×

### Let's log you in.

or

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

×

or

## Classical Analysis I

by: Elaina Osinski

7

0

0

# Classical Analysis I MATH 403

Elaina Osinski
Penn State
GPA 3.88

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
0
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 0 page Class Notes was uploaded by Elaina Osinski on Sunday November 1, 2015. The Class Notes belongs to MATH 403 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 7 views. For similar materials see /class/233009/math-403-pennsylvania-state-university in Mathematics (M) at Pennsylvania State University.

×

## Reviews for Classical Analysis I

×

×

### 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: 11/01/15
Math 403 Lecture Notes for Week 3 1 Subsequences De nition 1 Let Ingtn21be a sequence A subsequence of ltfngtn2118 a sequence lt1n1gti21 with ni lt ni1 for all i 2 1 De nition 2 A point y is called a cluster point or accumulation point of a sequence ltfngtn21lf for any 6 gt 0 and any N 2 1 there is an n 2 N such that ly 7 Inl lt e Example 3 The sequence 17 7117 717 H has two cluster points7 1 and 71 Note that the limsup and liminf of a sequence are both cluster points If a sequence has a limit7 then the limit is the unique cluster point of the sequence Proposition 4 The point y is a cluster point of the sequence Ingtn21if and only if there is a subsequence lt1n1gti21 wit liminmInl y Proof First7 suppose y is a cluster point of ltfngtn21 We will nd a subsequence whose limit is y First7 letting e E and N 1 we nd an n1 2 1 with lIn1 7 yl lt Next7 letting e i and N n1 17 we nd an n2 gt n1 with lIn2 7yl lt Continuing7 we nd an increasing sequence ni with lInl 7yl lt Since any 6 gt 0 has 6 lt 3 for some i we see that this subsequence satis es the limit de nition for converging to y Second7 suppose lt1n1gti21 is a subsequence whose limit is y we show y is a cluster point Let 6 gt 0 and N be given We know that there is an M such that for all i 2 M we have lInl 7 yl lt e Choosing 239 large enough so that ni 2 N7 we are done D Hence7 cluster points correspond to convergent subsequences Example 5 Let In n mod 017 ie if n h f where h is an integer and 0 S f lt 17 then In One can show that this sequence is dense in the interval 017 ie for any y E 01 and any 6 gt 0 there is an n with lIn 7 yl lt e Thus any point in 01 is a cluster point of the sequence ltfngtn21 Theorem 6 Bolzano Weierstrass Theorem Every bounded sequence in R has at least one cluster point Proof We will show that a bounded sequence has a convergent subsequence Let Ingtn21be bounded7 and say I gt 0 is such that 7b lt In lt b for all n 2 1 Consider the two intervals L1 7127 0 and R1 012 At least one of these two intervals must contain in nitely many terms of the sequence Choose that interval and let n1 be such that In1 is in that interval Next7 divide that interval in two eeg if the rst interval was L17 we divide it into L2 7127 and R2 7 0 Now one of these two intervals must contain in nitely many terms of the sequence beyond zml Choose this interval and let n2 gt n1 be such that mm is in the chosen intervall We repeat this procedure dividing intervals in two choosing one which contains in nitely many further terms of the sequence and picking one of them to add to our subsequencel At the end we have chosen a subsequence lt17 with the following property 212 for any i and anyj 2 i we have lznl 7 Injl lt since the two points are in the same subinterval obtained by dividing 7b 12 in halfi times This condition will imply that the subsequence mm is a Cauchy sequence de ned below and the Cauchy Convergence Criterion proved below will show that it therefore converges completing the proo l D Spaces which satisfy the Bolzano Weierstrass Theorem are called sequentially compact De nition 7 We say that a sequence lt1nn21is Cauchy if for any 6 gt 0 there is an N such that for any nm 2 N we have lzn 7 zml lt 6 The subsequence de ned above is Cauchy since for a given 6 we can choose N such that 227 lt 6 We start with a preliminary lemma Lemma 8 Nested Intervals Lemma Suppose ltInn21 is a nested sequence of closed bounded nonempty intervals ie each In ambn and In1 Q In for each n Suppose that the lengths of the intervals converges to 0 ie limnnmazn 7 an 0 Then there is a unique real number y with y 6 17211 ie y is in each of the intervals In Proof Since the intervals are nested for any n we have an S an1 S bn1 S bnl Hence the sequence ltan is a monotone increasing sequence bounded above by 121 so it converges say limnn00 an al Similarly the sequence 127 is monotone decreasing and bounded below by a1 so we have limnn00 bn b We claim that a 12 This follows since an S a S b S In for all n so I 7 a S n 7 an since limnnmazn 7 an 0 we must have b 7 a 0 et a 12 We see that y is in each of the intervals In since an S a y b S bnl We nally check that this y is unique Suppose we have another point 2 6 17211 say y lt 2 Again we have an S y lt 2 S In for each n so 27yltbn7anandweseethatz7y0sozyl D Theorem 9 Cauchy Convergence Criterion A sequence ltznn21converqes in R if and only it is Cauchy Proof We want to produce a nested sequence of intervals to which to apply the lemma Let lt1nn21be Cauchy and let 6k Applying the de nition we have numbers Mk such that for every mn 2 Mk we have lIm 7 Inl lt We replace the Mkls by Nkls de ned as Nk max Mj k 151516 The point is that Nk 2 Mk for each 16 and Nk lt Nk1i De ne the intervals Jk as 1 1 Jk INk PINc El Note that for every n 2 Nk we have In E Jki We can ont yet apply the lemma since the Jk7s are not necessarily nestedi So we let In g1JkJ1 J2 jn Note that each In is a nonempty closed interval since it contains INn in par ticular In1 Q In and the lengths of the In7s approaches 0 since In E Jn which has length By the Nested lntervals Lemma there is a point y with y E nleni We clai that limnH00 In yr Let 6 gt 0 be given and let k be large enough so that g lt 6 Let N Nki We then know that for every n 2 N we have both In and y in the interval IN and hence both in the interval JN so that lIniyllt2Elte D Thus we have also nished the proof of the Bolzano Weierstrass Theoremi Note 10 Convergent sequences in many other spaces besides R are Cauchy but the converse can fail For instance in the space Q the sequence In 7r rounded to n decimal digits is a Cauchy sequence in Q but it does not con verge in To be continued Math 403 Lecture Notes for Week 1 1 Ordered Fields De nition 1 Field Axioms A eld is a set X together with two binary functions and de ned on X which satisfy the following properties 1 zyyz E0 1y2zyz There is an element7 denoted 07 such that z 0 z for all 1 Pk For all I there is a w7 denoted 71 such that z w 0 51yyz 53gt 1y2Iyz gt1 There is an element7 denoted 17 such that z 1 z for all x 9 For all z 7amp0 there is a w7 denoted 1 1 such that 110 1 91y2zyzz The real numbers R with the usual addition and multiplication is a eld So are the rationals Q and the complex numbers C with the usual operations Example 2 We can de ne a somewhat more exotic eld R as follows ng 01 has two elements7 with and de ned as follows 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 De nition 3 A strictly totally ordered set consists of a set X and a binary relation lt which satis es 111 2 For all z and y exactly one of the three possibilities holds 1 lt y z y y lt z 3 lfzltyandyltzthenzlt2n De nition 4 Order Axioms An ordered eld consists of a eld X7 and a set P Q X of positive elements satisfying the following 1 lszPandy6P7thenzy P 21f16Pandy6P7thenzyEP 3i lszPthenizg Pi Example 5 2 4 For each I exactly one of the three possibilities holds 1 0 z E P P 1 The reals with the set P z z z gt 0 is an ordered eld The reals with the set P z z z lt 0 is not an ordered eld since 71 E P but 71 71 Pl 3 The rationals Q with P z E Q z gt 0 is an ordered eld 4 The eld ng cannot be ordered ie there is no set P which makes it an ordered eld To see this note that 71 1 so if 1 E P then we need 71 P iiei l P whereas if 1 P we need 716 P ie 16 Pl De nition 6 We can de ne a strict totally ordering on an ordered eld by setting zltyifandonlyifyizEP WewritezSyfor zyorzltyi Note 7 We can restate the ordering axioms in terms of the order lt as follows 1 2i 3 2 lfzltyandzltwthenzzltywi lf0ltzltyand0ltzltwthenzzltywi Exactly one of z lt y z y y