Mathematics Asked on January 5, 2022

Let $f,g$ be holomorphic function in $mathbb{D}$ that are continuous in $overline{mathbb{D}}$. Show that if $f=g$ on $|z|=1$, then $f=g$

It seems like identity theorem. But they have to be equal on an open connected set. $overline{mathbb{D}}$ is not open so I cannot use the identity theorem. At least directly.

Let $varphi=f-g$, then $varphi$ is holomorphic on $mathbb{D}$ and thus $$ forall zinmathbb{D},|varphi(z)|leqslantmax_{|u|=1}|varphi(u)|=0 $$ and thus $varphi=0$ on $mathbb{D}$ and $f=g$ on $overline{mathbb{D}}$.

Answered by Tuvasbien on January 5, 2022

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