# Cauchy Theorem for a triangle

Jump to navigation Jump to search

## Theorem

Let ${\displaystyle D\subseteq \mathbb {C} }$ be a domain, ${\displaystyle f:D\to \mathbb {C} }$ a differentiable function. Let ${\displaystyle T}$ be a triangle such that ${\displaystyle {\bar {T}}\subset D}$. Then

${\displaystyle \left|\int \limits _{\partial T}f\right|=0}$

## Proof

Assume

${\displaystyle \left|\int \limits _{\partial T}f\right|=c\geq 0}$.

It will be shown that ${\displaystyle c=0}$.

First, subdivide ${\displaystyle T}$ into four triangles, marked ${\displaystyle T^{1}}$, ${\displaystyle T^{2}}$, ${\displaystyle T^{3}}$, ${\displaystyle T^{4}}$ by joining the midpoints on the sides. Then it is true that

${\displaystyle \int \limits _{\partial T}f=\sum \limits _{r=1}^{4}\left(\int \limits _{T^{r}}f\right)}$.

Giving that

${\displaystyle c=\left|\int \limits _{\partial T}f\right|\leq \sum \limits _{r=1}^{4}\left|\int \limits _{T^{r}}f\right|}$

Choose ${\displaystyle r}$ such that

${\displaystyle \left|\int \limits _{\partial T^{r}}f\right|\geq {\frac {1}{4}}c}$

Defining ${\displaystyle T^{r}}$ as ${\displaystyle T_{1}}$, then

${\displaystyle \left|\int \limits _{\partial T_{1}}f\right|\geq {\frac {1}{4}}c}$ and ${\displaystyle L\left(\partial T_{1}\right)={\frac {1}{2}}L\left(\partial T\right)}$

(where ${\displaystyle L\left(\gamma \right)}$ describes length of curve).

Repeat this process of subdivision to get a sequence of triangles

${\displaystyle T\supset T_{1}\supset T_{2}\supset \ldots \supset \ldots T_{n}\supset \ldots }$

satisfying that

${\displaystyle \left|\int \limits _{\partial T_{n}}f\right|\geq \left({\frac {1}{4}}\right)^{n}c}$ and ${\displaystyle L\left(\partial T_{n}\right)=\left({\frac {1}{2}}\right)^{n}L\left(\partial T\right)}$.

Claim: The nested sequence ${\displaystyle {\bar {T}}\supset {\bar {T_{1}}}\supset {\bar {T_{2}}}\supset \ldots \supset {\bar {T_{n}}}\supset \ldots }$ contains a point ${\displaystyle z_{0}\in \bigcap \limits _{n=1}^{\infty }{\bar {T_{n}}}}$. On each step choose a point ${\displaystyle z_{n}\in T_{n}}$. Then it is easy to show that ${\displaystyle \left(z_{n}\right)}$ is a Cauchy sequence. Then ${\displaystyle \left(z_{n}\right)}$ converges to a point ${\displaystyle z_{0}\in \bigcap \limits _{n=1}^{\infty }{\bar {T_{n}}}}$ since each of the ${\displaystyle {\bar {T_{n}}}}$s are closed, hence, proving the claim.

We can generate another estimate of ${\displaystyle c}$ using the fact that ${\displaystyle f}$ is differentiable. Since ${\displaystyle f}$ is differentiable at ${\displaystyle z_{0}}$, for a given ${\displaystyle \varepsilon >0}$ there exists ${\displaystyle \delta >0}$ such that

${\displaystyle 0<\left|z-z_{0}\right|<\delta }$ implies ${\displaystyle \left|{\frac {f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}}-f'\left(z_{0}\right)\right|<\varepsilon }$

which can be rewritten as

${\displaystyle 0<\left|z-z_{0}\right|<\delta }$ implies ${\displaystyle \left|f\left(z\right)-f\left(z_{0}\right)-f'\left(z_{0}\right)\left(z-z_{0}\right)\right|<\varepsilon \left|z-z_{0}\right|}$

For ${\displaystyle z\in T_{n}}$ we have ${\displaystyle \left|z-z_{0}\right|, and so, by the Estimation Lemma we have that

${\displaystyle \left|\int \limits _{\partial T_{n}}\left\{f\left(z\right)-\left(f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)\right)\right\}dz\right|\leq \varepsilon L^{2}\left(\partial \right)}$

As ${\displaystyle f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)}$ is of the form ${\displaystyle \alpha z+\beta }$ it has an antiderivative in D, and so ${\displaystyle \int \limits _{\partial T_{n}}f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)=0}$, and the above is then just

${\displaystyle \left|\int \limits _{\partial T_{n}}f\left(z\right)dz\right|\leq \varepsilon L^{2}\left(\partial \right)}$

Notice that

${\displaystyle \left({\frac {1}{4}}\right)^{n}c\leq \left|\int \limits _{\partial T_{n}}f\left(z\right)dz\right|\leq \varepsilon L^{2}\left(\partial \right)=\left({\frac {1}{4}}\right)^{n}\varepsilon L^{2}\left(\partial \right)}$

Giving

${\displaystyle c\leq \varepsilon L^{2}\left(\partial \right)}$

Since ${\displaystyle \varepsilon >0}$ can be chosen arbitrary small, then ${\displaystyle c=0}$.