2 Standard tools

Let us give some standard tools which will be useful in the sequel.

The binomial series

$$ \begin{align*}\frac{1}{(1-x)^{\alpha}}=\sum_{k=0}^{\infty}\frac{\Gamma(k+\alpha)}{\Gamma(\alpha)k!}x^k,\end{align*} $$

where $|x|<1$ is a complex number and $\alpha $ is a nonnegative real number. The asymptotic behavior of the $\Gamma $ -function is the following: $\Gamma (k+\alpha )\asymp (k-1)!k^{\alpha }$ , where the symbol $\asymp $ denotes that the ratio of the two quantities either tends to a constant as k tends to infinity or it is rather two sides bound by constants.

The multinomial formula

$$ \begin{align*}(x_1+\cdots+x_n)^k=\sum_{|j|=k}\frac{k!}{j!}x_1^{j_1}\cdots x_n^{j_n},\end{align*} $$

where $j=(j_1,\ldots ,j_n)$ is an n-tuple index of nonnegative integers and $x_i$ are complex numbers.

The Stirling formula that describes the asymptotic behavior of the gamma function

$$ \begin{align*}k!\asymp k^{1/2}k^k/e^k.\end{align*} $$

Denote the normalized area measure on ${\mathbb C}^n={\mathbb R}^{2n}$ by $du(z)$ and the normalized rotation-invariant positive Borel measure on $\mathbb S_n$ by $d\sigma (\zeta )$ (see [Reference Rudin18, Reference Zhu22]). The measures $du(z)$ and $d\sigma (\zeta )$ are related by the formula

$$ \begin{align*}\int_{{\mathbb C}^n}f(z)du(z)=2n\int_{0}^{\infty}\int_{\mathbb S_n}\epsilon^{2n-1}f(\epsilon\zeta)d\sigma(\zeta)d\epsilon.\end{align*} $$

The holomorphic monomials are orthogonal to each other in $L^2(\sigma )$ , that is, if k and l are multiindices such that $k\neq l$ , then

$$ \begin{align*}\int_{\mathbb S_n}\zeta^k \bar{\zeta}^l d\sigma(\zeta)=0.\end{align*} $$

Moreover,

$$ \begin{align*}\int_{\mathbb S_n}|\zeta^k|^2d\sigma(\zeta)=\frac{(n-1)!k!}{(n-1+|k|)!} \quad \text{and} \quad \int_{{\mathbb B_n}}|z^k|^2du(z)=\frac{n!k!}{(n+|k|)!}.\end{align*} $$

3 Relation among Dirichlet-type spaces and an equivalent characterization

We study the structure of Dirichlet-type spaces. Note that

$$ \begin{align*}R(f)(z)=z_1\partial_{z_1}f(z)+\cdots+z_n\partial_{z_n}f(z)\end{align*} $$

is the radial derivative of a function f. The radial derivative plays a key role in the function theory of the unit ball. A crucial relation among these spaces is the following.

Proposition 1 Let $f\in \mathrm {Hol}({\mathbb B_n})$ and $\alpha \in {\mathbb R}$ be fixed. Then

$$ \begin{align*} f\in D_{\alpha}({\mathbb B_n})\quad \text{if and only if} \quad n^q f+R^q (f)+q\sum_{i=1}^{q-1}n^iR^{q-i}(f)\in D_{\upsilon}({\mathbb B_n}), \end{align*} $$

where $\alpha =2q+\upsilon $ , $q\in \mathbb N$ , and $R^q$ is the q-image of the operator R.

Proof Indeed, it is enough to check that

$$ \begin{align*}||nf+R(f)||^2_{\alpha-2}=\sum_{|k|=0}^{\infty}(n+|k|)^{\alpha-2}\frac{(n-1)!k!}{(n-1+|k|)!}(n+|k|)^2|a_{k}|^2=||f||_{\alpha}^2.\\[-40pt]\end{align*} $$

We continue by giving an equivalent characterization of Dirichlet-type spaces. In Dirichlet-type spaces in the unit ball, one of the Dirichlet-type integrals is achieved in a limited range of parameters.

Lemma 2 (See [Reference Michalska16])

If $\alpha \in (-1,1)$ , then $f\in D_\alpha(\mathbb B_n)$ if and only if

$$\begin{align*}|f|^2_{\alpha}:=\int_{{\mathbb B_n}} \frac{||\nabla(f)(z)||^2 - |R(f)(z)|^2}{ (1-||z||^2)^{\alpha}} du(z)<\infty.\end{align*}$$

Above, $\nabla (f)(z)=(\partial _{z_1}f(z),\ldots ,\partial _{z_n}f(z))$ denotes the holomorphic gradient of a holomorphic function f. Note that Proposition 1 allows us to use Lemma 2 whenever $\upsilon \in (-1,1)$ . Let $\gamma ,t\in {\mathbb R}$ be such that neither $n+\gamma $ nor $n+\gamma +t$ is a negative integer. If $f=\sum _{|k|=0}^{\infty }a_kz^k$ is the homogeneous expansion of a function $f\in \textrm {Hol}({\mathbb B_n})$ , then we may define an invertible continuous linear operator with respect to the topology of uniform convergence on compact subsets of ${\mathbb B_n}$ , denoted by $R^{\gamma ,t}: \textrm {Hol}({\mathbb B_n})\rightarrow \textrm {Hol}({\mathbb B_n})$ and having expression

$$ \begin{align*} R^{\gamma,t}f(z)=\sum_{|k|=0}^{\infty}C(\gamma,t,k)a_kz^k, \quad z\in {\mathbb B_n}, \end{align*} $$

where

(4) $$ \begin{align} C(\gamma,t,k)=\frac{\Gamma(n+1+\gamma)\Gamma(n+1+|k|+\gamma+t)}{\Gamma(n+1+\gamma+t)\Gamma(n+1+|k|+\gamma)}\asymp |k|^t. \end{align} $$

See [Reference Zhu22] for more information regarding these fractional radial derivatives.

