利用Freitag与Busam习题III.4.15证明开映射定理

Hey there, let's walk through how to use the given exercise result to prove the Open Mapping Theorem. First, let's restate the key exercise clearly:

Exercise III.4.15 (Freitag & Busam) #

Let $g$ be an analytic function on an open set containing the closed disk $\bar{U}_r(a)$. Suppose that for every $z$ on the boundary of the disk, $|g(a)| < |g(z)|$. Then $g$ has at least one zero inside the disk $\bar{U}_r(a)$.

Proving the Open Mapping Theorem #

Recall the Open Mapping Theorem states: If $f$ is a non-constant analytic function on a domain $D$, then the image $f(D)$ is an open set.

Here's how we use the exercise to prove this:

1. Setup: Pick any point $a \in D$. Since $D$ is open, there exists some $r > 0$ such that the closed disk $\bar{U}_r(a)$ is entirely contained in $D$.

2. Define the auxiliary function: For any complex number $w$, let $g_w(z) = f(z) - w$. Our goal is to show that for all $w$ sufficiently close to $f(a)$, $g_w$ has a zero inside $U_r(a)$ (which means $w$ is in $f(U_r(a))$, hence in $f(D)$).

3. Find a minimum distance on the boundary: Since $f$ is continuous on the compact set $\partial U_r(a)$, the function $|f(z) - f(a)|$ attains a minimum value here. Let $\delta = \min_{z \in \partial U_r(a)} |f(z) - f(a)|$. Since $f$ is non-constant, $\delta > 0$ (if $\delta$ were zero, $f$ would be constant on $D$ by the Identity Theorem, which contradicts our assumption).

4. Choose $\varepsilon$ and apply the exercise: Let $\varepsilon = \delta/2$. Take any $w$ such that $|w - f(a)| < \varepsilon$. Now:

   • $|g_w(a)| = |f(a) - w| < \varepsilon = \delta/2$
   • For every $z \in \partial U_r(a)$, by the reverse triangle inequality:
     $$|g_w(z)| = |f(z) - w| \geq | |f(z) - f(a)| - |f(a) - w| | \geq \delta - \varepsilon = \delta/2 > |g_w(a)|$$
     This satisfies the conditions of Exercise III.4.15 for $g_w$, so $g_w(z) = f(z) - w$ has a zero inside $U_r(a)$.

5. Conclusion: This means every $w$ in the open disk $U_\varepsilon(f(a))$ is in $f(U_r(a)) \subset f(D)$. Since $a$ was arbitrary, every point in $f(D)$ has an open neighborhood contained in $f(D)$, so $f(D)$ is open. That's exactly what the Open Mapping Theorem claims.

内容的提问来源于stack exchange，提问作者Jiu