|
|
|
|
proof of Banach fixed point theorem
|
(Proof)
|
|
|
Let be a non-empty, complete metric space, and let be a contraction mapping on with constant . Pick an arbitrary , and define the sequence
by
. Let
. We first show by induction that for any ,
For , this is obvious. For any , suppose that
. Then
by the triangle inequality and repeated application of the property
of . By induction, the inequality holds for all .
Given any
, it is possible to choose a natural number such that
for all , because
as
. Now, for any (we may assume that ),
so the sequence is a Cauchy sequence. Because is complete, this implies that the sequence has a limit in ; define to be this limit. We now prove that is
a fixed point of . Suppose it is not, then
. However, because converges to , there is a natural number such that
for all . Then
contradiction. So is a fixed point of . It is also unique. Suppose there is another fixed point of ; because
,
. But then
contradiction. Therefore, is the unique fixed point of .
|
"proof of Banach fixed point theorem" is owned by asteroid. [ full author list (2) | owner history (2) ]
|
|
(view preamble)
Cross-references: contradiction, converges, fixed point, limit, implies, Cauchy sequence, natural number, inequality, property, triangle inequality, obvious, induction, sequence, contraction mapping, metric space, complete
This is version 2 of proof of Banach fixed point theorem, born on 2002-11-10, modified 2007-11-10.
Object id is 3581, canonical name is ProofOfBanachFixedPointTheorem.
Accessed 7421 times total.
Classification:
| AMS MSC: | 54A20 (General topology :: Generalities :: Convergence in general topology ) | | | 47H10 (Operator theory :: Nonlinear operators and their properties :: Fixed-point theorems) | | | 54H25 (General topology :: Connections with other structures, applications :: Fixed-point and coincidence theorems) |
|
|
|
|
|
|
Pending Errata and Addenda
|
|
|
|
|
|
|
|
|
|
|