## Category: Complex Analysis

### Partial fractions and infinite product

1. Partial fractions

Theorem 1 (Mittag-Leffler) Let ${(a_n)_{n=1}^\infty}$ be a sequence of complex numbers with ${\lim_{n\rightarrow\infty}a_n=\infty}$ and let ${p_n(\cdot)}$ be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with poles at the points ${a_n}$ and the corresponding singular parts ${p_n\left(\frac{1}{z-a_n}\right)}$. Moreover, the most general meromorphic function of this kind can be written in the form

$\displaystyle f(z)=\sum_{n=1}^\infty\left[p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right]+g(z)$

where ${q_n(\cdot)}$‘s are suitably chosen polynomials and ${g(\cdot)}$ is analytic in the whole plane.

Proof: The idea is to subtract a certain portion of analytic part of ${p_n\left(\frac{1}{z-a_n}\right)}$ to ensure convergence without adding any poles. For each ${n\in{\mathbb N}}$, the singular part ${p_n\left(\frac{1}{z-a_n}\right)}$ is analytic in the ball ${B(0,\frac{|a_n|}{2})}$. So we can obtain Taylor expansion ${q_n(\cdot)}$ of ${p_n\left(\frac{1}{z-a_n}\right)}$ around the origin up to the degree ${d_n}$. Using estimate for the remainder term, we have

$\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq \sum_{d_n+1}^\infty\frac{M_n}{(|a_n|/2)^k}|z|^k=M_n\left(\frac{2|z|}{|a_n|}\right)^{d_n+1}\frac{1}{1-\frac{|z|}{|a_n|/2}}=M_n\frac{\beta_n^{d_n+1}}{1-\beta_n}$

where ${\beta_n=\frac{2|z|}{|a_n|}}$ and ${M_n}$ denotes the maximum of ${|p_n(\cdot)|}$ in ${|z|\leq\frac{|a_n|}{2}}$. In particular, this estimate implies that

$\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq 2M_n\beta_n^{d_n+1}\text{ whenever }|z|\leq\frac{|a_n|}{4}$

For each ${n}$, we may pick ${d_n}$ large enough to ensure ${2M_n\beta_n^{d_n+1}\leq 2^{-n}}$, which can be done by making ${\frac{M_n}{2^{d_n}}\leq 2^{-n}}$. In doing so, we obtain

$\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq 2^{-n}\text{ whenever }|z|\leq\frac{|a_n|}{4}$

Given ${R> 0}$, we shall show that the series converges uniformly in ${|z|\leq R}$. We may split the series ${\sum_{n=1}^\infty\left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|}$ into two categories: the first one with ${|a_n|\leq 4R }$ and the second one with ${|a_n|>4R}$. Then ${\sum_{|a_n|\leq 4R}\left[p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right]}$ represents a meromorphic function with poles at ${a_n}$ because it is a finite sum. In the latter category, since ${|a_n|>4R}$, we have ${|z|\leq\frac{|a_n|}{4}}$ in the disk.Thus above estimate implies

$\displaystyle \sum_{|a_n|> 4R}\left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq \sum 2^{-n}$

This completes the proof. $\Box$ Let us try to investigate several standard examples. Consider the function

$\displaystyle f(z)=\frac{\pi^2}{\sin^2\pi z}$

Recall that ${\sin\pi z=\pi z-\frac{(\pi z)^3}{3!}+\cdots}$ and so ${\sin^2\pi z =(\pi z)^2-\frac{2(\pi z)^4}{3!}+\cdots}$. We then can work out a first few terms in the Laurant expansion at the origin which is

$\displaystyle f(z)=\frac 1 {z^2}-\frac{2\pi^2}{3!}+\text{higher order terms}$

Thus the singular part of ${f}$ at 0 is ${\frac 1z^2}$. Moreover, as ${f}$ has period 1 namely ${f(z+n)=f(z)}$ for all ${n{\mathbb N}}$, we can see that the singular part at integers are ${\frac{1}{(z-n)^2}}$. Clearly ${\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}}$ converges on all compact subsets which do not contain integers. It follows that we may write

$\displaystyle f(z)=\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}+g(z)$

where ${g(z)}$ is entire. Write ${z=x+iy}$ we see that ${\frac{1}{(z-n)^2}=\frac{1}{(x-n)^2-y^2+2iy(x-n)}}$. From the uniform convergence of the series, we can see that ${\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}}$ converges to 0 as ${|y|\rightarrow 0}$. Using the idendity

