I am working on a problem in which we take some entire function $F$ which has zeros at every integer square root, i.e. $F(\sqrt{n}) = 0$ for every $n \in \mathbb{Z}^+$.
I need to show that, for any radius $r>0$, $\#\{z \in \overline{\mathbb{D}}_r(0) : F(z)=0 \} \geq \frac{r^2}{2}$. That is, the number of zeros of $F$ in a closed disc of radius $r$ is greater than or equal to $r^2/2$.
I am totally stumped on how to show this – any hints would be great.
Following this, I am to show that if $F \not \equiv 0$, $F(0)\neq 0$, and $|F| \leq \exp(\alpha |z|^{\beta})$ holds for $|z| \geq 1$ then $$r^2 \leq 16\alpha r^\beta.$$
For this, I have used Jensen's Formula with the above inequality for the number of zeros, and have obtained
$$\frac{r^2}{4} \leq \alpha r^{\beta} – \log|F(0)|$$
How would I deal with the $\log|F(0)|$ term? Is there a way I can justify $r$ getting larger rendering this term obsolete?
EDIT – added further question
If, by the answer below, we have $$r^2 – 1 \leq 4\alpha r ^{\beta} – 4\log|f(0)|$$
then is the following argument valid to reach the conclusion we are instructed to find:
Divide both sides by $r^\beta$ to obtain:
$$r^{2-\beta} – \frac{1}{r^\beta} \leq 4A – \frac{4\log|f(0)|}{r^\beta}$$
then as $r \to \infty$ we have $r^{2-\beta} \leq 4A$ and so $r^2 \leq 4Ar^\beta \leq 16Ar^\beta$ as required.
Then, for proving that $\beta<2 \implies F\equiv 0$, we go via contradiction. Let $\beta < 2$ and then assuming $F \not \equiv 0$ and WLOG $F(0)\neq 0$, the above result tells us for large $r$ we have $$r^2 \leq 16Ar^\beta$$
$$\implies r^{2-\beta} \leq 16A.$$
Since $\beta < 2$, we have that $r^{2-\beta}$ will tend to $\infty$ as $r \to \infty$. This contradicts the fact that $r^{2-\beta} \leq 16A$ and so we must have $F \equiv 0$.
Best Answer
Let $n(r)$ denote the number of zeros of $F$ in the closed disk $D_r(0)$.
If $r > 0$ and $k = \lfloor r^2 \rfloor $ is the largest integer with $k^2 \le r$ then $f$ has zeros at $\sqrt 1, \sqrt 2, \cdots, \sqrt k \in D_r(0)$, so that $$ n(r) \ge k = \lfloor r^2 \rfloor \, . $$ If $r \ge 1$ then $\lfloor r^2 \rfloor \ge r^2/2$ and therefore $n(r) \ge r^2/2$.
Now we use Jensen's formula in the form $$ \frac{1}{2\pi} \int_0^{2\pi} \log |F(re^{i\theta})| \, d\theta - \log |F(0)| = \int_0^r \frac{n(t)}{t} \, dt $$ If $r \ge 1$ then the right-hand side is $$ \ge \int_1^r \frac{n(t)}{t} \; dt \ge \int_1^r \frac{t^2/2}{t} \, dt = r^2 - 1 \, . $$ And if $|F(z)| \leq \exp(\alpha |z|^{\beta})$ then the left-hand side is $$ \le \alpha r^\beta - \log |F(0)| \, . $$ It follows that $$ r^2 - 1 \le \alpha r^\beta - \log |F(0)| $$ or $$ 1 + \frac{\log |F(0)|-1}{r^2} \le \alpha r^{\beta -2} $$ for $r \ge 1$. If $\beta < 2$ then the right-hand side converges to zero for $r \to \infty$ whereas the left-hand side converges to one. That is not possible, so we necessarily have $$ \beta \ge 2 \, . $$