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I want to solve this question:

Is there a function $f(z)$ analytic on $\mathbb C\setminus \{0\}$ that satisfies $|f(z)|\geq |z|^{-1/2}$ for every $z \neq 0$?

I can solve it with a long way (there isn't exist), but I have a big feeling that Picard big theorem can help me to solve it in a few raws. Can someone show me how to do it? (if this is true)

I haven't used this theorem ever so I will be happy (if it's not hard) to see a good explanation.


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up vote 2 down vote accepted

Indeed no such function exists.

a) If $f$ can be extended holomorphically through $0$, the inequality will fail for small $|z|$ since $|z|^{-1/2}$ tends to $\infty$ when $z$ tends to $0$, whereas $f(z)$ tends to $f(0)$.

b) If $f$ can be extended meromorphically through $0$ with a pole of order $k\gt 0$ the function $g(z):=z^kf(z)$ will be entire and satisfy $|g(z) |\geq |z|^{k-1/2} $ so that $h(z)=1/g(z)$ satisfies $|h(z)|\leq |z|^{−k+1/2 }$.
This implies that $h(z)$ can be extended holomorphically at $\infty $ by $h(\infty )=0$ and thus that $g(z)=z^kf(z)=1/h(z)$ has a pole ay $\infty$.
Hence $f(z)$ is meromorphic on the whole extended plane and is thus a rational function , necessarily of the form $f(z)= \frac {1}{z^k} P(z)$ with $P$ a polynomial satisfying $P(0)\neq 0 $.
But then:
$\bullet$ if $P$ is not constant our inequality is false at any zero of $P$.
$\bullet$ if $P$ is constant our inequality is false for large $|z|$.

c) If $f(z)$ has an essential singularity at $0$, then in the pointed disk $D^* $ defined by $0\lt |z|\lt 1$ the inequality implies $|f(z)| \gt 1$ .
As you very correctly conjectured, this contradicts the big Picard theorem according to which $f$ takes all complex values (except maybe one) in that pointed disk,.

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Georges, I was not able to rule out the case where $0$ was a pole, but $\infty$ was essential with lacunary value $0.$ Maybe you have that covered. – Will Jagy Dec 11 '11 at 0:48
I get it, when $0$ is a pole of order $k,$ once again Picard and $|z^k f(z)| \geq |z|^{k - 1/2}$ says $z^k f(z)$ cannot be essential at $\infty.$ So $f(z)$ is not essential at $\infty$ either. Little Picard would have sufficed. – Will Jagy Dec 11 '11 at 2:21
Dear @Will, yes what you say is quite correct. Another proof (without even little Picard) is to consider $h(z)=1/[z^k.f(z)]$. Since $|h(z)|\leq |z|^{-k+1/2}$, we see that $h$ tends to $0$ at infinity. Hence $h$ can holomorphically be extended (by zero) at infinity, so that $h(z)$ is meromorphic everywhere on the extended plane. Hence, so is $1/h(z)=z^k.f(z)$. I'll write an edit. Thanks for your interest. – Georges Elencwajg Dec 11 '11 at 7:35

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