Ramadanov Convergence Theorem (Theorem # 3709)
Theorem
Let $\Omega_0\subset\mathbb C^n$ be a bounded pseudoconvex domain and let $(\Omega_j)_{j\ge 1}$ be a sequence of bounded pseudoconvex domains in $\mathbb C^n$ satisfying: (i) all $\Omega_j$ and $\Omega_0$ are contained in a common fixed bounded [open set](/page/Open%20Set) $W\subset\mathbb C^n$; (ii) every compact subset of $\Omega_0$ is contained in $\Omega_j$ for all sufficiently large $j$; (iii) every compact subset of $\mathbb C^n\setminus\overline{\Omega_0}$ is eventually disjoint from $\Omega_j$; (iv) the domains are uniformly hyperconvex in the following quantitative sense: there exist constants $M>0$ and $\tau>0$ such that for every $j\geq 0$ there is a plurisubharmonic exhaustion function $\rho_j:\Omega_j\to[-M,0)$ satisfying $\{z\in\Omega_j:\rho_j(z)<-\varepsilon\}\subset\{z\in\Omega_j:\operatorname{dist}(z,\partial\Omega_j)>\tau\varepsilon\}$ for every $0<\varepsilon<1$. Then
\begin{align*}
K_{\Omega_j}(z,w)\longrightarrow K_{\Omega_0}(z,w)
\end{align*}
locally uniformly on $\Omega_0\times\Omega_0$.
Discussion
This theorem says that under suitable geometric convergence of bounded pseudoconvex domains, the Bergman kernels converge locally uniformly on the limit domain. It is useful for proving continuity of analytic invariants under domain approximation.
Proof
[proofplan]
We prove convergence through the extremal characterisation of the Bergman kernel and a normal-family compactness argument. The upper semicontinuity follows by taking norm-one holomorphic functions on the moving domains, extracting locally uniform limits on $\Omega_0$, and applying Fatou's lemma to their $L^2$ norms. The lower semicontinuity is the only place where the quantitative uniform hyperconvexity assumption is used: it gives the Ramadanov-Ligocka uniform $L^2$ approximation theorem for the Cauchy-Riemann operator $\bar{\partial}$, where for a smooth function $u:U\subset\mathbb{C}^n\to\mathbb{C}$ one has $\bar{\partial}u=\sum_{k=1}^n(\partial u/\partial \bar z_k)d\bar z_k$, extending holomorphic $L^2$ functions from $\Omega_0$ to the moving domains with asymptotically no loss of norm. Once the diagonal kernels converge locally uniformly, the full two-variable kernels converge locally uniformly by polarization and [Cauchy estimates](/theorems/2571).
[/proofplan]
[step:Declare the Bergman spaces and the extremal formula for the diagonal kernel]
For a bounded domain $D \subset \mathbb{C}^n$, define the Bergman space
\begin{align*}
A^2(D) := \left\{ f: D \to \mathbb{C} : f \text{ is holomorphic and } \int_D |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) < \infty \right\}.
\end{align*}
Its norm is
\begin{align*}
\|f\|_{A^2(D)} := \left(\int_D |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta)\right)^{1/2}.
\end{align*}
Let $K_D: D \times D \to \mathbb{C}$ denote the Bergman kernel of $D$. We use the Bergman kernel extremal characterisation in the following form: for every $z \in D$,
\begin{align*}
K_D(z,z) = \sup\left\{|f(z)|^2 : f \in A^2(D),\ \|f\|_{A^2(D)} \leq 1\right\}.
\end{align*}
Because $D$ is bounded, point evaluation on $A^2(D)$ is bounded by the submean-value inequality for holomorphic functions on a ball compactly contained in $D$, so the displayed supremum is finite.
[guided]
The proof will compare the Bergman kernels by comparing their extremal problems. For any bounded domain $D \subset \mathbb{C}^n$, the relevant [Hilbert space](/page/Hilbert%20Space) is
\begin{align*}
A^2(D) := \left\{ f: D \to \mathbb{C} : f \text{ is holomorphic and } \int_D |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) < \infty \right\},
\end{align*}
with norm
\begin{align*}
\|f\|_{A^2(D)} := \left(\int_D |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta)\right)^{1/2}.
