Notation 8 Let $k=(k_1,\dots ,k_n)\in \mathbb {Z}_{+}^n$ be a multi-index and let $z=(z_1,\dots ,z_n)\in \mathbb {C}^n$ . We write

$$\begin{align*}|k| = k_1 + \cdots + k_n, \qquad k! = k_1! \ldots k_n!, \qquad z^k = z_1^{k_1} \ldots z_n^{k_n}. \end{align*}$$

If $\gamma =(\gamma _1,\dots ,\gamma _n)\in \mathbb {R}^n$ , then with a slight abuse of notation, we write

$$\begin{align*}|z^\gamma| = |z_1|^{\gamma_1} \ldots |z_n|^{\gamma_n}. \end{align*}$$

Proof If we define $\Gamma _\zeta : A^2_{u,\alpha } \rightarrow \mathbb {C}$ by $\Gamma _\zeta (f) = f(\zeta )$ , then $\Gamma _\zeta $ is a bounded linear functional and consequently has a closed kernel. For any polynomial p, we have ${pf \in \mathrm {Ker} \Gamma _\zeta }$ . This implies

$$ \begin{align*} [f] \subseteq \mathrm{Ker} \Gamma_\zeta \subsetneq A^2_{u,\alpha}.\\[-37pt] \end{align*} $$

Proof Since $1/f_r$ is holomorphic on a neighborhood of $\overline {\Omega }$ , it is a multiplier by Remark 4, and (9) gives

$$\begin{align*}\left\{\frac{f}{f_r}:0<r<1\right\}\subseteq [f]. \end{align*}$$

By the hypothesis, this family is bounded in $A^2_{u,\alpha }(\Omega )$ . Choose $r_j\uparrow 1$ . The Banach–Alaoglu theorem and the reflexivity of $A^2_{u,\alpha }$ yield a weakly convergent sequence $\{ \frac {f}{f_{r_{j_k}}}\}_{k \geq 0}$ . Since point evaluations are bounded and

$$\begin{align*}\frac{f(z)}{f(r_{j_k}z)}\longrightarrow1 \qquad (z\in\Omega), \end{align*}$$

the weak limit is $1$ . Thus,

$$\begin{align*}1\in \overline{\left\{\frac{f}{f_r}:0<r<1\right\}}^{\,\mathrm{weak}}. \end{align*}$$

Finally, $[f]$ is a norm-closed linear subspace, and hence it is weakly closed by Mazur’s lemma. Therefore, $1\in [f]$ , and f is cyclic.

Proof Since g is cyclic, there exist polynomials $p_j$ such that $p_jg\to 1$ in $A^2_{u,\alpha }(\Omega )$ . Thus,

$$\begin{align*}\|p_jfg-f\|_{A^2_{u,\alpha}} \leq \|f\|_{\mathrm{Mult}(A^2_{u,\alpha})} \|p_jg-1\|_{A^2_{u,\alpha}}\longrightarrow0. \end{align*}$$

Hence, $f\in [fg]$ , and therefore $A^2_{u,\alpha }(\Omega )=[f]\subseteq [fg]$ .

Proof Fix $r<0$ . Following [Reference Demailly8, Chapter III, Example 6.11], we work in local logarithmic coordinates. At $z\in S_r(u)$ , only the terms satisfying $p_j(z)=e^r$ contribute locally to $u^{(r)}$ . If the corresponding vectors $\beta ^j$ have rank less than n, a linear change of coordinates shows that $u^{(r)}$ depends on fewer than n complex variables, so $(dd^c u^{(r)})^n=0$ there.

Thus, $\mu _{u^{(r)}}$ is supported on finitely many tori $S_{r,q}$ , each determined by n linearly independent equations $p_j=e^r$ . By rotation invariance, $\left .\mu _{u^{(r)}}\right |{}_{S_{r,q}}=C_q(r)\mu _q^r$ , where $\mu _q^r$ is the uniform probability measure on $S_{r,q}$ . Since $\sum _i\beta _i^j=1$ , we have $p_j(tz)=tp_j(z)$ . Consequently, for $r,s<0$ , the dilation $T(z)=e^{s-r}z$ maps $S_{r,q}$ onto $S_{s,q}$ and satisfies

$$\begin{align*}u^{(s)}\circ T=u^{(r)}+s-r \end{align*}$$

near $S_r(u)$ . Since the Monge–Ampère operator commutes with biholomorphic pullback, the definition of $\mu _{u^{(r)}}$ gives

$$\begin{align*}T_*\mu_{u^{(r)}}=\mu_{u^{(s)}},\qquad T_*\mu_q^r=\mu_q^s. \end{align*}$$

Hence, $C_q(r)=C_q(s)$ , and we may write

(11) $$ \begin{align} \mu_{u^{(r)}}=\sum_{q=1}^N C_q\mu_q^r, \end{align} $$

where $C_q>0$ is independent of r. The fact that $\mu _{u^{(r)}}(S_r(u))=1$ implies

$$\begin{align*}\sum_{q=1}^N C_q=1. \end{align*}$$

Given $S_{r,q}$ , choose indices $\sigma _q(j)$ , $j=1,\ldots ,n$ , corresponding to linearly independent equations defining this torus. To determine $S_{r,q}$ , we solve the following linear system:

$$\begin{align*}\begin{bmatrix} \beta_1^{\sigma_q(1)} & \ldots & \beta_n^{\sigma_q(1)} \\ \vdots & \ddots & \vdots \\ \beta_1^{\sigma_q(n)} & \ldots & \beta_n^{\sigma_q(n)} \end{bmatrix} \begin{bmatrix} \log |z_1| \\ \vdots \\ \log |z_n| \end{bmatrix} = \begin{bmatrix} r \\ \vdots \\ r \end{bmatrix} - \begin{bmatrix} \log \lambda_{\sigma_q(1)} \\ \vdots \\ \log \lambda_{\sigma_q(n)} \end{bmatrix}. \end{align*}$$

