[복소해석학] 5.1. Jensen's Formula

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우리는 가장 먼저, 주어진 전해석함수의 근의 분포에 대한 질문을 던진다. 지금까지 공부한 복소해석학의 결과를 이용하면 전해석함수 $f$의 근이 정확히 어떤 값인지 모두 찾아내기는 어려워도, $f$의 근이 원점을 중심으로하는 주어진 반지름 $R$ 안의 근방에서 몇 개나 존재하는지는 알 수 있다. 이를 위해 다음 정리를 증명하고, 그 따름정리로 zero-counting function $\mathfrak{n}(r)$을 도입한다.

Theorem. Let $\Omega$ be an open set containing the closure of a disc $D_R$ - the open disc of radius $R$ centered at the origin. Suppose $f$ is holomorphic in $\Omega$ such that $f(0)\ne 0$ and vanishes nowhere on $C_R$ - the circle of radius $R$ centered at the origin. If $z_1, \cdots, z_N$ are the zeros of $f$ inside $D_R$(counted with multiplicities) then the following holds :
    $$\log|f(0)|=\sum_{k=1}^{N}\log\left(\frac{|z_k|}{R}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|d\theta $$
 
먼저, 증명을 위해 다음 보조정리들을 도입한다. [1], Chapter 3의 정리들을 증명없이 소개한다.

Lemma. (1) Theorem 6.2 If $f$ is a nowhere vanishing holomorphic function in a simply connected region $\Omega$ then there exists a holomorphic function $g$ on $\Omega$ such that $$f(z)=e^{g(z)}$$ and the function $g$ can be denoted by $g(z)=\log f(z)$.

(2) Corollary 7.2 If $f$ is holomorphic in a disc $D_R(z_0)$, then $$f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})\;d\theta$$ for any $0<r<R$.

다음 증명은 Theorem의 증명이다. 증명 전체에서 $f$는 상수함수 $0$이 아니라고 가정한다.
 
Step 1. Let $f_1, f_2$ be two functions satisfying the hypotheses of \textbf{Theorem 2.1} then we obvserve that their product, $f_1f_2$ also satisfies the hypotheses of the given theorem - since $f_1f_2$ is also holomorphic in $\Omega$, $(f_1f_2)(0)\ne0$ and $\log|(f_1f_2)(0)|=\log|f_1(0)|+\log|f_2(0)|$.\\

Step 2. Now suppose $f$ is the function satisfying the hypotheses of the theorem. Define $$g(z)=\frac{f(z)}{(z-z_1)\cdots(z-z_N)}$$ If we show that $g$ is bounded near $z_j$ then by Riemann's theorem, each $z_j$ is a removable singularity of $g$. From \textbf{Theorem 1.1} of Chapter 3, $f(z_j)=0$ implies the existence of small neighborhood $U_j$ of $z_j$ and integer $m_j\ge 1$ such that $f(z)=(z-z_j)^{m_j}h(z)$ for all $z\in U_j$ where $h$ is non-vanishing in $U_j$. Then for that $z\in U_j$ we have
    $$g(z)=\frac{(z-z_j)^{m_j}h(z)}{(z-z_1)\cdots(z-z_N)}=(z-z_j)^{m_j-1}\frac{h(z)}{(z-z_1)\cdots(z-z_{j-1})(z-z_{j+1})\cdots(z-z_N)}$$
so by the continuity of each factor, $g$ is bounded in the small neighborhood of each $z_j$. Then $g$ is also holomorphic in the whole $\Omega$ and vanishes nowhere in $\overline{D_R}$. These first two steps, combined with the \textbf{Theorem 1.1} of Chapter 3, implies that it suffices to consider only the cases when $f(z)=z-w$ and nowhere vanishing function $g$ in $\overline{D_R}$\\

