Fall 2013 - Problem 3

Solution.

Suppose $f$ is such a function. Certainly $f$ is non-constant, since it is unbounded near the boundary. For each $z_0 \in \partial\D$, we $\delta_{z_0} > 0$ such that if $\abs{z - z_0} < \delta_{z_0}$, then $\abs{f\p{z}} \geq 1$, since $f$ is unbounded on the boundary. By compactness, there exist $z_1, \ldots, z_n \in \partial\D$ such that $\set{B\p{z_k, \delta_{z_k}} \mid 1 \leq k \leq n}$ covers $\partial\D$. Thus, if we let $\delta = \max_{1 \leq k \leq n} \delta_{z_k}$, then

Hence, the zeroes of $f$ are contained in the compact set $\cl{B\p{0, 1 - \delta}}$. Since $f$ is non-constant, $f$ can only vanish at finitely many points, or else they accumulate $\cl{B\p{0, 1 - \delta}}$, which implies that $f$ is identically zero. Let $\set{a_1, \ldots, a_m}$ be the zeroes of $f$ with corresponding orders $\set{k_1, \ldots, k_m}$. Thus,

where $g$ vanishes nowhere on $\D$. Thus,

and sending $\abs{z} \to 1$ and noting that polynomials are bounded on $\cl{\D}$, we see that

But $g$ vanishes nowhere, so $\frac{1}{g}$ is holomorphic in $\D$. By the maximum principle, this implies that $\frac{1}{g}$ is identically zero, which implies that $f$ is identically $\infty$, a contradiction. Thus, no $f$ could have existed to begin with.