lt I Bounded Sets De nition 8 Let A be a subset of an ordered eld 1 2i 3 H 539 533 We say that b is an upper bound of A if I S b for all z 6 Al We say that c is a lower bound of A if c S I for all z 6 Al i We say that b is a least upper bound of A if b is an upper bound of A and for any other upper bound 5 of A we have b S c We say that c is a greatest lower bound of A if c is a lower bound of A and for any other lower bound d of A we have d S c We say that b is the greatest element of A if b is an upper bound of A and b e A We say that c is the least element of A if c is a lower bound of A and c 6 Al Example 9 1 In R the set I z 0 S I lt 1 has 1 as an upper bound and 0 as a lower bound These are respectively a least upper bound and a greatest lower bound The set has a least element 0 but no greatest element 2 In R the set I z z lt 0 does not have a lower bound 3 ln Q the set I E Q lt z lt has an upper bound for instance 2 It does not have a least upper bound since any upper bound 1 must be rational so V3 lt b and hence there is another rational c with V3 lt c lt b This 5 is an upper bound which is less than 12 De nition 10 Completeness Axiom We say an ordered eld X is com plete or satis es the least upper bound property if whenever S is a nonempty subset of X which has an upper bound then S has a least upper bound We denote this least upper bound sup 5 Example 11 The reals are complete The rationals Q is not complete To see this consider the set I E Q lt This has an upper bound but no least upper bound Note 12 The least upper bound property implies the corresponding property for lower bounds Any nonempty set which has a lower bound has a greatest lower bound Note 13 The reals satisfy the least upper bound property In fact the reals are the unique complete ordered eld We will derive many properties of the reals just from the assumption that they form a complete ordered eld Proposition 14 Let L and U be nonempty subsets ofR with R L U U and such that for all l E L and u E U we have l lt u Then either L has a greatest element or U has a least element Proof Let u be any element of U Then u is an upper bound of L Hence by completeness L has a least upper bound b If I E L then L has a greatest element and we are done So suppose not then b E L We claim that b is the least element of U If not there is some u E U with u lt I But then u is an upper bound for L which is less than I contradicting that b was the least upper bound So I is the least element of U in this case B De nition 15 In an ordered eld an integer is a number that is either 0 or of the form 1 1 or of the form 71 1 A rational is an element of the form p q 1 where p and q are integers with q f 0 Proposition 16 Axiom of Archimedes In a complete ordered eld for any I there is an integer n with z lt n Proof If I lt 0 we can take n to be 0 So suppose I 2 0 Let S be the set S h h is an integer with h S I Then S is nonempty since it includes 0 and has an upper bound namely 1 Hence 5 has a least upper bound y The element y 7 is then not an upper bound for 5 so there is some integer k E S with y 7 lt kl But then y lt k lt k 1 so k 1 is not in 5 Since k 1 is an integer this means that z lt k 1 so setting n k 1 works D Corollary 17 The rationals are dense in a complete ordered eld ie ifz lt y then there is a rational r with z lt r lt y Proof First suppose I 2 0 By the Axiom of Archimedes there is an integer q with q gt y 7 z 1 so y 7 z gt 5 Consider the set S with n S n z n is an integer with y S 7 q S is nonempty and bounded below by y 4 so it has a greatest lower bound pi Since 5 consists only of integers p is in fact an element of 5 so p is the least element of 5 This means that 71 Lltygg q q Wealso havezy7y7zlt 7 Lgli Solettingrwehave z lt r lt y If I lt 0 we can nd an integer n with n gt 7r so n z gt 0 As above we then nd a rational r with n z lt r lt n y The number r 7 n is then a rational with z lt r 7 n lt y D 3 In nite Decimals For de niteness we will de ne the reals to be in nite decimals By an in nite decimal we mean a number of the form nidodldg or the negation of such a number where n and each di is an integer and 0 S di S 9 for each i 2 1 This