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Let $D$ be the open disk in $\mathbb{C}$ with origin $0$ and radius $1$.

Let $f,g: \overline{D} \to \mathbb{C}$ be continuous functions such that $f$ and $g$ are analytic on $D$ and such that $f=g$ on $S^1= \{z \in \mathbb{C}: |z| = 1\}$. Can I conclude that $f=g$ on $D$ as well?

It is enough to show that $\{z\in D: f(z) = g(z)\}$ has a limit point in $D$ but I can't see why this should hold.

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  • $\begingroup$ It even suffices to have the equality on a boundary arc. $\endgroup$ – Moishe Kohan Feb 22 at 17:32
  • $\begingroup$ Any boundary set of nonzero measure (lebesgue on the circle) will do, doesn't need to be arc $\endgroup$ – Conrad Feb 22 at 18:14
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Suppose that $f\neq g$. Let $M=\max_{z\in\overline D}\bigl\lvert f(z)-g(z)\bigr\rvert$, which is greater than $0$. Then there is some $z_0\in\overline D$ such that $\bigl\lvert f(z_0)-g(z_0)\bigr\rvert=M$. Then $z_0\in D$ since, if $z_0\in S^1$, then $f(z_0)-g(z_0)=0$. Now, apply the maximum modulus principle to deduce that $f=g$.

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  • $\begingroup$ Here is the version of the maximum modulus principle I know: if $G$ is an open connected set and $f: G \to \mathbb{C}$ is analytic on $G$ such that there is a point $a\in G$ such that $|f(a)| \geq |f(z)|$ for all $z \in G$. Then $f$ is constant. I guess we apply this to the analytic function $h: D \to \mathbb{C}: z \mapsto f(z)-g(z)$, right? Because we have $|h(z_0)| \geq |h(z)|$ for all $z \in D$ and thus $h$ is constant on $D$. By continuity and because $f=g$ on $S^1$, we then get that this constant must be $0$. Is this the reasoning to finish your argument? $\endgroup$ – user745578 Feb 22 at 17:31
  • $\begingroup$ Yes, that is correct. $\endgroup$ – José Carlos Santos Feb 22 at 17:56
  • $\begingroup$ Thank you for your help! $\endgroup$ – user745578 Feb 22 at 18:10

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