# New sharp bounds for the Jacobi heat kernel via
an extension of the Dijksma-Koornwinder formula

Adam Nowak  0000-0003-0814-6795 Institute of Mathematics, Polish Academy of
Sciences, Śniadeckich 8, 00-656 Warszawa, Poland<EMAIL_ADDRESS>,  Peter
Sjögren  0000-0001-7862-8584 Mathematical Sciences, University of Gothenburg
and Mathematical Sciences, Chalmers University of Technology, SE-412 96
Göteborg, Sweden<EMAIL_ADDRESS>and  Tomasz Z. Szarek
0000-0003-0821-5607 Department of Mathematics, University of Georgia, Athens,
GA 30602, USA and Instytut Matematyczny, Uniwersytet Wrocławski, Plac
Grunwaldzki 2, 50-384 Wrocław, Poland<EMAIL_ADDRESS>
###### Abstract.

We obtain sharp estimates for the Jacobi heat kernel in a range of parameters
where the result has not been established before. This extends and completes
an earlier result due to the authors. The proof is based on a generalization
of the Dijksma-Koornwinder product formula for Jacobi polynomials.

††footnotetext:  _2020 Mathematics Subject Classification:_ primary 35K08;
secondary 33C45, 42C05.
_Key words and phrases:_ Jacobi heat kernel, sharp estimate, product formula,
Jacobi polynomial, Jacobi expansions.

##  1\. Introduction

Let $\alpha,\beta>-1$. The Jacobi heat kernel is defined by

(1)
$G_{t}^{\alpha,\beta}(x,y):=\sum_{n=0}^{\infty}e^{-tn(n+\alpha+\beta+1)}\frac{P_{n}^{\alpha,\beta}(x)P_{n}^{\alpha,\beta}(y)}{h_{n}^{\alpha,\beta}},\qquad
x,y\in[-1,1],\quad t>0,$

where $P_{n}^{\alpha,\beta}$ are the classical Jacobi polynomials and
$h_{n}^{\alpha,\beta}$ are suitable normalizing constants. This is the kernel
of the Jacobi semigroup generated by the Jacobi operator; see Section 2 for
precise definitions. The oscillatory series displayed above cannot be
computed.

The kernel $G_{t}^{\alpha,\beta}(x,y)$ is an important object in analysis and
beyond, in particular due to its deep connections with, in a sense, more
elementary heat kernels in numerous classical frameworks. These include, among
others, the contexts of Euclidean spheres (cf. [12]), compact rank-one
symmetric spaces (cf. [13]), Fourier-Bessel expansions (cf. [9]) and algebraic
contexts related to multi-dimensional balls, simplices, cones, conical
surfaces and other domains, see e.g. [13, 6, 4, 2] and references therein.
Thus advancing knowledge about $G_{t}^{\alpha,\beta}(x,y)$ is widely
motivated.

In this paper we complete our study [13] of the behavior of the Jacobi heat
kernel by proving genuinely sharp two-sided global estimates in the whole
range of the parameters $\alpha$ and $\beta$ indicated above. In order to
state the main result, we define for $t>0$ and $\theta,\varphi\in[0,\pi]$ the
quantity

$Z_{t}^{\alpha,\beta}(\theta,\varphi)=\big{[}t+\theta\varphi\big{]}^{-\alpha-1/2}\,\big{[}t+(\pi-\theta)(\pi-\varphi)\big{]}^{-\beta-1/2}\;\frac{1}{\sqrt{t}}\exp\bigg{(}{-\frac{(\theta-\varphi)^{2}}{4t}}\bigg{)}.$

We will prove the following theorem, which concludes the long-standing open
problem of finding sharp bounds for the Jacobi heat kernel.

######  Theorem A.

Let $\alpha,\beta>-1$ and $T>0$ be fixed. There exists a constant
$C=C(\alpha,\beta,T)>1$ such that

$C^{-1}Z_{t}^{\alpha,\beta}(\theta,\varphi)\leq
G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)\leq
CZ_{t}^{\alpha,\beta}(\theta,\varphi)$

for $\theta,\varphi\in[0,\pi]$ and $0<t\leq T$.

In the restricted range of parameters $\alpha,\beta\geq-1/2$, Theorem A was
proved by the authors in [12, 13], see [13, Theorem 3.1]. We remark that only
qualitatively sharp, short-time bounds for $G_{t}^{\alpha,\beta}(x,y)$ were
earlier obtained in [1] and independently by other methods in [10]; in the
latter paper under the restriction $\alpha,\beta\geq-1/2$. The difference
between genuinely sharp and qualitatively sharp bounds is that in the latter
case the constants appearing in the exponential factors in the lower and upper
estimates (which are $1/4$ in our Theorem A) differ from each other and from
the optimal one. This exponential gap swallows various polynomial factors,
which in case of genuinely sharp bounds must be determined precisely.

On the other hand, for any $\alpha,\beta>-1$ showing the long-time sharp
estimates

(2) $C^{-1}\leq G_{t}^{\alpha,\beta}(x,y)\leq C,\qquad x,y\in[-1,1],\quad
t\geq T,$

with $T>0$ fixed and some $C=C(\alpha,\beta,T)>1$ independent of $x,y,t$, is
much more straightforward, see [1, 10].

The new contribution of Theorem A pertains essentially (see [13, Remark 3.2])
to the range of parameters when $\alpha<-1/2$ or $\beta<-1/2$, in particular
to the tricky case when also $\alpha+\beta\leq-3/2$. These cases are more
difficult to handle than the previously solved case $\alpha,\beta\geq-1/2$. In
fact, some crucial tools and techniques elaborated and used earlier in [10,
13] break down for the parameters below $-1/2$ and hence the proof of Theorem
A requires some new ideas and tools. One of them is an extended variant of a
_product formula_ for Jacobi polynomials due to Dijksma and Koornwinder [3].
This extension, obtained by means of analytic continuation, is contained in
Theorem 3.2, see also Remark 3.3; we believe that this result is of
independent interest. For the statement of the original Dijksma-Koornwinder
formula and further details, see Section 3. The extended product formula leads
to an extension of a _reduction formula_ for the Jacobi heat kernel from [10,
Theorem 3.1]; see Theorem 4.3, which is the key to proving the desired bounds.

The paper is organized as follows. In Section 2 we introduce basic notation
and gather various facts and results needed in the sequel. Section 3 is
devoted to the extension of the Dijksma-Koornwinder product formula. In
Section 4 we extend the reduction formula for the Jacobi heat kernel. Section
5 contains some preparations for the proof of Theorem A, given in Section 6.

##  2\. Preliminaries

Throughout the paper we use a standard notation. We will frequently write
$X\lesssim Y$ to indicate that $X\leq CY$ with a positive constant $C$
independent of significant quantities. We shall write $X\simeq Y$ when
simultaneously $X\lesssim Y$ and $Y\lesssim X$.

### Varia

For various facts below, see e.g. [15]. We use the following notation.

$\displaystyle P_{n}^{\alpha,\beta}$ Jacobi polynomials $\displaystyle
C_{n}^{\lambda}\;\;$ Gegenbauer/ultraspherical polynomials $\displaystyle
P_{n}^{\lambda}\;\;$ $\displaystyle\qquad\textrm{Jacobi
polynomial}\;\;P_{n}^{\lambda,\lambda}.$

The Gegenbauer and Jacobi polynomials are linked by

$C_{n}^{\lambda}(x)=\frac{\Gamma(\lambda+1/2)}{\Gamma(2\lambda)}\,\frac{\Gamma(n+2\lambda)}{\Gamma(n+\lambda+1/2)}\,P_{n}^{\lambda-1/2}(x),\qquad\lambda>-1/2,\quad\lambda\neq
0.$

We will use a similar shortened notation for the Jacobi heat kernel,
$G_{t}^{\lambda}(x,y):=G_{t}^{\lambda,\lambda}(x,y)$, and for other objects
defined below, e.g. $h_{n}^{\lambda}:=h_{n}^{\lambda,\lambda}$,
$J^{\lambda}:=J^{\lambda,\lambda}$.

For $\alpha,\beta>-1$ the Jacobi orthogonality measure on the interval
$[-1,1]$ is

$d\varrho_{\alpha,\beta}(x)=(1-x)^{\alpha}(1+x)^{\beta}\,dx.$

For complex $\alpha$ and $\beta$ such that $\Re\alpha,\Re\beta>-1$, we let

$h_{n}^{\alpha,\beta}=\frac{2^{\alpha+\beta+1}\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{(2n+\alpha+\beta+1)\Gamma(n+\alpha+\beta+1)\Gamma(n+1)},\qquad
n\geq 0,$

where in the case $n=0$ the product
$(2n+\alpha+\beta+1)\Gamma(n+\alpha+\beta+1)$ is replaced by
$\Gamma(\alpha+\beta+2)$. For real $\alpha,\beta>-1$, one has
$h_{n}^{\alpha,\beta}=\|P_{n}^{\alpha,\beta}\|^{2}_{L^{2}(d\varrho_{\alpha,\beta})}$.

