Math 508 Exam 2 Jerry L. Kazdan December 9, :00 10:20

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1 Math 58 Eam 2 Jerry L. Kazda December 9, 24 9: :2 Directios This eam has three parts. Part A has 8 True/False questio (2 poits each so total 6 poits), Part B has 5 shorter problems (6 poits each, so 3 poits), while Part C has 5 traditioal problems (2 poits each, so total is 6 poits). Maimum score is thus 6 poits. Closed book, o calculators or computers but you may use oe 3 5 card with otes o both sides. Please silece your cellphoe before the eam ad keep it out of sight for the duratio of the test period. This eam begis promptly at 9: ad eds at :2. Ayoe who cotiues workig after time is called may be deied the right to submit his or her eam or may be subject to other gradig pealties. Please idicate what work you wish to be graded ad what is scratch. Clarity ad eatess cout. Part A 8 True/False questios (2 poits each, so 6 poits). Aswer oly, o reasos eed be give. Circle your True/False choice. [If you wish to give a reaso or eample, your work will be read.] A. T F The quotiet of two irratioal umbers (with the deomiator ozero) is irratioal. Solutio FALSE 2/ 2 or, say, 2/ 8 A 2. T F If A is a oempty compact subset of the real lie R, the R A is ever coected. (The set R A cosists of all real umbers which are ot i A.) Solutio TRUE A 3. T F If A ad B are compact subsets of a metric space, the A B is also compact. Solutio TRUE A 4. T F A series of comple umbers coverges if ad oly if the correspodig sequece of partial sums is bouded. Solutio FALSE. For istace the series ( ) A 5. T F If X is ay metric space ad f : X R is ay cotiuous real-valued fuctio, the the fuctio g : X R defied by g() = (f()) 2 is always cotiuous.

2 Solutio TRUE The product of two cotiuous real (or comple-valued) fuctios is cotiuous. A 6. T F Let X ad Y be metric spaces, ad let A ad B be two subsets of X whose uio is X. If f : X Y is cotiuous o A ad cotiuous o B, the it is cotiuous o X. Solutio FALSE. Eample: Let A = { < }, B = { 2}, X = { 2} with f() = o A, ad f() = o B. A 7. T F If f : X Y is a cotiuous map betwee metric spaces, ad f(x) is compact, the X is compact. Solutio FALSE. Eample: Say f maps all of X to oe poit p Y. A set cosistig of oe poit is compact. This gives o iformatio about X. A 8. T F A closed ad bouded subset of a metric space must be compact. Solutio FALSE (although true i R ). Eample: the uit sphere i l 2. Part B 5 shorter problems (6 poits each, so 3 poits) B. For each of the followig give a eample of a sequece of cotiuous fuctios f (). If you prefer, a clear sketch of a graph will be adequate. a) f () for all [, ] but f ()d for all =, 2,.... Solutio Let f () be the bump fuctio i the figure o the left below. b) g () for all [, ] ad uiformly to zero o [, ]. g ()d but the g do ot coverge Solutio Let g () be the bump fuctio i the figure o the right below. f () g () 2/ 2/ 2

3 B 2. Show that there is some real > so that =. Solutio Let f() = The f() = ( + 5)/(3 + ) > while f(2) = (4 + 5)/(3 + 64) < so the assertio follows from the Itermediate Value Theorem. B 3. Say f(t)dt = si(+ 2 )+C, assumig that f(t) is cotiuous ad C is a costat, fid both C ad f. Solutio Let = o both sides to fid that = si() + C so C = si(). Take the derivative of both sides to get: f() = 2cos( + 2 ) B 4. Let f() = =2 2 + cos a) Prove that the series coverges uiformly for all real. Solutio Sice cos, the 2 + cos 2 Therefore 2 + cos 2. Because the series =2 coverges, by the Weierstrass M test, the origial 2 series coverges absolutely ad uiformly for all. b) Where (if aywhere) is f cotiuous? Why? Solutio Sice the uiform limit of a sequece of cotiuous fuctios is cotiuous, this f() is cotiuous for all real. B 5. Let f : R R be a smooth fuctio with the properties that f () ad f() cost. for all R. Show that f() = costat. Solutio That f () for all implies the graph of f is cove. Therefore it lies above every taget lie. Say there is a poit p where f (p) >. Sice the graph of f lies above this taget lie at p this would imply that as the f(), cotradictig that f() cost. Similarly, if there were a poit q where f (q), the the graph of f lies above the taget lie at q. Cosequetly as the f(), agai cotradictig the boudedess of f. Remark: Note that the assumptio f () does ot imply that f () is positive somewhere. A eample is e. 3

