# Inequalities via Number of Zeros & Jensen's Formula

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?

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 $$F \equiv 0$$.

• for the first part count how many given zeroes (square roots) are in the disc of radius $r$ - there are entire functions which have only such as zeroes so the bound you want must follow from them only Mar 18 at 19:27
• Would you propose that this is $\frac{r^2}{2}$? And if so, would it be a proof by induction that would give the desired result?
– 2307
Mar 18 at 19:41
• Just count them - enumerate the first few roots and see what happens for small $r$; as noted in general you cannot assume you have other roots unless more is given about the function Mar 18 at 19:48
• If it's not an integer than you get $[r^2]$ Mar 18 at 20:14
• Note that $\frac{r^2}{4} \leq \alpha r^{\beta} - \log|F(0)|$ is sufficient to prove that $\beta \ge 2$. Perhaps that is what you really need? Mar 19 at 9:37

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 $$\beta \ge 2 \, .$$
• Thank you so much - very clear! A couple small follow-up questions: I have made a small edit to my question - is the argument I have taken valid? It feels a bit circular, especially for the contradiction as I feel like I am assuming $\beta \geq 2$ when I am first dividing through to prove the bound holds? Also, I can see that the inequality for floor function holds, but is there anywhere I could see a proof of how this holds?
• @2307: That looks a bit too complicated. You must assume $F \not\equiv 0$ from the beginning, otherwise you can not apply Jensen's formula. So you show that if $F$ is not identically zero and has zeros at all those integer square roots and has a growth condition then $\beta \ge 2$. Mar 22 at 19:04