June 03, 2025
The Fantappiè and Laplace transforms realize isomorphisms between analytic functionals supported on a convex compact set \(K\subset{\mathbb{C}^n}\) and certain spaces of holomorphic functions associated with \(K\). Viewing the Bergman space of a bounded domain in \({\mathbb{C}^n}\) as a subspace of the space of analytic functionals supported on its closure, the images of the restrictions of these transforms have been studied in the planar setting. For the Fantappiè transform, this was done for simply connected domains (Napalkov Jr–Yulmukhametov, 1995), and for the Laplace transform, this was done for convex domains (Napalkov Jr–Yulmukhametov, 2004). In this paper, we study this problem in higher dimensions for strongly convex domains, and establish duality results analogous to the planar case. We also produce examples to show that the planar results cannot be generalized to all convex domains in higher dimensions.
For a compact convex subset \(K\) of \(\mathbb{C}^n\), the following spaces are known to be isomorphic:
\(\mathcal{O}'(K)\), the space of analytic functionals on \(K\),
\(\mathcal{O}(K^*)\), the space of holomorphic functions on \(K^*\), the dual complement of \(K\), given by \[K^*=\left\{z\in\mathbb{C}^n: \left\langle\zeta,z\right\rangle\neq 1,\,\forall \zeta\in K\right\},\] where \(\left\langle\zeta,z\right\rangle=\zeta_1z_1+\zeta_2z_2+\cdots\zeta_nz_n\), and
\(\mathcal{O}_{\operatorname{\operatorname{exp}}}(K)\), the space of entire functions \(F\) on \(\mathbb{C}^n\) such that, for every \(\epsilon>0\), there exists a \(C_\epsilon>0\) so that \[|F(z)|\leq C_\epsilon e^{(1+\epsilon)H_K(z)},\quad z\in \mathbb{C}^n,\] where \(H_K\) is defined as \[\begin{align} \label{eq:supprt32def} H_K(z)=\sup\limits_{\zeta\in K}\operatorname{Re}\langle\zeta,z\rangle,\,z\in\mathbb{C}^n. \end{align}\tag{1}\] Note that \(H_K(\overline{z})\) coincides with the classical support function of \(K\).
The duality between \((a)\), and \((b)\) was established by Aizenberg and Martineau in the case of convex sets, and by Gindikin and Henkin in the more general case of \(\mathbb{C}\)-convex sets; see [1]–[3]. The Fantappiè transform on \(\mathcal{O}'(K)\), given by \[\mathcal{F}_k\left(\mu\right)(z)=\mu\left(\frac{1}{\left(1-\left\langle\cdot,z\right\rangle\right)^k}\right),\quad \mu\in\mathcal{O}'(K), z\in K^*,\] for any fixed integer \(k\geq 1\), gives an explicit isomorphism between the two spaces (see Znamenskii [4]). Since \(\mathcal{O}'\left(K\right)\) is a large space, it is of interest to study the restrictions of \(\mathcal{F}_k\) to certain important subclasses of \(\mathcal{O}'\left(K\right)\). For instance, when \(K\) is the closure of a bounded domain \(\Omega\subset{\mathbb{C}^n}\), then under different assumptions on \(\Omega\), the restriction of \(\mathcal{F}_{1}\) to the Bergman space of \(\Omega\), the restriction of \(\mathcal{F}_n\) to the holomorphic Hardy space of \(\Omega\), and the restriction of \(\mathcal{F}_n\) to the space of holomorphic functions with polynomial growth on \(\Omega\), has been considered in [5], [6], and [7], respectively.
The Paley-Wiener type duality between \((a)\) and \((c)\) was established by Pólya-Ehrenpreis-Martineau, via the Laplace transform \[\mathcal{L}(\mu)(z)=\mu\left(e^{\langle\cdot,z\rangle}\right),\quad \mu\in\mathcal{O}'(K), z\in \mathbb{C}^n.\] See [8]. As is the case for \(\mathcal{F}_k\), the restriction of \(\mathcal{L}\) to different subspaces of \(\mathcal{O}'(K)\) has been considered. For instance, the restriction of \(\mathcal{L}\) to holomorphic Hardy spaces of a bounded domain has been studied for planar domains, strongly convex domains in \(\mathbb{C}^n\), and some weakly convex domains in \(\mathbb{C}^2\) in [9], [6], and [10], respectively.
In this note, we consider the restriction of the Fantappiè and the Laplace transforms on \(p\)-Bergman spaces on certain bounded domains \(\Omega\subset\mathbb{C}^n\). For a bounded domain \(\Omega\subset\mathbb{C}^n\), and \(p\in(1,\infty)\), the \(p\)-Bergman space on \(\Omega\) is defined as \[\begin{align} \label{eq:def32berg} A^p\left(\Omega\right)=\left\{f\in \mathcal{O}\left(\Omega\right):\|f\|_{L^p(\Omega)}=\left(\int_\Omega\left|f(\zeta)\right|^p dV(\zeta)\right)^{1/p}<\infty\right\}, \end{align}\tag{2}\] where \(V\) is the Lebesgue measure on \(\Omega\). Each \(A^p(\Omega)\) is a Banach space when endowed with the \(L^p\)-norm, and can be viewed as a subset of \(\mathcal{O}'\left(\overline{\Omega}\right)\) via the (possibly non-injective) inclusion \[\label{eq:inclusion} A^p(\Omega)\overset{\iota}{\hookrightarrow} L^p(\Omega)\xrightarrow{\theta} \mathcal{O}'\left(\overline{\Omega}\right),\tag{3}\] where \(\theta:L^p(\Omega)\rightarrow\mathcal{O}'\left(\overline{\Omega}\right)\) is given by \[\label{eq:32theta} \theta:f\mapsto \left(\mu_f:g\mapsto\frac{n!}{\pi^n}\int_\Omega\overline{f(\zeta)} g(\zeta) dV(\zeta)\right).\tag{4}\] The composition \(\mathcal{F}_k\circ\theta\circ\iota\), denoted simply by \(\mathcal{F} _k\) for convenience, is referred to as the Fantappiè transform on the \(p\)-Bergman space, and is explicitly given by \[\begin{align} \label{eq:Fan32berg} \mathcal{F}_{k}(g)(z)=\frac{n!}{\pi^n}\int_{\Omega} \frac{\overline{ g(\zeta)}}{\left(1-\langle \zeta,z\rangle\right)^{k}}dV(\zeta),\quad g\in A^p(\Omega),\, z\in \text{int}\,\Omega^*=\overline{\Omega}^*. \end{align}\tag{5}\]
First, we study the question of when \(\mathcal{F}_k\) is a normed space isomorphism between \(A^p(\Omega)\) and \(A^p\left(\Omega^*\right)\), where, in an abuse of notation, we denote the open set \(\overline{\Omega}^*\) by \(\Omega^*\). The range of \(\mathcal{F}_k\) can be sensitive to the choice of \(k\), and we focus on the case \(k=n+1\). This is motivated by the following observation: if \(\mathbb{B}^n\) denotes the unit ball in \({\mathbb{C}^n}\), then \[\mathcal{F}_{n+1}\left(f\right)(z)=\overline{f(\overline{z})},\quad f\in A^2\left(\mathbb{B}^n\right), z\in \mathbb{B}^n,\] and thus, \(\mathcal{F}_{n+1}\) is an isometry between \(A^2(\mathbb{B}^n)\) and \(A^2\left(\mathbb{B}^{n^*}\right)\). This follows from the fact that \(\frac{n!}{\pi^n}\left(1-\left\langle\overline{\zeta},z\right\rangle\right)^{-(n+1)}\), \(\left(\zeta,z\right)\in \mathbb{B}^n\times\mathbb{B}^n\), is the reproducing kernel of \(A^2(\mathbb{B}^n)\), and \(\overline{\mathbb{B}^n}^*=\mathbb{B}^n\). Note that \(\mathbb{B}^n\) is a strongly convex circled domain. Recall that a domain \(\Omega\subseteq\mathbb{C}^n\), is said to be circled, if \(e^{i\theta}z \in \Omega\), for all \(z \in \Omega\) and \(\theta \in [0, 2\pi)\). The above observation generalizes to the class of bounded strongly convex circled domains. In fact, we prove the following theorem in the general setting of any bounded strongly \(\mathbb{C}\)-convex domain with a strongly convex dual complement.
Theorem 1. Let \(\Omega\subset\mathbb{C}^n\) be a bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domain containing the origin, such that \(\Omega^*\) is strongly convex. Then, for any \(p\in(1,\infty)\), \(\mathcal{F}_{n+1}\) is a normed space isomorphism between \(A^p\left(\Omega\right)\) and \(A^p\left(\Omega^*\right)\).
We are not aware of any characterization of the class of bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domains whose dual complements are strongly convex. However, it is easy to check that the dual complement of a bounded \(\mathcal{C}^2\)-smooth circled strongly convex (and, therefore, strongly \(\mathbb{C}\)-convex) domain is always strongly convex. For examples of non-circled strongly \(\mathbb{C}\)-convex (but not strongly convex) domains with strongly convex dual complements; see Remark 4. In general, the dual complement of a bounded \(\mathcal{C}^2\)-smooth strongly convex (hence, strongly \(\mathbb{C}\)-convex) domain is not necessarily strongly convex; see Lemma 13. We do not know if strong convexity of the dual complement is a necessary condition for the duality in Theorem 1 to hold; see Remark 5 for more details.
Theorem 1 can now be used to describe the range of the Laplace transform on \(A^2(\Omega)\), given by \[\begin{align} \label{eq:lap32berg} \mathcal{L}(f)(z)=\int_\Omega\overline{f(\zeta)} e^{\left\langle\zeta,z\right\rangle} dV(\zeta),\quad z\in\mathbb{C}^n, \end{align}\tag{6}\] under an additional assumption of convexity on \(\Omega\).
Theorem 2. Let \(\Omega\subset\mathbb{C}^n\) be a bounded \(\mathcal{C}^2\)-smooth strongly convex domain containing the origin, such that \(\Omega^*\) is strongly convex. Then the Laplace transform \(\mathcal{L}\) on \(A^2(\Omega)\) is a normed space isomorphism between \(A^2(\Omega)\), and the weighted Bergman space \[A^2(\mathbb{C}^n,\omega_\Omega)=\left\{F\in\mathcal{O}(\mathbb{C}^n):\Vert F\Vert_{\omega_\Omega}^2=\int_{\mathbb{C}^n}|F(z)|^2 \omega_\Omega(z)<\infty\right\},\] with norm \(\|\cdot\|_{\omega_\Omega}\), where \[\begin{align} \omega_\Omega(z)=e^{-2H_\Omega(z)}\|z\|^{n+\frac{1}{2}} \left(dd^cH_\Omega\right)^n(z),\quad z\in\mathbb{C}^n, \end{align}\] and \(H_\Omega(\overline{z})\) is the support function of \(\Omega\).
Similar to Theorem 1, the hypothesis of this theorem holds true for any bounded \(\mathcal{C}^2\)-smooth strongly convex circled domain.
Furthermore, when \(\Omega\subset\mathbb{C}^n\) is a bounded strongly convex domain, the weighted Bergman space \(A^2(\mathbb{C}^n,\omega_\Omega)\) is isomorphic to \(A^2(\mathbb{C}^n,\mu_\Omega)\) via the identity map, where \[\mu_\Omega(z)=\|e^{\langle\cdot,z\rangle}\|^{-2}_{L^2(\Omega)} \left(dd^cH_\Omega\right)^n(z);\] see Lemma 8. In the planar case, similar results as Theorem 1 and Theorem 2 are known to hold for \(p=2\), under a much less restrictive hypothesis on the domain. In particular, for every bounded convex domain \(\Omega\subset\mathbb{C}\), it is known that \(\mathcal{F}_2\) is a normed space isomorphism between \(A^2(\Omega)\) and \(A^2(\Omega^*)\), and \(\mathcal{L}\) is a normed space isomorphism between \(A^2(\Omega)\) and \(A^2(\mathbb{C},\mu_\Omega)\); see [11]–[13]. However, we show that these results cannot be extended to higher dimensions in their full generality.
Theorem 3. Let \(\Omega=\left\{\left(\zeta_1,\zeta_2\right)\in\mathbb{C}^2: |\zeta_1|+|\zeta_2|<1\right\}\). Then,
\(\mathcal{F}_3\) is a bounded injective operator from \(A^2(\Omega)\) onto a proper subspace of \(A^2(\Omega^*)\),
\(\mathcal{L}\) is a bounded injective operator from \(A^2(\Omega)\) onto a proper subspace of both \(A^2(\mathbb{C}^2,\omega_{\Omega})\), and \(A^2(\mathbb{C}^2,\mu_\Omega)\). Moreover, \[\mathcal{L}(A^2(\Omega))\subsetneq A^2(\mathbb{C}^2,\omega_\Omega)\subsetneq A^2(\mathbb{C}^2,\mu_\Omega).\]
We provide an outline of the proofs. Let \(\Omega\) be as in Theorem 1. Then, it admits a strongly \(\mathbb{C}\)-convex neighborhood basis, and hence, \(\overline{\Omega}\) is \(\mathbb{C}\)-convex; see [8]. Thus, by the duality result stated earlier, \(\mathcal{F}_{n+1}\left(\mathcal{O}'\left(\overline{\Omega}\right)\right)=\mathcal{O}\left(\Omega^*\right)\). In particular, \(\mathcal{F}_{n+1}\) maps \(A^p(\Omega)\) to \(\mathcal{O}\left(\Omega^*\right)\). Thus, to prove Theorem 1, it suffices to establish that, for each \(p\in(1,\infty)\),
\(\mathcal{F}_{n+1}\) is an \(L^p(\Omega)\)-\(L^p(\Omega^*)\)-bounded operator,
\(\mathcal{F}_{n+1}\) is injective on \(A^p(\Omega)\), and
\(\mathcal{F}_{n+1}\) maps \(A^p(\Omega)\) onto \(A^p(\Omega^*)\).
For \(1\), we use the domain’s strong \(\mathbb{C}\)-convexity to convert \(\mathcal{F}_{n+1}\) into a singular integral operator on \(\Omega^*\). This operator is then compared with a projection operator, \(\mathcal{B}_{\Omega^*}\), on \(L^p(\Omega^*)\). This projection operator is a simpler version of an operator used in the literature to study the regularity properties of the Bergman projection on strictly pseudoconvex domains via a Kerzman–Stein-type approach; see [14], [15], and [16]. Lemma 4 is the key boundedness property of \(\mathcal{B}_{\Omega^*}\) that we need, and this result may be of independent interest. For both \(2\) and \(3\), we use that \(A^p(\Omega)'\) is isomorphic to \(A^q(\Omega)\) when \(p^{-1}+q^{-1}=1\); see Lemma 7. This duality allows us to show that \(\theta\circ\iota\) in 3 is injective, following which, the injectivity of \(\mathcal{F}_{n+1}\) on \(\mathcal{O}'\left(\overline{\Omega}\right)\) gives \(2\). For \(3\), we show that each function \(g\) in \(A^p(\Omega^*)\) induces a bounded functional on \(A^q(\Omega)\), and thus, can be mapped to a function \(\phi_g\) in \(A^p(\Omega)=A^q(\Omega)'\). Using a Cauchy–Fanttapiè-type reproducing formula for \(A^p(\Omega)\) — see Lemma 6 — we show that \(\mathcal{F}_{n+1}(\phi_g)\) is, in fact, \(g\).
To prove Theorem 2, we relate \(\mathcal{F}_{n+1}\) with \(\mathcal{L}\) via the following commutative diagram: \[\begin{tikzcd} A^2(\Omega) \arrow{dr}{\mathcal{L}} \arrow{r}{\mathcal{F}_{n+1}} & A^2(\Omega^*) \\& A^2({\mathbb{C}^n},\omega_\Omega) \arrow[swap]{u}{\mathfrak B_n} & \end{tikzcd}\] where \(\mathfrak B_n:F\mapsto\left(z\mapsto \int_0^\infty F(tz)t^ne^{-t}dt\right)\) is known as the Borel transform. Owing to Theorem 1, it suffices to show that \(\mathfrak B_n\) is a normed space isomorphism between \(A^2({\mathbb{C}^n},\omega_\Omega)\) and \(A^2(\Omega^*)\).
Let \(\Omega\) be as in Theorem 3. Then, both \(\Omega\) and \(\Omega^*\) are complete Reinhardt domains (in fact, \(\Omega^*\) is the unit polydisc in \(\mathbb{C}^2\)). This allows us to give characterizations of Bergman-space functions on both domains using power-series representations. We produce an explicit element in \(A^2(\Omega^*)\) that is not in the image of \(\mathcal{F}_3\), and an element in \(A^2(\mathbb{C}^2,\omega_\Omega)\), and hence in \(A^2(\mathbb{C}^2,\mu_\Omega)\) which is not in the image of \(\mathcal{L}\).
