# Doubt about the proof of Goursat theorem

Goursat theorem:

$$f\colon A \subseteq \mathbb{C} \to \mathbb{C}$$ holomorphic function in $$A$$ open set.

$$\Delta_0(z_1^{(0)},z_2^{(0)},z_3^{(0)})$$ is a triangle contained in $$A$$ of vertices $$z_1^{(0)},z_2^{(0)},z_3^{(0)} \in A$$.

Then the complex integral of $$f$$ over the perimeter of the triangle is $$\oint_{\partial\Delta_0} f(z)\,\text{d}z=0$$.

In order to show that, split $$\Delta_0$$ into four triangles $$\Delta^1,\Delta^2,\Delta^3,\Delta^4$$ like in the picture:

where $$E,D,F$$ are the midpoints of the respective sides.

Obviously $$\oint_{\partial\Delta_0} f(z)\,\text{d}z=\sum_{i=1}^4 \oint_{\partial\Delta^i} f(z)\,\text{d}z$$, and then $$|\oint_{\partial\Delta_0} f(z)\,\text{d}z| \le \sum_{i=1}^4 |\oint_{\partial\Delta^i} f(z)\,\text{d}z|$$.

So there is a triangle $$\Delta_1 \in \{\Delta^1,\Delta^2,\Delta^3,\Delta^4\}$$ such that $$|\oint_{\partial\Delta_1} f(z)\,\text{d}z| \ge \frac{1}{4}|\oint_{\partial\Delta_0} f(z)\,\text{d}z|$$.

Now repeat the above procedure to $$\Delta_1$$, and so on...

By mathematical induction we are able to find a sequence of nested triangles $$\Delta_0 \supseteq \Delta_1 \supseteq \dots \supseteq \Delta_n \supseteq \dots$$ such that:

$$|\oint_{\partial\Delta_n} f(z)\,\text{d}z| \ge \frac{1}{4^n}|\oint_{\partial\Delta_0} f(z)\,\text{d}z|$$.

This is what my textbook states (I'm using Rudin, but this approach is also used in Lang and Ahlfors, though triangles are replaced by rectangles).

My question is: How can we RIGOROUSLY use mathematical induction here to show that such a sequence exists? This approach doesn't seem enough precise to me.

Here is my attempt to "improve" this proof:

Let's define recursively $$\Delta_{n+1} \in \Big\{\Delta\left(z_1^{(n)},\frac{z_1^{(n)}+z_2^{(n)}}{2},\frac{z_1^{(n)}+z_3^{(n)}}{2}\right),\Delta\left(\frac{z_1^{(n)}+z_2^{(n)}}{2},z_2^{(n)},\frac{z_2^{(n)}+z_3^{(n)}}{2}\right),$$

$$\Delta\left(\frac{z_1^{(n)}+z_3^{(n)}}{2},\frac{z_2^{(n)}+z_3^{(n)}}{2},z_3^{(n)}\right),\Delta\left(\frac{z_1^{(n)}+z_2^{(n)}}{2},\frac{z_2^{(n)}+z_3^{(n)}}{2},\frac{z_1^{(n)}+z_3^{(n)}}{2}\right)\Big\}$$

such that

$$|\oint_{\partial\Delta_{n+1}} f(z)\,\text{d}z| \ge \frac{1}{4}|\oint_{\partial\Delta_n} f(z)\,\text{d}z|$$.

NOW (after we have explicitly defined the nested triangles) we can use mathematical induction showing that :

$$\forall \,n \in \mathbb{N} \quad |\oint_{\partial\Delta_n} f(z)\,\text{d}z| \ge \frac{1}{4^n}|\oint_{\partial\Delta_0} f(z)\,\text{d}z|$$.

Am I totally wrong? Thank you!

• I think this is one of the cases where absolutely strict mathematical rigor does nothing to improve the proof, but rather detracts from the actual content of the proof. For instance, I didn't bother actually going through your set of four triangles to find out wether their corners are correct. I just know that they are either correct, or can be easily corrected by a majority of undergrad students. And most mathematicians would just skip the induction with the words "it follows by induction" for the same reason. Aug 10, 2020 at 14:48
• You are absolutely right. I'm only trying to make all the "obscure" steps explicit as a pure exercise. However, I'm glad to hear that my method is correct. Thank you! Aug 10, 2020 at 15:15