Lemma 3 Let $t\in {\mathbb R}$ be such that $n-1+t \geq 0$ . If $f\in A({\mathbb B_n})$ , then

$$ \begin{align*}R^{-1,t}f(z)=\int_{\mathbb S_n}\frac{f(\zeta)}{(1-\langle z,\zeta \rangle)^{n+t}}d\sigma(\zeta), \quad z\in {\mathbb B_n}.\end{align*} $$

Proof The continuous linear operator $R^{\gamma ,t}$ (see [Reference Zhu22]) satisfies

$$ \begin{align*}R^{\gamma,t}\Big(\frac{1}{(1-\langle z,w \rangle)^{n+1+\gamma}}\Big)=\frac{1}{(1-\langle z,w \rangle)^{n+1+\gamma+t}}\end{align*} $$

for all $w\in {\mathbb B_n}$ . Next, define $f_{\epsilon }$ for $\epsilon \in (0,1)$ by

$$ \begin{align*}f_{\epsilon}(z)=\int_{\mathbb S_n}\frac{f(\zeta)}{(1-\langle z,\epsilon\zeta \rangle)^n}d\sigma(\zeta), \quad z\in {\mathbb B_n}.\end{align*} $$

The Cauchy formula holds for $f\in A({\mathbb B_n})$ and hence $f=\lim _{\epsilon \rightarrow 1^{-}}f_{\epsilon }$ . It follows that

$$ \begin{align*} R^{-1,t}f(z)&=R^{-1,t}\Big(\lim_{\epsilon\rightarrow 1^{-}}\int_{\mathbb S_n}\frac{f(\zeta)}{(1-\langle z,\epsilon\zeta \rangle)^n}d\sigma(\zeta) \Big)\\ &=\lim_{\epsilon\rightarrow 1^{-}}R^{-1,t}\Big(\int_{\mathbb S_n}\frac{f(\zeta)}{(1-\langle z,\epsilon\zeta \rangle)^n}d\sigma(\zeta) \Big)\\ &=\lim_{\epsilon\rightarrow 1^{-}}\int_{\mathbb S_n}f(\zeta)R^{-1,t}\Big(\frac{1}{(1-\langle z,\epsilon\zeta \rangle)^{n}}\Big)d\sigma(\zeta)\\ &=\lim_{\epsilon\rightarrow 1^{-}}\int_{\mathbb S_n}\frac{f(\zeta)}{(1-\langle z,\epsilon\zeta \rangle)^{n+t}}d\sigma(\zeta)\\ &=\int_{\mathbb S_n}\frac{f(\zeta)}{(1-\langle z,\zeta \rangle)^{n+t}}d\sigma(\zeta), \end{align*} $$

and the assertion follows.

Theorem 4 Let $\alpha \in {\mathbb R}$ be such that $n-1+\alpha /2\geq 0$ and $f\in A({\mathbb B_n})$ . Then $f\in D_{\alpha }({\mathbb B_n})$ if and only if

$$ \begin{align*}\int_{{\mathbb B_n}}(1-||z||^2)\Big|\int_{\mathbb S_n}\frac{f(\zeta)\bar{\zeta}_p}{(1-\langle z,\zeta \rangle)^{n+\alpha/2+1}}d\sigma(\zeta)\Big|^2du(z)<\infty\end{align*} $$

and

$$ \begin{align*}\int_{{\mathbb B_n}}\Big|\int_{\mathbb S_n}\frac{(\overline{z_p\zeta_q-z_q\zeta_p})f(\zeta)}{(1-\langle z,\zeta \rangle)^{n+\alpha/2+1}}d\sigma(\zeta)\Big|^2du(z)<\infty,\end{align*} $$

where $p,q=1,\ldots ,n$ .

Proof Choose t so that $\alpha =2t$ . Note that $n,t$ are fixed and hence

$$ \begin{align*} ||f||^2_{\alpha}\asymp\sum_{|k|=0}^{\infty}\frac{(n-1)!k!}{(n-1+|k|)!}||k|^ta_{k}|^2. \end{align*} $$

Thus, (4) implies that $||R^{-1,t}f||_{0}\asymp ||f||_{\alpha }$ . One can apply then the equivalent integral representation of Dirichlet-type norms to $R^{-1,t}f\in \textrm {Hol}({\mathbb B_n})$ , i.e., $R^{-1,t}f\in D_0(\mathbb B_n)$ if and only if $|R^{-1,t}f|_{0}<\infty$ . According to Lemma 3, we get that

$$ \begin{align*}\partial_{z_p}(R^{-1,t}f)(z)=\int_{\mathbb S_n}\frac{f(\zeta)\bar{\zeta}_p}{(1-\langle z,\zeta \rangle)^{n+t+1}}d\sigma(\zeta), \quad z\in {\mathbb B_n},\end{align*} $$

where $p=1,\ldots ,n$ . Expand the term $||\nabla (f)||^2-|R(f)|^2$ as follows:

$$ \begin{align*} ||\nabla(f)||^2-|R(f)|^2=(1-||z||^2)||\nabla(f)||^2+\sum_{p,q}|\bar{z}_p\partial_{z_q}f-\bar{z}_q\partial_{z_p}f|^2. \end{align*} $$

The assertion follows by Lemma 2.

4 Diagonal subspaces

In [Reference Bénéteau, Condori, Liaw, Seco and Sola3], a method of construction of optimal approximants via determinants in Dirichlet-type spaces in the unit disk is provided. Similarly, we may define optimal approximants in several variables (see [Reference Sargent and Sola20]).