$\displaystyle |\sin \pi z|^2=\cosh^2\pi y-\cos^2\pi x$

we see that ${f(z)\rightarrow 0}$ uniformly as ${|y|\rightarrow 0}$. In particular, this implies ${g(z)}$ is bounded in the strip ${0\leq x\leq 1}$. By periodicity, ${g(z)}$ is bounded in the entire complex plane. Then Liouville’s theorem says the ${g(z)}$ reduces to a constant. In this case, ${g(z)}$ must vanish as the limit is zero. Therefore, we have the identity

$\displaystyle \frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2}$

Integrate on both sides, we obtain that

$\displaystyle -\pi\cot\pi z=\sum\frac{-1}{z-n}$

In order to make the right-hand side converge, we need to substract extra terms from the Taylor series of ${\frac{1}{z-n}}$. In this case, subtracting all constant term is enough. That is ${\sum_{n=-N}^N\frac{1}{z-n}+\frac{1}{n}=\sum_{n=-N}^N\frac{z}{n(z-n)}}$ which is comparable to ${\frac 1 {n^2}}$. Thus we have

$\displaystyle \pi\cot\pi z=\frac{1}{z}+\sum_{n\neq 0}\left(\frac{1}{z-n}+\frac{1}{n}\right)+c$

We may bracket terms of ${n}$ and ${-n}$ together, and obtain

$\displaystyle \pi\cot\pi z=\frac{1}{z}+\sum_{n\neq 0}\frac{2z}{z^2-n^2}+c$

Now ${c=0}$ is necessary because both sides are odd.

Next we exploit this fact to investigate the sum

$\displaystyle \lim_{m\rightarrow\infty}\sum_{-m}^m\frac{(-1)^n}{z-n}=\frac 1z+\sum_{n=1}^\infty(-1)^n\frac{2z}{z^2-n^2}$

We may split into odd terms and even terms

$\displaystyle \sum_{-2k-1}^{2k+1}\frac{(-1)^n}{z-n}=\sum_{n=-k}^k\frac{1}{z-2n}-\sum_{n=-k-1}^{k}\frac{1}{z-1-2n}$

Clearly we know that ${\frac{\pi }{2}\cot\frac{\pi z}{2}=\lim_{m\rightarrow\infty}\sum_{m}^{-m}\frac{1}{z-2n}}$ and ${\frac{\pi }{2}\cot\frac{\pi (z-1)}{2}=\lim_{m\rightarrow\infty}\sum_{m}^{-m}\frac{1}{z-1-2n}}$. Thus in the limit we have

$\displaystyle \frac{\pi }{2}\cot\frac{\pi z}{2}-\frac{\pi }{2}\cot\frac{\pi (z-1)}{2}=\frac{\pi}{\sin\pi z}$

2. Product representation

An infinite product of complex numbers

$\displaystyle p_1p_2\cdots p_n=\prod_{n=1}^\infty p_n$

is evaluated by taking limits of the partial products ${P_n=p_1\cdots p_n}$. WLOG, we say that ${P_n}$ converges if and only if at most a finite number of the factors are zero and if the partial products formed by the nonvanishing factors tend to a finite nonzero limit. By writing ${p_n=\frac{P_n}{P_{n-1}}}$, we see that a necessary condition is that ${p_n\rightarrow 1}$. Thus WLOG, we may write

$\displaystyle \prod_{n=1}^\infty(1+a_n)$

so that ${a_n\rightarrow 0}$ is a necessary condition for convergence. If no factor is zero, it is natural to compare with

$\displaystyle \sum_{n=1}^\infty\log(1+a_n)$

We then have the following result:

Lemma 2 The infinite product ${\prod_{n=1}^\infty(1+a_n)}$ converges simultaneously with the series ${\sum_{n=1}^\infty\log(1+a_n)}$ whose terms represent the values of the principal branch of the logarithm.

Proof: Suppose ${\sum_{n=1}^\infty\log(1+a_n)}$ converges. Denote the partial sum ${S_n}$. Then clearly ${P_n=e^{S_n}}$ and by continuity of ${\exp}$ we have ${P_n\rightarrow e^S}$. $\Box$

Lemma 3 The product ${\prod_{n=1}^\infty (1+a_n)}$ converges absolutely if and only if ${\sum_{n=1}^\infty |a_n|}$ converges.

Proof: Recall that ${\lim_{z\rightarrow 0}\frac{\lim(1+z)}{z}=1}$. Then if either the product of the series converges, we see that ${a_n\rightarrow 0}$. Then we employ the following inequality

$\displaystyle (1-\epsilon)|a_n|\leq |\log(1+a_n)|\leq (1+\epsilon)|a_n|$

to derive the result. $\Box$ Next we prove Weierstrass products formula.

