Proof. It suffices to show that if $\mathcal {D}_k f=0$ in $\mathbb {B}^{m}\times \mathbb {B}^{m}$, then $f$ has a power series expansion converging to $f$ in a neighbourhood of $0$. The main idea is to use the Poisson integral formula and the series expansion of the Poisson kernel. Recall that if $f(\boldsymbol {x},\boldsymbol {u})\in C^{2}(\Omega \times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ and $\mathcal {D}_k f=0$ in $\Omega \times \mathbb {B}^{m}$, the Poisson integral formula [Reference Ding, Nguyen and Ryan11, Section 5.2] is given by

\begin{align*} f(\boldsymbol{x},\boldsymbol{v})=\displaystyle\frac{c_{m,k}}{2}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}Z_k\left(\frac{(\boldsymbol{x}-\boldsymbol{\zeta})\boldsymbol{u}(\boldsymbol{x}-\boldsymbol{\zeta})}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{2}},\boldsymbol{v}\right)f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta}), \end{align*}

where $Z_k(\boldsymbol {u},\boldsymbol {v})$ is the reproducing kernel of $k$-homogeneous harmonic polynomials in the following sense

\begin{align*} g(\boldsymbol{v})=\int_{\mathbb{S}^{m-1}}Z_k(\boldsymbol{u},\boldsymbol{v})g(\boldsymbol{u})dS(\boldsymbol{u}),\quad \text{for all}\ g\in\mathcal{H}_k(\mathbb{C}). \end{align*}

On the one hand, we already knew that there is a series expansion, which converges absolutely, for the Poisson kernel of Laplacian $\frac {1-|\boldsymbol {x}|^{2}}{|\boldsymbol {x}-\boldsymbol {\zeta }|^{m}}=\sum _{|\boldsymbol {\alpha }|=0}^{\infty }p_{\boldsymbol {\alpha }}(\boldsymbol {\zeta })\boldsymbol {x}^{\boldsymbol {\alpha }}$, where $|\boldsymbol {x}|<\sqrt {2}-1$ and $\boldsymbol {\zeta }\in \mathbb {S}^{m-1}$, see [Reference Axler, Bourdon and Ramey1, Theorem 1.28]. On the other hand, let $\boldsymbol {\eta }=\frac {(\boldsymbol {x}-\boldsymbol {\zeta })\boldsymbol {u}(\boldsymbol {x}-\boldsymbol {\zeta })}{|(\boldsymbol {x}-\boldsymbol {\zeta })|^{2}}$, and $\{\varphi _j\}_{j=1}^{d_k}$ be an orthonormal basis for $\mathcal {H}_k(\mathbb {C})$, where $d_k=\dim \mathcal {H}_k(\mathbb {C})$. Then we have [Reference Axler, Bourdon and Ramey1, (5.28)] $Z_k(\boldsymbol {\eta },\boldsymbol {v})=\sum _{j=1}^{d_k}\overline {\varphi _j(\boldsymbol {\eta })}\varphi _j(\boldsymbol {v})$. Since $\varphi _j$ is a homogeneous harmonic polynomial of degree $k$, and $\boldsymbol {\eta }$ has no singular point for $\boldsymbol {x}\in U$, $\boldsymbol {\zeta }\in \mathbb {S}^{m-1}$, we find that $\varphi _j(\boldsymbol {\eta })$ is analytic for $\boldsymbol {x}\in \mathbb {B}^{m}$. More specifically, the local series expansion of $\varphi _j(\boldsymbol {\eta })$ can be obtained by plugging the series expansion for $\boldsymbol {\eta }=\frac {(\boldsymbol {x}-\boldsymbol {\zeta })\boldsymbol {u}(\boldsymbol {x}-\boldsymbol {\zeta })}{|\boldsymbol {x}-\boldsymbol {\zeta }|^{2}}$ into $\varphi _j$. Notice that

\begin{align*} \boldsymbol{\eta}& =\frac{(\boldsymbol{x}-\boldsymbol{\zeta})\boldsymbol{u}(\boldsymbol{x}-\boldsymbol{\zeta})}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{2}}\\& =\boldsymbol{u}-2\langle\boldsymbol{x}-\boldsymbol{\zeta},\boldsymbol{u}\rangle|\boldsymbol{x}-\boldsymbol{\zeta}|^{{-}2}=\boldsymbol{u}-2\langle\boldsymbol{x}-\boldsymbol{\zeta},\boldsymbol{u}\rangle\sum_{s=0}^{\infty}c_s(|\boldsymbol{x}|^{2}-2\langle\boldsymbol{x},\boldsymbol{\zeta}\rangle)^{s}, \end{align*}

where $\sum _{s=0}^{\infty }c_s(t-1)^{s}$ is the Taylor series of $t^{-2}$ on $(0,2)$ at point $t=1$. This series converges to $\boldsymbol {\eta }$ absolutely in $|\boldsymbol {x}|<\sqrt {2}-1, \boldsymbol {u}\in \overline {\mathbb {B}^{m}}$ with a similar argument as in [Reference Axler, Bourdon and Ramey1, Theorem 1.28]. Now, we plug the expansion of $\boldsymbol {\eta }$ into $\varphi _j(\boldsymbol {\eta })$ and then rearrange it to obtain $\varphi _j(\boldsymbol {\eta })=\sum _{|\boldsymbol {\beta }|=0}^{\infty }q_{\boldsymbol {\beta }}(\boldsymbol {\zeta },\boldsymbol {u})\boldsymbol {x}^{\boldsymbol {\beta }}$. Now, we have series expansions for $\frac {1-|\boldsymbol {x}|^{2}}{|\boldsymbol {x}-\boldsymbol {\zeta }|^{m}}$ and $\varphi _j(\boldsymbol {\eta })$, and both series converge absolutely for $|\boldsymbol {x}|<\sqrt {2}-1, \boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$. This allows us to calculate

\begin{align*} & \frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}Z_k(\boldsymbol{\eta},\boldsymbol{v})=\sum_{j=1}^{d_k}\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}\varphi_j(\boldsymbol{\eta})\varphi_j(\boldsymbol{v})\\ & \quad=\sum_{j=1}^{d_k}\sum_{\boldsymbol{|\alpha|=0}}^{\infty}p_{\boldsymbol{\alpha}}(\boldsymbol{\zeta})\boldsymbol{x}^{\boldsymbol{\alpha}}\sum_{|\boldsymbol{\beta}|=0}^{\infty}q_{\boldsymbol{\beta}}(\boldsymbol{\zeta},\boldsymbol{u})\boldsymbol{x}^{\boldsymbol{\beta}}\varphi_j(\boldsymbol{v})=\sum_{j=1}^{d_k}\sum_{\boldsymbol{|\gamma|=0}}^{\infty}h_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u})\boldsymbol{x}^{\boldsymbol{\gamma}}\varphi_j(\boldsymbol{v})\\ & \quad=:\sum_{\boldsymbol{|\gamma|=0}}^{\infty}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}, \end{align*}

where $h_{\boldsymbol {\gamma }}(\boldsymbol {\zeta },\boldsymbol {u})=\sum _{|\boldsymbol {\alpha }|+|\boldsymbol {\beta }|=|\boldsymbol {\gamma }|}p_{\boldsymbol {\alpha }}(\boldsymbol {\zeta })q_{\boldsymbol {\beta }}(\boldsymbol {\zeta },\boldsymbol {u})$ and $g_{\boldsymbol {\gamma }}(\boldsymbol {\zeta },\boldsymbol {u},\boldsymbol {v})$ is a $k$-homogeneous harmonic polynomial with respect to $\boldsymbol {v}$. Further, this series expansion for the Poisson kernel also converges absolutely when $|\boldsymbol {x}|<\sqrt {2}-1, \boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$. Now, we plug this series expansion back into the Poisson integral formula to obtain

\begin{align*} f(\boldsymbol{x},\boldsymbol{v})& =\displaystyle\frac{c_{m,k}}{2}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\displaystyle\sum_{\boldsymbol{|\gamma|=0}}^{\infty}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\\ & =\Big[\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\sum_{\boldsymbol{|\gamma|=0}}^{\infty}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\Big]\boldsymbol{x}^{\boldsymbol{\gamma}}\\ & =\sum_{\boldsymbol{|\gamma|=0}}^{\infty}\left[\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\right]\boldsymbol{x}^{\boldsymbol{\gamma}} =:C_{\boldsymbol{\gamma}}(\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}, \end{align*}

where $C_{\boldsymbol {\gamma }}(\boldsymbol {v})\in \mathcal {H}_k(\boldsymbol {v},\mathbb {C})$. In the last second step, we interchange the integral and infinite sum because of the facts that $\sum _{\boldsymbol {|\gamma |=0}}^{\infty }g_{\boldsymbol {\gamma }}(\boldsymbol {\zeta },\boldsymbol {u},\boldsymbol {v})\boldsymbol {x}^{\boldsymbol {\gamma }}$ converges absolutely when $|\boldsymbol {x}|<\sqrt {2}-1, \boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$ and $f$ is a polynomial, which allows us to apply Fubini's Theorem.

Proof. Without loss of generality, we assume that $\boldsymbol {a}$ is the origin, because for arbitrary $\boldsymbol {a}\in \Omega$, one can obtain the result by a translation. We denote $f_l(\boldsymbol {x},\boldsymbol {u})=\sum _{|\boldsymbol {\alpha }|=l}\frac {\partial ^{\boldsymbol {\alpha }}f(0,\boldsymbol {u})}{\partial \boldsymbol {x}^{\boldsymbol {\alpha }}}\boldsymbol {x}^{\boldsymbol {\alpha }},$ and it is easy to see that $f_l\in \mathcal {H}_k(\boldsymbol {u},\mathbb {C})$. From the previous theorem, we know that there exists a neighbourhood $U$ of the origin, such that $f(\boldsymbol {x},\boldsymbol {u})=\sum _{l=0}^{\infty }f_l(\boldsymbol {x},\boldsymbol {u})$ for $\boldsymbol {x}\in U$.