Fix $N\in \mathbb N$ . We define the space of polynomials $p\in {\mathbb C}[z_1,\ldots ,z_n]$ with degree at most $nN$ as follows:

$$ \begin{align*}P_N^n:=\Big\{p(z)=\sum_{k_1=0}^{N}\cdots \sum_{k_n=0}^{N}a_{k_1,\ldots,k_n}z_1^{k_1}\cdots z_n^{k_n}\Big\}.\end{align*} $$

Remark 5 Let $(X,||\cdot ||)$ be a normed space and fix $x\in X$ , $C\subset X$ . The distance between x and the set C is the following:

$$ \begin{align*}\mathrm{dist}_X(x,C):=\inf\{||x-c||:c\in C\}.\end{align*} $$

It is well known that if X is a Hilbert space and $C\subset X$ a convex closed subset, then for any $x\in X$ , there exists a unique $y\in C$ such that $||x-y||=\mathrm {dist}_X(x,C)$ . Let $f\in D_{\alpha }({\mathbb B_n})$ be nonzero constant. We deduce that for any $N\in \mathbb N$ , there exists exactly one $p_N\in P_N^n$ satisfying

$$ \begin{align*}||p_Nf-1||_{\alpha}=\mathrm{dist}_{D_{\alpha}({\mathbb B_n})}(1,f\cdot P_N^n).\end{align*} $$

Let $f\in D_{\alpha }({\mathbb B_n})$ . We say that a polynomial $p_N\in P^n_N$ is an optimal approximant of order N to $1/f$ if $p_N$ minimizes $||pf-1||_{\alpha }$ among all polynomials $p\in P_N^n$ . We call $||p_Nf-1||_{\alpha }$ the optimal norm of order N associated with f.

Let $M=(M_1,\ldots ,M_n)$ be a multiindex, where $M_i$ are nonnegative integers, and $m\in \{1,\ldots ,n\}$ . Setting

$$ \begin{align*}\mu(m):=\frac{(M_1+\cdots+M_m)^{M_1+\cdots+M_m}}{M_1^{M_1}\cdots M_m^{M_m}},\end{align*} $$

we see that

(5) $$ \begin{align} \mu(m)^{1/2}|z_1|^{M_1}\cdots |z_m|^{M_m}\leq 1, \quad z\in {\mathbb B_n}. \end{align} $$

Using (5), we may construct polynomials that vanish in the closed unit ball along analytic subsets of the unit sphere.

Remark 6 Let $\tilde {f}\in \mathrm {Hol}({\mathbb D}(\mu (m)^{-1/4}))$ , where

$$ \begin{align*} {\mathbb D}(\mu)=\{z\in {\mathbb C}:|z|<\mu\}, \quad \mu> 0. \end{align*} $$

According to (5), we define the following function:

$$ \begin{align*} f(z)=f(z_1,\ldots,z_n)=\tilde{f}(\mu(m)^{1/4}z_1^{M_1}\cdots z_m^{M_m}), \quad z\in {\mathbb B_n}. \end{align*} $$

Then $f\in \mathrm {Hol}({\mathbb B_n})$ and it depends on m variables. Note that we may change the variables $z_1,\ldots ,z_m$ by any other m variables. For convenience, we choose the m first variables. The power $1/4$ will be convenient in the sequel.

Thus, the question that arises out is if we may define closed subspaces of $D_{\alpha }({\mathbb B_n})$ passing through one variable functions. We shall see that these subspaces are called diagonal subspaces due to the nature of the power series expansion of their elements.

Instead of the classical one variable Dirichlet-type spaces of the unit disk, we may consider spaces $d_{\beta }$ , $\beta \in {\mathbb R}$ , consisting of holomorphic functions $\tilde {f}\in \mathrm {Hol}({\mathbb D}(\mu ^{-1/4}))$ . Moreover, such functions with power series expansion $\tilde {f}(z)=\sum _{l=0}^{\infty }a_lz^l$ are said to belong to $d_{\beta }$ if

$$ \begin{align*} ||\tilde{f}||^2_{d_{\beta}}:= \sum_{l=0}^{\infty}\mu^{-l/2}(l+1)^{\beta}|a_l|^2<\infty. \end{align*} $$

There is a natural identification between the function theories of $D_{\beta }({\mathbb D})$ : one variable Dirichlet-type spaces of the unit disk, and $d_{\beta }$ , and one verifies that the results in [Reference Bénéteau, Condori, Liaw, Seco and Sola3] are valid for $d_{\beta }$ .

We are ready to define diagonal closed subspaces. Set

$$ \begin{align*} \beta(\alpha):=\alpha-n+\frac{m+1}{2}. \end{align*} $$

Let $\alpha $ , M, m be as above. The diagonal closed subspace of $D_{\alpha }({\mathbb B_n})$ is the following:

$$ \begin{align*} J_{\alpha,M,m}:=\{f\in D_{\alpha}({\mathbb B_n}):\exists \tilde{f}\in d_{\beta(\alpha)}, f(z)= \tilde{f}(\mu(m)^{1/4}z_1^{M_1}\cdots z_m^{M_m})\}. \end{align*} $$

The existence of a holomorphic function $\tilde {f}$ is unique by identity principle, and hence there is no any amiss in the definition. Any function $f\in J_{\alpha ,M,m}$ has an expansion of the form

$$ \begin{align*} f(z)=\sum_{l=0}^{\infty}a_l(z_1^{M_1}\cdots z_m^{M_m})^l. \end{align*} $$

The relation of norms between one variable and diagonal subspaces follows.

Proposition 7 If $f\in J_{\alpha ,M,m}$ , then $||f||_{\alpha }\asymp ||\tilde {f}||_{d_{\beta (\alpha )}}$ .

