Hadamard products

by tzy9393

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


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}


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


\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


\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