Acknowledgements. I am deeply grateful to my thesis advisor, Purvi Gupta, for her insightful guidance, many helpful ideas, and for engaging in regular discussions throughout the course of this project. I would also like to thank her for helping me with the writing of this paper. I also thank Mihai Putinar for valuable discussions and for suggesting that the Fantappiè kernel with exponent \(n+1\) might be effective in our context, as well as for highlighting connections of the Fantappiè transform to various other topics that may lead to promising future directions. Finally, I am grateful to the anonymous referee for providing many constructive comments that significantly improved this paper.
This work is supported by a scholarship from the Indian Institute of Science, and the DST-FIST programme (grant no. DST FIST-2021 [TPN-700661]).
In this section, we collect some preparatory results and observations for our main objects of study.
The following notation will be used throughout the paper.
\(\mathbb{B}^n\) denotes the unit Euclidean ball in \(\mathbb{C}^n\).
\(d=\partial+\overline{\partial}\) denotes the standard exterior derivative.
\(d^c={i}\left({\overline{\partial}-\partial}\right).\)
Given a bounded domain \(\Omega\), \(\sigma_\Omega\) denotes the Euclidean surface area measure on \(b\Omega\).
Given two \(\mathbb{R}\)-valued functions \(f\) and \(g\) on set a \(X\),
\(g\lesssim f\) on \(X\) denotes the existence of \(C_1>0\) such that \(C_1 g(x)\leq f(x)\), for all \(x\in X\).
\(g\approx f\) on \(X\) denotes the existence of \(C_1,C_2>0\) such that \(C_1 g(x)\leq f(x)\leq C_2 g(x)\), for all \(x\in X\).
\(\mathbb{N}\) denotes the set of all natural numbers, i.e., the set of all positive integers union \(\{0\}\).
Given a bounded \(\mathcal{C}^2\)-smooth domain \(\Omega\) that is star-convex with respect to the origin, \(m_\Omega\) denotes the Minkowski functional of \(\Omega\), defined in 7 .
Given a bounded domain \(\Omega\), \(H_\Omega\) denotes the function as defined in 1 .
Given a bounded \(\mathcal{C}^2\)-smooth convex domain \(\Omega\) containing the origin, \(\rho_\Omega\) denotes the defining function as given in 12 .
Given a domain \(\Omega\), and \(p\in(1,\infty)\), \(A^p(\Omega)\) denotes the \(p\)-Bergman space of \(\Omega\), defined in 2 .
Given a domain \(\Omega\), \(\mathcal{O}_{exp}(\Omega)\) denotes a subspace of exponential entire functions, defined in 38 .
Given a \(g\in A^p(\Omega)\) and \(k\in\mathbb{Z}_+\), \(\mathcal{F}_k(g)\) denotes the Fantappiè transform of \(g\), defined in 5 .
Given a \(f\in A^2(\Omega)\), \(\mathcal{L}(f)\) denotes the Laplace transform of \(f\), defined in 6 .
Given a \(F\in\mathcal{O}_{exp}(\Omega)\), \(\mathfrak B_n(F)\) denotes the Borel transform of \(F\), defined in 39 .
Let \(D\subset \mathbb{C}^n\) be a bounded \(\mathcal{C}^2\)-smooth domain that is star-convex with respect to the origin. The Minkowski functional of \(D\) is given by \[\begin{align} \label{eq:mink32def} m_D(z)=\inf\left\{t>0:t^{-1}z\in D\right\},\quad z\in\mathbb{C}^n. \end{align}\tag{7}\] Due to the star-convexity of \(D\), \(m_D\) is a positively \(1\)-homogeneous function on \(\mathbb{C}^n\), i.e., \(f(tz)=tf(z)\) for all \(z\in\mathbb{C}^n\) and \(t>0\). Moreover, \(D=\{z:m_D(z)<1\}\), and \(bD=\{z:m_D(z)=1\}\). We note the following result on the regularity of \(m_D\).
Lemma 1. Let \(D\subset\mathbb{C}^n\) be a \(\mathcal{C}^2\)-smooth bounded domain that is star-convex with respect to the origin. Further, assume that \(D\) has only non-radial tangents, i.e., for any \(\zeta\in bD\), \(\operatorname{Re}\langle\zeta,\overline{\eta}(\zeta)\rangle\neq 0\), where \(\eta(\zeta)\) is a normal to \(bD\) at \(\zeta\). Then \(m_D\) is \(\mathcal{C}^2\)-smooth on \(\mathbb{C}^n\setminus\{0\}\).
Proof. Since \(bD\) is \(\mathcal{C}^2\)-smooth, there exists a defining function of \(D\), \(\rho\), which is \(\mathcal{C}^2\)-smooth in \(\mathbb{C}^n\). We define \(\phi:\mathbb{C}^n\times \left(\mathbb{R}\setminus\{0\}\right)\rightarrow\mathbb{R}\) as \[\phi\left(z,t\right)=\rho\left(t^{-1}z_1,t^{-1}z_2,\cdots,t^{-1}z_n\right),\quad \left(z,t\right)\in\mathbb{C}^n\times\left(\mathbb{R}\setminus\{0\}\right).\] As \(\rho\in \mathcal{C}^2\left(\mathbb{C}^n\right)\), it follows that \(\phi\in\mathcal{C}^2\left(\mathbb{C}^n\times\left(\mathbb{R}\setminus\{0\}\right)\right)\). Also, since \(m_D\) is positively \(1\)-homogeneous, for any \(z\in\mathbb{C}^n\setminus\{0\}\), \(\frac{z}{m_D(z)}\in bD\). Consequently, \[\phi\left(z,m_D(z)\right)=0,\quad \forall z\in\mathbb{C}^n\setminus\{0\}.\] Let us now fix a \(\tilde{z}\in\mathbb{C}^n\setminus\{0\}\). We claim that \(\frac{\partial \phi}{\partial t}(\tilde{z},m_D(\tilde{z}))\neq 0\). Suppose \(\frac{\partial \phi}{\partial t}(\tilde{z},m_D(\tilde{z}))=0\). Then, using the chain rule, we get that \[\begin{align} \frac{\partial \phi}{\partial t}(\tilde{z},m_D(\tilde{z}))=\frac{-1}{m_D(\tilde{z})}\operatorname{Re}\left\langle2\partial \rho\left(\frac{\tilde{z}}{m_D\left(\tilde{z}\right)}\right),\frac{\tilde{z}}{m_D\left(\tilde{z}\right)}\right\rangle=0. \end{align}\] As \(m_D\left(\tilde{z}\right)\neq 0\), it must be that \(\operatorname{Re}\left\langle2\partial \rho\left(\frac{\tilde{z}}{m_D\left(\tilde{z}\right)}\right),\frac{\tilde{z}}{m_D\left(\tilde{z}\right)}\right\rangle=0\). This is a contradiction, since \(D\) has only non-radial tangents by assumption. Thus, we may apply the implicit function theorem to \(\phi\) and conclude that \(m_D\) is a \(\mathcal{C}^2\)-smooth function on \(\mathbb{C}^n\setminus\{0\}\). ◻
The next result allows us to freely move between \(D\) and \(D^*\) under the assumption of strong \(\mathbb{C}\)-convexity.
Lemma 2. Let \(D\subset\mathbb{C}^n\) be a bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domain that is star-convex with respect to the origin. Further, assume that \(D\) has only non-radial tangents, i.e., for any \(\zeta\in bD\), \(\operatorname{Re}\langle\zeta,\overline{\eta}(\zeta)\rangle\neq 0\), where \(\eta(\zeta)\) is a normal to \(bD\) at \(\zeta\). Then, the map \(T_{D}:D\rightarrow D^*\), defined as \[\begin{align} \label{eq:cov32map} T_{D}\left(\zeta\right)=\begin{cases} m_{D}^2(\zeta)\dfrac{\partial m_{D}(\zeta)}{\left\langle\partial m_{D}(\zeta),\zeta\right\rangle},&\quad \zeta\in D\setminus\{0\},\\ 0,&\quad \zeta=0, \end{cases} \end{align}\tag{8}\]
is invertible, with \(T_{D^*}\) as the inverse map,
is a homeomorphism between \(D\) and \(D^*\), and
is a \(\mathcal{C}^1\)-diffeomorphism between \(D\setminus\{0\}\) and \(D^*\setminus\{0\}\).
Moreover, if \(f\) is a measurable function on \(D^*\), then \[\begin{align} \label{eq:cov} \int_{D^*} f\left(\zeta\right)dV(\zeta)=\int_{D} f\left(T_{D}(\zeta)\right)h_{D}(\zeta) dV(\zeta), \end{align}\tag{9}\] where \(h_{D}\) is a positive function on \(D\), which is bounded above and below on \(D\) by positive constants.
Proof. Let \(S_{D}:bD\rightarrow bD^*\) be given by \[S_{D}\left(\eta\right)=\frac{\partial m_{D}(\eta)}{\left\langle\partial m_{D}(\eta),\eta\right\rangle},\qquad \eta\in bD.\] The following implications follow from [6].
\(D^*\) is also a bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domain that is star-convex with respect to the origin, and \(m_{D^*}\) is \(\mathcal{C}^1\)-smooth on \(\mathbb{C}^n\setminus\{0\}\); see [6]. As \(m_{D^*}\) is positively \(1\)-homogeneous, by Euler’s identity, for any \(\xi\in bD^*\), \[\label{eq:euler} \operatorname{Re}\left <\partial m_{D^*}(\xi),\xi\right>= \frac{m_{D^*}(\xi)}{2}= \frac{1}{2}.\tag{10}\] Hence, \(D^*\) has only non-radial tangents.
\(S_{D}\) is a \(\mathcal{C}^1\)-diffeomorphism from \(bD\) onto \(bD^*\), whose inverse is \(S_{D^*}\); see [6].
If \(g\) is a measurable function with respect to \(\sigma_{D^*}\), then, \[\int_{bD^*} g\left(\xi\right)d\sigma_{D^*}(\xi)=\int_{bD} g\left(S_{D}(\eta)\right)\widetilde{h}(\eta) d\sigma_{D}(\eta),\] for some positive continuous function \(\widetilde{h}\) on \(bD\); see [6].
If \(f\) is a Lebesgue integrable function on \(D^*\), then \[\begin{align} \label{eq:polar32coordinate} \int_{D^*} f\left(\zeta\right)dV(\zeta)&=&\int_0^1\int_{bD}f\left(rS_{D}(\eta)\right)r^{2n-1} {j_D(\eta)} dr\, d\sigma_{D}(\eta)\nonumber\\ &=&\int_0^1\int_{bD^*}f\left(r\xi\right)r^{2n-1} {\ell_{D^*}(\xi)} dr\, d\sigma_{D^*}(\xi), \end{align}\tag{11}\] for some positive continuous functions \(j_D\) and \(\ell_{D^*}\) on \(bD\) and \(bD^*\), respectively; see [6].
We write an arbitrary \(\zeta\in D\) as \(\zeta=r\eta\), where \(r\in(0,1)\), and \(\eta\in bD\). Then using the \(1\)-homogenity of \(m_{D}\), it follows that \(T_{D}(\zeta)=rS_{D}\left(\eta\right)\). From (ii), it is immediate that \(T_{D}\) is a homeomorphism between \(D\) and \(D^*\). Furthermore, since \(m_{D}\) is \(\mathcal{C}^2\)-smooth on \(\mathbb{C}^n\setminus\{0\}\), \(T_{D}\) is a \(\mathcal{C}^1\)-diffeomorphism between \(D\setminus\{0\}\) and \(D^*\setminus\{0\}\). Finally, using (iv), we get that \[\begin{align} \int_{D^*} f\left(\zeta\right)dV(\zeta) &=&\int_0^1\int_{bD}f\left(rS_D\left(\eta\right)\right)r^{2n-1} {j_D(\eta)} d\sigma_{D}(\eta)\,dr\\ &=&\int_0^1\int_{bD}f\left(T_{D}(r\eta)\right)r^{2n-1} {j_D(\eta)} d\sigma_{D}(\eta)\,dr\\ &=&\int_0^1\int_{bD}(f\circ T_{D})(r\eta) r^{2n-1} {h_{D}(r\eta)} \ell_D(\eta)d\sigma_{D}(\eta)\,dr\\ &=&\int_{D} (f\circ T_{D})(\zeta) h_{D}(\zeta) dV(\zeta), \end{align}\] where \(h_{D}(r\eta)=j_D(\eta){\ell_D(\eta)}^{-1}\), for \(r\in(0,1), \eta\in bD\). We obtained the last equality by applying (iv) to \(D\), instead of \(D^*\). This concludes the proof. ◻
Note that if the domain \(\Omega\subset\mathbb{C}^n\) satisfies the hypothesis of Theorem 1, then \(\Omega^*\) satisfies the hypothesis of Lemma 2 due to strong convexity, and hence \((\Omega^*)^*=\Omega\) also satisfies the hypothesis of Lemma 2.
Now, let \[\label{eq:defn32fns} \rho_D(\zeta)= m_D^2\left(\zeta\right)-1,\quad\zeta\in\mathbb{C}^n.\tag{12}\] Since \(m_D\) is a \(C^2\)-smooth function on \(\mathbb{C}^n\setminus\{0\}\), \(\rho_D\) is a \(C^2\)-smooth defining function of \(D\). In fact, the following lemma shows that \(\rho_D\) is also strongly convex on \(\mathbb{C}^n\setminus\{0\}\).
Lemma 3. Let \(D\subset\mathbb{C}^n\) be a bounded \(\mathcal{C}^2\)-smooth strongly convex domain containing the origin. Then \(\rho_D(\zeta)=m_D^2(\zeta)-1\) is a strongly convex function on \(\mathbb{C}^n\setminus\{0\}\), i.e., \(\operatorname{Hess}(\rho_D)(\zeta)\) is positive definite, for all \(\zeta\in\mathbb{C}^n\setminus\{0\}\).
Proof. Let \(\left(\zeta,w\right)\in \left(\mathbb{C}^n\setminus\{0\}\right)^2\). Writing \(w\) in real co-ordinates as \(w=\left(u_1,u_2,\cdots,u_{2n}\right)\), we get that, \[\begin{align} w^T \cdot \operatorname{Hess}(\rho_D)(\zeta)\cdot w &=& \sum_{j,k=1}^{2n} \frac{\partial^2\rho_D}{\partial x_j \partial x_k}(\zeta) u_ju_k\\ &=&2\sum_{j,k=1}^{2n} \left(\frac{\partial m_D}{\partial x_j}(\zeta)\frac{\partial m_D}{\partial x_k}(\zeta)+m_D(\zeta) \frac{\partial^2 m_D}{\partial x_j \partial x_k}(\zeta)\right)u_ju_k \\ &=&2\left(\sum_{j=1}^{2n}\frac{\partial m_D}{\partial x_j}(\zeta) u_j\right)^2+2m_D(\zeta) \sum_{j,k=1}^{2n}\frac{\partial^2 m_D}{\partial x_j \partial x_k}(\zeta) u_ju_k. \end{align}\] Since \(m_D\) is positively \(1\)-homogeneous, the right-hand side is \(0\)-homogeneous in \(\zeta\), i.e., \(f(t\zeta)=f(\zeta),\) for all \(\zeta\in\mathbb{C}^n\setminus\{0\}\) and \(t>0\). Thus, it suffices to show that \[\label{eq:claim32str32cvx} 2\left(\sum_{j=1}^{2n}\frac{\partial m_D}{\partial x_j}(\zeta) u_j\right)^2+2m_D(\zeta) \sum_{j,k=1}^{2n}\frac{\partial^2 m_D}{\partial x_j \partial x_k}(\zeta) u_ju_k>0,\quad\forall \zeta\in bD\,\text{and}\,\forall w\in\mathbb{C}^n\setminus\{0\}.\tag{13}\] Now, as \(D\) is convex, \(m_D\) is a convex function on \(\mathbb{C}^n\), which implies that \[\label{eq:mink32cvx} \sum_{j,k=1}^{2n}\frac{\partial^2 m_D}{\partial x_j \partial x_k}(\zeta) u_ju_k\geq 0,\quad\text{for all } \zeta\in bD\,\text{and } w\in\mathbb{C}^n\setminus\{0\}.\tag{14}\] Furthermore, as \(\rho_D\) is a defining function of \(D\), it follows that for any \(\zeta\in bD\) and \(w\in \mathbb{C}^n\setminus\{0\}\), \[\begin{align} \label{eq:tangent} \text{if } \sum_{j=1}^{2n}\frac{\partial m_D}{\partial x_j}(\zeta) u_j=0, \quad \text{then } \sum_{j,k=1}^{2n}\frac{\partial^2 m_D}{\partial x_j \partial x_k}(\zeta)u_ju_k>0. \end{align}\tag{15}\] Thus, 13 follows from 14 and 15 . ◻
We now inspect some singular integral operators that are closely related to our problem. Given a bounded strongly convex domain \(D\subset{\mathbb{C}^n}\), let \[\begin{align} B_D\left(\zeta,z\right)&=& m_D(\zeta)\left\langle 2\partial m_D(\zeta),\zeta-z\right\rangle+\left(1-m_D^2(\zeta)\right),\quad \left(\zeta,z\right)\in\overline{D}\times\overline{D}, \tag{16}\\ \mathcal{B}_D(f)(z)&=&\int_{D} f(\zeta) \left(B_D\left(\zeta,z\right)\right)^{-n-1} dV(\zeta),\quad f\in L^p(D), z\in D.\tag{17} \end{align}\] We first establish the \(L^p\)-regularity of \(\mathcal{B}_D\).