A direct application of Cramer’s rule shows that for every $z \in S_{r,q}$ ,

$$\begin{align*}|z_i|=C^{\prime}_{i,q}e^r, \end{align*}$$

where each $C^{\prime }_{i,q}$ is independent of r. Thus, for a monomial $z^L$ and $\alpha>-1$ ,

$$\begin{align*}\begin{aligned} \int_{-\infty}^{0}|r|^\alpha e^{2nr} \mu_q^r(|z^L|^2)\,dr &= \prod_{i=1}^n(C^{\prime}_{i,q})^{2l_i} \int_{-\infty}^{0}|r|^\alpha e^{2(n+|L|)r}\,dr\\ &= \prod_{i=1}^n(C^{\prime}_{i,q})^{2l_i} \frac{\Gamma(\alpha+1)} {[2(|L|+n)]^{\alpha+1}}. \end{aligned} \end{align*}$$

Consequently, recalling (6) and (11),

$$\begin{align*}\|z^L\|_{A^2_{u,\alpha}(\Omega)}^2 =C_L\left(\frac{n}{|L|+n}\right)^{\alpha+1}, \qquad C_L=\sum_{q=1}^N C_q\prod_{i=1}^n(C^{\prime}_{i,q})^{2l_i}. \end{align*}$$

Since $\|z^L\|_{H^2_u}^2=C_L$ , this formula also holds for $\alpha =-1$ . As $C_L$ is independent of $\alpha $ , taking ratios establishes the PSE property.

Proof of Theorem 23

Consider the mapping $L:\Omega \cap (\mathbb C\setminus \{0\})^n \to \Omega '\subset \mathbb R^n$ defined by

$$ \begin{align*} L(z_1,\ldots,z_n) = (\log |z_1|, \ldots, \log |z_n|). \end{align*} $$

For $x = (x^{(1)},\ldots ,x^{(n)}) \in \Omega '$ , let

$$ \begin{align*} F_0(x) = \max \{x^{(1)}, \ldots, x^{(n)} \}. \end{align*} $$

Given a dense sequence $\{x_i\}_{i \in \mathbb {N}}$ in $\partial \Omega '$ , define the iterative sequence

$$ \begin{align*} F_i = \max \{ F_{i-1}, f_i \}, \end{align*} $$

where $f_i(x) = x \cdot v_i + \lambda _i$ with $(1,\ldots ,1) \cdot v_i = 1$ , and $v_i \in [0,1]^n$ is such that ${f_i(x_i) = 0}$ and $f_i(x) < 0$ for all $x \in \Omega '$ . Then, the sequence $g_i = F_i \circ L$ consists of increasing plurisubharmonic functions on $\Omega $ converging to an exhaustion function u for $\Omega $ . So, by Theorem 3.6.1 in [Reference Klimek14], we have

$$ \begin{align*} \| z^L \|_{A^2_{g_i,\alpha}} \to \| z^L \|_{A^2_{u,\alpha}} \end{align*} $$

for any multi-index L and $\alpha \in [-1, 0]$ . Applying Proposition 28 completes the proof.

Notation 29 Theorem 23 allows us to define the Poletsky–Stessin weighted Bergman spaces on a wide range of domains and parameter values of $\alpha $ . From now on, when there is no ambiguity in the choice of the exhaustion function, we denote the space $A^2_{u,\alpha }(\Omega )$ by $D_{\beta }(\Omega )$ , where $\beta = -(\alpha +1)$ .

Proof Since $\alpha +\alpha '=-2$ , (10) gives

$$\begin{align*}\|z^L\|_{A^2_{u,\alpha}}^2 \|z^L\|_{A^2_{u,\alpha'}}^2=\|z^L\|_{H^2_u}^4. \end{align*}$$

For $f=\sum _L a_Lz^L\in A^2_{u,\alpha }(\Omega )$ and $g=\sum _L b_Lz^L\in A^2_{u,\alpha '}(\Omega )$ , define

$$\begin{align*}\langle f,g\rangle_u =\sum_L a_L\overline{b_L}\,\|z^L\|_{H^2_u}^2. \end{align*}$$

By Cauchy–Schwarz,

$$\begin{align*}|\langle f,g\rangle_u| \leq \|f\|_{A^2_{u,\alpha}}\|g\|_{A^2_{u,\alpha'}}. \end{align*}$$

Conversely, let $\Lambda \in (A^2_{u,\alpha }(\Omega ))^*$ . By the Riesz representation theorem, there exists ${h=\sum _L c_Lz^L\in A^2_{u,\alpha }(\Omega )}$ such that

$$\begin{align*}\Lambda(f)=\sum_L a_L\overline{c_L} \|z^L\|_{A^2_{u,\alpha}}^2, \qquad \|h\|_{A^2_{u,\alpha}}=\|\Lambda\|. \end{align*}$$

Set

$$\begin{align*}b_L=c_L\frac{\|z^L\|_{A^2_{u,\alpha}}^2} {\|z^L\|_{H^2_u}^2}. \end{align*}$$

Then,

$$\begin{align*}\sum_L |b_L|^2\|z^L\|_{A^2_{u,\alpha'}}^2 =\sum_L |c_L|^2\|z^L\|_{A^2_{u,\alpha}}^2 =\|\Lambda\|^2. \end{align*}$$

Hence, $g=\sum _L b_Lz^L\in A^2_{u,\alpha '}(\Omega )$ represents $\Lambda $ under the above pairing.

