[proofplan] We show that a separately holomorphic, locally bounded function on an [open set](/page/Open%20Set) $\Omega \subset \mathbb{C}^n$ is jointly holomorphic by representing it as a convergent [power series](/page/Power%20Series) on each polydisc $\mathbb{D}^n(a, r) \subset \Omega$. The strategy is to iterate the one-variable [Cauchy integral formula](/theorems/345): first in $z_1$ (holding the other variables fixed) to obtain a representation whose coefficients are functions of $(z_2, \dots, z_n)$, then show these coefficients are themselves holomorphic in the remaining variables by differentiation under the integral sign (justified by the uniform bound), and finally iterate for each subsequent variable to build the full iterated Cauchy integral. The resulting integral representation coincides with the [Cauchy Integral Formula on Polydiscs](/theorems/3398), from which [power series](/page/Power%20Series) expansion and joint holomorphicity follow. [/proofplan] [step:Fix a polydisc and establish a uniform bound on the torus] Let $a \in \Omega$. Since $f$ is locally bounded, there exist $r = (r_1, \dots, r_n)$ with $r_j > 0$ and $M > 0$ such that $\overline{\mathbb{D}^n(a, r)} \subset \Omega$ and $|f(z)| \leq M$ for all $z \in \overline{\mathbb{D}^n(a, r)}$. In particular, $|f(\zeta)| \leq M$ for all $\zeta$ on the distinguished boundary (torus) $\mathbb{T}^n(a, r) = \{(\zeta_1, \dots, \zeta_n) : |\zeta_j - a_j| = r_j \text{ for all } j\}$. [/step] [step:Represent $f$ via the one-variable Cauchy formula in $z_1$] Fix $(z_2, \dots, z_n) \in \mathbb{D}^{n-1}(a', r')$ where $a' = (a_2, \dots, a_n)$ and $r' = (r_2, \dots, r_n)$. Since $f$ is holomorphic in $z_1$ for each fixed value of the remaining variables (by the separate holomorphicity hypothesis), the function $\zeta_1 \mapsto f(\zeta_1, z_2, \dots, z_n)$ is holomorphic on a neighbourhood of $\overline{\mathbb{D}(a_1, r_1)}$. By the one-variable [Cauchy Integral Formula](/theorems/???), for each $z_1 \in \mathbb{D}(a_1, r_1)$: \begin{align*} f(z_1, z_2, \dots, z_n) &= \frac{1}{2\pi i} \oint_{|\zeta_1 - a_1| = r_1} \frac{f(\zeta_1, z_2, \dots, z_n)}{\zeta_1 - z_1}\, d\zeta_1. \end{align*} [guided] The separate holomorphicity hypothesis tells us that for each fixed $(z_2, \dots, z_n)$, the map $z_1 \mapsto f(z_1, z_2, \dots, z_n)$ is holomorphic on $\{z_1 : (z_1, z_2, \dots, z_n) \in \Omega\}$. Since $\overline{\mathbb{D}^n(a, r)} \subset \Omega$, this domain contains $\overline{\mathbb{D}(a_1, r_1)}$ whenever $(z_2, \dots, z_n) \in \mathbb{D}^{n-1}(a', r')$. Therefore the one-variable Cauchy formula applies. Note that we have not yet used the local boundedness hypothesis. It will be needed in the next step to justify differentiation under the integral sign, which will show that the result of the integration is holomorphic in the remaining variables. [/guided] [/step] [step:Show that the Cauchy integral defines a function holomorphic in $(z_2, \dots, z_n)$ by differentiation under the integral sign] Define \begin{align*} I(z_1, z_2, \dots, z_n) &:= \frac{1}{2\pi i} \oint_{|\zeta_1 - a_1| = r_1} \frac{f(\zeta_1, z_2, \dots, z_n)}{\zeta_1 - z_1}\, d\zeta_1. \end{align*} We claim $I$ is holomorphic in each variable $z_j$ for $j \geq 2$ on $\mathbb{D}^{n-1}(a', r')$. Fix $j \in \{2, \dots, n\}$ and fix all variables except $z_j$. For each $\zeta_1$ on the circle $|\zeta_1 - a_1| = r_1$, the function $z_j \mapsto f(\zeta_1, z_2, \dots, z_n)$ is holomorphic in $z_j$ (by separate holomorphicity of $f$). The integrand satisfies the uniform bound \begin{align*} \left|\frac{f(\zeta_1, z_2, \dots, z_n)}{\zeta_1 - z_1}\right| \leq \frac{M}{r_1 - |z_1 - a_1|} \end{align*} for all $\zeta_1$ on the circle and all $(z_2, \dots, z_n) \in \overline{\mathbb{D}^{n-1}(a', s')}$ for any $s' < r'$ componentwise, since $|f| \leq M$ on $\overline{\mathbb{D}^n(a, r)}$ and $|\zeta_1 - z_1| \geq r_1 - |z_1 - a_1| > 0$. This uniform bound justifies differentiation under the integral sign: for the Wirtinger derivative $\partial_{\bar{z}_j}$ with $j \geq 2$, \begin{align*} \partial_{\bar{z}_j} I &= \frac{1}{2\pi i} \oint_{|\zeta_1 - a_1| = r_1} \frac{\partial_{\bar{z}_j} f(\zeta_1, z_2, \dots, z_n)}{\zeta_1 - z_1}\, d\zeta_1 = 0, \end{align*} since $f$ is holomorphic in $z_j$ for each fixed $(\zeta_1, z_2, \dots, \widehat{z_j}, \dots, z_n)$ by hypothesis, so $\partial_{\bar{z}_j} f = 0$. [guided] This is the critical step where the local boundedness hypothesis is consumed. Without the bound $|f| \leq M$, we cannot justify passing the derivative under the integral sign, and the argument breaks down. Define \begin{align*} I(z_1, z_2, \dots, z_n) &:= \frac{1}{2\pi i} \oint_{|\zeta_1 - a_1| = r_1} \frac{f(\zeta_1, z_2, \dots, z_n)}{\zeta_1 - z_1}\, d\zeta_1. \end{align*} To show $I$ is holomorphic in $z_j$ (for $j \geq 2$), we verify that $\partial_{\bar{z}_j} I = 0$ by differentiating under the integral. The justification requires a dominating function. For any compact sub-polydisc $\overline{\mathbb{D}^{n-1}(a', s')} \subset \mathbb{D}^{n-1}(a', r')$ and any $z_1$ with $|z_1 - a_1| \leq \rho_1 < r_1$, the integrand is bounded by \begin{align*} \left|\frac{f(\zeta_1, z_2, \dots, z_n)}{\zeta_1 - z_1}\right| \leq \frac{M}{r_1 - \rho_1}, \end{align*} which is independent of $\zeta_1$ and of $(z_2, \dots, z_n)$. Since the circle $|\zeta_1 - a_1| = r_1$ has finite length $2\pi r_1$, we may differentiate under the integral sign with respect to $\bar{z}_j$: \begin{align*} \partial_{\bar{z}_j} I &= \frac{1}{2\pi i} \oint_{|\zeta_1 - a_1| = r_1} \frac{\partial_{\bar{z}_j} f(\zeta_1, z_2, \dots, z_n)}{\zeta_1 - z_1}\, d\zeta_1. \end{align*} Now $f$ is holomorphic in $z_j$ for each fixed value of the other variables, so $\partial_{\bar{z}_j} f(\zeta_1, z_2, \dots, z_n) = 0$ for each $\zeta_1$. Therefore $\partial_{\bar{z}_j} I = 0$, and $I$ is holomorphic in $z_j$. Since this holds for every $j \geq 2$, the function $I$ is separately holomorphic in $(z_2, \dots, z_n)$ and bounded (by the same bound $M / (r_1 - |z_1 - a_1|)$ on the integral). If $n = 2$, this already shows $I$ is holomorphic in $z_2$, which is a single complex variable, so separate holomorphicity is the same as holomorphicity. [/guided] [/step] [step:Iterate in each remaining variable to obtain the