Suppose that the function $f$ is defined by
$f(z) = \sum\_{j=0}^\infty a\_{2j} z^{2j}$ where $a\_{2j} \ge 0, z \in \mathbb{C}$. My questions are the following: 


1. First I want to check this point: if we have a bound$f(z) \le e^{z^2}$ for real $z$, then does this imply that $f$ is entire? I think $f$ should be entire, since for any complex $z \in \mathbb{C}$ we have $|f(z)| \le e^{|z|^2}$ so the power series converges. Is this correct?
2. What can we say about the growth of $a\_{2j}$? By Cauchy integral formula we can say $a\_{2j} \le \frac{e^{R^2}}{2\pi} (2j)! \frac{1}{R^{2j}}$ for $R > 0$, but can we find a sharper bound, possibly using the evenness of $f$ and $a\_{2j}\ge 0$?


Thank you for reading.



Bounds for the coefficients of the even entire function with positive coefficients
The answer to the question 1 is yes, and this is called Pringsheim's theorem:
For a series with non-negative coefficients the singular point which is closest to
the origin lies on the positive ray. Thus, if the series converges for all positive $z$, it converges everywhere and the function is entire. (Your estimate
$f(z)\leq e^{z^2}$ is not needed here). 


2. For a function with positive coefficients the maximum modulus is attained on the positive ray. So you have $|f(z)|\leq e^{|z|^2}$, and Cauchy's theorem gives you an estimate. Then you can optimize this estimate by finding the minimum of the RHS in $R$. This is the best you can do from your assumptions. Evenness will not give any improvement.


Reference. B. Levin, Distribution of roots of entire functions, Ch. 1, section 2.


