Strict positive definiteness under axial symmetry on the sphere

Abstract

Axial symmetry for covariance functions defined over spheres has been a very popular assumption for climate, atmospheric, and environmental modeling. For Gaussian random fields defined over spheres embedded in a three-dimensional Euclidean space, maximum likelihood estimation techiques as well kriging interpolation rely on the inverse of the covariance matrix. For any collection of points where data are observed, the covariance matrix is determined through the realizations of the covariance function associated with the underlying Gaussian random field. If the covariance function is not strictly positive definite, then the associated covariance matrix might be singular. We provide conditions for strict positive definiteness of any axially symmetric covariance function. Furthermore, we find conditions for reducibility of an axially symmetric covariance function into a geodesically isotropic covariance. Finally, we provide conditions that legitimate Fourier inversion in the series expansion associated with an axially symmetric covariance function.

Appendix: Mathematical Proofs and Lemmas

1.1 Proof of Proposition 1

Proof

We start by making use of the assumption $c_m(n,n')=c(n)\delta _{nn'}$ to write

$$\begin{aligned}&K(L,L^{\prime },\Delta \ell ) \\&\quad = \sum _{m=-\infty }^\infty \sum _{n,n'=\left| {m}\right| }^\infty e^{{\mathsf {i}} m \Delta \ell }\bar{P}^m_{n}(\sin L)\bar{P}^m_{n'}(\sin L^{\prime })c(n)\delta _{nn'} \\&\quad =\sum _{m=-\infty }^\infty \sum _{n=\left| {m}\right| }^\infty e^{{\mathsf {i}} m\Delta \ell }\bar{P}^m_{n}(\sin L)\bar{P}^m_n(\sin L^{\prime }) c(n). \end{aligned}$$

(23)

The two sums in (23) can be interchanged provided

$$\begin{aligned} \sum _{n=0}^\infty \sum _{m=-n}^n \left| {\bar{P}^m_{n}(\sin L)\bar{P}^m_n(\sin L^{\prime })}\right| c(n)<\infty . \end{aligned}$$

(24)

To show it, we note that Cauchy-Schwartz inequality in concert with (12) show that, for every $n\ge 0$,

$$\begin{aligned}&\sum _{m=-n}^n \left| {\bar{P}^m_{n}(\sin L)\bar{P}^m_n(\sin L^{\prime })}\right| \\&\quad \le \sqrt{\sum _{m=-n}^n (\bar{P}^m_{n}(\sin L))^2 \sum _{m=-n}^n (\bar{P}^m_n(\sin L^{\prime }))^2} =n+\frac{1}{2}. \end{aligned}$$

(25)

The fact that $\textstyle \sum _{n=0}^\infty (n+1/2)c(n)<\infty$ proves that (24) holds true, so that the interchange is legitimate. Therefore, we can swap the two sums in (23) and write:

$$\begin{aligned} K(L,L^{\prime },\Delta \ell )=\sum _{n=0}^\infty c(n) \sum _{m=-n}^{n}e^{{\mathsf {i}} m \Delta \ell } \bar{P}^m_n(\sin L)\bar{P}^m_n(\sin L^{\prime }). \end{aligned}$$

(26)

By (10),

$$\begin{aligned} \bar{P}^{-m}_n(\sin L )\bar{P}^{-m}_n(\sin L^{\prime })= \bar{P}^m_n(\sin L )\bar{P}^m_n(\sin L^{\prime } ) \end{aligned}$$

(27)

so that:

$$\begin{aligned}&c(n)\{e^{-{\mathsf {i}} m\Delta \ell }\bar{P}^{-m}_n(\sin L )\bar{P}^{-m}_n(\sin L^{\prime } )\\&\qquad + e^{{\mathsf {i}} m\Delta \ell }\bar{P}^m_n(\sin L )\bar{P}^m_n(\sin L^{\prime } )\}\\&\quad = 2c(n)\bar{P}^m_n(\sin L )\bar{P}^m_n(\sin L^{\prime } )\cos (m\Delta \ell ), \end{aligned}$$

and therefore (26) yields:

