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I have some trouble solving this problem: Does there exist an 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 appreciate! Thanks!

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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 $\leqslant \lfloor \alpha\rfloor$.

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  • $\begingroup$ Can you explain better why such an estimate forces g to be constant $\endgroup$ – user62138 Aug 9 '13 at 16:02
  • $\begingroup$ I have just started learning complex analysis. From what you showed, I see that there's few entire functions in some sense. Your addendum just gave me great enlightenment to what an entire function is. $\endgroup$ – user165633 Aug 5 '14 at 20:19
  • $\begingroup$ @user144542 That's right. Not in terms of cardinality or dimension of the vector space of entire functions; in both these senses, there are as many entire functions as continuous functions. But holomorphic functions are quite constrained, and entire functions even more so. You probably have already seen Liouville's theorem that a bounded entire function is constant, but Picard's (little) theorem is still a huge leap from that: a non-constant entire function attains every complex number with at most one exception as a value. If it's not a polynomial, even infinitely often. $\endgroup$ – Daniel Fischer Aug 5 '14 at 20:30
  • $\begingroup$ Like the function $f(x)=x \sin(x)$ on $\mathbb{R}$ ? Are these powerful properties of entire/holomorphic functions exist bacause $\mathbb{C}$ is equipped with a product inlike $\mathbb{R}^2$ or because the imagenary number $i$ is related to the number $\pi$ ? or both ? That Picard's theorem tells me that it has to do with the fact that $i$ is related to $\pi$, but I don't now. $\endgroup$ – user165633 Aug 5 '14 at 20:39
  • $\begingroup$ Good question, @user144542. If a function is complex differentiable, and the derivative is not zero, then the derivative, viewed as a map $\mathbb{R}^2\to\mathbb{R}^2$ is a rotation composed with a rescaling (well, that's what multiplication with a nonzero complex number is). And that means that locally, a holomorphic function cannot distort the domain (except in the zeros of the derivative), which prevents a lot of behaviour. Holomorphic functions are - except where the derivative vanishes - conformal mappings, they preserve angles. $\endgroup$ – Daniel Fischer Aug 5 '14 at 21:08

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