Prove that there is no function $f$ that is analytic.

Question

Prove that there is no function $f$ that is analytic in $\mathbb{C}\setminus\{0\}$ and satisfies $$|f(z)|\geq\frac{1}{\sqrt{|z|}},\quad \operatorname{for all}\quad z\in\mathbb{C}\setminus\{0\}$$

I am studying for a final exam. Not quite sure how to tackle this one. I was thinking maybe use Maximum Modulus Theorem somehow? Or Extended Liouville? Should I define a new function $g=\sqrt{z}f(z)$?

Answer 1

Proof:

1. Note that the function $f$ has no zeros because of the given inequality.

2. Define a holomorphic function $g:\mathbb{C}\backslash\{0\}\to \mathbb{C}$ as $$ g(z)~:=~\frac{1}{zf(z)^2}.\tag{1}$$

3. Note that $|g|\leq 1$ is bounded because of the given inequality.

4. The function $g$ has a removable singularity at $z=0$, cf. Riemann's theorem.

5. So the extension $\tilde{g}:\mathbb{C}\to \mathbb{C}$ is a bounded entire function.

6. Liouville's theorem then shows that $g=c$ is constant.

7. The constant $c\neq 0$ cannot be zero, cf. the definition (1) of $g$.

8. In other words, $$f(z)~=~\frac{1}{\sqrt{cz}},\tag{2}$$ which has a branch cut. Contradiction.