求证：全纯自同构$f:\mathbb{C} \to \mathbb{C}$满足$\vert f(z) \vert \to \infty$当$\vert z \vert \to \infty$

Hey there! Let's work through this problem step by step—first proving that key limit property, then extending to the Riemann sphere, and finally showing the map must be affine.

We can use proof by contradiction here:

• Suppose there exists a sequence ${z_n}$ where $\vert z_n \vert \to \infty$, but $\vert f(z_n) \vert$ stays bounded (say, $\vert f(z_n) \vert \leq M$ for some constant $M$).
• Since $f$ is a surjective entire function, there's a unique $z_0 \in \mathbb{C}$ such that $f(z_0) = w_0$, where $w_0$ is a limit point of ${f(z_n)}$ (we know such a $w_0$ exists because bounded sequences in $\mathbb{C}$ have limit points).
• Define the entire function $g(z) = f(z) - w_0$. This function has a single zero at $z_0$ (because $f$ is injective—only one $z$ maps to $w_0$). But the sequence ${z_n}$ satisfies $g(z_n) = f(z_n) - w_0 \to 0$ as $n \to \infty$, meaning we have infinitely many points approaching $\infty$ where $g(z)$ gets arbitrarily close to 0.
• A non-zero entire function can't have non-isolated zeros (or a sequence of points approaching $\infty$ where it tends to 0 without being identically zero). This contradicts the fact that $g(z)$ only has one zero at $z_0$. So our initial assumption is wrong, and $\vert f(z) \vert \to \infty$ as $\vert z \vert \to \infty$.

The Riemann sphere adds the point $\infty$ to $\mathbb{C}$, with neighborhoods of $\infty$ being sets of the form ${z \in \mathbb{C} \mid \vert z \vert > R} \cup {\infty}$ for some $R > 0$.

• Define the extended map $F: \mathbb{C}{\infty} \to \mathbb{C}{\infty}$ by $F(z) = f(z)$ for $z \in \mathbb{C}$, and $F(\infty) = \infty$.
• Continuity check:
  • For $z \to \infty$ in $\mathbb{C}$, we already proved $\vert f(z) \vert \to \infty$, so $F(z) \to F(\infty)$, which satisfies continuity at $\infty$.
  • The inverse map $f^{-1}: \mathbb{C} \to \mathbb{C}$ is also an entire automorphism (since biholomorphic maps have biholomorphic inverses). By the same logic as step 1, $\vert f^{-1}(w) \vert \to \infty$ as $\vert w \vert \to \infty$, so $F^{-1}(\infty) = \infty$, and $w \to \infty$ implies $F^{-1}(w) \to \infty$, ensuring continuity for the inverse.
• Biholomorphicity at $\infty$: Use local coordinates around $\infty$—take the chart $\phi(z) = 1/z$ for $z \neq \infty$, with $\phi(\infty) = 0$. The composition $\phi \circ F \circ \phi^{-1}$ maps $t \to 1/f(1/t)$ near $t=0$. Since $f(1/t) \to \infty$ as $t \to 0$, $1/f(1/t) \to 0$, and this function is holomorphic at $t=0$ (the singularity is removable). It's also bijective because $F$ is bijective, so the local map is biholomorphic. This means $F$ is a biholomorphic map on $\mathbb{C}_{\infty}$.

Biholomorphic maps on the Riemann sphere are exactly Möbius transformations, which have the general form:
$$F(z) = \frac{az + b}{cz + d}, \quad ad - bc \neq 0, , a,b,c,d \in \mathbb{C}$$

• Our extended map $F$ satisfies $F(\infty) = \infty$. Let's compute the limit as $z \to \infty$ for the Möbius transformation: $F(z) = \frac{a + b/z}{c + d/z} \to \frac{a}{c}$. For this limit to be $\infty$, we must have $c = 0$ (otherwise the limit is the finite value $a/c$).
• Setting $c=0$, the transformation simplifies to $F(z) = \frac{az + b}{d} = Az + B$, where $A = \frac{a}{d} \neq 0$ (since $ad - bc = ad \neq 0$) and $B = \frac{b}{d}$. This is exactly an affine map (linear function plus a constant), and it's clearly an entire automorphism of $\mathbb{C}$.

You can find detailed proofs and context in standard complex analysis textbooks:

• Complex Analysis by Lars Ahlfors: Covers entire automorphisms, Riemann sphere geometry, and Möbius transformations in depth.
• Functions of One Complex Variable I by John B. Conway: Includes rigorous arguments about extending entire maps to the Riemann sphere and characterizing the automorphism group.

内容的提问来源于stack exchange，提问作者GouldBach