Let $M_1,M_2,M_3,M_4$ be the suprema of $|f|$ on the edges of a square. Show that $|f(0)|\le \sqrt[4]{M_1M_2M_3M_4}$

Question

Let $G$ denote the interior of the square with vertices $1,i,-1,-i$. Suppose $f$ is holomorphic on $G$ extends continuously to $\overline{G}$, and $M_1,M_2,M_3,M_4$ are the suprema of $|f|$ on the edges of $G$. Show that $$|f(0)|\le \sqrt[4]{M_1M_2M_3M_4}$$ I have an idea if we assume $f$ is never zero. By Schwarz-Christoffel formula, we can map the unit cicle to the square ,fixing the origin and the vertices. Then define $h=f(g)$ and log $|h|$, we have log$|h(0)|={1\over{2\pi}}\int_0^{2\pi}$log$|h(e^{i\theta})|d\theta$ and then we can get the inequality. However, if $f$ is zero somewhere, we can't define log $h$ then I'm stuck.
Any help would be appreciated!

Answer 1

Consider the function

$$g(z) = f(z)\cdot f(iz)\cdot f(-z) \cdot f(-iz).$$

$g$ is holomorphic on $G$ and extends continuously to $\overline{G}$, and the maximum of $g$ on each of the edges is at most $M_1\cdot M_2\cdot M_3\cdot M_4$.

$g(0) = f(0)^4$. The maximum modulus principle does the rest.