×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

## ANALYSIS II

by: Cassidy Grimes

22

0

6

# ANALYSIS II MATH 555

Cassidy Grimes

GPA 3.51

Staff

These notes were just uploaded, and will be ready to view shortly.

Either way, we'll remind you when they're ready :)

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
6
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 6 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 555 at University of South Carolina - Columbia taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/229550/math-555-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

×

## Reviews for ANALYSIS II

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/26/15
Connectedness in Metric Space lof2 httpwwwmath 39 ANALYSIS II Metric Spaces Connectedness Defn A disconnection of a set A in a metric space Xd consists of two nonempty sets A1 A2 whose disjoint union is A and each is open relative to A A set is said to be connected if it does not have any disconnections Example The set 012 Ul2l is disconnected in the real number system Theorem Each interval open closed halfopen I in the real number system is a connected set Pf Let A1A2 be a disconnection for I Let a e A j 12 We may assume WLOG that a1 lt a2 otherwise relabel A1 and A2 Consider E1 X 6 A1 X S a2 then E1 is nonempty and bounded from above Let a supEl But a1 S a S a2 implies a e I since I is an interval First note that by the lemma to the least upper bound property either a 6 A1 or a is a limit point of A1 In either case a 6 A1 since A1 is closed relative to I Since A1 is also open relative to the interval I then there is an 8 gt 0 so that Nga 6 A1 But then a82 6 A1 and is less than a2 which contradicts that a is the sup ofEl 1 Theorem If A is a connected subset of real numbers with the standard metric then A is an interval Pf Otherwise there would be a1 lt a lt a2 with aj e A and a does not belong to A But then 01 ooamA and O2 aoomA form a disconnection of A L Note Each open subset of IR is the countable disjoint union of open intervals This is seen by looking at open components maximal connected sets and recalling that each open interval contains a rational Relatively with respect to A g IR open sets are just restrictions of these Theorem The continuous image of a connected set is connected Pf If C is a connected set in a metric space X and there is a disconnection of the image fC then it can be drawn back39 to form a disconnection of C if 01 O2 forms a disconnection for fC then f101f102 forms a disconnection for C 1 Corollary Intermediate Value Theorem Suppose f is a realvalued function which is continuous on an interval I If a1 a2 6 I and y is a number between fa1 and fa2 then there exists a between a1 and a2 such that fa y 04242000 1031 PM Connectedness in Metric Space httpWWW math d 39 html Pf We may assume WLOG that I a1 a2 We know that fI is a closed interval say 11 Any number y between fa1 and fa2 belongs to I1 and so there is an a e a1a2 such that fa y L Corollary The continuous image of a closed and bounded interval ab is an interval cd where cminaSXSbfX dmaXaSXsb fX Corollary Fixed Point Theorem Suppose that fab gt ab is continuous then f has a fixed point ie there is an x 6 ab such that foc x Pf Consider the function gX X fX then ga S 0 S gb g is continuous on ab so by the Intermediate Value Theorem there is an 0L 6 ab such that g0c 0 This implies that f0c 0L 1 Robert Sharpley Feb 21 1 998 2 of2 04242000 1031 PM 10f4 httpWWW ma isceduNsharpleyma l 5LecturesRSiExistence html ANALYSIS II RiemannStieltjes Integration Conditions for Existence In the previous section we saw that it was possible for on to be discontinuous but for the Reiemann Stieltjes integral of f to still exist The following example shows that the integral may not exist however if both f and or are discontinuous at a point Exam ple Let f or where fx is one for nonnegative x and zero otherwise In this case if P is any partition UP39foc 1 while LP39foc 0 This shows that the RiemannStieltjes integral for this pair does not exist Theorem A necessary and sufficient condition for f to be Riemann Stieltjes integrable with respect to x is for each given 8 gt 0 that one can obtain a partition P of ab such that UPfoc LPfoc lt 8 Pf First we show that is a sufficient condition This follows immediately since for each 8 gt 0 that there is a partition P such that holds b b U f fxd0cx L Jr fx d0cx s UPf0c LPf0c lt 2 Since 8 gt 0 was arbitrary then the upper and lower RiemannStieltjes integrals of f must coincide To prove that is a necessary condition for f to be Riemann integrable we let 8 gt 0 By the definition of the upper RiemannStieltjes integral as a in mum of upper sums we can find a partition P1 of ab such that r b r b J fxd0cx s UP1foc lt J fxd0cx e2 Similarly we have r J b b fx d0Lx 82 lt LP2foc S r x detx J 04242000 1054 PM httpWWW ma rsceduNshm pleymathSSSLecturesRSiExistence 1mm Let P be a common re nement of P1 and P2 then subtracting the two previous inequalities implies UPfoc LPfoo s UP1foc LP2f00 lt e 3 Theorem If f is continuous on ab then f is RiemannStieltjes integrable with respect to x onab Pf We use the condition to establish the proof If 8 gt 0 we set 80 810Lb 0La Since fis continuous on ab f is uniformly continuous Hence there is a 5 gt 0 such that fyfX lt 80 if yX lt 5 Suppose that P lt 5 then it follows that Mi mi lt 80 l S i S n Hence II Unmet Loam 2 Mi mi Aoci lt 80 0Lb0La lt a 1 i 1 Theorem If f is monotone and x is continuous on ab then f is RiemannStieltjes integrable with respect to x on ab Pf We prove the case for f monotone increasing and note that the case for monotone decreasing is similiar We again use the condition to prove the theorem If 8 gt 0 we set 80 8lfbfa Since 0c is continuous and ab is compact 0c is uniformly continuous So for 80 we can determine a 5 gt 0 so that if P is a partition with lt 5 then Adi lt 80 all i The function f is monotone increasing on ab so Mi fXi and mi fXi1 Hence n UPf0 LPf0 2 Mi 39 mi A08 i l n 2 xi 39 fXi1 Adi I l 1 H lt 20 2 fxifxi1 il g 80 fb fa lt a I Defn A RiemannStieltjes sum for f with respect to x for a partition P of an interval ab is defined by 2 of4 04242000 1054 PM httpWWW mathsceduNsharpleymatl155 5LecturesRSiExistence html 1391 RPE 2 f53 406 j 1 where the gj satisfying Xj1 S j S Xj l Sj S n are arbitrary Corollary Suppose that f is RiemannStieltjes integrable on ab then there is a unique number 7 la b f doc such that for every 8 gt 0 there eXists a partition P of ab such that if P S P1P2 then i 0 S UP1fx ylt 8 ji 0 S y LP2fcc lt a m yRP1 lt e where RP1 is any Riemann Stieltjes sum of f with respect to x for the partition P1 In this case we can interpret the integral as b Jr f doc lim RP a quotPM gt 0 although a careful proof is somewhat involved Pf Since LP2f CL S y S UP1f 0L for all partitions we see that parts i and ii follow from the de nition of the integral To see part iii we observe that mJ S fEJ S MJ and hence that LP1foc s RP1Es UP1foc But we also know that both LP1foc s v s UP1foc and condition hold from which part iii follows 1 Robert Sharpley Feb 25 I 998 3 of4 04242000 1054 PM httpWWW ma 1sceduNsharpleyma 155 5LecturesRSiExistence html 4 of4 04242000 1054 PM

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Amaris Trozzo George Washington University

#### "I made \$350 in just two days after posting my first study guide."

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Parker Thompson 500 Startups

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com