Does there exist an holomorphic function such that $|f(z)|\geq \frac{1}{\sqrt{|z|}}$? I have some trouble solving this problem:
Does there exist a holomorphic function $f$ on $\mathbb C\setminus \{0\}$ such that
$$|f(z)|\geq \frac{1}{\sqrt{|z|}}$$
for all $z\in\mathbb C \setminus \{0\}$?
I don't know where to start. My intuition is that you would get a problem with the singularity near $0$, but I am not sure how to prove it. Any help would be appreciated! Thanks!
 A: Such an inequality guarantees that $f$ has no zeros, hence we can invert the inequality and find
$$\left\lvert\frac{1}{f(z)}\right\rvert \leqslant \sqrt{\lvert z\rvert},$$
and that says that $1/f$ has a removable singularity (with value $0$) in the origin, thus, without loss of generality, $g = 1/f$ is an entire funtion that grows at most as fast as $\sqrt{\lvert z\rvert}$. But such an estimate forces $g$ to be constant, hence $g \equiv 0$. That on the other hand means $f \equiv \infty$, so $f$ was not an analytic function.

Addendum: The Cauchy integral for the derivative of $g$ yields, for $\lvert z\rvert \leqslant R$:
$$\begin{align}
\lvert g'(z)\rvert &= \left\lvert \frac{1}{2\pi i} \int_{\lvert\zeta\rvert = 2R} \frac{g(\zeta)}{(\zeta-z)^2}\,d\zeta\right\rvert\\
&\leqslant \frac{1}{2\pi} \int_0^{2\pi} \frac{\lvert g(2Re^{it})\rvert}{\lvert2Re^{it}-z\rvert^2}2R\,dt\\
&\leqslant \frac{1}{2\pi}\int_0^{2\pi}\frac{2R\sqrt{2R}}{R^2}\,dt = \frac{2\sqrt{2}}{\sqrt{R}},
\end{align}$$
and letting $R \to \infty$ shows $g' \equiv 0$.
Similarly, when $h$ is an entire function, and you have an estimate $\lvert h(z)\rvert \leqslant c\cdot \lvert z\rvert^\alpha$ for all $\lvert z\rvert \geqslant K$, the Cauchy integral for the $n$-th derivative of $h$ yields an estimate $\lvert h^{(n)}(z)\rvert \leqslant C\cdot R^{\alpha - n}$ for all $\lvert z\rvert \leqslant R/2$, where the constant $C$ depends on $n$ but not on $R$, and thus $h^{(n)} \equiv 0$ if $n > \alpha$, i.e. if you have such an estimate, then $h$ is a polynomial of degree $\leqslant \lfloor \alpha\rfloor$.