$$\begin{aligned}&K(L,xL^{\prime }, \Delta \ell )\\&\quad =\sum _{n=0}^\infty c(n) \left\{ 2\sum _{m=1}^{n} \bar{P}^m_n(\sin L)\bar{P}^m_n(\sin L^{\prime } )\cos (m \Delta \ell )\right. \\&\qquad \qquad \qquad \left. + \bar{P}^0_n(\sin L )\bar{P}^0_n(\sin L^{\prime } ) \right\} , \end{aligned}$$

which by (27) and recalling that the cosine is an even function, yields:

$$\begin{aligned}&K(L,L^{\prime },\Delta \ell )\\&\quad =\sum _{n=0}^\infty c(n) \sum _{m=-n}^{n} \bar{P}^m_n(\sin L )\bar{P}^m_n(\sin L^{\prime } )\cos (m\Delta \ell )\\&\quad =\sum _{n=0}^\infty c(n)(2n+1)\frac{1}{2} \sum _{m=-n}^{n} \frac{(n-m)!}{(n+m)!}\\&\qquad {P}^m_n(\sin L ){P}^m_n(\sin L^{\prime })\cos (m \Delta \ell ), \end{aligned}$$

which by the spherical harmonics addition theorem (3) yields

$$\begin{aligned} K(L,L^{\prime }, \Delta \ell )= \sum _{n=0}^\infty c(n)\left( n+\frac{1}{2}\right) P_n(\langle \varvec{s}, \varvec{s}^{\prime } \rangle ), \end{aligned}$$

(28)

where $\varvec{s},\varvec{s}^{\prime }$ belong to ${\mathbb {S}}^2$, so that $\varvec{s}=(L,\ell )$ and $\varvec{s}^{\prime }=(L^{\prime }, \ell ^{\prime })$. $\square$

1.2 Auxiliary Lemmas

To provide a constructive proof of Theorem 2, two auxiliary results are required.

Lemma 4

Fix$m\in {\mathbb {Z}}$and let$w_1,\ldots ,w_h$behdistinct points in$(-1,1)$such that

$$\begin{aligned} \sum _{r=1}^h z_{r}\bar{P}^m_n(w_r)=0 \end{aligned}$$

(29)

for every$n\in \{\left| {m}\right| ,\left| {m}\right| +1, \ldots \}$and for some complexh-ple$(z_{1}, \ldots , z_{h})$. Then, $z_{r}=0$for every$r=1,\ldots ,h$.

Proof

The identity (29) implies that

$$\begin{aligned} \sum _{r=1}^h {\text {Re}}(z_{r})\bar{P}^m_n(w_r)&=0 \end{aligned}$$

(30)

$$\begin{aligned} \sum _{r=1}^h {\text {Im}}(z_{r})\bar{P}^m_n(w_r)&=0 \end{aligned}$$

(31)

for every $n\in \{\left| {m}\right| ,\left| {m}\right| +1, \ldots \}$.

At this stage, recall that

$$\begin{aligned} \bar{P}^m_n(x)=k_{m,n}(1-x^2)^{\left| {m}\right| /2}\frac{d^{\left| {m}\right| }}{d x^{\left| {m}\right| }}P_n(x) \end{aligned}$$

where $k_{m,n}$ is a non zero constant, for each $n\in \{\left| {m}\right| ,\left| {m}\right| +1,\ldots \}$ and $x\in (-1,1)$.

Consider now a polynomial p of degree at most $h-1$ such that

$$\begin{aligned} p(w_r)={\text {Re}}(z_r)/\{(1-w_r^2)^{\left| {m}\right| /2}\}, \end{aligned}$$

(32)

for $r=1,\ldots ,h$, and let q be a polynomial of degree at most $h+\left| {m}\right| -1$ such that the $\left| {m}\right|$-th partial derivative of q is p. Since the ordinary Legendre polynomials $P_0,\ldots , P_{\left| {m}\right| +h-2}$ are known to generate the $(\left| {m}\right| +h-1)$-dimensional polynomial space, there exist $d_0,\ldots ,d_{\left| {m}\right| +h-2}\in {\mathbb {R}}$ such that

$$\begin{aligned} d_0P_0+\cdots + d_{\left| {m}\right| +h-2}P_{\left| {m}\right| +h-2}=q, \end{aligned}$$

which taking the $\left| {m}\right|$–th derivative of both sides and multiplying by $(1-x^2)^{\left| {m}\right| /2}$, yields that:

$$\begin{aligned} \frac{d_{\left| {m}\right| }}{k_{m,\left| {m}\right| }} \bar{P}^m_{\left| {m}\right| }(x)+\cdots + \frac{d_{\left| {m}\right| +h-2}}{k_{m,\left| {m}\right| +h-2}} \bar{P}^m_{\left| {m}\right| +h-2}(x)=(1-x^2)^{\left| {m}\right| /2}p(x) \end{aligned}$$