Proof If $f\in J_{\alpha ,M,m}$ , then

$$ \begin{align*}||f||^2_{\alpha} \asymp \sum_{l=0}^{\infty}(l+1)^{\alpha}\frac{(M_1l)!\cdots (M_ml)!}{(n-1+(M_1+\cdots M_m)l)!}|a_l|^2.\end{align*} $$

By Stirling’s formula, we obtain

$$ \begin{align*}||f||^2_{\alpha} \asymp \sum_{l=0}^{\infty}(l+1)^{\alpha-n+m/2+1/2}\mu(m)^{-l}|a_l|^2.\end{align*} $$

On the other hand, define the function $f'(z)=\sum _{l=0}^{\infty }\mu (m)^{-l/4}a_lz^l$ . Then $f'(\mu (m)^{1/4} z_1^{M_1}\cdots z_m^{M_m})=f(z_1,\ldots ,z_n)$ and

$$\begin{align*}||f'||^2_{d_{\beta(\alpha)}}\asymp \sum_{l=0}^{\infty}(l+1)^{\alpha-n+m/2+1/2}\mu(m)^{-l}|a_l|^2.\end{align*}$$

The assertion follows since $f'$ coincides with $\tilde {f}$ .

The corresponding Lemma 3.4 of [Reference Bénéteau, Condori, Liaw, Seco and Sola4] in our case is the following.

Lemma 8 Let $f\in J_{\alpha ,M,m}$ , where $\alpha ,M,m$ be as above. Let $r_N\in P_N^n$ with expansion

$$ \begin{align*}r_N(z)=\sum_{k_1=0}^{N}\cdots \sum_{k_n=0}^{N}a_{k_1,\ldots,k_n}z_1^{k_1}\ldots z_n^{k_n},\end{align*} $$

and consider its projection onto $J_{\alpha ,M,m}$

$$ \begin{align*}\pi(r_N)(z)=\sum_{\{l:M_1l,\ldots,M_ml\leq N\}}c_{M_1l,\ldots,M_ml,0,\ldots,0}z_1^{M_1l}\cdots z_m^{M_ml}.\end{align*} $$

Then

$$ \begin{align*}||r_Nf-1||_{\alpha}\geq ||\pi(r_N)f-1||_{\alpha}.\end{align*} $$

Moreover, just as in Proposition 7, there is a relation of optimal approximants between one variable and diagonal subspaces.

Proposition 9 If $f\in J_{\alpha ,M,m}$ , then

$$ \begin{align*}\mathrm{dist}_{D_{\alpha}({\mathbb B_n})}(1,f\cdot P_N^n)\asymp \mathrm{dist}_{d_{\beta}(\alpha)}(1,\tilde{f}\cdot P^1_N).\end{align*} $$

Proof Let $r_N$ , $\pi (r_N)$ be as in Lemma 8. Then $\pi (r_N)f-1\in J_{\alpha ,M,m}$ . It follows that

$$ \begin{align*} ||r_Nf-1||_{\alpha}\geq ||\pi(r_N)f-1||_{\alpha} \asymp ||\tilde{\pi}(r_N)\tilde{f}-1||_{d_{\beta(\alpha)}} \geq \mathrm{dist}_{d_{\beta(\alpha)}}(1,\tilde{f}\cdot P^1_N), \end{align*} $$

since $\tilde {\pi }(r_N)\in P^1_N$ . On the other hand, let

$$ \begin{align*}\mathrm{dist}_{d_{\beta(\alpha)}}(1,\tilde{f}\cdot P^1_N)=||q_N\tilde{f}-1||_{d_{\beta(\alpha)}}, \quad q_N(z)=\sum_{l=0}^{N}a_lz^l.\end{align*} $$

Then, the polynomial

$$ \begin{align*}q^{\prime}_N(z_1,\ldots,z_n)=\sum_{l=0}^{N}\mu(m)^{-l/4}a_lz_1^{M_1l}\cdots z_m^{M_ml}\end{align*} $$

satisfies $q_N^{\prime }\in J_{\alpha ,M,m}\cap P_N^n$ and $q_N^{\prime }f-1\in J_{\alpha ,M,m}$ . Thus,

$$ \begin{align*} ||q_N\tilde{f}-1||_{d_{\beta(\alpha)}}=||\tilde{q}_N^{\prime}\tilde{f}-1||_{d_{\beta(\alpha)}} \asymp ||q_N^{\prime}f-1||_{\alpha} \geq \mathrm{dist}_{D_{\alpha}({\mathbb B_n})}(1,f\cdot P_N^n) \end{align*} $$

and the assertion follows.

Define the function $\phi _{\beta }:[0,\infty )\rightarrow [0,\infty )$ by

$$ \begin{align*}\phi_{\beta}(t)= \begin{cases} t^{1-\beta}, & \beta<1, \\ \log^+(t), & \beta=1, \end{cases} \end{align*} $$

where $\log ^+(t):=\max \{\log t,0\}$ . We have the following.

Theorem 10 Let $\alpha \in {\mathbb R}$ be such that $\beta (\alpha )\leq 1$ . Let $f\in J_{\alpha ,M,m}$ be as above, and suppose that the corresponding $\tilde {f}$ has no zeros inside its domain, has at least one zero on the boundary, and admits an analytic continuation to a strictly bigger domain. Then f is cyclic in $D_{\alpha }({\mathbb B_n})$ whenever $\alpha \leq \frac {2n-m+1}{2}$ and

$$ \begin{align*}\mathrm{dist}^2_{D_{\alpha}({\mathbb B_n})}(1,f\cdot P_N^n)\asymp \phi_{\beta(\alpha)}(N+1)^{-1}.\end{align*} $$

Proof It is an immediate consequence of the identification between $D_{\beta }({\mathbb D})$ and $d_{\beta }$ and previous lemmas and propositions.

