### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# ANALYSIS I MATH 554

GPA 3.51

### View Full Document

## 13

## 0

## Popular in Course

## Popular in Mathematics (M)

This 38 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 13 views. For similar materials see /class/229529/math-554-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

## Reviews for ANALYSIS I

### What is Karma?

#### Karma is the currency of StudySoup.

#### 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

COMPACTNESS Handout 6 7 31196 Defn Suppose that K Q B A collection Q of open subsets such that K Q U O 069 is called an open cover of K K has a nite subcover from Q if there exist 917027 7 On in Q for which K Q U 97 J H Defn K is called compact7 if each open cover Q of K has a nite subcover Theorem The continuous image of a compact set is compact Proof Suppose f K gt B is continuous and K is compact Each open cover C of f K can be drawn back to an open cover C of K7 by considering the sets 6 f 1O7O e C K compact implies that we may draw a nite subcover from C Each of these members is the inverse image under f from a member of C These form the desired subcover of fK D Theorem Heine Borel Suppose that a 3 b7 then the interval a7 b is compact Proof Let C be an open cover for a7 b and consider the set A z mac has an open cover from C Note that A y since a E A Let 7 lubA It is enough to show that 7 gt b7 since if 951 E A and a S x S 9517 then x E A Suppose instead that 7 3 b7 then there must be some 90 E C such that 7 E 90 But 90 is open7 so there exists 6 gt 0 so that Ng Q 90 Since 7 is the least upper bound for A7 then there is an x E A such that 7 6 lt x S 7 But 05 E A implies there are members 917 7 9n of C whose union covers cum The collection 907 917 7 9n covers any 62 Contradiction7 since 7 is the least upper bound for the set A D Theorem Each closed subset C of a compact set K is compact Proof Let Q be an open cover for C Let 90 be the complement of C7 then 90 is open and Q Q U 90 is an open cover for K There is a nite subcover of Q which covers K and hence C This subcover dropping 90 if it appears is the desired nite subcover for C D Defn Suppose an is a sequence A sequence bk is called a subsequence of an if there exists a strictly increasing sequence of natural numbers n1ltn2ltltnklt such that bkank kl2 Theorem Suppose that K Q B then TFAE a K is compact b K is closed and bounded c each sequence in K has a subsequence which converges to a member of K d Bolzanno Weierstrass each in nite subset of K has a limit point in K Proof a gt b To show that K is bounded consider the open cover of K con sisting of the collection of nested open intervals 9 nn n E W To show that K is closed let me be a limit point of K Assume to the contrary that 950 K Consider the open cover of K consisting of the collection of nested open sets 9 z x E BHm x0 gt 111 71 E W Any nite subcollection which would cover K would have union whose complement would be a neighborhood of 950 not intersecting K This shows that 950 could not be a limit point of K b gt d We use the divide and conquer7 method better known as the bisection method Let A be an in nite subset of K Since K is bounded there is an interval a b such that K Q a b lnductively de ne the closed subintervals as follows Let a0 b0 a b Either the left or right half of a0 b0 contains an in nite number of members of K In the case that it is the right half set ah b1 bo a02 b0 Set a1b1 equal to the left half of a0 be otherwise lnductively let an1bn1 be the half of am bn which contains an in nite number of members of A Notice that the length of this interval is b 60271 that the an s satisfy an S anH S lt b and so must converge to some real number a S 950 S b Each neigborhood of 950 will contain one of the intervals am bn and hence will contain an in nite number of members of A ie x0 is a limit point of A This also shows that 950 is a limit point of the closed set K and must therefore belong to K d gt 0 Let x f be a sequence in K If the sequences image is nite then we may construct a constant subsequence which has the value which we may choose as any of the values of x f which is repeated in nitely often Otherwise let A be the range of the sequence Then A is an in nite subset of K By the Bolzanno Weierstrass property A must have a limit point me say which belongs to K For each k E W we may nd an integer nk larger than those previously picked ie 111 nkq so that xnk x0 lt lk This is the desired subsequence 0 gt b lf K were not bounded then there would exist a sequence xn E K such that gt n If this sequence had a subsequence which converged then it would have to be bounded But each subsequence of is clearly unbounded To show that K is closed we let me be a limit point of K which is not in K We can then nd a sequence from K which converges to 950 By condition c this has to have a subsequence which converges to a member of K Contradiction Each subsequence of a convergent sequence converges to the same limit in this case 050 which does not belong to K D Corollary Each continuous function f on a compact set K is bounded Proof The set f K is compact and is therefore bounded D Corollary Extreme Value Theorem Each continuous function on a compact set attains its maximum resp minimum Proof The set f K is compact and is therefore bounded and closed Hence the least upper bound 7 for fK must belong to Therefore there is an 050 E K such that 7 fx0 and so S fx0 for all x E K Similary the greatest lower bound of f K is attained by some member of K D Defn A function f is called uniformly continuous if for each 6 gt 0 El 6 gt 0 such that whenever 951952 6 domf and x1 x2 lt 6 then lt 6 Corollary Each continuous function on a b is uniformly continuous Proof Suppose not then negating the de nition implies that there exist an 60 gt 0 such that for each n 6 EV we can nd acmyn E K with xn yn lt ln but Z 60 K is compact so we can nd a subsequence xnk il of x f which converges to some 950 belonging to K Notice that ynk il also converges to me use an 62 proof But f is continuous at 950 so 60 S ffrm fynk S Wm fo lftro fynk gt 0 as k gt 00 which is a contradiction D MATH 554 i DIFFERENTIATION Handout 8 Defn A function f is said to be dz erentz39able at 0 if hm 0 712 f0 haO exists In this case the limit is called the derivative of f at 0 and is denoted f z0 Note 1 This de nition is equivalent to the requirement that the following lirnit exist hm we 7 mo 14mg 7 0 f ol 2 This7 in turn7 is equivalent to the following statement about how fast f converges to fz0 as x a 0 there exists a function 77 such that whine 77W 0 and f95 f96o 96 i 960 if960 7790 Note that if we de ne 77x0 07 then without loss of generality we can assume that 77 is continuous at 0 Examples 1 If fx 2 then f 2x 2 If g m then