11
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Great explanation of your progress so far, by the way! $\endgroup$ – Greg Martin Aug 3 '15 at 22:55
  • 5
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
1
$\begingroup$

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 $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.