Show the following:
Let $\Delta \subset \mathbb C$, be a triangle and let $f:\Delta \rightarrow \mathbb C$ be continous. Futhermore, assume that $f$ is holomorphic in the interior of $\Delta$. Then $$\int _{\partial \Delta}f = 0$$
My attempt: By Goursat Theorem the result is valid for any triangle in the interior of $\Delta$. But, I do not know how to extend this result in everywhere $\Delta$.
¡Any help would be awesome!
I would probably complete your argument like this:
Since $f$ is continuous on $\Delta$, and $\Delta$ is compact, we know that $f$ is uniformly continuous on $\Delta$. So for any given $\epsilon > 0$, there exists a $\delta > 0$ such that $$ |x - x_0 | < \delta \implies |f(x) - f(x_0)| < \epsilon $$ for all $x , x_o \in \Delta$.
If $\Delta_{\delta / 2}$ is a "slightly smaller version of $\Delta$" - more precisely, if $\Delta_{\delta/ 2}$ is a triangle in the interior of $\Delta$ whose centre coincides with the centre of $\Delta$ and whose corners are located at most a distance $\delta / 2$ from the corresponding corners of the original triangle $\Delta$ - then $$ | \ \int_{\partial \Delta} f - \frac L {L_{\delta/2}} \int_{\partial \Delta_{\delta / 2}} f \ | < L \epsilon,$$ where $L$ is the total perimeter of $\Delta$ and $L_{\delta/2}$ is the total perimeter of $\Delta_{\delta/2}$.
[The funny looking factor of $L/L_{\delta/2}$ is there to compensate for the fact that the lengths of the two contours are slightly different.]
But as you said, $\int_{\partial \Delta_{\delta / 2}} f = 0$ by Goursat's lemma. Therefore, $$ | \int_{\partial \Delta} f \ | < L \epsilon.$$
Since $\epsilon$ was chosen arbitrarily, we conclude that $$ \int_{\partial \Delta} f = 0.$$