Step 3. Let $g$ be the holomorphic function in $\Omega$ that nowhere vanishing in $\overline{D_R}$. Take small $\epsilon>0$ such that $D_{R+\epsilon}\subset\Omega$ and $g$ has no zero in $D_{R+\epsilon}$. This is possible since $\Omega$ is open and $g$ is continuous. Since $D_{R+\epsilon}$ is a simply connected region, we have ae holomorphic function $h$ in $D_{R+\epsilon}$ such that $g(z)=e^{h(z)}$. Then $$|g(z)|=|e^{h(z)}|=e^{\operatorname{Re}h(z)}$$ so we have $\log|g(z)|=\operatorname{Re}h(z)$ and this is holomorphic so by (2) of \textbf{Lemma 2.2} we have
$$\log|g(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|g(Re^{i\theta})|\;d\theta$$ as desired. Note that $g$ is nowhere vanishing in $D_R$.\\

Step 4. Take $w\in D_R$ and let $f(z)=z-w$. Then we claim that
    \begin{align*}
        \log|w|&=\log\left(\frac{|w|}{R}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|Re^{i\theta}-w|\;d\theta\\
        &=\log|w|-\log R+\log R+\frac{1}{2\pi}\int_0^{2\pi}\log|e^{i\theta}-w/R|\;d\theta
    \end{align*}
so if we prove that $$\frac{1}{2\pi}\int_0^{2\pi}\log|e^{i\theta}-w/R|\;d\theta=0$$ then the proof is over. But we have
    $$\int_0^{2\pi}\log|e^{i\theta}-a|\;d\theta=0$$
whenever $|a|<1$. Let us prove this. Changing variables $\theta\mapsto -\theta$ yields
    $$\int_0^{2\pi}\log|e^{i\theta}-a|\;d\theta=\int_0^{2\pi}\log|1-ae^{i\theta}|\;d\theta$$
and letting $F(z)=1-az$ which vanishes outside of $\overline{D_1}$ we have a small $\delta>0$ such that $F$ does not vanish inside $D_{1+\delta}$. Then (2) of \textbf{Lemma 2.2} implies
    $$\log|F(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|F(e^{i\theta)}|\;d\theta=\frac{1}{2\pi}\int_0^{2\pi}\log|1-ae^{i\theta}|\;d\theta=0$$ as desired. This finishes the proof.

Definition. If $f$ is a holomorphic function on $\overline{D_R}$, we denote the number of zeros of $f$ inside the disc $D_r$ for $0<r<R$, counted with multiplicities by $\mathfrak{n}(r)$.

앞으로의 논의는 [3]의 p.102, 103을 참고하였다. Stieltjes 적분의 정의는 [4]의 p.122에 있다. 앞으로 편의를 위하여 $\mathfrak{n}(0)=0$이라 하자.

