Complex Analysis/Goursat's Lemma

Introduction

Goursat's Lemma is a crucial result in the proof of the Cauchy's integral theorem.It restricts the integration paths to triangles, making it provable via a geometric subdivision argument.

Statement

Let $D\subseteq \mathbb {C}$ be a closed triangle, $G\supseteq D$ an open set, and $f\colon U\to \mathbb {C}$ a holomorphic function. Then: $\int _{\partial D}f(z),dz=0.$

Proof

Set $\Delta _{0}:=D$ . We inductively construct a sequence $(\Delta _{n})_{n\geq 0}$ with the properties:

1. $\Delta _{n}\subseteq \Delta _{n-1}$

2. ${\mathcal {L}}(\partial \Delta _{n})=2^{-n}{\mathcal {L}}(\partial D)$ , where ${\mathcal {L}}$ represents the length of a curve

3. $\left|\int _{\partial D}f(z),dz\right|\leq 4^{n}\left|\int _{\partial \Delta _{n}}f(z),dz\right|$

For $n\geq 0$ and $\Delta _{n}$ already constructed, we subdivide $\Delta _{n}$ by connecting the midpoints of its sides, forming four subtriangles $\Delta _{n+1}^{i}$ , $1\leq i\leq 4$ . Since the contributions of the midpoints cancel out in the integration, we have:

{\begin{array}{rl}\displaystyle \left|\int _{\partial \Delta _{n}}f(z)\,dz\right|&=\displaystyle \left|\sum _{i=1}^{4}\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\\&\leq \displaystyle \sum _{i=1}^{4}\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\\&\leq \displaystyle \max _{i}\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\end{array}}

Choose $1\leq i\leq 4$ such that $\left|\int _{\partial \Delta _{n+1}^{i}}f(z),dz\right|=\max _{i}\left|\int _{\partial \Delta _{n+1}^{i}}f(z),dz\right|$ and set $\Delta _{n+1}:=\Delta _{n+1}^{i}$ . Then, by construction: $\Delta _{n+1}\subseteq \Delta _{n}$ , ${\mathcal {L}}(\partial \Delta _{n+1})={\frac {1}{2}}{\mathcal {L}}(\partial \Delta _{n})=2^{-(n+1)}{\mathcal {L}}(\partial D)$ , and $\left|\int _{\partial D}f(z),dz\right|\leq 4^{n}\left|\int _{\partial \Delta _{n}}f(z),dz\right|\leq 4^{n+1}\left|\int _{\partial \Delta _{n+1}}f(z),dz\right|.$

This ensures $\Delta _{n+1}$ has the required properties.

Since all $\Delta _{n}$ are compact, $\bigcap _{n\geq 0}\Delta _{n}\neq \emptyset$ . Let $z_{0}\in \bigcap _{n\geq 0}\Delta _{n}$ . As $f$ is holomorphic at $z_{0}$ , there exists a neighborhood $V$ of $z_{0}$ and a continuous function $A\colon V\to \mathbb {C}$ with $A(z_{0})=0$ such that: $f(z)=f(z_{0})+(z-z_{0})f'(z_{0})+A(z)(z-z_{0}),\qquad z\in V.$

Since the function $z\mapsto f(z_{0})+(z-z_{0})f'(z_{0})$ has a primitive, it follows for $n\geq 0$ with $\Delta _{n}\subseteq V$ that: $\int _{\partial \Delta _{n}}f(z),dz=\int _{\partial \Delta _{n}}f(z_{0})+(z-z_{0})f'(z_{0})+A(z)(z-z_{0}),dz=\int _{\partial \Delta _{n}}A(z)(z-z_{0}),dz.$

Thus, due to the continuity of $A$ and $A(z_{0})=0$ , we obtain:

{\begin{array}{rl}\displaystyle \left|\int _{\partial D}f(z)\,dz\right|&\leq \displaystyle 4^{n}\left|\int _{\partial \Delta _{n}}f(z)\,dz\right|\\&=\displaystyle 4^{n}\left|\int _{\partial \Delta _{n}}A(z)(z-z_{0})\,dz\right|\\&\leq \displaystyle 4^{n}\cdot {\mathcal {L}}(\partial \Delta _{n})\max _{z\in \partial \Delta _{n}}|z-z_{0}||A(z)|\\&\leq \displaystyle 4^{n}\cdot {\mathcal {L}}(\partial \Delta _{n})^{2}\max _{z\in \partial \Delta _{n}}|A(z)|\\&=\displaystyle {\mathcal {L}}(\partial D)\max _{z\in \partial \Delta _{n}}|A(z)|\\&\to \displaystyle {\mathcal {L}}(\partial D)|A(z_{0})|=0,\qquad n\to \infty .\end{array}}

Notation in the Proof

$\Delta _{n}$ is the $n$ -th subtriangle of the original triangle, with side lengths scaled by a factor of ${\frac {1}{2^{n}}}$ .

$\partial \Delta _{n}$ is the integration path along the boundary of the $n$ -th subtriangle, with perimeter ${\mathcal {L}}(\partial \Delta _{n})={\frac {1}{2^{n}}}\cdot {\mathcal {L}}(\partial \Delta _{0})$ .

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