g 0 does not exist 3 If h xlzl then h exists and equals Theorem If f is differentiable at 0 then f is continuous at 0 Proof Use and let x a 0 E Theorem Basic rules of differentiation surns7 products7 quotients Suppose that f and g are differentiable at 0 then 1 f gym f Wo Mm 2 f9 96o f o9xo f96o9 96o 3 fQ Wo 9960f 96o fo9 o 996o27 if 9 7 0 Theor39em Chain rule If f is differentiable at 0 and g is differentiable at yo flt07 then h g o f is differentiable at 0 and hT950 9f950 f 950 Proof Use for f at 0 and for g at yo flt0gt3 M96 7 Wm 99 7 Mo y 7yo f o f o f o 91210 7729 7 mo 7 my 7 f zo 771969 yo 7729 where y The proof is completed by using this equation7 letting xn 7 0 and noticing that gm 7 yo where yn 7 rm 5 Theorem Rolle7s Theorem Suppose that b is differentiable on 177 is continuous on 17 and vanishes at the endpoints7 then there exists 0 strictly between a and b such that Mm 0 Proof If b is constant7 then any point can be selected for 0 Otherwise7 we may assume WLOG that b has positive values By the Extreme Value Theorem7 let x0 be such that Ms S z0 for all a S x S b First7 let xn 1 0 then since x0 gives a max7 we have 45 7 zo zni o 0 2 WWO and so7 by the Squeeze Theorem7 Mn 3 0 Similarly7 Mn 2 0 D Note Within the proof we actually established the critical point procedure of calculus local max and min can only occur at critical points Corollary Suppose that f is a differentiable function on cab and is continuous on 17 Then f vanishes identically if and only if f is a constant function Corollary Mean Value Theorem Suppose that f is differentiable on cab and is continuous on 17 then there exists x0 strictly between a and b such that N 7 at fSO b 7 a Proof Let W 7 we 7 We 7 a m and apply Rolle7s theorem D Defn F is called an anti derivative of f if F is differentiable and F x fx Corollary If both F and G are anti derivatives of f7 then they differ by a constant7 ie there exists a constant c such that 7 Gz 07 for all z E d0mf MATH 554 FALL 08 Lecture Note Set 1 Defn From the introductory lectures7 an ordered set is a set S with a relation lt7 which satis es two properties 1 Trichotomy property for any two elements ab 6 S7 exactly one of the following hold altb7ab7 orblta 2 Transitive property for any three elements a7b7 c 6 S7 if a lt b and b lt 07 then a lt 0 In this case the relation lt7 is called an order Defn Suppose that S is an ordered set and A Q S An element 6 E S is said to be an upper bound for A if a S 67 Va 6 A An element a is said to be a least upper bound for A if 1 a is an upper bound for A 2 if B is any upper bound for A7 then a S 6 In this case7 the supremum of A sup A is de ned as a The de nitions are similar for lower bound7 greatest lower bound and inf A7 respectively Note that we have already shown that the least upper bound for a nonempty set bounded from above is unique Defn A set S is said to have the least upper bound property if each nonempty subset of S which is bounded from above7 has a least upper bound Theorem Suppose the ordered set S has the least upper bound property7 then it has the greatest lower bound property ie each nonempty subset of S which is bounded from below has a greatest lower bound Proof Suppose that a nonempty set A has a lower bound7 call it 6 De ne L as the set of all lower bounds of A7 then L is nonempty Z 6 L Observe that each member of the nonempty set A is an upper bound of L so by the least upper bound property7 L has a least upper bound Call this element a First observe that a is a lower bound for A Otherwise7 there exists an element b E A with b lt 047 but each element of A is an upper bound for L7 so this element b is an upper bound of L which is smaller than 047 the least upper bound of L This would be a contradiction Therefore7 a E L Also7 if 6 is any lower bound of A7 then 6 S 04 since a sup L Hence a is the greatest lower bound of A Defn The real numbers are de ned to be a set R with two binary operations 1L7 which satisfy the following properties Given any a7 be in IR abcabc abba 3061R9a0a7Va 1R for each aEIR7 37a 6B7 so thata7a 0 r5905 5 abc abc 6 a b b a 7 31 B910anda1aVa 1R 8 for each a 6 R with a 31 0 3a 1 6 1R so that a 1 1 1 9 abc abac Note These properties just say that R is a nontrivial eld Moreover there is a distinguished subset P the positive cone of R with the following proper ties Given any ab in JP a a b 6 JP b a I 6 1F c For each a in IR exactly one of the following properties holds i a 6 JP ii 7a 6 JP iii a 0 Finally 1R must satisfy the least upper bound property that is each nonernpty subset of R which has an upper bound has a least upper bound These terms are de ned shortly Defn For the real numbers de ne a lt b as b 7 a 6 JP Lemma Using the eld properties of IR the following properties hold and are all assigned horne work problems see p 3 of this lecture set 1 The additive and rnultiplicative identities are unique 2 The additive and rnultiplicative inverses are unique 3 lfaElRthena00 Theorem The positive cone 1P induces an order on the real numbers ie 1R equipped with the relation lt7 is an ordered set Proof Assigned homework Problem 25 D Notation b 7 a7 is de ned as b 7a a S b7 means either a lt b or a b The fraction 7 means a b l Lemma For each ab c 6 1R i 7a 71 a ii 0 lt 1 iii if 0 lt a then 7a lt 0 E 1v ifaltband0ltcthenacltbc V v ifaltbthenacltbc vi if 0 lt a then the multiplicative inverse of a is positive ie 0 lt 1 1 v1i the product of two negative real numbers is positive while the product of a negative real number and a positive real number is negative Proof Since altlt71gt agt 7lt1agtltlt71gt agt 7alt1lt71gtgt 7M it follows from the fact that a 0 0 Homework Problem 24 that 04 71 a is an additive inverse for a But additive inverses are unique from your Homework Problem 227 so the conclusion of part follows To prove part ii7 we assume to the contrary7 ie that 1 Z 1P By the de nition7 1 31 07 so 71 6 JP and therefore 0 lt 71 But 71 71 771 1 by part and the HW Problem that additive inverses are unique This shows that 1 6 P by property b of the positive cone and the assumption that 71 6 JP Contradiction7 by the trichotomy property For part iii7 observe that 0 lt a means a 6 JP Since additive inverses are unique7 then 77a a7 and so 0 7 7a 77a a 6 JP This is equivalent to the statement 7a lt 0 To prove iv7 use the de nition of lt to show both b 7 a and c are in JP The positive cone is closed under multiplication7 so b7ac 6 JP Using the property shows then that be 7 ac 6 JP and so ac lt b0 Property v7 is proved similar to showing that 0 lt 1 lndeed7 suppose that b a 1 lt 07 then 7b is positive and so 71 7b a is positive Contradiction7 since 71 is negative The proof of property vi is left for additional practice D Lemma Suppose that A is a nonempty subset of R7 with least upper bound M7 