(33)

for every $x\in (-1,1)$. Hence, we can consider the following function

$$\begin{aligned} f(x)=\frac{d_{\left| {m}\right| }}{k_{m,\left| {m}\right| }}P^m_{\left| {m}\right| }(x)+\cdots + \frac{d_{\left| {m}\right| +h-2}}{k_{m,\left| {m}\right| +h-2}}P^m_{\left| {m}\right| +h-2}(x) \end{aligned}$$

knowing that by (32) and (33) we have that

$$\begin{aligned} f(w_r)={\text {Re}}(z_r) \end{aligned}$$

(34)

as $r=1,\ldots ,h$. By (30), we have that

$$\begin{aligned} \sum _{r=1}^h {\text {Re}}(z_{r})f(w_r)=0, \end{aligned}$$

which by (34) yields that

$$\begin{aligned} \sum _{r=1}^h {\text {Re}}(z_{r})^2=0, \end{aligned}$$

(35)

We have just proved that (30) implies (35). Similarly, we can prove that (31) implies

$$\begin{aligned} \sum _{r=1}^h {\text {Im}}(z_{r})^2=0 \end{aligned}$$

and the proof is complete. $\square$

Lemma 5

LetKbe an axially symmetric covariance function as in (17) such that for each$m\ge 0$, $c_m:\{m,m+1,\ldots \}\times \{m,m+1,\ldots \}\rightarrow {\mathbb {C}}$is a strictly positive definite function. Then, for$a=(a_1,\ldots ,a_k)\in {\mathbb {R}}^k$and$\varvec{s}_1,\ldots ,\varvec{s}_k$distinct points in${\mathbb {S}}^2$such that the longitude and the latitude of$\varvec{s}_j$are$\ell _j\in (-\pi ,\pi ]$and$L_j\in [-\pi /2,\pi /2]$, respectively ($j=1,\ldots ,k$), the following assertions are equivalent:

• (i) $a^TKa=0$, that is,

  $$\begin{aligned} \sum _{j,h=1}^k a_jK(L_j,L_h, \Delta \ell _{jh})a_h=0, \qquad (\Delta \ell _{jh} := \ell _j -\ell _h). \end{aligned}$$

  (36)

• (ii) The equality

  $$\begin{aligned} \sum _{j=1}^k a_j e^{ {\mathsf {i}} m \ell _j}\bar{P}^m_n(\sin L_j )=0. \end{aligned}$$

  (37)

  holds for each$m\in {\mathbb {Z}}$and each$n\in \{\left| {m}\right| ,\left| {m}\right| +1,\ldots \}$,

Proof

Note that

$$\begin{aligned} a^TKa&:= \sum _{j,h=1}^k a_jK(L_j,L_h, \Delta \ell _{jh})a_h \\&= \sum _{j,h=1}^k a_ja_h \sum _{m=-\infty }^\infty \sum _{n,n'=\left| {m}\right| }^\infty \\&\quad e^{{\mathsf {i}} m \Delta \ell _{jh}} \bar{P}^m_n(\sin L_j )\bar{P}^m_{n'}(\sin L_h ) c_m(n,n') \\&=\sum _{m=-\infty }^\infty \sum _{n,n'=\left| {m}\right| }^\infty c_m(n,n')\left\{ \sum _{j=1}^k a_j e^{ {\mathsf {i}} m \ell _j}\bar{P}^m_n(\sin L_j) \right\} \\&\quad \left\{ \sum _{h=1}^k a_h e^{- {\mathsf {i}} m \ell _h}\bar{P}^m_n(\sin L_h ) \right\} . \end{aligned}$$

(38)

