[proofplan] Since $\Omega$ is bounded, $\overline{\Omega}$ is compact, and the continuous function $|f|$ attains its maximum on $\overline{\Omega}$ at some point $z^*$. If $z^* \in \partial\Omega$, the estimate follows immediately. If $z^* \in \Omega$, the Strong Maximum Modulus Principle forces $f$ to be constant on the connected component of $\Omega$ containing $z^*$, and by continuity the maximum of $|f|$ on this component equals the maximum on its boundary, which is bounded by the boundary maximum on $\partial\Omega$. In either case, every interior value is bounded by $\max_{\partial\Omega} |f|$. [/proofplan] [step:Establish that $|f|$ attains its maximum on $\overline{\Omega}$] Since $\Omega$ is bounded, $\overline{\Omega}$ is a closed and bounded subset of $\mathbb{C}$, hence compact by the Heine-Borel theorem. The function $|f|: \overline{\Omega} \to \mathbb{R}$ is continuous (as the composition of $f \in C(\overline{\Omega})$ with the continuous modulus $|\cdot|: \mathbb{C} \to \mathbb{R}$). By the Extreme Value Theorem, $|f|$ attains its maximum on $\overline{\Omega}$: there exists $z^* \in \overline{\Omega}$ such that \begin{align*} |f(z^*)| &= \max_{z \in \overline{\Omega}} |f(z)|. \end{align*} [guided] The boundedness of $\Omega$ is consumed here. We need the maximum of $|f|$ on $\overline{\Omega}$ to exist and be attained — not merely a supremum. Since $\Omega$ is bounded, $\overline{\Omega} \subset \mathbb{C}$ is closed and bounded, hence compact by the Heine-Borel theorem (which applies in $\mathbb{C} \cong \mathbb{R}^2$). The function $|f|: \overline{\Omega} \to \mathbb{R}$ is continuous: $f$ is continuous on $\overline{\Omega}$ by hypothesis ($f \in C(\overline{\Omega})$), and the modulus $|\cdot|: \mathbb{C} \to \mathbb{R}$ is continuous, so the composition $|f| = |\cdot| \circ f$ is continuous. By the Extreme Value Theorem (a continuous real-valued function on a compact set attains its maximum), there exists $z^* \in \overline{\Omega}$ with \begin{align*} |f(z^*)| &= \max_{z \in \overline{\Omega}} |f(z)|. \end{align*} In particular, for any $z_0 \in \Omega$ we have $|f(z_0)| \le |f(z^*)|$, so the task reduces to showing $|f(z^*)| \le \max_{\partial\Omega} |f|$. [/guided] [/step] [step:If the maximum is attained on $\partial\Omega$, the estimate follows immediately] If $z^* \in \partial\Omega$, then for every $z_0 \in \Omega$: \begin{align*} |f(z_0)| &\le \max_{z \in \overline{\Omega}} |f(z)| = |f(z^*)| \le \max_{z \in \partial\Omega} |f(z)|, \end{align*} and the desired estimate holds. [/step] [step:If the maximum is attained in the interior, apply the Strong Maximum Modulus Principle to show $f$ is constant] Suppose $z^* \in \Omega$. Let $\Omega_0$ denote the connected component of $\Omega$ containing $z^*$. Since $\Omega$ is open, $\Omega_0$ is an open connected subset of $\mathbb{C}$, i.e., a domain. The restriction $f|_{\Omega_0} \in \mathcal{O}(\Omega_0)$, and $z^*$ is an interior maximum of $|f|$ on $\Omega_0$: \begin{align*} |f(z^*)| &\ge |f(z)| \quad \text{for all } z \in \Omega_0, \end{align*} since $|f(z^*)| = \max_{\overline{\Omega}} |f| \ge |f(z)|$ for all $z \in \Omega \supseteq \Omega_0$. By the [Strong Maximum Modulus Principle](/theorems/3345), which requires $\Omega_0$ to be a domain and $f|_{\Omega_0} \in \mathcal{O}(\Omega_0)$ (both satisfied), $f$ is constant on $\Omega_0$: there exists $c \in \mathbb{C}$ with $f(z) = c$ for all $z \in \Omega_0$. [guided] Now suppose $z^* \notin \partial\Omega$, so $z^* \in \Omega$. We need a more delicate argument. Let $\Omega_0$ be the connected component of $\Omega$ containing $z^*$. Since $\Omega$ is open, each connected component of $\Omega$ is open (connected components of an open set in $\mathbb{C}$ are open), so $\Omega_0$ is a domain. The restriction $f|_{\Omega_0}$ is holomorphic on $\Omega_0$. Moreover, $z^*$ is a global maximum of $|f|$ on all of $\overline{\Omega}$, so in particular $|f(z^*)| \ge |f(z)|$ for every $z \in \Omega_0$. This means $|f|$ attains an interior maximum on the domain $\Omega_0$. We apply the [Strong Maximum Modulus Principle](/theorems/3345): if $\Omega_0 \subset \mathbb{C}$ is a domain, $f \in \mathcal{O}(\Omega_0)$, and there exists a point in $\Omega_0$ where $|f|$ attains a maximum over $\Omega_0$, then $f$ is constant on $\Omega_0$. The hypotheses are satisfied ($\Omega_0$ is a domain by construction, $f|_{\Omega_0} \in \mathcal{O}(\Omega_0)$, and $z^* \in \Omega_0$ is a maximum point). Therefore $f$ is constant on $\Omega_0$: there exists $c \in \mathbb{C}$ with $f(z) = c$ for all $z \in \Omega_0$. [/guided] [/step] [step:Extend to the boundary of $\Omega_0$ by continuity and conclude the estimate] Since $f$ is constant on $\Omega_0$ with value $c$, and $f \in C(\overline{\Omega})$, continuity gives $f(z) = c$ for all $z \in \overline{\Omega_0}$. In particular, $|f(z)| = |c|$ for every $z \in \partial\Omega_0$. Since $\Omega_0 \subset \Omega$, we have $\partial\Omega_0 \subset \overline{\Omega_0} \subset \overline{\Omega}$. Moreover, $\partial\Omega_0 \cap \Omega = \varnothing$ (since $\Omega_0$ is a connected component of $\Omega$, hence clopen in $\Omega$, so no boundary point of $\Omega_0$ lies in $\Omega$). Therefore $\partial\Omega_0 \subset \overline{\Omega} \setminus \Omega = \partial\Omega$. It follows that $|c| = |f(z)|$ for some $z \in \partial\Omega_0 \subset \partial\Omega$, so \begin{align*} |f(z_0)| &\le \max_{z \in \overline{\Omega}} |f(z)| = |f(z^*)| = |c| \le \max_{z \in \partial\Omega} |f(z)|, \end{align*} for every $z_0 \in \Omega$. This establishes the main estimate. For points $z_0 \in \partial\Omega$ the inequality $|f(z_0)| \le \max_{z \in \partial\Omega} |f(z)|$ is immediate. The "in particular" statement follows by setting $M = \max_{\partial\Omega} |f|$: if $|f(z)| \le M$ on $\partial\Omega$, then $|f(z_0)| \le \max_{\partial\Omega} |f| \le M$ for all $z_0 \in \overline{\Omega}$. [guided] We know $f \equiv c$ on $\Omega_0$. By continuity of $f$ on $\overline{\Omega}$: for any $z \in \overline{\Omega_0}$, there exists a sequence $(z_n) \subset \Omega_0$ with $z_n \to z$, and $f(z) = \lim f(z_n) = c$. So $f \equiv c$ on $\overline{\Omega_0}$, and in particular $|f| = |c|$ on $\partial\Omega_0$. Where does $\partial\Omega_0$ sit relative to $\partial\Omega$? The set $\Omega_0$ is a connected component of the open set $\Omega$. Connected components of an open subset of $\mathbb{C}$ are clopen in the subspace topology of $\Omega$. So $\Omega_0$ is both open and closed in $\Omega$. A point $w \in \partial\Omega_0$ satisfies: every neighbourhood of $w$ meets both $\Omega_0$ and $\Omega_0^c$. Since $\Omega_0$ is closed in $\Omega$, any limit point of $\Omega_0$ that lies in $\Omega$ already belongs to $\Omega_0$, so $\partial\Omega_0 \cap \Omega = \varnothing$. Since $\partial\Omega_0 \subset \overline{\Omega_0} \subset \overline{\Omega}$, we get $\partial\Omega_0 \subset \overline{\Omega} \setminus \Omega = \partial\Omega$. Since $\Omega_0$ is non-empty and open, $\partial\Omega_0 \neq \varnothing$ (a non-empty proper clopen subset of $\mathbb{C}$ cannot exist, so $\Omega_0 \neq \mathbb{C}$, and any non-empty open proper subset of $\mathbb{C}$ has non-empty boundary). Pick any $w \in \partial\Omega_0 \subset \partial\Omega$. Then $|f(w)| = |c|$, so \begin{align*} |c| = |f(w)| &\le \max_{z \in \partial\Omega} |f(z)|. \end{align*} Combining with $|f(z_0)| \le |f(z^*)| = |c|$ for any $z_0 \in \Omega$: \begin{align*} |f(z_0)| &\le |c| \le \max_{z \in \partial\Omega} |f(z)|. \end{align*} This completes the proof of the main estimate. For the "in particular" clause: if $|f(z)| \le M$ for all $z \in \partial\Omega$, then $\max_{\partial\Omega} |f| \le M$, so $|f(z_0)| \le M$ for all $z_0 \in \Omega$. On $\partial\Omega$ we already have $|f| \le M$, so $|f(z)| \le M$ on all of $\overline{\Omega}$. [/guided] [/step]