[proofplan] We prove that every $L^2(\Omega,dV)$-limit of Bergman functions has a holomorphic representative. The key estimate is a local mean-value bound for holomorphic functions, which converts $L^2$ convergence into [uniform convergence](/page/Uniform%20Convergence) on compact subsets. The locally uniform limit is holomorphic by the Weierstrass theorem for holomorphic functions, and uniqueness of the $L^2$ limit identifies this holomorphic limit with the original $L^2$ limit almost everywhere. [/proofplan] [step:Reduce closedness to an $L^2$ convergent sequence in $A^2(\Omega)$] Let $(f_j)_{j=1}^{\infty}$ be a sequence in $A^2(\Omega)$ and suppose there exists $f \in L^2(\Omega,dV)$ such that \begin{align*} \lim_{j \to \infty}\|f_j-f\|_{L^2(\Omega)} = 0. \end{align*} It suffices to prove that the $L^2$ equivalence class of $f$ contains a holomorphic representative $g:\Omega \to \mathbb{C}$ satisfying \begin{align*} \int_\Omega |g(z)|^2 \, dV(z) < \infty. \end{align*} Indeed, this will show that the $L^2$ limit of every convergent sequence in $A^2(\Omega)$ again belongs to $A^2(\Omega)$, hence that $A^2(\Omega)$ is closed in $L^2(\Omega,dV)$. Since $L^2(\Omega,dV)$ is complete by the Riesz-Fischer theorem (citing a result not yet in the wiki: completeness of $L^2$ spaces), the same argument applies to every [Cauchy sequence](/page/Cauchy%20Sequence) in $A^2(\Omega)$ after taking its $L^2$ limit in the ambient space. [guided] We start with the exact property needed for closedness. A subspace $M$ of a [metric space](/page/Metric%20Space) $X$ is closed precisely when every sequence in $M$ that converges in $X$ has its limit in $M$. Here the ambient space is $L^2(\Omega,dV)$ and the subspace is $A^2(\Omega)$. Thus let $(f_j)_{j=1}^{\infty}$ be a sequence of holomorphic square-integrable functions on $\Omega$, and assume that it converges in the ambient norm to some $f \in L^2(\Omega,dV)$: \begin{align*} \lim_{j \to \infty}\|f_j-f\|_{L^2(\Omega)} = 0. \end{align*} The function $f$ is only an $L^2$ equivalence class, so it need not be defined pointwise and need not visibly be holomorphic. Our task is to find a representative $g:\Omega \to \mathbb{C}$ of this equivalence class that is holomorphic and square-integrable: \begin{align*} \int_\Omega |g(z)|^2 \, dV(z) < \infty. \end{align*} Once such a representative exists, the equivalence class $f$ belongs to $A^2(\Omega)$. The completeness of $L^2(\Omega,dV)$ itself is the Riesz-Fischer theorem (citing a result not yet in the wiki: completeness of $L^2$ spaces). We use it only to justify that a [Cauchy sequence](/page/Cauchy%20Sequence) in $A^2(\Omega)$ has an ambient $L^2$ limit. Since this step begins with an already convergent sequence, the rest of the proof establishes closedness directly. [/guided] [/step] [step:Convert $L^2$ control into uniform control on compact subsets] Let $K \subset \Omega$ be compact. Choose $r_K>0$ such that \begin{align*} \overline{B}(z,r_K) \subset \Omega \end{align*} for every $z \in K$, where $B(z,r_K)$ is the open Euclidean ball in $\mathbb{C}^n \cong \mathbb{R}^{2n}$. Let $\alpha_{2n}:=\mathcal{L}^{2n}(B(0,1))$ denote the volume of the unit ball in $\mathbb{R}^{2n}$. For $j,\ell \in \mathbb{N}$, define \begin{align*} h_{j\ell}:\Omega &\to \mathbb{C} \\ z &\mapsto f_j(z)-f_\ell(z). \end{align*} Then $h_{j\ell}$ is holomorphic on $\Omega$. By the mean-value inequality for holomorphic functions applied to $|h_{j\ell}|^2$ on $B(z,r_K)$, \begin{align*} |h_{j\ell}(z)|^2 &\leq \frac{1}{\alpha_{2n}r_K^{2n}}\int_{B(z,r_K)} |h_{j\ell}(w)|^2 \, dV(w) \\ &\leq \frac{1}{\alpha_{2n}r_K^{2n}}\int_{\Omega} |h_{j\ell}(w)|^2 \, dV(w). \end{align*} The [second inequality](/theorems/2136) uses the inclusion $B(z,r_K)\subset \Omega$ and non-negativity of $|h_{j\ell}|^2$. Therefore \begin{align*} \sup_{z\in K}|f_j(z)-f_\ell(z)| \leq C_K\|f_j-f_\ell\|_{L^2(\Omega)}, \qquad C_K := \left(\alpha_{2n}r_K^{2n}\right)^{-1/2}. \end{align*} Since $(f_j)_{j=1}^{\infty}$ converges in $L^2(\Omega,dV)$, it is Cauchy in $L^2(\Omega,dV)$. The estimate proves that $(f_j)_{j=1}^{\infty}$ is uniformly Cauchy on $K$. [guided] The goal of this step is to upgrade an $L^2$ estimate into a pointwise estimate on compact subsets. This is where holomorphicity is used. Fix a compact set $K \subset \Omega$. Because $K$ is compact and $\Omega$ is open, there is a radius $r_K>0$ such that every closed ball of radius $r_K$ centered at a point of $K$ remains inside $\Omega$: \begin{align*} \overline{B}(z,r_K) \subset \Omega \end{align*} for every $z \in K$. Let $\alpha_{2n}:=\mathcal{L}^{2n}(B(0,1))$ be the Euclidean volume of the unit ball in $\mathbb{R}^{2n}$, so \begin{align*} \mathcal{L}^{2n}(B(z,r_K))=\alpha_{2n}r_K^{2n}. \end{align*} For indices $j,\ell \in \mathbb{N}$, define the difference function \begin{align*} h_{j\ell}:\Omega &\to \mathbb{C} \\ z &\mapsto f_j(z)-f_\ell(z). \end{align*} Since $f_j$ and $f_\ell$ are holomorphic on $\Omega$, their difference $h_{j\ell}$ is holomorphic on $\Omega$. The mean-value inequality for holomorphic functions says that, for each $z \in K$, \begin{align*} |h_{j\ell}(z)|^2 &\leq \frac{1}{\mathcal{L}^{2n}(B(z,r_K))} \int_{B(z,r_K)} |h_{j\ell}(w)|^2 \, dV(w) \\ &= \frac{1}{\alpha_{2n}r_K^{2n}} \int_{B(z,r_K)} |h_{j\ell}(w)|^2 \, dV(w). \end{align*} The ball $B(z,r_K)$ is contained in $\Omega$, and the integrand $|h_{j\ell}|^2$ is non-negative. Enlarging the integration domain from $B(z,r_K)$ to $\Omega$ therefore gives \begin{align*} |h_{j\ell}(z)|^2 \leq \frac{1}{\alpha_{2n}r_K^{2n}} \int_{\Omega} |h_{j\ell}(w)|^2 \, dV(w). \end{align*} Taking square roots and then the supremum over $z\in K$ yields \begin{align*} \sup_{z\in K}|f_j(z)-f_\ell(z)| \leq C_K\|f_j-f_\ell\|_{L^2(\Omega)}, \qquad C_K := \left(\alpha_{2n}r_K^{2n}\right)^{-1/2}. \end{align*} Since $f_j \to f$ in $L^2(\Omega,dV)$, the sequence $(f_j)$ is Cauchy in $L^2(\Omega,dV)$. The displayed estimate transfers this Cauchy property to the uniform norm on $K$. Thus $(f_j)$ is uniformly Cauchy on every compact subset of $\Omega$. [/guided] [/step] [step:Construct the locally uniform holomorphic limit] For each compact $K\subset \Omega$, the uniform Cauchy property gives a continuous function $g_K:K\to \mathbb{C}$ such that $f_j|_K \to g_K$ uniformly on $K$. If $K_1\subset K_2\subset \Omega$ are compact, uniqueness of uniform limits gives $g_{K_2}|_{K_1}=g_{K_1}$. Hence these local limits define a function \begin{align*} g:\Omega &\to \mathbb{C} \end{align*} such that $f_j \to g$ uniformly on every compact subset of $\Omega$. By the Weierstrass theorem for holomorphic functions of several complex variables (citing a result not yet in the wiki: locally uniform limits of holomorphic functions are holomorphic), the locally uniform limit $g$ is holomorphic on $\Omega$. [guided] We now turn the compact uniform [Cauchy estimates](/theorems/2571) into an actual function. Fix a compact set $K\subset \Omega$. Since $(f_j|_K)$ is uniformly Cauchy in the [complete metric space](/page/Complete%20Metric%20Space) $C(K)$ with the supremum norm, there exists a continuous function $g_K:K\to \mathbb{C}$ such that \begin{align*} \lim_{j\to\infty}\sup_{z\in K}|f_j(z)-g_K(z)|=0. \end{align*} These compactly defined limits are compatible. If $K_1\subset K_2\subset \Omega$ are compact, then $f_j|_{K_1}$ converges uniformly both to $g_{K_1}$ and to $g_{K_2}|_{K_1}$. Uniform limits are unique, so \begin{align*} g_{K_2}|_{K_1}=g_{K_1}. \end{align*} Therefore the definitions agree on overlaps, and they determine a single function \begin{align*} g:\Omega &\to \mathbb{C} \end{align*} with the property that $f_j \to g$ uniformly on every compact subset of $\Omega$. The remaining point is holomorphicity. Each function $f_j:\Omega\to\mathbb{C}$ is holomorphic, and the convergence to $g$ is locally