Further, we notice that

(3.1)\begin{equation} 0=\mathcal{D}_k f=\mathcal{D}_k \displaystyle\sum_{l=0}^{\infty}f_l= \sum_{l=0}^{\infty}\mathcal{D}_k f_l. \end{equation}

Since $\mathcal {D}_k f_l\in \mathcal {P}_{l-2}(\Omega \times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, which implies that if $\{\varphi _j(\boldsymbol {u})\}_{j=1}^{d_{k}}$ is an orthonormal basis for $\mathcal {H}_{k}(\boldsymbol {u})$, then we can rewrite $\mathcal {D}_k f_l(\boldsymbol {x},\boldsymbol {u})=\sum _{j=1}^{d_{k}}a_{j,l-2}(\boldsymbol {x})\varphi _j(\boldsymbol {u})$, where $a_{j,l-2}(\boldsymbol {x})$ is a homogeneous polynomial of $\boldsymbol {x}$ with degree $l-2$. Then (3.1) gives $\sum _{l=0}^{\infty }\sum _{j=1}^{d_{k}}a_{j,l-2}(\boldsymbol {x})\varphi _j(\boldsymbol {u})=0.$ However, since $\{\varphi _j(\boldsymbol {u})\}_{j=1}^{d_{k}}$ is an orthonormal basis, we obtain $\sum _{l=0}^{\infty }a_{j,l-2}(\boldsymbol {x})=0,\; \text{for all}\; j,$ which implies that $a_{j,l-2}(\boldsymbol {x})=0$ for all $j,l$. This is equivalent to $\mathcal {D}_k f_l=0$ for all $l$, which completes the proof.

Proof. When $l=0,1$, the result is obviously true with $g=0$. Now, we assume that $l\geq 2$. Firstly, it is easy to observe that $(1-|\boldsymbol {x}|^{2})g+f=f$ when $\boldsymbol {x}\in \mathbb {S}^{m-1}$. Hence, with the boundary data $f\big \vert _{\mathbb {S}^{m-1}}$, if we can find a polynomial $g(\boldsymbol {x},\boldsymbol {u})$ satisfying $\deg _{\boldsymbol {x}}g\leq l-2$, $g(\boldsymbol {x},\cdot )\in \mathcal {H}_k(\boldsymbol {u},\mathbb {C})$ for each fixed $\boldsymbol {x}$ and $\mathcal {D}_k ((1-|\boldsymbol {x}|^{2})g+f)=0$, i.e., $\mathcal {D}_k(1-|\boldsymbol {x}|^{2})g=-\mathcal {D}_k f$, then from the uniqueness of the Dirichlet problem for bosonic Laplacians on the unit ball [Reference Ding, Nguyen and Ryan11, Section 4], we immediately have $P[f\big \vert _{\mathbb {S}^{m-1}}]=(1-|\boldsymbol {x}|^{2})g+f$.

Now, let $W$ stand for the vector space of polynomials $f(\boldsymbol {x},\boldsymbol {u})$ satisfying $\deg _{\boldsymbol {x}}f\leq l-2$ and $f(\boldsymbol {x},\cdot )\in \mathcal {H}_k(\boldsymbol {u},\mathbb {C})$ for each fixed $\boldsymbol {x}$. We have the following linear map

\begin{align*} T:\ W& \longrightarrow W,\\ g& \mapsto \mathcal{D}_k(1-|\boldsymbol{x}|^{2})g. \end{align*}

Suppose that $T g=0$, which implies that $\mathcal {D}_k(1-|\boldsymbol {x}|^{2})g=0$. However, $(1-|\boldsymbol {x}|^{2})g=0$ when $\boldsymbol {x}\in \mathbb {S}^{m-1}$, then according to the Poisson integral formula, $(1-|\boldsymbol {x}|^{2})g=0$ for all $\boldsymbol {x}\in \mathbb {R}^{m}$, i.e., $g=0$ for all $\boldsymbol {x}\in \mathbb {B}^{m}$. This implies that the linear map $T$ is injective. Since $W$ is finite-dimensional, then the linear map $T$ is also surjective. Therefore, we just proved that given an $f\in \mathcal {P}_l(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, which implies $\mathcal {D}_k f\in W$, there exists a function $g\in W$, such that $\mathcal {D}_k(1-|\boldsymbol {x}|^{2})g=-\mathcal {D}_k f$ as desired.

Proof. Suppose that there exists a function $f(\boldsymbol {x},\boldsymbol {u})\in \mathcal {P}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ with $\deg _{\boldsymbol {x}} f= l$ such that $\mathcal {D}_k|\boldsymbol {x}|^{2}f= 0$ in $\mathbb {B}^{m}\times \mathbb {B}^{m}$. One can see that $|\boldsymbol {x}|^{2}f=f$ when $\boldsymbol {x}\in \mathbb {S}^{m-1}$. According to the uniqueness for the Dirichlet problem given in [Reference Ding, Nguyen and Ryan11, Section 4], we must have $P[f]=|\boldsymbol {x}|^{2}f$ in $\mathbb {B}^{m}\times \mathbb {B}^{m}$. However, from the previous proposition, we notice that $\deg _{\boldsymbol {x}}P[f]= l$ but $\deg _{\boldsymbol {x}}|\boldsymbol {x}|^{2}f=l+2$, which is a contradiction. This completes our proof.

Proof. Given an $f\in \mathcal {P}_l(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, with Proposition 3.3, we know that there exists $g\in \mathcal {P}(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ with $\deg _{\boldsymbol {x}}g\leq l-2$, such that

\begin{align*} f=P[f]+(1-|\boldsymbol{x}|^{2})g, \text{ for all }(\boldsymbol{x},\boldsymbol{u})\in\mathbb{R}^{m}\times\mathbb{B}^{m}. \end{align*}

Now, we take the $l$-homogeneous terms with respect to $\boldsymbol {x}$ on the right-hand side above to obtain $f=f_l-|\boldsymbol {x}|^{2}g_{l-2}$, where $f_l$ is the $l$-homogeneous term for $P[f]$ and $g_{l-2}$ is the $(l-2)$-homogeneous term of $g$. Further, we know that $\mathcal {D}_k f_l=0$ by Proposition 3.2. This completes the proof for the existence of the decomposition. To prove the uniqueness for this decomposition, we assume that there exist $f_l,f'_l\in \ker \mathcal {D}_k$ and $g_{l-2},g'_{l-2}\in \mathcal {P}_{l-2}(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, such that $f_l-|\boldsymbol {x}|^{2}g_{l-2}=f'_l-|\boldsymbol {x}|^{2}g'_{l-2}$, i.e., $f_l-f'_l=|\boldsymbol {x}|^{2}g_{l-2}-|\boldsymbol {x}|^{2}g'_{l-2}\in \ker \mathcal {D}_k$. Then Proposition 3.4 tells that $f_l=f'_l$ and $g_{l-2}=g'_{l-2}$, which completes the proof.

Proof. Here, we only need to show that $\bigoplus _{l=0}^{\infty }\mathcal {B}_l(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$ is dense in $L^{2}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$. Let $f\in L^{2}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$ and $f=\sum _{j=1}^{d_k}f_j(\boldsymbol {x})\varphi _j(\boldsymbol {u})$ as explained above. The key to prove this proposition is to observe that $f\in L^{2}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$ is equivalent to $f_j\in L^{2}(\mathbb {S}^{m-1})$ for all $j=1,\cdots,d_k$. Indeed,

\begin{align*} \|f\|^{2}_{L^{2}(\mathbb{S}^{m-1}\times\mathbb{S}^{m-1},\mathcal{H}_k(\mathbb{S}^{m-1}))}:& =\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\Bigg\vert\sum_{j=1}^{d_k}f_j(\boldsymbol{x})\varphi_j(\boldsymbol{u})\Bigg\vert^{2}dS(\boldsymbol{u})dS(\boldsymbol{x})\\ & =\sum_{j=1}^{d_k}\int_{\mathbb{S}^{m-1}}|f_j(\boldsymbol{x})|^{2}dS(\boldsymbol{x}), \end{align*}

due to $\{\varphi _j\}_{j=1}^{d_k}$ is an orthonormal basis for $\mathcal {H}_k(\mathbb {S}^{m-1})$. As in classical harmonic analysis, there exists a sequence of polynomials $\{p_{l,j}\}_{l=1}^{\infty }$ such that $\|p_{l,j}-f_j\|_{L^{2}(\mathbb {S}^{m-1})}\rightarrow 0$ as $l\rightarrow \infty$ for all $j$. This is equivalent to

(3.4)\begin{equation} \displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\bigg\vert\displaystyle\sum_{j=1}^{d_k}p_{l,j}(\boldsymbol{x})\varphi_j(\boldsymbol{u})-\sum_{j=1}^{d_k}f_j(\boldsymbol{x})\varphi_j(\boldsymbol{u})\bigg\vert^{2}dS(\boldsymbol{u})dS(\boldsymbol{x})\rightarrow 0 \end{equation}

as $l\rightarrow \infty$. Further, Theorem 3.7 tells us that for each $l$, $\sum _{j=1}^{d_k}p_{l,j}(\boldsymbol {x})\varphi _j(\boldsymbol {u})\in \mathcal {P}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$ with $\deg _{\boldsymbol {x}}\leq l$, so that there exists a sequence $\{q^{l}_s\}_{s=0}^{\infty }$, where $q^{l}_s\in \mathcal {B}_s(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$ such that $\sum _{j=1}^{d_k}p_{l,j}(\boldsymbol {x})\varphi _j(\boldsymbol {u})=\sum _{s=0}^{[\frac {l}{2}]}q^{l}_s(\boldsymbol {x},\boldsymbol {u}).$ Combining this with (3.4), we have $\|\sum _{s=0}^{[\frac {l}{2}]}q^{l}_s-f\|_{L^{2}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))}\rightarrow 0$ as $l\rightarrow \infty$, which completes the proof.

Proof. Here we only provide an outline proof. To prove $(1)$, we assume $f\in \mathcal {B}_l(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$ is real valued. Then

\begin{align*} 0& =\mathrm{Im}f(\boldsymbol{x},\boldsymbol{v})=\mathrm{Im}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\overline{J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})}f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{\zeta})dS(\boldsymbol{u})\\ & ={-}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\mathrm{Im}J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{\zeta})dS(\boldsymbol{u}). \end{align*}

Now, we let $f=\mathrm {Im}J_{l,k}(\boldsymbol {\zeta },\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})$ to immediately obtain $\mathrm {Im}J_{l,k}=0$.

To prove $(2)$, let $\{\phi _1,\cdots,\phi _{d_{l,k}}\}$ be an orthonormal basis of $\mathcal {B}_l(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$, where $d_{l,k}=\dim \mathcal {B}_l(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {S}^{m-1}))$. Then, one has

(3.6)\begin{equation} J_{l,k}J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})=\displaystyle\sum_{j=1}^{d_{l,k}}\langle \phi_j\ |\ J_{l,k}({\cdot},\boldsymbol{x},\cdot,\boldsymbol{v}) \rangle\phi_j(\boldsymbol{\zeta},\boldsymbol{u})=\sum_{j=1}^{d_{l,k}}\overline{\phi_j(\boldsymbol{x},\boldsymbol{v})}\phi_j(\boldsymbol{\zeta},\boldsymbol{u}). \end{equation}

Since $J_{l,k}$ is real valued, the equation above is unchanged after taking complex conjugation, which implies $(2)$. One can obtain $(3)$ immediately from the definition of the reproducing kernel and the fact that the surface area element $dS$ is rotationally invariant.

In $(3)$, if we let $\boldsymbol {x}=T(\boldsymbol {\zeta })$ and $\boldsymbol {v}=T(\boldsymbol {u})$ in $J_{l,k}(T(\boldsymbol {\zeta }),\boldsymbol {x},T(\boldsymbol {u}),\boldsymbol {v})$, then we obtain $J_{l,k}(T(\boldsymbol {\zeta }),T(\boldsymbol {\zeta }),T(\boldsymbol {u}),T(\boldsymbol {u}))=J_{l,k}(\boldsymbol {\zeta },\boldsymbol {\zeta },\boldsymbol {u},\boldsymbol {u})$, i.e., $J_{l,k}(\boldsymbol {\zeta },\boldsymbol {\zeta },\boldsymbol {u},\boldsymbol {u})$ is invariant under rotation. Letting $\boldsymbol {x}=\boldsymbol {\zeta }$ and $\boldsymbol {v}=\boldsymbol {u}$ in 3.6 immediately gives us $(4)$. Taking absolute value on both sides of 3.6 and applying Cauchy–Schwarz inequality can easily provide us $(5)$.

Proof. Recall that in Proposition 3.1, we have shown that the Poisson kernel has the following series expansion

(3.7)\begin{equation} P(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})=\displaystyle\sum_{\boldsymbol{|\gamma|=0}}^{\infty}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}} \end{equation}

which converges absolutely when $|\boldsymbol {x}|<\sqrt {2}-1, \boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$, and $g_{\boldsymbol {\gamma }}(\boldsymbol {\zeta },\boldsymbol {u},\boldsymbol {v})\in \mathcal {H}_k(\boldsymbol {v},\mathbb {C})$. We claim that it also converges uniformly when $|\boldsymbol {x}|\leq \sqrt {2}-1-\epsilon,\boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$, where $\epsilon >0$ can be arbitrarily small. This can be observed from

\begin{align*} \bigg\vert\sum_{\boldsymbol{|\gamma|=0}}^{\infty}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}\bigg\vert\leq\sum_{\boldsymbol{|\gamma|=0}}^{\infty}\bigg\vert g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}\bigg\vert\leq\sum_{\boldsymbol{|\gamma|=0}}^{\infty}\bigg\vert g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\bigg\vert |\boldsymbol{x}|^{|\boldsymbol{\gamma}|}<{+}\infty, \end{align*}

for all $|\boldsymbol {x}|\leq \sqrt {2}-1-\epsilon, \boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$. The last step comes from the fact that $\sum _{\boldsymbol {|\gamma |=0}}^{\infty }g_{\boldsymbol {\gamma }}(\boldsymbol {\zeta },\boldsymbol {u},\boldsymbol {v})\boldsymbol {x}^{\boldsymbol {\gamma }}$ converges absolutely when $|\boldsymbol {x}|=\sqrt {2}-1-\epsilon$, particularly with $\boldsymbol {x}=(x_1,0,\cdots,0)$ and $\boldsymbol {\gamma }=(|\boldsymbol {\gamma }|,0,\cdots,0)$. Therefore, we have the result that the series expansion for the Poisson kernel given in (3.7) converges absolutely and uniformly when $|\boldsymbol {x}|\leq \sqrt {2}-1-\epsilon, \boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$ with an arbitrarily small $\epsilon >0$. Hence, given a function $f\in \mathcal {B}_l(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, with the Poisson integral formula, we have

\begin{align*} f(\boldsymbol{x},\boldsymbol{v})& =\displaystyle\frac{c_{m,k}}{2}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\displaystyle\sum_{\boldsymbol{|\gamma|=0}}^{\infty}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\\ & =\frac{c_{m,k}}{2}\sum_{\boldsymbol{|\gamma|=0}}^{\infty}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\boldsymbol{x}^{\boldsymbol{\gamma}} =:\sum_{\boldsymbol{|\gamma|=0}}^{\infty}f_{\boldsymbol{\gamma}}(\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}, \end{align*}

where $f_{\boldsymbol {\gamma }}(\boldsymbol {v})\in \mathcal {H}_k(\boldsymbol {v},\mathbb {C})$ and $|\boldsymbol {x}|\leq \sqrt {2}-1-\epsilon,\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$. We remind the reader that the uniform and absolute convergence of the series expansion of the Poisson kernel when $|\boldsymbol {x}|\leq \sqrt {2}-1-\epsilon, \boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$, $f$ is a polynomial and Fubini's Theorem allow us to interchange the integral and infinite sum in the last second step above. However, since $f\in \mathcal {B}_l(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, we take the degree $l$ (with respect to $\boldsymbol {x}$) terms on the right- hand side above, we have $f=\sum _{\boldsymbol {|\gamma |=l}}f_{\boldsymbol {\gamma }}(\boldsymbol {v})\boldsymbol {x}^{\boldsymbol {\gamma }}$. This implies that $f_{\boldsymbol {\gamma }}(\boldsymbol {v})=0$ for all $|\boldsymbol {\gamma }|\neq l$. In other words, we indeed have

\begin{align*} f(\boldsymbol{x},\boldsymbol{v})=\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\sum_{\boldsymbol{|\gamma|=l}}g_{\boldsymbol{\gamma}}(\boldsymbol{\zeta},\boldsymbol{u},\boldsymbol{v})\boldsymbol{x}^{\boldsymbol{\gamma}}f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta}), \end{align*}

for all $f\in \mathcal {B}_l(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$. Hence, we have the expression for our reproducing kernel as $J_{l,k}(\boldsymbol {\zeta },\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})=\frac {c_{m,k}}{2}\sum _{\boldsymbol {|\gamma |=l}}g_{\boldsymbol {\gamma }}(\boldsymbol {\zeta },\boldsymbol {u},\boldsymbol {v})\boldsymbol {x}^{\boldsymbol {\gamma }},$ which completes the proof as well.

Proof. Firstly, it is easy to observe that $\langle f\ |\ g\rangle =0$ when $k\neq l$, this is because of the already known orthogonality between functions in $\mathcal {H}_k(\mathbb {C})$ and functions in $\mathcal {H}_l(\mathbb {C})$ with respect to the $L^{2}$ inner product on $\mathbb {S}^{m-1}$. Hence, we assume $k=l$ in the rest of the proof. With the help of the previous proposition, we have

\begin{align*} \langle f\ |\ g\rangle& =\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\overline{f(\boldsymbol{x},\boldsymbol{v})}g(\boldsymbol{x},\boldsymbol{v})dS(\boldsymbol{v})dS(\boldsymbol{x})\\ & =\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\overline{\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\overline{J_{s,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})}f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})}\\ & \quad\cdot\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\overline{J_{t,l}(\boldsymbol{\eta},,\boldsymbol{x},\boldsymbol{\omega},\boldsymbol{v})}g(\boldsymbol{\eta},\boldsymbol{\omega})dS(\boldsymbol{\omega})dS(\boldsymbol{\eta})dS(\boldsymbol{v})dS(\boldsymbol{x})\\ & =\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}J_{s,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})\overline{J_{t,l}(\boldsymbol{\eta},\boldsymbol{x},\boldsymbol{\omega},\boldsymbol{v})}dS(\boldsymbol{v})dS(\boldsymbol{x})\\ & \quad\times\overline{f(\boldsymbol{\zeta},\boldsymbol{u})}g(\boldsymbol{\eta},\boldsymbol{\omega})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})dS(\boldsymbol{\omega})dS(\boldsymbol{\eta}). \end{align*}

