Introductory Analysis I Fall 2014 Homework #9 Due: Wednesday, November 19 1. Here is an easy one, to serve as warmup. Assume M is a compact metric space and N is a metric space. Assume that fn : M ! N for each n 2 M. Assume that for every p 2 M there exists a neighborhood Vp of p such that (fnjVp ) converges uniformly on Vp. Prove: (fn) converges uniformly on M. Proof. By compactness, there exist p1; : : : ; pn 2 M such that M = Sn V . Let > 0 be given. By hypothesis (f j ) is Cauchy in the j=1 pj n Vpj uniform norm for each j = 1; : : : ; n; thus there exist N1;:::;Nj such that jfn(p) − fm(p)j < if n; m ≥ Nj, p 2 Vpj , j = 1; : : : ; n. Clearly, if n; m ≥ max(N1;:::;Nn), then jfn(p) − fm(p)j < for all p 2 M. The sequence (fn) is thus Cauchy in the uniform norm, hence converges uni- formly. 2. Here is an exercise from the second set of pre-lim problems in the textbook; it is #70 on page 266. The book phrases it as follows: Let A be the set of all positive integers that do not contain the digit 9 in their decimal expansion. Prove that X 1 < 1: a a2A P1 In other words, we consider the series n=1 an where the sequence (1=an) is 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, . At first it is the almost the same as the sequence of positive integers, but eventually bigger and bigger gaps occur. k k+1 Solution. For k = 0; 1; 2;:::, let Ak = fa 2 A : 10 ≤ a < 10 g. These are numbers with k + 1 digits and some elementary counting tech- niques show that there are 8·9k numbers with k digits that do not contain the digit 9. Since all are ≥ 10k we see that k X 1 9 a ≤ 8 · 9k = 8 : 10k 10 a2Ak Assume now F is a finite subset of A. There is then k 2 N [ f0g such that Sk A ⊂ j=0 Aj, hence k k j 1 j X X X a X 9 X 9 a ≤ 8 ≤ 8 = 80 < 1: ≤ 10 10 a2F j=0 a2Aj j=0 j=0 The set of all finite sums is bounded; the result follows. 2 3. Prove the following result first proved by Euler: Let p1; p2;::: be the X 1 sequence of primes. Then = 1. In view of this result and the p j=1 j result of the previous exercise, what do you think appears more frequently in the number sequence: A prime, or an integer whose decimal expansion has no digit equal to 9? Hints: For a given integer N select n as in the solution of Exercise 10 of Homework 8; that is, n is such that fp1; : : : ; png is the set of all primes ≤ N. Recall also formula (1) from that solutions set, which is also valid if s = 1. The terms of the summation on the right side of (1) comprise all numbers that decompose into a product of (powers of) the primes p1; : : : ; pn; in particular every number from 1 to N. Thus N n X 1 Y 1 ≤ n 1 − 1 n=1 k=1 pk Prove that if 0 ≤ x ≤ 1=2 then 1 − x ≤ e2x. Use this to conclude that N X 1 Pn 2 ≤ e k=1 pk : n n=1 Solution. In the hints, \0 ≤ x ≤ 1=2 then 1 − x ≤ e2x," should have been \0 ≤ x ≤ 1=2 then (1 − x)−1 ≤ e2x. Let N 2 N and let p − 1; : : : ; pn be all the primes ≤ N. If k ≤ N, then j1 jn k = p1 ··· pn for some (unique) choice of j1; : : : ; jn 2 N [ f0g. Thus 1 = p−j1 ··· p−jn k 1 n and is thus a member of the summation appearing on the right hand side of equation (1) of the solution pages of Homework 8, with s = 1. It follows, by that formula, that N n X 1 Y 1 ≤ : (1) n 1 − 1 n=1 k=1 pk Claim: If 0 ≤ x ≤ 1, then (1 − x)−1 ≤ e2x. Letting '(x) = e2x(1 − x), the claim is equivalent to claiming that '(x) ≥ 1 for 0 ≤ x ≤ 1=2. Now '0(x) = e2x(1 − 2x) ≥ 0 for 0 ≤ x ≤ 1, thus ' is increasing in that interval, hence '(x) ≥ '(0) = 1 for all x 2 [0; 1=2], establishing the claim. −1 It follows that (1 − p−1)−1 ≤ e2p for all primes p and (1) implies N n X 1 Y 1 Pn 1 ≤ e pk = e k=1 pk : n n=1 k=1 3 Taking logarithms, n N X 1 X 1 ≥ : pk n k=1 n=1 Letting N ! 1, we also have n ! 1 and the right hand side of the last displayed inequality diverges to 1. This implies that the left hand side also has to diverge to 1 as n ! 1. The answer to the last question is that the frequency at which primes appear (i.e., on the average, the number of primes in any interval of in- tegers) is higher than that of integers with a decimal expansion with all digits different from 9. 4. More on monotone functions. In the previous homework you learned (if you did not know it already) that a monotone function can only have a countable set of discontinuities. The surprising fact is that every countable subset of an interval [a; b] can be the set of discontinuities of a monotone function. For example, there are monotone functions that are discontinu- ous at all rational points, continuous elsewhere. Here you are required to fill in the details of the constructions of such a function. Let −∞ < a < b < 1 and let (xn) be a sequence of distinct points of the interval (a; b). Define f :[a; b] ! R as follows. Select cn for n 2 N so that P1 cn > 0 and cn < 1. For x 2 [a; b] let S(x) = fn 2 N : xn < xg. Pn=1 Define f(x) = n2S(x) cn. Prove: (a) f is increasing in [a; b]. (b) f has a jump at xn for each n 2 N; in fact f(xn+) − f(xn−) = cn for each n 2 N, (c) f is continuous at all x 6= xn, n 2 N. Comments and hints. The points xn could be all over the place so it is quite possible (as it would be if (xn) is an ordering of the rational numbers in the interval) that S(x) is an infinite set for all x 2 (a; b). Of P course S(a) = ; and one defines cn = 0. If S(x) is a finite set there P n2; is no problem in how n2S(x) cn is defined. If it is infinite it is countably infinite and one can either define X X cn = supf cn : F is a finite subset of S(x)g n2S(x) n2F or, equivalently, one can order S(x) into a sequence fn1; n2;:::g and define P P1 n2S(x) cn = k=1 cnk . The way one does this ordering is irrelevant; the end result is always the same. Proving that f is increasing is easy, I think. For the rest, a direct proof can be sort of messy. Here is a more elegant way of proceeding. Let H be the Heaviside function; that is the function 4 defined by 0; if x < 0; H(x) = 1; if x ≥ 0: P1 Prove that f(x) = n=1 cnH(x − xn); in fact, prove the series P1 n=1 cnH(· − xn) converges uniformly to f. Use this to complete the proof. Solution. The hints were a bit muddled. That the function f defined in the statement of the exercise is the same as the function defined by f(x) = P1 P1 n=1 cnH(x − xn) is, I think, obvious. I will write f = n=1 gn, where gn(x) = cnH(x − xn) for n 2 N, x 2 (a; b). The series converges uniformly because jgn(x)j ≤ cn for all n 2 N, x 2 (a; b), so uniform convergence is P guaranteed by the Weierstrass M-test, since cn < 1. By a result seen in class (and in the textbook), the limit of a uniformly convergent sequence (or series) of functions from a metric space to another is continuous at each point where all the terms of the sequence (or series) are continuous. Now x 7! H(x − xn) is continuous at all points x 6= xn. Thus, if x 6= xn for n 2 N, then gn is continuous at x for all n, and f is continuous at all points x 6= xn, n 2 N. Let n 2 N. Then gm is continuous at xn for P m 6= m. Defining G(x) = m6=n gm(x), we have f = gn + G and since G is continuous at x we see that f(xn+) − f(xn−) = gn(x+) − gn(x−) = cn. At this point we proved (a) f has a jump at xn for each n 2 N; in fact f(xn+) − f(xn−) = cn for each n 2 N, (b) f is continuous at all x 6= xn, n 2 N. It remains to be seen that f is increasing. This is obvious from the first definition of f. 3. Textbook, Chapter 4, Exercise 8, p.