\end{align*}
Here $\mathcal{L}^{2n}$ is Lebesgue measure on $\mathbb{C}^n \cong \mathbb{R}^{2n}$. Let $K_D: D \times D \to \mathbb{C}$ denote the Bergman kernel of $D$. The Bergman kernel extremal characterisation states that for every $z \in D$,
\begin{align*}
K_D(z,z) = \sup\left\{|f(z)|^2 : f \in A^2(D),\ \|f\|_{A^2(D)} \leq 1\right\}.
\end{align*}
This formula is the bridge between domain convergence and kernel convergence: instead of trying to compare reproducing kernels directly, we compare all norm-one holomorphic functions on the varying domains. The boundedness of point evaluation follows from the submean-value inequality on any Euclidean ball contained in $D$; hence the extremal quantity is finite.
[/guided]
[/step]
[step:Prove the upper bound by extracting limits of extremal functions]
Fix $z_0 \in \Omega_0$. Choose $r>0$ such that the closed ball $\overline{B}(z_0,r)$ is contained in $\Omega_0$. The compact-inclusion hypothesis gives $\overline{B}(z_0,r) \subset \Omega_j$ for all sufficiently large $j$. Let $(j_k)_{k\geq 1}$ be a strictly increasing sequence such that
\begin{align*}
K_{\Omega_{j_k}}(z_0,z_0) \to \limsup_{j\to\infty}K_{\Omega_j}(z_0,z_0).
\end{align*}
For each sufficiently large $k$, choose $f_k \in A^2(\Omega_{j_k})$ with $\|f_k\|_{A^2(\Omega_{j_k})} \leq 1$ and
\begin{align*}
|f_k(z_0)|^2 \geq K_{\Omega_{j_k}}(z_0,z_0) - k^{-1}.
\end{align*}
The submean-value inequality applied to $|f_k|^2$ on $B(z_0,r)$ gives
\begin{align*}
|f_k(z_0)|^2 \leq \frac{1}{\mathcal{L}^{2n}(B(z_0,r))}\int_{B(z_0,r)} |f_k(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \frac{1}{\mathcal{L}^{2n}(B(z_0,r))}.
\end{align*}
The same estimate on balls contained in compact subsets of $\Omega_0$ gives local boundedness of $(f_k)$ on $\Omega_0$. By Montel's theorem, the sequence $(f_k)_{k\geq 1}$ has a subsequence, still denoted $(f_k)_{k\geq 1}$, converging locally uniformly on $\Omega_0$ to a holomorphic map $f: \Omega_0 \to \mathbb{C}$.
Let $K \subset \Omega_0$ be compact. The compact-inclusion hypothesis gives $K \subset \Omega_{j_k}$ eventually, and Fatou's lemma applied to the non-negative functions $|f_k|^2\mathbb{1}_K$ gives
\begin{align*}
\int_K |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \liminf_{k\to\infty}\int_K |f_k(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq 1.
\end{align*}
Choose compact sets $K_m \subset \Omega_0$, $m\geq 1$, such that $K_m \subset K_{m+1}$ and $\bigcup_{m=1}^{\infty}K_m=\Omega_0$. Applying the [monotone convergence theorem](/theorems/509) to the increasing sequence $|f|^2\mathbb{1}_{K_m}$ yields $f \in A^2(\Omega_0)$ and $\|f\|_{A^2(\Omega_0)} \leq 1$. Since $f_k(z_0)\to f(z_0)$ and $(j_k)$ realizes the limsup,
\begin{align*}
\limsup_{j\to\infty} K_{\Omega_j}(z_0,z_0) \leq |f(z_0)|^2 \leq K_{\Omega_0}(z_0,z_0).
