Page 87 - DMTH401_REAL ANALYSIS

Unit 6: Completeness

6.4 Contraction Mapping Theorem Notes

Definition: f : M  M contraction if  k < 1 s.t.
d(f(x), f(y))  kd(x, y) " x, y  M

Theorem: Banach
If f contraction on complete metric M then f has unique fixed point.
Proof: Uniqueness: If f(x) = x, f(y) = y then
d(x, y) = d(f(x), f(y))  kd(x, y)  d(x, y) = 0

Existence: Pick x  M, x = f(f ). By repeated application of the contraction property we get that
0 n+1 n
j
j
d(x, x )  k d(x , x ). å  j 1 k d(x , x )  so (x ) Cauchy.
=
j j+1 0 1 0 1 n
M complete so x  x  M, so f(x )  f(x). But also f(x ) = x  x so f(x) = x.
n n n n+1
6.5 Total Boundedness
Definition: Metric M totally bounded if "  > 0  finite set F  M s.t. M   x F B(x, ) .

Proposition: Subspace M of metric N is totally bounded iff "  > 0  finite H  N s.t. M 
 z H B(z, ) .

Proof: () Obvious.

 
() Given  > 0 let H  N be finite set s.t. M   z H B (z, ) . From each non-empty M  B(z, )

2 2
pick one point. Let F be set of these points.
F M finite.
  
If y  M then y in one of B (z, ) so M B (z, )  so  x M  B (z, ). Hence y  B(x, ) and
2 2 2
M   z F B(z, ) .

Corollary: Subspace of totally bounded metric space is totally bounded.

Theorem: Metric M totally bounded iff every sequence in M has Cauchy subsequence.
Proof: () Let x  M. M covered by finitely many balls radius 1/2 so  B s.t. N = {n   : x  B }
n 1 1 n 1
has |N | = .
1
Suppose inductively have defined infinite N  . Since M covered by finitely many balls of
k–1
1
radius  one ball B s.t. N = {n  N : x  B } is infinite.
2k k k k–1 n k
Let n(1) be least element of N , n(k) least element of N s.t. n(k) > n(k – 1).
1 k
1

Then (x (k) )   ( )  s.t. " k x  B for i k. Hence d(x , x ) < " i, j k so (x ) Cauchy.

x
n n 1 n n 1= n(i) k n(i) n(j) k n(k)
=
() Suppose M not totally bounded. Then for some  > 0 /  finite F with all points of M within
 of it. Choose x  M, inductively x s.t. d(x , x )  " i < k. x exists by assumption M not totally
1 k k i k
bounded.
This gives sequence (x )  = s.t. d(x , x)   " i  j. Then no subsequence of (x ) Cauchy.
k k 1
k
i
j
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