Let $T\subset \mathbb{C}$ be compact and $T^\circ$ be connected subsets of $\mathbb{C}$ where $T^\circ$ denotes interior of $T.$ Let $f: T \to\mathbb{C}$ be continuous and non constant. Also let $f\big|_{T^\circ}$, i.e, its restriction on interior of $T$ be analytic. Prove that $|f|$ attains its maximum over $T$ on the boundary $\partial T$
Let $f = u + iv $ be continuous on $T$
Then, $|f| = \sqrt{u^2 + v^2}$ is a continuous function $|f|:T \to \mathbb{R}$
Since, $T$ is compact, $|f| $ will attain maximum value, say M, somewhere on $T$.
Hence, $\exists z_0 \in T $ such that $|f(z_0)| = M$ and $|f(z)| \leq M \;\forall z\in T $
All that is left to prove is that $z_0 \notin T^\circ$.
But this follows directly from Maximum Modulus Principle which states that
If a function $f$ is analytic and not constant in a given domain $D$, then $| f (z)|$ has no maximum value in $D$. That is, there is no point $z_0$ in the domain such that $| f (z)| \leq | f (z_0)|$ for all points $z$ in it.