Proof Put $E=\mathcal Z(f)\cap \mathrm {supp}(\mu _u)$ . By the definition of capacity, there exists ${\mu \in \mathcal P(E)}$ such that

$$\begin{align*}I_\alpha[\mu] =\iint_{E\times E}k_\alpha(z,\zeta)\, d\mu(z)\,d\mu(\zeta)<\infty. \end{align*}$$

Expanding the kernel, we obtain

$$\begin{align*}I_\alpha[\mu] =\sum_L \frac{\left|\int_E z^L\,d\mu(z)\right|^2} {\|z^L\|_{A^2_{u,\alpha}}^2}. \end{align*}$$

Thus, by Cauchy–Schwarz, for every polynomial q,

$$\begin{align*}\left|\int_E q\,d\mu\right|^2 \leq \|q\|_{A^2_{u,\alpha}}^2 I_\alpha[\mu]. \end{align*}$$

By polynomial density, $\Lambda (q)=\int _E q\,d\mu $ extends to a bounded linear functional on $A^2_{u,\alpha }(\Omega )$ , with $\Lambda (1)=1$ .

Let p be a polynomial and put $h=pf$ . For each $0<r<1$ , we have $\Lambda (h_r)=\int _E h_r\,d\mu $ . Since $h\in \mathcal C(\overline \Omega )$ and $h=0$ on E,

$$\begin{align*}\Lambda(pf) =\lim_{r\to1^-}\Lambda(h_r) =\lim_{r\to1^-}\int_E h(r\zeta)\,d\mu(\zeta) =0. \end{align*}$$

Hence, $[f]\subseteq \ker \Lambda \neq A^2_{u,\alpha }(\Omega )$ , and f is not cyclic.

Proof Let J be a multi-index $f(z) = \sum _J a_J z^J \in D_{\beta }(\Omega _p)$ ; and applying the Hölder inequality on $|\sum _J a_J \zeta ^J|^2$ , we will have

$$ \begin{align*} \left| \sum_J a_J \zeta^J \right|^2 &\leq \left( \sum_J |a_J|^2 \frac{\Gamma\left(\frac{j_1}{p}+1\right) \Gamma\left(\frac{j_2}{p}+1\right)}{\Gamma\left(\frac{j_1 + j_2}{p}+2\right)} (|J|+2)^\beta \right) \\ &\quad \cdot \left( \sum_J \frac{1}{(|J|+2)^\beta} \frac{\Gamma\left(\frac{j_1 + j_2}{p}+2\right)}{\Gamma\left(\frac{j_1}{p}+1\right) \Gamma\left(\frac{j_2}{p}+1\right)} |\zeta^J|^2 \right) \\ &= \|f\|^2_{D_{\beta}(\Omega_p)} \left( \sum_J \frac{1}{(|J|+2)^\beta} \frac{\Gamma\left(\frac{j_1 + j_2}{p}+2\right)}{\Gamma\left(\frac{j_1}{p}+1\right) \Gamma\left(\frac{j_2}{p}+1\right)} |\zeta^J|^2 \right). \\ & \asymp \|f\|^2_{D_{\beta}(\Omega_p)} \left( \sum_{j \geq 0} \frac{1}{(j+2)^{\beta - 1}} \sum_{j_1+j_2 = j}\frac{\Gamma\left(\frac{j_1 + j_2}{p}+1\right)}{\Gamma\left(\frac{j_1}{p}+1\right) \Gamma\left(\frac{j_2}{p}+1\right)} |\zeta_1|^{2j_1} |\zeta_2|^{2j_2} \right). \end{align*} $$

Now, the claim is that

(12) $$ \begin{align} S(j) = \sum_{j_1+j_2 = j} \frac{\Gamma\left(\frac{j_1 + j_2}{p}+1\right)}{\Gamma\left(\frac{j_1}{p}+1\right) \Gamma\left(\frac{j_2}{p}+1\right)} |\zeta_1|^{2j_1} |\zeta_2|^{2j_2} &\leq C, \end{align} $$

where $C>0$ is a positive constant independent of j. Note that this finishes the proof as we will have

$$ \begin{align*} \left| \sum_J a_J \zeta^J \right|^2 &\lesssim \|f\|^2_{D_{\beta}(\Omega_p)} \left( \sum_{j \geq 0} \frac{1}{(j+2)^{\beta - 1}} \right). \end{align*} $$

And the sum is finite when $\beta> 2$ .

To prove the claim, we compare the sum in (12) with the following integral:

$$ \begin{align*} I(\lambda) & = \int_0^{\lambda} \frac{\Gamma(\lambda+1)}{\Gamma(x+1) \Gamma(\lambda - x+1)} r^x (1-r)^{\lambda - x} \,dx \\ &\lesssim \int_0^{\lambda} \frac{\lambda^\lambda}{x^x (\lambda - x)^{\lambda - x}} \sqrt{\frac{\lambda}{x (\lambda - x)}} r^x (1-r)^{\lambda - x} \,dx \\ &= \sqrt{\lambda} \int_0^{1} \left( \frac{1}{y^y (1-y)^{1 - y}} r^y (1-r)^{1-y} \right)^{\lambda} \sqrt{\frac{1}{y (1-y)}} \,dy \\ &= \sqrt{\lambda} \int_0^{1} e^{-\lambda h(y)} g(y) \,dy, \end{align*} $$

where $\lambda = \frac {j}{p}$ , $r = |\zeta _1|^{2p}$ , $h(y) = y \log (\frac {y}{r}) + (1-y) \log (\frac {1-y}{1-r})$ , and $g(y) = \sqrt {\frac {1}{y(1-y)}}$ . Note that in the case of $r = 0$ or $r = 1$ , $S(j) = 1$ and we get what we want; so we assume $r \in (0,1)$ . Let $a_\lambda $ denote the integrand defining $I(\lambda )$ . By convexity of $\log \Gamma $ , $a_\lambda $ is unimodal, and so

$$\begin{align*}\frac1p\sum_{k=0}^j a_\lambda(k/p) \leq \int_0^\lambda a_\lambda(x)\,dx +\frac2p\max_{[0,\lambda]}a_\lambda. \end{align*}$$

By Stirling’s inequalities, uniformly for $\lambda \geq 1$ ,

$$\begin{align*}a_\lambda(x)\lesssim \sqrt{\frac{\lambda+1}{(x+1)(\lambda-x+1)}} e^{-\lambda h(x/\lambda)}\leq1 \end{align*}$$

since $h\geq 0$ , with its continuous endpoint values. Thus, $S(j)\lesssim I(\lambda )+1$ .

