^{\mathcal{H}}f(x)\gtrsim\lambda^{\frac{d-2}{6}}$

if $x\in B(x_{0},c\lambda^{-\frac{1}{6}})$ with a sufficiently small $c>0$. We
therefore get (4.4). ∎

We now turn to the proof of Lemma 4.5.

###### Proof of Lemma 4.5.

To show (4.8) we may assume
$\|h\|_{L^{\infty}(B(y_{0},2(\lambda\mu)^{-\frac{1}{2}}))}=1$. Let $\varphi$
be a smooth cutoff function such that $\operatorname{supp}\varphi\subset
B(0,2)$ and $\varphi\equiv 1$ on $B(0,1)$. Let us set
$\phi_{y_{0}}(y)=\varphi(\sqrt{\lambda\mu}(y-y_{0}))$. Then by inversion we
write

$\displaystyle\partial_{x}^{\alpha}h(y_{0})=\partial_{x}^{\alpha}(\varphi_{y_{0}}h)(y_{0})=(2\pi)^{-d}\iint(i\xi)^{\alpha}$
$\displaystyle\varphi_{y_{0}}(y)e^{i(y_{0}-y)\cdot\xi}h(y)dyd\xi.$

Setting $\phi_{K}(\xi)=\varphi({\xi}/{K\sqrt{\lambda\mu}})$ with a large
positive constant $K$, we split the integral

$\displaystyle\partial_{x}^{\alpha}h(y_{0})=\mathrm{I}+\mathrm{I\\!I},$

where

$\displaystyle\mathrm{I}$
$\displaystyle:=(2\pi)^{-d}\iint(i\xi)^{\alpha}{\phi_{K}}(\xi)\phi_{y_{0}}(y)e^{i(y_{0}-y)\cdot\xi}h(y)dyd\xi,$
$\displaystyle\mathrm{I\\!I}$
$\displaystyle:=(2\pi)^{-d}\iint(i\xi)^{\alpha}(1-\phi_{K}(\xi))\phi_{y_{0}}(y)e^{i(y_{0}-y)\cdot\xi}h(y)dyd\xi.$

For $\mathrm{I}$ we have

$|{\mathrm{I}}|\lesssim\iint|\xi|^{|\alpha|}|\varphi_{K}(\xi)\phi_{y_{0}}(y)h(y)|dyd\xi\leq
C(\lambda\mu)^{\frac{|\alpha|}{2}}.$

Since $(|\xi|^{2}+\Delta_{y})h(y)=|\xi|^{2}+{|y|^{2}}-\lambda$, we may write

$\mathrm{I\\!I}=C\iint{(i\xi)^{\alpha}}\frac{(1-\varphi_{K}(\xi))\phi_{y_{0}}(y)}{|\xi|^{2}+y^{2}-\lambda}e^{i(y_{0}-y)\cdot\xi}(\xi^{2}+\Delta_{y})h(y)dyd\xi.$

By integration by parts, this is equal to

$C\iint\xi^{\alpha}(1-\varphi_{K}(\xi))A_{1}(y,\xi)e^{i(y_{0}-y)\cdot\xi}h(y)dyd\xi,$

where

$A_{1}(y,\xi):=\Delta_{y}\bigg{(}\frac{\phi_{y_{0}}(y)}{\xi^{2}+y^{2}-\lambda}\bigg{)}-2i\nabla_{y}\bigg{(}\frac{\phi_{y_{0}}(y)}{\xi^{2}+y^{2}-\lambda}\bigg{)}\cdot\xi.$

We now note that
$\xi^{2}+y^{2}-\lambda>\frac{K}{2}(\lambda\mu)^{\frac{1}{2}}+\frac{\xi^{2}}{2}$
for
$(y,\xi)\in{\operatorname{supp}(\phi_{y_{0}})\times\operatorname{supp}(\phi_{K})}$
with a sufficiently large $K$ and
$|\partial_{y}^{\alpha}\phi_{y_{0}}(y)|\lesssim(\lambda\mu)^{\frac{|\alpha|}{2}}$
for any $\alpha\in\mathbb{N}_{0}^{d}$. Thus, we have
$|A_{1}(y,\xi)|\lesssim(\lambda\mu)^{\frac{1}{2}}{|\xi|^{-1}}.$ Repeating this
by $N$ times, we have