uniform. The Weierstrass theorem for holomorphic functions of several complex variables states that a locally uniform limit of holomorphic functions is holomorphic (citing a result not yet in the wiki: locally uniform limits of holomorphic functions are holomorphic). Applying that theorem gives \begin{align*} g \in \mathcal{O}(\Omega). \end{align*} This is precisely why we needed [uniform convergence](/page/Uniform%20Convergence) on compact subsets rather than merely pointwise convergence. [/guided] [/step] [step:Identify the locally uniform limit with the ambient $L^2$ limit] Let $K\subset \Omega$ be compact. Since $f_j\to g$ uniformly on $K$ and $K$ has finite $2n$-dimensional Lebesgue measure, \begin{align*} \|f_j-g\|_{L^2(K)}^2 = \int_K |f_j(z)-g(z)|^2 \, dV(z) \leq \mathcal{L}^{2n}(K)\sup_{z\in K}|f_j(z)-g(z)|^2 \to 0. \end{align*} Also, since $f_j\to f$ in $L^2(\Omega,dV)$, \begin{align*} \|f_j-f\|_{L^2(K)} \leq \|f_j-f\|_{L^2(\Omega)} \to 0. \end{align*} By [uniqueness of limits](/theorems/742) in the normed space $L^2(K,dV)$, the restrictions of $f$ and $g$ to $K$ agree in $L^2(K,dV)$. Choose an exhaustion of $\Omega$ by compact sets \begin{align*} K_m := \{z\in \Omega : |z|\leq m \text{ and } \overline{B}(z,1/m)\subset \Omega\}, \qquad m\in\mathbb{N}. \end{align*} Then $\Omega=\bigcup_{m=1}^{\infty}K_m$. Since $f=g$ in $L^2(K_m,dV)$ for every $m$, it follows that $f=g$ $dV$-almost everywhere on $\Omega$. [guided] We have two candidates for the limit of $(f_j)$. The first is the ambient $L^2$ limit $f$. The second is the locally uniform holomorphic limit $g$. We must prove that they represent the same $L^2$ function. Fix a compact set $K\subset \Omega$. Because $K$ is compact in $\mathbb{C}^n\cong\mathbb{R}^{2n}$, its Lebesgue measure is finite: \begin{align*} \mathcal{L}^{2n}(K)<\infty. \end{align*} [Uniform convergence](/page/Uniform%20Convergence) of $f_j$ to $g$ on $K$ gives \begin{align*} \|f_j-g\|_{L^2(K)}^2 = \int_K |f_j(z)-g(z)|^2 \, dV(z) \leq \mathcal{L}^{2n}(K)\sup_{z\in K}|f_j(z)-g(z)|^2 \to 0. \end{align*} Thus $f_j|_K \to g|_K$ in $L^2(K,dV)$. On the other hand, convergence in $L^2(\Omega,dV)$ restricts to convergence on $K$ because $K\subset \Omega$ and the integrand is non-negative: \begin{align*} \|f_j-f\|_{L^2(K)} \leq \|f_j-f\|_{L^2(\Omega)} \to 0. \end{align*} Therefore the same sequence $f_j|_K$ converges in the normed space $L^2(K,dV)$ both to $f|_K$ and to $g|_K$. Norm limits are unique, so $f=g$ in $L^2(K,dV)$. To pass from compact subsets to all of $\Omega$, define compact sets \begin{align*} K_m := \{z\in \Omega : |z|\leq m \text{ and } \overline{B}(z,1/m)\subset \Omega\}, \qquad m\in\mathbb{N}. \end{align*} Each $K_m$ is compact, and every point of $\Omega$ belongs to some $K_m$ because $\Omega$ is open. Hence \begin{align*} \Omega=\bigcup_{m=1}^{\infty}K_m. \end{align*} Since $f=g$ in $L^2(K_m,dV)$ for every $m$, the set on which representatives of $f$ and $g$ differ is contained in a countable union of $dV$-null sets. Therefore $f=g$ $dV$-almost everywhere on $\Omega$. [/guided] [/step] [step:Conclude that the Bergman space is a Hilbert space] Since $g$ is holomorphic on $\Omega$ and $g=f$ $dV$-almost everywhere, we have \begin{align*} \int_\Omega |g(z)|^2 \, dV(z) = \int_\Omega |f(z)|^2 \, dV(z) <\infty. \end{align*} Thus $g\in A^2(\Omega)$ and the $L^2$ limit $f$ belongs to the same equivalence class as an element of $A^2(\Omega)$. Hence $A^2(\Omega)$ is closed in $L^2(\Omega,dV)$. The space $A^2(\Omega)$ is a linear subspace because sums and scalar multiples of holomorphic square-integrable functions are holomorphic and square-integrable. A closed linear subspace of a [Hilbert space](/page/Hilbert%20Space) is Hilbert with the restricted inner product. Since $L^2(\Omega,dV)$ is Hilbert, $A^2(\Omega)$ is Hilbert with \begin{align*} (f,h)_{A^2(\Omega)} = \int_\Omega f(z)\overline{h(z)} \, dV(z). \end{align*} [/step]