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If a function $f$ is analytic on the open disk $D=\{z:|z|<1\}$ and continuous on the closure of $D$, and if $f(z)=0$ for every $z$ in $\partial D$ with $\operatorname{Im} z \geq 0$, then I would like to show $f(z)=0$ for every $z$ in $D$.

It seems I should be using the symmetry but I am not sure how. If I can show that in fact $f(z)=0$ on all of $\partial D$, then I know I can conclude by maximum principle $f \equiv 0$.

Diffusion
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    Rotate $f$ to a $g$ that is zero on the other half circle and look at $fg$ – Conrad Aug 08 '24 at 22:13
  • OK, then $fg\equiv 0$ on $\overline{D}$, where $g(z)=f(-z)$. So $f(z)$ has an infinite number of zero in the disk, and there must be an accumulation point, implying that $f\equiv 0$? – Diffusion Aug 08 '24 at 22:47
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    if $fg\equiv 0$ then either $f\equiv 0$ or $g \equiv 0$ (or both - this is true on any connected open set and any two analytic functions and one can argue with accumulation points as noted) so yes that's a simple way of doing this - more generally if $f$ would be zero on a boundary arc rather than half circle, same proof applies just that you have to do enough rotations to cover the circle and multiply them; a considerably more difficult result is to show the same result if $f$ is $0$ on a set on the boundary of nonzero (real) Lebesgue measure which is not an arc – Conrad Aug 08 '24 at 23:13

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You can use Schwarz reflection principle to analytically continue $f$ into $D \cup H$, where $H$ is the upper half plane. After that, clearly, $f \equiv 0$, as $f$ has non-discrete set of zeros.

Below I outline the construction of the analytic $g:D \cup H \rightarrow \mathbb C$ such that $g|_D = f$.

It is a variation of the classical Schwarz reflection principle. In the classical case one has a domain $U \subset H$ and a holomorphic $h:U \rightarrow \mathbb C$ such that $h$ can be continuously extended to an open interval $J \subset \overline{U}\cap \mathbb R$ such that $h(J) \subset \mathbb R$. Then it is possible to continue $h$ analytically into $U \cup J \cup U^c$, where $U^c = \{z \in \mathbb C: \overline{z} \in U\}$ - the domain that is symmetric to $U$ with respect to the real line. Moreover, for $z \in U^c$ the analytic continuation is given as $h(z) = \overline{h(\overline{z})}$.

This construction is known to work with more general symmetries than conjugation. In particular, symmetries with respect to circles. Consider following symmetry with respect to the unit circle $T = \partial D$: $z \rightarrow \hat{z} = 1/\overline{z}$. I claim that if a holomorphic function $f:D \rightarrow \mathbb C$ extends to a segment $J$ of $T$ and maps it into the real line, then $f$ can be analytically continued through $J$ outside $D$. The formula for the analytic continuation is $g(z) = \overline{f(\hat{z})}$. Moreover, this works if $f$ maps $J$ into some other circle. In this case conjugation should be replaced with the corresponding symmetry.

Matsmir
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A simpler argument would be to consider function $g(z) = f(z)*f(e^{\pi i}z)$. Then $g$ is zero on boundary of the whole circle so it must be $0$ and hence $f$ is $0$ on the disk.

Salcio
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