then for every 6 gt 07 there exists a E A such that M 7 E lt a S M Proof Since 0 lt 6 then M 7 E lt M This shows that M 7 6 cannot be an upper bound for A Hence there is a member of A7 call it a7 so that M 7 E lt a D Theorem Archimedean Property Suppose a7b are positive real numbers7 then there exists n E N such that b lt n a Here IN is the set of natural numbers7 ie 1 and all its successors7 111111 Proof Suppose to the contrary that na lt b for all n 6 N7 then it follows that 04 ba is an upper bound for the natural numbers Let M be the least upper bound By the lemma7 12 gt 07 so there exists a natural number N so that M 712 lt N But then7 M lt N 12 lt N 17 which shows that M is not an upper bound for 1N Contradiction D Corollary The natural numbers N are not bounded Corollary Given any 6 gt 07 there exists n E N such that 0 lt 171 lt 6 Homework 2 Due Tuesday Sept 9 H Show that the additive or multiplicative identity is unique 3 Show that additive or multiplicative inverses are unique Prove that 7a c 7a 7c Prove that a 0 0 for each a in IR Prove that lt7 is an order for IR 99 Prove that a lt b and c lt 0 implies that b c lt a c 7 Provethat if0ltaltb7then0lt lt a Notation Next we de ne intervals of real numbers a7b z E lRla lt z lt b is called the open interval with endpoints a7b a7 b z E lRla S x S b is called the closed interval with endpoints a7b a7bl z E lRla lt x S b and ab z E lRla S x lt b are called the half open intervals with endpoints a7 b The common length7 or measure7 of these intervals is de ned to be b 7 a Theorem Suppose that I is an interval with endpoints ab and a lt b7 then I contains a rational number Proof De ne the length of by Z b7 a By the previous corollary7 there exists no E N such that 0 lt 1n0 lt Z Let A k an integer and kno lt a A is nonempty7 since the negative integers are not bounded from below Let k0 belong to A Set B k an integer and k 2 1 A Also7 A is bounded from above by a no7 which shows that B is in fact a nite set of integers Let K be the largest member of B and therefore of A7 then K 1 Z A Let r K 1n07 then K1 K lt7 a lt7TZ abiab7 no which shows that the rational r E a7b Q I D Corollary Each interval with nonzero length contains an in nite number of rationals Defn A real number is said to be irrational if it is not rational Remark Each interval with nonzero length contains an uncountably in nite number of irrationals Proved later We establish a few other facts about irrational numbers and also prove directly that each interval of positive length contains an in nite number of irrationals Lemma The product of a nonzero rational with an irrational is irrational Proof Suppose that ql 04 12 where 11412 are rational and 04 is irrational Since ql 31 07 then 04 q2q1 and it follows that 04 is rational Contradiction D Lemma If m is an odd integer7 then m2 is odd Proof If m is odd7 then there exists an integer k such that m 2k 1 In this case in2 22k2 2k 1 a Lemma xi is irrational Proof Suppose that xE mn where m7n are integers with n gt 0 We may assume that the rational is in lowest terms ie m and n have no common factors Squaring the equation and multiplying by n27 we obtain that m2 2n2 This shows that m2 is even By the lemma m must be even and equivalently that it contains 2 as a factor This shows 4k2 2n2 for some integer k Consequently n is even and 2 appears as one of its factors Contradiction since mn was supposed to be in lowest terms D Theorem Each interval with nonzero length contains an in nite number of irrationals Proof Let a b be the endpoints of the interval I Consider the interval axi It has length b 7 a gt 0 and so contains a nonzero rational number Q It follows that qxi is between a and b and hence belongs to I D Defn The absolute value of a real number a is de ned by a i a ifaZO 397 7a ifalt0 Lemma The absolute value function has the following properties 1 lal 2 0 for all a 6 1R 2 ia S lal 3 l 7 aw 7 W 4 labl7lallbl 5 Suppose Oz 2 0 then lal S 04 if and only if 704 S a S a 6 labl S lal lbl 7 llbl lall S lbial Proof To prove property 1 case it out7 Either a 2 0 or a lt 0 In the rst case lal a 2 0 In the second case lal 7a gt 0 since a lt 0 Property 2 follows similarly Properties 3 and 4 are left for the student To prove property 5 notice that lal S 04 is equivalent to the statement ia S a Using the ele mentary order properties it is easy to see that this last statement is equivalent to both inequalities a S 04 and 7o S a This is equivalent to the statement on the right hand side of property 5 To prove property 6 set 04 lal lbl and apply property 5 after verifying that ia b S a To prove property 7 use the fact that lbl lb 7 a al 3 lb 7 al lal and subtract lal from each side This shows that lbl 7 lal 3 lb 7 al By symmetry in a and b one proves that M 7w la7bl lb7al a CONNECTEDNESS Handout 6 Defn 1 A disconnection of a set A is two nonempty sets A1A2 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 1 1 Example The set 0 U E 1 is disconnected Theorem 1 Each interval open closed half open I is a connected set Proof Let A1A2 be a disconnection for I Let aj E A j 12 We may assume WLOG that a1 lt a2 otherwise relabel A1 and A2 Consider E1 x E A1x 3 a2 then E1 is nonempty and bounded from above Let a sup E1 But a1 3 a 3 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 e gt 0 so that NEW 6 A1 But then a 62 6 A1 and is less than a2 which contradicts that a is the sup of E1 D Theorem 2 If A is a connected set then A is an interval Proof Otherwise there would be a1 lt a lt a2 with aj E A and a A But then 91 ooa m A and 92 z a 00 m A form a disconnection of A D Theorem 3 The continuous image of a connected set is connected The continu ous image of a b is an interval 0 d where c min and d max agng aSISb Proof Any disconnection of the image f a bl can be drawn back7 to form a discon nection of a b if 91 92 forms a disconnection for fI then f 191 f 192 forms a disconnection for I a b D Corollary 1 Intermediate Value Theorem Suppose f is a real valued function which is continuous on an interval I If a1 a2 6 I and y is a number between f a1 and f02 then there exists a between a1 and a2 such that fa 3 Proof We may assume WLOG that I 01a2l We know that fI is a closed interval say I1 Any number 3 between fa1 and f02 belongs to I1 and so there is an a 6 0102 such that fa y D Theorem 4 Suppose that f a b gt a b is continuous then f has a xed point ie there is an oz 6 a b such that fa oz Proof Consider the function 995 x fx then 9a S 0 3 91 9 is continuous on a b so by the Intermediate Value Theorem there is an oz 6 a b such that 9a 0 This implies that fa a D Note There are some immediate consequences of these ideas 0 First we can get a better idea of the structure of general open sets in the real line Each open subset of B 