\end{align*}
[guided]
We first prove that the moving domains cannot create too large a diagonal Bergman kernel inside $\Omega_0$. Fix $z_0 \in \Omega_0$. Since $\Omega_0$ is open, there exists $r>0$ such that $\overline{B}(z_0,r) \subset \Omega_0$. The compact-inclusion hypothesis says that every compact subset of $\Omega_0$ is eventually contained in $\Omega_j$, so $\overline{B}(z_0,r) \subset \Omega_j$ for all sufficiently large $j$.
To prove an upper bound for the full limsup, we first choose the sequence on which that limsup is attained. Let $(j_k)_{k\geq 1}$ be a strictly increasing sequence such that
\begin{align*}
K_{\Omega_{j_k}}(z_0,z_0) \to \limsup_{j\to\infty}K_{\Omega_j}(z_0,z_0).
\end{align*}
For each sufficiently large $k$, choose a nearly extremising function $f_k \in A^2(\Omega_{j_k})$ satisfying $\|f_k\|_{A^2(\Omega_{j_k})} \leq 1$ and
\begin{align*}
|f_k(z_0)|^2 \geq K_{\Omega_{j_k}}(z_0,z_0) - k^{-1}.
\end{align*}
The submean-value inequality for holomorphic functions applies to $|f_k|^2$ on the Euclidean ball $B(z_0,r)$ because $f_k$ is holomorphic on $\Omega_{j_k}$ and $B(z_0,r) \subset \Omega_{j_k}$. It gives
\begin{align*}
|f_k(z_0)|^2 \leq \frac{1}{\mathcal{L}^{2n}(B(z_0,r))}\int_{B(z_0,r)} |f_k(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \frac{1}{\mathcal{L}^{2n}(B(z_0,r))}.
\end{align*}
Repeating this estimate on balls whose closures lie inside $\Omega_0$ proves local boundedness of $(f_k)$ on $\Omega_0$. Montel's theorem applies to locally bounded holomorphic functions on a domain in $\mathbb{C}^n$, so $(f_k)_{k\geq 1}$ has a subsequence, still denoted $(f_k)_{k\geq 1}$, converging locally uniformly on $\Omega_0$ to a holomorphic map $f: \Omega_0 \to \mathbb{C}$.
It remains to verify that the limit still has $A^2$ norm at most $1$. Let $K \subset \Omega_0$ be compact. By the compact-inclusion hypothesis, $K \subset \Omega_{j_k}$ eventually. Since $f_k \to f$ locally uniformly, $|f_k|^2 \to |f|^2$ pointwise on $K$. Let $\mathcal{B}(K)$ denote the Borel $\sigma$-algebra of the compact [metric space](/page/Metric%20Space) $K$. Fatou's lemma for the finite [measure space](/page/Measure%20Space) $(K,\mathcal{B}(K),\mathcal{L}^{2n}|_K)$ gives
\begin{align*}
\int_K |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \liminf_{k\to\infty}\int_K |f_k(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq 1.
\end{align*}
Choose compact sets $K_m \subset \Omega_0$, $m\geq 1$, such that $K_m \subset K_{m+1}$ and $\bigcup_{m=1}^{\infty}K_m=\Omega_0$. Applying the [monotone convergence theorem](/theorems/509) to the increasing sequence $|f|^2\mathbb{1}_{K_m}$ gives $f \in A^2(\Omega_0)$ and $\|f\|_{A^2(\Omega_0)} \leq 1$. Since $f_k(z_0)\to f(z_0)$ and $(j_k)$ realizes the limsup, the extremal formula for $K_{\Omega_0}$ then yields
\begin{align*}
\limsup_{j\to\infty} K_{\Omega_j}(z_0,z_0) \leq |f(z_0)|^2 \leq K_{\Omega_0}(z_0,z_0).
\end{align*}
[/guided]
[/step]
[step:Use uniform hyperconvexity to obtain the lower bound]
We use the following precise approximation input. Let $\bar{\partial}$ denote the Cauchy-Riemann operator: for a smooth map $u:U\subset\mathbb{C}^n\to\mathbb{C}$,
\begin{align*}
\bar{\partial}u := \sum_{k=1}^{n}\frac{\partial u}{\partial \bar z_k}\,d\bar z_k.
