Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly

Suppose convergence is locally uniform in the stated sense, and let $K\subseteq \mathrm{\Omega }$ be compact. Let $\mathcal{V}$ be the family of all open $V\subseteq \mathrm{\Omega }$ on which convergence is uniform. Local uniformity says this family covers $\mathrm{\Omega }$, without selecting a neighbourhood for every point. If $K$ is nonempty, [L10] supplies a finite list $V_1,\dots ,V_q$ covering $K$. For fixed $\epsilon >0$, let $N_j$ be the least natural threshold valid for $\epsilon$ on $V_j$, which exists by uniform convergence and [L11]. The natural $N=N_1+\cdots +N_q$ is at least each $N_j$, so $n\ge N$ gives $\mathrm{\mid }{f}_n-f\mathrm{\mid }<\epsilon$ on every $V_j$ and hence on $K$. For $K=\mathrm{\varnothing }$ the conclusion is vacuous. Thus convergence is uniform on every compact subset.

Conversely suppose convergence is uniform on every compact subset, and fix $a\in \mathrm{\Omega }$. By [L9] choose $s>0$ with $D(a,s)\subseteq \mathrm{\Omega }$ and let $r=s\mathrm{/}2$. The closed disc $\bar{D}(a,r)$ is closed and bounded in the complex metric; by [L9] its coordinate image is closed and bounded in ${\mathbb{R}}^2$, hence compact by [L7], and [L9] transports compactness back. Convergence is therefore uniform on that closed disc and on its open subdisc $D(a,r)$, an open neighbourhood of $a$ by [L9]. This proves the converse and fixes the local/compact equivalence for this proof.

If $\mathrm{\Omega }$ is nonempty, fix $a\in \mathrm{\Omega }$ and choose $s>0$ with $D(a,s)\subseteq \mathrm{\Omega }$ and put $S=s\mathrm{/}2>0$, so $\bar{D}(a,S)\subseteq \mathrm{\Omega }$; [L7] and [L9] make this closed disc compact, and step 1.1 makes convergence uniform there, and [L1] makes $f$ continuous on it, so $f$ is continuous throughout $\mathrm{\Omega }$.

Let $\mathrm{\Delta }\subseteq \mathrm{\Omega }$ be any filled triangle. By [L6], its boundary trace is the union of three affine edge images of $[0,1]$, each compact by [L7] and [L8]. An ambient open cover of this union covers each edge, so [L10] supplies a finite subcover for each of the three edges; concatenating these three finite lists covers the union, and the converse in [L10] proves its compactness. Step 1.1 therefore makes convergence uniform on the trace, [L3] makes every ${\int }_{\mathrm{\partial }\mathrm{\Delta }}{f}_n$ zero, and [L2] gives ${\int }_{\mathrm{\partial }\mathrm{\Delta }}f=0$.

The continuity from step 2.1 and the vanishing triangle integrals from step 2.2 satisfy [L4], so $f$ is holomorphic throughout $\mathrm{\Omega }$; on the empty open set this conclusion is vacuous.

Fix a natural derivative order $k$ and a point $a\in \mathrm{\Omega }$, and choose radii $0<r<R<S$ with $\bar{D}(a,S)\subseteq \mathrm{\Omega }$; by step 3.1 every difference $h_n:=f_n-f$ is holomorphic on $D(a,S)$.

Given $\epsilon >0$, the circle $\mathrm{\mid }\zeta -a\mathrm{\mid }=R$ is closed by the reverse triangle inequality and bounded, hence compact by [L7] and [L9]. Step 1.1 gives uniform convergence there and a natural $N$ such that $\mathrm{\mid }{h}_n(\zeta )\mathrm{\mid }<\epsilon (R-r{)}^{k+1}\mathrm{/}(2k!R)$ for $n\ge N$; applying [L5] with $M=\epsilon (R-r{)}^{k+1}\mathrm{/}(2k!R)$ then gives $\mathrm{\mid }{f}_n^{(k)}(z)-f^{(k)}(z)\mathrm{\mid }\le \epsilon \mathrm{/}2<\epsilon$ for every $\mathrm{\mid }z-a\mathrm{\mid }\le r$.

Step 5.1 proves uniform convergence of the $k$th derivatives on a neighbourhood of every point, hence locally uniform convergence by the explicit definition in the given data and [L9]; when $k=0$ it recovers the original convergence, and zero or eventually constant sequences require no exception.