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 Q IR open sets are just restrictions of these sets Connectedness is the basis of root nding for example with the Bisection method Consider the example of solving for polynomial roots or sinx x in the interval 0 It also permits us to study inverse functions of continuous strictly monotone functions We see that the continuous image under a monotone map f of a closed interval a b is a closed interval fa f That is any continu ous strictly monotone increasing function f maps a b one to one and onto fa Using compactness in the next notes we will show that in this settings inverse functions are also continuous COMPACTNESS Handout 7 Defn 1 Suppose that K Q IR A collection Q of open subsets such that K Q U O 069 is called an open cover of K K has a nite subcover from Q if there exist 91 92 7 On in Q for which K Q U Oj j H Defn 2 K is called compact if each open cover Q of K has a nite subcover Theorem 1 The continuous image of a compact set is compact Proof Suppose f K gt B is continuous and K is compact Each open cover C of flK can be drawn back to an open cover C of K by considering the sets 6 f 1OO e C K compact implies that we may draw a nite subcover from Each of these members is the inverse image under f from a member of C These form the desired subcover of K D Theorem 2 Heine Borel Suppose that a S b then the interval a b is compact Proof Let C be an open cover for a b and consider the set A z a S x S b 1 mac has a nite open cover from C Note that A is bounded and nonemtpy since a E A Let 7 lubA It is enough to show that 7 gt b since if 051 E A and a S x S 051 then x E A Suppose instead that 7 S b then there must be some 90 E C such that 7 E 90 But 90 is open so there exists 6 gt 0 so that 357 Q 90 Since 7 is the least upper bound for A then there is an x E A such that 7 6 lt x S 7 But 05 E A implies there are members 91 7 9n of C whose union covers a The collection 90 91 0 covers any 62 Contradiction since 7 is the least upper bound for the set A El Theorem Each closed subset C of a compact set K is compact Proof Let Q be an open cover for C Let 90 be the complement of C then 90 is open and Q Q U 90 is an open cover for K There is a nite subcover of Q which covers K and hence C This subcover dropping 90 if it appears is the desired nite subcover for C D Defn 3 Suppose an is a sequence A sequence bk is called a subsequence of an if there exists a strictly increasing sequence of natural numbers n1ltn2ltmltnkltm such that bk ank k 12 Theorem 4 Suppose that K Q B then TFAE a K is compact b K is closed and bounded c each sequence in K has a subsequence which converges to a member of K d Bolzanno Weierstrass each in nite subset of K has a limit point in K Proof a gt b To show that K is bounded consider the open cover of K con sisting of the collection of nested open intervals 9 nn n E W To show that K is closed let me be a limit point of K Assume to the contrary that 950 K Consider the open cover of K consisting of the collection of nested open sets 9 z x E BHm x0 gt 171 71 E W Any nite subcollection which would cover K would have union whose complement would be a neighborhood of 950 not intersecting K This shows that 050 could not be a limit point of K b gt d We use the divide and conquer7 method better known as the bisection method Let A be an in nite subset of K Since K is bounded there is an interval a b such that K Q a b lnductively de ne the closed subintervals as follows Let a0 b0 a b Either the left or right half of a0 b0 contains an in nite number of members of K In the case that it is the right half set ah b1 bo a02 b0 Set ah b1 equal to the left half of a0 be otherwise lnductively let an1bn1 be the half of am bn which contains an in nite number of members of A Notice that the length of this interval is b 60271 that the an s satisfy an S anH S lt b and so must converge to some real number a S 050 S b Each neigborhood of 950 will contain one of the intervals am bn and hence will contain an in nite number of members of A ie x0 is a limit point of A This also shows that 050 is a limit point of the closed set K and must therefore belong to K d gt 0 Let x f be a sequence in K If the sequences image is nite then we may construct a constant subsequence which has the value which we may choose as any of the values of x f which is repeated in nitely often Otherwise let A be the range of the sequence Then A is an in nite subset of K By the Bolzanno Weierstrass property A must have a limit point me say which belongs to K For each k E W we may nd an integer nk larger than those previously picked ie n1 nk1 so that xnk x0 lt lk This is the desired subsequence 0 gt b lf K were not bounded then there would exist a sequence 05 E K such that gt n If this sequence had a subsequence which converged then it would have to be bounded But each subsequence of is clearly unbounded To show that K is closed we let me be a limit point of K which is not in K We can then nd a sequence from K which converges to 950 By condition c this has to have a subsequence which converges to a member of K Contradiction Each subsequence of a convergent sequence converges to the same limit in this case 950 which does not belong to K D Corollary 1 Each continuous function f on a compact set K is bounded Proof The set f K is compact and is therefore bounded D Corollary 2 Extreme Value Theorem Each continuous function on a compact set K attains its maximum resp minimum Proof The set f K is compact and is therefore bounded and closed Hence the least upper bound 7 for fK must belong to Therefore there is an 050 E K such that 7 fx0 and so S fx0 for all x E K Similarly the greatest lower bound of f K is attained by some member of K D Defn 4 A function f is called uniformly continuous if for each 6 gt 0 36 gt 0 such that whenever 951952 6 domf and x1 x2 lt 6 then lt 6 Corollary 3 Each continuous function on a b is uniformly continuous Proof Suppose not then negating the de nition implies that there exist an 60 gt 0 such that for each n 6 EV we can nd acmyn E K with xn yn lt ln but Z 60 K is compact so we can nd a subsequence xnk il of x f which converges to some 950 belonging to K Notice that ynk il also converges to 050 use an 62 proof But f is continuous at 950 so 60 S ffrnk fynk S ffrnk fo lftro fynk gt 0 as k gt 00 which is a contradiction D MATH 554 i RIEMANN INTEGRATION Handout Qa Nov 29 Defn A collection of n 1 distinct points of the interval a b Px0alt1ltltzi1ltpiltltbpn is called a partition of the interval In this case we de ne the norm of the partition by 121 Am where Ax x 7 xi1 is the length of the i th subinterval pl1 Defn For a given partition P we de ne the Riemann upper sum of a function f by i1 where M denotes the supremum of f over each of the subintervals plAmi Similarly we