[proofplan]
We apply the maximum modulus principle to the reciprocal $1/f$, which is holomorphic on $\Omega$ because $f$ is non-vanishing. The maximum of $|1/f|$ on $\overline{\Omega}$ is attained on the boundary, which translates to the minimum of $|f|$ being attained on the boundary.
[/proofplan]
[step:Construct the reciprocal function $1/f$ and verify it is holomorphic]
Define the function
\begin{align*}
g: \overline{\Omega} &\to \mathbb{C} \\
z &\mapsto \frac{1}{f(z)}.
\end{align*}
Since $f(z) \neq 0$ for all $z \in \overline{\Omega}$, the function $g$ is well-defined on $\overline{\Omega}$. On $\Omega$, the function $g$ is holomorphic: it is the composition of the holomorphic function $f: \Omega \to \mathbb{C} \setminus \{0\}$ with the holomorphic function $w \mapsto 1/w$ on $\mathbb{C} \setminus \{0\}$. Since $f \in C(\overline{\Omega})$ and $f$ is non-vanishing on $\overline{\Omega}$, the function $g$ is continuous on $\overline{\Omega}$. Hence $g \in \mathcal{O}(\Omega) \cap C(\overline{\Omega})$.
[guided]
The idea is to convert a minimum problem for $|f|$ into a maximum problem. Since $|1/f(z)| = 1/|f(z)|$, maximising $|1/f|$ is equivalent to minimising $|f|$. But we need $1/f$ to be holomorphic, which requires $f$ to be non-vanishing — and this is precisely what the hypothesis $f(z) \neq 0$ for all $z \in \overline{\Omega}$ provides.
Define
\begin{align*}
g: \overline{\Omega} &\to \mathbb{C} \\
z &\mapsto \frac{1}{f(z)}.
\end{align*}
We verify that $g \in \mathcal{O}(\Omega) \cap C(\overline{\Omega})$:
- **Well-defined:** $f(z) \neq 0$ for all $z \in \overline{\Omega}$, so division by $f(z)$ is valid everywhere on $\overline{\Omega}$.
- **Holomorphic on $\Omega$:** The function $f: \Omega \to \mathbb{C} \setminus \{0\}$ is holomorphic by hypothesis. The function $h: \mathbb{C} \setminus \{0\} \to \mathbb{C}$, $h(w) = 1/w$, is holomorphic on its domain. The composition $g = h \circ f$ is holomorphic on $\Omega$.
- **Continuous on $\overline{\Omega}$:** Since $f \in C(\overline{\Omega})$ and $f$ is non-vanishing on the compact set $\overline{\Omega}$, the reciprocal $g = 1/f$ is continuous on $\overline{\Omega}$.
[/guided]
[/step]
[step:Apply the maximum modulus principle to $g = 1/f$ on $\overline{\Omega}$]
Since $\Omega$ is a bounded domain and $g \in \mathcal{O}(\Omega) \cap C(\overline{\Omega})$, the set $\overline{\Omega}$ is compact (closed and bounded in $\mathbb{C}$). The continuous function $|g|: \overline{\Omega} \to \mathbb{R}$ attains its maximum on $\overline{\Omega}$. By the [Maximum Modulus Principle](/theorems/???), the maximum of $|g|$ on $\overline{\Omega}$ is attained on the boundary $\partial\Omega$:
\begin{align*}
\max_{z \in \overline{\Omega}} |g(z)| &= \max_{z \in \partial\Omega} |g(z)|.
\end{align*}
[guided]
The maximum modulus principle (in its boundary form for bounded domains) states: if $h \in \mathcal{O}(\Omega) \cap C(\overline{\Omega})$ where $\Omega$ is a bounded domain, then $\max_{z \in \overline{\Omega}} |h(z)| = \max_{z \in \partial\Omega} |h(z)|$. We verify the hypotheses for $h = g$:
- $\Omega$ is a bounded domain by hypothesis.
- $g \in \mathcal{O}(\Omega) \cap C(\overline{\Omega})$ was established in the previous step.
Therefore
\begin{align*}
\max_{z \in \overline{\Omega}} |g(z)| &= \max_{z \in \partial\Omega} |g(z)|.
\end{align*}
This is the key step: the maximum principle, which is a statement about maxima, will translate into a statement about minima once we invert back.
[/guided]
[/step]
[step:Translate the maximum of $|1/f|$ back to the minimum of $|f|$]
Since $|g(z)| = 1/|f(z)|$ for all $z \in \overline{\Omega}$, the identity $\max_{z \in \overline{\Omega}} |g(z)| = \max_{z \in \partial\Omega} |g(z)|$ becomes
\begin{align*}
\max_{z \in \overline{\Omega}} \frac{1}{|f(z)|} &= \max_{z \in \partial\Omega} \frac{1}{|f(z)|}.
\end{align*}
Since $|f(z)| > 0$ for all $z \in \overline{\Omega}$, the map $t \mapsto 1/t$ is a strictly decreasing bijection on $(0, \infty)$, so $\max(1/|f|) = 1/\min(|f|)$. Applying this to both sides:
\begin{align*}
\frac{1}{\min_{z \in \overline{\Omega}} |f(z)|} &= \frac{1}{\min_{z \in \partial\Omega} |f(z)|}.
\end{align*}
Inverting yields
\begin{align*}
\min_{z \in \overline{\Omega}} |f(z)| &= \min_{z \in \partial\Omega} |f(z)|.
\end{align*}
This is the desired conclusion.
[guided]
We now translate back from $g$ to $f$. The relation $|g(z)| = |1/f(z)| = 1/|f(z)|$ holds for all $z \in \overline{\Omega}$ (valid since $f$ never vanishes). The maximum principle for $g$ gives
\begin{align*}
\max_{z \in \overline{\Omega}} \frac{1}{|f(z)|} &= \max_{z \in \partial\Omega} \frac{1}{|f(z)|}.
\end{align*}
Why does maximising $1/|f|$ give the minimum of $|f|$? Because $|f(z)| > 0$ on $\overline{\Omega}$, the function $t \mapsto 1/t$ is strictly decreasing on $(0, \infty)$, so
\begin{align*}
\max_{z \in S} \frac{1}{|f(z)|} &= \frac{1}{\min_{z \in S} |f(z)|}
\end{align*}
for any non-empty set $S \subset \overline{\Omega}$. Applying this with $S = \overline{\Omega}$ on the left and $S = \partial\Omega$ on the right:
\begin{align*}
\frac{1}{\min_{z \in \overline{\Omega}} |f(z)|} &= \frac{1}{\min_{z \in \partial\Omega} |f(z)|}.
\end{align*}
Both sides are positive and finite (the minima exist since $|f|$ is continuous on the compact set $\overline{\Omega}$ and $|f| > 0$). Taking reciprocals:
\begin{align*}
\min_{z \in \overline{\Omega}} |f(z)| &= \min_{z \in \partial\Omega} |f(z)|.
\end{align*}
[/guided]
[/step]
