I'm stuck on the following problem:

Let $f_n$ be a sequence of injective analytic functions on the unit disc $D$ such that $f_n$ converges uniformly to $f$ on compact subsets of $D$. Show that $f$ is either injective or constant.

Already $f$ is analytic as an almost uniform limit. If $f'$ is identically zero, then $f$ is constant. Otherwise $f'$ has isolated zeroes. If $c$ is such an isolated zero, then $f_n'$ converges uniformly to $f'$ on a small disc near $c$. By Hurwitz's theorem, for almost all $n$ the functions $f_n'$ should also have a zero in that small disc.

But an analytic function is locally injective at a point if and only if its derivative at the point is nonzero. And injective implies locally injective. So the $f_n'$ shouldn't have any zeros in $D$.

So, we can at least conclude that $f'$ is never zero on $D$. This shows that $f$ is locally injective. Is there an easy way to conclude injectivity from local injectivity here?

  • $\begingroup$ Great explanation of your progress so far, by the way! $\endgroup$ – Greg Martin Aug 3 '15 at 22:55
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    $\begingroup$ Idea: use the argument principle to write down a contour integral that equals the number of times that $f(z)=0$ (and, easy generalization, $f(z)=c$; compare to the analogous contour integrals for the $f_n$. Alternately, use Rouché's theorem for similar contour integrals. $\endgroup$ – Greg Martin Aug 3 '15 at 22:59
  • $\begingroup$ @D_S: note that a holomorphic function $h: D \to D$ that is locally injective is not necessarily injective. if you take $h(z)=e^{-4\pi} e^{4\pi z}$, then $h(D) \subset D$ and $h$ is locally injective, but $h(0)=h(i/2)$. it would be true however if you were to know eg that $f$ is proper $\endgroup$ – Glougloubarbaki Oct 5 '17 at 14:59
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    $\begingroup$ I don't see an easy way to conclude injectivity from the local injectivity. The only ways I could think of are badly concealed versions of the direct proof of injectivity via Hurwitz's theorem. (Suppose $f$ were neither constant nor injective. Let $f(z_1) = w = f(z_2)$ for $z_1 \neq z_2$. Take two disjoint small disks $D_k$ around $z_k$, such that $f(z) \neq w$ on $\partial D_k$. By Hurwitz, $f_n$ attains the value $w$ in $D_1$ and in $D_2$ for large enough $n$, contradicting the assumed injectivity of $f_n$.) $\endgroup$ – Daniel Fischer Nov 17 '17 at 15:27
  • $\begingroup$ Any reason you did not accept the posted answer? $\endgroup$ – Moishe Kohan Jul 6 at 15:45

Uniform limit of injective analytic functions is injective, if this limit IS NOT CONSTANT.

Proof. Suppose that $f(z)=a$ at two points $z_1,z_2$. Surround these two points by a Jordan curve $\gamma$ such that $f(z)\neq a$ on $\gamma$. Then according to the argument principle $$\frac{1}{2\pi}\int\frac{f'(z)dz}{f(z)-a}\geq 2.$$ (The right hand side equals to the number of solutions of $f(z)=a$ inside $\gamma$, counting multiplicity). On the other hand, $f_n(z)\neq a$ on $\gamma$ when $n$ is sufficiently large, and the similar integral with $f_n$ instead of $f$ is $\leq 1$, because $f_n$ are injective. But this is a contradiction because $f_n\to f$ uniformly, so integral with $f_n$ must converge to the integral with $f$.


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