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

Proof

Set ${\mathrm{\Delta }}_0:=D$. We inductively construct a sequence $({\mathrm{\Delta }}_n{)}_{n\ge 0}$ with the properties:

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

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

3. $|\underset{\partial D}{\int }f(z),dz|\le 4^n|\underset{\partial {\mathrm{\Delta }}_n}{\int }f(z),dz|$

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

$\begin{array}{cc}|\underset{\partial {\mathrm{\Delta }}_n}{\int }f(z)dz|& =|\sum _{i=1}^4\underset{\partial {\mathrm{\Delta }}_{n+1}^i}{\int }f(z)dz|\\ & \le \sum _{i=1}^4|\underset{\partial {\mathrm{\Delta }}_{n+1}^i}{\int }f(z)dz|\\ & \le \underset{i}{\max }|\underset{\partial {\mathrm{\Delta }}_{n+1}^i}{\int }f(z)dz|\end{array}$

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

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

Since all ${\mathrm{\Delta }}_n$ are compact, $\bigcap _{n\ge 0}{\mathrm{\Delta }}_n\ne \mathrm{\varnothing }$. Let $z_0\in \bigcap _{n\ge 0}{\mathrm{\Delta }}_n$. As $f$ is holomorphic at $z_0$, there exists a neighborhood $V$ of $z_0$ and a continuous function $A:V\to ℂ$ with $A(z_0)=0$ such that: $f(z)=f(z_0)+(z-z_0)f^{\prime }(z_0)+A(z)(z-z_0),z\in V.$

Since the function $z\mapsto f(z_0)+(z-z_0)f^{\prime }(z_0)$ has a primitive, it follows for $n\ge 0$ with ${\mathrm{\Delta }}_n\subseteq V$ that: $\underset{\partial {\mathrm{\Delta }}_n}{\int }f(z),dz=\underset{\partial {\mathrm{\Delta }}_n}{\int }f(z_0)+(z-z_0)f^{\prime }(z_0)+A(z)(z-z_0),dz=\underset{\partial {\mathrm{\Delta }}_n}{\int }A(z)(z-z_0),dz.$

Thus, due to the continuity of $A$ and $A(z_0)=0$, we obtain:

$\begin{array}{cc}|\underset{\partial D}{\int }f(z)dz|& \le 4^n|\underset{\partial {\mathrm{\Delta }}_n}{\int }f(z)dz|\\ & =4^n|\underset{\partial {\mathrm{\Delta }}_n}{\int }A(z)(z-z_0)dz|\\ & \le 4^n\cdot ℒ(\partial {\mathrm{\Delta }}_n)\underset{z\in \partial {\mathrm{\Delta }}_n}{\max }|z-z_0||A(z)|\\ & \le 4^n\cdot ℒ(\partial {\mathrm{\Delta }}_n{)}^2\underset{z\in \partial {\mathrm{\Delta }}_n}{\max }|A(z)|\\ & =ℒ(\partial D)\underset{z\in \partial {\mathrm{\Delta }}_n}{\max }|A(z)|\\ & \to ℒ(\partial D)|A(z_0)|=0,n\to \mathrm{\infty }.\end{array}$

Notation in the Proof

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

$\partial {\mathrm{\Delta }}_n$ is the integration path along the boundary of the $n$-th subtriangle, with perimeter $ℒ(\partial {\mathrm{\Delta }}_n)=\frac{1}{2^n}\cdot ℒ(\partial {\mathrm{\Delta }}_0)$.

See Also

• Goursat's Lemma (Details)
• rectifiable curve

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