\end{align*}
The Ramadanov-Ligocka uniform $L^2$ $\bar{\partial}$ approximation theorem for uniformly hyperconvex domain convergence states that if the domains lie in a common bounded [open set](/page/Open%20Set), compact subsets of $\Omega_0$ are eventually contained in $\Omega_j$, compact subsets of $\mathbb{C}^n\setminus\overline{\Omega_0}$ are eventually disjoint from $\Omega_j$, and there exist constants $M>0$ and $\tau>0$ with plurisubharmonic exhaustion functions $\rho_j:\Omega_j\to[-M,0)$ satisfying
\begin{align*}
\{\zeta\in\Omega_j:\rho_j(\zeta)<-\varepsilon\}\subset\{\zeta\in\Omega_j:\operatorname{dist}(\zeta,\partial\Omega_j)>\tau\varepsilon\}
\end{align*}
for every $0<\varepsilon<1$, then for every compact set $K \subset \Omega_0$ and every $g \in A^2(\Omega_0)$ there exist holomorphic maps $g_j: \Omega_j \to \mathbb{C}$, defined for all sufficiently large $j$, such that $g_j \to g$ uniformly on $K$ and
\begin{align*}
\limsup_{j\to\infty}\|g_j\|_{A^2(\Omega_j)} \leq \|g\|_{A^2(\Omega_0)}.
\end{align*}
The common bounded ambient set, the compact-inclusion and compact-exclusion assumptions, and the quantitative uniform hyperconvexity assumption in the theorem statement verify exactly these hypotheses.
Fix $z_0 \in \Omega_0$ and $\varepsilon>0$. Choose $g \in A^2(\Omega_0)$ with $\|g\|_{A^2(\Omega_0)} \leq 1$ and
\begin{align*}
|g(z_0)|^2 \geq K_{\Omega_0}(z_0,z_0)-\varepsilon.
\end{align*}
Applying the approximation theorem to $K=\{z_0\}$ gives $g_j(z_0)\to g(z_0)$ and $\limsup_j\|g_j\|_{A^2(\Omega_j)}\leq 1$. For every $\delta>0$, for all sufficiently large $j$ with $\|g_j\|_{A^2(\Omega_j)}>0$, the function $h_j:=g_j/\|g_j\|_{A^2(\Omega_j)}$ belongs to $A^2(\Omega_j)$ and has norm $1$, hence
\begin{align*}
K_{\Omega_j}(z_0,z_0) \geq |h_j(z_0)|^2 = \frac{|g_j(z_0)|^2}{\|g_j\|_{A^2(\Omega_j)}^2}.
\end{align*}
Taking the lower limit gives
\begin{align*}
\liminf_{j\to\infty}K_{\Omega_j}(z_0,z_0) \geq |g(z_0)|^2 \geq K_{\Omega_0}(z_0,z_0)-\varepsilon.
\end{align*}
Since $\varepsilon>0$ was arbitrary,
\begin{align*}
\liminf_{j\to\infty}K_{\Omega_j}(z_0,z_0) \geq K_{\Omega_0}(z_0,z_0).
\end{align*}
[guided]
The lower bound is the analytic core of Ramadanov's theorem. We need to show that holomorphic $L^2$ functions on $\Omega_0$ can be approximated by holomorphic $L^2$ functions on the moving domains without increasing the norm in the limit. The uniform hyperconvexity hypothesis is included precisely to guarantee this extension-and-correction step.
First define the operator that appears in the approximation theorem. For a smooth map $u:U\subset\mathbb{C}^n\to\mathbb{C}$, the Cauchy-Riemann operator $\bar{\partial}$ is the $(0,1)$-form-valued operator
\begin{align*}
\bar{\partial}u := \sum_{k=1}^{n}\frac{\partial u}{\partial \bar z_k}\,d\bar z_k.