Note that $P_{0}^{\alpha,\beta}\equiv 1$. We state some other useful formulas,
valid for $n\geq 0$:

(3) $\displaystyle P_{n}^{\alpha,\beta}(1)$
$\displaystyle=\frac{(\alpha+1)_{n}}{n!}=\frac{\Gamma(n+\alpha+1)}{\Gamma(n+1)\Gamma(\alpha+1)},$
(4) $\displaystyle\frac{d}{dx}P_{n}^{\alpha,\beta}(x)$
$\displaystyle=\frac{1}{2}(n+\alpha+\beta+1)P^{\alpha+1,\beta+1}_{n-1}(x).$

By convention, $P_{k}^{\alpha,\beta}\equiv 0$ for $k<0$. The Pochhammer symbol
is defined as $(a)_{k}=a(a+1)\cdot\ldots\cdot(a+k-1)$ for $k\geq 1$, with the
convention that $(a)_{0}=1$.

### Jacobi polynomials and complex values of the parameters

In this subsection we consider $\alpha,\beta\in\mathbb{C}$, see [15, Chapter
IV, Section 4.22].

Jacobi polynomials have natural extensions to all complex parameters, since,
defined a priori as orthogonal polynomials for $\alpha,\beta>-1$, they are
polynomials not only in their argument, but also in the parameters. One has
$P^{\alpha,\beta}_{0}\equiv 1$, and for $n\geq 1$

$\displaystyle P_{n}^{\alpha,\beta}(x)$
$\displaystyle=\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}(n+\alpha+\beta+1)_{k}\;(\alpha+k+1)_{n-k}\;\Big{(}\frac{x-1}{2}\Big{)}^{k}.$

Note that $P_{n}^{\alpha,\beta}$ is a polynomial of degree at most $n$ in
$\alpha$ and $\beta$. And $P^{\alpha,\beta}_{n}(x)$ is a polynomial of degree
at most $n$ in $x$ (the degree is equal to $n$ if e.g. $\alpha,\beta>-1$; for
more details, see [15, Chapter IV, Section 4.22]).

The formulas (3) and (4) hold for all complex parameters (understood as
appropriate limits whenever necessary). Moreover, the $P_{n}^{\alpha,\beta}$
are eigenfunctions of the Jacobi differential operator given by

$J^{\alpha,\beta}=-\big{(}1-x^{2}\big{)}\frac{d^{2}}{dx^{2}}-\big{[}\beta-\alpha-(\alpha+\beta+2)x\big{]}\frac{d}{dx}.$

More precisely, one has

(5)
$J^{\alpha,\beta}P_{n}^{\alpha,\beta}=n(n+\alpha+\beta+1)P_{n}^{\alpha,\beta},\qquad
n\geq 0.$

### Measures and some integrals

We observe that

$\varrho_{\alpha,\beta}\big{(}[-1,1]\big{)}=h_{0}^{\alpha,\beta}=2^{\alpha+\beta+1}\,\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)},\qquad\alpha,\beta>-1.$

For complex $\alpha$ such that $\Re\alpha>-1$ and $\alpha\neq-1/2$, define

$\Pi_{\alpha}(u)=\frac{\Gamma(\alpha+1)}{\sqrt{\pi}\Gamma(\alpha+1/2)}\int_{0}^{u}\big{(}1-w^{2}\big{)}^{\alpha-1/2}\,dw.$

Notice that $\Pi_{\alpha}$ is an odd function in $-1<u<1$, increasing if
$\alpha>-1/2$ and decreasing if $\alpha<-1/2$.

For real $\alpha>-1/2$ the measure $d\Pi_{\alpha}$ is a probability measure in
the interval $[-1,1]$, as seen by changing variable to $v=w^{2}$ in the
integral above. As $\alpha\searrow-1/2$, the weak limit of $d\Pi_{\alpha}$ is

$d\Pi_{-1/2}:=\frac{\delta_{-1}+\delta_{1}}{2},$

where $\delta_{\pm 1}$ is a unit point mass at $\pm 1$.

In general, for all complex $\alpha$ such that $\Re\alpha>-1$,
$\alpha\neq-1/2$, the distribution derivative $d\Pi_{\alpha}$ is a local
complex measure in $(-1,1)$. For real $\alpha\in(-1,-1/2)$ its density is
negative, even, and not integrable in $(-1,1)$. Furthermore, we have the
following lemma.

######  Lemma 2.1 ([11, Lemma 2.2]).

Let $\alpha\in(-1,-1/2)$ be fixed. Then the density $|\Pi_{\alpha}(u)|$
defines a finite measure on $[-1,1]$ and

$|\Pi_{\alpha}(u)|\simeq|u|\big{(}1-|u|\big{)}^{\alpha+1/2}\simeq|u|\frac{d\Pi_{\alpha+1}(u)}{du},\qquad
u\in(-1,1).$

We define

$\mathcal{A}:=\big{\\{}(\alpha,\beta)\in\mathbb{C}^{2}:\Re\alpha>-1,\Re\beta>-1\big{\\}}.$

Observe that, see [11, Section 2], if $\phi$ is a continuous function in
$[0,1]$ satisfying $|\phi(u)|\leq C(1-u)$, with a constant $C=C(\alpha)$ that
is locally bounded, then the integral

$I(\alpha)=\int_{(0,1]}\phi(u)\,d\Pi_{\alpha}(u)$

is well defined and analytic in $\\{\alpha:\Re\alpha>-1\\}$. Moreover,
$I(-1/2)=0$, as seen from the bound
$|I(\alpha)|\lesssim|\alpha+1/2|\int_{0}^{1}(1-u^{2})^{\Re\alpha+1/2}du$. This
observation generalizes as follows, cf. [11, Section 2].

######  Proposition 2.2.

  * (a)

Assume that $\phi_{\alpha,\beta}(u)$ is continuous in
$(u,\alpha,\beta)\in[0,1]\times\mathcal{A}$ and analytic in
$(\alpha,\beta)\in\mathcal{A}$ for each $u\in(0,1)$ and satisfies
$|\phi_{\alpha,\beta}(u)|\leq C(1-u)$, where the constant $C=C(\alpha,\beta)$
is locally bounded in $\mathcal{A}$. Then

$I(\alpha,\beta)=\int_{(0,1]}\phi_{\alpha,\beta}(u)\,d\Pi_{\alpha}(u)$

is well defined and analytic in $(\alpha,\beta)\in\mathcal{A}$. Moreover,
$I(-1/2,\beta)=0$.

  * (b)

Let $\phi_{\alpha,\beta}(u,v)$ be continuous in
$(u,v,\alpha,\beta)\in[0,1]^{2}\times\mathcal{A}$ and analytic in
$(\alpha,\beta)\in\mathcal{A}$ for each $u,v\in(0,1)$, and assume that
$|\phi_{\alpha,\beta}(u,v)|\leq C(1-u)(1-v)$ with $C$ as in (a). Then the
double integral

$I(\alpha,\beta)=\iint_{(0,1]^{2}}\phi_{\alpha,\beta}(u,v)\,d\Pi_{\alpha}(u)\,d\Pi_{\beta}(v)$

is well defined and analytic in $(\alpha,\beta)\in\mathcal{A}$. Furthermore,
$I(-1/2,\beta)=I(\alpha,-1/2)=0$.

##  3\. Dijksma-Koornwinder formula

In [3] Dijksma and Koornwinder found a symmetric _product formula_ for two
Jacobi polynomials, which was originally stated as follows (cf. [3, (2.5)]):

(6) $\displaystyle
P_{n}^{\alpha,\beta}\big{(}1-2s^{2}\big{)}P_{n}^{\alpha,\beta}\big{(}1-2t^{2}\big{)}=\frac{\Gamma(\alpha+\beta+1)\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\pi
n!\Gamma(n+\alpha+\beta+1)\Gamma(\alpha+1/2)\Gamma(\beta+1/2)}$
$\displaystyle\quad\times\int_{-1}^{1}\int_{-1}^{1}C_{2n}^{\alpha+\beta+1}\Big{(}ust+v\sqrt{1-s^{2}}\sqrt{1-t^{2}}\Big{)}\big{(}1-u^{2}\big{)}^{\alpha-1/2}\big{(}1-v^{2}\big{)}^{\beta-1/2}\,du\,dv.$

Here $-1\leq s,t\leq 1$, and the formula is valid for $n\geq 0$ and all
complex $\alpha,\beta$ such that $\Re\alpha>-1/2$ and $\Re\beta>-1/2$. The
proof in [3] is group theoretic. An analytic proof was given slightly later by
Koornwinder [7]. Yet another proof can be found in the more recent paper [8].

