Some Math

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)

3. Hadamard product

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<h+1} will show that {\sum_{n=1}^\infty|a_n|^{-h-1}<\infty}. \Box

Hadamard products

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

3. Hadamard product

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

5. Hadamard products

Theorem 8 Suppose {f} is entire and has order of growth {\rho_0}. Let {k} be the integer such that {k\leq \rho_0<k+1}. 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<s<k+1}, 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<L_0+1} 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