\end{align*}
We use the Ramadanov-Ligocka uniform $L^2$ $\bar{\partial}$ approximation theorem for uniformly hyperconvex domain convergence. It requires four inputs: the domains lie in a common bounded [open set](/page/Open%20Set); compact subsets of $\Omega_0$ are eventually contained in $\Omega_j$; compact subsets of $\mathbb{C}^n\setminus\overline{\Omega_0}$ are eventually disjoint from $\Omega_j$; and there exist constants $M>0$ and $\tau>0$ with plurisubharmonic exhaustion functions $\rho_j:\Omega_j\to[-M,0)$ satisfying
\begin{align*}
\{\zeta\in\Omega_j:\rho_j(\zeta)<-\varepsilon\}\subset\{\zeta\in\Omega_j:\operatorname{dist}(\zeta,\partial\Omega_j)>\tau\varepsilon\}
\end{align*}
for every $0<\varepsilon<1$. Under these hypotheses, for every compact set $K \subset \Omega_0$ and every $g \in A^2(\Omega_0)$ there exist holomorphic maps $g_j: \Omega_j \to \mathbb{C}$, defined for all sufficiently large $j$, such that $g_j \to g$ uniformly on $K$ and
\begin{align*}
\limsup_{j\to\infty}\|g_j\|_{A^2(\Omega_j)} \leq \|g\|_{A^2(\Omega_0)}.
\end{align*}
The common bounded ambient set, the compact-inclusion and compact-exclusion assumptions, and the quantitative uniform hyperconvexity assumption in the theorem statement verify precisely these four inputs. This theorem is the technical mechanism that converts compact convergence of domains into norm-controlled approximation of Bergman-space functions.
Fix $z_0 \in \Omega_0$ and $\varepsilon>0$. By the extremal characterisation of the diagonal Bergman kernel, choose $g \in A^2(\Omega_0)$ with $\|g\|_{A^2(\Omega_0)} \leq 1$ and
\begin{align*}
|g(z_0)|^2 \geq K_{\Omega_0}(z_0,z_0)-\varepsilon.
\end{align*}
Apply the approximation theorem with the compact set $K=\{z_0\}$. Then $g_j(z_0)\to g(z_0)$ and $\limsup_j\|g_j\|_{A^2(\Omega_j)}\leq 1$. For all sufficiently large $j$ with $g_j$ not identically zero, define the normalised function $h_j: \Omega_j \to \mathbb{C}$ by $h_j:=g_j/\|g_j\|_{A^2(\Omega_j)}$. Then $h_j \in A^2(\Omega_j)$ and $\|h_j\|_{A^2(\Omega_j)}=1$, so the extremal formula gives
\begin{align*}
K_{\Omega_j}(z_0,z_0) \geq |h_j(z_0)|^2 = \frac{|g_j(z_0)|^2}{\|g_j\|_{A^2(\Omega_j)}^2}.
\end{align*}
Passing to the lower limit and using $g_j(z_0)\to g(z_0)$ together with $\limsup_j\|g_j\|_{A^2(\Omega_j)}\leq 1$ gives
\begin{align*}
\liminf_{j\to\infty}K_{\Omega_j}(z_0,z_0) \geq |g(z_0)|^2 \geq K_{\Omega_0}(z_0,z_0)-\varepsilon.
\end{align*}
Because $\varepsilon>0$ was arbitrary, the lower bound follows.