Hence, it is clear that (ii) implies (i). Now suppose that (i) holds. Combining (36) and (38), we have that:

$$\begin{aligned} \sum _{m=-\infty }^\infty \sum _{n,n'=\left| {m}\right| }^\infty c_m(n,n')b_{n,m}b^*_{n',m}=0 \end{aligned}$$

(39)

where

$$\begin{aligned} b_{n,m}=\sum _{j=1}^k a_j e^{ {\mathsf {i}} m \ell _j}\bar{P}^m_n(\sin L_j ). \end{aligned}$$

(40)

Since $c_m$ is positive definite (for every m), then

$$\begin{aligned} \sum _{n,n'=\left| {m}\right| }^N c_m(n,n')b_{n,m}b^*_{n',m}\ge 0, \end{aligned}$$

for every N and every m. Thus, letting N diverge to infinity we have that

$$\begin{aligned} \sum _{n,n'=\left| {m}\right| }^\infty c_m(n,n')b_{n,m}b^*_{n',m}\ge 0, \end{aligned}$$

for every m. Therefore, by (39), we have that for every $m\in {\mathbb {Z}}$,

$$\begin{aligned} \sum _{n,n'=\left| {m}\right| }^\infty c_m(n,n')b_{n,m}b^*_{n',m}=0. \end{aligned}$$

(41)

At this stage, for each $m,N\in {\mathbb {Z}}$ with $m,N> 0$, let $C_m(N)$ be the $N\times N$ matrix whose $(n-m+1,n'-m+1)$ entry is $c_m(n,n')$ ($n,n'= m,\ldots , N+m-1$).

For each $m,N\in {\mathbb {Z}}$ with $N, m>0$, the matrix $C_m(N)$ must be positive definite since by assumption, for each $m\in {\mathbb {Z}}$, $c_m(n,n')$ is strictly positive definite. Hence, by Cholesky decomposition, there is a complex $N\times N$ lower triangular matrix $A_m(N)$ such that $C_m(N)=A_m(N) A_m(N)^*$ the diagonal entries of $A_m(N)$ are real and positive and the matrix $A_0(N)$ is real. Denoting by $a^{(m,N)}_{nr}$ the $(n-m+1,r-m+1)$ entry of the matrix $A_m(N)$ if $N,m> 0$, the Cholesky decomposition of the matrix $C_m(N)$ can be written in the following form:

$$\begin{aligned} c_m(n,n')=\sum _{r=m}^{n \wedge n'} a^{(m,N)}_{nr} a^{(m,N)*}_{n'r}, \end{aligned}$$

for $n,n'=m, \ldots , N+m-1$, where $x\wedge y$ denotes the minimum among x and y. If $M \ge N$, then by the Cholesky decomposition of the matrix $C_m(M)$ we have that:

$$\begin{aligned} \sum _{r=m}^{n \wedge n'} a^{(m,M)}_{nr} a^{(m,M)*}_{n'r}=\sum _{r=m}^{n \wedge n'} a^{(m,N)}_{nr} a^{(m,N)*}_{n'r}, \end{aligned}$$

for $n,n'=m, \ldots , N+m-1$. Note the Cholesky decomposition of $C_m(N)$ is unique being $C_m(N)$ positive definite and therefore $a^{(m,N)}_{nr}=a^{(m,M)}_{nr}$ if $m\le r \le n \le N+m-1 \le M+m-1$. In particular, we have that $a^{(m,n-m+1)}_{nr}=a^{(m,M)}_{nr}$ for any quartet (m, r, n, M) where $m\le r \le n \le M-m+1$. Therefore, we can write $a^{(m)}_{nr}$ for $a^{(m,n-m+1)}_{nr}$ and

$$\begin{aligned} c_m(n,n')=\sum _{r=\left| {m}\right| }^{n \wedge n'} a^{(m)}_{nr} a^{(m)*}_{n'r}, \end{aligned}$$

