Cauchy Theorem for a triangle

Theorem

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

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

Assume

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

It will be shown that $c=0$ .

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

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

Giving that

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

$\left|\int \limits _{\partial T^{r}}f\right|\geq {\frac {1}{4}}c$ Defining $T^{r}$ as $T_{1}$ , then

$\left|\int \limits _{\partial T_{1}}f\right|\geq {\frac {1}{4}}c$ and $L\left(\partial T_{1}\right)={\frac {1}{2}}L\left(\partial T\right)$ (where $L\left(\gamma \right)$ describes length of curve).

Repeat this process of subdivision to get a sequence of triangles

$T\supset T_{1}\supset T_{2}\supset \ldots \supset \ldots T_{n}\supset \ldots$ satisfying that

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

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

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

$0<\left|z-z_{0}\right|<\delta$ implies $\left|{\frac {f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}}-f'\left(z_{0}\right)\right|<\varepsilon$ which can be rewritten as

$0<\left|z-z_{0}\right|<\delta$ implies $\left|f\left(z\right)-f\left(z_{0}\right)-f'\left(z_{0}\right)\left(z-z_{0}\right)\right|<\varepsilon \left|z-z_{0}\right|$ For $z\in T_{n}$ we have $\left|z-z_{0}\right| , and so, by the Estimation Lemma we have that

$\left|\int \limits _{\partial T_{n}}\left\{f\left(z\right)-\left(f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)\right)\right\}dz\right|\leq \varepsilon L^{2}\left(\partial \right)$ As $f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)$ is of the form $\alpha z+\beta$ it has an antiderivative in D, and so $\int \limits _{\partial T_{n}}f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)=0$ , and the above is then just

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

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

$c\leq \varepsilon L^{2}\left(\partial \right)$ Since $\varepsilon >0$ can be chosen arbitrary small, then $c=0$ .