de ne the Riemann lower sum of a function f by Lf P Am i1 where m denotes the in mum of f over each of the subintervals pl1 Since in 3 Mi we note that LOUD S UMP for any partition P Defn Suppose 131132 are both partitions of ab then P2 is called a re nement of P1 denoted by P1 4 P2 if as sets P1 Q P2 Note If P1 4 P2 it follows that HPle 3 HRH since each of the subintervals formed by P2 is contained in a subinterval arising from P1 Lemma lf P1 4 P2 then 1103131 E 107132 and Uf7P2SUf7P1 Proof Suppose rst that P1 is a partition of a b and that P2 is the partition obtained from P1 by adding an additional point z The general case follows by induction adding one point at at time In particular we let P10altz1ltltzi1ltxiltltbn and P20altz1ltltzi1ltzltxiltltbn for some xed i We focus on the upper Riemann sum for these two partitions7 noting that the inequality for the lower sums follows similarly Observe that lf7 P1 3 Z Mj Aij j1 and iil n Uf7P2 3 Aij 7 M71 7 Z Z Mj Aij 71 739i1 where M supmiw f and M supmmi It then follows that UfP2 S Uf7 P1 since MJVIgMi n Defn lf P1 and P2 are arbitrary partitions of all7 then the common re nement of P1 and P2 is de ned as the formal union of the two Corollary Suppose P1 and P2 are arbitrary partitions of a7b7 then 103131 E Uf7P2 Proof Let P be the common re nement of P1 and P2 then Lf7P1 S Lf7P lt UMP S Uf7P2 D Defn The lower Riemann integral of f over a7b is de ned to be 17 fzd sup f7 P 4a all partitions P of ab Similarly7 the upper Riemann integral of f over a7b is de ned to be 717 afxdx Uf P inf all partitions P of ab By the de nitions of least upper bound and greatest lower bound7 it is evident that for any function f there holds Zfzdx S Defn A function f is Riemann integrable over a7b if the upper and lower Riemann integrals coincide We denote this common value by fa f dx Theorem A necessary and suf cient condition for f to be Riemann integrable is given 6 gt 07 there exists a partition P of ab such that UfPiLfPlte Note that in this case the unique number between these two values is fx dz Proof First we show that is a suf cient condition This follows immediately since for each 6 gt 0 that there is a partition P such that holds 1 b afxdx 7 fzdz Uf P i Lf P lt 6 Since 6 gt 0 was arbitrary then the upper and lower Riemann integrals of f must coincide To prove that is a necessary condition for f to be Riemann integrable we let 6 gt 0 By the de nition of the upper Riemann integral as a in mum of upper sums we can nd a partition P1 of ab such that b b fltzgtdz w P1 lt fltzgtdz 62 Similarly we have b b fzdz 7 62 lt LfP2 fzdz Let P be a common re nement of P1 and P2 then subtracting the two previous inequalities implies Uf7PLf7P Uf7P1Lf7P2 lt6 D MATH 554 703 I ANALYSIS 1 EXISTENCE OF SQUARE ROOTS Theorem If a is a nonnegative real number then there exists a unique positive real number oz such that a2 a We use the notation E oz Lemma Positive square roots are unique Proof Suppose not If x lt y and x y are both positive square roots of a gt 0 then x2 lt my lt 32 But 952 a 32 Contradiction ltgt Proof of the Theorem First notice that we may assume without loss of generality that 0 lt a lt 1 If a 1 then oz 1 is the unique square root of a If 1 lt a then b 1a is less than 1 and we denote its square root by B We set oz 1 then a2 152 1b a Also notice that in the case 0 lt a lt 1 the Lemma and its proof imply that 0 lt oz lt 1 For 0 lt a lt 1 we de ne the set Axgt0x2 a Notice that A is nonempty a E A and bounded from above by 1 so let oz lub A Suppose that a2 y a Case 1 If a lt a2 then we observe that e Q2270 is positive We claim that B oz e is an upper bound for A which would contract the statement that oz is the least upper bound By the de nition of e we see that 2gtoz2 2 ozgtoz2 2 a2x2 for each x E A Hence 5 is greater than all x E A Case 2 If a2 lt a then set x oz e where a a2 e min 7 1 2a 1 We claim that x E A which would contradict that oz is the least upper bound of A lndeed using the de nition of e we see that 0520z22 oz62 a220z1e a ltgt hdATH 5544INTEGRATH3N Handout 9 7 41296 Defn A collection of n 1 distinct points of the interval a b Px0altx1ltltxi1ltxilt ltbxn is called a partition of the interval In this case we de ne the norm of the partition W Inax Am 1991 where Axl xi ail1 is the length of the i th subinterval ml1 Defn For a given partition P we de ne the Riemann upper sum of a function f W i1 where Ml denotes the suprernurn of f over each of the subintervals ml1 Sim ilarly we de ne the Riemann lower sum of a function f by LP7f 3 imiAxi i1 where m2 denotes the in rnurn of f over each of the subintervals xi47 Since m2 3 Mi we note that M1371 S UP7f for any partition P Defn Suppose P1 P2 are both partitions of a b then P2 is called a re nement of P1 denoted by P1 lt P2 if as sets P1 Q P2 Note If P1 lt P2 it follows that 132 3 131 since each of the subintervals formed by P2 is contained in a subinterval which arises from P1 Lemma lf P1 lt P2 then and UP27fSUP17f Proof Suppose rst that P1 is a partition of ab and that P2 is the partition obtained from P1 by adding an additional point z The general case follows by induction adding one point at at time In particular we let P1x0altx1ltltxiS1ltxiltltbxn and P2x0altx1ltltxiS1ltzltxiltltbxn for some xed i We focus on the upper Riemann sum for these two partitions noting that the inequality for the lower sums follows similarly Observe that UP17f 3 ZMj Al j1 and 23971 n UP27f1 ZMjAjMZ i71M95i Z Z MjAfrj j1 jz 1 where M supmim and M supwgi It then follows that UP2 f S UP1 f since MM 3 Mi D Defn lf P1 and P2 are arbitrary partitions of a b then the common re nement of P1 and P2 is the formal union of the two Corollary Suppose P1 and P2 are arbitrary partitions of a b then M13171 S UP27f Proof Let P be the common re nement of P1 and P2 then LP17fS LP7f S UP7f S UP27f D Defn The lower Riemann integral of f over a b is de ned to be b fxdx sup LP a all partitions P of ab Similarly the upper Riemann integral of f over a b is de ned to be b fxdx in H UP a all partitlons P of ab By the de nitions of least upper bound and greatest lower bound it is evident that for any function f there holds xwx S Defn A function f is Riemann integrable over a b if the upper and lower Riemann integrals coincide We denote this common value by if f dx Examples 1 if k dx Mb a 2 05 dx ab a2 Hintz Use 21139 nn l2 Theorem A necessary and su icient condition for f to be Riemann integrable is given 6 gt 0 there exists a partition P of a b such that UP7f LPflt Proof First we show that is a su icient condition This follows immediately since for each 6 gt 0 that there is a partition P such that holds b b afxdfr afrrdx UltPfgt LPf lt 6 Since 6 gt 0 was arbitrary then the upper and lower Riemann integrals of f must coincide To prove that is