0

Why is a contraction not defined as a function from one metric space $M$ to another one $N$ with the usual property, i.e.: $ \exists k \in (0,1): $

$$ d_{N}(f(x), f(y)) \le k \, d_M(x,y), \quad \forall x,y \in M $$

Arthur11
  • 1,006
gen
  • 1,518
  • It is, see, e.g., Wikipedia (the second definition). – Arthur11 Jun 06 '17 at 15:54
  • 1
    Primary use of contraction is in Banach fixed point theorem. Formulation of the theorem assumes, that contraction composition is well defined. Function can be composed with itself only when image is subset of domain. – Przemek Jun 06 '17 at 15:59
  • 1
    @Przemek that answers my question -- could you post it as an answer? – gen Jun 06 '17 at 16:04

1 Answers1

0

Primary use of contraction is in Banach fixed point theorem. Formulation of the theorem assumes, that contraction composition is well defined. Function can be composed with itself only when image is subset of domain. Note, that if you let $k$ in your definition to be arbitrary large, this will give definition of Lipschitz continuity between metric spaces $M$ and $N$: $$\exists k\in [0,+\infty) \ \forall x,y \in M: \ d_N(f(x),f(y))\leq kd_M(x,y) $$ According to @mTur11 your definition is mentioned on Wikipedia, but it is uncommon. When speaking about contraction in general, $N=M$ is given unless otherwise stated.

Przemek
  • 855