Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

thanks.

share|improve this question
add comment

1 Answer

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,.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.