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By $\mathbb D$ denote the open unit disc in $\mathbb C$. Suppose that $f : \overline{\mathbb D}\to\mathbb C$ is analytic on $\mathbb D$ and continuous on $\overline{\mathbb D}$. Assume now that there are infinitely many distinct points $z_n\in\mathbb D$ which accumulate to the boundary of $\mathbb D$ such that $f(z_n) = 0$ for all $n\in\mathbb N$. Does it then follow that $f\equiv 0$?

The point is, we cannot make use of the usual identity theorem because the accumulation point of the zeros is not in $\mathbb D$. So, is there any "improvement" of that theorem covering the above case?

Friedrich Philipp
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  • the starting point is $g(z) = \sin\left( \frac{1}{1+z}\right)$ or $g(z) = \exp\left( \frac{1}{1+z}\right)$, hence you ask if the continuity (at the accumulation point, on the $|z| \le 1$ part) is enough to forbid the essential singularity at the accumulation point ? – reuns Mar 25 '16 at 01:48
  • if you ask $f$ and also each of its derivatives are continuous on $\overline{\mathbb{D}}$, hence uniformely continuous on $\overline{\mathbb{D}}$, aren't they also continuous on the neighborhood outside ? – reuns Mar 25 '16 at 01:52
  • I doubt that $f$ can be continuously extended to the exterior. Note that the Schwarz reflection principle can only be applied when $f$ maps the unit circle to itself (i.e., $|f(z)|=1$ for $|z|=1$). – Friedrich Philipp Mar 25 '16 at 02:00

1 Answers1

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Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be the series expansion of $f$ at zero, and suppose that this series has radius of convergence $R$. Then it's a general fact about holomorphic functions that $f$ must have a singularity on the circle $|z|=R$ (for a proof, see Big Rudin Chapter 16).

Since by assumption $f$ is continuous on $|z|=1$, this means that $R>1$, so we can still apply the identity theorem to conclude that $f\equiv 0$.

Edit: It appears I mis-remembered Rudin's definition of singularity here. He defines a singular point to be a boundary point which does not admit an analytic continuation. So just continuity on the boundary is not enough for my argument above.

For a construction relevant to your question, see this answer: Zeros of a holomorphic function on the boundary of a closed region

Community
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carmichael561
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  • Thanks a lot for this enlightening answer. :-) This is very interesting. – Friedrich Philipp Mar 25 '16 at 02:05
  • I'm not convinced. with the correct branch of $arg(z)$, $(z-1)^{1/2}$ is continuous at $z=1$ on the $Re(z) \le 1$ part, it doesn't mean it doesn't have a singularity there – reuns Mar 25 '16 at 02:05
  • what you say works if you ask $f$ being continuous in the complex sense on $\overline{\mathbb{D}}$, right ? not in the real sense ($f$ is continuous on $|z|\le 1$ in the real sense means we don't consider any value of $f$ outside $|z|\le 1$, while $f(z)$ is continuous at $z= a$ in the complex sense means $\lim_{z\to a} f(z) = f(a)$ from any direction/path) – reuns Mar 25 '16 at 02:08
  • @user1952009 How would you construct an $f$ as in the question from your example? – Friedrich Philipp Mar 25 '16 at 02:12
  • from an entire function $g(z)$ converging to $0$ at $\infty$ on every ray of the (closed) upper half plane ? then $f(z) = g(1/(1+z))$ or something like that, with $g(-1) \overset{def}{=} 0$ – reuns Mar 25 '16 at 02:15
  • user1952009 has a point. I've edited my post. – carmichael561 Mar 25 '16 at 02:31
  • I also just saw this in Rudin. So, question still open... – Friedrich Philipp Mar 25 '16 at 02:49
  • I think the post I linked answers it. – carmichael561 Mar 25 '16 at 02:50