Let $\mathbb{D}$ be the open unit disk and let $f,g: \mathbb{D} \rightarrow \mathbb{C}$ be holomorphic functions and be continuous on $\overline {\mathbb{D}}$.
Show when $f(z)=g(z)$ for all $z\in \partial \mathbb{D} $, then $f \equiv g $.
I thought that because $f(z)-g(z)=0$ on $\partial \mathbb{D} $ you can find a sequence in $\mathbb{D}$ which converges against these $z\in \partial \mathbb{D} $. So by identity theorem it follows $f \equiv g $.
for $z\in \mathbb{D}$. Am I right ?
Best Answer
The maximum modulus principle states that a holomorphic function takes its maximum on the boundary. So consider $h(z)=f(z)-g(z)$. We have, on $\partial\Bbb D$, $0\leq|h(z)|\leq0$, where the first inequality is by definition and the second inequality follows from $f(z)=g(z)$ on $\partial\Bbb D$. Thus $|h(z)|=0$ for all $z\in\Bbb D$. This means $h(z)=0$ for all $z\in\Bbb D$, giving $f(z)=g(z)$.