Question: If we have a limit point for the set of zeros which is on the boundary of our domain $D$, does the (complex analysis) identity theorem still hold?
Motivation: The identity theorem states that if two holomorphic functions agree at a limit point in $D$, then they are equivalent. I'm fairly sure that if the limit point is on the boundary, then one could use something like the Riemann Removable Singularity Theorem to "get rid of" all of the points converging to our limit point on the boundary. I just wanted to check.
Example: Let $D = B(1,1)$, the ball of radius 1, centered at 1. Suppose that $f$ is holomorphic on $D$ and has zeros at $\frac{1}{n}$ for each $n\in {\mathbb N}$. Is it the case that $f\equiv 0$?
Best Answer
It is not enough for the limit point to be on the boundary. Consider the analytic function $\sin\left(\frac{1}{x}\right)$ on the domain $(0, 1)$. Then we can choose a sequence $x_n = \frac{1}{\pi n} \rightarrow 0$ such that $f(x_n) = 0\ \forall n$ but the function is clearly non-zero.