In the last step above we interchange the order of integrations because all terms in the integrand are polynomials. Now, we take a look at

\begin{align*} & \displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}J_{s,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})\overline{J_{t,l}(\boldsymbol{\eta},\boldsymbol{x},\boldsymbol{\omega},\boldsymbol{v})}dS(\boldsymbol{v})dS(\boldsymbol{x})\\ & \quad=\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\overline{J_{t,l}(\boldsymbol{\eta},\boldsymbol{x},\boldsymbol{\omega},\boldsymbol{v})}J_{s,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})S(\boldsymbol{v})dS(\boldsymbol{x}), \end{align*}

where the last step above uses $(2)$ in Proposition 3.10. We notice that $J_{s,k}(\boldsymbol {\zeta },\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})\in \mathcal {B}_s(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ and $J_{t,l}(\boldsymbol {\eta },\boldsymbol {x},\boldsymbol {\omega },\boldsymbol {v})$ is the reproducing kernel in $\mathcal {B}_t(\mathbb {R}^{m}\times \mathbb {B}^{m},\mathcal {H}_l(\mathbb {C}))$, with the orthogonality property given in Theorem 3.13, we have

\begin{align*} \int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\overline{J_{t,l}(\boldsymbol{\eta},\boldsymbol{x},\boldsymbol{\omega},\boldsymbol{v})}J_{s,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})S(\boldsymbol{v})dS(\boldsymbol{x})=0,\quad \text{when}\ t\neq s. \end{align*}

Therefore, we have $\langle f\ |\ g\rangle =0$, when $t\neq s$ as desired.

Proof. Firstly, we assume that $f\in C^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))\cap C(\overline {\mathbb {B}^{m}}\times \overline {\mathbb {B}^{m}},\mathcal {H}_k(\mathbb {C}))$ and $\mathcal {D}_k f=0$ in $\mathbb {B}^{m}\times \mathbb {B}^{m}$. Then, with the Poisson integral formula and Proposition 3.11, for $\boldsymbol {x}\in B(0,\epsilon ),\ \boldsymbol {v}\in \mathbb {B}^{m}$, with $0<\epsilon <1$, we have

\begin{align*} f(\boldsymbol{x},\boldsymbol{v})& =\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\\ & =\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\sum_{l=0}^{\infty}J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\\ & =\frac{c_{m,k}}{2}\sum_{l=0}^{\infty}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta}) =\sum_{l=0}^{\infty}f_l(\boldsymbol{x},\boldsymbol{v}), \end{align*}

where $f_l(\boldsymbol {x},\boldsymbol {v})=\int _{\mathbb {S}^{m-1}}\int _{\mathbb {S}^{m-1}}J_{l,k}(\boldsymbol {\zeta },\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})f(\boldsymbol {\zeta },\boldsymbol {u})dS(\boldsymbol {u})dS(\boldsymbol {\zeta })$ and $\mathcal {D}_k f_l=0$ in $\mathbb {R}^{m}\times \mathbb {B}^{m}$. Notice that $(5)$ of Proposition 3.10 tells us that

\begin{align*} |J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})|& \leq\dim\mathcal{B}_l(\mathbb{R}^{m}\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C})) |\boldsymbol{x}|^{l}|\boldsymbol{v}|^{k}=\dim\mathcal{H}_k\cdot\dim\mathcal{H}_l|\boldsymbol{x}|^{l}|\boldsymbol{v}|^{k}\\ & \leq C_m(kl)^{m-2}|\boldsymbol{x}|^{l}|\boldsymbol{v}|^{k}, \end{align*}

where $C_m$ is a constant only depending on $m$. Hence, one can see that

\begin{align*} \sum_{l=0}^{\infty}|f_l(\boldsymbol{x},\boldsymbol{v})|\leq C_mk^{m-2}|\boldsymbol{v}|^{k}\sum_{l=0}^{\infty}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}l^{m-2}|\boldsymbol{x}|^{l}|f(\boldsymbol{\zeta},\boldsymbol{u})|dS(\boldsymbol{u})dS(\boldsymbol{\zeta})<{+}\infty, \end{align*}

when $(\boldsymbol {x},\boldsymbol {v})\in \overline {B(0,\epsilon )}\times \overline {\mathbb {B}^{m}}$. This implies that $\sum _{l=0}^{\infty }f_l$ converges absolutely and uniformly to $f$ in $\overline {B(0,\epsilon )}\times \overline {\mathbb {B}^{m}}$. Applying a dilation and a translation to the argument above can immediately give us the result on $B(\boldsymbol {a},r)\times \mathbb {B}^{m}$ as desired.

Proof. Let $0< r_1< d(\boldsymbol {a},\partial \Omega )$ and $0< r_2< d(\boldsymbol {v},\mathbb {S}^{m-1})$. We firstly apply the volume version of the mean-value property [Reference Ding, Nguyen and Ryan11, Proposition 5.3] with respect to $\boldsymbol {x}$ on $B(\boldsymbol {x},r_1)$ to obtain

(4.1)\begin{equation} |f(\boldsymbol{a},\boldsymbol{v})|^{p}\leq\left(\displaystyle\frac{m+2k-2}{m-2}\right)^{p}V(B(\boldsymbol{a},r_1))^{{-}1}\displaystyle\int_{B(\boldsymbol{a},r_1)} |f(\boldsymbol{x},\boldsymbol{\omega})|^{p}d\boldsymbol{x}, \end{equation}

where $\boldsymbol {\omega }=\frac {(\boldsymbol {x}-\boldsymbol {a})\boldsymbol {v}(\boldsymbol {x}-\boldsymbol {a})}{|\boldsymbol {x}-\boldsymbol {a}|^{2}}$ and Jensen's inequality is applied above. Further, we notice that $\boldsymbol {\omega }$ is obtained from $\boldsymbol {v}$ by a rotation, which implies that $f(\boldsymbol {x},\boldsymbol {\omega })\in \mathcal {H}_k(\boldsymbol {\omega },\mathbb {C})$. This motivates us to apply the volume version of the mean-value property to $f(\boldsymbol {x},\boldsymbol {\omega })$ with respect to $\boldsymbol {\omega }$ on $B(\boldsymbol {\omega },r_2)$ to have

(4.2)\begin{equation} |f(\boldsymbol{x},\boldsymbol{\omega})|^{p}\leq V(B(\boldsymbol{\omega},r_2))^{{-}1}\displaystyle\int_{B(\boldsymbol{\omega},r_2)}|f(\boldsymbol{x},\boldsymbol{u})|^{p}d\boldsymbol{u}. \end{equation}

Plugging (4.2) into (4.1), we obtain

\begin{align*} |f(\boldsymbol{a},\boldsymbol{v})|^{p}& \leq\left(\displaystyle\frac{m+2k-2}{m-2}\right)^{p}V(B(\boldsymbol{a},r_1))^{{-}1}\displaystyle\int_{B(\boldsymbol{a},r_1)}V(B(\boldsymbol{\omega},r_2))^{{-}1}\\ & \quad \times\int_{B(\boldsymbol{\omega},r_2)}|f(\boldsymbol{x},\boldsymbol{u})|^{p}d\boldsymbol{u} d\boldsymbol{x}\\ & \leq \left(\displaystyle\frac{m+2k-2}{m-2}\right)^{p}r_1^{{-}m}r_2^{{-}m}V(\mathbb{B}^{m})^{{-}2}\|f\|_{{b^{p}(\Omega\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C}))}}^{p}. \end{align*}

Now, we let $r_1\rightarrow d(\boldsymbol {a},\partial \Omega )$, $r_2\rightarrow d(\boldsymbol {v},\mathbb {S}^{m-1})$ and take $p$th root on both sides above to complete the proof.

Proof. Let $f\in L^{2}(\Omega \times \mathbb {B}^{m},\mathbb {C})$, $\boldsymbol {x}\in \Omega$ and $\boldsymbol {v}\in \mathbb {B}^{m}$. Then

\begin{align*} B_{k,\Omega\times\mathbb{B}^{m}}[f](\boldsymbol{x},\boldsymbol{v})=\langle R_{k,\Omega\times\mathbb{B}^{m}}({\cdot},\boldsymbol{x},\cdot,\boldsymbol{v}),B_{k,\Omega\times\mathbb{B}^{m}}[f]\rangle_{b^{2}}. \end{align*}

Since $B_{k,\Omega \times \mathbb {B}^{m}}$ is an orthogonal projection, it is also self-adjoint and $R_{k,\Omega \times \mathbb {B}^{m}}(\boldsymbol {x},\cdot,\boldsymbol {v},\cdot )\in b^{2}(\Omega \times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ tells us that $B_{k,\Omega \times \mathbb {B}^{m}}[R_{k,\Omega \times \mathbb {B}^{m}}(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v})]=R_{k,\Omega \times \mathbb {B}^{m}}(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v})$. Therefore, the equation above is equal to

\begin{align*} \langle R_{k,\Omega\times\mathbb{B}^{m}}({\cdot},\boldsymbol{x},\cdot,\boldsymbol{v}),f\rangle_{b^{2}}=\int_{\Omega}\int_{\mathbb{B}^{m}}\overline{R_{k,\Omega\times\mathbb{B}^{m}}(\boldsymbol{y},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})}f(\boldsymbol{y},\boldsymbol{u})d\boldsymbol{u} d\boldsymbol{y}, \end{align*}

which completes the proof since $R_{k,\Omega \times \mathbb {B}^{m}}$ is real valued.

