In Ahlfors' Complex Analysis text, page 227 the author claims that if the expression $$\rho(f)=\frac{2 |f'(z)|}{1+|f(z)|^2}$$ is locally bounded (here $f \in \mathfrak F$ is a family of meromorphic functions in some domain $\Omega$), then the family $\mathfrak F$ is equicontinuous (with respect to the spherical metric) on compact subsets of $\Omega$.
In particular, he states that
If $\rho(f) \leq M$ on the line segment between $z_1$ and $z_2$ we conclude that $d(f(z_1),f(z_2)) \leq M|z_1- z_2|$, and this immediately proves the equicontinuity when $\rho(f)$ is locally bounded.
I believe that he intended that we start the formula for the length
$$L=\int_\gamma \rho(f) |dz| $$ where $\gamma$ is the line segment between $z_1$ and $z_2$, and apply the triangle inequality.
My problem with this proof is that $f$ might not even be defined on that line segment (if $\Omega$ is not convex), and in that case the integral formula is invalid.
Is this proof correct after all? If so, what am I missing here?
Thanks.
The proof is correct, but curt. Some things remain implicit.
The integral estimate gives you for every $z \in K \subset \Omega$ a Lipschitz constant on a small disk around $z$. Since $K$ is compact, you can cover it by finitely many such disks. For small enough $\delta_0 > 0$, for any two points $z,w$ in $K$ with $\lvert z - w\rvert \leqslant \delta_0$, there is one disk containing both, and hence the segment connecting the two points. So for equicontinuity ($(\forall \varepsilon > 0)(\exists \delta > 0)(\dotsc)$), choose the $\delta$ always smaller than $\delta_0$.