University of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!

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1 Uiversity of Colorado Dever Dept. Math. & Stat. Scieces Applied Aalysis Prelimiary Exam 13 Jauary 01, 10:00 am :00 pm Name: The proctor will let you read the followig coditios before the exam begis, ad you will have time for questios. Oce the exam begis, you will have 4 hours to do your best. This is a closed book exam. Please put your ame o each sheet of paper that you tur i. Exam coditios: Submit as may solutios as you ca. All solutios will be graded ad your fial grade will be based o your six best solutios. Each problem is worth 0 poits; parts of problems have equal value uless oted otherwise. Justify your solutios: cite theorems that you use, provide couter-examples for disproof, give explaatios, ad show calculatios for umerical problems. If you are asked to prove a theorem, do ot merely quote that theorem as your proof; istead, produce a idepedet proof. Ask the proctor if you have ay questios. Good luck! Total DO NOT TURN THE PAGE UNTIL TOLD TO DO SO.

2 1. Let a ad b be real umbers with a 0. Determie the radius of covergece of (ax) ( ) 1 + b. =1 Solutio: This is a power series so we use the root test o the series of absolute values. lim ( ax ( 1 + b ) ) 1 = lim Hece the radius of covergece is eb a. ax ( 1 + b ) = ax e b < 1 x < eb a.. Give two real, bouded sequeces {a } ad {b }, put a = lim sup {a } ad b = lim sup {b }. Also, put s = lim sup{a + b }. Prove that Is it always valid? s a + b whe it is valid. Sice both sequeces are bouded, we clearly have that a < + ad b < +. Choose real umbers a ad b such that a < a ad b < b. Sice a > a, we kow that a is evetually a upper boud, i.e., there is a iteger N a for which a a for N a. Similarly, there is a iteger N b for which b b for N b. Put N = max{n a, N b }. The for N we have a + b a + b. Hece a + b is evetually a upper boud, which implies that s = lim sup{a + b } a + b. Sice this holds for all a > a ad all b > b, it must be that s a + b. I detail we have the followig. Suppose that s = a + b + ɛ, where ɛ > 0, the just put a = a + ɛ/3 ad b = b + ɛ/3 i the argumet above to see that s a + b + ɛ/3 < a + b + ɛ = s, a impossibility. 3. A real umber x is said to be a dyadic ratioal provided there is a iteger k ad a oegative iteger for which x = k. For each x [0, 1] ad each N, put { 1, if x = k for some k N; f (x) = 0, otherwise. (i) Prove that the dyadic ratioals are dese i R. Solutio: Let a, b R with a < b. We wat to show that there is a dyadic ratioal i the ope iterval (a, b). By the Archimedea priciple we ca choose a positive iteger such that (b a) > 1, so that we kow there is a iteger k with a < k < b, hece a < k < b, as desired.

3 (ii) Let f : [0, 1] R be the fuctio to which the sequece {f } coverges poitwise. Show that 1 0 f does ot exist. Solutio: If x = k [0, 1] for some k, N, the x = kr for all r 0. +r This says that f m (x) = 1 for all m. Ad if x is ot a dyadic ratioal, the f (x) = 0 for all N. Hece the limit fuctio f is give by { 1, if x is a ozero dyadic ratioal; f(x) = 0, otherwise. Cosider ay partitio P = {x 0,..., x } of [0, 1]. By the desity of the dyadic ratioals sup{f(x): x [x i 1, x i ]} = 1 for each subiterval of the partitio. This says that upper itegral of f o [0, 1] equals 1. Similarly, sice the irratioals are dese i [0, 1], it follows that the lower itegral of f o [0.1] equals 0. Hece 1 0 f does ot exist. (iii) Show that the covergece {f } f is ot uiform. Solutio: There are (at least) two ways to prove this. Our first way is the more simplistic. I order to show that the covergece is NOT uiform, we will show that for fixed ɛ = 1, o matter how large a we cosider, there is some x [0.1] for which f(x ) f (x ) > 1. For give put x = The x is a dyadic ratioal, so f(x ) = 1. But x caot be writte i the form k for ay iteger k, so f (x ) = 0, implyig f(x ) f (x ) = 1 > 1. Hece the covergece is ot uiform. Our secod way to prove this is to quote the theorem that says that if {f } is a sequece of itegrable fuctios from [a, b] to R, ad if {f } coverges uiformly to f, the b a f exists (ad eve equals the limit of the itegrals). Each f has the value 1 at oly fiitely may values of x, so f is itegrable o [0, 1]. As f is ot itegrable o [0, 1], the covergece could ot have bee uiform. 4. Show that if {f } ad {g } are two sequeces of real bouded fuctios that coverge uiformly o some set E R, the (i) the fuctios f ad g are uiformly bouded; (ii) {f g } also coverges uiformly o E. Solutio From the Cauchy criterio for uiformly coverget sequeces of fuctios there is a N such that m, > N = x E we have f m (x) f (x) < 1. Sice each f is bouded, for each k N ad for all x E we have f k (x) c k for some costat c k. The usig the triagle iequality we see that for > N we have f (x) f (x) f N+1 (x) + f N+1 (x) < c N

