Our first task is to define an appropriate distance function on the surface of
the cone. Unlike the sphere, the surface ${\mathbb{V}}_{0}^{d+1}$ has a
boundary at $t=1$ and a singularity at the apex $t=0$. Our distance function
should measure the distance between points near the boundary or the apex and
the distance between interior points differently. This is a well-known
phenomenon as we have already seen for the interval $[-1,1]$. More generally,
the distance on the interval $[a,b]$ is given by the change of variables
$x\in[a,b]\mapsto y\in[-1,1]$,

${\mathsf{d}}_{[a,b]}(x_{1},x_{2})=\frac{b-a}{2}\arccos\left(y_{1}y_{2}+\sqrt{1-y_{1}^{2}}\sqrt{1-y_{2}^{2}}\right),$

where $y_{i}=-1+2\frac{x_{i}-a}{b-a}$. In particular, the distance function
for $[0,1]$ is given by

$\displaystyle{\mathsf{d}}_{[0,1]}(x_{1},x_{2})\,$
$\displaystyle=\frac{1}{2}\arccos\left((2x_{1}-1)(2y_{1}-1)+4\sqrt{x_{1}(1-x_{1})}\sqrt{y_{1}(1-y_{1})}\right)$
$\displaystyle=\arccos\left(\sqrt{x_{1}x_{2}}+\sqrt{(1-x_{1})(1-x_{2})}\right),$

where the second identity follows from
$\arccos({\alpha})=\frac{1}{2}\arccos(2{\alpha}^{2}-1)$ . In particular,
setting $x_{i}=\cos^{2}\frac{{\theta}_{i}}{2}\in[0,1]$,
$0\leq{\theta}_{i}\leq\pi$, we obtain
${\mathsf{d}}_{[0,1]}(x_{1},x_{2})=\frac{1}{2}|{\theta}_{1}-{\theta}_{2}|$.

######  Definition 4.1.

For $(x,t)$ and $(y,s)$ on ${\mathbb{V}}_{0}^{d+1}$, define

(4.1)
${\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s)):=\arccos\left(\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}+\sqrt{1-t}\sqrt{1-s}\right).$

######  Proposition 4.2.

The function
${\mathsf{d}}_{{\mathbb{V}}_{0}^{d+1}}(\cdot,\cdot)={\mathsf{d}}_{{\mathbb{V}}_{0}}(\cdot,\cdot)$
defines a distance on the surface of the cone ${\mathbb{V}}_{0}^{d+1}$.

###### Proof.

Evidently ${\mathsf{d}}_{{\mathbb{V}}_{0}}(\cdot,\cdot)$ is symmetric. Since
$\|x\|=t$ and $\|y\|=s$, it follows readily that
$0\leq\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}+\sqrt{1-t}\sqrt{1-s}\leq 1$,
so that ${\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\geq 0$ and, furthermore,
${\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(x,t))=\arccos 1=0$. Hence, we only
need to prove that it satisfies the triangle inequality.

For $d=2$, we write $(x,t)=(x_{1},x_{2},t)\in{\mathbb{V}}_{0}^{3}$ with
$x_{1}^{2}+x_{2}^{2}=t^{2}$. For $(x,t),(y,s)\in{\mathbb{V}}_{0}^{3}$, it is
easy to verify the identity

$\left(\sqrt{(t+x_{1})(s+y_{1})}+\mathrm{sign}(x_{2}y_{2})\sqrt{(t-x_{1})(s-y_{1})}\right)^{2}=\frac{ts+x_{1}y_{1}+x_{2}y_{2}}{2}.$

Hence, setting
$z_{x,t}=(\sqrt{t+x_{1}},\mathrm{sign}(x_{2})\sqrt{t-x_{1}},\sqrt{1-t}$) and
define $z_{y,s}$ similarly, then they are elements in $\SS^{2}$ and it follows
that

${\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))=\arccos({\langle}z_{x,t},z_{y,s}{\rangle})={\mathsf{d}}_{\SS^{2}}(z_{x,t},z_{y,s}),$

where ${\mathsf{d}}_{\SS^{2}}(\cdot,\cdot)$ is the geodesic distance of the
unit sphere $\SS^{2}$. In particular,
${\mathsf{d}}_{{\mathbb{V}}_{0}}(\cdot,\cdot)$ satisfies the triangle
inequality. For $d>2$, given three distinct point in ${\mathbb{V}}_{0}^{d+1}$,
written as $(t_{i}\xi_{i},t_{i})$, $1\leq i\leq 3$, where
$\xi\in{\mathbb{S}^{d-1}}$, we can find a rotation in ${\mathbb{R}}^{d}$ so
that $\xi_{i}=(\eta_{i},0)$ with $\eta_{i}\in\SS^{2}$. Hence, the triangle
inequality for $d>2$ follows from the triangle inequality for $d=2$. This
completes the proof. ∎

The distance function ${\mathsf{d}}_{{\mathbb{V}}_{0}}(\cdot,\cdot)$ is
closely related to the distance function ${\mathsf{d}}_{[0,1]}(\cdot,\cdot)$
of the interval $[0,1]$ and the geodesic distance
${\mathsf{d}}_{\SS}(\cdot,\cdot)$ of the unit sphere ${\mathbb{S}^{d-1}}$.

######  Proposition 4.3.

For $d\geq 2$ and $(x,t),(y,s)\in{\mathbb{V}}_{0}^{d+1}$, write $x=t\xi$ and
$y=s\eta$ with $\xi,\eta\in{\mathbb{S}^{d-1}}$. Then

(4.2)
$1-\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))=1-\cos{\mathsf{d}}_{[0,1]}(t,s)+\sqrt{t}\sqrt{s}\left[1-\cos\left(\tfrac{1}{2}{\mathsf{d}}_{\SS}(\xi,\eta)\right)\right].$

In particular,

(4.3)
$c_{1}{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\leq{\mathsf{d}}_{[0,1]}(t,s)+(ts)^{\frac{1}{4}}{\mathsf{d}}_{\SS}(\xi,\eta)\leq
c_{2}{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s)).$

###### Proof.

Using $\arccos({\alpha})=\frac{1}{2}\arccos(2{\alpha}^{2}-1)$, we deduce

$\sqrt{\frac{ts+{\langle}x,y{\rangle}}{2}}=\sqrt{ts}\sqrt{\frac{1+{\langle}\xi,\eta{\rangle}}{2}}=\sqrt{ts}\cos\left(\tfrac{1}{2}\arccos{\langle}\xi,\eta{\rangle}\right).$

Consequently, in terms of the geodesic distance on the unit sphere, we can
write

${\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))=\arccos\left[\sqrt{t}\sqrt{s}\cos\left(\tfrac{1}{2}{\mathsf{d}}_{\SS}(\xi,\eta)\right)+\sqrt{1-t}\sqrt{1-s}\right].$

In particular, it follows that

$1-\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))=1-\sqrt{t}\sqrt{s}-\sqrt{1-t}\sqrt{1-s}+\sqrt{t}\sqrt{s}\left[1-\cos\left(\tfrac{1}{2}{\mathsf{d}}_{\SS}(\xi,\eta)\right)\right],$

which is the identity (4.2). From this identity, (4.3) follows from
$1-\cos{\theta}=2\sin^{2}\frac{{\theta}}{2}$,
$\frac{1}{\pi}{\theta}\leq\sin\frac{{\theta}}{2}\leq\frac{{\theta}}{2}$ for
$0\leq{\theta}\leq\pi$, and $(a+b)^{2}/2\leq a^{2}+b^{2}\leq(a+b)^{2}$ for
$a,b\geq 0$. ∎

The line segment from the apex to a point $(\xi,1)$,
$\xi\in{\mathbb{S}^{d-1}}$, on the top boundary of the cone
${\mathbb{V}}_{0}^{d+1}$ can be parametrized by $l_{\xi}=\\{(t\xi,t):0\leq
t\leq 1\\}$. For two points $(t\xi,t)$ and $(s\xi,s)$ on $l_{\xi}$, the
identity (4.2) shows that

${\mathsf{d}}_{{\mathbb{V}}_{0}}((t\xi,t),(s\xi,s))=\tfrac{|{\theta}-\phi|}{2}={\mathsf{d}}_{[0,1]}(t,s)$

if $t=\cos^{2}\frac{{\theta}}{2}$ and $s=\cos^{2}\frac{\phi}{2}$. Moreover,
the top boundary of ${\mathbb{V}}_{0}^{d+1}$ is the unit sphere
${\mathbb{S}^{d-1}}$, or $\\{(\xi,1):\xi\in{\mathbb{S}^{d-1}}\\}$. For
$(\xi_{1},1)$ and $(\xi_{2},1)$ on this boundary, (4.2) gives

${\mathsf{d}}_{{\mathbb{V}}_{0}}((\xi_{1},1),(\xi_{2},1))=\frac{1}{2}{\mathsf{d}}_{\SS}(\xi_{1},\xi_{2}),$

or half of the geodesic distance on the unit sphere.

######  Remark 4.1.

It is well-known that the geodesic distance ${\mathsf{d}}_{\SS}(\cdot,\cdot)$
of ${\mathbb{S}^{d-1}}$ is proportional to the Euclidean distance; that is,
$\|\xi-\eta\|\sim{\mathsf{d}}_{\SS}(\xi,\eta)$ for all
$\xi,\eta\in{\mathbb{S}^{d-1}}$. For the surface of the cone
${\mathbb{V}}_{0}^{d+1}$, however, this is no longer true when the points are
near the apex. Indeed, for $(x,t),(y,s)\in{\mathbb{V}}_{0}^{d+1}$ with
$x=t\xi$, $y=s\eta$, $\xi,\eta\in{\mathbb{S}^{d-1}}$,

$\|(x,t)-(y,s)\|^{2}=2(t-s)^{2}+2ts\big{(}1-{\langle}\xi,\eta{\rangle}\big{)}=2(t-s)^{2}+2ts(1-\cos{\mathsf{d}}_{\SS}(\xi,\eta)).$

In particular, if $t=s$, we see that $\|(t\xi,t)-(t\eta,t)\|\sim
t{\mathsf{d}}_{\SS}(\xi,\eta)$, whereas we have
$d_{{\mathbb{V}}_{0}}\big{(}(t\xi,t),(t\eta,t)\big{)}\sim\sqrt{t}{\mathsf{d}}_{\SS}(\xi,\eta)$
by (4.3). Hence the two distances are not compatible when $t$ is small.

We will also need the following lemma in the estimate of the kernels.

######  Lemma 4.4.

For $(x,t),(y,s)\in{\mathbb{V}}_{0}^{d+1}$,

$\big{|}\sqrt{t}-\sqrt{s}\big{|}\leq{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\quad\hbox{and}\quad\big{|}\sqrt{1-t}-\sqrt{1-s}\big{|}\leq{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s)).$

###### Proof.

Let $t=\cos^{2}\frac{{\theta}}{2}$ and $s=\cos^{2}\frac{\phi}{2}$,
$0\leq{\theta},\phi\leq\pi$. Since $|{\langle}x,y{\rangle}|\leq ts$, we obtain

$\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\leq\sqrt{ts}+\sqrt{1-t}\sqrt{1-s}=\cos\tfrac{{\theta}-\phi}{2},$

so that
${\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\geq\frac{1}{2}|{\theta}-\phi|$.
Elementary trigonometric identities shows that

$\big{|}\sqrt{t}-\sqrt{s}\big{|}=\big{|}\cos\tfrac{{\theta}}{2}-\cos\tfrac{\phi}{2}\big{|}\leq
2\sin\tfrac{|{\theta}-\phi|}{4}\leq\tfrac{|{\theta}-\phi|}{2}\leq{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s)).$

The inequality for $\big{|}\sqrt{1-t}-\sqrt{1-s}\big{|}$ follows from
$\big{|}\sin\tfrac{{\theta}}{2}-\sin\tfrac{\phi}{2}\big{|}\leq
2\sin\tfrac{|{\theta}-\phi|}{4}$. ∎

###  4.2. A family of doubling weights

For the conic surface, balls are conic caps. For $r>0$ and $(x,t)$ on
${\mathbb{V}}_{0}^{d+1}$, we denote the conic cap centered at $(x,t)$ with
radius $r$ by

${\mathsf{c}}((x,t),r):=\left\\{(y,s)\in{\mathbb{V}}_{0}^{d+1}:{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x,t),(y,s)\big{)}\leq
r\right\\}.$

A weight function ${\mathsf{w}}$ is a doubling weight if it satisfies

${\mathsf{w}}\big{(}{\mathsf{c}}((x,t),2r)\big{)}\leq
L\,{\mathsf{w}}\big{(}{\mathsf{c}}((x,t),r)\big{)},\quad r>0.$

In comparison with the spherical cap on ${\mathbb{S}^{d-1}}$, the geometry of
${\mathsf{c}}(x,t)$ is more complicated. Denote the surface measure on
${\mathbb{S}^{d-1}}$ by $\mathrm{d}\sigma_{\SS}$.

######  Lemma 4.5.

For $r>0$, $t,s\in[0,1]$, define $\tau_{r}(t,s)=(\cos
r-\sqrt{1-t}\sqrt{1-s})/\sqrt{ts}$ and
${\theta}_{r}(t,s)=\arccos\tau_{r}(t,s)$. Then, for
$(x,t)\in{\mathbb{V}}_{0}^{d+1}$ with $x=t\xi$, $\xi\in{\mathbb{S}^{d-1}}$,

$\displaystyle{\mathsf{w}}\big{(}{\mathsf{c}}((x,t),r)\big{)}$
$\displaystyle=\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}s^{d-1}\int_{{\mathsf{d}}_{\SS}(\xi,\eta)\leq\tfrac{1}{2}{\theta}_{r}(t,s)}{\mathsf{w}}(s\eta,s)\mathrm{d}\sigma_{\SS}(\eta)\mathrm{d}s.$

###### Proof.

From ${\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\leq r$, we obtain
${\mathsf{d}}_{[0,1]}(t,s)\leq r$ by (4.2) and, with $x=t\xi$ and $y=s\eta$,
it follows from (4.1) that

${\mathsf{d}}_{\SS}(\xi,\eta)\leq\arccos\left(2[\tau_{r}(t,s)]^{2}-1\right)=\tfrac{1}{2}\arccos\tau_{r}(t,s)=\tfrac{1}{2}{\theta}_{r}(t,s).$

Hence, the stated identity follows from
$\mathrm{d}\sigma(y,s)=s^{d-1}\mathrm{d}\sigma_{\SS}(\eta)\mathrm{d}s$. ∎

For ${\beta}>-d$ and ${\gamma}>-1$, consider the Jacobi weight function
defined on the cone

${\mathsf{w}}_{{\beta},{\gamma}}(t)=t^{\beta}(1-t)^{\gamma},\quad 0<t<1.$

Let ${\mathsf{b}}_{{\beta},{\gamma}}$ be the normalization constant so that
${\mathsf{b}}_{{\beta},{\gamma}}{\mathsf{w}}_{{\beta},{\gamma}}$ has unit
integral on ${\mathbb{V}}_{0}^{d+1}$. Setting $y=s\eta$,
$\eta\in{\mathbb{S}^{d-1}}$, then

${\mathsf{b}}_{{\beta},{\gamma}}^{-1}=\int_{0}^{1}s^{d+{\beta}-1}(1-s)^{\gamma}\mathrm{d}s\int_{{\mathbb{S}^{d-1}}}\mathrm{d}\sigma_{\mathbb{S}^{d-1}}(\xi)=\omega_{d}\frac{\Gamma({\beta}+d)\Gamma({\gamma}+1)}{\Gamma({\beta}+{\gamma}+d+1)},$

where $\omega_{d}$ is the surface are of ${\mathbb{S}^{d-1}}$.

######  Proposition 4.6.

Let $r>0$ and $(x,t)\in{\mathbb{V}}_{0}^{d+1}$. Then for ${\beta}>-d$ and
${\gamma}>-1$,

(4.4)
$\displaystyle{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}:=\,$
$\displaystyle{\mathsf{b}}_{{\beta},{\gamma}}\int_{{\mathsf{c}}((x,t),r)}{\mathsf{w}}_{{\beta},{\gamma}}(s)\mathrm{d}\sigma(y,s)$
$\displaystyle\sim$
$\displaystyle\,r^{d}(t+r^{2})^{{\beta}+\frac{d}{2}}(1-t+r^{2})^{{\gamma}+\frac{1}{2}}.$

In particular, ${\mathsf{w}}_{{\beta},{\gamma}}$ is a doubling weight and the
doubling index ${\alpha}({\mathsf{w}}_{{\beta},{\gamma}})$, defined in (2.1),
is give by
${\alpha}({\mathsf{w}}_{{\beta},{\gamma}})=d+2\max\\{0,{\beta}+\frac{d}{2}\\}+2\max\\{0,{\gamma}+\frac{1}{2}\\}$.

###### Proof.

Without loss of generality, we assume $r$ is bounded by a small positive
number $r\leq\delta$; for example, $\delta=\frac{\pi}{12}$ will do. By
rotation symmetry, we could choose $x=te_{1}$, where $e_{1}=(1,0,\ldots,0)$.
Then, by Lemma 4.5,

$\displaystyle{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}={\omega}_{d-1}\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}s^{d-1}{\mathsf{w}}_{{\beta},{\gamma}}(s)\int_{0}^{\tfrac{1}{2}{\theta}_{r}(t,s)}(\sin{\theta})^{d-2}\mathrm{d}{\theta}\mathrm{d}s,$

where we have used the identity (cf. [12, (A.5.1)])

(4.5)
$\int_{{\mathbb{S}^{d-1}}}g({\langle}\xi,\eta{\rangle})d\sigma(\eta)={\omega}_{d-1}\int_{0}^{\pi}g(\cos{\theta})(\sin{\theta})^{d-2}\mathrm{d}{\theta}$

with ${\omega}_{d-1}$ being the surface are of $\SS^{d-2}$. Since
${\theta}\sim\sin{\theta}\sim\sqrt{1-\cos{\theta}}$, it follows that

(4.6)
$\displaystyle{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}\,$
$\displaystyle\sim\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}s^{d-1}{\mathsf{w}}_{{\beta},{\gamma}}(s)\big{(}1-\tau_{r}(t,s)\big{)}^{\frac{d-1}{2}}\mathrm{d}s.$

By its definition, $\tau_{r}(t,s)\geq 0$ and, furthermore, $\tau_{r}(t,s)\leq
1$ since we can write

(4.7) $\displaystyle 1-\tau_{r}(t,s)=\frac{\cos{\mathsf{d}}_{[0,1]}(t,s)-\cos
r}{\sqrt{t}\sqrt{s}}.$

We need to consider three cases.

Case 1. Assume $3r^{2}\leq t\leq 1-3r^{2}$. By Lemma 4.4, this implies that
$s\sim t+r^{2}$ and $1-s\sim 1-t+r^{2}$, which allows us to conclude that

$\displaystyle{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}$
$\displaystyle\,\sim(t+r^{2})^{{\beta}+\frac{d}{2}}(1-t+r^{2})^{{\gamma}+\frac{1}{2}}$
$\displaystyle\times\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}\big{(}\cos({\mathsf{d}}_{[0,1]}(t,s))-\cos
r\big{)}^{\frac{d-1}{2}}\frac{\mathrm{d}s}{\sqrt{s(1-s)}}.$

Setting $t=\sin^{2}\frac{{\theta}}{2}$ and $s=\sin^{2}\frac{\phi}{2}$ so that
${\mathsf{d}}_{[0,1]}(t,s)=|{\theta}-\phi|/2$ and the last integral is easily
seen to be

$\int_{|{\theta}-\phi|\leq 2r}\big{(}\cos\tfrac{{\theta}-\phi}{2}-\cos
r\big{)}^{\frac{d-1}{2}}\mathrm{d}\phi=c\int_{|\zeta|\leq
2r}\big{(}\sin\tfrac{\zeta-r}{2}\sin\tfrac{\zeta+r}{2}\big{)}^{\frac{d-1}{2}}\mathrm{d}\zeta\sim
r^{d}.$

This completes the proof of the first case.

Case 2. $0\leq t\leq 3r^{2}$. For $(y,s)\in{\mathsf{c}}((x,t),r)$, we also
have $\sqrt{s}\leq\sqrt{t}+r\leq(1+\sqrt{3})r$ by Lemma 4.4, which shows, in
particular, that $s\leq c(t+r^{2})$. Evidently $1-s\sim 1-t\sim 1$ in this
case. Furthermore, let $t=\sin^{2}\frac{{\theta}}{2}$ and
$s=\sin^{2}\frac{\phi}{2}$; then

$|s-t|=\left|\sin\frac{{\theta}-\phi}{2}\sin\frac{{\theta}+\phi}{2}\right|\leq
cr^{2}$

since $|{\theta}-\phi|\leq 2{\mathsf{d}}_{[0,1]}(t,s)\leq r$ and ${\theta}\leq
cr$. Now, we have a trivial upper bound $1-\tau_{r}(t,s)\leq 2$, which leads
to, by (4.6),

$\displaystyle{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}\leq
c\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r/2}s^{d+{\beta}-1}\mathrm{d}s\sim\int_{0}^{r^{2}}\phi^{d+{\beta}-1}d\phi\sim
r^{2{\beta}+2d},$

which proves the upper bound in (4.4). For the lower bound, we consider a
subset of ${\mathsf{c}}((x,t),r)$ with $\mathrm{d}_{[0,1]}(t,s)\leq r/2$.
Using the upper bound of $s$ and $t$, we then deduce

$1-\tau_{r}(t,s)=\frac{\cos d_{[0,1]}(t,s)-\cos
r}{\sqrt{t}\sqrt{s}}\geq\frac{\cos\tfrac{r}{2}-\cos
r}{(3+\sqrt{3})r^{2}}\geq\frac{2}{\pi^{2}},$

where in the last step we have used the monotonicity of the function over
$0\leq r\leq\pi/12$, which shows then

$\displaystyle{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}\,$
$\displaystyle\geq c\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r/2}s^{d+{\beta}-1}\mathrm{d}s\sim r^{2{\beta}+2d}.$

This completes the proof of this second case.

Case 3. $1-t\leq 3r^{2}$. In this case, we clearly have $s\sim t\sim 1$ for
$(y,s)\in{\mathsf{c}}((x,t),r)$. Since
${\mathsf{d}}_{[0,1]}(t,s)={\mathsf{d}}_{[0,1]}(1-t,1-s)$, changing variable
$s\mapsto 1-s$ in (4.6), we obtain

$\displaystyle{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}\sim
c\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}(1-s)^{\gamma}\mathrm{d}s=c\int_{{\mathsf{d}}_{[0,1]}(1-t,s)\leq
r}s^{\gamma}\mathrm{d}s,$

where the last integral can be estimated as in Case 2. This completes the
proof of (4.4). ∎

It is worthwhile to mention that the proof relies on the geometry of
${\mathsf{c}}((x,t),r)$ when $t\leq cr^{2}$, which we describe in the
following remark.