[/guided]
[/step]
[step:Upgrade diagonal convergence to locally uniform two-variable convergence]
The upper and lower bounds show pointwise convergence
\begin{align*}
K_{\Omega_j}(z,z) \to K_{\Omega_0}(z,z)
\end{align*}
for every $z \in \Omega_0$. We first obtain the uniform local bounds needed for Montel's theorem. Let $E \subset \Omega_0$ be compact. Choose $\rho>0$ such that the compact set
\begin{align*}
E_\rho:=\{\zeta\in\mathbb{C}^n:\operatorname{dist}(\zeta,E)\leq \rho\}
\end{align*}
is contained in $\Omega_0$. By the compact-inclusion hypothesis, $E_\rho\subset\Omega_j$ for all sufficiently large $j$, and therefore $B(z,\rho)\subset\Omega_j$ for every $z\in E$ and all sufficiently large $j$. For every $u\in A^2(\Omega_j)$ with $\|u\|_{A^2(\Omega_j)}\leq 1$, the submean-value inequality gives
\begin{align*}
|u(z)|^2 \leq \frac{1}{\mathcal{L}^{2n}(B(z,\rho))}\int_{B(z,\rho)} |u(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \frac{1}{\mathcal{L}^{2n}(B(0,\rho))}
\end{align*}
for all $z\in E$. Taking the supremum over such $u$ gives
\begin{align*}
K_{\Omega_j}(z,z)\leq \frac{1}{\mathcal{L}^{2n}(B(0,\rho))}.
\end{align*}
The reproducing-kernel [Cauchy-Schwarz inequality](/theorems/432) then yields
\begin{align*}
|K_{\Omega_j}(z,w)|^2 \leq K_{\Omega_j}(z,z)K_{\Omega_j}(w,w) \leq \frac{1}{\mathcal{L}^{2n}(B(0,\rho))^2}
\end{align*}
for $z,w\in E$ and all large $j$. Montel's theorem for holomorphic functions of several variables, applied to $z$ and to the conjugate variable $\overline{w}$, implies that every subsequence has a further subsequence converging locally uniformly to a kernel $L: \Omega_0\times\Omega_0\to\mathbb{C}$ that is holomorphic in the first variable and anti-holomorphic in the second variable. Since each Bergman kernel satisfies $K_{\Omega_j}(w,z)=\overline{K_{\Omega_j}(z,w)}$, the locally uniform limit satisfies $L(w,z)=\overline{L(z,w)}$. The diagonal convergence gives
\begin{align*}
L(z,z)=K_{\Omega_0}(z,z)
\end{align*}
for every $z \in \Omega_0$.
Define the map $H:\Omega_0\times\Omega_0\to\mathbb{C}$ by
\begin{align*}
H(z,w):=L(z,w)-K_{\Omega_0}(z,w).
\end{align*}
The function $H$ is sesqui-holomorphic, Hermitian in the sense that $H(w,z)=\overline{H(z,w)}$, and satisfies $H(z,z)=0$ for every $z\in\Omega_0$. Around any point $a\in\Omega_0$, choose a polydisc $P\subset\Omega_0$ centered at $a$. For $z,w\in P$, write $z=a+\xi$ and $w=a+\eta$. The convergent expansion has the form
\begin{align*}
H(a+\xi,a+\eta)=\sum_{\alpha,\beta} c_{\alpha\beta}\xi^\alpha\overline{\eta}^\beta.
\end{align*}
For each integer $m\geq 0$, let
\begin{align*}
B_m(\xi,\eta):=\sum_{|\alpha|+|\beta|=m} c_{\alpha\beta}\xi^\alpha\overline{\eta}^\beta.
\end{align*}
The identity $H(a+\xi,a+\xi)=0$ for all sufficiently small $\xi$ implies that every homogeneous part $B_m(\xi,\xi)$ is zero. Because $H$ is Hermitian, each $B_m$ is a Hermitian sesquilinear form after polarization on the finite-dimensional space of monomials of total degree at most $m$; the complex polarization identity therefore gives $B_m(\xi,\eta)=0$ for all sufficiently small $\xi,\eta$. Hence all coefficients $c_{\alpha\beta}$ vanish, so $H=0$ on $P\times P$. The identity theorem for holomorphic functions in $(z,\overline{w})$ propagates this equality to $\Omega_0\times\Omega_0$. Thus every subsequential locally uniform limit is $K_{\Omega_0}$, and the full sequence converges locally uniformly on $\Omega_0\times\Omega_0$.
