Suppose $f$ is an entire function, $f(\sqrt{n}) = 0$ for $n =1,2,3,\ldots$, and there is a positive constant $\alpha$ such that $|f(z)|<\exp(|z|^\alpha)$ for all large enough $|z|$. For which $\alpha$ does it follow that $f(z) = 0$ for all $z$? [Consider $\sin(\pi z^2)$.]
Since $f$ is entire, using a Weierstrass product, the primary model of $f$ is
$$\prod_{n=1}^\infty\left(1-{z\over\sqrt{n}}\right)$$
Since $\sum \left({1\over\sqrt{n}}\right)^3<\infty$ and $\sum \left({1\over\sqrt{n}}\right)^2 = \infty$, we conclude that $f$ is of the form
$$e^{g(z)}\prod_{n=1}^\infty\left(1-{z\over\sqrt{n}}\right)\exp\left({z\over\sqrt{n}}+{z^2\over 2n}\right).$$
for some entire function $g$.
Since the problem is asking to consider $\sin(\pi z^2)$, we consider the order of growth of it.
$$|\sin(\pi z^2)| = \left|{e^{i\pi z^2}-e^{-i\pi z^2}\over 2i}\right|\leq e^{i\pi z^2} = e^{\pi |z|^2}.$$
So, it has the order of growth at most $2$. If we let $z = x-ix$ for $x\in\Bbb R$,
\begin{align*}
\left|{e^{i\pi z^2}-e^{-i\pi z^2}\over 2i}\right|&\geq {1\over 2}\left(|e^{i\pi z^2}|-|e^{-i\pi z^2}|\right)\\
& = {1\over 2}\left(e^{\operatorname{Re}(i\pi z^2)} – e^{\operatorname{Re}(-i\pi z^2)}\right)\\
& = {1\over 2}\left(e^{2\pi x^2}-e^{-2\pi x^2}\right)\\
&\geq C\cdot e^{2\pi x^2}.\\
\end{align*}
for some constant $C$. Hence the order of growth of $\sin(\pi z^2)$ is $2$.
So $2$ may be a candidate of $\alpha$ but I don't know what to do next. Could you help?
Best Answer
Following @Conrad's hint, suppose $\alpha =2-\epsilon$ for some $\epsilon>0$. Suppose $f(\sqrt{n}) =0$ for $n\in\Bbb N$ and $|f(z)|\leq\exp(|z|^{2-\epsilon})$. Suppose $f$ is nonconstant entire. Replacing $f(z)/z^m$ for some $m$ if necessary, we may assume $f(0)\neq 0$. Then by Jensen's formula, we get $$\log|f(0)|+\sum_{|a_n|<R}\log{R\over |a_n|} = {1\over 2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|\ d\theta\leq R^{2-\epsilon},$$ where $a_n$ are roots of $f$. Since $\sqrt{n}$ is a root of $f$ for each $n$, \begin{align*} R^{2-\epsilon}& \geq\sum_{\sqrt{n}<R}\log{R\over\sqrt{n}}\\ & \geq\sum_{n<R^2}\log{R\over\sqrt{R}}\\ & = \sum_{n<R^2}{1\over 2}\log R\\ & = O(R^2\log R).\\ \end{align*} And this is a contradiction for large $R$ which is allowed as $f$ is entire. Hence, $\alpha =2$.