detials

Question 1) Let (X, d) be a non-empty complete metric space, and T : X →X a contraction of X. (That is, there exists a nonnegative real number r < 1 such that d(T(x), T(y)) ≤ r d(x, y) for all x, y in X) Prove that T has exactly one fixed point. (That is, there exists a unique x ϵ X such that T(x) =x).
Solution:
As T:X ( X is a contraction mapping, therefore there exist r, 01.
Step 1 We construct a sequence
Consider any (X
Let,
=
T()
=
T()
=
T()
.
.
.
=
T()
Then, is a sequence in X
Step 2 - We show is a Cauchy sequence :-
Let, ( >0 be given,
To show, ( , such that,
Without loss of generality, let n m,
Consider,
2.
Now by using equation 1 m times,
=
.
.
.
=
(
.
.
.
Substituting these in 2
=
=
=
3
Now, as 0,
(
( corresponding to
i.e.
(
[by 3]
Step 3- As X is complete, ( ( p(X, such that
We assert that T (p) = p
If possible, suppose T(p) p, then d[T(p), p] > 0
Choose, ( such that 0<(As, , such that
Consider
=
<
Which contradicts our choice of (.
( T(p) = p
Step 4
Suppose ( p, q ( X, such that T(p)=p,
T(q)=q
Consider d(p, q) = d[T(p), T(q)]
r d(p,q)
((1-r) d(p, q) 0
Dividing by 1-r >0, we get, d(p,q) 0
Also d(p, q)0
( d(p, q)=0
i.e. p=q
Thus T has exactly one fixed point. This is also called banach’s fixed point theorem.
Question 2) Let (X, d) be a non-empty complete metric...