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.

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Ngô Quốc Anh

August 1, 2009