Solutions for problems in the 9 th International Mathematics Competition for University Students

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Lambert Barrett
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1 Solutios for problems i the 9 th Iteratioal Mathematics Competitio for Uiversity Studets Warsaw, July 9 - July 25, 2002 First Day Problem. A stadard parabola is the graph of a quadratic polyomial y = x 2 + ax + b with leadig coefficiet. Three stadard parabolas with vertices V, V 2, V 3 itersect pairwise at poits A, A 2, A 3. Let A s (A be the reflectio of the plae with respect to the x axis. Prove that stadard parabolas with vertices s (A, s (A 2, s (A 3 itersect pairwise at the poits s (V, s (V 2, s (V 3. Solutio. First we show that the stadard parabola with vertex V cotais poit A if ad oly if the stadard parabola with vertex s(a cotais poit s(v. Let A = (a, b ad V = (v, w. The equatio of the stadard parabola with vertex V = (v, w is y = (x v 2 + w, so it cotais poit A if ad oly if b = (a v 2 + w. Similarly, the equatio of the parabola with vertex s(a = (a, b is y = (x a 2 b; it cotais poit s(v = (v, w if ad oly if w = (v a 2 b. The two coditios are equivalet. Now assume that the stadard parabolas with vertices V ad V 2, V ad V 3, V 2 ad V 3 itersect each other at poits A 3, A 2, A, respectively. The, by the statemet above, the stadard parabolas with vertices s(a ad s(a 2, s(a ad s(a 3, s(a 2 ad s(a 3 itersect each other at poits V 3, V 2, V, respectively, because they cotai these poits. Problem 2. Does there exist a cotiuously differetiable fuctio f : R R such that for every x R we have f(x > 0 ad f (x = f(f(x? Solutio. Assume that there exists such a fuctio. Sice f (x = f(f(x > 0, the fuctio is strictly mootoe icreasig. By the mootoity, f(x > 0 implies f(f(x > f(0 for all x. Thus, f(0 is a lower boud for f (x, ad for all x < 0 we have f(x < f(0 + x f(0 = ( + xf(0. Hece, if x the f(x 0, cotradictig the property f(x > 0. So such fuctio does ot exist.

2 Problem 3. Let be a positive iteger ad let a = (, b = 2, for =, 2,...,. Show that a b + a 2 b a b = 0. ( Solutio. Sice ( ( = for all, ( is equivalet to [ ] ( 0 ( ( = (2 We prove (2 by iductio. For =, both sides are equal to 2. Assume that (2 holds for some. Let [ ] x = ; ( 0 ( ( the x + = 2+ + =0 ( ( = =0 ( ( + ( + + = = 2 + =0 + + ( This implies (2 for = 2 =0 ( = x Problem 4. Let f : [a, b] [a, b] be a cotiuous fuctio ad let p [a, b]. Defie p 0 = p ad p + = f(p for = 0,, 2,... Suppose that the set T p = {p : = 0,, 2,... } is closed, i.e., if x / T p the there is a δ > 0 such that for all x T p we have x x δ. Show that T p has fiitely may elemets. Solutio. If for some > m the equality p m = p holds the T p is a fiite set. Thus we ca assume that all poits p 0, p,... are distict. There is a coverget subsequece p ad its limit q is i T p. Sice f is cotiuous p + = f(p f(q, so all, except for fiitely may, poits p are accumulatio poits of T p. Hece we may assume that all of them are accumulatio poits of T p. Let d = sup{ p m p : m, 0}. Let δ be 2