Fix $0<\epsilon <\min \{r,1-r\}$ . Since h has its unique minimum at r, with $h(r)=0$ and $h"(r)=g(r)^2$ , Lemma 35 gives

$$\begin{align*}\sqrt{\lambda}\int_\epsilon^{1-\epsilon} e^{-\lambda h(y)}g(y)\,dy\longrightarrow\sqrt{2\pi}. \end{align*}$$

On the remaining intervals, $h\geq c:=\min \{h(\epsilon ),h(1-\epsilon )\}>0$ ; so,

$$\begin{align*}\sqrt{\lambda} \left(\int_0^\epsilon e^{-\lambda h(y)}g(y)\,dy + \int_{1-\epsilon}^1 e^{-\lambda h(y)}g(y)\,dy \right) \leq\sqrt{\lambda}e^{-c\lambda}\int_0^1g(y)\,dy \longrightarrow0. \end{align*}$$

Hence, $I(\lambda )$ , and therefore $S(j)$ remains bounded.

Proof

$$ \begin{align*} \begin{aligned} \sup_{0<r<1} \left\| \frac{f}{f_r} \right\|^2_{D_{\beta}(\Omega_p)} & =\sup_{0<r<1} \big\| (1 - r_1^{2p-1} z_1 - r_2^{2p-1} z_2) (1 + r(r_1^{2p-1} z_1 \\ &\quad + r_2^{2p-1} z_2) + r^2(r_1^{2p-1} z_1 + r_2^{2p-1} z_2)^2 + \cdots) \big\|^2_{D_{\beta}(\Omega_p)} \\ & =\sup_{0<r<1} \left\| 1 + (1-r) \sum_{j \geq 1} r^{j-1} (r_1^{2p-1} z_1 + r_2^{2p-1} z_2)^j \right\|^2_{D_{\beta}(\Omega_p)} \\ & =\sup_{0<r<1} \left\| 1 + (1-r) \sum_{j \geq 1} r^{j-1} \sum_{j_1 + j_2 = j} \binom{j}{j_1} r_1^{(2p-1)j_1} z^{j_1}_1 r_2^{(2p-1)j_2} z_2^{j_2} \right\|^2_{D_{\beta}(\Omega_p)} \\ & = 1 + \sup_{0<r<1}(1-r)^2 \sum_{j \geq 1} r^{2(j-1)} \sum_{j_1 + j_2 = j} \binom{j}{j_1}^2 \frac{\Gamma\left(\frac{j_1}{p}+1\right) \Gamma\left(\frac{j_2}{p}+1\right)} {\Gamma\left(\frac{j}{p}+2\right)} \\ & \quad \times (j+2)^{\beta} r_1^{2(2p-1)j_1} r_2^{2(2p-1)j_2}. \end{aligned} \end{align*} $$

After applying Stirling’s approximation to

$$\begin{align*}\binom{j}{j_1} \frac{\Gamma\left(\frac{j_1}{p}+1\right)\Gamma\left(\frac{j_2}{p}+1\right)} {\Gamma\left(\frac{j}{p}+2\right)} (j+2), \end{align*}$$

we continue as follows:

$$ \begin{align*} \begin{aligned} \sup_{0<r<1} \left\| \frac{f}{f_r} \right\|^2_{D_{\beta}(\Omega_p)} &\asymp 1 + \sup_{0<r<1}(1-r)^2 \sum_{j \geq 1} r^{2(j-1)} (j+2)^{\beta - 1} \\ & \quad \times \sum_{j_1 + j_2 = j} \binom{j}{j_1} r_1^{2p j_1} r_2^{2p j_2} \frac{j^{j\left(1-\frac{1}{p}\right)}} {j_1^{j_1\left(1-\frac{1}{p}\right)} j_2^{j_2\left(1-\frac{1}{p}\right)}} r_1^{2(p-1)j_1} r_2^{2(p-1)j_2} \\ &= 1 + \sup_{0<r<1}(1-r)^2 \sum_{j \geq 1} r^{2(j-1)} (j+2)^{\beta - 1} \\ & \quad \times \sum_{j_1 + j_2 = j} \binom{j}{j_1} r_1^{2p j_1} r_2^{2p j_2} \left( \frac{r_1^{2(p-1)\frac{j_1}{j}} r_2^{2(p-1)\frac{j_2}{j}}} {\left(\frac{j_1}{j} \right)^{\frac{j_1}{j}\left(1-\frac{1}{p}\right)} \left(\frac{j_2}{j} \right)^{\frac{j_2}{j}\left(1-\frac{1}{p}\right) }} \right)^{j}. \end{aligned} \end{align*} $$

Note that the function

$$\begin{align*}F(x) = \frac{r_1^{2(p-1)x} r_2^{2(p-1)(1-x)}} {x^{x\left(1-\frac{1}{p}\right)} (1-x)^{(1-x)\left(1-\frac{1}{p}\right)}} \end{align*}$$

in $[0,1]$ attains its maximum at $x = r_1^{2p}$ . Therefore,

$$ \begin{align*} \begin{aligned} &\sup_{0<r<1} \left\| \frac{f}{f_r} \right\|^2_{D_{\beta}(\Omega_p)} \\ &\lesssim 1 + \sup_{0<r<1}(1-r)^2 \sum_{j \geq 1} r^{2(j-1)} (j+2)^{\beta - 1} \sum_{j_1 + j_2 = j} \binom{j}{j_1} r_1^{2p j_1} r_2^{2p j_2} \\ &= 1 + \sup_{0<r<1}(1-r)^2 \sum_{j \geq 1} r^{2(j-1)} (j+2)^{\beta - 1} \\ &\leq 1+\sup_{0<r<1}(1-r)^2 \sum_{j\geq1}r^{2(j-1)}(j+2) =1+\sup_{0<r<1}\frac{3-2r^2}{(1+r)^2}<+\infty, \qquad \beta\leq2. \end{aligned} \end{align*} $$

Now, Theorem 21 finishes the proof.

