Given a meromorphic function $f:\mathbb{C}\to\mathbb{S}^2$ satisfies $$\int_{\mathbb{R}^2}\frac{|f'(z)|^2}{(1+|f(z)|^2)^2}dxdy<+\infty,$$ prove that $f$ is a rational function.
My attempts:
It suffices to check that $\infty$ is not an essential singularity of $f$. Otherwise, it will cover the extended complex plane infinitely many times (possibly except for one point). How to go on and use the integrable condition? Thanks.
On
My attempt is just so close to the conclusion. If we observe that $$A(r)=\int_{|z|<r}\frac{|f'(z)|^2}{(1+|f(z)|^2)^2}\text{d}x\text{d}y$$ is the area (w.r.t. the ordinary metric on the unit sphere) of the image of $\{|z|<r\}$ under the map $f:\mathbb{C}\to\mathbb{C}\cup\{\infty\}\hookrightarrow\mathbb{S}^2$. If $\infty$ is an essential singularity, then the image of $\mathbb{C}$ covers $\mathbb{S}^2$ for infinitely many times, which is impossible.
Define on $\mathbb{C}\cup \{\infty\}$ the measure $d\mu:=\frac{1}{(1+|z|^2)^2}dA$, where $dA$ is the usual area measure. It is easy to see that $\mu$ is a finite measure. Now let $\Omega, \Omega_1$ be open sets and let $f:\Omega \to \Omega_1$ be a conformal map. It follows from the change of variable formula that $$\int_\Omega \frac{|f'|^2}{(1+|f|^2)^2}dA=\int_{f(\Omega)}d\mu$$ More generally, by the Banach indicatrix theorem we have $$\int_\Omega \frac{|f'|^2}{(1+|f|^2)^2} dA=\int_{f(\Omega)} n(w)d\mu(z)$$ where $n(w):=\#\{z\in \Omega:f(z)=w\}$. If $\infty$ is an essential singularity and $\Omega=\mathbb{C}$, $n(w)=\infty$ for almost every $w\in \mathbb{C}$ and the integral diverges. On the other hand, if $f$ is rational then $n(w)$ is bounded and so the integral is finite since $\mu$ is a finite measure.