# Clarify about Weierstrass and Hadamard factorization theorems

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?

• I’m not sure I fully understand your question, but notice that Weierstrass factorisation is not unique. Apr 27 '20 at 15:46
• @ThomasShelby my question, in a few words, is: I take an entire function $f$, of finite order, that vanishes in the points $\{a_n\}_n$; now, if I use Weierstrass theorem on $\{a_n\}_n$, how is possible to obtain $f$? I mean, it must be possible because every entire function vanishing in $\{a_n\}_n$ can be obtained applying Weierstrass to $\{a_n\}_n$ and eventually multiplying for some $e^{h(z)}$; however I explained in the paragraph starting with "My problem is" why I can't understand how we can actually obtain our initial $f$ using Weierstrass on the set $\{a_n\}_n$. Thanks a lot Apr 27 '20 at 20:00

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:
Weierstrass Factorization Theorem. Let $$f$$ be an entire function and let $$\{a_n\}$$ be the non-zero zeros of $$f$$ repeated according to multiplicity. Suppose $$f$$ has a zero of order $$m\geq 0$$ at $$z=0$$. Then there exists a sequence of integers $$\{p_n\}$$ such that $$f(z)= e^{h(z)}z^m \prod_{n=1}^\infty E_{p_n}\Bigl(\frac z {a_n}\Bigl)$$ for some entire function $$h$$.
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.