We will rewrite and extend the product formula to all $\alpha,\beta$ such that
$\Re\alpha>-1$, $\Re\beta>-1$, i.e., $(\alpha,\beta)\in\mathcal{A}$; see
Theorem 3.2 below. For this purpose we define, to begin with for
$\alpha,\beta>-1$ such that $\alpha+\beta+1/2>-1$ and for $\lambda>-1$,

$\mathfrak{C}_{n}^{\alpha,\beta}:=\frac{h_{0}^{\alpha,\beta}}{(\alpha+\beta+3/2)h_{0}^{\alpha+\beta+1/2}h_{n}^{\alpha,\beta}},\quad
n\geq
0,\qquad\quad\mathrm{and}\qquad\quad\;\mathfrak{D}_{k}^{\lambda}:=\frac{P_{k}^{\lambda}(1)}{(\lambda+1)h_{k}^{\lambda}},\quad
k\geq 0.$

In what follows we use implicitly the duplication formula for the gamma
function, written as

$\Gamma(2\lambda+2)=\Gamma(\lambda+1)\Gamma(\lambda+3/2)2^{2\lambda+1}/\sqrt{\pi}.$

By a direct computation we see that

(7)
$\mathfrak{C}_{n}^{\alpha,\beta}=\begin{cases}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)(2n+\alpha+\beta+1)\Gamma(n+\alpha+\beta+1)\Gamma(n+1)}{\sqrt{\pi}\Gamma(\alpha+\beta+5/2)\Gamma(n+\alpha+1)\Gamma(n+\beta+1)},&n\geq
1,\\\
\frac{\Gamma(\alpha+\beta+2)}{\sqrt{\pi}\Gamma(\alpha+\beta+5/2)},&n=0,\end{cases}$

and this provides an analytic extension of each
$\mathfrak{C}_{n}^{\alpha,\beta}$, $n\geq 0$, to
$(\alpha,\beta)\in\mathcal{A}$. Similarly for $\lambda>-1$

(8)
$\mathfrak{D}_{k}^{\lambda}=\begin{cases}\frac{(2k+2\lambda+1)\Gamma(k+2\lambda+1)}{2^{2\lambda+1}\Gamma(\lambda+2)\Gamma(k+\lambda+1)},&k\geq
1,\\\
\frac{\Gamma(\lambda+3/2)}{\sqrt{\pi}\Gamma(\lambda+2)},&k=0.\end{cases}$

The last expression defines an analytic extension of each
$\mathfrak{D}_{2n}^{\lambda}$, $n\geq 0$, to $\Re\lambda>-3/2$. Thus, from now
on, we consider $\mathfrak{C}_{n}^{\alpha,\beta}$ and
$\mathfrak{D}_{2n}^{\alpha+\beta+1/2}$ as analytic functions of
$(\alpha,\beta)\in\mathcal{A}$ given by (7) and (8), respectively. Note that
$\mathfrak{C}_{n}^{\alpha,\beta},\mathfrak{D}_{2n}^{\alpha+\beta+1/2}>0$ for
$\alpha,\beta\in(-1,\infty)$.

We denote

(9)
$\Phi_{k}^{\lambda}(\theta,\varphi,u,v):=\mathfrak{D}_{k}^{\lambda}{P}_{k}^{\lambda}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\Big{)},$

and let $s=\sin\frac{\theta}{2}$ and $t=\sin\frac{\varphi}{2}$ in (6). By
means of (7) and (8), one then finds the following more compact form of the
Dijksma-Koornwinder formula (6):

(10)
$\mathfrak{C}_{n}^{\alpha,\beta}{P}_{n}^{\alpha,\beta}(\cos\theta){P}_{n}^{\alpha,\beta}(\cos\varphi)=\iint\Phi_{2n}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)\,d\Pi_{\alpha}(u)\,d\Pi_{\beta}(v),\quad
0\leq\theta,\varphi\leq\pi.$

This holds for $\Re\alpha>-1/2$ and $\Re\beta>-1/2$, and since Jacobi
polynomials are continuous functions of their parameters, it is also valid
when $\alpha=-1/2$ or $\beta=-1/2$.

We now proceed essentially as in [11, Section 2]. Taking the even parts of
$\Phi_{k}^{\lambda}(\theta,\varphi,u,v)$ in $u$ and $v$, we define

$\Phi_{k,E}^{\lambda}(\theta,\varphi,u,v):=\frac{1}{4}\sum_{\xi,\eta=\pm
1}\Phi_{k}^{\lambda}(\theta,\varphi,\xi u,\eta v).$

Then by (10) and for symmetry reasons, one has for $(\alpha,\beta)$ such that
$\Re\alpha,\Re\beta>-1/2$

(11)
$\mathfrak{C}_{n}^{\alpha,\beta}{{P}_{n}^{\alpha,\beta}(\cos\theta){P}_{n}^{\alpha,\beta}(\cos\varphi)}=4\iint_{(0,1]^{2}}\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)\,d\Pi_{\alpha}(u)\,d\Pi_{\beta}(v).$

######  Lemma 3.1.

For all $(\alpha,\beta)\in\mathcal{A}$, $n\geq 0$ and
$\theta,\varphi\in[0,\pi]$

$\displaystyle\mathfrak{C}_{n}^{\alpha,\beta}{{P}_{n}^{\alpha,\beta}(\cos\theta){P}_{n}^{\alpha,\beta}(\cos\varphi)}$
$\displaystyle=4\iint_{(0,1]^{2}}\Big{[}\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)-\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,1)$
$\displaystyle\qquad-\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,v)+\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,1)\Big{]}\,d\Pi_{\alpha}(u)\,d\Pi_{\beta}(v)$
$\displaystyle\quad+2\int_{(0,1]}\Big{[}\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,1)-\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,1)\Big{]}\,d\Pi_{\alpha}(u)$
$\displaystyle\quad+2\int_{(0,1]}\Big{[}\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,v)-\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,1)\Big{]}\,d\Pi_{\beta}(v)$
$\displaystyle\quad+\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,1).$

###### Proof.

When $\Re\alpha,\Re\beta>-1/2$, the claimed identity is a straightforward
consequence of (11), since the integrals can be evaluated term by term. The
left-hand side of the identity is analytic in $(\alpha,\beta)\in\mathcal{A}$.
It suffices to verify that the right-hand side is analytic in $\mathcal{A}$,
since then the conclusion will follow by analytic continuation.

The last term of the right-hand side is analytic in
$(\alpha,\beta)\in\mathcal{A}$, as easily seen from (9). The second term of
the right-hand side, i.e., the integral in $u$, is of the form
$I(\alpha,\beta)$ from item (a) of Proposition 2.2, with

$\phi_{\alpha,\beta}(u,v)=\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,1)-\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,1).$

Note that $|\phi_{\alpha,\beta}(u)|\leq C(1-u)$, since the derivative
$\partial_{u}\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,1)$ is bounded
locally uniformly in $\alpha$ and $\beta$, see (4). The third term of the
right-hand side, i.e., the integral in $v$, is analogous.

Finally, the first term of the right-hand side, i.e., the double integral, is
of the form $I(\alpha,\beta)$ from item (b) of Proposition 2.2, with

$\displaystyle\phi_{\alpha,\beta}(u,v)$
$\displaystyle\quad=\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)-\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,1)-\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,v)+\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,1,1),$

and the condition $|\phi_{\alpha,\beta}(u,v)|\leq C(1-u)(1-v)$ is satisfied
since the second order derivatives of
$\Phi_{2n,E}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)$ in $u,v$ are bounded
locally uniformly in $\alpha$ and $\beta$, cf. (4).

Altogether, it follows that the right-hand side of the identity is analytic in
$(\alpha,\beta)\in\mathcal{A}$. This finishes the proof. ∎

The next result is the announced extension of the Dijksma-Koornwinder formula.
We state it for real values of the parameters. A variant with complex
parameters is indicated in Remark 3.3 below. For the sake of completeness, the
original Dijksma-Koornwinder formula is included as item (i). By convention,
$\Phi_{k}^{\lambda}(\theta,\varphi,u,v)\equiv 0$ when $k<0$.

######  Theorem 3.2.

Let $n\geq 0$ and $\theta,\varphi\in[0,\pi]$. Then, with all integrations
taken over $[-1,1]^{2}$,

  * (i)

If $\alpha,\beta\geq-1/2$, then

$\mathfrak{C}_{n}^{\alpha,\beta}{{P}_{n}^{\alpha,\beta}(\cos\theta){P}_{n}^{\alpha,\beta}(\cos\varphi)}=\iint\Phi_{2n}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)\,d\Pi_{\alpha}(u)\,d\Pi_{\beta}(v).$

  * (ii)

If $-1<\beta<-1/2\leq\alpha$, then

$\displaystyle\mathfrak{C}_{n}^{\alpha,\beta}{{P}_{n}^{\alpha,\beta}(\cos\theta){P}_{n}^{\alpha,\beta}(\cos\varphi)}$
$\displaystyle\quad=\iint\Big{\\{}-2(\alpha+\beta+5/2)\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\,\Phi_{2n-1}^{\alpha+\beta+3/2}(\theta,\varphi,u,v)\,d\Pi_{\alpha}(u)\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad+\Phi_{2n}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)\,d\Pi_{\alpha}(u)\,d\Pi_{-1/2}(v)\Big{\\}}.$

  * (iii)

If $-1<\alpha<-1/2\leq\beta$, then

$\displaystyle\mathfrak{C}_{n}^{\alpha,\beta}{{P}_{n}^{\alpha,\beta}(\cos\theta){P}_{n}^{\alpha,\beta}(\cos\varphi)}$
$\displaystyle\quad=\iint\Big{\\{}-2(\alpha+\beta+5/2)\sin\frac{\theta}{2}\sin\frac{\varphi}{2}\,\Phi_{2n-1}^{\alpha+\beta+3/2}(\theta,\varphi,u,v)\,\Pi_{\alpha}(u)\,du\,d\Pi_{\beta}(v)$
$\displaystyle\qquad\qquad+\Phi_{2n}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)\,d\Pi_{-1/2}(u)\,d\Pi_{\beta}(v)\Big{\\}}.$

  * (iv)

