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 $L\left(\partial T_{1} \right) = \left( \frac{1}{2} \right)^{n} L\left(\partial T\right)$ .

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 $z_{0} \in T_{n}$ . 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 $\left| z-z_{0} \right| < L\left(\partial\right)$ , 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 $c = 0$ .

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