Weierstrass infinite products

August 1, 2009

Weierstrass infinite products

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:56

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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31

	Han on The implicit function theorem:…
	Alan de Gois on Variation of the Riemann tenso…
	Jet Foncannon on A generalization of the Morera…
	ibrahim tuet (@yiuyu… on In a normed space, finite line…
	Tran Minh Nguyet on Uniformly boundedness implies…
	Ngô Quốc Anh on Uniformly boundedness implies…
	Tran Minh Nguyet on Uniformly boundedness implies…
	An on Tính tích phân suy rộng loại…
	An on Tính tích phân suy rộng loại…
	BỔ ĐỀ SCHWARZ VÀ ĐỊN… on Schwarz’s Lemma, Schwarz…

Ngô Quốc Anh

August 1, 2009