If $-1<\alpha,\beta<-1/2$, then

$\displaystyle\mathfrak{C}_{n}^{\alpha,\beta}{{P}_{n}^{\alpha,\beta}(\cos\theta){P}_{n}^{\alpha,\beta}(\cos\varphi)}$
$\displaystyle\quad=\iint\Big{\\{}(\alpha+\beta+5/2)(\alpha+\beta+7/2)\sin\theta\sin\varphi\,\Phi_{2n-2}^{\alpha+\beta+5/2}(\theta,\varphi,u,v)\,\Pi_{\alpha}(u)\,du\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad-2(\alpha+\beta+5/2)\sin\frac{\theta}{2}\sin\frac{\varphi}{2}\,\Phi_{2n-1}^{\alpha+\beta+3/2}(\theta,\varphi,u,v)\,\Pi_{\alpha}(u)\,du\,d\Pi_{-1/2}(v)$
$\displaystyle\qquad\qquad-2(\alpha+\beta+5/2)\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\,\Phi_{2n-1}^{\alpha+\beta+3/2}(\theta,\varphi,u,v)\,d\Pi_{-1/2}(u)\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad+\Phi_{2n}^{\alpha+\beta+1/2}(\theta,\varphi,u,v)\,d\Pi_{-1/2}(u)\,d\Pi_{-1/2}(v)\Big{\\}}.$

###### Proof.

Item (i) is a restatement of (10). The proofs of the remaining items rely on
combining Lemmas 3.1 and 2.1 with symmetries of the quantity
$\Phi_{k,E}^{\lambda}(\theta,\varphi,u,v)$ and the measures involved. Further,
the following formulas for derivatives of $\Phi_{k}^{\lambda}$ are useful;
they follow from (4) and the identity
$\frac{1}{2}\,(k+2\lambda+1)\,\mathfrak{D}_{k}^{\lambda}=2(\lambda+2)\,\mathfrak{D}_{k-1}^{\lambda+1},\;k\geq
1$:

$\displaystyle\partial_{u}\Phi_{k}^{\lambda}(\theta,\varphi,u,v)$
$\displaystyle=2(\lambda+2)\sin\frac{\theta}{2}\sin\frac{\varphi}{2}\,\Phi_{k-1}^{\lambda+1}(\theta,\varphi,u,v),$
$\displaystyle\partial_{v}\Phi_{k}^{\lambda}(\theta,\varphi,u,v)$
$\displaystyle=2(\lambda+2)\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\,\Phi_{k-1}^{\lambda+1}(\theta,\varphi,u,v),$
$\displaystyle\partial_{u}\partial_{v}\Phi_{k}^{\lambda}(\theta,\varphi,u,v)$
$\displaystyle=(\lambda+2)(\lambda+3)\sin\theta\sin\varphi\,\Phi_{k-2}^{\lambda+2}(\theta,\varphi,u,v).$

The details are analogous to those in the proof of [11, Proposition 2.3] and
thus omitted. ∎

######  Remark 3.3.

To state Theorem 3.2 for complex values of the parameters, let

$\mathcal{B}_{1}:=\\{\alpha\in\mathbb{C}:\Re\alpha>-1/2\\}\cup\\{-1/2\\},\qquad\mathcal{B}_{2}:=\\{\alpha\in\mathbb{C}:\Re\alpha>-1\\}\setminus\mathcal{B}_{1}.$

Then the cases should be defined, respectively, by (i)
$\alpha,\beta\in\mathcal{B}_{1}$, (ii) $\alpha\in\mathcal{B}_{1}$,
$\beta\in\mathcal{B}_{2}$, (iii) $\alpha\in\mathcal{B}_{2}$,
$\beta\in\mathcal{B}_{1}$, (iv) $\alpha,\beta\in\mathcal{B}_{2}$, with the
formulas unchanged.

##  4\. Reduction formula

Recall that for $\alpha,\beta>-1$, $x,y\in[-1,1]$ and $t>0$ the Jacobi heat
kernel $G_{t}^{\alpha,\beta}(x,y)$ is given by (1). For basic information
concerning this kernel, see e.g. [10]. We will use the following lemma from
our earlier paper [13].

######  Lemma 4.1 ([13, Lemma 3.3]).

Let $\alpha,\beta>-1$. Then

$\displaystyle\frac{d}{dx}\,G_{t}^{\alpha,\beta}(x,1)=2(\alpha+1)\,e^{-t(\alpha+\beta+2)}\,G_{t}^{\alpha+1,\beta+1}(x,1),\qquad
x\in[-1,1],\quad t>0.$

In particular, the function $x\mapsto G_{t}^{\alpha,\beta}(x,1)$ is strictly
increasing on $[-1,1]$.

We now introduce an auxiliary function $H_{t}^{\lambda}(x)$ which can be
regarded as an extension for $-3/2<\lambda\leq-1$ of (a constant times) the
even part in $x$ of the Jacobi-ultraspherical heat kernel
$G_{t}^{\lambda}(x,1)$. Let

$H_{t}^{\lambda}(x):=\sum_{n=0}^{\infty}e^{-t2n(2n+2\lambda+1)}\,\mathfrak{D}_{2n}^{\lambda}\,P_{2n}^{\lambda}(x),\qquad
x\in[-1,1],\quad t>0,\quad-3/2<\lambda\leq-1.$

Note that $H_{t}^{\lambda}(x)$ is well defined. Indeed, the constants
$\mathfrak{D}_{2n}^{\lambda}$ are well defined by (8). By Stirling’s formula,
one has $\mathfrak{D}_{2n}^{\lambda}=\mathcal{O}(n^{\lambda+1})$ as
$n\to\infty$. Furthermore, the Jacobi polynomials satisfy (see [15, Chapter
VII, Section 7.32, Theorem 7.32.2])

$\big{|}P_{k}^{\alpha,\beta}(x)\big{|}\lesssim(k+1)^{\max\\{\alpha,\beta,-1/2\\}},\qquad
k\geq 0,\quad x\in[-1,1],$

for any fixed $\alpha,\beta\in\mathbb{R}$. Consequently, we see that the
series defining $H_{t}^{\lambda}(x)$ converges uniformly in
$(t,x)\in[T,\infty)\times[-1,1]$ for each fixed $T>0$. Using (4) we also see
that the series can be differentiated term by term in $x$ and $t$ arbitrarily
many times. Thus $H_{t}^{\lambda}(x)$ is a smooth function of
$(t,x)\in(0,\infty)\times[-1,1]$.

Next, we observe that $H_{t}^{\lambda}$ satisfies the heat equation based on
the Jacobi-ultraspherical operator $J^{\lambda}$,

(12) $\Big{(}\frac{d}{dt}+J^{\lambda}\Big{)}H_{t}^{\lambda}(x)=0,\qquad
x\in[-1,1],\quad t>0.$

This follows from the fact that each term of the defining series satisfies
this equation, see (5). Moreover, we have the following.

######  Lemma 4.2.

Let $-3/2<\lambda\leq-1$. Then

$\frac{d}{dx}\,H_{t}^{\lambda}(x)=2e^{-t(2\lambda+2)}\Big{[}G_{t}^{\lambda+1}(x,1)\Big{]}_{\emph{odd}},\qquad
x\in[-1,1],\quad t>0,$

where the subscript ‘odd’ indicates the odd part of the function in $x$.
Similarly,

$\frac{d^{2}}{dx^{2}}\,H_{t}^{\lambda}(x)=4(\lambda+2)e^{-t(4\lambda+6)}\Big{[}G_{t}^{\lambda+2}(x,1)\Big{]}_{\emph{even}},\qquad
x\in[-1,1],\quad t>0,$

where the subscript ‘even’ indicates the even part of the function in $x$.

###### Proof.

To verify the first identity, differentiate the series defining
$H_{t}^{\lambda}(x)$ term by term and use (4) together with the fact that
ultraspherical polynomials of odd orders are odd functions, and those of even
orders are even functions. The second identity follows from the first one and
Lemma 4.1. We leave the straightforward computations to the reader. ∎

The next result extends a _reduction formula_ for the Jacobi heat kernel that
was obtained in [10, Theorem 3.1] by means of the Dijksma-Koornwinder product
formula. For the sake of completeness, the result from [10] is stated as item
(i) of Theorem 4.3.

######  Theorem 4.3.

Let $t>0$ and $\theta,\varphi\in[0,\pi]$. Then, with all integrations taken
over $[-1,1]^{2}$,

  * (i)

If $\alpha,\beta\geq-1/2$, then

$\displaystyle\frac{h_{0}^{\alpha,\beta}}{h_{0}^{\alpha+\beta+1/2}}\,G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)$
$\displaystyle\quad=\iint
G_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{\alpha}(u)\,d\Pi_{\beta}(v).$

  * (ii)

If $-1<\beta<-1/2\leq\alpha$, then

$\displaystyle\frac{h_{0}^{\alpha,\beta}}{h_{0}^{\alpha+\beta+1/2}}\,G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)$
$\displaystyle\quad=\iint\Big{\\{}-2(\alpha+\beta+3/2)e^{-t(\alpha+\beta+3/2)/2}\cos\frac{\theta}{2}\cos\frac{\varphi}{2}$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{\alpha}(u)\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad+G_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{\alpha}(u)\,d\Pi_{-1/2}(v)\Big{\\}}.$

  * (iii)

If $-1<\alpha<-1/2\leq\beta$, then

$\displaystyle\frac{h_{0}^{\alpha,\beta}}{h_{0}^{\alpha+\beta+1/2}}\,G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)$
$\displaystyle\quad=\iint\Big{\\{}-2(\alpha+\beta+3/2)e^{-t(\alpha+\beta+3/2)/2}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,\Pi_{\alpha}(u)\,du\,d\Pi_{\beta}(v)$
$\displaystyle\qquad\qquad+G_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{-1/2}(u)\,d\Pi_{\beta}(v)\Big{\\}}.$

  * (iv)

