# $f$ is either univalent or constant

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?

• Rouche's theorem. – Shubhodip Mondal Nov 27 '14 at 21:16

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.

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.