######  Remark 4.2.

By (4.3), $(y,s)\in{\mathsf{c}}((x,t),r)$ in equivalent to

$2\sin^{2}\frac{{\mathsf{d}}_{\SS}(\xi,\eta)}{2}=\frac{\cos
d_{[0,1]}(t,s)-\cos
d_{{\mathbb{V}}_{0}}\big{(}(x,t),(y,s)\big{)}}{\sqrt{t}\sqrt{s}}\leq\frac{\cos
d_{[0,1]}(t,s)-\cos r}{\sqrt{t}\sqrt{s}}.$

If $t\leq cr^{2}$, then the proof of the Case 2 shows, by
$\sin{\theta}\leq{\theta}$, that the above inequality holds whenever
${\mathsf{d}}_{\SS}(\xi,\eta)\leq\frac{4}{\pi^{2}}$. In other words, for any
$s$ that satisfies ${\mathsf{d}}_{[0,1]}(t,s)\leq r$, the set
${\mathsf{c}}((x,t),r)$ contains a large spherical cap
$\\{\eta:{\mathsf{d}}_{\SS}(\xi,\eta)\leq\frac{4}{\pi^{2}}\\}$.

######  Corollary 4.7.

For $d\geq 2$, ${\beta}>-d$ and ${\gamma}>-1$, the space
$({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}},{\mathsf{d}}_{{\mathbb{V}}_{0}})$
is a homogeneous space.

For convenience, we will introduce and use the function
${\mathsf{w}}_{{\beta},{\gamma},d}(n;t)$ defined by

(4.8)
${\mathsf{w}}_{{\beta},{\gamma},d}(n;t):=n^{d}{\mathsf{w}}_{{\beta},{\gamma}}\big{(}{\mathsf{c}}((x,t),n^{-1})\big{)}=\big{(}t+n^{-2}\big{)}^{{\beta}+\frac{d}{2}}\big{(}1-t+n^{-2}\big{)}^{{\gamma}+\frac{1}{2}}.$

For any $\xi$ in the unit sphere $\SS^{m-1}$, a spherical cap
${\mathsf{c}}(\xi,r)$ has $\sigma({\mathsf{c}}(\xi,r))=cr^{m-1}$ so that, for
a doubling weight $w$ on the unit sphere $\SS^{m-1}$, the function
$w(n;\xi)=n^{m-1}\sigma({\mathsf{c}}(\xi,\tfrac{1}{n}))$ is an approximation
to $w(\xi)$ by the Lebesgue differentiation theorem. The function
${\mathsf{w}}_{{\beta},{\gamma},d}(n;t)$ in (4.8), however, is not an
approximation to ${\mathsf{w}}_{{\beta},{\gamma}}(t)$ on the conic surface
since for the Lebesgue measure $\mathrm{d}\sigma$ on ${\mathbb{V}}_{0}^{d+1}$,

$\sigma\big{(}{\mathsf{c}}((x,t),n^{-1})\big{)}\sim
n^{-d}(t+n^{-2})^{\frac{d}{2}}(1-t+n^{-2})^{\frac{1}{2}}$

by (4.4), whereas it is the ratio

${\mathsf{w}}_{n}(x,t)=\frac{{\mathsf{w}}({\mathsf{c}}((x,t),n^{-1}))}{\sigma({\mathsf{c}}((x,t),n^{-1}))}$

that provides an approximation to ${\mathsf{w}}(t)$ on
${\mathbb{V}}_{0}^{d+1}$.

For the unit sphere, it can be verified that $w_{n}(\xi)=w(n;\xi)$ is a
doubling weight on $\SS^{m-1}$ whenever $w$ is, and this property has been
used to show, for example, that $\|f\|_{p,w}\sim\|f\|_{p,w_{n}}$ for all
polynomials of degree at most $n$. For a doubling measure ${\mathsf{w}}$ on
the conic surface, it is not clear, however, if ${\mathsf{w}}_{n}$ defined
above is itself a doubling weight on ${\mathbb{V}}_{0}^{d+1}$.

###  4.3. Orthogonal polynomials on the conic surface

Orthogonal structure with respect to ${\mathsf{w}}_{{\beta},{\gamma}}$ on the
conic surface was studied in [53]. For ${\beta}>-d$ and ${\gamma}>-1$, define
the inner product

${\langle}f,g{\rangle}_{{\mathsf{w}}}={\mathsf{b}}_{{\beta},{\gamma}}\int_{{\mathbb{V}}_{0}^{d+1}}f(x,t)g(x,t){\mathsf{w}}_{{\beta},{\gamma}}\mathrm{d}\sigma(x,t),$

where $\mathrm{d}\sigma$ denote the surface measure on
${\mathbb{V}}_{0}^{d+1}$, which is well-defined for all polynomials restricted
on the conic surface. Let
${\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}})$ be
the space of orthogonal polynomials of degree $n$. Since
${\mathbb{V}}_{0}^{d+1}$ is a quadratic surface in ${\mathbb{R}}^{d+1}$, the
dimension of the space
${\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}})$ is
the same as the dimension of the spherical harmonics of degree $n$ on
$\SS^{d}$; that is,
$\dim{\mathcal{V}}_{0}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}})=1$
and

$\dim{\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}})=\binom{n+d-1}{n}+\binom{n+d-2}{n-1},\quad
n=1,2,3,\ldots.$

Furthermore, let $\Pi_{n}({\mathbb{V}}_{0}^{d+1})$ denote the space of
polynomials of degree at most $n$ restricted on ${\mathbb{V}}_{0}^{d+1}$, then
it is the union of
${\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}})$
and

$\dim\Pi_{n}({\mathbb{V}}_{0}^{d+1})=\binom{n+d}{n}+\binom{n+d-1}{n-1}.$

An orthogonal basis of
${\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},\varphi_{{\beta},{\gamma}})$ can be
given in terms of the Jacobi polynomials and spherical harmonics. Let
${\mathcal{H}}_{m}({\mathbb{S}^{d-1}})$ be the space of spherical harmonics of
degree $m$ in $d$ variables. Let
$\\{Y_{\ell}^{m}:1\leq\ell\leq\dim{\mathcal{H}}_{m}({\mathbb{S}^{d-1}})\\}$
denote an orthonormal basis of ${\mathcal{H}}_{m}({\mathbb{S}^{d-1}})$. Then
the polynomials

${\mathsf{S}}_{m,\ell}^{n}(x,t)=P_{n-m}^{(2m+{\beta}+d-1,{\gamma})}(1-2t)Y_{\ell}^{m}(x),\quad
0\leq m\leq n,\,\,1\leq\ell\leq\dim{\mathcal{H}}_{m}({\mathbb{S}^{d-1}}),$

consist of an orthogonal basis of
${\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},\varphi_{{\beta},{\gamma}})$. More
precisely,

${\langle}{\mathsf{S}}_{m,\ell}^{n},{\mathsf{S}}_{m^{\prime},\ell^{\prime}}^{n^{\prime}}{\rangle}_{{\mathsf{w}}_{b,{\gamma}}}={\mathsf{H}}_{m,n}^{{\beta},{\gamma}}\delta_{n,n^{\prime}}\delta_{m,m^{\prime}}\delta_{\ell,\ell^{\prime}}$

where the norm ${\mathsf{H}}_{m,n}^{{\beta},{\gamma}}$ of
${\mathsf{S}}_{m,\ell}^{n}$ is given by

${\mathsf{H}}_{m,n}^{{\beta},{\gamma}}=\frac{c_{{\beta}+d-1,{\gamma}}}{c_{2m+{\beta}+d-1,{\gamma}}}h_{n-m}^{(2m+{\beta}+d-1,{\gamma})},$

where $c_{{\alpha},{\beta}}$ is the normalization constant in (2.14) and
$h_{m}^{({\alpha},{\beta})}$ is the norm square of the Jacobi polynomial. We
call $S_{m,\ell}^{n}$ the Jacobi polynomials on the conic surface. The
reproducing kernel of the space
${\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}})$ is
denoted by ${\mathsf{P}}_{n}({\mathsf{w}}_{{\beta},{\gamma}};\cdot,\cdot)$,
which can be written in terms of the above basis,

${\mathsf{P}}_{n}\big{(}{\mathsf{w}}_{{\beta},{\gamma}};(x,t),(y,s)\big{)}=\sum_{m=0}^{n}\sum_{k=1}^{\dim{\mathcal{H}}_{m}^{d}}\frac{{\mathsf{S}}_{m,\ell}^{n}(x,t){\mathsf{S}}_{m,\ell}^{n}(y,s)}{{\mathsf{H}}_{m,n}^{{\beta},{\gamma}}},$

and it is the kernel of the orthogonal projection operator
$\operatorname{proj}_{n}({\mathsf{w}}_{{\beta},{\gamma}}):L^{2}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{{\beta},{\gamma}})\to{\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},\varphi_{{\beta},{\gamma}})$,

$\operatorname{proj}_{n}({\mathsf{w}}_{{\beta},{\gamma}};f)=\int_{{\mathbb{V}}_{0}^{d+1}}f(y,s){\mathsf{P}}_{n}\big{(}{\mathsf{w}}_{{\beta},{\gamma}};\,\cdot,(y,s)\big{)}{\mathsf{w}}_{{\beta},{\gamma}}(s)\mathrm{d}\sigma(y,s).$

The case ${\beta}=-1$ turns out to be the most interesting case, for which
there is a second order differential operator that has orthogonal polynomials
as eigenfunctions, akin the Laplace-Beltrami operator for the spherical
harmonics.

######  Theorem 4.8.

Let $\Delta_{0}^{(\xi)}$ denote the Laplace-Beltrami operator in
$\xi\in{\mathbb{S}^{d-1}}$. Define

$\Delta_{0,{\gamma}}:=\left(t(1-t)\partial_{t}^{2}+\big{(}d-1-(d+{\gamma})t\big{)}\partial_{t}+t^{-1}\Delta_{0}^{(\xi)}\right)$

for ${\gamma}>-1$. Then the polynomials in
${\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{-1,{\gamma}})$ are
eigenfunctions of $\Delta_{0,{\gamma}}$,

$\Delta_{0,{\gamma}}u=-n(n+{\gamma}+d-1)u,\qquad\forall
u\in{\mathcal{V}}_{n}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{-1,{\gamma}}).$

The reproducing kernel enjoys an addition formula that is a mixture of the
addition formula for the spherical harmonics and the Jacobi polynomials. The
formula has the most elegant form when ${\beta}=-1$, which is stated below.

######  Theorem 4.9.

Let $d\geq 2$ and ${\gamma}\geq-\frac{1}{2}$. Then, for
$(x,t),(y,s)\in{\mathbb{V}}_{0}^{d+1}$,

(4.9)
$\displaystyle{\mathsf{P}}_{n}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}=b_{{\gamma},d}\int_{[-1,1]^{2}}$
$\displaystyle Z_{2n}^{{\gamma}+d-1}\big{(}\zeta(x,t,y,s;v)\big{)}$
$\displaystyle\times(1-v_{1}^{2})^{\frac{d-4}{2}}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v,$

where $b_{{\gamma},d}$ is a constant so that
${\mathsf{P}}_{0}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}=1$ and

$\zeta(x,t,y,s;v)=v_{1}\sqrt{\tfrac{st+{\langle}x,y{\rangle}}{2}}+v_{2}\sqrt{1-t}\sqrt{1-s};$

moreover, the identity holds under the limit (2.28) when
${\gamma}=-\frac{1}{2}$ and/or $d=2$.

In particular, the orthogonal structure for the weight function
${\mathsf{w}}_{-1,{\gamma}}$ on the conic surface satisfies both
characteristic properties specified in Definition 3.1 and Definition 3.2. We
show that ${\mathsf{w}}_{-1,{\gamma}}$ admits highly localized kernels in the
next subsection.

###  4.4. Highly localized kernels

Let $\widehat{a}$ be an admissible cut-off function. For $(x,t)$,
$(y,s)\in{\mathbb{V}}_{0}^{d+1}$, define the kernel
${\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}})$ by

${\mathsf{L}}_{n}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}=\sum_{j=0}^{\infty}\widehat{a}\left(\frac{j}{n}\right){\mathsf{P}}_{j}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}.$

We show that this kernel is highly localized. It is worthwhile to point out
that ${\mathsf{w}}_{-1,{\gamma}}$ does not include the constant weight
function, or the Lebesgue measure on the conic surface.

We shall need ${\mathsf{w}}_{-1,{\gamma}}\big{(}{\mathsf{c}}((x,t),r)\big{)}$
with $r=n^{-2}$. Following the notation (4.8), we define

(4.10)
${\mathsf{w}}_{{\gamma},d}(n;t)=\big{(}1-t+n^{-2}\big{)}^{{\gamma}+\frac{1}{2}}\big{(}t+n^{-2}\big{)}^{\frac{d-2}{2}}.$

######  Theorem 4.10.

Let $d\geq 2$ and ${\gamma}\geq-\frac{1}{2}$. Let $\widehat{a}$ be an
admissible cutoff function. Then for any ${\kappa}>0$ and
$(x,t),(y,s)\in{\mathbb{V}}_{0}^{d+1}$,

$\left|{\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s))\right|\leq\frac{c_{\kappa}n^{d}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\big{)}^{-{\kappa}}.$

###### Proof.

We only prove the case when $d>2$ and ${\gamma}>-\frac{1}{2}$. The remaining
cases ${\gamma}=-\frac{1}{2}$ and/or $d=2$ follow similarly and are easier.
Using the addition formula for the reproducing kernel and, by the quadratic
transform (2.21),

(4.11)
$Z_{2n}^{\lambda}(x)=\frac{P_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(1)P_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(2x^{2}-1)}{h_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}},$

we can write ${\mathsf{L}}({\mathsf{w}}_{{\beta},{\gamma}})$ in terms of the
kernel $L_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}$ of the Jacobi
polynomials, where ${\lambda}={\gamma}+d-1$. Then

(4.12)
$\displaystyle{\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s))=c_{{\gamma}}\int_{[-1,1]^{2}}$
$\displaystyle
L_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}\big{(}2\zeta(x,t,y,s;v)^{2}-1\big{)}$
$\displaystyle\times(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v.$

Let $\theta(x,t,y,s;v)=\arccos(2\zeta(x,t,y,s;v)^{2}-1)$. Then

$1-\zeta(x,t,y,s;t)^{2}=\frac{1}{2}(1-\cos\theta(x,t,y,s;v))=\sin^{2}\frac{\theta(x,t,y,s;v)}{2}\sim\theta(x,t,y,s;v)^{2}.$

We apply the estimate (2.17) for $L_{n}^{{\alpha},{\beta}}$ with
$\alpha={\lambda}-1/2$, $\beta=-1/2$ to obtain

$\displaystyle\left|{\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s))\right|\leq
cn^{2{\lambda}+1}\int_{[-1,1]^{2}}$
$\displaystyle\frac{1}{\left(1+n\sqrt{1-\zeta(x,t,y,s;v)^{2}}\right)^{{\kappa}+2{\gamma}+d+1}}$
$\displaystyle\times(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v.$

From its definition, it is easy to verify that $|\zeta(x,t,y,s;v)|\leq 1$ and

$\displaystyle 1-\zeta(x,t,y,s;v)=\,$ $\displaystyle
1-\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))$ $\displaystyle+$
$\displaystyle\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}(1-v_{1})+\sqrt{1-s}\sqrt{1-t}(1-v_{2}).$

Consequently, since $ts+{\langle}x,y{\rangle}\geq 0$, it follows that

$\displaystyle 1-\zeta(x,t,y,s;v)\,$ $\displaystyle\geq
1-\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))$
$\displaystyle=2\sin^{2}\frac{{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))}{2}\geq\frac{2}{\pi^{2}}[{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))]^{2}.$

Applying this inequality, we obtain the estimate

$\displaystyle\left|{\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s))\right|\,$
$\displaystyle\leq
cn^{2{\lambda}+1}\frac{1}{\left(1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\right)^{\kappa}}$
$\displaystyle\times
b_{{\gamma},d}\int_{[-1,1]^{2}}\frac{(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}}{\left(1+n\sqrt{1-\zeta(x,t,y,s;v)}\right)^{2{\gamma}+d+1}}\mathrm{d}v,$

where
$b_{{\gamma},d}=c_{{\gamma}-\frac{1}{2},{\gamma}-\frac{1}{2}}c_{\frac{d-4}{2},\frac{d-4}{2}}$.
The estimate of the last expression is the crux of the proof and it is
contained in the lemma below. ∎

######  Lemma 4.11.

Let $d\geq 2$ and ${\gamma}>-\frac{1}{2}$. Then, for ${\beta}\geq
2{\gamma}+d+1$,

$\displaystyle b_{{\gamma},d}\int_{[-1,1]^{2}}$
$\displaystyle\frac{(1-v_{1})^{\frac{d-2}{2}-1}(1-v_{2})^{{\gamma}-\frac{1}{2}}}{\big{(}1+n\sqrt{1-\zeta(x,t,y,s;v)}\,\big{)}^{{\beta}}}\mathrm{d}v$
$\displaystyle\qquad\leq\frac{cn^{-(2{\gamma}+d-1)}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\big{)}^{{\beta}-3{\gamma}-\frac{3d+1}{2}}}.$

###### Proof.

Using the lower bound of $1-|\zeta(x,t,y,s;v)|$ in the proof of the previous
theorem, the left-hand side of the stated inequality has the upper bound

$\frac{c}{\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\big{)}^{{\beta}-2{\gamma}-d-1}}\int_{[-1,1]^{2}}\frac{(1-v_{1})^{\frac{d-2}{2}-1}(1-v_{2})^{{\gamma}-\frac{1}{2}}}{(1+n\sqrt{1-\zeta(x,t,y,s;v)})^{2{\gamma}+d+1}}\mathrm{d}v.$

Denote the above integral by $I(x,t,y,s)$. To complete the proof, we need to
show that

$\frac{I(x,t,y,s)}{\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\big{)}^{{\gamma}+\frac{d-1}{2}}}\leq\frac{cn^{-(2{\gamma}+d-1)}}{\sqrt{{\mathsf{w}}_{{\gamma}}(n;t)}\sqrt{{\mathsf{w}}_{{\gamma}}(n;s)}}.$

We need a lower bound for $1-\zeta$. Let
${\alpha}=\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}$ and
${\beta}=\sqrt{1-t}\sqrt{1-s}$. Then

$\displaystyle 1-\zeta(x,t,y,s;v)\,$
$\displaystyle=1-{\alpha}+(1-v_{1}){\alpha}-v_{2}{\beta}\geq
1-{\alpha}+\frac{1}{2}(1-v_{1}){\alpha}-v_{2}{\beta}$
$\displaystyle=1-\frac{1+v_{1}}{2}{\alpha}-\frac{1-v_{1}}{2}\sqrt{ts}+\frac{1-v_{1}}{2}\sqrt{ts}-v_{2}{\beta}.$

Using ${\alpha}\leq\sqrt{ts}$ and $1-\sqrt{ts}-\sqrt{1-t}\sqrt{1-s}\geq 0$, we
then obtain

(4.13) $\displaystyle 1-\zeta(x,t,y,s;v)\,$ $\displaystyle\geq
1-\sqrt{ts}-{\beta}+\frac{1}{2}(1-v_{1})\sqrt{ts}+(1-v_{2}){\beta}$
$\displaystyle\geq\frac{1}{2}(1-v_{1})\sqrt{t}\sqrt{s}+(1-v_{2})\sqrt{1-s}\sqrt{1-t}.$

Using this inequality in $I(x,t,y,s)$ and then making a change of variable
$v_{1}\mapsto 2u_{1}-1$ and $v_{2}\mapsto 2u_{2}-1$, we obtain

$\displaystyle I(x,t,y,s)\leq c\int_{[0,1]^{2}}$
$\displaystyle\frac{(1-v_{1})^{\frac{d-2}{2}-1}(1-v_{2})^{{\gamma}-\frac{1}{2}}}{\big{(}1+n\sqrt{1-(1-v_{1})\sqrt{t}\sqrt{s}-2(1-v_{2})\sqrt{1-s}\sqrt{1-t}}\,\big{)}^{2{\gamma}+d+1}}\mathrm{d}v.$

This integral can be estimated by using the inequality [12, (13.5.8)]

(4.14)
$\int_{0}^{1}\frac{(1-t)^{a-1}\mathrm{d}t}{(1+n\sqrt{B+A(1-t)})^{b}}\leq
c\frac{n^{-2a}}{A^{a}(1+n\sqrt{B})^{b-2a-1}},$

which holds for $A>0$, $B\geq 0$, $a>0$ and $b\geq 2a+1$. Applying this
inequality with $A=\sqrt{1-t}\sqrt{1-s}$, $a={\gamma}+\frac{1}{2}$, we obtain

$I(x,t,y,s)\leq\frac{cn^{-2{\gamma}-1}}{(\sqrt{1-t}\sqrt{1-s})^{{\gamma}+\frac{1}{2}}}\int_{0}^{1}\frac{(1-v_{1}^{2})^{\frac{d-2}{2}-1}}{\left(1+n\sqrt{\sqrt{t}\sqrt{s}(1-v_{1})}\right)^{d-1}}\mathrm{d}v_{1}.$

Applying (4.14) one more time with $B=0$, $a=\frac{d-2}{2}$ and $b=d-1$, we
see that the last integral is bounded by $cn^{-(d-2)}$, hence,

$\displaystyle I(x,t,y,s)\,$
$\displaystyle\leq\frac{cn^{-(2{\gamma}+d-1)}}{(\sqrt{1-t}\sqrt{1-s})^{{\gamma}+\frac{1}{2}}(\sqrt{t}\sqrt{s})^{\frac{d-2}{2}}}$
$\displaystyle\leq\frac{cn^{-(2{\gamma}+d-1)}}{(\sqrt{1-t}\sqrt{1-s}+n^{-2})^{{\gamma}+\frac{1}{2}}(\sqrt{t}\sqrt{s}+n^{-2})^{\frac{d-2}{2}}},$

where the second inequality follows since $I(x,t,y,s)\leq 1$ holds trivially
by the choice of $b_{{\gamma},\mathrm{d}}$. Now, using the elementary identity
[12, (11.5.13)]

(4.15) $(a+n^{-1})(b+n^{-1})\leq 3(ab+n^{-2})(1+n|b-a|)$

with $a=\sqrt{t}$ and $b=\sqrt{s}$ as well as with $a=\sqrt{1-t}$ and
$b=\sqrt{1-s}$, we obtain

$I(x,t,y,s)\leq\frac{cn^{-(2{\gamma}+d-1)}\big{(}1+n\big{|}\sqrt{t}-\sqrt{s}\big{|}\big{)}^{\frac{d-2}{2}}\big{(}1+n\big{|}\sqrt{1-t}-\sqrt{1-s}\big{|}\big{)}^{{\gamma}+\frac{1}{2}}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}}.$

Finally, by Lemma 4.4, we conclude that

$I(x,t,y,s)\leq\frac{cn^{-(2{\gamma}+d-1)}(1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s)))^{{\gamma}+\frac{d-1}{2}}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}},$

which is what we need to complete the proof. ∎

The following corollary, following immediately from (4.13), will be used
later.