3 positive umbers such that =0 δ < d. Let I 2 be a iterval of legth less tha δ cetered at p such that there are there are ifiitely may s such that p / I j, this ca be doe by iductio. Let 0 = 0 ad m+ be the j=0 smallest iteger > m such that p / m j=0 I j. Sice T p is closed the limit of the subsequece (p m must be i T p but it is impossible because of the defiitio of I s, of course if the sequece (p m is ot coverget we may replace it with its coverget subsequece. The proof is fiished. Remar. If T p = {p, p 2,... } ad each p is a accumulatio poit of T p, the T p is the coutable uio of owhere dese sets (i.e. the sigle-elemet sets {p }. If T is closed the this cotradicts the Baire Category Theorem. Problem 5. Prove or disprove the followig statemets: (a There exists a mootoe fuctio f : [0, ] [0, ] such that for each y [0, ] the equatio f(x = y has ucoutably may solutios x. (b There exists a cotiuously differetiable fuctio f : [0, ] [0, ] such that for each y [0, ] the equatio f(x = y has ucoutably may solutios x. Solutio. a. It does ot exist. For each y the set {x: y = f(x} is either empty or cosists of poit or is a iterval. These sets are pairwise disjoit, so there are at most coutably may of the third type. b. Let f be such a map. The for each value y of this map there is a x 0 such that y = f(x ad f (x = 0, because a ucoutable set {x: y = f(x} cotais a accumulatio poit x 0 ad clearly f (x 0 = 0. For every ε > 0 ad every x 0 such that f (x 0 = 0 there exists a ope iterval I x0 such that if x I x0 the f (x < ε. The uio of all these itervals I x0 may be writte as a uio of pairwise disjoit ope itervals J. The image of each J is a iterval (or a poit of legth < ε legth(j due to Lagrage Mea Value Theorem. Thus the image of the iterval [0, ] may be covered with the itervals such that the sum of their legths is ε = ε. This is ot possible for ε <. Remars.. The proof of part b is essetially the proof of the easy part of A. Sard s theorem about measure of the set of critical values of a smooth map. 2. If oly cotiuity is required, there exists such a fuctio, e.g. the first co-ordiate of the very well ow Peao curve which is a cotiuous map from a iterval oto a square. 3

4 Mx 2 Problem 6. For a matrix M with real etries let M = sup, x R \{0} x 2 where 2 deotes the Euclidea orm o R. Assume that a matrix A with real etries satisfies A A for all positive itegers Prove that A 2002 for all positive itegers. Solutio. Lemma. Let (a 0 be a sequece of o-egative umbers such that a 2 a 2+ a 2, a 2+ a 2+2 a a + for ay 0 ad lim sup a < /4. The lim sup a <. Proof. Let c l = sup 2 l( + a for l 0. We will show that c l+ 4c 2 l. Ideed, for ay iteger 2 l+ there exists a iteger 2 l such that = 2 or = 2 +. I the first case there is a 2 a 2+ a 2 4c 2 l 4c2 l 2+ c2 l ( , whereas i the secod case there is a 2+ a 2+2 a a + c 2 l (+(+2 4c2 l 2+2 4c2 l 2+3. Hece a sequece (a 4c2 l + 2l+ is o-decreasig ad its terms are o-positive sice it coverges to zero. Therefore a 4c2 l for + 2l+, meaig that c 2 l+ 4c2 l. This implies that a sequece ((4c l 2 l l 0 is oicreasig ad therefore bouded from above by some umber q (0, sice all its terms except fiitely may are less tha. Hece c l q 2l for l large eough. For ay betwee 2 l ad 2 l+ there is a c l ( q + q2l yieldig lim sup a q <, yieldig lim sup a q <, which eds the proof. Lemma 2. Let T be a liear map from R ito itself. Assume that lim sup T + T < /4. The lim sup T + T / <. I particular T coverges i the operator orm ad T is power bouded. Proof. Put a = T + T. Observe that T +m+ T +m = (T +m+2 T +m+ (T + T (T m+ T m implyig that a +m a +m+ + a a m. Therefore the sequece (a m m 0 satisfies assumptios of Lemma ad the assertio of Propositio follows. Remars.. The theorem proved above holds i the case of a operator T which maps a ormed space X ito itself, X does ot have to be fiite dimesioal. 2. The costat /4 i Lemma caot be replaced by ay greater umber sice a sequece a = satisfies the iequality a 4 +m a +m+ a a m for ay positive itegers ad m whereas it does ot have expoetial decay. 3. The costat /4 i Lemma 2 caot be replaced by ay umber greater that /e. Cosider a operator (T f(x = xf(x o L 2 ([0, ]. Oe ca easily 4

5 chec that lim sup T + T = /e, whereas T does ot coverge i the operator orm. The questio whether i geeral lim sup T + T < implies that T is power bouded remais ope. Remar The problem was icorrectly stated durig the competitio: istead of the iequality A ( A, the iequality 2002 ( A A ε ε was assumed. If A = the A =. Therefore ( 0 0 ε A A =, so for sufficietly small ε the coditio is satisfied 0 0 although the sequece ( A is clearly ubouded. 5

SAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx

SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval