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:

enter image description here

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


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!

  • 1
    $\begingroup$ 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. $\endgroup$ Aug 10, 2020 at 14:48
  • $\begingroup$ 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! $\endgroup$
    – Leonardo
    Aug 10, 2020 at 15:15

1 Answer 1


Unless there is some typo, your proof is absolutely correct and indeed more rigorous that the original proof.

However, often times in mathematics, we only give the very first step, and then (in a not so rigorous manner) say that we can inductively continue this process. Mathematicians believe that the reader of the proof has the capacity to interpret his proof in a more detailed way. This "sloppy" way of proof-writing improves readability drastically. Even though your proof is correct, it is much harder to read and does not contain any new idea.

As a student myself, I believe it is indeed hard to find out when it is alright to give up on rigorosity for readability. But it gets better over time!

  • $\begingroup$ Thank you so much! Yes, I'm aware that an excess of rigor could make things even more difficult. However, I like to try to make all the "obscure" steps explicit, in order to better understand the proof. $\endgroup$
    – Leonardo
    Aug 10, 2020 at 15:03
  • $\begingroup$ @Leonardo I understand you. I often do so myself. $\endgroup$
    – Zuy
    Aug 10, 2020 at 15:04

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