######  Corollary 4.12.

Let $d\geq 2$ and ${\gamma}>-\frac{1}{2}$. Then, for ${\beta}\geq
2{\gamma}+d+1$,

$\displaystyle b_{{\gamma},d}\int_{[-1,1]^{2}}$
$\displaystyle\frac{(1-v_{1})^{\frac{d-2}{2}-1}(1-v_{2})^{{\gamma}-\frac{1}{2}}}{\left(1+n\sqrt{1-(1-v_{1})\sqrt{t}\sqrt{s}-(1-v_{2})\sqrt{1-s}\sqrt{1-t}}\,\right)^{2{\gamma}+\mathrm{d}+1}}\mathrm{d}v$
$\displaystyle\qquad\leq\frac{cn^{-(2{\gamma}+d-1)}\big{(}1+n{\mathsf{d}}_{[0,1]}((x,t),(y,s))\big{)}^{{\gamma}+\frac{d-1}{2}}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}}.$

By (4.8), we have shown that ${\mathsf{w}}_{-1,{\gamma}}$ admits Assertion 1
of the highly localized kernel. Our next result shows that it also admits
Assertion 2.

######  Theorem 4.13.

Let $\widehat{a}$ be an admissible cutoff function. Let $d\geq 2$ and
${\gamma}\geq-\frac{1}{2}$. For $(x_{i},t_{i}),(y,s)\in{\mathbb{V}}_{0}^{d+1}$
and $(x_{1},t_{1})\in{\mathsf{c}}\big{(}(x_{2},t_{2}),c^{*}n^{-1}\big{)}$ with
$c^{*}$ small and for ${\kappa}>0$,

(4.16)
$\displaystyle\left|{\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}};(x_{1},t_{1}),(y,s))-{\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}};(x_{2},t_{2}),(y,s))\right|$
$\displaystyle\qquad\leq
c_{\kappa}\frac{n^{d+1}{\mathsf{d}}_{{\mathbb{V}}_{0}}((x_{1},t_{1}),(x_{2},t_{2}))}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t_{2})}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((y,s),(x_{2},t_{2}))\big{)}^{{\kappa}}}.$

###### Proof.

Denote the left-hand side of (4.17) by $K$. Let $\partial L(u)=L^{\prime}(u)$.
Using the integral expression (4.12) of
${\mathsf{L}}_{n}({\mathsf{w}}_{-1,{\gamma}})$, we obtain

(4.17) $\displaystyle K$ $\displaystyle\leq 2\int_{[-1,1]^{2}}\big{\|}\partial
L_{n}^{{\lambda}-\frac{1}{2},-\frac{1}{2}}\big{(}2(\cdot)^{2}-1\big{)}\big{\|}_{L^{\infty}(I_{v})}\big{|}\zeta_{1}(v)^{2}-\zeta_{2}(v)^{2}|$
$\displaystyle\qquad\qquad\qquad\times(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v,$

where $\zeta_{i}(v)=\zeta(x_{i},t_{i},y,s;v)$, and $I_{v}$ is the interval
with end points $\zeta_{1}(v)$ and $\zeta_{2}(v)$. Since $|\zeta(\cdot)|\leq
1$, $|\zeta_{1}(v)^{2}-\zeta_{2}(v)^{2}|\leq 2|\zeta_{1}(v)-\zeta_{2}(v)|$. We
claim that

(4.18) $\displaystyle|\zeta_{1}(v)-\zeta_{2}(v)|\leq
c\,{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x_{1},t_{1}),(x_{2},t_{2})\big{)}\big{(}\Sigma_{1}+\Sigma_{2}(v_{1})+\Sigma_{3}(v_{2})\big{)},$

where

$\displaystyle\Sigma_{1}\,$
$\displaystyle={\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x_{i},t_{i}),(y,s)\big{)}+{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x_{1},t_{1}),(x_{2},t_{2})\big{)},$
$\displaystyle\Sigma_{2}(v_{1})\,$ $\displaystyle=(1-v_{1})\sqrt{s},$
$\displaystyle\Sigma_{3}(v_{2})\,$ $\displaystyle=(1-v_{2})\sqrt{1-s}.$

To see this, we use (4.2) and writing $x_{i}=t_{i}\xi_{i}$ and $y=s\eta$ to
obtain

$\displaystyle\zeta_{1}(v)-\zeta_{2}(v)\,$
$\displaystyle=\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x_{1},t_{1}),(y,s)\big{)}-\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x_{2},t_{2}),(y,s)\big{)}$
$\displaystyle+(1-v_{1})\left(\sqrt{t_{2}s}\cos\frac{{\mathsf{d}}_{\SS}(\xi_{2},\eta)}{2}-\sqrt{t_{1}s}\cos\frac{{\mathsf{d}}_{\SS}(\xi_{1},\eta)}{2}\right)$
$\displaystyle+(1-v_{2})\left(\sqrt{1-t_{2}}-\sqrt{1-t_{1}}\right)\sqrt{1-s}.$

Denote temporarily
${\alpha}_{i}={\mathsf{d}}_{{\mathbb{V}}_{0}}((x_{1},t_{1}),(y,s))$ for
$i=1,2$. Hence, using the identity

$\displaystyle\cos{\alpha}_{1}-\cos{\alpha}_{2}\,$
$\displaystyle=2\sin\frac{{\alpha}_{1}-{\alpha}_{2}}{2}\sin\frac{{\alpha}_{1}+{\alpha}_{2}}{2}$
$\displaystyle=2\sin\frac{{\alpha}_{1}-{\alpha}_{2}}{2}\left(2\sin\frac{{\alpha}_{i}}{2}+\sin\frac{|{\alpha}_{1}-{\alpha}_{2}|}{2}\right),$

it follows readily that

$|\cos{\alpha}_{1}-\cos{\alpha}_{2}|\leq|{\alpha}_{1}-{\alpha}_{2}|\left(|{\alpha}_{1}|+\tfrac{1}{2}|{\alpha}_{1}-{\alpha}_{2}|\right).$

By the triangle inequality of ${\mathsf{d}}_{{\mathbb{V}}_{0}}$,
$|{\alpha}_{1}-{\alpha}_{2}|\leq{\mathsf{d}}_{{\mathbb{V}}_{0}}((x_{1},t_{1}),(x_{2},t_{2}))$,
this gives the estimate for the $\Sigma_{1}$ term. Moreover, assuming
$t_{2}\geq t_{1}$, for example, and applying similar argument for
${\mathsf{d}}_{\SS}$, we then obtain, using Lemma 4.4 and (4.3), that

$\displaystyle\bigg{|}\sqrt{t_{1}}\cos\frac{{\mathsf{d}}_{\SS}(\xi_{1},\eta)}{2}-$
$\displaystyle\sqrt{t_{2}}\cos\frac{{\mathsf{d}}_{\SS}(\xi_{2},\eta)}{2}\bigg{|}$
$\displaystyle\leq\left|\sqrt{t_{2}}-\sqrt{t_{1}}\right|+\sqrt{t_{1}}\left|\cos\frac{{\mathsf{d}}_{\SS}(\xi_{1},\eta)}{2}-\cos\frac{{\mathsf{d}}_{\SS}(\xi_{2},\eta)}{2}\right|$
$\displaystyle\leq{\mathsf{d}}_{{\mathbb{V}}_{0}}((x_{1},t_{1}),(x_{2},t_{2}))+\frac{1}{2}(t_{1}t_{2})^{\frac{1}{4}}{\mathsf{d}}_{\SS}(\xi_{1},\xi_{2})$
$\displaystyle\leq
c\,{\mathsf{d}}_{{\mathbb{V}}_{0}}((x_{1},t_{1}),(x_{2},t_{2})),$

which verifies the $\Sigma_{2}(v_{1})$ term. The third term with
$\Sigma_{3}(v_{2})$ follows from using Lemma 4.4 one more time. This verifies
the claim (4.18).

Since $\max_{r\in I_{v}}|1+n\sqrt{1-r}|^{-\sigma}$ is attained at one of the
end points of the interval, it follows from (2.18) with $m=1$.

$\big{\|}\partial
L_{n}^{{\lambda}-\frac{1}{2},-\frac{1}{2}}\big{(}2(\cdot)^{2}-1\big{)}\big{\|}_{L^{\infty}(I_{v})}\leq
c\left[\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\zeta_{1}(v)^{2}}\big{)}^{{\kappa}}}+\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\zeta_{2}(v)^{2}}\big{)}^{{\kappa}}}\right].$

Consequently, we see that $K$ is bounded by a sum of integrals

$\displaystyle K\leq
c\,{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x_{1},t_{1}),(x_{2},t_{2})\big{)}$
$\displaystyle\int_{[-1,1]^{2}}\left[\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\zeta_{1}(v)^{2}}\big{)}^{{\kappa}}}+\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\zeta_{2}(v)^{2}}\big{)}^{{\kappa}}}\right]$
$\displaystyle\quad\times\big{(}\Sigma_{1}+\Sigma_{2}(v)+\Sigma_{3}(v)\big{)}(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v.$

Since $(x_{1},t_{1})\in{\mathsf{c}}\big{(}(x_{2},t_{2}),c^{*}n^{-1}\big{)}$,
$\Sigma_{1}$ is bounded by $\Sigma_{1}\leq
cn^{-1}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x_{i},t_{i}),(y,s)\big{)}\big{)}$.
Hence, we obtain from Lemma 4.11 that

$\displaystyle\int_{[-1,1]^{2}}$
$\displaystyle\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\zeta_{i}(v)^{2}}\big{)}^{{\kappa}}}\Sigma_{1}(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v$
$\displaystyle\leq
c\frac{n^{d+1}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t_{i})}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((y,s),(x_{i},t_{i}))\big{)}^{{\kappa}({\gamma},d)-1}},$

where ${\kappa}({\gamma},d)={\kappa}-3{\gamma}-\frac{3d+1}{2}$ for either
$i=1$ or $i=2$. Since
${\mathsf{w}}_{{\gamma},d}(n,t_{1})\sim{\mathsf{w}}_{{\gamma},d}(n,t_{2})$ and
${\mathsf{d}}_{{\mathbb{V}}_{0}}((x_{1},t_{1}),(y,s))+n^{-1}\sim{\mathsf{d}}_{{\mathbb{V}}_{0}}((x_{2},t_{2}),(y,s))+n^{-1}$
by Lemma 4.4, we can replace $(x_{1},t_{1})$ in the right-hand side by
$(x_{2},t_{2})$. This shows that the integral containing $\Sigma_{1}$ has the
desired estimate.

For the remaining integrals, the same consideration shows that we only need to
consider those containing $\zeta_{2}(v)$. For the integral that contains
$\Sigma_{2}(v_{1})=(1-v_{1})\sqrt{s}$, the factor $(1-v_{1})$ increases the
power of the weight to $(1-v_{1})^{\frac{d}{2}}$, so that we can apply Lemma
4.11 with $\frac{d-2}{2}$ replaced by $\frac{d}{2}$, which leads to

$\displaystyle\int_{-1}^{1}$
$\displaystyle\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\zeta_{2}(v)^{2}}\big{)}^{{\kappa}}}\Sigma_{2}(v_{1})(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v$
$\displaystyle\leq
c\frac{n^{d+1}n^{-1}\sqrt{s}}{\sqrt{{\mathsf{w}}_{{\gamma},d+1}(n;s)}\sqrt{{\mathsf{w}}_{{\gamma},d+1}(n;t_{2})}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((y,s),(x_{2},t_{2}))\big{)}^{{\kappa}({\gamma},d+1)}}$
$\displaystyle\leq
c\frac{n^{d+1}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t_{2})}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((y,s),(x_{2},t_{2}))\big{)}^{{\kappa}({\gamma},d+1)}},$

where the last step follows from the inequality
$n^{-1}\sqrt{s}\leq(\sqrt{t_{2}}+n^{-1})(\sqrt{s}+n^{-1})$. The integral that
contains $\Sigma_{3}(v_{2})$ can be estimated similarly by applying Lemma 4.11
with ${\gamma}$ replaced by ${\gamma}+1$ and using
$n^{-1}\sqrt{1-s}\leq(\sqrt{1-t_{2}}+n^{-1})(\sqrt{1-s}+n^{-1})$. This
completes the proof. ∎

The first two assertions for the highly localized kernels are established for
${\mathsf{w}}_{-1,{\gamma}}$ when ${\gamma}\geq-\frac{1}{2}$. The case of
$p=1$ of the following lemma establishes Assertion 3. Recall that
${\mathsf{w}}_{{\beta},{\gamma},d}(n;t)$ is defined in (4.8).

######  Lemma 4.14.

Let $d\geq 2$, ${\beta}>-d$ and ${\gamma}>-1$. For $0<p<\infty$, assume
${\kappa}>\frac{2d}{p}+({\gamma}+{\beta}+\frac{d+1}{2})|\frac{1}{p}-\frac{1}{2}|$.
Then for $(x,t)\in{\mathbb{V}}_{0}^{d+1}$,

(4.19)
$\displaystyle\int_{{\mathbb{V}}_{0}^{d+1}}\frac{{\mathsf{w}}_{{\beta},{\gamma}}(s)\mathrm{d}\sigma(y,s)}{{\mathsf{w}}_{{\beta},{\gamma},d}(n;s)^{\frac{p}{2}}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\big{)}^{{\kappa}p}}\leq
cn^{-d}{\mathsf{w}}_{{\beta},{\gamma},d}(n;t)^{1-\frac{p}{2}}.$

###### Proof.

Let $J_{p}$ denote the left-hand side of (4.19). By Lemma 2.4, it is
sufficient to estimate $J_{2}$. Let $x=t\xi$ and $y=s\eta$. The definition of
${\mathsf{d}}_{{\mathbb{V}}_{0}}(\cdot,\cdot)$ shows that it is a function of
${\langle}\xi,\eta{\rangle}$. Hence, by (4.5), we deduce that

$\displaystyle J_{2,{\kappa}}\,$ $\displaystyle\leq
c\int_{0}^{1}\int_{-1}^{1}\frac{s^{d-1}{\mathsf{w}}_{{\beta},{\gamma}}(s)(1-u^{2})^{\frac{d-3}{2}}}{{\mathsf{w}}_{{\beta},{\gamma},d}(n;s)\big{(}1+n\arccos\big{(}\sqrt{ts}\sqrt{\frac{1+u}{2}}+\sqrt{1-t}\sqrt{1-s}\big{)}\big{)}^{2{\kappa}}}\mathrm{d}u\mathrm{d}s$
$\displaystyle\leq
c\int_{0}^{1}\int_{0}^{1}\frac{s^{d-1}{\mathsf{w}}_{{\beta},{\gamma}}(s)v^{d-2}(1-v^{2})^{\frac{d-3}{2}}}{{\mathsf{w}}_{{\beta},{\gamma},d}(n;s)\left(1+n\sqrt{1-\sqrt{ts}v-\sqrt{1-t}\sqrt{1-s}}\right)^{2k}}\mathrm{d}v\mathrm{d}s,$

where the second step follows from changing variable
$\sqrt{\frac{1+u}{2}}\mapsto v$ and the relation
${\theta}\sim\sin\frac{{\theta}}{2}\sim\sqrt{1-\cos{\theta}}$. Making a
further changing of variable $v\mapsto z/\sqrt{s}$ gives

$\displaystyle J_{2}\,$ $\displaystyle\leq
c\int_{0}^{1}\int_{0}^{\sqrt{s}}\frac{s\,{\mathsf{w}}_{{\beta},{\gamma}}(s)z^{d-2}(s-z^{2})^{\frac{d-3}{2}}}{{\mathsf{w}}_{{\beta},{\gamma},d}(n;s)\left(1+n\sqrt{1-\sqrt{t}\,z-\sqrt{1-t}\sqrt{1-s}}\right)^{2{\kappa}}}\mathrm{d}z\mathrm{d}s$
$\displaystyle\leq
c\int_{0}^{1}\int_{0}^{\sqrt{s}}\frac{(s-z^{2})^{\frac{d-3}{2}}}{(1-s+n^{-2})^{\frac{1}{2}}\left(1+n\sqrt{1-\sqrt{t}\,z-\sqrt{1-t}\sqrt{1-s}}\right)^{2{\kappa}}}\mathrm{d}z\mathrm{d}s,$

where we have used $sz^{d-2}\leq s^{\frac{d}{2}}\leq(s+n^{-2})^{\frac{d}{2}}$.
One more change of variable $s\mapsto 1-w^{2}$ with $\mathrm{d}s=w\mathrm{d}w$
and $w\leq(w^{2}+n^{-2})^{\frac{1}{2}}$, we obtain

$\displaystyle J_{2}\leq
c\int_{0}^{1}\int_{0}^{\sqrt{1-w^{2}}}\frac{(1-w^{2}-z^{2})^{\frac{d-3}{2}}}{\left(1+n\sqrt{1-\sqrt{t}z-\sqrt{1-t}\,w}\right)^{2{\kappa}}}\mathrm{d}z\mathrm{d}w,$

which is an integral over the positive quadrant
$\\{(z,w)\in{\mathbb{B}}^{2}:w\geq 0,z\geq 0\\}$ of the unit disk
${\mathbb{B}}^{2}$. Setting $p=\sqrt{t}z+\sqrt{1-t}\,w$ and
$q=-\sqrt{1-t}z+\sqrt{t}w$ in the integral, which is an orthogonal
transformation, and enlarging the integral domain while taking into account
that $p\geq 0$, it follows that

$\displaystyle J_{2}\,$ $\displaystyle\leq
c\int_{0}^{1}\frac{1}{\left(1+n\sqrt{1-p}\right)^{2{\kappa}}}\int_{-\sqrt{1-p^{2}}}^{\sqrt{1-p^{2}}}(1-p^{2}-q^{2})^{\frac{d-3}{2}}\mathrm{d}q\mathrm{d}p$
$\displaystyle\leq
c\int_{0}^{1}\frac{(1-p^{2})^{\frac{d-2}{2}}}{\left(1+n\sqrt{1-p}\right)^{2{\kappa}}}\leq
cn^{-d}\int_{0}^{n}\frac{r^{d-1}}{(1+r)^{\kappa}}\mathrm{d}r\leq cn^{-d}$

by setting $r=n\sqrt{1-p}$ and recalling that ${\kappa}>d$. This completes the
proof. ∎

######  Proposition 4.15.

Let $d\geq 2$ and ${\gamma}\geq-\frac{1}{2}$. For $0<p<\infty$ and
$(x,t)\in{\mathbb{V}}_{0}^{d+1}$,

$\int_{{\mathbb{V}}_{0}^{d+1}}\left|{\mathsf{L}}_{n}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}\right|^{p}{\mathsf{w}}_{-1,{\gamma}}(s)\mathrm{d}\sigma(y,s)\leq
c\left(\frac{n^{d}}{{\mathsf{w}}_{{\gamma},d}(n;t)}\right)^{p-1}.$

The proof is immediate by (i) of Theorem 4.10 and Lemma 4.14.

######  Corollary 4.16.

For ${\gamma}\geq-\frac{1}{2}$, the space
$({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{-1,{\gamma}},{\mathsf{d}}_{{\mathbb{V}}_{0}})$
is a localizable homogeneous space.

###  4.5. Maximal ${\varepsilon}$-separated sets and MZ inequality

Let ${\varepsilon}>0$. An ${\varepsilon}$-separated set is defined in
Definition 2.13. In this subsection, we provide a construction of some
examples of maximal ${\varepsilon}$-separated points on the conic surface.

For our construction, we shall need ${\varepsilon}$-separated points on the
unit sphere ${\mathbb{S}^{d-1}}$. We adopt the following notation. For
${\varepsilon}>0$, we denote by $\Xi_{\SS}({\varepsilon})$ a maximal
${\varepsilon}$-separated set on the unit sphere ${\mathbb{S}^{d-1}}$ and we
let $\SS_{\xi}({\varepsilon})$ be the subsets in ${\mathbb{S}^{d-1}}$ so that
the collection $\\{\SS_{\xi}({\varepsilon}):\xi\in\Xi_{\SS}({\varepsilon})\\}$
is a partition of ${\mathbb{S}^{d-1}}$, and we assume

(4.20)
${\mathsf{c}}_{\SS}(\xi,c_{1}{\varepsilon})\subset\SS_{\xi}({\varepsilon})\subset{\mathsf{c}}_{\SS}(\xi,c_{2}{\varepsilon}),\qquad\xi\in\Xi_{\SS}({\varepsilon}),$

where ${\mathsf{c}}_{\SS}(\xi,{\varepsilon})$ denotes the spherical cap
centered at $\xi$ with radius ${\varepsilon}$, $c_{1}$ and $c_{2}$ depending
only on $d$. Such a $\Xi_{\SS}({\varepsilon})$ exists for all
${\varepsilon}>0$, see for example [12, Section 6.4], and its cardinality
satisfies

(4.21) $c_{d}^{\prime}{\varepsilon}^{-d+1}\leq\\#\Xi_{\SS}({\varepsilon})\leq
c_{d}{\varepsilon}^{-d+1}.$

We now consider the subsets of points on the cone ${\mathbb{V}}_{0}^{d+1}$.
For ${\varepsilon}>0$, we denote by
$\Xi_{{\mathbb{V}}_{0}}=\Xi_{{\mathbb{V}}_{0}}({\varepsilon})$ an
${\varepsilon}$-separated set and denote by
$\\{{\mathbb{V}}_{0}(\xi,t):(t\xi,t)\in\Xi_{{\mathbb{V}}_{0}}\\}$ a partition
of ${\mathbb{V}}_{0}^{d+1}$. We start with an explicit construction of such
$\Xi_{{\mathbb{V}}_{0}}$ and ${\mathbb{V}}_{0}(\xi,t)$.