$\mathrm{I\\!I}=C_{N}\iint\xi^{\alpha}(1-{\phi_{K}}(\xi))A_{N}(y,\xi)e^{i(y_{0}-y)\cdot\xi}h(y)dyd\xi,$

where $A_{N}$ is a smooth function such that
$\operatorname{supp}A_{N}\subset\operatorname{supp}(\phi_{y_{0}}\otimes\varphi_{K})$
and $|A_{N}(y,\xi)|\lesssim|\xi|^{-N}(\lambda\mu)^{\frac{N}{2}}$. Since
$\|h\|_{L^{\infty}(B(y_{0},2(\lambda\mu)^{-\frac{1}{2}}))}=1$, we have

$\displaystyle|\mathrm{I\\!I}|\lesssim(\lambda\mu)^{\frac{N}{2}}\int_{K\sqrt{\lambda\mu}\leq|\xi|}|\xi|^{|\alpha|-N}\int_{|y-y_{0}|\leq
2/\sqrt{\lambda\mu}}dyd\xi\lesssim(\lambda\mu)^{\frac{|\alpha|}{2}}.$

Therefore, we obtain (4.8). ∎

###  4.3. Proof of (4.5)

Let us set $\widetilde{x}_{0}=\lambda^{-1/2}x_{0}$. Changing variables, we
note that the estimate (4.5) is equivalent to

(4.11)
$\|\chi_{\widetilde{Q}}\Pi_{\lambda}^{\mathcal{H}}(\lambda^{\frac{1}{2}}\widetilde{x}_{0},\lambda^{\frac{1}{2}}\cdot)\|_{p}\lesssim\lambda^{-\frac{1}{3}-\frac{d-1}{3p}},$

where

$\widetilde{Q}=\big{\\{}x\in\mathbb{R}^{d}:1/200\leq|x_{1}|\leq 1/100,\
|x_{i}|\leq 10^{2}\lambda^{-\frac{1}{3}},\ 2\leq i\leq d\,\big{\\}}.$

We recall the representation formula

$\Pi_{\lambda}^{\mathcal{H}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(\lambda-\mathcal{H})t}dt,$

which can be easily shown by utilizing the Hermite expansion and the fact that
$\lambda\in 2\mathbb{N}_{0}+d$. For example, see [13, p.11] for the detail.
Using the cutoff functions $\nu^{l}$ (see (2.20)) and scaling, we can write

$\Pi_{\lambda}^{\mathcal{H}}(\lambda^{\frac{1}{2}}\widetilde{x}_{0},\lambda^{\frac{1}{2}}x)=\sum_{l=0}^{\infty}[\nu^{l}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y).$

Thus, the desired estimate (4.11) follows if we show

$\displaystyle\|[\nu^{0}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},\cdot)\|_{L^{p}(\widetilde{Q})}\lesssim\lambda^{-\frac{1}{3}-\frac{d-1}{3p}},$
$\displaystyle\|[\nu^{l}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},\cdot)\|_{L^{p}(\widetilde{Q})}\lesssim\lambda^{-\frac{d-1}{3p}-N}2^{l(\frac{d-2}{2}-N)},\quad
l\geq 1.$

We recall (2.20). Then, using the properties (2.18) and (2.19), we see that
the above estimates follow if we obtain

(4.12)
$\displaystyle\|[\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},\cdot)\|_{L^{p}(\widetilde{Q})}\lesssim\lambda^{-\frac{1}{3}-\frac{d-1}{3p}},$
(4.13)
$\displaystyle\|[\psi(2^{l}\cdot)]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},\cdot)\|_{L^{p}(\widetilde{Q})}\lesssim\lambda^{-\frac{d-1}{3p}-N}2^{l(\frac{d-2}{2}-N)},\quad
l\geq 1.$

Since $\widetilde{x}_{0}\in
B((1-10^{2}\lambda^{-\frac{2}{3}})\mathbf{e}_{1},10^{2}\lambda^{-\frac{2}{3}})$,
for $y\in\widetilde{Q}$ we have

(4.14)
$\displaystyle|\mathcal{D}(\widetilde{x}_{0},y)|=(1-|\widetilde{x}_{0}|^{2})(1-|y|^{2})+|\widetilde{x}_{0}|^{2}|y|^{2}-\langle\widetilde{x}_{0},y\rangle^{2}\leq
10^{3}\lambda^{-\frac{2}{3}}.$

