We now turn to the construction of an entire function with prescribed zeros
Theorem (Weierstrass). Given any sequence of complex numbers with as , there exists an entire function that vanishes at all and nowhere else. Any other such entire function is of the form , where is entire.
To begin the proof, note first that if and are two entire functions that vanish at all and nowhere else, then has removable singularities at all the points an. Hence is entire and vanishes nowhere, so that there exists an entire function with . Therefore and the last statement of the theorem is verified.
Hence we are left with the task of constructing a function that vanishes at all the points of the sequence and nowhere else. A naive guess, suggested by the product formula for , is the product
The problem is that this product converges only for suitable sequences , so we correct this by inserting exponential factors. These factors will make the product converge without adding new zeros.
For each integer we define canonical factors by
for . The integer is called the degree of the canonical factor.
Lemma. If , then for some .
Suppose that we are given a zero of order m at the origin, and that are all non-zero. Then we define the Weierstrass product by
We claim that this function has the required properties; that is, is entire with a zero of order at the origin, zeros at each point of the sequence , and vanishes nowhere else.
Fix , and suppose that belongs to the disc . We shall prove that has all the desired properties in this disc, and since is arbitrary, this will prove the theorem.
We can consider two types of factors in the formula defining , with the choice depending on whether or . There are only finitely many terms of the first kind (since ), and we see that the finite product vanishes at all with . If , we have , hence the previous lemma implies
Note that by the above remark, does not depend on . Therefore, the product
defines a holomorphic function when , and does not vanish in that disc. This shows that the function has the desired properties, and the proof of Weierstrass’s theorem is complete.
Question (Harvard’s QE). Given a sequence of complex numbers such that , does there exist an entire function with ?
Proof. By the Weierstrass theorem, we can construct an entire function with simple zeros Then we define
where is chosen such that when ,
Because , for any , there exists such that when . Hence
holds for all when . In other words,
converges uniformly for all , so that in analytic in . Since can be arbitrarily large, is an entire function. It is easy to see that
while for ,
which implies that is an entire function satisfying the required condition.