Renato Caccioppoli

Sometimes it’s better to express with no words but listen to the music.

Let $(\mathcal{H}, d)$ a non null complete metric space and $T: (\mathcal{H}, d)\rightarrow (\mathcal{H}, d)$ such as $\forall x,y\in (\mathcal{H}, d),\ d(T(x), T(y)) with $0 < L < 1,$ then $\exists!\ x^* \in (\mathcal{H}, d): x^* =T(x^*).$

In other words$,T$ is a particular map over the metric space $(\mathcal{H}, d)$ known as contraction. The theorem ensures the contraction of a metric space admits only one fixed point$,x^*.$ It is known as Banach-Caccioppoli fixed-point theorem.

The trick is to define a particular succession of elements in the metric space. Starting from a generic element $x_0\in(\mathcal{H}, d)$ define a recursive succession: $x_1=T(x_0),x_2=T(x_1),\ldots x_{n+1}=T(x_n).$

First we measure the distance between two consecutive element:

$\forall n\in \mathbb N :d(x_{n+1},x_n)=d(T(x_n),T(x_{n-1}))

As you can see the distance is going smaller and smaller as $n\rightarrow \infty$, it suggests the succession is converging somewhere. (We implicitly use in this step the induction theorem, to prove the previous relation is true $\forall n\in \mathbb N$)

Now let measure the distance between two generic elements of the succession:

$\forall m>n \in \mathbb N :d(x_m,x_n)

This relation holds because of the previous one and the triangular inequality.
The term  $\sum \limits_{i=0}^{m-1-n}L^i$ is the partial sum of the geometric series, so sending $m\rightarrow \infty$ we have: $d(x_m,x_n) $\frac{L^n}{1-L}.$ If also $n\rightarrow \infty$ we finally have:

$d(x_m,x_n) $\frac{L^n}{1-L}$ $\rightarrow 0$

We’ve shown that the distance between two generic elements of the succession is approaching to zero as we send the elements far away to infinity. Infact the succession converges to an element $x^* \in (\mathcal{H}, d)$ thanks to the Cauchy criterion.

$\lim \limits_{n \to \infty}x_n=x^*$

The transformed succession $T(x_n)$ converges to the same point, from the definition of the map we take the limit of both sides:

$x^*=\lim \limits_{n \to \infty}T(x_n)=T(\lim \limits_{n \to \infty}x_n)=T(x^*)$

(Note: you can pass the limit through  $T$ because it is uniformly continuous).

The last relation prove that  $x^* \in (\mathcal{H}, d)$ is a fixed point.

Suppose exists another fixed point  $y^* \in (\mathcal{H}, d)$ so:

$d(x^*,y^*)=d(T(x^*),T(y^*))

It is a contradiction because $0 < L < 1$. So the fixed point $x^*$ is unique.

In order to visualize the fixed-point theorem, let make a simple example. Let  $(\mathbb{R}, \mid \ \mid)$ the metric space of the real number where the distance is defined by the absolute value of the difference and $T(x) =x/2$ is contraction over the metric space (it could be any linear transformation with an angular coefficient minor than 1).

You can see “the contraction” of the space in itself. The fixed point is  $x=0$ and it seems like the vanishing point used in the perspective.

Renato Caccioppoli was a gifted neapolitan mathematician. This post is a little tribute to him.