We also have $|\langle\widetilde{x}_{0},y\rangle|\leq 10^{-2}$ for
$y\in\widetilde{Q}$. Thus, from (2.23) we have

$\partial_{t}\mathcal{P}_{\mathcal{H}}(t,\widetilde{x}_{0},y)=\frac{(\cos
t-\langle\widetilde{x}_{0},y\rangle)^{2}-\mathcal{D}(\widetilde{x}_{0},y)}{2\sin^{2}t}\gtrsim
2^{2l}$

if $t\in\operatorname{supp}\psi(2^{l}\cdot)$, $l\geq 1$ since $\lambda$ is
assumed to be large. Recalling (2.13) we have

$[{\psi(2^{l}\cdot)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)=C_{d}\int\psi(2^{l}t)(\sin
t)^{-\frac{d}{2}}e^{i\lambda\mathcal{P}_{\mathcal{H}}(t,\widetilde{x}_{0},y)}dt.$

Since ${|\partial_{t}^{n}(\psi(2^{l}t)(\sin t)^{-\frac{d}{2}})|\lesssim
2^{l(n+\frac{d}{2})}}$ for any $n\geq 1$, applying Lemma 2.5 to
$[\psi^{l}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)$, we obtain

$|[\psi^{l}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)|\lesssim\lambda^{-N}2^{j(\frac{d-2}{2}-N)},\quad
N\in\mathbb{N}.$

Therefore, combining this with
$|\widetilde{Q}|\sim\lambda^{-\frac{(d-1)}{3}}$, we get (4.13).

Let $t_{*}$ be a number in $[0,\pi]$ such that

$\cos t_{*}=\langle\widetilde{x}_{0},y\rangle.$

The point $t_{*}$ can be contained in the support of $\nu^{0}\chi_{(0,\pi)}$.
To show the estimate (4.12), we make additional decomposition away from
$t_{\ast}$. Let us set

$\psi_{l}(t)=\psi(2^{l}(t-t_{\ast}))+\psi(2^{l}(t_{\ast}-t)).$

and also set

$\psi_{\lambda}(t):=\sum_{2^{-l}\leq C\lambda^{-1/3}}\psi_{l}(t),$

where $C$ is a sufficiently large constant. We make a further decomposition on
$[\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)$ as
follows. Then we may write

$\displaystyle[\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)$
$\displaystyle=[\psi_{\lambda}\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)+\sum_{2^{-l}>C\lambda^{-1/3}}[\psi_{l}\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y).$

Since $\sin t\sim 1$ on
$t\in\operatorname{supp}(\nu^{0}\chi_{(0,\pi)})\subset[2^{-2},\pi-2^{-2}]$,

$|[\psi_{\lambda}\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)|\lesssim\int|\psi_{\lambda}(t)|dt\lesssim\lambda^{-\frac{1}{3}}.$

Now we note that $(\cos t-\cos t_{\ast})^{2}\sim 2^{-2l}$ on the support of
$\psi_{l}\nu^{0}\chi_{(0,\pi)}$. Since
$\partial_{t}\mathcal{P}_{\mathcal{H}}(t,\widetilde{x}_{0},y)=\frac{(\cos
t-\cos t_{\ast})^{2}-\mathcal{D}(\widetilde{x}_{0},y)}{2\sin^{2}t}$, by (4.14)
we have

$\partial_{t}\mathcal{P}_{\mathcal{H}}(t,\widetilde{x}_{0},y)\gtrsim 2^{-2l}$

for $t\in\operatorname{supp}(\psi_{l}\nu^{0}\chi_{(0,\pi)})$. So, Lemma 2.5
gives
$[\psi_{l}\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)\lesssim\lambda^{-N}2^{l(-1+3N)}$,
for any $N\in\mathbb{N}$. Thus,
$\sum_{2^{-l}>C\lambda^{-1/3}}|[\psi_{l}\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)|\lesssim\lambda^{-1/3}$.
Combining these estimate, we now get

$|[\nu^{0}\chi_{(0,\pi)}]^{\mathcal{H}}_{\lambda}(\widetilde{x}_{0},y)|\lesssim\lambda^{-\frac{1}{3}}.$

Therefore, we get (4.12) because
$|\widetilde{Q}|\sim\lambda^{-\frac{(d-1)}{3}}$. This completes the proof of
(4.5). ∎

### Acknowledgement

This work was supported by the National Research Foundation of Korea (NRF)
grant number 2021R1A2B5B02001786.

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