Lemma 4. Let \(D\subset\mathbb{C}^n\) be a bounded strongly convex domain in \(\mathbb{C}^n\) containing the origin, and \(p\in(1,\infty)\). The integral operator denoted by \(\left|\mathcal{B}_D\right|\) defined on \(L^p\left(D\right)\) by \[\left|\mathcal{B}_D\right|(f)(z)=\int_{D} f(\zeta) \left|B_D\left(\zeta,z\right)\right|^{-n-1} dV(\zeta),\quad z\in D,\] is a bounded operator on \(L^p\left(D\right)\).
Proof. We claim that for \((\zeta,z)\in\overline{D}\times\overline{D}\), \(\operatorname{Re}{B_D}(\zeta,z)\) vanishes if and only if \(\left(\zeta,z\right)\in bD\times bD\) and \(\zeta=z\). When \(\zeta,z\in bD\), and \(\zeta=z\), clearly, \(\operatorname{Re}B_D\) vanishes. For the converse, suppose there is a \((\zeta_0, z_0)\in\overline{D}\times\overline{D}\) such that \(\operatorname{Re}B_D(\zeta_0, z_0)=0\). As \(m_D\) is positively \(1\)-homogeneous, by Euler’s identity, \[\label{veutfmnh} \operatorname{Re}\left\langle2\partial m_D(\zeta),\zeta\right\rangle= m_D(\zeta),\quad\forall\zeta\in \mathbb{C}^n.\tag{18}\] Using this, it follows that \(m_D^2(\zeta_0)-m_D(\zeta_0)\operatorname{Re}\left\langle2\partial m_D(\zeta_0),z_0\right\rangle=m_D^2(\zeta_0)-1\). In other words, \[\label{eq:constraint} m_D(\zeta_0)\operatorname{Re}\left\langle2\partial m_D(\zeta_0),z_0\right\rangle=1.\tag{19}\] Let us consider the polar set of \(D\), denoted as \(D^\circ\), and defined by \[D^\circ=\{z\in\mathbb{C}^n:H_D(z)<1\},\] where \(H_D(\overline{z})\) is the support function of \(D\). It is known that \(2\partial m_D(\zeta)\in bD^\circ\) for all \(\zeta\in bD\); see [6], and as \(\partial m_D\) is \(0\)-homogeneous, \(2\partial m_D(\zeta)\in bD^\circ\) for all \(\zeta\in\mathbb{C}^n\setminus\{0\}.\) Hence, by the definition of \(D^\circ\), we have that \(\operatorname{Re}\left\langle2\partial m_D(\zeta_0),z_0\right\rangle\leq 1\). Since \(m_D(\zeta)<1\), whenever \(\zeta\in D\), it follows from 19 that \(\left(\zeta_0,z_0\right)\notin{D}\times \overline{D}\). Thus, \(\left(\zeta_0,z_0\right)\in bD\times \overline{D}\). However, this means that \(\operatorname{Re}\left\langle2\partial m_D(\zeta_0),\zeta_0-z_0\right\rangle=0\), which implies that the real tangent space to \(bD\) at \(\zeta_0\) intersects the point \(z_0\). As \(D\) is strongly convex, the real tangent space to \(bD\) at \(\zeta_0\) does not intersect any points in \(\overline{D}\setminus\{\zeta_0\}\). Hence, \(\zeta_0=z_0\), which proves our claim.
Writing \(B_D\) in terms of \(\rho_D\), we have that \[B_D\left(\zeta,z\right)=\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D\left(\zeta\right).\] We claim that \(B_D\) satisfies the following estimate \[\begin{align} \label{eq:key32est32LS32ker} \left|B_D\left(\zeta,z\right)\right|\approx \left|\rho_D(\zeta)\right|+\left|\rho_D(z)\right|+\left|\operatorname{Im}\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle\right|+\|\zeta-z\|^2, \quad \left(\zeta,z\right)\in \overline{D}\times \overline{D}. \end{align}\tag{20}\] To prove this, first choose a small positive number \(\delta>0\), and consider the region \(G_{\delta}=\{\zeta\in\overline{D}: -\rho_D(\zeta)\leq\delta\}\). Applying Taylor’s theorem for each \(\zeta\in G_\delta\), we get that \[\rho_D(z)=\rho_D(\zeta)-2\operatorname{Re}\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle+\operatorname{Re}Q_\zeta(\rho_D,(\zeta-z))+R_\zeta(\rho_D,(\zeta-z))+E(\zeta,z),\quad \forall z\in\overline{D},\] where \[\begin{align} Q_\zeta(\rho_D,(\zeta-z))&=&\sum_{j,k=1}^{n}\frac{\partial^2 \rho_D}{\partial \zeta_j \partial \zeta_k}(\zeta) (\zeta_j-z_j)(\zeta_k-z_k),\\ R_\zeta(\rho_D,(\zeta-z))&=&\sum_{j,k=1}^{n}\frac{\partial^2 \rho_D}{\partial \zeta_j \partial \overline{\zeta_k}}(\zeta) (\zeta_j-z_j)\overline{(\zeta_k-z_k)}, \end{align}\] and \(E(\zeta,z)\) is a continuous function on \(G_\delta\times\overline{D}\), satisfying \(\lim_{\zeta\rightarrow z}E(\zeta,z)/\|\zeta-z\|^2=0\). Then \[\widetilde{E}\left(\zeta,z\right)=\begin{cases} \dfrac{E(\zeta,z)}{\|\zeta-z\|^2},&\quad \text{when}\,\zeta\neq z,\\ 0,&\quad \text{when}\,\zeta=z, \end{cases}\] is a uniformly continuous function on the compact set \(G_\delta\times\overline{D}\). Writing \(\zeta\) and \(z\) in real coordinates as \(\zeta=(u_1,u_2\cdots,u_{2n})\) and \(z=(x_1,x_2,\cdots,x_{2n})\), the real Hessian of \(\rho_D\) at \(\zeta\) is related to \(Q_\zeta,R_\zeta\) in the following way: \[\begin{align} \frac{1}{2}\sum_{j,k=1}^{2n}\frac{\partial^2 \rho_D}{\partial u_j \partial u_k}(\zeta) (u_j-x_j)(u_k-x_k)=\operatorname{Re}Q_\zeta(\rho_D,(\zeta-z))+R_\zeta(\rho_D,(\zeta-z)). \end{align}\] Furthermore, by Lemma 3, as \(\rho_D\) is a strongly convex function on \(\mathbb{C}^n\setminus\{0\}\), we get that \[C_1\|\zeta-z\|^2\leq\operatorname{Re}Q_\zeta(\rho_D,(\zeta-z))+R_\zeta(\rho_D,(\zeta-z))\leq C_2\|\zeta-z\|^2,\quad (\zeta,z)\in G_\delta\times\overline{D},\] where \(C_1,C_2\) are positive constants that are independent of both \(\zeta\) and \(z\). Combining this with the Taylor expansion of \(\rho_D\) above, we get that for a fixed \(\zeta\in G_\delta\), \[C_1\|\zeta-z\|^2+ E(\zeta,z)\leq2\operatorname{Re}\left(\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D(\zeta)\right)+\rho_D(\zeta)+\rho_D(z)\leq C_2\|\zeta-z\|^2+ E(\zeta,z),\quad z\in \overline{D}.\] Now, due to the uniform continuity of \(\widetilde{E}\), there exists a \(\epsilon>0\), such that for any \(\left(\zeta,z\right)\in G_\delta\times\overline{D}\) satisfying \(\|\zeta-z\|<\epsilon\), \[\widetilde{C_1}\leq\frac{C_1\|\zeta-z\|^2+E\left(\zeta,z\right)}{\|\zeta-z\|^2}\leq C_2+\widetilde{E}\left(\zeta,z\right)\leq \widetilde{C_2},\] where \(\widetilde{C_1}, \widetilde{C_2}\) are constants that do not depend on \(\zeta,z\). As a result, we have that \[\label{eq:est} 2\operatorname{Re}\left(\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D(\zeta)\right)\approx-\rho_D(\zeta)-\rho_D(z)+\|\zeta-z\|^2, \quad (\zeta,z)\in G_{\epsilon,\frac{\delta}{2}},\tag{21}\] where \(G_{\epsilon,\frac{\delta}{2}}=\{\left(\zeta,z\right)\in\overline{D}\times\overline{D}; -\rho_D(\zeta)<\delta/2,\|\zeta-z\|<\epsilon\}\). Note that both the right-hand side and left-hand side in 21 vanish if and only if \(\zeta\) and \(z\) are in \(bD\), and \(\zeta=z\). Hence, by the compactness of \(\overline{D}\times\overline{D}\setminus G_{\epsilon,\frac{\delta}{2}}\), it follows that \[\begin{align} \label{eq:real32part32est} 2\operatorname{Re}\left(\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D(\zeta)\right)\approx-\rho_D(\zeta)-\rho_D(z)+\|\zeta-z\|^2,\quad\forall\left(\zeta,z\right)\in\overline{D}\times\overline{D}. \end{align}\tag{22}\] Thus, \[\begin{align} \left|B_{D}\left(\zeta,z\right)\right|&\approx& \left|\operatorname{Re}\left(\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D(\zeta)\right)\right|+\left|\operatorname{Im}\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle\right|,\\ &\approx& \left|\rho_D(\zeta)\right|+\left|\rho_D(z)\right|+\left|\operatorname{Im}\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle\right|+\|\zeta-z\|^2,\quad \left(\zeta,z\right)\in \overline{D}\times\overline{D}. \end{align}\] This proves 20 . Now, using 20 , and following the steps presented in [16], we conclude the \(L^p\)-boundedness of \({\mathcal{B}_D}\). ◻
Next, we consider a kernel that appears when one pushes forward the Fantappiè transform on \(D^*\) under the diffeomorphism \(T_{D^*}\) considered in Lemma 2. Let \[\begin{align} K_{D}\left(\zeta,z\right)&=& 1-\langle T_D(\zeta),z\rangle,\quad (\zeta,z)\in\overline{D}\times\overline{D} \tag{23}\\ &=& \begin{cases} \dfrac{\left\langle \partial m_{D}(\zeta),\zeta-m_{D}^2(\zeta)z\right\rangle}{\left\langle \partial m_{D}(\zeta),\zeta\right\rangle},&\quad\left(\zeta,z\right)\in\left(\overline{D}\setminus\{0\}\right)\times\overline{D},\\ 1,&\quad \left(\zeta,z\right)\in\{0\}\times\overline{D}, \end{cases} \tag{24},\\ \mathcal{K}_D(f)(z)&=&\int_{D} f(\zeta) \left(K_D\left(\zeta,z\right)\right)^{-n-1} dV(\zeta),\quad f\in L^p(D), z\in D. \tag{25} \end{align}\]
Lemma 5. Let \(D\subset{\mathbb{C}^n}\) be a bounded strongly convex domain containing the origin. Then, \[\begin{align} \label{eq:comp32of} \left|B_D\left(\zeta,z\right)\right|\leq C \left|K_D\left(\zeta,z\right)\right|,\quad\forall\left(\zeta,z\right)\in\overline{D}\times\overline{D}. \end{align}\tag{26}\] where \(C>0\) are constants not depending on \(\zeta,z\).
Proof. Writing \(K_D\left(\zeta,z\right)\) and \(B_D(\zeta, z)\), with respect to \(\rho_D\), we get that \[\begin{align} K_D\left(\zeta,z\right)&=&\frac{1+\rho_D(\zeta)}{\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}\left(\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\frac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle\right),\\ B_D\left(\zeta,z\right)&=&\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D\left(\zeta\right). \end{align}\] We write \(\zeta\in \overline{D}\) as \(\zeta=r\xi\), where \(r\in[0,1]\) and \(\xi\in bD\). As \(m_D\) is positively \(1\)-homogeneous, \(\partial m_D\) is positively \(0\)-homogeneous. Thus, \[\label{eq:rho32m} \frac{1+\rho_D(\zeta)}{\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}=\frac{m_D(\zeta)}{\left\langle2\partial m_D(\zeta),\zeta\right\rangle}=\frac{1}{\left\langle2\partial m_D(\xi), \xi\right\rangle}.\tag{27}\] Now, from 10 , we have that \(\operatorname{Re}\left\langle2\partial m_D(\xi),\xi\right\rangle=1,\,\forall\xi\in bD\). Consequently, \(\left|\left\langle2\partial m_D(\xi),\xi\right\rangle\right|\approx1\) on \(bD\), which in turn implies that \(\left|\dfrac{1+\rho_D(\zeta)}{\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}\right|\approx 1\) on \(\overline{D}\). Thus, if we set \[\label{eq:tildeK} \widetilde{K}_D\left(\zeta,z\right)=K_D\left(\zeta,z\right)\frac{\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}{1+\rho_D(\zeta)},\quad(\zeta,z)\in\overline{D}\times\overline{D},\tag{28}\] we have that \[\begin{align} \label{eq:equiv32est} \left|{K_D}\left(\zeta,z\right)\right|\approx \left|\widetilde{K}_D\left(\zeta,z\right)\right|,\quad\left(\zeta,z\right)\in \overline{D}\times\overline{D}. \end{align}\tag{29}\] Now, observe that \[\begin{align} \left|\frac{{B_D\left(\zeta,z\right)}}{\widetilde{K}_D\left(\zeta,z\right)}\right|&=&\left|\frac{\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D(\zeta)}{\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\dfrac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}\right|\\ &=&\left|\frac{\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\dfrac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle+\rho_D(\zeta)\left(\dfrac{\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}{1+\rho_D(\zeta)}-1\right)}{\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\dfrac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}\right|\\ &\leq&1+\frac{\left|\rho_D(\zeta)\right|\left|\dfrac{\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}{1+\rho_D(\zeta)}-1\right|}{\left|\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\dfrac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle\right|}\\ &\lesssim& 1+\frac{\left|\rho_D(\zeta)\right|}{\left|\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\dfrac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle\right|}\\ &\leq&1+\frac{\left|\rho_D(\zeta)\right|}{\left|\operatorname{Re}\left(\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\dfrac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle\right)\right|}\\ (\text{by \eqref{eq:rho32m}}) &=& 1+\frac{\left|\rho_D(\zeta)\right|}{\left|\operatorname{Re}\left(\left\langle\partial \rho_D(\zeta),\zeta-z\right\rangle-\rho_D(\zeta)\right)\right|}\\ (\text{by \eqref{eq:real32part32est}}) &\leq& 1+\frac{\left|\rho_D(\zeta)\right|}{C_1\left(\left|\rho_D(\zeta)\right|+\left|\rho_D(z)\right|+\|\zeta-z\|^2\right)}\\ &\leq& 1+\frac{1}{C_1}. \end{align}\] Finally, combining this with 29 , we conclude the lemma. ◻
We now discuss a reproducing kernel on \(p\)-Bergman spaces of strongly convex domains. Define a \((1,0)\)-form as follows. For \(\left(\zeta,z\right)\in\overline{D}\times D\), consider \[\label{eq:repr32ker} G\left(\zeta,z\right)=\dfrac{\partial\rho_D(\zeta)}{\widetilde{K}_D\left(\zeta,z\right)},\tag{30}\] where \(\widetilde{K}_D\) is as in 28 . Since \(\rho_D\) is \(\mathcal{C}^2\) smooth on \(\mathbb{C}^n\setminus\{0\}\), it follows that for each \(z\in D\), \(G\) is a \(C^1\)-smooth form on \(\overline{D}\setminus\{0\}\). We show that the \((n,n)\)-form \(\left(\overline{\partial}_{\zeta}G\right)^{n}(\zeta,z)\) reproduces functions in \(A^p( D)\). For this, we exploit the reproducing property of the Cauchy–Leray operator for functions in the dense subclass \(\mathcal{O}\left(D\right)\cap C^1\left(\overline{D}\right)\).