(42)

where for each r, $a^{(m)}_{rr}>0$. Note that (42) holds also for $m<0$ if we let $a^{(-m)}_{nr}=a^{(m)*}_{nr}$. Substituting (42) in (41) we obtain that, for every $m\in {\mathbb {Z}}$,

$$\begin{aligned} \sum _{n,n'=\left| {m}\right| }^\infty \sum _{r=\left| {m}\right| }^{n \wedge n'} a^{(m)}_{nr} a^{(m)*}_{n'r}b_{n,m}b^*_{n',m}=0, \end{aligned}$$

that is equivalent to:

$$\begin{aligned} \sum _{r=\left| {m}\right| }^{\infty } \sum _{n,n'= r}^\infty a^{(m)}_{nr} a^{(m)*}_{n'r}b_{n,m}b^*_{n',m}=0, \end{aligned}$$

which in turn is equivalent to:

$$\begin{aligned} \sum _{r=\left| {m}\right| }^{\infty } \left| {\sum _{n=r}^\infty a^{(m)}_{nr} b_{n,m}}\right| ^2=0, \end{aligned}$$

which implies that for every $r\ge \left| {m}\right|$

$$\begin{aligned} \sum _{n=r}^\infty a^{(m)}_{nr} b_{n,m}=0. \end{aligned}$$

(43)

Thus, for every fixed $m\in {\mathbb {Z}}$ and every fixed $r\ge \left| {m}\right|$ we can write:

$$\begin{aligned} \sum _{n=r+s}^\infty a^{(m)}_{n,r+s} b_{n,m}=0, \end{aligned}$$

(44)

for $s=0,1,2,\ldots$ Recalling that $a^{(m)}_{r+s,r+s}>0$ for each s, we can define a sequence $(d_s)_{s=0}^\infty$ according to the recursion:

$$\begin{aligned} d_0=1,\, d_1=-\frac{a^{(m)}_{r+1,r}}{a^{(m)}_{r+1,r+1}}, \ldots , d_s=-\frac{1}{a^{(m)}_{r+s,r+s}}\sum _{l=0}^{s-1} d_l a^{(m)}_{r+s,r+l}, \ldots \end{aligned}$$

This ensures that:

$$\begin{aligned} \sum _{s=0}^{t} d_s a_{r+t,r+s}=0 \end{aligned}$$

for $t=1,2,\ldots$ and therefore by (44),

$$\begin{aligned} 0&= \sum _{s=0}^\infty d_s \sum _{n=r+s}^\infty a^{(m)}_{n,r+s} b_{n,m} =\sum _{n=r}^\infty b_{n,m} \sum _{s=0}^{n-r} d_s a^{(m)}_{n,r+s}\\&= b_{r,m} d_0 a^{(m)}_{r,r} + \sum _{n=r+1}^\infty b_{n,m} \sum _{s=0}^{n-r} d_s a^{(m)}_{n,r+s}\\&= b_{r,m} a^{(m)}_{r,r} + \sum _{t=1}^\infty b_{r+t,m} \sum _{s=0}^{t} d_s a^{(m)}_{r+t,r+s} = b_{r,m} a^{(m)}_{r,r}. \end{aligned}$$

We have just proved that $b_{r,m} a^{(m)}_{r,r}=0$ for any $m\in {\mathbb {Z}}$ and any $r\ge \left| {m}\right|$. At this stage, recall once again that $a^{(m)}_{r,r}>0$ for any (m, r). Therefore, for any $m\in {\mathbb {Z}}$ and any $r\ge \left| {m}\right|$, $b_{r,m}=0$, namely by (40)

$$\begin{aligned} \sum _{j=1}^k a_j e^{ {\mathsf {i}} m \ell _j}\bar{P}^m_n(\sin L_j )= 0 \end{aligned}$$

(45)

for each $m\in {\mathbb {Z}}$ and each $n\in \{\left| {m}\right| ,\left| {m}\right| +1,\ldots \}$, proving (ii) . $\square$

Now we are ready to prove our main result.

1.3 Proof of Theorem 2

Proof

We aim at proving that if $(a_1,\ldots ,a_k)\in {\mathbb {R}}^k$, and $\varvec{s}_1,\ldots ,\varvec{s}_k$ are k distinct points in ${\mathbb {S}}^2$ such that the longitude and the latitude of $\varvec{s}_j$ are $\ell _j\in (-\pi ,\pi ]$ and $L_j\in [-\pi /2,\pi /2]$, respectively ($j=1,\ldots ,k$), and

$$\begin{aligned} \sum _{j,h=1}^k a_jK(L_j,L_h, \Delta \ell _{jh})a_h=0, \end{aligned}$$

(46)

then $a_j=0$ for $j=1,\ldots ,k$.

