I don't understand how the two theorems in the title conciliate. Namely, suppose that $f$ is entire with growth order $\rho$, and $f$ vanish in the points $a_1, a_2 \dots$ , and eventually in $0$; then Hadamard factorization theorem states that we can write $f$ as $$e^{P(z)}z^m \prod_{n=1}^\infty E_k\Bigl(\frac z {a_n}\Bigl)\ ,$$ where $k$ is the largest integer lower than $\rho$, and $P(z)$ is a polynomial of degree $\le k$. Obviously, $m$ is the order of the zero in $0$, and $E_k$ is the canonical factor.
Weierstrass theorem states that, given the sequence $a_0=0,a_1, a_2 \dots$ (we can consider the same $a_n$ as above since, excluding that $f$ is identically zero, it's true that $|a_n|\to \infty$) there is an entire function $g$ vanishing only in the $a_n$. This function $g$ is obtained as $$(*) \ \ \ \ \ \ \ \ \ e^{h(z)}z^m \prod_{n=1}^\infty E_n\Bigl(\frac z {a_n}\Bigl)\ ,$$ with $h(z)$ an holomorphic function.
My problem is that I don't understand how we can construct our function $f$, using Weierstrass theorem, from the sequence of the $a_n$ (namely, to have $f=g$); it should be possible, since every entire function vanishing in the $a_n$ is of the form $(*)$, right? I intuitively understand that, if we want $g$ to have growth order equal to $\rho$ (necessary to obtain $f=g$), it is reasonable to take $h(z)=P(z)$, with $\deg P \lt k$; I say "intuitively" and "reasonable" because we didn't study the proof of Hadamard theorem, only the statement. However I don't understand how the two products $\prod_n E_k(z/a_n)$ and $\prod_n E_n(z/a_n)$ can be equal, since in one product the degrees are fixed and in the other they increase with $n$. Can you clarify my ideas? Thanks a lot
I make an edit, since I gave a look also to the product formula for the sine function, namely $$\frac {\sin \pi z} \pi = z\prod_{n=1}^\infty \Bigl(1-\frac{z^2}{n^2}\Bigl)\ .$$ It is clear that this infinite product converges and that the zeros of this product are exactly the zeros of $\sin \pi z$; however I still don't understand how we could obtain it using Weierstrass theorem with the sequence $\{0,1,-1,2,-2,\dots \}$. Maybe we could take the $h(z)$ equal to the inverse of the product of the exponential part of the canonical factors?
Best Answer
Given a sequence $\{a_n\}$ of complex numbers with $a_n\neq 0$ for all $n\geq1$ and $|a_n|\to\infty$ as $n\to\infty$, you probably know that the function $\tilde f(z)=\prod_{n=1}^\infty E_n\Bigl(\frac z {a_n}\Bigl)$ is an entire function vanishing only at the $a_n$. But there's nothing special about the product $\prod_{n=1}^\infty E_n\Bigl(\frac z {a_n}\Bigl)$. Suppose $\{a_n\}$ is a sequence of complex numbers with $a_n\neq 0$ for all $n\geq1$ and $|a_n|\to\infty$ as $n\to\infty.$ If $\{p_n\}$ is any sequence of integers such that $\sum_{n=1}^{\infty}\left(\frac{1}{|a_n|}\right)^{p_n+1}< \infty$, then $\prod_{n=1}^\infty E_{p_n}\Bigl(\frac z {a_n}\Bigl)$ is an entire function with zeros only at $a_n$(for a proof, see Conway). Since $|a_n|\to\infty$, there is an $N$ such that $|a_n|> 2$ for all $n\geq N$. Thus we see that $\sum_{n= 1}^{\infty}\left(\frac{1}{|a_n|}\right)^{n}$ converges and therefore $p_n=n-1$ is a choice. Notice that the same holds for $p_n\geq n-1$.
Now, if we want the function $\tilde f$ to have a zero (of order $m\geq 0$) at $z=0$ as well, $f(z)=z^m\prod_{n=1}^\infty E_{p_n}\Bigl(\frac z {a_n}\Bigl)$ will do the job. If $g$ is another entire function which vanishes only at the above prescribed points, it will be of the form $f(z)e^{h(z)}$, where $h$ is some entire function(consider the nowhere vanishing entire function $\frac g{f}$). This is summarized as follows:
Now if $f$ is an entire function with growth order $\rho$, then one can show that $\sum_{n=1}^{\infty}\left(\frac{1}{|a_n|}\right)^{k+1}< \infty$, where $k$ is the largest integer less than or equal to $\rho$(again, see Conway or Stein-Shakarchi for a proof). Thus, we have $f(z)= e^{h(z)}z^m \prod_{n=1}^\infty E_{k}\Bigl(\frac z {a_n}\Bigl)$ for some entire function $h$. Again, it takes some effort to establish that the entire function $h$ is indeed a polynomial.
I hope this helps.