[proofplan] We choose a Euclidean ball centered at $a$ and compactly contained in $\Omega$. For each holomorphic $f \in A^2(\Omega)$, the function $|f|^2$ is plurisubharmonic, so the mean-value inequality bounds $|f(a)|^2$ by the average of $|f|^2$ over that ball. Since the ball lies inside $\Omega$, this average is bounded by the global $A^2(\Omega)$ norm, giving a uniform operator norm bound for evaluation. [/proofplan] [step:Choose an interior ball centered at the evaluation point] For every center $c \in \mathbb{C}^n$ and radius $s > 0$, write \begin{align*} B(c,s) &:= \{z \in \mathbb{C}^n : |z-c| < s\}, \\ \overline{B}(c,s) &:= \{z \in \mathbb{C}^n : |z-c| \leq s\}. \end{align*} Since $\Omega \subset \mathbb{C}^n$ is open and $a \in \Omega$, there exists $\rho > 0$ such that $B(a,\rho) \subset \Omega$. Choose $r := \rho/2$. Then $\overline{B}(a,r) \subset B(a,\rho) \subset \Omega$. Define the geometric constant \begin{align*} \alpha_{2n} := \mathcal{L}^{2n}(B(0,1)). \end{align*} By translation and scaling invariance of Lebesgue measure on $\mathbb{C}^n \cong \mathbb{R}^{2n}$, \begin{align*} \mathcal{L}^{2n}(B(a,r)) = \alpha_{2n} r^{2n}. \end{align*} [/step] [step:Bound the value of a holomorphic function by its local square integral] Let $f \in A^2(\Omega)$. Define \begin{align*} u: \Omega &\to [0,\infty) \\ z &\mapsto |f(z)|^2. \end{align*} Since $f$ is holomorphic, $f \in C^\infty(\Omega;\mathbb{C})$. For every $z \in \Omega$ and $\xi \in \mathbb{C}^n$, the Levi form of $u$ satisfies \begin{align*} \mathcal{L}_u(z;\xi) = \left|\sum_{j=1}^n \frac{\partial f}{\partial z_j}(z)\xi_j\right|^2 \geq 0. \end{align*} By the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403), $u = |f|^2$ is a [plurisubharmonic function](/page/Plurisubharmonic%20Function) on $\Omega$. Since $\overline{B}(a,r) \subset \Omega$, the [Sub-Mean Value Inequality for Plurisubharmonic Functions](/theorems/TEMP-9), in its Euclidean-ball form, applies to $u$ on the ball $B(a,r)$ and gives \begin{align*} |f(a)|^2 = u(a) \leq \frac{1}{\mathcal{L}^{2n}(B(a,r))} \int_{B(a,r)} u(z)\, d\mathcal{L}^{2n}(z). \end{align*} Substituting the definition of $u$ and the value of the ball measure gives \begin{align*} |f(a)|^2 \leq \frac{1}{\alpha_{2n} r^{2n}} \int_{B(a,r)} |f(z)|^2\, d\mathcal{L}^{2n}(z). \end{align*} [guided] Fix $f \in A^2(\Omega)$. The goal is to control the single value $f(a)$ using an integral norm. The bridge between pointwise values and averages is the mean-value inequality. To use it, define the nonnegative function \begin{align*} u: \Omega &\to [0,\infty) \\ z &\mapsto |f(z)|^2. \end{align*} Since $f$ is holomorphic, $f \in C^\infty(\Omega;\mathbb{C})$. To verify plurisubharmonicity of $u=|f|^2$, we use the smooth Levi-form criterion. For every $z \in \Omega$ and every direction $\xi \in \mathbb{C}^n$, differentiating $u=f\overline{f}$ gives \begin{align*} \mathcal{L}_u(z;\xi) = \left|\sum_{j=1}^n \frac{\partial f}{\partial z_j}(z)\xi_j\right|^2 \geq 0. \end{align*} Thus the hypotheses of the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403) are satisfied, and $u$ is a [plurisubharmonic function](/page/Plurisubharmonic%20Function) on $\Omega$. The previous step chose $r > 0$ so that $\overline{B}(a,r) \subset \Omega$. This verifies the closed-ball domain condition needed for the [Sub-Mean Value Inequality for Plurisubharmonic Functions](/theorems/TEMP-9), in its Euclidean-ball form, so the inequality applies to $u$ on $B(a,r)$ and yields \begin{align*} u(a) \leq \frac{1}{\mathcal{L}^{2n}(B(a,r))} \int_{B(a,r)} u(z)\, d\mathcal{L}^{2n}(z). \end{align*} Now substitute $u(z)=|f(z)|^2$. Since $\mathcal{L}^{2n}(B(a,r))=\alpha_{2n}r^{2n}$, this becomes \begin{align*} |f(a)|^2 \leq \frac{1}{\alpha_{2n} r^{2n}} \int_{B(a,r)} |f(z)|^2\, d\mathcal{L}^{2n}(z). \end{align*} This is the local estimate: the value at $a$ is bounded by the $L^2$ mass of $f$ on any ball around $a$ that remains inside $\Omega$. [/guided] [/step] [step:Enlarge the integral to the whole domain] Since $B(a,r) \subset \Omega$ and $|f|^2 \geq 0$, monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_{B(a,r)} |f(z)|^2\, d\mathcal{L}^{2n}(z) \leq \int_{\Omega} |f(z)|^2\, d\mathcal{L}^{2n}(z) = \|f\|_{A^2(\Omega)}^2. \end{align*} Combining this with the previous estimate yields \begin{align*} |f(a)|^2 \leq \frac{1}{\alpha_{2n} r^{2n}} \|f\|_{A^2(\Omega)}^2. \end{align*} Taking square roots gives \begin{align*} |\operatorname{ev}_a(f)| = |f(a)| \leq \frac{1}{(\alpha_{2n} r^{2n})^{1/2}} \|f\|_{A^2(\Omega)}. \end{align*} Thus $\operatorname{ev}_a$ is bounded with operator norm at most $(\alpha_{2n} r^{2n})^{-1/2}$. [/step] [step:Verify linearity of the evaluation map] Let $f,g \in A^2(\Omega)$ and let $\lambda,\mu \in \mathbb{C}$. Since sums and scalar multiples of holomorphic square-integrable functions are holomorphic and square-integrable, $\lambda f+\mu g \in A^2(\Omega)$. By pointwise linearity of functions, \begin{align*} \operatorname{ev}_a(\lambda f+\mu g) = (\lambda f+\mu g)(a) = \lambda f(a)+\mu g(a) = \lambda \operatorname{ev}_a(f)+\mu \operatorname{ev}_a(g). \end{align*} Therefore $\operatorname{ev}_a$ is linear. Together with the uniform bound from the previous step, this proves that $\operatorname{ev}_a: A^2(\Omega)\to\mathbb{C}$ is a bounded linear functional. [/step]