Page 86 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Proof:

ì ì n , 1 ü x 0
>
ý
f (x) = ï í min í x þ
n î
ï n x = 0
î
so (f )  C[0, 1].
n
1 1 1 n æ 1 ö
-
0 ò |f (x) f (x)|dx = ( m - n ) dx + m è x - n dx
m
0 ò
1 ç ò
m
n
÷
ø
1 2
 +
m n
3
 ¾¾¾ 0
n n
so (f ) Cauchy.
n
Let f  C[0, 1]. Find k  s.t. |f|  k. Then for n > k
1 1 n æ 1 ö
1
0 ò |f (x) f (x)|dx = ( m - n ) dx + 1 ç ò m è x - n dx
m
-
0 ò
÷
m
n
ø
æ 1 1 ö 1
 2 ç è k - n ø ÷ - k
1 2
= - ¾¾¾ > 0
n
k n
6.3 Completion
Definition: S  M is dense in M if S = M.
Definition: A completion of metric space M is:

 Complete metric space N s.t. M dense subset of N.
 Complete metric space N and isometry i: M  A  N s.t i(M) is dense in N.
Theorem: Any metric M can be isometrically embedded into complete metric space.

Proof: Find isometry of M onto subset of B(M), complete. Fix a  M, define F : M  B(M) by F(x)(z)
= d(z,x) – d(z, a). |F(x)(z)|  d(x, a) so F(x)  B(M).

|F(x)(z) - F(y)(z)| = |d(z, x) – d(z, y)|
 d(x, y)

Equality occurs when z = y. Then ||F(x) – F(y)|| = d(x, y) so F isometry.
Corollary: Any metric space M has a completion.

Proof: Embet M into complete N. Then M (closure taken in N) is complete by 6.2, M dense in M .
Then M completion of M.

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