Theorem 4 There exists an entire function with prescribed zeros ${a_n}$ provided in the infinitely many zeros case that ${a_n\rightarrow\infty}$. Every entire function with these zeros and no other zeros can be written in the form

$\displaystyle f(z)=z^me^{g(z)}\prod_{n=1}^\infty \left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}}$

where the product is taken over all ${a_n\neq 0}$, ${g(z)}$ is an entire function.

Proof: First we shall show that the product converges. This is direct because

$\displaystyle \begin{array}{rcl} \log\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}}& = \log(1-z/a_n)+\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}\\ & =-\sum_{k=m_n+1}^\infty\frac{(z/a_n)^{k}}{k} \end{array}$

By taking ${m_n=n}$, we see that for ${|z|\leq R}$, the series ${\sum_{k=n+1}^\infty\frac{(z/a_n)^{k}}{k}\leq \sum_{k=n+1}^\infty\left(\frac{R}{|a_n|}\right)^{k}<\infty}$ as we may take ${|a_n|}$ sufficiently large enough to make ${\frac{R}{|a_n|}<1}$. $\Box$

Corollary 5 Every function which is meromorphic in the whole plane is the quotient of two entire function.

Proof: Suppose ${f(z)}$ is meromorphic in the whole plane. Let ${g(z)}$ be an entire function with poles of ${f(z)}$ for zeros. Then ${h(z)=f(z)g(z)}$ is an entire function. Then ${f(z)=\frac{h}{g}}$ as desired. $\Box$ The product representation becomes more interesting if it is possible to pick a uniform ${m_n=h}$. The previous proof shows that the product

$\displaystyle \prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{z^2}{2a_n^2}+\cdots+\frac{z^h}{ha_n^h}}$

converges to an entire function if the series ${\sum_{n=1}^\infty\frac{1}{h+1}\left(\frac{R}{a_n}\right)^{h+1}}$ converges for all ${R\in{\mathbb R}^+}$. That is to say ${\sum_{n=1}^\infty\frac{1}{|a_n|^{h+1}}<\infty}$. Assume that ${h}$ is the smallest integer for which this series converges; the previous product expression is called the canonical product associated with the sequence ${(a_n)_{n=1}^\infty}$ and ${h}$ is the genus of the canonical product. If the exponent ${g(z)}$ in the product representation of an entire function ${f(z)}$ with

$\displaystyle z^me^{g(z)}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{z^2}{2a_n^2}+\cdots+\frac{z^h}{ha_n^h}}$

reduces to a polynomial, then ${f}$ is of finite genus. The genus of ${f}$ is equal to ${\max(h,\deg(g))}$. For instance, an entire function of genus zero is of the form

$\displaystyle Cz^m\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)$

with ${\sum_{n=1}^\infty 1/|a_n|<\infty}$. The canonical representation of an entire function of genus 1 is either of the form

$\displaystyle Cz^me^{\alpha z}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)e^{\frac{z}{a_n}}$

with ${\sum_{n=1}^\infty\frac{1}{|a_n|^2}<\infty, \sum_{n=1}^\infty\frac{1}{|a_n|}=\infty}$; or of the form

$\displaystyle Cz^m e^{\alpha z}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)$

with ${\sum_{n=1}^\infty 1/|a_n|<\infty}$.

Next we discuss an example: ${\sin\pi z}$. It has zeros at ${z=\pm n}$. Since ${\sum\frac 1n=\infty }$adn ${\sum\frac 1{n^2}<\infty}$, we must take ${h=1}$ so that

$\displaystyle \sin\pi z=ze^{g(z)}\prod_{n\neq 0}\left(1-\frac{z}{n}\right)e^{\frac{z}{n}}$

In order to determine ${g(z)}$ we form logarithmic derivative on both sides and find

$\displaystyle \pi\cot\pi z=\frac{1}{z}+g'(z)+\sum_{n\neq 0}\left(\frac{1}{z-n}+\frac 1n\right)=g'(z)+\pi\cot\pi z$

This shows that ${g'(z)=0}$ and ${g(z)}$ reduces to a constant. Since ${\frac{\sin\pi z}{z}\rightarrow \pi}$ as ${z\rightarrow 0}$, we must have ${e^{g(z)}=\pi}$ identically. Hence we can write

$\displaystyle \sin\pi z=\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)$

Recall that an entire function is said to of order ${\lambda}$ is ${\lambda}$ is the smallest number such that

$\displaystyle M(r)\leq e^{r^{\lambda+\epsilon}}$

for all ${\epsilon>0}$ as soon as ${r}$ is sufficiently large. Here ${M(r)}$ denotes the maximum modulus of ${f(z)}$ on ${|z|\leq r}$. The genus and the order are closely related as seen by the following theorem.

