What does holomorphic function on the Riemann sphere mean? Is that just $f\in H(\mathbb{C}\cup \{\infty\})$, so $f:\mathbb{C\cup \{\infty\}}\to \mathbb{C}$. And is $f$ entire? Since $f$ is holomorhpic over the whole complex plane.

I am attempting to prove the following:

All holomorphic functions on the Riemann shpere are constant

The way to do this is by Liouville's theorem, which said every bounded entire function is constant. So I just need to show that entire function on Riemann sphere is bounded.

  • 2
    $\begingroup$ The restriction of $f$ to $\mathbb{C}$ is an entire function. $\endgroup$ Sep 8, 2015 at 19:37
  • $\begingroup$ It does mean that f restricted to complex plane is holomorphic together with the fact that the map f(1/z) as a function of z is holomorphic in z possibly after removing a removable singularity. Using this it is easy to see that f is a continuous map from riemann spehere to complex plane is a continuous function and hence its image must be compact and thus bounded. You can find this in beginning of any text on riemann surface. $\endgroup$
    – random123
    Sep 9, 2015 at 6:16

1 Answer 1


$f$ is holomorphic at $\infty$. So $f$ if bounded as well in an open ball (say $B$) at $\infty$. Also $f$ is continuous imply $f$ is bounded in $\Bbb{C_\infty}\backslash\,B$ (as this set is compact).

So $f$ if bounded in ehole of $\Bbb{C_\infty}$.


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