Lagmental Vicfred

Banach's Fixed-Point Theorem Gives a Unique Contractive Fixed Point by Vicfred

Last updated: Thu 25 April 2019

A contraction on a complete metric space has one fixed point reached geometrically by iteration. The formulas are more useful when each symbol has a job rather than merely decorating the theorem.

Notation

A Banach space \(X\) is complete in its norm, and a bounded linear operator \(T:X\to Y\) has norm \(\|T\|=\sup_{\|x\|\le1}\|Tx\|\). Completeness powers the major structural theorems.

$$ d(Tx,Ty)\le q\,d(x,y),\qquad0<q<1 $$

A reliable calculation names domain and codomain. The notation \(\mathsf{data}\mapsto\mathsf{claim}\) is harmless only after both \(\operatorname{dom}\) and \(\operatorname{cod}\) have been fixed.

$$ \exists!\,x^\ast,\quad Tx^\ast=x^\ast $$

Stress the formula

Here is a concrete symbolic test. Reading it from left to right reveals which transformation is reversible and which is only an implication.

$$ d(x_n,x^\ast)\le\frac{q^n}{1-q}d(x_1,x_0),\qquad x_{n+1}=Tx_n $$

Interpretation

The two-row display is also a debugging tool: if the conclusion changes when only notation changes, some hidden choice has entered the argument.

$$ \begin{aligned} \mathsf{D}\;&:\quad d(Tx,Ty)\le q\,d(x,y),\qquad0<q<1,\\[5pt] \mathsf{C}\;&:\quad \exists!\,x^\ast,\quad Tx^\ast=x^\ast. \end{aligned} $$

Limit of the argument

Finite-dimensional intuition can fail badly in infinite dimensions. Closed, bounded sets need not be compact, and linear maps need not be bounded automatically.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \exists!\,x^\ast,\quad Tx^\ast=x^\ast \end{gathered}} $$

A symbolic summary is trustworthy only because the example and limitation remain visible beside it. The box compresses the conclusion without hiding its origin.

This article was posted on Thu 09 January 2014. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.