a necessary condition for f to be Riemann integrable we let 6 gt 0 By the de nition of the upper Riemann integral as a in mum of upper sums we can nd a partition P1 of a b such that b b afxdrr UP1f lt afrrdrr e2 Similarly we have b b afxdrr 62 lt L032 f afrrdfr Let P be a common re nement of P1 and P2 then subtracting the two previous inequalities implies UPf LPf s UP1f LP2f lt e a Defn A Riemann sum for f for a partition P of an interval a b is de ned by RltPfo mm where the Q satisfying xj1 S Q 3 xj l S j S n are arbitrary Corollary Suppose that f is Riemann integrable on cg b then there is a unique number 7 fffxdx such that for every 6 gt 0 there exists a partition P of cg b such that if P lt P171327 then 0 UP1f 7lte 0 S 7 LP2f lt e where RP1 f g is any Riemann sum of f for the partition P1 In this sense we can interpret b fxdx lim RP g 1 HPHHO although we would actually need to show a little more to be entirely correct Proof Since LP2 f S 7 S UP1 f for all partitions we see that parts i and ii follow from the de nition of the Riemann integral To see part iii we observe that mj S S Mj and hence that LPl7f S RPl7f7 S But we also know that both LPl7f S and condition hold from which part iii follows E Theorem If f is continuous on cg b then f is Riemann integrable on m b Proof We use the condition to prove that f is Riemann integrable If e gt 0 we set 60 eb a Since f is continuous on cg b f is uniformly continuous Hence there is a 6 gt 0 such that lt 60 if y x lt 6 Suppose that lt 6 then it follows that mi S 60 l S 139 S Hence UP f LP f m sz 3 601 a e D 39l N Theorem If f is monotone on m b then f is Riemann integrable on cg b Proof If f is constant then we are done We prove the case for f monotone increas ing The case for monotone decreasing is similiar We again use the condition to prove that f is Riemann integrable If e gt 0 we set 6 efb fa and consider any partition P with P lt 6 Since f is monotone increasing on cub then Ml and m2 ag1 Hence UltP7 f LltP f git mam 2ftrz39 fzgt1mi 2 MPH gm fol1 6W fa e a Theorem Properties of the Riemann Integral Suppose that f and g are Riemann integrable and k is a real number then 1 ffk rr doc k LEM drr ii fffgdxfffdxffgdx iii 9 S f implies fig drr S fffdm iv Iii de srflfldx Proof To prove part i we observe that in case k 2 0 then supmihxi MM and infmihxi kmi Hence UPkf kUPf and LPkf kLP In the case that k lt 0 then supmihxi kmi and infmiwi kM It follows in this case that UP kf kLP f and LP kf kUP f and so fxdx k xm 1 b kfxdx k fxdx To prove property ii we notice that sup1f g S sup f sup 9 and inf f infjg S inf1f g for any interval I for example I ml1 Hence 1 LP7f LP79 S LP7fg S UP7f9 S UP7f UP79 Let 6 gt 0 then since f and g are Riemann integrable there exist partitions P1 P2 such that 2 UP17f LP17f lt627 UP279 LP279 lt62 If we let P be a common re nement of P1 and P2 then by combining inequalities 1 and 2 we see that see that UP17f LP17f UP279 LP279 62 62 e Property iii follows directly from the de nition of the upper and lower integrals using for example the inequality sup 995 3 sup f Property iv is proved by applying property iii to the inequality f S f S If I from which it follows that ff dx 3 ff dx S if dx But this inequality implies property iv D Defn We extend the de nition of the integral to include general limits of integra tion These are consistent with our earlier de nition 1 5 dx 0 2 lbw doc ism doc Theorem If f is Riemann integrable on a b then it is Riemann integrable on each subinterval 0 d Q a b Moreover if c E a b then 3 4 d9 m dx Abm dag Proof We show rst that condition holds for the interval 0 d Suppose e gt 0 then by applied to f over the interval a b we have that there exists a partition P of a b such that condition holds Let P be the re nement obtained from P which contains the points 0 and d Let P be the partition obtained by restricting the partition P to the interval 0 d then and so f is Riemann integrable over 0 d To prove the identity 3 we use the fact that condition holds when f is Riemann integrable Let 6 gt 0 then for 63 gt 0 we may apply to each of the intervals I ab ac and 0 b respectively to obtain partitions P1 which satisfy 4 0 s UIltP1fgt f doc s UIPI7 f L1P17f lt 63 We let P be the partition of ab formed by the union of the two partitions PMC PM and P be the common re nement of P and PM Observing that 5 Uab57 f UacPl7f Ucb527f7 we can combine with inequality 4 to obtain fifdrr fffdrr fffdxl Uaclt15fgt fif dxl Meta f If dxl lUabp7f fffdf l lt 360 6 Since 6 gt 0 was arbitrary then equality 3 must hold D Theorem lntermediate Value Theorem for lntegrals If f is continuous on a b then there exists 5 between a and I such that rm doc flt gtltb a fffdx Proof Since f is continuous on a b and for 77 z b there holds a Iglib l fr S 77 S 1118th then by the Intermediate Value Theorem for continuous functions there exists a 6 ab such that 77 D Theorem Fundamental Theorem of Calculus l Derivative of an lntegral Sup pose that f is continuous on a b and set f fydy then F is differentiable and Fx for a lt x lt b Proof Notice that Fxh Fx ff hfdx h h for some 5 between 950 and x0 h Hence as h gt 0 then 5 Eh converges to 950 and so the displayed difference quotient has a limit of aco as h gt 0 D Theorem Fundamental Theorem of Calculus Part ll lntegral of a Derivative Suppose that F is function with a continuous derivative on a b then b my dy Fe 2 Fltbgt 1 Proof De ne C05 f Fy dy and set H F G Since the derivative of H is identically zero by Part I of the Fundamental Theorem of Calculus then the Mean Value Theorem implies that H b H a Expressing this in terms of F and G gives b Fltbgt Fltygt dy Fm which establishes the theorem D MATH 554703 1 FALL 08 Lecture Note Set 2 Please read Chapter 2 of Rudin which deals with the concepts of countability pages 24730 and the metric space topology pages 30 36 Some portions are outlined again here for convenience Defn a Two sets are said to have the same cardinality and are called equiualent if there exists a 1 1 onto mapping from one to the other This determines an equivalence relation b Consider the set Jn 1 2 7 A set A which is equivalent to Jn for some n is called nite with cardinality n This is the number of elements of A c A set is called in nite if it is not nite d A is called countable if it has the same cardinality as the natural numbers W e A is said to be at most countable if it is countable or nite We showed in class that the integers and the rationals were countable but that the irrationals were not We only indicated the proof of the last statement since it needed some elementary properties of sequences which we cover in Lecture Set 4 Moreover we showed Theorem The union of at most a countable number of at most countable sets is at most countable recall to use a diagonal