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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?

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1 Answer 1

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User Ted Shifrin proved it:

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.

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    $\begingroup$ 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 $\endgroup$
    – Conrad
    Jul 29, 2020 at 21:49

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