[복소해석학] 5.3. Infinite product

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

Theorem. If $\sum|a_n|<\infty$, then the product $\prod(1+a_n)$ converges. Moreover, the product converges to $0$ if and only if one of its factors is $0$ and is rearrangeable.

The theorem above is a necessary condition that guarantees the existence of a product.

proof. If $1+a_k=0$ for some $k$ then the product is $0$, obviously. Now suppose $1+a_n\ne0$ for all $n$. Since $\sum|a_n|<\infty$ we have $|a_n|\to0$ as $n\to\infty$ so there exists large $N$ such that $n>N\;\rightarrow\;|a_n|<1/2$. Ignoring the first finite terms, we may consider $|a_n|<1/2$ for all $n$. Then we have
 
$$\operatorname{Log}(1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-\cdots=\sum_{n=1}^\infty\frac{(-1)^{n+1}z^n}{n}$$
 
for $|z|<1$. Then we have $|\operatorname{Log}(1+z)|\le 2|z|$ for $|z|<1/2$ since

\begin{align*}
        |\operatorname{Log}(1+z)|&=\Bigg|\sum_{n=1}^\infty\frac{(-1)^{n+1}z^n}{n}\Bigg|=|z|\cdot\Bigg|\sum_{n=1}^\infty\frac{(-1)^{n+1}z^{n-1}}{n}\Bigg|\\
        &\le|z|\cdot\sum_{n=1}^\infty\frac{z^{n-1}}{n}\le|z|\cdot\sum_{n=1}^\infty\frac{1}{n\cdot2^{n-1}}<|z|\cdot\sum_{n=1}^\infty\frac{1}{2^{n-1}}=2|z|
\end{align*}
 
Thus we have $|\operatorname{Log}(1+a_n)|\le 2|a_n|$ for all $n$. We also have that $1+z=e^{\operatorname{Log}(1+z)}$ when $|z|<1$ so
$$\prod_{k=1}^n(1+a_k)=\prod_{k=1}^ne^{\operatorname{Log}(1+a_k)}=\exp\left(\sum_{k=1}^n\operatorname{Log}(1+a_k)\right)$$

and from the continuity of exponential function we have
 
$$\lim_{n\to\infty}\prod_{k=1}^n(1+a_k)=\lim_{n\to\infty}\exp\left(\sum_{k=1}^n\operatorname{Log}(1+a_k)\right)$$
 
Since $\sum_{n=1}^\infty\operatorname{Log}(1+a_n)$ absolutely converges, the limit of the product above converges and thus rearrangeable by the rearrangement theorem.

Here, $\operatorname{Log}z$ stands for the principal branch of the complex logarithm.
 
Lemma 1. Let $f$ be a complex function defined on a set $E$. Denote $\sup_{x\in E}=\|f\|_E$. If the sequence of functions $\{f_n\}$ satisfies
    $$\sum_{n=1}^\infty\|f_n\|_E<\infty$$
then $\sum f_n(x)$ absolutely converges for all $x\in E$ and uniformly converges in $E$.

proof. That $\sum f_n(x)$ absolutely converges for all $x\in E$ is obvious since $f_n(x)\le\|f_n\|_E$ for all $x\in E$. For the partial sum $S_n=\sum_{k=1}^n f_k$ and positive integers $n>m$ we have
    \begin{align*}
        |S_n-S_m|&=\Bigg|\sum_{k=m+1}^n f_k\Bigg|\le\sum_{k=m+1}^n\|f_k\|_E
    \end{align*}
Since $\sum_{n=1}^\infty\|f_n\|_E$ converges, the above implies that $\{S_n\}$ is an uniform Cauchy sequence so $\sum f_n$ uniformly converges in $E$.

Lemma 2. Let $\Omega$ be open and $\{f_n\}$ be the sequence of holomorphic functions in $\Omega$. If
    $$\sum_{n=1}^\infty \|f_n\|_\Omega<\infty$$
then $\sum f_n'(x)$ absolutely converges for all $x\in\Omega$ and uniformly converges in every compact subset of $\Omega$.
 
proof. Let $K\subset\Omega$ be compact then we have an uniform $r>0$ such that $a\in K\;\Rightarrow\;\overline{D_r(a)}\subset\Omega$. For $a\in K$ and holomorphic function $f$ in $\Omega$ we have
$$|f'(a)|=\Bigg|\frac{1}{2\pi i}\int_{|z-a|=r}\frac{f(z)}{(z-a)^2}dz\Bigg|\le\frac{1}{2\pi}\cdot 2\pi r\cdot\frac{\|f\|_\Omega}{r^2}=\frac{\|f\|_\Omega}{r}$$
 
Thus $\|f_n'\|_K\le r^{-1}\|f_n\|_\Omega$ for the constant $r>0$, the remaining proof is almost the copy of the proof of Lemma 1.

Theorem. Suppose $\{F_n\}$ is a sequence of holomorphic functions on the open set $\Omega$. If $\sum_{n=1}^\infty\|F_n-1\|_\Omega<\infty$ then the product $\prod F_n$ uniformly converges in $\Omega$ to a holomorphic function $G$ in $\Omega$. Moreover, If $F_n$ does not vanish for any $n$ in $\Omega$ then $G$ does not vanish in $\Omega$ and
    $$\sum_{n=1}^\infty\frac{F'_n(z)}{F_n(z)}=\frac{G'(z)}{G(z)}$$
for all $z\in\Omega$. In fact, the convergence $F'_n/F_n \to G'/G$ is uniform in every compact subset of $\Omega$.

proof. Let $G_n=\prod_{k=1}^nF_k=F_1F_2\cdots F_n$. Then $G_n$ is holomorphic in $\Omega$ for all $n$.