If we focus on polynomials, then the following is true.

Note that the Theorem 11 is not a characterization. We shall study the case $\alpha>\frac {2n+1-m}{2}$ .

5 Cyclicity for model polynomials via radial dilation

The radial dilation of a function $f:{\mathbb B_n}\rightarrow {\mathbb C}$ is defined for $r\in (0,1)$ by $f_r(z)=f(rz)$ . To prove Theorem 11, it is enough to prove the following lemma.

We follow the arguments of [Reference Knese, Kosiński, Ransford and Sola13, Reference Kosiński and Vavitsas14]. Indeed, if Lemma 12 holds, then $\phi _r\cdot p\rightarrow 1$ weakly, where $\phi _r:=1/p_r$ . This is a consequence of a crucial property of Dirichlet-type spaces: if $\{f_n\}\subset D_{\alpha }({\mathbb B_n})$ , then $f_n\rightarrow 0$ weakly if and only if $f_n\rightarrow 0$ pointwise and $\sup _{n}\{||f_n||_{\alpha }\}<\infty $ . Since $\phi _r$ extends holomorphically past the closed unit ball, $\phi _r$ are multipliers, and hence, $\phi _r\cdot p\in [p]$ . Finally, $1$ is weak limit of $\phi _r\cdot p$ and $[p]$ is closed and convex or, equivalently, weakly closed. It is clear that $1\in [p]$ , and hence p is cyclic.

Moreover, it is enough to prove that $||p/p_r||_{\alpha }<\infty $ , as $r\rightarrow 1^{-}$ , for $\alpha _0=\frac {2n+1-m}{2}$ . Then the case $\alpha <\alpha _0$ follows since the inclusion $D_{\alpha _0}({\mathbb B_n})\hookrightarrow D_{\alpha }({\mathbb B_n})$ is a compact linear map and weak convergence in $D_{\alpha _0}({\mathbb B_n})$ gives weak convergence in $D_{\alpha }({\mathbb B_n})$ .

Proof of Lemma 12

By Theorem 4, it is enough to show the following:

$$ \begin{align*}I_p:=\int_{{\mathbb B_n}}(1-||z||^2)\Big|\int_{\mathbb S_n}\frac{(1-\lambda \zeta_1\cdots \zeta_m)\bar{\zeta}_p}{(1-r^m\lambda \zeta_1\cdots \zeta_m)(1-\langle z,\zeta \rangle)^{\beta}}d\sigma(\zeta)\Big|^2du(z)\end{align*} $$

and

$$ \begin{align*}I_{p,q}:=\int_{{\mathbb B_n}}\Big|\int_{\mathbb S_n}\frac{(\overline{z_p\zeta_q-z_q\zeta_p})(1-\lambda \zeta_1\cdots \zeta_m)}{(1-r^m\lambda \zeta_1\cdots \zeta_m)(1-\langle z,\zeta \rangle)^{\beta}}d\sigma(\zeta)\Big|^2du(z)\end{align*} $$

are finite, as $r\rightarrow 1^{-}$ , where $\beta =n+t+1$ , $t=\frac {2n+1-m}{4}$ , and $\lambda =m^{m/2}$ .

Denote

$$ \begin{align*}S_p:=\int_{\mathbb S_n}\frac{(1-\lambda \zeta_1\cdots \zeta_m)\bar{\zeta}_p}{(1-r^m\lambda \zeta_1\cdots \zeta_m)(1-\langle z,\zeta \rangle)^{\beta}}d\sigma(\zeta),\end{align*} $$

where the last integral is equal to

$$ \begin{align*}\frac{1}{2\pi}\int_{\mathbb S_n}\int_{0}^{2\pi}\frac{(1-\lambda e^{im\theta}\zeta_1\cdots \zeta_m)e^{-i\theta}\bar{\zeta}_p}{(1-r^m\lambda e^{im\theta}\zeta_1\cdots \zeta_m)(1-e^{-i\theta}\langle z,\zeta \rangle)^{\beta}}d\theta d\sigma(\zeta).\end{align*} $$

Let $z,\zeta $ be fixed. Then

$$ \begin{align*}\int_{0}^{2\pi}\frac{e^{-i\theta}}{(1-e^{-i\theta}\langle z,\zeta \rangle)^{\beta}}d\theta=0.\end{align*} $$

Thus, replacing $p(e^{i\theta }\zeta )/p(re^{i\theta }\zeta )$ by $p(e^{i\theta }\zeta )/p(re^{i\theta }\zeta )-1$ , we obtain

$$ \begin{align*}S_p=\frac{\lambda(r^m-1)}{2\pi}\int_{\mathbb S_n}\int_{0}^{2\pi}\frac{ \bar{\zeta}_p\zeta_1\cdots\zeta_me^{i(m-1)\theta}}{(1-r^m\lambda e^{im\theta}\zeta_1\cdots \zeta_m)(1-e^{-i\theta}\langle z,\zeta \rangle)^{\beta}}d\theta d\sigma(\zeta).\end{align*} $$

Next, expand the binomials

$$ \begin{align*} \int_{0}^{2\pi}&\frac{e^{i(m-1)\theta}}{(1-r^m\lambda e^{im\theta}\zeta_1\cdots \zeta_m)(1-e^{-i\theta}\langle z,\zeta \rangle)^{\beta}}d\theta\\ &=\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{\Gamma(k+\beta)}{\Gamma(\beta)k!}(r^m\lambda\zeta_1\cdots \zeta_m)^l\langle z,\zeta \rangle^k\int_{0}^{2\pi}e^{i(m(l+1)-k-1)\theta}d\theta\\ &=2\pi\sum_{k=0}^{\infty}\frac{\Gamma(m(k+1)-1+\beta)}{\Gamma(\beta)(m(k+1)-1)!}(r^m\lambda\zeta_1\cdots \zeta_m)^k\langle z,\zeta \rangle^{m(k+1)-1}\\ &=2\pi \sum_{k=0}^{\infty}\sum_{|j|=m(k+1)-1}\frac{\Gamma(m(k+1)-1+\beta)}{\Gamma(\beta)j!}(r^m\lambda\zeta_1\cdots \zeta_m)^kz^j\bar{\zeta}^j. \end{align*} $$