Proof of Theorem 34

Let $\{\zeta _j\}_{j=1}^k=\mathcal Z(q)\cap \partial \Omega _p$ , and let

$$\begin{align*}s_j(z)=1-\sum_{\ell=1}^2 |\zeta_{j,\ell}|^{2p-2}\overline{\zeta_{j,\ell}}z_\ell, \end{align*}$$

with zero coefficients at zero coordinates. Up to coordinatewise rotations, these are the polynomials in Proposition 37, vanishing on $\partial \Omega _p$ only at $\zeta _j$ . Since $s_j(\zeta _j)=0$ , Cauchy–Schwarz gives

$$\begin{align*}\mathrm{dist}(z,\zeta_j)\geq C_j|s_j(z)|, \qquad C_j=\left(\sum_{\ell=1}^2 |\zeta_{j,\ell}|^{4p-2}\right)^{-1/2}. \end{align*}$$

On the other hand, the Łojasiewicz’s inequality [Reference Krantz and Parks17] gives a positive integer m and a constant $C>0$ such that

$$\begin{align*}|q(z)|\geq C\mathrm{dist} (z,\mathcal Z(q)\cap\partial\Omega_p)^m, \end{align*}$$

for any $z\in \partial \Omega _p$ . Choosing $\zeta _j$ nearest to z and bounding the remaining factors by their supremum norms, we obtain

$$\begin{align*}|q(z)|\geq C'\prod_{j=1}^k|s_j(z)|^m, \qquad C'=C\left(\min_{1\leq j\leq k} \frac{C_j}{\prod_{i\ne j}\|s_i\|_{\infty}}\right)^m>0, \end{align*}$$

for any $z\in \partial \Omega _p$ , where the supremum norms are taken over $\partial \Omega _p$ . By increasing m sufficiently, we may assume that

$$\begin{align*}Q(z) = \frac{\prod_{j = 1}^k s_j(z)^m}{q(z)} \in D_\beta(\Omega_p). \end{align*}$$

Assume $\beta \leq 2$ . By Proposition 37, each $s_j$ is cyclic in $D_\beta (\Omega _p)$ . Since the $s_j$ are cyclic polynomials and hence multipliers, Lemma 22 implies that $qQ=\prod _{j=1}^k s_j^m$ is cyclic. Fix $\epsilon> 0$ , and let $p_1$ and $p_2$ be polynomials such that

$$\begin{align*}\| p_1 q Q - 1 \|_{D_\beta} < \frac{\epsilon}{2} \end{align*}$$

and

$$\begin{align*}\| p_2 - Q \|_{D_\beta} < \| p_1 q \|^{-1}_{\mathrm{Mult}(D_\beta)} \frac{ \epsilon }{2}, \end{align*}$$

where $\| p_1 q \|_{\mathrm {Mult}(D_\beta )}$ denotes the multiplier norm of $p_1 q$ , and the inequality follows from the density of polynomials in $D_\beta $ . Then, we have

$$ \begin{align*} \| p_1 p_2 q - 1 \|_{D_\beta} \leq \| p_1 p_2 q - p_1 q Q \|_{D_\beta} + \| p_1 q Q - 1 \|_{D_\beta} < \epsilon. \end{align*} $$

Conversely, if $\beta>2$ , choose $\zeta \in \mathcal Z(q)\cap \partial \Omega _p$ . By Proposition 36, evaluation at $\zeta $ is bounded on $D_\beta (\Omega _p)$ . Therefore, q is not cyclic by Remark 6.

Proof Let $\epsilon> 0$ and take $u(z) = \log \max \{|z_1|,\ldots ,|z_n| \}$ . Also, let $f(z) = \sum _{|L|\geq 0} (l_1+1)^{\frac {\alpha -\epsilon }{2}} \ldots (l_n+1)^{\frac {\alpha -\epsilon }{2}} z^L$ . Then, one can write

$$ \begin{align*} \begin{aligned} ||f||^2_{A^2_{\alpha}(\mathbb{D}^n)} = \sum_{L} \frac{1}{(l_1+1)^{1+\epsilon}} \ldots \frac{1}{(l_n+1)^{1+\epsilon}} < \infty. \end{aligned} \end{align*} $$

So, $f \in A^2_{\alpha }(\mathbb {D}^n)$ . Also,

(13) $$ \begin{align} \begin{aligned} ||f||^2_{A^2_{u,\alpha}(\mathbb{D}^n)} = \sum_L \frac{(l_1+1)^{\alpha-\epsilon} \ldots (l_n+1)^{\alpha-\epsilon}}{(|L|+n)^{\alpha + 1}} &\geq \\ \sum_{l \geq 0} \frac{(l+1)^{\alpha-\epsilon} \ldots (l+1)^{\alpha-\epsilon}}{(n l+n)^{\alpha + 1}} &\gtrsim \\ \sum_{l \geq 0} (l+1)^{(n-1)\alpha - n \epsilon - 1}&. \end{aligned} \end{align} $$

Note that the right-hand side of (13) diverges whenever $\alpha>0$ and $\epsilon = \frac {n-1}{n} \alpha $ ; so, for such a case, $f \not \in A^2_{u,\alpha }(\mathbb {D}^n)$ . Now, Proposition 39 finishes the proof.

