Proof of Hopf's theorem using Liouville

Question

Hopf Theorem

A topological sphere immersed as a constant mean curvature surface in $\mathbb{E}^3$ is a round sphere

In Heinz Hopf's Differential Geometry in the Large, a proof is given of the above theorem using Riemann surfaces and Liouville's theorem. (See section 4 of the following paper: http://www.ime.usp.br/~eufrasio/matcont/edicoes/35/35_10.pdf)

One argument has troubled me: given a local parametrisation $x(u,v)$ of a point on the surface $S$, which is conformally equivalent to the Riemann sphere $\mathbb{C}_\infty$, one can define the coefficients of the second fundamental form $L, M, N$. What is done is the proof, is that we define a function $$ \Phi(u,v) = \frac{L-N}{2}-iM$$ and we do a change of variables to $z = u+iv, \bar{z}=u-iv$. From the Codazzi equations we find that $\Phi$ is holomorphic and thus depends only on $z$.

It is then shown that $\Phi(z) = -2 \langle \bar{x}_z,x_z\rangle$ where $\bar{x}$ is a unit normal on the surface $S$. For different parameters $u',v'$ we can define $w=u' + iv'$ and we define $$\Psi(w) = -2 \langle \bar{x}_w,x_w\rangle$$ Now we assume $z$ covers all of the Riemann sphere except $\infty$, and $w = 1/z$ is the coordinate that covers all except $0$.

Then using the rules of transition between two coordinates, we define the Hopf Differential $$\Phi(z)(dz)^2 = \Psi(w)(dw)^2$$

The idea of the proof is now to apply Liouville's theorem to $\Phi$. Now, I agree with the proof that $\Phi$ is an entire function of $z$, but why is it bounded? If $\Phi$ were defined on all of $S = \mathbb{C}_\infty$, then clearly it would be. But since we only defined $\Phi$ for the complex coordinate $z$, can we really assume this?

(Note: I know that one could simply apply Poincaré's theorem that the only quadratic differential on $\mathbb{C}_\infty$ is the trivial one, but the point of this proof was specifically to avoid that)

Answer 1

After stating the relation $\Phi(w) = \Psi(z)z^4$ (where $w=1/z$), the author says:

Being $\Phi$ an entire function of $w$ and being $\Psi$ regular for $z = 0$, then $\Phi = 0$ for $w = \infty$...

Expressed in a more precise form: $$ \lim_{w\to \infty}\Phi(w) = \lim_{z\to 0} \Psi(z)z^4 = 0 $$ Since $\Phi$ has a finite limit at infinity, it is bounded.