Show that if $f$ is holomorphic in the unit disk, is bounded, and converges uniformly to zero in the sector $\theta < \arg z < \phi $ as $|z| \rightarrow 1$, then $f=0.$

I'm having trouble proving this fact, I don't even know where to begin.

Any hints are appreciated.

  • $\begingroup$ I didn't think on the problem too much, but I'd start with $f(z)+f(1/z)$. $\endgroup$
    – ajotatxe
    Feb 17, 2018 at 14:07
  • $\begingroup$ How does that help? $\endgroup$
    – Burrrrb
    Feb 17, 2018 at 15:24
  • $\begingroup$ Here's an interesting generalization: Suppose $f\in H(\mathbb D)$ and $A\subset \partial \mathbb D$ is an arc of positive length. Suppose that for each $w\in A$ that $$\lim_{r\to 1^-} f(rw) = 0.$$ Then $f\equiv 0.$ Note that we don't need any boundedness condition on $f$ or any uniform convergence hypothesis. $\endgroup$
    – zhw.
    Feb 17, 2018 at 19:36

1 Answer 1


Choose $n$ so $n\phi/2>2\pi$. Define $$g(z)=\prod_{j=0}^nf(e^{ij\phi/2}z).$$

Since $f$ is bounded it follows that $g(re^{it})\to0$ uniformly as $r\to1$. Hence $g=0$.

If $Z$ is the zero set of $f$ this shows that the disk is the union of finitely many rotations of $Z$. So $Z$ is uncountable, hence $f=0$.


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