Proof Let J be a multi-index $f(z) = \sum _J a_J z^J \in D_{\beta }(\mathbb {D}^n)$ , applying the Hölder inequality on $|\sum _J a_J \zeta ^J|^2$ , we will have

$$ \begin{align*} \left| \sum_J a_J \zeta^J \right|^2 &\leq \left( \sum_J |a_J|^2 (|J|+n)^\beta \right) \cdot \left( \sum_J \frac{1}{(|J|+n)^\beta} |\zeta^J|^2 \right) \\ &= \|f\|^2_{D_{\beta}(\mathbb{D}^n)} \left( \sum_J \frac{1}{(|J|+n)^\beta} \right). \\ \end{align*} $$

The last sum on the right-hand side is finite. Indeed, grouping the terms according to $j=|J|$ , we obtain

$$\begin{align*}\sum_{J\in\mathbb N^n}\frac{1}{(|J|+n)^\beta} = \sum_{j\geq0} \binom{j+n-1}{n-1}(j+n)^{-\beta} \lesssim \sum_{j\geq1}j^{n-1-\beta}<+\infty, \end{align*}$$

since $\beta>n$ . This completes the proof.

Proof Let $f(z) = \sum _{|L| \geq 0} a_L z^L$ and $g(z) = \sum _{|L| \geq 0} b_L z^L$ be holomorphic functions in $D_{\beta }(\mathbb {D}^n)$ . Their product is

$$\begin{align*}f(z)g(z) = \sum_{|L| \geq 0} \sum_{L_1 + L_2 = L} a_{L_1} b_{L_2} z^L. \end{align*}$$

The squared norm of $ f \cdot g $ is

$$\begin{align*}\|f \cdot g\|_{D_{\beta}(\mathbb{D}^n)}^2 = \sum_{|L|\geq 0} (|L| + n)^{\beta} \left| \sum_{L_1 + L_2 = L} a_{L_1} b_{L_2} \right|^2. \end{align*}$$

Applying Hölder’s inequality,

$$ \begin{align*} \left| \sum_{L_1 + L_2 = L} a_{L_1} b_{L_2} \right|^2 &\leq \left( \sum_{L_1 + L_2 = L} \frac{1}{(|L_1| + n)^{\beta} (|L_2| + n)^{\beta}} \right) \\ &\quad \cdot \left( \sum_{L_1 + L_2 = L} |a_{L_1}|^2 (|L_1| + n)^{\beta} |b_{L_2}|^2 (|L_2| + n)^{\beta} \right). \end{align*} $$

Thus,

$$ \begin{align*} \begin{aligned} \|f \cdot g\|^2_{D_{\beta}(\mathbb{D}^n)} &\leq \sum_{L} (|L| + n)^{\beta} \left( \sum_{L_1 + L_2 = L} \frac{1}{(|L_1| + n)^{\beta} (|L_2| + n)^{\beta}} \right) \\ &\quad \cdot\left( \sum_{L_1 + L_2 = L} |a_{L_1}|^2 (|L_1| + n)^{\beta} |b_{L_2}|^2 (|L_2| + n)^{\beta} \right)\\&\leq \sum_{|L| \geq 0} \left( \frac{|L| + n}{|L| + 2n} \right)^{\beta} \sum_{L_1 + L_2 = L} \left( \frac{1}{(|L_1| + n)} + \frac{1}{(|L_2| + n)} \right)^\beta \\& \quad\cdot \left( \sum_{L_1 + L_2 = L} |a_{L_1}|^2 (|L_1| + n)^{\beta} |b_{L_2}|^2 (|L_2| + n)^{\beta} \right) \\ &\leq \sum_{|L| \geq 0} \sum_{L_1 + L_2 = L} \left( \frac{1}{(|L_1| + n)} + \frac{1}{(|L_2| + n)} \right)^\beta \\ &\quad \cdot \left( \sum_{L_1 + L_2 = L} |a_{L_1}|^2 (|L_1| + n)^{\beta} |b_{L_2}|^2 (|L_2| + n)^{\beta} \right) \\ &\leq C_{\beta} \sum_{|L| \geq 0} \sum_{L_1 + L_2 = L} \left( \frac{1}{(|L_1| + n)^\beta} + \frac{1}{(|L_2| + n)^\beta} \right) \\ & \quad \cdot \left( \sum_{L_1 + L_2 = L} |a_{L_1}|^2 (|L_1| + n)^{\beta} |b_{L_2}|^2 (|L_2| + n)^{\beta} \right) \end{aligned} \end{align*} $$

$$ \begin{align*} \begin{aligned} &\leq 2 C_{\beta} \left( \sum_{|L| \geq 0} \frac{1}{(|L| + n)^\beta} \right) \\ & \quad \cdot \left( \sum_{|L| \geq 0} \sum_{L_1 + L_2 = L} |a_{L_1}|^2 (|L_1| + n)^{\beta} |b_{L_2}|^2 (|L_2| + n)^{\beta} \right) \\ & \leq 2 C_{\beta} \sum_{|L| \geq 0} \frac{1}{(|L| + n)^\beta} \|f\|_{D_{\beta}(\mathbb{D}^n)}^2 \|g\|_{D_{\beta}(\mathbb{D}^n)}^2, \end{aligned} \end{align*} $$

where $C_\beta = 2^{\beta - 1}$ is obtained from Jensen’s inequality. So, for $\beta> n$ , $ f \cdot g \in A^2_{u,\alpha }(\mathbb {D}^n) $ .

Proof Proposition 41 implies that for $\beta> 2$ , the point evaluation functional is bounded at any point of $\overline {\mathbb D}^2$ . Therefore, by Remark 6, a cyclic function in $D_\beta (\mathbb D^2)$ cannot have any zero on $\overline {\mathbb D}^2$ .

Conversely, if f does not vanish on $\overline {\mathbb D}^2$ , then f and $1/f$ are multipliers. Thus, ${1=(1/f)f\in [f]}$ , and f is cyclic (see Remark 7).