Proof. We firstly notice that for a function $f\in L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, $f_l(\boldsymbol {x},\boldsymbol {v}):=f(\tfrac {l-1}{l}\boldsymbol {x},\boldsymbol {v})\in L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ for $l=1,2,\cdots$. Further, we can see that $f_l$ converges to $f$ in $L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ when $l\rightarrow \infty$. This can be observed from the case $f\in C(\overline {\mathbb {B}^{m}}\times \overline {\mathbb {B}^{m}},\mathcal {H}_k(\mathbb {C}))$ and the fact that $C(\overline {\mathbb {B}^{m}}\times \overline {\mathbb {B}^{m}},\mathcal {H}_k(\mathbb {C}))$ is dense in $L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$. More specifically, for $f\in L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, there exists a sequence $\{g_s\}_{s=1}^{\infty }\in C(\overline {\mathbb {B}^{m}}\times \overline {\mathbb {B}^{m}},\mathcal {H}_k(\mathbb {C}))$ such that for arbitrary $\epsilon >0$, there exists $N>0$, such that $\|f-g_s\|_{L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))}<\epsilon$ when $s>N$. Let $g_{s,l}(\boldsymbol {x},\boldsymbol {v})=g_s(\tfrac {l-1}{l}\boldsymbol {x},\boldsymbol {v})$, then we can easily check that $\|f_l-g_{s,l}\|_{L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))}<\frac {l}{l-1}\epsilon$ when $s>N$. Therefore, we have

(4.4)\begin{align} \|f-f_l\|_{L^{2}(\mathbb{B}^{m}\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C}))}& \leq \|f-g_s\|_{L^{2}(\mathbb{B}^{m}\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C}))}+\|g_s-g_{s,l}\|_{L^{2}(\mathbb{B}^{m}\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C}))}\nonumber\\ & \quad+\|g_{s,l}-f_l\|_{L^{2}(\mathbb{B}^{m}\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C}))}\leq \epsilon+\epsilon+\displaystyle\frac{l}{l-1}\epsilon=\frac{3l-1}{l}\epsilon, \end{align}

when $l>N'$ and $s>N$, where $N'$ is sufficiently large so that $\|g_s-g_{s,l}\|_{L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))}<\epsilon$, which comes from the continuity of $g_s$.

Since $\ker \mathcal {D}_k$ is invariant with respect to dilations, any function $f\in b^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ can be approximated by a sequence of functions $\{f_l\}_{l=1}^{\infty }$ satisfying $\mathcal {D}_k f_l=0$ in $\overline {\mathbb {B}^{m}}\times \overline {\mathbb {B}^{m}}$. Further, notice that $f_l\in C^{2}(\tfrac {l}{l-1}\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, then Proposition 3.16 tells us that each $f_l$ can be approximated by a sequence of homogeneous polynomial null solutions to $\mathcal {D}_k$, which converges absolutely and uniformly in $\overline {\mathbb {B}^{m}}\times \overline {\mathbb {B}^{m}}$. Hence, with a similar argument as applied in (4.4), there exists a sequence of homogeneous polynomials null solutions to $\mathcal {D}_k$ which converges to $f$ in $L^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ as desired.

The claim that the decomposition is an orthogonal decomposition can be observed by changing to polar coordinates and applying Theorem 3.13.

Proof. For $\boldsymbol {x},\boldsymbol {y},\boldsymbol {u},\boldsymbol {v}\in \mathbb {B}^{m}\backslash {\{0\}}$, we have

\begin{align*} |J_{l,k}(\boldsymbol{y},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})|& =|\boldsymbol{x}\boldsymbol{y}|^{l}|\boldsymbol{u}\boldsymbol{v}|^{k}|J_{l,k}\left(\displaystyle\frac{\boldsymbol{y}}{|\boldsymbol{y}|},\frac{\boldsymbol{x}}{|\boldsymbol{x}|},\frac{\boldsymbol{u}}{|\boldsymbol{u}|},\frac{\boldsymbol{v}}{|\boldsymbol{v}|}\right)|\leq|\boldsymbol{x}\boldsymbol{y}|^{l}|\boldsymbol{u}\boldsymbol{v}|^{k}|\dim\mathcal{H}_k\dim\mathcal{H}_l\\ & \leq C_m(lk)^{m-2}|\boldsymbol{x}|^{l}|\boldsymbol{y}|^{l}|\boldsymbol{u}|^{k}|\boldsymbol{v}|^{k}. \end{align*}

This implies that $\sum _{l=0}^{\infty }(m+2l)(m+2k)J_{l,k}(\boldsymbol {y},\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})$ converges absolutely and uniformly in $\mathbb {B}^{m}\times K\times \mathbb {B}^{m}\times \mathbb {B}^{m}$, where $K$ is a compact subset of $\mathbb {B}^{m}$. Now, if we denote $F(\boldsymbol {y},\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})=\sum _{l=0}^{\infty }(m+2l)(m+2k)J_{l,k}(\boldsymbol {y},\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})$, then $F(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v})\in \ker \mathcal {D}_k$ is bounded in $\mathbb {B}^{m}\times \mathbb {B}^{m}$ for fixed $\boldsymbol {x},\boldsymbol {v}\in \mathbb {B}^{m}$ and hence $F(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v})\in b^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$. Now, if $f$ is a polynomial solution to $\mathcal {D}_k$, then with the calculation in (4.3) and the orthogonality given in Proposition 3.13, we can easily obtain that $\langle F(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v}),f \rangle _{b^{2}}=f$ in $\mathbb {B}^{m}\times \mathbb {B}^{m}$. Further, the previous proposition tells us that a function in $b^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C})$ can be approximated by polynomial solutions to $\mathcal {D}_k$. This immediately gives us that for any $f\in b^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C})$, we have $\langle F(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v}),f \rangle _{b^{2}}=f$ in $\mathbb {B}^{m}\times \mathbb {B}^{m}$. This implies that $F(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v})$ is the reproducing kernel of $b^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C})$, which completes the proof.

Proof. This result can be observed from the previous proposition and Proposition 4.4. The orthogonality given in Corollary 3.15 and the decomposition given in Theorem 3.7 explain the disappearance of the terms in the series with $l>s$.

Proof. For $0\leq l\leq s$ and $\boldsymbol {x},\boldsymbol {v}\in \mathbb {B}^{m}$, one has

\begin{align*} & \displaystyle\int_{\mathbb{B}^{m}}\int_{\mathbb{B}^{m}}J_{l,k}(\boldsymbol{y},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})f(\boldsymbol{y},\boldsymbol{u})d\boldsymbol{u} d\boldsymbol{y}\\ & \quad=\int_0^{1}\int_0^{1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}r_1^{m-1}r_2^{m-1}J_{l,k}(r_1\boldsymbol{\zeta},\boldsymbol{x},r_2\boldsymbol{\eta},\boldsymbol{v})f(r_1\boldsymbol{\zeta},r_2\boldsymbol{\eta})dS(\boldsymbol{\eta})dS(\boldsymbol{\zeta})dr_1dr_2\\ & \quad=\int_0^{1}\int_0^{1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}r_1^{m+l+s-1}r_2^{m+2k-1}J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{\eta},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{\eta})dS(\boldsymbol{\eta})dS(\boldsymbol{\zeta})dr_1dr_2\\ & \quad=(m+l+s)^{{-}1}(m+2k)^{{-}1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}J_{l,k}(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{\eta},\boldsymbol{v})f(\boldsymbol{\zeta},\boldsymbol{\eta})dS(\boldsymbol{\eta})dS(\boldsymbol{\zeta})\\ & \quad=(m+l+s)^{{-}1}(m+2k)^{{-}1}f_l(\boldsymbol{x},\boldsymbol{v}), \end{align*}

where the last equation is obtained by applying the Poisson integral formula (3.2), the expression of the Poisson kernel given in Proposition 3.11, the decomposition of $f(\boldsymbol {\zeta },\boldsymbol {\eta })$ given in Theorem 3.7 and the orthogonality given in Theorem 3.13 . Plugging the equation above into the previous proposition gives us the desired result.

Proof. Indeed, the difference between the isometrical isomorphism given in (5.2) and the isomorphism above is the extra condition $\mathcal {H}_k(\mathbb {C})$ added to the second variable $\boldsymbol {u}$. Therefore, to prove the lemma above, we only need to show that for the variable $\boldsymbol {u}$,

(5.3)\begin{align} M(\mathbb{S}^{m-1},\mathcal{H}_k) & \longrightarrow \mathcal{H}_k(\mathbb{C})^{*},\nonumber\\ \mu & \mapsto \Lambda_{\mu},\quad (\Lambda_{\mu}(f)=\displaystyle\int_{\mathbb{S}^{m-1}}fd\mu,\quad \text{for}\ f\in \mathcal{H}_k(\mathbb{C})), \end{align}

is an isometrical isomorphism, where $M(\mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ stands for the space of finite complex Borel measures given by $d\mu =hdS$, and $h$ is a $k$-homogeneous harmonic polynomial. To prove this, it suffices to show the linear map (5.3) is into, because $\dim M(\mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))=\dim \mathcal {H}_k(\mathbb {C})^{*}=\dim \mathcal {H}_k(\mathbb {C})$ is finite. This is equivalent to prove that if $\Lambda _{\mu }(f)=0$ for all $f\in \mathcal {H}_k(\mathbb {C})$, then $\mu =0$. Assume that $\mu =hdS$, where $h\in \mathcal {H}_k(\mathbb {C})$, then

\begin{align*} \Lambda_{\mu}(f)=\int_{\mathbb{S}^{m-1}}fd\mu=\int_{\mathbb{S}^{m-1}}fhdS. \end{align*}

Since $f\in \mathcal {H}_k(\mathbb {C})$ is arbitrary, then $h=0$, which means $\mu =0$. This completes the proof for (5.3). Hence, our lemma is true.