By Lemma 5, letting $m=0$, we obtain:

$$\begin{aligned} \sum _{j=1}^k a_j \bar{P}_n(\sin (L_j))= 0 \end{aligned}$$

(47)

for every $n\in \{0,1,2,\ldots \}$, where $\bar{P}_n$ denotes the ordinary normalized Legendre polynomial of degree n.

By denoting $L_{(1)},\ldots , L_{(h)}$ the distinct elements among $L_1,\ldots ,L_k$ and letting $a_{(r)}=\sum _{j\in B_r} a_j$ where $B_r=\{j=1,\ldots , k: L_j=L_{(r)}\}$, as $r=1,\ldots , h$, the equation (47) becomes:

$$\begin{aligned} \sum _{j=1}^h a_{(j)} \bar{P}_n(\sin L_{(j)})= 0 \end{aligned}$$

(48)

for every $n\in \{0,1,2,\ldots \}$.

At this stage, consider the following function:

$$\begin{aligned} p(L)=\sum _{n\in E} d_{n} \bar{P}_n(\sin L ), \end{aligned}$$

where E is a finite or infinite subset of $\{0,1,2,\ldots \}$, so that by (48),

$$\begin{aligned} \sum _{j=1}^h a_{(j)} p(L_{(j)})=0. \end{aligned}$$

(49)

There exist $\{d_{n}\in {\mathbb {R}}: n\in E \}$ for some set E such that $p(L_{(j)})=a_{(j)}$ for $j=1,\ldots ,h$. Indeed, we can just take $E=\{0,1,\ldots ,h-1\}$ recalling that $\bar{P}_0$,..., $\bar{P}_{h-1}$ generate the h–dimensional polynomial space and that interpolation of arbitrary data at h nodes is possible.

Hence, by (49), $\textstyle \sum _{j=1}^h a_{(j)}^2=0$ and therefore $a_{(j)}=0$, namely

$$\begin{aligned} \sum _{j\in B_r} a_j=0, \quad j=1,\ldots ,h. \end{aligned}$$

(50)

Now note that if $L_j\in \{-\pi /2,\pi /2\}$ then $\varvec{s}_j$ is a pole and therefore $L_i\ne L_j$ for any $i\ne j$ since $\varvec{s}_1,\ldots , \varvec{s}_n$ are distinct points (i.e., $B_r=\{j\}$ if $L_j\in \{-\pi /2,\pi /2\}$). Hence, by (50), we have just proved that if $L_j\in \{-\pi /2,\pi /2\}$ then $a_j=0$.

Hence , the equation (45) becomes:

$$\begin{aligned} \sum _{r=1}^h \sum _{j\in B_r} a_j e^{ {\mathsf {i}} m \ell _j} \bar{P}^m_n(\sin L_{(r)} )= 0, \end{aligned}$$

(51)

for each $m\in {\mathbb {Z}}$ and each $n\in \{\left| {m}\right| ,\left| {m}\right| +1,\ldots \}$. Since $\overline{P}_n^m(\pm 1) =0$, (51) becomes:

$$\begin{aligned} \sum _{r\in A} \sum _{j\in B_r} a_j e^{{\mathsf {i}} m \ell _j}\bar{P}^m_n(\sin L_{(r)} )= 0 \end{aligned}$$

(52)

where $A=\{r=1,\ldots ,h: L_{r}\notin \{-\pi /2,\pi /2\}\}$, for each $m\in {\mathbb {Z}}$ and each $n\in \{\left| {m}\right| ,\left| {m}\right| +1,\ldots \}$. Moreover, since we have just proved that $a_j=0$ if $L_j=\pm \pi /2$, then we need to show that $a_j=0$ for all $j\in B_r$ and for all $r\in A$.

By Lemma 4, (52) implies that:

$$\begin{aligned} \sum _{j\in B_r} a_j e^{{\mathsf {i}} m \ell _j}=0 \end{aligned}$$

(53)

for every $m\in {\mathbb {Z}}$ and $r\in A$. At this stage, observe that $\ell _i \ne \ell _j$ when $i,j\in B_r$ because in this case $L_i=L_j$ and $\varvec{s}_i\ne \varvec{s}_j$.

