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


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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Have you considered $1/f$? – Jonas Meyer Aug 15 '11 at 8:37
@Jones Meyer $1/f$ won't work, just check that for $z=4$ you have $1/4 = |f(z)| < \frac{1}{\sqrt{|z|}} = 1/2$. – Thomas Klimpel Aug 15 '11 at 9:16
@Thomas, why do you think Jonas was proposing $f(z)=1/z$, when what Jonas wrote was "Have you considered $1/f$"? In any event, have you considered $1/f$? – Gerry Myerson Aug 15 '11 at 9:59
@Thomas: Thanks for clearing that up (I said $1/f$, not $1/z$). No, there is no such function, and I would approach it by contradiction, supposing such $f$ exists, and considering $g(z)=1/f(z)$ as the next step. I wonder if movesonthemove had any thoughts on this. – Jonas Meyer Aug 15 '11 at 10:00
@robjohn: Since $1/f(z)$ vanishes at 0, it’s of the form $zg(z)$ with $g$ entire. – Pierre-Yves Gaillard Aug 15 '11 at 10:35

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. – Pierre-Yves Gaillard Aug 15 '11 at 10:55

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 $\frac{1}{f(z)}=g(z)$.

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