Therefore,

$$ \begin{align*}S_p=\lambda(r^m-1)\sum_{k=0}^{\infty}\sum_{|j|=m(k+1)-1}\frac{\Gamma(m(k+1)-1+\beta)}{\Gamma(\beta)j!}(r^m\lambda)^kz^jc(k),\end{align*} $$

where $c(k)=\int _{\mathbb {S}_n}\zeta ^{\alpha (k)}\bar {\zeta }^{b(k)}d\sigma (\zeta )$ , $\alpha (k)=(k+1,\ldots ,k+1\text {(m-comp.)},0,\ldots ,0)$ and $b(k)=(j_1,\ldots ,j_{p-1},j_p+1,j_{p+1},\ldots ,j_n)$ . Whence, $1\leq p\leq m$ . Since the holomorphic monomials are orthogonal to each other in $L^2(\sigma )$ , we get that

$$ \begin{align*}|S_p|\asymp(1-r^m)\Big|z^{\prime}_p\sum_{k=0}^{\infty}(k+1)^{\beta-n}(r^m\lambda z_1\cdots z_m)^k\Big|,\end{align*} $$

where $z^{\prime }_p=z_1\cdots z_{p-1}z_{p+1}\cdots z_m$ . Hence, we obtain

$$ \begin{align*}I_p\asymp(1-r^m)^2\sum_{k=0}^{\infty}(k+1)^{2(\beta-n)}(r^m\lambda)^{2k}\int_{{\mathbb B_n}}(1-||z||^2)|z^{\prime}_p|^2|z_1\cdots z_m|^{2k}du(z),\end{align*} $$

where has been used again the orthogonality of the holomorphic monomials in $L^2(\sigma )$ . To handle the integral above, we use polar coordinates

$$ \begin{align*} \int_{{\mathbb B_n}}&(1-||z||^2)|z^{\prime}_p|^2|z_1\cdots z_m|^{2k}du(z)\\ &\asymp \int_{0}^{1}\int_{\mathbb S_n}\epsilon^{2n-1}(1-\epsilon^2)\epsilon^{2km+2m-2}|\zeta_p^{\prime}|^2|\zeta_1\cdots \zeta_m|^{2k}d\sigma(\zeta)d\epsilon\\ &\asymp \frac{[(k+1)!]^{m-1}k!}{(n+m(k+1)-2)!}\cdot\frac{1}{(k+1)^2}. \end{align*} $$

If we recall that $\beta =n+t+1$ , $t=\frac {2n+1-m}{4}$ and $\lambda ^{2k}=m^{mk}$ , then applying the Stirling formula more than one time, we see that

$$ \begin{align*}I_p\asymp(1-r^m)^2 \sum_{k=0}^{\infty}(k+1)r^{2mk}.\end{align*} $$

This proves the assertion made about $I_p$ .

It remains to estimate the following term:

$$ \begin{align*}I_{p,q}=\int_{{\mathbb B_n}}\Big|\int_{\mathbb S_n}\frac{(\overline{z_p\zeta_q-z_q\zeta_p})(1-\lambda \zeta_1\cdots \zeta_m)}{(1-r^m\lambda \zeta_1\cdots \zeta_m)(1-\langle z,\zeta \rangle)^{\beta}}d\sigma(\zeta)\Big|^2du(z).\end{align*} $$

We shall show that $I_{p,q}\asymp I_p$ . Denote again the inner integral by $S_{p,q}$ which is convenient to expand it as $S_{p,q}=\bar {z}_pS_q-\bar {z}_qS_p$ . Recall that $z_p^{\prime }=z_1\cdots z_{p-1}z_{p+1}\cdots z_m$ . Similar calculations to the one above lead to

$$ \begin{align*}|S_{p,q}|\asymp(1-r^m)|\bar{z}_pz_q^{\prime}-\bar{z}_qz_p^{\prime}|\Big|\sum_{k=0}^{\infty}(k+1)^{\beta-n}(r^m\lambda z_1\cdots z_m)^k\Big|.\end{align*} $$

Moreover, the orthogonality of the holomorphic monomials in $L^2(\sigma )$ gives the following estimation:

$$ \begin{align*}I_{p,q}\asymp (1-r^m)^2\sum_{k=0}^{\infty}(k+1)^{2\beta-2n}(r^m\lambda)^{2k}\int_{{\mathbb B_n}}|\bar{z}_pz_q^{\prime}-\bar{z}_qz_p^{\prime}|^2|z_1\cdots z_m|^{2k}du(z).\end{align*} $$

It is easy to see that $|\bar {z}_pz_q^{\prime }-\bar {z}_qz_p^{\prime }|^2=|z_p|^2|z^{\prime }_q|^2+|z_q|^2|z^{\prime }_p|^2-2|z_1\cdots z_m|^2$ . Let us estimate the integral

$$ \begin{align*}\int_{{\mathbb B_n}}(|z_p|^2|z_q^{\prime}|^2-|z_1 \cdots z_m|^2)|z_1\cdots z_m|^{2k}du(z).\end{align*} $$