Lemma 6. Let \(D\subset{\mathbb{C}^n}\) be a bounded strongly convex domain. Given \(p\in(1,\infty)\), let \(f\in A^p\left(D\right)\). Then, for each \(z\in D\), \[\label{eq:repr32prop} f(z)=\frac{1}{(2\pi i)^n}\int_D f(\zeta)\left(\overline{\partial}G\right)^n\left(\zeta,z\right).\tag{31}\]
Proof. Consider the modified \((1,0)\)-form \[\begin{align} L\left(\zeta,z\right)=\dfrac{\partial\rho_D(\zeta)}{\widetilde{K}_D\left(\zeta,z\right)+\frac{\rho_D(\zeta)}{1+\rho_D(\zeta)}\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}=\dfrac{\partial\rho_D(\zeta)}{\left\langle\partial\rho_D(\zeta),\zeta-z\right\rangle},\quad \left(\zeta,z\right)\in\overline{D}\times D. \end{align}\] This form is a generating form for a convex domain with \(\mathcal{C}^2\)-smooth boundary; see[15]. Thus the \((n,n-1)\)-form, \(L\wedge\left(\overline{\partial} L\right)^{n-1}\) is a Cauchy–Fanatppiè form on \(bD\), and reproduces functions in \(\mathcal{O}\left(D\right)\cap\mathcal{C}\left(\overline{D}\right)\) from its boundary values. As a consequence, we get that for any \(f\in\mathcal{O}\left(D\right)\cap\mathcal{C}^1\left(\overline{D}\right)\), \[\begin{align} \label{eq:repr32larey}\nonumber f(z)&=\frac{1}{(2\pi i)^n}\int_{bD} f(\zeta)j^*\left(L\wedge\left(\overline{\partial} L\right)^{n-1}\right)(\zeta,z)\\ &=\frac{1}{(2\pi i)^n}\int_{bD} \dfrac{f(\zeta)}{\left\langle\partial\rho_D(\zeta),\zeta-z\right\rangle^n} j^*\left(\partial \rho_D\wedge\left(\overline{\partial}\partial\rho_D\right)^{n-1}\right)(\zeta), \quad z\in D, \end{align}\tag{32}\] where \(j:bD\rightarrow\mathbb{C}^n\) is the inclusion map. It is easy to verify that \[j^*\left(L\wedge\left(\overline{\partial} L\right)^{n-1}\right)(\zeta,z)=j^*\left(G\wedge\left(\overline{\partial} G\right)^{n-1}\right)(\zeta,z),\quad \left(\zeta,z\right)\in bD\times D.\] Hence by 32 , \[\label{eq:repr32bdry32our} f(z)=\frac{1}{(2\pi i)^n}\int_{bD} f(\zeta)j^*\left(G\wedge\left(\overline{\partial} G\right)^{n-1}\right)(\zeta,z),\quad z\in D.\tag{33}\] Now, to conclude 31 for \(f\), we wish to apply Stokes’ theorem to the term on the right-hand side of the above equality. However, since the form \(\left(\overline{\partial} G\right)^{n}\) is not \(\mathcal{C}^1\)-smooth at the origin, we cannot apply Stokes’ theorem directly. To get around this issue, let us first fix a \(z\in D\). Consider \(\lambda\in(0,1)\), such that \(z\in D\setminus \overline{\lambda D}\), where \(\lambda D=\{\lambda\zeta:\zeta\in D\}\). Applying Stokes’ theorem on the domain \(D\setminus \overline{\lambda D}\), we obtain that \[\label{eq:int32decomp} \int_{ D\setminus \overline{\lambda D}}f(\zeta) \left(\overline{\partial} G\right)^n\left(\zeta,z\right)=\int_{b D} f(\zeta)j^*\left(G\wedge\left(\overline{\partial} G\right)^{n-1}\right)(\zeta,z)\minus\int_{\lambda b D} f(\zeta)j^*\left(G\wedge\left(\overline{\partial} G\right)^{n-1}\right)(\zeta,z).\tag{34}\] By a straightforward computation, we can verify that \[\begin{align} \label{eq:comp32vol} \left(\overline{\partial} G\right)^n(\zeta,z) &=& -\frac{\left(\left(\overline{\partial}\partial\rho_D\right)^{n-1}\wedge\overline{\partial}\widetilde{K}_ D\wedge\partial\rho_D+\widetilde{K}_ D\left(\overline{\partial}\partial\rho_D\right)^n\right)}{\widetilde{K}_ D\left(\zeta,z\right)^{n+1}}\nonumber\\ &=&\dfrac{\mathfrak h_ D\left(\zeta\right)}{\widetilde{K}_ D\left(\zeta,z\right)^{n+1}}dV(\zeta),\quad \left(\zeta,z\right)\in\left(D\setminus\{0\}\right)\times D, \end{align}\tag{35}\] where \[\renewcommand\arraystretch{2.5} \mathfrak h_ D(\zeta)= c_n\det \begin{pmatrix} \dfrac{\rho_D(\zeta){\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}}{1+\rho_D(\zeta)} & \dfrac{\partial\rho_D}{\partial \zeta_j}\\ \dfrac{\partial}{\partial \bar \zeta_j}\left(\dfrac{\rho_D(\zeta)\left\langle\partial\rho_D(\zeta),\zeta\right\rangle}{1+\rho_D(\zeta)}\right)& \dfrac{\partial^2\rho_D}{\partial \zeta_j\partial \bar \zeta_k} \end{pmatrix}_{1\leq j,k\leq n}, \quad \zeta\in D.\] Similarly, using [15], we have that \[\begin{align} \label{eq:leray32surf32area} j^*\left(G\wedge\left(\overline{\partial} G\right)^{n-1}\right)(\zeta,z) =\frac{j^*\left(\partial \rho_D\wedge\left(\overline{\partial}\partial\rho_D\right)^{n-1}\right)}{\widetilde{K}_ D\left(\zeta,z\right)^{n}}\nonumber=\dfrac{\mathfrak h_{bD}\left(\zeta\right)}{\widetilde{K}_ D\left(\zeta,z\right)^{n}}\sigma_ D, \end{align}\tag{36}\] where \(\sigma_ D\) is the surface area measure on \(b D\), and \[\mathfrak{h}_{b D}\left(\zeta\right)=\frac{(-1)^n}{\pi^n}\det\begin{pmatrix} 0 & \dfrac{\partial\rho_D}{\partial \zeta_j}\\ \dfrac{\partial\rho_D}{\partial \bar \zeta_j} & \dfrac{\partial^2\rho_D}{\partial \zeta_j\partial \bar \zeta_k} \end{pmatrix}_{1\leq j,k\leq n}, \quad \zeta\in bD.\] Since \(D\) is strongly convex, \(\mathfrak{h}_{b D}\) is bounded above and below by positive constants. Furthermore, since \(m_D\) is \(1\)-homogeneous, the second order partial derivatives of \(\rho_D\) are \(0\)-homogeneous, which implies that \(\sup_{\zeta\in\overline{D}} \frac{\partial^2\rho_D}{\partial \zeta_j\partial \bar \zeta_k}(\zeta)=\sup_{\zeta\in b D}\frac{\partial^2\rho_D}{\partial \zeta_j\partial \bar \zeta_k}(\zeta)\). Consequently, as \(\rho_D\) is \(\mathcal{C}^2\)-smooth on \(bD\), it follows that \(\frac{\partial^2\rho_D}{\partial \zeta_j\partial \bar \zeta_k}\) is bounded above on \(\overline{D}\). Hence \[\label{eq:density32bdd32abv} \left|\mathfrak h_ D(\zeta)\right|\leq C,\quad \zeta\in\overline{D},\tag{37}\] where \(C\) is a positive constant. As mentioned earlier, since \(z\in D\) is fixed \(\widetilde{K}_ D\left(\zeta,z\right)\neq 0,\, \forall\zeta\in\overline{D}\), and hence by continuity \(\left|\frac{1}{\widetilde{K}_ D\left(\zeta,z\right)}\right|\leq C_z\), where \(C_z\) is a constant depending on \(z\). Thus combining this with 35 , 37 , and the fact \(f\in \mathcal{C}\left(\overline{D}\right)\), we get that \[\begin{align} \int_{ D}\left|f(\zeta)\right| \left|\left(\overline{\partial} G\right)^n\right|\left(\zeta,z\right)\lesssim \text{vol}(D). \end{align}\] Thus, applying the dominated convergence theorem, we get that, as \(\lambda\rightarrow 0\), \[\int_{ D\setminus \overline{\lambda D}}f(\zeta) \left(\overline{\partial} G\right)^n\left(\zeta,z\right)\rightarrow \int_{ D}f(\zeta) \left(\overline{\partial} G\right)^n\left(\zeta,z\right).\] Applying the change of variable \(\zeta=\lambda\zeta'\), where \(\zeta'\in b D\), we get that \[\begin{align} \left|\int_{\lambda b D} f(\zeta)j^*\left(G\wedge\left(\overline{\partial} G\right)^{n-1}\right)(\zeta,z)\right|&\lesssim&\int _{\lambda b D}j^*\left(\partial \rho_D\wedge\left(\overline{\partial}\partial\rho_D\right)^{n-1}\right)(\zeta),\\ &\lesssim& \lambda^{2n}\int_{b D} j^*\left(\partial \rho_D\wedge\left(\overline{\partial}\partial\rho_D\right)^{n-1}\right)(\zeta'),\\ &\lesssim& \lambda^{2n} \sigma\left(b D\right). \end{align}\] The right-hand side tends to \(0\), as \(\lambda\) tends to \(0\). Hence, \(\lim_{\lambda\rightarrow 0}\int_{\lambda b D} f(\zeta)j^*\left(G\wedge\left(\overline{\partial} G\right)^{n-1}\right)(\zeta,z)\rightarrow 0\). Combining these observations with 34 and 33 , we get that \[f(z)=\frac{1}{(2\pi i)^n}\int_ D f(\zeta)\left(\overline{\partial}G\right)^n\left(\zeta,z\right),\quad z\in D.\]
Since \(D\) is a convex domain, the class \(\mathcal{O}\left(D\right)\cap\mathcal{C}^1\left(\overline{D}\right)\) is dense in \(A^p\left(D\right)\). Hence to prove 31 for \(A^p( D)\) functions, where \(p\in\left(1,\infty\right)\), it is enough to show that for every \(p\in\left(1,\infty\right)\), the integral operator \[L^p( D)\ni f\mapsto\frac{1}{(2\pi i)^n}\int_ D f\left(\zeta\right)\left(\overline{\partial} G\right)^n\left(\zeta,z\right),\] is a bounded operator on \(L^p\left(D\right)\). To see this, for \(f\in L^p\left(D\right)\), \[\begin{align} \left\|\frac{1}{(2\pi i)^n}\int_ D f\left(\zeta\right)\left(\overline{\partial} G\right)^n\left(\zeta,\cdot\right)\right\|_{L^p(D)}^p&=&\left\|\frac{1}{(2\pi i)^n}\int_ D f\left(\zeta\right)\frac{\mathfrak h_D(\zeta)}{\widetilde{K}_D(\zeta,\cdot)^{n+1}}dV(\zeta)\right\|_{L^p(D)}^p\\ (\text{by}\,\eqref{eq:comp32vol},\eqref{eq:density32bdd32abv})&\lesssim&\left\|\int_ D\left|f\left(\zeta\right)\right|\left|K_ D\left(\zeta,\cdot\right)\right|^{-n-1} dV(\zeta)\right\|_{L^p(D)}^p\\ (\text{by Lemma~\ref{le:comp32of32two32ker}})&\lesssim&\left\|\int_ D\left|f\left(\zeta\right)\right|\left|B_ D\left(\zeta,\cdot\right)\right|^{-n-1} dV(\zeta)\right\|_{L^p(D)}^p\\ (\text{by Lemma~\ref{le:bdd32of32LS32ker}})&\lesssim& \|f\|_{L^p(D)}^p. \end{align}\] This proves our claim, and hence completes the proof of the lemma. ◻
Next, we note the following duality result on strongly convex domains, which is implicitly present in the literature. In fact, the result holds for all \(\mathcal{C}^2\)-smooth strongly pseudoconvex domains.
Lemma 7. Let \(D\subset\mathbb{C}^n\) be a bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domain. Given \(p\in(1,\infty)\), let \(q\in(1,\infty)\) be the conjugate exponent of \(p\), i.e., \(p^{-1}+q^{-1}=1\). Then, the map \(\Phi_p:A^q\left(D\right)\rightarrow A^p( D)'\), defined as \[\Phi_p(f)(h)=\int_{D} h(\zeta)\overline{f(\zeta)} dV(\zeta), \quad f\in A^q\left(D\right),\, h\in A^p\left(D\right),\] is an isomorphism.
Proof. Fix \(p\in(1,\infty)\). According to Theorem 2.15 and Remark 2.6 in [17], the surjectivity of \(\Phi_p\) is guaranteed if the following two conditions are satisfied:
For \(p \in(1, \infty)\), the positive Bergman operator \(|P|\), defined by \[|P|(f)(z)=\int_D f(\zeta) | \widetilde{B}_D(\zeta,z)| dV(\zeta),\quad z\in D, \,f\in L^p(D),\] where \(\widetilde{B}_D\) is the Bergman kernel of the domain \(D\), is a bounded operator on \(L^p(D)\).
The Bergman projection acts as the identity operator on \(A^p(D)\).
Both conditions are known to hold for bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domains. Specifically, condition (i) was established in [16], while by [17] and condition (i), condition (ii) is equivalent to \(A^2(\Omega)\cap A^p(\Omega)\) being dense in \(A^p(\Omega)\), which follows from [16].
Following the same argument but switching the roles of \(p\) and \(q\), we obtain that \(\Phi_q\) is also surjective. Hence, from [17], it follows that \(\Phi_p\) is injective, which completes the proof. ◻
Next, we introduce the Borel transform on \(A^2(\mathbb{C}^n,\omega_D)\), which allows us to relate \(\mathcal{F}_{n+1}\) with \(\mathcal{L}\), to establish Theorem 2.
Let \(D\subset\mathbb{C}^n\) be a bounded convex domain. Let \(\mathcal{O}_{\operatorname{\operatorname{exp}}}(D)\) denote the space of entire functions \(F\in\mathcal{O}(\mathbb{C}^n)\) such that, for every \(\epsilon>0\), there exist a \(C_\epsilon>0\) so that \[\label{eq:Borel32dfn32cond} |F(z)|\leq C_\epsilon e^{(1+\epsilon)H_D(z)},\quad z\in\mathbb{C}^n.\tag{38}\] The Borel transform on \(\mathcal{O}_{\operatorname{\operatorname{exp}}}(D)\) is defined as \[\label{eq:def32borel} \mathfrak B_n\left(F\right)(z)=\int_0^\infty F(tz)e^{-t}t^n dt,\quad F\in \mathcal{O}_{\operatorname{\operatorname{exp}}}(D).\tag{39}\] It follows from the definition that \(\mathfrak B_n(F)\in\mathcal{O}(D^\circ)\), where \(D^{\circ}\) is the polar set of \(D\). Furthermore, it is known that \(\mathfrak B_n(F)\) has a holomorphic extension to \(D^*\); see [2], or [8]. Next, we show that for planar strongly convex domain, \(\mathcal{B}_1\) is an isomorphism between \(A^2(\mathbb{C},\omega_D)\) and \(A^2(D^*)\), where recall that \(\omega_D=e^{-2H_\Omega(z)}\|z\|^{\frac{3}{2}}\left(dd^cH_D\right)\). This result essentially follows from [13]. However, the weighted Bergman space appearing in [13] is different from \(\omega_D\). Thus, in order to establish the isomorphism between \(A^2(\mathbb{C},\omega_D)\), and \(A^2(D^*)\), we first show that the weighted Bergman space appearing in [13], and \(A^2(\mathbb{C},\omega_D)\) are normed space isomorphic.
Lemma 8. Let \(D\subset\mathbb{C}^n\) be a bounded strongly convex domain with \(0\in D\). Then the identity map is a normed space isomorphism between \(A^2(\mathbb{C}^n,\omega_D)\) and \(A^2(\mathbb{C}^n,\mu_D)\), where \[\begin{align} A^2(\mathbb{C}^n,\mu_D)=\left\{F\in\mathcal{O}(\mathbb{C}^n):\|F\|_{\mu_D}^2=\int_{\mathbb{C}^n}{|F(z)|^2}{\left\|e^{\langle\cdot, z\rangle}\right\|^{-2}_{L^2(D)}}\left(dd^cH_D\right)^n(z)<\infty\right\}, \end{align}\] equipped with the norm \(\|.\|_{\mu_D}\).
