MHB Why is the contraction constant important in the Banach contraction principle?

ozkan12
Messages
145
Reaction score
0
It was important in the proof of BCP that the contraction constant h be strictly less than 1. That gave us control over the rate of convergence of f^n (x0) to the fixed point since h^n goes to 0 as n goes to infinity...If we consider f is contractive mapping instead of a contraction, then we lose that control and indeed a fixed point need not exist...But I didnt understand this case...how is contraction constant h important ? I don't understand...please talk about this
 
Physics news on Phys.org
ozkan12 said:
It was important in the proof of BCP that the contraction constant h be strictly less than 1. That gave us control over the rate of convergence of f^n (x0) to the fixed point since h^n goes to 0 as n goes to infinity...If we consider f is contractive mapping instead of a contraction, then we lose that control and indeed a fixed point need not exist.
As far as I know, the term "contractive mapping", as distinct from "contraction", or "contraction mapping", does not have a universally accepted meaning in English. You'll have to give us its definition. Do you mean a "nonexpansive map"?

ozkan12 said:
how is contraction constant h important ?
You said it yourself (I am guessing): if $h\ge 1$, a fixed point need not exist.

ozkan12 said:
I don't understand...
What exactly?
 
no I say contractive mapping...the definition of contractive mapping: let f:X to X be self mapping on (X,d) metric space
if d(fx,fy)<d(x,y) then f is contractive mapping and note that in this concept 'h' contraction constant equal to 1 and inequality is strict inequality
 
ozkan12 said:
let f:X to X be self mapping on (X,d) metric space
if d(fx,fy)<d(x,y) then f is contractive mapping
OK. So, what is your question?
 
It was important in the proof of BCP that the contraction constant h be strictly less than 1. That gave us control over the rate of convergence of f^n (x0) to the fixed point since h^n goes to 0 as n goes to infinity...If we consider f is contractive mapping instead of a contraction, then we lose that control and indeed a fixed point need not exist...

how this happened ? that is why if we choose contractive map we lost control of rate of convergence of f^n(x0)

x0 is arbitrary point in (X,d) metric space
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K