Let ${\varepsilon}>0$ and let
$N=\lfloor\frac{\pi}{2}{\varepsilon}^{-1}\rfloor$. For $1\leq j\leq N$ we
define

${\theta}_{j}:=\frac{(2j-1)\pi}{2N},\qquad{\theta}_{j}^{-}:={\theta}_{j}-\frac{\pi}{2N}\quad\hbox{and}\quad{\theta}_{j}^{+}:={\theta}_{j}+\frac{\pi}{2N}.$

Let $t_{j}=\sin^{2}\frac{{\theta}_{j}}{2}$ and define $t_{j}^{-}$ and
$t_{j}^{+}$ accordingly. In particular, $t_{1}^{-}=0$ and $t_{N}^{+}=1$. Then
${\theta}_{j+1}^{-}={\theta}_{j}^{+}$ and ${\mathbb{V}}_{0}^{d+1}$ can be
partitioned by

${\mathbb{V}}_{0}^{d+1}=\bigcup_{j=1}^{N}{\mathbb{V}}_{0}^{(j)},\quad\hbox{where}\quad{\mathbb{V}}_{0}^{(j)}:=\left\\{(x,t)\in{\mathbb{V}}_{0}^{d+1}:t_{j}^{-}<t\leq
t_{j}^{+}\right\\}.$

Let ${\varepsilon}_{j}:=(2\sqrt{t_{j}})^{-1}\pi{\varepsilon}$. Then
$\Xi_{\SS}({\varepsilon}_{j})$ is the maximal ${\varepsilon}_{j}$-separated
set of ${\mathbb{S}^{d-1}}$ such that
$\\{\SS_{\xi}({\varepsilon}_{j}):\xi\in\Xi_{\SS}({\varepsilon}_{j})\\}$ is a
partition
${\mathbb{S}^{d-1}}=\bigcup_{\eta\in\Xi_{\SS}({\varepsilon}_{j})}\SS_{\eta}({\varepsilon}_{j})$,
and

$\\#\Xi_{\SS}({\varepsilon}_{j})\sim{\varepsilon}_{j}^{-d+1}.$

For each $j=1,\ldots,N$, we decompose ${\mathbb{V}}_{0}^{(j)}$ by

${\mathbb{V}}_{0}^{(j)}=\bigcup_{\xi\in\Xi_{\SS}({\varepsilon}_{j})}{\mathbb{V}}_{0}(\xi,t_{j}),\quad\hbox{where}\quad{\mathbb{V}}_{0}(\xi,t_{j}):=\left\\{(t\eta,t):t_{j}^{-}<t\leq
t_{j}^{+},\,\eta\in\SS_{\xi}({\varepsilon}_{j})\right\\}.$

Finally, we define the subset $\Xi_{{\mathbb{V}}_{0}}$ of
${\mathbb{V}}_{0}^{d+1}$ by

$\Xi_{{\mathbb{V}}_{0}}=\big{\\{}(t_{j}\xi,t_{j}):\,\xi\in\Xi_{\SS}({\varepsilon}_{j}),\,1\leq
j\leq N\big{\\}}.$

######  Proposition 4.17.

Let ${\varepsilon}>0$ and $N=\lfloor\frac{\pi}{2}{\varepsilon}^{-1}\rfloor$.
Then $\Xi_{{\mathbb{V}}_{0}}$ is a maximal ${\varepsilon}$-separated set of
${\mathbb{V}}_{0}^{d+1}$ and
$\\{{\mathbb{V}}_{0}(\xi,t_{j}):\xi\in\Xi_{\SS}({\varepsilon}_{j}),\,1\leq
j\leq N\\}$ is a partition

${\mathbb{V}}_{0}^{d+1}=\bigcup_{j=1}^{N}\bigcup_{\xi\in\Xi_{\SS}({\varepsilon}_{j})}{\mathbb{V}}_{0}(\xi,t_{j}).$

Moreover, there are positive constants $c_{1}$ and $c_{2}$ depending only on
$d$ such that

(4.22)
${\mathsf{c}}\big{(}(t_{j}\xi,t_{j}),c_{1}{\varepsilon}\big{)}\subset{\mathbb{V}}_{0}(\xi,t_{j})\subset{\mathsf{c}}\big{(}(t_{j}\xi,t_{j}),c_{2}{\varepsilon}\big{)},$

and $c_{d}^{\prime}$ and $c_{d}$ such that

$c_{d}^{\prime}{\varepsilon}^{-d}\leq\\#\Xi_{{\mathbb{V}}_{0}}\leq
c_{d}{\varepsilon}^{-d}.$

###### Proof.

Let $(t_{j}\xi,t_{j})$ and $(t_{k}\eta,t_{k})$ be two distinct points in
$\Xi_{{\mathbb{V}}_{0}}$. If $t_{j}\neq t_{k}$, then

${\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(t_{j}\xi,t_{j}),(t_{k}\eta,t_{k})\big{)}\geq{\mathsf{d}}_{[0,1]}(t_{j},t_{k})=\frac{1}{2}|{\theta}_{j}-{\theta}_{k}|\geq\frac{\pi}{2N}\geq{\varepsilon}.$

If $j=k$, then $\xi$ and $\eta$ are both elements of $\SS({\varepsilon}_{j})$,
so that ${\mathsf{d}}_{\SS}(\xi,\eta)\geq{\varepsilon}_{j}$. Hence, using
$\frac{2}{\pi}\phi\leq\sin\phi\leq\phi$, we deduce from (4.2) that

${\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(t_{j}\xi,t_{j}),(t_{j}\eta,t_{j})\big{)}\geq\frac{2}{\pi}\sqrt{t_{j}}{\mathsf{d}}_{\SS}(\xi,\eta)\geq\frac{2}{\pi}\sqrt{t_{j}}{\varepsilon}_{j}={\varepsilon}.$

Hence, $\Xi_{{\mathbb{V}}_{0}}$ is ${\varepsilon}$-separated. Moreover, since
$\\#\Xi_{\SS}({\varepsilon}_{j})\sim{\varepsilon}_{j}^{-d+1}$ and
${\varepsilon}_{j}\sim{\varepsilon}/{\theta}_{j}$, it follows that

$\\#\Xi_{{\mathbb{V}}_{0}}=\sum_{j=1}^{N}\\#\Xi_{\SS}({\varepsilon}_{j})\sim\sum_{j=1}^{N}{\varepsilon}_{j}^{-d+1}\sim{\varepsilon}^{-d+1}\sum_{j=1}^{N}{\theta}_{j}^{d-1}\sim{\varepsilon}^{-d+1}N\sim{\varepsilon}^{-d}.$

For the proof of (4.22), we first consider the ball
${\mathsf{c}}_{[0,1]}(t_{j},r)=\\{s:{\mathsf{d}}_{[0,1]}(s,t_{j})\leq r\\}$ on
$[0,1]$. For $0<\delta<\pi/2$, it is easy to see that

${\mathsf{c}}_{[0,1]}(t_{j},\delta/N)\subset\\{s:(s\eta,s)\in{\mathbb{V}}_{0}(\xi,t_{j})\\}\subset{\mathsf{c}}_{[0,1]}(t_{j},\pi/N).$

We further choose $\delta$ so that $4\delta(1+\delta)<\frac{1}{2}$. For
$s\in{\mathsf{c}}_{[0,1]}(t_{j},\delta/N)$, write $s=\sin^{2}\frac{\phi}{2}$,
then

$|t_{j}-s|=\frac{1}{2}|\cos{\theta}-\cos\phi|=\left|\sin\frac{{\theta}_{j}-\phi}{2}\sin\frac{{\theta}_{j}+\phi}{2}\right|\leq\frac{\delta}{N}\left(2\sqrt{t_{j}}+\frac{\delta}{N}\right),$

where we have used $|\sin\frac{{\theta}_{j}+\phi}{2}|\leq
2\sin\frac{{\theta}_{j}}{2}+|\sin\frac{{\theta}_{j}-\phi}{2}|\leq
2\sqrt{t_{j}}+{\mathsf{d}}_{[0,1]}(s,t_{j})$. For $j\geq 1$,
$N^{-1}=\frac{2}{(2j-1)\pi}{\theta}_{j}\leq
2\sin\frac{{\theta}_{j}}{2}=2\sqrt{t_{j}}$, it follows that

$|t_{j}-s|\leq 2\delta(2+\delta)t_{j}\leq\tfrac{1}{2}t_{j},$

which implies in particular that $\frac{1}{2}t_{j}\leq s\leq\frac{3}{2}t_{j}$.
Furthermore, the same proof shows if $s\in{\mathsf{c}}_{[0,1]}(t_{j},\pi/N)$,
then $s\leq c_{*}t_{j}$. By definition, there are constants $b_{1}>0$ and
$b_{2}>0$ such that
${\mathsf{c}}_{\SS}(\xi,b_{1}{\varepsilon}_{j})\subset\SS_{\xi}({\varepsilon}_{j})\subset{\mathsf{c}}_{\SS}(\xi,b_{2}{\varepsilon}_{j})$.
We claim that (4.22) holds for some $c_{1}<\delta$ and some $c_{2}>b_{2}$.
Indeed, if
$(y,\eta)\subset{\mathsf{c}}\big{(}(t_{j}\xi,t_{j}),c_{1}{\varepsilon}\big{)}$,
then ${\mathsf{d}}_{[0,1]}(s,t_{j})\leq c_{1}{\varepsilon}\leq\delta/N$ so
that $s\geq t_{j}/2$, and $(st)^{\frac{1}{4}}{\mathsf{d}}_{\SS}(\xi,\eta)\leq
cc_{1}{\varepsilon}$ by (4.3) so that ${\mathsf{d}}_{\SS}(\xi,\eta)\leq
2^{\frac{1}{4}}cc_{1}{\varepsilon}/\sqrt{t_{j}}\leq b_{1}{\varepsilon}_{j}$ by
choosing $c_{1}$ small. This establishes the left-hand side inclusion of
(4.22). The right-hand side inclusion can be similarly established. The proof
is completed. ∎

The above construction establishes the existence of maximal
${\varepsilon}$-separated set on the conic surface. Since
$(\Omega,{\mathsf{w}}_{-1,{\gamma}},{\mathsf{d}}_{{\mathbb{V}}_{0}})$ is a
localizable homogeneous space, we can then deduce the Marcinkiewicz-Zygmund
inequality on such sets for all doubling weights on the conic surface. The
weight ${\mathsf{w}}$ on ${\mathbb{V}}_{0}^{d+1}$ is in general a function of
both $x$ and $t$.

######  Theorem 4.18.

Let ${\mathsf{w}}$ be a doubling weight on ${\mathbb{V}}_{0}^{d+1}$. Let
$\Xi_{{\mathbb{V}}_{0}}$ be a maximal $\frac{\delta}{n}$-separated subset of
${\mathbb{V}}_{0}^{d+1}$ and $0<\delta\leq 1$.

  1. $(i)$

For all $0<p<\infty$ and $f\in\Pi_{m}({\mathbb{V}}_{0}^{d+1})$ with $n\leq
m\leq cn$,

$\sum_{z\in\Xi_{{\mathbb{V}}_{0}}}\Big{(}\max_{(x,t)\in{\mathsf{c}}((z,r),\frac{\delta}{n})}|f(x,t)|^{p}\Big{)}{\mathsf{w}}\\!\left({\mathsf{c}}((z,r),\tfrac{\delta}{n})\right)\leq
c_{{\mathsf{w}},p}\|f\|_{p,{\mathsf{w}}}^{p}$

where $c_{{\mathsf{w}},p}$ depends on the doubling constant $L({\mathsf{w}})$
and on $p$ when $p$ is close to $0$.

  2. $(ii)$

For $0<r<1$, there is a $\delta_{r}>0$ such that for $\delta\leq\delta_{r}$,
$r\leq p<\infty$ and $f\in\Pi_{n}({\mathbb{V}}_{0}^{d+1})$,

$\displaystyle\|f\|_{p,{\mathsf{w}}}^{p}\leq
c_{{\mathsf{w}},r}\sum_{z\in\Xi}\Big{(}\min_{(x,t)\in{\mathsf{c}}\bigl{(}(z,r),\tfrac{\delta}{n}\bigr{)}}|f(x,t)|^{p}\Big{)}{\mathsf{w}}\bigl{(}{\mathsf{c}}((z,r),\tfrac{\delta}{n})\bigr{)}$

where $c_{{\mathsf{w}},r}$ depends only on $L({\mathsf{w}})$ and on $r$ when
$r$ is close to $0$.

###  4.6. Cubature rules and localized tight frames

We now turn our attention to Assertion 4 in Subsection 2.5 and construct fast
decaying polynomials on the conic surface.

######  Lemma 4.19.

Let $d\geq 2$. For each $(x,t)\in{\mathbb{V}}_{0}^{d+1}$, there is a
polynomial $T_{x,t}$ of degree $n$ that satisfies

  1. (1)

$T_{x,t}(x,t)=1$, $T_{x,t}(y,s)\geq c>0$ if
$(y,s)\in{\mathsf{c}}((x,t),\frac{\delta}{n})$, and for every ${\kappa}>0$,

$0\leq T_{x,t}(y,s)\leq
c_{\kappa}\left(1+{\mathsf{d}}_{{\mathbb{V}}_{0}}\big{(}(x,t),(y,s)\big{)}\right)^{-{\kappa}},\quad(y,s)\in{\mathbb{V}}_{0}^{d+1}.$

  2. (2)

there is a polynomial $q(t)$ of degree $n$ such that $q(t)T_{x,t}$ is a
polynomial of degree $2n$ in $(x,t)$ variables and $1\leq q_{n}(t)\leq c$.

###### Proof.

Let $r$ be a positive integer such that ${\kappa}\leq 2r$. For positive
integer $n$, let $m=\lfloor\frac{n}{r}\rfloor+1$ and define

$S_{n}(\cos{\theta})=\left(\frac{\sin(m+\frac{1}{2})\frac{{\theta}}{2}}{(m+\frac{1}{2})\sin\frac{{\theta}}{2}}\right)^{2r},\qquad
0\leq{\theta}\leq\pi.$

By considering $m{\theta}\geq 1$ and $m{\theta}\leq 1$ separately if
necessary, it follows readily that

(4.23) $S_{n}(1)=1,\qquad 0\leq S_{n}(\cos{\theta})\leq
c\big{(}1+n{\theta}\big{)}^{-2r},\quad 0\leq{\theta}\leq\pi.$

Moreover, $S_{n}(z)$ is an even algebraic polynomial of degree at most $2n$.
For a fixed $(x,t)\in{\mathbb{V}}_{0}^{d+1}$, we define

$T_{(x,t)}(y,s):=\frac{S_{n}\Big{(}\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}+\sqrt{1-t}\sqrt{1-s}\Big{)}+S_{n}\Big{(}\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}-\sqrt{1-t}\sqrt{1-s}\Big{)}}{1+S_{n}(2t-1)}.$

As an even polynomial, $S_{n}(z)$ is a sum of even monomials. Since, by
binomial formula, $(\sqrt{a}+\sqrt{b})^{2k}+(\sqrt{a}-\sqrt{b})^{2k}$ is a
polynomial of degree $k$ in $a$ and $b$, it follows that $T_{(x,t)}$ is indeed
a polynomial in $(y,s)$ of degree at most $n$. Moreover, it satisfies
$T_{(x,t)}(x,t)=1$ since ${\langle}x,x{\rangle}=t^{2}$. Furthermore, using
$\frac{2}{\pi}{\theta}\leq\sin{\theta}\leq{\theta}$ for
$0\leq{\theta}\leq\pi/2$, we see that $S_{n}$ satisfies

$S_{n}(\cos{\theta})\geq\left(\frac{2}{\pi}\right)^{2r},\qquad\hbox{if \,
$0\leq{\theta}\leq\frac{2\pi}{2m+1}$}.$

Hence, since $0\leq S_{n}(2t-1)\leq 1$, it follows that

$T_{(x,t)}(y,s)\geq\frac{S_{n}\Big{(}\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}+\sqrt{1-t}\sqrt{1-s}\Big{)}}{1+S_{n}(2t-1)}\geq\frac{1}{2}S_{n}(\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s)))\geq\frac{1}{2}\left(\frac{2}{\pi}\right)^{2r}$

for $(y,s)\in{\mathsf{c}}((x,t),\frac{2\pi}{2m+1})$. Finally, we can use the
fast decay (4.23) of $S_{n}$ with
${\theta}\sim\sin\frac{{\theta}}{2}=\sqrt{\frac{1}{2}(1-\cos{\theta})}$ to
derive an upper bound of $T_{(x,t)}$. Indeed, since evidently
$1-\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}+\sqrt{1-t}\sqrt{1-s}\geq
1-\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}-\sqrt{1-t}\sqrt{1-s},$ we obtain
then

$\displaystyle 0\leq T_{(x,t)}(y,s)\,$ $\displaystyle\leq
c\frac{1}{\left(1+n\sqrt{1-\sqrt{\frac{{\langle}x,y{\rangle}+ts}{2}}-\sqrt{1-t}\sqrt{1-s}}\right)^{2r}}$
$\displaystyle=c\frac{1}{\left(1+n\sqrt{1-\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))}\right)^{2r}}.$

Hence, by the definition of ${\mathsf{d}}_{{\mathbb{V}}_{0}}$ and
$1-\cos{\theta}\sim{\theta}^{2}$, we obtain

(4.24) $0\leq T_{(x,t)}(y,s)\leq
c\frac{1}{\left(1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\right)^{2r}}.$

This completes the proof of (1). Finally, $q_{n}(t)=1+S_{n}(2t-1)$ is a
polynomial of degree $n$ and it satisfies $1\leq q_{n}(t)\leq c$ for all
$t\in[0,1]$. Moreover, $q_{n}(t)T_{x,t}$ is a polynomial of degree $2n$ in
$(x,t)$ variable. This completes the proof. ∎

From Propositions 2.17 and 2.18, we have established the following result.

######  Corollary 4.20.

Let ${\mathsf{w}}$ be a doubling weight function on ${\mathbb{V}}_{0}^{d+1}$.
Then

${\lambda}_{n}\big{(}{\mathsf{w}};(x,t)\big{)}\leq
c\,{\mathsf{w}}\left({\mathsf{c}}\left((x,t),\tfrac{1}{n}\right)\right).$

Moreover, for ${\gamma}\geq-\frac{1}{2}$,

${\lambda}_{n}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t)\big{)}\geq
c\,{\mathsf{w}}_{-1,{\gamma}}\left({\mathsf{c}}\left((x,t),\tfrac{1}{n}\right)\right)=cn^{-d}{\mathsf{w}}_{-1,{\gamma}}(n;t).$

This implies, in particular, that the cubature rule in Theorem 2.20 holds for
all doubling weights on the conic surface.

######  Theorem 4.21.

Let ${\mathsf{w}}$ be a doubling weight on ${\mathbb{V}}_{0}^{d+1}$. Let $\Xi$
be a maximum $\frac{\delta}{n}$-separated subset of ${\mathbb{V}}_{0}^{d+1}$.
There is a $\delta_{0}>0$ such that for $0<\delta<\delta_{0}$ there exist
positive numbers ${\lambda}_{z,r}$, $(z,r)\in\Xi$, so that

(4.25)
$\int_{{\mathbb{V}}_{0}^{d+1}}f(x,t){\mathsf{w}}(x,t)\mathrm{d}\sigma(x,t)=\sum_{(z,r)\in\Xi}{\lambda}_{z,r}f(z,r),\qquad\forall
f\in\Pi_{n}({\mathbb{V}}_{0}^{d+1}).$

Moreover,
${\lambda}_{z,r}\sim{\mathsf{w}}\\!\left({\mathsf{c}}((z,r),\tfrac{\delta}{n})\right)$
for all $(z,r)\in\Xi$.

We can now construct localized frame on the conic surface. Let us recapitulate
necessary definitions in the setting of ${\mathbb{V}}_{0}^{d+1}$. For
$j=0,1,\ldots,$ let $\Xi_{j}$ be a maximal $\frac{\delta}{2^{j}}$-separated
subset in ${\mathbb{V}}_{0}^{d+1}$, so that

$\int_{{\mathbb{V}}_{0}^{d+1}}f(x,t){\mathsf{w}}(x,t)\mathrm{d}\sigma(x,t)=\sum_{(z,r)\in\Xi_{j}}{\lambda}_{(x,r),j}f(z,r),\qquad
f\in\Pi_{2^{j}}({\mathbb{V}}_{0}^{d+1}).$

Let $\widehat{a}$ be an admissible cut-off function satisfying (2.43). Let
${\mathsf{L}}_{2^{j}}({\mathsf{w}})*f$ denote the near best approximation
operator defined by

${\mathsf{L}}_{2^{j}}({\mathsf{w}})*f(x):=\int_{{\mathbb{V}}_{0}^{d+1}}f(y,s){\mathsf{L}}_{2^{j}}({\mathsf{w}};(x,t),(y,s)){\mathsf{w}}(y,s)\mathrm{d}\sigma(y,s).$

For $j=1,\ldots,$ we introduce the notation
$F_{j}({\mathsf{w}})={\mathsf{L}}_{2^{j}}({\mathsf{w}})$ for both the kernel
and the operator. More precisely, $F_{0}({\mathsf{w}};\cdot,\cdot)=1$ and for
$j\geq 1$,

$F_{j}({\mathsf{w}};\cdot,\cdot)={\mathsf{L}}_{2^{j-1}}({\mathsf{w}};\cdot,\cdot)\quad\hbox{and}\quad
F_{j}({\mathsf{w}})*f=F_{j}({\mathsf{w}})*f$

and define the frame elements $\psi_{(z,r),j}$ for $(z,r)\in\Xi_{j}$ by

$\psi_{(z,r),j}(x,t):=\sqrt{{\lambda}_{(z,r),j}}F_{j}({\mathsf{w}};(x,t),(z,r)),\qquad(x,t)\in{\mathbb{V}}_{0}^{d+1},$

which are well defined for all doubling weight by Theorem 4.21. By Theorem
2.21, the system $\Phi=\\{\psi_{(z,r),j}:(z,r)\in\Xi_{j},\,j=1,2,3,\ldots\\}$
is a tight frame of $L^{2}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}})$.

######  Theorem 4.22.

Let ${\mathsf{w}}$ be a doubling weight on ${\mathbb{V}}_{0}^{d+1}$. If $f\in
L^{2}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}})$, then

$f=\sum_{j=0}^{\infty}\sum_{(z,r)\in\Xi_{j}}\langle
f,\psi_{(z,r),j}\rangle_{\mathsf{w}}\psi_{(z,r),j}\qquad\mbox{in
$L^{2}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}})$}$

and

$\|f\|_{2,{\mathsf{w}}}=\Big{(}\sum_{j=0}^{\infty}\sum_{(z,r)\in\Xi_{j}}|\langle
f,\psi_{(z,r),j}\rangle_{\mathsf{w}}|^{2}\Big{)}^{1/2}.$

For the weight function ${\mathsf{w}}_{-1,{\gamma}}$, the frame elements are
highly localized.

######  Theorem 4.23.

For ${\gamma}\geq-\frac{1}{2}$, the frame for ${\mathsf{w}}_{-1,{\gamma}}$ is
highly localized in the sense that, for every ${\kappa}>0$, there exists a
constant $c_{\kappa}>0$ such that

(4.26) $|\psi_{(z,r),j}(x,t)|\leq
c_{\sigma}\frac{2^{jd/2}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(2^{j};t)}\left(1+2^{j}{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(z,r))\right)^{\kappa}},\quad(x,t)\in{\mathbb{V}}_{0}^{d+1}.$

###### Proof.

The inequality (4.26) follows from the highly localized estimate of the kernel
in Theorem 4.10 and ${\lambda}_{(z,r),j}\sim
2^{-jd}{\mathsf{w}}_{{\gamma}}(2^{j};r)$ which holds for
${\mathsf{w}}_{-1,{\gamma}}$ according to Corollary 4.20. ∎

It is worth pointing out that the localization of the frame elements is
established for ${\mathsf{w}}_{-1,{\gamma}}(t)=t^{-1}(1-t)^{{\gamma}}$ but not
for the Lebesgue measure on ${\mathbb{V}}_{0}^{d+1}$.

###  4.7. Characterization of best approximation

For $f\in L_{p}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}})$, $1\leq p<\infty$, and
$f\in C({\mathbb{V}}_{0}^{d+1})$ if $p=\infty$, the error of best
approximation by polynomials of degree at most $n$ is defined by

${\mathsf{E}}_{n}(f)_{p,{\mathsf{w}}}=\inf_{g\in\Pi_{n}({\mathbb{V}}_{0}^{d+1})}\|f-g\|_{p,{\mathsf{w}}},\quad
1\leq p\leq\infty.$

Following the study in Section 3, we can give a characterization for this
quantity by the modulus of smoothness defined via the operator
${\mathsf{S}}_{{\theta},{\mathsf{w}}}$ and the $K$-functional defined via the
differential operator $\Delta_{0,{\gamma}}$ for ${\mathsf{w}}_{-1,{\gamma}}$.