Proof. To prove $(1)$, firstly, we have

\begin{align*} & f_{r_1,r_2}(\boldsymbol{\xi},\boldsymbol{\eta})\\ & \quad=f(r_1\boldsymbol{\xi},r_2\boldsymbol{\eta})=P[\mu](r_1\boldsymbol{\xi},r_2\boldsymbol{\eta})\\ & \quad =\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}Z_k\left(\frac{(r_1\boldsymbol{\xi}-\boldsymbol{\zeta})\boldsymbol{u}(r_1\boldsymbol{\xi}-\boldsymbol{\zeta})}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{2}},r_2\boldsymbol{\eta}\right)h(\boldsymbol{u})dS(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\\ & \quad =\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}Z_k\Bigg(\boldsymbol{u},\phi(\boldsymbol{\eta})\Bigg)h(\boldsymbol{u})dS(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\\ & \quad =\omega_m^{{-}1}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}h(\phi(\boldsymbol{\eta}))d\mu_1(\boldsymbol{\zeta}), \end{align*}

where $\phi (\boldsymbol {\eta })=\frac {(r_1\boldsymbol {\xi }-\boldsymbol {\zeta })r_2\boldsymbol {\eta }(r_1\boldsymbol {\xi }-\boldsymbol {\zeta })}{|r_1\boldsymbol {\xi }-\boldsymbol {\zeta }|^{2}},$ and $\omega _m$ is the area of $\mathbb {S}^{m-1}$. The last equation above is obtained from the facts that $Z_k(\boldsymbol {a}\boldsymbol {u} \boldsymbol {a},\boldsymbol {v})=Z_k(\boldsymbol {u},\boldsymbol {a}\boldsymbol {v} \boldsymbol {a})$ with $\boldsymbol {a}\in \mathbb {R}^{m}$ and $Z_k$ is the reproducing kernel of $k$-homogeneous harmonic polynomials and [Reference Ding, Nguyen and Ryan11, Lemma 3.6].

Now, we have

\begin{align*} \|f_{r_1,r_2}\|_{1}& =\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\bigg\vert\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}h(\phi(\boldsymbol{\eta}))d\mu_1(\boldsymbol{\zeta})\bigg\vert dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})\\ & \leq\frac{c_{m,k}}{2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}\int_{\mathbb{S}^{m-1}}|h(\phi(\boldsymbol{\eta}))|dS(\boldsymbol{\eta})dS(\boldsymbol{\xi})d|\mu_1|(\boldsymbol{\zeta}). \end{align*}

Further, noticing that $\phi (\boldsymbol {\eta })$ is a reflection of $r_2\boldsymbol {\eta }$, hence, the integral above becomes

\begin{align*} & \displaystyle\frac{c_{m,k}}{2}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}\int_{\mathbb{S}^{m-1}}|h(r_2\boldsymbol{\eta})|dS(\boldsymbol{\eta})dS(\boldsymbol{\xi})d|\mu_1|(\boldsymbol{\zeta})\\ & \quad=\frac{m+2k-2}{m-2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|h(r_2\boldsymbol{\eta})|dS(\boldsymbol{\eta})d|\mu_1|(\boldsymbol{\zeta})\\ & \quad\leq \frac{m+2k-2}{m-2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|h(\boldsymbol{\eta})|dS(\boldsymbol{\eta})d|\mu_1|(\boldsymbol{\zeta})\\ & \quad=\frac{m+2k-2}{m-2}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}d|\mu_2|(\boldsymbol{\eta})d|\mu_1|(\boldsymbol{\zeta})\leq\frac{m+2k-2}{m-2}\|\mu\|, \end{align*}

where the following facts are used in the last two steps above

\begin{align*} \omega_m^{{-}1}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}dS(\boldsymbol{\xi})=\omega_m^{{-}1}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1|^{2}}{|\boldsymbol{\xi}-r_1\boldsymbol{\zeta}|^{m}}dS(\boldsymbol{\xi})=1, \end{align*}

and $\int _{\mathbb {S}^{m-1}}|h(r_2\boldsymbol {\eta })|dS(\boldsymbol {\eta })$ is increasing with respect to $r_2$, since $h$ is harmonic, see [Reference Axler, Bourdon and Ramey1, Corollary 6.6].

For $(2)$, we firstly assume that $1\leq p<\infty$, with similar argument applied to $(1)$, we obtain

\begin{align*} |f_{r_1,r_2}(\boldsymbol{\xi},\boldsymbol{\eta})| \leq\displaystyle\frac{c_{m,k}}{2}\displaystyle\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}|g(\boldsymbol{\zeta},\phi(\boldsymbol{\eta}))|dS(\boldsymbol{\zeta}). \end{align*}

Then, with the Jensen's integral inequality, we have

\begin{align*} & \left(\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|f_{r_1,r_2}(\boldsymbol{\xi},\boldsymbol{\eta})|^{p}dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})\right)^{1/p}\\ & \quad\leq\displaystyle\frac{c_{m,k}}{2}\left(\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}|g(\boldsymbol{\zeta},\phi(\boldsymbol{\eta}))|^{p}dS(\boldsymbol{\zeta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})\right)^{1/p}\\ & \quad=\frac{c_{m,k}}{2}\left(\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|r_1\boldsymbol{\xi}|^{2}}{|r_1\boldsymbol{\xi}-\boldsymbol{\zeta}|^{m}}|g(\boldsymbol{\zeta},\boldsymbol{\eta})|^{p}dS(\boldsymbol{\eta})dS(\boldsymbol{\zeta})dS(\boldsymbol{\xi})\right)^{1/p}\\ & \quad=\frac{m+2k-2}{m-2}\left(\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|g(\boldsymbol{\zeta},\boldsymbol{\eta})|^{p}dS(\boldsymbol{\eta})dS(\boldsymbol{\zeta})\right)^{1/p}=\frac{m+2k-2}{m-2}\|g\|_p. \end{align*}

Similar argument as above can be applied to $p=\infty$.

Proof. This can be obtained immediately from the previous theorem and [Reference Ding, Nguyen and Ryan11, Theorem 3.7] as follows

\begin{align*} \|f_{r_1,r_2}\|_p=\|P[g_{s_1,s_2}]_{\frac{r_1}{s_1},\frac{r_2}{s_2}}\|_p\leq\frac{m+2k-2}{m-2}\|g_{s_1,s_2}\|_p. \end{align*}

Proof. Suppose $1\leq p<\infty$ and a function $g\in L^{p}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$, let $f=P[g]$. For a fixed $\epsilon >0$, there exists a function $h\in C(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$, such that $\|g-h\|_p<\epsilon$. Let $f^{{\dagger} }=P[h]$, then $\|f_{r_1,r_2}-g\|_p\leq \|f_{r_1,r_2}-f^{{\dagger} }_{r_1,r_2}\|_p+\|f^{{\dagger} }_{r_1,r_2}-h\|_p+ \|h-g\|_p.$ According to Theorem 5.2, we observe that

\begin{align*} \|f_{r_1,r_2}-f^{{{\dagger}}}_{r_1,r_2}\|_p=\|P[g-h]_{r_1,r_2}\|_p\leq\frac{m+2k-2}{m-2}\|g-h\|_p\leq\frac{m+2k-2}{m-2}\epsilon. \end{align*}

Further, with [Reference Ding, Nguyen and Ryan11, Theorem 3.7] and [Reference Axler, Bourdon and Ramey1, Theorem 6.7], we obtain $\|f^{{\dagger} }_{r_1,r_2}-h\|_p\leq \epsilon$ when $r_1,r_2\rightarrow 0$. Therefore, we have $\|f_{r_1,r_2}-f^{{\dagger} }_{r_1,r_2}\|_p\leq (\frac {m+2k-2}{m-2}+2)\epsilon.$ Since $\epsilon$ is arbitrary, we complete the proof.

Proof. Recall that $M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))=C(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))^{*}$, let $f=P[\mu ]$ with $\mu =\mu _1\times \mu _2\in M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$. To prove $(1)$, we only need to show that for all $g\in C(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$,

\begin{align*} \lim_{r_1,r_2\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}f_{r_1,r_2}(\boldsymbol{\xi},\boldsymbol{\eta})g(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})= \int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}g(\boldsymbol{\zeta},\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})d\mu_2(\boldsymbol{u}). \end{align*}

Now, we look at the integral on the left side, with a similar argument as in $(1)$ in Theorem 5.2,

\begin{align*} & \lim_{r_1,r_2\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}f_{r_1,r_2}(\boldsymbol{\xi},\boldsymbol{\eta})g(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})\\ & \quad=\lim_{r_1,r_2\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P(\boldsymbol{\zeta},r_1\boldsymbol{\xi},\boldsymbol{u},r_2\boldsymbol{\eta})d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})g(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta}). \end{align*}

Recall that $P(\boldsymbol {\zeta },r_1\boldsymbol {\xi },\boldsymbol {u},r_2\boldsymbol {\eta })=\tfrac {c_{m,k}}{2}\frac {1-r_1^{2}}{|r_1\boldsymbol {\xi }-\boldsymbol {\zeta }|^{m}}Z_k(\tfrac {(r_1\boldsymbol {\xi }-\boldsymbol {\zeta })\boldsymbol {u}(r_1\boldsymbol {\xi }-\boldsymbol {\zeta })}{|r_1\boldsymbol {\xi }-\boldsymbol {\zeta }|^{2}},r_2\boldsymbol {\eta }).$ Since $Z_k(\boldsymbol {u},\boldsymbol {v})$ is a homogeneous polynomial, when $r_1,r_2\rightarrow 1$, the singularities only happen in $\frac {1-r_1^{2}}{|r_1\boldsymbol {\xi }-\boldsymbol {\zeta }|^{m}}$. Therefore, we can let $r_1=r_2=1$ in $Z_k,\ \boldsymbol {\omega }$ above. Further, since $\boldsymbol {\zeta },\boldsymbol {\eta }\in \mathbb {S}^{m-1}$, we have $\frac {1-r_1^{2}}{|r_1\boldsymbol {\xi }-\boldsymbol {\zeta }|^{m}}=\frac {1-r_1^{2}}{|\boldsymbol {\xi }-r_1\boldsymbol {\zeta }|^{m}}.$ Hence, the integral above becomes

\begin{align*} & \lim_{r_1,r_2\rightarrow 1}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P(r_1\boldsymbol{\zeta},\boldsymbol{\xi},\boldsymbol{u},r_2\boldsymbol{\eta})g(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\\ & \quad=\lim_{r_1\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P(r_1\boldsymbol{\zeta},\boldsymbol{\xi},\boldsymbol{u},\boldsymbol{\eta})g(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\\ & \quad=\lim_{r_1\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P[g]_{r_1,1}(\boldsymbol{\zeta},\boldsymbol{u})d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\\ & \quad=\lim_{r_1,r_2\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P[g]_{r_1,r_2}(\boldsymbol{\zeta},\boldsymbol{u})d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\\ & \quad=\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}g(\boldsymbol{\zeta},\boldsymbol{u})d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta}), \end{align*}

which completes the proof of $(1)$.

