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