If $-1<\alpha,\beta<-1/2$ and $\alpha+\beta>-3/2$, then

$\displaystyle\frac{h_{0}^{\alpha,\beta}}{h_{0}^{\alpha+\beta+1/2}}\,G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)$
$\displaystyle\quad=\iint\Big{\\{}(\alpha+\beta+3/2)(\alpha+\beta+5/2)e^{-t(\alpha+\beta+2)}\sin\theta\sin\varphi$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+5/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,\Pi_{\alpha}(u)\,du\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad-2(\alpha+\beta+3/2)e^{-t(\alpha+\beta+3/2)/2}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,\Pi_{\alpha}(u)\,du\,d\Pi_{-1/2}(v)$
$\displaystyle\qquad\qquad-2(\alpha+\beta+3/2)e^{-t(\alpha+\beta+3/2)/2}\cos\frac{\theta}{2}\cos\frac{\varphi}{2}$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{-1/2}(u)\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad+G_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{-1/2}(u)\,d\Pi_{-1/2}(v)\Big{\\}}.$

  * (v)

If $-1<\alpha,\beta<-1/2$ and $\alpha+\beta\leq-3/2$, then

$\displaystyle
h_{0}^{\alpha,\beta}\mathfrak{C}_{0}^{\alpha,\beta}G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)$
$\displaystyle\quad=\iint\Big{\\{}(\alpha+\beta+5/2)e^{-t(\alpha+\beta+2)}\sin\theta\sin\varphi$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+5/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,\Pi_{\alpha}(u)\,du\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad-2e^{-t(\alpha+\beta+3/2)/2}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,\Pi_{\alpha}(u)\,du\,d\Pi_{-1/2}(v)$
$\displaystyle\qquad\qquad-2e^{-t(\alpha+\beta+3/2)/2}\cos\frac{\theta}{2}\cos\frac{\varphi}{2}$
$\displaystyle\qquad\qquad\qquad\times
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{-1/2}(u)\,\Pi_{\beta}(v)\,dv$
$\displaystyle\qquad\qquad+H_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\Big{)}\,d\Pi_{-1/2}(u)\,d\Pi_{-1/2}(v)\Big{\\}}.$

###### Proof.

To verify items (i)–(iv) above, one starts by multiplying the identities in
(i)–(iv) of Theorem 3.2 by the factor $\exp(-tn(n+\alpha+\beta+1))$. The
resulting equations are summed over $n=0,1,2,\ldots$. Then one uses the simple
relations

$\displaystyle tn(n+\alpha+\beta+1)$
$\displaystyle\quad=(t/4)2n[2n+(\alpha+\beta+1/2)+(\alpha+\beta+1/2)+1]$
$\displaystyle\quad=(t/4)(2n-1)[2n-1+(\alpha+\beta+3/2)+(\alpha+\beta+3/2)+1]+t(\alpha+\beta+3/2)/2$
$\displaystyle\quad=(t/4)(2n-2)[2n-2+(\alpha+\beta+5/2)+(\alpha+\beta+5/2)+1]+t(\alpha+\beta+2)$

and the fact that Jacobi-ultraspherical polynomials of even (odd) orders are
even (odd) functions. It follows that the series obtained represents in each
case either the even or the odd part with respect to the first variable of the
Jacobi-ultraspherical heat kernel with appropriate parameter and the second
variable fixed at the right endpoint. Since, for symmetry reasons, the
corresponding complementary odd/even parts give no contributions to the
integrals, we end up with the desired conclusions.

Dealing with the last term in item (v) is even more straightforward, since it
requires only some of the above arguments. ∎

##  5\. Preliminaries for the proof of Theorem A

### Initial estimates

The next result is a special case of Theorem A.

######  Lemma 5.1.

Let $\lambda>-1$. Then

$G_{t}^{\lambda}(\cos\theta,1)\simeq
t^{-\lambda-1}\exp\bigg{(}-\frac{\theta^{2}}{4t}\bigg{)},$

uniformly in $0\leq\theta\leq\pi/2$ and $0<t\leq 1$.

######  Remark 5.2.

This estimate is new for $\lambda<-1/2$, but for $\lambda\geq-1/2$ it is
contained in the known part of Theorem A.

###### Proof of Lemma 5.1.

We first verify that $\lim_{t\to 0^{+}}G_{t}^{\lambda}(0,1)=0$. The heat
equation yields

$\frac{\partial}{\partial
t}\,G_{t}^{\lambda}(0,1)=-J^{\lambda}\,G_{t}^{\lambda}(x,1)\Big{|}_{x=0}=\frac{d^{2}}{dx^{2}}\,G_{t}^{\lambda}(x,1)\Big{|}_{x=0}.$

To estimate this second derivative, we apply Lemma 4.1 twice and then use
Remark 5.2, getting

(13) $\frac{\partial}{\partial t}\,G_{t}^{\lambda}(0,1)\simeq
G_{t}^{\lambda+2}(0,1)\simeq
t^{-\lambda-3}\,\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)}>0,\qquad 0<t\leq 1.$

Thus $G_{t}^{\lambda}(0,1)$ is a non-negative, increasing function of $t$ in
$(0,1]$, and so the limit

$\eta:=\lim_{t\to 0^{+}}G_{t}^{\lambda}(0,1)$

exists and is non-negative.

Aiming at a contradiction, we assume that $\eta>0$. Since
$G_{t}^{\lambda}(x,1)$ is increasing in $x$ by Lemma 4.1, we then have
$G_{t}^{\lambda}(x,1)\geq G_{t}^{\lambda}(0,1)>\eta$ for $0\leq x\leq 1$ and
$0<t\leq 1$, and by symmetry $G_{t}^{\lambda}(1,y)>\eta$ for $0\leq y\leq 1$.
Now let $\big{(}T_{t}^{\lambda}\big{)}$ be the Jacobi (ultraspherical)
semigroup and take a nontrivial continuous function $f\geq 0$ supported in a
compact subinterval of $(0,1)$. Then we get

$T^{\lambda}_{t}f(1)=\int_{-1}^{1}G_{t}^{\lambda}(1,y)f(y)\,d\varrho_{\lambda}(y)>\eta\,\|f\|_{L^{1}(d\varrho_{\lambda})}>0$

for $0<t\leq 1$. But $\big{(}T^{\lambda}_{t}\big{)}$ is a symmetric diffusion
semigroup in the sense of [14, Chapter 3] and its maximal operator is
$L^{\infty}$-bounded. By standard arguments, see e.g. [5, Chapter 2, Section
2], it follows that $T^{\lambda}_{t}f$ converges to $f$ as $t\to 0^{+}$,
uniformly in $[-1,1]$, since this convergence certainly holds for the dense
subset of $C([-1,1])$ consisting of finite linear combinations of Jacobi-
ultraspherical polynomials. This is a contradiction which proves that
$\eta=0$.

Then (13) and Remark 5.2 imply for $0<t\leq 1$

$\displaystyle G_{t}^{\lambda}(0,1)$
$\displaystyle\simeq\int_{0}^{t}\tau^{-\lambda-3}\,\exp\bigg{(}-\frac{\pi^{2}}{16\tau}\bigg{)}\,d\tau\simeq\int_{1/t}^{\infty}\sigma^{\lambda+1}\,\exp\bigg{(}-\frac{\pi^{2}}{16}\,\sigma\bigg{)}\,d\sigma$
(14) $\displaystyle\simeq
t^{-\lambda-1}\,\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)},$

where we made the change of variable $\sigma=1/\tau$.

For $x=\cos\theta\in(0,1]$, we apply first Lemma 4.1 and then Remark 5.2, to
get

$\displaystyle G_{t}^{\lambda}(x,1)-G_{t}^{\lambda}(0,1)$
$\displaystyle=\int_{0}^{x}\frac{d}{dy}\,G_{t}^{\lambda}(y,1)\,dy\simeq\int_{0}^{x}G_{t}^{\lambda+1}(y,1)\,dy$
$\displaystyle\quad=\int_{\theta}^{\pi/2}G_{t}^{\lambda+1}(\cos\varphi,1)\,\sin\varphi\,d\varphi\simeq
t^{-\lambda-2}\,\int_{\theta}^{\pi/2}\,\exp\bigg{(}-\frac{\varphi^{2}}{4t}\bigg{)}\,\varphi\,d\varphi$
$\displaystyle\quad=2\,t^{-\lambda-1}\,\bigg{[}\exp\bigg{(}-\frac{\theta^{2}}{4t}\bigg{)}-\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)}\bigg{]}.$

From this and (5), we can finish the proof by separating the case
$((\pi/2)^{2}-\theta^{2})/t>1$ and the contrary case. ∎

######  Lemma 5.3.

Let $\lambda\in(-3/2,-1]$. Then

$H_{t}^{\lambda}(\cos\theta)\simeq
t^{-\lambda-1}\,\exp\bigg{(}-\frac{\theta^{2}}{4t}\bigg{)},$

uniformly in $0\leq\theta\leq\pi/2$ and $0<t\leq 1$.

###### Proof.

We first claim that

$H_{t}^{\lambda}(0)\to 0\qquad\mathrm{as}\qquad t\to 0^{+}.$

In the formula of Theorem 4.3 (v), we take $\alpha=\beta=\lambda/2-1/4>-1$,
$\varphi=0$ and $\theta=\pi$. Then the fourth integral in the right-hand side
of this formula is precisely $H_{t/4}^{\lambda}(0)$. Moreover, the first three
integrals in the right-hand side all vanish because of the trigonometric
factors. In the left-hand side, we have constant times $G_{t}^{\alpha}(-1,1)$,
and this quantity is non-negative and, in view of Lemma 4.1, no larger than
$G_{t}^{\alpha}(0,1)$. Lemma 5.1 shows that $G_{t}^{\alpha}(0,1)\to 0$ as
$t\to 0^{+}$, and the claim follows.