Proof. Following the same idea as in the proof of [10], it is sufficient to prove that for any \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq1\), \[\label{eq:comp32weight} \left\|e^{\langle\cdot, z\rangle}\right\|^2_{L^2(D)}\approx {e^{2H_D(z)}}\|z\|^{-n-\frac{1}{2}}.\tag{40}\] To show this, by applying 11 to \(D\), we get that \[\begin{align} {e^{-2H_D(z)}}\left\|e^{\langle\cdot, z\rangle}\right\|^2_{L^2(D)}&\approx& { \int_0^1\int_{bD} e^{2r\operatorname{Re}\langle\xi,z\rangle-2H_D(z)}} r^{2n-1} d\sigma_D(\xi) dr\nonumber\\ \label{eq:berg32eq32key32est} &=& \int_0^1 e^{2(r-1)H_D(z)} r^{2n-1}\int_{bD} e^{2r\left(\operatorname{Re}\langle\xi,z\rangle-H_D(z)\right)} d\sigma_D(\xi) dr. \end{align}\tag{41}\] Also, applying [10], we obtain that for any \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq1\), and \(r\in (0,1)\), \[\begin{align} \int_{b\mathbb{B}^n} e^{-2C_1\|z\|\left(1-\operatorname{Re}\langle\eta,z\rangle\right)} d\sigma_{\mathbb{B}^n}(\eta)\lesssim \int_{bD} e^{2r\left(\operatorname{Re}\langle\xi,z\rangle-H_D(z)\right)} d\sigma_D(\xi)\lesssim \int_{b\mathbb{B}^n} e^{-2rC_2\|z\|\left(1-\operatorname{Re}\langle\eta,z\rangle\right)} d\sigma_{\mathbb{B}^n}(\eta), \end{align}\] where \(C_1,C_2\) are positive constants not depending on either \(r\) or \(z\). Furthermore, following the same computations as in the proof of [10] to obtain [10], we get that for any \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq1\), and \(r\in (0,1)\), \[\begin{align} \|z\|^{n-\frac{1}{2}}\int_{b\mathbb{B}^n} e^{-2C_1\|z\|\left(1-\operatorname{Re}\langle\eta,z\rangle\right)} d\sigma_{\mathbb{B}^n},(\eta)&\gtrsim& 1\\ r^{n-\frac{1}{2}}\|z\|^{n-\frac{1}{2}}\int_{b\mathbb{B}^n} e^{-2rC_2\|z\|\left(1-\operatorname{Re}\langle\eta,z\rangle\right)} d\sigma_{\mathbb{B}^n}(\eta)&\lesssim& 1. \end{align}\] From 41 , it now follows that for any \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq1\), \[\begin{align} \label{eq:berg32eq32prel32est} \|z\|\int_0^1 e^{-{(1-r)H_D(z)}}r^{2n-1} dr\lesssim\|z\|^{n+\frac{1}{2}}{e^{-2H_D(z)}}\|e^{\cdot z}\|^2_{L^2(D)}\lesssim \|z\|\int_0^1 e^{-{(1-r)H_D(z)}} r^{n-\frac{1}{2}} dr. \end{align}\tag{42}\] Moreover, as there exists \(r_1,r_2>0\), such that \(\mathbb{B}^n(r_1)\subset D\subset \mathbb{B}^n(r_2)\), for any \(z\in \mathbb{C}^n\setminus\{0\}\), \[\begin{align} \label{eq:support32radial32comp} r_1 \|z\|\leq H_D(z)\leq r_2\|z\|. \end{align}\tag{43}\] Combining 43 with the change of variable \(r'=(1-r)H_D(z)\) yields that for any \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq1\), \[\begin{align} \|z\| \int_0^1 e^{-{(1-r)H_D(z)}}r^{2n-1} dr&\approx& H_D(z) \int_0^1 e^{-{(1-r)H_D(z)}}r^{2n-1} dr\\ &=& \int_0^{H_D(z)} e^{-r'}\left(1-\frac{r'}{H_D(z)}\right)^{2n-1} dr'\\ &\leq& \int_0^{H_D(z)} e^{-r'} dr'\\ &=&1-e^{-H_D(z)}\lesssim 1, \end{align}\] Also, by 43 , \(\|z\|\geq 1\), implies \(H_D(z)\geq r_1\). Thus, for \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq 1\), \[\begin{align} \int_0^{H_D(z)} e^{-r'}\left(1-\frac{r'}{H_D(z)}\right)^{2n-1} dr'&\geq& \left(\frac{1}{2}\right)^{2n-1}\int_0^{\frac{r_1}{2}} e^{-r'} dr'\approx 1. \end{align}\] Hence, we get that for any \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq1\), \[\begin{align} \|z\| \int_0^1 e^{-{(1-r)H_D(z)}}r^{2n-1} dr&\approx& 1. \end{align}\] A similar computation yields, for any \(z\in\mathbb{C}^n\), satisfying \(\|z\|\geq1\), \[\begin{align} \|z\| \int_0^1 e^{-{(1-r)H_D(z)}}r^{n-\frac{1}{2}} dr &\approx& 1. \end{align}\] Finally, combining these with 42 , we obtain 40 . ◻
Lemma 9. Let \(D\subset\mathbb{C}\) be a bounded strongly convex domain with \(0\in D\). Then \(\mathfrak B_1\) is a normed space isomorphism between \(A^2(\mathbb{C},\omega_D)\) and \(A^2(D^*)\).
Proof. Let us consider the following space: \[B_2^1\left(\mathbb{C}\setminus D\right)=\left\{f\in\mathcal{O}( \widehat{\mathbb{C}}\setminus D): f(\infty)=0,\,\int_{\mathbb{C}\setminus D}|f'(\lambda)|^2 dV(\lambda)<\infty\right\},\] equipped with the norm \(\|f\|_{B_2^1\left(\mathbb{C}\setminus D\right)}^2=\left(\int_{\mathbb{C}\setminus D}|f'(\lambda)|^2 dV(\lambda)\right)\). Since \(D^*\) is biholomorphic to \(\widehat{\mathbb{C}}\setminus D\) via the map \(\tau:\lambda\rightarrow\frac{1}{\lambda}\), the following map \[\begin{align} \mathfrak T: &B_2^1\left(\mathbb{C}\setminus D\right)\rightarrow A^2(D^*),\\ & F\rightarrow (F\circ \tau)', \end{align}\] is a normed space isomorphism between \(B_2^1\left(\mathbb{C}\setminus D\right)\), and \(A^2(D^*)\). From [13], it follows that \(\widetilde{\mathfrak B}:A^2(\mathbb{C},\mu_D)\rightarrow B_2^1(\mathbb{C}\setminus D)\), defined as \[\widetilde{\mathfrak B}(F)(z)=\int_0^\infty F(t) e^{-tz} dt,\] is a normed space isomorphism between \(A^2(\mathbb{C},\mu_D)\) and \(B_2^1(\mathbb{C}\setminus D)\). Note that, in [13], the inverse of \(\widetilde{\mathfrak{B}}\) has been considered as the Borel transform. For \(z\in D^*\), applying a change of variable \(t'=\frac{t}{z}\) we get that \[\begin{align} \left(\mathfrak T\circ\widetilde{\mathfrak B}(F)\right)(z) &=& \frac{1}{z^2}\int_0^\infty F(t)e^{-t/z}t dt\\ &=&\int_0^\infty F(t'z) e^{-t'} t'dt'\\ &=& \mathfrak B_{1}(F)(z). \end{align}\] Combining all these with Lemma 8, it follows that, \(\mathfrak B_1\) is a normed space isomorphism between \(A^2(\mathbb{C},\omega_D)\) and \(A^2(D^*)\). ◻
Proof of Theorem 1 Fix \(p\in(1,\infty)\). As discussed in the introduction, the proof splits into three parts: the \(L^p\)-boundedness, the injectivity, and the surjectivity of \(\mathcal{F}_{n+1}\).
By Lemma 2, applied to \(D=\Omega^*\), \[\begin{align} \mathcal{F}_{n+1}\left(f\right)(z)&=&\frac{n!}{\pi^n}\int_{\Omega} \frac{\overline{ f(\zeta)}}{\left(1-\langle \zeta,z\rangle\right)^{n+1}}dV(\zeta)\\ &=& \frac{n!}{\pi^n}\int_{\Omega^*} \dfrac{\overline{\left(f\circ T_{\Omega^*}\right)}(\zeta)}{\left(1-\langle T_{\Omega^*}(\zeta),z\rangle\right)^{n+1}} h_{\Omega^*}(\zeta) dV(\zeta) \\ &=& \frac{n!}{\pi^n}\int_{\Omega^*}\dfrac{\overline{\left(f\circ T_{\Omega^*}\right)}(\zeta)\left\langle \partial m_{\Omega^*}(\zeta),\zeta\right\rangle^{n+1} h_{\Omega^*}(\zeta)}{\left(\left\langle \partial m_{\Omega^*}(\zeta),\zeta\right\rangle-m_{\Omega^*}^2(\zeta)\left\langle \partial m_{\Omega^*}(\zeta),z\right\rangle\right)^{n+1}} dV(\zeta)\\ &=&\frac{n!}{\pi^n}\int_{\Omega^*}\dfrac{\overline{\left(f\circ T_{\Omega^*}\right)}(\zeta)\left\langle \partial m_{\Omega^*}(\zeta),\zeta\right\rangle^{n+1} h_{\Omega^*}(\zeta)}{\left\langle \partial m_{\Omega^*}(\zeta),\zeta-m_{\Omega^*}^2(\zeta)z\right\rangle^{n+1}} dV(\zeta)\\ &=& \frac{n!}{\pi^n}\int_{\Omega^*} \overline{\left(f\circ T_{\Omega^*}\right)}(\zeta) h_{\Omega^*}(\zeta) K_{\Omega^*}\left(\zeta,z\right)^{-n-1} dV(\zeta), \quad f\in A^p\left(\Omega\right), z\in\Omega^*, \end{align}\] where \(K_{\Omega^*}\) is as given by 24 .
Now for a bounded strongly convex domain \(D\subset{\mathbb{C}^n}\) containing the origin, let \[|\mathcal{K}_D|(f)(z)=\int_{\Omega} f(\zeta) |K_D\left(\zeta,z\right)|^{-n-1} dV(\zeta),\quad z\in D.\] By Lemma 5, we have that for any \(g\in L^p\left(D\right)\), \[\label{eq:int32comp} \left|\mathcal{K}_{D}\right|\left(\left|g\right|\right)(z)\leq \tilde{C}\left|\mathcal{B}_{D}\right|\left(\left|g\right|\right)(z), \quad \forall z\in D,\tag{44}\] where \(\tilde{C}\) is a positive constant, which is independent of \(z\) and \(g\).
Applying 44 to \(D=\Omega^*\), we get for \(g\in L^p\left(\Omega\right)\), \[\begin{align} \|\mathcal{F}_{n+1}(g)\|_{L^p(\Omega^*)}&=&\left\|\mathcal{K}_{\Omega^*}\left(\overline{\left(g\circ T_{\Omega^*}\right)} h_{\Omega^*}\right)\right\|_{L^p(\Omega^*)}\\ &\leq& \left\|\left|\mathcal{K}_{\Omega^*}\right|\left(\left|\overline{\left(g\circ T_{\Omega^*}\right)}\right|\left|h_{\Omega^*}\right|\right)\right\|_{L^p(\Omega^*)}\\ &\lesssim& \left\|\left|\mathcal{B}_{\Omega^*}\right|\left(\left|\overline{\left(g\circ T_{\Omega^*}\right)}\right|\left|h_{\Omega^*}\right|\right)\right\|_{L^p(\Omega^*)}\\ &\lesssim&\left\| \overline{\left(g\circ T_{\Omega^*}\right)}h_{\Omega^*}\right\|_{L^p\left(\Omega^*\right)}\\ &\lesssim& \left\| g\right\|_{L^p\left(\Omega\right)} \end{align}\] where the last inequality follows due to the boundedness of \(h_{\Omega^*}\). This shows that \(\mathcal{F}_{n+1}\) is a bounded operator from \(A^p\left(\Omega\right)\) to \(L^p\left(\Omega^*\right)\), \(\forall p\in(1,\infty)\).
Recall that \(\mathcal{F}_{n+1}\) on \(A^p(\Omega)\) is, in fact, the composition \[A^p(\Omega)\overset{\iota} {\hookrightarrow}L^p(\Omega)\xrightarrow\theta\mathcal{O}'(\overline{\Omega})\xrightarrow{\mathcal{F}_{n+1}} \mathcal{O}(\Omega^*),\] where \(\iota\) is the inclusion map, and \(\theta\) is as in 4 . By the Martineau–Aizenberg duality theorem, the right-most map is already known to be injective. Thus, we must show that, for \(g\in A^p(\Omega)\), if \((\theta\circ\iota)(g)=\mu_g\) given by \[\mu_g:f\mapsto \int_{\Omega}\overline{g(\zeta)} f(\zeta)dV(\zeta),\,\quad f\in \mathcal{O}\left(\overline{\Omega}\right)\] is the zero map, then \(g\equiv 0\). Let \(g\in A^p\left(\Omega\right)\), such that \(\mu_{g}=0\). For any \(f\) in the dense subspace \(\mathcal{O}\left(\overline{\Omega}\right)\subset A^q\left(\Omega\right)\), \(\mu_g(f)=\overline{\Phi_g(f)}\), where \(\Phi_g\) is as in Lemma 7. Thus, \(\Phi_g\equiv 0\), which implies that \(g\equiv 0\). This completes the proof of injectivity.
Let \(g\in A^p\left(\Omega^*\right)\), where \(p\in\left(1,\infty\right)\). We define a linear functional \(F_g\) on \(A^q\left(\Omega\right)\), where \(p^{-1}+q^{-1}=1\), as follows \[F_g(\psi)=\frac{1}{(2i)^nn!}\int_{\Omega} \psi\left(\zeta\right)\left(g\circ T_\Omega\right)\left(\zeta\right)\left(\mathfrak H\circ T_\Omega\right)\left(\zeta\right)dV(\zeta),\quad \forall \psi\in A^q\left(\Omega\right),\] where \(\mathfrak H:\Omega^*\rightarrow\mathbb{C}\), defined as \[\label{eq:mathfrak32H} \mathfrak H\left(\eta\right)= {\mathfrak h_{\Omega^*}(\eta)}{h_{\Omega}\left(T_{\Omega^*}\left(\eta\right)\right)}\left(\frac{1+\rho_{\Omega^*}(\eta)}{\left\langle\partial \rho_{\Omega^*}(\eta),\eta\right\rangle}\right)^{n+1},\tag{45}\] where \(\mathfrak h_{\Omega^*}\), \(h_{\Omega}\) are as in 35 and 9 , respectively, and \(T_\Omega, T_{\Omega^*}\) are as in 8 . Since \(h_{\Omega}(T_{\Omega^*}(\eta))\), \(\left|\mathfrak h_{\Omega^*}(\eta)\right|\), and \(\left|\frac{1+\rho_{\Omega^*}(\eta)}{\left\langle\partial \rho_{\Omega^*}(\eta),\eta\right\rangle}\right|\) are bounded above by positive constants on \(\Omega^*\), \(\left|\mathfrak H\right|\) is bounded above on \(\Omega^*\). Thus, it follows that \[\begin{align} \left|F_g\left(\psi\right)\right|&\lesssim&\left(\int_\Omega\left| g\circ T_\Omega\right|^p\left(\zeta\right)dV\left(\zeta\right)\right)^{\frac{1}{p}}\|\psi\|_{L^q\left(\Omega\right)},\\ &\lesssim& \|g\|_{L^p\left(\Omega^*\right)}\|\psi\|_{L^q\left(\Omega\right)}. \end{align}\] This shows that \(F_g\) is a bounded linear functional on \(A^q\left(\Omega\right)\). Thus, by Lemma 7, there exists a \(\phi_g\in A^p\left(\Omega\right)\), such that \[F_g(\psi)=\int_\Omega\overline{\phi_g\left(\zeta\right)} \psi\left(\zeta\right)dV\left(\zeta\right).\] Now, for a fixed \(z\in\Omega^*\), taking \(\psi(\zeta)=\frac{n!}{\pi^n}\left(1-\left\langle\zeta,z\right\rangle\right)^{-n-1}\), we obtain that \[\begin{align} \nonumber\mathcal{F}_{n+1}(\phi_g)(z)=\frac{n!}{\pi^n}\int_\Omega\frac{\overline{\phi_g\left(\zeta\right)}}{\left(1-\left\langle\zeta,z\right\rangle\right)^{n+1}} dV\left(\zeta\right)&=& \frac{1}{(2\pi i)^n}\int_\Omega\dfrac{\left(g\circ T_\Omega\right)\left(\zeta\right)}{\left(1-\left\langle\zeta,z\right\rangle\right)^{n+1}}\left(\mathfrak H\circ T_\Omega\right)\left(\zeta\right)dV(\zeta)\\ \nonumber(\text{since }T_\Omega^{-1}=T_{\Omega^*})\quad &=&\frac{1}{(2\pi i)^n}\int_\Omega\dfrac{g(T_\Omega(\zeta))\mathfrak H(T_\Omega(\zeta))}{\left(1-\left\langle T_{\Omega^*}(T_\Omega(\zeta)),z\right\rangle\right)^{n+1}} \frac{h_\Omega(\zeta)}{h_\Omega(T_{\Omega^*}(T_\Omega(\zeta)))}dV(\zeta)\\ \nonumber(\text{by Lemma \ref{le:cov32dom}})\quad &=&\frac{1}{(2\pi i)^n}\int_{\Omega^*} \dfrac{g(\eta)\mathfrak H(\eta)}{\left(1-\left\langle T_{\Omega^*}(\eta),z\right\rangle\right)^{n+1}} \frac{1}{h_\Omega(T_{\Omega^*}(\eta))} dV(\eta)\\ \nonumber(\text{by \eqref{eq:def32of32FLD32kera}, \eqref{eq:tildeK}, and \eqref{eq:mathfrak32H}})\quad &=&\frac{1}{(2\pi i)^n}\int_{\Omega^*}\dfrac{g\left(\eta\right)}{\widetilde{K}_{\Omega^*}\left(\eta,z\right)^{n+1}}\mathfrak h_{\Omega^*}(\eta) dV(\eta)\\ \nonumber(\text{by \eqref{eq:comp32vol}}) \quad &=& \frac{1}{(2\pi i)^n}\int_{\Omega^*}g(\eta)\left(\overline{\partial}_{\eta} G \right)^n(\eta,z)\\ (\text{by Lemma~\ref{le:repr}})\quad \label{eq:repr32fant}&=& g(z). \end{align}\tag{46}\] This completes the proof of Theorem 1.