Theorem 6 The genus ${h}$ and the order ${\lambda}$ of an entire function satisfies

$\displaystyle h\leq \lambda\leq h+1$

Proof: Suppose that ${f}$ is an entire function of order ${\lambda}$. First we claim that ${\sum_{n=1}^\infty|a_n|^{-(\lambda+\epsilon)}<\infty}$ for any given ${\epsilon>0}$. This will address the issue of convergence of the canonical product. Given ${r>0}$. Recall Jensen’s formula

$\displaystyle \log|f(0)|=\sum_{n=1}^\infty\log\left|\frac{z_n}{r}\right|+\frac{1}{2\pi}\int_0^{2\pi}\log|f(re^{i\theta})|d\theta$

and the fact

$\displaystyle \int_0^{r}\frac{n(x)}{x}dx=-\sum_{n=1}^\infty\log\left|\frac{z_n}{r}\right|$

Apply these formulae with ${2r}$ and we will get

$\displaystyle \begin{array}{rcl} n(r)\log 2&=\int_r^{2r}\frac{n(r)dx}{x}\\ & \leq \int_0^{2r}\frac{n(x)dx}{x}\\ & \leq \frac{1}{2\pi}\int_0^{2\pi}\log|f(2re^{i\theta})|d\theta-\log|f(0)| \end{array}$

This implies ${n(r)r^{-\lambda-\epsilon}\leq \frac{2}{\log 2}r^{\lambda-\lambda-\epsilon}-\frac{\log|f(0)|}{\log 2}r^{-\lambda-\epsilon}\rightarrow 0}$ as ${r\rightarrow\infty}$. Moreover, note we also have the relation

$\displaystyle k\leq n(|z_k|)<|a_n|^{\lambda+\epsilon}$

This implies

$\displaystyle \sum\frac{1}{|a_n|^{h+1}}\leq\sum\frac{1}{n^{\frac{h+1}{\lambda+\epsilon}}}$

Pick ${\epsilon>0}$ so that ${\lambda+\epsilon will show that ${\sum_{n=1}^\infty|a_n|^{-h-1}<\infty}$. $\Box$

In this post, we will discuss the product representation of entire functions. In order to do so, we first discuss the convergence of infinite products.

1. Infinite products

Given a sequence of complex numbers ${(a_n)_{n=1}^\infty}$, we say the product ${\prod_{n=1}^\infty(1+a_n)}$ converges if the limit ${\lim_{N\rightarrow\infty}\prod_{n=1}^\infty(1+a_n)}$ exists.

Theorem 1 If ${\sum_{n=1}^\infty|a_n|<\infty}$, then ${\prod_{n=1}^\infty(1+a_n)}$ converges. Moreover, the product converges to 0 if and only if one of its factors is 0.

Proof: WLOG, we may assume that ${|a_n|\leq \frac 12}$ for all ${n\in{\mathbb N}}$. Thus we may use power series to define logarithm ${\log(1+a_n)}$ so that ${1+a_n=e^{\log(1+a_n)}}$. Then we obtain the following partial products

$\displaystyle \prod_{n=1}^N(1+a_n)=e^{\sum_{n=1}^N\log(1+a_n)}$

For each ${n\in{\mathbb N}}$, we make the estimate

$\displaystyle |\log(1+a_n)|\leq\sum_{k=1}^\infty\frac{|a_n|^k}{k}\leq \sum_{k=1}^\infty|a_n|(1+\frac 12+\frac 14+\cdots)\leq 2|a_n|$

Using continuity of ${e}$, we obtain the first result. To prove the second assertion, we note that if ${1+a_n\neq 0}$ for all ${0}$, then clearly the products converges to ${e^{\text{something}}}$ which is nonzero. $\Box$ Similarly we have convergence theorem for functions.

