Ngô Quốc Anh

August 1, 2009

Weierstrass infinite products

We now turn to the construction of an entire function with prescribed zeros

Theorem (Weierstrass). Given any sequence \{a_n\} of complex numbers with |a_n| \to \infty as n\to \infty, there exists an entire function f that vanishes at all z = a_n and nowhere else. Any other such entire function is of the form f(z)e^{g(z)}, where g is entire.

To begin the proof, note first that if f_1 and f_2 are two entire functions that vanish at all z = a_n and nowhere else, then \frac{f_1}{f_2} has removable singularities at all the points an. Hence \frac{f_1}{f_2} is entire and vanishes nowhere, so that there exists an entire function g with \frac{f_1(z)}{f_2(z)} = e^{g(z)}. Therefore f_1(z) = f_2(z)e^{g(z)} 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 \{a_n\} and nowhere else. A naive guess, suggested by the product formula for \sin \pi z, is the product

\displaystyle\prod\limits_n {\left( {1 - \frac{z} {{{a_n}}}} \right)} .

The problem is that this product converges only for suitable sequences \{a_n\}, so we correct this by inserting exponential factors. These factors will make the product converge without adding new zeros.

For each integer k \geq 0 we define canonical factors by

\displaystyle E_0(z) = 1-z

and

\displaystyle E_k(z) = (1-z)e^{z+\frac{z^2}{2}+ \cdots +\frac{z^k}{k}},

for k \geq 1. The integer k is called the degree of the canonical factor.

Lemma
. If |z| \leq \frac{1}{2}, then |1-E_k(z)| \leq c|z|^{k+1} for some c >0.

Suppose that we are given a zero of order m at the origin, and that a_1, a_2,... are all non-zero. Then we define the Weierstrass product by

\displaystyle f\left( z \right) = {z^m}\prod\limits_{n = 1}^\infty {{E_n}\left( {\frac{z} {{{a_n}}}} \right)} .

We claim that this function has the required properties; that is, f is entire with a zero of order m at the origin, zeros at each point of the sequence \{a_n\}, and f vanishes nowhere else.

Fix R>0, and suppose that z belongs to the disc |z|<R. We shall prove that f has all the desired properties in this disc, and since R is arbitrary, this will prove the theorem.

We can consider two types of factors in the formula defining f, with the choice depending on whether |a_n| \leq 2R or |a_n| > 2R. There are only finitely many terms of the first kind (since |a_n| \to \infty), and we see that the finite product vanishes at all z = a_n with |a_n| < R. If |a_n| \geq 2R, we have \left|\frac{z}{a_n}\right| \leq \frac{1}{2}, hence the previous lemma implies

\displaystyle\left| {1 - {E_n}\left( {\frac{z}{{{a_n}}}} \right)} \right| \leqslant c{\left| {\frac{z}{{{a_n}}}} \right|^{n + 1}} \leqslant \frac{c}{{{2^{n + 1}}}}.

Note that by the above remark, c does not depend on n. Therefore, the product

\displaystyle\prod\limits_{\left| {{a_n}} \right| \geqslant 2R} {{E_n}\left( {\frac{z} {{{a_n}}}} \right)}

defines a holomorphic function when |z| < R, and does not vanish in that disc. This shows that the function f has the desired properties, and the proof of Weierstrass’s theorem is complete.

Question (Harvard’s QE). Given a sequence of complex numbers z_1, z_2,... such that |z_n| \to \infty, does there exist an entire function f with f(z_j) = a_j?

Proof. By the Weierstrass theorem, we can construct an entire function g(z) with simple zeros z_1, z_2,... Then we define

\displaystyle f\left( z \right) = \sum\limits_{n = 1}^\infty {{u_n}\left( z \right)} = \sum\limits_{n = 1}^\infty {{e^{{\gamma _n}(z - {z_n})}}\frac{{g(z)}}{{z - {z_n}}}\frac{{{a_n}}}{{g'({z_n})}}}

where \gamma_n is chosen such that when |z| < \frac{|z_n|}{2},

\displaystyle\left| {{u_n}\left( z \right)} \right| = \left| {{e^{{\gamma _n}(z - {z_n})}}\frac{{g(z)}}{{z - {z_n}}}\frac{{{a_n}}}{{g'({z_n})}}} \right| \leqslant \frac{1}{{{n^2}}}.

Because |z_n| \to \infty, for any R>0, there exists N>0 such that |z_n| > 2R when n \geq N. Hence

\displaystyle\left| {{u_n}\left( z \right)} \right| \leqslant \frac{1} {{{n^2}}}

holds for all |z| \leq R when n \geq N. In other words,

\displaystyle\sum\limits_{n = 1}^\infty {{u_n}\left( z \right)}

converges uniformly for all |z| \leq R, so that f(z) in analytic in \{ |z| < R\}. Since R can be arbitrarily large, f(z) is an entire function. It is easy to see that

\displaystyle\mathop {\lim }\limits_{x \to {z_n}} {u_n}\left( z \right) = \mathop {\lim }\limits_{x \to {z_n}} {e^{{\gamma _n}(z - {z_n})}}\frac{{g(z)}}{{z - {z_n}}}\frac{{{a_n}}}{{g'({z_n})}} = {a_n}

while for k \ne n,

\displaystyle u_k(z_n)=0,

which implies that f(z) is an entire function satisfying the required condition.

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