sequence represents the real number a Em 13921 These are added and multiplied with the expected carrying of digits The one point to be careful about is that we must identify eventually 9 sequences and eventually 0 sequences such as 0999 1 000 a With this understanding the collection of in nite decimals forms a complete ordered e di 4 Limits De nition 18 A sequence is a list of real numbers indexed by the positive integers We use the notation ltfngtn21 to denote the sequence 111213 De nition 19 Ne de nition of limit Let Ingtn21be a sequence and y a real number We say that the sequence has limit y and write limnH00 In y if For any 6 gt 0 there is a natural number N such that lIn 7 yl lt e for all n 2 N De nition 20 Let Ingtn21be a sequence We say that the sequence has limit 007 and write limnH00 In 00 if For any h there is a natural number N such that In gt h for all n 2 N We de ne limit 700 similarly Note 21 We shall try to be careful about what we mean when we say a sequence has a limit We shall generally say a sequence has a limit as a real number to indicate that there is a real number y which is the limit of the sequence We will say a sequence has a limit as an eItended real number if we also allow the limit to be ioo Example 22 Consider the sequence with I n Then limnH00 0 To see this7 let 6 gt 0 be given We need to nd an N such that lIn 7 0 lt e for all n 2 N7 is i lt s This is equivalent to ensuring n gt so choosing N gt suf ces De nition 23 We say that a sequence ltIngtn21is bounded above if the set In n 2 l is bounded above We de ne bounded below similarly De nition 24 We say a sequence is monotone increasing if In S In1 for all n 2 1 We de ne monotone decreasing similarly Proposition 25 Let ltfngtn21 be a sequence which is bounded above and mono tone increasing Then ltfngtn21hll8 a limit as a real number Proof Consider the set In n 2 1 It is nonempty and bounded above7 so it has a least upper bound y We claim that limnH00 In y et 6 gt 0 be given We want to nd an N as required in the de nition of limit Suppose there is none Then for any N there is an n 2 N with lIn 7 yl 2 s This means In S y7 s Since the sequence is monotone increasing7 we have IN S In S y7e as well Since this holds for any N7 we have that y 7 e is an upper bound for the sequence7 contradicting that y was a least upper bound D Note 26 The same is true for a sequence which is bounded below and monotone decreasing Note 27 Both conditions in the proposition are necessary For instance7 the sequence 17 7117 717 is bounded7 but has no limit The sequence with In n is monotone but not bounded above7 and has no limit as a real number Note7 though7 that a monotone increasing sequence always has a limit as an extended real number7 since if it is not bounded above it has limit 00 Math 403 Lecture Notes for Week 2 1 Limsup and Liminf De nition 1 Let A Q R If A has an upper bound7 we let supA be the least upper bound of A When A is unbounded we sometimes write supA 00 If A has a lower bound7 we let ian be the greatest lower bound of A7 and write ian 700 when A has no lower bound For a sequence ltfngtn21 we let supngt1 In supIn n 2 l and similarly for in De nition 2 For a sequence ltfngtn21 we de ne the limit superior limsup to limsup E In inf sup In nnm ngtoo n21 nZn We de ne the limit inferior to be liminf lim In sup inf In 400 ngt n21 an Example 3 1 Consider the sequence with In We compute 7 1 1 lim In inf supi inf 7 0 naoo n 1n2n k n21 n 1 limI supinf7sup00 nix 21an k 721 Note that these both are equal to the limit of the sequence to Consider the sequence 17 711 711 H We have T 39 f 39 f 1 1 121212 31 liim In sup inf In supil 71 naoo nzlk n 7121 If a sequence is bounded both above and below then the limsup and liminf both exist as real numbers We can think of the limsup as the largest number which the sequence approaches in nitely often7 and limsup the smallest Proposition 4 liimn Do In g End In Proof Set 39 f yn klrzln 9 2n sup In kgn so that lim inf In sup yn n21 l39 39 f 1m sup In 7131 2n Note that yn S 2n for all n and if n S m then yn S ym and 2m S zni From this we see that yn S 2m for any two numbers n and m so that lim inf In S lim