4 Part C 5 traditioal problems (2 poits each, so total is 6 poits) C. Let a be a sequece of comple umbers with a A as ad let Show that S A as. Solutio S A = a + a a S = a + a a. A = (a A) + (a 2 A) + + (a A). We reduce to the special case where A = by lettig b = a A. The b ad we eed to show that T := b + b b. () Sice b, Give ǫ > there is a N so that if > N the b < ǫ. Rewrite () as the sum of two terms: T = b + b b N + b N+ + b N b = I + J. By our choice of N, J ( N)ǫ/ < ǫ. It is importat to ote that this will remai true if we choose a eve larger value of. To estimate I, we use that sice the sequece b coverges, it is bouded. Thus for some M we kow that b < M. Cosequetly, I NM/. Now choose so large that NM/ < ǫ. The T I + J < 2ǫ. C 2. Let f() C([, ]) have the property that f()h()d = for it all fuctios h C([, ]) with the additioal property that h() = h() =. Prove that f() o all of [, ]. Solutio By cotradictio, say that f(p) > at some p [, ]. Because f is cotiuous, it will be positive at all earby poits. Thus we may assume that p is a iterior poit ad also that f > i a small iterval J cotaiig p. Let h() be a cotiuous bump fuctio with h() > i J, ad h() = i the remaider of [, ] The f()h()d = f()h()d >, h() which is a cotradictio. The idetical costructio works if f(p) < at some p [, ]. J p 4

5 C 3. Let M be a metric space ad let B(M; R) be the set of all bouded real-valued fuctios o M with the uiform orm: f = sup f(). M Sice f is assumed to be bouded, the f <. Defie the distace betwee f ad g to be f g. This makes B ito a metric space. Show that this is a complete metric space. There are two steps: (i). Get a cadidate for the limit fuctio f(), ad (ii). Prove that this f() is bouded. [Where does your proof use that R is complete?] Solutio Let f () be a sequece of bouded fuctio that is a Cauchy sequece i this orm. I particular, at ay poit = p M, f (p) is a Cauchy sequece of real umbers. Because of the completeess of the real umbers, the f (p) coverge of some real umber, say q so we will defie f(p) to be this umber q. This gives us the fuctio to which the Cauchy sequece f coverges. We ow eed oly show that f() is a bouded fuctio. This follows from the fact that a Cauchy sequece is always bouded. I fact, lettig ǫ =, there is a N so that if k > N the f k f N <. Thus, for ay poit p f(p) f(p) f k (p) + f k f N + f N < f(p) f k (p) + + f N. Sice f k (p) f(p) the first term o the right ca be made less tha, say,, by choosig k large, while the last term i bouded because the fuctios we were workig with were assumed to be bouded. Cosequetly, the space of bouded cotiuous fuctios if complete. C 4. Suppose that G : R R is a cotiuous fuctio with the property that for some real M G() G(y) M y for all, y R. (2) Here is the stadard Euclidea distace i R. If λ > is small eough, show that the fuctio F : R R defied by F() = λg() is oe-to-oe ad oto, so for every z R the equatio F() = z has oe ad oly oe solutio R. Note that a solutio is a fied poit of some map. Solutio We use a cotractig mappig. For our complete metric space we use R with the Euclidea orm. The equatio we wat to solve is λg() = z. We seek a fied poit of the map T() := λg() + z. 5

6 Clearly T maps R ito itself so we oly eed to verify the cotractig coditio usig the special property (2) of G: T() T(y) = λ G() G(y) λm y. It is ow clear that pickig λ so that λm < the cotractig assumptio is satisfied. Thus,as desired, T has a uique fied poit. [Remark: This problem is the essece of the Iverse Fuctio Theorem]. C 5. Let f() : R R be cotiuous for all with f() = for ad let g () be the sequece of fuctios i the figure. Let h (t) = f(t )g ()d a) Show that h is uiformly cotiuous. Solutio h (t) h (s) = [f(t ) f(s )]g ()d / / g () Because f() is cotiuous o R ad zero for, it is uiformly cotiuous o R. Give ay ǫ > there is a δ so that if t s < δ the f(t) f(s) < ǫ. But (t ) (s ) = t s so f(t ) f(s ) < ǫ. Therefore, h (t) h (s) This shows that h is uiformly cotiuous. b) Show that lim h (t) = f(t) uiformly. Solutio We use f(t) = f(t)g ()d. The h (t) f(t) = / / ǫg ()d = ǫ. [f(t ) f(t)]g ()d. With δ from part (a), pick so that / < δ. The (t ) t = < / < δ so that f(t ) f(t) < ǫ. Therefore h (t) f(t) / / ǫg ()d = ǫ. Because the right side is idepedet of t we have proved that h coverges to f uiformly. 6

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges