Show that $f$ is either injective or a constant function.

Let $\Omega$ be a domain in $\mathbb{C}$ and let $\{f_n\}_{n \in \mathbb{N}}$ be a sequence of injective functions that converge in $O(\Omega)$ to $f$ . Show that $f$ is either injective or a constant function.

How does the conclusion change if, instead of a domain, we allow $\Omega$ to be an arbitrary open set ?

I know that $f$ is holomorphic as an almost uniform limit. But I dont know how to proceed.

• Hint: the derivative of an analytic function can be recovered from the function's values via the Cauchy integral formula. What can you say about the derivatives of all the functions involved? – Greg Martin Apr 11 '16 at 4:24
• @GregMartin: I don't see how that would work. Even if you show that the derivative of the limit function vanishes nowhere, that would not be sufficient to prove that an analytic function is injective. Or what did you have in mind? – Martin R Apr 12 '16 at 8:07

Assume that the limit function $f$ is neither constant nor injective. Then $f$ takes some value $a$ in disjoint disks $B_1, B_2 \subset \Omega$. As in

it follows from Rouché's theorem that there are $n_1, n_2$ such that for $n \ge n_j$, $f_n$ takes the value $a$ in $B_j$ (at least once). For $n \ge \max(n_1, n_2)$ this is a contradiction to $f_n$ being injective.

Rouché's theorem is applied separately to the two disks in $\Omega$, so the same conclusion holds if $\Omega$ is a (not necessarily connected) open set.

$\textbf{Sketch of proof:}$

$\textbf{1:}$ If holomorphic function series {$f_n$} converges uniformly to f on a compact set A $\subset$ $\Omega$,then {$f_n^{'}$} converges uniformly to $f^{'}$.

$\textbf{Proof of 1:}$ $\forall$ $\epsilon$ $>$0 $\exists$ $N\in N^{*}$,when n>N,$\mid f_n-f \mid$$<$$\epsilon$.$\forall$ $z\in A$ choose a compact circle C contained in $\Omega$ with its ratio R.Then use Cauchy's formula: $(f-f_n)^{'}(z)={\frac{1}{2{\pi}i}}\oint_C\frac{(f-f_n)(\varepsilon)}{(\varepsilon-z)^{2}}d\varepsilon$.Then we have estimate:

$\mid (f-f_n)^{'}(z) \mid$$<=$$\epsilon$/R. $\textbf{QED}$

$\textbf{2}$.If f is injective on compact set A,then $f^{'}$$\neq 0 on any point in A. \textbf{Proof of 2:} Using Rouche's theorem.If there is a z_0 \in A,f^{'}(z_0)=0.By \textbf{Isolated zero theorem},there is a region 0<\mid z-z_0 \mid <\delta contained in \Omega ,in this region f^{'}(z) \neq 0.Denote \Delta=\mid z-z_0 \mid=\delta,let m:=inf_{\Delta}\mid f(z)-f(z_0) \mid(because \Delta is compact,the definition is reasonable.) \textbf{By Rouche Theorem}: h(z):=f(z)-f(z_0)-\frac{m}{2} has same zero as g(z):=f(z)-f(z_0).since g(z) obviouly has a zero z_0 and its multiple is at least 2,h(z) has 2 zeros in \mid z-z_0 \mid<\delta.They are different from z_0,so the derivation on them is not 0,so the zero is simple,and there is z_1 \neq z_2 such that h(z)=0,then f(z_1)=f(z_2)=f(z_0)+\frac{m}{2},then f is not injective on A.\textbf{QED} \textbf{3} If there is a z_0 such that f^{'}(z_0)$$\neq$$0$,then there is a neighbourhood contained in $\Omega$,in this neighbourhood f is holomorphic and injective.

$\textbf{Proof of 3}$:Consider f=u(x,y)+$\textbf{i}$v(x,y).Proof that $|f^{'}|$ is the Jacobian $\frac{\partial{(u,v)}}{\partial{(x,y)}}$ and it is not zero.Use implicit function theorem to show that there is a neighbourhood in which f is holomorphic and injective.$\textbf{QED}$

$\textbf{4}$ if {$f_n$} is nowhere vanishing in $\Omega$ and it converges uniformly to f on any compact set of $\Omega$,then f is zero,or nowhere vanishing.

$\textbf{Proof of 4:}$.Google $\textbf{Hurwitz theorem}$.$\textbf{QED}$

$\quad$ Then according to conditions of your problem,$f_n^{'}$ is $\textbf{Nowhere vanishing}$ in $\Omega$ and it converges uniformly on any compact set in $\Omega$.(By 1,2).By 4 $f^{'}$ is zero or nowhere vanishing.If $f^{'}$ is 0,then f=$\textbf{CONST}$,and otherwise by 3 on any point belonged to $\Omega$ there is a neighborhood such that f is injective.By connectness of $\Omega$（In Euclidean space a connected open set is path-connected,so we choose a path from x and y.The path is compact in $E^2$,so we can useuse Heine-Borel theorem.),f is injective in $\Omega$. $\textbf{QED}$

• locally injectivity doesn't imply injectivity. – Mathillda Oct 6 '18 at 14:14