Moreover, the matrix of the system (53) is Vandermonde–like associated to the distinct points $e^{{\mathsf {i}} \ell _j}$, $j\in B_r$ for each $r\in A$. Therefore, (53) implies that $a_j=0$ for each $j=1,\ldots ,k$ such that $L_j\notin \{-\pi /2,\pi /2\}$ and the proof is complete. $\square$

1.4 Proof of Theorem 3

Proof

In order to evaluate the integral in the right hand side of (19), we can plug in (17) and swap the double series and the triple integral if the following is verified:

$$\begin{aligned}&\int _{-\pi /2}^{\pi /2}\int _{-\pi /2}^{\pi /2} \int _{-\pi }^\pi \sum _{n,n'=0}^\infty \sum _{m=-(n\wedge n')}^{n\wedge n'} \\&\quad g^{(m,n,n')}_{m_0,n_0,n'_0}(\Delta \ell ,L,L^{\prime }) {\mathrm {d}}\Delta \ell {\mathrm {d}}L {\mathrm {d}}L^{\prime } <\infty , \end{aligned}$$

(54)

where:

$$\begin{aligned}&g^{(m,n,n')}_{m_0,n_0,n'_0}(\Delta \ell ,L,L^{\prime })\\&\quad =\left| {e^{im \Delta \ell } \bar{P}^m_{n}(\sin L)\bar{P}^m_{n'}(\sin L^{\prime })c_m(n,n') e^{- {\mathsf {i}} m_0 \Delta \ell }\bar{P}^{m_0}_{n_0}(\sin L) \bar{P}^{m_0}_{n'_0}(\sin L^{\prime }) \cos L\cos L^{\prime }}\right| \end{aligned}$$

for every $m_0\in {\mathbb {Z}}$ and $n_0,n_0'\ge \left| {m_0}\right|$. Trivially,

$$\begin{aligned}&g^{(m,n,n')}_{m_0,n_0,n'_0}(\Delta \ell ,L,L^{\prime }) \\&\quad \le \left| {\bar{P}^m_{n}(\sin L)\bar{P}^m_{n'}(\sin L^{\prime })c_m(n,n') \bar{P}^{m_0}_{n_0}(\sin L) \bar{P}^{m_0}_{n'_0}(\sin L^{\prime })}\right| . \end{aligned}$$

(55)

At this stage, note that by (12),

$$\begin{aligned} \left| {\bar{P}^{m}_{n}(\sin L)}\right| \le \sqrt{n+\frac{1}{2}}, \end{aligned}$$

for every $m,n\in {\mathbb {Z}}$ with $n\ge \left| {m}\right|$, and therefore (55) yields:

$$\begin{aligned}&g^{(m,n,n')}_{m_0,n_0,n'_0}(\Delta \ell ,L,L^{\prime })\\&\quad \le \left| {\bar{P}^m_{n}(\sin L)\bar{P}^m_{n'}(\sin L^{\prime })c_m(n,n') \sqrt{(n_0+1/2)(n'_0+1/2)}}\right| , \end{aligned}$$

which, by the Cauchy–Schwartz inequality and (12), in turn yields:

$$\begin{aligned}&\sum _{m=-( n\wedge n')}^{n\wedge n'} g^{(m,n,n')}_{m_0,n_0,n'_0}(\Delta \ell ,L,L^{\prime })\\&\quad \le \sqrt{\left( n_0+\frac{1}{2}\right) \left( n'_0+\frac{1}{2}\right) \left( n+\frac{1}{2}\right) \left( n'+\frac{1}{2}\right) } \\&\qquad \times \sup _{m\in \{-(n\wedge n'), \ldots , n\wedge n'\}} \left| {c_m(n,n')}\right| , \end{aligned}$$

which by (18) implies that (54) is verified. It is therefore possible to plug (17) into the integral in (19) and swap the double sum with the triple integral. In this way, due to the orthogonality of the associated Legendre polynomials $\{\bar{P}^m_{n}(\sin L): n\in {\mathbb {N}}\}$ given by (6) and of $\{e^{im \ell }: m\in {\mathbb {Z}}\}$, and due to (7), (19) is obtained. $\square$