Proof Let $F:\mathbb {D}^2 \rightarrow \mathbb {D}$ be $(z_1,z_2) \mapsto \frac {z_1+z_2}{2}$ and define $C_F:D_{\beta - \frac {1}{2}}(\mathbb {D}) \rightarrow D_{\beta }(\mathbb {D}^2)$ as $f \mapsto f \circ F$ . Then, for $f(z) = \sum _{j \geq 0} a_j z^j$ , we may write

(14) $$ \begin{align} \begin{aligned} \left\| C_F(f) \right\|^2_{D_{\beta }(\mathbb{D}^2)} &= \sum_{j \geq 0} |a_j|^2 \frac{1}{2^{2j}} \| (z_1 + z_2)^j \|_{D_{\beta }(\mathbb{D}^2)}^2 \\ &= \sum_{j \geq 0} |a_j|^2 \frac{1}{2^{2j}} (j+2)^\beta \sum_{j_1 = 0}^j \binom{j}{j_1}^2 \\ &= \sum_{j \geq 0} |a_j|^2 \frac{1}{2^{2j}} \binom{2j}{j} (j+2)^\beta \\ &\asymp \sum_{j \geq 0} |a_j|^2 \frac{1}{2^{2j}} \frac{2^{2j}}{(j+2)^{\frac{1}{2}}} (j+2)^\beta \\ &= \sum_{j \geq 0} |a_j|^2 (j+2)^{\beta - \frac{1}{2}} \\ &\asymp \| f \|_{D_{\beta - \frac{1}{2}}(\mathbb{D})}^2. \end{aligned} \end{align} $$

This proves that $C_F$ is bounded. Moreover, $B = \mathrm {Im}(C_F)$ is a closed subspace of $D_{\beta }(\mathbb {D}^2)$ . This implies the cyclicity of $f_0$ whenever $\beta \leq \frac {3}{2}$ since it is the image of the function $g(z) = 1 - z$ under $C_F$ , which is cyclic in $D(\mathbb {D})$ . In this case, there exists a sequence of polynomials $\{p_n\}$ such that $p_n g \to 1$ in $D(\mathbb {D})$ , and the corresponding sequence $\{q_n = C_F(p_n)\}$ can be used to approximate $1$ via $q_n f_0$ in $D_{\beta }(\mathbb {D}^2)$ .

Proof of Theorem 38 (a)

Let f be a polynomial with no zeros in $\mathbb D^2$ and a nonempty finite zero set $\{\zeta _j\}_{j=1}^k$ on $\mathbb T^2$ . Put

$$\begin{align*}s_j(z)=1-\frac{\overline{\zeta_{j,1}}z_1+ \overline{\zeta_{j,2}}z_2}{2}. \end{align*}$$

By rotation invariance and Proposition 44, each $s_j$ is cyclic for $\beta \leq \frac 32$ .

As explained in [Reference Bénéteau, Knese, Kosiński, Liaw, Seco and Sola2, Section 2.2], a zero of f on $\partial \mathbb D^2\setminus \mathbb T^2$ would force a factor $z_i-\zeta $ , with $\zeta \in \mathbb T$ , contradicting the finiteness of $\mathcal Z(f)\cap \mathbb T^2$ . Thus,

$$\begin{align*}\mathcal Z(f)\cap\overline{\mathbb D^2} =\{\zeta_j\}_{j=1}^k. \end{align*}$$

Łojasiewicz’s inequality [Reference Krantz and Parks17] gives

$$\begin{align*}|f(z)|\gtrsim \mathrm{dist}(z,\{\zeta_j\}_{j=1}^k)^N, \qquad z\in\overline{\mathbb D^2}, \end{align*}$$

for some $N>0$ . Since $|s_j(z)|\lesssim |z-\zeta _j|$ , choosing m sufficiently large ensures that

$$\begin{align*}Q=\frac{\prod_{j=1}^k s_j^m}{f} \end{align*}$$

and its first partial derivatives are bounded on $\mathbb D^2$ , by the quotient rule and the preceding estimate. Hence, $Q\in D_2(\mathbb D^2)\subseteq D_\beta (\mathbb D^2)$ for $\beta \leq \frac 32$ .

By Lemma 22, $fQ=\prod _j s_j^m$ is cyclic. Choose polynomials $p_n\to Q$ in $D_\beta (\mathbb D^2)$ . Since f is a multiplier, $fp_n\to fQ$ , and so

$$\begin{align*}D_\beta(\mathbb D^2)=[fQ]\subseteq[f]. \end{align*}$$

For $\beta>2$ , evaluation at any $\zeta _j$ is a bounded nonzero functional annihilating $[f]$ , by Proposition 41. Thus, f is not cyclic.

Proof Without loss of generality, we can prove the statement only for $f(z) = 1 - z_1$ . Assume that $\beta \leq 1$ . To prove the cyclicity of f, as in Proposition 37, we need to estimate the norm of $f/f_r$ and then apply Theorem 21:

$$ \begin{align*} \sup_{0<r<1} \left\| \frac{f}{f_r} \right\|^2_{D_{\beta }(\mathbb{D}^2)} = 1 + \sup_{0<r<1} (1-r)^2 \sum_{j \geq 1} (j+2)^{\beta} r^{2j - 2} \asymp 1 + \sup_{0<r<1} (1-r)^{1 - \beta} < +\infty. \end{align*} $$

Now let $\beta> 1$ ; our goal is to show that the capacity of $\mathcal {Z}(f) \cap \mathbb {T}^2$ is positive to apply Theorem 33. Assuming that $\mu $ is the uniform probability measure on ${\{1\}\times \mathbb {T} = \mathcal {Z}(f) \cap \mathbb {T}^2}$ , we will have

$$ \begin{align*} I_{\beta}[\mu] &= \iint_{\mathbb{T}^2 \times \mathbb{T}^2} k_{\beta}(z,w) d\mu(z)d\mu(w) \\ &= \iint_{\mathbb{T}\times\mathbb{T}} \sum_{|J| \geq 0} \frac{z_2^{j_2} \bar{w}_2^{j_2}}{(|J|+2)^\beta} d\mu(z_2) d\mu(w_2) = \sum_{j_1 \geq 0} \frac{1}{(j_1+2)^{\beta}} < +\infty. \end{align*} $$

This implies that the capacity of $\mathcal {Z}(f) \cap \mathbb {T}^2$ is positive.

