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.

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.

share|improve this question
4  
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
3  
@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
1  
@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
3  
@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

2 Answers 2

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.

share|improve this answer
1  
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 $\frac{1}{f(z)}=g(z)$.

share|improve this answer

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.