Step 1. Suppose $\|F_n-1\|_\Omega<1/2$ and prove the theorem. In this case, $F_n$ does not vanish in $\Omega$ for all $n$ so $\operatorname{Log}F_1,\;\operatorname{Log}F_2, \cdots$ are holomorphic in $\Omega$ since $F_n$ does not intersects with the negative real axis. We also have $\|\operatorname{Log}F_n\|_\Omega\le 2\|F_n-1\|_\Omega$ and
    $$G_n=\prod_{k=1}^n\exp(\operatorname{Log}F_k)=\exp\left(\sum_{k=1}^n \operatorname{Log}F_k\right)$$
From the hypotheses we have $\sum\|\operatorname{Log}F_n\|_\Omega\le 2\sum\|F_n-1\|_\Omega<\infty$ so by Lemma 1 $\sum\operatorname{Log}F_n(z)$ absolutely converges for all $z\in\Omega$ and $\sum\operatorname{Log}F_n$ uniformly converges in $\Omega$. Letting $\sum\|\operatorname{Log}F_n\|_\Omega=R$ we have
    $$\Bigg|\sum_{k=1}^n\operatorname{Log}F_k(z)\Bigg|\le R$$
for all $n$ and $z\in\Omega$ obviously. Now we want to prove that $\{G_n\}$ uniformly converges. To see this, suppose $n>m$. Since $\sum\operatorname{Log}F_n$ uniformly converges, letting $S_n=\sum_{k=1}^n\operatorname{Log}F_k$, $\{S_n\}$ is a uniform Cauchy sequence. Then
    $$|S_n-S_m|=\Bigg|\sum_{k=m+1}^n \operatorname{Log}F_k\Bigg|<\epsilon$$
for any $\epsilon>0$ so we can control $|G_n-G_m|$ by
    \begin{align*}
        |G_n-G_m|&=\Bigg|\exp\left(\sum_{k=1}^n \operatorname{Log}F_k-\sum_{k=1}^m \operatorname{Log}F_k\right)\Bigg|\\&=\Bigg|\exp\left(\sum_{k=1}^m \operatorname{Log}F_k\right)\Bigg|\cdot\Bigg|\exp\left(\sum_{k=m+1}^n \operatorname{Log}F_k\right)-1\Bigg|\\
        &\le e^R\cdot |e^\epsilon-e^0|
    \end{align*}
Then since exponential function is continuous, $\{G_n\}$ uniformly converges to $G$ on $\Omega$ and since each $G_n$ is holomorphic on $\Omega$, $G$ is also holomorphic on $\Omega$.

Step 2. $G=\prod F_n=\exp\left(\sum\operatorname{Log}F_n\right)$ so $G$ does not vanish on $\Omega$. Then we can easily observe that
    $$\frac{G'}{G}=\left(\operatorname{Log}G\right)'=\left(\sum_{n=1}^\infty\operatorname{Log}F_n\right)'=\sum_{n=1}^\infty(\operatorname{Log}F_n)'=\sum_{n=1}^\infty\frac{F_n'}{F_n}$$
Note that the third equal sign, $(\sum\operatorname{Log}F_n)'=\sum(\operatorname{Log}F_n)'$ came from the uniform convergence of $\sum\operatorname{Log}F_n$.

Step 3. Now we deal with the general case. Since $\sum_{n=1}^\infty\|F_n-1\|_\Omega<\infty$, we have a large $N$ such that if $n>N$ then $\|F_n-1\|_\Omega<1/2$. Then the product $\prod_{n=N+1}^\infty F_n$ uniformly converges on $\Omega$. That is, the sequence of functions $\{F_{N+1}F_{N+2}\cdots F_n\}_{n=N+1}^\infty$ uniformly converges. Since $\sum\|F_n-1\|_\Omega<\infty$, $F_1F_2\cdots F_n$ is bounded on $\Omega$ hence $\{(F_1F_2\cdots F_n)F_{N+1}F_{N+2}\cdots F_n\}_{n=N+1}^\infty$ uniformly converges and thus $\prod F_n$ uniformly converges on $\Omega$. Moreover, if $F_1, \cdots, F_n$ do not vanish on $\Omega$ then $G=\prod F_n=(F_1\cdots F_n)\prod_{n=N+1}^\infty F_n$ does not vanish on $\Omega$ and we have
    $$\frac{G'}{G}=\frac{F_1'}{F_1}+\cdots+\frac{F_n'}{F_n}+\sum_{n=N+1}^\infty\frac{F_n'}{F_n}$$
then we can easily handle the finite terms so the proof is over.

For example, [1] deals with the two identity,
\begin{align}
    \frac{\sin\pi z}{\pi}&=z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)\\
    \pi\cot\pi z=\sum_{n=-\infty}^\infty&\frac{1}{z+n}=\frac{1}{z}+\sum_{n=1}^\infty\frac{2z}{z^2-n^2}
\end{align}
But here we omit the proof of these two identities. Check p.143, p.144.

 

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[2], 김영원, 계승혁. 2014. 기초복소해석. 서울대학교출판문화원.
[3], 김영원, 2023. 복소해석학 강의노트. 한빛아카데미.
[4], Rudin, Walter 1976. Principles of Mathematical Analysis 3ed. McGraw-Hill Press.
[5], Munkres, J.R., 2000. Topology 2ed. Pearson Education.