Theorem 2 Suppose ${(F_n)_{n=1}^\infty}$ is a sequence of holomorphic functions in an open set ${\Omega}$. If for all ${z\in\Omega}$ we have

$\displaystyle |1-F_n(z)|\leq c_n\text{ and }\sum_{n=1}^\infty c_n<\infty$

then ${\prod_{n=1}^\infty F_n(z)}$ converges to a holomorphic function. Moreover, if ${F_n}$ does not vanish in ${\Omega}$ for any ${n\in{\mathbb N}}$, then

$\displaystyle \frac{F'(z)}{F(z)}=\sum_{n=1}^\infty\frac{F_n'(z)}{F_n(z)}$

2. Weisterass product

Theorem 3 Given any sequence ${(a_n)_{n=1}^\infty}$ of complex numbers with ${|a_n|\rightarrow\infty}$ as ${n\rightarrow\infty}$, there exists an entire function ${f}$ that vanishes at all ${z=a_n}$ and nowhere else. Any other entire function is of the form ${f(z)e^{g(z)}}$ where ${g}$ is entire.

In general the function ${\prod_{n=1}^\infty(1-z/a_n)}$ may not converge, so we need to insert extra factors to make it converge without adding any zeros. For each integer ${k\geq 0}$, we define ${E_0=(1-z)}$ and ${E_k(z)=(1-z)e^{\sum_{n=1}^k\frac{z^k}{n}}}$. Then we prove the following estimate:

Lemma 4 If ${|z|\leq\frac 12}$, then ${|1-E_k(z)|\leq c|z|^{k+1}}$ for some constant ${c>0}$.

Proof: Since ${|z|\leq \frac 12}$, we obtain the taylor expansion ${\log(1-z)=-\sum_{n=1}^\infty \frac{z^n}{n}}$. Thus we have

$\displaystyle E_k(z)=e^{\log(1-z)+\sum_{n=1}^k\frac{z^n}{n}}=e^{-\sum_{n=k+1}^\infty\frac{z^n}{n}}=e^w$

Clearly ${|w|\leq|z|^{n+1}|1+z^2+z^3+\cdots|\leq 2|z|^{n+1}}$. In particular, ${|w|\leq 1}$, thus

$\displaystyle |1-E_k(z)|=|1-e^w|\leq c_1|w|\leq 2c_1|z|^{n+1}$

This provides a desired bound. $\Box$ Proof: Define ${h(z)=\prod_{k=0}^\infty E_k(z/a_n)}$. Fix ${R>0}$, we will show that ${h(z)}$ converges uniformly on ${|z|\leq R}$. Decompose ${h(z)=\prod_{|a_n|\leq 2R} E_k(z/a_n)\prod_{|a_n|>2R}E_n(z/a_n)}$ and we only need to consider the latter factor. Clearly ${|z/a_n|\leq \frac 12}$ for all ${|z|\leq R}$ and all ${|a_n|\geq 2R}$ and by previous lemma we see that the product converges uniformly in ${|z|\leq R}$. Moreover, they never vanish in ${|z|\leq R}$ which implies the limit does not vanish also. Hence the function ${z^m\prod_{n=1}^\infty E_n(z/a_n)}$ provides the desired function. $\Box$

In order to refine the result, we need some elementary results. First tool is Jensen’s formula.

Theorem 5 Let ${\Omega}$ be an open set contains the closure of a disk ${D_R}$ and suppose ${f}$ is holomorphic in ${\Omega}$, ${f(0)\neq 0}$, and never vanishes on the circle ${C_R}$. If ${z_1,...,z_N}$ denote the zeros of ${f}$ inside the disc, then

$\displaystyle \log|f(0)|=\sum_{k=1}^N\log\left(\frac{|z_n|}{R}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|d\theta$

Proof: First note that if the theorem holds true for ${f_1}$ and ${f_2}$ then it holds true for ${f_1f_2}$. Consider ${g(z)=\frac{f(z)}{(z-a_1)\cdots(z-a_N)}}$ which is holomorphic function in ${D_R}$ that vanishes nowhere. So we only need to show that the theorem is true for ${g}$ and ${z-w}$. Since ${g}$ has no zeros we need to show ${\log|g(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|g(Re^{i\theta})|d\theta}$. This is clearly true by mean-value property of the harmonic function ${\log|g(z)|}$. It remains to show

$\displaystyle \begin{array}{rcl} \log|w|&=\log\left(\frac{|w|}{R}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|Re^{i\theta}-w|d\theta\\ & = \log\left(|w|\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|e^{i\theta}-\frac{w}{R}|d\theta\\ \end{array}$

This boils down to show ${\frac{1}{2\pi}\int_0^{2\pi}\log|e^{i\theta}-a|d\theta=0}$ for any ${|a|<1}$. Making change of variables ${\theta\rightarrow-\theta}$ shows this is equivalent to ${\frac{1}{2\pi}\int_0^{2\pi}\log|e^{-i\theta}-a|d\theta=\frac{1}{2\pi}\int_0^{2\pi}\log|1-ae^{i\theta}|d\theta=0}$. Consider the function ${F(z)=1-az}$ which is holomorphic and vanishes nowhere in the unit disk. Thus applying mean-value property to ${\log|F(z)|}$ we have

