Let $U = B(0,1)\subset \mathbb{C}$ be the unit ball, and $f\in A(U)= H(U)\cap C^0(\overline{U})$ non constant where $H(\Omega)$ are the holomorphic functions over $\Omega$. We know that if $z_0 \in U$ is a zero of $f$ than it must be isolated since $f$ is holomorphic in $U$ and non constant. But apparently nothing is against $f$ having a non isolated zero on the boundary $\partial U$(where $f$ can be not-holomorphic).

It is possible that $f(e^{i \theta})= 0$ for any $\theta\in [0,\alpha] \subset [0,2\pi]$ for some $\alpha$?

We know that if $\alpha = 2\pi, f$ would be constant thanks to maximum modulus principle so $0 <\alpha < 2\pi$ is the interesting case, but what else can we say?

P.S. I know that it has been asked this: Zeros of a holomorphic function on the boundary of a closed region but my question is more specific, I'm asking for a set of zeros not only infinite but with 1-dimensional Lebesgue measure positive ($m_1([0,\alpha))>0$ ), and it has not been answered there.

Thank you.

  • 1
    $\begingroup$ The accepted answer in the linked question ends "It is worth noting that by the F. And M. Riesz theorem, the set of zeros on the boundary has zero one-dimensional Lebesgue measure." $\endgroup$ – Hagen von Eitzen Jun 8 '17 at 9:59

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