Passing through polar coordinates, we get, for $p\neq q$ , that

$$ \begin{align*} \int_{{\mathbb B_n}}|z_p|^2|z_q^{\prime}|^2&|z_1\cdots z_m|^{2k}du(z)\\ &\asymp 2n(n-1)!\frac{[(k+1)!]^{m-1}k!}{(mk+n+m-1)!}\frac{k+2}{2km+2n+2m} \end{align*} $$

and

$$ \begin{align*} \int_{{\mathbb B_n}}|z_1\cdots z_m|^{2(k+1)}&du(z)\\ &=2n(n-1)!\frac{[(k+1)!]^{m-1}k!}{(mk+n+m-1)!}\frac{k+1}{2km+2n+2m}. \end{align*} $$

Hence, we obtain

$$ \begin{align*} \int_{{\mathbb B_n}}(|z_p|^2|z_q^{\prime}|^2-|z_1 \cdots z_m|^2)&|z_1\cdots z_m|^{2k}du(z)\\ &\asymp \frac{[(k+1)!]^{m-1}k!}{(mk+n+m-2)!(k+1)^2}. \end{align*} $$

Again, applying the Stirling formula to the one above estimates, we obtain

$$ \begin{align*}I_{p,q}\asymp(1-r^m)^2 \sum_{k=0}^{\infty}(k+1)r^{2mk}.\end{align*} $$

This proves the assertion made about $I_{p,q}$ .

6 Sufficient conditions for noncyclicity via Cauchy transforms and $\alpha $ -capacities

We consider the Cauchy transform of a complex Borel measure $\mu $ on the unit sphere by

$$ \begin{align*}C_{[\mu]}(z)=\int_{\mathbb S_n}\frac{1}{(1-\langle z,\bar{\zeta}\rangle )^n}d\mu(\zeta), \quad z\in {\mathbb B_n}.\end{align*} $$

Note that this definition differs from the classical one.

Let $f\in D_{\alpha }({\mathbb B_n})$ and put a measure $\mu $ on $\mathcal {Z}(f^*)$ : the zero set in the sphere of the radial limits of f. The results in [Reference Sola21] about Cauchy transforms and noncyclicity are valid in our settings. We deduce that $[f] \neq D_{\alpha }({\mathbb B_n})$ , and hence noncyclicity, whenever $C_{[\mu ]}\in D_{-\alpha }({\mathbb B_n})$ . Thus, it is important to compute the Dirichlet-type norm of the Cauchy transform.

Let $\mu $ be a Borel measure on $\mathbb {S}_n$ and set $\mu ^*(j)=\int _{\mathbb S_n}\zeta ^jd\mu (\zeta )$ , $\bar {\mu }^*(j)=\int _{\mathbb S_n}\bar {\zeta }^jd\mu (\zeta )$ . We have the following.

Lemma 13 Let $\mu $ be a Borel measure on $\mathbb {S}_n$ . Then

$$ \begin{align*}||C_{[\mu]}||_{-\alpha}^2\asymp \sum_{k=0}^{\infty}\sum_{|j|=k}\frac{(k+1)^{n-1-\alpha}k!}{j!}|\bar{\mu}^*(j)|^2.\end{align*} $$

Proof Our Cauchy integral of $\mu $ on ${\mathbb B_n}$ has the following expansion:

$$ \begin{align*}C_{[\mu]}(z)=\sum_{k=0}^{\infty}\sum_{|j|=k}\frac{\Gamma(k+n)}{\Gamma(n)j!}\bar{\mu}^*(j)z^j.\end{align*} $$

Therefore, one can compute the norm of $C_{[\mu ]}$ in the space $D_{-\alpha }({\mathbb B_n})$ . The assertion follows.

The following lemma is crucial in the sequel. It is probably known, but we were not able to locate it in the literature, and hence we include its proof.

Lemma 14 Let $j_1,\ldots ,j_n,k$ be nonnegative integers satisfying $j_1+\cdots +j_n=nk$ . Then

$$ \begin{align*}j_1!\cdots j_n!\geq (k!)^n.\end{align*} $$

Proof The $\Gamma $ -function is logarithmically convex, and hence we may apply the Jensen inequality to it:

$$ \begin{align*}\log \Gamma\Big(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\Big)\leq \frac{\log \Gamma(x_1)}{n}+\cdots+\frac{\log \Gamma(x_n)}{n}.\end{align*} $$

Set $x_i:=j_i+1$ , $i=1,\ldots ,n$ . Since $j_1+\cdots +j_n=nk$ , the assertion follows.

We may identify noncyclicity for model polynomials via Cauchy transforms.

Lemma 15 Consider the polynomial $p(z)=1-m^{m/2} z_1\cdots z_m$ , where $1\leq m\leq n$ . Then p is not cyclic in $D_{\alpha }({\mathbb B_n})$ whenever $\alpha>\frac {2n+1-m}{2}$ .

Proof Recall that the model polynomials vanish in the closed unit ball along analytic sets of the form:

$$ \begin{align*}\mathcal{Z}(p)\cap \mathbb{S}_n=\{1/\sqrt{m}(e^{i\theta_1},\ldots,e^{i\theta_{m-1}}, e^{-i(\theta_1+\cdots+\theta_{m-1})},0,\ldots,0):\theta_i\in \mathbb{R}\}.\end{align*} $$

It is easy to see that for a proper measure $\mu $ , $\mu ^*(j)$ is nonzero, when $mj_m=k$ and $\mu ^*(j)\asymp m^{-k/2}$ . By Stirling’s formula and Lemma 14, we get that

$$ \begin{align*} ||C_{[\mu]}||_{-\alpha}^2&\leq C \sum_{k=0}^{\infty}\frac{(mk+1)^{n-1-\alpha}(mk)!}{(k!)^mm^{mk}}\asymp \sum_{k=0}^{\infty}(k+1)^{1/2(2n-m-1)-\alpha}. \end{align*} $$

Thus, p is not cyclic in $D_{\alpha }({\mathbb B_n})$ for $\alpha>\frac {2n+1-m}{2}$ .