More specifically, for $f\in
L^{p}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{-1,{\gamma}})$ and $r>0$, the
modulus of smoothness is defined by

${\omega}_{r}(f;\rho)_{p,{\mathsf{w}}_{-1,{\gamma}}}=\sup_{0\leq{\theta}\leq\rho}\left\|\left(I-{\mathsf{S}}_{{\theta},{\mathsf{w}}_{-1,{\gamma}}}\right)^{r/2}f\right\|_{p,{\mathsf{w}}_{-1,{\gamma}}},\quad
1\leq p\leq\infty,$

where the operator ${\mathsf{S}}_{{\theta},{\mathsf{w}}_{-1,{\gamma}}}$ is
defined by, for $n=0,1,2,\ldots$ and ${\lambda}={\gamma}+d-1$,

$\operatorname{proj}_{n}({\mathsf{w}}_{-1,{\gamma}};{\mathsf{S}}_{{\theta},{\mathsf{w}}_{-1,{\gamma}}}f)=R_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(\cos{\theta})\operatorname{proj}_{n}({\mathsf{w}}_{-1,{\gamma}};f).$

Moreover, in terms of the fractional differential operator
$(-\Delta_{0,{\gamma}})^{\frac{r}{2}}$, the $K$-functional is defined for a
weight ${\mathsf{w}}$ by

${\mathsf{K}}_{r}(f,t)_{p,{\mathsf{w}}}:=\inf_{g\in
W_{p}^{r}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}})}\left\\{\|f-g\|_{p,{\mathsf{w}}}+t^{r}\left\|(-\Delta_{0,{\gamma}})^{\frac{r}{2}}f\right\|_{p,{\mathsf{w}}}\right\\}.$

Both these quantities are well defined as shown in Section 3. Since Assertions
1–4 hold for ${\mathsf{w}}_{-1,{\gamma}}$, we only have to verify that the
Assertion 5 in Subsection 3.5 holds for the differential operator
${\mathfrak{D}}_{\varpi}=\Delta_{0,{\gamma}}$. By Theorem 4.8, the kernel
$L_{n}^{(r)}(\varpi)$ in Assertion 5 becomes

${\mathsf{L}}_{n}^{(r)}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}=\sum_{k=0}^{\infty}\widehat{a}\left(\frac{k}{n}\right)(k(k+{\gamma}+d-1))^{\frac{r}{2}}{\mathsf{P}}_{k}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}.$

######  Lemma 4.24.

Let ${\gamma}\geq-\frac{1}{2}$ and ${\kappa}>0$. Then, for $r>0$ and
$(x,t),(y,s)\in{\mathbb{V}}_{0}^{d+1}$,

$|{\mathsf{L}}_{n}^{(r)}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s)\big{)}|\leq
c_{\kappa}\frac{n^{d+r}}{\sqrt{{\mathsf{w}}_{{\gamma},d}(n;t)}\sqrt{{\mathsf{w}}_{{\gamma},d}(n;s)}}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((x,t),(y,s))\big{)}^{-{\kappa}}.$

###### Proof.

By (4.9), the kernel can be written as

$\displaystyle{\mathsf{L}}_{n}^{(r)}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s))=b_{{\gamma},d}\int_{[-1,1]^{2}}$
$\displaystyle L_{n,r}\left(2\zeta(x,t,y,s;v)^{2}-1\right)$
$\displaystyle\times(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v,$

in which $L_{n,r}$ is defined by, with ${\lambda}={\gamma}+d-1$,

$L_{n,r}(t)=\sum_{k=0}^{\infty}\widehat{a}\left(\frac{k}{n}\right)(k(k+{\gamma}+d-1))^{\frac{r}{2}}\frac{P_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(1)P_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(t)}{h^{({\lambda}-\frac{1}{2},-\frac{1}{2})}}.$

Applying (2.18) with
$\eta(t)=\widehat{a}(t)\left(t(t+n^{-1}({\gamma}+d-1))\right)^{\frac{r}{2}}$
and $m=0$, it follows that

$\left|L_{n,r}(t)\right|\leq
cn^{r}\frac{n^{2{\lambda}+1}}{(1+n\sqrt{1-t})^{\ell}}.$

Consequently, it follows readily that

$\displaystyle|{\mathsf{L}}_{n}^{(r)}\big{(}{\mathsf{w}}_{-1,{\gamma}};(x,t),(y,s))|\leq
cn^{2{\lambda}+r+1}\int_{[-1,1]^{2}}$
$\displaystyle\frac{1}{\left(1+n\sqrt{1-\zeta(x,t,y,s;v)^{2}}\right)^{{\kappa}+2{\gamma}+d+1}}$
$\displaystyle\times(1-v_{1}^{2})^{\frac{d-2}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v.$

Apart from $n^{r}$, this estimate is the same as that of
$L_{n}({\mathsf{w}}_{-1,{\gamma}})$ appeared in the proof of Theorem 4.10, so
that the desired upper bound follows from the estimate there. ∎

With all assertions hold for the ${\mathsf{w}}_{-1,{\gamma}}$, the
characterization of the best approximation by polynomials in Subsection 3.3
holds on the conic surface, which we state below.

######  Theorem 4.25.

Let $f\in L^{p}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}})$ if $1\leq p<\infty$ and
$f\in C({\mathbb{V}}_{0}^{d+1})$ if $p=\infty$. Le $r>0$ and $n=1,2,\ldots$.
For ${\mathsf{w}}={\mathsf{w}}_{-1,{\gamma}}$ with ${\gamma}\geq-\frac{1}{2}$,
there holds

  1. (i)

direct estimate

${\mathsf{E}}_{n}(f)_{p,{\mathsf{w}}_{-1,{\gamma}}}\leq
c{\mathsf{K}}_{r}(f;n^{-1})_{p,{\mathsf{w}}_{-1,{\gamma}}};$

  2. (ii)

inverse estimate

${\mathsf{K}}_{r}(f;n^{-1})_{p,{\mathsf{w}}_{-1,{\gamma}}}\leq
cn^{-r}\sum_{k=0}^{n}(k+1)^{r-1}{\mathsf{E}}_{k}(f)_{p,{\mathsf{w}}_{-1,{\gamma}}}.$

Furthermore, the characterization can be given via the modulus of smoothness,
since it is equivalent to the $K$-functional.

######  Theorem 4.26.

Let ${\gamma}\geq-\frac{1}{2}$ and $f\in
L_{p}^{r}({\mathbb{V}}_{0}^{d+1},{\mathsf{w}}_{-1,{\gamma}})$, $1\leq
p\leq\infty$. Then for $0<{\theta}\leq\pi/2$ and $r>0$

$c_{1}{\mathsf{K}}_{r}(f;{\theta})_{p,{\mathsf{w}}_{-1,{\gamma}}}\leq{\omega}_{r}(f;{\theta})_{p,{\mathsf{w}}_{-1,{\gamma}}}{\mathsf{K}}_{r}(f;{\theta})_{p,{\mathsf{w}}_{-1,{\gamma}}}.$

##  5\. Homogeneous space on solid cones

In this section we work in the setting of homogeneous space on the solid cone

${\mathbb{V}}^{d+1}=\\{(x,t):\|x\|\leq t,\,x\in{\mathbb{B}}^{d},\,0\leq t\leq
1\\},$

which is bounded by ${\mathbb{V}}_{0}^{d+1}$ and the hyperplane $t=1$ in
${\mathbb{R}}^{d+1}$. As in the previous section, we shall verify that the
framework we developed for approximation and tight frame on homogeneous space
is applicable on this domain with respect to the Jacobi weight function
$W_{{\gamma},\mu}(x,t)=(1-t)^{{\gamma}}(t^{2}-\|x\|^{2})^{\mu-\frac{1}{2}}$.

The structure of the section is parallel to that of the conic surface. We
again need to define and understand an intrinsic distance and doubling weights
on the cone and do so in the first two subsections. The orthogonal structure
with respect to the Jacobi weight is reviewed in the third subsection. The
highly localized kernels are established in the fourth subsection, where the
Assertions 1 and 3 are verified for $W_{{\gamma},\mu}$ for all $\mu\geq 0$ but
the Assertion 2 is established only for $W_{{\gamma},0}$. We also provide
construction of $ve$-separated set on the cone and state the Marcinkiewicz-
Zygmund inequality in the fifth subsection. Assertion 4 is verified in the
sixth subsection, which allows us to state the positive cubature rules and the
tight frame on the cone. While the tight localized frame holds for the Jacobi
wight with $\mu=0$, the characterization of the best approximation by
polynomials works out for all $\mu\geq 0$ in the seventh subsection.

###  5.1. Distance on solid cone

The distant function ${\mathsf{d}}_{{\mathbb{B}}}(\cdot,\cdot)$ on the unit
ball ${\mathbb{B}}^{d}$ can be deduced from regarding ${\mathbb{B}}^{d}$ as
the projection of the upper hemisphere of $\SS^{d+1}$, or the distant function
${\mathsf{d}}_{\SS}(\cdot,\cdot)$ applied on the points
$X=(x,\sqrt{1-\|x\|^{2}})$ and $Y=(y,\sqrt{1-\|y\|^{2}})$ with
$x\in{\mathbb{B}}^{d}$. The same holds for the solid cone
${\mathbb{V}}^{d+1}$.

######  Definition 5.1.

For $(x,t)$ and $(y,s)$ on ${\mathbb{V}}^{d+1}$, let
$X=\big{(}x,\sqrt{t^{2}-\|x\|^{2}}\big{)}$ and
$Y=\big{(}y,\sqrt{s^{2}-\|y\|^{2}}\big{)}$. Define

(5.1)
${\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)):=\arccos\left(\sqrt{\frac{{\langle}X,Y{\rangle}+ts}{2}}+\sqrt{1-t}\sqrt{1-s}\right).$

######  Proposition 5.2.

The function ${\mathsf{d}}_{{\mathbb{V}}}(\cdot,\cdot)$ defines a distance
function on the solid cone ${\mathbb{V}}^{d+1}$.

###### Proof.

Since $\|X\|=t$ and $\|Y\|=s$, both $(X,t)$ and $(Y,s)$ are elements of
${\mathbb{V}}_{0}^{d+2}$. Furthermore,

(5.2)
${\mathsf{d}}_{{\mathbb{V}}^{d+1}}((x,t),(y,s))={\mathsf{d}}_{{\mathbb{V}}_{0}^{d+2}}((X,t),(Y,s)),$

from which it is easy to see that ${\mathsf{d}}_{{\mathbb{V}}}(\cdot,\cdot)$
is a distance function on ${\mathbb{V}}^{d+1}$. ∎

The distance function ${\mathsf{d}}_{{\mathbb{V}}}(\cdot,\cdot)$ is closely
related to the distance function ${\mathsf{d}}_{[0,1]}(\cdot,\cdot)$ of the
interval $[0,1]$ and the distance ${\mathsf{d}}_{{\mathbb{B}}}(\cdot,\cdot)$
of the unit ball ${\mathbb{B}}^{d}$.

######  Proposition 5.3.

For $d\geq 2$ and $(x,t),(y,s)\in{\mathbb{V}}^{d+1}$, write $x=tx^{\prime}$
and $y=sy^{\prime}$ with $x^{\prime},y^{\prime}\in{\mathbb{B}}^{d}$. Then

$1-\cos{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))=1-\cos{\mathsf{d}}_{[0,1]}(t,s)+\sqrt{t}\sqrt{s}\left[1-\cos\left(\tfrac{1}{2}{\mathsf{d}}_{{\mathbb{B}}}(x^{\prime},y^{\prime})\right)\right].$

In particular,

$c_{1}{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\leq{\mathsf{d}}_{[0,1]}(t,s)+(ts)^{\frac{1}{4}}{\mathsf{d}}_{{\mathbb{B}}}(x^{\prime},y^{\prime})\leq
c_{2}{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)).$

###### Proof.

With $x=tx^{\prime}$, we see that $X=t(x^{\prime},\sqrt{1-\|x^{\prime}|^{2}})$
and $Y=s(y^{\prime},\sqrt{1-\|y^{\prime}|^{2}})$. Hence, it follows as in the
case of ${\mathbb{V}}_{0}^{d+1}$ that

(5.3)
${\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))=\arccos\left[\sqrt{t}\sqrt{s}\cos\left(\tfrac{1}{2}{\mathsf{d}}_{{\mathbb{B}}}(x^{\prime},y^{\prime})\right)+\sqrt{1-t}\sqrt{1-s}\right].$

The rest of the proof follows from that of Proposition 4.3 almost verbatim. ∎

Like the distance function on the surface, we see that the distance on the
line segment from the apex to $(x^{\prime},1)$ on the top boundary of the cone
${\mathbb{V}}^{d+1}$ is exactly the distance function on $[0,1]$; that is, for
$x^{\prime}\in{\mathbb{B}}^{d}$,

${\mathsf{d}}_{{\mathbb{V}}}((x^{\prime},t),(x^{\prime},s))={\mathsf{d}}_{[0,1]}(t,s),\qquad
0\leq t,s\leq 1.$

The following lemma is an analog of Lemma 4.4 and will be needed in the next
subsection.

######  Lemma 5.4.

For $(x,t),(y,s)\in{\mathbb{V}}^{d+1}$,

$\big{|}\sqrt{t}-\sqrt{s}\big{|}\leq{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\quad\hbox{and}\quad\big{|}\sqrt{1-t}-\sqrt{1-s}\big{|}\leq{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)),$

and

$\big{|}\sqrt{t^{2}-\|x\|^{2}}-\sqrt{s^{2}-\|y\|^{2}}\big{|}\leq
c\max\\{\sqrt{s},\sqrt{t}\\}\,{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)).$

###### Proof.

The first two inequalities are immediate consequence of (5.2) and Lemma 4.4.
Let $x=tx^{\prime}$ and $y=sy^{\prime}$ with
$x^{\prime},y^{\prime}\in{\mathbb{B}}^{d}$. We know [12, (A.1.4)] that

(5.4)
$\left|\sqrt{1-\|x^{\prime}\|^{2}}-\sqrt{1-\|y^{\prime}\|^{2}}\right|\leq\sqrt{2}{\mathsf{d}}_{\mathbb{B}}(x^{\prime},y^{\prime}).$

Without loss of generality, assuming $t\geq s$. Then $s\leq\sqrt{s}\sqrt{t}$,
so that

$\displaystyle\left|\sqrt{t^{2}-\|x\|^{2}}-\sqrt{s^{2}-\|y\|^{2}}\right|$
$\displaystyle=\left|t\sqrt{1-\|x^{\prime}\|^{2}}-s\sqrt{1-\|y^{\prime}\|^{2}}\right|$
$\displaystyle\leq|t-s|\sqrt{1-\|x^{\prime}\|^{2}}+s\left|\sqrt{1-\|x^{\prime}\|^{2}}-\sqrt{1-\|y^{\prime}\|^{2}}\right|$
$\displaystyle\leq\sqrt{2}\max\\{\sqrt{s},\sqrt{t}\\}\left(\big{|}\sqrt{t}-\sqrt{s}\big{|}+(ts)^{\frac{1}{4}}\,{\mathsf{d}}_{\mathbb{B}}(x^{\prime},y^{\prime})\right),$

which is bounded by
$c\max\\{\sqrt{s},\sqrt{t}\\}\,{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))$ by
the first inequality in the statement and the inequality in Proposition 5.3. ∎

###  5.2. A family of doubling weights

For $d\geq 1$, $\mu\geq 0$, ${\beta}+2\mu>-d$ and ${\gamma}>-1$, we consider
the Jacobi weight function defined on the solid cone by

$W_{{\beta},{\gamma},\mu}(x,t):=(t^{2}-\|x\|^{2})^{\mu-\frac{1}{2}}t^{\beta}(1-t)^{\gamma},\quad(x,t)\in{\mathbb{V}}^{d+1}.$

Let ${\mathbf{b}}_{{\beta},{\gamma},\mu}$ be the normalization constant so
that ${\mathbf{b}}_{{\beta},{\gamma},\mu}W_{{\beta},{\gamma},\mu}$ has unit
integral on ${\mathbb{V}}^{d+1}$. Setting $x=tx^{\prime}$ with
$x^{\prime}\in{\mathbb{B}}^{d}$, then

${\mathbf{b}}_{{\beta},{\gamma},\mu}=\bigg{(}\int_{0}^{1}t^{d+{\beta}+2\mu-1}(1-t)^{\gamma}\mathrm{d}t\int_{{\mathbb{B}}^{d}}(1-\|x^{\prime}\|^{2})^{\mu-\frac{1}{2}}\mathrm{d}x^{\prime}\bigg{)}^{-1}=c_{{\beta}+2\mu+d-1,{\gamma}}b_{\mu}^{\mathbb{B}},$

where $c_{{\alpha},{\beta}}$ is defined in (2.14) and $b_{\mu}^{\mathbb{B}}$
is the normalization constant of the classical weight function on the unit
ball. The case ${\beta}=0$ is of particular interest and will be denoted by
$W_{{\gamma},\mu}$; that is,

$W_{{\gamma},\mu}(x,t):=W_{0,{\gamma},\mu}(x,t)=(t^{2}-\|x\|^{2})^{\mu-\frac{1}{2}}(1-t)^{\gamma},\quad(x,t)\in{\mathbb{V}}^{d+1}.$

We show that this is a doubling weight with respect to the distance
${\mathsf{d}}_{\mathbb{V}}$. For $r>0$ and $(x,t)$ in ${\mathbb{V}}^{d+1}$, we
denote the ball centered at $(x,t)$ with radius $r$ by

${\mathbf{c}}((x,t),r):=\big{\\{}(y,s)\in{\mathbb{V}}^{d+1}:{\mathsf{d}}_{{\mathbb{V}}}\big{(}(x,t),(y,s)\big{)}\leq
r\big{\\}}.$

######  Proposition 5.5.

Let $r>0$ and $(x,t)\in{\mathbb{V}}^{d+1}$. Then for ${\beta}>-d$,
${\gamma}>-1$ and $\mu\geq 0$,

(5.5) $\displaystyle
W_{{\beta},{\gamma},\mu}\big{(}{\mathbf{c}}((x,t),r)\big{)}$
$\displaystyle:=\int_{{\mathbf{c}}((x,t),r)}W_{{\beta},{\gamma},\mu}(y,s)\mathrm{d}y\mathrm{d}s$
$\displaystyle\sim
r^{d+1}(t+r^{2})^{{\beta}+\frac{d-1}{2}}(1-t+r^{2})^{{\gamma}+\frac{1}{2}}(t^{2}-\|x\|^{2}+r^{2})^{\mu}.$

In particular, $W_{{\beta},{\gamma},\mu}$ is a doubling weight and the
doubling index ${\alpha}(W_{{\beta},{\gamma},\mu})$ is give by
${\alpha}(W_{{\beta},{\gamma},\mu})=2\mu+d+1+2\max\\{0,{\beta}+\frac{d-1}{2}\\}+2\max\\{0,{\gamma}+\frac{1}{2}\\}$.

###### Proof.

Let $\tau_{r}(t,s)$ and ${\theta}_{r}(t,s)=\arccos\tau_{r}(t,s)$ be as in the
proof of Proposition 4.6. From ${\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\leq
r$, we obtain ${\mathsf{d}}_{[0,1]}(t,s)\leq r$ and, by (5.1),

${\mathsf{d}}_{{\mathbb{B}}}(x^{\prime},y^{\prime})\leq\arccos\left(2[\tau_{r}(t,s)]^{2}-1\right)=\tfrac{1}{2}\arccos\tau_{r}(t,s)=\tfrac{1}{2}{\theta}_{r}(t,s),$

where ${\mathsf{d}}_{\mathbb{B}}(\cdot,\cdot)$ is the distance on the unit
ball ${\mathbb{B}}^{d}$. Hence, it follows that

$\displaystyle
W_{{\beta},{\gamma},\mu}\big{(}{\mathbf{c}}((x,t),r)\big{)}=\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}s^{d}\int_{{\mathsf{d}}_{{\mathbb{B}}}(x^{\prime},y^{\prime})\leq\tfrac{1}{2}{\theta}_{r}(t,s)}W_{{\beta},{\gamma},\mu}(sy^{\prime},s)\mathrm{d}y^{\prime}\mathrm{d}s$
$\displaystyle\qquad\sim\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}s^{d+{\beta}+2\mu-1}(1-s)^{\gamma}\int_{{\mathsf{d}}_{{\mathbb{B}}}(x^{\prime},y^{\prime})\leq\tfrac{1}{2}{\theta}_{r}(t,s)}(1-\|y^{\prime}\|^{2})^{\mu-\frac{1}{2}}\mathrm{d}y^{\prime}\mathrm{d}s$

For $\mu\geq 0$ and $0<\rho\leq 1$, it is known [38, Lemma 5.3] or [12, p.
107] that

(5.6)
$\int_{{\mathsf{d}}_{{\mathbb{B}}}(x^{\prime},y^{\prime})\leq\rho}(1-\|y^{\prime}\|^{2})^{\mu-\frac{1}{2}}\mathrm{d}y^{\prime}\sim(1-\|x^{\prime}\|^{2}+\rho^{2})^{\mu}\rho^{d},$

which can also be established by following the approach in Proposition 4.6.
Consequently, using ${\theta}_{r}(t,s)\sim\sqrt{1-\tau_{r}(t,s)}$, we conclude
that

$\displaystyle
W_{{\beta},{\gamma},\mu}\big{(}{\mathbf{c}}((x,t),r)\big{)}\sim\,$
$\displaystyle\int_{{\mathsf{d}}_{[0,1]}(t,s)\leq
r}s^{d+{\beta}+2\mu-1}(1-s)^{\gamma}$
$\displaystyle\quad\times\big{(}1-\|x^{\prime}\|^{2}+1-\tau_{r}(t,s)\big{)}^{\mu}\big{(}1-\tau_{r}(t,s)\big{)}^{\frac{d}{2}}\mathrm{d}s.$

If $t\geq 3r^{2}$, then $s\sim t+r^{2}$ and, by (4.7), it follows that

$s^{\mu}\big{(}1-\|x^{\prime}\|^{2}+1-\tau_{r}(t,s)\big{)}^{\mu}\sim(t^{2}-\|x\|^{2}+r^{2})^{\mu}.$

With this term removed, the integral of the remaining integrand can be
estimated by following the estimates of Case 1 and Case 3 of the proof of
Proposition 4.6. If $t\leq 3r^{2}$, then $t^{2}-\|x\|^{2}+r^{2}\sim r^{2}$. We
use $1-\|x^{\prime}\|^{2}+1-\tau_{r}(t,s)\leq 2$ for the upper bound and
$1-\tau_{r}(t,s)\geq 2/\pi^{2}$ on the subset ${\mathsf{d}}_{[0,1]}(t,s)\leq
r/2$, proved in the Case 2 of the proof of Proposition 4.6, to remove the term
$\big{(}1-\|x^{\prime}\|^{2}+1-\tau_{r}(t,s)\big{)}^{\mu}$ from the integral.
The rest of the proof then follows from that of the Case 2 of the proof of
Proposition 4.6. This completes the proof. ∎

######  Corollary 5.6.

For $d\geq 2$, ${\beta}>-d$, ${\gamma}>-1$ and $\mu\geq 0$,
$({\mathbb{V}}^{d+1},W_{{\beta},{\gamma},\mu},{\mathsf{d}}_{{\mathbb{V}}})$ is
a homogeneous space.

