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Suppose $f$ is continuous on the closed unit disc $\overline{\mathbb{D}}$ and is holomorphic on the open unit disc $\mathbb{D}$. Must $f$ have finitely many zeros in $\mathbb{D}$? I know that this is true if $f$ is holomorphic in $\overline{\mathbb{D}}$ (by compactness of the closed unit disc), but I'm not sure of what happens when I just consider $\mathbb{D}$.

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  • $\begingroup$ Consider the function 0. $\endgroup$ – mez Aug 20 '14 at 1:20
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If $f$ is continuous on $\overline{\mathbb D}$ then it is bounded. In that case there is a strong consequence of Jensen's formula. If $|f| \leq M$, $f(0) \neq 0$ and $S$ is the (possibly infinite) multiset of roots of $f$ on $\mathbb D$ then

$$\prod_{z \in S} |z| \geq \frac{|f(0)|}{M} > 0.$$

This implies that $$\sum_{z \in S}(1-|z|) < \infty.$$

So, loosely speaking, the roots of $f$ must quickly converge to the unit circle. The condition $f(0) \neq 0$ can be circumvented. There is a unique integer $m \geq 0$ such that $$g(z) = \frac{f(z)}{z^m}$$ is holomorphic and $g(0) \neq 0$. If $S$ is the multiset of non-zero roots of $f$ then the inequality above becomes

$$\prod_{z \in S} |z| \geq \frac{|f^{(m)}(0)|}{m! \, M}.$$

An explicit example of a function $f$ that is holomorphic on $\mathbb D$, continuous on $\overline{\mathbb D}$ and has infinitely many roots is

$$ f(z) = (1-z^2) \sin(\operatorname{atanh}(z)).$$

To understand this example, note that $\operatorname{atanh}$ maps $\mathbb D$ biholomorphically onto the strip $\{ z \mid -\frac{\pi}{4} < \operatorname{Im}(z) < \frac{\pi}{4} \}$ where $\pm 1$ maps to $\pm \infty$. Now note that $\sin$ is bounded and has infinitely many roots on this strip. The factor $1-z^2$ makes sure that it is continuous at $\pm 1$ (and has roots there as it should).

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  • $\begingroup$ How about the branch cuts of artanh? $\endgroup$ – user64494 Nov 10 '13 at 9:50
  • $\begingroup$ @user64494 I assume $\operatorname{atanh}(0) = 0$. It is holomorphic (single valued) on $\mathbb D$. $\endgroup$ – WimC Nov 10 '13 at 9:57
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An example of a function on the open unit disk with infinitely many zeros would be the Fourier expansion of a modular form. As mentioned in WimC's answer, the zeros accumulate along the unit disk - in the picture below of the normalized Eisenstein series of weight 4 (a basic example of a modular form), the blue areas surround the zeros, and unfortunately they get hard to see as you get toward the unit circle, but they are there. Follow their arc and you can make them out for a while. This function is given by $$E_4(z)=1 + 240\sum\limits_{n=1}^\infty \sigma_{3}(n)e^{2\pi i z},$$ where $\sigma_3(n)=\sum\limits_{d\vert n} d^3$ is the sum of divisors function.

Modulus of Eisenstein series of weight 4

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Any accumulation point of the zeros must be on the boundary, of course. You can show in this case that the series centered at 0 has radius of convergence 1. For a specific example, I think the following works:

For $|z| \leq 1$, $$f(z) = (z-1)\prod_{n=1}^\infty\frac{z-1+1/n^2}{1-(1-1/n^2)z}.$$

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  • $\begingroup$ Could you explain why $f$ is continuous in $ \overline{\mathbb{D}}$? $\endgroup$ – user64494 Nov 10 '13 at 7:06

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