We consider Riesz $\alpha $ -capacity for a fixed parameter $\alpha \in (0,n)$ with respect to the anisotropic distance in $\mathbb S_n$ given by

$$ \begin{align*}d(\zeta,\eta)=|1-\langle \zeta,\eta \rangle|^{1/2}\end{align*} $$

and the nonnegative kernel $K_{\alpha }:(0,\infty )\rightarrow [0,\infty )$ given by

$$ \begin{align*}K_{\alpha}(t)= \begin{cases} t^{\alpha-n}, & \alpha\in(0,n), \\ \log(e/t), & \alpha=n. \end{cases} \end{align*} $$

Note that we may extend the definition of K to $0$ by defining $K(0):=\lim _{t\rightarrow 0^+}K(t)$ .

Let $\mu $ be any Borel probability measure supported on some Borel set $E\subset \mathbb {S}_n$ . Then the Riesz $\alpha $ -energy of $\mu $ is given by

$$ \begin{align*} I_{\alpha}[\mu]=\iint_{\mathbb S_n}K_{\alpha}(|1-\langle \zeta,\eta \rangle|)d\mu(\zeta)d\mu(\eta) \end{align*} $$

and the Riesz $\alpha $ -capacity of E by

$$ \begin{align*} \mathrm{cap}_{\alpha}(E)=\inf\{I_{\alpha}[\mu]:\mu\in \mathcal{P}(E)\}^{-1}, \end{align*} $$

where $\mathcal {P}(E)$ is the set of all Borel probability measures supported on E. Note that if $\text {cap}_{\alpha }(E)>0$ , then there exist at least one probability measure supported on E having finite Riesz $\alpha $ -energy. Moreover, any $f\in D_{\alpha }({\mathbb B_n})$ has finite radial limits $f^*$ on $\mathbb {S}_n$ , except possibly, on a set E having $\text {cap}_{\alpha }(E)=0$ . Theory regarding to the above standard construction in potential theory can be found in [Reference Ahern and Cohn1, Reference Cohn and Verbitsky9, Reference El-Fallah, Kellay, Mashreghi and Ransford11, Reference Pestana and Rodríguez17].

The relation between noncyclicity of a function and the Riesz $\alpha $ -capacity of the zeros of its radial limits follows.

Theorem 16 Fix $\alpha \in (0,n]$ and let $f\in D_{\alpha }({\mathbb B_n})$ . If $\mathrm {cap}_{\alpha }(\mathcal {Z}(f^*))>0$ , then f is not cyclic in $D_{\alpha }({\mathbb B_n})$ .

Proof Let $\mu $ be a probability measure supported in $\mathcal {Z}(f^*)$ , with finite Riesz n-energy. If $r\in (0,1)$ , then

$$ \begin{align*} \log\frac{e}{|1-r\langle \zeta,\eta\rangle|}&=1+\textrm{Re}\Big(\log\frac{1}{1-r\langle \zeta,\eta\rangle}\Big)\\ &=1+\textrm{Re} \sum_{k=1}^{\infty}\sum_{|j|=k}\frac{r^kk!}{kj!}\zeta^j\overline{\eta}^j. \end{align*} $$

Note that $\mu $ is a probability measure, and hence

$$ \begin{align*} \iint_{\mathbb S_n}\log\frac{e}{|1-r\langle \zeta,\eta\rangle|}d\mu(\zeta)d\mu(\eta) =1+ \sum_{k=1}^{\infty}\sum_{|j|=k}\frac{r^kk!}{kj!}|\mu^*(j)|^2. \end{align*} $$

Since $|1-w|/|1-rw|\leq 2$ for $r\in (0,1)$ and $w\in \overline {{\mathbb D}}$ , the dominated convergence theorem and Lemma 13 give

$$ \begin{align*}||C_{[\mu]}||^2_{-n}\asymp\sum_{k=1}^{\infty}\sum_{|j|=k}\frac{k!}{kj!}|\mu^*(j)|^2<\infty.\end{align*} $$

The assertion follows.

We continue setting a probability measure $\mu $ , supported in $\mathcal {Z}(f^*)$ , with finite Riesz $\alpha $ -energy, where $\alpha \in (0,n)$ . If $r\in (0,1)$ , then

$$ \begin{align*} \frac{1}{(1-r\langle \zeta,\eta\rangle)^{n-\alpha}}&=\sum_{k=0}^{\infty}\sum_{|j|=k} \frac{\Gamma(k+n-\alpha)k!r^k}{k!\Gamma(n-\alpha)j!}\zeta^j\overline{\eta}^j. \end{align*} $$

Similar arguments to the one above show that

$$ \begin{align*} I_{\alpha}[\mu]&\geq\Big|\iint_{\mathbb S_n}\textrm{Re}\Big(\frac{1}{(1-r\langle \zeta,\eta\rangle)^{n-\alpha}}\Big)d\mu(\zeta)d\mu(\eta)\Big|\\ &=\Big|\sum_{k=0}^{\infty}\sum_{|j|=k} \frac{\Gamma(k+n-\alpha)k!r^k}{k!\Gamma(n-\alpha)j!}\iint_{\mathbb S_n}\zeta^j\overline{\eta}^jd\mu(\zeta)d\mu(\eta)\Big|\\ &\asymp \sum_{k=0}^{\infty}\sum_{|j|=k}\frac{(k+1)^{n-1-\alpha}k!}{j!}r^k|\mu^*(j)|^2. \end{align*} $$

Again, letting $r\rightarrow 1^{-}$ by Lemma 13, we obtain that $C_{[\mu ]}\in D_{-\alpha }({\mathbb B_n})$ . The assertion follows.