###  5.3. Orthogonal polynomials on the solid cone

These polynomials are studied in [53] and they are orthogonal with respect to
the inner product

${\langle}f,g{\rangle}_{{\beta},{\gamma},\mu}={\mathbf{b}}_{{\beta},{\gamma},\mu}\int_{{\mathbb{V}}^{d+1}}f(x,t)g(x,t)W_{{\beta},{\gamma},\mu}(x,t)\mathrm{d}x\mathrm{d}t.$

Let ${\mathcal{V}}_{n}({\mathbb{V}}^{d+1},W_{{\beta},{\gamma},\mu})$ be the
space of orthogonal polynomials of degree $n$, which has dimension
$\binom{n+d+1}{n}$. An orthogonal basis of
${\mathcal{V}}_{n}({\mathbb{V}}^{d+1},W_{{\beta},{\gamma},\mu})$ can be given
in terms of the Jacobi polynomials and the orthogonal polynomials on the unit
ball. For $m=0,1,2,\ldots$, let
$\\{P_{\mathbf{k}}^{m}(W_{\mu}):|{\mathbf{k}}|=m,{\mathbf{k}}\in{\mathbb{N}}_{0}^{d}\\}$
be an orthonormal basis of ${\mathcal{V}}_{n}({\mathbb{B}}^{d},W_{\mu})$ on
the unit ball. Let $\alpha:=\mu+\frac{{\beta}+d-1}{2}$ and

(5.7)
${\mathbf{J}}_{m,{\mathbf{k}}}^{n}(x,t):=P_{n-m}^{(2{\alpha}+2m,{\gamma})}(1-2t)t^{m}P_{\mathbf{k}}^{m}\left(W_{\mu};\frac{x}{t}\right),\quad$

Then $\\{{\mathbf{J}}_{m,{\mathbf{k}}}^{n}(x,t):|{\mathbf{k}}|=m,\,0\leq m\leq
n\\}$ is an orthogonal basis of
${\mathcal{V}}_{n}({\mathbb{V}}^{d+1},W_{{\beta},{\gamma},\mu})$ and the norm
of ${\mathbf{J}}_{m,{\mathbf{k}}}^{n}$ is given by

(5.8)
${\mathbf{H}}_{m,n}^{({\alpha},{\gamma})}:={\langle}{\mathbf{J}}_{m,{\mathbf{k}}}^{n},{\mathbf{J}}_{m,{\mathbf{k}}}^{n}{\rangle}_{W_{\mu,{\beta},{\gamma}}}=\frac{c_{2{\alpha},{\gamma}}}{c_{2{\alpha}+2m,{\gamma}}}h_{n-m}^{(2{\alpha}+2m,{\gamma})},$

where $c_{{\alpha},{\beta}}$ is as in (2.14) and $h_{m}^{({\alpha},{\gamma})}$
is the norm square of the Jacobi polynomial. We call
${\mathbf{J}}_{m,{\mathbf{k}}}^{n}$ the Jacobi polynomials on the cone.

The reproducing kernel of the space
${\mathcal{V}}_{n}({\mathbb{V}}^{d+1},W_{{\beta},{\gamma},\mu})$, denoted by
${\mathbf{P}}_{n}(W_{{\beta},{\gamma},\mu};\cdot,\cdot)$, can be written in
terms of the above basis,

${\mathbf{P}}_{n}\big{(}W_{{\beta},{\gamma},\mu};(x,t),(y,s)\big{)}=\sum_{m=0}^{n}\sum_{|{\mathbf{k}}|=n}\frac{{\mathbf{J}}_{m,{\mathbf{k}}}^{n}(x,t){\mathbf{J}}_{m,{\mathbf{k}}}^{n}(y,s)}{{\mathbf{H}}_{m,n}^{{\beta},{\gamma}}}.$

It is the kernel of the projection
$\operatorname{proj}_{n}(W_{{\beta},{\gamma},\mu}):L^{2}({\mathbb{V}}^{d+1},W_{{\beta},{\gamma},\mu})\to{\mathcal{V}}_{n}({\mathbb{V}}^{d+1},W_{{\beta},{\gamma},\mu})$,

$\operatorname{proj}_{n}(W_{{\beta},{\gamma},\mu};f)=\int_{{\mathbb{V}}^{d+1}}f(y,s){\mathbf{P}}_{n}\big{(}W_{{\beta},{\gamma},\mu};\,\cdot,(y,s)\big{)}W_{{\beta},{\gamma},\mu}(s)\mathrm{d}y\mathrm{d}s.$

In contrast to the conic surface, it is the case ${\beta}=0$ that turns out to
be the most interesting one. For ${\beta}=0$, there is a second order
differential operator that has orthogonal polynomials as eigenfunctions.
Recall that $W_{0,{\gamma},\mu}=W_{{\gamma},\mu}$.

######  Theorem 5.7.

Let $\mu>-\tfrac{1}{2}$, ${\gamma}>-1$ and $n\in{\mathbb{N}}_{0}$. Define the
second order differential operator ${\mathfrak{D}}_{\mu,{\gamma}}$ by

$\displaystyle{\mathfrak{D}}_{{\gamma},\mu}:=$
$\displaystyle\,t(1-t)\partial_{t}^{2}+2(1-t){\langle}x,\nabla_{x}{\rangle}\partial_{t}+\sum_{i=1}^{d}(t-x_{i}^{2})\partial_{x_{i}}^{2}-2\sum_{i<j}x_{i}x_{j}\partial_{x_{i}}\partial_{x_{j}}$
$\displaystyle+(2\mu+d)\partial_{t}-(2\mu+{\gamma}+d+1)({\langle}x,\nabla_{x}{\rangle}+t\partial_{t}),$

where $\nabla_{x}$ and $\Delta_{x}$ denote the gradient and the Laplace
operator in $x$-variable. Then the polynomials in
${\mathcal{V}}_{n}({\mathbb{V}}^{d+1},W_{{\gamma},\mu})$ are eigenfunctions of
${\mathfrak{D}}_{{\gamma},\mu}$,

(5.9) ${\mathfrak{D}}_{{\gamma},\mu}u=-n(n+2\mu+{\gamma}+d)u,\qquad\forall
u\in{\mathcal{V}}_{n}({\mathbb{V}}^{d+1},W_{{\gamma},\mu}).$

The reproducing kernel enjoys an addition formula that is a mixture of the
addition formula for orthogonal polynomials with respect to the classical
weight function $W_{\mu}$ on the unit ball and the Jacobi polynomials. The
addition formula is complicated and has the most elegant form when
${\beta}=0$.

######  Theorem 5.8.

Let $d\geq 2$. For $\mu\geq 0$ and ${\gamma}\geq-\frac{1}{2}$, let
${\alpha}=\mu+\frac{d-1}{2}$; then

(5.10)
$\displaystyle{\mathbf{P}}_{n}\big{(}W_{{\gamma},\mu};(x,t),(y,s)\big{)}=\,$
$\displaystyle
c_{\mu,{\gamma},d}\int_{[-1,1]^{3}}Z_{2n}^{2{\alpha}+{\gamma}+1}(\xi(x,t,y,s;u,v))$
$\displaystyle\times(1-u^{2})^{\mu-1}(1-v_{1}^{2})^{{\alpha}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}u\mathrm{d}v,$

where $c_{\mu,{\gamma},d}$ is a constant, so that ${\mathbf{P}}_{0}=1$ and
$\xi(x,t,y,s;u,v)\in[-1,1]$ is defined by

(5.11) $\displaystyle\xi(x,t,y,s;u,v)=$
$\displaystyle\,v_{1}\sqrt{\tfrac{1}{2}\left(ts+{\langle}x,y{\rangle}+\sqrt{t^{2}-\|x\|^{2}}\sqrt{s^{2}-\|y\|^{2}}\,u\right)}$
$\displaystyle+v_{2}\sqrt{1-t}\sqrt{1-s}.$

When $\mu=0$ or ${\gamma}=-\frac{1}{2}$, the identity (5.10) holds under the
limit (2.28). In particular,

(5.12) $\displaystyle{\mathbf{P}}_{n}\big{(}W_{{\gamma},0};$
$\displaystyle(x,t),(y,s)\big{)}=c_{0,{\gamma},d}$
$\displaystyle\times\int_{[-1,1]^{2}}\left[Z_{2n}^{{\gamma}+d}(\xi(x,t,y,s;1,v))+Z_{2n}^{{\gamma}+d}(\xi(x,t,y,s;-1,v))\right]$
$\displaystyle\qquad\qquad\quad\times(1-v_{1}^{2})^{\frac{d-1}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v.$

As a consequence, the orthogonal structure for the weight function
$W_{{\gamma},\mu}$ on the cone ${\mathbb{V}}^{d+1}$ satisfies both
characteristic properties specified in Definition 3.1 and Definition 3.2. In
the next subsection, we use the addition formula to establish Assertions 1–3
for highly localized kernels.

###  5.4. Highly localized kernels

Let $\widehat{a}$ be an admissible cut-off function. For $(x,t)$,
$(y,s)\in{\mathbb{V}}^{d+1}$, the localized kernel
${\mathbf{L}}_{n}(W_{{\gamma},\mu};\cdot,\cdot)$ is defined by

${\mathbf{L}}_{n}\left(W_{{\gamma},\mu};(x,t),(y,s)\right)=\sum_{j=0}^{\infty}\widehat{a}\left(\frac{j}{n}\right){\mathbf{P}}_{j}\left(W_{{\gamma},\mu};(x,t),(y,s)\right).$

For ${\gamma}>-1$ and $\mu\geq 0$, we need the function
$n^{d+1}W_{{\gamma},\mu}\big{(}{\mathbf{c}}((x,t),n^{-2})\big{)}$, which we
denote by $W_{{\gamma},\mu}(n;x,t)$; that is,

$W_{{\gamma},\mu}(n;x,t):=\big{(}t+n^{-2}\big{)}^{\frac{d-1}{2}}\big{(}1-t+n^{-2}\big{)}^{{\gamma}+\frac{1}{2}}\big{(}t^{2}-\|x\|^{2}+n^{-2}\big{)}^{\mu}.$

######  Theorem 5.9.

Let $d\geq 2$, ${\gamma}\geq-\frac{1}{2}$ and $\mu\geq 0$. Then for any $k>0$
and $(x,t),(y,s)\in{\mathbb{V}}^{d+1}$,

$\left|{\mathbf{L}}_{n}\left(W_{{\gamma},\mu};(x,t),(y,s)\right)\right|\leq\frac{c_{\kappa}n^{d+1}}{\sqrt{W_{{\gamma},\mu}(n;x,t)}\sqrt{W_{{\gamma},\mu}(n;y,s)}}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\big{)}^{-{\kappa}}.$

###### Proof.

The proof follows along the line of the estimate on the surface of the cone,
but it will be more involved as can be seen in the lower bound of
$1-\xi(x,t,y,s;u,v)$ at (5.15) below.

The kernel ${\mathbf{L}}_{n}(W_{\mu,{\gamma}})$ can be written in terms of the
kernel of the Jacobi polynomials by the addition formula in Theorem 5.8, which
gives, with ${\alpha}=\mu+\frac{d-1}{2}$ and ${\lambda}=2\mu+{\gamma}+d$, that

$\displaystyle{\mathbf{L}}_{n}\left(W_{{\gamma},\mu};(x,t),(y,s)\right)=c_{\mu,{\gamma},d}\int_{[-1,1]^{3}}$
$\displaystyle
L_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}\big{(}2\xi(x,t,y,s;u,v)^{2}-1\big{)}$
$\displaystyle\times(1-u^{2})^{\mu-1}(1-v_{1}^{2})^{{\alpha}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}u\mathrm{d}v.$

Then, as in the proof of Theorem 4.10, we can apply the estimate (2.17) for
$L_{n}^{{\alpha},{\beta}}$ with $\alpha={\lambda}-1/2$, $\beta=-1/2$ to obtain

$\displaystyle\left|{\mathbf{L}}_{n}\left(W_{{\gamma},\mu};(x,t),(y,s)\right)\right|\leq
cn^{2{\lambda}+1}\int_{[-1,1]^{3}}$
$\displaystyle\frac{1}{\left(1+n\sqrt{1-\xi(x,t,y,s;u,v)^{2}}\right)^{{\kappa}+5\mu+3{\gamma}+\frac{3d}{2}+1}}$
$\displaystyle\times(1-u^{2})^{\mu-1}(1-v_{1}^{2})^{{\alpha}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}u\mathrm{d}v.$

We first need an lower bound for $\xi(x,t,y,s;u,v)$. We simplify the notation
and write

$\xi(x,t,y,s;u,v)=v_{1}\sqrt{\eta(x,t,y,s;u)}+v_{2}\sqrt{1-t}\sqrt{1-s}$

by introducing the notation

$\eta(x,t,y,s;u)=\tfrac{1}{2}\left(ts+{\langle}x,y{\rangle}+\sqrt{t^{2}-\|x\|^{2}}\sqrt{s^{2}-\|y\|^{2}}\,u\right).$

Let $x=tx^{\prime}$ and $y=sy^{\prime}$, where
$x^{\prime},y^{\prime}\in{\mathbb{B}}^{d}$. Then by $|u|\leq 1$, it is easy to
verify that $0\leq\eta(x,t,y,s;u)\leq ts$ and, consequently,
$|\xi(x,t,y,s;u,v)|\leq\sqrt{ts}+\sqrt{1-s^{2}}\sqrt{1-t^{2}}\leq 1$.
Furthermore, we can write

(5.13) $\displaystyle 1-\xi(x,t,y,s;u,v)=\,$ $\displaystyle
1-\sqrt{\eta(x,t,y,s;u)}-\sqrt{1-t}\sqrt{1-s}$
$\displaystyle+(1-v_{1})\sqrt{\eta(x,t,y,s;u)}+(1-v_{2})\sqrt{1-t}\sqrt{1-s}$
$\displaystyle\geq\,$ $\displaystyle
1-\sqrt{\eta(x,t,y,s;1)}-\sqrt{1-t}\sqrt{1-s}$ $\displaystyle=\,$
$\displaystyle
1-\cos{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))=2\sin^{2}\frac{{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))}{2}$
$\displaystyle\geq\,$
$\displaystyle\frac{2}{\pi^{2}}[{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))]^{2}.$

As in the case of ${\mathbb{V}}_{0}^{d+1}$, we use this inequality to obtain
the estimate

$\displaystyle\left|{\mathbf{L}}_{n}\left(W_{{\gamma},\mu};(x,t),(y,s)\right)\right|\leq
cn^{2{\lambda}+1}\frac{1}{(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)))^{{\kappa}+\mu+{\gamma}+\frac{d}{2}}}I(x,t,y,s)$

with the integral $I(x,t,y,s)$ defined by

$I(x,t,y,s)=c_{{\gamma},\mu}\int_{[-1,1]^{3}}\frac{(1-u^{2})^{\mu-1}(1-v_{1}^{2})^{{\alpha}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}}{(1+n\sqrt{1-\xi(x,t,y,s;u,v)})^{4\mu+2{\gamma}+d+1}}\mathrm{d}u\mathrm{d}v,$

where $c_{{\gamma},\mu}$ is the normalization constant so that $I(x,t,y,s)=1$
when $n=0$. To complete the proof, we need to show that

$\frac{I(x,t,y,s)}{(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)))^{\mu+{\gamma}+\frac{d}{2}}}\leq\frac{cn^{-(4\mu+2{\gamma}+d)}}{\sqrt{W_{{\gamma},\mu}(n;x,t)}\sqrt{W_{{\gamma},\mu}(n;y,s)}}.$

This is a special case of Lemma 5.10 established below. ∎

######  Lemma 5.10.

Let $d\geq 2$ and ${\gamma}\geq-\frac{1}{2}$ and $\mu\geq 0$. Then for
${\beta}\geq 4\mu+2{\gamma}+d+1$,

(5.14) $\displaystyle c_{{\gamma},\mu}$
$\displaystyle\int_{[-1,1]^{3}}\frac{(1-u^{2})^{\mu-1}(1-v_{1}^{2})^{{\alpha}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}}{(1+n\sqrt{1-\xi(x,t,y,s;u,v)})^{{\beta}}}\mathrm{d}u\mathrm{d}v$
$\displaystyle\leq\frac{cn^{-(4\mu+2{\gamma}+d)}}{\sqrt{W_{{\gamma},\mu}(n;x,t)}\sqrt{W_{{\gamma},\mu}(n;y,s)}(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)))^{{\beta}-5\mu-3{\gamma}-\frac{3d}{2}-1}}.$

###### Proof.

Let $I(x,t,y,s)$ be defined as in the proof of the previous proof. Using the
lower bound of $\xi$ in (5.13), we see that the left-hand side of (5.14) is
bounded by

$\frac{1}{(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)))^{{\beta}-4\mu-2{\gamma}-d-1}}I(x,t,y,s).$

In order to estimate $I(x,t,y,s)$ we need a refined lower bound for
$1-\xi(x,t,y,s;u,v)$. Let $x=tx^{\prime}$ and $y=sy^{\prime}$ with
$x^{\prime},y^{\prime}\in{\mathbb{B}}^{d}$. We claim that for $-1\leq
v_{1},v_{2},u\leq 1$

(5.15) $\displaystyle 1-\xi(x,t,y,s;u,v)\,$
$\displaystyle\geq\frac{1}{4}\sqrt{ts}(1-v_{1})+\sqrt{1-s}\sqrt{1-t}(1-v_{2})$
$\displaystyle+\frac{1}{4}\sqrt{ts}\sqrt{1-\|x^{\prime}\|^{2}}\sqrt{1-\|y^{\prime}\|^{2}}(1-u).$

To prove this lower bound, we write
$A(x^{\prime},y^{\prime})=\sqrt{1-\|x^{\prime}\|^{2}}\sqrt{1-\|y^{\prime}\|^{2}}$
so that the notation can be further simplified as

$\displaystyle\eta(x,t,y,s;u)=\tfrac{1}{2}ts\big{(}1+{\langle}x^{\prime},y^{\prime}{\rangle}+A(x^{\prime},y^{\prime})u\big{)}.$

We shall use $\xi$ and $\eta$ without their variables below. Our first step is
to write

$\displaystyle 1-\xi\,$
$\displaystyle=1-\sqrt{ts}-\sqrt{1-t}\sqrt{1-s}+\sqrt{ts}-\sqrt{\eta}+(1-v_{1})\sqrt{\eta}+(1-v_{2})\sqrt{1-t}\sqrt{1-s}$
$\displaystyle\geq\sqrt{ts}-\sqrt{\eta}+(1-v_{1})\sqrt{\eta}+(1-v_{2})\sqrt{1-t}\sqrt{1-s}.$

Next, since $\eta(x,t,y,s;u)\leq ts$ by the Cauchy’s inequality, we obtain

$\displaystyle\sqrt{ts}-\sqrt{\eta}\,$
$\displaystyle=\frac{ts-\eta}{\sqrt{ts}+\sqrt{\eta}}\geq\frac{ts}{4\sqrt{ts}}\left(1-{\langle}x^{\prime},y^{\prime}{\rangle}-A(x^{\prime},y^{\prime})\,u\right)$
$\displaystyle=\frac{\sqrt{ts}}{4}\big{(}1-{\langle}x^{\prime},y^{\prime}{\rangle}-A(x^{\prime},y^{\prime})+(1-u)A(x^{\prime},y^{\prime})\big{)}.$

Since $1-{\langle}x^{\prime},y^{\prime}{\rangle}-A(x^{\prime},y^{\prime})\geq
0$ and $1\geq 1-v_{1}$, this leads to

$\displaystyle 1-\xi\,$ $\displaystyle\geq
B\sqrt{ts}(1-v_{1})+\sqrt{1-s}\sqrt{1-t}(1-v_{2})+\frac{\sqrt{ts}}{4}A(x^{\prime},y^{\prime})(1-u),$

where $B$ is given by

$\displaystyle
B=\frac{1}{4}\big{(}1-{\langle}x^{\prime},y^{\prime}{\rangle}-A(x^{\prime},y^{\prime})\big{)}+\sqrt{\tfrac{1}{2}\big{(}1+{\langle}x^{\prime},y^{\prime}{\rangle}+A(x^{\prime},y^{\prime})u\big{)}},$

which can be rewritten to give

$\displaystyle B\,$
$\displaystyle=1-\frac{1}{4}(1-u)A(x^{\prime},y^{\prime})-\frac{1}{2}\left(1-\sqrt{\tfrac{1}{2}\big{(}1+{\langle}x^{\prime},y^{\prime}{\rangle}+A(x^{\prime},y^{\prime})u\big{)}}\right)^{2}$
$\displaystyle\geq 1-\frac{1}{4}-\frac{1}{2}=\frac{1}{4}.$

This proves (5.15). We are now ready to prove (5.14). Denote its left-hand
side by $J(x,t,u,v)$. Using the inequality (5.15), we obtain

$J(x,t,y,s)\leq
c\int_{[0,1]^{3}}\frac{(1-u^{2})^{\mu-1}(1-v_{1}^{2})^{\mu+\frac{d-1}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}}{\left(1+n\sqrt{B_{1}(1-v_{1})+B_{2}(1-v_{2})+B_{3}(1-u)}\right)^{4\mu+2{\gamma}+d+1}}\mathrm{d}u\mathrm{d}v,$

where $B_{1}=\frac{1}{4}\sqrt{t}\sqrt{s}$, $B_{2}=\sqrt{1-s}\sqrt{1-t}$ and
$B_{3}=\frac{1}{4}\sqrt{ts}\sqrt{1-\|x^{\prime}\|^{2}}\sqrt{1-\|y^{\prime}\|^{2}}$.
We can estimate the integral as in the proof of Theorem 4.10 by applying
(4.14), which we now need to apply three times, and we obtain

$\displaystyle J(x,t,y,s)\,$ $\displaystyle\leq
c\frac{n^{-2\mu}}{B_{3}^{\mu}}\int_{[0,1]^{2}}\frac{(1-v_{1}^{2})^{\mu+\frac{d-1}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}}{\left(1+n\sqrt{B_{1}(1-v_{1})+B_{2}(1-v_{2})}\right)^{2\mu+2{\gamma}+d}}\mathrm{d}v$
$\displaystyle\leq
c\frac{n^{-2\mu-2{\gamma}-1}}{B_{3}^{\mu}B_{2}^{{\gamma}+\frac{1}{2}}}\int_{0}^{1}\frac{(1-v_{1}^{2})^{\mu+\frac{d-1}{2}-1}}{\left(1+n\sqrt{B_{1}(1-v_{1})}\right)^{2\mu+d-1}}\mathrm{d}v_{1}$
$\displaystyle\leq
c\frac{n^{-4\mu-2{\gamma}-d}}{B_{3}^{\mu}B_{2}^{{\gamma}+\frac{1}{2}}B_{1}^{\mu+\frac{d-1}{2}}}.$

Combining part of $B_{1}$ and $B_{3}$, it follows from
$\sqrt{t}\sqrt{s}B_{3}=\frac{1}{4}\sqrt{t^{2}-\|x\|^{2}}\sqrt{s^{2}-\|y\|^{2}}$
that