Proof of Theorem 2 We begin by proving that \(\mathfrak B_n\) is an normed space isomorphism between \(A^2(\mathbb{C}^n,\omega_\Omega)\), and \(A^2(\Omega^*)\).
Lemma 10.
Let \(\Omega\subset\mathbb{C}^n\) be a domain which satisfies the hypothesis of Theorem 2. Then, \[\begin{align} \label{eq:norm32distor32borel} C_1 \|f\|_{A^2\left(\mathbb{C}^n,\omega_\Omega\right)} \leq \|\mathfrak B_n(f)\|_{A^2(\Omega^*)}\leq C_2 \|f\|_{A^2\left(\mathbb{C}^n,\omega_\Omega\right)},\quad f\in A^2\left(\mathbb{C}^n,\omega_\Omega\right), \end{align}\tag{47}\] where \(C_1,C_2\) are positive constants not depending on \(f\).
Proof. Let \(F\in A^2(\mathbb{C}^n,\omega_\Omega)\). For a fixed \(\zeta\in \Omega^*\), consider the sets \[\Omega^*_\zeta=\left\{\eta\in\mathbb{C}:\eta\zeta\in \Omega^*\right\},\quad M_\zeta=\left(\Omega^*_\zeta\right)^*,\] and the function \[F_\zeta(w)=F(w\zeta) w^{n-1},\quad w\in\mathbb{C}.\] Following the same computations as in [6], we obtain the following.
If \(\varphi\) is a \(\sigma_{\Omega^*}\)-measurable function on \(b\Omega^*\), then \[\label{eq:bdry32int32equiv} \int_{b\Omega^*}\varphi(\zeta)d\sigma_{\Omega^*}(\zeta)\approx\int_{b\Omega^*}\int_{b\Omega^*_\zeta}\varphi(\lambda\zeta)d\sigma_{\Omega^*_\zeta}(\lambda) d\sigma_{\Omega^*}(\zeta).\tag{48}\] See [6].
For any \(F\in A^2(\mathbb{C}^n,\omega_\Omega)\), \[\begin{align} \label{eq:eqiv32norm32PW32aux} \|F\|^2_{A^2(\mathbb{C}^n,\omega_\Omega)}\approx\int_{b\Omega^*}\|F_\zeta\|^2_{A^2\left(\mathbb{C},\omega_{M_\zeta}\right)}d\sigma_{\Omega^*}(\zeta). \end{align}\tag{49}\] See [6].
\(M_\zeta\) is a bounded \(\mathcal{C}^2\)-smooth strongly convex domain in \(\mathbb{C}\), and \[\label{eq:support32fns32reln} H_{M_\zeta}(w)=H_\Omega(w\zeta),\quad w\in\mathbb{C}.\tag{50}\] See [6].
The function \(F_\zeta\) satisfies \[\label{eq:one32d32PW} \int_{\mathbb{C}}|F_\zeta(w)|^2 |w|^{\frac{3}{2}}e^{-2H_{M_\zeta}(w)} \left(dd^cH_{M_\zeta}\right)(w)<\infty.\tag{51}\] See [6].
\(F\) satisfies 38 , and hence, \(\mathfrak B_n F\) is well-defined on \(\Omega^\circ\), the polar set of \(\Omega\), and has an analytic continuation to \(\Omega^*\). See [6].
For \(\eta>0\), \[\begin{align} \eta^{(n+1)}\mathfrak B_n F\left(\eta\zeta\right)&=&\eta^{(n+1)}\int_0^\infty F\left({t\eta\zeta}\right)t^n e^{-t}dt\\ &=&\eta^2\int_0^\infty F_{\zeta}\left({t\eta}\right){t} e^{-t}{dt},\\ &=& \eta^{2}\mathfrak B_1 F_\zeta\left({\eta}\right). \end{align}\] Note that the LHS is convergent for all \(\eta\) satisfying \({\eta}\zeta\in \Omega^\circ\). Equivalently by 50 , \(H_{\Omega}(\eta\zeta)=H_{{M_\zeta}}(\eta)<1\), i.e., \(\eta\in {M_\zeta}^{\circ}\). Similar to \(\mathfrak B_nF\) , \(\mathfrak B_1 F_\zeta\) has a holomorphic extension to \({M_\zeta}^*=\Omega^*_\zeta\). From 51 it follows that, \(F_\zeta\in A^2\left(\mathbb{C},\omega_{{M_\zeta}}\right)\), and hence, applying Lemma 9 we get that \[\begin{align} \|F_\zeta\|^2_{A^2\left(\mathbb{C},\omega_{M_\zeta}\right)}&\approx& \left\|\mathfrak B_1F_\zeta\right\|^2_{A^2(\Omega^*_\zeta)}\\ &=&\int_{\Omega^*_\zeta}\left|\mathfrak B_n F_\zeta(\eta\zeta)\right|^2|\eta|^{2n-2} dV(\eta)\\ &\approx& \int_0^1\int_{b\Omega^*_\zeta} \left|\mathfrak B_n F_\zeta(r\lambda\zeta)\right|^2 r^{2n-1} d\sigma_{\Omega^*_\zeta}(\lambda) dr. \end{align}\] Now integrating both side with respect to \(\zeta\) on \(b\Omega^*\), and applying 49 we get that \[\begin{align} \nonumber\|F\|^2_{A^2(\mathbb{C}^n,\omega_\Omega)}&\approx&\int_{b\Omega^*}\|F_\zeta\|^2_{A^2\left(\mathbb{C},\omega_{M_\zeta}\right)} d\sigma_{\Omega^*}(\zeta)\\ \nonumber&\approx& \int_0^1\int_{b\Omega^*}\int_{b\Omega^*_\zeta} \left|\mathfrak B_n F_\zeta(r\lambda\zeta)\right|^2 r^{2n-1} d\sigma_{\Omega^*_\zeta}(\lambda) d\sigma_{\Omega^*}(\zeta)dr\\ \nonumber(\text{by}\,\eqref{eq:bdry32int32equiv})&\approx&\int_0^1\int_{b\Omega^*} \left|\mathfrak B_n F(r\zeta)\right|^2 r^{2n-1} d\sigma_{\Omega^*}(\zeta)dr\\ \label{eq:borel32norm32equiv}(\text{by}\,\eqref{eq:polar32coordinate})&\approx& \|\mathfrak B_n(F)\|^2_{A^2(\Omega^*)}. \end{align}\tag{52}\] This shows 47 , and hence, Lemma 10 is proved. ◻
Let \(f\in A^2\left(\Omega\right)\). Then by definition, for any \(z\in\mathbb{C}^n\), \[\begin{align} \left|\mathcal{L}(f)(z)\right|&\leq&\int_\Omega|f(\zeta)| e^{\operatorname{Re}\langle\zeta,z\rangle} dV(\zeta),\\ &\leq& e^{H_\Omega(z)} \int_\Omega|f(\zeta)| dV(\zeta), \\ &\leq& C e^{H_\Omega(z)}, \end{align}\] where \(C\) is a positive constant does not depends on \(f\) or \(z\). This shows \(\mathcal{L}(f)\) satisfies the condition in 38 , and hence, \(\mathcal{L}(f)\in \mathcal{O}_{\operatorname{exp}}(\Omega)\). Thus, \(\mathfrak B_n\left(\mathcal{L}(f)\right)\) has a holomorphic extension to \(\Omega^*\). Furthermore, for \(z\in\Omega^*\), \[\begin{align} \label{eq:Lap32borel32reln} \nonumber\mathfrak B_n\left(\mathcal{L}(f)\right)(z)&=&\int_0^\infty \int_{\Omega} \overline{f(\zeta)} e^{-t\left(-\langle\zeta,z\rangle+1\right)} t^n dV(\zeta) dt\\ \nonumber&=& \int_\Omega\overline{f(\zeta)}\int_0^\infty e^{-t\left(-\langle\zeta,z\rangle+1\right)} t^n dt dV(\zeta)\\ \nonumber&=&n!\int_\Omega\frac{\overline{f(\zeta)}}{\left(1-\langle\zeta,z\rangle\right)^{n+1}} dV(\zeta)\\ &=& \pi^n \mathcal{F}_{n+1}(f)(z). \end{align}\tag{53}\] By Theorem 1, \(\mathcal{F}_{n+1}\) is a normed space isomorphism between \(A^2(\Omega)\) and \(A^2(\Omega^*)\). Thus, proof of the theorem reduces to showing \(\mathfrak B_n\) is a normed space isomorphsim between \(A^2(\mathbb{C}^n,\omega_\Omega)\), and \(A^2(\Omega^*)\). However, from Lemma 10, we have already obtained the injectivity and the \(L^2\)-boundedness of \(\mathfrak B_n\). Hence, to conclude the theorem, it suffices to show that \(\mathfrak B_n\) is surjective.
Let \(g\in A^2(\Omega^*)\). Consider the following analytic functional on \(\mathcal{O}'\left(\overline{\Omega}\right)\). \[\mu_g(\psi)=\int_\Omega\psi(\zeta)\left(g\circ T_\Omega\right)(\zeta) \left(\mathfrak H\circ T_\Omega\right)(\zeta) dV(\zeta),\quad\forall \psi\in \mathcal{O}(\overline{\Omega}),\] where \(\mathfrak H:\Omega^*\rightarrow\mathbb{C}\) is defined as \[\label{eq:pairing} \mathfrak H\left(\eta\right)= {\mathfrak h_{\Omega^*}(\eta)}{h_{\Omega}\left(T_{\Omega^*}\left(\eta\right)\right)}\left(\frac{1+\rho_{\Omega^*}(\eta)}{\left\langle\partial \rho_{\Omega^*}(\eta),\eta\right\rangle}\right)^{n+1},\tag{54}\] for \(\mathfrak h_{\Omega^*}\), \(h_{\Omega}\) as in 35 and 9 , respectively, and \(T_\Omega, T_{\Omega^*}\) as in 8 . Let \(\widehat\mu_g\) be the Laplace transform of \(\mu_g\) defined as \[\widehat\mu_g(z)=\mu_g\left(e^{\langle\cdot,z\rangle}\right),\quad z\in\mathbb{C}^n.\] Then, \(\widehat\mu_g\in\mathcal{O}(\mathbb{C}^n)\), and satisfies the condition in 38 . Hence, \(\mathfrak B_n\left(\widehat \mu_g\right)\) has an analytic continuation to \(\Omega^*\). \[\begin{align} \mathfrak B_n\left(\widehat \mu_g\right)=\pi^n{\mathcal{F}_{n+1}(\mu_g)}. \end{align}\]
Following the same computation as in the last paragraph of the previous section, we get that \[\mathcal{F}_{n+1}(\mu_g)(z)=g(z),\quad z\in \Omega^*.\] Also, as \(g\in A^2(\Omega^*)\), using the same computations in 52 backwards, we get, \(\widehat \mu_g\in A^2(\mathbb{C}^n,\omega_\Omega)\). This shows \(\frac{\widehat \mu_g}{\pi^n}\) is in the pre-image of \(g\) under \(\mathfrak B_n\), which shows \(\mathfrak B_n\) is surjective, and the proof is complete.
Proof of Theorem 3 Let us recall \[\begin{align} \Omega=\left\{(\zeta_1,\zeta_2)\in\mathbb{C}^2: |\zeta_1|+|\zeta_2|<1\right\}. \end{align}\] By definition, \(\Omega\) is a circled domain, in fact, it is a Reinhardt domain. Consequently, it follows from [6] that (the interior of) \(\Omega^*\) can be given as \[\Omega^*=\left\{z\in\mathbb{C}^2: H_\Omega(z)<1\right\},\] where \(H_\Omega(\overline{z})\) is the support function of \(\Omega\). In this case, \(H_\Omega(z)=\max\{|z_1|,|z_2|\}\). Hence, \[\Omega^*=\left\{\left(z_1,z_2\right)\in\mathbb{C}^2: |z_1|<1,|z_2|<1\right\}.\] Consider the following parameterization of \(\Omega\). \[\begin{align} \vartheta:(0,1)\times[0,1]\times [0,2\pi)^2&\rightarrow& \Omega,\notag \\ (r,s,\theta_1,\theta_2)&\mapsto&\left(rse^{i\theta_1},r(1-s)e^{i\theta_2}\right). \label{eq:param32of32diamond} \end{align}\tag{55}\] Under this change of co-ordinates, the pull-back of the Lebesgue measure is given by \[\begin{align} \label{eq:pullback32of32lbge} \vartheta^*\left(dV\right)=-\frac{1}{4}\vartheta^*\left(dz_1 dz_2 d\overline{z_1} d\overline{z_2}\right)=\frac{1}{2}r^3s(1-s) dr ds d\theta_1 d\theta_2. \end{align}\tag{56}\] Since \(\Omega\) is a complete Reinhardt domain, any holomorphic function has a global power series expansion on \(\Omega\).
Lemma 11. Let \(\Omega=\{(\zeta_1,\zeta_2)\in\mathbb{C}^2: |\zeta_1|+|\zeta_2|<1\}\). Then, the following holds.
\(f\in A^2\left(\Omega\right)\) if and only if \[\label{eq:series32exp32diamond} f(\zeta_1,\zeta_2)=\sum\limits_{(m_1,m_2)\in \mathbb{N}^2} a_{m_1m_2}\zeta_1^{m_1}\zeta_2^{m_2},\quad \left(\text{in} \,L^2\left(\Omega\right)\right),\tag{57}\] where \(\left(a_{m_1,m_2}\right)_{\mathbb{N}^2}\) are complex numbers that satisfy \[\label{eq:condition32on32coeff32diamond} \sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{|a_{m_1,m_2}|^2}{(m_1+m_2+2)}\frac{{\left(2m_1+1\right)!}{\left(2m_2+1\right)!}}{{\left(2m_1+2m_2+3\right)!}}<\infty.\tag{58}\] Moreover, \[\label{eq:norm32eq32diamond} \|f\|_{A^2\left(\Omega\right)}^2\approx \sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{|a_{m_1,m_2}|^2}{(m_1+m_2+2)}\frac{{\left(2m_1+1\right)!}{\left(2m_2+1\right)!}}{{\left(2m_1+2m_2+3\right)!}}.\tag{59}\]
\(F\in A^2\left(\Omega^*\right)\) if and only if \[\label{eq:series32exp32polydisc} F(z_1,z_2)=\sum\limits_{(m_1,m_2)\in \mathbb{N}^2} t_{m_1m_2}z_1^{m_1}z_2^{m_2},\quad \left(\text{in} \,L^2\left(\Omega^*\right)\right),\tag{60}\] where \(\left(t_{m_1,m_2}\right)_{\mathbb{N}^2}\) are complex numbers that satisfy \[\label{eq:condition32on32coeff32polydisc} \sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{|t_{m_1,m_2}|^2}{(m_1+1)(m_2+1)}<\infty.\tag{61}\] Moreover, \[\label{eq:norm32eq32polydisc} \|F\|_{A^2\left(\Omega^*\right)}^2\approx \sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{|t_{m_1,m_2}|^2}{(m_1+1)(m_2+1)}.\tag{62}\]
Let \(G\in \mathcal{O}(\mathbb{C}^2)\) with the power series expansion \[\label{eq:series32exp32PW} G(z_1,z_2)=\sum\limits_{(m_1,m_2)\in \mathbb{N}^2} \ell_{m_1m_2}z_1^{m_1}z_2^{m_2},\quad \left(\text{uniformly on compact subsets of } \mathbb{C}^2 \right).\tag{63}\]
Proof. Let \(f\in A^2(\Omega)\). Due to the Reinhardtness of \(\Omega\), there exists a sequence \((a_{m_1,m_2})_{\mathbb{N}^2}\) such that \[p_k(\zeta_1,\zeta_2)\rightarrow f(\zeta_1,\zeta_2)\quad \text{in}\, L^2(\Omega),\] where \(p_k(\zeta_1,\zeta_2)=\sum\limits_{m_1,m_2=0}^k a_{m_1,m_2} \zeta_1^{m_1}\zeta_2^{m_2}\). Now \[\begin{align} \nonumber \|p_k\|_{A^2(\Omega)}^2&=&\int_\Omega\left|{\sum\limits_{m_1,m_2=0}^k a_{m_1,m_2}} \zeta_1^{m_1}\zeta_2^{m_2}\right|^2 dV(\zeta_1,\zeta_2)\\ \nonumber \text{(by \eqref{eq:param32of32diamond}, \eqref{eq:pullback32of32lbge})}&=& 4\pi^2 \sum\limits_{m_1,m_2=0}^k |a_{m_1,m_2}|^2\int_0^1r^{2m_1+2m_2+3}dr\int_0^1 s^{2m_1+1}(1-s)^{2m_2+1} ds\\ \nonumber&\approx& \sum\limits_{m_1,m_2=0}^k \frac{|a_{m_1,m_2}|^2}{m_1+m_2+2}\frac{(2m_1+1)!(2m_2+1)!}{(2m_1+2m_2+3)!} \end{align}\] As \(\lim\limits_{k\rightarrow\infty}\|p_k\|_{A^2(\Omega)}^2=\|f\|_{A^2(\Omega)}^2\), it follows that 58 , and 59 holds.