We will next estimate $H_{t}^{\lambda}(x)$ for $0\leq x\leq 1$ and $0<t\leq
1$, by integrating its derivatives. Letting first $x=0$, we conclude from (12)
and Lemma 4.2 that

$\frac{\partial}{\partial
t}\,H_{t}^{\lambda}(0)=-J^{\lambda}H_{t}^{\lambda}(x)\Big{|}_{x=0}=\frac{d^{2}}{dx^{2}}\,H_{t}^{\lambda}(x)\Big{|}_{x=0}\simeq
G_{t}^{\lambda+2}(0,1)\simeq
t^{-\lambda-3}\,\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)},$

the last relation by Lemma 5.1. Integrating as in (5), we get for $0<t\leq 1$

(15) $H_{t}^{\lambda}(0)\simeq
t^{-\lambda-1}\,\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)}.$

Let now $0\leq x\leq 1$ and $0<t\leq 1$. We write, using Lemma 4.2 and then
Lemma 4.1,

$\displaystyle H_{t}^{\lambda}(x)$
$\displaystyle=H_{t}^{\lambda}(0)+\int_{0}^{x}\frac{d}{dy}\,H_{t}^{\lambda}(y)\,dy=H_{t}^{\lambda}(0)+e^{-t(2\lambda+2)}\,\int_{0}^{x}\Big{[}G_{t}^{\lambda+1}(y,1)-G_{t}^{\lambda+1}(-y,1)\Big{]}\,dy$
$\displaystyle\simeq
H_{t}^{\lambda}(0)+\int_{0}^{x}\,\int_{-y}^{y}G_{t}^{\lambda+2}(z,1)\,dz\,dy\simeq
H_{t}^{\lambda}(0)+\int_{0}^{x}\,\int_{0}^{y}G_{t}^{\lambda+2}(z,1)\,dz\,dy,$

the last step by the monotonicity of $G_{t}^{\lambda+2}(\cdot,1)$. Now let
$x=\cos\theta,\;y=\cos\varphi$ and $z=\cos\kappa$. From Lemma 5.1, we see that
the last double integral is of order of magnitude

(16)
$\displaystyle\int^{\pi/2}_{\theta}\int^{\pi/2}_{\varphi}G_{t}^{\lambda+2}(\cos\kappa,1)\sin\kappa\,d\kappa\,\sin\varphi\,d\varphi\simeq
t^{-\lambda-3}\,\int^{\pi/2}_{\theta}\int^{\pi/2}_{\varphi}\exp\bigg{(}-\frac{\kappa^{2}}{4t}\bigg{)}\kappa\,d\kappa\,\varphi\,d\varphi$
$\displaystyle\qquad=2\,t^{-\lambda-2}\,\int^{\pi/2}_{\theta}\bigg{[}\exp\bigg{(}-\frac{\varphi^{2}}{4t}\bigg{)}-\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)}\bigg{]}\,\varphi\,d\varphi$
$\displaystyle\qquad=4\,t^{-\lambda-1}\,\bigg{[}\exp\bigg{(}-\frac{\theta^{2}}{4t}\bigg{)}-\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)}\bigg{]}-t^{-\lambda-2}\,\bigg{[}\Big{(}\frac{\pi}{2}\Big{)}^{2}-\theta^{2}\bigg{]}\,\exp\bigg{(}-\frac{\pi^{2}}{16t}\bigg{)}$
$\displaystyle\qquad=4\,t^{-\lambda-1}\,\exp\bigg{(}-\frac{\theta^{2}}{4t}\bigg{)}\bigg{[}1-\bigg{(}1+\frac{(\pi/2)^{2}-\theta^{2}}{4t}\bigg{)}\,\exp\bigg{(}-\frac{(\pi/2)^{2}-\theta^{2}}{4t}\bigg{)}\bigg{]}.$

Now if ${((\pi/2)^{2}-\theta^{2})}/{4t}>C$ for some suitably large constant
$C$, the product inside the square bracket in the last expression will be
small, and then the whole expression is of order of magnitude
$t^{-\lambda-1}\exp(-\theta^{2}/{4t})$. Together with (15), this implies the
conclusion of the lemma. And in the contrary case
${((\pi/2)^{2}-\theta^{2})}/{4t}\leq C$, one has
$t^{-\lambda-1}\exp(-\theta^{2}/{4t})\simeq
t^{-\lambda-1}\exp(-{\pi^{2}}/{16t})\simeq H_{t}^{\lambda}(0)$ because of
(15). Since the expressions in (16) are non-negative and no larger than a
constant times $t^{-\lambda-1}\exp(-\theta^{2}/{4t})$, the conclusion follows
from (15). ∎

### Further preparations

Let $0\leq\theta,\varphi\leq\pi$. For $0\leq u,v\leq 1$ we define

$F(u,v):=\bigg{[}\arccos\bigg{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\bigg{)}\bigg{]}^{2}.$

Then $0\leq F(1,1)=(\theta-\varphi)^{2}/4\leq F(u,v)\leq\pi^{2}/4=F(0,0)$.

######  Lemma 5.4.

There exist absolute constants $c_{1}>c_{2}>0$ such that the bounds

$\displaystyle\exp\bigg{(}-c_{1}\,\frac{(1-u)\theta\varphi}{t}\bigg{)}\exp\bigg{(}-c_{1}\,\frac{(1-v)(\pi-\theta)(\pi-\varphi)}{t}\bigg{)}$
$\displaystyle\qquad\qquad\leq\exp\bigg{(}-\frac{F(u,v)}{t}\bigg{)}\,\exp\bigg{(}\frac{F(1,1)}{t}\bigg{)}$
$\displaystyle\qquad\qquad\qquad\qquad\leq\exp\bigg{(}-c_{2}\,\frac{(1-u)\theta\varphi}{t}\bigg{)}\exp\bigg{(}-c_{2}\,\frac{(1-v)(\pi-\theta)(\pi-\varphi)}{t}\bigg{)}$

hold for $\theta,\varphi\in[0,\pi]$, $u,v\in[0,1]$ and $t>0$.

###### Proof.

We have
$\cos\sqrt{F(u,v)}=u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2}$.
Differentiating this equation with respect to $u$ and $v$, we get

$\displaystyle-\partial_{u}F(u,v)$
$\displaystyle=2\,\frac{\sqrt{F(u,v)}}{\sin\sqrt{F(u,v)}}\,\sin\frac{\theta}{2}\sin\frac{\varphi}{2}\simeq\sin\frac{\theta}{2}\sin\frac{\varphi}{2}\simeq\theta\varphi;$
$\displaystyle-\partial_{v}F(u,v)$
$\displaystyle=2\,\frac{\sqrt{F(u,v)}}{\sin\sqrt{F(u,v)}}\,\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\simeq\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\simeq(\pi-\theta)(\pi-\varphi).$

Thus for any $0\leq u,v\leq 1$ and $0\leq\theta,\varphi\leq\pi$

$F(u,v)-F(1,1)\simeq(1-u)\theta\varphi+(1-v)(\pi-\theta)(\pi-\varphi).$

Dividing by $t$ and exponentiating, we obtain the desired conclusion. ∎

We will also need the following.

######  Lemma 5.5.

Let $\gamma\geq-1/2$ be fixed. Then uniformly in $\xi\geq 0$

$\int_{[0,1]}\exp\big{(}-\xi(1-s)\big{)}\,d\Pi_{\gamma}(s)\simeq\frac{1}{(1+\xi)^{\gamma+1/2}}.$

###### Proof.

The case $\gamma=-1/2$ is trivial, so we assume $\gamma>-1/2$. Using the
definition of $\Pi_{\gamma}$ and making the transformation $\sigma=\xi(1-s)$,
we get

$\displaystyle\int_{[0,1]}\exp\big{(}-\xi(1-s)\big{)}\,d\Pi_{\gamma}(s)$
$\displaystyle\simeq\int_{0}^{1}\exp\big{(}-\xi(1-s)\big{)}\,(1-s)^{\gamma-1/2}\,ds$
$\displaystyle=\xi^{-\gamma-1/2}\,\int_{0}^{\xi}\exp(-\sigma)\,\sigma^{\gamma-1/2}\,d\sigma$
$\displaystyle\simeq\xi^{-\gamma-1/2}\,\min\big{(}\xi^{\gamma+1/2},1\big{)}\simeq\frac{1}{(1+\xi)^{\gamma+1/2}}.$

∎

##  6\. Proof of Theorem A

In this proof we may assume that $T=1$, since in compact time intervals
separated from $0$ the short-time and long-time bounds coincide, and the long-
time bounds (2) are a simple consequence of the short-time bounds (see [10, p.
233]). In what follows all relations $\simeq$ and $\gtrsim$ are uniform with
respect to $\theta,\varphi\in[0,\pi]$ and $0<t\leq 1$.

The proof of Theorem A is based on the five parts of Theorem 4.3, which are
defined in terms of $\alpha$ and $\beta$. In Part (i) of Theorem 4.3 one has
$\alpha,\beta\geq-1/2$. This case is contained in the known part of Theorem A
and will not be considered here.

### STEP 1: Part (ii) of Theorem 4.3

Here $-1<\beta<-1/2\leq\alpha$.