$\displaystyle 0=\log|F(z)|=\frac{1}{2\pi}\int_0^{2\pi}\log|F(e^{i\theta})|d\theta=\frac{1}{2\pi}\int_0^{2\pi}\log|1-e^{i\theta}|d\theta$

$\Box$ Next we claim that if ${f(0)\neq 0}$ and ${f}$ does not vanish on the circle ${C_R}$, then

$\displaystyle \int_0^R\frac{n(r)}{r}dr=\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|d\theta-\log|f(0)|$

This is immediate if we prove

Lemma 6 If ${z_1,...,z_N}$ are the zeros of ${f}$ inside ${D_R}$ then

$\displaystyle \int_0^R\frac{n(r)}{r}dr=-\sum_{k=1}^N\log\left(\frac{|z_k|}{R}\right)$

Proof: We start from the right-hand side.

$\displaystyle -\sum_{k=1}^N\log\left(\frac{|z_k|}{R}\right)=-\sum_{k=1}^N\int_{R}^{|z_k|}\frac{dr}{r}=\sum_{k=1}^N\int_{|z_k|}^{R}\frac{dr}{r}=\int_{0}^{R}\sum_{k=1}^N1_{(|z_k|,R]}(r)\frac{dr}{r}=LHS$

$\Box$

4. Functions of finite order

Let ${f}$ be an entire function. We call ${f}$ has order of growth ${\leq\rho}$ if

$\displaystyle |f(z)|\leq Ae^{B|z|^\rho}\text{ }\forall z\in\mathbb C$

In particular, the order of ${f}$ is ${\rho_0=\inf \rho}$.

Theorem 7 If ${f}$ is an entire function that has order of growth ${\leq\rho}$, then ${n(r)\leq Cr^\rho}$ for sufficiently large ${r}$. Moreover if ${z_1, z_2, ...}$ are zeros of ${f}$, then ${\sum_{n=1}^\infty|a_n|^{-s}}$ converges for any ${s>\rho}$.

Proof: We shall use Jensen’s formula. Note that ${n(R)}$ is an increasing function of ${R}$. Now choose ${R=2r}$, then clearly we have

$\displaystyle \int_{r}^{2r}\frac{n(x)dx}{x}\geq n(r)\int_{r}^{2r}\frac{dx}{x}=n(r)\log 2$

By Jensen’s formula we have

$\displaystyle \begin{array}{rcl} n(r)&\leq\frac{1}{\log 2}\int_0^{2r}\frac{n(x)dx}{x}\\ & \leq \frac{1}{2\pi\log 2}\int_0^{2\pi}\log|f(2re^{i\theta})|d\theta\\ & \leq \frac{1}{2\pi\log 2}\int_0^{2\pi}\log Ae^{Br^\rho}d\theta\leq Cr^\rho \end{array}$

This proves the first part. Then we directly estimate

$\displaystyle \begin{array}{rcl} \sum_{|a_k|\geq 1}^\infty|a_n|^{-s}&=\sum_{j=0}^\infty\sum_{2^j\leq|a_n|\leq 2^{j+1}}|a|^{-s}\\ &\leq \sum_{j=0}^\infty n(2^{j+1})2^{-js}\\ & \leq C\sum_{j=0}^\infty 2^{\rho(j+1)-js}\\ &\leq C'\sum_{j=0}^\infty 2^{(\rho-s)j}<\infty \end{array}$

$\Box$

Theorem 8 Suppose ${f}$ is entire and has order of growth ${\rho_0}$. Let ${k}$ be the integer such that ${k\leq \rho_0. If ${a_1, a_2,...}$ are zeros of ${f}$, then

$\displaystyle f(z)=z^m e^{g(z)}\prod_{n=1}^\infty E_k(z/a_n)$

where ${g(z)}$ is a polynomial of degree ${\leq k}$, ${m}$ is the order of zero at ${0}$.

Proof: The convergence of the canonical products is now easily solved. Using Lemma 2.2, we have ${|1-E_k(z/a_n)|\leq c_1|z/a_n|^{k+1}\leq c_2|a_n|^{-k-1}}$. Thus the convergence of ${\sum_{n=1}^\infty |a_n|^{-k-1}}$ implies the convergence of the product. In order to show that ${g(z)}$ is a polynomial of degree ${\leq k}$, we need more efforts.