The proof of $(2)$ is similar as in $(1)$. Firstly, we prove the case $1< p<\infty$. We notice that $L^{p}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))^{*}=L^{q}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$, where $q$ is the conjugate number of $p$, i.e., $\frac {1}{p}+\frac {1}{q}=1$. Let $g\in L^{p}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ and $f=P[g]$, we need to show that for all $h\in L^{q}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$,

\begin{align*} \lim_{r_1,r_2\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\Bigg(f_{r_1,r_2}(\boldsymbol{\xi},\boldsymbol{\eta})-g(\boldsymbol{\xi},\boldsymbol{\eta})\Bigg)h(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})= 0. \end{align*}

According to Hölder's inequality, the integral above can be estimated by

\begin{align*}{\leq}\|f_{r_1,r_2}-g\|_p\cdot\|h\|_q\longrightarrow 0,\quad \text{as}\ r_1,r_2\to 1, \end{align*}

where Corollary 5.5 and $h\in L^{q}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ are applied above. This proves $(2)$ when $1< p<\infty$. The argument is the same for the case $p=1$ except that we do not need to use Hölder's inequality.

For $p=\infty$, we notice that $L^{1}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))^{*}=L^{\infty }(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$. With $g\in L^{\infty }(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ and $f=P[g]$, we need to show that for all $h\in L^{1}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$,

\begin{align*} & \lim_{r_1,r_2\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}f_{r_1,r_2}(\boldsymbol{\xi},\boldsymbol{\eta})h(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta})\\ & \quad = \int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}g(\boldsymbol{\xi},\boldsymbol{\eta})h(\boldsymbol{\xi},\boldsymbol{\eta})dS(\boldsymbol{\xi})dS(\boldsymbol{\eta}). \end{align*}

With a similar argument as in $(1)$, the left-hand side of the equation above becomes

\begin{align*} \lim_{r_1,r_2\rightarrow 1}\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P[g]_{r_1,r_2}(\boldsymbol{\zeta},\boldsymbol{u})h(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta}). \end{align*}

The Corollary 5.5 tells us that $P[g]_{r_1,r_2}(\boldsymbol {\zeta },\boldsymbol {u})$ converges to $g(\boldsymbol {\zeta },\boldsymbol {u})$ in $L^{1}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ as $r_1,r_2\rightarrow 1$, since $g\in L^{\infty }(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$, we also have that $P[g]_{r_1,r_2}(\boldsymbol {\zeta },\boldsymbol {u})h(\boldsymbol {\zeta },\boldsymbol {u})$ converges to $g(\boldsymbol {\zeta },\boldsymbol {u})h(\boldsymbol {\zeta },\boldsymbol {u})$ in $L^{1}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ as $r_1,r_2\rightarrow 1$. This completes the proof of the theorem.

Proof. Let $\mu =\mu _1\times \mu _2\in M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$, $p>1$ and $q$ is the conjugate number of $p$. For convenience, all $\lambda _{m,k}$ in this proof are positive constants only depending on $m$ and $k$ but not necessarily the same. We have

\begin{align*} |P[\mu](\boldsymbol{x},\boldsymbol{v})|& =\bigg\vert\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\bigg\vert\\ & =\bigg\vert\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{c_{m,k}}{2}\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}Z_k\left(\frac{(\boldsymbol{x}-\boldsymbol{\zeta})\boldsymbol{u}(\boldsymbol{x}-\boldsymbol{\zeta})}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{2}},\boldsymbol{v}\right)d\mu_2(\boldsymbol{u})d\mu_1(\boldsymbol{\zeta})\bigg\vert\\ & \leq\lambda_{m,k}\bigg\vert\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}d|\mu_2|(\boldsymbol{u})d|\mu_1|(\boldsymbol{\zeta})\bigg\vert\\ & \leq\lambda_{m,k}\left[\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\left(\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}\right)^{p}d|\mu_2|(\boldsymbol{u})d|\mu_1|(\boldsymbol{\zeta})\right]^{\frac{1}{p}}\\ & \quad\times\left[\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}d|\mu_2|(\boldsymbol{u})d|\mu_1|(\boldsymbol{\zeta})\right]^{\frac{1}{q}}, \end{align*}

where the fact that $|Z_k(\boldsymbol {u},\boldsymbol {v})|\leq \dim \mathcal {H}_k(\mathbb {C})$ for $\boldsymbol {u},\boldsymbol {v}\in \mathbb {B}^{m}$ [Reference Axler, Bourdon and Ramey1, Proposition 5.27] is applied above. Notice that

\begin{align*} \frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}\leq \frac{1+|\boldsymbol{x}|}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m-1}}\leq 2|\boldsymbol{x}-\boldsymbol{\zeta}|^{1-m},\; \text{for all}\; \boldsymbol{x}\in\mathbb{B}^{m},\boldsymbol{\zeta}\in\mathbb{S}^{m-1}. \end{align*}

Therefore, we obtain $|P[\mu ](\boldsymbol {x},\boldsymbol {v})|\leq \lambda _{m,k}|\boldsymbol {x}-\boldsymbol {\zeta }|^{1-m}\|\mu \|^{\frac {1}{q}+\frac {1}{p}}=\lambda _{m,k}|\boldsymbol {x}-\boldsymbol {\zeta }|^{1-m}\|\mu \|.$ Thus,

\begin{align*} \|P[\mu]\|_{L^{p}(\mathbb{B}^{m}\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C}))}\leq\lambda_{m,k}\left(\int_{\mathbb{B}^{m}}|\boldsymbol{x}-\boldsymbol{\zeta}|^{p(1-m)}d\boldsymbol{x}\right)^{1/p}\|\mu\|\leq\lambda_{m,k}\|\mu\|, \end{align*}

where the integral in the parentheses above exists if $p(1-m)>-m$, i.e., $p<\frac {m}{m-1}$, which completes the proof for $p>1$. The proof for the case $p=1$ is much easier, one can adapt the same argument without using the Hölder's inequality.

Proof. $(1)$. Firstly, it is obvious that the map $\mu \mapsto P[\mu ]$ is linear. Hence, we only need to prove it is bijective. We will prove this by showing it is an injective and surjective map, respectively. To prove it is an injective map, we only need to prove the inequalities in $(1)$. Let $f\in h^{1}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, on the one hand, Theorem 5.2 already tells us that $\|P[\mu ]\|_{h^{1}}\leq \frac {m+2k-2}{m-2}\|\mu \|$. On the other hand, since $\{P[\mu ]_{r_1,r_2}\}$ converges weak$^{*}$ to $\mu$ as in Theorem 5.6, we have

\begin{align*} \|\mu\|\leq\underset{r_1,r_2\rightarrow 1}{\lim\inf}\|P[\mu]_{r_1,r_2}\|_1\leq\sup_{0\leq r_1,r_2<1}\|P[\mu]_{r_1,r_2}\|_1=\|P[\mu]\|_{h^{1}}, \end{align*}

which completes the proof of the inequalities in $(1)$. Hence, the map in $(1)$ is injective.

To prove the map is surjective, let $f\in h^{1}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, according to the definition, there exists a family of functions $\{f_{r_1,r_2},r_1,r_2\in [0,1)\}$, which is norm-bounded in $L^{1}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$, and hence in $M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))=C(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))^{*}$. Further, since $M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1})$ is a separable metric space and $M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ is a subspace of $M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1})$. This implies that $M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ is also separable. Therefore, we have that $\{f_{r_1,r_2},r_1,r_2\in [0,1)\}$ is a norm-bounded sequence in a separable normed linear space $M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$. According to Proposition 5.10, we know that there exists a subsequence converges weak$^{*}$ to some $\mu \in M(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$. Now, to complete the proof of $(1)$, we only need to show that $f=P_B[\mu ]$. Recall that, for fixed $\boldsymbol {x},\boldsymbol {v}\in \mathbb {B}^{m}$, the Poisson integral formula given in [Reference Ding, Nguyen and Ryan11] tells us that

\begin{align*} f(r_1\boldsymbol{x},r_2\boldsymbol{v})=P_B[f](r_1\boldsymbol{x},r_2\boldsymbol{v})=\displaystyle\frac{c_{m,k}}{2}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P_B(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},r_2\boldsymbol{v})f(r_1\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta}). \end{align*}

Since $P_B(\boldsymbol {\zeta },\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})$ is continuous on $\mathbb {S}^{m-1}$ for fixed $\boldsymbol {x},\boldsymbol {v}\in \mathbb {B}^{m}$, then by the continuity of $f$, we let $r_1,r_2\rightarrow 1$ to immediately obtain that

\begin{align*} f(\boldsymbol{x},\boldsymbol{v})=\lim_{r_1,r_2\rightarrow 1}P_B[f](r_1\boldsymbol{x},r_2\boldsymbol{v})=\displaystyle\frac{c_{m,k}}{2}\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P_B(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})d\mu (\boldsymbol{u},\boldsymbol{\zeta}). \end{align*}

In other words, $f=P_B[\mu ]$.