By Theorem 4.3 (ii)

(17) $G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)\simeq I_{1}+I_{2},$

where

$\displaystyle I_{1}$ $\displaystyle=-(\pi-\theta)(\pi-\varphi)\iint
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{\alpha}(u)\,\Pi_{\beta}(v)\,dv;$
$\displaystyle I_{2}$ $\displaystyle=\iint
G_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{\alpha}(u)\,d\Pi_{-1/2}(v).$

Observe that $I_{1}$ and $I_{2}$ are positive, in the case of $I_{1}$ since
$G_{t}^{\alpha+\beta+3/2}(\cdot,1)$ is increasing and $\Pi_{\beta}$ is odd and
decreasing. We will estimate $I_{1}$ and $I_{2}$ separately.

For the simpler $I_{2}$ we have (the relations will be justified in a moment)

$\displaystyle I_{2}$
$\displaystyle\simeq\iint_{[0,1]^{2}}G_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{\alpha}(u)\,d\Pi_{-1/2}(v)$
$\displaystyle\simeq
t^{-\alpha-\beta-3/2}\int_{[0,1]}\exp\bigg{(}-\frac{F(u,1)}{t}\bigg{)}\,d\Pi_{\alpha}(u).$

Here we first restrict the set of integration by means of the monotonicity of
$G_{t/4}^{\alpha+\beta+1/2}(\cdot,1)$ (see Lemma 4.1) and the symmetry of the
measures involved. The second relation follows from plugging in the bound of
Lemma 5.1. Applying Lemma 5.4 with $v=1$, we conclude

$I_{2}\gtrsim
t^{-\alpha-\beta-3/2}\exp\bigg{(}-\frac{F(1,1)}{t}\bigg{)}\int_{[0,1]}\exp\bigg{(}-c_{1}\,\frac{(1-u)\theta\varphi}{t}\bigg{)}\,d\Pi_{\alpha}(u),$

and this inequality can be reversed if we replace $c_{1}$ by $c_{2}$. But for
any fixed $c>0$, Lemma 5.5 implies

$\int_{[0,1]}\exp\bigg{(}-c\,\frac{(1-u)\theta\varphi}{t}\bigg{)}\,d\Pi_{\alpha}(u)\simeq\frac{1}{(1+\theta\varphi/t)^{\alpha+1/2}}=t^{\alpha+1/2}(t+\theta\varphi)^{-\alpha-1/2}.$

It follows that

(18)
$I_{2}\simeq(t+\theta\varphi)^{-\alpha-1/2}\,t^{-\beta-1}\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)}.$

To treat $I_{1}$, we recall that $\Pi_{\beta}(v)$ is an odd function which is
negative for $v>0$. Thus we have

$I_{1}=(\pi-\theta)(\pi-\varphi)\int_{[-1,1]}\int_{[0,1]}\Big{[}G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\\\
-G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}-v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\Big{]}\,|\Pi_{\beta}(v)|\,dv\,d\Pi_{\alpha}(u).$

We write the difference here as the integral from $-v$ to $v$ of the
derivative and use Lemma 4.1, to get

$\displaystyle I_{1}$ $\displaystyle\simeq(\pi-\theta)^{2}(\pi-\varphi)^{2}$
$\displaystyle\qquad\times\int_{[-1,1]}\int_{[0,1]}\int_{-v}^{v}G_{t/4}^{\alpha+\beta+5/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v^{\prime}\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,dv^{\prime}\,|\Pi_{\beta}(v)|\,dv\,d\Pi_{\alpha}(u).$

Since the integrand here is increasing in $u$ and in $v^{\prime}$, the
symmetries of the measures $d\Pi_{\alpha}$ and $dv^{\prime}$ show that one can
restrict the integration in these two variables to $[0,1]$ and $[0,v]$,
respectively. Then we use Fubini’s theorem and integrate first in $v$, which
produces a factor
$\int_{v^{\prime}}^{1}|\Pi_{\beta}(v)|\,dv\simeq(1-v^{\prime})^{\beta+3/2}$.
As a result,

$I_{1}\simeq(\pi-\theta)^{2}(\pi-\varphi)^{2}\int_{[0,1]}\int_{[0,1]}G_{t/4}^{\alpha+\beta+5/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v^{\prime}\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,(1-v^{\prime})^{\beta+3/2}\,dv^{\prime}\,d\Pi_{\alpha}(u).$

Now Lemma 5.1 tells us that

$I_{1}\simeq(\pi-\theta)^{2}(\pi-\varphi)^{2}\,t^{-\alpha-\beta-7/2}\int_{[0,1]}\int_{[0,1]}\exp\bigg{(}-\frac{F(u,v^{\prime})}{t}\bigg{)}\,(1-v^{\prime})^{\beta+3/2}\,dv^{\prime}\,d\Pi_{\alpha}(u).$

From Lemma 5.4 we see that this expression can be estimated from above and
below by constant times

$\displaystyle(\pi-\theta)^{2}(\pi-\varphi)^{2}\,t^{-\alpha-\beta-7/2}\exp\bigg{(}-\frac{F(1,1)}{t}\bigg{)}$
$\displaystyle\;\;\times\int_{[0,1]}\exp\bigg{(}-c\,\frac{(1-u)\theta\varphi}{t}\bigg{)}\,d\Pi_{\alpha}(u)\int_{[0,1]}\exp\bigg{(}-c\,\frac{(1-v^{\prime})(\pi-\theta)(\pi-\varphi)}{t}\bigg{)}\,(1-v^{\prime})^{\beta+3/2}\,dv^{\prime},$

though with different values of $c>0$ in the upper and lower estimates. In the
last integral here, we can replace $(1-v^{\prime})^{\beta+3/2}\,dv^{\prime}$
by the comparable measure $d\Pi_{\beta+2}(v^{\prime})$. Then Lemma 5.5 implies
that the product of the two integrals is of order of magnitude

$t^{\alpha+1/2}(t+\theta\varphi)^{-\alpha-1/2}\,t^{\beta+5/2}\big{[}t+(\pi-\theta)(\pi-\varphi)\big{]}^{-\beta-5/2}.$

Summing up, we have

$I_{1}\simeq(t+\theta\varphi)^{-\alpha-1/2}(\pi-\theta)^{2}(\pi-\varphi)^{2}\,\,\big{[}t+(\pi-\theta)(\pi-\varphi)\big{]}^{-\beta-5/2}\,t^{-1/2}\,\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)}.$

Adding this and (18), we obtain from (17)

(19)
$G_{t}^{\alpha,\beta}(\cos\theta,\cos\varphi)\simeq(t+\theta\varphi)^{-\alpha-1/2}\,\big{[}t+(\pi-\theta)(\pi-\varphi)\big{]}^{-\beta-1/2}\,t^{-1/2}\,\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)},$

as desired.

### STEP 2: Part (iii) of Theorem 4.3

Here $-1<\alpha<-1/2\leq\beta$.

This part is completely analogous to Part (ii). We need only swap $\alpha$
with $\beta$, $u$ with $v$ and $\theta,\;\varphi$ with
$\pi-\theta,\;\pi-\varphi$. The result will coincide with (19).

### STEP 3: The first three integrals in Parts (iv) and (v) of Theorem 4.3

Here $-1<\alpha,\beta<-1/2$.

Neglecting irrelevant positive factors, these three integrals in Part (iv)
coincide with those in Part (v), and they are

$\displaystyle J_{1}$ $\displaystyle=\,\sin\theta\sin\varphi\,\iint
G_{t/4}^{\alpha+\beta+5/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,\Pi_{\alpha}(u)\,du\,\Pi_{\beta}(v)\,dv;$
$\displaystyle J_{2}$
$\displaystyle=\,-\sin\frac{\theta}{2}\sin\frac{\varphi}{2}\,\iint
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,\Pi_{\alpha}(u)\,du\,d\Pi_{-1/2}(v);$
$\displaystyle J_{3}$
$\displaystyle=\,-\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\,\iint
G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{-1/2}(u)\,\Pi_{\beta}(v)\,dv.$

Each of these three quantities will be seen to be positive in the estimates
that follow.