iterated Cauchy integral representation] Apply the one-variable Cauchy formula in $z_2$ to the integrand: for each fixed $\zeta_1$ on $|\zeta_1 - a_1| = r_1$ and each fixed $(z_3, \dots, z_n) \in \mathbb{D}^{n-2}(a'', r'')$, the function $z_2 \mapsto f(\zeta_1, z_2, z_3, \dots, z_n)$ is holomorphic in $z_2$ (by separate holomorphicity). Therefore \begin{align*} f(\zeta_1, z_2, z_3, \dots, z_n) &= \frac{1}{2\pi i} \oint_{|\zeta_2 - a_2| = r_2} \frac{f(\zeta_1, \zeta_2, z_3, \dots, z_n)}{\zeta_2 - z_2}\, d\zeta_2. \end{align*} Substituting into the formula from the previous steps and interchanging the order of integration (justified by continuity of $f$ and the uniform bound $M$ on the compact torus, via [Fubini's theorem](/theorems/2961) for continuous functions on compact sets): \begin{align*} f(z) &= \frac{1}{(2\pi i)^2} \oint_{|\zeta_1 - a_1| = r_1} \oint_{|\zeta_2 - a_2| = r_2} \frac{f(\zeta_1, \zeta_2, z_3, \dots, z_n)}{(\zeta_1 - z_1)(\zeta_2 - z_2)}\, d\zeta_2\, d\zeta_1. \end{align*} Repeating for $z_3, \dots, z_n$ yields after $n$ iterations: \begin{align*} f(z) &= \frac{1}{(2\pi i)^n} \oint_{|\zeta_1 - a_1| = r_1} \cdots \oint_{|\zeta_n - a_n| = r_n} \frac{f(\zeta)}{(\zeta_1 - z_1)\cdots(\zeta_n - z_n)}\, d\zeta_n \cdots d\zeta_1. \end{align*} [/step] [step:Expand the Cauchy kernel as a power series and conclude joint holomorphicity] For each $j = 1, \dots, n$, since $|z_j - a_j| < r_j = |\zeta_j - a_j|$, the geometric series expansion \begin{align*} \frac{1}{\zeta_j - z_j} &= \frac{1}{(\zeta_j - a_j) - (z_j - a_j)} = \sum_{k_j = 0}^{\infty} \frac{(z_j - a_j)^{k_j}}{(\zeta_j - a_j)^{k_j + 1}} \end{align*} converges absolutely and uniformly in $\zeta_j$ on $|\zeta_j - a_j| = r_j$ for each fixed $z_j$ with $|z_j - a_j| < r_j$. Substituting all $n$ such expansions into the iterated Cauchy integral and using the uniform bound $|f(\zeta)| \leq M$ on the torus to justify the interchange of summation and integration: \begin{align*} f(z) &= \sum_{\alpha \in \mathbb{N}_0^n} c_\alpha (z - a)^\alpha, \end{align*} where $\alpha = (\alpha_1, \dots, \alpha_n)$, $(z - a)^\alpha = \prod_{j=1}^n (z_j - a_j)^{\alpha_j}$, and the coefficients are given by \begin{align*} c_\alpha &= \frac{1}{(2\pi i)^n} \oint_{|\zeta_1 - a_1| = r_1} \cdots \oint_{|\zeta_n - a_n| = r_n} \frac{f(\zeta)}{(\zeta_1 - a_1)^{\alpha_1 + 1} \cdots (\zeta_n - a_n)^{\alpha_n + 1}}\, d\zeta_n \cdots d\zeta_1. \end{align*} The series converges absolutely on $\mathbb{D}^n(a, r)$: the general term satisfies $|c_\alpha (z - a)^\alpha| \leq M \prod_{j=1}^n (|z_j - a_j|/r_j)^{\alpha_j}$, and $\sum_\alpha \prod_{j=1}^n (|z_j - a_j|/r_j)^{\alpha_j} = \prod_{j=1}^n (1 - |z_j - a_j|/r_j)^{-1} < \infty$ since $|z_j - a_j| < r_j$. Therefore $f$ is represented by a convergent [power series](/page/Power%20Series) on $\mathbb{D}^n(a, r)$. A convergent [power series](/page/Power%20Series) in $n$ complex variables defines a [holomorphic function](/page/Holomorphic%20Function) (its partial sums are polynomials, hence holomorphic, and the convergence is uniform on compact subsets). Since $a \in \Omega$ was arbitrary, $f \in \mathcal{O}(\Omega)$. [/step]