# Complex Analysis/Lemma of Goursat

The Goursat Lemma is an important intermediate result in the proof of the Cauchy Integral Theorem. It restricts the integration paths to triangles, and its proof is based on a geometrical subdivision argument.

## Statement

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

## Proof

$\Delta _{n}\subseteq \Delta _{n-1}$
${\mathcal {L}}(\partial \Delta _{n})=2^{-n}{\mathcal {L}}(\partial D)$ (where $L$ denotes the length of a curve)
$\left|\int _{\partial D}f(z),dz\right|\leq 4^{n}\left|\int _{\partial \Delta _{n}}f(z),dz\right|$
So, for some $n\geq 0$, suppose $\Delta _{n}$ is already constructed. We subdivide $\Delta _{n}$ by connecting the midpoints of the sides, creating four smaller triangles $\Delta _{n+1}^{i}$, $1\leq i\leq 4$. Since the connections of the midpoints cancel each other out during integration, we have:
${\begin{array}{rl}\displaystyle \left|\int _{\partial \Delta _{n}}f(z)\,dz\right|&=\displaystyle \left|\sum _{i=1}^{4}\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\\&\leq \displaystyle \sum _{i=1}^{4}\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\\&\leq \displaystyle \max _{i}\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\end{array}}$
Now, choose $1\leq i\leq 4$ with $\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|=\max _{i}\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|$ and set $\Delta _{n+1}:=\Delta _{n+1}^{i}$. Then, by construction, we have $\Delta _{n+1}\subseteq \Delta _{n}$, and also:
${\mathcal {L}}(\partial \Delta _{n+1})={\frac {1}{2}}{\mathcal {L}}(\partial \Delta _{n})=2^{-(n+1)}{\mathcal {L}}(\partial D)$
and
$\left|\int _{\partial D}f(z)\,dz\right|\leq 4^{n}\left|\int _{\partial \Delta _{n}}f(z)\,dz\right|\leq 4^{n+1}\left|\int _{\partial \Delta _{n+1}}f(z)\,dz\right|$
Thus, $\Delta _{n+1}$ has exactly the desired properties. Since all $\Delta _{n}$ are compact, the intersection $\bigcap _{n\geq 0}\Delta _{n}\neq \emptyset$, and let $z_{0}\in \bigcap _{n\geq 0}\Delta _{n}$. Since $f$ is holomorphic at $z_{0}$, there exists a continuous function $A\colon V\to \mathbb {C}$ with $A(z_{0})=0$ in a neighborhood $V$ of $z_{0}$ such that:
$f(z)=f(z_{0})+(z-z_{0})f'(z_{0})+A(z)(z-z_{0}),\qquad z\in V$
Since $z\mapsto f(z_{0})+(z-z_{0})f'(z_{0})$ has an antiderivative, for all $n\geq 0$ with $\Delta _{n}\subseteq V$, we have:
$\int _{\partial \Delta _{n}}f(z)\,dz=\int _{\partial \Delta _{n}}f(z_{0})+(z-z_{0})f'(z_{0})+A(z)(z-z_{0})\,dz=\int _{\partial \Delta _{n}}A(z)(z-z_{0})\,dz.$
Thus, using the continuity of $A$ and $A(z_{0})=0$, we get:
${\begin{array}{rl}\displaystyle \left|\int _{\partial D}f(z)\,dz\right|&\leq \displaystyle 4^{n}\left|\int _{\partial \Delta _{n}}f(z)\,dz\right|\\&=\displaystyle 4^{n}\left|\int _{\partial \Delta _{n}}A(z)(z-z_{0})\,dz\right|\\&\leq \displaystyle 4^{n}\cdot {\mathcal {L}}(\partial \Delta _{n})\max _{z\in \partial \Delta _{n}}|z-z_{0}||A(z)|\\&\leq \displaystyle 4^{n}\cdot {\mathcal {L}}(\partial \Delta _{n})^{2}\max _{z\in \partial \Delta _{n}}|A(z)|\\&=\displaystyle {\mathcal {L}}(\partial D)\max _{z\in \partial \Delta _{n}}|A(z)|\\&\to \displaystyle {\mathcal {L}}(\partial D)|A(z_{0})|=0,\qquad n\to \infty .\end{array}}$
==Notation in the Proof==
$\Delta _{n}$ is the $n$-th similar subtriangle of the original triangle with side lengths shortened by a factor of ${\frac {1}{2^{n}}}$.
$\partial \Delta _{n}$ is the integration path along the boundary of the $n$-th similar subtriangle, with a perimeter ${\mathcal {L}}(\partial \Delta _{n})={\frac {1}{2^{n}}}\cdot {\mathcal {L}}(\partial \Delta _{0})$.