Lemma 9 We have the following estimates

$\displaystyle |E_k(z)|\geq e^{-c|z|^{k+1}}\text{ if }|z|\leq\frac 12$

$\displaystyle |E_k(z)|\geq |1-z| e^{-c'|z|^{k}}\text{ if }|z|\geq\frac 12$

Proof of Lemma. If ${|z|\leq\frac 12}$, we directly have

$\displaystyle \begin{array}{rcl} |E_k(z)|&= |e^{\log(1-z)+\sum_{n=1}^k\frac{z^k}{n}}|\\ & = |e^{-\sum_{n=k+1}^\infty\frac{z^k}{n}}|\\ & \geq e^{-|\sum_{n=k+1}^\infty\frac{z^k}{n}|}\\& \geq e^{-c|z|^{k+1}} \end{array}$

If ${|z|\geq \frac 12}$, then

$\displaystyle \begin{array}{rcl} |E_k(z)|&=|1-z|||e^{\sum_{n=1}^k\frac{z^k}{n}}|\\ &\geq|1-z| e^{-\sum_{n=1}^k\frac{|z|^k}{n}}\\ &\geq |1-z|e^{-c|z|^k} \end{array}$

Lemma 10 For any ${\rho_0, we have

$\displaystyle \prod_{n=1}^\infty|E_k(z/a_n)|\geq e^{-c|z|^s}$

except possibly for ${z\in B_{|a_n|^{-k-1}}(a_n)}$.

Proof of Lemma. Again we use usual decomposition

$\displaystyle \begin{array}{rcl} \prod_{n=1}^\infty|E_k(z/a_n)|=\prod_{|a_n|\leq 2|z|}|E_k(z/a_n)|\prod_{|a_n|>2|z|}|E_k(z/a_n)| \end{array}$

We first estimate the second product. Note that ${|z/a_n|\leq \frac 12}$ for all ${n\in{\mathbb N}}$. By previous lemma, we have

$\displaystyle \begin{array}{rcl} \prod_{|a_n|>2|z|}|E_k(z/a_n)|&\geq \prod_{|a_n|>2|z|}e^{-c|z/a_n|^{k+1}}\\ & = e^{-c|z|^{k+1}\sum_{|a_n|>2|z|}|a_n|^{-k-1}} \end{array}$

Note that ${|a_n|^{-k-1}=|a_n|^{-s}|a_n|^{s-k-1}\leq C|a_n|^{-s}|z|^{s-k-1}}$ for some constant ${C>0}$ as ${|a_n/z|^{s-k-1}\leq 2^{s-k-1}\leq C}$. Thus we have

$\displaystyle e^{-c|z|^{k+1}\sum_{|a_n|>2|z|}|a_n|^{-k-1}}\geq e^{-c'|z|^{k+1}\sum_{|a_n|>2|z|}|a_n|^{-s}|z|^{s-k-1}}\geq e^{-c''|z|^s}$

because ${\sum |a_n|^{-s}<\infty}$. To estimate the first prouct we use previous lemma to obtain

$\displaystyle \prod_{|a_n|\leq 2|z|}|E_k(z/a_n)|\geq \prod_{|a_n|\leq 2|z|}|1-z/a_n|e^{-c|z/a_n|^{k}}=|1-z/a_n|e^{-c|z|^k\sum_{|a_n|\leq 2|z|}|a_n|^{-k}}$

Using a similar trick we have ${|a_n|^{-k}=|a_n|^{-s}|a_n|^{s-k}\leq C|a_n|^{-s}|z|^{s-k}}$. Thus

$\displaystyle \prod_{|a_n|<2|z|}|E_k(z/a_n)|\geq \prod_{|a_n|\leq 2|z|}|1-z/a_n|e^{-c|z|^s}$

It remains to estimate ${|1-z/a_n|}$. But using the property that ${|z-a_n|\geq |a_n|^{-k-1}}$, we have

$\displaystyle |1-z/a_n|=|\frac{a_n-z}{a_n}|\geq \frac{|a_n|^{-k-1}}{|a_n|}=|a_n|^{-k-2}$

Now we consider

$\displaystyle \begin{array}{rcl} (k+2)\sum\log|a_n|&\leq (k+1)n(2|z|)\log 2|z|\leq C|z|^{s} \end{array}$

This completes the proof of the lemma.

