As title goes, one knows that for a holomorphism $f(z)$, zeros of $f(z)$ cannot have limit point unless $f(z)$ vanishes everywhere. But I am confused at the behaviors of holomorphism at boundary. My question is
Let $\mathbb{D}=\{z\in \mathbb{C}: |z|<1\}$ to be the unit disc. If $f: \mathbb{D}\to \mathbb{C}$ is continuous in $\overline{\mathbb{D}}$ and holomorphic inside $\mathbb{D}$. Can the zeros of $f$ in $\partial \mathbb{Z}$, $Z_0=\{z\in \partial\mathbb{D}: f(z)=0\}$, have limit point for nonzero $f$?
Here are some remark on it.
- Firstly, if $Z_0$ is the complete unit circle, then by maximal module principle, $f$ equals to zero inside $\mathbb{D}$.
- And, if $Z_0$ consists a piece of arc of unit circle, then $f$ vanishes everywhere. Since one can take $F=\prod_n f(\mathrm{e}^{i\theta_n}z)$ for some suitable $\theta_n$ such that $F$ vanishes all over the unit circle.
- Note that given a boundary value of $f(z)$, then by Cauchy's formula, $f(z)$ is given by $$f(z)=\frac{1}{2\pi i}\int_{C} \frac{f(\zeta)\textrm{d}\zeta}{\zeta-z}$$ Naive thought is to give a continuous function at boundary whose zeros have limit point. But there are two problems, firstly, the integral may fails to be continuous at boundary; secondly, the function should at least satisfy $\int_C f(\zeta)\textrm{d}\zeta=0$, so not all continuous functions can derive a holomorphism from this method.
- Poisson's integral determines $f(z)$ by the real part of boundary value of $f(z)$, which may not help in this problem.
Even more, I am confused that
What is the condition of a function on unit circle being the boundary value of some holomorphism?
Of course, it is essentially a PDE problem.