4 Hece > N ad k N ad x E we have f k (x) max{c 1,..., c N, c N+1 + 1}. This shows that the f are uiformly bouded. The same argumet works for the fuctios g. So we kow there is a costat C such that This proves part (i). k N ad x E we have f k (x) C ad g k (x) C. We ow show that {f g } satisfies the uiform Cauchy criterio. Let ɛ > 0 be give. From the uiform Cauchy criterio for {f } ad {g }, there are N 1 ad N such that m, > N 1 = x E : f m (x) f (x) < ɛ C ; m, > N 1 = x E : g m (x) g (x) < ɛ C. Now let m, > N = max{n 1, N } ad compute that f m (x)g m (x) f (x)g (x) f m (x)g m (x) f (x)g m (x) + f (x)g m (x) f (x)g (x) = f m (x) f (x) g m (x) + f (x) g m (x) g (x) < ɛ C C + ɛ C C = ɛ. 5. Let A be a subset of the metric space (X, d). Suppose that x is a limit poit of A. Show that x must be a limit poit of A. Solutio Suppose that x is a limit poit of A but ot a limit poit of A. Sice x is ot a limit poit of A there must be a ope ball B r (x), r > 0, with B r (x) A =. But sice x is a limit poit of A, each ope set cotaiig x must cotai a poit y of A \ {x}. But sice y A, each ope set cotaiig y must have a poit of A. This meas B r (x) must cotai a poit of A, givig a cotradictio. Hece x must be a limit poit of A. Alterative Solutio: We have see that A is closed, ad that a closed set cotais all its limit poits. Hece x A. We suppose that x is NOT a limit poit of A ad work for a cotradictio. It follows that x A ad is ot a limit poit of A. But sice x is a limit poit of A, for each 1 there is a poit y B 1 (x) (A \ {x}). Put r = d(y, x), so 0 < r < 1. Sice x is ot a limit poit of A, there must be some r > 0 such that B r (x) A = {x}. Choose N large eough so that > N = 1 < r. The > N = y A \ A = y is a limit poit of A. Hece B r (y ) cotais a poit x A. But d(x, y ) < r = d(y, x) = x x. Ad d(x, x) d(x, y ) + d(y, x) < r + r < 1.

5 We have ow established a sequece {x } of poits of A differet from x with d(x, x) < 1 for all > N. Let U be ay ope set cotaiig x. There must be some s > 0 such that x B s (x) U. Now pick N > N such that > N = 1 < s. The > N implies that x B 1 (x) B s (x) U, x A, x x. This shows that x is a limit poit of A, cotradictig our hypothesis that x is ot a limit poit of A. Hece it must be that x is ideed a limit poit of A. 6. Let (X, d) ad (Y, ρ) be metric spaces ad let = A X. If f ad g are cotiuous mappigs of X ito Y for which f(x) = g(x) for all x A, show that f(x) = g(x) for all x A. Solutio: Let x A \ A. We wat to show that f(x) = g(x). Let ɛ > 0 be give. Sice f ad g are cotiuous at x, there are δ g > 0 ad δ f > 0 such that the followig hold: d(x, y) < δ f = ρ(f(x), f(y)) < ɛ/; ad d(x, y) < δ g = ρ(g(x), g(y)) < ɛ/. Sice x is a limit poit of A by hypothesis, for each positive iteger, the ope ball B 1/ (x) cotais a poit a A. Choose large eough so that 1 < mi{δ f, δ g }. The we have ρ(f(x), g(x)) ρ(f(x), f(a )) + ρ(f(a ), g(a )) + ρ(g(a ), g(x)) ɛ/ ɛ/ = ɛ. Sice ɛ > 0 is arbitrary, we must have f(x) = g(x). 7. Fid 1 lim i=1 i, ad justify your aswer. Hit: Thik of the sum as a Riema sum. Solutio: For each the sum is a Riema sum, i=1 f(t i) x i, where t i = i/ ad the partitio is uiform, i.e., x i = 1/. By Theorem 6.14 i Rudi (or ay such theorem), the sum coverges to 1 0 xdx = /3. 8. Let f(x) be cotiuous, real-valued fuctio o [a, b]. (a) Prove that for ay ɛ > 0 there is a polyomial, p(x), such that b a f(x) p(x) dx < ɛ.

6 Solutio: By the Weierstrass theorem (e.g. thm 7.4 Rudi), there exists p(x) so that for every x [a, b], f(x) p(x) < ɛ/(b a). So, b a f(x) p(x) dx < b a ɛ/(b a)dx = ɛ. (b) Prove that there is a sequece p k (x) of polyomials havig the property that f(x) = p k (x) for all x [a, b]. k=1 Solutio: By the Weierstrass Theorem, for ɛ = 1/ there is a polyomial p 1 (x) such that for each x [a, b] there holds f(x) p 1 (x) < 1/. The for ɛ = 1/( ), there is a polyomial p (x) such that for all x [a, b] there holds (f(x) p 1 (x)) p (x) < 1/( ). We ow proceed by iductio. Suppose that p 1 (x),..., p j (x) have bee chose so that f(x) (p 1 (x) + + p k (x)) < 1, for 1 k j. k The by the Weierstrass Theorem there is a polyomial p j (x) such that for all x [a, b] we have f(x) (p 1 (x) + + p j (x)) p j+1 (x) < 1 j+1. By iductio we may assume that p (x) has bee costructed for each N so that m f(x) p i (x) < 1 m. i=1 Hece f(x) = k=1 p k(x) for all x [a, b].

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