# Does entire function extend to the holomorphic function of Riemann sphere?

Let $f\colon \mathbb{C}\to \mathbb{C}$ be a nonconstant entire function.

Is it true that there exists $\bar{f}\colon S^2\to S^2$ with $\bar{f}|_\mathbb{C}=f$?

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possible duplicate of Nomenclature in complex analysis –  t.b. Nov 15 '11 at 7:22
Isn't it true that $f\colon \mathbb{C}\to \mathbb{C}$ defined by $z\mapsto e^z$ can be extended to the holomorphic function of $S^2$? –  complexanalysisnewbie Nov 15 '11 at 7:26
No, since $\exp{1/z}$ has an essential singularity at $0$, there is no way to extend this function to a continuous function in $0$, let alone a holomorphic one (the image of every punctured neighborhood of $0$ is dense in $\mathbb{C}$ by Casorati-Weierstrass). –  t.b. Nov 15 '11 at 7:30
I got ya. Thank you man. –  complexanalysisnewbie Nov 15 '11 at 7:32
What do you mean by "there exists $\bar f:\ S^2\to S^2\$"? You could define $\bar f(\infty):=0$. –  Christian Blatter Nov 15 '11 at 9:40

Proposition. Let $f:\mathbb{C}\to\mathbb{C}$ be an entire function. The following are equivalent:

1. $f$ extends continuously (and hence holomorphically) to a function $f:S^2\to S^2$;
2. $f$ is either constant or $z\mapsto f(1/z)$ has a pole at $0$;
3. $f$ is either constant or a proper map.
4. $f$ is a polynomial;

Proof. Let us assume that $f$ is non-constant (otherwise all statements are true); then by Liouville's Theorem, $f$ is unbounded.

The equivalence of the first three statements is easy. Indeed, as $f$ is unbounded, $f$ extending continuously to $\infty$ is equivalent to $\lim_{z\to\infty} f(z)=\infty$, which in turn is equivalent to $f$ being a proper map.

If $f$ is a non-constant polynomial, then clearly again $\lim_{z\to\infty} f(z)=\infty$, so 4. implies 1. To prove that 1. implies 4., assume 1. and observe that the set of zeros of $f$ is a discrete and closed subset of $S^2$, so this set is finite. Let $g$ be a polynomial with the same zeros (and same multiplicities) as $f$ and consider the function $h:=f/g$. It is not difficult to show that the function $h$ is constant, and hence $f$ is a polynomial.

(Alternatively, if you know about Laurent series, just observe the relationship between the Laurent series of $f(1/z)$ around zero and the Taylor series that represents $f$ in the complex plane. So the singularity of the former function is non-essential if and only if the Taylor series is finite; i.e., if and only if $f$ is a polynomial.) $\blacksquare$

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No, the function $\exp$, for example has an essential singularity at $\infty$ and has no holomorphic extension. The same is true for the simple function $f(z) = z$ which has a pole of first order at $\infty$ and so on. Basically, the behavior of $f(z)$ at $\infty$ is the same than the behavior of $f(1/z)$ at zero.

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Thank you guys, then is it true that the entire function is proper map? –  complexanalysisnewbie Nov 15 '11 at 7:30
The function $f(z)=z$ does extend continuously to a map from the Riemann sphere to itself, with $f(\infty)=\infty$. –  mathstribble Nov 16 '11 at 9:43