Spring 2016 - Problem 10

Solution.

Suppose $f$ is such a function. For each $e^{i\theta} \in \partial\D$, there exists $\delta_\theta > 0$ so that on $B\p{e^{i\theta}, \delta_\theta}$, we have $\abs{f\p{z}} \geq 1$. By compactness, there exist finitely many $B\p{e^{i\theta_k}, \delta_k}$ whose union covers $\partial\D$. If $r = d\p{\partial\D, \p{\bigcup_{k=1}^n B\p{e^{i\theta_k}, \delta_k}}^\comp}$, which is positive since it is the distance between two disjoint closed sets. Then if $\abs{z} > 1 - r$, we have $\abs{f\p{z}} > 1$, i.e., if $f$ has any zeroes, they must lie in the compact set $\cl{B\p{0, r}}$. Since $f$ is not identically zero, this means that $f$ has only finitely many zeroes $a_1, \ldots, a_n$ counting multiplicity. Let

so that $g$ has no zeroes in $\D$, but we still have $\lim_{\abs{z}=1} \abs{g\p{z}} = \infty$ since linear functions are bounded. This means that $\frac{1}{g}$ is a holomorphic function on $\D$ such that $\abs{g\p{z}} = 0$ on $\partial\D$. By the maximum principle, this implies that $g$ is identically zero, but this implies that $f$ is identically infinity, which is impossible. Hence, no such $f$ can exist.