Cauchy Theorem for a triangle

Theorem

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

$\left|\underset{\partial T}{\int }f\right|=0$

Proof

Assume

$\left|\underset{\partial T}{\int }f\right|=c\ge 0$.

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

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

Giving that

$c=\left|\underset{\partial T}{\int }f\right|\le \sum _{r=1}^4\left|\underset{T^r}{\int }f\right|$

Choose $r$ such that

$\left|\underset{\partial T^r}{\int }f\right|\ge \frac{1}{4}c$

Defining $T^r$ as $T_1$, then

$\left|\underset{\partial T_1}{\int }f\right|\ge \frac{1}{4}c$ 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

$T\supset T_1\supset T_2\supset \dots \supset \dots T_n\supset \dots$

satisfying that

$\left|\underset{\partial T_n}{\int }f\right|\ge {\left(\frac{1}{4}\right)}^{n}c$ and $L\left(\partial T_n\right)={\left(\frac{1}{2}\right)}^{n}L\left(\partial T\right)$.

Claim: The nested sequence $\overline{T}\supset \overline{T_1}\supset \overline{T_2}\supset \dots \supset \overline{T_n}\supset \dots$ contains a point $z_0\in \bigcap _{n=1}^{\mathrm{\infty }}\overline{T_n}$. 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 $z_0\in \bigcap _{n=1}^{\mathrm{\infty }}\overline{T_n}$ since each of the $\overline{T_n}$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 $\epsilon >0$ there exists $\delta >0$ such that

$0<|z-z_0|<\delta$ implies $|\frac{f\left(z\right)-f\left(z_0\right)}{z-z_0}-f^{\prime }\left(z_0\right)|<\epsilon$

which can be rewritten as

$0<|z-z_0|<\delta$ implies $|f\left(z\right)-f\left(z_0\right)-f^{\prime }\left(z_0\right)(z-z_0)|<\epsilon |z-z_0|$

For $z\in T_n$ we have $|z-z_0|<L\left(\partial \right)$, and so, by the Estimation Lemma we have that

$\left|\underset{\partial T_n}{\int }\{f\left(z\right)-(f\left(z_0\right)+f^{\prime }\left(z_0\right)(z-z_0))\}dz\right|\le \epsilon L^2\left(\partial \right)$

As $f\left(z_0\right)+f^{\prime }\left(z_0\right)(z-z_0)$ is of the form $\alpha z+\beta$ it has an antiderivative in D, and so $\underset{\partial T_n}{\int }f\left(z_0\right)+f^{\prime }\left(z_0\right)(z-z_0)=0$, and the above is then just

$\left|\underset{\partial T_n}{\int }f\left(z\right)dz\right|\le \epsilon L^2\left(\partial \right)$

Notice that

${\left(\frac{1}{4}\right)}^{n}c\le \left|\underset{\partial T_n}{\int }f\left(z\right)dz\right|\le \epsilon L^2\left(\partial \right)={\left(\frac{1}{4}\right)}^n\epsilon L^2\left(\partial \right)$

Giving

$c\le \epsilon L^2\left(\partial \right)$

Since $\epsilon >0$ can be chosen arbitrary small, then $c=0$.