Let $U\subset \mathbb C$ be a bounded domain and $f:\overline{U}\to\mathbb C$ continuous and holomorphic $U$.
- Show that $|f(z)|\leq\max\{|f(w)|:w\in\partial U\}$ for all $z\in U$.
- Show that if there is one $z\in U$ with $|f(z)|<\min\{|f(w)|:w\in\partial U\}$ then $f$ has a zero inside $U$.
If we consider the ball $B_r(z)\subset U$ with $r>0$ then we get via the mean value theorem and the estimation lemma that $$|f(z)|\leq\frac{1}{2\pi}\int_0^{2\pi}|f(z+r\mathrm e^{\mathrm it})|\,\mathrm dt\overset{(*)}{\leq}\max_{t\in[0,2\pi]}|f(z+r\mathrm e^{\mathrm it})|$$ and therefore there exists a $w\in\partial B_r(z)$ with $|f(z)|\leq |f(w)|$.
I might have come very quickly to my conclusion since I couldn't find a more decent explanation for $(*)$. Is it reasonable to let $r=1$ and thus the length of the curve be $2\pi$ due to homotopy which cancels out nicely?
Since $f$ will usually attain its minumum at the boundary one value smaller than everything inside of $U$ might be zero thus making $f$ have a zero inside $U$. Unfortunately I don't know how to show this more thoroughly and formally.
$\overline{U}$ is compact because $U$ is bounded, hence $f$ attains its maximum on $\overline{U}$. If $f$ is non-constant then it is an open mapping, hence does not have a local maximum in $U$. Therefore $$|f(z)|\leq \max\{|f(w)|:w\in\partial U\}$$ for all $z\in U$, since this inequality certainly holds if $f$ is constant.
If $f(w)=0$ for some $w\in\partial U$ then the inequality $$|f(z)|<\min\{|f(w)|:w\in \partial U\}=0$$ is impossible, so we may assume that $f$ does not vanish on $\partial U$. If $f$ has no zeros in $U$, then $\frac{1}{f}$ is holomorphic in $U$ and continuous on $\overline{U}$, and it follows from part (1) applied to $\frac{1}{f}$ that $$ |f(z)|\geq \min\{|f(w)|:w\in \partial U\} $$ for all $z\in U$. Therefore if $|f(z)|<\min\{|f(w)|:w\in \partial U\}$ for some $z\in U$, then $f$ must have a zero in $U$.