# Uniqueness of holomorphic functions up to the boundary

Suppose that $$f : \{ z \in \mathbb C : \mathrm{Re}(z) > 0 \} \to \mathbb C$$ is a holomorphic function in the right half plane with the property that $$\lim_{z \to 0} f(z) \in \mathbb C$$ i.e. the limit exists. Is it true that if $$(z_n)$$ is a convergent sequence with limit in $$\{ z \in \mathbb C : \mathrm{Re}(z) > 0 \} \cup \{ 0 \}$$ such that $$f(z_n)=0$$ for all $$n$$, then $$f$$ vanishes identically? Clearly this holds if the limit lies only in $$\{ z \in \mathbb C : \mathrm{Re}(z) > 0 \}$$ in view of the uniqueness theorem.

• Your use of the term "limit point" is quite confusing. I assume you mean a convergent sequence, with limit in $\{z: \textrm{Re}\: z>0\} \cup \{ 0 \}$. I've taken the liberty to edit accordingly, but of course feel free to edit further if that wasn't your intention. Jun 26 at 16:52

This is false. We can take any function $$g\not\equiv 0$$ with $$\lim_{z\to 0} g(z)=0$$ such as $$g(z)=z$$ and then $$f(z)=B(z)g(z)$$, where $$B$$ is a Blaschke product with zeros $$z_n$$ (the linked page discusses Blaschke products on the unit circle, but of course we can move things over to the half plane by a Cayley transform).
The $$z_n=x_n+iy_n$$ must satisfy $$\sum \frac{x_n}{x_n^2+y_n^2+1}<\infty$$, but are otherwise arbitrary, so we can definitely fix a sequence $$z_n\to 0$$.
Then $$f(z_n)=0$$ and also still $$\lim_{z\to 0}f(z)=0$$ since $$|B|\le 1$$, but $$f$$ is not identically zero.