2
$\begingroup$

Let $\mathrm{\Omega}$ be an open connected set in $\mathbb{C}$. Let ${f_n}$ be a sequence of univalent analytic functions on $\mathrm{\Omega}$ and assume that ${f_n}$ converges locally uniformly to a function $f$. Show that $f$ is either univalent or a constant.

How to attack this problem?

$\endgroup$
  • $\begingroup$ Rouche's theorem. $\endgroup$ – Shubhodip Mondal Nov 27 '14 at 21:16
2
$\begingroup$

Notice that $f$ is folomorphic by uniform convergence on compact parts (equivalent locally uniform convergence).

Suppose that is $f(z)$ non-constant and not injective.

Let $a\neq b$ such that $f(a)=f(b)=p$

Let $\overline{D(a,r)}$ and $\overline{D(b,r)}$ disjoint compact discs such that:

$z=a$ is the only solution of equation $f(z)=f(a)$ on $\overline{D(a,r)}$

$z=b$ is the only solution of equation $f(z)=f(b)$ on $\overline{D(b,r)}$

Let $K=\partial D(a,r)\bigcup \partial D(b,r)$

Consider $d=\inf \{|f(z)-p|:z\in K\}$ , $d>0$

By uniform convergence on compact parts:

for all $n$ large enough , $|f_n(z)-f(z)|<d$ for all $z\in K$ then

$|f_n(z)-f(z)|<|f(z)-f(a)|$ on $\partial D(a,r)$ then $f_n(z_0)-f(a)=0$ by Rouche's Theorem

$|f_n(z)-f(z)|<|f(z)-f(b)|$ on $\partial D(b,r)$ then $f_n(z_1)-f(b)=0$ by Rouche's Theorem

For some $z_0\in D(a,r)$ , $z_1 \in D(b,r)$ therefore $f_n(z_0)=f_n(z_1)$ and $z_0\neq z_1$ contradiction.

$\endgroup$
1
$\begingroup$

It is enough to show that for any $c \in \mathbb C$, $f(z) - c$ has at most one root in $\mathrm {\Omega}$.On the contrary, assume that there are two points $z_0,z_1 \in \mathrm{\Omega}$ such that $f(z_0) = f(z_1) = c$. If $f$ is non-constant, zeros of $f(z) - c$ must be an isolated set. Hence,there exists $r_0$ such that $f(z) \ne c$ for $0 < |z-z_0| < r_0$ and similarly $r_1$ for the point $z_1$. We can assume that $r_0, r_1$'s are small enough such that $\overline{B_{r_0}(z_0)} \cap \overline{B_{r_1}(z_1)} = \emptyset$.

Now let us define $h_n(z) = f_n(z) - f(z)$, then $f_n(z)$ converges to $0$ uniformly (on every compact subset) which means that, for every $\epsilon >0$ there is some $M$ such that if $n \ge M$, then $|h_n(z)| \le \epsilon$. So, in particular if we consider the circle of radius $r_0$ i.e $C_{r_0} (z_0)$, which is a closed curve in $\mathrm{\Omega}$, we have that $|h_n(z)| < |f(z) - c| $ on $C_{r_0} (z_0)$ for all large enough $n$.(By compactness of the circle, $\text{inf}_{z \in C_{r_0}(z_0)} |f(z) - c| > 0$.)

Hence, by Rouche's theorem $h_n(z) + f(z) - c = f_n(z) - c$ and $f(z) - c$ has the same number of zeros inside $C_{r_0} (z_0)$, which means $f_n(z) - c$ has a root inside $C_{r_0}(z_0)$ for all large $n$. Similarly $f_n(z) - c$ has a root inside $C_{r_1}(z_1)$ for all large $n$. Which means, $f_n(z) - c$ has at least $2$ roots for all large enough n, contradicting the fact that they are univalent.

$\endgroup$

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.