Proof of Theorem 38 (b)

Recall that the classical spaces $\mathfrak D_\alpha (\mathbb D^2)$ have squared monomial norms

$$\begin{align*}\|z_1^{j_1}z_2^{j_2}\|_{\mathfrak D_\alpha}^2 =(j_1+1)^\alpha(j_2+1)^\alpha. \end{align*}$$

By [Reference Bénéteau, Knese, Kosiński, Liaw, Seco and Sola2, Main Theorem], an irreducible polynomial depending on both variables, non-vanishing in $\mathbb D^2$ and vanishing at infinitely many points of $\mathbb T^2$ , is cyclic in $\mathfrak D_\alpha $ if and only if $\alpha \leq \frac 12$ . We adapt their argument to $D_\beta $ .

Assume $\beta \leq 1$ . Factors with a nonempty finite zero set on $\mathbb T^2$ are cyclic by part (a). Factors with no zeros on $\mathbb T^2$ have no zeros on $\overline {\mathbb D^2}$ , by [Reference Bénéteau, Knese, Kosiński, Liaw, Seco and Sola2, Section 2.2], and are cyclic by part (c) and the inclusions between the spaces. Thus, by Lemma 22, it remains to consider irreducible factors with infinitely many zeros on $\mathbb T^2$ . The case $\zeta -z_i$ has already been treated, so assume that f depends on both variables.

Let $(n,m)$ be the bidegree of f. By [Reference Bénéteau, Knese, Kosiński, Liaw, Seco and Sola2, Section 2.2 and Proposition 4.2], there exist a unitary $(n+m)\times (n+m)$ matrix U and a column vector polynomial ${B\in \mathbb C^{n+m}[z_1,z_2]}$ such that

$$\begin{align*}(I-A(z))B(z)\in f(z)\mathbb C^{n+m}[z_1,z_2], \qquad A(z)=U \begin{bmatrix} I_nz_1&0\\ 0&I_mz_2 \end{bmatrix}. \end{align*}$$

Moreover, there exists a row vector polynomial a such that $p(z_2)=a(z)B(z)$ is a one-variable polynomial with no zeros in $\mathbb D$ .

Let $g\in [f]^\perp $ and let $v\in \mathbb C^{n+m}[z_1,z_2]$ be a row vector polynomial. Since

$$\begin{align*}(I-A^k)B=\sum_{\ell=0}^{k-1}A^\ell(I-A)B \in f\mathbb C^{n+m}[z_1,z_2], \end{align*}$$

we obtain

$$\begin{align*}\langle g,vB\rangle=\langle g,vA^kB\rangle, \qquad k\geq0. \end{align*}$$

Put $d=\deg v+\deg B+1$ , where degrees are total degrees, taking the maximum over the entries. Since A is homogeneous of degree one, the monomials in $vA^{kd}B$ have degrees in $[kd,(k+1)d)$ . Hence, $\{vA^{kd}B\}_{k\geq 0}$ is orthogonal. Moreover, A is unitary on $\mathbb T^2$ ; so,

$$\begin{align*}\|vA^{kd}B\|_{D_\beta}^2 \lesssim (k+1)^\beta \|vA^{kd}B\|_{H^2(\mathbb D^2)}^2 \lesssim (k+1)^\beta, \end{align*}$$

where the last estimate follows from $|vA^{kd}B|\leq |v||B|$ on $\mathbb T^2$ , with Euclidean norms on the vectors.

If some $vA^{kd}B=0$ , then $\langle g,vB\rangle =0$ . Otherwise, Bessel’s inequality gives

$$\begin{align*}\|g\|_{D_\beta}^2 \geq\sum_{k\geq0} \frac{|\langle g,vA^{kd}B\rangle|^2} {\|vA^{kd}B\|_{D_\beta}^2} \gtrsim|\langle g,vB\rangle|^2 \sum_{k\geq0}\frac1{(k+1)^\beta}. \end{align*}$$

Since the series diverges for $\beta \leq 1$ , $\langle g,vB\rangle =0$ . Taking $v=z_1^{j_1}z_2^{j_2}a$ , we obtain $g\perp [p]$ . Since the restriction of the norm to functions of $z_2$ alone is equivalent to the $D_\beta (\mathbb D)$ norm, one-variable cyclicity gives polynomials $p_j(z_2)$ such that $p_jp\to 1$ in $D_\beta (\mathbb D^2)$ . Thus, p is cyclic, and $g=0$ .

Conversely, assume $\beta>1$ . Some irreducible factor h of f has infinitely many zeros on $\mathbb T^2$ . If h is a constant multiple of $\zeta -z_i$ , it is noncyclic by the preceding proposition. Otherwise, by [Reference Bénéteau, Knese, Kosiński, Liaw, Seco and Sola2, Main Theorem], h is noncyclic in $\mathfrak D_{\beta /2}(\mathbb D^2)$ . Since

$$\begin{align*}(j_1+1)^{\beta/2}(j_2+1)^{\beta/2} \leq(j_1+j_2+2)^\beta, \qquad \|\cdot\|_{\mathfrak D_{\beta/2}} \leq\|\cdot\|_{D_\beta}, \end{align*}$$

h is also noncyclic in $D_\beta (\mathbb D^2)$ . Finally, $[f]\subseteq [h]\neq D_\beta (\mathbb D^2)$ , and so f is noncyclic.