Locally uniform limits of holomorphic functions are holomorphic, with locally uniform convergence of all derivatives

Each $f_n$ is continuous by [L6], and locally uniform convergence makes $f$ continuous at every point by [L4] applied on a neighbourhood where the convergence is uniform.

Fix $a\in U$ and $k<m$, and let $V$ be the $k$th slice domain of $U$ through $a$, an open subset of $\mathbb{C}$ by [L5]. The slices of the $f_n$ are holomorphic on $V$ by [L6], and they converge to the slice of $f$ locally uniformly on $V$, since a neighbourhood in $U$ of a point of the slice meets the slice in a neighbourhood there. So [L3] makes the slice of $f$ holomorphic on $V$, and $f$ is separately holomorphic by [L5].

By steps 1.1 and 1.2 the limit $f$ is continuous and separately holomorphic on $U$, so [L1] makes it holomorphic.

Fix $a\in U$ and a multi-index $\alpha$. By [L8] choose $\epsilon >0$ with the ball $B(a,\epsilon )\subseteq U$ and put ${\rho }_k=\epsilon \mathrm{/}(2\sqrt{m})$, so ${\mathrm{\Delta }}_{\rho }(a)\subseteq U$ by [L7] and [L9], and put $r_k={\rho }_k\mathrm{/}2$. Shrinking $\epsilon$ if necessary, the convergence $f_n\to f$ is uniform on ${\bar{\mathrm{\Delta }}}_{\rho }(a)$.

Fix $b\in {\mathrm{\Delta }}_r(a)$ and put ${\sigma }_k:={\rho }_k-\mathrm{\mid }{b}_k-a_k\mathrm{\mid }$ for each $k<m$. Then ${\sigma }_k>r_k$ because $\mathrm{\mid }{b}_k-a_k\mathrm{\mid }<r_k={\rho }_k\mathrm{/}2$, and if $\mathrm{\mid }{\zeta }_k-b_k\mathrm{\mid }<{\sigma }_k$ then $\mathrm{\mid }{\zeta }_k-a_k\mathrm{\mid }\le \mathrm{\mid }{\zeta }_k-b_k\mathrm{\mid }+\mathrm{\mid }{b}_k-a_k\mathrm{\mid }<{\rho }_k$, so ${\mathrm{\Delta }}_{\sigma }(b)\subseteq {\mathrm{\Delta }}_{\rho }(a)\subseteq U$ by [L7] and [L9]. Also ${\mathrm{\Gamma }}_r(b)\subseteq {\bar{\mathrm{\Delta }}}_{\rho }(a)$ because $\mathrm{\mid }{b}_k-a_k\mathrm{\mid }<r_k$ and $r_k+r_k={\rho }_k$. The difference $f_n-f$ is holomorphic on $U$ by [L6] and step 2.1, so [L2] applied on ${\mathrm{\Delta }}_{\sigma }(b)$ with inner polyradius $r$ gives $\mathrm{\mid }{\mathrm{\partial }}_z^{\alpha }{f}_n(b)-{\mathrm{\partial }}_z^{\alpha }f(b)\mathrm{\mid }\le \alpha !({\sup }_{{\bar{\mathrm{\Delta }}}_{\rho }(a)}\mathrm{\mid }{f}_n-f\mathrm{\mid }){\prod }_{k<m}{r}_k^{-{\alpha }_k}$.

The right-hand side of step 4.1 does not depend on $b$ and tends to $0$ by the uniform convergence of step 3.1, so ${\mathrm{\partial }}_z^{\alpha }{f}_n\to {\mathrm{\partial }}_z^{\alpha }f$ uniformly on the neighbourhood ${\mathrm{\Delta }}_r(a)$ of $a$. Since $a\in U$ and $\alpha$ were arbitrary, the convergence is locally uniform for every multi-index.