Conversely, if \((a_{m_1,m_2})_{\mathbb{N}^2}\) is a sequence of complex numbers satisfying 58 , then the function \(f\) defined as in 57 is in \(L^2(\Omega)\), and is approximable by holomorphic polynomials of the form \(p_k(\zeta_1,\zeta_2)=\sum\limits_{m_1,m_2=0}^ka_{m_1,m_2}\zeta_1^{m_1} \zeta_2^{m_2}\), \(k\in\mathbb{N}\). Thus, \(f\in A^2(\Omega)\). This completes the proof of (i).
Suppose \(q_k\) is a polynomial of the form \(q_k(z_1,z_2)=\sum\limits_{m_1,m_2=0}^kt_{m_1,m_2}z_1^{m_1} z_2^{m_2}\). Then \[\begin{align} \nonumber \|q_k\|_{A^2(\Omega^*)}^2&=&\int_{\Omega^*}\left|{\sum\limits_{m_1,m_2=0}^k t_{m_1,m_2}} z_1^{m_1}z_2^{m_2}\right|^2 dV(z_1,z_2)\\ \nonumber &=& 4\pi^2 \sum\limits_{m_1,m_2=0}^k |t_{m_1,m_2}|^2\int_0^1r^{2m_1+1}dr\int_0^1 r^{2m_2+1} dr\\ \nonumber &\approx& \sum\limits_{m_1,m_2=0}^k \frac{|t_{m_1,m_2}|^2}{(m_1+1)(m_1+1)}. \end{align}\] Now, following the same argument as in (i), we get (ii).
From [18], it follows that the measure \((dd^c H_\Omega)^2\) is supported on the real hypersurface \(M=\{(z_1,z_2)\in\mathbb{C}^2:|z_1|=|z_2|\}\), and if we parameterize \(M\) via \(\vartheta_M \left(r,\psi_1,\psi_2\right)=\left(re^{i\psi_1},re^{i\psi_2}\right)\), we obtain that \[\label{eq:MA32bi32disc} \vartheta_M^* (dd^cH_\Omega)^2\left(r,\psi_1,\psi_2\right)\approx dr d\psi_1 d\psi_2,\quad (r,\psi_1,\psi_2)\in(0,\infty)\times[0,2\pi)^2.\tag{64}\] For an explicit computation, see [10]. Next, if we take a polynomial of the form \(\mathcal{P}_k(z_1,z_2)=\sum\limits_{m_1,m_2=0}^k\ell_{m_1,m_2}z_1^{m_1} z_2^{m_2}\). Then from 64 , it follows that \[\begin{align} \nonumber \|\mathcal{P}_k\|_{A^2(\mathbb{C}^2,\omega_\Omega)}^2&=&\int_{\mathbb{C}^2}\left|{\sum\limits_{m_1,m_2=0}^k \ell_{m_1,m_2}} z_1^{m_1}z_2^{m_2}\right|^2 e^{-2\max\{|z_1|,|z_2|\}}\|z\|^{\frac{5}{2}}(dd^cH_\Omega)^2(z_1,z_2)\\ \nonumber &\approx& \sum\limits_{m_1,m_2=0}^k |\ell_{m_1,m_2}|^2\int_0^\infty r^{2m_1+2m_2+\frac{5}{2}}e^{-2r} dr\\ \nonumber &\approx& \sum\limits_{m_1,m_2=0}^k \frac{|\ell_{m_1,m_2}|^2}{2^{2m_1+2m_2}}\Gamma\left(2m_1+2m_2+\frac{7}{2}\right). \end{align}\] Now, following the same steps as in (i) and (ii), we obtain (iii)(a).
To show (iii)(b), we first show that for any \(z\in\mathbb{C}^n\), satisfying \(H_\Omega(z)>1\), \[\left\|e^{\langle\cdot, z\rangle}\right\|^2_{L^2(\Omega)}\approx {e^{2H_\Omega(z)}}\|z\|^{-2}.\] Parameterizing the set \(\{z\in\mathbb{C}^2: H_\Omega(z)>1\}\) as \(\{(te^{i\psi_1},te^{i\psi_2}): t>1, \psi_1,\psi_2\in[0,2\pi]\}\) we get \[\begin{align} \label{eq:main32est32counter} \nonumber e^{-2H_\Omega(z)}&\|z\|^2\left\|e^{\langle\cdot, z\rangle}\right\|^2_{L^2(\Omega)}\\ \nonumber&\approx t^2\int_0^1\int_0^1\int_0^{2\pi}\int_0^{2\pi} e^{2tr\left(s\cos(\theta_1-\psi_1)+(1-s)\cos(\theta_2-\psi_2)-1\right)} e^{2t(r-1)}r^3s(1-s) d\theta_1 d\theta_2 ds dr\\ &=t^2\int_0^1\int_0^1\int_0^{2\pi}\int_0^{2\pi} e^{2tr\left(s\cos\theta_1+(1-s)\cos\theta_2-1\right)} e^{2t(r-1)}r^3s(1-s) d\theta_1 d\theta_2 ds dr. \end{align}\tag{65}\] For \(t>1\) and \(r>0\), let us denote \[\begin{align} Q(r,t)&=rt\int_0^1\int_0^{2\pi}\int_0^{2\pi} e^{2rt\left(s\cos\theta_1+(1-s)\cos\theta_2-1\right)} s(1-s) d\theta_1 d\theta_2 ds\\ &= 4\pi^2 e^{-2rt}rt\int_0^1 I_0(2rts) I_0(2rt(1-s)) s(1-s) ds, \end{align}\] where \(I_0\) is the modified Bessel function of the first kind of order \(0\), given by \[I_0(x)=\frac{1}{2\pi}\int_0^{2\pi} e^{{x} \cos\theta} d\theta,\quad \text{for } x\in\mathbb{R}.\] By [19], as \(x\rightarrow\infty\), \[I_0(x) \frac{(1+x)^{\frac{1}{2}}}{e^x}\rightarrow 1.\] This implies that for \(x\geq 0\), \[I_0(x)\approx \frac{e^x}{(1+x)^{\frac{1}{2}}}.\] Using this estimate on \(Q(r,t)\), we obtain that for \(t>1\) and \(r>0\), \[\begin{align} Q(r,t)&\approx rt\int_0^1 (1+2rts)^{-\frac{1}{2}}(1+2rt(1-s))^{-\frac{1}{2}} s(1-s) ds\\ &= rt\int_0^1 (1+2rt+(2rt)^2 s(1-s))^{-\frac{1}{2}} s(1-s) ds. \end{align}\] From this expression, for \(t>1\) and \(r>0\) such that \(rt\leq 1\), \[\begin{align} Q(r,t)\approx rt \int_0^1 s(1-s) ds\approx rt. \end{align}\] Furthermore, for \(t>1\) and \(r>0\) such that \(rt\geq 1\), \[\begin{align} \nonumber Q(r,t)&\approx \int_0^1 \left(s(1-s)+\frac{1}{rt}\right)^{-\frac{1}{2}} s(1-s) ds\\ &\approx \int_0^1 s^{\frac{1}{2}}(1-s)^{\frac{1}{2}} ds\approx 1. \end{align}\] Consequently, as \(t\rightarrow\infty\), \[\begin{align} \label{eq:Qt321} \nonumber t\int_0^{\frac{1}{t}} e^{2t(r-1)}Q(r,t) r^2 dr&\leq t^2 e^{2-2t}\int_0^{\frac{1}{t}}r^3 dr\\ &= \frac{ e^{2-2t}}{4t^2}\rightarrow 0. \end{align}\tag{66}\] Furthermore, by applying a change of variable \(u=t(1-r)\), and applying the dominated convergence theorem, as \(t\rightarrow \infty\), \[\begin{align} \label{eq:Qt322} \nonumber t \int_{\frac{1}{t}}^1 e^{2t(r-1)}Q(r,t) r^2 dr&= t \int_{\frac{1}{t}}^1 e^{2t(r-1)} r^2 dr\\ &=\int_0^{t-1} e^{-2u} \left(1-\frac{u}{t}\right)^2 du \rightarrow \int_0^\infty e^{-2u} du=\frac{1}{2}. \end{align}\tag{67}\] Combining 66 and 67 , we get that as \(t\rightarrow\infty\), \[\begin{align} \label{eq:Qt32final} t\int_0^1 e^{2t(r-1)}Q(r,t) r^2 dr\rightarrow \frac{1}{2}. \end{align}\tag{68}\] Finally by 65 and 68 , for \(t>1\), \[\begin{align} e^{-2H_\Omega(z)}\|z\|^2\left\|e^{\langle\cdot, z\rangle}\right\|^2_{L^2(\Omega)}\approx t\int_0^1 e^{2t(r-1)}Q(r,t) r^2 dr\approx 1. \end{align}\] This completes the proof of our claim.
Combining this with 64 , we get \[\begin{align} \nonumber \|\mathcal{P}_k\|_{A^2(\mathbb{C}^2,\mu_\Omega)}^2&\approx&\int_{\mathbb{C}^2}\left|{\sum\limits_{m_1,m_2=0}^k \ell_{m_1,m_2}} z_1^{m_1}z_2^{m_2}\right|^2 e^{-2\max\{|z_1|,|z_2|\}}\|z\|^{2}(dd^cH_\Omega)^2(z_1,z_2)\\ \nonumber &\approx& \sum\limits_{m_1,m_2=0}^k |\ell_{m_1,m_2}|^2\int_0^\infty r^{2m_1+2m_2+2}e^{-2r} dr\\ \nonumber &\approx& \sum\limits_{m_1,m_2=0}^k \frac{|\ell_{m_1,m_2}|^2}{2^{2m_1+2m_2}}(2m_1+2m_2+2)!. \end{align}\] Now, following the same steps as in (i) and (ii), we obtain (iii)(b). ◻
Next, we expand the Fantappiè and Laplace transforms of \(A^2\left(\Omega\right)\)-functions in terms of power series.
Lemma 12. Let \(\Omega\) be as above, and \(f\in A^2\left(\Omega\right)\), which admits the expansion in 57 .
The Fantappiè transform of \(f\) is given by, \[\begin{align} \label{eq:Fan32expand} \mathcal{F}_3\left(f\right)(z)=\sum\limits_{(m_1,m_2)\in\mathbb{N}^2} t_{m_1,m_2} z_1^{m_1}z_2^{m_2},\quad z=(z_1,z_2)\in\Omega^*, \end{align}\tag{69}\] where \[\begin{align} \label{eq:Fan32coeff} t_{m_1,m_2}=2\overline{a_{m_1,m_2}}\frac{\left(m_1+m_2+1 \right)!}{m_1! m_2!}\dfrac{(2m_1+1)!(2m_2+1)!}{(2m_1+2m_2+3)!}. \end{align}\tag{70}\]
The Laplace transform of \(f\) is given by, \[\begin{align} \label{eq:Lap32expand} \mathcal{L}\left(f\right)(z)=\sum\limits_{(m_1,m_2)\in\mathbb{N}^2} \ell_{m_1,m_2} z_1^{m_1}z_2^{m_2},\quad z=(z_1,z_2)\in\mathbb{C}^n, \end{align}\tag{71}\] where \[\begin{align} \label{eq:Lap32coeff} \ell_{m_1,m_2}=\pi^2\frac{\overline{a_{m_1,m_2}}}{m_1! m_2!}\dfrac{(2m_1+1)!(2m_2+1)!}{(m_1+m_2+2)(2m_1+2m_2+3)!}. \end{align}\tag{72}\]
Proof. Let \(z\in\Omega^*\). By the Cauchy–Schwarz inequality, there exists a \(C_z>0\) such that, \[\begin{align} \left|\mathcal{F}_3\left(f\right)(z)\right|\leq C_z \|f\|_{A_2\left(\Omega\right)}, \end{align}\] and \[\begin{align} \left|\mathcal{L}\left(f\right)(z)\right|\leq C_z \|f\|_{A_2\left(\Omega\right)},\quad f\in A^2(\Omega) \end{align}\] Thus, it suffices to establish the claim for polynomials of the form \(p_k\left(z_1,z_2\right)=\sum\limits_{m_1,m_2=0}^k a_{m_1,m_2}z_1^{m_1}\,z_2^{m_2}.\) We first begin by showing that, \(\left|\left\langle\zeta,z\right\rangle\right|< 1,\,\forall\zeta\in\overline{\Omega}\). To see this, if there exists \(\widetilde{\zeta}\in\overline{\Omega}\) such that \(\left|\left\langle\widetilde{\zeta},z\right\rangle\right|=1\). Since \(\Omega\) is Reinhardt, \(\frac{\widetilde{\zeta}}{\left\langle\widetilde{\zeta},z\right\rangle}\in\overline{\Omega}\), and \(\left\langle\frac{\widetilde{\zeta}}{\left\langle\widetilde{\zeta},z\right\rangle},z\right\rangle=1\), which is not possible since, \(z\in \Omega^*\), which proves our claim. Thus, using the series of \(\left(1-\left\langle\zeta,z\right\rangle\right)^{-3}\), and 57 , we get that for each \(z=(z_1,z_2)\in\Omega^*\), \[\begin{align} \mathcal{F}_3\left(p_k\right)(z)&=&\frac{2}{\pi^2}\sum_{(k_1,k_2)\in\mathbb{N}^2} \dfrac{\left(k_1+k_2+2\right)!}{k_1!k_2!} \left(\int_\Omega\overline{f\left(\zeta\right)} \zeta_1^{k_1}\zeta_2^{k_2} dV(\zeta)\right)z_1^{k_1} z_2^{k_2}\\ &=&\frac{2}{\pi^2}\sum_{(k_1,k_2)\in\mathbb{N}^2}\sum\limits_{m_1,m_2=0}^k \dfrac{\left(k_1+k_2+2\right)!}{k_1!k_2!} \left(\overline{a_{m_1,m_2}}\int_\Omega\overline{\zeta_1^{m_1}}\overline{\zeta_2^{m_2}}\zeta_1^{k_1}\zeta_2^{k_2} dV(\zeta)\right)z_1^{k_1} z_2^{k_2}. \end{align}\] Similarly for \(\mathcal{L}\), we get that for each \(z=(z_1,z_2)\in\mathbb{C}^2\), \[\begin{align} \mathcal{L}(p_k)(z)&=&\sum\limits_{(k_1,k_2)\in\mathbb{N}^2} \left(\int_\Omega\overline{f\left(\zeta\right)} \zeta_1^{k_1}\zeta_2^{k_2} dV(\zeta)\right)\frac{z_1^{k_1}z_2^{k_2}}{k_1!k_2!}\\ &=&\sum\limits_{(k_1,k_2)\in\mathbb{N}^2}\left(\overline{a_{m_1,m_2}}\int_\Omega\overline{\zeta_1^{m_1}}\overline{\zeta_2^{m_2}}\zeta_1^{k_1}\zeta_2^{k_2} dV(\zeta) \right)\frac{z_1^{k_1}z_2^{k_2}}{k_1!k_2!}. \end{align}\] Finally, by combining the orthogonality of monomials (due to Reinhardtness) with the computation in the proof establishing 58 , we deduce that \[\begin{align} \int_{\Omega}\overline{\zeta_1^{m_1}}\overline{\zeta_2^{m_2}}\zeta_1^{k_1}\zeta_2^{k_2} dV(\zeta)=\begin{cases} \frac{2\pi^2}{\left(2m_1+2m_2+4\right)}\frac{{\left(2m_1+1\right)!}{\left(2m_2+1\right)!}}{{\left(2m_1+2m_2+3\right)!}},&\quad \text{for}\,(k_1,k_2)=(m_1,m_2),\\ 0,&\quad \text{otherwise}. \end{cases} \end{align}\] Replacing this in the above equation, the claim follows for \(p_k\), and hence for \(f\in A^2\left(\Omega\right)\). ◻
Let \(f\in A^2\left(\Omega\right)\), with power series as in 57 . Due to Lemma 12, we get that for \(z\in\Omega^*\), \(\mathcal{F}_3\left(f\right)(z)=\sum\limits_{(m_1,m_2)\in\mathbb{N}^2}t_{m_1,m_2} z_1^{m_1}z_2^{m_2}\), where \(t_{m_1,m_2}\) is as in 70 . Now, using 59 , 62 , and computations we get that \[\begin{align} \|\mathcal{F}_3\left(f\right)\|_{A^2\left(\Omega^*\right)}^2 &\approx& \sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{|t_{m_1,m_2}|^2}{(2m_1+2)(2m_2+2)}\\ &\approx&\sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{\left|a_{m_1,m_2}\right|^2}{\left(2m_1+1\right)\left(2m_2+1\right)}\frac{{\left(m_1+m_2+1\right)!}^2}{{m_1!}^2 {m_2!}^2}\frac{{\left(2m_1+1\right)!}^2{\left(2m_2+1\right)!}^2}{{\left(2m_1+2m_2+3\right)!}^2} \\ &\lesssim& \sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{|a_{m_1,m_2}|^2}{(m_1+m_2+2)}\frac{{\left(2m_1+1\right)!}{\left(2m_2+1\right)!}}{{\left(2m_1+2m_2+3\right)!}}\\ &\lesssim& \|f\|_{A^2\left(\Omega\right)}^2, \end{align}\] where the second last inequality follows from Stirling’s approximation applied as follows: \[\begin{align} \frac{m_1+m_2+2}{\left(2m_1+1\right)\left(2m_2+1\right)}\frac{{\left(m_1+m_2+1\right)!}^2}{{m_1!}^2 {m_2!}^2}\frac{{\left(2m_1+1\right)!}{\left(2m_2+1\right)!}}{{\left(2m_1+2m_2+3\right)!}}\approx{m_1^{-\frac{1}{2}}+m_2^{-\frac{1}{2}}} \lesssim 1. \end{align}\] This shows that \(\mathcal{F}_3\) is a \(L^2\)-bounded operator from \(A^2\left(\Omega\right)\) to \(A^2\left(\Omega^*\right)\).
