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## ANALYSIS I

by: Cassidy Grimes

27

0

2

# ANALYSIS I MATH 554

Cassidy Grimes

GPA 3.51

R. Sharpley

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COURSE
PROF.
R. Sharpley
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 554 at University of South Carolina - Columbia taught by R. Sharpley in Fall. Since its upload, it has received 27 views. For similar materials see /class/229529/math-554-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

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Date Created: 10/26/15
MORE ON COMPACTNESS Handout 7 part B Defn 1 A function f is called Lipschz39tz if there is an M gt 0 such that lflt1 952M S M951 9527 for 8L119517952 E d0mltfgt If M lt 1 then f is called a contraction Theorem 1 Each LipschitZ function is uniformly continuous Theorem 2 Suppose that K is compact and f K gt K is a contraction then f has a xed point in K Proof Let me be an arbitrary point in K De ne inductively an acn n 012 We claim that the sequence xni 1 is convergent to some oz 6 K First note that for each n E W lfrn1 Inl ffrn ffrn71 S M lfrn frn71 Hence by induction for each n 6 EV xn1 xn S Mnx1 x0 We then see that if m gt n then m n k Where k 6 W and low MI E 95nk nk71l lfrnk71 nk72 lmn1 95 ltMnkil AJTLJrk 2 l lxl xol M 1MMk71l951x0l l1 71 l n M and so xni 1 is Cauchy It must converge to some limit oz which will belong to K since K is closed But f is continuous so an gt fa Notice also that an gt oz so oz is our xed point D Theorem 3 Suppose that f a b gt K is one to one onto and continuous then f 1 is continuous Proofl Suppose that g f 1 and yn gt yo 6 K There exists unique 05 6 la b such that yn or equivalently 05 gyn If 05 74gt 950 then there exists 60 gt 0 and a subsequence xnk such that xnk x0 Z 60 This sequence in turn has a subsequence which converges in K to some 2 E K We may as well assume that the subsequence is the sequence f is continuous so ynk gt But then yo fx0 f is one to one so 2 950 which is a contradiction since xnk x0 Z 60 D Pr00f2 Let 9 Q cub be relatively open then f 1 19 Let C be the complement in cg b of 9 then C is closed and hence compact Therefore f C is compact in K and consequently it is closed lts complement in K must then be open That complement however is f D

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