Are two analytic functions equal if they are equal on the boundary of an open disk?

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.

• It even suffices to have the equality on a boundary arc. – Moishe Kohan Feb 22 at 17:32
• Any boundary set of nonzero measure (lebesgue on the circle) will do, doesn't need to be arc – Conrad Feb 22 at 18:14

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 $$f=g$$.
• 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? – user745578 Feb 22 at 17:31