Fall 2012 - Problem 9

Solution.

Replacing $f$ with $f - f\p{0}$ if necessary, we may assume without loss of generality that $f\p{0} = 0$. Set $g\p{z} = f\p{z}\conj{f\p{\conj{z}}}$. Notice that if we write $f = u + iv$, then $\conj{f\p{\conj{z}}} = u\p{x, -y} - iv\p{x, -y}$. Thus,

Since $f$ is holomorphic, $u$ and $v$ satisfying the Cauchy-Riemann equations, and so

so $\conj{f\p{\conj{z}}}$ satisfies the Cauchy-Riemann equations, hence holomorphic, and so $g$ is holomorphic.

For $x \in \p{-1, 1}$, we have $\conj{x} = x$ and so $g\p{x} = \abs{f\p{x}}^2 \in \R$. Hence, we may view $\func{g}{\p{-1, 1}}{\R}$. For $n \in \N$, we have $g\p{x_{2n}} = g\p{x_{2n+1}}$ so by the mean value theorem, there exists $y_n \in \p{x_{2n+1}, x_{2n}}$ with $g'\p{y_n} = 0$. Since $x_n \to 0$, we have $y_n \to 0$, too. In other words, the zeroes of $g'$ have an accumulation point at the origin, so $g'$ is identically $0$. Thus, $g$ must be constant on $\D$, since $\D$ is connected. Since $f\p{0} = 0$, it follows that $g\p{0} = 0$, so $g$ is identically $0$.

If $f\p{z_0} \neq 0$ for some $z_0 \in \D$, then $f\p{z_0} \neq 0$ on a neighborhood $U$ of $z_0$. Thus, since $f\p{z}\conj{f\p{\conj{z}}} = 0$ on $U$, it follows that $\conj{f\p{\conj{z}}} = 0$ on $U$. But this implies that $f\p{z}$ is $0$ on the open set $\set{\overline{z} \mid z \in U}$, so $f$ must be identically zero on all of $\D$, a contradiction. Thus, $f\p{z} \equiv 0$ to begin with, which completes the proof.