$\displaystyle J(x,t,y,s)\leq
c\frac{n^{-4\mu-2{\gamma}-d}}{\left(\sqrt{t^{2}-\|x\|^{2}}\sqrt{s^{2}-\|y\|^{2}}\right)^{\mu}\left(\sqrt{1-t}\sqrt{1-s}\right)^{{\gamma}+\frac{1}{2}}\left(\sqrt{t}\sqrt{s}\right)^{\frac{d-1}{2}}}$
$\displaystyle\leq
c\frac{n^{-4\mu-2{\gamma}-d}}{\left(\sqrt{t^{2}-\|x\|^{2}}\sqrt{s^{2}-\|y\|^{2}}+n^{-2}\right)^{\mu}\left(\sqrt{1-t}\sqrt{1-s}+n^{-2}\right)^{{\gamma}+\frac{1}{2}}\left(\sqrt{t}\sqrt{s}+n^{-2}\right)^{\frac{d-1}{2}}},$

where the second inequality follows since $I(x,t,y,s)\leq 1$ holds trivially.
We can now apply (4.15) three times to obtain

$\displaystyle J(x,t,y,s)\leq$
$\displaystyle\,c\frac{n^{-4\mu-2{\gamma}-d}}{\sqrt{W_{\mu,{\gamma}}(n;x,t)}\sqrt{W_{\mu,{\gamma}}(n;y,s)}}\big{(}1+n\big{|}\sqrt{t^{2}-\|x\|^{2}}-\sqrt{s^{2}-\|y\|^{2}}\big{|}\big{)}^{\mu}$
$\displaystyle\times\big{(}1+n\big{|}\sqrt{t}-\sqrt{s}\big{|}\big{)}^{\frac{d-1}{2}}\big{(}1+n\big{|}\sqrt{1-t}-\sqrt{1-s}\big{|}\big{)}^{{\gamma}+\frac{1}{2}}.$

Finally, by Lemma 5.4, we conclude

$J(x,t,y,s)\leq
c\frac{n^{-4\mu-2{\gamma}-d}(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)))^{\mu+{\gamma}+\frac{d}{2}}}{\sqrt{W_{{\gamma},\mu}(n;x,t)}\sqrt{W_{{\gamma},\mu}(n;y,s)}}$

which is what we need to complete the proof. ∎

The theorem we just proved verifies Assertion 1 for $W_{{\gamma},\mu}$. Our
next theorem verifies Assertion 2 but only for $W_{{\gamma},0}$. Recall that
$W_{{\gamma},0}$ depends only on $t$,

$W_{{\gamma},0}(n;x,t)=\big{(}t+n^{-2}\big{)}^{\frac{d-1}{2}}\big{(}1-t+n^{-2}\big{)}^{{\gamma}+\frac{1}{2}},$

######  Theorem 5.11.

Let $d\geq 2$, ${\gamma}\geq-\frac{1}{2}$. For
$(x_{i},t_{i}),(y,s)\in{\mathbb{V}}^{d+1}$, and
$(x_{1},t_{1})\in{\mathsf{c}}((x_{2},t_{2}),c^{*}n^{-1})$ with $c^{*}$ small,
and any ${\kappa}>0$,

(5.16)
$\displaystyle\left|{\mathbf{L}}_{n}\left(W_{{\gamma},0};(x_{1},t_{1}),(y,s)\right)-{\mathbf{L}}_{n}\left(W_{{\gamma},0};(x_{2},t_{2}),(y,s)\right)\right|$
$\displaystyle\qquad\qquad\leq\frac{c_{\kappa}n^{d+2}{\mathsf{d}}_{{\mathbb{V}}}((x_{1},t_{1}),(x_{2},t_{2}))}{\sqrt{W_{{\gamma},0}(n;x_{2},t_{2})}\sqrt{W_{{\gamma},0}(n;y,s)}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}}((x_{2},t_{2}),(y,s))\big{)}^{{\kappa}}}.$

###### Proof.

Let $K$ denote the left-hand side of (5.16). As in the proof of Theorem 4.13,
we use the integral expression of ${\mathbf{L}}_{n}(W_{{\gamma},0})$ and
(5.12) to obtain $K\leq K_{1}+K_{-1}$, where

$\displaystyle K_{u}$ $\displaystyle\leq 2\int_{[-1,1]^{2}}\big{\|}\partial
L_{n}^{{\gamma}+d-\frac{1}{2},-\frac{1}{2}}\big{(}2(\cdot)^{2}-1\big{)}\big{\|}_{L^{\infty}(I_{u,v})}\big{|}\xi_{1}(u,v)^{2}-\xi_{2}(u,v)^{2}|$
$\displaystyle\qquad\qquad\qquad\times(1-v_{1}^{2})^{\frac{d-1}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v,$

where $u=1$ or $-1$, $\partial f=f^{\prime}$,
$\xi_{i}(u,v)=\xi(x_{i},t_{i},y,s;u,v)$ and $I_{u,v}$ is the interval with end
points $\xi_{1}(u,v)$ and $\xi_{2}(u,v)$. We first consider the case $u=1$.
Clearly $|\xi_{1}(1,v)^{2}-\xi_{2}(1,v)^{2}|\leq
2|\xi_{1}(1,v)-\xi_{2}(1,v)|$. We claim that it has the upper bound

(5.17) $\displaystyle|\xi_{1}(u,v)-\xi_{2}(u,v)|\leq
c\,{\mathsf{d}}_{{\mathbb{V}}}\big{(}(x_{1},t_{1}),(x_{2},t_{2})\big{)}\big{[}\Sigma_{1}+\Sigma_{2}(v_{1})+\Sigma_{3}(v_{2})\big{]},$

where

$\displaystyle\Sigma_{1}\,$
$\displaystyle={\mathsf{d}}_{{\mathbb{V}}}\big{(}(x_{i},t_{i}),(y,s)\big{)}+{\mathsf{d}}_{{\mathbb{V}}}\big{(}(x_{1},t_{1}),(x_{2},t_{2})\big{)},$
$\displaystyle\Sigma_{2}(v_{1})\,$ $\displaystyle=(1-v_{1})\sqrt{s},$
$\displaystyle\Sigma_{3}(v_{2})\,$ $\displaystyle=(1-v_{2})\sqrt{1-s}.$

To see this, we follow the notation used in the proof of Lemma 5.10, writing
$\xi_{i}=v_{1}\sqrt{\eta_{i}(1)}+v_{2}\sqrt{1-t}\sqrt{1-s}$, where
$\eta_{i}(1)=\eta(x_{i},t_{i},s_{i};1)$. Using (5.3) and writing
$x_{i}=t_{i}x_{i}^{\prime}$ and $y=sy^{\prime}$, we obtain

(5.18) $\displaystyle\xi_{1}(1,v)-\xi_{2}(1,v)\,$
$\displaystyle=\xi_{1}(1,\mathbf{1})-\xi_{2}(1,\mathbf{1})$
$\displaystyle+(1-v_{1})\big{(}\sqrt{\eta_{2}(1)}-\sqrt{\eta_{1}(1)}\big{)}$
$\displaystyle+(1-v_{2})\big{(}\sqrt{1-t_{2}}-\sqrt{1-t_{1}}\big{)}\sqrt{1-s},$

where $\mathbf{1}=(1,1)$. For $u=1$, as in Proposition 5.3,

$\eta_{i}(1)=t_{i}s\cos\frac{{\mathsf{d}}_{{\mathbb{B}}}(x_{i}^{\prime},y^{\prime})}{2}\quad\hbox{and}\quad\xi_{i}(1,\mathbf{1})=\cos{\mathsf{d}}_{{\mathbb{V}}}\big{(}(x_{i},t_{i}),(y,s)\big{)}.$

Substituting these into (5.18), the resulted identity is similar to the
corresponding expression in the proof of Theorem 4.13 and can be analogously
estimated using Proposition 5.3 and Lemma 5.4, so that (5.17) follows.

As in the proof of Theorem 4.13, it follows from (2.18) with $m=1$ that

$\displaystyle\big{\|}\partial
L_{n}^{{\lambda}-\frac{1}{2},-\frac{1}{2}}\big{(}2(\cdot)^{2}-1\big{)}\big{\|}_{L^{\infty}(I_{1,v})}$
$\displaystyle\qquad\qquad\leq
c\left[\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\xi_{1}(1,v)^{2}}\big{)}^{{\kappa}}}+\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\xi_{2}(1,v)^{2}}\big{)}^{{\kappa}}}\right].$

Hence, $K_{1}$ is bounded by, with ${\lambda}={\gamma}+d$,

$\displaystyle K_{1}\leq
c\,{\mathsf{d}}_{{\mathbb{V}}}\big{(}(x_{1},t_{1}),(x_{2},t_{2})\big{)}$
$\displaystyle\int_{[-1,1]^{2}}\left[\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\xi_{1}(1,v)^{2}}\big{)}^{{\kappa}}}+\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\xi_{2}(1,v)^{2}}\big{)}^{{\kappa}}}\right]$
$\displaystyle\times\big{(}\Sigma_{1}+\Sigma_{2}(v_{1})+\Sigma_{3}(v_{2})\big{)}(1-v_{1}^{2})^{\frac{d-1}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v.$

Using the assumption that
$(x_{1},t_{1})\in{\mathsf{c}}((x_{2},t_{2}),c^{*}n^{-1})$, the integral that
contains $\Sigma_{1}$ term can be handled as in the proof of Theorem 4.13,
which leads to

$\displaystyle\int_{[-1,1]^{2}}$
$\displaystyle\frac{n^{2{\lambda}+3}}{\big{(}1+n\sqrt{1-\xi_{i}(1,v)^{2}}\big{)}^{{\kappa}}}\Sigma_{1}(1-v_{1}^{2})^{\frac{d-1}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}v$
$\displaystyle\leq\frac{n^{2{\lambda}+2}}{\big{(}1+n{\mathsf{d}}_{\mathbb{V}}((x,t),(y,s))\big{)}^{{\kappa}-2{\gamma}-d-2}}\int_{[-1,1]^{2}}\frac{(1-v_{1}^{2})^{\frac{d-1}{2}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}}{\big{(}1+n\sqrt{1-\xi_{i}(1,v)^{2}}\big{)}^{2{\gamma}+d+2}}\mathrm{d}v.$

Now, the inequality (5.15) shows that

$1-\xi_{i}(1,v)\geq\frac{1}{4}\sqrt{t_{i}s}(1-v_{1})+\sqrt{1-t_{i}}\sqrt{1-s}(1-v_{2}).$

Hence, applying Corollary 4.12 with $d-1$ replaced by $d$ and Lemma 5.4, we
conclude that the integral in the right-hand side is bound by

$\displaystyle\frac{c\,n^{d+2}}{\sqrt{W_{{\gamma},0}(n;x_{i},t_{i})}\sqrt{W_{{\gamma},0}(n;y,s)}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}}((x_{i},t_{i}),(y,s))\big{)}^{{\kappa}({\gamma},d)}},$

where ${\kappa}({\gamma},d)={\kappa}-3{\gamma}-\frac{3d+4}{2}$, since
$W_{{\gamma},0}(n;x,t)={\mathsf{w}}_{{\gamma},d+1}(n;t)$. Putting together, we
see that the integral containing $\Sigma_{1}$ has the desired upper bound. The
two integrals containing $\Sigma_{2}(v_{1})$ and $\Sigma_{3}(v_{2})$ can be
estimated similarly and more straightforwardly by using Corollary 4.12.
Moreover, the estimate is parallel to the two corresponding terms in the proof
of Theorem 4.13 and can be carried out similarly. This completes the proof of
the case $u=1$.

For $u=-1$, we observe that if we define $Y_{*}=(y,-\sqrt{s^{2}-\|y\|^{2}})$
and

${\mathsf{d}}_{*}((x,t),(y,s))=\arccos\left(\sqrt{\frac{{\langle}X,Y_{*}{\rangle}+ts}{2}}+\sqrt{1-t}\sqrt{1-s}\right),$

then for $u=-1$ and $v=\mathbf{1}=(1,1)$,

$\xi(x,t,y,s;-1,\mathbf{1})=\cos{\mathsf{d}}_{*}((x,t),(y,s)).$

Since $(Y_{*},s)\in{\mathbb{V}}_{0}^{d+2}$ and
${\mathsf{d}}_{*}((x,t),(y,s))={\mathsf{d}}_{{\mathbb{V}}_{0}^{d+2}}((X,t),(Y_{*},s))$,
it follows that ${\mathsf{d}}_{*}$ is also a distance function and satisfies,
in particular, the triangle inequality. As a result, we can repeat the proof
for $u=1$ to estimate $K_{-1}$, so that $K_{-1}$ is bounded by the right-hand
side of (5.16) with $(1+n{\mathsf{d}}_{*}((x,t),(y,s)))^{\kappa}$ in place of
$(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)))^{\kappa}$ in the denominator.
Since ${\langle}X,Y{\rangle}\geq{\langle}X,Y_{*}{\rangle}$,
$\cos{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\geq\cos{\mathsf{d}}_{*}((x,t),(y,s))$
and, consequently,
${\mathsf{d}}_{\mathbb{V}}((x,t),(y,s))\leq{\mathsf{d}}_{*}((x,t),(y,s))$.
Hence, the upper bound of $K_{-1}$ is bounded by the right-hand side of
(5.16). This completes the proof. ∎

######  Remark 5.1.

It is not clear if this theorem also holds for $W_{{\gamma},\mu}$ when
$\mu>0$. The main task lies in establishing an analog of (5.15), perhaps with
a multiple of $1-u$ as one more term. This turns out to be elusive, not
because of lack of trying, and there appears to be a real obstacle for
$(x_{i},t_{i})$ around the apex of the cone.

Our next lemma is an analog of Lemma 4.14, which establishes Assertion 3 when
$p=1$ for the weight function $W_{{\gamma},\mu}$ on the solid cone.

######  Lemma 5.12.

Let $d\geq 2$, ${\gamma}>-1$ and $\mu>-\frac{1}{2}$. For $0<p<\infty$, assume
${\kappa}>\frac{2d+2}{p}+({\gamma}+\mu+\frac{d}{2})|\frac{1}{p}-\frac{1}{2}|$.
Then for $(x,t)\in{\mathbb{V}}^{d+1}$,

(5.19)
$\displaystyle\int_{{\mathbb{V}}^{d+1}}\frac{W_{{\gamma},\mu}(s)}{W_{{\gamma},\mu}(n;s)^{\frac{p}{2}}\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\big{)}^{{\kappa}p}}\mathrm{d}y\mathrm{d}s\leq
cn^{-d-1}\,W_{{\gamma},\mu}(n;t)^{1-\frac{p}{2}}.$

###### Proof.

Let $J_{p}$ denote the left-hand side of (5.19). Using the doubling property
of $W_{{\beta},{\gamma}}$ and by Lemma 2.4, it is sufficient to estimate
$J_{2,{\kappa}}$. Setting $y=sy^{\prime}$, $y^{\prime}\in{\mathbb{B}}^{d}$, we
obtain

$\displaystyle
J_{2}=\int_{0}^{1}s^{d}\int_{{\mathbb{B}}^{d}}\frac{W_{{\gamma},\mu}(sy^{\prime},s)}{W_{{\gamma},\mu}(n;sy^{\prime},s)(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(sy^{\prime},s)))^{2{\kappa}}}\mathrm{d}y^{\prime}\mathrm{d}s.$

Using ${\mathsf{w}}_{{\gamma},d}(n;t)$ defined in (4.10), it follows readily
that

$\frac{W_{{\gamma},\mu}(y,s)}{W_{{\gamma},\mu}(n;y,s)}=c\frac{s{\mathsf{w}}_{-1,{\gamma}}(s)}{(s^{2}-\|y\|^{2}+n^{-2})^{\frac{1}{2}}{\mathsf{w}}_{{\gamma},d+1}(n;s)}\leq
c\frac{{\mathsf{w}}_{-1,{\gamma}}(n;s)}{{\mathsf{w}}_{{\gamma},d+1}(n;s)\sqrt{1-\|y^{\prime}\|^{2}}},$

which leads to

$\displaystyle J_{2}\leq
c\int_{0}^{1}s^{d}\int_{{\mathbb{B}}^{d}}\frac{{\mathsf{w}}_{-1,{\gamma}}(n;s)}{{\mathsf{w}}_{{\gamma},d+1}(n;s)(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(sy^{\prime},s)))^{2{\kappa}}}\frac{\mathrm{d}y^{\prime}}{\sqrt{1-\|y^{\prime}\|^{2}}}\mathrm{d}s.$

Setting $x=tx^{\prime}$, $X=(x^{\prime},\sqrt{1-\|x^{\prime}\|^{2}})$ and
$Y=(y^{\prime},\sqrt{1-\|y^{\prime}\|^{2}})$, so that
${\mathsf{d}}_{{\mathbb{B}}^{d}}(x^{\prime},y^{\prime})={\mathsf{d}}_{\SS^{d}}(X,Y)$,
we use the identity

(5.20)
$\int_{{\mathbb{B}}^{d}}g\left(y^{\prime},\sqrt{1-\|y^{\prime}\|^{2}}\right)\frac{dy^{\prime}}{\sqrt{1-\|y^{\prime}\|^{2}}}=\int_{\SS_{+}^{d}}g(y)\mathrm{d}\sigma(y),$

where $\SS_{+}^{d}$ denotes the upper hemisphere of $\SS^{d}$, which allows us
to follow the proof of Proposition 4.15 to obtain

$\displaystyle J\,$ $\displaystyle\leq
c\int_{0}^{1}\int_{-1}^{1}\frac{s^{d}{\mathsf{w}}_{-1,{\gamma}}(s)(1-u^{2})^{\frac{d-2}{2}}}{{\mathsf{w}}_{{\gamma},d+1}(n;s)\left(1+n\arccos\left(\sqrt{ts}\sqrt{\frac{1+u}{2}}+\sqrt{1-t}\sqrt{1-s}\right)\right)^{2{\kappa}}}\mathrm{d}u\mathrm{d}s.$

Comparing with the proof of Proposition 4.15, we see that the above integral
with $d$ replaced by $d-1$ has already appeared in that proof, so that it is
bounded by $cn^{-d-1}$ as the proof of Proposition 4.15 shows. The proof is
completed. ∎

######  Proposition 5.13.

Let $d\geq 2$, $\mu\geq 0$ and ${\gamma}\geq-\frac{1}{2}$. For $0<p<\infty$
and $(x,t)\in{\mathbb{V}}^{d+1}$,

$\int_{{\mathbb{V}}^{d+1}}\left|{\mathbf{L}}_{n}\big{(}W_{{\gamma},\mu};(x,t),(y,s)\big{)}\right|^{p}W_{{\gamma},\mu}(y,s)\mathrm{d}y\mathrm{d}s\leq
c\left(\frac{n^{d+1}}{W_{{\gamma},\mu}(n;x,t)}\right)^{p-1}.$

This follows immediately from applying Lemma 5.12 on (i) of Theorem 5.9.

######  Corollary 5.14.

For $d\geq 2$ and ${\gamma}\geq-\frac{1}{2}$, the space
$({\mathbb{V}}^{d+1},W_{{\gamma},0},{\mathsf{d}}_{{\mathbb{V}}})$ is a
localizable homogeneous space.

###  5.5. Maximal ${\varepsilon}$-separated sets and MZ inequality

We provide a construction of maximal ${\varepsilon}$-separated sets on the
solid cone as defined in Definition 2.13.

Our construction follows the approach in Subsection 4.5. Instead of starting
with ${\varepsilon}$-separated sets on the unit sphere, we now need such sets
on the unit ball ${\mathbb{B}}^{d}$. We adopt the following notation. For
${\varepsilon}>0$, we denote by $\Xi_{{\mathbb{B}}}({\varepsilon})$ a maximal
${\varepsilon}$-separated set on the unit ball ${\mathbb{B}}^{d}$ and we let
${\mathbb{B}}_{u}({\varepsilon})$ be the subsets in ${\mathbb{B}}^{d}$ so that
the collection
$\\{{\mathbb{B}}_{u}({\varepsilon}):u\in\Xi_{\mathbb{B}}({\varepsilon})\\}$ is
a partition of ${\mathbb{B}}^{d}$, and we assume

(5.21)
${\mathbf{c}}_{{\mathbb{B}}}(u,c_{1}{\varepsilon})\subset{\mathbb{B}}_{u}({\varepsilon})\subset{\mathbf{c}}_{{\mathbb{B}}}(u,c_{2}{\varepsilon}),\qquad
u\in\Xi_{{\mathbb{B}}}({\varepsilon}),$

where ${\mathbf{c}}_{{\mathbb{B}}}(u,{\varepsilon})$ denotes the ball centered
at $u$ with radius ${\varepsilon}$ in ${\mathbb{B}}^{d}$, $c_{1}$ and $c_{2}$
depend only on $d$. It is known (see, for example, [38]) that such a
$\Xi_{\mathbb{B}}({\varepsilon})$ exists for all ${\varepsilon}>0$ and its
cardinality satisfies

(5.22)
$c_{d}^{\prime}{\varepsilon}^{-d}\leq\\#\Xi_{{\mathbb{B}}}({\varepsilon})\leq
c_{d}{\varepsilon}^{-d}.$

On the solid cone ${\mathbb{V}}^{d+1}$ we denote by
$\Xi_{{\mathbb{V}}}=\Xi_{{\mathbb{V}}}({\varepsilon})$ a maximum
${\varepsilon}$-separated set and denote by
$\\{{\mathbb{V}}(u,t):(tu,t)\in\Xi_{{\mathbb{V}}}\\}$ a partition of
${\mathbb{V}}^{d+1}$. We give a construction of these sets.