Let $f$ be holomorphic in $\Omega$ such that $D_R\subset\overline{D_R}\subset\Omega$ and $z_1, \cdots, z_N$ be the zeros of $f$ in $D_R$ aligned by the ascending order of absolute value. Now define the characteristics $\eta_k(r)$ by
 $$\eta_k(r)=\begin{cases}
     1\;\;\;\;\;\operatorname{if}\;r\ge|z_k|\\
     0\;\;\;\;\;\operatorname{if}\;r<|z_k|
 \end{cases}$$
Then it is obvious that $\sum_{k=1}^N\eta_k(r)=\mathfrak{n}(r)$. Since $\mathfrak{n}(r)$ and $\eta_k(r)$ is a monotone increasing function, the Stieltjes integral along those $\mathfrak{n}$ and $\eta_k$ is well-defined. If $\phi$ is a continuous function in $[a, R]$ for $a=|z_1|$ then we have
$$\int_a^R\phi(r)d\eta_k(r)=\phi(|z_k|)$$
by the definition of Stieltjes integral. Then we also have
$$\int_a^R \phi(r)d\mathfrak{n}(r)=\int_a^R\phi(r)d\sum_{k=1}^N\eta_k(r)=\sum_{k=1}^{N}\int_a^R\phi(r)d\eta_k(r)=\sum_{k=1}^{N}\phi(|z_k|)$$
by \textbf{Theorem 6.12} of \cite{[4]}. If $\phi$ is continuously differentiable, that is, of $C^1$ then the partial integral of $\phi$ along $\mathfrak{n}(r)$ yields
\begin{align*}
    \int_a^R\phi(r)d\mathfrak{n}(r)&=\phi(R)\mathfrak{n}(R)-\phi(a)\mathfrak{n}(a)-\int_a^R\mathfrak{n}(r)d\phi(r)\\
    &=\phi(R)\mathfrak{n}(R)-\int_a^R\mathfrak{n}(r)\phi'(r)dr\\
    &=\phi(R)\mathfrak{n}(R)-\int_a^R\mathfrak{n}(r)\phi'(r)dr-\int_0^a\mathfrak{n}(r)\phi'(r)dr\\
    &=\phi(R)\mathfrak{n}(R)-\int_0^R\mathfrak{n}(r)\phi'(r)dr
\end{align*}
by \textbf{Theorem 6.17} of \cite{[4]} and the fact that $\mathfrak{n}(r)=0$ when $0\le r\le a$. Finally we have
$$\sum_{k=1}^{N}\phi(|z_k|)=\phi(R)\mathfrak{n}(R)-\int_0^R\mathfrak{n}(r)\phi'(r)dr$$
Lemma. If $z_1, \cdots,z_N$ are the zeros of $f$ inside the disc $D_R$, then
    $$\int_0^R\mathfrak{n}(r)\frac{dr}{r}=\sum_{k=1}^N\log\frac{R}{|z_k|}=\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|\;d\theta-\log|f(0)|$$

proof. The proof is immediate from the argument above. Let $\phi(z)=\log(R/z)=\log R-\log z$ then we have $\phi(R)=0$ and the proof is over.

영점의 개수와 함수의 성장 속도의 관계는 가장 먼저 다항함수의 예시를 통해 생각해 볼 수 있다. 대수학의 기본정리에 의하면 $n$차 다항함수는 정확히 $n$개의 복소근을 가지고, 다항식의 차수가 커질수록 함수의 성장 속도가 빠르다는 것을 생각해보자. 이 관찰을 일반적인 전해석함수에 대해 자연스럽게 일반화한다면 영점의 개수가 많을수록 성장 속도가 빠를 것이라고 추측할 수 있다. Jensen's formula는 함수의 영점과 그 성장 속도의 관계에 대한 이 추측을 수식의 형태로 제시할 뿐 아니라 함수의 성장 속도가 그 영점의 분포와도 관계있음을 알려준다.\\

Jensen's formula의 형태를 조금 바꾸어 아래와 같은 식으로 생각하자.
$$\log|f(0)|+\sum_{k=1}^N\log\left(\frac{R}{|z_k|}\right)=\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|\;d\theta$$
식의 좌변은 $f$의 $0$에서의 값 그리고 크기가 $R$인 원판 안에 위치한 영점들의 위치에 관련되어 있다. 영점 $z_k$가 원점에서 가까울수록, 그리고 영점이 많을수록 좌변의 값은 커진다. 반면 식의 우변은 크기가 $R$인 원판 둘레에서의 $f$의 값과 관련되어 있다. 우변의 적분은, 정확히 원 $C_R$을 따르는 $f$의 로그평균이다. 즉, 함수의 영점이 많을수록 그리고 그 영점의 절댓값이 작을수록 함수의 성장 속도가 빨라진다.\\

여기서 한 걸음 나아가 함수의 성장 속도를 정확히 정의한다면, 영점의 개수와 함수의 성장 속도 사이의 관계를 조금 더 명확한 수식의 형태로 나타낼 수 있다.

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[1], Stein, Elias M., Shakarachi, Rami 2003. Complex Analysis. Princeton University Press.
[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.