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Let $f : \Bbb C^* \to \Bbb C$ be a holomorphic function such that $\lvert f(z) \rvert \leq \sqrt{\lvert z \rvert} + \frac{1}{\sqrt{\lvert z \rvert}}$. Show that $f$ is constant.

I cannot use Liouville theorem because the function is not entire and we do not know what happens near $0$.

I noticed that $\lvert f(\frac1z)\rvert \leq \sqrt{\lvert z \rvert} + \frac{1}{\sqrt{\lvert z \rvert}}$. I also noticed that for $z$ such that $\lvert z \rvert \leq 1$, we have that $\lvert f(z) \rvert \leq 2$. I tried to look at what happens for $1/f$ or $e^f$ but no interesting result. How can I do it ?

I wrote $\lvert f(z) \rvert \leq \sqrt{\lvert z \rvert} + \frac{1}{\sqrt{\lvert z \rvert}}=\frac{\lvert z \rvert +1}{\sqrt{\lvert z \rvert} }$. For $\lvert z \rvert \geq 1,$ $\lvert f(z) \rvert \leq \frac{2 \lvert z \rvert}{\sqrt{\lvert z \rvert}}=2\sqrt{\lvert z \rvert}$ but it does not help a lot.

Kilkik
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1 Answers1

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Note that $|zf(z)| \to 0$ as $z\to 0$, so $f$ has a removable singularity at $z = 0$. Let $F$ be a holomorphic extension of $f$ to an entire function. Use Cauchy’s estimates to obtain $|F^{(n)}(0)| \le n!\, (r^{1/2-n}+r^{-n-1/2})$ for all $r > 0$ and $n\ge 1$. Let $r\to \infty$ to deduce $F$ is constant. Then $f$ is constant.

kobe
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