# Is a meromorphic function determined by its boundary values?

Let $$f: \mathbb D \to \widehat {\mathbb{C}}$$ be a meromorphic function inside the unit disk.

Assume that $$f$$ is zero on the boundary and continuous in the closed disk (as a function into $$\widehat {\mathbb{C}}$$).

Is $$f$$ necessarily identically zero?

If $$f$$ is not surjective then we can take $$a\not \in \text{Image}(f)$$ and reduce to the holomorphic case via $$\frac 1 {f(z)-a}$$, where the maximum modulus finishes.

What if $$f$$ is not surjective?

By the continuity on the boundary, there must be finitely many poles, and we may multiply by suitably many factors of the form $$z-a_j$$ to cancel them out and reduce to the holomorphic case again.
• note that the result is true even if meromorphic $f$ has zero non-tangential limits on an arc or even more generally on a set of non-zero Lebesgue measure on the circle; in general, it is called Privalov theorem (the proof being a clever reduction - called the ice-cream cone construction - to the bounded holomorphic case when the theorem is known as Fatou's theorem); in the case here the proof is easy as noted Jul 29, 2020 at 21:49