argument Theorem Every in nite subset of a countable set is countable Defn A sequence in a set A is a mapping f from the natural numbers to A which may be written as a17a27a37 where aj fj jE V We recall the concept of mathematical induction which is applied to a sequence of statements Theorem Principle of Mathematical Induction Suppose that a statement pn is de ned for each natural number n lf 1 p1 is true 2 pn true gt pn1 true is a true statement for each n E W then pn is a true statement for each natural number n Proof Suppose to the contrary that the set E n E lNl pn false is not empty Notice that 1 Z B Let N be the smallest element of B possible since you can take a minimum of a nite set of integers7 then set 71 N 7 1 Observe by assumption 1 that n 6 N7 and by the de nition of B that pn is true By assumption 27 it follows that pn1 is true But 71 1 N Contradiction Hence B must be empty D MATH 554 LIMITS AND CONTINUOUS FUNCTIONS Handout 5 7 Fall 2008 We need to recall the following Earlier De nitions 0 A point 950 is called a limit point of a set A if each nbhd of 950 contains a member of A different from 950 ie for each 6 gt 0 N6x0x0 m A y 0 o A point 050 E A is called an isolated point ofA if x0 belongs to A but is not a limit point Earlier Theorems 0 A set F is closed if and only if it contains all its limit points 0 x0 is a limit point of a set A if and only if there exists a sequence Q A such that 05 gt 950 but 05 y x0 V71 6 With these ideas we begin the study continuity of functions which is very nat urally framed in a general metric space Defn Suppose that 950 is a limit point of the domain of a function f A gt B then f is said to have a limit L as x approaches 050 if Vegt 0EISgt 0 3 x Ed0mf amp 0lt dAxx0 lt6 gtdBfxL lt 6 In this case we use the notation lim L zaxo Defn Suppose f A gt B metric spaces lf 050 E A then f is said to be continuous at 950 if for each 6 gt 0 there is a 6 gt 0 so that if x E A and dAxx0 lt 6 then dBf957f950 lt 6 Note Remember that a point 050 E A is either an isolated point of A or it is a limit point of A Considering these two cases separately the de nition for continuity of f at a point can be seen to be equivalent to the following in each of the respective situations 1 if x0 is an isolated point of A then f is automatically continuous or 2 if x0 is a limit point of the domain A then the condition lim aco must hold Defn Consider a set B Q B A set C Q B is called open relative to B or briefly pelatiuely open if 0 9 m B for some open set 9 Q B That is to say 0 is just the restriction to B of an open set in the Whole space Theorem Let f A gt B Where A B are metric spaces then TFAE a f is continuous at each point of its domain for each limit point 950 of the domain A lim f exists amp equals oco maze b c for each sequence xn gt 950 then gt aco must hold d f 1 9 is open for each open subset 9 of B We will be concentrating on the metric space of real numbers Note The same proof above for continuity at a point 050 can be used to show the corresponding result for limits holds The only difference is that f is not required to be de ned at 950 The statement reads as Suppose that f A gt B is a peal ualued function of a real uapiable ie A B Q R If x0 is a limit point of the domain of f then TFAE 39 a lim L maze b For every sequence in the domain of f if xn gt 950 then gt L Corollary The nite sum product or the quotient of continuous functions is each continuous on their respective domains Corollary All polynomials are continuous Rational functions are continuous on their domains Theorem The composition of continuous functions is continuous Example Each of the following are examples of continuous functions on their respective domains 1 M m 2 995 2 25 339 WmW Homework 5 due Tuesday Nov 11 1 Suppose that 050 is a limit point of the domain of f g and that both f and 9 have limits at 950 then prove that 351133 f 9W 1133 it hm 905 maze 2 Suppose that f is de ned by 3x2 if lgx 95quot 2x1ifxlt 1 Determine at each point Whether or not f is continuous Justify your answer 3 2 2 Determine the domain of F 5 3 and carefully show that F x 1 is continuous at each point of its domain MATH 554 7 RIEMANN INTEGRATION Handout Qb Dec 4 Corollary If f is Riemann integrable on 17 then so is 7f and b b Hm dz e we dz Proof We use the condition and note that Uf7P Lf7P7 M4713 Uf7P and so Uf7P Lf7P Uf7P Lf7P lt 6 Since this holds for each positive 6 it follows that dx 7 dx D Note This result shows that for a function to be Riemann integrable it is enough to nd7 for each positive 6 a partition ne enough that the corresponding upper and lower sums are within 6 units of one another In this case the Riemann integral is within 6 units of either approximating sum Examples You should go through the following two examples on your own to make sure you understand the mechanics 1 gig dz kb i a 2 ssdx b2 712 Hint Use the result proved earlier 21239 nn12 Theorem If f is continuous on 17 then f is Riemann integrable on 17 Proof We use the condition to prove that f is Riemann integrable lf 6 gt 07 we set 60 Eb 7 1 Since f is continuous on the compact set 17 f is uniformly continuous Hence there is a 6 gt 0 such that 7 lt 60 if y 7 xi lt 6 Suppose that lt 6 then it follows that Mi 7 g 60 1 S 239 g Hence lf7 7 7 S 6003 a E D 11 The following indented material de nition and resulting corollary are included for completeness and are not required for the further development You will not be responsible for these two on the Final Defn A Riemann sum for f for a partition P of an interval 17 is de ned by Ru P s 7 Z mam 71 where the Q satisfying 791 S Q 3 7 1 Sj S n7 are arbitrary Corollary Suppose that f is Riemann integrable on 17 then there is a unique number 7 such that for every 6 gt 0 there exists a partition P of 17 such that if P 4 Png7 then 239 0 UfP17 ylte 2d 0 S 77 LfP2 lt 6 ma w 7 RltfP1sgtilt e where Rf7 P17 5 is any Riemann sum of f for the partition P1 In this sense7 we can interpret bfxdx li mORf 135 HPquot although we would actually need to show a little more to be precise Proof Since Lf7 P2 3 y S Uf7 P1 for all partitions7 we see that parts i and ii follow from the de nition of the Riemann integral To see part iii7 we observe that 771 3 g 3 M and hence that Lf7P1 S Rf7P17 S Uf7P1 But we also know that both 103131 S Y S Uf7P1 and condition hold7 from which part iii follows E Note The following theorem is also included for completeness7 but will not be needed for our development Theorem If f is monotone on 17 then f is Riemann integrable on 17 Proof If f is constant7 then we are done We prove the case for f monotone increasing The case for monotone decreasing is similar We again use the condition to prove that f is Riemann integrable lf 6 gt 07 we set 6 6fb 7 fa and consider any partition