We start with $J_{1}$. Imitating the first step of the treatment of $I_{1}$
above, now for both the variables $u$ and $v$, we have

(20)
$J_{1}=\,\sin\theta\sin\varphi\,\iint_{[0,1]^{2}}\Delta(u,v)\,|\Pi_{\alpha}(u)|\,du\,|\Pi_{\beta}(v)|\,dv,$

where $\Delta(u,v)$ is defined as

$\displaystyle
G_{t/4}^{\alpha+\beta+5/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}-G_{t/4}^{\alpha+\beta+5/2}\Big{(}-u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}$
$\displaystyle-\,G_{t/4}^{\alpha+\beta+5/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}-v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}+G_{t/4}^{\alpha+\beta+5/2}\Big{(}-u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}-v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}.$

This second difference is also given by

(21)
$\Delta(u,v)=\iint_{|u^{\prime}|<u,\,|v^{\prime}|<v}\partial_{u^{\prime}}\partial_{v^{\prime}}\Big{[}G_{t/4}^{\alpha+\beta+5/2}\Big{(}u^{\prime}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v^{\prime}\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\Big{]}\,du^{\prime}\,dv^{\prime}.$

Applying Lemma 4.1 twice, we have

$\displaystyle\partial_{u^{\prime}}\partial_{v^{\prime}}\Big{[}G_{t/4}^{\alpha+\beta+5/2}\Big{(}u^{\prime}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v^{\prime}\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\Big{]}$
$\displaystyle\quad\simeq\sin\theta\sin\varphi\>G_{t/4}^{\alpha+\beta+9/2}\Big{(}u^{\prime}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v^{\prime}\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}.$

By combining this with (20) and (21), we obtain

$J_{1}\simeq\,\sin^{2}\theta\,\sin^{2}\varphi\iint_{[0,1]^{2}}\iint_{|u^{\prime}|<u,\,|v^{\prime}|<v}\\\
G_{t/4}^{\alpha+\beta+9/2}\Big{(}u^{\prime}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v^{\prime}\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,du^{\prime}\,dv^{\prime}\,|\Pi_{\alpha}(u)|\,du\,|\Pi_{\beta}(v)|\,dv.$

As in the case of $I_{1}$ in Step 1, we can restrict the inner integration
here to $0<u^{\prime}<u$ and $0<v^{\prime}<v$, then apply Lemma 5.1, and
finally swap the order of integration. The result will be

$J_{1}\simeq\sin^{2}\theta\,\sin^{2}\varphi\;t^{-\alpha-\beta-11/2}\iint_{[0,1]^{2}}\exp\bigg{(}-\frac{F(u^{\prime},v^{\prime})}{t}\bigg{)}\,\int_{u^{\prime}}^{1}|\Pi_{\alpha}(u)|\,du\,\int_{v^{\prime}}^{1}|\Pi_{\beta}(v)|\,dv\,du^{\prime}\,dv^{\prime}.$

From Lemma 2.1 we see that the two inner integrals here are of order of
magnitude $(1-u^{\prime})^{\alpha+3/2}$ and $(1-v^{\prime})^{\beta+3/2}$,
respectively. The same lemma allows us to replace
$(1-u^{\prime})^{\alpha+3/2}\,du^{\prime}$ and
$(1-v^{\prime})^{\beta+3/2}\,dv^{\prime}$ by $d\Pi_{\alpha+2}(u^{\prime})$ and
$d\Pi_{\beta+2}(v^{\prime})$. Then Lemma 5.4 implies that $J_{1}$ is bounded
above and below by expressions of type

$\displaystyle\sin^{2}\theta\,\sin^{2}\varphi\;t^{-\alpha-\beta-11/2}\exp\bigg{(}-\frac{F(1,1)}{t}\bigg{)}\,$
$\displaystyle\quad\times\int_{[0,1]}\,\exp\bigg{(}-c\,\frac{(1-u^{\prime})\theta\varphi}{t}\bigg{)}\,d\Pi_{\alpha+2}(u^{\prime})\;\int_{[0,1]}\exp\bigg{(}-c\,\frac{(1-v^{\prime})(\pi-\theta)(\pi-\varphi)}{t}\bigg{)}\,d\Pi_{\beta+2}(v^{\prime}),$

with constants $c>0$. Because of Lemma 5.5, the integrals in this expression
are of orders of magnitude $(1+\theta\varphi/t)^{-\alpha-5/2}$ and
$[1+(\pi-\theta)(\pi-\varphi)/t]^{-\beta-5/2}$, respectively. Thus

(22) $\displaystyle J_{1}$
$\displaystyle\simeq\theta^{2}\varphi^{2}\,(t+\theta\varphi)^{-\alpha-5/2}\,(\pi-\theta)^{2}(\pi-\varphi)^{2}\,\big{[}t+(\pi-\theta)(\pi-\varphi)\big{]}^{-\beta-5/2}$
$\displaystyle\qquad\times
t^{-1/2}\,\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)}.$

Next we consider $J_{2}$. As in the argument for $I_{1}$ above but with $u$
and $v$ swapped, we write

$\displaystyle J_{2}$
$\displaystyle=\,\sin\frac{\theta}{2}\sin\frac{\varphi}{2}\,\int_{v\in[-1,1]}\int_{u\in[0,1]}\Big{[}G_{t/4}^{\alpha+\beta+3/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}$
$\displaystyle\qquad\qquad\qquad-
G_{t/4}^{\alpha+\beta+3/2}\Big{(}-u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\Big{]}\,|\Pi_{\alpha}(u)|\,du\,d\Pi_{-1/2}(v)$
$\displaystyle\simeq\,\sin^{2}\frac{\theta}{2}\,\sin^{2}\frac{\varphi}{2}\,\int_{v\in[-1,1]}\int_{u\in[0,1]}\int_{|u^{\prime}|<u}$
$\displaystyle\qquad\qquad\qquad
G_{t/4}^{\alpha+\beta+5/2}\Big{(}u^{\prime}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,du^{\prime}\,|\Pi_{\alpha}(u)|\,du\,d\Pi_{-1/2}(v),$

where we also wrote the difference as an integral of the derivative and used
Lemma 4.1. Since $G_{t/4}^{\alpha+\beta+5/2}(\cdot,1)$ is increasing and the
measures are symmetric, the integrations in $u^{\prime}$ and $v$ can be
restricted to $0<u^{\prime}<u$ and $v=1$, respectively. As in the case of
$J_{1}$ above, we switch the order of integration in $u^{\prime}$ and $u$, and
then replace $\int_{u^{\prime}}^{1}|\Pi_{\alpha}(u)|\,du\,du^{\prime}$ by
$d\Pi_{\alpha+2}(u^{\prime})$. The result will be

$J_{2}\simeq\,\theta^{2}\,\varphi^{2}\,\int_{[0,1]}G_{t/4}^{\alpha+\beta+5/2}\Big{(}u^{\prime}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{\alpha+2}(u^{\prime}).$

Apply now first Lemma 5.1 and then Lemma 5.4, to conclude that

$J_{2}\simeq\,\theta^{2}\,\varphi^{2}\;t^{-\alpha-\beta-7/2}\,\exp\bigg{(}-\frac{F(1,1)}{t}\bigg{)}\,\int_{[0,1]}\,\exp\bigg{(}-c\,\frac{(1-u^{\prime})\theta\varphi}{t}\bigg{)}\,d\Pi_{\alpha+2}(u^{\prime}).$

Using finally Lemma 5.5, we get

(23)
$J_{2}\simeq\,\theta^{2}\,\varphi^{2}\;(t+\theta\varphi)^{-\alpha-5/2}\,t^{-\beta-1}\,\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)}.$

The integral $J_{3}$ is obtained from $J_{2}$ by the same swapping that led
from Part (ii) to Part (iii) above. It follows that

(24)
$J_{3}\simeq\,(\pi-\theta)^{2}(\pi-\varphi)^{2}\,t^{-\alpha-1}\,\big{[}t+(\pi-\theta)(\pi-\varphi)\big{]}^{-\beta-5/2}\,\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)}.$

### STEP 4: The fourth integral in Part (iv) of Theorem 4.3

Here $-1<\alpha,\beta<-1/2$ and $\alpha+\beta>-3/2$.

This integral is

$\displaystyle\iint
G_{t/4}^{\alpha+\beta+1/2}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)}\,d\Pi_{-1/2}(u)\,d\Pi_{-1/2}(v)$
$\displaystyle\quad\simeq\,G_{t/4}^{\alpha+\beta+1/2}\Big{(}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+\cos\frac{\theta}{2}\cos\frac{\varphi}{2},1\Big{)},$

since $G_{t/4}^{\alpha+\beta+1/2}(\cdot,1)$ is increasing. Lemma 5.1 tells us
that this is of order of magnitude

(25)
$t^{-\alpha-\beta-3/2}\,\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)}.$

We can now finish the argument for Part (iv) of Theorem 4.3. The quantities in
(25) and those in (22), (23) and (24) are all dominated by
$Z_{t}^{\alpha,\beta}(\theta,\varphi)$. Moreover, depending on the relative
sizes of $\theta\varphi$, $(\pi-\theta)(\pi-\varphi)$ and $t$, one of these
quantities is in each case of the same order of magnitude as
$Z_{t}^{\alpha,\beta}(\theta,\varphi)$.

### STEP 5: The fourth integral in Part (v) of Theorem 4.3

Here $-1<\alpha,\beta<-1/2$ and $\alpha+\beta\leq-3/2$.

The integral is

(26) $\iint
H_{t/4}^{\lambda}\Big{(}u\sin\frac{\theta}{2}\sin\frac{\varphi}{2}+v\cos\frac{\theta}{2}\cos\frac{\varphi}{2}\Big{)}\,d\Pi_{-1/2}(u)\,d\Pi_{-1/2}(v),$

where $\lambda:=\alpha+\beta+1/2\in(-3/2,-1]$. We can now use Lemma 5.3,
together with the facts that $H_{t}^{\lambda}$ is even in $[-1,1]$ and
increasing in $[0,1]$, to estimate the integral in (26). This integral equals
one-half of

$H_{t/4}^{\lambda}\bigg{(}\cos\frac{\theta-\varphi}{2}\bigg{)}+H_{t/4}^{\lambda}\bigg{(}\cos\frac{\theta+\varphi}{2}\bigg{)}\simeq
H_{t/4}^{\lambda}\bigg{(}\cos\frac{\theta-\varphi}{2}\bigg{)}\simeq
t^{-\lambda-1}\,\exp\bigg{(}-\frac{(\theta-\varphi)^{2}}{4t}\bigg{)}.$

By means of these estimates together with those in (22), (23) and (24), we can
now finish the proof for Part (v) of Theorem 4.3, in the same way as we ended
that of Part (iv) in Step 4.

Theorem A is completely proved. ∎

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