Spring 2017 - Problem 8

Solution.

Let $\func{\phi_a}{\D}{\D}$ be the Blaschke factor which maps $a \in \D$ to $0$. Notice that

which gives the identity $\phi_a\p{z}\conj{\phi_a\p{1/\conj{z}}} = 1$. Thus, $B\p{z}\conj{B\p{1/\conj{z}}} = 1$ by multiplying all the terms together. Taking the derivative on both sides, we obtain

which tells us that if $B\p{z} \neq 0$, then $B'\p{z} = 0 \iff B'\p{1/\conj{z}} = 0$. Another calculation gives us

First, assume that $B\p{0} \neq B'\p{0} \neq 0$ and that $B$ has distinct zeroes. Multiplying the above by $\prod_{j=1}^n \p{z - a_j}\p{1 - \conj{a_j}z}$, we get a polynomial equation $P\p{z} = 0$ of degree $2n - 2$. Since the zeroes of $B$ are precisely $a_1, \ldots, a_n$, $P$ does not vanish at any of these points, so $P\p{z} = 0 \iff B'\p{z} = 0$.

Given a zero $z_j$ of $B'$ in $\D$, however, we know that $B'\p{1/\conj{z_j}}$ is a zero of $B'$ outside of $\D$. Since $B'$ has $2n - 2$ zeroes, it follows that half of them are in the disk and half are outside the disk, i.e., $B'$ has $n - 1$ zeroes in $\D$.

In the general case, by perturbing $a$ in $\phi_a$, we may approximate $B$ uniformly on compact sets by a Blaschke product $B_\epsilon$ such that $B_\epsilon\p{0} \neq B_\epsilon'\p{0} \neq 0$ and the roots of $B_\epsilon$ are all distinct. By the Cauchy integral formula,

So, it follows that $B_\epsilon'$ converges uniformly to $B'$ on compact sets and similarly, $B_\epsilon''$ converges uniformly to $B''$ on compact sets. By uniform convergence,

Since $B$ is holomorphic on $\D$, it follows that $B'$ has no poles inside of $\D$. Thus, by the argument principle, $B'$ has precisely $n - 1$ zeroes in $\D$, which completes the proof.