P with lt 6 Since f is monotone increasing on 17 then Ml and ml fzi1 Hence M Uf7P Lf7P Mi miAi H H M f71A39 H MPH ice 7 foo1 lt 5fbfa6 a Theorem Monotone Property of the Riemann Integral Suppose that f and g are Riemann integrable and k is a real number7 then 1 g s f implies fig dz 3 f f d96 ngt lfffdzl ifmdz Proof Property i follows directly from the de nition of the upper and lower integrals using the inequalities supI g S supI f and ian g S ian f for each subinterval I Property ii is proved by applying property i to the inequality lfl S f S lfl to obtain 7 lfl dx 3 ff dx 3 lfl dx But this inequality is equivalent to property ii D Defn We extend the de nition of the integral to include general limits of integration These are consistent with our earlier de nition 1 f dx 0 2 If we dz 7 7 Lime dz Theorem If f is Riemann integrable on 17 then it is Riemann integrable on each subinterval 07d Q 17 Moreover7 ifc E 17 then lt3 fltzgtdz7 Mm ow Proof We show rst that condition holds for the interval 07d Suppose E gt 07 then by applied to f over the interval 17 we have that there exists a partition P of 17 such that condition holds Let P be the re nement obtained from P which contains the points 0 and d Let P be the partition obtained by restricting the partition P to the interval 07d7 then U f7 13 7 L02 13 Uf713 M1213 Uf7 P 7 Lu P lt e and so f is Riemann integrable over 07d To prove the identity 37 we use the fact that condition holds when f is Riemann integrable Let E gt 07 then for 63 gt 07 we may apply to each of the intervals I 17 10 and 07b respectively7 to obtain partitions P1 which satisfy 4 o 11102131 7 I f dx U1ltf7PIgt 7 L102 PI lt 63 We let P be the partition of 17 formed by the union of the two partitions PacPcb and P be the common re nement of P and PPM Observing that 5 Uabf7 13 Uacf7 151 Ucbf7 152 we can combine with inequality 4 to obtain moo fffdz 7 fffdzl WM 7 rsfdzl New 7 rffdzl lUmf P 7 L f dzl lt 360 6 Since 6 gt 0 was arbitrary then equality 3 must hold D Corollary Each bounded piece wise continuous function with left and right hand limits at each point of an interval ab is Riemann integrable Moreover its integral is the sum of the integrals of the 77pieces Theorem lntermediate Value Theorem for lntegrals If f is continuous on 1 b then there exists between a and b such that pf om7fww7a b d fgfi there holds Proof Since f is continuous on ab and for 77 7 a mums n lt maxf96 i 11 11 then by the Intermediate Value Theorem for continuous functions there exists a 6 ab such that f 77 Theorem Fundamental Theorem of Calculus l Derivative of an lntegral Suppose that f is continuous on ab and set fy dy then F is differentiable and F z f for a lt z lt b Proof Notice that Fm h i Fm 500 f dz f f 5 for some between 0 and 0141 Hence as h 7 0 then converges to 0 and so the displayed difference quotient has a limit of fz0 as h 7 0 D Theorem Fundamental Theorem of Calculus Part ll lntegral of a Derivative Suppose that F is function with a continuous derivative on 1 b then rmwmo ww7mm Proof De ne Gz F y dy and set H F 7 G Since the derivative of H is identically zero Part I of the Fundamental Theorem of Calculus then the Mean Value Theorem implies that Hb 7 Ha 0b 7 a 0 Expressing this in terms of F and G gives 17 Hw7mm7mw7mm7wawo which establishes the theorem D Defn For a function f we call any function F whose derivative is f an antiderioatwe off Theorem Linearity Property of the Riemann lntegral Suppose that f and g are Riemann inte grable and k is a real number then 1 Li ka dz 7 k Lime dz 11 ff9dffdavf9d Proof It is a good exercise to prove these directly from the de nition and to use the condition The serious student should go through this in detail just for additional practice We present sirnpler proofs using the Fundamental Theorerns of Calculus For part ii we let F be an antiderivative of f and G be an antiderivative of 97 then H F G is an antiderivative of f 9 Therefore f5 f 9 dz 7 Hb 7 Ha 7 m 7 Fa Gb 7 Ga fffdzffgdm Part i is proved sirnilarly using the corresponding property of di erentiation D MATH 554 i DIFFERENTIATION Handout 8 7 32596 Defn A function f is said to be di erentiable at 950 if 950 h 950 h lim hgt0 exists In this case the limit is called the derivative of f at 950 and is denoted fxo Note 1 This de nition is equivalent to the requirement that the following limit exist hm rm mo maze x x0 fmol 2 This in turn is equivalent to the following statement about how fast converges to fx0 as x gt 950 there exists a function 77 such that 7795 0 and f0 fe m 060 f o 77015 Examples 1 If 952 then fx 2x 2 If 995 x7 then gO does not exist 3 If Mac xx7 then hx exists and equals Theorem If f is differentiable at 050 then f is continuous at 950 Proof Use and let x gt 950 D Theorem Basic rules of differentiation sums products quotients Suppose that f and g are differentiable at 950 then 1 f glfrO JCT150 9950 2 fQVWO f09950 T few950 3 WW950 9950V950 flt090 995027 11390950 7f 0 Theorem Chain rule If f is differentiable at 950 and g is differentiable at yo aw then h g o f is differentiable at 950 and hfro JUL 50 f 950 Proof Use for f at 050 and for g at yo are 0 0 0 fm NCO LCyo 772G hm hose 7 ND 9110 wyo 77210 ffro 77195 LCyo 772G where y z The proof is completed by using this equation letting 05 gt 050 and noticing that yn gt yo where yn D Theorem Rolle s Theorem Suppose that q is differentiable on a b is contin uous on a b and vanishes at the endpoints then there exists are strictly between a and I such that q xo 0 Proof If q is constant then any point can be selected for 950 Otherwise we may assume WLOG that q has positive values By the Extreme Value Theorem let 950 be such that 3 re for all a S x S I First let 05 i 950 then since 950 gives a max we have 0 Z gt x0 7 0 and so by the Squeeze Theorem q xo S 0 Similarly q xo Z 0 D Note Within the proof we actually established the critical point procedure of calculus local max and min can only occur at critical points Corollary Mean Value Theorem Suppose that f is differentiable on a b and is continuous on a b then there exists x0 strictly between a and I such that N ax fxo b a Proof Let we M We a m and apply Rolle s theorem D Defn F is called an anti derivative of f if F is differentiable and F f Corollary If both F and G are anti derivatives of f then they differ by a constant ie there exists a constant c such that C05 c for all x E d0mf

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

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

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "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!"

#### "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."

### 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

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.