Lower bounding the derivative of a simple zero of an analytic function

Question

Let $f : \mathbb C \to \mathbb C$ be an entire function with a separated zero set, i.e. there is a $\delta>0$ s.t. $|z-z'| > \delta$ for every distinct zeros of $f$. Further, suppose that all zeros of $f$ are simple and $$ |f(z)| \geq C_\varepsilon e^{a|z|^2}, \quad z \in Z_\varepsilon $$ where $Z_\varepsilon$ denotes all complex numbers with distant $\geq \varepsilon$ to the zero set of $f$. Hence, on such a set, $f$ grows (up to constants) precisely like $e^{a|z|^2}$.

I was wondering if this implies a lower bound on $|f'(z)|$ for zeros $z$ of $f$. Intuitively, $|f'(z)|$ should grow rather fast if the zero $z$ is far away from the origin. Are there results in this direction or is there a way to show a growth estimate?

Answer 1

Sure enough, the idea is that if $f(z_0) = 0$ then $g(z) = \frac{f(z)-f(z_0)}{z-z_0}$ is an entire function which is non-vanishing and for which we know the lower bound on the circle $|z-z_0| =\frac{\delta}{2} = r$. Moreover, $g(z_0) = f'(z_0)$. But then on this disk the function $u(z) = \log g(z)$ is harmonic, thus we have the mean-value property:

$$\log |f'(z_0)| = \frac{1}{2\pi r}\int_{|z-z_0| = r} \log |g(z)||dz| \ge \frac{1}{2\pi r} \int_{|z-z_0|=r} a|z|^2 + \log (C_r) - \log |z-z_0||dz|.$$

Last two terms will give us a constant independent of $z_0$, so we are left with the integral

$$\frac{1}{2\pi r}\int_{|z-z_0|=r} |z|^2|dz| = \frac{1}{2\pi r}\int_{|w| = r} |w|^2 + |z_0|^2 + 2Re(w\bar{z_0})|dz| = \frac{r}{2\pi} + |z_0|^2,$$ where the integral $\int_{|w| = r}2Re(w\bar{z_0})|dz|$ is zero by, say, symmetry.

Collecting everything we have $\log |f'(z_0)| \ge c + a|z_0|^2$ for some absolute constant $c$, thus $|f'(z_0)| \ge e^c e^{a|z_0|^2}$. This should be optimal up to possibly a power of $z_0$.

Although I used some shortcuts based on the chosen function $|z|^2$, this method works in a very big generality, say for all functions of finite order, you just have to play with the radius of the circle you consider.