$(2)$. The proof of $(2)$ is similar. Let $1< p\leq \infty$, and $f\in h^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$. To prove the map is injective, one can apply a similar argument as in $(1)$, with the help of Theorem 5.2 and Theorem 5.6, one can obtain the inequalities in $(2)$, which also implies that the map in $(2)$ is injective. To prove it is surjective, let $q$ is the conjugate number of $p$, then $\{f_{r_1,r_2},r_1,r_2\in [0,1)\}$ is norm-bounded in $L^{p}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))=L^{q}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))^{*}$. Theorem 5.10 tells us that there exists a subsequence of $\{f_{r_1,r_2},r_1,r_2\in [0,1)\}$, which converges weak$^{*}$ to some function $g\in L^{p}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$. To complete the proof of $(2)$, we need to show that $f=P_B[g]$. This can be derived from a similar argument as in $(1)$ if we can prove that $P_B(\cdot,\boldsymbol {x},\cdot,\boldsymbol {v})\in L^{q}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$. This can be observed from the fact that, for fixed $\boldsymbol {x}\in \mathbb {B}^{m}$, the classical Poisson kernel $\frac {1-|\boldsymbol {x}|^{2}}{|\boldsymbol {x}-\boldsymbol {\zeta }|^{m}}\in L^{q}(\mathbb {S}^{m-1})$ with respect to $\boldsymbol {\zeta }$ and the reproducing kernel $Z_k$ is bounded.

Proof. We firstly consider the case $1< p<\infty$. Suppose $f\in h^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, then the previous theorem tells us that there exists a function $g\in L^{p}(\mathbb {S}^{m-1}\times \mathbb {S}^{m-1},\mathcal {H}_k(\mathbb {C}))$ such that $f=P[g]$. Let $q$ be the conjugate number of $p$, then we have

(5.4)\begin{align} |f(\boldsymbol{x},\boldsymbol{v})|& =\bigg\vert\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}P(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})g(\boldsymbol{\zeta},\boldsymbol{u})dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\bigg\vert\nonumber\\ & \leq \left(\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|P(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})|^{q}dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\right)^{\frac{1}{q}}\nonumber\\& \quad\times\left(\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|g(\boldsymbol{\zeta},\boldsymbol{u})|^{p}dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\right)^{\frac{1}{p}}\nonumber\\ & =\left(\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|P(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})|^{q}dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\right)^{\frac{1}{q}}\|g\|_p. \end{align}

Now, recall that $P(\boldsymbol {\zeta },\boldsymbol {x},\boldsymbol {u},\boldsymbol {v})=\frac {c_{m,k}}{2}\frac {1-|\boldsymbol {x}|^{2}}{|\boldsymbol {x}-\boldsymbol {\zeta }|^{m}}Z_k(\frac {(\boldsymbol {x}-\boldsymbol {\zeta })\boldsymbol {u}(\boldsymbol {x}-\boldsymbol {\zeta })}{|\boldsymbol {x}-\boldsymbol {\zeta }|^{2}},\boldsymbol {v}),\ \boldsymbol {x},\boldsymbol {v}\in \mathbb {B}^{m},\boldsymbol {\zeta },\boldsymbol {u}\in \mathbb {S}^{m-1},$ and $Z_k(\boldsymbol {u},\boldsymbol {v})\leq \dim \mathcal {H}_k(\mathbb {C})$ for $\boldsymbol {u},\boldsymbol {v}\in \overline {\mathbb {B}^{m}}$, see [Reference Axler, Bourdon and Ramey1, Proposition 5.27]. Then,

(5.5)\begin{align} & \left(\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}|P(\boldsymbol{\zeta},\boldsymbol{x},\boldsymbol{u},\boldsymbol{v})|^{q}dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\right)^{\frac{1}{q}}\nonumber\\ & \quad\leq\displaystyle\frac{c_{m,k}}{2}\dim\mathcal{H}_k(\mathbb{C})\left[\displaystyle\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\left(\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}\right)^{q}dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\right]^{\frac{1}{q}}\nonumber\\ & \quad\leq\frac{c_{m,k}}{2}\dim\mathcal{H}_k(\mathbb{C})\sup_{\boldsymbol{\zeta}\in\mathbb{S}^{m-1}}\left(\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}\right)^{\frac{q-1}{q}}\left(\int_{\mathbb{S}^{m-1}}\int_{\mathbb{S}^{m-1}}\frac{1-|\boldsymbol{x}|^{2}}{|\boldsymbol{x}-\boldsymbol{\zeta}|^{m}}dS(\boldsymbol{u})dS(\boldsymbol{\zeta})\right)^{\frac{1}{q}}\nonumber\\ & \quad=\frac{m+2k-2}{m-2}(\omega_m)^{\frac{p-2}{p}}\dim\mathcal{H}_k(\mathbb{C})\left(\frac{1+|\boldsymbol{x}|}{(1-|\boldsymbol{x}|)^{m-1}}\right)^{\frac{1}{p}}. \end{align}

Further, Theorem 5.2 tells us that $\|P[g]_{r_1,r_2}\|_p\leq \frac {m+2k-2}{m-2}\|g\|_p$ and we also know that $\lim _{r_1,r_2\rightarrow 1}P[g]_{r_1,r_2}=P[g]=g$ on $\mathbb {S}^{m-1}\times \mathbb {S}^{m-1}$. This implies that

\begin{align*} \|g\|_p\leq \|P[g]\|_{h^{p}}=\sup_{0\leq r_1,r_2<1}\|P[g]_{r_1,r_2}\|_p\leq \frac{m+2k-2}{m-2}\|g\|_p, \end{align*}

in other words, with $f=P[g]$, it is equivalent to $\frac {m-2}{m+2k-2}\|f\|_{h^{p}}\leq \|g\|_p\leq \|f\|_{h^{p}}.$ Plugging this together with (5.5) into (5.4) completes our proof immediately.

Proof. From the proof of the proposition above, we have that

\begin{align*} & \|f\|_{L^{p}(\mathbb{B}^{m}\times\mathbb{B}^{m},\mathcal{H}_k(\mathbb{C}))}^{p}\leq C_{m,k,p}\int_{\mathbb{B}^{m}}\int_{\mathbb{B}^{m}}\int_{\mathbb{S}^{m-1}}|\boldsymbol{x}-\boldsymbol{\zeta}|^{p(1-m)}dS(\boldsymbol{\zeta})d\boldsymbol{x} d\boldsymbol{v}\|f\|_{h^{p}}^{p}\\ & \quad\leq C'_{m,k,p}\int_{\mathbb{B}^{m}}\int_{\mathbb{S}^{m-1}}|\boldsymbol{x}-\boldsymbol{\zeta}|^{p(1-m)}dS(\boldsymbol{\zeta})d\boldsymbol{x} \|f\|_{h^{p}}^{p}\leq C''_{m,k,p}\|f\|^{p}_{h^{p}}, \end{align*}

when $1\leq p<\frac {m}{m-1}$. This completes the proof.

Proof. Suppose that $\{f_j\}_{j=1}^{\infty }$ converges to $f$ in $L^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathbb {C})$ and $\{f_j\}_{j=1}^{\infty }$ is a Cauchy sequence in $h^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$. We will show that $f\in h^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ up to a modification on a set of measure zero on $\mathbb {B}^{m}\times \mathbb {B}^{m}$.

Let $K_1\times K_2\subset \mathbb {B}^{m}\times \mathbb {B}^{m}$ be a compact subset. Proposition 5.13 tells us that there exists a constant $C>0$ only depending on $m,k$ and $p$ such that $|f_j(\boldsymbol {x},\boldsymbol {v})-f_i(\boldsymbol {x},\boldsymbol {v})|\leq C\|f_j-f_i\|_{h^{p}},$ for all $(\boldsymbol {x},\boldsymbol {v})\in K_1\times K_2$ and all $j,i$. Since $\{f_j\}_{j=1}^{\infty }$ is a Cauchy sequence in $h^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, $\{f_j\}_{j=1}^{\infty }$ is also a Cauchy sequence in $C(K_1\times K_2)$. Hence, $\{f_j\}_{j=1}^{\infty }$ converges uniformly on $K_1\times K_2$. According to [Reference Ding, Nguyen and Ryan11, Proposition 5.8], $\{f_j\}_{j=1}^{\infty }$ converges uniformly to a function $g\in C^{2}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$ and $\mathcal {D}_k g=0$ on $\mathbb {B}^{m}\times \mathbb {B}^{m}$. Since $\{f_j\}_{j=1}^{\infty }$ converges to $f$ in $L^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathbb {C})$, some subsequence of $\{f_{j}\}_{=1}^{\infty }$ converges to $f$ pointwise almost everywhere on $\mathbb {B}^{m}\times \mathbb {B}^{m}$. Therefore, $f=g$ almost everywhere on $\mathbb {B}^{m}\times \mathbb {B}^{m}$ and thus $f\in h^{p}(\mathbb {B}^{m}\times \mathbb {B}^{m},\mathcal {H}_k(\mathbb {C}))$, which completes the proof.