Let ${\varepsilon}>0$ and let
$N=\lfloor\frac{\pi}{2}{\varepsilon}^{-1}\rfloor$. We define
$t_{j}=\sin^{2}\frac{{\theta}_{j}}{2}$, $t_{j}^{+}$ and $t_{j}^{-}$, $1\leq
j\leq N$, as in Subsection 4.5, so that

${\mathbb{V}}^{d+1}=\bigcup_{j=1}^{N}{\mathbb{V}}^{(j)},\quad\hbox{where}\quad{\mathbb{V}}^{(j)}:=\left\\{(x,t)\in{\mathbb{V}}^{d+1}:t_{j}^{-}<t\leq
t_{j}^{+}\right\\}.$

Let ${\varepsilon}_{j}:=(2\sqrt{t_{j}})^{-1}\pi{\varepsilon}$. Then
$\Xi_{\mathbb{B}}({\varepsilon}_{j})$ is the maximal
${\varepsilon}_{j}$-separated set of ${\mathbb{B}}^{d}$ such that, for each
$j\geq 1$,
$\\{{\mathbb{B}}_{u}({\varepsilon}_{j}):u\in\Xi_{\mathbb{B}}({\varepsilon}_{j})\\}$
is a partition of ${\mathbb{B}}^{d}$ and
$\\#\Xi_{\mathbb{B}}({\varepsilon}_{j})\sim{\varepsilon}_{j}^{-d}.$ For each
$j=1,\ldots,N$, we decompose ${\mathbb{V}}^{(j)}$ by

${\mathbb{V}}^{(j)}=\bigcup_{u\in\Xi_{\mathbb{B}}({\varepsilon}_{j})}{\mathbb{V}}(u,t_{j}),\quad\hbox{where}\quad{\mathbb{V}}(u,t_{j}):=\left\\{(tv,t):t_{j}^{-}<t\leq
t_{j}^{+},\,v\in{\mathbb{B}}_{u}({\varepsilon}_{j})\right\\}.$

Finally, we define the set $\Xi_{{\mathbb{V}}}$ of ${\mathbb{V}}^{d+1}$ by

$\Xi_{{\mathbb{V}}}=\big{\\{}(t_{j}u,t_{j}):\,u\in\Xi_{\mathbb{B}}({\varepsilon}_{j}),\,1\leq
j\leq N\big{\\}}.$

######  Proposition 5.15.

Let ${\varepsilon}>0$ and $N=\lfloor\frac{\pi}{2}{\varepsilon}^{-1}\rfloor$.
Then $\Xi_{{\mathbb{V}}}$ is a maximal ${\varepsilon}$-separated set of
${\mathbb{V}}^{d+1}$ and
$\\{{\mathbb{V}}(t_{j}u,t_{j}):u\in\Xi_{\mathbb{B}}({\varepsilon}_{j}),\,1\leq
j\leq N\\}$ is a partition

${\mathbb{V}}^{d+1}=\bigcup_{j=1}^{N}\bigcup_{u\in\Xi_{\mathbb{B}}({\varepsilon}_{j})}{\mathbb{V}}(u,t_{j}).$

Moreover, there are positive constants $c_{1}$ and $c_{2}$ depending only on
$d$ such that

(5.23)
${\mathbf{c}}\big{(}(t_{j}u,t_{j}),c_{1}{\varepsilon}\big{)}\subset{\mathbb{V}}(u,t_{j})\subset{\mathbf{c}}\big{(}(t_{j}u,t_{j}),c_{2}{\varepsilon}\big{)},\qquad(t_{j}u,t_{j})\in\Xi_{\mathbb{V}},$

and $c_{d}^{\prime}$ and $c_{d}$ depending only on $d$ such that

$c_{d}^{\prime}{\varepsilon}^{-d-1}\leq\\#\Xi_{{\mathbb{V}}}\leq
c_{d}{\varepsilon}^{-d-1}.$

###### Proof.

The proof is parallel to that of Proposition 4.17. In fact, it can be carried
out almost verbatim under obvious modifications such as replacing
${\mathsf{d}}_{\SS}$ by ${\mathsf{d}}_{\mathbb{B}}$ and ${\mathbb{V}}_{0}$ by
${\mathbb{V}}$. We shall omit it. ∎

Even though we only established Assertion 2 for $W_{{\gamma},0}$ for
${\gamma}\geq 0$, it is sufficient for applying Theorem 5.16 to conclude that
the Marcinkiewicz-Zygmund inequality on a maximal ${\varepsilon}$-separated
set holds for any doubling weight on ${\mathbb{V}}^{d+1}$.

######  Theorem 5.16.

Let $W$ be a doubling weight on ${\mathbb{V}}^{d+1}$. Let $\Xi_{{\mathbb{V}}}$
be a maximal $\frac{\delta}{n}$-separated subset of ${\mathbb{V}}^{d+1}$ and
$0<\delta\leq 1$.

  1. $(i)$

For all $0<p<\infty$ and $f\in\Pi_{m}^{d+1}$ with $n\leq m\leq cn$,

$\sum_{z\in\Xi_{{\mathbb{V}}}}\Big{(}\max_{(x,t)\in{\mathbf{c}}((z,r),\frac{\delta}{n})}|f(x,t)|^{p}\Big{)}W\\!\left({\mathbf{c}}((z,r),\tfrac{\delta}{n})\right)\leq
c_{W,p}\|f\|_{p,W}^{p}$

where $c_{W,p}$ depends on $p$ when $p$ is close to $0$ and on the doubling
constant ${\alpha}(W)$.

  2. $(ii)$

For $0<r<1$, there is a $\delta_{r}>0$ such that for $\delta\leq\delta_{r}$,
$r\leq p<\infty$ and $f\in\Pi_{n}^{d+1}$,

$\displaystyle\|f\|_{p,W}^{p}\leq
c_{W,r}\sum_{z\in\Xi}\Big{(}\min_{(x,t)\in{\mathbf{c}}\bigl{(}(z,r),\tfrac{\delta}{n}\bigr{)}}|f(x,t)|^{p}\Big{)}W\bigl{(}{\mathbf{c}}((z,r),\tfrac{\delta}{n})\bigr{)}$

where $c_{W,r}$ depends only on $L(W)$ and on $r$ when $r$ is close to $0$.

###  5.6. Cubature rules and localized tight frames

We first verify Assertion 4 in Subsection 2.5 by constructing fast decaying
polynomials on the solid cone.

######  Lemma 5.17.

Let $d\geq 2$. For each $(x,t)\in{\mathbb{V}}^{d+1}$, there is a polynomial
${\mathcal{T}}_{x,t}$ of degree $n$ that satisfies

  1. (1)

${\mathcal{T}}_{x,t}(x,t)=1$, ${\mathcal{T}}_{x,t}(y,s)\geq c>0$ if
$(y,s)\in{\mathbf{c}}((x,t),\frac{\delta}{n})$, and for every ${\kappa}>0$,

$0\leq{\mathcal{T}}_{x,t}(y,s)\leq
c_{\kappa}\left(1+{\mathsf{d}}_{{\mathbb{V}}}\big{(}(x,t),(y,s)\big{)}\right)^{-{\kappa}},\quad(y,s)\in{\mathbb{V}}^{d+1}.$

  2. (2)

there is a polynomial $q$ of degree $n$ such that $q(x,t){\mathcal{T}}_{x,t}$
is a polynomial of degree $3n$ in $(x,t)$ and $1\leq q(x,t)\leq c$.

###### Proof.

For $(x,t),(y,s)\in{\mathbb{V}}^{d+1}$, we introduce the notation
$X=(x,\sqrt{t^{2}-\|x\|^{2}})$ and $Y=(y,\sqrt{s^{2}-\|y\|^{2}})$. Moreover,
we also denote $Y_{*}=(y,-\sqrt{t^{2}-\|y\|^{2}})$. Then $(X,t),(Y,s)$ and
$(Y_{*},s)$ are all elements of ${\mathbb{V}}_{0}^{d+2}$. Let $T_{(X,t)}$
denote the polynomial of degree $n$ on ${\mathbb{V}}_{0}^{d+2}$ defined in
Lemma 4.19. We now define

${\mathcal{T}}_{x,t}(y,s):=\frac{T_{(X,t)}(Y,s)+T_{(X,t)}(Y_{*},s)}{1+T_{(X,t)}(X_{*},t)}.$

Since $T_{(X,t)}(X,t)=1$, it follows that ${\mathcal{T}}_{(x,t)}(x,t)=1$.
Since $\sqrt{\frac{{\langle}X_{*},X{\rangle}+t^{2}}{2}}=\|x\|$, we have

$T_{(X,t)}(X_{*},t)=\frac{S_{n}(\|x\|+(1-t))+S_{n}(\|x\|-(1-t))}{1+S_{n}(2t-1)}.$

By $0\leq S_{n}(t)\leq c$, we see that $0\leq T_{(X,t)}(X_{*},s)\leq c$. In
particular, it follows that

${\mathcal{T}}_{x,t}(y,s)\geq c\,T_{(X,t)}(Y,s)\geq
c>0,\qquad(y,s)\in{\mathbf{c}}\big{(}(x,t),\tfrac{\delta}{n}\big{)},$

since
${\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))={\mathsf{d}}_{{\mathbb{V}}_{0}^{d+2}}((X,t),(Y,s))$.
Furthermore, since
$\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((X,t),(Y,s))\geq\cos{\mathsf{d}}_{{\mathbb{V}}_{0}}((X,t),(Y_{*},s))$,
we obtain

${\mathsf{d}}_{{\mathbb{V}}_{0}}((X,t),(Y_{*},s))\geq{\mathsf{d}}_{{\mathbb{V}}_{0}}((X,t),(Y,s))={\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s)).$

Hence, using the estimate of $T_{(X,t)}$ in Lemma 4.19, we conclude that

$\displaystyle 0\leq{\mathcal{T}}_{(x,t)}(y,s)\,$ $\displaystyle\leq
c\left[\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((X,t),(Y,s))\big{)}^{-{\kappa}}+\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}_{0}}((X,t),(Y_{*},s))\big{)}^{-{\kappa}}\right]$
$\displaystyle\leq
c\big{(}1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\big{)}^{-{\kappa}}.$

Finally, let $q(x,t)=(1+S_{n}(2t-1))T_{(X,t)}(X_{*},t)$. Then $q(x,t)$ is a
polynomial of degree at most $2n$, so that $q(x,t){\mathcal{T}}_{x,t}$ is a
polynomial of degree at most $3n$ and $1\leq q(x,t)\leq c$. This completes the
proof. ∎

By Propositions 2.17 and 2.18 and using Theorem 5.9, we have established the
following result on the Christoffel function ${\lambda}_{n}(W)$.

######  Corollary 5.18.

Let $W$ be a doubling weight function on ${\mathbb{V}}^{d+1}$. Then

$\lambda_{n}\big{(}W;(x,t)\big{)}\leq
c\,W\\!\left({\mathbf{c}}\left((x,t),\tfrac{1}{n}\right)\right).$

Moreover, for ${\gamma}\geq-\frac{1}{2}$ and $\mu\geq 0$,

$\lambda_{n}\big{(}W_{{\gamma},\mu};(x,t)\big{)}\geq
c\,W_{{\gamma},\mu}\left({\mathbf{c}}\left((x,t),\tfrac{1}{n}\right)\right)=cn^{-d-1}W_{{\gamma},\mu}(n;t).$

Since the lower bound of the Christoffel function holds for all doubling
weight, we see that the cubature rules in Theorem 2.20 holds for all doubling
weights on the solid cone.

######  Theorem 5.19.

Let $W$ be a doubling weight on ${\mathbb{V}}^{d+1}$. Let $\Xi$ be a maximum
$\frac{\delta}{n}$-separated subset of ${\mathbb{V}}^{d+1}$. There is a
$\delta_{0}>0$ such that for $0<\delta<\delta_{0}$ there exist positive
numbers ${\lambda}_{z,r}$, $(z,r)\in\Xi$, so that

(5.24)
$\int_{{\mathbb{V}}^{d+1}}f(x)W(x,t)\mathrm{d}x\mathrm{d}t=\sum_{(z,r)\in\Xi}{\lambda}_{z,r}f(z,r),\qquad\forall
f\in\Pi_{n}^{d+1}.$

Moreover, ${\lambda}_{z,r}\sim
W\\!\left({\mathbf{c}}((z,r),\tfrac{\delta}{n})\right)$ for all $(z,r)\in\Xi$.

We can now state our local frame on the solid cone. For $j=0,1,\ldots,$ let
$\Xi_{j}$ be a maximal $\frac{\delta}{2^{j}}$-separated subset in
${\mathbb{V}}^{d+1}$, so that

$\int_{{\mathbb{V}}^{d+1}}f(x,t)W(x,t)\mathrm{d}x\mathrm{d}t=\sum_{(z,r)\in\Xi_{j}}{\lambda}_{(z,r),j}f(z,r),\qquad
f\in\Pi_{2^{j}}^{d+1}.$

Denote by ${\mathbf{L}}_{n}(W)*f$ the near best approximation operator defined
by

${\mathbf{L}}_{n}(W)*f(x,t)=\int_{{\mathbb{V}}^{d+1}}f(y,s){\mathbf{L}}_{n}(W;(x,t),(y,s))W(y,s)\mathrm{d}y\mathrm{d}s.$

For $j=0,1,\ldots,$ define the operator $F_{j}(W)$ by

$F_{j}(W)*f={\mathbf{L}}_{2^{j-1}}(W)*f$

and define the frame elements $\psi_{(z,r),j}$ for $(z,r)\in\Xi_{j}$ by

$\psi_{(z,r),j}(x,t):=\sqrt{{\lambda}_{(z,r),j}}F_{j}((x,t),(z,r)),\qquad(x,t)\in{\mathbb{V}}^{d+1}.$

Then $\Phi=\\{\psi_{(z,r),j}:(z,r)\in\Xi_{j},\,j=1,2,3,\ldots\\}$ is a tight
frame.

######  Theorem 5.20.

Let $W$ be a doubling weight on ${\mathbb{V}}^{d+1}$. If $f\in
L^{2}({\mathbb{V}}^{d+1},W)$, then

$f=\sum_{j=0}^{\infty}\sum_{(z,r)\in\Xi_{j}}\langle
f,\psi_{(z,r),j}\rangle_{W}\psi_{(z,r),j}\qquad\mbox{in
$L^{2}({\mathbb{V}}^{d+1},W)$}$

and

$\|f\|_{2,W}=\Big{(}\sum_{j=0}^{\infty}\sum_{(z,r)\in\Xi_{j}}\left|\langle
f,\psi_{(z,r),j}\rangle_{W}\right|^{2}\Big{)}^{1/2}.$

Furthermore, for $W_{{\gamma},\mu}$ with ${\gamma}\geq-\frac{1}{2}$ and
$\mu\geq 0$, the frame is highly localized in the sense that, for every
${\kappa}>0$, there exists a constant $c_{\kappa}>0$ such that

(5.25) $|\psi_{(z,r),j}(x,t)|\leq
c_{\sigma}\frac{2^{j(d+1)/2}}{\sqrt{W_{{\gamma},\mu}(2^{j};x,t)}(1+2^{j}{\mathsf{d}}_{{\mathbb{V}}}((x,t),(z,r)))^{\kappa}},\quad(x,t)\in{\mathbb{V}}^{d+1}.$

The frame elements are well defined for all doubling weight by Theorem 5.19.
The decomposition is the consequence of Theorem 2.21. Moreover, the
localization (5.25) follows from Theorem 5.9 and ${\lambda}_{(z,r),j}\sim
2^{-j(d+1)}W_{{\gamma},\mu}(2^{j};t)$ that holds for $W_{{\gamma},\mu}$ as see
from Corollary 5.18 and (5.5).

###  5.7. Characterization of best approximation

For $f\in L^{p}({\mathbb{V}}^{d+1},W)$, we denote by
${\mathbf{E}}_{n}(f)_{p,W}$ the error of best approximation to $f$ from
$\Pi_{n}^{d+1}$, the space of polynomials of degree at most $n$, in the norm
$\|\cdot\|_{p,W}$,

${\mathbf{E}}_{n}(f)_{p,W}:=\inf_{g\in\Pi_{n}^{d+1}}\|f-g\|_{p,W},\qquad 1\leq
p\leq\infty.$

We give a characterization of this quantity in terms of the modulus of
smoothness defined via the operator ${\mathsf{S}}_{{\theta},W}$ and the
$K$-functional defined via the differential operator
${\mathfrak{D}}_{{\gamma},\mu}$ for $W_{{\gamma},\mu}$.

For $f\in L^{p}({\mathbb{V}}^{d+1},W_{{\gamma},\mu})$ and $r>0$, the modulus
of smoothness is defined by

${\omega}_{r}(f;\rho)_{p,W_{{\gamma},\mu}}=\sup_{0\leq{\theta}\leq\rho}\left\|\left(I-{\mathbf{S}}_{{\theta},W_{{\gamma},\mu}}\right)^{r/2}f\right\|_{p,W_{{\gamma},\mu}},\quad
1\leq p\leq\infty,$

where the operator ${\mathbf{S}}_{{\theta},W_{{\gamma},\mu}}$ is defined by,
for $n=0,1,2,\ldots$ and ${\lambda}=2\mu+{\gamma}+d$,

$\operatorname{proj}_{n}(W_{{\gamma},\mu};{\mathbf{S}}_{{\theta},W_{{\gamma},\mu}}f)=R_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(\cos{\theta})\operatorname{proj}_{n}(W_{{\gamma},\mu};f).$

Moreover, in terms of the fractional differential operator
$(-{\mathfrak{D}}_{{\gamma},\mu})^{\frac{r}{2}}$, the $K$-functional is
defined for a weight $W$ on ${\mathbb{V}}^{d+1}$ by

${\mathbf{K}}_{r}(f,\rho)_{p,W}:=\inf_{g\in{\mathcal{W}}_{p}^{r}({\mathbb{V}}^{d+1},W)}\left\\{\|f-g\|_{p,W}+\rho^{r}\left\|(-{\mathfrak{D}}_{{\gamma},\mu})^{\frac{r}{2}}f\right\|_{p,W}\right\\},$

where ${\mathcal{W}}_{p}^{r}({\mathbb{V}}^{d+1},W)$ denotes the Sobolev space
consisting of functions in $L^{p}({\mathbb{V}}^{d+1},W)$ with finite
$\left\|(-{\mathfrak{D}}_{{\gamma},{\beta}})^{\frac{r}{2}}f\right\|_{p,W}$.

The weight function $W_{{\gamma},\mu}$ admits Assertion 1 and 3 by Theorem 5.9
and Lemma 5.12. We now verify that the Assertion 5 in Subsection 3.5 holds. By
Theorem 4.8, the kernel $L_{n}^{(r)}(\varpi)$ in Assertion 5 becomes

${\mathbf{L}}_{n}^{(r)}\big{(}W_{{\gamma},\mu};(x,t),(y,s)\big{)}=\sum_{k=0}^{\infty}\widehat{a}\left(\frac{k}{n}\right)(k(k+2\mu+{\gamma}+d))^{\frac{r}{2}}{\mathbf{P}}_{k}\big{(}W_{{\gamma},\mu};(x,t),(y,s)\big{)}.$

######  Lemma 5.21.

Let ${\gamma}\geq-\frac{1}{2}$ and $\mu\geq 0$. Let ${\kappa}>0$. Then, for
$r>0$ and $(x,t),(y,s)\in{\mathbb{V}}^{d+1}$,

$|{\mathbf{L}}_{n}^{(r)}\big{(}W_{{\gamma},\mu};(x,t),(y,s)\big{)}|\leq
c_{\kappa}\frac{n^{r+d}}{\sqrt{W_{{\gamma},\mu}(n;x,t)}\sqrt{W_{{\gamma},\mu}(n;y,s)}\left(1+n{\mathsf{d}}_{{\mathbb{V}}}((x,t),(y,s))\right)^{{\kappa}}}.$

###### Proof.

By (5.10), the kernel can be written as

$\displaystyle{\mathbf{L}}_{n}^{(r)}\big{(}W_{{\gamma},\mu};(x,t),(y,s))=\,$
$\displaystyle
c_{\mu,{\gamma},d}\int_{[-1,1]^{3}}L_{n,r}\left(2\xi(x,t,y,s;u,v)^{2}-1\right)$
$\displaystyle\times(1-u^{2})^{\mu-1}(1-v_{1}^{2})^{{\alpha}-1}(1-v_{2}^{2})^{{\gamma}-\frac{1}{2}}\mathrm{d}u\mathrm{d}v,$

in which ${\alpha}=\mu+\frac{d-1}{2}$ and $L_{n,r}$ is defined by, with
${\lambda}=2{\alpha}+{\gamma}+1$,

$L_{n,r}(t)=\sum_{k=0}^{\infty}\widehat{a}\left(\frac{k}{n}\right)(k(k+{\gamma}+d-1))^{\frac{r}{2}}\frac{P_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(1)P_{n}^{({\lambda}-\frac{1}{2},-\frac{1}{2})}(t)}{h^{({\lambda}-\frac{1}{2},-\frac{1}{2})}}.$

Applying (2.18) with
$\eta(t)=\widehat{a}(t)\left(t(t+n^{-1}(2\mu+{\gamma}+d))\right)^{\frac{r}{2}}$
and $m=0$, it follows that

$\left|L_{n,r}(t)\right|\leq
cn^{r}\frac{n^{2{\lambda}+1}}{(1+n\sqrt{1-t})^{\ell}}.$

Using this estimate, we can then deduce the proof to one that has already
appeared for ${\mathbf{L}}_{n}(W_{{\gamma},\mu})$ in the proof of Theorem
4.10. ∎

With Assertions 1, 3 and 5 verified for $W_{{\gamma},\mu}$, the
characterization of the best approximation by polynomials in Subsection 3.3
holds on the solid cone, which we state below.

######  Theorem 5.22.

Let $f\in L^{p}({\mathbb{V}}^{d+1},W)$ if $1\leq p<\infty$ and $f\in
C({\mathbb{V}}^{d+1})$ if $p=\infty$. Le $r>0$ and $n=1,2,\ldots$. For
$W=W_{{\gamma},\mu}$ with ${\gamma}\geq-\frac{1}{2}$ and $\mu\geq 0$, there
holds

  1. (i)

direct estimate

${\mathbf{E}}_{n}(f)_{p,W_{{\gamma},\mu}}\leq
c\,{\mathbf{K}}_{r}(f;n^{-1})_{p,W_{{\gamma},\mu}}.$

  2. (ii)

inverse estimate, for $\mu=0$,

${\mathbf{K}}_{r}(f;n^{-1})_{p,W_{{\gamma},0}}\leq
cn^{-r}\sum_{k=0}^{n}(k+1)^{r-1}{\mathbf{E}}_{k}(f)_{p,W_{{\gamma},0}}.$

###### Proof.

The direct estimate follows from Theorem 3.14, which requires only Assertions
1, 3, 5 and holds for $W_{{\gamma},\mu}$ by Theorem 5.9 and Lemma 5.12. The
inverse estimate follows from Theorem 3.12, which requires one weigh function
that admits all Assertions 1–3 and 5 and that holds for $W_{{\gamma},0}$ on
the cone. ∎

Both the direct and the inverse estimates hold for the weight function
$W_{{\gamma},\mu}$ in the above theorem. However, it should be noted that the
inverse estimate uses the $K$-functional

${\mathbf{K}}_{r}(f,\rho)_{p,W}=\inf_{g\in{\mathcal{W}}_{p}^{r}({\mathbb{V}}^{d+1},W)}\left\\{\|f-g\|_{p,W}+\rho^{r}\left\|(-{\mathfrak{D}}_{{\gamma},0})^{\frac{r}{2}}f\right\|_{p,W}\right\\}$

defined via the operator ${\mathfrak{D}}_{{\gamma},0}$ for the weight
$W_{{\gamma},0}$.

For $W=W_{{\gamma},\mu}$, both direct and inverse estimates can be given via
the modulus of smoothness, since it is equivalent to the $K$-functional.

######  Theorem 5.23.

Let ${\gamma}\geq-\frac{1}{2}$, $\mu\geq 0$ and $f\in
L_{p}^{r}({\mathbb{V}}^{d+1},W_{{\gamma},\mu})$, $1\leq p\leq\infty$. Then for
$0<{\theta}\leq\pi/2$ and $r>0$

$c_{1}{\mathbf{K}}_{r}(f;{\theta})_{p,W_{{\gamma},\mu}}\leq{\omega}_{r}(f;{\theta})_{p,W_{{\gamma},\mu}}{\mathbf{K}}_{r}(f;{\theta})_{p,W_{{\gamma},\mu}}.$

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