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Does there exist a function $f(z)$ holomorphic in $\mathbb{C}\backslash\{0\}$, such that

$\left|f(z)\right|\geq\frac{1}{\sqrt{\left|z\right|}}$

for all $z\in\mathbb{C}\backslash\{0\}?$

I'm not really sure on how to proceed or which particular theorems I should look at.

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    @robjohn: Since $1/f(z)$ vanishes at 0, it’s of the form $zg(z)$ with $g$ entire.2011-08-15

2 Answers 2

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It has gotten to the point where the main ideas for a solution have already appeared in the comments, so I figured an answer collecting some of these might as well be posted.

Suppose such $f$ exists. Define $h(z)=1/f(z)$ for $z\neq 0$, and $h(0)=0$. As Pierre-Yves Gaillard commented, $h$ has the form $h(z)=zg(z)$ for some entire $g$. Rearranging the original inequality in terms of $g$ shows that $g(z)\to 0$ as $z\to \infty$, and I strongly suspect that you have seen a theorem that will tell you what possible entire functions go to $0$ at infinity.

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    Dear Jonas, thanks for mentioning my name, but it was clearly **your** argument.2011-08-15
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I will flesh out my comment.

Let $g(z)=\frac{1}{f(z)}$. Since $|f(z)|\ge\sqrt{|z|}$, $f(z)\neq 0$ on $\mathbb{C}\backslash \{0\}$. Thus, $g(z)$ is holomorphic on $\mathbb{C}\backslash \{0\}$. Furthermore, $|g(z)|=\left|\frac{1}{f(z)}\right|\le\sqrt{|z|}$, so $\lim_{z\to 0}\;g(z)=0$. Therefore, $g(z)$ has a removable singularity at $0$, and so $g(z)$ is entire with $g(0)=0$.

By Cauchy's Integral Formula, g'(z)=\frac{1}{2\pi i}\int_\gamma \frac{g(w)\;\mathrm{d}w}{(w-z)^2} Where $\gamma$ is any curve circling $z$ once counterclockwise. Let $\gamma$ be a circle of radius $R+|z|$ centered at the origin. Then |g'(z)|\le\frac{1}{2\pi}\frac{\sqrt{R+|z|}\;2\pi(R+|z|)}{R^2} Since $R$ is arbitrary, we get that g'(z)=0 for all z. Since $g(0)=0$, we get that $g(z)=0$ for all $z\in\mathbb{C}$. Thus, there can be no $f$ so that $\frac{1}{f(z)}=g(z)$.