Lemma 11 There exists a sequence of complex numbers ${(r_m)_{m=1}^\infty}$ with ${|r_m|\rightarrow\infty}$ such that

$\displaystyle |E_k(z/a_n)|\geq e^{-c|z|^s}\;\forall |z|=r_m$

Proof of Lemma. First we note that ${\sum|a_n|^{-k-1}}$ converges so we may take ${\sum_{N}^\infty|a_n|<\frac {1}{ 10}}$. For each positive integer ${L}$ we may find ${L\leq r\leq L+1}$ such that the the disk ${|z|=r}$ does not intersect the forbidden areas in previous lemma. Suppose it is not the case, then there is a particular ${L_0}$ such that all balls of radius ${r}$ with ${L_0\leq r will intersect the forbidden area. In particular, by rotating the those balls on the real line so that the diameter will lie in the positive real axis, we see that ${[L_0, L_0+1]\subset\cup_n [|a_n|-|a_n|^{-k-1},|a_n|+|a_n|^{-k-1}]}$. This implies ${2\sum_{N}^{\infty}|a_n|^{-k-1}\geq 1}$ which is a contradiction.

Now we prove the Hadamard’s theorem. Let ${E(z)=z^m\prod_{n=1}^\infty E_k(z/a_n)}$. Previously we’ve shown that ${E(z)}$ converges to a holomorphic which has zeros as ${f}$ does and does not vanish everywhere else. Thus we have ${\frac{f(z)}{E(z)}=e^{g(z)}}$ and in particular

$\displaystyle e^{\Re g(z)}=|e^{g(z)}|=\left|\frac{f(z)}{E(z)}\right|\leq \frac{A_1e^{B_1|z|^s}}{A_2e^{B_2|z|^{-s}}}=Ae^{B|z|^s}\text{ when }|z|=r_m$

This implies ${\Re g(z)\leq B|z|^s}$ on ${|z|=r_m}$. Finally we show that this is enough to guarantee that ${g}$ is a polynomial of degree less than ${s}$.

Lemma 12 Suppose ${g}$ is entire and ${u=\Re g}$ satisfies

$\displaystyle u(z)\leq Cr^s \text{ whenever }|z|=r$

for a sequence of positive real numbers ${r}$ that tends to infinity. Then ${g}$ is a polynomial of degree ${\leq s}$.

Proof of Lemma. First we write ${g(z)=\sum_{n=0}^\infty a_n z^n}$. Let ${C_r}$ denote the circle centerer at the origin with radius ${r}$. By Cauchy’s integral formula, we have ${a_n=\frac{1}{2\pi i}\int_{C_r}\frac{g(\zeta)}{\zeta^{n+1}}d\zeta}$ whenver ${n\geq 0}$. This implies

$\displaystyle a_n=\frac{1}{2\pi i}\int_0^{2\pi}\frac{g(re^{i\theta})ire^{i\theta}}{r^{n+1}e^{i(n+1)\theta}}d\theta=\frac{1}{2\pi r^n}\int_0^{2\pi }\frac{g(re^{i\theta})}{e^{in\theta}}d\theta\text{ whenever }n\geq 0$

and

$\displaystyle \int_0^{2\pi }\frac{g(re^{i\theta})}{e^{in\theta}}d\theta=0 \text{ whenever }n<0$

which further implies

$\displaystyle 0=\overline{\int_0^{2\pi }\frac{g(re^{i\theta})}{e^{in\theta}}d\theta}=\int_0^{2\pi}\overline{g(re^{i\theta})}e^{-in\theta}d\theta\text{ whenever }n\geq 0$

Thus

$\displaystyle a_n=\frac{1}{2\pi r^n}\int_0^{2\pi}(g(re^{i\theta})+\overline{g(re^{i\theta})})e^{-in\theta}d\theta=\frac{1}{\pi r^n}\int_0^{2\pi} u(re^{i\theta})e^{-in\theta}d\theta$

Recall that ${\int_0^{2\pi}e^{-in\theta}d\theta=0}$ whenever ${n> 0}$. Thus for ${n\geq 1}$, we have

$\displaystyle \begin{array}{rcl} |a_n|&=\left|\frac{1}{\pi r^n}\int_0^{2\pi}(u(re^{i\theta})-Cr^s)e^{-in\theta}d\theta\right|\\ & \leq \frac{1}{\pi r^n}\int_0^{2\pi}|u(re^{i\theta}-Cr^s)|d\theta\\ & = \frac{1}{\pi r^n}\int_0^{2\pi}(Cr^s-u(re^{i\theta}))d\theta\\ & =2Cr^{s-n}-\frac{2u(0)}{r^n} \end{array}$

Letting ${r\rightarrow\infty}$ shows that ${a_n=0}$ if ${n>s}$. Hence ${g}$ is a polynomial of degree ${\leq s}$. $\Box$