[guided]
We have proved the convergence of the diagonal quantities $K_{\Omega_j}(z,z)$ at each point $z\in\Omega_0$. To obtain convergence of $K_{\Omega_j}(z,w)$ for two variables, we first need a uniform bound on compact subsets. Let $E \subset \Omega_0$ be compact. Choose $\rho>0$ such that the compact set
\begin{align*}
E_\rho:=\{\zeta\in\mathbb{C}^n:\operatorname{dist}(\zeta,E)\leq \rho\}
\end{align*}
is contained in $\Omega_0$. The compact-inclusion hypothesis puts this single compact neighborhood $E_\rho$ inside $\Omega_j$ for all sufficiently large $j$, so $B(z,\rho)\subset\Omega_j$ for every $z\in E$ and all sufficiently large $j$.
For every $u\in A^2(\Omega_j)$ with $\|u\|_{A^2(\Omega_j)}\leq 1$, the submean-value inequality gives
\begin{align*}
|u(z)|^2 \leq \frac{1}{\mathcal{L}^{2n}(B(z,\rho))}\int_{B(z,\rho)} |u(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \frac{1}{\mathcal{L}^{2n}(B(0,\rho))}
\end{align*}
for every $z\in E$. Taking the supremum over all such $u$ in the extremal formula gives
\begin{align*}
K_{\Omega_j}(z,z)\leq \frac{1}{\mathcal{L}^{2n}(B(0,\rho))}.
\end{align*}
The reproducing-kernel [Cauchy-Schwarz inequality](/theorems/432) gives
\begin{align*}
|K_{\Omega_j}(z,w)|^2 \leq K_{\Omega_j}(z,z)K_{\Omega_j}(w,w) \leq \frac{1}{\mathcal{L}^{2n}(B(0,\rho))^2}
\end{align*}
for $z,w\in E$ and all sufficiently large $j$. This is the local boundedness needed for Montel's theorem. Applying Montel's theorem in the variables $(z,\overline{w})$ shows that every subsequence has a further subsequence converging locally uniformly to a kernel $L: \Omega_0\times\Omega_0\to\mathbb{C}$ that is holomorphic in $z$ and anti-holomorphic in $w$. Since each Bergman kernel is Hermitian, meaning $K_{\Omega_j}(w,z)=\overline{K_{\Omega_j}(z,w)}$, the locally uniform limit also satisfies $L(w,z)=\overline{L(z,w)}$.
The diagonal convergence already proved gives
\begin{align*}
L(z,z)=K_{\Omega_0}(z,z)
\end{align*}
for every $z\in\Omega_0$. Define the map $H:\Omega_0\times\Omega_0\to\mathbb{C}$ by
\begin{align*}
H(z,w):=L(z,w)-K_{\Omega_0}(z,w).
\end{align*}
Then $H$ is sesqui-holomorphic, Hermitian, and vanishes on the diagonal. Around any point $a\in\Omega_0$, choose a polydisc $P\subset\Omega_0$ centered at $a$. On $P\times P$, write $z=a+\xi$ and $w=a+\eta$, so
\begin{align*}
H(a+\xi,a+\eta)=\sum_{\alpha,\beta} c_{\alpha\beta}\xi^\alpha\overline{\eta}^\beta.
\end{align*}
The condition $H(a+\xi,a+\xi)=0$ makes every homogeneous diagonal part vanish. For each total degree $m$, the Hermitian homogeneous part is recovered from its diagonal values by the complex polarization identity on the finite-dimensional space spanned by the relevant monomials. Therefore each homogeneous part is zero, all coefficients $c_{\alpha\beta}$ vanish, and $H=0$ on $P\times P$. The identity theorem in the holomorphic variables $(z,\overline{w})$ then gives $H=0$ on $\Omega_0\times\Omega_0$. Therefore every subsequential locally uniform limit is $K_{\Omega_0}$, and the whole sequence satisfies
\begin{align*}
K_{\Omega_j}(z,w)\longrightarrow K_{\Omega_0}(z,w)
\end{align*}
locally uniformly on $\Omega_0\times\Omega_0$.
[/guided]
[/step]
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