A similar computation yields that \[\begin{align} \|\mathcal{L} (f)\|_{A^2(\mathbb{C}^2,\omega_\Omega)}^2&\approx&\sum\limits_{(m_1,m_2)\in\mathbb{N}^2}\frac{|\ell_{m_1,m_2}|^2}{2^{2m_1+2m_2}}{\Gamma\left(2m_1+2m_2+\frac{7}{2}\right)}\\ &\approx&\sum\limits_{(m_1,m_2)\in\mathbb{N}^2}{|a_{m_1,m_2}|^2}\dfrac{\Gamma\left(2m_1+2m_2+\frac{7}{2}\right)(2m_1+1)!^2(2m_2+1)!^2}{2^{2m_1+2m_2}(m_1+m_2+2)^2m_1!^2m_2!^2(2m_1+2m_2+3)!^2}\\ &\lesssim& \sum_{(m_1,m_2)\in \mathbb{N}^2}\frac{|a_{m_1,m_2}|^2}{(m_1+m_2+2)}\frac{{\left(2m_1+1\right)!}{\left(2m_2+1\right)!}}{{\left(2m_1+2m_2+3\right)!}}\\ &\lesssim& \|f\|_{A^2\left(\Omega\right)}^2, \end{align}\] where the second last inequality follows from Stirling’s approximation applied as follows. \[\begin{align} \dfrac{\Gamma\left(2m_1+2m_2+\frac{7}{2}\right)(2m_1+1)!(2m_2+1)!}{2^{2m_1+2m_2}(m_1+m_2+2)m_1!^2m_2!^2(2m_1+2m_2+3)!}&\approx& \frac{m_1^{\frac{1}{2}} m_2^{\frac{1}{2}}}{(m_1+m_2)^{\frac{3}{2}}}\\ \text{(by AM-GM)} &\lesssim& \frac{1}{(m_1+m_2)^{\frac{1}{2}}}\lesssim 1. \end{align}\] This shows \(\mathcal{L}\) is \(L^2\)-bounded operator form \(A^2(\Omega)\) to \(A^2( \mathbb{C}^2,\omega_\Omega)\).
Next, to show that \(\mathcal{F}_3\) is not an onto map, consider \(F\left(z_1,z_2\right)=\sum\limits_{(m_1,m_2)\in \mathbb{N}^2} t_{m_1,m_2} z_1^{m_1} z_2^{m_2}\), where \[\begin{align} t_{m_1,m_2}=\begin{cases} (2k+1)^{\frac{1}{4}},& \text{when } m_1=m_2=k>0,\\ 0,&\text{otherwise}. \end{cases} \end{align}\] Then, by 62 , it follows that \(\|F\|_{A^2\left(\Omega^*\right)}^2\approx \sum\limits_{k=1}^\infty \frac{1}{k^{\frac{3}{2}}}<\infty\). Hence, \(F\in A^2\left(\Omega^*\right)\). If there exists a \(f\in A^2\left(\Omega\right)\), such that \(\mathcal{F}_3(f)=F\). Then, \(f\) admits an expansion, as in 57 for some sequence \(\left(a_{m_1,m_2}\right)_{\mathbb{N}^2}\) satisfying 58 . Furthermore, from 70 it follows that \[\begin{align} a_{m_1,m_2}=\begin{cases} \frac{(2k+1)^{\frac{1}{4}}k!^2(4k+3)!}{2(2k+1)!^3},& \text{when } m_1=m_2=k>0,\\ 0,&\text{otherwise}. \end{cases} \end{align}\] Then, by Stirling’s approximation, we get that \[\begin{align} \|f\|_{A^2\left(\Omega\right)}^2&\approx& \sum_{k=1}^\infty \frac{(2k+1)^{\frac{1}{2}}k!^4(4k+3)!}{(2k+1)!^4(2k+2)}\\ &\approx& \sum_{k=1}^\infty \frac{1}{k}=\infty. \end{align}\] This is a contradiction, since \(f\in A^2\left(\Omega\right)\), which proves that \(\mathcal{F}_3\) is not onto from \(A^2\left(\Omega\right)\) to \(A^2\left(\Omega^*\right)\).
Finally, to show that \(\mathcal{L}\) is not onto, consider \(G(z_1,z_2)=\sum\limits_{(m_1,m_2)\in\mathbb{N}^2}\ell_{m_1,m_2}z_1^{m_1}z_2^{m_2}\), where \[\begin{align} \ell_{m_1,m_2}=\begin{cases} \frac{2^{2k}}{k^{\frac{3}{4}} \Gamma\left(4k+\frac{7}{2}\right)^{\frac{1}{2}}},& \text{when } m_1=m_2=k>0,\\ 0,&\text{otherwise}. \end{cases} \end{align}\] Then \(G\in\mathcal{O}(\mathbb{C}^2)\), and by [eq:norm32eq32PW], it follows that \(\|G\|_{A^2(\mathbb{C}^2,\omega_\Omega)}^2\approx \sum\limits_{k=1}^\infty \frac{1}{k^\frac{3}{2}}<\infty.\) Hence, \(G\in A^2(\mathbb{C}^2,\omega_\Omega)\). If there exists a \(f\in A^2(\Omega)\), such that \(\mathcal{L}(f)=G\), then by the relation 72 we have that \[\begin{align} a_{m_1,m_2}=\begin{cases} \frac{2^{2k} k!^2(2k+2)(4k+3)!}{(2k+1)!^2k^{\frac{3}{4}} \Gamma\left(4k+\frac{7}{2}\right)^{\frac{1}{2}}},& \text{when } m_1=m_2=k>0,\\ 0,&\text{otherwise}. \end{cases} \end{align}\] Now, by 58 , and Stirling’s approximation, it follows that \[\begin{align} \|f\|^2_{A^2(\Omega)}&\approx&\sum\limits_{k=1}^\infty \frac{2^{4k} k!^4(2k+2)(4k+3)!}{(2k+1)!^2k^{\frac{3}{2}} \Gamma\left(4k+\frac{7}{2}\right)}\\ &\approx& \sum\limits_{k=1}^\infty \frac{1}{k}=\infty. \end{align}\] This is a contradiction, since \(f\in A^2(\Omega)\).
As for every \((m_1,m_2)\in \mathbb{N}^2\), \((2m_1+2m_2+2)!\leq \Gamma\left(2m_1+2m_2+\frac{7}{2}\right)\), by [eq:condition32on32coeff32PW], and [eq:norm32condition32on32co32eff32PW32YL] it follows that \[A^2(\mathbb{C}^2,\omega_\Omega)\subseteq A^2(\mathbb{C}^2,\mu_\Omega).\] Finally, to show that the inclusion is strict, let us consider \(\widetilde{G}(z_1,z_2)=\sum\limits_{(m_1,m_2)\in\mathbb{N}^2}\widetilde{\ell}_{m_1,m_2}z_1^{m_1}z_2^{m_2}\), where \[\begin{align} \widetilde{\ell}_{m_1,m_2}=\begin{cases} \frac{2^{2k}}{k^{\frac{3}{4}} (4k+2)!^{\frac{1}{2}}},& \text{when } m_1=m_2=k>0,\\ 0,&\text{otherwise}. \end{cases} \end{align}\] Then \(G\in\mathcal{O}(\mathbb{C}^2)\), and by [eq:norm32eq32PW], it follows that \(\|\widetilde{G}\|_{A^2(\mathbb{C}^2,\mu_\Omega)}^2\approx \sum\limits_{k=1}^\infty \frac{1}{k^\frac{3}{2}}<\infty.\) Hence, \(\widetilde{G}\in A^2(\mathbb{C}^2,\mu_\Omega)\). However, by Stirling’s approximation, it follows that \[\begin{align} \|\widetilde{G}\|_{A^2(\mathbb{C}^2,\omega_\Omega)}^2\approx \sum\limits_{k=1}^\infty \frac{1}{k}=\infty. \end{align}\] Hence, \(\widetilde{G}\notin A^2(\mathbb{C}^2,\omega_\Omega)\), which completes the proof of the theorem.
Further remarks In this section, we expand on the scope of the geometric conditions presented in our theorems. We begin by giving an example of a class of bounded \(\mathcal{C}^2\)-smooth strongly convex domains, whose dual complement may not be strongly convex.
Let us consider the following class of real ellipsoids in \(\mathbb{C}^n\). \[\begin{align} \label{eq:counter32nD} \Omega_n=\left\{z=(z_1,z_2,\cdots,z_n)\in\mathbb{C}^n: \sum_{j=1}^n a_j(\operatorname{Re}z_j)^2 +\sum_{j=1}^n b_j(\operatorname{Im}z_j)^2<1\right\}, \end{align}\tag{73}\] where \(a_j\) and \(b_j\) are positive constants such that there exists at least one index \(j\in\{1,2,\cdots,n\}\), with \(a_j\neq b_j\).
Lemma 13. Let \(\Omega_n\) be a real ellipsoid as defined in 73 , where \(a_j \neq b_j\), for some \(j \in \{1, 2, \dots, n\}\). Then its dual complement \(\Omega_n^*\) is not strongly convex when \(a_j = 2b_j\) or \(b_j = 2a_j\), and it fails to be even convex when \(\max\left\{\frac{a_j}{b_j}, \frac{b_j}{a_j}\right\}>2\).
Proof. When \(n=1\), by the definition of the dual complement it follows that \(\zeta\in \Omega_1^*\setminus\{0\}\) if and only if \(\zeta^{-1}\in \overline{\Omega_1}\). Thus, applying the transformation \(\zeta\rightarrow \zeta^{-1}\) on \(\overline{\Omega_1}\setminus\{0\}\), we can verify that \[\Omega_1^*=\{z\in\mathbb{C}: ((\operatorname{Re}z)^2+(\operatorname{Im}z)^2)^2< a_1(\operatorname{Re}z)^2+b_1(\operatorname{Im}z)^2\}.\] It is straightforward to see that \(\Omega_1^*\) is a bounded \(\mathcal{C}^2\)-smooth domain. By analyzing the hessian of the defining function we get that \(\Omega_1^*\) is convex but not strongly convex when \(2a_1=b_1\) or \(2b_1=a_1\), and is not convex when \(\max\{\frac{a_1}{b_1},\frac{b_1}{a_1}\}> 2\).
When \(n\geq2\), it is difficult to explicitly compute the dual complement \(\Omega_n^*\). However, we can determine the failure of convexity or strong convexity by analyzing its linear slices. Without loss of generality, let us assume \(a_1\neq b_1\) and consider the complex line \(\ell_{z_1} =\{ (z_1, 0, \dots, 0) : z_1 \in \mathbb{C}\}\). The slice \(\Omega_n^*\cap \ell_{z_1}\) is a planar domain that is identical to \(\Omega_1^*\). To see this, note that a point \(w = (w_1, 0, \dots, 0)\) belongs to \(\Omega_n^*\) if and only if \(w_1 z_1 \neq 1\) for all \(z \in \Omega_n\). If \(w \in \Omega_n^*\) and \(z_1 \in \Omega_1\), then the point \((z_1, 0, \dots, 0)\) lies in \(\Omega_n\). Therefore, \(w_1 z_1 \neq 1\) for any \(z_1 \in \Omega_1\), which implies \(w_1 \in \Omega_1^*\). Conversely, suppose \(w_1 \in \Omega_1^*\) and \(z = (z_1, z_2,\dots,z_n) \in\Omega_n\). Then \(z_1\in\Omega_1\), which guarantees that the pairing \(\sum_{j=1}^n w_j z_j = w_1 z_1 \neq 1\). This implies \((w_1, 0, \dots, 0) \in \Omega_n^* \cap \ell_{z_1}\). Finally, based on the convexity properties of \(\Omega_1^*\), it follows that the slice \(\Omega_n^* \cap \ell_{z_1}\), and consequently \(\Omega_n^*\) is not strongly convex when \(a_1 = 2b_1\) or \(b_1 = 2a_1\), and fails to be convex when \(\max\left\{\frac{a_1}{b_1}, \frac{b_1}{a_1}\right\}>2\), which completes the proof. ◻
Remark 4. As mentioned in the introduction, we are unaware of any general conditions on bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domains such that their dual complements are strongly convex. However, we can construct a class of domains satisfying this property. Let \(E\) be a bounded \(\mathcal{C}^2\)-smooth strongly convex domain containing the origin, and \(\Omega= E^*\). Then, \(\Omega\) is a strongly \(\mathbb{C}\)-convex domain, and due to \(\mathbb{C}\)-convexity, \(\Omega^*= E\), which is strongly convex. This shows that the hypothesis of Theorem 1 is satisfied for the dual complement of bounded \(\mathcal{C}^2\)-smooth strongly convex domains containing the origin. For example, \(\Omega_n^*\), where \(\Omega_n\) is as defined in 73 , satisfies the hypothesis of Theorem 1.
Remark 5. The hypothesis in Theorem 1 of \(\Omega^*\) being strongly convex is used only to show that \(\mathcal{F}_{n+1}\) is a \(L^p\)-bounded operator between \(A^p(\Omega)\) and \(A^p(\Omega^*)\), which essentially follows from the fact that when \(D\) is a bounded \(\mathcal{C}^2\)-smooth strongly convex domain containing the origin, the operator \(|\mathcal{B}_{D}|\), as given in Lemma 4 is bounded on \(L^p(D)\). Consequently, the requirement of \(\Omega^*\) being strongly convex can be relaxed if one can answer the following question.
Question. Let \(D\subset\mathbb{C}^n\) be a bounded \(\mathcal{C}^2\)-smooth strongly \(\mathbb{C}\)-convex domain containing the origin, and \(|\mathcal{B}_D|\) is the operator as given in Lemma 4. Is \(|\mathcal{B} _ {D}|\) a \(L^p(D)\)-bounded operator?
Remark 6. Note that the domain considered in Theorem 3 is a bounded convex Reinhardt domain which lacks \(\mathcal{C}^1\)-smoothness, which raises the following question.
Question. Does there exist a bounded \(\mathcal{C}^1\)-smooth convex domain \(\Omega\subset \mathbb{C}^n\) such that \(\mathcal{F}_{n+1}\)
is not a normed space isomorphism from \(A^2(\Omega)\) and \(A^2(\Omega^*)\), and \(\mathcal{L}\) is not a normed space isomorphism from \(A^2(\Omega)\) to both \(A^2(\mathbb{C}^n,\omega_\Omega)\) and \(A^2(\mathbb{C}^n,\mu_\Omega)\)?