sup In D Note 5 This proposition is also true for unbounded sequences with the con vention that 700 S 700 lt 7 lt 00 S 00 for any real number Ti Proposition 6amp7 a sequence ltfngtn21 we have limnH00 In y and only if himn x In limnaoo In y Proof Suppose lim In y we will show that liimnH00 In Mm y We consider lim sup rst For any 6 gt 0 there is an N such that lIn 7 yl lt e for all n 2 Ni Hence supngtN In S y 6 so limsupIn inf supk 2 nIn S ye n21 Since this is true for any 6 gt 0 we have limsupI S y Similarly we nd lim inf In 2 yr Since we know liminf In S lim sup In we conclude that they both equal yr For the other direction suppose liimnH00 In End In y we will show that limIn yr Suppose not so there is some 6 gt 0 so that for all N there is an n 2 n with lIn 7 yl 2 6 ie either In 2 y e or In S y 7 6 Hence there are in nitely many n where this happens so there are either in nitely many n for which In 2 y e or there are in nitely many n for which In S y 7 e or both In the rst case we see that limsup In 2 y e and in the second case we have liminf In S y 7 e contradicting that lim sup In lim inf In yr Proposition 7 For a sequence ltfngtn21 lim inf In 7 lim sup 7In Proof We compute l39 39 f 39 f 1m1n In In sup infIn k 2 n n21 sup 7 sup7In k 2 n from homework n21 7 39 f 7 z k gt iglsup In S n 7 39 f 7 251212 3 7 lim sup 7In 2 Properties of Limits Proposition 8 Let Ingtn21zmd ltyngtn21 be sequences with limIn a and limyn en 1 limInyn ab 21imcIncaforc lR 5 limInynab Proof 1 Let 6 gt 0 be given We then know that there is an N1 such that lIn 7 al lt 3 for all n 2 N1 and an N2 such that lyn 7 bl lt 5 for all n 2 N2 Letting N maxN1N2 we have yn 7 a bl S lIn 7 al lyn 7 bl lt 3 3 e for all n 2 N so the de nition of limit is satls edi 2 This is a special case of 3 when yn c for all mi 3 Assume for simplicity that a gt 0 and b gt 0 the other cases are similar Let 6 gt 0 be given Note that lznyn7abl lrnyn7znbznb7abl S lznyn7bllbzn7al We choose N1 such that lIn 7 al lt i for n 2 Nli We choose N2 so that lIn 7 al lt for n 2 N2 iiei lt In lt gal We then choose N3 so that lyn 7 bl lt 37 for n 2 Ngi Then letting N maxN1N2N3 we have that for n 2 N 3 lInyn7ablSEaia4rbi e 3 2b and hence the limit de nition is satis ed D Note 9 We must be careful when dealing with in nite limitsi When lim In 00 and limyn 700 the sequence ltIn yn may or may not have a limit Similarly if lim In 00 and lim yn 0 the sequence ltIn may or may not have a limit Also the converses of the above can fail For instance if In n and yn 7n then limIn yn 0 but limIn and limyn do not exist as real numbers Similarly when In n and yn i we have limIn yn 1 but lim In 00 and limyn 0 Theorem 10 Comparison Theorem Let ltfngtn21 and yngtn21 be se quences such that In S yn for all n and suppose the limits of both sequences eIz39st Then lim In S lim yn Proof We note that supkgtn 1k S supkgtnyk for all n7 since 1k S yk for all kl Hence inf su I lt inf su 72 k 7 nzlkgayk so that MIn S Eynl Since these are equal to the respective limits7 the theorem follows E The same is true for lim sup and liminfl Note that we do not always have strict inequality The sequence with In 0 for all n and yn 1 satis es n In lt yn for all n7 but the limits are both equal to 0 Theorem 11 Squeeze Theorem Let ltfngtn21 ltyngtn2h and lt2ngtn21 be sequences with In S yn S 2n for all n Suppose limIn lim 2n Then lim yn lim In lim 2n Proof Let limIn lim 2n ll Let 6 gt 0 be given We can choose an N large enough so that lIn 71 lt e and lzn 71 lt 6 whenever n 2 Ni Then 7eltIn7lSyn7lSzn7llte so lyn 7 ll lt 6 whenever n 2 N7 satisfying the limit de nition D Example 12 We can use the squeeze theorem to compute the limit of the sequence with yn COS by letting In 7 2n i and noting that In S yn S 2n and that limIn lim 2n 0i

×

×

### 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!"

Anthony Lee UC Santa Barbara

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

Jim McGreen Ohio University

Forbes

#### "Their 'Elite Notetakers' are making over \$1,200/month in sales by creating high quality content that helps their classmates in a time of need."

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