I just worked on a problem and was able to solve it pretty easily, using Hadamard's product representation. But I wonder whether the solution that I compared my work to doesn't actually use the Hadamard factorization.
The question asks to determine the class of all functions $f$ such that
i) $f$ is analytic on the whole complex plane
ii) $f$ is zero if and only if z=0
iii) $|f(re^{i\theta}|\le e^{\delta r^3}$ for any $\delta >0$ and for all large $r$.
My work:
By part iii), we know that $f$ is a function of order m = 3, by definition.
By part i), $f$ is entire.
Then since $f$ is an entire function of finite order, it has a Hadamard canonical product representation:
$$f(z) = z^me^{g(z)} \prod \big(1- \frac {z}{z_n}\big)e^{\large \frac{z}{z_n}+...+\frac{1}{p}(\frac{z}{z_n})^p}$$
where we know that g(z) is a polynomial, since $f$ is of finite order.
Now by part ii) we know that $f(z)$ has no non-zero roots. Then I think that the infinite product factor simply does not appear in the Hadamard factorization, and the class of functions is easily seen to be of the form
$$f(z) = z^me^{g(z)}$$
Where g(z) is a polynomial of degree at most d=3, in order to satisfy the growth condition of part iii).
So, my question is: the solution simply says that $\large\frac {f}{z^m}$ is entire and never zero, and so it must have the form $e^{g(z)}$, where $g(z)$ is a polynomial of degree at most 3. Is this solution saying the same thing as me? Or is it saying something more basic?
...not that Hadamard's factorization is super difficult (although I remember there was essentially only one way to prove the factorization), but it is usually a topic that shows up much later in any complex variables book, and also what I have seen early in the textbooks is the rewriting of functions in the form of $e^g$, or $\frac{1}{f}$, etc to prove fundamental results -- e.g., maximum / minimum modulus principle stuff.
So, I wonder whether I have overlooked a basic result in solving this problem. If so, I would love to be reminded of it :-)
Any comments are welcome.
Thanks,