Existence of analytic function on punctured disk

Question

For background, this post asks if there exists a non-constant analytic function on the open unit disc in $\mathbb{C}$ such that $$n^{-1/2} < \left| f(1/n) \right| < 2n^{-1/2}, \quad \forall n \in \mathbb{N},$$ in which case there is no such analytic function.

My question is whether or not there exists a (non-constant) analytic function on the punctured unit disk that satisfies the above inequality. In my attempts, since $f$'s domain no longer contains the accumulation point, I've tried splitting the proof by cases based on the type of singularity that $f$ possesses at the origin; if it is either removable or a pole, I can show that there is no such analytic function, but I am stuck on the essential singularity case.

Any help would be appreciated.

Let $g(z)=f(1/z)$ and note that $g(z)$ must satisfy $\frac{1}{\sqrt{n}}<g(n)<\frac{2}{\sqrt{n}}$. Now, use the result of this posting to find an analytic function with $g(n)=\frac{1.5}{\sqrt{n}}$ for all $n\in\mathbb{N}_1$. — Sangchul Lee, Apr 05 '25 at 09:27
Theorem 15.13 in Rudin's RCA shows that you can arbitrarily assign values at the points $\frac 1n$. — Kavi Rama Murthy, Apr 05 '25 at 09:29

score 1 · Accepted Answer · answered Apr 05 '25 at 09:38

1

If $(z_n)$ is a sequence without limit points in a region $\Omega$ and $c_n$ is any sequence of complex numbers then there is an analytic fucnion $f$ on $\Omega$ such that $f(z_n)=c_n$ for all $n$. This follows from a theorem in Rudin's RCA. [Theorem 15.13 in my copy of the book, which is not an international edition!].

answered Apr 05 '25 at 09:38

Kavi Rama Murthy

359,332

It is Theorem 15.13 in my old English edition as well (reprint from 1986). – Martin R Apr 05 '25 at 14:49

Existence of analytic function on punctured disk

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