Locally uniform convergence (Tag D-G42SR)
Let \(\Omega\subseteq\CC\) be open and let \(f_k,f\colon\Omega\to\CC\) for \(k\ge1\). The sequence \((f_k)\) converges locally uniformly to \(f\) on \(\Omega\) if \(f_k\to f\) uniformly on every compact subset of \(\Omega\).
Let \(\Omega\subseteq\CC\) be open and let \(f_k,f\colon\Omega\to\CC\), with the convention \(\operatorname{dist}(A,\varnothing)=+\infty\) when \(\Omega=\CC\). Then \(f_k\to f\) locally uniformly on \(\Omega\) if and only if \(f_k\to f\) uniformly on every bounded set \(A\subseteq\Omega\) with \(\operatorname{dist}(A,\partial\Omega)>0\).
Let \(A\subseteq\Omega\) be bounded with \(\operatorname{dist}(A,\partial\Omega)=d>0\). Its closure \(\overline A\) is closed and bounded, hence compact, and every point of \(\overline A\) has distance at least \(d\) from \(\partial\Omega\). A point of \(\overline A\) lies in \(\overline\Omega\) but not in \(\partial\Omega\), so it lies in \(\Omega\); thus \(\overline A\) is a compact subset of \(\Omega\), and uniform convergence on \(\overline A\) gives uniform convergence on \(A\).
Conversely, let \(K\subseteq\Omega\) be compact. Then \(K\) is bounded, and \(z\mapsto\operatorname{dist}(z,\partial\Omega)\) is continuous and positive on \(K\) when \(\partial\Omega\neq\varnothing\), because \(\partial\Omega\) is closed and disjoint from \(K\), so it has a positive minimum on \(K\). Hence \(K\) is one of the sets \(A\).
Univalent functions and biholomorphisms onto their images (Tag D-OHFRH)
Let \(\Omega\subseteq\CC\) be open. A function \(f\colon\Omega\to\CC\) is univalent if \(f\) is holomorphic and injective.
Let \(\Omega\subseteq\CC\) be open and let \(f\colon\Omega\to\CC\) be univalent. Then \(f'(z)\neq0\) for every \(z\in\Omega\), the image \(f(\Omega)\) is open, and \(f\colon\Omega\to f(\Omega)\) is a biholomorphism. Conversely, every biholomorphism from \(\Omega\) onto an open subset of \(\CC\) is univalent.
Suppose \(f'(z_0)=0\) for some \(z_0\in\Omega\). An injective function is not constant on any disc, so on a disc \(D\subseteq\Omega\) about \(z_0\), \(f-f(z_0)\) has a zero of order $n$ at \(z_0\) with \(n\ge2\): \(f(z)-f(z_0)=(z-z_0)^n g(z)\) with \(g\) holomorphic and \(g(z_0)\neq0\). Shrinking \(D\), \(g\) has no zeros on \(D\), so it has a holomorphic \(n\)-th root \(k\) on \(D\), and \(f(z)-f(z_0)=\varphi(z)^n\) with \(\varphi(z)\coloneqq(z-z_0)k(z)\). Since \(\varphi(z_0)=0\) and \(\varphi'(z_0)=k(z_0)\neq0\), the holomorphic inverse function theorem gives a neighborhood \(V\subseteq D\) of \(z_0\) mapped by \(\varphi\) bijectively onto a disc \(D_\varepsilon(0)\). For \(0<\delta<\varepsilon\), the \(n\ge2\) distinct points \(\varphi^{-1}(\delta e^{2\pi ij/n})\), \(j=0,\ldots,n-1\), all have image \(f(z_0)+\delta^n\) under \(f\), contradicting injectivity. Hence \(f'\) has no zeros.
By the open mapping theorem applied on each connected component of \(\Omega\), \(f(\Omega)\) is open. For \(w_0=f(z_0)\), the holomorphic inverse function theorem gives open neighborhoods \(V\ni z_0\) and \(W\ni w_0\) with \(f\colon V\to W\) bijective and holomorphic inverse; since \(f\) is injective, \(f^{-1}\) agrees with this local inverse on \(W\). Thus \(f^{-1}\colon f(\Omega)\to\Omega\) is holomorphic. The converse holds because a biholomorphism is holomorphic and bijective onto its image.
Hurwitz's theorem (Tag T-FZWEC)
Let \(\Omega\subseteq\CC\) be open and let \((f_k)_{k\geq1}\) be a sequence of holomorphic functions on \(\Omega\) converging locally uniformly on \(\Omega\) to \(f\). Suppose \(f\) has a zero of order \(n\) at \(z_0\in\Omega\). Then there is \(\delta>0\) such that for all sufficiently large \(k\), \(f_k\) has exactly \(n\) zeros in the disc \(\{z : \abs{z-z_0}<\delta\}\), counted with multiplicity. Moreover, for every \(\varepsilon\in(0,\delta)\) there is \(K\) such that for all \(k\geq K\) these \(n\) zeros lie in \(\{z : \abs{z-z_0}<\varepsilon\}\).
Near a zero of order \(n\) of the limit, the \(n\) zeros of \(f_k\) for large \(k\) converge to that zero.
Let \(\Omega\) be a connected open set and let \(f_k\to f\) locally uniformly on \(\Omega\), with each \(f_k\) holomorphic and nowhere zero on \(\Omega\). Then \(f\) is nowhere zero on \(\Omega\) or identically zero.
If \(f\) is not identically zero and \(f(z_0)=0\), then \(z_0\) is a zero of finite order \(n\geq 1\) by the identity principle, and Hurwitz's theorem gives zeros of \(f_k\) near \(z_0\) for large \(k\), a contradiction.
In the proof of the Riemann mapping theorem, Montel's theorem produces a locally uniform limit of univalent maps, and Locally uniform limits of univalent functions are univalent or constant shows the limit is univalent because its derivative at the base point is nonzero, so it is not constant.
Locally uniform limits of univalent functions are univalent or constant (Tag T-SULVA)
Let \(\Omega\subseteq\CC\) be a connected open set and let \((f_k)_{k\geq1}\) be univalent functions on \(\Omega\) converging locally uniformly on \(\Omega\) to \(f\). Then \(f\) is either univalent or constant.